A polygonal line with a finite number of segments has all its vertices on a parabola. Any two adjacent segments make equal angles with the tangent to the parabola at their point of intersection. One end of the polygonal line is also on the axis of the parabola. Show that the other vertices of the polygonal line are all on the same side of the axis.


What is the smallest n for which there is a solution to \(\sin x_{1} + \sin x_{2} + \ldots +\sin x_{n} = 0\) , \(\sin x_{1} + 2 \sin x_{2} + \ldots + n \sin x_{n} = 100\) ?

Solution: Put \(x_{1} = x_{2} = \ldots = x_{10} = 3\pi /2\) , \(x_{11} = x_{12} = \ldots = x_{20} = \pi /2\) . Then \(\sin x_{1} + \sin x_{2} + \ldots +\sin x_{20} = (- 1 - 1 - 1 - \ldots - 1) + (1 + 1 + \ldots + 1) = 0\) , and \(\sin x_{1} + 2 \sin x_{2} + \ldots + 20 \sin x_{20} = - (1 + 2 + \ldots + 10) + (11 + 12 + \ldots + 20) = 100\) . So there is a solution with \(\mathrm{n} = 20\) . If there is a solution with \(\mathrm{n} < 20\) , then there must be a solution for \(\mathrm{n} = 19\) (put any extra \(x_{i} = 0\) ). But then \(100 = (\sin x_{1} + 2 \sin x_{2} + \ldots + 19 \sin x_{19}) - 10 (\sin x_{1} + \sin x_{2} + \ldots + \sin x_{19}) = - 9 \sin x_{1} - 8 \sin x_{2} - 7 \sin x_{3} - \ldots - \sin x_{9} + \sin x_{11} + 2 \sin x_{12} + \ldots + 9 \sin x_{19}\) . But \(\mathrm{l} \mathrm{rhs} \mathrm{l} \leq (9 + 8 + \ldots + 1) + (1 + 2 + \ldots + 9) = 90\) . Contradiction. So there is no solution for \(\mathrm{n} < 20\) .


The sequence of integers \(\mathrm{a}_{\mathrm{n}}\) is given by \(\mathrm{a}_{0} = 0\) , \(\mathrm{a}_{\mathrm{n}} = \mathrm{p}(\mathrm{a}_{\mathrm{n} - 1})\) , where \(\mathrm{p}(\mathrm{x})\) is a polynomial whose coefficients are all positive integers. Show that for any two positive integers \(\mathrm{m}\) , \(\mathrm{k}\) with greatest common divisor d, the greatest common divisor of \(\mathrm{a}_{\mathrm{m}}\) and \(\mathrm{a}_{\mathrm{k}}\) is \(\mathrm{a}_{\mathrm{d}}\) .


Prove that for any tetrahedron the radius of the inscribed sphere \(\mathrm{r} < \mathrm{ab} / (2(\mathrm{a} + \mathrm{b}))\) , where a and b are the lengths of any pair of opposite edges.


7 boys each went to a shop 3 times. Each pair met at the shop. Show that 3 must have been in the shop at the same time.


Can 77 blocks each \(3 \times 3 \times 1\) be assembled to form a \(7 \times 9 \times 11\) block?


The incircle of ABC touches AB at M. N is any point on the segment BC. Show that the incircles of AMN, BMN, ACN have a common tangent.


A positive integer n has exactly 12 positive divisors \(1 = \mathrm{d}_1 < \mathrm{d}_2 < \mathrm{d}_3 < \ldots < \mathrm{d}_{12} = \mathrm{n}\) . Let \(\mathrm{m} = \mathrm{d}_4\) - 1. We have \(\mathrm{d}_{\mathrm{m}} = (\mathrm{d}_1 + \mathrm{d}_2 + \mathrm{d}_4) \mathrm{d}_8\) . Find n.


Eight pawns are placed on a chessboard, so that there is one in each row and column. Show that an even number of the pawns are on black squares.


ABC is a triangle. A', B', C' are points on the segments BC, CA, AB respectively. Angle B'A'C' = angle A and AC'/C'B = BA'/A'C = CB'/B'A. Show that ABC and A'B'C' are similar.


One bird lives in each of n bird- nests in a forest. The birds change nests, so that after the change there is again one bird in each nest. Also for any birds A, B, C, D (not necessarily distinct), if the distance \(\mathrm{AB} < \mathrm{CD}\) before the change, then \(\mathrm{AB} > \mathrm{CD}\) after the change. Find all possible values of n.


Show that the 120 five digit numbers which are permutations of 12345 can be divided into two sets with each set having the same sum of squares.


We are given 1998 normal coins, 1 heavy coin and 1 light coin, which all look the same. We wish to determine whether the average weight of the two abnormal coins is less than, equal to, or greater than the weight of a normal coin. Show how to do this using a balance 4 times or less.


A triangle with perimeter 1 has side lengths a, b, c. Show that \(\mathrm{a}^2 +\mathrm{b}^2 +\mathrm{c}^2 +4\mathrm{abc}< 1 / 2\) .


ABCD is a convex quadrilateral. X lies on the segment AB with AX/XB = m/n. Y lies on the segment CD with CY/YD = m/n. AY and DX intersect at P, and BY and CX intersect at Q. Show that area XQYP/area ABCD < mn/(m² + mn + n²).


A 23 x 23 square is tiled with 1 x 1, 2 x 2 and 3 x 3 squares. What is the smallest possible number of 1 x 1 squares?


Do there exist two reals whose sum is rational, but the sum of their nth powers is irrational for all \(n > 1\) ? Do there exist two reals whose sum is irrational, but the sum of whose nth powers is rational for all \(n > 1\) ?


An insect is on a square ceiling side 1. The insect can jump to the midpoint of the segment joining it to any of the four corners of the ceiling. Show that in 8 jumps it can get to within 1/100 of any chosen point on the ceiling.


ABCD has AB = CD, but AB not parallel to CD, and AD parallel to BC. The triangle is ABC is rotated about C to A'B'C. Show that the midpoints of BC, B'C and A'D are collinear.


Show that for each integer \(n > 0\) , there is a polygon with vertices at lattice points and all sides parallel to the axes, which can be dissected into \(1 \times 2\) (and/or \(2 \times 1\) ) rectangles in exactly \(n\) ways.


Find the smallest positive integer \(n\) for which we can find an integer \(m\) such that \([10^n /m] = 1989\) .


ABC is a triangle. Points D, E, F are chosen on BC, CA, AB such that B is equidistant from D and F, and C is equidistant from D and E. Show that the circumcenter of AEF lies on the bisector of EDF.


S and S' are two intersecting spheres. The line BXB' is parallel to the line of centers, where B is a point on S, B' is a point on S' and X lies on both spheres. A is another point on S, and A' is another point on S' such that the line AA' has a point on both spheres. Show that the segments AB and A'B' have equal projections on the line AA'.


Two walkers are at the same altitude in a range of mountains. The path joining them is piecewise linear with all its vertices above the two walkers. Can they each walk along the path until they have changed places, so that at all times their altitudes are equal?


Find the least possible value of \((x + y)(y + z)\) for positive reals satisfying \((x + y + z)xyz = 1\) .


A polyhedron has an even number of edges Show that we can place an arrow on each edge so that each vertex has an even number of arrows pointing towards it (on adjacent edges).


N is the set of positive integers. Does there exist a function f: \(\mathrm{N}\to \mathrm{N}\) such that \(\mathrm{f(n + 1)} = \mathrm{f}(\mathrm{f(n)}\) ) + f( \(\mathrm{f(n + 2)}\) ) for all n.


A convex polygon is such that any segment dividing the polygon into two parts of equal area which has at least one end at a vertex has length \(< 1\) . Show that the area of the polygon is \(< \pi /4\) .


Show that \(\mathrm{x}^4 > \mathrm{x} - 1 / 2\) for all real x.

Solution: \(\mathrm{x}^4 - \mathrm{x} + 1 / 2 = (\mathrm{x}^2 - 1 / 2)^2 +(\mathrm{x} - 1 / 2)^2\geq 0\) . We could only have equality if \(\mathrm{x}^2 = \mathrm{x} = 1 / 2\) , which is impossible, so the inequality is strict.


The line joining the midpoints of two opposite sides of a convex quadrilateral makes equal angles with the diagonals. Show that the diagonals are equal.

Solution:
![md5:54360b517d323d1521aef21360afdede](54360b517d323d1521aef21360afdede.jpeg)

Let L, M, N be the midpoints of BC, CD, DA. Assume that NL makes equal angles with AC and BD, so \(\angle \mathrm{NLM} = \angle \mathrm{BEL} = \angle \mathrm{AFN} = \angle \mathrm{LNM}\) , so \(\mathrm{LM} = \mathrm{MN}\) and hence \(\mathrm{BD} = \mathrm{AC}\) .


A graph has 30 points and each point has 6 edges. Find the total number of triples such that each pair of points is joined or each pair of points is not joined.

## Answer

1990

Solution: There are 30. 29. \(28 / 6 = 4060\) triples in all. Let m be the number of triples with 0 or 3 edges, and let n be the number of triples with 1 or 2 edges. So \(\mathrm{m} + \mathrm{n} = 4060\) . Each point is joined to 6 others, so it is in 6. \(5 / 2 = 15\) triples where it is joined to both the other points, and it is in 23. \(22 / 2 = 253\) triples where it is not joined to either of the other points. So the total number of triples (a,b,c), where a is joined to b and c or not joined to b or c is \(30(15 + 253) = 8040\) . This counts the m triples 3 times each, and the n triples once each, so \(3\mathrm{m} + \mathrm{n} = 8040\) . Hence \(\mathrm{m} = 1990\) .


Does there exist a rectangle which can be dissected into 15 congruent polygons which are not rectangles? Can a square be dissected into 15 congruent polygons which are not rectangles?

## Answer

yes, yes

Solution:
![md5:fe9e789413e4c2d406e979a3b5cc493e](fe9e789413e4c2d406e979a3b5cc493e.jpeg)

By stretching vertically we get a square.


The point P lies inside the triangle ABC. A line is drawn through P parallel to each side of the triangle. The lines divide AB into three parts length c, c', c" (in that order), and BC into three parts length a, a', a" (in that order), and CA into three parts length b, b', b" (in that order). Show that abc = a'b'c' = a"b"c".

Solution:
![md5:fca7c846c76e89d2f888faaeee6233c1](fca7c846c76e89d2f888faaeee6233c1.jpeg)

The three small triangles are similar, so a/a" = c'/c = b"/b' and a/a' = c'/c" = b"/b. Hence (a/a")(b/b") = (c'/c)(c"/c') = c"/c, so abc = a"b"c". Similarly, (a/a')(c/c') = (b"/b)(b"/b") = b/b, so abc = a'b'c'.


Find three non- zero reals such that all quadratics with those numbers as coefficients have two distinct rational roots.

Answer 1,2,- 3

Solution: If a + b + c = 0, then 1 is a root of ax² + bx + c, and so the other root is - b/a - 1, which is rational.


What is the largest possible value of \(|\ldots |\mathrm{la}_1 - \mathrm{a}_2| - \mathrm{a}_3| - \ldots - \mathrm{a}_{1990}|\) , where \(a_1, a_2, \ldots , a_{1990}\) is a permutation of 1, 2, 3, ..., 1990?

Answer 1989

Solution: Since \(\mathrm{la - bl} \leq \max (\mathrm{a,b})\) , a trivial induction shows that the expression does not exceed \(\max (\mathrm{a}_1, \mathrm{a}_2, \ldots , \mathrm{a}_{1990}) = 1990\) . But for integers \(\mathrm{la - bl}\) has the same parity as \(\mathrm{a + b}\) , so a trivial induction shows that the expression has the same parity as \(\mathrm{a}_1 + \mathrm{a}_2 + \ldots + \mathrm{a}_{1990} = 1990\) . 1991/2, which is odd. So it cannot exceed 1989. That can be attained by the permutation 2, 4, 5, 3, 6, 8, 9, 7, ... , \(4\mathrm{k} + 2\) , \(4\mathrm{k} + 4\) , \(4\mathrm{k} + 5\) , \(4\mathrm{k} + 3\) , ..., \(1984 + 2\) , \(1984 + 4\) , \(1984 + 5\) , \(1984 + 3\) , 1990, 1. Because we get successively 2, 3, 0; 6, 2, 7, 0; 10, 2, 11, 0; ... ; \(4\mathrm{k} + 2\) , 2, \(4\mathrm{k} + 3\) , 0; ... ; 1986, 2, 1987, 0; 1990, 1989.


An equilateral triangle of side n is divided into \(\mathrm{n}^2\) equilateral triangles of side 1. A path is drawn along the sides of the triangles which passes through each vertex just once. Prove that the path makes an acute angle at at least n vertices.

Solution:
![md5:58aed129ef4e27f6b77023add17b0c63](58aed129ef4e27f6b77023add17b0c63.jpeg)

The diagram has \(1 + 2 + \ldots + \mathrm{n} = \mathrm{n}(\mathrm{n} + 1) / 2\) upright triangles and \(1 + 2 + \ldots + \mathrm{n} - 1 = \mathrm{n}(\mathrm{n} - 1) / 2\) upside down triangles. It has \(1 + 2 + \ldots + \mathrm{n} + 1 = (\mathrm{n} + 1)(\mathrm{n} + 2) / 2\) vertices. So the path must be \((\mathrm{n} + 1)(\mathrm{n} + 2) / 2 - 1 = (\mathrm{n}^2 +3\mathrm{n}) / 2\) units long. Each unit length of the path is in just one upright triangle. The path cannot contain all three sides of a small triangle, or it would pass through a vertex more than once. So it must contain two sides of \((\mathrm{n}^2 +3\mathrm{n}) / 2 - \mathrm{n}(\mathrm{n} + 1) / 2 = \mathrm{n}\) triangles. But if it contains two sides of a triangle, then it must make an acute angle at the vertex where they meet.


Can the squares of a 1990 x 1990 chessboard be colored black or white so that half the squares in each row and column are black and cells symmetric with respect to the center are of opposite color?

Answer no

Solution:
![md5:ba89446342b07ea6059c12a58ecc2d7f](ba89446342b07ea6059c12a58ecc2d7f.jpeg)

Suppose it can be done. Divide the board into 4 quadrants. Suppose there are b black and \(995^{2}\) - b white squares in the top left quadrant. Then there are \(995^{2}\) - b black and b white squares in the bottom right quadrant (by the symmetry property).

If half the squares in each row are black, then half the squares in the first 995 rows are black, so the number of black squares in the top right quadrant is 995. \(1990 / 2 - \mathrm{b} = 995^{2} - \mathrm{b}\) . So if half the squares in each column are black, then half the squares in the right- hand half of the board are black, so \((995^{2} - \mathrm{b}) + (995^{2} - \mathrm{b}) = 995^{2}\) , in other words, \(\mathrm{b} = 995^{2} / 2\) , which is impossible.


Let \(\mathbf{x}_1\) , \(\mathbf{x}_2\) , ..., \(\mathbf{x}_n\) be positive reals with sum 1. Show that \(\mathbf{x}_1^{2} / (\mathbf{x}_1 + \mathbf{x}_2) + \mathbf{x}_2^{2} / (\mathbf{x}_2 + \mathbf{x}_3) + \ldots + \mathbf{x}_n\) . \(\frac{1}{2} /(\mathbf{x}_{\mathrm{n - 1}} + \mathbf{x}_{\mathrm{n}}) + \mathbf{x}_{\mathrm{n}}^{2} / (\mathbf{x}_{\mathrm{n}} + \mathbf{x}_{\mathrm{1}}) \geq 1 / 2\) .

Solution: \[\sum \mathbf{x}_1^{2} / (\mathbf{x}_1 + \mathbf{x}_{i + 1}) - \sum \mathbf{x}_{i + 1}^{2} / (\mathbf{x}_i + \mathbf{x}_{i + 1}) = \sum (\mathbf{x}_i - \mathbf{x}_{i + 1}) = 0.\] \[\mathbf{x}_i^{2} + \mathbf{x}_{i + 1}^{2} / (\mathbf{x}_i + \mathbf{x}_{i + 1}) \geq \frac{1}{4} \sum (\mathbf{x}_i + \mathbf{x}_{i + 1}) = \frac{1}{2}.\]


ABCD is a convex quadrilateral. X is a point on the side AB. AC and DE intersect at Y. Show that the circumcircles of ABC, CDY and BDX have a common point.


Two grasshoppers sit at opposite ends of the interval [0, 1]. A finite number of points (greater than zero) in the interval are marked. A move is for a grasshopper to select a marked point and jump over it to the equidistant point the other side. This point must lie in the interval for the move to be allowed, but it does not have to be marked. What is the smallest n such that if each grasshopper makes n moves or less, then they end up with no marked points between them?


Find all integers n such that \([\mathrm{n} / 1!] + [\mathrm{n} / 2!] + \ldots + [\mathrm{n} / 10!] = 1001\) .


A, B, C are adjacent vertices of a regular 2n- gon and D is the vertex opposite to B (so that BD passes through the center of the 2n- gon). X is a point on the side AB and Y is a point on the side BC so that angle XDY = π/2n. Show that DY bisects angle XYC.


A graph has n points and n(n- 1)/2 edges. Each edge is colored with one of k colors so that there are no closed monochrome paths. What is the largest possible value of n (given k)?


Given a point X and n vectors \(\mathbf{x}_{\mathrm{i}}\) with sum zero in the plane. For each permutation of the vectors we form a set of n points, by starting at X and adding the vectors in order. For example, with the original ordering we get \(\mathbf{X}_{1}\) such that \(\mathbf{X}\mathbf{X}_{1} = \mathbf{x}_{1}\) , \(\mathbf{X}_{2}\) such that \(\mathbf{X}_{1}\mathbf{X}_{2} = \mathbf{x}_{2}\) and so on. Show that for some permutation we can find two points Y, Z with angle \(\mathbf{Y}\mathbf{X}\mathbf{Z} = 60\) deg, so that all the points lie inside or on the triangle XYZ.


Two unequal circles intersect at X and Y. Their common tangents intersect at Z. One of the tangents touches the circles at P and Q. Show that ZX is tangent to the circumcircle of PXQ.


Given 1990 piles of stones, containing 1, 2, 3, ..., 1990 stones. A move is to take an equal number of stones from one or more piles. How many moves are needed to take all the stones?


A quadratic polynomial \(\mathrm{p(x)}\) has positive real coefficients with sum 1. Show that given any positive real numbers with product 1, the product of their values under p is at least 1.


A cube side 100 is divided into a million unit cubes with faces parallel to the large cube. The edges form a lattice. A prong is any three unit edges with a common vertex. Can we decompose the lattice into prongs with no common edges?


For which positive integers n is \(3^{2n + 1} - 2^{2n + 1} - 6^n\) composite?

Answer all \(n\neq 1\)

Solution: \(3^{2n + 1} - 2^{2n + 1} - 6^n = (3^n - 2^n)(3^{n + 1} + 2^{n + 1})\) , so it is certainly composite for \(n > 1\) . For \(n = 1\) , it is \(27 - 8 - 6 = 13\) , which is prime.


If every altitude of a tetrahedron is at least 1, show that the shortest distance between each pair of opposite edges is more than 2.


A game is played in three moves. The first player picks any real number, then the second player makes it the coefficient of a cubic, except that the coefficient of \(x^{3}\) is already fixed at 1. Can the first player make his choices so that the final cubic has three distinct integer roots?


Given 2n genuine coins and 2n fake coins. The fake coins look the same as genuine coins but weigh less (but all fake coins have the same weight). Show how to identify each coin as genuine or fake using a balance at most 3n times.


Solution:
\(11 = (\mathrm{ab} - 2\mathrm{cd})^{2} + 2(\mathrm{ac} + \mathrm{bd})^{2} = (\mathrm{a}^{2} + 2\mathrm{d}^{2})(\mathrm{b}^{2} + 2\mathrm{c}^{2})\) , so we must have either (1) \(\mathrm{a}^{2} + 2\mathrm{d}^{2} = 1\) , \(\mathrm{b}^{2} + 2\mathrm{c}^{2} = 11\) , or (2) \(\mathrm{a}^{2} + 2\mathrm{d}^{2} = 11\) , \(\mathrm{b}^{2} + 2\mathrm{c}^{2} = 1\) .

(1) gives \(\mathrm{a} = \pm 1\) , \(\mathrm{d} = 0\) , \(\mathrm{b} = \pm 3\) , \(\mathrm{c} = \pm 1\) . If \(\mathrm{a} = 1\) and \(\mathrm{d} = 0\) , then \(\mathrm{ac} + \mathrm{bd} = 1\) implies \(\mathrm{c} = 1\) , and \(\mathrm{ab} - 2\mathrm{cd} = 3\) implies \(\mathrm{b} = 3\) . Similarly, if \(\mathrm{a} = -1\) , then \(\mathrm{c} = -1\) , and \(\mathrm{b} = -3\) . Similarly, (2) gives \((\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}) = (3,1,0,1),(-3, - 1,0, - 1)\) .


n numbers are written on a blackboard. Someone then repeatedly erases two numbers and writes half their arithmetic mean instead, until only a single number remains. If all the original numbers were 1, show that the final number is not less than \(1 / \mathrm{n}\) .

Solution: Put \(\mathrm{c} = (\mathrm{a} + \mathrm{b}) / 4\) . We have \(1 / \mathrm{c} = 4 / (\mathrm{a} + \mathrm{b}) \leq 1 / \mathrm{a} + 1 / \mathrm{b}\) , so each move does not increase the sum of the reciprocals of the numbers. If the final number is \(\mathrm{k}\) , then the final sum of reciprocals is \(1 / \mathrm{k}\) . The initial sum is \(\mathrm{n}\) , so \(1 / \mathrm{k} \leq \mathrm{n}\) , or \(\mathrm{k} \geq 1 / \mathrm{n}\) .


Four lines in the plane intersect in six points. Each line is thus divided into two segments and two rays. Is it possible for the eight segments to have lengths 1, 2, 3, ..., 8? Can the lengths of the eight segments be eight distinct integers?

Answer no, yes

Solution:
![md5:386ceaf84dd3e55dc0eef7999b32a956](386ceaf84dd3e55dc0eef7999b32a956.jpeg)

If a triangle has integer sides, one of which is 1, then it must be isosceles. So the only candidates for the segment length 1 are AB and AE. wlog \(\mathrm{AB} = 1\) , so \(\mathrm{BF} = \mathrm{AF}\) . Hence \(\cos\)

DFE = 1 - 1/(2 AF²). Hence ED² = DF² + EF² + 2DF. EF(1 - 1/2AF²) = DF² + EF² + 2DF. EF - DF. EF/AF². But the first three terms are integers and the last term is < 1. Contradiction. (Careful, looking at the figure one is tempted to conclude that ED < AB, but a more realistic figure shows that is false.).

![md5:de556ee4d753e91963e7e8be0bcc6baf](de556ee4d753e91963e7e8be0bcc6baf.jpeg)

Building on the 3,4,5 triangle we get the figure above.


A lottery ticket has 50 cells into which one must put a permutation of 1, 2, 3, ..., 50. Any ticket with at least one cell matching the winning permutation wins a prize. How many tickets are needed to be sure of winning a prize?

Answer 26

Solution:
Take the tickets:

1 2 3 ... 25 26 27 ... 50 2 3 4 ... 26 1 27 ... 50 3 4 5 ... 1 2 27 ... 50

26 1 2 ... 24 25 27 ... 50 Each of the numbers 1, 2, ..., 26 occurs in each of the places 1, 2, ..., 26, but the winning ticket cannot have all these numbers in the last 24 places. So there must be at least one match. So 26 tickets suffice.

Now given any 25 tickets we show that they could all fail to match the winning permutation. In other words, we construct a permutation which fails to match any of the 25 tickets in any cell. We place the numbers 1, 2, 3, ..., 50 in turn. We start by placing 1. Clearly at most 25 places are ruled out, so we can place the 1. Now suppose we have placed 1, 2, ..., a. There must be at least 25 places where a+1 is not ruled out. If any of them are still unoccupied, then we are done. If not, they must be occupied by numbers \(\mathbf{x}_1\) , \(\mathbf{x}_2\) , ..., \(\mathbf{x}_{25}\) already placed. Take any empty place. 26 numbers cannot be ruled out for it, and we know that a+1 is ruled out, so at least one of the \(\mathbf{x}_i\) is not ruled out. So we can move that \(\mathbf{x}_i\) to it and then place a+1 where the \(\mathbf{x}_i\) came from.


Find unequal integers m, n such that mn + n and mn + m are both squares. Can you find such integers between 988 and 1991?

Answer no

Solution:
For example, \(49 = 7^{2}\) , \(50 = 2\) . \(5^{2}\) , \(8 = 2\) . \(2^{2}\) , \(9 = 3^{2}\) , so \(49\) . \(8 + 8 = 20^{2}\) , \(49\) . \(8 + 49 = 21^{2}\) .

\(\mathrm{wlog m}< \mathrm{n}\) . Then \(\mathrm{mn} + \mathrm{m} = (\mathrm{m} + \mathrm{h})^{2}\) , \(\mathrm{mn} + \mathrm{n} = (\mathrm{m} + \mathrm{k})^{2}\) , with \(\mathrm{k} > \mathrm{h}\) . So \(\mathrm{n} - \mathrm{m} = (\mathrm{m} + \mathrm{k})^{2} - (\mathrm{m} + \mathrm{h})^{2} = (\mathrm{k} - \mathrm{h})(2\mathrm{m} + \mathrm{k} + \mathrm{h}) > 2\mathrm{m}\) , so \(\mathrm{n} > 3\mathrm{m}\) . Hence we cannot have \(\mathrm{m}\) and \(\mathrm{n}\) between 988 and 1991.


ABCD is a rectangle. Points K, L, M, N are chosen on AB, BC, CD, DA respectively so that KL is parallel to MN, and KM is perpendicular to LN. Show that the intersection of KM and LN lies on BD.

Solution:
![md5:289e0cda6fd1182ce46a641ac3fcffd1](289e0cda6fd1182ce46a641ac3fcffd1.jpeg)

Let LN and KM meet at O. \(\angle \mathrm{NOM} = \angle \mathrm{NDM} = 90^{\circ}\) , so OMDN is cyclic. Hence \(\angle \mathrm{NOD} = \angle \mathrm{NMD}\) . Similarly, BLOK is cyclic and \(\angle \mathrm{LOB} = \angle \mathrm{LKB}\) . But NM is parallel to LK and AB is parallel to CD, so \(\angle \mathrm{LKB} = \angle \mathrm{NMD}\) . Hence \(\angle \mathrm{NOD} = \angle \mathrm{LOB}\) , so DOB is a straight line.


An investigator works out that he needs to ask at most 91 questions on the basis that all the answers will be yes or no and all will be true. The questions may depend upon the earlier answers. Show that he can make do with 105 questions if at most one answer could be a lie.

Solution:
Suppose he asks n questions as usual, and then asks "did you lie to any of the last n questions?" If the reply is a truthful no, then the n answers were correct. If the reply is a lying no, then the n answers were still correct. On the other hand if the answer is yes, then the n answers might have been correct and might not. However, a lie has certainly been told, so all future answers must be truthful and so he could ask the n questions again.

\(91 = 7\) . 13, so the obvious candidates for n are 7 and 13. If we take \(\mathrm{n} = 7\) , then the worst case is 13 check questions and 7 repeat questions. That does not work because he needs 20 extra questions and only has 14. A little thought suggests reducing n each time. So the first batch of questions is 13, followed by a check question. If the check answer is yes, then he knows a lie has been told and asks the 13 questions again. No further check questions are needed, and he has used exactly 14 extra questions. If the check answer is no, then the lie may not have been told, so the next batch of questions is 12, followed by a check question, and so on. That allows him to ask \(13 + 12 + \ldots +1 = 91\) questions. If he gets a yes to the check question after the batch of i, then he ignores the answers to that batch and asks them again, thus asking a total of 14 extra questions, but thereafter asks no check questions.


A minus sign is placed on one square of a \(5 \mathrm{x} 5\) board and plus signs are placed on the remaining squares. A move is to select a \(2 \mathrm{x} 2\) , \(3 \mathrm{x} 3\) , \(4 \mathrm{x} 4\) or \(5 \mathrm{x} 5\) square and change all the signs in it. Which initial positions allow a series of moves to change all the signs to plus?

Answer only the central square

Solution:
![md5:02463e47ef2122f6f333c91ae48786ab](02463e47ef2122f6f333c91ae48786ab.jpeg)

We take the 5x5 square, the two yellow 3x3 squares, which overlap at the center, and the two blue 2x2 squares. Then every square except the center square is changed an even number of times. So this works if the central square was selected.

![md5:35c41272386d9da0a2235f7d237dfecf](35c41272386d9da0a2235f7d237dfecf.jpeg)

It is easy to check that any 2x2, 3x3, 4x4 or 5x5 square has an even number of green squares, so if the selected square was green, and we change it an odd number of times, then some other green square must also be changed an odd number of times and hence end up with a minus. So if all the squares end up plus, then the selected square was not green, so it must belong to the central white column. Similarly, it must belong to the central row and hence must be the center square.


Show that \((\mathrm{x} + \mathrm{y} + \mathrm{z})^{2} / 3 \geq \mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})} + \mathrm{y}\sqrt{(\mathrm{z}\mathrm{x})} + \mathrm{z}\sqrt{(\mathrm{x}\mathrm{y})}\) for all non- negative reals \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) .

Solution:
By AM/GM xy + yz \(\geq 2\mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})}\) . Adding the similar results gives \(2(\mathrm{xy} + \mathrm{yz} + \mathrm{zx})\geq 2(\mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})} +\) \(\mathrm{y}\sqrt{(\mathrm{z}\mathrm{x})} + \mathrm{z}\sqrt{(\mathrm{x}\mathrm{y})}\) ).

By AM/GM \(\mathrm{x}^{2} + \mathrm{x}^{2} + \mathrm{y}^{2} + \mathrm{z}^{2}\geq 4\mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})}\) . Adding the similar results gives \(\mathrm{x}^{2} + \mathrm{y}^{2} + \mathrm{z}^{2}\geq \mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})}\) \(+\mathrm{y}\sqrt{(\mathrm{z}\mathrm{x})} +\mathrm{z}\sqrt{(\mathrm{x}\mathrm{y})}\) . Adding the first result gives \((\mathrm{x} + \mathrm{y} + \mathrm{z})^{2} / 3\geq \mathrm{x}\sqrt{(\mathrm{y}\mathrm{z})} +\mathrm{y}\sqrt{(\mathrm{z}\mathrm{x})} +\mathrm{z}\sqrt{(\mathrm{x}\mathrm{y})}\)


Does there exist a triangle in which two sides are integer multiples of the median to that side? Does there exist a triangle in which every side is an integer multiple of the median to that side?

Answer yes, no

Solution:
The obvious approach is to make the triangle isosceles. So suppose the sides are a, b, b. Then the length m of a median to one of the sides length b satisfies: \(\mathrm{a}^2 +\mathrm{b}^2 = 2\mathrm{m}^2 +\mathrm{b}^2 /2\) . The simplest possibility is to take \(\mathrm{m} = \mathrm{b}\) , so \(\mathrm{a}^2 = 3\mathrm{b}^2 /2\) . Thus if \(\mathrm{b} = 2\) , \(\mathrm{a} = \sqrt{6}\) .

Suppose we have a triangle ABC, with medians AD, BE, CF, and BC/AD, CA/BE, AB/CF all integers. If \(\mathrm{AD} = \mathrm{BC} / 2\) , then \(\angle A = 90^{\circ}\) . If \(\mathrm{AD}< \mathrm{BC} / 2\) , then \(\angle A\) is obtuse, so at least two of the medians must be equal to the corresponding sides. So wlog we have \(\mathrm{b}^2 +\mathrm{c}^2 = 5\mathrm{a}^2 /2\) , \(\mathrm{c}^2 +\mathrm{a}^2 = 5\mathrm{b}^2 /2\) . Subtracting, \(\mathrm{b}^2 -\mathrm{a}^2 = (5 / 2)(\mathrm{a}^2 -\mathrm{b}^2)\) , so \(\mathrm{a} = \mathrm{b}\) . Hence \(\mathrm{c} / \mathrm{a} = \sqrt{(3 / 2)}\) . So the third median has length m where \(\mathrm{a}^2 +\mathrm{a}^2 = (3 / 4)\mathrm{a}^2 +2\mathrm{m}^2\) , so \(\mathrm{a} / \mathrm{m} = \sqrt{(8 / 5)}\) , which is not integral. Contradiction.


The numbers 1, 2, 3, ..., n are written on a blackboard (where \(\mathrm{n} \geq 3\) ). A move is to replace two numbers by their sum and non- negative difference. A series of moves makes all the numbers equal k. Find all possible k.

Answer all powers of \(2 \geq \mathrm{n}\)

Solution:
If a prime p divides \(\mathrm{a} + \mathrm{b}\) and a- b, then it divides 2a and 2b, so if p is odd, it divides a and b. Thus if an odd prime p divides k, then it must divide all the original numbers including 1. So k must be a power of 2. Note that \(\mathrm{k}, \mathrm{k} \rightarrow 0, 2\mathrm{k} \rightarrow 2\mathrm{k}, 2\mathrm{k}\) and \(\mathrm{k}, \mathrm{k}, \mathrm{k} \rightarrow 0, \mathrm{k}, 2\mathrm{k} \rightarrow \mathrm{k}, \mathrm{k}, 2\mathrm{k} \rightarrow 0, 2\mathrm{k}, 2\mathrm{k} \rightarrow 2\mathrm{k}, 2\mathrm{k}, 2\mathrm{k}\) . So (by a trivial induction) if we get all the numbers equal to k, then we can get them all to equal 2k. Finally, note that we can never decrease the largest number on the board, so the answer must be all powers of 2 greater than some minimum, which must be at least n.

We use induction to show that if \(2^{\mathrm{m}}\) is the smallest power of 2 which is \(\geq \mathrm{n}\) , then we can get all numbers equal to \(2^{\mathrm{m}}\) . Note that \(0, \mathrm{k} \rightarrow \mathrm{k}, \mathrm{k} \rightarrow 0, 2\mathrm{k}\) , so with a zero we can double each member of any set of numbers as often as we wish and finally convert the zero. For example, we could convert 0, 2, 4 to 8, 8, 8. It is convenient to take the induction hypothesis as \(\mathrm{S}_{\mathrm{n}}\) : we can convert 1, 2, ..., n to 0, \(2^{\mathrm{k}}, 2^{\mathrm{k}}, \ldots , 2^{\mathrm{k}}\) , where \(2^{\mathrm{k}}\) is the smallest power of 2 which is \(\geq \mathrm{n}\) .

We show first that \(\mathrm{S}_{\mathrm{n}}\) is true for \(\mathrm{n} \leq 8\) . For \(\mathrm{n} = 3\) , we take \(1,3 \rightarrow 2,4\) , then \(2,2 \rightarrow 0,4\) . For \(\mathrm{n} = 4\) , we ignore the 4 and use the case \(\mathrm{n} = 3\) . For \(\mathrm{n} = 5\) , we take \(3,5 \rightarrow 2,8\) . Then \(2,2 \rightarrow 0,4\) . Then we use the 0 to convert the remaining powers of 2 (1,4,4) to 8. For \(\mathrm{n} = 6\) , we take \(2,6 \rightarrow 4,8\) and \(3,5 \rightarrow 2,8\) , then \(4,4 \rightarrow 0,8\) . Finally, we use the 0 to convert 1 and 2 to 8. For \(\mathrm{n} = 7\) , we take \(1,7 \rightarrow 6,8\) , then \(2,6 \rightarrow 4,8\) , then \(3,5 \rightarrow 2,8\) , then \(4,4 \rightarrow 0,8\) , then \(2,6 \rightarrow 4,8\) and finally use the 0 to convert the remaining 4 to 8.

Let \(\mathrm{n} = 2^{\mathrm{a}} + \mathrm{b}\) , where \(0 < \mathrm{b} \leq 2^{\mathrm{a}}\) and assume \(\mathrm{S}_{\mathrm{m}}\) is true for all \(\mathrm{m} < \mathrm{n}\) . If \(\mathrm{b} = 1\) , we convert the pair \(2^{\mathrm{a}} - 1\) , \(2^{\mathrm{a}} + 1\) to 2, \(2^{\mathrm{a} + 1}\) . We have \(2^{\mathrm{a}} - 2 > 2\) , so by induction we can convert 1, 2, ..., \(2^{\mathrm{a}} - 2\) to 0, \(2^{\mathrm{a}}\) , ..., \(2^{\mathrm{a}}\) . Now all the numbers except 0 are powers of 2 and we can use the 0 to convert them each to \(2^{\mathrm{a} + 1}\) . Similarly, if \(\mathrm{b} = 2\) , we convert \(2^{\mathrm{a}} - 1\) , \(2^{\mathrm{a}} + 1\) to 2, \(2^{\mathrm{a} + 1}\) and \(2^{\mathrm{a}} - 2\) , \(2^{\mathrm{a}} + 2\) to 4, \(2^{\mathrm{a} + 1}\) and then proceed as in the previous case. If \(3 \leq \mathrm{b} < 2^{\mathrm{a}}\) , then we start by converting the pairs \((2^{\mathrm{a}} + \mathrm{b}, 2^{\mathrm{a}} - \mathrm{b})\) , \((2^{\mathrm{a}} + \mathrm{b} - 1, 2^{\mathrm{a}} - \mathrm{b} + 1)\) , \((2^{\mathrm{a}} + \mathrm{b} - 2, 2^{\mathrm{a}} - \mathrm{b} + 2)\) , ..., \((2^{\mathrm{a}} + 1, 2^{\mathrm{a}} - 1)\) . That gives some \(2^{\mathrm{a} + 1}\) s and 2, 4, ..., 2b. Now by \(\mathrm{S}_{\mathrm{b}}\) we can convert 2, 4, ..., 2b to 0, \(2^{\mathrm{a} + 1}\) , ..., \(2^{\mathrm{a} + 1}\) . The remaining

numbers 1, 2, ..., \(2^{\mathrm{a}}\) - b- 1 can either be converted to powers of 2 by \(S^{2\mathrm{a - b - 1}}\) (if \(2^{\mathrm{a}}\) - b- 1 \(\geq 3\) ) or are already powers of 2. Finally we use the 0 to bring all powers of 2 up to \(2^{\mathrm{a + 1}}\) . In the case \(\mathrm{b} = 2^{\mathrm{a}}\) , we ignore \(2^{\mathrm{a}} + \mathrm{b} (= 2^{\mathrm{a + 1}})\) and use the case b- 1 to convert the others.


The figure below is cut along the lines into polygons (which need not be convex). No polygon contains a 2 x 2 square. What is the smallest possible number of polygons?

![md5:04fbaa7e3df54e301a41791ce24d25b9](04fbaa7e3df54e301a41791ce24d25b9.jpeg)

Answer 12

Solution:
We can clearly cut the polygon into 12 strips width 1, so the smallest number is \(\leq 12\) .

There are 84 unit squares in the figure. Each cut along the edge of a unit square not already cut (and not on the boundary) increases the number of pieces by at most 1. So it is sufficient to show that at most 72 edges remain uncut (after cutting into polygons). Because then cutting the remaining edges would increase the total number of pieces by at most 72. But the final number of pieces is 84, so we would have to start with at least 12.

Initially, there are 144 edges, so we have to show that at least 72 of them are cut to make the polygons. An interior vertex has 4 edges. At least two of them must be cut, or the vertex would be the center of an uncut 2 x 2 square. If we take alternate interior vertices (36 in total, as shown below), then each has at least two cut edges, so in total at least 72 edges are cut to make the polygons.


ABC is an acute- angled triangle with circumcenter O. The circumcircle of ABO intersects AC and BC at M and N. Show that the circumradii of ABO and MNC are the same.

Solution: It is sufficient to show that \(\angle \mathrm{MBN} = \angle \mathrm{C}\) . But \(\angle \mathrm{MBN} = \angle \mathrm{MBO} + \angle \mathrm{OBN} = \angle \mathrm{MAO} + \angle \mathrm{OBN} = \angle \mathrm{MCO} + \angle \mathrm{OCN} = \angle \mathrm{C}\) .


A polygon can be transformed into a new polygon by making a straight cut, which creates two new pieces each with a new edge. One piece is then turned over and the two new edges are reattached. Can repeated transformations of this type turn a square into a triangle?


An \(h \times k\) minor of an n x n table is the hk cells which lie in h rows and k columns. The semiperimeter of the minor is \(\mathrm{h} + \mathrm{k}\) . A number of minors each with semiperimeter at least n together include all the cells on the main diagonal. Show that they include at least half the cells in the table.


(1) \(\mathbf{r}_1, \mathbf{r}_2, \ldots , \mathbf{r}_{100}, \mathbf{c}_1, \mathbf{c}_2, \ldots , \mathbf{c}_{100}\) are distinct reals. The number \(\mathbf{r}_i + \mathbf{c}_j\) is written in position i, j of a \(100 \times 100\) array. The product of the numbers in each column is 1. Show that the product of the numbers in each row is -1. (2) \(\mathbf{r}_1, \mathbf{r}_2, \ldots , \mathbf{r}_{2n}, \mathbf{c}_1, \mathbf{c}_2, \ldots , \mathbf{c}_{2n}\) are distinct reals. The number \(\mathbf{r}_i + \mathbf{c}_j\) is written in position i, j of a \(2n \times 2n\) array. The product of the numbers in each column is the same. Show that the product of the numbers in each row is also the same.


A sequence of positive integers is constructed as follows. If the last digit of \(\mathbf{a}_{\mathrm{n}}\) is greater than 5, then \(\mathbf{a}_{\mathrm{n} + 1}\) is \(9\mathbf{a}_{\mathrm{n}}\) . If the last digit of \(\mathbf{a}_{\mathrm{n}}\) is 5 or less and \(\mathbf{a}_{\mathrm{n}}\) has more than one digit, then \(\mathbf{a}_{\mathrm{n} + 1}\) is obtained from \(\mathbf{a}_{\mathrm{n}}\) by deleting the last digit. If \(\mathbf{a}_{\mathrm{n}}\) has only one digit, which is 5 or less, then the sequence terminates. Can we choose the first member of the sequence so that it does not terminate?


\(\mathrm{p(x)}\) is the cubic \(\mathrm{x}^3 - 3\mathrm{x}^2 +5\mathrm{x}\) . If h is a real root of \(\mathrm{p(x)} = 1\) and k is a real root of \(\mathrm{p(x)} = 5\) find \(\mathrm{h} + \mathrm{k}\)

Solution:
Put \(\mathrm{y} = 2\mathrm{- h}\) , where \(\mathrm{p(h)} = 1\) , then \((2 - \mathrm{y})^{3} - 3(2 - \mathrm{y})^{2} + 5(2 - \mathrm{y}) - 1 = 0\) , so \(8 - 12\mathrm{y} + 6\mathrm{y}^{2} - \mathrm{y}^{3} - 12 + 12\mathrm{y} - 3\mathrm{y}^{2} + 10 - 5\mathrm{y} - 1 = 0\) , or \(\mathrm{y}^3 - 3\mathrm{y}^2 +5\mathrm{y} = 5\) , or \(\mathrm{p(y)} = 5\) . So if h is a root of \(\mathrm{p(h)} = 1\) , then there is a root k of \(\mathrm{p(k)} = 5\) such that \(\mathrm{h} + \mathrm{k} = 2\) . To complete the proof we have to show that \(\mathrm{p(x)} = 5\) has only one real root.

But \(\mathrm{x}^3 - 3\mathrm{x}^2 +5\mathrm{x} = (\mathrm{x} - 1)^3 +2(\mathrm{x} - 1) + 3\) which is a strictly increasing function of \(\mathrm{x} - 1\) and hence of \(\mathrm{x}\) . So \(\mathrm{p(x)} = \mathrm{k}\) has only one real root.


The chords AB and CD of a sphere intersect at X. A, C and X are equidistant from a point Y on the sphere. Show that BD and XY are perpendicular.


Do there exist 4 vectors in the plane so that none is a multiple of another, but the sum of each pair is perpendicular to the sum of the other two? Do there exist 91 non- zero vectors in the plane such that the sum of any 19 is perpendicular to the sum of the others?


ABCD is a square. The points X on the side AB and Y on the side AD are such that AX·AY = 2 BX·DY. The lines CX and CY meet the diagonal BD in two points. Show that these points lie on the circumcircle of AXY.


X is a set with 100 members. What is the smallest number of subsets of X such that every pair of elements belongs to at least one subset and no subset has more than 50 members? What is the smallest number if we also require that the union of any two subsets has at most 80 members?


The real numbers \(\mathbf{x}_1,\mathbf{x}_2,\ldots ,\mathbf{x}_{1991}\) satisfy \(\left|\mathbf{x}_1 - \mathbf{x}_2\right| + \left|\mathbf{x}_2 - \mathbf{x}_3\right| + \ldots +\left|\mathbf{x}_{1990} - \mathbf{x}_{1991}\right| = 1991\) . What is the maximum possible value of \(\left|\mathbf{s}_1 - \mathbf{s}_2\right| + \left|\mathbf{s}_2 - \mathbf{s}_3\right| + \ldots +\left|\mathbf{s}_{1990} - \mathbf{s}_{1991}\right|\) , where \(\mathbf{s}_n = (\mathbf{x}_1 + \mathbf{x}_2 + \ldots +\mathbf{x}_n) / n\) ?