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.

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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

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Problems

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First Round | 1 holds for 1. Let be a polynomial of degree 2 such that | 1,0,1 . Show that for every 1,1 5 | | 4 2. (Meysam Aghighi) We have a 50 50 garden which is divided into 1 1 pieces and in some pieces an apple, orange or peach tree is planted. We know that there is at least one apple tree adjacent to every orange tree. There are at least one apple tree and one orange tree adjacent to every peach tree and at least one apple tree, one orange tree and one peach tree adjacent to every empty piece (a piece without any tree). (We say that two pieces are adjacent if they have a common side.) Show that the number of empty pieces is not greater than 1000. 3. (Meysam Aghighi) Let the inner bisector of the angle of triangle intersects and the circumcircle of in and respectively. We draw a line through so that it intersects the rays and (with the startpoint) in and . Show that . 4. (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. When commander commands each soldier either doesn’t move or takes a step in one of the four directions! After the command, the soldiers are standing in 2 equal columns, such that the first and last rows are omitted and two columns are added in left and right. Prove that is even.

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5. (Mohsen Jamali) Natural numbers for every distinct and , is divisible by

have this property that . Prove that for every ,

6. (Omid Naghshineh) There are 11 men sitting around a circular table with equal distances and 11 cards with numbers 1,2, … ,11 on them are dealt among them. It is possible that one has no cards and the other has more than one. 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, the places of the cards with numbers 1, and 1 are not the vertices of an acute-angled triangle. (card 0 is the same as card 11, and card 12 is the same as card 1) At the beginning the cards 1 to 11 are dealt among them counter-clockwise. (everyone has exactly one card) Prove that there will never be a man who has all the cards.

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Second Round 1. (Ali Khezeli) Let 2 and , , … , be points on the plane that no three of them are collinear. , ,…, be points on the segments , ,…, a. Let , ,…, be points in the triangles respectively. Show that if , ,…, respectively, then the following holds | | | | | | | | | | | | Where | | means the length of the segment . b. 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 half-planes , ,…, respectively, then, | | | | | | | | | | | | Suggestion: for part (b), first solve the problem assuming that is on the outer bisector of the corresponding angle.

2. (Mohsen Jamali) We call the permutation of 1,2, … , consistent if the set | 1,2, … , has 2 members. Prove that the total number of consistent permutations is , where is the sum of the positive divisors of and is the number of positive divisors of .

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3. We have partitioned a triangle into similar triangles with itself, such that every side of a small triangle is parallel to one of the sides of the main triangle. In

this case every small triangle is derived by a homothety (with a positive or negative ratio) from the main triangle. Prove that the sum of all of these ratios is 1.

4. (Amir Ja'fari) Do there exist two functions , : the following inequality holds? | | |

such that for all |

1

5. Consider a spherical ball on a plane and a point marked on it. 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. Note that the ball must not roll in its place (it means rolling without moving on the plane). Prove that this is possible. 6. Let , be two relatively prime integers. Prove that the following equation has an infinite number of solutions:

7.

is a circumscribed polyhedral. Its faces are colored with black and white such that there isn't any pair of black faces with an edge in common.

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Prove that sum of the black areas is not greater than sum of the white areas.

8. (Omid Naghshineh) Consider that we have omitted some of the points of a 2dlattice (same as ). Look at these points as a graph, two points (vertices) are connected if they are the same in one of their coordinates and differ 1 in another. Each connected component is called a cluster. Suppose that for all , the number of omitted points inside the horizontal square with side length 2 1 and the origin as its center, be less than /2. Prove that the nonomitted points contain exactly one infinite cluster.

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Third Round 1. (Naser Talebizadeh) Prove that for every natural number , there exists a natural number such that for every natural number that 2 1389, the sum of digits of in base , is more than . 2. (Sepehr Ghazinezami) In triangle , is the circumcenter and is the orthocenter. and are the midpoints of and respectively and is a diameter of the circumcircle. If be an inscribed quadrilateral, prove that 1 2 3. (Morteza Saghafian) There is a board divided into unit squares, and we have drawn one of the two diagonals of each unit square. 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. 4. (Morteza Saghafian) We have drawn a polygon with 2n sides without picking up the pencil from the paper. If we name the sides from 1 to 2 with the order of drawing them, then the odd sides are all vertical and are drawn bottom-up. Prove that this polygon intersects itself. 5. (Davood Vakili) In the isosceles triangle , we have and . and are the midpoints of and respectively. is a point that and . is the intersection of and . If be the intersection of with the circumcircle of (not ), prove that the tangent line in to the circumcircle of is parallel to . 6. (Ali Khezeli) We call a sequence , … , of real numbers concave if for every 0

1389 we have

. Find the maximum number such

that for every concave sequence of nonnegative numbers, we have

7. (Mehdi E'tesami) is an arbitrary point on the side of the triangle . is a circle tangent to the segments and in and respectively, also is tangent to the circumcircle of in . If , prove that the circumcircles of and are tangent to each other.

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8. (Mir Omid Hajimirsadeghi) and are two trees without having a vertex of degree 2. Each edge of them has a positive number named the length of the edge. The distance between two vertices is the sum of the length of the edges of the path between them. We call the vertices with degree 1, leaf. 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 , the distance between and in is equal to the distance between and in . 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 , the distance between and in , is equal to the distance between and in . 0 0 , 9. (Mohammad Ja'fari) Find all increasing functions : 0 we have such that for every , 2 2 (note that is not necessarily strictly increasing.) 10. (Ali Khezeli & Morteza Saghafian) Find all polynomials , with real coefficients such that for every , , , , 1 , 1 , 1 0 11. (Ehsan Azarmsa) : is an increasing function and is a natural number. and natural numbers Suppose that there exist prime numbers , , … , , , … , such that for every 1 the set | 1,2, … be an infinite arithmetic progression. Prove that there exists natural number such that 1 , 2 ,…, is an arithmetic progression. and intersect at and . is the 12. (Davood Vakili) Two circles and is on . common tangent of them near to such that is on for the second time at , and intersects for the second time intersects at . is the intersection of and . If be the second intersection point of circumcircle and circumcircle , prove that

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Solutions

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First Round 1. Notice that 1 2

If 0

1

1

1, we see that 1 1 | | 2 2 If 1 0, 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, 1 . The conditions for equality 1

are a.

1 at

, and

b.

1 at

.

2. Let , , , be the total number of apple, orange and peach trees, and the total number of empty pieces respectively. We now want to count the total number of adjacencies of apple trees to an orange tree, a peach tree, or an empty piece. We know that these adjacencies are at most 4 , and since every orange tree, every peach tree, and every empty piece has at least one adjacent , hence apple tree; 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. Note that these adjacencies are at most 3 , because every orange tree must have an adjacent apple tree, and it can have 3 adjacent peach tree or empty piece. On the other hand since every peach tree and every empty piece has an adjacent orange tree, so the total number of , so we have adjacencies are at least 3 Similarly, we have the following inequality 2

12

50

Now, noting that inequalities, we have

50

2 3

3 2

2

4

2

2500, 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. 3. Since is the bisector, we have

So the circumcircle of is tangent to arc in this circle be , we have

180°

180°

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1000

2 in . Hence if the measure of the

2 2 . Thus

Similarly we have

2

4. Suppose that it can be done for . We claim that it can be done for 2, and since it can't be done for 1, must be even. Let up, down, left and right be the four directions. Note that if columns have moved to 2 columns, then the first row must've moved down, the last row must've moved up, the leftmost column must've moved left, and the rightmost column must've moved right. Now, consider the remaining soldiers, they are 2 columns with soldiers in each column, that have moved in columns with 2 soldiers in each. Thus, if it can be done for , then must be even. | , and 5. For all distinct , that , we have On the other hand all of the above terms are divisors of in decreasing order be , then know that

, so if the divisors of . Since we

, we have

1 1 1 And the problem is solved. 6. First divide the table into 11 equal arcs. Now if the cards , be on the points , on the table, we define the distance between these two cards, the number of arcs between , on the table. (the smaller one) For example, the distance between , is 5 in the following figure:

Now, after each step we sum the distances between every two cards with consecutive numbers. Easily it can be proved that after each step this sum either remains invariant or varies by 2 . Because if the place of card be between 1 and 1, this sum doesn’t change, otherwise changes by 2. So the parity of this sum remains invariant. At first it is odd ( 11), and if they are all in one place this sum would be even, so it is impossible.

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Second Round 1. a. Lemma. If be a point in the triangle , then: . PROOF. We extend to meet in . We have: (by triangle inequality) Now we return to the problem, from the triangle inequality we have:

(according to lemma)

b. Suppose that for all 1 and length of this case the vector angle .

, we have

|

is the

is the unit vector codirectional with . In is codirectional with the inner bisector of the , hence from the definition of 0 , plus we have: (consider

Also suppose that · that: same as , , ) Note that |

in which

we can say , , the

1, we can say: ·

· ·

·

Hence we have: · From the other hand, with respect to

, we have: 0

·

· So we deduce that:

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·

· ·

|

|

2. Since ∑ 0 , so from the two members of 1,2, … , , one should be positive and the other one negative. Let

| be the

positive and – be the negative members. Also let | and | . First note that if then . Because if and , then and hence , and so , it means that . Similarly, we can say that if (and 1) then . Also, note that for every 1 , we have . So considering the above paragraph, if for 1 , the remainder of upon division by , be less than or equal to , then . Similarly for , we have . Thus from every consecutive numbers, there are at least of them in and hence if the remainder of the division of by , be more than , then . So easily we deduce that | . For each divisor of , there are 1 different pairs of , whose sum is . So the total number of states is equal to: 1 |

1 |

|

3. Suppose that the bottom side of the main triangle is horizontal, 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, hence: (let be the length of the horizontal side of the triangle , and let Δ be the main triangle)

1 But in the above summation, every horizontal segment (that contains no smaller horizontal segment) once appears with positive sign for its above

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triangle and once appears with negative sign for its below triangle, except the segments of , and these segments appear with the positive sign; so

and so sum of the ratios with respect to their signs is 1. 4. For all we have: 1 | 2

|

|

|

1 2

For every , consider the point , in the plane. The above inequality shows that the distance between every two of these points is more than

√

. For every point

and a radius of

√

,

, consider a circle with center at this point

, so not two of these circles intersect. On the other hand the

number of these points are uncountable, so we have an uncountable number of circles in the plane that no two of them intersect and this is impossible; 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, this is a contradiction; so no two functions with this property exist. 5. Let be the radius of the ball and be its first position on the plane. First, we roll the ball (sphere) on its great circle that passes through the marked point and the top point of the ball, until the mark be on top of the ball. Let be the current position of the ball on the plane. It is trivial that 2 . So there exists a point on the plane so that 2 . Now we roll the ball from to on the segment , and since 2 , the mark will be at the top of the ball. 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. 6. First note that if , , , , , be a solution, then . , . , . , . , . , . is another solution. So it is enough to find a primary solution. We know that 2 3 5 2 3 5 2 3 5 2 3 5 Now since gcd , 1 , hence there exists 1 such that 1 . So 1 , , and hence 1 1 . So we have

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2

3

5

2

2

3

5

3

2 3 5 5 Thus, a group of infinite solutions for the mentioned equation is: , , , , , 3 5 ,2 ,2 3 5 ,3 2 ,2 3 5 ,5 7. For every face of the polyhedral, consider the point of tangency of the inscribed sphere with that face, and draw segments from that point to the vertices of the face. Thus, we have divided each face of the polyhedral to some triangles. Now, consider a black face … , and let be the point of tangency of this face with the inscribed sphere, consider the triangle , from the assumption there is a white face containing the edge that the inscribed circle is tangent to it at . It is trivial that and (the length of the tangents from a point to a sphere are equal), so the triangles and are congruent and have equal areas. Thus, every black triangle has a corresponding white triangle and these white triangles are distinct. So sum of the black areas is not greater than sum of the white areas. 8. We suppose that the origin is not omitted without loss of generality. Consider the cluster having the origin, if the cluster is finite, its border contains some omitted vertices that are adjacent to at least one of the vertices of the cluster. Among all of the points of the cluster having origin, consider the point having maximum horizontal distance from the origin, and then consider the point having maximum vertical distance from the origin, let be the maximum of these two distances. So all of the points of the cluster having origin, are in the square whose center is on the origin and has side-length 2 1. But since, the points on the cluster's border have rounded around the origin, hence for every

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0 there exists a point on the border that its horizontal or vertical distance from the origin is equal to . Thus, at least 1 points are omitted which is a contradiction. So the cluster having the origin is infinite. Now assume that there are two infinite clusters. Obviously their border contains a path of omitted points which is infinite from both directions. We extend the origin-centered square until it contains a piece of this path. From now on, by increasing one by one, the square's side will extend by 2 and there will be 2 more vertices of the path in the square, but the value of /2 will increase by 1/2. So but increasing , 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. This contradiction shows that there is exactly one infinite cluster.

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Third Round 1. We claim that 1389! 1 has the desired properties. Let 2 1389 , we know that there are at least 1 zeros in the rightmost of in base , because 1389! is divisible by . Hence, the 1389! 1 rightmost digits of 1389! 1 is 1, then sum of the digits of in base is at least 1 1 which is greater than or equal to 1. 2. Let be the circumcircle of the quadrilateral . A 2 homothety at , takes , to , respectively. So this homothety maps to the circumcircle of . So is tangent to the circumcircle of at and its radius is half of the radius of the circumcircle of . We know that the radius of the circumcircle of is equal to the radius of the circumcircle of ( ). Hence passes through the center of the circumcircle of and its radius is . So

is tangent to the circumcircle of

(at ). The circumcircles of

and are symmetric across the line , as well as , (the points of tangency of these to circles with ) and we deduce that . On the other hand, is a diameter of ( ), hence and so . Hence the line is perpendicular to , and since , this line is the . Now let be the midpoint of perpendicular bisector of , so , are equal in the isosceles triangle . Now , then the medians since and are both perpendicular to , they are parallel. Similarly, and are both perpendicular to , and are parallel. So is a parallelogram and its diagonals bisect each other, from this we determine that is the midpoint of

and hence:

.

3. A connected component of diagonals is some of the diagonals that are connected to each other by a path of diagonals, and if we add another diagonal to it, it will no longer be connected. Easily, it can be seen that the diagonals that are not in this component but are on the border of this component, are connected. (until they don’t cross the table sides.) Now, 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. If the height of this path be , the problem is solved, otherwise suppose that the height be less than . Consider the connected component having this path. If this component is not connected to the right side of the table, then its

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rightmost border's height will be more, and this is a contradiction. Similarly if this connected component is not connected to the left side of the table, then its leftmost border's height will be more, and a contradiction. Thus, this connected component is connected to both right and left sides of the table, and hence they're connected to each other. 4. First Solution: If a polygon does not intersect itself, then by moving clockwise (or counterclockwise) along its sides, the inside (or outside) of the polygon must always be at our right-side. But if the said polygon doesn’t intersect itself, then the right-side of its rightmost side (because it is drawn upward) is its outside, and the right-side of its leftmost side is its (because it is drawn upward) is its inside, which is a contradiction. Second Solution: Suppose that the leftmost side of the polygon connects the vertex upward to the vertex , and the rightmost side connects the vertex upward to the vertex . Then all of the polygon must be between the lines and . Clearly, the remaining of the polygon contains a path that connects to , and another one that connects to ; but these two paths intersect. Third Solution: suppose that this polygon has 2 sides and doesn’t intersect itself. In this case, total sum of the angles of this polygon is equal to 2 2 180° 1 360° . From the other hand each non-vertical side of this polygon is adjacent to two vertical sides that are drawn upward. So the sum of the two adjacent angles to each non-vertical side is 360° , and since we have non-vertical sides, the total sum of all angles equals 360° , which is a contradiction. 5. Obviously , , , are collinear. Since is an inscribed quadrilateral, then , and since is inscribed ( is the foot of the altitude from on ), then . Hence, . Now if we draw ray parallel to such that , then in the circumcircle of : (let be the measure of the arc ) 2 Hence we deduce that is tangent to circumcircle of . 6. First consider the concave sequence 1389 ,0 Note that the maximum possible value of for this sequence is .

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1389, ∑ ∑

0.

Now we claim that for every concave sequence , we have ∑ ∑ , and since we have the above equality, must be the answer. , we define Consider a concave sequence , , … , 1389

1389 Indeed we are trying to make the sequence linear. Now we must prove that:

Considering the definition of and its linearity, and the fact that , easily it can be proved that if then then . So we have:

, , and if

.

So the desired equals to

7. It is obvious that the bisector of is perpendicular to and so to the (by assumption). Thus , and the circumcenter of is on the bisector of (the perpendicular bisector of ). Since the center of is on the bisector of , and is on the line connecting the centers of these two circles, it is on the bisector of , too. Hence 180° . 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 , then this ray is tangent to the circumcircles of and , and according to the above equation such ray exists. Thus the circles and are tangent in . 8. It is obvious that the total number of the leaves of and are equal. We solve the problem using induction. The base case for 2 leaves is trivial because there is no vertex of degree 2. Now suppose that , have leaves. Let ,

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In which is a leaf. It means that we omitted leaf from , and leaf from . Assume the induction hypothesis holds for , , so there is a correspondence between their vertices, like . Now let be the neighbor of in , and be the neighbor of in . It is enough that we prove that and the length of the edge is equal to the length of the edge . First note that if , be two leaves in and the path between , , intersects the path between , for the first time in vertex , then we have: (let denote the length of the path between and ) 2 doesn’t have a 2-degree vertex:

And since

,

min

,

2

min

2

Now do the same for . Thus, because these two values for a pair , and its correspondence , , minimizes, so and the length of the edge is equal to the length of the edge , hence the induction and proof are complete. 9. Suppose that 0 . Let 0, we have 2 Let

,

0, we have 2

Let

,

2

2

0, we have 2

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2

Since

is increasing, so for every 2 . From the other hand by letting

Hence

2 Hence we have 2 2 Letting 0, Now letting

2

2 .

we have , we have

2

in the main equation results in

0 0 in the main equation results in

Then let

0

0

0

, and we have 1 3 2 2 2 2 So is surjective. Hence in the equation , can be all values in domain, so is the identity function. 10. Let , , 1 . So we must find all polynomials , such that for all , , : , , 0 , Suppose that , be the polynomial with the least degree that satisfies the above equation, we reach the contradiction by finding a polynomial with lesser degree that satisfies the above equation, concluding that can only be zero. Let 0 , so 0,0 0 . Now let 0 , it results that 0, 0, hence we can say that for every , we have 0, 0. Then by letting 0 we conclude that ,0 0 . Thus for all , we have , , . By applying it to the ,0 0, so we can consider above equation, we have: 0 , , , , , , 0 If we let 0, it concludes that 0, 0. So for all , 0, 0, , . It is seen that satisfies the , so we it means that , reached the contradiction, and hence 0.

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11. Suppose that | be an arithmetic progression with common difference . Since is increasing, must be positive. (even if for one , we 0 , the problem is easy) so we know that have . Now by using Chinese Remainder Theorem, we such that and . conclude that exists This means that can be displayed as both forms and . Now it is seen that is in both progressions, so:

So

. And hence

|

for every , and also

is constant for every .

). From the other hand we can find Name this constant value ( , it means appears in all of the progressions. such that : , it means : | .( , Also we can find such that : are found using the Chinese Remainder Theorem) so for every , and progressions, hence: are in both : And this means that if , then 1 , 2 ,… form an arithmetic progression. 12. Since is on both circumcircles of and , so there is a spiral similarity about carrying one of these circles to another such that it carries to , and to . Suppose this similarity carries to a point . It is enough that we prove that the points , , are collinear, because then, considering the similarity of and we have: ° 180° 180 Now we prove that , , are collinear: The spiral similarity about , carries , , to , , respectively. So the triangles and are similar. From the other hand: (let denote the measure of the arc ) 2 2

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Thus the triangles similar. So deduce that

and

are similar, and hence

, also and

and

are

. From which we are similar. So

, and: 180

26

°

, ,