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Vertices $M$ and $N$ of the equilateral triangle $B M N$ lie on the sides $A D$ and $C D$ of the square $A B C D$ with side length $a$. Find $M N$.

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## Perin way.

Right triangles $ABM$ and $CBN$ are equal by leg and hypotenuse, therefore

$$
\angle ABM = \angle CBN = \frac{1}{2}(\angle ABC - \angle MBN) = \frac{1}{2}\left(90^{\circ} - 60^{\circ}\right) = 15^{\circ}
$$

Thus,

$$
\angle MND = 180^{\circ} - (\angle BNC + \angle BNM) = 180^{\circ} - 75^{\circ} - 60^{\circ} = 45^{\circ} = \angle ACD
$$

Therefore, $MN \parallel AC$. Let the diagonal $BD$ of the square intersect the segment $MN$ at point $K$. Then $BK$ and $DK$ are the altitudes of triangles $BMN$ and $DMN$. Denote $MN = BM = BN = x$. From the isosceles right triangle $MDN$ and the equilateral triangle $BMN$, we find that

$$
DK = \frac{1}{2} MN = \frac{x}{2}, \quad BK = \frac{x \sqrt{3}}{2}
$$

Since $DK + BK = BD = a \sqrt{2}$, we have the equation

$$
\frac{x}{2} + \frac{x \sqrt{3}}{2} = a \sqrt{2}
$$

from which we find that $x = (\sqrt{6} - \sqrt{2}) a$.

## Answer

$(\sqrt{6} - \sqrt{2}) a$.