\begin{aligned} & \text { The Law of Sines } \\ & {[\text { Area of a triangle (using two sides and the included angle).] }}\end{aligned}

Two circles with centers $O_{1}$ and $O_{2}$, lying on the side $M N$ of triangle $M P N$, touch each other and intersect sides $M P$ and $P N$ at points $M, D$, and $N, C$ respectively, such that $M O_{1}=O_{1} D=3$ and $N O_{2}=C O_{2}=6$. Find the area of triangle $M N P$, given that the ratio of the area of triangle $M C O_{2}$ to the area of triangle $O_{1} D N$ is $8 \sqrt{2-\sqrt{3}}$.
Compose a system of trigonometric equations with respect to angles $P M N$ and $P N M$.

## Solution

Let $\angle P M N=\alpha, \angle P N M=\beta$. Then $\angle D O_{1} N=2 \alpha, \angle C O_{2} M=2 \beta$. Since

$$
\frac{S_{\triangle M C O_{2}}}{S_{\triangle O_{1} D N}}=\frac{O_{2} C \cdot O_{2} M \sin 2 \beta}{O_{1} D \cdot O_{1} N \sin 2 \alpha}=\frac{8 \sin 2 \beta}{5 \sin 2 \alpha}=\frac{8 \sqrt{3}}{5}
$$

then

$$
\frac{\sin \beta \cos \beta}{\sin \alpha \cos \alpha}=\sqrt{3}
$$

By the Law of Sines,

$$
\frac{\sin \beta}{\sin \alpha}=\frac{P M}{P N}=\frac{1}{\sqrt{2-\sqrt{3}}}=\sqrt{2+\sqrt{3}}
$$

Substituting the found ratio into the previous equation, we get that

$$
\frac{\cos \beta}{\cos \alpha}=\sqrt{2+\sqrt{3}}
$$

Therefore,

$$
\left\{\begin{array}{l}
\sin \alpha=\sqrt{2-\sqrt{3}} \sin \beta \\
\cos \alpha=\frac{\sqrt{2+\sqrt{3}} \cos \beta}{\sqrt{3}}
\end{array}\right.
$$

Squaring both sides of these equations and adding the resulting equations term by term, we get the equation

$$
1=(2-\sqrt{3}) \sin ^{2} \beta+\frac{2+\sqrt{3}}{3} \cos ^{2} \beta
$$

Since $\cos ^{2} \beta=1-\sin ^{2} \beta$, this equation can be written as

$$
1=(2-\sqrt{3}) \sin ^{2} \beta+\frac{2+\sqrt{3}}{3}\left(1-\sin ^{2} \beta\right)
$$

From this, we find that $\sin ^{2} \beta=\frac{1}{4}$.

Since $\alpha$ and $\beta$ are acute angles of right triangles $M D K$ and $N C K$ (K is the point of tangency of the circle),

then $\beta=30^{\circ}$, and since $\sin \alpha=\frac{\sqrt{2-\sqrt{3}}}{2}$, then $\alpha=15^{\circ}$. Therefore, $\angle M P N=135^{\circ}$. Then

$$
\begin{gathered}
P N=\frac{M N \sin 15^{\circ}}{\sin 45^{\circ}}=\frac{\frac{18 \sqrt{2-\sqrt{3}}}{2}}{\frac{\sqrt{2}}{2}}=\frac{18 \sqrt{2-\sqrt{3}}}{\sqrt{2}} \\
S_{\triangle M N P}=\frac{1}{2} M N \cdot P N \sin 30^{\circ}=\frac{81 \sqrt{2-\sqrt{3}}}{\sqrt{2}}=\frac{81(\sqrt{3}-1)}{2}
\end{gathered}
$$

## Answer

$\frac{81(\sqrt{3}-1)}{2}$