Difference between revisions of "2002 IMO Problems/Problem 2"

(Problem)
(Problem)
Line 3: Line 3:
 
:<math>\text{EF is the chord which is the perpendicular bisector of AO. D is the midpoint of the minor arc AB. The line through}</math>
 
:<math>\text{EF is the chord which is the perpendicular bisector of AO. D is the midpoint of the minor arc AB. The line through}</math>
 
:<math>\text{O parallel to AD meets AC at J. Show that J is the incenter of triangle CEF.}</math>
 
:<math>\text{O parallel to AD meets AC at J. Show that J is the incenter of triangle CEF.}</math>
 +
:
 +
==Solution==
 +
:<math>\text{By construction, AEOF is a rhombus with } 60^\circ - 120^\circ \text{angles}</math>
 +
:<math>\text{ Consequently, we may set } s = AO = AE = AF = EO = EF</math>

Revision as of 10:27, 7 October 2022

Problem

$\text{BC is a diameter of a circle center O. A is any point on the circle with } \angle AOC \not\le 60^\circ$
$\text{EF is the chord which is the perpendicular bisector of AO. D is the midpoint of the minor arc AB. The line through}$
$\text{O parallel to AD meets AC at J. Show that J is the incenter of triangle CEF.}$

Solution

$\text{By construction, AEOF is a rhombus with } 60^\circ - 120^\circ \text{angles}$
$\text{ Consequently, we may set } s = AO = AE = AF = EO = EF$