APMO/md/en-apmo1993_sol.md

APMO 1993 - Problems and Solutions

Problem 1

Let $A B C D$ be a quadrilateral such that all sides have equal length and angle $\angle A B C$ is 60 degrees. Let $\ell$ be a line passing through $D$ and not intersecting the quadrilateral (except at $D)$. Let $E$ and $F$ be the points of intersection of $\ell$ with $A B$ and $B C$ respectively. Let $M$ be the point of intersection of $C E$ and $A F$. Prove that $C A^{2}=C M \times C E$.

Solution

Triangles $A E D$ and $C D F$ are similar, because $A D | C F$ and $A E | C D$. Thus, since $A B C$ and $A C D$ are equilateral triangles,

$$\frac{AE}{CD}=\frac{AD}{CF}⟺\frac{AE}{AC}=\frac{AC}{CF}\backslash frac\{A\; E\}\{C\; D\}=\backslash frac\{A\; D\}\{C\; F\}\; \backslash Longleftrightarrow\; \backslash frac\{A\; E\}\{A\; C\}=\backslash frac\{A\; C\}\{C\; F\}$$

The last equality combined with

$$\mathrm{\angle }EAC={180}^{\circ }-\mathrm{\angle }BAC={120}^{\circ }=\mathrm{\angle }ACF\backslash angle\; E\; A\; C=180^\{\backslash circ\}-\backslash angle\; B\; A\; C=120^\{\backslash circ\}=\backslash angle\; A\; C\; F$$

shows that triangles $E A C$ and $A C F$ are also similar. Therefore $\angle C A M=\angle C A F=\angle A E C$, which implies that line $A C$ is tangent to the circumcircle of $A M E$. By the power of a point, $C A^{2}=C M \cdot C E$, and we are done.

Problem 2

Find the total number of different integer values the function

$$f(x)=[x]+[2x]+\left[\frac{5x}{3}\right]+[3x]+[4x]f(x)=[x]+[2\; x]+\backslash left[\backslash frac\{5\; x\}\{3\}\backslash right]+[3\; x]+[4\; x]$$

takes for real numbers $x$ with $0 \leq x \leq 100$. Note: $[t]$ is the largest integer that does not exceed $t$. Answer: 734.

Solution

Note that, since $[x+n]=[x]+n$ for any integer $n$,

$$f(x+3)=[x+3]+[2(x+3)]+\left[\frac{5(x+3)}{3}\right]+[3(x+3)]+[4(x+3)]=f(x)+35,f(x+3)=[x+3]+[2(x+3)]+\backslash left[\backslash frac\{5(x+3)\}\{3\}\backslash right]+[3(x+3)]+[4(x+3)]=f(x)+35,$$

one only needs to investigate the interval $[0,3)$. The numbers in this interval at which at least one of the real numbers $x, 2 x, \frac{5 x}{3}, 3 x, 4 x$ is an integer are

• $0,1,2$ for $x$;
• $\frac{n}{2}, 0 \leq n \leq 5$ for $2 x$;
• $\frac{3 n}{5}, 0 \leq n \leq 4$ for $\frac{5 x}{3}$;
• $\frac{n}{3}, 0 \leq n \leq 8$ for $3 x$;
• $\frac{n}{4}, 0 \leq n \leq 11$ for $4 x$.

Of these numbers there are

• 3 integers $(0,1,2)$;
• 3 irreducible fractions with 2 as denominator (the numerators are $1,3,5$ );
• 6 irreducible fractions with 3 as denominator (the numerators are $1,2,4,5,7,8$ );
• 6 irreducible fractions with 4 as denominator (the numerators are $1,3,5,7,9,11,13,15$ );
• 4 irreducible fractions with 5 as denominator (the numerators are 3,6,9,12).

Therefore $f(x)$ increases 22 times per interval. Since $100=33 \cdot 3+1$, there are $33 \cdot 22$ changes of value in $[0,99)$. Finally, there are 8 more changes in [99,100]: 99, 100, $99 \frac{1}{2}, 99 \frac{1}{3}, 99 \frac{2}{3}, 99 \frac{1}{4}$, $99 \frac{3}{4}, 99 \frac{3}{5}$. The total is then $33 \cdot 22+8=734$. Comment: A more careful inspection shows that the range of $f$ are the numbers congruent modulo 35 to one of

$$0,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,300,1,2,4,5,6,7,11,12,13,14,16,17,18,19,23,24,25,26,28,29,30$$

in the interval $[0, f(100)]=[0,1166]$. Since $1166 \equiv 11(\bmod 35)$, this comprises 33 cycles plus the 8 numbers in the previous list.

Problem 3

Let

$$f(x)=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0\text{ and }g(x)=c_{n+1}{x}^{n+1}+c_n{x}^n+\cdots +c_{0}f(x)=a\_\{n\}\; x^\{n\}+a\_\{n-1\}\; x^\{n-1\}+\backslash cdots+a\_\{0\}\; \backslash quad\; \backslash text\; \{\; and\; \}\; \backslash quad\; g(x)=c\_\{n+1\}\; x^\{n+1\}+c\_\{n\}\; x^\{n\}+\backslash cdots+c\_\{0\}$$

be non-zero polynomials with real coefficients such that $g(x)=(x+r) f(x)$ for some real number $r$. If $a=\max \left(\left|a_{n}\right|, \ldots,\left|a_{0}\right|\right)$ and $c=\max \left(\left|c_{n+1}\right|, \ldots,\left|c_{0}\right|\right)$, prove that $\frac{a}{c} \leq n+1$.

Solution

Expanding $(x+r) f(x)$, we find that $c_{n+1}=a_{n}, c_{k}=a_{k-1}+r a_{k}$ for $k=1,2, \ldots, n$, and $c_{0}=r a_{0}$. Consider three cases:

• $r=0$. Then $c_{0}=0$ and $c_{k}=a_{k-1}$ for $k=1,2, \ldots, n$, and $a=c \Longrightarrow \frac{a}{c}=1 \leq n+1$.
• $|r| \geq 1$. Then

$$\begin{array}{c}{\displaystyle \mid a_0\mid =\mid \frac{c_0}{r}\mid \le c}\\ {\displaystyle \mid a_1\mid =\mid \frac{c_1-a_0}{r}\mid \le \mid c_1\mid +\mid a_0\mid \le 2c}\end{array}\backslash begin\{gathered\}\; \backslash left|a\_\{0\}\backslash right|=\backslash left|\backslash frac\{c\_\{0\}\}\{r\}\backslash right|\; \backslash leq\; c\; \backslash \backslash \; \backslash left|a\_\{1\}\backslash right|=\backslash left|\backslash frac\{c\_\{1\}-a\_\{0\}\}\{r\}\backslash right|\; \backslash leq\backslash left|c\_\{1\}\backslash right|+\backslash left|a\_\{0\}\backslash right|\; \backslash leq\; 2\; c\; \backslash end\{gathered\}$$

and inductively if $\left|a_{k}\right| \leq(k+1) c$

$$\mid a_{k+1}\mid =\mid \frac{c_{k+1}-a_k}{r}\mid \le \mid c_{k+1}\mid +\mid a_k\mid \le c+(k+1)c=(k+2)c\backslash left|a\_\{k+1\}\backslash right|=\backslash left|\backslash frac\{c\_\{k+1\}-a\_\{k\}\}\{r\}\backslash right|\; \backslash leq\backslash left|c\_\{k+1\}\backslash right|+\backslash left|a\_\{k\}\backslash right|\; \backslash leq\; c+(k+1)\; c=(k+2)\; c$$

Therefore, $\left|a_{k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c \Longleftrightarrow \frac{a}{c} \leq n+1$.

• $0<|r|<1$. Now work backwards: $\left|a_{n}\right|=\left|c_{n+1}\right| \leq c$,

and inductively if $\left|a_{n-k}\right| \leq(k+1) c$

Therefore, $\left|a_{n-k}\right| \leq(k+1) c \leq(n+1) c$ for all $k$, and $a \leq(n+1) c$ again.

Problem 4

Determine all positive integers $n$ for which the equation

$$x^n+(2+x{)}^n+(2-x{)}^n=0x^\{n\}+(2+x)^\{n\}+(2-x)^\{n\}=0$$

has an integer as a solution.

Answer: $n=1$.

Solution

If $n$ is even, $x^{n}+(2+x)^{n}+(2-x)^{n}>0$, so $n$ is odd. For $n=1$, the equation reduces to $x+(2+x)+(2-x)=0$, which has the unique solution $x=-4$. For $n>1$, notice that $x$ is even, because $x, 2-x$, and $2+x$ have all the same parity. Let $x=2 y$, so the equation reduces to

$$y^n+(1+y{)}^n+(1-y{)}^n=0.y^\{n\}+(1+y)^\{n\}+(1-y)^\{n\}=0\; .$$

Looking at this equation modulo 2 yields that $y+(1+y)+(1-y)=y+2$ is even, so $y$ is even. Using the factorization

$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+\cdots +b^{n-1})\text{ for }n\text{ odd, }a^\{n\}+b^\{n\}=(a+b)\backslash left(a^\{n-1\}-a^\{n-2\}\; b+\backslash cdots+b^\{n-1\}\backslash right)\; \backslash quad\; \backslash text\; \{\; for\; \}\; n\; \backslash text\; \{\; odd,\; \}$$

which has a sum of $n$ terms as the second factor, the equation is now equivalent to

$$y^n+(1+y+1-y)((1+y{)}^{n-1}-(1+y{)}^{n-2}(1-y)+\cdots +(1-y{)}^{n-1})=0y^\{n\}+(1+y+1-y)\backslash left((1+y)^\{n-1\}-(1+y)^\{n-2\}(1-y)+\backslash cdots+(1-y)^\{n-1\}\backslash right)=0$$

or

$$y^n=-2((1+y{)}^{n-1}-(1+y{)}^{n-2}(1-y)+\cdots +(1-y{)}^{n-1})\mathrm{.}y^\{n\}=-2\backslash left((1+y)^\{n-1\}-(1+y)^\{n-2\}(1-y)+\backslash cdots+(1-y)^\{n-1\}\backslash right)\; .$$

Each of the $n$ terms in the second factor is odd, and $n$ is odd, so the second factor is odd. Therefore, $y^{n}$ has only one factor 2 , which is a contradiction to the fact that, $y$ being even, $y^{n}$ has at least $n>1$ factors 2 . Hence there are no solutions if $n>1$.

Problem 5

Let $P_{1}, P_{2}, \ldots, P_{1993}=P_{0}$ be distinct points in the $x y$-plane with the following properties: (i) both coordinates of $P_{i}$ are integers, for $i=1,2, \ldots, 1993$; (ii) there is no point other than $P_{i}$ and $P_{i+1}$ on the line segment joining $P_{i}$ with $P_{i+1}$ whose coordinates are both integers, for $i=0,1, \ldots, 1992$.

Prove that for some $i, 0 \leq i \leq 1992$, there exists a point $Q$ with coordinates $\left(q_{x}, q_{y}\right)$ on the line segment joining $P_{i}$ with $P_{i+1}$ such that both $2 q_{x}$ and $2 q_{y}$ are odd integers.

Solution

Call a point $(x, y) \in \mathbb{Z}^{2}$ even or odd according to the parity of $x+y$. Since there are an odd number of points, there are two points $P_{i}=(a, b)$ and $P_{i+1}=(c, d), 0 \leq i \leq 1992$ with the same parity. This implies that $a+b+c+d$ is even. We claim that the midpoint of $P_{i} P_{i+1}$ is the desired point $Q$. In fact, since $a+b+c+d=(a+c)+(b+d)$ is even, $a$ and $c$ have the same parity if and only if $b$ and $d$ also have the same parity. If both happen then the midpoint of $P_{i} P_{i+1}, Q=\left(\frac{a+c}{2}, \frac{b+d}{2}\right)$, has integer coordinates, which violates condition (ii). Then $a$ and $c$, as well as $b$ and $d$, have different parities, and $2 q_{x}=a+c$ and $2 q_{y}=b+d$ are both odd integers.