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

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minor edits by lpieleanu
 
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<math>\textbf{Second solution:}</math>
 
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Firstly, call <math>x=\angle PBC + \angle PCB =\angle PBA + \angle PCA</math>
 
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Then, by the triangle sum of <math>\triangle ABC</math> and <math>\triangle BPC</math>, we have:
 
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<math>\angle BAC +2x=180°</math> and <math>\angle BPC +x=180°</math>
 
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<math>\Rightarrow \angle BPC = 90°+\frac{\angle BAC}{2} </math>
 
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Therefore, since <math>\angle BAC</math> is fixed and looks at fixed segment <math>BC</math>, <math>P</math> is contained within a circle <math>\pi</math> that passes through <math>B</math>, <math>C</math> and <math>I</math> (since we can quickly access that it satisfies <math>\angle BIC = 90°+\frac{\angle BAC}{2} </math>).
 
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Hence, to prove <math>AP \geq AI</math> it suffices to show that <math>AI</math> meets the center <math>O</math> of circle <math>\pi</math>, since that would directly imply that <math>I</math> is the closest point to <math>A</math> on the circle.
 
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<cmath>\angle BOC=2(180°-\angle BPC)=180°-\angle BAC</cmath>
 
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It follows that <math>O</math> is contained within the circumcircle of <math>\triangle ABC</math>.But since <math>O</math> is the center of circle <math>\pi</math>, <math>OB=OC=Radius</math>, meaning <math>O</math> sits at the middle point of arc <math>BC</math> of the circumcircle, therefore proving that it is contained in the angle bissector of <math>A</math> <math>\boxed{}</math>.
 
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By Pietro Leão Baruffato(a.k.a Spacephysics) :P
 
  
 
==See Also==
 
==See Also==
  
 
{{IMO box|year=2006|before=First Problem|num-a=2}}
 
{{IMO box|year=2006|before=First Problem|num-a=2}}

Latest revision as of 19:04, 20 May 2025

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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