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

(Solution)
(Solution)
Line 7: Line 7:
 
:<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{article}  
+
:<math>\documentclass{imosol}  
 
\usepackage[pdftex]{graphicx}  
 
\usepackage[pdftex]{graphicx}  
\begin{document}  
+
\begin{imosol}  
 
\includegraphics{imosol.png}  
 
\includegraphics{imosol.png}  
\end{document}</math>
+
\end{imosol}</math>

Revision as of 11:04, 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{imosol}

\usepackage[pdftex]{graphicx} Unknown environment 'imosol'$ (Error compiling LaTeX. Unknown error_msg)