Three vertices are chosen randomly from the seven vertices of a regular 7-sided polygon. The probability that hey form the vertices of an isosceles triangle is:

I know that the this problem deals with geometry and combinations. I just drew its picture and from it I think that from a single vertex of the polygon 5 isosceles triangles can be drawn. So, is 7*5=35 the no. of isosceles triangles that can be drawn?

  • $\begingroup$ No. You correctly found that for every vertex there are five isosceles triangles containing that vertex. However, for two different vertices $v, w$, the set of triangles containing $v$ and the set containing $w$ might overlap, so you can't say that the total will be 35. In fact, it happens that 35 counts the total number of possible triangles in your heptagon, isosceles or not. $\endgroup$ – Rick Decker Aug 23 '13 at 16:20

There are seven choices for the apex. For each apex there are three choices for the base line. Hence there are $21$ isosceles triangles. In total, there are $7\choose 3$ triangles, so the probability is $\frac{3}{5}$.

  • $\begingroup$ But what about the isosceles triangles formed by three consecutive vertices of the heptagon? $\endgroup$ – Rajath Radhakrishnan Aug 23 '13 at 15:50
  • $\begingroup$ That's accounted for in Hagen's count. $\endgroup$ – Rick Decker Aug 23 '13 at 16:22
  • $\begingroup$ +1 As a side note, you should mention that there is no double counting, since we can't form an equilateral triangle. $\endgroup$ – Calvin Lin Aug 23 '13 at 16:59

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