Let $D,E,F$ be the feet of the altitudes on $BC,CA,AB.$ $M$ is the midpoint of $BC$ and $EF,FD,DE,$ cut $BC,CA,AB$ at $P,Q,R.$ Since $(B,C,D,P)=-1$ $\Longrightarrow$ $PB \cdot PC=PM \cdot PD$ $\Longrightarrow$ $P$ has equal power WRT $(O_1)$ and 9-point circle $\odot(DEF) \equiv (O_2)$ and similarly $Q$ and $R$ $\Longrightarrow$ $\overline{PQR}$ is radical axis of $(O_1),(O_2).$ Since $PH$ is the polar of $A$ WRT the circle $(M)$ with diameter $BC,$ then $PH$ is perpendicular to $AM$ at $T,$ the 2nd intersection of $\odot(AEHF)$ and $(O_3).$ Hence $PH \cdot PT=PE \cdot PF$ $\Longrightarrow$ $P$ has equal power WRT $(O_2)$ and $(O_3)$ and similarly $Q,R$ $\Longrightarrow$ $\overline{PQR}$ is radical axis of $(O_2),(O_3).$ The conclusion follows.

In general, given two circles $ \odot (O_1, R_1) $ and $ \odot (O_2, R_2) $ . Let $ P, Q $ be the insimilicenter, exsimilicenter of $ \odot (O_1) \sim \odot (O_2) $, respectively . Then $ \odot (O_1), \odot (O_2), \odot (O_3) \equiv \odot (PQ) $ are coaxial . ____________________________________________________________ Proof : Since $ \odot (O_3) $ is the locus of the point $ T $ such that $ TO_1:TO_2=R_1:R_2 $ , so for any point $ T $ on $ \odot (O_3) $ we have $ \frac{{TO_1}^2-{R_1}^2}{{TO_2}^2-{R_2}^2}=\left ( \frac{R_1}{R_2} \right )^2=\text{Const} \Longrightarrow \odot (O_1), \odot (O_2), \odot (O_3) $ are coaxial . Done

Does this work? Note that the centres of the circles are collinear on some line $l$. Then, their axis' are perpendicular to $l$ and they are concurrent by the Radical Axis Theorem. Thus, they must coincide i.e. They have the same axis.

Does the solution above work? It seems a little short for the problem...

MathPanda1 wrote: Does the solution above work? It seems a little short for the problem... All you did was prove that the radical axii are parallel

aops wrote: Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center. See http://www.artofproblemsolving.com/wiki/index.php/Radical_axis Is parallel radical axii a possibility?

MathPanda1 wrote: aops wrote: Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center. See http://www.artofproblemsolving.com/wiki/index.php/Radical_axis Is parallel radical axii a possibility? Yes, but just because they're parallel doesn't mean that they're the same line...

Well, don't they have to meet? The only way parallel lines meet is if they coincide. Unless point of infinity exists...

MathPanda1 wrote: Well, don't they have to meet? The only way parallel lines meet is if they coincide. Unless point of infinity exists... Parallel radical axii coincide at the point at infinity; a valid way to do this problem would be to show that there exists some point with equal power wrt to all three circles, which would force the conclusion.

Thank you for pointing out the errors in my proof, EulerMacaroni. I really appreciate it!

here's my solution let $D,E,F$ be the feet of altitudes and define $X_a=EF \cap BC$ and so define $X_b,X_c$ I'll claim that $X_aX_bX_c$ is the radical axis of $O_1,O_2,O_3$ claim(1): $X_a$ is on the radical axis of $O_1,O_2$ proof: easily $X_aE.X_aF=X_aB.X_aC$ $\blacksquare$ claim(2): $X_a$ is on the radical axis of $O_1,O_3$ proof: we have that $A_{HM}$ point , $H$and $X_a$ are collinear (since the image of $X_a$ is the $A_{HM}$ point after inversion around $H$ with radius $-\sqrt{AH.HD}$ ) and of course the $A_{HM}$ point is on $O_3$ so $X_aA_{HM}.X_aH=X_aB.X_aC$ since $HBCA_{HM}$ is cyclic $\blacksquare$ and we win

Clearly centers of three circles are lies on Euler line ,so it suffice to find a point $P$ with equal power with respect to all three circles. ( Then the line perpendicular to Euler line and passing from $P$ would be their radical axis. ) Take $P$ on Euler line such that $H$ be between $P$ and $O$ and $PH = x$. Let $l=HO$ , $N =$ center of nine point circle and $K$ be the center of $O_3$. we want to choose x such that these three numbers are equal : \[ (I): P^P_{O_1} = PO^2 - R^2 = (x+l)^2 - R^2 = x^2 + 2xl + l^2 - R^2 \]\[ (II): P^P_{O_2} = PN^2 - \frac{R^2}{4} = (x + \frac{l}{2})^2 - \frac{R^2}{4} = x^2 + xl + \frac{l^2}{4} - \frac{R^2}{4}\]\[ (III): P^P_{O_3} = PK^2 - KH^2 = (x+ \frac{l}{3})^2 - \frac{l^2}{9} = x^2 + \frac{2l}{3}x\]Easily you could check that $x= \frac{\frac{3}{4}(R^2-l^2)}{l}$ works.

Let $D$ be the midpoint of $BC$ and $H_A$ be the $A$-Humpty point. Then \[ \frac{\text{Pow}(D,O_1)}{\text{Pow}(D,O_3)} = \frac{ - DB^2}{DH_A \cdot DG} = \frac{-3 DB^2}{DH_A \cdot DA} = -3 \]hence we finish by coaxiality lemma.