Let $\Gamma_1$ and $\Gamma_2$ be two non-overlapping circles. $A,C$ are on $\Gamma_1$ and $B,D$ are on $\Gamma_2$ such that $AB$ is an external common tangent to the two circles, and $CD$ is an internal common tangent to the two circles. $AC$ and $BD$ meet at $E$. $F$ is a point on $\Gamma_1$, the tangent line to $\Gamma_1$ at $F$ meets the perpendicular bisector of $EF$ at $M$. $MG$ is a line tangent to $\Gamma_2$ at $G$. Prove that $MF=MG$.
Problem
Source: CGMO 2015 Q6
Tags: geometry, perpendicular bisector, power of a point, radical axis
TelvCohl
20.08.2015 17:01
Let $ X, $ $ Y $ be the midpoint of $ AB, $ $ CD $, respectively. Obviously, $ AC $ is perpendicular to $ BD $, so we get $ {XA}^2 $ $ = $ $ {XB}^2 $ $ = $ $ {XE}^2 $ and $ {YC}^2 $ $ = $ $ {YD}^2 $ $ = $ $ {YE}^2 $ $ \Longrightarrow $ $ \Gamma_1, $ $ \Gamma_2 $ and the degenerate circle $ E $ are coaxial with common radical axis $ XY $ $ \equiv $ $ \tau $, hence from $ {ME} ^2 $ $ = $ $ {MF}^2 $ $ \Longrightarrow $ $ M $ lie on $ \tau $ $ \Longrightarrow $ $ {MF}^2 $ $ = $ $ {MG}^2 $ $ \Longrightarrow $ $ MF $ $ = $ $ MG $.
pi37
21.08.2015 00:39
We claim that an inversion about $E$ maps $\Gamma_1$ and $\Gamma_2$ to concentric circles. It is clear that this will show every circle orthogonal to $\Gamma_1$ passing through $E$ will also be orthogonal to $\Gamma_2$. Let $E'$ be the center of this inversion; we aim to show $E'$ lies on $AC$ and $BD$. After the inversion (for convenience we will omit primes), the circles map to a circle with common center $O$. Line $AB$ maps to some circle $\omega_1$ externally tangent to $\Gamma_1$ and internally tangent to $\Gamma_2$, while $CD$ maps to a circle $\omega_2$ internally tangent to both. Suppose (after inversion) $CD$ intersects $\Gamma_1$ again at $X$. Then $AX\parallel CD$ and $CX$ is a diameter of $\Gamma_1$. Then $AC\perp BD$, so circles $\omega_1$ and $\omega_2$ with diameters $AB$ and $CD$ intersect at $E'$, which is also the intersection of $AC$ and $BD$, as desired.
AlastorMoody
12.04.2020 18:10
China Girls MO 2015 P6 wrote:
Let $\Gamma_1$ and $\Gamma_2$ be two non-overlapping circles. $A,C$ are on $\Gamma_1$ and $B,D$ are on $\Gamma_2$ such that $AB$ is an external common tangent to the two circles, and $CD$ is an internal common tangent to the two circles. $AC$ and $BD$ meet at $E$. $F$ is a point on $\Gamma_1$, the tangent line to $\Gamma_1$ at $F$ meets the perpendicular bisector of $EF$ at $M$. $MG$ is a line tangent to $\Gamma_2$ at $G$. Prove that $MF=MG$.
By Converse of Reim's, $\odot (AO_1C)$ $\cap$ $\odot (BO_2D)$ $\in$ $AB,O_1O_2$. Let them be $K,L$. Let $E'$ $=$ $O_1O_2$ $\cap$ $AC$. Since, $BD$ $\perp$ $AC$
\begin{align*}\angle CE'O_2=\angle LO_2K=\angle CDL \end{align*}Thus, $CE'DL$ is cyclic $\implies$ $DE'||KO_1$ $\implies$ $E'$ $\in$ $DB$. Thus, $E \equiv E'$. By Definition $M$ lies on Radical Axes of $\odot (O_1)$ and $\odot (E)$. By La Hire's Theorem, $E$ and $L$ are Inverses of each other under $\odot (O_1)$ and $\odot (O_2)$ $\implies$ the perpendicular bisector of $\overline{EL}$ is the radical axis of $\odot (E)$ and $\odot (O_2)$ & $\odot (E)$ and $\odot (O_1)$ $\implies$ $M$ lies on radical axis $\odot (O_1)$ and $\odot (O_2)$ which implies the desired conclusion $\qquad \blacksquare$
Mogmog8
30.04.2022 20:39
Notice $$\measuredangle CED=\measuredangle CDE+\measuredangle ECD=\measuredangle EBA+\measuredangle BAE=\measuredangle BEA=-\measuredangle CED,$$so $\measuredangle CED=90.$ Letting $P$ and $Q$ be the midpoints of $\overline{AB}$ and $\overline{CD},$ respectively, we have $$QC^2=QD^2=QE^2.$$Hence, $Q$ lies on the radical axis of $\omega_E$ and $\Gamma_1,$ where $\omega_E$ is the circle with center $E$ and radius zero. Similarly, $P$ lies on the radical axis of $\Gamma_1$ and $\omega_E$ and $M$ lies on $\overline{PQ}$ since $ME^2=MF^2.$ But $\overline{PQ}$ is the radical axis of $\Gamma_1$ and $\Gamma_2$ so $MF^2=MG^2.$ $\square$ Remarks: (Motivation) The desired result made me suspect $M$ was on the radical axis. I first constructed $P,Q$ in order to draw the radical axis, and the $\omega_E$ was motivated by the radical axes and $ME=MF.$