P,Q,B are collinear

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MNJ2357

644 posts

MNJ2357

#1 h Nov 21, 2020, 6:43 PM • 7 Y

Y by Congruentisogonal44, chrono223, mathematicsy, centslordm, MathLuis, itslumi, Rounak_iitr

$H$ is the orthocenter of an acute triangle $ABC$ , and let $M$ be the midpoint of $BC$ . Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$ , and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$ . Prove that $P,Q,B$ are collinear.

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jacoporizzo

615 posts

jacoporizzo

#2 h Nov 21, 2020, 7:08 PM • 4 Y

Y by jbala, centslordm, Mango247, Efesc128e968

Since $\angle{ADH}=\angle{AEH}=90$ , $E,D$ are the feet of the $B, C$ altitudes. Additionally, it is well known that $MD, ME$ are tangents to $(AH)$ . Then as the center of $(AH)$ lies on $AH$ , $Q=DD \cap HH$ , and Pascal's on $DDAHHE$ gives the desired collinearity.

This post has been edited 1 time. Last edited by jacoporizzo, Nov 21, 2020, 7:13 PM

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Kamran011

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Kamran011

#3 h Nov 21, 2020, 7:19 PM • 1 Y

Y by centslordm

Here goes my solution , notice some super well known facts such as $D,E$ being feet of the altitudes and $MD$ being tangent to $(AH)$ at $D$ .
Then just to make the finish different than the previous solution , observe that by La Hire $Q\in{BP}\hspace{3 mm}\textbf{Q.E.D.}$

This post has been edited 2 times. Last edited by Kamran011, Aug 23, 2021, 4:57 AM

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RodwayWorker

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RodwayWorker

#4 h Nov 21, 2020, 7:45 PM • 2 Y

Y by centslordm, hodu

2020 Korea National Olympiad P2 wrote:

$H$ is the orthocenter of an acute triangle $ABC$ , and let $M$ be the midpoint of $BC$ . Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$ , and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$ . Prove that $P,Q,B$ are collinear.

Great problem! Though trivialized by projective geometry. This is extremely detailed solution.
2020 Korea National Olympiad P2 Solution

2020 Korea National Olympiad P2 Alternate Solution

RIP to my Polars bash.

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Pluto04

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Pluto04

#6 h Nov 21, 2020, 9:44 PM • 1 Y

Y by centslordm

MNJ2357 wrote:

$H$ is the orthocenter of an acute triangle $ABC$ , and let $M$ be the midpoint of $BC$ . Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$ , and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$ . Prove that $P,Q,B$ are collinear.

Solution similar to above few solutions.

By three tangents lemma , $MD$ and $ME$ are tangent to $(ADHE)$ and by pascal's theorem on $DDAHHE$ we get $P,Q,B$ are collinear.

Attachments:

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Flash_Sloth

230 posts

Flash_Sloth

#7 h Nov 22, 2020, 3:46 AM • 3 Y

Y by Anthraquinone, RodwayWorker, centslordm

The orthocenter $H$ is unnecessary. The problem can be generalized as follows:
Generalization: Let $D,E$ be arbitrary points on $AB$ and $AC$ . Let $H = CD \cap BE$ , $P=AH \cap DE$ . The line through $H$ parallel to $BC$ meets $DM$ at $Q$ . Prove that $P,Q,B$ are collinear.

One line Proof:Click to reveal hidden text

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unknownpi

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unknownpi

#8 h Feb 5, 2021, 12:58 PM • 2 Y

Y by centslordm, Mango247

any solution for people who do not know about pascal's theorem or la hire's theorem?

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bopappu

23 posts

bopappu

#9 h Feb 6, 2021, 8:37 PM • 1 Y

Y by centslordm

Tragic Problem.

All poles and polars will be taken wrt $(AH)$ . By Brokard on $ADHE$ we see that $BP$ is the polar of $C$ , and since $C \in DH$ is the polar of $Q$ , by La Hire $Q$ lies on the polar of $C$ , hence $B, Q, P$ collinear as desired.

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A.L.E.X

133 posts

A.L.E.X

#10 h Feb 7, 2021, 10:44 PM • 1 Y

Y by centslordm

R.I.P

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Spacesam

597 posts

Spacesam

#11 h May 2, 2021, 3:20 AM • 1 Y

Y by centslordm

Let $\omega$ be the circle through $(ADEH)$ . Recall that $M \in \overline{DD}$ by Three Tangents lemma. As a result, $Q = \overline{DD} \cap \overline{HH}$ . Now, Pascal's on $DDAHHE$ gives us $Q$ , $B = \overline{DA} \cap \overline{HE}$ , and $P = \overline{AH} \cap \overline{DE}$ are collinear, as desired.

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MathLuis

1471 posts

MathLuis

#12 h May 30, 2021, 1:23 AM • 1 Y

Y by centslordm

MNJ2357 wrote:

$H$ is the orthocenter of an acute triangle $ABC$ , and let $M$ be the midpoint of $BC$ . Suppose $(AH)$ meets $AB$ and $AC$ at $D,E$ respectively. $AH$ meets $DE$ at $P$ , and the line through $H$ perpendicular to $AH$ meets $DM$ at $Q$ . Prove that $P,Q,B$ are collinear.

Even if is a bit known, is cute :3. (Love projections :3). It is know that $D,E$ are feets of altitude.
We will take polars w.r.t. $(AH)$
Let $BP \cap DH=S$ and $BP \cap AE=T$ and let $CP \cap EH=U$ and $CP \cap AD=V$
$(D, S; H, C) \overset{B}{=} (A, T; E, C) \overset{P}{=} (H, S; D, C) \implies -1=(D, S; H, C) \overset{B}{=} (A, T; E, C) \implies B \in \mathcal P_P$ $(E, U; H, B) \overset{C}{=} (A, V; D, B) \overset{P}{=} (H, U; E, B) \implies -1=(E, U: H, B) \overset{C}{=} (A, V; D, B) \implies C \in \mathcal P_P$ And both of them implies that $\mathcal P_P=BC$ . We also have:
$-1=(D, S; H, C) \overset{B}{=} (A, T; E, C) \implies T \in \mathcal P_C$ $-1=(E, U; H, B) \overset{C}{=} (A, V; D, B) \implies V \in \mathcal P_B$ But otherside by the first ratio chasing $P \in \mathcal P_B$ and $P \in \mathcal P_C$ and then $\mathcal P_B=PV$ and $\mathcal P_C=PT$
We know that $\mathcal P_Q=DH$ and $\mathcal P_P=BC$ . So proving that $B,P,Q$ are colinear is equivalent to prove that its polars are concurrent but:
$\mathcal P_B \cap \mathcal P_Q \cap \mathcal P_P=VC \cap DH \cap BC=C$
Thus we are done :blush:

This post has been edited 1 time. Last edited by MathLuis, May 30, 2021, 1:28 AM

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bora_olmez

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bora_olmez

#13 h Aug 22, 2021, 8:45 PM

Y by

Same as #9 - posting for storage.

Notice that both $QH$ and $QD$ are tangent to the circle with diameter $AH$ , then $DH$ is the polar of $Q$ with respect to the aforementioned circle which contains $C$ and consequently by La Hire, we get that the polar of $C$ which by Brokard is $BP$ and contains $Q$ , as desired. $\blacksquare$

This post has been edited 2 times. Last edited by bora_olmez, Aug 22, 2021, 8:46 PM

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rafaello

1079 posts

rafaello

#14 h Aug 22, 2021, 9:16 PM

Y by

Bruh, absolutely trivial.

Note that $QH,QD$ are tangent to $(AH)$ by trivial angle chase.
Hence, by Pascal's theorem on $DDEHHA$ , we get that $P,Q,B$ are collinear.

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Mogmog8

1080 posts

Mogmog8

#15 h Oct 27, 2021, 4:20 AM • 1 Y

Y by centslordm

We present two solutions, the second written on 6/6/23 in Korea.

Solution One: Three Tangents Lemma and Pascal. $\square$

Solution Two: Let $Q'=\overline{BP}\cap\ell$ where $\ell$ is the line through $H$ perpendicular to $\overline{AH}$ . Then, letting $M'=\overline{BC}\cap\overline{DQ}$ , it suffices to show $M'=M$ . Indeed, letting $P_{\infty}$ be the point at infinity along $\ell$ , we have $(B,C;M,P_{\infty})\stackrel{Q}=(\overline{BP}\cap\overline{CD},C;D,H)\stackrel{P}=(B,\overline{AB}\cap\overline{CP};D,A)=-1$ $\square$

This post has been edited 2 times. Last edited by Mogmog8, Jun 6, 2023, 12:49 PM
Reason: added second solution

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primesarespecial

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primesarespecial

#16 h Oct 27, 2021, 11:31 AM

Y by

Bruh,
$MD$ is tangent to $(MH)$ .
Now $Q$ is pole of $DH$ ,and it lies on $PB$ ,the polar of $C$ by Brocard.

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Albert123

204 posts

Albert123

#17 h Dec 4, 2021, 1:36 AM

Y by

Note that in $\triangle PBC$ : $P \equiv c ; B \equiv c$
By chasing angles: $MD ; ME$ are tangents to $(AH)$
$\implies GD ; GH$ are tangents to $(AH)$
$\implies G \equiv c$
$\implies P,Q,B$ are collinear.

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Rexaria112

48 posts

Rexaria112

#18 h Mar 30, 2023, 10:34 AM • 1 Y

Y by Om245

One linear solution

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ismayilzadei1387

219 posts

ismayilzadei1387

#19 h Dec 12, 2023, 12:18 PM

Y by

very easy

$D-H-C$ and $E-H-B$ are obvious
$MD$ and $ME$ are tangent to $(ADHE)$
Thus Pascal yields $DDAHHED$ and we are done

This post has been edited 1 time. Last edited by ismayilzadei1387, Dec 12, 2023, 12:19 PM

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john0512

4176 posts

john0512

#20 h Dec 14, 2023, 11:37 PM • 1 Y

Y by Creeper1612

oops.

Rename $D$ to $F$ for the rest of this solution since this problem has strange naming conventions. Furthermore, let $D$ then be the foot from $A$ to $BC$ . Employ barycentric coodinates with reference triangle $\triangle DEF$ , and let $EF=a,DF=b,DE=c$ .

Our strategy will be to let $Q'$ denote the intersection of $BP$ and $MF$ , and show that $HQ'$ is parallel to $BC$ , which would solve the problem.

Note that $A=(-a:b:c),B=(a:-b:c),C=(a:b:-c).$ We will first compute $M$ . Multiply $B$ by $a+b-c$ and $C$ by $a-b+c$ to make them the same sum to get $B=(a^2+ab-ac:-ab-b^2+bc:ac+bc-c^2)$ and $C=(a^2-ab+ac:ab-b^2+bc:-ac+bc-c^2).$ Averaging these (this is allowed now since $B$ and $C$ have the same sum) gives $M=(a^2:-b^2+bc:-c^2+bc).$ For now, we also know that $\overrightarrow{BC}$ is a scalar multiple of $B-C=(2ab-2ac:-2ab:2ac)$ so it is also a multiple of $(b-c:-b:c),$ which we will use later.

Then, cevian $MF$ is described by $(a^2:-b^2+bc:t).$ We can find that the equation of line $BP$ is $2bcx+acy-abz=0,$ so if $Q'=(a^2:-b^2+bc:t)$ then we can solve to get $t=2ac-bc+c^2,$ which means that $Q'=(a^2:-b^2+bc:2ac-bc+c^2).$ The sum of this is $a^2+2ac+c^2-b^2=(a+c)^2-b^2=(a+b+c)(a-b+c),$ so we multiply $H=(a:b:c)$ by $a-b+c$ to get $H=(a^2-ab+ac:ab-b^2+bc:ac-bc+c^2),$ so $\overrightarrow{Q'H}$ is a scalar multiple of $Q'-H=(ab-ac:-ab:ac)$ so it is also a multiple of $(b-c:-b:c).$ Since this is also true of $\overrightarrow{BC}$ , they are parallel, so we are done.

This post has been edited 1 time. Last edited by john0512, Dec 14, 2023, 11:38 PM

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cursed_tangent1434

565 posts

cursed_tangent1434

#21 h Jan 13, 2024, 1:08 AM • 1 Y

Y by GeoKing

First of all, it is well known that $D$ and $E$ are the feet of the altitudes from $C$ and $B$ respectively. Further, let $F= \overline{AH} \cap \overline{BC}$ and $H' = \overline{HM} \cap (AH)$ . Also, let $R= \overline{BE} \cap \overline{DM}$ .

We observe that $QH$ is tangent to $(AH)$ (since the center of $(AH)$ lies on the segment $AH$ ). It is well known that $MD$ and $ME$ are tangent to $(AH)$ . Thus,
$-1 = (DE;HH') \overset{H}{=}(DR;QM)$
Now, let $X_C = \overline{EF} \cap \overline{AB}$ and $Q'= \overline{BP} \cap \overline{DM}$ . Then,
$-1=(AH;PF) \overset{B}{=}(DR;Q'M)$
But this means that $Q'=Q$ and indeed $B-Q-P$ as desired.

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lelouchvigeo

175 posts

lelouchvigeo

#22 h Jan 13, 2024, 7:01 AM

Y by

Has anybody done it by trig bash?

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dolphinday

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dolphinday

#23 h Feb 15, 2024, 9:58 PM

Y by

By the Three Tangents Lemma, $ME$ , $MD$ are both tangents to $(ADHE)$ . Then $HQ$ is also a tangent since $\angle QHA = \angle HEA$ . So Pascal's on $DDAHHE$ finishes.

This post has been edited 1 time. Last edited by dolphinday, Mar 18, 2024, 1:32 AM
Reason: wrong pascal

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GrantStar

815 posts

GrantStar

#24 h Mar 25, 2024, 6:17 PM

Y by

cute

I claim that the three points lie on the polar of $B$ with respect to $(AH)$ under renamed points. This is since

$B$ is the pole on line $AC$ by Brokard
Since $MF$ and the line through $A$ parallel to $BC$ are tangents to $(AH)$ , we find $Q$ is the pole of line $AFB$ .

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Aiden-1089

277 posts

Aiden-1089

#25 h Mar 26, 2024, 4:16 AM

Y by

Note that $D$ and $E$ are feet of the altitudes from $B$ and $C$ respectively. It is well-known that QH and MD are tangent to $(AEHF)$ .
Pascal on $DDAHHE$ yields the desired result.

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Mrcuberoot

26 posts

Mrcuberoot

#26 h Jun 24, 2024, 8:09 PM • 1 Y

Y by Hamzaachak

Let $T=AB \cap HQ$

Since $HT$ is parallel to $BC$

$\triangle DQT \sim \triangle DMB \implies \frac{DQ}{DM}=\frac{QT}{MB}$

And $\triangle DHQ \sim \triangle DCM \implies \frac{DQ}{DM}=\frac{HQ}{MC}$

Hence $\frac{HQ}{MC}=\frac{QT}{MB}$ and since $M$ is the midpoint of $BC$ we get $HQ=QT$ $\star$

Denote $\angle CAB=\alpha$ $\angle ABC =\beta$ and $\angle ACB=\gamma$

By Law of sine In triangle $PHD$ we get $\frac{\sin(\angle PDH)}{PH}=\frac{\sin(\angle HPD)}{HD}$

It is well Known that $EDBC$ is cyclic hence $\angle PEC=180-\beta$

Thus

$180-\angle HPD =\angle EPH =360-(\angle PEC + \gamma +90 )= 360-(180-\beta +90+\gamma)$

$=90+\beta -\gamma$

Thus $\sin(\angle HPD)=\sin(90+\beta-\gamma)$

And also we have $90-\gamma=\angle EBC =\angle EDC=\angle PDH$ so

$\sin(\angle PDH)=\sin(90-\gamma)=\cos(\gamma)$

Hence $PH=\frac{HD\cos(\gamma)}{\sin(90+\beta+\gamma)}$

Similarly in triangle $APD$ by law of sine we get

$\frac{\sin(\angle APD)}{AD}=\frac{\sin(\angle ADP)}{PA}$

And we have that $\angle APD =\angle EPH=90+\beta-\gamma$ (got it before)

And also by cyclic quad $EDBC$ $\gamma =\angle EDA=\angle PDA$

Hence $PA=\frac{AD\sin(\gamma)}{\sin(90+\beta-\gamma)}$

$\frac{PH}{PA}=\frac{HD\cot(\gamma)}{AD}$ and since $\frac{HD}{AD}=\cot(\beta)$ we get $\frac{PH}{PA}=\cot(\beta)\cot(\gamma)$ $\star$

We know that $BT=AB-AT$

In right Triangle $AHT$ $\frac{AH}{AT}=\sin(\angle ATH)=\sin(\beta)$

hence $AT=\frac{AH}{\sin(\beta)}$

So $\frac{BT}{BA}=1-\frac{AH}{AB\sin(\beta)}$

And by Law of sine in triangle $AHB$

$\frac{\sin(\angle HBA)}{AH}=\frac{\sin(AHB)}{AB} \implies \frac{AH}{AB}=\frac{\sin(\angle HBA)}{\sin(AHB)}$

by angle chasing we get that $\angle AHB =180-\gamma$ and $\angle HBA=90-\alpha$

Hence $\frac{AH}{AB}=\frac{\cos(\alpha)}{\sin(\gamma)}$

So $\frac{BT}{BA}=1-\frac{\cos(\alpha)}{\sin(\beta)\sin(\gamma)}$

$=\frac{\sin(\beta)\sin(\gamma)-cos(180-(\beta+\gamma))}{\sin(\beta)\sin(\gamma)}$

$=\frac{\sin(\beta)\sin(\gamma)+cos(\beta)\cos(\gamma)-\sin(\beta)\sin(\gamma)}{\sin(\beta)\sin(\gamma)}$

$=\cot(\beta)\cot(\gamma)$

Thus $\frac{BA}{BT}=\tan(\beta)\tan(\gamma)$ $\star$

Finally Using the 3 $\star$

By Menelaus Theorem

$\frac{PH}{PA}$ × $\frac{BA}{BT}$ × $\frac{QT}{QH}$ $=\cot(\beta)\cot(\gamma)$ × $\tan(\beta)\tan(\gamma)$ × $\frac{QT}{QT}=1$ $\iff$ $P,Q,B$ collinear

This post has been edited 3 times. Last edited by Mrcuberoot, Jun 24, 2024, 8:15 PM
Reason: .

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Maths_Girl

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Maths_Girl

#27 h Oct 17, 2024, 8:41 AM

Y by

Cute yet easy, here is a solution that doesn't involve projective geo:

Let $O_1$ be the midpoint of $AH$ and let $E$ , $D$ be the feet of the altitudes from $B, C$ respectively. Define $Q'$ to be the intersection of $BP$ with the line through $O_1$ , parallel to $AB$ and let $R'$ be the intersection of $PC$ with the line through $O_1$ , parallel to $AC$ (in other words, let $Q', R'$ be the images of $B, C$ of a homothety centered at $P$ that sends $A$ to $O_1$ . It is well-known (and follows directly from PoP) that $\frac{PO_1}{PA} = \frac{PH}{PH_a}$ , so $H_a$ is sent to $H$ with this homothety. Hence $Q',R',H$ are collinear and $Q'R'\perp AH_a$ and $Q'R'\parallel BC$ . Note that by trivial angle-chase $\angle O_1DH = \beta = \angle ABC = \angle O_1Q'R'$ , so $DO_1HQ'$ is cyclic, so $\angle Q'DO_1 = 90^{\circ}$ . However, as mentioned in pervious posts and also by angle-chase, $\angle MDO_1 = 90^{\circ}$ , so $D, Q', M$ are collinear, hence $Q'\equiv Q$ . $\square$

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Eka01

204 posts

Eka01

#28 h Oct 26, 2024, 7:17 AM

Y by

By Brokard's, we see that $PB$ is the polar of $C$ with respect to $(AH)$ so if $Q$ lies on this, La Hire's Implies that $C$ lies on the polar of $Q$ . Now it is well known that $MD$ is tangent to $(AH)$ and obviously $QH$ is tangent to $(AH)$ so this implies that $Q$ is pole of $DH$ or that $DH$ is polar of $Q$ but $C$ obviously lies on $DH$ .

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Ianis

400 posts

Ianis

#29 h Oct 26, 2024, 2:56 PM

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Let $F=AH\cap BC$ , $G=DM\cap BH$ and $Q'=BP\cap DM$ . Then $\begin{align*}\{G,D;M,Q\} & \underset{H}{=}\{B,C;M,\infty \} \\ & =-1 \\ & =\{H,A;F,P\} \\ & \underset{B}{=}\{G,D;M,Q'\}, \end{align*}$ so $Q=Q'$ , and therefore $P,Q,B$ are collinear.

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