What is the proof that this holds for any isogonal conjugates o and h?

I think it's the same using similar triangles $ABX$ and $ANC$ and similar triangles $ABY$ and $AMC$. I found an inversive solution that kinda overcomplicates everything as well.

It is not difficult to show that $BX=XC$ and let $I$ be the midpoint of $BC$, $L$ is the intersection of $AX$ and $BC$. Since $XBMC$ is cyclic so $\angle BMX= \angle BCX= \angle XBC= \angle XMC= \angle BAC.$ From here we can obtain that $\triangle XBL \sim \triangle XMB \; ( \text{A.A})$ then $\frac{BL}{BM}= \frac{BX}{XM} \qquad (1)$. We also have $\angle BMX= \angle MBA+ \angle MAB, \angle BAC= \angle MAB+ \angle MAC$ and $\angle BMX= \angle BAC$ so $\angle MAC= \angle MBA$. From this we obtain $\triangle AMB \sim \triangle CMA \; ( \text{A.A})$. Hence $\frac{BM}{AM}= \frac{AB}{AC} \qquad (2)$. From $(1)$ and $(2)$ we get $\frac{AM}{MX}= \frac{AC}{AB} \cdot \frac{BL}{BX}$. Similarly, we also obtain $\frac{AM}{MX}= \frac{AB}{AC} \cdot \frac{CL}{XC}$. This follows $\frac{AB^2}{AC^2}= \frac{CX}{BX} \cdot \frac{BL}{CL}= \frac{BL}{CL}$ or $BL= \frac{AB^2 \cdot BC}{AB^2+AC^2}$. Thus, \[\frac{AM}{MX}= \frac{AC}{AB} \cdot \frac{BL}{BX}= \frac{AB \cdot BC \cdot CA}{BX \cdot (AB^2+AC^2)}.\] Note that $\triangle XBI \sim \triangle OBI \; ( \text{A.A})$ so $BX= \frac{BO \cdot BC}{2OI}= \frac{BO \cdot BC}{AH}$, hence $\frac{AM}{MX}= \frac{AH \cdot AB \cdot CA}{BO \cdot (AB^2+AC^2)}$. It is not difficult to prove that $HY$ is the diameter of $\omega_2$. Therefore $BH \perp BY$. Hence $BY \parallel AC$ since $BH \perp AC$. Similarly, we also have $CY \parallel AB$. We obtain $ACYB$ is a parallelogram which means $I$ is the midpoint of $AY$. Since $BNCY$ is cyclic so $\angle NCI= \angle NYB$ and since $AC \parallel BY$ so $\angle NYB= \angle YAC$. Thus, $\angle YAC= \angle NCI$. This follows $\triangle ICN \sim \triangle IAC \; ( \text{A.A})$ then $IN= \frac{BI^2}{AI}= \frac{BC^2}{4AI}$. We have \[\frac{AN}{NY}= \frac{AI-IN}{AI+IN}= \frac{(2AI)^2-BC^2}{(2AI)^2+BC^2}= \frac{AB^2+AC^2-BC^2}{AB^2+AC^2}.\] We need to prove $MN \parallel XY$. This remain to show that $\frac{AN}{NY}= \frac{AM}{MX}$ or \[\frac{BO}{2OI}=\frac{BO}{AH}= \frac{bc}{b^2+c^2-a^2}.\] We have $\frac{BO}{2OI} = \frac{AB}{2AP}$ with $AP$ is the altitude of $\triangle ABC$ since $\triangle BOI \sim \triangle BAP \; ( \text{A.A})$. Hence, we just need to prove $2AC \cdot AP= c^2+b^2-a^2$, which is obviously right since by Power of a Point, $2AC \cdot AP= 2 (AI^2-BI^2)= c^2+b^2-a^2$. Thus, $\frac{AM}{MX}= \frac{AN}{NY}$ or $MN \parallel XY$. $\blacksquare$

ABCDE wrote: I think it's the same using similar triangles $ABX$ and $ANC$ and similar triangles $ABY$ and $AMC$. I found an inversive solution that kinda overcomplicates everything as well. May be if we prove $\triangle YCN \sim \triangle ALC$ then we can prove $\triangle ABY \sim \triangle AMC$. $L$ in here is the intersection of $AX$ and $BC$.

Here is a less messy solution: Note that since $\angle OMX = \angle HNY = 90$, $OX$ and $HY$ are diameters of $\omega_1 , \omega_2$, respectively. Thus, $X$ is the intersection of the tangents off of $B$ and $C$ of the circumcircle of $ABC$. By well known properties of symmedians, $AX$ is a symmedian. In addition, if we call the midpoint of $BC$ to be $P$, $P$ is the center of a homothety from circumcircle $ABC$ and $\omega_2$ with ratio 1 to 1. We also know that the reflection of $H$ over $P$ is the (antipode?) of $ABC$ with respect to $A$. Thus, since $HY$ is also the diameter of $\omega_2$, we get that $Y$ is the image of $A$ under the homothety and $A$, $P$, and $Y$ are collinear, giving that $AY$ is the median. Now we just angle chase given that both $AX$, $AY$ and $AH$, $AO$ are isogonal. This eventually gives $OAM$ similar to $HAN$ as well as $AOX$ similar to $AHY$, or that $\frac{AM}{AN} = \frac {AX}{AY}$

Generalization: Let $P_1,P_2$ be isogonal conjugates wrt $\triangle ABC$. Point $Q_1$ is on the circumcircle of $BCP_1$ such that $P_1,Q_1$ are diametrically opposite, $Q_2$ is contructed similarly. Prove that $Q_1,Q_2$ are also isogonal conjugates wrt $\triangle ABC$. The Proof the straight forward angle chasing, note that $BQ_1\perp BP_1, BQ_2\perp BP_2$, hence $\angle (AB,BQ_2)=180-\angle (BQ_2,BC)-\angle B=$ $90-(\angle B-\angle (BC,BP_2))=90-\angle (BC,BP_1)=\angle (BQ_1,BC)$, and $BQ_1,BQ_2$ are isogonals wrt $\angle B$. Analogously $CQ_1,CQ_2$ are also isogonals and the result follows. Furthermore we can prove $\triangle AP_1Q_2\sim \triangle AP_2Q_2$(This is the key to the original problem), we've already proven $\angle P_1AQ_1=\angle P_2AQ_2$, it suffices to prove $\angle AP_1Q_1=\angle AP_2Q_2$ which is just more direct angle chasing.

Claim 1:$AX$ is the $A$-symmedian. Proof of Claim 1:We show that $X$ is the midpoint of the arc $BC$ of $(OBC)$ (other than $O$). The rest is a well-known symmedian property. Indeed, \[ \angle{XBC} = \angle{XMC} \]\[= \angle{OMC} - 90^{\circ} = 90^{\circ}- \angle{OBC} = 90^{\circ} - \angle{OCB} \]\[= 90^{\circ}- \angle{OMB} = \angle{BMX} = \angle{BCX} \] Claim 2:$ABYC$ is a parallelogram. Proof of Claim 2:We show that $CY \parallel AB$ and $BY \parallel AC$. Angle chasing gives \[\angle{BCY} = \angle{BNY}=\angle{BNH} - 90^{\circ} =90^{\circ} - \angle{HCB}=B\]The analogous argument for the other side proves our claim. Therefore, clearly, $AY$ is median and therefore, $AY$ and $AX$ are isogonal. We show that $\triangle{ABN} \sim \triangle{AXC}$. Obviously, $\angle{BAN} = \angle{XAC}$. Also, $\angle{ANB} = 180^{\circ} - \angle{BNY} = 180^{\circ} - \angle{BCY} = 180^{\circ} - B = A+C = \angle{ACX}$. Similarly, $\triangle{ABM} \sim \triangle{AYC}$ because $\angle{AMB} = 180^{\circ} - \angle{BMX} = 180^{\circ} - \angle{BMX} = 180^{\circ} - A = \angle{ACY}$. Comparing side ratios, we get $AB.AC = AN.AX = AM.AY$. Therefore, $\frac{AN}{AY} = \frac{AM}{AX} \implies MN \parallel XY$.

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.597504213007954, xmax = 14.43681578743231, ymin = -4.751664516057952, ymax = 6.526108524306414; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((1.273064938604060,5.675797660006203)--(-0.6600000000000008,1.060000000000001)--(4.380000000000004,0.1800000000000002)--cycle, zzttqq); /* draw figures */ draw((1.273064938604060,5.675797660006203)--(-0.6600000000000008,1.060000000000001), zzttqq); draw((-0.6600000000000008,1.060000000000001)--(4.380000000000004,0.1800000000000002), zzttqq); draw((4.380000000000004,0.1800000000000002)--(1.273064938604060,5.675797660006203), zzttqq); draw((1.273064938604060,5.675797660006203)--(2.201245458045489,2.574405805169606)); draw((0.5905740225130863,1.766986049666991)--(1.273064938604060,5.675797660006203)); draw(circle((1.746955949823893,-0.02743410555407675), 2.641202354670093)); draw(circle((1.518754541954515,-1.334405805169606), 3.237306058116431)); draw(circle((1.737155198324775,4.125101732587905), 1.618653029058216)); draw((1.292666441602292,-2.629274016277759)--(1.273064938604060,5.675797660006203)); draw(circle((0.9318194805585732,3.721391854836597), 1.983973415627672)); draw((1.273064938604060,5.675797660006203)--(2.446935061395944,-4.435797660006202)); draw((1.280389925769249,2.572232416043008)--(1.711734543879979,1.897143240222432)); draw((1.292666441602292,-2.629274016277759)--(2.446935061395944,-4.435797660006202)); label("$\omega_1$",(3.577650043326026,-0.8118311395114484),SE*labelscalefactor); label("$\omega_2$",(-1.248645842943432,-2.535508241750543),SE*labelscalefactor); /* dots and labels */ dot((1.273064938604060,5.675797660006203),dotstyle); label("$A$", (1.361493769018622,5.812013724807360), NE * labelscalefactor); dot((-0.6600000000000008,1.060000000000001),dotstyle); label("$B$", (-0.5591750020477951,1.207333465968635), NE * labelscalefactor); dot((4.380000000000004,0.1800000000000002),dotstyle); label("$C$", (4.488736511652404,0.3208709562456713), NE * labelscalefactor); dot((2.201245458045489,2.574405805169606),dotstyle); label("$O$", (2.297204195948415,2.734018899380405), NE * labelscalefactor); dot((0.5905740225130863,1.766986049666991),dotstyle); label("$H$", (0.6966468867264007,1.921428265467688), NE * labelscalefactor); dot((2.201245458045489,2.574405805169606),dotstyle); dot((1.280389925769249,2.572232416043008),dotstyle); label("$M$", (1.386117727622038,2.709394940776989), NE * labelscalefactor); dot((1.292666441602292,-2.629274016277759),dotstyle); label("$X$", (1.386117727622038,-2.486260324543712), NE * labelscalefactor); dot((1.711734543879979,1.897143240222432),dotstyle); label("$N$", (1.804725023880103,2.044548058484767), NE * labelscalefactor); dot((2.446935061395944,-4.435797660006202),dotstyle); label("$Y$", (2.543443781982571,-4.283809302593054), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Since $M$ lies on the circle with diameter $AO,$ $AM \perp OM \implies XM \perp OM,$ which implies $OX$ is a diameter of $\omega_1.$ Similar arguments show that $HY$ is a diameter of $\omega_2.$ Since $XC \perp OC$ and $XB \perp OB,$ $BX$ and $CX$ are both tangents to the circumcircle of $\triangle ABC,$ implying $X$ lies on the $A$-symmedian of $\triangle ABC.$ Note that \[\angle YCB = \angle YNB = \angle HNY - \angle HNB = 90^{\circ} - \angle HCB \\ = 90^{\circ} - (90^{\circ} - \angle ABC) = \angle ABC,\] so $AB \parallel CY.$ Similarly, $AC \parallel BY,$ implying $ABYC$ is a parallelogram. Since the diagonals of a parallelogram bisect each other, $AY$ must pass through the midpoint of $BC.$ It follows that $\angle BAX = \angle CAY$ and $\angle CAX = \angle BAY.$ We also have that \[\angle ACX = \angle ACB + \angle BCX = \angle ACB + \angle BAC = 180^{\circ} - \angle BNY = \angle BNA,\] so $\triangle ABN \sim \triangle AXC.$ Similarly, $\triangle ABM \sim \triangle ACY.$ Now, we have that \[\dfrac{BN}{CY} = \dfrac{AB}{AX} = \dfrac{AN}{AC} \quad \dfrac{BM}{CY} = \dfrac{AB}{AY}= \dfrac{AM}{AC} \\ \implies \dfrac{AM}{AX} = \dfrac{AN}{AY} \implies MN \parallel XY. \quad \blacksquare\]

Outline: We first show that $AX$ and $AY$ are the $A$-symmedian and $A$-median in $\triangle ABC$, respectively. Then, by two angle chases, it follows that $\triangle AOX \sim \triangle AHY.$ Then, since $\overline{OM}$ and $\overline{HN}$ are simply altitudes in a pair of similar triangles, it follows that the associated ratios in which each altitude divides the opposite side are equal. We then have that $\frac{AM}{AX} = \frac{AN}{AY}.$ This ratio relation implies that a homothety centered at $A$ maps $M$ to $X$ and $N$ to $Y$, showing that $MN \parallel XY.$ More detail follows: Claim: $AX$ is the $A$-symmedian in $\triangle ABC.$ Proof. Since $AO$ is a diameter of $\odot (AMO)$, we have $\angle OBX = \angle OMX = 180^{\circ} - \angle OMA = 90^{\circ}.$ Therefore $BX$ is a tangent to $\odot (ABC).$ Similarly, since $AH$ is a diameter of $\odot (ANH)$, we have $\angle OCX = 180^{\circ} - \angle OMX = \angle OMA = 90^{\circ}.$ Therefore $CX$ is a tangent to $\odot (ABC).$ Hence, $X$ is the intersection of the tangents to $\odot (ABC)$ at $B, C.$ It is a well known fact that this implies $AX$ is the $A$-symmedian. $\quad$ $\blacksquare$ See http://www.scribd.com/doc/201353908/MR4-Symmedians or http://yufeizhao.com/olympiad/geolemmas.pdf for a proof of this. Claim: $AY$ is the $A$-median in $\triangle ABC.$ Proof. Since $\angle YNH = 180^{\circ} - \angle ANH = 90^{\circ}$, it follows that $\overline{YH}$ is a diameter of $\odot (BHNCY).$ Therefore \[\angle YBC = \angle YBH - \angle CBH = 90^{\circ} - \left(90^{\circ} - C\right) = \angle BCA.\] This angle equality implies that $BY \parallel AC.$ Similarly, we find that $CY \parallel AB.$ Therefore quadrilateral $ABYC$ is a parallelogram. It follows that the diagonals of $ABYC$ bisect one another. Therefore $AY$ bisects $\overline{BC}$, and so $AY$ is the $A$-median in $\triangle ABC.$ $\quad$ $\blacksquare$ We now show two angle equalities between $\triangle AHY$ and $\triangle AOX.$ Firstly, since the pairs of lines $AX, AY$ and $AO, AH$ are isagonal with respect to angle $BAC$, it follows that $\angle OAX = \angle HAY.$ (We have used the well known fact that $H$ and $O$ are isagonal conjugates) $\quad$ $\blacksquare$ For the second angle chase, denote the midpoint of $\overline{BC}$ by $P$ and let $Z = AX \cap BC.$ Then since $\angle OMZ = 180^{\circ} - \angle OMA = 90^{\circ}.$ In addition, since $P$ is the midpoint of $\overline{BC}$, we have $OP \perp BC.$ Due to these pairs of right angles, it follows that quadrilateral $OMZP$ is cyclic. However, since $\angle OMX = 90^{\circ}$, it follows that $OX$ is a diameter of $\odot (BMOCX.)$ Therefore, we have $OX \perp BC \implies OX = OP.$ We use these relations, as well as the facts that $AB \parallel CY$ and $AC \parallel BY$, as proved earlier. It follows that \[\angle MOX = \angle MOP = 180^{\circ} - \angle MZP = \angle MZB = 180^{\circ} - B - \angle BAZ.\] In addition, we have \[\angle NHY = \angle NBY = \angle NBC + \angle CBY = \angle NYC + \angle CBY = \angle AYC + C\]\[\left(180^{\circ} - \angle YCA - \angle CAY\right) + C = 180^{\circ} - \angle YCB - \angle BCA + C = 180^{\circ} - B - \angle CAY.\] However, Since $AY$ and $AZ$ are isagonal with respect to $\angle BAC$, it follows that $\angle BAZ = \angle CAY.$ Therefore, we conclude that $\angle MOX = \angle NHY.$ We then have \[\angle AOX = \angle AOM + \angle MOX = \left(90^{\circ} - \angle OAM\right) + \angle MOX\]\[= \left(90^{\circ} - \angle HAN\right) + \angle NHY = \angle AHY. \quad \blacksquare\] Thus, $\angle HAY = \angle OAX$ and $\angle AOX = \angle AHY \implies \triangle AOX \sim \triangle AHY.$ Since $\overline{OM}$ and $\overline{HN}$ are just altitudes in similar triangles, it follows that the associated ratios with which each altitude divides the opposite side are equal. Therefore $\frac{AM}{AX} = \frac{AN}{AY}.$ It follows that a homothety centered at $A$ takes $M$ to $X$ and $N$ to $Y.$ This implies that $MN \parallel XY$, as desired. $%Error. "qedsymbol" is a bad command. $

Dear Mathlinkers, a question : did the lines MX and NY intersect on BC? Sincerely Jean-Louis

Generalization: Let $ P_1, Q_1 $ be the isogonal conjugate of $ \triangle ABC $ . Let $ P_2\equiv \odot (AP_1) \cap \odot (BP_1C), Q_2\equiv \odot (AQ_1) \cap \odot (BQ_1C) $ . Let $ P_3\equiv AP_2 \cap \odot (BP_1C), Q_3 \equiv AQ_2 \cap \odot (BQ_1C) $ . Then $ P_2Q_2 \parallel P_3Q_3 $ Proof: Invert with center $ A $ (factor $ AB \cdot AC $) and then reflect in the bisector of $ \angle BAC $ . Denote $ T' $ as the image of $ T $ under this transformation $ \psi $ . Easy to see $ P_1' \in AQ_1, Q_1' \in AP_1 $ . Since $ AB \cdot AC=AP_1 \cdot AP_1'=AQ_1 \cdot AQ_1' $ , so we get $ \triangle ABP_1 \sim \triangle AP_1'C $ and $ \triangle ABQ_1 \sim \triangle AQ_1'C $ . Since $ \angle CP_1'Q_1=\angle P_1BA=\angle CBQ_1 $ , so $ P_1' $ is the second intersection of $ AQ_1 $ and $ \odot (BQ_1C) $ . Similarly, $ Q_1' $ is the second intersection of $ AP_1 $ and $ \odot (BP_1C) $ , so $ \psi (\odot (BP_1C))=\odot (BQ_1C) $ and $ \psi (\odot (BQ_1C))=\odot (BP_1C) $ . ... $ (\star) $ Since $ \psi (\odot (AP_1)) $ is a line passing through $ P_1' $ and perpendicular to $ AP_1' $ , so combine with $ (\star) $ we get $ P_2' $ is the antipode of $ Q_1 $ in $ (BQ_1C) $ . i.e. $ P_2' \equiv Q_3 $ Similarly, $ Q_2' \equiv P_3 \Longrightarrow AP_2 \cdot AQ_3 = AQ_2 \cdot AP_3 $ . i.e. $ P_2Q_2 \parallel P_3Q_3 $ Q.E.D jayme wrote: Dear Mathlinkers, a question : did the lines MX and NY intersect on BC? Sincerely Jean-Louis Yes , this can be generalized : $ P_2Q_3 \cap Q_2P_3 \in BC $ Let $ X\equiv BQ_3 \cap CP_2, Y \equiv CP_3 \cap BQ_2 $ . From my proof above $\Longrightarrow \triangle ABQ_2 \sim \triangle AP_3C ,\triangle ACP_2 \sim \triangle AQ_3B $ , so we get $ A, B, X, P_2 $ are concyclic and $ A, C, Y, Q_2 $ are concyclic . Since $ \angle XAB+\angle CAY=(180^{\circ}-\angle BP_2C)+(180^{\circ}-\angle BQ_2C) $ $ =(180^{\circ}-\angle BP_1C)+(180^{\circ}-\angle BQ_1C)=360^{\circ}-(180^{\circ}+\angle BAC)=180^{\circ}-\angle BAC $ , so $ X, A, Y $ are collinear $ \Longrightarrow BC, P_2Q_3, P_3Q_2 $ are concurrent (Desargue theorem ($ \triangle BQ_2Q_3 , \triangle CP_3P_2 $)). i.e. $ P_2Q_3 \cap P_3Q_2 \in BC $ Done

Dear Mathlinkers, I have this reference Yiu P., Reflections, Message Hyacinthos du 24/09/2004. for a possible the origine of a point of view of the problem Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, I have this reference Yiu P., Reflections, Message Hyacinthos du 24/09/2004. for a possible the origine of a point of view of the problem Sincerely Jean-Louis The problem: https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/10528 The response: https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/10529

One liner Apply $\sqrt{bc}$ inversion with center $A$. We are done. EDIT: Not yet.......

First we claim that $X$ is the intersection of the tangents to $(ABC)$ from $B$ and $C$, implying that line $AM$ is a symmedian. This is obvious, since $\angle OMX=\angle OCX=\angle OBX=90^{\circ}$. We further claim that $AN$ is a median. To prove this, let $Y'$ be the point such that $ABY'C$ is a parallelogram, then $\angle BY'C=\angle A=180^{\circ}-\angle BHC$, so that $BHCY'$ is cyclic, and we need to show that $\angle HNY'=90^{\circ}$, but $\angle HBY'=\angle HBC+\angle Y'BC=90^{\circ}-\angle C+\angle C=90^{\circ}$. Now, perform a transformation about $A$ which composes an inversion with radius $\sqrt{AB \cdot AC}$ and reflection about the $A$-angle bisector. Then $(BHC)$ and $(BOC)$ clearly swap from $H,O$ isogonal, so that $X$ swaps with $N$ and $M$ swaps with $Y$. Hence $AX \cdot AN=AM\cdot AY\Longrightarrow \frac{AX}{AM}=\frac{AY}{AN}$ which implies the result.

Begin with a preliminary angle chasing. $$\angle CMX = \angle OMC - \angle OMX = 180-\angle OBC - 90 = \angle A$$and since $\angle BOC = 2\angle A$, we have $X$ as a midpoint of arc $BC$ in the circumcircle of $BOC$. Since $\angle BXC = 180-2\angle A$, we get that $AX$ is a symmedian. Also, $\angle BYC = 180-\angle BHC = \angle BAC$ and $\angle ABY = \angle ABH + \angle HBY = 90-\angle A + 90 = 180-\angle A$. This gives us that $ABYC$ is a parallelogram. This gives us that $AX, AY$ are isogonal. We have $$\angle BNA = 180-\angle BNY = 180-\angle BCY = 180-\angle ABC = \angle BCA + \angle A = \angle BCA + \angle BCX = \angle ACX$$so combine this with $\angle BAN = \angle CAX$ to get $\triangle ABN \sim \triangle AXC$. We also have $$\angle AMC = 180-\angle CMX = 180-\angle A = \angle ABY$$so similarly we have $\triangle ABY \sim \triangle AMC$. Now we have $AB \cdot AC = AM \cdot AY = AN \cdot AX$, giving $\frac{AM}{AX}=\frac{AN}{AY}$, which gives us $MN \parallel XY$, as required. $\blacksquare$

It is just an exercise in applied algebra. We set $\odot (ABC)$ to be the unit circle. Notice that $X$ is the intersection of the tangents at $B$ and $C$ to $\odot (ABC)$, so $x=\frac{2bc}{b+c}$. Also see that $Y$ is the antipode of $H$ in $\omega_2$, whose center is the reflection of $O$ across $BC$. So $y=b+c$. Now we intersect $AX$ and $AY$ with the $\odot (ABC)$ to obtain points $F$ and $G$. We see that $m=\frac{a+f}{2}$ and $n=a+g+ag\overline{h}$. Now we let $x'=x-a$. It remains to show that $$\frac{m'}{x'}=\frac{n'}{y'}$$which is a straightforward computation.

A bit easy for a #5? It's just angle and length chasing with some well known stuff. We can start by noticing that the points $M,N,X,Y$ are all well known. It is clear that $X,Y$ are the antipodes of $O,H$ on $\omega_1,\omega_2$, and $M,N$ are the second intersections of $AX, AY$ with the circles $\omega_1\omega_2$. It suffices to show $AMN\sim AXY$. We see that $X$ is the intersection of the $B,C$ tangents to the circle $ABC$, and $Y$ is the reflection of $A$ over the midpoint of $BC$, which we will denote $P$. Also, let $AX$ intersect the circle $ABC$ again at $Q$. Also, $M$ is the midpoint of $AQ$. Reflect $N$ over $BC$ to $N'$, and $H$ over $BC$ to $H'$. $\angle H'N'P = 90$, so $N'P$ passes through $A'$, where $AA'\parallel BC$ and $A'$ on the circle $ABC$. This yields $AN', AP$ are isogonal, so $N' = Q$. Hence, $QN\parallel PX$ is the perpendicular bisector of $BC$. So, $\frac{AQ}{AX} = \frac{AN}{AP}$, so $\frac{AM}{AX} = \frac{AN}{AY}$. This implies the conclusion.

EulerMacaroni wrote: Now, perform a transformation about $A$ which composes an inversion with radius $\sqrt{AB \cdot AC}$ and reflection about the $A$-angle bisector. Then $(BHC)$ and $(BOC)$ clearly swap from $H,O$ isogonal, so that $X$ swaps with $N$ and $M$ swaps with $Y$. Can someone explain why $(BHC)$ and $(BOC)$ are swapped?

Achillys wrote: EulerMacaroni wrote: Now, perform a transformation about $A$ which composes an inversion with radius $\sqrt{AB \cdot AC}$ and reflection about the $A$-angle bisector. Then $(BHC)$ and $(BOC)$ clearly swap from $H,O$ isogonal, so that $X$ swaps with $N$ and $M$ swaps with $Y$. Can someone explain why $(BHC)$ and $(BOC)$ are swapped? Let the performed operation (the $\sqrt {\text {bc}} $-flip) be denoted by $\Phi $. Then $\Phi (BC)\equiv \odot (ABC) $. If $A'$ is the antipode of $A $ and $D $ be the foot of perpendicular from $A $ to $BC $. Then $\Phi (A')\equiv D\implies \Phi (O) $ is the reflection of $A $ in $D $, denoted by $A_0$. Then $A_0$ is the reflection of $A$ in $BC $. Since $B,C $ remain invariant under $\Phi $, we conclude that $\Phi (\odot (OBC))\equiv\odot (A_0BC) $. But $\angle BA_0C+\angle BHC=\angle BAC+(180^{\circ}-\angle BAC)=180^{\circ} $, so $A_0BHC $ is cyclic $\implies \odot (A_0BC)\equiv \odot (HBC) $. Hence $\Phi (\odot (OBC))\equiv \odot (HBC) $. Since $\Phi $ is an involution we conclude that $\odot (HBC) $ and $\odot (OBC) $ are swapped under it.

Solution with 2 inversions: First, let $\Psi$ be the composition of inversion with radius $\sqrt{\frac{1}{2} \cdot AB \cdot AC}$ and reflection over the bisector of $\angle BAC$. Also, let $A_{1}$ and $A_{2}$ be the midpoint of $BC$ and the foot of the altitude from $A$ to $BC$. It's easy to see that $\Psi$ takes $\odot BOC$ to the nine point circle of $\triangle ABC$ because it takes $B$ and $C$ to the midpoints of $AB$ and $AC$ and takes $O$ to the foot of the altitude from $A$. That also means that $\Psi$ takes the circe with diameter $AO$ to the line $BC$. The intersection of these two circles, point $M$ must go to the intersection of $BC$ and the nine point circle, so it must go to $A_{1}$ because $O$ goes to $A_{2}$. From this we get $AM \cdot AA_{1} = \frac{1}{2} \cdot AB \cdot AC$. Now let $S$ be the intersection of $AA_{1}$ and the nine point circle other than $A_{1}$. It's easy to see that $\Psi$ takes $X$ to $S$, so $AX \cdot AS = AM \cdot AA_{1}$. Now let $\psi$ be the inversion with center $A$ and radius $\sqrt{AH \cdot AA_{2}}$. Let $AN$ intersect $BC$ at $T$. Because $\angle AA_{2}T = \angle ANH = 90^{\circ}$ we get that $HA_{2}TN$ is cyclic so $\psi$ takes $N$ to $T$. But $\psi$ also takes $B$ and $C$ to the feet of the altitudes from $C$ and $B$, and takes $H$ to $A_{2}$, so it takes $\odot BHNC$ to the nine point circle. It also takes the circle with diameter $AH$ to line $BC$ so $N$ must go to the midpoint of $BC$, or $T = A_{2}$. Also $\psi$ takes $Y$ to the intersection of $AA_{1}$ and the nine point circle other than $A_{1}$, which means that $Y$ goes to $S$ so $AY \cdot AS = AN \cdot AA_{1}$. Finally we have $\frac{AY}{AN} = \frac{AA_{1}}{AS} = \frac{AX}{AM}$ so $XY \parallel MN$.

K6160 wrote: It is just an exercise in applied algebra. We set $\odot (ABC)$ to be the unit circle. Notice that $X$ is the intersection of the tangents at $B$ and $C$ to $\odot (ABC)$, so $x=\frac{2bc}{b+c}$. Also see that $Y$ is the antipode of $H$ in $\omega_2$, whose center is the reflection of $O$ across $BC$. So $y=b+c$. Now we intersect $AX$ and $AY$ with the $\odot (ABC)$ to obtain points $F$ and $G$. We see that $m=\frac{a+f}{2}$ and $n=a+g+ag\overline{h}$. Now we let $x'=x-a$. It remains to show that $$\frac{m'}{x'}=\frac{n'}{y'}$$which is a straightforward computation. $y=b+c-a$

Solution: It's pretty easy to see that $M$ and $N$ are Humpty-Dumpty points of $\triangle{ABC}$. As a matter of fact, $HN \perp AN$ and $M$ lies on the $(\triangle{CHB}) \implies$ its Humpty point, and similarly for the other point. Since $M$ and $N$ are isogonal conjugates, chasing angles gives us that $\triangle{AMB} \sim \triangle{AYC}$ and also $\triangle{ABN} \sim \triangle{AXC}$. Ergo, using this, we have $AY*AM = AB*AC = AN*NX \implies \frac{AM}{AX} = \frac{AN}{YA}$, as claimed. $\square$ Oh sorry, I meant to say $AH \perp HN$, thanks for pointing it out.

v_Enhance wrote: Let $ABC$ be a triangle with circumcenter $O$ and orthocenter $H$. Let $\omega_1$ and $\omega_2$ denote the circumcircles of triangles $BOC$ and $BHC$, respectively. Suppose the circle with diameter $\overline{AO}$ intersects $\omega_1$ again at $M$, and line $AM$ intersects $\omega_1$ again at $X$. Similarly, suppose the circle with diameter $\overline{AH}$ intersects $\omega_2$ again at $N$, and line $AN$ intersects $\omega_2$ again at $Y$. Prove that lines $MN$ and $XY$ are parallel. Proposed by Sammy Luo [asy][asy] import graph; size(15cm); real labelscalefactor = 0.25; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 3.58381736170553, xmax = 43.73260011935113, ymin = 3.6343787400109595, ymax = 22.482001756794613; /* image dimensions */ pen qqzzcc = rgb(0,0.6,0.8); pen ffqqtt = rgb(1,0,0.2); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((1.3199280930795845,15.437543520442091)--(2.497749113398948,-0.6992304297136127)--(26.128668035621757,6.174194271510027)--cycle, linewidth(2) + qqzzcc); /* draw figures */ draw((1.3199280930795845,15.437543520442091)--(2.497749113398948,-0.6992304297136127), linewidth(2) + qqzzcc); draw((2.497749113398948,-0.6992304297136127)--(26.128668035621757,6.174194271510027), linewidth(2) + qqzzcc); draw((26.128668035621757,6.174194271510027)--(1.3199280930795845,15.437543520442091), linewidth(2) + qqzzcc); draw(circle((12.736146630144013,8.15944034229615), 13.538863746923209), linewidth(2) + ffqqtt); draw(circle((17.26929645608864,-7.425584686073689), 16.230910402050043), linewidth(2) + qqzzcc); draw(circle((15.890270518876696,-2.6844765004997266), 13.538863746923202), linewidth(2) + yellow); draw(circle((7.028037361611799,11.79849193136912), 6.769431873461604), linewidth(2) + qqzzcc); draw((xmin, -1.8771204243636868*xmin + 17.915207502653193)--(xmax, -1.8771204243636868*xmax + 17.915207502653193), linewidth(2) + qqzzcc); /* line */ draw(circle((2.8969900374459296,10.015585099044152), 5.64665896789744), linewidth(2) + yellow); draw((xmin, -0.9774330368450098*xmin + 16.727684844877913)--(xmax, -0.9774330368450098*xmax + 16.727684844877913), linewidth(2) + qqzzcc); /* line */ draw((xmin, 2.370626568829112*xmin-74.69606813610105)--(xmax, 2.370626568829112*xmax-74.69606813610105), linewidth(2) + wrwrwr); /* line */ draw((xmin, 2.3706265688291026*xmin-11.240422487466294)--(xmax, 2.3706265688291026*xmax-11.240422487466294), linewidth(2) + wrwrwr); /* line */ /* dots and labels */ dot((1.3199280930795845,15.437543520442091),linewidth(4pt) + dotstyle); label("$A$", (5.033634516842732,15.567489170755639), NE * labelscalefactor); dot((2.497749113398948,-0.6992304297136127),linewidth(4pt) + dotstyle); label("$B$", (4.476012534097654,-0.3804995357536054), NE * labelscalefactor); dot((26.128668035621757,6.174194271510027),linewidth(4pt) + dotstyle); label("$C$", (26.557843050802738,7.091635033030446), NE * labelscalefactor); dot((12.736146630144013,8.15944034229615),linewidth(4pt) + dotstyle); label("$O$", (13.174915464920868,9.09907417091273), NE * labelscalefactor); dot((4.474051981812274,4.593626677646214),linewidth(4pt) + dotstyle); label("$H$", (4.922110120293716,5.530293481344226), NE * labelscalefactor); dot((7.028037361611799,11.79849193136912),linewidth(4pt) + dotstyle); label("$D$", (7.487171240921074,12.667854860481231), NE * labelscalefactor); dot((6.863786858502337,5.031053002079388),linewidth(4pt) + dotstyle); label("$M$", (7.264122447823043,5.976391067540289), NE * labelscalefactor); dot((21.802446282033245,-23.01060971444354),linewidth(4pt) + dotstyle); label("$X$", (3.58381736170553,22.482001756794613), NE * labelscalefactor); dot((2.8969900374459296,10.015585099044152),linewidth(4pt) + dotstyle); label("$D_{1}$", (3.918390551352576,10.660415722598948), NE * labelscalefactor); dot((8.353527304276588,8.562671283491138),linewidth(4pt) + dotstyle); label("$N$", (8.825463999509262,9.433647360559775), NE * labelscalefactor); dot((27.30648905594112,-9.962579678645643),linewidth(4pt) + dotstyle); label("$Y$", (27.784611412841908,-9.079402466576829), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] I begin by stating well known facts. $N$ is the $A$-Humpty point or $A_H$ of $\triangle ABC$. $M$ is the $A$-Dumpty point or $A_D$ of $\triangle ABC$. $M, N$ are isogonal conjugates $AN \perp HN$ Hence, using angle chasing, we establish that $\triangle AMB \sim \triangle AYC$ and $\triangle ABN \sim \triangle \triangle AXC$, which put together yields that $\dfrac{AM}{AX} = \dfrac{AN}{AY}$ and we're done by converse of Basic Proportionality Theorem. @above $AN \perp HN$ as $N$ is $A$-Humpty Point and also because of $N$ lies on circle with diameter $AH$. Clearly $AM \not\perp HM$ except for $1-2$ special cases.

$M$ is $D_A$ (Dumpty point), $X$ is the intersection point of tangents at $B$ and $C$. $N$ is $H_A$ (Humpty point), $Y$ is the reflection of $A$ across midpoint of $BC$. Change the labeling as in the diagram. $$\frac{AH_A}{AA'}=\frac{1}{2} \cdot \frac{AH_A}{AM}=\frac{1}{2} \cdot (1-\frac{MH_A}{MA})=\frac{1}{2} \cdot (1-\frac{MD}{MA})=\frac{1}{2} \cdot (1-\frac{MD \cdot MA}{MA^2})=\frac{1}{2} \cdot (1-\frac{MB^2}{MA^2})=\frac{1}{2} \cdot (1-\frac{CE^2}{CA^2})=\frac{1}{2} \cdot (1-\frac{TE}{TA})=\frac{1}{2} \cdot(\frac{AE}{AT})=\frac{AD_A}{AT}$$

Attachments:

It is well known that $M$ is the midpoint of the A-symmedian chord, i.e Dumpty point and $N$ is the Humpty point of $\triangle ABC$. Let $T = AM \cap (ABC)$ and $D$ be the midpoint of $BC$. Now since $\triangle BAT \sim \triangle DAC$, we have $AT = \frac{AB\cdot AC}{AD} \implies AM = \frac{AB \cdot AC}{2\cdot AD}$. Note also that $Y$ is the reflection of $A$ across $D$, so $AY = 2\cdot AD$. So, $AM \cdot AY = AB \cdot AC$ and $AM$ , $AN$ are isogonal w.r.t $\angle BAC$. Thus, this motivates us to take an $\sqrt{AB \cdot AC}$ inversion centered at $A$ followed by a reflection across the angle bisector of $\angle BAC$. Because $AM \cdot AY = AB \cdot AC$ and $AM$, $AN$ are isogonal, we have $M \leftrightarrow Y$ under inversion. Also since $B$ and $C$ swap places, this gives that $(BOC) \leftrightarrow (BHC)$. So, $N$ and $X$ also swap places and hence, $AN \cdot AX = AB \cdot AC = AM \cdot AY \implies \frac{AM}{AX} = \frac{AN}{AY}$, we are done.

Let $AN$ meet $BC$ at $T$ and $AM$ meet $(ABC)$ again at $K$. It's well-known that $M$ is the $A$-Dumpty point, i.e. the midpoint of $AK$, which coincides with the $A$-symmedian of $ABC$. As a result, it's easy to see that $X = BB \cap CC$, where tangents are taken wrt $(ABC)$. It's also well-known that $N$ is the $A$-Humpty Point of $ABC$, i.e. the projection of $H$ onto the $A$-median, so $T$ is the midpoint of $BC$. Because $(ABC)$ and $(BHC)$ are symmetric about $BC$, we know $TA = TY$, i.e. $Y$ is the reflection of $A$ over $T$. Now, we consider $\sqrt{bc}$-inversion. It's well-known that $N, X$ and $T, K$ swap under this transformation. Now, we have $$AY \cdot AM = (2 \cdot AT) \cdot \frac{AK}{2} = AT \cdot AK = AB \cdot AC = AN \cdot AX$$so $$\frac{AM}{AX} = \frac{AN}{AY}$$which clearly suffices. $\blacksquare$ Remarks: I solved this problem mentally, as nearly every line in my solution is well-known. Observing that $\sqrt{bc}$-inversion relates $M$ and $Y$ is the only non-trivial insight required.

Notice how $M$ is the Dumpty Point and $N$ is the Humpty point. This motivates us to try $\sqrt{bc}$ inversion, so if we succed on showing that this maps $(BHC)$ and $(BOC)$ then we would have that it maps $N$ with $X$ and $M$ with $Y,$ ending the problem. In order to do this let $D$ be the foot of the $A-$altitude and $A'$ the reflection of $A$ wrt $D.$ Now the inversion maps $BC$ with $(ABC),$ meaning that $D$ is sent to $AO\cap (ABC),$ which implies that $A'$ goes to $O.$ A we have that $A'\in(BHC)$ trivially, we're done.

We notice that $N$ and $M$ are simply the Humpty/Dumpty points, respectively, which are known to be isogonal conjugates. They induce the following similar triangles, as \begin{align*} \triangle AMB \sim \triangle ACY: \quad &\angle MAB = \angle CAN = \angle CAY \\ &\angle ABM = \angle NBC = \angle AYC \\ \\ \triangle ANB \sim \triangle ACX: \quad &\angle BAN = \angle MAC = \angle XAC \\ &\angle ABN = \angle MBC = \angle AXC \end{align*} These two give us similarity ratios, which can be rewritten as \[AM \cdot AY = AN \cdot AX = AB \cdot AC \implies \frac{AM}{AX} = \frac{AN}{AY}.~\blacksquare\]

Solved with GrantStar, ihatemath123, alsk, SenorSloth. We see $N$ is the humpty point so $Y$ is the point with $ABYC$ parallelogram. Then $\sqrt{bc}$ invert swaps $(BOC),(BHC)$ and $M,Y$ so it swaps $N,X$ then $AM\cdot AY=AB\cdot AC=AN\cdot AX$ implies parallel.

We characterise each of the points. $M=A$ Dumpty point. $X=$ intersection of tangents at $B$ and $C$. $N=A$ Humpty point. $Y=$ Reflection of $A$ over midpoint of $BC$. Now note that under $\sqrt{bc}$ inversion $M \longleftrightarrow Y$ $N \longleftrightarrow X$ so $AB.AC=AN.AX=AM.AY$ so rearranging we get $\frac{AM}{AN}=\frac{AX}{AY}$ and the required follows by converse of Thales' theorem.