## Three Tangent Circles

A circle is inscribed in an equilateral triangle with a side of length $2$. Three circles are drawn externally tangent to this circle and internally tangent to two sides of the triangle. Three more circles are drawn externally tangent to these circles and internally tangent to two sides of the triangle, as shown. If this process continued forever, what would be the sum of the areas of all the circles?

Source: NCTM Mathematics Teacher, February 2006

SOLUTION

If the equilateral triangle has a side of length $2$, the altitude $AE=\sqrt{3}$.
Stage 0: circle $D$
radius = $\sqrt{3}/3$ (in an equilateral triangle the incenter $D$ is also the centroid)
area = $\pi(\sqrt{3}/3)^2=\pi/3$
Stage 1: three circle $C$ (only one is drawn)
radius = $\sqrt{3}/3\div 3=\sqrt{3}/9$ ($C$ is the centroid of the smaller equilateral triangle)
area of $3$ circles $C$ = $3\left [\pi(\sqrt{3}/9)^2\right ]=\pi/9$
Stage 2: three circles $B$ (only one is drawn)
radius = $\sqrt{3}/9\div 3=\sqrt{3}/27$
area of $3$ circles $B$ = $3\left [\pi(\sqrt{3}/27)^2\right ]=\pi/81$
If we do the calculations for Stage 3 (circles too small to draw), we observe that the radius equals $1/3$ of the Stage 2 radius and that the area of the circles equals $1/9$ of the Stage 2 area.
Sum of the circles = $\dfrac{\pi}{3}+\dfrac{\pi}{9}+\dfrac{\pi}{81}+\dfrac{\pi}{729}+\cdots$
$\dfrac{\pi}{9}+\dfrac{\pi}{81}+\dfrac{\pi}{729}+\cdots$ is a geometric series with first term equal $\pi/9$ and ratio equal $1/9$
$\dfrac{\pi}{3}+\dfrac{\pi}{9}+\dfrac{\pi}{81}+\dfrac{\pi}{729}+\cdots=\dfrac{\pi}{3}+\dfrac{\pi/9}{1-1/9}=\dfrac{11\pi}{24}$

Answer: $11\pi/24$