8.5. From the right angle $C$ of triangle $ABC$, the median $CM$ is drawn. The circle inscribed in triangle $ACM$ touches side $CM$ at its midpoint. Find the angles of triangle $ABC$.
Answer: $30^{\circ}, 60^{\circ}, 90^{\circ}$.

Solution 1. Angle $\angle C$ is a right angle, i.e., $\angle C=90^{\circ}$. Let's find angles $\angle A$ and $\angle B$. Let $K, L$, and $N$ be the points of tangency of the circle inscribed in triangle $A C M$ with sides $C M, A M$, and $A C$ of this triangle, respectively. By the condition, $M K=C K$. Since the segments of tangents drawn from one point to a circle are equal, then $M L=M K=C K=C N$ and $A L=A N$. Therefore, $A C=A N+N C=A L+L M=A M$. But $C M=A M$, since $A M$ and $M C$ are radii of the circumcircle of triangle $A B C$. Thus, $A C=C M=A M$, i.e., triangle $A C M$ is equilateral. Hence, $\angle A=60^{\circ}, \angle B=180^{\circ}-\angle A-\angle C=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4dc87fb8be38105fe1c5g-3.jpg?height=420&width=294&top_left_y=1343&top_left_x=1646)

In Solution 2. Let $O$ be the center of the circle inscribed in triangle $A C M$, and $K$ be the point of tangency of this circle with side $C M$. Segment $O K$ is both a median (by the condition, $C K=C M$) and an altitude (by the property of a tangent to a circle, $O K \perp C M$) in triangle $C O M$. Therefore, triangle $C O M$ is isosceles, $C O=O M$, and the angles at its base are equal: $\angle O M C=\angle O C M$. Since $O$ is the center of the inscribed circle, $C O$ and $M O$ are the angle bisectors of $\angle A C M$ and $\angle A M C$, respectively, so $\angle A M C=2 \angle O M C=2 \angle O C M=\angle A C M$. Thus, triangle $A C M$ is also isosceles, $A C=A M=\frac{1}{2} A B$. Therefore, triangle $A B C$ is a right triangle with angles $30^{\circ}, 60^{\circ}, 90^{\circ}$.

8.8, 9.7. Petya and Vanya built columns of cubes, each with four cubes, but they followed different rules: in each column, Petya has red, yellow, green, and blue cubes, while Vanya has only red, yellow, and green cubes. It turned out that all the columns they built are different, and neither Petya nor Vanya can build new columns following their rules. Who built more columns and by how many times?

Answer: Vanya built one and a half times more.

Solution. From each column built by Petya, we draw three threads to such columns built by Vanya, where instead of a blue cube, there is a red, yellow, or green cube, and the other cubes in the columns match. Since in each column built by Vanya, there are two cubes of the same color, it will be connected to two columns built by Petya. Therefore, three times the number of Petya's columns equals two times the number of Vanya's columns, so Vanya built one and a half times more columns than Petya.