The
circumcircle of the hexagon PQRSTU has a radius of length 1. It is known that PQ = RS = TU = 1.
Let J, K and L correspond respectively to the mid points of the sides QR, ST and UP.
Is it always the case that the triangle JKL is equilateral?
Let cis(theta) denote cos(theta) + i sin(theta).
WOLOG let the vertices be
P = cis(alpha+30)
Q = cis(alpha+90)
R = cis([beta+90])
S = cis([beta+30])
T = cis(30)
U = cis(30)
The midpoints then become
J = (Q + R)/2
K = (S + T)/2
L = (U + P)/2
It is easy, but tedious to show that
(K  J) cis(60) = L  J
Therefore, JKL is an equilateral triangle.

Posted by Bractals
on 20070802 03:48:07 