Let $ABC$ be an acute triangle, and let $X$ be a variable interior point on the minor arc $BC$ of its circumcircle. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to lines $CA$ and $CB$, respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. (Specifically: prove that there is a point $F$, determined by triangle $ABC$, such that no matter where $X$ is on arc $BC$, line $\ell$ passes through $F$.) Robert Simson et al.
Problem
Source: USA December TST for IMO 2014, Problem 1
Tags: geometry, geometric transformation, parallelogram, reflection, Asymptote, xooks
24.12.2013 09:57
I have a very ugly solution .Let $\angle BCX= \alpha$ and let $l$ intersect $BR$ at $M .$ $RM =XP =CXsin(C+\alpha)=\frac{a}{sinA}.sin(A-\alpha).sin(C+\alpha)$ and from law of sines to $\Delta BQR$, $BR=\frac{BQ}{sin(\alpha)}.cos(A-\alpha)=\frac{BX.cos(C+\alpha)}{sin(\alpha)}.cos(A-\alpha)= \frac{\frac{a}{sinA}sin(\alpha)cos(C+\alpha)}{sin(\alpha)}.cos(A-\alpha)=\frac{a.cos(C+\alpha).cos(A-\alpha)}{sinA}$ $BM=RM-RB=\frac{a}{sinA}(.sin(A-\alpha).sin(C+\alpha)-cos(C+\alpha).cos(A-\alpha))=\frac{a}{sinA}(-cos(C+A))$ which is independent from $\alpha$ .Hence Done!!
24.12.2013 10:28
Dear Mathlinkers, the fix point seems to be the orthocenter of ABC. Sincerely Jean-Louis
24.12.2013 10:45
Dear Mathlinkers, the midpoint of XH is on PQ (Simson's line of X wrt ABC) and HR and XP are parallels. I think we are done... Sincerely Jean-Louis
24.12.2013 13:01
Dilate about $X$ with factor $2$, then $P$ gets mapped to the line for which $X$ is the anti-steiner point, which is parallel to simson line (I forgot what that line was called, but u can see a proof on my blog). Let $P$ go to $P' \implies P'PRH$ is a parallelogram. So, $HR = PP' = PX$ and we are done ($HPXR$ is a parallelogram so $HP \parallel RX$). Of course, what Jayme did was practically the same. Some inspiration despite my terrible drawing is looking at the concurrence point of the lines. It appeared to lie on $BH$, but $X$ being the reflection of $H$ over the midpoint of $BC$ tells the point must then be $H$ and then the rest follows by just trying to prove it was $H$.
28.12.2013 23:40
I'll prove the lemma referred to in previous posts.
Lemma: The Simson line of $X$ bisects $XH$.
Now, all you have to do is construct a parallelogram. Indeed, let $M$ be the midpoint of $XH$, which lies on $PR$ by the lemma. But since $PX\parallel RH$, $XPHR$ is a parallelogram, so $PH\parallel XR$ as desired.
29.12.2013 02:04
And as luck would have it, I was writing a section about Simson lines when I saw this problem... The lemma is also a standard example of complex numbers; just check the reflections of $X$ over $P$ and $Q$ are colliear with $H$. Since this lemma essentially trivializes the problem, that means complex numbers can kill the problem pretty rapidly too, without even mentioning the Simson line. I also know of people who successfuly coordinate bashed it. Love the authorship information
26.04.2014 12:30
The theorem that the Simson line of a point wrt a triangle bisects the segment joining that point and the orthocenter of the triangle is undoubtedly the main key. Let $BT \perp CA$ with $T$ on $CA$. Let $\el$ intersect $BT$ at $H$. Now clearly $PXRH$ is a parallelogram. So $PR$ bisects $XH$. Now let the orthocenter of $\triangle ABC$ be $H'\neq H$. Then $PR$ also bisects $XH'$. Let the midpoints of $XH$ and $XH'$ be $M,M'$ respectively. Then $M,M'$ both lie on $PQ$ ,and $H,H'$ both lie on $BT$ But $MM'\parallel HH'$. So, $PQ\parallel BT$ which is impossible. So, $H$ must be the orthocenter of $\triangle ABC$. Thus $H$ is our desired fixed point.
15.10.2014 07:25
We proceed with complex numbers. Let $ H $ be the orthocenter of $ \triangle{ABC} $ and let $ A, B, C, X, P, Q, R, H $ have complex coordinates $ a, b, c, x, p, q, r, h $ respectively. WLOG assume that the circumcircle of $ \triangle{ABC} $ is the unit circle. It is clear that $ p = \frac{1}{2}\left(a + x + c - \frac{ac}{x}\right) $ and that $ q = \frac{1}{2}\left(b + x + c - \frac{bc}{x}\right) $. Therefore line $ PQ $ has equation $ \frac{z - p}{\overline{z} - \overline{p}} = \frac{p - q}{\overline{p} - \overline{q}} = \frac{abc}{x} \Longrightarrow z = \left(\frac{abc}{x}\right)\overline{z} - \frac{abx + bcx + cax + abc}{2x^2} + \frac{a + b + c + x}{2} $. Since $ BR \perp AC $ we find that line $ BR $ has equation $ \frac{z - b}{\overline{z} - \frac{1}{b}} = -\frac{a - c}{\frac{1}{a} - \frac{1}{c}} = ac \Longrightarrow z = ac\overline{z} + b - \frac{ac}{b} $. Computing the coordinates of $ R $, the intersection of these two lines, we find that $ r = \frac{1}{2}\left(a + x + c + \frac{ac}{x}\right) + b $. But since $ h = a + b + c $ this implies that $ x + h = p + r $ so quadrilateral $ XPHR $ is a parallelogram and so $ PH \parallel XR $, hence, $ H $ is the desired fixed point.
03.04.2015 01:51
Hey guys, what's the motivation for the Lemma pi37 used? I got to the conjecture tha H is the common point but died from there ._.
07.08.2015 21:12
DrMath wrote: Hey guys, what's the motivation for the Lemma pi37 used? I got to the conjecture tha H is the common point but died from there ._. The motivation is that from the diagram, it's clear that $XPHR$ is a parallelogram, so going after the midpoint is a standard method of proving the existence of the parallelogram.. In any case, it's a well-known lemma
07.08.2015 21:17
Ooops LOL I forgot I had posted in this... and I've actually used this lemma on many a geo problem. Thanks anyways
29.10.2015 06:22
pi37 wrote: Lemma: The Simson line of $X$ bisects $XH$. Proof Let $X_A,H_A$ be the reflections of $X,H$ over $BC$, and let $X_B$ be the reflection of $X$ over $AC$. It suffices to show that $X_A,X_B$, and $H$ are collinear. Note that $C$ is the circumcenter of $XX_AX_B$, so $\angle XX_AX_B=\frac{1}{2}\angle XCX_B=\angle ACX$. Meanwhile, since $HH_AXX_A$ is an isosceles trapezoid, $\angle HX_AX=\angle AH_AX=180-\angle XX_AX_B$, so the three points are collinear. Why does it suffice to show that $X_A,X_B$, and $H$ are collinear? Thanks!
29.10.2015 06:49
The Simson line of $X$ is the line through the projections of $X$ onto $BC,AC$. These are the midpoints of $XX_A,XX_B$, so by homothety, if $X_AX_B$ passes through $H$, then the line through the midpoints of $XX_A$ and $XX_B$ passes through the midpoint of $XH$.
17.09.2016 02:54
I realize that posts of mine like these are not contributing, but this is my way of marking whether I did the problem or not so sorry for this post and everything else in advance.
19.09.2016 10:38
Since the simson line bisects HX it follows that the point is the orthocenter of ABC.
06.04.2017 02:06
Interesting that they put this on the 2014 TST when the 2013 TSTST had a problem with a similar idea...
18.04.2017 04:25
What does the problem mean by "interior point"?
18.04.2017 22:26
Recognizing that $PQ$ is a Simpson Line, we recall that (thank you EGMO!) quadrilateral $RXPH$ is a parallelogram, where $H$ is the orthocenter of triangle $ABC$. It follows that since line $\ell$ is parallel to $RX$, the intersection of $\ell$ and the altitude from $B$ to $AC$ is the orthocenter. Thus $\ell$ always passes through the orthocenter. $\square$
26.12.2017 19:26
Simple
29.07.2022 19:22
We claim that $\ell$ passes through $H$, the orthocenter of $\triangle ABC$. Let $M$ be the midpoint of $\overline{HX}$ and identify line $PQ$ as the Simson Line of $X$. It is well-known that $PQ$ passes through $M$. Now reflect $P$ over $M$ to get $R’$. By construction, $HPXR’$ is a parallelogram with $HP\parallel XR’$ in particular. It suffices to show that $BR’\perp AC$ to establish $R’=R$ which finishes the problem. To do this, toss on the complex plane with $(ABC)$ as the unit circle so that $A=a,B=b,C=c,X=x,H=a+b+c.$ Clearly, $M=\frac{a+b+c+x}{2}$ and $P=\frac12(x+a+c-\tfrac{ac}{x})$ so $R’=2M-P= \frac{a+2b+c+x+\frac{ac}{x}}{2}$. To show that $BR’\perp AC$, we need to check $\frac{r’-h}{c-a}+\overline{\left(\frac{r’-h}{c-a}\right)}=0$. Note that $\frac{r’-h}{c-a}=\frac{\frac{a+2b+c+x+\frac{ac}{x}}{2}-(a+b+c)}{c-a}=\frac{-a-c+x+\frac{ac}{x}}{2c-2a}$ and its conjugate is $\overline{\left(\frac{r’-h}{c-a}\right)}=\overline{\left(\frac{-a-c+x+\frac{ac}{x}}{2c-2a}\right)}=\frac{-\frac1a-\frac1c+\frac1x+\frac{x}{ac}}{\frac2c-\frac2a}=\frac{\frac{-cx-ax+ac+x^2}{acx}}{\frac{2x(a-c)}{acx}}=-\frac{-a-c+x+\frac{ac}{x}}{2c-2a}$ so $BR’\perp AC$ and we are done.
20.01.2023 04:04
We claim that the fixed point is the orthocenter. Let the altitude intersect line $\ell$ at $S$. Note that $RXPS$ is a parallelogram since $BS$ and $XP$ are both perpendicular to $AC$ and we are given that the other side. It is well known that $PQ$, the Simson line, passes through the midpoint of $XH$, where $H$ is the orthocenter. Therefore, $RXPH$ is also a parallelogram, since we already know that $RH$ is parallel to $XC$. Note that the orthocenter is the only point that completes the paralleogram with three vertices $R,X,P$ in the correct order, so $S$ must be the orthocenter.
20.01.2023 04:17
srijonrick wrote: [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 = -10.933, xmax = 10.825, ymin = -8.036, ymax = 5.692; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen qqzzff = rgb(0,0.6,1); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen ffqqff = rgb(1,0,1); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-4.34,4.07)--(-7.38,-2.53)--(3.756,-2.456)--cycle, linewidth(0.8) + ttffqq); draw((-8.915714298550334,-4.4351705426085655)--(-4.321356994370616,1.2644795852858564)--(2.0590632643892013,-1.0881381995311175)--(-2.5352940397905166,-6.78778832742554)--cycle, linewidth(1.2) + ffqqff); /* draw figures */ draw((-4.34,4.07)--(-7.38,-2.53), linewidth(0.8) + ttffqq); draw((-7.38,-2.53)--(3.756,-2.456), linewidth(0.8) + ttffqq); draw((3.756,-2.456)--(-4.34,4.07), linewidth(0.8) + ttffqq); draw(circle((-1.821321502814693,-1.090239792642929), 5.74210902802787), linewidth(0.8) + wvvxds); draw((xmin, 1.2405761569108182*xmin-3.642562990903317)--(xmax, 1.2405761569108182*xmax-3.642562990903317), linewidth(0.8) + linetype("0 3 4 3") + sexdts); /* line */ draw((xmin, -150.4864864864868*xmin-388.3152805856316)--(xmax, -150.4864864864868*xmax-388.3152805856316), linewidth(0.8) + linetype("0 3 4 3") + sexdts); /* line */ draw((xmin, 1.2405761569108182*xmin + 6.62545203800184)--(xmax, 1.2405761569108182*xmax + 6.62545203800184), linewidth(0.8) + qqzzff); /* line */ draw((xmin, 0.3049749595271946*xmin-1.7161009352521472)--(xmax, 0.3049749595271946*xmax-1.7161009352521472), linewidth(1.2) + dtsfsf); /* line */ draw((xmin, -0.36872458073384967*xmin-7.7226135592843255)--(xmax, -0.36872458073384967*xmax-7.7226135592843255), linewidth(0.8) + dbwrru); /* line */ draw((xmin, -0.36872458073384967*xmin-0.32891096066473735)--(xmax, -0.36872458073384967*xmax-0.32891096066473735), linewidth(0.8) + dbwrru); /* line */ draw((-8.915714298550334,-4.4351705426085655)--(-4.321356994370616,1.2644795852858564), linewidth(1.2) + ffqqff); draw((-4.321356994370616,1.2644795852858564)--(2.0590632643892013,-1.0881381995311175), linewidth(1.2) + ffqqff); draw((2.0590632643892013,-1.0881381995311175)--(-2.5352940397905166,-6.78778832742554), linewidth(1.2) + ffqqff); draw((-2.5352940397905166,-6.78778832742554)--(-8.915714298550334,-4.4351705426085655), linewidth(1.2) + ffqqff); draw((-4.321356994370616,1.2644795852858564)--(-2.5352940397905166,-6.78778832742554), linewidth(0.8) + rvwvcq); draw((xmin, -150.4864864864868*xmin-649.0413513513527)--(xmax, -150.4864864864868*xmax-649.0413513513527), linewidth(0.8) + dotted + wrwrwr); /* line */ draw((xmin, -0.46060606060606063*xmin-0.7259636363636363)--(xmax, -0.46060606060606063*xmax-0.7259636363636363), linewidth(0.8) + dotted + wrwrwr); /* line */ /* dots and labels */ dot((-4.34,4.07),linewidth(2pt) + dotstyle); label("$A$", (-4.245,4.152), NE * labelscalefactor); dot((-7.38,-2.53),linewidth(2pt) + dotstyle); label("$B$", (-7.281,-2.448), NE * labelscalefactor); dot((3.756,-2.456),linewidth(2pt) + dotstyle); label("$C$", (3.851,-2.36), NE * labelscalefactor); dot((-2.5352940397905166,-6.78778832742554),linewidth(2pt) + dotstyle); label("$X$", (-2.441,-6.694), NE * labelscalefactor); dot((2.0590632643892013,-1.0881381995311175),linewidth(2pt) + dotstyle); label("$P$", (2.157,-0.996), NE * labelscalefactor); dot((-2.5638002042230275,-2.4979957987708783),linewidth(2pt) + dotstyle); label("$Q$", (-2.485,-2.404), NE * labelscalefactor); dot((-8.915714298550334,-4.4351705426085655),linewidth(2pt) + dotstyle); label("$R$", (-8.821,-4.34), NE * labelscalefactor); dot((-4.321356994370616,1.2644795852858564),linewidth(2pt) + dotstyle); label("$H$", (-4.223,1.358), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let the line $l$ through $P$ parallel to $RX$ meet the altitude from $B$ at $H -(\bigstar)$. Now note that $\overline{RBH} \perp CA$ and $XP \perp CA$ $\implies RH \parallel XP - (\bigstar\bigstar)$ also $PH \parallel RX$ (from $\bigstar$),so, $RHPX$ is a parallelogram. Also $\overline{QP}(\equiv \overline{RQP})$ is the Simson line of point $X$ w.r.t $\triangle ABC$. Next observe that $RP$ and $HX$ bisects each other, but we know that the line joining the orthocentre of $\triangle ABC$ and the point $X$ bisects the Simson line $(\overline{QP})$. Hence from all these we can conclude that $H$ here, is the orthocentre of $\triangle ABC$, and it's clearly a fixed point, through which the line $l$ passes as $X$ varies along the minor arc $\overarc{BC}$. $\quad \blacksquare$ how did you draw that
14.04.2023 12:41
USA TST 2014 P1. $X$ is a variable point on the minor arc $BC$ of the circumcircle $(ABC)$ of acute $\triangle ABC$. Let $P$ and $Q$ be the feet of the perpendiculars from $X$ to $CA$ and $CB$ respectively. Let $R$ be the intersection of line $PQ$ and the perpendicular from $B$ to $AC$. Let $\ell$ be the line through $P$ parallel to $XR$. Prove that as $X$ varies along minor arc $BC$, the line $\ell$ always passes through a fixed point. Solution. Let $H$ be the orthocenter of $\triangle ABC$. By the Simson line bisection we know that $X$, $Q$, $H$ are collinear and $XQ=QH$. Also it is obvious that $RD\parallel PX$, so $PHRX$ is a parallelogram, forcing $\ell=PH$. [asy][asy] import graph; size(6cm); 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 = -6.406197132008787, xmax = 12.583124554125094, ymin = -8.666773136629851, ymax = 0.6899984410813239; /* image dimensions */ /* draw figures */ draw(circle((-0.6672559936935142,-3.6198789087014167), 2.8728045589712123), linewidth(0.8)); draw((-1.542454397044784,-0.8836344406728074)--(-3.4119890967645343,-4.468080669458789), linewidth(0.8)); draw((-3.4119890967645343,-4.468080669458789)--(2.098218439251572,-4.3977974100708295), linewidth(0.8)); draw((2.098218439251572,-4.3977974100708295)--(-1.542454397044784,-0.8836344406728074), linewidth(0.8)); draw((-1.6776086209705285,-6.309152714508519)--(-1.7013699878880846,-4.4462615481721), linewidth(0.8)); draw((-1.6776086209705285,-6.309152714508519)--(1.0986428914133959,-3.4329561476787687), linewidth(0.8)); draw((-4.297964579554645,-5.385951269629345)--(-3.4119890967645343,-4.468080669458789), linewidth(0.8)); draw((-4.297964579554645,-5.385951269629345)--(1.0986428914133959,-3.4329561476787687), linewidth(0.8)); draw((-4.297964579554645,-5.385951269629345)--(-1.6776086209705285,-6.309152714508519), linewidth(0.8)); draw((-3.4119890967645343,-4.468080669458789)--(-0.7191652101887513,-1.6783151229662738), linewidth(0.8)); draw((-1.5217130671707177,-2.5097547027995923)--(1.0986428914133959,-3.4329561476787687), linewidth(0.8)); /* dots and labels */ dot((-1.542454397044784,-0.8836344406728074),linewidth(2pt) + dotstyle); label("$A$", (-1.5012789786401808,-0.846481944311248), NE * labelscalefactor); dot((-3.4119890967645343,-4.468080669458789),linewidth(2pt) + dotstyle); label("$B$", (-3.37263329418242,-4.431602843560583), NE * labelscalefactor); dot((2.098218439251572,-4.3977974100708295),linewidth(2pt) + dotstyle); label("$C$", (2.133088086807642,-4.362658210882711), NE * labelscalefactor); dot((-1.6776086209705285,-6.309152714508519),linewidth(2pt) + dotstyle); label("$X$", (-1.6391682439959248,-6.27340945938373), NE * labelscalefactor); dot((-1.7013699878880846,-4.4462615481721),linewidth(2pt) + dotstyle); label("$Q$", (-1.6588667104753168,-4.402055143841495), NE * labelscalefactor); dot((1.0986428914133959,-3.4329561476787687),linewidth(2pt) + dotstyle); label("$P$", (1.1383155295983463,-3.3974333533925054), NE * labelscalefactor); dot((-4.297964579554645,-5.385951269629345),linewidth(2pt) + dotstyle); label("$R$", (-4.25906428575506,-5.347581534852308), NE * labelscalefactor); dot((-1.5217130671707177,-2.5097547027995923),linewidth(2pt) + dotstyle); label("$H$", (-1.4815805121607888,-2.471605428861084), NE * labelscalefactor); dot((-0.7191652101887513,-1.6783151229662738),linewidth(2pt) + dotstyle); label("$D$", (-0.6837926197454132,-1.634420603486926), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]
12.06.2023 23:42
Let $H$ be the orthocenter. Then, $R$, $B$, and $H$ are collinear and $PQR$ is the simson line of $X$ wrt to $\triangle ABC$. Then, since $HR \parallel XP$ and since the simson line bisects $XH$ it follows that $HPRX$ is a parallelogram, and the fixed point is thus $H$.
Attachments:

19.06.2023 00:19
Claim: $\ell$ always passes through $H$ when $X$ is on minor arc $BC$ Proof: Let us say that the intersection of $RH$ and $\ell$ is $H'$. We can say that $RH\parallel PX$, so $PXRH'$ is a parallelogram. This means that $PR$ bisects $XH'$. We can say that $PR$ is a Simson Line of $\triangle{ABC}$, and since the Simson Line bisects $XH$, $H=H'$. This means that the fixed point has to be $H$
07.08.2023 07:37
I thought i posted here, but apparently I haven't.. The line always passes through H, the orthocenter. Since RH perp. AC and XP perp. AC, RH//XP; it's well known that the Simson line PQ(R) bisects XH, hence RXPH is a parallelogram, with HP//XR, so H is the fixed point independent of everything else
03.12.2023 03:52
0 MOHS Geo exists? (As per Evan's website) I claim that the fixed point is $H,$ the orthocenter of the triangle. Actually, I will prove that $HPXR$ is a parallelogram. Obviously $RH \parallel DP$ since both are perpendicular to $AC.$ However, line $PQ$ bisects $XH$ by the well-known Simson line configuration, and we are done.
13.12.2023 18:17
We begin our solution by citing( and not proving) a very well known lemma Lemma: Let $H$ be the orthocenter of $\triangle ABC$. Then we have that the Simson line of $X$ w.r.t $\triangle ABC$ bisects $XH$. Now the problem is quite trivial. Claim: We claim that $H$ is the fixed problem referred to in the problem. Proof: It suffices to show that $XR \parallel PH$. We will instead prove that $HPXR$ is a parallelogram which also gives the desired conclusion. It is easy to observe that $RE \parallel XP$ as both are perpendicular to $AC$. Now let $K = PR \cap XH$. We must have $KH=KX$. Now it is clear that $\triangle BKH \cong \triangle PKX$ (by AAS congruency). This gives $KR=KP$ which indeed gives $\triangle XKR \cong \triangle HKR$ (SAS congruency). This finally gives $\angle PRX= \angle RPH$ giving away the fact that $PH \parallel XR$. Our proof is thus complete.
01.01.2024 23:44
I claim that the desired fixed point is the orthocenter $H$ of $ABC$. We first use a lemma that trivializes this problem: Lemma: $RHXP$ is a parallelogram. Proof: Let $K = XP \cap (ABC)$ and $K'$ be the reflection of $K$ over $\overline{BC}$. Then we observe in this order that first $LBKP$ is a parallelogram, $K'$ is the orthocenter of $\triangle XAC$, and finally that $RX \parallel HP$: We first have $LBKP$ is a parallelogram as \[\measuredangle XPQ = \measuredangle XCB = \measuredangle XKB.\]Now as $XP \perp AC$ and $K'$ is the reflection of $K$, it is obvious that $K'$ is the orthocenter of $\triangle XAC$. Now in order to show $RX \parallel HP$, it suffices to show $BH = XK'$ as $LBKP$ is already a parallelogram. Now as $K', H$ are the orthocenters of $\triangle BAC$ and $\triangle XAC$ respectively, we have that $HK' \parallel BX \implies BH = XP$, as desired. $\square$ Now as $RHXP$ is a parallelogram, we have that $\ell$ always passes through $H$, so that $H$ becomes our desired fixed point. $\blacksquare$
03.01.2024 22:12
09.03.2024 17:35
Indeed $0$ MOHS. Let $H$ be the orthocenter. See that $XP \parallel RH$ and it is well known that the Simson Line of $X$ with respect to $ABC$ (basically $PQR$) bisects $\overline{XH}$ and $RH \parallel XP$ and hence $RHPX$ is a parallelogram. So we get $RX \parallel HP \implies H \in \ell$, as desired.
16.04.2024 07:10
We claim that $\ell$ always passes through the orthocenter $H$ of $\Delta ABC$. Lemma. The Simson line of $\Delta ABC$ bisects $XH$. Proof. Well-known result. Since we have $RH \parallel PX$, applying the lemma on the Simson line $PR$ implies that $RHPX$ is a parallelogram. Thus, $PH \parallel RX$, so $H \in \ell$, as desired. Is it a coincidence that the key claim is after the problem proposer?
02.05.2024 08:10
Let $H$ be the orthocenter of $\triangle ABC$, we claim it is the desired point. Obviously, $\overline{RH} \parallel \overline{PX}$, and it is well-known that the Simson line, or $\overline{PR}$, bisects $\overline{HX}$. This is enough information to determine that $RHPX$ is a parallelogram, completing the proof. $\square$
02.09.2024 18:03
mmm yes robert simson ... i wonder why