APMO/md/en-apmo1989_sol.md

APMO 1989 - Problems and Solutions

Problem 1

Let $x_{1}, x_{2}, \ldots, x_{n}$ be positive real numbers, and let

$$S=x_1+x_2+\cdots +x_{n}S=x\_\{1\}+x\_\{2\}+\backslash cdots+x\_\{n\}$$

Prove that

$$(1+x_1)(1+x_2)\cdots (1+x_n)\le 1+S+\frac{S^2}{2!}+\frac{S^3}{3!}+\cdots +\frac{S^n}{n!}\backslash left(1+x\_\{1\}\backslash right)\backslash left(1+x\_\{2\}\backslash right)\; \backslash cdots\backslash left(1+x\_\{n\}\backslash right)\; \backslash leq\; 1+S+\backslash frac\{S^\{2\}\}\{2!\}+\backslash frac\{S^\{3\}\}\{3!\}+\backslash cdots+\backslash frac\{S^\{n\}\}\{n!\}$$

Solution 1

Let $\sigma_{k}$ be the $k$ th symmetric polynomial, namely

$${\sigma }_k=\sum _{\begin{array}{c}\mathrm{\mid }S\mathrm{\mid }=k\\ S\subseteq \{1,2,\dots ,n\}\end{array}}\prod _{i\in S}{x}_i,\backslash sigma\_\{k\}=\backslash sum\_\{\backslash substack\{|S|=k\; \backslash \backslash \; S\; \backslash subseteq\backslash \{1,2,\; \backslash ldots,\; n\backslash \}\}\}\; \backslash prod\_\{i\; \backslash in\; S\}\; x\_\{i\},$$

and more explicitly

$${\sigma }_1=S,{\sigma }_2=x_1{x}_2+x_1{x}_3+\cdots +x_{n-1}{x}_n,\text{ and so on. }\backslash sigma\_\{1\}=S,\; \backslash quad\; \backslash sigma\_\{2\}=x\_\{1\}\; x\_\{2\}+x\_\{1\}\; x\_\{3\}+\backslash cdots+x\_\{n-1\}\; x\_\{n\},\; \backslash quad\; \backslash text\; \{\; and\; so\; on.\; \}$$

Then

$$(1+x_1)(1+x_2)\cdots (1+x_n)=1+{\sigma }_1+{\sigma }_2+\cdots +{\sigma }_n\backslash left(1+x\_\{1\}\backslash right)\backslash left(1+x\_\{2\}\backslash right)\; \backslash cdots\backslash left(1+x\_\{n\}\backslash right)=1+\backslash sigma\_\{1\}+\backslash sigma\_\{2\}+\backslash cdots+\backslash sigma\_\{n\}$$

The expansion of

$$S^k={(x_1+x_2+\cdots +x_n)}^k=\underset{k\text{ times }}{\underbrace{(x_1+x_2+\cdots +x_n)(x_1+x_2+\cdots +x_n)\cdots (x_1+x_2+\cdots +x_n)}}S^\{k\}=\backslash left(x\_\{1\}+x\_\{2\}+\backslash cdots+x\_\{n\}\backslash right)^\{k\}=\backslash underbrace\{\backslash left(x\_\{1\}+x\_\{2\}+\backslash cdots+x\_\{n\}\backslash right)\backslash left(x\_\{1\}+x\_\{2\}+\backslash cdots+x\_\{n\}\backslash right)\; \backslash cdots\backslash left(x\_\{1\}+x\_\{2\}+\backslash cdots+x\_\{n\}\backslash right)\}\_\{k\; \backslash text\; \{\; times\; \}\}$$

has at least $k$ ! occurrences of $\prod_{i \in S} x_{i}$ for each subset $S$ with $k$ indices from ${1,2, \ldots, n}$. In fact, if $\pi$ is a permutation of $S$, we can choose each $x_{\pi(i)}$ from the $i$ th factor of $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{k}$. Then each term appears at least $k$ ! times, and

$$S^k\ge k!{\sigma }_k⟺{\sigma }_k\le \frac{S^k}{k!}S^\{k\}\; \backslash geq\; k!\backslash sigma\_\{k\}\; \backslash Longleftrightarrow\; \backslash sigma\_\{k\}\; \backslash leq\; \backslash frac\{S^\{k\}\}\{k!\}$$

Summing the obtained inequalities for $k=1,2, \ldots, n$ yields the result.

Solution 2

By AM-GM,

$$(1+x_1)(1+x_2)\cdots (1+x_n)\le {\left(\frac{(1+x_1)+(1+x_2)+\cdots +(1+x_n)}{n}\right)}^n={(1+\frac{S}{n})}^n\backslash left(1+x\_\{1\}\backslash right)\backslash left(1+x\_\{2\}\backslash right)\; \backslash cdots\backslash left(1+x\_\{n\}\backslash right)\; \backslash leq\backslash left(\backslash frac\{\backslash left(1+x\_\{1\}\backslash right)+\backslash left(1+x\_\{2\}\backslash right)+\backslash cdots+\backslash left(1+x\_\{n\}\backslash right)\}\{n\}\backslash right)^\{n\}=\backslash left(1+\backslash frac\{S\}\{n\}\backslash right)^\{n\}$$

By the binomial theorem,

$${(1+\frac{S}{n})}^n=\sum _{k=0}^n\left(\genfrac{}{}{0px}{}{n}{k}\right){\left(\frac{S}{n}\right)}^k=\sum _{k=0}^n\frac{1}{k!}\frac{n(n-1)\dots (n-k+1)}{n^k}{S}^k\le \sum _{k=0}^n\frac{S^k}{k!}\backslash left(1+\backslash frac\{S\}\{n\}\backslash right)^\{n\}=\backslash sum\_\{k=0\}^\{n\}\backslash binom\{n\}\{k\}\backslash left(\backslash frac\{S\}\{n\}\backslash right)^\{k\}=\backslash sum\_\{k=0\}^\{n\}\; \backslash frac\{1\}\{k!\}\; \backslash frac\{n(n-1)\; \backslash ldots(n-k+1)\}\{n^\{k\}\}\; S^\{k\}\; \backslash leq\; \backslash sum\_\{k=0\}^\{n\}\; \backslash frac\{S^\{k\}\}\{k!\}$$

and the result follows. Comment: Maclaurin's inequality states that

$$\frac{{\sigma }_1}{n}\ge \sqrt{\frac{{\sigma }_2}{\left(\genfrac{}{}{0px}{}{n}{2}\right)}}\ge \cdots \ge \sqrt[k]{\frac{{\sigma }_k}{\left(\genfrac{}{}{0px}{}{n}{k}\right)}}\ge \cdots \ge \sqrt[n]{\frac{{\sigma }_n}{\left(\genfrac{}{}{0px}{}{n}{n}\right)}}\backslash frac\{\backslash sigma\_\{1\}\}\{n\}\; \backslash geq\; \backslash sqrt\{\backslash frac\{\backslash sigma\_\{2\}\}\{\backslash binom\{n\}\{2\}\}\}\; \backslash geq\; \backslash cdots\; \backslash geq\; \backslash sqrt[k]\{\backslash frac\{\backslash sigma\_\{k\}\}\{\backslash binom\{n\}\{k\}\}\}\; \backslash geq\; \backslash cdots\; \backslash geq\; \backslash sqrt[n]\{\backslash frac\{\backslash sigma\_\{n\}\}\{\backslash binom\{n\}\{n\}\}\}$$

Then $\sigma_{k} \leq\binom{ n}{k} \frac{S^{k}}{n^{k}}=\frac{1}{k!} \frac{n(n-1) \ldots(n-k+1)}{n^{k}} S^{k} \leq \frac{S^{k}}{k!}$.

Problem 2

Prove that the equation

$$6(6a^2+3b^2+c^2)=5n^{2}6\backslash left(6\; a^\{2\}+3\; b^\{2\}+c^\{2\}\backslash right)=5\; n^\{2\}$$

has no solutions in integers except $a=b=c=n=0$.

Solution

We can suppose without loss of generality that $a, b, c, n \geq 0$. Let $(a, b, c, n)$ be a solution with minimum sum $a+b+c+n$. Suppose, for the sake of contradiction, that $a+b+c+n>0$. Since 6 divides $5 n^{2}, n$ is a multiple of 6 . Let $n=6 n_{0}$. Then the equation reduces to

$$6a^2+3b^2+c^2=30n_0^{2}6\; a^\{2\}+3\; b^\{2\}+c^\{2\}=30\; n\_\{0\}^\{2\}$$

The number $c$ is a multiple of 3 , so let $c=3 c_{0}$. The equation now reduces to

$$2a^2+b^2+3c_0^2=10n_0^{2}2\; a^\{2\}+b^\{2\}+3\; c\_\{0\}^\{2\}=10\; n\_\{0\}^\{2\}$$

Now look at the equation modulo 8:

$$b^2+3c_0^2\equiv 2(n_0^2-a^2)(8)b^\{2\}+3\; c\_\{0\}^\{2\}\; \backslash equiv\; 2\backslash left(n\_\{0\}^\{2\}-a^\{2\}\backslash right)\; \backslash quad(\backslash bmod\; 8)$$

Integers $b$ and $c_{0}$ have the same parity. Either way, since $x^{2}$ is congruent to 0 or 1 modulo 4 , $b^{2}+3 c_{0}^{2}$ is a multiple of 4 , so $n_{0}^{2}-a^{2}=\left(n_{0}-a\right)\left(n_{0}+a\right)$ is even, and therefore also a multiple of 4 , since $n_{0}-a$ and $n_{0}+a$ have the same parity. Hence $2\left(n_{0}^{2}-a^{2}\right)$ is a multiple of 8 , and

$$b^2+3c_0^2\equiv 0(8)b^\{2\}+3\; c\_\{0\}^\{2\}\; \backslash equiv\; 0\; \backslash quad(\backslash bmod\; 8)$$

If $b$ and $c_{0}$ are both odd, $b^{2}+3 c_{0}^{2} \equiv 4(\bmod 8)$, which is impossible. Then $b$ and $c_{0}$ are both even. Let $b=2 b_{0}$ and $c_{0}=2 c_{1}$, and we find

$$a^2+2b_0^2+6c_1^2=5n_0^{2}a^\{2\}+2\; b\_\{0\}^\{2\}+6\; c\_\{1\}^\{2\}=5\; n\_\{0\}^\{2\}$$

Look at the last equation modulo 8:

$$a^2+3n_0^2\equiv 2(c_1^2-b_0^2)(8)a^\{2\}+3\; n\_\{0\}^\{2\}\; \backslash equiv\; 2\backslash left(c\_\{1\}^\{2\}-b\_\{0\}^\{2\}\backslash right)\; \backslash quad(\backslash bmod\; 8)$$

A similar argument shows that $a$ and $n_{0}$ are both even. We have proven that $a, b, c, n$ are all even. Then, dividing the original equation by 4 we find

$$6(6(a\mathrm{/}2{)}^2+3(b\mathrm{/}2{)}^2+(c\mathrm{/}2{)}^2)=5(n\mathrm{/}2{)}^{2}6\backslash left(6(a\; /\; 2)^\{2\}+3(b\; /\; 2)^\{2\}+(c\; /\; 2)^\{2\}\backslash right)=5(n\; /\; 2)^\{2\}$$

and we find that $(a / 2, b / 2, c / 2, n / 2)$ is a new solution with smaller sum. This is a contradiction, and the only solution is $(a, b, c, n)=(0,0,0,0)$.

Problem 3

Let $A_{1}, A_{2}, A_{3}$ be three points in the plane, and for convenience,let $A_{4}=A_{1}, A_{5}=A_{2}$. For $n=1,2$, and 3 , suppose that $B_{n}$ is the midpoint of $A_{n} A_{n+1}$, and suppose that $C_{n}$ is the midpoint of $A_{n} B_{n}$. Suppose that $A_{n} C_{n+1}$ and $B_{n} A_{n+2}$ meet at $D_{n}$, and that $A_{n} B_{n+1}$ and $C_{n} A_{n+2}$ meet at $E_{n}$. Calculate the ratio of the area of triangle $D_{1} D_{2} D_{3}$ to the area of triangle $E_{1} E_{2} E_{3}$. Answer: $\frac{25}{49}$. Solution Let $G$ be the centroid of triangle $A B C$, and also the intersection point of $A_{1} B_{2}, A_{2} B_{3}$, and $A_{3} B_{1}$ 。

By Menelao's theorem on triangle $B_{1} A_{2} A_{3}$ and line $A_{1} D_{1} C_{2}$,

$$\frac{A_1{B}_1}{A_1{A}_2}\cdot \frac{D_1{A}_3}{D_1{B}_1}\cdot \frac{C_2{A}_2}{C_2{A}_3}=1⟺\frac{D_1{A}_3}{D_1{B}_1}=2\cdot 3=6⟺\frac{D_1{B}_1}{A_3{B}_1}=\frac{1}{7}\backslash frac\{A\_\{1\}\; B\_\{1\}\}\{A\_\{1\}\; A\_\{2\}\}\; \backslash cdot\; \backslash frac\{D\_\{1\}\; A\_\{3\}\}\{D\_\{1\}\; B\_\{1\}\}\; \backslash cdot\; \backslash frac\{C\_\{2\}\; A\_\{2\}\}\{C\_\{2\}\; A\_\{3\}\}=1\; \backslash Longleftrightarrow\; \backslash frac\{D\_\{1\}\; A\_\{3\}\}\{D\_\{1\}\; B\_\{1\}\}=2\; \backslash cdot\; 3=6\; \backslash Longleftrightarrow\; \backslash frac\{D\_\{1\}\; B\_\{1\}\}\{A\_\{3\}\; B\_\{1\}\}=\backslash frac\{1\}\{7\}$$

Since $A_{3} G=\frac{2}{3} A_{3} B_{1}$, if $A_{3} B_{1}=21 t$ then $G A_{3}=14 t, D_{1} B_{1}=\frac{21 t}{7}=3 t, A_{3} D_{1}=18 t$, and $G D_{1}=A_{3} D_{1}-A_{3} G=18 t-14 t=4 t$, and

$$\frac{G{D}_1}{G{A}_3}=\frac{4}{14}=\frac{2}{7}\backslash frac\{G\; D\_\{1\}\}\{G\; A\_\{3\}\}=\backslash frac\{4\}\{14\}=\backslash frac\{2\}\{7\}$$

Similar results hold for the other medians, therefore $D_{1} D_{2} D_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $-\frac{2}{7}$. By Menelao's theorem on triangle $A_{1} A_{2} B_{2}$ and line $C_{1} E_{1} A_{3}$,

$$\frac{C_1{A}_1}{C_1{A}_2}\cdot \frac{E_1{B}_2}{E_1{A}_1}\cdot \frac{A_3{A}_2}{A_3{B}_2}=1⟺\frac{E_1{B}_2}{E_1{A}_1}=3\cdot \frac{1}{2}=\frac{3}{2}⟺\frac{A_1{E}_1}{A_1{B}_2}=\frac{2}{5}\backslash frac\{C\_\{1\}\; A\_\{1\}\}\{C\_\{1\}\; A\_\{2\}\}\; \backslash cdot\; \backslash frac\{E\_\{1\}\; B\_\{2\}\}\{E\_\{1\}\; A\_\{1\}\}\; \backslash cdot\; \backslash frac\{A\_\{3\}\; A\_\{2\}\}\{A\_\{3\}\; B\_\{2\}\}=1\; \backslash Longleftrightarrow\; \backslash frac\{E\_\{1\}\; B\_\{2\}\}\{E\_\{1\}\; A\_\{1\}\}=3\; \backslash cdot\; \backslash frac\{1\}\{2\}=\backslash frac\{3\}\{2\}\; \backslash Longleftrightarrow\; \backslash frac\{A\_\{1\}\; E\_\{1\}\}\{A\_\{1\}\; B\_\{2\}\}=\backslash frac\{2\}\{5\}$$

If $A_{1} B_{2}=15 u$, then $A_{1} G=\frac{2}{3} \cdot 15 u=10 u$ and $G E_{1}=A_{1} G-A_{1} E_{1}=10 u-\frac{2}{5} \cdot 15 u=4 u$, and

$$\frac{G{E}_1}{G{A}_1}=\frac{4}{10}=\frac{2}{5}\backslash frac\{G\; E\_\{1\}\}\{G\; A\_\{1\}\}=\backslash frac\{4\}\{10\}=\backslash frac\{2\}\{5\}$$

Similar results hold for the other medians, therefore $E_{1} E_{2} E_{3}$ and $A_{1} A_{2} A_{3}$ are homothetic with center $G$ and ratio $\frac{2}{5}$. Then $D_{1} D_{2} D_{3}$ and $E_{1} E_{2} E_{3}$ are homothetic with center $G$ and ratio $-\frac{2}{7}: \frac{2}{5}=-\frac{5}{7}$, and the ratio of their area is $\left(\frac{5}{7}\right)^{2}=\frac{25}{49}$.

Problem 4

Let $S$ be a set consisting of $m$ pairs $(a, b)$ of positive integers with the property that $1 \leq a<$ $b \leq n$. Show that there are at least

$$4m\frac{(m-\frac{n^2}{4})}{3n}4\; m\; \backslash frac\{\backslash left(m-\backslash frac\{n^\{2\}\}\{4\}\backslash right)\}\{3\; n\}$$

triples $(a, b, c)$ such that $(a, b),(a, c)$, and $(b, c)$ belong to $S$.

Solution

Call a triple $(a, b, c)$ good if and only if $(a, b),(a, c)$, and $(b, c)$ all belong to $S$. For $i$ in ${1,2, \ldots, n}$, let $d_{i}$ be the number of pairs in $S$ that contain $i$, and let $D_{i}$ be the set of numbers paired with $i$ in $S$ (so $\left|D_{i}\right|=d_{i}$ ). Consider a pair $(i, j) \in S$. Our goal is to estimate the number of integers $k$ such that any permutation of ${i, j, k}$ is good, that is, $\left|D_{i} \cap D_{j}\right|$. Note that $i \notin D_{i}$ and $j \notin D_{j}$, so $i, j \notin D_{i} \cap D_{j}$; thus any $k \in D_{i} \cap D_{j}$ is different from both $i$ and $j$, and ${i, j, k}$ has three elements as required. Now, since $D_{i} \cup D_{j} \subseteq{1,2, \ldots, n}$,

$$\mid D_i\cap D_j\mid =\mid D_i\mid +\mid D_j\mid -\mid D_i\cup D_j\mid \le d_i+d_j-n\backslash left|D\_\{i\}\; \backslash cap\; D\_\{j\}\backslash right|=\backslash left|D\_\{i\}\backslash right|+\backslash left|D\_\{j\}\backslash right|-\backslash left|D\_\{i\}\; \backslash cup\; D\_\{j\}\backslash right|\; \backslash leq\; d\_\{i\}+d\_\{j\}-n$$

Summing all the results, and having in mind that each good triple is counted three times (one for each two of the three numbers), the number of good triples $T$ is at least

$$T\ge \frac{1}{3}\sum _{(i,j)\in S}(d_i+d_j-n)T\; \backslash geq\; \backslash frac\{1\}\{3\}\; \backslash sum\_\{(i,\; j)\; \backslash in\; S\}\backslash left(d\_\{i\}+d\_\{j\}-n\backslash right)$$

Each term $d_{i}$ appears each time $i$ is in a pair from $S$, that is, $d_{i}$ times; there are $m$ pairs in $S$, so $n$ is subtracted $m$ times. By the Cauchy-Schwartz inequality

$$T\ge \frac{1}{3}(\sum _{i=1}^n{d}_i^2-mn)\ge \frac{1}{3}(\frac{{\left(\sum _{i=1}^n{d}_i\right)}^2}{n}-mn)\mathrm{.}T\; \backslash geq\; \backslash frac\{1\}\{3\}\backslash left(\backslash sum\_\{i=1\}^\{n\}\; d\_\{i\}^\{2\}-m\; n\backslash right)\; \backslash geq\; \backslash frac\{1\}\{3\}\backslash left(\backslash frac\{\backslash left(\backslash sum\_\{i=1\}^\{n\}\; d\_\{i\}\backslash right)^\{2\}\}\{n\}-m\; n\backslash right)\; .$$

Finally, the sum $\sum_{i=1}^{n} d_{i}$ is $2 m$, since $d_{i}$ counts the number of pairs containing $i$, and each pair $(i, j)$ is counted twice: once in $d_{i}$ and once in $d_{j}$. Therefore

$$T\ge \frac{1}{3}(\frac{(2m{)}^2}{n}-mn)=4m\frac{(m-\frac{n^2}{4})}{3n}\mathrm{.}T\; \backslash geq\; \backslash frac\{1\}\{3\}\backslash left(\backslash frac\{(2\; m)^\{2\}\}\{n\}-m\; n\backslash right)=4\; m\; \backslash frac\{\backslash left(m-\backslash frac\{n^\{2\}\}\{4\}\backslash right)\}\{3\; n\}\; .$$

Comment: This is a celebrated graph theory fact named Goodman's bound, after A. M. Goodman's method published in 1959. The generalized version of the problem is still studied to this day.

Problem 5

Determine all functions $f$ from the reals to the reals for which (1) $f(x)$ is strictly increasing, (2) $f(x)+g(x)=2 x$ for all real $x$, where $g(x)$ is the composition inverse function to $f(x)$. (Note: $f$ and $g$ are said to be composition inverses if $f(g(x))=x$ and $g(f(x))=x$ for all real x.)

Answer: $f(x)=x+c, c \in \mathbb{R}$ constant.

Solution

Denote by $f_{n}$ the $n$th iterate of $f$, that is, $f_{n}(x)=\underbrace{f(f(\ldots f}{n \text { times }}(x)))$. Plug $x \rightarrow f{n+1}(x)$ in (2): since $g\left(f_{n+1}(x)\right)=g\left(f\left(f_{n}(x)\right)\right)=f_{n}(x)$,

$$f_{n+2}(x)+f_n(x)=2f_{n+1}(x)f\_\{n+2\}(x)+f\_\{n\}(x)=2\; f\_\{n+1\}(x)$$

that is,

$$f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_n(x)f\_\{n+2\}(x)-f\_\{n+1\}(x)=f\_\{n+1\}(x)-f\_\{n\}(x)$$

Therefore $f_{n}(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find

$$f_n(x)-x=n(f(x)-x)\mathrm{.}f\_\{n\}(x)-x=n(f(x)-x)\; .$$

Since $g$ has the same properties as $f$,

$$g_n(x)-x=n(g(x)-x)=-n(f(x)-x)\mathrm{.}g\_\{n\}(x)-x=n(g(x)-x)=-n(f(x)-x)\; .$$

Finally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y) \Longrightarrow f(g(x))>$ $f(g(y)) \Longrightarrow x>y$. An induction proves that $f_{n}$ and $g_{n}$ are also increasing functions. Let $x>y$ be real numbers. Since $f_{n}$ and $g_{n}$ are increasing,

$$x+n(f(x)-x)>y+n(f(y)-y)⟺n[(f(x)-x)-(f(y)-y)]>y-xx+n(f(x)-x)>y+n(f(y)-y)\; \backslash Longleftrightarrow\; n[(f(x)-x)-(f(y)-y)]>y-x$$

and

Summing it up,

Suppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$,

which is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x \in \mathbb{R}$, that is, $f(x)=x+c$. It is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.