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Parallel Lines and Q Point

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Source: India EGMO TST 2025 Day 1 P3

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29 posts

#1 h Dec 13, 2024, 8:34 AM • 3 Y

Y by GeoKing, Rounak_iitr, SatisfiedMagma

Let $\Delta ABC$ be an acute angled scalene triangle with circumcircle $\omega$ . Let $O$ and $H$ be the circumcenter and orthocenter of $\Delta ABC,$ respectively. Let $E,F$ and $Q$ be points on segments $AB,AC$ and $\omega$ , respectively, such that
$\angle BHE=\angle CHF=\angle AQH=90^\circ.$ Prove that $OQ$ and $AH$ intersect on the circumcircle of $\Delta AEF$ .

Proposed by Antareep Nath

This post has been edited 2 times. Last edited by taptya17, Dec 17, 2024, 6:08 AM

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1603 posts

#2 h Dec 13, 2024, 9:29 AM • 5 Y

Y by bin_sherlo, Rounak_iitr, mxlcv, GeoKing, HoRI_DA_GRe8

nice problem
solution

This post has been edited 1 time. Last edited by starchan, Dec 13, 2024, 1:24 PM
Reason: unriddling the typos

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6 posts

LP088

#3 h Dec 13, 2024, 9:52 AM • 1 Y

Y by GeoKing

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

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696 posts

#4 h Dec 13, 2024, 10:11 AM • 2 Y

Y by GeoKing, egxa

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. $\blacksquare$

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596 posts

#5 h Dec 13, 2024, 12:12 PM • 17 Y

Y by bin_sherlo, GeoKing, iamnotgentle, LP088, starchan, alexanderhamilton124, ehuseyinyigit, Supercali, MrOreoJuice, SatisfiedMagma, Aryan-23, Combe2768, kamatadu, EpicBird08, BVKRB-, SilverBlaze_SY, ihategeo_1969

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

starchan wrote:

nice problem
solution

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.

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32 posts

#6 h Dec 15, 2024, 4:25 PM • 2 Y

Y by GeoKing, iliya8788

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

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163 posts

Om245

#7 h Dec 17, 2024, 4:30 AM • 3 Y

Y by taptya17, GeoKing, HoRI_DA_GRe8

Congratulations HoRI_DA_GRe8 :coolspeak:

:coolspeak:

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 $(AEF)$ .

hehe Comments

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#8 h Dec 17, 2024, 6:04 AM • 2 Y

Y by Om245, GeoKing

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 $P\in AH$ .

This post has been edited 4 times. Last edited by taptya17, Dec 17, 2024, 6:21 AM

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#9 h Dec 20, 2024, 9:30 AM • 1 Y

Y by GeoKing

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.

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517 posts

#10 h Jan 1, 2025, 9:42 AM • 1 Y

Y by Om245

Solved with sanyalarnab
Note that $AEHF$ is a parallelogram . Let $D$ be the common midpoint of $AH,EF$ . Since (AQ,AH;AB,AC)=-1 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.

https://cdn.discordapp.com/attachments/1247512024687181896/1323949288837087242/image.png?ex=67765f5c&is=67750ddc&hm=b14eee297feabbb9f3dfd1e113e4b9f9fc244ab953ec98c7ef1b41eb3c3dfd3e&

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338 posts

#11 h Jan 4, 2025, 9:49 AM • 1 Y

Y by GeoKing

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 $CH=CH'$ .

This post has been edited 2 times. Last edited by TestX01, Jan 4, 2025, 9:51 AM

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#12 h Jan 25, 2025, 11:44 AM

Y by

An Apollonius Circle

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193 posts

#13 h Feb 23, 2025, 1:09 AM • 2 Y

Y by babarazamtruefan152-0, HoRI_DA_GRe8

Let $\triangle XYZ$ be orthic triangle of $\triangle ABC$ .

Claim: $(AEF)$ , $(AH)$ , $(ABC)$ are coaxial.
Proof: Using the OG coaxiality lemma, we need to prove $\frac{\text{Pow}(E,(AH))}{\text{Pow}(E,(ABC))}=\frac{\text{Pow}(F,(AH))}{\text{Pow}(F,(ABC))} \iff \frac{EZ}{EB}=\frac{FY}{FC}$ Which is true since $\triangle HEB \cup Z \overset{-}{\sim} \triangle HFC \cup Y$ because see that $\angle EZH=\angle FYH=90^{\circ}$ . $\square$

Let $T=\overline{AH} \cap (AEF)$ . See that $(Q,T;E,F) \overset{A}=(\overline{AQ} \cap \overline{BC},X;B,C)=-1$ and so all we need to prove is that $O$ is pole of $\overline{EF}$ in $(AEF)$ .

Now since $\angle BHE+\angle CHF=180 ^{\circ}$ , it is well known that this is equivalent to $H$ having an isogonal conjugate in $BEFC$ and it is obvious that this must be $O$ .

To finish just look at this angle chase $\angle OFE=\angle HFC=90 ^{\circ}-\angle ACH=\angle BAC=\angle EAF$ And so $\overline{OF}$ is tangent to $(AEF)$ and so is $\overline{OE}$ and done.

Remarks

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everythingpi3141592

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everythingpi3141592

#14 h Apr 5, 2025, 1:57 PM • 1 Y

Y by HoRI_DA_GRe8

Note that $BEFC$ is convex, and $H$ satisfies $\angle BHE + \angle CHF = 180^{\circ}$ , thus its isogonal conjugate in said quadrilateral exists. This must be O, due to its reflections in the angle bisectors of $B$ , $C$ . This means $\angle OEF = \angle HEC = 90^{\circ} - \angle HCE = \angle BAC = \angle EAF$ , thus $OE$ and analogously $OF$ are both tangents to the circumcircle of $\triangle AEF$ .

This means that $OA$ is the symmedian in $\triangle AEF$ , and thus $AH$ being its reflection in angle bisector must be the median, so, $(AE, AF; AH, A\infty) = -1$ . We project at $A$ to get that $AH$ intersects $(AEF)$ at the harmonic conjugate of $A'$ , reflection of $A$ in the perpendicular bisector of $EF$ . Reflecting back, about this line we see that in fact, this point is reflection of the isogonal conjugate of $A$ in the perpendicular bisector of $EF$ , i.e. angle bisector of $\angle EOF$ . Thus, it suffices to show that $OA$ , $OQ$ are isogonal in $\angle EOF$ , which we can do by proving that $AQ$ and $EF$ share a perpendicular bisector. For this, we prove $(AQEF)$ is concyclic, which would finish as the line joining $O$ with the centre of this circle works ( $O$ is not the centre itself as it is intersection of tangents at $E$ , $F$ and thus lies outside the circle)

To prove this, note that $Q$ is $(AE'F') \cap (ABC)$ where $E'$ , $F'$ are the feet of perpendiculars from $B$ , $C$ respectively. This is the centre of spiral similarity sending $BF'$ to $CE'$ . To finish, note that $BF = \frac{BH}{\cos(90^{\circ}-A)}$ and $BF' = BH\cos(90^{\circ}-A)$ , and thus $\frac{BF'}{BF}$ can be expressed in trigonometric ratios in $\angle BAC$ , and thus it is equal to $\frac{CE'}{CE}$ which we can calculate analogously, thus the spiral similarity also sends $F$ to $E$ , and we are done.

Remark/light-hearted rant: This does indeed give TST Day 4 P3 vibes as author has mentioned above, in some ways the opposite since you are isogonal conjugating perpendiculars at H and not O. But unlike TST, I did not fail this time ~~redemption~~. We can only hope that holds for this edition of TST as well ~~true redemption~~, that being said, I did find this problem interesting, with its 'config geo' flavor, thanks @HoRI_DA_GRe8 for giving me the idea to try this problem

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Haris1

#15 h Apr 6, 2025, 11:38 AM

Y by

Just $\sqrt(AE*AF)$ invert then the whole problem turns into a very famous $Hm$ , $Dm$ config which is very easy to prove.

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