Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
m
Line 11: Line 11:
 
By (1), (2), and (3), we get <math>\angle IBP = \angle ICP</math>; hence <math>B,I,P,C</math> are concyclic.
 
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\phantom{0}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>.  
  
 
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.

Revision as of 04:06, 23 December 2022

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 \[\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA)\] (1) and similarly \[\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle PBA - \angle PBC)\] (2). Since $\angle PBA + \angle PCA = \angle PBC + \angle PCB$, we have $\angle PBA - \angle PBC = \angle PCB - \angle PCA$ (3).

By (1), (2), and (3), we get $\angle IBP = \angle ICP$; hence $B,I,P,C$ 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 :)

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_IMO_Problems/Problem_1&oldid=184509"