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

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==Problem==
 
==Problem==
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:<math>\text{BC is a diameter of a circle center O. A is any point on the circle with } \angle AOC \not\le 60^\circ</math>
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<math>BC</math> is a diameter of a circle center <math>O</math>. <math>A</math> is any point on the circle with <math>\angle AOC \not\le 60^\circ</math>. <math>EF</math> is the chord which is the perpendicular bisector of <math>AO</math>. <math>D</math> is the midpoint of the minor arc <math>AB</math>. The line through <math>O</math> parallel to <math>AD</math> meets <math>AC</math> at <math>J</math>. Show that <math>J</math> is the incenter of triangle <math>CEF</math>.
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:<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>
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:<math>\text{O parallel to AD meets AC at J. Show that J is the incenter of triangle CEF.}</math>
 
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==Solution==
 
==Solution==
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:<math>\text{By construction, AEOF is a rhombus with } 60^\circ - 120^\circ \text{angles}</math>
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{{solution}}
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:<math>\text{ Consequently, we may set } s = AO = AE = AF = EO = EF</math>
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:<math>\documentclass{article} \usepackage[pdftex]{graphicx} \begin{document} \begin{center} \includegraphics{myimage.png} \end{center} \end{document}}</math>
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==See Also==
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{{IMO box|year=2002|num-b=1|num-a=3}}

Latest revision as of 08:32, 5 July 2024

Problem

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

Solution

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See Also

2002 IMO (Problems) • Resources
Preceded by
Problem 1 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 3
All IMO Problems and Solutions
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