Given 12 vertices and 16 edges arranged as follows:

![md5:21705a320698046255d6dc2ebb5e015a](21705a320698046255d6dc2ebb5e015a.jpeg)

Draw any curve which does not pass through any vertex. Prove that the curve cannot intersect each edge just once. Intersection means that the curve crosses the edge from one side to the other. For example, a circle which had one of the edges as tangent would not intersect that edge.

Solution:
A- - - - B- - - - C  |  |  |  D- - E- - F- - G- - H  |  |  |  I- - J- - - - K- - L

If a curve intersects the boundary of a region R (such as ABFED), then it moves from inside R to outside or vice versa. Hence if R has an odd number of edges (like ABFED) then a curve intersecting all of them just once must have one endpoint inside R. But there are four such regions (ABFED, BCHGF, EFGKJ and the outside of ABCHLKJID) and only two endpoints. Note that we can easily intersect all edges but one. For example, start above AB, then cross successively AB, AD, DI, DE, EF, EJ, IJ, JK, GK, KL, HL, GH, CH, BC, FG.


Given a rectangle ABCD with AC length e and four circles centers A, B, C, D and radii a, b, c, d respectively, satisfying \(\mathrm{a + c = b + d< e}\) . Prove you can inscribe a circle inside the quadrilateral whose sides are the two outer common tangents to the circles center A and C, and the two outer common tangents to the circles center B and D.

Solution: Let O be the center of the rectangle. Let \(\mathrm{r} = (\mathrm{a} + \mathrm{c}) / 2 = (\mathrm{b} + \mathrm{d}) / 2\) . The required circle has center O, radius r. Let an outer common tangent touch the circle center A at W, and the circle center C at X. Let P be the midpoint of WX, then OP is parallel to AW and CX and has length r, hence the circle center O touches AW at P. Similarly for the other common tangents.


Prove that any 39 successive natural numbers include at least one whose digit sum is divisible by 11.

Solution: Let n be the smallest number in the sequence and m the smallest with last digit 0. m and m+10 have different digit sums unless (possibly) the penultimate digit of m is 9, but in that case m+10 and m+20 have different digit sums. So two of m, m+10, m+20 are sure to have different digit sums. Hence at least one has a digit sum not congruent to 1 mod 11. Adding the appropriate final digit gives a number whose digit sum is divisible by 11. This number lies in the range m to m+29 and m<=n+9. Hence the result. n=999981 shows it is best possible.


(a) Arrange 7 stars in the 16 places of a 4 x 4 array, so that no 2 rows and 2 columns contain all the stars.
(b) Prove this is not possible for <7 stars.

Solution:
(a) \(* * .\) \(* .* .\) \(* * .\) \(* .* .\)

Pick any two rows. The unpicked stars lie in different columns.

(b) If there is a row with at least 3 stars, pick it. That leaves at most 3 stars, pick the row for one and the columns for the others. Now assume no row has more than 2 stars. 6 stars in <6 rows, so we can pick a row with 2 stars. That leaves 4 stars in 3 rows, so we can pick another row with 2 stars. That leaves 2 stars. Pick their columns. [This glosses over the case of <6 stars. In this case we can add extra stars to make the number up to 6. Now the procedure above deals with the original stars and the extra stars, and in particular with the original stars.]


(a) Given a quadruple (a, b, c, d) of positive reals, transform to the new quadruple (ab, bc, cd, da). Repeat arbitarily many times. Prove that you can never return to the original quadruple unless a=b=c=d=1.
(b) Given n a power of 2, and an n-tuple (a₁, a₂, ..., aₙ) transform to a new n-tuple (a₁a₂, a₂a₃, ..., aₙ-₁aₙ, aₙa₁). If all the members of the original n-tuple are 1 or -1, prove that with sufficiently many repetitions you obtain all 1s.

Solution:
(a) Let \(Q_0\) be the original quadruple (a, b, c, d) and \(Q_n\) the quadruple after n transformations. If abcd>1, then the products form a strictly increasing sequence, so return is impossible. Similarly if abcd<1. So we must have abcd=1. Let the largest of the four values of a quadruple \(Q\) be \(M(Q)\) . If a member of \(Q_1\) is not 1, then \(M(Q_1) > 1\) . \(Q_3\) consists of the elements of \(Q_1\) squared and permuted, so \(M(Q_3) = M(Q_1)^2\) . Hence the sequence \(M(Q_1)\) , \(M(Q_3)\) , \(M(Q_5)\) , ... increases without limit. This means no return is possible, because a return would lead to the values cycling.

(b) After r<n transformations, the first number of the n-tuple is the product \(\mathbf{a}_{1}^{(\mathrm{rl}0)}\mathbf{a}_{2}^{(\mathrm{rl}1)}\ldots\) \(\mathbf{a}_{\mathrm{r}+1}^{(\mathrm{rl}\mathrm{r})}\) , where (rli) denotes the binomial coefficient. [This is an easy induction.] Hence after \(\mathrm{n = 2^{k}}\) transformations it is \(\mathbf{a}_{1}^{2}\) times the product \(\mathbf{a}_{2}^{(\mathrm{nl}1)}\ldots \mathbf{a}_{\mathrm{n}}^{(\mathrm{nl}1)}\) . So it is sufficient to prove that (nli) is even for n a power of 2 and \(0< \mathrm{i}< \mathrm{n}\) . But observe that (nli) = (n-1li) n/(n-i) and n is divisible by a higher power of 2 than n-i.


(a) A and B move clockwise with equal angular speed along circles center P and Q respectively. C moves continuously so that \(\mathrm{AB = BC = CA}\) . Establish C's locus and speed. \(\ast (\mathrm{b})\) ABC is an equilateral triangle and P satisfies \(\mathrm{AP} = 2\) , \(\mathrm{BP} = 3\) . Establish the maximum possible value of CP.

Solution:
(a) Represent A, B as complex numbers \(\mathrm{z}_{1} + \mathrm{w}_{1}\mathrm{e}^{\mathrm{i}}\) , \(\mathrm{z}_{2} + \mathrm{w}_{2}\mathrm{e}^{\mathrm{i}}\) . Then C is \((\mathrm{z}_{1} + \mathrm{w}_{1}\mathrm{e}^{\mathrm{i}}) + (\mathrm{z}_{2} + \mathrm{w}_{2}\mathrm{e}^{\mathrm{i}} - \mathrm{z}_{1} - \mathrm{w}_{1}\mathrm{e}^{\mathrm{i}})\mathrm{e}^{\mathrm{i}\pi /3}\) , which is also of the form \(\mathrm{z} + \mathrm{w}\mathrm{e}^{\mathrm{i}}\) .

However, there is one subtlety. There are actually two circles possible for C depending on which side of AB we place it. The continuity requirement means that C is normally confined to one of the circles. However, if A and B ever coincide then C may be able to switch to the other circle.

If we regard "moves continuously" as allowing a discontinuous velocity, then a switch is always possible (provided A and B coincide).

(b) Answer: 5.

P must be the opposite side of AB to C (or we could increase CP, whilst keeping AP and BP the same, by reflecting in AB). Similarly it must be on the same side of AC as B, and on the same side of BC as A. For any P in this region the quadrilateral APBC is convex and hence satisfies Ptolemy's inequality CP. \(\mathrm{AB}\leq \mathrm{AP}\) . BC + BP. AC, with equality iff APBC is cyclic. But \(\mathrm{AB} = \mathrm{BC} = \mathrm{CA}\) , so we have \(\mathrm{CP}\leq \mathrm{AP} + \mathrm{BP} = 5\) with equality iff P lies on the arc AB of the circle ABC. Note that there is just one such point, because the locus of P such that \(\mathrm{BP} = 1.5\) AP is a circle which cuts the arc just once.

Ptolemy's inequality for 4 points A, B, C, D: AB: \(\mathrm{CD} + \mathrm{BC}\) . \(\mathrm{AD}\geq \mathrm{AC}\) . BD with equality iff ABCD is a cyclic quadrilateral (meaning A, B, C, D lie on a circle in that order).

## Proof

Take E inside ABCD such that \(\angle \mathrm{DAE} = \angle \mathrm{CAB}\) and \(\angle \mathrm{ADE} = \angle \mathrm{ACB}\) . Then ADE and ACB are similar, so \(\mathrm{DE / CB} = \mathrm{AD / AC}\) and hence BC. \(\mathrm{AD} = \mathrm{AC}\) . DE. It also follows that AE/AB = AD/AC. But we also have \(\angle \mathrm{EAB} = \angle \mathrm{DAC}\) and hence AEB and ADC are also similar. So \(\mathrm{EB / AB} = \mathrm{DC / AC}\) , and hence AB: \(\mathrm{CD} = \mathrm{AC}\) . EB. Adding, we have: AB: \(\mathrm{CD} + \mathrm{BC}\) . \(\mathrm{AD} = \mathrm{AC}(\mathrm{BE} + \mathrm{ED})\geq \mathrm{AC}\) . BD with equality iff E lies on BD, or equivalently ABCD is cyclic.

This glosses over one point. It only follows that \(\angle \mathrm{EAB} = \angle \mathrm{DAC}\) if ABCD is convex. For the convex case, we have that \(\angle \mathrm{EAB} = \angle \mathrm{CAB} + \angle \mathrm{EAC}\) and \(\angle \mathrm{DAC} = \angle \mathrm{DAE} + \angle \mathrm{EAC}\) , or \(\angle \mathrm{EAB} = \angle \mathrm{CAB} - \angle \mathrm{EAC}\) and \(\angle \mathrm{DAC} = \angle \mathrm{DAE} - \angle \mathrm{EAC}\) . Either way \(\angle \mathrm{EAB} = \angle \mathrm{DAC}\) . But in the non- convex case, we can have \(\angle \mathrm{EAB} = \angle \mathrm{CAB} + \angle \mathrm{EAC}\) and \(\angle \mathrm{DAC} = \angle \mathrm{DAE} - \angle \mathrm{EAC}\) (or - ... +) and hence the angles \(\angle \mathrm{EAB}\) and \(\angle \mathrm{DAC}\) are not necessarily equal.


Given an m x n array of real numbers. You may change the sign of all numbers in a row or of all numbers in a column. Prove that by repeated changes you can obtain an array with all row and column sums non- negative.

Solution: The array has mn entries. Call an array that can be obtained by repeated changes a reachable array. A reachable array differs from the original only in that some or all of the signs of its mn entries may be different. There are at most 2 possibilities for each sign and hence at most \(2^{\mathrm{mn}}\) different reachable arrays. For each reachable array calculate the sum of all its entries. Take the reachable array with the largest such sum. It must have non- negative row and column sums, because if any such sum was negative, changing the sign of that row or column would give another reachable array with strictly greater total sum.


Given n<1 points, some pairs joined by an edge (an edge never joins a point to itself). Given any two distinct points you can reach one from the other in just one way by moving along edges. Prove that there are n- 1 edges.

Solution:
Every point must have at least one edge. We show that there is a point with just one edge. Suppose the contrary, that every point has at least two edges. We now construct a path in which the same edge or point never appears twice. Starting from any point b, move along an edge to c. c is not already on the path, because otherwise the edge would join b to itself. Now suppose we have reached a point x not previously on the path. x has at least two edges, so it must have another one besides the one we used to reach it. Suppose this joins x to y. If y is already on the path, then we have two distinct ways of moving along edges from x to y: directly, or by backtracking along the path from x to y. But this is impossible, so y is not already on the path and we may extend the path to it. But this procedure allows us to construct a path containing more than the n distinct points available. Contradiction.

The result is now easy. Induction on n. Take a point with just one edge. Remove it and the edge. Then the remaining n- 1 points satisfy the premise and hence have just n- 2 edges.


Given any natural numbers m, n and k. Prove that we can always find relatively prime natural numbers r and s such that rm+sn is a multiple of k.

Solution:
Care is needed. Although easy, this is more awkward than it looks.

Let \(\mathrm{d} = (\mathrm{m},\mathrm{n})\) , the greatest common divisor of m and n. Let \(\mathrm{r} = \mathrm{n} / \mathrm{d}\) , \(\mathrm{s} = \mathrm{nhk} - \mathrm{m} / \mathrm{d}\) , where h is any integer sufficiently large to ensure that \(\mathrm{s} > 0\) . Now \(\mathrm{rm} + \mathrm{sn} = \mathrm{mn} / \mathrm{d} + \mathrm{nnhk} - \mathrm{mn} / \mathrm{d} = \mathrm{nnhk}\) , which is a multiple of k. If e divides r, then it also divides \(\mathrm{rdhk} = \mathrm{nhk}\) . So if e divides r and s, then it

also divides s - nhk = - m/d. But n/d and m/d are relatively prime, so e must be 1. Hence r and s are relatively prime.


A and B play the following game with N counters. A divides the counters into 2 piles, each with at least 2 counters. Then B divides each pile into 2 piles, each with at least one counter. B then takes 2 piles according to a rule which both of them know, and A takes the remaining 2 piles. Both A and B make their choices in order to end up with as many counters as possible. There are 3 possibilities for the rule:

\(R1\) B takes the biggest heap (or one of them if there is more than one) and the smallest heap (or one of them if there is more than one).

\(R2\) B takes the two middling heaps (the two heaps that A would take under \(R1\) ).

\(R3\) B has the choice of taking either the biggest and smallest, or the two middling heaps. For each rule, how many counters will A get if both players play optimally?

Solution:
Answers: [N/2], [(N+1)/2], [N/2].

Suppose A leaves piles n, m with n≤m.

Under \(R1\) , B can certainly secure m by dividing the larger pile into 1 and m- 1. He cannot do better, because if b is the biggest of the 4 piles, then the smallest is at most m- b. Hence A's best strategy is to leave [N/2], [(N+1)/2].

Under \(R2\) , if A leaves a=2, b=N- 2, then B cannot do better than [N/2], because if he divides the larger pile into a,b with a≤b, then he takes a+1. A cannot do better, because if he leaves a,b with 3≤a≤b, then B can divide to leave 1, a- 1, [b/2], [(b+1)/2]. Now if a- 1≥[(b+1)/2], then B takes b≥[(N+1)/2]. If a- 1<[(b+1)/2], then B takes a- 1+[b/2]. But a- 1≥2 and [b/2]≥[(b+1)/2]- 1, so a- 1+[b/2]≥1+[(b+1)/2], or B takes at least as many as A, so B takes at least [(N+1)/2].

Under \(R3\) , A's best strategy is to divide into [N/2],[(N+1)/2]. We have already shown that B can secure [(N+1)/2] and no more by following \(R1\) . He cannot do better under R2, for if he divides so that the biggest pile comes from [N/2], then the smallest does too and so he gets [(N+1)/2]. If he divides so that the biggest and smallest piles come from [(N+1)/2], then he gets only [N/2]. But one of these must apply, because if he divided so that the smaller from [N/2] was smaller than the smaller from [(N+1)/2], and the bigger from [N/2] was smaller than the bigger from [(N+1)/2], then [N/2] would be at least 2 less than [(N+1)/2] (which it is not).


Given three arbitary infinite sequences of natural numbers, prove that we can find unequal natural numbers m, n such that for each sequence the mth member is not less than the nth member.

Solution:
Given any infinite sequence of natural numbers we can find a non- decreasing subsequence (proof below). So suppose the three sequences are a, b, and c. Take a non- decreasing

subsequence of \(\mathbf{a}_i\) . Suppose it is \(\mathbf{a}_{i1}\) , \(\mathbf{a}_{i2}\) , \(\mathbf{a}_{i3}\) , ... . Now consider the infinite sequence \(\mathbf{b}_{11}\) , \(\mathbf{b}_{12}\) , ... . It must have a non- decreasing subsequence. Suppose it is \(\mathbf{b}_{11}\) , \(\mathbf{b}_{12}\) , ... . Now consider the infinite sequence \(\mathbf{c}_{j1}\) , \(\mathbf{c}_{j2}\) , ... . It must have a non- decreasing subsequence \(\mathbf{c}_{k1}\) , \(\mathbf{c}_{k2}\) , ... . Each of the three sub- sequences \(\mathbf{a}_{k1}\) , \(\mathbf{a}_{k2}\) , ... , \(\mathbf{b}_{k1}\) , \(\mathbf{b}_{k2}\) , ... , \(\mathbf{c}_{k1}\) , \(\mathbf{c}_{k2}\) , ... is non- decreasing. So we may take, for example, \(\mathrm{m} = \mathrm{k}2\) and \(\mathrm{n} = \mathrm{k}1\) .

[Proof that any infinite sequence of natural numbers has a non- decreasing subsequence: if the original sequence is unbounded, then we can take a strictly increasing subsequence. If not, then since there are only finitely many possible numbers not exceeding the bound, at least one of them must occur infinitely often.]


120 unit squares are arbitrarily arranged in a \(20 \times 25\) rectangle (both position and orientation is arbitrary). Prove that it is always possible to place a circle of unit diameter inside the rectangle without intersecting any of the squares.

Solution:
If a circle with unit diameter intersects a unit square, then its center must lie inside an area \(3 + \pi /4\) , namely an oval centered on the square and comprising: the original square, area 1; four \(1 \times 1 / 2\) rectangles on the sides, total area 2; and four quarter circles at the corners, total area \(\pi /4\) . So if it does not intersect any of the 120 unit squares, then it must avoid ovals with a total area of \(120 \times (3 + \pi /4) = 454.2\) . Of course, for many arrangements of the squares, these ovals might overlap substantially, but the worst case would be no overlap.

The circle is also required to lie inside the rectangle, so its center must lie outside a strip \(1 / 2\) wide around the edge, and hence inside an inner \(19 \times 24\) rectangle, area 456. The total area of ovals is less, so they cannot cover it completely and it must be possible to place a circle as required.


ABCD is any convex quadrilateral. Construct a new quadrilateral as follows. Take A' so that A is the midpoint of DA'; similarly, B' so that B is the midpoint of AB'; C' so that C is the midpoint of BC'; and D' so that D is the midpoint of CD'. Show that the area of A'B'C'D' is five times the area of ABCD.

Solution: Compare the triangles A'B'A and ADB. The base of A'B'A can be taken as A'A, which is the same length as AD. The height of A'B'A is AB' times sin B'A'A', which is twice AB times sin BAD. So area A'B'A = 2 area ADB. Similarly, area B'C'B = 2 area BAC, area C'D'C = 2 area CBD, and area D'A'D = 2 area DCA. So adding, the area A'B'A + area C'D'C = 2 area ABCD, and area B'C'B + area D'A'D = 2 area ABCD. But ABCD = A'B'A + B'C'B + C'D'C + D'A'D + ABCD. Hence result.


Given a fixed circle C and a line L throught the center O of C. Take a variable point P on L and let K be the circle center P through O. Let T be the point where a common tangent to C and K meets K. What is the locus of T?

Solution: Let the common tangent meet C at S. Let X be the intersection of C and OP lying between O and P. PT = PO, hence \(\angle \mathrm{POT} = \angle \mathrm{PTO}\) , so \(\angle \mathrm{OPT} = 180^{\circ} - 2\angle \mathrm{POT}\) . But PT and OS are parallel, because both are perpendicular to the common tangent. Hence \(\angle \mathrm{POS} = 2\angle \mathrm{POT}\) , so \(\angle \mathrm{SOT} = \angle \mathrm{XOT}\) . Hence TX is tangent to C, in other words T lies on the (fixed) tangent to C at X. Conversely, it is easy to see that any such point can be obtained (just take P such that PO = PT). Thus the required locus is the pair of tangents to C which are perpendicular to L.


Solution: An easy induction gives \(\mathrm{a_r} = (2^{\mathrm{r}} - 1)\mathrm{a_1} - (2^{\mathrm{r}} - 2)\mathrm{a_0}\) for \(\mathrm{r} = 2\) , 3, ..., 100. Hence, in particular, \(\mathrm{a_{100} = (2^{100} - 2)(a_1 - a_0) + a_1}\) . But \(\mathrm{a_1}\) and \((\mathrm{a_1} - \mathrm{a_0})\) are both at least 1. Hence result.


Prove that there are no integers a, b, c, d such that the polynomial \(\mathrm{ax^3 + bx^2 + cx + d}\) equals 1 at \(\mathrm{x = 19}\) and 2 at \(\mathrm{x = 62}\) .

Solution: If there were such values, then subtract the equation with \(\mathrm{x} = 19\) from the equation with \(\mathrm{x} = 62\) to get: \(\mathrm{a}(62^{3} - 19^{3}) + \mathrm{b}(62^{2} - 19^{2}) + \mathrm{c}(62 - 19) = 1\) . But the left hand side is divisible by 62 - \(19 = 43\) , contradiction.


Given an n x n array of numbers. n is odd and each number in the array is 1 or - 1. Prove that the number of rows and columns containing an odd number of - 1s cannot total n.

Solution: If we change a - 1 to 1, we affect the total number of rows and columns (containing an odd number of - 1s) by 0, 2 or - 2. After changing all the - 1s we have total of 0. Hence the starting total must be even. So it cannot be n.


Given the lengths AB and BC and the fact that the medians to those two sides are perpendicular, construct the triangle ABC.

Solution: Let M be the midpoint of AB and X the midpoint of MB. Construct the circle center B, radius BC/2 and the circle diameter AX. If they do not intersect (so BC<AB/2 or BC>AB) then the construction is not possible. If they intersect at N, then take C so that N is the midpoint of BC. Let CM meet AN at O. Then AO/AN = AM/AX =2/3, so the triangles AOM and ANX are similar. Hence \(\angle AOM = \angle ANX = 90^{\circ}\) .


Given four positive real numbers a, b, c, d such that \(\mathrm{abcd} = 1\) , prove that \(\mathrm{a}^{2} + \mathrm{b}^{2} + \mathrm{c}^{2} + \mathrm{d}^{2} + \mathrm{ab} + \mathrm{ac} + \mathrm{ad} + \mathrm{bc} + \mathrm{bd} + \mathrm{cd} \geq 10\) .

Solution: Applying the arithmetic/geometric mean result to the 10 numbers gives the result immediately.


Given a fixed regular pentagon ABCDE with side 1. Let M be an arbitary point inside or on it. Let the distance from M to the closest vertex be \(\mathbf{r}_1\) , to the next closest be \(\mathbf{r}_2\) and so on, so that the distances from M to the five vertices satisfy \(\mathbf{r}_1 \leq \mathbf{r}_2 \leq \mathbf{r}_3 \leq \mathbf{r}_4 \leq \mathbf{r}_5\) . Find (a) the locus of M which gives \(\mathbf{r}_3\) the minimum possible value, and (b) the locus of M which gives \(\mathbf{r}_3\) the maximum possible value.

Solution:
Let X be the midpoint of AB and O the center of ABCDE. Suppose M lies inside AXO. Then \(\mathrm{ME} = \mathbf{r}_3\) . So we maximise \(\mathbf{r}_3\) by taking M at X, with distance 1.5590, and we minimise \(\mathbf{r}_3\) by taking M as the intersection of AO and EB with distance 0.8090. AXO is one of 10 congruent

areas, so the required loci are (a) the 5 midpoints of the diagonals, and (b) the 5 midpoints of the sides.


Given a number with 1998 digits which is divisible by 9. Let x be the sum of its digits, let y be the sum of the digits of x, and z the sum of the digits of y. Find z.

Solution: \(\mathrm{x}\leq 9\) . \(1998 = 17982\) . Hence \(\mathrm{y}\leq\) the greater of \(1 + 7 + 9 + 9 + 9 = 35\) and \(9 + 9 + 9 + 9 = 36\) . But 9 divides the original number and hence also x, y and z. Hence \(\mathrm{z} = 9\) .


\(\mathrm{AB} = \mathrm{BC}\) and M is the midpoint of AC. H is chosen on BC so that MH is perpendicular to BC. P is the midpoint of MH. Prove that AH is perpendicular to BP.

Solution: Take X on AH so that BX is perpendicular to AH. Extend to meet HM at P'. Let N be the midpoint of AB. A, B, M and X are on the circle center N radius NA (because angles AMB and AXB are 90). Also MN is parallel to BC (because AMN, ACB are similar), so NM is perpendicular to MH, in other words HM is a tangent to the circle. hence P'M. \(\mathrm{P'M} = \mathrm{P'X}\) . P'B. Triangles P'XH and P'HB are similar (angles at P' same and both have a right angle), so \(\mathrm{P'H / P'X} = \mathrm{P'B / P'H}\) , so \(\mathrm{P'H}\) . \(\mathrm{P'H} = \mathrm{P'X}\) . \(\mathrm{P'B}\) . Hence \(\mathrm{P'H} = \mathrm{P'M}\) and P' coincides with P.


The triangle ABC satisfies \(0\leq \mathrm{AB}\leq 1\leq \mathrm{BC}\leq 2\leq \mathrm{CA}\leq 3\) . What is the maximum area it can have?

Solution: If we ignore the restrictions of CA, then the maximum area is 1, achieved when AB is perpendicular to BC. But in this case CA satisfies the restrictions.


Given unequal integers x, y, z prove that \(\left(\mathrm{x - y}\right)^{5} + \left(\mathrm{y - z}\right)^{5} + \left(\mathrm{z - x}\right)^{5}\) is divisible by \(5(\mathrm{x - y})(\mathrm{y - z})(\mathrm{z - x})\) .

Solution: Put \(\mathrm{x - y = r}\) , \(\mathrm{y - z = s}\) . Then \(\mathrm{z - x} = - (\mathrm{r + s})\) , and \(\left(\mathrm{x - y}\right)^{5} + \left(\mathrm{y - z}\right)^{5} + \left(\mathrm{z - x}\right)^{5} = \mathrm{r}^{5} + \mathrm{s}^{5} - (\mathrm{r + s})^{5} = - 5\mathrm{r}^{4}\mathrm{s} - 10\mathrm{r}^{3}\mathrm{s}^{2} - 10\mathrm{r}^{2}\mathrm{s}^{3} - 5\mathrm{rs}^{4} = - 5\mathrm{rs}(\mathrm{r + s})(\mathrm{r}^{2} + \mathrm{rs} + \mathrm{s}^{2})\) .


Given \(\mathrm{a_0}\) , \(\mathrm{a_1}\) , ..., \(\mathrm{a_n}\) , satisfying \(\mathrm{a_0} = \mathrm{a_n} = 0\) , and and \(\mathrm{a_{k - 1}} - 2\mathrm{a_k} + \mathrm{a_{k + 1}}\geq 0\) for \(\mathrm{k = 0}\) , 1, ..., n- 1. Prove that all the numbers are negative or zero.


Solution: The essential point is that if we plot the values \(\mathbf{a}_{\mathrm{r}}\) against r, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it. There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine: \(\mathbf{a}_{\mathrm{r} + 1} - \mathbf{a}_{\mathrm{r}} \geq \mathbf{a}_{\mathrm{r}} - \mathbf{a}_{\mathrm{r} - 1}\) . Hence (easy induction) if \(\mathbf{a}_{\mathrm{s}} - \mathbf{a}_{\mathrm{s} - 1} > 0\) , then \(\mathbf{a}_{\mathrm{n}} > \mathbf{a}_{\mathrm{s}}\) . Take \(\mathbf{a}_{\mathrm{s}}\) to be the first positive, then certainly \(\mathbf{a}_{\mathrm{s}} > \mathbf{a}_{\mathrm{s} - 1}\) , so \(\mathbf{a}_{\mathrm{n}} > 0\) . Contradiction.


Given two sets of positive numbers with the same sum. The first set has m numbers and the second n. Prove that you can find a set of less than m+n positive numbers which can be arranged to part fill an m x n array, so that the row and column sums are the two given sets.

Example: row sums 1, 5, 3; column sums 2, 7. Array is:

x5 x1 21

Solution:
Induction on m+n. Trivial for m+n=2.

Let \(\mathbf{x}\) be the largest number in the two given sets. Suppose it is a row total; let y be the largest column total. If \(\mathbf{y}< \mathbf{x}\) , then replace \(\mathbf{x}\) by \(\mathbf{x} - \mathbf{y}\) in the set of row totals and remove y from the col totals. By induction find \(\leq = \mathrm{m} + \mathrm{n} - 2\) positive numbers in an m x (n- 1) array with the new totals. Adding a col empty except for y in the row totalling \(\mathbf{x} - \mathbf{y}\) gives the required original set.

If \(\mathbf{y} = \mathbf{x}\) , then drop \(\mathbf{x}\) from the row totals and y from the col totals and argue as before.

If \(\mathbf{x}\) was a col total we interchange rows and cols in the argument.


Given 5 circles. Every 4 have a common point. Prove that there is a point common to all 5.

Solution: Let the circles be a, b, c, d, e. Let A be a point common to b, c, d, e, let B be a point common to a, c, d, e and so on. If any two of A, B, C, D, E coincide then the coincident point is on all 5 circles. Suppose they are all distinct. Then A, B, C are on d and e. Hence d and e coincide (3 points determine a circle). Hence D is on all 5 circles.


8 players compete in a tournament. Everyone plays everyone else just once. The winner of a game gets 1, the loser 0, or each gets 1/2 if the game is drawn. The final result is that everyone gets a different score and the player placing second gets the same as the total of the four bottom players. What was the result of the game between the player placing third and the player placing seventh?

Solution: The bottom 4 played 6 games amongst themselves, so their scores must total at least 6. Hence the number 2 player scored at least 6. The maximum score possible is 7, so if the number 2 player scored more than 6, then he must have scored 6 1/2 and the top player 7. But then the top player must have won all his games, and hence the number 2 player lost at least one game and could not have scored 6 1/2. Hence the number 2 player scored exactly 6, and the bottom 4 players lost all their games with the top 4 players. In particular, the number 3 player won against the number 7 player.


(a) The two diagonals of a quadrilateral each divide it into two parts of equal area. Prove it is a parallelogram.

(b) The three main diagonals of a hexagon each divide it into two parts of equal area. Prove they have a common point. [If ABCDEF is a hexagon, then the main diagonals are AD, BE and CF.]

Solution:
(a) Let the quadrilateral be ABCD and let the diagonals AC, BD meet at E. Then area ABC = AC.EB.sin CEB/2, and area ADC = AC.ED.sin CEB/2, so E is the midpoint of BD. Similarly, it is the midpoint of AC. Hence the triangles AEB and CED are congruent, so angle CDE = angle ABE, and hence AB is parallel to CD. Similarly, AD is parallel to BC.

(b) Let the hexagon be ABCDEF. Let BE, CF meet at J, let AD, CF meet at K, and let AD, BE meet at L. Let AK=a, BJ=b, CJ=c, DL=d, EL=e, FK=f. Also let KL=x, JL=y and JK=z. Consider the pair of diagonals AD, BE. They divide the hexagon into 4 parts: the triangles

ALB and DLE, and the quadrilaterals AFEL and BCDL. Since area ALB + area AFEL = area DLE + area BCDL, and area ALB + area BCDL = area DLE + area AFEL, the two triangles must have the same area (add the two inequalities). But area ALB = 1/2 AL.BL.sin ALB, and area DLE = 1/2 DL.EL.sin DLE = 1/2 DL.EL.sin ALB, so AL.BL = DL.EL or de = (a+x)(b+y). Similarly, considering the other two paris of diagonals, we get bc = (e+y)(f+z) and af = (c+z)(d+x). Multiplying the three inequalities gives: abcdef = (a+f)(b+y)(c+z)(d+x)(e+y)(f+z). But x, y, z are non- negative, so they must be zero and hence the three diagonals pass through a common point.


The natural numbers m and n are relatively prime. Prove that the greatest common divisor of \(\mathrm{m} + \mathrm{n}\) and \(\mathrm{m}^2 +\mathrm{n}^2\) is either 1 or 2.

Solution: If d divides \(\mathrm{m} + \mathrm{n}\) and \(\mathrm{m}^2 +\mathrm{n}^2\) , then it also divides \((\mathrm{m} + \mathrm{n})^2 - (\mathrm{m}^2 +\mathrm{n}^2) = 2\mathrm{mn}\) and hence also \(2\mathrm{m}(\mathrm{m} + \mathrm{n}) - 2\mathrm{mn} = 2\mathrm{m}^2\) and \(2\mathrm{n}(\mathrm{m} + \mathrm{n}) - 2\mathrm{mn} = 2\mathbf{n}^2\) . But m and n are relatively prime, so \(\mathrm{m}^2\) and \(\mathrm{n}^2\) are also. Hence d must divide 2.


Given a circle c and two fixed points A, B on it. M is another point on c, and K is the midpoint of BM. P is the foot of the perpendicular from K to AM.

(a) prove that KP passes through a fixed point (as M varies);
(b) find the locus of P.

Solution:
(a) Take Y on the circle so that angle ABY=90. Then AY is a diameter and so angle AMY=90. Take X as the midpoint of BY. Then triangles BXK and BYM are similar, so XK is parallel to YM. Hence XK is perpendicular to AM, and so P is the intersection of XK and AM. In other words, KP always passes through X.

(b) P must lie on the circle diameter AX, and indeed all such points can be obtained (given a point P on the circle, take M as the intersection of AP and the original circle). So the locus of P is the circle diameter AX.


Find the smallest value x such that, given any point inside an equilateral triangle of side 1, we can always choose two points on the sides of the triangle, collinear with the given point and a distance x apart.

Solution:
Answer: 2/3.

Let O be the center of ABC. Let AO meet BC at D, let BO meet CA at E, and let CO meet AB at F. Given any point X inside ABC, it lies in one of the quadrilaterals AEOF, CDOE, BFOD. Without loss of generality, it lies in AEOF. Take the line through X parallel to BC. It meets

AB in P and AC in Q. Then PQ is shorter than the parallel line MON with M on AB and N on AC, which has length 2/3. If we twist the segment PXQ so that it continues to pass through X, and P remains on AB and Q on AC, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than 2/3. So at some intermediate position its length will be 2/3.

To show that no value smaller than 2/3 is possible, it is sufficient to show that any segment POQ with P and Q on the sides of the triangle has length at least 2/3. Take P on MB and Q on AN with P, O, Q collinear. Then PQ cos POM = MN - QN cospi/3 + PM cos pi/3. But PM>QN (using the sine rule, PM = OM sinPOM/sinOPM and QN = ON sinQON/sinOQN, but OM=ON, angle POM = angle QON, and angle OQN = angle OPM + pi/3 > angle OPM), and hence PQ > MN sec POM > MN.


(a) A \(6 \times 6\) board is tiled with \(2 \times 1\) dominos. Prove that we can always divide the board into two rectangles each of which is tiled separately (with no domino crossing the dividing line).

#(b) Is this true for an \(8 \times 8\) board?

Solution:
(a) We say a domino bridges two columns if half the domino is in each column. We show that for \(0 < n < 6\) the number of dominoes bridging columns n and \(n+1\) must be at least 2 and even.

Consider first \(n = 1\) . There cannot be 3 dominos entirely in column 1, or it would be separately tiled. So there must be at least one domino bridging columns 1 and 2. The number must be even, because it must equal the number of squares in column 1 (even) less twice the number of dominoes (entirely) in column 1.

Now suppose it is true for \(n < 5\) and consider column \(n + 1\) . There must be at least one domino bridging columns \(n + 1\) and \(n + 2\) , or columns 1 thru \(n + 1\) would be separately tiled. The number must be even, because it must equal the number of squares in column \(n + 1\) (even) less the number bridging n and \(n + 1\) (even) less twice the number entirely in the column.

So in total there are at least \(5 \times 2 = 10\) dominos bridging columns. By the same argument there are at least another 10 bridging rows, but there are only 18 dominoes in total.

(b) No. For example:

1 2 3 3 1 1 2 2 1 2 1 2 2 3 3 1 3 3 1 3 1 2 4 1 1 2 2 3 1 2 4 3 1 3 3 2 2 1 2 3 3 2 1 1 4 1 2 1 3 2 3 2 4 3 3 1 1 1 3 2 1 1 2 2


Given a set of n different positive reals \(\{a_{1}, a_{2}, \ldots , a_{n}\}\) . Take all possible non- empty subsets and form their sums. Prove we get at least \(\mathrm{n(n + 1)} / 2\) different sums.

Solution:
Assume \(a_{1} < a_{2} < \ldots < a_{n}\) . We have the following collection of increasing sums:

\(a_{1} < a_{2} < \ldots < a_{n}\) n sums \(a_{1} + a_{n} < a_{2} + a_{n} < \ldots < a_{n - 1} + a_{n}\) n- 1 sums \(a_{1} + a_{n - 1} + a_{n} < a_{2} + a_{n - 1} + a_{n} < \ldots < a_{n - 2} + a_{n - 1} + a_{n}\) n- 2 sums

\(a_{1} + a_{2} + \ldots + a_{n}\) 1 sum

A total of \(1 + 2 + \ldots + n = \mathrm{n(n + 1)} / 2\) .


Given a triangle ABC. Let the line through C parallel to the angle bisector of B meet the angle bisector of A at D, and let the line through C parallel to the angle bisector of A meet the angle bisector of B at E. Prove that if DE is parallel to AB, then \(\mathrm{CA} = \mathrm{CB}\) .

Solution:
The idea is to find an expression for the perpendicular distance h from D to AB. Let \(\gamma = \angle ACB\) , \(\alpha = \frac{1}{2}\angle CAB\) , and \(\beta = \frac{1}{2}\angle ABC\) . We have \(\mathrm{h} = \mathrm{AP}\) sin \(\alpha\) .

Using the sine rule on APC, we have \(\mathrm{AP} = \mathrm{AC} \sin (\gamma + \beta) / \sin (\alpha + \beta)\) , so \(\mathrm{h} = \mathrm{AC} \sin \alpha \sin (\gamma + \beta) / \sin (\alpha + \beta)\) . Similarly, the perpendicular distance k from E to AB is BC sin \(\beta \sin (\gamma + \alpha) / \sin (\alpha + \beta)\) .

We also have that \(\mathrm{AC} / \mathrm{BC} = \sin 2\beta /\sin 2\alpha\) , and hence \(\mathrm{h} / \mathrm{k} = \sin 2\beta \sin \alpha \sin (\gamma + \beta) / (\sin 2\alpha \sin \beta \sin (\gamma + \alpha))\) . Using the fact that \(\sin (\gamma + \beta) = \sin (2\alpha + \beta)\) , and the expression for \(\sin 20\) , we get \(\mathrm{h} / \mathrm{k} = (\sin (2\alpha + 2\beta) + \sin 2\alpha) / (\sin (2\alpha + 2\beta) + \sin 2\beta)\) and hence \(\mathrm{h} = \mathrm{k}\) iff the triangle is isosceles.

For some reason the geometric solution took me longer to find. Let ED meet BC at X. Then XCD and XBE are isosceles, so \(\mathrm{BC} = \mathrm{BX} + \mathrm{XC} = \mathrm{DX} + \mathrm{XE} = \mathrm{DE}\) . Similarly, \(\mathrm{AC} = \mathrm{DE}\) . Hence \(\mathrm{AC} = \mathrm{BC}\) .


An infinite arithmetic progression contains a square. Prove it contains infinitely many squares.

Solution: Let the square be \(a^{2}\) and the difference d, so that all numbers of the form \(a^{2} + \mathrm{nd}\) belong to the arithmetic progression (for n a natural number). Take n to be \(2a + \mathrm{dr}^{2}\) , then \(a^{2} + \mathrm{nd} = (a + \mathrm{dr})^{2}\) .


Can we label each vertex of a 45- gon with one of the digits 0, 1, ..., 9 so that for each pair of distinct digits i, j one of the 45 sides has vertices labeled i, j?

Solution: 10 x 5 > 45, so some digit \(\mathrm{i}_{0}\) must appear less than 5 times. But each occurrence can give at most 2 edges \(\mathrm{i}_{0}\) , j, so there are at most 8 edges \(\mathrm{i}_{0}\) , j, which is one too few.


Find all real p, q, a, b such that we have \((2\mathrm{x} - 1)^{20} - (\mathrm{ax} + \mathrm{b})^{20} = (\mathrm{x}^{2} + \mathrm{px} + \mathrm{q})^{10}\) for all x.

Solution:
Comparing coefficients of \(\mathrm{x}^{20}\) , we must have \(\mathrm{a} = (2^{20} - 1)^{1 / 20}\) (note that we allow either the positive or the negative root).

Set \(\mathrm{x} = 1 / 2\) . Then we must have \((\mathrm{ax} + \mathrm{b})^{20} = 0 = (\mathrm{x}^{2} + \mathrm{px} + \mathrm{q})^{10}\) , and hence \(\mathrm{ax} + \mathrm{b} = 0\) and \(\mathrm{x}^{2} + \mathrm{px} + \mathrm{q} = 0\) . So \(\mathrm{b} = - \mathrm{a} / 2\) , and \(1 / 4 + \mathrm{p} / 2 + \mathrm{q} = 0\) .

Set \(\mathrm{x} = 0\) . Then we get \(\mathrm{q}^{10} = 1 - \mathrm{b}^{20} = 1 / 2^{20}\) , so \(\mathrm{q} = 1 / 4\) or \(- 1 / 4\) , and \(\mathrm{p} = - 1\) or 0 respectively. Comparing the coefficients of \(\mathrm{x}^{19}\) , we must have \(\mathrm{p} = - 1\) and \(\mathrm{q} = 1 / 4\) . So, if there is a solution, then it must be: \(\mathrm{a} = (2^{20} - 1)^{1 / 20}\) , \(\mathrm{b} = - \mathrm{a} / 2\) , \(\mathrm{p} = - 1\) , \(\mathrm{q} = 1 / 4\) . This is indeed a solution because with these values, the \(\mathrm{lhs} = 2^{20}(\mathrm{x} - 1 / 2)^{20} - (\mathrm{x} - 1 / 2)^{20}\mathrm{a}^{20} = (\mathrm{x} - 1 / 2)^{20} = (\mathrm{x}^{2} - \mathrm{x} + 1 / 4)^{10} = \mathrm{rhs}\) .


We place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both 1. At step \(\mathrm{n} > 1\) , place a point at the midpoint of each arc created at step \(\mathrm{n} - 1\) and label it with the sum of the labels at the two adjacent points. What is the total sum of the labels after step \(\mathrm{n}\) ?

For example, after step 4 we have: 1, 4, 3, 5, 2, 5, 3, 4, 1, 4, 3, 5, 2, 5, 4, 3, 4.

Solution:
Answer: \(2.3^{\mathrm{n} - 1}\) .

True for \(\mathrm{n} = 1\) . The new points added at step \(\mathrm{n} + 1\) have twice the sum of the points after step \(\mathrm{n}\) , because each old point contributes to two new points. hence the total after step \(\mathrm{n} + 1\) is three times the total after step \(\mathrm{n}\) .


Given an isosceles triangle, find the locus of the point P inside the triangle such that the distance from P to the base equals the geometric mean of the distances to the sides.

Solution:
Let the triangle be ABC, with AB=AC. Take the circle through B and C which has AB and AC as tangents. The required locus is the arc BC.

Suppose P lies on the arc. Let the perpendiculars from P meet BC in L, AB in N and AC in M. Join PB and PC. The triangles PNB and PLC are similar (PNB and PLC are both 90, and NBP \(=\) LCP because NB is tangent to the circle). Hence \(\mathrm{PN / PL} = \mathrm{PB / PC}\) . Similarly, triangles PMC and PLB are similar and hence \(\mathrm{PM / PL} = \mathrm{PC / PB}\) . Multiplying gives the required result \(\mathrm{PL}^2 =\) PM.PN.

If P is inside the circle and not on it, take P' as the intersection of the line AP and the arc. We have \(\mathrm{PL}< \mathrm{PL}\) , but \(\mathrm{PM} > \mathrm{PM}\) and \(\mathrm{PN} > \mathrm{PN}\) , hence \(\mathrm{PL}^2 < \mathrm{PM.PN}\) . Similarly, if P is outside the circle and not on it, then \(\mathrm{PL}^2 > \mathrm{PM.PN}\) .


In the triangle ABC, the length of the altitude from A is not less than BC, and the length of the altitude from B is not less than AC. Find the angles.

Solution: Let k be twice the area of the triangle. Then \(\mathrm{k}\geq \mathrm{BC}^2\) , \(\mathrm{k}\geq \mathrm{AC}^2\) and \(\mathrm{k}\leq \mathrm{AC}.\mathrm{BC}\) , with equality in the last case only if AC is perpendicular to BC. Hence AC and BC have equal lengths and are perpendicular. So the angles are 90, 45, 45.


If m, k, n are natural numbers and \(\mathrm{n} > 1\) , prove that we cannot have \(\mathrm{m(m + 1)} = \mathrm{k^n}\) .

Solution: m and \(\mathrm{m + 1}\) have no common divisors, so each must separately be an nth power. But the difference between the two nth powers is greater than 1 (for \(\mathrm{n} > 1\) ).


Reduce each of the first billion natural numbers (billion \(= 10^9\) ) to a single digit by taking its digit sum repeatedly. Do we get more 1s than 2s?

Solution: Taking digit sums repeatedly gives the remainder after dividing the number by 9, or 9 if the number is exactly divisible by 9. \(10^9 - 1 = 9\mathrm{n}\) , and for any \(\mathrm{r} > = 0\) the nine consecutive numbers \(9\mathrm{r} + 1\) , \(9\mathrm{r} + 2\) , ..., \(9\mathrm{r} + 9\) include just one number giving remainder 1 and one number giving remainder 2. Hence the numbers up to \(10^9 - 1\) give equal numbers of 1s and 2s. \(10^9\) itself gives 1, so there is just one more of the 1s than the 2s.


Given n odd and a set of integers \(\mathrm{a_1}\) , \(\mathrm{a_2}\) , ..., \(\mathrm{a_n}\) , derive a new set \((\mathrm{a_1} + \mathrm{a_2}) / 2\) , \((\mathrm{a_2} + \mathrm{a_3}) / 2\) , ..., \((\mathrm{a_{n - 1}} + \mathrm{a_n}) / 2\) , \((\mathrm{a_n} + \mathrm{a_1}) / 2\) . However many times we repeat this process for a particular starting set we always get integers. Prove that all the numbers in the starting set are equal.

For example, if we started with 5, 9, 1, we would get 7, 5, 3, and then 6, 4, 5, and then 5, 4, 5, 5.5. The last set does not consist entirely of integers.

Solution:
Let the smallest value be s and suppose it occurs m times (with \(\mathrm{m}< \mathrm{n}\) ). Then the values in the next stage are all at least s, and at most \(\mathrm{m} - 1\) equal s. So after at most m iterations the smallest value is increased.

We can never reach a stage where all the values are equal, because if \((a_{1} + a_{2}) / 2 = (a_{2} + a_{3}) / 2 =\) ... \(= (a_{\mathrm{n - 1}} + a_{\mathrm{n}}) / 2 = (a_{\mathrm{n}} + a_{1}) / 2\) , then \(a_{1} + a_{2} = a_{2} + a_{3}\) and hence \(a_{1} = a_{3}\) . Similarly, \(a_{3} = a_{5}\) , and so \(a_{1} =\) \(a_{3} = a_{5} = \ldots = a_{\mathrm{n}}\) (n odd). Similarly, \(a_{2} = a_{4} = \ldots = a_{\mathrm{n - 1}}\) . But we also have \(a_{\mathrm{n}} + a_{1} = a_{1} + a_{2}\) and so \(a_{2} = a_{\mathrm{n}}\) , so that all \(a_{i}\) are equal. In other words, if all the values are equal at a particular stage, then they must have been equal at the previous stage, and hence at every stage.

Thus if the values do not start out all equal, then the smallest value increases indefinitely. But that is impossible, because the sum of the values is the same at each stage, and hence the smallest value can never exceed \((a_{1} + \ldots +a_{\mathrm{n}}) / \mathrm{n}\) .

Note that for n even the argument breaks down because a set of unequal numbers can iterate into a set of equal numbers. For example: 1, 3, 1, 3, ..., 1, 3.


The convex hexagon ABCDEF has all angles equal. Prove that \(\mathrm{AB - DE = EF - BC = CD - }\) FA.

(b) Given six lengths \(a_{1}\) , ..., \(a_{6}\) satisfying \(a_{1} - a_{4} = a_{5} - a_{2} = a_{3} - a_{6}\) , show that you can construct a hexagon with sides \(a_{1}\) , ..., \(a_{6}\) and equal angles.

Solution:
(a) Extend AB, CD, EF. We get an equilateral triangle with sides AF + AB + BC, BC + CD + DE, ED + EF + FA. Hence AB - DE = CD - FA = EF - BC, as required.

(b) Take an equilateral triangle with sides s, t, u lengths \(a_{2} + a_{3} + a_{4}\) , \(a_{4} + a_{5} + a_{6}\) , and \(a_{6} + a_{1} + a_{2}\) respectively. Construct BC length \(a_{2}\) parallel to t with B on u and C on s. Construct DE length \(a_{4}\) parallel to u with D on s and E on t. Construct FA length \(a_{6}\) parallel to s with F on t and A on u. Then ABCDEF is the required hexagon, with AB = \(a_{1}\) , BC = \(a_{2}\) etc.


Find all possible integer solutions for \(\mathrm{sqrt(x + sqrt(x\ldots(x + sqrt(x)) \ldots)) = y}\) , where there are 1998 square roots.

Solution:
Let \(s_{1} = \mathrm{sqrt(x)}\) , \(s_{2} = \mathrm{sqrt(x + s_{1})}\) , \(s_{3} = \mathrm{sqrt(x + s_{2})}\) and so on. So the equation given is \(y = s_{1998}\) . We show first that all \(s_{\mathrm{n}}\) must be integral for \(1 \ll n \ll 1998\) . \(y\) is integral, so \(s_{1998}\) is integral. Now suppose \(s_{\mathrm{n}}\) is integral. Then \(s_{\mathrm{n - 1}} = s_{\mathrm{n}}^{2} - x\) is integral, proving the claim.

So in particular \(s_{1}\) and \(s_{2}\) are integers and \(s_{2}^{2} = s_{1}^{2} + s_{1}\) . But if \(s_{1} > 0\) , then \(s_{1}^{2} < s_{1}^{2} + s_{1} < (s_{1} + 1)^{2}\) , which is impossible. Similarly \(s_{1} < 0\) is impossible. So the only possible solution is \(s_{1} = 0\) and hence \(x = 0\) and \(y = 0\) .


ABCD is a convex quadrilateral. A' is the foot of the perpendicular from A to the diagonal BD, B' is the foot of the perpendicular from B to the diagonal AC, and so on. Prove that A'B'C'D' is similar to ABCD.

Solution: Let the diagonals meet at O. Then \(\mathrm{CC^{\prime}O}\) is similar to AA'O (because \(\mathrm{CC^{\prime}O = AA^{\prime}O = 90}\) , and COC', AOA' are opposite angles), so \(\mathrm{A^{\prime}O / C^{\prime}O = AO / CO}\) . Similarly, \(\mathrm{B^{\prime}O / D^{\prime}O = BO / DO}\) . AA'O is also similar to BB'O, so \(\mathrm{A^{\prime}O / B^{\prime}O = AO / BO}\) . Thus \(\mathrm{OA^{\prime}:OB^{\prime}:OC^{\prime}:OD^{\prime} =}\) OA:OB:OC:OD. Hence triangles OA'B' and OAB are similar. Likewise OB'C' and OBC, OC'D' and OCD, and OD'A' and ODA. Hence result.


Find all natural numbers n such that \(\mathrm{n}^2\) does not divide n!.

Solution:
Answer: \(\mathrm{n} = 4\) or prime.

If \(\mathrm{n} = \mathrm{rs}\) , with \(1< \mathrm{r}< \mathrm{s}\) , then \(\mathrm{r}< \mathrm{s}< \mathrm{n}\) , and hence \(\mathrm{rsn} = \mathrm{n}^2\) divides n!. Similarly, if \(\mathrm{n} = \mathrm{r}^2\) with \(\mathrm{r} > 2\) , then \(\mathrm{r}< 2\mathrm{r}< \mathrm{n}\) , and hence \(\mathrm{n}^2\) divides n!. This covers all possibilities except \(\mathrm{n} = 4\) or \(\mathrm{n} =\) prime, and it is easy to see that in these cases \(\mathrm{n}^2\) does not divide n!.


Given a lattice of regular hexagons. A bug crawls from vertex A to vertex B along edges of the hexagons, taking the shortest possible path (or one of them). Prove that it travels a distance at least AB/2 in one direction. If it travels exactly AB/2 in one direction, how many edges does it traverse?

Solution:
![md5:675f77deace78f718b28d352bbac53d9](675f77deace78f718b28d352bbac53d9.jpeg)

Suppose vertex A is that marked \* at the bottom left. Without loss of generality, B is in a 60 degree sector as shown. Assume the edges have unit length. The vertices can be partitioned into two sets (marked ° and . in the diagram). Each set forms a skewed lattice with axes at 60 degrees. Any path must alternate between the two lattices.

If B is on the same lattice as A, then we can give B coordinates (m,n) relative to A and the shortest path from A to B must move m units east and n units east of north. The shortest path between a lattice point and the next lattice point east is evidently one edge in direction 3 followed by one edge in direction 2. Similarly, the shortest path between a lattice point and the next lattice point east of north is one edge in direction 1, followed by one edge in direction 2. So a shortest path from A to B must have \(\mathrm{m + n}\) edges in direction 2.

B is a distance \(\sqrt[3]{(m + n / 2)}\) east of A and a distance \(3\mathrm{n} / 2\) north of A, so \(\mathrm{AB}^2 = (3\mathrm{m}^2 + 3\mathrm{mn} + 3\mathrm{n}^2)\) \(< (4\mathrm{m}^2 + 8\mathrm{mn} + 4\mathrm{n}^2) = 4(\mathrm{m} + \mathrm{n})^2\) . So in this case the bug must travel more than AB/2 in direction 2.

Now suppose B is on the other lattice. Let C be the lattice point immediately north of A and D the lattice point in direction 3 from A. Then a shortest path from A to B must either be A to C and then a shortest path from C to B, or A to D and then a shortest path from D to B. Take B to have coordinates (m, n) relative to C or D.

In the first case, \(\mathrm{AB}^2 = (\sqrt{3} (\mathrm{m} + \mathrm{n} / 2))^2 + (3\mathrm{n} / 2 + 1)^2 = (3\mathrm{m}^2 + 3\mathrm{mn} + 3\mathrm{n}^2) + 3\mathrm{n} + 1\) and a shortest path has \(\mathrm{m} + \mathrm{n}\) units in direction 2. But \(4(\mathrm{m} + \mathrm{n})^2 > (3\mathrm{m}^2 + 3\mathrm{mn} + 3\mathrm{n}^2) + 3\mathrm{n} + 1\) , if \(\mathrm{m}^2 + \mathrm{n}^2 + 5\mathrm{mn} > 3\mathrm{n} + 1\) , which is true for m, n at least 1. If \(\mathrm{m} = 0\) and \(\mathrm{n} = 1\) , then a shortest path has 2 units in direction 1 and \(\mathrm{AB} = \sqrt{7} < 4\) . If \(\mathrm{m} = 1\) and \(\mathrm{n} = 0\) , then \(\mathrm{AB} = 2\) and a shortest path has 1 unit in each direction. So in this case (the only one so far) we have equality.

It remains to consider the case where the path starts out towards D. In this case \(\mathrm{AB}^2 = (\sqrt{3} (\mathrm{m} + \mathrm{n} / 2) + \sqrt{3} /2)^2 + (3\mathrm{n} / 2 - 1 / 2)^2 = (3\mathrm{m}^2 + 3\mathrm{mn} + 3\mathrm{n}^2) + 3\mathrm{m} + 1\) and a path has \(\mathrm{m} + \mathrm{n}\) units in direction 2. But \(4(\mathrm{m} + \mathrm{n})^2 > (3\mathrm{m}^2 + 3\mathrm{mn} + 3\mathrm{n}^2) + 3\mathrm{m} + 1\) for \(\mathrm{m}^2 + \mathrm{n}^2 + 5\mathrm{mn} > 3\mathrm{m} + 1\) , which is true for m, n at least 1. If \(\mathrm{m} = 1\) , \(\mathrm{n} = 0\) , then a shortest path has 2 units in direction 3 and \(\mathrm{AB} = \sqrt{7} < 4\) . Finally, if \(\mathrm{m} = 0\) and \(\mathrm{n} = 1\) , then a shortest path has 1 unit in each direction and \(\mathrm{AB} = 2\) .

Thus the answer to the final question is 3, because the only cases where the bug travels exactly \(\mathrm{AB} / 2\) in one direction are where it goes to the opposite vertex of a hexagon it is on.


A circle center O is inscribed in ABCD (touching every side). Prove that angle AOB + angle COD equals 180 degrees.

Solution: Let AB touch the circle at W, BC at X, CD at Y, and DA at Z. Then AO bisects angle ZOW and BO bisects angle XOW. So angle AOB is half angle ZOX. Similarly angle COD is half angle XOZ and hence angle AOB + angle COD equals 180.


The natural numbers a, b, n are such that for every natural number k not equal to b, b - k divides a - \(\mathrm{k^n}\) . Prove that \(\mathrm{a} = \mathrm{b^n}\) .

Solution: We have \(\mathrm{k^n - a = b^n - a}\) (mod b - k). Hence \(\mathrm{b^n - a = 0}\) (mod b - k) for every k not equal to b. But if \(\mathrm{b^n}\) does not equal a, then by taking \(\mathrm{k - b > b^n - a}\) we could render the equation false.


How many (algebraically) different expressions can we obtain by placing parentheses in \(\mathrm{a_1 / a_2 / ... / a_n?}\)

Solution: Answer \(2^{\mathrm{n} - 2}\) . \(\mathrm{a_1}\) must be in the numerator, and \(\mathrm{a_2}\) must be in the denominator, but the other symbols can be in either. This is easily proved by induction.


What is the smallest number of tetrahedrons into which a cube can be partitioned?

Solution:
Answer: 5.

Tetrahedral faces are triangular, so each cube face requires at least two tetrahedral faces. So at least 12 tetrahedral faces are needed in all. At most three faces of a tetrahedron can be mutually orthogonal (and no two faces can be parallel), so at most 3 faces from each tetrahedron can contribute towards these 12. So we require at least 4 tetrahedra to provide the cube faces. But these tetrahedra each have volume at most \(1 / 6\) ( \(1 / 3\) x face area x 1, and face area is at most \(1 / 2\) ). So if we have only 4 tetrahedra in total then their total volume is less than the cube's volume. Contradiction. Hence we need at least 5 tetrahedra.

It can be done with 5: lop off 4 non- adjacent corners to leave a tetrahedron. More precisely, take the cube as ABCDA'B'C'D' with ABCD horizontal, A' directly under A, B' directly under B and so on. Then the five tetrahedra are AA'BD, CC'BC, DD'AC', BB'AC', BDA'C'.


a) Find the smallest square with last digit not 0 which becomes another square (not zero) by the deletion of its last two digits. #(b) Find all squares, not containing the digits 0 or 5, such that if the second digit is deleted the resulting number divides the original one.

Solution:
(a) This one must have slipped through: 121!

(b) Answer: 16,36,121,484. Suppose the number has more than 2 digits. Write it as \((10\mathrm{m}+\) \(\mathrm{n})10^{\mathrm{r}} + \mathrm{s}\) , where \(1< = \mathrm{m}< = 9\) \(0< = \mathrm{n}< = 9\) \(0< = \mathrm{s}< 10^{\mathrm{r}}\) . Then we have \(\mathrm{k(m.10^{r} + s) = (10m}\) \(+\mathrm{n})10^{\mathrm{r}} + \mathrm{s}\) , for some \(\mathrm{k} > 1\)

s does not contain the digits 0 or 5, so 5 does not divide s. Hence 5 divides k- 1, and so k must be 6, 11, or 16 (if k was 21 or more, then the rhs would be negative). Since 25 does not divide k- 1, we must have \(\mathrm{r} = 1\) and s is a single digit.

We look at each possibility for k in turn. \(\mathrm{k} = 6\) gives no solutions. \(\mathrm{k} = 11\) gives about two dozen multiples of 11 from 121 to 891. By inspection the only squares are 121 and 484. \(\mathrm{k} =\) 16 gives 192, which is not a square.

In addition, there is the possibility of 2 digit solutions, which I had overlooked. It is easiest to check each of the 2 digit squares, thus finding the additional solutions 16, 36.


A circle is inscribed in ABCD. AB is parallel to CD, and BC = AD. The diagonals AC, BD meet at E. The circles inscribed in ABE, BCE, CDE, DAE have radius \(\mathrm{r}_1\) , \(\mathrm{r}_2\) , \(\mathrm{r}_3\) , \(\mathrm{r}_4\) respectively. Prove that \(1 / \mathrm{r}_1 + 1 / \mathrm{r}_3 = 1 / \mathrm{r}_2 + 1 / \mathrm{r}_4\) .

Solution:
A necessary and sufficient condition for ABCD to have an inscribed circle is \(\mathrm{AB} + \mathrm{CD} = \mathrm{BC}\) \(+\mathrm{AD}\) . So we have \(\mathrm{AB} + \mathrm{CD} = 2\mathrm{AD}\) , which we use repeatedly. Extend DC to X so that BX is parallel to EC. Then \(\mathrm{DX} = \mathrm{AB} + \mathrm{CD} = 2\mathrm{AD}\) and the triangles DEC, AEB, DBX are similar. Let h be the perpendicular distance from AB to CD. The similar triangles give us the heights of DEC and AEB in terms of h.

\(1 / \mathrm{r}_1 =\) perimeter ABE/(2 area ABE) \(= (\mathrm{AB} + 2\mathrm{EB}) / (\mathrm{AB}.\mathrm{height}) = (\mathrm{AB}+\) 2. BD.AB/(AB+CD))/(AB.h.AB/(AB+CD)) = 2(AD + BD)/(AB.h). Similarly, \(1 / \mathrm{r}_3 = 2(\mathrm{AD}+\) BD)/(CD.h).

The area of AED \(=\) area ABD - area ABE \(= 1 / 2\) AB.h.CD/(2AD), so \(1 / \mathrm{r}_2 = 1 / \mathrm{r}_4 =\) perimeter ADE/(2 area ADE) \(= (\mathrm{AD} + \mathrm{BD}) / (\mathrm{h}.\mathrm{AB}.\mathrm{CD} / 2\mathrm{AD})\) , and \(1 / \mathrm{r}_2 + 1 / \mathrm{r}_4 = 2(\mathrm{AD} + \mathrm{BD}) / \mathrm{h}\) \(2\mathrm{AD} / (\mathrm{AB}.\mathrm{CD}) = 2(\mathrm{AB} + \mathrm{BD}) / \mathrm{h}\) \((\mathrm{AB} + \mathrm{CD}) / (\mathrm{AB}.\mathrm{CD}) = 1 / \mathrm{r}_1 + 1 / \mathrm{r}_3\)


(a) Each of \(\mathbf{x}_{1}, \ldots , \mathbf{x}_{\mathrm{n}}\) is -1, 0 or 1. What is the minimal possible value of the sum of all \(\mathbf{x}_{i}\mathbf{x}_{j}\) with \(1 \leq = \mathrm{i} < \mathrm{j} \leq = \mathrm{n}\) ? (b) Is the answer the same if the \(\mathbf{x}_{i}\) are real numbers satisfying \(0 \leq = |\mathbf{x}_{i}| \leq = 1\) for \(1 \leq = \mathrm{i} \leq = \mathrm{n}\) ?

Solution:
(a) Answer: \([- \mathrm{n} / 2]\) .

Let \(\mathbf{A} = (\mathbf{x}_{1} + \ldots + \mathbf{x}_{\mathrm{n}})^{2}\) , \(\mathbf{B} = \mathbf{x}_{1}^{2} + \ldots + \mathbf{x}_{\mathrm{n}}^{2}\) . Then we must minimize \(\mathbf{A} - \mathbf{B}\) . For \(\mathrm{n}\) even, we separately minimize \(\mathbf{A}\) and maximize \(\mathbf{B}\) by taking half the \(\mathbf{x}\) 's to be \(+1\) and half to be - 1. For \(\mathrm{n}\) odd we can take \([\mathrm{n} / 2]\) \(\mathbf{x}\) 's to be \(+1\) , \([\mathrm{n} / 2]\) to be - 1, and one to be 0. That minimizes \(\mathbf{A}\) and gives \(\mathbf{B}\) one less than its maximum. That is the best we can do if we fix \(\mathbf{A} = 0\) , since \(\mathbf{A} = 0\) requires an even number of \(\mathbf{x}\) 's to be non- zero and hence at least one to be zero. If we do not minimize \(\mathbf{A}\) , then since its value must be an integer, its value will be at least 1. In that case, even if \(\mathbf{B}\) is maximized we will not get a lower total.

(b) Answer: \(-[\mathrm{n} / 2]\) . For \(\mathrm{n}\) even, the same argument works. For \(\mathrm{n}\) odd we can clearly get \(-[\mathrm{n} / 2]\) , so it remains to prove that we cannot get a smaller sum. Suppose otherwise, so that \(\mathbf{x}_{i}\) is a minimal sum with sum less than \(-[\mathrm{n} / 2]\) . Let \(\mathbf{x}_{\mathrm{n}} = \mathbf{x}\) , then the sum is \(\mathbf{x}(\mathbf{x}_{1} + \ldots + \mathbf{x}_{\mathrm{n - 1}}) + \mathbf{sum}\) of terms \(\mathbf{x}_{i}\mathbf{x}_{j}\) with \(1 \leq = \mathrm{i}, \mathrm{j} < \mathrm{n}\) . But this is less than the sum for \(\mathrm{n - 1}\) , so \(\mathbf{x}(\mathbf{x}_{1} + \ldots + \mathbf{x}_{\mathrm{n - 1}})\) must be negative, and since it is minimal we must have \(|\mathbf{x}| = 1\) . But the same argument shows that all the terms have modulus 1. We now have a contradiction since we know that the minimum in this case is \(-[\mathrm{n} / 2]\) .


Two players have a 3 x 3 board. 9 cards, each with a different number, are placed face up in front of the players. Each player in turn takes a card and places it on the board until all the cards have been played. The first player wins if the sum of the numbers in the first and third rows is greater than the sum in the first and third columns, loses if it is less, and draws if the sums are equal. Which player wins and what is the winning strategy?

Solution:
The first player always wins.

Let the board be:

. F . S . S . F .

We call the squares marked F the F- squares, the squares marked S the S- squares, and the remaining squares the neutral squares. The first player wins if the sum of the two cards on the F- squares exceeds the sum of the two cards on the S- squares. We also call the first player F and the second player S.

Let the cards be \(\mathrm{a_1 > a_2 > \ldots > a_9}\) . Let \(\mathrm{t_1 = a_1 + a_9}\) , \(\mathrm{t_2 = a_2 + a_8}\) , \(\mathrm{t_3 = a_3 + a_7}\) , \(\mathrm{t_4 = a_4 + a_6}\) .

If \(\mathrm{t_1 > t_2}\) , or \(\mathrm{t_1 = t_2 > t_3}\) , or \(\mathrm{t_1 = t_2 = t_3 > = t_4}\) (\*), then F's strategy is to get a total of \(\mathrm{t_1}\) or better on the F- squares and to force S to a lower score on the S- squares. If (\*) does not hold, then F's strategy is to force S to \(\mathrm{t_1}\) or lower, and to get a higher score.

If (\*) holds, then F starts by playing \(\mathrm{a_1}\) to an F- square. S must play to the remaining F- square, otherwise F will play \(\mathrm{a_3}\) or better to it on his next move and win. So S must play \(\mathrm{a_9}\) to the remaining F- square, giving F a total of \(\mathrm{t_1}\) .

Now if \(\mathrm{t_1 > t_2}\) , then F forces S to \(\mathrm{t_2}\) or worse by playing \(\mathrm{a_8}\) to an S- square.

If \(\mathrm{t_1 = t_2 > t_3}\) , then F forces S to \(\mathrm{t_3}\) or worse by playing \(\mathrm{a_2}\) to a neutral square. If S plays to an S- square, then he cannot do better than \(\mathrm{a_3 + a_8}\) , which loses. So he plays \(\mathrm{a_8}\) to a neutral square. But now F plays \(\mathrm{a_3}\) to an S- square, and S cannot do better than \(\mathrm{t_3}\) .

If \(\mathrm{t_1 = t_2 = t_3 > t_4}\) , then F forces S to \(\mathrm{t_4}\) or worse. He starts by playing \(\mathrm{a_2}\) to a neutral square. If does not prevent F playing \(\mathrm{a_8}\) to an S- square on his next move, then he cannot do better than \(\mathrm{a_3 + a_8}\) , which loses. So he must play \(\mathrm{a_8}\) to a neutral square. Now F plays \(\mathrm{a_3}\) to a neutral square. If S does not prevent F playing \(\mathrm{a_7}\) to an S- square on the following move, then he cannot do better than \(\mathrm{a_4 + a_7}\) which loses, so he plays \(\mathrm{a_7}\) to a neutral square. F now plays \(\mathrm{a_4}\) to an S- square. S cannot now do better than \(\mathrm{t_4}\) , which loses.

Finally, if \(\mathrm{t_1 = t_2 = t_3 = t_4}\) , then F proceeds as in the last case except that at the end he plays \(\mathrm{a_4}\) to the last neutral square instead of to an S- square. S now gets \(\mathrm{a_5 + a_6}\) on the S- squares, which loses.

If (\*) does not hold, then F starts by playing \(\mathrm{a_9}\) to an S- square. If S does not play to the other S- square, then F will play \(\mathrm{a_7}\) or \(\mathrm{a_8}\) there on his next move and S will lose. So S must play \(\mathrm{a_1}\) to the other square, and gets a total of \(\mathrm{t_1}\) . F now plays to get \(\mathrm{t_2}\) , \(\mathrm{t_3}\) or \(\mathrm{t_4}\) on the F- squares.

If \(\mathrm{t_1 < t_2}\) , then F plays \(\mathrm{a_2}\) to an F- square and so gets at least \(\mathrm{t_2}\) and wins.

If \(\mathrm{t_1 = t_2 < t_3}\) , then F plays \(\mathrm{a_8}\) to a neutral square. If S does not prevent F playing \(\mathrm{a_2}\) to an F- square on his next move, then F will get at least \(\mathrm{a_2 + a_7}\) and win. So S must play \(\mathrm{a_2}\) to a neutral square. Now F plays \(\mathrm{a_3}\) to an F- square and so gets at least \(\mathrm{t_3}\) on the F- squares and wins.

Finally, if \(\mathrm{t_1 = t_2 = t_3 < t_4}\) , then F plays as in the previous case, except that at the end he plays \(\mathrm{a_7}\) to a neutral square instead of \(\mathrm{a_3}\) to an F- square. S must prevent F playing \(\mathrm{a_3}\) to an F- square the following move, or F gets at least \(\mathrm{a_3 + a_6}\) and wins. So S plays \(\mathrm{a_3}\) to a neutral square. F now plays \(\mathrm{a_4}\) to an F square and so must get at least \(\mathrm{t_4}\) , which wins.


A circle is circumscribed about the triangle ABC. X is the midpoint of the arc BC (on the opposite side of BC to A), Y is the midpoint of the arc AC, and Z is the midpoint of the arc AB. YZ meets AB at D and YX meets BC at E. Prove that DE is parallel to AC and that DE passes through the center of the inscribed circle of ABC.

Solution:
ZY bisects the angle AYB, so AD/BD = AY/BY. Similarly, XY bisects angle BYC, so CE/BE = CY/BY. But AY = CY. Hence AD/BD = CE/BE. Hence triangles BDE and BAC are similar and DE is parallel to AC.

Let BY intersect AC at W and AX at I. I is the in- center. AI bisects angle BAW, so WI/IB = AW/AB. Now consider the triangles AYW, BYA. Clearly angle AYW = angle BYA. Also angle WAY = angle CAY = angle ABY. Hence the triangles are similar and AW/AY = AB/BY. So AW/AB = AY/BY. Hence WI/IB = AY/BY = AD/BD. So triangles BDI and BAW are similar and DI is parallel to AW and hence to DE. So DE passes through I.


Bus numbers have 6 digits, and leading zeros are allowed. A number is considered lucky if the sum of the first three digits equals the sum of the last three digits. Prove that the sum of all lucky numbers is divisible by 13.

Solution: The total is made up of numbers of the form abcabc, and pairs of numbers abcxyz, xyzabc. The former is abc.1001 and the sum of the pair is 1001(abc + xyz). So the total is divisible by 1001 and hence by 13.


The beam of a lighthouse on a small rock penetrates to a fixed distance d. As the beam rotates the extremity of the beam moves with velocity v. Prove that a ship with speed at most v/8 cannot reach the rock without being illuminated.

Solution: Let the lighthouse be at L. Take time \(t = 0\) at the moment the boat starts its run, so that at \(t = 0\) it is at S a distance d from L, and thereafter it is at a distance less than d. Take A and B a distance d from L so that ALBS is a semicircle with diameter AB and S the midpoint of the arc AB. During the period to \(t = 2.5\) pi.d/v the boat has traveled a distance less than d, so it cannot reach AB. But it is a distance less than d from L, so it must be inside the semicircle. But during this period the beam sweeps across from LA to LB and so it must illuminate the boat.


A group of 100 people is formed to patrol the local streets. Every evening 3 people are on duty. Prove that you cannot arrange for every pair to meet just once on duty.

Solution: Every time a person is on duty he is paired with two other people, so if the arrangement were possible the number of pairs involving a particular person would have to be even. But it is 99.


A tangent to the inscribed circle of a triangle drawn parallel to one of the sides meets the other two sides at X and Y. What is the maximum length XY, if the triangle has perimeter p?

Solution: Let BC be the side parallel to XY, h the length of the altitude from A, and r the radius of the in- circle. Then XY/BC = (h - 2r)/h. But r.p = h.BC. So XY = (p - 2BC)BC/p = (p²/8 - 2(BC - p/4)²)/p. So the maximum occurs when BC = p/4 and has value p/8.


The \(\mathrm{n}^2\) numbers \(\mathbf{x}_{\mathrm{ij}}\) satisfy the \(\mathrm{n}^3\) equations: \(\mathbf{x}_{\mathrm{ij}} + \mathbf{x}_{\mathrm{jk}} + \mathbf{x}_{\mathrm{ki}} = 0\) . Prove that we can find numbers \(\mathbf{a}_1\) , ..., \(\mathbf{a}_{\mathrm{n}}\) such that \(\mathbf{x}_{\mathrm{ij}} = \mathbf{a}_{\mathrm{i}} - \mathbf{a}_{\mathrm{j}}\) .

Solution: Taking \(\mathrm{i} = \mathrm{j} = \mathrm{k}\) , we have that \(\mathbf{x}_{\mathrm{ii}} = 0\) . Now taking \(\mathrm{j} = \mathrm{k}\) , we have that \(\mathbf{x}_{\mathrm{ij}} = - \mathbf{x}_{\mathrm{ji}}\) . Define \(\mathbf{a}_{\mathrm{i}} = \mathbf{x}_{\mathrm{i1}}\) . Then we have \(\mathbf{x}_{\mathrm{i1}} + \mathbf{x}_{\mathrm{ij}} + \mathbf{x}_{\mathrm{ji}} = 0\) . Hence \(\mathbf{x}_{\mathrm{ij}} = \mathbf{a}_{\mathrm{i}} - \mathbf{a}_{\mathrm{j}}\) .


Can 1965 points be arranged inside a square with side 15 so that any rectangle of unit area placed inside the square with sides parallel to its sides must contain at least one of the points?

Solution:
Yes. Place a grid of 900 points in 30 equally spaced rows and columns, so that each point is a distance 15/31 from its nearest neighbours (or 15/31 from the edge). This blocks all rectangles except those slimmer than 1/2. Those slimmer than 1/2 must have length at least 2, so we can block them with a smaller set of rows and columns containing more finely spaced points.

Label the rows 1- 30. In each of the 7 rows 3, 7, 11, 15, 19, 23, 27 place an additional 31 points, so that each of these rows has 61 equally spaced points at a spacing of 15/62. Similarly for the columns. So in total we are placing an additional \(2.7.31 = 434\) points. Any rectangle of length \(>2\) must encounter one of these rows (or columns) and hence must have width less than 1/4. This blocks any rectangle except those with width \(< 1 / 4\) .

In each of the 3 rows 7, 15, 23 place an additional 62 points, so that each of these rows has 123 equally spaced points at a spacing of 15/124. Similarly for the columns. So in total we are placing an additional \(2.3.62 = 372\) points. Any rectangle of length \(>4\) must encounter one of these rows (or columns) and hence must have width less than 1/8. This blocks any rectangle except those with width \(< 1 / 8\) and hence length \(>8\) .

In row 15 place an additional 124 points, so that it has a total of 247 equally spaced points at a spacing of 15/247. Similarly for column 15. This requires an additional 248 points. Any rectangle which can fit through these gaps has area at most \(15 \times 15 / 247 < 1\) . So we have blocked all rectangles with area 1 or more and used \(900 + 434 + 372 + 248 = 1954\) points.

Ilan Mayer, who seems to solve these problems effortlessly, came up with a neater arrangement of points. He used narrowly spaced points along widely spaced diagonals: (k/15, k/15) for \(\mathrm{k} = 1,2,\ldots ,224\) ; ((28\*n+k)/15, k/15) for \(\mathrm{n} = 1,2,\ldots ,7\) , \(\mathrm{k} = 1,2,\ldots ,224 - 28\*n\) ; (k/15,

\((28^{*}\mathrm{n} + \mathrm{k}) / 15\) for \(\mathrm{n} = 1,2,\dots,7\) \(\mathrm{k} = 1,2,\dots,224 - 28^{*}\mathrm{n}\) . The diagonals are spaced 28/15 apart, so the biggest rectangle that can be fitted between two diagonals has sides 15/15 less epsilon and 15/15 less epsilon. For example, take the vertices as \((14 / 15 + \mathrm{e},\mathrm{e})\) , \((29 / 15 - \mathrm{e},\mathrm{e})\) , \((14 / 15 + \mathrm{e}\) , \(15 / 15 - \mathrm{e})\) , \((29 / 15 - \mathrm{e},15 / 15 - \mathrm{e})\) . If one allows a rectangle to touch points (in other words if one took the rectangles to exclude their boundaries) then this does not work - many \(15\mathrm{x}1 / 15\) rectangles will fit. But one can add an additional point on each of the 15 lines, keeping the points on each line evenly spaced. That blocks rectangles without boundary and still has only 1821 points.


Given n real numbers \(\mathrm{a_1,a_2,\ldots,a_n}\) , prove that you can find n integers \(\mathrm{b_1,b_2,\ldots,b_n}\) , such that \(\mathrm{|a_i - b_i| < 1}\) and the sum of any subset of the original numbers differs from the sum of the corresponding \(\mathrm{b_i}\) by at most \((\mathrm{n} + 1) / 4\) .

Solution:
We can take all \(\mathrm{a_i}\) to lie in the range (0,1) and all \(\mathrm{b_i}\) to be 0 or 1. The largest positive value of the sum of \((\mathrm{a_i - b_i})\) for any subset is achieved by taking the subset of those i for which \(\mathrm{b_i = 0}\) . Similarly, the largest negative value is achieved by taking those i for which \(\mathrm{b_i = 1}\) . So the worst subset will be one of those two.

If \(\mathrm{a_i< a_j}\) , then we cannot have \(\mathrm{b_i = 1}\) and \(\mathrm{b_j = 0}\) if the set of \(\mathrm{b_i}\) is to minimise the maximum sum, because swapping them would reduce the sum of a's with \(\mathrm{b = 0}\) and the sum of \((1 - \mathrm{a})\) 's with \(\mathrm{b = 1}\) . So if we order the a's so that \(\mathrm{a_1< = a_2< = \ldots < = a_n}\) , then a best set of b's is \(\mathrm{b_i = 0}\) for \(\mathrm{i< =}\) some k, and \(\mathrm{b_i = 1}\) for \(\mathrm{i > k}\) . [If some of the \(\mathrm{a_i}\) are equal, then we can find equally good sets of b's do not have this form, but we cannot get a lower maximum sum by departing from this form.]

Let \(\mathrm{L_i = a_1 + a_2 + \ldots + a_i}\) , and \(\mathrm{R_i = a_{i + 1} + a_{i + 2} + \ldots + a_n}\) . As we increase i the sums \(\mathrm{L_i}\) increase and the sums \(\mathrm{R_i}\) decrease, so for some k we must have \(\mathrm{L_k< R_k}\) , \(\mathrm{L_{k + 1} > = R_{k + 1}}\) . Either k or \(\mathrm{k + 1}\) must correspond to the optimum choice of b's to minimise the maximum sum.

Now assume that the a's form a maximal set, in other words they are chosen so that the minimum is as large as possible. We show first that in this case \(\mathrm{L_{k + 1} = R_k}\) . Suppose \(\mathrm{L_{k + 1}< R_k}\) . Then we could increase each of \(\mathrm{a_{k + 1}}\) , \(\mathrm{a_{k + 2}}\) , ..., \(\mathrm{a_n}\) by epsilon. This would leave \(\mathrm{L_k}\) unaffected, but slightly increase \(\mathrm{L_{k + 1}}\) and slightly reduce \(\mathrm{R_k}\) . For small epsilon this does not change the value of k, but increases the smaller of \(\mathrm{L_{k + 1}}\) and \(\mathrm{R_k}\) , thus increasing the minimum and contradicting the maximality of the original a's. Similarly, if \(\mathrm{L_{k + 1} > R_k}\) , we could decrease each of \(\mathrm{a_1}\) , \(\mathrm{a_2}\) , ..., \(\mathrm{a_{k + 1}}\) by epsilon, thus slightly increasing \(\mathrm{R_k}\) and reducing \(\mathrm{L_{k + 1}}\) .

Suppose not all of \(\mathrm{a_1,a_2,\ldots,a_{k + 1}}\) are equal. Take i so that \(\mathrm{a_i< a_{i + 1}}\) . Now increase each of \(\mathrm{a_1}\) , \(\mathrm{a_2,\ldots,a_i}\) by epsilon and reduce each of \(\mathrm{a_{i + 1}}\) , \(\mathrm{a_{i + 2},\ldots,a_{k + 1}}\) by epsilon', with epsilon and epsilon' sufficiently small that we do not upset the ordering or change the value of k, and with their relative sizes chosen so that \(\mathrm{L_{k + 1}}\) is increased. \(\mathrm{R_k}\) is also increased, so we contradict the maximality of the a's. Hence all \(\mathrm{a_1,a_2,\ldots,a_{k + 1}}\) are equal. Similarly, we show that all of \(\mathrm{a_{k + 1}}\) , ..., \(\mathrm{a_n}\) are equal. For if not we can increase slightly \(\mathrm{a_{k + 1}}\) , ..., \(\mathrm{a_j}\) and reduce slightly \(\mathrm{a_{j + 1}}\) , ..., \(\mathrm{a_n}\) to get a contradiction.

So we have established that all the a's must be equal. Suppose n is odd \(= 2\mathrm{m} + 1\) and that all the a's equal x. Then for the optimum k we have \((\mathrm{k + 1})\mathrm{x} = (2\mathrm{m} + 1 - \mathrm{k})(1 - \mathrm{x})\) , hence \(\mathrm{k + 1} = (2\mathrm{m} + 2)(1-\)

x) and the maximum difference is \((k + 1)x = (2m + 2)(1 - x)x\) . This is maximised by taking \(\mathrm{x} =\) \(1 / 2\) , \(\mathrm{k} = \mathrm{m}\) , and is \((m + 1) / 2 = (n + 1) / 4\) . If n is even \(= 2\mathrm{m}\) , then for the optimum \(\mathrm{k}\) we have \((k + 1)x = (2m - k)(1 - x)\) , so \(\mathrm{k} + 1 = (2m + 1)(1 - x)\) , and the maximum difference is \((k + 1)x =\) \((2m + 1)(1 - x)x\) . However, in this case we cannot take \(\mathrm{x} = 1 / 2\) , because that would give \(\mathrm{k} = \mathrm{m}\) - \(1 / 2\) which is non- integral, so we take \(\mathrm{k} = \mathrm{m} - 1\) or m, both of which give a maximum difference of \(\mathrm{m(m + 1)} / (2\mathrm{m + 1}) = \mathrm{n(n + 2)} / (4\mathrm{n + 4}) < (\mathrm{n + 1}) / 4\) .


A tourist arrives in Moscow by train and wanders randomly through the streets on foot. After supper he decides to return to the station along sections of street that he has traversed an odd number of times. Prove that this is always possible. [In other words, given a path over a graph from A to B, find a path from B to A consisting of edges that are used an odd number of times in the first path.]

Solution:
Disregard all edges except those used in the path from A to B, and for each of those let the multiplicity be the number of times it was traversed. Let the degree of a vertex be the sum of the multiplicities of its edges. The key is to notice that the degree of every vertex except A and B must be even. For as we traverse the path from A to B we increase the degree by 2 each time we pass through a vertex. But at the start of the path, as we leave A, we only increase its degree by 1. Similarly as we arrive at B for the last time.

Now construct a path from B as follows. Since B has odd degree it must have an edge of odd multiplicity. Suppose the edge connects B to C. Follow that edge and reduce its multiplicity by one, so that B's degree and C's degree are each reduced by one. Now C has odd degree, so it must have an edge of odd multiplicity. Repeat. Since there are only finitely many edges we must eventually be unable to continue the path. But the only way that can happen is if we reach A.


(a) A committee has met 40 times, with 10 members at every meeting. No two people have met more than once at committee meetings. Prove that there are more than 60 people on the committee.

(b) Prove that you cannot make more than 30 subcommittees of 5 members from a committee of 25 members with no two subcommittees having more than one common member.

Solution:
(a) Each meeting involves \(10.9 / 2 = 45\) pairs. So after 40 meetings, there have been 1800 pairs. We are told that these are all distinct. But if there are N people on the committee, then there are only N(N-1)/2 pairs available. For N=60, this is only 1770.

(b) A subcommittee of 5 has \(5.4 / 2 = 10\) pairs. So 31 subcommittees have 310 pairs, and these are all distinct, since no two people are on more than one subcommittee. But a committee of 25 only has \(25.24 / 2 = 300\) pairs available.


Given two relatively prime natural numbers r and s, call an integer good if it can be represented as mr + ns with m, n non- negative integers and bad otherwise. Prove that we can find an integer c, such that just one of k, c - k is good for any k. How many bad numbers are there?

Solution:
Notice that 0 is good and all negative numbers are bad. Take \(\mathrm{c} = \mathrm{rs} - \mathrm{r} - \mathrm{s}\) . First c, is bad. For suppose otherwise: \(\mathrm{c} = \mathrm{mr} + \mathrm{ns}\) . Then \(\mathrm{mr} + \mathrm{ns} = (\mathrm{s - 1})\mathrm{r} - \mathrm{s}\) . Hence \((\mathrm{s - 1 - m})\mathrm{r} = (\mathrm{n + 1})\mathrm{s}\) , so r divides \(\mathrm{n + 1}\) . Say \(\mathrm{n + 1 = kr}\) , and then \(\mathrm{s - 1 - m = ks}\) , so \(\mathrm{m} = (1 - \mathrm{k})\mathrm{s} - 1\) . But \(\mathrm{n + 1}\) is positive, so \(\mathrm{k} > = 1\) , and hence m is negative. Contradiction.

If k is good, then c- k must be bad (otherwise c would be good). Suppose k is bad. Since r and s are relatively prime we can find integers a and b with \(\mathrm{ar} + \mathrm{bs} = 1\) and hence integers m and n with \(\mathrm{mr} + \mathrm{ns} = \mathrm{k}\) . Adding a multiple of sr - rs to both sides if necessary, this gives a pair m, n with \(\mathrm{mr} + \mathrm{ns} = \mathrm{k}\) and m non- negative. Now take the pair with the smallest possible nonnegative m. Then \(\mathrm{m} < = \mathrm{s - 1}\) (for otherwise \(\mathrm{m}' = \mathrm{m - s}\) , \(\mathrm{n}' = \mathrm{n + r}\) would be a pair with smaller non- negative m). Also \(\mathrm{n} < = - 1\) , otherwise k would be good. Now \(\mathrm{c - k} = (\mathrm{s - 1 - m})\mathrm{r} + (\mathrm{- n - 1})\mathrm{s}\) and the coefficients \(\mathrm{s - 1 - m}\) and \(\mathrm{- n - 1}\) are both non- negative, so \(\mathrm{c - k}\) is good.

So exactly \((\mathrm{rs} - \mathrm{r} - \mathrm{s} + 1) / 2\) integers are bad.


A spy- plane circles point A at a distance \(10\mathrm{km}\) with speed \(1000\mathrm{km / h}\) . A missile is fired towards the plane from A at the same speed and moves so that it is always on the line between A and the plane. How long does it take to hit?

Solution:
Answer: 18pi sec.

Let C be the position of the spy- plane at the moment the missile is fired. Let B be the point a quarter of the way around the circle from C (in the direction the spy- plane is moving). Then the missile moves along the semi- circle on diameter AB and hits the plane at B.

To see this take a point P on the quarter circle and let the line AX meet the semi- circle at Q. Let O be the center of the semicircle. The angle BOQ is twice the angle BAQ, so the arc BP is the same length as the arc BQ. Hence also the arc AQ is the same length as the arc CP.


Prove that the sum of the lengths of the edges of a polyhedron is at least 3 times the greatest distance between two points of the polyhedron.

Solution:
If A and B are at the greatest distance, then they must be vertices. For suppose A is not a vertex. Then there is a segment XY entirely contained in the polyhedron with A as an interior point. But now at least one of angles BAX, BAY must be at least 90. Suppose it is BAX. Then BX is longer than BA. Contradiction.

Take a plane through A perpendicular to the line AB. Then the polyhedron must lie entirely on one side of the plane, for if Z lay on the opposite side to B, then BZ would be longer than BA. Now move the plane slightly towards B keeping it perpendicular to AB. The intersection of the plane and the polyhedron must be a small polygon. The polygon must have at least 3 vertices, each of which must lie on an edge of the polyhedron starting at A. Select three of these edges.

As the plane is moved further towards B, the selected vertices may sometimes split into multiple vertices or they may sometimes coalesce. In the former case, just choose one of the daughter vertices. In the latter case, let O be the point of intersection of the plane and AB. Let O' be the point of intersection at the last coalescence (or A if there was none). Then we have three paths along edges, with no edges in common, each of which projects onto O'O and hence has length at least O'O. Now select one or more new vertices to replace any lost through coalescence and repeat.


An alien moves on the surface of a planet with speed not exceeding u. A spaceship searches for the alien with speed v. Prove the spaceship can always find the alien if \(v > 10u\) .

Solution:
The spacecraft flies at a constant height, so that it can see a circular spot on the surface. It starts at the north pole and spirals down to the south pole, overlapping its previous track on each circuit. The alien cannot move fast enough to cross the track before the next circuit, so it is trapped inside a reducing area surrounding the south pole.

The value of 10 is not critical, so we do not have to optimise the details. Take the height above the surface to be half the radius. Then a diameter of the spot subtends an angle 2 cos' \((1 / 1.5)\) at the center of the planet. \(1 / 1.5< 1 / \sqrt{2}\) , so the angle is more than 90 degrees. The critical case is evidently when the spacecraft is circling the equator. Using suitable units, we may take the radius of the planet to be 1 and the spaceship speed to be 1. Then the diameter of the spot is \(\mathrm{pi} / 2\) . We take the overlap to be \(2 / 3\) , so that each revolution the track advances \(\mathrm{pi} / 6\) . If the planet flew in a circle above the equator, the distance for a revolution would be \(2\mathrm{pi}1.5 = 3\mathrm{pi}\) . The helical distance must be less than \(3\mathrm{pi} + \mathrm{pi} / 6 = 19\mathrm{pi} / 6\) . So the alien can travel a distance \(19\mathrm{pi} / 60< 2 / 3\mathrm{pi} / 2\) and is thus trapped as claimed.


There are an odd number of soldiers on an exercise. The distance between every pair of soldiers is different. Each soldier watches his nearest neighbour. Prove that at least one soldier is not being watched.

Solution:
The key is to notice that no loops of size greater than two are possible. For suppose we have \(\mathrm{A_1}\) \(\mathrm{A}_2\) ... \(\mathrm{A_n}\) with \(\mathrm{A_i}\) watching \(\mathrm{A_{i + 1}}\) for \(0< \mathrm{i}< \mathrm{n}\) , and \(\mathrm{A_n}\) watching \(\mathrm{A_1}\) . Then the distance \(\mathrm{A_i}\) \(\mathrm{A_i}\) is greater than the distance \(\mathrm{A_iA_{i + 1}}\) for \(1< \mathrm{i}< \mathrm{n}\) , and the distance \(\mathrm{A_1A_n}\) is less than the distance \(\mathrm{A_1A_2}\) . Hence the distance \(\mathrm{A_1A_n}\) is less than the distance \(\mathrm{A_{n - 1}A_n}\) and so \(\mathrm{A_{n - 1}}\) is closer to \(\mathrm{A_n}\) than \(\mathrm{A_1}\) . Contradiction.

Pick any soldier. Now pick the soldier he is watching, and so on. The total number of soldiers is finite so this process must terminate with some soldier watching his predecessor. If the process terminates after more than two soldiers have been picked, then the penultimate soldier is watched by more than one soldier. But in that case there must be another soldier who is unwatched, because the number of soldiers equals the number of soldiers watching.

If the process terminates after just two soldiers, then we have a pair of soldiers watching each other. Now repeat on the remaining soldiers. Either we find a soldier watched twice (in which case some other soldier must be unwatched) or all the soldiers pair off, except one, since the total number is odd. But that soldier must be unwatched.


(a) B and C are on the segment AD with \(\mathrm{AB} = \mathrm{CD}\) . Prove that for any point P in the plane: \(\mathrm{PA} + \mathrm{PD} \geq \mathrm{PB} + \mathrm{PC}\) .

(b) Given four points A, B, C, D on the plane such that for any point P on the plane we have \(\mathrm{PA} + \mathrm{PD} \geq \mathrm{PB} + \mathrm{PC}\) . Prove that B and C are on the segment AD with \(\mathrm{AB} = \mathrm{CD}\) .

Solution:
(a) Suppose the points lie in the order A, B, C, D. If P lies on AD, then the result is trivial, and we have equality if P lies outside the segment AD. So suppose P does not lie on AD.

Let M be the midpoint of AD. Take P' so that P, M, P' are collinear and \(\mathrm{PM} = \mathrm{MP}\) . Then we wish to prove that \(\mathrm{PA} + \mathrm{AP}' > \mathrm{PB} + \mathrm{BP}'\) . Extend P'B to meet PA at Q. Then \(\mathrm{P'A} + \mathrm{AQ} > \mathrm{P'Q}\) , so \(\mathrm{P'A} + \mathrm{AP} > \mathrm{P'Q} + \mathrm{QP}\) . But \(\mathrm{QP} + \mathrm{BQ} > \mathrm{PB}\) , so \(\mathrm{QP} + \mathrm{QP}' > \mathrm{PB} + \mathrm{PB}'\) . Hence result.

(b) Let the foot of the perpendicular from B, C onto AD be X, Y respectively. Suppose that N, the midpoint of XY, is on the same side of M, the midpoint of AD, as D. Then take P to be a remote point on the line AD, the opposite side of A to D, so that A, D, M and N are all on the same side of the line PAD from P. Then \(\mathrm{PA} + \mathrm{PD} = 2\mathrm{PM} < 2\mathrm{PN} \leq \mathrm{PB} + \mathrm{PC}\) .

Contradiction. So we must have N coincide with M. But we still have \(\mathrm{PA} + \mathrm{PD} = 2\mathrm{PM} = 2\mathrm{PN} < \mathrm{PB} + \mathrm{PC}\) , unless both B and C are on the line AD. So we must have B and C on the line AD

and \(\mathrm{AB} = \mathrm{CD}\) . It remains to show that B and C are between A and D. Take \(\mathrm{P} = \mathrm{B}\) . Then if C is not between A and D, we have \(\mathrm{PC} > \mathrm{PD}\) (or PA), contradiction.


Can both \(\mathrm{x}^2 +\mathrm{y}\) and \(\mathrm{x} + \mathrm{y}^2\) be squares for \(\mathrm{x}\) and \(\mathrm{y}\) natural numbers?

Solution: No. The smallest square greater than \(\mathrm{x}^2\) is \((\mathrm{x} + 1)^2\) , so we must have \(\mathrm{y} > 2\mathrm{x}\) . Similarly \(\mathrm{x} > 2\mathrm{y}\) . Contradiction.


A group of children are arranged into two equal rows. Every child in the back row is taller than the child standing in front of him in the other row. Prove that this remains true if each row is rearranged so that the children increase in height from left to right.

Solution:
Rearrange the children in the back row into order, and rearrange the front row in the same way, so that each child stays in front of the same child in the back row. Denote heights in the back row by \(\mathrm{a_i}\) and heights in the front row by \(\mathrm{b_i}\) . So we have \(\mathrm{a_1}\leq \mathrm{a_2}\leq \ldots \leq \mathrm{a_n}\) , and \(\mathrm{a_i} > \mathrm{b_i}\) for \(i = 1,2,\ldots ,\mathrm{n}\) .

Now if \(\mathrm{i}< \mathrm{j}\) , but \(\mathrm{b_i} > \mathrm{b_j}\) , then we may swap \(\mathrm{b_i}\) and \(\mathrm{b_j}\) and still have each child taller than the child in front of him. For \(\mathrm{b_i}< \mathrm{a_i}< = \mathrm{a_j}\) , and \(\mathrm{b_j}< \mathrm{b_i}< \mathrm{a_i}\) . By repeated swaps we can get the front row into height order. [For example, identify the shortest child and swap him to the first position, then the next shortest and so on.]


A rectangle ABCD is drawn on squared paper with its vertices at lattice points and its sides lying along the gridlines. \(\mathrm{AD} = \mathrm{k}\) AB with \(\mathrm{k}\) an integer. Prove that the number of shortest paths from A to C starting out along AD is \(\mathrm{k}\) times the number starting out along AB.

Solution: Let ABCD have n lattice points along the side AB. Then it has kn lattice points along the side AD. Let X be the first lattice point along AB after leaving A. A shortest path from X to C must involve a total of \(\mathrm{kn} + \mathrm{n} - 1\) moves between lattice points, \(\mathrm{n} - 1\) in the direction AB and \(\mathrm{kn}\) in the direction BC. Hence the total number of such paths is \((\mathrm{kn} + \mathrm{n} - 1)! / ((\mathrm{kn})! (\mathrm{n} - 1)!)\) . Similarly, the number of paths starting out along AD is \((\mathrm{kn} + \mathrm{n} - 1)! / ((\mathrm{kn} - 1)! \mathrm{n}!)\) . Let \(\mathrm{m} = (\mathrm{kn} + \mathrm{n} - 1)! / ((\mathrm{kn} - 1)! (\mathrm{n} - 1)!)\) . Then the number starting along AB is \(\mathrm{m} / (\mathrm{kn})\) and the number starting along AD is \(\mathrm{m} / \mathrm{n}\) , which is \(\mathrm{k}\) times larger, as required.


Given non- negative real numbers \(\mathrm{a_1}\) , \(\mathrm{a_2}\) , ..., \(\mathrm{a_n}\) , such that \(\mathrm{a_{i - 1}} \leq \mathrm{a_i} \leq 2\mathrm{a_{i - 1}}\) for \(i = 2,3,\ldots ,\mathrm{n}\) . Show that you can form a sum \(\mathrm{s} = \mathrm{b_1a_1} + \ldots +\mathrm{b_na_n}\) with each \(\mathrm{b_i} + 1\) or - 1, so that \(0 \leq \mathrm{s} \leq \mathrm{a_1}\) .

Solution: We show that you can pick \(\mathbf{b}_{\mathrm{n}}\) \(\mathbf{b}_{\mathrm{n - 1}}\) , ..., \(\mathbf{b}_{\mathrm{r}}\) so that \(\mathrm{s}_{\mathrm{r}} = \mathrm{b}_{\mathrm{n}}\mathrm{a}_{\mathrm{n}} + \mathrm{b}_{\mathrm{n - 1}}\mathrm{a}_{\mathrm{n - 1}} + \ldots +\mathrm{b}_{\mathrm{r}}\mathrm{a}_{\mathrm{r}}\) satisfies \(0< = \mathrm{s}_{\mathrm{r}}\) \(\leq = \mathrm{a}_{\mathrm{r}}\) . Induction on r. Trivial for \(\mathrm{r} = \mathrm{n}\) . Suppose true for r. Then \(- \mathrm{a}_{\mathrm{r - 1}}< = \mathrm{s}_{\mathrm{r}} - \mathrm{a}_{\mathrm{r - 1}}< = \mathrm{a}_{\mathrm{r}} - \mathrm{a}_{\mathrm{r - 1}}< =\) \(\mathrm{a}_{\mathrm{r - 1}}\) . So with \(\mathbf{b}_{\mathrm{r - 1}} = - 1\) we have \(\left|\mathbf{s}_{\mathrm{r - 1}}\right|< = \mathbf{a}_{\mathrm{r - 1}}\) . If necessary, we change the sign of all \(\mathbf{b}_{\mathrm{n}}\) \(\mathbf{b}_{\mathrm{n - 1}}\) , ..., \(\mathbf{b}_{\mathrm{r - 1}}\) and obtain \(\mathbf{s}_{\mathrm{r - 1}}\) as required. So the result is true for all \(\mathrm{r} > = 1\) and hence for \(\mathrm{r} = 1\)


Prove that you can always draw a circle radius A/P inside a convex polygon with area A and perimeter P.

Solution: Draw a rectangle width A/P on the inside of each side. The rectangles at each vertex must overlap since the angle at the vertex is less than 180. The total area of the rectangles is A, so the area covered must be less than A. Hence we can find a point not in any of the rectangles. But this point must be a distance more than A/P from each side, so we can use it as the center of the required circle.


A graph has at least three vertices. Given any three vertices A, B, C of the graph we can find a path from A to B which does not go through C. Prove that we can find two disjoint paths from A to B.

[A graph is a finite set of vertices such that each pair of distinct vertices has either zero or one edges joining the vertices. A path from A to B is a sequence of vertices \(\mathrm{A_1}\) , \(\mathrm{A_2}\) , ..., \(\mathrm{A_n}\) such that \(\mathrm{A = A_1}\) , \(\mathrm{B = A_n}\) and there is an edge between \(\mathrm{A_i}\) and \(\mathrm{A_{i + 1}}\) for \(\mathrm{i} = 1\) , 2, ..., n- 1. Two paths from A to B are disjoint if the only vertices they have in common are A and B.]

Solution:
Take any path from A to B. Suppose it is \(\mathrm{A = A_0}\) , \(\mathrm{A_1}\) , ..., \(\mathrm{A_n = B}\) . We show by induction on r that we can find two disjoint paths from A to \(\mathrm{A_r}\) . If \(\mathrm{r} = 1\) , then take any vertex C distinct from A and \(\mathrm{A_1}\) . Take any path from \(\mathrm{A_1}\) to C which does not go through A. Now take any path from C to A which does not go through \(\mathrm{A_1}\) . Joining these two paths together gives a path p from A to \(\mathrm{A_1}\) which does not involve the edge \(\mathrm{AA_1}\) . Then p and the edge \(\mathrm{AA_1}\) are the required disjoint paths.

Suppose now we have two disjoint paths \(\mathrm{A}\) , \(\mathrm{B_1}\) , \(\mathrm{B_2}\) , ..., \(\mathrm{B_s}\) , \(\mathrm{A_r}\) and \(\mathrm{A}\) , \(\mathrm{B_t}\) , \(\mathrm{B_{t - 1}}\) , ..., \(\mathrm{B_{s + 1}}\) , \(\mathrm{A_r}\) and we wish to find two disjoint paths joining A and \(\mathrm{A_{r + 1}}\) . Take a path between A and \(\mathrm{A_{r + 1}}\) which does not include \(\mathrm{A_r}\) . If it also avoids all of \(\mathrm{B_1}\) , ..., \(\mathrm{B_t}\) , then we are home, because it is disjoint from the alternative path \(\mathrm{A}\) , \(\mathrm{B_1}\) , \(\mathrm{B_2}\) , ..., \(\mathrm{B_s}\) , \(\mathrm{A_r}\) , \(\mathrm{A_{r + 1}}\) . If not, let \(\mathrm{B_i}\) be the first of the B's on the path as we move from \(\mathrm{A_{r + 1}}\) to A. This allows us to construct two disjoint paths from A to \(\mathrm{A_{r + 1}}\) . One path goes from A to \(\mathrm{B_i}\) and then from \(\mathrm{B_i}\) to \(\mathrm{A_{r + 1}}\) . The other path goes around the other way to \(\mathrm{A_r}\) and then along the edge to \(\mathrm{A_{r + 1}}\) . [Explicitly, if \(\mathrm{i} \leq = \mathrm{s}\) , then the paths are A, \(\mathrm{B_1}\) , \(\mathrm{B_2}\) , ..., \(\mathrm{B_i}\) , ..., (new path) ... \(\mathrm{A_{r + 1}}\) and A, \(\mathrm{B_t}\) , \(\mathrm{B_{t - 1}}\) , ..., \(\mathrm{A_r}\) , \(\mathrm{A_{r + 1}}\) . If \(\mathrm{i} > \mathrm{s}\) , then the paths are A, \(\mathrm{B_t}\) , \(\mathrm{B_{t - 1}}\) , ..., \(\mathrm{B_i}\) , ..., (new path) ... \(\mathrm{A_{r + 1}}\) and A, \(\mathrm{B_1}\) , ..., \(\mathrm{B_s}\) , \(\mathrm{A_r}\) , \(\mathrm{A_{r + 1}}\) .] Hence, by induction, there are two disjoint paths from A to B.


Given a triangle ABC. Suppose the point P in space is such that PH is the smallest of the four altitudes of the tetrahedron PABC. What is the locus of H for all possible P?

Solution:
Answer: the triangle DEF with FAE parallel to BC, DBF parallel to CA and DCE parallel to AB.

Let \(\alpha\) be the angle between planes ABC and PBC. Let h be the perpendicular distance from H to the line BC, and let \(\mathrm{h}_{\mathrm{A}}\) be the perpendicular distance from A to the line BC. Then \(\mathrm{PH} = \mathrm{h}\) tan \(\alpha\) , and the altitude from A to PBC is \(\mathrm{h}_{\mathrm{A}}\) sin \(\alpha\) . Hence if PH is shorter than the altitude from A we require that \(\mathrm{h}< \mathrm{h}_{\mathrm{A}}\cos \alpha < \mathrm{h}_{\mathrm{A}}\) . Similar arguments apply for B and C. So if PH is the shortest then H lies within triangle DEF.

If H does lie within DEF, then if we make \(\alpha\) sufficiently small we will have \(\mathrm{h}< \mathrm{h}_{\mathrm{A}}\cos \alpha\) and hence PH will be shorter than the altitude from A. Similarly we can make PH sufficiently short that PH is less than the altitudes from B and C. Hence the inside of DEF is the required locus.


Given 100 points on the plane. Prove that you can cover them with a collection of circles whose diameters total less than 100 and the distance between any two of which is more than 1. [The distance between circles radii r and s with centers a distance d apart is the greater of 0 and d - r - s.]

Solution: If we have two circles diameters d and \(\mathrm{d}'\) , the distance between which is less than 1, then they are contained in a circle diameter \(\mathrm{d} + \mathrm{d}' + 1\) . [If the line through the centers cuts the circles in A, B, A', B', then take a circle diameter AB'.] So start with 100 circles of diameter 1/1000 each. If any pair is a distance \(\leq = 1\) apart, then replace them by a single circle, increasing the total diameter by 1. Repeat until all the circles are a distance \(>1\) apart. We must end up with at least one circle, so the total increase is at most 99. Hence the final total diameter is at most 99 1/10.


The distance from A to B is d kilometers. A plane P is flying with constant speed, height and direction from A to B. Over a period of 1 second the angle PAB changes by \(\alpha\) degrees and the angle PBA by \(\beta\) degrees. What is the minimal speed of the plane?

Solution:
Answer: \(20\pi \mathrm{d}\sqrt{(\alpha\beta)}\) kilometers per hour.

Let the plane be at height h and a (horizontal) distance y from A. Let the angle PAB be \(\theta + \alpha\) and the angle PBA be \(\phi\) . After 1 second, the angle PAB is \(\theta\) and the angle PBA is \(\phi + \beta\) . We have immediately that:

\[\mathrm{h / y = tan(\theta + \alpha),h / (d - y) = tan\phi,h / (y + x) = tan\theta,h / (d - y - x) = tan(\phi + \beta).}\]

Eliminating \(\theta\) , we obtain: \(\mathrm{h / y = (tan\alpha + tan\theta) / (1 - tan\alpha tan\theta) = (a(y + x) + h) / (y + x - ah)}\) , where \(\mathrm{a} = \mathrm{tan}\alpha\) . Hence \(\mathrm{x} = \mathrm{a}(\mathrm{h}^{2} + \mathrm{y}^{2}) / (\mathrm{h} - \mathrm{ay})\) . Similarly, eliminating \(\phi\) , we obtain \(\mathrm{x} = \mathrm{b}(\mathrm{h}^{2} + (\mathrm{d} - \mathrm{y})^{2}) / (\mathrm{h} + (\mathrm{d} - \mathrm{y})\mathrm{b})\) .

At this point I do not see how to make further progress without approximating. But approximating seems reasonable, since \(\alpha\) and \(\beta\) , are certainly small, at least when expressed in radians. For example, typical values might be 10,000 ft for h and more than 10 miles for y or d- y and 500 mph for the aircraft speed. That gives \(\mathrm{x} = 0.14\) miles, so \(\mathrm{x / y} = 0.014\) and \(\mathrm{x / h} = 0.07\) . So, let us neglect \(\mathrm{a / h}\) , \(\mathrm{b / h}\) , \(\mathrm{a / y}\) etc. Then we get the simplified expressions: \(\mathrm{x} = \mathrm{a}(\mathrm{h}^{2} + \mathrm{y}^{2}) / \mathrm{h} = \mathrm{b}(\mathrm{h}^{2} + (\mathrm{d} - \mathrm{y})^{2}) / \mathrm{h}\) .

If \(\mathrm{a} = \mathrm{b}\) , then we quickly obtain \(\mathrm{y} = \mathrm{d} / 2\) , \(\mathrm{h} = \mathrm{d} / 2\) , \(\mathrm{x} = \mathrm{ad}\) . Assume \(\mathrm{a} > \mathrm{b}\) . Then we can solve for \(\mathrm{h}\) , substitute back in and obtain an expression for \(\mathrm{x}\) in terms of \(\mathrm{y}\) . It is convenient to divide through by \(\mathrm{d}\) and to write \(\mathrm{X} = \mathrm{x} / \mathrm{d}\) , \(\mathrm{Y} = \mathrm{y} / \mathrm{d}\) . Note that since we are assuming \(\mathrm{a} > \mathrm{b}\) , we require \(\mathrm{Y} < 1 / (1 + \sqrt{(\mathrm{a} / \mathrm{b})})\) . After some manipulation we obtain: \(\mathrm{X} = \mathrm{ab}(1 - 2\mathrm{Y}) / \sqrt{((\mathrm{a} - \mathrm{b})(\mathrm{b}(1 - \mathrm{Y})^{2} - \mathrm{a}\mathrm{Y}^{2})}\) . Differentiating, we find that there is a minimum at \(\mathrm{Y} = \mathrm{b} / (\mathrm{a} + \mathrm{b})\) , which is in the allowed range, and that the minimum value of \(\mathrm{X}\) is \(\sqrt{(\mathrm{ab})}\) . By symmetry, we obtain the same result for \(\mathrm{a} < \mathrm{b}\) and we notice that it is also true for \(\mathrm{a} = \mathrm{b}\) . So in all cases we have that the minimum value of \(\mathrm{x}\) is \(\mathrm{d}\sqrt{(\mathrm{ab})}\) .

We are assuming \(\alpha\) and \(\beta\) are small, so we may take \(\mathrm{a} = \alpha\) , \(\mathrm{b} = \beta\) . However, the question specified that \(\alpha\) and \(\beta\) were measured in degrees, so to obtain the final answer we must convert, giving: \(\mathrm{x} = \mathrm{d}(\pi / 180) \sqrt{(\alpha\beta)}\) , and hence \(\mathrm{speed} = 20\pi \mathrm{d}\sqrt{(\alpha\beta)}\) kilometers per hour.