Circumcircle of triangle PQR passes through midpoint of BC

orl

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 $BC.$

**Posted:** Sat Aug 09, 2008 7:24 pm

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The QuattoMaster 6000

**Posted:** Sat Aug 09, 2008 8:23 pm

jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 640

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

**Posted:** Sun Aug 10, 2008 3:28 am

orl

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.

**Posted:** Sun Aug 10, 2008 4:42 am

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jayme · Yang-Mills Theory Offline Joined: 20 Nov 2007 Posts: 640

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

**Posted:** Sun Aug 10, 2008 6:29 am

dgreenb801

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

**Posted:** Sun Aug 10, 2008 6:40 am

The QuattoMaster 6000

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, $OP\perp YX$ .

**Posted:** Sun Aug 10, 2008 5:31 pm

dgreenb801

Thank you

!

**Posted:** Sun Aug 10, 2008 6:06 pm

April

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) Mr. Green

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 $PQR$ .

**Posted:** Wed Sep 10, 2008 5:18 pm

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limes123

**Posted:** Tue Nov 11, 2008 11:36 am

sunken rock

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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luisgeometria

Lines $PR$ and $PQ$ cut again the circumference $T = RCQB$ at $X$ and $Y$ . Notice that $P$ is radical center of $T$ , the Feurebach circle and the circumfrence with diameter $BC$ . An inversion with center $P$ and power $PB.PC$ transforms them on themselves, takes $D$ $\to$ $M$ , $R$ $\to$ $X$ and $Q$ $\to$ $Y$ . In other words the circumference $PQR$ is taken to the line $XY$ . Notice that $(B,C,D,P)$ are harmonically separeted and $P = QY \cap XR$ belows to the polar of the meetpoint of the diagonals of the quadrilateral $RXQY$ WRT $T$ , hence $XY$ passes through $D$ . $If$ $XY$ passes through $D$ ,its inverse circumference $(PQR)$ passes through the inverse of $D$ , wich is $M$ ......Done..

**Posted:** Fri Mar 20, 2009 4:35 pm

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