## Task B-2.3.

In a triangle with side lengths $a, b, c$ and opposite angles $\alpha, \beta, \gamma$ respectively, the following equality holds:

$$
a^{3}+b^{3}+c^{3}=a b(a+b)-b c(b+c)+a c(a+c)
$$

If $\cos \beta=\frac{24}{25}$, determine the ratio in which the height from the vertex of angle $\alpha$ divides the opposite side.
## Solution.

From the equality $a^{3}+b^{3}+c^{3}=a b(a+b)-b c(b+c)+a c(a+c)$, it follows that

$$
\begin{gathered}
a^{3}+b^{3}+c^{3}-a^{2} b-a b^{2}+b^{2} c+b c^{2}-a^{2} c-a c^{2}=0 \\
a^{3}-a b^{2}-a c^{2}+b^{3}-a^{2} b+b c^{2}+c^{3}+b^{2} c-a^{2} c=0 \\
a\left(a^{2}-b^{2}-c^{2}\right)+b\left(b^{2}-a^{2}+c^{2}\right)+c\left(c^{2}+b^{2}-a^{2}\right)=0 \\
\left(b^{2}+c^{2}-a^{2}\right)(-a+b+c)=0
\end{gathered}
$$

Due to the triangle inequality, $b+c-a>0$, so it must be that $b^{2}+c^{2}-a^{2}=0$, or $b^{2}+c^{2}=a^{2}$. Therefore, the given triangle is a right triangle with the hypotenuse of length $a$. Since $\cos \beta=\frac{24}{25}$, the lengths of the sides of this triangle are of the form

$$
a=25 k, \quad c=24 k, \quad b=\sqrt{a^{2}-c^{2}}=\sqrt{(25 k)^{2}-(24 k)^{2}}=7 k
$$

Let's denote the parts of the given triangle as shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_30_d18d2bbdee068dba95dfg-08.jpg?height=557&width=857&top_left_y=1509&top_left_x=608)

From the similarity of triangles $A B D$ and $C B A$ and triangles $C A D$ and $C B A$ (or from Euclid's theorem), we get

$$
\frac{p}{b}=\frac{b}{a} \Rightarrow p=\frac{b^{2}}{a}, \quad \frac{q}{c}=\frac{c}{a} \Rightarrow q=\frac{c^{2}}{a}
$$

The desired ratio is $\frac{p}{q}=\frac{b^{2}}{c^{2}}=\left(\frac{7 k}{24 k}\right)^{2}=\frac{49}{576}$ (or its reciprocal).