# Thread: dividing an n-gon into rhombuses

1. ## dividing an n-gon into rhombuses

I'm trying to prove that any regular and convex 2n-polygon can be divided into rhombuses.

I proceed by induction.
BASE CASE: For n=2 this is obviously true because a square is a rhombus.
INDUCTIVE HYPOTHESIS: suppose the statement is true for a polygon with 2n sides.

Now to make a polygon with 2(n+1) sides, we take the 2n-polygon which has already been divided into rhombuses. We take any n sides and replace them with n+2 sides, thus 'elongating' the 2n-polygon (so to speak) in such a way that we draw n lines parallel to those n sides, at some distance. And then join them with the 2n-polygon with a line at each end. Hence n more rhombuses are added to the 2n-polygon and this new shape is a 2n+2-polygon.

This completes my proof.
I have a feeling there's some loophole in this argument :s Can someone kindly help?

2. Sorry, I can't visualize your procedure because there are too many possible variants.

Now to make a polygon with 2(n+1) sides, we take the 2n-polygon which has already been divided into rhombuses. We take any n sides
Absolutely any n sides? Not necessarily connected?
we draw n lines parallel to those n sides, at some distance
This looks suspicious because if those lines are not drawn at a precise distance, then the resulting figure will not all have sides equal, i.e., will not be a regular polygon.
And then join them with the 2n-polygon with a line at each end.
There are so many ways to "join them with the 2n-polygon". It is not cleat what you mean.

Could you draw a picture how you transform a hexagon into an octagon?

3. I just tried doing it on paper and I realized that this way the polygon i come up with is extremely lopsided its not regular at all.

To answer your questions first, I take any n consecutive sides and draw lines parallel to them at a distance that is equal to any one side of the 2n-gon.

The main thing that I missed before is, that when we go from a 2n-gon to 2(n+1)-gon.... EACH interior angle must change too. So we can't replace just half the sides with new ones like I was doing.

So how then should I proceed? Any hints?

4. From google search, the answer is here

http://mathcircle.berkeley.edu/newsi...002examsol.pdf

I haven't looked at it though.

5. Thanks. I think I understood solution #2 on p. 4.

There is also a Youtube video, but I have not figured out yet how to turn into a precise proof.

6. @undefined: thank you for that link! I'm glad to see that my initial attempt at the proof was pretty close!

@emakarov: So the outline of the proof is that at nth step we add n new rhombuses starting from the top, in clockwise direction. But to complete the proof we need to:
1) prove that at every step the polygon is regular.
2) We need to give the angle at which we construct the new rhombus at every step.

7. So the outline of the proof is that at nth step we add n new rhombuses starting from the top, in clockwise direction. But to complete the proof we need to:
1) prove that at every step the polygon is regular.
2) We need to give the angle at which we construct the new rhombus at every step.
Hmm, I am not sure what solution you are talking about. Both solutions on p. 4 of the PDF document the link to which appears above start with the bottom side of the polygon. Further, solution #1 does not use induction (or, if it does, its statement is different from the problem). Solution #2, which I found clearer, does not mention steps. Finally, solution #2 works with more general EP-gons, which are not necessarily regular. It often happens in proofs by induction that one has to generalize the induction statement.