10.409 A triangle with sides 13, 14, and 15 is divided into three equal areas by lines perpendicular to the longest side. Find the distances from these lines to the nearest vertices of the triangle located on the longest side.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_21_f024bf2ff7725246f3bfg-248.jpg?height=277&width=349&top_left_y=104&top_left_x=85)

Let in $\triangle ABC$, $AC=15$, $BC=14$, $AB=13$, $MN \perp AC$, $KL \perp AC$,

$S_{AMN}=S_{B K L M N}=S_{K C L}$. By the cosine rule for $\triangle ABC$, $\cos \angle A=\frac{13^{2}+15^{2}-14^{2}}{2 \cdot 13 \cdot 15}=\frac{33}{65}$,

$\operatorname{tg} \angle A=\frac{56}{33}$, $\cos \angle C=\frac{14^{2}+15^{2}-13^{2}}{2 \cdot 14 \cdot 15}=\frac{3}{5}$,

$\sin \angle C=\frac{4}{5}$, $\operatorname{tg} \angle C=\frac{4}{3}$,

$S_{ABC}=\frac{AC \cdot BC}{2} \cdot \sin \angle C=84$, $S_{KCL}=28$ (by the condition). On the other hand,

$S_{KCL}=\frac{KL \cdot LC}{2}=\frac{(LC)^{2} \operatorname{tg} \angle C}{2} \Rightarrow LC=\sqrt{42}$.

Similarly, $S_{AMN}=28=\frac{(AN)^{2} \operatorname{tg} \angle A}{2} \Rightarrow AN=\sqrt{33}$.

Answer: $\sqrt{33}, \sqrt{42}$.