Recreational Mathematics 2003

Recreational MathematicsPaul Yiu Department of Mathematics Florida Atlantic University Summer 2003 Chapters 1–44 Version 031209 ii Contents 1 Lattice polygons 101 1.1 Pick’s Theorem: area of lattice polygon . . . . . . . . . 102 1.2 Counting primitive triangles . . . . . . . . . . . . . . . 103 1.3 The Farey sequence . . . . . . . . . . . . . . . . . . . 104 2 Lattice points 109 2.1 Counting interior points of a lattice triangle . . . . . . . 110 2.2 Lattice points on a circle . . . . . . . . . . . . . . . . . 111 3 Equilateral triangle in a rectangle 117 3.1 Equilateral triangle inscribed in a rectangle . . . . . . . 118 3.2 Construction of equilateral triangle inscribed in a rectangle . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 4 Basic geometric constructions 4.1 Geometric mean . . . . . . . . . 4.2 Harmonic mean . . . . . . . . . 4.3 Equal subdivisions of a segment . 4.4 The Ford circles . . . . . . . . . 123 124 125 126 127 201 202 203 204 205 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Greatest common divisor 5.1 gcd(a, b) as an integer combination of a and b . 5.2 Nonnegative integer combinations of a and b . . 5.3 Cassini formula for Fibonacci numbers . . . . . 5.4 gcd of generalized Fibonacci and Lucas numbers . . . . . . . . . . . . . . . . 6 Pythagorean triples 209 6.1 Primitive Pythagorean triples . . . . . . . . . . . . . . 210 6.2 Primitive Pythagorean triangles with square perimeters 211 iv CONTENTS 6.3 6.4 6.5 Lewis Carroll’s conjecture on triples of equiareal Pythagorean triangles . . . . . . . . . . . . . . . . . . . . . . . . . 212 Points at integer distances from the sides of a primitive Pythagorean triangle . . . . . . . . . . . . . . . . . . . 213 Dissecting a rectangle into Pythagorean triangles . . . . 214 7 The tangrams 225 7.1 The Chinese tangram . . . . . . . . . . . . . . . . . . 226 7.2 A British tangram . . . . . . . . . . . . . . . . . . . . 227 7.3 Another British tangram . . . . . . . . . . . . . . . . . 228 8 The classical triangle centers 8.1 The centroid . . . . . . . . . . . . . . 8.2 The circumcircle and the circumcircle 8.3 The incenter and the incircle . . . . . 8.4 The orthocenter and the Euler line . . . 8.5 The excenters and the excircles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 232 233 234 235 236 9 The area of a triangle 301 9.1 Heron’s formula for the area of a triangle . . . . . . . . 302 9.2 Heron triangles . . . . . . . . . . . . . . . . . . . . . . 303 9.3 Heron triangles with consecutive sides . . . . . . . . . 304 10 The golden section 10.1 The golden section ϕ . . . . . . . . . . 10.2 Dissection of a square . . . . . . . . . . 10.3 Dissection of a rectangle . . . . . . . . . 10.4 The golden right triangle . . . . . . . . 10.5 What is the most non-isosceles triangle? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 310 311 313 314 316 321 322 323 325 327 328 329 11 Constructions with the golden section 11.1 Construction of golden rectangle . . . . . . . . . . . . 11.2 Hofstetter’s compass-only construction of the golden section . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Hofstetter’s 5-step division of a segment in the golden section . . . . . . . . . . . . . . . . . . . . . . . . . . 11.4 Construction of regular pentagon . . . . . . . . . . . . 11.5 Ahlburg’s parsimonious construction of the regular pentagon . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.6 Construction of a regular 17-gon . . . . . . . . . . . . CONTENTS v 12 Cheney’s card trick 335 12.1 Principles . . . . . . . . . . . . . . . . . . . . . . . . 336 12.2 Examples . . . . . . . . . . . . . . . . . . . . . . . . . 337 13 Digit problems 13.1 When can you cancel illegitimately and yet get the correct answer? . . . . . . . . . . . . . . . . . . . . . . . 13.2 A Multiplication problem . . . . . . . . . . . . . . . . 13.3 A division problem . . . . . . . . . . . . . . . . . . . 13.4 The most popular Monthly problem . . . . . . . . . . . 13.5 The problem of 4 n’s . . . . . . . . . . . . . . . . . . 14 Numbers with many repeating digits 14.1 A quick multiplication . . . . . . . 14.2 The repunits . . . . . . . . . . . . 14.3 Squares of repdigits . . . . . . . . 14.4 Sorted numbers with sorted squares 15 Digital sum and digital root 15.1 Digital sum sequences . . . . . . . 15.2 Digital root . . . . . . . . . . . . . 15.3 The digital roots of the powers of 2 15.4 Digital root sequences . . . . . . . 16 3-4-5 triangles in the square 17 Combinatorial games 17.1 Subtraction games . . . . . . . . . . 17.1.1 The Sprague-Grundy sequence 17.1.2 Subtraction of square numbers 17.1.3 Subtraction of square numbers 17.2 The nim sum of natural numbers . . 17.3 The game Nim . . . . . . . . . . . . 17.4 Northcott’s variation of Nim . . . . . 17.5 Wythoff’s game . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 402 403 404 407 408 415 416 417 418 419 423 424 425 426 427 431 501 502 503 504 505 509 510 511 512 18 Repunits 517 18.1 k -right-transposable integers . . . . . . . . . . . . . . 518 18.2 k -left-transposable integers . . . . . . . . . . . . . . . 519 18.3 Sam Yates’ repunit riddles . . . . . . . . . . . . . . . . 520 vi CONTENTS 18.4 18.5 Recurring decimals . . . . . . . . . . . . . . . . . . . 522 The period length of a prime . . . . . . . . . . . . . . 523 531 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 535 536 537 537 538 539 540 601 602 603 604 611 612 613 614 621 622 623 624 627 628 629 630 633 701 702 703 704 19 More digital trivia 20 The shoemaker’s knife 20.1 Archimedes’ twin circles . . . . 20.2 Incircle of the shoemaker’s knife 20.2.1 Archimedes’ construction 20.2.2 Bankoff’s constructions . 20.2.3 Woo’s three constructions 20.3 More Archimedean circles . . . . 21 Infinitude of prime numbers 21.1 Proofs by construction of sequence of relatively prime numbers . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Somos sequences . . . . . . . . . . . . . . . . . . . . 21.3 F¨ urstenberg’s topological proof made easy . . . . . . . 22 Strings of prime numbers 22.1 The prime number spiral . . . . . . . . . . . . . . . . . 22.2 The prime number spiral beginning with 17 . . . . . . . 22.3 The prime number spiral beginning with 41 . . . . . . . 23 Strings of composites 23.1 Strings of consecutive composite numbers . . . . . . . 23.2 Strings of consecutive composite values of n 2 + 1 . . . 23.3 Consecutive composite values of x2 + x + 41 . . . . . 24 Perfect numbers 24.1 Perfect numbers . . . . . . . . . . . . . . . . . . . . . 24.2 Charles Twigg on the first 10 perfect numbers . . . . . 24.3 Abundant and deficient numbers . . . . . . . . . . . . 24.3.1 Appendix: Two enumerations of the rational numbers in (0,1) . . . . . . . . . . . . . . . . . . . . 25 Routh and Ceva theorems 25.1 Routh theorem: an example . . . . . . . . . . . . . . . 25.2 Routh theorem . . . . . . . . . . . . . . . . . . . . . . 25.3 Ceva Theorem . . . . . . . . . . . . . . . . . . . . . . CONTENTS vii 26 The excircles 26.1 Feuerbach theorem . . . . . . . . . . . . . . . . . 26.2 A relation among the radii . . . . . . . . . . . . . 26.3 The circumcircle of the excentral triangle . . . . . 26.4 The radical circle of the excircles . . . . . . . . . 26.5 Apollonius circle: the circular hull of the excircles 27 Figurate numbers 27.1 Triangular numbers . . . . . . . . . . . . . 27.2 Special triangular numbers . . . . . . . . . 27.3 Pentagonal numbers . . . . . . . . . . . . . 27.4 The polygonal numbers P n,k . . . . . . . . . 27.4.1 Appendix: Solution of Pell’s equation . . . . . . . . . . . . . . . 711 712 713 714 715 716 719 719 719 721 722 723 725 726 727 727 728 730 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 Polygonal triples 28.1 Double ruling of S . . . . . . . . . . . . . . . . . . . . 28.2 Primitive Pythagorean triple associated with a k -gonal triple . . . . . . . . . . . . . . . . . . . . . . . . . . . 28.3 Triples of triangular numbers . . . . . . . . . . . . . . 28.4 k -gonal triples determined by a Pythagorean triple . . . 28.5 Correspondence between (2h + 1)-gonal and 4h-gonal triples . . . . . . . . . . . . . . . . . . . . . . . . . . 29 Sums of consecutive squares 801 29.1 Sum of squares of natural numbers . . . . . . . . . . . 802 29.2 Sums of consecutive squares: odd number case . . . . . 803 29.3 Sums of consecutive squares: even number case . . . . 805 30 Sums of powers of natural numbers 31 A high school mathematics contest 807 809 32 Mathematical entertainments 817 32.1 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 818 32.2 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 820 33 Maxima and minima without calculus 34 A British test of teachers’ mathematical background 903 911 1003 38 Permutations 1005 38. . . . . . k ) .4 Edge-magic heptagon . . . . . . . . . . . . 1001 37. . . . . . . . . . . . . . 43 Lewis Carroll’s unused geometry pillow problem 44 Japanese Temple Geometry . . . . . . . . . . . . .3 Volume of a tetrahedron . . 42. . 39. . . . . . . . . . . . . . . . . . . .1 The universal sequence of permutations . . . 1009 . . . . . . . . 1005 38. . . . . . . .1 The Josephus problem . . . . . . . . . . . . . . . . . . .5 Edge-magic pentagram 40. . . . . . . . . . . . . . 1010 . . . . . . . . . . . . . . . . . 40 Graph labelling 40.1 The disjoint cycle decomposition of a permutation 39. . .4 The game of ropes and rungs . . . . . 39. .3 Dudeney’s Canterbury puzzle 3 . . . . . . . . . . . . . . . . . . . 42. . . . . . . . . . . . . . . . . . . . . . .1 The isosceles tetrahedron . . . . . . . .2 Generalized Josephus problem J(n. . . . . .2 Face-magic tetrahedron 40. . . . . . . 1107 1107 1108 1109 1110 1113 1115 41 Card tricks from permutations 42 Tetrahedra 42. . . 40.2 The position of a permutation in the universal sequence 1007 39 Cycle decompositions 39. .6 A perfect magic circle . . . . . . . . . . . 1017 1017 1018 1019 1020 1021 1022 1101 . . . . 1015 . 40. . . . . . . 40. .2 The volume of an isosceles tetrahedron 42. . . . . . .viii CONTENTS 35 A mathematical contest 36 Some geometry problems from recent journals 915 919 37 The Josephus problem and its generalization 1001 37. . . . . . . . 1013 . 1009 . . . . .2 2 .4 2 .3 Magic cube . . . . . . . . . . . . . . . . . . . . . . . . .1 Edge-magic triangle . . . . . . . Chapter 1 Lattice polygons 1 2 3 Pick’s Theorem: area of lattice polygon Counting primitive triangles The Farey sequences Appendix: Regular solids Exercise Project: A cross number puzzle . y3 ). we can partition it into primitive lattice triangles. If the vertices are at the points (x1 .102 Lattice polygons 1. if one of the vertices is at the origin (0. otherwise negative. let I = the number of interior points. 1 This formula is valid for arbitrary vertices. (x3 . y1). y1). there is a simple formula to find its area in terms of the coordinates of its vertices. 2 1 2 Given a lattice polygon. This follows (2) The area of a primitive lattice triangle is always 1 2 from a study of the Farey sequences. then the points are collinear. then the area is 1 x y 1 1 1 1 x2 y2 1 . each triangle contains no lattice point apart from its three vertices.. A lattice polygon is one whose vertices are lattice points (and whose sides are straight line segments).1 (Pick). 0). and B = the number of boundary points. i. It is positive if the vertices are traversed counterclockwise. (x2 . then the area is 1 |x y − x2 y1 |. Two wonderful things happen that make it easy to find the area of the polygon as given by Pick’s theorem. This is an application of Euler’s polyhedral formula.e. If the polygon is a triangle. The area of a lattice polygon is I + B 2 − 1. y2 ). (x2 . y2 ). Theorem 1. . If it is zero. and the other two have coordinates (x1 . For a lattice polygon. (1) There are exactly 2I + B − 2 such primitive lattice triangles no matter how the lattice points are joined. . 2 x y 1 3 3 In particular.1 Pick’s Theorem: area of lattice polygon A lattice point is a point with integer coordinates. then V − E + F = 2. . From this. we take two identical copies and glue them along their common boundary. Theorem 1.2. Euler’s polyhedron formula gives (2I + B ) − 3T + 2T = 2.2 Counting primitive triangles 103 1. Since every face is a triangle. Imagine the 2-sheet polygon blown into a closed polyhedron. Now. If a closed polyhedron has V vertices.2 Counting primitive triangles We shall make use of the famous Euler polyhedron formula. and each edge is contained in exactly two faces. It follows that E = 3T . we have 2E = 3F . E edges and F faces. Then there are F = 2T faces of the polyhedron. Given a lattice polygon with a partition into primitive lattice triangles. The number of vertices is V = 2I + B . we have T = 2I + B − 2. Suppose there are T primitive triangles in each sheet.1. 4 h k 3 . . 4 1 . we can record the successive terms of the Farey sequence Fn by rotating a ruler about 0 counterclockwise from the positive x-axis. 1 0 . 5 1 .104 Lattice polygons 1. We write 0 = 0 and 1 = 1 . .3 The Farey sequence Let n be a positive integer. The Farey sequence of order n is the sequence of rational numbers in [0. and h k are three successive terms of Fn . 6 3 . 5 3 . 3 1 . 1 1 . 5 1 . 7 1 . 2 1 . 5 1 . If h k h k . If and are successive terms of Fn . 1 0 . 1 0 . 3 1 . 1] of denominators ≤ n arranged in increasing order. then h+h h = . kh − hk = 1 . 3 3 . Restricting to the left side of the vertical line x = n. 2 1 . 5 1 . . . 4 1 .3. 3 7 4 5 6 7 1 Theorem 1. 2 2 . 5 4 . 1 2 . . 5 1 . 6 1 . 6 2 . 1 1 . 5 3 . Corollary 1. 1 0 . 1. 3 3 . 1 0 . A primitive lattice triangle has area 1 2 . 1 1 F1 F2 F3 F4 F5 F6 F7 : : : : : : : 0 . The sequence of “visible” lattice points swept through corresponds to Fn . 1 0 . 2 1 . 3 1 . 4 1 . 7 5 . k k+k The rational numbers in [0. 1 1 . then and k + k > n. 7 4 . 5 2 . . 3 1 . 1 4 . then the rational numbers corresponding to P and Q are successive terms in a Farey sequence (of order ≤ their denominators). 1] can be represented by lattice points in the first quadrant (below the line y = x). 4 1 . If P and Q are two lattice points such that triangle OP Q contains no other lattice points in its interior or boundary.4. 4 2 . 7 1 . 1 3 . 3 2 . 3 1 . 5 1 . 5 2 . 1 2 5 3 4 5 6 1 . 4 1 . 2. 2 1 . 3 h k 2 . 2 1 . . Lehmer [Lehmer]. 2 n→∞ n 2π lim .3 The Farey sequence 105 The Farey polygons P10 and P20 Let φ(n) be the number of integers m satisfying 1 ≤ m ≤ n and gcd(m. By Pick’s formula. By a calcula2 tion of D. The Farey sequence Fn has 1 + n k =1 φ(k ) terms. this area is about 3 of the (smallest) square containing it: 2π 2 3 An = 2. N. for large values of n.1. its area is An = 1 k =1 φ(k ). n) = 1. The n polygon Pn contains 2 + k=1 φ(k ) boundary points and no interior n points. edges. say. Note that m ≥ 3 and n ≥ 3. and each vertex belongs to the same number of faces. Since m 2E 2E V − E + F = 2. n ≥ 3. say. E . we have m − E + n = 2. m 3 3 3 4 5 n 3 4 5 3 3 E V = 2E m F = 2E n regular solid tetrahedron cube duodecahedron octahedron icosahedron . and faces reE E spectively. the only possibilities are as follows. and we require 2(m + n) > mn. From this. Then nF = 2E = mV . n-gons. 2(m + n) − mn Since m.106 Lattice polygons Appendix: Regular solids A regular solid is one whose faces of regular polygons of the same type. m faces. and F be the numbers of vertices. E= 2mn . Let V . F = 2n . and V = 2 . 1. 2 4. are the origin and the points (k. Can a lattice triangle be equilateral? Why? 2. Find the area of P6 and that of Pn for a general n. 6. taken in the Farey sein order. 2 In [Weaver].3 The Farey sequence 107 Exercise 1. The Farey polygon Pn is the lattice polygon whose vertices. 5. Can a lattice polygon be regular? Why? [You may make use of the nontrivial fact that the only values of n for which sin π is rational n is n = 6. 8. Here is P6 . . h) for h k quence Fn . How many terms does the Farey sequence Fn have? [Hint: Give the answer in terms of the Euler ϕ-function]. For B = 3.] 3. 9. 6. it is shown that these are the only possible values of B if I = 1. give an example of a lattice triangle with exactly one interior point and B boundary points. Give an example of an equilateral lattice hexagon. 4. a permutation of the digits 0 through 9 Second smallest prime of the form n2 + 2n + 1. n > 0 A gross number 19-down minus 18-down plus 26-down Number of 2-digit primes The product of the digits of this number is 78125 The sum of this number’s positive divisors is 91 332 + 32 This number is the sum of the factorials of its digits A power of 6 The sum of the fourth power of 5 consecutive triangular numbers A Mersenne prime A power of 2 The number of the beast . and their product is 84 Palindromic square The only even prime number 20-across minus 28-down A perfect square A palindromic integer This many degrees Fahrenheit is 265 degrees Centigrade With 3-down.108 Lattice polygons A cross number puzzle 9 8 7 6 5 4 3 2 1 0 0 Across 1 4 8 9 11 13 14 17 18 20 21 22 24 27 30 31 32 20 22 18 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 19 21 23 24 25 26 27 28 29 30 31 32 1 2 3 4 5 6 7 Down 1 2 3 4 5 6 7 10 12 15 16 18 19 20 23 25 26 28 29 8 9 A multiple of 11 The product of the positive divisors of the number is 2022 See 9-across A Fermat prime Product of the first 3 primes Colleague of 1-across In base 2. this number is written 11010001110001 Yet another perfect square! The first prime year after 1950 π radians This many degrees is 25 6 The 17th Fibonacci number 2102 + 1112 The least common multiple of 36 and 1631 The number of positive perfect squares less than 105 The number of positive integers less than 625 which are not divisible by 5 The sum of these digits is 15. Chapter 2 Lattice points 1 Counting interior points of a lattice triangle 2 Lattice points on a circle Appendix: The floor and the ceiling Appendix: Number of lattice points in a circle Appendix: Number of lattice points under a hyperbola Exercise Project: Cross number puzzle . b) = 2 m=1 ma − (a − 1)(b − 1) + 1.110 Lattice points 2. b) + 1 2 lattice points in the interior of the triangle. if gcd(a. Since the rectangle has (a − 1)(b − 1) interior points. 0). b) = 1. b) − 1 points in the latter group. we obtain as a−1 n=1 nb ma = . and (a. Note that we have an algorithm for computing the gcd of two integers: b−1 a gcd(a. then the triangle has 1 (a − 1)(b − 1) 2 interior points. a b m=1 b−1 Note that each of these is the total number of interior points and those on the hypotenuse (except the two vertices). b In particular. If we put these two copies together to form a rectangle. we see that the interior points along the diagonal are counted twice. There are gcd(a. we have exactly (a − 1)(b − 1) − gcd(a. a b a Counting in two ways. b . 0). b). (a.1 Counting interior points of a lattice triangle A lattice triangle has vertices at (0. the circle of radius 100 has lattice points.2. On the other hand.2 Lattice points on a circle 111 2. ±a). ±b).2 Lattice points on a circle How many lattice points are there on the circle x2 + y 2 = n2 ? This number is usually denoted by r 2 (n2 ). Four of these are on the coordinate axes. i Thus. The first few values can be read from the following figure. for example. if the number n is factored into a product of prime powers n = 2a · i i pb i · qj j . j c where pi and qj are prime numbers of the forms 4k + 1 and 4k + 3 respectively. then r2 (n2 ) = 4 (2bi + 1). (±b. The other 8 are of the form (±a. if p is a prime number of the form 4n + 1. then the circle of radius p has 12 lattice points. . n 1 2 3 4 5 6 7 8 9 10 r2 (n 2 ) In general. and depend on how the prime p is written as a sum of two squares. 112 Lattice points Appendix: The floor and the ceiling The floor of a real number x is the greatest integer not exceeding x: 1 x := max{n ∈ Z : n ≤ x}. If x is not a half-integer. we denote by {x} the integer nearest x. the ceiling of x is the least integer not exceeded by x: x := min{n ∈ Z : n ≥ x}. Find all integers√ n for which n dividing n? How about 1 It is sometimes called the greatest integer function of x and denoted by [x]. . On the other hand. Project √ n divides n. center at the origin? Write a computer program to find out exactly how many lattice points are inside or on the circle radius 100. the number K (R) of lattice points inside and on the circle of radius R satisfies √ √ √ √ πR − π (2 2 R − 2) < K (R) < πR + π (2 2 R + 2). H (R) = R log R + O (R). Appendix: Number of lattice points under a hyperbola Given a real number R.2.2 Lattice points on a circle 113 Appendix: Number of lattice points inside a circle Given a real number r . how many lattice points in the first quadrant are under the hyperbola xy = R (but not on the axes)? This number is H (R ) = 1≤n≤ R d(n). A better estimate was given by Dirichlet √ H (R) = R log R + (2γ − 1)R + O ( R). As a crude estimate. γ is the Euler constant n→∞ lim 1+ 1 1 1 + +···+ 2 3 n − log n ≈ 0. . For large values of √ R. how many lattice points are there inside or on the circle of radius r . Here.5772157 · · · . This is often expressed by writing √ K (R) = πR + O ( R). 5. 15) and C = (16. . Calculate the lengths of the sides of the triangle.114 Lattice points Exercise 1. . 55). Write 97 as a sum of 2 squares and find the lattice points on the circle of radius 97. 0). the k -th statement reads: The number of false statements in this list is greater than k . . 0). 14. Determine the truth value of each of these statements. What is the area? How many interior and boundary points does the triangle have? 4. . B = (36. For k = 1. 45) and C = (48. Consider the lattice triangle ABC with A = (0. B = (24. Repeat the same for the lattice triangle ABC with A = (0. 30). There is a list of n statements. What is the area of the triangle? How many boundary and interior points does it have? 2. 3. Give an example of a lattice triangle whose side lengths are 13. 15. n. Solve the equation x+1 2 x+2 3 x+3 = 819. . 4 6. 2. 6 5 4 3 2 1 0 0 1 2 3 4 5 6 .2 Lattice points on a circle 115 Project: Cross number puzzle Fill in the accompanying square with distinct 2-. and 4-digit numbers which are perfect squares. 3-.2. none of which begins with 0. 116 Lattice points . Chapter 3 Equilateral triangle in a rectangle 1 Equilateral triangle inscribed in a rectangle 2 Construction Exercise . These satisfy a2 + y 2 = b2 + x2 = (a − x)2 + (b − y )2. 2(ax + by ) = a2 + y 2 = b2 + x2 . This can be rewritten as (x2 + b2 )((x2 + b2 ) − 4ax − 4b2 ) = 0. and BP = y . from which Similarly. BC = AD = b. DQ = x.1 Equilateral triangle inscribed in a rectangle Given a rectangle ABCD. x = 2a − √ 3b. and we have (x2 − 2ax + b2 )2 = 4b2 (b2 + x2 ) − 4a2 b2 . y = 2b − √ 3a. . From these.118 Equilateral triangle in a rectangle 3. how can we choose a point P on BC and a point Q on CD such that triangle AP Q is equilateral? D x Q a−x C b−y b P y A a B Suppose AB = DC = a. 3. What is the range of the ratio rectangle? a b for X and Y to be in the interior of the .1. Then AXY is equilateral.2 Construction of equilateral triangle inscribed in a rectangle 119 3. Construct equilateral triangles BCY and CDX so that X and Y are in the interior of the rectangle.2 Construction of equilateral triangle inscribed in a rectangle What is more interesting is that the above calculation leads to a very easy construction of the equilateral triangle AXY . 1 Join AX to intersect BC at P and AY to intersect CD at Q. Construction 3. D Q C N Y X P A M B 1 This is not always possible. Show that Area(ABP ) + Area(ADQ) = Area(CP Q). Suppose AP = a. A pavement of width d is constructed across a rectangular field of dimensions a by b. 3.120 Equilateral triangle in a rectangle Exercise 1. P is a point in the interior of a rectangle ABCD. D Q C P A B 2. A D P B C 4. BP = b. CP = c. What is the area of the pavement? A a P B b d D Q C . b. Take a 9 × 10 rectangle ABCD and obtain the equilateral triangle AP Q by folding. c. Find the distance DP in terms of a. and the percentage of the paper wasted. Find the largest cube that can be obtained. and the width is 8 units? A .3. What is the length of the paper if the corner A is on the edge. A piece of 8 × 11 paper is to be cut into a pattern that can be folded into a cube. A 6.2 Construction of equilateral triangle inscribed in a rectangle 121 5. 122 Equilateral triangle in a rectangle . Chapter 4 Basic geometric constructions 1 2 3 4 Geometric mean Harmonic mean Equal subdivisions of a segment The Ford circles Appendix: Some basic construction principles Appendix: The Geometer’s Sketchpad Exercise . Then P Q2 = P A · P B . P . and A.124 Basic geometric constructions 4. B are on the same side of P . A. Q P A B Construction 4.1. Describe a semicircle with P B as diameter. Construction 4. Describe a semicircle with AB as diameter. Given two segments of length a < b.2. These are based on Euclid’s proof of the Pythagorean theorem. Q A P B . B on a line such that P A = a. Given two segments of length a. P B = b. P B = b.1 Geometric mean We present two ruler-and-compass constructions of the geometric means of two quantities given as lengths of segments. mark three points A. B on a line (P between A and B ) such that P A = a. mark three points P . and let the perpendicular through P intersect the semicircle at Q. b. and let the perpendicular through A intersect the semicircle at Q. so that the length of P Q is the geometric mean of a and b. Then P Q2 = P A · P B . so that the length of P Q is the geometric mean of a and b. PQ AB CD A b B K P Q harmonic mean D a C Another construction b harmonic mean a . and the line through K parallel to AB intersect AD and BC at P and Q respectively.2 Harmonic mean Let ABCD be a trapezoid with AB//CD.4.2 Harmonic mean 125 4. then P Q is the harmonic mean of AB and CD: 2 1 1 = + . If the diagonals AC and BD intersect at K . 1 For example. construct the segments Pj −1D and AQj . . Qm . 2i i = 1. Here is the case for n = 13 with binary representation 11012 . . . we relabel C as Q . . Let n < 2k+1 . Ck such that DCi = 1 DC. Along the direction CD. . and mark the projection of their intersection on AB as the point P j . D Q2 Q1 C = Q0 A P2 P1 B = P0 It would be interesting if this procedure can be modified to give a simple construction of subdivision points with ratio m : n instead of 1 : n. . which has binary representation 1101 .126 Basic geometric constructions 4. . Q1 . By repeated bisections of CD. Its binary representation has no more than k + 1 digits. . 2. relabel those points corresponding to the nonzero digits. 1 Let P0 be the (orthogonal) projection of Q0 on AB . For each j = 1. . . Suppose it has m + 1 nonzero binary digits. which contains not more than k + 1 digits. . We also put C0 = C . introduce the points C 1 . . if n = 13. k. making use of the binary representation of n. C as Q . . . as Q0 .3 Equal subdivisions of a segment Here is an algorithm to divide a given segment into n < 2 k+1 equal parts. Then Pm is the point which divides AB into n equal parts. Construct a square ABCD on the given segment. C2 . and 2 0 0 2 1 C2 as Q2 . m. . . Here are the Ford circles corresponding to F5 . 0).4.1. what is the point of tan1. Theorem 4. In the latter case we say that two rational numbers are adjacent.4 The Ford circles 127 4. 0 1 1 5 4 1 3 2 5 1 2 3 5 2 3 3 4 4 5 1 Exercise and P are adjacent rational numbers. there is a sequence of tangent Ford circles. Two rational numbers are adjacent if and only if they are consecutive terms in a Farey sequence. Corresponding to each Farey sequence. The Ford circles of two distinct rational numbers are either disjoint or tangent to each other externally. If p q Q gency of their Ford circles? .4 The Ford circles The Ford circle of a rational number r = p is the circle of radius 21 in q q2 the upper half plane tangent to the x-axis at the point (r. Theorem 4. the line joining their centers passes through the point of tangency. . A point P is equidistant from two given intersecting lines if and only if it lies on the bisector of an angle between the two lines.128 Basic geometric constructions Appendix: Some basic construction principles Theorem 4. The distance between their centers is the sum (respectively difference) of their radii if the tangency is external (respectively internal).2 (Perpendicular bisector locus). a point P is equidistant from A and B if and only if P lies on the perpendicular bisector of the segment AB . If two circles are tangent to each other.4. Theorem 4. Given two distinct A and B on a plane. Note that two intersecting lines have two angle bisectors.3 (Angle bisector locus). • Given a line and a point P . • Hide an object. choose Document options. construct lines perpendicular and parallel to through P . (If the sketch you are making becomes too complicated. On this blank page. • Translate. square. Change the name of the object. Add page. and type a name for the tool. Select Create New Tool. Remember to save the file. right triangle.gsp file contains the following tools: equilateral triangle. equilateral triangle. Blank page. • Create New Tool. . Construct a circle given its center and radius. parallelogram. From the File menu. say. and dilate. rectangle. • Construct a segment.4. rotate. Toolbox Open a new sketch. A new tool will now appear. Select Create new tool and name this square. circles. segments. construct a square on a segment.gsp tool by opening new blank pages and creating new tools. Save the file. • Construct a circle given its center and a point on it. lines. ray. Select a segment and build an equilateral triangle on it. • Construct the midpoint of a segment. • Construct the intersection of two objects. • Save a Sketch. You can extend this basic shapes. Save this as a file basic shapes. Select everything by dragging the mouse from top left to bottom right. The current basic shape. You may of course add your own tools to the same file. Note that this is not the same as “delete”). • Label an object. you may wish to hide some of the objects to make it simpler.gsp in the folder tool folder. Hide everything except the vertices and sides of the equilateral triangle. or line through two specified points.4 The Ford circles 129 The Geometer’s Sketchpad Basic commands • Construct points. rhombus. 2 a+b 4. B A C D 5. Construct the following diagram. Given triangle ABC . CAY and ABZ externally on the sides of the triangle. a+b √ 2ab ≥ ab ≥ . Make a sketch to show that for two given positive quantities a and b. BY . CZ . What can you say about the intersections. construct the equilateral triangles BCX . Show that the 90◦ angle of a right triangle is bisected by the line joining it to the center of the square on the hypotenuse. 3. and directions of these lines (segments)? 2. lengths.130 Basic geometric constructions Exercise 1. Construct the following diagram. Join AX . B A C D . and the √ line joining their centers. If the inradii of the quadrilateral and the isosceles triangle are √ √ equal. 8.4. an isosceles triangle. . Construct the circle. Consider the circle tangent to one of them internally. Construct the partition. An equilateral triangle of side 2a is partitioned symmetrically into a quadrilateral. the common inradius is ( 3 − 2)a. Outline a simple procedure to divide a segment into 123 equal parts. and two other congruent triangles.4 The Ford circles 131 6. 7. the other externally. It is known that this circle has radius 43 a. Two congruent circles of radii a have their centers on each other. . b) as an integer combination of a and b Nonnegative integer combinations of a and b Cassini formula for Fibonacci numbers gcd of generalized Fibonacci and Lucas numbers Appendix: The Euler φ-function Exercise Project k −1 0 1 2 3 .Chapter 5 Greatest common divisor 1 2 3 4 gcd(a. n−3 n−2 n−1 n rk qk Fn+1 ∗ ∗ ∗ Fn 1 Fn−1 1 Fn−2 1 Fn−3 1 F3 F2 F1 0 xk 1 0 F1 −F2 F3 yk 0 1 −F2 F3 −F4 1 (−1)n−2 Fn−3 (−1)n−1 Fn−2 1 (−1)n−1 Fn−2 (−1)n Fn−1 1 (−1)n Fn−1 (−1)n+1 Fn ∗∗∗ . . We form two sequences rk and qk as follows.1. −yn ) or (b + xn . Beginning with r −1 = a and r0 = b. These divisions eventually terminate when some r n divides rn−1 . b) as an integer combination of a and b It is well known that the gcd of two (positive) integers a and b can be calculated efficiently by repeated divisions. xn and yn are opposite in sign. Assume a > b. b) as an integer combination of a and b: 1 gcd(a. a − yn ) according as xn > 0 or < 0.2. Let p be a prime number. Clearly. k < a such that ak − bh = 1. Corollary 5. y 0 = 1. namely. For every integer a not divisible by p. there are unique integers h. there exists a positive integer b < p such that ab − 1 is divisible by p. In that case. along with these divisions. x0 = 0. rk ) := rk−1 − qk rk . . b) = rn = axn + byn . we introduce two more sequences (x k ) and (yk ) with the same rule but specific initial values. then we obtain gcd(a. Proof. let qk = rk − 1 . xk+1 =xk−1 − qk xk . Take (k. . If. for k ≥ 0. 1 In each of these steps. y −1 = 0. h) = (xn . rk rk+1 = mod(rk−1 . Theorem 5.202 Greatest common divisor 5. x−1 = 1. b) = rn . gcd(a. . n−1 n n+1 rk qk a ∗∗∗ a b b a− a b rb1 b rn − 1 rn 0 xk 1 0 x1 yk 0 1 y1 qn−1 xn−1 yn−1 qn xn yn It can be proved that |xn | < b and |yn | < a. Given relatively prime integers a > b.1 gcd(a. yk+1 =yk−1 − qk yk . k −1 0 1 . rk = axk + byk . (ii) Among the first 11 columns. . . is in S . n = a(b − u) + bv . Write ab − n = au − bv for 0 < u < b and 0 ≤ v < a. Then ab = a(x + 1) + b(y + 1). Note that a|b(y + 1). there is a unique entry (with asterisk) which is a multiple of 11. 0 ≤ v < a. b − 1} such that au ≡ t (mod b). . .. every positive integer < a + b is of the form au − bv . This entry with asterisk. y nonnegative integers}. This is clearly a contradiction. . This shows that n ∈ S .5.. (v) The rightmost entry with an asterisk is 66... Every integer greater than ab − a − b can be written as ax + by for nonnegative integers x and y . . being a positive multiple of 7. along each of the first 6 rows. . (iii) None of the entries on the left of an entry with asterisk is in S . for a contradiction.. Since gcd(a.. It follows that au − t = bv for some 1 ≤ v < a. From this. (Choose u ∈ {1.3. b) = 1.. (iv) The entries with asterisks are on different columns. we must have a|y + 1.2 Nonnegative integer combinations of a and b 203 5. Every integer t in the range 0 < t < a can be written as t = au − bv for 0 < u < b and 0 ≤ v < a. . Theorem 5. x.. 0 < u < b... From this. y nonnegative integers}.. . the largest integer not in S is 66 − 7 = 59. Then ab − n < a + b.. But y + 1 is a positive integer smaller than a.. y ≥ 0. Let S := {7x + 11y : x. Observations: (i) Every number in the bottom row. y ≥ 0. Let S := {ax + by : x. Then 0 < au − t < ab. Thus. Arrange the positive integers in the form 1 2 3 4 5 6 7 8 9 10 11∗ 12 13 14 15 16 17 18 19 20 15 22∗ 23 24 25 26 27 28 29 30 31 32 33∗ 34 35 36 37 38 39 40 41 42 43 44∗ 45 46 47 48 49 50 51 52 53 54 55∗ 56 57 58 59 60 61 62 63 64 65 66∗ 67 68 69 70 71 72 73 74 75 76 77 . Proof.2 Nonnegative integer combinations of a and b Find the largest positive integer which cannot as 7x + 11y for integers x.. Suppose (a − 1)(b − 1) ≤ n < ab. ab − a − b = ax + by . Suppose. Let a and b be given relatively prime positive integers. . . From this ab − a − b ∈ / S. and those on its right along the row. are in S . 2. following the euclidean algorithm. . n−3 n−2 n−1 n rk qk Fn+1 ∗ ∗ ∗ Fn 1 Fn−1 1 Fn−2 1 Fn−3 1 F3 F2 F1 0 xk 1 0 F1 −F2 F3 yk 0 1 −F2 F3 −F4 F0 = 0. . . . Fn−1 ) = gcd(Fn−1 . 1 (−1)n−2 Fn−3 (−1)n−1 Fn−2 1 (−1)n−1 Fn−2 (−1)n Fn−1 1 (−1)n Fn−1 (−1)n+1 Fn ∗∗∗ . . = gcd(F2 .204 Greatest common divisor 5.3 Cassini formula for Fibonacci numbers The Fibonacci numbers Fn are defined recursively by Fn+1 = Fn + Fn−1 . Fn+1 Fn−1 − Fn k −1 0 1 2 3 . Fn−2 ) . F1 ) = 1. . However. It is easy to see that gcd(Fn+1 . The first few Fibonacci numbers are n 0 1 2 3 4 5 6 7 8 9 10 11 12 . . F1 = 1. . Fn ) = gcd(Fn . Fn 0 1 1 2 3 5 8 13 21 34 55 89 144 . we obtain the Cassini formula 2 = (−1)n . b) Lucas Fibonacci Lucas periodic periodic natural 2 constant n Mersenne 2 + 1 Fermat repunits 10k−11 Ln Ln (a.4 gcd of generalized Fibonacci and Lucas numbers Given a and b. b) Ln+1 = aLn + bFn−1 . F0 = 0.n) .4 gcd of generalized Fibonacci and Lucas numbers 205 5. un ) = ugcd(m. −2) 2 − 1 (11. gcd(1m . Examples (a. L1 = a. gcd(Fm . Theorem 5. b) Fn+1 = aFn + bFn−1 .n) . 1) Fn (1.b) 1. −10) 1n Corollary 5.4. L0 = 2. −1) n n (3. .n) − 1. Let a and b be relatively prime integers. gcd(2m − 1. 3.n) . 2n − 1) = 2gcd(m. Fn ) = Fgcd(m. (a.5. The associated Fibonacci and Lucas sequences satisfy gcd(um . 2. the generalized Fibonacci sequence F (a. −1) (2. b) Fn (1.5. F1 = 1. There is also an accompanying Lucas sequence L(a. 1n ) = 1gcd(m. and vice versa. Find the immediate neighbors of 13 31 in F31 . Express the gcd as an integer combination of the given numbers by completing the last two columns. Later. the teller made a mistake and paid him the amount which was written as cents. calling for a certain amount of money in dollars and cents. in dollars. When he went to cash the check. he suddenly realized that he had twice the amount of the money the check called for.50. 1 2 3. Somebody received a check. Find the immediate neighbors of the fraction quence Fn . rk 54321 12345 qk ∗∗∗ xk 1 0 yk 0 1 rk 267914296 196418 qk ∗∗∗ xk 1 0 yk 0 1 2. after spending $ 3. Find the gcd of the following pairs of numbers by completing the second column of each table. in the Farey se- 4. What was the amount on the check ? .206 Greatest common divisor Exercise 1. 2 2 3 9 + 2 7 3 9 =3 × + 2 7 3 9 =3 × × . 2 7 3+ 3 9 81 + + 2 7 67 3 9 81 =3 × + + 2 7 67 3 9 81 =3 × × + 2 7 67 3 9 81 =3 × × × 2 7 67 3+ 3 9 81 6561 + + + 2 7 67 5623 3 9 81 6561 + =3 × + + 2 7 67 5623 3 9 81 6561 + =3 × × + 2 7 67 5623 3 9 81 6561 + =3 × × × 2 7 67 5623 3 9 81 6561 × =3 × × × 2 7 67 5623 3+ .5.4 gcd of generalized Fibonacci and Lucas numbers 207 Project Generalize 3+ 3 3 =3× . then a 2 between them. How many n’s are there in the sequence? . then a 4 between any two numbers whose sum is 4. then a 3 between any two numbers whose sum is 3. and so forth.208 Greatest common divisor Project Write down two 1’s. Chapter 6 Pythagorean triples 1 2 3 4 5 Primitive Pythagorean triples Primitive Pythagorean triples with square perimeter Lewis Carroll’s conjecture on infinity of triples of equiareal Pythagorean triangles Points at integer distances from the sides of a Pythagorean triangle Dissecting a rectangle into Pythagorean triangles Appendix: Primitive Pythagorean triples < 1000 Appendix: Dissection of a square Exercise Project: Factorable x2 + px ± q Project: 64 primitive Pythagorean triangles with a common hypotenuse Project: Primitive Pythagorean triangles with equal perimeters Project: Two pairs of primitive Pythagorean triples with almost equal perimeters Project: Cross number puzzle on Pythagorean triples . c = m2 + n2 for relatively prime integers m and n of different parity. b = 2mn. The area is divisible by 6. 4. Indeed. Fermat has proved that the area of a Pythagorean triangle can never be a square. c) is of the form a = m2 − n2 . 3. there is no Pythagorean triangle with two sides whose lengths are square (numbers). b.1 Primitive Pythagorean triples It is well known that every primitive Pythagorean triple (a. . Exactly one leg is divisible by 3. B m 2 + n 2 m2 − n 2 A 2mn C Some basic properties of Pythagorean triples: 1. Exactly one leg is even. Exactly one side is divisible by 5. 2.210 Pythagorean triples 6. q ) = 1. we must have m = 2q 2 and m + n = p2 for some integers p and q . they generate a primitive Pythagorean triple (a. p q m n a b 3 2 8 1 63 16 5 3 18 7 275 252 7 4 32 17 735 1088 9 5 50 31 1539 3100 11 6 72 49 2783 7056 11 7 98 23 9075 4508 13 7 98 71 4563 13916 13 8 128 41 14703 10496 15 8 128 97 6975 24832 13 9 162 7 26195 2268 17 9 162 127 10115 41148 17 10 200 89 32079 35600 19 10 200 161 14079 64400 . p2 − 2q 2 ).2 Primitive Pythagorean triangles with square perimeters 211 6. c =m2 + n2 = p4 − 4p2 q 2 + 8q 4 . From these.6. . b. and gcd(p. If this perimeter is a square (number). a =m2 − n2 = p2 (4q 2 − p2 ). (m. b =2mn = 4q 2 (p2 − 2q 2 ). c 65 373 1313 3461 7585 10133 14645 18065 25793 26293 42373 47921 65921 2pq 12 30 56 90 132 154 182 208 240 234 306 340 380 . √ 2q ≤ p < 2q . n) = (2q 2 . . . Here are the first few of such triangles. m2 + n2 ) with perimeter p = 2m(m + n). 2mn. . The perimeNote that p is odd.2 Primitive Pythagorean triangles with square perimeters If m > n are relatively prime integers of opposite parity. c) = (m2 − n2 . The last column gives the square root of the perimeter. . ter is 4p2 q 2 = (2pq )2. r ) satisfy p2 + pq + q 2 = r 2 . r ) have the same area pqr (p + q ). then the Pythagorean triangles generated by (r. Since there is an infinity of such triangles. This is essentially in Diophantus. there conjecture is established. q ). q. (r. .212 Pythagorean triples 6. and (p + q. p).3 Lewis Carroll’s conjecture on triples of equiareal Pythagorean triangles Lewis Carroll conjectured that there is an infinity of Pythagorean triangles with equal areas. If (p. 1).2) we have the triangle (21.1) we have the Pythagorean triangle (35. (6. (1.11). b. 8. 6. (5. 12 5 4 3 2 1 0 35 Another example: with paramters (5. with vertices (a. (0. 3. 4. (3.13). (4.12.6). The arrangement is not as regular as the previous example. 4.4 Points at integer distances from the sides of a primitive Pythagorean triangle 213 6.20. 20 3 1 8 6 11 4 2 0 21 .1) at distances 1. Here the five points (29.37). and (0. (11. With the parameters (6. 5 from the hypotenuse. b). We c seek interior points which are at integer distances from the hypotenuse. (17. The distance of an interior point (x.2).29). (16. 0). 3. y ) to the hypotenuse is 1 (ab − bx − ay ). 11 from the hypotenuse.5) are at distances 1. 0). c) be a primitive Pythagorean triangle.2). The hypotenuse is the line bx + ay = ab. Here we have the interior points (8. 2. (11.6.4 Points at integer distances from the sides of a primitive Pythagorean triangle Let (a.3).4). (23.4).8). 2. If we put CP = a2 . Indeed. b. we want two Pythagorean triangles with the same hypotenuse. AQD and AP Q are similar. the three right triangles QP C . 65 2 = 632 + 162 = 562 + 332 . then the lengths of the various segments are as shown below. (ii) the smallest such rectangle with AP = AQ. Here.214 Pythagorean triples 6. 5). This also answers (i) since it is clearly also the smallest rectangle (with any restrictions). From this it is easy to complete the data. Note that ABP now is a Pythagorean triangle with parameters b and a. With (a. (iv) the smallest square. D ab Q ab ac a2 b2 bc C 2 2 +b 2 =a c P b2 − a2 B A 2ab (ii) is also tractable without the help of a computer. . The answer to (iv) is given in the Appendix.5 Dissecting a rectangle into Pythagorean triangles How many matches (of equal lengths) are required to make up the following figure? D x Q a−x w C b−y P b A v u a y B This is Problem 2237 of the Journal of Recreational Mathematics. c) = (3. It is well known that the smallest such hypotenuse is 65. which asks for the (i) the smallest such rectangle. 4. we obtain the smallest rectangle satisfying (iv). If they are similar to the Pythagorean triangle a : b : c. In (iii). the ratios of similarity are also a : b : c. (iii) the smallest such rectangle with AP Q Pythagorean. c 15. 13 21. 829 775. 17 19. 13 22. 504. 865 825. 305 31. 52. 16. 445 345. 901 925. 269 171. 2 17. 13 25. 85 91. 2 13. 9 20. 5 15. 6 8. 28. n 4. 17 21. 397 165. 1 7. 780. 15 23. 125 21. 457 41. 16 22. 541 475. 265 135. 864. 925 465. 997 m. 120. 12 14. 1 13. 19 25. 557 391. 785 551. 409 231. 308. 13 29. 16 26. 10 18. 6 23. 7 26. 352. 3 14. 461 37. 673 129. 264. 325 93. 540. 6 14. 12 18. 617 351. 15 17. 265 105. c 3. 257 175. 19 23. 6 17. 468. 60. 8 30. 865 589. 449 111. 2 7. 101 117. 485 403. 7 19. 420. 12 31. 24. 785 663. 137 143. 10 19. 4 23. 157 119. 2 19. 4 29. 900. 8 14. 208. 601 287. 185 95. 509 259. 2 23. 421 207. 14 26. 7 24. 36. 612. 877 899. 20 24. 7 22. 124. 5 22.5 Dissecting a rectangle into Pythagorean triangles 215 Appendix: Primitive Pythagorean triples < 1000 m. 221 23. 793 615. 84. 8. 252. 857 697. 8 17. 840. 232. 7 a. 12 20. 152. 144. 593 273. 709 525. 580. 14 16. 48. 120. 845 493. 9 15. 396. 193 133. 11 25. 8 25. 684. 420. 641 429. 65 77. 17 27. 8 22. 3 11. 340. 228. 2 11. 2 27. 1 18. 521 437. 701 451. 280. 2 29. 5 18. 14 18. 281 87. 220. 11 29. 481 225. 7 10. 312. 629 205. 725 387. 677 595. 440. 156. 8 12. 132. 10 28. 7 16. 104. 456. 532. 360. 965 629. 9 11. b. n 3. 1 31. 881 651. 41 33. 85 15. 60. 60. 9 22. 661 369. 5 16. 377 273. 760. 18 24. 1 24. 112. 7 11. 13 15. 16 28. 380. 600. 216. 221 221. 84. 17 23. 368. 204. 797 713. 697 621. 308. 348. 884. 260. 4 7. 205 27. 6 12. 8 16. 13 9. 92. 901 665. 241 247. 4 12. 416. 252. 1 11. 4 a. 277 209. 160. 761 341. 569 425. 15 27. 528. 28. 644. 6 30. 96. 949 m. 40. 289 231. 224. 608. 11 15. b. 5 24. 145 165. 2 10. 1 26. b. 2 a. 56. 12 23. 728. 276. 533 195. 169 153. 836. 4 17. 680. 11 21. 353 33. 505 399. n 2. 20. c 7. 5 9. 1 28. 5 29. 493 325. 6 19. 565 333. 16 18. 181 57. 16 20. 56. 757 725. 140. 10 12. 4 19. 248. 11 27. 365 29. 168. 52. 4 10. 985 703. 700. 4 16. 756. 3 16. 24. 180. 17 25. 425 105. 2 8. 72. 460. 7 20. 3 6. 400. 88. 505 483. 168. 13 19. 912. 10 23. 120. 12. 337 285. b. 336. 36. 97 51. 468. 8 19. 3 28. 9 29. 76. 280. 748. 624. 1 5. 4 25. 493 315. 229 255. 941 957. 480. 3 9. 4. 364. 925 759. 317 161. 25 11. 365 261. 32. 349 35. 288. 7 13. 11 17. 14 24. 4 8. 8 28. 3 22. 5 21. 5 13. 53 55. 396. 745 609. 545 203. 68. 220. 433 299. 136. 672. 325 155. 16 24. 109 105. 853 527. 920. 184. 149 85. 420. 696. c 5. 464. 48. 10 14. 305 145. 140. 733 533. 168. 173 69. 821 667. 11 13. 953 777. 1 14. 905 805. 61 63. 197 75. 304. 565 123. 18 20. 929 455. 44. 20. 364. 3 26.6. 240. 29 45. 736. 401 279. 965 . 577 407. 520. 689 377. 9 27. 313 115. 572. 2 5. 613 297. 80. 100. 73 65. 985 945. 1 22. 336. 1 9. n 4. 10 31. 924. 445 185. 949 627. 132. 14 20. 372. 689 301. 19 21. 180. 905 851. 4 21. 616. 377 217. 185 25. 272. 10 22. 84. 1 6. 816. 572. 2 16. 629 481. 745 473. 240. 233 187. 769 675. 2 21. 18 26. 13 17. 793 783. 552. 116. 8 10. 481 39. 937 561. 725 575. 1 16. 6 25. 37 13. 12 26. 389 323. 89 17. 373 357. 293 189. 545 385. 21 23. 300. 9 17. 841 435. 108. 4 27. 4 13. 697 459. 205 195. 828. 533 429. 485 275. 44. 113 99. 408. 476. 5 26. 653 43. 773 513. 176. 2 25. 156. 685 555. 145 19. 800. 5 8. 660. 40. 20 22. 60. 8 23. 260. 809 837. 1 20. 544. 3 20. 625 215. 685 319. 425 253. 65 39. 845 741. 432. 17 35. 6 a. 11 19. 977 m. 14 28. 200. 12. 216 Pythagorean triples Appendix: Dissection of a square 255 136 289 375 424 224 360 105 360 . P .5 Dissecting a rectangle into Pythagorean triangles 217 Exercise 1. If the distance between P and Q is 91 feet. What is the least number of matches of equal lengths to make up the following configuration? 3. Do there exist Pythagorean triangles whose sides are Fibonacci numbers? 5. Q. A man has a square field. and that from P to C is an exact number of feet. at A. Give an example of a primitive Pythagorean triangle in which the even leg is a square. What is the least number of matches of equal lengths to make up the following configuration? 4. Give an example of a primitive Pythagorean triangle in which the hypotenuse is a square. . with other property adjoining the highway. He put up a straight fence in the line of 3 trees. 6. what is this distance? A 60 B P 60 ? D C 91 Q 2.6. 60 feet by 60 feet. Find an integer-sided right angled triangle with sides x 2 − 1. triangle to be a square? 8. where x. 12. z are integers. how many Pythagorean triangles are there such that the area is n times the perimeter ? How many of these are primitive ? . 9. Give an example of a primitive Pythagorean triangle in which the odd leg is a square. Find the shortest perimeter common to two different primitive Pythagorean triangles.218 Pythagorean triples 7. Show that there are an infinite number of Pythagorean triangles whose hypotenuse is an integer of the form 3333 · · · 3. The number of primitive Pythagorean triangle with a fixed inradius is always a power of 2. y 2 − 1 z 2 − 1. For each natural number n. 11. 10. y . 1. The polynomials x2 + px ± q are both factorable (over integers) if and only if |p| and |q | are respectively the hypotenuse and area of a primitive Pythagorean triangle.6. Let p and q be relatively prime integers.5 Dissecting a rectangle into Pythagorean triangles 219 Project: Factorable x2 + px ± q Theorem 6. . What are the 64 triples? . and 53. 17. 37. 29. 13. 41. Their product is 2576450045. The first 7 primes of the form 4k + 1 are 5. repeatedly by making use of the identity 2 2 2 2 2 (x2 1 + y1 )(x2 + y2 ) = (x1 x2 − y1 y2 ) + (x1 y2 + x2 y1 ) .220 Pythagorean triples Project: 64 primitive Pythagorean triangles with a common hypotenuse A product of n + 1 distint prime numbers of the form 4k + 1 can be written a sum of two squares in 2n different ways. 5 Dissecting a rectangle into Pythagorean triangles 221 Project: Primitive Pythagorean triangles with equal perimeters perimeter 1716 14280 m+n 33 39 85 105 119 407 429 481 555 899 957 1023 1131 1209 1785 1955 1995 2185 2261 2415 3135 3315 3553 3705 3927 4199 4389 2m m n a b c 317460 1542684 6240360 19399380 .6. B = m2 − n2 . Can you find at least one such set ? perimeter 117390 117392 313038 313040 339150 m + n 2m m 273 301 253 319 459 527 455 559 425 475 525 451 517 469 603 437 475 575 n a b c 339152 371448 371450 . not both odd. The four students produced four entirely different primitive triangles. This interested the class greatly. while the other two also had the same perimeter. C = m2 + n2 . this perimeter differing from the first one by 2.222 Pythagorean triples Project: Two pairs of primitive Pythagorean triples with almost equal perimeters In a class in Number Theory the professor gave four students the assignment of finding a fairly large primitive Pythagorean triangle using the well known formula for the legs: A = 2mn. but on comparing them it was found that two of them had the same perimeter.000. only to discover that there were only four such sets with perimeters less than 500. and much time was spent in an effort to find other such sets. where m and n are coprime integers. 9U. in increasing size. 12 (1986) 139. 21D 29B. 2D. 11U is a 3-digit number with its tens and units digits in squares 16 and 11 respectively. 21A. 1B has its tens and units digits in the squares labelled 2 and 1 respectively. 3U 4U. 34A. 15D. B = back. A = across. 10A 12B. 11U. 7D 20A. 7A. 35A. 33A The answers are distinct 2. For example. 25B 28D. Guy. 3D. K. 23B 22A. 17D. 6D. 3A. 15U 16A. . D = down.6.. 26D 5D. so that the third member is the hypotenuse. 9A 30U. U = up.and 3-digit decimal numbers. 9B 2D. 8A 30D. Problem 1153. 21D 19U. 8D. 13D 19U.5 Dissecting a rectangle into Pythagorean triangles 223 Project: Cross number puzzle on primitive Pythagorean triples 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 1B. 14A. 18B. 31A. Each of the above sets of answers is a primitive Pythagorean triple. 20U 22D. none beginning with zero. 32U. 24D. Crux Math. 32D 32U. 33A 16B. 5B 27A. 1 R. 224 Pythagorean triples . Chapter 7 The tangrams 1 The Chinese tangram 2 A British tangram 3 Another British tangram Project . cˇ ı q¯ ı qiˇ ao zh¯ ı huˇ ajˇ ıng yˇ e! (what a sublimity of the tangram!) 1 F. 2 pentagons. are 1 triangle. I finally know what the mysterious Chinese character on p. to drink.1 The Chinese tangram It is known that exactly 13 convex polygons can be formed from the tangram. The Chinese inscription reads Liˇ ang r´ en du` ı zhu´ o (Two people having a drink together). 2 A very interesting book on the Chinese tangram has recently appeared: J. . and 4 hexagons. Sterling. American Math. 49 (1942) 596–599. The Tangram Book. Wang and C.226 The tangrams 7. 6 quadrilaterals.45 of Dudeney’s Amusements in Mathematics is. The undeciphered Chinese character is zhu´ o. Monthly. 2003. C. 1 Show them. Slocum. Hsiung. 2 There T. A theorem on the tangram. New York. . 3 3 Singmaster. All sloping lines are at 45◦ .2 A British tangram The seven pieces of the puzzle can be fitted together as shown in the figure to make a perfect square. Find all ways of making a rectangle using all of these pieces.. 6 (1973) 152–153. Problem 267.7. and the lines which intersect the outer square do so at the midpoints of the sides.2 A British tangram 227 7. Journal of Recreational Math. .. All sloping lines are at 45◦ . 4 (1) Are there any solution with the green square in a corner or not touching a side? (2) Prove of disprove: any solution has the two gray trapezoids and the green square in the above relative positions. 4 Singmaster. Journal of Recreational Math.3 Another British tangram Ten pieces of the puzzle can be fitted together as shown in the figure to make 4 × 5 rectangle. Problem 812. 13 (1980–1981) 62–63. (3) Find all solutions.228 The tangrams 7. 7.3 Another British tangram 229 Project Show how to dissect a 3-4-5 triangle into 4 pieces that may be rearranged into a square. 5 5 S. Rabinowitz, Problem 1299, Journal Recreational Math., 16 (1983–1984) 139. 230 The tangrams Chapter 8 The classical triangle centers 1 2 3 4 5 The centroid The cirumcenter and the circumcircle The incenter and the incircle The orthocenter and the Euler line The excenters and the excircles Exercise 232 The classical triangle centers The following triangle centers have been known since ancient times. We shall adopt the following notations. Let ABC be a given triangle. The lengths of the sides BC , CA, AB opposite to A, B , C are denoted by a, b, c. 8.1 The centroid The centroid G is the intersection of the three medians. It divides each median in the ratio 2 : 1. A F G E B D C The triangle DEF is called the medial triangle of ABC . It is the ). image of ABC under the homothety h(G, − 1 2 The lengths of the medians are given by Apollonius’ theorem: 1 2 2 2 m2 a = (2b + 2c − a ), 4 etc. Exercise Calculate the lengths of the medians of a triangle whose sidelengths are 136, 170, and 174. 8.2 The circumcircle and the circumcircle 233 8.2 The circumcircle and the circumcircle The perpendicular bisectors of the three sides of a triangle are concurrent at the circumcenter of the triangle. This is the center of the circumcircle, the circle passing through the three vertices of the triangle. A A F O E O B D C B D C Theorem 8.1 (The law of sines). Let R denote the circumradius of a triangle ABC with sides a, b, c opposite to the angles A, B , C respectively. b c a = = = 2R. sin A sin B sin C Since the area of a triangle is given by dius can be written as abc R= . 4 bc sin A, the circumra= 1 2 234 The classical triangle centers 8.3 The incenter and the incircle The internal angle bisectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. If the incircle touches the sides BC , CA and AB respectively at X , Y , and Z , AY = AZ = s − a, BX = BZ = s − b, A CX = CY = s − c. s−a s−a Y Z I s−c s−b B C s−b X s−c Denote by r the inradius of the triangle ABC . r= 2 = . a+b+c s 8.4 The orthocenter and the Euler line 235 8.4 The orthocenter and the Euler line The orthocenter H is the intersection of the three altitudes of triangle ABC . These altitudes can be seen as the perpendicular bisectors of the antimedial triangle XY Z of ABC , which is bounded by the three lines each passing through A, B , C parallel to their respective opposite sides. A Z Y O H G B C X XY Z is the image of triangle ABC under the homothety h(G, −2). It follows that H is the image of O under the same homothety. We conclude that O , G, and H are collinear, and OG : GH = 1 : 2. The line containing O , G, H is the famous Euler line of triangle ABC . 236 The classical triangle centers 8.5 The excenters and the excircles The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. An excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle. Ib A Ic X B C Y Z Ia The exradii of a triangle with sides a, b, c are given by ra = s−a , rb = s−b , rc = s−c . 8.5 The excenters and the excircles 237 Exercise 1. Given a triangle ABC , construct a triangle whose sides have the same lengths as the medians of ABC . 2. Construct the incircle of triangle ABC , and mark the points of contact X on BC , Y on CA, and Z on AB . Are the lines AX , BY , CZ concurrent? If so, is their intersection the incenter of triangle ABC ? 3. Given a rectangle ABCD, construct points P on BC and Q on CD such that the triangle AP Q is equilateral. 4. Let D , E , F be the midpoints of BC , CA, AB of triangle ABC . Construct the circumcircle of DEF . This is called the nine-point circle of triangle ABC . Construct also the incircle of triangle ABC . What do you observe about the two circles? How would you justify your observation? 5. Construct the circle through the excenters of triangle ABC . How is its center related to the circumcenter and incenter of triangle ABC ? 6. Given three non-collinear points as centers, construct three circles mutually tangent to each other externally. Chapter 9 The area of a triangle 1 2 3 Heron’s formula for the area of a triangle Heron triangles Heron triangles with consecutive sides Appendix: Heron triangles with sides < 1000 Appendix: A famous unsolved problem Project: Triangles whose sides and one altitude are in arithmetic progression 13 12 15 14 302 The area of a triangle 9.1 Heron’s formula for the area of a triangle Theorem 9.1. The area of a triangle of sidelengths a, b, c is given by = (a + b + c ). where s = 1 2 s(s − a)(s − b)(s − c), B Ia ra I r Y s−b C s−c Y s−a A Proof. Consider the incircle and the excircle on the opposite side of A. From the similarity of triangles AIZ and AI Z , r s−a . = ra s From the similarity of triangles CIY and I CY , r · ra = (s − b)(s − c). From these, (s − a)(s − b)(s − c) , s and the area of the triangle is r= = rs = s(s − a)(s − b)(s − c). 9.2 Heron triangles 303 9.2 Heron triangles A Heron triangle is one whose sidelengths and area are both integers. It can be constructed by joining two integer right triangles along a common leg. For example, by joining the two Pythagorean triangles (9, 12, 15) and (5, 12, 13), we obtain the Heron triangle (13, 14, 15) with area 84. 13 12 15 5 9 Some properties of Heron triangles 1. The semiperimeter is an integer. 2. The area is always a multiple of 6. Exercise 1. Construct four Heron triangles by joining suitable multiples of (3,4,5) and (5,12,13) along common legs. The Heron triangles you obtain should be primitive, i.e., the sidelengths of each should be relatively prime. 2. Can the Heron triangle (25,34,39,420) be obtained by joining two Pythagorean triangles along a common leg? 304 The area of a triangle 9.3 Heron triangles with consecutive sides If (b − 1, b, b + 1, ) is a Heron triangle, then b must be an even integer. We write b = 2m. Then s = 3m, and 2 = 3m2 (m − 1)(m + 1). This requires m2 − 1 = 3k 2 for an integer k , and = 3km. The solutions of m2 − 3k 2 = 1 can be arranged in a sequence mn+1 kn+1 = 2 3 1 2 mn , kn m1 k1 = 2 . 1 From these, we obtain the side lengths and the area. The middle sides form a sequence (bn ) given by bn+2 = 4bn+1 − bn , b0 = 2, b1 = 4. The areas of the triangles form a sequence n+2 = 14 n+1 − Tn 0 6 84 n, T0 = 0, T1 = 6. Heron triangle (1, 2, 3, 0) (3, 4, 5, 6) (13, 14, 15, 84) n 0 1 2 3 4 5 Exercise bn 2 4 14 1. There is a uniquely Heron triangle with sides (b − 1, b, b + 1) in which b is an 8-digit number. What is the area of this Heron triangle? 2280) (60. 97. 29. ) (3. 1386) (34. 288) (29. 630) (14. 51. 420) (65. 51. 30. 35. 37. 37. 132) (19. 456) (51. 65. 56. 1386) (65. 70. 65. 75. 61. 65. 41. 29.3 Heron triangles with consecutive sides 305 Appendix: Heron triangles with sides < 100 (a. 52. 62. 25. 37. ) (5. 85. 660) (22. 28. 1680) (38. 336) (20. 68. 60. 12. 51. 924) (41. 87. 20. 252) (19. 73. 420) (48. 264) (21. 390) (41. 1020) (25. 85. 51. 2640) (57. 744) (37. 52. 85. b. 58. 21. 420) (17. 73. 39. 720) (28. 55. 60. 86. 264) (36. 12) (9. 17. 35. 72) (15. 40. 85. 40. 168) (9. 60) (15. 2340) (11. 82. 41. 34. 75. 85. 68. 85. 78. 198) (32. 1800) (17. 210) (25. 61. 25. 360) (24. 396) . 87. 96. 30) (8. 330) (39. 61. 8. 720) (40. 52. 15. 65. 89. 41. 1848) (57. 65. 420) (29. 61. 52. 70. 65. 55. 40. 528) (17. 75. 42. 2394) (39. 1656) (73. 80. 930) (41. 24) (11. 36. 34. 50. 420) (78. 68. 52. 156) (5. 75. 546) (52. 48. 51. 92. 624) (26. 408) (16. ) (5. 96. 72. 50. 17. 39. 15. 73. 75. 75. 37. 114) (16. 20. 75. 39. 61. 63. 3420) (a. 74. 924) (9. 65. 41. 13. 85. 95. 51. 13. 37. 61. 1080) (21. 69.9. 1020) (25. 28. 45. 35. 210) (5. 13. 84) (14. 120) (13. 924) (39. 780) (27. 65. 25. 330) (12. 37. 58. 40. 660) (41. 91. 21. 5. 82. 95. 24. 3120) (26. 36) (17. 85. 91. 25. 420) (29. c. 420) (15. 26. 92. c. 1260) (29. 120) (25. 75. 63. 91. 20. 840) (75. 1260) (22. 60. 660) (33. 13. 252) (12. 65. 52. 360) (25. 2310) (35. 420) (15. 84) (7. 68. 74. 41. 87. 24. 73. 25. 936) (41. c. 53. 73. 168) (6. 420) (78. 1290) (37. 97. 14. 2220) (28. 55. 26. 75. 522) (41. 1680) (35. 630) (25. 840) (34. 180) (17. 1140) (53. 90) (17. 33. 60) (17. 336) (13. 68. 39. 1116) (26. 76. 44. 65. 28. 87. 924) (35. 5. 41. 89. 1008) (7. 84. 30. 240) (18. ) (5. 41. 52. 504) (43. 91. 95. 35. 85. 75. 420) (25. 77. 89. 26. 61. 17. 72. 89. 75. 1656) (32. 48. 65. 73. 95. 1560) (65. 61. 2184) (39. 90) (53. 53. 29. 55. 97. 12) (13. 38. 29. 270) (25. 37. 462) (33. 6) (4. 35. 35. c. 84) (13. 51. 65. 77. 97. 204) (11. 1320) (35. 87. 55. 3096) (a. 61. 53. 690) (35. 65. 360) (33. 17. 53. 84. 66. 120) (13. 58. 29. 51. 21. 52. 51. 60. 97. 73. b. 1260) (a. 17. 168) (20. 88. 60) (13. 1092) (44. 74. 55. 41. 97. 966) (68. 35. 210) (8. 156) (17. 85. 87. 52. 396) (61. 39. 51. 80. 52. 73. 840) (25. 89. 37. b. 77. 74. 126) (51. 53. 80. 87. 15. 52. 40. 45. 66) (12. 25. 25. c. 61. 87. 41. 51. 66. 756) (48. 51. 1260) (11. 90. 65. 2016) (41. 65. 924) (39. 25. 53. 89. 25. 420) (13. 4. 984) (50. 40. 30. 61. 546) (52. 42) (7. 210) (10. 468) (29. 53. 25. 2772) (31. 456) (29. 55. b. 29. 69. 456) (15. 44. 34. 73. 53. 58. 65. 15. 630) (29. 462) (13. 90. 91. 17. 85. 40. 68. 5. 65. 1584) (a. 77. 1710) (26. 85. 180) (25. 924) (29. 840) (37. ) (10. 89. 41. 65. 37. 51. 29. 216) (39. 6. 25. 1680) (68. 39. 60. 126) (3. 756) (56. 1170) (33. 84) (20. 65. b. 660) (13. 80. 1320) (72. 798) (65. 85. 13. 234) (4. 73. 82. 44. 52. 1980) (21. 300) (34. 252) (26. 30. 29. 126) (15. 2850) (26. 504) (25. 75. 56. 546) (53. 50. 39. 93. 97. 20. 50. 88. 330) (24. 2640) (44. 89. 97. 85. 87. 65. 17. 91. 63. 82. 306) (37. 53. 52. 336) (20. 10. 56. 37. 36) (10. 1560) (17. 39. 60) (16. 420) (25. 90. 73. 13. 68. 210) (9. 66. 1092) (36. 131. See [Guy. and c = 146. Verify that this triangle is Heron and find the lengths of the medians. . 174 have medians 158. b = 102. It has an area . 170. 127.306 The area of a triangle Appendix: A famous unsolved problem Find a Heron triangle whose medians have integer lengths. But it is not a Heron triangle. Here is the first one. The triangle with sides 136. Let a = 52. Buchholz and Rathbun have found an infinite set of Heron triangles with two integer medians. This is a famous unsolved problem. Problem D21]. P. 2. 1 + 2x. 15 with altitude 12 on the side 14. . the sides of the triangles are 1 − x. . he obtained 16x4 − 80ax3 + 144a2 x2 − 10a3 x + 25a4 = 0. by the Heron formula. and x > 0 for the common difference. equations = · 1 · (1 + kx) for k = 1. Newton did not solve this equation nor did he give any numerical example. Setting AC = a. These lead to the . 1+ x. and AB = 2a − x. and the altitude on the shortest side is 1. and the altitude on the longest side to be The angles of the triangles are . we realize that these lengths can be in A. . .3 Heron triangles with consecutive sides 307 Project: Triangles whose sides and one altitude are in arithmetic progression This is an extension of Problem 29 of Isaac Newton’s Lectures on Algebra ([Whiteside.9. DC = 2x − a. we may assume the sides of the triangles to be . pp. (B) Recalling the Heron triangle with sides 13. . 3. In (1). (†) can be rewritten as √ (3 − 3 2). (A) Newton considered a triangle ABC with an altitude DC . Assuming unit length for this altitude. Note that the altitude in question is either the first or the second terms of the A. or 2. and 1 + 3x. (2x − 3a)3 + 2a3 = 0. 14. and 1+2x. DC is shorter than AC and BC . the other two roots being complex.234 – 237]). the three sides of the triangles are 1 + x. By taking so that x = a 2 a = 2. Actually. (†) “Divide this equation by 2x − a and there will result 8x 3 − 36ax2 + 54a2 x − 25a3 = 0”. we have either 1. is given by 2 = 1 2 3 (1 + 2x)2 (1 + 4x). Clearly. 16 On the other hand. in some other order. (in increasing order). BC = x.P. the area of the triangle. we seek one such triangle with sides 15. we obtain the cubic equation z 3 − 120z + 16 = 0. There are altogether five similarity classes of triangles whose three sides and one altitude. By considering the area of the triangle in two different ways. • for k = 2: 48x3 + 44x2 + 8x − 1 = 0. . . Now we consider (2). . in some order. . that the polynomial 48x3 +56x2 +16x − 1 has exactly one real root. the polynomial factors as so that we have x = . .P. 14. it is easy to see. we reduce this to sin 3θ = so that the positive roots of (∗) are the two numbers z= . (∗) √ This can be solved by writing z = 4 10 sin θ for an angle θ. Using the trigonometric identity sin 3θ = 3 sin θ − 4 sin3 θ. . The case k = 3 has been dealt with in Newton’s solution. using elementary calculus. . and the altitude 16 + z on the shortest side. are in arithmetic progression. Instead of constructing an equation in x.308 The area of a triangle • for k = 1: 48x3 + 56x2 + 16x − 1 = 0. For k = 2. . We obtain two similarity classes of triangles. 17 + 2z . which is positive. respectively with angles . 18 + 3z . when the altitude in question is the second term of the A. . For k = 1. and . This leads to the Heronian triangle with sides 13.P. More detailed calculation shows that the angles of such triangles are . This gives a similarity class of triangle with the three sides and the altitude on the shortest side in A. . • for k = 3: 48x3 + 24x − 1 = 0. 15. The angles of the triangles are . and altitude 12 on the side 14. 6180339887 · · · 2 The golden triangle Dissection of a square Dissection of a rectangle Characterizations of the golden triangle What is the most non-isosceles triangle? Appendix: Cutting a square into three similar parts Exercise .Chapter 10 The golden section √ ϕ= 1 2 3 4 5 5+1 = 1. 4. 1 ϕ M A P B AB =ϕ= AP Some basic properties of ϕ 1.. The diagonal and the side of a regular pentagon are in the golden ratio. 2 = ϕ − 1. 1 ϕ √ 5+1 . ϕ2 = 1 + ϕ. 3. . . . The simple continued fraction expansion of ϕ is 1+ 1+ 1+ 1 1 1 1+ 1 .1 The golden section ϕ A segment AB is said to be divided into the golden ratio by a point P if AP 2 = AB · P B . 2. The ratio of successive Fibonacci numbers tends to ϕ.310 The golden section 10. . D C Q Y P A X B Let AB = BC = a.2 Dissection of a square 311 10. We want to find x = AX = P C such that the areas of the rectangle and the square are the same.2 Dissection of a square Show how the square should be dissected so that it can be reassembled into a rectangle.10. 312 The golden section Resolve the paradox: How can the area increase after reassembling? D 8 C 3 Q 3 8 5 Y P 5 3 5 5 5 5 A 3 X 5 B 5 8 . then P and Q divide the sides in the golden ratio.3 Dissection of a rectangle Let P and Q be points on the sides AB and BC of rectangle ABCD.10. then the rectangle is golden. and ∠DP Q is a right angle. DP = P Q. BP Q and CDQ are equal. If the areas of triangles AP D .3 Dissection of a rectangle 313 10. in addition. D C T3 Q T1 T2 A P B . If. then the triangle is a golden right triangle. If one side of a right triangle is the geometric mean of the other two sides. C A ϕ 1 B Characterizations of the golden triangle 1. geometric.314 The golden section 10.4 The golden right triangle The golden right triangle is the one whose hypotenuse and one leg are in the golden ratio. 4. If the sides of a right triangle are in geometric progression. If the altitude of a right triangle divides the hypotenuse in the golden ratio. then the triangle is golden. . If the sides of a right triangle are respectively the harmonic. then it is the golden triangle. 2. and arithmetic means of two numbers. then this is a golden right triangle. 3. If θ is the base angle.10. the perimeter is 2(tan θ + cot θ + csc θ). We want to find the θ that minimizes this perimeter.4 The golden right triangle 315 Find the isosceles triangle of smallest perimeter that circumscribed a given semicircle. Suppose the semicircle has radius 1. C A B . The one which is closest to 1 is called the ratio of non-isoscelity of the triangle. b b η 1 >0 It follows that η 2 + η > 1. 1.1. Instead. is one For each number t in this range. there are 6 ratios obtained by comparing the lengths of a pair of sides. Theorem 10. the triangle with sides t. we have c 1 a η+1≥ +1> ≥ . the most non-isosceles triangles almost degenerate to a segment divided in the golden ratio. There is therefore no “most non-isosceles” triangle.5 What is the most non-isosceles triangle? Given a triangle. 1 t with ratio of non-isoscelity t. . 1]. 1]. Since a + b > b c c.316 The golden section 10. then so is 1 r 2 r r follows that the ratio of non-isoscelity η ≤ 1. This shows that η ∈ ( ϕ . we must have η > ϕ − 1. A number η is the ratio of non-isoscelity of a triangle if 1 and only if it lies in the interval ( ϕ . First note that if r < 1 is the ratio of the length of two sides of . Since 1 (r + 1 ) > 1. r is closer to 1 than 1 . Proof. Since the roots of x2 + x − 1 = 0 are ϕ 1 and −ϕ < 0. then η = max( a . Determine the range of the ratio of nonisoscelity of triangles. b ). If a ≤ b ≤ c are the side lengths. It a triangle. no two congruent.5 What is the most non-isosceles triangle? 317 Project: Cutting a square into three similar parts Cut a square into three similar parts.10. D 1 x2 C x x2 − x + 1 A x2 + 1 B . Show that cos 36◦ = ϕ . Which of the two equilateral triangles inscribed in a regular pentagon has larger area? 3.318 The golden section Exercise 1. Which of the two squares inscribed in a regular pentagon has larger area? . 2 2. the first one has length and the second one has shorter length between the two is approx 0.0328 · · · . 2 which is smaller only by 0. This indeed is not exact.0324786 · · · . but it gives an approximate rectangle with base √ √ 1 4 5(2 − 2 5 + 5).10. The first square exceeds the second one by 1 8 √ √ √ √ 20 + 5 2 + 10 − 4 5 − 4 50 − 10 5 ≈ 0.08%! As for the triangles. the sidelengths of these squares are respectively √ 10 √ 5+1 8 and √ 5 2 √ √ 10 − 2 5 − 5 + 1 . √ ( 5−1) 5− 5+2 6 The difference . 5 2 4+ √ 5− √ 15 + 6 5 . √ √ 10( 5+1) √ √ √ . One of you [GC] suggested an “easy” construction of the square as follows.5 What is the most non-isosceles triangle? 319 If we assume the regular pentagon inscribed in a unit circle. 320 The golden section . Chapter 11 Constructions with the golden section 1 2 3 4 5 6 Construction of golden rectangle Hofstetter’s compass-only construction of the golden section Hofstetter’s 5-step division of a segment in the golden section Construction of a regular pentagon Ahlburg’s parsimonious construction of the regular pentagon Construction of a regular 17-gon Exercise Project . 1 Construction of golden rectangle .322 Constructions with the golden section 11. It is clear that C . √ If we assume AB of length 2. CX 2 √ This is easy√ to verify.1. Figure 1 shows two circles A(B ) and B (A) intersecting at C and D . construct 1 K.11. to replace the line AB by a circle. we need four circles and one line.2 Hofstetter’s compass-only construction of the golden section Kurt Hofstetter has found the following construction of the golden section by striking the compass only five times. however. √ 5−1 CD = . 2003. X . . 1 We denote by P (Q) the circle with P as center and P Q as radius. Given a segment AB . A simple construction of the golden section.2 Hofstetter’s compass-only construction of the golden section 323 11. say C (D ). Here is a simple application: to divide a given segment into golden section. Thus.e.. Hofstetter on December 9. in 5 steps. See Figure 2. The line AB intersects the circles again at E and F . i. 2 Communicated Hofstetter. = CX 2 15 + 3 5+1 X X D E A B F A D B E F C C Y Y This shows that to construct three collinear points in golden section. D . Forum Geometricorum. 2 (2002) 65–66. Y are on a line. It is much more interesting to note that D divides the segment CX in the golden ratio. √ √ 2 3 5−1 CD 2 √ =√ =√ . the golden section can be constructed with compass only. It is possible. then CD = 2 3 and CX = 15 + 3. 2 Construction 11. From these. The circles A(F ) and B (E ) intersect at two points X and Y . by K. EF and extend to intersect AB at G. The point G divides the segment AB in the golden section. CD to intersect AB at (their common midpoint) M . then A divides G B in the golden section. 3. C1 C C2 C3 A M F G B F C4 E D Proof. C2 = B (A). 6. If the lines EF and AB intersect at G . intersecting C1 at C and D .324 Constructions with the golden section 1. By [1]. F divides F B in the golden section. G divides AB in the golden section as well. C1 = A(B ). Remark. 4. F closer to M then G . C3 = A(M ) to intersect C2 at E . Since EF is parallel to F A. . C4 = E (A) to intersect C3 at F and F . 2. 5. 4. Lemoine. C3 = C (A).11.. After the publication of this paper. 5. Hofstetter. 3 (2003) 205–206. Given a segment AB . C1 = A(B ).51. the segment CD to intersect C3 at F . (Teubner 1904). C4 = E (F ) to intersect AB at G. G´ eom´ etrographie ou art des Constructions g´ eom´ etriques. Naud. 2. . The point G divides the segment AB in the golden section. and J. A 5-step division of a segment in the golden section. p. 1902. Forum Geom. intersecting C1 again at E . 3 K. Paris. Hofstetter has also found the following parsimonious division of a given segment in the golden section. C2 = B (A).3 Hofstetter’s 5-step division of a segment in the golden section 325 11. C. Reusch. pg 37. it was found that the same construction has been given previously by E. 3 C3 C4 E C G H A G B F C2 C1 D Construction 11. construct 1. 3.2. intersecting C1 at C and D . Planimetrische Konstruktionen in geometrographischer Ausfuhrung.3 Hofstetter’s 5-step division of a segment in the golden section K. Since . and HG2 = EG2 − EH 2 = 2 − 3 = 5 . Let H be the orthogonal projection of E on the line AB . Suppose AB has unit length. This shows that G divides AB in the golden section. Remark. The other √ intersection G of C4 and the line AB is such that ( 5 + 1) : 1. G A : AB = 1 2 . Then CD = 3 and EG = EF = √ 2. we have AG = HA = 1 2 4 4 √ 1 HG − HA = 2 ( 5 − 1).326 Constructions with the golden section √ Proof. Mark the midpoint M of OY and bisect angle OMA to intersect OA at P . center O . 2. E are three adjacent vertices of a regular pentagon inscribed in the circle.4 Construction of regular pentagon Let OA and OY be two perpendicular radii of a circle. . Y B C M O P A D E Then A. Construct the perpendicular to OA at P to intersect the circle at B and E . B . The remaining two vertices can be easily constructed. 1.11.4 Construction of regular pentagon 327 11. 6 (1980) 50. Then ABCDE is the required regular pentagon.328 Constructions with the golden section 11. (Euclidean operations include (i) setting a compass. . (4) Strike an arc B (CA). (iii) drawing a line. (1) Strike an arc B (B C ).. (2) Strike an arc A(AP ). (ii) striking an arc. Here is Hayo Ahlburg’s construction. (3) Strike an arc C (AP ).3. 12 euclidean construction in all.5 Ahlburg’s parsimonious construction of the regular pentagon Make use of a given right triangle AB C with AC = 2B C to construct a regular pentagon in the fewest number of euclidean operations. meeting AB at P . striking 4 arcs. there is √ the relation a = 52−1 d. The construction requires 3 compass settings. 4 Crux Math. 4 A E B P D B C Construction 11. meeting arc A(AP ) at B . meeting arcs A(AP ) and C (AP ) at D and E. Between the side a and the diagonal d of a regular pentagon. with center B and radius B C . and drawing 5 lines for the sides of the pentagon. that is. On the radius OB perpendicular to OA. Mark the intersections of the circle E (K ) with the diameter of O (A) through A. 5 Then A4 . The polygon can be completed by successively laying off arcs equal to A4 A6 . . Construct the perpendicular through P4 and P6 to intersect the circle O (A) at A4 and A6 . mark a point J such that OJ = 1 OA. 2. Mark a point E on the segment OA such that ∠OJE = 1 4 3. 4. With AF as diameter. A1 = A. 5. A16 . .11. . and the other and P6 . 1.6 Construction of a regular 17-gon 329 11.6 Construction of a regular 17-gon To construct two vertices of the regular 17-gon inscribed in a given circle O (A). . A5 . A10 . construct a circle intersecting the radius OB at K . A3 . 5 Note that P4 is not the midpoint of AF . . leading to A8 . 6. . . Mark a point F on the diameter through A such that O is between E and F and ∠EJF = 45◦ . A15 . . . A6 are two vertices of a regular 17-gon inscribed in O (A). 4 ∠OJA. A17 . A2 . Label the one between O and A points P 4 . i. √ DE 5−1 = .. Then E divides DF in the golden section. P B : P A = ( 5 + 1) : 2.e.. The line joining the midpoints D . i.e. P inside and Q outside the square. E of two sides intersects the circumcircle at F . Show that √ the rectangle AP BQ is a golden rectangle. M is the midpoint of the side AB of a square ABCD. The line DM intersects the circle with diameter AB at two points. Let ABC be an equilateral triangle. DF 2 A D E F B C 2.330 Constructions with the golden section Exercise 1. D C P A M B Q . Calculate ∠BEF . A 20◦ E F 60◦ B 50◦ C . E and F are points on AC and AB such that ∠EBC = 60◦ and ∠F CB = 50◦ . While it is not difficult to guess what the answer is. it has been quite baffling to most people to arrive at the answer logically. ABC is an isosceles triangle with angle A = 20 ◦ .6 Construction of a regular 17-gon 331 Project This is a classic problem you should know if you teach geometry.11. Gazette. C be points on the circumference such that AX = AY = radius.332 Constructions with the golden section Appendix: Another construction of regular pentagon 6 Given a circle. Nelson. (iii) P = AX ∩ OB and Q = AY ∩ OC . (ii) B . (iv) the line P Q intersect the circumference at B and C . . Math. A regular pentagon construction. A B C P B Q C O X Y 6 D. center O . 61 (1977) 215–216. let (i) AXY be an isosceles triangle whose height is 5 of the radius of the 4 circle. Then AB and AC are two sides of a regular pentagon inscribed in the circle. P and Q are the midpoints of the sides BC and DE .11.6 Construction of a regular 17-gon 333 ABCDE is a pentagon symmetrically inscribed in a circle such that AB = CD = AE = the radius of the circle. Show that triangle AP Q is equilateral. A B E O P Q C D . 334 Constructions with the golden section . Chapter 12 Cheney’s card trick 1 Principles 2 Examples Exercise A ♠ 2 ♣ 3 ♣ 4 ♠ ? ? . 1. he will tell everybody what the 5th card is. What he will do is to ask a spectator to give you any 5 cards from a deck of 52. 2. and K are respectively 1. and you have to do things properly by making use of the following three basic principles. and 13 respectively. Among 5 cards at least 2 must be of the same suit. From one of these. The distance of two points on a 13-hour clock is no more than 6.1 Principles You are the magician’s assistant. . 1 Now rearrange them properly to tell the magician what number he should add (clockwise) to the first card to get the number on the secret card. and large. medium. for cards with the same number. Here are some examples. We decide which of the two cards to be shown as the first. The pigeonhole principle. 12. You show him 4 of the cards. and in no time. J. There are 6 arrangements of three objects. and which to be kept secret. you get to the other by travelling this distance on the 13-hour clock. Now you can determine the distance between these two cards. So you and the magician agree that the secret card has the same suit as the first card. The remaining three cards can be ordered as small. This of course depends on you. Let’s agree on this: arrangement distance sml 1 slm 2 msl 3 mls 4 lsm 5 lms 6 1 First by numerical order. 11. we treat A. For calculations.336 Cheney’s card trick 12. order by suits: ♣ < ♦ < ♥ < ♠. going clockwise. Keep the latter as the secret card. Q. hours distance clockwise 2 and 7 5 2 to 7 3 and 10 6 10 to 3 2 and J 4 J to 2 A and 8 6 8 to A 3. ♣4.2 Examples 337 12. 2. ♦J. The second card is ♦J. Suppose you have ♠5. 4 ♣ J ♦ 5 ♠ Q ♠ ? ? The secret card is ♣7. ♦J.12. 1. for example. and ♠Q. the large as the third. and arrange the other three cards. and the small as the fourth card. large. the third ♠5. want to tell the magician that he should add 4 to the number (clockwise) on the first card. in the order medium. small. you. ♠5. Now to the magician. and you decide to use the ♣ cards for the first and the secret ones. You therefore show ♣4 as the first card. Going clockwise. . which is 6. and the fourth ♠Q. small. medium. deal the medium as the second card.2 Examples If. clockwise from ♣4 to ♣7. and ♠Q. and you get the number by adding to Q the number determined by the order large. The secret card is ♠5. Here are two examples. this is 5. ♣7. Suppose your assistant show you these four cards in order: Q ♠ J ♦ 7 ♣ 4 ♣ ? ? Then you know that the secret card is a ♠. the assistant. The distance between ♣7 and ♣4 is of course 3. ♣5. For the magician: what is the secret card? 5 ♠ 7 ♠ 6 ♦ 5 ♣ ? ? 2 ♥ J ♠ 2 ♣ 8 ♠ ? ? . 2. For the assistant: (a) ♠5.338 Cheney’s card trick Exercise 1. ♦6. ♠J. ♠7. ♣2. ♣Q. ♥K. (b) ♥2. ♠8. Chapter 13 Digit problems 1 2 3 4 5 When can you cancel illegitimately and yet get the correct answer? A multiplication problem A division problem The most famous Monthly problem The problem of 4 n’s Appendix: Squares with 9 distinct nonzero digits Appendix: Squares with 10 distinct digits Exercise Charles Twigg’s square number puzzle Cross number puzzle x x x x) x x x x x x x x x x x 8 x x x x x x x x x x x x x x x x x x . b cd d allowing perhaps further simplifications of a ? d How about . When do such illegimate cancellations as a b a ab = = bc bc c give correct results in lowest terms? a a b c = .1 When can you cancel illegitimately and yet get the correct answer? Let ab and bc be 2-digit numbers.402 Digit problems 13. 2 A Multiplication problem 403 13.13. Reconstruct the multiplication. .2 A Multiplication problem A multiplication of a three-digit number by 2-digit number has the form p × p p p p p p p p p p p p p p p p p in which all digits involved are prime numbers. 404 Digit problems 13. . the answer is also unique. then there are two solutions. If the dividend is prefixed by another asterisk.3 A division problem Reconstruct the division problem ∗ ∗ ∗ ∗) ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ Charles Twigg: If the digit 2 is replaced by 9. Since 8d is a 3-digit number 9xx.13. Consider the 3-digit number in the seventh line. and therefore 992 = 8 × 124. This cannot be the same as the 3-digit number in the fifth line. 1 2 4) 1 2 1 1 1 1 9 8 1 9 2 6 6 6 0 9 7 8 0 9 8 3 1 6 8 8 0 9 1 1 3 2 1 1 6 1 1 6 . This must be 9xx. since the difference between the 3-digit numbers in the fourth and fifth lines is a 3-digit number. the digit after 7 is a larger one. From this 8d must be 99x. which must be smaller than the first and the last digits. It follows that the quotient is 97809. Its difference from the 4-digit number in the sixth line is a 2-digit number. the last second digit of the quotient is 0. Let the divisor be the 3-digit number d.3 A division problem 405 Another division problem: Problem E1 of the Monthly x x x) x x x x x x x x x x x x x x x x 7 x x x x x x x x x x x x x x x x x x x x x Clearly. in the quotient. Therefore. since these give 4-digit multiples of d. which is a multiple of d. the 4-digit number in the third and bottom lines is 9d = 10xx or 11xx. 406 Digit problems One more division problem: not even a single known digit This is Problem E10 of the Monthly. by Fitch Cheney: x x x) x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x . has been busily engaged testing on his desk calculator the 81 · 109 possible solutions to the problem of reconstructing the following exact long division in which the digits indiscriminately were each replaced by x save in the quotient where they were almost entirely omitted. It was proposed by P.4 The most popular Monthly problem The following problem.13. Professor Euclide Paracelso Bombasto Umbugio. Chessin and appeared in the April issue of 1954. L. reduce the possibilities to (18 · 109 )0 . E 1111. is said to be the most popular problem in the American Mathematical Monthly. making it the third digit of a five-digit answer. x x x) x x x x x x x x x x x x 8 x x x x x x x x x x x x x x x x x x Deflate the Professor! That is. Martin Gardner’s remark: Because any number raised to the power of zero is one. yielding readily to a few elementary insights. the reader’s task is to discover the unique reconstruction of the problem.4 The most popular Monthly problem 407 13. . The problem is easier than it looks. Our good friend and eminent numerologist. The 8 is in correct position above the line. .n .n + n n n .n n n − .n .n .n · n n n . Proof. . Let n be any positive number different from 1 and let p be any integer greater than 3.n n n n . loglog√n n logq√ log n+n logq√ n √ ··· n √ ··· n n =k.n .n n+.n .n n n .n ! ! ! Theorem 13.n + n+n−.408 Digit problems 13.n .n 13 = 14 = 15 16 17 18 19 20 21 .n n+n+.n n+ n+ n n n−. = = = = = = = n n + .n+. Every integer may be expressed by using p n’s and a finite number of operator symbols used in high school texts. These settle the cases of 4 and 5 numbers.5 The problem of 4 n’s 1 2 3 4 5 6 7 8 9 10 11 12 = = = = = = = = = = = = n+ n n+ n n n n + n n+ n+ n n n−. . where the base of the second logarithms contains k square root signs.n + n . It is easily verified that for every positive integer k .n n n .n n .n + .n . n =k. add an appropriate numbers of (n − n) + · · · + (n − n).n n n + .n n .n√ n(n+n) .n+.n+.n n .1 (Hoggatt and Moser). For higher values of p.n−.n + n n+.n + .n . 5 The problem of 4 n’s 409 Appendix: Squares with 9 distinct nonzero digits 11826 15681 19377 22887 24237 25059 26733 29106 139854276 245893761 375468129 523814769 587432169 627953481 714653289 847159236 12363 15963 19569 23019 24276 25572 27129 30384 152843769 254817369 382945761 529874361 589324176 653927184 735982641 923187456 12543 18072 19629 23178 24441 25941 27273 157326849 326597184 385297641 537219684 597362481 672935481 743816529 14676 19023 20316 23439 24807 26409 29034 215384976 361874529 412739856 549386721 615387249 697435281 842973156 Appendix: Squares with 10 distinct digits 32043 35337 37905 40545 45567 49314 54918 55626 58455 61575 63051 66105 68781 75759 78072 81222 85353 87639 89355 90198 94695 98055 1026753849 1248703569 1436789025 1643897025 2076351489 2431870596 3015986724 3094251876 3416987025 3791480625 3975428601 4369871025 4730825961 5739426081 6095237184 6597013284 7285134609 7680594321 7984316025 8135679204 8967143025 9614783025 32286 35757 38772 42744 45624 49353 55446 56532 58554 61866 63129 66276 69513 76047 78453 81945 85743 88623 89523 91248 95154 98802 1042385796 1278563049 1503267984 1827049536 2081549376 2435718609 3074258916 3195867024 3428570916 3827401956 3985270641 4392508176 4832057169 5783146209 6154873209 6714983025 7351862049 7854036129 8014367529 8326197504 9054283716 9761835204 33144 35853 39147 43902 46587 50706 55524 57321 59403 62679 65634 67677 71433 76182 80361 83919 85803 89079 90144 91605 96702 99066 1098524736 1285437609 1532487609 1927385604 2170348569 2571098436 3082914576 3285697041 3528716409 3928657041 4307821956 4580176329 5102673489 5803697124 6457890321 7042398561 7362154809 7935068241 8125940736 8391476025 9351276804 9814072356 35172 37176 39336 44016 48852 53976 55581 58413 60984 62961 65637 68763 72621 77346 80445 84648 86073 89145 90153 92214 97779 1237069584 1382054976 1547320896 1937408256 2386517904 2913408576 3089247561 3412078569 3719048256 3964087521 4308215769 4728350169 5273809641 5982403716 6471398025 7165283904 7408561329 7946831025 8127563409 8503421796 9560732841 .13. 8. Find A if the digits of C are those of A in reverse order. a mathematics student whose plausible mistakes in computation always results in correct answers. and let C = B + A . Find digits m.410 Digit problems Exercise 1. What specific number did Lucky Larry write? 2. f edcba 9m + 1 9. and 407. we find A = 12 or 69. if C is the reverse of A. Lucky Larry. This quickly implies k ≤ 2. b. c. What is the remaining one? 8. What if we also include the digit 0? 7. 4. let B = A + N . 7. . once wrote an answer in the form ab ca = abca where abca represents a four-digit integer whose digits a. and by at least 99 · 10 2 if k is odd. which when increased by 1. 3. 371. There are exactly four 3-digit numbers each equal to the sum of the cubes of its own digits. Find the three 3-digit numbers each of which is equal to the product of the sum of its digits by the sum of the squares of its digits. then C cannot exceed A by more than 18k . d. Find a number of the form aaabbbccc. From this. Find all 4-digit numbers abcd such that 3 abcd = a + b + c + d. 6. 3. 6. b. √ 5. f such that 9m abcdef = . a. Find all natural numbers whose square (in base 10) is represented by odd digits only. 2. If A has k digits. On the other hand. Solution. gives a square. let A be the sum of the digits of the number B . 9 exactly once to form prime numbers whose sum is smallest possible. 4. Three of them are 153. 5. then C exceeds A k −2 k −3 by at least 9 · 10 2 is k is even. e. c are all different. Let N be the sum of the digits of a natural number A. Use each digit 1. 855552 − 1 = 7319658024. Find the smallest 3-digit number N such that the three numbers obtained by cyclic permutations of its digits are in arithmetic progression. 169. 1. Here are seven consecutive squares for each of which its decimal digits sum to a square: 81. There are three pandigital perfect squares whose square roots are palindromes.13. Form a square of 8 digits which is transformed into a second square when the second digit from the left is increased by 1. 2 1 2 These are the only possibilities even if we consider more generally numbers consisting of two consecutive blocks of repeating digits. whose squares. Find another set of seven consecutive squares with the same property. one having 4 digits and the other 8 digits. 225. A pandigital number is one whose decimal representation contains all digits 0. all different. 121. . 196. 1 11. to within ±1. Find them. Find the number. . 144. . contain all ten digits without repetition. 12. 100. The seven numbers beginning with 9999. 977772 − 1 = 9560341728.5 The problem of 4 n’s 411 10. 14. 15. 9. 13. Find a perfect square of 12 digits formed from the juxtaposition of two squares. The number (abbbb)2 − 1 has 10 digits. . . are palindromes. 10. . 5. form reversal pairs. 8. 7. 9. are three permutations of the same digit set. the sum of the digits is 19. are permutations of consecutive digits. 4. 6. 2. the central digit is perfect.412 Digit problems Charles Twigg’s square number trivia What three-digit squares have the following characteristics? 1. three of its permutations are prime. is also a cube. the central digit is a nonzero cube. are composed of even digits. 3. 13. (The solution is unique). .5 The problem of 4 n’s 413 Cross number puzzle Place 13 three .digit square numbers in the spaces in the accompanying grid. c. 390. 5390. . . . The only choice is c = 6 and d = 0. c + d + 3 is divisible by 3. 1. Necessarily. 7) must satisfy 2 mod (e. . 7) + mod(10f + g. b. 3 respectively. 1. Since ef g is a multiple of 7. 4. f must be 0 or 5. Now it remains to arrange 0. and f ghi is only one of 5286. 4 or 4. mod(e. bcd cde def ef g f gh ghi hij 2 3 5 7 11 13 17 Solution: 1460357289 or 4160357289 Since def is divisible by 5. Now we want to find a multiple of 17 beginning with the last two digits of these. cd3 are 3 digit numbers divisible by 2. Clearly. 1. and 832. e = 2. a and b can be 1. 728. 3. 5728. 6 as a. no repetition) in a row abcdef ghij so that the following 3-digit numbers in the table below are divisible by the prime below them. 52867. 57289. 1. f = 5. The only number that can be formed from the first two digits of 13m by appending a 0 or a 5 on the left to form a multiple of 11 are 286. 7) ≡ 0 mod 7.414 Digit problems Project Arrange the ten digits (0. 9. the number f gh must be divisible by 11. Now. Now we must have ef ghij = 357289. d such that the three digit numbers bcd. 5832. 53901. This eliminates the last case. In particular. We have ab60357289. . Chapter 14 Numbers with many repeating digits 1 2 3 4 A quick multiplication The repunits Squares of repdigits Sorted numbers with sorted squares Appendix: Factorization of 10n + 1 for n ≤ 50 Exercise . extended if necessary.1 A quick multiplication What is the smallest positive integer with the property that when the digit 1 is appended to both ends. The first time this occurs is at the 22nd zero. the new number is 99 times the original? Suppose the number X has n digits. 89 For this value of X we have 99 × 112359550561797752809 = 11123595505617977528091. We require 10n +10X +1 = 99X or 89X = 10n +1. See Appendix for the factorization of numbers of the form 10n−1 1. Note that 89 is prime. until a remainder 80 occurs. To find the smallest n we divide 89 into the number 1 followed by a string of zeros. the smallest possible value of X is 1022 + 1 = 112359550561797752809.416 Numbers with many repeating digits 14. Then add 1 and obtain an integer quotient which is the smallest possible X . . Thus. (Rp )2 = Ak mk Bk 1. Notations A :=123456790. for 3 ≤ m ≤ 9. Theorem 14. Thus. 2k :=12k . B :=098765432.14.2 The repunits The repunit Rn is the number whose decimal representation consists of a string of n ones. Rp Rq = Ak mq−p+1 Bk 1.2 The repunits 417 14. mk :=12 · (m − 1)mk (m − 1) · · · 2.1. 1k :=1k−1 . k ≥ 0. Let p = 9k + m. For q ≥ p. In particular. . Rn = 1n . 1 ≤ m ≤ 9. B and c are given by a 1 2 4 5 7 8 and m is given by a 1 2 4 5 7 8 1 2 12 48 193 302 592 774 3 1232 4928 19713 30802 60372 78854 4 123432 493728 1974913 3085802 6048172 7899654 5 12345432 49381728 197526913 308635802 604926172 790107654 6 1234565432 4938261728 19753046913 30864135802 60493706172 79012187654 7 123456765432 493827061728 1975308246913 3086419135802 6049381506172 7901232987654 8 12345678765432 49382715061728 197530860246913 308641969135802 604938259506172 790123440987654 9 1234567898765432 4938271595061728 19753086380246913 30864197469135802 60493827039506172 79012345520987654 A 123456790 493827160 197530864 308641975 604938271 790123456 B 098765432 395061728 580246913 469135802 839506172 320987654 c 1 4 5 6 9 4 1 2 4 6 . 2. 5. k ≥ 0.2.3. where A. Theorem 14. then the square of 6n by multiplication by 4. Proof.418 Numbers with many repeating digits 14. (3n )2 =1n−1 08n−1 9. Let n = 9k + m. (9n )2 =(10n − 1)2 =102n − 2 · 10n + 1 =10n (10n − 2) + 1 =9n−1 80n−1 1. The last one is easiest. 8. Theorem 14. For n ≥ 2. 4. From this we obtain the square of 3n by division by 9. For a = 1. we write an for a string of n digits each equal to a. (9n )2 =9n−1 89n−1 1.3 Squares of repdigits In the decimal representations of integers. (aRn )2 = Ak mBk c. 1 ≤ m ≤ 9. 7. (6n )2 =4n−1 35n−1 6. 59 (1986) 1.4 Sorted numbers with sorted squares 419 14. Mag. (3m 6n 7)2 = More generally. solution. 16.. • 3n 4. Math. then 3x = 10m−1 10n 1.4 Sorted numbers with sorted squares A number is sorted if its digits are nondecreasing from left to right. 66 (1993) 257–262. 117. Mag. 116. if n + 1 ≥ m.14. Monotonic numbers. . 6. 1m 3m 4n−m+1 6m 8n 9. • 3m 6n 7. It is strongly sorted if its square is also sorted. 15. 2. 60 (1987)1. (3n 51 )2 =(10 · 3n + 5)2 =100 · (3n )2 + 100 · (3n ) + 25 =1n−108n−1 91 02 + 3n 25 =1n−112n−1 225 =1n 2n+1 5. if n + 1 < m. Math. Blecksmith and C. and it is easy to find its square. 12. 38. Nicol. 1 Problem 1234. See also R. 1m 3n+1 5m−n−1 6n+1 8n 9. 3. 1 • 1. 13. the product of any two numbers of the form 3 m 6n 7 is sorted. If x = 3m 6n 7. It is known that the only strongly sorted integers are given in the table below. • 16n 7. • 3n 5.. 420 Numbers with many repeating digits Appendix: Factorization of 10n + 1 for 1 ≤ n ≤ 50 n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Factorization of 10n + 1 11 101 7 × 11 × 13 73 × 137 11 × 9091 101 × 9901 11 × 909091 17 × 5882353 7 × 11 × 13 × 19 × 52579 101 × 3541 × 27961 112 × 23 × 4093 × 8779 73 × 137 × 99990001 11 × 859 × 1058313049 29 × 101 × 281 × 121499449 7 × 11 × 13 × 211 × 241 × 2161 × 9091 353 × 449 × 641 × 1409 × 69857 11 × 103 × 4013 × 21993833369 101 × 9901 × 999999000001 11 × 909090909090909091 73 × 137 × 1676321 × 5964848081 72 × 11 × 13 × 127 × 2689 × 459691 × 909091 89 × 101 × 1052788969 × 1056689261 11 × 47 × 139 × 2531 × 549797184491917 17 × 5882353 × 9999999900000001 11 × 251 × 5051 × 9091 × 78875943472201 101 × 521 × 1900381976777332243781 7 × 11 × 13 × 19 × 52579 × 70541929 × 14175966169 73 × 137 × 7841 × 127522001020150503761 11 × 59 × 154083204930662557781201849 61 × 101 × 3541 × 9901 × 27961 × 4188901 × 39526741 11 × 909090909090909090909090909091 19841 × 976193 × 6187457 × 834427406578561 7 × 112 × 13 × 23 × 4093 × 8779 × 599144041 × 183411838171 101 × 28559389 × 1491383821 × 2324557465671829 11 × 9091 × 909091 × 4147571 × 265212793249617641 73 × 137 × 3169 × 98641 × 99990001 × 3199044596370769 11 × 7253 × 422650073734453 × 296557347313446299 101 × 722817036322379041 × 1369778187490592461 7 × 11 × 132 × 157 × 859 × 6397 × 216451 × 1058313049 × 388847808493 17 × 5070721 × 5882353 × 19721061166646717498359681 11 × 2670502781396266997 × 3404193829806058997303 29 × 101 × 281 × 9901 × 226549 × 121499449 × 4458192223320340849 11 × 57009401 × 2182600451 × 7306116556571817748755241 73 × 137 × 617 × 16205834846012967584927082656402106953 7 × 11 × 13 × 19 × 211 × 241 × 2161 × 9091 × 29611 × 52579 ×3762091 × 8985695684401 101 × 1289 × 18371524594609 × 4181003300071669867932658901 11 × 6299 × 4855067598095567 × 297262705009139006771611927 97 × 353 × 449 × 641 × 1409 × 69857 × 206209 × 66554101249 × 75118313082913 11 × 197 × 909091 × 5076141624365532994918781726395939035533 101 × 3541 × 27961 × 60101 × 7019801 × 14103673319201 × 1680588011350901 . 24 and 74 . What was the number? 5. 3 Answer: . “Multiply that huge number by 8 in my head? You’ve got to be kidding. 9. 3. “It’s the serial number on that clock you brought back from Kaloat. and it’s the smallest number that works that way. If you take the last two digits and put them in front. Dad.14. 9n 8 8 2. and it is the smallest number for which it works.” Doug replied. 2 7. “You just shift its last digit to the front.” Mike told him. Show that an integer can always be found which contains only digits 0 and 1 (in the decimal scale) and which is divisible by n. “That must be an interesting number. (3n 4)2 = 1n+1 5n 6.” “But it’s easy. What other numbers besides nine are possible ? 8. “Homework?” “Just fun. Show that (16n 7)2 = 27n 8n+1 9. you get exactly four times the original number.” he said. Determine an n-digit number such that the number formed by reversing the digits is nine times the original number.4 Sorted numbers with sorted squares 421 Exercise 1. Show that 16n 1 = . There are only two repdigits an whose squares have digital sum 37. Write 52 9n−2 893n−1 9 as a sum of three squares of natural numbers. Given an integer n. 6n 4 4 19n 1 = . What are these? 3 2 AMM Problem 4281.” The boy was right. Dad. John shook its head. 9n 5 5 26n 2 = . 6n 5 5 49n 4 = . John looked over his son’s shoulder. and I’ve just noticed something special about it. 4.” What was the serial number? 6. 422 Numbers with many repeating digits . Chapter 15 Digital sum and digital root 1 2 3 4 Digital sum sequences Digital root Digital roots of the powers of 2 Digital root sequences Exercise . 4.35. . 109. 95. 83. 17. 103. 3. 45. 40. 73. 31. 47. 10. . 99. 81. Here are the numbers below 100 which are not of this form: 1. 64. It also traces back to 91. 28. 72. . .130. 2. 97. .46. 16. What is the smallest number that does not appear in any of these digit sum sequences? Find the first 10 terms of the digital sum sequence beginning with this number. 75. and eventually 1.134. 53. .34. . 135. 77. 52. 13. Find the next smallest number which is not in any of the 6 digit sum sequences and generate a new digit sum sequence from it. 96.113. 144. 94. . n ≥ 1. 14. 114. . 122. 137. Show that S(3) = R(3) and R(9) = S(9).142. . 18. . 27. . 3. .26.106. 119.109.424 Digital sum and digital root 15. 33. .139. The number 101 is n + d(n) for n = 91 and 100. 31. 11. . 107. . .67.119. We denote this by d(n). An infinite sequence of “starters”: 10n 122. 20. 51. 15. 130. 6.89. 5. 90. 126. 30. The number 101 traces back to 100. 7. . 70. 111.76. 9. 117.55. . 79. 86 which is a starter. 37. . . Here are the first few digit sum sequences: S(1) S(3) S(5) S(7) S(9) 1.50. 44.41.167. 8. 101. 9. 23. There are infinitely many digit sum sequences because there are infinitely many numbers which are not of the form n + d(n). . a1 = a. 84. 36.152. 24. 59.56. 29. 91.128. 54.88.104. 5. 38.65. 71. 49. 32. 62. 153.1 Digital sum sequences The digital sum of a positive integer n is the sum of the digits of n. every number n ≥ 10n 115 has d(n) ≥ 10n 123. Every number n ≤ 10n 114 has n + d(n) ≤ 10n 121. 127. 19. Given a positive integer a. 86. 63.118.43.160. 58. 7. the digit sum sequence S(a) = (a n ) is defined recursively by an+1 = an + d(an ).80. 20.22. 25. 39. 42. 115. 12. 69. Note that they are quite similar to the digital root sequences. 57. 107. 21. D(m + n) = D(D(m) + D(n)). 5. 4. D(D(n)) = D(n). which we call the digital root of n.1. D(n + 9) = D(n). Amusements. 3. 2. If the operation d is repeated indefinitely. Proof. (6) D(9n) = D(9D(n)) = 9 since D(n) has one single digit. 6.2 Digital root 425 15. denoted D(n). D(mn) = D(D(m)D(n)). See [Dudeny. it stabilizes after a finite number of steps and yield a number between 1 and 9.2 Digital root Given a positive integer n. Theorem 15. 1.157]. let d(n) be the sum of the digits of n. p. D(mn ) = D(D(m)n ).15. (5) D(n + 9) = D(D(n) + D(9)) = D(D(n) + 9) = D(n) since D(n) is a single-digit number. D(9n) = 9. . 7 1. 8.3 The digital roots of the powers of 2 n 0 1 2 3 4 5 6 n 1 2 4 8 16 32 64 2 D(2n ) 1 2 4 8 7 5 1 It follows that the sequence of digital roots of 2 n is periodic with period 6. 2k ≡ 4 or mod6 according as k is even or odd. . 9 The sequence of the digital roots of 2n +1 is also periodic with period 2. 5 1. 6. 6. 2 1. 7. 2. 2. 4 1. 1. we need only find 2k mod 6. 5. 4. Now. . 8. 5. 6. n 2 3n 4n 5n 6n 7n 8n 9n 6: sequence of digital roots 1. it is clear that 2k ≡ 1 or 2 mod 3 according as k is even or odd. 9. 7. 9. 5. 8. 4. 2. 4.. 9 1. . 7. 1. 8. 9 1.426 Digital sum and digital root 15. 3. Digital roots of Fermat numbers To find the digital root of the Fermat number Fk = 22 + 1. 1. 4. 9. 3. 5. Therefore. 1. . . From this. k The digital roots of the Fibonacci numbers form a sequence of period 24: 1. 8. 8. 1. 7. 7. 8 1. . 3. 8. 4. 5 if k ≡ 1 mod 2. 3. . we have D(Fk ) = 8 if k ≡ 0 mod 2. . 6. 8. 1. 55. 86. 35. 9. 42. 21. . 79. 68. . 26. 62. Here are five digital root sequences with initial terms 1. . 57. 102. 15. 53. 4. 104. 1. 2. 81. 92. 58. 35. 88. 1. 45.4 Digital root sequences A digit root sequence (an ) is defined recursively by an+1 = an + D (an ). The digital roots of these sequences are 2. 7. 3. 56. 71. 98. 84. 70. 8. 25. 8 9 . . 40. 30. 69. 46. 75. 106. 87. . 29. . 76. 135. 39. . 100. . 23. 1. 47. 61. 153. 18. Obviously. 91. 7. 27. 52. 73. 58. 59. 28. . 72. 29. . 10. The digital root sequence beginning with 1 is 1. 82. 31. 70. By taking a term not in this sequence we generate a new one. 7. 48. 55. 14. 34. 2. the digital root sequence beginning with any of these terms is a subsequence of this. 117. 17. 16. 56. 6. 3. 33. 5. 20. 8. 8. 108. 90. 126. 13. 85. . . . 36. 49. 63. 144. 4. 65. 83. 23. 60. 64. . 95. 77. 93. 37. 66. 8. 7 7. 78. 5. 4. 74. 22. 43. 50. 80. R(1) R(3) R(5) R(7) R(9) 1. 99. 6 5. 54. 43. The digital roots of these sequences are R1 R3 R5 R7 R9 1. 4. 16. 19. 2. 67. 97. These five sequences partition the natural numbers. 44. 4. 28. 31.2 (Kumar). a1 = a. 51. . 89. Theorem 15. 2. 96. . 62. 50. 2. 32. 9. 5 3. 24. periodic. 94. 11.4 Digital root sequences 427 15. 41.15. . 12. 5. . 38. 6. Find the digital root sequence of n2 . Find all integers n such that the sum of the digits of n 2 is equal to n. Find the digital root of n!. the other known one is 135. 144 = 1 · 4 · 4(1 + 4 + 4). 7.428 Digital sum and digital root Exercise 1. A Filzian number is one which is the product of its digits and digial sum.. What is the digital root of a number of the form 2 n−1(2n − 1)? 1 Answer: 135. Find one more such number. It is known that there are only finitely many Filzian numbers. d(n2 ) = n. The cubes. 4.e. i. 1 2. 3. The triangular numbers. 5. . For example. Apart from 1 and 144. f (12) = 5. For example. . f (123) = 14 etc. the iterations of f beginning with 4 leads to the cycle 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4.15. f (10) = 1. Study the iterations of f .4 Digital root sequences 429 Project: Sum of squares of digits Let f (n) be sum of the digits of n. For example. f (1) = 1. f (2) = 4. 430 Digital sum and digital root . Chapter 16 3-4-5 triangles in the square . 432 3-4-5 triangles in the square . 433 . 434 3-4-5 triangles in the square θ tan θ = 1 . ϕ . Show that the shaded triangle is a 3:4:5 triangle. Find the ratio of the areas of ABY . and AXD. BCXY . D X C Y A B . 2.435 Exercise 1. Chapter 17 Combinatorial games 1 2 2 3 4 Subtraction games The nim sum Nim Northcott’s variation of nim Wythoff’s game Appendix: Beatty’s theorem Exercise Project: Another subtraction game . d) is a winning position. the first player has a winning strategy if and only if N is not divisible by d. Theorem 17.502 Combinatorial games 17.1 Subtraction games Starting with a given positive integer N . The one who gets to a specified N wins. An equivalent version: the battle of numbers Starting with 0. The one who gets down to 0 wins.1. The winning positions are the terms of the arithmetic progression of common difference d containing N . two players alternately add positive integers less than a given limit d. Specifically. the small number mod (N. The player who secures a multiple of d has a winning strategy. two players alternately subtract a positive amount less than a given positive number d < N . . Therefore. 1. . m2 . . The player who secures a position n with g (n) = 0 has a winning strategy. 2. . Theorem 17.1 The Sprague-Grundy sequence Let G be a 2-person counter game in which two players alternately remove a positive amount of counters according to certain specified rules. The values of n for which g (n) = 0 are precisely the multiples of d. . then g (n) is the smallest nonnegative integer different from g (m1 ).1. Example 1: the trivial counter game If G is the game which the players may subtract any positive amount. (1) g (n) = 0 for all n which have no legal move to another number. . d − 1. . g (mk ). g (0) = 0. mk . . (2) Suppose from position n it is possible to move to any of positions m1 . g (m2 ). .2. . . the Sprague-Grundy sequence is the natural sequence 0. . . . (all < n). 1. .1 Subtraction games 503 17. . n. . . . then the Sprague-Grundy sequence is periodic 0. . In particular. The Sprague-Grundy sequence of G is the sequence (g (n)) of nonnegative integers defined recursively as follows. Example 2: removing not more than d counters If G is the game which subtracts numbers < d. 3. . 2. .17. This means no matter how B moves. p. 6. 0 =⇒ g (9) = 2 10 → 9. since now B can only move to 4 or 1. 2 =⇒ g (6) = 1 7 → 6. 5. Exercise How would you win if the starting number is 200? or 500? 1 [Smith. . with period 5. 1 =⇒ g (5) = 0 6 → 5. 1 =⇒ g (10) = 0 The values of n ≤ 1000 for which g (n) = 0 are as follows: 1 Suppose we start with 74. g (0) = 0 1 → 0 =⇒ g (1) = 1 g (2) = 0 3 → 2 =⇒ g (3) = 1 4 → 3.2 Subtraction of square numbers Two players alternately subtract a positive square number.68] incorrectly asserts that this sequence is periodic. This is clear if B moves to 9 or 1.1.504 Combinatorial games 17. Player A can subtract 64 to get 10. 4 =⇒ g (8) = 1 9 → 8. We calculate the Sprague-Grundy sequence. 3 =⇒ g (7) = 0 8 → 7. 0 =⇒ g (4) = 2 5 → 4. A can always win. then A can move to 5 which again has value 0. which has value 0. But if B moves to 6. 1.1 Subtraction games 505 17.3 Subtraction of square numbers The first 100 terms of the Sprague-Grundy sequence are 1 1 1 1 2 2 5 5 3 2 2 2 0 0 0 3 3 0 0 0 3 6 3 1 1 1 4 2 1 1 1 4 4 4 2 2 2 0 0 3 3 5 5 5 5 0 0 3 1 1 4 0 6 0 0 6 1 1 2 2 2 5 1 2 1 1 7 0 0 3 3 3 0 0 3 6 6 8 1 1 4 2 2 1 1 4 3 4 9 10 2 0 2 0 5 3 0 1 3 4 3 4 2 4 5 6 2 4 2 4 The winning positions within 500 are as follows.17. 0 62 150 257 369 2 65 170 260 377 5 67 177 262 390 7 72 180 267 392 10 85 182 272 397 12 95 187 275 425 15 109 192 312 430 17 119 197 317 437 20 124 204 322 442 22 127 210 327 447 34 130 215 332 449 39 132 238 335 464 44 137 243 340 52 142 249 345 57 147 255 350 . 3. By way of example. second player may subtract either 1. 2. 2. or 6 (but not 12). or 5. if the original number is 12. Winner is the last player able to perform such a subtraction. first player may subtract either 1. 4. The first 100 terms of the Sprague-Grundy sequence are 1 0 0 0 0 0 0 0 0 0 0 2 1 2 1 5 1 2 1 3 1 2 3 0 0 0 0 0 0 0 0 0 0 4 2 1 3 1 2 1 6 1 2 1 5 0 0 0 0 0 0 0 0 0 0 6 1 4 1 2 1 3 1 2 1 5 7 0 0 0 0 0 0 0 0 0 0 8 3 1 2 1 4 1 2 1 3 1 9 10 0 1 0 2 0 1 0 3 0 1 0 2 0 1 0 4 0 1 0 2 The winning positions within 500 are as follows.506 Combinatorial games Subtraction of aliquot parts Two players start with a positive integer and alternately subtract any aliquot part (divisor) with the exception of the number itself from the number left by the opponent. leaving 10. If he subtract 2. 0 29 59 89 119 149 179 209 239 269 299 329 359 389 419 449 479 1 31 61 91 121 151 181 211 241 271 301 331 361 391 421 451 481 3 33 63 93 123 153 183 213 243 273 303 333 363 393 423 453 483 5 35 65 95 125 155 185 215 245 275 305 335 365 395 425 455 485 7 37 67 97 127 157 187 217 247 277 307 337 367 397 427 457 487 9 39 69 99 129 159 189 219 249 279 309 339 369 399 429 459 489 11 41 71 101 131 161 191 221 251 281 311 341 371 401 431 461 491 13 43 73 103 133 163 193 223 253 283 313 343 373 403 433 463 493 15 45 75 105 135 165 195 225 255 285 315 345 375 405 435 465 495 17 47 77 107 137 167 197 227 257 287 317 347 377 407 437 467 497 19 49 79 109 139 169 199 229 259 289 319 349 379 409 439 469 499 21 51 81 111 141 171 201 231 261 291 321 351 381 411 441 471 23 53 83 113 143 173 203 233 263 293 323 353 383 413 443 473 25 55 85 115 145 175 205 235 265 295 325 355 385 415 445 475 27 57 87 117 147 177 207 237 267 297 327 357 387 417 447 477 . 1 Subtraction games 507 Subtraction of proper divisors The rules are the same with the exception that only proper divisors may be subtracted. Consider 1 an improper divisor. The first 100 terms of the Sprague-Grundy sequence are 1 0 0 0 0 0 0 0 0 0 0 2 0 3 1 0 1 2 1 4 1 2 3 0 0 0 0 0 0 0 0 0 0 4 1 1 4 1 2 1 5 1 2 1 5 0 0 0 0 0 0 0 0 0 0 6 2 2 1 2 1 4 1 2 1 6 7 0 0 0 0 0 0 0 0 0 0 8 0 1 2 1 3 1 2 1 4 1 9 10 0 1 0 3 0 1 0 4 0 1 0 2 0 1 0 3 0 1 0 2 The winning positions within 500 are as follows. 0 25 53 83 113 141 171 201 231 261 291 321 351 381 411 441 471 1 27 55 85 115 143 173 203 233 263 293 323 353 383 413 443 473 2 29 57 87 117 145 175 205 235 265 295 325 355 385 415 445 475 3 31 59 89 119 147 177 207 237 267 297 327 357 387 417 447 477 5 32 61 91 121 149 179 209 239 269 299 329 359 389 419 449 479 7 33 63 93 123 151 181 211 241 271 301 331 361 391 421 451 481 8 35 65 95 125 153 183 213 243 273 303 333 363 393 423 453 483 9 37 67 97 127 155 185 215 245 275 305 335 365 395 425 455 485 11 39 69 99 128 157 187 217 247 277 307 337 367 397 427 457 487 13 41 71 101 129 159 189 219 249 279 309 339 369 399 429 459 489 15 43 73 103 131 161 191 221 251 281 311 341 371 401 431 461 491 17 45 75 105 133 163 193 223 253 283 313 343 373 403 433 463 493 19 47 77 107 135 165 195 225 255 285 315 345 375 405 435 465 495 21 49 79 109 137 167 197 227 257 287 317 347 377 407 437 467 497 23 51 81 111 139 169 199 229 259 289 319 349 379 409 439 469 499 .17. 508 Combinatorial games Subtraction of primes The first 100 terms of the Sprague-Grundy sequence 1 0 0 5 6 3 4 6 6 4 6 2 0 1 6 3 3 1 3 8 8 0 3 4 5 1 1 2 1 2 2 6 7 7 4 7 0 4 8 5 5 2 6 4 7 5 4 7 5 5 10 6 4 1 5 6 2 3 0 0 7 0 8 8 0 2 7 8 9 10 3 3 4 0 3 4 4 5 4 1 5 2 1 1 2 2 6 8 9 0 4 1 5 2 4 10 5 7 6 10 9 7 4 1 5 2 6 3 4 7 The winning positions within 500 are as follows. 0 126 290 404 1 134 296 416 2 146 302 426 10 156 310 446 11 170 311 452 26 176 320 470 35 188 326 476 36 196 336 482 50 206 344 486 56 218 356 494 86 236 362 92 248 376 101 254 386 116 260 392 122 266 396 . and has 0 as identity element. If we write 0 0 = 0. Here are the nim sums of numbers ≤ 15: 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 1 1 0 3 2 5 4 7 6 9 8 11 10 13 12 15 14 2 2 3 0 1 6 7 4 5 10 11 8 9 14 15 12 13 3 3 2 1 0 7 6 5 4 11 10 9 8 15 14 13 12 4 4 5 6 7 0 1 2 3 12 13 14 15 8 9 10 11 5 5 4 7 6 1 0 3 2 13 12 15 14 9 8 11 10 6 6 7 4 5 2 3 0 1 14 15 12 13 10 11 8 9 7 7 6 5 4 3 2 1 0 15 14 13 12 11 10 9 8 8 8 9 10 11 12 13 14 15 0 1 2 3 4 5 6 7 9 9 8 11 10 13 12 15 14 1 0 3 2 5 4 7 6 10 10 11 8 9 14 15 12 13 2 3 0 1 6 7 4 5 11 11 10 9 8 15 14 13 12 3 2 1 0 7 6 5 4 12 12 13 14 15 8 9 10 11 4 5 6 7 0 1 2 3 13 13 12 15 14 9 8 11 10 5 4 7 6 1 0 3 2 14 14 15 12 13 10 11 8 9 6 7 4 5 2 3 0 1 15 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Theorem 17. .2 The nim sum of natural numbers 509 17. . . .3. commutative. . Gk which have Sprague-Grundy sequences (g1 (n)). . . .17. a b = (a1 a2 · · · an ) (b1 b2 · · · bn ) = (a1 b1 )(a2 b2 ) · · · (an bn ).2 The nim sum of natural numbers The nim sum of two nonnegative integers is the addition in their base 2 notations without carries. The player who secures a position (n1 . Suppose two players alternately play one of the counter games G1 . n2 . . (gk (n)) respectively. 0 1=1 0 = 1. nk ) with g1 (n1 ) g2 (n2 ) · · · gk (nk ) = 0 has a winning strategy. then in terms of the base 2 expansions of a and b. The nim sum is associative. . In particular. a a = 0 for every natural number a. . . . 1 1 = 0. Therefore. . b. players A and B alternately remove a positive amount of marbles from any pile. Since 12 9 = 5.510 Combinatorial games 17. suppose the initial position is (12. the first player can remove 2 marbles from the second pile to maintain a balance of the nim sum equation. c marbles respectively.3 The game Nim Given three piles of marbles. b. This theorem indeed generalizes to an arbitrary number of piles. c) does not satisfy a b c = 0. 12 5 9=0 thereby securing a winning position. 9). In the game nim.4. provided that the initial position (a. the first player has a winning strategy. For example. 7. Theorem 17. The player who makes the last move wins. with a. the player who can balance the nim sum equation has a winning strategy. For example. This is equivalent to nim if one considers a number of piles of counters corresponding to the number of spaces between the counters on the rows. the numbers of spaces have nim sum 3 2 4 2 2 = 5.4 Northcott’s variation of Nim 511 17. (If a player tries to increase the number of spaces. . Therefore the player who can balance the nim sum equation has a winning strategy. the only restriction being not moving onto or beyond the opponent’s counter. It can be made 0 by moving 3 spaces in row 3.17. The one who cannot move loses.4 Northcott’s variation of Nim Two players alternately move their counters on one of the rows. in the above arrangement. the other player can force the same distance by pursueing the same number of spaces). bi . Here are the 18 smallest winning positions for Wythoff’s game: 1 2 3 5 4 7 6 10 8 13 9 15 11 18 12 20 14 23 16 26 17 28 19 31 21 34 22 36 24 39 25 41 27 44 29 47 Theorem 17. i < k }.512 Combinatorial games 17. b1 ) = (1.5. to forbid your opponent to get to this position. construct (ak . 1). bk ) by setting ak := min{c : c > ai .5 Wythoff’s game Wythoff’s game is a variant of Nim. The sequence of winning positions: starting with (a 1 . Given two piles of marbles. nϕ2 ). 5). We describe the position of the game by the amounts of marbles in the two piles. If you can make (2. you should occupy (3. where ϕ = 5+1 is the golden ratio. 2 . then you will surely win no matter how your opponent moves. or an equal (positive) amount of marbles from both piles. a player either removes an arbitrary positive amount of marbles from any one pile. The winning positions of Wythoff’s game are the pairs √ ( nϕ . Now. 2). The player who makes the last move wins. bk :=ak + k. s (r − 1)n .5 Wythoff’s game 513 Appendix: Beatty’s Theorem If α and β are positive irrational numbers satisfying sequences α . 2. . 3α . . form a partition of the sequence of positive integers. 3β .. . . then the is the natural sequence 1.. 62 (1989) 203. . . r 2n . . r and n . solution. s 1 α 1 +β = 1. See Russian problem book. s 2n . then the union of the two sequences n . and β .. 2β . 2α .. . r (s − 1)n . . n − 2 if and only if r and s are relatively prime.. . If r and s are positive integers and n = r + s.. . .17. . MG Problem 1300. All numbers are reduced modulo 100. The golden ratio ϕ has the property that ϕ 2 and ϕ have exactly the same digits after the decimal point. What choices has A for the initial number in order to insure that B will at some time announce 00 ? 2 2 56.514 Combinatorial games Exercise 1. so that only two digit numbers are announced. 3. B reverses the digits of this number and adds to it the sum of its digits and then announces his result. A continues in the same pattern. . 80 or 68. What are the winning positions in the battle of numbers with N = 100 and d = 7? 2. A announces a two digit number from 01 to 99. Find all numbers θ which have exactly the same digits as θ 2 after the decimal point. two players alternately subtract numbers not more than the predecessor used. The one who gets down to zero is the winner. Under what condition does the first player have a winning strategy? .17.5 Wythoff’s game 515 Project: Another subtraction game Starting with a given positive integer. 516 Combinatorial games . Project: A number whose cube root divides all numbers obtained by cyclic permutations of its digits. .Chapter 18 Repunits 1 2 3 4 5 k -right-transposable integers k -left-transposable integers Sam Yates’ repunit riddles Recurring decimals The period length of a prime Appendix: Factorizations of repunits Rn . n ≤ 50 Appendix: Primitive divisors of repunits Appendix: Multiplicative digital root Exercise Project: Period lengths of primes < 100. From this. Note that X is a repdigit if and only if k = 1. The equation will reduce to X = a repdigit. 10 − k can only have prime divisors 2. Now. we have 7X = a(10n − 1). If k = 3. If a = 7. .1 k -right-transposable integers Let k be a given positive integer. then again X is a repdigit. (10 − k )X = a(10n − 1) = 9a · Rn . which is clearly impossible. the number is multiplied by k . A positive integer X is k -transposable if in moving the leftmost digit to the rightmost. Suppose the number X has n digits. Therefore. Therefore X = a · 10 7 −1 and has first digit a. This is possible 6m only if n is a multiple of 6. We shall assume k > 1.518 Repunits 18. 5. Therefore. the only k -transposable numbers are (142857)m and (285714)m with k = 3. we must have 7 dividing 10 n − 1. 7 It is easy to see that a can only be 1 or 2. 3. 106m − 1 = (142857)m . For k = 3. with leftmost digit a. We have 10(X − a · 10n−1) + a = kX. . This shows that b ≥ k . =1012658227848. b · 10n + X − b = 10kX . b · Xk for k = b. and b > 1 k − 10 . k 2 19X = b(10n − 1) 3 29X = b(10n − 1) 4 39X = b(10n − 1) 5 49X = b(10n − 1) 6 59X = b(10n − 1) 7 69X = b(10n − 1) 8 79X = b(10n − 1) 9 89X = b(10n − 1) These lead to the following numbers: X2 X3 X4 X5 X6 X7 X8 X9 n 18m 28m 12m 42m 58m 22m 13m 44m 10n−1 (10k −1) 10n −1 = =105263157894736842. =102040816326530612244897959183673469387755.2 k -left-transposable integers 519 18. b(10n −1) > (10k −1)10n−1. and its rightmost digit is b. We have X −b b · 10n−1 + = kX. =10112359550561797752808988764044943820224719. and (10k − 1)X = b(10n − 1). =1034482758620689655172413793.18. Every Each of these Xk can be replaced by k k -left-transposable number is of the form (Xk )m for Xk given above and m ≥ 1. A positive integer X is k -left-transposable if in moving the rightmost digit to the leftmost. Note that X is a repdigit if and only if k = 1. =1014492753623188405797. the number is multiplied by k . We shall assume k > 1. Suppose the number X has n digits. . 10 From this. Since X is has n digits. . =1016949152542372881355932203389830508474576271186440677966. . =102564102564.2 k -left-transposable integers Let k be a given positive integer. . 9. 520 Repunits 18. 3 Yes. and gcd(R . Are two repunits with consecutive numbers as their subscripts relatively prime? 3 4. Aside from 3. An old car dealer’s record in the 1960’s shows that the total receipts for the sale of new cars in one year came to 1. 6 and the sum is R9 .111. Consecutive numbers are relatively prime.n) . R ) = R m n gcd(m.3 Sam Yates’ repunit riddles 1. 6R × R = R .00 dollars. Repunits with even subscripts are divisible by 11. 7 7 13 None . 5 . Consecutive odd numbers are relatively prime. how many cars did he sell? 1 REPUNITS=12345679 2 No. what prime p divides R p ? 5 6. If each car had eight cylinders and was sold for the same price as each other car. 4 Yes. What digit does each letter of this multiplication represent? R R R R R R R × R R R R R R R R E P U N I T I N U P E R 7.111. What digits should be substituted for the letters so that the sum of the nine identical addends will be a repunit? 1 R R R R R R R R R E E E E E E E E E P P P P P P P P P U U U U U U U U U N N N N N N N N N I I I I I I I I I T T T T T T T T T S S S S S S S S S + 2. Are two repunits with consecutive even numbers as their subscripts relatively prime? 2 3. Are two repunits with consecutive odd numbers as their subscripts relatively prime? 4 5. Since the p q product is a palindrome. . where N = nRn . . 8M 3 = 23 − 1 = 7 is prime but R3 = 111 = 3 × 37. is the corresponding repunit Rp also prime? 8 10.3 Sam Yates’ repunit riddles 521 8. 7 If R · R has 100 digits. What is the smallest repunit divisible by the square of 11? What is the smallest repunit divisible by the square of R 11 ? In general. Suppose p ≤ q and p = 9k + m for 1 ≤ m ≤ 9. p + q = 101. We must have k = 0 and p ≤ 9. . For any p = 2. the smallest repunit divisible by the square of R n is RN . 7 9. it cannot contain A and B. . 9. p= 12 · · · (p − 1)(p − 1)32 if 3 ≤ p ≤ 9.18. . the product Rp · Rq is the palindrome p102−2p 1. where ( 12k if p = 2. . Find a pair of repunits whose product is a 100-digit palindrome. If a Mersenne number Mp = 2p − 1 is prime. For p = 3.4 Recurring decimals The decimal expansion of a rational number is eventually periodic. 3 with period length 1.333 · · · = 0. It is finite if and only if the denominator has no prime divisors other than 2 and 5. 1 The decimal expansion of n is purely periodic if and only if n is a prime. Here are the periods of the reciprocals of the first few primes.3. p 3 7 11 13 17 19 23 29 31 37 41 43 47 period 3 142857 09 076923 0588235294117647 052631578947368421 0434782608695652173913 0344827586206896551724137931 032258064516129 027 0243902439 023255813953488372093 0212765957446808510638297872340425531914893617 λ(p) 1 6 2 6 16 18 22 28 15 3 10 21 46 . together with their period lengths. this is the most well known recurring decimal 1 = 0.522 Repunits 18. 5. Theorem 18. Clearly. The period length of p is the smallest divisor n of p − 1 such that p divides R n . for a prime p > 5. Thus. p cannot be 2 or 5. p Then moving the decimal places n places to the right. We say that p is a primitive prime divisor the repunit n.2. A table of primitive prime divisors of repunits is given in an Appendix.a1 · · · aλ . .1.18. p and 10λ 1 − = a1 · · · aλ p p is an integer. Theorem 18. p 1 Suppose p has period length λ: 1 = 0.a1 · · · aλ . the period length of p is the smallest divisor λ of p − 1 such that p divides 10λ − 1. 5. If p = 3. Note that 10n − 1 = 9n = 9 × Rn . we have 10λ = a1 · · · aλ . then 10 p−1 − 1 is divisible by p. Proof. and any number λ for which 10λ − 1 is divisible by p divides p − 1. This means that p divides 10 λ − 1. It is known that if p = 2. Let p > 5 be a prime.5 The period length of a prime The period length of a prime number p means the length of the shortest repeating block of digits in the decimal expansion of 1 . then p divides Rn . If p = 2. the period length of p is the number n for which p is primitive prime divisor.5 The period length of a prime 523 18. 11 × 47 × 139 × 2531 × 549797184491917 × R23 . 11 × 73 × 101 × 137. 2028119 × 247629013 × 2212394296770203368013. 11 × 53 × 79 × 859 × 265371653 × 1058313049. 32 × 7 × 11 × 13 × 19 × 37 × 52579 × 333667. 2791 × 6943319 × 57336415063790604359. 11 × 103 × 4013 × 2071723 × 5363222357 × 21993833369. 32 × 31 × 37 × 41 × 271 × 238681 × 333667 × 2906161 × 4185502830133110721. 32 × 7 × 11 × 13 × 19 × 37 × 101 × 9901 × 52579 × 333667 × 999999000001. 3 × 7 × 11 × 13 × 37. 239 × 4649 × 505885997 × 1976730144598190963568023014679333.524 Repunits Appendix: Factorizations of repunits Rn for n ≤ 50 n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Rn 11. 11 × 17 × 73 × 101 × 137 × 353 × 449 × 641 × 1409 × 69857 × 5882353. 112 × 23 × 4093 × 8779 × 21649 × 513239. 173 × 1527791 × 1963506722254397 × 2140992015395526641. 3 × 7 × 11 × 13 × 17 × 37 × 73 × 101 × 137 × 9901 × 5882353 × 99990001 × 9999999900000001. 11 × 909090909090909091 × R19 . 3 × 7 × 11 × 13 × 31 × 37 × 41 × 211 × 241 × 271 × 2161 × 9091 × 2906161. 11 × 41 × 251 × 271 × 5051 × 9091 × 21401 × 25601 × 182521213001 × 78875943472201. 2071723 × 5363222357. prime 11 × 41 × 101 × 271 × 3541 × 9091 × 27961. prime 3 × 7 × 11 × 13 × 37 × 73 × 101 × 137 × 9901 × 99990001. 3 × 72 × 11 × 13 × 37 × 43 × 127 × 239 × 1933 × 2689 × 4649 × 459691 × 909091 × 10838689. 112 × 23 × 89 × 101 × 4093 × 8779 × 21649 × 513239 × 1052788969 × 1056689261. 239 × 4649. 3 × 7 × 11 × 13 × 37 × 101 × 9901. 53 × 79 × 265371653. 83 × 1231 × 538987 × 201763709900322803748657942361. 11 × 101. 41 × 271 × 21401 × 25601 × 182521213001. 11 × 29 × 101 × 239 × 281 × 4649 × 909091 × 121499449. 11 × 239 × 4649 × 909091. 33 × 37 × 757 × 333667 × 440334654777631. 3191 × 16763 × 43037 × 62003 × 77843839397. 35121409 × 316362908763458525001406154038726382279. 11 × 41 × 271 × 9091. 3 × 37 × 67 × 21649 × 513239 × 1344628210313298373. 41 × 71 × 239 × 271 × 4649 × 123551 × 102598800232111471. 41 × 271. 3 × 31 × 37 × 41 × 271 × 2906161. 3 × 37. . 11 × 17 × 73 × 101 × 137 × 5882353. 3 × 37 × 53 × 79 × 265371653 × 900900900900990990990991. 21649 × 513239. 32 × 37 × 333667. 3 × 37 × 43 × 239 × 1933 × 4649 × 10838689. 11 × 41 × 73 × 101 × 137 × 271 × 3541 × 9091 × 27961 × 1676321 × 5964848081. 127. 9999999900000001. 123551. Exercise Find all prime numbers p for which 1 p has period length 11.. 909090909090909091. 2161. 4649. 21993833369. 43037. 9091. 52579. 316362908763458525001406154038726382279. 1409. 9901. 2906161.18. 99990001. 6943319. 538987. 69857. 1527791. 2028119. 247629013. 13. 9 9 Problem 2207. 71. 2071723. 4013. 440334654777631. 77843839397. 4185502830133110721. 459691. 333667. 271. 909091. 5363222357. 1052788969. 7. 73. 121499449. 173. 4093. 27 (1995) 59. 999999000001. 641. 79. 29. 1676321. 17. 102598800232111471. 859. 1344628210313298373. 241. 1231. . 8779. 19. 238681. 900900900900990990990991. 1963506722254397. 3541. 21649. 57336415063790604359. 211. 201763709900322803748657942361. 10838689. 513239. 3. 251. 21401. 239. 89. 353. 2791. 5051. 1976730144598190963568023014679333. 2140992015395526641. 5882353. 83. 182521213001. 53. 41. 505885997. 2689. 35121409. 62003. 37. Recreational Math. 31. 43.5 The period length of a prime 525 Appendix: Primitive prime divisors of repunits n 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 Primitive prime divisors of Rn 11. 1933. 549797184491917. 281. 449. 5964848081. J. 78875943472201. 11111111111111111111111. 1058313049. 265371653. 137. 139. 27961. 16763. 1056689261. 2212394296770203368013. 47. 1111111111111111111. 101. 25601. 103. 2531. 67. 3191. 757. 23. The multiplicative digital root of n is 1 if and only if n is a repunit.526 Repunits Appendix: The multiplicative digital root The multiplicative digital root of a number is the single digit number obtained by iterating the digital product operation. Their product cannot be 11 modulo 100. and its prime divisors can only be 3 and 7.3 (Kuczma). Those of 3 modulo 100 are given in the first row of the table below. The powers of 7 modulo 100 are 1. If so. 1 7 49 43 3 21 47 29 9 63 41 87 27 89 23 61 81 67 69 83 43 1 7 49 29 3 21 47 87 9 63 41 61 27 89 23 83 81 67 69 49 43 1 7 47 29 3 21 41 87 9 63 23 61 27 89 69 83 81 67 7 49 43 1 21 47 29 3 63 41 87 9 89 23 61 27 67 69 83 813 . this would mean that R n factors into one digit numbers. 7. 43. Theorem 18. n ≥ 2. The digital product of a number is 1 if and only if it is a repunit. for digital product. 6244 → 6 · 2 · 4 · 4 = 192 → 1 · 9 · 2 = 18 → 1 cdot8 = 8. 49. Thus. Proof. We claim that no number can have a repunit Rn . k ≥ 0. R n = 3h · 7k for some integers h. In other words. If the digits are reversed. What is the smallest integer of 2n identical digits which is the product of two n-digit numbers? Clearly n ≥ 2. Find the most general such number N . . find the smallest integer whose cube terminates in seven sevens. Calculate the period of p = 67. 4. show that the decimal expansion of the frac1 will repeat in p− digits or some factor thereof it and only if tion 1 p 2 k p ≡ ±3 (mod 40). 2k 5k and 1 2h ·5k ? 3. the quotient is 48. if the units digits is transposed from right to left. There are four ways of rearranging it as a product of two 3-digit numbers: 143 × 777 = 231 × 481 = 259 × 429 = 273 × 407.5 The period length of a prime Exercise 527 1. If the last digit is removed and placed before the remaining p − 1 digits. 6. 10 7. If p is any odd prime. R6 = 3 × 7 × 11 × 13 × 37. M = 714285 = 5N . Now. Without the use of tables. 5. What are the decimal expansions of 2. N = 142857. A certain 3-digit number yields a quotient of 26 when divided by the sum of its digits. Since 1111 = 11 × 101. 11 10 11 96607533 = 901639512372747777777. 1 . a number M is obtained where M = 5N . (a) Find the smallest integer N such that. a new number of p digits is 1 formed which is n th of the original number. Let N be an integer of p digits. 1.18. What is the smallest 3-digit number for which this is possible ? 8. we seek 6 digit numbers. and 97 are the only primes < 100 that cannot be found in the table above. Prime Period lengths 7 13 19 29 37 43 53 61 71 79 89 Prime Period lengths 11 17 23 31 41 47 59 67 73 83 97 .528 Project: Period lengths of primes < 100 Repunits Find the period lengths of the primes < 100 by using the factorization of repunits given above. 61. Note that 59. i 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 Ni 19000458461599776807277716631 90004584615997768072777166311 45846159977680727771663119 458461599776807277716631190 4584615997768072777166311900 45846159977680727771663119000 58461599776807277716631190004 84615997768072777166311900045 46159977680727771663119000458 61599776807277716631190004584 15997768072777166311900045846 59977680727771663119000458461 99776807277716631190004584615 97768072777166311900045846159 77680727771663119000458461599 76807277716631190004584615997 68072777166311900045846159977 80727771663119000458461599776 7277716631190004584615997768 72777166311900045846159977680 27771663119000458461599776807 77716631190004584615997768072 77166311900045846159977680727 71663119000458461599776807277 16631190004584615997768072777 66311900045846159977680727771 63119000458461599776807277716 31190004584615997768072777166 11900045846159977680727771663 Ni ÷ M . Find its cube root M . Verify that N = N0 the twentyeight numbers Ni . 1 ≤ i ≤ 28 which are formed by cyclic permutations of its digits are all divisible by the cube root of N .5 The period length of a prime 529 Project It is known that the number N = 19000458461599776807277716631 is a perfect cube.18. 530 Repunits . 1. the reverse of 86). Curly.Chapter 19 More digital trivia 1. 5. 6. (a) N = 112. and Moe had an unusual combination of ages. 4. contain the digits 0. Let n be a nonnegative integer. considered jointly. Y. y. 4. z by inserting an extra digit (the same for all) on the left. (a) Find the smallest positive integer N having the property that the sum of its digits does not divide the sum of the cubes of its digits. 8 and 9 each once. Z are derived from x. Modulo 3. what was the youngest he could be? 2.. solution: . (b) 370 and 371. The number formed by placing 2 n and 2n+1 side by side in any order is divisible by 3. 4 1 Suppose 2n has k digits. Is the solution unique ? 3 6. Putting 2n+1 on the left hand side of 2n gives the number 2n+1 · 10k + 2n . The sum of any two of the three ages was the reverse of the third age (e. 16 + 52 = 68. Find two perfect cubes which. 2. Larry.g. 3. 2 4 3 Unique 812222 = 6597013284. (b) Find the two consecutive positive integers each of which equals the sum of the cubes of its digits. 2 5. 7. this is (−1)n+1 + (−1)n ≡ 0. Show that there is but one five-digit integer whose last three digits are alike and whose square contains no duplicate digits. All were under 100 years old. 213 = 9261 and 933 = 804357. 1 3. (a) What was the sum of the ages? (b) If Larry was older than either of the others. y. such that x 2 + y 2 = z 2 and X 2 + Y 2 = Z 2 where X. z (less than 100). Find positive integers x. 2. 4 5 . 7. 8.532 More digital trivia 7. 10. 8 9 10 11 12 16922 = 2862864. 3. Find a four-digit square which remains a square when two zeros are intercalated between the thousands digit and the hundreds digit. 9 15. 8. 9. 4. 6. 7 13. used once each in both cases. 7. Find a number and its fourth power. 8. 5. 6. Find a perfect square of 7 digits with all digits even and positive. (See E116 and Problem following E602). 6 12. Determine all binomial coefficients whose decimal expansions consist of the ten digits 0. 1522 = 23104 and 1792 = 32041. Find two perfect squares. once each. `595´ ` ´ = 5169738420 and 253 = 8301429675. 10 16. 578). 7 . 3. even without restriction to multiples of 7. 8. all different. 5. Determine the largest and smallest perfect squares which can be written with the ten digits 0. 4. n = 32. which together contain all the digits 0. 3. 29782 = 8868484. 200704 = 4482 and 872 = 7569. 7. 1. 2. What is the largest prime whose square contains no duplicate digits ? 8 14. 4. 18 and 69 are the only numbers with two distinct powers which together contain the ten digits. and 9 without repetition. 213972 = 457831609. 1. 5. 2. 8. 2. 700569 = 8372 . once each. of five digits each. which together have nine digits. 28782 = 8282884. 2. 7. Find a multiple of 7 whose square has eight digits of the form ababbbcc. How many solutions are there ? 9. Determine all numbers of the form 1 k + 2k + · · · + nk whose decimal expansions consist of the ten digits 0. and n4 = 1048576. and 9 without repetition. Find a number whose cube and fourth power together contain the ten digits. 11 17. Find a perfect square whose digits form one of the permutations of five consecutive digits. 5. Let k ≥ 3. 522 = 2704. 5 11. 75882 = 57577744 18 is the only possible number:183 = 5832 and 184 = 104976. 9. 6. 1. 6. 1. 4. 3. 12 5 6 In fact. (See also E538. (See also E714). Find a number of the form ab0cd whose square contains the nine digits 1. squares: 363652 = 1322413225. 326597184.842973156. 9. Find a perfect cube whose digits form a permutation of consecutive digits. Find the smallest four-digit number such that the sum of products of pairs of digits is equal to the sum of products of sets of three. 6. (See also E538). 13 The 14 243. 8. 5. Find three three-digit numbers in geometrical progression which can be derived from one another by cyclic permutation of digits. 636362 = 4049540496. Find a square of ten digits such that the two numbers formed by the first five and last five digits are consecutive.361874529.533 18.529874361. 14 20. 15 21. 22.627953481.576432081. 3. 7. 4. 2. 15 2033 = 8365427. 13 19. The only higher powers satisfying the same condition are the fifth powers of 32 and . 534 More digital trivia . Chapter 20 The shoemaker’s knife 1 Archimedes’ twin circles 2 Incircle of the shoemaker’s knife Archimedes’ construction Bankoff’s constructions Woo’s constructions 3 More Archimedean circles Exercise . 1 Archimedes’ twin circles Let P be a point on a segment AB . The two circles each tangent to CP . Suppose the smaller semicircles have radii a and b respectively. Q H U R V K A O1 O P O2 B A O1 O P O2 B Theorem 20. given by ab t= . AP and P B is called a shoemaker’s knife. The region bounded by the three semicircles (on the same side of AB ) with diameters AB . This perpendicular is an internal common tangent of the smaller semicircles.536 The shoemaker’s knife 20.1 (Archimedes). the largest semicircle AB and one of the smaller semicircles have equal radii t. a+b Q A O1 O P O2 B A O1 O P O2 B . Let Q be the intersection of the largest semicircle with the perpendicular through P to AB . The circle tangent to each of the three semicircles has radius given by ρ= a2 ab(a + b) .2 Incircle of the shoemaker’s knife 20.2 Incircle of the shoemaker’s knife 537 20. + ab + b2 C X Y A O1 O P O2 B Construction Z G Y X H A G C H B .1 Archimedes’ construction Theorem 20.20.2 (Archimedes).2. then the circle through the points P . the common radius of Archimedes’ twin circles. Y has the same radius as the Archimedean circles.3 (Leon Bankoff). The semiperimeter of the triangle CO1 O2 is a + b + ρ = (a + b) + (a + b)3 ab(a + b) = . = 3 (a + b) a+b This is the same as t. First construction Z Q1 X C Q2 C3 Y A O1 O P O2 B . Z C X Y A O1 O P O2 B Proof. The circle through P .2. O1 X = O1 P = a. If the incircle C (ρ) of the shoemaker’s knife touches the smaller semicircles at X and Y . X . O2 Y = O2 P = b.538 The shoemaker’s knife 20. Y is clearly the incircle of the triangle CO1 O2 .2 Bankoff’s constructions Theorem 20. since CX = CY = ρ. a2 + ab + b2 a2 + ab + b2 The inradius of the triangle is given by abρ = a+b+ρ ab · ab(a + b) ab . X . 20.2 Incircle of the shoemaker’s knife Second construction Z Q Y X P 539 A C B 20.3 Woo’s three constructions Z Z Q Y S P X B M Y A C X B A C Z Y X C B A .2. ab a+b O2 V = b. C5 M N V K C4 A O1 O P O2 B H U Let M be the midpoint of the chord HK . radius t = aab +b S A O1 O P O2 B Proof. The incircle of the curvilinear triangle bounded by the semicircle O (a + b) and the circles A(2a) and B (2b) has .4 (Thomas Schoch). Since O1 U = a. Theorem 20. From this. . (2a + x)2 + (2b + x)2 = 2[(a + b)2 + (a + b − x)2 ].3 More Archimedean circles Let UV be the external common tangent of the semicircles O 1 (a) and O2 (b). the circle tangent to HK and the minor arc HK of O (a + b) has radius t. Denote this circle by S (x). Let C4 be the intersection of O1 V and O2 U . it follows that (a + b) − QM = P N = 2t. = t. This means that the circle C4 (t) passes through P and C4 P = touches the common tangent HK of the semicircles at N . x = ab a+b = t. Since O and P are symmetric (isotomic conjugates) with respect to O 1 O2 . From this. OM + P N = O1 U + O2 V = a + b. By Apollonius theorem. This circle touches the minor arc at the point Q. which extends to a chord HK of the semicircle O (a + b).540 The shoemaker’s knife 20. Note that SO is a median of the triangle SO1 O2 . and O1 P : P O2 = a : b. similarly for the circles . We call the semicircle with diameter O1 O2 the midway semicircle of the shoemaker’s knife.20.3 More Archimedean circles 541 Exercise 1. the outer semicircle. Show that the circle tangent to the line P Q and with center at the intersection of (O1 ) and the midway semicircle has radius t = aab . Show that the radius of the circle tangent to the midway semicircle. +b Q C C A O1 O P O2 B 3. Show that they have equal radii t = aab +b C1 Q1 C1 C2 Q2 C2 A O1 O P O2 B 2. The circles (C1 ) and (C1 ) are each tangent to the outer semicircle of the shoemaker’s knife. and to OQ1 at Q1 . (C2 ) and (C2 ). and with center on the line P Q has radius . t = aab +b Q C A O1 O P O2 B . 000 Exercise Project: Prime links Project: Truncatable primes .Chapter 21 Infinitude of prime numbers 1 Fibonacci and Fermat numbers 2 Somos sequences 3 F¨ urstenberg’s topological proof made easy Appendix: Euclid’s proof Appendix: The prime numbers below 20. it follows that there are infinitely primes. . Fermat numbers The Fermat numbers are Fn := 22 + 1. Wunderlich. American Math. It is well known that Fermat’s conjecture of the primality of Fn is wrong. Note that Fn − 2 = 22 − 1 = 22 By induction. n n−1 +1 22 n−1 − 1 = Fn−1 (Fn−1 − 2). 72 (1965) 305. pk .1 Proofs by construction of sequence of relatively prime numbers Fibonacci numbers 1 Since gcd(Fm . . 1 M. From this.602 Infinitude of prime numbers 21. . While F0 = 3. F1 = 5. From this.n) . . F4 = 65537 n are all primes. Euler found that F5 = 232 + 1 = 4294967297 = 641 × 6700417. . Fpk are distinct. F3 = 257. F2 = 17. F1 . . and each one of them has only one prime divisors. . the Fermat numbers are pairwise relatively prime. . . we see that Fn does not contain any factor of F0 . . . Fn ) = Fgcd(m. Fn = Fn−1 Fn−2 · · · F1 · F0 + 2. This contradicts F 19 = 4181 = 37 × 113. . if there are only finitely many primes p1 . n ≥ 1. Fn−1 . Hence. . then the primes in Fp1 . Another proof of the infinite primes theorem. Monthly. . = b2 .2 Somos sequences Define two sequences (an ) and (bn ) with initial values a0 = a1 = 1. namely. b2n−1 . Then. inductively. . a2n+1 =b2 b4 · · · b2n . b2n . bn =an−1 + an . Here are the first few terms 7 8 9 ··· n 0 1 2 3 4 5 6 an 1 1 1 2 3 10 39 490 20631 10349290 · · · bn 0 1 2 3 5 13 49 529 21121 10369921 · · · Theorem 21. Proof. b2 . . . . . . b1 = 1. . . . b2 . that b1 . and for n ≥ 2 recursively by an =an−2 bn−1 . If we calculate the first few terms a2 a3 a4 a5 a6 a7 =a0 b1 =a1 b2 =a2 b3 =a3 b4 =a4 b5 =a5 b6 . . b2 . . n ≥ 1 are relatively prime. we see a pattern. = b1 . = b2 b4 . The terms of the sequence (bn ). . . contain a common divisor. b2 . a2n =b1 b3 · · · b2n−1 . . . b2n−1 are relatively prime.21. . = b1 b3 b5 . Suppose. . b0 = 0. b2n = a2n−1 + a2n = b2 b4 · · · b2n−2 + b1 b3 · · · b2n−1 does not contain any divisor of b1 .1. bn .2 Somos sequences 603 21. = b2 b4 b6 . . . . nor does b2n+1 = a2n + a2n+1 = b1 b3 · · · b2n−1 + b2 b4 · · · b2n contain any divisor of b1 . = b1 b3 . . It follows by induction that no two of the terms b1 . being finite. If A is periodic. Each arithmetic progression is closed as well as open. avoiding the language of topology. 10. The most recent issue of Mathematics Magazine 2 contains a paraphrase of this proof. For example. under this topology. a period set is a (finite) union of doubly infinite arithmetic progressions of the same common difference. A cannot be closed. then A3 ∪ A5 is the union of the arithmetic progressions of common difference 15. here is the entire article of [F¨ urstenberg]: We introduce a topology into the space of integers S . if A3 := {3n : n ∈ Z } and A5 := {5n : n ∈ Z }. by using the arithmetic progressions (from −∞ to +∞) as a basis. This contradiction shows that there are indeed infinitely many primes. It can be found in many books.604 Infinitude of prime numbers ¨ 21.. since its complement is the union of other arithmetic progressions (having the same difference). The following are clear.3 Furstenberg’s topological proof made easy There is a famous proof of the infinitude of primes using topology. 6. 1} is clearly not an open set. 2 Math. . 3. Hence A is not a finite union of closed sets which proves that there are an infinity of primes. 9. let Ap := {np : n ∈ Z} consists of the multiples of p. For each prime number p. If A and B are periodic sets. 76 (2003) number 3. In fact. so is their union. then so is its complement Z \ A. n ∈ A if and only if n + k ∈ A. Then. We say that a subset A ⊂ Z is periodic if there is an integer k such that for every integer n ∈ Z. which. and since the set {−1. 5. the union of S any finite number of arithmetic progression is closed. In other words. 1. 2. Mag. This is clearly periodic. the (finite) union A := p prime Ap is periodic. Consider now the set A = Ap . As a result. cannot be periodic. It is not difficult to verify that this actually yields a topological space. This complement is clearly the set {−1. The only numbers not belonging to A are −1 and 1. 3. the period of the union being the lcm of the periods of the sets. Suppose there are only finitely many primes. where Ap consists of all multiples of p. Let Z be the set of integers. containing the terms 0. and p runs through the set of primes ≥ 2. (2) extends to finite unions. and so is its complement. S may be shown to be normal and hence metrizable. Apart from an introductory sentence. 1}. . and reductio ad absurdum. is one of a mathematician’s finest weapon. 29. . 11. 5. on this hypothesis. P . The proof is by reductio ad absurdum. and therefore there is a prime (which may be Q itself) greater than any of them. . and let us. H. 3. and that 2. for it leaves the remainder 1 when divided by any one of these numbers. . But.. that there is no prime greater than P . 3. it is divisible by some prime.3 Furstenberg’s ¨ topological proof made easy 605 Appendix: Euclid’s proof 3 The prime numbers or primes are the numbers (A) 2. which cannot be resolved into smaller factors. Let us suppose that it does. P is the complete series (so that P is the largest prime). which Euclid loved so much. that the series (A) never comes to an end. . . . [Hardy. 23. 3 G. 13. 5. We have to prove that there are infinitely many primes. . . It is a far finer gambit than any chess gambit: a chess player may offer the sacrifice of a pawn or even a prize. This contradicts our hypothesis. 5. . Hardy’s paraphrase. . 3. 19. consider the number Q defined by the formula Q = (2 · 3 · 5 · · · P ) + 1.95–96]. if not itself prime. . . It is plain that Q is not divisible by any of 2. i.21. . 17. . and therefore this hypothesis is false.e. pp. but a mathematician offers the game. 7. 606 Infinitude of prime numbers Appendix: The prime numbers below 20000 10 20 30 40 50 60 70 80 90 100 100 200 300 400 500 600 700 800 900 1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 . 21.3 Furstenberg’s ¨ topological proof made easy 607 Exercise 1. Let pn denote the n-th prime number. Find the smallest value of n for which p1 p2 · · · pn + 1 is not a prime number. 2. Find the smallest value of n for which p1 p2 · · · pn − 1 is not a prime number. 3. Find the smallest value of n > 3 for which n!+1 is a prime number. 4. Find the smallest value of n > 7 for which n! − 1 is a prime number. 5. Find a shortest sequence of prime numbers p1 < p2 < · · · < pn satisfying the following conditions. (i) p1 = 2, (ii) pk+1 < 2pk for k = 1, . . . , n − 1, (iii) pn > 10000. 608 Infinitude of prime numbers Project: Prime links A prime link of length n is a permutation of 1, 2, . . . n beginning with 1 and ending with n such that the sum of each pair of adjacent terms is prime. This was proposed and solved by Morris Wald [157]. For n ≤ 6, the link is unique. For n = 7 there are two links: 1, 4, 3, 2, 5, 6, 7 and and 1, 6, 5, 2, 3, 4, 7. Wald suggested working backwards. Start with n and precede it with the greatest remaining member of the set whose sum with n is a prime, and repeat in like fashion. Here are the first 10 links: 1. 1, 2. 1, 2, 3. 1, 2, 3, 4. 1, 4, 3, 2, 5. 1, 4, 3, 2, 5, 6. 1, 4, 3, 2, 5, 6, 7. 1, 2, 3, 4, 7, 6, 5, 8. 1, 2, 3, 4, 7, 6, 5, 8, 9. 1, 2, 3, 4, 7, 6, 5, 8, 9, 10. Continue with larger values of n. 21.3 Furstenberg’s ¨ topological proof made easy 609 Project: Tuncatable primes Consider the prime number 73939133. The numbers obtained by truncations from the right are all primes: 73939133, 7393913, 739391, 73939, 7393, 739, 73, 7. More generally, we call a number N a right-truncatable prime if every number obtained from truncating N from the right is a prime or 1. A complete list of right-truncatable primes can be found in [Waltrom and Berg]. Write a computer program to find all right-truncatable primes. Similarly, define left-truncatable primes, and bi-truncatable primes as those which are both left- and right-truncatable primes. 53 317 599 797 2393 3793 3797 73331 7331 373393 23333 593993 23339 719333 31193 739397 31379 739399 37397 2399333 7393931 7393933 23399339 29399999 37337999 59393339 73939133 Here are all the bi-truncatable primes: 1 11 37 131 317 2 13 53 137 373 3 17 71 173 1373 5 23 73 311 3137 7 31 113 313 610 Infinitude of prime numbers Chapter 22 Strings of prime numbers 1 2 3 The prime number spiral The prime number spiral beginning with 17 The prime number spiral beginning with 41 Appendix: The number spiral Appendix: Long chains of primes Appendix: Consecutive primes with consecutive prime digital sums Projects 612 Strings of prime numbers 22.1 The prime number spiral The first 1000 prime numbers arranged in a spiral. = prime of the form 4n + 1; = prime of the form 4n + 3. 2 22.2 The prime number spiral beginning with 17 613 22.2 The prime number spiral beginning with 17 17 The numbers on the 45 degree line are n2 + n + 17. Let f (n) = n2 + n + 17. The numbers f (0), f (1), . . . f (15) are all prime. n 0 4 8 12 f (n) 17 37 89 173 n 1 5 9 13 f (n) 19 47 107 199 n 2 6 10 14 f (n) 23 59 127 227 n 3 7 11 15 f (n) 29 73 149 257 f (n) = n2 + n + 41 is prime for 0 ≤ n ≤ 39. n 0 5 10 15 20 25 30 35 f (n) 41 71 151 281 461 691 971 1301 n 1 6 11 16 21 26 31 36 f (n) 43 83 173 313 503 743 1033 1373 n 2 7 12 17 22 27 32 37 f (n) 47 97 197 347 547 797 1097 1447 n 3 8 13 18 23 28 33 38 f (n) 53 113 223 383 593 853 1163 1523 n 4 9 14 19 24 29 34 39 f (n) 61 131 251 421 641 911 1231 1601 .3 The prime number spiral beginning with 41 41 The numbers on the 45 degree line are f (n) = n2 + n + 41.614 Strings of prime numbers 22. 3 The prime number spiral beginning with 41 615 Prime number spiral beginning with 41: A closer look 41 .22. −(n − 1)) = (2n − 1)2 . . m) if 2m − 1 ≤ q ≤ 4m − 1. n) =  4n2 − n − m    2 4n − 3n + m if if if if m > |n|. if n > 0.616 Strings of prime numbers Appendix: The number spiral Beginning with the origin. let (2m − 1)2 be the largest odd square ≤ N .  (−m.    (−7m + q + 1. n). 2. n) = 2n(2n − 1).    (3m − q − 1. n) = 4n2 and More generally. f (n. − n = |n| > |m|. Then the number N appears at the lattice point  (m. we label the lattice points 0. −m) if 6m − 1 ≤ q ≤ 8m − 1. Along with this. f (n. 30 16 12 4 2 0 1 9 6 20 25 Given a positive integer N . n > |m|. 1. 5m − q − 1) if 4m − 1 ≤ q ≤ 6m − 1. n) the number at the lattice point (m.  4m2 − 3m + n    4m2 − m − n f (m. and write N = (2m − 1)2 + q. consecutively. . It is clear that along the 45-degree line. − m = |m| > |n|. f (−n. we trace out a spiral clockwise through the lattice points. if n ≥ 0. 0 ≤ q < 8m. Also. . −m + q + 1) if q ≤ 2m − 1. Denote by f (m. .22.3 The prime number spiral beginning with 41 617 Exercise 1. Label the vertices of a cube with the numbers 1 through 8 in such a way that the sum of the endpoints of each edge is prime. we only have 12 distinct primes. . 39. . f (−1). we still get two long chains of primes: f (−14). . 10. . . 183: n n3 + n2 + 17 173 5147771 174 5237731 175 5328733 176 5420783 177 5513887 178 5608051 179 5703281 180 5799583 181 5896963 182 5995427 183 6094981 184 6195631 185 6297383 factorization 683 × 7537 prime prime prime prime prime prime prime prime prime prime 13 × 476587 prime Higgins: 40 primes from g (x) = 9x2 − 231x + 1523. or h(x) = 9x2 − 471x + 6203 give the same primes in reverse order. f (−13). . x = 0. f (−2). f (10). . f (0). . . . .220] also considers the cubic function f (n) = n3 + n2 + 17. −13. Note that on the negative side. . The first of these begin with n = 717. This is true only when we take −1 as a prime. there is a string of 10 consecutive primes from −183 to −174. the string of 25 values are all primes. . f (−4) a chain of 11 primes. the longest strings of primes have 6 members. But f (0) = f (−1) = 17. and noted that for n = −14. . Replacing n by −n we consider n3 − n − 17 for n = 174. · · · . since f (−3) = −1. .618 Strings of prime numbers Appendix: Long chains of primes Beiler [p. . Beyond these. Even if we break the string into two. . 23). though only with 5 different values: 2442113. 29 3512051. 19 3511999. 13). 23 2442353. 11 14347. 17. 31 2442151. 19 5919937. (1307. 7. 31 3512053. 17). 313. But within this range he had overlooked the better records (102251. 17 2442191. And then five. 23 2442179. are consecutive primes. 29 Another “long” chain of 9 consecutive primes with 5 different consecutive prime digital sums can be found among “small” primes: 14293. 19). 3. 5). 5. and 199. 11 14369. The first sequence of five consecutive primes who digital sums form another sequence of 5 consecutive primes is (1291. 43 5920043. in some order. 197. 13 14387. 17 2442173..22. though these are the same for 313 and 331. 23 5919959. 337 all have prime digital sums. 11. 47 5920049. (102293. 31 Within the same range. 331. 23 2442359. 37 5920003. 37 3512057. 17 . 19 14321. with digit sums 11. 193. Problem 228. 23 14327. 11). 13. 29 2442263. (102301. for four consecutive primes having digital sums that. (1303. 7). (102253. 23 14323. 317. (102259. 19 2442197. 29 5919971. or 3. 19. there are also 15 consecutive primes whose digital sums are primes. 19). 41 5920069. 13 A little bit beyond these we find the best record for eight up to the first 1 million primes: 5919931. 19 2442199. the first quadruple is 191. Beyond these. The beginning of the prime number sequence provides an easy answer: just consider the primes 2. 7) and 3511973. 29 2442133. 23 2442287. 13).3 The prime number spiral beginning with 41 619 Appendix: Consecutive primes with consecutive prime digital sums Charles Twigg asked. (1301. 13 14303. 5. 29 2442289. in Crux Math. (1297. 17 3511993. Twigg listed such quadruples and quintuples up to primes around 5 million (about 350000 primes). 23 3512011. The five consecutive primes 311. (102299. 31 2442311. 19 14341. 7. 23 2442227. 11). 620 Strings of prime numbers Project Find the coordinates of the n-th point of the path. What is the position of the lattice point (a. beginning with the origin. b) in the sequence? . Chapter 23 Strings of composites 1 Strings of consecutive composite numbers 2 Strings of consecutive composite values of n 2 + 1 3 Strings of consecutive composite values of n 2 + n + 41 Appendix: 50 consecutive composite values of x 2 + x + 41 Project . (n + 1)! + 3. we give the first string of n consecutive composite numbers. with 111 composite numbers. The string beginning with 396734 just misses by 1. 1 These are significantly less than 101!. (n + 1)! + (n + 1) are all composites. the n numbers (n + 1)! + 2.1 Strings of consecutive composite numbers It is well known that there are strings of consecutive composite numbers of arbitrary lengths. . very large. n 3 5 7 13 17 19 21 33 35 43 51 71 85 95 111 113 117 131 147 153 179 209 first string of n composite numbers 8 · · · 10 24 · · · 28 90 · · · 96 114 · · · 126 524 · · · 540 888 · · · 906 1130 · · · 1150 1328 · · · 1360 9552 · · · 9586 15684 · · · 15726 19610 · · · 19660 31398 · · · 31468 155922 · · · 156006 360654 · · · 360748 370262 · · · 370372 492114 · · · 492226 1349534 · · · 1349650 1357202 · · · 1357332 2010734 · · · 2010880 4652354 · · · 4652506 17051708 · · · 17051886 20831324 · · · 20831532 The first string of 100 consecutive composite numbers begins with 370262. For example.622 Strings of composites 23. 1 It actually extends to 370372. In the table below. however. it gives 99 consecutive composites. · · · . These numbers are. 2 Strings of consecutive composite values of n2 + 1 m 3 9 13 15 19 33 39 45 87 99 111 129 151 211 n···n + m − 1 7···9 27 · · · 35 41 · · · 53 95 · · · 109 185 · · · 203 351 · · · 383 497 · · · 535 3391 · · · 3435 3537 · · · 3623 45371 · · · 45469 82735 · · · 82845 99065 · · · 99193 357165 · · · 357315 840905 · · · 841115 .23.2 Strings of consecutive composite values of n 2 + 1 623 23. L.3 Consecutive composite values of x2 + x + 41 Problem 142 of Crux Mathematicorum asks for 40 consecutive positive integer values of x for which f (x) = x2 + x + 41 are all composites. The string of 38 consecutive numbers beginning with this all give composite f (x).000. . Several solutions were published. then f (xn ) = x2 n + xn + 41 = (1601!) · (1601! + 2b + 1) + f (n) is a multiple of f (n) which is greater than f (n). A near miss is 176955. beginning with 2561526. 2 For example. then (and now) editor of Journal of Recreational Mathematics. 191. which contains 13 primes. beginning with 1204431. with factorization of the corresponding f (x). Nelson. There are three such strings. 39. There are longer strings. f (n) ≤ f (39) = 1601. It begins with 1081296.000. The only strings of ≥ 10 consecutive primes begin with 66. we have a string of 45 composites. H.000. the longest string of composites has 50 numbers. . For example. 3033715. found this smallest string. if we set xn = 1601! + n. 2 Up to 5. and 3100535 respectively. How about long strings of primes? They are relatively few. and no more up to 5. See Appendix 2 for the first of these strings. Unfortunately these numbers were quite large. The Crux editor wrote that “[i]t would be interesting if some computer nut were to make a search and discover the smallest set of 40 consecutive integers x for which f (x) is composite”.000. . Since for n = 0. 534. being constructed by a method similar to the one above.624 Strings of composites 23. 219. except the one beginning with 219. . . These numbers are therefore composite. These numbers are quite large since 1601! has 4437 digits. here is one. Each of these strings contains 10 primes. and 179856. 23.3 Consecutive composite values of x 2 + x + 41 625 Appendix: 50 consecutive composite values of x2 + x + 41 x 2561525 2561526 2561527 2561528 2561529 2561530 2561531 2561532 2561533 2561534 2561535 2561536 2561537 2561538 2561539 2561540 2561541 2561542 2561543 2561544 2561545 2561546 2561547 2561548 2561549 2561550 2561551 2561552 2561553 2561554 2561555 2561556 2561557 2561558 2561559 2561560 2561561 2561562 2561563 2561564 2561565 2561566 2561567 2561568 2561569 2561570 2561571 2561572 2561573 2561574 2561575 2561576 x2 + x + 41 6561412887191 6561418010243 6561423133297 6561428256353 6561433379411 6561438502471 6561443625533 6561448748597 6561453871663 6561458994731 6561464117801 6561469240873 6561474363947 6561479487023 6561484610101 6561489733181 6561494856263 6561499979347 6561505102433 6561510225521 6561515348611 6561520471703 6561525594797 6561530717893 6561535840991 6561540964091 6561546087193 6561551210297 6561556333403 6561561456511 6561566579621 6561571702733 6561576825847 6561581948963 6561587072081 6561592195201 6561597318323 6561602441447 6561607564573 6561612687701 6561617810831 6561622933963 6561628057097 6561633180233 6561638303371 6561643426511 6561648549653 6561653672797 6561658795943 6561663919091 6561669042241 6561674165393 Factorization prime 1693357 × 3874799 1097 × 5981242601 167 × 971 × 40463429 499 × 13149165089 167 × 797 × 853 × 57793 773 × 8488284121 71 × 133261 × 693487 379 × 7591 × 2280667 1512947 × 4336873 39233 × 167243497 347 × 2339 × 8084281 367 × 17878676741 3049 × 2152010327 53 × 83 × 661 × 2256559 3947 × 1662399223 10501 × 624844763 3557 × 1844672471 71 × 92415564823 47 × 139606600543 722317 × 9083983 53 × 4973 × 24894887 2927 × 2241723811 419 × 15659977847 472 × 2970364799 2003 × 3275856697 43 × 919 × 1039 × 159811 83 × 12577 × 6285667 151 × 43454015453 43 × 152594452477 653 × 10048340857 41 × 160038334213 41 × 160038459167 2053 × 3196094471 9049 × 725117369 1601 × 8933 × 458797 1994669 × 3289567 691 × 23689 × 400853 4111 × 1596109843 131 × 419 × 1259 × 94951 238363 × 27527837 4783 × 1371863461 2039 × 3218061823 61 × 97 × 1108945949 694367 × 9449813 5417 × 6529 × 185527 347 × 18909649999 4933 × 1330154809 5839 × 1123764137 151 × 397 × 109457753 313 × 1999 × 10487143 prime . 626 Strings of composites Project: Strings of consecutive composite values of n2 + n + 1 Find the first strings of consecutive composite values of n 2 + n + 1. The string with more than 100 composite values of n 2 + 1 has 111 members beginning with 82735. What is the longest of such a string you can find? . m n···n + m − 1 3 9 · · · 11 5 28 · · · 32 Project The first string of 99 consecutive composite values of n 2 + 1 begins with n = 45371. Chapter 24 Perfect numbers 1 Perfect numbers 2 Charles Twigg on the first 10 perfect numbers 3 Abundant and deficient numbers Appendix: Mersenne primes Appendix: Three important number theoretic functions Exercise . 13 (1991) 40. If 1 + 2 + 22 + · · · + 2k−1 = 2k − 1 is a prime number. Theorem 24. 1 Proof.628 Perfect numbers 24. Note: 2k − 1 is usually called the k -th Mersenne number and denoted by Mk . Theorem 24. Look at the odd divisors of n. a contradiction: perfect squares don’t exist. .2 (Euler).1 (Euclid). They all divide the largest of them. then k must be prime. In particular. Intelligencer. Open problem Does there exist an odd perfect number? Theorem-joke 24. then 2k−1 (2k − 1) is a perfect number. This shows that the odd divisors of n come in pairs a. Every even perfect number is of the form given by Euclid. Therefore the number of odd divisors of n is also odd. Suppose n is a perfect square. Hence n is not perfect. If Mk is prime. 1 Math.1 (Hendrik Lenstra).1 Perfect numbers A number is perfect is the sum of its proper divisors (including 1) is equal to the number itself. which is itself a square. b where a · b = d 2 . say d 2 . Perfect squares do not exist. Only d is paired to itself. it is not 2n. the prime 19. 2 • Three repdigits are imbedded in P5 . The first and last digits of P4 are like cubes. • P4 terminates both P11 and P14 . • P10 is the smallest perfect number to contain each of the ten decimal digits at least once.24. Let Pn be the n-th perfect number. In P2 . The sums of the digits of P 3 and P4 are the same. n 1 2 3 4 5 6 7 8 9 10 k 2 3 5 7 13 17 19 31 61 89 Mk 3 7 31 127 8191 131071 524287 2147483647 2305843009213693951 618970019642690137449562111 Pn = 2k−1 Mk 6 28 496 8128 33550336 8589869056 137438691328 2305843008139952128 2658455991569831744654692615953842176 191561942608236107294793378084303638130997321548169216 • P1 is the difference of the digits of P2 .2 Charles Twigg on the first 10 perfect numbers 629 24. • P7 contains each of the ten decimal digits except 0 and 5. and therefore 39 known perfect numbers.2 Charles Twigg on the first 10 perfect numbers There are only 39 known Mersenne primes. • P3 and P4 are the first two perfect numbers prefaced by squares. the units digit is the cube of the of tens digit. The first two digits of P3 are consecutive squares. See Appendix. . • P9 is the smallest perfect number to contain each of the nine nonzero digits at least once. It is zerofree. 2 These contain respectively 65 and 366 digits. namely. equal to. 5775 and 5776 are the first pair of abundant numbers. 43065 61424. The first triple of abundant numbers 3 n 171078830 171078831 171078832 3 Discovered factorization 2 · 5 · 13 · 23 · 29 · 1973 33 · 7 · 11 · 19 · 61 · 71 24 · 31 · 344917 σ (n) 358162560 342835200 342158656 σ (n) − 2n 16004900 677538 992 in 1975 by Laurent Hodges and Reid. then n is abundant. including 1 and n itself. 7425 26144. 39376 58695. 61425 77175. 76545 81675. 79696 84524. or less than 2n. 58696 76544.3 Abundant and deficient numbers A number n is abundant. 945 is the first odd abundant number. 5776 21735. 89776 5984. 20 is abundant. 70785 81080. 21945 43064. But not conversely. 77176 82004. equal to. 5985 21944. perfect. The advantage of using σ (n) is that it can be easily computed if we know how n factors into primes: Abundant numbers up to 200: 12 78 138 200 18 80 140 20 84 144 24 88 150 30 90 156 36 96 160 40 100 162 42 102 168 48 104 174 54 108 176 56 112 180 60 114 186 66 120 192 70 126 196 72 132 198 Deficient even numbers up to 200: 2 58 122 172 4 62 124 178 8 64 128 182 10 68 130 184 14 74 134 188 16 76 136 190 22 82 142 194 26 86 146 32 92 148 34 94 152 38 98 154 44 106 158 46 110 164 50 116 166 52 118 170 All multiples of 6 are abundant. 82005 91664. 11025 27404. 69616 79695. 364]. 49665 69615. or deficient if the sum of its proper divisors (including 1 but excluding n itself) is greater than.. 98176 11024. 56925 70784. 27405 56924. Pairs of consecutive abundant numbers up to 10. perfect. 84645 . 84525 98175.630 Perfect numbers 24. 81081 84644. or less than n. 81676 89775. 21736 39375. 26145 49664. 91665 7424. If we denote by σ (n) the sum of all divisors of n. p.000: 5775. or deficient according as σ (n) is greater than.. [Pickover. . Robinson H.Slowinski W.Robinson R.Slowinski Armengaud. Woltman.Robinson R.Robinson A.B.Pervushin E. Kurowski et.M.Gillies D.M.Fauquembergue R.B.Euler R.Riesel A.3 Abundant and deficient numbers 631 Appendix: Mersenne primes Primes of the form Mk = 2k − 1 are called Mersenne prime.N.Slowinski.Gillies B.Tuckerman C.Noll.Slowinski D.24.M. Woltman. The only known Mersenne primes are listed below. Cameron. Kurowski et.B. Woltman.Lucas R.Robinson R.Hurwitz D.Cataldi I.A. al. al.E.Noll D.Slowinski D.Nelson. Woltman et al. .Gillies C. Clarkson. L.M. Hajratwala. et.Powers E. k 17 31 89 127 607 2203 3217 4423 9941 19937 23209 86243 132049 756839 1257787 2976221 6972593 Year 1588 1750 1911 1876 1952 1952 1957 1961 1963 1971 1979 1982 1983 1992 1996 1997 1999 Discoverer P.Slowinski D. M13466917 has 4053946 digits and is the largest known prime.al.Colquitt.P.M.Hurwitz D. Woltman. L.Welsch D.Nickel H.A.Cataldi L. Kurowski et. k 19 61 107 521 1279 2281 4253 9689 11213 21701 44497 110503 216091 859433 1398269 3021377 13466917 Year 1588 1883 1913 1952 1952 1952 1961 1963 1963 1978 1979 1988 1985 1993 1996 1998 2001 Discoverer P.M. al. D.Gage Slowinski and Gage Spence. The number of divisors of n is k d(n) = i=1 (1 + ai ). 2. . pi i=1 These functions are all multiplicative in the sense that f (mn) = f (m)f (n) whenever gcd(m. 1. The number of positive integers < n which are relatively prime to n is given by k 1 φ(n) = n 1− . n) = 1. pi − 1 3. is k σ (n) = i=1 i +1 pa i . including 1 and n itself.632 Perfect numbers Appendix: Three important number theoretic functions Let n be a positive integer with prime factorization k n= i=1 i pa i There are several important numbers associated with n. The sum of divisors of n. This was Computer Challenge 511 of Journal of Recreational Mathematics. . E1 lists them by increasing denominator.... . the rational number m n gcd(m.24. . 1 1 1 2 1 1 2 3 1 3 1 . .1). χ(m. . n) = 1.1 Appendix: Two enumerations of the rational numbers in (0. . . Thus. . .3 Abundant and deficient numbers 633 24. . . . ... 1 1 2 1 3 1 2 3 4 1 5 . . and for equal denominators. by increasing numerator. E2 : 2 3 4 3 5 6 5 4 7 5 8 . otherwise. . Thus.. and for equal sums of terms. by increasing numerator. . lists them by increasing sum of numerator and denominator. n) in enumeration E1 and position 1 2 m+n−1 m φ (k ) + k =3 k =1 χ(k. . .1) Consider two enumerations of the rational numbers in (0. 4 1 2 9 30 59 67 97 197 513 Fraction 1 2 3 5 23 73 143 163 235 477 1238 Position 1 2 7 158 1617 6211 8058 16765 69093 465988 What is the next match? For m + n ≤ 20000. m + n − k ) in enumeration E2 . 10 (1977–78) 122–123. . and 2 occupy the same positions in both seThe fractions 1 2 3 5 (with m < n and quences. Here. .. The solution lists 10 of these. on the other hand. 1 . n) = 1) occupies position E1 : n−1 m φ (k ) + k =2 k =1 χ(k. . . I found four more: Fraction Position 4 Journal 1729 4175 1922 4641 3239 7820 4646 11217 5297983 6546724 18588348 38244610 of Recreational Math. n) = 1 0 if gcd(m. 2 3 3 4 4 5 5 5 5 6 6 E2 . .3. More generally. All the people. Starting with the 6th in line. The athletic young warden was ordered to free a certain number of these prisoners at his discretion. During his third walk. each third person was given three jewels. Starting with the 8th in line. On the 6th day of the 28th anniversary of his accession to the throne. he shuts every second door. Starting at the beginning again. The prisoners in cells whose doors were still open were freed. until at last he had completed the full hundred walks. closing it if it was open and opening it if it was shut. while very sparsely populated. each second person was given two jewels. he stopped at every third door: if it was open he shut it. including the king the advisors. Starting with the second in line. opening closed doors and closing open doors. . How many people were there in Pythagora ? 2. show that if σ (n) is prime. The isle of Pythagora. if it was shut he opened it. The man at the extreme end of the line noticed that the number of jewels he received corresponded to his position in line. and this is how he did it. Which were the lucky cells? 3. And so on. each fourth person was given four jewels. On his fourth walk he did the same. Starting with the 4th in line. the king of the island called a meeting of his 496 advisors to decide how to celebrate the auspicious occasion.634 Exercise Perfect numbers 1. For a positive integer n. and so on. each person was given one jewel. They decided to divide the regal jewels among the people of the land. were lined up in a single file. is capable of supporting a population of thirty million. except he did it for every fourth door. A minimum security prison contains 100 cells with one prisoner in each. then so is d(n). First he walked along the row of cells opening every door. starting at the beginning. On his fifth walk he stopped at every fifth door. and the jewels were distributed as follows. Chapter 25 Routh and Ceva theorems 1 Routh theorem: an example 2 Routh theorem 3 Ceva theorem and coordinates of triangle centers centroid incenter Gergonne point Nagel point Exercise . Y . Q. CY : Y A = 5 : 3. R can be worked out easily: P = BY ∩ CZ Q = CZ ∩ AX R = AX ∩ BY Y = (5 : 0 : 3) Z = (2 : 3 : 0) X = (0 : 1 : 2) Z = (2 : 3 : 0) X = (0 : 1 : 2) Y = (5 : 0 : 3) P = (10 : 15 : 6) Q = (2 : 3 : 6) R = (10 : 3 : 6) This means that the absolute barycentric coordinates of X . The lines AX . AZ : ZB = 3 : 2. 6479 .1 Routh theorem: an example Given a triangle ABC . X = (0 : 1 : 2). Find the area of triangle P QR. X . 2 X 1 C Those of P .702 Routh and Ceva theorems 25. BY . 11 R= 1 (10A + 3B + 6C ). Z = (2 : 3 : 0). Y = (5 : 0 : 3). Z are P = 1 (10A + 15B + 6C ). CZ bound a triangle P QR. A 3 3 Y R Z 5 P Q 2 B We make use of homogeneous barycentric coordinates with respect to ABC . Y . 19 The area of triangle P QR = 10 15 6 1 2 3 6 · 31 · 11 · 19 10 3 6 = 576 . 31 Q= 1 (2A + 3B + 6C ). Suppose triangle ABC has area . Z are points on the side lines specified by the ratios of divisions BX : XC = 2 : 1. µν + µ + 1 1 (A + νB + νλC ).25. λµ + λ + 1 P = From these.2 Routh theorem 703 25. . Z = (1 : ν : 0). Y . Q= νλ + ν + 1 1 R= (λµA + B + λC ). Those of P . R can be worked out easily: P = BY ∩ CZ Q = CZ ∩ AX R = AX ∩ BY Y = (µ : 0 : 1) Z = (1 : ν : 0) X = (0 : 1 : λ) Z = (1 : ν : 0) X = (0 : 1 : λ) Y = (µ : 0 : 1) P = (µ : µν : 1) Q = (1 : ν : νλ) R = (λµ : 1 : λ) This means that the absolute barycentric coordinates of X . µ µν 1 1 ν νλ λµ 1 λ · Area(P QR) = (µν + µ + 1)(νλ + ν + 1)(λµ + λ + 1) (λµν − 1)2 = · (µν + µ + 1)(νλ + ν + 1)(λµ + λ + 1) . Q. Y = (µ : 0 : 1).2 Routh theorem A 1 ν R Z Y µ P Q 1 B λ X 1 C We make use of homogeneous barycentric coordinates with respect to ABC . X = (0 : 1 : λ). Z are 1 (µA + µνB + C ). BY . Example: centroid If AX . Y . The lines AX . CZ can be written down by combining the coordinates of X . XC Y A ZB If this condition is satisfied. the intersection is the centroid G: A 1 1 Z Y G 1 1 B 1 X 1 C X Y Z G = = = = (0 : 1 : 1) (1 : 0 : 1) (1 : 1 : 0) (1 : 1 : 1) .704 Routh and Ceva theorems 25. Z . Let X . Z be points on the lines BC . AB respectively. CZ are the medians.3 Ceva Theorem Theorem 25. the homogeneous barycentric coordinates of the common point of AX . BY . BY . CZ are concurrent if and only if BX CY AZ · · = 1.1 (Ceva). Y . CA. 3 Ceva Theorem 705 Example: incenter If AX . CZ are the angle bisectors. BY .25. the intersection is the incenter I: A c b Y Z I a a B c X b C X Y Z I = = = = (0 : b : c) (a : 0 : c) (a : b : 0) (a : b : c) . Y . CZ intersect at the Gergonne point. A s−a s−a Y Z P s−c s−b B s−b X s−c C X Y Z P = (0 : s − c : s − b) = (s − c : 0 : s − a) = (s − b : s − a : 0) = = = = = 0: 1 s −a 1 s −a 1 s −a 1 s −b 1 : s− c 1 : 0 : s− c 1 : s− :0 b 1 1 : s− : s− b c . the lines AX .706 Routh and Ceva theorems Example: Gergonne point If X . BY . Z are the points of tangency of the incircle with the sidelines. Z are the points of tangency of the excircles with the respective sidelines. Y . BY .3 Ceva Theorem 707 Example: Nagel point If X . A s−b Z s−c Y s−a Q s−a B s−c X s−b C X Y Z Q = = = = (0 : s − b : s − c) (s − a : 0 : s − c) (s − a : s − b : 0) (s − a : s − b : s − c) . the lines AX . CZ intersect at the Nagel point.25. Project In the Routh formula. ν = 6. (c) λ = 3. how should one choose integer values for λ. 2. (b) λ = 1. Calculate the area of triangle P QR given (a) λ = µ = ν = 2. µ = 7. 1 of that of triangle ABC for and ν so that the area of triangle P QR is n an integer n? . µ. ν = 4. Calculate the homogeneous barycentric coordinates of the orthocenter of triangle ABC . µ = 7.708 Routh and Ceva theorems Exercise 1. µ.25. ν with numerators and denominators < 10 for which the area of triangle P QR is k times that of ABC with the numerator and denominator of k less than 10. λ 1 9 1 7 1 6 1 6 1 4 2 7 1 3 3 8 4 9 1 2 5 9 5 8 3 4 5 6 7 8 µ 4 5 1 3 1 5 7 5 1 4 1 6 4 3 5 6 1 8 3 7 ν 9 8 1 6 6 5 2 7 1 4 7 5 1 4 1 2 9 5 2 3 9 4 k 1 9 1 2 2 9 1 6 3 7 1 6 1 8 1 9 1 9 1 8 1 9 1 7 1 5 2 9 1 9 1 9 1 4 1 6 1 9 1 6 2 9 1 6 1 3 1 8 1 9 1 8 1 8 3 7 1 9 1 6 1 7 λ 1 8 1 7 1 6 1 5 1 4 1 3 1 3 2 5 1 2 1 2 4 7 2 3 3 4 6 7 7 8 µ 1 3 2 5 1 2 6 7 1 3 1 6 8 5 1 4 1 8 1 2 1 4 1 6 ν 8 5 7 6 3 4 1 2 4 3 1 7 1 8 5 8 3 8 1 2 k 1 7 1 6 1 5 1 6 1 8 1 2 1 7 2 9 1 3 1 7 1 9 1 8 1 8 1 9 1 9 1 8 1 9 1 9 1 9 1 8 1 6 1 9 1 6 2 9 1 3 1 2 1 4 1 6 1 5 1 6 1 3 λ 1 8 1 7 1 6 1 5 1 4 1 3 3 8 2 5 1 2 1 2 3 5 2 3 4 5 6 7 8 9 µ 3 8 ν 1 2 1 4 6 7 1 6 2 5 3 5 8 7 1 7 6 7 1 3 1 2 3 7 1 9 1 5 k 1 3 1 4 1 9 2 9 2 9 1 6 1 9 1 6 1 6 1 6 1 6 1 8 1 9 1 6 1 9 1 9 1 9 1 6 1 8 1 8 1 9 1 8 1 5 1 6 1 6 1 7 1 9 2 9 1 8 1 4 1 9 λ 1 8 1 6 1 6 1 4 1 4 1 3 3 8 3 7 1 2 1 2 5 8 5 7 5 6 6 7 µ 9 5 1 7 ν 4 9 1 3 2 3 k 1 9 1 2 1 8 1 4 1 9 1 9 1 3 1 8 1 9 1 5 2 9 1 6 1 9 1 6 1 4 1 8 1 9 2 9 1 5 1 7 1 9 1 7 1 9 1 9 1 9 1 6 2 9 2 9 1 2 1 2 1 9 1 7 8 6 5 5 8 1 2 1 3 7 6 1 5 3 5 1 3 1 2 9 8 1 2 5 4 1 1 7 1 4 7 3 8 1 8 1 2 5 6 1 6 1 4 7 2 3 8 1 8 7 1 2 2 3 3 8 3 4 2 5 8 3 1 6 8 1 2 3 1 6 1 1 4 7 8 8 3 1 6 4 5 6 1 2 5 2 1 4 5 6 7 4 7 6 1 6 1 4 7 1 7 3 2 7 6 1 5 3 2 3 1 9 8 7 9 7 4 1 3 1 1 7 6 6 5 7 5 8 5 1 9 8 7 6 5 4 3 2 5 3 1 8 7 7 6 4 3 3 2 7 4 4 3 8 1 1 8 7 6 5 4 3 8 5 9 5 7 1 7 8 3 2 7 4 2 5 6 6 1 6 6 8 9 5 1 4 7 3 2 1 3 2 1 6 5 2 4 7 6 8 3 9 1 2 6 5 2 1 8 4 9 6 1 3 1 8 6 3 8 3 5 3 1 3 2 2 4 7 3 4 3 6 7 2 2 7 3 8 3 5 8 3 2 6 5 3 4 2 2 5 2 8 3 3 8 5 2 2 5 2 2 9 4 8 3 2 5 9 7 8 8 3 5 7 5 2 5 6 3 4 4 6 6 8 2 2 4 3 4 4 7 6 7 2 5 8 3 4 8 7 5 7 3 3 4 5 6 7 8 8 7 1 2 4 3 6 7 4 2 6 7 7 6 6 7 5 3 2 8 3 7 8 2 5 2 8 3 2 3 3 4 5 6 7 8 2 5 8 3 7 2 8 7 4 6 8 5 1 5 6 6 3 2 4 9 4 1 1 5 9 4 6 6 7 9 3 6 8 9 5 7 3 5 4 .3 Ceva Theorem 709 Appendix We give those values of λ. 710 Routh and Ceva theorems . Chapter 26 The excircles 1 2 3 4 5 Feuerbach theorem A relation among radii The circumcircle of the excentral triangle The radical circle of the excircles Apollonius circle: the circular hull of the excircles Appendix: Three mutually orthogonal circles with given centers . A F Fb Fc I N B Fa C .712 The excircles 26.1 Feuerbach theorem The nine-point circle is tangent internally to the incircle and externally to each of the excircles. Ib M A Ic rb O I rc r D B C M ra Ia ra − r =2DM . .2 A relation among the radii ra + rb + rc = 4R + r.26.2 A relation among the radii 713 26. ra + rb + rc − r =4R. rb + rc =2MD = 2(2R − DM ). O Ib = O Ic = 2R. and radius twice the circumradius. Ib A Ic O I O B D X C Ia O Ia =ra + O X =ra + 2OD − r =ra + 2(R − DM ) − r =ra + 2R − (ra − r ) − r =2R.3 The circumcircle of the excentral triangle The circle through the excenters has center at the reflection of the incenter in the circumcenter. Similarly.714 The excircles 26. (from previous page) . A I I B C .26.4 The radical circle of the excircles 715 26. the incenter of the medial triangle. Its radius is 2 r 2 + s2 .4 The radical circle of the excircles The circle orthogonal to each of the excircles has center √ at the Spieker 1 point. 5 Apollonius circle: the circular hull of the excircles Fc A Fa Fc I N B Fa C Fb fb .716 The excircles 26. . CA = b. then 2 2 + rc = a2 . C that form an acute-angled triangle. rc = (a2 + b2 − c2 ). 2 2 2 These are all positive since ABC is an acute triangle. construct a semicircle externally of the triangle. rb 2 2 rc + ra = b2 . construct three circles with these points as centers that are mutually orthogonal to each other. 2 2 ra + rb = c2 . A Y E Z F B D H C X Solution Let BC = a. Label these X . (1) With each side as diameter.26. B (Z ) and C (X ) are mutually orthogonal to each other. 1 1 1 2 2 2 ra = (b2 + c2 − a2 ).5 Apollonius circle: the circular hull of the excircles 717 Appendix: Three mutually orthogonal circles with given centers Given three points A. Z on the semicircles on BC . AB respectively. It follows if we extend BE ra 2 to intersect at Y the semicircle constructed externally on the side AC as 2 . rb . rb = (c2 + a2 − b2 ). so that 2 =1 (b2 + c2 − a2 ) = bc cos A = AC · AE . and CX = CY . Consider the perpendicular foot E of B on AC . then. Y . (3) The circles A(Y ). From these. rc respectively. These satify AY = AZ . Therefore we have the followdiameter. AY 2 = AC · AE = ra ing simple construction of these circles. B . (2) Extend the altitudes of the triangle to intersect the semicircles on the same side. and AB = c. BZ = BX . CA. If these circles have radii ra . Note that AE = c cos A. 718 The excircles . 1 = + = . nk+1 mk+1 3 4 2 3 nk mk 1 . 6. 21. 28. 55. 27. 45. 1 n1 m1 1 . 3. 36. . 2 The first few of these are 1. 10. . 15. m2 if and The n-th triangular number Tn = 1 2 only if (2n + 1)2 − 2(2m)2 = 1.2 Special triangular numbers Triangular numbers which are squares n(n + 1) is a square. . .Chapter 27 Figurate numbers 27. say.1 Triangular numbers The nth triangular number is 1 Tn = 1 + 2 + 3 + · · · + n = n(n + 1). 720 Figurate numbers Find the first few triangular numbers which are squares: k nk mk Tn 1 2 3 1 8 49 1 6 35 1 36 1225 4 5 6 7 Palindromic triangular numbers n Tn 1 1 2 3 3 6 10 55 11 66 18 171 34 595 36 666 77 3003 n Tn n Tn 109 5995 3185 5073705 132 8778 3369 5676765 173 15051 3548 6295926 363 66066 8382 35133153 1111 617716 11088 61477416 1287 828828 18906 178727871 1593 1269621 57166 1634004361 1833 1680861 102849 5289009825 2662 3544453 111111 6172882716 T11111111 = 61728399382716. . 3 Pentagonal numbers The pentagonal numbers are the sums of the arithmetic progression 1 + 4 + 7 + · · · + (3n − 2) + · · · n(3n − 1).3 Pentagonal numbers 721 27. The nth pentagonal number is Pn = 1 2 Palindromic pentagonal numbers n Pn n Pn 1 1 101 12521 2 5 693 720027 4 22 2173 7081807 26 1001 2229 7451547 44 2882 4228 26811862 n 6010 26466 26906 31926 44059 Pn 54177145 1050660501 1085885801 1528888251 2911771192 .27. the k -gonal numbers are the sums of the arithmetic progression 1 + (k − 1) + (2k − 3) + · · · .n = 1 2 .722 Figurate numbers 27.k More generally.4 The polygonal numbers Pn. The nth k -gonal number is Pk. for a fixed k . n((k − 2)n − (k − 4)). b) the smallest positive solution of x2 − dy 2 = 1) but with (x1 . b (2) If the equation x2 − dy 2 = −1 has a solution in nonzero integers.4. yn x1 y1 = a . y ) = (a. b) is the smallest positive solution. If (x. . y1) given by its smallest positive solution. The positive integer solutions of the equation x2 − dy 2 = 1 can be arranged in a sequence as follows. then xn+1 yn+1 = a db b a xn .k 723 27.27. its integer solutions can be arranged in the form a sequence satisfying the same recurrence relation above (with (a.4 The polygonal numbers Pn.1 Appendix: Solution of Pell’s equation (1) Let d be a positive integer which is not a square. 724 Figurate numbers Exercise 1. sequentially from one. the Government’s mathematical whiz-kid informs the Minister that there is precisely a 50-50 chance that the other two numbers will both be less than the one just displayed. Ramanujan’s house number. . An unidentified country has 7-digit population – and everyone has been given a National ID Number. Prove that every hexagonal number is a triangular number. When the first number appears on the screen. What is the population. and what is the first number? 1 1 Problem 2585. 2. What is this value? 5. 31 (2002–2003) 71. 3. 4. 6. allocated by no identifiable logic. JRM. Find a pentagonal numbers (apart from 1) that is also a square. It is known that there is only one value of n for which 12 + 22 + 32 + · · · + n2 is a square. The Censure Minister has chosen three names at random. and is finding their ID number on the computer. Find two triangular numbers (apart from 1) that are squares. c) are all odd integers exceeding 1. c) satisfying Pk. by dividing throughout by (k − 4) 2 . This is always an integer point for k = 3.5) −2) . 4. we rewrite (28. 6. By completing squares. 6. 2c + 1). that this determines a rational point on the surface S: x2 + y 2 = z 2 + 1.2) as (2(k − 2)a − (k − 4))2 + (2(k − 2)b − (k − 4))2 = (2(k − 2)c − (k − 4))2 + (k − 4)2 . 5.a + Pk. For a given positive integer k . a. b.4) (28. (28. b. 2b + 1. c) := (ga − 1. a. a. (28. c) := (2a + 1.3) and note. we shall change signs. The coordinates of P (3. b.b = Pk.n := A 4-gonal triple is simply a Pythagorean triple satisfying a 2 + b2 = c2 . b. (28. and consider instead the point P (3. the sequence of k -gonal numbers consists of the integers 1 (k − 2)n2 − (k − 4)n . gc − 1). 8. 3. we mean a triple of positive integers (a.Chapter 28 Polygonal triples We consider polygonal numbers of a fixed shape. P (k .2) Pk.1) 2 By a k -gonal triple.6) . (28. with where g = 2(kk− 4 corresponding g = −2. We shall assume that k = 4. For k = 3 (triangular numbers). namely. (28. gb − 1.c . Figure 28. z0 ) be a point on the surface S. (28. We summarize this in the following proposition. y = y0 + qt.7) and (28.8) qx0 − py0 = r. being the surface of revolution of a rectangular hyperbola about its conjugate axis. It has a double ruling. through each point on the surface. is a rectangular hyperboloid of one sheet. = ±1. y0. It follows that r2 = = = = This means 2 2 r 2 (x2 0 + y0 − z0 ) 2 2 r 2 (x2 0 + y0 ) − (px0 + qy0 ) 2 2 2 2 (p + q )(x0 + y0 ) − (px0 + qy0 )2 (qx0 − py0 )2 . z = z0 + rt. we determine the direction numbers of the line. A line through P with direction numbers p : q : r has parametrization : x = x0 + pt. .. i. p2 + q 2 = r 2 . (28.1: Let P (x0 .e.7) (28.9) Solving equations (28. there are two straight lines lying entirely on the surface.726 Polygonal triples 28.1 Double ruling of S The surface S.4) shows that the line is entirely contained in the surface S if and only if px0 + qy0 = rz0 . Substitution of these expressions into (28.9). 2 Primitive Pythagorean triple associated with a k -gonal triple 727 Proposition 28. r y0 = qz0 − p . there are odd integers u and v such that qu + rv = 1. q. we obtain. We study the converse question of determining k -gonal triples from (primitive) Pythagorean triples.8). The two lines lying entirely on the hyperboloid S : x2 + y 2 = z 2 + 1 and passing through P (x 0 . q = 2mn. r ) for these direction numbers. As is well known. b. from (28. we . r ) as in (28. it follows from the fact that g = 2(kk− > 2. b.3 Triples of triangular numbers Given a primitive Pythagorean triple (p. 4 The direction numbers of the ruling lines on S through the point P . since q is even. r (28. c) corresponding to it. if P is a rational point. c). as given by (28. Given an odd integer z0 > 1.7) and (28. In view of (28. By the euclidean algorithm. for k ≥ 5 and (28.11) We claim that it is possible to choose z 0 > 1 so that x0 and y0 are also odd integers > 1. x0 = pz0 + q . (Note that v must be odd.2 Primitive Pythagorean triple associated with a k gonal triple Let P be the rational point determined by a k -gonal triple (a.28. z0 ) have direction numbers 2 x0 z0 − y0 : y0 z0 + x0 : x2 0 + y0 for = ±1.10) for relatively prime integers m > n of different parity. r = m2 + n2 (28.10).6) for k = 3 (triangular numbers). every such triple is given by p = m2 − n2 . y0. are all positive.5). If u is even. 28. q.9). we may therefore choose a primitive Pythagorean triple (p. In particular. This is clear for −2) k = 3. as given in Proposition 1. We first note that the coordinates of P all exceed 1.1. and for k ≥ 5. these direction numbers are rational. we want to determine a triangular triple (a. 28. 24. 2. 8) (6. b+ (0). 37) (21. 41) (8. the integers z0 . 9.4 k -gonal triples determined by a Pythagorean triple Now. 2) (4. c (t)). This makes y0 an integer. 10) (4. the integer (k − 2) > 1 cannot divide both n and m − n. c+ (0)) (2. = ±1. v + q ). From (28. 29) (7. 12. 20. 35. 36) (35. Theorem 28. Triangular triples from primitive Pythagorean triples (m. we consider k ≥ 5. q. v ) by (u − r. 5. −4) Theorem 28. Let (p. 4) (p. 4.728 Polygonal triples replace (u. h if h is even. 21) (6. 28. 5) (15. 3) (9.10) by relatively prime integers m > n with different parity corresponds to a k -gonal triple if and only if one of 2gn h if h is odd. 15) (14. 2) (5. 14. The primitive Pythagorean 2 triple (p. the integers x0 and y0 are both odd.2. 13) (35. 6) (5. This means that a primitive Pythagorean triple . 1) (7. 28) (9. and y0 can be chosen greater than 1. since p and q are of different parity and z0 is odd. r ) defined in (28. 40. There are two infinite families of triangular triples (a (t). 2 and 2(m−n) g is an integer. 27. 7. 1) (4.3. We shall adopt the notation h := for an integer h. 36) (a− (0). n) (2. 1) (3. 6. 6. Clearly. in which both entries are odd). 12. q. 41) (a+ (0). has direction numbers p : q : r . 8. 4. 25) (63. 17) (5. 8. 20) (7. 3. 2) (6. 16. 5. 28) (27. We summarize this in the following theorem. r ) (3. 6) (5. r ) be a primitive Pythagorean triple. a (t). 21) (35. 14. Since m and n are relatively prime. 45) 28. b− (0). 1) (5. 5. c (t)). 44. Let k ≥ 5 and g = 2(kk− . the integer z0 = pu is such that qz0 − p = p(qu − 1) is divisible by r . 11) (9. The corresponding x0 is also an integer.11). x0 . b (t). 15) (14. Replacing z0 by z0 + rt for a positive integer t if necessary. 21. 20. P = P (3. 10) (20. such that one of the lines (P ). 65) (45. q. c− (0)) (3. 53) (9. b (t). 3) (8. 116. Indeed. 97) (17. 41) (45. 8) (11. 8 ) 3 (3. (4h + 2) − gonal triples (h. 78) (68. b0 = 8h2 + 2h + 1. 215. 21) (11. 4) (7. 38.28. and cannot divide the odd integer m − n. with n a multiple of 2h give (4h + 2)-gonal pairs. 4) (11. 44. 75. 40. b. 2) (7. 125) (38. 122) (21. n). 176. 88. 28. 144. 132. 39) (18. 185) (21. bt = 8h2 + 2h + 1 + (8h2 + 4h)t. 14. 29) (9. It follows that only those pairs (m. 125) (105. 105. 12) (14. 84. 117) (57. 82) (11. 145) (117. 176. 84. c) (5. 4) (9. 132. 140. 56. 11. 77) (17. 10) (5. 145) (105. 141) (104. For example. 5. 41) (33. 4) (7. 4) (7. 72. 10. 4) (m. These give an infinite family of (4h + 2)-gonal triples: at = (4h + 1)(t + 1).4 k -gonal triples determined by a Pythagorean triple 729 (p. 220. 144. 12 ) 5 . 6) (11. 2) (11. 55. (k − 2) is the even number 2h. 21) (13. 11. k. r = 8h2 + 4h + 1. 221) (9. 85) (65. 216) (9. 185) (13. 40. 6) (p. 8) (7. 50. a0 = 4h + 1. r ) corresponds to at most one line on S associated with k -gonal triples (for k ≥ 5). 8) (11. 37. 12. 157) (a. 88. ct = 8h2 + 2h + 2 + (8h2 + 4h + 1)t. 131. we have p = 4h + 1. 4) (9. 2) (9. 2) (5. q = 8h2 + 4h. 6) (11. 6) (9. by choosing m = 2h + 1. 137) (57. 137) (85. 72. n = 2h. 4) (11. n) (3. 56. 36. 64) (52. 20. 81. r ) (5. 2) (5. if k = 4h + 2. 174. 183) (13. 80) (85. 53) (33. 156) (2. 57. q. 79. q. 65) (65. 39. 97) (17. 139) (90. 8) (11. 12) (13. 18. 6. 157) (57. 85) (77. 14. c0 = 8h2 + 2h + 2. 4) (9. 111) (60. 13. 19) (9. g ) (1. 13) (21. 65) (13. 138. 85) (85. 38) (33. 19) (17. 4h) (2. 84. 3) (10. c ) are related by (m − n)(c − c ) ≡ n 2h − 1 (mod m2 + n2 ). 64. 6) (8. 85) (91. 16. 23) (10. 10. 31) (40. 50. 180. n) (4. 5) (7. 80. 24) (5. 5. 173. 23. 47) (19. 7) (9. 140. 181) (2h + 1) − gonal (a. 57. 1) (10. 10. 5. (k1 − 2) = (k2 − 2) = 2h − 1. 109) (51. 28. 10. 88) (24. 22. 4) (8. q. 16) (13. 180. 16) (7. while (k1 − 4) = 2h − 3. 72. 27. 8) (13. 20. 17) (7. 49) (26. 5) (10. 149) (19. 83) (22. 1) (4. 21. 12. 29) (33. 60. 72. The (2h + 1)-gonal triple (a. 36) (29. for some h ≥ 2. 110. 9) (p. 1) (10. 20. 97) (63. 24. 20) (75. Theorem 2 suggests that there is a one-to-one correspondence between k1 -gonal triples and k2 -gonal triples. 60. 32.5 Correspondence between (2h + 1)-gonal and 4hgonal triples Let k1 < k2 be two positive integers ≥ 5. 113) (77. 11. 8. c ) (14. 3) (8. 36. 48) (60. 147) (95. 174) 4h − gonal (a . (2h + 1) − gonal and 4h − gonal triples (h. 97) (19. 7. 58) (33. 30) (15. 1) (6. 25) (21. 79. 80. 71) (96. 175) (33. 8) (7. 107. 5) (9. provided (k1 − 2) = (k2 − 2) . 3) (5. 3) (10. 8) (m. 89) (99. 57) (44. 52) (36. (28. 107) (37. 14) (8. 140. 2) (7. 48) (26. 4) (7. 16. 82. 36. 36) (39. 108) (51. 177. 73) (39. 18. 6. 81) (44. 34. 62) (29. 3) (8. 20) . 48. 41) (90. 12. 4. 8. 65) (13. 45. 37) (11. 61) (45. 35) (11. 20. 172. 58) (31. 20. 62) (15. 149) (99. 101) (91. 7) (10. 17. 9. 28.12) In this case. 174. 9) (6. 85) (55. 181) (35. 48) (46. c) and a 4h-gonal triple (a . 60. 90) (17. r ) (15. 24. 56. 9. 108) (18. 106. 178) (16. 53) (55. 73) (39. b. 8. 89) (65. k2 = 4h. 48. 98) (52. 80) (48. 138. 12) (4. This is the case if and only if k 1 = 2h + 1. 29) (35. 2) (10. 20. b. 109) (51. 80. 56. 52. b . 173) (3. 17) (11. 19. 1) (8. 65) (15. 14. 7) (10. 16) (5. 39) (27. 7. c) (7. 117) (19. 2h + 1. 38. 7) (4. 85) (19. 58) (32. b .730 Polygonal triples 28. 101) (19. 60. 2) (8. 112. and (k2 − 4) = 2h − 2. 80. Chapter 29 Sums of consecutive squares 1 Sum of squares of natural numbers 2 Sums of consecutive squares: odd number case 3 Sums of consecutive squares: even number case . . Let Tn = 1 + 2 + 3 · · · + n = 1 2 Sn = 12 + 22 + 32 + · · · + n2 . Find n so that n2 + (n + 1)2 is a square. 1 12 + 22 + 32 + · · · + n2 = n(n + 1)(2n + 1). 23 = 13 33 = 23 3 33 4 = . 2. Since Tn is known. we have (n + 1)3 = 13 + 3Sn + 3Tn + n. 6 n(n + 1) and Proof.1. Find 12 + 32 + 52 + · · · + (2n − 1)2 . 3 n = (n − 1)3 n3 (n + 1)3 = + + + 3 · 12 3 · 22 3 · 32 + + + 3·1 3·2 3·3 + 1 + 1 + 1 + 3(n − 1)2 + 3(n − 1) + 1 + 3 · n2 + 3·n + 1 Combining these equations.802 Sums of consecutive squares 29. . we have Sn = 1 3 (n + 1)3 − n − 1 − n(n + 1) 3 2 1 = n(n + 1)(2n + 1).1 Sum of squares of natural numbers Theorem 29. . 6 Exercise 1. 3 (Ek ) 1. 3. the equation (Ek ) has no solution: . .2 Sums of consecutive squares: odd number case Suppose the sum of the squares of 2k + 1 consecutive positive integers is a square. 0 1 1 2 7 3 5 3 3 10 . . Number of solutions of (Ek ) when 2k + 1 is a square 2k + 1 25 49 121 169 289 361 529 625 841 961 . . In each case. Suppose 2k + 1 is a square. then the equation (E k ) has no solution. b ± 1.29. Show that (Ek ) has solution only when k = 6m(m + ) for some integers m > 1. . If the integers are b. Find the unique sequence of 49 (respectively 121) consecutive positive integers whose squares sum to a square. . . Suppose 2k + 1 is not a square. From this we obtain the equation 1 a2 − (2k + 1)b2 = k (k + 1)(2k + 1).2 Sums of consecutive squares: odd number case 803 29. b ± k . We require 1 (2k + 1)b2 + k (k + 1)(2k + 1) = a2 3 for an integer a. the number of solutions is finite. . 4. Verify that for the following values of k < 50. . 2. Find the two sequences of 169 consecutive squares whose sums are squares. and = ±1. If k + 1 is divisible 9 = 3 2 or by any prime of the form 4k + 3 ≥ 7. 44. 30. 2181933022). 8. 18. 25. 39. 22. 29 are the equations (Ek ) solvable. 98478182. 16. we obtain two sequences of solutions of (E73 ): Answer: (4088. 9. . 3. Work out 5 sequences of 23 consecutive integers whose squares add up to a square in each case. for example. 44. Suppose p = 2k +1 is a prime. (23360. (18642443912. 2734). 21. . 16. Consider the equation (E36 ) : u2 −73v 2 = 12 · 37 · 73. only for k = 5. 45. v ) = (4088. 10. (85056113063608088. . 31. 21. 48. b) = (2281249. the equation (Ek ) has no solution: − 1 k (k +1) 1. 11. 267000). 34. . . 43. 15. 42. The fundamental solution of the Pell equation x 2 − 73y 2 = 1 being (a. the sum of the squares of the 73 numbers with center 478 (respectively 2734) is equal to the square of 4088 (respectively 23360). 56912849921762734). 9955065049008478). 478). (486263602888235360. This means. 29. 17. 35. 12474054766). 8. Verify that for the following values of k < 50. 7. 42782 . 26.804 Sums of consecutive squares k = 6. 38. . . 28. 23. 2052522. . 2734). 478). 20. 13. 32. 3 = 5. it remains to consider (Ek ) for the following values of k: 5. 40. 19. . 36. 6. (106578370640. This equation does in fact have solutions (u. (23360. 47. 37. 20. 7. 46. 8. then the equation (Ek ) has no solution. 14. For k ≤ 50. 26. 2. Among these. If the Legendre symbol p −1. . 27. 23. Answer: 72 + 82 + · · · + 292 881 + 8822 + · · · + 9032 427872 + 427882 + · · · + 428092 20534012 + 20534022 + · · · + 20534232 ········· 2 = = = = 922 . 11. 41. 33. 49. 17. 11. the equation (Ek ) has no solution for the following values of k : k = 3. 23. This means (2a)2 − 2k (2b + 1)2 = 2k (4k 2 − 1). 33. or by any prime of the form 4k + 1. The equation (Ek ) can be written as (2b + 1)2 − 2ky 2 = − 4k 2 − 1 . 31. 40. Show that the equation (Ek ) has no solution if 2k is a square. 29. is equal to a2 . the equation (E k ) has solutions only for k = 1. 38. 1. 31. 9. 6. 5.3 Sums of consecutive squares: even number case Suppose the sum of the squares of the 2k consecutive numbers b − k + 1. 4k 2 − 1 are relatively prime. 5. Show that (34. 47. then the equation (E k ) has no solution. 13. . Excluding square values of 2k < 100. 3 (Ek ) Note that the numbers 2k . b + k. . . 7. Suppose 2k is not a square. 39.29. 2. 3. 12.3 Sums of consecutive squares: even number case 805 29. 19. 37. 41. 29. 7) are solutions of (E ” 12 ). . 49. b. (50. 15. . . 27. 43. 3). 41. 24. b + k − 1. 3 By considering Legendre symbols. 17. Let k be a prime. 35. 4. b − k + 2. 21. the equation (E k ) has no solution for the following values of k ≤ 50: k = 5. 0). . 45. . 44. (38. . For k ≤ 50. Construct from them three infinite sequences of expressions of the sum of 24 consecutive squares as a square. Show that if 2k + 1 is divisible by 9. 4. 8. 2942 + 2952 + · · · + 3672 = 28492 . 330).235). The equation (E37 ) has solutions (185.806 Sums of consecutive squares 7. 59252 + 59262 + · · · 60122 = 559902. 130962 + 130972 + · · · + 131792 = 7638652 . 4) and (2222. The equation (E44 ) has solutions (242. and (2849. Answer: 2252 + 2262 + · · · + 2982 = 22572 . 1922 + 1932 + · · · + 2792 = 22222.261). 2). . (2257. From these we construct three infinite sequences of expressions of the sum of 74 consecutive squares as a square. From these we obtain two infinite sequences of expressions of the sum of 88 consecutive squares as a square. 1 (Bernoulli). 5 2 3 where c = 1 − 1 5 +1 + 2 1 3 = −1 . where c is determined by the condition that the sum of the coefficients is 1. (1) S3 (n) = 13 + 23 + · · · + n3 = 1 4 4 3 (2) Since 4S3 (n) = n + 2n + n2 . 5 2 3 30 .Chapter 30 Sums of powers of natural numbers Notation Sk (n) := 1k + 2k + · · · + nk . we have 1 1 1 S4 (n) = n5 + n4 + n3 + cn. Sk (n) is a polynomial in n of degree k + 1 without constant term. Examples n2 (n + 1)2 . It can be obtained recursively as Sk+1 (n) = (k + 1)Sk (n)dn + cn. 30 Therefore. 1 1 1 1 14 + 24 + · · · + n4 = n5 + n4 + n3 − n. Theorem 30. 3. 2. Find the sum of the series 1 · 2 · 3 + 2 · 3 · 4 + 3 · 4 · 5 + · · · + n(n + 1)(n + 2). 4. Find the sum of the first n triangular numbers. Find the sum of the first n odd numbers. Find S5 (n) and S6 (n). 5.808 Sums of powers of natural numbers Exercise 1. . Find the sum of the cubes of the first n odd numbers. 29 (2003) 193–195. . The total. Randy and Hannah are eating at a restaurant.Chapter 31 A high school mathematics contest Christopher Newport University Regional High School Mathematics Contest. November. Hannah leaves a tip of 20% of her items. paid by the pair is exactly 70 dollars. 2002 1 1. The items ordered by Randy cost twice as much as the items ordered by Hannah. including tips.. How much was the cost of the items Hannah ordered? 1 Crux Math. Randy leaves a tip of 15% of the price of what he has ordered. 810 A high school mathematics contest 2. but 34534 is not. 3. How many mountain numbers are greater than 70000? . 34541 is a mountain number. If ap = q and aq = p. Let (an ) be an arithmetic sequence. For example. Solve the equation x2 − |x| − 1 = 0. find ap+q . 4. A five-digit number is called a mountain number if the first three digits are increasing and the last three are decreasing. while Wai’s score on Wednesday was 20% less than his score on Tuesday. However. Strangely. while Wai’s score on Tuesday was 20 points higher than on Monday. Hai and Wai separately play a video game and compare scores. on Wednesday. Hai’s score on Tuesday was 10% less than his score on Monday.811 5. Each day. What were their scores on Wednesday? . Hai’s score plus Wai’s score turned out to be the same on all three days. Hai’s score was 10 points higher than on Tuesday. What is the total area of the two triangles P AD and P BC (shaded in the figure)? A B P D C 7. In the year after that. What is the smallest whole number of years for which this is possible? . the stock price fell by 99%. Samantha bought a stock for 1000 dollars whose price then doubled every year for the next n years. the stock was still worth than 1000 dollars. Nevertheless. A point P is given in the interior of a rectangle ABCD with AB = CD = 24 and AD = BD = 5.812 A high school mathematics contest 6. Determine the shape of the triangle.813 8. as shown in the diagram. What is the area of the region inside the larger circle. In triangle ABC . cos(A − B ) + sin(A + B ) = 2. 9. but outside all the smaller circles? . Four small circles of radius 1 are tangent to each other and to a larger circle containing them. The two circles touch each other. as indicated. and each touches two adjacent sides of the rectangle. Two circles of radii 9 and 17 centimeters are enclosed in a rectangle with one side of length 50 centimeters. Find the area of the rectangle.814 A high school mathematics contest 10. . Find three different prime numbers a. What is her average speed. She did this by running for 10 minutes. and so on until she crossed the finish line.815 11. 12. 49 minutes. while running? . Karen ran a 42 kilometer marathon in 3 hours. c so that their sum a + b + c and their product abc both end in the digit 7. walking for 2 minutes. in kilometers per hour. She runs twice as fast as she walks. walking for 2 minutes. then running for 10 minutes. b. 816 A high school mathematics contest . M.Chapter 32 Mathematical entertainments 1 David Wells’ survey of Beauty in Mathematics 2 T. Apostol’s mathematical acrostic . 10:4 (1988) 31. c. 4 1 2×3×4 1 4×5×6 1 6×7×8 I π is transcendental. K The maximum area of a quadrilateral with sides a. Intelligencer. b. J Every number greater than 77 is the sum of integers. 6 π −3 . C If p(n) is the number of partitions of n. F Every prime of the form 4n + 1 is the sum of two integral squares in exactly one way. A F K P U B G L Q V C H M R W D I N S X E J O T A Euler’s formula for a polyhedron: V − E + F = 2. where s is half the perimeter. then 5((1 − x2 )(1 − x10 )(1 − x15 ) · · · )5 ((1 − x)(1 − x2 )(1 − x3 )(1 − x4 ) · · · )6 =p(4) + p(9)x + p(14)x2 + · · · .1 Beauty in mathematics: David Wells’ survey Give each of the following theorems a score for beauty between 0 (the least) and 10 (the most beautiful). 1 Math. E There is no rational number whose square is 2.818 Mathematical entertainments 1 32. D The number of primes is infinite. B Any square matrix satisfies its own characteristic equation. . the sum of whose reciprocal is 1. d is (s − a)(s − b)(s − c)(s − d). G 1+ H 1 22 + − 1 32 +···+ + 1 n2 +··· = −··· = π2 . N The number of partitions of an integer into odd integers is equal to the number of partitions into distinct integers.1 2 819 L There is no equilateral triangle whose vertices are plane lattice points. there is a pair of people who have the same number of friends present. T The number of representations of an odd number as the sum of 4 squares is 8 times the sum of its divisors. O If the points of the plane are each colored red. X There are 5 regular polyhedra. there is a pair of points of the same color of mutual distance unity. Q A continuous mapping of the closed unit disk into itself has a fixed point. or blue. of an even number. U The word problem for groups is unsolvable. V The order of a subgroup divides the order the group. yellow. 24 times the sum of its odd divisors. The difference is 2n in the n-th place. S A regular icosahedron inscribed in a regular octahedron divides the edges in the golden ratio. M At any party. . . √ R Write down the multiples of 2. W eiπ = −1. . 3 6 10 13 17 20 23 27 30 . . P Every plane map can be colored with 4 colors. igonoring fractional parts. .32. and underneath the number missing from the first sequence: 1 2 4 5 7 8 9 11 12 . 2 Math. M. Intelligencer. When completely filled in. the diagram contains a passage from a published work concerning mathematics. The initial letters of the WORDS spell out the author and title of the work. Apostol’s mathematical acrostic 2 Guess as many WORDS as you can. then write each letter in the correspondingly numbered square in the diagram. . 10:3 (1988) 43.820 Mathematical entertainments 32. All the WORDS are related to mathematics or mathematicians.2 T. 2 2 821 A B C D E F G H I J K L M N O P Unit of speed 131 The second cervical vertebra In opposition Countless 61 Pallet of an escapement 180 Successive volumes (2 words) Pupil of Gauss.32. last name followed by initials One way to describe Pythagoras Exact opposite Make a mosaic of 148 First to prove e transcendental Logical and sophistical reasoning A type of polynomial Providing pleasure or delight Added Contour connecting points of equal depth below a water surface One of the first to use the method of successive approximations Equal to the end proposed Three-digit code 98 44 106 23 112 51 65 152 99 38 174 95 19 134 140 105 128 130 168 57 40 78 107 16 37 77 79 135 161 54 94 6 56 126 96 123 118 166 178 91 173 68 165 173 117 13 73 181 167 88 113 116 M 2 = 0 (two words) 80 Distribution 157 French mathematician (1765–1843) Directed Lowness in pitch 83 One of Cayley’s hyperdeterminants German geometer (1833–1872) 114 108 46 22 154 5 14 67 31 81 17 26 92 21 122 176 59 35 7 48 102 10 58 138 127 71 60 32 18 100 149 147 164 179 29 53 89 146 50 69 97 66 141 139 109 85 158 34 111 1 182 30 184 15 52 3 159 45 144 133 137 41 155 121 12 90 86 27 82 142 170 119 125 33 25 75 70 55 101 76 64 185 150 43 49 110 93 63 163 42 183 115 24 87 160 9 4 2 153 177 84 62 20 28 145 171 104 Q 74 11 124 103 156 R S T U V W X Y Z 129 169 39 136 132 120 162 72 151 143 172 36 8 47 . 32.2 2 901 17 1 J 2 C 3 E 4 B 5 P 6 Z 7 O 8 Y 9 A 16 10 K 11 Q 12 I 13 Z 14 J 15 H 16 Y 17 N 18 K 19 V 20 B 21 J 15 22 N 23 U 24 D 25 I 26 O 27 G 28 C 29 N 30 F 31 L 32 J 14 33 H 34 L 35 N 36 W 37 Z 38 S 39 R 40 V 41 F 42 D 13 43 G 44 S 45 G 46 M 47 Z 48 P 49 D 50 N 51 W 52 I 53 O 12 54 X 55 H 56 T 57 U 58 L 59 M 60 P 61 D 62 A 63 D 64 I 11 65 X 66 N 67 K 68 V 69 N 70 G 71 O 72 R 73 Y 74 Q 75 F 10 76 I 77 T 78 W 79 U 80 T 81 M 82 H 83 X 84 C 85 J 86 F 9 87 B 88 U 89 P 90 E 91 T 92 P 93 D 94 Y 95 U 96 V 97 N 98 S 8 99 Z 100 L 101 I 102 J 103 Q 104 C 105 Y 106 T 107 X 108 L 7 109 L 110 D 111 L 112 V 113 T 114 K 115 D 116 T 117 Y 118 X 119 F 120 R 6 121 H 122 K 123 W 124 Q 125 G 126 U 127 N 128 Z 129 R 130 S 5 131 A 132 R 133 I 134 W 135 V 136 R 137 E 138 M 139 K 140 X 141 J 4 142 I 143 R 144 H 145 C 146 P 147 N 148 J 149 M 150 G 151 R 152 Y 3 153 A 154 O 155 G 156 Q 157 U 158 K 159 F 160 C 161 W 162 R 163 D 2 164 O 165 W 166 Y 167 T 168 T 169 R 170 E 171 C 172 V 173 X 1 174 T 175 U 176 L 177 B 178 Z 179 P 180 E 181 Y 182 J 183 D 184 G 185 F 0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 . 902 Mathematical entertainments . We have 1000 feet of fencing and wish to make with it a rectangular pen. at one side of a long wall.Chapter 33 Maxima and minima without calculus 1. How can we arrange to surround the maximum area? 1000 − 2x x x . A Norman window has a fixed perimeter a. Find the largest possible area. A right pyramid has a square base and a given surface area A.904 Maxima and minima without calculus 2. r h 2r 3. What is the largest possible volume? h l 2b . 905 4. What is the largest possible capacity of the tray? 5. The perimeter of a triangle is a constant 2s. What is the largest possible area of the triangle? . A tray is constructed from a square metal sheet by dimension a × a by cutting squares of the same size from the four corners and folding up the sides. what is the least possible surface area? 7. 2r h r .906 Maxima and minima without calculus 6. Inscribe in a given cone a cylinder whose volume is largest possible. If the volume is a constant V . The volume of a cylindrical can is given by V = πr 2 l and the surface area by A = 2πr (l + r ). 907 8. R r h . Find the largest cylinder contained in a sphere of radius R. h R r .908 Maxima and minima without calculus 9. Find the largest right circular cone contained in a sphere of radius R. 909 10. Two corridors of widths a and b meet at a right-angled corner. What is the length of the longest ladder which may be carried round the corner? Assume the workman sufficiently unimaginative to keep the ladder horizontal). v b a u . 910 Maxima and minima without calculus . If β < 7. P is false. If β < 7. Answer: 2. None of these. Which one of the following statements must be true? A B C D E If Q is true. If Q is true. β < 0. If Q is false. β ≥ 0. P is true. Gazette. β ≥ 0. A sufficient condition for the truth a statement Q is that β < 0. 53 (1969) 357–362. . Which one of the following statement must be true? A B C D E If β ≥ 7. 1 1. P is true. None of these. A necessary condition for the truth of a statement P is that β ≥ 7. β < 0. If Q is false. If β ≥ 7. P is false.Chapter 34 A British test of teachers’ mathematical background Samples from a test on the mathematical background of (British) teachers in training in the late 1960’s. Answer: 1 Math. Answer: x 2 −9 x −3 when x = 3? . What is the value of A 0 B 1 C 6 D ∞ E Do not know Answer: 5. D The converse is false but requires dis-proof. Which of the following statements is necessarily true? A The converse is true and does not require proof. B The converse is true but requires proof.912 A British test of teachers’ mathematical background 3. E None of the above. A certain theorem in Euclidean geometry has been proved. Which of the words in boldfont should be altered? A The B quadrilateral C cyclic D supplementary E None of them Answer: 4. C The converse is false and does not require dis-proof. “The condition for a quadrilateral to be cyclic is that the opposite angles must be supplementary”. if any. Answer: 7. Answer: . Which of the following is necessarily correct? A S is true. B S is true. Statement S: All triangles possess property Q. It has been proved that an infinite number of triangles possess a property Q. does the first mistake in the calculation occur? A (I) B (II) C (III) D (IV) E None of these: it is correct. but proof is required. Consider the following calculation (all logarithms to base e): −3 −5 dx 3 = [log(x + 1)]− −5 x+1 (I) (II) (III) (IV) 3 [log(x + 1)]− −5 = log(−2) − log(−4) −2 log(−2) − log(−4) = log −4 −2 = − log 2 log −4 In which line. E None of the above. C S is more likely to be true than false.913 6. and no further proof is required. D S is more likely to be false than true. B The best order is 1M . An algebraic problem involves a. and we have to write the sum of the products.914 A British test of teachers’ mathematical background 8. N in some order. D The best order is 1L. We wish to letter the points 1. two at a time. 2N . 3M . D The best order is cb + ab + ac. 2M . C The best order is 1N . Answer: 9. 3L. 2L. 3M . With which of the following statements do you agree? A The best order is ab + ac + bc. 2. E There is no “best” order. Answer: . B The best order is ac + ba + cb. b. E There is no “best” order. 3 in the diagram with the letter L. 2N . M . C The best order is bc + ca + ab. c. 3N . A 1 3 B 2 C With which of the following statements do you agree¿ A The best order is 1L. the ratio of Xavior’s age to Yolanda’s age is 6:7. . the ages of Xaviere. What is Zo¨ e’s present age? 1 Grades 10 and 11.. Yolanda. 29 (2003) 129–132. and Zo¨ e were in the ratio 1:2:5. Today.Chapter 35 A mathematical contest 2001 Canadian Invitational Mathematics Challenge 1 1 Thirty years ago. Crux Math. x2 − y 2 + z 2 =2. . x − 3y 2 + z =0.916 A mathematical contest 2 Solve the system of equations x + y + z =2. 917 3(a) A flat mirror is perpendicular to the xy -plane and stands on the line y = x + 4. A laser beam from the origin strikes the mirror at P (−1. 3(b) A flat mirror is perpendicular to the xy -plane and stands on a line L. Determine the equation of the line L. 0). Determine the coordinates of the point Q. A laser beam from the origin strikes the mirror at P (−1. 3) and is reflected to the point Q on the x-axis. . 5) and is reflected to the point Q(24. . (a) Show that f (n) can be written as the product of two quadratic polynomials with integer coefficients.918 A mathematical contest 4 Determine all pairs of nonnegative integers (m. n) which are solutions to the equation 3(2m ) + 1 = n2 . 5 Let f (n) = n4 + 2n3 − n2 + 2n + 1. (b) Determine all integers n for which |f (n)| is a prime number. Chapter 36 Some geometry problems from recent journals 1 Crux Mathematicorum Jim Totten Department of Mathematics and Statistics University College of the Cariboo Kamloops. Monthly Problems Department of Mathematics Texas A&M University 3368 TAMU College Station. Problem Editor Department of Mathematics Iowa State University Ames. Canada. IA 50011 American Mathematical Monthly Doug Hensley. BC. PA 16214 2 3 4 . TX 77843-3368 Pi Mu Epsilon Journal Michael McConnell 840 Wood Street Mathematics Department Clarion University Clarion. V2C 4Z9 Mathematics Magazine Elgin Johnston. 920 Some geometry problems from recent journals Crux 2813. Regina. NB and J. Let P and Q be the points of intersections of the line MD with the circle. Prove that the rectangle AP BQ is a golden rectangle. √ P B : P A = ( 5 + 1) : 2. Chris Fisher. University of Regina. Fredericton. proposed by Barry Monson. radius MA(= MB ). SK. center M . where P is inside the square and Q is outside. . Canada Suppose that M is the midpoint of side AB of the square ABCD. University of New Brunswick. that is. proposed by Peter Woo. and that F and F are the foci of the ellipse Λ that is tangent to the four sides of Π at their midpoints. La Mirada. CA Suppose that Π is a parallelogram with sides of lengths 2a and 2b and with acute angle α. (b) Find a straight-edge and compass construction for F and F . .921 Crux 2822. b and α. (a) Find the major and minor semi-axes of Π in terms of a. Biola University. LC λ CM 1−µ = . Clifton COllege. proposed by Christopher J. UK Suppose that L. AB respectively. Bradley. N are points on BC . and are distinct from A. BNL. and CLM meet at the Miquel point P . C . Suppose further that 1−λ BL = . µ.922 Some geometry problems from recent journals Crux 2823. NB ν and that the circles AMN . ν and the side lengths of triangle ABC . M . . MA µ AN 1−ν = . Find [BCP ] : [CAP ] : [ABP ] in terms of λ. Britol. CA. B . J. If Ω is the center of the circumcircle of triangle Ia Ib Ic .923 Crux 2830. Zaltbommel. proposed by D. R) is the circumcircle of triangle ABC . the Netherlands Suppose that Γ(O. Smeenk. Suppose that Ia . Ic are the centers of the excircles of triangle ABC opposite A. Ib . B . determine the locus of Ω as C varies. C respectively. Suppose that side AB is fixed and that C varies on Γ (always on the same side of AB ). . We conclude that y = z . and P D = x+x y +z AD 2 = Therefore. E . The point P has coordinates x : y : z = 1 : 1 : 1. P E = P F if and only if (x + y )2 y 2(z 2 + zx + x2 ) = (z + x)2 z 2 (x2 + xy + y 2). F respectively. Suppose P has homogeneous barycentric coordinates (x : y : z ) with respect to triangle ABC . or (y − z )(x + y + z )(x3 (y + z )+ x2 (y 2 + yz + z 2 )+ xyz (y + z )+ y 2z 2 ) = 0. BP . (x + y + z )2 (x + y )2 z y yz y 2 + yz + z 2 2 2 a2 + a2 − a = ·a . CP intersect the opposite sides at D . (x + y + z )2 (y + z )2 Therefore. Suppose that P D = P E = P F . Tsintsifas. Then. D divides BC in the ratio · AD . Determine the locus of P . Thessaloniki. By Stewart’s theorem. y+z y+z (y + z )2 (y + z )2 x2 (y 2 + yz + z 2 ) · a2 . . Greece Suppose that triangle ABC is equilateral and that P is an interior point. proposed by G. (x + y + z )2 (z + x)2 z 2 (x2 + xy + y 2 ) PF2 = · a2 . Since P is an interior point. it is the center of the equilateral triangle. BP : P C = z : y . y . we obtain P E2 = y 2 (z 2 + zx + x2 ) · a2 . z are positive.924 Some geometry problems from recent journals Crux 2836. The lines AP . P F = P D if and only if z = x. x. Similarly. P D2 = Similarly. . BE . Valldolid. and CD are concurrent. University of Valladolid. The interior bisectors of ∠BA A and ∠CA A intersect AB and CA at D and E respectively. Spain Let A be an interior point of the line segment BC in triangle ABC . Prove that AA .925 Crux 2840. proposed by Juan-Bosco Romero M´ arquez. Tsintsifas. Greece The inscircle inscribed in a tetrahedron is a circle of maximum radius inscribed in the tetrahedron.926 Some geometry problems from recent journals Crux 2847. proposed by G. considering every possible orientations in E3 . Thessaloniki. . Find the radius of the inscircle of a regular tetrahedron. Kawasaki. Prove that ABCD has an inscribed circle. Suppose that AB 2 + BC 2 + CD2 = AD 2 . proposed by Toshio Seimiya. .927 Crux 2849. Japan In a convex quadrilateral ABCD. we have ∠ABC = ∠BCD = 120◦ . A circle passing through A and C intersects BA and BC in points G and E respectively. Problem 1669.928 Some geometry problems from recent journals Mathematics Magazine. A line through D and perpendicular to BC intersects AC at F . N. Turkey Let ABC be a triangle and let E be the midpoint of BC . Duman. with 3AF = F C . proposed by A. . Prove that triangle F DG is similar to triangle ABC . Bilkent University. Let D be the midpoint of EC . and ∠ACD = 1 2 3 (a) Characterize the sets {a. N. . Nagpur.929 Mathematics Magazine. proposed by M. DC . b. AD . Problem 1671. Deshpande. (b) Find the triangle of minimum area in T . BD . Institute of Science. c} that are sets of side lengths of triangles in T . and AC have integral lengths ∠ABC = 1 ∠ADB . India Let T be the set of triangles ABC for which there is a point D on BC such that segments AB . K and L are concyclic. (b) G. and let E be the midpoint of AB . Prove that (a) KH//AC . . H . and K lies on the Euler circle (nine-point circle) of triangle ABC . Suceav˘ a. CA Let ABC be an acute triangle. let H and L be the projections of B and C on AT . California State University. Problem 11006. Let G and K be the projections of A and T respectively on BC . GL//BT . T the midpoint of arc BC of the circle circumscribing ABC . GH//T C . (c) The center of the circle through G. proposed by B. Fullerton.930 Some geometry problems from recent journals American Mathematical Monthly. and LK//AB . H . |BC | = y .931 Pi Mu Epsilon Journal. A. then a + b + c = 1 (perimeter) of triangle ABC . Let D . |CA| = z . |AB | = a. Lindstrom. and |CF | = c. F be points on sides AB . NY Suppose that triangle ABC has an interior point P . Batavia. 2. 1. Show that (x − a)2 + (y − b)2 + (z − c)2 = a2 + b2 + c2 . CA respectively. BC . so that P D ⊥ AB . Problem 1058. P F ⊥ CA. Let |AB | = x. |BE | = b. Show that if ABC is an equilateral triangle. proposed by P. E . P E ⊥ BC . 2 . CA respectively such that |AD | = |BE = |CF |. Ayoub. Problem 1060. Abington. The points D . . College. B.932 Some geometry problems from recent journals Pi Mu Epsilon Journal. Abington. Show that the circumcircles of triangles ABC and DEF are concentric. Pennsylvania State University. proposed by A. BC . PA Suppose ABC is an equilateral triangle. E . and F are on AB . .Chapter 37 The Josephus problem and its generalization 37.1 The Josephus problem n prisoners are arranged in a circle. n = 21. 5. In succession. Who is the survivor? Examples 1. 21. n = 10: 2 * 5 3 6 7 7 4 8 9 9 10 5 4 3 2 1 8 6 1 2. and the last one is set free. which is 11. . there are the 10 odd numbers 3. . After the removal of the 10 even numbered ones and then the first. 19. . The survivor is the 5-th of this list. every second one is removed from the circle and executed. . The binary expansion of f (n) is obtained by transferring the leftmost digit 1 of n to the rightmost.2. Let f (n) be the survivor in the Josephus problem of n prisoners. then f (n) = 2(n − 2m ) + 1. f (100) = f (11001002) = 10010012 = 64 + 8 + 1 = 73. f (2n + 1) =2f (n) + 1.1. Corollary 37. There is an almost explicit expression for f (n): if 2 m is the largest power of 2 ≤ n. Example f (100) =2f (50) − 1 =2(2f (25) − 1) − 1 = 4f (25) − 3 =4(2f (12) + 1) − 3 = 8f (12) + 1 =8(2f (6) − 1) + 1 = 16f (6) − 7 =16(2f (3) − 1) − 7 = 32f (3) − 23 =32(2f (1) + 1) − 23 = 64f (1) + 9 =73.1002 The Josephus problem and its generalization Theorem 37. f (2n) =2f (n) − 1. . . k ) J(10.37.2 Generalized Josephus problem J(n. k ) for various values of k For n = 10. here are the sequences of execution depending on the values of k .2 Generalized Josephus problem J(n. k ) 1003 37. k ∗ 1 1 2 3 4 5 6 7 8 9 10 2 2 4 6 8 10 3 7 1 9 5 3 3 6 9 2 7 1 8 5 10 4 4 4 8 2 7 3 10 9 1 6 5 5 5 10 6 2 9 8 1 4 7 3 6 6 2 9 7 5 8 1 10 4 3 7 7 4 2 1 3 6 10 5 8 9 8 8 6 5 7 10 3 2 9 4 1 9 9 8 10 2 5 3 4 1 6 7 10 10 1 3 6 2 9 5 7 4 8 Positions 2 and 6 are deadly positions for the Josephus problem of 10 prisoners and random choice of k . The last column gives the survivors. 3): * 4 8 5 2 3 4 1 2 6 1 6 7 5 8 7 9 3 10 9 J(10. This means that one other position is most likely to survive? Which one is it? 5. 12. . 13. For what values of n is f (n) = n? 2. . there is only one deadly position 1. 22. For what values of n is f (n) = n − 1? 3. 24 are 5. k ). 9. . 11). . 3 9 10 1 11 8 7 6 5 4 3 2 1 24 23 12 13 14 15 16 17 18 19 20 21 4 22 2 6. . 4.1004 The Josephus problem and its generalization Exercise 1. Make a list of the deadly positions of the Josephus problem for n = 4. The deadly positions for J(24. . Find out the survivor in the Josephus problem J(24. k = 1. . What is the one with the best chance of survival? . 18. . 5. 16. For n = 7. 19. . f (m) = m for m > n. (qi . Let L1 = (1). For i ≥ 2. 0).. Then Lk is the permutation corresponding to m. i. . r1 ) = (m − 1.Chapter 38 Permutations 38. These can be calculated recursively as follows.e. . ri ) = qi−1 . . form Li by inserting i into Li−1 so that there are exactly ri members smaller than i on its right hand side. we write m − 1 = r2 × 1! + r3 × 2! + · · · + rk × (k − 1)! for 0 ≤ ri < i. In other words. we construct a sequence of lists which ends at the desired permutation. qi−1 = i × qi + ri . we set. The permutations of 1. . Beginning with (q1 . . n as a bijection π : N → N which is “eventually” constant. for each i ≥ 2. . The enumeration begins with the identity permutation.1 The universal sequence of permutations For convenient programming we seek an enumeration of the permutations. Regard each permutation of 1. i Along with these. and each of the first (n − 1)! permutations ends with n. mod(qi−1 . Given an integer m. 2. i) . 2. n are among the first n! of them. 1. 2. 4. 4. 3. 6) (5. 1. 2) (1. 3. 4. 3. 2) (1.4.1006 Permutations Example To find the 12345th permutation.8.2. 6) The permutation is (5. 2) (5. The corresponding sequences are k 1 2 3 4 5 6 7 8 Lk (1) (1.3. .1. 2) (5. 2. 4. 1.7. 3.6). 3. 1. 7. 4. 3. 2. 7. 8. we write 12344 = 0 × 1! + 1 × 2! + 1 × 3! + 4 × 4! + 0 × 5! + 3 × 6! + 2 × 7!. 6) (5. 2. 7. .2 The position of a permutation in the universal sequence Given a permutation Ln . 4. . 4. 7. 2. 3. 8. j 9 8 7 6 5 4 3 2 1 Lj (1. 3. 5) (1.2 The position of a permutation in the universal sequence 1007 38. 5) (1. 3) (1. This number can be computed recursively as follows. 7. s1 =s2 × 1 + 1. s2 =s2 × 2 + r2 . 5) (1. 4. 6. 3) (1. we want to determine its position in the enumeration scheme above. sj −1 =sj × (j − 1) + rj −1 . . n − 1. 2. . 3. For j = n. 4. . 6. Then. 2. 5). 3. . 5) (1. 7. . let (i) rj be the number of elements in Lj on the right hand side of j . the position number of the permutation L n is 1 + r2 × 1! + r3 × 2! + · · · + rn × (n − 1)!. 8. . Example Consider the permutation L = (1. 2. 3. 4.38. 4. 2. 5) (1. 6. 3. . 2) (1) rj 2 1 1 3 0 2 0 0 1 sj 2 17 120 723 3615 14462 43386 86772 86773 This permutation appears as the 86773-th in the universal sequence. 9. 2. 6. 2. 9. sn =rn . sn−1 =sn × (n − 1) + rk−1 . . 4. 6. (ii) Lj −1 be the sequence Lj with j deleted. 8. . 2. 1n ) be a permutation of (1.6) is (2. . .1. 2. .8. Find the one-billionth permutation in the universal sequence. . . . 2. . .8. The inverse permutation of (5. . . . . . n) such that the sums a1 − a2 + · · · − an−1 + an . Project: Nice odd integers An odd integer n is said to be nice if and only if there is a permutation (a1 . . (i) cyclic |ai − ai+1 | ≥ 2n − 2. Find all nice integers. .1008 Permutations Exercise 1. .7. . .4.6). (ii) For how many distinct permutations of (1. .4. .5. . 2.1.3. n) does equality hold? Answer: n · 2n . . a2 . .7.2. n).3. . Let (a1 . a2 − a3 + · · · − an + a1 . an − a1 + · · · − an−2 + an−1 are all positive. What is its position number in the universal sequence? 3. 2. an ) of (1. 4.3.2. For example. Thus. Corollary 39.Chapter 39 Cycle decompositions 39. Even (respectively odd) permutations are said to have parity +1 (respectively −1).1 The disjoint cycle decomposition of a permutation Theorem 39. . More generally. Theorem 39.1. the fixed points are counted as 1-cycles. the permutation 1 2 3 4 5 6 7 8 9 X J Q K 6 9 8 3 4 2 K 7 Q 5 1 J X decomposes into the two disjoint cycles (1629QJ )(387KX 54). Every cycle decomposes into a product of transpositions. The parity (of the number of transpositions) of the permutation is independent of the decomposition. A cycle of length k has parity (−1) k−1 . a permutation of n objects has parity (−1)n− number of disjoint cycles in a decomposition . Theorem 39. A permutation can be decomposed into a product of transpositions. permutations are classified into even and odd permutations. In using this formula. Every permutation can be decomposed into a product of disjoint cycles. though we usually do not write them in the cycle decomposition of a permutation. • 1 2 3 4 5 6 7 8 9 6 5 3 9 2 7 1 8 4 • 1 2 3 4 5 6 7 8 9 7 1 4 6 5 3 2 8 9 • 1 2 3 4 5 6 7 8 9 7 4 3 8 1 6 5 2 9 1 Puzzle 128 of [Dudeney] .2 Dudeney’s puzzle 1 Take nine counters numbered 1 to 9. 26733. But it can be done in much fewer moves. and place them in a row in the natural order. As an example in six moves we give the following: (7846932). The square of 12543 can be found in four moves. and 27273 can also be obtained in four moves. (25) (37) (38) (34) 1 1 1 1 1 2 5 5 5 5 3 3 7 7 7 4 4 4 4 3 5 2 2 2 2 6 6 6 6 6 7 7 3 8 8 8 8 8 3 4 9 9 9 9 9 The squares of 25572. which give the number 139854276. It is required in as few exchanges of pairs as possible to convert this into a square number. as follows. which is the square of 11826.1010 Cycle decompositions 39. 2 2 1011 However. Can you find it? 1 2 3 4 5 6 7 8 9 .39. there is one which can be made in three moves. and 39147. the square of 35757. There are three other ways of making 4 moves to make a pandigital square. there are 4 ways to move 1 (37) 1 (48) 1 (4X ) 1 (49) 1 2 2 2 2 2 3 7 7 7 7 4 4 8 8 8 5 5 5 5 5 6 6 6 6 6 7 3 3 3 3 8 8 4 0 0 9 9 9 9 4 0 0 0 4 9 This gives the 1278563049. 38772. These are the squares of 35853.1012 Cycle decompositions The pandigital case of Dudeney’s puzzle In the pandigital case. • 1 2 3 4 5 6 7 8 9 0 1 2 8 5 4 3 7 6 0 9 • 1 2 3 4 5 6 7 8 9 0 1 2 7 8 5 6 3 0 4 9 • 1 2 3 4 5 6 7 8 9 0 1 5 3 2 4 8 7 6 0 9 . .39. Move as few counters as possible to make the equations valid. 2 2 (27)(495) to 2 × 78 = 156 = 39 × 4.3 Dudeney’s Canterbury puzzle 3 1013 39. While 7 × 28 = 196.3 Dudeney’s Canterbury puzzle 3 7 × 2 8 = 1 9 6 = 3 4 × 5. it is not true that 34 × 5 = 196. 3 How about 1 2 × 3 4 5 = 6 7 8 9 0? 3 (348965) to 12 × 483 = 5796. Move as few counters as possible to make the equation valid.1014 Cycle decompositions Project The multiplication 1 2 × 3 4 5 = 6 7 8 9 is clearly not valid. . Who wins the prize ? * A B C D E F G * Suppose you are at position A.39. and are permitted to add any number of rungs in any positions. (provided that no two rungs are at the same horizontal level).4 The game of ropes and rungs Each of the seven players starts by sliding down his own vertical rope. How would you add rungs to claim the prize ? * .4 The game of ropes and rungs 1015 39. and makes a turn every time he encounters a rung. 1016 Cycle decompositions . 6 without repitition so that the sum of the three number along each edge is constant. .Chapter 40 Graph labelling 40. .1 Edge-magic triangle Label the vertices and edges of a triangle with the numbers 1. . . . 2. . . . 10 so that the four faces all have the same sum 32.1018 Graph labelling 40.2 Face-magic tetrahedron The vertices and edges of a tetrahedron with consecutive integers 1. 2. . 9 4 7 5 1 10 6 3 8 2 Can you label them with a smaller common face sum? . . . 8 without repetition such that the sum of the numbers at the four vertices of each face is a constant.3 Magic cube 1019 40.40.3 Magic cube Label the vertices of a cube by the numbers 1. . 8 2 7 1 3 5 4 6 8 2 7 1 5 3 6 4 8 3 6 1 5 2 7 4 . 2. . . . . . . 14 (without repetition) so that the sum of the three numbers along each edge is constant. 2.1020 Graph labelling 40.4 Edge-magic heptagon Label the vertices and edges with the fourteen numbers 1. However.5 Edge-magic pentagram 1021 40. 10 without repetition such that the sum of the four numbers along each line is constant. 2 4 1 5 3 . it is possible to label the five outer vertices such that the sum of the four vertices along each of the five edges is constant. .5 Edge-magic pentagram It is known that the pentagram cannot be labelled with the numbers 1. . 2. . . Find such a labelling with minimum edge sum.40. given the labelling of the five inner vertices below. 6 A perfect magic circle 73 15 64 22 66 20 67 21 65 23 72 14 12 13 74 26 29 60 59 19 36 69 50 17 38 71 48 24 31 75 62 57 61 12 27 68 18 70 16 63 25 58 42 44 35 28 51 37 45 43 52 49 39 56 30 53 33 55 40 46 34 54 32 47 41 .1022 Graph labelling 40. rearranges the cards according to some simple rule.Chapter 41 Card tricks from permutations Consider mn cards arranged in order. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Rearrangements with simple cycle structure lead to interesting card puzzles. . and from top to bottom. the magician asks an audience to note the card at a specific spot. We call this the standard order. For example. then asks the audience to tell the new card at the spot. from left to right. The magician is able to tell the card that originally occupies this position. from top to bottom. For m = n. and from left to right. this rearrangement is the reflection in the main diagonal. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 .n Pick up the cards along the columns.1102 Card tricks from permutations The rearrangement ω = ωm. Then rearrange them in the standard order. this rearrangement is the cyclic permutation of the vertices of 4 squares: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 The cycle structure is simpler after one more application: it is simply rotation through 180◦ about the center of the square. Then rearrange them in the standard order. from bottom to top. For m = n = 4. .n Pick up the cards along the columns. and from left to right.1103 The rearrangement ρ = ρm. For m = n = 4.n 1 1 2 5 3 2 4 9 5 6 6 3 7 13 8 10 9 7 10 4 11 14 12 11 13 8 14 15 15 12 16 16 with cycle structure 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 The cycle structure is simpler after one more application: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 .1104 Card tricks from permutations The repeated diagonal rearrangement Pick up the cards along the diagonals. this is the permutation δ = δm. and from left to right. Then rearrange them in the standard order. from from bottom to top. and alternately.1105 The repeated snake rearrangement Pick up the card along the columns. Rearrange in the standard order. first from top to bottom.n 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 The cycle structure after one more application: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 . then from bottom to top. This is the permutation σ = σm. 1106 Card tricks from permutations . forming an isosceles tetrahedron ABCD. construct its anticomplimentary triangle A B C by drawing lines through the vertices parallel to their opposite sides. D C =D A B = D A B C L A = D B C To compute the volume of a tetrahedron.Chapter 42 Tetrahedra 42. Every isosceles tetrahedron arises from any one of its faces in this way. the position of this perpendicular foot is clearly the same for the four faces. we would drop the perpendicular from a vertex to its opposite face (of area ∆) to determine the ∆h. C into a point D . . The volume of the tetrahedron is then V = 1 3 For an isosceles tetrahedron. height h on this face.1 The isosceles tetrahedron An isosceles tetrehedron is one whose four faces are congruent triangles. Fold along the sides of the given triangle to bring the vertices A . B . We may therefore ask for the volume of the isosceles tetrahedron ABCD in terms of the side lengths of triangle ABC . Given a triangle ABC . C D C A A D B Z G L H O Z B C L B C A Proposition 42. The point L is the reflection of H in O . Proposition 42. . and the common pedal X of these points on the line BC .1108 Tetrahedra 42.3. This is the plane containing D . Since C Z is perpendicular to AB . B L. A B respectively.2.2 The volume of an isosceles tetrahedron Let L be the pedal of the vertex D on the face ABC . The same reasoning applied to the other two faces DBC and DCA shows that A L. C L is perpendicular to A B . C L are perpendicular to B C . 72 1 We use the word pedal for perpendicular foot or orthogonal projection.1. The point L is called the de Longchamps point of triangle ABC . L. The volume of the isosceles tetrahedron on triangle ABC is given by Viso = 1 2 (b + c2 − a2 )(c2 + a2 − b2 )(a2 + b2 − c2 ). OL2 = OH 2 = R2 (1 − 8 cos α cos β cos γ ). Theorem 42. triangle DZL becomes the segment C L intersecting AB at Z . It follows that L is the orthocenter of triangle A B C . C A . Upon unfolding the face DAB into triangle ABC .1 Consider the plane through D perpendicular to the faces ABC and DBC . . BD = b . CA = b.42.3 Volume of a tetrahedron 1109 42. The volume V of the tetrahedron is given by 0 1 1 1 1 1 0 a2 b2 c2 1 2 1 a 2 0 c2 b2 V = 288 1 b 2 c2 0 a2 1 c 2 b2 a2 0 = 1 144 a2 edges a2 a 2 − 2 a2 a 2 (a2 + a 2 ) − faces a2 b2 c2 . AD = a . AB = C.3 Volume of a tetrahedron Let ABCD be a tetrahedron with BC = a. CD = c . 6. Fold the diagonals AC and BD .. W.1110 Tetrahedra 42.4 Charles Twigg’s envelope model of the tetrahedron 2 Take an sealed envelope which is two copies of a rectangle ABCD glued along the perimeter. Assume AB < BC . Mag. Tuck D under AB (or A under DC ) and press up on B and C until DC and BA coincide. F is the midpoint of the side BC . A D E = E B F C 1. Underneath E is the point E 5. Their intersection is the double point E and E . thus forming a hexahedron. Cut along two half diagonals AE and DE to remove the sector containing one long side and the flap of the envelope. Separate E and E until EF E is a straight line. 3. Trigg. Fold back along the same lines and crease firmly again. Fold up around EF E until D meets A. 51 (1978) 66–67. Fold along the remaining portions of the half diagonals (BE and CE ) and crease firmly. Fold AB into DC to form the crease EF . . A = C E F B = D E 2 C. Here. 2. 4. Math. Tetrahedral models from envelopes. √ a2 + b2 . What is the volume of the resulting tetrahedron? 3 3 This is an isosceles tetrahedron with face 2a. Its volume is 4 2 a 9 √ b2 − a2 .42. In Twigg’s envelope model. suppose AB = 2a and BC = 2b.4 2 1111 Exercise 1. Find the volumes of the isosceles tetrahedra with face given below: a 11 33 69 21 b 20 65 91 99 c 21 72 100 100 V 2. . and √ a2 + b2 . What is the shape of the envelope in Twigg’s model for which the resulting tetrahedron is regular? 4. Find the volumes of the following tetrahedra: a b c a b c 32 33 35 76 70 44 21 32 47 58 76 56 V 3. 1112 Tetrahedra . . by paper folding. to find. It follows that A B = A W = A C . and that of C in A Z . a line which intersects AC and AB at Y and Z respectively) such that if A is the reflection of A in . one of the pillow problems Lewis Carroll had attempted but did not include in his collection of pillow problems was the following. and A is on the perpendicular bisector of BC . A Y W Z B C A The point W is both the reflection of B in A Y .Chapter 43 Lewis Carroll’s unused geometry pillow problem According to [Rowe]. then the reflections of B in A Z and of C in A Y coincide. Given a triangle ABC . In a standard proof of the Euler line theorem. . A O N H D B C A Let D be the midpoint BC and H the orthocenter of triangle ABC . which is necessarily the circumcenter O of triangle ABC . It follows that the midpoint of AA is the same as that of OH . This is ∠BA C =∠BA W + ∠W A C =2∠Y A W + 2∠W A Z =2∠Y A Z = − 2∠Y AZ since A Y AZ is a rhombus. The reflection of A in the side BC is therefore the point Q on the perpendicular bisector such that ∠BQC = 2∠BAC . and the reflection line is the perpendicular bisector of the line AA . where R is the circumradius of triangle ABC . We therefore conclude that A is the reflection of the circumcenter O in the side BC . the ninepoint center N of triangle ABC . 1 and that the midpoint of OH is the nine-point center of triangle ABC . 1 AH = 2 · OD = 2R cos A. it is established that AH = 2OD .1114 Lewis Carroll’s unused geometry pillow problem Consider the directed angle ∠BA C . The Lewis Carroll paper-folding line is the perpendicular to AN at N . This means that AH = OA . This means that ∠BA C = −2∠BAC . and AOA H is a parallelogram. Chapter 44 Japanese Temple Geometry . 1116 Japanese Temple Geometry . 1117 . 1118 Japanese Temple Geometry . 1119 . 13th edition. L.. R. [2] I. 1998. Bailey. W. 4-. 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