For a triangle with sides $a, b, c$, draw lines through an interior point that are parallel to the sides. If the segments of these lines within the triangle are of equal length, what is this length?
Solution. Let's denote the lengths of the segments of the lines inside the triangle by $x$.

All triangles visible in the diagram are similar, since their sides are parallel. Let's denote their areas as shown in the diagram, and the area of triangle $ABC$ by $T$. Notice that $t_{A}+t_{B}+t_{C}=T+t_{a}+t_{b}+t_{c}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c2022df27a2a64428548g-1.jpg?height=450&width=512&top_left_y=309&top_left_x=796)

Due to the similarities,

$$
t_{A}=\left(\frac{x}{a}\right)^{2} T, \quad t_{B}=\left(\frac{x}{b}\right)^{2} T, \quad t_{C}=\left(\frac{x}{c}\right)^{2} T
$$

and using the fact that the opposite sides of the parallelograms in the diagram are equal:

$$
t_{a}=\left(\frac{a-x}{a}\right)^{2} T, \quad t_{b}=\left(\frac{b-x}{b}\right)^{2} T, \quad t_{c}=\left(\frac{c-x}{c}\right)^{2} T
$$

Thus, we can write the following equation:

$$
\left(\frac{x}{a}\right)^{2} T+\left(\frac{x}{b}\right)^{2} T+\left(\frac{x}{c}\right)^{2} T=T+\left(\frac{a-x}{a}\right)^{2} T+\left(\frac{b-x}{b}\right)^{2} T+\left(\frac{c-x}{c}\right)^{2} T
$$

Simplifying by $T$ and rearranging:

$$
4-2 x\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0
$$

Therefore, the desired length is:

$$
x=\frac{2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}
$$

which is exactly $\frac{2}{3}$ of the harmonic mean of the sides.