The heights of a triangle are 10, 12, and 15. What are the lengths of its sides?
Solution. Let the sides of the triangle be $a, b$, and $c$. We can assume that $a > b > c$. Since the sides of the triangle are in the same ratio as the reciprocals of the corresponding altitudes, the ratio of the sides is $a: b: c = 1/10: 1/12: 1/15 = 6: 5: 4$. Therefore, the triangle we are looking for is similar to the triangle $\mathcal{H}$, whose sides are 6, 5, and 4.

According to Heron's formula, if the area of $\mathcal{H}$ is denoted by $T$, then

$$
T=\sqrt{\frac{(6+5+4)}{2} \cdot \frac{(6+5-4)}{2} \cdot \frac{(6+4-5)}{2} \cdot \frac{(5+4-6)}{2}}=\frac{15 \sqrt{7}}{4}
$$

Therefore, the altitude corresponding to the largest side of $\mathcal{H}$ is

$$
m=\frac{2 T}{6}=\frac{30 \sqrt{7}}{24}=\frac{5 \sqrt{7}}{4}
$$

Thus, the ratio of the corresponding segments in the two similar triangles is $a: 6 = b: 5 = c: 4 = 10: (5 \sqrt{7} / 4) = 8 / \sqrt{7}$.

Therefore, the sides we are looking for are

$$
a=\frac{48}{\sqrt{7}}=\frac{48 \sqrt{7}}{7}, \quad b=\frac{40}{\sqrt{7}}=\frac{40 \sqrt{7}}{7}, \quad c=\frac{32}{\sqrt{7}}=\frac{32 \sqrt{7}}{7}
$$

Such a triangle exists because the sides satisfy the triangle inequality.