Two circles $O_1$ and $O_2$ intersect at $M,N$. The common tangent line nearer to $M$ of the two circles touches $O_1,O_2$ at $A,B$ respectively. Let $C,D$ be the symmetric points of $A,B$ with respect to $M$ respectively. The circumcircle of triangle $DCM$ intersects circles $O_1$ and $O_2$ at points $E,F$ respectively which are distinct from $M$. Prove that the circumradii of the triangles $MEF$ and $NEF$ are equal.
Problem
Source: Italy TST 2009 p5
Tags: geometry, circumcircle, trapezoid, geometric transformation, reflection, power of a point, radical axis
WakeUp
10.03.2012 14:21
Clearly the result is equivalent to proving $\angle EMF+\angle ENF=180^{\circ}$. Let $MN$ meet $AB$ at $L$. Then $MN$ is the radical axis of $O_1$ and $O_2$ so $L$ has equal power with respect to $O_1$ and $O_2$ i.e. $LA^2=LB^2$ (since $AB$ is a common tangent) hence $L$ is the midpoint of $AB$. Now clearly $ABCD$ is a parallelogram. Since $M$ is the midpoint of $BD$, and $L$ is the midpoint of $AB$ it follows that $ML||AD||BC$. Next we show $B,C,F$ are collinear. Note that $\angle CFM=\angle CDM=\angle ABD=\angle ABM=\angle BFM$. It follows $B,C,F$ are collinear and hence so are $A,D,E$ by a symmetric argument. But now $AMNE$ and $BMNF$ are cyclic trapeziums, and hence they must be isosceles cyclic trapeziums. Then $\angle MNE=\angle NMA$ and $\angle MNP=\angle NMB$. Therefore $\angle AMB=\angle ENF$. So $\angle EMF+\angle ENF=180^{\circ}\iff \angle EMF=\angle BMC$. We will in fact prove $\triangle MBC$ is similar to $\triangle MEF$ which will prove the result. Indeed, $\angle MBC=\angle ADB=\angle MFE$ and of course $\angle MCB=\angle MEF$, so $\triangle MBC$ is similar to $\triangle MEF$.
Attachments:
Potla
08.06.2012 13:54
Claim.
We claim that $EF$ is the common tangent to the two circles.
Proof.
Let $E$ and $F$ be the points of tangency of the other common tangent to the two circles. Then, we note that $MN\cap AB=X\implies AX=BX,$ because $X$ lies on the radical axis of these two circles. Then, $MX\parallel AD,$ leading to $AD\perp O_1O_2.$ Since $A$ and $E$ are reflections wrt $O_1O_2,$ so we get $A,D,E$ are collinear. So are $B,C,F.$
Now, note that $\angle DCA=\angle CAB=\angle DEM,$ so that $E\in\odot(MCD),$ and our claim is proved.$\Box$
Again, observe that $\angle ENF=\pi-\angle NEF-\angle NFE=\pi-\angle NME-\angle NMF=\pi-\angle EMF.$ Now, let reflections of $M$ and $N$ wrt $EF$ be $M',N'.$ Note that $\angle ENF+\angle EMF=\pi$ leads to the fact that $M'ENF$ and $MEN'F$ are cyclic. Therefore, we can safely say that the circles $\odot(M'ENF)$ and $\odot(MEN'F)$ are reflections of each other wrt $EF.$ So, these two circles have the same radii.
Mahdi_Mashayekhi
08.01.2022 10:45
Note that if radius of $MEF$ and $NEF$ are equal then Reflection of $N$ must lie an $MEF$ so we need to prove $\angle EMF + \angle ENG = \angle 180$. we will prove $EF$ is common tangent of our two circles. Let $E,F$ be tangency points of common tangent we will prove $E,F$ lie on $MCD$. Step1 : $A,D,E$ and $B,C,F$ are collinear. $MN$ bisects both $AB$ and $CD$ so $MN || AD || BC$ and we have $MN || AE || BF$ so $A,D,E$ and $B,C,F$ are collinear. Step2 : $E,F$ lie on $MCD$. $\angle MED = \angle MAB = \angle MCD \implies MDEC$ is cyclic. $\angle MFC = \angle MBA = \angle MDC \implies MCFD$ is cyclic. so $MCFED$ is cyclic as wanted. Now that we have $EF$ is tangent to circles let's prove our main goal : $\angle MEF + \angle NEF = \angle 180$. $\angle EMF + \angle ENF = \angle FEN + \angle EFN + \angle EFN = \angle 180$. we're Done.
Mahdi_Mashayekhi
08.01.2022 11:07
Another way to show $EF$ is common tangent. Let $O_1$ and $O_2$ be circles of two circles. $\angle MED = \angle DCM = \angle MAB = \angle MEA \implies E,D,A$ are collinear. $\angle MFC = \angle MDC = \angle MBA = \angle MFB \implies F,C,B$ are collinear. Let $E'$ and $F'$ be tangency points of other common tangent we have $AE' \perp O_1O_2$ and $BF' \perp O_1O_2$ and $MN \perp O_1O_2$ so $AE' || MN || BF'$. But $MN$ bisects $AB$ and $CD$ so $AE || MN || BF$ so $E$ is $E'$ and $F$ is $F'$.