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

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m (→‎Solution)
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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> (1)
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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> (2).
+
 
−
Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PBA -
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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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\angle PBC = \angle PCB - \angle PCA</math> (3).
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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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Hence, <math>B,P,I,</math> and <math>C</math> are concyclic.
  
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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>\triangle  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>\triangle  ABC\, </math> at point <math>J</math>. Then, by the Incenter-Excenter Lemma, <math>JB=JC=JI=JP</math>.  
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latexed by tluo5458 :)
 
latexed by tluo5458 :)
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 +
minor edits by lpieleanu

Revision as of 19:48, 30 May 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

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