APMO/md/en-apmo1991_sol.md

APMO 1991 - Problems and Solutions

Problem 1

Let $G$ be the centroid of triangle $A B C$ and $M$ be the midpoint of $B C$. Let $X$ be on $A B$ and $Y$ on $A C$ such that the points $X, Y$, and $G$ are collinear and $X Y$ and $B C$ are parallel. Suppose that $X C$ and $G B$ intersect at $Q$ and $Y B$ and $G C$ intersect at $P$. Show that triangle $M P Q$ is similar to triangle $A B C$.

Solution 1

Let $R$ be the midpoint of $A C$; so $B R$ is a median and contains the centroid $G$.

It is well known that $\frac{A G}{A M}=\frac{2}{3}$; thus the ratio of the similarity between $A X Y$ and $A B C$ is $\frac{2}{3}$. Hence $G X=\frac{1}{2} X Y=\frac{1}{3} B C$. Now look at the similarity between triangles $Q B C$ and $Q G X$ :

$$\frac{QG}{QB}=\frac{GX}{BC}=\frac{1}{3}⟹QB=3QG⟹QB=\frac{3}{4}BG=\frac{3}{4}\cdot \frac{2}{3}BR=\frac{1}{2}BR\mathrm{.}\backslash frac\{Q\; G\}\{Q\; B\}=\backslash frac\{G\; X\}\{B\; C\}=\backslash frac\{1\}\{3\}\; \backslash Longrightarrow\; Q\; B=3\; Q\; G\; \backslash Longrightarrow\; Q\; B=\backslash frac\{3\}\{4\}\; B\; G=\backslash frac\{3\}\{4\}\; \backslash cdot\; \backslash frac\{2\}\{3\}\; B\; R=\backslash frac\{1\}\{2\}\; B\; R\; .$$

Finally, since $\frac{B M}{B C}=\frac{B Q}{B R}, M Q$ is a midline in $B C R$. Therefore $M Q=\frac{1}{2} C R=\frac{1}{4} A C$ and $M Q | A C$. Similarly, $M P=\frac{1}{4} A B$ and $M P | A B$. This is sufficient to establish that $M P Q$ and $A B C$ are similar (with similarity ratio $\frac{1}{4}$ ).

Solution 2

Let $S$ and $R$ be the midpoints of $A B$ and $A C$, respectively. Since $G$ is the centroid, it lies in the medians $B R$ and $C S$.

Due to the similarity between triangles $Q B C$ and $Q G X$ (which is true because $G X | B C$ ), there is an inverse homothety with center $Q$ and ratio $-\frac{X G}{B C}=\frac{X Y}{2 B C}$ that takes $B$ to $G$ and $C$ to $X$. This homothety takes the midpoint $M$ of $B C$ to the midpoint $K$ of $G X$.

Now consider the homothety that takes $B$ to $X$ and $C$ to $G$. This new homothety, with ratio $\frac{X Y}{2 B C}$, also takes $M$ to $K$. Hence lines $B X$ (which contains side $A B$ ), $C G$ (which contains the median $C S$ ), and $M K$ have a common point, which is $S$. Thus $Q$ lies on midline $M S$. The same reasoning proves that $P$ lies on midline $M R$. Since all homothety ratios are the same, $\frac{M Q}{M S}=\frac{M P}{M R}$, which shows that $M P Q$ is similar to $M R S$, which in turn is similar to $A B C$, and we are done.

Problem 2

Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points?

Solution

Embed the points in the cartesian plane such that no two points have the same $y$-coordinate. Let $P_{1}, P_{2}, \ldots, P_{997}$ be the points and $y_{1}<y_{2}<\ldots<y_{997}$ be their respective $y$-coordinates. Then the $y$-coordinate of the midpoint of $P_{i} P_{i+1}, i=1,2, \ldots, 996$ is $\frac{y_{i}+y_{i+1}}{2}$ and the $y$-coordinate of the midpoint of $P_{i} P_{i+2}, i=1,2, \ldots, 995$ is $\frac{y_{i}+y_{i+2}}{2}$. Since

there are at least $996+995=1991$ distinct midpoints, and therefore at least 1991 red points. The equality case happens if we take $P_{i}=(0,2 i), i=1,2, \ldots, 997$. The midpoints are $(0, i+j)$, $1 \leq i<j \leq 997$, which are the points $(0, k)$ with $1+2=3 \leq k \leq 996+997=1993$, a total of $1993-3+1=1991$ red points.

Problem 3

Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=b_{1}+b_{2}+$ $\cdots+b_{n}$. Show that

$$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+\cdots +\frac{a_n^2}{a_n+b_n}\ge \frac{a_1+a_2+\cdots +a_n}{2}\backslash frac\{a\_\{1\}^\{2\}\}\{a\_\{1\}+b\_\{1\}\}+\backslash frac\{a\_\{2\}^\{2\}\}\{a\_\{2\}+b\_\{2\}\}+\backslash cdots+\backslash frac\{a\_\{n\}^\{2\}\}\{a\_\{n\}+b\_\{n\}\}\; \backslash geq\; \backslash frac\{a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\}\{2\}$$

Solution

By the Cauchy-Schwartz inequality,

$$(\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+\cdots +\frac{a_n^2}{a_n+b_n})((a_1+b_1)+(a_2+b_2)+\cdots +(a_n+b_n))\ge {(a_1+a_2+\cdots +a_n)}^2\mathrm{.}\backslash left(\backslash frac\{a\_\{1\}^\{2\}\}\{a\_\{1\}+b\_\{1\}\}+\backslash frac\{a\_\{2\}^\{2\}\}\{a\_\{2\}+b\_\{2\}\}+\backslash cdots+\backslash frac\{a\_\{n\}^\{2\}\}\{a\_\{n\}+b\_\{n\}\}\backslash right)\backslash left(\backslash left(a\_\{1\}+b\_\{1\}\backslash right)+\backslash left(a\_\{2\}+b\_\{2\}\backslash right)+\backslash cdots+\backslash left(a\_\{n\}+b\_\{n\}\backslash right)\backslash right)\; \backslash geq\backslash left(a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\backslash right)^\{2\}\; .$$

Since $\left(\left(a_{1}+b_{1}\right)+\left(a_{2}+b_{2}\right)+\cdots+\left(a_{n}+b_{n}\right)\right)=2\left(a_{1}+a_{2}+\cdots+a_{n}\right)$,

$$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+\cdots +\frac{a_n^2}{a_n+b_n}\ge \frac{{(a_1+a_2+\cdots +a_n)}^2}{2(a_1+a_2+\cdots +a_n)}=\frac{a_1+a_2+\cdots +a_n}{2}\backslash frac\{a\_\{1\}^\{2\}\}\{a\_\{1\}+b\_\{1\}\}+\backslash frac\{a\_\{2\}^\{2\}\}\{a\_\{2\}+b\_\{2\}\}+\backslash cdots+\backslash frac\{a\_\{n\}^\{2\}\}\{a\_\{n\}+b\_\{n\}\}\; \backslash geq\; \backslash frac\{\backslash left(a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\backslash right)^\{2\}\}\{2\backslash left(a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\backslash right)\}=\backslash frac\{a\_\{1\}+a\_\{2\}+\backslash cdots+a\_\{n\}\}\{2\}$$

Problem 4

During a break, $n$ children at school sit in a circle around their teacher to play a game. The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and soon. Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.

Answer: All powers of 2 .

Solution 1

Number the children from 0 to $n-1$. Then the teacher hands candy to children in positions $f(x)=1+2+\cdots+x \bmod n=\frac{x(x+1)}{2} \bmod n$. Our task is to find the range of $f: \mathbb{Z}{n} \rightarrow \mathbb{Z}{n}$, and to verify whether the range is $\mathbb{Z}_{n}$, that is, whether $f$ is a bijection. If $n=2^{a} m, m>1$ odd, look at $f(x)$ modulo $m$. Since $m$ is odd, $m|f(x) \Longleftrightarrow m| x(x+1)$. Then, for instance, $f(x) \equiv 0(\bmod m)$ for $x=0$ and $x=m-1$. This means that $f(x)$ is not a bijection modulo $m$, and there exists $t$ such that $f(x) \not \equiv t(\bmod m)$ for all $x$. By the Chinese Remainder Theorem,

Therefore, $f$ is not a bijection modulo $n$. If $n=2^{a}$, then

$$f(x)-f(y)=\frac{1}{2}(x(x+1)-y(y+1))=\frac{1}{2}(x^2-y^2+x-y)=\frac{(x-y)(x+y+1)}{2}\mathrm{.}f(x)-f(y)=\backslash frac\{1\}\{2\}(x(x+1)-y(y+1))=\backslash frac\{1\}\{2\}\backslash left(x^\{2\}-y^\{2\}+x-y\backslash right)=\backslash frac\{(x-y)(x+y+1)\}\{2\}\; .$$

and

$$f(x)\equiv f(y)(2^a)⟺(x-y)(x+y+1)\equiv 0(2^{a+1})f(x)\; \backslash equiv\; f(y)\; \backslash quad\backslash left(\backslash bmod\; 2^\{a\}\backslash right)\; \backslash Longleftrightarrow(x-y)(x+y+1)\; \backslash equiv\; 0\; \backslash quad\backslash left(\backslash bmod\; 2^\{a+1\}\backslash right)$$

If $x$ and $y$ have the same parity, $x+y+1$ is odd and $(*)$ is equivalent to $x \equiv y\left(\bmod 2^{a+1}\right)$. If $x$ and $y$ have different parity,

$$(\ast )⟺x+y+1\equiv 0(2^{a+1})(*)\; \backslash Longleftrightarrow\; x+y+1\; \backslash equiv\; 0\; \backslash quad\backslash left(\backslash bmod\; 2^\{a+1\}\backslash right)$$

However, $1 \leq x+y+1 \leq 2\left(2^{a}-1\right)+1=2^{a+1}-1$, so $x+y+1$ is not a multiple of $2^{a+1}$. Therefore $f$ is a bijection if $n$ is a power of 2 .

Solution 2

We give a full description of $a_{n}$, the size of the range of $f$. Since congruences modulo $n$ are defined, via Chinese Remainder Theorem, by congruences modulo $p^{\alpha}$ for all prime divisors $p$ of $n$ and $\alpha$ being the number of factors $p$ in the factorization of $n, a_{n}=\prod_{p^{\alpha} | n} a_{p^{\alpha}}$. Refer to the first solution to check the case $p=2: a_{2^{\alpha}}=2^{\alpha}$. For an odd prime $p$,

$$f(x)=\frac{x(x+1)}{2}=\frac{(2x+1{)}^2-1}{8}f(x)=\backslash frac\{x(x+1)\}\{2\}=\backslash frac\{(2\; x+1)^\{2\}-1\}\{8\}$$

and since $p$ is odd, there is a bijection between the range of $f$ and the quadratic residues modulo $p^{\alpha}$, namely $t \mapsto 8 t+1$. So $a_{p^{\alpha}}$ is the number of quadratic residues modulo $p^{\alpha}$. Let $g$ be a primitive root of $p^{\alpha}$. Then there are $\frac{1}{2} \phi\left(p^{\alpha}\right)=\frac{p-1}{2} \cdot p^{\alpha-1}$ quadratic residues that are coprime with $p: 1, g^{2}, g^{4}, \ldots, g^{\phi\left(p^{n}\right)-2}$. If $p$ divides a quadratic residue $k p$, that is, $x^{2} \equiv k p$ $\left(\bmod p^{\alpha}\right), \alpha \geq 2$, then $p$ divides $x$ and, therefore, also $k$. Hence $p^{2}$ divides this quadratic residue, and these quadratic residues are $p^{2}$ times each quadratic residue of $p^{\alpha-2}$. Thus

$$a_{p^{\alpha }}=\frac{p-1}{2}\cdot p^{\alpha -1}+a_{p^{\alpha }-2}\mathrm{.}a\_\{p^\{\backslash alpha\}\}=\backslash frac\{p-1\}\{2\}\; \backslash cdot\; p^\{\backslash alpha-1\}+a\_\{p^\{\backslash alpha\}-2\}\; .$$

Since $a_{p}=\frac{p-1}{2}+1$ and $a_{p^{2}}=\frac{p-1}{2} \cdot p+1$, telescoping yields

$$a_{p^{2t}}=\frac{p-1}{2}(p^{2t-1}+p^{2t-3}+\cdots +p)+1=\frac{p(p^{2t}-1)}{2(p+1)}+1a\_\{p^\{2\; t\}\}=\backslash frac\{p-1\}\{2\}\backslash left(p^\{2\; t-1\}+p^\{2\; t-3\}+\backslash cdots+p\backslash right)+1=\backslash frac\{p\backslash left(p^\{2\; t\}-1\backslash right)\}\{2(p+1)\}+1$$

and

$$a_{p^{2t-1}}=\frac{p-1}{2}(p^{2t-2}+p^{2t-4}+\cdots +1)+1=\frac{p^{2t}-1}{2(p+1)}+1a\_\{p^\{2\; t-1\}\}=\backslash frac\{p-1\}\{2\}\backslash left(p^\{2\; t-2\}+p^\{2\; t-4\}+\backslash cdots+1\backslash right)+1=\backslash frac\{p^\{2\; t\}-1\}\{2(p+1)\}+1$$

Now the problem is immediate: if $n$ is divisible by an odd prime $p, a_{p^{\alpha}}<p^{\alpha}$ for all $\alpha$, and since $a_{t} \leq t$ for all $t, a_{n}<n$.

Problem 5

Given are two tangent circles and a point $P$ on their common tangent perpendicular to the lines joining their centres. Construct with ruler and compass all the circles that are tangent to these two circles and pass through the point $P$.

Solution

Throughout this problem, we will assume that the given circles are externally tangent, since the problem does not have a solution otherwise. Let $\Gamma_{1}$ and $\Gamma_{2}$ be the given circles and $T$ be their tangency point. Suppose $\omega$ is a circle that is tangent to $\Gamma_{1}$ and $\Gamma_{2}$ and passes through $P$. Now invert about point $P$, with radius $P T$. Let any line through $P$ that cuts $\Gamma_{1}$ do so at points $X$ and $Y$. The power of $P$ with respect to $\Gamma_{1}$ is $P T^{2}=P X \cdot P Y$, so $X$ and $Y$ are swapped by this inversion. Therefore $\Gamma_{1}$ is mapped to itself in this inversion. The same applies to $\Gamma_{2}$. Since circle $\omega$ passes through $P$, it is mapped to a line tangent to the images of $\Gamma_{1}$ (itself) and $\Gamma_{2}$ (also itself), that is, a common tangent line. This common tangent cannot be $P T$, as $P T$ is also mapped to itself. Since $\Gamma_{1}$ and $\Gamma_{2}$ have exactly other two common tangent lines, there are two solutions: the inverses of the tangent lines.

We proceed with the construction with the aid of some macro constructions that will be detailed later.

Step 1. Draw the common tangents to $\Gamma_{1}$ and $\Gamma_{2}$. Step 2. For each common tangent $t$, draw the projection $P_{t}$ of $P$ onto $t$. Step 3. Find the inverse $P_{1}$ of $P_{t}$ with respect to the circle with center $P$ and radius $P T$. Step 4. $\omega_{t}$ is the circle with diameter $P P_{1}$. Let's work out the details for steps 1 and 3 . Steps 2 and 4 are immediate. Step 1. In this particular case in which $\Gamma_{1}$ and $\Gamma_{2}$ are externally tangent, there is a small shortcut:

• Draw the circle with diameter on the two centers $O_{1}$ of $\Gamma_{1}$ and $O_{2}$ of $\Gamma_{2}$, and find its center $O$.
• Let this circle meet common tangent line $O P$ at points $Q, R$. The required lines are the perpendicular to $O Q$ at $Q$ and the perpendicular to $O R$ at $R$.

Let's show why this construction works. Let $R_{i}$ be the radius of circle $\Gamma_{i}$ and suppose without loss of generality that $R_{1} \leq R_{2}$. Note that $O Q=\frac{1}{2} O_{1} O_{2}=\frac{R_{1}+R_{2}}{2}, O T=O O_{1}-R_{1}=\frac{R_{2}-R_{1}}{2}$, so

$$\sin \mathrm{\angle }TQO=\frac{OT}{OQ}=\frac{R_2-R_1}{R_1+R_2}\backslash sin\; \backslash angle\; T\; Q\; O=\backslash frac\{O\; T\}\{O\; Q\}=\backslash frac\{R\_\{2\}-R\_\{1\}\}\{R\_\{1\}+R\_\{2\}\}$$

which is also the sine of the angle between $O_{1} O_{2}$ and the common tangent lines. Let $t$ be the perpendicular to $O Q$ through $Q$. Then $\angle\left(t, O_{1} O_{2}\right)=\angle(O Q, Q T)=\angle T Q O$, and $t$ is parallel to a common tangent line. Since

$$d(O,t)=OQ=\frac{R_1+R_2}{2}=\frac{d(O_1,t)+d(O_2,t)}{2}d(O,\; t)=O\; Q=\backslash frac\{R\_\{1\}+R\_\{2\}\}\{2\}=\backslash frac\{d\backslash left(O\_\{1\},\; t\backslash right)+d\backslash left(O\_\{2\},\; t\backslash right)\}\{2\}$$

and $O$ is the midpoint of $O_{1} O_{2}, O$ is also at the same distance from $t$ and the common tangent line, so these two lines coincide. Step 3. Finding the inverse of a point $X$ given the inversion circle $\Omega$ with center $O$ is a well known procedure, but we describe it here for the sake of completeness.

• If $X$ lies in $\Omega$, then its inverse is $X$.
• If $X$ lies in the interior of $\Omega$, draw ray $O X$, then the perpendicular line $\ell$ to $O X$ at $X$. Let $\ell$ meet $\Omega$ at a point $Y$. The inverse of $X$ is the intersection $X^{\prime}$ of $O X$ and the line perpendicular to $O Y$ at $Y$. This is because $O Y X^{\prime}$ is a right triangle with altitude $Y X$, and therefore $O X \cdot O X^{\prime}=O Y^{2}$.
• If $X$ is in the exterior of $\Omega$, draw ray $O X$ and one of the tangent lines $\ell$ from $X$ to $\Omega$ (just connect $X$ to one of the intersections of $\Omega$ and the circle with diameter $O X$ ). Let $\ell$ touch $\Omega$ at a point $Y$. The inverse of $X$ is the projection $X^{\prime}$ of $Y$ onto $O X$. This is because $O Y X^{\prime}$ is a right triangle with altitude $Y X^{\prime}$, and therefore $O X \cdot O X^{\prime}=O Y^{2}$. $X$ is inside $\Omega$