Let $(\mathrm{a}, \mathrm{b}, \ldots, \mathrm{k})$ denote the greatest common divisor of the integers $\mathrm{a}, \mathrm{b}, \ldots \mathrm{k}$ and $[\mathrm{a}, \mathrm{b}, \ldots, \mathrm{k}]$ denote their least common multiple. Show that for any positive integers $\mathrm{a}, \mathrm{b}, \mathrm{c}$ we have $(\mathrm{a}, \mathrm{b}, \mathrm{c})^{2}[\mathrm{a}, \mathrm{b}][\mathrm{b}, \mathrm{c}][\mathrm{c}, \mathrm{a}]=[\mathrm{a}, \mathrm{b}$, $c]^{2}(a, b)(b, c)(c, a)$.

Solution: If we express $\mathrm{a}, \mathrm{b}, \mathrm{c}$ as a product of primes then the $\mathrm{gcd}$ has each prime to the smallest power and the $1 \mathrm{~cm}$ has each prime to the largest power. So the equation given is equivalent to showing that $2 \min (\mathrm{r}, \mathrm{s}, \mathrm{t})+\max (\mathrm{r}, \mathrm{s})$ $+\max (\mathrm{s}, \mathrm{t})+\max (\mathrm{t}, \mathrm{r})=2 \max (\mathrm{r}, \mathrm{s}, \mathrm{t})+\min (\mathrm{r}, \mathrm{s})+\min (\mathrm{s}, \mathrm{t})+\min (\mathrm{t}, \mathrm{r})$ for non-negative integers $\mathrm{r}, \mathrm{s}, \mathrm{t}$. Assume $\mathrm{r} \leq \mathrm{s} \leq \mathrm{t}$. Then each side is $2 \mathrm{r}+\mathrm{s}+2 \mathrm{t}$.


A tetrahedron has opposite sides equal. Show that all faces are acute-angled.

Solution: Let the tetrahedron be $A B C D$. Let $M$ be the midpoint of $B C$. We have $A M+M D>A D(*)$. Now the triangles $\mathrm{ABC}$ and $\mathrm{DCB}$ are congruent because $\mathrm{AB}=\mathrm{DC}, \mathrm{BC}=\mathrm{CB}$ and $\mathrm{AC}=\mathrm{DB}$. Hence $\mathrm{AM}=\mathrm{DM}$. Also $\mathrm{AD}=\mathrm{BC}=2 \mathrm{MC}$. So $(*)$ implies that $\mathrm{AM}>\mathrm{MC}$. But that implies that angle $\mathrm{BAC}$ is acute. Similarly for all the other angles.


$\mathrm{n}$ digits, none of them 0 , are randomly (and independently) generated, find the probability that their product is divisible by 10 .

Solution:
Answer: $1-(8 / 9)^{n}-(2 / 9)^{n}+(4 / 9)^{n}$.

A number is divisible by 10 iff it has an even number and a 5 amongst its digits. The probability of no 5 is $(8 / 9)^{\mathrm{n}}$. The probability of no even number is $(5 / 9)^{\mathrm{n}}$. The probability of no 5 and no even number is $(4 / 9)^{\mathrm{n}}$. Hence result.


Let $k$ be the real cube root of 2. Find integers A, B, C, a, b, c such that $\left|\left(A x^{2}+B x+C\right) /\left(a x^{2}+b x+c\right)-k\right|<\mid$ $\mathrm{x}-\mathrm{k} \mid$ for all non-negative rational $\mathrm{x}$.

Solution:
Taking the limit, we must have $\left(\mathrm{Ak}^{2}+\mathrm{Bk}+\mathrm{C}\right)=\mathrm{k}\left(\mathrm{ak}^{2}+\mathrm{bk}+\mathrm{c}\right)$, so $\mathrm{A}=\mathrm{b}, \mathrm{B}=\mathrm{c}, \mathrm{C}=2 \mathrm{a}$. Now notice that $\left(b x^{2}+c x+2 a\right)-k\left(a x^{2}+b x+c\right)=(b-a k) x^{2}+(c-b k) x+(2 a-c k)=(x-k)\left((b-a k) x+c-a k^{2}\right)$. So we require $\left|(b-a k) x+c-\mathrm{ak}^{2}\right|<\left|a x^{2}+b x+c\right|$ for all $x \geq 0$.

There are many ways to satisfy this. For example, take $\mathrm{a}=1, \mathrm{~b}=\mathrm{c}=2$. Then ( $\mathrm{b}-\mathrm{ak}) \mathrm{x}$ is always positive and less than $\mathrm{bx}$ for positive $\mathrm{x}$, and $\mathrm{c}-\mathrm{ak}^{2}$ is positive and less than $\mathrm{c}$.


A pentagon is such that each triangle formed by three adjacent vertices has area 1. Find its area, but show that there are infinitely many incongruent pentagons with this property.

Solution:
Let the pentagon be ABCDE. Triangles BCD and ECD have the same area, so $\mathrm{B}$ and $\mathrm{E}$ are the same perpendicular distance from $\mathrm{CD}$, so $\mathrm{BE}$ is parallel to $\mathrm{CD}$. The same applies to the other diagonals (each is parallel to the side with which it has no endpoints in common). Let BD and CE meet at X. Then ABXE is a parallelogram, so area $\mathrm{BXE}=$ area $\mathrm{EAB}=1$. Also area $\mathrm{CDX}+$ area $\mathrm{EDX}=$ area $\mathrm{CDX}+$ area $\mathrm{BCX}=1$. Put
area $\mathrm{EDX}=\mathrm{x}$. Then $\mathrm{DX} / \mathrm{XB}=$ area $E D X /$ area $\mathrm{BXE}=\mathrm{x} / 1$ and also $=$ area $\mathrm{CDX} / \operatorname{area} \mathrm{BCX}=(1-\mathrm{x}) / \mathrm{x}$. So $\mathrm{x}^{2}+$ $x-1=0, x=\sqrt{5}-1) / 2$ (we know $x<0$, so it cannot be the other root). Hence area $\operatorname{ABCDE}=3+x=(\sqrt{ } 5+$ $5) / 2$.

Take any triangle XCD of area $(3-\sqrt{5}) / 2$ and extend DX to $B$, so that $B C D$ has area 1 , and extend CX to E so that CDE has area 1. Then take BA parallel to CE and EA parallel to BD. It is easy to check that the pentagon has the required property.


Show that if two points lie inside a regular tetrahedron the angle they subtend at a vertex is less than $\pi / 3$.

Solution:
Let the tetrahedron be $\mathrm{ABCD}$ and the points be $\mathrm{P}$ and $\mathrm{Q}$. Note that we are asked to prove the result for any vertex, not just some. So consider angle PAQ. Let the rays AP, AQ meet the plane BCD at $\mathrm{P}^{\prime}, \mathrm{Q}^{\prime}$ respectively. So we have to show that angle $\mathrm{P}^{\prime} \mathrm{AQ}^{\prime}<60^{\circ}$ for $\mathrm{P}^{\prime}$ and $\mathrm{Q}^{\prime}$ interior points of the triangle $\mathrm{BCD}$. Extend $\mathrm{P}^{\prime} \mathrm{Q}^{\prime}$ to meet the sides of the triangle at $\mathrm{X}$ and $\mathrm{Y}$. Without loss of generality, $\mathrm{X}$ lies $\mathrm{BC}$ and $\mathrm{Y}$ lies on CD. Obviously if is sufficient to show that angle XAY $<60^{\circ}$.

$\mathrm{X}$ and $\mathrm{Y}$ cannot both be vertices (or $\mathrm{P}^{\prime}, \mathrm{Q}^{\prime}$ would not have been interior points of the triangle and hence $\mathrm{P}$ and $\mathrm{Q}$ would not have been strictly inside the tetrahedron). So suppose $\mathrm{X}$ is not a vertex. We show that $\mathrm{XY}<=$ $\mathrm{XD}$. Consider triangle XYD. $\angle \mathrm{XDY}<\angle \mathrm{BDC}=60^{\circ}$, but $\angle \mathrm{XYD}=\angle \mathrm{XCD}+\angle \mathrm{CXY} \geq 60^{\circ}$, so $\angle \mathrm{XDY}<$ $\angle \mathrm{XYD}$. Hence $\mathrm{XY} \leq \mathrm{XD}$. But XD = AX (consider, for example, the congruent triangles AXB and DXB). Hence $X Y<$ AX. Similarly, XY $\leq$ AY. Hence angle XAY $<60^{\circ}$.


The sequence $a_{n}$ is defined by $a_{1}=a_{2}=1, a_{n+2}=a_{n+1}+2 a_{n}$. The sequence $b_{n}$ is defined by $b_{1}=1, b_{2}=7$, $b_{n+2}=2 b_{n+1}+3 b_{n}$. Show that the only integer in both sequences is 1 .

Solution:
We can solve the first recurrence relation to give $a_{n}=A(-1)^{n}+B 2^{n}$. Using $a_{1}$ and $a_{2}$, we get $a_{n}=\left(2^{n}+(-\right.$ $\left.1)^{n+1}\right) / 3$. Similarly, for the second recurrence relation we get $b_{n}=2 \cdot 3^{n-1}+(-1)^{n}$. So if $a_{m}=b_{n}$ then $2 \cdot 3^{n}+3(-$ $1)^{\mathrm{n}}=2^{\mathrm{m}}+(-1)^{\mathrm{m}+1}$ or $2^{\mathrm{m}}=2.3^{\mathrm{n}}+3(-1)^{\mathrm{n}}+(-1)^{\mathrm{m}}$.

If $m=1$ or 2 , then we find $n=1$ is the only solution, corresponding to the fact that the term 1 is in both sequences. If $\mathrm{m}>2$, then $2^{\mathrm{m}}=0 \bmod 8$. But $3^{\mathrm{n}}=(-1)^{\mathrm{n}} \bmod 4$, so $2.3^{\mathrm{n}}+3(-1)^{\mathrm{n}}+(-1)^{\mathrm{m}}=5(-1)^{\mathrm{n}}+(-1)^{\mathrm{m}} \bmod 8$ which cannot be $0 \bmod 8$. So there are no solutions for $\mathrm{m}>2$.


Three vertices of a regular $2 n+1$ sided polygon are chosen at random. Find the probability that the center of the polygon lies inside the resulting triangle.

Solution:
Answer: $(n+1) /(4 n-2)$.

Label the first vertex picked as 1 and the others as $2,3, \ldots, 2 n+1$ (in order). There are $2 n(2 n-1) / 2$ ways of choosing the next two vertices. If the second vertex is 2 (or $2 n+1$ ), then there is just one way of picking the third vertex so that the center lies in the triangle (vertex $n+2$ ). If the second vertex is 3 (or $2 n$ ), then there are two $(n+2, n+3)$ and so on. So the total number of favourable triangles is $2(1+2+\ldots+n)=n(n+1)$. Thus the required probability is $(n+1) /(4 n-2)$.


Find all complex numbers $x, y, z$ which satisfy $x+y+z=x^{2}+y^{2}+z^{2}=x^{3}+y^{3}+z^{3}=3$.

Solution:
Answer: $1,1,1$.

We have $(x+y+z)^{2}=\left(x^{2}+y^{2}+z^{2}\right)+2(x y+y z+z x)$, so $x y+y z+z x=3$.

$(x+y+z)^{3}=\left(x^{3}+y^{3}+z^{3}\right)+3\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+6 x y z$, so $8=\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x\right.$ $\left.+z x^{2}\right)+2 x y z$. But $(x+y+z)(x y+y z+z x)=\left(x^{2} y+x y^{2}+y^{2} z+y z^{2}+z^{2} x+z x^{2}\right)+3 x y z$, so $x y z=-1$. Hence $x, y, z$ are the roots of the cubic $w^{3}-3 w^{2}+3 w-1=(w-1)^{3}$. Hence $x=y=z=1$.


Show that the cube roots of three distinct primes cannot be terms in an arithmetic progression (whether consecutive or not).

Solution:
Suppose the primes are $p, q, r$ so that $q^{1 / 3}=p^{1 / 3}+m d, r^{1 / 3}=p^{1 / 3}+n d$, where $m$ and $n$ (but not necessarily $d$ ) are integers. Then $n q^{1 / 3}-m r^{1 / 3}=(n-m) p^{1 / 3}$. Cubing: $n^{3} q-3 n^{2} m q^{2 / 3} r^{1 / 3}+3 n m^{2} q^{1 / 3} r^{2 / 3}-m^{3} r=(n-m)^{3} p$, or $q^{1 / 3} r^{1 / 3}\left(m r^{1 / 3}-n q^{1 / 3}\right)=\left((n-m)^{3} p+m^{3} r-n^{3} q\right) /(3 m n)$. But $_{m r^{1 / 3}}-n q^{1 / 3}=(m-n) p^{1 / 3}$, so we have $(p q r)^{1 / 3}=((n-$ $\left.\mathrm{m})^{3} \mathrm{p}+\mathrm{m}^{3} \mathrm{r}-\mathrm{n}^{3} \mathrm{q}\right) /(3 \mathrm{mn}(\mathrm{m}-\mathrm{n}))(*)$.

It is now clear that we do not need $\mathrm{p}, \mathrm{q}, \mathrm{r}$ prime, just that $\mathrm{pqr}$ is not a cube, for then by the usual argument it must be irrational so that $(*)$ is impossible.


$p(x)$ is a polynomial with integral coefficients. Show that there are no solutions to the equations $p(a)=b$, $\mathrm{p}(\mathrm{b})=\mathrm{c}, \mathrm{p}(\mathrm{c})=\mathrm{a}$, with $\mathrm{a}, \mathrm{b}, \mathrm{c}$ distinct integers.

Solution:
Suppose there $\mathrm{a}, \mathrm{b}, \mathrm{c}$ satisfy the equations. Then $\mathrm{p}(\mathrm{x})=(\mathrm{x}-\mathrm{a}) \mathrm{q}(\mathrm{x})+\mathrm{b}=(\mathrm{x}-\mathrm{b}) \mathrm{r}(\mathrm{x})+\mathrm{c}=(\mathrm{x}-\mathrm{c}) \mathrm{s}(\mathrm{x})+\mathrm{a}$ for some polynomials $q(x), r(x), s(x)$ with integer coefficients. Hence $(b-a) q(b)+b=p(b)=c$, so $(b-a)$ divides $(\mathrm{c}-\mathrm{b})$. Similarly, $(\mathrm{c}-\mathrm{b})$ divides $(\mathrm{a}-\mathrm{c})$, and $(\mathrm{a}-\mathrm{c})$ divides $(\mathrm{b}-\mathrm{a})$. But $(\mathrm{b}-\mathrm{a})$ divides $(\mathrm{c}-\mathrm{b})$ divides $(\mathrm{a}-\mathrm{c})$ implies that $(\mathrm{b}-\mathrm{a})$ divides $(\mathrm{a}-\mathrm{c})$. So we have $(\mathrm{b}-\mathrm{a})$ and $(\mathrm{a}-\mathrm{c})$ dividing each other. Hence $(\mathrm{b}-\mathrm{a})= \pm(\mathrm{a}-\mathrm{c})$.

If $\mathrm{b}-\mathrm{a}=\mathrm{a}-\mathrm{c}$, then $\mathrm{b}-\mathrm{c}=(\mathrm{b}-\mathrm{a})+(\mathrm{a}-\mathrm{c})=2(\mathrm{a}-\mathrm{c})$. But $(\mathrm{b}-\mathrm{a})$ divides $2(\mathrm{a}-\mathrm{c})$ (and both are non-zero since $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are distinct), so that is impossible. If $\mathrm{b}-\mathrm{a}=-(\mathrm{a}-\mathrm{c})$, the $\mathrm{b}=\mathrm{c}$, contradicting the fact that they are distinct. So there are no solutions.


Show that for any positive reals $\mathrm{x}, \mathrm{y}, \mathrm{z}$ we have $\mathrm{x}^{\mathrm{x}} \mathrm{y}^{\mathrm{y}} \mathrm{z}^{\mathrm{z}} \geq(\mathrm{xyz})^{\mathrm{a}}$, where $\mathrm{a}$ is the arithmetic mean of $\mathrm{x}, \mathrm{y}, \mathrm{z}$.

Solution: Without loss of generality $x \geq y \geq z$. We have $x^{x} y^{y} \geq x^{y} y^{x}$, because that is equivalent to $(x / y)^{x} \geq(x / y)^{y}$ which is obviously true. Similarly $y^{y} z^{z} \geq y^{z} z^{y}$ and $z^{z} x^{x} \geq z^{x} x^{z}$. Multiplying these three together we get $\left(x^{x} y^{y} z^{z}\right)^{x} \geq$ $x^{y+z} y^{z+x} z^{x+y}$. Multiplying both sides by $x^{x} y^{y} z^{z}$ gives $\left(x^{x} y^{y} z^{z}\right)^{3} \geq(x y z)^{3 a}$. Taking cube roots gives the required result.


Two points in a thin spherical shell are joined by a curve shorter than the diameter of the shell. Show that the curve lies entirely in one hemisphere.

Solution: Suppose the shell has diameter 2. Let $\mathrm{M}$ be the midpoint of the curve. Let $\mathrm{O}$ be the center of the shell and $\mathrm{X}$ the midpoint of MO. Let $S$ be the circle center $X$ radius $\sqrt{3}) / 2$ in the plane normal to OM. Then $S$ lies in the shell and every point of $\mathrm{S}$ is a distance (in space) of 1 from M. Hence the curve cannot cross $\mathrm{S}$ (because if it crossed at $\mathrm{Y}$, we would have $\mathrm{SY} \geq 1$ along the curve, but the curve has length $<2$, so $\mathrm{SY}<1$. So we have a stronger result than required.


A, B, C play a series of games. Each game is between two players, The next game is between the winner and the person who was not playing. The series continues until one player has won two games. He wins the series. A is the weakest player, C the strongest. Each player has a fixed probability of winning against a given opponent. A chooses who plays the first game. Show that he should choose to play himself against B.

Solution:
It must be wrong to choose $\mathrm{B}$ against $\mathrm{C}$, for then after the first game (whatever its outcome) A would be playing one of the other players $(\mathrm{X})$, and that player would already have won a game. That is a worse position than playing that person as the first game, because if he loses the game then $\mathrm{X}$ has won the series, whereas if he lost to $\mathrm{X}$ on the first game, there is still a chance $\mathrm{A}$ could win the series. [If he wins the game as the second game, then he is certainly no better off than he would be after winning the match as the first game.]

Use $\mathrm{XbY}$ to denote that $\mathrm{X}$ beats $\mathrm{Y}$. If $\mathrm{A}$ chooses to play $\mathrm{B}$ in the first game, then he wins the series if either (1) $\mathrm{AbB}, \mathrm{AbC}$, (2) $\mathrm{AbB}, \mathrm{CbA}, \mathrm{BbC}, \mathrm{AbB}$, or (3) $\mathrm{BbA}, \mathrm{CbB}, \mathrm{AbC}$, $\mathrm{AbB}$. If $\mathrm{A}$ plays $\mathrm{C}$ in the first game, then he wins the series if (1') $\mathrm{AbC}, \mathrm{AbB},\left(2^{\prime}\right) \mathrm{AbC}, \mathrm{BbA}, \mathrm{CbB}, \mathrm{AbC},\left(3^{\prime}\right) \mathrm{CbA}, \mathrm{BbC}, \mathrm{AbB}, \mathrm{AbC}$. Evidently the probability of (1) is the same as the probability of (1'). If we compare (2) and (3) they are the same except that in (2) A must beat B and in (3') A must beat C. Similarly, if we compare (3) and (2') they are the same except that in (3) A must beat B and in (2') A must beat C. We assume that, since C is a stronger player than $\mathrm{B}, \mathrm{A}$ is more likely to beat $\mathrm{B}$ than $\mathrm{C}$. Hence $\operatorname{prob}(2)>\operatorname{prob}\left(3^{\prime}\right)$ and $\operatorname{prob}(3)>\operatorname{prob}\left(2^{\prime}\right)$. Thus A should choose to play B in the first game.


A point inside an equilateral triangle with side 1 is a distance $\mathrm{a}, \mathrm{b}, \mathrm{c}$ from the vertices. The triangle $\mathrm{ABC}$ has $\mathrm{BC}=\mathrm{a}, \mathrm{CA}=\mathrm{b}, \mathrm{AB}=\mathrm{c}$. The sides subtend equal angles at a point inside it. Show that sum of the distances of the point from the vertices is 1 .

Solution:
Let $\mathrm{D}$ be the point inside $\mathrm{ABC}$, so that $\angle \mathrm{ADB}=\angle \mathrm{BDC}=120^{\circ}$. The key is to start from $\mathrm{ABC}$ and to rotate the triangle $\mathrm{BDC}$ through $60^{\circ}$ away from the triangle $\mathrm{ADB}$. After that everything is routine.

Suppose $\mathrm{D}$ goes to $\mathrm{D}^{\prime}$ and $\mathrm{C}$ to $\mathrm{C}^{\prime}$. Then $\mathrm{BD}=\mathrm{BD}^{\prime}$ and $\angle \mathrm{DBD}^{\prime}=60^{\circ}$, so $\mathrm{BDD}^{\prime}$ is equilateral. Hence $\angle \mathrm{D}^{\prime} \mathrm{DB}$ $=60^{\circ} . \angle \mathrm{BDA}=120^{\circ}$, so $\mathrm{ADD}^{\prime}$ is a straight line. Also $\angle \mathrm{DD} ' \mathrm{~B}=60^{\circ}$ and $\angle \mathrm{C}^{\prime} \mathrm{D}^{\prime} \mathrm{B}=120^{\circ}$, so $\mathrm{DD}^{\prime} \mathrm{C}^{\prime}$ is a straight line. Thus $A C^{\prime}$ has length DA + DB + DC.

Note that $\mathrm{BC}=\mathrm{BC}^{\prime}$ and $\angle \mathrm{CBC}^{\prime}=60^{\circ}$, so $\mathrm{CBC}^{\prime}$ is equilateral. Hence $\angle \mathrm{CC} \mathrm{C}^{\prime} \mathrm{B}=60^{\circ}$. Now take $\mathrm{Y}$ such that $\mathrm{AC}^{\prime} \mathrm{Y}$ is equilateral, $\mathrm{Y}$ is on the opposite side of $\mathrm{AC}^{\prime}$ to $\mathrm{C}$. Then $\angle \mathrm{BC}^{\prime} \mathrm{Y}=60^{\circ}-\angle \mathrm{AC} \mathrm{C}^{\prime} \mathrm{B}=\angle \mathrm{CC}^{\prime} \mathrm{A}$. Also $\mathrm{BC}^{\prime}=\mathrm{CC}^{\prime}$ and $\mathrm{YC}^{\prime}=\mathrm{AC}^{\prime}$, so triangles $\mathrm{BC}^{\prime} \mathrm{Y}$ and $\mathrm{CC}^{\prime} \mathrm{A}$ are congruent. Hence $\mathrm{BY}=\mathrm{CA}=\mathrm{b}$. Also $\mathrm{BC}^{\prime}=\mathrm{BC}=$ $\mathrm{a}$ and $\mathrm{BA}=\mathrm{c}$. Thus $\mathrm{B}$ is a point inside an equilateral triangle and distances $\mathrm{a}, \mathrm{b}, \mathrm{c}$ from the vertices. Hence the triangle must have side 1 . So $\mathrm{DA}+\mathrm{DB}+\mathrm{DC}=\mathrm{AC}^{\prime}=1$.


Show that for any non-negative reals $\mathrm{x}, \mathrm{y},[5 \mathrm{x}]+[5 \mathrm{y}] \geq[3 \mathrm{x}+\mathrm{y}]+[\mathrm{x}+3 \mathrm{y}]$. Hence or otherwise show that (5a)! $(5 b)!/(a!b!(3 a+b)!(a+3 b)!)$ is integral for any positive integers $a, b$.

Solution:
If is obviously sufficient to prove that $[5 x]+[5 y] \geq[3 x+y]+[x+3 y]$ for $0<x<y<1$. If $2 x \geq y$, then $[5 x] \geq$ $[3 x+y]$ and $[5 y] \geq[x+3 y]$, so the result holds. So assume $2 x<y$. It is now a question of examining a lot of cases.

If $y<2 / 5$, then $3 x+y<5 y / 2<1$, so $[3 x+y]=0$, and $[3 y+x]<=[5 y]$, so the result holds. If $2 / 5 \leq y<3 / 5$ and $x<1 / 5$, then $[5 y]=2,[3 x+y]=0$ or 1 and $[x+3 y]=1$. If $2 / 5 \leq y<3 / 5$ and $x \geq 1 / 5$, the $[5 x]+[5 y]=3,[3 x+y]$
$=1,[x+3 y]=1$ or 2 . If $3 / 5 \leq y<4 / 5$, then $[5 y]=3,[3 x+y]=0$ or $1,[x+3 y]=2$ or 3 . If $y \geq 4 / 5$ and $x<1 / 5$, then $[5 x]+[5 y]=4,[3 x+y]=0$ or $1,[x+3 y]=2$ or 3 . If $y \geq 4 / 5$ and $1 / 5 \leq x$, then $[5 x]+[5 y]=5,[3 x+y]=2$, $[\mathrm{x}+3 \mathrm{y}]=2$ or 3 .

The highest power of a prime $p$ dividing $n!$ is $[n / p]+\left[n / p^{2}\right]+\left[n / p^{3}\right]+\ldots$. Thus it is sufficient to show that $[5 \mathrm{~m} / \mathrm{p}]+[5 \mathrm{n} / \mathrm{p}] \geq[(3 \mathrm{~m}+\mathrm{n}) / \mathrm{p}]+[(\mathrm{m}+3 \mathrm{n}) / \mathrm{p}]$, which follows immediately from the previous result, putting $\mathrm{x}$ $=\mathrm{m} / \mathrm{p}, \mathrm{y}=\mathrm{n} / \mathrm{p}$.


Show that for any tetrahedron the sum of the squares of the lengths of two opposite edges is at most the sum of the squares of the other four.

Solution: Let the vertices be A, B, C, D. We will show that $\mathrm{AB}^{2}+\mathrm{CD}^{2} \leq \mathrm{AC}^{2}+\mathrm{AD}^{2}+\mathrm{BC}^{2}+\mathrm{BD}^{2}$. Use vectors with origin A. Let the vectors from A to B, C, D be B, C, D respectively. We have to show that $\mathbf{B}^{2}+(\mathbf{C}-$ D) $)^{2} \leq \mathbf{C}^{2}+\mathbf{D}^{2}+(\mathbf{B}-\mathbf{C})^{2}+(\mathbf{B}-\mathbf{D})^{2}$. Rearranging, this is equivalent to $(\mathbf{B}-\mathbf{C}-\mathbf{D})^{2} \geq 0$.


A polynomial $p(x)$ of degree $n$ satisfies $p(0)=0, p(1)=1 / 2, p(2)=2 / 3, \ldots, p(n)=n /(n+1)$. Find $p(n+1)$.

Solution: The polynomial $(\mathrm{x}+1) \mathrm{p}(\mathrm{x})-\mathrm{x}$ has degree $\mathrm{n}+1$ and zeros at $0,1,2, \ldots, \mathrm{n}$, so it must be $\mathrm{k} \mathrm{x}(\mathrm{x}-1)(\mathrm{x}-2) \ldots(\mathrm{x}$ $-n)$. Also it has value 1 at $x=-1$, so $k(n+1)!(-1)^{n+1}$. Hence $(n+2) p(n+1)-(n+1)=(-1)^{n+1}$. So for $n$ odd, $p(n+1)=1$, and for $n$ even, $p(n+1)=n /(n+2)$.


Two circles intersect at two points, one of them $\mathrm{X}$. Find $\mathrm{Y}$ on one circle and $\mathrm{Z}$ on the other, so that $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are collinear and $\mathrm{XY} . \mathrm{XZ}$ is as large as possible.

Solution:
Let the circle through $\mathrm{XY}$ have center $\mathrm{O}$ and the other circle have center $\mathrm{O}$ '. $\mathrm{XY}=2 \mathrm{OX} \sin \mathrm{XOY}, \mathrm{XZ}=2$ $\mathrm{O}^{\prime} \mathrm{X} \sin \mathrm{XOZ}$, so we wish to maximise $2 \sin \mathrm{XOY} \sin \mathrm{XO}$ 'Z. But $1 / 2 \angle \mathrm{XOY}=90^{\circ}-\angle \mathrm{OXY}, 1 / 2 \angle \mathrm{XO} Z=$ $90^{\circ}-\angle \mathrm{OXZ}$, so $1 / 2 \angle \mathrm{XOY}+1 / 2 \angle \mathrm{XO} \mathrm{Z}=\angle \mathrm{OXO}$ ', which is fixed. Thus $\angle \mathrm{XOY}+\angle \mathrm{XO} ' \mathrm{Z}$ is fixed. But $2 \sin \mathrm{XOY} \sin \mathrm{XO} \mathrm{Z}=\cos \left(\mathrm{XOY}-\mathrm{XO}^{\prime} \mathrm{Z}\right)-\cos (\mathrm{XOY}+\mathrm{XO} \mathrm{O})$ ), so we maximise by taking $\mathrm{XOY}=\mathrm{XO} \mathrm{Z}$.

Note that in this case $\angle \mathrm{OYZ}=\angle \mathrm{O}$ 'ZY, so if we extend $\mathrm{YO}$ and $\mathrm{ZO}$ ' to meet at $\mathrm{C}$, then $\mathrm{CY}=\mathrm{CZ}$ and hence $\mathrm{C}$ is the center of a circle containing the two circles and touching them at $\mathrm{Y}$ and $\mathrm{Z}$. Also $\angle \mathrm{CZY}=\angle \mathrm{OXY}$, so O'Z is parallel to OX. Similarly OY is parallel to O'X, which shows how to construct $\mathrm{Y}$ and $\mathrm{Z}$.


A pack of $\mathrm{n}$ cards, including three aces, is well shuffled. Cards are turned over in turn. Show that the expected number of cards that must be turned over to reach the second ace is $(\mathrm{n}+1) / 2$.

Solution: For each arrangement A of the cards, let $\mathrm{A}^{\prime}$ be the reflection about the middle of the pack, so that if a card is in position $\mathrm{m}$ in $\mathrm{A}$, then it is in position ( $\mathrm{n}+1-\mathrm{m}$ ) in $\mathrm{A}^{\prime}$. Then all possible arrangements can be grouped into pairs ( $A, A^{\prime}$ ) (note that $A$ cannot equal $A^{\prime}$ ). If the position of the second ace in $A$ is $m$, then it is $n+1-m$ in $A^{\prime}$, so the average over $\mathrm{A}$ and $\mathrm{A}^{\prime}$ is $(\mathrm{n}+1) / 2$. Hence that is also the average over all the arrangements.


The squares of a $4 \times 7$ chess board are colored red or blue. Show that however the coloring is done, we can find a rectangle with four distinct corner squares all the same color. Find a counter-example to show that this is not true for a $4 \times 6$ board.

Solution:
A counter-example for $4 \mathrm{x} 6$ is:

| $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{B}$ |
| :--- | :--- | :--- | :--- | :--- | :--- |
| $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{R}$ |
| $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{B}$ | $\mathrm{R}$ |
| $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{B}$ | $\mathrm{R}$ | $\mathrm{R}$ | $\mathrm{B}$ |

Every column has two blue and two red squares and no two columns have the red squares in the same two rows or the blue squares in the same two rows, so there can be no rectangles.

Suppose there is a counter-example for a $4 \times 7$ rectangle. Suppose it has three red squares in the first column. Then in those rows each remaining column can have at most one red square, so four remaining columns each have at least two blue squares in those three columns. Hence two of those columns have blue squares in the same two rows and hence a blue cornered rectangle. Contradiction. Similarly if there are three blue squares in the first column. So each column must have two red and two blue squares. But there are only 6 ways of choosing 2 items from 4, so two columns must have red squares in the same rows. Contradiction.


$\mathrm{AB}$ is a fixed chord of a circle, not a diameter. $\mathrm{CD}$ is a variable diameter. Find the locus of the intersection of $\mathrm{AC}$ and $\mathrm{BD}$.

Solution:
Let the lines meet at $\mathrm{X}$ and suppose $\mathrm{X}$ lies outside the circle. $\angle \mathrm{AXB}=\angle \mathrm{AXD}$ (same angle) $=180 \mathrm{deg}-\angle$ $\mathrm{XAD}-\angle \mathrm{XDA}=90^{\circ}-\angle \mathrm{XDA}$ (CD is a diameter, so angle $\left.\mathrm{CAD}=90^{\circ}\right)=90^{\circ}-\angle \mathrm{BDA}$ (same angle). But $\angle$ BDA is constant, so $\angle \mathrm{AXB}$ is constant and hence $\mathrm{X}$ lies on a circle through $\mathrm{A}$ and $\mathrm{B}$.

Let $\mathrm{O}$ be the center of the circle $\mathrm{ABC}$ and $\mathrm{O}^{\prime}$ the center of the circle $\mathrm{ABX}$. We have $\angle \mathrm{AOB}=2 \angle \mathrm{ADB}, \angle$ $\mathrm{AO}^{\prime} \mathrm{B}=2 \angle \mathrm{AXB}$, so $\angle \mathrm{AOB}+\angle \mathrm{AO}^{\prime} \mathrm{B}=180^{\circ}$. Hence $\angle \mathrm{OAO}^{\prime}+\angle \mathrm{OBO}^{\prime}=180^{\circ}$. But $\angle \mathrm{OAO}^{\prime}=\angle \mathrm{OBO}^{\prime}$, so $\angle \mathrm{OAO}^{\prime}=90^{\circ}$. In other words, the circles are orthogonal.

If $\mathrm{X}$ lies inside the circle center $\mathrm{O}$, then $\angle \mathrm{AXB}=\angle \mathrm{XAD}+\angle \mathrm{XDA}=\angle \mathrm{CAD}+\angle \mathrm{ADB}$ (same angles) $=$ $90^{\circ}+\angle \mathrm{ADB}$ (CD diameter). So $\mathrm{X}$ lies on the same circle.

Conversely, suppose $\mathrm{X}$ lies on the circle O'. Extend XA, XB to meet the circle center O at C and D respectively. If $\mathrm{X}$ lies outside the circle center $\mathrm{O}$, assume $\mathrm{C}$ does not lie inside the circle center $\mathrm{O}^{\prime}$ (if not use D). Then $\angle \mathrm{CAD}=\angle \mathrm{AXD}+\angle \mathrm{ADX}=\angle \mathrm{AXB}+\angle \mathrm{ADB}$ (same angles) $=(\angle \mathrm{AO} \mathrm{B}+\angle \mathrm{AOB}) / 2=90^{\circ}$. Hence $\mathrm{CD}$ is a diameter.

If $\mathrm{X}$ lies inside the circle $\mathrm{O}^{\prime}$, then $\angle \mathrm{ABX}=90^{\circ}+\angle \mathrm{ADB}$. But $\angle \mathrm{AXB}=\angle \mathrm{ADB}+\angle \mathrm{XBD}$. So $\angle \mathrm{CBD}=\angle$ $\mathrm{XBD}=90^{\circ}$. Hence CD is a diameter.


Find all integral solutions to $a^{2}+b^{2}+c^{2}=a^{2} b^{2}$.

Solution:
Answer: $0,0,0$.

Squares must be 0 or $1 \bmod 4$. Since the rhs is a square, each of the squares on the 1 hs must be $0 \bmod 4$. So a, $\mathrm{b}, \mathrm{c}$ are even. Put $\mathrm{a}=2 \mathrm{a}_{1}, \mathrm{~b}=2 \mathrm{~b}_{1}, \mathrm{c}=2 \mathrm{c}_{1}$. Then $\mathrm{a}_{1}{ }^{2}+\mathrm{b}_{1}{ }^{2}+\mathrm{c}_{1}{ }^{2}=$ square. Repeating, we find that $\mathrm{a}, \mathrm{b}, \mathrm{c}$ must each be divisible by an arbitrarily large power of 2. So they must all be zero.


A tetrahedron ABCD has edges of total length 1. The angles at A (BAC etc) are all $90^{\circ}$. Find the maximum volume of the tetrahedron.

Solution:
Answer: $(2 \sqrt{2}-7) / 162$.

Let the edges at $A$ have lengths $x, y, z$. Then the volume is $x y z / 6$ and the perimeter is $x+y+z+\sqrt{ }\left(x^{2}+y^{2}\right)+$ $\sqrt{ }\left(y^{2}+z^{2}\right)+\sqrt{ }\left(z^{2}+x^{2}\right)=1$. By AM/GM we have $x+y+z \geq 3 k$, where $k=(x y z)^{1 / 3}$. Also $x^{2}+y^{2} \geq 2 x y$, so $\sqrt{ }\left(x^{2}+y^{2}\right)+\sqrt{ }\left(y^{2}+z^{2}\right)+\sqrt{ }\left(z^{2}+x^{2}\right) \geq \sqrt{ }(2 x y)+\sqrt{ }(2 y x)+\sqrt{ }(2 z x)$. By AM/GM that is $>=3 \sqrt{2}$ k. So we have $1 \geq$ $3(1+\sqrt{2}) \mathrm{k}$. Hence $\mathrm{k}^{3} \leq(5 \sqrt{2}-7) / 27$. Hence the volume is at most $(5 \sqrt{2}-7) / 162$. This is achieved if $\mathrm{x}=\mathrm{y}=\mathrm{z}$ $=(\sqrt{ } 2-1) / 3$. Then the other three sides are $(2-\sqrt{ } 2) / 3$ and the perimeter is 1 .


The polynomials $a(x), b(x), c(x), d(x)$ satisfy $a\left(x^{5}\right)+x b\left(x^{5}\right)+x^{2} c\left(x^{5}\right)=\left(1+x+x^{2}+x^{3}+x^{4}\right) d(x)$. Show that $\mathrm{a}(\mathrm{x})$ has the factor $(\mathrm{x}-1)$.

Solution:
Take $\mathrm{k}$ to be a complex 5 th root of 1 , so that $1+\mathrm{k}+\mathrm{k}^{2}+\mathrm{k}^{3}+\mathrm{k}^{4}=0$. Putting $\mathrm{x}=\mathrm{k}, \mathrm{k}^{2}, \mathrm{k}^{3}, \mathrm{k}^{4}$ in the given equation we get:

$\mathrm{a}(1)+\mathrm{k} b(1)+\mathrm{k}^{2} \mathrm{c}(1)=0$

$a(1)+k^{2} b(1)+k^{4} c(1)=0$

$\mathrm{a}(1)+\mathrm{k}^{3} \mathrm{~b}(1)+\mathrm{k} c(1)=0$

$a(1)+k^{4} b(1)+k^{3} c(1)=0$

Multiplying by $-k,-k^{2},-k^{3},-k^{4}$ respectively, we get

$-k a(1)-k^{2} b(1)-k^{3} c(1)=0$

$-k^{2} a(1)-k^{4} b(1)-k c(1)=0$

$-k^{3} a(1)-k b(1)-k^{4} c(1)=0$

$-k^{4} a(1)-k^{3} b(1)-k^{2} c(1)=0$

Adding all eight equations gives $5 \mathrm{a}(1)=0$. Hence $\mathrm{a}(\mathrm{x})$ has the root $\mathrm{x}=1$ and hence the factor $(\mathrm{x}-1)$.


For which positive integers $\mathrm{a}, \mathrm{b}$ does $\left(\mathrm{x}^{\mathrm{a}}+\ldots+\mathrm{x}+1\right)$ divide $\left(\mathrm{x}^{\mathrm{ab}}+\mathrm{x}^{\mathrm{ab}-\mathrm{b}}+\ldots+\mathrm{x}^{2 b}+\mathrm{x}^{\mathrm{b}}+1\right)$ ?

Solution:
Answer: $\mathrm{a}+1$ and $\mathrm{b}$ relatively prime.

The question is when $\left(\mathrm{x}^{\mathrm{a}+1}-1\right) /(\mathrm{x}-1)$ divides $\left(\mathrm{x}^{\mathrm{b}(a+1)}-1\right) /\left(\mathrm{x}^{\mathrm{b}}-1\right)$ or when $\left(\mathrm{x}^{\mathrm{a}+1}-1\right)\left(\mathrm{x}^{\mathrm{b}}-1\right)$ divides $\left(\mathrm{x}^{\mathrm{b}(\mathrm{a}+1)}-\right.$ 1)( $x-1)$. Now both $\left(x^{a+1}-1\right)$ and $\left(x^{b}-1\right)$ divide ( $\left.x^{b(a+1)}-1\right)$. They both have a factor $(x-1)$, so if that is their only common factor, then their product divides $\left(x^{b(a+1)}-1\right)(x-1)$. That is true if $a+1$ and $b$ are relatively prime, for the roots of $x^{k}-1$ are the kth roots of 1 . Thus if $a+1$ and $b$ are relatively prime, then the only (complex) number which is an (a+1)th root of 1 and a bth root of 1 is 1 .

But suppose $d$ is a common factor of $a+1$ and $b$, then $\exp (2 \pi i / d)$ is a root of both $x^{a+1}-1$ and $x^{b}-1$. It is a root of $\mathrm{x}^{\mathrm{b}(a+1)}-1$, but only with multiplicity 1 , so $\left(\mathrm{x}^{\mathrm{a+1}}-1\right)\left(\mathrm{x}^{\mathrm{b}}-1\right)$ does not divide $\left(\mathrm{x}^{\mathrm{b}(a+1)}-1\right)(\mathrm{x}-1)$.


The triangles ABC and DEF have AD, BE and CF parallel. Show that $[\mathrm{AEF}]+[\mathrm{DBF}]+[\mathrm{DEC}]+[\mathrm{DBC}]+$ $[\mathrm{AEC}]+[\mathrm{ABF}]=3[\mathrm{ABC}]+3[\mathrm{DEF}]$, where $[\mathrm{XYZ}]$ denotes the signed area of the triangle XYZ. Thus $[\mathrm{XYZ}]$ is + area XYZ if the order $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ is anti-clockwise and - area $\mathrm{XYZ}$ if the order $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ is clockwise. So, in particular, $[\mathrm{XYZ}]=[\mathrm{YZX}]=-[\mathrm{YXZ}]$.

Solution:
The starting point is that $[\mathrm{ABC}]=[\mathrm{XBC}]+[\mathrm{AXC}]+[\mathrm{ABX}](*)$ for any point $\mathrm{X}$. If $\mathrm{X}$ is inside the triangle, then all the rotations have the same sense. If $\mathrm{X}$ is outside, then they do not. But it is easy to check that $(*)^{*}$ ) always holds.

So $[\mathrm{ABC}]=[\mathrm{DBC}]+[\mathrm{ADC}]+[\mathrm{ABD}],[\mathrm{DEF}]=[\mathrm{AEF}]+[\mathrm{DAF}]+[\mathrm{DEA}]$. Now, ignoring sign, the triangles $\mathrm{ABD}$ and DEA have equal area, because they have a common base AD and the same height (since AD is parallel to $\mathrm{BE})$. But the sign is opposite, so $[\mathrm{ABD}]+[\mathrm{DEA}]=0$. Similarly, $[\mathrm{ADC}]+[\mathrm{DAF}]=0$, so $[\mathrm{ABC}]+$ $[\mathrm{DEF}]=[\mathrm{DBC}]+[\mathrm{AEF}]$. Adding the two similar equations (obtained from $\mathrm{E}$ with $[\mathrm{ABC}]$ and $\mathrm{B}$ with $[\mathrm{DEF}]$, and from $\mathrm{F}$ with $[\mathrm{ABC}]$ and $\mathrm{C}$ with $[\mathrm{DEF}]$ ) gives the required result.


Prove that the product of the two real roots of $x^{4}+x^{3}-1=0$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1=0$.

Solution:
Let the roots of the quartic be $\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}$. We show that ab, ac, ad, bc, bd, cd are the roots of the sextic. We have $a+b+c+d=-1, a b+a c+a d+b c+b d+c d=0,1 / a+1 / b+1 / c+1 / d=0, a b c d=-1$.

Let $x^{6}+a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$ be the sextic with roots $a b, a c, a d, b c$, bd, cd. Since their sum is zero, we have $\mathrm{a}_{5}=0$. Their product is (abcd) ${ }^{3}$, so $\mathrm{a}_{0}=-1$.


$\mathrm{ABCD}$ is a tetrahedron. The midpoint of $\mathrm{AB}$ is $\mathrm{M}$ and the midpoint of $\mathrm{CD}$ is $\mathrm{N}$. Show that $\mathrm{MN}$ is perpendicular to $\mathrm{AB}$ and $\mathrm{CD}$ iff $\mathrm{AC}=\mathrm{BD}$ and $\mathrm{AD}=\mathrm{BC}$.

Solution:
Use vectors. Take any origin $\mathrm{O}$ and write the vector $\mathrm{OX}$ as $\mathbf{X}$. Then $\mathrm{MN}$ perpendicular to $\mathrm{AB}$ and $\mathrm{CD}$ is equivalent to:

$(\mathbf{A}+\mathbf{B}-\mathbf{C}-\mathbf{D}) .(\mathbf{A}-\mathbf{B})=0$ and

$(\mathbf{A}+\mathbf{B}-\mathbf{C}-\mathbf{D}) .(\mathbf{C}-\mathbf{D})=0$.

Expanding and adding the two equations gives $(\mathbf{A}-\mathbf{D})^{2}=(\mathbf{B}-\mathbf{C})^{2}$ or $\mathrm{AD}=\mathrm{BC}$. Subtracting gives $\mathrm{AC}=\mathrm{BD}$.

Conversely, $\mathrm{AD}=\mathrm{BC}$ and $\mathrm{AC}=\mathrm{BD}$ gives:

$(\mathbf{A}-\mathbf{D})^{2}=(\mathbf{B}-\mathbf{C})^{2}$ and

$(\mathbf{A}-\mathbf{C})^{2}=(\mathbf{B}-\mathbf{D})^{2}$.

Adding gives $(\mathbf{A}+\mathbf{B}-\mathbf{C}-\mathbf{D}) .(\mathbf{A}-\mathbf{B})=0$, so MN is perpendicular to AB. Subtracting gives MN

perpendicular to $\mathrm{CD}$.


The positive reals $\mathrm{v}, \mathrm{w}, \mathrm{x}, \mathrm{y}, \mathrm{z}$ satisfy $0<\mathrm{h} \leq \mathrm{v}, \mathrm{w}, \mathrm{x}, \mathrm{y}, \mathrm{z} \leq \mathrm{k}$. Show that $(\mathrm{v}+\mathrm{w}+\mathrm{x}+\mathrm{y}+\mathrm{z})(1 / \mathrm{v}+1 / \mathrm{w}+1 / \mathrm{x}$ $+1 / \mathrm{y}+1 / \mathrm{z}) \leq 25+6(\sqrt{ }(\mathrm{h} / \mathrm{k})-\sqrt{ }(\mathrm{k} / \mathrm{h}))^{2}$. When do we have equality?

Solution:
Fix four of the variables and allow the other to vary. Suppose, for example, we fix all but $x$. Then the expression on the lhs has the form $(r+x)(s+1 / x)=(r s+1)+s x+r / x$, where $r$ and $s$ are fixed. But this is convex. That is to say, as $\mathrm{x}$ increases if first decreases, then increases. So its maximum must occur at $\mathrm{x}=\mathrm{h}$ or $\mathrm{x}=\mathrm{k}$. This is true for each variable.

Suppose all five are $\mathrm{h}$ or all five are $\mathrm{k}$, then the $\mathrm{lh}$ is 25 , so the inequality is true and strict unless $\mathrm{h}=\mathrm{k}$. If four are $h$ and one is $k$, then the $\mathrm{hs}$ is $17+4(\mathrm{~h} / \mathrm{k}+\mathrm{k} / \mathrm{h})$. Similarly if four are $\mathrm{k}$ and one is $\mathrm{h}$. If three are $\mathrm{h}$ and two are $\mathrm{k}$, then the lhs is $13+6(\mathrm{~h} / \mathrm{k}+\mathrm{k} / \mathrm{h})$. Similarly if three are $\mathrm{k}$ and two are $\mathrm{h}$.

$\mathrm{h} / \mathrm{k}+\mathrm{k} / \mathrm{h} \geq 2$ with equality iff $\mathrm{h}=\mathrm{k}$, so if $\mathrm{h}<\mathrm{k}$, then three of one and two of the other gives a larger lhs than four of one and one of the other. Finally, we note that the rhs is in fact $13+6(\mathrm{~h} / \mathrm{k}+\mathrm{k} / \mathrm{h})$, so the inequality is true with equality iff either (1) $\mathrm{h}=\mathrm{k}$ or (2) three of $\mathrm{v}, \mathrm{w}, \mathrm{x}, \mathrm{y} \mathrm{z}$ are $\mathrm{h}$ and two are $\mathrm{k}$ or vice versa.


The sum of 5 real numbers is 8 and the sum of their squares is 16 . What is the largest possible value for one of the numbers?

Solution:
Answer: $16 / 5$.

Let the numbers be $\mathrm{v}, \mathrm{w}, \mathrm{x}, \mathrm{y}, \mathrm{z}$. We have $(\mathrm{v}-6 / 5)^{2}+(\mathrm{w}-6 / 5)^{2}+\ldots+(\mathrm{z}-6 / 5)^{2}=\left(\mathrm{v}^{2}+\ldots+\mathrm{z}^{2}\right)-12 / 5(\mathrm{v}+\ldots$ $+z)+36 / 5=16-96 / 5+36 / 5=4$. Hence $|v-6 / 5| \leq 2$, so $v \leq 16 / 5$. This value can be realized by putting $v=$ $16 / 5$ and setting the other numbers to $6 / 5$.


Two square maps cover exactly the same area of terrain on different scales. The smaller map is placed on top of the larger map and inside its borders. Show that there is a unique point on the top map which lies exactly above the corresponding point on the lower map. How can this point be constructed?

Solution:
The point is obviously unique, because the two maps have different scales (but if $\mathrm{P}$ and $\mathrm{Q}$ where two fixed points the distance between them would be the same on both maps).

Let the small map square be $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and the large be $A B C D$, where $X$ and $X^{\prime}$ are corresponding points. We deal first with the special case where $A^{\prime} B^{\prime}$ is parallel to $A B$. In this case let $A A^{\prime}$ and $B B^{\prime}$ meet at $O$. Then triangles $\mathrm{OAB}$ and $\mathrm{OA}^{\prime} \mathrm{B}^{\prime}$ are similar, so $\mathrm{O}$ must represent the same point. So assume $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ is not parallel to AB.

Let the lines $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ and $\mathrm{AB}$ meet at $\mathrm{W}$, the lines $\mathrm{B}^{\prime} \mathrm{C}^{\prime}$ and $\mathrm{BC}$ meet at $\mathrm{X}$, the lines $\mathrm{C}^{\prime} \mathrm{D}^{\prime}$ and $\mathrm{CD}$ meet at $\mathrm{Y}$, and the lines D'A' and DA meet at Z. We claim that the segments WY and XZ meet at a point O inside the smaller square. W cannot lie between $\mathrm{A}^{\prime}$ and $\mathrm{B}^{\prime}$ (or one of the vertices $\mathrm{A}^{\prime}, \mathrm{B}^{\prime}$ of the smaller square would lie outside the larger square). If it lies on the opposite side of $\mathrm{A}^{\prime}$ to $\mathrm{B}^{\prime}$, then $\mathrm{Y}$ must lie on the opposite side of $\mathrm{C}^{\prime}$ to $\mathrm{D}^{\prime}$. Thus the segment WY must cut the side $\mathrm{A}^{\prime} \mathrm{D}^{\prime}$ at some point $\mathrm{Z}^{\prime}$ and the side $\mathrm{B}^{\prime} \mathrm{C}^{\prime}$ at some point $\mathrm{X}^{\prime}$. The same conclusion holds if $\mathrm{W}$ lies on the opposite side of $\mathrm{B}^{\prime}$ to $\mathrm{A}^{\prime}$, because then $\mathrm{Y}$ must lie on the opposite side of $\mathrm{D}^{\prime}$ to $\mathrm{C}^{\prime}$. Similarly, the segment XZ must cut the side $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ at some point $\mathrm{W}^{\prime}$ and the side $\mathrm{C}^{\prime} \mathrm{D}^{\prime}$ at some point $Y^{\prime}$. But now the segments $X^{\prime} Z^{\prime}$ and $W^{\prime} Y^{\prime}$ join pairs of points on opposite sides of the small square and so they must meet at some point $\mathrm{O}$ inside the small square.

Now the triangles WOW' and YOY' are similar (WW' and YY' are parallel). Hence OW/OY $=\mathrm{OW}^{\prime} / \mathrm{OY}^{\prime}$. So if we set up coordinate systems with $\mathrm{AB}$ as the $\mathrm{x}$-axis and $\mathrm{AD}$ as the $\mathrm{y}$-axis (for the large square) and $\mathrm{A}^{\prime} \mathrm{B}^{\prime}$ as the $x^{\prime}$-axis and $A^{\prime} D^{\prime}$ as the $y^{\prime}$-axis (for the small square) so that corresponding points have the same coordinates, then the $y$ coordinate of $\mathrm{O}$ equals the $\mathrm{y}^{\prime}$ coordinate of O. Similarly, XOX' and ZOZ' are similar, so $\mathrm{OX} / \mathrm{OZ}=\mathrm{OX}^{\prime} / \mathrm{OZ}$ ', so the $\mathrm{x}$-coordinate of $\mathrm{O}$ equals its $\mathrm{x}^{\prime}$-coordinate. In other words, $\mathrm{O}$ represents the same point on both maps.


You are told that all integers from 33 to 73 inclusive can be expressed as a sum of positive integers whose reciprocals sum to 1 . Show that the same is true for all integers greater than 73.

Solution: The trick is consider the integers $2 a_{1}, 2 a_{2}, \ldots, 2 a_{m}$ given that $a_{1}, a_{2}, \ldots, a_{m}$ is a solution for $n$. The sum of their reciprocals is $1 / 2$. So if we throw in two $4 \mathrm{~s}$, we get a solution for $2 n+8$. Similarly, adjoining 3 and 6 gives a solution for $2 \mathrm{n}+9$. It is now a simple induction. For the starter set gives 74 thru 155 , then those give 156 thru 319 , and so on. In general, $n$ thru $2 n+7$ gives $2 n+8$ thru $4 n+23=2(2 n+8)+7$.


Show that if the angle between each pair of faces of a tetrahedron is equal, then the tetrahedron is regular.

Does a tetrahedron have to be regular if five of the angles are equal?

Solution:
Answer: no.

Let the tetrahedron be $\mathrm{ABCD}$. Let the insphere have center $\mathrm{O}$ and touch the sides at $\mathrm{W}, \mathrm{X}, \mathrm{Y}, \mathrm{Z}$. The $\mathrm{OW}$, $\mathrm{OX}, \mathrm{OY}, \mathrm{OZ}$ are the normals to the faces. But the angle between each pair of normals is equal. So $\mathrm{OW}, \mathrm{OX}$,
$\mathrm{OY}$ and $\mathrm{OZ}$ are vectors of equal length at equal angles. Hence each side of $\mathrm{WXYZ}$ is equal (eg $\mathrm{WX}=2 \mathrm{OW}$ $\sin (\mathrm{WOX} / 2)$ ). So $\mathrm{WXYZ}$ is a regular tetrahedron. But the faces of $\mathrm{ABCD}$ are just the tangent planes at $\mathrm{W}$, $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$. So if we rotate through an angle $120^{\circ}$ about the line $\mathrm{OW}$, then $\mathrm{X}$ goes to $\mathrm{Y}, \mathrm{Y}$ to $\mathrm{Z}$ and $\mathrm{Z}$ to $\mathrm{X}$. Hence $\mathrm{AB}=\mathrm{AC}, \mathrm{AC}=\mathrm{AD}, \mathrm{AD}=\mathrm{AB}$ and $\mathrm{BC}=\mathrm{CD}=\mathrm{DB}$ (assuming appropriate labeling). Similarly, for rotation about the other axes. So $A B C D$ has equal edges and hence is regular.

Consider the four normals OW, OX, OY, OZ. We can move $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ slightly closer together so that $\mathrm{XYZ}$ remains an equilateral triangle. Then move $\mathrm{W}$ so that $\mathrm{WX}=\mathrm{WY}=\mathrm{XY}$. So five of the distances are equal, but the sixth is unequal. The reason for slightly is that all the angles between pairs of normals must remain less than $180^{\circ}$. The corresponding tetrahedron will now have the angles between only five pairs of faces equal.


There are 9 delegates at a conference, each speaking at most three languages. Given any three delegates, at least 2 speak a common language. Show that there are three delegates with a common language.

Solution: Suppose not. Then A can shares a language with at most 3 delegates, because if he shared a language with 4 , he would have to share the same language with 2 of them (since he can only speak 3 languages). So there are 5 delegates who do not share a language with A. Let one of them be B. By the same argument, there must be at least one of the other 4 (call her C) who does not share a language with B. But now no two of A, B, C share a language. Contradiction.


Find all sets of 14 or less fourth powers which sum to 1599 .

Solution:
Answer: none.

The only 4th powers less than 1599 are $1,16,81,256,625,1296\left(7^{4}=2401\right)$. Note that $1,81,625=1 \mathrm{mod}$ 16 and $16,256,1296=0 \bmod 16$. But $1599=15 \bmod 16$, so it cannot be expressed as a sum of 14 or less fourth powers.


$\mathrm{N}$ is the north pole. A and $\mathrm{B}$ are points on a great circle through $\mathrm{N}$ equidistant from $\mathrm{N} . \mathrm{C}$ is a point on the equator. Show that the great circle through $\mathrm{C}$ and $\mathrm{N}$ bisects the angle $\mathrm{ACB}$ in the spherical triangle $\mathrm{ABC}$ (a spherical triangle has great circle arcs as sides).

Solution:
Let $\mathrm{S}_{\mathrm{A}}, \mathrm{S}_{\mathrm{B}}, \mathrm{S}_{\mathrm{N}}$ be the great circles through $\mathrm{A}$ and $\mathrm{C}, \mathrm{B}$ and $\mathrm{C}$, and $\mathrm{N}$ and $\mathrm{C}$ respectively. Let $\mathrm{C}^{\prime}$ be the point directly opposite $\mathrm{C}$ on the sphere. Then any great circle through $\mathrm{C}$ also goes through $\mathrm{C}^{\prime}$. So, in particular, $\mathrm{S}_{\mathrm{A}}$, $\mathrm{S}_{\mathrm{B}}$ and $\mathrm{S}_{\mathrm{N}}$ go through $\mathrm{C}^{\prime}$.

Two great circles through $\mathrm{C}$ meet at the same angle at $\mathrm{C}$ and at $\mathrm{C}^{\prime}$, so the spherical angles $\mathrm{ACN}$ and $\mathrm{AC}^{\prime} \mathrm{N}$ are equal. Now rotate the sphere through an angle $180^{\circ}$ about the diameter through $\mathrm{N}$. Then great circles through $\mathrm{N}$ map into themselves, so $\mathrm{C}$ and $\mathrm{C}^{\prime}$ change places ( $\mathrm{C}$ is on the equator). Also $\mathrm{A}$ and $\mathrm{B}$ change places (they are equidistant from $\mathrm{N}$ ). $\mathrm{S}^{\mathrm{A}}$ must go into another great circle through $\mathrm{C}$ and $\mathrm{C}^{\prime}$. But since $\mathrm{A}$ maps to $\mathrm{B}$, it must be $\mathrm{S}_{\mathrm{B}}$. Hence the spherical angle $\mathrm{AC}^{\prime} \mathrm{N}=$ angle $\mathrm{BCN}$ (since one rotates into the other). Hence $\mathrm{ACN}$ and $\mathrm{BCN}$ are equal.


$a_{1}, a_{2}, \ldots, a_{n}$ is an arbitrary sequence of positive integers. A member of the sequence is picked at random. Its value is a. Another member is picked at random, independently of the first. Its value is $b$. Then a third, value c. Show that the probability that $\mathrm{a}+\mathrm{b}+\mathrm{c}$ is divisible by 3 is at least $1 / 4$.

Solution:
Let the prob of a value $=0,1,2 \bmod 3$ be $\mathrm{p}, \mathrm{q}, \mathrm{r}$ respectively. So $\mathrm{p}+\mathrm{q}+\mathrm{r}=1$ and $\mathrm{p}, \mathrm{q}, \mathrm{r}$ are non-negative. If $\mathrm{a}=\mathrm{b} \bmod 3$, then to get $\mathrm{a}+\mathrm{b}+\mathrm{c}=0 \bmod 3$, we require $\mathrm{c}=\mathrm{a} \bmod 3$. So the prob of such events is $\mathrm{p}^{3}+\mathrm{q}^{3}+$ $r^{3}$. If the first two values are different mod 3, then the third must be different again, so the prob. is 6 pqr. Thus we have to show that $\mathrm{p}^{3}+\mathrm{q}^{3}+\mathrm{r}^{3}+6 \mathrm{pqr}>=1 / 4$.

$1=(p+q+r)^{3}=p^{3}+q^{3}+r^{3}+6 p q r+3\left(p^{2} q+p q^{2}+p^{2} r+p r^{2}+q^{2} r+q^{2}\right)$. So we have to show that $\left(p^{2} q+\right.$ $\left.\mathrm{pq}^{2}+\mathrm{p}^{2} \mathrm{r}+\mathrm{pr}^{2}+\mathrm{q}^{2} \mathrm{r}+\mathrm{qr}^{2}\right)<=1 / 4$. Or $\mathrm{p}^{2}(\mathrm{q}+\mathrm{r})+\mathrm{p}\left(\mathrm{q}^{2}+\mathrm{r}^{2}\right)+\mathrm{qr}(\mathrm{q}+\mathrm{r})<=1 / 4$, or $\mathrm{p}^{2}(1-\mathrm{p})+\mathrm{p}(\mathrm{q}+\mathrm{r})^{2}+\mathrm{qr}(\mathrm{q}+$ $r-2 p) \leq 1 / 4$, or $p^{2}(1-p)+p(1-p)^{2}+q r(1-3 p)=p(1-p)+q r(1-3 p) \leq 1 / 4$.

If $p \geq 1 / 3$, then we maximise $p(1-p)+q r(1-3 p)$ by taking $q r=0$ and $p=1 / 2$ to maximise $p(1-p)$. Thus the maximum is $1 / 4$, achieved at $p=1 / 2, q=1 / 2, r=0$ and $p=1 / 2, q=0, r=1 / 2$. But because of the symmetry, there is no loss of generality in assuming that $p \geq q \geq r$, so $p$ must be $\geq 1 / 3$. So the maximum value is $1 / 4$.


$\mathrm{P}$ lies between the rays $\mathrm{OA}$ and $\mathrm{OB}$. Find $\mathrm{Q}$ on $\mathrm{OA}$ and $\mathrm{R}$ on $\mathrm{OB}$ collinear with $\mathrm{P}$ so that $1 / \mathrm{PQ}+1 / \mathrm{PR}$ is as large as possible.

Solution: Let the line through $\mathrm{P}$ parallel to $\mathrm{OA}$ meet $\mathrm{OB}$ at $\mathrm{C}$. Note that $\mathrm{C}$ is fixed. Let the line through $\mathrm{C}$ parallel to $\mathrm{QR}$ meet $\mathrm{OP}$ at $\mathrm{D}$. $\mathrm{D}$ varies as $\mathrm{Q}$ varies. Triangles $\mathrm{ODC}, \mathrm{OPR}$ are similar, so $\mathrm{CD} / \mathrm{PR}=\mathrm{OD} / \mathrm{OP}$. Also triangles OPQ and $\mathrm{PDC}$ are similar, so $\mathrm{CD} / \mathrm{PQ}=\mathrm{DP} / \mathrm{OP}$. Adding, $\mathrm{CD} / \mathrm{PR}+\mathrm{CD} / \mathrm{PQ}=1$. Hence we maximise $1 / \mathrm{PQ}+1 / \mathrm{PR}$ by making $\mathrm{CD}$ as small as possible. That is achieved by making angle $\mathrm{CDP}=90^{\circ}$ and hence QR perpendicular to OP.


$\mathrm{X}$ has $\mathrm{n}$ members. Given $\mathrm{n}+1$ subsets of $\mathrm{X}$, each with 3 members, show that we can always find two which have just one member in common.

Solution:
We use induction on $n$. The result is true for $n<5$, because there are at most $n$ distinct subsets of 3 elements. Suppose it is true for all sets $\mathrm{X}$ with $<\mathrm{n}$ members. Let $\mathrm{X}$ have $\mathrm{n}$ members and consider a collection of $\mathrm{n}+1$ subsets with 3 members. If every member of $X$ was in at most 3 of the subsets, there would be at most $n$ subsets, so some member A is in at least 4 of the subsets. Suppose one of them is $\{A, B, C\}$. There are at least three others, so B (say) must be in at least two of the others. Suppose they are $\{$ A, B, D $\}$ and $\{A, B, E\}$. Now any other subset containing A must contain B, because otherwise it would have to contain C, D and E, which is impossible. Similarly, any other subset containing B must contain A. So suppose there are $m$ subsets containing $A$ and $B$. If $\{A, B, K\}$ is such a subset, then $K$ cannot belong to any other subsets (because if it also belonged to $\mathrm{S}$, then $\mathrm{S}$ and $\{\mathrm{A}, \mathrm{B}, \mathrm{K}\}$ would have just one member in common). So consider the $n-(m+2)$ people other than those who belong to sets of the type $\{A, B, K\}$. There can be at most $\mathrm{n}-\mathrm{m}-2$ subsets involving them. Hence at most $\mathrm{n}-2$ in total. So the result is true for $\mathrm{n}$.

Note that we do worse than $n$ subsets if we allow any member to be in 4 subsets. But at least for $n=4 m$ we can achieve $n$ subsets. Just take $m$ groups of 4 and take all subsets with 3 members of each group.


A balance has unequal arms and pans of unequal weight. It is used to weigh two objects of unequal weight. The first object balances against a weight A, when placed in the left pan and against a weight a, when placed in the right pan. The corresponding weights for the second object are $\mathrm{B}$ and $\mathrm{b}$. A third object balances against a weight C, when placed in the left pan. What is its true weight?

Solution:
The effect of the unequal arms and pans is that if an object of weight $x$ in the left pan balances an object of weight $\mathrm{y}$ in the right pan, then $\mathrm{x}=\mathrm{hy}+\mathrm{k}$ for some constants $\mathrm{h}$ and $\mathrm{k}$. Thus if the first object has true weight
$\mathrm{x}$, then $\mathrm{x}=\mathrm{hA}+\mathrm{k}, \mathrm{a}=\mathrm{h} x+\mathrm{k}$. So $\mathrm{a}=\mathrm{h}^{2} \mathrm{~A}+(\mathrm{h}+1) \mathrm{k}$. Similarly, $\mathrm{b}=\mathrm{h}^{2} \mathrm{~B}+(\mathrm{h}+1) \mathrm{k}$. Subtracting gives $\mathrm{h}^{2}=(a-$ b) /(A - B). and hence $(h+1) k=a-h^{2} A=(b A-a B) /(A-B)$.

The true weight of the third object is thus $h C+k=\sqrt{ }((a-b) /(A-B)) C+(b A-a B) /(A-B) 1 /(\sqrt{ }(a-b) /(A-B))$ $+1)$.


Find the maximum possible number of three term arithmetic progressions in a monotone sequence of $n$ distinct reals.

Solution:
Answer: $\mathrm{m}^{2}$ for $\mathrm{n}=2 \mathrm{~m}+1, \mathrm{~m}(\mathrm{~m}-1)$ for $\mathrm{n}=2 \mathrm{~m}$.

Let the reals be $\mathrm{a}_{1}, \ldots, \mathrm{a}_{\mathrm{n}}$. Suppose $\mathrm{n}=2 \mathrm{~m}+1$ and the middle term is $\mathrm{a}_{\mathrm{k}}$. If $\mathrm{k}<\mathrm{m}+1$, then we are constrained by the shortage of first terms. If $k>m+1$ we are constrained by the shortage of third terms. Thus if $k=1$, $\mathrm{a}_{\mathrm{k}}$ cannot be the middle term. If $\mathrm{k}=2$, there is only one candidate for the first term. If $\mathrm{k}=3$, there are two candidates for the middle terms and so on. Thus the total number of possible progressions certainly cannot exceed: $1+2+\ldots+\mathrm{m}+\mathrm{m}-1+\mathrm{m}-2+\ldots+1=\mathrm{m}(\mathrm{m}+1) / 2+\mathrm{m}(\mathrm{m}-1) / 2=\mathrm{m}^{2}$. But this bound is achieved by the sequence $1,2,3, \ldots, \mathrm{n}$.

Similarly, if $\mathrm{n}=2 \mathrm{~m}$, then the upper bound is $1+2+\ldots+\mathrm{m}-1+\mathrm{m}-1+\ldots+1=\mathrm{m}(\mathrm{m}-1)$. Again, this is achieved by the sequence $1,2, \ldots, \mathrm{n}$.


A $+B+C$ is an integral multiple of $\pi . x, y, z$ are real numbers. If $x \sin A+y \sin B+z \sin C=x^{2} \sin 2 A+$ $y^{2} \sin 2 B+z^{2} \sin 2 C=0$, show that $x^{n} \sin n A+y^{n} \sin n B+z^{n} \sin n C=0$ for any positive integer $n$.

Solution: The juxtaposition of $\mathrm{x}^{2}$ and $\sin 2 \mathrm{~A}$ strongly suggests considering $\cos \mathrm{A}+\mathrm{i} \sin \mathrm{A}$. So put $\mathrm{u}=\mathrm{x}(\cos \mathrm{A}+\mathrm{i} \sin$ $A), v=y(\cos B+i \sin B), w=z(\cos C+i \sin C)$. Put $a_{n}=u^{n}+v^{n}+w^{n}$. So $a_{1}$ and $a_{2}$ are real. Hence also uv + $\mathrm{vw}+\mathrm{wu}=\left(\mathrm{a}_{1}{ }^{2}-\mathrm{a}_{2}\right) / 2$ is real. Also $u v w=\mathrm{xyz} \exp (\mathrm{i}(\mathrm{A}+\mathrm{B}+\mathrm{C}))= \pm \mathrm{xyz}$, since $\mathrm{A}+\mathrm{B}+\mathrm{C}$ is an integral multiple of $\pi$. Thus $\mathrm{u}, \mathrm{v}, \mathrm{w}$ are roots of some cubic $\mathrm{p}^{3}+\mathrm{ap}^{2}+\mathrm{bp}+\mathrm{c}=0$ with real coefficients. Putting $\mathrm{p}=\mathrm{u}, \mathrm{v}, \mathrm{w}$ and adding, we get $a_{3}+a a_{2}+b a_{1}+3 c=0$, so $a_{3}$ is real. Also multiplying through by $p^{n}$, then putting $p=u, v, w$ and adding, we get: $a_{n+3}+a a_{n+2}+b a_{n+1}+c a_{n}=0$. So by a trivial induction $a_{n}$ is real for all positive $n$. Hence result.


The insphere of a tetrahedron touches each face at its centroid. Show that the tetrahedron is regular.

Solution: Let the tetrahedron be $\mathrm{ABCD}$. Let $\mathrm{G}$ be the centroid of $\mathrm{ABC}$ and $\mathrm{H}$ the centroid of $\mathrm{ACD}$. Let $\mathrm{AM}$ be a median in $\mathrm{ABC}$ and $\mathrm{AN}$ a median in $\mathrm{ACD}$. Then $\mathrm{AG}$ and $\mathrm{AH}$ are tangents to the insphere, so they are equal. $\mathrm{CG}$ and $\mathrm{CH}$ are also tangents and hence equal. So the triangles $\mathrm{ACG}$ and $\mathrm{ACH}$ are congruent. Hence $\angle \mathrm{AGC}$ $=\angle \mathrm{AHC}$ and so $\angle \mathrm{CGM}=\angle \mathrm{CHN}$. But $\mathrm{GM}=\mathrm{AG} / 2=\mathrm{AH} / 2=\mathrm{GN}$, so the triangles $\mathrm{CGM}$ and $\mathrm{CHN}$ are also congruent. Hence $\mathrm{CM}=\mathrm{CN}$. Hence $\mathrm{CB}=\mathrm{CD}$. So every pair of adjacent edges is equal. Hence all the edges are equal and the tetrahedron is regular.


If $x, y, z$ are reals such that $0 \leq x, y, z \leq 1$, show that $x /(y+z+1)+y /(z+x+1)+z /(x+y+1) \leq 1-(1-$ $\mathrm{x})(1-\mathrm{y})(1-\mathrm{z})$.

Solution:
Consider $\mathrm{x} /(\mathrm{y}+\mathrm{z}+1)+\mathrm{y} /(\mathrm{z}+\mathrm{x}+1)+\mathrm{z} /(\mathrm{x}+\mathrm{y}+1)+(1-\mathrm{x})(1-\mathrm{y})(1-\mathrm{z})$ as a function of $\mathrm{x}$, with $\mathrm{y}$ and $\mathrm{z}$ fixed. Each term is convex, so the whole function is convex. Hence its maximum value occurs at its endpoints. The same is true for $\mathrm{x}$ and $\mathrm{y}$, so we need only check the eight possible values $\mathrm{x}, \mathrm{y}, \mathrm{z}=0$ or 1 . In fact, we easily find the expression has value 1 at all eight points. The result follows.

A function is convex if for any three points $\mathrm{a}<\mathrm{b}<\mathrm{c}$, the point $(\mathrm{b}, \mathrm{f}(\mathrm{b})$ ) lies on or below the chord joining the points (a, $\mathrm{f}(\mathrm{a})$ ) and $(\mathrm{c}, \mathrm{f}(\mathrm{c}))$. Analytically, this means that if $\mathrm{b}=\mathrm{ha}+(1-\mathrm{h}) \mathrm{c}$, where $0 \leq \mathrm{h} \leq 1$, then $\mathrm{f}(\mathrm{b}) \leq$ $h \mathrm{f}(\mathrm{a})+(1-\mathrm{h}) \mathrm{f}(\mathrm{c})$. The linear nature of this relation implies immediately that a sum of convex functions is convex and that a positive multiple of a convex function is convex. Linear functions are obviously convex. It is obvious from the graph that the function $\mathrm{a} /(\mathrm{b}+\mathrm{x})$ is convex. To prove it analytically we must show that $a /(b+h x+(1-h) y)) \leq h a /(b+x)+(1-h) a /(b+y)$ or $a\left(b^{2}+b x+b y+x y\right) \leq h a(b+y)(b+h x+(1-h) y)+(1-$

h)a(b+x)(b+hx+(1-h)y). After cancelling some terms, we have to show that $x y<=h(1-h) x^{2}+(1-h)^{2} x y+h^{2} x y$ $+h(1-h) y^{2}$. This is obviously true for $h=1$ or 0 . Otherwise we may divide by $1-h$, then $h$ to get $2 x y \leq x^{2}+y^{2}$, which is true.

To see that the maximum value of a convex function must occur at its endpoints just draw a chord between the endpoints. All other points of the curve must lie below the chord.


Prove that if $n$ is not a multiple of 3 , then the angle $\pi / n$ can be trisected with ruler and compasses.

Solution: The key is to use $\pi / 3$. Since $n$ and 3 are relatively prime we can find integers a and $b$ such that $3 a+n b=1$. Hence $a \pi / n+b \pi / 3=\pi /(3 n)$. So take a circle with arcs subtending $\pi / n$ and $\pi / 3$ at the center. Start at a point $X$ on the circumference, mark off $|a|$ arcs subtending $\pi / n$ in one direction, then mark off $|b|$ arcs subtending $\pi / 3$ in the other direction. We end up at a point $\mathrm{Y}$ with $\mathrm{XY}$ subtending $\pi / 3 \mathrm{n}$ at the center.


What is the largest number of towns that can meet the following criteria. Each pair is directly linked by just one of air, bus or train. At least one pair is linked by air, at least one pair by bus and at least one pair by train. No town has an air link, a bus link and a trian link. No three towns, A, B, C are such that the links between $\mathrm{AB}, \mathrm{AC}$ and $\mathrm{BC}$ are all air, all bus or all train.

Solution:
Answer: 4. [eg A and B linked by bus. C and D both linked to A and B by air and to each other by train.]

Suppose A is linked to three other towns by air. Let them be B, C, D. B has at most one other type of link. Suppose it is bus. Then $\mathrm{B}$ must be linked to $\mathrm{C}$ and $\mathrm{D}$ by bus. But now there is a problem with the $\mathrm{CD}$ link. If it is air, then A, C, D are all linked by air. If it is bus, then B, C, D are all linked by bus. If it is train, then C has links of all three types. So A cannot be linked to three others by air. Similarly it cannot be linked to three others by train, or to three others by bus. Since it has at most two types of link, it can be linked to at most 4 towns ( 2 by one type of link and 2 by another). So certainly there cannot be more than 5 towns.

Suppose 5 is possible. A must be linked to two towns by one type of link and two by another. Without loss of generality we may suppose $\mathrm{A}$ is linked to $\mathrm{B}$ and $\mathrm{C}$ by bus and to $\mathrm{D}$ and $\mathrm{E}$ by train. Now suppose $\mathrm{BE}$ is an air link. Then B cannot have a train link (or it will have all three). It cannot be linked to $C$ by bus (or $A B C$ is all bus), so $\mathrm{BC}$ must be air. Similarly, ED must be air (it cannot be train or ADE is all train, or bus because then $\mathrm{E}$ has air, bus and train). But now there is a problem with the $\mathrm{CE}$ link. It cannot be air (or $\mathrm{BCE}$ is all air). It cannot be train, because $\mathrm{C}$ already has air and bus links. It cannot be bus, because $\mathrm{E}$ already has train and air links. So BE cannot be an air link. So it must be a bus or train link. Without loss of generality, we may assume it is a bus link.

E has a bus and a train link, so it cannot have air links. It cannot be linked to $\mathrm{D}$ by train (or AED is all train). So ED must be bus. Now DB cannot be air (or D has air, train and bus). It cannot be bus (or DBE is all bus). So it must be train. So $\mathrm{BC}$ cannot be air (or $\mathrm{B}$ has train, bus and air). It cannot be bus (or $\mathrm{ABC}$ is all bus). So $\mathrm{BC}$ must be train. $\mathrm{CD}$ cannot be air (or $\mathrm{C}$ has air, bus and train). It cannot be train (or $\mathrm{BCD}$ is all train). So CD must be bus. Finally, $\mathrm{CE}$ cannot be air (or $\mathrm{C}$ has air, bus and train). It cannot be bus (or CDE is all bus). So CE must be train. So none of the links are air. Contradiction. Hence 5 towns is not possible. But 4 is, as given in the Answer above.


Show that for any triangle, $3(\sqrt{3}) / 2 \geq \sin 3 A+\sin 3 B+\sin 3 C \geq-2$. When does equality hold?

Solution:
Answer: 1st inequality, $\mathrm{A}=\mathrm{B}=20^{\circ}, \mathrm{C}=140^{\circ}$. 2nd inequality, $\mathrm{A}=0^{\circ}, \mathrm{B}=\mathrm{C}=90^{\circ}$ (which is degenerate).

For $\mathrm{x}$ between 0 and $180^{\circ}, \sin 3 \mathrm{x}$ is negative iff $60^{\circ}<\mathrm{x}<120^{\circ}$. So at most two angles $\mathrm{x}$ in a triangle can have $\sin 3 x$ negative. Obviously $\sin 3 x \geq-1$, so $\sin 3 A+\sin 3 B+\sin 3 C \geq-2$. We can only get equality in the degenerate case $\mathrm{A}=0^{\circ}, \mathrm{B}=\mathrm{C}=90^{\circ}$ (or with the angles relabeled).

We can certainly achieve $3(\sqrt{3}) / 2$ as shown above. But $3(\sqrt{3}) / 2>2$, so we must have all three sines positive. [If only two are positive, then their sum is at most 2.]. $\sin 3 \mathrm{x}$ is positive for $\mathrm{x}<60^{\circ}$ and $\mathrm{x}>120^{\circ}$. So one angle must be $>120^{\circ}$. Assume, for definiteness, that $\mathrm{A} \leq \mathrm{B}<60^{\circ}, \mathrm{C}>120^{\circ}$. Put $\mathrm{C}^{\prime}=\mathrm{C}-120^{\circ}$. Then $3 \mathrm{~A}+3 \mathrm{~B}$ $+3 C^{\prime}=180^{\circ}$. Now $\sin \mathrm{x}$ is a concave function for $0 \leq \mathrm{x} \leq 180^{\circ}$, or - $\sin \mathrm{x}$ is a convex function. So by Jensen's inequality $-\sin x-\sin y-\sin z \geq 3-\sin (x+y+z) / 3$. Hence $\sin 3 A+\sin 3 B+\sin 3 C \leq 3 \sin 60^{\circ}=3(\sqrt{3}) / 2$.


A convex polygon has $\mathrm{n}$ sides. Each vertex is joined to a point $\mathrm{P}$ not in the same plane. If $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are adjacent vertices of the polygon take the angle between the planes PBA and PBC. The sum of the $n$ such angles equals the sum of the $\mathrm{n}$ angles subtended at $\mathrm{P}$ by the sides of the polygon (such as the angle APB). Show that $\mathrm{n}=3$.

Solution:
$\mathrm{n}=3$ is certainly possible. For example, take $\angle \mathrm{APB}=\angle \mathrm{APC}=\angle \mathrm{BPC}=90^{\circ}$ (so that the lines $\mathrm{PA}, \mathrm{PB}, \mathrm{PC}$ are mutually perpendicular). Then the three planes through $\mathrm{P}$ are also mutually perpendicular, so the two sums are both $270^{\circ}$.

We show that $\mathrm{n}>3$ is not possible.

The sum of the $\mathrm{n}$ angles APB etc at $\mathrm{P}$ is less than $360^{\circ}$. This is almost obvious. Take another plane which meets the lines PA, PB, PC etc at $\mathrm{A}^{\prime}, \mathrm{B}^{\prime}, \mathrm{C}^{\prime}, \ldots$ and so that the foot of the perpendicular from $\mathrm{P}$ to the plane lies inside the $\mathrm{n}-\mathrm{g}$ - $\mathrm{A}^{\prime} \mathrm{B}^{\prime} \mathrm{C}^{\prime}$... then as we move $\mathrm{P}$ down the perpendicular the angles $\mathrm{A}^{\prime} \mathrm{PB}$ ' etc all increase. But when it reaches the plane their sum is $360^{\circ}$. However, I do not immediately see how to make that rigorous. Instead, take any point $\mathrm{O}$ inside the $\mathrm{n}$-gon $\mathrm{ABC} \ldots$... We have $\angle \mathrm{PBA}+\angle \mathrm{PBC}>\angle \mathrm{ABC}$. Adding the $\mathrm{n}$ such equations we get $\sum\left(180^{\circ}-\mathrm{APB}\right)>\sum \mathrm{ABC}=(\mathrm{n}-2) 180^{\circ}$. So $\sum \mathrm{APB}<360^{\circ}$.

The sum of the $\mathrm{n}$ angles between the planes is at least $(\mathrm{n}-2) 180^{\circ}$. If we take a sphere center $\mathrm{P}$. Then the lines $\mathrm{PA}, \mathrm{PB}$ intersect it at $\mathrm{A}^{\prime \prime}, \mathrm{B}^{\prime \prime}, \ldots$ which form a spherical polygon. The angles of this polygon are the angles between the planes. We can divide the polygon into $\mathrm{n}-2$ triangles. The angles in a spherical triangle sum to at least $180^{\circ}$. So the angles in the spherical polygon are at least $(n-2) 180^{\circ}$. So we have $(n-2) 180^{\circ}<$ $360^{\circ}$ and hence $\mathrm{n}<4$.


Show that for any positive real $\mathrm{x},[\mathrm{nx}] \geq \sum_{1}{ }^{\mathrm{n}}[\mathrm{kx}] / \mathrm{k}$.

Solution:
If $x$ is an integer, then we have equality. So it is sufficient to prove the result for $0<x<1$. The rhs only increases at $\mathrm{x}=\mathrm{a} / \mathrm{b}$, where $\mathrm{a}, \mathrm{b}$ are coprime positive integers with $1 \leq \mathrm{b} \leq \mathrm{n}$, and $0 \leq \mathrm{a} \leq \mathrm{b}$. So it is sufficient to consider $\mathrm{x}$ of this form. In fact we can assume $0<\mathrm{a}<\mathrm{b}$ since the equality is obvious for $\mathrm{x}=0$ or 1 . We may write:

$\mathrm{a}=\mathrm{q}_{1} \mathrm{~b}+\mathrm{r}_{1}$

$2 a=q_{2} b+r_{2}$

$3 \mathrm{a}=\mathrm{q}_{3} \mathrm{~b}+\mathrm{r}_{3}$

...

$\mathrm{na}=\mathrm{q}_{\mathrm{n}} \mathrm{b}+\mathrm{r}_{\mathrm{n}}$

with each $0 \leq r_{i}<b$.

Thus $\mathrm{kx}=\mathrm{q}_{\mathrm{k}}$ and we have to prove that $\mathrm{q}_{1}+\mathrm{q}_{2} / 2+\ldots+\mathrm{q}_{\mathrm{n}} / \mathrm{n} \leq \mathrm{q}_{\mathrm{n}}$.

We claim that $r_{1}, r_{2}, \ldots, r_{b-1}$ is just a permutation of $1,2, \ldots, b-1$. For if $r_{i}=r_{j}$ with $i<j$, then $(j-i) a=\left(q_{j}-\right.$ $\left.\mathrm{q}_{\mathrm{i}}\right) \mathrm{b}$, but a and $\mathrm{b}$ are coprime and $\mathrm{j}-\mathrm{i}<\mathrm{b}$, so that is impossible. So we may use the rearrangement inequality to give $\mathrm{r}_{1} / 1+\mathrm{r}_{2} / 2+\ldots+\mathrm{r}_{\mathrm{b}-1} / \mathrm{b}-1 \geq 1 / 1+2 / 2+\ldots+(\mathrm{b}-1) /(\mathrm{b}-1)=\mathrm{b}-1$. The inequality remains true if we add some positive terms to the lhs, so we have $r_{1} / 1+r_{2} / 2+\ldots+r_{n} / n \geq b-1$. Hence $r_{1} / b+r_{2} /(2 b)+\ldots+r_{n} /(n b) \geq(b-$ $1) / b$.

So $\left(\mathrm{q}_{1}+\mathrm{q}_{2} / 2+\ldots+\mathrm{q}_{\mathrm{n}} / \mathrm{n}\right)+(\mathrm{b}-1) / \mathrm{b} \leq\left(\mathrm{q}_{1}+\mathrm{r}_{1} / \mathrm{b}\right)+\left(\mathrm{q}_{2}+\mathrm{r}_{2} /(2 \mathrm{~b})\right)+\ldots\left(\mathrm{q}_{\mathrm{n}}+\mathrm{r}_{\mathrm{n}} /(\mathrm{nb})\right)=\mathrm{a} / \mathrm{b}+\mathrm{a} / \mathrm{b}+\ldots+\mathrm{a} / \mathrm{b}=$ $\mathrm{na} / \mathrm{b}=\left(\mathrm{q}_{\mathrm{n}} \mathrm{b}+\mathrm{r}_{\mathrm{n}}\right) / \mathrm{b} \leq \mathrm{q}_{\mathrm{n}}+(\mathrm{b}-1) / \mathrm{b}$. Subtracting $(\mathrm{b}-1) / \mathrm{b}$ from both sides gives the required result.


A graph has 1982 points. Given any four points, there is at least one joined to the other three. What is the smallest number of points which are joined to 1981 points?

Solution:
Answer: 1979.

Suppose there were 4 points not joined to 1981 points. Let one of them be A. Take a point B not joined to A. Now if $\mathrm{X}$ and $\mathrm{Y}$ are any two other points, $\mathrm{X}$ and $\mathrm{Y}$ must be joined, because otherwise none of the points $\mathrm{A}$, B, X, Y could be joined to the other 3. There must be two other points C and D not joined to 1981 points. We have just shown that $\mathrm{C}$ must be joined to every point except possibly $\mathrm{A}$ and $\mathrm{B}$. So it must be not joined to one of those. Similarly D. But now none of A, B, C, D is joined to the other 3. Contradiction. So there cannot be 4 points not joined to 1981 points. But there can be 3. Just take the graph to have all edges except AB and AC.


Show that if $m, n$ are positive integers such that $\left(x^{m+n}+y^{m+n}+z^{m+n}\right) /(m+n)=\left(x^{m}+y^{m}+z^{m}\right) / m\left(x^{n}+y^{n}+\right.$ $\left.\mathrm{z}^{\mathrm{n}}\right) / \mathrm{n}$ for all real $\mathrm{x}, \mathrm{y}, \mathrm{z}$ with sum 0 , then $\{\mathrm{m}, \mathrm{n}\}=\{2,3\}$ or $\{2,5\}$.

Solution:
Put $\mathrm{z}=-\mathrm{x}-\mathrm{y}$. If $\mathrm{m}$ and $\mathrm{n}$ are both odd, the lhs has a term in $\mathrm{x}^{\mathrm{m}+\mathrm{n}}$ but the rhs does not. So at least one of $\mathrm{m}$ and $n$ is even. Suppose both are even. Then comparing terms in $x^{m+n}$ we get $2 /(m+n)=4 / m n$. Put $m=2 M, n$ $=2 N$, then $M N=M+N$. So $M$ must divide $N$ and $N$ must divide $M$. Hence $M=N=2$. So $m=n=4$. But put $\mathrm{x}=\mathrm{y}=1, \mathrm{z}=-2$, the lhs is $(1+1+256) / 8=129 / 4$ and the rhs is $((1+1+16) / 4)^{2}=81 / 4$. So one of $\mathrm{m}, \mathrm{n}$ must be odd and the other even. Without loss of generality we may take m odd.

Comparing the terms in $x^{\mathrm{m}+\mathrm{n}-1} \mathrm{y}$, the coefficient on the lhs is 1 . On the rhs it is $1 \mathrm{x} 2 / \mathrm{n}$. So we must have $\mathrm{n}=2$. Put $x=y=1, z=-2$. We get lhs $=\left(1+1-2^{\mathrm{m}+2}\right) /(\mathrm{m}+2)$ and rhs $=(1+1+4) / 2 x\left(1+1-2^{\mathrm{m}}\right) / \mathrm{m}$. So $(6-\mathrm{m}) 2^{\mathrm{m}-}$ $1=2 m+6$. We cannot have $m>6$ or the lhs is negative. Trying $m=1,3,5$ we find that 3 and 5 are the only solutions.

Arguably, we still have to verify that $\mathrm{m}=3, \mathrm{n}=2$ and $\mathrm{m}=5, \mathrm{n}=2$ are solutions. That is just tedious algebra. [We do have equality for those values.]


$\mathrm{D}$ is a point inside the equilateral triangle $\mathrm{ABC}$. $\mathrm{E}$ is a point inside $\mathrm{DBC}$. Show that area $\mathrm{DBC} /($ perimeter $\mathrm{DBC})^{2}>$ area $\mathrm{EBC} /(\text { perimeter } \mathrm{EBC})^{2}$.

Solution:
Let us find an expression for $\mathrm{t}=($ area $\mathrm{DBC}) /(\text { perimeter } \mathrm{DBC})^{2}$. Let angle $\mathrm{B}=2 \mathrm{x}$, angle $\mathrm{C}=2 \mathrm{y}$ and angle $\mathrm{D}$ $=2 z$ and let the inradius be $r$. Then area $D B C=r / 2 x$ perimeter $D B C$. Also $B C=r \cot x+r \cot y$, and similarly for the other sides, so perimeter $=2 r(\cot x+\cot y+\cot z)$. Hence $t=1 /(4(\cot x+\cot y+\cot z))$. Putting $\mathrm{z}=90^{\circ}-\mathrm{x}-\mathrm{y}$, we get $1 / 4 \mathrm{t}=\cot \mathrm{x}+\cot \mathrm{y}+(\cot \mathrm{x}+\cot \mathrm{y}) /(\cot \mathrm{x} \cot \mathrm{y}-1)$. Now EBC has both $\mathrm{x}$ and $\mathrm{y}$ smaller than $\mathrm{DBC}$, and cot $\mathrm{x}$ is a decreasing function of $\mathrm{x}$ (certainly for $\mathrm{x}$ in the range 0 to $30^{\circ}$ ). So writing
$u=\cot x, v=\cot y$, it is sufficient to show that $u+v+(u+v) /(u v-1)$ is an increasing function of $u$. [It is symmetric, so it follows that it is also an increasing function of $v$.] We have $x, y<30^{\circ}$, and hence $u, v>\sqrt{3}$, so we need the result at least for $\mathrm{u}, \mathrm{v}>\sqrt{ }$.

We have $u+v+(u+v) /(u v-1)=u+v+1 / v(u v-1) /(u v-1)+(v+1 / v) /(u v-1)$. In considering the dependence on $u$, we can ignore the terms that do not depend on $u$, so we have $u+\left(1+1 / v^{2}\right) /(u-1 / v)$. Put $k$ $=\mathrm{u}-1 / \mathrm{v}$, then we have to consider $\mathrm{k}+\mathrm{h}^{2} / \mathrm{k}$, where $\mathrm{h}^{2}=1+1 / \mathrm{v}^{2}$. But this is an increasing function for $\mathrm{k}>\mathrm{h}$ (see below). Thus $u+v+(u+v) /(u v-1)$ is an increasing function of $u$ for $u-1 / v>\left(1+1 / v^{2}\right)^{1 / 2}$, or for $u>$ $1 / v+\left(1+1 / v^{2}\right)^{1 / 2}$. This bound is highest for the smallest $\mathrm{v}$ in other words for $\mathrm{v}=\sqrt{ }$, when it is $\sqrt{3}$. So for $\mathrm{v}>$ $\sqrt{3}, u+v+(u+v) /(u v-1)$ is an increasing function of $u$ for $u>\sqrt{3}$.

[To see that $\mathrm{k}+\mathrm{h}^{2} / \mathrm{k}$ is increasing for $\mathrm{k}>\mathrm{h}$, take $\mathrm{k}^{\prime}>\mathrm{k}>\mathrm{h}$. Then $\mathrm{k}^{\prime}+\mathrm{h}^{2} / \mathrm{k}^{\prime}-\mathrm{k}-\mathrm{h}^{2} / \mathrm{k}=\left(\mathrm{k}^{\prime}-\mathrm{k}\right)-\mathrm{h}^{2}\left(1 / k-1 / \mathrm{k}^{\prime}\right)$ $\left.=\left(\mathrm{k}^{\prime}-\mathrm{k}\right)\left(1-\mathrm{h}^{2} / \mathrm{kk}^{\prime}\right)>0.\right]$


Show that there is a positive integer $\mathrm{k}$ such that, for every positive integer $\mathrm{n}, \mathrm{k} 2^{\mathrm{n}}+1$ is composite.

Solution:
Answer: $\mathrm{k}=542258$ (for example).

Suppose $\mathrm{p}$ is a prime dividing $2^{\mathrm{b}}-1,0 \leq \mathrm{a}<\mathrm{b}$ and $\mathrm{k}=-2^{\mathrm{b}-\mathrm{a}}$ mod $\mathrm{p}$. Then if $\mathrm{n}=\mathrm{a}$ mod $\mathrm{b}$, we have $\mathrm{k} 2^{\mathrm{n}}=-2^{\mathrm{b}-}$ ${ }^{a} 2^{a+h b}=-2^{(h+1) b}=-1 \bmod p$, so $k 2^{n}+1$ is divisible by p. So we would like to find a collection of pairs (a, b), such that every positive integer $n$ satisfies $n=a \bmod b$ for at least one member of the collection. We need the corresponding $\mathrm{p}$ distinct so that we can be sure of finding $\mathrm{k}$ by the Chinese Remainder theorem which satisfies $k=-2^{b-a} \bmod p$ for all members of the collection. For the congruences $n=a \bmod b$ to cover all the integers, we need the $1 \mathrm{~cm}$ of the $\mathrm{b}$ to be small relative to their size, so we look for an $1 \mathrm{~cm}$ with many factors.

$6=2.3$ does not work because $2^{2}-1=3,2^{3}-1=7,2^{6}-1=3^{2} 7$, we cannot find distinct primes p. $10=2.5$ does not work because there are not enough factors to cover all integers. The mod 2 residue covers $1 / 2$, the $\bmod 5$ residue covers $1 / 5$ and the $\bmod 10$ residue covers $1 / 10$, but that adds up to less than 1 . Similarly 12 does not work. We must drop one of $2,3,6$. But then the rest cover at most 11 residue classes mod 12 . So we try 24 . Again we drop 6 , but we have:

```
\(2: 2^{2}-1=3\)
\(3: 2^{3}-1=7\)
\(4: 2^{5}-1\) has factor 5
\(8: 2^{8}-1\) has factor 17
12: \(2^{12}-1\) has factor 13
\(24: 2^{24}-1\) has factor 241
```

We now find, for example, the following covering set:

```
0 mod 2 covers the even residues
1 mod 3 covers 1, 7, 13, 19
3 mod 4 covers 3, 11, 15, 23
5 mod 8 covers 5, 21
5 mod 12 covers 17
9 mod 24 covers 9
```

So we now need $\mathrm{k}$ which is

```
-4 mod 3
-4 mod 7
-2 mod 5
-8 mod 17
-128 mod 13
-215 mod 241
```


$\mathrm{O}$ is the center of a sphere $\mathrm{S}$. Points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are inside $\mathrm{S}, \mathrm{OA}$ is perpendicular to $\mathrm{AB}$ and $\mathrm{AC}$, and there are two spheres through $A, B$, and $\mathrm{C}$ which touch $\mathrm{S}$. Show that the sum of their radii equals the radius of $\mathrm{S}$.

Solution: Let $\mathrm{D}$ be the circumcenter of $\mathrm{ABC}$. The triangle $\mathrm{ABC}$ is in the plane normal to OA. The two spheres both contain through the circumcircle of $\mathrm{ABC}$, so their centers must lie on a line $\mathrm{L}$ normal to the plane $\mathrm{ABC}$ and hence parallel to OA. Take the plane through OA and the line L. Suppose the centers are P and Q. The sphere center $\mathrm{P}$ must touch $\mathrm{S}$ at a point $\mathrm{X}$ on the line OP. Similarly, the sphere center $\mathrm{Q}$ must touch $\mathrm{S}$ at a point $\mathrm{Y}$ on the line $\mathrm{OQ}$. Since the spheres pass through $A$, we have $\mathrm{PA}=\mathrm{PX}$ and hence $\mathrm{OP}+\mathrm{PA}=\mathrm{R}$, the radius of the sphere $\mathrm{S}$. Similarly $\mathrm{OQ}+\mathrm{QA}=\mathrm{R}$. Indeed if $\mathrm{K}$ is any point on the line $\mathrm{L}$ such that $\mathrm{OK}+\mathrm{KA}=$ $R$, then the sphere center $K$ will touch $S$ and pass through A. Since it will also have KD perpendicular to DA, it will contain all points on the circumcircle of $A B C$. But the locus of points such that $O K+K A=R$ is an ellipse with foci $\mathrm{O}$ and $\mathrm{A}$. So it meets the line $\mathrm{L}$ in just two points, which must therefore be $\mathrm{P}$ and $\mathrm{Q}$. Moreover, since the line $\mathrm{L}$ is parallel to $\mathrm{OA}$ the points must be equidistant from the midpoint of OA (which is the center of the ellipse). Hence $\mathrm{OP}=\mathrm{AQ}$ and so $\mathrm{AQ}+\mathrm{AP}=\mathrm{R}$, as required.


If six points are chosen sequentially at random on the circumference of a circle, what is the probability that the triangle formed by the first three is disjoint from that formed by the second three.

Solution:
Answer: $3 / 10$.

Only the order is important. We are interested in permutations of 123456 where the 123 are together (allowing wrapping). wlog the 1 is in first position. So the triangles are disjoint in the cases $123 \mathrm{xxx}, 132 \mathrm{xxx}$, $1 \mathrm{xxx} 23,1 \mathrm{xxx} 32,12 \mathrm{xxx} 3,13 \mathrm{xxx} 2$. So the probability is $6 \cdot 6 / 5!=3 / 10$.


Solution: Let the roots be $\mathrm{r}_{\mathrm{i}}$. If the condition holds, then $2 \sum \mathrm{r}_{\mathrm{i}}<5 \sum \mathrm{r}_{\mathrm{i}} \mathrm{r}_{\mathrm{j}}$. Expanding, $2 \sum \mathrm{r}_{\mathrm{i}}^{2}+4 \sum \mathrm{r}_{\mathrm{i}} \mathrm{r}_{\mathrm{j}}<5 \sum \mathrm{r}_{\mathrm{i}} \mathrm{r}_{\mathrm{j}}$, or $2 \sum$ $r_{i}^{2}<\sum r_{i} r_{j}$. But if $r_{i}$ and $r_{j}$ are real we have we have $2 r_{i} r_{j} \leq r_{i}^{2}+r_{j}^{2}$. So if all the roots are real, adding the 10 similar equations gives $2 \sum r_{i} r_{j} \leq 4 \sum r_{i}^{2}$. Contradiction. Hence not all the roots are real.


$\mathrm{S}_{1}, \mathrm{~S}_{2}, \ldots, \mathrm{S}_{\mathrm{n}}$ are subsets of the real line. Each $\mathrm{S}_{\mathrm{i}}$ is the union of two closed intervals. Any three $\mathrm{S}_{\mathrm{i}}$ have a point in common. Show that there is a point which belongs to at least half the $\mathrm{S}_{\mathrm{i}}$.

Solution: We can write $\mathrm{S}_{\mathrm{i}}=\left[\mathrm{a}_{\mathrm{i}}, \mathrm{b}_{\mathrm{i}}\right] \cup\left[\mathrm{c}_{\mathrm{i}}, \mathrm{d}_{\mathrm{i}}\right]$, where $\mathrm{a}_{\mathrm{i}} \leq \mathrm{b}_{\mathrm{i}} \leq \mathrm{c}_{\mathrm{i}}<=\mathrm{d}_{\mathrm{i}}$. Put $\mathrm{a}=\max \mathrm{a}_{\mathrm{i}}, \mathrm{d}=\min \mathrm{d}_{\mathrm{i}}$. Then a belongs to some $\mathrm{S}_{\mathrm{h}}$, and d belongs to some $\mathrm{S}_{\mathrm{k}}$. Suppose there is some $\mathrm{S}_{\mathrm{i}}$ which does not contain a or $\mathrm{d}$. Then $\mathrm{b}_{\mathrm{i}}<\mathrm{a}$, so any point in $S_{i}$ and $S_{h}$ does not belong to $\left[a_{i}, b_{i}\right]$. Similarly $c_{i}>b$, so that any point in $S_{i}$ and $S_{k}$ does not belong to $\left[\mathrm{c}_{\mathrm{i}}, \mathrm{d}_{\mathrm{i}}\right]$. But that means that $\mathrm{S}_{\mathrm{i}}, \mathrm{S}_{\mathrm{h}}$ and $\mathrm{S}_{\mathrm{k}}$ cannot have a point in common. Contradiction. So every $\mathrm{S}_{\mathrm{i}}$ must contain a or $\mathrm{d}$. Hence either a or $\mathrm{d}$ belongs to at least half of them.


Show that one can construct (with ruler and compasses) a length equal to the altitude from $\mathrm{A}$ of the tetrahedron $A B C D$, given the lengths of all the sides. [So for each pair of vertices, one is given a pair of points in the plane the appropriate distance apart.]

Solution:
Let the altitude from $\mathrm{A}$ be $\mathrm{AH}$ with $\mathrm{H}$ in the plane $\mathrm{BCD}$. The plane normal to $\mathrm{BC}$ through $\mathrm{A}$ also contains $\mathrm{H}$.

Suppose it meets BC at X. Then HX and AX are both perpendicular to BC.

Since we have the side lengths we can construct a cardboard cutout of the tetrahedron: the base BCD and the face BCA next to it, also the face $\mathrm{CDA}^{\prime}$ (and the face BDA", although we do not need it. If we folded along the lines $\mathrm{BC}, \mathrm{CD}$ and $\mathrm{BD}$, then $\mathrm{A}, \mathrm{A}^{\prime}$ and $\mathrm{A}^{\prime \prime}$ would become coincident and we would get the tetrahedron.) We have just shown that in the plane $\mathrm{AH}$ is a straight line perpendicular to $\mathrm{BC}$ (and meeting it at $\mathrm{X}$ ). So we draw this line and also the line through $\mathrm{A}^{\prime}$ perpendicular to $\mathrm{CD}$, giving $\mathrm{H}$ as their point of intersection. Thus we have $\mathrm{AH}$ and $\mathrm{HX}$ and we know that in the tetrahedron AH is perpendicular to HX. So draw a circle diameter $\mathrm{AX}$ and take a circle center $\mathrm{H}$ radius HX meeting the circle at $\mathrm{K}$. Then $\mathrm{AK}$ is the required length.


Prove that an open interval of length $1 / \mathrm{n}$ in the real line contains at most $(\mathrm{n}+1) / 2$ rational points $\mathrm{p} / \mathrm{q}$ with $1 \leq$ $\mathrm{q} \leq \mathrm{n}$.

Solution:
This is a variant on the familiar result that $\mathrm{m}+1$ integers from $\{1,2, \ldots, 2 \mathrm{~m}\}$ must include one which divides another. To prove that, take the largest odd divisor of each of the $\mathrm{m}+1$ integers. That gives us $\mathrm{m}+1$ odd numbers from $\{1,3, \ldots, 2 \mathrm{~m}-1\}$, so by the pigeonhole principle we must have some odd integer $\mathrm{b}$ twice. If the corresponding integers are $2^{\mathrm{h}} \mathrm{b}$ and $2^{\mathrm{k}} \mathrm{b}$, then one must divide the other.

Now if $1 \leq \mathrm{q} \leq \mathrm{n}$ and $1 \leq \mathrm{kq} \leq \mathrm{n}$, then $\left|\mathrm{p} / \mathrm{q}-\mathrm{p}^{\prime} / \mathrm{kq}\right| \geq 1 / \mathrm{kq} \geq 1 / \mathrm{n}$. But we cannot have two such points in an open interval of length $1 / \mathrm{n}$. Obviously we cannot have two points with the same denominator, so if $\mathrm{n}=$ $2 \mathrm{~m}$, there are at most $\mathrm{m}$ points and if $\mathrm{n}=2 \mathrm{~m}+1$ there are at most $\mathrm{m}+1$ points.


Two roots of the real quartic $x^{4}-18 x^{3}+a x^{2}+200 x-1984=0$ have product -32 . Find a.

Solution: Let the two roots satisfy the quadratic $\mathrm{x}^{2}+\mathrm{hx}-32=0$ (we have not yet shown that $\mathrm{h}$ is real). The other two roots have product $-1984 /-32=62$. Let them satisfy the quadratic $x^{2}+k x+62$. So $x^{4}-18 x^{3}+a x^{2}+200 x-$ $1984=\left(\mathrm{x}^{2}+\mathrm{hx}-32\right)\left(\mathrm{x}^{2}+\mathrm{kx}+62\right)=\mathrm{x}^{4}+(\mathrm{h}+\mathrm{k}) \mathrm{x}^{3}+(\mathrm{hk}+30) \mathrm{x}^{2}+(62 \mathrm{~h}-32 \mathrm{k}) \mathrm{x}-1984=0$. Equating coefficients: $h+k=-18,62 h-32 k=200$. Solving, $h=-4, k=-14$. Hence $a=86$.


Can one find a set of $n$ distinct positive integers such that the geometric mean of any (non-empty, finite) subset is an integer? Can one find an infinite set with this property?

Solution:
Answer: yes, no.

Take each member to be an $n!$ power (for example, $\left.1^{n!}, 2^{n!}, \ldots, n^{n!}\right)$.

Suppose we could find an infinite set. Take any two members $\mathrm{m}$ and $\mathrm{n}$. Then for sufficiently large $\mathrm{k}$, $(m / n)^{1 / k}$ must be irrational. But now if we take any other $a_{1}, a_{2}, \ldots, a_{k-1}$ in the set, $\left(m a_{1} \ldots a_{k}\right)^{1 / k}$ and $\left(n a_{1} \ldots\right.$ $\left.a_{k}\right)^{1 / k}$ cannot both be integers. Contradiction.


$A, B, C, D, X$ are five points in space, such that $A B, B C, C D, D A$ all subtend the acute angle $\theta$ at $X$. Find the maximum and minimum possible values of $\angle \mathrm{AXC}+\angle \mathrm{BXD}$ (for all such configurations) in terms of $\theta$.

Solution:
Answer: minimum 0 (see below); maximum $2 \cos ^{-1}(2 \cos \theta-1)$, achieved by a square pyramid.

If we take $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ to lie in a plane so that $\mathrm{AC}$ and $\mathrm{BD}$ meet at $\mathrm{X}$ with the angle $\mathrm{AXC}=\theta$, then $\mathrm{AB}, \mathrm{BC}$, $\mathrm{CD}$, DA all subtend the angle $\theta$ at $\mathrm{X}$, but $\mathrm{AC}$ and $\mathrm{BD}$ subtend the angle zero. So the minimum is 0 .

A little playing around suggests that we should take $\mathrm{ABCD}$ to be a square, with $\mathrm{X}$ on the normal through the center, so that $\mathrm{XABCD}$ is a square pyramid. We calculate the result for this case. Suppose $\mathrm{XA}=\mathrm{XB}=\mathrm{XC}=$ $\mathrm{XD}=1, \angle \mathrm{AXB}=\angle \mathrm{BXC}=\angle \mathrm{CXD}=\angle \mathrm{DXA}=\theta$. Then $\mathrm{AB}=2 \sin \theta / 2$, so $\mathrm{AC}=2 \sqrt{2} \sin \theta / 2$. So $\sin$ $\mathrm{AXC} / 2=\sqrt{2} \sin \theta / 2$. Hence $\cos \mathrm{AXC}=1-2 \sin ^{2} A X C / 2=1-4 \sin ^{2} \theta / 2=2 \cos \theta-1$. Hence $\angle \mathrm{AXC}+\angle$ BXD $=2 \cos ^{-1}(2 \cos \theta-1)$. We note that this increases monotonically from 0 (at $\theta=0$ ) to $2 \pi$ (at $\theta=\pi / 2$ ). For $\theta>0$, we have $\cos \theta<1$, hence $2 \cos \theta-1<\cos \theta$ and hence $2 \cos ^{-1}(2 \cos \theta-1)>2 \theta$. In other words, except for $\theta=0$, we can certainly do better than $2 \theta$.

Take vectors origin $\mathrm{X}$. Write $\mathrm{XA}=\mathbf{A}$ etc. Since we are only interested in the angles, it is convenient to take a to be the unit vector in the direction $\mathbf{A}$. Then we have $\mathbf{a . b}=\mathbf{b} . \mathbf{c}=\mathbf{c} . \mathbf{d}=\mathbf{d} . \mathbf{a}=\cos \theta$. So $(\mathbf{a}-\mathbf{c}) .(\mathbf{b}-\mathbf{d})=$ 0 . So either $\mathbf{a}=\mathbf{c}$, or $\mathbf{b}=\mathbf{d}$ or $\mathbf{a}-\mathbf{c}$ is perpendicular to $\mathbf{b}-\mathbf{d}$. Suppose $\mathbf{a}=\mathbf{c}$. Then $X$ lies on the line $A C$. In this case we certainly have $A D$ and $C D$ subtending the same angle at $X$, and $A B$ and $B C$ subtending the same angle at $\mathrm{X}$. But $\angle \mathrm{AXC}=0 . \mathrm{XABD}$ is a tetrahedron with $\angle \mathrm{AXD}=\angle \mathrm{AXB}=\theta . \angle \mathrm{BXD} \leq \angle \mathrm{AXB}+\angle$ AXD (with equality only if A, B, D, $X$ lie in a plane). So $\angle B X D+\angle A X C \leq 2 \theta$ and we have just seen that this is not maximal. Similarly if $\mathbf{b}=\mathbf{d}$. So we may assume that $\mathbf{a}-\mathbf{c}$ is perpendicular to $\mathbf{b}-\mathbf{d}$ and both are non-zero.

We also have $(\mathbf{a}+\mathbf{c}) \cdot(\mathbf{a}-\mathbf{c})=\mathrm{a}^{2}-\mathbf{c}^{2}=0$ and $(\mathbf{a}+\mathbf{c}) .(\mathbf{b}-\mathbf{d})=0$. So $\mathbf{a}+\mathbf{c}$ is normal to the plane containing $\mathbf{a}-$ $\mathbf{c}$ and $\mathbf{b}-\mathbf{d}$. Similarly, $\mathbf{b}+\mathbf{d}$. Hence $\mathbf{a}+\mathbf{c}$ and $\mathbf{b}+\mathbf{d}$ are multiples of each other and $(\mathbf{a}+\mathbf{c}) .(\mathbf{b}+\mathbf{d})=|\mathbf{a}+\mathbf{c}|$ $|\mathbf{b}+\mathbf{d}| \cdot$ But lhs $=4 \cos \theta$ and rhs $=(2 \cos \mathrm{AXC} / 2)(2 \cos \mathrm{BXD} / 2)$. So $\cos \theta=\cos \mathrm{AXC} / 2 \cos \mathrm{BXD} / 2=1 / 2$ $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)+1 / 2 \cos (\mathrm{AXC} / 2-\mathrm{BXD} / 2)$. We want to maximise $\mathrm{AXC} / 2+\mathrm{BXD} / 2$ and hence to minimise $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)$, so we must maximise $\cos (\mathrm{AXC} / 2$ - $\mathrm{BXD} / 2)$ and hence take $\angle \mathrm{AXC}=\angle$ BXD. That then gives $\cos (\mathrm{AXC} / 2+\mathrm{BXD} / 2)=2 \cos \theta-1$, so $\mathrm{AXC}+\mathrm{BXD}=2 \cos ^{-1}(2 \cos \theta-1)$. So the square pyramid is indeed maximal.


A maths exam has two papers, each with at least one question and 28 questions in total. Each pupil attempted 7 questions. Each pair of questions was attempted by just two pupils. Show that one pupil attempted either nil or at least 4 questions in the first paper.

Solution:
Each pupil attempts 7 questions and hence 21 pairs of questions. There are $28 \cdot 27 / 2=378$ pairs of questions in total and each is attempted by 2 pupils. So there must be $378 \cdot 2 / 21=36$ pupils. Suppose n pupils solved question 1. Each solved 6 pairs involving question 1, so there must be $3 \mathrm{n}$ pairs involving question 1. But there are 27 pairs involving question 1 , so $\mathrm{n}=9$. The same applies to any other question. So each question was solved by 9 pupils.

Suppose the result is false. Suppose there are $\mathrm{m}$ questions in the first paper, that the number of pupils solving $1,2,3$ questions in the first paper is $\mathrm{a}, \mathrm{b}, \mathrm{c}$ respectively. So $\mathrm{a}+\mathrm{b}+\mathrm{c}=36, \mathrm{a}+2 \mathrm{~b}+3 \mathrm{c}=9 \mathrm{~m}$. Now consider pairs of problems in the first paper. There are $\mathrm{m}(\mathrm{m}-1) / 2$ such pairs. Pupils solving just 1 solve no pairs, those solving 2 solve 1 pair and those solving 3 solve 3 pairs, so we have $b+3 c=m(m-1)$. Solving for $b$ we get $b$ $=-2 m^{2}+29 m-108=-2(m-29 / 4)^{2}-23 / 8<0$. Contradiction. So the result must be true.


A polynomial of degree $3 n$ has the value 2 at $0,3,6, \ldots, 3 n$, the value 1 at $1,4,7, \ldots, 3 n-2$ and the value 0 at $2,5,8, \ldots, 3 n-1$. Its value at $3 n+1$ is 730 . What is $n$ ?

Solution:
Answer: $\mathrm{n}=4$.

The $(3 n+1)$ th differences of the polynomial are zero. Call it $p(x)$, so we have $p(3 n+1)-(3 n+1) C 1 p(3 n)+$ $(3 n+1) C 2 p(3 n-1)-\ldots+(-1)^{3 n+1} p(0)=0$, where $r C s$ is the binomial coefficient. Hence $p(3 n+1)=2($

$(3 n+1) C 1-(3 n+1) C 4+\ldots)+((3 n+1) C 3-(3 n+1) C 6+\ldots)$. Putting $n=1$, we get: $p(4)=2(4 C 1-4 C 4)+$ $4 C 3=6+4=10$. So $n$ is not 1 . Putting $n=2$, we get: $p(7)=2(7 C 1-7 C 4+7 C 7)+(7 C 3-7 C 6)=2(7-35$ $+1)+(35-7)=-26$. So $n$ is not 2. Putting $n=3$, we get: $p(10)=2(10 C 1-10 C 4+10 C 7-10 C 10)+(10 C 3$ $-10 C 6+10 C 9)=2(10-210+120-1)+(120-210+10)=-162-100=-262$. So $n$ is not 3 . Putting $n=4$, we get: $p(13)=2(13 C 1-13 C 4+13 C 7-13 C 10+13 C 13)+(13 C 3-13 C 6+13 C 9-13 C 12)=2(13-715+$ $1716-286+1)+(286-1716+715-13)=1458-728=730$. So $n=3$ works.


Do there exist 1985 distinct positive integers such that the sum of their squares is a cube and the sum of their cubes is a square?

Solution:
Answer: yes.

Take any $\mathrm{n}$ integers $\mathrm{a}_{\mathrm{i}}$. Suppose that $\sum \mathrm{a}_{\mathrm{i}}^{2}=\mathrm{b}$ and $\sum \mathrm{a}_{\mathrm{i}}^{3}=\mathrm{c}$. Now multiply each $\mathrm{a}_{\mathrm{i}}$ by $\mathrm{b}^{4} \mathrm{c}^{3}$. The sum of their squares becomes $b^{9} c^{6}$ which is a cube and the sum of their cubes becomes $b^{12} c^{10}$ which is a square.


Find all real roots of the quartic $\mathrm{x}^{4}-(2 \mathrm{~N}+1) \mathrm{x}^{2}-x+\mathrm{N}^{2}+N-1=0$ correct to 4 decimal places, where $\mathrm{N}=$ $10^{10}$.

Solution:
Answer: 99999.9984 and 100000.0016 .

We can write the equation as $\left(x^{2}-N-1 / 2\right)^{2}=x+5 / 4$. For $x<-5 / 4$ the lhs is positive and the rhs is negative, so there are no roots with $x<-5 / 4$. If $x$ lies between $-5 / 4$ and 0 , then the lhs is obviously much larger than the rhs, so again there is no root. Thus there are no negative roots. Descartes' rule of signs (see below) tells us that there are at most 2 positive roots.

If $x=10^{5}$, then the lhs is $1 / 4$ and the rhs is much larger (approx $10^{5}$ ). If $x=10^{5} \pm 1$, then the lhs is $\left( \pm 210^{5}+\right.$ $1 / 2)^{2}$ which is approximately $410^{10}$ and much larger than the rhs. So there is a root either side of $10^{5}$. Put $\mathrm{x}=$ $10^{5} \pm \mathrm{k}$. Then we want $\left( \pm 2 \mathrm{k} \cdot 10^{5}+\mathrm{k}^{2}-1 / 2\right)^{2}=10^{5}+5 / 4$, or $\left(4 \mathrm{k}^{2} 10^{5}-1\right) 10^{5} \pm 2 \mathrm{k}\left(1-2 \mathrm{k}^{2}\right) 10^{5}-5 / 4=0$. So evidently we need approximately $4 \mathrm{k}^{2}=10^{-5}$, or $\mathrm{k}= \pm 0.0016$. So it looks as though the roots are $10^{5} \pm 0.0016$ $=99999.9984$ and 100000.0016.

Put $\mathrm{x}=10^{5} \pm 0.00155$. Then $\left(\mathrm{x}^{2}-10^{10}-1 / 2\right)^{2}-\mathrm{x}-5 / 4=\left( \pm 310-1 / 2+0.00155^{2}\right)^{2}-10^{5} \pm 0.00155-5 / 4<$ $311^{2}-10^{5}<0$. Put $x=10^{5} \pm 0.00165$, then $\left(x^{2}-10^{10}-1 / 2\right)^{2}-x-5 / 4=\left( \pm 330-1 / 2+0.00165^{2}\right)^{2}-10^{5} \pm$ $0.00165-5 / 4>329^{2}-10^{5}-2>0$. So indeed one root lies between $10^{5}-0.00165$ and $10^{5}-0.00155$ and the other root lies between $10^{5}+0.00155$ and $10^{5}+0.00165$.

## Descartes' rule of signs.

This states that if the number of sign changes in the coefficients of the polynomial is $d$, then the number of positive roots is $\mathrm{d}$ or $\mathrm{d}$ less an even number. So, for example, if the polynomial is $\mathrm{x}^{5}+14.3 \mathrm{x}^{4}-34 \mathrm{x}^{2}-\mathrm{x}+$ 3.2, then there are two sign changes ( +14.3 to -34 and -1 to 3.2 ), so there are either 0 or 2 positive roots. Note that we ignore zero coefficients. If $r$ is a positive root of $p(-x)=0$, then $-r$ is a negative root of $p(x)=0$. So if we substitute - $\mathrm{x}$ for $\mathrm{x}$ in the polynomial and the number of sign changes is then $\mathrm{d}^{\prime}$, then we can conclude that the number of negative roots of the polynomial is either $\mathrm{d}^{\prime}$ or $\mathrm{d}^{\prime}$ less an even number. With the example above we get $-x^{5}+14.3 x^{4}-34 x^{2}+x+3.2$, which has 3 sign changes. So the polynomial has 1 or 3 negative roots.

The proof is not difficult. The key idea is to show that if $k$ is a positive root, so that $p(x)=(x-k) q(x)$, then (1) $p(x)$ has at least one more sign change than $q(x)$, and (2) the difference between the number of sign changes is odd (note that the signs of the constant coefficients of $p(x)$ and $q(x)$ are different).


A tetrahedron has at most one edge longer than 1 . What is the maximum total length of its edges?

Solution:
Answer: $5+\sqrt{3}$.

Suppose $\mathrm{AB}$ is the edge which may be longer than 1. Then if we rotate the triangle $\mathrm{ACD}$ about $\mathrm{CD}$ until $\mathrm{A}$ lies in the same plane as $B C D$ and is on the opposite side of $\mathrm{CD}$ to $\mathrm{B}$, then we maximise $\mathrm{AB}$ without changing any of the other side lengths. So the maximal configuration must be planar.

Now suppose we regard $\mathrm{C}$ and $\mathrm{D}$ as fixed and the other points as variable. Suppose $\mathrm{CD}=2 \mathrm{x}(<=1)$. Then $\mathrm{A}$ and $\mathrm{B}$ must both lie inside the circle center $\mathrm{C}$ radius 1 and inside the circle center $\mathrm{D}$ radius 1 and hence inside their intersection which is bounded by the two arcs XY (assuming they meet at $\mathrm{X}$ and $\mathrm{Y}$ ). Obviously we maximise $\mathrm{AC}+\mathrm{AD}+\mathrm{BC}+\mathrm{BD}$ by taking $\mathrm{A}$ at $\mathrm{X}$ and $\mathrm{B}$ at $\mathrm{Y}$ (or vice versa). We claim that that choice also maximises $\mathrm{AB}$. Suppose that is true. Then it also maximises $\mathrm{AC}+\mathrm{AD}+\mathrm{BC}+\mathrm{BD}+\mathrm{AB}$ at $4+2\left(1-\mathrm{x}^{2}\right)^{1 / 2}$. So we now have to vary $\mathrm{CD}$ to maximise $2 \mathrm{x}+4+2\left(1-\mathrm{x}^{2}\right)^{1 / 2}$. We show that $\mathrm{x}+\left(1-\mathrm{x}^{2}\right)^{1 / 2}$ is increasing for $\mathrm{x}$ $<=1 / 2$ and hence that the maximum is at $x=1 / 2$. Put $x=\sin$. Then we have $x+\left(1-x^{2}\right)^{1 / 2}=\sqrt{2} \sin (t+\pi / 4)$ which is indeed increasing for $\mathrm{x} \leq \pi / 6$.

It remains to prove the claim. Take the circle diameter XY. Then the two ares both lie inside this circle. [Two circles intersect in at most two points, so each arc must lie entirely inside or entirely outside the circle center $\mathrm{O}$ and it obviously does not lie outside. ] But AB lies inside a chord of this circle The length of the chord cannot exceed the diameter of the circle (which is $\mathrm{XY}$ ) and hence $\mathrm{AB} \leq \mathrm{XY}$.


A graph has $n>2$ points. Show that we can find two points $A$ and $B$ such that at least $[n / 2]-1$ of the remaining points are joined to either both or neither of $\mathrm{A}$ and $\mathrm{B}$.

Solution: Consider the number of pairs $(X,\{Y, Z\})$, where $\mathrm{X}, \mathrm{Y}, \mathrm{Z}$ are distinct points such that $\mathrm{X}$ is joined to just one of $Y, Z$. If $X$ is joined to just $k$ points, then there are just $k(n-1-k) \leq(n-1)^{2} / 4$ such pairs $(X,\{Y, Z\})$. Hence in total there are at most $n(n-1)^{2} / 4$ such pairs. But there are $n(n-1) / 2$ possible $\{\mathrm{Y}, \mathrm{Z}\}$. So we must be able to find one of them $\{A, B\}$ which belongs to at most $[(n-1) / 2]$ such pairs. Hence there are at least $n-2$ $-[(n-1) / 2]=[n / 2]-1$ points $X$ which are joined to both of A and B or to neither of A and B. [If confused by the [] , consider $n=2 m$ and $n=2 m+1$ separately! $]$


$0<a_{1} \leq a_{2} \leq a_{3} \leq \ldots$ is an unbounded sequence of integers. Let $b_{n}=m$ if $a_{m}$ is the first member of the sequence to equal or exceed $n$. Given that $a_{19}=85$, what is the maximum possible value of $a_{1}+a_{2}+\ldots+$ $\mathrm{a}_{19}+\mathrm{b}_{1}+\mathrm{b}_{2}+\ldots+\mathrm{b}_{8} ?$

Solution:
We show that the only possible value of the sum is $85.20=1700$.

That is certainly the value if all $a_{i}=85$, for then all $b_{j}=1$ and so the sum is $19 \cdot 85+85 \cdot 1=85 \cdot 20$. Now consider the general case. Suppose that we increase some $\mathrm{a}_{\mathrm{i}}$ by 1 from $\mathrm{k}$ to $\mathrm{k}+1$ (whilst preserving the property that $a_{i}$ is increasing, so we must have $a_{i}<a_{i+1}$ before the increase). The effect of the increase is to change $b_{k+1}$ from $i+1$ to $i$, but not to change any other $b_{j}$. This is obvious if $a_{i-1}<a_{i}$ and $a_{i+1}>a_{i}+1$. If $a_{i-1}=a_{i}$, then before the change $b_{k}=i-1$ (not i) and that is still true after the change. Equally, if $a_{i}=a_{i+1}$ after the change, then it is still true that $\mathrm{b}_{\mathrm{k}+1}$ changes from $\mathrm{i}+1$ to $\mathrm{i}$. Thus the overall effect of the increase is not to change the sum of the $\mathrm{a}_{\mathrm{i}}$ plus the sum of the $\mathrm{b}_{\mathrm{j}}$. But by a series of such changes we convert any initial sequence to all $\mathrm{a}_{\mathrm{i}}=85$.


Do there exist 14 consecutive positive integers each divisible by a prime less than 13 ? What about 21 consecutive positive integers each divisible by a prime less than 17 ?

Solution:
Answer: no, yes.

There are 7 odd numbers. At most one can be a multiple of 7 and one a multiple of 11. It is only possible to have two of them multiples of 5 if either the largest or smallest is a multiple of 5 . But it is only possible to have three of them multiples of 3 if both the largest and the smallest are multiples of 3 . So at most four numbers are multiples of 3 or 5 . That leaves one odd number unaccounted for.

A little juggling shows that we want the odd numbers to be divisible by: $3,7,5,3,11,13,3,5,7,3$ (or the 11 and 13 can be interchanged). We need the 6th odd to be 13 mod 143 to get the 11 and 13 correct, and hence the 1 st in the sequence to be 2 mod 143. But the 1st must be even, so it must be 2 mod 286 . The 2 nd must be divisible by 3 , so the 1 st must be $2 \bmod 858$. But we need the 4 th to be divisible by 7 , so the 1 st must be $3434 \bmod 6006$. Then the 6 th must be divisible by 5 , so the 1 st must be $9440 \bmod 30030$. Thus an example is $9440,9441, \ldots, 9460$.


Five professors attended a lecture. Each fell asleep just twice. For each pair there was a moment when both were asleep. Show that there was a moment when three of them were asleep.

Solution:
This is a slightly tricky application of the pigeonhole principle.

For each pair take the first moment when they are both asleep. There are ten pairs, so ten such moments. If two coincide, then we are done because at that moment at least three professors were asleep. So suppose they are all distinct and form a set S. Each such moment must also be one of the 10 occasions when a professor falls asleep. But consider the earliest member of S. Two professors were asleep at that moment so two fell asleep at or before that moment. Thus each of the remaining 9 members of $\mathrm{S}$ must be one of the 8 later occasons when a professor fell asleep. So they cannot all be distinct. Contradiction.


What is the smallest $\mathrm{n}>1$ for which the average of the first $\mathrm{n}$ (non-zero) squares is a square?

Solution:
Answer: 337.

We need $1 / 6(n+1)(2 n+1)$ a square. We need $n=6 m+1$ for it to be an integer. So $(3 m+1)(4 m+1)$ must be a square. But $3 m+1$ and $4 m+1$ are relatively prime, so each must be a square. Suppose $3 m+1=a^{2}$ and $4 m+1=$ $(a+k)^{2}$, then (subtracting) $m=2 a k+k^{2}$, so $a^{2}-6 a k-\left(3 k^{2}+1\right)=0$, so $a=3 k+\sqrt{ }\left(12 k^{2}+1\right)$. By inspection the smallest solution of this is $\mathrm{k}=2$, giving $\mathrm{a}=13$ and hence $\mathrm{m}=56$ and $\mathrm{n}=337$.