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.

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 $\blacksquare$

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.

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.

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

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.

Another angle chasing solution: [asy][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][/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.

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 $M.$ $\blacksquare$

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.

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 $CD$?

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][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][/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. Part 1 Part 2

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 $R$.

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.

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. $\square$

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][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][/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! $\blacksquare$

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 $\blacksquare$

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.

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

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.

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

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