Marion Walter 
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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  one tenth.

figure 9

Figure 9

Lemma 1.   In figure 10, the area of triangle b l c  is one fifth the area of triangle a b c.

Figure 10

Figure 10

Proof:  For simplicity, let's assume (throughout) that the area of triangle a b c  is 1.   The trick is to draw in the segment from E to H.  This forms a  trapezoid whose top base is  two thirds B C  and whose height is one third  the height from A to segment b c.  It follows that the area of the trapezoid is five ninths.  Here's why:  Suppose length of segment B C equals b  and the height to  segment B C  is h.  Then

BCHE equals five ninths (derivation)  

Now look just at the trapezoid:

figure 11

Figure 11

We know that the top base is two thirds  the bottom base.   Convince yourself that, if the area of triangle b l c is x,  and the area of, say, triangle E L B is y, then the other areas are as marked.   Notice that the area of triangle b e c is  one third (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  one third:

    system of equations

Solving, we get x equals one fifth, as advertised.  By the way, by symmetry, the same is true for triangles AJB and ANC;  they too have area one fifth.   

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

Figure 12


It has two properties:
    bullet Its top base is one third b   and its height is two  thirds h , so
    bullet its area is eight ninths.

Notice, too, that the area of  triangle a c d is two thirds .  Letting the area of triangle b q c be z and the area of triangle d q b be w, we have two more equations in two unknowns:

        system of equations
    


Which implies that z equals one half .  We have:

Lemma.  The area of triangle BQC  is one half.

The rest is easy:

figure 13

Figure 13


By Lemma 1, each herringbone triangle in Figure 13 has area one fifth.   By Lemma 2, the bricked-in triangle has area one half .  So, the white chevron has area
        
    one minus one half minus one fifth minus one fifth equals one tenth

figure 14

Figure 14

Now, look at the whole triangle, painted as in Figure 6:  It's composed of three checkered triangles of area one fifth, three chevrons of area one tenth, and the hexagon in the middle.  That leaves one tenth 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, det(X, Y) 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 det(X, Y) 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  (k equals one third in the original Marion Walter problem).  Then the vertices can be expressed as combinations of A and B as in Figure 15:

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

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 r A + s B and t A + t B and the "bricked-in'' triangle determined by x A + y B and j A + j B.

Figure 17

Figure 17


So, the ratio of the gray area to the area of the whole original triangle can be calculated like this:

very long chain of equalities showing that the ratio is r t minus s t minus x j plus y j

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 one third will give you one tenth.



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