1966 IMO Problems/Problem 2
Let 
, 
, and 
be the lengths of the sides of a triangle, and 
respectively, the angles opposite these sides. If,
![\[a+b=\tan{\frac{\gamma}{2}}(a\tan{\alpha}+b\tan{\beta})\]](http://latex.artofproblemsolving.com/a/7/0/a707cd7360756a01d3f6e17d4dff56bc3561a277.png)
Prove that the triangle is isosceles.
Solution
We'll prove that the triangle is isosceles with 
.
We'll prove that . Assume by way of contradiction WLOG that 
.
First notice that as 
then and the identity 
our equation becomes:
![\[a+b=\cot \frac{\alpha +\beta}{2}\left(a\tan \alpha + b\tan \beta \right)\]](http://latex.artofproblemsolving.com/a/1/c/a1c3a6e7ad9660df3cdb00e2a28d476ff3586e9c.png)
![\[\iff a\tan\frac{\alpha +\beta}{2}+b\tan \frac{\alpha +\beta}{2}=a\tan \alpha + b\tan \beta\]](http://latex.artofproblemsolving.com/b/d/2/bd278cc330f9bafbb88664474be840209a38ddeb.png)
![\[\iff a\left(\tan \alpha -\tan \frac{\alpha +\beta}{2}\right)+b\left(\tan \beta -\tan \frac{\alpha +\beta}{2} \right)=0\]](http://latex.artofproblemsolving.com/d/1/e/d1ea2bfcbcc0c9063ccf226363be2bbd7a3bb5ac.png)
Using the identity 
and inserting this into the above equation we get:
![\[\iff a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)+b\tan \frac{\beta -\alpha}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0\]](http://latex.artofproblemsolving.com/e/1/0/e104076d54e67e960085ca32e1f1fa83cb12c026.png)
![\[\underbrace{\iff}_{\tan -A=-\tan B}a\tan \frac{\alpha -\beta}{2}\left(1+\tan \alpha \tan \frac{\alpha +\beta}{2}\right)-b\tan \frac{\alpha -\beta}{2}\left(1+\tan \beta \tan \frac{\alpha +\beta}{2} \right)=0\]](http://latex.artofproblemsolving.com/0/0/6/0062a028412198b8dad1a7c5bbdc8094a34506c4.png)
![\[\iff \tan \frac{\alpha -\beta}{2}\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)=0\]](http://latex.artofproblemsolving.com/8/3/5/8353f65dee16d3a0cd3caad6b565e0c0293d54b0.png)
Now, since and the definitions of 
being part of the definition of a triangle, 
.
Now, 
(as 
and the angles are positive), 
, and furthermore, 
. By all the above, ![\[\left(a-b+\tan \frac{\alpha +\beta}{2}(a\tan\alpha -b\tan \beta) \right)>0\]](http://latex.artofproblemsolving.com/f/e/8/fe8ddf3fb8a97b25b77bc52f2ddb140f21ae5f4b.png)
Which contradicts our assumption, thus 
. By the symmetry of the condition, using the same arguments, 
. Hence .
See Also
| 1966 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
