Compose a system of three equations with respect to the sides of a triangle. Use the equality $\sqrt{6}-\sqrt{2}=4 \cos 75^{\circ}$ when solving it.

## Solution

Let $C D$ be the height of the triangle. Denote $A B=c, B C=a, A C=b$. Then $S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot C D=\frac{1}{2} c$. On the other hand, $S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A C \cdot \sin \angle A C B=\frac{1}{2} a b \sin 75^{\circ}$, so $c=a b \sin 75^{\circ}$. By the cosine theorem,

$$
A B^{2}=B C^{2}+A C^{2}-2 B C \cdot A C \cdot \cos \angle A C B, \quad
$$

or $c^{2}=a^{2}+b^{2}-2 a b \cos 75^{\circ}$. Finally, by the problem's condition, $a+b+c=4+\sqrt{6}-\sqrt{2}$. Thus, we have the system

$$
\left\{\begin{array}{l}
c=a b \sin 75^{\circ} \\
A^{2}=a^{2}+b^{2}-2 a b \cos 75^{\circ} \\
a+b+c=4+\sqrt{6}-\sqrt{2}
\end{array}\right.
$$

Noting that $\sqrt{6}-\sqrt{2}=4 \cos 75^{\circ}$, we rewrite the system as

$$
\left\{\begin{array}{l}
a b=\frac{c}{\sin 75^{\circ}} \\
A^{2}=(a+b)^{2}-2 a b\left(1+\cos 75^{\circ}\right) \\
a+b=4+4 \cos 75^{\circ}-A
\end{array}\right.
$$

Substituting $a b$ from the first equation and $a+b$ from the third into the second, we get an equation in terms of $c$:

$$
c^{2}=\left(4\left(1+\cos 75^{\circ}\right)-c\right)^{2}-\frac{2 c\left(1+\cos 75^{\circ}\right)}{\sin 75^{\circ}}
$$

from which we find that

$$
c=\frac{8\left(1+\cos 75^{\circ}\right) \sin 75^{\circ}}{4 \sin 75^{\circ}+1}=\frac{8 \sin 75^{\circ}+4 \sin 150^{\circ}}{4 \sin 75^{\circ}+1}=\frac{8 \sin 75^{\circ}+2}{4 \sin 75^{\circ}+1}=2 .
$$

Let $R$ be the radius of the circumscribed circle of the given triangle. Then

$$
R=\frac{A B}{2 \sin \angle A C B}=\frac{c}{2 \sin 75^{\circ}}=2 \cdot \frac{\sqrt{6}+\sqrt{2}}{4}=\sqrt{6}-\sqrt{2} \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_e3b1ddf947c7dd2094eeg-48.jpg?height=488&width=509&top_left_y=179&top_left_x=777)

## Answer

$\$|\operatorname{sqrt}\{6\}-| \operatorname{sqrt}\{2\} . \$-->\sqrt{6}-\sqrt{2}$.
[Law of Cosines] [Law of Sines]

In an acute triangle $ABC$, the angle $\angle ACB = 75^{\circ}$, and the altitude dropped from this vertex is 1.

Find the radius of the circumscribed circle if it is known that the perimeter of triangle $ABC$ is $4 + \sqrt{6} - \sqrt{2}$.