circumcenter of BJK lies on line AC, median, right angle, circumcircle related

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#1 h Jun 18, 2020, 9:20 PM

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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

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#2 h Jun 18, 2020, 10:38 PM • 1 Y

Y by Sourorange

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)

Note

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#3 h Jun 19, 2020, 5:47 AM • 2 Y

Y by AlastorMoody, Han1728

Beautiful problem ! :love:

:love:

Here is a solution
Remark

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#4 h Jul 1, 2020, 6:15 PM • 7 Y

Y by AlastorMoody, anser, jelena_ivanchic, SnowPanda, Aryan-23, srijonrick, tigerzhang

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 $\blacksquare$

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#5 h Jul 2, 2020, 4:15 AM

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Why the shortlist is not in the contest collection?

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#6 h Sep 13, 2020, 4:33 AM • 3 Y

Y by jelena_ivanchic, Mathematicsislovely, v4913

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 $APCQ$ .

$[asy] size(9cm); pair D = dir(90); pair T = dir(-90); pair C = dir(-20); pair A = -conj(C); pair P = 0.4*T+0.6*A; pair M = extension(A, C, D, T); pair B = conj(abs(A-C)**2/(4*M-4*P))+M; pair Q = A+C-P; pair E = extension(C, Q, D, A); pair K = extension(T, C, B, D); filldraw(circumcircle(M, B, C), invisible, blue); filldraw(A--T--C--D--cycle, invisible, red); filldraw(A--P--C--Q--cycle, invisible, orange); draw(P--B, orange); draw(A--T, red); draw(D--T, red); draw(D--K, blue); draw(A--C, red); draw(A--K, dotted+red); pair N = extension(Q, foot(Q, A, C), C, T); draw(C--K, red); draw(C--E, orange+dashed); draw(P--N--Q, blue); dot("$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$", N, dir(N)); /* TSQ Source: !size(9cm); D = dir 90 T = dir -90 C = dir -20 A = -conj(C) P = 0.4*T+0.6*A M = extension A C D T B = conj(abs(A-C)**2/(4*M-4*P))+M Q = A+C-P E = extension C Q D A K = extension T C B D circumcircle M B C 0.1 blue / blue A--T--C--D--cycle 0.1 lightred / red A--P--C--Q--cycle 0.1 yellow / orange P--B orange A--T red D--T red D--K blue A--C red A--K dotted red N = extension Q foot Q A C C T C--K red C--E orange dashed P--N--Q blue */ [/asy]$

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.

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#7 h Nov 8, 2020, 5:42 PM

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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. $\blacksquare$

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#8 h Nov 9, 2020, 9:32 PM

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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! $\Box$

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#10 h Dec 19, 2020, 11:49 AM • 1 Y

Y by Mango247

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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 $\frac{1}{2} \angle BOK = \angle BON \overset{(\clubsuit)}{=} \angle BMN \overset{(*)}{=} \angle AJM \overset{\text{ver.opp.}}{=} \angle BJK.\ \blacksquare$

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#11 h Jan 14, 2021, 3:15 AM • 3 Y

Y by Mango247, Mango247, Mango247

Amazing problem! :trampoline:

:trampoline:

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.

$\blacksquare$

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#12 h Jan 19, 2021, 7:18 AM

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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 $AC$ .

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#13 h Apr 10, 2021, 7:05 PM

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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 $ABC$ .
Proof: Well -known.Click to reveal hidden text

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.

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#14 h May 11, 2021, 3:12 PM

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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 $(BJXK).$

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#15 h Aug 29, 2021, 6:06 AM

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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.

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L567

#16 h Aug 29, 2021, 6:40 AM • 1 Y

Y by everythingpi3141592

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. $\blacksquare$

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#17 h Oct 9, 2021, 12:36 PM

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#18 h Oct 11, 2021, 8:05 PM • 2 Y

Y by centslordm, Bradygho

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.

This post has been edited 1 time. Last edited by tigerzhang, Nov 4, 2021, 3:20 AM

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OronSH

#19 h Feb 1, 2024, 11:16 PM

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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 $\angle JCK.$

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#20 h Mar 9, 2024, 5:07 PM

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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.

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#21 h Mar 20, 2024, 7:32 PM

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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.

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#22 h Jun 7, 2024, 3:51 AM

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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
$\measuredangle KBJ = 90 + \measuredangle CBJ = 90 + \measuredangle ACL = \measuredangle KLJ. \quad \blacksquare$

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#23 h Feb 17, 2025, 11:05 AM

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Huh, G1? I guess its bit harder for G1:

Here we go (HM Points):

Re-define $J$ to be $B$ HM point, and let $\omega$ denote the $B$ Apollonius circle. Let $X, B'$ denote the reflections of $J, B$ with respect to $AC$ respectively. Note that $AX$ is $A$ -symmedian.
Note that: $X, B' \in \omega$ . Refine $K = JA \cap \omega$ with $K \neq J$ . Note that the circumcenter of $BJK$ lies on $AC$ .
It suffice to show $KC$ to be tangent to $(BMC)$ and $\measuredangle KBC=90^\circ$ .

Claim: $KC$ is tangent to $(BMC)$
Note that: $\frac{AJ}{JC}=\frac{AK}{KC}$ as both of them lie of the $B$ -Apollonius circle. Thus, by Angle-Bisector theorem, $\angle JCA = \angle KCA$ . Notice that: $\angle KCA = \angle JCA = \angle JBC = \angle MBC$ , and we are done.

Claim: $\measuredangle KBC=90^\circ$
Pure-Angle-Chase:
$\measuredangle KBJ =\measuredangle KJB+ \measuredangle BKJ = \measuredangle CBA + \measuredangle XBB' = \measuredangle CBA + \measuredangle OBM = 90^\circ - \measuredangle JBC.$

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#25 h Today at 5:21 AM

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SO COOOLLLLLL

Let $L$ be second intersection of $(M)$ and $(BMC)$
Let $R$ be second intersection of $BK$ and $(BMC)$
Reflect $J$ across $AC$ to get $N$
Let $T = RM\cap CK$

Clearly:
1. $\angle RBC = 90$ since $\angle RBC = 180 - \angle CBK = 90$
2. $ALR$ collinear $\angle ALR = \angle ALC + \angle RLC = 180$
3. $RATC$ is a cyclic kite
Now, let $J' = BM\cap LC.$
By Pascal's on $LRBMCC,$ we get $J'AK$ are collinear. Thus, $J = J'$

Finally, note that $\angle KBM = \angle MCR = 90 - \angle MAR= 90 - \angle LCR = \angle CJN.$
Thus, since $AC$ is perp bisector of $JN,$ we are done.

In fact, one can get $AT\parallel LC$ , as $\angle CAT = \angle ACT = \angle ART = \angle LCM$

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