Six Incircles in an Equilateral Triangle

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)