# 1966 IMO Problems/Problem 2

Let $a$, $b$, and $c$ be the lengths of the sides of a triangle, and $\alpha,\beta,\gamma$ respectively, the angles opposite these sides. If,

$$a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta})$$

Prove that the triangle is isosceles.

## Solution

We'll prove that the triangle is isosceles with $a=b$. We'll prove that $a=b$. Assume by way of contradiction WLOG that $a>b$. First notice that as $\gamma = \pi -\alpha-\beta$ then and the identity $\tan\left(\frac \pi 2 - x \right)=\cot x$ our equation becomes: $$a+b=\cot \frac{\alpha +\beta}{2}\left(a\tan \alpha + b\tan \beta \right)$$$$\iff a\tan\frac{\alpha +\beta}{2}+b\tan \frac{\alpha +\beta}{2}=a\tan \alpha + b\tan \beta$$ $$\iff a\left(\tan \alpha -\tan \frac{\alpha +\beta}{2}\right)+b\left(\tan \beta -\tan \frac{\alpha +\beta}{2} \right)=0$$ Using the identity $\tan (A-B)=\frac {\tan A-\tan B}{1+\tan A\tan B}$ $\iff \tan A-\tan B=\tan(A-B)(1+\tan A\tan B)$ and inserting this into the above equation we get: $$\iff a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)+b\tan \frac{\beta -\alpha}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0$$ $$\underbrace{\iff}_{\tan -A=-\tan A}a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)-b\tan \frac{\alpha -\beta}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0$$ $$\iff \tan \frac{\alpha -\beta}{2}\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)=0$$ Now, since $a>b$ and the definitions of $a,b,\alpha,\beta$ being part of the definition of a triangle, $\alpha >\beta$. Now, $\pi >\alpha -\beta >0$ (as $\alpha+\beta +\gamma = \pi$ and the angles are positive), $\tan \frac{\alpha -\beta}{2}\neq 0$, and furthermore, $\tan \frac{\alpha+\beta}{2}>0$. By all the above, $$\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0$$ Which contradicts our assumption, thus $a\leq b$. By the symmetry of the condition, using the same arguments, $a\geq b$. Hence $a=b$.