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Difference between revisions of "2006 IMO Problems/Problem 1"

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(→‎Solution)
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Let ray <math>AI</math> meet the circumcircle of <math>\Delta ABC</math> at point <math>J</math>. Then, by the Incenter-Excenter Lemma, <math>JB=JC=JI=JP</math>.  
 
Let ray <math>AI</math> meet the circumcircle of <math>\Delta ABC</math> at point <math>J</math>. Then, by the Incenter-Excenter Lemma, <math>JB=JC=JI=JP</math>.  
  
−
Finally, <math>AP+JP \geq AJ = AI+IJ</math> (since triangle APJ can be degenerate), but <math>JI=JP</math>; hence <math>AP \geq AI</math> and we are done.
+
Finally, <math>AP+JP \geq AJ = AI+IJ</math> (since triangle APJ can be degenerate, which happens only when <math>P=I</math>), but <math>JI=JP</math>; hence <math>AP \geq AI</math> and we are done.
  
 
By Mengsay LOEM  , Cambodia IMO Team 2015
 
By Mengsay LOEM  , Cambodia IMO Team 2015
  
 
latexed by tluo5458 :)
 
latexed by tluo5458 :)

Revision as of 17:46, 18 August 2017

Problem

Let $ABC$ be triangle with incenter $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA = \angle PBC+\angle PCB$. Show that $AP \geq AI$, and that equality holds if and only if $P=I$

Solution

We have \[\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PBA - \angle PBC)\] (1) and similarly \[\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PCB - \angle PCA)\] (2). Since $\angle PBA + \angle PCA = \angle PBC + \angle PCB$, we have $\angle PBA - \angle PBC = \angle PCB - \angle PCA$ (3).

By (1), (2), and (3), we get $\angle IBP = \angle ICP$; hence $B,I,P,C$ are concyclic.

Let ray $AI$ meet the circumcircle of $\Delta ABC$ at point $J$. Then, by the Incenter-Excenter Lemma, $JB=JC=JI=JP$.

Finally, $AP+JP \geq AJ = AI+IJ$ (since triangle APJ can be degenerate, which happens only when $P=I$), but $JI=JP$; hence $AP \geq AI$ and we are done.

By Mengsay LOEM , Cambodia IMO Team 2015

latexed by tluo5458 :)

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