Let ABC be the triangle and A'B'C' its medial triangle.
Let the sides of triangle ABC be replaced with point masses (proportional to their length) located at the midpoints A', B', and C'.
If G is the center of gravity, then
(ka)A'A' + (kb)A'B' + (kc)A'C'
A'G = --------------------------------
ka + kb + kc
1
= ----------- [(b)A'B' + (c)A'C']
a + b + c
1 A'B' A'C'
= ----------- [(b|A'B'|)-------- + (c|A'C'|)--------]
a + b + c |A'B'| |A'C'|
1 A'B' A'C'
= ----------- [(bc/2)-------- + (cb/2)--------]
a + b + c |A'B'| |A'C'|
bc A'B' A'C'
= -------------- [-------- + --------]
2(a + b + c) |A'B'| |A'C'|
Clearly A'G bisects angle C'A'B'. A similar argument would show that B'G and C'G bisect angles A'B'C' and B'C'A' respectively.
Therefore, the center of gravity of the perimeter of a triangle is the incenter of its medial triangle.
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