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

(Solution)
(Solution)
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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>
:<math>\documentclass{imosol}</math>
+
:\documentclass{article}
\usepackage[pdftex]{graphicx}  
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\usepackage[pdftex]{graphicx}
\begin{imosol}  
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\begin{document}
\includegraphics{imosol.png}  
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\end{imosol}
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This is my first image.
 +
 
 +
\includegraphics{myimage.png}
 +
 
 +
That's a cool picture up above.
 +
\end{document}

Revision as of 11:07, 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

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

\usepackage[pdftex]{graphicx} \begin{document}

This is my first image.

\includegraphics{myimage.png}

That's a cool picture up above. \end{document}