APMO/md/en-apmo2012_sol.md

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

$$y:z=\mathrm{△}BCP:\mathrm{△}ACP=BF:AF=\mathrm{△}BPF:\mathrm{△}APF=(x-1):1y:\; z=\backslash triangle\; B\; C\; P:\; \backslash triangle\; A\; C\; P=B\; F:\; A\; F=\backslash triangle\; B\; P\; F:\; \backslash triangle\; A\; P\; F=(x-1):\; 1$$

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

$$C(1,b-1)\le 1,C(b+1,n)\le 1,1\le b\le nC(1,\; b-1)\; \backslash leq\; 1,\; \backslash quad\; C(b+1,\; n)\; \backslash leq\; 1,\; \backslash quad\; 1\; \backslash leq\; b\; \backslash leq\; n$$

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

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

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

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

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

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

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

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

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

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

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

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

$$\mathrm{\angle }BCK=\mathrm{\angle }ACK-\mathrm{\angle }ACB={90}^{\circ }-\mathrm{\angle }ACB=\mathrm{\angle }CBH\backslash angle\; B\; C\; K=\backslash angle\; A\; C\; K-\backslash angle\; A\; C\; B=90^\{\backslash circ\}-\backslash angle\; A\; C\; B=\backslash angle\; C\; B\; H$$

and

$$\mathrm{\angle }CBK=\mathrm{\angle }ABK-\mathrm{\angle }ABC={90}^{\circ }-\mathrm{\angle }ABC=\mathrm{\angle }BCH,\backslash angle\; C\; B\; K=\backslash angle\; A\; B\; K-\backslash angle\; A\; B\; C=90^\{\backslash circ\}-\backslash angle\; A\; B\; C=\backslash angle\; B\; C\; H,$$

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

$$n-a_i{a}_j\ge n-\frac{a_i^2+a_j^2}{2}\ge n-\frac{n}{2}=\frac{n}{2}>0.n-a\_\{i\}\; a\_\{j\}\; \backslash geq\; n-\backslash frac\{a\_\{i\}^\{2\}+a\_\{j\}^\{2\}\}\{2\}\; \backslash geq\; n-\backslash frac\{n\}\{2\}=\backslash frac\{n\}\{2\}>0\; .$$

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:

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:

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

$$\frac{a_i{a}_j}{n-a_i{a}_j}\le \frac{a_i{a}_j}{n-\frac{a_i^2+a_j^2}{2}}\le \frac{{\left(\frac{a_i+a_j}{2}\right)}^2}{n-\frac{a_i^2+a_j^2}{2}}=\frac{1}{2}\cdot \frac{{(a_i+a_j)}^2}{(n-a_i^2)+(n-a_j^2)}\backslash frac\{a\_\{i\}\; a\_\{j\}\}\{n-a\_\{i\}\; a\_\{j\}\}\; \backslash leq\; \backslash frac\{a\_\{i\}\; a\_\{j\}\}\{n-\backslash frac\{a\_\{i\}^\{2\}+a\_\{j\}^\{2\}\}\{2\}\}\; \backslash leq\; \backslash frac\{\backslash left(\backslash frac\{a\_\{i\}+a\_\{j\}\}\{2\}\backslash right)^\{2\}\}\{n-\backslash frac\{a\_\{i\}^\{2\}+a\_\{j\}^\{2\}\}\{2\}\}=\backslash frac\{1\}\{2\}\; \backslash cdot\; \backslash frac\{\backslash left(a\_\{i\}+a\_\{j\}\backslash right)^\{2\}\}\{\backslash left(n-a\_\{i\}^\{2\}\backslash right)+\backslash left(n-a\_\{j\}^\{2\}\backslash right)\}$$

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

$$(\frac{a_j^2}{n-a_i^2}+\frac{a_i^2}{n-a_j^2})((n-a_i^2)+(n-a_j^2))\ge {(a_i+a_j)}^2,\backslash left(\backslash frac\{a\_\{j\}^\{2\}\}\{n-a\_\{i\}^\{2\}\}+\backslash frac\{a\_\{i\}^\{2\}\}\{n-a\_\{j\}^\{2\}\}\backslash right)\backslash left(\backslash left(n-a\_\{i\}^\{2\}\backslash right)+\backslash left(n-a\_\{j\}^\{2\}\backslash right)\backslash right)\; \backslash geq\backslash left(a\_\{i\}+a\_\{j\}\backslash right)^\{2\},$$

from which it follows that

$$\frac{{(a_i+a_j)}^2}{(n-a_i^2)+(n-a_j^2)}\le (\frac{a_j^2}{n-a_i^2}+\frac{a_i^2}{n-a_j^2})\backslash frac\{\backslash left(a\_\{i\}+a\_\{j\}\backslash right)^\{2\}\}\{\backslash left(n-a\_\{i\}^\{2\}\backslash right)+\backslash left(n-a\_\{j\}^\{2\}\backslash right)\}\; \backslash leq\backslash left(\backslash frac\{a\_\{j\}^\{2\}\}\{n-a\_\{i\}^\{2\}\}+\backslash frac\{a\_\{i\}^\{2\}\}\{n-a\_\{j\}^\{2\}\}\backslash right)$$

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

which establishes the desired inequality (i).