Redefine $J$ as the $B-$Humpty point of $ABC$, and $K$ as the intersection of $AJ$ and the circle through $B, J$ and the reflection of $B$ in $AC$ (say $B'$). Note that the reflection of $J$ in $AC$ (say $X$), which is the endpoint of the $B-$symmedian chord, is also on this circle. This circle is in fact the $B-$Apollonian circle of $ABC$, so its center is on $AC$. Now we show that our definitions of $J, K$ are consistent with the previous definitions, i.e., we need to show that $\angle KBC = 90^\circ$ and $KC$ is tangent to $(BMC)$. For the first part, note that $\angle KBJ = 180^\circ - \angle BJK - \angle JKB = 180^\circ - 180^\circ + \angle BAC - \angle JB'B = \angle BAC - \angle XBB' = \angle BAC - \angle OBM$, where $O$ is the circumcenter of $ABC$. So $\angle KBC = \angle KBJ + \angle JBC = 90^\circ$. For the second part, we claim that $C, K, X$ are collinear, from where it would follow that $\angle KCM = \angle XCA = \angle XBA = \angle CBJ$, which would finish the proof. To this end, note that $\angle KXC = \angle KXB + \angle BXC = \angle AJB + \angle BAC = 180^\circ$, whence we are done. (The other configurations can be handled similarly)

Beautiful problem !

An easy solution which does not invoke HM points. Let $N$ be the midpoint of $KC$.Clearly $NB=NK$ Let the perpendicular bisector of $BK$ intersect $AC$ at point $O$.It is also easy to see that $AK$ is parallel to $MN$ Since $\angle NOC=\angle BCA=180-\angle MBN \Rightarrow MBNO$ is cyclic.Therefore $0.5\times \angle BOK=\angle BON=\angle BMN=\angle AJM=\angle BJK$.Thus $O$ is the circumcenter of $BNK$ and we are done$\blacksquare$

Why the shortlist is not in the contest collection?

The following solution was found jointly with Serena An, Sanjana Das, and Sunaina Pati. Let $D$ be the antipode of $C$ on $(BMC)$, so we have right triangle $DMC$. Complete it to cyclic kite $DCAT$ inscribed in a circle with diameter $\overline{DT}$ (i.e.\ with $\angle DCT = \angle DAT = 90^{\circ}$). We define $P = \overline{BM} \cap \overline{AT}$ and then construct parallelogram $APCQ$. [asy][asy] size(9cm); pair D = dir(90); pair T = dir(-90); pair C = dir(-20); pair A = -conj(C); pair P = 0.4*T+0.6*A; pair M = extension(A, C, D, T); pair B = conj(abs(A-C)**2/(4*M-4*P))+M; pair Q = A+C-P; pair E = extension(C, Q, D, A); pair K = extension(T, C, B, D); filldraw(circumcircle(M, B, C), invisible, blue); filldraw(A--T--C--D--cycle, invisible, red); filldraw(A--P--C--Q--cycle, invisible, orange); draw(P--B, orange); draw(A--T, red); draw(D--T, red); draw(D--K, blue); draw(A--C, red); draw(A--K, dotted+red); pair N = extension(Q, foot(Q, A, C), C, T); draw(C--K, red); draw(C--E, orange+dashed); draw(P--N--Q, blue); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$C$", C, dir(C)); dot("$A$", A, dir(A)); dot("$P$", P, dir(P)); dot("$M$", M, dir(M)); dot("$B$", B, dir(B)); dot("$Q$", Q, dir(Q)); dot("$E$", E, dir(E)); dot("$K$", K, dir(K)); dot("$N$", N, dir(N)); /* TSQ Source: !size(9cm); D = dir 90 T = dir -90 C = dir -20 A = -conj(C) P = 0.4*T+0.6*A M = extension A C D T B = conj(abs(A-C)**2/(4*M-4*P))+M Q = A+C-P E = extension C Q D A K = extension T C B D circumcircle M B C 0.1 blue / blue A--T--C--D--cycle 0.1 lightred / red A--P--C--Q--cycle 0.1 yellow / orange P--B orange A--T red D--T red D--K blue A--C red A--K dotted red N = extension Q foot Q A C C T C--K red C--E orange dashed P--N--Q blue */ [/asy][/asy] Claim: Lines $CQ$ and $AD$ meet on $(BMC)$, say at $E$. Proof. Because $\measuredangle QCM = \measuredangle PAM = \measuredangle TAM = \measuredangle ADM$. $\blacksquare$ Claim: The points $A$, $Q$, $K$ are collinear, so $Q=J$. Proof. Pascal on $DECCMB$. $\blacksquare$ Claim: Reflect $Q$ across $\overline{AC}$ to obtain $N$. Then we have $BKNQ$ concyclic. Proof. Because $\measuredangle KNQ = 90^{\circ} - \measuredangle PNT = 90^{\circ} - \measuredangle ACT = \measuredangle DCM = \measuredangle DBM = \measuredangle KBQ$. $\blacksquare$ Since $\overline{AC}$ is obviously the perpendicular bisector of $\overline{QN}$, the proof is complete.

Let the tangent to $(BMC)$ hit $AC$ at $T$. Since $\triangle KBC$ is a right triangle, $T$ must be the midpoint of $KC$. Let $N$ be the midpoint of $BK$. The perpendicular bisector $NT$ of $BK$ hits $AC$ at $O$. It suffices to show that $O$ is the center of $(BJK)$. First off, note that\[\angle NTB = \angle TBC = \angle TCB = \angle CMB = \angle OMB\]so $BTOM$ is cyclic. This is in fact, quite nice. It tells us that\[\angle BMT = \angle BDT = \frac12\angle BOK.\]Furthermore, by midpoints, $MT \parallel AK$ so $\angle BMT = \angle BJK$ hence $\angle BJK = \frac12\angle BOK$ and thus $J$ lies on the circle centered at $O$ with radius $OB = OK$, as desired. $\blacksquare$

Main Claim: $J$ is the $B$-HM point of $\triangle ABC$. Proof: Let $K'$ be the $\sqrt{ac}$ inverse of $K$; this inversion swaps $A$ and $C$ as well. Note \[\angle BK'A = \angle BCK = \angle BMC = 180^\circ - \angle BMA,\]so $K'BMA$ is cyclic. As \[ \angle K'MA = \angle K'BA = \angle CBK = 90^\circ, \]so $K'$ lies on the perpendicular bisector of $\overline{AC}$. Now, \begin{align*} \angle BKA &= \angle BCK' \\ &= C - \angle K'CA \\ &= C - \angle K'AM \\ &= C - (180^\circ - \angle K'BM) \\ &= C - (180^\circ - (90^\circ + \angle ABM)) \\ &= C + \angle ABM - 90^\circ. \end{align*}As a result, \begin{align*} \angle AKC &= \angle BKC - \angle BKA \\ &= (90^\circ - \angle BCK) - (C + \angle ABM - 90^\circ) \\ &= (90^\circ - \angle BMC) - (C + \angle ABM - 90^\circ) \\ &= (90^\circ - (180^\circ - \angle MBC - C)) - (C + \angle ABM - 90^\circ) \\ &= \angle MBC - \angle ABM. \end{align*}In particular, \begin{align*} \angle CAJ &= 180^\circ - \angle AKC - \angle KCA \\ &= 180^\circ - (\angle MBC - \angle ABM) - (180^\circ - \angle MBC) \\ &= \angle ABM, \end{align*}so $\overline{AC}$ is tangent to $(ABJ)$, implying the claim. $\blacksquare$ Now, let $T$ be the foot of the $B$-internal angle bisector, let $T'$ be the foot of the $B$-external angle bisector, and let $L = \overline{BB} \cap \overline{AC}$. We claim that $L$ is the circumcenter of $(TJBK)$, which is enough to solve the problem. Noting that $(J, B)$ are inverses under $(AC)$ inversion, it follows that $\frac{AT}{TC} = \frac{AB}{BC} = \frac{AJ}{JC}$, implying that $\overline{JT}$ bisects $\angle AJC$. As $(AC;TT') = -1$, it follows that $\angle TJT' = 90^\circ$ (by right angles/bisectors lemma), so $TJBT'$ is cyclic with diameter $TT'$. Noting that $AJHC$ is cyclic (this follows from $(AC)$ inversion), we find \begin{align*} \angle TJK &= 180^\circ - \angle AJT \\ &= 180^\circ - \frac{1}{2}\angle AJC \\ &= 180^\circ - \frac{1}{2}\angle AHC \\ &= 90^\circ + \frac{B}{2} \\ &= \angle TBK. \end{align*}Hence, $TJBKT'$ is cyclic. Yet \[ \angle LBT = \angle LBA - \angle TBA = 180^\circ - C - \frac{B}{2} = \angle LTB, \]so $L$ is the circumcenter of $TBT'$. Thus, $L$ is the circumcenter of $\triangle BJK$, and we are done! $\Box$

Storage parmenides51 wrote: Let $BM$ be a median in an acute-angled triangle $ABC$. A point $K$ is chosen on the line through $C$ tangent to the circumcircle of $\vartriangle BMC$ so that $\angle KBC = 90^\circ$. The segments $AK$ and $BM$ meet at $J$. Prove that the circumcenter of $\triangle BJK$ lies on the line $AC$. Aleksandr Kuznetsov, Russia Solution. Let $N$ be the midpoint of $KC.$ We know that $NB=NK$. Let the perpendicular bisector of $BK$, through $N$, intersect $AC$ at $O.$ Now, as $CB \perp KB$, and also $ON \perp BK$, so $BC \parallel ON$, and thus $$\angle NOC = \angle BCA \overset{(\spadesuit)}{=} 180^{\circ} - \angle MBN,$$thus $M,O,N,B$ are concyclic; and $(\spadesuit)$ follows as $\angle NBC = \angle NCB$ and $\measuredangle (\overline{AMC}, \overline{CC}) = \angle MBC$ by Alternate segment theorem, so $\angle MBN = \angle NCB + \measuredangle (\overline{AMC}, \overline{CC}) = 180^{\circ} - \angle BCA.\quad(\clubsuit)$ Now, by Midpoint theorem in $\triangle ACK$, we have $MN \parallel AK-(*).$ Lastly, we note that $$\frac{1}{2} \angle BOK = \angle BON \overset{(\clubsuit)}{=} \angle BMN \overset{(*)}{=} \angle AJM \overset{\text{ver.opp.}}{=} \angle BJK.\ \blacksquare$$

Amazing problem! We prove that $J$ is actually the $B$-humpty point WRT $ABC$. Let $X_B$ the $B$-humpty point WRT $ABC$, and let $K'$ the intersection of $AX_B$ with the line through $C$ tangent to $(BMC)$. Observe that from the ratio lemma, we have that $$\frac{sin \angle X_BBA}{sin \angle X_BBC}=\frac{BA}{BC} \qquad (\spadesuit)$$Now, notice that since $CK'$ is tangent to $(BMC)$, then $\measuredangle X_BAC= \measuredangle X_BBA \implies \measuredangle K'AC= \measuredangle X_BBA$. Hence, by Law of Sines on $K'AC$, $$\frac{K'A}{K'C}= \frac{sin \angle K'AC}{sin \angle ACK'}=\frac{sin \angle X_BBA}{sin \angle X_BBC} \underbrace{=}_{(\spadesuit)} \frac{BA}{BC}$$$\implies K'$ lies on the $B$-Apollonius circle of $ABC$. Let $L$ the centerof the $B$-Apollonius circle of $ABC \implies LK'^2=LB^2=LA.LC$ (it's well known that $L$ is the intersection of the tangent through $B$ to $(ABC)$ with $AC$) $\implies \angle LK'C= \angle CAK'= \angle CAX_B= \angle X_BBA (\star)$, and since $\angle ACK'= 180º- \angle MBC= 180º- \angle X_BBC \implies \angle LK'C= \angle X_BBC- \angle X_BBA \implies$ from $(\star)$, $\angle LK'X_B= \angle X_BB_C \implies$ if $P= K'L \cap BC \implies P \in (BX_BK')$, and since $PK$ is diameter of $(BX_BK')$, $\angle PBK'=90º \implies \angle CBK'= 90º \implies K=K'$ and $J=X_B$, so $(BJK)$ is the $B$-Apollonius circle of $ABC$, whose center lies on $AC$ (it's well known that the center of the $B$-Apollonius circle of $ABC$ is the intersection of the tangent through $B$ to $(ABC)$ with $AC$), as desired. $\blacksquare$

I present my solution which is a bit different from others (idea is still the same ig). We want to claim that $J$ is the $B$-humpty point of $\triangle ABC$. So let $I$ be the $B$-humpty point of $\triangle ABC$ (Nota bene: $I$ is not a incenter lol). Let $I'$ be the reflection of $I$ over $AC$ and let $K=I'C\cap AI$. Let $D$ be the centre of $(MBC)$. We have $\measuredangle IKI'=\measuredangle ACI'-\measuredangle CAK=\measuredangle ICA-\measuredangle CAI$ and $\measuredangle IBI'=\measuredangle IBC-\measuredangle I'BC=\measuredangle ICA-\measuredangle ABJ=\measuredangle ICA-\measuredangle CAI$, since it is kind of well-known that $BI'$ is the symmedian (can be proven by simple angle chase using properties of a humpty point). Hence, $I'IBK$ is cyclic. Now, $$\measuredangle IBK=\measuredangle II'K=\measuredangle II'C=\measuredangle CII'=\measuredangle ICA-90^\circ= \measuredangle IBC-90^\circ\implies \measuredangle CBK=90^\circ.$$ We have $$\measuredangle CMB+\measuredangle BCD=\frac{\measuredangle CDB}{2} +\measuredangle BCD=90^\circ$$Also, $$\measuredangle KCB=90^\circ-\measuredangle BKC=90^\circ+\measuredangle I'KB=90^\circ+\measuredangle I'IC+\measuredangle CIB=90^\circ+90^\circ-\measuredangle ICA+\measuredangle CIB=\measuredangle ACI+\measuredangle CIM=\measuredangle CMI=\measuredangle CMB,$$thus $\measuredangle KCD=90^\circ$. Thus, $K$ coincides with original $K$ and $J=I$. Since $AC$ is the perpendicular bisector of $JJ'$ ($J'$ is the reflection of $J$ over $AC$). And since $J'$ lies on $JBK$, we have $O$ lying on $AC$.

Took 1.5 bloody hours. WLOG $BA<BC$. Define $CK \cap \odot(ABC) = X$. Claim: $BX$ is symmedian of $\triangle ABC$. Proof: This is simple angle chasing $\angle ACX = \angle MBC =\angle ABX$ $\blacksquare$ The most difficult part of the problem is to realise that $\odot(BJK)$ is $B$ Apollonian circle. This simply motivates to add point $F$, which is interaction of $\angle ABC$ bisector and side $BC$. Then $\odot(BFX)$ is $B$ Apollonian circle (well-known fact). Claim: $XF$ intersects $\odot(ABC)$ at point $N$, which is midpoint of arc $ABC$.

Consequently this immediately gives us $\angle FXC =90^{\circ} - 0.5 \angle B = \angle KBF$. Thus $K$ lies on $\odot(BFX)$. From Apollonian circle definition it follows: $$ \frac{KA}{KC}=\frac{BA}{BC} = \frac{AF}{FC} $$This reveals that $KF$ is angle bisector $\angle AKC$. We conclude by: $$ \angle AKF = \angle FKX = \angle FBX = \angle JBF $$since this simply says that $J$ lies on $\odot(BFK)$. Now we are done since both points $J,K$ lies on $B$ Apollonian circle.

Invert around $B$ with radius $\sqrt{BA\cdot BC}$ and then reflect the diagram in the angle bisector of $\angle ABC.$ Let $X=\overline{CK}\cap(ABC),$ and let $T$ be the intersection of the tangents to $(ABC)$ at $A$ and $C.$ $\textbf{Claim: }$ $M^*=X$. $\emph{Proof: }$ Since $\angle ABX=\angle ACX=\angle MBC,$ we know $\overline{BM},\overline{BX}$ are isogonal with respect to $\angle ABC.$ Therefore, since the inverse of $\overline{AC}$ is $(ABC),$ we are done. $\blacksquare$ $\textbf{Claim: }$ $J^*=T$ $\emph{Proof: }$ By definition, $K\in\overline{CX},$ so $K^*\in (BMA).$ Furthermore, $\angle KBC=90^\circ,$ so $\angle K^*BA=90^\circ.$ This implies that $K^*$ is the antipode of $A$ in $(BMA).$ In particular, $K^*$ lies on the perpendicular bisector of $\overline{AC}.$ Since $J\in\overline{KA},$ quadrilateral $BK^{*}CJ^*$ is cyclic. Therefore, \begin{align*} \angle CJ^*B &=180^\circ-\angle BK^*C\\ &=180^\circ-(\angle BK^*A+2\angle AK^*M)\\ &= 180^\circ-(\angle BMA+2\angle ABM)\\ &= \angle CAB-\angle ABM.\end{align*}But we also know that $J\in\overline{BM},$ so $J^*\in\overline{BX}$ and thus $\angle J^*BC=\angle ABM.$ It follows that $\angle BCJ^*=180^\circ-\angle CAB,$ which implies the desired conclusion. $\blacksquare$ $\textbf{Claim: }$ $X\in (BJK)$ $\emph{Proof: }$ Just note that $J^*,X^*,K^*$ all lie on the perpendicular bisector of $\overline{BC}.$ $\blacksquare$ Now write \begin{align*} \angle CAJ &=\angle CAB-\angle JAB\\ &=\angle CAB-\angle CJ^*B\\ &=\angle ABM\\ &=\angle XBC\\ &=\angle XAC, \end{align*}and similarly, $\angle JCA=\angle ACX.$ Thus, $\overline{AC}$ is the perpendicular bisector of $\overline{JX},$ so it passes through the center of $(BJXK).$

nice problem Let $N$ denote the midpoint of $BK$ and $O$ the circumcenter of $BJK.$ The key claim is as follows: Claim: $BONM$ is cyclic. Proof: Note that $ON \parallel BC$ and $NM \parallel KA,$ so $\measuredangle ONM$ is equal to the complement of the angle formed by lines $KA$ and $BC.$ Let $\measuredangle JBC = \alpha$ and $\measuredangle JKB = \theta.$ Then $\measuredangle ONM = 90 + \theta.$ A quick angle chase gives $\measuredangle OBK = \alpha - \theta,$ so \[ \measuredangle OBJ = \measuredangle OBM = \measuredangle OBK + \measuredangle KBM = 90 - \alpha + \alpha - \theta = 90 - \theta. \]Therefore, $\measuredangle OBM + \measuredangle ONM = 180$ so the points are concyclic. Furthermore, note that $BN$ happens to be tangent to $(BMC)$ since $BN = NC$ (circumcenter of right triangle). Then, \[ \measuredangle OMB = \measuredangle ONB = \measuredangle CBN = \measuredangle NCB = \alpha + \measuredangle C = \measuredangle AMB \]by exterior angles. Hence, since $\measuredangle OMB = \measuredangle AMB,$ $O,A,$ and $M$ are collinear which implies that $O \in AC$ as desired.

Let $N$ be midpoint of $KC$ and $(BJK)$ meet $KC$ at $Z$. Let the circumcenter of $(BJK)$ be $O$ Since $\angle NZB = 180 - \angle KZB = 180 - \angle KJB = 180 - \angle NMB$, we have $BMNZ$ is cyclic Since $\angle ZOB = 2 \angle BKC = \angle BNC$, we also have $BMNZO$ is cyclic But now we have $\angle OMB = \angle ONB = \angle NBC = \angle AMB$ and so $O$ lies on $AC$, as desired. $\blacksquare$

my job is to overkill geo Let $C'$ be the antipode of $C$ with respect to the circumcircle of $\triangle BMC$. Notice that $AK$ passes through $C'$. We claim that $J$ is the $B$-humpty point of $\triangle ABC$, after which we are done by easy angle chasing. Rewrite the problem as the following: Let $M$ be the midpoint of $\overline{BC}$ in $\triangle ABC$. Let $B'$ be the antipode of $B$ with respect to the circumcircle of $\triangle ABM$. Let $J$ be the $A$-humpty point of $\triangle ABC$. Prove that $\overline{AB'}$, $\overline{CJ}$, and the line passing through $B$ tangent to the circumcircle of $\triangle ABM$ are concurrent. Let the circumcircle of $\triangle ABM$ intersect the circle centered at $B'$ passing through $B$ at a point $L \neq B$. Claim: $C$, $J$, and $L$ are collinear. Proof: We have $$\measuredangle BLJ=\measuredangle BAM=\measuredangle BB'M=\measuredangle BLC.$$ Claim: $AB'CL$ is cyclic. Proof: More angle chasing. Now, we use the radical axis theorem on the circumcircles of $\triangle ABM$, $\triangle BCL$, and $AB'CL$ to finish.

solved with orz @GrantStar!!! Let $N,P,O$ be the midpoint of $BC,$ the midpoint of $CK$ and the circumcenter of $BKJ.$ First we notice $BP$ is also tangent to $(BMC)$ so $PM$ is the $M$ symmedian of $\triangle BMC.$ Now we have $\angle CAJ=180-\angle CMP=\angle NMB=\angle ABJ$ so $J$ is the $B$ humpty point. Then by the tangency $\angle KCM=\angle CBM=\angle JCM,$ and we have $\angle JCK=2\angle MBC=2(90-\angle KBJ)=180-\angle KOJ$ so $KOJC$ is cyclic. However $O$ is on the perpendicular bisector of $JK$ so it is the fact $5$ point so it lies on $AC$ which bisects $\angle JCK.$

The HM point ends up being sort of a red herring... We let $D$ be the midpoint of $\overline{CK}$. It suffices to show that $O =\overline{D\infty_{BC}} \cap \overline{AC}$ satisfies $\angle BOD = \angle BJK$ as $\overline{OD}$ bisects $\angle AOK$. Observe that as $\angle MOD + \angle MBC + \angle CBD = 180^\circ$, it follows that $MBDO$ is cyclic. Then $\angle BOD = \angle BMD = \angle BJK$ as $\overline{MD} \parallel \overline{JK}$ by midline. Remark: If we instead wrote $\angle BOD = \angle BMD = \angle AJM$, the configuration actually follows from $J$ being the $B$-HM point. Proving this, however, is much more difficult than the problem itself, hence the irony.

Introduce $B'$ as the reflection of $B$ over $\overline{AC}$. The condition then naturally rewrites as $KBJB'$ being cyclic, which is equivalent to $\angle B'BM = \angle B'KA$. Now we complex bash with unit circle $(BCM)$. Then $a = 2m - c$, $k = \frac{3bc-c^2}{b+c}$, and $b' = c + m - \frac{cm}{b}$, so $\frac{b'-k}{a-k} = \frac{b-c}{2b}$ and $\frac{b'-b}{m-b} = \frac{b-c}{b}$, done.

The bulk of this proof is to show that $J$ is the $B$-Humpty point. Consider $\sqrt{ca}$ inversion about $B$ composed with a reflection over the $B$-bisector. $CK \mapsto (ABM)$, so $K$ maps to the point on $(ABM)$ such that $\angle K^*BA = 90$. Define $T = AA \cap CC$, which lies on the $B$-symmedian. Note \begin{align*} \measuredangle BTC &= \measuredangle BAC + \measuredangle ABM \\ &= \measuredangle BMC + 2 \measuredangle ABM \\ &= \measuredangle BK^*A + 2 \measuredangle AK^*M \\ &= \measuredangle BK^*C, \end{align*}so $T \in (K^*BC)$, implying $T$ is the inverse of $J$, making $J$ the Humpty Point. If we reflect $K$ over $AC$ to $L$, showing that $JBKL$ is cyclic suffices. From the Humpty Point classification, we get \[\measuredangle ACJ = \measuredangle CBM = \measuredangle KCA = \measuredangle ACL,\] so $J$, $C$, and $L$ are collinear. Finally, we the desired from \[\measuredangle KBJ = 90 + \measuredangle CBJ = 90 + \measuredangle ACL = \measuredangle KLJ. \quad \blacksquare\]