INMO 1992

This is the first question of the 1992 INMO (Indian National Maths Olympiad). People have essentially cursed me for doing this geometry problem by bashing. It has a plain geometric proof which is way more prettier than mine. But still, here's the question and my solution:

A triangle ABC has B\angle B half of A\angle A. Prove that a=b(b+c)a=b(b+c).

Proof: First take A=2x\angle A=2x. So, B=x\angle B=x.

By sine rule, asin2x\frac{a}{\sin2x}=bsinx\frac{b}{sinx}=csin3x\frac{c}{\sin3x}. From here you can get b and c in terms of a (and some trigonometric functions) and you'll be done. This proof requires considerably less geometric knowledge and basic algebraic manipulation, though one could argue that this destroys the beauty of the problem. Comments are extremely welcome.

Note by Rajdeep Ghosh
2 years, 2 months ago

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doing the problem is the main thing whether u bash or use plain geometry.but I had done it using plain geometry with the help of angular bisector theorem.

ALEKHYA CHINA - 2 years, 1 month ago

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why don't u add your own proof

Rajdeep Ghosh - 2 years, 1 month ago

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The only beauty of maths is to solve it in shorter and easier steps.

MAINAK CHAUDHURI - 1 year, 8 months ago

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