Problem A circle of radius 1 inch is inscribed in an equilateral triangle. A smaller circle is inscribed at each vertex, tangent to the circle and two sides of the triangle. The process is continued with progressively smaller circles. What is the sum of the circumference of all the circles.

Solution

$y = 2 ~ \text{inches}$

$\Sigma d = y - 1$

$\Sigma d = 1 ~ \text{inch}$

Sum of circumference $C = 3 \Sigma (\pi d) + \pi (2)$

$C = 3\pi \, \Sigma d + 2\pi$

$C = 3\pi \, (1) + 2\pi$

$C = 5\pi ~ \text{inches}$ ← answer

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