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Six Incircles in an Equilateral Triangle

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From an interior point P of an equilateral triangle ABC draw lines perpendicular to the sides. Let A', B', and C' be the points on the sides opposite A, B and C. Inscribe circles in the six subtriangles. Let (XYZ) denote the diameter of the incircle in triangle XYZ.

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Then the sum of the diameters of the red circles equals the sum of the diameters of the blue circles:

(APC') + (BPA') + (CPB') = (C' PB) + (A' PC) + (B' PA)

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Contributed by: Jay Warendorff (March 2011)
Open content licensed under CC BY-NC-SA

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