Let $(X,XA)$ meets $AB, AC$ again at $E, F$ $\angle AEX = \angle BAX = 2 \angle XBA \Rightarrow \angle EBX = \angle EXB$ so $EB=EX$ $\angle AFX = \angle CAX = 2 \angle XCA \Rightarrow \angle FCX = \angle FXC$ so $FC=FX=EX=EB$ Since $EB = FC,$ $(X, XA)$ passes through $M$ (See Prove that MN is parallel to AI post #4)

This problem was proposed by Burii.

Since $EB = FC,$ $(X, XA)$ passes through $M$ Can someone show how does that follow?

Mprog. wrote: Since $EB = FC,$ $(X, XA)$ passes through $M$ Can someone show how does that follow? Let $Q$ be the Miquel point of $BEFC$. We have that $\triangle QEB \sim \triangle QFC$ by spiral similarity so as $EB = FC$, the two triangles are in fact congruent so $QB = QC$. As $Q, A$ clearly lie on the same side of $\overline{BC}$, ($X$ lies inside $\triangle ABC$) we conclude that $Q = M$ as desired.

Amazing problem. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 1; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.931619015998317, xmax = 12.35268271700373, ymin = -6.2, ymax = 6.295618999209102; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pair A = (-3.023693176494155,4.80720091682328), B = (-5.559967711756501,-2.749982437357092), C = (3.9630069673681714,-2.8168117495423215); pair M = (-0.7418154036524869,5.291189386209763), P = (-2.5219236580982507,0.48914284080444176), Q = (0.07919666832617143,0.712700419228189); pair X = (-1.2118272753134591,1.8859515227576646), Y = (-3.6364372221499246,4.089377965597112), Z = (0.8344687186207644,4.067687193302564); pair O = (-0.78,-0.15), Op = (-1.407153798515055,2.8068644781947234); path omega = circle(Op, 2.5718758578637173); pair R = intersectionpoint(X--(2*X-A),omega); /* draw figures */ draw(Y--X--Z^^X--R,pink); draw(O--Op,gray(0.7)); draw(circle(O, 5.441323368446319), darkgreen); draw(B--X--C^^A--X,purple); draw(omega, red+linetype("4 4")); draw(circle((-0.8114130509364313,-4.168902949863692), 4.962180682439345), red+linetype("4 4")); draw(P--A--Q,orange); draw(A--B--C--cycle, rvwvcq); /* dots and labels */ dot((-3.023693176494155,4.80720091682328),dotstyle); label("$A$", A, NW * labelscalefactor); dot((-5.559967711756501,-2.749982437357092),dotstyle); label("$B$", B, SW * labelscalefactor); dot((3.9630069673681714,-2.8168117495423215),dotstyle); label("$C$", C, SE * labelscalefactor); dot(M,linewidth(4pt) + dotstyle); label("$M$", M, N * labelscalefactor); dot(X,dotstyle); label("$X$", X, W*1.5); dot(P,linewidth(4pt) + dotstyle); label("$P$", P, 1.7*S); dot(Q,linewidth(4pt) + dotstyle); label("$Q$", Q, 1.7*S); dot(Z,linewidth(4pt) + dotstyle); label("$S$", Z, NE * labelscalefactor); dot(Y,linewidth(4pt) + dotstyle); label("$T$", Y, W * labelscalefactor); dot(R,linewidth(4pt) + dotstyle); label("$R$", R, S * labelscalefactor); dot(O,linewidth(4pt) + dotstyle); dot(Op,linewidth(4pt) + dotstyle); dot(A^^B^^C,linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] The proof proceeds in four steps. Step 1: quadrilateral $\boldsymbol{BPQC}$ is cyclic. Let $P$ and $Q$ be the feet of the $A$-angle bisectors in triangles $ABX$ and $ACX$, respectively. Observe that \[ \angle PAX = \tfrac12\angle BAX = \angle ABX \]by the given condition, so $\triangle XPA\sim\triangle XAB$. This yields $XA^2 = XP\cdot XB$, and analogous reasoning with $\triangle XAC$ yields $XA^2 = XQ\cdot XC$. Therefore $XP\cdot XB = XQ\cdot XC$, and so quadrilateral $BPQC$ is cyclic. Step 2: quadrilateral $\boldsymbol{APQM}$ is cyclic. Suppose $PQ$ intersects $BC$ again at point $Y$ (not shown in the diagram). Menelaus and the Angle Bisector Theorem imply \[ \frac{BY}{YC} = \frac{BP}{PX}\cdot\frac{XQ}{QC} = \frac{BA}{AX}\cdot\frac{XA}{AC} = \frac{BA}{AC}. \]This means that $Y$ is the foot of the $A$-external angle bisector of $\triangle ABC$, meaning that $PQ$, $BC$, and $AM$ are concurrent. Now \[ YA\cdot YM = YB\cdot YC = YP\cdot YQ, \]and so quadrilateral $APQM$ is cyclic. Step 3: $\boldsymbol{X}$ is a center of homothety. Let $AX$, $BX$, and $CX$ intersect $\odot(APQM)$ again at $R$, $S$, and $T$ respectively. Then a bit of angle chasing yields \[ \angle RSP = \angle RAP \equiv \angle XAP = \angle ABX, \]and so $SR\parallel AB$. Analogous reasoning yields $RT\parallel AC$, and furthermore Reim's Theorem tells us $ST\parallel BC$. Therefore triangles $ABC$ and $RST$ are homothetic, and their center of homothety is $X$. Step 4: finishing touches. Observe that the homothety above sends $\odot(APQM)$ to $\odot(ABC)$, and so $X$ lies on the line connecting the centers of these two circles. But this line is precisely the perpendicular bisector of $\overline{AM}$, and so $XA=XM$ as desired.

okay well i guess this was already mentioned above, but i'll still post this here

djmathman wrote: Amazing problem.

Here's another way to finish the solution after the first 2 steps: Invert w.r.t circle $(X,XA)$, this swaps $P\leftrightarrow B$, $Q\leftrightarrow C$ and $A$ stays, hence swaps $(APQ)\leftrightarrow (ABC)$. Now $M$, which lies on both circles stays, thus $M$ lies on $(X,XA)$ and $XM=XA$.

djmathman wrote: Amazing problem. Only few could solve it during the contest.

Let $\ell$ be the perpendicular bisector of $AB$ and let $D=\overline{BX}\cap\ell$, which means $\overline{AD}$ is the angle bisector of $\angle BAX$. $\frac{AX}{AB}=\frac{DX}{DB}=\frac{d(X,\ell)}{d(B,\ell)}=\frac{d(X,\ell)}{\frac{AB}{2}}$. By symmetry, $X$ is equidistant from the perpendicular bisectors of $AB,AC$, as desired (sure it could be equidistant to the other arc midpoint but config issues don't matter).

Nice problem! Let $E$ and $F$ on $AB$ and $AC$ respectively; Such that $XA=XF=XE.$ By easy angle chasing we have $BE=EX=XA=XF=FC.$ Then we have $\triangle MEB= \triangle MFC$ because we know that $\angle EBM= \angle FCM,$ $BE=CF$ and $BM=CM.$ We conclude that $AMFE$ is cyclic and we are done. $\square$

Someone told me to do this problem. Let $\Gamma$ be the circle centered at $A$ with radius $\sqrt{AB \cdot AC}{}$. Consider the "force-overlay polarity" given by doing pole-polar duality wrt $\Gamma$, then reflecting over the angle bisector of $\angle BAC$. It is not hard to check that under this transformation, the problem is as follows (angle chase; the two angle conditions act independently): Dual problem wrote: Let $\triangle ABC{}$ be an acute triangle, and let the external angle bisector of $\angle BAC$ intersect line $BC$ at $X$. Let $\omega_B$ be the circle centered at $2B - A$ through $B$; define $\omega_C$ similarly. Suppose $\ell$ is a common tangent of $\omega_B$ and $\omega_C$. Show that $2X-A$ lies on $\ell$. But it is easy to check that $2X-A$ is the exsimilicenter of the two circles (compare radii and distances). $\blacksquare$