Point $D$ lies on side $A C$ of triangle $A B C$. The circle $S_{1}$, inscribed in triangle $A B D$, touches segment $B D$ at point $M$; the circle $S_{2}$, inscribed in triangle $B C D,$ at point $N$. The ratio of the radii of circles $S_{1}$ and $S_{2}$ is $\frac{7}{4}$. It is known that $B M=3, M N=N D=1$. Find the sides of triangle $A B C$.
Find $\cos \angle B D C$ and apply the cosine theorem (or use Heron's formula)

## Solution

First method.

Let $O_{1}$ and $O_{2}$ be the centers of circles $S_{1}$ and $S_{2}$, $r_{1}$ and $r_{2}$ their radii, $P$ and $Q$ the points of tangency with sides $A D$ and $D C$ respectively. Denote $\angle B D C=\alpha$. Since $\angle O_{1} D O_{2}=90^{\circ}$ and $\angle O_{2} D Q=\frac{\alpha}{2}$, then $\angle O_{1} D P=90^{\circ} -\frac{\alpha}{2}$. Therefore,

$$
r_{1}=D P \cot \frac{\alpha}{2}=D M \cot \frac{\alpha}{2}=2 \cot \frac{\alpha}{2}, r_{2}=D Q \tan \frac{\alpha}{2}=D N \tan \frac{\alpha}{2}=\tan \frac{\alpha}{2}
$$

On the other hand, $\frac{r_{1}}{r_{2}}=\frac{7}{4}$. From the obtained system of equations, we find that

$$
r_{2}=\sqrt{\frac{8}{7}}, r_{1}=\frac{7}{4} \sqrt{\frac{8}{7}}
$$

Therefore,

$$
\tan \frac{\alpha}{2}=r_{2}=\sqrt{\frac{8}{7}}, \cos \alpha=\frac{1-\tan^{2} \frac{\alpha}{2}}{1+\tan^{2} \frac{\alpha}{2}}=-\frac{1}{15}
$$

Let $C Q=x$. In triangle $B C D$, it is known that

$$
B D=5, C D=1+x, B C=x+4, \cos \angle B D C=-\frac{1}{15}
$$

By the cosine theorem,

$$
(x+4)^{2}=25+(x+1)^{2}+2 \cdot 5 \cdot(x+1) \cdot \frac{1}{15}
$$

From this equation, we find that $x=2$. Similarly, we find that $A P=7$. Therefore,

$$
\begin{gathered}
A C=A P+P D+D Q+Q C=7+2+1+2=12 \\
B C=C Q+B N=2+4=6, A B=A P+B M=7+3=10
\end{gathered}
$$

Second method.

Let $A P=x$ (where $P$ is the point of tangency of the first circle with side $A D$) and express the area of triangle $A B D$ using Heron's formula, as well as through the semiperimeter and $r_{1}$. Solving the obtained equation, we find that $x=7$. Similarly, we find $C Q$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f3f7c9453eb6f3f37ea7g-29.jpg?height=383&width=558&top_left_y=1372&top_left_x=750)

Answer

$A B=10, B C=6, A C=12$.