# Solutions to the 2005 CMO 

written March 30, 2005

1. Consider an equilateral triangle of side length $n$, which is divided into unit triangles, as shown. Let $f(n)$ be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) or revisits a triangle. An example of one such path is illustrated below for $n=5$. Determine the value of $f(2005)$.

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-1.jpg?height=325&width=371&top_left_y=824&top_left_x=926)

## Solution

We shall show that $f(n)=(n-1)$ !.

Label the horizontal line segments in the triangle $l_{1}, l_{2}, \ldots$ as in the diagram below. Since the path goes from the top triangle to a triangle in the bottom row and never travels up, the path must cross each of $l_{1}, l_{2}, \ldots, l_{n-1}$ exactly once. The diagonal lines in the triangle divide $l_{k}$ into $k$ unit line segments and the path must cross exactly one of these $k$ segments for each $k$. (In the diagram below, these line segments have been highlighted.) The path is completely determined by the set of $n-1$ line segments which are crossed. So as the path moves from the $k$ th row to the $(k+1)$ st row, there are $k$ possible line segments where the path could cross $l_{k}$. Since there are $1 \cdot 2 \cdot 3 \cdots(n-1)=(n-1)$ ! ways that the path could cross the $n-1$ horizontal lines, and each one corresponds to a unique path, we get $f(n)=(n-1)$ !.

Therefore $f(2005)=(2004)$ !.

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-1.jpg?height=322&width=384&top_left_y=1975&top_left_x=903)

2. Let $(a, b, c)$ be a Pythagorean triple, i.e., a triplet of positive integers with $a^{2}+b^{2}=c^{2}$.

a) Prove that $(c / a+c / b)^{2}>8$.

b) Prove that there does not exist any integer $n$ for which we can find a Pythagorean triple $(a, b, c)$ satisfying $(c / a+c / b)^{2}=n$.

## a) Solution 1

Let $(a, b, c)$ be a Pythagorean triple. View $a, b$ as lengths of the legs of a right angled triangle with hypotenuse of length $c$; let $\theta$ be the angle determined by the sides with lengths $a$ and $c$. Then

$$
\begin{aligned}
\left(\frac{c}{a}+\frac{c}{b}\right)^{2} & =\left(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\right)^{2}=\frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta}{(\sin \theta \cos \theta)^{2}} \\
& =4\left(\frac{1+\sin 2 \theta}{\sin ^{2} 2 \theta}\right)=\frac{4}{\sin ^{2} 2 \theta}+\frac{4}{\sin 2 \theta}
\end{aligned}
$$

Note that because $0<\theta<90^{\circ}$, we have $0<\sin 2 \theta \leq 1$, with equality only if $\theta=45^{\circ}$. But then $a=b$ and we obtain $\sqrt{2}=c / a$, contradicting $a, c$ both being integers. Thus, $0<\sin 2 \theta<1$ which gives $(c / a+c / b)^{2}>8$.

## Solution 2

Defining $\theta$ as in Solution 1, we have $c / a+c / b=\sec \theta+\csc \theta$. By the AM-GM inequality, we have $(\sec \theta+\csc \theta) / 2 \geq \sqrt{\sec \theta \csc \theta}$. So

$$
c / a+c / b \geq \frac{2}{\sqrt{\sin \theta \cos \theta}}=\frac{2 \sqrt{2}}{\sqrt{\sin 2 \theta}} \geq 2 \sqrt{2}
$$

Since $a, b, c$ are integers, we have $c / a+c / b>2 \sqrt{2}$ which gives $(c / a+c / b)^{2}>8$.

## Solution 3

By simplifying and using the AM-GM inequality,

$$
\left(\frac{c}{a}+\frac{c}{b}\right)^{2}=c^{2}\left(\frac{a+b}{a b}\right)^{2}=\frac{\left(a^{2}+b^{2}\right)(a+b)^{2}}{a^{2} b^{2}} \geq \frac{2 \sqrt{a^{2} b^{2}}(2 \sqrt{a b})^{2}}{a^{2} b^{2}}=8
$$

with equality only if $a=b$. By using the same argument as in Solution $1, a$ cannot equal $b$ and the inequality is strict.

## Solution 4

$$
\begin{aligned}
\left(\frac{c}{a}+\frac{c}{b}\right)^{2} & =\frac{c^{2}}{a^{2}}+\frac{c^{2}}{b^{2}}+\frac{2 c^{2}}{a b}=1+\frac{b^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}+1+\frac{2\left(a^{2}+b^{2}\right)}{a b} \\
& =2+\left(\frac{a}{b}-\frac{b}{a}\right)^{2}+2+\frac{2}{a b}\left((a-b)^{2}+2 a b\right) \\
& =4+\left(\frac{a}{b}-\frac{b}{a}\right)^{2}+\frac{2(a-b)^{2}}{a b}+4 \geq 8,
\end{aligned}
$$

with equality only if $a=b$, which (as argued previously) cannot occur.

## b) Solution 1

Since $c / a+c / b$ is rational, $(c / a+c / b)^{2}$ can only be an integer if $c / a+c / b$ is an integer. Suppose $c / a+c / b=m$. We may assume that $\operatorname{gcd}(a, b)=1$. (If not, divide the common factor from $(a, b, c)$, leaving $m$ unchanged.)

Since $c(a+b)=m a b$ and $\operatorname{gcd}(a, a+b)=1$, $a$ must divide $c$, say $c=a k$. This gives $a^{2}+b^{2}=a^{2} k^{2}$ which implies $b^{2}=\left(k^{2}-1\right) a^{2}$. But then $a$ divides $b$ contradicting the fact that $\operatorname{gcd}(a, b)=1$. Therefore $(c / a+c / b)^{2}$ is not equal to any integer $n$.

## Solution 2

We begin as in Solution 1, supposing that $c / a+c / b=m$ with $\operatorname{gcd}(a, b)=1$. Hence $a$ and $b$ are not both even. It is also the case that $a$ and $b$ are not both odd, for then $c^{2}=a^{2}+b^{2} \equiv 2(\bmod 4)$, and perfect squares are congruent to either 0 or 1 modulo 4. So one of $a, b$ is odd and the other is even. Therefore $c$ must be odd.

Now $c / a+c / b=m$ implies $c(a+b)=m a b$, which cannot be true because $c(a+b)$ is odd and mab is even.

3. Let $S$ be a set of $n \geq 3$ points in the interior of a circle.

a) Show that there are three distinct points $a, b, c \in S$ and three distinct points $A, B, C$ on the circle such that $a$ is (strictly) closer to $A$ than any other point in $S, b$ is closer to $B$ than any other point in $S$ and $c$ is closer to $C$ than any other point in $S$.

b) Show that for no value of $n$ can four such points in $S$ (and corresponding points on the circle) be guaranteed.

## Solution 1

a) Let $H$ be the smallest convex set of points in the plane which contains $S . \dagger$ Take 3 points $a, b, c \in S$ which lie on the boundary of $H$. (There must always be at least 3 (but not necessarily 4) such points.)

Since $a$ lies on the boundary of the convex region $H$, we can construct a chord $L$ such that no two points of $H$ lie on opposite sides of $L$. Of the two points where the perpendicular to $L$ at $a$ meets the circle, choose one which is on a side of $L$ not containing any points of $H$ and call this point $A$. Certainly $A$ is closer to $a$ than to any other point on $L$ or on the other side of $L$. Hence $A$ is closer to $a$ than to any other point of $S$. We can find the required points $B$ and $C$ in an analogous way and the proof is complete.

[Note that this argument still holds if all the points of $S$ lie on a line.]

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-4.jpg?height=428&width=433&top_left_y=1434&top_left_x=629)

(a)

![](https://cdn.mathpix.com/cropped/2024_04_17_af19c08dc8a5848f069fg-4.jpg?height=458&width=436&top_left_y=1403&top_left_x=1251)

(b)

b) Let $P Q R$ be an equilateral triangle inscribed in the circle and let $a, b, c$ be midpoints of the three sides of $\triangle P Q R$. If $r$ is the radius of the circle, then every point on the circle is within $(\sqrt{3} / 2) r$ of one of $a, b$ or $c$. (See figure (b) above.) Now $\sqrt{3} / 2<9 / 10$, so if $S$ consists of $a, b, c$ and a cluster of points within $r / 10$ of the centre of the circle, then we cannot select 4 points from $S$ (and corresponding points on the circle) having the desired property.[^0]

## Solution 2

a) If all the points of $S$ lie on a line $L$, then choose any 3 of them to be $a, b, c$. Let $A$ be a point on the circle which meets the perpendicular to $L$ at $a$. Clearly $A$ is closer to $a$ than to any other point on $L$, and hence closer than other other point in $S$. We find $B$ and $C$ in an analogous way.

Otherwise, choose $a, b, c$ from $S$ so that the triangle formed by these points has maximal area. Construct the altitude from the side $b c$ to the point $a$ and extend this line until it meets the circle at $A$. We claim that $A$ is closer to $a$ than to any other point in $S$.

Suppose not. Let $x$ be a point in $S$ for which the distance from $A$ to $x$ is less than the distance from $A$ to $a$. Then the perpendicular distance from $x$ to the line $b c$ must be greater than the perpendicular distance from $a$ to the line $b c$. But then the triangle formed by the points $x, b, c$ has greater area than the triangle formed by $a, b, c$, contradicting the original choice of these 3 points. Therefore $A$ is closer to $a$ than to any other point in $S$.

The points $B$ and $C$ are found by constructing similar altitudes through $b$ and $c$, respectively.

b) See Solution 1.

4. Let $A B C$ be a triangle with circumradius $R$, perimeter $P$ and area $K$. Determine the maximum value of $K P / R^{3}$.

## Solution 1

Since similar triangles give the same value of $K P / R^{3}$, we can fix $R=1$ and maximize $K P$ over all triangles inscribed in the unit circle. Fix points $A$ and $B$ on the unit circle. The locus of points $C$ with a given perimeter $P$ is an ellipse that meets the circle in at most four points. The area $K$ is maximized (for a fixed $P$ ) when $C$ is chosen on the perpendicular bisector of $A B$, so we get a maximum value for $K P$ if $C$ is where the perpendicular bisector of $A B$ meets the circle. Thus the maximum value of $K P$ for a given $A B$ occurs when $A B C$ is an isosceles triangle. Repeating this argument with $B C$ fixed, we have that the maximum occurs when $A B C$ is an equilateral triangle.

Consider an equilateral triangle with side length $a$. It has $P=3 a$. It has height equal to $a \sqrt{3} / 2$ giving $K=a^{2} \sqrt{3} / 4$. ¿From the extended law of sines, $2 R=a / \sin (60)$ giving $R=a / \sqrt{3}$. Therefore the maximum value we seek is

$$
K P / R^{3}=\left(\frac{a^{2} \sqrt{3}}{4}\right)(3 a)\left(\frac{\sqrt{3}}{a}\right)^{3}=\frac{27}{4} .
$$

## Solution 2

From the extended law of sines, the lengths of the sides of the triangle are $2 R \sin A$, $2 R \sin B$ and $2 R \sin C$. So

$$
P=2 R(\sin A+\sin B+\sin C) \text { and } K=\frac{1}{2}(2 R \sin A)(2 R \sin B)(\sin C),
$$

giving

$$
\frac{K P}{R^{3}}=4 \sin A \sin B \sin C(\sin A+\sin B+\sin C)
$$

We wish to find the maximum value of this expression over all $A+B+C=180^{\circ}$. Using well-known identities for sums and products of sine functions, we can write

$$
\frac{K P}{R^{3}}=4 \sin A\left(\frac{\cos (B-C)}{2}-\frac{\cos (B+C)}{2}\right)\left(\sin A+2 \sin \left(\frac{B+C}{2}\right) \cos \left(\frac{B-C}{2}\right)\right) .
$$

If we first consider $A$ to be fixed, then $B+C$ is fixed also and this expression takes its maximum value when $\cos (B-C)$ and $\cos \left(\frac{B-C}{2}\right)$ equal 1 ; i.e. when $B=C$. In a similar way, one can show that for any fixed value of $B, K P / R^{3}$ is maximized when $A=C$. Therefore the maximum value of $K P / R^{3}$ occurs when $A=B=C=60^{\circ}$, and it is now an easy task to substitute this into the above expression to obtain the maximum value of $27 / 4$.

## Solution 3

As in Solution 2, we obtain

$$
\frac{K P}{R^{3}}=4 \sin A \sin B \sin C(\sin A+\sin B+\sin C)
$$

From the AM-GM inequality, we have

$$
\sin A \sin B \sin C \leq\left(\frac{\sin A+\sin B+\sin C}{3}\right)^{3},
$$

giving

$$
\frac{K P}{R^{3}} \leq \frac{4}{27}(\sin A+\sin B+\sin C)^{4}
$$

with equality when $\sin A=\sin B=\sin C$. Since the sine function is concave on the interval from 0 to $\pi$, Jensen's inequality gives

$$
\frac{\sin A+\sin B+\sin C}{3} \leq \sin \left(\frac{A+B+C}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} .
$$

Since equality occurs here when $\sin A=\sin B=\sin C$ also, we can conclude that the maximum value of $K P / R^{3}$ is $\frac{4}{27}\left(\frac{3 \sqrt{3}}{2}\right)^{4}=27 / 4$.

5. Let's say that an ordered triple of positive integers $(a, b, c)$ is $n$-powerful if $a \leq b \leq c$, $\operatorname{gcd}(a, b, c)=1$, and $a^{n}+b^{n}+c^{n}$ is divisible by $a+b+c$. For example, $(1,2,2)$ is 5 -powerful.

a) Determine all ordered triples (if any) which are $n$-powerful for all $n \geq 1$.

b) Determine all ordered triples (if any) which are 2004-powerful and 2005-powerful, but not 2007-powerful.

[Note that $\operatorname{gcd}(a, b, c)$ is the greatest common divisor of $a, b$ and $c$.]

## Solution 1

Let $T_{n}=a^{n}+b^{n}+c^{n}$ and consider the polynomial

$$
P(x)=(x-a)(x-b)(x-c)=x^{3}-(a+b+c) x^{2}+(a b+a c+b c) x-a b c .
$$

Since $P(a)=0$, we get $a^{3}=(a+b+c) a^{2}-(a b+a c+b c) a+a b c$ and multiplying both sides by $a^{n-3}$ we obtain $a^{n}=(a+b+c) a^{n-1}-(a b+a c+b c) a^{n-2}+(a b c) a^{n-3}$. Applying the same reasoning, we can obtain similar expressions for $b^{n}$ and $c^{n}$ and adding the three identities we get that $T_{n}$ satisfies the following 3 -term recurrence:

$$
T_{n}=(a+b+c) T_{n-1}-(a b+a c+b c) T_{n-2}+(a b c) T_{n-3} \text {, for all } n \geq 3
$$

¿From this we see that if $T_{n-2}$ and $T_{n-3}$ are divisible by $a+b+c$, then so is $T_{n}$. This immediately resolves part (b) - there are no ordered triples which are 2004-powerful and 2005-powerful, but not 2007-powerful-and reduces the number of cases to be considered in part (a): since all triples are 1-powerful, the recurrence implies that any ordered triple which is both 2 -powerful and 3 -powerful is $n$-powerful for all $n \geq 1$.

Putting $n=3$ in the recurrence, we have

$$
a^{3}+b^{3}+c^{3}=(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)-(a b+a c+b c)(a+b+c)+3 a b c
$$

which implies that $(a, b, c)$ is 3 -powerful if and only if $3 a b c$ is divisible by $a+b+c$. Since

$$
a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(a b+a c+b c),
$$

$(a, b, c)$ is 2 -powerful if and only if $2(a b+a c+b c)$ is divisible by $a+b+c$.

Suppose a prime $p \geq 5$ divides $a+b+c$. Then $p$ divides $a b c$. Since $\operatorname{gcd}(a, b, c)=1, p$ divides exactly one of $a, b$ or $c$; but then $p$ doesn't divide $2(a b+a c+b c)$.

Suppose $3^{2}$ divides $a+b+c$. Then 3 divides $a b c$, implying 3 divides exactly one of $a$, $b$ or $c$. But then 3 doesn't divide $2(a b+a c+b c)$.

Suppose $2^{2}$ divides $a+b+c$. Then 4 divides $a b c$. Since $\operatorname{gcd}(a, b, c)=1$, at most one of $a, b$ or $c$ is even, implying one of $a, b, c$ is divisible by 4 and the others are odd. But then $a b+a c+b c$ is odd and 4 doesn't divide $2(a b+a c+b c)$.

So if $(a, b, c)$ is 2 - and 3 -powerful, then $a+b+c$ is not divisible by 4 or 9 or any prime greater than 3. Since $a+b+c$ is at least $3, a+b+c$ is either 3 or 6 . It is now a simple matter to check the possibilities and conclude that the only triples which are $n$-powerful for all $n \geq 1$ are $(1,1,1)$ and $(1,1,4)$.

## Solution 2

Let $p$ be a prime. By Fermat's Little Theorem,

$$
a^{p-1} \equiv \begin{cases}1(\bmod p), & \text { if } p \text { doesn't divide } a \\ 0(\bmod p), & \text { if } p \text { divides } a\end{cases}
$$

Since $\operatorname{gcd}(a, b, c)=1$, we have that $a^{p-1}+b^{p-1}+c^{p-1} \equiv 1,2$ or $3(\bmod p)$. Therefore if $p$ is a prime divisor of $a^{p-1}+b^{p-1}+c^{p-1}$, then $p$ equals 2 or 3 . So if $(a, b, c)$ is $n$-powerful for all $n \geq 1$, then the only primes which can divide $a+b+c$ are 2 or 3 .

We can proceed in a similar fashion to show that $a+b+c$ is not divisible by 4 or 9 .

Since

$$
a^{2} \equiv \begin{cases}0(\bmod 4), & \text { if } p \text { is even; } \\ 1(\bmod 4), & \text { if } p \text { is odd }\end{cases}
$$

and $a, b, c$ aren't all even, we have that $a^{2}+b^{2}+c^{2} \equiv 1,2$ or $3(\bmod 4)$.

By expanding $(3 k)^{3},(3 k+1)^{3}$ and $(3 k+2)^{3}$, we find that $a^{3}$ is congruent to 0,1 or -1 modulo 9. Hence

$$
a^{6} \equiv \begin{cases}0(\bmod 9), & \text { if } 3 \text { divides } a ; \\ 1(\bmod 9), & \text { if } 3 \text { doesn't divide } a .\end{cases}
$$

Since $a, b, c$ aren't all divisible by 3 , we have that $a^{6}+b^{6}+c^{6} \equiv 1,2$ or $3(\bmod 9)$.

So $a^{2}+b^{2}+c^{2}$ is not divisible by 4 and $a^{6}+b^{6}+c^{6}$ is not divisible by 9 . Thus if $(a, b, c)$ is $n$-powerful for all $n \geq 1$, then $a+b+c$ is not divisible by 4 or 9 . Therefore $a+b+c$ is either 3 or 6 and checking all possibilities, we conclude that the only triples which are $n$-powerful for all $n \geq 1$ are $(1,1,1)$ and $(1,1,4)$.

See Solution 1 for the (b) part.


[^0]:    ${ }^{\dagger}$ By the way, $H$ is called the convex hull of $S$. If the points of $S$ lie on a line, then $H$ will be the shortest line segment containing the points of $S$. Otherwise, $H$ is a polygon whose vertices are all elements of $S$ and such that all other points in $S$ lie inside or on this polygon.