2006 IMO Problems/Problem 1: Difference between revisions
| Line 4: | Line 4: | ||
==Solution== | ==Solution== | ||
We have | We have | ||
<cmath>\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle | <cmath>\angle IBP = \angle IBC - \angle PBC = \frac{1}{2} \angle ABC - \angle PBC = \frac{1}{2}(\angle PCB - \angle PCA)</cmath> (1) | ||
and similarly <cmath>\angle ICP = \angle PCB - \angle ICB = \angle PCB - \frac{1}{2} \angle ACB = \frac{1}{2}(\angle | 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 - | Since <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>, we have <math>\angle PBA - | ||
\angle PBC = \angle PCB - \angle PCA</math> (3). | \angle PBC = \angle PCB - \angle PCA</math> (3). | ||
Revision as of 06:31, 28 December 2017
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
(1)
and similarly
(2).
Since
, we have
(3).
By (1), (2), and (3), we get
; hence
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 :)