Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that $X,Y,T,K$ lie on $(ABC)$ i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics...

@above you are right

$\sqrt{bc}$ inversion + involution + pascals

a solution by cadaeibf posted yesterday , in a deleted afterwards post Quote: We have $\angle UAQ=\angle WAS$. By $\sqrt{bc}$-inversion+reflection, we have $\frac{AU}{AQ}=\frac{AQ^*}{AU^*}=\frac{AW}{AS}$, implying $AUQ\sim AWS$ and so $\angle AUQ=\angle AWS$, which implies that $UQ$ and $WS$ concur on $\Gamma$ (they form the same arc). Similarly, $X,Y,K,T$ lie on $\Gamma$. This problem was also 2020 Cono Sur Shortlist G5

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Solved with Mathslion. The converse of Pascal makes a lot of sense here, so apply it on $AVWXUZ$ (since $UZ||PQ||VW$ so they intersect at infinity) and we get $X$ lies on the circumcircle. By another application on $AVWKZU$ we get $K$ lies on the circle as well. Similarly we get $X,Y,K,T$ lie on $\Gamma$. Now by Reim's ($\leftrightarrow$ just angles) we get the following cyclic quadrilaterals: $XYQR$, $XYPS$, $KPST$,$KTRQ$. Complete the quadrilateral $XYKT$ by adding $XY \cap KT = D$. It's enough to have that $A$ lies on the polar of $D$ (the line $MN$), this would mean that $DA$ is tangent to $\Gamma$. This follows by looking at the radical center of the circles: $APS,KPST,XYPS,ABC$ (notice that the radical of $APS$ and $ABC$ is the tangent at $A$)

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By the angles, we have that $(B,C;Q,S) = (B, C; P, R) \stackrel{A}{=} (B, C; U, W) \implies X \in \Gamma$, since $X=UQ \cap SW$. Similarly, $Y$, $K$, $Y \in \Gamma$. Let $D$ be the intersection point of the tangent to $\Gamma$ by $A$ and $BC$. We will prove that $D$, $X$, $Y$ are collinear (Similarly, $D$, $T$, $K$ are collinear): By inversion $\sqrt{bc}$ + reflection, we have that $X'$ and $Y'$ lies on $BC$, and $D'$ lies on $\Gamma$ such that $BC \parallel AD'$, so, $D$, $X$, $Y$ are collinear $\iff$ $D'$, $X'$, $Y'$, $A$ are cyclic $\iff$ $X'C' = Y'B'$, by the isosceles trapezoids, $\iff \dfrac{XC}{XA}\cdot c=\dfrac{YB}{YA}\cdot b$. But, $(B,A;Y,C) \stackrel{V}{=} (B, Q; P, C) \stackrel{U}{=} (B, X; A, C) \implies \dfrac{YB}{BC} \div \dfrac{YA}{AC} = \dfrac{BA}{BC} \div \dfrac{XA}{XC} \implies \dfrac{XC}{XA}\cdot c=\dfrac{YB}{YA}\cdot b$. Now, by Brocard's Theorem in $XYKT$, we have that $MN$ is the polar of $D$, but, $D$ lies in the polar of $A$, so, $A$ lies in the polar of $D$, so, $M$, $N$, $A$ are collinear.

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MathLuis wrote: Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that $X,Y,T,K$ lie on $(ABC)$ i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics... I want revenge of this problem... After learning $\sqrt{bc}$ inversion this became trivial (sadly i did not think abt it on the test :c) Continuing my progress we do a $\sqrt{bc}$ inversion, since by Reim's we have $AKPQ, ARST, AYPR, AXQS$ cyclic, we get that $K$ is send to $BC \cap WZ=K'$, $T$ is send to $BC \cap UV=T'$, $Y$ is send to $BC \cap VZ=Y'$ and $X$ is send to $BC \cap UW=X'$. Now let $T$ the interesction of the tangent from $A$ to $\Gamma$ and let $A'$ be a point on $\Gamma$ such that $AA' \parallel BC$. Then $T$ is send to $A'$, now since by symetry w.r.t. the bisector line of $BC$ we have that $AA'K'T', AA'X'Y'$ isosceles trapezoids and this means that $KT,XY,BC$ are concurrent at $T$ and by duality we are done

MathLuis wrote: Applying Multiple converse of pascal and 4 conics theorem and 3 conics theorem should give that $X,Y,T,K$ lie on $(ABC)$ i believe. @below I wish i wasnt confused about that theorem during the test, at least i would've got 2 points for my proof using conics... to prove this, it is enough to note a simple involution from which it turns out that $\{R, C; S, B\} = \{U, C; V, B\} \Rightarrow T \in \Gamma$. Could you tell me what the theorem you mention is about (or failing that, provide a link to that topic)? i'm not familiar with it

almost trivial with projective Note that $(B,P;Q,C)=(B,R;S,C)=(B,W;Z,C)_\Gamma$ implies that $K \in \Gamma$, and similarly $T, X, Y \in \Gamma$. By Brocard it suffices to show that the tangent $\ell$ to $\Gamma$ through $A$ passes through $XY \cap KT$, but this is true because $$(B,Y;A,C)_\Gamma=(B,P;Q,C)=(B,R;S,C)=(B,A;X,C)_\Gamma=(C,X;A,B)_\Gamma$$ so $XY$ passes through $\ell \cap BC$, and similarly so does $KT$.

YLG_123 wrote: By the angles, we have that $(B,C;Q,S) = (B, C; P, R) \stackrel{A}{=} (B, C; U, W) \implies X \in \Gamma$, since $X=UQ \cap SW$. Similarly, $Y$, $K$, $Y \in \Gamma$. Let $D$ be the intersection point of the tangent to $\Gamma$ by $A$ and $BC$. We will prove that $D$, $X$, $Y$ are collinear (Similarly, $D$, $T$, $K$ are collinear): By inversion $\sqrt{bc}$ + reflection, we have that $X'$ and $Y'$ lies on $BC$, and $D'$ lies on $\Gamma$ such that $BC \parallel AD'$, so, $D$, $X$, $Y$ are collinear $\iff$ $D'$, $X'$, $Y'$, $A$ are cyclic $\iff$ $X'C' = Y'B'$, by the isosceles trapezoids, $\iff \dfrac{XC}{XA}\cdot c=\dfrac{YB}{YA}\cdot b$. But, $(B,A;Y,C) \stackrel{V}{=} (B, Q; P, C) \stackrel{U}{=} (B, X; A, C) \implies \dfrac{YB}{BC} \div \dfrac{YA}{AC} = \dfrac{BA}{BC} \div \dfrac{XA}{XC} \implies \dfrac{XC}{XA}\cdot c=\dfrac{YB}{YA}\cdot b$. Now, by Brocard's Theorem in $XYKT$, we have that $MN$ is the polar of $D$, but, $D$ lies in the polar of $A$, so, $A$ lies in the polar of $D$, so, $M$, $N$, $A$ are collinear. Really nice solution! I proved that $X\in\Gamma$ with something of a spiral similarity and was trying to prove that $X, Y$ and the point you named $D$ are collinear, but I was struggling, thaks for the help.