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[ Area of a triangle (using the semiperimeter and the radius of the inscribed or excircle),

## Heron's Formula

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Area of a triangle (using height and base)
Integer triangles

The radius of the circle inscribed in the triangle is $\frac{4}{3}$, and the lengths of the heights of the triangle are integers, the sum of which is 13. Calculate the lengths of the sides of the triangle.

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Let $a, b, c$ be the lengths of the sides of a triangle, $h_{\mathrm{a}}, h_{\mathrm{b}}, h_{\mathrm{c}}$ be the lengths of the altitudes dropped to these sides, and $S$ be its area. Then

$$
\frac{1}{r}=\frac{a+b+c}{2 S}=\frac{a}{2 S}+\frac{b}{2 S}+\frac{c}{2 S}=\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}
$$

Thus, we obtain the system of equations:

$$
\left\{\begin{array}{l}
\frac{1}{h_{a}}+\frac{1}{h_{b}}+\frac{1}{h_{c}}=\frac{3}{4} \\
h_{a}+h_{b}+h_{c}=13
\end{array}\right.
$$

From the second equation, it follows that $h_{\mathrm{a}}, h_{\mathrm{b}}, h_{\mathrm{c}} \leq 10$, and the length of one of the altitudes is not less than five. By enumeration, we find the unique (up to permutation) solution $h_{\mathrm{a}}=3, h_{\mathrm{b}}=4, h_{\mathrm{c}}=6$. Therefore, $3 a=4 b=6 c$, which means $a=4 x, b=3 x, c=2 x$. It remains to find $x$. On the one hand, $S=\frac{1}{2} a h_{\mathrm{a}}=6 x$. On the other hand, by Heron's formula $S$

$=\frac{\sqrt{135}}{4} x^{2}$. Therefore, $x=\frac{8}{\sqrt{15}}$.