Label
some points as in the diagram on the left. Let r be the radius of
the smallest circle. Then |AE| = |CD| = r,
and |EB| = 1/2 - r. Since
|BG| = 1/2, Pythagoras applied to the triangle ECB
gives Correct solutions were received from Steve King, Bill Webb, Juan Carlos Marivela, Philippe Fondanaiche, Alejandro Vellano, Dennis Muniz, Nancy Schwarzkopf.
Label
some points as in the diagram on the left. Let r be the radius of
the smallest circle. Then |AE| = |CD| = r,
and |EB| = 1/2 - r. Since
|BG| = 1/2, Pythagoras applied to the triangle ECB
gives
|EC| = Ö((1/2 + r)2 - (1/2 - r)2) = Ö(2r).
Applying Pythagoras to the triangle ECA gives |AC| = 1 - r = 2r + r2 and r = 1/4. The coordinates for the center of the circle are (1/4, Ö2/2).
Several people wrote to ask about the location of the next smaller circle which fits into the upper left hand corner, and the next, and the next, etc. Can you find this sequence?
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