Kazitsyna T.V.

Lёsha drew a geometric picture by tracing his plastic right-angled triangle four times, placing the shorter leg against the hypotenuse and aligning the vertex of the acute angle with the vertex of the right angle. It turned out that the "closing" fifth triangle is isosceles (see the figure, the marked sides are equal). Find the acute angles of Lёsha's triangle?

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-33.jpg?height=700&width=646&top_left_y=1371&top_left_x=709)
Let the smaller angle of Leshin's triangle be $\alpha$. Denote the points as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-34.jpg?height=760&width=726&top_left_y=0&top_left_x=677)

Notice that four angles of the white pentagon are equal to $90^{\circ}+\alpha$. Therefore, the fifth angle (at vertex $F$) is $540^{\circ}-4\left(90^{\circ}+\alpha\right)=180^{\circ}-4 \alpha$. Hence, $\angle E A F=\angle A F E=4 \alpha$.

Now notice that the isosceles triangles $A B C$ and $C D E$ with the angle at the vertex equal to $90^{\circ}+\alpha$ are equal by two sides and the angle between them. Therefore, $A C=C E$ and $\angle B A C=\angle C E D=45^{\circ}-\frac{\alpha}{2}$. This means that triangle $A C E$ is isosceles and $\angle C A E=1 / 2\left(180^{\circ}-\left(\left(90^{\circ}+\alpha\right)-\left(90^{\circ}-\alpha\right)\right)=90^{\circ}-\alpha\right.$.

On the other hand, $\angle C A E=\angle B A E-\angle B A C=\left(90^{\circ}+4 \alpha\right)-\left(45^{\circ}-\alpha / 2\right)=45^{\circ}+9 \alpha / 2$, from which $11 \alpha=90^{\circ}$.

## Answer

$90^{\circ} / 11,900^{\circ} / 11$.