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> |
− | and similarly <cmath>\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC)</cmath> | + | 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> | + | 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 <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. | |
− | 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 |
Revision as of 18:50, 30 May 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