Solution to the irregular hexagon problem

Three equilateral triangles share a common point which is the center of a circle on which the three sides opposite the point become three sides of a cyclic hexagon [a hexagon inscribed in the circle] ABCDEF, with, say, AB, CD, and EF as the sides of the equilateral triangles. Consider the triangle formed from the midpoints of sides BC, DE, and FA [G, H, and J, respectively]. Prove that triangle GHJ is equilateral.

Solution:

Let O be the origin of the circle.  Let $$\alpha =\angle BOC$$, $$\beta =\angle DOE$$, and $$\gamma =\angle FOA$$.  Note that $$\alpha +\beta +\gamma =\pi$$.

The length of a side of $$\Delta GHJ$$, say, $$\overline{GH}$$, is determined by the length of the bisectors, in this case $$\left|\overline{OG}\right|$$ and $$\left|\overline{OH}\right|$$; and the angle $$\angle GOH$$ between the bisectors, as follows:

$$\left|\overline{GH}\right|^2$$ = $$\left|\overline{OG}\right|^2$$ + $$\left|\overline{OH}\right|^2$$ – $$2 \left|\overline{OG}\right|$$ $$\left|\overline{OH}\right|$$ $$\cos \angle GOH$$

where $$\left|\overline{OG}\right|$$ = $$\cos \frac{\alpha}{2}$$ , $$\left|\overline{OH}\right|$$= $$\cos \frac{\beta}{2}$$, and $$\angle GOH=\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3}$$.

Then in terms of the parameters $$\alpha$$ and $$\beta$$, the length of the side $$\overline{GH}$$ is

$$\left|\overline{GH}\right|^2$$ = $$\cos^2 \frac{\alpha}{2}$$ + $$\cos^2 \frac{\beta}{2}$$ – $$2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})$$

The same reasoning applies to the other sides, e.g.,

$$\left|\overline{JG}\right|^2$$ = $$\cos^2 \frac{\alpha}{2}$$ + $$\cos^2 \frac{\gamma}{2}$$ – $$2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})$$

To prove that $$\Delta GHJ$$ is equilateral, we must show that the lengths of its sides are equal.  The problem then reduces to showing that

$$\cos^2 \frac{\alpha}{2}$$ + $$\cos^2 \frac{\beta}{2}$$ – $$2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \cos(\frac{\alpha }{2}+\frac{\beta }{2}+\frac{\pi }{3})$$ = $$\cos^2 \frac{\alpha}{2}$$ + $$\cos^2 \frac{\gamma}{2}$$ – $$2 \cos \frac{\alpha}{2} \cos \frac{\gamma}{2} \cos(\frac{\alpha }{2}+\frac{\gamma}{2}+\frac{\pi }{3})$$

where, again,  $$\alpha +\beta +\gamma =\pi$$.

Suffice it to say that this is an exercise in manipulating trig identities beyond the patience of most sane people, even those that consider math puzzles for recreation.  I will leave most of the details to the reader, assuring said reader that the identity to be proven does in fact hold true.  Your trusty problem poser has performed said manipulations himself, and can offer the following road map.

First, observe that $$2 \cos(z + \frac{\pi}{3})$$ = $$\cos z$$ – $$\sqrt{3}$$ $$\sin z$$.  Note that the term associated with the $$\sqrt{3}$$ will be treated separately.

Also note that $$\cos \frac{\gamma }{2}$$ = $$\sin \frac{\alpha +\beta }{2}$$ and $$\sin \frac{\gamma }{2}$$=$$\cos \frac{\alpha +\beta }{2}$$.

Keep organized, and enjoy!