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

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(latexed)
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==Solution==
 
==Solution==
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We have angle(IBP)= angle(IBC) - angle(PBC)=1/2 angle(ABC) - angle(PBC)=1/2 [ angle(PBA) - angle(PBC)]  (1)  
+
We have  
−
and similarly angle(ICP)=angle(PCB) - angle(ICB) = angle(PCB) - 1/2 angle(ACB)= 1/2[ angle(PCB) - angle(PCA)] (2)
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<cmath>\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PBA - \angle PBC)</cmath> (1)
−
since angle(PBA)+angle(PCA)=angle(PBC)+angle(PCB)  , then  angle(PBA) - angle(PBC) = angle(PCB) - angle(PCA)  (3)
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and similarly <cmath>\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PCB - \angle PCA)</cmath> (2).
−
(1) , (2) and (3) therefore : angle(IBP)=angle(ICP)  i.e  BIPC is con cyclic .
+
Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PBA -
−
Let J be a point of intersection between (AI) and the circumcise of triangle ABC , then IJB is isosceles traingle because of angle(BIJ)=angle(IBJ)
+
\angle PBC = \angle PCB - \angle PCA</math> (3).
−
so JB=JC=JI , then JI=JP  (because BIPC con cyclic) .  
 
  
−
In triangle APJ  :  AP+JP >=  AJ=AI+IJ   but JI=JP so  AP >= AI  .
+
By (1), (2), and (3), we get <math>\angle IBP = \angle ICP</math>; hence <math>B,I,P,C</math> are concyclic.
 +
 
 +
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.
  
 
By Mengsay LOEM  , Cambodia IMO Team 2015
 
By Mengsay LOEM  , Cambodia IMO Team 2015
 +
 +
latexed by tluo5458 :)

Revision as of 18:41, 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), 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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