Prove that H, O and D are collinear

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Prove that H, O and D are collinear

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Source: Chinese TST, Test 3, Problem 1 2012

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761 posts

#1 h Mar 25, 2012, 5:18 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

In an acute-angled $ABC$ , $\angle A>60^{\circ}$ , $H$ is its orthocenter. $M,N$ are two points on $AB,AC$ respectively, such that $\angle HMB=\angle HNC=60^{\circ}$ . Let $O$ be the circumcenter of triangle $HMN$ . $D$ is a point on the same side with $A$ of $BC$ such that $\triangle DBC$ is an equilateral triangle. Prove that $H,O,D$ are collinear.

This post has been edited 1 time. Last edited by bluecarneal, Mar 13, 2016, 4:31 PM
Reason: fixed latex error

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761 posts

#2 h Mar 25, 2012, 5:20 AM • 2 Y

Y by Adventure10, Mango247

I can't find a solution without analytical method with which I solved it in the Test.
Could someone find a geometric methgod?

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5 posts

#3 h Mar 25, 2012, 8:17 AM • 4 Y

Y by JRD, littletush, Adventure10, Mango247

I found a solution without analytical method in the exam.Let L be a point on AC satisfying LN=HL=HN,and let K be a point on AB satisfying KM=KH=MH .We can prove △DHC∽△MNL,△DHB∽△MNK,so DHC=OHC.So H,O,D are collinear.

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73 posts

#4 h Mar 25, 2012, 12:02 PM • 3 Y

Y by R8450932, Adventure10, Mango247

This is my solution , we know that for a point $P$ inside ( not necessary ) an equilateral triangle $\Delta XYZ$ , a triangle with angles $\angle YPZ - 60^o ~,~ \angle ZPX - 60^o ~,~ \angle XPY - 60^o$ has side ratio $PX : PY : PZ$ , which can be proved by rotation .

Now apply this result on $H$ and equilateral triangle $BDC$ . Since $\angle MHD = 120^o - \angle A = \angle BHC - 60^o$ and $HM:HN= BH:CH$ , we find that $\Delta MHN$ is similar to the triangle with sides $CH ~,~ DH ~,~ BH$ . Therefore , $\angle MHO = 90^o - \angle MNH = 90^o - ( \angle CHD - 60^o )$
$= 150^o - \angle CHD = \angle MHD$ .

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148 posts

cwein3

#5 h Mar 27, 2012, 12:53 AM • 2 Y

Y by Adventure10, Mango247

Construct $F$ on the same side of $HC$ as $A$ such that $\triangle FCH$ is equilateral. Then a rotation around $C$ maps $\triangle DCB$ to $\triangle FCH$ , so $FD = HB$ . Furthermore, $\angle DFH = \angle CHB - 60 = 120 - \angle CAB$ . Furthermore, $\angle MHN = 120 - \angle CAB$ also. In addition, $\triangle CMH \sim \triangle BNH$ , so $\dfrac {HM} {HN} = \dfrac {CH} {HB} = \dfrac {FH} {FD}$ . Therefore, $\triangle DFH \sim \triangle NHM$ . Then we have $\angle MHD$ $= 60 + \angle FHD - (\angle CAB + 30)$ $= \angle FHD + 120 - \angle CAB - 90$ $=\angle HMN + \angle MHN - 90$ $= 90 - \angle MNH$ $= \angle MHO$ , so $H$ , $O$ , and $D$ are collinear.

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38 posts

JRD

#6 h Dec 13, 2012, 2:45 PM • 2 Y

Y by Adventure10, Mango247

Hint: let HM intersects AC at point E. and HN intersects AB at F. it’s easy to see that points M,N,F,E lie on a same circle, so it’s enough to say that HD is perpendicular to EF. That means we want to show that DF²₋DE²=HF²₋HE². now use cosine law in triangles BDF and CDE and BHF and CHE.the rest is not difficult.

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2312 posts

#7 h Oct 21, 2014, 5:24 PM • 4 Y

Y by md0308, sabkx, Adventure10, and 1 other user

I generalize this problem at http://www.artofproblemsolving.com/community/c6h610521

This post has been edited 1 time. Last edited by TelvCohl, Aug 20, 2015, 8:43 PM

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314 posts

#8 h Dec 22, 2014, 11:05 PM • 1 Y

Y by Adventure10

My solution:Ok,after we look at the picture we see that we need to make some connection beetwen the angles in triangles $CHD$ and $HMN$ and now we get the intution to construct $G$ such that $GHC$ is equilateral(because of spiral similarity we will have $BGC$ is similar to $CHD$ and $BHG$ will be similar to $HMN$ ).Now,from this we have an easy angle chase.Let $<HMN=x$ and by similarity of $HMN$ and $HBG$ we have $<HBG=<HMN=x$ and since $<CBH=90-<BCA$ we get $<CBG=<<CDH=x+<ACB-90$ and from this we get $<HDB=150-x-<BCA$ and from this we get $<BHD=60+x$ and also $<BHO=<MHO+<MHB=30+x-<BAC+30+<BAC=60+x$ ,so we get $<BHO=<BHD$ and from this we are finished.

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355 posts

#9 h Feb 18, 2015, 6:59 PM • 2 Y

Y by Adventure10, Mango247

Say $<BHM = <CHN = x+30$ . Then $<MON = 240-2x$

Let $S_1$ be spiral similarity around $H$ with angle $x+30$ and $k_1=HM/HB$ ,
Let $R_O$ be rotation around $O$ with angle $240-2x$ ,
Let $S_2$ be spiral similarity around $H$ with angle $x+30$ and $k_2=HC/HN$ .
Let $R_D$ be rotation around $D$ with angle $60$ ,

Then composition $\mathcal{J} = \mathcal{R}_D\circ \mathcal{S}_2 \circ \mathcal{R}_O \circ \mathcal{S}_1$ is isometry since $k_1 *k_2=1$ . Since it's sum of rotational angles is $360$ and $B$ it is fixed point:

$\mathcal{J}: B\stackrel{\mathcal{S}_1}{\longmapsto}M\stackrel{\mathcal{R}_O}{\longmapsto}N\stackrel{\mathcal{S}_2}{\longmapsto}C \stackrel{\mathcal{R}_D}{\longmapsto}B$

$\mathcal{J}$ must be identity map. So if $\mathcal{R}_O: H \longmapsto H'$ and $\mathcal{S}_2: H'\longmapsto H''$ we have:

$\mathcal{J}: H\stackrel{\mathcal{S}_1}{\longmapsto}H\stackrel{\mathcal{R}_O}{\longmapsto}H'\stackrel{\mathcal{S}_2}{\longmapsto}H'' \stackrel{\mathcal{R}_D}{\longmapsto}H$

Now since $<H'HO = x-30$ and $<H'HH'' = x+30$ we have $<OHH'' = 60$ . But since $\mathcal{R}_C$ maps $H''$ to $H$ we have also $<DHH'' = 60$ so $O,D,H$ are collinear.

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160 posts

#10 h Sep 10, 2018, 6:19 AM • 2 Y

Y by Adventure10, Mango247

Nice question

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Arefe

#11 h Apr 20, 2020, 7:38 AM

Y by

We can find that sin(EHO)/sin(FHO) = sin(EHD)/sin(FHD) where E and F are the intersects of altitudes with AC and AB , the problem is solved .

This post has been edited 1 time. Last edited by Arefe, Apr 20, 2020, 7:38 AM

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862 posts

#12 h May 15, 2021, 3:53 PM • 1 Y

Y by Rounak_iitr

projective >>> synthetic :mad:

:mad:

$[asy] import graph; size(10cm); real labelscalefactor = 1.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(11); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.868953582044828, xmax = 19.051346935130493, ymin = -1.424030804570298, ymax = 10.436451535195745; /* image dimensions */ draw((4.1011591118959165,8.443093158764478)--(0,0)--(10,0)--cycle, linewidth(1.5)); /* draw figures */ draw((0,0)--(5,8.660254037844386), linewidth(0.75)); draw((5,8.660254037844386)--(10,0), linewidth(0.75)); draw((0,0)--(6.719869915485527,4.694895895949962), linewidth(1)); draw((1.9090225772392457,3.930121953830859)--(10,0), linewidth(1)); draw((2.523791648867181,5.195752572215299)--(4.1011591118959165,2.8653106868493228), linewidth(0.75)); draw((4.1011591118959165,2.8653106868493228)--(5.663558402505913,6.206809283332205), linewidth(0.75)); draw(circle((4.395606767138283,4.763652767795307), 1.9210419250937896), linewidth(1) + dotted); draw((5,8.660254037844386)--(4.1011591118959165,2.8653106868493228), linewidth(0.75) + linetype("4 4")); draw(shift((5,0))*xscale(5)*yscale(5)*arc((0,0),1,0,180), linewidth(1) + linetype("4 4") + heavygrey); draw((1.9090225772392457,3.930121953830859)--(4.542830594759458,5.712823808268292), linewidth(0.75) + heavygrey); draw((4.542830594759458,5.712823808268292)--(7.5,4.330127018922194), linewidth(0.75) + heavygrey); draw(circle((4.32199485332769,4.289067247558802), 1.4407814438203435), linewidth(0.75) + dotted); /* dots and labels */ dot((4.1011591118959165,8.443093158764478),dotstyle); label("$A$", (4.1011591118959165,8.443093158764478), dir(90) * labelscalefactor); dot((0,0),dotstyle); label("$B$", (0,0), dir(225) * labelscalefactor); dot((10,0),dotstyle); label("$C$", (10,0), dir(315) * labelscalefactor); dot((5,8.660254037844386),linewidth(4pt) + dotstyle); label("$D$", (5,8.660254037844386), dir(90) * labelscalefactor); dot((6.719869915485527,4.694895895949962),linewidth(4pt) + dotstyle); label("$E$", (6.719869915485527,4.694895895949962), dir(45) * labelscalefactor); dot((1.9090225772392457,3.930121953830859),linewidth(4pt) + dotstyle); label("$F$", (1.9090225772392457,3.930121953830859), dir(135) * labelscalefactor); dot((4.1011591118959165,2.8653106868493228),linewidth(4pt) + dotstyle); label("$H$", (4.1011591118959165,2.8653106868493228), dir(60) * labelscalefactor); dot((2.523791648867181,5.195752572215299),linewidth(4pt) + dotstyle); label("$M$", (2.523791648867181,5.195752572215299), dir(135) * labelscalefactor); dot((5.663558402505913,6.206809283332205),linewidth(4pt) + dotstyle); label("$N$", (5.663558402505913,6.206809283332205), dir(45) * labelscalefactor); dot((4.395606767138283,4.763652767795307),linewidth(4pt) + dotstyle); label("$O$", (4.395606767138283,4.763652767795307), dir(45) * labelscalefactor); dot((2.5,4.330127018922194),linewidth(4pt) + dotstyle); label("$K$", (2.5,4.330127018922194), dir(90) * labelscalefactor); dot((7.5,4.330127018922194),linewidth(4pt) + dotstyle); label("$L$", (7.5,4.330127018922194), dir(45) * labelscalefactor); dot((4.542830594759458,5.712823808268292),linewidth(4pt) + dotstyle); label("$P$", (4.542830594759458,5.712823808268292), dir(45) * labelscalefactor); dot((2.918133514624364,4.613142100873805),linewidth(4pt) + dotstyle); label("$Q$", (2.918133514624364,4.613142100873805), dir(90) * labelscalefactor); dot((5.2729585798534115,5.371434634211479),linewidth(4pt) + dotstyle); label("$R$", (5.2729585798534115,5.371434634211479), dir(30) * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Let $E=BH\cap AC$ , $F=CH\cap AB$ , and $\omega = (BCEF)$ . Let $K=BD\cap\omega$ and $L=CD\cap\omega$ . Let $P=EL\cap FK$ , $Q=FK\cap HM$ , and $R=EL\cap HN$ .

By Pascal's theorem, $D$ , $P$ , and $H$ are collinear, so it suffices to show that $P$ , $O$ , and $H$ are collinear. Note that $\measuredangle PEH = \measuredangle LCB = 60^{\circ}$ and $\measuredangle HFP = \measuredangle CBK = 60^{\circ}$ , thus $EP\perp HN$ and $FP\perp HM$ . Thus the center of $(PQHR)$ lies on $HP$ and it suffices to show that $(HQR)$ and $(HMN)$ are tangent at $H$ . Note that $\tfrac{HQ}{HM}=\tfrac{HR}{HN}=\tfrac{3}{4}$ , which implies a homothety centered at $H$ , as desired. $\blacksquare$

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139 posts

#13 h Mar 2, 2025, 4:13 PM

Y by

I post a different synthetic solution:
Let the circumcircle of $HMN$ intersects $AB and AC$ at $X, Y$ respectively. It is easy to see that $HXY$ is equilateral and the circumcircles of
$XHB and YHC$ have equal radii. Let $O_1$ , $O_2$ be their centers and let $T$ be their second intersection point. Then obviously $T, H, O$ are collinear so we just need to prove $D$ lies on this line. Take $D'$ the reflection of $T$ over $XY$ .
Then $D'O_1 = D'O_2$ and the triangle $O_1 D O_2$ is equilateral.
Then simple angle chase gives $\angle D'O_1 B = \angle D'O_2 C$ therefore these triangles are equal $\implies$ $O_1 D O_2$ is similar to $D'BC$ thus
$D=D'$ as desired.

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