Newsgroups: rec.puzzles
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: 07 Mar 1995 21:54:13 GMT
Subject: Re: Triangle Puzzle (spoiler)

gsg5501@Msu.oscs.montana.edu writes:

> The problem, stated Feb. 20 and then again Feb. 28, went something
> like this: take an equilateral triangle, trisect the angles.  What's
> the ratio of the area of the triangle to the area of the irregular
> hexagon formed in the middle?

Well, _that_ problem hasn't been stated before here.  All the previous
problems have involved trisecting a _side_ of the triangle with
lines from the opposite vertex.  I posted the answer to that this
morning.

Oddly enough, it is the answer you give.

But the problem you pose, of trisecting the _angles_ to form a
hexagon, is also interesting, though the answer is not rational.
Using the method of my previous message, it's easy to see that if the
ratio of the central segment of each side to the side is X, then the
area of the hexagon is (72 / (9 - X^2)) - 8 times the area of the
triangle.  For an equilateral triangle whose angles are trisected,
X=Sqrt[3] Tan[Pi/18], so the hexagon is

  (24 / (3 - Tan^2[Pi/18])) - 8 ~ 0.0837781

times the area of the triangle.

ObPuzzle:  Express that in radicals (i.e., without trig functions).

Dan
Hoey@AIC.NRL.Navy.Mil