4. One angle of a right triangle is $50^{\circ}$. Let $X, Y$ and $Z$ be the points where the inscribed circle touches the sides of the triangle. Calculate the sizes of the angles of triangle $\triangle X Y Z$.
First solution.

Sketch:

![](https://cdn.mathpix.com/cropped/2024_05_30_81794be492b317694280g-35.jpg?height=645&width=554&top_left_y=1045&top_left_x=288)

Since one acute angle of the right triangle is $50^{\circ}$, the other acute angle must be $40^{\circ}$ because the sum of the angles adjacent to the hypotenuse is $90^{\circ}$.

$1 \text{ POINT}$

Perpendiculars from the center $S$ intersect the sides of the triangle at points $X, Y$, and $Z$, and the lengths $\overline{S X}, \overline{S Y}$, and $\overline{S Z}$ are equal because the center of the inscribed circle is equidistant from all three sides of the triangle.

Since the sum of the angles in a quadrilateral is $360^{\circ}$, and in quadrilateral $Z A X S$ the known angles are $50^{\circ}, 90^{\circ}$, and $90^{\circ}$, it follows that the angle at vertex $S$ is $130^{\circ}$.

In quadrilateral $X B Y S$, the known angles are $40^{\circ}, 90^{\circ}$, and $90^{\circ}$, so the angle at vertex $S$ is $140^{\circ}$.

Quadrilateral $Y C Z S$ is a square, so all angles are right angles.

$1 \text{ POINT}$

We consider triangles $Z X S, X Y S$, and $Y Z S$, which are isosceles because the sides $\overline{S X}, \overline{S Y}$, and $\overline{S Z}$ are of equal length and thus serve as the legs of these triangles.

$1 \text{ POINT}$

Since the sum of the angles in a triangle is $180^{\circ}$, we conclude that the angles at the base of triangle $Z X S$ are $\left(180^{\circ}-130^{\circ}\right): 2=25^{\circ}$,

in triangle $X Y S$ they are $\left(180^{\circ}-140^{\circ}\right): 2=20^{\circ}$,

and in triangle $Y Z S$ they are $\left(180^{\circ}-90^{\circ}\right): 2=45^{\circ}$.

$1 \text{ POINT}$

Therefore, the angles of triangle $X Y Z$ are

$$
\begin{aligned}
& 20^{\circ}+25^{\circ}=45^{\circ} \\
& 20^{\circ}+45^{\circ}=65^{\circ} \text{ and} \\
& 25^{\circ}+45^{\circ}=70^{\circ}
\end{aligned}
$$

##