# 2012 UNCO Math Contest II Problems/Problem 7

## Problem

A circle of radius $1$ is externally tangent to a circle of radius $3$ and both circles are tangent to a line. Find the area of the shaded region that lies between the two circles and the line. $[asy] draw((-1,-1)--(7,-1),black); filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); filldraw(circle((0,0),1),white); filldraw(circle((2*sqrt(3),2),3),white); [/asy]$

## Solution $[asy] draw((-1,-1)--(7,-1),black); filldraw((0,-1)--(0,0)--(2*sqrt(3),2)--(2*sqrt(3),-1)--cycle,grey); filldraw(circle((0,0),1),white); filldraw(circle((2*sqrt(3),2),3),white); [/asy]$

If we draw in the second tangent between these circles, we get a point where the tangents intersect. Drawing a line through the centers of the circle and the point where the tangents meet, we get two similar triangles in ratio $1:3$. Let the hypotenuse of the smaller triangle be $x$. Since the distance between the centers of the circles is $1+3=4$, we can write the ratio $\frac{x}{x+4} = \frac{1}{3}$. Solving for $x$, we get $x=2$. Since the hypotenuse of the smaller triangle is $2$, and one of the legs is $1$, we see that it is a $30-60-90$ triangle. So the side lengths of the smaller triangle is $\sqrt{3},1,2$ and the side lengths of the larger triangle is $3\sqrt{3},3,6$. Finding the difference between the areas of both triangles, we get $4\sqrt3$ which is the area of the trapezoid. The trapezoid is the area of a $120^\circ$ sector of the smaller circle, a $60^\circ$ sector of the larger circle, and the shaded region. Subtracting the areas of both sectors from the area of the trapezoid, we get $4\sqrt{3} - \frac{120}{360} \cdot \pi - \frac{60}{360} \cdot 9\pi = 4\sqrt{3} - \frac{1}{3} \cdot \pi - \frac{1}{6} \cdot 9\pi = \boxed {4\sqrt{3} - \frac{11\pi}{6}}$

~Ultraman