In the right-angled triangle $ABC$, the circle drawn over the leg $AC$ intersects the hypotenuse $AB$ at point $D$. What are the angles of the triangle if the area of the circular segment cut off by the chord $AD$ is equal to the area of the region bounded by the segments $BC$ and $BD$ and the arc $CD$?
We attach to the two equal-area regions the region bounded by segments $AC$, $AD$, and arc $CD$. This slice, together with the other region, covers the semicircle $ADC$, and with the other region, it covers the triangle $ABC$, so the semicircle and the triangle have equal areas. With the notation $AC=b$, $BC=a$, and $\angle BAC = \alpha$,

$$
\begin{gathered}
\frac{1}{2} \cdot \frac{b^{2} \pi}{4}=\frac{a b}{2}, \text { from which } \frac{a}{b}=\tan \alpha=\frac{\pi}{4} \approx 0.7854 \\
\alpha \approx 38^{\circ} 15^{\prime}, \quad \text { and } \quad \beta \approx \angle ABC=51^{\circ} 45^{\prime}
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_f8aa2bae2207e1a8f913g-1.jpg?height=466&width=544&top_left_y=499&top_left_x=777)

Tamás Komor (Budapest, Fazekas M. High School, 2nd grade)