$\left[\begin{array}{l}{[\text { Equilateral (regular) triangle }]} \\ {[\quad \text { Area of a circle, sector, and segment }}\end{array}\right]$

In an equilateral triangle $ABC$, a circle is drawn passing through the center of the triangle and touching side $BC$ at its midpoint $D$. A line is drawn from point $A$, tangent to the circle at point $E$, and $\angle BAE < 30^{\circ}$. Find the area of triangle $ABE$ if the area of triangle $ABC$ is $\frac{10}{4-\sqrt{2}}$.
Find $\sin \angle D A E$ from the right triangle $E A Q(Q$ - the center of the given circle).

## Solution

Let $O$ be the center of triangle $A B C, Q$ be the center of the given circle. Denote the side of triangle $A B C$ by $a$. Then

$$
O D=\frac{1}{3} A D=\frac{a \sqrt{3}}{6}, O Q=Q D=\frac{a \sqrt{3}}{12}
$$

$$
A Q=A D-Q D=\frac{a \sqrt{3}}{2}-\frac{a \sqrt{3}}{12}=\frac{5 a \sqrt{3}}{12}
$$

From the right triangle $A E Q$, we find that

$$
\begin{gathered}
\sin \angle E A Q=\frac{E Q}{A Q}=\frac{1}{5}, \cos \angle E A Q=\frac{2 \sqrt{6}}{5} \\
A E=E Q \tan \angle E A Q=\frac{a \sqrt{2}}{2}
\end{gathered}
$$

Therefore,

$$
\begin{aligned}
& S_{\triangle \mathrm{ABE}}=\frac{1}{2} A E \cdot A B \sin \left(30^{\circ}-\angle E A Q\right)= \\
= & \frac{a^{2}}{\sqrt{2}} \cdot \frac{\frac{\sqrt{6}}{5}-\frac{\sqrt{3}}{15}}{4}=\frac{a^{2} \sqrt{6}(2 \sqrt{2}-1)}{40}
\end{aligned}
$$

According to the problem statement

$$
S_{\triangle \mathrm{ABC}}=\frac{10}{4-\sqrt{2}}=\frac{a^{2} \sqrt{3}}{4}
$$

Solving for $a^{2}$ from this equation and substituting it into the previous one, we get: $S_{\Delta \mathrm{ABE}}=1$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-07.jpg?height=392&width=606&top_left_y=1353&top_left_x=726)

## Answer

1. 

## Problem