Solution to Challenge Problem for April 10, 2007

Solution to Challenge Problem for April 10, 2007

Christopher G. Green April 15, 2007

Suppose that in a hexagon ABCDEF the diagonal AD divides the hexagon in two parts of equal area. Suppose the same is true for BE and CF. Show that the three diagonals intersect in one point.

Note. We are told that each diagonal divides the hexagon into two parts; this eliminates “pathological” cases where four vertices lie on one side of the diagonal determined by the other two vertices, for instance (see Figure 1, left). To avoid further difficulties, we will assume (a) the polygon is a simple polygon, and (b) all the diagonals are “interior” diagonals, i.e., they lie entirely within the polygon. (This is true for any convex hexagon, and for some concave hexagons as well (see Figure 1, right, for an example.)

It is not clear that the problem can be solved in the general case where diagonals may not lie entirely within the polygon;

indeed, it is not even clear that a hexagon meeting the problem description exists in this case.

F A

B

C

D

E

F

E

D

C

I G

H B

A B

E D

C F

A

Figure 1: (left) A hexagon not allowed by the problem statement since diagonal AD does not divide the interior into two regions. (middle) General (convex) hexagon satisfying the hypotheses of the problemand the condition that all diagonals lie entirely within the polygon. (right) An example of a concave hexagon for which each diagonal is an interior diagonal, divides the region into parts of equal area, and intersects the other two diagonals in a single point.

Solution. Suppose to the contrary that the three diagonals do not intersect in one point (see Figure 1, middle). By assumption, we have Area(ADEF) = Area(ABCD), Area(ABEF) = Area(BCDE), and Area(ABCF) = Area(CDEF).

Each of the two regions determined by a given diagonal can be decomposed into a triangle and a quadrilateral in two ways using each of the other two diagonals. This gives us the equations

Area(AF G) + Area(DEF G) = Area(ABCG) + Area(CDG) (from AD and CF) Area(DIE) + Area(AF EI) = Area(BCDI) + Area(ABI) (from AD and BE) Area(ABI) + Area(AF EI) = Area(BCDI) + Area(DIE) (from BE and AD) Area(BHC) + Area(HCDE) = Area(ABHF) + Area(F HE) (from BE and CF) Area(AF G) + Area(ABCG) = Area(DEF G) + Area(CDG) (from CF and AD) Area(BHC) + Area(HCDE) = Area(ABHF) + Area(F HE) (from CF and BE).

From these equations we determine the equalities

Area(AF G) = Area(CDG), Area(ABI) = Area(DIE), Area(BHC) = Area(F HE). (1) We also know that

• angle AIB is equal to angle DIE;

• angle AGF is equal to angle DGC; and

• angle FHE is equal to angle BHC,

since these are vertical angle pairs. This suggests that we use the Law of Sines to compute the areas of the triangles mentioned in (1).

Suppose we have two triangles that have an equal angle.

a

b

f e d c

The Law of Sines tells us that the area of the left triangle is ¹₂absinc, while the area of the right triangle is ¹₂desinf. If c=f, then we would conclude thatab=de.

The above observation, applied to our problem, tells us that

• EI·DI= (AG+GI)·(BH+HI);

• AG·F G= (DI+GI)·(CH+HG); and

• BH·CH= (F G+HG)·(EI+HI).

We can take the product of both sides to see that

AG·F G·EI·DI·CH·BH= (DI+GI)·(AG+GI)·(F G+HG)·(BH+HI)·(CH+HG)·(EI+HI).

All of the lengths involved are non-negative, and particular the left-hand side is positive. We can divide through by the left-hand side and rearrange to arrive at the equation

1 =

1 + GI

DI 1 + GI

AG 1 +HG

F G 1 + HI

BH 1 + HG

CH 1 +HI EI

.

This can only hold ifGI =HG=HI= 0. Hence the three diagonals must intersect in a single point.