Denote $F$ and $X$ as the midpoint of $DE$ and the midpoint of arc $BC$ that doesn't contain $A$. Let $EA$ intersects $\omega$ again at $G$. Let $K'$ be the intersection of $FX$ and $\omega$. Lemma 01. $K'$ lies on $PQ$. Proof. Notice that $GK' \perp FX$ (since $G$ is the antipode of $X$), moreover $G$ lies on the radical axis of $(ADE)$ and $(BIC)$. This is because of $\triangle GAB \sim \triangle GBE$ ($\measuredangle GAB = \measuredangle GCB = \measuredangle CBG = \measuredangle EBG$). Therefore, $GA \cdot GE = GB^2$. Since $F$ and $X$ are center of $(AED), (BIC)$, then $K'$ lies on radical axis of $(AED), (BIC)$ as well: $PQ$. It is now suffices to prove $K' = K$. Lemma 02. Let $K'D$ intersects $\omega $ at $T$, then $AT \parallel BC$. Proof. Notice that $FD^2 = FB \cdot FC = FK \cdot FX$, as we know $(E,D ; B,C) = -1$ and $F$ being the midpoint of $DE$. Therefore, \[ \measuredangle KDF = \measuredangle KXD = \measuredangle KXA = \measuredangle KTA \]Hence, $AT \parallel DF \equiv BC$. Lemma 03. $T,M,B$ are collinear. Proof. As $ATCB$ is an isosceles trapezoid, and $M \in AC$, and $MB = MC$. It follow that $T,M,B$ are collinear. Lemma 04. $K' = K$. Proof. Notice that Pascal Theorem on $AK'TBBC$ gives us $AK' \cap BB$, $K'T \cap BC = D$ and $TB \cap AC = M$ are collinear. This gives us $AK', DM$ and tangent of $\omega$ at $B$ are concurrent. But, we know $BB \cap DM = S$. So, $A , K', S$ are collinear. Since $K$ and $K'$ both lie on $\omega$, then $K = K'$.

Let $X=AA \cap BC$ and let $L$ be the midpoint of arc $\widehat{BC}$ of not containing $A$. $XL$ intersects $\omega$ for a second time at $K'$ and $J \neq A$ is the intersection of the other tangent from $X$ to $\omega$. Claim 1: $AJ,BL,CK'$ concur Proof: Let $V=BL \cap CK$ then by considering the cyclic quad $BKLC$, as $X=BC \cap KL$ by Brokard we have $V$ lies on the polar of $X$ but this is just line $AJ$. Claim 2: $V,M,D$ colinear on line $\ell$. Proof: Let $N$ be the midpoint of arc $\widehat{BAC}$ then applying Pascal to $JACBLN$ we get $M,V,BC \cap NJ$ colinear. It remains to show $NJ$ passes through $D$ but this follows from: $$\measuredangle LJD=\measuredangle EJD=90^{\circ}=\measuredangle LJN$$ Note by definition $S$ also lies on $\ell$. Claim 3: $S,BL \cap KC,D$ colinear. Proof: By applying Brokard to $BBLAKC$ the claim follows From Claim 3 we have $BL \cap KC \in \ell$. But then as by Claim 2 $V$ lies on $\ell$ we have: $$V \equiv \ell \cap BL \equiv BL \cap K'C \equiv BL \cap KC$$and so $K \equiv K'$ as they are both the second intersection of $CV$ with $\omega$. Claim 4: $N$ lies on $PQ$ Proof: An easy angle chase gives: $$\measuredangle EAB=\measuredangle NAB=\measuredangle NCB=\measuredangle CBN=\measuredangle EBN$$Hence $NB$ is tangent to $\odot ABE$ so $NB^2=NA \cdot NE$. Also as $\angle LBN=90^{\circ}$ and $L$ is the centre of $\odot BIC$ we have $NB$ tangent to $\odot BIC$. This combined with the previous relation shows $N$ is on the radical axis of $\odot ADE$ and $\odot BIC$ which is line $PQ$. Claim 5: $PQ \bot LX$ Proof: $L,X$ are the centres of $\odot BIC$, $\odot ADE$ respectively from which the claim follows. Now we have $\angle LKN=90^{\circ}$ (as $LN$ diameter of $\omega$) hence by Claim 4 and Claim 5 we have $K$ on $PQ$ as desired.

I have a feeling that this was posted in HSO few days ago in some other contest.

This is my problem

oh nice! (darn i had a sol but it got deleted ) just wondering - is there a good way of dealing with all that M and S and tangent stuff without using pascal? That was the only thing that really bothered me about this problem - that aside it was a very nice orthocenter config geo problem (albeit a bit easy for a 6), and so it was kind of disappointing that there wasn't a good way to deal with the other half of the config.

Pluto1708 wrote: I have a feeling that this was posted in HSO few days ago in some other contest. No, the problems have been posted recently for the same contest. We had to delete that topics because the user that posted them did that quite randomly, did not warn us that he was doing so, and left behind a few issues (not posting the proposer's name, having some typos, not adding a logo). khina wrote: oh nice! (darn i had a sol but it got deleted ) We're sorry for that, we couldn't handle the situation other way .

khina wrote: just wondering - is there a good way of dealing with all that M and S and tangent stuff without using pascal? That was the only thing that really bothered me about this problem - that aside it was a very nice orthocenter config geo problem (albeit a bit easy for a 6), and so it was kind of disappointing that there wasn't a good way to deal with the other half of the config.

Yes ;)

Nice problem! Congrats to rmtf1111. Let $T$ be the intersection of the tangent to $\omega$ at $A$ and $\overline{BC}$. Let $\overline{AD}$ cut $\omega$ again at $L$. It's well known that $T,L$ are centers of $\odot(ADE)$ and $\odot(ABC)$ respectively. Let $A',D'$ be the reflections of $A,D$ across the perpendicular bisector of $BC$. The key observation is $S,D'$ are isogonal conjugate w.r.t. $\triangle ABM$. This is obvious because $\angle ABS =$ $ 180^{\circ}-\angle B =$ $180^{\circ} - \angle ABB'$ and $\angle BMS = 180^{\circ} - \angle D'MA$. Thus the $\sqrt{bc}$ inversion maps $K\mapsto D'$. It also maps $L\mapsto D$ and $T\mapsto A'$. As $AA'D'D$ is isosceles trapezoid, which is cyclic, we deduce by inversion that $K,L,T$ are colinear. Finally, let $N$ be the midpoint of arc $BAC$. As $\angle NKL = 90^{\circ}$, it suffices to show that $N$ lie on the radical axis. WLOG $AB < AC$ and note that $\angle BAE = 90^{\circ} - \tfrac{\angle A}{2} = 180^{\circ} - \angle NBE$ thus $NB$ is tangent to $\odot(ABE)$ so $NA\cdot NE = NB^2$ hence we are done. Remark: The $\sqrt{bc}$ step can be replaced with a quick sine law bash: just show that $BK : CK = AB^2 : AC^2$ and conclude by Angle Bisector Theorem.

Let $N$ and $T$ be the midpoints of arc $BC$ and segment $DE$ respectively. let $\overline{BS}$ intersect $\overline{AT}$ at $J$. Then since $\triangle BJA\sim\triangle BMC$, it follows that $A,B,M,J$ are concyclic. Therefore, $\angle AMJ=\angle ABJ=ACT$, and so $JM\parallel BC$. Claim: $N,T,K$ are collinear. Proof. Let $\overline{NT}$ intersect $(ABC)$ at $K_1$. Then \[(A,K_1;B,C)=(A,N;C,B)\stackrel{A}{=}(T,D;C,B)\]where the first equality is by an inversion at $T$ with radius $TA$. Also, \[(A,K;B,C)\stackrel{A}{=}(J,S;B,\overline{AC}\cap\overline{BS})\stackrel{M}{=}(P_\infty,D;B,C)=\frac{CD}{BD}.\]Since $CT:BT=CD^2:BD^2$, these two cross ratios are equal and so it follows that $K_1=K$. $\blacksquare$ Let $R$ be the midpoint of arc $BC$. Then since $\angle FBE=\angle CAE=\angle FAB$, $\triangle FAB\sim\triangle FBE$, so $FA\cdot FE=FB^2$ and consequently $F$ lies on the radical axis of $(ADE)$ and $(BIC)$. Finally, notice that $K$ is the foot of perpendicular from $F$ to the line joining the centers of those two circles. Therefore, $\overline{FK}$ must be the radical axis of them, which coincides with $\overline{PQ}$. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(140); pair B = dir(210); pair C = dir(330); pair I = incenter(A,B,C); pair D = extension(A,I,B,C); pair E = extension(A,rotate(90,A)*I,B,C); pair O = circumcenter(A,B,C); pair F = 2*foot(O,A,E)-A; pair N = 2*foot(O,A,D)-A; pair M = extension(F,N,A,C); pair T = midpoint(D--E); pair K = 2*foot(O,N,T)-N; pair S = extension(A,K,D,M); pair J = extension(B,S,A,T); pair PQ[] = intersectionpoints(circumcircle(A,D,E),circumcircle(B,I,C)); draw(A--B--C--cycle, black+1); draw(circumcircle(A,B,C)); draw(circumcircle(B,I,C)); draw(circumcircle(A,D,E)); draw(E--F); draw(A--N); draw(A--T); draw(B--M); draw(A--S); draw(J--S); draw(M--S); draw(T--N, mydash); draw(F--PQ[1], mydash); draw(B--E); dot("$A$", A, dir(90)); dot("$B$", B, dir(B)); dot("$C$", C, dir(0)); dot("$D$", D, dir(315)); dot("$E$", E, dir(180)); dot("$F$", F, dir(90)); dot("$I$", I, dir(150)); dot("$J$", J, dir(135)); dot("$K$", K, dir(20)); dot("$M$", M, dir(30)); dot("$N$", N, dir(270)); dot("$S$", S, dir(225)); dot("$T$", T, dir(270)); dot(PQ[0]); dot(PQ[1]); [/asy][/asy]

A different solution I guess . $(ADE)$ is the $A-$apollonius circle of $ABC$ and hence $\frac{BQ}{QC} = \frac{BD}{DC} \implies QD$ is angle bisector of $\angle BQC$ and hence passes through mid point of arc $BC$ of circle $(BIC)$ Let $L$ be the midpoint of arc $BC$ of $(ABC)$ not containing $A$ and let $L'$ be the other midpoint, let $A'$ be midpoint of $BC$. It is known that $L$ is center of $(BIC)$ and that $L'B$ is tangent to it . So $Pow(L',(BIC)) = L'B^2 = L'A'\times L'L = \L'A \times L'E = Pow(L',(ADE))$ (As $L',A,E$ are collinear, $L,A',A,E$ are concyclic ) Hence we get that $L'$ lies on radical axis of $(BIC)$ and $(ADE)$ that is $PQ$. Let $PQ$ intersect $(ABC)$ again at $K'$ and $BC$ at $X$ , $K'X$ is angle bisector of $BK'C$ So $\frac{BK'}{K'C} = \frac{BX}{XC} = \frac{sin \angle BPQ \times sin \angle PCB} {sin \angle CPQ \times sin \angle PBC} = \frac{BQ\times BP}{CQ \times PC} = \frac{BP^2}{CP^2} $ In the last step we used that $BPCQ$ is harmonic as tangents at $B,C$ intersect on $PQ$ , $PD$ is angle bisector of $\angle BPC$ and hence $\frac{BK'}{K'C} =\frac{BP^2}{CP^2}=\frac{BD^2}{CD^2}$ Claim - Let $T$ be on $(ABC)$ such that $ATCB$ is trapezium. Then $K$ lies on $TD$ Proof - Note that $TB \cap AC = M, AK \cap BB = S $ now apply Pascal on $ABBKCT$ to get that $BC \cap TK$ lies on $MS$ and hence must be $D$. Now to finish problem we just compute $\frac{BK}{KC} = \frac{sin \angle BTD}{sin \angle DTC} = \frac{BD\times TC}{CD \times TB } = \frac{BD\times AB}{CD \times AC } = \frac{BD^2}{CD^2} $. Now simply note that both $K,K'$ are on opposite side of $A$ as $BC$ and hence we must have $K = K'$ Hence proved .

Not sure if this is new... Denote by $F,G$ the midpoint of arcs $(BAC) , (BC)$respectively. One can easily observe that $FA.FE=FB^2$ So $F$ lies on radical axis of $(BIC),(ADE)$. Let $T = AA \cap BC$ which is the center of $(ADE)$. Since $GT$ is the line connecting the centers of $(BIC),(ADE)$ , and $F$ is already on $PQ$ , It's enough to show that $K$ lies on $FT$ which implies the statement of problem by $\angle FKG = 90$. Note that by ratio lemma , if we redefine $K$ to be $TG \cap \omega$: $$\frac{sin \angle KAB}{sin \angle KAC}=\frac{KB}{KC} = \frac{TB}{TC}= (\frac{AB}{AC})^2$$So it's enough to show that : $$\frac{sin \angle{SAB}}{sin \angle{SAC}} = \frac{SB}{SC} \cdot \frac{sin \angle{C}}{sin \angle C+\alpha} =? (\frac{AB}{AC})^2*$$But we already know that: $$\frac{SB}{SC} \cdot \frac{sin \angle B}{sin \angle C+\alpha} = \frac{sin \angle SMB}{sin \angle SMC} = \frac{sin \angle DMB}{sin \angle DMC} = \frac{AB}{AC} **$$Now after multiplying $**$ by $\frac{sin \angle C}{sin \angle B} = \frac{AB}{AC}$ , we'll get that $*$ is indeed true ; and we're done.

Easy for P6? Let the tangent at $A$ intersect $BC$ at $T$, Let $N$ and $F$ be the arcmidpoints, let $BS \cap AT = L$ Key Claim$N-T-K$

Proof

Clearly $F$ belong to the radax of the two circles, also $FK$ perpendicular to line joining centers, hence done.

Diagram