[ Area of a circle, sector, and segment ] [ Midline of a triangle ]

The area of triangle $ABC$ is $1, AC=2 BC$, point $K$ is the midpoint of side $AC$. A circle centered at point $K$ intersects side $AB$ at points $M$ and $N$, such that $AM=MN=NB$. Find the area of the part of triangle $ABC$ enclosed within the circle.
Prove that triangle $ABC$ is a right triangle.

## Solution

Let $BC=a, R$ be the radius of the circle. Then $AC=2a$. Let $F$ be the projection of the center $K$ of the circle onto side $AB$. Then $F$ is the midpoint of $MN$, and thus the midpoint of $AB$. Therefore, $KF$ is the midline of triangle $ABC$. Consequently, triangle $ABC$ is a right triangle ( $\angle B=90^{\circ}$ ).

Since $AC=2BC$, then

\[
\begin{gathered}
\angle A=30^{\circ}, AB=a \sqrt{3}, MN=\frac{a \sqrt{3}}{3}, FM=\frac{1}{2} MN=\frac{a \sqrt{3}}{6} \\
KF=\frac{1}{2} BC=\frac{a}{2}, \angle FKM=30^{\circ}, R=KM=2 FM=\frac{a \sqrt{3}}{3} \\
S_{\triangle \mathrm{NKM}}=\frac{R^{2} \sqrt{3}}{4}=\frac{a^{2} \sqrt{3}}{12}
\end{gathered}
\]

Since $R\frac{a \sqrt{3}}{3}$ ). Therefore, the desired area is the difference between the area of the semicircle and the segment cut off from the semicircle by the chord $MN$, i.e.,

\[
\frac{\pi R^{2}}{2}-\left(\frac{\pi R^{2}}{6}-\frac{R^{2} \sqrt{3}}{4}\right)=R^{2}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right)=\frac{1}{3} a^{2}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right)
\]

Since $S_{\triangle \mathrm{ABC}}=\frac{a^{2} \sqrt{3}}{2}=1$, then $a^{2}=\frac{2}{\sqrt{3}}$. Therefore, the desired area is

\[
\left(\frac{2}{\sqrt{3}}\right)^{2}\left(\frac{\pi}{3}+\frac{\sqrt{3}}{4}\right)=\frac{2 \pi \sqrt{3}}{27}+\frac{1}{6}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-19.jpg?height=386&width=849&top_left_y=1459&top_left_x=610)

## Answer

$\frac{2 \pi \sqrt{3}}{27}+\frac{1}{6}$