In triangle $A B C$, it is known that $\angle B A C=\alpha, \angle A B C=\beta, B C=a, A D$ is the altitude. On side $A B$, a point $P$ is taken such that $\frac{A P}{P B}=\frac{1}{2}$. A circle is drawn through point $P$, touching side $B C$ at point $D$. Find the radius of this circle.
Let $O$ be the center of the given circle. Apply the Law of Cosines to triangle $A O P$.

## Solution

Note that the center $O$ of the given circle lies on the altitude $A D$. Let $A D = h$, and the radius of the circle be $r$. Suppose that point $O$ lies between points $A$ and $D$. Then,

$$
\begin{gathered}
A B = \frac{A D}{\sin \angle A B D} = \frac{h}{\sin \beta}, \quad A P = \frac{h}{3 \sin \beta}, \\
A O = A D - O D = h - r, \quad O P = r
\end{gathered}
$$

By the Law of Cosines in triangle $O A P$, we have

$$
O P^{2} = A P^{2} + A O^{2} - 2 A P \cdot A O \cos \left(90^{\circ} - \angle A B D\right)
$$

or

$$
r^{2} = \left(\frac{h}{3 \sin \beta}\right)^{2} + (h - r)^{2} - 2 h \cdot \frac{h - r}{3}
$$

From this, we find that

$$
r = \frac{h \left(3 \sin^{2} \beta + 1\right)}{12 \sin^{2} \beta}
$$

From triangle $A B C$ using the Law of Sines, we find that

$$
A B = \frac{a \sin (\alpha + \beta)}{\sin \alpha}
$$

Then,

$$
h = A B \sin \beta = \frac{a \sin \beta \sin (\alpha + \beta)}{\sin \alpha}
$$

Therefore,

$$
r = \frac{a \sin (\alpha + \beta) \left(3 \sin^{2} \beta + 1\right)}{12 \sin \alpha \sin \beta}
$$

If point $A$ is located between points $O$ and $D$, the solution is similar (in this case, $A O = r - h$, and $\cos \angle O A P = -\sin \beta$)

![](https://cdn.mathpix.com/cropped/2024_05_06_1b164f828743b46f847cg-33.jpg?height=363&width=688&top_left_y=139&top_left_x=685)

## Answer

$\frac{a \sin (\alpha + \beta) \left(3 \sin^{2} \beta + 1\right)}{12 \sin \alpha \sin \beta}$