Solution to Problem 272

Correct solutions came from John Snyder, USA; Brian Bradie, USA; Burkart Venzke, Germany; Farid Lian, Colombia; Eric Pittelkau, USA; Alejandro Vellano, Argentina; Bill Webb, USA; Jean Moreau de Saint-Martin, France; Ron Welch, USA; Kai Grinde Myrann, Norway; Lou Cairoli, USA; Aaron Lo Cascio, USA; Cee Ann Franklin, USA; Juan Balonga, Spain; Phillip Adkins, USA; Allen Druze, USA.

The circles in the figure-8 have radius sqrt(3)/10.

Let the radius be r. Notice that the center of the top circle, the point where the top circle meets one of the sides of the triangle, and the top vertex of the triangle form a 30-60-90 triangle with the side opposite the 30-degree angle having length r. The hypotenuse of this small triangle thus has length 2r. The altitude of the original triangle includes this hypotenuse, a radius of the upper circle and a diameter of the lower circle and thus measures 5r. But this altitude measures sqrt(3)/2.

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