Newsgroups: rec.puzzles
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Date: 07 Mar 1995 13:48:29 GMT
Subject: Re: Triangle Puzzle

Lee Shere <sh...@leland.stanford.edu> writes again:

> Start with an equilateral triangle.
> Draw two lines from each vertice [vertex] to trisect
> the opposite side, forming an irregular hexagon
> in the center of the triangle.
> What is the ratio of the area of triangle to
> the area of the hexagon (in whole numbers)?

Stitch <SWIL...@nauvax.ucc.nau.edu> admired the puzzle so much I guess
you might like seeing a solution.  I wrote most of this when Lee Shere
posed the problem here in 1992.

In addition to drawing two lines from each vertex to trisect the
opposite side, let us draw an altitude from each vertex to bisect the
opposite side.  The bisectors cut the central hexagon into six
congruent triangular pieces.  I will calculate the area of one of
those pieces.

Take such a piece that shares a side of the hexagon formed by a line
from vertex A of the original triangle.  The piece is the difference
of two long thin triangles with vertex A, both of which have as sides:
  1. the line from A to a point trisecting the opposite side,
  2. the line from A to the midpoint of the opposite side, and
  3. the altitude from a side adjacent to A.
By the symmetry of the original triangle, there are just two kinds of
such long triangles, and the difference of their areas will be the
area of a sixth of the hexagon.

Consider the ratio into which the triangle's altitude is cut by a line
from a vertex to the opposite side

         +B
        /:\
       / : \
      /  : ,\E
     /  D;'  \
    /  ,':    \
   / ,'  :     \
  /,'    :H     \
A+'------+-------+C

The line AE has a vertical rise 1-BE/BC as far as that of AB, and
covers a horizontal distance 1+BE/BC times as far as AB, so
DH/BH=(1-BE/BC)/(1+BE/BC), and the area of triangle ADH is
(1-BE/BC)/(1+BE/BC) times the area of triangle ABH, which is half of
ABC.  So as BE/BC assumes the values 1/3, 1/2, and 2/3, the area of
ADH assumes the values 1/4, 1/6, and 1/10 the area of ABC.  The two
interior triangles with bases on BH then have areas 1/4-1/6=1/12 and
1/6-1/10=1/15 of the area of ABC.  The difference of the two triangles
is then 1/60 of ABC, so the ratio of the hexagon to ABC is 1/10, QEI.*

Here are a few interesting features of this puzzle.

1. The triangle need not be equilateral: the shape of the triangle
   does not affect the outcome.  The equilateral case is just handier
   to solve, because of its symmetry.

2. Suppose we divide each sides of the triangle into 2n+1 parts: how
   big is the central hexagon then?  Prove that it's a unit fraction
   of the triangle.

ObPuzzle: What fractions of the original triangle can appear as
regions in a triangle cut by lines from vertices to evenly-spaced
points on the opposite side?

Dan Hoey
Hoey@AIC.NRL.Navy.Mil
* Look it up.
.
quit