Let $\ell$ be a line, and let $\gamma$ and $\gamma'$ be two circles. The line $\ell$ meets $\gamma$ at points $A$ and $B$, and $\gamma'$ at points $A'$ and $B'$. The tangents to $\gamma$ at $A$ and $B$ meet at point $C$, and the tangents to $\gamma'$ at $A'$ and $B'$ meet at point $C'$. The lines $\ell$ and $CC'$ meet at point $P$. Let $\lambda$ be a variable line through $P$ and let $X$ be one of the points where $\lambda$ meets $\gamma$, and $X'$ be one of the points where $\lambda$ meets $\gamma'$. Prove that the point of intersection of the lines $CX$ and $C'X'$ lies on a fixed circle. Gazeta Matematica
Problem
Source: Romania TST 5 2010, Problem 2
Tags: ratio, geometry proposed, geometry
Luis González
25.08.2012 23:54
Let $Q \equiv CX \cap C'X'.$ $CX$ cuts $\ell$ at $M$ and $\gamma$ again at $E.$ $C'X'$ cuts $\ell$ at $N$ and $\gamma'$ again at $F.$ Since $\ell$ is polar of $C$ and $C'$ WRT $\gamma$ and $\gamma',$ then the pencils $P(C,M,E,X)$ and $P(C',N,F,X')$ are harmonic $\Longrightarrow$ $P,E,F$ are collinear. Hence, by Menelaus' theorem for $\triangle QCC'$ cut by $XX'$ and $EF,$ we get $\frac{QX}{QX'}=\frac{PC'}{PC} \cdot \frac{CX}{C'X'} \ , \ \frac{QE}{QF}=\frac{PC'}{PC} \cdot \frac{CE}{C'F} \Longrightarrow$ $\frac{QX \cdot QE}{QX' \cdot QF}=\left (\frac{PC'}{PC} \right)^2 \cdot \frac{CX \cdot CE}{C'X' \cdot C'F}= \left ( \frac{PC' \cdot CA}{PC \cdot C'A'} \right)^2=\text{const}.$ Thus, the ratio of the powers of $Q$ WRT $\gamma$ and $\gamma'$ is constant $\Longrightarrow$ locus of $Q$ is a circle $\omega$ coxal with $\gamma$ and $\gamma'.$ This circle $\omega$ then passes through the intersections $CA \cap C'A',$ $CB \cap C'B',$ $CA \cap C'B'$ and $CB \cap C'A'.$
simplependulum
27.08.2012 05:55
Quote: Thus, the ratio of the powers of $Q$ WRT $\gamma$ and $\gamma'$ is constant $\Longrightarrow$ locus of $Q$ is a circle $\omega$ coxal with $\gamma$ and $\gamma'.$ I have learnt this theorem very recently but is there any name given to it?