# 2002 Canadian Mathematical Olympiad Solutions 

1. Let $S$ be a subset of $\{1,2, \ldots, 9\}$, such that the sums formed by adding each unordered pair of distinct numbers from $S$ are all different. For example, the subset $\{1,2,3,5\}$ has this property, but $\{1,2,3,4,5\}$ does not, since the pairs $\{1,4\}$ and $\{2,3\}$ have the same sum, namely 5 .

What is the maximum number of elements that $S$ can contain?

## Solution 1

It can be checked that all the sums of pairs for the set $\{1,2,3,5,8\}$ are different.

Suppose, for a contradiction, that $S$ is a subset of $\{1, \ldots, 9\}$ containing 6 elements such that all the sums of pairs are different. Now the smallest possible sum for two numbers from $S$ is $1+2=3$ and the largest possible sum is $8+9=17$. That gives 15 possible sums: $3, \ldots, 17$.

Also there are $\left(\begin{array}{l}6 \\ 2\end{array}\right)=15$ pairs from $S$. Thus, each of $3, \ldots, 17$ is the sum of exactly one pair. The only pair from $\{1, \ldots, 9\}$ that adds to 3 is $\{1,2\}$ and to 17 is $\{8,9\}$. Thus $1,2,8,9$ are in $S$. But then $1+9=2+8$, giving a contradiction. It follows that the maximum number of elements that $S$ can contain is 5 .

## Solution 2.

It can be checked that all the sums of pairs for the set $\{1,2,3,5,8\}$ are different.

Suppose, for a contradiction, that $S$ is a subset of $\{1, \ldots 9\}$ such that all the sums of pairs are different and that $a_{1}<a_{2}<\ldots<a_{6}$ are the members of $S$.

Since $a_{1}+a_{6} \neq a_{2}+a_{5}$, it follows that $a_{6}-a_{5} \neq a_{2}-a_{1}$. Similarly $a_{6}-a_{5} \neq a_{4}-a_{3}$ and $a_{4}-a_{3} \neq a_{2}-a_{1}$. These three differences must be distinct positive integers, so,

$$
\left(a_{6}-a_{5}\right)+\left(a_{4}-a_{3}\right)+\left(a_{2}-a_{1}\right) \geq 1+2+3=6 .
$$

Similarly $a_{3}-a_{2} \neq a_{5}-a_{4}$, so

$$
\left(a_{3}-a_{2}\right)+\left(a_{5}-a_{4}\right) \geq 1+2=3 .
$$

Adding the above 2 inequalities yields

$$
a_{6}-a_{5}+a_{5}-a_{4}+a_{4}-a_{3}+a_{3}-a_{2}+a_{2}-a_{1} \geq 6+3=9,
$$

and hence $a_{6}-a_{1} \geq 9$. This is impossible since the numbers in $S$ are between 1 and 9 .

2. Call a positive integer $n$ practical if every positive integer less than or equal to $n$ can be written as the sum of distinct divisors of $n$.

For example, the divisors of 6 are $\mathbf{1 , 2}, \mathbf{3}$, and $\mathbf{6}$. Since

$$
1=\mathbf{1}, \quad 2=\mathbf{2}, \quad 3=\mathbf{3}, \quad 4=\mathbf{1}+\mathbf{3}, \quad 5=\mathbf{2}+\mathbf{3}, \quad 6=\mathbf{6},
$$

we see that 6 is practical.

Prove that the product of two practical numbers is also practical.

## Solution

Let $p$ and $q$ be practical. For any $k \leq p q$, we can write

$$
k=a q+b \text { with } 0 \leq a \leq p, 0 \leq b<q .
$$

Since $p$ and $q$ are practical, we can write

$$
a=c_{1}+\ldots+c_{m}, \quad b=d_{1}+\ldots+d_{n}
$$

where the $c_{i}$ 's are distinct divisors of $p$ and the $d_{j}$ 's are distinct divisors of $q$. Now

$$
\begin{aligned}
k & =\left(c_{1}+\ldots+c_{m}\right) q+\left(d_{1}+\ldots+d_{n}\right) \\
& =c_{1} q+\ldots+c_{m} q+d_{1}+\ldots+d_{n} .
\end{aligned}
$$

Each of $c_{i} q$ and $d_{j}$ divides $p q$. Since $d_{j}<q \leq c_{i} q$ for any $i, j$, the $c_{i} q$ 's and $d_{j}$ 's are all distinct, and we conclude that $p q$ is practical.

3. Prove that for all positive real numbers $a, b$, and $c$,

$$
\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b} \geq a+b+c
$$

and determine when equality occurs.

Each of the inequalities used in the solutions below has the property that equality holds if and only if $a=b=c$. Thus equality holds for the given inequality if and only if $a=b=c$.

## Solution 1.

Note that $a^{4}+b^{4}+c^{4}=\frac{\left(a^{4}+b^{4}\right)}{2}+\frac{\left(b^{4}+c^{4}\right)}{2}+\frac{\left(c^{4}+a^{4}\right)}{2}$. Applying the arithmetic-geometric mean inequality to each term, we see that the right side is greater than or equal to

$$
a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2} \text {. }
$$

We can rewrite this as

$$
\frac{a^{2}\left(b^{2}+c^{2}\right)}{2}+\frac{b^{2}\left(c^{2}+a^{2}\right)}{2}+\frac{c^{2}\left(a^{2}+b^{2}\right)}{2} .
$$

Applying the arithmetic mean-geometric mean inequality again we obtain $a^{4}+b^{4}+c^{4} \geq$ $a^{2} b c+b^{2} c a+c^{2} a b$. Dividing both sides by $a b c$ (which is positive) the result follows.

## Solution 2.

Notice the inequality is homogeneous. That is, if $a, b, c$ are replaced by $k a, k b, k c, k>0$ we get the original inequality. Thus we can assume, without loss of generality, that $a b c=1$. Then

$$
\begin{aligned}
\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b} & =a b c\left(\frac{a^{3}}{b c}+\frac{b^{3}}{c a}+\frac{c^{3}}{a b}\right) \\
& =a^{4}+b^{4}+c^{4} .
\end{aligned}
$$

So we need prove that $a^{4}+b^{4}+c^{4} \geq a+b+c$.

By the Power Mean Inequality,

$$
\frac{a^{4}+b^{4}+c^{4}}{3} \geq\left(\frac{a+b+c}{3}\right)^{4}
$$

so $a^{4}+b^{4}+c^{4} \geq(a+b+c) \cdot \frac{(a+b+c)^{3}}{27}$.

By the arithmetic mean-geometric mean inequality, $\frac{a+b+c}{3} \geq \sqrt[3]{a b c}=1$, so $a+b+c \geq 3$.

Hence, $a^{4}+b^{4}+c^{4} \geq(a+b+c) \cdot \frac{(a+b+c)^{3}}{27} \geq(a+b+c) \frac{3^{3}}{27}=a+b+c$.

## Solution 3.

Rather than using the Power-Mean inequality to prove $a^{4}+b^{4}+c^{4} \geq a+b+c$ in Proof 2, the Cauchy-Schwartz-Bunjakovsky inequality can be used twice:

$$
\begin{aligned}
& \left(a^{4}+b^{4}+c^{4}\right)\left(1^{2}+1^{2}+1^{2}\right) \geq\left(a^{2}+b^{2}+c^{2}\right)^{2} \\
& \left(a^{2}+b^{2}+c^{2}\right)\left(1^{2}+1^{2}+1^{2}\right) \geq(a+b+c)^{2}
\end{aligned}
$$

So $\frac{a^{4}+b^{4}+c^{4}}{3} \geq \frac{\left(a^{2}+b^{2}+c^{2}\right)^{2}}{9} \geq \frac{(a+b+c)^{4}}{81}$. Continue as in Proof 2 .

4. Let $\Gamma$ be a circle with radius $r$. Let $A$ and $B$ be distinct points on $\Gamma$ such that $A B<\sqrt{3} r$. Let the circle with centre $B$ and radius $A B$ meet $\Gamma$ again at $C$. Let $P$ be the point inside $\Gamma$ such that triangle $A B P$ is equilateral. Finally, let $C P$ meet $\Gamma$ again at $Q$. Prove that $P Q=r$.

![](https://cdn.mathpix.com/cropped/2024_04_17_74e70b301e0241021e8cg-4.jpg?height=569&width=653&top_left_y=469&top_left_x=736)

## Solution 1.

Let the center of $\Gamma$ be $O$, the radius $\mathrm{r}$. Since $B P=B C$, let $\theta=\measuredangle B P C=\measuredangle B C P$.

Quadrilateral $Q A B C$ is cyclic, so $\measuredangle B A Q=180^{\circ}-\theta$ and hence $\measuredangle P A Q=120^{\circ}-\theta$.

Also $\measuredangle A P Q=180^{\circ}-\measuredangle A P B-\measuredangle B P C=120^{\circ}-\theta$, so $P Q=A Q$ and $\measuredangle A Q P=2 \theta-60^{\circ}$.

Again because quadrilateral $Q A B C$ is cyclic, $\measuredangle A B C=180^{\circ}-\measuredangle A Q C=240^{\circ}-2 \theta$.

Triangles $O A B$ and $O C B$ are congruent, since $O A=O B=O C=r$ and $A B=B C$.

Thus $\measuredangle A B O=\measuredangle C B O=\frac{1}{2} \measuredangle A B C=120^{\circ}-\theta$.

We have now shown that in triangles $A Q P$ and $A O B, \measuredangle P A Q=\measuredangle B A O=\measuredangle A P Q=\measuredangle A B O$.

Also $A P=A B$, so $\triangle A Q P \cong \triangle A O B$. Hence $Q P=O B=r$.

## Solution 2.

Let the center of $\Gamma$ be $O$, the radius $r$. Since $A, P$ and $C$ lie on a circle centered at $B$, $60^{\circ}=\measuredangle A B P=2 \measuredangle A C P$, so $\measuredangle A C P=\measuredangle A C Q=30^{\circ}$.

Since $Q, A$, and $C$ lie on $\Gamma, \measuredangle Q O A=2 \measuredangle Q C A=60^{\circ}$.

So $Q A=r$ since if a chord of a circle subtends an angle of $60^{\circ}$ at the center, its length is the radius of the circle.

Now $B P=B C$, so $\measuredangle B P C=\measuredangle B C P=\measuredangle A C B+30^{\circ}$.

Thus $\measuredangle A P Q=180^{\circ}-\measuredangle A P B-\measuredangle B P C=90^{\circ}-\measuredangle A C B$.

Since $Q, A, B$ and $C$ lie on $\Gamma$ and $A B=B C, \measuredangle A Q P=\measuredangle A Q C=\measuredangle A Q B+\measuredangle B Q C=2 \measuredangle A C B$.

Finally, $\measuredangle Q A P=180-\measuredangle A Q P-\measuredangle A P Q=90-\measuredangle A C B$.

So $\measuredangle P A Q=\measuredangle A P Q$ hence $P Q=A Q=r$.

5. Let $\mathbb{N}=\{0,1,2, \ldots\}$. Determine all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that

$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right)
$$

for all $x$ and $y$ in $\mathbb{N}$.

## Solution 1.

We claim that $f$ is a constant function. Suppose, for a contradiction, that there exist $x$ and $y$ with $f(x)<f(y)$; choose $x, y$ such that $f(y)-f(x)>0$ is minimal. Then

$$
f(x)=\frac{x f(x)+y f(x)}{x+y}<\frac{x f(y)+y f(x)}{x+y}<\frac{x f(y)+y f(y)}{x+y}=f(y)
$$

so $f(x)<f\left(x^{2}+y^{2}\right)<f(y)$ and $0<f\left(x^{2}+y^{2}\right)-f(x)<f(y)-f(x)$, contradicting the choice of $x$ and $y$. Thus, $f$ is a constant function. Since $f(0)$ is in $\mathbb{N}$, the constant must be from $\mathbb{N}$.

Also, for any $c$ in $\mathbb{N}, x c+y c=(x+y) c$ for all $x$ and $y$, so $f(x)=c, c \in \mathbb{N}$ are the solutions to the equation.

## Solution 2.

We claim $f$ is a constant function. Define $g(x)=f(x)-f(0)$. Then $g(0)=0, g(x) \geq-f(0)$ and

$$
x g(y)+y g(x)=(x+y) g\left(x^{2}+y^{2}\right)
$$

for all $x, y$ in $\mathbb{N}$.

Letting $y=0$ shows $g\left(x^{2}\right)=0$ (in particular, $g(1)=g(4)=0$ ), and letting $x=y=1$ shows $g(2)=0$. Also, if $x, y$ and $z$ in $\mathbb{N}$ satisfy $x^{2}+y^{2}=z^{2}$, then

$$
g(y)=-\frac{y}{x} g(x) .
$$

Letting $x=4$ and $y=3,(*)$ shows that $g(3)=0$.

For any even number $x=2 n>4$, let $y=n^{2}-1$. Then $y>x$ and $x^{2}+y^{2}=\left(n^{2}+1\right)^{2}$. For any odd number $x=2 n+1>3$, let $y=2(n+1) n$. Then $y>x$ and $x^{2}+y^{2}=\left((n+1)^{2}+n^{2}\right)^{2}$. Thus for every $x>4$ there is $y>x$ such that $(*)$ is satisfied.

Suppose for a contradiction, that there is $x>4$ with $g(x)>0$. Then we can construct a

sequence $x=x_{0}<x_{1}<x_{2}<\ldots$ where $g\left(x_{i+1}\right)=-\frac{x_{i+1}}{x_{i}} g\left(x_{i}\right)$. It follows that $\left|g\left(x_{i+1}\right)\right|>$ $\left|g\left(x_{i}\right)\right|$ and the signs of $g\left(x_{i}\right)$ alternate. Since $g(x)$ is always an integer, $\left|g\left(x_{i+1}\right)\right| \geq\left|g\left(x_{i}\right)\right|+1$. Thus for some sufficiently large value of $i, g\left(x_{i}\right)<-f(0)$, a contradiction.

As for Proof 1, we now conclude that the functions that satisfy the given functional equation are $f(x)=c, c \in \mathbb{N}$.

Solution 3. Suppose that $W$ is the set of nonnegative integers and that $f: W \rightarrow W$ satisfies:

$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right) .
$$

We will show that $f$ is a constant function.

Let $f(0)=k$, and set $S=\{x \mid f(x)=k\}$.

Letting $y=0$ in $(*)$ shows that $f\left(x^{2}\right)=k \quad \forall x>0$, and so

$$
x^{2} \in S \quad \forall x>0
$$

In particular, $1 \in S$.

Suppose $x^{2}+y^{2}=z^{2}$. Then $y f(x)+x f(y)=(x+y) f\left(z^{2}\right)=(x+y) k$. Thus,

$$
x \in S \quad \text { iff } \quad y \in S
$$

whenever $x^{2}+y^{2}$ is a perfect square.

For a contradiction, let $n$ be the smallest non-negative integer such that $f\left(2^{n}\right) \neq k$. By (l) $n$ must be odd, so $\frac{n-1}{2}$ is an integer. Now $\frac{n-1}{2}<n$ so $f\left(2^{\frac{n-1}{2}}\right)=k$. Letting $x=y=2^{\frac{n-1}{2}}$ in (*) shows $f\left(2^{n}\right)=k$, a contradiction. Thus every power of 2 is an element of $S$.

For each integer $n \geq 2$ define $p(n)$ to be the largest prime such that $p(n) \mid n$.

Claim: For any integer $n>1$ that is not a power of 2 , there exists a sequence of integers $x_{1}, x_{2}, \ldots, x_{r}$ such that the following conditions hold:
a) $x_{1}=n$.
b) $x_{i}^{2}+x_{i+1}^{2}$ is a perfect square for each $i=1,2,3, \ldots, r-1$.
c) $p\left(x_{1}\right) \geq p\left(x_{2}\right) \geq \ldots \geq p\left(x_{r}\right)=2$.

Proof: Since $n$ is not a power of $2, p(n)=p\left(x_{1}\right) \geq 3$. Let $p\left(x_{1}\right)=2 m+1$, so $n=x_{1}=$ $b(2 m+1)^{a}$, for some $a$ and $b$, where $p(b)<2 m+1$.

Case 1: $a=1$. Since $\left(2 m+1,2 m^{2}+2 m, 2 m^{2}+2 m+1\right)$ is a Pythagorean Triple, if $x_{2}=b\left(2 m^{2}+\right.$ $2 m)$, then $x_{1}^{2}+x_{2}^{2}=b^{2}\left(2 m^{2}+2 m+1\right)^{2}$ is a perfect square. Furthermore, $x_{2}=2 b m(m+1)$, and so $p\left(x_{2}\right)<2 m+1=p\left(x_{1}\right)$.

Case 2: $a>1$. If $n=x_{1}=(2 m+1)^{a} \cdot b$, let $x_{2}=(2 m+1)^{a-1} \cdot b \cdot\left(2 m^{2}+2 m\right), x_{3}=$ $(2 m+1)^{a-2} \cdot b \cdot\left(2 m^{2}+2 m\right)^{2}, \ldots, x_{a+1}=(2 m+1)^{0} \cdot b \cdot\left(2 m^{2}+2 m\right)^{a}=b \cdot 2^{a} m^{a}(m+1)^{a}$. Note that for $1 \leq i \leq a, x_{i}^{2}+x_{i+1}^{2}$ is a perfect square and also note that $p\left(x_{a+1}\right)<2 m+1=p\left(x_{1}\right)$.

If $x_{a+1}$ is not a power of 2 , we extend the sequence $x_{i}$ using the same procedure described above. We keep doing this until $p\left(x_{r}\right)=2$, for some integer $r$.

By (2), $x_{i} \in S$ iff $x_{i+1} \in S$ for $i=1,2,3, \ldots, r-1$. Thus, $n=x_{1} \in S$ iff $x_{r} \in S$. But $x_{r}$ is a power of 2 because $p\left(x_{r}\right)=2$, and we earlier proved that powers of 2 are in $\mathrm{S}$. Therefore, $n \in S$, proving the claim.

We have proven that every integer $n \geq 1$ is an element of $S$, and so we have proven that $f(n)=k=f(0)$, for each $n \geq 1$. Therefore, $f$ is constant, Q.E.D.