} \underline{\text { Folklore }}
$$

In a right triangle $ABC$, angle $A$ is $60^{\circ}$, and $M$ is the midpoint of the hypotenuse $AB$.

Find the angle $IMA$, where $I$ is the center of the inscribed circle of the given triangle.
Let $r$ be the radius of the circle inscribed in triangle $ABC$, and let $D$, $E$, and $F$ be the points of tangency with sides $AB$, $BC$, and $AC$ respectively (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_dc8ed1d6cdac2bf77445g-34.jpg?height=486&width=366&top_left_y=2201&top_left_x=846)

Obviously, quadrilateral $CEIF$ is a square, so $CE = r$.

By the given condition $AB = 2AC$, hence $AM = AC$. Therefore, $DM = AM - AD = AC - AF = r$.

Thus, triangle $IDM$ is a right isosceles triangle, so $\angle IMA = 45^{\circ}$.

Remark. The equality $DM = r$ can also be obtained from the formula for calculating the radius of the inscribed circle in a right triangle (see problem $\underline{52552}$) and the equality $BD = p - b$ (see problem $\underline{52554}$).

## Answer

$45^{\circ}$.