# SOLUTIONS 

## QUESTION 1

Solution

Note that

$$
f(1-x)=\frac{9^{1-x}}{9^{1-x}+3}=\frac{9}{9+3 \times 9^{x}}=\frac{3}{9^{x}+3},
$$

from which we get

$$
f(x)+f(1-x)=\frac{9^{x}}{9^{x}+3}+\frac{3}{9^{x}+3}=1 .
$$

Therefore,

$$
\begin{aligned}
& \sum_{k=1}^{1995} f\left(\frac{k}{1996}\right) \\
& \quad=\sum_{k=1}^{997}\left[f\left(\frac{k}{1996}\right)+f\left(\frac{1996-k}{1996}\right)\right]+f\left(\frac{998}{1996}\right) \\
& \quad=\sum_{k=1}^{997}\left[f\left(\frac{k}{1996}\right)+f\left(1-\frac{k}{1996}\right)\right]+f\left(\frac{1}{2}\right) \\
& \quad=997+\frac{3}{3+3}=997 \frac{1}{2} .
\end{aligned}
$$

## QUESTION 2

## Solution 1.

We prove equivalently that $a^{3 a} b^{3 b} c^{3 c} \geq(a b c)^{a+b+c}$. Due to complete symmetry in $a, b$ and $c$, we may assume, without loss of generality, that $a \geq b \geq c$. Then $a-b \geq 0, b-c \geq 0, a-c \geq 0$ and $\frac{a}{b} \geq 1, \frac{b}{c} \geq 1, \frac{a}{c} \geq 1$. Therefore,

$$
\frac{a^{3 a} b^{3 b} c^{3 c}}{(a b c)^{a+b+c}}=\left(\frac{a}{b}\right)^{a-b}\left(\frac{b}{c}\right)^{b-c}\left(\frac{a}{c}\right)^{a-c} \geq 1
$$

## Solution 2.

If we assign the weights $a, b, c$ to the numbers $a, b, c$, respectively, then by the wighted geometric-mean-harmonic-mean inequality followed by the arithmeticmean-geometric-mean inequality, we get

$$
\sqrt[a+b+c]{a^{a} b^{b} c^{c}} \geq \frac{a+b+c}{\frac{a}{a}+\frac{b}{b}+\frac{c}{c}}=\frac{a+b+c}{3} \geq \sqrt[3]{a b c}
$$

from which $a^{a} b^{b} c^{c} \geq(a b c)^{\frac{a+b+c}{3}}$ follows immediately.

## QUESTION 3

## Solution

For convenience, the interior angle in a boomerang which is greater than $180^{\circ}$ will be called a "reflex angle".

Clearly, there are $b$ reflex angles, each occurring in a different boomerang and each with the corresponding vertex in the interior of $C$. Angles around these vertices add up to $2 b \pi$. On the other hand, the sum of all the interior angles of $C$ is $(s-2) \pi$ and the sum of the interior angles of all the $q$ quadrilaterals is $2 \pi q$.

Therefore, $2 \pi q \geq 2 b \pi+(s-2) \pi$ from which $q \geq b+\frac{s-2}{2}$ follows.

## QUESTION 4

## Solution 1.

Since $1^{3}+2^{3}+\cdots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2}$, we see that when $k=0,\left(x_{1}, x_{2}, \cdots, x_{n} ; y\right)=$ $\left(1,2, \cdots, n ; \frac{n(n+1)}{2}\right)$ is a solution. To see that we can generate infinitely many solutions in general, set $c=\frac{n(n+1)}{2}$ and notice that for all positive integers $q$, we have:

$$
\begin{aligned}
& \left(c^{k} q^{3 k+2}\right)^{3}+\left(2 c^{k} q^{3 k+2}\right)^{3}+\cdots+\left(n c^{k} q^{3 k+2}\right)^{3} \\
& \quad=c^{3 k} q^{3(3 k+2)}\left(1^{3}+2^{3}+\cdots n^{3}\right) \\
& \quad=c^{3 k} q^{3(3 k+2)}\left(\frac{n(n+1)}{2}\right)^{2} \\
& \quad=c^{3 k+2} q^{3(3 k+2)}=\left(c q^{3}\right)^{3 k+2} .
\end{aligned}
$$

That is, $\left(x_{1}, x_{2}, \cdots, x_{n} ; y\right)=\left(c^{k} q^{3 k+2}, 2 c^{k} q^{3 k+2}, \cdots, n c^{k} q^{3 k+2} ; c q^{3}\right)$ is a solution. This completes the proof.

## Solution 2.

For any positive integer $q$, take $x_{1}=x_{2}=\cdots=x_{n}=n^{2 k+1} q^{3 k+2}, y=n^{2} q^{3}$. Then

$$
\sum_{i=1}^{n} x_{i}^{3}=n \cdot n^{6 k+3} \cdot q^{9 k+6}=\left(n^{2} q^{3}\right)^{3 k+2}=y^{3 k+2}
$$

## Solution 3.

If $n=1$, take $x_{1}=q^{3 k+2}, y=q^{3}$ as in solution 2. For $n>1$, we look for solutions of the form

$$
x_{1}=x_{2}=\cdots x_{n}=n^{p}, y=n^{q} .
$$

Then

$$
\sum_{i=1}^{n} x_{i}^{3}=y^{3 k+2} \Leftrightarrow n^{3 p+1}=n^{(3 k+2) q} \Leftrightarrow 3 p+1=(3 k+2) q \Leftrightarrow(3 k+2) q-3 p=1 .
$$

The last equation is satisfied if we take

$$
q=3 t+2 \text { and } p=(3 k+2) t+(2 k+1) \text { where } t
$$

is any nonnegative integer. Thus, infinetely many solutions in positive integers are given by

$$
x_{1}=x_{2}=\cdots x_{n}=n^{(3 k+2) t+(2 k+1)}, y=n^{3 t+2}
$$

## SOLUTIONS (Cont'd)

## QUESTION 5

## Solution

Note first that $u_{1}=1-u$. Since for all $x \in[u, 1], u \leq x$ and $1-x \leq 1-u$ we have

$$
\begin{aligned}
1- & (\sqrt{u x}+\sqrt{(1-u)(1-x)})^{2} \\
& =1-u x-(1-u)(1-x)-2 \sqrt{u x(1-u)(1-x)} \\
& =u+x-2 u x-2 \sqrt{u x(1-u)(1-x)} \\
& \leq u+x-2 u x-2 u(1-x)=x-u .
\end{aligned}
$$

Therefore,

$$
f(x)=0 \text { if } 0 \leq x \leq u
$$

and

$$
f(x) \leq x-u \text { if } u \leq x \leq 1
$$

From (2) we get $u_{2}=f\left(u_{1}\right) \leq u_{1}-u=1-2 u$ if $u_{1} \geq u$. An easy induction then yields

$$
u_{n+1}=f\left(u_{n}\right) \leq u_{n}-u \leq 1-(n+1) u \text { if } u_{i} \geq u \text { for all } i=1,2, \cdots, n
$$

Thus for sufficiently large $k$, we must have $u_{k-1}<u$ and then $u_{k}=f\left(u_{k-1}\right)=0$ by (1).