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

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:<math>\text{By construction, AEOF is a rhombus with } 60^\circ - 120^\circ \text{angles}</math>
 
:<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>
 
:<math>\text{ Consequently, we may set } s = AO = AE = AF = EO = EF</math>
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Revision as of 12:11, 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

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