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

Triangle divided


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


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

Note:

Equal means, equal in area.

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

splitted pentagon

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

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    $\begingroup$ +1 but how can i believe that their areas are equal $\endgroup$ – manshu Feb 24 '16 at 19:33
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    $\begingroup$ Haha, I was just in the process of drawing exactly this, then saw this answer appear. Nice one. $\endgroup$ – jhabbott Feb 24 '16 at 19:34
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    $\begingroup$ @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). $\endgroup$ – jhabbott Feb 24 '16 at 19:36
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    $\begingroup$ @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 $\endgroup$ – Paul Evans Feb 24 '16 at 19:42
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    $\begingroup$ "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. $\endgroup$ – user2357112 Feb 24 '16 at 23:14

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