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

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==Solution==
 
==Solution==
 
We have  
 
We have  
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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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<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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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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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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Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PCB - \angle PCA = \angle PBA - \angle PBC.</math> It follows that <math>\angle IBP = \frac{1}{2} (\angle PCB - \angle PCA) = \frac{1}{2} (\angle PBA - \angle PBC) = \angle ICP</math>.
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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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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.
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Hence, <math>B,P,I,</math> and <math>C</math> are concyclic.
 
  
  
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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.
 
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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By Mengsay LOEM  , Cambodia IMO Team 2015
 
By Mengsay LOEM  , Cambodia IMO Team 2015
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minor edits by lpieleanu
 
minor edits by lpieleanu
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==See Also==
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{{IMO box|year=2006|before=First Problem|num-a=2}}

Latest revision as of 00:02, 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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