it's just butterfly theorem $HLOD$ is a parallelogram. $LD$ is perpendicular to $HK$, deduce: $KL$ is perpendicular to $HO$ i.e $KL$ is parallel to $PQ$. Using butterfly theorem, it's easy to realize $HP=1/2HK$ so $QH=1/2HL=1/4HA$

solution by Ali Mirzaie define $N$ the midpoint of $AH$, $Q$ the midpoint of $NH$, $M$ the midpoint of $BC$, $S$ the intersection of $MN$ and $OH$ , $A'$ the ,intersection of $AO$ and circumcircle, $R$ midpoint of $FE$ we know that $AA'$ is perpendicular to $EF$ and $MN$ is parallel to $AA'$ hence $MN$ is perpendicular to $EF$ and because $M$ is centre of circle passing through $B,F,E,C$ hence R is the intersection of $EF,MN$ so $H,S,R,P$ lie on a circle so $$\widehat{HRP}=\widehat{HSP}$$$$\widehat{HRP}=\widehat{A'MC}$$$$\widehat{A'MC}=\widehat{HSP}$$also $HA'||QS$ so $\widehat{DHM}=\widehat{SQH}$ hence $\widehat{HPS}=\widehat{SQH}$ and that means $S,Q,P,H$ lie on a circle so$$\widehat{SQP}=90 , SQ||HM \Rightarrow PQ \perp HM$$

H.HAFEZI2000 wrote: for acute triangle $\triangle ABC$ with orthocenter $H$, and $E,F$ the feet of altitudes for $B,C$, we have $P$ on $EF$ such as that $HO \perp HP$. $Q$ is on segment $AH$ so $HM \perp PQ$. prove $QA=3QH$ Proposed by Tran Quang Hung (Vietnam)

Is O supposed to mean the circumcenter?

MathematicalPhysicist wrote: Is O supposed to mean the circumcenter? It is.

This problem is also susceptible to barycentric coordinates.

$\textrm{ I wish all Iran 3rd geometries were proposed by buratinogigle}$ Call the $A-$queue point as $V$ , midpoint of $\overline{AH}$ as $N$ , $\overline{PH} \cap \overline{BC} = K$ , $S =$ midpoint $HK$ , $T = \overline{AO} \cap \overline{EF}$ , $N_9$ is the nine point center ,$A' = $ antipode of $A$ in $(O)$ , $\overline{HP} \cap \overline{AV} = W$ and $\overline{AH} \cap (O) = H'$ , midpoint of $HC = C'$ and midpoint of $HB = B'$ The problem is equivalent to $Q$ being midpoint of $HN$. Since $\angle AVA' = 90 \implies AV \parallel QP \implies \frac{HQ}{HA} = \frac{HP}{HW} $ So it's enough to prove $\frac{HN}{2HA} = \frac{HP}{HW}$ by butterfly theorem in $(AA'H'V)$ we have $HW=2HK$ it's enough to prove $\frac{HN}{2HA} = \frac{HP}{2HK} \iff \frac{1}{2} = \frac{HN}{HA} = \frac {HP}{HK}$ Now notice $O,H,N_9$ are collinear and $N_9$ is circumcircle of $DEF$ so by butterfly theorem in $\odot(DEF)$ for $EFB'C'$ we get $HP=HS= \frac{HK}{2}$ $\boxed{\mathcal{Q.E.D}}$

Gaussian_cyber wrote: $\textrm{ I wish all Iran 3rd geometries were proposed by buratinogigle}$ Not only Iran Problems. I wish all Contest Geos were proposed by Buratinogigle.

$ N , S $ be the midpoint of $ AC , AB $ and $ T $ be the intersection of circle $AEF$ and $MH$ and $X$ be the intersection of $ SN , EF $ and $R$ be the intersection of $BC , EF$ and $D$ is foot of perpendicular of $A$ and $Q'$ be the intersection of $PQ , MH$ and $L$ be the intersection of $EF , AH$ . We know that reflection of $H$ in $M$ is $A'$ ( facing diameter of $A$ ) so we have $ T \in \omega_{ABC} $ In order to Radical axis Theorem in $\odot {AEF} , \odot {EFBC} , \odot {ABC} $ we have $ AT , EF , BC $ are congruent . So $R \in AT$ $\implies RA || PQ , PH || AX \implies \angle RAX = \angle HPQ $ $ EFSN $ is cyclic $\implies XE.XF=XS.XN $ $\implies AX $ is radical axis of $\odot AEF , \odot ASN $ $\implies AX \perp HO \implies AX || HP $ $\frac{HP}{AX} =\frac{HL}{LA}$ $\frac{HQ}{HA} = \frac{HQ'}{HT} = \frac{sin{RAX} . HP}{HT} = \frac {sin{RAX} . AX . HL}{HT . AL} = \frac{RX . sin{AXF}}{AR} . \frac{sin{ALF}}{sin{AXF}} . \frac{HL}{HT}$ $ = \frac{RX}{RA} . sin{ALF} . \frac{HL}{HT} = \frac{RX}{AR} . sin{ALF} . \frac{HL}{HF} . \frac{HF}{HT}$ $= \frac{RX}{RA} . sin{ALF} . \frac{cos{C}}{sin{ALF}} .\frac{cos{B}}{sin{HAR}} = \frac{RX . cosB .cos C}{AR . sin{HAR}} = \frac{RX . cosB .cosC}{RD} $ $=\frac{RF}{FB} . \frac{AB}{2} . \frac{cosB .cosC}{RD} = \frac{RF}{RD} . \frac{AB . cosB .cosC}{2.cosB . BC} =\frac{RF}{RD} .\frac{AB}{2BC} . cosC $ $=\frac{sinA}{sin2C} . \frac{AB}{2BC} . cosC = \frac{sinA}{2sinC . cosC} .\frac{sinC}{2sinA} . cosC = \frac{1}{4} $ $\implies \frac{HQ}{AH} =\frac{1}{4} \implies \frac{HQ}{AQ} = \frac{1}{3} \implies 3HQ = AQ \implies \blacksquare $

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.744904112133643cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ real xmin = -18.425425180874665, xmax = 7.06438304339262, ymin = -12.98485148669736, ymax = 3.9813592296935254; /* image dimensions */ pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen ffqqff = rgb(1.,0.,1.); pair A = (-10.861833247181348,3.227445735800793), B = (-14.18337817499907,-10.711387765864046), C = (4.261804462358381,-10.96792716276749), O = (-4.887375806404415,-5.561402975360461), M = (-4.960786856320345,-10.839657464315767), H = (-11.008655347013208,-7.329063242109822), D = (-11.056302308606892,-10.754879780695648), F = (-13.25011024314061,-6.794937259453172), P = (-11.505387725417677,-5.608914974775164), Q = (-10.971949822055242,-4.689935997632167), O_N = (-7.948015576708812,-6.445233108735142), G = (-12.127045060513268,-6.679873784298371); draw(A--B--C--cycle, linewidth(0.8) + ffdxqq); /* draw figures */ draw(A--B, linewidth(0.8) + ffdxqq); draw(B--C, linewidth(0.8) + ffdxqq); draw(C--A, linewidth(0.8) + ffdxqq); draw(circle(O, 10.627229336912594), linewidth(1.2) + linetype("0 3 4 3") + ffqqff); draw(A--D, linewidth(0.8)); draw(C--F, linewidth(0.8)); draw(B--(-5.672205013214733,-1.6436513772717791), linewidth(0.8)); draw(H--O, linewidth(0.8)); draw((-5.672205013214733,-1.6436513772717791)--F, linewidth(0.8)); draw(H--P, linewidth(0.8)); draw(O--M, linewidth(0.8)); draw(M--Q, linewidth(0.8)); draw(O_N--Q, linewidth(0.8)); draw(O_N--P, linewidth(0.8)); draw(O--P, linewidth(0.8)); draw(O--Q, linewidth(0.8)); draw(Q--P, linewidth(0.8)); draw(P--G, linewidth(0.8)); draw(M--G, linewidth(0.8)); draw(O--(-10.935244297097277,-2.0508087531545143), linewidth(0.8)); draw(M--(-10.935244297097277,-2.0508087531545143), linewidth(0.8)); /* dots and labels */ dot(A,blue); label("$A$", (-11.518806180403216,3.4617823543959334), NE * labelscalefactor,blue); dot(B,blue); label("$B$", (-15.150348692347071,-10.822284859249987), NE * labelscalefactor,blue); dot(C,blue); label("$C$", (4.6718208520131395,-11.336753381775369), NE * labelscalefactor,blue); dot(O,linewidth(4.pt) + blue); label("$O$", (-4.770189679040884,-5.314445382801771), NE * labelscalefactor,blue); dot(M,linewidth(4.pt) + blue); label("$M$", (-4.830715387573282,-10.610444879386593), NE * labelscalefactor,blue); dot(H,linewidth(4.pt) + blue); label("$H$", (-12.184588974259588,-7.826262286896287), NE * labelscalefactor,blue); dot(D,linewidth(4.pt) + blue); label("$D$", (-11.095126220676432,-11.91174761283315), NE * labelscalefactor,blue); dot((-5.672205013214733,-1.6436513772717791),linewidth(4.pt) + blue); label("$E$", (-5.557023889962053,-1.4105371824621025), NE * labelscalefactor,blue); dot(F,linewidth(4.pt) + blue); label("$F$", (-14.030623084497716,-6.615748116248327), NE * labelscalefactor,blue); dot(P,linewidth(4.pt) + blue); label("$P$", (-12.124063265727191,-5.34470823706797), NE * labelscalefactor,blue); dot(Q,linewidth(4.pt) + blue); label("$Q$", (-10.671446260949649,-4.346034046283403), NE * labelscalefactor,blue); dot(O_N,linewidth(4.pt) + blue); label("$O_N$", (-6.949115186207198,-7.13021663877371), NE * labelscalefactor,blue); dot(G,linewidth(4.pt) + blue); label("$G$", (-12.759583205317366,-6.8578509503779195), NE * labelscalefactor,blue); dot((-10.935244297097277,-2.0508087531545143),linewidth(4.pt) + blue); label("$N$", (-10.822760532280643,-1.8039542879226893), NE * labelscalefactor,blue); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Consider that $O_N$ is the midpoint of $OH$ so it is the nine point center of $ABC$ and $N$ is the midpoint of $AH$. now by angle chasing we have that $\angle PQO_N=90^{\circ}$ so $PQO_NH$ is cyclic now since $QGDM$ is cyclic we can see that: \[\angle OND=\angle DMH=90^{\circ} - \angle HMD=\angle HQO_N \quad \textbf{So} \quad QO_N||ON\]as desired

Here is a projective solution. Let $U=\overline{EF}\cap \overline{BC}$, $V=\overline{DE}\cap \overline{AB}$. Then $\overline{UV}$ is the orthic axis of $\triangle ABC$, and it is well known that this is perpendicular to the Euler line. Let $K=\overline{PH}\cap \overline{AU}$ and $L=\overline{PH}\cap \overline{BC}$. Then \[(K,L;P,P_\infty)\stackrel{U}{=}(A,B;F,V)=-1\]which shows that $KP=PL$. We also have \[(K,H;P,L)\stackrel{U}{=}(A,\overline{UH}\cap\overline{AB};F,B)=-1\]which shows that $\frac{PH}{HL}=-\frac{PK}{KL}=-\frac{1}{2}$. Therefore, $PK=PL=3PH$. As both $\overline{PQ}$ and $\overline{AV}$ are perpendicular to $\overline{HM}$, we have \[\frac{PH}{PK}=\frac{QH}{QA}=\frac{1}{3}\]as desired.

iman007 wrote: now by angle chasing we have that $\angle PQO_N=90^{\circ}$ can you explain this part? after that it's just using the fact $HM||ON$