You are provided with a circle, a triangle, and a square which each have the same length perimeter. If the circle is rotated, without slipping, around each of the other two, in which case will it rotate through the greatest number of degrees?

(In reply to solution by Larry)

I agree with Larry's conclusion, but his analysis assumes that the triangle is equilaterial.  This is not a condition of the problem.

If the triangle has angles of a, b, and c, then the extra turns at the three vertices are 180 -a, 180-b, and 180 - c respectively.  This sums to 540 - a -b -c = 360 degrees of extra turns.   When added to the normal  360 when it is not at a vertex, this yields 720 degrees, which is consistent with the special case solved by Larry. 

So, no matter what kind of triangle it is, the square yields the same 720 degrees. 

I believe that this will be true if the circle is rotated around any convex shape with the same perimeter, even if the shape is irregular and even if it is not all straight lines.  In particular, 720 degrees of turn results if the circle rotates around another circle of the same perimeter!  Try it with two quarters, if you don't believe me.

Posted by Steve Herman on 2005-07-03 01:46:27