In my opinion, this problem is significantly harder than the TST geometry from the past two years, but maybe that’s just me. This one took me 3.5 hours oops. Suppose $(XAM)$ hits $\overline{AC}$ again at $R$ and $(XAN)$ hits $\overline{AB}$ again at $S$, and let $T = \overline{RM}\cap \overline{SN}$ so that $X$ is the Miquel point of complete quadrilateral $AMTNRS$. We show that $T$ is the desired intersection point. [asy][asy] defaultpen(fontsize(10pt)); size(250); pair A, B, C, M, N, X, R, S, T; real r = 0.65; A = dir(120); B = dir(210); C = dir(330); M = midpoint(A--B); N = midpoint(A--C); T = r*B + (1-r)*C; R = extension(T, M, A, C); S = extension(T, N, A, B); X = IP(circumcircle(R, A, M), circumcircle(S, A, N), 0); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(T--R^^S--N, heavygreen); draw(circumcircle(R, A, M), heavycyan); draw(circumcircle(S, A, N), heavycyan); draw(R--A^^S--B, orange); draw(X--A, purple); draw(circumcircle(B, M, T), blue+dotted); draw(circumcircle(C, N, T), blue+dotted); draw(S--X--M--cycle, magenta+linewidth(0.9)); draw(R--X--N--cycle, magenta+linewidth(0.9)); dot("$A$", A, dir(70)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$M$", M, dir(0)); dot("$N$", N, dir(10)); dot("$R$", R, dir(80)); dot("$S$", S, dir(250)); dot("$T$", T, dir(270)); dot("$X$", X, dir(180)); [/asy][/asy] Claim: $T$ lies on $\overline{BC}$. Proof: Let $T_1 = \overline{RM}\cap \overline{BC}$ and $T_2 = \overline{SN}\cap \overline{BC}$. By Menelaus in $\triangle ABC$, we have $$\frac{RA}{RC} = \frac{BT_1}{T_1C} \quad \text{and} \quad \frac{SB}{SA} = \frac{BT_2}{T_2C},$$so it suffices to show that $\tfrac{RA}{RC} = \tfrac{SB}{SA}$. But this is equivalent to \begin{align*} 1 - \frac{RA}{RC} = 1 - \frac{SB}{SA} & \iff \frac{b}{RC} = \frac{c}{SA} \\ & \iff \frac{RC}{b} - \frac{1}{2} = \frac{SA}{c} - \frac{1}{2} \\ & \iff \frac{RN}{b} = \frac{SM}{c}. \end{align*}Since $X$ is the Miquel point, $X$ is the center of the spiral similarity sending $\overline{SM}$ to $\overline{RN}$, and hence we just need to show that $$\frac{RN}{SM} = \frac{XR}{XM} = \frac{b}{c}.$$We have $\angle XRM = \angle XAM = \angle C$ and $\angle RXM = \angle A$, so $\triangle XMR \sim \triangle ABC$, giving the desired ratio. Thus the claim is proven. $\blacksquare$ Claim: $T$ lies on $\omega_B$ and $\omega_C$. Proof: We just angle chase: $$\angle BTM = \angle C+\angle MRA = \angle XRM+\angle MRA = \angle XRA = \angle BMX,$$so $(BTM)$ is tangent to $\overline{MX}$. Hence $T$ lies on $\omega_B$, and similarly, $T$ lies on $\omega_C$. $\blacksquare$ Combining these two claims finishes the problem.

$\qquad \qquad$

This note is strictly for V_Enhance No offence intended in any means ,@V_Enhance why do you post all the official solutions to the problems very soon after the problem is being posted also none of the solution is found by you so you can wait for 1 week and then post the official solution rather than posting it right after posting the problem that does not make any sense ,Also please consider this Rather than deleting this post !

This actually wasn't too painful to complex bash. Let $ABC$ be inscribed in the unit circle, so that $$|a|=|b|=|c|=1$$$$m=\frac{a+b}2,\quad n=\frac{a+c}2$$$$\overline{x}=\frac{2a-x}{a^2}$$ Now, we want to find the coordinates of $O_B$ and $O_C$, the circumcenters of $\omega_B$ and $ \omega_C$. We have the two equations $$O_BB = O_BM$$$$O_BM \perp MX$$We deal with the one on top first. This becomes (letting $o$ represent the coordinate of point $O_B$ for brevity): $$|o-b|=|o-m|$$$$(o-b)\left(\overline{o}-\frac1b\right) = \left(o-\frac{a+b}2\right)\left(\overline{o}-\frac{\frac1a+\frac1b}2\right)$$$$-\frac{o}b - \overline{o}b + 1 = -\frac{o}{2a} - \frac{o}{2b} - \frac{\overline{o}a}2 - \frac{\overline{o}b}2 + \frac12 + \frac{a}{4b} + \frac{b}{4a}$$$$\frac{o}{2a} - \frac{o}{2b} + \frac{\overline{o}a}2 - \frac{\overline{o}b}2 + \frac12 - \frac{a}{4b} - \frac{b}{4a} = 0$$Dividing out by $a-b$: $$-\frac{o}{2ab} + \frac{\overline{o}}2 - \frac1{4b} + \frac1{4a}=0$$$$-2o + 2ab\overline{o} - a + b = 0$$$$\overline{o} = \frac{a-b+2o}{2ab}$$And now we plug this in to the second equation, which becomes $$\frac{m-o}{m-x} \in i\mathbb{R}$$$$\frac{a+b-2o}{a+b-2x} = -\frac{\frac1a+\frac1b-2\overline{o}}{\frac1a+\frac1b-2\overline{x}} = -\frac{\frac1a+\frac1b - \frac1b + \frac1a - \frac{2o}{ab}}{\frac1a + \frac1b - \frac4a + \frac{2x}{a^2}} = \frac{2a(o-b)}{a^2 +2bx - 3ab}$$Now we expand to get $$a^3 + 2abx - 3a^2b + a^2b + 2b^2x - 3ab^2 - 2a^2o - 4bxo + 6abo = 2a^2o + 2abo - 4axo - 2a^2b - 2ab^2 + 4abx$$$$a^3-2abx+2b^2x-ab^2=4a^2o+4bxo-4axo-4abo$$Again dividing out by $a-b$: $$a^2 + ab - 2bx = 4o(a-x)$$$$o=\frac{a^2+ab-2bx}{4(a-x)}$$After applying an identical derivation for the coordinate of $O_C$ (which consists simply of swapping $b$ with $c$ in the algebra), we obtain $$o_b=\frac{a^2+ab-2bx}{4(a-x)}$$$$o_c=\frac{a^2+ac-2cx}{4(a-x)}$$Now, we will show that the second intersections of $\omega_B$ and $\omega_C$ with $BC$ are in fact the same point. We know one of the intersections of $\omega_B$ with $BC$ is $B$. Calling $P_B$ the projection of $O_B$ onto $BC$, and $Q_B$ the second intersection of $\omega_B$ with $BC$, we observe that $P_B$ is the midpoint of $BQ_B$. Thus, $$p_b = \frac12(b+c+o_b-bc\overline{o}_b)$$$$q_b = 2p_b - b = c + o_b - bc\overline{o}_b$$Defining $P_C$ and $Q_C$ similarly, we observe $$q_c = b + o_c - bc\overline{o}_c$$We just need to show that $Q_B$ and $Q_C$ are the same point. Now we simplify the coordinate of $Q_B$: $$q_b = c + o_b - bc\overline{o}_b = c + \frac{a^2+ab-2bx}{4(a-x)} - bc\cdot\frac{\frac1{a^2}+\frac1{ab}-\frac4{ab}+\frac{2x}{a^2b}}{4\left(\frac1a-\frac2a + \frac{x}{a^2}\right)}$$$$q_b=\frac{4ac - 4cx + a^2 + ab - 2bx + bc + 2cx - 3ac}{4(a-x)} = \frac{(a+b)(a+c) - 2(b+c)x}{4(a-x)}$$And then, since swapping the variables $b$ and $c$ does not change the value of the expression, $q_c = q_b$ and we are done. $\blacksquare$

ZhangLi wrote: This note is strictly for V_Enhance No offence intended in any means ,@V_Enhance why do you post all the official solutions to the problems very soon after the problem is being posted also none of the solution is found by you so you can wait for 1 week and then post the official solution rather than posting it right after posting the problem that does not make any sense ,Also please consider this Rather than deleting this post ! he does this for the likes

The projective solution I think is a lot easier than is being described; the actual work of showing $X \to \omega_B \cap BC$ is a projective map is straightforward enough if you realize that this kills the problem. As v_Enhance mentions, noting $\angle XMD$ is fixed suffices, or you can also explicitly write it as a composition of perspectivities. In a sense this has to work--it's a theorem of projective geometry that any projective map can be written as the composition of (at most two) perspectivities.

By angle chasing/Miquel, it suffices to prove that $\omega_B$ and $\omega_C$ meet on $(AMN) = \omega$. Let the points of tangency from $X$ to $\omega$ be $A$ and $T$. The claim is that $T$ is the desired point of intersection. It suffices to show that $(BMT)$ is tangent to $XM$, and by analogous reasoning it will follow that $(CNT)$ is tangent to $XN$. Extend $XM$ to meet $\omega$ again at $Y$, and let $T'$ be the reflection of $T$ over $M$.

. Hence the similar triangles follow by SAS. Combining the similarity with the fact that $ATBT'$ is a parallelogram gives$$\angle TMY=\angle TAY =\angle T'AM = \angle MBT,$$so $XM$ is tangent to $(BMT)$, as desired. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -19.18, xmax = 11.02, ymin = -8.04, ymax = 9.44; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-7.9982181120824265,2.10509303247173), 2.9195469421660403), linewidth(1) + wrwrwr+red); draw(circle((-6.976436224164854,-0.6298139350565414), 5.839093884332081), linewidth(1) + wrwrwr+red); draw(circle((-11.295156128426985,-1.5969159111919922), 2.2793906765790553), linewidth(1) + wrwrwr+red); draw((-9.02,4.84)--(-11.88,-3.8), linewidth(1) + wrwrwr); draw((-11.88,-3.8)--(-2.06,-3.78), linewidth(1) + wrwrwr); draw((-2.06,-3.78)--(-9.02,4.84), linewidth(1) + wrwrwr); draw((-15.348436186036503,2.4756484697619165)--(-7.312234189408197,-0.7327195768003403), linewidth(1) + wrwrwr); draw((-9.28987652553386,-0.513183669955118)--(-15.348436186036503,2.4756484697619165), linewidth(1) + wrwrwr); draw((-15.348436186036503,2.4756484697619165)--(-9.02,4.84), linewidth(1) + wrwrwr); draw((-11.61012347446614,1.5531836699551196)--(-9.28987652553386,-0.513183669955118), linewidth(1) + wrwrwr); draw((-11.61012347446614,1.5531836699551196)--(-11.88,-3.8), linewidth(1) + wrwrwr); draw((-9.02,4.84)--(-11.61012347446614,1.5531836699551196), linewidth(1) + wrwrwr); draw((-9.02,4.84)--(-9.28987652553386,-0.513183669955118), linewidth(1) + wrwrwr); draw((-9.28987652553386,-0.513183669955118)--(-11.88,-3.8), linewidth(1) + wrwrwr); /* dots and labels */ dot((-9.02,4.84),dotstyle); label("$A$", (-8.94,5.04), NE * labelscalefactor); dot((-11.88,-3.8),dotstyle); label("$B$", (-12.38,-4.28), NE * labelscalefactor); dot((-2.06,-3.78),dotstyle); label("$C$", (-1.7,-4.32), NE * labelscalefactor); dot((-10.45,0.52),linewidth(3pt) + dotstyle); label("$M$", (-10.18,0.62), NE * labelscalefactor); dot((-5.54,0.53),linewidth(3pt) + dotstyle); label("$N$", (-5.24,0.5), NE * labelscalefactor); dot((-15.348436186036503,2.4756484697619165),dotstyle); label("$X$", (-15.26,2.68), NE * labelscalefactor); dot((-9.28987652553386,-0.513183669955118),linewidth(3pt) + dotstyle); label("$T$", (-9.14,-1.18), NE * labelscalefactor); dot((-11.61012347446614,1.5531836699551196),linewidth(3pt) + dotstyle); label("$T'$", (-11.96,1.64), NE * labelscalefactor); dot((-7.312234189408197,-0.7327195768003403),linewidth(3pt) + dotstyle); label("$Y$", (-7.24,-0.58), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

ZhangLi wrote: This note is strictly for V_Enhance No offence intended in any means ,@V_Enhance why do you post all the official solutions to the problems very soon after the problem is being posted also none of the solution is found by you so you can wait for 1 week and then post the official solution rather than posting it right after posting the problem that does not make any sense ,Also please consider this Rather than deleting this post ! Okay, I will stop posting the official TST(ST) solutions. They all get uploaded on http://web.evanchen.cc/problems.html eventually anyways and so they will still be accessible (albeit maybe with some delay).

Where I can find references to understand the moving point solution? I don't know what a projective map is v_Enhance wrote: Okay, I will stop posting the official TST(ST) solutions. . Please continue posting the solutions on AoPS too.

[asy][asy] size(12cm); defaultpen(fontsize(10pt)); pair A =dir(130); pair B = dir(220); pair C = dir(320); pair M = 0.5*A + 0.5*B; pair N = 0.5*A + 0.5*C; pair O = circumcenter(A,B,C); pair K = 0.8*O + 0.2*A; pair X = rotate(90,A)*K; pair Y = 2*X - A; pair P = foot(A, B, Y); pair Q = foot(A, C, Y); pair Z = intersectionpoints(circumcircle(A, M, N), circumcircle(A, P, Q))[0]; pair L = intersectionpoints(circumcircle(B, M, P), circumcircle(C, N, Q))[1]; draw(A--B--C--A); draw(B--X); draw(C--X); draw(A--Y); draw(B--Y); draw(C--Y); draw(A--P); draw(A--Q); draw(B--L); draw(circumcircle(B, M, P)); draw(circumcircle(C, N, Q)); draw(circumcircle(A, M, N)); draw(circumcircle(A, P, Q)); dot('$A$', A, dir(90)); dot('$B$', B, dir(B)); dot('$C$', C, dir(225)); dot('$M$', M, dir(M)); dot('$N$', N, dir(N)); dot('$X$', X, dir(X)); dot('$Y$', Y, dir(Y)); dot('$P$', P, dir(300)); dot('$Q$', Q, dir(315)); dot('$Z$', Z, dir(315)); dot('$L$', L, dir(L)); [/asy][/asy] Let $Y$ be the reflection of $A$ over $X$, and $P$ and $Q$ be the feet of the perpendiculars from $A$ to $\overline{BY}$ and $\overline{CY}$, respectively. As $X$, $M$, and $N$ are the midpoints of $\overline{AY}$, $\overline{AB}$, and $\overline{AC}$, respectively, we see that $\overline{MX} \parallel \overline{BY}$ and $\overline{NX} \parallel \overline{CY}$. Also, $MB=MP$ and $NC=NQ$, so $\overline{MX}$ is tangent to $(BMP)$ and $\overline{NX}$ is tangent to $(CNP)$. Thus, $\omega_B=(BMP)$ and $\omega_C=(CNP)$, so it suffices to show that $(BMP)$ and $(CNQ)$ intersect on $\overline{BC}$. As $\angle APY = \angle AQY = 90^{\circ}$, $A$, $P$, $Q$, and $Y$ lie on the circle with diameter $\overline{AY}$. Let this circle intersect $(AMN)$ at a second point $Z\neq A$. We claim that $Z$ lies on $(BMP)$ and $(CNQ)$. Since $\overline{AY}$ is tangent to $(ABC)$, it is also tangent to $(AMN)$, so $\measuredangle ZAY = \measuredangle ZMA$. We now have $\measuredangle ZPB=\measuredangle ZPY=\measuredangle ZAY=\measuredangle ZMA=\measuredangle ZMB$, and $Z \in (BMP)$. By symmetry, we also have $Z \in (CNQ)$. Let $L\neq Z$ be the other intersection of $(BMP)$ and $(CNQ)$. We have $\measuredangle ZLB=\measuredangle ZMB=\measuredangle ZMA=\measuredangle ZNA=\measuredangle ZNC=\measuredangle ZLC$, so $L \in \overline{BC}$, as desired.

Kayak wrote: Where I can find references to understand the moving point solution? I don't know what a projective map is There isn't anything public AFAIK... But to answer the question a projective map from a line $\mathbb{RP}^1$ to another line $\mathbb{RP}^1$ is any (bijective) map which preserves the cross ratio. It's a theorem that all such maps are Moebius transformations, i.e. can be written as $x \mapsto \frac{ax+b}{cx+d}$ for a suitable choice of coordinates on the source and target. Kayak wrote: v_Enhance wrote: Okay, I will stop posting the official TST(ST) solutions. . Please continue posting the solutions on AoPS too. Okay. (But I cannot promise they will be easy to find, given the delay. Lots of scrolling.)

This is extremely easy to complex bash. Let $G$ be the other point of tangency to $(AMN)$ from $X$ besides $X$. Note that the reflection of $A$ over $G$ (call this $D$) lies on $(ABC)$. Assume $D\not=B,C$. Let $Y$ be the reflection of $A$ over $X$, so if we let $a,b,c,d$ be on the complex unit circle we have that \[y=\frac{2ad}{a+d}\longrightarrow x=\frac{a}{2}+\frac{ad}{a+d},g=\frac{a+d}{2},m=\frac{a+b}{2}.\] We claim $XM$ is tangent to $MGB$. To prove this, it suffices to show that directed angles $\angle XMB=\angle MGB$, or that \[\frac{\frac{x-m}{m-b}}{\frac{m-g}{g-b}}\]is real. Note that this equals \[\frac{\frac{ad}{a+d}-\frac{b}{2}}{\frac{a-b}{2}}\cdot\frac{\frac{a+d-2b}{2}}{\frac{b-d}{2}}=\frac{(\frac{2ad}{a+d}-b)(a+d-2b)}{(a-b)(b-d)}=\frac{(\frac{2}{\frac{1}{a}+\frac{1}{d}}-b)(\frac{1}{b}-\frac{2}{a+d})b(a+d)}{(a-b)(b-d)}.\]$(\frac{2}{\frac{1}{a}+\frac{1}{d}}-b)(\frac{1}{b}-\frac{2}{a+d})$ is clearly self conjugate, and if we conjugate $\frac{b(a+d)}{(a-b)(b-d)}$ we get $\frac{\frac{1}{b}(\frac{1}{a}+\frac{1}{d})}{(\frac{1}{a}-\frac{1}{b})(\frac{1}{b}-\frac{1}{d})}=\frac{b(a+d)}{(b-a)(d-b)}$ so this is self conjugate so the tangency is proved. Thus $\omega_B=(MGB)$, and by symmetry $\omega_C=(NGC)$. Letting $K=BC\cap (MGB)$, we have that $G$ is the Miquel point of $K,M,N$ with respect to $ABC$, and thus $G$ lies on $(NGC)$ as desired. The $D=B,C$ case follows by continuity.

I will only consider the below config, as all the other ones are analogous. First, note that we wish to show the existence of a point $Y$ on $BC$ such that $\angle NMY=\angle MYB=\angle XMB$ and $\angle MNY=\angle ANX$. Suppose that the length $AX=d$. Then, after dropping a perpendicular from $X$ to $AC$, it is clear that $\cot ANX=\frac{d\cos B+\frac{b}{2}}{d\sin B}=\cot B+\frac{R}{d}$. Similarly, $\cot XMB=\cot C-\frac{R}{d}$, so we have that $\cot ANX+\cot XMB=\cot B+\cot C$. Therefore, if we construct a triangle with these two angles as base angles, the ratio of the base to the height will be just $\cot B+\cot C$, which happens to be the ratio of the base to the height of the medial triangle. Therefore, point $Y$ exists, as desired.

By angle chasing, it's enough to prove that there's a point $T\in (AMN)$ that $T\in \omega_B$ and $T\in \omega_C$. We'll prove the stronger proposition: let $T$ be the point where other tangent from $X$ intersects $(AMN)$, then $T\in \omega_B$. Similarly, we'll get that $T\in \omega_C$ and the prove will be done. Let $XM$ intersects $(AMN)$ for the second time at $M'\neq M$. From $AM=MB$, we get $\frac{AM}{MT}=\frac{MB}{MT}\implies \frac{\sin (\angle{ATM})}{\sin (\angle{MAT})}= \frac{\sin (\angle{MTB})}{\sin (\angle{MBT})}.$ But we've $(M,M'; A,T)=-1\implies \frac{MT}{TM'}=\frac{AM}{AM'}\implies \frac{\sin (\angle{MM'T})}{\sin (\angle{ TMM'})} =\frac{\sin (\angle{AM'M})}{\sin (\angle{AMM'})} .$ So, $\frac{\sin (\angle{MTB})}{\sin (\angle{MBT})}=\frac{\sin (\angle{ATM})}{\sin (\angle{MAT})}=\frac{\sin (\angle{AM'M})}{\sin (\angle{MM'T})} =\frac{\sin (\angle{AMM'})}{\sin (\angle{TMM'})}.$ Hence, $\frac{\sin (\angle{MTB})}{\sin (\angle{MBT})} =\frac{\sin (\angle{XMB})}{\sin (\angle{TMM'})}$ where $\angle{MTB} +\angle{MBT}=\angle{XMB}+\angle{TMM'}.$ From now, easy to prove that $\angle{MTB}=\angle{XMB}$ and $\angle{TMM'}=\angle{MBT}.$ This means $XM$ is tangent to $(MTB)$, so $(MTB)=\omega_B\implies T\in \omega_B$, done. The other side can be proved in similar way using $(A,T;N,N')=-1$.

I guess this was quite easy for a USA TST. Anyway, here's another approach: Invert the figure about $A$ with radius $\sqrt{AB \cdot AC}$ followed by reflection in the angle bisector of $\angle BAC$, to get the following equivalent problem:- Inverted problem wrote: Let $ABC$ be a triangle and let $M$ and $N$ denote the midpoints of $\overline{AB}$ and $\overline{AC}$ respectively. Let $X$ be a point such that $\overline{AX}$ is parallel to $\overline{BC}$. Denote by $\omega_B$ the circle through $M$ and $B$ tangent to $\odot (ABX)$, and by $\omega_C$ the circle through $N$ and $C$ tangent to $\odot (ACX)$. Show that $\omega_B$ and $\omega_C$ intersect on $\odot (AMN)$. PROOF: Let $P$ and $Q$ be the centers of $\odot (ABX)$ and $\odot (ACX)$. Then $\overline{PQ}$ is the perpendicular bisector of $\overline{AX}$. As $\overline{AX} \parallel \overline{BC}$, we get $\overline{PQ} \perp \overline{BC}$. Let $\overline{PQ} \cap \overline{BC}=U$. Then, as $\measuredangle PUB=\measuredangle PMB=90^{\circ}$, we get that $B,M,P,U$ lie on a circle which is tangent to $\odot (P,PB)$, which means that $\odot (BMPU) \equiv \omega_B$. Similarly, we get that $\odot (CNQU) \equiv \omega_C$. This gives that $\omega_B$ and $\omega_C$ meet on $\overline{BC}$. Then we are done by Miquel's Theorem. $\blacksquare$

Here is a seemingly different solution. Let $\odot(AXM), \odot(AXN)$ intersects $AC, AB$ again at $P, Q$. Claim : $MP, NQ, BC$ are concurrent. Proof : Since $AX$ is tangent to $\odot(AMN)$, it's easy to see that $\triangle XMP\sim\triangle XQN\sim\triangle ABC$. Thus by Spiral Similarity, we get $\triangle XMQ\sim\triangle XNP$. Thus $$\frac{MQ}{NP} = \frac{XM}{XP} = \frac{AB}{AC}$$which means if $P'$ is the reflection of $P$ across $N$, then $\overline{AMQB}\sim\overline{ANP'C}$. Let $K=MP\cap BC$ and $L=NQ\cap BC$. By Menelaus' Theorem, we deduce $$\frac{BK}{KC}\cdot \frac{CP}{PA}\cdot\frac{AM}{MB} = 1 \implies \frac{BK}{KC} = \frac{AP}{PC} = \frac{P'C}{P'A}$$Analogously, we get $\frac{BL}{LC} = \frac{BQ}{QA}$. Thus $\frac{BK}{KC} = \frac{BL}{LC}\implies K=L$ as desired. Now just notice that $\measuredangle KBM = \measuredangle CBA = \measuredangle CAX = \measuredangle PMX = \measuredangle KMX$. Thus $K\in\omega_B$. Similarly $K\in\omega_C$ so we are done. Motivation : Circles $\omega_B, \omega_C$ are annoying to deal with. So I worked backward and extend $MK, NK$ because it's nicely related to lines $MX, NX$ by angles. This naturally induces $P, Q$ (which will cause cyclic quad). Then many pairs of spiral similarities are found on $P, Q$. Thus I decided to stick with these two points and complete the solution using ratio chasing.

Let $P$ be the midpoint of $BC$ and $T$ be the second intersection of $\omega_{B}$ with $BC$. By angle chasing, we get $\angle MPT = \angle ACB = \angle XAB$ and $\angle PTM = 180 - \angle MTB = 180 - \angle XMB = \angle AMX$ Therefore, $\triangle PTM \sim \triangle AMX \rightarrow \frac{PT}{PM} = \frac{AM}{AX}$ From $MP = AN, AM = PN$, we get $\frac{PT}{PN} = \frac{AN}{AX}$ From $\angle XAN = 180 - \angle CBA = \angle NPT$, we get $\triangle XAN \sim \triangle NPT$ Therefore, we get $\angle ANX = \angle PTN$ From $\angle XNT = \angle XNM + \angle MNT = \angle XNM + \angle PTN = \angle XNM + \angle ANX = \angle ANM = \angle NCP$, we get $XN$ is tangent to the circle pass through $C, N, T$. So, $\omega_{B}$ and $\omega_{C}$ intersect on $BC$.

Lengths finish immediately. JMO 1? Let $M', N'$ be the intersections of $\omega_B, \omega_C$ with $MN$ respectively. It suffices to show $MM' = NN'$ since $MN \parallel BC$ and projecting onto $BC$, since it suffices for the distance between the midpoints of $M'M$ and $N'N$ to be $\frac{1}{2} BC = MN$. $\frac{NN'}{MM'} \cdot \frac{c}{b} = \frac{NN'}{MM'} \cdot \frac{MB}{NC} = \frac{\frac{MB}{MM'}}{\frac{NC}{NN'}} = \frac{\frac{\sin XMB}{\sin XMN}}{\frac{\sin XNC}{\sin XNM}} = \frac{XM \cdot \sin XMA}{XN \cdot \sin XNA} = \frac{XA \cdot \sin XAB}{XA \cdot \sin XAC} = \frac{\sin C}{\sin B} = \frac{c}{b}$ so we are done.

arajaram wrote: Lengths finish immediately. JMO 1? Apparently not.

Invert the figure at $A$; now the problem becomes the following: Inverted problem wrote: In $\triangle AMN$, $B$ is the midpoint of $\overline{AM}$ and $C$ is the midpoint of $\overline{AN}$. Let $X$ be located on the line through $A$ parallel to $BC$. Let $\omega_B$ be the circle through $M$ and $B$ tangent to the circle $(MAX)$, and $\omega_C$ be the circle through $N$ and $C$ tangent to the circle $(NAX)$. Prove that one of the intersections of $\omega_B$ and $\omega_C$ is concyclic with $A$, $B$, and $C$. Note that a homothety with factor 2 centered at $M$ sends $\omega_B$ to $(MAX)$, and similarly a homothety with factor 2 centered at $N$ sends $\omega_C$ to $(NAX)$. Let $Y$ be the point on $MN$ such that $YA = YX$. We claim that $Y \in \omega_B$. To show that, let $M'$ be the intersection of $MN$ with $(MAX)$ other than $M$. Then $MAXM'$ is an isosceles trapezoid, and $Y$ is the midpoint of $MM'$, so the homothety sending $(MAXM')$ to $\omega_B$ sends $M'$ to $Y$, so $Y$ lies on $\omega_B$, as desired. By similar reasoning, $Y \in \omega_C$. Now let $Z$ be the intersection of $\omega_B$ and $\omega_C$ other than $Y$. But by Miquel's theorem on $\triangle AMN$, with $\omega_B$ and $\omega_C$ passing through $B$ and $C$ respectively and through a common point $Y$ on $MN$, $ABCZ$ is cyclic, as desired.

By Miquel, it suffices to show that $\omega_B$, $\omega_C$, and $(AMN)$ have a common point. We claim that this common point is the tangent from $X$ to $(AMN)$ other than $A$, which we call $Y$. Thus, $XM$ is a symmedian in $\triangle AMY$. We want to show that $XM$ is tangent to $(BMY)$, so we want to show that $\angle AMX=\angle MYB.$ If $M_2$ is the midpoint of $AY$, then by isogonality, $\angle AMX=\angle YMM_2$. However, $MM_2\parallel BY$ since it is the mid-line of $\triangle ABY$, so $\angle AMX=\angle MYB$. Hence, $XM$ is tangent to $(BMY)$. Hence, $Y$ lies on $\omega_B$. Similarly, $Y$ also lies on $\omega_C$, and we are done.

$\text{Nice one. The problem turns out to be easy after we use the transformation}$ $\digamma=\mathcal{S}_{l_A}\circ \mathcal{I}_A^{\frac{1}{2}.\overline{AB}.\overline{AC}}$

what. It suffices to prove the converse is true: given a point $K$ on $\overline{BC}$, the tangents to $(BMK)$ and $(CNK)$ at $M$ and $N$ respectively intersect on the tangent to $(AMN)$ at $A$. This is because the intersection of the tangents at $M$ and $A$ can fall anywhere on the tangent at $A$: to see this, note that as $K$ approaches the midpoint of $\overline{BC}$, the intersection point goes to infinity in a direction (dependent on which side of $M$ that $K$ lies on), and as $K$ approaches infinity, the intersection point goes to $A$, and apply continuity. Let $D$ and $E$ be the second intersections of $(BMK)$ and $(CNK)$ with $\overline{MN}$ respectively, so $BMDK$ and $CNEK$ are isosceles trapezoids. By Ceva and Law of Sines it suffices to prove that

On the other hand, if $K$ is translated by some vector, clearly $D$ and $E$ are translated by that vector as well. Moreover, since $D=M$ and $C=N$ when $K$ is the foot of the $A$-altitude, it follows that $\frac{KD}{KE}=\frac{AB}{AC}$ and $\frac{NE}{MD}=1$ always hold (except when $D=M$, but this is handled by continuity), so we are done. $\blacksquare$

the "synthetic" solution above gives me nightmares so i will do the morally correct thing and kill the problem with coordinates instead Let $A=(0,2),B=(2b,0),C=(2c,0)$, so $M=(b,1),N=(c,1)$. It is easy to calculate the coordinates of the circumcenter as $(b+c,bc+1)$, so the tangent to $(ABC)$ at $A$ should have slope $\tfrac{b+c}{1-bc}$. Let $X=(p,q)$. Then the slope of $\overline{MX}$ is $\tfrac{q-1}{p-b}$, so the circumcenter of $\omega_B$ can be found by intersecting the perpendicular bisector of $\overline{BM}$, which has equation $y=bx-\tfrac{3b^2-1}{2}$, with the perpendicular to $\overline{MX}$ at $M$, which has equation $y=\tfrac{p-b}{1-q}(x-b)+1$. Some computation gives the $x$-coordinate of this center as $$\frac{bp-\frac{5b^2}{2}+\frac{3b^2q}{2}-\frac{1}{2}+\frac{q}{2}}{p-2b+bq}.$$In fact, this is all we need to find the second intersection of $\omega_B$ with $\overline{BC}$: the $x$-coordinate of this second intersection should be twice this expression minus $2b$, which works out to be $$\frac{-b^2+b^2q-1+q}{p-2b+bq}=\frac{(b^2+1)(q-1)}{p+b(q-2)}.$$We now calculate the denominator. Since the slope of $\overline{AX}$ is $\tfrac{b+c}{1-bc}$, we should have $$\frac{q-2}{p}=\frac{b+c}{1-bc} \iff p=\frac{(1-bc)(q-2)}{b+c} \implies p+b(q-2)=\frac{(1-bc+b^2+bc)(q-2)}{b+c}=\frac{(b^2+1)(q-2)}{b+c}.$$It follows that $$\frac{(b^2+1)(q-1)}{p+b(q-2)}=\frac{(b+c)(q-1)}{q-2},$$which is symmetric in $b$ and $c$, hence the analogously defined point for $\omega_C$ will be the same, i.e. $\omega_B$ and $\omega_C$ concur on $\overline{BC}$. Note that denominators equalling zero is not an issue for continuity reasons. We are done. $\blacksquare$

Just another solution using the method of moving points: Let $\omega_B$ intersect $BC$ at $S$ . Similarly, Let $\omega_C$ intersect $BC$ at $T$ . We shall define $P$ as the intersection of two lines $MS$ and $AC$ , and $Q$ as the intersection of the lines $NT$ and $AB$ . Claim 1: $PXMA$ and $QXNA$ are cyclic quadrilaterals. Proof 1: $\angle APM=\angle CPS =\angle BSP -\angle BCP=\angle BSM -\angle BAX=\angle BMX -\angle BAX=\angle AXM$. Therefore, we get the desired result for $PXMA$; utilizing the same approach, we come to prove the claim. Now, fix $A,B,C$ and move $X$ along the line tangent to $(ABC)$ at $A$ (let this line be $l$). Claim 2: $f$:$X\mapsto S$ and $g$:$X\mapsto T$ are projective. Proof 2: We will demonstrate the first one as the latter is similar. Let's consider inversion $\Phi$ centered at A with radius 1. Note that $\Phi(X),\Phi(M),\Phi(P)$ lie on a line. If we get $f$:$X\mapsto \Phi(X)\mapsto\Phi(X)\Phi(M)\mapsto\Phi(X)\Phi(M)\cap\Phi(AC)=\Phi(P)\mapsto P\mapsto PM\mapsto PM \cap BC=S$ , the function is projective. Hence, the result follows. Finally, we need to observe three cases. Our first point will be $MN\cap l$. It is obvious that both $S$ and $T$ are the foot of altitude from $A$. Second point will be at infinit along $l$. We can see that both $S$ and $T$ are the midpoint of the side $BC$. As third, we get the intersection $R$ of $l$ and $BC$. Since \[ \frac{RM^2}{RN^2} = \frac{RB}{RB} \](This follows from the similarity of $RAMB$ and $RCNA$.), again $S=T$. So, for all $X$ on line $l$, $S=T$.

Oops. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17.5cm); 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 = -4.123705830044876, xmax = 4.87995712658351, ymin = -5.28191760484135, ymax = 1.967229278661107; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); draw(circle((-1.182152078949922,-0.8414411514978412), 2.244662198512096), linewidth(0.9) + zzttff); draw(circle((-1.890651113165819,-1.6868604311144357), 1.4207720777081836), linewidth(0.9) + blue); draw(circle((0.23302527404884588,-1.102186043065105), 0.8519679649548042), linewidth(0.9) + blue); /* draw figures */ draw((0.15006136387206684,0.9651345397461043)--(-3.306349893889249,-1.5669010252580262), linewidth(0.9)); draw((-3.306349893889249,-1.5669010252580262)--(0.9366137566295555,-1.5826157054393228), linewidth(0.9)); draw((0.9366137566295555,-1.5826157054393228)--(0.15006136387206684,0.9651345397461043), linewidth(0.9)); draw((0.9366137566295555,-1.5826157054393228)--(-0.4741025953800833,-1.577390830063441), linewidth(0.9)); draw((3.3841645459585936,-1.4197728602213702)--(-1.578144265008591,-0.30088324275596096), linewidth(0.9)); draw((0.5433375602508111,-0.30874058284660927)--(1.5402304160674711,-0.6986200600090183), linewidth(0.9) + blue); draw((1.5402304160674711,-0.6986200600090183)--(0.7399756584401833,-0.9456781441429661), linewidth(0.9) + blue); draw((0.7399756584401833,-0.9456781441429661)--(0.23302527404884582,-1.102186043065105), linewidth(0.9) + blue); draw((0.23302527404884582,-1.102186043065105)--(0.23125558062473608,-1.580003267751382), linewidth(0.9) + blue); draw((0.15006136387206684,0.9651345397461043)--(3.3841645459585936,-1.4197728602213702), linewidth(0.9) + blue); draw((-0.4741025953800833,-1.577390830063441)--(0.5433375602508111,-0.30874058284660927), linewidth(0.9)); draw((1.5402304160674711,-0.6986200600090183)--(3.3841645459585936,-1.4197728602213702), linewidth(0.9) + blue); /* dots and labels */ dot((0.15006136387206684,0.9651345397461043),dotstyle); label("$A$", (0.1727188888025096,1.0600160913514725), NE * labelscalefactor); dot((0.9366137566295555,-1.5826157054393228),dotstyle); label("$C$", (0.9515528517808205,-1.7728854652474806), NE * labelscalefactor); dot((-1.578144265008591,-0.30088324275596096),linewidth(4pt) + dotstyle); label("$M$", (-1.7101764063098903,-0.22377615484140662), NE * labelscalefactor); dot((0.5433375602508111,-0.30874058284660927),linewidth(4pt) + dotstyle); label("$N$", (0.5835324077361241,-0.22377615484140662), NE * labelscalefactor); dot((3.3841645459585936,-1.4197728602213702),dotstyle); label("$X$", (3.459227040271426,-1.4733339411358088), NE * labelscalefactor); dot((-0.4741025953800833,-1.577390830063441),linewidth(4pt) + dotstyle); label("$D$", (-0.6232323041313685,-1.7215337753997655), NE * labelscalefactor); dot((-3.306349893889249,-1.5669010252580262),linewidth(4pt) + dotstyle); label("$B$", (-3.464692476755536,-1.6787407005266695), NE * labelscalefactor); dot((0.23302527404884582,-1.102186043065105),linewidth(4pt) + dotstyle); label("$O_C$", (0.16416027382472595,-0.9854928875825149), NE * labelscalefactor); dot((0.23125558062473608,-1.580003267751382),linewidth(4pt) + dotstyle); label("$T$", (0.2668636535581296,-1.5075684010342856), NE * labelscalefactor); dot((1.5402304160674711,-0.6986200600090183),linewidth(4pt) + dotstyle); label("$R$", (1.5592145152034587,-0.6260310586485087), NE * labelscalefactor); dot((0.7399756584401833,-0.9456781441429661),linewidth(4pt) + dotstyle); label("$S$", (0.6434427125806096,-1.1481065721002794), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\Gamma_B$ and $\Gamma_C$ denote the circles tangent to $\overline{XM}$ passing through $B$, and $\overline{XN}$ passing through $C$, and denote their centers by $O_B$ and $O_C$. Also, let $R$ be the pole of $\overline{CN}$ with respect to $\Gamma_C$, $S$ be the midpoint of $\overline{CN}$, and , and $T$ be the midpoint of $\overline{CD}$. Animate $X$ linearly along $\overline{AA}$. Define the map $f: AA \mapsto BC$, such that $X \in \overline{AA}$ is taken to $D \in \overline{BC}$ such that $(BMD)$ is tangent to $\overline{XM}$. To see that this map is projective consider the following compositions: Map $X \mapsto R$, by the negative homothety at $N$. Map $R \mapsto S$, by inversion with respect to $\Gamma_C$. Map $S \mapsto O_C$, by a projection at $R$. Map $O_C \mapsto T$, as a map from any point to a fixed line is projective. Map $T \mapsto D$, by a homothety with a fixed scale factor of $2$ at $C$. Similarly define the map $g: \overline{AA} \mapsto \overline{BC}$ such that $X \in \overline{AA}$ is taken to $E \in \overline{BC}$ such that $(CNE)$ is tangent to $\overline{CN}$. To show that $D \equiv E$, it suffices to show the two projective maps are the same. To do this we must show that there are equivalent for three such points $X$. Take $X = A$, for the first case. Then $D$ and $E$ are both sent to the point at infinity along $\overline{BC}$. Take $X = P_\infty$, where $P_\infty$ denotes the point at infinity along $\overline{AA}$. Then $D$ and $E$ are sent to the midpoint of $\overline{BC}$. Take $X = \overline{MN} \cap \overline{AA}$. Then $D$ and $E$ concur at the foot of the altitude from $A$ to $\overline{BC}$. Thus we are done.

We use complex numbers. I had to use so many hints.... Let $(ABC)$ be the unit circle. Then \[m = \frac{a + b}{2}, n = \frac{c + a}{2}, \overline{x} = \frac{2a - x}{a^2},\]as $x$ is on chord $AA$ of $(ABC)$. Now observe that $O_BM \perp MX$ and $BO_B = O_BM$. Now as $BO_B = O_BM$, we have that \begin{align*} (o_b - b)(\overline{o_b} - \overline{b}) &= (o_b - m)(\overline{o_b} - \overline{m}) \\ o_b \overline{b} - o_b \overline{b} - b \overline{o_b} + b \overline{b} &= -\overline{m}o_b - m \overline{o_b} + m \overline{m} \\ -\frac{o_b}{2b} + \frac{o_b}{2a} - \frac{b \overline{o_b}}{2} + \frac{a \overline{o_b}}{2} + \frac{1}{2} - \frac{a}{4b} - \frac{b}{4a} &= 0 \\ -\frac{o_b(a - b)}{2ab} + \frac{\overline{o_b}(a - b)}{2} - \frac{(a - b)^2}{4ab} &= 0 \\ \implies \overline{o_b} = \frac{2o_b + a - b}{2ab}; \implies \overline{o_c} &= \frac{2o_c + a - c}{2ac}. \\ \end{align*}Now as $O_BM \perp MX$ we have that \[\frac{m - o_b}{m - x} = -\left(\overline{\frac{m - o_b}{m - x}} \right) \iff \frac{a + b - 2o_b}{a + b - 2x} = - \frac{\overline{m} - \overline{o_b}}{\overline{m} - \overline{x}}.\]\begin{align*} \implies \frac{a + b - o_b}{a + b - 2x} &= \frac{2a(o_b - b)}{a^2 + 2bx - 3ab} \\ \implies a^3 + 2abx - 3a^2b + a^2b + 2b^x - 3b^2a - 2o_ba^2 - 4bxo_b + 6abo_b &= 2a^2o_b + 2abo_b - 4axo_b - 2a^2b - 2b^2a + 4abx \\ \implies a^3 - 2abx + 2b^2x - b^2a &= 4o_ba^2 + 4bxo_b - 4abo_b - 4axo_b \\ \implies a(a^2 - b^2) + 2bx(b - a) &= 4o_ba(a - b) + 4xo_b(b - a) \\ \implies a(a + b) - 2bx &= 4o_ba - 4xo_b \\ \implies o_b = \frac{a^2 + ab - 2bx}{4(a - x)}; \implies o_c &= \frac{a^2 + ac - 2cx}{4(a - x)} \\ \end{align*} Now let the foot from $O_B$ to $BC$ be $D$ and the foot from $O_C$ to $BC$ be $E$. Also let $P_1$ be the intersection of $\omega_B$ with $BC$ (not $B$), and let $P_2$ similarly be the intersection of $\omega_C$ with $BC$ (not $C$). Then we have that $d = 1/2(b + p_1)$ and $d = 1/2(b + c + o_b - bc \overline{o_b})$, so that $p_1 = c + o_b - bc \overline{o_b}$, and likewise $p_2 = b + o_c - bc \overline{o_c}$. Then we have that \[p_1 = c + \frac{a^2 + ab - 2bx}{4(a - x)} - bc \left(\frac{\frac{1}{a^2} + \frac{1}{ab} - \frac{4}{ab} + \frac{2x}{a^2b}}{4 \left(\frac{x - a}{a^2} \right)} \right) \implies p_1 = \frac{(a + b)(a + c) - 2x(b + c)}{4(a - x)}.\]However now observe that as $p_1$ is invariant if we interchange $b, c$, it must follow that $p_1 = p_2$, as desired. $\blacksquare$

solved with ihatemath123 orz . Under an inversion at $A$ with radius $\sqrt{\frac{AB\cdot AC}2}$ followed by reflection across the angle bisector of $A$ let $X$ get sent to a point $Y$ with $AY\parallel BC.$ We can see $\omega_B$ is sent to the image of $(AYC)$ under homothety at $C$ with scale factor $\frac12,$ and $\omega_C$ is sent to the image of $(AYB)$ under homothety at $B$ with scale factor $\frac12.$ Now let $T$ be a point on $BC$ with $AT=YT.$ Letting $B',C'$ be the reflections of $B,C$ over $T,$ we see $AYBB',AYCC'$ are isosceles trapezoids, so under the homothety, $T$ lies on the images of both $\omega_B$ and $\omega_C.$ Undoing the inversion, $T$ gets sent to a point on $(AMN),$ so by Miquel points the other intersection of $\omega_B$ and $\omega_C$ lies on $BC.$

Fix $\triangle ABC$ and animate $X$ along the tangent to $(ABC)$ at $A.$ Let $D$ be the point on the perpendicular to $\overline{MX}$ through $M$ where $\overline{BD} \perp \overline{BM}.$ Similarly denote $E$ the point on the perpendicular to $\overline{NX}$ through $N$ where $\overline{EC} \perp \overline{CN}.$ Let $T_1=\omega_B \cap \overline{BC}.$ Note $\angle DMT_1$ is fixed, since $\angle DT_1M = \angle XMD$ and $\angle MDT_1=180^\circ - \angle MBT_1,$ which is fixed. Now observe the map $X \mapsto D \mapsto T_1$ is projective, with both maps using fixed angle rotation about $M.$ On the other hand, denoting $T_2=\omega_C \cap \overline{BC},$ we obtain the projective map $X \mapsto E \mapsto T_2,$ this time by rotation about $N.$ It suffices to check these maps coincide for three positions of $X.$ When $X=A,$ circles $\omega_B$ and $\omega_C$ are simply $\overline{AMB}$ and $\overline{ANC},$ which both intersect $\overline{BC}$ again at $\infty_{BC}.$ When $X$ is $\infty$ along the the tangent to $(ABC)$ passing through $A,$ we have $MX$ and $NX$ are the lines through $M,N$ parallel to the tangent line to $(ABC)$ through $A.$ Then, by homotheties of ratio $1/2$ centered at $B$ and $C,$ we see that $\omega_B$ and $\omega_C$ intersect at the midpoint of $BC.$ When $X= \overline{MN} \cap \overline{AA},$ then $\overline{MD}$ and $\overline{NE}$ are parallel to the $A$-altitude. By homotheties at $B$ and $C$ with ratio $2$ we can see that $(MBD)$ and $(CNE)$ both pass through $\text{foot}(A,\overline{BC}).$ We have checked three cases, and may conclude.

The setup is perfect for applying a $\sqrt{\frac{bc}2}$ inversion. Under this inversion, $M,N\longleftrightarrow C,B$ $\text{line tangent to circumcircle }ABC\text{ at A}\longleftrightarrow \text{line through }A\text{ parallel to }BC$ $\text{line }BC\longleftrightarrow\text{circumcircle }AMN$. Claim: If we take a point $X^*$ such that $AX^*\parallel BC$ and consider circle $\tau_1$ passing through $M,B$ and tangent to $(ABX^*)$ and circle $\tau_2$ passing through $N,C$ and tangent to $(ACX^*)$, then the circles intersect on line $BC$. Proof. Let $O_1,O_2$ be the centres of $(ABX),(ACX)$, respectively. Then, line $O_1O_2$ is the perpendicular bisector and hence $O_1O_2\perp AX\parallel BC$. Let $E^*$ be the intersection of $O_1O_2$ and $BC$. Also, $O_1$ lies on perpendicular bisector of $\overline {AB}$ which means $O_1M\perp AB$. However we also know $O_1E^*\perp BC$. This means $\angle O_1MB=O_1E^*B=90^\circ$, so $O_1,M,B,E^*$ are cyclic with diameter $\overline{O_1B}$. So, this new circle $O_1MBE^*$ is just the circle $(ABX^*)$ but with homothety at $B$ with factor $\frac12$. Hence, this circle is tangent to $(ABX^*)$ and it also passes through $M,B$. Hence, it follows that $O_1,M,B,E^*\in\tau_1$. More importantly, $E^\star\in\tau_1$ and similarly we get $E^*\in\tau_2$. Since $E^*$ is a point common to $BC,\tau_1,\tau_2$, the claim is true. $\blacksquare$ Let $D^*$ be the second intersection of $\tau_1,\tau_2$. Note that $\angle CBA=\angle E^*BM=180^\circ-\angle MD^*E^*$ and similarly $\angle E^*D^*N=180^\circ-\angle ACB$. Adding both, we get $\angle ND^*M=180^\circ-\angle MAN$, which means $D^*\in (MAN)$. Apply $\sqrt{\frac{bc}2}$ inversion. Under this inversion, suppose $X*$ goes to $X$. Then, by definition, $D^*,E^*$ go to $D,E$. Since $D^*\in(MAN)$, it means $D\in BC$, as desired.

Untethered moving points. Move $X$ along $\overline{AA}.$ Let $\overline{XM}, \overline{XN}$ be moving lines. Note that both of these have degree $1$ by the converse of Zack's lemma. Then consider the rotations of these lines about $M,N$ by $90$ degrees, respectively. Note that these lines must also be degree $1.$ Thus, the intersection of these rotated lines and the perpendicular bisectors of $BM, CN,$ respectively, must also move with degree $1.$ Note that the aformentioned points are the centers of $\omega_B, \omega_C,$ respectively. Let $H$ be the orthocenter of $\triangle ABC.$ Note that $\omega_B \cap \overline{BC}, \omega_C \cap \overline{BC}$ also both move with degree $1$ because it's just a reflection of $B,C$ respectively over the centers of $\omega_B,\omega_C$ after projecting to $\overline{BC}$ from $\infty_{\overline{AH}}.$ Thus we simply need to check $2$ or $3$ cases for the problem to be complete. Let's just check $3.$ $XMN$ collinear. $X=A.$ $X=\infty_{\overline{AA}}.\blacksquare$

for the sake of memorization