In an isosceles right triangle $ABC$, from point $C$, which is the vertex of the right angle, a height $CC_1$ is dropped to the hypotenuse. From point $C_1$, two perpendicular lines are drawn, intersecting sides $BC$ and $AC$ at points $A_1$ and $B_1$ respectively. It is known that $\angle C_1 A_1 B = 60^\circ$, and the hypotenuse $AB = 2 \sqrt{5 + 2 \sqrt{6}}$. Find the length of the segment $A_1 B_1$. Provide its approximate value to two decimal places.
Prove that triangle $A_{1} B_{1} C_{1}$ is isosceles and find $A_{1} C_{1}$ in triangle $A_{1} C_{1} B$ using the Law of Sines.

## Solution

Notice that

$$
\begin{gathered}
\angle A_{1} C_{1} B=180^{\circ}-\angle C_{1} A_{1} B-\angle C_{1} B C=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}, \angle C C_{1} A_{1}=15^{\circ} \\
\angle A C_{1} B_{1}=180^{\circ}-\angle A_{1} C_{1} B_{1}-\angle A_{1} C_{1} B=180^{\circ}-90^{\circ}-75^{\circ}=15^{\circ}
\end{gathered}
$$

Triangles $A C_{1} B_{1}$ and $C C_{1} A_{1}$ are equal by side and two adjacent angles. Therefore, $A_{1} B_{1} C_{1}$ is an isosceles right triangle. Consequently, $A_{1} B_{1}=A_{1} C_{1} \sqrt{2} \cdot A_{1} C_{1}$. We will find $A_{1} C_{1}$ from triangle $C_{1} A_{1} B$ using the Law of Sines:

$$
\begin{aligned}
& A_{1} C_{1}=\frac{B C_{1} \sin 45^{\circ}}{\sin 60^{\circ}}=\sqrt{5+2 \sqrt{6}} \cdot \frac{\sqrt{2}}{\sqrt{3}}=\sqrt{2} \sqrt{\frac{5+2 \sqrt{6}}{3}}= \\
& =\sqrt{2} \sqrt{\frac{15+6 \sqrt{6}}{9}}=\sqrt{2} \sqrt{\frac{\sqrt{6}+2 \cdot 3 \sqrt{6}+9}{9}}=\frac{\sqrt{2}}{3}(\sqrt{6}+3)
\end{aligned}
$$

Therefore,

$$
A_{1} B_{1}=A_{1} C_{1} \sqrt{2}=\frac{2(\sqrt{6}+3)}{3}
$$

Answer

$2+\frac{2 \sqrt{6}}{3} ; 3.63$