7. In triangle $A B C$, the altitudes $A D, B E, C F$ are drawn. The length of side $A C$ is $\sqrt{6}+\sqrt{2}$. The distances from the center of the inscribed circle of triangle $D E F$ to points $A$ and $C$ are 2 and $2 \sqrt{2}$, respectively. Find the radius of the circumscribed circle around triangle $D E F$. (16 points)
Solution. $\quad A D, B E, C F$ are the altitudes of triangle $A B C, D A, E B, F C$ are the angle bisectors of angles $D, E, F$ of triangle $D E F, O$ is the point of intersection of the altitudes of triangle $A B C$, which is also the center of the inscribed circle of triangle $D E F$. Thus, $A O=2, C O=2 \sqrt{2}$. Let $O E=x, A E=y$. Then we arrive at the system $\left\{\begin{array}{c}x^{2}+y^{2}=4, \\ (\sqrt{2}+\sqrt{6}-y)^{2}+x^{2}=8 .\end{array}\right.$ Solving the system, we get $y=\sqrt{2}, x=\sqrt{2}$. Then $\quad \angle D A C=\angle B C A=45^{\circ}, \quad B C=2 \sqrt{3}$,

![](https://cdn.mathpix.com/cropped/2024_05_06_d66465e084edd0e7e96fg-11.jpg?height=414&width=620&top_left_y=564&top_left_x=1369)
$\angle F C A=\operatorname{arctg} \frac{1}{\sqrt{3}}=30^{\circ}, \quad \angle A B E=30^{\circ}, \quad A B=2 \sqrt{2}$. $\angle D F E=90^{\circ}, D E=A B \cos 45^{\circ}=2 . \quad R_{o n .}=D E / 2=1$.

## Answer: 1.