Circumcircle of triangle PQR passes through midpoint of BC

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orl

3647 posts

orl

#1 h Aug 10, 2008, 3:24 AM • 4 Y

Y by nguyendangkhoa17112003, Me.021, Adventure10, Mango247

The altitudes through the vertices $A,B,C$ of an acute-angled triangle $ABC$ meet the opposite sides at $D,E, F,$ respectively. The line through $D$ parallel to $EF$ meets the lines $AC$ and $AB$ at $Q$ and $R,$ respectively. The line $EF$ meets $BC$ at $P.$ Prove that the circumcircle of the triangle $PQR$ passes through the midpoint of $BC.$

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The QuattoMaster 6000

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The QuattoMaster 6000

#2 h Aug 10, 2008, 4:23 AM • 4 Y

Y by Adventure10, Doctorstrange07, Mango247, ddami

orl wrote:

The altitudes through the vertices $A,B,C$ of an acute-angled triangle $ABC$ meet the opposite sides at $D,E, F,$ respectively. The line through $D$ parallel to $EF$ meets the lines $AC$ and $AB$ at $Q$ and $R,$ respectively. The line $EF$ meets $BC$ at $P.$ Prove that the circumcircle of the triangle $PQR$ passes through the midpoint of $BC.$

Solution

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jayme

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jayme

#3 h Aug 10, 2008, 11:28 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
this problem has also being proposed at : O.M. IRAN 1998 and Hubei Math Contest 1994.
Thanks for your nice proof.
I saw another solution based on power in the book of Mohamed Assila.
Sincerely
Jean-Louis

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orl

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orl

#4 h Aug 10, 2008, 12:42 PM • 2 Y

Y by Adventure10, Mango247

Please post those further solutions you know for this problem in this thread. You are also invited to send your solutions for the other posted IMO Shortlist problems. Thanks.

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jayme

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jayme

#5 h Aug 10, 2008, 2:29 PM • 2 Y

Y by Adventure10, Mango247

Dear "Orl",
sorry, I made a confusion in the name. Here is the reference of the book that I have given.
Soulami Tarik Belhaj, Les Olympiades de mathématiques, Ellipses.
I will send my proof that I have in mind.
Sincerely
Jean-Louis

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dgreenb801

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dgreenb801

#6 h Aug 10, 2008, 2:40 PM • 2 Y

Y by Adventure10, Mango247

Quatto, what is point T? Do you mean F instead? Also, can you explain why $MH \perp AP$ ?

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The QuattoMaster 6000

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The QuattoMaster 6000

#7 h Aug 11, 2008, 1:31 AM • 2 Y

Y by Adventure10, Mango247

Yes, sorry dgreenb801, $F$ is the same as $T$ ; sorry for the error. [Moderator edit: Error fixed.]

Apply the following lemma to cyclic quadrilateral $ABCD$ to get that $MH\perp AP$ :

Consider cyclic quadrilateral $ABCD$ with center $O$ and intersection of diagonals being $P$ . Let $AB$ and $CD$ intersect at $X$ and $AD$ and $BC$ meet at $Y$ . Then, $OP\perp XY$ .

This can be proven as follows:

Let the intersection between the circumcircles of $\triangle ABY$ and $DCY$ be $Z$ . Let the feet of the perpendiculars from $P$ to $AD$ and $BC$ be $Q$ and $R$ respectively and the midpoints of $AD$ and $BC$ be $M$ and $N$ respectively. Now, $\angle ADZ = \angle ZCB$ and $\angle ZBC = 180 - \angle ZBY = 180 - \angle YAZ = \angle ZAD$ , so $\triangle ZAD\sim \triangle ZBC$ . Since $M$ and $N$ are the respective midpoints of $AD$ and $BC$ , we see that $\triangle ZAM\sim \triangle ZBN$ , so $\angle YMZ = \angle YNZ$ , so $YZNM$ is cyclic. Since $O$ is the circumcenter of $ABCD$ , we see that $\angle YMO + \angle YNO = 90 + 90 = 180$ , so $YNOM$ is cyclic, so $YZNOM$ is cyclic. Thus, $\angle YZO = \angle YNO = 90$ , so $OZ\perp YZ$ . By Miguel's Theorem, we see that $Y$ , $Z$ , and $X$ are collinear, so $OZ\perp YX$ . Now, notice that since $ABCD$ is cyclic, $\triangle PAD\sim \triangle PBC$ , so $\triangle PAQ\sim \triangle PBR$ , so $\frac {AQ}{BR} = \frac {AP}{PB} = \frac {AD}{BC}$ , which means that $\triangle ZAQ\sim \triangle ZBR$ , which implies that $\angle ZQY = \angle ZRY$ , so $ZRQY$ is cyclic. Since $\angle PRY + \angle PQY = 90 + 90 = 180$ , we see that $YZRPQ$ is cyclic, so $\angle PZY = \angle PRY = 90$ , which implies that $P$ and $O$ lie on the perpendicular through $Z$ to $YX$ . Thus, $OP\perp YX$ .

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dgreenb801

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dgreenb801

#8 h Aug 11, 2008, 2:06 AM • 1 Y

Y by Adventure10

Thank you

!

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April

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April

#9 h Sep 11, 2008, 1:18 AM • 6 Y

Y by AlastorMoody, Adventure10, Acrylic2005, Math_legendno12, Mango247, and 1 other user

I think this problem is posted many times on MathLinks, but I can't get the links, so let me present the simplest solution I have in mind (surely that it's not new)

We have quadrilateral $BCEF$ is cyclic and $EF\parallel QR$ , so the quadrilateral $BQCR$ is also cyclic. Therefore $DQ\cdot DR = DB\cdot DC$ $(1)$

On the other hand, $(PBDC)$ is a harmonic division, so $DB\cdot DC = DM\cdot DP$ $(2)$ , where $M$ is the midpoint of $BC$ .

From $(1)$ and $(2)$ , we conclude that $M$ lies on the circumcircle of triangle $PQR$ .

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limes123

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limes123

#10 h Nov 11, 2008, 7:36 PM • 2 Y

Y by Adventure10, Mango247

The QuattoMaster 6000 wrote:

Consider cyclic quadrilateral $ABCD$ with center $O$ and intersection of diagonals being $P$ . Let $AB$ and $CD$ intersect at $X$ and $AD$ and $BC$ meet at $Y$ . Then, $OP\perp XY$ .

We can prove, that O is orthocenter of XYP. We see that $(A,D;F,Y)=(B,C;G,Y)=(D,A;F,Y)$ => (A,D;F,Y)=(B,C;G,Y)=1. Hence PX is polar of Y => $PX\perp OY$ . Analogously $XO\perp YP$ QED (I think it's known as Brokard's theorem)

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sunken rock

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sunken rock

#11 h Mar 20, 2009, 7:42 PM • 2 Y

Y by Adventure10, Mango247

I have a simpler way to prove this lemma: it is well known (Pascal to the ‘hexagons’ ABCCDA and ABBCDD) that the tangents to circle (ABCD) at A and C concur on XY, let their common point be K, same the ones at B and D, which concur at L. Then since AP.PC = BP.PD it follows that P belongs to radical axis of the circles (K,KA) and (L,LB). But O belongs to that line too [the tangents from O to both circles are radii of (ABCD)], therefore OP and KL are parallel, but KL and XY are perpendicular.

Best regards,
sunken rock

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Luis González

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Luis González

#12 h Mar 21, 2009, 12:35 AM • 3 Y

Y by Pluto1708, Adventure10, Mango247

Lines $PR$ and $PQ$ cut circle $\mathcal{T} \equiv{} \odot (RCQB)$ again at $X$ and $Y.$ Notice that $P$ is radical center of $\mathcal{ T},$ the 9-point circle $(N)$ and the circle with diameter $\overline{BC}.$ Inversion with center $P$ and power $\overline{ PB} \cdot \overline{PC}$ takes $D \mapsto M,$ $R \mapsto X$ and $Q \mapsto Y.$ In other words, circle $\odot( PQR)$ is taken to the line $XY.$ But $(B,C,D,P)=-1$ and $P \equiv {} QY \cap XR$ is on the polar of the intersection of the diagonals of the quadrilateral $RXQY$ WRT $\mathcal{ T}$ $\Longrightarrow$ $XY$ passes through $D.$ If $XY$ passes through $D,$ then its inverse $\odot (PQR)$ passes through the inverse $M$ of $D$ and the proof is completed.

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sayantanchakraborty

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sayantanchakraborty

#13 h Jan 23, 2014, 5:34 AM • 1 Y

Y by Adventure10

Let O be the midpoint of BC.

We note that $\angle PFB=\angle AFE=\angle C$

$\angle FPB=\angle B-\angle C$

Applying the law of sines in triangle PFB.

$PB=\frac{BFsinC}{sin(B-C)}=\frac{acosBsinC}{sin(B-C)}$

$PD=PB+BD=\frac{acosBsinC}{sin(B-C)}+ccosB$

$OD=OB-BD=\frac{a}{2}-ccosB$

$\angle QFD=\angle BFD=\angle C$

$\angle FQD=\angle AFE=\angle C$

So $QD=FD=\frac{csin2B}{2sinC}$

Similarly applying the law of sines in triangle DRC.

$DR=\frac{DCsinC}{sinB}=\frac{bsin2C}{2sinB}$

Now points P,Q,O,R lie on a circle

$\Leftrightarrow PD*OD=QD*DR$

$\Leftrightarrow (\frac{acosBsinC}{sin(B-C)}+ccosB)(\frac{a}{2}-ccosB)=\frac{bcsin2Bsin2C}{4sinBsinC}$

$\Leftrightarrow 4sinBcosBsinCcosC=4sinBcosBsinCcosC$

which is true.Hence the proof.

Maths is the doctor of science.

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AnonymousBunny

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AnonymousBunny

#14 h Jun 5, 2014, 12:51 PM • 2 Y

Y by Adventure10, Mango247

Darn I failed to come up with a synthetic solution, and tried trig bash instead, which I believe is the more natural approach. My solution is probably similar to that posted by sayantanchakraborty, although I didn't read it fully.

Diagram
Solution

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jayme

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jayme

#15 h Nov 29, 2018, 10:21 AM • 1 Y

Y by Adventure10

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Regard%203.pdf p. 12-13.

Sincerely
Jean-Louis

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khanhnx

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khanhnx

#16 h Oct 1, 2019, 12:04 PM • 2 Y

Y by Adventure10, Mango247

Let $M$ is midpoint of $BC$
Since: $PQ$ $\parallel$ $EF$ then: $B$ , $R$ , $C$ , $Q$ lie on a circle
So: $\overline{DQ} . \overline{DR} = \overline{DB} . \overline{DC} = \overline{DP} . \overline{DM}$ or $M$ $\in$ $(PQR)$

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Kimchiks926

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Kimchiks926

#17 h Jun 6, 2020, 10:50 AM • 1 Y

Y by mkomisarova

Let $M$ be a midpoint of $BC$ .
Claim: $BRCQ$ is cyclic.
Proof: Note that:
$\angle BRQ = \angle PFB = \angle BCQ$ We will prove that $PRQM$ is cyclic. It is sufficient to show that $PD \cdot DM = RD \cdot DQ$ . Since $BRCQ$ is cyclic we have $RD \cdot DQ = BD \cdot DC$ . Therefore we are left to prove:
$PD \cdot DM =BD \cdot DC$ Note that:
$\begin{align*} BD \cdot DC =(BM-DM)(MC+DM) =BM^2-DM^2 = PD \cdot DM \\ BM^2=DM(PD+DM)=DM \cdot MP \end{align*}$ Since $(P,D;B,C)=-1$ and $M$ is midpoint of $BC$ by well -known lemma we get that $BM^2 = DM \cdot MP$ and we are done.

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dgreenb801

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dgreenb801

#18 h Jun 7, 2020, 12:17 AM

Y by

See my solution on my Youtube channel here:

https://www.youtube.com/watch?v=CryOVKdVZC0

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iman007

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iman007

#19 h Dec 8, 2020, 6:22 AM • 2 Y

Y by tarannom, kn07

orl wrote:

The altitudes through the vertices $A,B,C$ of an acute-angled triangle $ABC$ meet the opposite sides at $D,E, F,$ respectively. The line through $D$ parallel to $EF$ meets the lines $AC$ and $AB$ at $Q$ and $R,$ respectively. The line $EF$ meets $BC$ at $P.$ Prove that the circumcircle of the triangle $PQR$ passes through the midpoint of $BC.$

I think I have seen this one on the other exam before :huh:

harmonic

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snakeaid

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snakeaid

#20 h Dec 10, 2020, 1:32 PM

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We have that $\triangle RAQ \sim \triangle FAE \sim \triangle CAB \implies AQ \cdot AC =AR \cdot AB \implies BRCQ$ is cyclic $\implies RD \cdot QD=BD \cdot CD$ . Let $X$ be the second intersection of $(AEF)$ and $(ABC)$ . Then by the radical center $A,P,X$ are collinear. Also if $N$ is the reflection of $H$ w.r.t. $M$ , then $\angle BNC=\angle BHC=180^{\circ}-\angle BAC \implies N \in (ABC)$ and $\angle AXN=\angle ABN=\angle ABC+\angle CBN=\angle ABC+\angle HCB=90^{\circ}=\angle AHN \implies X,H,N$ are collinear. Since $H,M,N$ are collinear we have that $X,H,M$ are collinear. Thus $H$ is the orthocenter of $\triangle PAM$ . If $S$ is the reflection of $H$ over $\overline{BC}$ , then $\angle BSC=\angle BHC=180^{\circ}-\angle BAC \implies S\in (ABC)$ . In a similar way we prove that $S \in (ADM)$ . Hence $D$ lies on the radical axis of $(ABC)$ and $(ADM)$ , so it has equal powers w.r.t. these circles, which are $BD \cdot CD$ and $PD \cdot MD$ , respectively. Thus $RD \cdot QD=BD \cdot CD=PD \cdot MD \implies PRMQ$ is cyclic. Done.

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EulersTurban

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EulersTurban

#21 h Dec 12, 2020, 4:21 PM • 3 Y

Y by Mango247, Mango247, Mango247

Nice problem for practicing harmonic ratios

We easily see that $BQRC$ is a cyclic quadrilateral.
Let $M$ be the midpoint of $BC$ .
We have that $-1=(P,D;B,C)=(B,C;P,D)$ , this implies that $DM.DP=DB.DC$
But since we have that $BQCR$ is cyclic this implies that $BD.DC=QD.DR$ by PoP, thus we have that $DM.DP=DQ.DR$ , by PoP we have that $PQMR$ is cyclic.

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MathsLover04

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MathsLover04

#22 h Feb 14, 2021, 5:34 PM

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$AD,BE,CF$ concure at $H$ , thus $(P,D;B,C)=-1\Rightarrow BD\cdot PC=DC\cdot PB\Rightarrow BD\cdot DP+BD\cdot DC=DC\cdot PB\Rightarrow DP(BC-DC)+BD\cdot DC=DC\cdot PB\Rightarrow DP\cdot BM=DC\cdot PB\Rightarrow DP\cdot DM=DC\cdot PB-BD\cdot DP\Rightarrow DP\cdot DM=DC\cdot BD+DC\cdot PB-BD\cdot PC\Rightarrow DP\cdot DM=DC\cdot BD=RD\cdot DQ$ . So $PRMQ$ is cyclic.

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bora_olmez

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bora_olmez

#23 h May 19, 2021, 9:03 AM

Y by

For storage I guess...

We know that $(P,D;B,C) = -1$ and that $R, Q, B,C$ concyclic. Let $M$ be the midpoint of $BC$ .
Then $DM \cdot DP = DB \cdot DC = DQ \cdot DR$ thus $M \in circ(PQR)$

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lazizbek42

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lazizbek42

#24 h Dec 24, 2021, 8:03 PM

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$M$ is midpoint of $BC$
$PD \cdot DM=RD \cdot DQ$ By sinus Law.

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Evolhimar

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Evolhimar

#26 h Dec 26, 2021, 8:17 AM

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bora_olmez wrote:

For storage I guess...

Can u explain why the condition (P,D;B,C) = -1 implies
DM . DP = DB . DC ?

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Mogmog8

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Mogmog8

#27 h Jan 26, 2022, 4:55 AM • 2 Y

Y by centslordm, megarnie

We know $BQCR$ is cyclic as $\angle RQC=\angle FEC=180-\angle CBA=RBC$ so $DB\cdot DC=DQ\cdot DR.$ Since $(BC;DP)=-1,$ $D^*=P$ with respect to $(BC).$ Hence, letting $M$ be the midpoint of $\overline{BC},$ we see $MD\cdot MP=BM^2.$ Therefore, $\begin{align*}DB\cdot DC&=(BM-DM)(CM+DM)=BM^2-DM^2\\&=MP\cdot MD-DM^2=DM(MP-DM)=DM\cdot DP.\end{align*}$ $\square$

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JAnatolGT_00

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JAnatolGT_00

#28 h May 17, 2022, 9:31 PM

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Denote midpoint of $BC$ by $M.$ Since $BRCQ$ are concyclic and $P,D$ are inverse wrt circle with diameter $BC,$ we obtain $|PD|\cdot |DM|=|BD|\cdot |DC|=|RD|\cdot |DQ|.$

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Pyramix

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Pyramix

#29 h Jul 4, 2023, 8:08 AM

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storage

This post has been edited 1 time. Last edited by Pyramix, Jul 4, 2023, 8:09 AM
Reason: latex error

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Eka01

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Eka01

#31 h Sep 6, 2024, 12:29 PM

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By angle chasing, it is easy to show that $BRCQ$ is cyclic, so $RD.DQ=BD.DC$ and since $(PD;BC)=-1$ , it follows from the midpoint theorem for harmonic conjugates that $BD.CD=PD.DM=DR.DQ$ so the result follows from $PoP$ .

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