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

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==Problem==
 
==Problem==
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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>
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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==
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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)  
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We have  
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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 PCB - \angle PCA).</cmath>
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since angle(PBA)+angle(PCA)=angle(PBC)+angle(PCB)  , then  angle(PBA) - angle(PBC) = angle(PCB) - angle(PCA) (3)
+
 
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(1) , (2) and (3) therefore : angle(IBP)=angle(ICP)  i.e  BIPC is con cyclic .  
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and similarly <cmath>\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC).</cmath>
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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)
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Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PCB - \angle PCA = \angle PBA - \angle PBC.</math>
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so JB=JC=JI , then JI=JP (because BIPC con cyclic) .  
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 +
It follows that <cmath>\angle IBP = \frac{1}{2} (\angle PCB - \angle PCA) = \frac{1}{2} (\angle PBA - \angle PBC) = \angle ICP.</cmath> Hence, <math>B,P,I,</math> and <math>C</math> are concyclic.
 +
 
 +
 
 +
Let ray <math>AI</math> meet the circumcircle of <math>\triangle 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, which happens only when <math>P=I</math>), but <math>JI=JP</math>; hence <math>AP \geq AI</math> and we are done.
  
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In triangle APJ  :  AP+JP >=  AJ=AI+IJ  but JI=JP  so  AP >= AI  .
 
  
 
By Mengsay LOEM  , Cambodia IMO Team 2015
 
By Mengsay LOEM  , Cambodia IMO Team 2015
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 +
latexed by tluo5458 :)
 +
 +
minor edits by lpieleanu
 +
 +
 +
==See Also==
 +
 +
{{IMO box|year=2006|before=First Problem|num-a=2}}

Latest revision as of 01:02, 19 November 2023

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 \[\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA).\]

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

It follows that \[\angle IBP = \frac{1}{2} (\angle PCB - \angle PCA) = \frac{1}{2} (\angle PBA - \angle PBC) = \angle ICP.\] Hence, $B,P,I,$ and $C$ are concyclic.


Let ray $AI$ meet the circumcircle of $\triangle ABC\,$ 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 :)

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
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