APMO/md/en-apmo2003_sol.md

# XV APMO: Solutions and Marking Schemes 

1. Let $a, b, c, d, e, f$ be real numbers such that the polynomial

$$
p(x)=x^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f
$$

factorises into eight linear factors $x-x_{i}$, with $x_{i}>0$ for $i=1,2, \ldots, 8$. Determine all possible values of $f$.

## Solution.

From

$$
x^{8}-4 x^{7}+7 x^{6}+a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f=\left(x-x_{1}\right)\left(x-x_{2}\right) \ldots\left(x-x_{8}\right)
$$

we have

$$
\sum_{i=1}^{8} x_{i}=4 \quad \text { and } \quad \sum x_{i} x_{j}=7
$$

where the second sum is over all pairs $(i, j)$ of integers where $1 \leq i<j \leq 8$. Since this sum can also be written

$$
\frac{1}{2}\left[\left(\sum_{i=1}^{8} x_{i}\right)^{2}-\sum_{i=1}^{8} x_{i}^{2}\right]
$$

we get

$$
14=\left(\sum_{i=1}^{8} x_{i}\right)^{2}-\sum_{i=1}^{8} x_{i}^{2}=16-\sum_{i=1}^{8} x_{i}^{2}
$$

so

$$
\sum_{i=1}^{8} x_{i}^{2}=2 \quad \text { while } \quad \sum_{i=1}^{8} x_{i}=4 . \quad[3 \text { marks }]
$$

Now

$$
\sum_{i=1}^{8}\left(2 x_{i}-1\right)^{2}=4 \sum_{i=1}^{8} x_{i}^{2}-4 \sum_{i=1}^{8} x_{i}+8=4(2)-4(4)+8=0
$$

which forces $x_{i}=1 / 2$ for all $i$. [3 marks] Therefore

$$
f=\prod_{i=1}^{8} x_{i}=\left(\frac{1}{2}\right)^{8}=\frac{1}{256} . \quad[1 m x]
$$

Alternate solution: After obtaining (1) [3 marks], use Cauchy's inequality to get

$$
16=\left(x_{1} \cdot 1+x_{2} \cdot 1+\cdots+x_{8} \cdot 1\right)^{2} \leq\left(x_{1}^{2-}+x_{2}^{2}+\cdots+x_{8}^{2}\right)\left(1^{2}+1^{2}+\cdots+1^{2}\right)=8 \cdot 2=16
$$

or the power mean inequality to get

$$
\frac{1}{2}=\frac{1}{8} \sum_{i=1}^{8} x_{i} \leq\left(\frac{1}{8} \sum_{i=1}^{8} x_{i}^{2}\right)^{1 / 2}=\frac{1}{2} . \quad[2 \text { marks }]
$$

Either way, equality must hold, which can only happen if all the terms $x_{i}$ are equal, that is, if $x_{i}=1 / 2$ for all $i$. [1 mark] Thus $f=1 / 256$ as above. [ 1 mark]
2. Suppose $A B C D$ is a square piece of cardboard with side length $a$. On a plane are two parallel lines $\ell_{1}$ and $\ell_{2}$, which are also $a$ units apart. The square $A B C D$ is placed on the plane so that sides $A B$ and $A D$ intersect $\ell_{1}$ at $E$ and $F$ respectively. Also, sides $C B$ and $C D$ intersect $\ell_{2}$ at $G$ and $H$ respectively. Let the perimeters of $\triangle A E F$ and $\triangle C G H$ be $m_{1}$ and $m_{2}$ respectively. Prove that no matter how the square was placed, $m_{1}+m_{2}$ remains constant.

Solution 1.
Let $E H$ intersect $F G$ at $O$. The distance from $G$ to line $F D$ and line $E F$ are both $a$. So $F G$ bisects $\angle E F D$. Similarly, $E H$ bisects $\angle B E F$. So $O$ is an excentre of $\triangle A E F$. Similarly, $O$ is an excentre of $\triangle C G H$. [2 marks] Construct these excircles with centre $O$. Let $M, N, P, Q$ be on sides $A B, B C, C D, D A$ respectively, where these excircles touch the square. Then $O M \perp A B, O N \perp B C, O P \perp C D$, and $O Q \perp D A$. Since $A B \| C D$ and $A D \| B C, M, O, P$ are collinear and $N, O, Q$ are collinear. Now $M P=N Q=a$. [2 marks] Using the fact that the two tangents from a point to a circle have the same length, we get $E F=E M+F Q$ and $G H=G N+H P$. [1 mark] Then

$$
m_{1}=A E+A F+E F=A E+A F+(E M+F Q)=A M+A Q=O Q+O M
$$

and

$$
m_{2}=C G+C H+G H=C G+C H+(G N+H P)=C N+C P=O P+O N . \quad[1 \text { mark }]
$$

Therefore

$$
m_{1}+m_{2}=(O Q+O M)+(O P+O N)=M P+N Q=2 a . \quad[1 \mathrm{mark}]
$$

## Solution 2.

Extend $A B$ to $I$ and $D C$ to $J$ so that $A E=B I=C J$. Let $\ell_{2}$ intersect $I J$ at $M$, and let $K$ lie on $I J$ so that $G K \perp I J$. Then, since $A E=G K, \triangle A E F$ and $\triangle K G M$ are congruent. [1 mark] Thus, since $G K=C J$ and $G C=K J$,

$$
m_{1}+m_{2}=\operatorname{perimeter}(K G M)+\operatorname{perimeter}(C G H)=\operatorname{perimeter}(H M J) . \quad[\mathbf{2} \text { marks }]
$$

Let $L$ lie on $C D$ so that $E L \perp C D$. Then a circle with centre $E$ and radius $a$ will touch $D C$ at $L, I J$ at $I$, and the interior of $H M$ at some point $N$, so

$$
\operatorname{perimeter}(H M J)=J H+(H N+N M)+J M=(J H+H L)+(M I+J M)=J L+I J=a+a=2 a
$$

[4 marks] Thus $m_{1}+m_{2}=2 a$.
Solution 3.
Without loss of generality, assume the square has side $a=1$. Let $\theta$ be the acute angle between $\ell_{1}$ (or $\ell_{2}$ ) and the sides $A B$ and $C D$ of the square. Then, letting $E F=x$ and $G H=y$, we have

$$
E A=x \cos \theta, \quad A F=x \sin \theta, \quad C H=y \cos \theta, \quad C G=y \sin \theta
$$

Thus

$$
m_{1}+m_{2}=(x+y)(\sin \theta+\cos \theta+1) . \quad[2 \text { marks }]
$$

Draw lines parallel to $\ell_{1}, \ell_{2}$ through $A$ and $C$ respectively. The distance between these lines is $\sin \theta+\cos \theta$ [1 mark], as can be seen by drawing a mutual perpendicular to these lines through $B$, say. Also, the altitudes from $A$ to $E F$ and from $C$ to $G H$ have lengths $x \sin \theta \cos \theta$ and $y \sin \theta \cos \theta$ respectively [ 1 mark]. Therefore the distance between $\ell_{1}$ and $\ell_{2}$ must be

$$
(\sin \theta+\cos \theta)-x \sin \theta \cos \theta-y \sin \theta \cos \theta
$$

But we are given that this distance is $a=1$, so

$$
(x+y) \sin \theta \cos \theta+1=\sin \theta+\cos \theta
$$

or

$$
x+y=\frac{\sin \theta+\cos \theta-1}{\sin \theta \cos \theta} \cdot \quad[1 \text { mark }]
$$

Therefore, by (1),

$$
\begin{aligned}
m_{1}+m_{2} & =\frac{(\sin \theta+\cos \theta-1)(\sin \theta+\cos \theta+1)}{\sin \theta \cos \theta} \\
& =\frac{\left(\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta\right)-1}{\sin \theta \cos \theta} \\
& =\frac{1+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}=2 . \quad[2 \text { marks }]
\end{aligned}
$$

3. Let $k \geq 14$ be an integer, and let $p_{k}$ be the largest prime number which is strictly less than $k$. You may assume that $p_{k} \geq 3 k / 4$. Let $n$ be a composite integer. Prove:
(a) if $n=2 p_{k}$, then $n$ does not divide $(n-k)$ !;
(b) if $n>2 p_{k}$, then $n$ divides $(n-k)$ !.

Solution.
(a) Note that $n-k=2 p_{k}-k<2 p_{k}-p_{k}=p_{k}$, so $p_{k} \nmid(n-k)$ !, so $2 p_{k} \nless(n-k)$ !. [1 mark]
(b) Note that $n>2 p_{k} \geq 3 k / 2$ implies $k<2 n / 3$, so $n-k>n / 3$. So if we can find integers $a, b \geq 3$ such that $n=a b$ and $a \neq b$, then both $a$ and $b$ will appear separately in the product $(n-k)!=1 \times 2 \times \cdots \times(n-k)$, which means $n \mid(n-k)!$. Observe that $k \geq 14$ implies $p_{k} \geq 13$, so that $n>2 p_{k} \geq 26$.

If $n=2^{\alpha}$ for some integer $\alpha \geq 5$, then take $a=2^{2}, b=2^{\alpha-2}$. [ 1 mark] Otherwise, since $n \geq 26>16$, we can take $a$ to be an odd prime factor of $n$ and $b=n / a$ [1 mark], unless $b<3$ or $b=a$.

Case (i): $b<3$. Since $n$ is composite, this means $b=2$, so that $2 a=n>2 p_{k}$. As $a$ is a prime number and $p_{k}$ is the largest prime number which is strictly less than $k$, it follows that $a \geq k$. From $n-k=2 a-k \geq$ $2 a-a=a>2$ we see that $n=2 a$ divides into $(n-k)$. [ 2 marks]

Case (ii): $b=a$. Then $n=a^{2}$ and $a>6$ since $n \geq 26$. Thus $n-k>n / 3=a^{2} / 3>2 a$, so that both $a$ and $2 a$ appear among $\{1,2, \ldots, n-k\}$. Hence $n=a^{2}$ divides into $(n-k)!$. [2 marks]
4. Let $a, b, c$ be the sides of a triangle, with $a+b+c=1$, and let $n \geq 2$ be an integer. Show that

$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<1+\frac{\sqrt[n]{2}}{2}
$$

## Solution.

Without loss of generality, assume $a \leq b \leq c$. As $a+b>c$, we have

$$
\frac{\sqrt[n]{2}}{2}=\frac{\sqrt[n]{2}}{2}(a+b+c)>\frac{\sqrt[n]{2}}{2}(c+c)=\sqrt[n]{2 c^{n}} \geq \sqrt[n]{b^{n}+c^{n}} \quad \quad[2 \text { marks }]
$$

As $a \leq c$ and $n \geq 2$, we have

$$
\begin{aligned}
\left(c^{n}+a^{n}\right)-\left(c+\frac{a}{2}\right)^{n} & =a^{n}-\sum_{k=1}^{n}\binom{n}{k} c^{n-k}\left(\frac{a}{2}\right)^{k} \\
& \leq\left[1-\sum_{k=1}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n} \quad\left(\text { since } c^{n-k} \geq a^{n-k}\right) \\
& =\left[\left(1-\frac{n}{2}\right)-\sum_{k=2}^{n}\binom{n}{k}\left(\frac{1}{2}\right)^{k}\right] a^{n}<0
\end{aligned}
$$

Thus

$$
\sqrt[n]{c^{n}+a^{n}}<c+\frac{a}{2} . \quad[3 \text { marks }]
$$

Likewise

$$
\sqrt[n]{b^{n}+a^{n}}<b+\frac{a}{2} \quad \quad[1 \text { mark }]
$$

Adding (1), (2) and (3), we get

$$
\sqrt[n]{a^{n}+b^{n}}+\sqrt[n]{b^{n}+c^{n}}+\sqrt[n]{c^{n}+a^{n}}<\frac{\sqrt[n]{2}}{2}+c+\frac{a}{2}+b+\frac{a}{2}=1+\frac{\sqrt[n]{2}}{2} . \quad[1 \text { mark }]
$$

5. Given two positive integers $m$ and $n$, find the smallest positive integer $k$ such that among any $k$ people, either there are $2 m$ of them who form $m$ pairs of mutually acquainted people or there are $2 n$ of them forming $n$ pairs of mutually unacquainted people.

## Solution.

Let the smallest positive integer $k$ satisfying the condition of the problem be denoted $r(m, n)$. We shall show that

$$
r(m, n)=2(m+n)-\min \{m, n\}-1
$$

Observe that, by symmetry, $r(m, n)=r(n, m)$. Therefore it suffices to consider the case where $m \geq n$, and to prove that

$$
r(m, n)=2 m+n-1 . \quad[1 \text { mark }]
$$

First we prove that

$$
r(m, n) \geq 2 m+n-1
$$

by an example. Call a group of $k$ people, every two of whom are mutually acquainted, a $k$-clique. Consider a set of $2 m+n-2$ people consisting of a $(2 m-1)$-clique together with an additional $n-1$ people none of whom know anyone else. (Call such people isolated.) Then there are not $2 m$ people forming $m$ mutually acquainted pairs, and there also are not $2 n$ people forming $n$ mutually unacquainted pairs. Thus $r(m, n) \geq$ $(2 m-1)+(n-1)+1=2 m+n-1$ by the definition of $r(m, n)$. [1 mark]

To establish (1), we need to prove that $r(m, n) \leq 2 m+n-1$. To do this, we now show that

$$
r(m, n) \leq r(m-1, n-1)+3 \quad \text { for all } m \geq n \geq 2
$$

Let $G$ be a group of $t=r(m-1, n-1)+3$ people. Notice that

$$
t \geq 2(m-1)+(n-1)-1+3=2 m+n-1 \geq 2 m \geq 2 n
$$

If $G$ is a $t$-clique, then $G$ contains $2 m$ people forming $m$ mutually acquainted pairs, and if $G$ has only isolated people, then $G$ contains $2 n$ people forming $n$ mutually unacquainted pairs. Otherwise, there are three people in $G$, say $a, b$ and $c$, such that $a, b$ are acquainted but $a, c$ are not. Now consider the group $A$ obtained byremoving $a, b$ and $c$ from $G$. A has $t-3=r(m-1, n-1)$ people, so by the definition of $r(m-1, n-1)$, A either contains $2(m-1)$ people forming $m-1$ mutually acquainted pairs, or else contains $2(n-1)$ people forming $n-1$ mutually unacquainted pairs. In the former case, we add the acquainted pair $a, b$ to $A$ to form $m$ mutually acquainted pairs in $G$. In the latter case, we add the unacquainted pair $a, c$ to $A$ to form $n$ mutually unacquainted pairs in $G$. This proves (2). [3 marks]

Trivially, $r(s, 1)=2 s$ for all $s[\mathbf{1}$ mark], so $r(m, n) \leq 2 m+n-1$ holds whenever $n=1$. Proceeding by induction on $n$, by (2) we obtain

$$
r(m, n) \leq r(m-1, n-1)+3 \leq 2(m-1)+(n-1)-1+3=2 m+n-1
$$

which completes the proof. [1 mark]
Note. Give an additional 1 mark to any student who gets at most 5 marks by the above marking scheme, but in addition gives a valid argument that $r(2,2)=5$.