Prove that the given triangle is a right triangle.

## Solution

Multiply both sides of the given equation by 2, replace $\angle C=180^{\circ}-\angle A-\angle B$, and use trigonometric formulas:

$$
\begin{gathered}
2 \cos ^{2} A+2 \cos ^{2} B+2 \cos ^{2}(A+B)=2 \Leftrightarrow \\
\Leftrightarrow 1+\cos 2 A+1+\cos 2 B+2 \cos ^{2}(A+B)=2 \Leftrightarrow \\
\Leftrightarrow \cos 2 A+\cos 2 B+2 \cdot \cos ^{2}(A+B)=0 \Leftrightarrow \\
\Leftrightarrow 2 \cos (A+B) \cos (A-B)+2 \cos ^{2}(A+B)=0 \Leftrightarrow \\
\Leftrightarrow 2 \cos (A+B)(\cos (A-B)+\cos (A+B))=0 \Leftrightarrow \\
\Leftrightarrow 4 \cos (A+B) \cos A \cos B=0 .
\end{gathered}
$$

From the last equation, it follows that one of the angles of triangle $ABC$ is $90^{\circ}$, i.e., triangle $ABC$ is a right triangle.

Assume that $\angle C=90^{\circ}$. Let $R=3 \sqrt{2}$ be the radius of the circumscribed circle of triangle $ABC$, $r=\sqrt{3}$ be the radius of the inscribed circle, $O$ be its center, and $K, M$, and $N$ be the points of tangency with sides $AB, BC$, and $AC$ respectively. Then

$$
\begin{gathered}
AB=2 R=6 \sqrt{2}, S_{\Delta \mathrm{ABC}}=2 S_{\Delta \mathrm{AOK}}+2 S_{\Delta \mathrm{BOK}}+S_{\mathrm{OMCN}}= \\
=2\left(S_{\Delta \mathrm{AOK}}+S_{\Delta \mathrm{BOK}}\right)+S_{\mathrm{OMCN}}=2 S_{\Delta \mathrm{AOB}}+S_{\mathrm{OMCN}}=AB \cdot O K+O M^{2}= \\
=AB \cdot r+r^{2}=2 R r+r^{2}=6 \sqrt{2} \cdot \sqrt{3}+3=6 \sqrt{6}+3
\end{gathered}
$$

## Answer

$6 \sqrt{6}+3$.
8,"

The angles of triangle $ABC$ satisfy the equation

$$
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=1
$$

Find the area of this triangle if the radii of the inscribed and circumscribed circles are $\sqrt{3}$ and $3 \sqrt{2}$, respectively.