APMO/md/en-apmo1994_sol.md

APMO 1994 - Problems and Solutions

Problem 1

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that (i) For all $x, y \in \mathbb{R}$,

$$f(x)+f(y)+1\ge f(x+y)\ge f(x)+f(y)f(x)+f(y)+1\; \backslash geq\; f(x+y)\; \backslash geq\; f(x)+f(y)$$

(ii) For all $x \in[0,1), f(0) \geq f(x)$, (iii) $-f(-1)=f(1)=1$.

Find all such functions $f$. Answer: $f(x)=\lfloor x\rfloor$, the largest integer that does not exceed $x$, is the only function.

Solution

Plug $y \rightarrow 1$ in (i):

$$f(x)+f(1)+1\ge f(x+1)\ge f(x)+f(1)⟺f(x)+1\le f(x+1)\le f(x)+2f(x)+f(1)+1\; \backslash geq\; f(x+1)\; \backslash geq\; f(x)+f(1)\; \backslash Longleftrightarrow\; f(x)+1\; \backslash leq\; f(x+1)\; \backslash leq\; f(x)+2$$

Now plug $y \rightarrow-1$ and $x \rightarrow x+1$ in (i):

$$f(x+1)+f(-1)+1\ge f(x)\ge f(x+1)+f(-1)⟺f(x)\le f(x+1)\le f(x)+1f(x+1)+f(-1)+1\; \backslash geq\; f(x)\; \backslash geq\; f(x+1)+f(-1)\; \backslash Longleftrightarrow\; f(x)\; \backslash leq\; f(x+1)\; \backslash leq\; f(x)+1$$

Hence $f(x+1)=f(x)+1$ and we only need to define $f(x)$ on $[0,1)$. Note that $f(1)=$ $f(0)+1 \Longrightarrow f(0)=0$. Condition (ii) states that $f(x) \leq 0$ in $[0,1)$. Now plug $y \rightarrow 1-x$ in (i):

$$f(x)+f(1-x)+1\le f(x+(1-x))\le f(x)+f(1-x)⟹f(x)+f(1-x)\ge 0f(x)+f(1-x)+1\; \backslash leq\; f(x+(1-x))\; \backslash leq\; f(x)+f(1-x)\; \backslash Longrightarrow\; f(x)+f(1-x)\; \backslash geq\; 0$$

If $x \in(0,1)$ then $1-x \in(0,1)$ as well, so $f(x) \leq 0$ and $f(1-x) \leq 0$, which implies $f(x)+f(1-x) \leq 0$. Thus, $f(x)=f(1-x)=0$ for $x \in(0,1)$. This combined with $f(0)=0$ and $f(x+1)=f(x)+1$ proves that $f(x)=\lfloor x\rfloor$, which satisfies the problem conditions, as since $x+y=\lfloor x\rfloor+\lfloor y\rfloor+{x}+{y}$ and $0 \leq{x}+{y}<2 \Longrightarrow\lfloor x\rfloor+\lfloor y\rfloor \leq x+y<\lfloor x\rfloor+\lfloor y\rfloor+2$ implies

$$\lfloor x\rfloor +\lfloor y\rfloor +1\ge \lfloor x+y\rfloor \ge \lfloor x\rfloor +\lfloor y\rfloor \mathrm{.}\backslash lfloor\; x\backslash rfloor+\backslash lfloor\; y\backslash rfloor+1\; \backslash geq\backslash lfloor\; x+y\backslash rfloor\; \backslash geq\backslash lfloor\; x\backslash rfloor+\backslash lfloor\; y\backslash rfloor\; .$$

Problem 2

Given a nondegenerate triangle $A B C$, with circumcentre $O$, orthocentre $H$, and circumradius $R$, prove that $|O H|<3 R$.

Solution 1

Embed $A B C$ in the complex plane, with $A, B$ and $C$ in the circle $|z|=R$, so $O$ is the origin. Represent each point by its lowercase letter. It is well known that $h=a+b+c$, so

$$OH=\mathrm{\mid }a+b+c\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }+\mathrm{\mid }c\mathrm{\mid }=3R\mathrm{.}O\; H=|a+b+c|\; \backslash leq|a|+|b|+|c|=3\; R\; .$$

The equality cannot occur because $a, b$, and $c$ are not collinear, so $O H<3 R$.

Solution 2

Suppose with loss of generality that $\angle A<90^{\circ}$. Let $B D$ be an altitude. Then

$$AH=\frac{AD}{\cos ({90}^{\circ }-C)}=\frac{AB\cos A}{\sin C}=2R\cos AA\; H=\backslash frac\{A\; D\}\{\backslash cos\; \backslash left(90^\{\backslash circ\}-C\backslash right)\}=\backslash frac\{A\; B\; \backslash cos\; A\}\{\backslash sin\; C\}=2\; R\; \backslash cos\; A$$

By the triangle inequality,

Comment: With a bit more work, if $a, b, c$ are the sidelengths of $A B C$, one can show that

$$O{H}^2=9R^2-a^2-b^2-c^2\mathrm{.}O\; H^\{2\}=9\; R^\{2\}-a^\{2\}-b^\{2\}-c^\{2\}\; .$$

In fact, using vectors in a coordinate system with $O$ as origin, by the Euler line

$$\overrightarrow{OH}=3\overrightarrow{OG}=3\cdot \frac{\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}}{3}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\backslash overrightarrow\{O\; H\}=3\; \backslash overrightarrow\{O\; G\}=3\; \backslash cdot\; \backslash frac\{\backslash overrightarrow\{O\; A\}+\backslash overrightarrow\{O\; B\}+\backslash overrightarrow\{O\; C\}\}\{3\}=\backslash overrightarrow\{O\; A\}+\backslash overrightarrow\{O\; B\}+\backslash overrightarrow\{O\; C\}$$

so

$$O{H}^2=\overrightarrow{OH}\cdot \overrightarrow{OH}=(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\cdot (\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})O\; H^\{2\}=\backslash overrightarrow\{O\; H\}\; \backslash cdot\; \backslash overrightarrow\{O\; H\}=(\backslash overrightarrow\{O\; A\}+\backslash overrightarrow\{O\; B\}+\backslash overrightarrow\{O\; C\})\; \backslash cdot(\backslash overrightarrow\{O\; A\}+\backslash overrightarrow\{O\; B\}+\backslash overrightarrow\{O\; C\})$$

Expanding and using the fact that $\overrightarrow{O X} \cdot \overrightarrow{O X}=O X^{2}=R^{2}$ for $X \in{A, B, C}$, as well as $\overrightarrow{O A} \cdot \overrightarrow{O B}=O A \cdot O B \cdot \cos \angle A O B=R^{2} \cos 2 C=R^{2}\left(1-2 \sin ^{2} C\right)=R^{2}\left(1-2\left(\frac{c}{2 R}\right)^{2}\right)=R^{2}-\frac{c^{2}}{2}$, we find that

as required. This proves that $O H^{2}<9 R^{2} \Longrightarrow O H<3 R$, and since $a, b, c$ can be arbitrarily small (fix the circumcircle and choose $A, B, C$ arbitrarily close in this circle), the bound is sharp.

Problem 3

Let $n$ be an integer of the form $a^{2}+b^{2}$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \leq \sqrt{n}$, then $p$ divides $a b$. Determine all such $n$.

Answer: $n=2,5,13$.

Solution

A prime $p$ divides $a b$ if and only if divides either $a$ or $b$. If $n=a^{2}+b^{2}$ is a composite then it has a prime divisor $p \leq \sqrt{n}$, and if $p$ divides $a$ it divides $b$ and vice-versa, which is not possible because $a$ and $b$ are coprime. Therefore $n$ is a prime. Suppose without loss of generality that $a \geq b$ and consider $a-b$. Note that $a^{2}+b^{2}=(a-b)^{2}+2 a b$.

• If $a=b$ then $a=b=1$ because $a$ and $b$ are coprime. $n=2$ is a solution.
• If $a-b=1$ then $a$ and $b$ are coprime and $a^{2}+b^{2}=(a-b)^{2}+2 a b=2 a b+1=2 b(b+1)+1=$ $2 b^{2}+2 b+1$. So any prime factor of any number smaller than $\sqrt{2 b^{2}+2 b+1}$ is a divisor of $a b=b(b+1)$. One can check that $b=1$ and $b=2$ yields the solutions $n=1^{2}+2^{2}=5$ (the only prime $p$ is 2 ) and $n=2^{2}+3^{2}=13$ (the only primes $p$ are 2 and 3 ). Suppose that $b>2$. Consider, for instance, the prime factors of $b-1 \leq \sqrt{2 b^{2}+2 b+1}$, which is coprime with $b$. Any prime must then divide $a=b+1$. Then it divides $(b+1)-(b-1)=2$, that is, $b-1$ can only have 2 as a prime factor, that is, $b-1$ is a power of 2 , and since $b-1 \geq 2$, $b$ is odd. Since $2 b^{2}+2 b+1-(b+2)^{2}=b^{2}-2 b-3=(b-3)(b+1) \geq 0$, we can also consider any prime divisor of $b+2$. Since $b$ is odd, $b$ and $b+2$ are also coprime, so any prime divisor of $b+2$ must divide $a=b+1$. But $b+1$ and $b+2$ are also coprime, so there can be no such primes. This is a contradiction, and $b \geq 3$ does not yield any solutions.
• If $a-b>1$, consider a prime divisor $p$ of $a-b=\sqrt{a^{2}-2 a b+b^{2}}<\sqrt{a^{2}+b^{2}}$. Since $p$ divides one of $a$ and $b, p$ divides both numbers (just add or subtract $a-b$ accordingly.) This is a contradiction.

Hence the only solutions are $n=2,5,13$.

Problem 4

Is there an infinite set of points in the plane such that no three points are collinear, and the distance between any two points is rational? Answer: Yes.

Solution 1

The answer is yes and we present the following construction: the idea is considering points in the unit circle of the form $P_{n}=(\cos (2 n \theta), \sin (2 n \theta))$ for an appropriate $\theta$. Then the distance $P_{m} P_{n}$ is the length of the chord with central angle $(2 m-2 n) \theta \bmod \pi$, that is, $2|\sin ((m-n) \theta)|$. Our task is then finding $\theta$ such that (i) $\sin (k \theta)$ is rational for all $k \in \mathbb{Z}$; (ii) points $P_{n}$ are all distinct. We claim that $\theta \in(0, \pi / 2)$ such that $\cos \theta=\frac{3}{5}$ and therefore $\sin \theta=\frac{4}{5}$ does the job. Proof of (i): We know that $\sin ((n+1) \theta)+\sin ((n-1) \theta)=2 \sin (n \theta) \cos \theta$, so if $\sin ((n-1) \theta$ and $\sin (n \theta)$ are both rational then $\sin ((n+1) \theta)$ also is. Since $\sin (0 \theta)=0$ and $\sin \theta$ are rational, an induction shows that $\sin (n \theta)$ is rational for $n \in \mathbb{Z}{>0}$; the result is also true if $n$ is negative because $\sin$ is an odd function. Proof of (ii): $P{m}=P_{n} \Longleftrightarrow 2 n \theta=2 m \theta+2 k \pi$ for some $k \in \mathbb{Z}$, which implies $\sin ((n-m) \theta)=$ $\sin (k \pi)=0$. We show that $\sin (k \theta) \neq 0$ for all $k \neq 0$. We prove a stronger result: let $\sin (k \theta)=\frac{a_{k}}{5^{k}}$. Then

Since $a_{0}=0$ and $a_{1}=4, a_{k}$ is an integer for $k \geq 0$, and $a_{k+1} \equiv a_{k}(\bmod 5)$ for $k \geq 1$ (note that $a_{-1}=-\frac{4}{25}$ is not an integer!). Thus $a_{k} \equiv 4(\bmod 5)$ for all $k \geq 1$, and $\sin (k \theta)=\frac{a_{k}}{5^{k}}$ is an irreducible fraction with $5^{k}$ as denominator and $a_{k} \equiv 4(\bmod 5)$. This proves (ii) and we are done.

Solution 2

We present a different construction. Consider the (collinear) points

$$P_k=(1,\frac{x_k}{y_k}),P\_\{k\}=\backslash left(1,\; \backslash frac\{x\_\{k\}\}\{y\_\{k\}\}\backslash right),$$

such that the distance $O P_{k}$ from the origin $O$,

$$O{P}_k=\frac{\sqrt{x_k^2+y_k^2}}{y_k}O\; P\_\{k\}=\backslash frac\{\backslash sqrt\{x\_\{k\}^\{2\}+y\_\{k\}^\{2\}\}\}\{y\_\{k\}\}$$

is rational, and $x_{k}$ and $y_{k}$ are integers. Clearly, $P_{i} P_{j}=\left|\frac{x_{i}}{y_{i}}-\frac{x_{j}}{y_{j}}\right|$ is rational. Perform an inversion with center $O$ and unit radius. It maps the line $x=1$, which contains all points $P_{k}$, to a circle (minus the origin). Let $Q_{k}$ be the image of $P_{k}$ under this inversion. Then

$$Q_i{Q}_j=\frac{1^2{P}_i{P}_j}{O{P}_i\cdot O{P}_j}Q\_\{i\}\; Q\_\{j\}=\backslash frac\{1^\{2\}\; P\_\{i\}\; P\_\{j\}\}\{O\; P\_\{i\}\; \backslash cdot\; O\; P\_\{j\}\}$$

is rational and we are done if we choose $x_{k}$ and $y_{k}$ accordingly. But this is not hard, as we can choose the legs of a Pythagorean triple, say

$$x_k=k^2-1,y_k=2kx\_\{k\}=k^\{2\}-1,\; \backslash quad\; y\_\{k\}=2\; k$$

This implies $O P_{k}=\frac{k^{2}+1}{2 k}$, and then

$$Q_i{Q}_j=\frac{\mid \frac{i^2-1}{i}-\frac{j^2-1}{j}\mid }{\frac{i^2+1}{2i}\cdot \frac{j^2+1}{2j}}=\frac{\mathrm{\mid }4(i-j)(ij+1)\mathrm{\mid }}{(i^2+1)(j^2+1)}Q\_\{i\}\; Q\_\{j\}=\backslash frac\{\backslash left|\backslash frac\{i^\{2\}-1\}\{i\}-\backslash frac\{j^\{2\}-1\}\{j\}\backslash right|\}\{\backslash frac\{i^\{2\}+1\}\{2\; i\}\; \backslash cdot\; \backslash frac\{j^\{2\}+1\}\{2\; j\}\}=\backslash frac\{|4(i-j)(i\; j+1)|\}\{\backslash left(i^\{2\}+1\backslash right)\backslash left(j^\{2\}+1\backslash right)\}$$

Problem 5

You are given three lists $A, B$, and $C$. List $A$ contains the numbers of the form $10^{k}$ in base 10, with $k$ any integer greater than or equal to 1 . Lists $B$ and $C$ contain the same numbers translated into base 2 and 5 respectively:

$A$	$B$	$C$
10	1010	20
100	1100100	400
1000	1111101000	13000
$\vdots$	$\vdots$	$\vdots$

Prove that for every integer $n>1$, there is exactly one number in exactly one of the lists $B$ or $C$ that has exactly $n$ digits.

Solution

Let $b_{k}$ and $c_{k}$ be the number of digits in the $k$ th term in lists $B$ and $C$, respectively. Then

and, similarly

$$c_k=\lfloor k\cdot {\log }_{5}10\rfloor +1c\_\{k\}=\backslash left\backslash lfloor\; k\; \backslash cdot\; \backslash log\; \_\{5\}\; 10\backslash right\backslash rfloor+1$$

Beatty's theorem states that if $\alpha$ and $\beta$ are irrational positive numbers such that

$$\frac{1}{\alpha }+\frac{1}{\beta }=1\backslash frac\{1\}\{\backslash alpha\}+\backslash frac\{1\}\{\backslash beta\}=1$$

then the sequences $\lfloor k \alpha\rfloor$ and $\lfloor k \beta\rfloor, k=1,2, \ldots$, partition the positive integers. Then, since

$$\frac{1}{{\log }_{2}10}+\frac{1}{{\log }_{5}10}={\log }_{10}2+{\log }_{10}5={\log }_{10}(2\cdot 5)=1\backslash frac\{1\}\{\backslash log\; \_\{2\}\; 10\}+\backslash frac\{1\}\{\backslash log\; \_\{5\}\; 10\}=\backslash log\; \_\{10\}\; 2+\backslash log\; \_\{10\}\; 5=\backslash log\; \_\{10\}(2\; \backslash cdot\; 5)=1$$

the sequences $b_{k}-1$ and $c_{k}-1$ partition the positive integers, and therefore each integer greater than 1 appears in $b_{k}$ or $c_{k}$ exactly once. We are done. Comment: For the sake of completeness, a proof of Beatty's theorem follows. Let $x_{n}=\alpha n$ and $y_{n}=\beta n, n \geq 1$ integer. Note that, since $\alpha m=\beta n$ implies that $\frac{\alpha}{\beta}$ is rational but

$$\frac{\alpha }{\beta }=\alpha \cdot \frac{1}{\beta }=\alpha (1-\frac{1}{\alpha })=\alpha -1\backslash frac\{\backslash alpha\}\{\backslash beta\}=\backslash alpha\; \backslash cdot\; \backslash frac\{1\}\{\backslash beta\}=\backslash alpha\backslash left(1-\backslash frac\{1\}\{\backslash alpha\}\backslash right)=\backslash alpha-1$$

is irrational, the sequences have no common terms, and all terms in both sequences are irrational. The theorem is equivalent to proving that exactly one term of either $x_{n}$ of $y_{n}$ lies in the interval $(N, N+1)$ for each $N$ positive integer. For that purpose we count the number of terms of the union of the two sequences in the interval $(0, N)$ : since $n \alpha<N \Longleftrightarrow n<\frac{N}{\alpha}$, there are $\left\lfloor\frac{N}{\alpha}\right\rfloor$ terms of $x_{n}$ in the interval and, similarly, $\left\lfloor\frac{N}{\beta}\right\rfloor$ terms of $y_{n}$ in the same interval. Since the sequences are disjoint, the total of numbers is

$$T(N)=\lfloor \frac{N}{\alpha }\rfloor +\lfloor \frac{N}{\beta }\rfloor T(N)=\backslash left\backslash lfloor\backslash frac\{N\}\{\backslash alpha\}\backslash right\backslash rfloor+\backslash left\backslash lfloor\backslash frac\{N\}\{\backslash beta\}\backslash right\backslash rfloor$$

However, $x-1<\lfloor x\rfloor<x$ for nonintegers $x$, so

that is, $T(N)=N-1$. Therefore the number of terms in $(N, N+1)$ is $T(N+1)-T(N)=N-(N-1)=1$, and the result follows.