This problem was written by Aron Thomas, United Kingdom (me).

nice problem but i think a bit hard for P1

Using inversion in $A$ it can be seen that $X\mapsto X^*\mapsto B^*X^*\mapsto C^*Y^*\mapsto Y^*\mapsto Y$ is a projective map (the third map is a translation+reflection=a reflection). Furthermore, it can be checked that it also is an involution (for example by looking at the arcs formed by the angles $\measuredangle AB^*X^*,\measuredangle AB^*Y*$, etc). Therefore, it has a fixed point (since it isn't the Identity)

Nice problem. Pascal twice immediately finishes. Let $XA$, $YA$, $BA$, and $CA$ intersect $\omega$ at points $B', C', T, S$. Notice that $$\angle BXB' = \angle CYC'$$meaning that $BC$ is parallel to $B'C'$. We will show that the desired concurrency point is the intersection of $ST$ and the parallel through $A$ to $BC$, which is clearly fixed. Let $K$ be the intersection of $BX$ and $C'S$, notice that by Pascal's Theorem on $STBXYC'$ that $AK$ passes through the intersection of $ST$ and $XY$. Moreover, by Pascal's Theorem on $BXB'C'SC$, we have that $AK$ also passes through the intersection of $BC$ and $B'C'$ and therefore $AK$ is the parallel through $A$ to $BC$. We ultimately have that $XY$ passes through $AK \cap ST$ which as mentioned before is fixed. $\blacksquare$

Let $AB$ and $AC$ intersect $\omega$ again at $P$ and $Q$, and let $PQ$ intersect the line through $A$ parallel to $BC$ at $E$. We claim $E$ is the desired fixed point. If $XA$ and $YA$ intersect $\omega$ again at $F$ and $G$, the angle condition implies $BCFG$ is an isosceles trapezium. Hence, $FG||BC$, so: $$\angle{XYA}=\angle{AFG}=\angle{XAE}$$Which means $EA$ is tangent to $(AXY)$. It's easy to check that $(APQ)$ is also tangent to $EA$. Thus, by the radical axis theorem applied to $PQXY$, $PQAA$ and $XYAA$, $XY$ meets $EA$ at $E$ as desired.

Balkan MO 2021/1 wrote: Let $ABC$ be a triangle with $AB<AC$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle BXA= \angle AYC$. Suppose also that $X$ and $C$ lie on opposite sides of the line $AB$ and that $Y$ and $B$ lie on opposite sides of the line $AC$. Show that, as $X, Y$ vary on $\omega$, the line $XY$ passes through a fixed point. Cool problem! Congratulations to the author! [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -8.76, xmax = 6.2, ymin = -6.1, ymax = 6.1; /* image dimensions */ pen wwwwww = rgb(0.4,0.4,0.4); /* draw figures */ draw((-0.96,1.3)--(-2.52,-1.38), linewidth(1)); draw((-2.52,-1.38)--(2.52,-1.34), linewidth(1)); draw((2.52,-1.34)--(-0.96,1.3), linewidth(1)); draw(circle((-0.014197491462569661,0.4288839242837845), 3.090486573878745), linewidth(1) + wwwwww); draw((-1.4206770237557222,3.18077828791136)--(0.010329407318762973,-2.661505322164137), linewidth(1)); draw((-5.165887274981112,1.2666199422620552)--(-0.96,1.3), linewidth(1) + wwwwww); draw((-1.4206770237557222,3.18077828791136)--(-5.165887274981112,1.2666199422620552), linewidth(1) + wwwwww); draw((-5.165887274981112,1.2666199422620552)--(1.4677093318989498,3.140905205579223), linewidth(1)); draw((-2.6962246053781356,1.9644091841041895)--(3.00775294278384,-0.21835856217914973), linewidth(1) + wwwwww); draw((1.4677093318989498,3.140905205579223)--(-3.025494212321238,-0.2662414761085552), linewidth(1) + wwwwww); draw((-3.025494212321238,-0.2662414761085552)--(3.00775294278384,-0.21835856217914973), linewidth(1) + linetype("2 2") + wwwwww); /* dots and labels */ dot((-0.96,1.3),dotstyle); label("$A$", (-0.88,1.5), NE * labelscalefactor); dot((-2.52,-1.38),dotstyle); label("$B$", (-3,-1.3), NE * labelscalefactor); dot((2.52,-1.34),dotstyle); label("$C$", (2.65,-1.3), NE * labelscalefactor); dot((0.010329407318762973,-2.661505322164137),dotstyle); label("$M$", (0.1,-2.46), NE * labelscalefactor); dot((-1.4206770237557222,3.18077828791136),linewidth(4pt) + dotstyle); label("$Z$", (-1.34,3.34), NE * labelscalefactor); dot((-5.165887274981112,1.2666199422620552),linewidth(4pt) + dotstyle); label("$D$", (-5.08,1.42), NE * labelscalefactor); dot((1.4677093318989498,3.140905205579223),dotstyle); label("$Y$", (1.54,3.34), NE * labelscalefactor); dot((-2.6962246053781356,1.9644091841041895),linewidth(4pt) + dotstyle); label("$X$", (-2.82,2.12), NE * labelscalefactor); dot((-3.025494212321238,-0.2662414761085552),linewidth(4pt) + dotstyle); label("$Y'$", (-3.4,-0.3), NE * labelscalefactor); dot((3.00775294278384,-0.21835856217914973),linewidth(4pt) + dotstyle); label("$X'$", (3.08,-0.06), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $M$ be the midpoint of arc $BC$ that lies on different side with $A$ wrt $BC$ in $\omega$. We claim that if $Z = AM \cap \omega$, then $D$, which is the intersection of tangent of $\omega$ at $Z$ and parallel of $BC$ passing through $A$ is the desired fixed point. Before that, let us define $X' = XA \cap \omega$ and $Y' = YA \cap \omega$ and redefine $D$ as the intersection of $XY$ with the line passing through $A$ parallel to $BC$. Claim 01. $X'Y' \parallel BC \parallel AD$. Proof. Note that $\angle BXX' = \angle BXA = \angle AYC = \angle Y'YC$. Furthermore, we know that $BC \parallel AD$, which leads to the desired claim. We have $\angle XAD = \angle XX'Y = \angle XYY' = \angle XYA$, which proves that $\triangle DXA \sim \triangle DAY$, and therefore \[ DX \cdot DY = DA^2 \]Now, let us redefine $Z$ so that $Z \in \omega$ and $DZ$ tangent to $\omega$. Then, we must have \[ DZ^2 = DX \cdot DY = DA^2 \]It suffices to prove that as $X,Y$ varies, $Z$ is fixed. Claim 02. $Z,A,M$ is collinear. Proof. Let $ZA \cap \omega = M'$ and $BC \cap ZA = K$. Note that \[ \angle BKZ = \angle DAZ = \angle DZA = \angle DZM' = \angle ZCM' \ \text{and} \ \angle ZBK = \angle ZBC = \angle ZM'C \]This implies $\triangle ZBK \sim \triangle ZM'C$, which implies $\angle BZM' = \angle BZK = \angle M'ZC$, which forces $M'$ to be the midpoint arc of $BC$.

I hate who I've become, but a correct solution is a correct solution. Let $P = AX\cap \omega$ and $Q = AY\cap\omega$; the condition $\angle BXA = \angle AYC$ is equivalent to $PQ\parallel BC$, that is, $PQ$ passes through $\infty_{BC}$. Now the map $X\mapsto P\mapsto Q\mapsto Y$ is a projective involution (project through $A$, then $\infty_{BC}$, then $A$ again), so $XY$ passes through a fixed point. (And the condition that $AB < AC$ ensures that this point is not an infinity point.)

Let $X_1,Y_1$ be fixed points such that $\angle BX_1A=\angle AY_1C$ and let $X_2,Y_2$ vary on $\omega$ such that $\angle BX_2A=\angle AY_2C$. We will prove that as $X_2,Y_2$ vary on $\omega$, the point $X_1Y_1\cap X_2Y_2$ is fixed. Let $X_1Y_1\cap X_2Y_2=P$. Let $X_1A\cap \omega=\{X_1,X_1^*\}$. Define $X_2^*,Y_1^*,Y_2^*$ similarly. $\angle X_1^*X_1B=\angle AX_1B=AY_1C=Y_1^*Y_1C\Rightarrow X_1^*Y_1^*//BC$. Similarly, $X_2^*Y_2^*//BC$. Apply Pascal on $Y_2^*X_1Y_1Y_1^*X_2Y_2$. This gives us $A, P$ and $Y_2^*X_1\cap Y_1^*X_2$ are collinear. Let $Y_2^*X_1\cap Y_1^*X_2=K$, Then $A-P-K$. Apply Pascal on $X_1X_1^*Y_1^*X_2X_2^*Y_2^*$. This gives us, $AK//X_1^*Y_1^*$. Since $A-P-K$ and $AK//X_1^*Y_1^*$, we can find $AP//X_1^*Y_1^*$. Then, $P$ is intersection of the line passes through $A$ and parallel to $X_1^*Y_1^*$ with $X_1Y_1$. Also $P\in X_2Y_2$. Since $X_1$ and $Y_1$ is fixed, $P$ is fixed. Thus, as $X_2,Y_2$ vary on $\omega$, the line $X_2Y_2$ always passes through the point $P$.

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This works, right? Let $l$ be the line through $A$, parallel to $BC$. Let $AB$ and $AC$ meet $\omega$ at $D$ and $E$, and let $DE$ meet $l$ at $P$. We claim $P$ works. Also let $XA$ and $YA$ meet $\omega$ at $M$ and $N$. The angle condition implies $MN||BC$. Note that $<PAE=<ACB=<ADE$, so $l$ is tangent to $(ADE)$. Now note that $<PAX=<NMA=<NYX$, so $l$ is tangent to $(AXY)$. Now we use radical axis on $(AXY)$, $(EAD)$ and $\omega$ to finish.

This was probably quite technical for a p1 Solution: Let $A'$ be the reflection of $A$ over the perpendicular bisector. Then, let $h$ be the negative inversion at $A$ with radius $\sqrt{Pow_{\omega} A}$. It is easy to see that all lines $XY$ pass through $h(A')$

Let the perpendicular bisector of $BC$ meets $\omega$ at $I, J$($J,A$ lies on the same half-planes with respect to the line $BC$. ), $IA \cap \omega = G;JA \cap \omega = H;XA\cap \omega = E; YA \cap \omega =F$. Note that $\angle FXB = \angle FYB = \angle AYC - \angle BYC = \angle BXA - \angle BXC = \angle CXE$, so $EF \parallel BC.$ The inversion at $A$ with radius $k=\sqrt{Pow_{A/ \omega}}$ sends $(FE;IJ)$ to $(YX;HG)$. Thus $(YX;HG)=-1$, so $XY$ passes through the intersection of the two tangents at $G$ and $H$ of $\omega$, which is a fixed point.

naruto_uzumaki_06 wrote: Let the perpendicular bisector of $BC$ meets $\omega$ at $I, J$($J,A$ lies on the same half-planes with respect to the line $BC$. ), $IA \cap \omega = G;JA \cap \omega = H;XA\cap \omega = E; YA \cap \omega =F$. Note that $\angle FXB = \angle FYB = \angle AYC - \angle BYC = \angle BXA - \angle BXC = \angle CXE$, so $EF \parallel BC.$ The inversion at $A$ with radius $k=\sqrt{Pow_{A/ \omega}}$ sends $(FE;IJ)$ to $(YX;HG)$. Thus $(YX;HG)=-1$, so $XY$ passes through the intersection of the two tangents at $G$ and $H$ of $\omega$, which is a fixed point. Nice, this was my original solution!

This problem hard for p1

$l:$ line through $A$ parallel to $BC$. We claim that the fixed point $P$ is a point on the line $l$ and $-PA^2=Pow_{(ABC)}(P)$. $\textbf{Claim:}$ There exists exactly one such point $P$ on the $l$. $\textbf{Proof:}$ Assume that $P$ is on the same side with $B$ wrt $AC$. $U=l \cap \omega$ and $U$ is on the same side with $B$ wrt $AC$. As $P$ goes to infinity on $l$ from $U$ , $f(P)=PA^2-Pow_{(ABC)}(P)$ takes the all values between $UA^2$ and $-R^2$ so there exists at least one such point. If there exists two, then we have $R^2=PO^2-PA^2=P'O^2-P'A^2 \implies PP' \perp OA \implies AB=AC$ contradiction. Thus there is exactly one such $P$. $\square$ To prove that $P-X-Y$ are collinear, define $Y'=PX\cap (ABC)$. Because $PA^2=PX\cdot PY'$, $F=Y'A \cap (ABC)$ and $E=XA \cap (ABC)$. $\angle XEF=\angle XY'A=\angle XAP \implies EF \parallel BC \implies \angle BXA=\angle CY'A \implies Y'=Y$ as desired.

I wonder why has nobody posted a solution with complex numbers, as this problem is easily doable with them. Let $\omega$ be the unit circle.Then the hypothesis says $\frac{b-x}{a-x}:\frac{a-y}{c-y}\in\mathbb{R}\iff \frac{(b-x)(c-x)}{(a-x)(a-y)}=\frac{(\overline{b}-\overline{x})(\overline{b}-\overline{y})}{(\overline{a}-\overline{x})(\overline{a}-\overline{y})}=\frac{(b-x)(c-x)}{bc(\overline{a}x-1)(\overline{a}y-1)}\iff bc(\overline{a}x-1)(\overline{a}y-1)=(a-x)(a-y)\iff (xbc\overline{a}^2-bc\overline{a})y-bc(\overline{a}x-1)=(a-x)a-(a-x)y\iff y=\frac{(a-x)a+bc(\overline{a}x-1)}{(xbc\overline{a}^2-bc\overline{a})+a-x}=\frac{x-\frac{bc-a^2}{\overline{a}bc-a}}{x\frac{bc\overline{a}^2-1}{\overline{a}bc-a}-1}=\frac{x-p}{x\overline{p}-1}$, where $P$ is a fixed point of afix $p=\frac{bc-a^2}{\overline{a}bc-a}$. Then you have $p+\overline{p}xy=x+y\iff P\in XY$, so the problem is done.

@above I think it is a good idea to also check that $P$ is not a point at infinity, i.e. that $\overline{a}bc - a \neq 0$. In my opinion, this problem is a huge advantage for complex bash people even with not so much experience. Identify $A$, $B$, $C$, $X$ and $Y$ with their complex numbers $a$, $b$, $c$, $x$, $y$ and $\omega$ as the unit circle. Having in mind the locations of $X$ and $Y$ from the problem condition, the angles $\angle AXB$ and $\angle CYA$ with the same orientation are equal precisely when $\frac{a-x}{c-x}/\frac{c-y}{a-y}$ is real, i.e. (after conjugating, replacing $\overline{b}$ with $1/b$, etc. and cancelling $c-y$ and $b-x$ from both sides) $$(a-x)(a-y) = (xb\overline{a} - b)(yc\overline{a} - c) \Leftrightarrow xy\frac{bc\overline{a}^2-1}{\overline{a}bc-a} + \frac{bc-a^2}{\overline{a}bc-a} = x+y $$which is if and only if the fixed point $\frac{bc-a^2}{\overline{a}bc-a}$ lies on $XY$ (quick calculation shows that its conjugate is indeed $\frac{bc\overline{a}^2-1}{\overline{a}bc-a}$). We should also check that this fixed point is not a point at infinity, i.e. that $\overline{a}bc - a \neq 0$. If we suppose otherwise, then from $|a-b| = |a-c| \Leftrightarrow b\overline{a} + \overline{b}a = c\overline{a} + \overline{c}a \Leftrightarrow \left(\frac{1}{b}-\frac{1}{c}\right)a = \overline{a}(c-b) \Leftrightarrow a = \overline{a}bc $ we get that $AB = AC$, contradiction!

Let $XA,YA$ meet $\omega$ at $K,S$. Note that $\angle KXB = \angle SYC$ so $SK || BC$. Let $l$ be line parallel to $BC$ through $A$. Let $XY$ meet $l$ at $T$. Note that $\angle XAT = \angle XKS = \angle XYA$ so $XYA$ is tangent to $l$ so $TA^2 = pow_\omega(T)$. Now we should prove there exists only one point like $T$ such that $TA^2 = pow_\omega(T)$. Assume not so there exists $T'$ as well. Let $TA$ meet $\omega$ at $P,Q$ and Let $T'T = x, TP = y, PA = z, ZQ = k$. Now note that we have $(y+z)^2 = y(y+z+k) , (x+y+z)^2 = (x+y)(x+y+z+k)$ so $y^2 + z^2 + 2yz = y^2 + yz + yk \implies z^2 + yz = yk$ and $x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = x^2 + xy + xz + xk + yx + y^2 + yz + yk \implies z^2 + yz + xz = xk + yk \implies xz = xk \implies z = k$ so now from $z^2 + yz = yk$ we have that $y = k = 0$ which isn't true so $T'$ doesn't exists.

https://www.geogebra.org/m/vrypvn4w the figure is a little simplified P is AY U (BCXY) and Q is AQ U (BCXY) PQ II BC can easily be proven with simple angle chasing. Hence We are done

Many of the above posts seem to be overcomplicating this problem. Here is a very short solution. Extend $XA$ and $YA$ to meet $\omega$ again at $X'$ and $Y'$. The angle condition gives that $X'Y'\parallel BC$. Now, let $XY$ meet the line $\ell$ through $A$ parallel to $BC$ at $T$. As $\measuredangle XAT = \measuredangle XX'Y = \measuredangle XYY' = \measuredangle XYA$, we get that $TX\cdot TY = TA^2$. Thus, $T$ is the intersection of $\ell$ and the radical axis of $\odot(A,0)$ and $\omega$, which is fixed, done.

Let $\tau_P$ denotes projection $\omega \stackrel{P}{\mapsto} \omega.$ Note that $\overline{\tau_A (X)\tau_A (Y)}\parallel BC,$ hence $X\mapsto Y$ is the involution $\tau_A \circ \tau_{\infty_{BC}} \circ \tau_A,$ done.

Let $XA$ and $YA$ meet $\omega$ again at $X',Y'$, respectively. Note that $X'Y'\parallel BC$ and hence $A,X',Y',A_1$ are concyclic where $A_1$ is the image of $A$ under reflection across the perpendicular bisector of $BC$. Applying negative inversion centered at $A$ that preserves $\omega$, we have $AA_1$ is fixed and $(AX'Y')$ is mapped to be line $XY$. Hence, $XY$ and $AA_1$ meet each other at the image under the inversion of $A_1$ which is independent from $X,Y$ since $A_1$ is fixed.

Complex Bashing

VicKmath7 wrote: Let $ABC$ be a triangle with $AB<AC$. Let $\omega$ be a circle passing through $B, C$ and assume that $A$ is inside $\omega$. Suppose $X, Y$ lie on $\omega$ such that $\angle BXA=\angle AYC$. Suppose also that $X$ and $C$ lie on opposite sides of the line $AB$ and that $Y$ and $B$ lie on opposite sides of the line $AC$. Show that, as $X, Y$ vary on $\omega$, the line $XY$ passes through a fixed point. Proposed by Aaron Thomas, UK 1/ Let $A'$ is the second intersection of the line through $A$ with parallel $BC$ and circle $(ABC)$ Let $X'$, $Y'$ is second intersection of the line $AX$, $AY$ with circle $\omega$ Because $\angle BXA = \angle CYA$ $\Leftrightarrow $ $\angle X'Y'B = \angle Y'X'C $ so $X'Y' \parallel BC $ Therefore, consider three radical axis of circles $(ABC)$, $(AX'Y')$, $\omega$, we easy to see that $A$, $A'$, $X'$, $Y'$ are concylic 2/ Let $Z$ is intersection of $XY$ and the line through $A$ parallel $BC$ Consider inversion tranformation center $A$ with power equal power of $A$ respect to $\omega $ : $\star$ $f$ : $X \leftrightarrow X'$, $Y \leftrightarrow Y'$. $\Rightarrow$ $f$ : $Z \leftrightarrow A' $. (Cuz $A$, $A'$, $X'$, $Y'$ are concylic and $X$, $Y$, $Z$ are collinear) Which means $\overline{AA'} .\overline{AZ}$ equal to powar of $A$ respect to $\omega$. So $Z$ is a fixed point when $XY$ vary

Let $l$ be the line through $A$ parallel to $BC$. Let $P, Q$ be the intersections between $l$ and $\omega$. Let $T$ be the unique point on $l$ such that $TA^2=TP \cdot TQ$. (To show that this is unique, define $A'$ as the reflection of $A$ across $T$, then note that $(P,Q;A,A')=-1$.) First we show that $(XYA)$ is tangent to $l$. Let $X'$ and $Y'$ denote the second intersections between $\omega$ and the lines $XA$ and $YA$ respectively. From the angle condition we know that $BC // X'Y'$. $\measuredangle XYA = \measuredangle XYY' = \measuredangle XX'Y' = \measuredangle XAD$, so $(XYA)$ is tangent to $l$. Now note that $T$ has equal power to $(XYA)$ and $\omega$ by definition, so $T$ lies on the radical axis of the two circles, which is $XY$.

Let $AX, AY$ hit $\omega$ at $E,F$ respectively, let $\omega$ hit $AB,AC$ at $D,G$ respectively, let $A_1$ a point in $DG$ with $AA_1 \parallel BC$, and let $A'$ such that $AA'CB$ is an ISLsosceles Trapezoid, consider an inversion centered at $A$ with a radius that fixed $\omega$, then because of the definition of $A'$ we have that $A_1 \to A'$ but note that $AA'FE$ is also cyclic by symetry on the perpendicular bisector of $BC$ therefore by inverting back we get $X,Y,A_1$ colinear, but since $\omega$ and $\triangle ABC$ are fixed we get that $A_1$ is fixed so it is our desired point, thus we are done .