A hexagon is drawn with its vertices at

\((0,0),(1,0),(2,1),(2,2),(1,2), \text{ and } (0,1),\)

and all of its diagonals are also drawn, as shown below. The diagonals cut the hexagon into 24 regions of various shapes and sizes. These 24 regions are shown in pink and yellow below. If the smallest region (by area) has area a, and the largest has area b, then what is the ratio a:b? Give your answer in lowest terms.

Guest Nov 11, 2019

#1**+1 **

I have redrawn and recolored the image to make the following claims:

1) All regions of the same color in this image have the same area., and

2) the measure of the area of each green region is twice that of the area of any blue region.

If these claims are proved, the ratio of the smallest area to the largest area would of course be 1/2.

The proofs I give will be mostly based on the image, something that math people normally discourage for good reasons, but in this case should be justified since the facts are well known and the details could become overwhelming.

First we notice that regions 1 and 18 together make up half of the area of the rectangle ABGF, which in turn has area equal to half of the square CHFG that has area 1 square unit; so

Area of 1 +Area of 18 = 1/4.

Next consider regions 18 and 19, which again together make up a triangle with area 1/4 of the square. So

Area of 19 + Area of 18 = 1/4.

This forces not only regions 1 and 19 to have the same area, but all the othe other green regions other than 16, 24, 21 and 7 due to obvious congruence.

To show that the latter 4 regions have the same area, note that 16, 24, 15, and 17 together make up a triangle that has area 1/2, and that 24 and 15 together make up a triangle with area 1/4, which implies that 16 and 17 also add up to 1/4. Since 15 and 17 are congruent and therefore have the same area, we are forced to conclude that 16 and 24 (and therefore 21 and 7 by congruence) also have the same area.

Similar reasoning shows that all blue regions also have the same area.

To prove that each green region has area twice as large as each blue region, consider regions 24 and 18 and refer to the image. Region 18 is divided into two congruent triangles by means of the line segment CE, each of which is congruent to region 18 triangle.

Of course this proof consists of mainly 'hand-waving' and is mainly heuristic, but the reader may provide the details!!

Gadfly Nov 12, 2019