Draw an acute isosceles triangle inside a circle. Over one of the sides of this triangle, construct another isosceles triangle such that its apex also lies on the circumference of the circle. Continuing this process, determine the limit of the angles at the vertices of the triangles.
From the fact that triangles can always be constructed on the chords in two different ways, it follows that this problem is solvable as long as a certain law holds between the successive constructions.

1. First, we will examine the case where the constructed isosceles triangles are always obtuse.

If the angle at the vertex of the given $ABC$ triangle is $\alpha$, the other two angles are $\beta=\gamma$, and the vertices of the successively constructed obtuse isosceles triangles are $A_{1}, A_{2}, \ldots A_{n}, \ldots$, and the angles at these vertices are $\alpha_{1}, \alpha_{2}, \ldots \alpha_{n} \ldots$, and the center of the circle is $O$, then from the cyclic quadrilateral $ABCA_{1}$:

$$
\alpha_{1}+\beta=2R
$$

and since

$$
\beta=R-\frac{\alpha}{2}
$$

therefore

$$
\alpha_{1}=R+\frac{\alpha}{2}
$$

and similarly, in general,

$$
\alpha_{n}=R+\frac{\alpha_{n-1}}{2}
$$

thus

$$
\begin{gathered}
\alpha_{n}=R+\frac{R}{2}+\frac{\alpha_{n-2}}{4}=R+\frac{R}{2}+\frac{R}{4}+\frac{\alpha_{n-3}}{8}=\ldots= \\
=R+\frac{R}{2}+\ldots+\frac{R}{2^{n-1}}+\frac{\alpha}{2^{n}}=R \cdot \frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}+\alpha \cdot \frac{1}{2^{n}}= \\
=2R-(2R-\alpha) \cdot \frac{1}{2^{n}}
\end{gathered}
$$

and since for $n=\infty$

$$
\lim _{n=\infty} \cdot \frac{1}{2^{n}}=\frac{1}{\lim _{n=\infty} \cdot 2^{n}}=\frac{1}{\infty}=0
$$

therefore

$$
\lim _{n=\infty} \alpha_{n}=2R
$$

2. Now let the constructed isosceles triangle always be acute, and denote the vertices by $A^{\prime}, A^{\prime \prime}, \ldots A^{(n)} \ldots$; the angles at these vertices by $\alpha^{\prime}, \alpha^{\prime \prime}, \ldots \alpha^{(n)} \ldots$; from the cyclic quadrilateral $AA^{\prime}CA_{1}$:

$$
\alpha^{\prime}+\alpha_{1}=2R
$$

and since

$$
\alpha_{1}=R+\frac{\alpha}{2}
$$

therefore

$$
\alpha^{\prime}=R-\frac{\alpha}{2}
$$

and in general

$$
\alpha^{(n)}=R-\frac{\alpha^{(n-1)}}{2}
$$

and thus

$$
\begin{aligned}
& \alpha^{(n)}=R-\frac{R}{2}+\frac{R^{2}}{2}+\ldots+(-1)^{n-1} \frac{R}{2^{n-1}}+(-1)^{n} \frac{\alpha}{2^{n}}= \\
& =R \cdot \frac{1+\frac{1}{2^{n}}}{1+\frac{1}{2}}+(-1)^{n} \alpha \cdot \frac{1}{2^{n}}=\frac{2}{3} R+\left(\frac{2}{3} R+(-1)^{n} \alpha\right) \frac{1}{2^{n}}
\end{aligned}
$$

thus for $n=\infty$

$$
\lim _{n=\infty} \alpha^{(n)}=\frac{2R}{3}=60^{\circ}
$$

In this case, the limiting triangle is the inscribed regular triangle.

(Sándor Pichler, Budapest.)

The problem was also solved by: Bartók I., Jánosy Gy., Neidenbach E., Székely J. They focused on one case: Dömény I., Havas E., Haar A., Kertész G., Kiss J., Szúcs A.