| 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 to two sides of the triangle. If this process is continued indefinitely, what is the sum of the circumferences of all the circles? |  |
Solution 1 by Ben Miller, Benzie Central
High School |
 |
Solution 2 by Rob Moran, Ann Arbor Huron H.S.
The big circle has a radius of 1 and circumference of 
;
the radii of the remaining
circles are 1/3, 1/9, 1/27 . . . So, the sum of all the circumferences will
be:
 |
Also solved by: Nathaniel Sotuyo (Charlevoix H.S.); Andrew Jeanguenat
(De La Salle Collegiate H.S.); Pranav Moudgil (Detroit Country Day H.S.); Allison
Bryan, Kyle DeBoer, Max Dixon, Lindsey Westerhof (Forest Hills Eastern H.S.);
David Danks, Chantel Dowden, Stefanie Ells, Katelyn Evans, Allison Hogan, Carol
Jiang, Steve Lenio, Visakh Ponnuru, Andrew Post, Maja Redzic, Tom Wakefield,
Danny Walker (Forest Hills Northern H.S.); Meghan Borzenski, Robert Harrison
(Lansing Catholic Central H.S.); Cathy Krueger (Grosse Pointe South H.S.); Matt
DeYoung (Mattawan H.S.); Dominick Good, Kim Lebioda (Plymouth H.S.); Matthew
Brownell, Armila Francis, Kristen Kitti, Melissa Rich, Sara Ross, Matt Saagman
(Plymouth Christian Academy); and Craig Rogers (St. Louis H.S.).
right triangle,
I found that DC = 2. Since the large circle has a radius of 1
inch, it “cuts” the hypotenuse DC in half leaving
1 inch in the hypotenuse. The second half of the hypotenuse is the sum
of all the diameters of the remaining circles. I then used the circumference
formula to compute the sum of the circumferences of all the circles in
each vertex of the triangle. In each vertex the sum of the circumferences
is 
. Adding all of the circumferences,
we get 
inches.