[ Equilateral (regular) triangle ]

The side of an equilateral triangle is $a$. Find the radius of the excircle.

#
Prove that the radius of the desired circle is equal to the height of the given triangle.

## Solution

Let $O$ be the center of the circle that touches side $AB$ of the equilateral triangle $ABC$ at point $M$ (the midpoint of $AB$) and the extensions of sides $CA$ and $CB$ at points $L$ and $K$ respectively. If $R$ is the desired radius, then

$$
CO = 2OK \left(\angle OCK = 30^{\circ}\right)
$$

Therefore, $CM + R = 2R$. Hence,

$$
R = CM = \frac{a \sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-14.jpg?height=458&width=504&top_left_y=1649&top_left_x=777)

Answer

$\frac{a \sqrt{3}}{2}$