Difference between revisions of "2006 IMO Problems/Problem 1"
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Latest revision as of 00:02, 19 November 2023
Problem
Let 
be triangle with incenter 
. A point 
in the interior of the triangle satisfies 
. Show that 
, and that equality holds if and only if 
Solution
We have
![\[\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA).\]](http://latex.artofproblemsolving.com/f/4/2/f421455e7a91f3c575a372c25191d800b9d050b0.png)
and similarly ![\[\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC).\]](http://latex.artofproblemsolving.com/9/e/8/9e877bd39e2a567854a33848d7b67b84525f2179.png)
Since 
, we have 
It follows that ![\[\angle IBP = \frac{1}{2} (\angle PCB - \angle PCA) = \frac{1}{2} (\angle PBA - \angle PBC) = \angle ICP.\]](http://latex.artofproblemsolving.com/7/c/1/7c12a2c53931996bdeaf3f9e8bb2d2b800bef328.png)
Hence, 
and 
are concyclic.
Let ray 
meet the circumcircle of 
at point 
. Then, by the Incenter-Excenter Lemma, 
.
Finally, 
(since triangle APJ can be degenerate, which happens only when 
), but 
; hence and we are done.
By Mengsay LOEM , Cambodia IMO Team 2015
latexed by tluo5458 :)
minor edits by lpieleanu
See Also
| 2006 IMO (Problems) • Resources | ||
| Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
| All IMO Problems and Solutions | ||
