SOLUTIONS FOR 2011 APMO PROBLEMS
Problem 1.
Solution: Suppose all of the 3 numbers $a^{2}+b+c, b^{2}+c+a$ and $c^{2}+a+b$ are perfect squares. Then from the fact that $a^{2}+b+c$ is a perfect square bigger than $a^{2}$ it follows that $a^{2}+b+c \geq(a+1)^{2}$, and therefore, $b+c \geq 2 a+1$. Similarly we obtain $c+a \geq 2 b+1$ and $a+b \geq 2 c+1$.
Adding the corresponding sides of the preceding 3 inequalities, we obtain $2(a+b+c) \geq 2(a+b+c)+3$, a contradiction. This proves that it is impossible to have all the 3 given numbers to be perfect squares. Alternate Solution: Since the given conditions of the problem are symmetric in $a, b, c$, we may assume that $a \geq b \geq c$ holds. From the assumption that $a^{2}+b+c$ is a perfect square, we can deduce as in the solution above the inequality $b+c \geq 2 a+1$. But then we have
a contradiction, which proves the assertion of the problem.
Problem 2.
Solution: We will show that $36^{\circ}$ is the desired answer for the problem. First, we observe that if the given 5 points form a regular pentagon, then the minimum of the angles formed by any triple among the five vertices is $36^{\circ}$, and therefore, the answer we seek must be bigger than or equal to $36^{\circ}$.
Next, we show that for any configuration of 5 points satisfying the condition of the problem, there must exist an angle smaller than or equal to $36^{\circ}$ formed by a triple chosen from the given 5 points. For this purpose, let us start with any 5 points, say $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$, on the plane satisfying the condition of the problem, and consider the smallest convex subset, call it $\Gamma$, in the plane containing all of the 5 points. Since this convex subset $\Gamma$ must be either a triangle or a quadrilateral or a pentagon, it must have an interior angle with $108^{\circ}$ or less. We may assume without loss of generality that this angle is $\angle A_{1} A_{2} A_{3}$. By the definition of $\Gamma$ it is clear that the remaining 2 points $A_{4}$ and $A_{5}$ lie in the interior of the angular region determined by $\angle A_{1} A_{2} A_{3}$, and therefore, there must be an angle smaller than or equal to $\frac{1}{3} \cdot 108^{\circ}=36^{\circ}$, which is formed by a triple chosen from the given 5 points, and this proves that $36^{\circ}$ is the desired maximum.
Problem 3.
Solution: Since $\angle B_{1} B B_{2}=90^{\circ}$, the circle having $B_{1} B_{2}$ as its diameter goes through the points $B, B_{1}, B_{2}$. From $B_{1} A: B_{1} C=B_{2} A: B_{2} C=B A: B C$, it follows that this circle is the Apolonius circle with the ratio of the distances from the points $A$ and $C$ being $B A: B C$. Since the point $P$ lies on this circle, we have
from which it follows that $P A \sin A=P C \sin C$. Similarly, we have $P A \sin A=$ $P B \sin B$, and therefore, $P A \sin A=P B \sin B=P C \sin C$.
Let us denote by $D, E, F$ the foot of the perpendicular line drawn from $P$ to the line segment $B C, C A$ and $A B$, respectively. Since the points $E, F$ lie on a circle having $P A$ as its diameter, we have by the law of sines $E F=P A \sin A$. Similarly, we have $F D=P B \sin B$ and $D E=P C \sin C$. Consequently, we conclude that $D E F$ is an equilateral triangle. Furthermore, we have $\angle C P E=\angle C D E$, since the quadrilateral $C D P E$ is cyclic. Similarly, we have $\angle F P B=\angle F D B$. Putting these together, we get
which proves the assertion of the problem. Alternate Solution: Let $O$ be the midpoint of the line segment $B_{1} B_{2}$. Then the points $B$ and $P$ lie on the circle with center at $O$ and going through the point $B_{1}$. From
it follows that the triangles $O C D$ and $O B A$ are similar, and therefore we have that $O C \cdot O A=O B^{2}=O P^{2}$. Thus we conclude that the triangles $O C P$ and $O P A$ are similar, and therefore, we have $\angle O P C=\angle P A C$. Using this fact, we obtain
from which we conclude that $\angle P A C+\angle P B C=\angle P B A+\angle P C A$. Similarly, we get $\angle P A B+\angle P C B=\angle P B A+\angle P C A$. Putting these facts together and taking into account the fact that
we conclude that $\angle P B A+\angle P C A=60^{\circ}$, and finally that $\angle B P C=(\angle P B A+\angle P A B)+(\angle P C A+\angle P A C)=\angle B A C+(\angle P B A+\angle P C A)=90^{\circ}$, proving the assertion of the problem.
Problem 4.
Solution: We will show that the desired maximum value for $m$ is $n(n-1)$. First, let us show that $m \leq n(n-1)$ always holds for any sequence $P_{0}, P_{1}, \cdots, P_{m+1}$ satisfying the conditions of the problem.
Call a point a turning point if it coincides with $P_{i}$ for some $i$ with $1 \leq i \leq m$. Let us say also that 2 points ${P, Q}$ are adjacent if ${P, Q}=\left{P_{i-1}, P_{i}\right}$ for some $i$ with $1 \leq i \leq m$, and vertically adjacent if, in addition, $P Q$ is parallel to the $y$-axis.
Any turning point is vertically adjacent to exactly one other turning point. Therefore, the set of all turning points is partitioned into a set of pairs of points using the relation of "vertical adjacency". Thus we can conclude that if we fix $k \in{1,2, \cdots, n}$, the number of turning points having the $x$-coordinate $k$ must be even, and hence it is less than or equal to $n-1$. Therefore, altogether there are less than or equal to $n(n-1)$ turning points, and this shows that $m \leq n(n-1)$ must be satisfied.
It remains now to show that for any positive odd number $n$ one can choose a sequence for which $m=n(n-1)$. We will show this by using the mathematical induction on $n$. For $n=1$, this is clear. For $n=3$, choose
It is easy to see that these points satisfy the requirements (See fig. 1 below).
figure 1
Let $n$ be an odd integer $\geq 5$, and suppose there exists a sequence satisfying the desired conditions for $n-4$. Then, it is possible to construct a sequence which gives a configuration indicated in the following diagram (fig. 2), where the configuration inside of the dotted square is given by the induction hypothesis: figure 2 By the induction hypothesis, there are exactly $(n-4)(n-5)$ turning points for the configuration inside of the dotted square in the figure 2 above, and all of the lattice points in the figure 2 lying outside of the dotted square except for the 4 points $(n, 2),(n-1, n-2),(2,3),(1, n-1)$ are turning points. Therefore, the total number of turning points in this configuration is
showing that for this $n$ there exists a sequence satisfying the desired properties, and thus completing the induction process.
Problem 5.
Solution: By substituting $x=1$ and $y=1$ into the given identity we obtain $f(f(1))=f(1)$. Next, by substituting $x=1$ and $y=f(1)$ into the given identity and using $f(f(1))=f(1)$, we get $f(1)^{2}=f(1)$, from which we conclude that either $f(1)=0$ or $f(1)=1$. But if $f(1)=1$, then substituting $y=1$ into the given identity, we get $f(x)=x$ for all $x$, which contradicts the condition (1). Therefore, we must have $f(1)=0$.
By substituting $x=1$ into the given identity and using the fact $f(1)=0$, we then obtain $f(f(y))=2 f(y)$ for all $y$. This means that if a number $t$ belongs to the range of the function $f$, then so does $2 t$, and by induction we can conclude that for any non-negative integer $n, 2^{n} t$ belongs to the range of $f$ if $t$ does. Now suppose that there exists a real number $a$ for which $f(a)>0$, then for any non-negative integer $n 2^{n} f(a)$ must belong to the range of $f$, which leads to a contradiction to the condition (1). Thus we conclude that $f(x) \leq 0$ for any real number $x$.
By substituting $\frac{x}{2}$ for $x$ and $f(y)$ for $y$ in the given identity and using the fact that $f(f(y))=2 f(y)$, we obtain
from which it follows that $x f(y)-f(x f(y))=f(y) f\left(\frac{x}{2}\right)-f\left(\frac{x}{2} f(y)\right) \geq 0$, since the values of $f$ are non-positive. Combining this with the given identity, we conclude that $y f(x) \geq f(x y)$. When $x>0$, by letting $y$ to be $\frac{1}{x}$ and using the fact that $f(1)=0$, we get $f(x) \geq 0$. Since $f(x) \leq 0$ for any real number $x$, we conclude that $f(x)=0$ for any positive real number $x$. We also have $f(0)=f(f(1))=2 f(1)=0$.
If $f$ is identically 0 , i.e., $f(x)=0$ for all $x$, then clearly, this $f$ satisfies the given identity. If $f$ satisfies the given identity but not identically 0 , then there exists a $b<0$ for which $f(b)<0$. If we set $c=f(b)$, then we have $f(c)=f(f(b))=2 f(b)=$ $2 c$. For any negative real number $x$, we have $c x>0$ so that $f(c x)=f(2 c x)=0$, and by substituting $y=c$ into the given identity, we get
from which it follows that $f(x)=2 x$ for any negative real $x$. We therefore conclude that if $f$ satisfies the given identity and is not identically 0 , then $f$ is of the form $f(x)=\left{\begin{array}{ll}0 & \text { if } x \geq 0 \ 2 x & \text { if } x<0 .\end{array}\right.$ Finally, let us show that the function $f$ of the form shown above does satisfy the conditions of the problem. Clearly, it satisfies the condition (1). We can check that $f$ satisfies the condition (2) as well by separating into the following 4 cases depending on whether $x, y$ are non-negative or negative.
- when both $x$ and $y$ are non-negative, both sides of the given identity are 0 .
- when $x$ is non-negative and $y$ is negative, we have $x y \leq 0$ and both sides of the given identity are $4 x y$.
- when $x$ is negative and $y$ is non-negative, we have $x y \leq 0$ and both sides of the given identity are $2 x y$.
- when both $x$ and $y$ are negative, we have $x y>0$ and both sides of the given identity are $2 x y$. Summarizing the arguments above, we conclude that the functions $f$ satisfying the conditions of the problem are