You need to be sure!
From a regular polygon of 50 - sides , you have to select 3 vertices ( at random) .You selected the triplets times .
What is the value of to surely say that you have at least a triplet which form obtuse- angled triangle upon joining.
Assumptions :
-
After selecting first triplet, for the second triplet you have to select again from the 50- vertices (not from the left 47 vertices)
-
No repetition of the same triplet is allowed
-
You are selecting the vertices with eyes closed. (You can't check , whether the condition is satisfied)
Try my set : Let's play with polygons
The answer is 5801.
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1 solution
X=50(50/2-2)(50/2-1)/2 are the no of obtuse angled triangle
you may get the the no of obtuse triangle as follow
choose any of the vertex of the obtuse angled triangle this would be our side containing obtuse angled triangle let it be A
now join this vertex to the (diam)opposite point.
now we have two regions take the point just right of A. it has s/2-2 choice on left vertex to be tria
the second right has s/2-3
and so on till 1
so no of obtuse angled triangle on single vertex=(s/2-2)+(s/2-3)+(s/2-4)++++1
=(s/2-2)(s/2-1)/2
total on n vertex=s(s/2-2)(s/2-1)
this one way is original ;p
so n=(50C3)-X+1