## Task 4 - 260924

For a right triangle $A B C$ with the right angle at $C$, it is required that this right angle be divided into four equal angles by the median of side $A B$, the angle bisector of $\angle A C B$, and the altitude perpendicular to side $A B$.

Investigate whether there exists a triangle $A B C$ that meets these requirements, and whether all such triangles are similar to each other!

Determine, if this is the case, the measures of the angles $\angle B A C$ and $\angle A B C$!
Let $W$ be the intersection of the angle bisector with $AB$, $H$ the foot of the perpendicular from $C$ to $AB$, and $M$ the midpoint of the segment $AB$.

Since the angle bisector divides the right angle at $C$ into the two angles $\angle A C W = \angle W C B = 45^{\circ}$, the altitude and the median must bisect these two sub-angles.

Without loss of generality, assume that the median bisects the angle $\angle A C W$ and the altitude bisects the angle $\angle W C B$. (The case where the roles of the two points $A$ and $B$ are swapped is obtained by interchanging these points.)

Thus, we have

$$
\angle A C M = \angle M C W = \angle W C H = \angle H C B = 22.5^{\circ}
$$

By definition, $\angle B H C = 90^{\circ}$, so due to the sum of the interior angles in the triangle $\triangle B H C$, we have

$$
\angle C B A = \angle C B H = 180^{\circ} - \angle H C B - \angle B H C = 67.5^{\circ}
$$

Due to the sum of the interior angles in the triangle $\triangle A B C$, we then have $\angle B A C = 180^{\circ} - \angle A C B - \angle C B A = 22.5^{\circ}$. Thus, the two remaining interior angles of the triangle $\triangle A B C$ are uniquely determined (up to swapping), provided such a triangle exists.

Now, let us consider a triangle $\triangle A B C$ with $\angle B A C = 22.5^{\circ}$, $\angle C B A = 67.5^{\circ}$, and $\angle A C B = 90^{\circ}$, and let the points $H$, $W$, and $M$ be defined as above. Then, due to the definition of the angle bisector, $\angle A C W = \angle W C B = \frac{90^{\circ}}{2} = 45^{\circ}$.

By the Thales' theorem, $C$, $A$, and $B$ lie on the circle centered at $M$ passing through $A$, so $|A M| = |B M| = |C M|$. In particular, the triangle $\triangle A M C$ is isosceles, and we have

$$
\angle A C M = \angle M A C = \angle B A C = 22.5^{\circ} = \frac{90^{\circ}}{4} \quad \text{and} \quad \angle M C W = \angle A C W - \angle A C M = 22.5^{\circ} = \frac{90^{\circ}}{4}
$$

In the triangle $\triangle H C B$, we have $\angle B H C = 90^{\circ}$ by the definition of the altitude, so from the sum of the interior angles in this triangle and $\angle C B H = \angle C B A$, we get

$$
\angle H C B = 180^{\circ} - \angle B H C - \angle C B H = 180^{\circ} - 90^{\circ} - 67.5^{\circ} = 22.5^{\circ} = \frac{90^{\circ}}{4}
$$

Since the triangle $\triangle A B C$ has no obtuse interior angle, $H$ lies on the segment $AB$, and we finally have

$$
\angle W C H = \angle W C B - \angle H C B = 45^{\circ} - 22.5^{\circ} = 22.5^{\circ} = \frac{90^{\circ}}{4}
$$

Thus, as required in the problem statement, the median, angle bisector, and altitude divide the right angle into four equal parts.

Therefore, there is exactly one choice of the two remaining interior angles of $\triangle A B C$, namely the ones given above, $22.5^{\circ}$ and $67.5^{\circ}$, up to swapping the angles at $A$ and $B$ (which preserves similarity).