8,9}

Point $M$ lies inside an equilateral triangle $ABC$. Calculate the area of this triangle, given that $AM = BM = 2$, and $CM = 1$.
Apply the Law of Cosines to triangle $B C M$.

## Solution

Since $A M=C M$, point $M$ lies on the perpendicular bisector of side $A B$. Therefore, $C M$ is the angle bisector of angle $B C A$. Let $A B=B C=A C=x$. By the Law of Cosines in triangle $B C M$, we find that

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

From this, we find that

$$
x=\frac{\sqrt{3}+\sqrt{15}}{2}
$$

Therefore,

$$
s_{\Delta \mathrm{ABC}}=\frac{1}{2} x^{2} \sin 60^{\circ}=\frac{9 \sqrt{3}+3 \sqrt{15}}{8} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-40.jpg?height=441&width=598&top_left_y=2112&top_left_x=730)

## Answer

$\frac{9 \sqrt{3}+3 \sqrt{15}}{8}$

## Problem