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Topic: Rhombi Tiling (Read 2229 times) |
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william wu
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Rhombi Tiling
« on: Oct 24th, 2004, 6:10pm » |
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A regular polygon with 100 sides of length 1 is subdivided into rhombi with sides of length 1. Show that exactly 25 of these rhombi are squares.
Notes:
1. Recall that a rhombus is a 4-sided polygon with equal side lengths.
2. Disclaimer: Really neat result but I don't know how to solve this.
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| « Last Edit: Oct 24th, 2004, 6:24pm by william wu » |
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Barukh
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I've attached a visualization of the possible tiling of the 12-gon. Really amazing!
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John_Gaughan
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Re: Rhombi Tiling
« Reply #2 on: Oct 26th, 2004, 6:34am » |
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Maybe this is being naive but Barukh's drawing has 3/12 or 25% of the rhombi as squares... maybe there is a more general relation to prove?
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Grimbal
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Re: Rhombi Tiling
« Reply #3 on: Oct 26th, 2004, 7:26am » |
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It seems obvious to me.
>>
If you start from one side, you can stack rhombi up to the opposite side. All of them have the "top" side parallel to the starting side (the "base"). If you do the same from the side that is perpendicular to the first one, you get another pile crossing the polygon at roughly a right angle. The crossing of these 2 piles is bound to be a square rhombus. Considering that in a 100-sided polygon there are 25 sets of 4 sides on which you can do that construction, there must be at least 25 squares.<<
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Barukh
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Re: Rhombi Tiling
« Reply #4 on: Oct 27th, 2004, 4:47am » |
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That's really clever, Grimbal! Bravo! Of course, the same is true for the general case (of 4n-gon).
By the way, every regular 2n-gon is covered by exactly n(n-1)/2 rhombi.
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Obob
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Nice work showing that there must be at least 25 squares, Grimbal. But why can't there be more than 25?
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Rezyk
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Re: Rhombi Tiling
« Reply #6 on: Oct 27th, 2004, 8:06am » |
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The "parallel stack" path is unique in either direction. Each square must trace out to 4 unique sides.
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