APMO/md/en-apmo2013_sol.md

Solutions of APMO 2013

Problem 1. Let $A B C$ be an acute triangle with altitudes $A D, B E$ and $C F$, and let $O$ be the center of its circumcircle. Show that the segments $O A, O F, O B, O D, O C, O E$ dissect the triangle $A B C$ into three pairs of triangles that have equal areas.

Solution. Let $M$ and $N$ be midpoints of sides $B C$ and $A C$, respectively. Notice that $\angle M O C=\frac{1}{2} \angle B O C=\angle E A B, \angle O M C=90^{\circ}=\angle A E B$, so triangles $O M C$ and $A E B$ are similar and we get $\frac{O M}{A E}=\frac{O C}{A B}$. For triangles $O N A$ and $B D A$ we also have $\frac{O N}{B D}=\frac{O A}{B A}$. Then $\frac{O M}{A E}=\frac{O N}{B D}$ or $B D \cdot O M=A E \cdot O N$.

Denote by $S(\Phi)$ the area of the figure $\Phi$. So, we see that $S(O B D)=\frac{1}{2} B D \cdot O M=$ $\frac{1}{2} A E \cdot O N=S(O A E)$. Analogously, $S(O C D)=S(O A F)$ and $S(O C E)=S(O B F)$.

Alternative solution. Let $R$ be the circumradius of triangle $A B C$, and as usual write $A, B, C$ for angles $\angle C A B, \angle A B C, \angle B C A$ respectively, and $a, b, c$ for sides $B C, C A, A B$ respectively. Then the area of triangle $O C D$ is

$$S(OCD)=\frac{1}{2}\cdot OC\cdot CD\cdot \sin (\mathrm{\angle }OCD)=\frac{1}{2}R\cdot CD\cdot \sin (\mathrm{\angle }OCD)S(O\; C\; D)=\backslash frac\{1\}\{2\}\; \backslash cdot\; O\; C\; \backslash cdot\; C\; D\; \backslash cdot\; \backslash sin\; (\backslash angle\; O\; C\; D)=\backslash frac\{1\}\{2\}\; R\; \backslash cdot\; C\; D\; \backslash cdot\; \backslash sin\; (\backslash angle\; O\; C\; D)$$

Now $C D=b \cos C$, and

$$\mathrm{\angle }OCD=\frac{{180}^{\circ }-2A}{2}={90}^{\circ }-A\backslash angle\; O\; C\; D=\backslash frac\{180^\{\backslash circ\}-2\; A\}\{2\}=90^\{\backslash circ\}-A$$

(since triangle $O B C$ is isosceles, and $\angle B O C=2 A$ ). So

$$S(OCD)=\frac{1}{2}Rb\cos C\sin ({90}^{\circ }-A)=\frac{1}{2}Rb\cos C\cos AS(O\; C\; D)=\backslash frac\{1\}\{2\}\; R\; b\; \backslash cos\; C\; \backslash sin\; \backslash left(90^\{\backslash circ\}-A\backslash right)=\backslash frac\{1\}\{2\}\; R\; b\; \backslash cos\; C\; \backslash cos\; A$$

A similar calculation gives

so $O C D$ and $O A F$ have the same area. In the same way we find that $O B D$ and $O A E$ have the same area, as do $O C E$ and $O B F$.

Problem 2. Determine all positive integers $n$ for which $\frac{n^{2}+1}{[\sqrt{n}]^{2}+2}$ is an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.

Solution. We will show that there are no positive integers $n$ satisfying the condition of the problem.

Let $m=[\sqrt{n}]$ and $a=n-m^{2}$. We have $m \geq 1$ since $n \geq 1$. From $n^{2}+1=\left(m^{2}+a\right)^{2}+1 \equiv$ $(a-2)^{2}+1\left(\bmod \left(m^{2}+2\right)\right)$, it follows that the condition of the problem is equivalent to the fact that $(a-2)^{2}+1$ is divisible by $m^{2}+2$. Since we have

we see that $(a-2)^{2}+1=k\left(m^{2}+2\right)$ must hold with $k=1,2$ or 3 . We will show that none of these can occur.

Case 1. When $k=1$. We get $(a-2)^{2}-m^{2}=1$, and this implies that $a-2= \pm 1, m=0$ must hold, but this contradicts with fact $m \geq 1$.

Case 2. When $k=2$. We have $(a-2)^{2}+1=2\left(m^{2}+2\right)$ in this case, but any perfect square is congruent to $0,1,4 \bmod 8$, and therefore, we have $(a-2)^{2}+1 \equiv 1,2,5(\bmod 8)$, while $2\left(m^{2}+2\right) \equiv 4,6(\bmod 8)$. Thus, this case cannot occur either.

Case 3. When $k=3$. We have $(a-2)^{2}+1=3\left(m^{2}+2\right)$ in this case. Since any perfect square is congruent to 0 or $1 \bmod 3$, we have $(a-2)^{2}+1 \equiv 1,2(\bmod 3)$, while $3\left(m^{2}+2\right) \equiv 0$ $(\bmod 3)$, which shows that this case cannot occur either.

Problem 3. For $2 k$ real numbers $a_{1}, a_{2}, \ldots, a_{k}, b_{1}, b_{2}, \ldots, b_{k}$ define the sequence of numbers $X_{n}$ by

$$X_n=\sum _{i=1}^k[a_{i}n+b_i](n=1,2,\dots )X\_\{n\}=\backslash sum\_\{i=1\}^\{k\}\backslash left[a\_\{i\}\; n+b\_\{i\}\backslash right]\; \backslash quad(n=1,2,\; \backslash ldots)$$

If the sequence $X_{n}$ forms an arithmetic progression, show that $\sum_{i=1}^{k} a_{i}$ must be an integer. Here $[r]$ denotes the greatest integer less than or equal to $r$.

Solution. Let us write $A=\sum_{i=1}^{k} a_{i}$ and $B=\sum_{i=1}^{k} b_{i}$. Summing the corresponding terms of the following inequalities over $i$,

we obtain $A n+B-k<X_{n}<A n+B$. Now suppose that $\left{X_{n}\right}$ is an arithmetic progression with the common difference $d$, then we have $n d=X_{n+1}-X_{1}$ and $A+B-k<X_{1} \leq A+B$ Combining with the inequalities obtained above, we get

or

from which we conclude that $|A-d|<\frac{k}{n}$ must hold. Since this inequality holds for any positive integer $n$, we must have $A=d$. Since $\left{X_{n}\right}$ is a sequence of integers, $d$ must be an integer also, and thus we conclude that $A$ is also an integer.

Problem 4. Let $a$ and $b$ be positive integers, and let $A$ and $B$ be finite sets of integers satisfying: (i) $A$ and $B$ are disjoint; (ii) if an integer $i$ belongs either to $A$ or to $B$, then $i+a$ belongs to $A$ or $i-b$ belongs to $B$.

Prove that $a|A|=b|B|$. (Here $|X|$ denotes the number of elements in the set $X$.) Solution. Let $A^{}={n-a: n \in A}$ and $B^{}={n+b: n \in B}$. Then, by (ii), $A \cup B \subseteq A^{} \cup B^{}$ and by (i),

$$\mathrm{\mid }A\cup B\mathrm{\mid }\le \mid A^{\ast }\cup B^{\ast }\mid \le \mid A^{\ast }\mid +\mid B^{\ast }\mid =\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }=\mathrm{\mid }A\cup B\mathrm{\mid }|A\; \backslash cup\; B|\; \backslash leq\backslash left|A^\{*\}\; \backslash cup\; B^\{*\}\backslash right|\; \backslash leq\backslash left|A^\{*\}\backslash right|+\backslash left|B^\{*\}\backslash right|=|A|+|B|=|A\; \backslash cup\; B|$$

Thus, $A \cup B=A^{} \cup B^{}$ and $A^{}$ and $B^{}$ have no element in common. For each finite set $X$ of integers, let $\sum(X)=\sum_{x \in X} x$. Then

which implies $a|A|=b|B|$. Alternative solution. Let us construct a directed graph whose vertices are labelled by the members of $A \cup B$ and such that there is an edge from $i$ to $j$ iff $j \in A$ and $j=i+a$ or $j \in B$ and $j=i-b$. From (ii), each vertex has out-degree $\geq 1$ and, from (i), each vertex has in-degree $\leq 1$. Since the sum of the out-degrees equals the sum of the in-degrees, each vertex has in-degree and out-degree equal to 1. This is only possible if the graph is the union of disjoint cycles, say $G_{1}, G_{2}, \ldots, G_{n}$. Let $\left|A_{k}\right|$ be the number of elements of $A$ in $G_{k}$ and $\left|B_{k}\right|$ be the number of elements of $B$ in $G_{k}$. The cycle $G_{k}$ will involve increasing vertex labels by $a$ a total of $\left|A_{k}\right|$ times and decreasing them by $b$ a total of $\left|B_{k}\right|$ times. Since it is a cycle, we have $a\left|A_{k}\right|=b\left|B_{k}\right|$. Summing over all cycles gives the result.

Problem 5. Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$, and let $P$ be a point on the extension of $A C$ such that $P B$ and $P D$ are tangent to $\omega$. The tangent at $C$ intersects $P D$ at $Q$ and the line $A D$ at $R$. Let $E$ be the second point of intersection between $A Q$ and $\omega$. Prove that $B, E, R$ are collinear.

Solution. To show $B, E, R$ are collinear, it is equivalent to show the lines $A D, B E, C Q$ are concurrent. Let $C Q$ intersect $A D$ at $R$ and $B E$ intersect $A D$ at $R^{\prime}$. We shall show $R D / R A=R^{\prime} D / R^{\prime} A$ so that $R=R^{\prime}$.

Since $\triangle P A D$ is similar to $\triangle P D C$ and $\triangle P A B$ is similar to $\triangle P B C$, we have $A D / D C=$ $P A / P D=P A / P B=A B / B C$. Hence, $A B \cdot D C=B C \cdot A D$. By Ptolemy's theorem, $A B \cdot D C=B C \cdot A D=\frac{1}{2} C A \cdot D B$. Similarly $C A \cdot E D=C E \cdot A D=\frac{1}{2} A E \cdot D C$.

Thus

$$\frac{DB}{AB}=\frac{2DC}{CA}\backslash frac\{D\; B\}\{A\; B\}=\backslash frac\{2\; D\; C\}\{C\; A\}$$

and

$$\frac{DC}{CA}=\frac{2ED}{AE}\backslash frac\{D\; C\}\{C\; A\}=\backslash frac\{2\; E\; D\}\{A\; E\}$$

Since the triangles $R D C$ and $R C A$ are similar, we have $\frac{R D}{R C}=\frac{D C}{C A}=\frac{R C}{R A}$. Thus using (4)

$$\frac{RD}{RA}=\frac{RD\cdot RA}{R{A}^2}={\left(\frac{RC}{RA}\right)}^2={\left(\frac{DC}{CA}\right)}^2={\left(\frac{2ED}{AE}\right)}^2\backslash frac\{R\; D\}\{R\; A\}=\backslash frac\{R\; D\; \backslash cdot\; R\; A\}\{R\; A^\{2\}\}=\backslash left(\backslash frac\{R\; C\}\{R\; A\}\backslash right)^\{2\}=\backslash left(\backslash frac\{D\; C\}\{C\; A\}\backslash right)^\{2\}=\backslash left(\backslash frac\{2\; E\; D\}\{A\; E\}\backslash right)^\{2\}$$

Using the similar triangles $A B R^{\prime}$ and $E D R^{\prime}$, we have $R^{\prime} D / R^{\prime} B=E D / A B$. Using the similar triangles $D B R^{\prime}$ and $E A R^{\prime}$ we have $R^{\prime} A / R^{\prime} B=E A / D B$. Thus using (3) and (4),

$$\frac{R^{\mathrm{\prime }}D}{R^{\mathrm{\prime }}A}=\frac{ED\cdot DB}{EA\cdot AB}={\left(\frac{2ED}{AE}\right)}^2\backslash frac\{R^\{\backslash prime\}\; D\}\{R^\{\backslash prime\}\; A\}=\backslash frac\{E\; D\; \backslash cdot\; D\; B\}\{E\; A\; \backslash cdot\; A\; B\}=\backslash left(\backslash frac\{2\; E\; D\}\{A\; E\}\backslash right)^\{2\}$$

It follows from (5) and (6) that $R=R^{\prime}$.