4. In a given acute triangle $ABC$, let $A_{1}, B_{1}$ be the feet of the altitudes from vertices $A, B$. Determine the sizes of its internal angles at vertices $B$ and $C$, if the size of angle $BAC$ is $40^{\circ}$ and the radii of the inscribed circles of triangles $A_{1}B_{1}C$ and $ABC$ are in the ratio $1: 2$.

II. round of category B takes place

## on Tuesday, March 31, 1998

so that it starts in the morning and the contestants have 4 hours of pure time to solve the problems. For each problem, a contestant can earn 6 points, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the start of the competition.
4. Let $\alpha, \beta, \gamma$ be the measures of the interior angles of triangle $ABC$. The feet of the altitudes $A_{1}$ and $B_{1}$ (Fig.3) lie on the Thales circle with diameter $AB$, so quadrilateral $AB A_{1} B_{1}$ is cyclic, and for its opposite angles we have $|\angle CAB| + |\angle B A_{1} B_{1}| = 180^{\circ}$. Therefore, $|\angle B_{1} A_{1} C| = \alpha$, so triangles $ABC$ and $A_{1} B_{1} C$ are similar by the AA similarity criterion.

If we denote by $\varrho$ and $\varrho_{1}$ the radii of the incircles of triangles $ABC$ and $A_{1} B_{1} C$ respectively, it follows from the mentioned similarity that

$$
\cos \gamma = \frac{|B_{1} C|}{|BC|} = \frac{|A_{1} C|}{|AC|} = \frac{\varrho_{1}}{\varrho} = \frac{1}{2},
$$

so $\gamma = 60^{\circ}$. Given that $\alpha = 40^{\circ}$, we finally get $\beta = 80^{\circ}$.

Thus, the problem is solved.

Award 6 points for a complete solution, and 3 points for proving the similarity of triangles $ABC$ and $A_{1} B_{1} C$.

![](https://cdn.mathpix.com/cropped/2024_04_17_95cd66dd41f619cc2db4g-4.jpg?height=363&width=442&top_left_y=496&top_left_x=1427)

Fig. 3