Given 1000 square plates in the plane with their sides parallel to the coordinate axes (but possibly overlapping and possibly of different sizes). Let S be the set of points covered by the plates. Show that you can choose a subset T of plates such that every point of S is covered by at least one and at most four plates in T.


You are given a set of scales and a set of n different weights. R represents the state in which the right pan is heavier, L represents the state in which the left pan is heavier and B represents the state in which the pans balance. Show that given any n- letter string of Rs and Ls you can put the weights onto the scales one at a time so that the string represents the successive states of the scales. For example, if the weights were 1, 2 and 3 and the string was LRL, then you would place 1 in the left pan, then 2 in the right pan, then 3 in the left pan.


A polynomial is monic if its leading coefficient is 1. Two polynomials \(\mathrm{p(x)}\) and \(\mathrm{q(x)}\) commute if \(\mathrm{p(q(x)) = q(p(x))}\) .

(1) Find all monic polynomials of degree 3 or less which commute with \(\mathrm{x}^2 - \mathrm{k}\) .

(2) Given a monic polynomial \(\mathrm{p(x)}\) , show that there is at most one monic polynomial of degree n which commutes with \(\mathrm{p(x)^2}\) .

(3) Find the polynomials described in (2) for \(\mathrm{n} = 4\) and \(\mathrm{n} = 8\) .

(4) If \(\mathrm{q(x)}\) and \(\mathrm{r(x)}\) are monic polynomials which both commute with \(\mathrm{p(x)^2}\) , show that \(\mathrm{q(x)}\) and \(\mathrm{r(x)}\) commute.

(5) Show that there is a sequence of polynomials \(\mathrm{p}_2(\mathrm{x}), \mathrm{p}_3(\mathrm{x}), \ldots\) such that \(\mathrm{p}_2(\mathrm{x}) = \mathrm{x}^2 - 2\) , \(\mathrm{p}_n(\mathrm{x})\) has degree n and all polynomials in the sequence commute.


\(\mathbf{a}_{\mathrm{n}}\) is the nearest integer to \(\sqrt{\mathrm{n}}\) . Find \(1 / \mathrm{a}_{1} + 1 / \mathrm{a}_{2} + \ldots +1 / \mathrm{a}_{1980}\) .


ABCD is a quadrilateral. M is a point inside it such that ABMD is a parallelogram. \(\angle \mathrm{CBM} =\) \(\angle \mathrm{CDM}\) . Show that \(\angle \mathrm{ACD} = \angle \mathrm{BCM}\) .


Show that there is no positive integer n for which \(1000^{\mathrm{n}} - 1\) divides \(1978^{\mathrm{n}} - 1\) .


If P, Q are points in space the point [PQ] is the point on the line PQ on the opposite side of Q to P and the same distance from Q. \(\mathrm{K}_{0}\) is a set of points in space. Given \(\mathrm{K}_{\mathrm{n}}\) we derive \(\mathrm{K}_{\mathrm{n} + 1}\) by adjoining all the points [PQ] with P and Q in \(\mathrm{K}_{\mathrm{n}}\) .

(1) \(\mathrm{K}_{0}\) contains just two points A and B, a distance 1 apart, what is the smallest n for which \(\mathrm{K}_{\mathrm{n}}\) contains a point whose distance from A is at least 1000?

(2) \(\mathrm{K}_{0}\) consists of three points, each pair a distance 1 apart, find the area of the smallest convex polygon containing \(\mathrm{K}_{\mathrm{n}}\) .

(3) \(\mathrm{K}_{0}\) consists of four points, forming a regular tetrahedron with volume 1. Let \(\mathrm{H}_{\mathrm{n}}\) be the smallest convex polyhedron containing \(\mathrm{K}_{\mathrm{n}}\) . How many faces does \(\mathrm{H}_{1}\) have? What is the volume of \(\mathrm{H}_{\mathrm{n}}\) ?


Two players play a game. There is a heap of m tokens and a heap of \(\mathrm{n}< \mathrm{m}\) tokens. Each player in turn takes one or more tokens from the heap which is larger. The number he takes must be a multiple of the number in the smaller heap. For example, if the heaps are 15 and 4, the first player may take 4, 8 or 12 from the larger heap. The first player to clear a heap wins. Show that if \(\mathrm{m} > 2\mathrm{n}\) , then the first player can always win. Find all k such that if \(\mathrm{m} > \mathrm{k}\mathrm{n}\) , then the first player can always win.


Show that there is an infinite sequence of reals \(\mathrm{x}_{1}\) , \(\mathrm{x}_{2}\) , \(\mathrm{x}_{3}\) , ... such that \(\mathrm{l}\mathrm{x}_{\mathrm{n}}\mathrm{l}\) is bounded and for any \(\mathrm{m} > \mathrm{n}\) , we have \(\mathrm{l}\mathrm{x}_{\mathrm{m}} - \mathrm{x}_{\mathrm{n}}\mathrm{l} > 1 / (\mathrm{m} - \mathrm{n})\) .


Let \(\mathrm{p}(\mathrm{x}) = \mathrm{x}^{2} + \mathrm{x} + 1\) . Show that for every positive integer n, the numbers n, \(\mathrm{p}(\mathrm{n})\) , \(\mathrm{p}(\mathrm{p}(\mathrm{n}))\) , \(\mathrm{p}(\mathrm{p}(\mathrm{p}(\mathrm{n})))\) , ... are relatively prime.


Show that for some k, you can find 1978 different sizes of square with all its vertices on the graph of the function \(\mathrm{y} = \mathrm{k}\sin \mathrm{x}\) .


The set \(\mathrm{S_0}\) has the single member (5, 19). We derive the set \(\mathrm{S_{n + 1}}\) from \(\mathrm{S_n}\) by adjoining a pair to \(\mathrm{S_n}\) . If \(\mathrm{S_n}\) contains the pair (2a, 2b), then we may adjoin the pair (a, b). If \(\mathrm{S}\) contains the pair (a, b) we may adjoin (a+1, b+1). If \(\mathrm{S}\) contains (a, b) and (b, c), then we may adjoin (a, c). Can we obtain (1, 50)? (1, 100)? If We start with (a, b), with a < b, instead of (5, 19), for which n can we obtain (1, n)?


An n- gon area A is inscribed in a circle radius R. We take a point on each side of the polygon to form another n- gon. Show that it has perimeter at least 2A/R.


Two players play a game by moving a piece on an n x n chessboard. The piece is initially in a corner square. Each player may move the piece to any adjacent square (which shares a side with its current square), except that the piece may never occupy the same square twice. The first player who is unable to move loses. Show that for even n the first player can always win, and for odd n the second player can always win. Who wins if the piece is initially on a square adjacent to the corner?


Given a set of n non- intersecting segments in the plane. No two segments lie on the same line. Can we successively add n- 1 additional segments so that we end up with a single non- intersecting path? Each segment we add must have as its endpoints two existing segment endpoints.


a and b are positive real numbers. \(\mathbf{x}_i\) are real numbers lying between a and b. Show that \((\mathbf{x}_1 + \mathbf{x}_2 + \ldots + \mathbf{x}_n)(1 / \mathbf{x}_1 + 1 / \mathbf{x}_2 + \ldots + 1 / \mathbf{x}_n)\leq \mathrm{n}^2 (\mathrm{a} + \mathrm{b})^2 /4\mathrm{ab}\) .


\(\mathrm{n} > 3\) is an integer. Let S be the set of lattice points (a, b) with \(0\leq \mathrm{a},\mathrm{b}< \mathrm{n}\) . Show that we can choose n points of S so that no three chosen points are collinear and no four chosen points from a parallelogram.


Given any tetrahedron, show that we can find two planes such that the areas of the projections of the tetrahedron onto the two planes have ratio at least \(\sqrt{2}\) .


\(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots , \mathbf{a}_{\mathrm{n}}\) are real numbers. Let \(\mathbf{b}_{\mathrm{k}} = (\mathbf{a}_{1} + \mathbf{a}_{2} + \ldots +\mathbf{a}_{\mathrm{k}}) / \mathbf{k}\) for \(\mathbf{k} = 1, 2, \ldots , \mathbf{n}\) . Let \(\mathbf{C} = (\mathbf{a}_{1} - \mathbf{b}_{1})^{2} + (\mathbf{a}_{2} - \mathbf{b}_{2})^{2} + \ldots +(\mathbf{a}_{\mathrm{n}} - \mathbf{b}_{\mathrm{n}})^{2}\) , and \(\mathbf{D} = (\mathbf{a}_{1} - \mathbf{b}_{\mathrm{n}})^{2} + (\mathbf{a}_{2} - \mathbf{b}_{\mathrm{n}})^{2} + \ldots +(\mathbf{a}_{\mathrm{n}} - \mathbf{b}_{\mathrm{n}})^{2}\) . Show that \(\mathbf{C} \leq \mathbf{D} \leq 2\mathbf{C}\) .


Let \(\mathbf{x}_{\mathrm{n}} = (1 + \sqrt{2} + \sqrt{3})^{\mathrm{n}}\) . We may write \(\mathbf{x}_{\mathrm{n}} = \mathbf{a}_{\mathrm{n}} + \mathbf{b}_{\mathrm{n}}\sqrt{2} + \mathbf{c}_{\mathrm{n}}\sqrt{3} + \mathbf{d}_{\mathrm{n}}\sqrt{6}\) , where \(\mathbf{a}_{\mathrm{n}}, \mathbf{b}_{\mathrm{n}}, \mathbf{c}_{\mathrm{n}}, \mathbf{d}_{\mathrm{n}}\) are integers. Find the limit as \(\mathbf{n}\) tends to infinity of \(\mathbf{b}_{\mathrm{n}} / \mathbf{a}_{\mathrm{n}}, \mathbf{c}_{\mathrm{n}} / \mathbf{a}_{\mathrm{n}}, \mathbf{d}_{\mathrm{n}} / \mathbf{a}_{\mathrm{n}}\) .


T is an isosceles triangle. Another isosceles triangle T' has one vertex on each side of T. What is the smallest possible value of area T'area T?


A grasshopper hops about in the first quadrant \((\mathrm{x},\mathrm{y}\geqslant = 0)\) . From \((\mathrm{x},\mathrm{y})\) it can hop to \((\mathrm{x} + 1,\mathrm{y} - 1)\) or to \((\mathrm{x} - 5,\mathrm{y} + 7)\) , but it can never leave the first quadrant. Find the set of points \((\mathrm{x},\mathrm{y})\) from which it can never get further than a distance 1000 from the origin.


In a group of people every person has less than 4 enemies. Assume that A is B's enemy iff B is A's enemy. Show that we can divide the group into two parts, so that each person has at most one enemy in his part.


Let S be the set {0, 1}. Given any subset of S we may add its arithmetic mean to S (provided it is not already included - S never includes duplicates). Show that by repeating this process we can include the number 1/5 in S. Show that we can eventually include any rational number between 0 and 1.


The real sequence \(\mathrm{x}_1\geq \mathrm{x}_2\geq \mathrm{x}_3\geq \ldots\) satisfies \(\mathrm{x}_1 + \mathrm{x}_4 / 2 + \mathrm{x}_9 / 3 + \mathrm{x}_{16} / 4 + \ldots +\mathrm{x}_\mathrm{N} / \mathrm{n}\leq 1\) for every square \(\mathrm{N} = \mathrm{n}^2\) . Show that it also satisfies \(\mathrm{x}_1 + \mathrm{x}_2 / 2 + \mathrm{x}_3 / 3 + \ldots +\mathrm{x}_\mathrm{n} / \mathrm{n}\leq 3\) .


Given a finite set X of points in the plane. S is a set of vectors AB where (A, B) are some pairs of points in X. For every point A the number of vectors AB (starting at A) in S equals the number of vectors CA (ending at A) in S. Show that the sum of the vectors in S is zero.


What is the smallest number of pieces that can be placed on an \(8\mathrm{x}8\) chessboard so that every row, column and diagonal has at least one piece? [A diagonal is any line of squares parallel to one of the two main diagonals, so there are 30 diagonals in all.] What is the smallest number for an n x n board?


a and b are real numbers. Find real x and y satisfying: \((\mathrm{x} - \mathrm{y}(\mathrm{x}^2 - \mathrm{y}^2)^{1 / 2} = \mathrm{a}(1 - \mathrm{x}^2 + \mathrm{y}^2)^{1 / 2}\) and \((\mathrm{y} - \mathrm{x}(\mathrm{x}^2 - \mathrm{y}^2)^{1 / 2} = \mathrm{b}(1 - \mathrm{x}^2 + \mathrm{y}^2)^{1 / 2}\) .


A set of square carpets have total area 4. Show that they can cover a unit square.


\(\mathbf{x}_{\mathrm{i}}\) are real numbers between 0 and 1. Show that \(\left(\mathbf{x}_{1} + \mathbf{x}_{2} + \ldots +\mathbf{x}_{\mathrm{n}} + 1\right)^{2}\geq 4(\mathbf{x}_{1}^{2} + \mathbf{x}_{2}^{2} + \ldots +\mathbf{x}_{\mathrm{n}}^{2})\) .


m and n are relatively prime positive integers. The interval [0, 1] is divided into m + n equal subintervals. Show that each part except those at each end contains just one of the numbers 1/m, 2/m, 3/m, ..., (m- 1)/m, 1/n, 2/n, ..., (n- 1)/n.


Given a point P in space and 1979 lines \(\mathbf{L}_{1}\) , \(\mathbf{L}_{2}\) , ..., \(\mathbf{L}_{1979}\) containing it. No two lines are perpendicular. \(\mathbf{P}_{1}\) is a point on \(\mathbf{L}_{1}\) . Show that we can find a point \(\mathbf{A}_{\mathrm{n}}\) on \(\mathbf{L}_{\mathrm{n}}\) (for \(\mathrm{n} = 2\) , 3, ..., 1979) such that the following 1979 pairs of lines are all perpendicular: \(\mathbf{A}_{\mathrm{n - 1}}\mathbf{A}_{\mathrm{n + 1}}\) and \(\mathbf{L}_{\mathrm{n}}\) for \(\mathrm{n} = 1\) , ..., 1979. [We regard \(\mathbf{A}_{\mathrm{- 1}}\) as \(\mathbf{A}_{1979}\) and \(\mathbf{A}_{1980}\) as \(\mathbf{A}_{1}\) .]


Find a sequence \(\mathbf{a}_{1}\) , \(\mathbf{a}_{2}\) , ..., \(\mathbf{a}_{25}\) of 0s and 1s such that the following sums are all odd:

\(\begin{array}{r l} & {\mathrm{a}_{1}\mathrm{a}_{1} + \mathrm{a}_{2}\mathrm{a}_{2} + \ldots +\mathrm{a}_{25}\mathrm{a}_{25}}\\ & {\mathrm{a}_{1}\mathrm{a}_{2} + \mathrm{a}_{2}\mathrm{a}_{3} + \ldots +\mathrm{a}_{24}\mathrm{a}_{25}}\\ & {\mathrm{a}_{1}\mathrm{a}_{3} + \mathrm{a}_{2}\mathrm{a}_{4} + \ldots +\mathrm{a}_{23}\mathrm{a}_{25}}\\ & {\ldots}\\ & {\mathrm{a}_{1}\mathrm{a}_{24} + \mathrm{a}_{2}\mathrm{a}_{25}} \end{array}\)

Show that we can find a similar sequence of n terms for some \(\mathrm{n} > 1000\) .


A convex quadrilateral is divided by its diagonals into four triangles. The incircles of each of the four are equal. Show that the quadrilateral has all its sides equal.


All two digit numbers from 19 to 80 inclusive are written down one after the other as a single number \(\mathrm{N} = 192021\ldots 7980\) . Is N divisible by 1980?

## Answer

Yes

Solution: \(1980 = 2^{2}3^{2}5\) . 11. N is obviously divisible by \(2^{2}\) and 5. The digits in odd position are \(9+\) \((0 + 1 + 2 + \ldots +9) + (0 + 1 + 2 + \ldots +9) + \ldots +(0 + 1 + 2 + \ldots +9) + 0 = 9 + 6\cdot 45 = 279\) . The digits in even position are \(1 + (2 + 2 + \ldots +2) + (3 + 3 + \ldots +3) + \ldots +(7 + \ldots +7) + 8 = 9 + 10(2 + 3 + \ldots +7) = 279\) . So the sum of the digits of N is 2. 279 which is divisible by 9. Hence N is divisible by 9. The difference between the odd and even sums is 0, which is divisible by 11, so N is divisible by 11.


A square is divided into n parallel strips (parallel to the bottom side of the square). The width of each strip is integral. The total width of the strips with odd width equals the total width of the strips with even width. A diagonal of the square is drawn which divides each strip into a left part and a right part. Show that the sum of the areas of the left parts of the odd strips equals the sum of the areas of the right parts of the even strips.

Solution: Let LO be the total area of the left parts of the odd strips, LE the total area of the left parts of the even strips, and RE the total area of the right parts of the even trips. Since the diagonal bisects the square, \(\mathrm{LO} + \mathrm{LE} = \mathrm{A} / 2\) , where A is the area of the square. Also \(\mathrm{RE} + \mathrm{LE} = \mathrm{A} / 2\) (because the total width of the even strips equals the total width of the odd strips). Subtracting, \(\mathrm{LO} = \mathrm{RE}\) , as required.


35 containers of total weight 18 must be taken to a space station. One flight can take any collection of containers weighing 3 or less. It is possible to take any subset of 34 containers in 7 flights. Show that it must be possible to take all 35 containers in 7 flights.

Solution:
Let the lightest container be S weighing s. The other containers can be taken in 7 flights. The smallest load on these flights must be \(\leq (18 - \mathrm{s}) / 7\) , and so that flight has spare capacity of at least \(3 - (18 - \mathrm{s}) / 7 = (3 + \mathrm{s}) / 7\) . Thus it can accomodate S provided that \((3 + \mathrm{s}) / 7 \geq \mathrm{s}\) , or \(\mathrm{s} \leq 1 / 2\) .

At least one of the 7 flights takes \(\leq 4\) containers. These weigh at most the weight of the 4 heaviest. Since the 31 lightest weigh at least 31s, the 4 heaviest weigh at most 18- 31s. Thus

this flight has spare capacity of at least \(3 - (18 - 31s) = 31s - 15\) and can accomodate S provided that \(31s - 15 \geq s\) , or \(s \geq 1 / 2\) .


ABCD is a convex quadrilateral. M is the midpoint of BC and N is the midpoint of CD. If \(k = \mathrm{AM} + \mathrm{AN}\) show that the area of ABCD is less than \(k^2 /2\) .


Are there any solutions in positive integers to \(\mathrm{a}^4 = \mathrm{b}^3 + \mathrm{c}^2\) ?

Solution: We have \(\mathrm{b}^3 = (\mathrm{a}^2 - \mathrm{c})(\mathrm{a}^2 + \mathrm{c})\) , so one possibility is that \(\mathrm{a}2 \pm \mathrm{c}\) are both cubes. So we want two cubes whose sum is twice a square. Looking at the small cubes, we soon find \(8 + 64 = 2\) . 36 giving \(6^4 = 28^2 + 8^3\) . Multiplying through by \(k^{12}\) gives an infinite family of solutions. Note that the question does not ask for all solutions.


Given a point P on the diameter AC of the circle K, find the chord BD through P which maximises the area of ABCD.


There are several settlements around Big Lake. Some pairs of settlements are directly connected by a regular shipping service. For all \(\mathrm{A} \neq \mathrm{B}\) , settlement A is directly connected to X iff B is not directly connected to Y, where B is the next settlement to A counterclockwise and Y is the next settlement to X counterclockwise. Show that you can move between any two settlements with at most 3 trips.

Solution:
Suppose there are n settlements \(\mathrm{A}_1, \mathrm{A}_2, \ldots , \mathrm{A}_n\) in counterclockwise order around the lake. We will use cyclic indices, so that \(\mathrm{A}_{n + 1}\) means \(\mathrm{A}_1\) , and so on. wlog \(\mathrm{A}_1\) has a direct service to \(\mathrm{A}_2\) . Then it follows that \(\mathrm{A}_2\) does not have a direct service to \(\mathrm{A}_3\) , so \(\mathrm{A}_3\) does have a direct service to \(\mathrm{A}_4\) , and so on. So \(\mathrm{A}_i\) has direct service to \(\mathrm{A}_{i + 1}\) iff i is odd. But \(\mathrm{A}_1 = \mathrm{A}_{n + 1}\) has direct service to \(\mathrm{A}_2\) , so n must be even.

Now suppose we want to get from \(\mathrm{A}_i\) to \(\mathrm{A}_j\) . If there is direct service we are done. So suppose not. Then there must be direct service from \(\mathrm{A}_{i + 1}\) to \(\mathrm{A}_{j + 1}\) . If i and j are both odd, then there is direct service from \(\mathrm{A}_i\) to \(\mathrm{A}_{i + 1}\) and from \(\mathrm{A}_j\) to \(\mathrm{A}_{j + 1}\) , so we can make the journey in 3 trips. If i and j are both even, then we can go \(\mathrm{A}_i\) to \(\mathrm{A}_{i - 1}\) to \(\mathrm{A}_{j - 1}\) to \(\mathrm{A}_j\) . So suppose i and j have opposite parity. wlog i is odd and j is even. If there is direct service from \(\mathrm{A}_i\) to \(\mathrm{A}_{j - 1}\) , then we can make the journey in two trips: \(\mathrm{A}_i\) to \(\mathrm{A}_{j - 1}\) to \(\mathrm{A}_j\) . If not, then we can go \(\mathrm{A}_i\) to \(\mathrm{A}_{i + 1}\) to \(\mathrm{A}_j\) .


A six digit (decimal) number has six different digits, none of them 0, and is divisible by 37. Show that you can obtain at least 23 other numbers which are divisible by 37 by permuting the digits.

Solution:
Suppose the digits are a1, a2, a3, a4, a5, a6 and that a1a2a3a4a5a6 is divisible by 37. We claim that a2a3a4a5a6a1 is also divisible by 37. Put \(\mathrm{n} = \mathrm{a}_2\mathrm{a}_3\mathrm{a}_4\mathrm{a}_5\mathrm{a}_6\) and \(\mathrm{m} = \mathrm{a}_1\) . The original number is \(10^{5}\mathrm{m} + \mathrm{n}\) and the derived number is \(10\mathrm{n} + \mathrm{m}\) . But 37 divides \(10^{3} - 1\) and hence also \(10^{6} - 1\) and \(\mathrm{m}(10^{6} - 1)\) . So it also divides \(10(10^{5}\mathrm{m} + \mathrm{n}) - \mathrm{m}(10^{6} - 1) = 10\mathrm{n} + \mathrm{m}\) , which proves the claim.

Iterating, we get the original number and 5 others:

a1a2a3a4a5a6 a2a3a4a5a6a1 a3a4a5a6a1a2 a4a5a6a1a2a3 a5a6a1a2a3a4 a6a1a2a3a4a5

Similarly, we have that a1a2a3a4a5a6 - a1a2a6a4a5a3 = (a6- a3)999, so a1a2a6a4a5a3 is also divisible by 37. Iterating we get:

a1a2a6a4a5a3 a2a6a4a5a3a1 a6a4a5a3a1a2 a4a5a3a1a2a6 a5a3a1a2a6a4 a3a1a2a6a4a5 a5a3a1a2a6a4 a3a1a2a3a4a5

Similarly, we could swap the first term and the fourth to get:

a4a2a3a1a5a6 a2a3a1a5a6a4 a3a1a5a6a4a2 a1a5a6a4a2a3 a5a6a4a2a3a1 a6a4a2a3a1a5

or the second and the fifth to get:

a1a5a3a4a2a6 a5a3a4a2a6a1 a3a4a2a6a1a5 a4a2a6a1a5a3 a2a6a1a5a3a4 a6a1a5a3a4a2


Find all real solutions to: \(\sin x + 2\sin (x + y + z) = 0\) \(\sin y + 3\sin (x + y + z) = 0\) \(\sin z + 4\sin (x + y + z) = 0\)


Given 1980 vectors in the plane. The sum of every 1979 vectors is a multiple of the other vector. Not all the vectors are multiples of each other. Show that the sum of all the vectors is zero.

Solution: Let the vectors be \(\mathbf{x}_{\mathrm{i}}\) and their sum \(\mathbf{s}\) . Then we have \(\mathbf{s} - \mathbf{x}_{\mathrm{i}} = \mathbf{n}_{\mathrm{i}}\mathbf{x}_{\mathrm{i}}\) for some scalar \(\mathbf{n}_{\mathrm{i}}\) . Hence \((\mathbf{n}_{\mathrm{i}} + 1)\mathbf{x}_{\mathrm{i}} = \mathbf{s}\) . If \(\mathbf{s}\) is non- zero, then it follows that every vector is a multiple of \(\mathbf{s}\) and hence all the vectors are multiples of each other. But we are told that is not true. Hence \(\mathbf{s}\) is zero.


Let \(\mathbf{f}(\mathbf{n})\) be the sum of \(\mathbf{n}\) and its digits. For example, \(\mathbf{f}(34) = 41\) . Is there an integer such that \(\mathbf{f}(\mathbf{n}) = 1980\) ? Show that given any positive integer \(\mathbf{m}\) we can find \(\mathbf{n}\) such that \(\mathbf{f}(\mathbf{n}) = \mathbf{m}\) or \(\mathbf{m} + 1\) .

## Answer

\[\mathrm{f}(1962) = 1962 + 18 = 1980\]

Solution: If the last digit of \(\mathbf{n}\) is not 9, then \(\mathbf{f}(\mathbf{n} + 1) = \mathbf{f}(\mathbf{n}) + 2\) . If the last digit of \(\mathbf{n}\) is 9, then \(\mathbf{f}(\mathbf{n} + 1)< \mathbf{f}(\mathbf{n})\) . On the other hand \(\mathbf{f}\) clearly achieves arbitrarily large values. Also \(\mathbf{f}(1) = 1\) . Now consider any \(\mathbf{m} > 1\) . Let \(\mathbf{M}\) be the smallest integer such that \(\mathbf{f}(\mathbf{M}) > \mathbf{m}\) . Then \(\mathbf{f}(\mathbf{M} - 1)\leq \mathbf{m}\) . Since \(\mathbf{f}(\mathbf{M}) > \mathbf{f}(\mathbf{M} - 1)\) we must have \(\mathbf{f}(\mathbf{M}) = \mathbf{f}(\mathbf{M} - 1) + 2\) . Hence either \(\mathbf{f}(\mathbf{M}) = \mathbf{m} + 1\) or \(\mathbf{f}(\mathbf{M} - 1) = \mathbf{m}\) .


Some unit squares in an infinite sheet of squared paper are colored red so that every \(2 \times 3\) and \(3 \times 2\) rectangle contains exactly two red squares. How many red squares are there in a \(9 \times 11\) rectangle?

## Answer

33

Solution:
![md5:5f00730fde8fe7df4a9585aa390cc519](5f00730fde8fe7df4a9585aa390cc519.jpeg)

There cannot be two red squares with a common side. For consider as the diagram shows we can immediately conclude that the squares with a \(\ast\) are not red, but now the bold rectangle has at most 1 red square. Contradiction.

![md5:b97a810dbe5c47c0e3729869585fd247](b97a810dbe5c47c0e3729869585fd247.jpeg)

Consider a red square. One of the two diagonally adjacent squares marked \* must be red. But it is now easy to show that all red squares on that diagonal are red and that the other red squares are those on every third parallel diagonal line. Any \(9 \times 11\) rectangle must have just three such diagonals on a 9 cell border row, and hence just 3 red cells in that border row. But the remaining \(9 \times 10\) rectangle can easily be partitioned into fifteen \(3 \times 2\) rectangles, each with 2 red squares.


There is a flu epidemic in elf city. The course of the disease is always the same. An elf is infected one day, he is sick the next, recovered and immune the third, recovered but not immune thereafter. Every day every elf who is not sick visits all his sick friends. If he is not immune he is sure to catch flu if he visits a sick elf. On day 1 no one is immune and one or more elves are infected from some external source. Thereafter there is no further external infection and the epidemic spreads as described above. Show that it is sure to die out (irrespective of the number of elves, the number of friends each has, and the number infected on day 1). Show that if one or more elves is immune on day 1, then it is possible for the epidemic to continue indefinitely.

Solution:
This is curiously easy. Write S for sick, N for not sick and not immune, and I for immune. Suppose group A are S on day 1 and group B are N on day 1. Then on day 2, B are S, and A are I. So on day 3 no one is sick, A are N and B are I. Thereafter no one can get sick, so the epidemic has died out.

Suppose on day 1, there is also group C who are I. Then on day 2, B are S, A are I and C are N. So on day 3, C are S, A are N and B are I. On day 4, A are S, B are N and C are I, the same as day 1, so the epidemic continues indefinitely.


Define the sequence \(\mathbf{a}_{\mathrm{n}}\) of positive integers as follows. \(\mathbf{a}_{1} = \mathbf{m}\) . \(\mathbf{a}_{\mathrm{n} + 1} = \mathbf{a}_{\mathrm{n}}\) plus the product of the digits of \(\mathbf{a}_{\mathrm{n}}\) . For example, if \(\mathrm{m} = 5\) , we have 5, 10, 10, ... . Is there an m for which the sequence is unbounded?

## Answer

No.

Solution:
Put \(\mathrm{p(n)}\) for the product of the digits of n. We show that, for sufficiently large n, a sequence starting below it cannot get past the "gap" from \(10_{\mathrm{n}}\) to \(10_{\mathrm{n}} + 10_{\mathrm{n} - 1}\) . For suppose N is the last member of the sequence below the gap. Then N has at most n digits, so \(\mathrm{p(N)} \leq 9^{\mathrm{n}}\) . But for sufficiently large n (in fact for \(\mathrm{n} \geq 21\) ) we have \(9^{\mathrm{n}} < 10^{\mathrm{n} - 1}\) . So \(\mathrm{N} + \mathrm{p(N)} < 10^{\mathrm{n}} + 10^{\mathrm{n} - 1}\) . But \(\mathrm{N} + \mathrm{p(N)} > 10^{\mathrm{n}}\) by assumption. Hence \(\mathrm{N} + \mathrm{p(N)}\) is sure to have second digit (from the left) zero.

So all further terms of the sequence are the same. But for any m there is certainly a gap above m, and, as shown, the sequence will not be able to get beyond it. So it is bounded.


ABC is equilateral. A line parallel to AC meets AB at M and BC at P. D is the center of the equilateral triangle BMP. E is the midpoint of AP. Find the angles of DEC.

## Answer

\[\mathrm{D} = 60^{\circ},\mathrm{E} = 90^{\circ}\]

Solution:
![md5:a902c5010ecb112e54485bddc3d17aca](a902c5010ecb112e54485bddc3d17aca.jpeg)

Let K be the midpoint of BP and L the midpoint of AC. EL is parallel to BC, so \(\angle ELC = 120^{\circ}\) . EK is parallel to AB, so \(\angle EKC = 60^{\circ}\) , so ELCK is cyclic. But \(\angle DKC = \angle DLC = 90^{\circ}\) , so DLCK is cyclic. Hence D, K, C, L, E all lie on a circle. Hence \(\angle DEC = \angle DLC = 90^{\circ}\) , and \(\angle EDC = \angle EKC = 60^{\circ}\) .


A rectangular box has sides \(\mathbf{x}< \mathbf{y}< \mathbf{z}\) . Its perimeter is \(\mathrm{p} = 4(\mathrm{x} + \mathrm{y} + \mathrm{z})\) , its surface area is \(\mathrm{s} =\) \(2(\mathrm{xy} + \mathrm{yz} + \mathrm{zx})\) and its main diagonal has length \(\mathrm{d} = \sqrt{(\mathrm{x}^2 + \mathrm{y}^2 + \mathrm{z}^2)}\) . Show that \(3\mathrm{x}< (\mathrm{p} / 4-\) \(\sqrt{(\mathrm{d}^2 - \mathrm{s} / 2)}\) and \(3\mathrm{z} > (\mathrm{p} / 4 + \sqrt{(\mathrm{d}^2 - \mathrm{s} / 2)}}\)

Solution:
We have \(3(\mathrm{y - x})(\mathrm{z - x}) > 0\) , so \(3\mathrm{x}^2 +3\mathrm{yz} > 3\mathrm{xy} + 3\mathrm{xz}\) . Hence \(\mathrm{y}^2 +\mathrm{z}^2 +4\mathrm{x}^2 +2\mathrm{yz} - 4\mathrm{xy} - 4\mathrm{xz} > \mathrm{x}^2 + \mathrm{y}^2 +\mathrm{z}^2 - \mathrm{xy} - \mathrm{yz} - \mathrm{xz}\) or \((\mathrm{y} + \mathrm{z} - 2\mathrm{x})^2 >(\mathrm{d}^2 - \mathrm{s} / 2)\) . Hence \((\mathrm{x} + \mathrm{y} + \mathrm{z}) > 3\mathrm{x} + \sqrt{(\mathrm{d}^2 - \mathrm{s} / 2)}\) . So \(3\mathrm{x}< \mathrm{p} / 4 - \sqrt{(\mathrm{d}^2 - \mathrm{s} / 2)}\) .

Similarly, \(3(\mathrm{z - x})(\mathrm{z - y}) > 0\) , so \(\mathrm{x}^2 +\mathrm{y}^2 +4\mathrm{z}^2 >\mathrm{x}^2 +\mathrm{y}^2 +\mathrm{z}^2 +3\mathrm{zx} + 3\mathrm{zy} - 3\mathrm{xy}\) , so \((2\mathrm{z} - \mathrm{x} - \mathrm{y})^2 >\mathrm{x}^2 +\mathrm{y}^2 +\mathrm{z}^2 - \mathrm{xy} - \mathrm{yz} - \mathrm{zx}\) or \((3\mathrm{z} - \mathrm{p} / 4)^2 >(\mathrm{d}^2 - \mathrm{s} / 2)\) . Hence \(3\mathrm{z} > \mathrm{p} / 4 + \sqrt{(\mathrm{d}^2 - \mathrm{s} / 2)}\) .


S is a set of integers. Its smallest element is 1 and its largest element is 100. Every element of S except 1 is the sum of two distinct members of the set or double a member of the set. What is the smallest possible number of integers in S?

## Answer

Solution:
Let \(\mathrm{n} = \mathrm{M}(\mathrm{n}) + \mathrm{m}(\mathrm{n})\) , where \(\mathrm{M}(\mathrm{n})\geq \mathrm{m}(\mathrm{n})\) . Put \(\mathrm{M}^{1}(\mathrm{n}) = \mathrm{M}(\mathrm{n})\) , \(\mathrm{M}^{2}(\mathrm{n}) = \mathrm{M}(\mathrm{M}(\mathrm{n}))\) etc. Then \(\mathrm{M}(100)\geq 50\) , \(\mathrm{M}^{2}(100)\geq 25\) , \(\mathrm{M}^{3}(100)\geq 13\) , \(\mathrm{M}^{4}(100)\geq 7\) , \(\mathrm{M}^{5}(100)\geq 4\) , \(\mathrm{M}^{6}(100)\geq 2\) (and obviously \(\mathrm{n} > \mathrm{M}(\mathrm{n})\) ), so we need at least 8 numbers. There are several ways of using 9 numbers. For example, {1, 2, 4, 8, 16, 32, 36, 64, 100}, where \(36 = 4 + 32\) , \(100 = 36 + 64\) and the others are double another number.

Doubling every time does not work: 1, 2, 4, 8, 16, 32, 64, 128. But if we do not double every time, then we cannot get a number larger than 96 with 8 numbers: the best we can do is \(1\cdot 2^{6}\) . \((3 / 2) = 96\) (on the occasion when we do not double the best we can do is to the largest plus the next largest, or \(3 / 2\) x the largest). Hence we need at least 9 numbers. [To be more formal, write the elements as \(1 = \mathrm{a}_{1}< \mathrm{a}_{2}< \ldots < \mathrm{a}_{\mathrm{n}}\) , then each \(\mathrm{a}_{\mathrm{i}}\) must be a sum of preceding elements. The largest possible \(\mathrm{a}_{\mathrm{i}}\) is \(2\mathrm{a}_{\mathrm{i - 1}}\) and the next largest \(\mathrm{a}_{\mathrm{i - 1}} + \mathrm{a}_{\mathrm{i - 2}}\) and so on.]


Show that there are infinitely many positive integers n such that \([\mathrm{a}^{3 / 2}] + [\mathrm{b}^{3 / 2}] = \mathrm{n}\) has at least 1980 integer solutions.

Solution: Consider all a, b in the range 1, 2, 3, ..., \(\mathrm{N}^{2}\) . There are \(\mathrm{N}^{4}\) possible pairs of values. But \([\mathrm{a}^{3 / 2}]\) and \([\mathrm{b}^{3 / 2}]\) are in the range 1, 2, ..., \(\mathrm{N}^{3}\) , so their sum is in the range 1, 2, ..., \(2\mathrm{N}^{3}\) . Hence one of these values has at least \(\mathrm{N} / 2\) solutions. By taking \(\mathrm{N}\) sufficiently large we can get \(\mathrm{a}_{1} > 1980\) solutions for some \(\mathrm{N}_{1}\leq 2\mathrm{N}^{3}\) . But now by taking \(\mathrm{N}\) sufficiently large we can get \(\mathrm{a}_{2} > \mathrm{a}_{1}\) solutions for some \(\mathrm{N}_{2}\) . Since \(\mathrm{a}_{2}\neq \mathrm{a}_{1}\) , we must have \(\mathrm{N}_{2}\neq \mathrm{N}_{1}\) . In other words, we have a different n, also with \(>1980\) solutions. Continuing, we get an infinite sequence of distinct n each with at least 1980 solutions.


ABCD is a tetrahedron. Angles ACB and ADB are 90 deg. Let k be the angle between the lines AC and BD. Show that \(\cos \mathrm{k}< \mathrm{CD} / \mathrm{AB}\) .


\(\mathrm{x}_{0}\) is a real number in the interval (0, 1) with decimal representation \(0. \mathrm{d}_{1} \mathrm{d}_{2} \mathrm{d}_{3} \ldots\) . We obtain the sequence \(\mathrm{x}_{\mathrm{n}}\) as follows. \(\mathrm{x}_{\mathrm{n} + 1}\) is obtained from \(\mathrm{x}_{\mathrm{n}}\) by rearranging the 5 digits \(\mathrm{d}_{\mathrm{n} + 1}\) , \(\mathrm{d}_{\mathrm{n} + 2}\) , \(\mathrm{d}_{\mathrm{n} + 3}\) , \(\mathrm{d}_{\mathrm{n} + 4}\) , \(\mathrm{d}_{\mathrm{n} + 5}\) . Show that the sequence \(\mathrm{x}_{\mathrm{n}}\) converges. Can the limit be irrational if \(\mathrm{x}_{0}\) is rational? Find a number \(\mathrm{x}_{0}\) so that every member of the sequence is irrational, no matter how the rearrangements are carried out.


A chess board is placed on top of an identical board and rotated through 45 degrees about its center. What is the area which is black in both boards?


AB is a diameter of the circle C. M and N are any two points on the circle. The chord MA' is perpendicular to the line NA and the chord MB' is perpendicular to the line NB. Show that AA' and BB' are parallel.


Find an example of m and n such that m is the product of n consecutive positive integers and also the product of \(\mathrm{n} + 2\) consecutive positive integers. Show that we cannot have \(\mathrm{n} = 2\) .


Write down a row of arbitrary integers (repetitions allowed). Now construct a second row as follows. Suppose the integer n is in column k in the first row. In column k in the second row write down the number of occurrences of n in row 1 in columns 1 to k inclusive. Similarly, construct a third row under the second row (using the values in the second row), and a fourth row. An example follows:

7	1	2	1	7	1	1

1	1	1	2	2	3	4

1	2	3	1	2	1	1

1	1	1	2	2	3	4

Show that the fourth row is always the same as the second row.


Let S be the set of points (x, y) given by \(\mathrm{y}\leq - \mathrm{x}^2\) and \(\mathrm{y}\geq \mathrm{x}^2 - 2\mathrm{x} + \mathrm{a}\) . Find the area of the rectangle with sides parallel to the axes and the smallest possible area which encloses S.


ABC, CDE, EFG are equilateral triangles (not necessarily the same size). The vertices are counter- clockwise in each case. A, D, G are collinear and \(\mathrm{AD} = \mathrm{DG}\) . Show that BFD is equilateral.


1000 people live in a village. Every evening each person tells his friends all the news he heard during the day. All news eventually becomes known (by this process) to everyone. Show that one can choose 90 people, so that if you give them some news on the same day, then everyone will know in 10 days.


The reals a and b are such that a cos \(x + b\) cos \(3x > 1\) has no real solutions. Show that \(|b| \leq 1\) .


ABCD is a convex quadrilateral. K is the midpoint of AB and M is the midpoint of CD. L lies on the side BC and N lies on the side AD. KLMN is a rectangle. Show that its area is half that of ABCD.


The sequence \(\mathbf{a}_{\mathrm{n}}\) of positive integers is such that (1) \(\mathbf{a}_{\mathrm{n}} \leq \mathbf{n}^{3 / 2}\) for all n, and (2) m- n divides \(\mathbf{k}_{\mathrm{m}} - \mathbf{k}_{\mathrm{n}}\) (for all \(\mathbf{m} > \mathbf{n}\) ). Find \(\mathbf{a}_{\mathrm{n}}\) .


Is it possible to color half the cells in a rectangular array white and half black so that in each row and column more than 3/4 of the cells are the same color?


ACPH, AMBE, AHBT, BKXM and CKXP are parallelograms. Show that ABTE is also a parallelogram (vertices are labeled anticlockwise).


Find all solutions (x, y) in positive integers to \(x^{3} - y^{3} = xy + 61\) .

## Answer

(6,5)

Solution: Put \(x = y + a\) . Then \((3a - 1)y^{2} + a(3a - 1)y + (a^{3} - 61) = 0\) . The first two terms are positive, so the last term must be negative, so \(a = 1, 2, 3\) . Trying each case in turn, we get \((y + 6)(y - 5) = 0\) , \(5y^{2} + 10y - 53 = 0\) , \(4y^{2} + 12y - 17 = 0\) . The last two equations have no integers solutions.


Eighteen teams are playing in a tournament. So far, each team has played exactly eight games, each with a different opponent. Show that there are three teams none of which has yet played the other.


ABC is a triangle. A' lies on the side BC with \(\mathrm{BA} / \mathrm{BC} = 1 / 4\) . Similarly, B' lies on the side CA with \(\mathrm{CB} / \mathrm{CA} = 1 / 4\) , and C' lies on the side AB with \(\mathrm{AC} / \mathrm{AB} = 1 / 4\) . Show that the perimeter of A'B'C' is between 1/2 and 3/4 of the perimeter of ABC.


The positive reals \(\mathbf{x}\) , y satisfy \(\mathbf{x}^{3} + \mathbf{y}^{3} = \mathbf{x} - \mathbf{y}\) . Show that \(\mathbf{x}^{2} + \mathbf{y}^{2} < 1\) .

Solution: Since \(\mathbf{x}\) , y are positive, so is \(\mathbf{x}^{3} + \mathbf{y}^{3}\) , and hence \(\mathbf{x} > \mathbf{y}\) . So \((\mathbf{x}^{2} + \mathbf{y}^{2})(\mathbf{x} - \mathbf{y}) = (\mathbf{x}^{3} - \mathbf{y}^{3}) - \mathbf{xy}(\mathbf{x} - \mathbf{y})\) \(< \mathbf{x}^{3} - \mathbf{y}^{3} = \mathbf{x} - \mathbf{y}\) . Hence \(\mathbf{x}^{2} + \mathbf{y}^{2} < 1\) .


A convex polygon is drawn inside the unit circle. Someone makes a copy by starting with one vertex and then drawing each side successively. He copies the angle between each side and the previous side accurately, but makes an error in the length of each side of up to a factor \(1 \pm \mathbf{p}\) . As a result the last side ends up a distance \(\mathbf{d}\) from the starting point. Show that \(\mathbf{d} < 4 \mathbf{p}\) .


An integer is initially written at each vertex of a cube. A move is to add 1 to the numbers at two vertices connected by an edge. Is it possible to equalise the numbers by a series of moves in the following cases? (1) The initial numbers are (1) 0, except for one vertex which is 1. (2) The initial numbers are 0, except for two vertices which are 1 and diagonally opposite on a face of the cube. (3) Initially, the numbers going round the base are 1, 2, 3, 4. The corresponding vertices on the top are 6, 7, 4, 5 (with 6 above the 1, 7 above the 2 and so on).


Find 21 consecutive integers, each with a prime factor less than 17.


Each of the numbers from 100 to 999 inclusive is written on a separate card. The cards are arranged in a pile in random order. We take cards off the pile one at a time and stack them into 10 piles according to the last digit. We then put the 1 pile on top of the 0 pile, the 2 pile on top of the 1 pile and so on to get a single pile. We now take them off one at a time and stack them into 10 piles according to the middle digit. We then consolidate the piles as before. We then take them off one at a time and stack them into 10 piles according to the first digit and finally consolidate the piles as before. What can we say about the order in the final pile?


Given 6 points inside a \(3 \times 4\) rectangle, show that we can find two points whose distance does not exceed \(\sqrt{5}\) .


What is the smallest value of \(4 + \mathbf{x}^{2}\mathbf{y}^{4} + \mathbf{x}^{4}\mathbf{y}^{2} - 3\mathbf{x}^{2}\mathbf{y}^{2}\) for real \(\mathbf{x}\) , \(\mathbf{y}\) ? Show that the polynomial cannot be written as a sum of squares. [Note the candidates did not know calculus.]


ABCDEF is a prism. Its base ABC and its top DEF are congruent equilateral triangles. The side edges are AD, BE and CF. Find all points on the base which are equidistant from the three lines AE, BF and CD.


The circle C has center O and radius r and contains the points A and B. The circle C' touches the rays OA and OB and has center O' and radius r'. Find the area of the quadrilateral OAOB.


The sequence \(\mathbf{a}_{\mathrm{n}}\) is defined by \(\mathbf{a}_{1} = 1\) , \(\mathbf{a}_{2} = 2\) , \(\mathbf{a}_{\mathrm{n} + 2} = \mathbf{a}_{\mathrm{n} + 1} + \mathbf{a}_{\mathrm{n}}\) . The sequence \(\mathbf{b}_{\mathrm{n}}\) is defined by \(\mathbf{b}_{1} = 2\) , \(\mathbf{b}_{2} = 1\) , \(\mathbf{b}_{\mathrm{n} + 2} = \mathbf{b}_{\mathrm{n} + 1} + \mathbf{b}_{\mathrm{n}}\) . How many integers belong to both sequences?

## Answer

1,2,3 only

Solution:
The first few terms are:

n 1 2 3 4 5 6 7 \(\mathbf{a}_{\mathrm{n}}\) 1 2 3 5 8 13 21 \(\mathbf{b}_{\mathrm{n}}\) 2 1 3 4 7 11 18

Note that for \(\mathbf{n} = 4\) , 5 we have \(\mathbf{a}_{\mathrm{n - 1}} < \mathbf{b}_{\mathrm{n}} < \mathbf{a}_{\mathrm{n}}\) . So by a trivial induction, the inequality holds for all \(\mathbf{n} \geq 4\) .


N is a sum of n powers of 2. If N is divisible by \(2^{\mathrm{m}} - 1\) , prove that \(\mathrm{n} \geq \mathrm{m}\) . Does there exist a number divisible by 11...1 (m 1s) which has the sum of its digits less than m?


A non- negative real is written at each vertex of a cube. The sum of the eight numbers is 1. Two players choose faces of the cube alternately. A player cannot choose a face already chosen or the one opposite, so the first player plays twice, the second player plays once. Can the first player arrange that the vertex common to all three chosen faces is \(\leq 1 / 6\) ?


A library is open every day except Wednesday. One day three boys, A, B, C visit the library together for the first time. Thereafter they visit the library many times. A always makes his next visit two days after the previous visit, unless the library is closed on that day, in which case he goes the following day. B always makes his next visit three days after the previous visit (or four if the library is closed). C always makes his next visit four days after the previous visit (or five if the library is closed). For example, if A went first on Monday, his next visit would be Thursday, then Saturday. If B went first on Monday, his next visit would be on Thursday. All three boys are subsequently in the library on a Monday. What day of the week was their first visit?


ABCD is a parallelogram and AB is not equal to BC. M is chosen so that (1) \(\angle \mathrm{MAC} =\) \(\angle \mathrm{DAC}\) and M is on the opposite side of AC to D, and (2) \(\angle \mathrm{MBD} = \angle \mathrm{CBD}\) and M is on the opposite side of BD to C. Find AM/BM in terms of \(\mathrm{k} = \mathrm{AC} / \mathrm{BD}\) .


3n points divide a circle into 3n arcs. One third of the arcs have length 1, one third have length 2 and one third have length 3. Show that two of the points are at opposite ends of a diameter.


M is a point inside a regular tetrahedron. Show that we can find two vertices A, B of the tetrahedron such that \(\cos \mathrm{AMB} \leq - 1 / 3\) .


\(0 < \mathrm{x}, \mathrm{y}, \mathrm{z} < \pi / 2\) . We have \(\cos \mathrm{x} = \mathrm{x}, \sin (\cos \mathrm{y}) = \mathrm{y}, \cos (\sin \mathrm{z}) = \mathrm{z}\) . Which of \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is the largest and which the smallest?


P is a polygon with \(2\mathrm{n} + 1\) sides. A new polygon is derived by taking as its vertices the midpoints of the sides of P. This process is repeated. Show that we must eventually reach a polygon which is homothetic to P.


\(\mathrm{a}_1, \mathrm{a}_2, \ldots , \mathrm{a}_{1982}\) is a permutation of 1, 2, ..., 1982. If \(\mathrm{a}_1 > \mathrm{a}_2\) , we swap \(\mathrm{a}_1\) and \(\mathrm{a}_2\) . Then if (the new) \(\mathrm{a}_2 > \mathrm{a}_3\) we swap \(\mathrm{a}_2\) and \(\mathrm{a}_3\) . And so on. After 1981 potential swaps we have a new permutation \(\mathrm{b}_1, \mathrm{b}_2, \ldots , \mathrm{b}_{1982}\) . We then compare \(\mathrm{b}_{1982}\) and \(\mathrm{b}_{1981}\) . If \(\mathrm{b}_{1981} > \mathrm{b}_{1982}\) , we swap them. We then compare \(\mathrm{b}_{1980}\) and (the new) \(\mathrm{b}_{1981}\) . So we arrive finally at \(\mathrm{c}_1, \mathrm{c}_2, \ldots , \mathrm{c}_{1982}\) . We find that \(\mathrm{a}_{100} = \mathrm{c}_{100}\) . What value is \(\mathrm{a}_{100}\) ?


Cucumber River has parallel banks a distance 1 meter apart. It has some islands with total perimeter 8 meters. It is claimed that it is always possible to cross the river (starting from an arbitrary point) by boat in at most 3 meters. Is the claim always true for any arrangement of islands? [Neglect the current.]


The parabola \(\mathrm{y} = \mathrm{x}^2\) is drawn and then the axes are deleted. Can you restore them using ruler and compasses?


An integer is put in each cell of an \(\mathrm{n} \times \mathrm{n}\) array. The difference between the integers in cells which share a side is 0 or 1. Show that some integer occurs at least \(\mathrm{n}\) times.


\(\mathbf{x}\) is a positive integer. Put \(\mathbf{a} = \mathbf{x}^{1 / 12}\) , \(\mathbf{b} = \mathbf{x}^{1 / 4}\) , \(\mathbf{c} = \mathbf{x}^{1 / 6}\) . Show that \(2^{\mathrm{a}} + 2^{\mathrm{b}} \geq 2^{1 + \mathrm{c}}\) .

Solution: Put \(\mathbf{x} = \mathbf{r}^{12}\) . Since \(\mathbf{x}\) is a positive integer, we have \(\mathbf{r} \geq 1\) . We have to show that \((2^{\mathrm{r}} + 2^{\mathrm{r}^{3}}) / 2 \geq 2^{\mathrm{r}^{2}}\) . But this follows immediately from AM/GM.


What is the largest subset of \(\{1, 2, \ldots , 1982\}\) with the property that no element is the product of two other distinct elements.


A real number is assigned to each unit square in an infinite sheet of squared paper. Show that some cell contains a number that is less than or equal to at least four of its eight neighbors.


Given a real sequence \(\mathbf{a}_{1}, \mathbf{a}_{2}, \ldots , \mathbf{a}_{\mathrm{n}}\) , show that it is always possible to choose a subsequence such that (1) for each \(\mathrm{i} \leq \mathrm{n} - 2\) at least one and at most two of \(\mathbf{a}_{\mathrm{i}}, \mathbf{a}_{\mathrm{i} + 1}, \mathbf{a}_{\mathrm{i} + 2}\) are chosen and (2) the sum of the absolute values of the numbers in the subsequence is at least \(1 / 6 \sum \mathrm{n}^{\mathrm{i}} \mathrm{la}_{\mathrm{i}}\) .


An \(\mathrm{n} \times \mathrm{n}\) array has a cross in \(\mathrm{n} - 1\) cells. A move consists of moving a row to a new position or moving a column to a new position. For example, one might move row 2 to row 5, so that row 1 remained in the same position, row 3 became row 2, row 4 became row 3, row 5 became row 4, row 2 became row 5 and the remaining rows remained in the same position. Show that by a series of moves one can end up with all the crosses below the main diagonal.


Let \(\{\mathbf{a}\}\) denote the difference between a and the nearest integer. For example \(\{3.8\} = 0.2, \{- 5.4\} = 0.4\) . Show that \(\mathrm{la} \mathrm{la} - 11 \mathrm{la} - 2 \mathrm{l} \ldots \mathrm{la} - \mathrm{nl} \geq = \{\mathrm{a} \} \mathrm{n}!/2^{\mathrm{n}}\) .


Do there exist polynomials \(\mathrm{p}(\mathrm{x}), \mathrm{q}(\mathrm{x}), \mathrm{r}(\mathrm{x})\) such that \(\mathrm{p}(\mathrm{x} - \mathrm{y} + \mathrm{z})^{3} + \mathrm{q}(\mathrm{y} - \mathrm{z} - 1)^{3} + \mathrm{r}(\mathrm{z} - 2\mathrm{x} + 1)^{3} = 1\) for all \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) ? Do there exist polynomials \(\mathrm{p}(\mathrm{x}), \mathrm{q}(\mathrm{x}), \mathrm{r}(\mathrm{x})\) such that \(\mathrm{p}(\mathrm{x} - \mathrm{y} + \mathrm{z})^{3} + \mathrm{q}(\mathrm{y} - \mathrm{z} - 1)^{3} + \mathrm{r}(\mathrm{z} - \mathrm{x} + 1)^{3} = 1\) for all \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) ?


A tetrahedron T' has all its vertices inside the tetrahedron T. Show that the sum of the edge lengths of T' is less than 4/3 times the corresponding sum for T.


A \(4 \mathrm{x} 4\) array of unit cells is made up of a grid of total length 40. Can we divide the grid into 8 paths of length 5? Into 5 paths of length 8?


Three positive integers are written on a blackboard. A move consists of replacing one of the numbers by the sum of the other two less one. For example, if the numbers are 3, 4, 5, then one move could lead to 4, 5, 8 or 3, 5, 7 or 3, 4, 6. After a series of moves the three numbers are 17, 1967 and 1983. Could the initial set have been 2, 2, 2? 3, 3, 3?


\(\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}\) are circles, none of which lie inside either of the others. \(\mathrm{C}_{1}\) and \(\mathrm{C}_{2}\) touch at Z, \(\mathrm{C}_{2}\) and \(\mathrm{C}_{3}\) touch at X, and \(\mathrm{C}_{3}\) and \(\mathrm{C}_{1}\) touch at Y. Prove that if the radius of each circle is increased by a factor \(2 / \sqrt{3}\) without moving their centers, then the enlarged circles cover the triangle XYZ.


Find all real solutions x, y to \(\mathrm{y}^{2} = \mathrm{x}^{3} - 3\mathrm{x}^{2} + 2\mathrm{x}, \mathrm{x}^{2} = \mathrm{y}^{3} - 3\mathrm{y}^{2} + 2\mathrm{y}.\)


The positive integer k has n digits. It is rounded to the nearest multiple of 10, then to the nearest multiple of 100 and so on (n- 1 roundings in all). Numbers midway between are rounded up. For example, 1474 is rounded to 1470, then to 1500, then to 2000. Show that the final number is less than 18k/13.


M is the midpoint of BC. E is any point on the side AC and F is any point on the side AB. Show that area \(\mathrm{MEF} \leq\) area \(\mathrm{BMF} +\) area CME.


\(\mathrm{a}_{\mathrm{n}}\) is the last digit of \([10^{n / 2}]\) . Is the sequence \(\mathrm{a}_{\mathrm{n}}\) periodic? \(\mathrm{b}_{\mathrm{n}}\) is the last digit of \([2^{n / 2}]\) . Is the sequence \(\mathrm{b}_{\mathrm{n}}\) periodic?


A and B are acute angles such that \(\sin^{2} \mathrm{A} + \sin^{2} \mathrm{B} = \sin (\mathrm{A} + \mathrm{B})\) . Show that \(\mathrm{A} + \mathrm{B} = \pi / 2\) .


The projection of a tetrahedron onto the plane P is ABCD. Can we find a distinct plane P' such that the projection of the tetrahedron onto P' is A'B'C'D' and AA', BB', CC' and DD' are all parallel?


Given a quadratic equation \(\mathrm{ax}^2 +\mathrm{bx} + \mathrm{c}\) . If it has two real roots \(\mathrm{A}\leq \mathrm{B}\) , transform the equation to \(\mathrm{x}^2 +\mathrm{Ax} + \mathrm{B}\) . Show that if we repeat this process we must eventually reach an equation with complex roots. What is the maximum possible number of transformations before we reach such an equation?


a, b, c are positive integers. If ab divides ba and ca divides ac, show that cb divides bc.


A word is a finite string of As and Bs. Can we find a set of three 4- letter words, ten 5- letter words, thirty 6- letter words and five 7- letter words such that no word is the beginning of another word. [For example, if ABA was a word, then ABAAB could not be a word.]


Can you place an integer in every square of an infinite sheet of squared paper so that the sum of the integers in every 4 x 6 (or 6 x 4) rectangle is (1) 10, (2) 1?


A point is chosen on each of the three sides of a triangle and joined to the opposite vertex. The resulting lines divide the triangle into four triangles and three quadrilaterals. The four triangles all have area A. Show that the three quadrilaterals have equal area. What is it (in terms of A)?


A group of children form two equal lines side- by- side. Each line contains an equal number of boys and girls. The number of mixed pairs (one boy in one line next to one girl in the other line) equals the number of unmixed pairs (two girls side- by- side or two boys side- by- side). Show that the total number of children in the group is a multiple of 8.


A 1 x k rectangle can be divided by two perpendicular lines parallel to the sides into four rectangles, each with area at least 1 and one with area at least 2. What is the smallest possible k?


O is a point inside the triangle ABC. \(\mathrm{a} =\) area OBC, \(\mathrm{b} =\) area OCA, \(\mathrm{c} =\) area OAB. Show that the vector sum \(\mathrm{aOA} + \mathrm{bOB} + \mathrm{cOC}\) is zero.


Show that given any \(2\mathrm{m} + 1\) different integers lying between - (2m- 1) and 2m- 1 (inclusive) we can always find three whose sum is zero.


Interior points D, E, F are chosen on the sides BC, CA, AB (not at the vertices). Let k be the length of the longest side of DEF. Let a, b, c be the lengths of the longest sides of AFE, BDF, CDE respectively. Show that \(\mathrm{k}\geq \sqrt{3}\) min(a, b, c) /2. When do we have equality?


X is a union of k disjoint intervals of the real line. It has the property that for any \(\mathrm{h}< 1\) we can find two points of X which are a distance h apart. Show that the sum of the lengths of the intervals in X is at least \(1 / \mathrm{k}\)


x is a real. The decimal representation of x includes all the digits at least once. Let f(n) be the number of distinct n- digit segments in the representation. Show that if for some n we have \(\mathrm{f(n)}\leq \mathrm{n} + 8\) , then x is rational.


Show that we can find n integers whose sum is 0 and whose product is n iff n is divisible by 4.


Show that \((a + b)^2 /2 + (a + b) / 4 \geq a\sqrt{b} + b\sqrt{a}\) for all positive a and b.

## Answer

By AM/GM \(\sqrt{(ab)} \leq (a + b) / 2\) , so \(\sqrt{2} (a + b) + \sqrt{(ab)} \leq a + b\) . Hence \(\sqrt{(2a + 2b)} \geq \sqrt{a} + \sqrt{b} (*)\) .

By AM/GM \((a + b) \geq 2\sqrt{(ab)}\) and \(2(a + b) + 1 \geq 2\sqrt{(2a + 2b)}\) . Multiplying, \((a + b)(2a + 2b + 1) \geq 4\sqrt{(ab)}\sqrt{(2a + 2b)}\) . Then using \((*)\geq 4\sqrt{(ab)}(\sqrt{a} +\sqrt{b})\) .


ABC and A'B'C' are equilateral triangles and ABC and A'B'C' have the same sense (both clockwise or both counter- clockwise). Take an arbitrary point O and points P, Q, R so that OP is equal and parallel to AA', OQ is equal and parallel to BB', and OR is equal and parallel to CC'. Show that PQR is equilateral.


Take a large number of unit squares, each with one edge red, one edge blue, one edge green, and one edge yellow. For which m, n can we combine mn squares by placing similarly colored edges together to get an m x n rectangle with one side entirely red, another entirely blue, another entirely green, and the fourth entirely yellow.


Let \(\mathrm{A} = \cos^2 \mathrm{a}\) , \(\mathrm{B} = \sin^2 \mathrm{a}\) . Show that for all real a and positive x, y we have \(\mathrm{x}^\mathrm{A}\mathrm{y}^\mathrm{B} < \mathrm{x} + \mathrm{y}\) .


Two players play a game. Each takes it in turn to paint three unpainted edges of a cube. The first player uses red paint and the second blue paint. So each player has two moves. The first player wins if he can paint all edges of some face red. Can the first player always win?


\(\mathrm{n} > 3\) positive integers are written in a circle. The sum of the two neighbours of each number divided by the number is an integer. Show that the sum of those integers is at least 2n and less than 3n. For example, if the numbers were 3, 7, 11, 15, 4, 1, 2 (with 2 also adjacent to 3), then the sum would be \(14 / 7 + 22 / 11 + 15 / 15 + 16 / 4 + 6 / 1 + 4 / 2 + 9 / 3 = 20\) and \(14 \leq 20 < 21\) .


The incircle of the triangle ABC has center I and touches BC, CA, AB at D, E, F respectively. The segments AI, BI, CI intersect the circle at D', E', F respectively. Show that DD', EE', FF' are collinear.


Find all integers m, n such that \((5 + 3\sqrt{2})^{\mathrm{m}} = (3 + 5\sqrt{2})^{\mathrm{n}}\) .


\(\mathrm{x_1< x_2< x_3< \ldots < x_n}\) . \(\mathrm{y_i}\) is a permutation of the \(\mathrm{x_i}\) . We have that \(\mathrm{x_1 + y_1< x_2 + y_2< \ldots < x_n + }\) \(\mathrm{y_n}\) . Prove that \(\mathrm{x_i = y_i}\) .


ABC is a triangle and P is any point. The lines PA, PB, PC cut the circumcircle of ABC again at A'B'C' respectively. Show that there are at most eight points P such that A'B'C' is congruent to ABC.


The positive reals x, y, z satisfy \(\mathrm{x^2 + xy + y^2 / 3 = 25}\) , \(\mathrm{y^2 / 3 + z^2 = 9}\) , \(\mathrm{z^2 + zx + x^2 = 16}\) . Find the value of \(\mathrm{xy + 2yz + 3zx}\) .


Starting with the polynomial \(\mathrm{x^2 + 10x + 20}\) , a move is to change the coefficient of \(\mathrm{x}\) by 1 or to change the coefficient of \(\mathrm{x^0}\) by 1 (but not both). After a series of moves the polynomial is changed to \(\mathrm{x^2 + 20x + 10}\) . Is it true that at some intermediate point the polynomial had integer roots?

## Answer

Yes.

Solution: We have \(\mathrm{x^2 + (n + 1)x + n = (x + n)(x + 1)}\) , so \(\mathrm{x^2 + ax + b}\) has integer roots if \(\mathrm{a = b + 1}\) (and a and b are integers). But initially a- b is - 10 and it ends up as +10. Each move changes a- b by \(\pm 1\) , so it must pass through all values between - 10 and +10.


The center of a coin radius r traces out a polygon with perimeter p which has an incircle radius \(\mathrm{R > r}\) . What is the area of the figure traced out by the coin?


Each weight in a set of n has integral weight and the total weight of the set is 2n. A balance is initially empty. We then place the weights onto a pan of the balance one at a time. Each time we place the heaviest weight not yet placed. If the pans balance, then we place the weight

onto the left pan. Otherwise, we place the weight onto the lighter pan. Show that when all the weights have been placed, the scales will balance. [For example, if the weights are 2, 2, 1, 1. Then we must place 2 in the left pan, followed by 2 in the right pan, followed by 1 in the left pan, followed by 1 in the right pan.]


A number is prime however we order its digits. Show that it cannot contain more than three different digits. For example, 337 satisfies the conditions because 337, 373 and 733 are all prime.


Find all pairs of digits (b, c) such that the number b ... b6c ... c4, where there are n bs and n cs is a square for all positive integers n.


A, B, C and D lie on a line in that order. Show that if X does not lie on the line then \(\vert \mathrm{XA}\vert +\) \(\vert \mathrm{XD}\vert + \vert \vert \mathrm{AB}\vert - \vert \mathrm{CD}\vert \vert >\vert \mathrm{XB}\vert +\vert \mathrm{XC}\vert\)


The real sequence \(\mathbf{x}_{\mathrm{n}}\) is defined by \(\mathbf{x}_{1} = 1\) \(\mathbf{x}_{2} = 1\) \(\mathbf{x}_{\mathrm{n} + 2} = \mathbf{x}_{\mathrm{n} + 1}{}^{2} - \mathbf{x}_{\mathrm{n}} / 2\) . Show that the sequence converges and find the limit.


The squares of a 1983 x 1984 chess board are colored alternately black and white in the usual way. Each white square is given the number 1 or the number - 1. For each black square the product of the numbers in the neighbouring white squares is 1. Show that all the numbers must be 1.


A 3 x 3 chess board is colored alternately black and white in the usual way with the center square white. Each white square is given the number 1 or the number - 1. A move consists of simultaneously changing each number to the product of the adjacent numbers. So the four corner squares are each changed to the number previously in the center square and the center square is changed to the product of the four numbers in the corners. Show that after finitely many moves all numbers are 1.


\(\mathbf{C}_{1}\) , \(\mathbf{C}_{2}\) , \(\mathbf{C}_{3}\) are circles with radii \(\mathbf{r}_{1}\) , \(\mathbf{r}_{2}\) , \(\mathbf{r}_{3}\) respectively. The circles do not intersect and no circle lies inside any other circle. \(\mathbf{C}_{1}\) is larger than the other two. The two outer common tangents to \(\mathbf{C}_{1}\) and \(\mathbf{C}_{2}\) meet at A ("outer" means that the points where the tangent touches the two circles

lie on the same side of the line of centers). The two outer common tangents to \(C_1\) and \(C_3\) intersect at B. The two tangents from A to \(C_3\) and the two tangents from B to \(C_2\) form a quadrangle. Show that it has an inscribed circle and find its radius.


Show that any cross- section of a cube through its center has area not less than the area of a face.


ABC is an acute angled triangle. The midpoints of BC, CA and AB are D, E, F respectively. Perpendiculars are drawn from D to AB and CA, from E to BC and AB, and from F to CA and BC. The perpendiculars form a hexagon. Show that its area is half the area of the triangle.


Is there an integer n such that the sum of the (decimal) digits of n is 1000 and the sum of the squares of the digits is \(1000^{2}\) ?


An \(8 \times 8\) chess- board is colored in the usual way. What is the largest number of pieces can be placed on the black squares (at most one per square), so that each piece can be taken by at least one other? A piece A can take another piece B if they are (diagonally) adjacent and the square adjacent to B and opposite to A is empty.


Call a side or diagonal of a regular n- gon a segment. How many colors are required to paint all the segments of a regular n- gon, so that each segment has a single color and every two segments with a vertex in common have different colors.


Given a line L and a point O not on the line, can we move an arbitrary point X to O using only rotations about O and reflections in L?


The quadratic \(\mathrm{p}(\mathrm{x}) = \mathrm{ax}^{2} + \mathrm{bx} + \mathrm{c}\) has a \(>100\) . What is the maximum possible number of integer values x such that \(\mathrm{lp}(\mathrm{x})< 50\) ?


In the diagram below a, b, c, d, e, f, g, h, i, j are distinct positive integers and each (except a, e, h and j) is the sum of the two numbers to the left and above. For example, \(\mathrm{b} = \mathrm{a} + \mathrm{e}\) , \(\mathrm{f} = \mathrm{e} + \mathrm{h}\) , \(\mathrm{i} = \mathrm{h} + \mathrm{j}\) . What is the smallest possible value of d?

h i e f g a b c d


\(\mathrm{a}_{1}< \mathrm{a}_{2}< \ldots < \mathrm{a}_{\mathrm{n}}< \ldots\) is an unbounded sequence of positive reals. Show that there exists k such that \(\mathrm{a}_{1} / \mathrm{a}_{2} + \mathrm{a}_{2} / \mathrm{a}_{3} + \ldots + \mathrm{a}_{\mathrm{h}} / \mathrm{a}_{\mathrm{h} + 1}< \mathrm{h} - 1\) for all \(\mathrm{h} > \mathrm{k}\) . Show that we can also find a k such that \(\mathrm{a}_{1} / \mathrm{a}_{2} + \mathrm{a}_{2} / \mathrm{a}_{3} + \ldots + \mathrm{a}_{\mathrm{h}} / \mathrm{a}_{\mathrm{h} + 1}< \mathrm{h} - 1985\) for all \(\mathrm{h} > \mathrm{k}\) .


Find all pairs (x, y) such that \(\mathrm{Isin x - sin y| + sin x sin y< = 0}\)


ABCDE is a convex pentagon. A' is chosen so that B is the midpoint of AA', B' is chosen so that C is the midpoint of BB' and so on. Given A', B', C', D', E', how do we construct ABCDE using ruler and compasses?


The sequence \(\mathrm{a_1,a_2,a_3,\ldots}\) satisfies \(\mathrm{a_{4n + 1} = 1}\) \(\mathrm{a_{4n + 3} = 0}\) \(\mathrm{a_{2n} = a_n}\) . Show that it is not periodic.


n lines are drawn in the plane. Some of the resulting regions are colored black, no pair of painted regions have a boundary line in common (but they may have a common vertex). Show that at most \((\mathrm{n}^2 +\mathrm{n}) / 3\) regions are black.


Each face of a cube is painted a different color. The same colors are used to paint every face of a cubical box a different color. Show that the cube can always be placed in the box, so that every face is a different color from the box face it is in contact with.


The points A, B, C, D, E, F are equally spaced on the circumference of a circle (in that order) and AF is a diameter. The center is O. OC and OD meet BE at M and N respectively. Show that \(\mathrm{MN} + \mathrm{CD} = \mathrm{OA}\) .


A move replaces the real numbers a, b, c, d by a- b, b- c, c- d, d- a. If a, b, c, d are not all equal, show that at least one of the numbers can exceed 1985 after a finite number of moves.


\(\mathrm{a_1< a_2< \ldots < a_n}\) and \(\mathrm{b_1 > b_2 > \ldots > b_n}\) . Taken together the \(\mathrm{a_i}\) and \(\mathrm{b_i}\) constitute the numbers 1, 2, ..., 2n. Show that \(\mathrm{la_1 - b_1| + la_2 - b_2| + \ldots + la_n - b_n| = n^2}\) .


An r x s x t cuboid is divided into rst unit cubes. Three faces of the cuboid, having a common vertex, are colored. As a result exactly half the unit cubes have at least one face colored. What is the total number of unit cubes?


ABCD is a parallelogram. A circle through A and B has radius R. A circle through B and D has radius R and meets the first circle again at M. Show that the circumradius of AMD is R.


A regular hexagon is divided into 24 equilateral triangles by lines parallel to its sides. 19 different numbers are assigned to the 19 vertices. Show that at least 7 of the 24 triangles have the property that the numbers assigned to its vertices increase counterclockwise.


\(\mathrm{x}\) is a real number. Define \(\mathrm{x}_0 = 1 + \sqrt{(1 + \mathrm{x})}\) , \(\mathrm{x}_1 = 2 + \mathrm{x} / \mathrm{x}_0\) , \(\mathrm{x}_2 = 2 + \mathrm{x} / \mathrm{x}_1\) , ..., \(\mathrm{x}_{1985} = 2 + \mathrm{x} / \mathrm{x}_{1984}\) . Find all solutions to \(\mathrm{x}_{1985} = \mathrm{x}\) .

## Answer

3

Solution: If \(\mathrm{x} = 0\) , then \(\mathrm{x}^{1985} = 2 \neq \mathrm{x}\) . Otherwise we find \(\mathrm{x}_1 = 2 + \mathrm{x} / (1 + \sqrt{(1 + \mathrm{x})}) = 2 + (\sqrt{(1 + \mathrm{x})} - 1) = 1 + \sqrt{(1 + \mathrm{x})}\) . Hence \(\mathrm{x}_{1985} = 1 + \sqrt{(1 + \mathrm{x})}\) . So \(\mathrm{x} - 1 = \sqrt{(1 + \mathrm{x})}\) . Squaring, \(\mathrm{x} = 0\) or 3. We have already ruled out \(\mathrm{x} = 0\) . It is easy to check that \(\mathrm{x} = 3\) is a solution.


\(\mathrm{x}\) is a real number. Define \(\mathrm{x}_0 = 1 + \sqrt{(1 + \mathrm{x})}\) , \(\mathrm{x}_1 = 2 + \mathrm{x} / \mathrm{x}_0\) , \(\mathrm{x}_2 = 2 + \mathrm{x} / \mathrm{x}_1\) , ..., \(\mathrm{x}_{1985} = 2 + \mathrm{x} / \mathrm{x}_{1984}\) . Find all solutions to \(\mathrm{x}_{1985} = \mathrm{x}\) .


A regular pentagon has side 1. All points whose distance from every vertex is less than 1 are deleted. Find the area remaining.


Given a large sheet of squared paper, show that for \(\mathrm{n} > 12\) you can cut along the grid lines to get a rectangle of more than \(\mathrm{n}\) unit squares such that it is impossible to cut it along the grid lines to get a rectangle of \(\mathrm{n}\) unit squares from it.


The cube ABCDA'B'C'D' has unit edges. Find the distance between the circle circumscribed about the base ABCD and the circumcircle of ABC.


The quadratic \(\mathbf{x}^{2} + \mathbf{ax} + \mathbf{b} + 1\) has roots which are positive integers. Show that \((\mathbf{a}^{2} + \mathbf{b}^{2})\) is composite.

Solution: Let the roots be c, d, so c + d = - a, cd = b+1. Hence a² + b² = (c² + 1)(d² + 1).


Two equal squares, one with blue sides and one with red sides, intersect to give an octagon with sides alternately red and blue. Show that the sum of the octagon's red side lengths equals the sum of its blue side lengths.


ABC is acute- angled. What point P on the segment BC gives the minimal area for the intersection of the circumcircles of ABP and ACP?


Given n points can one build n- 1 roads, so that each road joins two points, the shortest distance between any two points along the roads belongs to {1, 2, 3, ..., n(n- 1)/2 }, and given any element of {1, 2, 3, ..., n(n- 1)/2 } one can find two points such that the shortest distance between them along the roads is that element?


Prove that there is no convex quadrilateral with vertices at lattice points so that one diagonal has twice the length of the other and the angle between them is 45 degrees.


Prove that we can find an m x n array of squares so that the sum of each row and the sum of each column is also a square.


Two circles intersect at P and Q. A is a point on one of the circles. The lines AP and AQ meet the other circle at B and C respectively. Show that the circumradius of ABC equals the distance between the centers of the two circles. Find the locus of the circumcircle as A varies.


A regular hexagon has side 1000. Each side is divided into 1000 equal parts. Let S be the set of the vertices and all the subdividing points. All possible lines parallel to the sides and with endpoints in S are drawn, so that the hexagon is divided into equilateral triangles with side 1.

Let X be the set of all vertices of these triangles. We now paint any three unpainted members of X which form an equilateral triangle (of any size). We then repeat until every member of X except one is painted. Show that the unpainted vertex is not a vertex of the original hexagon.


Let \(\mathrm{d(n)}\) be the number of (positive integral) divisors of n. For example, \(\mathrm{d}(12) = 6\) . Find all n such that \(\mathrm{n} = \mathrm{d}(\mathrm{n})^2\) .


Show that for all positive reals \(\mathbf{x}_{\mathrm{i}}\) we have \(1 / \mathbf{x}_{1} + 1 / (\mathbf{x}_{1} + \mathbf{x}_{2}) + \ldots +\mathbf{n} / (\mathbf{x}_{1} + \ldots +\mathbf{x}_{\mathrm{n}}) < 4 / \mathbf{a}_{1} + 4 / \mathbf{a}_{2} + \ldots +4 / \mathbf{a}_{\mathrm{n}}\) .


ABC is a triangle with \(\mathrm{AB} \neq \mathrm{AC}\) . Show that for each line through A, there is at most one point X on the line (excluding A, B, C) with \(\angle \mathrm{ABX} = \angle \mathrm{ACX}\) . Which lines contain no such points X?


An n x n x n cube is divided into \(\mathrm{n}^3\) unit cubes. Show that we can assign a different integer to each unit cube so that the sum of each of the \(3\mathrm{n}^2\) rows parallel to an edge is zero.


Find all positive integers a, b, c so that \(\mathrm{a}^2 + \mathrm{b} = \mathrm{c}\) and a has \(\mathrm{n} > 1\) decimal digits all the same, b has n decimal digits all the same, and c has 2n decimal digits all the same.


Two points A and B are inside a convex 12- gon. Show that if the sum of the distances from A to each vertex is a and the sum of the distances from B to each vertex is b, then \(\mathrm{la} - \mathrm{bl} < 10\) |AB|.


There are 30 cups each containing milk. An elf is able to transfer milk from one cup to another so that the amount of milk in the two cups is equalised. Is there an initial distribution of milk so that the elf cannot equalise the amount in all the cups by a finite number of such transfers?


A \(99 \times 100\) chess board is colored in the usual way with alternate squares black and white. What fraction of the main diagonal is black? What if the board is \(99 \times 101\) ?


\(\mathrm{A_1A_2\ldots A_n}\) is a regular n- gon and P is an arbitrary point in the plane. Show that if n is even we can choose signs so that the vector sum \(\pm \mathrm{PA}_1\pm \mathrm{PA}_2\pm \ldots \pm \mathrm{PA}_\mathrm{n} = 0\) , but if n is odd, then this is only possible for finitely many points P.


A 1 or a - 1 is put into each cell of an n x n array as follows. A - 1 is put into each of the cells around the perimeter. An unoccupied cell is then chosen arbitrarily. It is given the product of the four cells which are closest to it in each of the four directions. For example, if the cells below containing a number or letter (except x) are filled and we decide to fill x next, then x gets the product of a, b, c and d.

- 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 a 1 -1 c x d -1 -1 -1 b -1 -1 -1 -1

What is the minimum and maximum number of 1s that can be obtained?


Prove that \(\mathrm{Isin} 1\mathrm{l} + \mathrm{Isin} 2\mathrm{l} + \ldots + \mathrm{Isin} 3\mathrm{nl} > 8\mathrm{n} / 5\)


Let S be the set of all numbers which can be written as 1/mn, where m and n are positive integers not exceeding 1986. Show that the sum of the elements of S is not an integer.


The incircle of a triangle has radius 1. It also lies inside a square and touches each side of the square. Show that the area inside both the square and the triangle is at least 3.4. Is it at least 3.5?


How many polynomials \(\mathrm{p(x)}\) have all coefficients 0, 1, 2 or 3 and take the value n at \(\mathrm{x} = 2\) ?


A and B are fixed points outside a sphere S. X and Y are chosen so that S is inscribed in the tetrahedron ABXY. Show that the sum of the angles AXB, XBY, BYA and YAX is independent of X and Y.


Ten players play in a tournament. Each pair plays one match, which results in a win or loss. If the ith player wins \(\mathbf{a}_{\mathrm{i}}\) matches and loses \(\mathbf{b}_{\mathrm{i}}\) matches, show that \(\sum \mathbf{a}_{\mathrm{i}}^{2} = \sum \mathbf{b}_{\mathrm{i}}^{2}\) .


Find all sets of 6 weights such that for each of \(\mathrm{n} = 1,2,3,\ldots ,63\) , there is a subset of weights weighing n.


ABCDEFG is a regular 7- gon. Prove that \(1 / \mathrm{AB} = 1 / \mathrm{AC} + 1 / \mathrm{AD}\) .


Your opponent has chosen a 1 x 4 rectangle on a 7 x 7 board. At each move you are allowed to ask whether a particular square of the board belongs to his rectangle. How many questions do you need to be certain of identifying the rectangle. How many questions are needed for a 2 x 2 rectangle?


An L is an arrangement of 3 adjacent unit squares formed by deleting one unit square from a 2 x 2 square. How many Ls can be placed on an 8 x 8 board (with no interior points overlapping)? Show that if any one square is deleted from a 1987 x 1987 board, then the remaining squares can be covered with Ls (with no interior points overlapping).


Squares ABC'C", BCA'A", CAB'B" are constructed on the outside of the sides of the triangle ABC. The line A'A" meets the lines AB and AC at P and P'. Similarly, the line B'B" meets the lines BC and BA at Q and Q', and the line C'C" meets the lines CA and CB at R and R'. Show that P, P', Q, Q', R and R' lie on a circle.


\(\mathrm{A_1,A_2,\ldots,A_{2m + 1}}\) and \(\mathbf{B}_1,\mathbf{B}_2,\ldots ,\mathbf{B}_{2n + 1}\) are points in the plane such that the \(2\mathrm{m} + 2\mathrm{n} + 2\) lines \(\mathrm{A_1A_2,A_2A_3,\ldots,A_{2m}A_{2m + 1},A_{2m + 1}A_1,B_1B_2,B_2B_3,\ldots,B_{2n}B_{2n + 1},B_{2n + 1}B_1}\) are all different and no three of them are concurrent. Show that we can find i and j such that \(\mathrm{A_iA_{i + 1}}\) \(\mathrm{B_iB_{j + 1}}\) are opposite sides of a convex quadrilateral (if \(\mathrm{i} = 2\mathrm{m} + 1\) , then we take \(\mathrm{A_{i + 1}}\) to be \(\mathrm{A_1}\) . Similarly for \(\mathrm{j} = 2\mathrm{n} + 1\) ).


Find 5 different relatively prime numbers, so that the sum of any subset of them is composite.


ABCDE is a convex pentagon with \(\angle ABC = \angle ADE\) and \(\angle AEC = \angle ADB\) . Show that \(\angle BAC = \angle DAE\) .


Show that there is a real number x such that all of cos x, cos 2x, cos 4x, ... cos \((2^{n}x)\) are negative.


The positive reals a, b, c, x, y, z satisfy a + x = b + y = c + z = k. Show that ax + by + cz ≤ k².


A real number with absolute value at most 1 is put in each square of a 1987 x 1987 board. The sum of the numbers in each 2 x 2 square is 0. Show that the sum of all the numbers does not exceed 1987.


AB is a chord of the circle center O. P is a point outside the circle and C is a point on the chord. The angle bisector of APC is perpendicular to AB and a distance d from O. Show that BC = 2d.


Players take turns in choosing numbers from the set {1, 2, 3, ..., n}. Once m has been chosen, no divisor of m may be chosen. The first player unable to choose a number loses. Who has a winning strategy for n = 10? For n = 1000?


What is the smallest number of subsets of S = {1, 2, ..., 33}, such that each subset has size 9 or 10 and each member of S belongs to the same number of subsets?


Some lattice points in the plane are marked. S is a set of non- zero vectors. If you take any one of the marked points P and add place each vector in S with its beginning at P, then more vectors will have their ends on marked points than not. Show that there are an infinite number of points.


A convex pentagon is cut along all its diagonals to give 11 pieces. Show that the pieces cannot all have equal areas.


The set \(\mathrm{S}_0 = \{1,2!4!8!16!,\ldots \}\) . The set \(\mathrm{S}_{\mathrm{n + 1}}\) consists of all finite sums of distinct elements of \(\mathrm{S}_{\mathrm{n}}\) . Show that there is a positive integer not in \(\mathrm{S}_{1987}\) .


If the graph of the function \(\mathrm{f} = \mathrm{f}(\mathrm{x})\) is rotated through 90 degrees about the origin, then it is not changed. Show that there is a unique solution to \(\mathrm{f}(\mathrm{b}) = \mathrm{b}\) . Give an example of such a function.


A convex polyhedron has all its faces triangles. Show that it is possible to color some edges red and the others blue so that given any two vertices one can always find a path between them along the red edges and another path between them along the blue edges.


Show that \((2\mathrm{n} + 1)^{\mathrm{n}}\geq (2\mathrm{n})^{\mathrm{n}} + (2\mathrm{n} - 1)^{\mathrm{n}}\) for every positive integer n.


A book contains 30 stories. Each story has a different number of pages under 31. The first story starts on page 1 and each story starts on a new page. What is the largest possible number of stories that can begin on odd page numbers?

Solution:
Answer: 23.

Call stories with an odd number of pages odd stories and stories with an even number of pages even stories. There are 15 odd stories and 15 even stories. The odd stories change the parity of the starting page (in the sense that the following story starts on a page of opposite parity), whereas the even stories do not. So the odd stories must start alternately on odd and even pages. Hence 8 of them must start on odd pages and 7 on even pages (irrespective of how the stories are arranged). We can, however, control the even stories. In particular, if we put each of them after an even number of odd stories, then they will all begin on odd pages. For example, we could put them all first (before any of the odd stories).


ABCD is a convex quadrilateral. The midpoints of the diagonals and the midpoints of AB and CD form another convex quadrilateral Q. The midpoints of the diagonals and the midpoints of BC and CA form a third convex quadrilateral Q'. The areas of Q and Q' are equal. Show that either AC or BD divides ABCD into two parts of equal area.

Solution:
Note that Q is a parallelogram because each side is formed by joining the midpoints of two sides of a triangle, so it is parallel to and half the length of the base of the triangle. But the triangles corresponding to opposite sides have the same base. Hence opposite sides of Q are parallel and equal. Similarly Q'.

Let the midpoints of the diagonals be X, Y. Take two adjacent side midpoints which are on the same side of the line XY. Suppose they are M, the midpoint of AB, and N, the midpoint of BC. Suppose also that X is the midpoint of BD, and Y the midpoint of AC. If X does not lie on AC, then we may assume it lies on the same side of AC as M and N (if not just consider the other two midpoints instead of M and N). So the line parallel to XY through M cuts the altitude from N of NXY. So XYM has the same base XY as XYN, but smaller height, so it has smaller area. Hence the two parallelograms also have different areas. Contradiction. So X must lie on AC. But AX bisects ABD and CX bisects CBD, so AC bisects ABD and CBD and hence ABCD.


Show that there are infinitely many triples of distinct positive integers a, b, c such that each divides the product of the other two and a + b = c + 1.

Solution: \[\mathrm{n(n + 1), n(n^2 + n - 1), (n + 1)(n^2 + n - 1).}\]


Given a sequence of 19 positive integers not exceeding 88 and another sequence of 88 positive integers not exceeding 19. Show that we can find two subsequences of consecutive terms, one from each sequence, with the same sum.

Solution:
We prove the general case. Let the first sequence be \(\mathbf{a}_1,\mathbf{a}_2,\ldots ,\mathbf{a}_{\mathrm{m}}\) and the second sequence be \(\mathbf{b}_1,\mathbf{b}_2,\ldots ,\mathbf{b}_{\mathrm{n}}\) , were \(0< \mathbf{a}_{\mathrm{i}}\leq \mathbf{n}\) and \(0< \mathbf{b}_{\mathrm{j}}\leq \mathbf{m}\) . Put \(\mathbf{s}_{\mathrm{k}} = \mathbf{a}_{1} + \mathbf{a}_{2} + \ldots +\mathbf{a}_{\mathrm{k}}\) , \(\mathbf{t}_{\mathrm{k}} = \mathbf{b}_{1} + \mathbf{b}_{2} + \ldots +\mathbf{b}_{\mathrm{k}}\) . Assume \(\mathbf{s}_{\mathrm{m}} > \mathbf{t}_{\mathrm{n}}\) (if they are equal, then we are done).

Let \(\mathbf{f}(\mathbf{i})\) be the smallest \(\mathbf{k}\) such that \(\mathbf{s}_{\mathrm{k}} > = \mathbf{t}_{\mathrm{i}}\) . If it is equal, we are done, so assume \(\mathbf{s}_{\mathrm{k}} > \mathbf{t}_{\mathrm{i}}\) . Now consider the n numbers \(\mathbf{s}_{\mathrm{f}(\mathrm{i})} - \mathbf{t}_{\mathrm{i}}\) . Each is at least 1 and at most n- 1 (if it was n or more then \(\mathbf{s}_{\mathrm{f}(\mathrm{i}) - 1} \geq \mathbf{s}_{\mathrm{f}(\mathrm{i})} - \mathbf{n} \geq \mathbf{t}_{\mathrm{i}}\) , contradicting the minimality of \(\mathbf{f}(\mathrm{i})\) ). So there must be two the same. So we have \(\mathbf{s}_{\mathrm{f}(\mathrm{i})} - \mathbf{t}_{\mathrm{i}} = \mathbf{s}_{\mathrm{f}(\mathrm{i})} - \mathbf{t}_{\mathrm{j}}\) for some \(\mathrm{i} > \mathrm{j}\) and hence \(\mathbf{a}_{\mathrm{f}(\mathrm{j}) + 1} + \mathbf{a}_{\mathrm{f}(\mathrm{j}) + 2} + \ldots + \mathbf{a}_{\mathrm{f}(\mathrm{i})} = \mathbf{b}_{\mathrm{j} + 1} + \mathbf{b}_{\mathrm{j} + 2} + \ldots + \mathbf{b}_{\mathrm{i}}\) .


The quadrilateral ABCD is inscribed in a fixed circle. It has AB parallel to CD and the length AC is fixed, but it is otherwise allowed to vary. If h is the distance between the midpoints of AC and BD and k is the distance between the midpoints of AB and CD, show that the ratio h/k remains constant.

Solution: Let the center of the circle be O and its radius be R. Let \(\angle \mathrm{AOB} = 2\mathrm{x}\) (variable) and let \(\angle \mathrm{AOC} = 2\mathrm{y}\) (fixed). Then \(\mathrm{AC} = 2\mathrm{R}\) sin y. We find \(\angle \mathrm{COD} = 180^{\circ} - \mathrm{x} - 2\mathrm{y}\) , so \(2\mathrm{h} = 2\mathrm{R}\sin \mathrm{x} + 2\mathrm{R}\) \(\sin (\mathrm{x} + 2\mathrm{y})\) . Angle \(\mathrm{ACD} = \mathrm{x} + \mathrm{y}\) , so \(\mathrm{k} = \mathrm{AB}\sin (\mathrm{x} + \mathrm{y}) = 2\mathrm{R}\sin (\mathrm{x} + \mathrm{y})\) sin y. Hence the ratio \(\mathrm{h / k} = (\sin \mathrm{x} + \sin (\mathrm{x} + 2\mathrm{y})) / (2\sin (\mathrm{x} + \mathrm{y})\sin \mathrm{y})\) . We have \(\sin \mathrm{x} = \sin (\mathrm{x} + \mathrm{y} - \mathrm{y}) = \sin (\mathrm{x} + \mathrm{y})\cos \mathrm{y} - \cos (\mathrm{x} + \mathrm{y})\sin \mathrm{y}\) , and \(\sin (\mathrm{x} + 2\mathrm{y}) = \sin (\mathrm{x} + \mathrm{y} + \mathrm{y}) = \sin (\mathrm{x} + \mathrm{y})\cos \mathrm{y} + \cos (\mathrm{x} + \mathrm{y})\sin \mathrm{y}\) , so \((\sin \mathrm{x} + \sin (\mathrm{x} + 2\mathrm{y}) = 2\sin (\mathrm{x} + \mathrm{y})\cos \mathrm{y}\) . Hence \(\mathrm{h / k} = \cot \mathrm{y}\) , which is constant.


The numbers 1 and 2 are written on an empty blackboard. Whenever the numbers m and n appear on the blackboard the number \(\mathrm{m} + \mathrm{n} + \mathrm{mn}\) may be written. Can we obtain (1) 13121, (2) 12131?

Solution:
(1) \(13121 = 2 + 4373 + 2\) . \(4373,4373 = 2 + 1457 + 2\) . \(1457,1457 = 2 + 485 + 2\) . \(485,485 = 2+\) \(161 + 2\) . \(161,161 = 2 + 53 + 2\) . \(53,53 = 2 + 17 + 2\) . \(17,17 = 2 + 5 + 2\) . \(5,5 = 2 + 1 + 2\) . 1.

Put \(\mathrm{M} = \mathrm{m} + 1\) , \(\mathrm{N} = \mathrm{n} + 1\) . Then we the number derived from m and n is MN - 1. So if M and N are of the form \(2^{a}3^{b}\) then so is MN. Thus we can only ever write up numbers of the form \(2^{a}3^{b} - 1\) . But \(12131 = 2^{2}3^{2}337 - 1\) , which is not of the required form.

Note that it makes no difference whether the two numbers m, n are allowed to be the same (which is ambiguous).


If rationals x, y satisfy \(\mathbf{x}^{5} + \mathbf{y}^{5} = 2\mathbf{x}^{2}\mathbf{y}^{2}\) show that 1 - x y is the square of a rational.

Solution: Put \(\mathbf{y} = \mathbf{k}\mathbf{x}\) , then \(\mathbf{x}^{5}(1 + \mathbf{k}^{5}) = 2\mathbf{k}^{2}\mathbf{x}^{4}\) , so \(\mathbf{x} = 2\mathbf{k}^{2} / (1 + \mathbf{k}^{5})\) , \(\mathbf{y} = 2\mathbf{k}^{3} / (1 + \mathbf{k}^{5})\) and \(1 - \mathbf{x}\mathbf{y} = (1-\) \(\mathbf{k}^{5})^{2} / (1 + \mathbf{k}^{5})^{2}\) . x and y are rational, so \((1 - \mathbf{k}^{5}) / (1 + \mathbf{k}^{5})\) is rational.


There are 21 towns. Each airline runs direct flights between every pair of towns in a group of five. What is the minimum number of airlines needed to ensure that at least one airline runs direct flights between every pair of towns?

Solution:
Answer: 21.

There are 210 pairs of towns. Each airline serves 10 pairs, so we certainly need at least 21 airlines. The following arrangement shows that 21 is possible:

1	2	3	4	5

1	6	7	8	9

1	10	11	12	13

1	14	15	16	17

1	18	19	20	21

2	6	10	14	18

2	7	11	15	19

2	8	12	16	20

2	9	13	17	21

3	6	11	16	21

3	7	10	17	20

3	8	13	14	19

3	9	12	15	18

4	6	12	17	19

4	7	13	16	18

4	8	10	15	21

4	9	11	14	20

5	6	13	15	20

5	7	12	14	21

5	8	11	17	18

5	9	10	16	19


Find all positive integers n satisfying \((1 + 1 / \mathrm{n})^{n + 1} = (1 + 1 / 1998)^{1998}\) .

Solution:
Answer: no solutions.

We have \((1 + 1 / \mathrm{n})^{\mathrm{n} + 1} > \mathrm{e} > (1 + 1 / \mathrm{n})^{\mathrm{n}}\) .


A, B, C are the angles of a triangle. Show that \(2(\sin \mathrm{A}) / \mathrm{A} + 2(\sin \mathrm{B}) / \mathrm{B} + 2(\sin \mathrm{C}) / \mathrm{C} \leq (1 / \mathrm{B} + 1 / \mathrm{C}) \sin \mathrm{A} + (1 / \mathrm{C} + 1 / \mathrm{A}) \sin \mathrm{B} + (1 / \mathrm{A} + 1 / \mathrm{B}) \sin \mathrm{C}\) .

Solution: Assume \(\mathrm{A} \leq \mathrm{B} \leq \mathrm{C}\) . Then \(\sin \mathrm{A} \leq \sin \mathrm{B}\) . Also \(\mathrm{A} \leq \mathrm{C} < 180^{\circ} - \mathrm{A}\) , so \(\sin \mathrm{A} \leq \sin \mathrm{C}\) . Similarly \(\sin \mathrm{B} \leq \sin \mathrm{C}\) . Hence \((1 / \mathrm{A} - 1 / \mathrm{B})(\sin \mathrm{B} - \sin \mathrm{A})\) , \((1 / \mathrm{B} - 1 / \mathrm{C})(\sin \mathrm{C} - \sin \mathrm{B})\) and \((1 / \mathrm{C} - 1 / \mathrm{A})(\sin \mathrm{A} - \sin \mathrm{C})\) are all non- negative. Hence their sum is also non- negative, which gives the result.


Form 10A has 29 students who are listed in order on its duty roster. Form 10B has 32 students who are listed in order on its duty roster. Every day two students are on duty, one from form 10A and one from form 10B. Each day just one of the students on duty changes and is replaced by the following student on the relevant roster (when the last student on a roster is replaced he is replaced by the first). On two particular days the same two students were on duty. Is it possible that starting on the first of these days and ending the day before the second, every pair of students (one from 10A and one from 10B) shared duty exactly once?

Solution:
Answer: no.

Suppose such an arrangement is possible. Suppose that it includes m cycles through the form 10A roster and n cycles through the 10B roster. Then the total number of changes is \(29\mathrm{m}+\) \(32\mathrm{n} = 29\mathrm{x}32\) (since each pair occurs once). But that means 29 divides n and 32 divides m. Both m and n are at least 1, so that means \(\mathrm{n}\geq 29\) and \(\mathrm{m}\geq 32\) , but then \(29\mathrm{m} + 32\mathrm{n} > 29\) . 32. Contradiction.


In the triangle ABC, the angle C is obtuse and D is a fixed point on the side BC, different from B and C. For any point M on the side BC, different from D, the ray AM intersects the circumcircle S of ABC at N. The circle through M, D and N meets S again at P, different from N. Find the location of the point M which minimises MP.

Solution: Take A' on the circle S such that AA' is parallel to BC. Let the ray AD meet S again at P'. Then \(\angle \mathrm{MNP}' = \angle \mathrm{ANP}'\) (same angle) \(= \angle \mathrm{AA'P}'\) (A'AP'N cyclic) \(= \angle \mathrm{A'DB}\) (BC parallel to AA') \(=\) angle MDP (opposite angles). So MDNP' is cyclic, so P must be P'. Since P is a fixed point, independent of M, we minimise MP by taking M as the foot of the perpendicular from P to BC.


Solution:
We show that infinitely many odd numbers in the sequence are divisible by 3.

If \(\mathrm{n} = 14\) mod 36, then \(\mathrm{n} = 36\mathrm{m} + 14\) for some m. So there are \(18\mathrm{m} + 7\) odd numbers in the sum and \(18\mathrm{m} + 7\) even numbers. Hence the sum is odd. There are \(12\mathrm{m} + 5\) numbers equal to 1 mod 3, \(12\mathrm{m} + 5\) equal to - 1 mod 3 and \(12\mathrm{m} + 4\) equal to 0 mod 3. Any product of numbers equal to 1 mod 3 equals 1 mod 3, so if \(\mathrm{k} = 1\) mod 3, then \(\mathrm{k}^{\mathrm{k}} = 1\) mod 3. Similarly, if \(\mathrm{k} = 0\) mod 3, then \(\mathrm{k}^{\mathrm{k}} = 0\) mod 3. If \(\mathrm{k} = - 1\) mod 3 then \(\mathrm{k}^{\mathrm{k}} = - 1\) mod 3 if \(\mathrm{k}\) is odd and 1 mod if \(\mathrm{k}\) is even. Of the \(12\mathrm{m} + 5\) numbers equal to - 1 mod 3, \(6\mathrm{m} + 3\) are even and \(6\mathrm{m} + 2\) are odd. Hence the sum \(= (12\mathrm{m} + 5)\cdot 1 + (6\mathrm{m} + 3)\cdot 1 + (6\mathrm{m} + 2)\cdot - 1 + (12\mathrm{m} + 4)\cdot 0 = 12\mathrm{m} + 6 = 0\) mod 3. So the sum is divisible by 3.


ABC is an acute- angled triangle. The tangents to the circumcircle at A and C meet the tangent at B at M and N. The altitude from B meets AC at P. Show that BP bisects the angle MPN.

Solution:
If the tangent at B is parallel to AC, then angle NBC \(=\) angle BCA (parallel lines) and angle NBC \(=\) angle BAC (NB tangent), so BCA is isosceles and \(\mathrm{BC} = \mathrm{BA}\) . Hence the figure is symmetrical about the line PB and so BP bisects MPN.

So assume AC is not parallel to the tangent at B. Assume it meets it at L on the same side of B as N. Take A' on the line AC so that \(\mathrm{MA} = \mathrm{MA}'\) . We show that LMA' and LNC are similar. Obviously the angles at L are the same. \(\angle \mathrm{MA'L} = \angle \mathrm{MAL}\) ( \(\mathrm{MA} = \mathrm{MA}'\) ) \(= \angle \mathrm{ABC}\) (MA tangent) \(= \angle \mathrm{LCN}\) (NC tangent). So the triangles are similar. Hence \(\mathrm{LN} / \mathrm{NC} = \mathrm{LM} / \mathrm{MA}'\) . But \(\mathrm{NC} = \mathrm{NB}\) and \(\mathrm{MA}' = \mathrm{MA} = \mathrm{MB}\) , so \(\mathrm{LN} / \mathrm{NB} = \mathrm{LM} / \mathrm{MB}\) and hence \(\mathrm{LM} / \mathrm{LN} = \mathrm{MB} / \mathrm{NB}\) . So the circle on LB as diameter has all points Q on it satisfying \(\mathrm{QM} / \mathrm{QN} = \mathrm{BM} / \mathrm{BN}\) . But \(\angle \mathrm{LPB} = 90^{\circ}\) , so P must lie on the circle and hence \(\mathrm{PM} / \mathrm{PN} = \mathrm{BM} / \mathrm{BN}\) . Hence PB is the angle bisector of MPN.


What is the minimal value of \(\mathrm{b} / (\mathrm{c} + \mathrm{d}) + \mathrm{c} / (\mathrm{a} + \mathrm{b})\) for positive real numbers b and c and nonnegative real numbers a and d such that \(\mathrm{b} + \mathrm{c} \geq \mathrm{a} + \mathrm{d}\) ?

Solution:
Answer: \(\sqrt{2} - 1 / 2\) .

Obviously \(\mathrm{a} + \mathrm{d} = \mathrm{b} + \mathrm{c}\) at the minimum value, because increasing a or d reduces the value. So we may take \(\mathrm{d} = \mathrm{b} + \mathrm{c} - \mathrm{a}\) . We also take \(\mathrm{b} \gg \mathrm{c}\) (interchanging b and c if necessary). Dividing through by \(\mathrm{b} / 2\) shows that there is no loss of generality in taking \(\mathrm{b} = 2\) , so \(0 < \mathrm{c} \ll = 2\) . Thus we have to find the minimum value of \(2 / (2\mathrm{c} - \mathrm{a} + 2) + \mathrm{c} / (\mathrm{a} + 2)\) . We show that it is \(\sqrt{2} - 1 / 2\) .

This is surprisingly awkward. Note first that \((\mathrm{c} - (\mathrm{h} - \mathrm{k}))^{2} > = 0\) , so \(\mathrm{c}^{2} + \mathrm{c}(2\mathrm{k} - 2\mathrm{h}) + \mathrm{h}^{2} - 2\mathrm{h}\mathrm{k} + \mathrm{k}^{2} > = 0\) . Hence \(\mathrm{c}^{2} + \mathrm{ck} + \mathrm{h}^{2} > = (2\mathrm{h} - \mathrm{k})(\mathrm{c} + \mathrm{k})\) . Hence \(\mathrm{c} / \mathrm{h}^{2} + 1 / (\mathrm{c} + \mathrm{k}) > = 2 / \mathrm{h} - \mathrm{k} / \mathrm{h}^{2}\) with equality iff \(\mathrm{c} = \mathrm{h} - \mathrm{k}\) . Applying this to \(\mathrm{c} / (\mathrm{a} + 2) + 1 / (\mathrm{c} + 1 - \mathrm{a} / 2)\) where \(\mathrm{h} = \sqrt{(\mathrm{a} + 2)}\) , \(\mathrm{k} = 1 - \mathrm{a} / 2\) , we find that \(\mathrm{c} / (\mathrm{a} + 2) + 2 / (2\mathrm{c} + 2 - \mathrm{a}) > = 2 / \sqrt{(\mathrm{a} + 2)} + (\mathrm{a} - 2) / (2\mathrm{a} + 4)\) .

The allowed range for c is \(0\leq \mathrm{c}\leq 2\) and \(0\leq \mathrm{a}\leq \mathrm{c} + 2\) , hence \(0\leq \mathrm{a}\leq 4\) . Put \(\mathrm{x} = 1 / \sqrt{(\mathrm{a} + 2)}\) , so \(1 / \sqrt{6} < = \mathrm{x}< = 1 / \sqrt{2}\) . Then \(2 / \sqrt{(\mathrm{a} + 2)} + (\mathrm{a} - 2) / (2\mathrm{a} + 4) = 2\mathrm{x} + 1 / 2 - 2\mathrm{x}^{2} = 1 - (2\mathrm{x} - 1)^{2} / 2\) . We have - \(0.184 = (2 / \sqrt{6} - 1)\leq 2\mathrm{x} - 1\leq \sqrt{2} - 1 = 0.414\) . Hence \(\mathrm{c} / (\mathrm{a} + 2) + 2 / (2\mathrm{c} + 2 - \mathrm{a})\geq 1 - (\sqrt{2} - 1)^{2} / 2 = \sqrt{2} - 1 / 2\) .

We can easily check that the minimum is achieved at \(\mathrm{b} = 2\) , \(\mathrm{c} = \sqrt{2} - 1\) , \(\mathrm{a} = 0\) , \(\mathrm{d} = \sqrt{2} + 1\) .


\(\mathrm{n}^{2}\) real numbers are written in a square \(\mathrm{n}\times \mathrm{n}\) table so that the sum of the numbers in each row and column equals zero. A move is to add a row to one column and subtract it from another (so if the entries are \(\mathrm{a_{ij}}\) and we select row i, column h and column k, then column h becomes \(\mathrm{a_{1h} + a_{i1}}\) , \(\mathrm{a_{2h} + a_{i2}}\) , ..., \(\mathrm{a_{nh} + a_{in}}\) , column k becomes \(\mathrm{a_{1k} - a_{i1}}\) , \(\mathrm{a_{2k} - a_{i2}}\) , ..., \(\mathrm{a_{nk} - a_{in}}\) , and the other entries are unchanged). Show that we can make all the entries zero by a series of moves.


In the acute- angled triangle ABC, the altitudes BD and CE are drawn. Let F and G be the points of the line ED such that BF and CG are perpendicular to ED. Prove that \(\mathrm{EF} = \mathrm{DG}\) .

Solution: \(\angle \mathrm{BDC} = \angle \mathrm{BEC} = 90^{\circ}\) , so BCDE is cyclic, so \(\angle \mathrm{BDE} = \angle \mathrm{BCE} = 90^{\circ} - \angle \mathrm{B}\) . Hence \(\angle \mathrm{DGC} = 90^{\circ} - \angle \mathrm{CDG} = \angle \mathrm{BDE} = 90^{\circ} - \mathrm{B}\) . So \(\mathrm{DG} = \mathrm{CD}\) sin \(\mathrm{DGC} = \mathrm{BC}\) sin CBD sin \(\mathrm{DGC} = \mathrm{BC}\) cos C cos B. Similarly EF.


Find the minimum value of \(\mathrm{xy / z + yz / x + zx / y}\) for positive reals \(\mathrm{x}\) , \(\mathrm{y}\) , \(\mathrm{z}\) with \(\mathrm{x}^{2} + \mathrm{y}^{2} + \mathrm{z}^{2} = 1\) .

Solution:
Answer: min \(\sqrt{3}\) when all equal.

Let us consider \(\mathrm{z}\) to be fixed and focus on \(\mathrm{x}\) and \(\mathrm{y}\) . Put \(\mathrm{f(x,y,z) = xy / z + yz / x + zx / y}\) . We have \(\mathrm{f(x,y,z) = p / z + z(1 - z^{2}) / p = (p + k^{2} / p) / z}\) , where \(\mathrm{p} = \mathrm{xy}\) , and \(\mathrm{k} = z\sqrt{(1 - z^{2})}\) . Now \(\mathrm{p}\) can take any value in the range \(0< \mathrm{p}\leq (1 - z^{2}) / 2\) . The upper limit is achieved when \(\mathrm{x} = \mathrm{y}\) .

We have \(\mathrm{p} + \mathrm{k}^{2} / \mathrm{p} = (\mathrm{p} - \mathrm{k})^{2} / \mathrm{p}\) . For \(\mathrm{p}\leq \mathrm{k}\) , \((\mathrm{p} - \mathrm{k})\) and \(1 / \mathrm{p}\) are both decreasing functions of \(\mathrm{p}\) , so \(\mathrm{p} + \mathrm{k}^{2} / \mathrm{p}\) is a decreasing function of \(\mathrm{p}\) . Thus if \(\mathrm{p}\) is restricted to the interval \((0,\mathrm{h}]\) , then for \(\mathrm{k}\leq\) \(\mathrm{h}\) the minimum value of \(\mathrm{p} + \mathrm{k}^{2} / \mathrm{p}\) is \(2\mathrm{k}\) and occurs at \(\mathrm{p} = \mathrm{k}\) . For \(\mathrm{k}\geq \mathrm{h}\) the minimum is \(\mathrm{h} + \mathrm{k}^{2} / \mathrm{h}\) and occurs at \(\mathrm{p} = \mathrm{h}\) .

We have \(\mathrm{h} = (1 - \mathrm{z}^{2}) / 2\) , \(\mathrm{k} = z\sqrt{(1 - \mathrm{z}^{2})}\) . So \(\mathrm{k}\leq \mathrm{h}\) iff \(\mathrm{z}\leq = 1 / \sqrt{5}\) . So if \(\mathrm{z}\leq 1 / \sqrt{5}\) , then \(\mathrm{f(x,y,z)}\geq 2\mathrm{k} / \mathrm{z}\) \(= 2\sqrt{(1 - \mathrm{z}^{2})}\geq 27\sqrt{(1 - 1 / 5)} = 4 / \sqrt{5} >\sqrt{3}\) .

If \(z > 1 / \sqrt{5}\) , then the minimum of \(\mathrm{f}(\mathrm{x}, \mathrm{y}, \mathrm{z})\) occurs at \(\mathrm{x} = \mathrm{y}\) and is \(\mathrm{x}^{2} / \mathrm{z} + \mathrm{z} + \mathrm{z} = (1 - \mathrm{z}^{2}) / (2\mathrm{z}) + 2\mathrm{z} = 3\mathrm{z} / 2 + 1 / (2\mathrm{z}) = (\sqrt{3}) / 2 (\mathrm{z} \sqrt{3} + 1 / (\mathrm{z} \sqrt{3}) \geq \sqrt{3}\) with equality at \(\mathrm{z} = 1 / \sqrt{3}\) (and hence \(\mathrm{x} = \mathrm{y} = 1 / \sqrt{3}\) also).


A polygonal line connects two opposite vertices of a cube with side 2. Each segment of the line has length 3 and each vertex lies on the faces (or edges) of the cube. What is the smallest number of segments the line can have?

Answer 6

Solution:
![md5:8cf996e790873a262ab31b6ce4f02ca0](8cf996e790873a262ab31b6ce4f02ca0.jpeg)

Suppose one endpoint of a segment length 3 is at A. Evidently the other end could be at the edge midpoints B, C, D. It could also be on the circular arc connecting B and C (with center O and radius \(\sqrt{5}\) ). Similarly, it could be on arcs connecting C and D, or B and D. We claim that if X is a point of one of these arcs other than its endpoints, then the only possible segment length 3 with an endpoint at X (and the other endpoint on the surface of the cube) is AX. wlog we can consider X to be on the arc BC. Take axes with origin O, so that A is (0,0,2). Suppose X is (a,b,0) and that the other endpoint of the segment is Y (x,y,z). Then \(\mathrm{XY}^{2} = (\mathrm{x} - \mathrm{a})^{2} + (\mathrm{y} - \mathrm{b})^{2} + \mathrm{z}^{2} = \mathrm{a}^{2} + \mathrm{b}^{2} + \mathrm{z}^{2} - \mathrm{x}(2\mathrm{a} - \mathrm{x}) - \mathrm{y}(2\mathrm{b} - \mathrm{y}) = 5 + \mathrm{z}^{2} - \mathrm{x}(2\mathrm{a} - \mathrm{x}) - \mathrm{y}(2\mathrm{b} - \mathrm{y})\) . But a, \(\mathrm{b} > 1\) since X is not an endpoint of the arc, so (2a- x) and (2b- y) are both positive. Hence - x(2a- x) - y(2b- y) \(\leq 0\) with equality iff \(\mathrm{x} = \mathrm{y} = 0\) . Similarly, \(\mathrm{z}^{2} \leq 4\) with equality iff \(\mathrm{z} = 2\) . Hence \(\mathrm{XY}^{2} \leq 9\) with equality iff \(\mathrm{Y} = \mathrm{A}\) , which proves the claim.

Thus if the next link of the polygonal line goes from A to anywhere except B, C, D, then it has to go back to A. So a minimal line must go to B, C, or D.

Now from D the line can only go to A or O. For if it goes to Z (x,y,z), then we have \(\mathrm{DZ}^{2} = (\mathrm{x} - 2)^{2} + (\mathrm{y} - 2)^{2} + (\mathrm{z} - 1)^{2} \leq 2^{2} + 2^{2} + 1^{2} = 3^{2}\) with equality iff \(\mathrm{x} = 0\) , \(\mathrm{y} = 0\) and \(\mathrm{z} = 0\) or 2.

So let us take A as the starting point of the polygonal line. wlog the first segment is AD. Then the second segment must be DO (for a minimal line). Thus the best we can do with 2 segments is to move along an edge. It takes three such moves to get to the opposite corner, and hence at least 6 segments. But it is obvious that it can be done with 6 segments.


Let m, n, k be positive integers with \(\mathrm{m} \geq \mathrm{n}\) and \(1 + 2 + \ldots + \mathrm{n} = \mathrm{mk}\) . Prove that the numbers 1, 2, ..., n can be divided into k groups in such a way that the sum of the numbers in each group equals m.

Solution:
Induction on n, then m. For \(\mathrm{n} = 1\) , 2 there is nothing to prove. Assume the result is proved for \(< \mathrm{n}\) and consider the case n. If n is odd, we have \(\mathrm{n} = \mathrm{n} - 1 + 1 = \mathrm{n} - 2 + 2 = \ldots = (\mathrm{n} + 1) / 2 + (\mathrm{n} - 1) / 2\) , so the result is true for \(\mathrm{m} = \mathrm{n}\) , \(\mathrm{k} = (\mathrm{n} + 1) / 2\) . If n is even, we have \(\mathrm{n} + 1 = \mathrm{n} - 1 + 2 = \ldots = (\mathrm{n} / 2 + 1) + (\mathrm{n} / 2 - 1)\) , so the result is true for \(\mathrm{m} = \mathrm{n} + 1\) and \(\mathrm{k} = \mathrm{n} / 2\) . Now suppose it is true for \(< \mathrm{m}\) .

If \(2\mathrm{n} > \mathrm{m} > \mathrm{n} + 1\) , then for m odd we can take the sums \(\mathrm{m} = \mathrm{n} + \mathrm{m} - \mathrm{n} = \mathrm{n} - 1 + \mathrm{m} - \mathrm{n} + 1 = \ldots = (\mathrm{m} + 1) / 2 + (\mathrm{m} - 1) / 2\) . These use up the numbers \(\mathrm{m} - \mathrm{n}\) , \(\mathrm{m} - \mathrm{n} + 1\) , ..., n and give some sums of m. By induction the remaining numbers 1, 2, ..., \(\mathrm{m} - \mathrm{n} - 1\) will give the remaining sums of m (obviously \(\mathrm{m} > \mathrm{m} - \mathrm{n} - 1\) ). If m is even, we can take the sums \(\mathrm{m} = \mathrm{n} + \mathrm{m} - \mathrm{n} = \mathrm{n} - 1 + \mathrm{m} - \mathrm{n} + 1 = \ldots = (\mathrm{m} / 2 + 1) + (\mathrm{m} / 2 - 1)\) . That gives some sums of m and leaves us with the integers 1, 2, ..., \(\mathrm{m} - \mathrm{n} - 1\) and \(\mathrm{m} / 2\) . But since \(\mathrm{m} < 2(\mathrm{n} + 1)\) , \(\mathrm{m} / 2 > \mathrm{m} - \mathrm{n} - 1\) and hence we can use the integers 1, 2, ..., \(\mathrm{m} - \mathrm{n} - 1\) to form sums of \(\mathrm{m} / 2\) . With the integer \(\mathrm{m} / 2\) that gives us sums of m (we know that the parity must come out right because we know that the sum of all the remaining numbers is divisible by m).

Finally, consider \(\mathrm{m} \geq 2\mathrm{n}\) . In that case we can form k sums of \(2\mathrm{n} - 2\mathrm{k} + 1\) : \(\mathrm{n} + (\mathrm{n} - 2\mathrm{k} + 1)\) , \((\mathrm{n} - 1) + (\mathrm{n} - 2\mathrm{k} + 2)\) , ..., \((\mathrm{n} - \mathrm{k} + 1) + (\mathrm{n} - \mathrm{k})\) . So we are home provided the remaining integers 1, 2, ..., \(\mathrm{n} - 2\mathrm{k}\) can be used to form k sums of \(\mathrm{m} - (2\mathrm{n} - 2\mathrm{k} + 1)\) . That follows by induction provided that \(\mathrm{m} - (2\mathrm{n} - 2\mathrm{k} + 1) \geq \mathrm{n} - 2\mathrm{k}\) , or \(\mathrm{m} + 4\mathrm{k} - 1 \geq 3\mathrm{n}\) , or \(\mathrm{m} + 2\mathrm{n}(\mathrm{n} + 1) / \mathrm{m} - 1 \geq 3\mathrm{n}\) or \((\mathrm{m} - 2\mathrm{n})(\mathrm{m} - \mathrm{n} - 1) \geq 0\) , which is true.