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 2007-08-02 03:48:07