MN bisects segment CH

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MN bisects segment CH

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#1 h Jul 22, 2019, 3:27 AM • 2 Y

Y by Adventure10, Mango247

Let $H$ be the orthocentre of an acute triangle $ABC$ with $BC > AC$ , inscribed in a circle $\Gamma$ . The circle with centre $C$ and radius $CB$ intersects $\Gamma$ at the point $D$ , which is on the arc $AB$ not containing $C$ . The circle with centre $C$ and radius $CA$ intersects the segment $CD$ at the point $K$ . The line parallel to $BD$ through $K$ , intersects $AB$ at point $L$ . If $M$ is the midpoint of $AB$ and $N$ is the foot of the perpendicular from $H$ to $CL$ , prove that the line $MN$ bisects the segment $CH$ .

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#2 h Aug 16, 2019, 10:34 PM • 2 Y

Y by Adventure10, Mango247

Related topics:
- Bosnia and Herzegovina Regional Olympiad 2018 : https://artofproblemsolving.com/community/c6h1709471p11016808
- European Mathematical Cup 2018 , Junior P3 : https://artofproblemsolving.com/community/c6h1758655p11490969
- Danube Mathematical Competition 2014 , Junior P3 : https://artofproblemsolving.com/community/c6h1703980p10958876

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#5 h Feb 14, 2020, 9:44 AM • 5 Y

Y by Euiseu, fastlikearabbit, Grumpah, augustin_p, Adventure10

Let $P$ is the midpoint of $CH$ . We need to prove that $P-N-M$ are colliniar. We perform homotety from $H$ with ratio $2$ . $P$ goes to $C$ . $M$ goes to antipode of $C$ wrt $\circ ABC$ , and let $N$ goes to $Q$ . We need now to prove that $Q$ lies on diameter from $C$ . This can be easily proved with angle chasing and some similarities.
Proof with angle chasing and some similarities
$[asy] import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -12.009090909090915, xmax = 10.172727272727274, ymin = -5.835454545454546, ymax = 7.346363636363635; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-2.2200116993272885,0.740298332845862), 5.706489632560346), linewidth(0.5) + wrwrwr); draw((-6.3,4.73)--(-7.22,-2.01), linewidth(0.5) + wrwrwr); draw((-7.22,-2.01)--(2.98,-1.61), linewidth(0.5) + wrwrwr); draw((2.98,-1.61)--(-6.3,4.73), linewidth(0.5) + wrwrwr); draw((-6.3,4.73)--(0.2460429784135627,-4.405826252986779), linewidth(0.5) + wrwrwr); draw((2.98,-1.61)--(0.2460429784135627,-4.405826252986779), linewidth(0.5) + wrwrwr); draw((-2.3379346965024017,-0.7995604283541823)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); draw((-6.3,4.73)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); draw((-6.3,4.73)--(-6.099976601345422,-0.370596665691723), linewidth(0.5) + wrwrwr); draw((-6.099976601345422,-0.370596665691723)--(-3.374106427962564,-1.859180644233826), linewidth(0.5) + wrwrwr); draw((-6.099976601345422,-0.370596665691723)--(-4.37519129392871,0.3952858493872016), linewidth(0.5) + wrwrwr); draw((-6.199988300672711,2.1797016671541387)--(-2.12,-1.81), linewidth(0.5) + linetype("2 2") + wrwrwr); draw((-3.374106427962564,-1.859180644233826)--(-0.6831768080801006,0.8926445003478276), linewidth(0.5) + wrwrwr); draw((-7.22,-2.01)--(-0.6831768080801006,0.8926445003478276), linewidth(0.5) + wrwrwr); draw((-6.099976601345422,-0.370596665691723)--(-2.650405986511998,1.1611683644661264), linewidth(0.5) + wrwrwr); draw((-2.650405986511998,1.1611683644661264)--(-6.3,4.73), linewidth(0.5) + wrwrwr); /* dots and labels */ dot((-6.3,4.73),dotstyle); label("$C$", (-6.736363636363641,4.819090909090908), NE * labelscalefactor); dot((-7.22,-2.01),dotstyle); label("$A$", (-7.918181818181823,-2.1809090909090916), NE * labelscalefactor); dot((2.98,-1.61),dotstyle); label("$B$", (3.1545454545454534,-1.8354545454545461), NE * labelscalefactor); dot((0.2460429784135627,-4.405826252986779),linewidth(4pt) + dotstyle); label("$D$", (0.0454545454545434,-5.180909090909091), NE * labelscalefactor); dot((-2.3379346965024017,-0.7995604283541823),linewidth(4pt) + dotstyle); label("$K$", (-2.0818181818181847,-0.8718181818181826), NE * labelscalefactor); dot((-3.374106427962564,-1.859180644233826),linewidth(4pt) + dotstyle); label("$L$", (-3.4090909090909123,-2.308181818181819), NE * labelscalefactor); dot((-6.099976601345422,-0.370596665691723),linewidth(4pt) + dotstyle); label("$H$", (-6.5,-0.3990909090909099), NE * labelscalefactor); dot((-4.37519129392871,0.3952858493872016),linewidth(4pt) + dotstyle); label("$N$", (-4.3,0.5463636363636355), NE * labelscalefactor); dot((-2.12,-1.81),linewidth(4pt) + dotstyle); label("$M$", (-2.0818181818181847,-2.1990909090909097), NE * labelscalefactor); dot((-6.199988300672711,2.1797016671541387),linewidth(4pt) + dotstyle); label("$P$", (-6.118181818181823,2.3281818181818172), NE * labelscalefactor); dot((-0.6831768080801006,0.8926445003478276),linewidth(4pt) + dotstyle); label("$W$", (-0.8272727272727296,1.0736363636363628), NE * labelscalefactor); dot((-2.650405986511998,1.1611683644661264),linewidth(4pt) + dotstyle); label("$Q$", (-2.572727272727276,1.31), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

This post has been edited 3 times. Last edited by microsoft_office_word, Feb 14, 2020, 11:01 PM

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#7 h May 6, 2020, 8:08 PM

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What was the morality of this act?

This post has been edited 2 times. Last edited by Nymoldin, Jan 8, 2022, 10:18 PM
Reason: Cmon,who cares this post?

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#8 h Jun 15, 2021, 6:53 AM

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Let $O$ be circumcenter of $\triangle ABC$ and $T$ midpoint of $CH$ . We know that $TM\parallel CO$ . After writing $\angle CBA=\alpha$ and $\angle ABD=2\beta$ and doing some angle chasing we can find that $\angle HTN=\angle HCO$ . It means that $TN\parallel CO$ $\implies$ $T-N-M$ are collinear.

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#9 h Sep 25, 2021, 7:11 PM

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$\angle LKD=\angle KDB=\angle CDB=\angle CAB=\angle CAL$
$\implies ALKC$ cyclic
let $\angle BAC=x$ and $\angle ABC=y$
$\angle ACD=\angle ACB-\angle DCB=180-x-y-(180-2x)=x-y$
$\angle CAK=\angle CKA=90-\frac{\angle ACD}{2}=90+\frac{y}{2}-\frac{x}{2}$
$\angle ACL=180-\angle LAC-\angle CLA=180-x-\angle CKA=180-x-90+\frac{x}{2}-\frac{y}{2}=90-\frac{x}{2}-\frac{y}{2}=\frac{\angle ACB}{2}$
let $A_1$ and $B_1$ be foot of perpendiculars from $A$ and $B$ to $BC$ and $AC$ respectively
let $E$ be midpoint of $A_1B_1$
let $O$ be midpoint of $HC$
$\angle HB_1C=90=\angle HNC=\angle HA_1C$
$\implies HNA_1CB_1$ cyclic with center $O$
$\angle NB_1A_1=\angle NCA_1=\angle NCB_1=\angle NA_1B_1$
$\implies NB_1=NA_1$
$M, E, O$ collinear because they lie on Gauss line of $B_1HA_1C$
$\angle AB_1B=90=\angle AA_1B$
$\implies AB_1A_1B$ cyclic with center $M$
$A_1B_1$ is radical axis of circles $(B_1HNA_1C)$ and $(AB_1A_1B)$
so $OM$ is perpendicular to $A_1B_1$ , but because it passes through $E$ , midpoint of $A_1B_1$ , it is perpendicular bisector of $A_1B_1$
but $N$ passes through perpendicular bisector of $A_1B_1$ because $NB_1=NA_1$
$\implies M, N, O$ collinear

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#10 h Jan 25, 2025, 10:09 AM

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let $X,Y,$ and $Z$ be the foot of perpendiculars from $C,A,$ and $B$ to $AB, BC,$ and $AC$ respectively. Also, let $P$ and $E$ be the midpoint of $CH$ and $BC$ respectively, which results in $P$ and $E$ are the circumcenters of $(CHN)$ and $(CXB)$ respectively, along with $PE||BZ$ and $EM||AC$

Claim 1: $ALKC$ is cyclic
since $LK||BD$ , we have that $\angle CAB = \angle CDB = \angle LKD = 180 -\angle LKC$ , as desired. Furthermore, since $AC=CK$ , then $\angle ALC = \angle CKA = \angle CAK = \angle CLK$ , making $\angle CLA = \frac{\angle KLA}{2}$

Claim 2: $PXME$ is cyclic
From $\angle PEM = 180 - \angle MEB - \angle PEC = 180 - \angle ACB -\angle ZBC = 90$ , we have that $\angle PXM = \angle PEM = 90$ , therefore $PXME$ is cyclic

For finishing up, just notice that $\angle XPN = 2\angle HCN = 2(90 - \angle HLC) = 180 - \angle ALK = \angle KLM = \angle ABD$ , and $\angle XPM =\angle XEM = \angle XEB - \angle MEB = 2\angle XCB - \angle ACB = 2(90 -\angle XBC) - (180 - \angle CAB - \angle CBA) = 180 -2(\angle CBD - \angle ABD) - (180 - \angle CDB - (\angle (CBD - \angle (ABD)) = 180 -2\angle CBD +2\angle ABD - 180 + \angle CBD + \angle CBD -\angle ABD = \angle ABD$ , which reveals that $\angle XPM = \angle XPN$ , so $M,N,P$ are colinear, thus done :-D

:-D

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