Invert through $A$ with radius $\sqrt{AB\cdot AC}$ and then reflect everything across the bisector of angle $\angle{BAC}$. Denote this transformation by $\Phi$. Let $\Phi(P)=P’$ for every point $P$ in the plane. Let $D$ be the antipodal of $ A$ in $\odot(ABC)$. Points $K’, T’, S’, L’$ are just the intersection points of $AX$ and $AY$ with $BD$ and $CD.$ Observe that $\angle{DS’L’}=\angle{BS’A}=\angle{AT’C}$ because $\triangle{ABS’} \sim \triangle{ACT’}$. The last equality of angles implies that $K’T’L’S’$ is cyclic. We need to prove that $ \odot(ABC)$, $\odot(AS’T’)$, and $\odot(AK’L’)$ have a common point, different from $A$. The last is equivalent with proving that $$ \dfrac{pow_{\odot(ABC)} (K’)}{pow_{\odot(ABC)}(L’)}=\dfrac{pow_{\odot(AS’T’)}(K’)}{pow_{\odot(AS’T’)}(L’)}.$$Notice that $\dfrac{pow_{\odot(ABC)} (K’)}{pow_{\odot(ABC)}(L’)}=\dfrac{K’A \cdot K’X’}{L’A\cdot L’Y’}$, and $\dfrac{pow_{\odot(AS’T’)}(K’)}{pow_{\odot(AS’T’)}(L’)}=\dfrac{K’A\cdot K’T’}{L’A \cdot L’S’}.$ So, we have to prove that $\dfrac{K’T’}{K’X’}=\dfrac{L’S’}{L’Y’} \iff \dfrac{X’T’}{X’K’}=\dfrac{Y’S’}{Y’L’}$, or, in other words, that $X’$ and $Y’$ are corresponding points in the similar triangles $\triangle{DK’T’}$ and $\triangle{DS’L’}$, respectively. The problem is solved as $X’$ is the foot of the perpendicular from $D$ to $S’L’$, and $Y’$ is the foot of the perpendicular from $D$ to $K’T’$.

Invert in A with any radius. Consider cyclic complete quadrilateral KTSL with extensions of opposite sides. Let Q be Miquel point of this complete quadrilateral. From well known lemma it lies on AM where M is other intersection of opposite sides of KTSL. Let R be intersection of diagonals of KTSL. From another lemma O,R and Q are collinear where O is center of KTSL. Also OR is perpendicular to AM so Q is on circle with diameter AR. B and C are obviously also on this circle so we are done.

Note that $\odot{AKL},\odot{AST}$ are concyclic.We will show $\odot{KLST}$ cyclic and then the result follows from radical axis theorm.$AS\cdot AL=AK\cdot AT \Longleftrightarrow AB.AC.\cos CAY \cdot \cos BAY=AB.AC \cos BAX \cos CAX$ which is obviously true.$\blacksquare$ EDIT-This doesn't complete it I'll post the correct solution later

Let $H$ be the feet of $A$ on $BC$. Then $A, L, T, H, C$ concyclic and $A, B, S, K, T$ concyclic. It is clear from the assumption that $\triangle ATC \sim \triangle ASB$ and $\triangle AKB \sim \triangle ALC$, which implies $AL \cdot AS = AK \cdot AT$, thus $L, S, K, T$ is concyclic. Furthermore, let $M$ be the midpoint of BC, then by simple reflexion argument $ML =MS$ and $MT =MK$, which means $M$ is the center of $\odot (LSKT)$. Let $P= ST \cap LK$, $Q = LT \cap SK$. By radical axis theorem, $AH$, $LT$, $SK$ are concurrent , thus $Q$ lies on $AH$, which is perpendicular to BC. Moreover, by Brocard's thm, $MP \perp AQ$, therefore $P$ lies on $BC$ since $M$ lies on $BC$.

No body tempted to overkill this thing yet? Let \(BK \cap CL = M, BS\cap CT= N, LT \cap KS = O\). By Desargues' on \(\Delta BKS\) and \(\Delta CLT\), it suffices to prove that \(\overline{M,O,N}.\) Let \(M'\) be the reflection of \(M\) in \(KL\). Since \(\Delta OLK \sim \Delta OST\), it suffices to prove that \(OM, OM'\) are isogonal lines wrt \(\angle O.\) Note that \(\angle MLO = \angle CAX = \angle BAY = \angle MKO\) so by isogonality lemma we are done.

Well, since no one bashed it by using Menelaus and Steiner yet, here it goes : Notice that $KL$ and $ST$ both cut $BC$ in the exterior of segment $[XY]$, and denote by $U$ and $V$ these intersections. We have to prove that $U = V$, and because both $U$ and $V$ lie in the exterior of $[XY]$, it is enough to show that : $\frac{UX}{UY} = \frac{VX}{VY},$ or after applying Menelaus' Theorem in $\triangle AXY$, $\frac{KX}{KA} \cdot \frac{LA}{LY} = \frac{TX}{TA} \cdot \frac{SA}{SY},$ which rewrites as : $\frac{BX}{CY} \cdot \frac{AC}{AB} \cdot \frac{cos \angle CAY}{cos \angle BAX} \cdot \frac{cos \angle AXB}{cos \angle CYA} = \frac{CX}{BY} \cdot \frac{AB}{AC} \cdot \frac{cos \angle BAY}{cos \angle CAX} \cdot \frac{cos \angle AXB}{cos \angle CYA}$ As the cosines cancel out by isogonality, this rewrites as : $\frac{BX \cdot BY}{CX \cdot CY} = \frac{AB^2}{AC^2},$ which is true by Steiner's Theorem. Done.

fastlikearabbit wrote: After inverting through $A$ with radius $\sqrt{AB\cdot AC}$ and then reflecting everything across the bisector of angle $\angle{BAC}$, you need this to finish: https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2015-05/article_1_lema_coaxalitate.pdf After invert , then how ?

This problem was proposed by Greece and the author is Raphael Tsiamis.

Solution. A bit more detailed approach. Clearly, $ASKB$ and $ATLC$ are cyclic, from which we can obtain that $KSLT$ is also cyclic and notice that $TL\perp AB$ and $KS\perp AC$. Let $E$ be the midpoint of $\overline{BC}$, thus $EM\parallel AB$ and then $EM\perp TL$; hence, being $E$ the center of $(ATLC)$, we must have that $MT=ML$. Similarly, we get $MS=MK$. Since $K,\ S,\ L$ and $X$ all lie on the same circle, this forces $MS=MK=MT=ML$ to happen. Let $H$ be the foot of the $A$-altitude. $H$ is the second intersection point of $(ASKHB)$ and $(AHTLC)$; therefore, applying the radical axis theorem to these circles and $(KSLT)$ we infer that $AH,\ SK$ and $TL$ concur at a point, say $P$. Let $Z=\overline{ST}\cap \overline{KL}$. By Brockard's theorem, $MZ\perp AP$, so $H,\ Z$ and $M$ are collinear. We are done!

Let $BK$$\cap$$AS$$=$$F$, $BS$$\cap$$AK$$=$$G$,$CL$$\cap$$AX$$=$$W$ and $CT$$\cap$$AY$$=$$M$.By simple angle chasing we find that $KSLT$ is cyclic. And the way we defined points gives us that $FKWL$,$MTGS$ are cyclics.And also $FKGS$ and $TWLM$ are cyclics.We use Power of point wrt to $FKGS$, $FKWL$ and $MTGS$ and we can deduce that B is a point in the radical axis of $FKWL$ and $MTGS$.WE use POP again wrt to $FKWL$, $MTGS$ and $TWLM$ and get that point $C$ is a point on the radical axis of the circles $FKWL$ and $MTGS$. So $BC$ is the radical axis of circles $FKWL$ and $MTGS$. Now use POP wrt to $KSLT$,$MTGS$ and $FKWL$ and get that $KL$, $TS$ and $BC$ are concurrent.

Let $KS\cap TL =F$ , $BS\cap TC=D$ , $BK\cap CL=E$ . We will prove that points $D$ , $F$ and $E$ are collinear . Let $D'$ be the reflexion of $D$ over the midpoint of $TS$ . $ ATDS$ is cyclic because $\angle CTA=90$ and $\angle BSA=90$ . $D'$ is the orthocenter of $\Delta ATS$ because it's reflexion over the midpoint of $TS$ is $D$ . $AD$ and $AD'$ are isogonal in $\Delta ATS$ , because $AD$ is diameter of $(ATDS)$ . $\Delta DST$$ \sim $$\Delta EKL$ , because $\angle EKL=90-(90-\angle TSD)=\angle TSD$ and $\angle ELK=90-(90-\angle DTS)=\angle DTS$ . $\Delta TD'S\sim \Delta DST\sim \Delta EKL$ . $\Delta TFS \sim \Delta KFL$ $=>$$FTD'S \sim EKFL$$=>$ $\angle EFL = \angle D'FS$ . If points $D$ , $F$ and $E$ are collinear then $\angle EFL =\angle DFT$ or $\angle D'FS = \angle DFT$ . Now using trig Ceva two times in quadrilateral $ATFS$ , with point $D$ and with point $D'$ yields $\angle D'FS=\angle DFT=>$ points $D$ , $F$ and $E$ are collinear$=>$$BC$ , $TS$ and $KL$ are concurrent because of Desargues theorem in $\Delta BSK$ and $\Delta CTL$ .

After we invert around $A$ with power $\sqrt{bc}$ problem is equivalent to this: Given triangle $ABC$ and let antipod of $A$ be $A'$.$X,Y$ are on $(ABC)$ (on shorter arc $BC$) s.t. $<BAX = <CAY $. $AX$ intersects $BA'$ and $CA'$ at $P$,$Q$, and $AY$ intersects $BA'$ and $CA'$ at $R$,$S$. Prove that center of spiral similarity that sends $QR$ to $PS$ lies on $(ABC)$. First we compute $p$ as intersection of $b(-a)$ and $ax$. We get $$p=\frac{ax(b-a)+ba(a+x)}{ax+ba}$$or $$p=\frac{2bx+a(b-x)}{b+x}$$analogously $$q=\frac{2cx+a(c-x)}{c+x}$$(noticing that we just need to exchange $b$ with $c$ in intersection formula to get $q$ from $p$) And now we see that we will get $r,s$ by substituting $x$ with $\frac{bc}{x}$.So $$r=\frac{2bc+a(x-c)}{c+x}$$and $$s=\frac{2bc+a(x-b)}{b+x}$$And now the only real computation is finding $M$ , so we get $$m=\frac{qs-pr}{q+s-p-r}$$.Now we just bash, but first we do denominator because it seems easier. $$q+s=\frac{(2cx+a(c-x))(b+x)+(2bc+a(x-b))(c+x)}{(b+x)(c+x)}$$$$q+s=\frac{4xbc+2xa(c-b)+2cx^2+2bc^2}{(b+x)(c+x)}$$By symmetry (again we can just substitute $b$ with $c$ and vice versa) $$p+r=\frac{4xbc+2xa(b-c)+2bx^2+2cb^2}{(b+x)(c+x)}$$Finishing we get $$q+s-p-r=\frac{2(c-b)(x^2+bc+2xa)}{(b+x)(c+x)}$$Now we see something quite nice, we see the term $x^2+bc+2xa$ which will after conjugation become $\frac{2bcx+ax^2+abc}{x^2abc}$, so this part ($2bcx+ax^2+abc$) needs to be in the numerator if problem statement is true,or else after conjugation there will be no term equal to its conjugate(or it can be both in numerator and denominator so it cancels).It is hard for me to explain this part but if we have $\frac{t}{q}$ and we wish to show that it lies on unit circle we need to have $\frac{q}{t}=\frac{\bar t}{\bar q}$ then $q=\bar t k$ with $k$ having modulus $1$. This way if we have trouble factoring we can notice that on of factors will be $2bcx+ax^2+abc$(also it is safe to assume one of them will be $b-c$). Here is the harder part: $$qs=\frac{(2cx+a(c-x))(2bc+a(x-b))}{(b+x)(c+x)}$$or $$qs=\frac{4c^2bx+2acx(x-b)+2abc(c-x)+a^2(c-x)(x-b)}{(b+x)(c+x)}$$and again by symmetry $$pr=\frac{4b^2cx+2abx(x-c)+2abc(b-x)+a^2(b-x)(x-c)}{(b+x)(c+x)}$$$$qs-pr=\frac{2(c-b)(2bcx+ax^2+abc)}{(b+x)(c+x)}$$and finaly $$m=\frac{qs-pr}{q+s-p-r}=\frac{2bcx+ax^2+abc}{x^2+bc+2xa}$$and we see $m=\frac{1}{\bar m}$ which finishes problem. Motivation: We want to bash, thus we invert after seeing that is is immpossible in given problem and that there are isogonal conjugates. Now, $p$ is not that bad,and by symmetry we can get others. And finally we are able to handle $m$ because $M$ is center of spiral similarity. We see that it is possible to bash this problem,which leads to solution above. (note:this should be the method of last resort in a competition,but its nice for practicing bashing). This takes about 45min-1h.

Sorry for reviving, but tought i'd share this Similarity based solution, as nobody seems to have mentioned it. The result is equivalent with: $(AXKT)=(AYLS)$, which follows from: - $\triangle AKB \sim \triangle ALC$ ; - $\triangle ASB \sim \triangle ATC$ ; - $\frac{BK}{TC}=\frac{BX}{XC}$ and $\frac{BS}{CL}=\frac{BY}{YC}$ ; (Thales)

Here's a solution using moving points: Fix $\triangle ABC$, and animate $X$ linearly on $BC$. By isogonality, $X \mapsto Y$ is a projection. Also, note that $K,S$ simply lie on the circle with $AB$ as diameter and $L,T$ lie on the circle with $AC$ as diameter. Thus, by perspectivity from $A$, we get that $X \mapsto K,T$ is a projective map. Similarly, $Y \mapsto L,S$ is also a projective map. This gives that the degree of each of the points $K,L,S,T$ is two. Now, when $X=A \infty_{\perp_{AB}} \cap BC$, it's easy to see that $Y=A \infty_{\perp_{AC}} \cap BC$ (since, in this case, $\angle XAB=\angle YAC=90^{\circ}$). Then, in this situation, we will have $K=L=A$ (as $AX$ is tangent to $\odot (AB)$ and $AY$ is tangent to $\odot (AC)$). This gives that the degree of line $KL$ is atmost $2+2-1=3$. Similarly, we have that degree of $ST$ is also atmost $3$. Thus it suffices to show that $KL,ST,BC$ are concurrent for $(3+3+0)+1=7$ positions of $X$. When $X$ is the foot of the $A$-altitude, we have $K=S=X$, which directly gives the desired result. When $X$ is taken so that $Y$ is the foot of the $A$-altitude, we again have $L=T=Y$, and hence the result. For $X=B$, we get $Y=C$. This gives $K=B$ and $L=C$, and so $KL,BC$ only coincide. The case $X=C$ and $Y=B$ follows similarly as above. Take $X$ as the foot of the $A$-internal angle bisector. Then $Y=X$, and we have that lines $KL,ST$ themselves coincide. The case when $X$ is the foot of the $A$-external angle bisector follows in a similar fashion as above. The seventh position is a bit of a problem, as I cannot seem to find a nice position which would not require any solving per se. For now, you can consider any point you like ( I'd recommend $X=\infty_{BC}$). I'll try to find a better point later on.

Dear mathlinkers, I don't know if I'm doing it right or not. Can you check it for me, please! Here is my solution: Let H is the feet of A on BC. We have: (HT,HA)= (CT,CA) (since (A,T,H,C) cyclic))= π/2 - (AC,AT)=π/2-(AC,AX) =π/2-(AY,AB) (since aX,AY are isogonal in angle BAC) =π/2-(AS,AB)= (BA,BS)= (HA,HS) (because (A,S,H,B) cyclic) (mod π). Then HT, HS are reflect about HA. On the other hand: (HK,HB)=(AK,AB)=(AX,AB)=(AC,AY)=(AC,AL)=(HC,HL). So that HK, HL are reflect about HA. Now we use double ratio of division on circle: $\odot (AB)$ (AB,KS)= H(AB,KS)= H(AC,LT) (since pencil reflection) = $\odot (AC)$ (AC,LT) (?! ) So that (AB,KS)=(AC,LT) as known as BC, KL, ST converge.

Pretty straightforward by cross ratio. We have $\dfrac{LY}{SY}=\dfrac{CL}{BS}$ and $\dfrac{LA}{SA}=\dfrac{CL \cot \angle{CAY}}{BS \cot \angle{BAY}}$. So that $\{L,S;Y,A\}=-\dfrac{ \cot \angle{BAY}}{\cot \angle{CAY}}$. Similarly $\{K,T;X,A\}=-\dfrac{\cot \angle{CAX}}{\cot \angle{BAX}}$. Since $\angle{BAX}=\angle{CAY}$ we get that cross ratios are equal and hence $ST,XY,KL$ concur. Done

Child's play for desergaues. 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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Note that by desergaues theorem we just need to prove that $\Delta NBX$ and $\Delta YCM$ are perspective with respect to a line or that if $H=BN \cap CY,I=BX\cap CM$ then $A-H-I$ are collinear.Let $K=BY \cap CN$ Firstly observe that since $\angle ANK+\angle AYK\equiv \angle ANC +\angle AYB=90+90=180$ we have that $ANKY$ is cyclic. Similiarly since $\angle AXI+\angle AMI=360-(\angle AXB+\angle AMC)=180$ we have $AXMI$ is cyclic .Note that by isogonal line lemma we have that $AH,AK$ are isogonal. So we just need to prove $AI,AK$ as isogonal. Now note that $\angle BAX=\angle MAC$ since $AP,AQ$ are isogonal and also $\angle BXA=\angle CMA=90$ so by $AA$ similiarity we have$$\Delta ABX\sim \Delta ACM\implies \frac{AB}{AC}=\frac{AX}{AM}.......(1)$$. Also note that as $\angle NAC\equiv \angle PAC=\angle QAB \equiv \angle YAB$ since $AP,AQ$ are isogonal and also as $\angle AYB=\angle ANB=90$ so by $AA$ similiarity we have$$\Delta AYB \sim \Delta ANC \implies \frac{AY}{AN}=\frac{AB}{AC}....(2)$$Now (1), (2) imply that $\Delta AYN \sim \Delta AXM$ by $SAS$ similiarity $\implies \angle ANY=\angle AMX$ and since $AKNY$ and $AXIM$ are cyclic we have $\angle AKY=\angle ANY=\angle AMX=\angle AIX \implies \angle YAK=90-\angle AKY=90-\angle AIX=\angle IAX \implies AI,AK$ are isogonal w.r.t $\angle PAQ \implies AI,AK$. Now note that by the isogonality we obtained we have $\angle BAK=\angle BAP+\angle XAK=\angle CAM+\angle MAO = \angle CAM\implies AK,AI $ are isogonal w.r.t $\angle BAC$ . We are done. $\blacksquare$

Menelaus solution: Let $P = LK\cap BC$. By menelaus, we have $\frac{AL}{LY} \cdot\frac{YP}{PX}\cdot\frac{XK}{KA} = -1$. If $S,T,P$ were collinear, then we would have $\frac{AS}{SY}\cdot\frac{YP}{PX}\cdot\frac{XT}{TA} = -1$, thus we need to prove that these two expressions are equal. Setting these equal to each other and simplifying, we have to prove that $$\frac{AL}{LY}\cdot\frac{XK}{KA} = \frac{XT}{TA}\cdot\frac{AS}{SY}$$Let $\angle BAY = \angle CAX = \gamma$ and $\angle YAX = \alpha$. Then, since $\angle ALC = 90$, we have $AL = AC\cos(\alpha+\gamma)$. Similarly, we also have $KA = AB\cos(\alpha+\gamma)$, thus the LHS of what we want to prove becomes $\frac{AC}{LY}\cdot\frac{XK}{AB}$. In a similar vein, the RHS becomes $\frac{AB}{SY}\cdot \frac{XT}{AC}$. Since $BK$ is parallel to $CT$, we let $\angle KBC = \angle BCT = \theta$ and $\angle CBS = \angle BCL = \mu$. Then, we have $$LY = CY\sin(\mu), XK = BX\sin(\theta), XT = XC\sin(\theta), SY = BY\sin(\mu)$$Thus, plugging this is, our equation becomes $$\frac{AC}{AB} \cdot \frac{BX\sin(\theta)}{CY\sin(\mu)} = \frac{AB}{AC}\cdot\frac{CX\sin(\theta)}{BY\sin(\mu)}$$$$\Leftrightarrow \frac{AC}{AB}\cdot\frac{BX}{CX} = \frac{AB}{AC}\cdot\frac{CY}{BY}$$However, by the ratio lemma, $\frac{BX}{CX} = \frac{AB}{AC}\cdot\frac{\sin(\gamma+\alpha)}{\sin(\gamma)}$ and $\frac{CY}{BY} = \frac{AC}{AB}\cdot\frac{\sin(\gamma+\alpha)}{\sin(\gamma)}$. Plugging this in our equation becomes $$\frac{AC}{AB}\cdot\frac{AB}{AC}\cdot\frac{\sin(\gamma+\alpha)}{\sin(\gamma)}= \frac{AB}{AC}\cdot\frac{AC}{AB}\cdot\frac{\sin(\gamma+\alpha)}{\sin(\gamma)}$$$$\Leftrightarrow 1 = 1$$Since all of our steps are reversible, we have proven that the original expression is true. Thus, $S,T,P$ are collinear, so $KL$ and $ST$ must intersect at some point on $BC$.

Flash_Sloth wrote: Furthermore, let $M$ be the midpoint of BC, then by simple reflexion argument $ML =MS$ and $MT =MK$, which means $M$ is the center of $\odot (LSKT)$. Where do $ML=MS$ and $MT=MK$ come from?

First by some angle chasing (or trig) it's easy to see $KLST$ is cyclic. Let $KS\cap LT=P$ and, consider the Miquel point $M$ of $KLST$ ,and by Brokard theorem $\angle PMA=90$ . Now invert around $A$ with radius $\sqrt{AS.AL}$ , let $X'$ denote the inverse of some point $X$ .Then $\angle AC'P=\angle AC'S=\angle ALC=90$ , and similarly $\angle AB'P=90$ and so $\{ C',B'\}$ lie on the circle with diameter $AP$ , inverting back we're done. $\blacksquare$

Is this really $G6$ it must be $G1$ or $G2$ let $M,N,P$ the midpoints of $BC,AB,AC$ let $KL \cap ST=Z$ and $KS \cap LT=W$ main calim(2): $KLST$ si cyclic $\angle ASK=\angle ABK=90-\angle BAK=90-\angle CAS$ so $SK \perp AC$ so $MN \perp SK$ since $NA=NB=NK=NS$ so $MS=MK$ in a similar way $ML=MT$ it's obvious that $ML=MS$ so $ML=MS=MK=MT$ $\blacksquare$ radical axis on $(AKSB),(ALTC),(KLST)$ we have $W$ is on the altitude from $A$ to $BC$ Brokard on $KLST$ gives $AW \perp MZ$ so $Z \in BC$

Bro it's even easy for G2 Note that ignoring trivial special cases we could $WLOG$ suppose that $X$ is between $Y$ and $C$ and $T,S$ are outside $\triangle ABC$. Let $KL\cap BC=M_1 , ST\cap BC=M_2$ It's easy to se by Menelaus and law of sins that $\frac{XM_1}{M_1Y}=\frac{YL}{LA}.\frac{AK}{KX}$ $=\frac{CY.cos(\angle{BAY}+\hat{B})}{AC.cos\angle{CAY}}.\frac{AB.cos\angle{BAX}}{BX.cos(\angle{CAX}+\hat{C})}$ $=\frac{AB.CY}{AC.BX}.\frac{cos(\angle{BAY}+\hat{B})}{cos(\angle{CAX}+\hat{C})}$ Also $\frac{XM_2}{M_2Y}=\frac{YS}{SA}.\frac{AT}{TX} =\frac{BY.AC}{CX.AB}.\frac{cos(\angle{BAY}+\hat{B})}{cos(\angle{CAX}+\hat{C})}$ So $\frac{XM_1}{M_1Y}=\frac{XM_2}{M_2Y} \Leftrightarrow \left( \frac{AB}{AC}\right)^2=\frac{BX}{XC}.\frac{BY}{YC}$ Which is a well known relation so $M_1=M_2$ , done

Simple angle chasing gives us that $KTLS$ is cyclic. Let $H$ be the foot of the $A$-altitude. Since $AKSH$ and $ATLH$ are cyclic, we can see that $H$ is the Miquel Point of the cyclic quadrilateral $KTLS$. Let $KL\cup ST=E$. From a well-known lemma, $\angle APE=90^\circ$. Hence, $E$ lies on $BC$.

Let $H$ be the foot of the $A$ altidue. By angle chasing: Lemma: $BC$ is the external bisector of both $\angle SHL$ and $\angle KHT$ So both $SL\cap BC$ and $KT\cap BC$ are the point $P$ such that $(X, Y; H, P)=-1$, the end. $\blacksquare$ Remark: Alternatively, after showing that $KTLS$ is cyclic, you can prove the result by fixing $\triangle ALS$ and animating $T\in AX$ linearly (the condition has degree 4, which is now easy).

Let $AH$ be altitude and $KL, ST$ meet at $P$. Claim1 : $LKTS$ is cyclic. Proof : Let $\angle BAY = \angle CAX = x$ and $\angle BAX = \angle CAY = y$. $AL.AS = b.cos{y} . c.cos{x} = AK.AT$ so $LKTS$ is cyclic. Note that $AKHS$ and $ALHT$ are cyclic so $AKS$ and $ALT$ meet at $H$ and it's well known that $\angle AHP = \angle 90$ so $P$ lies on $BC$.

BY SIMPLE ANGLE CHASING, WE GET LKST IS CYCLIC THEN BY RADICAL AXIS THEOREM, AD( LET THIS BE THE A-ALTITUDE OF ABC WITH D ON BC), LT AND KS ALL ARE CONCURRENT. ( LET THIS MEETING POINT BE P) LET F=LK U ST AND LET AP U ST =E THEN WE KNOW (F, E, S, T) MUST BE HARMONIC ALDTC AND AKDSB ARE BOTH CYCLIC, THUS ANGLE ADK=ANGLE ABK = ANGLE ACL = ANGLE ADL THUS AD BISECTS ANGLE LDK . LET LK U AD = O AND LET BC U LK = Q WE GET (Q, O, L, K) IS HARMONIC ( TRIVIAL) SINCE ALS, ADE AND AKT ARE ALL COLLINEAR, THUS A ,Q, AND F MUST BE COLLINEAR, BUT SINCE A DOES NOT LIE ON LK, THERE IS ONLY ONE POSSIBILITY THAT Q=F. HENCE PROVED

Let $P = KL \cap ST$. Some basic angle chasing yields $ABSK \sim ACTL$. Thus, $$\angle LSK = 180^{\circ} - \angle ASK = 180^{\circ} - \angle ATL = 180^{\circ} - \angle LTK$$which means $KSLT$ is cyclic. Now, parallel lines, Steiner's Ratio Theorem, and similarity give $$\frac{LY}{YS} \cdot \frac{TX}{XK} = \frac{CY}{YB} \cdot \frac{CX}{XB} = \left( \frac{AC}{AB} \right)^2 = \left( \frac{TL}{SK} \right)^2 = \left( \frac{PL}{PS} \right)^2.$$Thus, properties of anti-parallel lines and Phantom Points imply $P, X, Y$ are collinear, which clearly finishes. $\blacksquare$ Remark: Some details are left out of this write-up, as I am short on time. Also, how did I miss the Brocard's solution after noticing that Radical Axes imply $KS$ and $LT$ meet on the $A$-altitude?

Let $P=KL \cap ST$. Firstly, note that $$AS \cdot AL = (AB \cos{\angle BAY})(AC \cos{\angle CAX}) = (AB \cos{\angle BAX})(AC \cos{\angle CAY})=AK \cdot AT,$$so points $K,S,T,L$ are concyclic by pop. Let $O$ be the circumcenter of $(KSTL)$. Next, consider the radical centre $Q$ of circles $(ABKS), (ACTL), (KSTL)$. Let $A'$ be the foot of the $A$-altitude of $\Delta ABC$. Since points $A$ and $A'$ lie on circles $(ABKS)$ and $(ACTL)$, $Q=KS \cap TL$ lies on the $A$-altitude of $\Delta ABC$. By Brocard's theorem, it suffices to show that $O$ lies on $BC$. Let $\perp_{UV}$ denote the perpendicular bisector of segment $UV$, so that $O=\perp_{KS} \cap \perp_{TL}$. Let $D=SK \cap BC, E=SK \cap AC$, and let $M$ be the midpoint of $BC$. $\measuredangle KDB = \measuredangle KBD + \measuredangle DKB = \measuredangle KAA' + \measuredangle YAB = \measuredangle XAA' + \measuredangle CAX = \measuredangle CAA'$, so $\measuredangle DEC = \measuredangle EDC + \measuredangle DCE = \measuredangle CAA' + \measuredangle A'CA = \measuredangle CA'A = 90^{\circ}$. It follows that $KS \perp AC$. Consider circle $(ABKS)$, since $\perp_{KS}$ passes through the midpoint of $AB$ and is parallel to $AC$, it passes through $M$. Similarly, $\perp_{TL}$ passes through $M$. It follows that $O=M$, so we are done. $\square$

The statement after taking pole-polar duality at $A$ with arbitrary radius is quite interesting (and arguably easier). Let $\ell_P$ denote the polar of $P$, and $P_{\ell}$ denote the pole of $\ell$. Obtaining the dual statement. Note that $\ell_X, \ell_Y, \ell_B, \ell_C$ all pass through $P_{BCXY}$ with $(\ell_X, \ell_Y)$ being equally inclined with respect to $(\ell_B, \ell_C)$ (basically $\measuredangle (\ell_B, \ell_X)=\measuredangle (\ell_Y, \ell_C)$). Now \[ K \in AX, AX \perp KB \Leftrightarrow P_{AX} \in \ell_K, AP_{AX} \perp AP_{KB} \]which can be rephrased to $\ell_K \parallel \ell_X$ and $\ell_X \perp AP_{KB}$. Basically, $P_{KB}$ is the point on $\ell_B$ such that $AP_{BK} \perp \ell_X$. The other points are obtained in a similar manner. The finished diagram is as shown, and we want to show that $P_{KL}, P_{BCXY}, P_{ST}$ are collinear. Solving the problem. Note that \[ \measuredangle(\ell_B) + \measuredangle(\ell_C) = \measuredangle(\ell_X) + \measuredangle(\ell_Y) = \measuredangle(\overline{AP_{BK}P_{CT}}) + \measuredangle(\overline{AP_{CL}P_{BS}}) \]implies $P_{BK}P_{CL}P_{BS}P_{CT}$ is cyclic (and allows us to delete $\ell_X$ and $\ell_Y$ from the diagram). Let $O$ be the center of the circle. Let $P_{BK}P_{KL}$ and $P_{CL}P_{KL}$ meet the circle again at $M$ and $N$ respectively. Then $P_{CT}M$ and $P_{BS}N$ are diameters, so they intersect at $O$. Hence by Pascal on $P_{CT}P_{CL}NP_{BS}P_{BK}M$, we get $O, P_{BCXY}, P_{KL}$ collinear. Similarly, $O, P_{BCXY}, P_{ST}$ collinear, so the result follows.

why 3 n on imo. anyways geo grind never stops: original solution. Let $P=XL\cap YK$. It suffices to prove that $P$ lies on the $A$-altitude, as then, letting $D$ be the foot of $A$ to $BC$, by complete quad we have $(KL\cap BC,D;X,Y)=-1$, and a similar result for $ST$, and since $D$ is fixed we can get the concurrence being shared. We use Ceva. Let $\angle XAB=\alpha,\angle BXA=\beta, \angle AYC=\theta$. Clearly, \[\frac{XK}{KA}\times\frac{AL}{LY}\times \frac{YD}{DX}=\frac{\tan \alpha}{\tan \beta}\times \frac{\tan \theta}{\tan \alpha}\times \frac{\tan \beta}{\tan \theta}=1\]from the right triangles, and isogonality.

Suppose that $M, N, P$ are midpoint of $BC, CA, AB;$ $KS$ intersects $TL$ at $U;$ $KL$ intersects $TS$ at $V$. It's easy to see that $K, S \in \bigodot(AB)$ and $L, T \in \bigodot(CA)$. We have $\angle{ASK} = 180^{\circ} - \angle{ABK} = 180^{\circ} - \angle{ACL} = \angle{ATL}$. Then $K, S, L, T$ lie on a circle. We also have $\angle{AKS} = \angle{ABS} = \angle{ACT} = 90^{\circ} - \angle{CAX}$. Hence $KS \perp CA$ or $KS \perp MN$. Similarly, $LT \perp MP$. Then $M$ is center of $(KSLT)$. Note that the radical axis of $\bigodot(AB)$ and $\bigodot(CA)$ is $A$ - altitude of $\triangle ABC$ and $\mathcal{P}_{U / \bigodot(AB)} = \overline{UK} \cdot \overline{US} = \overline{UL} \cdot \overline{UT} = \mathcal{P}_{U / \bigodot(CA)},$ we have $AU \perp BC$. But by Brocard theorem, $MV \perp AU$ then $V \in BC$

Quite a nice and simple problem, with a bit of bashing lol. Notice that $$\angle AXY = \angle ABY = 90 - \angle BAY = 90 - \angle CAN = \angle ACN = \angle AMN$$so $XYMN$ is cyclic, we get $\frac{\cos \angle NYM}{\cos \angle XNY} = \frac{\cos \angle NXM}{\cos \angle XMY}$ and notice that $\triangle BXY \sim \triangle CMN$, so $\frac{BY}{CN} = \frac{BX}{CM}$. Combining the two gives $$\frac{BY}{CN} \cdot \frac{\cos \angle NYM}{\cos \angle XNY} = \frac{BX}{CM} \cdot \frac{\cos \angle NXM}{\cos \angle XMY}$$$$\Rightarrow \frac{BY}{CN} \cdot \frac{\sin \angle BYN}{\sin \angle CNY} = \frac{BX}{CM} \cdot \frac{\sin \angle BXM}{\sin \angle CMX} \quad (\dag)$$Let $D = \overline{BC} \cap \overline{NY}, E = \overline{BC} \cap \overline{MX}$, we have $$\frac{BD}{\sin \angle BYD} \cdot \frac{\sin \angle CND}{CD} = \frac{BY}{\sin \angle BDY} \cdot \frac{\sin \angle CDN}{CN} = \frac{BY}{CN}$$$$\Rightarrow \frac{BD}{CD} = \frac{BY}{CN} \cdot \frac{\sin \angle BYN}{\sin \angle CNY}$$Similarly, $$\frac{BE}{CE} = \frac{BX}{CM} \cdot \frac{\sin \angle BXM}{\sin \angle CMX}$$And by $(\dag)$, we have $\frac{BD}{CD} = \frac{BE}{CE}$, i.e. $D = E$, so $\overline{BC}, \overline{NY}, \overline{MX}$ are concurrent, as desired.

$\textbf{This was way too easy for a G6/P3}$ $\text{Define}$ $\omega_{B}$ $\text{and}$ $\omega_{C}$ $\text{as the circles with diameter}$ $AC$, $\text{and}$ $AB$, $\text{respectively}$. $\newline$ $\text{Further define}$ $D$ $\text{as the foot of the perpendicular from }$ $A$ $\newline$ $\text{Lastly, define}$ $X=KS\cap LT$, $\text{and}$ $Y=KL\cap TS$ $\newline$ $\text{Now, obviously}$ $K, S, D \in \omega_{C},$ $\text{and}$ $L, T, D \in \omega_{B}$ $\newline$ $\textbf{Claim:}$ $K,L,S,T$ $\text{are cyclic}$ $\newline$ $\textbf{Proof:}$ $\angle BSK = \angle BAK = \angle CAL = \angle LTC$ $\angle LTK=\frac{\pi}{2}+\angle LTC=\frac{\pi}{2}+\angle BSK=\pi -(\frac{\pi}{2}-\angle BSK)=\pi-\angle LSK$ $\newline$ $\textbf{Claim:}$ $X$ $\text{lies on the radical axis of}$ $\omega_{B}$ $\text{and}$ $\omega_{C}$ $\newline$ $\textbf{Proof:}$ $\newline$ $\text{Radical axis theorem on }$ $\omega_{B},$ $\omega_{C}$ $\text{and}$ $(KSLT)$ $\text{suffices.}$ $\newline$ $\textbf{Finishing:}$ $\newline$ $D$ $\text{must be the Miquel point of the complete quadrilateral}$ $\mathcal{Q} \{KT,TL,LS,SK\},$ $\text{as}$ $D=(ADTLC)\cap (ADSKB)$ $\text{It is well-known that the Miquel point of a convex, complete, cyclic quadrilateral is the foot of the perpendicular}$ $\text{ from the intersection of the diagonals to the line connecting the intersections of the sides}$ $\newline$ $\Longrightarrow$ $YD\perp AX$. $\newline$ $\text{But this finishes the problem, as}$ $\overline{AX}$ $\text{is the same as}$ $\overline{AD}$ $\text{and}$ $AD\perp BC$ $\newline$ $\Longrightarrow$ $Y\in BC,$ $\text{which is exactly what we wanted.}$ $\blacksquare$

Main claim is to use Desargue's Theorem on $\triangle BSK$ and $\triangle CTL$ for $BC, TS$ and $KL$ concurrence. For this one needs to show $\overline{BK\cap CL-LT\cap SK-BS\cap CT}$ is collinear.

Main claim is to use Desargue's theorem on triangles $BTK$ and $CSL$. So we just need to show that $D=BT\cap CS$, $E=BK\cap CL$ and $A=KT \cap SL$ are collinear. If we denote $F=BS \cap CT$ by the parallelogram isogonolaty lemma $AD, AF$ and $AE, AF$ are isogonal as we wanted.