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.$

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

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.

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

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

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$.

Thank you !

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$.

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)

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

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.

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.

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.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Regard%203.pdf p. 12-13. Sincerely Jean-Louis

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 $M$ $\in$ $(PQR)$

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.

See my solution on my Youtube channel here: https://www.youtube.com/watch?v=CryOVKdVZC0

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

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.

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.

$AD,BE,CF$ 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.

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 $M \in circ(PQR)$

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

bora_olmez wrote: For storage I guess... Can u explain why the condition (P,D;B,C) = -1 implies DM . DP = DB . DC ?

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*}$\square$

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

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 $PoP$.