Two identical equilateral triangles $A B C$ and $C D E$ with side length 1 are positioned such that they share only one common point $C$ and the angle $B C D$ is less than $60^{\circ}$. Point $K$ is the midpoint of $A C$, point $L$ is the midpoint of $C E$, and point $M$ is the midpoint of $B D$. The area of triangle $K L M$ is $\frac{\sqrt{3}}{5}$. Find $B D$.
Since $C M$ is the median of the isosceles triangle $B C D$, then $C M$ is the height of this triangle. The segment $C D$ is seen from points $M$ and $L$ at a right angle. Therefore, points $D, L, C$, and $M$ lie on a circle with diameter $C D$. Hence,

$$
\angle D M L = \angle D C L = 60^{\circ}.
$$

Similarly, $\angle B M K = 60^{\circ}$. Then $\angle K M L = 60^{\circ}$, and since $M K = M L$ (from the congruence of triangles $M K C$ and $M L C$), triangle $K L M$ is equilateral. Since its area is $\frac{\sqrt{3}}{5}$, the side length is $\frac{2}{\sqrt{5}}$.

Let $B C = 2 x$. Then, applying the cosine theorem to triangle $B M K$, we get the equation

$$
x^{2} + \frac{4}{5} - \frac{2 x}{\sqrt{5}} = \frac{3}{4}, \text{ or } x^{2} - \frac{2 x}{\sqrt{5}} + \frac{1}{20} = 0
$$

The problem condition is satisfied by only one root of this equation:

$$
x = \frac{2 - \sqrt{3}}{2 \sqrt{5}} < \frac{1}{2}
$$

Therefore, $B D = 2 x = \frac{2 - \sqrt{3}}{\sqrt{5}}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_0bf3a2f278e52693111ag-23.jpg?height=372&width=734&top_left_y=0&top_left_x=665)

## Answer

$\frac{2 - \sqrt{3}}{\sqrt{5}}$.

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