$A C$ and $B D$ are diagonals of the inscribed quadrilateral $A B C D$. Angles $D A C$ and $A B D$ are equal to $\gamma$ and $\delta$ respectively, side $C D=a$. Find the area of triangle $A C D$
Apply the Law of Sines to triangle $A C D$.

## Solution

Since the inscribed angles $A C D$ and $A B D$ subtend the same arc, then

$$
\angle A C D=\angle A B D=\delta
$$

Applying the Law of Sines to triangle $A C D$, we get:

$$
\frac{A C}{\sin \left(180^{\circ}-\angle C A D-\angle A C D\right)}=\frac{C D}{\sin \angle C A D} \text {, or } \frac{A C}{\sin \left(180^{\circ}-\gamma-\delta\right)}=\frac{a}{\sin \gamma} \text {. }
$$

From this, we find that

$$
A C=a \cdot \frac{\sin \left(180^{\circ}-\gamma-\delta\right)}{\sin \gamma}=a \cdot \frac{\sin (\gamma+\delta)}{\sin \gamma}
$$

Therefore,

$$
S_{\triangle \mathrm{ACD}}=\frac{1}{2} \cdot C D \cdot A C \cdot \sin \angle A C D=\frac{1}{2} \cdot a \cdot a \cdot \frac{\sin (\gamma+\delta)}{\sin \gamma} \cdot \sin \delta=\frac{1}{2} a^{2} \cdot \frac{\sin \delta \cdot \sin (\gamma+\delta)}{\sin \gamma} .
$$

## Answer

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-40.jpg?height=565&width=552&top_left_y=2130&top_left_x=753)

$\frac{1}{2} a^{2} \cdot \frac{\sin \delta \cdot \sin (\gamma+\delta)}{\sin \gamma}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-41.jpg?height=160&width=2044&top_left_y=109&top_left_x=13)

In the center of a circular pool, a student is swimming. Suddenly, a teacher approaches the pool. The teacher cannot swim but runs 4 times faster than the student can swim. The student runs faster. Can he escape?

## Solution

Let $R$ be the radius of the pool. The student can swim along a circle of radius $r$, where $(1-\pi / 4) R<r<R / 4$, then his angular velocity is greater than the teacher's angular velocity, and after some time, the student and the teacher will be at opposite ends of the diameters. Then the student has to swim less than ${ }^{\pi R} / 4$ to the nearest point on the larger circle, while the teacher has to run a distance of $\pi R$. Therefore, the student will reach the edge of the pool before the teacher and will have time to escape.