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An equilateral triangle $ABC$, with side length $a$, has a circle inscribed in it, touching side $AB$ at point $D$. A second circle, located inside triangle $ABC$, is externally tangent to the first (inscribed) circle at point $K$, touches side $AB$ at point $M$, and side $BC$. Find the area of the figure $DKM$, bounded by the smaller arc $DK$, the smaller arc $KM$, and the segment $MD$.

#
Triangle $M K D-$ is a right triangle.

## Solution

Let $R$ and $r$ be the radii of the given circles $(R>r)$, $O$ and $Q$ be their centers, and $P$ be the point of intersection of the common tangent passing through point $K$ with side $A B$. Since $P K = P D = P M$, triangle $M K D$ is a right triangle.

The desired area is the difference between the area of triangle $M K D$ and the areas of two segments. The area of the first segment is the difference between the area of sector $D O K$ and triangle $D O K$, i.e.,

$$
\frac{\pi R^{2}}{6} - \frac{R^{2} \sqrt{3}}{4}
$$

and the second segment is the difference between the area of sector $M Q K$ and triangle $M Q K$, i.e.,

$$
\frac{\pi r^{2}}{3} - \frac{r^{2} \sqrt{3}}{4}
$$

Since $B K = 3 R - 2 R = R$, then $r = \frac{1}{3} R$.

From triangles $M Q K$ and $D O K$, we find that

$$
M K = r \sqrt{3} = \frac{R \sqrt{3}}{3}, \quad D K = R \left( R = \frac{a \sqrt{3}}{6} \right)
$$

$$
\begin{gathered}
\frac{1}{2} M K \cdot D K - \left( \frac{\pi R^{2}}{6} - \frac{R^{2} \sqrt{3}}{4} \right) - \left( \frac{\pi r^{2}}{3} - \frac{r^{2} \sqrt{3}}{4} \right) = \\
= \frac{R^{2} \sqrt{3}}{6} - \frac{R^{2}}{2} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) - \frac{R^{2}}{9} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) = \\
= \frac{R^{2} (48 \sqrt{3} - 22 \pi)}{108} = \frac{R^{2} (24 \sqrt{3} - 11 \pi)}{54} = \frac{a^{2} (24 \sqrt{3} - 11 \pi)}{648}.
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-28.jpg?height=388&width=548&top_left_y=595&top_left_x=755)

Answer

$\frac{a^{2} (24 \sqrt{3} - 11 \pi)}{648}$