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How many triangles can be formed using the vertices of a regular dodecagon (a 12-sided polygon)?

Logic  Oct 18, 2018
 #1
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For each side of the 12-gon, there are 12-4=8 non-adjacent vertices which you can use to form a triangle, so there are 12*8=96 such triangles. There are also those triangles which share two sides with the 12-gon, of which there are 12 (one for each vertex of the 12-gon--choose the two sides adjacent to the vertex). So there are 108 total excluded triangles, and thus the answer is (123)−108=220−108=112(123)−108=220−108=112.

Guest Oct 18, 2018
 #2
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That sounds like good logic :)

Melody  Oct 18, 2018
edited by Melody  Oct 18, 2018
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The guest found the number of triangles that don't share a side with the dodecagon, I don't think we need to do that.

Guest Oct 18, 2018
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Well offer an alternate answer then. 

Melody  Oct 18, 2018
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The answer is 12 choose 3=220, but you could also add 108 to the guest's answer to get that (because he subtracted 108 from the total number)

Guest Oct 18, 2018
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Well, that is correct. Read here, for more info: https://cs.uwaterloo.ca/journals/JIS/sommars/newtriangle.html

We have N choose 3, in this case, 12 choose 3, which is 12!/9!*3!=12*11*10/3*2*1=220 triangles. You can add 108 triangles to that.

tertre  Oct 19, 2018

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