A triangle has sides of 13, 17, and 24 units. The 24-unit side is divided into three equal parts. What is the distance from the opposite vertex to the division points?
We denote the height of the triangle in question by $m$, the projection of the side of length 13 on the 24-unit side by $x$, so the projection of the 17-unit side is $24-x$, because the foot of the perpendicular from the longest side lies on the side segment. The distance from the vertices opposite the 13 and 17-unit sides to the points that trisect these sides are denoted by $p$ and $q$. The distances from the trisection points to the foot of the perpendicular are $|x-8|$ and $|24-x-8|$. The 13 and 17-unit sides, along with the segments $p$ and $q$, form right triangles with the height.

![](https://cdn.mathpix.com/cropped/2024_05_02_700ff852980fd3bcd20ag-1.jpg?height=320&width=637&top_left_y=385&top_left_x=733)

We apply the Pythagorean theorem to these triangles:

$$
\begin{gathered}
x^{2}+m^{2}=13^{2} \\
(x-8)^{2}+m^{2}=p^{2} \\
(16-x)^{2}+m^{2}=q^{2} \\
(24-x)^{2}+m^{2}=17^{2}
\end{gathered}
$$

From these, we can eliminate $m$ by subtracting (1) from the other equations:

$$
\begin{gathered}
64-16 x=p^{2}-13^{2} \\
256-32 x=q^{2}-13^{2} \\
576-48 x=17^{2}-13^{2}=4 \cdot 30=120
\end{gathered}
$$

From the last equation, we get $x=19 / 2$, and substituting this into (5) and (6) gives:

$$
p^{2}=81 ; \quad q^{2}=121
$$

Thus, the sought distances are $p=9, q=11$.

Katalin Balla (Budapest, Radnóti M. Practical Grammar School II. grade)