Inside an acute angle domain, 2 points are given. Construct that isosceles triangle whose base lies on one side of the angle, its apex is on the other side of the angle, and one of its legs passes through each of the given points.
I. solution. Let the given angle be $\angle M O N = \omega \angle P_{1} P^{\prime \prime} P = 2 \angle O P^{\prime \prime} P = \angle P_{1} A Q$.

If $P^{\prime \prime}$ coincides with $Q^{\prime \prime}$, that is, $P Q \perp O M$, then we have equality instead of inequality, and $A$ coincides with $P^{\prime \prime}$, making the solution degenerate into a line segment.

We get another solution if we look for the base of our isosceles triangle on the $O N$ side. The degeneracy mentioned can only occur in one case.

From the work of András Benczúr (Budapest, Fazekas M. g.), with additional discussion.

Many computational solutions were received. We outline one such solution, leaving the discussion to the reader.

![](https://cdn.mathpix.com/cropped/2024_05_02_1ed52d168b88346b5d63g-2.jpg?height=602&width=770&top_left_y=122&top_left_x=664)

II. solution. Draw a line parallel to $O M$ through $A$ and $P$, and let the projections of $P$ and $Q$ on the former be $S$ and $T$, and the projection of $Q$ on the latter be $U$. From the obvious similarity of the triangle pairs $A P S, A Q T$ and $A P^{\prime \prime} S, A Q^{\prime \prime} T$:

$$
\frac{P S}{Q T} = \frac{A S}{A T} = \frac{P^{\prime \prime} S}{Q^{\prime \prime} T}
$$

Therefore, with the notation $P S = x, \quad P^{\prime \prime} P = p, \quad Q^{\prime \prime} Q = q, \quad Q U = r$ (where $r$ can also be negative):

$$
\begin{gathered}
\frac{x}{x-r} = \frac{x-p}{q+r-x}, \quad 2 x^{2} - (p+q+2 r) x + p r = 0 \\
x = \frac{1}{4} \left( p + q + 2 r \pm \sqrt{(p+q+2 r)^{2} - 8 p r} \right)
\end{gathered}
$$

Zoltán Sebestyén (Celldömölk, Berzsenyi D. g.)