Collinearity with orthocenter

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95 posts

#1 h Jan 4, 2016, 9:38 PM • 26 Y

Y by Davi-8191, tenplusten, Arthur., Tawan, Wizard_32, Arshia.esl, OlympusHero, Instance, HWenslawski, donotoven, centslordm, jhu08, TheCollatzConjecture, mathlearner2357, tiendung2006, megarnie, RedFlame2112, rayfish, ImSh95, Lamboreghini, newinolympiadmath, Adventure10, Mango247, Rounak_iitr, farhad.fritl, ItsBesi

Let $ABC$ be an acute triangle with orthocenter $H$ , and let $W$ be a point on the side $BC$ , lying strictly between $B$ and $C$ . The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$ , respectively. Denote by $\omega_1$ is the circumcircle of $BWN$ , and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$ . Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$ , and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$ . Prove that $X,Y$ and $H$ are collinear.

Proposed by Warut Suksompong and Potcharapol Suteparuk, Thailand

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High

876 posts

High

#2 h Jan 24, 2016, 5:03 PM • 9 Y

Y by Tawan, RedFlame2112, ImSh95, megarnie, Danielzh, lian_the_noob12, Adventure10, Mango247, Rounak_iitr

Just happened to stumble upon this problem recently; I guess I'll post a quick solution. Also, I think this problem needs to be added as the 2013 ISL G1 in the contest section.
Solution

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rkm0959

1721 posts

rkm0959

#3 h Feb 25, 2016, 3:55 PM • 6 Y

Y by Tawan, JUSTemom, RedFlame2112, ImSh95, Adventure10, Mango247

As $\angle BNC = \angle BMC = 90$ , $B, N, M, C$ is cyclic. Let this circle be $\omega_3$ . Let $Z(\not= W) = \omega_1 \cap \omega_2$ . Let $D = AH \cap BC$ .
Now $BN$ is the radical axis of $\omega_1$ and $\omega_3$ and $CM$ is the radical axis of $\omega_2$ and $\omega_3$ .
This implies that $A=BN \cap CM$ is the radical center of these three circles, so $A, Z, W$ are colinear.
Now by Power of a Point, we have $AN \cdot AB = AH \cdot AD= AZ \cdot AW = AM \cdot AC$ , so we have $\angle AEW = \angle AZH = \angle XZW = \angle WZY = \angle AZX = \angle AZY = 90$ by simple angle chasing - so $X, Y, Z, H$ are colinear.

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PROF65

2016 posts

PROF65

#4 h Feb 26, 2016, 2:57 PM • 4 Y

Y by RedFlame2112, ImSh95, Adventure10, Mango247

Let $Z$ the second point of intersection of $\omega _1$ and $\omega _2$ .
$(BCMN)$ are concyclic and the radical axes of its circumcircle , $\cal{C}$ $\ \text{and}\ \omega _1,$ $\cal{C}$ $\text{and} \ \omega _2,\omega _1\ \text{and}\ \omega _2,$ are concurrent at $A$ . $\widehat{WZY}=\frac{\pi}{2},\widehat{WZX}=\frac{\pi}{2}$ thus $X,Y$ and $Z$ are colinear .if $H'$ is the intersection of $XY$ and $AH$ then $H'H_1ZW$ is cyclic where $H_1$ is the foot of $AH$ .hence $AZ\cdot AW=AH_1\cdot AH'$ but $AZ\cdot AW=AN\cdot AB$ then $AH'\cdot AH_1=AB\cdot AN$ besides $BHH_1N$ is cyclic implies $AN\cdot AB=AH\cdot AH_1$ so $H=H'$ and the result follows.
R.HAS

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liberator

95 posts

liberator

#5 h Feb 26, 2016, 4:23 PM • 27 Y

Y by rkm0959, jam10307, vlohani, A_Math_Lover, Vfire, FadingMoonlight, Polynom_Efendi, zuss77, hara-huri, OlympusHero, sabrinamath, myh2910, mathlearner2357, prMoLeGend42, RedFlame2112, rayfish, ImSh95, megarnie, newinolympiadmath, Danielzh, Dimanas23, aqwxderf, Adventure10, Mango247, saltamonte, Math_.only., raffigm

The easiest way to do this is with Reim's theorem.
$[asy] /* IMO 2014 Question 4, free script by liberator, 12 August 2014 */ unitsize(2cm); defaultpen(fontsize(10pt)); /* Initialize objects */ pair A = (-2.5, 2.5); pair B = (-3.5, -0.5); pair C = (0.5, -0.5); pair H = orthocenter(A,B,C); pair M = foot(B,C,A); pair N = foot(C,A,B); pair W = (-1.5, -0.5); pair X = rotate(180, circumcenter(B,W,N))*W; pair Y = rotate(180, circumcenter(C,W,M))*W; pair P = intersectionpoint(H--Y, circumcircle(B,W,N)); /* Draw objects */ draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(X--Y, rgb(0.4,0.6,0.8)+dashed+linewidth(1)); draw(A--H, rgb(0.4,0.6,0.8)); draw(B--X, rgb(0.4,0.6,0.8)); draw(C--Y, rgb(0.4,0.6,0.8)); draw(circumcircle(B,W,N), red); draw(circumcircle(C,W,M), red); draw(circumcircle(A,M,N), red); /* Place dots on and label each point, label circles */ dot(A); label("$A$", A, dir(90)); dot(B); label("$B$", B, dir(200)); dot(C); label("$C$", C, dir(340)); dot(H); label("$H$", H, dir(-90)); dot(M); label("$M$", M, dir(0)); dot(N); label("$N$", N, dir(180)); dot(P); label("$P$", P, dir(-70)); dot(W); label("$W$", W, 2.5*dir(-80)); dot(X); label("$X$", X, dir(180)); dot(Y); label("$Y$", Y, dir(60)); label("$\omega_1$", B--W, 15*dir(-90)); label("$\omega_2$", C--Y, 10*dir(0)); [/asy]$
Let $P$ be the second intersection of $\omega_1, \omega_2$ . Then $P$ is the Miquel point of $\triangle MNW$ w.r.t $\triangle ABC$ , so $AMPHN$ is cyclic. $BX \parallel AH$ , hence $X, H, P$ are collinear by Reim's theorem; similarly, $H,P,Y$ are collinear, and the result follows.

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K6160

276 posts

K6160

#6 h Mar 20, 2016, 6:25 PM • 4 Y

Y by Mr.Mister, RedFlame2112, Adventure10, Mango247

Let $Z$ be the second intersection of $\omega_1$ and $\omega_2$ . Let $A'$ be the foot of the $A$ -altitude. We have that $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ because $BCMN$ is cyclic. Also by power of a point, $AZ\cdot AW=AN\cdot AB = AH\cdot AA'$ , so $A'WZH$ is cyclic. Therefore, $\angle XZW=\angle YZW=\angle HZW=90^{\circ}$ and the conclusion follows. $\boxed{}$

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Kezer

986 posts

Kezer

#7 h Mar 29, 2016, 8:31 PM • 7 Y

Y by MSTang, S117, RedFlame2112, Adventure10, Mango247, Rounak_iitr, Math_.only.

Different angle chasing approach without the radical axis/Reim

Was too noob after drawing in the Miquel point to think of proving that it lies on $XH$ , so I needed Evan's hint from his book to do so.

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MSTang

6012 posts

MSTang

#8 h Jun 17, 2016, 3:58 AM • 5 Y

Y by donotoven, RedFlame2112, Adventure10, Mango247, Natrium

Spiral sims!

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Reynan

634 posts

Reynan

#9 h Jun 17, 2016, 5:21 AM • 3 Y

Y by RedFlame2112, Adventure10, Mango247

$AN\cdot AB=AM\cdot AC = AP\cdot AW\cdots (1)$

we have $\angle XPW=90=\angle WPX$ so $XPY$ collinear

let $D$ the perpendicular from $A$ to BC $we have$ NHDB $and HDPW$ cylich from $(1)$
$\angle XHD=\angle DWP=180-\angle DHP$ so $XHP$ collinear

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the_executioner

533 posts

the_executioner

#10 h Jul 1, 2016, 3:03 AM • 3 Y

Y by Tawan, RedFlame2112, Adventure10

Let $Z$ be the second intersection point $\omega_1$ and $\omega_2$ . Now $\angle YZW$ $+$ $\angle XZW$ $=$ $90^o+90^o=180^o$ . Therefore $Y$ , $Z$ and $X$ are collinear. So we are left to prove that $Z$ , $H$ and $X$ are collinear. Now since $Z$ is the miquel point, we have $A$ , $N$ , $H$ , $Z$ , $M$ are concyclic. So， $\angle AZM$ $+$ $\angle MZW$ $= C+(A+B)=180^o$ .
Therefore $A$ , $Z$ and $W$ are collinear. Now, $\angle AZH$ $=90^o$ and thus $\angle HZW$ $=90^o$ . But $\angle WZX$ $=90^o$ , therefore $Z$ , $H$ and $X$ are collinear and we are done.

This post has been edited 2 times. Last edited by the_executioner, Jul 1, 2016, 3:06 AM

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anantmudgal09

1980 posts

anantmudgal09

#11 h Jul 5, 2016, 3:31 PM • 4 Y

Y by Tawan, Aryan-23, RedFlame2112, Adventure10

A non standard way to do it...

Let $\ell_b,\ell_c$ be the lines perpendicular to line $BC$ at points $B,C$ . Notice that as $W$ varies, $\angle WMY=\angle WNX=90^{\circ}$ and points $X,Y$ vary on lines $\ell_b,\ell_c$ respectively. Therefore, since this implies that $N$ and $NW$ are related projectively, i.e., the pencil and range have same cross ratio and essentially this is the same cross ratio for $X$ as it varies on the line $\ell_b$ we see that points $X$ and $W$ are projectively related to each other. Similarly, points $W$ and $Y$ are projectively related to each other. Thus, points $X$ and $Y$ are projectively related, i.e., there exist a projectivity mapping $X$ to $Y$ . We prove that it is in fact a perspectivity. Thus, we only need to consider three positions of $W$ for which $X,Y,H$ are collinear. These are clearly true for $W=B,C,D$ where $D$ is the feet of altitude from $A$ . Hence it holds for all $W$ .

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Ghost_rider

35 posts

Ghost_rider

#12 h Jul 6, 2016, 2:42 AM • 2 Y

Y by RedFlame2112, Adventure10

Let $\omega_1 \cap \omega_2$ = $G$ . $A$ has the same power with respect $\omega_1$ and $\omega_2$ so $A,G$ and $W$ are collinear.
Then $\angle NGA$ = $\angle ABW$ = $\angle NHA$ , so $H,N,A$ and $G$ are concyclic ( $\angle HGA$ = $90^\circ$ ).
Finally $X,H,G$ and $Y$ are collinear.

This post has been edited 1 time. Last edited by Ghost_rider, Jul 6, 2016, 2:43 AM
Reason: type

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jayme

9792 posts

jayme

#13 h Jul 6, 2016, 6:29 AM • 6 Y

Y by AlastorMoody, Kanep, Siddharth03, RedFlame2112, Adventure10, Mango247

Dear Mathlinkers,
just for history, this result comes from Mannhein in 1893... actually a rediscovery...

Sincerely
Jean-Louis

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Vexation

49 posts

Vexation

#14 h Aug 10, 2016, 5:59 AM • 2 Y

Y by RedFlame2112, Adventure10

$K$ is the foot of the altitudes from $A$ .

We need to check that $\frac{XB \cdot KC+YC \cdot KB}{BC} = {HK}$ ... (*)

From $NXB \sim NWC$ and $MYC \sim MWB$ , (*) becomes $KH \cdot KA = KB \cdot KC$ , which is true when $H$ is reflected in $BC$ by power of $K$ wrt the circumcircle.

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CountofMC

838 posts

CountofMC

#15 h Dec 6, 2016, 5:15 PM • 4 Y

Y by RedFlame2112, Adventure10, Mango247, Rounak_iitr

Solution

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peace09

5419 posts

peace09

#184 h Aug 4, 2024, 6:27 PM

Y by

peace09 wrote:

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Silly. Here's an actual solution: let $P$ be the Miquel point of $(AMN)$ , $(BNW)$ , and $(CMW)$ . Clearly $A$ is the radical center of $(BCMN)$ , $(BNPW)$ , and $(CMPW)$ ; in particular $A,P,W$ are collinear. Then

$HP\perp AW$ since $P\in(AMN)$ of diameter $AH$ ,
$XP\perp AW$ since $P\in(BNW)$ of diameter $WX$ , and
$YP\perp AW$ since $P\in(CMW)$ of diameter $WY$ ;

and so $H,X,Y$ are collinear. $\square$

Attachments:

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kinnikuma

9 posts

kinnikuma

#185 h Aug 23, 2024, 6:50 AM

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Consider $D = (BNW) \cap (WMC)$ , and let us demonstrate how powerful this point is.

- $X, D, Y$ are collinear : it's because $\angle XDY = \angle XDW + \angle WDY = \frac{\pi}{2} + \frac{\pi}{2} = \pi$ .
- $A, D, W$ are collinear : we just need to show that $A$ lies on the radical axis of $(BNM)$ and $(WMC)$ . That's true because $AN \cdot AB = AM \cdot AC$ by power of point (here, $B, N, M, C$ are concyclic).

Hence $\angle ADH = \angle HDW = \frac{\pi}{2}$ . Consider finally $(AHD)$ and $(HDW)$ . If we show that $(XY)$ is their radical axis than we are done. Let us remember that a radical axis is the line perpendicular to the line formed by the centers, and, if a common point exist between the circles, passing through this common point. Here, we already showed that $(XY)$ contains $D$ , a common point between the circles. In addition, it is perpendicular to $(AW)$ , which is parallel to the line formed by the centers (why : the centers, since there are only right triangles, are midpoints, so the parallelism is not mysterious anymore, by Thales) $\huge \blacksquare$ .

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fasttrust_12-mn

118 posts

fasttrust_12-mn

#186 h Aug 31, 2024, 1:20 AM

Y by

let $\omega_3$ be the circumcircle of $\triangle ANM$ donate $E$ be the miquel point now we have that $A-E-W$ as a fast result we get $X-E-H-Y$

$[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 = -42.64193212363468, xmax = 37.62142919514159, ymin = -12.456287979858324, ymax = 25.85389619279325; /* image dimensions */ /* draw figures */ draw((-10.4950997206116,22.60018473975595)--(15.812330657344743,-5.12607334397776), linewidth(0.4)); draw((15.812330657344743,-5.12607334397776)--(-20.190422376757308,-4.594012954311232), linewidth(0.4)); draw((-20.190422376757308,-4.594012954311232)--(-10.4950997206116,22.60018473975595), linewidth(0.4)); draw((-10.4950997206116,22.60018473975595)--(-10.899013648982507,-4.731324413342385), linewidth(0.4)); draw((-16.298569222789236,6.322160526330897)--(15.812330657344743,-5.12607334397776), linewidth(0.4)); draw((-20.190422376757308,-4.594012954311232)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw((-5.47591954759328,-4.811468661146167)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw(circle((5.274466321965228,2.2215409038274205), 12.84655673265849), linewidth(0.4)); draw((-16.298569222789236,6.322160526330897)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); draw(circle((-12.779695673279125,-1.0842462590880841), 8.199837347688193), linewidth(0.4)); draw((16.024852191523735,9.254550468801007)--(-1.5099931362833348,13.130488137396515), linewidth(0.4)); draw((-5.47591954759328,-4.811468661146167)--(16.024852191523735,9.254550468801007), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(-16.298569222789236,6.322160526330897), linewidth(0.4)); draw((-20.083471798964982,2.6429761429699887)--(16.024852191523735,9.254550468801007), linewidth(0.4)); draw(circle((-10.629958786198769,13.474721301690662), 9.126459879221372), linewidth(0.4)); draw((-10.4950997206116,22.60018473975595)--(-5.47591954759328,-4.811468661146167), linewidth(0.4)); /* dots and labels */ dot((-10.4950997206116,22.60018473975595),dotstyle); label("$A$", (-11.030154386078584,23.32730429183617), NE * labelscalefactor); dot((15.812330657344743,-5.12607334397776),dotstyle); label("$B$", (16.057261110228783,-4.523964569876786), NE * labelscalefactor); dot((-20.190422376757308,-4.594012954311232),dotstyle); label("$C$", (-21.78285945294246,-5.052786130542223), NE * labelscalefactor); dot((-10.899013648982507,-4.731324413342385),linewidth(4pt) + dotstyle); label("$D$", (-10.677606678968292,-4.288932765136593), NE * labelscalefactor); dot((-10.76481785178594,4.34925786362537),linewidth(4pt) + dotstyle); label("$H$", (-10.501332825413147,4.818549668545914), NE * labelscalefactor); dot((-16.298569222789236,6.322160526330897),linewidth(4pt) + dotstyle); label("$M$", (-16.083338187992755,6.816320008837561), NE * labelscalefactor); dot((-1.5099931362833348,13.130488137396515),linewidth(4pt) + dotstyle); label("$N$", (-1.27633448936053,13.57348439511813), NE * labelscalefactor); dot((-5.47591954759328,-4.811468661146167),dotstyle); label("$W$", (-5.21311721875878,-4.230174813951544), NE * labelscalefactor); dot((16.024852191523735,9.254550468801007),linewidth(4pt) + dotstyle); label("$X$", (16.233534963783928,9.695459616904934), NE * labelscalefactor); dot((-20.083471798964982,2.6429761429699887),linewidth(4pt) + dotstyle); label("$Y$", (-21.665343550572363,2.1156839140336863), NE * labelscalefactor); dot((-7.270440253264129,4.989091970055546),linewidth(4pt) + dotstyle); label("$E$", (-7.034613705495285,5.464887131581447), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

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MagicalToaster53

159 posts

MagicalToaster53

#187 h Sep 3, 2024, 1:43 AM

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We use directed angles $\measuredangle$ modulo $\pi$ . Let $Z$ be the second point of intersection between $\omega_1$ and $\omega_2$ . Then observe that $X, Z, Y$ are collinear as $\measuredangle XZW = \measuredangle YZW = 90^{\circ}$ . Let $D$ be the point where the altitude from $A$ meets $\overline{BC}$ . Then make the observation that $BX \parallel HD$ , as $\measuredangle XBW = \measuredangle 90^{\circ} = \measuredangle HDW$ . We now make a claim:

Claim: $X, H, Z$ are collinear.
Proof: We have the following angle chase: $\measuredangle BXH = \measuredangle DHZ = \measuredangle DWZ = \measuredangle BWZ = \measuredangle BXZ. \square$
Hence as $XZ$ is the same line passing simultaneously through both $H$ and $Y$ , we find that $X, H, Z, Y$ are collinear, as desired. $\blacksquare$

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ezpotd

1269 posts

ezpotd

#188 h Sep 8, 2024, 1:34 AM

Y by

Draw in $(BWN) \cap (CWM) = K$ , then Miquel tells us $(AMNHK)$ is cyclic. Direct angles. We then see $\angle NKX = \angle NBX = 90 - \angle ABC = \angle NAH = \angle NKH$ , so $K,H,X$ are collinear, and so are $K,H,Y$ , finishing.

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legogubbe

19 posts

legogubbe

#189 h Sep 18, 2024, 6:27 PM

Y by

Let $Z$ be the second intersection point of $(BWN)$ and $(CWN)$ .

Claim 1. Points $X$ , $Y$ and $Z$ are collinear.
Proof. We see by constructions that $BXZW$ and $CYZW$ are cyclic. This implies
$\measuredangle WZX = \measuredangle WBX = 90^{\circ} = \measuredangle WCY = \measuredangle WZY,$ which means $X$ , $Y$ and $Z$ are indeed collinear.

Claim 2. Quadrilateral $AHZM$ is cyclic.
Proof. By Miquel's theorem, $ANZM$ is cyclic. However, by the properties of the orthocenter, $ANHM$ is cyclic. From this we deduce that in fact, $ANHZM$ is cyclic, thereby $AHZM$ is cyclic.

Let $K$ be the intersection point of lines $AH$ and $BC$ .

Claim 3. Quadrilateral $KHZW$ is cyclic.
Proof. From cyclic quadrilaterals,
$\measuredangle KHZ = \measuredangle AHZ = \measuredangle AMZ = \measuredangle CMZ = \measuredangle CWZ = \measuredangle KWZ,$ proving our claim.

Finally, $\measuredangle WZH = \measuredangle WKH = \measuredangle CKA = 90^{\circ}$ . Hence $X$ , $Y$ , $Z$ and $H$ are collinear.

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cursed_tangent1434

625 posts

cursed_tangent1434

#190 h Oct 6, 2024, 2:16 PM

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We let $T$ denote the second intersection of circles $(BNW)$ and $(CMW)$ . We start ff by noting that since,
$\measuredangle WTX = \measuredangle WBX = \frac{\pi}{2}$ and
$\measuredangle WTY = \measuredangle WCY = \frac{\pi}{2}$ which implies that points $X$ , $T$ and $Y$ are collinear. Next, by Radical Center Theorem on $(BNTW)$ , $(CMTW)$ and $(BNMC)$ , lines $\overline{BN}$ , $\overline{TW}$ and $\overline{MC}$ concur. It is clear that $A= BN \cap CM$ , so
$AT \cdot AW = AN \cdot AB = AH \cdot AD$ which implies that $DHTW$ is also cyclic. Now this finishes since,
$\measuredangle WTH = \measuredangle WDH = \frac{\pi}{2} = \measuredangle WTX$ which implies that $X$ , $H$ and $T$ are collinear. Combining this with our previous observation concludes that points $X$ , $Y$ and $H$ are collinear, as desired.

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Saucepan_man02

1340 posts

Saucepan_man02

#191 h Nov 18, 2024, 5:59 AM

Y by

Storage

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cj13609517288

1916 posts

cj13609517288

#192 h Nov 19, 2024, 4:14 PM • 1 Y

Y by peace09

Let $Z$ be the other intersection of $\omega_1$ and $\omega_2$ . By radax on $\omega_1,\omega_2,(BNMC)$ we get that $A,Z,W$ are collinear. Since $\angle XZW=\angle YZW=90^{\circ}$ , we get that $X,Y,Z$ are collinear. Thus it suffices to show that $\angle HZW=90^{\circ}$ , which is equivalent to $Z$ being on $(AH)$ .

Let $f$ denote an inversion centered at $A$ with radius $\sqrt{AN\cdot AB}$ . Then $f$ swaps $Z$ and $W$ . Unsurprisingly, $f$ also swaps $(AH)$ and $BC$ , so we win. $\blacksquare$

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ItsBesi

146 posts

ItsBesi

#194 h Jan 12, 2025, 1:42 PM

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Nothing new but posting it for storage.

Let $D$ be the feet of the altitude from $A$ to $BC$ , let $\omega_1 \cap \omega_2=\{Z\}$ .

Claim: Points $\overline{X-Z-Y}$ are collinear.

Proof:

$\angle XZW \stackrel{\omega_1}{=}=90$ and $\angle YZW \stackrel{\omega_1}{=}=90$

Hence $\angle XZY=\angle XZW+\angle YZW=180 \implies \angle XZY=180 \implies$ Points $\overline{X-Z-Y}$ are collinear. $\square$

Claim: Points $\overline{A-Z-W}$ are collinear.

Proof: First note that $\omega_1 \cap \omega_2 =\{W,Z\}$ so $ZW$ is the radical axis of $\omega_1$ and $\omega_2$

Also its well known that points $B,N,M$ and $C$ are concyclic so by Power of the Point Theorem we get:

$Pow(A,\omega_1)=AN \cdot AB=Pow(A, \odot(BNMC))=AM \cdot AC=Pow(A,\omega_2) \implies Pow(A,\omega_1)=Pow(A,\omega_2)$

So $A$ lies on the radical axis of $\omega_1$ and $\omega_2$ hence $A$ lies on the line $WZ \therefore$ Points $\overline{A-Z-W}$ are collinear. $\square$

Claim: Points $W,D,H$ and $Z$ are concyclic.

Proof: It's also well known that points $D,N,B$ and $H$ are concyclic

So by POP we get:
$AH \cdot AD=Pow(A,\odot(DNBH))=AN \cdot AB=Pow(A,\omega_1)=AZ \cdot AW \implies AH \cdot AD= AZ \cdot AW$

which by the converse of POP we have that: Points $W,D,H$ and $Z$ are concyclic. $\square$ Let $\odot(WDHZ)=\Gamma$ .

Claim: Points $\overline{X-H-Y}$ are collinear.

Proof:

From $\Gamma \implies \angle HZW \stackrel{\Gamma}{=} 180-\angle HDZ=180-90=90=\angle XZW \implies \angle HZW=\angle XZW \implies$
Points $\overline{X-H-Z}$ are collinear, combining with the first claim that Points $\overline{X-Z-Y}$ are collinear we get that $\overline{X-H-Z-Y}$ are all collinear

Hence points $\overline{X-H-Y}$ are collinear $\blacksquare$

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Maximilian113

575 posts

Maximilian113

#195 h Feb 24, 2025, 9:47 PM

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Let $P$ be the second intersection of $w_1, w_2.$ Note that by Miquel's Theorem, $AMPN$ is cyclic, but clearly $H$ also lies on this circle so $AMPHN$ is cyclic. Therefore, $\angle NPH = \angle NAH = 90^\circ - \angle ABC = \angle XBN = \angle XPN,$ so $X, H, P$ are collinear. Similarly, $H, P, Y$ are collinear as well and we are done. QED

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Ilikeminecraft

627 posts

Ilikeminecraft

#196 h Mar 14, 2025, 11:53 PM

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let $P$ be second intersection of $(BNW), (MWC).$
By Radax on those two circles and $(BNMC),$ we get $A,P,W$ are collinear.
furthermore, miquel theorem on $ABC$ with $M, P, W$ as the points, we get $ANHPM$ is cyclic. Hence, $\angle APH = 90.$
However, $\angle BHP = 90 = \angle HPW$ so $X, H, P$ are collinear.

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endless_abyss

47 posts

endless_abyss

#197 h Mar 23, 2025, 3:15 PM

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Let $P$ denote the second intersection of the circumcircles $B N W$ and $C M W$ ,

Claim we claim that $X - P - H$ and $Y - P - H$ are collinear

Let $Q$ denote the foot of $A$ onto $B C$ , then we know that -

$\angle B H Q = C$
$\angle B H X = \angle P W B - C$
$\angle Q H P = \angle 180 - \angle P W B$

So, $X - P - H$ are collinear, similarly, $Y - P - H$ are collinear.
$\square$
:starwars:

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bjump

1026 posts

bjump

#198 h Mar 27, 2025, 3:01 AM

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Let $D$ be the second interssection of $\omega_1$ , and $\omega_2$ . Since $\angle XDW = \angle YDW = 90^\circ$ , $X$ , $D$ , and $Y$ are collinear. Let $E$ be the foot from $A$ , since $AM \cdot AC = AN \cdot AB$ . $A$ lies on $WD$ by radical axis. Converse of Radical axis with lines $NB$ , $MC$ and $HE$ , and $\omega_1$ , and $\omega_2$ gives $HEWD$ cyclic, which means that $\angle WDH = 90^\circ$ . So the points are collinear.

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joshualiu315

2534 posts

joshualiu315

#199 h Apr 26, 2025, 11:56 PM

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not necessarily the fastest solution + i took a ridiculous amount of time to find the miquel point

Let $P$ be the second intersection of $\omega_1$ and $\omega_2$ . Since $\angle WPX = \angle WPY$ , we have that $P$ lies on $\overline{XY}$ . Thus, it suffices to show that $X, H, P$ are collinear since showing that $Y,H,P$ are collinear is analogous.

Note that $P$ is the Miquel point of $(WBN)$ , $(WCM)$ and $(AMN)$ . Then, note that

$\angle NPX = \angle NWX = 90^\circ - \angle NXW = 90^\circ - \angle NBW = \angle NAH = \angle NPH,$
so we are done. $\blacksquare$

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