4. Around the triangle $ABC$ with angle $\beta=2 \delta<\frac{\pi}{3}$ is described a circle $k$ with radius $R$ whose diameter bisects the angle $\beta$. Find the area of triangle $ABM$ if $M=t \cap BC$, where $t$ is the tangent to $k$ drawn at point $A$.
Solution. Since the diameter of the circumscribed circle is the angle bisector of angle $\beta$ (see diagram), it follows that triangle $A B C$ is isosceles. Therefore, we have

$$
\alpha=90^{\circ}-\delta, \measuredangle C A M=2 \delta, \measuredangle A C M=90^{\circ}+\delta, \measuredangle A M C=90^{\circ}-3 \delta
$$

Applying the Law of Sines to $\triangle A B C$, we get

$$
\overline{A C}=2 R \sin 2 \delta, \overline{A B}=2 R \sin \left(90^{\circ}-\delta\right)=2 R \cos \delta
$$

If we apply it to triangle AMC, we will get:

$$
\overline{A M}=\overline{A C} \cdot \frac{\sin \left(90^{\circ}+\delta\right)}{\sin \left(90^{\circ}-3 \delta\right)}=2 R \frac{\cos \delta}{\cos 3 \delta} \sin 2 \delta
$$

For the area of triangle $\triangle A B M$:

$$
P=\frac{1}{2} \overline{A B} \cdot \overline{A M} \sin \left(90^{\circ}+\delta\right)=4 R^{2} \frac{\sin \delta \sin ^{4} \delta}{\cos 3 \delta}
$$

## IV year

![](https://cdn.mathpix.com/cropped/2024_06_05_9925ad5ba00e33a79b8cg-6.jpg?height=551&width=466&top_left_y=210&top_left_x=1014)