Medium geometry with AH diameter circle

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#1 h Jun 28, 2016, 2:25 PM • 25 Y

Y by doxuanlong15052000, john111111, anantmudgal09, Davi-8191, AlastorMoody, Vietjung, rashah76, itslumi, v4913, tenebrine, sotpidot, HWenslawski, innout, anonman, megarnie, HamstPan38825, Jc426, Lamboreghini, CyclicISLscelesTrapezoid, tiendung2006, mathmax12, Adventure10, Mango247, geobo, Rounak_iitr

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$ . Denote by $M$ , $N$ the midpoints of $\overline{AH}$ , $\overline{BC}$ . Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$ , and meets line $AN$ at a point $Q \neq A$ . The tangent to $\gamma$ at $G$ meets line $OM$ at $P$ . Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$ .

Proposed by Evan Chen

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v_Enhance

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v_Enhance

#2 h Jun 28, 2016, 2:27 PM • 25 Y

Y by anantmudgal09, jlammy, math4444, baladin, Davi-8191, Iyerie, Anar24, AlastorMoody, Omeredip, amar_04, mathlogician, v4913, tenebrine, sotpidot, TheCollatzConjecture, innout, megarnie, hsiangshen, HamstPan38825, Jc426, Lamboreghini, mathmax12, EpicBird08, Adventure10, Mango247

Denote by $\triangle DEF$ the orthic triangle. Note that $\overline{AG}$ , $\overline{EF}$ , $\overline{BC}$ are concurrent at $R$ by radical axis, and that $\overline{PA}$ and $\overline{PG}$ are tangents to $\gamma$ .

Now, consider circles $(PAGM)$ , $(MFDNE)$ , and $(MBC)$ . They intersect at $M$ but have radical center $R$ , so are coaxial; assume they meet again at $T \in \overline{RM}$ , say. Then $\angle PTM$ and $\angle MTN$ are both right angles, hence $T$ lies on $\overline{PN}$ .

Finally $H$ is the orthocenter of $\triangle ARN$ , and thus the circle with diameter $\overline{RN}$ passes through $G$ , $Q$ , $N$ .

This post has been edited 2 times. Last edited by v_Enhance, Jun 29, 2016, 6:48 AM
Reason: Un-hide now that people have had chance to try it

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LeVietAn

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LeVietAn

#3 h Jun 28, 2016, 2:59 PM • 3 Y

Y by AllanTian, Adventure10, Mango247

My solutions:
We have $\overline{G, H, N}$ (well-known). Let $AG\cap BC=S, AH\cap BC=D$ , and let $MK\perp PN at K$ .
We have $AG, KM, DN$ are concurent at point $S$ is the radical center of ${\odot (APGKM), \odot (MKDN), \odot (AGDN)}$ .
Becaues $AH\perp SN, NH\perp AS$ $\Rightarrow H$ is othorcenter of $\triangle ANS$ $\Rightarrow Q\in SH$ .
We have $SK.SM=SG.SA=SB.SC$ $\Rightarrow K\in \odot (BCM)$ .
We have $K\in \odot (SN)\equiv (GQN)$ .
So, $(GNQ)\cap (MBC)\cap PN=K$ . DONE

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ABCDE

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ABCDE

#4 h Jun 28, 2016, 3:11 PM • 4 Y

Y by char2539, amuthup, Adventure10, Mango247

I always overcomplicate Evan's medium geometry proposals lol

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houssam9990

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houssam9990

#6 h Jun 28, 2016, 4:15 PM • 2 Y

Y by Adventure10, Mango247

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ThE-dArK-lOrD

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ThE-dArK-lOrD

#7 h Jun 28, 2016, 5:26 PM • 3 Y

Y by talkon, Adventure10, Mango247

Let $U,V$ be the feet of altitudes from $B,C$ respectively.
Let $UV\cap BC=X$ and let $XM \cap \mathcal{N} =T$ where $\mathcal{N}$ is the nine-point circle of $\triangle{ABC}$ .
Considering radical axes of $\mathcal{N},(ABC),(MBC)$ we found that $X$ is the radical center and so $T \in (MBC)$ .
Considering those of $(PAG),(ABC),(MBC)$ we found that $X$ is the radical center and so $T\in (PAG)$ .
Then, we have $\angle{PTM}=90^{\circ}$ and $\angle{NTM}=\angle{NUM}=90^{\circ}$ . So $T\in PN$ .
From $\angle{XTN}=90^{\circ}=\angle{XQN}=\angle{XGN}=90^{\circ}$ we get $T\in (GQN)$ .
Hence $(GQN)\cap(MBC)=T\in PN$ .

This post has been edited 4 times. Last edited by ThE-dArK-lOrD, Jun 13, 2019, 6:04 AM

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talkon

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talkon

#8 h Jun 28, 2016, 5:31 PM • 3 Y

Y by ThE-dArK-lOrD, Adventure10, Mango247

My solution.

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liberator

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liberator

#9 h Jun 28, 2016, 5:32 PM • 12 Y

Y by abk2015, v_Enhance, PARISsaintGERMAIN, BobaFett101, JasperL, rashah76, parola, Abidabi, sotpidot, ike.chen, Adventure10, Mango247

$[asy] unitsize(3.5cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); path carc(pair A, pair B, pair C, real d=0, bool dir=CW) { pair O=circumcenter(A,B,C); return arc(O,circumradius(A,B,C),degrees(A-O)+d,degrees(C-O)-d,dir); } pair A=dir(120), B=dir(205), C=dir(-25), E=foot(B,C,A), F=foot(C,A,B), H=A+B+C, O=(0,0), M=(A+H)/2, N=(B+C)/2, R=extension(E,F,B,C), G=foot(N,A,R), P=extension(O,M,A,A+C-B), Q=extension(R,H,A,N), T=foot(M,P,N); D(unitcircle,heavygreen); DPA(CP(M,A)^^A--P--G,red); D(carc(R,Q,N),purple); D(carc(B,M,C),orange); D(A--B--C--cycle,1); DPA(A--R--E^^B--R--M^^P--N); DPA(H--A--N^^P--O,dashed+pathpen); D("A",A,dir(A)); D("B",B,SW); D("C",C,SE); D("E",E,dir(0)); D("F",F,dir(F)); D("G",G,NW); D("H",H); D("M",M,NE); D("N",N); D("O",O); D("P",P,dir(P)); D("Q",Q); D("R",R,dir(R)); D("T",T); [/asy]$
Let $\triangle DEF$ be the orthic triangle, and note that $AG,EF,BC$ are concurrent at $R$ by radical axis. Let $T=MR\cap PN$ . With respect to $\gamma$ , $\overline{AG}$ is the polar of $P$ and $\overline{EF}$ is the polar of $N$ , thus $\overline{PN}$ is the polar of $R$ (La Hire), so $R,T$ are inverses, and $\angle RTN=90^{\circ}$ .

$H$ is the orthocenter of $\triangle ARN$ (Brokard), and thus the circle on diameter $\overline{RN}$ passes through $G,T,Q,N$ . But also $RT\cdot RM=RE\cdot RF=RB\cdot RC$ , so $T$ lies on $(MBC)$ , and the result follows.

Comment. This is a nice, novel problem based off a popular configuration. Well done Evan!

This post has been edited 1 time. Last edited by liberator, Jun 28, 2016, 5:37 PM
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EulerMacaroni

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EulerMacaroni

#10 h Jun 28, 2016, 6:58 PM • 2 Y

Y by Adventure10, Mango247

Let $DEF$ be the orthic triangle, $V$ be the harmonic conjugate of $D$ with respect to segment $\overline{BC}$ , and $T'$ be the harmonic conjugate of $D$ with respect to $EF$ in the nine-point circle. Projecting from $M$ onto line $EF$ implies that $M,T',V$ are collinear, so $VT'\cdot VM=VF\cdot VE=VB\cdot VC$ so $T'\in\odot(MBC)$ , and since $\angle MT'N=\pi-\angle VT'N=\frac{\pi}{2}$ , $T'\in \odot(GQN)$ , hence $T'\equiv T$ . Finally, $VT\cdot VM=VB\cdot VC=VG\cdot VA$ so pentagon $APGTM$ is cyclic, and $\angle PTM=\pi-\angle PAM=\frac{\pi}{2}$ .

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First

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First

#11 h Jun 28, 2016, 7:39 PM • 1 Y

Y by Adventure10

V_Enhance, how did you come up with this problem?

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PROF65

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PROF65

#12 h Jun 28, 2016, 11:51 PM • 2 Y

Y by Adventure10, Mango247

Let $S= AG\cap BC$ we know that $NG\cap SQ=H$ is the orthocenter of $ASN$ if J is the center of $NSGQ$ then $\widehat{JGM}=\widehat{GNS}=\widehat{GQS}=\widehat{GQH}$ thus $GJ$ is tangent to $(M)$ i.e. $\overline{JGP} \perp GM$ besides $OM \perp AG,NG\perp AG$ then $OM \parallel NG$ implies $\widehat{GPM}=\widehat{JGN}=\widehat{GQH}=\widehat{GAH} =\widehat{GAM}$ hence $GPMA$ s cyclic but $GJ\perp GM$ thus $PA \perp AM=AH$ implies $PA\parallel BC$ , applying the converse of Reim's to $N-R-P , S-G-A$ where $R=NP\cap (GMAP)$ yields $GRSB$ is cyclic .In the other hand $RM\perp PR, SR\perp NR$ we deduce $S,R,M$ are collinear ; further $GA,RM,BC$ are concurrent so by radical axis 's we conclude $RMBC$ is cyclic.
R HAS.

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DrMath

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DrMath

#13 h Jun 29, 2016, 4:22 AM • 2 Y

Y by Adventure10, Mango247

Nice problem!

Let $\triangle DEF$ be the orthic triangle. First, $\angle AQH=\angle AQA' = 90$ where $A'$ is the antipode of $A$ . Thus $Q, H, N, A'$ are collinear. An easy angle chase gives $\angle MFN =\angle MEN =90$ , so $QH$ , the tangent to $(AEF)$ at $E$ , and the tangent at $F$ concur. Thus $QFHE$ is harmonic. A perspectivity at $A$ gives $XBDC$ harmonic where $X=AQ\cap BC$ . Thus $X\in EF$ by a well known lemma. On the other hand, since $NH\perp AQ$ and $AH\perp XN$ , $XH\perp AG$ and $X, H, G$ are collinear. Moreover, $(XQGN)$ is cyclic.

Let $T$ be the intersection of circles $(XQGN)$ and $(MFDNE)$ other than $N$ . Note $XT\cdot XM = XF\cdot XE=XB\cdot XC$ , so $T$ lies on $(MBC)$ . By Radical Axis on $(AQFHGE), (XQGN), (MFDNE)$ , $GQ, EF, TN$ concur. Let $R$ be the point of concurrence. On the other hand, by Radical Axis on $(MQDG), (XQGN), (MFDNE)$ , we have $MD, GQ, TN$ concur. Thus $R$ lies on $AH$ as well. We claim that $P, R, N$ are collinear.

Note $MO\parallel QHN\perp QA$ , so $PM\perp QA$ . Thus $PQ=PA$ , so $P$ is the intersection of the tangents to $(AQFHGE)$ at $A, Q$ . But recall $N$ was the intersection of the tangents at $E, F$ . Thus by Pascal's on $AQHQAG$ we deduce $P, R, N$ are collinear, as desired. $\square$

This post has been edited 4 times. Last edited by DrMath, Jun 29, 2016, 5:15 PM

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MS_Kekas

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MS_Kekas

#14 h Jul 3, 2016, 4:26 PM • 4 Y

Y by Mindstormer, Didier, Adventure10, Mango247

Niche ne ponel no pooral

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v_Enhance

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v_Enhance

#15 h Jul 4, 2016, 2:10 AM • 8 Y

Y by niraekjs, smy2012, AlastorMoody, v4913, HamstPan38825, Adventure10, Mango247, Rounak_iitr

First wrote:

V_Enhance, how did you come up with this problem?

Well, the original proposal was to show that the nine-point circle and $(MBC)$ intersected on line $PN$ . The points $G$ and $Q$ were added after the fact, to prevent people from using straight coordinates.

Here is also my original solution: Let $L$ be diametrically opposite $A$ on the circumcircle. Denote by $\triangle DEF$ the orthic triangle. Let $X = \overline{AH} \cap \overline{EF}$ . Finally, let $T$ be the second intersection of $(MFDNE)$ and $(MBC)$ .

We begin with a few easy observations. First, points $H$ , $G$ , $N$ , $L$ are collinear and $\angle AGL = 90^\circ$ . Also, $Q$ is the foot from $H$ to $\overline{AN}$ . Consequently, lines $AG$ , $EF$ , $HQ$ , $BC$ , $TM$ concur at a point $R$ (radical axis). Moreover, we already know $\angle MTN = 90^\circ$ . This implies $T$ lies on the circle with diameter $\overline{RN}$ , which is exactly the circumcircle of $\triangle GQN$ .

Note by Brokard's Theorem on $AFHE$ , the point $X$ is the orthocenter of $\triangle MBC$ . But $\angle MTN = 90^\circ$ already, and $N$ is the midpoint of $\overline{BC}$ . Consequently, points $T$ , $X$ , $N$ are collinear.

Finally, we claim $P$ , $X$ , $N$ are collinear, which solves the problem. Note $P = \overline{GG} \cap \overline{AA}$ . Set $K = \overline{HNL} \cap \overline{AP}$ . Then by noting $-1 = (D,X;A,H) \overset{N}{=} (\infty, \overline{NX} \cap \overline{AK}; A, K)$ we see that $\overline{NX}$ bisects segment $\overline{AK}$ , as desired. (A more projective finish is to show that $\overline{PXN}$ is the polar of $R$ to $\gamma$ ).

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Aiscrim

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Aiscrim

#17 h Jul 5, 2016, 5:06 AM • 1 Y

Y by Adventure10

Let $A^\star$ be the antipode of $A$ in $(ABC)$ and let $QN\cap (ABC)=\{A,Q^\star\}$ . It is well known that $A^\star$ is the reflection of $H$ over $N$ . As $HQ\parallel Q^\star A^\star$ (they are both perpendicular to $AN$ ) and $NH=NA^\star$ , we infer that $HQA^\star Q^\star$ is a parallelogram, so $NQ=NQ^\star$ . Also note that $G-H-N$ are collinear.

Let $\{T\}=AG\cap HQ,\ \{R\}=AH\cap GQ$ . Then $T$ lies on the polar of $P$ wrt $\gamma$ , so $P$ lies on the polar of $T$ wrt to $\gamma$ , i.e. $P\in NR$ . Observe that by Power of Point we have
$NG\cdot NH=NA\cdot NQ=NA\cdot NQ^\star=NB\cdot NC$
The inversion $\mathcal{I} (N,NB^2)$ swaps $G$ with $H$ , $Q$ with $A$ , and leaves $B$ and $C$ invariant. Let $\mathcal{I}(M)=M^\star$ . Under inversion, the concurrency of $(NGQ),(MBC)$ and $PN$ is equivalent to the concurrency of $AH,(M^\star BC)$ and $NP$ , i.e. to proving that $R \in (M^\star BC)$ .

A little computation shows us that $BR\perp MC$ , whence $R$ is the orthocenter of $\triangle{MBC}$ , so $\widehat{BRC}=180^\circ-\widehat{BMC}=\widehat{BM^\star C}\Rightarrow R \in (M^\star BC)$

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Pyramix

419 posts

Pyramix

#85 h Mar 5, 2024, 7:49 AM

Y by

We have that $Q$ is the $A-$ Humpty Point and $G$ is the $A-$ Queue Point.
Let $X=EF\cap BC$ and let points $T,N$ be the intersection of $(GNQ)$ and $(DEF)$ . It is well-known that $H$ is the orthocenter of $AXN$ and $AD,NG,XQ$ are the altitudes. Hence, $NGQX$ is cyclic with diameter $\overline{NX}$ , which gives $\angle XTN=90^\circ$ . Moreover, $\overline{MN}$ is the diameter of $(DEF)$ (nine point circle). So, $\angle MTN=90^\circ=\angle XTN$ . So, $M,T,X$ are collinear.

Note that $MG=MA$ and $OG=OA$ , as $M$ is the center of $(AGH)$ and $O$ is center of $(ABC)$ . So, $OM$ is the perpendicular bisector of $\overline{AG}$ . It follows that $PA=PG$ , which means that $\overline{PA}$ is tangent to $(AGH)$ at $A$ . Hence, $P,A,M,G$ are concyclic.

Claim: $T\in(MBC)$
Proof. Let $K,G$ be the intersection of $(GNQ)$ and $(ABC)$ . It is well-known that $K$ is the reflection of $Q$ in $\overline{BC}$ and that $AK$ is the $A-$ symmedian. Let $A,H'$ be the intersection of line $AH$ with $(ABC)$ . It is well-known that $H'$ is the reflection of $H$ in $\overline{BC}$ . Finally, note that $\angle XQN=\angle HQN=\angle HDN=90^\circ$ , which means $H,Q,N,D$ are cyclic. Reflecting about $\overline{BC}$ , we get that $H',K,N,D$ are cyclic.
We now perform $\sqrt{-HA\cdot HD}$ inversion, which is the inversion about the circle with center $H$ and radius $\sqrt{HA\cdot HD}$ , and then reflection about $H$ . Under this inversion, the nine-point circle $(DEF)$ goes to $(ABC)$ and the circle $(GNQ)$ goes to itself as $G$ goes to $N$ , $Q$ goes to $X$ , $N$ goes to $G$ and $X$ goes to $Q$ . This can be proved using the fact that $H$ is the orthocenter of $AXN$ with altitudes $AD,XQ,NG$ . So, $T$ goes to an intersection of $(GNQ)$ and $(ABC)$ . So, $T$ goes to either $G$ or $K$ . However, $N$ goes to $G$ , so $T$ must go to $K$ . Finally, note that $D$ goes to $A$ and $H'$ goes to $M$ , as $HM=\frac{HA}{2}$ and $HH'=2HD$ , while $N$ goes to $G$ . So, circle $(H'KND)$ goes to $(MTGA)$ . Hence, $T\in(PAMG)$ .
So, $XG\cdot XA=XT\cdot XM$ . But since $A,G,B,C$ are concyclic, $XG\cdot XA=XB\cdot XC$ . So, $XT\cdot XM=XB\cdot XC$ which means $T\in(MBC)$ . $\blacksquare$

To conclude, note that $\overline{MP}$ is the diameter of $(PMT)$ , so $\angle MTP=\angle MTN=90^\circ$ , which means $T\in\overline{PN}$ . So, $T\in\overline{PN}$ , $T\in(GNQ)$ and $T\in(MBC)$ . Hence, $T$ is the required point of intersection. $\blacksquare$

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ihatemath123

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ihatemath123

#86 h Apr 21, 2024, 10:01 PM

Y by

Since $\overline{GH} \perp \overline{GA}$ and $\overline{OM} \parallel \overline{NH}$ , it follows that $OM$ is the perpendicular bisector of $\overline{AG}$ . So, $\overline{AP}$ is also tangent to $\gamma$ ; this means $\overline{AP} \parallel \overline{BC}$ .

Let $X$ be the intersection of lines $AG$ and $BC$ . Then, it's well known that $X$ , $H$ and $Q$ are collinear, so $XGQN$ is cyclic.

Claim: lines $PN$ and $XM$ are perpendicular.

Proof: Let $P'$ be the reflection of $A$ over $P$ , so that $P'$ lies on line $NHG$ . We will in fact show that $\triangle AXH \sim \triangle P'NA$ , which implies our claim (since the segments we're trying to show perpendicular are the medians of the respective triangles). We already have $\angle XAH = \angle NP'A$ since $\overline{XA} \perp \overline{HP'}$ ; furthermore,
$\angle AHX = 180^{\circ} - \angle ANX = \angle P'AN,$ where the step is because $H$ is the orthocenter of $\triangle AXN$ . So, our claim is proved.

Let $U$ be the intersection of $\overline{PN}$ and $\overline{XM}$ . As stated earlier, $PM$ is the perpendicular bisector of $\overline{AG}$ , so $(PAMG)$ is cyclic with diameter $PM$ . Since $\angle PUM = 90^{\circ}$ , it follows that $GUMA$ is cyclic. So, by the radical axis theorem, $BUMC$ is cyclic. Since $\angle XUN = 90^{\circ}$ , we also have $XQUN$ cyclic, finishing.

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Aiden-1089

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Aiden-1089

#87 h May 13, 2024, 7:53 AM

Y by

Note that $G$ is the $A$ -Queue point, $Q$ is the $A$ -Humpty point. Let $X$ be the $A$ -Ex point.
Reflecting about line $OM$ , we see that $G$ goes to $A$ , so $PA$ is tangent to $(AH)$ . Thus, $P,A,M,G$ are concyclic.

Let $T' \neq M$ be the intersection between $(PAMG)$ and $(MBC)$ .
By radax on $(PAMG)$ , $(ABC)$ , $(MBC)$ , we get that $AG, BC, T'M$ are concurrent. Thus $T'$ lies on $XM$ .
Let $D$ be the foot of the $A$ -altitude. It is well-known that we can take a suitable inversion about $X$ that swaps the following pair of points: $(A,G), (M,T'), (H,Q), (D,N)$ . Thus the $A$ -altitude $AMHD$ is taken to the circle $(XGT'QN)$ .

Now, since $\measuredangle MT'N = \measuredangle XT'N = \measuredangle XGN = \measuredangle AGN = 90^{\circ} = \measuredangle MAP = \measuredangle MT'P$ , we get that $P,N,T'$ are collinear.

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EpicBird08

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EpicBird08

#88 h Jun 20, 2024, 2:52 AM

Y by

For some reason I love overcomplicating things.

Humpty point properties imply that $B,H,Q,C$ are concyclic. Thus applying the radical center theorem on $\gamma, (BHQC),$ and $(ABC)$ gives that $AG, QH,$ and $BC$ are concurrent, say at point $X.$

Claim 1: $XGQN$ is cyclic with diameter $XN.$
Proof: It is well-known that $G,H,N$ are collinear, so we see that $\angle XGN = \angle AGH = \angle AQH = \angle XQN = 90^\circ,$ as claimed.

We now get rid of the point $O$ from the problem:

Claim 2: $PA$ is tangent to $\gamma$ as well.
Proof: Note that $AM = MH$ and $AO = OG,$ so $MO$ is none other than the perpendicular bisector of $AG.$ Since $PG$ is tangent to $\gamma,$ this implies that $PA$ is as well.

Thus $P$ is just the intersection of the tangents to $\gamma$ at $A$ and $G.$

Now let $Z$ be the intersection of the circumcircle of $GQN$ with $PN.$ By Claim 1, we see that $Z$ is just the foot of the altitude from $X$ to $PN.$ However,

Claim 3: $XM \perp PN.$
Proof: We in fact show the stronger statement that $PM$ is the polar of $X$ with respect to $\gamma.$ First, we show that $P$ lies on the polar of $X.$ By La-Hire this is equivalent to showing that $X$ lies on the polar of $P,$ which is just $AG,$ and this holds by the definition of $X.$ Now we show that $N$ lies on the polar of $X,$ which proves the claim. Again by La-Hire this is equivalent to showing that $X$ lies on the polar of $N,$ which is just $EF$ where $E$ the foot of the altitude from $B$ to $AC$ and $F$ is the foot of the altitude from $C$ to $AB.$ But by the radical center theorem on $\gamma, (BFEC),$ and $(ABC),$ we see that $X,E,F$ are collinear. Therefore, $X$ lies on the polar of $N,$ which proves our claim.

Hence $Z$ is the foot of the altitude from $N$ to $XM.$

Now, let $AH$ intersect $BC$ at point $D.$ Then $\angle MZN = \angle MDN = 90^\circ,$ so $MZDN$ is cyclic. Thus by power of a point at $X,$ we get $XZ \cdot XM = XD \cdot XN.$ Since $\angle AGN = \angle ADN = 90^\circ,$ we get that $AGDN$ is cyclic, so $XD \cdot XN = XG \cdot XA = XB \cdot XC$ by power of a point on $(ABC).$ Combining everything together, we get
$XZ \cdot XM = XD \cdot XN = XG \cdot XA = XB \cdot XC,$ which implies that $ZMBC$ is cyclic, and we are done.

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Clew28

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Clew28

#89 h Jun 20, 2024, 3:50 PM • 1 Y

Y by duckman234

Given that the intersection of $QG$ with $BC$ is at $K$ , and $N$ is the midpoint of the arc $BC$ that does not contain $A$ , we start with the known result that $ND$ and $IP$ intersect at $Q$ . Given that $DP \parallel AN$ , by Reim's theorem, $Q$ , $P$ , $D$ , and $G$ are concyclic. It's apparent that they all lie on the $D$ -Apollonius circle with respect to $BC$ , leading to the relation $QB \cdot CG = QC \cdot BG$ . This implies the cross-ratio $-1 = (Q, G; B, C)$ , indicating that $GK$ must align with the $G$ -symmedian of triangle $BGC$ . Therefore, $DG$ bisects the angle $\angle QGM$ .

Furthermore, $QK$ aligns with the $Q$ -symmedian of triangle $QBC$ . Consequently, the tangents to $(ABC)$ at $B$ and $C$ intersect at $T$ on line $QK$ , establishing that the cross-ratio $(Q, G; T, K) = -1$ . This conclusion leads to the result that $DM$ bisects $\angle GMQ$ .

Hence, it follows that $D$ serves as the incenter of triangle $GQM$ , as required.

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kiemsibongtoi

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kiemsibongtoi

#90 h Jul 1, 2024, 7:18 AM

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v_Enhance wrote:

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$ . Denote by $M$ , $N$ the midpoints of $\overline{AH}$ , $\overline{BC}$ . Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$ , and meets line $AN$ at a point $Q \neq A$ . The tangent to $\gamma$ at $G$ meets line $OM$ at $P$ . Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$ .

Proposed by Evan Chen

Let $D$ , $E$ , $F$ be the foot of altitude from vertex $A$ , $B$ , $C$ in triangle $ABC$ respectively
$\hspace{0.5cm}$ $S$ be the intersection of lines $EF$ and $BC$ ; $T$ be the intersection of lines $PN$ and $MS$
Examine radical axis of circles $\gamma$ , circle with diameter $BC$ , $(ABC)$ we see that lines $EF$ , $BC$ , $HQ$ concur at $S$ , so $\angle SQN = \angle HQA = 90^\circ$
Cuz $OM \perp AG$ so $P$ is the pole of $AG$ respect to $\gamma$ , and we ez to see that $N$ is the pole of $EF$ respect to $\gamma$
Therefore, $PN$ is the polar of $S$ respect to $\gamma$ (Follow La Hire theorem)
Lead to $SM \perp PN$ at $T$ and $\overline{ST}.\overline{SM} = \overline{SE}.\overline{SF} = \overline{SB}.\overline{SC}$
Combine with $\angle SPN = \angle SQN = 90^\circ$ , we see that $T$ lie on circles $(TQN)$ , circle $(BMC)$ , done

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usatstst-2016.pdf (73kb)

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bin_sherlo

730 posts

bin_sherlo

#91 h Aug 12, 2024, 5:07 PM • 2 Y

Y by egxa, ehuseyinyigit

Let $D,E,F$ be the altitudes from $A,B,C$ to $BC,CA,AB$ . $EF\cap BC=T$ . Let $M^*$ be the reflection of $A$ with respect to $BC$ and $K$ be the point on $TM^*$ such that $GK\perp BC$ . $L$ is on $TM^*$ such that $GL\perp AD$ .
$OM$ is the perpendicular bisector of $AG$ hence $PA$ is tangent to $(AGH)$ which means $AP\parallel BC$ .
Invert from $A$ with radius $\sqrt{AH.AD}$ . $PN\leftrightarrow (AQP^*)$ where $TP^*=TP$ and $P^*$ is on $AP$ . $(MBC)\leftrightarrow (M^*EF)$ and $(TQGN)\leftrightarrow (TQNG)$ .
When we invert from $T$ with radius $\sqrt{TB.TC},$ $A,Q,K,P^*$ swap with $G,H,M^*,L$ which are cyclic since $\angle GLM^*=\angle AP^*M^*=\angle TAP^*=\angle GHA$ . Thus, $(AQP^*)$ pass through $K$ . Also $TK.TM^*=TG.TA=TE.TF$ hence $K$ is on $(M^*EF)$ . $\angle TKG=\angle KGT=\angle TNG$ so $K$ also lies on $(TNGQ)$ which gives that $(AQP^*),(M^*EF),(TNQG)$ are concurrent as desired. $\blacksquare$

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Eka01

204 posts

Eka01

#92 h Aug 20, 2024, 1:34 PM • 1 Y

Y by AaruPhyMath

Let $X$ be the $A$ expoint. We make a few observations:-
$P$ is the pole of $AG$ in $\gamma$ since it is the intersection of one tangent and perpendicular bisector of $AG$ $\implies (PAGM)$ is cyclic.
$Q$ is the $A$ humpty point and it lies on $HX$ so $(GXNQ)$ is cyclic and $\Delta AXN$ has orthocenter $H$ .

Now, by radax on $(ABC)$ , $(PAMG)$ and $(MBC)$ , $X$ lies on the radical axis of $(MBC)$ and $(PAMG)$ . By radical axis on the nine point circle, $(MBC)$ and $(PAMG)$ , we get that they are coaxial and that the second intersection $T$ lies on $PN$ as well as the circle with diamter $XN$ which is what we needed to prove.

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L13832

268 posts

L13832

#93 h Sep 5, 2024, 1:29 PM

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Finally got time to latex this!

v_Enhance wrote:

Let $ABC$ be a scalene triangle with orthocenter $H$ and circumcenter $O$ . Denote by $M$ , $N$ the midpoints of $\overline{AH}$ , $\overline{BC}$ . Suppose the circle $\gamma$ with diameter $\overline{AH}$ meets the circumcircle of $ABC$ at $G \neq A$ , and meets line $AN$ at a point $Q \neq A$ . The tangent to $\gamma$ at $G$ meets line $OM$ at $P$ . Show that the circumcircles of $\triangle GNQ$ and $\triangle MBC$ intersect at a point $T$ on $\overline{PN}$ .

Proposed by Evan Chen

Let $PN\cap RM=T$ . Note that $G$ is the $A-$ Queue Point, so we have the following claims
Claim I: ${G-H-N}$ is collinear. [Coxeter]
Claim II: $\odot(GAND)$ [a**]
Claim III: $EF, BC, AG$ concur at $R$ , the radical center of $\odot(AGHQ) \odot(AGBC), \odot(BHQC)$ .
Claim IV: $\odot(RGQN)$ with $RN$ as the diameter.
Proof: We need to show $\angle RQN=90^{\circ}$ , which is true since $\overline{R-H-Q}$ by Brocard's Theorem on $BCEF$ , we get $RH \perp AN$ .
Claim V: $\odot(AMGP)$
Proof: $P$ is the pole of $AG$ w.r.t $\gamma$ and it lies on the perpendicular bisector of $AG$ .
So we also have $PN$ is the polar of $R$ w.r.t $\gamma\implies PN\perp RM$ where $PN\cap RM=T$ .

Claim VI: $\overline{R-T-M}\iff \angle MTN=90^{\circ}\iff T \in (AMGP)$
Proof: $\angle APT=\angle RNT=\angle TGA\implies \boxed{T \in \odot (RGTQN)}$ .
Claim VII: $\boxed{T \in \odot(MBC)}$
Proof: $RM \cdot RT=RE\cdot RF=RC\cdot RB$

Figure

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EeEeRUT

82 posts

EeEeRUT

#94 h Jan 6, 2025, 6:17 AM

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Let $S,E,F$ be feel of altitudes from $A,B,C$ , respectively.
Denote $X : EF \cap AH$ and $I$ be midpoint of $EF$ . Also, let $\gamma_1$ be circle with diameter $MX$ .
Note that $MIN$ collinear and $E,F$ lies on $\gamma$ with their tangent intersect at $N$ .
By Brokard, $X$ is orthocenter of $\triangle MBC$ .
Note that $\angle MIX = 90^{\circ}$ , so $I \in \gamma_1$ .But $I$ also lies on the median of $\triangle MBC$ , so $I$ is humpty point of $\triangle MBC$ .
Let $Y$ be midpoint of $GA$ , $R$ be $EF \cap BC$ . It is well-known that $AGR$ are collinear.
and $PA$ is tangent to the circumcircle of $ABC$ ( the line $OM$ is coaxis of $\gamma$ and the circumcircle $ABC$ , so reflecting $G$ across perpendicular bisector of this line will give point $A$ .)
Note that $\angle MRY = 180^{\circ} - \angle ASR$ , so $MYR$ is cyclic.
Also, it is well known that $(R,S;B,C) = -1 \rightarrow NR \times NS = NB^2 = NI \times NM$
Thus, $SRYMI$ is cyclic.
Inverting this circle about $\gamma$ gives $PXN$ collinear.
Let $T’$ be $PN \cap (MBC)$ , so $T’$ is queue point of $\triangle MBC$ , that is $NX \times NT = NB^2 = NI \times NM$ .
Inverting about $(BCEF)$ map $(MBC)$ to $(IBC)$ and $(GNQ)$ to $AH$ ( $Q$ is humpty point so we have $NQ \times NA = NB^2$ ). Also, it maps $T’$ to $X$ .
Since $X$ already lies on $AH$ , it is suffice to show that $X \in (IBC)$ , which is a well known property of $M$ humpty point that it lies on the circle with orthocenter and the remaining vertex of triangle $MBC$ .
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iStud

268 posts

iStud

#95 h Jan 6, 2025, 12:29 PM

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It's a really nice problem! Evan Chen's medium geometry problems are the best!

Let $D,E,F$ be the foot of altitudes in $\triangle{ABC}$ .

Firstly, note that $G$ is the $A$ -queue point and $Q$ is the $A$ -humpty point. It's well known that $\overline{G,H,N}$ and $BHQC$ is cyclic. Radical Axis Theorem at $(ABC),(BCEF),(AGFHQE),(BHQC)$ tells us that lines $AG,EF,QH,BC$ are concurrent, say the intersection point to be $R$ . Notice that $BDHF$ and $AGFH$ cyclic easily implies $RGFB$ is cyclic by Miquel Point. Moreover, $P$ lies on the perpendicular bisector of $AG$ which is $OM$ and $PG$ is tangent to $(AGFHQE)$ at $G$ , so $PA$ must be tangent to the same circle at $A$ and therefore is parallel to $BC$ .

One can show that $R$ lies on $(GQN)$ easily. Let $RM$ hits $(MBC)$ at $J$ . Observe that $AD,BE,CF$ are concurrent at $H$ and $R=EF\cap BC$ , so by Ceva-Menelause, $(R,D;B,C)=-1$ . Since that $N$ is the midpoint of $BC$ , we have $RD\times RN=RB\times RC$ . But by Power of Point's Theorem, we have $RJ\times RN=RB\times RC=RD\times RN$ . This results in $J$ lying on the nine-point circle of $\triangle{ABC}$ , so $\angle{RJN}=180^\circ-\angle{MJN}=180^\circ-\angle{MDN}=90^\circ$ , so $J$ lies on $(RGQN)$ .

Lastly, we claim that $J$ lies on $(APGM)$ . Indeed,
$\begin{align*} \angle{GJM}&=180^\circ-\angle{GJR}\\ &=180^\circ-\angle{GNR}\\ &=180^\circ-\angle{HND}\\ &=90^\circ+\angle{GHA}\\ &=90^\circ+\frac{180^\circ-2\angle{GPM}}{2}\\ &=180^\circ-\angle{GPM} \end{align*}$
For the final act, simply see that $\angle{MJN}+\angle{PJM}=\angle{MDN}+\angle{PGM}=90^\circ+90^\circ=180^\circ$ , so $\overline{P,J,N}$ , as desired. $\blacksquare$

P.S. I was about to invert when I suddenly realized that synthetic approach is more than enough for this problem

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Cali.Math

128 posts

Cali.Math

#96 h Mar 10, 2025, 4:59 PM

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We uploaded our solution https://calimath.org/pdf/USATSTST2016-2.pdf on youtube https://youtu.be/FA4RxRnKcFk.

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waterbottle432

8 posts

waterbottle432

#97 h Apr 21, 2025, 10:36 AM

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Let $E=AH \cap ABC$ , $D$ be reflection of $E$ w.r.t. $ON$ and $L=AH \cap BC.$ Since $AH\perp BC$ and $ON\perp BC \Longrightarrow AH\parallel ON.$ Since $ON\perp ED \Longrightarrow AH\perp ED$ Since $\angle ECB = \angle EAB = \angle HCB$ it follows that $BC$ bisects segment $HE.$ Therefore, $NH=NE=ND.$ Since $\angle HED = 90^\circ \Longrightarrow H,N,D$ collinear and $A,O,D$ collinear. Since $AM=MH,AO=OD \Longrightarrow MO\parallel HD.$ Since $\angle AGH = 90^\circ = \angle AGD \Longrightarrow G,H,N,D$ collinear. Let $F=HQ\cap BC.$ By P.O.P. $FH \cdot HQ=AH\cdot HL=GH\cdot HN \Longrightarrow G,F,Q,N$ concyclic. $\angle AGH+\angle FGH = 90^\circ+90^\circ=180^ \circ \Longrightarrow A,G,F$ collinear. Let $B'$ be a point that lies on $AC$ s.t. $BB'\perp AC.$ Define $C'$ similarly. Clearly $A,G,H,Q,B',C'$ concyclic. Since $\angle GAM=\angle GAH = \angle HGS = \angle OPS = \angle MPG \Longrightarrow P,G,M,A$ concyclic.

Let $T=(PAMG)\cap(MBC)$ . By Radical Axis concurrence theorem on $(GTMA),(BTMC),(AGBC) \Longrightarrow F,T,M$ collinear. By Radical Axis concurrence theorem on $(BCB'C'),(AGHQ),(AGBC) \Longrightarrow F,C',B'$ collinear. By P.O.P. $FT\cdot FM=FB\cdot FC=FB'\cdot FC' \Longrightarrow T$ lies on the nine point circle of $\triangle ABC \Longrightarrow M,T,C',L,N,B'$ comcyclic. Therefore, $\angle FTN = 180^\circ -\angle MTN = 180^\circ -\angle MLN=90^\circ \Longrightarrow T$ lies on $(GNQ).$ Thus, $T=(GNQ)\cap(MBC)$ .

Since $OM\parallel NH \Longrightarrow OM\perp AG$ and since $O$ is center of $(AGBC) \Longrightarrow OM$ is perpendicular bisector of $AG$ . Thus, $PM$ is diameter of $(PGTMA)$ . Since $\angle FTN = 90^\circ = \angle PTM$ and $F,T,M$ collinear $\Longrightarrow T$ lies on $PN$ and we are done.

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alexanderchew

10 posts

alexanderchew

#98 h May 1, 2025, 1:38 AM

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Solution. Let $X=EF \cap BC$ , where E and F are such that BE and CF are altitudes of ABC. We claim that T is the intersection of $PN$ and $MX$ .
First, we show that $T$ lies on $(XGQN)$ . Note that since $B$ is the pole of $PTN$ , the conclusion follows.
Next, we show that $T$ lies on $(MBC)$ . By Power of a Point theorem, $XT\times XM = XE\times XF$ since T lies on the nine-point circle. Since $XE\times XF = XB\times XC$ , we have $XT\times XM = XB\times XC$ which implies that $T$ lies on $(MBC)$ . Thus we are done.

This post has been edited 1 time. Last edited by alexanderchew, May 1, 2025, 1:48 AM
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Markas

150 posts

Markas

#99 h May 11, 2025, 5:48 PM

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Let DEF be the orthic triangle in $\triangle ABC$ . Let $(ABC) = \omega_1$ , $(BFEC) = \omega_2$ , $(AGFHE) = \gamma$ . Now $rad(\omega_1,\omega_2) = BC$ , $rad(\omega_1,\gamma) = AG$ , $rad(\omega_2,\gamma) = FE$ $\Rightarrow$ their radical center is $AG \cap BC \cap FE = T$ . By Brokard on BFEC we get that H is the orthocenter of $\triangle ATN$ $\Rightarrow$ G, H, N lie on one line, also T, H, Q lie on one line and $\angle TGN = \angle TQN = 90^{\circ}$ since $\angle HGA = \angle HQA = 90^{\circ}$ from AH diameter. Let $PN \cap TM = R$ . It is enough to prove that $R \in (GNQ)$ and $R \in (MBC)$ . To show that $R \in (GNQ)$ we need to show $R \in (TGQN)$ - it is enough to prove $\angle TRN = 90^{\circ}$ . Wrt. $\gamma$ AG is the polar of P and EF is the polar of N $\Rightarrow$ $T \in AG$ , T $\in$ polar of P and $T \in EF$ , T $\in$ polar of N and from La Hire P $\in$ polar of T and N $\in$ polar of T $\Rightarrow$ PN $\equiv$ polar of T $\Rightarrow$ $\angle TRN = 90^{\circ}$ $\Rightarrow$ $R \in (GNQ)$ . It is left to show that $R \in (MBC)$ . Now $R \in (GPAM)$ from $\angle PGM = \angle PRM = \angle PAM = 90^{\circ}$ $\Rightarrow$ TR.TM = TG.TA = TB.TC $\Rightarrow$ TR.TM = TB.TC $\Rightarrow$ RMBC is cyclic $\Rightarrow$ $R \in (MBC)$ $\Rightarrow$ $R \in (GNQ)$ , $R \in (MBC)$ and $R \in PN$ $\Rightarrow$ we are ready.

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