Common chord bisects segment

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Common chord bisects segment

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Source: China TSTST 3 Day 1 Q2

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#1 h Mar 18, 2017, 8:12 AM • 10 Y

Y by rkm0959, anantmudgal09, baopbc, ThE-dArK-lOrD, guangzhou-2015, buratinogigle, nguyendangkhoa17112003, Ya_pank, Adventure10, Rounak_iitr

Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$ . Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$ .

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#2 h Mar 18, 2017, 9:54 AM • 16 Y

Y by baopbc, Ankoganit, smy2012, dagezjm, Saro00, guangzhou-2015, zed1969, anantmudgal09, Mosquitall, enhanced, nguyendangkhoa17112003, Gaussian_cyber, AllanTian, JG666, Adventure10, thinkcow

Let $\widetilde{A},$ $\widetilde{D}$ be the isogonal conjugate of $A,$ $D$ WRT $\triangle BCD,$ $\triangle ABC,$ respectively. Clearly, $\widetilde{A}$ and $\widetilde{D}$ are symmetry WRT $BC,$ so one of the intersection $U$ of $\odot (PQR)$ and $\odot (XYZ)$ lies on $BC.$ Let $V$ be the second intersection of these two circles, then note that $( A,B,P,Q )$ and $( B,D,Y,Z )$ are concyclic we get $\measuredangle ZVQ = \measuredangle ZYU + \measuredangle UPQ = \measuredangle ZDB + \measuredangle BAQ = 2\measuredangle ABD \ ,$ hence $V$ lies on the 9-point circle of $\triangle ABD.$ Let $M$ be the midpoint of $BH.$ Since $M,$ $Q,$ $V,$ $Z$ are concyclic (the 9-point circle of $\triangle ABD$ ), so we conclude that $\measuredangle MVQ = \measuredangle BAQ = \measuredangle UPQ = \measuredangle UVQ \Longrightarrow M \ \text{lies on} \ UV \ .$

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bobthesmartypants

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bobthesmartypants

#3 h Aug 31, 2017, 7:01 PM • 3 Y

Y by anantmudgal09, Adventure10, Mango247

solution

EDIT: 4321st post woo

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#5 h Oct 27, 2017, 4:40 PM • 2 Y

Y by Adventure10, Mango247

I think that this solution is a bit different from others posted here. Not that hard a problem with the knowledge of circumrectangular hyperbolas, especially with geogebra by your side

mofumofu wrote:

Let $ABCD$ be a non-cyclic convex quadrilateral. The feet of perpendiculars from $A$ to $BC,BD,CD$ are $P,Q,R$ respectively, where $P,Q$ lie on segments $BC,BD$ and $R$ lies on $CD$ extended. The feet of perpendiculars from $D$ to $AC,BC,AB$ are $X,Y,Z$ respectively, where $X,Y$ lie on segments $AC,BC$ and $Z$ lies on $BA$ extended. Let the orthocenter of $\triangle ABD$ be $H$ . Prove that the common chord of circumcircles of $\triangle PQR$ and $\triangle XYZ$ bisects $BH$ .

First we clear up our notation a bit. Consider the equivalent problem:

China TST 2017 #2, morally correct notation wrote:

Let $ABCP$ be a non-cyclic quadrilateral and $\mathcal{H}$ denote the rectangular hyperbola circumscribing them. Let $H_B$ be the orthocenter of triangle $ABP$ . Let $\omega_1$ be the pedal circle of $P$ wrt $\triangle ABC$ and $\omega_2$ be the pedal circle of $A$ wrt $\triangle BCP$ . Prove that the radical axis (or common chord) of $\omega_1, \omega_2$ passes through the mid-point of $\overline{BH_B}$ .

For convenience, suppose $P$ lies in the interior of $\triangle ABC$ . The general case can be accounted for by directing angles accordingly. Now let $Z$ be the center of $\mathcal{H}$ and $P_A, P_B, P_C$ be the projections of $P$ on $\overline{BC}, \overline{CA}, \overline{AB}$ respectively. Let $\odot(P_AP_BP_C)$ meet the sides $\overline{BC}, \overline{CA}, \overline{AB}$ again at $Q_A,Q_B,Q_C$ respectively.

Let $M$ be the midpoint of $\overline{BH_B}$ and $P_B'=\overline{AP} \cap \overline{BH_B}$ . Then $H_B \in \mathcal{H} \implies Z \in \odot(P_BMP_B')$ .

Claim. $\overline{ZQ_A}$ is the common chord of $\omega_1, \omega_2$ (in fact these two points lie on both the circles).

(Proof) Observe that $Z$ is a common point due to the fundamental theorem and $Q_A \in \omega_1$ . So let $A'$ be the isogonal conjugate of $A$ wrt $\triangle BPC$ ; and $P'$ be the isogonal conjugate of $P$ wrt $\triangle ABC$ . Then $A', P'$ are symmetric in $\overline{BC}$ . Consequently, $Q_A$ lies on the pedal circle $\omega_2$ of $A'$ wrt $\triangle BPC$ , as desired. $\blacksquare$

Finally, we see $\angle P_BZQ_A=\angle P_BP_AB=90^{\circ}-\angle ABP$ and $\angle P_BZM=\angle P_BP_B'B=90^{\circ}-\angle ABP$ hence $M$ lies on $\overline{ZQ_A}$ , as desired. $\blacksquare$

P.S. 1400th post!

This post has been edited 1 time. Last edited by anantmudgal09, Oct 27, 2017, 4:42 PM

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#6 h Jul 18, 2019, 5:15 PM • 3 Y

Y by top1csp2020, JG666, Adventure10

ZR-BC=U then XZU=XAR=XYC so UXYZ cyclic, similarly QXU collinear so RQU=RZX=RAC=RPC so UPQR cyclic hence ZVQ=ZVU-UVQ=ZYU-UPQ=180-ZDB-QB=ZMQ so MVQ+QVU=180-MZQ+QAB=180 so q.e.d

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#7 h Jul 18, 2020, 11:24 PM

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Notice that $Z,Q,X,R$ all lies on the circle with diameter $AD$ . Denote this circle by $\omega$ .Let $N$ be the midpoint of $AD$ . Let $M$ be the midpoint of $BH$ . Suppose $ZX$ and $QR$ meet at $G$ , and $ZR$ and $QX$ meet at $I$ .
CLAIM. Both $I$ and $G$ lies on the common chord
Proof. Applying radical axis theorem to $(PQR)$ , $(XYZ)$ and $\omega$ we see that $G$ is the radical center of the three circles hence lying on the common chord. Now
$\angle QIR=\angle ZRQ-\angle XQR=\angle BAQ-(90^{\circ}-\angle ACD)=\angle QPC-\angle RPC=\angle QPR$ hence $I$ lies on $(PQR)$ . By symmetry it lies on $(XYZ)$ as well. This justfies our claim.

Now notice that $\angle MQB+\angle NQD=\angle MBQ+\angle NQD=90^{\circ}$ . Therefore $MZ$ and $MQ$ are both tangent to $\omega$ . Let $ZQ$ and $DX$ meet at $J$ . Then $J$ lies on $ZQ$ , the polar of $M$ w.r.t. $\omega$ . Hence by La Hire's theorem, $M$ lies on the polar of $J$ . w.r.t. $\omega$ , that is, $GI$ by Brokard's theorem.
This shows that the common chord of the two circles bisect $BH$ .

This post has been edited 2 times. Last edited by mathaddiction, Jul 18, 2020, 11:47 PM

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Plops

#8 h Apr 29, 2021, 7:29 AM

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Pascal's on $\odot (AZRDXQ)$ , $ZR \cap QX \cap PY=I'_2$ . Let $U=AD \cap BC$ , $V=ZQ \cap RX$ , and assume $AB<CD$ . By simple angle chasing, we see

$\angle RI_2B=\angle RAD-\angle ZDA+\angle DUC=\angle ADC-\angle DAB+\angle DAB+\angle CBA-\pi=\angle ADC+\angle CBA-\pi=\pi-\angle AQD+\pi-\angle PQA-\pi=\pi-\angle AQD-\angle PQA=\pi-\angle PQR$
so $\odot (I'_2PQR)$ is cyclic, and similarly, $\odot (I'_2YXZ)$ is cyclic. Let $M$ be the midpoint of $BH$ . Then, $MH=MZ=MQ=MB$ , and

$\angle MZB= \angle MBZ=\frac{\pi}{2}-\angle ZAD=\angle ZDA$
so $MZ, MQ$ are tangent to circle $\odot (AZRDXQ)$ , and $M$ lies on the polar of $V$ w.r.t. $\odot (AZRDXQ)$ , which, by Brokard's theorem, is $I_1I_2$ , the radical axis of $\odot (XYZ)$ and $\odot (PQR)$ .

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#10 h Jun 6, 2021, 6:28 AM

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Let $K$ be the orthocenter of $\triangle ACD$ , and let $M$ and $N$ be the midpoints of $BH$ and $CK$ .

Claim: $\overline{HK}, \overline{BC}$ and $\overline{ZR}$ are concurrent one of the intersections of $(PQR)$ and $(XYZ)$ , say $S$ .

Proof. Let $U=\overline{KR}\cap\overline{HZ}$ , and $V=\overline{CR}\cap\overline{BZ}$ . Then as $D$ is the orthocenter of $\triangle AUV$ , $UV\parallel CK\parallel BH$ . Hence $\triangle KRC$ and $\triangle HZB$ are perspective and the concurrency follows. Now it is left to show that this point lies on both circles. Since $ZAQR$ and $BAQP$ are cyclic, by Miquel's theorem it follows that $S$ lies on $(PQR)$ . Similarly, $S$ lies on $(XYZ)$ so we are done. $\square$

Now as $BH\parallel CK$ it follows by homothety that $\overline{MN}$ passes through $S$ . Consider a rectangular circumhyperbola $\mathcal{H}$ passing through $A,B,C,D,H$ . Obviously, this passes through $K$ as well. Let $T$ be the center of this hyperbola. Then by the fundamental theorem, this must be the other intersection point of $(PQR)$ and $(XYZ)$ (Configuration issues can be dealt without much difficulty). Since $BH$ and $CK$ are parallel chords of a conic, it follows that $\overline{MN}$ passes through $T$ , so it must be the radical axis of $(PQR)$ and $(XYZ)$ as desired. $\square$

This post has been edited 1 time. Last edited by KST2003, Jun 6, 2021, 6:29 AM

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#11 h Mar 20, 2022, 2:26 PM

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casey theorem for distance $H,B$ with radical axis of circles $\odot XYZ$ , $\odot PQR$ .

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#12 h Sep 27, 2022, 2:45 PM

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Denote by $\Omega$ the circle passing through $A,Z,R,D,X,Q$ .
By Pascal's theorem on $(ZRDQXA)$ , we have that $L:=ZR \cap QX$ lies on line $BC$ . Simple angle chasing indicates that $L$ is one of the intersections of circles $(PQR)$ and $(XYZ)$ .
Consider the radical axis of $\Omega$ , $(PQR)$ and $(XYZ)$ , we have $S:=XZ\cap RQ$ lies on the radical axis of $(PQR)$ and $(XYZ)$ , which we denote by $l$ . Thus now we conclude that $l$ is line $SL$ , so we only have to prove that the $M$ , the midpoint of segment $BH$ , lies on $SL$ .
To finish, it is well-known that $M$ is the pole of line $QZ$ with respect to $\Omega$ , therefore apply Pascal's theorem on $(ZZRQQX)$ , and then we are done!

This post has been edited 2 times. Last edited by ChanandlerBong, Dec 10, 2022, 3:05 AM
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#13 h Aug 9, 2023, 11:21 AM • 1 Y

Y by Rounak_iitr

$\text{a good problem for training the \textit{Pascal theorem}, but it is a bit easy for China TST}$

https://i.ibb.co/340Wpbq/2017-China-TST-R3-D1-P2.png

geogebra solution link

This post has been edited 5 times. Last edited by trinhquockhanh, Aug 9, 2023, 1:49 PM

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#14 h Apr 19, 2025, 3:39 PM

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Allow me to disagree with the above comment! This is a very beautiful problem, well placed as a P2. It's difficulty relies mostly on drawing a good diagram on paper (sth you may haven't noticed through geogebra) as well as keeping it clear from the many lines/circles that seem tempting.

This post has been edited 1 time. Last edited by sttsmet, Apr 19, 2025, 3:40 PM

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