[Subrata Ghosh]

Hello Swastik,

The image below can give you some hint you are looking for.

[Subrata Ghosh]

Solved by Shinjini Ghosh

[Subrata Ghosh]

There seems to be nothing special about the given expression since the cyclic symmetry is not present. So it is not an identity with constant value for all a,b,c.

So, we can have variable value of the expression depending on the values of a,b and c. For example, for a=0,b=1 and c=2 the sum is -1.5 and that for a=1, b=4 and c=2 is 1.167.

Restoring cyclic symmetry in the above expression will give us the following identity :
a/(a−c)(a−b)+b/(b−c)(b−a)+c/(c−a)(c−b) = 0

This can be proved by simple addition of the LHS.

[Subrata Ghosh]

Using the same formula can we predict the probability of generating an obtuse angled triangle from any 3 randomly chosen points on a circle.

Hint : A circle has infinite number of points ( n ---> infinity)

[Subrata Ghosh]

When 2n = 90, n =45 then using the formula derived earlier,

Number of obtuse angled triangles = 45 x 44 x 43 = 85,140

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