Circumcircles and intersections madness

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Circumcircles and intersections madness

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Source: Germany VAIMO 2015 - #3

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986 posts

Kezer

#1 h Jul 11, 2015, 11:05 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be an acute triangle with $|AB| \neq |AC|$ and the midpoints of segments $[AB]$ and $[AC]$ be $D$ resp. $E$ . The circumcircles of the triangles $BCD$ and $BCE$ intersect the circumcircle of triangle $ADE$ in $P$ resp. $Q$ with $P \neq D$ and $Q \neq E$ .
Prove $|AP|=|AQ|$ .

(Notation: $|\cdot|$ denotes the length of a segment and $[\cdot]$ denotes the line segment.)

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#2 h Jul 11, 2015, 2:58 PM • 3 Y

Y by Kezer, AlastorMoody, Adventure10

It is the Problem6 in 2014 China Girls Math Olympiad，here is the link:http://artofproblemsolving.com/community/c6h603726p3585065

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#3 h Jul 24, 2016, 9:53 PM • 3 Y

Y by AlastorMoody, Adventure10, Mango247

\documentclass{article}
\usepackage {amsmath}
% Uncomment the following line to allow the usage of graphics (.png, .jpg)
%\usepackage[pdftex]{graphicx}
% Comment the following line to NOT allow the usage of umlauts
\usepackage[utf8]{inputenc}

% Start the document
\begin{document}
\begin {align}
% Create a new 1st level heading
\section{Main Heading}
First of all we easily see that the lines EQ,PD,BC are congruent because they are the radical axises of the circumcircles of \bigtriangleup ADE, \bigtriangleup BCD and \bigtriangleup BEC.
Let PD \cap BC \cap EQ=O.
We have to prove that AP=AQ
\leftrightarrow \angle ADP=\angle BDQ,but \angle ADE=\angle AEP and \angle BQD=\angle APE= 180-\angle ABC.So it'senough to prove that \bigtriangle APE \sim \bigtriangleup BQD ,so we must prove \frac{AP}{PE}=\frac{BQ}{QD}
From similar triangle we get BQ=\frac{BO.EC}{EO} and DQ=\frac{OD.PE}{EO}.Therefore we shall prove OD.AP=OB.EC
\Leftrightarrow \frac{OB}{OD}=\frac{AP}{EC} but from similar triangles \frac{OB}{OD}=\frac{OP}{OC}.So we must prove OP.EC=OC.AP
but from DE||BC and the cyclic quadraterial ADEP we get \angle APO=\angle ACO.Therefore AOCP is cyclic and moreover \angle EPC=\angle APO=\angle ABC
PO is symmedian in \bigtriangleup APC and APCO is a harmonic quadraterial.Therefore AO.PC=AP.CO.Finally we apply the Ptolemys theorem and we get OP.EC=OC.AP.Quite a nice problem.

% Uncomment the following two lines if you want to have a bibliography
%\bibliographystyle{alpha}
%\bibliography{document}

\end{document}

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