# CMO 1996 SOLUTIONS 

## QUESTION 1

## Solution .

If $f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$ has roots $\alpha, \beta, \gamma$ standard results about roots of polynomials give $\alpha+\beta+\gamma=0, \alpha \beta+\alpha \gamma+\beta \gamma=-1$, and $\alpha \beta \gamma=1$.

Then

$$
S=\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=\frac{N}{(1-\alpha)(1-\beta)(1-\gamma)}
$$

where the numerator simplifies to

$$
\begin{aligned}
N & =3-(\alpha+\beta+\gamma)-(\alpha \beta+\alpha \gamma+\beta \gamma)+3 \alpha \beta \gamma \\
& =3-(0)-(-1)+3(1) \\
& =7
\end{aligned}
$$

The denominator is $f(1)=-1$ so the required sum is -7 .

## QUESTION 2

## Solution 1.

For any $t, \quad 0 \leq 4 t^{2}<1+4 t^{2}$, so $0 \leq \frac{4 t^{2}}{1+4 t^{2}}<1$. Thus $x, y$ and $z$ must be non-negative and less than 1 .

Observe that if one of $x y$ or $z$ is 0 , then $x=y=z=0$.

If two of the variables are equal, say $x=y$, then the first equation becomes

$$
\frac{4 x^{2}}{1+4 x^{2}}=x
$$

This has the solution $x=0$, which gives $x=y=z=0$ and $x=\frac{1}{2}$ which gives $x=y=z=\frac{1}{2}$.

Finally, assume that $x, y$ and $z$ are non-zero and distinct. Without loss of generality we may assume that either $0<x<y<z<1$ or $0<x<z<y<1$. The two proofs are similar, so we do only the first case.

We will need the fact that $f(t)=\frac{4 t^{2}}{1+4 t^{2}}$ is increasing on the interval $(0,1)$.

To prove this, if $0<s<t<1$ then

$$
\begin{aligned}
f(t)-f(s) & =\frac{4 t^{2}}{1+4 t^{2}}-\frac{4 s^{2}}{1+4 s^{2}} \\
& =\frac{4 t^{2}-4 s^{2}}{\left(1+4 s^{2}\right)\left(1+4 t^{2}\right)}
\end{aligned}
$$

$$
>0 .
$$

So $0<x<y<z \Rightarrow f(x)=y<f(y)=z<f(z)=x$, a contradiction.

Hence $x=y=z=0$ and $x=y=z=\frac{1}{2}$ are the only real solutions.

## Solution 2.

Notice that $x, y$ and $z$ are non-negative. Adding the three equations gives

$$
x+y+z=\frac{4 z^{2}}{1+4 z^{2}}+\frac{4 x^{2}}{1+4 x^{2}}+\frac{4 y^{2}}{1+4 y^{2}} .
$$

This can be rearranged to give

$$
\frac{x(2 x-1)^{2}}{1+4 x^{2}}+\frac{y(2 y-1)^{2}}{1+4 y^{2}}+\frac{z(2 z-1)^{2}}{1+4 z^{2}}=0 .
$$

Since each term is non-negative, each term must be 0 , and hence each variable is either 0 or $\frac{1}{2}$. The original equations then show that $x=y=z=0$ and $x=y=z=\frac{1}{2}$ are the only two solutions.

## Solution 3.

Notice that $x, y$, and $z$ are non-negative. Multiply both sides of the inequality

$$
\frac{y}{1+4 y^{2}} \geq 0
$$

by $(2 y-1)^{2}$, and rearrange to obtain

$$
y-\frac{4 y^{2}}{1+4 y^{2}} \geq 0
$$

and hence that $y \geq z$. Similarly, $z \geq x$, and $x \geq y$. Hence, $x=y=z$ and, as in Solution 1, the two solutions follow.

## Solution 4.

As for solution 1 , note that $x=y=z=0$ is a solution and any other solution will have each of $x, y$ and $z$ positive.

The arithmetic-geometric mean inequality (or direct computation) shows that $\frac{1+4 x^{2}}{2} \geq \sqrt{1 \cdot 4 x^{2}}=2 x$ and hence $x \geq \frac{4 x^{2}}{1+4 x^{2}}=y$, with equality if and only if $1=4 x^{2}-$ that is, $x=\frac{1}{2}$. Similarly, $y \geq z$ with equality if and only if $y=\frac{1}{2}$ and $z \geq x$ with equality if and only if $z=\frac{1}{2}$. Adding $x \geq y, y \geq z$ and $z \geq x$ gives $x+y+x \geq x+y+z$. Thus equality must occur in each inequality, so $x=y=z=\frac{1}{2}$.

## QUESTION 3

## Solution.

Let $a_{1}, a_{2}, \ldots, a_{n}$ be a permutation of $1,2, \ldots, n$ with properties (i) and (ii).

A crucial observation, needed in Case II (b) is the following: If $a_{k}$ and $a_{k+1}$ are consecutive integers (i.e. $a_{k+1}=a_{k} \pm 1$ ), then the terms to the right of $a_{k+1}$ (also to the left of $a_{k}$ ) are either all less than both $a_{k}$ and $a_{k+1}$ or all greater than both $a_{k}$ and $a_{k+1}$.

Since $a_{1}=1$, by (ii) $a_{2}$ is either 2 or 3 .

CASE I: Suppose $a_{2}=2$. Then $a_{3}, a_{4}, \ldots, a_{n}$ is a permutation of $3,4, \ldots, n$. Thus $a_{2}, a_{3}, \ldots, a_{n}$ is a permutation of $2,3, \ldots, n$ with $a_{2}=2$ and property (ii). Clearly there are $f(n-1)$ such permutations.

CASE II: Suppose $a_{2}=3$.

(a) Suppose $a_{3}=2$. Then $a_{4}, a_{5}, \ldots, a_{n}$ is a permutation of $4,5, \ldots, n$ with $a_{4}=4$ and property (ii). There are $f(n-3)$ such permutations.

(b) Suppose $a_{3} \geq 4$. If $a_{k+1}$ is the first even number in the permutation then, because of (ii), $a_{1}, a_{2}, \ldots, a_{k}$ must be $1,3,5, \ldots, 2 k-1$ (in that order). Then $a_{k+1}$ is either $2 k$ or $2 k-2$, so that $a_{k}$ and $a_{k+1}$ are consecutive integers. Applying the crucial observation made above, we deduce that $a_{k+2}, \ldots, a_{n}$ are all either greater than or smaller than $a_{k}$ and $a_{k+1}$. But 2 must be to the right of $a_{k+1}$. Hence $a_{k+2}, \ldots, a_{n}$ are the even integers less than $a_{k+1}$. The only possibility then, is

$$
1,3,5, \ldots, a_{k-1}, a_{k}, \ldots, 6,4,2
$$

Cases I and II show that

$$
f(n)=f(n-1)+f(n-3)+1, \quad n \geq 4 .
$$

Calculating the first few values of $f(n)$ directly gives

$$
f(1)=1, f(2)=1, f(3)=2, f(4)=4, f(5)=6 \text {. }
$$

Calculating a few more $f(n)$ 's using $\left(^{*}\right)$ and mod 3 arithmetic, $f(1)=1, f(2)=1, f(3)=$ $2, f(4)=1, f(5)=0, f(6)=0, f(7)=2, f(8)=0, f(9)=1, f(10)=1, f(11)=2$. Since $f(1)=f(9), f(2)=f(10)$ and $f(3)=f(11) \bmod 3,\left({ }^{*}\right)$ shows that $f(a)=f(a \bmod 8), \bmod 3, a \geq$ 1 .

Hence $f(1996) \equiv f(4) \equiv 1(\bmod 3)$ so 3 does not divide $f(1996)$.

## QUESTION 4

## Solution 1.

Let $B E=B D$ with $E$ on $B C$, so that $A D=E C$ :

![](https://cdn.mathpix.com/cropped/2024_04_17_481f12c4dfc2227ae8efg-5.jpg?height=423&width=897&top_left_y=499&top_left_x=294)

By a standard theorem, $\frac{A B}{C B}=\frac{A D}{D C} ; \quad$ so in

$\triangle C E D$ and $\triangle C A B$ we have a common angle and

$$
\frac{C E}{C D}=\frac{A D}{C D}=\frac{A B}{C B}=\frac{C A}{C B}
$$

Thus $\triangle C E D \sim \triangle C A B$, so that $\angle C D E=\angle D C E=\angle A B C=2 x$.

Hence $\angle B D E=\angle B E D=4 x$, whence $9 x=180^{\circ}$ so $x=20^{\circ}$.

Thus $\angle A=180^{\circ}-4 x=100^{\circ}$.

## Solution 2 .

Apply the law of sines to $\triangle A B D$ and $\triangle B D C$ to get

$$
\frac{A D}{B D}=\frac{\sin x}{\sin 4 x} \quad \text { and } \quad 1+\frac{A D}{B D}=\frac{B C}{B D}=\frac{\sin 3 x}{\sin 2 x} .
$$

Now massage the resulting trigonometric equation with standard identities to get

$$
\sin 2 x(\sin 4 x+\sin x)=\sin 2 x(\sin 5 x+\sin x) .
$$

Since $0<2 x<90^{\circ}$, we get

$$
5 x-90^{\circ}=90^{\circ}-4 x,
$$

so that $\angle A=100^{\circ}$.

## QUESTION 5

## Solution.

Let

$$
\begin{aligned}
f(n) & =n-\sum_{k=1}^{m}\left[r_{k} n\right] \\
& =n \sum_{k=1}^{m} r_{k}-\sum_{k=1}^{m}\left[r_{k} n\right] \\
& =\sum_{k=1}^{m}\left\{r_{k} n-\left[r_{k} n\right]\right\} .
\end{aligned}
$$

Now $0 \leq x-[x]<1$, and if $c$ is an integer, $(c+x)-[c+x]=x-[x]$.

Hence $0 \leq f(n)<\sum_{k=1}^{m} 1=m$. Because $f(n)$ is an integer, $0 \leq f(n) \leq m-1$.

To show that $f(n)$ can achieve these bounds for $n>0$, we assume that $r_{k}=\frac{a_{k}}{b_{k}}$ where $a_{k}, b_{k}$ are integers; $a_{k}<b_{k}$.

Then, if $n=b_{1} b_{2} \ldots b_{m},\left(r_{k} n\right)-\left[r_{k} n\right]=0, k=1,2, \ldots, m$ and thus $f(n)=0$.

Letting $n=b_{1} b_{2} \ldots b_{n}-1$, then

$$
\begin{aligned}
r_{k} n & =r_{k}\left(b_{1} b_{2} \ldots b_{m}-1\right) \\
& \left.=r_{k}\left\{\left(b_{1} b_{2} \ldots b_{m}-b_{k}\right)+b_{k}-1\right)\right\} \\
& =\text { integer }+r_{k}\left(b_{k}-1\right)
\end{aligned}
$$

This gives

$$
\begin{aligned}
r_{k} n-\left[r_{k} n\right] & =r_{k}\left(b_{k}-1\right)-\left[r_{k}\left(b_{k}-1\right)\right] \\
& =\frac{a_{k}}{b_{k}}\left(b_{k}-1\right)-\left[\frac{a_{k}}{b_{k}}\left(b_{k}-1\right)\right] \\
& =\left(a_{k}-\frac{a_{k}}{b_{k}}\right)-\left[a_{k}-\frac{a_{k}}{b_{k}}\right] \\
& =\left(a_{k}-\frac{a_{k}}{b_{k}}\right)-\left(a_{k}-1\right) \\
& =1-\frac{a_{k}}{b_{k}}=1-r_{k} .
\end{aligned}
$$

Hence

$$
f(n)=\sum_{k=1}^{m}\left(1-r_{k}\right)=m-1 .
$$