Fixed Point on QR

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Fixed Point on QR

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Source: ELMO SL 2018 G1

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223 posts

#1 h Jun 28, 2018, 8:38 PM • 7 Y

Y by Amir Hossein, Durjoy1729, Mathuzb, cosmicgenius, Adventure10, Mango247, lian_the_noob12

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $P$ be a point on the nine-point circle of $ABC$ . Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$ , respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$ , respectively. Show that as $P$ varies along the nine-point circle of $ABC$ , the line $QR$ passes through a fixed point.

Proposed by Brandon Wang

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1979 posts

#2 h Jun 28, 2018, 9:16 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Funny how I used the same key words for G4

a1267ab wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $P$ be a point on the nine-point circle of $ABC$ . Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$ , respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$ , respectively. Show that as $P$ varies along the nine-point circle of $ABC$ , the line $QR$ passes through a fixed point.

Proposed by Brandon Wang

Invert at $H$ . Then $P' \in \odot(ABC)$ and $Q'=\overline{BP'} \cap \overline{CH}, R'=\overline{CP'} \cap \overline{BH}$ . We claim that $\odot(HQ'R')$ passes through a fixed point. Let $M$ be the midpoint of $\overline{BC}$ and $K=\overline{MH} \cap \odot(AH) \cap \odot(ABC)$ . Then we prove $K$ is the fixed point. Animate $P'$ on $\odot(ABC)$ ; then $Q' \mapsto P' \mapsto R'$ is projective; hence we just need to check three choices of $P'$ . Now $P'=A$ is obvious; we show $P' \in \overline{CH}$ as the other case follows analogously. Note that $\angle P'KH=\angle HCA'=\angle BAC=\angle P'HB$ proving the claim.

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515 posts

#3 h Jun 29, 2018, 12:07 AM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

We rewrite the problem in terms of orthic triangle as :
$\textbf{REFORMULATED PROBLEM:}$
In a $\Delta ABC$ let $I$ be the incenter and let $P$ be a point on $\odot (ABC)$ and $\odot (BPI)\cap CI=X,\odot (CPI)\cap BI=Y$ then $XY$ passes through a fixed point .

Now invert at $I$ and the problem becomes,
$\textbf{INVERTED PROBLEM:}$
In a $\Delta ABC$ let $H$ be the orthocenter and let $P$ be a point on $\odot (ABC)$ and $BH\cap CP=X$ and $CH\cap BP=Y$ then $\odot (HXY)$ passes through a fixed point.

$\textbf{PROOF:}$
Let circle with diameter $AH$ meet $\odot (ABC)$ at $M$ ,we will show that $M$ is the fixed point now let $E$ be the foot of $B-$ altitude ,Now simple angle chasing tells us that $P\in \odot (HXY)$ and we have $\angle PMH=\angle PMA-\angle AMH=90-\angle ACB-\angle PCB$ on the other hand we have $\angle PXH=\angle PXE=90-\angle ACB-\angle PCB=\angle PMH$ , $\implies M$ lies on $\odot (HXY)$ so we are done $\blacksquare$

This post has been edited 3 times. Last edited by enhanced, Jun 29, 2018, 12:11 AM

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#4 h Jun 29, 2018, 12:56 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the midpoint of $\overline{BC}$ and let $Q'=PM\cap CH$ and $R'=PM\cap BH$ . Also let $D$ be the foot of $A$ -altitude on $BC$ . Since the quadrilaterals $BFHD$ and $CEHD$ is cyclic, $\measuredangle{FHB}=\measuredangle{FDB}=\measuredangle{FPR'}$ $\measuredangle{Q'HE}=\measuredangle{CDE}=\measuredangle{Q'PE}.$
Therefore we have $Q'=Q$ and $R'=R$ , which means that $QR$ passes through $M$ which is fixed.

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2487 posts

#5 h Jun 30, 2018, 4:50 PM • 1 Y

Y by Adventure10

Invert at $H$ with power $\sqrt{-HB \cdot HE}$ . Then we need to show the following

Inverted ELMO SL 2018 G1 wrote:

Let $ABC$ be a triangle with orthocenter $H$ . A point $P$ is chosen on $(ABC)$ and $PC \cap HB = Y, PB \cap HC = Z$ . Prove that the circumcircle of $HYZ$ passes through a fixed point.

We claim that the fixed point is $(AH)\cap (ABC) = X$ .
To see this, first note that $P$ is on $(HYZ)$ , as $\angle BHC = 180^\circ - \angle A$ . Now see that $\angle HXP = 90^\circ - \angle PXA = \angle PQH$ , so we are done.

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PROF65

#7 h Jun 30, 2018, 10:13 PM • 2 Y

Y by Adventure10, Mango247

remark that $P$ is the center of the similarity that send $QE\mapsto FR$ then $\angle QPE =\angle FPR =\angle A$ so $P$ is on $QR$ moreover recall that if $M$ is the midpoint of $BC$ then $\angle EFM=\angle A$ so $\angle FPQ= \angle FPM$ hence $QR$ pass through $M$ .
RH HAS

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695 posts

#8 h Jul 1, 2018, 2:58 AM • 4 Y

Y by AlastorMoody, zuss77, itslumi, Adventure10

Let $M$ be the midpoint of $BC.$ Let $FH$ meet the nine-point circle for a second time at $X.$

As $MX$ is the $C$ -midline of $\triangle HBC$ , we see that $MX \parallel RH.$ By Reim's theorem for the nine-point circle and $\odot(FHP)$ cut by $MR$ and $XH$ , it follows that $R \in PM.$ Analogously, $Q \in PM$ , so $M$ is the fixed point.

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1179 posts

#9 h Jul 9, 2018, 10:15 PM • 3 Y

Y by Amir Hossein, Adventure10, Mango247

Another angle chasing solution:

$[asy] size(7cm); pair A,B,C; A = (1.8,6.2); B = (0,0); C = (7,0); pair H = orthocenter(A,B,C); pair D,E,F; D = extension(A,H,B,C); E = extension(B,H,A,C); F = extension(C,H,A,B); draw(A--B--C--A, blue); draw(A--D, blue); draw(B--E, blue); draw(C--F, blue); pair M = (B+C)/2; draw(circumcircle(D,E,F),red); pair L = (2.5,-18); pair M = (2.5,31); path q = L--M; path f = circumcircle(D,E,F); pair [] P = intersectionpoints(q,f); pair P = P[0]; pair H = extension(A,D,B,E); draw(circumcircle(F,H,P),red); draw(circumcircle(E,H,P),red); path q = circumcircle(E,H,P); path w = H -- C; dot(P); label("$P$",P,S); pair [] Q = intersectionpoints(q,w); pair Q = Q[1]; pair R = extension(Q,P,B,H); draw(R--P--Q,blue+1.2bp); pair M = 0.5 * B + 0.5 * C; dot(M); label("$M$",M,S); draw(B--R,blue+dashed); label("$R$",R,W); label("$Q$",Q,S); label("$H$", H, NNW); label("$A$",A,N); label("$F$",F,NW); label("$B$",B,W); label("$E$",E,N); pair X = E; label("$C$", C, S); [/asy]$

Define $D$ to be the foot of the altitude from $A$ . We claim $R,P,Q$ are collinear. Indeed, we have
$\begin{align*} \measuredangle RPQ &= \measuredangle RPF + \measuredangle FPE + \measuredangle EPQ\\ &= \measuredangle RHF + \measuredangle FDE + \measuredangle EHQ\\ &= \measuredangle BHF + \measuredangle FDE + \measuredangle EHC\\ &= \measuredangle BDF + \measuredangle FDE + \measuredangle EDC\\ &= 180^\circ \end{align*}$ Let $M$ be the midpoint of $BC$ . We claim $M$ is the fixed point. Since $M$ lies on the nine-point circle and $R,P,Q$ are collinear,
$\begin{align*} \measuredangle RPM &= \measuredangle RPF + \measuredangle FPM\\ &= \measuredangle BHF + \measuredangle FDM\\ &= \measuredangle BHF + \measuredangle FDH + 90^\circ\\ &= \measuredangle BDF + \measuredangle FDH + 90^\circ\\ &= 180^\circ, \end{align*}$ as desired.

This post has been edited 1 time. Last edited by winnertakeover, Jul 10, 2018, 8:51 PM
Reason: small typo

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#10 h Dec 23, 2018, 7:16 AM • 3 Y

Y by AlastorMoody, Aryan-23, Adventure10

It's wonderful how simple tools work so well.

a1267ab wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $P$ be a point on the nine-point circle of $ABC$ . Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$ , respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$ , respectively. Show that as $P$ varies along the nine-point circle of $ABC$ , the line $QR$ passes through a fixed point.

Define $\Omega$ to be the nine-point circle of $\triangle ABC.$ We claim that $M,$ the midpoint of side $BC$ always lies on $QR.$

Claim: $P$ lies on $QR.$
Proof: Note that $P$ is the center of spiral similarity taking $FQ$ to $RE$ and so $\angle EPR=\angle QPF.$ Also, $P$ is the center of spiral similarity taking $FR$ to $QE$ and so $\angle EPQ=\angle RPF.$ Thus, $\angle EPR=\angle EPQ,$ as desired. $\square$

As shown above, $\angle EPR=\angle QPF$ implies that $PR \cap \Omega$ is the midpoint of arc $EF$ of $\Omega,$ which is precisely the point $M.$ $\blacksquare$

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#11 h May 5, 2019, 4:43 AM • 4 Y

Y by Pathological, Pluto1708, Adventure10, Mango247

What a nice problem!

We use the old lemma: Let $\ell_1, \ell_2$ be two lines, and let $X\to Y\colon \ell_1\to \ell_2$ be projective so that $\ell_1\cap \ell_2$ is mapped to itself. Then, for all $X\to Y$ , $XY$ passes through a fixed point $P$ .

Proof is easy, say $T = \ell_1\cap \ell_2$ , say $X_1\to Y_1$ , $X_2\to Y_2$ and $X_1Y_1\cap X_2Y_2 = P$ , then $P$ is the desired fixed point.

Clearly $P \to Q$ , $P\to R$ is projective (e.g. after inversion in $H$ ) Also, $\measuredangle EPF = 2\measuredangle CAB$ . So, $Q = H$ if and only if $(EHP)$ tangent to $CH$ if and only if $\measuredangle EPH = \measuredangle EHF = \measuredangle CAB$ , and $R = H$ if and only if $(FHP)$ tangent to $BH$ if and only if $\measuredangle HPF = \measuredangle EHF = \measuredangle CAB$ , so $Q = H$ if and only if $R = H$ so the projective map $Q\to R$ takes $H\to H$ indeed.

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178 posts

#12 h Jun 29, 2019, 3:57 PM • 2 Y

Y by Adventure10, Mango247

I have two questions:
In Wizard_32 s solution, from the part:
Claim: $P$ lies on $QR.$
Proof: Note that $P$ is the center of spiral similarity taking $FQ$ to $RE$ and so $\angle EPR=\angle QPF.$ Also, $P$ is the center of spiral similarity taking $FR$ to $QE$ and so $\angle EPQ=\angle RPF.$ Thus, $\angle EPR=\angle EPQ,$ as desired. $\square$
How do we conclude that $P,Q,R$ are collinear?
Also, can someone explain how Dukejukem used Reim s Theorem to solve the problem? Isnt Reim s Theroem the fact that in a cyclic quadrilateral $ABCD$ , where $P,Q$ are on $AC,BD$ , $ABPQ$ is cyclic if and only if $PQ$ is parallel to $CD$ ?

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#13 h Jan 26, 2020, 3:34 PM • 5 Y

Y by Crystal1011, strawberry_circle, gamerrk1004, Adventure10, Mango247

Nothing different from others.

ELMO SL 2018 G1 wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $P$ be a point on the nine-point circle of $ABC$ . Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$ , respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$ , respectively. Show that as $P$ varies along the nine-point circle of $ABC$ , the line $QR$ passes through a fixed point.

Proposed by Brandon Wang

Claim:- The fixed point is the midpoint of $BC$ .

Now Invert around $H$ with radius $-\sqrt{HA\cdot HD}$ . Now the problem becomes equivalent to this problem.

Inverted Problem wrote:

$D'E'F'$ is a triangle with altitudes $A'D,B'E',C'F'$ and orthocenter $H$ . $P'$ be an arbitary point on $\odot(D'E'F')$ and let $E'P'\cap F'H=Q'$ and $F'P'\cap E'H=R'$ . Then Prove that $\odot(R'Q'H)$ passes through the $D'-\text{Queue Point}(Q_A)$ of $\triangle D'E'F'$ .

Here's the diagram of the Inverted Image.

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.553333333333331, xmax = 7.633333333333336, ymin = -2.29, ymax = 5.616666666666658; /* image dimensions */ pen qqttcc = rgb(0,0.2,0.8); pen ffxfqq = rgb(1,0.4980392156862745,0); pen ttffqq = rgb(0.2,1,0); /* draw figures */ draw((-2.22,3.15)--(-3.74,-0.59), linewidth(1.2) + qqttcc); draw((1.36,-0.57)--(-2.22,3.15), linewidth(1.2) + qqttcc); draw((-3.74,-0.59)--(1.36,-0.57), linewidth(1.2) + qqttcc); draw((-2.22,3.15)--(-2.1998668225017686,-0.5839602620490265), linewidth(0.4)); draw((-3.74,-0.59)--(-1.07267223914642,1.9578046730795204), linewidth(0.4)); draw((1.36,-0.57)--(-3.010086636397104,1.2059710393913365), linewidth(0.4)); draw(circle((-1.1944483185952233,0.5543212417820158), 2.7909325443471795), linewidth(0.4) + red); draw((-3.74,-0.59)--(1.3133333333333357,4.243333333333324), linewidth(0.4)); draw((1.3133333333333357,4.243333333333324)--(1.36,-0.57), linewidth(0.4)); draw((1.337720621853682,1.7279587173774786)--(-3.74,-0.59), linewidth(0.4)); draw((-1.19,-0.58)--(-3.266609880522185,2.4239337493269977), linewidth(0.4)); draw(circle((-0.8295605302108242,2.9647518251412333), 2.4953486296519722), linewidth(0.8) + linetype("4 4") + ffxfqq); draw(circle((-2.218742289111523,2.038937562880504), 1.1489382512713515), linewidth(0.4) + ttffqq); /* dots and labels */ dot((-2.22,3.15),dotstyle); label("$D'$", (-2.3,3.296666666666657), NE * labelscalefactor); dot((-3.74,-0.59),dotstyle); label("$E'$", (-3.98,-0.8366666666666775), NE * labelscalefactor); dot((1.36,-0.57),dotstyle); label("$F'$", (1.4733333333333356,-0.77), NE * labelscalefactor); dot((-2.22,0.89),dotstyle); label("$H$", (-2.18,1.07), NE * labelscalefactor); dot((-2.1998668225017686,-0.5839602620490265),dotstyle); label("$A'$", (-2.26,-0.8633333333333443), NE * labelscalefactor); dot((-1.07267223914642,1.9578046730795204),dotstyle); label("$B'$", (-1.1266666666666645,2.1633333333333233), NE * labelscalefactor); dot((-3.010086636397104,1.2059710393913365),dotstyle); label("$C'$", (-3.3,1.123333333333323), NE * labelscalefactor); dot((1.3133333333333357,4.243333333333324),dotstyle); label("$R'$", (1.3666666666666691,4.376666666666657), NE * labelscalefactor); dot((1.337720621853682,1.7279587173774786),dotstyle); label("$P'$", (1.5133333333333356,1.6166666666666565), NE * labelscalefactor); dot((-1.3148867565367968,0.517054287138188),dotstyle); label("$Q'$", (-1.393333333333331,0.23), NE * labelscalefactor); dot((-1.19,-0.58),dotstyle); label("$M$", (-1.2333333333333312,-0.85), NE * labelscalefactor); dot((-3.266609880522185,2.4239337493269977),dotstyle); label("$Q_A$", (-3.6866666666666648,2.456666666666657), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Note that $\begin{cases} \angle E'P'F'=\angle E'D'F'=\angle R'HF'\implies P'\in\odot(Q'HR')\\ \angle Q_AHC'=\angle Q_AD'E'=\angle Q_AF'E'=\angle Q_AP'Q'\implies Q_A\in\odot(H'Q'P')\end{cases}\implies Q_A\in\odot(R'Q'H)$ So Inverting back we get that $QR$ passes through the midpoint of $BC$ . $\blacksquare$

To know more about Queue Points (Name given by the User Math-Pi-Rate). Visit these two Blogposts of his.

This post has been edited 2 times. Last edited by amar_04, Jan 26, 2020, 3:41 PM

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#14 h Jul 3, 2020, 8:52 PM • 2 Y

Y by Kanep, nixon0630

Trivial.

Let $AD$ be an altitude of $\triangle ABC$ , and $M$ be the midpoint of $BC$ . Redefine $Q$ as the intersection of lines $PM$ and $CH$ , and redefine $R$ similarly. It suffices to show that $PQEH$ and $PRFH$ are cyclic. But this follows by easy angle chasing since both $\angle EHQ$ and $\angle EPQ$ are easily computable, and similarly for $R$ .

This post has been edited 1 time. Last edited by mathlogician, Jul 3, 2020, 8:54 PM

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#15 h Sep 30, 2020, 12:28 PM

Y by

Posted for storage.Here is a solution using $-\sqrt{HA\cdot HD}$ .We know that during such inverzion $N_9 <---> (ABC)$ and $A <-->D$ ,etc.

1)Guess the fixed point.
From a good diagram we see that $M-Q-R$ Collinear,so we guess that $M$ (the midpoint of BC) is the fixed point.

Now prove that $M-Q-R$ collinear.Denote $\phi$ (X) the inverze of X during $-\sqrt{HA\cdot HD}$ inverzion.
Since p is on $N_9$ thus $\phi$ (P) lies on $(ABC)$ and P-H- $\phi$ (P) collinear.

Also we know that $\phi$ (Q) will be of ray $HF$ And will satisfy that $\phi$ (P)-B- $\phi$ (Q) collinear.
NOTE;The midpoint of $BC$ is send to the miquel point of $(BFEC)$ , let denote that point as $M_q$

similarly define $\phi$ (R),Now we wanted to show that $Q-R-M$ collinear,but in our inverted problem is enough to show that $($ \phi $(Q),$ M_q $,$ H $,$ \phi $(R))$ Are concyclic.

For easier interpretation let $\phi$ (Q)=X and $\phi$ (R)=Y, $\phi$ (P)=Z

We see that from $\angle YBZ=\angle XBH$ Also $\angle BZY=\angle XHB$ ,Which gieves us that
$\angle BYZ=\angle HXB$ ,which gives that $(XYZH)$ - cyclic.
Now we have $\angle XZM_q=\angle BZM_q$ ,which means that is equal to $\angle M_qAB=\angle M_qHF=\angle M_qHX$ Which gives us that $(XM_qHY)$ cycli,and we are done.

This post has been edited 3 times. Last edited by itslumi, Sep 30, 2020, 12:30 PM

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#16 h Dec 19, 2020, 7:16 PM

Y by

Very easy. Let $M$ be the midpoint of $\overline{BC}$ . Notice that $CDHE$ is cyclic, so $\angle EDM=\angle EDC=\angle EHC$ . From the cyclic quad $EPHQ$ we have $\angle EPQ=\angle EHQ=\angle EHC$ . From the cyclic quad $EPDM$ we have $\angle EPM=\angle EDM$ . Thus $\angle EPQ=\angle EPM \implies P,Q,M$ are collinear. Similarly $P,R,M$ are collinear. Thus $Q,R,M$ are collinear, as desired. $\square$

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656 posts

#17 h Jul 29, 2021, 10:34 AM

Y by

Below is a different solution (which is not very elegant, but very simple and easily motivated):

If we consider the inversion at $H$ which swaps nine-point circle of $\triangle ABC$ with $\odot(ABC)$ , we get the following equivalent problem:

Inverted ELMO SL 2018 G1 wrote:

Let $H$ be the orthocenter of a $\triangle ABC$ and let $P$ be a variable point on $\odot(ABC)$ . Let $Q = \overline{BP} \cap \overline{CH} ~,~ R = \overline{CP} \cap \overline{BH}$ . Show that as $P$ varies on $\odot(ABC)$ , the circle $\odot(HQR)$ passes through a fixed point.

Let $T = \odot(AH) \cap \odot(ABC) \ne A$ be the $A \text{-Queue}$ point wrt $\triangle ABC$ . We will prove that $\odot(HQR)$ always passes through $T$ .
$[asy] pair A=dir(110),B=dir(-150),C=dir(-30),H=A+B+C,P=dir(10),Q=extension(B,P,C,H),R=extension(C,P,B,H),HB=2*foot(H,A,C)-H,HC=2*foot(H,A,B)-H,T=2*foot(A,1/2*(A+H),(0,0))-A; fill(unitcircle,cyan); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(-90)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$H_B$",HB,dir(HB)); dot("$H_C$",HC,dir(HC)); draw(unitcircle,red); draw(A--B--C--A^^B--R^^C--HC^^B--P^^C--R^^T--H,magenta); dot("$T$",T,dir(T)); draw(circle(1/2*(A+H),abs(A-H)/2),red); draw(HC--B^^C--HB,magenta); draw(arc(circumcenter(T,H,R),T,R),dashed); [/asy]$
Let $H_B,H_C$ be the reflections of $H$ in sides $\overline{CA},\overline{AB}$ , respectively. Then $H_B,H_C \in \odot(ABC)$ .
It is well known that $T$ is the center of spiral similarity which maps $\overline{H_CH} \mapsto \overline{HH_B}$ . So it suffices to show that $Q \mapsto R$ under this spiral similarity, or equivalently, $\frac{QH}{QH_C} = \frac{RH_B}{RH}$ .

Note that
$\frac{QH}{QH_C} = \frac{QH \cdot \sin \angle BQH}{QH_C \cdot \sin \angle BQH_C} = \frac{BH \cdot \sin \angle QBH}{BH_C \cdot \sin \angle QBH_C} = \frac{\sin \angle PBH_B}{\sin \angle PBH_C}$ where we used $BH = BH_C$ . Similarly, we obtain that $\frac{RH_B}{RH} = \frac{\sin \angle PCH_B}{\sin \angle PCH_C}$ . Since $\angle PBH_B = \angle PCH_B$ and $\angle PBH_C = \angle PCH_B$ , so we are done! $\blacksquare$

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#18 h Oct 7, 2021, 12:34 PM

Y by

Wow this turned out to be so simple at the end ... ~~assuming i did not fakesolve~~
We claim that $QR$ must always pass through the midpoint ' $M$ ' of $BC$ . Let $D,E,F$ be the points on $BC, CA, AB$ respectively such that $AD, BE, CF$ are the altitudes in $\triangle{ABC}$ . Let $I,G$ be the midpoint of $CH, BH$ respectively. Let $PM \cap BE = R'$ . By midpoint theorem we have that $BE \parallel MI$ . Therefore, $\angle{MIF}=\angle{GHF}$ . Further, we also have $\angle{FPR'}=\angle{FPM}=\angle{MIF}$ . Therefore we have that $FPR'H$ is cyclic, so we get that $R' \equiv{R}$ . In a very similar way we get $Q' \equiv{Q}$ , so we are done $\blacksquare$

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593 posts

#20 h Jan 12, 2022, 6:10 PM

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Solution

This post has been edited 4 times. Last edited by HoRI_DA_GRe8, Jan 15, 2022, 6:43 PM

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#21 h Dec 30, 2022, 3:54 PM

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Hm, why is everything so complicated?
Let $D$ be the foot from $A$ and $E = \tfrac{B+C}{2}$ , note that $\measuredangle EPM = \measuredangle EDM = \measuredangle EDC = \measuredangle EHC = \measuredangle EHQ = \measuredangle EPQ$ so $P-M-Q$ and similarly $P-M-R$ so gg.

This post has been edited 1 time. Last edited by MrOreoJuice, Dec 30, 2022, 3:55 PM

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#22 h Sep 6, 2023, 12:38 AM • 1 Y

Y by centslordm

The fixed point is the midpoint $M$ of $\overline{BC}$ .

Let $\overline{PM} \cap \overline{CH}=Q'$ . It is well-known that $\overline{ME}$ and $\overline{MF}$ are tangent to $(AEFH)$ . Then,
$\measuredangle EHQ'=\measuredangle EHF=\measuredangle EFD=\measuredangle EPD=\measuredangle EPQ',$ hence $EPHQ'$ is cyclic, hence $Q'=Q$ so $Q$ lies on $\overline{PM}$ . Likewise, $R$ lies on $\overline{PM}$ , so we're done. $\blacksquare$

Remark: You can just straight angle chase without phantom points but I didn't think about it this way. I think philosophically this solution focuses on $\overline{PM}$ anyways

This post has been edited 3 times. Last edited by IAmTheHazard, Sep 6, 2023, 12:40 AM

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#23 h Sep 6, 2023, 1:01 AM

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Let $M$ be the midpoint of $\overline{BC}$ and $D$ be the foot of the $A$ -altitude. I claim that $M$ is the desired point. Since
$\measuredangle EPQ = \measuredangle EHQ = \measuredangle EHC = \measuredangle EDC = \measuredangle EDM = \measuredangle EPM,$ we find that $P$ , $Q$ , and $M$ are collinear. Similarly $P$ , $M$ , and $R$ are collinear, which implies the conclusion.

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#24 h Sep 6, 2023, 1:30 AM

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Let $M$ the midpoint of $BC$ then $\angle FPR=\angle EHC=\angle BAC=\angle FEM$ so $P,R,M$ are colinear and in a similar way u can prove $P,Q,M$ colinear so u get $P,Q,R,M$ colinear hence $QR$ goes through $M$ , thus we are done

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#25 h Apr 2, 2024, 3:34 PM

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a1267ab wrote:

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $P$ be a point on the nine-point circle of $ABC$ . Lines $BH, CH$ meet the opposite sides $AC, AB$ at $E, F$ , respectively. Suppose that the circumcircles $(EHP), (FHP)$ intersect lines $CH, BH$ a second time at $Q,R$ , respectively. Show that as $P$ varies along the nine-point circle of $ABC$ , the line $QR$ passes through a fixed point.

Proposed by Brandon Wang

Let $Q=CH\cap MP$ then $<EPQ=<EPM=<A=<EHQ$ done

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