Difference between revisions of "2006 IMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
| − | Let <math>ABC</math> be triangle with incenter <math>I</math>. A point <math>P</math> in the interior of the triangle satisfies <math>\angle PBA+\angle PCA = \angle PBC+\angle PCB</math>. Show that <math>AP \geq AI</math>, and that equality holds if and only if <math>P=I</math> | + | Let <math>ABC</math> be triangle with incenter <math>I</math>. A point <math>P</math> in the interior of the triangle satisfies <math>\angle PBA+\angle PCA = \angle PBC+\angle PCB</math>. Show that <math>AP \geq AI</math>, and that equality holds if and only if <math>P=I.</math> |
==Solution== | ==Solution== | ||
Revision as of 20:42, 23 July 2019
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/a/9/2/a928a15d342a0b7c66d3cd1820552fb3c4495813.png)
(1)
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/7/5/5/755e61e57474a4a57009c18bac156836ae9434fc.png)
(2).
Since 
, we have 
(3).
By (1), (2), and (3), we get 
; hence 
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 :)
