nice problem

Nice Sketch of my solution : Hc and Hb are reflections of H respect to AB and AC respectively, E is circumcenter of QHHc and F is circumcenter of HQHb now note that with angle chasing you can prove AEFQ(w) is concylice and OE and OF are tangent to w then by harmonic bundles you can finish the problem

Let $D,K,L$ be the feet of the altitudes from $A,B,C$ to $BC,CA,AB$ respectively. Note that $Q$ is $A-$queue point. Let $(AEF)\cap AH=P$. Claim: $A,E,F,Q$ are concyclic. Proof: Let's apply coaxiality lemma on $(AQHKL)$ and $(AQBC)$. \[\frac{EL.EA}{EB.EA}\overset{?}{=}\frac{FK.FA}{FC.FA}\iff \frac{HL.HK}{HB.HC}=\frac{EA.\frac{LH}{HC}}{EA.\frac{HB}{HK}}=\frac{EL}{EB}\overset{?}{=}\frac{FK}{FC}=\frac{FA.\frac{HK}{HB}}{FA.\frac{HC}{HL}}=\frac{HK.HL}{HB.HC}\]Which proves the result.$\square$ Claim: $OE=OF$. Proof: Work on the complex plane. \[\overline{e}=\frac{1}{a}+\frac{1}{b}-\frac{e}{ab}, \ \ \frac{h-b}{h-e}=-\frac{\overline{h}-\overline{b}}{\overline{h}-\overline{e}}\implies \frac{1}{e-a-b-c}=\frac{1}{a+\frac{ec}{b}}\]Thus, $e=\frac{(2a+b+c)b}{b-c}$. Similarily $f=\frac{(2a+b+c)c}{(c-b)}$. \[e.\overline{e}=\frac{(2a+b+c)b}{b-c}.\frac{\overline{(2a+b+c)}.\frac{1}{b}}{c-b}=\frac{(2a+b+c)c}{c-b}.\frac{\overline{(2a+b+c)}.\frac{1}{c}}{b-c}=f.\overline{f}\]So $OE=OF$.$\square$ Let $AO\cap (AEFQ)=S$. $\measuredangle FAS=90-\measuredangle B=\measuredangle PAE$ thus, $PS\parallel EF$. We have $OE=OF,OA=OQ$ and $A,F,E,Q$ are concyclic and $O$ is not the circumcenter of $(AEFQ)$ hence the perpendicular bisectors of $AQ,EF$ intersect at another point than circumcenter of $(AEF)$ which implies their perpendicular bisectors coincide. We see that $PS \parallel EF\parallel AQ$. Since $ASPQ$ is an isosceles trapezoid and $OA=OQ$, also $OS=OP$ must hold. $\measuredangle SPO=\measuredangle OSP=\measuredangle OAQ=\measuredangle AQO$ which proves the collinearity of $O,P,Q$ as desired.$\blacksquare$

My first problem proposed to any contest and I am so happy it is selected. starchan wrote: nice problem

This was more or less the solution I sent in.The problem was inspired by the last paragraph .I learnt about this isogonal conjugate thing in quadrilaterals from D4P3 of this year India TST .I had changed the conditions of the original problem from circumcentre to orthocentre and then played in geogebra to finally land up here ! Thanks everyone.

One can also show that $QM$ and $AO$ intersect on $(AEF)$ where $M$ is the midpoint of $AH$. Anyways, very nice configuration!

Congratulations HoRI_DA_GRe8 Indeed cool problem Consider $E$ on $AC$ and $F$ on $AB$ (mainly cuz after writing whole solution I realize that it's opposite (weird)) Let $D = AH \cap BC$, $R = AQ \cap BC$ and $N=AH \cap EF$. It's well known that $(D,Q;B,C)=-1$. Now we project cross ratio from $A$ to get $$(AQ,AN;AE,AF)=-1$$As $N$ is midpoint of $EF$, we get $AQ \parallel EF$. As $N$ is center of circle $(AQH)$ we get $AQEF$ is cyclic trapezium. Let $A'$ is point on $(ABC)$ such that $\angle A'BA = \angle A'CA = 90$. It's well known that $Q-H-M-A'$ where $M$ is midpoint of $BC$. Notice by angle chase we get \[\Box AEHF \sim \Box CHBA'\] Destination is now within reach; it's just an angle chase that will lead us there. Let $X = OQ \cap AH$ then assuming $AC > AB$ we get $$\angle AXQ = \angle XAO + \angle XOA = \angle B - \angle C + 2 \angle QBA$$ By parallelograms similarity $$\angle AEF = \angle CA'H = \angle CA'Q = \angle CBQ$$Thus $$\angle AEQ = \angle AEF - \angle QEF = \angle QBC - \angle QAB = (\angle B + \angle QBA) + (\angle QBA - \angle C) = \angle AXQ$$Hence $X$ lie on circle $(AEF)$.

Some part of it was done with Om245. Since $\angle AFH+\angle AEH=180$, $H$ and $O$ are isogonal conjugates in $FECB$. Thus, $\angle OFE=\angle OEF=\angle BAC \implies OF=OE$ are tangents to $AEF$. We know $OA=OQ$. As shown in Om245's solution above, parallelograms $AEHF$ and $BHCA'$ are similar and thus an angle chase tells us $AQ || EF$. Thus, since $O$ lies on the perpendicular bisectors of both $AQ$ and $EF$, we conclude that $AQEF$ must be an isosceles trapezium and hence cyclic. Further, $AO$ is the symmedian in $\Delta AEF$ and thus if $P=OQ\cap\odot AEF$ then by symmetry $AP$ must bisect $EF$ and thus $P\in AH$.

Let $P=OQ\cap AH$. Let $E',F'$ be the feet from $B,C$ to $AC, AB$ respectively. By similar triangles we have $\frac{BF'}{F'E}=\frac{CE'}{E'F}$, so by spiral similarity we get $AQEF$ cyclic. Let the center of $(AQEF)$ be $O'$. Therefore $O'$ lies on the perpendicular bisectors of $AQ$ and $EF$. Let $N=AH\cap EF$. Since $AFHE$ is clearly a parallelogram, $N$ is the midpoint of both $EF$ and $AH$. $\angle AQH=90^\circ\implies N$ is the center of $(AQH)\implies N$ lies on the perpendicular bisector of $AQ$. Therefore $N$ also lies on the perpendicular bisectors of $AQ$ and $EF$. (Clearly $N$ and $O'$ are distinct because $\angle A$ cannot be $90^\circ$.) Since two points lie on both perpendicular bisectors, the perpendicular bisectors must be the same implying $AQEF$ is an isosceles trapezoid. Let $QF$ intersect $(ABC)$ again at $Y$. By Reims', $BY\parallel EF\parallel AQ$. Therefore $QY$ and $AB$ subtend equal arcs in $(ABC)$. Now $$\angle PQF = \angle OQY=\frac{180^\circ-\angle QOY}{2}=90^\circ-\angle QCY=90^\circ-C=\angle PAF.$$This implies $P\in (AQEF)$, as desired.

Solved with sanyalarnab

and $AH$ bisects $EF$ we have $AQ \parallel EF$ .Note that $(AQH)$ is a right triangle with circumcenter $D$ ,thus $EF$ is the perpendicular bisector of $QH$.Thus $Q$ is the reflection of $H$ across $EF$ which implies $AQEF$ is cyclic isosceles trapezoid. Let $AH$ meet $(AEF)$ again at $G$ and $QE$ meet $(ABC)$ again at $Y$. By reims theorem $YC \parallel EF$ which implies $AQYC$ is an isosceles trapezoid. $\measuredangle YQO=90^\circ-\measuredangle QCY=90^\circ-\measuredangle CYA=90^\circ-\measuredangle CBA=\measuredangle EAG=\measuredangle EQG=\measuredangle YQG$. Thus $Q-O-G$ are collinear.

no proj i want copic markers but im too poor Let $G$ be the intersection point. firstly, $BH\perp AC$ so $EH\parallel AC$. Similarly, $FH\parallel AB$. Using the Forgotten Coaxiality Lemma on $E,F$ with $(AH)$ and $(ABC)$, we want if $E'$ is foot of $H$ to $AB$ and $F'$ foot of $H$ to $AC$, that $\frac{EE'}{EB}=\frac{FF'}{FC}$ but this is because of $\triangle HBE\sim \triangle HCF$ and similarly defined altitude. now let $AH$ intersect $(ABC)$ at $H'$. Forgotten coaxiality lemma makes us want $\frac{EE'}{EB}=\frac{GH}{GH'}$. But note that \[\frac{EE'}{EB}=\frac{EE'\times EB}{EB^2}=\frac{EH^2}{EB^2}=\sin^2{90^\circ-\angle A}=\cos^2{\angle A}\]Now use ratio lemma on $\triangle QHH'$ so \[\frac{GH}{GH'}=\frac{QH}{QH'}\times\frac{\sin HQO}{\sin H'QO}\]now $\angle HQO=90^\circ-\angle OQA=\angle AH'Q$ (because angle at centre theorem), and $\angle H'QO=90^\circ-\angle QAH'=\angle AHQ$. Sine law in $\triangle QHH'$ means this relation is $\frac{QH}{QH'}$. Hence, $\frac{GH}{GH'}=\left(\frac{QH}{QH'}\right)^2$, so all we want now is \[\cos{\angle A}=\frac{QH}{QH'}\]Now, doing more trig, \[\frac{QH}{QH'}=\frac{AH \sin \angle QAH}{\sin\angle QAH}\times \frac{\sin \angle AQH'}{AH'}=\frac{AH}{AH'}\times \sin\angle AQH'\]hence we just need \[\frac{AH}{AH'}=\frac{\sin \angle ACH}{\sin \angle ACH'}\]This is a direct consequence of ratio lemma on $\triangle ACH'$ and the well known fact $CH=CH'$.

An Apollonius Circle