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Six Incircles in an Equilateral Triangle

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d(APC')
≈
0.58
d(BPA')
≈
0.27
d(CPB')
≈
0.35
d(C'PB)
≈
0.51
d(A'PC)
≈
0.25
d(B'PA)
≈
0.44
d(APC')
+
d(BPA')
+
d(CPB')
≈
1.20
d(C'PB)
+
d(A'PC)
+
d(B'PA)
≈
1.20
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
d
(XYZ) denote the diameter of the incircle in triangle XYZ.
Then the sum of the diameters of the red circles equals the sum of the diameters of the blue circles:
d
(APC') +
d
(BPA') +
d
(CPB') =
d
(C' PB) +
d
(A' PC) +
d
(B' PA)

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