Start with an equilateral triangle.

geometry/hexagon.in.triangle New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

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

[ Editor: posed by Lee Shere <shere@leland.stanford.edu>. ]

Show solution

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 side 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?

* Look it up.

-- Dan Hoey <Hoey@AIC.NRL.Navy.Mil>