I have come across a problem two days ago which I could not solve.It's a basic problem but i need a procedure for the problem.

A triangle of sides of lengths a,b,c and the angles opposite to the sides are A,B,C respectively , satisfies the following condition :

a + b = tan (C/2) [ atan (B) + btan (A) ]

The perimeter of the triangle is 36 cm...and the side length c = 16 cm...then the area of triangle is ** ____** .

[WHEN I ATTEMPTED A QUESTION IN I.M.O. ,THEY SAID IT IS ACTUALLY AN ISOSCELES TRIANGLE.......BUT I DON'T KNOW THE PROOF. ]

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TopNewestYou can format your equation by putting \ [ on the left and \ ] on the right (without the spaces). – Tim Vermeulen · 3 years, 7 months ago

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– Vaibhav Reddy · 3 years, 7 months ago

you could understand the question right :) now did you solve the problemLog in to reply

– Tim Vermeulen · 3 years, 7 months ago

I haven't tried it yet. The problem as it is might be understandable, but if you typeset your problem, it looks better and thus users are way more likely to try the problem. :)Log in to reply

– Vaibhav Reddy · 3 years, 7 months ago

Let a; b; c be the lengths of the sides of a triangle, and A; B;C, respectively, the angles opposite these sides. Prove that if a + b = tan C/2(a tan B + b tan A); the triangle is isosceles. SORRY,my bad......the upper one is the original question...it appeared in 1966 I.M.O. Now is it easy to solve???Log in to reply

– Chris Quinones · 3 years, 7 months ago

I see you on every discussion, whether it be a answer to a question or your own question.Log in to reply