Track / Stage 5 / 47 of 400 #647 of 2000

Problem 647

AIME late

Geometry Difficulty 5.2 Find the answer

[ Auxiliary circle ] $[$ Sine theorem $\quad]$

On the side $A B$ of triangle $A B C$ , an equilateral triangle is constructed outward. Find the distance between its center and vertex $C$ , if $A B=c$ and $\angle C=120^{\circ}$ .

Your answer

Official solution

## Solution

Point $C$ lies on the circumcircle of the constructed equilateral triangle ( $60^{\circ}+120^{\circ}=$ $\left.180^{\circ}\right)$ , so the desired distance is equal to the radius of this circle, i.e.,

$R=\frac{c}{2 \sin 60^{\circ}}=\frac{c}{\sqrt{3}}=\frac{c \sqrt{3}}{3}$

## Answer

$\frac{c \sqrt{3}}{3}$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.