Home > #MTBoS, Maths, SSM, Starters, Teaching > Circles, hexagons and a lot of dead ends

## Circles, hexagons and a lot of dead ends

Today’s puzzle comes from @Catriona Shearer (@cshearer41) and is kinda great:

Have a go, before reading on.

I know sometimes when I type up my solutions it looks like the answer comes quickly and easily to me. And this is sometimes the case, but not always. And this was one of those time where it really didn’t. Here are some of the pages of dead end

And there were many more over a couple of sittings.

Today I looked with fresh eyes and considered what I had. I knew I had 2 overlapping congruent regular hexagons to make the whole shape (I will referto a hexagon this size as H1). I also knew that the white intersection was made up of a regular hexagon around a circle (H2) and 2 trapezia around semi-circles that would also make a regular hexagon (H3). (If you can’t see that these are regular consider symmetry, the angles you know are 120 and equal tangents).

I knew that the side lengths of H1 was equal to the sum of the side lengths of H2 and H3, so I figured if I could express the side length of H2 in terms of H3 or vice versa I would be able to solve it quite easily. I then had the realisation that side length of H3 had to be half of the side length of H2. Although given that I’d inferred something wrongly the other day I thought I’d better justify it fully to make sure:

Once it was justified the problem was fairly straightforward.

Side length ratio H1:H2:H3 is 3/2:1:1/2

So area ratio is: 9/4:1:1/4

The whole shape area is 2H2 – H2 -H3 so 13/4.

The white area is H2+H3 so 5/4

So the white area as a fraction of the whole is 5/13 hence the yellow area as a fraction of the whole is 8/13.

A lovely solution in the end, but one that came after a lot of dead ends. A very enjoyable puzzle, as all Catriona’s are

Categories: #MTBoS, Maths, SSM, Starters, Teaching