7. Determine the triangle with the minimum perimeter whose side lengths are integers, and one of the medians is equal to the corresponding side.
Solution. Let $A_{1}$ be the midpoint of side $BC$ of triangle $ABC$ (diagram on the right) and let $a, b, c$ be natural numbers such that $BC = AA_{1} = a$, $CA = b$, and $AB = c$. Then

$$
\begin{aligned}
& b^{2} = \frac{5 a^{2}}{4} - a^{2} \cos \measuredangle A A_{1} C, \\
& c^{2} = \frac{5 a^{2}}{4} + a^{2} \cos \measuredangle A A_{1} C,
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_faf575a75202fd173f3cg-12.jpg?height=267&width=281&top_left_y=756&top_left_x=1162)

from which it follows that $\frac{5 a^{2}}{4} - b^{2} = c^{2} - \frac{5 a^{2}}{4}$, i.e., $5 a^{2} = 2(b^{2} + c^{2})$. From the last equation, it follows that $a$ is an even number. It is easily verified that for $a \in \{2, 4, 6, 8, 12, 14\}$, the equation $5 a^{2} = 2(b^{2} + c^{2})$ has no solution in the set of natural numbers for which $b + c > a$, $c + a > b$, and $a + b > c$. For $a = 10$, the only solution for which the given conditions hold is $a = 10$, $\{b, c\} = \{9, 13\}$. The perimeter of the corresponding triangle is 32. If for natural numbers $a, b, c$ that can be the lengths of the sides of triangles, $5 a^{2} = 2(b^{2} + c^{2})$ and $a \geq 16$, then $a + b + c > a + a \geq 32$. According to this, the lengths of the sides of the desired triangle are 9, 10, and 13.