I've done too much config geo in my life. Let $X'_A=\overline{AB}\cap (BOC)$ and let $Y'_A=\overline{AC}\cap (BOC)$. Then since $B,C,X'_A,Y'_A$ are cyclic and $\overline{X_AY_A}||\overline{BC}$, $X_A,X'_A,Y_A,Y'_A$ are cyclic. Thus $K_A-O-\overline{X_AY_A}\cap \overline{X'_AY'_A}$. Let $P=\overline{AO}\cap (BOC)$. Then angle chasing gives us that $P$ is the reflection of $A$ over $\overline{X'_AY'_A}$. Thus $\overline{X_AY_A}\cap \overline{X'_AY'_A}$ is the circumcenter of $(AHP)$. More angle chasing gives that $H,P$ swap under $\sqrt{bc}$ inversion. If $A'$ is the reflection of $A$ across $BC$, $D=\overline{AH}\cap \overline{BC}$, and $H_A=\overline{AH}\cap (ABC)$, then by this inversion, $(A,\overline{AO}\cap \overline{BC};O,P)=(P_{\infty},H_A;A',H)=\frac{HH_A}{A'H_A}=2\cdot \frac{HD}{AH}=(A,D;A',H)$, so $\overline{HP},\overline{BC},\overline{OA'}$ concur. If $N_9$ is the $9$-point center of $\triangle ABC$, then $2\cdot N_9-A=B+C-O$, so $\overline{AN_9},\overline{OA'}$ are symmetric across $\overline{BC}$. Thus $\overline{AN_9},\overline{HP},\overline{BC}$ concur. By $\sqrt{bc}$ inverting, the radical axis of $(AHP)$ and $(ABC)$ is isogonal to $\overline{AN_9}$. Thus $\overline{K_AO}$ is perpendicular to the line isogonal to $\overline{AN_9}$ in $\angle BAC$. Let $B'$ be the reflection of $B$ across $\overline{AC}$ and define $C'$ similarly. Under a homothety of factor $\frac{1}{4}$ around the centroid of $\triangle ABC$, $B'$ goes to the foot from $N_9$ to $\overline{AC}$, and $C'$ goes to the foot from $N_9$ to $\overline{AB}$. Thus by some angle chasing, $\overline{B'C'}$ is perpendicular to the line isogonal to $\overline{AN_9}$ in $\angle ABC$. Thus $\overline{K_AO}||\overline{B'C'}$. Let $T$ be the antipode of $O$ on $(BOC)$. By some angle chasing, $B'-C-P$ and $C'-B-P$, and $BB'=BC=CC'$, so the spiral center sending $B'$ to $C'$ is $T$. Thus the line through $T$ perpendicular to $\overline{B'C'}$ goes through $O'$, the center of $(A'B'C')$. This line is also perpendicular to $\overline{K_AO}$, so it is the line through $K_A$ perpendicular to $\overline{K_AO}$. Thus $K_A$ lies on the circle with diameter $OO'$, and so does $K_B$ and $K_C$, as desired.

Seems a bit easy for USA TSTST #6. Let $O_A, O_B, O_C$ the centers of $(BOC), (COA), (AOB)$ and $O_A'$ the centre of $(OX_AY_A)$. Claim. $AO=AK_A$ Proof. From $AX_A=HX_A$, $AO=CO$ and $\angle X_AAH=\angle OAC$ it follows that $\triangle X_AAH\sim\triangle OAC$, so there is a spiral similarity centred in $A$ sending $X_A$ to $O$ and $H$ to $C$, which implies $\triangle AX_AO\sim\triangle AHC$. Therefore $\angle AOX_A=\angle ACH=90^\circ-\angle A$. Similarly $\angle AOY_A=90^\circ-\angle A$, which implies $\angle X_AOY_A=180^\circ-2\angle A=\angle BOC$, so $\triangle X_AO_A'Y_A\sim\triangle BO_AC$. Hence there is a homothety with center $A$ which maps $\triangle X_AO_A'Y_A$ to $\triangle BO_AC$, which implies that $A, O_A', O_A$ are collinear, which implies $A$ is on the perpendicular bisector of $OK_A$ $\blacksquare$ [asy][asy] import olympiad; import geometry; size(10cm); defaultpen(fontsize(11pt)+linewidth(0.65)); dotfactor *= 0.83; settings.tex="pdflatex"; settings.prc=false; settings.render=0; pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(A--B--C--cycle); pair H = orthocenter(A, B, C); pair O = circumcenter(A,B,C); pair OA=circumcenter(B,O,C); pair [] KA = intersectionpoints(circle(A,1), circumcircle(B,O,C)); pair [] U = intersectionpoints(circle(A, abs(A-H)), circle(H, abs(A-H))); pair KAs = extension(O, KA[0], B, C); pair XA = extension(U[0], U[1], A, B); pair YA = extension(U[0], U[1], A, C); pair OAp = circumcenter(O,XA,YA); draw(circumcircle(XA, YA, O)); draw(circle(A, 1)); pair [] L = intersectionpoints(circle(A,1), circumcircle(A,B,C)); draw(KAs--L[0]--L[1]); draw(XA--YA); draw(XA--H); draw(XA--O--YA); draw(C--H); draw(B--KAs--O); draw(circumcircle(A,B,C)); draw(circumcircle(B,O,C)); draw(A--H); draw(A--OA); draw(KA[0]--A--O); dot("$A$", A, dir(100)); dot("$B$", B, dir(-130)); dot("$C$", C, dir(0)); dot("$O$", O, dir(-60)); dot("$H$", H, dir(40)); dot("$O_A$", OA, dir(0)); dot("$O_A'$", OAp, dir(0)); dot("$K_A$", KA[0], dir(-70)); dot("$K_A^*$", KAs, dir(180)); dot("$X_A$", XA, dir(180)); dot("$Y_A$", YA, dir(0)); [/asy][/asy] Now consider the inversion $\Phi$ respect to $(ABC)$ and denote the image of $\bullet$ by $\bullet^*$. This inversion fix $A, B, C$ and maps $(BOC)$ to the line $BC$, so $K_A^*\in OK_A\cap BC$, which implies that $K_A^*$ is the radical center of the circles $(ABC)$, $(BOC)$ and the circle with center in $A$ and radius $AO$, which implies that $K_A^*$ is on the perpendicular bisector of $OA$. Therefore $K_A^*\in BC\cap O_BO_C$, and similarly $K_B^*\in AC\cap O_AO_C$, $K_C^*\in AB\cap O_AO_B$. Because $AO_A, BO_B$ and $CO_C$ are concurrent in the Kosnita point $K$ of $\triangle ABC$ it follows that $\triangle ABC$ and $\triangle O_AO_BO_C$ are perspective, so by Desargues theorem it follows that $K_A^*, K_B^*$ and $K_C^*$ are collinear, which implies $K_A, K_B, K_C$ and $O$ are concyclic.

This year USA TSTST seems to like center spam. Extend $AO$ to meet $\odot(BOC)$ again at $A_1$. Notice that $\triangle AX_AH \sim \triangle AOC$, inducing the similarity $\triangle AX_AO\sim\triangle AHC$. However, by an easy angle chasing, $\triangle AHC\sim\triangle ABA_1$, so $\triangle A_1BC$ and $\triangle OX_AY_A$ are homothetic. Let $O_A$, $O_B$, $O_C$ be the centers of $\odot(BOC)$, $\odot(COA)$, $\odot(AOB)$. Let $O_A'$, $O_B'$, $O_C'$ be the centers of $\odot(OX_AY_A)$, $\odot(OX_BY_B)$, and $\odot(OX_CY_C)$. It suffices to show that $O_AO_A'$, $O_BO_B'$, and $O_CO_C'$ are concurrent (at the center of $\odot(OK_AK_BK_C)$). However, notice that from the homothety in the above paragraph, $A\in O_AO_A'$, Thus, these three lines concur at the Kosnita's point of $\triangle ABC$, done. Remark: I'm sure I have seen the first part (figuring the shape of $\triangle OX_AY_A$) before. However, I can't recall where exactly I've seen it.

@above the same idea appears here

Let $P,Q$ be the reflections of $H$ across $AC, AB$. Note that $X_A, Y_A$ are circumcenters of $\triangle AHP$ and $\triangle AHQ$. So $OX_A, OY_A$ are perpendicular bisectors of $AP, AQ$ so $\angle X_AOY_A = \frac{1}{2} \angle POQ = 180 - 2A$. If $Z$ is the circumcenter of $AX_AY_A$, then since $X_AY_A || BC$, $A,Z,O$ are collinear and by the previous sentence, we have $(ZX_AOY_A)$ cyclic, so taking a homothety from $A$, we have that if $O_A, O_A'$ are the circumcenters of $(X_AY_AO)$ and $(BOC)$, then $A,O_A, O_A'$ are collinear. So if $K = AO_A' \cap BO_B' \cap CO_C'$ is the Kosnita point, then $K$ lies on the perpendicular bisectors of $OK_A, OK_B$ and $OK_C$, so the points $K_A, K_B, K_C, O$ are concyclic with center $K$, as desired. $\blacksquare$

Let $UVW$ be the triangle formed by the lines $X_AY_A,X_BY_B,X_CY_C$. Claim. $(OVW),(BOC),(X_AY_AO)$ are coaxial. Proof: Let $DEF$ be the orthic triangle of $ABC$. $R=(OBC)\cap AB,AC=R,S$; $UV\cap AC=K, UW\cap AB=L$. By coaxility lemma it is enough to prove that $$\frac{XV}{YV} \cdot \frac{XW}{YW}=\frac{XS}{YR} \cdot \frac{XC}{YB}$$ Note that $R-O-V$ and $S-O-W$ are collinear. One can get the following equations by parallel lines and sinus law: $$\frac{XV}{XY}=\frac{XK}{XA},\frac{YW}{YX}=\frac{YL}{YA} \implies \frac{XV}{YW}=\frac{AY}{AX} \cdot \frac{XK}{YL}=\frac{c}{b} \cdot \frac{XK}{YL}$$ $$\frac{XS}{XW} \cdot \frac{YV}{YR}=\frac{\cos B}{\cos A}\cdot \frac{\cos A}{\cos C}, \frac{XC}{YB}=\frac{b}{c}$$ Thus we need to prove that $\frac{XK}{YL}=\frac{\cos B}{\cos C} \cdot \frac{b^2}{c^2}$. Because $U,V,W$ are the reflections of $O$ across the sides of $ABC$, $X_AY_AX_BY_BX_CY_C$ is a hexagon with opposite sides equal. Hence $$\frac{AH}{2AD}=\frac{XY}{a} \text{ similar ones for the other sides} \implies $$ $$\frac{XK}{YL}=\frac{\frac{BH \cdot b}{2BE}}{\frac{CH \cdot c}{2CF}}=\frac{BH}{CH} \cdot \frac{b^2}{c^2} \cdot \frac{c \cdot CF}{b \cdot BF}=\frac{\cos B}{\cos C} \cdot \frac{b^2}{c^2}$$$\square$ Now let $O_A$ be the center of $BOC; O_B, O_C$ defined similarly. By inversion with center $O$ radius $OA$, we get $K_A*=O_BO_C\cap BC$. Because $O_AO_BO_C$ and $ABC$ are perspective; $K_A*,K_B*, K_C*$ are collinear. Therefore $O, K_A, K_B, K_C$ are concyclic.

Let $T_A,P_A$ be the antipodes of $O$ wrt $\odot(BOC),\odot(X_AOY_A)$. Define $T_B,T_C,P_B,P_C$ similarly. We'll prove that $T_AP_A,T_BP_B,T_CP_C$ are concurrent at a point called $Q$. Suppose that we can prove, we will get that $K_A,K_B,K_C$ lie on a circle with diameter $OQ$, as desired. Let $I_A,O_A$ be centers of $\odot(X_AY_AO),\odot(BOC)$. Note that $\triangle AX_AH \overset{+}{\sim} \triangle AOC$. By spiral similarity, $\triangle AX_AO \overset{+}{\sim} \triangle AHC$. Hence, $\measuredangle X_AOY_A = \measuredangle X_AOA + \measuredangle AOY_A = \measuredangle HCA + \measuredangle ABH = \measuredangle BOC$. Therefore, a homothety at $A$ can send $\odot(X_AY_AO)$ to $\odot(BOC)$, and $I_A,O_A,A$ are collinear. Let $R_A$ be the reflection of $O$ across $A$. Notice that $R_A,R_B$ are isogonal wrt $\angle T_AT_CT_B$. Therefore, by Jacobi's theorem, $T_AR_A,T_BR_B,T_CR_C$ are concurrent at a point called $Q'$. Dilation with ratio 2 gives us that $R_A,T_A,P_A,Q'$ are collinear, as desired. [asy][asy] size(300); pair A,B,C,X,Y,I,OA,O,K,R,T,P,K,H; A = dir(110); B = dir(212); C = dir(-32); H = A+B+C; O = 0; X = extension(A,B,A/2+H/2,rotate(90,A/2+H/2)*A); Y = extension(A,C,A/2+H/2,rotate(90,A/2+H/2)*A); I = circumcenter(X,Y,O); T = 2*B*C / (B+C); OA = O/2+T/2; P = 2*I-O; R = 2*A-O; fill(A--B--C--cycle,0.1*cyan+0.9*white); draw(unitcircle); draw(A--B--C--cycle); draw(circumcircle(B,O,C)); draw(circumcircle(O,X,Y)); draw(O--T,green); draw(O--P,green); draw(O--R,green); draw(R--T,magenta); draw(A--OA,blue); dot("$A$",A,dir(80)); dot("$B$",B,dir(200)); dot("$C$",C,dir(-20)); dot("$X_A$",X,dir(X)); dot("$Y_A$",Y,dir(Y)); dot("$O$",O,dir(-45)); dot("$T_A$",T,dir(T)); dot("$P_A$",P,dir(P)); dot("$O_A$",OA,dir(30)); dot("$I_A$",I,dir(250)); dot("$R_A$",R,dir(R)); [/asy][/asy]

Second problem accepted at TSTST! (oh no, my real name is disclosed) Not really expect it to be a P6.

thanks for the complex bashable problems

I dont have so much time now, so i will post only the second sol i got (the first one was MMP and got it in an hour) then after some minutes i got this sol in like, ummm....5 minutes Claim 1: $AK_A=BK_B=CK_C=R$ (where $R$ is the radius of $(ABC)$) Proof: Its enough to show that $AK_A=AO$ so we start. Since $H,A$ are symetric w.r.t. $X_AY_A$ we have that $\angle X_AHY_A+\angle BHC=180$ so $H$ has an isogonal conjugate in $BX_AY_A$, and we clearly see its $O$, now let $O_A$ the center of $(BOC)$ and let $Z_A$ the center of $(X_AY_AO)$, by isogonals we have $\angle X_AOY_A+\angle BOC=180$ so that means $\angle BO_AC=\angle X_AZ_AY_A$, and since $X_AY_A \parallel BC$ we have that there is an homothety sending $\triangle X_AZ_AY_A$ to $\triangle BO_AC$ hence $A,Z_A,O_A$ are colinear and since $Z_AO_A$ is the perpendicular bisector of $K_AO$ we have $AK_A=AO$ as desired yay. Finishing: Let $K$ the Kosnita point of $\triangle ABC$ then spamming Claim 1 u actualy have that $K_AK_BK_CO$ is cyclic with center $K$ becuase the perpendicular bisectors of $K_AO,K_BO,K_CO$ are concurrent at $K$. Hence the problem is done (man, i'm finally unpostbanned WOOOOO)

too lazy to post full solution, but here's the gist: Step 1: Prove $AO = AK_A$. This isn't hard, let $AO \cap (BOC)$ again at $A'$ and then prove $OX_AY_A$ and $A'BC$ are homothetic, which shows $A$ is the center of homothety sending $(OX_AY_A)$ to $(BOC)$ from which the result follows. Step 2: Prove that the perpendicular bisectors of $OK_A, OK_B, OK_C$ concur. From this, we are done, since the concurrence point $T$ would simply be the center of our desired cyclic quadrilateral. This is equivalent to showing that if $P_A, P_B, P_C$ are the circumcenters of $(BOC)$, etc, then $AP_A, BP_B, CP_C$ concur. This is well known, though. Done.

Solution 1: Claim: $AO=AK_A$. Proof: Let $\triangle DEF$ be determined by $X_AY_A, X_BY_B, X_CY_C$ in the usual order. Then a dilation at $H$ of factor $2$ sends $\triangle DEF$ to $\triangle A'B'C'$, such that $\triangle ABC$ is the medial triangle of $\triangle A'B'C'$. Thus $\triangle DEF$ is obtained from $\triangle ABC$ by a rotation about $N$ by $180^{\circ}$, where $N$ is the nine-point centre of $\triangle ABC$. Thus, if the arc midpoint opposite $O$ in $(OX_AY_A)$ is $Z_A$, then $$\angle X_AZ_AY_A = 180^{\circ}-\angle X_AOY_A=180^{\circ}-\angle Y_CHX_B=360^{\circ}-\angle CY_CH-\angle BX_BH=2\angle BAC$$so $Z_A$ is the centre of $(AX_AY_A)$. Thus, there exists dilation at $A$ sends $(OX_AY_A)$ to $(BOC)$ so the perpendicular bisector of $OK_A$ is $AN'$, where $N'$ is the isogonal conjugate of $N$ in $\triangle ABC$. Thus, the perpendicular bisectors of $OK_A, OK_B, OK_C$ concur at $N'$, implying that $OK_AK_BK_C$ is cyclic. Solution 2: A more complicated way of finishing, albeit "cooler", is to observe that $O$ is the orthocentre of $\triangle DEF$ and $OK_AK_BK_C$ is in fact the Hagge circle of $\triangle DEF$ with respect to $D'E'F'$, its circumcevian triangle with respect to $N$. This solution involves proving: $K_A'\in (DEF)$ where $K_A$ is the reflection of $K_A$ over $EF$: this is true since $AK_A=AO=AE=AF$ so $K_A\in (EFO)$; and $K_A'$ lies on line $AND$: this is true by some angle chasing and Reim's with $Z_AK_AH$.

Never thought this day would ever come in my life.What a splendid problem !! 2022 TSTST #6 wrote: Let $O$ and $H$ be the circumcenter and orthocenter, respectively, of an acute scalene triangle $ABC$. The perpendicular bisector of $\overline{AH}$ intersects $\overline{AB}$ and $\overline{AC}$ at $X_A$ and $Y_A$ respectively. Let $K_A$ denote the intersection of the circumcircles of triangles $OX_AY_A$ and $BOC$ other than $O$. Define $K_B$ and $K_C$ analogously by repeating this construction two more times. Prove that $K_A$, $K_B$, $K_C$, and $O$ are concyclic. Hongzhou Lin Let $I$ and $G$ be circumcentres of $\triangle CHA$ and $\triangle BHA$.It is well known that $AI=AG=AO=R$ and $I,G$ are reflections of $O$ over $AC,AB$ respectively. Claim : Let $K,J$ be the inverses of $Y_A,X_A$ under an inversion centered at $A$ with radius $AO=R$.$K,J \in \odot(\triangle BOC)$ Proof : Clearly $KJX_AY_A$ is cyclic and $K,J$ lie on $AC,AB$.Again $X_AY_A \parallel BC$ which implies $KJ$ is antiparallel to $BC$ therefore $BCKJ$ is cyclic.Now since $AIKJG$ is cyclic and $AI=AG$ we have by fact 5, $\angle IKA=\frac{1}{2}\angle IKG=90-\frac{1}{2}\angle IAG=90-\angle BAC$.(It could be angle chased to get $\angle IAG=2\angle BAC$).Now as $$K \in AC \implies KO=KI \implies \angle OKA=\angle IKA=90-\angle BAC=\angle OBC \implies \text{ OBCK is cyclic. }$$Similarly we get that $OBJC$ is cyclic as well $\implies K,J \in \odot(\triangle BOC)\text{ } \blacksquare$ Claim : $AK_A=BK_B=CK_C=R=AO=BO=CO$ Proof : Use radical axis on $\odot(\triangle OX_AY_A),\odot(X_AY_AJK),\odot(\triangle BOC)$ this gives that $IG,OK_A,KJ$ are concurrent. Now let $\odot(A,AO) \cap \odot(\triangle BOC)=K'$.Using radical axis on $\odot(A,AO),\odot(\triangle BOC) $ and $\odot(AIKJG)$ gives $IG,OK',KJ$ are concurrent.This gives that $K'=K_A \implies K_A \in \odot (A,AO)$ which gives $AO=AK_A=R$ . Similarly other cases hold as well $\blacksquare$. Now we invert at $O$ with radius $R$ and take a homothety of scale factor $2$,the problem becomes Quote: Let the tangents to the circumcircle of $\triangle ABC$ at $A,B,C$ meet $BC,CA,AB$ at $E,F,G$ respectively .Prove that $E,F,G$ are collinear Using Pascal on the degenerate hexagon $AABBCC$ proves the result $\blacksquare$

Let $O_A$ denote the center of $(BOC)$, and analogously define $O_B$, $O_C$. Claim: $AO_A \perp OK_A$ Proof. By the homothety centered at $A$ sending $(OX_AY_A)$ to $(BOC)$, it suffices to show that if $O' = AO \cap (BOC)$, then $\tfrac{AO}{AO'}=\tfrac{AX_A}{AB} = \tfrac{AY_A}{AC}$. Let $AB$ intersect $(BOC)$ again at $B'$. Then $\triangle AB'C$ is isosceles with $AB'=B'C$ and $\angle AB'C = 180^{\circ} - 2A$. Thus, $AB'=\tfrac{b}{2\cos A}$, so the power of $A$ with respect to $(BOC)$ is $AB \cdot AB' = \tfrac{bc}{2\cos A}$. Thus \[ \frac{AO}{AO'} = \frac{R}{\frac{bc}{2 \cos A \cdot R}} = \frac{2R^2 \cos A}{bc}. \]On the other hand, note that $\tfrac{AX_A}{AB}$ is the ratio of $AH/2$ to the length of the $A$-altitude in $\triangle ABC$, which is \[ \frac{\frac{AH}{2}}{\frac{2[ABC]}{a}} = \frac{R \cos A}{\frac{bc}{2R}} = \frac{2R^2 \cos A}{bc} = \frac{AO}{AO'}, \]as desired. Perform an inversion at $O$ with radius $R$, and denote mapped points with a $'$. It suffices to show that $K_A'$, $K_B'$, and $K_C'$ are collinear, where we can redefine $K_A'$ and cyclic variants as $OK_A \cap BC$. It is well-known that $AO_A$, $BO_B$, and $CO_C$ are concurrent, so by Ceva-Menelaus, it follows that the perpendicular lines from $O$ to each of these lines concur at points on $BC$, $AC$, and $AB$ respectively that are collinear, concluding. EDIT: Actually, a much cleaner finish is to note that it suffices to show that the perpendicular bisectors of $OK_A$, $OK_B$, and $OK_C$ concur. By the claim, $AO_A$ is also the perpendicular bisector of $OK_A$, so we are done by the concurrence of $AO_A$, $BO_B$, and $CO_C$.

Quick sketch. Solved with @Lemmas Claim 1: $AK_{A}=AO$. Proof: Its enough to prove that $A$ is a center of Homotety which sends $(OX_{A}Y_{A})$ to $(OBC)$. Let $AO$ intersect $(BOC)$ at point $T$. Its obvious that $X_{A}Y_{A}$ is parallel to $BC$. We show that $OY_{A}$ is parallel to $TC$ or that $\frac{AY_{A}}{AC}= \frac{AO}{AC}$ which is doable by sines. Claim 2: If $O_{A}$ is circumcenter of $BOC$, similary define $O_{B}$ and $O_{C}$ then $AO_{A}$, $BO_{B}$, $CO_{C}$, are concurrent(Kosnita Point). Perpendecular bisector to $OK_{A}$ is line $O_{A}A$. It follows that perpendicular bisectors of $OK_{A}$, $OK_{B}$, $OK_{C}$ share a common point which completes the proof.(P6 -_-)

huh??????? Let $A'$ be the $A$-antipode and let $H_B$ and $H_C$ the reflections of $H$ over $AC$ and $AB$ respectively. Take the intersections of $A'H_B$ and $A'H_C$ with $AC$ and $AB$ respectively to be $E$, $F$. Pascal gives that $H$, $E$, $F$ collinear. Moreover, this also implies that $A'A$ is the angle bisector of $EA'F$, so this circle and $BOC$ are homothetic. From a homothety at $A$ with scale factor $1/2$, It follows that $K_A$ is the reflection of $O$ over $AX_{54}$, the A-Kosnita Cevian. It follows that $X_{54}$ is the center of this circle.

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