Collinearity and points on the circumcenter

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Collinearity and points on the circumcenter

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#1 h Jul 27, 2011, 6:33 AM • 2 Y

Y by Adventure10, Mango247

Let $M$ be a point on the cicumcircle of $\triangle ABC$ . Given a point $R$ , denote by $A_1,B_1,C_1$ the intersection points of the lines $AR,BR,CR$ with the circumcircle. Prove that the intersection points of the lines $MA_1$ and $BC$ , $MB_1$ and $CA$ , $MC_1$ and $AB$ are collinear.

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#2 h Jul 27, 2011, 2:44 PM • 2 Y

Y by Adventure10, Mango247

This is a particular case of the following configuration (discussed before):

Proposition. $P,Q$ are two arbitrary points in the plane of $\triangle ABC$ with circumcircle $(O).$ $PA,$ $PB,$ $PC$ cut $(O)$ again at $P_1,P_2,P_3$ and $QA,$ $QB,$ $QC$ cut $(O)$ again at $Q_1,Q_2,Q_3.$ $P_1Q_1,$ $P_2Q_2,$ $P_3Q_3$ cut $BC,CA,AB$ at $A_1,B_1,C_1.$ Then $A_1,B_1,C_1$ are collinear.

$\frac{A_1B}{A_1C}=\frac{BP_1}{CP_1} \cdot \frac{\sin \widehat{A_1P_1B}}{\sin \widehat{A_1P_1C}}=\frac{ \sin \widehat{BAP_1}}{\sin \widehat{CAP_1}} \cdot \frac{ \sin \widehat{BAQ_1}}{\sin \widehat{CAQ_1}}$

Multiplying cyclic expressions together, keeping in mind that

$\frac{ \sin \widehat{BAP_1}}{\sin \widehat{CAP_1}} \cdot \frac{ \sin \widehat{CBP_2}}{\sin \widehat{ABP_2}} \cdot \frac{ \sin \widehat{ACP_3}}{\sin \widehat{BCP_3}}=1 \ , \ \frac{ \sin \widehat{BAQ_1}}{\sin \widehat{CAQ_1}} \cdot \frac{ \sin \widehat{CBQ_2}}{\sin \widehat{ABQ_2}} \cdot \frac{ \sin \widehat{ACQ_3}}{\sin \widehat{BCQ_3}}=1$

We get $\frac{A_1B}{A_1C} \cdot \frac{B_1C}{B_1A} \cdot \frac{C_1A}{C_1B}=1$ $\Longrightarrow$ $A_1,B_1,C_1$ are collinear.

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#3 h Jul 1, 2012, 4:04 AM • 2 Y

Y by Adventure10, Mango247

I am not sure if it is correct, as I managed to come up with extremely simple solution.
Let the intersections $A_2,B_2,C_2$ . Pascal's theorem wrt $B_1AC_1BMC$ implies $B_2,C_2,P$ collinear. Similarly $A_2,C_2,P$ collinear. Thus we get the conclusion.

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jayme

#4 h Jul 1, 2012, 7:16 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
for the proof and more, see also

http://perso.orange.fr/jl.ayme vol. 3 La P-transversale de Q

Sincerely
Jean-Louis

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