## 3rd grade - high school - A category,  Primošten,  April 7, 2008.

## Solutions

Problem 1. The lengths of the sides of a triangle are three consecutive natural numbers, and one of the angles of the triangle is twice as large as one of the remaining two angles. Determine the lengths of the sides of the triangle.
Solution. Let $a=n-1, b=n, c=n+1$. Then $\alpha<\beta<\gamma$. There are three possible cases.

$1^{\circ} \beta=2 \alpha$

Using the sine and cosine rules, we get:

$$
\begin{gathered}
\cos \alpha=\frac{\sin 2 \alpha}{2 \sin \alpha}=\frac{\sin \beta}{2 \sin \alpha}=\frac{b}{2 a}=\frac{n}{2(n-1)} \\
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{n+4}{2(n+1)}
\end{gathered}
$$

From the equation $\frac{n}{2(n-1)}=\frac{n+4}{2(n+1)}$, we get $n=2$. The sides of the triangle would be $1, 2,$ and $3$, which is not possible.

$2^{\circ} \gamma=2 \alpha$

Similarly, we would get:

$$
\begin{gathered}
\cos \alpha=\frac{\sin 2 \alpha}{2 \sin \alpha}=\frac{\sin \gamma}{2 \sin \alpha}=\frac{n+1}{2(n-1)} \\
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{n+4}{2(n+1)}
\end{gathered}
$$

From the equation $\frac{n+1}{2(n-1)}=\frac{n+4}{2(n+1)}$, we get $n=5$ and the side lengths of the triangle are 4, 5, and 6.

$3^{\circ} \gamma=2 \beta$

Repeating the process from the previous two cases, we get:

$$
\begin{aligned}
& \cos \beta=\frac{\sin 2 \beta}{2 \sin \beta}=\frac{\sin \gamma}{2 \sin \beta}=\frac{n+1}{2 n} \\
& \cos \beta=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{n^{2}+2}{2\left(n^{2}-1\right)}
\end{aligned}
$$

From the equation $\frac{n+1}{2 n}=\frac{n^{2}+2}{2\left(n^{2}-1\right)}$, we get the quadratic equation $n^{2}-3 n-1=0$, whose solutions $n_{1,2}=\frac{3 \pm \sqrt{13}}{2}$ are irrational.

Thus, the only solution is a triangle with side lengths 4, 5, and 6.

Problem 2. Let $x_{1}, x_{2}, \ldots, x_{n-1}, x_{n}$ be positive real numbers such that $\sum_{i=1}^{n} x_{i}=1$. Prove the inequality

$$
\frac{x_{1}^{2}}{x_{1}+x_{2}}+\frac{x_{2}^{2}}{x_{2}+x_{3}}+\ldots+\frac{x_{n-1}^{2}}{x_{n-1}+x_{n}}+\frac{x_{n}^{2}}{x_{n}+x_{1}} \geq \frac{1}{2}
$$

Solution. With the notation $x_{n+1}=x_{1}$, we have:

$$
\begin{aligned}
\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+x_{i+1}} & =\sum_{i=1}^{n} \frac{x_{i}^{2}+x_{i} x_{i+1}-x_{i} x_{i+1}}{x_{i}+x_{i+1}} \\
& =\sum_{i=1}^{n}\left(x_{i}-\frac{x_{i} x_{i+1}}{x_{i}+x_{i+1}}\right) \geq \sum_{i=1}^{n} x_{i}-\sum_{i=1}^{n} \frac{\left(\frac{x_{i}+x_{i+1}}{2}\right)^{2}}{x_{i}+x_{i+1}} \\
& =1-\frac{1}{4} \sum_{i=1}^{n}\left(x_{i}+x_{i+1}\right)=\frac{1}{2}
\end{aligned}
$$

## Second Solution.

With the notation $x_{n+1}=x_{1}$, we get:

$$
\begin{aligned}
\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+x_{i+1}} & =\sum_{i=1}^{n} \frac{x_{i}^{2}-x_{i+1}^{2}+x_{i+1}^{2}}{x_{i}+x_{i+1}} \\
& =\sum_{i=1}^{n}\left(x_{i}-x_{i+1}\right)+\sum_{i=1}^{n} \frac{x_{i+1}^{2}}{x_{i}+x_{i+1}} \\
& =\sum_{i=1}^{n} \frac{x_{i+1}^{2}}{x_{i}+x_{i+1}}
\end{aligned}
$$

From this, we get

$$
\sum_{i=1}^{n} \frac{x_{i}^{2}}{x_{i}+x_{i+1}}=\frac{1}{2} \sum_{i=1}^{n} \frac{x_{i}^{2}+x_{i+1}^{2}}{x_{i}+x_{i+1}} \geq \frac{1}{4} \sum_{i=1}^{n}\left(x_{i}+x_{i+1}\right)=\frac{1}{2}
$$

where we used the inequality between the arithmetic and quadratic means.

Problem 3. Among all numbers of the form $36^{m}-5^{n}$, where $m$ and $n$ are natural numbers, find the smallest in absolute value.

Solution. Since the number $36^{m}$ ends in 6 for any natural number $m$, and the number $5^{n}$ ends in 5 for any natural number $n$, the number $N=\left|36^{m}-5^{n}\right|$ ends in 1 or 9. Therefore, the smallest possible values of $N$ could be $1, 9, 11, \ldots$

For $m=1$ and $n=2$, we get $N=11$. We will show that $N$ cannot be 1 or 9, i.e., that $N=11$ is the smallest desired number.

From the equation $36^{m}-5^{n}= \pm 9$, it would follow that the number $5^{n}$ is divisible by 9, which is impossible.

From the equation $36^{m}-5^{n}=1$, it would follow that $5^{n}=36^{m}-1=\left(6^{m}-1\right)\left(6^{m}+1\right)$, i