A T-square allows you to construct a straight line through two points and a line perpendicular to a given line through a given point. Circles $\mathrm{C}$ and $\mathrm{C}^{\prime}$ intersect at $\mathrm{X}$ and $\mathrm{Y}$. $\mathrm{XY}$ is a diameter of $\mathrm{C}$. $\mathrm{P}$ is a point on $\mathrm{C}^{\prime}$ inside

C. Using only a $T$-square, find points $\mathrm{Q}, \mathrm{R}$ on $\mathrm{C}$ such that $\mathrm{QR}$ is perpendicular to $\mathrm{XY}$ and $\mathrm{PQ}$ is perpendicular to PR.

Solution:
$\mathrm{C}^{\text {' }}$ seems slightly odd. Why not just say that $\mathrm{XY}$ is a diameter of $\mathrm{C}, \mathrm{P}$ is a point inside $\mathrm{C}$ and we want points $\mathrm{Q}, \mathrm{R}$ on $\mathrm{C}$ such that $\mathrm{QR}$ is perpendicular etc.? So presumably it is there to help.

Note that PQ perpendicular to PR means that P lies on the circle diameter QR. So take M to be the midpoint of $\mathrm{QR}$. Then $M Q=M R=M P$. Guided by the hint about $C^{\prime}$, we extend $P M$ to meet $C^{\prime}$ at $S$. Then $Q M \cdot M R=$ $\mathrm{XM} \cdot \mathrm{MY}($ circle $\mathrm{C})=\mathrm{PM} \cdot \mathrm{MS}$ (circle $\left.\mathrm{C}^{\prime}\right)$. So M is also the midpoint of PS. So we want to find $\mathrm{S}$ on $\mathrm{C}^{\prime}$ so that XY bisects SP.

Let the line XP meet $\mathrm{C}$ at U. Draw the line through U perpendicular to XY to meet $\mathrm{C}$ again at V. Draw XV. Draw the line through $\mathrm{P}$ perpendicular to $\mathrm{XY}$ to meet $\mathrm{XV}$ at $\mathrm{T}$. Then $\mathrm{XY}$ bisects $\mathrm{PT}$, which is parallel to $\mathrm{QR}$. Now draw the line through T perpendicular to PT (and hence parallel to XY) to meet $\mathrm{C}^{\prime}$ at $\mathrm{S}$. XY must also bisect ST. So we have the required point M. Draw the line through M perpendicular to XY to meet the circle $\mathrm{C}$ at $\mathrm{Q}$ and $\mathrm{R}$.


A partition of $n$ is an increasing sequence of integers with sum $n$. For example, the partitions of 5 are: $1,1,1$, 1,$1 ; 1,1,1,2 ; 1,1,3 ; 1,4 ; 5 ; 1,2,2 ;$ and 2,3 . If $\mathrm{p}$ is a partition, $\mathrm{f}(\mathrm{p})=$ the number of $1 \mathrm{~s}$ in $\mathrm{p}$, and $\mathrm{g}(\mathrm{p})=$ the number of distinct integers in the partition. Show that $\sum f(p)=\sum g(p)$, where the sum is taken over all partitions of $n$.

Solution:
Let $\pi(n)$ be the number of partitions of $n$. If $p=a_{1}, a_{2}, \ldots, a_{m}$ is a partition of $n$, then $h(p)=1, a_{1}, \ldots, a_{m}$ is $a$ partition of $n+1$ which contains a 1 . Moreover, $h$ is obviously a bijection between partitions of $n$ and partitions of $\mathrm{n}+1$ which contain a 1. But $h(p)$ also contains one more 1 than $p$, so $\sum f(p)$ for $n+1$ is $\sum f(p)$ for $\mathrm{n}$ plus $\pi(\mathrm{n})$. Thus $\sum \mathrm{f}(\mathrm{p})$ for $\mathrm{n}$ is $1+\pi(1)+\pi(2)+\ldots+\pi(\mathrm{n}-1)$. [Obviously $\sum \mathrm{f}(\mathrm{p})$ for $\mathrm{n}=1$ is 1 .]

Checking, we find $\pi(1)=1, \pi(2)=2, \pi(3)=3, \pi(4)=5, \pi(5)=7$. So this formula gives $\sum f(p)$ for $n=5$ is $1+$ $1+2+3+5=12$, which is correct.

Now fix $n$ and consider the number of pairs $(p, m)$, where $p$ is a partition of $n$ containing $m$. For each $p$ the number of such pairs is $g(p)$. So the total number of such pairs $\sum g(p)$. But the number of pairs $(p, m)$ for fixed $m$ is just $\pi(n-m)($ taking $\pi(0)=0)$. So the total is also $\sum \pi(n-m)$. So $\sum g(p)=1+\pi(1)+\ldots+\pi(n-1)=$ $\sum \mathrm{f}(\mathrm{p})$.


Find all solutions to $\left(m^{2}+n\right)\left(m+n^{2}\right)=(m-n)^{3}$, where $m$ and $n$ are non-zero integers.

Solution:
Answer: $(m, n)=(-1,-1),(8,-10),(9,-6),(9,-21)$.

If $m, n>0$, then lhs is certainly positive, so we must have $m>n$ to make the rhs positive. But then $m^{2}+n>$ $\mathrm{m}^{2}$ and $\mathrm{n}^{2}+\mathrm{m}>\mathrm{m}$, so $\mathrm{lh} \mathrm{p}>\mathrm{m}^{3}>$ rhs. Contradiction. So there are no solutions with $\mathrm{m}$ and $\mathrm{n}$ both positive.

Put $M=|m|, N=|n|$. Consider next $m=M, n=-N$. So we have $\left(M^{2}-N\right)\left(N^{2}+M\right)=(M+N)^{3}$. Hence $M^{2} N^{2}+$ $M^{3}-N^{3}-M N=M^{3}+3 M^{2} N+3 M N^{2}+N^{3}$. $N$ is non-zero, so we can divide to get: $2 N^{2}+N\left(3 M-M^{2}\right)+$ $3 M^{2}+M=0$. Regarding this as a quadratic in $N$ we solve, getting $N=\left(\left(M^{2}-3 M\right) \pm \sqrt{ }\left(M^{4}-6 M^{3}+9 M^{2}-\right.\right.$ $\left.\left.24 M^{2}-8 M\right)\right) / 4$. Thus $M^{4}-6 M^{3}-15 M^{2}-8 M=M(M-1)^{2}(M-8)$ must be a square. Hence $M(M-8)$ must be a square. We have $(M-4)^{2}=M(M-8)+16$ and for $M \geq 13$, we have $(M-5)^{2}=M^{2}-10 M+25<M^{2}-8 M$. So we need only consider $M \leq 12$. But obviously we cannot have $M<8$, or $M(M-8)$ is negative. Checking the remaining values, we find $M=8$ and 9 are the only solutions. They give the solutions $(m, n)=(8,-10)$, $(9,-6),(9,-21)$.

Next, consider the case $m=-M, n=N$. That clearly does not work. We get $\left(M^{2}+N\right)\left(N^{2}-M\right)=-(M+N)^{3}$. So $2 N^{2}+\left(M^{2}+3 M\right) N+3 M^{2}-M=0$, which has no solutions because $3 M^{2}>M$.

Finally, consider the case $m=-M, n=-N$. Then we get $\left(M^{2}-N\right)\left(N^{2}-M\right)=(N-M)^{3}$, so $2 N^{2}-\left(M^{2}+3 M\right) N+$ $\left(3 M^{2}-M\right)=0$. Solving for $N$, we get $N=\left(\left(M^{2}+3 M\right) \pm \sqrt{ }\left(M^{4}+6 M^{3}+9 M^{2}-24 M^{2}+8 M\right)\right) / 4$. So we require
$M\left(M^{3}+6 M^{2}-15 M+8\right)=M(M+8)(M-1)^{2}$ to be a square. Hence $M(M+8)$ is a square. We have $(M+$ $4)^{2}>M(M+8)$ and for $M \geq 5$, we have $(M+3)^{2}=M^{2}+6 M+9<M^{2}+8 M$. So we need only check $M=1$, $2,3,4$. We find $M=1$ gives the only square, and that gives the solution ( $m, n)=(-1,-1)$.


The feet of the angle bisectors of the triangle $\mathrm{ABC}$ form a right-angled triangle. If the right-angle is at $\mathrm{X}$, where $\mathrm{AX}$ is the bisector of angle $\mathrm{A}$, find all possible values for angle $\mathrm{A}$.

Solution:
Answer: $120^{\circ}$.

Use vectors origin $A$. Write the vector $A B$ as $\mathbf{B}$ etc. Using the familiar $B X / C X=A B / A C$ etc, we have $\mathbf{Z}=$ $\mathrm{b} \mathbf{B} /(\mathrm{a}+\mathrm{b}), \mathbf{Y}=\mathrm{c} \mathbf{C} /(\mathrm{a}+\mathrm{c}), \mathbf{X}=(\mathrm{b} \mathbf{B}+\mathrm{c} \mathbf{C}) /(\mathrm{b}+\mathrm{c})$. Hence $\mathbf{Z}-\mathbf{X}=\mathrm{b} \mathbf{B}(\mathrm{c}-\mathrm{a}) /((\mathrm{a}+\mathrm{b})(\mathrm{b}+\mathrm{c}))-\mathrm{c} \mathbf{C} /(\mathrm{b}+\mathrm{c}), \mathbf{Y}-\mathbf{X}=-$ $b \mathbf{B} /(b+c)+c \mathbf{C}(b-a) /((a+c)(b+c))$

We have (Z-X). $(\mathbf{Y}-\mathbf{X})=0$, so after multiplying through by $(a+b)(a+c)(b+c)^{2}$, we get $b^{2} \mathbf{B}^{2}\left(a^{2}-c^{2}\right)+c^{2} \mathbf{C}^{2}\left(a^{2}-b^{2}\right)$ $+2 b c \mathbf{B} \cdot \mathbf{C}\left(a^{2}+b c\right)=0$. But $\mathbf{B}^{2}=c^{2}, \mathbf{C}^{2}=b^{2}$, so $b c\left(2 a^{2}-b^{2}-c^{2}\right)+2$ B. $C\left(a^{2}+b c\right)=0$.

But $a^{2}=b^{2}+c^{2}-2$ B. $\mathbf{C}$ (cosine rule), so $2 a^{2}-b^{2}-c^{2}=a^{2}-2$ B.C. Hence $b c\left(a^{2}-2\right.$ B.C. $)+2$ B. $\mathbf{C}\left(a^{2}+b c\right)=0$, so $\mathbf{B} \cdot \mathbf{C}=$ -bc/2. Hence $\cos \mathrm{A}=-1 / 2$, so $\angle \mathrm{A}=120^{\circ}$.


$X$ is the smallest set of polynomials $p(x)$ such that: (1) $p(x)=x$ belongs to $X$; and (2) if $r(x)$ belongs to $X$, then $x r(x)$ and $(x+(1-x) r(x))$ both belong to $X$. Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $\mathrm{r}(\mathrm{x}) \neq \mathrm{s}(\mathrm{x})$ for any $0<\mathrm{x}<1$.

Solution:
If they are never equal, then we must be able to order them, so that $r(x)>s(x)$ for all $x$ in $(0,1)$ or $s(x)>r(x)$ for all $\mathrm{x}$ in $(0,1)$. Let us use the notation $[+--+\ldots]$. The operation $r(x)$ to $x r(x)$ is denoted as - , and the operation $r(x)$ to $x+(1-x) r(x)$ is denoted as + . Then the operations are listed in reverse order with the last carried out put first. So for example $[+-]$ means we first apply - to get $x^{2}$, then + to get $x+(1-x) x^{2}=x+$ $x^{2}-x^{3}$. We claim that we have the ordering $+>$ nothing $>-$, which we apply starting with the first term. So looking at the first few polynomials we have: $[++]>[+]$, because we compare the first terms which are equal, then we compare the second terms. We regard $[+]$ as having nothing for the second term, so $+>$ nothing. Then $[+]>[+-]$, because they have equal first terms, but unequal second terms (nothing $>-$ ). The starting polynomial is represented as [ ]. So we have $[+-]>[]$ because $+>$ nothing. Then []$>[-+]>$ $[-]>[--]$. Clearly this ordering, which is a type of lexicographic ordering is a total order, that is, for any two distinct polynomials in $\mathrm{X}$ we will find that one is larger than the other.

So suppose $r(x)$ belongs to the set $X$. We have to establish that + > nothing $>-$. In other words, that $x+(1-$ x) $r(x)>x$ and that $x>x r(x)$. But that is obvious, because by a trivial induction we have $0<r(x)<1$ for all $r(x)$ in $X$ and $x$ in $(0,1)$. So, if follows that if $r(x)$ and $s(x)$ are in $X$ then $x+(1-x) r(x)>x ~ s(x)$. It is also obvious that if $[\mathrm{a}]>[\mathrm{b}]$, where $\mathrm{a}$ and $\mathrm{b}$ are some strings of + and,- then $[+\mathrm{a}]>[+b]$ and $[-\mathrm{a}]>[-b]$. But that is sufficient to establish the ordering.


$\mathrm{M}$ is the midpoint of $\mathrm{XY}$. The points $\mathrm{P}$ and $\mathrm{Q}$ lie on a line through $\mathrm{Y}$ on opposite sides of $\mathrm{Y}$, such that $|\mathrm{XQ}|=$ $2|\mathrm{MP}|$ and $|\mathrm{XY}| / 2<|\mathrm{MP}|<3|\mathrm{XY}| / 2$. For what value of $|\mathrm{PY}| /|\mathrm{QY}|$ is $|\mathrm{PQ}|$ a minimum?

Solution:
Let the angle between the line through $\mathrm{Y}$ and $\mathrm{XY}$ be $\theta$. Take $\mathrm{Y}^{\prime}$ on the line such that $\mathrm{X} \mathrm{Y}^{\prime}=\mathrm{XY}$. If $\mathrm{P}$ is on the opposite side of $\mathrm{Y}$ to $\mathrm{Y}^{\prime}$, then $\mathrm{Q}$ is on the opposite side of $\mathrm{Y}^{\prime}$ to $\mathrm{Y}$. As $\mathrm{P}$ approaches $\mathrm{Y}, \mathrm{Q}$ approaches $\mathrm{Y}^{\prime}$ so the minimum value of $\mathrm{PQ}$ is $\mathrm{Y} \mathrm{Y}^{\prime}$, corresponding to $\mathrm{PY} / \mathrm{QY}=0$. But it is unrealised, since the problem requires $\mathrm{MY}<\mathrm{MP}$. If $\mathrm{P}$ is on the same side of $\mathrm{Y}$ as $\mathrm{Y}^{\prime}$, then as $\mathrm{P}$ approaches the midpoint of $\mathrm{Y} \mathrm{Y}^{\prime}, \mathrm{Q}$

![](https://cdn.mathpix.com/cropped/2024_05_09_ff3994484212c24fa6bfg-24.jpg?height=48&width=1665&top_left_y=2620&top_left_x=181)
the problem requires MY <MP. $\mathrm{P}$ is allowed to be on either side of $\mathrm{Y}$, so the unrealised minimum value of $\mathrm{PQ}$ is $\mathrm{Y}^{\prime} / 2$ as $\mathrm{PY/QY}$ approaches infinity.


$\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}$ is a sequence of $0 \mathrm{~s}$ and $1 \mathrm{~s}$. $\mathrm{T}$ is the number of triples $\left(\mathrm{a}_{i}, \mathrm{a}_{j}, \mathrm{a}_{\mathrm{k}}\right)$ with $\mathrm{i}<\mathrm{j}<\mathrm{k}$ which are not equal to $(0,1,0)$ or $(1,0,1)$. For $1 \leq i \leq n, f(i)$ is the number of $j<i$ with $a_{j}=a_{i}$ plus the number of $j>i$ with $a_{j} \neq a_{i}$. Show that $\mathrm{T}=\mathrm{f}(1)(\mathrm{f}(1)-1) / 2+\mathrm{f}(2)(\mathrm{f}(2)-1) / 2+\ldots+\mathrm{f}(\mathrm{n})(\mathrm{f}(\mathrm{n})-1) / 2$. If $\mathrm{n}$ is odd, what is the smallest value of T?

Solution:
For $n$ odd, the smallest value of $T$ is $n(n-1)(n-3) / 8$ achieved by $01010 \ldots 010$.

Suppose a particular $a_{i}=0$. Let $S_{i}$ be the set of $a_{j}$ with $j<i$ and $a_{j}=a_{i}$ and $a_{j}$ with $j>i$ and $a_{j} \neq a_{i}$. Suppose we take any two members of $S_{i}$ and consider the triple formed with $a_{i}$ itself. Surprisingly perhaps, they are all of the required form ( $\mathrm{a}_{\mathrm{i}}$ is bold $)$ :

$$
\begin{array}{llllll}
\ldots & 0 & \ldots & 0 & \ldots & 0 \\
\ldots & 0 & \ldots & 0 & \ldots & 1 \\
\ldots & 0 & \ldots & 1 & \ldots & 1
\end{array}
$$

Similarly, if $\mathrm{a}_{\mathrm{i}}=1$ :

. . $\quad 1$. . 1 . . 1

. . 1 . . . 1 ... 0

... 1 . . . 0 . . . 0

Conversely, any triple of the required form can be considered (uniquely) to be an $\mathrm{a}_{\mathrm{i}}$ with members of the triple before it equal to it and members of the triple after it unequal to it. Since $\mathrm{f}(\mathrm{m})(\mathrm{f}(\mathrm{m})-1)$ is the number of ways of choosing two items from $f(m)$, we have established that $T=\sum f(m)(f(m)-1)$.

We show that $\sum f(m)=n(n-1) / 2$ (and is hence independent of the details of the particular sequence $a_{i}-$ it depends only on its length). In the case of a sequence of all $0 \mathrm{~s}$, we have $\mathrm{f}(1)=0, \mathrm{f}(2)=1, \ldots, \mathrm{f}(\mathrm{n})=\mathrm{n}-1$, so $\sum$ $f(m)=n(n-1) / 2$. Now suppose we have a particular sequence and we change $a_{i}$ from 0 to 1 . Suppose there are a 0s before $a_{i}$, $b 1 s$ before $a_{i}, c 0 s$ after $a_{i}$ and $d 1 s$ after $a_{i}$. Then the value of $f(i)$ is changed from a $+d$ to $b+$ $c$, an increase of $(b-a)+(c-d)$. The value of $f(j)$ with $j<i$ is decreased by 1 if $f(j)$ is 0 and increased by 1 if $f(j)$ is 0 . So in total there is an increase of $(a-b)$ for such $j$. Similarly, for $j>i$, the value of $f(j)$ is decreased by 1 if $f(j)$ is 0 and increased by 1 if $f(j)$ is 1 , a net increase of $(d-c)$. Thus overall $\sum f(m)$ is increased by (b a) $+(c-d)+(a-b)+(d-c)=0$. But by a sequence of such changes we can get to any sequence $a_{i}$, so all sequences (of length $n$ ) have the same value for $\sum \mathrm{f}(\mathrm{m})$.

Thus $T$ is minimised when $\sum \mathrm{f}(\mathrm{m})^{2}$ is minimised. But since $\sum \mathrm{f}(\mathrm{m})$ is fixed, that is achieved when the $\mathrm{f}(\mathrm{m})$ are as equal as possible. If $n=2 k+1$, then $\sum f(m)=n(n-1) / 2$, so the average value of $f(m)$ is exactly $k$, so we look for an arrangement with all $\mathrm{f}(\mathrm{m})=\mathrm{k}$. It is not hard to see that $01010 \ldots 1010$ works. So this gives $\mathrm{T}=\sum$ $\mathrm{k}(\mathrm{k}-1) / 2=\mathrm{n}(\mathrm{n}-1)(\mathrm{n}-3) / 8$.

The question does not ask for it, but the even case is also straightforward. If $n=2 k$, then the best we can hope for is $k$ values of $k$ and $k$ values of $k-1$, so that $\sum f(m)=k \cdot k+k .(k-1)=k(2 k-1)=n(n-1) / 2$. That is achieved by $0101 \ldots 0101$ (all the 0s have $\mathrm{f}(\mathrm{m})=\mathrm{k}$ and all the $1 \mathrm{~s}$ have $\mathrm{f}(\mathrm{m})=\mathrm{k}-1)$. Hence $\mathrm{T}=\mathrm{k} \cdot \mathrm{k}(\mathrm{k}-1) / 2+(\mathrm{k}-$ 1). $(k-1)(k-2) / 2=(k-1)\left(2 k^{2}-3 k+2\right) / 2=(n-2)\left(n^{2}-3 n+4\right) / 8$.


The repeating decimal $0 . \mathrm{ab} \ldots \mathrm{kpq} \ldots \mathrm{u}=\mathrm{m} / \mathrm{n}$, where $\mathrm{m}$ and $\mathrm{n}$ are relatively prime integers, and there is at least one decimal before the repeating part. Show that $\mathrm{n}$ is divisible by 2 or 5 (or both). [For example, $0.011 \underline{36}=0.01136363636 \ldots=1 / 88$ and 88 is divisible by 2.]

Solution: Note that $\mathrm{k}$ and $\mathrm{u}$ are not equal (otherwise we should have regarded the repeating part as starting at $\mathrm{k}$ ). We have $\mathrm{m} / \mathrm{n}=\mathrm{ab} \ldots \mathrm{k} / 10^{\mathrm{r}} \mathrm{pq} \ldots \mathrm{u} /\left(10^{\mathrm{r}}\left(10^{\mathrm{s}}-1\right)\right)=\left(\mathrm{ab} \ldots \mathrm{k}\left(10^{\mathrm{s}}-1\right)+\mathrm{pq} \ldots \mathrm{u}\right) /\left(10^{\mathrm{r}}\left(10^{\mathrm{s}}-1\right)\right)$. The numerator $=\mathrm{u}-\mathrm{k}$ $\bmod 10$, which is non-zero, so the numerator is not divisible by 10 . But the denominator is divisible by 10 . Hence after reduction to lowest terms the denominator is divisible by 2 or 5 or both.


The cubic $x^{3}+a x^{2}+b x+c$ has real coefficients and three real roots $r \geq s \geq t$. Show that $k=a^{2}-3 b \geq 0$ and that $\sqrt{ } \mathrm{k} \leq \mathrm{r}-\mathrm{t}$.

Solution:
$\mathrm{a}^{2}-3 \mathrm{~b}=(\mathrm{r}+\mathrm{s}+\mathrm{t})^{2}-3(\mathrm{rs}+\mathrm{st}+\mathrm{tr})=\mathrm{r}^{2}+\mathrm{s}^{2}+\mathrm{t}^{2}-(\mathrm{rs}+\mathrm{st}+\mathrm{tr})$. By Cauchy-Schwartz we have $(\mathrm{rs}+\mathrm{st}+\mathrm{tr})^{2} \leq$ $\left(r^{2}+s^{2}+t^{2}\right)^{2}$, so $r^{2}+s^{2}+t^{2} \geq|r s+s t+\operatorname{tr}| \geq(r s+s t+$ tr $)$. Hence $a^{2}-3 b \geq 0$.

$a^{2}-3 b \leq(r-t)^{2}$ is the same as $r^{2}+s^{2}+t^{2}-r s-s t-t r \leq r^{2}-2 r t+t^{s}$ or $s^{2}+r t-r s-s t \leq 0$ or $(r-s)(s-t) \geq 0$, which is true since $r \geq s$ and $s \geq t$. So $a^{2}-3 b \leq(r-t)^{2}$. Taking the non-negative square roots, we get the required result.


Let $X$ be the set $\{1,2, \ldots, 20\}$ and let $\mathrm{P}$ be the set of all 9-element subsets of $\mathrm{X}$. Show that for any map f: $\mathrm{P}$ $\rightarrow X$ we can find a 10 -element subset $Y$ of $X$, such that $f(Y-\{k\}) \neq k$ for any $k$ in $Y$.

Solution: Consider pairs $(S, k)$ with $S$ in $P$ and $k$ in $X$ such that $f(S)=k$. There are evidently $20 \mathrm{C} 9$ such pairs, since we can choose any $S$ and $k$ is then fixed. Now consider the pairs $(\mathrm{Y}, \mathrm{k})$ such that $\mathrm{Y}$ is a 10-element subset of $\mathrm{X}$ containing $k$ and $f(Y-\{k\})=k$. The map $(Y, k)$ to $(Y-\{k\}, k)$ is an injection because if $(Y-\{k\}, k)=\left(Y^{\prime}-\right.$ $\left.\left\{\mathrm{k}^{\prime}\right\}, \mathrm{k}^{\prime}\right)$, then $\mathrm{k}=\mathrm{k}^{\prime}$ and hence $\mathrm{Y}=\mathrm{Y}^{\prime}$. It is not necessarily a bijection because if there are any pairs ( $\mathrm{S}, \mathrm{k}$ ) with $\mathrm{k}$ in $\mathrm{S}$ then they do not correspond to any $(\mathrm{Y}, \mathrm{k})$. But certainly the number of pairs $(\mathrm{Y}, \mathrm{k})$ is at most the number of pairs $(\mathrm{S}, \mathrm{k})$. So there are at most $20 \mathrm{C} 9$ pairs $(\mathrm{Y}, \mathrm{k})$. But there are $20 \mathrm{C} 10$ subsets $\mathrm{Y}$ with 10 elements, so at least 20C10 - 20C9 of them (more than 16000) do not belong to any pairs ( $\mathrm{Y}, \mathrm{k}$ ), in other words they are such that $f(Y-\{k\})$ is not $k$ for any $k$ in $Y$.


$A B C$ is a triangle with incenter I. Show that the circumcenters of IAB, IBC, ICA lie on a circle whose center is the circumcenter of $\mathrm{ABC}$.

Solution:
In fact they lie on the circumcircle of $\mathrm{ABC}$.

Extend AI to meet the circumcircle again at $\mathrm{A}^{\prime}$. We show that $\mathrm{A}^{\prime}$ is the circumcenter of $\mathrm{BCI}$. Angle $\mathrm{A}^{\prime} \mathrm{AC}=$ angle $\mathrm{A}^{\prime} \mathrm{AB}$, so $\mathrm{A}^{\prime}$ is the midpoint of the arc $\mathrm{BC}$, so $\mathrm{A}^{\prime} \mathrm{B}=\mathrm{A}^{\prime} \mathrm{C}$. Also $\angle \mathrm{A}^{\prime} \mathrm{CB}=\angle \mathrm{A}^{\prime} \mathrm{AB}=\mathrm{A} / 2$, so $\angle \mathrm{A}^{\prime} \mathrm{CI}=$ $\mathrm{A} / 2+\mathrm{B} / 2$. But $\angle \mathrm{A}^{\prime} \mathrm{IC}=\angle \mathrm{IAC}+\angle \mathrm{ICA}=\mathrm{A} / 2+\mathrm{B} / 2$, so $\mathrm{A}^{\prime} \mathrm{CI}$ is isosceles, so $\mathrm{A}^{\prime} \mathrm{C}=\mathrm{A}^{\prime} \mathrm{I}$.


Let $p(x)$ be the polynomial $(1-x)^{a}\left(1-x^{2}\right)^{b}\left(1-x^{3}\right)^{c} \ldots\left(1-x^{32}\right)^{k}$, where $a, b, \ldots, k$ are integers. When expanded in powers of $x$, the coefficient of $x^{1}$ is -2 and the coefficients of $x^{2}, x^{3}, \ldots, x^{32}$ are all zero. Find $\mathrm{k}$.

Solution:
Answer: $2^{27}-2^{11}$

We have $p(x)=1-2 x+O\left(x^{33}\right)$. Hence $p(-x)=1+2 x+O\left(x^{33}\right)$. Multiplying $p(x) p(-x)=1-2^{2} x^{2}+O\left(x^{33}\right)$. Now $p(x) p(-x)$ cannot have any odd terms, so we can write it as a polynomial in $x^{2}, q\left(x^{2}\right)$. Hence $q\left(x^{2}\right)=1-$ $2^{2} x^{2}+O\left(x^{34}\right)$. Similarly, $r\left(x^{4}\right)=q\left(x^{2}\right) q\left(-x^{2}\right)=1-2^{4} x^{4}+O\left(x^{36}\right), s\left(x^{8}\right)=r\left(x^{4}\right) r\left(-x^{4}\right)=1-2^{8} x^{8}+O\left(x^{40}\right)$, and $t\left(x^{16}\right)=1-2^{16} x^{16}+O\left(x^{48}\right)$.

Now go back to the definition of $p(x)$. When we take $p(x) p(-x)$, the term $(1-x)^{a}$ becomes $\left(1-x^{2}\right)^{a}$. All the even terms just double their exponent, so $\left(1-x^{2}\right)^{b}$ becomes $\left(1-x^{2}\right)^{2 b},\left(1-x^{4}\right)^{d}$ becomes $\left(1-x^{4}\right)^{2 d}$ and so on. The odd terms all keep the same exponent, so $\left(1-\mathrm{x}^{3}\right)^{\mathrm{c}}$ becomes $\left(1-\mathrm{x}^{6}\right)^{\mathrm{c}}$ and so on. Thus we get $\mathrm{t}\left(\mathrm{x}^{16}\right)=(1-$ $\left.x^{16}\right)^{n}\left(1-x^{32}\right)^{16 k} \ldots$. The first exponent is a sum of several exponents from $p(x)$, but the details are unimportant. We know that $\mathrm{t}\left(\mathrm{x}^{16}\right)=1-2^{16} \mathrm{x}^{16}+\mathrm{O}\left(\mathrm{x}^{48}\right)$. The $\mathrm{x}^{16}$ term can only come from $\left(1-\mathrm{x}^{16}\right)^{\mathrm{n}}$, so $\mathrm{n}=2^{16}$. Now there is no $\mathrm{x}^{32}$ term, so putting $\mathrm{N}=2^{16}$ we have $\mathrm{NC} 2=16 \mathrm{k}$, were $\mathrm{NC} 2$ is the binomial coefficient $\mathrm{N}(\mathrm{N}-$ $1) / 2=2^{31}-2^{15}$. Hence $\mathrm{k}=2^{27}-2^{11}$.


A license plate has six digits from 0 to 9 and may have leading zeros. If two plates must always differ in at least two places, what is the largest number of plates that is possible?

Solution:
Answer: $10^{2}$.

We show by induction that we can find a set of $10^{\mathrm{n}-1}$ plates for $\mathrm{n}>1$ digits. It is true for $\mathrm{n}=2$ : take the plates $00,11,22, \ldots, 99$. Suppose it is true for $\mathrm{n}$. If $\mathrm{d}$ is a digit from 0 to 9 and $\mathrm{s}$ is a plate of $\mathrm{n}$ digits, let $[\mathrm{d}, \mathrm{s}]$ be the plate of $\mathrm{n}+1$ digits which has a as its first digit, and the remaining digits the same as those of $\mathrm{s}$, except that the last digit is that for $\mathrm{s}$ plus $\mathrm{d}$ (reduced mod 10 if necessary). Let $\mathrm{S}$ be a set of plates for $\mathrm{n}$ digits. We claim that the set $\mathrm{S}^{\prime}=\{[\mathrm{d}, \mathrm{s}]: \mathrm{d}=0,1, \ldots$ or 9 and $\mathrm{s}$ belongs to $\mathrm{S}\}$ is a set of plates for $\mathrm{n}+1$ digits. It obviously has 10 times as many members as $\mathrm{S}$, so this claim is sufficient to establish the induction.

We have to show that $[\mathrm{a}, \mathrm{s}]$ and $[\mathrm{b}, \mathrm{t}]$ differ in at least two places. If $\mathrm{a}=\mathrm{b}$, then $\mathrm{s} \neq \mathrm{t}$, so $\mathrm{s}$ and $\mathrm{t}$ differ in at least two places. The same change is made to their last digits, so $[\mathrm{a}, \mathrm{s}]$ and $[\mathrm{a}, \mathrm{t}]$ differ in at least two places. If $\mathrm{a} \neq \mathrm{b}$ and $\mathrm{s}=\mathrm{t}$, then $[\mathrm{a}, \mathrm{s}]$ and $[\mathrm{b}, \mathrm{s}]$ differ in both their first and last places. If $\mathrm{a} \neq \mathrm{b}$ and $\mathrm{s} \neq \mathrm{t}$, then $\mathrm{s}$ and $\mathrm{t}$ differ in at least two places and so the modified $\mathrm{s}$ and $\mathrm{t}$, differ in at least one place. But $[\mathrm{a}, \mathrm{s}]$ and $[\mathrm{b}, \mathrm{t}]$ also differ in the first place, so they differ in at least two places.

So we have established that the largest number is at least $10^{n-1}$ for $\mathrm{n}$ digits.

But any two plates which differ only in the last digit cannot both be chosen. So at most $1 / 10$ of the $10^{\mathrm{n}}$ possible plates can be chosen. That shows that $10^{\mathrm{n}-1}$ is best possible.


Define $f_{1}(x)=\sqrt{ }\left(x^{2}+48\right)$ and $f_{n}(x)=\sqrt{ }\left(x^{2}+6 f_{n-1}(x)\right)$. Find all real solutions to $f_{n}(x)=2 x$.

Solution:
Answer: For each $\mathrm{n}, \mathrm{x}=4$ is the only solution.

Obviously $\mathrm{x}=4$ is a solution. Since $\mathrm{f}_{\mathrm{n}}(\mathrm{x})>=0$, any solution must be non-negative. So we restrict attention to $\mathrm{x} \geq 0$.

Suppose $x<4$. We show by induction that $f_{n}(x)>2 x$. For $n=1$, the claim is equivalent to $4 x^{2}<x^{2}+48$, or $x^{2}<16$, which is true. So suppose the result is true for $n$. Then $x^{2}+6 f_{n}(x)>x^{2}+12 x$. But $x<4$, so $3 x^{2}<$ $12 \mathrm{x}$, so $4 \mathrm{x}^{2}<\mathrm{x}^{2}+12 \mathrm{x}$. Hence $\mathrm{f}_{\mathrm{n}+1}(\mathrm{x})>2 \mathrm{x}$, as required.

An exactly similar argument shows that $\mathrm{f}_{\mathrm{n}}(\mathrm{x})<2 \mathrm{x}$ for $\mathrm{x}>4$. Hence $\mathrm{x}=4$ is the only solution.


Show that for any odd positive integer we can always divide the set $\{\mathrm{n}, \mathrm{n}+1, \mathrm{n}+2, \ldots, \mathrm{n}+32\}$ into two parts, one with 14 numbers and one with 19 , so that the numbers in each part can be arranged in a circle, with each
number relatively prime to its two neighbours. For example, for $\mathrm{n}=1$, arranging the numbers as $1,2,3, \ldots$, 14 and $15,16,17, \ldots, 33$, does not work, because 15 and 33 are not relatively prime.

Solution:
Suppose we use $\mathrm{n}, \mathrm{n}+1, \ldots, \mathrm{n}+13$ for the first circle. That certainly works for $\mathrm{n}$ not divisible by 13 , since consecutive numbers are always relatively prime and any common divisor of $n$ and $n+13$ must also divide their difference 13. We could then take the second circle to be $n+15, n+14, n+16, n+17, \ldots, n+32$ for $n \neq 2$ $\bmod 17$ or $n+14, n+15, \ldots, n+29, n+30, n+32, n+31$ if $n=2 \bmod 17$. Note that any common factor of $n+14$ and $n+16$ must divide their difference 2 , but $n$ is odd, so they are relatively prime. Similarly, $n+30$ and $n+32$.

If $\mathrm{n}$ is divisible by 13 , then $\mathrm{n}+19$ is not, so we can take $n+19, n+20, \ldots, n+32$ for the first circle. Then we can take the second circle to be $n+1, n, n+2, n+3, \ldots, n+18$ for $n \neq 16 \bmod 17$ or $n, n+1, \ldots, n+15, n+16, n+18$, $\mathrm{n}+17$ if $\mathrm{n}=16 \bmod 17$.


How many positive integers can be written in base $\mathrm{n}$ so that (1) the integer has no two digits the same, and (2) each digit after the first differs by one from an earlier digit? For example, in base 3 , the possible numbers are $1,2,10,12,21,102,120,210$.

Solution:
Answer: $2^{n+1}-2 n-2$.

We use a more elaborate induction hypothesis. We claim that for base $n+1$, the following numbers of integers satisfy the two conditions: $2^{n+1}-2 n-2$ not using the digit $n ; 2^{n}-1$ with the digit $n$ is the last position; $2^{\mathrm{n}-1}$ with the digit $\mathrm{n}$ in the last but 1 position; $2^{\mathrm{n}-2}$ with the digit $\mathrm{n}$ having 2 following digits; $2^{\mathrm{n}-3}$ with the digit $\mathrm{n}$ having 3 following digits; $\ldots ; 1$ with the digit $\mathrm{n}$ having $\mathrm{n}$ following digits.

Thus for $\mathrm{n}=2$, the claim is that there are 2 numbers only involving 0 and $1(1,10), 3$ numbers with 2 as the last digit $(2,12,102), 2$ numbers with one digit after $2(21,120)$ and 1 number with two digits after $2(210)$. Suppose this holds for base $n+1$.

If we add up the various possibilities we get $2^{n+1}-2 n-2+\left(2^{n}-1+2^{n-1}+2^{n-2}+\ldots+1\right)=2^{n+1}-2 n-2+2^{n+1}-$ $2=2^{n+2}-2(n+1)-2$. The number of base $n+2$ integers not involving the digit $n+1$ is the same as the number of base $n+1$ numbers, which is (by induction and the addition above) $2^{n+2}-2(n+1)-2$. If a base $n+2$ number has the digit $\mathrm{n}+1$ in the last place, then either that is the only digit in the number or the earlier digits must form a base $n+1$ number with the digit $n$ in it. There are $\left(2^{n}-1+2^{n-1}+\ldots+1\right)=2^{n+1}-2$ such numbers, so in total we have $2^{n+1}-1$ base $n+2$ numbers with $n+1$ in the last place. If a base $n+2$ number has the digit $n+1$ in the penultimate place, then either the number has $n+1$ as the first digit, in which case it must be $n+1 n$, or the other digits form a base $\mathrm{n}+1$ number with the digit $\mathrm{n}$ in the penultimate place or earlier. There are $2^{\mathrm{n}-1}+\ldots+$ $1=2^{\mathrm{n}}-1$ such numbers. So $2^{\mathrm{n}}$ in total. Similarly for the other possibilities.

Finally, we need to check the answer for $n=2\left(1,10\right.$, so two numbers and $\left.2^{2+1}-2 \cdot 2-2=8-4-2=2\right)$.


$\mathrm{ABC}$ is acute-angled. The circle diameter $\mathrm{AB}$ meets the altitude from $\mathrm{C}$ at $\mathrm{P}$ and $\mathrm{Q}$. The circle diameter $\mathrm{AC}$ meets the altitude from B at R and S. Show that P, Q, R and S lie on a circle.

Solution: Use vectors. Take $\mathrm{A}$ as the origin. Let $\mathrm{AB}=\mathbf{b}, \mathrm{AC}=\mathbf{c}, \mathrm{AR}=\mathbf{r}$. $\mathrm{AR}$ is perpendicular to $\mathrm{RC}$, so $\mathbf{r} .(\mathbf{c}-\mathbf{r})=0$. $\mathrm{BR}$ is perpendicular to $\mathrm{AC}$, so $(\mathbf{b}-\mathbf{r}) . \mathbf{c}=0$. Hence $\mathbf{r . r}=\mathbf{r . c}=$ b.c. Thus $|\mathrm{AR}|=\sqrt{ }(|\mathrm{AB}| \cdot|\mathrm{AC}| \cos \mathrm{A})$. But the identical argument gives the same value for $|\mathrm{AS}|$. The situtation is symmetrical between $\mathrm{B}$ and $\mathrm{C}$, so we get the same result for $|\mathrm{AP}|$ and $|\mathrm{AQ}|$. Hence all four points lie on a circle center $\mathrm{A}$.


An obtuse angled triangle has integral sides and one acute angle is twice the other. Find the smallest possible perimeter.

Solution:
Answer: $77=16+28+33$.

Let the triangle be $\mathrm{ABC}$ with angle $\mathrm{A}$ obtuse and angle $\mathrm{B}=2$ angle $\mathrm{C}$. Let the sides be $\mathrm{a}, \mathrm{b}, \mathrm{c}$ as usual. Note that $\mathrm{a}>\mathrm{b}>\mathrm{c}$. We have $\mathrm{b} \sin \mathrm{C}=\mathrm{c} \sin 2 \mathrm{C}$ and $\mathrm{c}^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}-2 \mathrm{ab} \sin \mathrm{C}$. Hence, $\mathrm{b} / 2 \mathrm{c}=\sin 2 \mathrm{C} / \sin \mathrm{C}=\cos \mathrm{C}=$ $\left(a^{2}+b^{2}-c^{2}\right) / 2 a b$. So $a b^{2}=a^{2} c+b^{2} c-c^{3}$. Hence $b^{2}(a-c)=c\left(a^{2}-c^{2}\right)$. Dividing by $a-c$ we get $b^{2}=c(a+c)$.

Now the triangle with smallest perimeter will have $\mathrm{a}, \mathrm{b}, \mathrm{c}$ relatively prime (otherwise we could divide by the common factor). Hence $\mathrm{c}$ must be a square. For if $\mathrm{c}$ and $\mathrm{a}+\mathrm{c}$ have a common factor, then so do a and $\mathrm{c}$ and hence $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$, which means they cannot be the minimal set. Clearly $\mathrm{c}$ is not 1 (or the triangle would have nil area). $\mathrm{c}=4$ gives $\mathrm{a}=5$, which is too short, or $\mathrm{a} \geq 21$, which is too long. $\mathrm{c}=9$ gives $\mathrm{a}=7$ (too short), 18 (3 divides $\mathrm{a}, \mathrm{b}, \mathrm{c}$ ), or $\geq 40$ (too long). $\mathrm{c}=16$ gives $\mathrm{a}=20$ (too short) or $\mathrm{a}=33$ which works. Larger $\mathrm{c}$ gives $\mathrm{a}$ larger perimeter. $\operatorname{Eg} \mathrm{c}=25$ gives $\mathrm{a}=56, \mathrm{~b}=45$ (perimeter 126).


For each non-empty subset of $\{1,2, \ldots, n\}$ take the sum of the elements divided by the product. Show that the sum of the resulting quantities is $n^{2}+2 n-(n+1) s_{n}$, where $s_{n}=1+1 / 2+1 / 3+\ldots+1 / n$.

Solution:
This is a straightforward induction. For $\mathrm{n}=1$ the only term in the sum is $1 / 1$ with sum 1 . The formula gives $1^{2}+2.1-2.1=1$. So it is true for $n=1$.

Note first that for $n$ the sum of the inverses of the products, including 1 for the empty set, is $(1+1 / 1)(1+$ $1 / 2)(1+1 / 3) \ldots(1+1 / n)=n+1$ (it telescopes). Now the sum for $n+1$ is the sum for $n$ plus the sum of terms involving $n+1$. But if a term involves $n+1$, then the sum is increased by $n+1$ and the product is increased by a factor $n+1$. So the sum of all the terms involving $n+1$ is $(1 / n+1 \times$ sum for $n)+$ (sum of inverse products for $n$ ) $=\left(n^{2}+2 n-(n+1) s_{n}\right) /(n+1)+(n+1)=n+1-1 /(n+1)-s_{n}+(n+1)=2(n+1)-s_{n+1}$. Hence the sum for $n+1$ is $\left(n^{2}+2 n\right)-(n+1) s_{n}+2(n+1)-s_{n+1}=(n+1)^{2}+2(n+1)-1-(n+1) s_{n}-s_{n+1}=(n+1)^{2}+2(n+1)-(n+1) /(n+1)-$ $(n+1) s_{n}-s_{n+1}=(n+1)^{2}+2(n+1)-(n+2) s_{n+1}$.


Define the function $\mathrm{f}$ on the natural numbers by $\mathrm{f}(1)=2, \mathrm{f}(\mathrm{n})=2^{\mathrm{f}(\mathrm{n}-1)}$. Show that $\mathrm{f}(\mathrm{n})$ has the same residue mod $m$ for all sufficiently large $n$.

Solution:
This is the so-called tower of exponents, but HTML is not up showing it! The trick is to use induction on $\mathrm{m}$ (which is not at all obvious). The result is trivial for $\mathrm{m}=1$.

If $\mathrm{m}$ is even, then we can write $\mathrm{m}=2^{\mathrm{a}} \mathrm{b}$, where $\mathrm{b}$ is odd. Then $\mathrm{f}(\mathrm{n})$ is eventually constant mod $\mathrm{b}$. Obviously $\mathrm{f}(\mathrm{n})$ is eventually $0 \bmod 2^{\mathrm{a}}$, so the residue mod $\mathrm{m}$ is eventually constant.

If $\mathrm{m}$ is odd, then $2^{\varphi(\mathrm{m})}=1 \bmod \mathrm{m}$, where $\varphi(\mathrm{m})<\mathrm{m}$. So by induction $\mathrm{f}(\mathrm{n})$ is eventually constant $\bmod \varphi(\mathrm{m})$. Hence $\mathrm{f}(\mathrm{n}+1)=2^{\mathrm{f}(\mathrm{n})}$ is eventually constant $\bmod m$.


$\mathrm{a}$ and $\mathrm{b}$ are positive integers and $\mathrm{c}=\left(\mathrm{a}^{\mathrm{a}+1}+\mathrm{b}^{\mathrm{b}+1}\right) /\left(\mathrm{a}^{\mathrm{a}}+\mathrm{b}^{\mathrm{b}}\right)$. By considering $\left(\mathrm{x}^{\mathrm{n}}-\mathrm{n}^{\mathrm{n}}\right) /(\mathrm{x}-\mathrm{n})$ or otherwise, show that $c^{a}+c^{b} \geq a^{a}+b^{b}$.

Solution:
$\left(x^{n}-n^{n}\right) /(x-n)=x^{n-1}+x^{n-2} n+\ldots+n^{n-1}$. If $x>n$, then each term is $>n^{n-1}$, so $\left(x^{n}-n^{n}\right) /(x-n)>n^{n}$. If $x<n$, then each term is $<=n^{n}$ with equality only for the last term, so $\left(x^{n}-n^{n}\right) /(x-n)<n^{n}$. So multiplying by $x-n$, which is positive for $x>n$ and negative for $x<n$, we get $\left(x^{n}-n^{n}\right)>(x-n) n^{n}$ except for $x=n$, and hence $\left(x^{n}-n^{n}\right) \geq$ $(\mathrm{x}-\mathrm{n}) \mathrm{n}^{\mathrm{n}}$ for all $\mathrm{x} \geq 0$.

Putting $x=c, n=a$, we get $\left(c^{a}-a^{a}\right) \geq(c-a) a^{a}$. Similarly, putting $x=c, n=b$, we get $\left(c^{b}-b^{b}\right) \geq(c-b) b^{b}$. Adding $\left(c^{a}+c^{b}\right)-\left(a^{a}+b^{b}\right) \geq(c-a) a^{a}+(c-b) b^{b}=c\left(a^{a}+b^{b}\right)-a^{a+1}-b^{b+1}=0$, which is the required result.


$\mathrm{X}$ is a point on the side $\mathrm{BC}$ of the triangle $\mathrm{ABC}$. Take the other common tangent (apart from $\mathrm{BC}$ ) to the incircles of ABX and ACX which intersects the segments AB and AC. Let it meet AX at Y. Show that the locus of $\mathrm{Y}$, as $\mathrm{X}$ varies, is the arc of a circle.

Solution:
We show that $\mathrm{AY}=(\mathrm{AB}+\mathrm{AC}-\mathrm{BC}) / 2=$ constant.

Let the common tangent meet the incircles of $\mathrm{ABX}, \mathrm{ACX}$ at $\mathrm{R}, \mathrm{S}$ respectively. Let $\mathrm{AX}$ meet them at $\mathrm{P}, \mathrm{Q}$ respectively and let $\mathrm{BC}$ meet them at $\mathrm{U}, \mathrm{V}$ respectively. Let $\mathrm{AB}$ meet the incircle of $\mathrm{ABX}$ at $\mathrm{K}$ and let $\mathrm{AC}$ meet the incircle of $\mathrm{ACX}$ at $\mathrm{L}$. We have $\mathrm{AY}=\mathrm{AQ}-\mathrm{QY}$ and $\mathrm{AY}=\mathrm{AP}-\mathrm{PY}$. So adding $2 \mathrm{AY}=\mathrm{AP}+\mathrm{AQ}-$ $(\mathrm{YQ}+\mathrm{YP})=\mathrm{AP}+\mathrm{AQ}-(\mathrm{YS}+\mathrm{YR})=\mathrm{AP}+\mathrm{AQ}-\mathrm{RS}$. If we reflect about the line of centers of the two incircles $R$ goes to $U$ and $S$ to $V$. Hence $R S=U V$. So $2 A Y=A P+A Q-U V$. We have $A P=A K=A B-B K$ $=A B-B U$ and $A Q=A L=A C-C L=A C-C V$. Hence $2 A Y=A B+A C-B U-U V-C V=A B+A C-B C$.


Let $a_{n}$ be the number written with $2^{n}$ nines. For example, $a_{0}=9, a_{1}=99, a_{2}=9999$. Let $b_{n}=\Pi_{0}{ }^{n} a_{i}$. Find the sum of the digits of $b_{n}$.

Solution:
Answer: $9 \cdot 2^{\text {n }}$.

Induction on $\mathrm{n}$. We have $\mathrm{b}_{0}=9$, digit sum 9 , and $\mathrm{b}_{1}=891$, digit sum 18 , so the result is true for $\mathrm{n}=0$ and 1 . Assume it is true for n-1. Obviously $\mathrm{a}_{\mathrm{n}}<10$ to the power of $2^{\mathrm{n}}$, so $\mathrm{b}_{\mathrm{n}-1}<10$ to the power of $\left(1+2+2^{2}+\ldots+\right.$ $\left.2^{n-1}\right)<10$ to the power of $2^{n}$. Now $b_{n}=b_{n-1} 10^{N}-b_{n-1}$, where $N=2^{n}$. But $b_{n-1}<10^{N}$, so $b_{n}=\left(b_{n-1}-1\right) 10^{N}+$ $\left(10^{N}-b_{n-1}\right)$ and the digit sum of $b_{n}$ is just the digit sum of $\left(b_{n-1}-1\right) 10^{N}$ plus the digit sum of $\left(10^{N}-b_{n-1}\right)$.

Now $b_{n-1}$ is odd and so its last digit is non-zero, so the digit sum of $b_{n-1}-1$ is one less than the digit sum of $b_{n-}$ 1, and hence is $9 \cdot 2^{n-1}-1$. Multiplying by $10^{N}$ does not change the digit sum. $\left(10^{N}-1\right)-b_{n-1}$ has $2^{n}$ digits, each 9 minus the corresponding digit of $b_{n-1}$, so its digit sum is $9 \cdot 2^{n}-9 \cdot 2^{n-1} \cdot b_{n-1}$ is odd, so its last digit is not 0 and hence the last digit of $\left(10^{N}-1\right)-b_{n-1}$ is not 9 . So the digit sum of $10^{N}-b_{n-1}$ is $9 \cdot 2^{n}-9 \cdot 2^{n-1}+1$. Hence $b_{n}$ has digit sum $\left(9 \cdot 2^{n-1}-1\right)+\left(9 \cdot 2^{n}-9 \cdot 2^{n-1}+1\right)=9 \cdot 2^{n}$.


Let $\mathrm{k}=1^{0}$. Show that $\Sigma_{0}^{88} 1 /(\cos \mathrm{nk} \cos (\mathrm{n}+1) \mathrm{k})=\cos \mathrm{k} / \sin ^{2} \mathrm{k}$.

Solution: $\tan (n+1) k-\tan n k=(\sin (n+1) k \cos n k-\sin n k \cos (n+1) k) /(\cos n k \cos (n+1) k)=\sin k /(\cos n k \cos (n+1) k$ ). Using this expression the sum telescopes and we get $\sum_{0}^{88} 1 /(\cos n k \cos (\mathrm{n}+1) \mathrm{k})=(\tan 89 \mathrm{k}-\tan 0) / \sin \mathrm{k}$. But $\tan 0=0$ and $\tan 89 \mathrm{k}=\cot (\pi / 2-89 \mathrm{k})=\cot \mathrm{k}$.


A set of 11 distinct positive integers has the property that we can find a subset with sum $n$ for any $n$ between 1 and 1500 inclusive. What is the smallest possible value for the second largest element?

Solution:
Answer: 248.

By taking the integers to be $1,2,4,8, \ldots, 1024$ we can generate all integers up to 2047 . But by taking some integers smaller, we can do better. For example, 1, 2, 4, .., 128, 247, 248, 750 gives all integers up to 1500 . We can obviously use the integers $1,2,4, \ldots, 128$ to generate all integers up to 255 . Adding 248 gives all integers from 256 up to 503 . Then adding 247 gives all integers from 504 to 750 . So adding 750 gives all integers up to 1500 . We show that we cannot do better than this.

Let the integers be $a_{1}<a_{2}<\ldots<a_{11}$. Put $s_{n}=a_{1}+\ldots+a_{n}$. Take $N$ such that $s_{N-1}<1500 \leq s_{N}$. Note that 1500 must be a sum of some of the integers so certainly $1500 \leq s_{11}$. Equally we obviously have $\mathrm{a}_{1}=1, \mathrm{a}_{2}=2$, so $\mathrm{N}$ is well-defined and we have $1<\mathrm{N} \leq 11$. Now if $1<\mathrm{k} \leq \mathrm{N}$, we have $\mathrm{s}_{\mathrm{k}-1}<1500$ and hence $\mathrm{s}_{\mathrm{k}-1}+1 \leq 1500$. So some sum of distinct $\mathrm{a}_{\mathrm{i}}$ must equal $\mathrm{s}_{\mathrm{k}-1}+1$ and it must involve an $\mathrm{a}_{\mathrm{i}}$ with $\mathrm{i}>\mathrm{k}-1$ since $\mathrm{s}_{\mathrm{k}-1}<\mathrm{s}_{\mathrm{k}-1}+1$. Hence $\mathrm{a}_{\mathrm{k}} \leq \mathrm{s}_{\mathrm{k}-1}+1$ for $1<\mathrm{k} \leq \mathrm{N}$.

Now an easy induction gives $\mathrm{s}_{\mathrm{k}} \leq 2_{\mathrm{k}}-1$. We must have $\mathrm{a}_{1}=1$ and hence $\mathrm{s}_{1}=1$, so it is true for $\mathrm{k}=1$.

Suppose it is true for $\mathrm{k}-1$. Then $\mathrm{a}_{\mathrm{k}} \leq \mathrm{s}_{\mathrm{k}-1}+1 \leq 2^{\mathrm{k}-1}$, so $\mathrm{s}_{\mathrm{k}}=\mathrm{s}_{\mathrm{k}-1}+\mathrm{a}_{\mathrm{k}} \leq 2^{\mathrm{k}-1}-1+2^{\mathrm{k}-1}=2^{\mathrm{k}}-1$, so it is true for $\mathrm{k}$. Hence for all $\mathrm{k} \leq \mathrm{N}$. But $2^{10}-1=1023<1500$, so if $\mathrm{N}<11$, then $\mathrm{s}_{\mathrm{N}}<1500$. Contradiction. Hence $\mathrm{N}=11$.

We have $\mathrm{s}_{\mathrm{k}}=\mathrm{s}_{\mathrm{k}-1}+\mathrm{a}_{\mathrm{k}} \leq 2 \mathrm{~s}_{\mathrm{k}-1}+1$. Hence $\mathrm{s}_{\mathrm{k}-1} \geq\left(\mathrm{s}_{\mathrm{k}}-1\right) / 2$. So $\mathrm{s}_{11} \geq 1500$ implies $\mathrm{s}_{10} \geq 750$. But $\mathrm{s}_{8} \leq 255$, so $\mathrm{a}_{9}+$ $\mathrm{a}_{10} \geq 495$, so $\mathrm{a}_{10} \geq 248$.


Three chords of a sphere are meet at a point $\mathrm{X}$ inside the sphere but are not coplanar. A sphere through an endpoint of each chord and $\mathrm{X}$ touches the sphere through the other endpoints and $\mathrm{X}$. Show that the chords have equal length.

Solution:
Let two of the chords be $\mathrm{AB}$ and $\mathrm{CD}$. Take the plane containing them. In this plane we have a circle through A, B, C, D, and a circle through A, C, X which touches a circle through B, D, X at X. We show that AX = CX. Let the common tangent at X meet the larger circle at $\mathrm{E}$ and $\mathrm{F}$. [Assume the points are in the order $\mathrm{A}, \mathrm{C}$, $\mathrm{F}, \mathrm{B}, \mathrm{D}, \mathrm{E}$ as we go round the circle.] We have $\angle \mathrm{XAC}=\angle \mathrm{BAC}$ (same angle $)=\angle \mathrm{BDC}($ circle $\mathrm{ABCD})=$ $\angle \mathrm{BXF}(\mathrm{XF}$ tangent to $\mathrm{BXD})=\angle \mathrm{EXA}$ (opposite angle) $=\angle \mathrm{XCA}$ (EX tangent to AXC). So XAC is isosceles. So $\mathrm{XA}=\mathrm{XC}$. Similarly $\mathrm{XB}=\mathrm{XD}$. Hence $\mathrm{AB}=\mathrm{CD}$.

Similarly, the other pairs of chords are equal.


A complex polynomial has degree 1992 and distinct zeros. Show that we can find complex numbers $\mathrm{z}_{\mathrm{n}}$, such that if $\mathrm{p}_{1}(\mathrm{z})=\mathrm{z}-\mathrm{z}_{1}$ and $\mathrm{p}_{\mathrm{n}}(\mathrm{z})=\mathrm{p}_{\mathrm{n}-1}(\mathrm{z})^{2}-\mathrm{z}_{\mathrm{n}}$, then the polynomial divides $\mathrm{p}_{1992}(\mathrm{z})$.

Solution:
Let the polynomial of degree 1992 be $\mathrm{q}(\mathrm{z})$. Suppose its roots are $\mathrm{w}_{1}, \mathrm{w}_{2}, \ldots, \mathrm{w}_{1992}$. Let $\mathrm{S}_{1}=\left\{\mathrm{w}_{1}, \ldots, \mathrm{w}_{1992}\right\}$.

We now define $S_{2}$ as follows. Let $z_{1}=\left(w_{1}+w_{2}\right) / 2$ and take $S_{2}$ to be the set of all numbers $\left(w-z_{1}\right)^{2}$ with $w$ in $\mathrm{S}_{1}$. Note that $\mathrm{w}=\mathrm{w}_{1}$ and $\mathrm{w}=\mathrm{w}_{2}$ give the same number. It is possible that other pairs may also give the same number. But certainly $\left|\mathrm{S}_{2}\right|<=\left|\mathrm{S}_{1}\right|-1$. We now repeat this process until we get a set with only one member. Thus if $\mathrm{S}_{\mathrm{i}}$ has more than one member, then we take we take $\mathrm{z}_{\mathrm{i}}$ to be an average of any two distinct members. Then we take $\mathrm{S}_{\mathrm{i}+1}$ to be the set of all $\left(\mathrm{w}-\mathrm{z}_{\mathrm{i}}\right)^{2}$ with $\mathrm{w}$ in $\mathrm{S}_{\mathrm{i}}$. So after picking at most 1991 elements $\mathrm{z}_{\mathrm{i}}$ we have a set $\mathrm{S}_{\mathrm{N}}$ with only one member. Take $\mathrm{Z}_{\mathrm{N}}$ to be that one member, so that $\mathrm{S}_{\mathrm{N}+1}=\{0\}$. Now if $\mathrm{N}<1991$, take the remaining $\mathrm{z}_{\mathrm{i}}$ to be 0 until we reach $\mathrm{z}_{1992}$.

Now as we allow $z$ to take the values in $S$ :

$\mathrm{p}_{1}(\mathrm{z})=\left(\mathrm{z}-\mathrm{z}_{1}\right)$ takes 1992 possible values;

$\mathrm{p}_{2}(\mathrm{z})=\left(\mathrm{z}-\mathrm{z}_{1}\right)^{2}-\mathrm{z}_{2}$ takes at most 1991 possible values;

$\mathrm{p}_{3}(\mathrm{z})=\left(\left(\mathrm{z}-\mathrm{z}_{1}{ }^{2}-\mathrm{z}_{2}\right)^{2}-\mathrm{z}_{3}\right.$ takes at most 1990 possible values;

$\mathrm{p}_{4}(\mathrm{z})=\left(\left(\left(\mathrm{z}-\mathrm{z}_{1}\right)^{2}-\mathrm{z}_{2}\right)^{2}-\mathrm{z}_{3}\right)^{2}-\mathrm{z}_{4}$ takes at most 1989 possible values;

...

$\mathrm{p}_{1992}(\mathrm{z})$ takes only the value 0 .

So every root of $\mathrm{q}(\mathrm{z})$ is also a root of $\mathrm{p}_{1992}(\mathrm{z})$. Hence $\mathrm{q}(\mathrm{z})$ must divide $\mathrm{p}_{1992}(\mathrm{z})$.


$\mathrm{n}>1$, and $\mathrm{a}$ and $\mathrm{b}$ are positive real numbers such that $\mathrm{a}^{\mathrm{n}}-\mathrm{a}-1=0$ and $\mathrm{b}^{2 \mathrm{n}}-\mathrm{b}-3 \mathrm{a}=0$. Which is larger?

Solution:
Answer: $\mathrm{a}>\mathrm{b}$.

Note that $\mathrm{a}^{\mathrm{n}}=\mathrm{a}+1>1$ (since $\mathrm{a}$ is positive). Hence $\mathrm{a}>1$. So $\mathrm{a}^{2 \mathrm{n}}=(\mathrm{a}+1)^{2}=\mathrm{a}^{2}+2 \mathrm{a}+1$. Put $\mathrm{a}=1+k$, then $a^{2}=1+2 k+k^{2}>1+2 k$, so $a^{2}+1>2+2 k=2 a$. Hence $a^{2 n}>4 a$. So $(b / a)^{2 n}=(b+3 a) / a^{2 n}<(b+3 a) / 4 a$. If $\mathrm{b} / \mathrm{a} \geq 1$, then $(\mathrm{b}+3 \mathrm{a}) / 4 \mathrm{a} \leq(\mathrm{b}+3 \mathrm{~b}) / 4 \mathrm{a}=\mathrm{b} / \mathrm{a}$, so $(\mathrm{b} / \mathrm{a})^{2 \mathrm{n}}<\mathrm{b} / \mathrm{a}$. Contradiction. Hence $\mathrm{b} / \mathrm{a}<1$.


The diagonals of a convex quadrilateral meet at right angles at $X$. Show that the four points obtained by reflecting $\mathrm{X}$ in each of the sides are cyclic.

Solution:
If we shrink the reflected points by $1 / 2$ about the point $X$, then we get the feet of the perpendiculars from $X$ to the sides. So it is sufficient to show that these four points are cyclic. Let the quadrilateral be ABCD. Let the feet of the perpendiculars from $\mathrm{X}$ to $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}, \mathrm{DA}$ be $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ respectively.

$\mathrm{XPBQ}$ is cyclic, so $\angle \mathrm{XQP}=\angle \mathrm{XBP}$. Similarly, $\mathrm{XPAS}$ is cyclic, so $\angle \mathrm{XSP}=\angle \mathrm{XAP}$. But $\mathrm{XBP}$ and $\mathrm{XAP}$ are two angles in the triangle $\mathrm{XAB}$ and the third is $\angle \mathrm{AXB}=90^{\circ}$. Hence $\angle \mathrm{XQP}+\angle \mathrm{XSP}=90^{\circ}$. Similarly $\angle$ $\mathrm{XQR}+\angle \mathrm{XSR}=90^{\circ}$. Adding, $\angle \mathrm{PQR}+\angle \mathrm{PSR}=180^{\circ}$, so $\mathrm{PQRS}$ is cyclic.


Let $\mathrm{S}$ be the set of functions $\mathrm{f}$ defined on reals in the closed interval $[0,1]$ with non-negative real values such that $\mathrm{f}(1)=1$ and $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \leq \mathrm{f}(\mathrm{x}+\mathrm{y})$ for all $\mathrm{x}, \mathrm{y}$ such that $\mathrm{x}+\mathrm{y} \leq 1$. What is the smallest $\mathrm{k}$ such that $\mathrm{f}(\mathrm{x}) \leq$ $\mathrm{kx}$ for all $\mathrm{f}$ in $\mathrm{S}$ and all $\mathrm{x}$ ?

Solution:
Answer: $\mathrm{k}=2$.

Consider the function $f(x)=0$ for $0 \leq x \leq 1 / 2,1$ for $1 / 2<x \leq 1$. If $x+y \leq 1$, then at least one of $x, y$ is $\leq 1 / 2$, so at least one of $f(x), f(y)$ is 0 . But $f$ is obviously increasing, so the other of $f(x), f(y)$ is $\leq f(x+y)$. Thus $f$ satisfies the conditions. But $\mathrm{f}(1 / 2+\varepsilon)=1$, so $\mathrm{k}$ cannot be smaller than 2 .

So now let $\mathrm{f}$ be any function satisfying the conditions. We wish to show that $\mathrm{f}(\mathrm{x}) \leq 2 \mathrm{x}$. Putting $\mathrm{y}=1$ - $\mathrm{x}$, we get $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y})<=\mathrm{f}(1)=1$. But $\mathrm{f}(\mathrm{y})$ is non-negative, so $\mathrm{f}(\mathrm{x}) \leq 1$. Put $\mathrm{y}=\mathrm{x} \leq 1 / 2$. Then $2 \mathrm{f}(\mathrm{x}) \leq \mathrm{f}(2 \mathrm{x})$. A simple induction gives $2^{n} f(x) \leq f\left(2^{n} x\right)$ for $x \leq 1 / 2^{n}$. Now take any $x$ in $[0,1]$. If $x>1 / 2$, then $2 x>1$, so $f(x)<2 x$. If $x$ $\leq 1 / 2$, choose $\mathrm{n} \geq 1$, so that $1 / 2^{n+1}<\mathrm{x} \leq 1 / 2^{n}$. Then $2^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \leq \mathrm{f}\left(2^{n^{n}} \mathrm{x}\right) \leq 1$, so $\mathrm{f}(\mathrm{x}) \leq 1 / 2^{\mathrm{n}} \leq 2 \mathrm{x}$.


The sequence $a_{n}$ of odd positive integers is defined as follows: $a_{1}=r, a_{2}=s$, and $a_{n}$ is the greatest odd divisor of $a_{n-1}+a_{n-2}$. Show that, for sufficiently large $n, a_{n}$ is constant and find this constant (in terms of $r$ and $s$ ).

Solution:
This is awkward to get started. Note that if $a_{n-1}=a_{n-2}$, then $a_{n-1}+a_{n-2}=2 a_{n-1}$, whose greatest odd divisor is just $a_{n-1}$, so $a_{n}=a_{n-1}$. So once two consecutive terms are constant the following terms are constant.

Both $a_{n-1}$ and $a_{n-2}$ are odd, so $a_{n-1}+a_{n-2}$ is even and hence $a_{n} \leq\left(a_{n-1}+a_{n-2}\right) / 2$. So if $a_{n-1}$ and $a_{n-2}$ are unequal, then $a_{n}<\max \left(a_{n-1}, a_{n-2}\right)$. As already noted, if they are equal, then $a_{n}=\max \left(a_{n-1}, a_{n-2}\right)$.

Put $b_{n}=\max \left(a_{n}, a_{n-1}\right)$. If $a_{n}<a_{n-1}$, then $a_{n+1}<\max \left(a_{n}, a_{n-1}\right)=a_{n-1}$. Also $a_{n+2} \leq \max \left(a_{n+1}, a_{n}\right)<a_{n-1}$. Hence $\max \left(a_{n+2}, a_{n+1}\right)<\max \left(a_{n}, a_{n-1}\right)$ or $b_{n+2}<b_{n}$. If $a_{n}>a_{n-1}$, then $a_{n+1}<\max \left(a_{n}, a_{n-1}\right)=a_{n}$, and $a_{n+2}<\max \left(a_{n+1}, a_{n}\right)=$
$a_{n}$. Hence $b_{n+2}<b_{n}$. Thus if $a_{n}$ and $a_{n-1}$ are unequal, then $b_{n+2}<b_{n}$. But obviously $a_{n}>0$ for all $n$, and so $b_{n}>0$ for all $n$. Also it is integral, and $b_{1}=\max (r, s)$, so we can only have $b_{2 n+1}<b_{2 n-1}$ for at most $\max (r, s)$ values of $n$. Hence for some $n$ we must have $a_{n}=a_{n-1}$ and then $a_{n}$ is constant from that point on.

We show that the constant is the greatest common divisor of $\mathrm{r}$ and $\mathrm{s}$. Use parentheses to denote the greatest common divisor, so the greatest common divisor of $r$ and $s$ is $(r, s)$. We have $\left(a_{n-1}, a_{n-2}\right)=\left(a_{n-1}, a_{n-1}+a_{n-2}\right)=$ $\left(a_{n}, a_{n-1}\right)$. So if $a_{n}$ is constant for $n \geq N$, we have $(r, s)=\left(a_{1}, a_{2}\right)=\left(a_{2}, a_{3}\right)=\ldots=\left(a_{N}, a_{N+1}\right)=a_{N}$. We claim that $\mathrm{a}_{\mathrm{n}}=\mathrm{d}$ for $\mathrm{n}>3$. Clearly $\mathrm{d}$ divides $\mathrm{r}+\mathrm{s}$ and hence $\mathrm{d}$ divides $\mathrm{a}_{3}$. But $\mathrm{d}$ divides $\mathrm{a}_{2}=\mathrm{s}$, so $\mathrm{d}$ also divides $\mathrm{a}_{2}+$ $\mathrm{a}_{3}$ and hence d divides $\mathrm{a}_{4}$. Now suppose some odd $\mathrm{k}$ divides $\mathrm{r}+\mathrm{s}$, but does not divide $\mathrm{s}$. So $\mathrm{k}$ divides $\mathrm{a}_{3}$, but not $\mathrm{a}_{2}$. Hence it does not divide $\mathrm{a}_{4}$.


A sequence $\mathrm{x}_{\mathrm{n}}$ of positive reals satisfies $\mathrm{x}_{n-1} \mathrm{x}_{\mathrm{n}+1} \leq \mathrm{x}_{\mathrm{n}}{ }^{2}$. Let $\mathrm{a}_{\mathrm{n}}$ be the average of the terms $\mathrm{x}_{0}, \mathrm{x}_{1}, \ldots, \mathrm{x}_{\mathrm{n}}$ and $b_{n}$ be the average of the terms $x_{1}, x_{2}, \ldots, x_{n}$. Show that $a_{n} b_{n-1} \geq a_{n-1} b_{n}$.

Solution:
Put $\mathrm{k}=\mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}-1}$. We have to show that $\left(\mathrm{x}_{0}+\mathrm{k}+\mathrm{x}_{\mathrm{n}}\right) /(\mathrm{n}+1) \mathrm{k} /(\mathrm{n}-1) \geq\left(\mathrm{x}_{0}+\mathrm{k}\right) / \mathrm{n} \quad\left(\mathrm{k}+\mathrm{x}_{\mathrm{n}}\right) / \mathrm{n}$ or $\mathrm{n}^{2}(\mathrm{k}+$ $\left.x_{0}+x_{n}\right) \geq\left(n^{2}-1\right)\left(k+x_{0}\right)\left(k+x_{n}\right)$. Simplifying slightly, this is equivalent to $k\left(k+x_{0}+x_{n}\right) \geq\left(n^{2}-1\right) x_{0} x_{n}$. So it is evidently sufficient to show that $k \geq(n-1)\left(x_{0} x_{n}\right)^{1 / 2}$. [Then AM/GM gives $x_{0}+x_{n} \geq 2\left(x_{0} x_{n}\right)^{1 / 2}$, so $\left(k+x_{0}+\right.$ $\left.\left.\mathrm{x}_{\mathrm{n}}\right) \geq(\mathrm{n}+1)\left(\mathrm{x}_{0} \mathrm{x}_{\mathrm{n}}\right)^{1 / 2}.\right]$

We have $\mathrm{x}_{0} / \mathrm{x}_{1} \leq \mathrm{x}_{1} / \mathrm{x}_{2} \leq \ldots \leq \mathrm{x}_{\mathrm{n}-1} / \mathrm{x}_{\mathrm{n}}$. Hence (apparently weakening) $\mathrm{x}_{0} \mathrm{x}_{\mathrm{n}} \leq \mathrm{x}_{1} \mathrm{x}_{\mathrm{n}-1} \leq \mathrm{x}_{2} \mathrm{x}_{\mathrm{n}-2} \leq \ldots$. So $\mathrm{k}=\left(\mathrm{x}_{1}+\mathrm{x}_{\mathrm{n}-}\right.$ $\left.{ }_{1}\right)+\left(x_{2}+x_{n-2}\right)+\ldots \geq(n-1)\left(x_{0} x_{n}\right)^{1 / 2}$, using first AM/GM, then the relation just established.


$a_{1}, a_{2}, a_{3}, \ldots$ are positive integers such that $a_{n}>a_{n-1}+1$. Put $b_{n}=a_{1}+a_{2}+\ldots+a_{n}$. Show that there is always a square in the range $b_{n}, b_{n}+1, b_{n}+2, \ldots, b_{n+1}-1$.

Solution:
If the result fails then for some $m$ we have $b_{n}>m^{2}$ and $b_{n+1} \leq(m+1)^{2}$. So $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2}<1$. Thus it is sufficient to prove that $b_{n+1}{ }^{1 / 2}-b_{n}{ }^{1 / 2} \geq 1$. Squaring, that is equivalent to $a_{n+1} \geq 2 b_{n}{ }^{1 / 2}+1$ or $b_{n}{ }^{1 / 2} \leq\left(a_{n+1}-1\right) / 2$.

We have $a_{n-1}<=a_{n}-2$. So $b_{n} \leq a_{n}+\left(a_{n}-2\right)+\left(a_{n}-4\right)-\ldots-\left(a_{n}-2(n-1)\right)$. Adding extra terms to the rhs if necessary we get $b_{n}=1+3+5+\ldots+a_{n}$ for $a_{n}$ odd, or $2+4+6+\ldots+a_{n}$ for $a_{n}$ even. In the odd case there are $\left(a_{n}+1\right) / 2$ terms of average size $\left(a_{n}+1\right) / 2$ (group them in pairs working from the outside in), so the sum is $\left(a_{n}+1\right)^{2} / 4$. In the even case there are $a_{n} / 2$ terms of average size $\left(a_{n}+2\right) / 2$, so the sum is $a_{n}\left(a_{n}+2\right) / 4$. So in either case we have $b_{n} \leq\left(a_{n}+1\right)^{2} / 4$ and hence $b_{n}{ }^{1 / 2} \leq\left(a_{n}+1\right) / 2 \leq\left(a_{n+1}-1\right) / 2$.


The sequence $a_{1}, a_{2}, \ldots, a_{99}$ has $a_{1}=a_{3}=a_{5}=\ldots=a_{97}=1, a_{2}=a_{4}=a_{6}=\ldots=a_{98}=2$, and $a_{99}=3$. We interpret subscripts greater than 99 by subtracting 99 , so that $\mathbf{a}_{100}$ means $\mathrm{a}_{1}$ etc. An allowed move is to change the value of any one of the $a_{n}$ to another member of $\{1,2,3\}$ different from its two neighbors, $a_{n-1}$ and $a_{n+1}$. Is there a sequence of allowed moves which results in $\mathrm{a}_{\mathrm{m}}=\mathrm{a}_{\mathrm{m}+2}=\ldots=\mathrm{a}_{\mathrm{m}+96}=1, \mathrm{a}_{\mathrm{m}+1}=\mathrm{a}_{\mathrm{m}+3}=\ldots=\mathrm{a}_{\mathrm{m}+95}=2$, $\mathrm{a}_{\mathrm{m}+97}=3, \mathrm{a}_{\mathrm{n}+98}=2$ for some $\mathrm{m}$ ? [So if $\mathrm{m}=1$, we have just interchanged the values of $\mathrm{a}_{98}$ and $\mathrm{a}_{99}$.]

Solution:
This is a classic invariant problem. We strongly suspect that there is no sequence of moves (otherwise it would be too easy to find), so that we must prove there is no sequence. The standard approach is to look for some invariant which is not changed by the allowed moves, but which is different for the initial and desired final positions.

For each member $a_{i}$ of the sequence let $b_{i}=$

0 if $\mathrm{a}_{\mathrm{i}}=\mathrm{a}_{\mathrm{i}+1}$,

1 if $\left(\mathrm{a}_{\mathrm{i}}, \mathrm{a}_{\mathrm{i}+1}\right)=(1,2),(2,3)$ or $(3,1)$

-1 if $\left(\mathrm{a}_{\mathrm{i}}, \mathrm{a}_{\mathrm{i}+1}\right)=(1,3),(2,1)$ or $(3,2)$.

We hope that $b_{1}+b_{2}+\ldots+b_{9 g}$ will be a suitable invariant. Suppose we make an allowed move by changing $a_{i}$. That has the effect of changing just $b_{i-1}$ and $b_{i}$. If $a_{i-1}=a_{i+1}$, then $b_{i-1}+b_{i}=0$, so the total of the $b_{j}$ does not change. If $a_{i-1}$ does not equal $a_{i+1}$, then we cannot change $a_{i}$ since it must be different from both. Thus allowed moves do not change $b_{1}+\ldots+b_{g 9}$. So it is an invariant. We now check it is suitable. The initial value is +3 (there are 49 pairs $(1,2), 48$ pairs $(2,1), 1$ pair $(2,3)$ and 1 pair $(3,1)$ total $49-48+1+1=3$. But the desired final position has value -3 (it has 48 pairs $(1,2), 49$ pairs $(2,1), 1$ pair $(1,3)$ and 1 pair $(3,2)$, total 48 - 49 - 1 $-1=-3)$. So we cannot get from one to the other by allowed moves.


The hexagon $\mathrm{ABCDEF}$ has the following properties: (1) its vertices lie on a circle; (2) $\mathrm{AB}=\mathrm{CD}=\mathrm{EF}$; and (3) the diagonals $\mathrm{AD}, \mathrm{BE}, \mathrm{CF}$ meet at a point. Let $\mathrm{X}$ be the intersection of $\mathrm{AD}$ and $\mathrm{CE}$. Show that $\mathrm{CX} / \mathrm{XE}=$ $(\mathrm{AC} / \mathrm{CE})^{2}$.

Solution:
Let the diagonals $\mathrm{AD}, \mathrm{BE}, \mathrm{CF}$ meet at $\mathrm{Y}$. We show first that the triangles $\mathrm{AEC}$, $\mathrm{YED}$ are similar. $\angle \mathrm{ACE}=$ $\angle \mathrm{ADE}(\mathrm{ACDE}$ circle) $=\mathrm{YDE}$ (same angle). $\angle \mathrm{AEB}=\angle \mathrm{CED}(\mathrm{AB}=\mathrm{CD}$ ), so $\angle \mathrm{AEB}+\angle \mathrm{BEC}=\angle \mathrm{CED}+$ $\angle \mathrm{BEC}$ or $\angle \mathrm{AEC}=\angle \mathrm{YED}$. So $\mathrm{AEC}$ and $\mathrm{YED}$ are similar, so $\mathrm{AC} / \mathrm{CE}=\mathrm{YD} / \mathrm{DE}$.

We next show that $\mathrm{AEC}$ and $\mathrm{CDY}$ are similar. $\angle \mathrm{AEC}=\angle \mathrm{ADC}$ (circle) $=\angle \mathrm{CDY}$ (same angle). $\angle \mathrm{EAC}=$ $\angle \mathrm{EAD}+\angle \mathrm{DAC}=\angle \mathrm{ECD}$ (circle) $+\angle \mathrm{ECF}(\mathrm{EF}=\mathrm{CD})=\angle \mathrm{DCY}$. So $\mathrm{AEC}$ and $\mathrm{CDY}$ are similar. So $\mathrm{AC} / \mathrm{CE}=\mathrm{CY} / \mathrm{YD}$.

Hence $(\mathrm{AC} / \mathrm{CE})^{2}=\mathrm{CY} / \mathrm{DE}$. Finally we show that $\mathrm{CXY}$ and EXD are similar. That is almost obvious because $\mathrm{CF}$ is parallel to $\mathrm{DE}$ since $\mathrm{CD}=\mathrm{EF}$. Hence $\mathrm{CY} / \mathrm{DE}=\mathrm{CX} / \mathrm{XE}$, which gives the required result.


$\mathrm{x}_{\mathrm{i}}$ is a infinite sequence of positive reals such that for all $\mathrm{n}, \mathrm{x}_{1}+\mathrm{x}_{2}+\ldots+\mathrm{x}_{\mathrm{n}} \geq \sqrt{ }$. Show that $\mathrm{x}_{1}{ }^{2}+\mathrm{x}_{2}{ }^{2}+\ldots+$ $\mathrm{x}_{\mathrm{n}}^{2}>(1+1 / 2+1 / 3+\ldots+1 / \mathrm{n}) / 4$ for all $\mathrm{n}$.

Solution:
Note that this is rather a weak inequality. Taking $\mathrm{n}=1$, we get $\mathrm{x}_{1}>=1$, but $(1+1 / 2+1 / 3+\ldots+1 / 30)<4$, so it is only for $\mathrm{n}>30$ that we need to consider $\mathrm{x}_{2}$ ! Of course, weak inequalities can be awkward to prove.

For $\mathrm{a}+\mathrm{b}$ constant, we minimise $\mathrm{a}^{2}+\mathrm{b}^{2}$ by taking $|\mathrm{a}-\mathrm{b}|$ as small as possible. So we suspect that the minimum value of $x_{1}{ }^{2}+\ldots+x_{n}{ }^{2}$ is when $x_{1}=1, x_{2}=\sqrt{2}-1, x_{3}=\sqrt{3}-\sqrt{2}, x_{4}=\sqrt{4}-\sqrt{3}, \ldots(*)$. Note that $\sqrt{n}-$ $\sqrt{ }(n-1)=1 /\left(\sqrt{n}+\sqrt{ }(n-1)>1 /(2 \sqrt{n})\right.$. So for the values $(*)$ we have $x_{1}{ }^{2}+\ldots+x_{n}{ }^{2}>(1+1 / 2+1 / 3+\ldots+1 / n) / 4$. So it remains to show that the values $\left(^{*}\right)$ do indeed give the minimum.

We use Abel's partial summation formula (whose proof is trivial). Put $y_{n}=V_{n}-\sqrt{ }(n-1), s_{n}=x_{1}+x_{2}+\ldots+x_{n}$, $t_{n}=y_{1}+\ldots+t_{n}$. So we assume $s_{n} \geq t_{n}$. Note also that $y_{1}>y_{2}>y_{3}>\ldots$. Then the partial summation formula is: $\sum x_{i} y_{i}=\sum s_{i}\left(y_{i}-y_{i+1}\right)+s_{n} y_{n+1}$ (where the sum is taken from 1 to $n$ ). We also have $\sum y_{i}^{2}=\sum t_{i}\left(y_{i}-y_{i+1}\right)+$ $\mathrm{t}_{\mathrm{n}} \mathrm{y}_{\mathrm{n}+1}$. But each term is not more than the corresponding term in the first equality, so $\sum \mathrm{y}_{\mathrm{i}}{ }^{2} \leq \sum \mathrm{x}_{\mathrm{i}} \mathrm{y}_{\mathrm{i}}$. Now Cauchy-Schwartz gives $\left(\sum \mathrm{x}_{\mathrm{i}} \mathrm{y}_{\mathrm{i}}\right)^{2} \leq \sum \mathrm{x}_{\mathrm{i}}^{2} \sum \mathrm{y}_{\mathrm{i}}^{2} \cdot$ Hence $\sum \mathrm{y}_{\mathrm{i}}^{2} \leq \sum \mathrm{x}_{\mathrm{i}}^{2}$, as required.


$\mathrm{X}$ is a set of $\mathrm{n}$ positive integers with sum $\mathrm{s}$ and product $\mathrm{p}$. Show for any integer $\mathrm{N}>=\mathrm{s}, \Sigma$ ( parity( $\mathrm{Y}$ ) ( $\mathrm{N}-$ sum(Y))Cs $)=\mathrm{p}$, where $\mathrm{aCb}$ is the binomial coefficient $\mathrm{a}!/(\mathrm{b}!(\mathrm{a}-\mathrm{b})$ ! ), the sum is taken over all subsets $\mathrm{Y}$ of $\mathrm{X}$, parity $(\mathrm{Y})=1$ if $\mathrm{Y}$ is empty or has an even number of elements, -1 if $\mathrm{Y}$ has an odd number of elements, and $\operatorname{sum}(\mathrm{Y})$ is the sum of the elements in $\mathrm{Y}$.

Solution:
$\left(\sum(-1)^{n}\right.$ binomial) with the sum over all subsets $\mathrm{Y}$ of a set $\mathrm{X}$ strongly suggests the principle of inclusion and exclusion.

Consider all sequences of length $\mathrm{N} \geq \mathrm{s}$ of $0 \mathrm{~s}$ and $1 \mathrm{~s}$ with a total of $\mathrm{s} 1 \mathrm{~s}$. We can regard positions $1,2, \ldots$, $a_{1}$ as block $a_{1}$, positions $a_{1}+1, a_{2}+2, \ldots, a_{1}+a_{2}$ as block $a_{2}$, positions $a_{1}+a_{2}+1, a_{1}+a_{2}+2, \ldots, a_{1}+a_{2}+a_{3}$ as block $a_{3}$, and so on up to $a_{1}+\ldots+a_{n-1}+1, \ldots a_{1}+\ldots+a_{n}$ as block $a_{n}$. We are interested in sequences which have just one 1 in each block. Counting directly there are obviously just p such sequences.

But we can also take $\mathrm{N}^{\prime}$ to be the total number of sequences of $0 \mathrm{~s}$ and $1 \mathrm{~s}$ with $\mathrm{s} 1 \mathrm{~s}$ and $\mathrm{N}_{\mathrm{Y}}$ to be the number with no 1s in block $\mathrm{k}$ for $\mathrm{k}$ in $\mathrm{Y}$, where $\mathrm{Y}$ is a subset of $\mathrm{X}$. Requiring one 1 in each block is equivalent to requiring no block to have no 1s. So the PIE gives that $\mathrm{p}=\mathrm{N}^{\prime}-\mathrm{N}_{\mathrm{Y}}$ for $\mathrm{Y}$ with one element $+\mathrm{N}_{\mathrm{Y}}$ for $\mathrm{Y}$ with 2 elements and so on. $\mathrm{N}^{\prime}=\mathrm{NCs}$ and $\mathrm{N}_{\mathrm{Y}}=(\mathrm{N}-\operatorname{sum}(\mathrm{Y}))$ Cs, so we have $\mathrm{p}=\Sigma(\operatorname{parity}(\mathrm{Y})(\mathrm{N}-\operatorname{sum}(\mathrm{Y}))$ Cs $)$, which is the required relation.


The sequence $\mathrm{a}_{0} \mathrm{a}_{1}, \mathrm{a}_{2}, \ldots$ of non-negative integers is defined as follows. The first $\mathrm{p}-1$ terms are $0,1,2,3, \ldots$, $p-2$. Then $a_{n}$ is the least positive integer so that there is no arithmetic progression of length $p$ in the first $\mathrm{n}+1$ terms. If $\mathrm{p}$ is an odd prime, show that $\mathrm{a}_{\mathrm{n}}$ is the number obtained by writing $\mathrm{n}$ in base $\mathrm{p}-1$, then treating the result as a number in base p. For example, if p is 5 , to get the 5 th term one writes 5 as 11 in base 4 , then treats this as a base 5 number to get 6 .

Solution:
Let $b_{n}$ be the number obtained by writing $n$ in base $p-1$ and then treating the result as a number in base $p$. The resulting sequence $\mathrm{b}_{\mathrm{n}}$ is all those non-negative integers whose base $\mathrm{p}$ representation does not have a digit $\mathrm{p}-1$. We show that $\mathrm{b}_{\mathrm{n}}$ cannot contain an arithmetic progression of length $\mathrm{p}$. For suppose there was such a progression with difference $\mathrm{d}$. Suppose the last non-zero digit of $\mathrm{d}$ in base $\mathrm{p}$ is $\mathrm{k}$. Suppose the first term of the progression has digit $\mathrm{h}$ in that position, then the terms of the progression have digit $\mathrm{h}, \mathrm{h}+\mathrm{k}, \mathrm{h}+2 \mathrm{k}, \ldots \mathrm{h}+(\mathrm{p}-1) \mathrm{k}$ $\bmod p$ in that position. But these must be a complete set of residues $\bmod p$, so one of them must be $p-1$ mod p. So the corresponding term has a digit p-1 in this position. Contradiction.

Now to show that $\mathrm{a}_{\mathrm{n}}=\mathrm{b}_{\mathrm{n}}$ we use induction on $\mathrm{n}$. Evidently, it is true for $\mathrm{n}<\mathrm{p}-1$. Suppose it is true for all $\mathrm{m}<$ $\mathrm{n}$. It is sufficient to show that if $\mathrm{b}_{\mathrm{n}}<\mathrm{m}<\mathrm{b}_{\mathrm{n}+1}$, then $\left\{\mathrm{b}_{1}, \mathrm{~b}_{2}, \ldots, \mathrm{b}_{\mathrm{n}}, \mathrm{m}\right\}$ contains an arithmetic progression. $\mathrm{m}$ must contain a digit p-1, for otherwise it would be $\mathrm{b}_{\mathrm{k}}$ for some $\mathrm{k}>\mathrm{n}$. Let $\mathrm{m}_{1}$ be the number obtained from $\mathrm{m}$ by reducing every digit $\mathrm{p}-1$ in $\mathrm{m}$ by 1 . Then $\mathrm{m}$ has no digit $\mathrm{p}-1 \mathrm{~s}$, so it must be some $\mathrm{b}_{\mathrm{k}}$ and hence one of $\mathrm{b}_{1}, \ldots$ , $b_{n}$. Now take $m_{2}$ to be the number obtained by reducing the same digits by another 1 . Similarly, define $m_{3}$, $\ldots, \mathrm{m}_{p-1}$. Then each $\mathrm{m}_{\mathrm{i}}$ is in $\left\{\mathrm{b}_{1}, \ldots, \mathrm{b}_{\mathrm{n}}\right\}$ and $\mathrm{m}_{\mathrm{p}-1}, \ldots, \mathrm{m}_{1}, \mathrm{~m}$ is a progression of length $\mathrm{p}$.


A trigonometric map is any one of sin, cos, tan, arcsin, arccos and arctan. Show that given any positive rational number $\mathrm{x}$, one can find a finite sequence of trigonometric maps which take 0 to $\mathrm{x}$. [So we need to show that we can always find a sequence of trigonometric maps $\mathrm{t}_{\mathrm{i}}$ so that: $\mathrm{x}_{1}=\mathrm{t}_{0}(0), \mathrm{x}_{2}=\mathrm{t}_{1}\left(\mathrm{x}_{1}\right), \ldots, \mathrm{x}_{\mathrm{n}}=\mathrm{t}_{\mathrm{n}}$. $\left.1\left(\mathrm{x}_{\mathrm{n}-1}\right), \mathrm{x}=\mathrm{t}_{\mathrm{n}}\left(\mathrm{x}_{\mathrm{n}}\right) \cdot\right]$

Solution:
We have $\cos ^{2} t+\sin ^{2} t=1$. Hence $\cos t=\cos t /\left(\sqrt{ }\left(\cos ^{2} t+\sin ^{2} t\right)\right)=1 /\left(\sqrt{ }\left(1+\tan ^{2} t\right)\right)$. So if we put $\tan t=\sqrt{x}$, then $\cos \tan ^{-1} \sqrt{x}=1 / \sqrt{ }(1+x)$. We also have $\cos (\pi / 2-x)=\sin x$ and $\tan (\pi / 2-x)=1 / \tan x$. So $\tan ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-}$ ${ }^{1} x=1 / x(1)$. Hence also $\tan ^{-1} \cos ^{-1} \sin ^{-1} \tan ^{-1} \cos \tan ^{-1} \sqrt{x}=\sqrt{ }(x+1)$ (2). These two relations solve the problem.

Using (2) and iterating we can get $\sqrt{ }$ for any positive integer $n$. Hence in particular we can get $\mathrm{n}$ for any positive integer $\mathrm{n}$. Now suppose we want $\mathrm{m} / \mathrm{n}$ with $\mathrm{m}$ and $\mathrm{n}$ relatively prime. We show that $\mathrm{m} / \mathrm{n}$ can be achieved by induction on $n$. We have just dealt with the case $\mathrm{n}=1$. Suppose we have dealt with all $\mathrm{a} / \mathrm{b}$ with $\mathrm{b}$ $<\mathrm{n}$. If $\mathrm{m}>\mathrm{n}$, then we can write $\mathrm{m}=\mathrm{qn}+\mathrm{r}$ with $0<\mathrm{r}<\mathrm{n}$ and use (2) to reduce the problem to getting $\mathrm{r} / \mathrm{n}$. If $\mathrm{m}<\mathrm{n}$, then put $\mathrm{r}=\mathrm{m}$. Now use (1) to reduce the problem to $\mathrm{n} / \mathrm{r}$, which is solved by induction.


The circumcenter $\mathrm{O}$ of the triangle $\mathrm{ABC}$ does not lie on any side or median. Let the midpoints of $\mathrm{BC}, \mathrm{CA}$, $\mathrm{AB}$ be $\mathrm{L}, \mathrm{M}, \mathrm{N}$ respectively. Take $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ on the rays $\mathrm{OL}, \mathrm{OM}, \mathrm{ON}$ respectively so that $\angle \mathrm{OPA}=\angle \mathrm{OAL}$, $\angle \mathrm{OQB}=\angle \mathrm{OBM}$ and $\angle \mathrm{ORC}=\angle \mathrm{OCN}$. Show that $\mathrm{AP}, \mathrm{BQ}$ and $\mathrm{CR}$ meet at a point.

Solution:
We show that the circumcircle $A B C$ is the incircle of $P Q R$. Then $(A R / A Q)(C Q / C P)(B P / B R)=1$ since $A R$ $=\mathrm{BR}, \mathrm{AQ}=\mathrm{CQ}, \mathrm{BP}=\mathrm{CP}$ (equal tangents) and hence $\mathrm{PA}, \mathrm{QB}, \mathrm{RC}$ are concurrent by Ceva's theorem.

So let the tangents to the circumcircle at $\mathrm{B}$ and $\mathrm{C}$ meet at $\mathrm{P}^{\prime}$. It is sufficient to show that $\angle \mathrm{OP} \mathrm{A}=\angle \mathrm{OAL}$, for then it follows that $\mathrm{P}^{\prime}=\mathrm{P}$ (since there is obviously a unique point on the ray OL at which the segment OA subtends the $\angle \mathrm{OAL})$.

This is curiously difficult to prove. Let $\mathrm{LP}^{\prime}$ meet the circle at $\mathrm{K}$. Then $\angle \mathrm{KCP}^{\prime}=\angle \mathrm{KBC}$ ( $\mathrm{P}^{\prime} \mathrm{C}$ tangent) $=\angle$ $\mathrm{KCB}$ (KO perpendicular to $\mathrm{BC}$, since $\mathrm{L}$ midpoint) $=\angle \mathrm{KCL}$ (same angle). So $\mathrm{KC}$ bisects $\angle \mathrm{P}^{\prime} \mathrm{CL}$. Hence $\mathrm{KP}^{\prime} / \mathrm{KL}=\mathrm{CP}^{\prime} / \mathrm{CL}$. But obviously $\mathrm{CP}^{\prime} / \mathrm{CL}=\mathrm{BP}^{\prime} / \mathrm{BL}$. So $\mathrm{K}, \mathrm{C}$ and $\mathrm{B}$ lie on the circle of Apollonius, the points $\mathrm{X}$ such that $\mathrm{XP}^{\prime} / \mathrm{XL}$ is constant. But A also lies on the circle of Apollonius. Hence KA bisects $\angle \mathrm{P}^{\prime} \mathrm{AL}$. (See Canada 71/9 if you are not familiar with the circle of Apollonius.) But $\angle \mathrm{OP}^{\prime} \mathrm{A}+\angle \mathrm{KAP}^{\prime}=\angle \mathrm{OKA}=\angle$ $\mathrm{OAK}(\mathrm{OA}$ and $\mathrm{OK}$ radii $)=\angle \mathrm{OAL}+\angle \mathrm{KAL}$. So $\angle \mathrm{OP}^{\prime} \mathrm{A}=\angle \mathrm{OAL}$.


$a_{0}, a_{1}, a_{2}, \ldots$ is an infinite sequence of integers such that $a_{n}-a_{m}$ is divisible by $n-m$ for all (unequal) $n$ and $m$. For some polynomial $\mathrm{p}(\mathrm{x})$ we have $\mathrm{p}(\mathrm{n})>\left|\mathrm{a}_{\mathrm{n}}\right|$ for all $\mathrm{n}$. Show that there is a polynomial $\mathrm{q}(\mathrm{x})$ such that $\mathrm{q}(\mathrm{n})=$ $a_{n}$ for all $n$.

Solution:
Clearly for any finite $N$, we can find a polynomial $q(n)$ of degree $N$ such that $q(n)=a_{n}$ for $n=0, \ldots, N$. Also, once $\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots, \mathrm{a}_{\mathrm{N}}$ are fixed, $\mathrm{a}_{\mathrm{m}}$ is somewhat constrained for $\mathrm{m}>\mathrm{N}$, because we require $\mathrm{a}_{\mathrm{m}}$ to be $\mathrm{a}_{0}$ mod $\mathrm{m}$, $a_{1} \bmod m-1, \ldots, a_{N} \bmod m-N$. Put $M=1 c m(m, m-1, \ldots, m-N)$. Then $a_{m}$ is determined mod M.

We would like to argue that $\mathrm{q}(\mathrm{m})$ is known to satisfy the congruences (because $m-n$ divides $q(m)-q(n)$ ), that $\mathrm{q}(\mathrm{m})$ and $\mathrm{a}_{\mathrm{m}}$ are bounded so that they cannot differ by as much as $\mathrm{M}$ and hence must be equal. The snag is that $\mathrm{q}(\mathrm{x})$ does not necessarily have integer coeffcients, so it is not necessarily true that $\mathrm{m}$ - $\mathrm{n}$ divides $\mathrm{q}(\mathrm{m})$ $\mathrm{q}(\mathrm{n})$. [The argument is that $\mathrm{m}-\mathrm{n}$ divides $\mathrm{m}^{\mathrm{k}}-\mathrm{n}^{\mathrm{k}}$ for any integer $\mathrm{k}$ and hence divides $\mathrm{q}(\mathrm{m})-\mathrm{q}(\mathrm{n})$ for any polynomial with integer coefficients.]

However, the coefficients of $\mathrm{q}(\mathrm{x})$ are rational. So let $\mathrm{Q}$ be the $1 \mathrm{~cm}$ of the denominators. Then $\mathrm{Q} q(\mathrm{x})$ does have integer coefficients and $m-n$ does divide $Q(q(m)-q(n))$. So $Q q(m)=Q q(0)=Q a_{0}=Q a_{m} m o d m$, and similarly $Q \mathrm{q}(\mathrm{m})=Q \mathrm{a}_{\mathrm{m}}$ mod $\mathrm{m}-1$ and so on. Hence $\mathrm{Q} q(\mathrm{~m})=Q \mathrm{a}_{\mathrm{m}}$ mod $M$.

But we know that $|\mathrm{q}(\mathrm{m})|<\mathrm{c}^{\mathrm{N}}$ for some constant $\mathrm{c}$, since $\mathrm{q}(\mathrm{x})$ has degree N. Similarly, we know that $\left|\mathrm{a}_{\mathrm{m}}\right|$ is bounded by a polynomial of degree $\mathrm{N}$ and hence $\left|\mathrm{a}_{\mathrm{m}}\right|<\mathrm{c}^{\prime} \mathrm{m}^{\mathrm{N}}$ for some constant $\mathrm{c}^{\prime}$. Now $\mathrm{M}$ certainly exceeds the product of the $\mathrm{N}+1$ numbers $\mathrm{m}, \mathrm{m}-1, \ldots, \mathrm{m}-\mathrm{N}$ divided by the greatest common divisor for each of the $\mathrm{N}(\mathrm{N}+1) / 2$ pairs ( $\mathrm{m}-\mathrm{i}, \mathrm{m}-\mathrm{j})$. But each gcd is at most $\mathrm{N}$, so $\mathrm{M}$ is more than $\mathrm{c}^{\prime \prime} \mathrm{m}^{\mathrm{N}+1}$ for some (small) constant $c^{\prime \prime}$ which does not depend on $m$. Hence for $m \geq$ some $N^{\prime}$ we have $M>|Q q(m)|+\left|Q a_{m}\right|$ and hence $q(m)=a_{m}$.

So we have $\mathrm{q}(\mathrm{m})=\mathrm{a}_{\mathrm{m}}$ for $\mathrm{m} \leq \mathrm{N}$ and for $\mathrm{m} \geq \mathrm{N}^{\prime}$. Suppose $\mathrm{N}<\mathrm{m}<\mathrm{N}^{\prime}$. Then for any $\mathrm{m}^{\prime}>\mathrm{N}^{\prime}$ we have that ( $\mathrm{m}^{\prime}$ $-m)$ divides $Q q\left(m^{\prime}\right)-Q q(m)$ and $Q a_{m^{\prime}}-Q a_{m}$. So it must divide their difference. But $q\left(m^{\prime}\right)=a_{m^{\prime}}$, so it divides $Q\left(q(m)-a_{m}\right)$. But we can choose $m^{\prime}$ so that $\left(m^{\prime}-m\right)$ is prime to $Q$ and larger than $\left(q(m)-a_{m}\right)$. Hence $q(m)=a_{m}$. So we have established that $q(m)=a_{m}$ for all $m$.


A graph with $\mathrm{n}$ points and $\mathrm{k}$ edges has no triangles. Show that it has a point $\mathrm{P}$ such that there are at most $\mathrm{k}(1$ $-4 k / \mathrm{n}^{2}$ ) edges between points not joined to P (by an edge).

Solution:
Given a point $\mathrm{P}$, the edges of the graph can be divided into three categories: (1) edges $\mathrm{PQ}$, (2) edges $\mathrm{QQ}^{\prime}$, where Q is joined to $\mathrm{P}$, and (3) edges $\mathrm{Q}^{\prime} \mathrm{Q}^{\prime \prime}$, where $\mathrm{Q}^{\prime}$ and $\mathrm{Q}^{\prime \prime}$ are not joined to $\mathrm{P}$. Note that if $\mathrm{Q}$ is joined to $\mathrm{P}$ and there is an edge $\mathrm{QQ}^{\prime}$, then $\mathrm{Q}^{\prime}$ cannot be joined to $\mathrm{P}$, or $\mathrm{PQQ}$ ' would be a triangle. So the total number of edges in categories (1) and (2) is $\sum^{\prime} \operatorname{deg} Q$, where $\sum^{\prime}$ indicates that the sum is taken over points $\mathrm{Q}$ which are joined to $\mathrm{P}$ (by an edge). Thus the total number of edges in category (3) is $\mathrm{k}-\Sigma^{\prime}$ deg $\mathrm{Q}$. So we have to show that for some $\mathrm{P}, \Sigma^{\prime} \operatorname{deg} \mathrm{Q} \geq 4 \mathrm{k}^{2} / \mathrm{n}^{2}$.

The total for all points $\mathrm{P}$ is $\sum_{\mathrm{P}} \sum^{\prime} \operatorname{deg} \mathrm{Q}$. If we invert the order of the summations, this becomes $\sum \operatorname{deg}^{2} \mathrm{Q}$, where the sum is over all points $Q$. Now by Cauchy, $\left(\sum \operatorname{deg} Q\right)^{2} \leq\left(\sum 1^{2}\right)\left(\sum \operatorname{deg}^{2} Q\right)=n \sum \operatorname{deg}^{2} Q$. But $\sum$ deg $\mathrm{Q}=2 \mathrm{k}$, so $\sum \operatorname{deg}^{2} \mathrm{Q} \geq 4 \mathrm{k}^{2} / \mathrm{n}$ and hence $\sum_{\mathrm{P}} \sum^{\prime} \operatorname{deg} \mathrm{Q} \geq 4 \mathrm{k}^{2} / \mathrm{n}$. But if a sum of $\mathrm{n}$ terms is at least $4 \mathrm{k}^{2} / \mathrm{n}$, then at least one for the terms must be at least $4 \mathrm{k}^{2} / \mathrm{n}^{2}$. In other words, there is a point $P$ such that $\sum^{\prime} \operatorname{deg} Q \geq 4 \mathrm{k}^{2} / \mathrm{n}^{2}$, as required.