# SOLUTIONS FOR 2012 APMO PROBLEMS 

## Problem 1.

Solution: Let us denote by $\triangle X Y Z$ the area of the triangle $X Y Z$. Let $x=\triangle P A B, y=\triangle P B C$ and $z=\triangle P C A$.

From

$$
y: z=\triangle B C P: \triangle A C P=B F: A F=\triangle B P F: \triangle A P F=(x-1): 1
$$

follows that $z(x-1)=y$, which yields $(z+1) x=x+y+z$. Similarly, we get $(x+1) y=x+y+z$ and $(y+1) z=x+y+z$. Thus, we obtain $(x+1) y=(y+1) z=$ $(z+1) x$.

We may assume without loss of generality that $x \leq y, z$. If we assume that $y>z$ holds, then we get $(y+1) z>(z+1) x$, which is a contradiction. Similarly, we see that $y<z$ leads to a contradiction $(x+1) y<(y+1) z$. Therefore, we must have $y=z$. Then, we also get from $(y+1) z=(z+1) x$ that $x=z$ must hold. We now obtain from $(x-1): 1=y: z=1: 1$ that $x=y=z=2$ holds. Therefore, we conclude that the area of the triangle $A B C$ equals $x+y+z=6$.

## Problem 2.

Solution: If we insert numbers as in the figure below ( 0 's are to be inserted in the remaining blank boxes), then we see that the condition of the problem is satisfied and the total number of all the numbers inserted is 5 .

| 0 | 1 | 0 |  |
| :--- | :--- | :--- | :--- |
| 1 | 1 | 1 |  |
| 0 | 1 | 0 |  |
|  |  |  |  |

We will show that the sum of all the numbers to be inserted in the boxes of the given grid cannot be more than 5 if the distribution of the numbers has to satisfy the requirement of the problem. Let $n=2012$. Let us say that the row number (the column number) of a box in the given grid is $i(j$, respectively) if the box lies on the $i$-th row and the $j$-th column. For a pair of positive integers $x$ and $y$, denote by $R(x, y)$ the sum of the numbers inserted in all of the boxes whose row number is greater than or equal to $x$ and less than or equal to $y$ (assign the value 0 if $x>y$ ).

First let $a$ be the largest integer satisfying $1 \leq a \leq n$ and $R(1, a-1) \leq 1$, and then choose the smallest integer $c$ satisfying $a \leq c \leq n$ and $R(c+1, n) \leq 1$. It is possible to choose such a pair $a, c$ since $R(1,0)=0$ and $R(n+1, n)=0$. If $a<c$, then we have $a<n$ and so, by the maximality of $a$, we must have $R(1, a)>1$, while from the minimality of $c$, we must have $R(a+1, n)>1$. Then by splitting the grid into 2 rectangles by means of the horizontal line bordering the $a$-th row and the $a+1$-th row, we get the splitting contradicting the requirement of the problem. Thus, we must have $a=c$.

Similarly, if for any pair of integers $x, y$ we define $C(x, y)$ to be the sum of the numbers inserted in all of the boxes whose column number is greater than or equal to $x$ and less than or equal to $y(C(x, y)=0$ if $x>y)$, then we get a number $b$ for which

$$
C(1, b-1) \leq 1, \quad C(b+1, n) \leq 1, \quad 1 \leq b \leq n
$$

If we let $r$ be the number inserted in the box whose row number is $a$ and the column number is $b$, then since $r \leq 1$, we conclude that the sum of the numbers inserted into all of the boxes is

$$
\leq R(1, a-1)+R(a+1, n)+C(1, b-1)+C(b+1, n)+r \leq 5
$$

## Problem 3.

## Solution

For integers $a, b$ and a positive integer $m$, let us write $a \equiv b(\bmod m)$ if $a-b$ is divisible by $m$. Since $\frac{n^{p}+1}{p^{n}+1}$ must be a positive integer, we see that $p^{n} \leq n^{p}$ must hold. This means that if $p=2$, then $2^{n} \leq n^{2}$ must hold. As it is easy to show by induction that $2^{n}>n^{2}$ holds if $n \geq 5$, we conclude that if $p=2$, then $n \leq 4$ must be satisfied. And we can check that $(p, n)=(2,2),(2,4)$ satisfy the condition of the problem, while $(2,3)$ does not.

Next, we consider the case where $p \geq 3$.
Suppose $s$ is an integer satisfying $s \geq p$. If $s^{p} \leq p^{s}$ for such an $s$, then we have

$$
\begin{aligned}
(s+1)^{p} & =s^{p}\left(1+\frac{1}{s}\right)^{p} \leq p^{s}\left(1+\frac{1}{p}\right)^{p} \\
& =p^{s} \sum_{r=0}^{p}{ }_{p} C_{r} \frac{1}{p^{r}}<p^{s} \sum_{r=0}^{p} \frac{1}{r!} \\
& \leq p^{s}\left(1+\sum_{r=1}^{p} \frac{1}{2^{r-1}}\right) \\
& <p^{s}(1+2) \leq p^{s+1}
\end{aligned}
$$

Thus we have $(s+1)^{p}<p^{s+1}$, and by induction on $n$, we can conclude that if $n>p$, then $n^{p}<p^{n}$. This implies that we must have $n \leq p$ in order to satisfy our requirement $p^{n} \leq n^{p}$.

We note that since $p^{n}+1$ is even, so is $n^{p}+1$, which, in turn implies that $n$ must be odd and therefore, $p^{n}+1$ is divisible by $p+1$, and $n^{p}+1$ is also divisible by $p+1$. Thus we have $n^{p} \equiv-1(\bmod (p+1))$, and therefore, $n^{2 p} \equiv 1(\bmod (p+1))$.

Now, let $e$ be the smallest positive integer for which $n^{e} \equiv 1(\bmod (p+1))$. Then, we can write $2 p=e x+y$, where $x, y$ are non-negative integers and $0 \leq y<e$, and we have

$$
1 \equiv n^{2 p}=\left(n^{e}\right)^{x} \cdot n^{y} \equiv n^{y} \quad(\bmod (p+1))
$$

which implies, because of the minimality of $e$, that $y=0$ must hold. This means that $2 p$ is an integral multiple of $e$, and therefore, $e$ must equal one of the numbers $1,2, p, 2 p$.

Now, if $e=1, p$, then we get $n^{p} \equiv 1(\bmod (p+1))$, which contradicts the fact that $p$ is an odd prime. Since $n$ and $p+1$ are relatively prime, we have by Euler's Theorem that $n^{\varphi(p+1)} \equiv 1(\bmod (p+1))$, where $\varphi(m)$ denotes the number of integers $j(1 \leq j \leq m)$ which are relatively prime with $m$. From $\varphi(p+1)<p+1<2 p$ and the minimality of $e$, we can then conclude that $e=2$ must hold.

From $n^{2} \equiv 1(\bmod (p+1))$, we get

$$
-1 \equiv n^{p}=n^{\left(2 \cdot \frac{p-1}{2}+1\right)} \equiv n \quad(\bmod (p+1))
$$

which implies that $p+1$ divides $n+1$. Therefore, we must have $p \leq n$, which, together with the fact $n \leq p$, show that $p=n$ must hold.

It is clear that the pair $(p, p)$ for any prime $p \geq 3$ satisfies the condition of the problem, and thus, we conclude that the pairs $(p, n)$ which satisfy the condition of the problem must be of the form $(2,4)$ and $(p, p)$ with any prime $p$.

Alternate Solution. Let us consider the case where $p \geq 3$. As we saw in the preceding solution, $n$ must be odd if the pair $(p, n)$ satisfy the condition of the problem. Now, let $q$ be a prime factor of $p+1$. Then, since $p+1$ divides $p^{n}+1, q$ must be a prime factor of $p^{n}+1$ and of $n^{p}+1$ as well. Suppose $q \geq 3$. Then, from $n^{p} \equiv-1(\bmod q)$, it follows that $n^{2 p} \equiv 1(\bmod q)$ holds. If we let $e$ be the smallest positive integer satisfying $n^{e} \equiv 1(\bmod q)$, then by using the same argument as we used in the preceding solution, we can conclude that $e$ must equal one of the numbers $1,2, p, 2 p$. If $e=1, p$, then we get $n^{p} \equiv 1(\bmod q)$, which contradicts the assumption $q \geq 3$. Since $n$ is not a multiple of $q$, by Fermat's Little Theorem we get $n^{q-1} \equiv 1(\bmod q)$, and therefore, we get by the minimality of $e$ that $e=2$ must hold. From $n^{2} \equiv 1(\bmod q)$, we also get

$$
n^{p}=n^{\left(2 \cdot \frac{p-1}{2}+1\right)} \equiv n \quad(\bmod q),
$$

and since $n^{p} \equiv-1(\bmod q)$, we have $n \equiv-1(\bmod q)$ as well.
Now, if $q=2$ then since $n$ is odd, we have $n \equiv-1(\bmod q)$ as well. Thus, we conclude that for an arbitrary prime factor $q$ of $p+1, n \equiv-1(\bmod q)$ must hold.

Suppose, for a prime $q, q^{k}$ for some positive integer $k$ is a factor of $p+1$. Then $q^{k}$ must be a factor of $n^{p}+1$ as well. But since

$$
\begin{gathered}
n^{p}+1=(n+1)\left(n^{p-1}-n^{p-2}+\cdots-n+1\right) \quad \text { and } \\
n^{p-1}-n^{p-2}+\cdots-n+1 \equiv(-1)^{p-1}-(-1)^{p-2}+\cdots-(-1)+1 \not \equiv 0 \quad(\bmod q)
\end{gathered}
$$

we see that $q^{k}$ must divide $n+1$. By applying the argument above for each prime factor $q$ of $p+1$, we can then conclude that $n+1$ must be divisible by $p+1$, and as we did in the preceding proof, we can conclude that $n=p$ must hold.

## Problem 4.

Solution: If $A B=A C$, then we get $B F=C F$ and the conclusion of the problem is clearly satisfied. So, we assume that $A B \neq A C$ in the sequel.

Due to symmetry, we may suppose without loss of generality that $A B>A C$. Let $K$ be the point on the circle $\Gamma$ such that $A K$ is a diameter of this circle. Then, we get

$$
\angle B C K=\angle A C K-\angle A C B=90^{\circ}-\angle A C B=\angle C B H
$$

and

$$
\angle C B K=\angle A B K-\angle A B C=90^{\circ}-\angle A B C=\angle B C H,
$$

from which we conclude that the triangles $B C K$ and $C B H$ are congruent. Therefore, the quadrilateral $B K C H$ is a parallelogram, and its diagonal $H K$ passes through the center $M$ of the other diagonal $B C$. Therefore, the 3 points $H, M, K$ lie on the same straight line, and we have $\angle A E M=\angle A E K=90^{\circ}$.

From $\angle A E D=90^{\circ}=\angle A D M$, we see that the 4 points $A, E, D, M$ lie on the circumference of the same circle, from which we obtain $\angle A M B=\angle A E D=$ $\angle A E F=\angle A C F$. Putting this fact together with the fact that $\angle A B M=\angle A F C$, we conclude that the triangles $A B M$ and $A F C$ are similar, and we get $\frac{A M}{B M}=\frac{A C}{F C}$. By a similar argument, we get that the triangles $A C M$ and $A F B$ are similar, and
therefore, that $\frac{A M}{C M}=\frac{A B}{F B}$ holds. Noting that $B M=C M$, we also get $\frac{A C}{F C}=\frac{A B}{F B}$, from which we can conclude that $\frac{B F}{C F}=\frac{A B}{A C}$, proving the assertion of the problem.

## Problem 5.

Solution: Let us note first that if $i \neq j$, then since $a_{i} a_{j} \leq \frac{a_{i}^{2}+a_{j}^{2}}{2}$, we have

$$
n-a_{i} a_{j} \geq n-\frac{a_{i}^{2}+a_{j}^{2}}{2} \geq n-\frac{n}{2}=\frac{n}{2}>0 .
$$

If we set $b_{i}=\left|a_{i}\right|(i=1,2, \ldots, n)$, then we get $b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=n$ and $\frac{1}{n-a_{i} a_{j}} \leq$ $\frac{1}{n-b_{i} b_{j}}$, which shows that it is enough to prove the assertion of the problem in the case where all of $a_{1}, a_{2}, \cdots, a_{n}$ are non-negative. Hence, we assume from now on that $a_{1}, a_{2}, \cdots, a_{n}$ are all non-negative.

By multiplying by $n$ the both sides of the desired inequality we get the inequality:

$$
\sum_{1 \leq i<j \leq n} \frac{n}{n-a_{i} a_{j}} \leq \frac{n^{2}}{2}
$$

and since $\frac{n}{n-a_{i} a_{j}}=1+\frac{a_{i} a_{j}}{n-a_{i} a_{j}}$, we obtain from the inequality above by subtracting $\frac{n(n-1)}{2}$ from both sides the following inequality:

$$
\sum_{1 \leq i<j \leq n} \frac{a_{i} a_{j}}{n-a_{i} a_{j}} \leq \frac{n}{2}
$$

We will show that this inequality (i) holds.
If for some $i$ the equality $a_{i}^{2}=n$ is valid, then $a_{j}=0$ must hold for all $j \neq i$ and the inequality (i) is trivially satisfied. So, we assume from now on that $a_{i}^{2}<n$ is valid for each $i$.

Let us assume that $i \neq j$ from now on. Since $0 \leq a_{i} a_{j} \leq\left(\frac{a_{i}+a_{j}}{2}\right)^{2} \leq \frac{a_{i}^{2}+a_{j}^{2}}{2}$ holds, we have

$$
\frac{a_{i} a_{j}}{n-a_{i} a_{j}} \leq \frac{a_{i} a_{j}}{n-\frac{a_{i}^{2}+a_{j}^{2}}{2}} \leq \frac{\left(\frac{a_{i}+a_{j}}{2}\right)^{2}}{n-\frac{a_{i}^{2}+a_{j}^{2}}{2}}=\frac{1}{2} \cdot \frac{\left(a_{i}+a_{j}\right)^{2}}{\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)}
$$

Since $n-a_{i}^{2}>0, n-a_{j}^{2}>0$, we also get from the Cauchy-Schwarz inequality that

$$
\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right)\left(\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)\right) \geq\left(a_{i}+a_{j}\right)^{2},
$$

from which it follows that

$$
\frac{\left(a_{i}+a_{j}\right)^{2}}{\left(n-a_{i}^{2}\right)+\left(n-a_{j}^{2}\right)} \leq\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right)
$$

holds. Combining the inequalities (ii) and (iii), we get

$$
\begin{aligned}
\sum_{1 \leq i<j \leq n} \frac{a_{i} a_{j}}{n-a_{i} a_{j}} & \leq \frac{1}{2} \sum_{1 \leq i<j \leq n}\left(\frac{a_{j}^{2}}{n-a_{i}^{2}}+\frac{a_{i}^{2}}{n-a_{j}^{2}}\right) \\
& =\frac{1}{2} \sum_{i \neq j} \frac{a_{j}^{2}}{n-a_{i}^{2}} \\
& =\frac{1}{2} \sum_{i=1}^{n} \frac{n-a_{i}^{2}}{n-a_{i}^{2}} \\
& =\frac{n}{2}
\end{aligned}
$$

which establishes the desired inequality (i).