We draw an isosceles right triangle $ABC$ over the hypotenuse $AB$. Over the hypotenuse $BC$, we draw the isosceles right triangles $BCC_{1}$, then $BC_{1}C_{2}$, $BC_{2}C_{3}$, and so on.

Calculate the length of the broken line $AC C_{1} C_{2} C_{3} \ldots C_{n}$, as well as the sum of the areas of the triangles, if $AB=a$. What will be the result of the obtained results when $n$ becomes infinite?
The length of the broken line $A C C_{1} C_{2} C_{3} \ldots C_{n}$

$$
\begin{gathered}
S=A C+C C_{1}+C_{1} C_{2}+C_{2} C_{3}+\cdots+C_{n-1} C_{n}= \\
\frac{a}{\sqrt{2}}+\frac{a}{2}+\frac{a}{2 \sqrt{2}}+\frac{a}{4}+\cdots+\frac{a}{(\sqrt{2})^{n+1}}= \\
\frac{a}{\sqrt{2}}\left[1+\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2})^{2}}+\frac{1}{(\sqrt{2})^{3}}+\cdots+\frac{1}{(\sqrt{2})^{n}}\right]= \\
=\frac{a}{\sqrt{2}} \frac{1-\frac{1}{(\sqrt{2})^{n+1}}}{1-\frac{1}{\sqrt{2}}}
\end{gathered}
$$

The sum of the areas of the triangles

$$
\begin{gathered}
T=A B C \Delta+B C C_{1} \Delta+B C_{1} C_{2} \Delta+B C_{2} C_{3} \Delta+\cdots+B C_{n-1} C_{n} \Delta= \\
=\frac{a^{2}}{4}+\frac{a^{2}}{8}+\frac{a^{2}}{16}+\frac{a^{2}}{32}+\cdots+\frac{a^{2}}{2^{n+2}}= \\
=\frac{a^{2}}{4}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{n}}\right)=\frac{a^{2}}{4} \frac{2^{n+1}-1}{2^{n}}=a^{2} \frac{2^{n+1}-1}{2^{n+2}}
\end{gathered}
$$

If $n$ becomes infinite, then

$$
S=a(\sqrt{2}+1)
$$

and

$$
T=a^{2}\left(\frac{1}{2}-\frac{1}{2^{n+2}}\right)=\frac{a^{2}}{2}
$$

(Miksa Rosenthal, Pécs.)

Number of solutions: 38.