Iran 2010-2011.pdf

Iranian Team Members (from left to right): • Mina Dalirrooyfard • Mohammad Pedramfar • Erfan Tavakoli • Mojtaba Shokrian Zini • Sahand Seifnashri • Alireza Fallah This booklet is prepared by Meysam Aghighi and Masoud Shafaei with special thanks to Morteza Saghaan. Copyright c Young Scholars Club 2010-2011. All rights reserved. 1 Iranian Mathematical Olympiad 2010-2011 28th Contents First Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Second Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Third Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Solutions Problems 3 4 5 8 First Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Second Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Third Round . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 11 2 Problems 3 . Let E and F be points on the extensions of AB and AC such that BE = CF = BC . A and B are two points on W1 and W2 respectively such that AB is their common tangent. (Arashk Hamidi) The coecients of P (x) = ax3 + bx2 + cx + d are real and min{d. then the number of their participants are dierent. 5. (Mohsen Jamali) There are n points in the plane that no three of them are collinear. is not greater than 3 3. 6. b + d} > max{|c|. ∠A = 60◦ . The circumcircle of ACE intersects EF in K (dierent from E ). We also know that if two dierent classes have two common students.First Round 1. |a + c|} prove that the equation P (x) = 0 does not have a solution in [−1. prove that: ∠DP M = ∠BDC 4. 1]. (Davood Vakili) In triangle ABC . prove that (ab + 1)2 + (a − b)2 is not a perfect square. (Davood Vakili) Two circles W1 and W2 intersect at D and P . Suppose D is closer to AB than P . If AD intersects W2 again in C . 2. prove that K lies on the bisector of ∠BAC . 4 . and so ab + 1 and a − b. Prove that the total number of classes is not greater than (n − 1)2 . and having area equal to 1. Prove that the total number of triangles whose vertices are from these n points 2 (n2 − n). If ab − 1 and a + b are relatively prime. We know that there are at least two participants in any class. (Morteza Adl) a and b are two integers such that a > b. and M be the midpoint of BC . (Mohammad Ali Karami) A school has n students and there are some extra classes provided for them so that each student can participate in any number of them. Prove that if the numbers 2a + 1 . (Note that A does not contain a parallelogram with zero angle. or a parallelogram that its opposite vertices coincide. y ) be a polynomial in two variables with real coecients.Second Round 1. homothety or any combination of these transformations. b and c are positive integers. y ) = 3x4 y − 2x2 y 3 . A version of A is a set of points that can be reached from A by translation. y ) with highest degree. (A subset S of the plane is bounded if there exists a positive number M such that the distance of the elements of S from origin is less than M . (For example if P (x.) (b) Prove that we can always do this using above transformations. Let Q(x. and the intersection of every three of them be empty. The degree of a monomial is sum of the powers of x and y in it. 5 . (Sepehr Ghazi Nezami) A is a set of n points in the plane. If this black square lies on a square of lattice. y ) = 3x4 y − 2x2 y 3 + 5xy 2 + x − 5 then Q(x.) 2. We can roll the parallelepiped in every four directions.) Suppose there exists real numbers x1 . and a. (a) Prove that if no four points of A form a parallelogram. Q(x2 . y2 ) < 0 prove that the set {(x. y ) be sum of the expressions in P (x. x2 . rotation. where the vertices of its bottom face are lattice points. 3. (Mostafa Eynollahzadeh) Let P (x. then it would become black as well. We want to place n versions of A in the plane so that the intersection of every two of them be a single point. y1 ) > 0 . y1 and y2 such that Q(x1 . and its faces are divided into unit squares while the square in the middle of each face is black colored. then we can make every square of the lattice black by rolling the parallelepiped. (Sina Jalali) We have put a rectangular parallelepiped with dimensions (2a + 1) × (2b + 1) × (2c + 1) on an innite lattice plane with unit squares. 2b + 1 and 2c + 1 are pairwise coprime. y ) = 0} is not bounded. y )|P (x. we can do this using only translations. . a sequence of 1 and -1 with these properties: • is periodic with 2n − 1 as its fundamental period. 0) and (0. (i. for every positive integer j we have xj +2n −1 = xj and 2n − 1 is the minimum number with this property. −1). 5. x2 .) • there exist distinct integers 0 ≤ t1 < t2 < . (It is clear that the centers of these 12-gons coincide and is the origin of the 3D-space. × xj +tk prove that for every positive integer s < 2n − 1 we have 2n −1 xi xi + s = − 1 i=1 6. 1) . Find the number of faces of the polyhedral with maximum volume that its projections on xy .Shania. yz and zx planes are good 12-gons... (Sepehr Ghazi Nezami) Suppose S be a gure in the plane that its border does not contain any lattice point and let x. Prove that the area of S is equal to the number of lattice points in it. • the origin be its center.Mansouri..e. 0) . (1.. M.Dashmiz) Let n be a positive integer and x1 . (Sepehr Ghazi Nezami) We dene a 12-gon in the plane to be good if it has these conditions: • be regular. (−1. y ∈ S be two lattice points with unit distance from one another..4.H.. S. Assume that there exists a tiling of the plane with copies of S such that x and y lie on lattice points for every copy of S in this tiling.) 6 . A copy of S is a combination of rotations and reections of S . (M. < tk < n such that for every j we have xj +n = xj +t1 × xj +t2 × . • be lled. • its vertices contain points (0. prove that the number of elements in the range of f is not greater than n. f (B )) ≥ f (A ∪ B ). 7 .7. (Bijan Aghdasi) S is a set with n elements. • For every two A. 8. and P (S ) is the set of all subsets of S . B ⊆ S we have max(f (A). (Mohsen Jamali) Show that there exist innitely many positive integers n such that n2 + 1 has no divisor of the form k2 + 1 except for n2 + 1. Let f : P (S ) −→ N be a function with these characteristics: • For each A ⊆ S we have f (A) = f (S − A). and tangent to AC at H ... the incenter of triangle OBC . Let m denote the number of maximal elements of F . Let F be a family of these intervals such that every element of F is a subset of at most one other element of F . b. (Mohsen Jamali) There are n points on a circle (n > 1). • For every two distinct b. (Hesam Rajabzadeh) In non-isosceles triangle ABC . c ∈ A we have gcd(a. . Consider circle ω1 inside triangle ABC which is tangent to ω . nd all good sets. Consider circle ω with center O and radius OB .. 5.Third Round 1.. and T be a point on LK such that AT ||BC . prove that T A = T M . y ∈ R we have: f (x + f (x) + 2f (y )) = f (2x) + f (2y ) 6. (Sepehr Ghazi Nezami) Let OBC be an isosceles triangle (OB = OC ). (Morteza Saghaan) For what integer values of n > 2. b. tangent to ω1 at J .. Circle ω2 is also inside triangle ABC and is tangent to ω . h2 . 8 . . If L and K be the midpoints of M D and M E . h3 are the distances of P from sides of the triangle. and let D and E be the foot of perpendiculars from C and B to AB and AC respectively. 7. Tangents to this circle from B and C intersect at A. an ∈ N such that the following sequence is a nonconstant arithmetic progression? a1 a2 .. 2. c) = 1. prove that: m+ a ≤n 2 4. Prove that the bisector of ∠KJH passes through I . c and a|bc. and a be the number of its non-maximal elements. let M be the midpoint of BC . a2 . and tangent to AB at K . an a1 3. there exist numbers a1 . c ∈ A. An interval is called maximal if it is not a subset of any other interval. a2 a3 . there exists a ∈ A that a = b. (Ali Khezeli) Find the locus of all points P inside an equilateral triangle having the following property h1 + h2 = h3 where h1 . (Mohammad Jafari) Find all surjective functions f : R −→ R such that for every x. We dene interval to be an arc between any two of these n points. (Mohammad Mansouri) We dene a nite subset of N like A to be good if it has the following properties: • For every three distinct a. Ak (x) with integer coecients such that k f (x) = i=0 (xp − x)i pk−i Ai (x) also prove that the statement is not correct for k > p. b. 9 . . 10. A line l in the plane is called wise if one can divide these n points into two sets A and B (A ∩ B = ∅) such that sum of distances of points of A from l is equal to the sum of distances of points of B from l. (Mahyar Sedgaran) p is a prime number and k ≤ p.8. (Mohammad Jafari) Find the smallest real k such that for every a. Prove that there exist polynomials A0 (x).. (Morteza Saghaan) There are n points in the plane. Prove that the line P P passes through a xed point. B and C denote the midpoints of sides BC . AC and AB of triangle ABC respectively. (Sepehr Ghazi Nezami) Let A .. af (a) + bf (b) + 2ab is a perfect square.. f (x) is divisible by pk . P B = P B and P C = P C . (Mohsen Jamali) Let f : N −→ N be a function such that for every a. c. f (x) is a polynomial with integer coecients that for every integer x. b ∈ N. d ∈ R we have: (a2 + 1)(b2 + 1)(c2 + 1) + + (c2 + 1)(d2 + 1)(a2 + 1) + (b2 + 1)(c2 + 1)(d2 + 1) (d2 + 1)(a2 + 1)(b2 + 1) ≥ 2(ab + bc + cd + da + ac + bd) − k 11. Assume that P and P are two variable points such that P A = P A . not all on the same line. 9. Prove that there are innite points in the plane such that n + 1 wise lines pass through them. A1 (x). 12. Prove that for every a ∈ N we must have f (a) = a. 10 . Solutions 11 . proof: If there are more than 4 triangles with this property. lemma 2: The number of triangles with vertices A and B having area 1 is at most 4. N is the midpoint of AB (because N A2 = N D × N P = N B 2 ). i.First Round 1. hence there exists a prime number p such that: p|a2 + 1. The case p|a + b similarly reaches contradiction. it is enough to prove that ∠DP M = ∠BP C or equivalently ∠DP B = ∠M P C . Now that a2 + 1 and b2 + 1 are relatively prime and their multiplication is a perfect square. then the distance of C from line AB is 2 2 . Now we have ∠P BC = ∠P DC = ∠P CB = AP = ∠P AB 2 PB = ∠P BA 2 hence P CB and P BA are similar. b2 + 1) = 1. two lines at distance d Now we count the number of triangles (T ) in this way that for each pair of points there can be at most 4 triangles and we have counted each point at least 3 times. Since ∠BDC = ∠BP C . the locus of such points consist of two lines parallel to AB at a d distance d from it. Let points A and B in the plane (with distance d). hence n lemma 1: T ≤ 2 ×4 2 =⇒ T ≤ × n(n − 1) 3 3 3. if there is a point C such that the area of triangle ABC is 1. 12 . Suppose the contrary. We know that (a + b)2 + (ab − 1)2 = a2 b2 + a2 + b2 + 1 =⇒ c2 = (a2 + 1)(b2 + 1) now we prove that (a2 + 1. p|b2 + 1 =⇒ p|a2 − b2 =⇒ p|a − b or p|a + b If p|a − b. a − b) = 1. Let the extension of P D intersect AB at N .e. then p|ab − b2 and since p|b2 + 1 then p|ab + 1 which is a contradiction as we know (ab + 1. and since P M and P N are their medians. 2. both should be perfect squares which is impossible because a2 + 1 can not be one and the proof is complete. proof: trivial. then from lemma 1 and pigeonhole principle it concludes that 3 of the points must lie on one of the 2 from AB which is a contradiction. the angles ∠DP B = ∠N P B = ∠M P C . Suppose that α be a real root for P (x) such that |α| ≤ 1. b + d} hence max{|c|. Since every 2-element subset of students are together in show that |Ai | ≤ n i(i−1) at most one class of each Ai . 1] in one of its boundary points. Let T be the intersection of BF and CE .e. similarly ∠CBF = ∠ACB . and from BE = F C we conclude that KBE and KF C are congruent. so there is no α with such property.4. 6. hence P (α) = 0 =⇒ aα3 + bα2 + cα + d = 0 =⇒ aα3 + cα = −(bα2 + d) =⇒ |α||aα2 + c| = |bα2 + d| =⇒ |aα2 + c| ≥ |bα2 + d| (∗) on the other hand f (x) = |ax + c| takes its maximum over the interval [0. then the number of 2-element subsets of Ai multiply the elements of Ai must be less than or equal to the number of 2-element subsets n (n−1) 2) of students. Hence 2 2 ∠CT F = ∠BCE + ∠CBF = ∠ABC + ∠ACB = 60◦ 2 and the quadrilateral ABT C is cyclic. i i(i−1) (2 ) hence we have m ≤ n(n − 1)( 1 1 1 + . thus ABKF is cyclic. so |aα2 + c| ≤ max{|c|. so ∠EBF = ∠ACE = ∠AKE and then ∠ABF = 180 − ∠EBF = 180 − ∠AKE = ∠AKF . f (x) ≤ max{|c|. Suppose Ai (2 ≤ i ≤ n) be the set of all classes with i participants. b + d} and this is a contradiction. |b + d|} ≥ min{d. since BC = BE we have ∠BCE = ∠ABC . |b + d|} from these inequalities and (∗) we have max{|c|. |a + c|} it follows from the hypothesis that d and b + d are positive. |a + c|}. i. so KC = KE =⇒ KC = KE . 5. Having ∠EBK = ∠CF K and ∠BEK = ∠F CK .. hence |Ai | ≤ ( = n .. so bx + d > 0 for each 0 ≤ x ≤ 1. We will (n−1) .. We have ∠BCE + ∠BEC = ∠ABC . |a + c|} ≥ min{d. thus AK is the bisector of ∠BAC . hence similarly we can say that (considering minimum of g (x) = |bx + d|) |bα2 + d| ≥ min{|d|. |a + c|} ≥ |aα2 + c| ≥ |bα2 + d| ≥ min{|d|. + |An |. + ) = n(n − 1)(1 − ) = (n − 1)2 2(2 − 1) n(n − 1) n 13 .. On the other hand m = |A2 | + . so when A → +∞ the sign of L(A) is the sign of the coecient of Ak which is Q(x.) and with respect to the fact that the sign of Q(x. and point (t1 . we mean the ray with origin as its start point) • We can suppose that (0. .. • We know that Q is homogeneous so if we put k = deg(Q) we have Q(tx. Ay ) . y ). y ) out of this line such that Q(x. so P has a root on this line and the problem is solved. + Q0 let L(A) = P (Ax. now we can replace (x2 . otherwise choose another point (x. Qk−2 .. m2 ) < 0. m2 ) that P (l1 .Second Round 1. y ) and we have tk > 0 for t > 0 so Q(tx. t2 ) on the second ray such that the distance of origin from the line passing through these points is greater than an arbitrary M ∈ R+ . now since Q and Qi are homogeneous we have L(A) = P (Ax. + Q0 (x.. (x1 . l2 ) > 0.) lemma. For every (x. y1 ) or Q(x2 . y ) ∈ R2 that Q(x. proof of lemma.. ty ) and Q(x. Ay ) = Ak Q(x. Note that if degQ = k then there exist homogeneous polynomials Q0 . P (m1 . hence it must be zero overall. then there exists point (s1 . y ) + . y1 ) and (x2 . y ) + Ak−1 Qk−1 (x. y1 ). and the distance of origin from the line passing through these points is greater than an arbitrary M ∈ R+ .. y ). (we assumed that Q(x. y ) is dierent from the sign of either Q(x1 . Now since the restriction of P to this line is a single variable polynomial (by replacing x = ay + b or y = ax + b) that is negative in a point and positive in another. ty ) = tk Q(x. y ) that Qi (x. y ) have the same sign for t > 0. 0) . y ) are of the same sign.. and another point on the ray passing through origin and (x2 .. there exists S ∈ R+ such that if A > S then P (Ax. y ) are constant numbers. First note some points: (in the proof by the ray passing through origin and (x. y ). y ) = 0 (if such point doesn't exist then Q must be zero on all plane except a line. let it be Q(x1 . y ) = 0) So the lemma is proved. Qk−1 such that degQi = i (0 ≤ i ≤ k − 1) and P = Q + Qk−1 + Qk−2 + . Ay ) and Q(x. (this is trivial. 14 . Ay ) and Fi (A) = Qi (Ax. y1 ) like (l1 . y2 ). 0). l2 ). y2 ) are not collinear. y2 ) by (x. y ) = (0. y2 ) like (m1 . F (A) = Q(Ax. Ay ). • If two rays passing through origin doesn't form a line together. Now with respect to above notes there is a point on the ray passing through origin and (x1 . s2 ) on the rst ray. y ) (1 ≤ i ≤ k − 1) and Q(x. an be origin. so xij = ai + aj ∈ Bi ∩ Bj . lying on the same face as well. As 2a + 1 and 2c + 1 are coprime. Clearly. we can color these cells as well (x + 2a + 1. 2b + 1.. translating by the vector (2b + 1. an are complex numbers in this plane. Now with respect to the symmetry between a and c. and A + an have this property.. because if x ∈ Bi1 ∩ Bi2 ∩ Bi3 . Similar to the previous part we dene 15 . (b) Let an arbitrary point apart from a1 . if we can color a cell of the lattice plane with (x. then rolling it downward. so ai2 = ai3 and i2 = i3 a contradiction. y ) cells where x + y is even. therefore Bi ∩ Bj has a single element. up. . . y ) coordinates. and with respect to what is said. Suppose that at the beginning the cuboid lies on the plane like the gure. 2c + 1) = 1. then we must have ai1 + ai2 = ai1 + ai3 . By rolling it in the directions down. y + 2b + 1) (x + 2c + 1. now we can write (x + 1. up. Let Bi = A + ai (1 ≤ i ≤ n).. b ∈ R2 . We dene S + b = {x + b|x ∈ S } where S ⊆ R2 . . since aj + ai ∈ Bi and ai + aj ∈ Bj . it is translated by the vector (2a +1.. Now by rolling the cuboid in the shown direction we can color a cell that sum of its coordinates are dierent from the previous cell. A + a2 . Without loss of generality assume that a and b have the same parity and a + b is even. 2a +1) and lies on the same face.2. Now since the center of the cuboid face is of integer values. Now we prove that the sets A + a1 . an }. Therefore part (a) is complete.ai1 . s ∈ Z such that r(2a + 1) + s(2c + 1) = 1. there exists r. left. Assume that a1 . hence x − xij = aj1 − aj = ai1 − ai . y + 1) = (x + r(2a + 1) + s(2c + 1). the intersection of every three of them is empty. 0) cell of the plane. y + 2a + 1) (x + 2b + 1. then there must exist j1 = j and i1 = i that ai1 + aj = x = aj1 + ai . we can color all (x.. Remark. If the center of the bottom face of the cuboid lies on the (0. we can translate it by the vector (2c + 1. y + r(2a + 1) + s(2c + 1)) so this cell can be colored. y + 1) too. a2 . Now we show that this can also be achieved for b by rolling it upward. so a + b + 1 is odd. a contradiction. right respectively.aj and aj1 form a parallelogram.... and suppose that the points of the set A be {a1 . with respect to previous part. .... so ai . Now if x ∈ Bi ∩ Bj and x = xij . y + 2c + 1) now we prove that we can color (x + 1. we can color all of the plane cells. (a) Let an arbitrary point of the plane be origin. right. First note that these are n translations of A. 3. 2b + 1). 2c + 1). The statement also holds for the hypothesis (2a + 1. Consider the following squares of size t × t and (t + 2d) × (t + 2d). z . Note that each one of the B1 .i2 .aj = aj . let this number be k. and all of them are lattice points... As said in the hypothesis. on the k other hand none of the tiles has any point outside the (t + 2d) × (t + 2d) square.S. Also let x .ai . there are two points x. similar to part (a) we have ai . and let S be a copy of S . since 2 every tile has at most k lattice points of that square we have u ≥ (t+1) .i3 and i4 that at least three of them are distinct we must a4 i1 have ai1 ai2 = ai3 ai4 . y with unit distance on a tile that their corresponding points on other tiles are lattice points. we can choose O in such a way that this won't happen. otherwise a = ai and it means that the origin is ai3 i2 the center of a rotational homothety that brings the segment ai1 ai3 to the segment ai2 ai4 . therefore the total number of lattice points of each of the copies of S are equal. (1 ≤ i ≤ n). and the Bi ∩ Bj would have one element in this case. since it is obtained from S by a rotational homothety about the origin. (note that it is required that no two of ai s are collinear with origin for taking it as center of this rotational homothety) 4. Let z be another lattice point in S . . y . Let Bi = A. so 16 . Assume that u be the number of tiles that has intersection with t × t square.i2 . Let d ∈ N be greater than the maximum distance between points of S . z ∈ S be the corresponding points to x.b = {xb|x ∈ S } where b ∈ C and S ⊆ C is arbitrary.i3 and i4 are nite. There are four possible places for z shown in below gure. y. Clearly since the ways of choosing i1 . Bn is a version of S ..ai ∈ Bi ∩ Bj now for every i1 . 1. Also note that xi xi+s can be written in the form of product of xi−1 . xi−n . this summation is equal to -1. k . · · · . xi+s−n−1 and so xi+s can be written in the form of xi−1 . Lemma. 1)} and X is the set of all x∈ X Proof of lemma. because i can be changed from n + 1 to n + 1 + 2n − 2 and xi xi+s can be evaluated by n members before xi . · · · . 5. and these n members form a n-tuple and we know that by changing i these sequences will contain all n-tuples of {−1. n-tuples of {−1. It is trivial that this polyhedron is in the intersection of three 12-gonal prisms that their projections on the axis planes are mentioned 12-gonals. now xi+s can be written in the form of product of xi+s−1 . · · · . · · · . then we have [x]i1 [x]i2 · · · [x]ik = −1 where X = X − {(1.+1) . xi−n . Now let s be the area of S . 1}. 0 < i1 < i2 < · · · < ik ≤ n be integers. then we have [x]i1 [x]i2 · · · [x]ik = 0. 1) so using lemma. · · · . · · · . since the inner square is totally we have u ≤ (t+2d k covered we have us ≥ t2 and us ≤ (t + 2d)2 . Because 2n − 1 is the minimum period of the sequence. We dene [x]i = xi where x = (x1 . the n-tuples (xi . xi−n . · · · . there exist 0 < u1 < u2 < · · · < up ≤ n such that for every l ∈ Z we have xp = xl−u1 × xl−u2 × · · · × xl−up 2n −1 n hence i=1 xi xi+2 is in the form of one of the summations stated in the lemma. xi+1 . xi+s−n . there exists a n-tuple y such that [x]i1 = [y ]i1 and [x]s = [y ]s for all s = i1 . Let n. n 6. because from the hypothesis we have xi = xi−n+t1 × · · · × xi−n+tk . n Since for every n-tuple x. · · · . 1. · · · . Indeed every member of the sequence can be evaluated using n consecutive members of this sequence. xi+n−1 ) for 1 ≤ i ≤ 2n − 1 are distinct. · · · . now since x2 l = 1 for every l ∈ Z. 1. Now for constructing the shape with maximum 17 . 1) is not among them) Note that by dening xi = x2n −1+i we can expand the sequence to negative indices. according to what we said. 1} except (1. x∈X This proves the lemma. ((1. hence 2 (t + 1)2 ≤ uk ≤ (t + 2d + 1)2 t2 ≤ us ≤ (t + 2d)2 ⇒ (t + 1)2 (t + 2d + 1)2 k ≤ ≤ (t + 2d)2 s t2 (t + 2d + 1)2 (t + 1)2 k ⇒ lim ≤ ≤ lim ⇒k=s t→∞ (t + 2d)2 s t→∞ t2 so the proof is complete. We show that the desired polyhedron has 36 faces. xi can be written in the form of product of xi−1 . xn ). xi+s−1 can be written in the form of product of xi+s−2 . First Solution.. . rstly.Am−1 . Hence.. −1).A2 .. then the function f : P ((S − {i. A2 . Suppose that f (A1 ) < f (A2 ) < . (because the projection of xy prism on xz plane is a nite strip. y. Second Solution. we will show that each face of the recent prism. construct a new element by joining i and l).. secondly since the polyhedron is convex.. so {i} = A1 ∪ A2 ∪ . if we consider the intersection of this polyhedron and the yz prism. f (B )}. Am−1 ∈ Domain(f )... if we restrict f to the subsets of S that either have both i and l or none of them. z ) = (0. f (B c )} = max{f (A).. For showing this it is sucient to show that each face of the xz prism. f (B ) ≤ max{f (A1 ). 1. adds exactly one face to the shape and the intersection of this two prisms forms a polyhedron with 24 faces. (0.. 1. (0. then lemma. from the hypothesis of induction we have m − 1 ≤ n − 1 so m ≤ n and the proof is complete. one face adds to the polyhedron. because it is the intersection of two convex shapes. On the other hand this polyhedron contains the points (x. every face of this prism... because the xy prism and every face are convex.. it contains the 12-gon on xz plane) So the intersection of these two prisms is a polyhedron with 24 faces which is convex. By replacing some of Aj s (1 ≤ j ≤ m − 1) with their complement we can assume that i ∈ A1 ∪ A2 ∪ .. 1). so there exists i ∈ Am such that {i} is not obtained from A1 .. intersects the xy prism. A2 ... First consider the prism corresponding to xy plane. Am . ∪ Am . now consider the intersection of this shape with the prism corresponding to xz plane.volume we should consider the intersection of these three prisms. ∪ Am .. and since i is not obtained from A1 . the resulting shape is a 12-gon. 1). Similarly to the previous part. (by intersection and complement) Since Am is obtained from single-element sets of itself.. Now let a = {i.Am−1 .A2 . so it is connected and only adds one face. f (A2 ). . and the projection of this polyhedron on the yz plane totally contains the 12-gon corresponding to yz plane. therefore the resulting shape has 24 + 12 = 36 faces. because A1 . every face intersects the polyhedron in one region and adds one face to the shape. intersects the polyhedron... It concludes from the lemma that f (A∆B ) ≤ max{f (A).A2 . l} (indeed. f (B )} and the proof is complete. < f (Am )... ∪ Am−1 . −1. . so it contains a square with vertices at these points... −1. (0. −1). using lemma. It is sucient to prove that f (A ∩ B ) ≤ max{f (A). l}) ∪ {a}) −→ N is obtained and clearly has the hypothesis properties... f (A ∩ B ) = f ((A ∪ B )c ) = f (Ac ∩ B c ) ≤ max{f (Ac ). f (Am )} proof. 18 .. So for every face of the 12-gonal prism. If B is obtained by intersection and complement from A1 . therefore there exists l = i such that l ∈ A1 ∪ A2 ∪ . it concludes that Am is not obtained from A1 . f (B )}... 7. the desired intersection is also convex.Am−1 . . 19 . We claim that for every positive integer n. . Now it is sucient to consider an innite sequence an of positive integers of the form k2 + 1 such that (am . let f (A) := max{min(A). If k2 + 1 has no divisor of this form we are done.now note that P (S ) with operation ∆ forms a group (∅ as identity and A itself as its inverse). an ) = 1 whenever m = n. . Otherwise l2 + 1|k 2 + 1. m} be the range of f ).. If n2 + 1 has no divisor of the form k 2 + 1 (except n2 + 1) we are done. |B2 | ≥ 22 . Note that there is a f that its range has exactly n elements. As an example consider n an = 22 + 1... with respect to this the sets Bi = {x|f (x) ≤ i} (1 ≤ i ≤ m) are subgroups of P (S ) (let {1. by continuing this way our claim is proved. 8. Therefore by Lagrange's Theorem |Bi |||P (S )|. |Bm | ≥ 2m =⇒ 2n ≥ 2m =⇒ m ≤ n therefore the proof is complete. so |Bi | is a power of 2. and since B1 B2 . min(Ac )} and by min{X } we mean the minimum element of X . n2 + 1 has a divisor of the form k2 + 1 that satises the statement of the problem.. Bm = P (S ) and B1 has at least two elements (x ∈ B1 ⇒ xc ∈ B1 ) hence |B1 | ≥ 21 . 2.. Otherwise suppose that k 2 + 1|n2 + 1 for some positive integer k.. [xm . . Let n be even. b2k+1 b1 = 2k + 1 3 1 .. If xi = xj then one of their corresponding intervals must contain the other one. j ≤ m : xi = xj .. in this case no such numbers exist because if n = 2k we have a1 a2 < a2 a3 < . Let the points be 1. y ] be a non-maximal interval of F . Let A = {[x1 . assume b1 = x and b1 b2 = 1. j ∈ {1.... and so the powers of T with respect to M and ω are equal.. We denote an interval by [i. y1 ].··· b1 b2 = 1 ⇒ b2 = ⇒ b3 = 2x ⇒ b4 = x 2x b2i = b2i+1 = 1×3×···×(2i−1) 1 ×x 2×4×···×(2i−2) 2×4×···×(2i) x 1×3×···×(2i−1) 2≤i≤k 1≤i≤k but b2k+1 b1 = 2k + 1 so 2 × 4 × · · · × (2k ) x2 = 2k + 1 ⇒ x = 1 × 3 × · · · × (2k − 1) 1 × 3 × · · · × (2k + 1) 2 × 4 × · · · × (2k ) now let ai = bi (2k + 1)!. thus it is trivial that a1 a2 . a2k+1 ∈ R+ . < a2k−1 < a1 which is a contradiction.. j ] where i.. hence M E is tangent to ω and so is ∠M BE = 90 − ∠C = ∠HAC = EH 2 M D.. On the other hand a2i+1 = a2i = 2 × 4 × · · · × (2i) ×x× 1 × 3 × · · · × (2i − 1) (2k + 1)! = 2 ×· · ·× (2i) (2i +1) ×· · ·× (2k +1) 1 × 3 × · · · × (2i − 1) 1 × × 2 × 4 × · · · × (2i − 2) x (2k + 1)! = 1 × 3 × · · · × (2i − 1) (2i) × · · · × (2k ) therefore all of ai 's are distinct natural numbers. Let X.. < a2k−1 a2k < a2k a1 hence a1 < a3 < . ym ]} be the set of maximal intervals of F . and let ω be the circle with diameter AH . b2 b3 = 2. 3. Similarly ∀1 ≤ i. Let H be the orthocenter of ABC . . Thus LK is the radical axis of point M and circle ω . n. Y denote the multi-set (the members of a multi-set need not to be distinct) of all lower and upper endpoints of all intervals of F respectively.. . which means T A2 = T M 2 =⇒ T A = T M 2. . Now let n be odd. we show that either x ∈ / X − { x} 20 .. . T A is tangent to ω in A.. and let [x.. n} and i < j ..Third Round 1. we claim that ∀1 ≤ i... and the statement is true for all n = 2k + 1. Since T A ⊥ AH . a2 a3 . On other hand BM = EM ⇒ ∠M EB = (of ω ). an a1 forms an arithmetic progression too. y2 ]. j ≤ m : yi = yj . 2. so the claim is proved. we construct such progression with a1 . [x2 . · · · .. 2.. e be distinct. y ]. in this case a ∈ A exists such that for two pairs {b. hence p|d or p|e. If B ⊆ C. and let p be a prime factor of a. a|de because we can choose these pairs in n ways. suppose 21 . hence d|c. hence if e = c then c = d which can't be true. Now we prove that if a. a that b|a and this means that 1 < b < a which is a contradiction. we claim that b|ac. c} and {d. a] and [b. c. a) = 1. b} so e|d. C . B ⊆ D then two intervals contain B . b} and (d. a contradiction. e} we have a|bc. a contradiction (note that the existence of e needs A to have at least 4 elements. d. hence m + |A1 | ≤ n. If x ∈ X − {x} and y ∈ Y − {y }. y ]. and if n ≥ 4 we have n > n. C ⊆ B . Similarly there exists e ∈ A that e|ad. This results that the upper endpoints of intervals in A and A1 are distinct. c) = e > 1. then there must exists b ∈ A. d) = a ⇒ b|a ⇒ a = b  b|ab which is a contradiction. D are not maximal and since B is not maximal either there exists E ∈ A such that D ⊆ B ⊆ E . Let B = [x. b = 1. Similarly B C. so D is subset of two intervals. y ] if B ⊆ C. So assume that |A| ≥ 4. hence two intervals contain D. otherwise e|d|c which means (e. B D. y ] in F dierent from [x. and let A2 denote the set of non-maximal intervals that their lower endpoints are dierent from the lower endpoints of other elements of F . then for every c ∈ A − {a. so our claim is proved. B C which means D ⊆ B. a contradiction. and since or |A2 | ≥ a . so m + a ≤ n. b ∈ A and |A| ≥ 4. c) = 1 now there exists d ∈ A such that d|ac. D = [b. C = [x. since d ∈ A − {a. then a b. And if B D. c. either |A1 | ≥ a 2 2 2 4. Assume that a|b. so we have |A| + |A1 | ≤ n. B ⊆ D is not possible. so |A1 | + |A2 | ≥ a. b. if e = b then b|ad|ac which is contradiction. a contradiction.or y ∈ / Y − {y }. a]. b} we have  b|ac  b|ad ⇒ b|a(b. Now let A1 denote the set of non-maximal intervals that their upper endpoints are dierent from the upper endpoints of other elements of F . d. there must exist intervals [x. Similarly m + |A2 | ≤ n. if b ac. |A1 | + |A2 | ≥ a. therefore e ∈ A − {a. b} we must have (a. Now 2 2 if b. then we must have D ⊆ B and so D ⊆ B. therefore C. for |A| = 4 there isn't such e and we lead to contradiction sooner) Thus b|ac so if c and d be two elements of A − {a. First note that 1 ∈ / A because if 1 ∈ A and a be the minimum element of A − {1}. c) = (a. b. yet c|ab so c = ab. a) = b  a|ba which can't be true. Therefore one condition is that c = ab and (a.g we must have b = d. so (a. therefore p|(a. r) = 1 5. b are two coprime factors of c. d) which is a contradiction. c unless q = 1 and it means a|c. Let b = qr and c = pr . we have c|ab ⇒ pr |pq. so f (x + f (x) + 2f (y )) = f (2x) + f (2f (y )). b. thus a = pq such that q |b and p|c.p|d. Therefore |A| ≤ 3. similarly suppose that p|c. q. so r = r which means that a = pq . c} be good. c). b) = 1 OR A = {pq. so do a. b) = (a. rp} (p. so for every c ∈ R we have f (f (0) + c) − f (t + c) = f (0) on the other hand c = t ⇒ f (f (0) + t) = c=0 ⇒ f (f (0)) = c = f (0) ⇒ x = 0. c. there exists t ∈ R such that f (t) = 0. now because b|ac we have b|c which means that a. hence  a|bc  a|be ⇒ a|b(c. but in this situation we must have (p. y = f (0) in the rst equation. y = 0 ⇒ (2) (1) f (0) f (0) (2) (3) f (2f (0)) f (3f (0)) = 2f (0) (4) = 2f (0) (5) now by putting x = 0. Thus we all sets are of the form A = {a. now let x = y = t ⇒ f (t) = 2f (2t) ⇒ f (2t) = 0 (1) x = 0 ⇒ f (f (0) + 2f (y )) = f (0) + f (2y ) x = t ⇒ f (t + 2f (y )) = f (2y ) (1) f (f (0) + 2f (y )) ⇒ −f (t + 2f (y )) = f (0) but 2f (y ) takes all real values. now assume that A = {a. b = qr and c = pr. (4) that (5) f (3f (0)) = 3f (0) ⇒ f (0) = 0 now if we put x = 0. Since f (x) is surjective. b|ac and c|ab. b. we obtain from (3). b. b = 1. it results that f (2f (y )) = f (2y ). so e. b.qr ⇒ r |q 2 r note that if r and q have common factors. Since 2f (y ) takes all values in R we can conclude that f (x + f (x) + y ) = f (2x) + f (y ) (6) 22 . if q > 1 we must have (r . qr. a|bc. ab} (a. b. c) = 1. q ) = 1 so r |r. so (a. q. q. e. r) = 1 because if a prime number s|(p. similarly r|r . r) then s|(a. c) = 1. N is a point on BE and P is a point on CF such that K P and HN are perpendicular to BC . = K Q HS QB similarly the condition that P lies on C H . In the next gure BB and CC are diagonals of ω and E and F are midpoints of arcs B C and BC respectively. Lemma. . BT QB = C U SC the lemma is Now. HC intersects line CF at P and circle at Y . =1 K Q B T NB but TN NB = CH HS . = K Q HS SC and considering the identity proved. Also 23 . Using Menelaus' theorem in triangle BT Q and secant secant K B : Proof of lemma. BK QB T N . leads to BK CH CU .by putting y = 0 we have f (x + f (x)) = f (2x) (7) also putting y = f (x + f (x)) = f (2x) in (6) results f (x + f (x) + y ) = f (x + f (x) + f (x + f (x))) = f (2x + 2f (x)) = f (x + f (x) + (x + f (x))) = f (2x) + f (x + f (x)) = 2f (2x) on the other hand f (x + f (x) + y ) = f (x + f (x) + f (2x)) = f (2x) + f (f (2x)) so f (f (2x)) = f (2x). draw a perpendicular from H to BC such that intersects BE at N and B N intersects with circle at X and with AB at K . According to above lemma. so BK CH BT . and since f (2x) takes all real values we have ∀a ∈ R : f (a) = a 6. We prove that N lies on K B if and only if P lies on HC . K P is perpendicular to BC . B . Suppose that J is the second intersection point of circumcircles of these pentagons. We claim that the answer is the incircle. B. So circumcircle of J Y H is ω1 and circumcircle of K J X is ω2 . without loss of generality we assume that the point is in the quadrilateral AEOD. On the other hand we know that if a circle is tangent to AC at H and to ω at Y . So circumcircle of J Y H is tangent to ω . 7. Also J = J and K = K and ∠IJH = 180◦ − ∠ICH = 180◦ − ∠IBK = ∠IJK . therefore Y = Y . because all are equal to half of arc B C . We have ∠J HA = ∠J IC = ∠J Y H so circumcircle of J Y H is tangent to AC . Hence this circle is exactly the circumcircle of J Y H . a homothety with center A and AX as its coecient. C respectively. 24 . maps X . according to previous AX √ √ √ part since the incircle of A B C passes through X . so IJ is bisector. let X denote the point of intersection of AX with the incircle (closer to A). it follows that circumcircles of J Y H and K J X are also tangent to each other.∠CHN = ∠CXN = ∠CIN . So the pentagon XHCN I and similarly the pentagon BP IY K are cyclic. Similarly circumcircle of K XJ is also tangent to AB and ω . We have ∠XDP + ∠XEQ = ∠XF R − ∠XDP = = ◦ 60 so if ∠XDP = α we have ∠XEQ = 60 − α and ∠XF R = 60◦ + α. We have ∠K J H = ∠ACI + ∠ABI = 180◦ − = ∠BIC = ∠AHJ + ∠AK J ∠BAC 2 and since AH and AK are tangent to circumcircles of J Y H and K J X . then Y is the inverse homothetic center of these two circles and since C and H are homothetic so Y should be on C H . C to X. we have XR = XQ+ XP . According to the law of sines DX + XE = ∠DOE = 60◦ 2 2 2 F D+DX D − DX = F2 = ∠F OD 2 2 ◦ PX RX QX √ ⇒ RX = XD sin α = (2r sin α) sin α = 2r sin2 α = XF sin ∠XF R = XF sin(60 + α) = 2r sin2 (60 + α) = XE sin ∠XEQ = XE sin(60 − α) = 2r sin2 (60 − α) √ √ √ = 2r sin(60 + α) = 2r(sin α + sin(60 − α)) = P X + QX Now if this property holds for a point not on the incircle. e. we have Ai (x) = pBi (x) (III) in which the coecients of Bi (x) are integer. For an arbitrary triangle ABC ... Hence p|(h(x) − a)k−1 Ak−1 (x) + · · · .. By replacement of (III) in (I) the statement results. thus its coecients are divisible by p. . + yAn = 0 n so the proof is complete. .. Remark. Gn 25 . Let G be the origin of the plane and l be the x-axis. hence X and X coincide. therefore p|y k−1 Ak−1 (x) + · · · + A0 (x) (II) now let x be xed. lemma. which completes the proof. so we have a polynomial in terms of y whose degree is at most k − 1 ≤ p − 1.. Suppose that we have a set of n points is the plane and G be their centroid.. Let A = {A1 . if a line l passes through G then sum of the signed distances of these n points from l is zero. (i. 8..for k = 1 the problem is well-known. A2 . For k > p the counterexample f (x) = (xp − x)k−1 − pp−1 (xp − x)k−1 works. Now we add the constraint that deg(Ai ) ≤ p − 1 for i ≤ k − 1. Now we dene n + 1 points G... the desired locus is an ellipse tangent to sides of the triangle. hence ∀x. G1 . deg(g ) ≤ kp − 1 √ √ √ now the hypothesis of induction for k − 1 on g (x) results that f (x) = (xp − x)k Ak (x) + (xp − x)k−1 Ak−1 (x) + · · · + pk−1 A0 (x) ⇒ p|( xp − x k − 2 xp − x k−1 ) Ak−1 (x) + ( ) Ak−2 (x) + · · · + A0 (x) p p p (I) −x if we let h(x) = x p . 9. i : Ai (x) ≡ 0 (mod p). We prove the rst part by induction . a takes all values module p. sum of the distances of points in each side of the line are equal) proof of lemma. Now since the degree of Ai is less than or equal to p − 1. An } be the points. + yAn ⇒ yA1 + yA2 + .on the other hand we know that XR = XQ + XP which suces that XR = XR. it is obvious that h has integer value but it does not have integer coecients. since G is the average of all of the points we have 0 = yG = yA1 + yA2 + .. Using division algorithm we can write f (x) = (xp − x)k Ak (x) + g (x). We have (x + ap)p − (x + ap) xp − x − ap g (x + ap) ≡ ≡ ≡ h(x) − a (mod p) p p and Ai (x + ap) ≡ Ai (x) (mod p). so for a constant x. Ai2 . . Aik } be one of the sets and Y = A − X be the other set. and assume that the center of the homothety taking OAi Aj to GGi Gj is T ... Ai−1 .. · · · . Aik }. 26 √ . we have A1 +A2 +. Ai . Ai+1 . thus it is sucient that O does not lie on lines A1 G. we dene Gi to be the centroid of A1 .in this way: let G be the centroid of these n points and assume that O be an arbitrary point in the plane.G and Gi are not collinear is correct. Suppose that for a point O..+A −A +A Gi = 1 2 n n i i G= ⇒ G − Gi = Ai − Ai 2 = (Ai − O) n n this means that O. therefore O TS − − → − − → GO GO/GT n/2 + 1 1 2 = = 2 = = GS GS/GT n /4 − 1 n/2 − 1 n−2 2 in this case a homothety with center G takes O to S and we have OG = n− . . thus T =n .+An n A +A +. so the minimum value for k is 4... Hence the 2 triangles GGi Gj and OAi Aj are similar with ratio n . A2 G. suppose that O.Gi and Gj are collinear. G and Gi are collinear with O. Ai1 . (we can assign the points on OG to each of the sets) We claim that OGi is a wise line. Now we have to prove that there exists innite points like O that these lines are pairwise distinct for them. According to Cauchy-Schwartz inequality we can write (a2 + 1)(b2 + 1)(c2 + 1) = ((a + b)2 + (ab − 1)2 )(c2 + 1) ≥ (a + b)c + ab − 1 2 2 2 (a + 1)(b + 1)(c + 1) ≥ 2 ab − 4 cyc sym ⇒ and equality holds for a = b = c = d = 3. Ai and G are collinear. Now let the extension of Ai Aj TG 2 O 2 intersects OG in S . . and GGi = n Ai O. then let X = {Ai . 2 Now.. −2 10. GS 2 Therefore it is sucient that O does not lie on the lines that are resulted by 2 homothety over G with ratio n− from the lines Ai Aj . hence we have found n + 1 wise lines.... if Ai be in one side of OGi with points {Ai1 . According to what we said. if we divide the points into two sets by OG... the line OG is a wise line for the point O. . Ai2 .T =n .. An . Therefore the assumption that O. An G... let Ai be the reection of Ai over O.. P = P + 4 + B. so the A B C .BA (1). 2 XP1 1 the locus of the points X with the property that XP = 2 ). p s. and ratio − 2 AP BP C P circumcircle of A B C is the apollonius circle about P1 and P with ratio 1 (i. so the locus of P is a circle centered at the circumcenter of ABC . and P1 maps to P with a homothety with ratio − . Now let X P be a point on P P such that XP = 2.P 2 2 (P − C )2 = (P − C )2 ⇒ P 2 + C 2 − 2C. First note that if p be an odd prime. (2) ⇒ 4 − − → 3 (2). we have (S and T are intersection points of the apollonius circle with P P ) P1 T = x 3 P1 S = x ⇒ NS = P 1 S + P1 T 2x x 4x = ⇒ N P1 = . Let af (a) = p2k−1 s where s ∈ N. Now since P maps to P1 with a homothety centered at N 1 1 . therefore these two homotheties P maps to P with a homothety with ratio − 2 P P passes through the center of this homothety which is a xed point.11. on the other hand a homothety with respect to G takes P to P1 . N P = 2 3 3 3 note that N T = 23x is the radius of the apollonius circle. so x is constant. and let x = |P P1 |.C = 2 2 2 − − → the hypothesis results that (P − A)2 = (P − A )2 ⇒ P 2 + A2 − 2A.P 4 2 B2 2 2 2 2 (P − B ) = (P − B ) ⇒ P + B − 2B.e. therefore for a prime p we have p2k−1 af (a) that k ∈ N. Let G be the origin. Let the homothety with center at G. then we have P =P =P =2 . af (a) is not a perfect square.B = . centroid of ABC and 1 1 1A 1B 1C maps P to P1 . Now suppose that for a ∈ N.P = P + C + C. so p|p + f (p) and p|f (p). First Solution. so 2P + P is determined uniquely. Second Solution.P 4 2 2 (1) (2) (3) − → 3 (A2 − B 2 ) = (2P + P ). we have (so by K we mean GK ) A = −B −C −A . thus X = 2P + . 12. by combining having ratio 4 2 1 . hence the locus of P1 and P are two circles with centers at N . Let N denote the center of this circle. (3) ⇒ 4 (B 2 − C 2 ) = (2P + P ). putting a = b = p results that 2p(p + f (p)) is a perfect square.CB − → − − → note that the vectors BA and CB are independent vectors and the dot product of them with 2P + P is a xed value. hence X is the xed point XP 3 that P P passes through.P = P + A + A. now put b = p2k so p2k−1 s + p2k (f (p2k ) + 2a) is 27 . 28 . hence [af (a) + p2 + 2ap] − (a + p)2 = af (a) − a2 . so for every odd prime p we have f (p) = p.a perfect square. Now assume that p > f (1) be an odd prime. so af (a) is a perfect square. hence f (1) + 2p = 2 pf (p) + 1 now since p|f (p) and pf (p) is a perfect square so f (p) = k2 p that k ∈ N. so (b = p in which p is and odd prime) (a + p)2 = af (a) + p2 + 2ap. Now suppose for a ∈ N we have f (a) = a. because p ≤ f (p) so p ≤ pf (p). which means p2k−1 (s + p(f (p2k ) + 2a)) is a perfect square but the term in the parenthesis does not have p factor which is a contradiction. therefore 4 pf (p) ≥ 4p > 2p + f (1). hence pf (p) + f (1) + 2p < ( pf (p) + 2)2 ⇔ f (1) + 2p < 4 pf (p) + 4. So we have ( pf (p))2 < pf (p) + f (1) + 2p < ( pf (p) + 2)2 so pf (p) + f (1) + 2p = ( pf (p) + 1)2 . so for any odd prime we have |af (a) − a2 | ≥ 2(a + p) − 1 which can't be true for a constant a. but the absolute value of left side of the equation is greater than or equal to 2(a + p) − 1 because the dierence of (a + p)2 with any perfect square except itself is at least this value. since p|f (p) so p ≤ f (p). but above equation simultaneously proves that f (1) = 1. Now if k ≥ 2 we have 3p > f (1) + 2p = 2 pf (p) + 1 ≥ 4p + 1 so f (p) = p for any odd prime p greater than f (1). so f (a) = a. 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