Line passes through a fixed point

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math154

4302 posts

math154

#1 h Dec 24, 2013, 5:50 AM • 8 Y

Y by narutomath96, Davi-8191, HWenslawski, geometrylover123, centslordm, jhu08, Adventure10, Mango247

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.

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MMEEvN

252 posts

MMEEvN

#2 h Dec 24, 2013, 7:57 AM • 6 Y

Y by Anish_S, geometrylover123, centslordm, jhu08, Adventure10, Mango247

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!!

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jayme

9782 posts

jayme

#3 h Dec 24, 2013, 8:28 AM • 5 Y

Y by lebathanh, geometrylover123, jhu08, Adventure10, Mango247

Dear Mathlinkers,
the fix point seems to be the orthocenter of ABC.
Sincerely
Jean-Louis

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jayme

9782 posts

jayme

#4 h Dec 24, 2013, 8:45 AM • 6 Y

Y by haitienbg, xdiegolazarox, Kanep, jhu08, Adventure10, Mango247

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

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IDMasterz

1412 posts

IDMasterz

#5 h Dec 24, 2013, 11:01 AM • 3 Y

Y by jhu08, Adventure10, Mango247

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 $H$ .

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pi37

2079 posts

pi37

#6 h Dec 28, 2013, 9:40 PM • 9 Y

Y by Anish_S, yojan_sushi, tapir1729, AllanTian, jhu08, Adventure10, Mango247, and 2 other users

I'll prove the lemma referred to in previous posts.
Diagram
Lemma: The Simson line of $X$ bisects $XH$ .
Proof
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.

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v_Enhance

6874 posts

v_Enhance

#7 h Dec 29, 2013, 12:04 AM • 10 Y

Y by Anish_S, Kezer, claserken, alex_g, alkumus, HamstPan38825, jhu08, sabkx, Adventure10, Mango247

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

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Mikasa

56 posts

Mikasa

#8 h Apr 26, 2014, 9:30 AM • 4 Y

Y by jhu08, Adventure10, Mango247, and 1 other user

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.

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Wolstenholme

543 posts

Wolstenholme

#9 h Oct 15, 2014, 4:25 AM • 4 Y

Y by yojan_sushi, jhu08, Adventure10, Mango247

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.

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DrMath

2130 posts

DrMath

#12 h Apr 2, 2015, 10:51 PM • 3 Y

Y by jhu08, Adventure10, Mango247

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 ._.

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EulerMacaroni

851 posts

EulerMacaroni

#13 h Aug 7, 2015, 6:12 PM • 3 Y

Y by jhu08, Adventure10, Mango247

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

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DrMath

2130 posts

DrMath

#14 h Aug 7, 2015, 6:17 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Ooops LOL I forgot I had posted in this... and I've actually used this lemma on many a geo problem. Thanks anyways

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MathPanda1

1135 posts

MathPanda1

#15 h Oct 29, 2015, 3:22 AM • 3 Y

Y by jhu08, Adventure10, Mango247

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!

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pi37

2079 posts

pi37

#16 h Oct 29, 2015, 3:49 AM • 3 Y

Y by jhu08, Adventure10, Mango247

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 $XH$ .

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bobthesmartypants

4337 posts

bobthesmartypants

#17 h Sep 16, 2016, 11:54 PM • 3 Y

Y by jhu08, Adventure10, Mango247

solution

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.

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isaacmeng

113 posts

isaacmeng

#44 h Apr 14, 2023, 9:41 AM

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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 $\ell=PH$ .

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YaoAOPS

1519 posts

YaoAOPS

#45 h Jun 12, 2023, 8:42 PM

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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:

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trk08

614 posts

trk08

#46 h Jun 18, 2023, 9:19 PM

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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 $H$

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huashiliao2020

1292 posts

huashiliao2020

#47 h Aug 7, 2023, 4:37 AM

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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

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EpicBird08

1749 posts

EpicBird08

#48 h Dec 3, 2023, 12:52 AM

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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.

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Jishnu4414l

154 posts

Jishnu4414l

#49 h Dec 13, 2023, 3:17 PM

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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.

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MagicalToaster53

159 posts

MagicalToaster53

#50 h Jan 1, 2024, 8:44 PM

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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. $\blacksquare$

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peace09

5417 posts

peace09

#51 h Jan 3, 2024, 7:12 PM • 1 Y

Y by GrantStar

Ew

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ihategeo_1969

205 posts

ihategeo_1969

#52 h Mar 9, 2024, 2:35 PM

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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.

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blueberryfaygo_55

340 posts

blueberryfaygo_55

#53 h Apr 16, 2024, 4:10 AM

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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?

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joshualiu315

2513 posts

joshualiu315

#54 h May 2, 2024, 5:10 AM

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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. $\square$

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AngeloChu

470 posts

AngeloChu

#55 h Sep 2, 2024, 3:03 PM

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mmm yes
robert simson ... i wonder why

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cherry265

16 posts

cherry265

#56 h Feb 14, 2025, 1:20 AM

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can you quote egmo for this problem? like if i just said "by egmo chapter 4"

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Tony_stark0094

65 posts

Tony_stark0094

#57 h Apr 1, 2025, 2:19 AM

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RHPX is a parallelogram :w00t:

This post has been edited 1 time. Last edited by Tony_stark0094, Apr 1, 2025, 2:20 AM
Reason: jjjjj

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shendrew7

794 posts

shendrew7

#58 h Apr 1, 2025, 2:54 AM

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We claim the desired point is the orthocenter $H$ . Let $H' = \ell \cap BR$ . Then $RXPH'$ is a parallelogram, but $RXPH'$ is also a paralleogram by Simson lines, and thus $H = H'$ . $\blacksquare$

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