        ## Results on Marion Walter's Theorem

Two proofs are available; a proof using classical geometry is followed by a proof using vector geometry

#### Proof of Marion Walter's Theorem Using Classical Geometry

We want to show that ratio of the area of the hexagon to the area of the triangle is . Figure 9

Lemma 1.   In figure 10, the area of is the area of . Figure 10

Proof:  For simplicity, let's assume (throughout) that the area of is 1.   The trick is to draw in the segment from E to H.  This forms a  trapezoid whose top base is and whose height is the height from A to .  It follows that the area of the trapezoid is .  Here's why:  Suppose and the height to is h.  Then Now look just at the trapezoid: Figure 11

We know that the top base is the bottom base.   Convince yourself that, if the area of is x,  and the area of, say, is y, then the other areas are as marked.   Notice that the area of is (look at the original picture).  This leads to two equations in two unknowns;  the first one   says that the sum of all four pieces is the area of the trapezoid, and the second one says that the sum of the bottom triangle and the left little one is : Solving, we get , as advertised.  By the way, by symmetry, the same is true for triangles AJB and ANC;  they too have area .

This argument can be recycled:  Look at the trapezoid you get by joining up the top trisection points rather than the bottom ones: Figure 12

It has two properties: Its top base is and its height is , so its area is .

Notice, too, that the area of is .  Letting the area of be z and the area of be w, we have two more equations in two unknowns: Which implies that .  We have:

Lemma.  The area of triangle BQC  is .

The rest is easy: Figure 13

By Lemma 1, each herringbone triangle in Figure 13 has area .   By Lemma 2, the bricked-in triangle has area .  So, the white chevron has area  Figure 14

Now, look at the whole triangle, painted as in Figure 6:  It's composed of three checkered triangles of area , three chevrons of area , and the hexagon in the middle.  That leaves for the hexagon, which is what we wanted to show.

#### Proof of Marion Walter's Theorem Using Vector Geometry

In what follows, if X and Y are points in the plane, is the determinant of the matrix whose columns are the coordinates of X and Y.  It turns out (see one of the warm-up problem number 4 that equals twice the area of the triangle whose vertices are the origin, X and Y.

Suppose we move in from each endpoint by a factor of k  ( in the original Marion Walter problem).  Then the vertices can be expressed as combinations of A and B as in Figure 15: Figure 15

This looks awful, but the intersection points are quite easy to derive (using the vector equations of lines).  The important thing to notice is that there's a certain symmetry in the figure that can be expressed like this: Figure 16

All the coefficients are expressions in k (which is independent of A and B).  Notice how the middle points and the origin are collinear.  This allows us to "split" the hexagon in half, and to operate with the difference of the areas of two triangles in each half.

For example, in the upper half, the gray area is the difference between the area  of the triangle determined by and and the "bricked-in'' triangle determined by and . Figure 17

So, the ratio of the gray area to the area of the whole original triangle can be calculated like this: and this is independent of A and B.

The bottom half of the hexagon can be calculated similarly, and the answer comes out the same, because of the transpose rule.  Replacing r, s, j, t, x, and y by the expressions in k gives a formula for the ratio of the areas of the hexagon and  triangle.  Replacing k by will give you .  