JBMO/md/en-official/en-solutions-english-jbmo2015.md

$19^{\text {th }}$ Junior Balkan Mathematical Olympiad June 24-29, 2015, Belgrade, Serbia

Problem 1. Find all prime numbers $a, b, c$ and positive integers $k$ which satisfy the equation

$$a^2+b^2+16\cdot c^2=9\cdot k^2+1a^\{2\}+b^\{2\}+16\; \backslash cdot\; c^\{2\}=9\; \backslash cdot\; k^\{2\}+1$$

Solution:

The relation $9 \cdot k^{2}+1 \equiv 1(\bmod 3)$ implies

$$a^2+b^2+16\cdot c^2\equiv 1(3)\iff a^2+b^2+c^2\equiv 1(3)a^\{2\}+b^\{2\}+16\; \backslash cdot\; c^\{2\}\; \backslash equiv\; 1(\backslash bmod\; 3)\; \backslash Leftrightarrow\; a^\{2\}+b^\{2\}+c^\{2\}\; \backslash equiv\; 1(\backslash bmod\; 3)$$

Since $a^{2} \equiv 0,1(\bmod 3), \quad b^{2} \equiv 0,1(\bmod 3), c^{2} \equiv 0,1(\bmod 3)$, we have:

$a^{2}$	0	0	0	0	1	1	1	1
$b^{2}$	0	0	1	1	0	0	1	1
$c^{2}$	0	1	0	1	0	1	0	1
$a^{2}+b^{2}+c^{2}$	0	1	1	2	1	2	2	0

From the previous table it follows that two of three prime numbers $a, b, c$ are equal to 3 .

Case 1. $a=b=3$

We have

Case 2. $c=3$

If $\left(3, b_{0}, c, k\right)$ is a solution of the given equation, then $\left(b_{0}, 3, c, k\right)$ is a solution too.

Let $a=3$. We have

$$a^2+b^2+16\cdot c^2=9\cdot k^2+1\iff 9\cdot k^2-b^2=152\iff (3k-b)\cdot (3k+b)=152.a^\{2\}+b^\{2\}+16\; \backslash cdot\; c^\{2\}=9\; \backslash cdot\; k^\{2\}+1\; \backslash Leftrightarrow\; 9\; \backslash cdot\; k^\{2\}-b^\{2\}=152\; \backslash Leftrightarrow(3\; k-b)\; \backslash cdot(3\; k+b)=152\; .$$

Both factors shall have the same parity and we obtain only 2 cases:

• $\left{\begin{array}{l}3 k-b=2, \ 3 k+b=76,\end{array} \Leftrightarrow\left{\begin{array}{l}b=37, \ k=13,\end{array}\right.\right.$ and $(a, b, c, k)=(3,37,3,13)$;
• $\left{\begin{array}{l}3 k-b=4, \ 3 k+b=38,\end{array} \Leftrightarrow\left{\begin{array}{l}b=17, \ k=7,\end{array}\right.\right.$ and $(a, b, c, k)=(3,17,3,7)$.

So, the given equation has 5 solutions:

${(37,3,3,13),(17,3,3,7),(3,37,3,13),(3,17,3,7),(3,3,2,3)}$.

Problem 2. Let $a, b, c$ be positive real numbers such that $a+b+c=3$. Find the minimum value of the expression

$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}A=\backslash frac\{2-a^\{3\}\}\{a\}+\backslash frac\{2-b^\{3\}\}\{b\}+\backslash frac\{2-c^\{3\}\}\{c\}$$

$19^{\text {th }}$ Junior Balkan Mathematical Olympiad June 24-29, 2015, Belgrade, Serbia

Solution:

We can rewrite $A$ as follows:

Recall now the well-known inequality $(x+y+z)^{2} \geq 3(x y+y z+z x)$ and set $x=a b, y=b c, z=c a$, to obtain $(a b+b c+c a)^{2} \geq 3 a b c(a+b+c)=9 a b c$, where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain:

$$ab+bc+ca\ge 3\sqrt{abc}a\; b+b\; c+c\; a\; \backslash geq\; 3\; \backslash sqrt\{a\; b\; c\}$$

Also by using AM-GM inequality we get that

$$\frac{1}{abc}+1\ge 2\sqrt{\frac{1}{abc}}\backslash frac\{1\}\{a\; b\; c\}+1\; \backslash geq\; 2\; \backslash sqrt\{\backslash frac\{1\}\{a\; b\; c\}\}$$

Multiplication of (1) and (2) gives:

$$(ab+bc+ca)(\frac{1}{abc}+1)\ge 3\sqrt{abc}\cdot 2\sqrt{\frac{1}{abc}}=6(a\; b+b\; c+c\; a)\backslash left(\backslash frac\{1\}\{a\; b\; c\}+1\backslash right)\; \backslash geq\; 3\; \backslash sqrt\{a\; b\; c\}\; \backslash cdot\; 2\; \backslash sqrt\{\backslash frac\{1\}\{a\; b\; c\}\}=6$$

So $A \geq 2 \cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is 3.

Problem 3. Let $\triangle A B C$ be an acute triangle. The lines $l_{1}, l_{2}$ are perpendicular to $A B$ at the points $A, B$ respectively. The perpendicular lines from the midpoint $M$ of $A B$ to the sides of the triangle $A C, B C$ intersect the lines $l_{1}, l_{2}$ at the points $E, F$, respectively. If $D$ is the intersection point of $E F$ and $M C$, prove that

$$\mathrm{\angle }ADB=\mathrm{\angle }EMF\backslash angle\; A\; D\; B=\backslash angle\; E\; M\; F$$

Solution:

Let $H, G$ be the points of intersection of $M E, M F$ with $A C, B C$ respectively. From the similarity of triangles $\triangle M H A$ and $\triangle M A E$ we get $\frac{M H}{M A}=\frac{M A}{M E}$, thus

$$M{A}^2=MH\cdot MEM\; A^\{2\}=M\; H\; \backslash cdot\; M\; E$$

Similarly, from the similarity of triangles $\triangle M B G$ and $\triangle M F B$ we get $\frac{M B}{M F}=\frac{M G}{M B}$, thus

$$M{B}^2=MF\cdot MGM\; B^\{2\}=M\; F\; \backslash cdot\; M\; G$$

Since $M A=M B$, from (1), (2), we conclude that the points $E, H, G, F$ are concyclic.

Therefore, we get that $\angle F E H=\angle F E M=\angle H G M$. Also, the quadrilateral $C H M G$ is cyclic, so $\angle C M H=\angle H G C$. We have

$$\mathrm{\angle }FEH+\mathrm{\angle }CMH=\mathrm{\angle }HGM+\mathrm{\angle }HGC={90}^{\circ }\backslash angle\; F\; E\; H+\backslash angle\; C\; M\; H=\backslash angle\; H\; G\; M+\backslash angle\; H\; G\; C=90^\{\backslash circ\}$$

thus $C M \perp E F$. Now, from the cyclic quadrilaterals $F D M B$ and $E A M D$, we get that $\angle D F M=\angle D B M$ and $\angle D E M=\angle D A M$. Therefore, the triangles $\triangle E M F$ and $\triangle A D B$ are similar, so $\angle A D B=\angle E M F$.

Problem 4.

An $L$-figure is one of the following four pieces, each consisting of three unit squares:

A $5 \times 5$ board, consisting of 25 unit squares, a positive integer $k \leq 25$ and an unlimited supply $L$-figures are given. Two players, $\boldsymbol{A}$ and $\boldsymbol{B}$, play the following game: starting with $\boldsymbol{A}$ they alternatively mark a previously unmarked unit square until they marked a total of $k$ unit squares.

We say that a placement of $L$-figures on unmarked unit squares is called good if the $L$-figure do not overlap and each of them covers exactly three unmarked unit squares of the board. $\boldsymbol{B}$ wins if every $\boldsymbol{g o o d}$ placement of $L$-figures leaves uncovered at least three unmarked unit squares. Determine the minimum value of $k$ for which $\boldsymbol{B}$ has a winning strategy.

Solution:

We will show that player $\boldsymbol{A}$ wins if $k=1,2,3$, but player $\boldsymbol{B}$ wins if $k=4$. Thus the smallest $k$ for which $\boldsymbol{B}$ has a winning strategy exists and is equal to 4 .

If $k=1$, player $\boldsymbol{A}$ marks the upper left corner of the square and then fills it as follows.

$19^{\text {th }}$ Junior Balkan Mathematical Olympiad

June 24-29, 2015, Belgrade, Serbia

If $k=2$, player $\boldsymbol{A}$ marks the upper left corner of the square. Whatever square player $\boldsymbol{B}$ marks, then player $\boldsymbol{A}$ can fill in the square in exactly the same pattern as above except that he doesn't put the $L$-figure which covers the marked square of $\boldsymbol{B}$. Player $\boldsymbol{A}$ wins because he has left only two unmarked squares uncovered.

For $k=3$, player $\boldsymbol{A}$ wins by following the same strategy. When he has to mark a square for the second time, he marks any yet unmarked square of the $L$-figure that covers the marked square of $\boldsymbol{B}$.

Let us now show that for $k=4$ player $\boldsymbol{B}$ has a winning strategy. Since there will be 21 unmarked squares, player $\boldsymbol{A}$ will need to cover all of them with seven $L$-figures. We can assume that in his first move, player $\boldsymbol{A}$ does not mark any square in the bottom two rows of the chessboard (otherwise just rotate the chessboard). In his first move player $\boldsymbol{B}$ marks the square labeled 1 in the following figure.

If player $\boldsymbol{A}$ in his next move does not mark any of the squares labeled 2,3 and 4 then player $\boldsymbol{B}$ marks the square labeled 3 . Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure.

If player $\boldsymbol{A}$ in his next move marks the square labeled 2, then player $\boldsymbol{B}$ marks the square labeled 5. Player $\boldsymbol{B}$ wins as the square labeled 3 is left unmarked but cannot be covered with an $L$-figure.

Finally, if player $\boldsymbol{A}$ in his next move marks one of the squares labeled 3 or 4, player $\boldsymbol{B}$ marks the other of these two squares. Player $\boldsymbol{B}$ wins as the square labeled 2 is left unmarked but cannot be covered with an $L$-figure.

Since we have covered all possible cases, player $\boldsymbol{B}$ wins when $k=4$.