Triangle from 17 sided polygon
Probability
Level
5
In a 17 sided regular polygon find total number of obtuse angle triangles that can be formed from 3 vertices ..?
The answer is 476.
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1 solution
Consider the general case of an n-gon. Consider any one of the n points. From this point, if we join two other arbitrary points, we get a triangle. Our goal is to find the number of such triangles such that atleast one of the angles are obtuse.
From this one point, if one of the two other vertices are directly opposite to it (Only possible in even sided regular polygons, hence mentioned at the beginning) , then the angle subtended will be a right angle. thales' theorem
If the two vertices are on either side of this diagonal, then an acute angle is formed. Obtuse triangles will only be formed when both points are on the same side of the semi-polygon.
Hence, the total number of ways to choose the remaining points is (n/2 - 1)/2. Therefore considering for each of the n vertices, in total there would be n x (n/2 - 1)/2 or n×( (n/2)(upper bound) - 1)((n/2)(upper bound) - 2)/2. Substituting n=17 yields 476.