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?
Three Tangent Circles Cabri2D v2c.html
Source: NCTM Mathematics Teacher, February 2006

SOLUTION
image
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

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About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , , , . Bookmark the permalink.

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