Cute orthocenter geometry

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MarkBcc168

1595 posts

MarkBcc168

#1 h Jul 28, 2020, 7:49 AM • 7 Y

Y by KKSingla2003, Amir Hossein, mira74, giveandtake, centslordm, brickmaster8, Rounak_iitr

Let acute scalene triangle $ABC$ have orthocenter $H$ and altitude $AD$ with $D$ on side $BC$ . Let $M$ be the midpoint of side $BC$ , and let $D'$ be the reflection of $D$ over $M$ . Let $P$ be a point on line $D'H$ such that lines $AP$ and $BC$ are parallel, and let the circumcircles of $\triangle AHP$ and $\triangle BHC$ meet again at $G \neq H$ . Prove that $\angle MHG = 90^\circ$ .

Proposed by Daniel Hu.

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GeoMetrix

924 posts

GeoMetrix

#2 h Jul 28, 2020, 7:49 AM • 4 Y

Y by amar_04, centslordm, brickmaster8, PRMOisTheHardestExam

Let $O$ denote the circumcenter of $\triangle{BHC}$ and let $E$ denote the circumcenter of $\triangle{AHP}$ . We begin with a crucial claim.

Claim: If $H'$ is the antipode of $H$ w.r.t $\odot(BHC)$ then $\overline{AM} \cap \overline{PG}=H'$

Proof: Let $H_A$ denote the $A$ -humpty point. Then notice that $\angle HH_AH'=\angle HH_AM=90^\circ$ and hence $H' \in \overline{AM}$ . Now just note that $\angle HGH'+\angle HGP=180^\circ$ and so $H' \in \overline{PG}$ $\qquad \blacksquare$

Now notice that its well known that $H'$ is the reflection of $A$ in $M$ (just angle chase) Also clearly $\overline{AP} \parallel \overline{BC}$ and hence we have that $\overline{AM}=\tfrac{1}{2}\overline{AP}$ . But notice that clearly if $X=\overline{MH} \cap \overline{AP}$ then $X$ Is the midpoint of $\overline{AP}$ and hence we get that $MXPF$ is a parallelogram which implies that $\overline{MH} \parallel \overline{PG}$ with which we are done $\qquad \blacksquare$

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Omega18

161 posts

Omega18

#3 h Jul 28, 2020, 7:49 AM • 7 Y

Y by zephyr7723, Amir Hossein, indulged, Braman, thedragon01, PRMOisTheHardestExam, Mathlover_1

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(25.15639756385678cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.693141418188027, xmax = 10.463256145668753, ymin = -10.172937052451232, ymax = -0.3526248182641615; /* image dimensions */ pen qqttcc = rgb(0.,0.2,0.8); /* draw figures */ draw((-2.685950600670045,-2.6026295254967415)--(-3.624326490091042,-5.644785490957004), linewidth(0.8)); draw((-3.624326490091042,-5.644785490957004)--(0.33825647657857444,-5.690281506463871), linewidth(0.8)); draw((0.33825647657857444,-5.690281506463871)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw(circle((-1.63070852463975,-4.593929414274443), 2.2536217708548194), linewidth(0.8)); draw(circle((-1.655361488872718,-6.741137583146432), 2.253621770854819), linewidth(0.8)); draw(circle((-5.254956026365177,-3.6468793463522564), 2.77312938133218), linewidth(0.8)); draw((-7.799308487827341,-2.543920998335784)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw((-1.6430350067562338,-5.667533498710437)--(-5.242629544248693,-2.5732752619162627), linewidth(0.8) + qqttcc); draw((-0.6001194128424232,-8.732437471924131)--(-7.799308487827341,-2.543920998335784), linewidth(0.8) + qqttcc); draw((-0.6001194128424232,-8.732437471924131)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); draw((-0.5650721123368723,-5.67991002608484)--(-7.799308487827341,-2.543920998335784), linewidth(0.8)); draw((-2.720997901175595,-5.655156971336034)--(-2.685950600670045,-2.6026295254967415), linewidth(0.8)); /* dots and labels */ dot((-2.685950600670045,-2.6026295254967415),linewidth(2.pt) + dotstyle); label("$A$", (-2.6739736932342324,-2.4396740976359865), NE * labelscalefactor); dot((-3.624326490091042,-5.644785490957004),linewidth(2.pt) + dotstyle); label("$B$", (-3.885207650012522,-5.551613648127904), NE * labelscalefactor); dot((0.33825647657857444,-5.690281506463871),linewidth(2.pt) + dotstyle); label("$C$", (0.38206275156022085,-5.6261511223911835), NE * labelscalefactor); dot((-2.710603564903013,-4.749837694368731),linewidth(2.pt) + dotstyle); label("$H$", (-2.6367049561025926,-4.675798325534371), NE * labelscalefactor); dot((-2.720997901175595,-5.655156971336034),linewidth(2.pt) + dotstyle); label("$D$", (-2.6553393246684127,-5.588882385259544), NE * labelscalefactor); dot((-1.6430350067562338,-5.667533498710437),linewidth(2.pt) + dotstyle); label("$M$", (-1.574545947850862,-5.588882385259544), NE * labelscalefactor); dot((-0.5650721123368723,-5.67991002608484),linewidth(2.pt) + dotstyle); label("$D'$", (-0.4937525710333115,-5.607516753825364), NE * labelscalefactor); dot((-7.799308487827341,-2.543920998335784),linewidth(2.pt) + dotstyle); label("$P$", (-7.910231260229606,-2.4396740976359865), NE * labelscalefactor); dot((-3.782515686424972,-5.996808283772401),linewidth(2.pt) + dotstyle); label("$G$", (-4.220626284197279,-6.0920103365366804), NE * labelscalefactor); dot((-5.242629544248693,-2.5732752619162627),linewidth(2.pt) + dotstyle); label("$T$", (-5.1709790810540905,-2.4955772033334465), NE * labelscalefactor); dot((-0.6001194128424232,-8.732437471924131),linewidth(2.pt) + dotstyle); label("$Z$", (-0.549655676730771,-8.551746987224902), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $T = HM \cap AP$
Now $\triangle AHP \sim \triangle DHD'$ , consider the homothety $\Omega$ centered at $H$ that maps $DD' \mapsto AP$
Then $\Omega : M \mapsto T$ . Now as $M$ is midpoint of $DD' \Rightarrow \boxed{ T \mbox{ is midpoint of } AP }$

Now it is well known that $( BHC ) :=$ reflection of $(ABC)$ about $BC$
Let $AM \cap ( BHC ) = Z$ , then $Z$ is reflection of $A$ about $M$ .
Thus, we have $\boxed{ M \mbox{ is midpoint of } AZ }$

Hence by midpoint theorem $\boxed{ MT \parallel PZ }$

Claim : $G,P,Z$ are collinear.
Let $\measuredangle :=$ directed angle mod $180^{\circ}$
Now $B,C,Z,G$ are concyclic and $ABZC$ is a parallelogram thus
$\measuredangle BGZ = \measuredangle BCZ = \measuredangle CBA$ Also $\measuredangle PGB + \measuredangle BGH = \measuredangle PGH = \measuredangle PAH = 90^{\circ}$
$\begin{align*} \Rightarrow \measuredangle PGB & = 90^{\circ} - \measuredangle BGH \\ & = 90^{\circ} - \measuredangle BCH \\ & = \measuredangle ABC \\ & = - \measuredangle BGZ \end{align*}$
$\therefore \measuredangle BGP = \measuredangle BGZ$
Thus we get $\boxed{ G,P,Z \mbox{ are collinear. } }$
Hence $PG \parallel HM$
$\Rightarrow \measuredangle MHG = \measuredangle PGH = \measuredangle PAH = 90^{\circ}$
Hence, proved.

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lilavati_2005

357 posts

lilavati_2005

#4 h Jul 28, 2020, 7:50 AM • 1 Y

Y by char2539

Introduce the following points :
$O_1$ is the center of $\odot(ABC)$
$O_2$ is the center of $\odot(HBC)$
$O_3$ is the center of $\odot(AHP)$
$O_4$ is the center of $\odot(HDD')$

We know that $O_3$ and $O_4$ are the midpoints of $PH$ and $HD'$ respectively.

Lemma(Well Known)
Proof

By Lemma $O_2$ is the reflection of $O_1$ over $BC$ .

Since $M$ is the midpoint of $DD'$ and $O_1O_2 \perp DD',O_1, O_4, O_2$ are collinear.
$O_2O_3 \perp GH$ since $GH$ is the radical axis of $\odot(AHP)$ and $\odot(BHC)$
Hence it suffices to show that $MH \parallel O_2O_3$

$AP \parallel BC \Longrightarrow \frac{DH \div 2}{HP \div 2} = \frac{O_4H}{HO_3} = \frac{D'H}{HP} = \frac{AH}{HD}$

$O_4M = HD \div 2$ since $O_4M$ is the $D'$ midline wrt $\triangle D'HD$
$O_2M =O_1M = O_1B \cos A = AH \div 2$
The last equality above follows from LOS in $\triangle AHB$ .

Hence, $\frac{O_4H}{HO_3} = \frac{O_4M}{MO_2} \Longrightarrow MH \parallel O_2O_3$

This post has been edited 1 time. Last edited by lilavati_2005, Jul 28, 2020, 7:56 AM
Reason: added a diagram

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Severus

742 posts

Severus

#5 h Jul 28, 2020, 7:54 AM • 1 Y

Y by Imayormaynotknowcalculus

A four page coordinate bash also does the trick.

Sketch

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Reason: added sketch

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MP8148

888 posts

MP8148

#6 h Jul 28, 2020, 7:55 AM • 8 Y

Y by Mathasocean, Amir Hossein, Muaaz.SY, Siddharth03, SSaad, PRMOisTheHardestExam, vrondoS, Radin.AmirAslani

$[asy] size(8cm); defaultpen(fontsize(10pt)); pair A = dir(105), B = dir(210), C = dir(330), H = orthocenter(A,B,C), D = foot(A,B,C), M = (B+C)/2, R = dir(75), D1 = 2M-D, A1 = 2M-H, Q = 2*D1-R, P = extension(A,R,H,D1), G = foot(H,P,Q); draw(unitcircle^^circumcircle(H,G,Q)); draw(B--C^^A--P--Q--R--A--D^^A1--H--D1^^P--H--G); dot("$A$", A, dir(95)); dot("$B$", B, dir(180)); dot("$C$", C, dir(0)); dot("$H$", H, dir(50)); dot("$D$", D, dir(270)); dot("$M$", M, dir(240)); dot("$D'$", D1, dir(315)); dot("$A'$", A1, dir(320)); dot("$P$", P, dir(150)); dot("$G$", G, dir(180)); dot("$Q$", Q, dir(285)); dot("$R$", R, dir(75)); [/asy]$
Let $A'$ be the antipode of $A$ on $(ABC)$ , $R$ be the point on $(ABC)$ such that $\overline{AR} \parallel \overline{BC}$ , and $Q$ be the reflection of $A$ over $M$ . It is well known that $M$ is the midpoint of $\overline{A'H}$ , so by symmetry $Q$ is the antipode of $H$ on $(BHC)$ , and $R$ , $D'$ , $A'$ , $Q$ are collinear with $D$ the midpoint of $\overline{RQ}$ .

We can redefine $G \ne Q = \overline{PQ} \cap (BHC)$ , and it suffices to prove $\overline{HA'} \parallel \overline{PQ}$ . But this is immediate from $\dfrac{D'A'}{D'H} = \dfrac{DH}{D'H} = \dfrac{D'R}{D'P} = \dfrac{D'Q}{D'P}.$ $\blacksquare$

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Idio-logy

206 posts

Idio-logy

#7 h Jul 28, 2020, 8:03 AM • 1 Y

Y by Nathanisme

Solution

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Mathematicsislovely

245 posts

Mathematicsislovely

#8 h Jul 28, 2020, 8:19 AM

Y by

In the following solution $(XYZ)$ denote the circumcircle of $XYZ$ .

Let $AH$ cut $(BHC)$ again at $X$ .Let perpendicular on $BC$ at $D'$ cut $(BHC)$ at $R$ and $Y$ where $R,H$ lies same side on $BC$ .As the perpendicular line on $BC$ at $D$ and $D'$ are symmetric about the diametre of $(BHC)$ passing through $M$ , so $HRYX$ ; $HRD'D$ ; $D'DXY$ are rectangle.

$\textcolor{blue}{CLAIM1:}$ $P,G,Y$ are collinear.
$\textcolor{red}{Proof:}$ Let $PG$ cut $(BHC)$ again at $Y'$ .As $G,H$ are common point of $(BHC)$ and $(PAH)$ and $AH\cap (BHC)= X$ ,so by Reim's theorem $AP||XY'$ .On the other hand $XY|| DD'||AP$ and $Y$ lies on $(BHC)$ .So $Y'\equiv Y$ . $\square$

Now as $PAHG$ is concyclic so $\angle PGH=180^{\circ}-\angle PAH=90^{\circ}$ .So it is enough to show that $PG||HM$ which will prove that $90^{\circ}=\angle PGH=\angle MHG$ .

We will prove that $PY||HM$

Let the perpendicular line on $BC$ at $D'$ cut $PA$ at $S$ .Also let $MH\cap PS=Q$ and $HM\cap YS=V$

$\textcolor{blue}{CLAIM2:}$ $D'S=D'Y$
$\textcolor{red}{Proof:}$ Let $X'$ be the reflection of $A$ over $BC$ .Then $\angle BXC=\angle BAC=180^{\circ}-\angle BHC$ So $B,H,C,X'$ are concyclic.So $X'\equiv X$
Now as $ASD'D$ and $DD'YX$ are rectangle so
$SD'=AD=DX=D'Y$ . $\square$

$\textcolor{blue}{CLAIM3:}$ $RD'=D'V$
$\textcolor{red}{Proof:}$ $DH||D'V$ and $M= HV\cap DD'$ is the midpoint of $DD'$ .SO $DHD'V$ is a parallegram.So $HD=D'V$ .But since $HDD'R$ is rectangle $HD=RD'$ $\square$

$\textcolor{blue}{CLAIM4:}$ $QV||PY$ .
$\textcolor{red}{Proof:}$ On one hand we have,
$\frac{SY}{SP}=2\frac {SD'}{SP}=2\frac{RD'}{RG}$ .[The first equality by $D'$ =midpoint of $SY$ ]
On the other hand ,
$\frac{SV}{SQ}=\frac{RV}{RH}=2\frac{RD'}{RH}$ [AS, $D'$ is the midpoint of $RV$ ]

SO $\frac{SY}{SP}=\frac{SV}{SQ}$ which implies
$PY||QV$ .So $PG||HM$ .SO $\angle MHG=90^{\circ}$ $\blacksquare$ .

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srijonrick

168 posts

srijonrick

#9 h Jul 28, 2020, 8:28 AM • 3 Y

Y by Aritra12, Amir Hossein, A-Thought-Of-God

MarkBcc168 wrote:

Let acute scalene triangle $ABC$ have orthocenter $H$ and altitude $AD$ with $D$ on side $BC$ . Let $M$ be the midpoint of side $BC$ , and let $D'$ be the reflection of $D$ over $M$ . Let $P$ be a point on line $D'H$ such that lines $AP$ and $BC$ are parallel, and let the circumcircles of $\triangle AHP$ and $\triangle BHC$ meet again at $G \neq H$ . Prove that $\angle MHG = 90^\circ$ .

Proposed by Daniel Hu.

Another $MH$ line configuration!

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.7655891980115, xmax = 9.257249900918303, ymin = -8.548927726358093, ymax = 5.1113621322226335; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen zzttff = rgb(0.6,0.2,1); pen ffvvqq = rgb(1,0.3333333333333333,0); pen ccqqqq = rgb(0.8,0,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((-3.394872238111926,2.523862673503301)--(-5.9650112473135035,-4.858490859924286)--(7.0349887526864965,-4.858490859924286)--cycle, linewidth(0.8) + zzttqq); draw((-5.9650112473135035,-4.858490859924286)--(-3.394872238111926,-1.2273734294705232)--(7.0349887526864965,-4.858490859924286)--cycle, linewidth(0.8) + ffvvqq); draw((-5.264643222971541,-3.250976410057223)--(-3.394872238111926,-1.2273734294705232)--(0.5349887526864965,-4.858490859924286)--cycle, linewidth(1.2) + fuqqzz); /* draw figures */ draw((-3.394872238111926,2.523862673503301)--(-5.9650112473135035,-4.858490859924286), linewidth(0.8) + zzttqq); draw((-5.9650112473135035,-4.858490859924286)--(7.0349887526864965,-4.858490859924286), linewidth(0.8) + zzttqq); draw((7.0349887526864965,-4.858490859924286)--(-3.394872238111926,2.523862673503301), linewidth(0.8) + zzttqq); draw((-3.394872238111926,2.523862673503301)--(-3.394872238111926,-4.858490859924286), linewidth(0.8) + blue); draw((4.464849743484919,-4.858490859924286)--(-3.394872238111926,-1.2273734294705232), linewidth(0.8)); draw((xmin, 0*xmin + 2.5238626735033014)--(xmax, 0*xmax + 2.5238626735033014), linewidth(0.8) + zzttqq); /* line */ draw((-11.514596653914866,2.5238626735033014)--(-3.394872238111926,-1.2273734294705232), linewidth(0.8)); draw(circle((-7.454734446013396,0.6482446220163888), 4.472183384233048), linewidth(0.8) + zzttff); draw(circle((0.5349887526864965,-6.7341089114112), 6.765200889483147), linewidth(0.8) + blue); draw((-5.9650112473135035,-4.858490859924286)--(-3.394872238111926,-1.2273734294705232), linewidth(0.8) + ffvvqq); draw((-3.394872238111926,-1.2273734294705232)--(7.0349887526864965,-4.858490859924286), linewidth(0.8) + ffvvqq); draw((7.0349887526864965,-4.858490859924286)--(-5.9650112473135035,-4.858490859924286), linewidth(0.8) + ffvvqq); draw((-7.454734446013397,0.6482446220163874)--(0.5349887526864965,-6.734108911411201), linewidth(0.8) + ccqqqq); draw((-5.264643222971541,-3.250976410057223)--(-3.394872238111926,-1.2273734294705232), linewidth(0.8) + fuqqzz); draw((-3.394872238111926,-1.2273734294705232)--(0.5349887526864965,-4.858490859924286), linewidth(0.8) + fuqqzz); draw((0.5349887526864965,-4.858490859924286)--(-5.264643222971541,-3.250976410057223), linewidth(0.8) + fuqqzz); draw((0.5349887526864956,-2.9828728084373735)--(0.5349887526864965,-6.734108911411201), linewidth(1.2)); draw((-7.454734446013397,2.5238626735033014)--(-7.454734446013397,0.6482446220163874), linewidth(0.8)); draw((-7.454734446013397,2.5238626735033014)--(-3.394872238111926,-1.2273734294705232), linewidth(0.8) + fuqqzz); /* dots and labels */ dot((-3.394872238111926,2.523862673503301),linewidth(4pt) + dotstyle); label("$A$", (-3.3084654497633146,2.7032982148446205), NE * labelscalefactor); dot((-5.9650112473135035,-4.858490859924286),linewidth(4pt) + dotstyle); label("$B$", (-5.869769798247198,-4.6741339683952905), NE * labelscalefactor); dot((7.0349887526864965,-4.858490859924286),linewidth(4pt) + dotstyle); label("$C$", (7.111883865436075,-4.6741339683952905), NE * labelscalefactor); dot((-3.394872238111926,-1.2273734294705232),linewidth(3pt) + dotstyle); label("$H$", (-3.3084654497633146,-1.1058210726442357), NE * labelscalefactor); dot((-3.394872238111926,-4.858490859924286),linewidth(3pt) + dotstyle); label("$D$", (-3.3084654497633146,-4.717916948711254), NE * labelscalefactor); dot((0.5349887526864965,-4.858490859924286),linewidth(3pt) + dotstyle); label("$M$", (0.6320027786734296,-4.717916948711254), NE * labelscalefactor); dot((4.464849743484919,-4.858490859924286),linewidth(3pt) + dotstyle); label("$D'$", (4.550579516952192,-4.717916948711254), NE * labelscalefactor); dot((-11.514596653914866,2.5238626735033014),linewidth(3pt) + dotstyle); label("$P$", (-11.49588276884855,1.9808790396312168), NE * labelscalefactor); dot((-5.264643222971541,-3.250976410057223),linewidth(3pt) + dotstyle); label("$G$", (-5.760312347457289,-3.64523393097014), NE * labelscalefactor); dot((-7.454734446013397,0.6482446220163874),linewidth(3pt) + dotstyle); label("$X$", (-7.358391128989968,0.7768470809422106), NE * labelscalefactor); dot((0.5349887526864965,-6.734108911411201),linewidth(3pt) + dotstyle); label("$O'$", (0.6320027786734296,-6.600585102297701), NE * labelscalefactor); dot((0.5349887526864956,-2.9828728084373735),linewidth(3pt) + dotstyle); label("$O$", (0.52254532788352,-2.879031775440772), NE * labelscalefactor); dot((-2.3648272351425224,-4.054733634990755),linewidth(3pt) + dotstyle); label("$E$", (-2.695503725339821,-4.411436086499507), NE * labelscalefactor); dot((-7.454734446013397,2.5238626735033014),linewidth(3pt) + dotstyle); label("$F$", (-7.358391128989968,2.659515234528657), NE * labelscalefactor); dot((-4.329757730541734,-2.2391749197638733),linewidth(3pt) + dotstyle); label("$L$", (-4.2497995265565365,-2.1128296199114045), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$

Claim 1: $\overline{MH}$ passes through the midpoint of $AP.$
Proof

Now, let us denote the centers of $\odot(APH)$ and $\odot(ABC)$ by $X$ and $O$ respectively. Also, we note $X$ is the midpoint of $PH$ , since $\triangle APH$ is right-angled at $A$ .

Claim 2: $O'$ , the reflection of $O$ over $BC$ is the circumcenter of $\triangle BHC.$
Proof

Now, let us denote the midpoint of $GH$ by $L$ . Since $\overline{GH}$ is the radical axis of $\odot(APH)$ and $\odot(BHC)$ , so $\overline{O'X}$ must pass through $L$ and $\overline{O'X} \perp \overline{GH}.$

Next, in $\triangle APH$ , we have $XF \parallel AH$ and $XF = \frac{AH}{2}$ by Midpoint theorem. Also, we have $O'M \parallel AH$ and $O'M = \frac{AH}{2}.$

Hence, $XFMO'$ is a parallelogram. So, $XO' \parallel FM\implies \angle GLO' = \angle GHM.$ By Radical Axis theorem, $\angle GLO' = 90^{\circ} \implies \angle MHG = 90^{\circ}. \quad\blacksquare$

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Kimchiks926

256 posts

Kimchiks926

#10 h Jul 28, 2020, 9:52 AM • 2 Y

Y by Amir Hossein, mkomisarova

Let $O$ denote the circumcenter of $\odot(ABC)$ and $O'$ the circumcenter of $\odot(BHC)$ . It is well - known that $\odot(BHC)$ is reflection of $\odot(ABC)$ over $BC$ , therefore $O'$ is reflection of $O$ over the midpoint of $BC$ . Let $J$ denote the midpoint of $HP$ , which is obviously the center of $\odot(AHP)$ .

Claim: $HM \parallel JO'$
Proof: Assume that $OO'$ intersects $HD'$ at point $K$ . Clearly $KM$ is the midline in $\triangle HDD'$ . It is well - known that $AH=OO'$ (this can be simply proved using complex numbers). Also note that $\triangle APH \sim \triangle DD'H$ . Therefore:
$\frac{PH}{HD'}=\frac{2JH}{2HK} =\frac{AH}{HD}=\frac{OO'}{HD}=\frac{2MO'}{2KM}$ As a result:
$\frac{JH}{HK}=\frac{O'M}{MK}$ This implies that $HM \parallel JO'$ as desired.

Now to finish note that $GH$ is radical axis of $\odot(BHC)$ and $\odot(AHP)$ . Therefore $GH \perp JO'$ . Since $HM \parallel JO'$ , we conclude that $GH \perp HM$ , which implies that $\angle MHG=90$ as desired.

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Steve12345

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#11 h Jul 28, 2020, 11:01 AM • 1 Y

Y by ehuseyinyigit

The $A-Queue$ point is the orthocenter of triangle $APG$ . The problem statement follows easily from this fact.

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Googolplexian

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#12 h Jul 28, 2020, 11:13 AM • 3 Y

Y by Amir Hossein, ihatemath123, Mango247

I didn't find a synthetic solution until after the time, so I will write the (fairly horrible) barybash I did (I am sorry).

We use Barycentric coordintates wrt triangle $ABC$ .
Use Conway's Notation: $S_A=\frac {b^2+c^2-a^2}{2}, S_B= \frac {a^2+c^2-b^2}{2}, S_C= \frac {a^2+b^2-c^2}{2}$ , and let $S_{AB}=S_AS_B$ and $S$ be twice the area of triange ABC.
We have $B=(0, 1, 0), C=(0, 0, 1), H=(S_{BC} : S_{CA} : S_{AB})$ .
Let circle $BHC$ have equation $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$
From points $B$ and $C$ , we get $v=0, w=0$ respectively, and then using $H$ , we obtain $-S_AS_BS_C(a^2S_A+b^2S_B+c^2S_C)+uS_BS_C(S_{BC}+S_{CA}+S_{AB})=0$ Now $S_B+S_C=a^2$ , with cyclic varations, and so it follows that $\frac{1}{2} \cdot (a^2S_A+b^2S_B+c^2S_C)=S_{BC}+S_{CA}+S_{AB}$ , (and also by the Conway identites, both are equal to $S^2$ ) and we get $u=2S_A$ .

Since $H=(S_{BC} : S_{CA} : S_{AB}), D(0: S_{CA} : S_{AB})=(0: S_C: S_B)$ , meaning $D'=(0: S_B: S_C)$ .

Let $D'H$ have equation $ux+vy+wz$ =0. By using $D'$ and $H$ , we can find the ratio $u: v: w$ and find that the line has equation $S_A(S_C^2-S_B^2)x-S_BS_C^2y+S_B^2S_Cz=0$ .

The equation of $BC$ is $x=0$ , so the point at infinity on $BC$ is $(0: -1: 1)$ .
Since $AP$ is parallel to $BC$ , it meets it at $(0: -1: 1)$ . Additionally using $A=(1, 0, 0)$ , we find that $AP$ has equation $y+z=0$

P is the intersection of $S_A(S_C^2-S_B^2)x-S_BS_C^2y+S_B^2S_Cz=0$ and $y+z=0$ and we can solve this to find that $P=(-S_{BC} : S_A(S_B-S_C): S_A(S_C-S_B))$

Let circle $AHP$ have equation $-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0$
Using $A=(1, 0, 0)$ gives $u=0$ . Using $H=(S_{BC} : S_{CA} : S_{AB})$ and the identity $\frac{1}{2} \cdot (a^2S_A+b^2S_B+c^2S_C)=S_{BC}+S_{CA}+S_{AB}$ gives $S_Cv+S_Bw=0$ .
Using $P=(-S_{BC} : S_A(S_B-S_C): S_A(S_C-S_B))$ , we get (after some simplifications) that $v-w=\frac{a^2S_A(S_B-S_C)}{S_{BC}}+c^2-b^2$ .
We can solve these equations to find that $v=\frac{S_AS_B}{S_C}-S_A+S_B$ and $w=S_C-S_A+\frac{S_AS_C}{S_B}$ .

We have that circle $BHC$ is $-a^2yz-b^2zx-c^2xy+2S_A(x+y+z)=0$ and that cirlce $AHP$ is $-a^2yz-b^2zx-c^2xy+((\frac{S_AS_B}{S_C}-S_A+S_B)y+(S_C-S_A+\frac{S_AS_C}{S_B})z))(x+y+z)=0$ .

These circles meet at $H$ and $G$ , so we have that $GH$ is their radical axis.

By the Barycentric Radical Axis theorem the two circles $-a^2yz-b^2zx-c^2xy+(u_1x+v_1y+w_1z)(x+y+z)=0, -a^2yz-b^2zx-c^2xy+(u_2x+v_2y+w_2z)(x+y+z)=0$ have radical axis $(u_1-u_2)x+(v_1-v_2)y+(w_1-w_2)z=0$
So GH is $2S_Ax+(S_A-S_B-\frac{S_AS_B}{S_C})y+(S_A-S_C-\frac{S_AS_C}{S_B})z=0$ which we write as $2S_AS_BS_Cx+(S_AS_BS_C-S_B^2S_C-S_AS_B^2)y+(S_AS_BS_C-S_BS_C^2-S_AS_C^2)z=0$
Choose the point $P'$ on this with $x=0$ . We find $P'=(0 : S_AS_C^2+S_BS_C^2-S_AS_BS_C : S_AS_BS_C-S_B^2S_C-S_AS_B^2)$ $=(0 , \frac{S_AS_C^2+S_BS_C^2-S_AS_BS_C}{(S_A+S_B)S_C^2-(S_A+S_C)S_B^2} , \frac{S_AS_BS_C-S_B^2S_C-S_AS_B^2)}{(S_A+S_B)S_C^2-(S_A+S_C)S_B^2}=(0 : \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2} , \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2})$ .

Shift the circumcenter of the triangle to the zero vector so that $\vec H= \vec A+\vec B+ \vec C$ . We have $\overrightarrow{P'H}=\vec A+ \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2}\vec B + \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2}\vec C$

We also have $\vec H= \frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}\vec A + \frac{S_{CA}}{S_{BC}+S_{CA}+S_{AB}}\vec B+ \frac{S_{AB}}{S_{BC}+S_{CA}+S_{AB}}\vec C, \vec M =\frac{1}{2}\vec B+\frac{1}{2}\vec C$ giving $\overrightarrow{MH}=\frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}\vec A+\frac{S_{CA}-S_{BC}-S_{AB}}{2S_{BC}+2S_{CA}+2S_{AB}}\vec B+\frac{S_{AB}-S_{BC}-S_{CA}}{2S_{BC}+2S_{CA}+2S_{AB}}\vec C$
The sum of the coefficients of this vector is $0$ , so by the Generalised perpendicularity criterion, $MH\perp HP' \Leftrightarrow a^2(z_1y_2+y_1z_2)+b^2(x_1z_2+z_1x_2)+c^2(y_1x_2+x_1y_2)=0$ where $(x_1,y_1,z_1)=(1, \frac{S_AS_BS_C-b^2S_B^2}{c^2S_C^2-b^2S_B^2}, \frac{c^2S_C^2-S_AS_BS_C}{c^2S_C^2-b^2S_B^2}), (x_2,y_2.z_2)=(\frac{S_{BC}}{S_{BC}+S_{CA}+S_{AB}}, \frac{S_{CA}-S_{BC}-S_{AB}}{2S_{BC}+2S_{CA}+2S_{AB}},\frac{S_{AB}-S_{BC}-S_{CA}}{2S_{BC}+2S_{CA}+2S_{AB}})$ .

It is sufficient to show that $a^2((c^2S_C^2-S_AS_BS_C)(S_{CA}-S_{BC}-S_{AB})+(S_AS_BS_C-b^2S_B^2)(S_{AB}-S_{BC}-S_{CA}))$ $+b^2((c^2S_C^2-b^2S_B^2)(S_{AB}-S_{BC}-S_{CA})+(c^2S_C^2-S_AS_BS_C)(2S_{BC}))$ $+c^2((S_AS_BS_C-b^2S_B^2)(2S_{BC})+(c^2S_C^2-b^2S_B^2)(S_{CA}-S_{BC}-S_{AB}))=0$
Using $S_B+S_C=a^2$ and cyclic variations, we can express everything in terms of $S_A, S_B, S_C$ , and we show that this expression is equal to 0:

Replace $S_A, S_B, S_C$ with $A, B, C$ respectively.

The above is $(B+C)(((A+B)C^2-ABC)(CA-BC-AB)+(ABC-(A+C)B^2)(AB-BC-CA))$ $+(A+C)(((A+B)C^2-(A+C)B^2)(AB-BC-CA)+((A+B)C^2-ABC)(2BC))$ $+(A+B)((ABC-(A+C)B^2)(2BC)+((A+B)C^2-(A+C)B^2)(CA-BC-AB))$ $=(B+C)(C(AC+BC-AB)(CA-BC-AB)+B(AC-AB-BC)(AB-BC-CA))$ $+(A+C)((AC^2+BC^2-AB^2-B^2C)(AB-BC-CA)+2BC^2(AC+BC-AB))$ $+(A+B)((2B^2C(AC-AB-BC)+((AC^2+BC^2-AB^2-B^2C)(CA-BC-AB))$
Using $AC^2+BC^2-AB^2-B^2C=C(AC+BC-AB)+B(AC-AB-BC)$ , we can factorise the first line in the above expression as $(B+C)(AC+BC-AB)(AC-BC-AB)(C-B)$ , the second line as $(A+C)(AC+BC-AB)(AC-BC-AB)(-B-C)$ and the third as $(A+B)(AC+BC-AB)(AC-BC-AB)(B+C)$ .

So the expression is the sum of these, which is $(B+C)(AC+BC-AB)(AC-BC-AB)(C-B-A-C+A+B)=0$

This implies $MH\perp HP'$ which shows $MH\perp HG$ as required.

This post has been edited 2 times. Last edited by Googolplexian, Jul 29, 2020, 8:39 AM

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#13 h Jul 28, 2020, 12:19 PM • 1 Y

Y by TheRealGraceWu

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Let $A'$ be the antipode of $A$ and let $T'$ be the reflection of $T$ wrt $M$ .

Claim. If $HG\perp HT$ , $HG=AT$ .
pf)
note that $(BHC)$ has the same radius with $(ABC)$ . Since $HT'=TA'$ , by Pythagorean thm, we get $AT=HG$ .

Now, let $N=MH\cap AP, L=MH\cap (AHP)$

note that $N$ is the midpoint of $AP$ since $\triangle AHP\sim \triangle DHD'$

Since $\angle ATN=\angle NLP=90$ and $AN=NP$ , $ATPL$ is a parallelogram.

Redefine $G$ as the intersection of $(AHP)$ and a line perpendicular to $AP$ passing through $T$ .

Note that $AN^2=NP^2=NT\cdot NH$ . Therefore $\angle ATP=180-\angle AHP=180-\angle AGP$ .

Since $GT\perp AP$ , $T$ is the orthocenter of $\triangle AGP$ .

Let $O'$ be the midpoint of $HP$ . (the circumcenter of $(AHP)$ )

It is well-known that $GT=2O'N$ . Since $AH=2O'N$ ( $N, O'$ are midpoints of $AP, PH$ ), $AH=GT$ .

Hence, $AHGT$ is a parallelogram.

Then, we obtain that $AT=HG$ .

Note that since $L$ is the reflection of $T$ wrt $N$ , $L$ is the antipode of $G$ .

$\therefore$ $GL$ is the diameter of $(AHP)$ . $\angle GHL=90$

Now by Claim, we obtain that $B, H, C, G$ are cyclic. We are done.

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#14 h Jul 28, 2020, 12:41 PM

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Phantom Points

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#15 h Jul 28, 2020, 12:48 PM

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Complex numbers work here very well

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One-liner (excluding constructions) $[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.028184276513091, xmax = 13.651231623973653, ymin = -9.68127833226805, ymax = 5.291831456807111; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((1.5,2.51)--(-0.52,-2.77)--(9.84,-2.77)--cycle, linewidth(2) + zzttff); /* draw figures */ draw((1.5,2.51)--(-0.52,-2.77), linewidth(0.8) + zzttff); draw((-0.52,-2.77)--(9.84,-2.77), linewidth(0.8) + zzttff); draw((9.84,-2.77)--(1.5,2.51), linewidth(0.8) + zzttff); draw((-2.6384543058622527,2.51)--(1.5,2.51), linewidth(0.8)); draw((-2.6384543058622527,2.51)--(7.82,-2.77), linewidth(0.8)); draw((-0.52,-2.77)--(1.5,0.4206818181818154), linewidth(0.8)); draw((1.5,0.4206818181818154)--(9.84,-2.77), linewidth(0.8)); draw((1.5,2.51)--(1.5,-2.77), linewidth(0.8)); draw(circle((-0.5692271529311265,1.4653409090909082), 2.3179761919930204), linewidth(0.8) + ccqqqq); draw(circle((4.66,-3.8146590909090947), 5.284289225261901), linewidth(0.8) + ccqqqq); draw((4.66,-2.77)--(1.5,0.4206818181818154), linewidth(0.8) + ffvvqq); draw((1.5,0.4206818181818154)--(0.45538967495481386,-0.6138834613557778), linewidth(0.8)); draw((-2.6384543058622527,2.51)--(4.66,-2.77), linewidth(0.8)); draw((4.66,-2.77)--(0.45538967495481386,-0.6138834613557778), linewidth(0.8)); draw((0.45538967495481386,-0.6138834613557778)--(-2.6384543058622527,2.51), linewidth(0.8) + ffvvqq); draw((xmin, -1.670886075949367*xmin + 5.0163291139240505)--(xmax, -1.670886075949367*xmax + 5.0163291139240505), linewidth(0.8)); /* line */ draw((-0.52,-2.77)--(-2.6384543058622527,2.51), linewidth(0.8)); draw((9.84,-2.77)--(7.82,-8.05), linewidth(0.8)); draw((7.82,-8.05)--(-0.52,-2.77), linewidth(0.8)); draw((1.5,0.4206818181818154)--(2.420659623917179,0.9716826537080058), linewidth(0.8)); draw((0.45538967495481386,-0.6138834613557778)--(7.82,-8.05), linewidth(0.8) + ffvvqq); draw((1.5,-8.05)--(7.82,-8.05), linewidth(0.8)); draw(circle((4.66,-1.7253409090909093), 5.284289225261899), linewidth(0.8) + zzttff); draw((4.66,-2.77)--(7.82,-5.960681818181815), linewidth(0.8) + ffvvqq); draw((7.82,-5.960681818181815)--(7.82,-2.77), linewidth(0.8)); draw((7.82,-5.960681818181815)--(7.82,-8.05), linewidth(0.8)); draw((1.5,-2.77)--(1.5,-8.05), linewidth(0.8)); /* dots and labels */ dot((1.5,2.51),linewidth(1pt) + dotstyle); label("$A$", (1.5793801724103085,2.544820998024576), NE * labelscalefactor); dot((-0.52,-2.77),linewidth(1pt) + dotstyle); label("$B$", (-0.4431706239291322,-2.7330734609945218), NE * labelscalefactor); dot((9.84,-2.77),linewidth(1pt) + dotstyle); label("$C$", (9.919994408743431,-2.7330734609945218), NE * labelscalefactor); dot((1.5,0.4206818181818154),linewidth(1pt) + dotstyle); label("$H$", (1.5793801724103085,0.4644830360754426), NE * labelscalefactor); dot((1.5,-2.77),linewidth(1pt) + dotstyle); label("$D$", (1.5793801724103085,-2.7330734609945218), NE * labelscalefactor); dot((4.66,-2.77),linewidth(1pt) + dotstyle); label("$M$", (4.738411892407149,-2.7330734609945218), NE * labelscalefactor); dot((7.82,-2.77),linewidth(1pt) + dotstyle); label("$D'$", (7.89744361240399,-2.7330734609945218), NE * labelscalefactor); dot((-2.6384543058622527,2.51),linewidth(1pt) + dotstyle); label("$P$", (-2.5620333629514036,2.544820998024576), NE * labelscalefactor); dot((0.45538967495481386,-0.6138834613557778),linewidth(1pt) + dotstyle); label("$G$", (0.539211191435739,-0.5756859448991241), NE * labelscalefactor); dot((7.82,-8.05),linewidth(1pt) + dotstyle); label("$A'$", (7.89744361240399,-8.010967920013618), NE * labelscalefactor); dot((1.5,-8.05),linewidth(1pt) + dotstyle); label("$A''$", (1.5793801724103085,-8.010967920013618), NE * labelscalefactor); dot((2.420659623917179,0.9716826537080058),linewidth(1pt) + dotstyle); label("$H_A$", (2.5039748221654814,1.003829915099292), NE * labelscalefactor); dot((7.82,-5.960681818181815),linewidth(1pt) + dotstyle); label("$A'''$", (7.89744361240399,-5.930629958064485), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Let $A'''$ be the antipode of $A$ on $(ABC)$
Let $A'$ be the "parallelogram point"
Let $A''$ be the reflection of $A$ in $BC$
By Reim's we have $P-G-A'$
now $\frac{DH}{HP}=\frac{DH}{HA}=\frac{DA'''}{DA'}$
implying $PG || HM$ implying the result.

This post has been edited 2 times. Last edited by math_comb01, Nov 24, 2023, 11:35 AM

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bjump

#69 h Dec 23, 2023, 4:57 AM

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12AM solves sus
Let $O$ , $O_{1}$ , and $O_{2}$ denote the centers of $(ABC)$ , $(AHP)$ , and $(BHC)$ respectively. Let $M_{1}$ denote the midpoint of $HD'$ .

Claim: $O_{1}O_{2}$ and $HM$ are homothetic at $M_{1}$ .
Proof: It is well known that $2OM=AH$ . But note that by considering a triangle $A'BC$ where $A$ is reflected about $M$ to $A'$ , by orthocenter reflections $H$ lies on $(A'BC)$ . So $O_2$ is the center of $(A'BC)$ and thus the reflection of $O$ over $BC$ . So $2O_{2}M=2OM=AH$ .
Now
$\frac{HO_1}{HM_1}=\frac{PH}{HD'}=\frac{AH}{HD}=\frac{2O_{2}M}{2M_{1}M}=\frac{O_{2}M}{M_{1}M}$ Thus our claim is proven $\square$

Now Since $HG$ is the radical axis of $(APH)$ and $(HBC)$ . $O_{1}O_{2} \perp HG$ , but since $HM$ and $O_{1}O_{2}$ are homothetic, $HM \perp HG$
And we are finished

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Leo.Euler

577 posts

Leo.Euler

#70 h Feb 11, 2024, 8:42 PM

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Still can't believe I tried to linpop this.

Let $O_1$ and $O_2$ denote the centers of $(APH)$ and $(BHC)$ , respectively. It is equivalent to show that $\overline{O_1O_2} \parallel \overline{MH}$ . We prove the following stronger claim:

Claim: Let $M'$ denote the midpoint of $AP$ . Then $M'O_1O_2M$ is a parallelogram.
Proof. First, note that $\overline{MO_2} \parallel \overline{M'O_1}$ as both lines are parallel to $\overline{AH}$ . Furthermore, since $AHO_2O$ is a parallelogram (well-known), it follows that $M'O_1 = \frac{AH}{2} = \frac{OO_2}{2} = O_2M,$ so $M'O_1O_2M$ is a parallelogram, as desired.
:yoda:

Done.

This post has been edited 2 times. Last edited by Leo.Euler, Feb 11, 2024, 8:47 PM

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ihatemath123

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ihatemath123

#71 h Mar 15, 2024, 1:52 AM

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Let $A'$ be the reflection of $A$ across $M$ . Then, since the foot from $H$ to $\overline{PA'}$ lies on both $(AHP)$ and $(BHC)$ , it follows that $G$ is the foot from $H$ to $\overline{PA'}$ . A homothety of scale factor $\tfrac{1}{2}$ centered at $A$ sends line $PA'$ to line $HM$ , so from $\overline{HG} \perp \overline{A'P}$ it follows that $\overline{HG} \perp \overline{HM}$ .

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blueprimes

326 posts

blueprimes

#72 h Mar 25, 2024, 5:11 PM

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Extend rays $AD$ and $PG$ to intersect $(BHC)$ at $S$ and $T$ respectively. Suppose $AS$ and $PT$ intersect at $N$ . First, we claim that $\triangle HDM \sim \triangle NST$ . Clearly we have $\angle NST = 90^\circ = \angle HDM$ . Since $BC \parallel ST$ , we get $TD' \perp BC$ , so $DD' = ST$ and $\frac{DM}{ST} = \frac{DM}{DD'} = \frac{1}{2}$ . Now by similar triangles from parallel lengths, we have
$\frac{ST}{AP} = \frac{DD'}{AP} \implies \frac{SN}{SA} = \frac{DH}{DA}.$ But $SA = 2(DA)$ since it is well known that $(BHC)$ is a reflection of $(ABC)$ about $D$ , then $\frac{DH}{SN} = \frac{1}{2}$ . Combining this with the earlier facts, by SAS Similarity yields $\triangle HDM \sim \triangle NST.$ Then $PT \parallel HM$ , and $\angle MHG = \angle HGT = 90^\circ$ and we are finished.

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GrantStar

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GrantStar

#73 h May 1, 2024, 4:30 PM

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By homothety $MH$ bisects $AP$ and since $PG \perp GH$ line $PG$ passes through $B+C-A$ and we are done by homothety centered at $A$ with scale factor $2$ .

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Clew28

45 posts

Clew28

#74 h May 21, 2024, 1:31 AM • 1 Y

Y by duckman234

Solution, I think

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Clew28

45 posts

Clew28

#75 h May 21, 2024, 1:51 AM • 1 Y

Y by duckman234

GrantStar wrote:

By homothety $MH$ bisects $AP$ and since $PG \perp GH$ line $PG$ passes through $B+C-A$ and we are done by homothety centered at $A$ with scale factor $2$ .

Very slick solution :coolspeak:

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Pyramix

419 posts

Pyramix

#76 h Jun 2, 2024, 2:19 PM • 1 Y

Y by GeoKing

Let $O_1,O_2$ be the centers of circles $(AHP)$ and $(BHC)$ . Let $N$ be the mid-point of $\overline{DH'}$ . Then, we have $\triangle HAP \sim \triangle HDD'$ . Since both triangles are right-angled, $O_1\in HP$ and $N$ is the circumcenter of $HDD'$ . Hence, the homothety at $H$ with ratio $-\frac{AH}{HD}$ sends $HDD'$ to $HAP$ . Hence, $\frac{AH}{HD}=\frac{HO_1}{NH}=\frac{AH/2}{HD/2}=\frac{OM}{NM}=\frac{MO_2}{NM}$ and the conclusion follows by Thales' Theorem. $\blacksquare$ .
Note, we have $AH=2OM$ and $OM=MO_2$ as $O_2$ is the reflection of $O$ in $BC$ .

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bjump

999 posts

bjump

#77 h Jul 9, 2024, 4:32 AM • 1 Y

Y by GeoKing

7 minute speedrun b4 mom make me sleep

Let $F=\frac{P+A}{2}$ , $FHM$ collinear. $FM \perp HG \perp FD$ , let $PG$ intersect $(AGM)$ at $I$ , and $(BHC)$ at $J$ . Let $FM$ intersect $(BHC)$ at $G'$ (sus why i named it this ur boutta see why). By reim $GHG'J$ rectangle so $\angle GHM = \angle GHG' = 90^{\circ}$ QED

This post has been edited 2 times. Last edited by bjump, Jul 9, 2024, 4:36 AM

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Om245

163 posts

Om245

#78 h Jul 9, 2024, 10:29 AM • 1 Y

Y by GeoKing

Interesting....

Let $X$ be reflection of $A$ above midpoint $M$ with $H' = HM \cap XD'$ . Observe $XD' \perp BC$ and as everything is just reflection above midpoint $M$ , we get $H'$ is orthocenter of $\triangle XBC$ . Hence $\frac{D'H}{PH} = \frac{DH}{HA} = \frac{H'D'}{H'X} \Rightarrow HM \parallel PX$ Now if $G = PX \cap (BHC)$ .
Key observation is $XH$ is diameter in $(BHC)$ . Because If $H_A$ be $A$ -humpty point, then we know $X \in AH_A$ and $\measuredangle XH_AH = 90$ .

$HG\perp GX \parallel HM \Rightarrow GH \perp HM \Rightarrow \measuredangle GHM = 90$

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fearsum_fyz

48 posts

fearsum_fyz

#79 h Oct 16, 2024, 2:52 PM

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https://i.imgur.com/09gkjV4_d.webp?maxwidth=1520&fidelity=grand

Let $H'$ be the antipode of $H$ in $(BHC)$ .

Claim 1: $H'$ is the reflection of $A$ over $M$ .
Proof. Angle chasing shows that $\square{ABH'C}$ is a parallelogram.

By a homothety centered at $M$ , $MH$ bisects $AP$ . Hence by midpoint theorem, $MH$ is parallel to $H'P$ . Since $HP$ and $HH'$ are both diameters, $H'P$ is in turn parallel to the line joining the centers of $(BHC)$ and $(AHP)$ . Therefore $MH$ is parallel to the line joining the centers of $(BHC)$ and $(AHP)$ .
However, it is well known that the line joining the centers of two circles is perpendicular to their radical axis. Hence $MH \perp HG$ and we are done!

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Maximilian113

544 posts

Maximilian113

#80 h Jan 14, 2025, 1:50 AM

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Let $A', D', H'$ be the reflections of $A, D, H$ respectively over $M.$ It is well-known that $(ABC)$ passes through $H',$ therefore by homothety (reflections are basically homothety) $(\triangle BHC)$ passes through $A'.$ But it is also well-known that $A'H$ is its diameter because $AH'$ is the diameter of $(\triangle ABC),$ therefore $\angle PGA'=\angle PGH+\angle HGA'=180^\circ,$ so $G=(\triangle BHC) \cap PA'.$

It thus suffices to show that $PA' \parallel HM.$ Clearly by our definitions $HDH'D'$ is a parallelogram so $HM$ passes through $H'.$ Therefore $\frac{D'H'}{D'H} = \frac{DH}{HD'} = \frac{AH}{PH} = \frac{H'A}{PH},$ so $\triangle D'H'H \sim \triangle D'A'P$ and the desired result follows. QED

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pi271828

3363 posts

pi271828

#81 h Mar 18, 2025, 4:29 PM • 1 Y

Y by peace09

Let the reflection of $A$ over $M$ and $H$ to be $X$ and $Y$ respectively. Define the intersection of $HM$ and $AP$ to be $M'$ . It is clear by homothety that $M'$ is the midpoint of $AP$ .

Claim: $X$ lies on $\overline{PG}$ .

Proof. Note that $A$ is the orthocenter of $\triangle BHC$ so by orthocenter reflections, $X$ is the antipode of $H$ in $(BHC)$ . Therefore $\begin{align*} \angle PGH + \angle XGH = 90^{\circ} + 90^{\circ} = 180^{\circ}\end{align*}$ so we are done. $\square$

Claim: $Y$ lies on $\overline{PGX}$

Proof. Note that by midlines that $\overline{HM} = \overline{HM'} \parallel \overline{PY}$ and $\overline{HM} \parallel \overline{XY}$ . Therefore $Y$ lies on $\overline{PX}$ . $\square$

To finish note that we must have $\overline{HM} \parallel \overline{PGXY}$ , so $\angle MHG = \angle PGH = 90^{\circ}$ .

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ErTeeEs06

49 posts

ErTeeEs06

#82 h Mar 29, 2025, 4:11 PM

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Let $AG$ intersect $BC$ at $E$ . Then from Reim we get $D'EGH$ cyclic. Now let $A'$ be the antipode of $A$ on $(ABC)$ , $X$ the intersection of $GH$ and $BC$ and $F\neq A'$ the intersection of $XA'$ and $(ABC)$ . Then by PoP we have $XF\cdot XA'=XB\cdot XC=XG\cdot XH=XD'\cdot XE$ So $A'FGH$ and $A'FED'$ are also cyclic. Now $\angle A'FE=180^\circ-\angle A'DE=90^\circ=\angle A'FA$ so $F$ is on line $AGE$ . This implies $\angle A'HG=180^\circ-\angle A'FG=90^\circ$ so $\angle MHG=90^\circ$ which is what we needed to prove.

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