Difference between revisions of "2006 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
We have | We have | ||
− | <cmath>\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA)</cmath> | + | <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> | |
+ | Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PCB - \angle PCA = \angle PBA - \angle PBC.</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>. | + | 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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+ | 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. | 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 | ||
latexed by tluo5458 :) | latexed by tluo5458 :) | ||
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+ | 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 be triangle with incenter . A point in the interior of the triangle satisfies . Show that , and that equality holds if and only if
Solution
We have
and similarly Since , we have
It follows that Hence, and are concyclic.
Let ray meet the circumcircle of at point . Then, by the Incenter-Excenter Lemma, .
Finally, (since triangle APJ can be degenerate, which happens only when ), but ; hence 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 |