Sine Theorem  [ Two tangents drawn from one point ]

One of the angles of a triangle is $\frac{3 \pi}{4}$, the radius of the inscribed circle is 4, and the perimeter of the triangle is $16(6+\sqrt{2})$. Find the radius of the circle circumscribed around this triangle.
Let the inscribed circle with center $O$ touch the sides $B C, A C$, and $A B$ of triangle $A B C$ at points $A 1, B 1$, and $C 1$ respectively, with $\angle B A C=\frac{3 \pi}{4}$. Denote $B A 1=B C 1=x, C A 1=C B 1=y$. Then $B C=x+y$. From the right triangle $A O B 1$, we find that

$$
A B 1=\frac{O B_{1}}{\lg \frac{1}{2} \angle B A C}=\frac{4}{\lg \frac{3 \pi}{8}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-15.jpg?height=111&width=1992&top_left_y=1814&top_left_x=-1)
$\frac{2 t}{1-t^{2}}$. Since $\frac{3 \pi}{5}<\frac{\pi}{2}$, the condition of the problem is satisfied only by the positive root of this equation: $t=1+\sqrt{2}$. Then

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-15.jpg?height=111&width=618&top_left_y=2008&top_left_x=720)

According to the problem, $2 x+2 y+8(\sqrt{2}-1)=16(6+\sqrt{2})$, from which we find that $x+y=4 \sqrt{2}+52$. Let $R$ be the radius of the circumcircle of triangle $A B C$. By the sine rule,

$$
R=\frac{B C}{2 \sin \angle B A C}=\frac{x+y}{2 \sin \frac{\operatorname{in}}{1}}=\frac{4 \sqrt{2}+52}{\sqrt{2}}=4+26 \sqrt{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_08de36b0d140fd8f2c0dg-16.jpg?height=543&width=1886&top_left_y=0&top_left_x=126)

## Answer

$26 \sqrt{2}+4$.