The base of a triangle is $a$, and its height is $m$. Draw a square on the base such that two of its vertices fall on the sides of the triangle. In the smaller triangle that is formed, inscribe a square in the same manner, and continue this process to infinity. What is the sum of the areas of the squares?
Let the sides of the consecutive squares be: $x_{1}, x_{2}, x_{3}, \ldots$ Then, from the similar triangles formed:

$$
x_{1}: a=\left(m-x_{1}\right): m
$$

from which

$$
\begin{gathered}
x_{1}=\frac{a m}{m+a} \\
x_{2}: a=\left(m-x_{1}-x_{2}\right): m_{1}
\end{gathered}
$$

or

$$
\begin{aligned}
& x_{2}: a=\left(m-\frac{a m}{m+a}-x_{2}\right): m \\
& m x_{2}=a\left(\frac{m^{2}+a m-a m}{m+a}-x_{2}\right)
\end{aligned}
$$

thus

$$
x_{2}(m+a)=\frac{a m^{2}}{m+a}
$$

or

$$
x_{2}=\frac{a m^{2}}{(m+a)^{2}}
$$

Continuing this procedure, we see that the areas of the consecutive squares are:

$$
x_{1}^{2}=\frac{a^{2} m^{2}}{(m+a)^{2}}, x_{2}^{2}=\frac{a^{2} m^{4}}{(m+a)^{4}}, \text { etc. }
$$

These areas form an infinite geometric progression whose first term is:

$$
\frac{a^{2} m^{2}}{(m+a)^{2}}
$$

and common ratio is:

$$
\frac{m^{2}}{(m+a)^{2}}
$$

Since the common ratio is less than 1, the sum of the areas is:

$$
S=\frac{a^{2} m^{2}}{(m+a)^{2}} \cdot \frac{1}{1-\frac{m^{2}}{(m+a)^{2}}}=\frac{a^{2} m^{2}}{a^{2}+2 a m+m^{2}-m^{2}}=\frac{a m^{2}}{a+2 m}
$$

(Miklós Dénes, Budapest.)