# Dividing the pentagon

It is easy to divide an equilateral triangle into three equal, though not equilateral, triangles.

It is even simpler to divide a square into four equal squares.

The difficult part is, whether you can divide a regular pentagon into five equal pentagons?

Note:

Equal means, equal in area.

• can we have 1 extra pentagon in the end or the question is about exactly 5? – manshu Feb 24 '16 at 19:21
• If all are equal in terms of area, then I'm ok with it :-) – ABcDexter Feb 24 '16 at 19:24

Easy, the only definition of a pentagon is that is must have 5 sides :)

You could replicate it for orders 7,9, 11 and so on. (Just figured it won't work for even numbers)

• +1 but how can i believe that their areas are equal – manshu Feb 24 '16 at 19:33
• Haha, I was just in the process of drawing exactly this, then saw this answer appear. Nice one. – jhabbott Feb 24 '16 at 19:34
• @manshu if you add 5 points somewhere on the straight lines from the centre to the 5 vertices (doesn't matter where, somewhere near the middle), then rotate all 5 of those elbow points around the centre, then all 5 shapes will be identical (and therefore have the same area). – jhabbott Feb 24 '16 at 19:36
• @manshu because all of the areas are symmetrical - if all the red lines are drawn in the exact same way and are simply $72\unicode{xb0}$ rotations of each other – Paul Evans Feb 24 '16 at 19:42
• "Just figured it won't work for even numbers" - then you make the cuts in the middle of the original polygon's edges, like the example with the square, instead of at the vertices. – user2357112 Feb 24 '16 at 23:14