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Collinearity with orthocenter
liberator   180
N Apr 26, 2025 by joshualiu315
Source: IMO 2013 Problem 4
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
180 replies
liberator
Jan 4, 2016
joshualiu315
Apr 26, 2025
Collinearity with orthocenter
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G H BBookmark kLocked kLocked NReply
Source: IMO 2013 Problem 4
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liberator
95 posts
#1 • 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
#2 • 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
#3 • 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
#4 • 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
#5 • 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
#6 • 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
#7 • 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. :blush:
This post has been edited 2 times. Last edited by Kezer, Mar 29, 2016, 8:39 PM
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MSTang
6012 posts
#8 • 5 Y
Y by donotoven, RedFlame2112, Adventure10, Mango247, Natrium
Spiral sims!
This post has been edited 2 times. Last edited by MSTang, Jun 18, 2016, 3:03 AM
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Reynan
634 posts
#9 • 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
#10 • 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
#11 • 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
#12 • 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
#13 • 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
#14 • 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.
This post has been edited 2 times. Last edited by Vexation, Aug 10, 2016, 6:44 AM
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CountofMC
838 posts
#15 • 4 Y
Y by RedFlame2112, Adventure10, Mango247, Rounak_iitr
Solution
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