Iranian_Math 2010

Iranian Team Members (from left to right): • Saeed Tayyebi • Javad Abedi • Morteza Ashrafi-jou • Sina Rezaei • Hessamoddin Rajab-Zadeh • Masoud Shafaei This booklet is prepared by Meysam Aghighi and Morteza Saghafian. Copyright © Young Scholars Club 2009-2010. All rights reserved. 1 27th Iranian Mathematical Olympiad 2009-2010 Contents Problems FIRST ROUND................................................................................................................ 4 SECOND ROUND .......................................................................................................... 6 THIRD ROUND .............................................................................................................. 9 Solutions FIRST ROUND.............................................................................................................. 12 SECOND ROUND ........................................................................................................ 15 THIRD ROUND ............................................................................................................ 20 2 Problems 3 . orange or peach tree is planted.) Show that the number of empty pieces is not greater than 1000. 3.First Round | 1 holds for 1. There are at least one apple tree and one orange tree adjacent to every peach tree and at least one apple tree.1 5 | | 4 2. We know that there is at least one apple tree adjacent to every orange tree. When commander commands each soldier either doesn’t move or takes a step in one of the four directions! After the command. (Erfan Salavati) There are 2 soldiers standing in equal columns next to each other such that the distance between every two of them is one step. (Meysam Aghighi) We have a 50 50 garden which is divided into 1 1 pieces and in some pieces an apple. Show that for every 1. Prove that is even. We draw a line through so that it intersects the rays and (with the startpoint) in and . Show that . 4 . such that the first and last rows are omitted and two columns are added in left and right. Let be a polynomial of degree 2 such that | 1. the soldiers are standing in 2 equal columns. one orange tree and one peach tree adjacent to every empty piece (a piece without any tree).0. (Meysam Aghighi) Let the inner bisector of the angle of triangle intersects and the circumcircle of in and respectively. 4.1 . (We say that two pieces are adjacent if they have a common side. (Mohsen Jamali) Natural numbers for every distinct and . (card 0 is the same as card 11.11 on them are dealt among them.2. … .5. and 1 are not the vertices of an acute-angled triangle. (Omid Naghshineh) There are 11 men sitting around a circular table with equal distances and 11 cards with numbers 1. 5 . Prove that for every . It is possible that one has no cards and the other has more than one. the places of the cards with numbers 1. is divisible by have this property that . and card 12 is the same as card 1) At the beginning the cards 1 to 11 are dealt among them counter-clockwise. 6. (everyone has exactly one card) Prove that there will never be a man who has all the cards. In each step one can give one of his cards to his adjacent individual if the card number has the following property: before and after this stage. | | | | | | | | | | | | Suggestion: for part (b). consistent if the set | 1. Prove that the total number of consistent permutations is . be points on the segments . Show that if . .….…. Let .….2.…. . first solve the problem assuming that is on the outer bisector of the corresponding angle. … . then the following holds | | | | | | | | | | | | Where | | means the length of the segment . … . . 2. .2. (Ali Khezeli) Let 2 and . then. where is the sum of the positive divisors of and is the number of positive divisors of . . respectively. b. a. … .…. We define the half-plane with the outer bisector of angle as its boundary which does not contain the inner bisector of . . Show that if . be points in the triangles respectively. (Mohsen Jamali) We call the permutation of 1. . respectively. be points in the half-planes . has 2 members.….Second Round 1. 6 . be points on the plane that no three of them are collinear. . Let . (Amir Ja'fari) Do there exist two functions . such that every side of a small triangle is parallel to one of the sides of the main triangle. 7 .3. In this case every small triangle is derived by a homothety (with a positive or negative ratio) from the main triangle. is a circumscribed polyhedral. We want to roll the ball on a closed polygon of the plane to bring the mark to the top of the ball while the ball is on its first place. Prove that the sum of all of these ratios is 1. Prove that the following equation has an infinite number of solutions: 7. 4. 6. We have partitioned a triangle into similar triangles with itself. Note that the ball must not roll in its place (it means rolling without moving on the plane). be two relatively prime integers. : the following inequality holds? | | | such that for all | 1 5. Prove that this is possible. Consider a spherical ball on a plane and a point marked on it. Its faces are colored with black and white such that there isn't any pair of black faces with an edge in common. two points (vertices) are connected if they are the same in one of their coordinates and differ 1 in another. the number of omitted points inside the horizontal square with side length 2 1 and the origin as its center. 8. Suppose that for all . Look at these points as a graph. 8 . Each connected component is called a cluster. (Omid Naghshineh) Consider that we have omitted some of the points of a 2dlattice (same as ).Prove that sum of the black areas is not greater than sum of the white areas. Prove that the nonomitted points contain exactly one infinite cluster. be less than /2. 5. (Morteza Saghafian) We have drawn a polygon with 2n sides without picking up the pencil from the paper. and are the midpoints of and respectively and is a diameter of the circumcircle. … . (Naser Talebizadeh) Prove that for every natural number . we have 7. is a circle tangent to the segments and in and respectively. Prove that there is a path using only these diagonals that either connects the upper side of the board to the lower side or connects the left side of it to the right side. If we name the sides from 1 to 2 with the order of drawing them. we have and . there exists a natural number such that for every natural number that 2 1389. of real numbers concave if for every 0 1389 we have . If . (Mehdi E'tesami) is an arbitrary point on the side of the triangle . prove that the circumcircles of and are tangent to each other. (Davood Vakili) In the isosceles triangle . also is tangent to the circumcircle of in . Find the maximum number such that for every concave sequence of nonnegative numbers. prove that 1 2 3. If be the intersection of with the circumcircle of (not ). Prove that this polygon intersects itself. If be an inscribed quadrilateral. is more than . 6. (Morteza Saghafian) There is a board divided into unit squares. then the odd sides are all vertical and are drawn bottom-up. the sum of digits of in base . 2. is a point that and .Third Round 1. prove that the tangent line in to the circumcircle of is parallel to . (Ali Khezeli) We call a sequence . (Sepehr Ghazinezami) In triangle . and we have drawn one of the two diagonals of each unit square. 9 . is the intersection of and . and are the midpoints of and respectively. is the circumcenter and is the orthocenter. 4. …. is an arithmetic progression. (Ehsan Azarmsa) : is an increasing function and is a natural number. The distance between two vertices is the sum of the length of the edges of the path between them. . and natural numbers Suppose that there exist prime numbers . 1 . 1 . . . prove that 10 . 2 . (Mir Omid Hajimirsadeghi) and are two trees without having a vertex of degree 2. 9. and intersects for the second time intersects at . the distance between and in is equal to the distance between and in . … be an infinite arithmetic progression. . . and intersect at and . leaf. . We call the vertices with degree 1. the distance between and in . (Ali Khezeli & Morteza Saghafian) Find all polynomials . is equal to the distance between and in . 0 0 . 2 2 (note that is not necessarily strictly increasing. such that for every 1 the set | 1. Prove that there exists an injective and surjective function from the set of vertices of to the set of vertices of such that for every two vertices and in . is the 12. Prove that there exists natural number such that 1 . (Mohammad Ja'fari) Find all increasing functions : 0 we have such that for every . (Davood Vakili) Two circles and is on . Each edge of them has a positive number named the length of the edge. … . If be the second intersection point of circumcircle and circumcircle .2. 1 0 11. is the intersection of and .) 10. … . common tangent of them near to such that is on for the second time at .8. with real coefficients such that for every . is an injective and surjective function from the set of leaves of to the set of leaves of with this property that for every two leaves and in . Solutions 11 . so we have adjacencies are at least 3 Similarly. and every empty piece has at least one adjacent . so the total number of adjacencies are at least we have 4 Now we do the same for counting the total number of adjacencies of orange trees to a peach tree or an empty space. We know that these adjacencies are at most 4 . Notice that 1 2 If 0 1 1 1. and since every orange tree. we see that 1 1 | | 2 2 If 1 0. we have the following inequality 2 12 . orange and peach trees. We now want to count the total number of adjacencies of apple trees to an orange tree. because every orange tree must have an adjacent apple tree. Let . 1 at . 1 . . or an empty piece. 1 at . be the total number of apple. hence apple tree. On the other hand since every peach tree and every empty piece has an adjacent orange tree. and it can have 3 adjacent peach tree or empty piece. and the total number of empty pieces respectively. a peach tree. so the total number of . every peach tree. we see that 1 1 | | 2 2 | This proves that | for all 2 1 1 5 4 1 1 2 0 5 4 1 5 5 2 4 4 1. 2. Note that these adjacencies are at most 3 . The conditions for equality 1 are a.First Round 1. and b. . 3. we have 180° 180° 13 1000 2 in . we have 50 2 3 3 2 2 4 2 2500. noting that inequalities. we have So the circumcircle of is tangent to arc in this circle be .50 Now. Since is the bisector. and using the above 2 2 2 2 5 2500 2 2 2 2 So the total number of empty pieces are not greater than 1000. Thus Similarly we have 2 . Hence if the measure of the 2 2 . 14 . consider the remaining soldiers. and if they are all in one place this sum would be even. the last row must've moved up. Because if the place of card be between 1 and 1. and the rightmost column must've moved right. must be even. be on the points . then know that . they are 2 columns with soldiers in each column. the leftmost column must've moved left. Since we . First divide the table into 11 equal arcs. Let up. and 5. and since it can't be done for 1. Thus. so it is impossible. | . Easily it can be proved that after each step this sum either remains invariant or varies by 2 . that . (the smaller one) For example. otherwise changes by 2. down. We claim that it can be done for 2. At first it is odd ( 11).4. we have On the other hand all of the above terms are divisors of in decreasing order be . the number of arcs between . on the table. so if the divisors of . the distance between . left and right be the four directions. is 5 in the following figure: Now. Now if the cards . that have moved in columns with 2 soldiers in each. Now. if it can be done for . we have 1 1 1 And the problem is solved. this sum doesn’t change. 6. For all distinct . on the table. Note that if columns have moved to 2 columns. we define the distance between these two cards. So the parity of this sum remains invariant. then must be even. then the first row must've moved down. Suppose that it can be done for . after each step we sum the distances between every two cards with consecutive numbers. Second Round 1. Lemma. with respect to . Suppose that for all 1 and length of this case the vector angle . a. In is codirectional with the inner bisector of the . from the triangle inequality we have: (according to lemma) b. hence from the definition of 0 . We extend to meet in . plus we have: (consider Also suppose that · that: same as . then: . we have: 0 · · So we deduce that: 15 · · · . If be a point in the triangle . the 1. . . we can say: · · · · Hence we have: · From the other hand. ) Note that | in which we can say . PROOF. we have | is the is the unit vector codirectional with . We have: (by triangle inequality) Now we return to the problem. . Similarly for . so from the two members of 1. and let Δ be the main triangle) 1 But in the above summation.2. we have . . every horizontal segment (that contains no smaller horizontal segment) once appears with positive sign for its above 16 . For each divisor of . So easily we deduce that | . then . So considering the above paragraph. hence: (let be the length of the horizontal side of the triangle . Because if and . the remainder of upon division by . Since ∑ 0 . So the total number of states is equal to: 1 | 1 | | 3. Also let | and | . we can say that if (and 1) then . hence sum of the ratios is sum of the ratios of the horizontal sides of the small triangles (with respect to the sign of the ratio) to the horizontal side of the main triangle. whose sum is . then . be more than . there are 1 different pairs of . Also. and so . there are at least of them in and hence if the remainder of the division of by . Suppose that the bottom side of the main triangle is horizontal. First note that if then . Let | be the positive and – be the negative members. be less than or equal to . … .| | 2. if for 1 . Thus from every consecutive numbers. Similarly. one should be positive and the other one negative. we have . then and hence . it means that . note that for every 1 . . hence there exists 1 such that 1 . so we have an uncountable number of circles in the plane that no two of them intersect and this is impossible. consider the point . 1 . and these segments appear with the positive sign. . . . . . . so and so sum of the ratios with respect to their signs is 1. First. . First note that if . so no two functions with this property exist. Let be the radius of the ball and be its first position on the plane. So it is enough to find a primary solution. For every point and a radius of √ . So we have 17 . except the segments of . because there is a point with rational coordinates in every one of the circles. hence we would have an uncountable number of points with rational coordinates and since we know that the total number of the points with rational coordinates is countable. we roll the ball (sphere) on its great circle that passes through the marked point and the top point of the ball. . then . . . 5. Now we roll the ball from to on the segment . . until the mark be on top of the ball. The above inequality shows that the distance between every two of these points is more than √ . so not two of these circles intersect. . . is another solution. So there exists a point on the plane so that 2 . and since 2 . Let be the current position of the ball on the plane. 4. Then we roll the ball from to on and the mark is at the top of the ball while it is on its first position. We know that 2 3 5 2 3 5 2 3 5 2 3 5 Now since gcd . For all we have: 1 | 2 | | | 1 2 For every . and hence 1 1 .triangle and once appears with negative sign for its below triangle. in the plane. . this is a contradiction. consider a circle with center at this point . the mark will be at the top of the ball. On the other hand the number of these points are uncountable. So 1 . 6. It is trivial that 2 . . be a solution. consider a black face … . .2 3 5 . Thus. hence for every 18 . Consider the cluster having the origin. . But since. We suppose that the origin is not omitted without loss of generality.2 3 5 . Now. consider the point having maximum horizontal distance from the origin. consider the point of tangency of the inscribed sphere with that face. and let be the point of tangency of this face with the inscribed sphere. Among all of the points of the cluster having origin. So sum of the black areas is not greater than sum of the white areas. its border contains some omitted vertices that are adjacent to at least one of the vertices of the cluster. and then consider the point having maximum vertical distance from the origin. For every face of the polyhedral.5 7. the points on the cluster's border have rounded around the origin. 8. . . So all of the points of the cluster having origin. and draw segments from that point to the vertices of the face.3 2 . let be the maximum of these two distances. we have divided each face of the polyhedral to some triangles. It is trivial that and (the length of the tangents from a point to a sphere are equal). Thus.2 . if the cluster is finite. are in the square whose center is on the origin and has side-length 2 1. consider the triangle . 3 5 . a group of infinite solutions for the mentioned equation is: . so the triangles and are congruent and have equal areas.2 3 5 2 2 3 5 3 2 3 5 5 Thus. from the assumption there is a white face containing the edge that the inscribed circle is tangent to it at . every black triangle has a corresponding white triangle and these white triangles are distinct. by increasing one by one. at least 1 points are omitted which is a contradiction. We extend the origin-centered square until it contains a piece of this path. So the cluster having the origin is infinite. 19 . From now on. Now assume that there are two infinite clusters. Obviously their border contains a path of omitted points which is infinite from both directions. Thus. the square's side will extend by 2 and there will be 2 more vertices of the path in the square. the total number of vertices of the path which is in the square are growing faster than /2. and hence it cannot be always less than it.0 there exists a point on the border that its horizontal or vertical distance from the origin is equal to . This contradiction shows that there is exactly one infinite cluster. but the value of /2 will increase by 1/2. So but increasing . So is a parallelogram and its diagonals bisect each other. from this we determine that is the midpoint of and hence: . Now let be the midpoint of perpendicular bisector of . the problem is solved. Hence. and since . We claim that 1389! 1 has the desired properties. So is tangent to the circumcircle of at and its radius is half of the radius of the circumcircle of . Hence the line is perpendicular to . On the other hand. so . hence and so . We know that the radius of the circumcircle of is equal to the radius of the circumcircle of ( ). (the points of tangency of these to circles with ) and we deduce that . because 1389! is divisible by . So is tangent to the circumcircle of (at ). and are parallel. If the height of this path be . (until they don’t cross the table sides. respectively. Similarly. and if we add another diagonal to it. to . it can be seen that the diagonals that are not in this component but are on the border of this component. they are parallel. then its 20 . then the medians since and are both perpendicular to . A connected component of diagonals is some of the diagonals that are connected to each other by a path of diagonals.) Now. Let be the circumcircle of the quadrilateral . the 1389! 1 rightmost digits of 1389! 1 is 1. Consider the connected component having this path. 3. otherwise suppose that the height be less than . Now . then sum of the digits of in base is at least 1 1 which is greater than or equal to 1. Hence passes through the center of the circumcircle of and its radius is . as well as . 2. takes . The circumcircles of and are symmetric across the line . A 2 homothety at . is a diameter of ( ). we know that there are at least 1 zeros in the rightmost of in base . it will no longer be connected. Let 2 1389 . are connected.Third Round 1. So this homothety maps to the circumcircle of . consider a path of diagonals that starts from the upper side of the table and has the maximum height among all the paths starting from the upper side. Easily. and are both perpendicular to . this line is the . are equal in the isosceles triangle . If this component is not connected to the right side of the table. then in the circumcircle of : (let be the measure of the arc ) 2 Hence we deduce that is tangent to circumcircle of . but these two paths intersect. then the right-side of its rightmost side (because it is drawn upward) is its outside. In this case.0 Note that the maximum possible value of for this sequence is . . ∑ ∑ 0. the remaining of the polygon contains a path that connects to . and a contradiction. Similarly if this connected component is not connected to the left side of the table. this connected component is connected to both right and left sides of the table. which is a contradiction. Third Solution: suppose that this polygon has 2 sides and doesn’t intersect itself. Hence. are collinear. which is a contradiction. and the rightmost side connects the vertex upward to the vertex . From the other hand each non-vertical side of this polygon is adjacent to two vertical sides that are drawn upward. and another one that connects to . and since is inscribed ( is the foot of the altitude from on ). 6. and this is a contradiction. and hence they're connected to each other. then . . First consider the concave sequence 1389 . Clearly. and the right-side of its leftmost side is its (because it is drawn upward) is its inside. Since is an inscribed quadrilateral. then . So the sum of the two adjacent angles to each non-vertical side is 360° . and since we have non-vertical sides. 5. First Solution: If a polygon does not intersect itself. But if the said polygon doesn’t intersect itself. then by moving clockwise (or counterclockwise) along its sides. Obviously . Now if we draw ray parallel to such that . Second Solution: Suppose that the leftmost side of the polygon connects the vertex upward to the vertex . Thus. total sum of the angles of this polygon is equal to 2 2 180° 1 360° . .rightmost border's height will be more. 4. the inside (or outside) of the polygon must always be at our right-side. . Then all of the polygon must be between the lines and . 21 1389. the total sum of all angles equals 360° . then its leftmost border's height will be more. 1389 1389 Indeed we are trying to make the sequence linear. . easily it can be proved that if then then . and the circumcenter of is on the bisector of (the perpendicular bisector of ). Now suppose that . Since the center of is on the bisector of . It is obvious that the total number of the leaves of and are equal. 22 . 8. we have ∑ ∑ . then this ray is tangent to the circumcircles of and . Thus the circles and are tangent in . The base case for 2 leaves is trivial because there is no vertex of degree 2. and according to the above equation such ray exists. Let . and if . … . . We solve the problem using induction. Now we must prove that: Considering the definition of and its linearity. So we have: . and the fact that . must be the answer. Thus . too. So is on the bisector of and hence is the incenter of . So we have: 90 2 2 2 So if we draw the ray such that and . and since we have the above equality. .Now we claim that for every concave sequence . have leaves. it is on the bisector of . It is obvious that the bisector of is perpendicular to and so to the (by assumption). we define Consider a concave sequence . Hence 180° . So the desired equals to 7. and is on the line connecting the centers of these two circles. so and the length of the edge is equal to the length of the edge . and leaf from . because these two values for a pair . . Let 0. First note that if . for the first time in vertex . we have 2 Let . intersects the path between . we have 2 Let . . It is enough that we prove that and the length of the edge is equal to the length of the edge . Suppose that 0 . and be the neighbor of in . 9. be two leaves in and the path between . 0. It means that we omitted leaf from . Thus. minimizes. so there is a correspondence between their vertices. 2 min 2 Now do the same for . 2 2 0. min .In which is a leaf. then we have: (let denote the length of the path between and ) 2 doesn’t have a 2-degree vertex: And since . hence the induction and proof are complete. and its correspondence . Now let be the neighbor of in . . Assume the induction hypothesis holds for . we have 2 23 2 . like . 1 . 0. 24 . we have . By applying it to the . .Since is increasing. . hence we can say that for every . and we have 1 3 2 2 2 2 So is surjective. So we must find all polynomials . . 10. it results that 0. . Now let 0 . So for all . be the polynomial with the least degree that satisfies the above equation. such that for all . . . 0 . and hence 0. Let 0 . so 0. . Let .0 0. so for every 2 . . . 0. we have: 0 . 0 If we let 0. it concludes that 0. so we can consider above equation. Then by letting 0 we conclude that .0 0 . so is the identity function. we have 2 in the main equation results in 0 0 in the main equation results in Then let 0 0 0 . so we it means that . : . . It is seen that satisfies the . 0. 0.0 0 . . Hence in the equation . From the other hand by letting Hence 2 Hence we have 2 2 Letting 0. we have . we have 0. can be all values in domain. Now letting 2 2 . Suppose that . . concluding that can only be zero. Thus for all . reached the contradiction. we reach the contradiction by finding a polynomial with lesser degree that satisfies the above equation. 0. From the other hand: (let denote the measure of the arc ) 2 2 25 . respectively. Since is on both circumcircles of and . . . ). and progressions. Since is increasing. the problem is easy) so we know that have . Now it is seen that is in both progressions. 12. And hence | for every .( . 2 . Suppose this similarity carries to a point . and also is constant for every . we 0 . Also we can find such that : are found using the Chinese Remainder Theorem) so for every . and to .11.… form an arithmetic progression. are collinear: The spiral similarity about . so there is a spiral similarity about carrying one of these circles to another such that it carries to . it means appears in all of the progressions. conclude that exists This means that can be displayed as both forms and . then 1 . hence: are in both : And this means that if . must be positive. such that : . we such that and . are collinear. carries . Suppose that | be an arithmetic progression with common difference . It is enough that we prove that the points . From the other hand we can find Name this constant value ( . . so: So . considering the similarity of and we have: ° 180° 180 Now we prove that . So the triangles and are similar. to . . it means : | . Now by using Chinese Remainder Theorem. because then. (even if for one . and: 180 26 ° . . So deduce that and are similar. also and and are . and hence . From which we are similar.Thus the triangles similar. So . .