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I am looking for a easy way for dividing a regular hexagon into 5 equal parts and preferably equal shapes or continuing shapes to make it easy to see the regions.

The way that I found is dividing each triangle in the hexagon in five parts (Splitting equilateral triangle into 5 equal parts). But it is difficult to see the 5 final regions.

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chema989
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  • You might look for an easy way, but that doesn't mean it exists. I can think of three strategies, but if you didn't think of them directly, they're almost certainly going to be what you consider hard. – Nij Sep 13 '16 at 06:12
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    Dissection of triangle in that fashion can't be extended to hexagon (I mean, you may dissect each triangle into equal shapes, but that won't give you a dissection of your hexagon into equal shapes). – Ivan Neretin Sep 13 '16 at 06:28
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    Your "easy way to divide" and "easy way to see the parts" are at odds with each other. The easiest way to divide any convex polygon into equal parts is to divide the perimeter into equal parts and to join the segments to the polygons center. To see that these are equal you break them into triangles and see they all have the same basis (as the perimeter was equally divided) and same height (because distance from the center to perimeter is equal). The tiny amount of work to "see" the parts are equal more than make up for the extreme difficulty it would take to divide them into the same shape. – fleablood Sep 13 '16 at 06:46
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    Okay, the restriction on the linked page that the parts be equal to rotation is not necessary and made the problem exceedingly hard to solve and to see. Without the restriction dividing the triangle or hexagon is exceedingly easy. Just divide the perimeters into equal parts and cut like a pizza from the center. Will work for all convex polygons. It will not be symmetric to rotation but who the heck cares about that? – fleablood Sep 13 '16 at 06:50
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    What are "continuining shapes"? I mean, I suppose that's a typo for "continuing shapes", but I don't know what that means either. – Gerry Myerson Sep 13 '16 at 07:14

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In the linked page the restriction "obtainable from each other by a rigid motion" makes the question exceedingly hard. You do not need that restriction and the answer becomes very easy without it.

Simply divide the hexagons perimeter into 5 equal parts and connect to the center point.

For example if each side is 1 inch, start at a vertex go to the next vertex of one inch and go 1/5 of an inch into the next side. This forms a triangle of base 1 inch and height $\sqrt{3}/2$ and a second triangle of base 1/5 and height $\sqrt{3}/2$ so the area of this piece is $1.2*\sqrt{3}/4$: 1/5 of the area of the hexigon.

For the second piece go along the side 4/5 of an inch and along the next side 2/5 of an inch. This forms a piece the shape of two triangles glued together. Both have the same height $\sqrt{3}/2$ and the sum of their two bases is $1.2$. So it has the same area as the first piece.

Keep going.

This will work for all regular polygons of course. It will work for all polygons but that might not be immediately obvious.

fleablood
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    How do you extend this construction to non-convex polygons? I don't see a natural way to make it work outside of the convex (or at least star-shaped) case. – RavenclawPrefect Dec 08 '20 at 22:02