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Given a circle with 24 evenly spaced points, how would you find the number of possible isosceles triangles (which includes equilateral) that can by drawn using the points?

My attempt was to say that the number of ways to pick the vertex where the congruent sides meet is 24. For each of the vertexes there are 11 possible pairs of other vertices. This gives 264 possible triangles, but I don't think I did it right. Can anyone help?

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You have to treat equilateral triangles separately, to avoid counting them more than once.

There are eight equilateral triangles. And for each of the 24 points, there are ten possible pairs of vertices that form non-equilateral isosceles triangles. So I get 8 + 240 = 248 triangles.

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  • $\begingroup$ How did you get that there are 10 possible pairs of vertices that form non-equilateral triangles? And how did you find there were 8 equilateral triangles? $\endgroup$ – 1110101001 Dec 4 '13 at 0:38

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