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A circle containing nine-point cener

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Source: 2019 Belarus Team Selection Test 2.2

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#1 h Sep 2, 2019, 7:46 AM • 6 Y

Y by HWenslawski, jhu08, Adventure10, Mango247, lian_the_noob12, ItsBesi

Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$ . Point $T$ is the midpoint of the segment $AO$ . The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$ .
Prove that the circumcircle of the triangle $AST$ bisects the segment $OH$ .

(M. Berindeanu, RMC 2018 book)

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#2 h Sep 2, 2019, 7:55 AM • 4 Y

Y by A-Thought-Of-God, HWenslawski, jhu08, Adventure10

Vlados021 wrote:

Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$ . Point $T$ is the midpoint of the segment $AO$ . The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$ .
Prove that the circumcircle of the triangle $AST$ bisects the segment $OH$ .

Let $O'$ be the reflection of $O$ in $BC$ . We get $AOO'H$ is a parallelogram, so $N$ is midpoint of $AO'$ where $N$ is midpoint of $OH$ . Further, $SA = SO = SO'$ and as $N$ is midpoint of $AO'$ , $SN\perp AO'$ . This gives $\angle ATS = 90^{\circ} = \angle ANS$ and we are done.

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#3 h Sep 27, 2019, 3:58 PM • 2 Y

Y by jhu08, Adventure10

Let $AD$ be altitude of $\triangle ABC$ and $N$ the midpoint of $OH$ .
Then $\angle ADS = 90^{\circ} = \angle ATS$ .
So $ASTD$ is cyclic.
$T$ and $N$ are midpoints of $OA$ and $OH$ . $\therefore$ $TN \parallel AH$ .
Also $N$ is the centre of NPC of $\triangle ABC$ and $D$ lies on NPC.
$\therefore$ $ND = \frac{R}{2} = AT$ .
Also $AD > TN$ as $\triangle ABC$ is acute.
So $ADNT$ is an isosceles trapezium $\implies ADNT$ is cyclic.
So $ATNDS$ is cyclic $\implies N$ lies on $\odot AST$ as desired.
Q.E.D.

This post has been edited 4 times. Last edited by Euler365, Sep 27, 2019, 4:00 PM

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brainiacmaniac31

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brainiacmaniac31

#4 h Sep 28, 2019, 12:01 AM • 3 Y

Y by jhu08, Adventure10, Mango247

Could there be a solution with 9-point circle?

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#5 h Dec 26, 2020, 12:13 AM • 1 Y

Y by jhu08

$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -23.696792103092832, xmax = 22.46131210341322, ymin = -14.856517958715257, ymax = 13.984754996461037; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); draw((1.4152989376448384,5.869067483969411)--(-6.331542670121041,-3.331876842159191)--(4.938046654327134,-9.660184693580092)--cycle, linewidth(2) + zzttqq); /* draw figures */ draw(circle((1.6982712469743642,-2.2309279852220922), 8.104936762786627), linewidth(0.8) + red); draw((1.4152989376448384,5.869067483969411)--(-6.331542670121041,-3.331876842159191), linewidth(0.4) + blue); draw((-6.331542670121041,-3.331876842159191)--(4.938046654327134,-9.660184693580092), linewidth(0.4) + blue); draw((4.938046654327134,-9.660184693580092)--(1.4152989376448384,5.869067483969411), linewidth(0.4) + blue); draw((1.4152989376448384,5.869067483969411)--(1.6982712469743642,-2.2309279852220922), linewidth(0.4) + blue); draw((-6.331542670121041,-3.331876842159191)--(-14.50520038116374,1.2579463340417105), linewidth(0.4) + blue); draw((-14.50520038116374,1.2579463340417105)--(1.5567850923096014,1.8190697493736592), linewidth(0.4) + blue); draw(circle((-6.544950721759452,3.5635069090055604), 8.287411188355128), linewidth(0.8) + red); draw((1.4152989376448384,5.869067483969411)--(-4.369931193388714,-4.433397132939652), linewidth(0.4) + blue); draw((-3.374739572097797,-2.6611380813256886)--(1.6982712469743642,-2.2309279852220922), linewidth(0.4) + blue); draw((-2.4581218662381015,1.2685953209051097)--(3.176672795985986,-1.8955586048053408), linewidth(0.4) + blue); draw(circle((1.5567850923096014,1.8190697493736592), 4.052468381393314), linewidth(0.8) + red); draw((1.5567850923096014,1.8190697493736592)--(-0.8382341625617165,-2.44603303327389), linewidth(0.4) + blue); /* dots and labels */ dot((1.4152989376448384,5.869067483969411),dotstyle); label("$A$", (1.5242373849065534,6.171062846470806), NE * labelscalefactor); dot((-6.331542670121041,-3.331876842159191),dotstyle); label("$B$", (-6.198948678404263,-3.030389299270584), NE * labelscalefactor); dot((4.938046654327134,-9.660184693580092),dotstyle); label("$C$", (5.0539747654040745,-9.36581536683023), NE * labelscalefactor); dot((-3.374739572097797,-2.6611380813256886),linewidth(4pt) + dotstyle); label("$H$", (-3.242416513543091,-2.427015388074427), NE * labelscalefactor); dot((1.6982712469743642,-2.2309279852220922),linewidth(4pt) + dotstyle); label("$O$", (1.8259243405046321,-2.0046536502371173), NE * labelscalefactor); dot((-0.8382341625617165,-2.44603303327389),linewidth(4pt) + dotstyle); label("$N$", (-0.7082460865192295,-2.2158345191557722), NE * labelscalefactor); dot((1.5567850923096014,1.8190697493736592),linewidth(4pt) + dotstyle); label("$T$", (1.6750808627055926,2.0681202503369405), NE * labelscalefactor); dot((-14.50520038116374,1.2579463340417105),linewidth(4pt) + dotstyle); label("$S$", (-14.374665175112197,1.4949150347005917), NE * labelscalefactor); dot((-4.369931193388714,-4.433397132939652),linewidth(4pt) + dotstyle); label("$D$", (-4.237983467016751,-4.2069684261030895), NE * labelscalefactor); dot((3.176672795985986,-1.8955586048053408),linewidth(4pt) + dotstyle); label("$M$", (3.304190422935218,-1.6426293035194235), NE * labelscalefactor); dot((-2.4581218662381015,1.2685953209051097),linewidth(4pt) + dotstyle); label("$E$", (-2.337355646748855,1.4949150347005917), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

I will post this somewhat different solution. BTW the problem is ultra-cool

$\color{black}\rule{25cm}{1pt}$
Let's add the midpoints of $AC$ and $AB$ , we denote them with $M$ and $E$ respectively. Also let's denote with $D$ the foot of the A-altitude of $ABC$ .

Since we have that $\angle ATS=90$ we also have that $D$ belongs on the circle $(ATS)$ , this holds since we have that $90=\angle BDA=\angle SDA$ .
Now I present to you a nice little lemma, which used kills the problem.
$\color{red}\rule{25cm}{1pt}$
Lemma: The center of the nine-point circle of $ABC$ , when reflected across $ME$ gives us $T$ .
Proof:
Lets call the center of the nine-point circle $N$ , and let's denote its reflection across $ME$ with $T'$ .
Since we have that $NE=NM$ , we must have that $T'E=T'M$ .
But notice that when we reflect $D$ across $ME$ we must get $A$ , since $ME$ cuts the A-altitude in half because of the midpoints $M$ and $E$ . Thus we have that $ND=T'A$ .
This implies that $T'$ is the circumcenter of $(AME)$ .
Now let's do a homothety centered at $A$ and having a coefficient of $\frac{1}{2}$ , we easily see that $O$ gets sent into $T$ , since homothety preserves circumcenters we have that $T' \equiv T$ .
$\color{red}\rule{25cm}{1pt}$

Now back to our problem at hand.

Because we have that $NT \parallel AD$ and because of the lemma we have that $TNDA$ is an isoceles trapezoid. This implies that $N\in (ATD)$ .
But this implies that $N \in (ATS)$ and since $N$ is the midpoint of $OH$ we come to the conclusion of the problem.

This post has been edited 1 time. Last edited by EulersTurban, Dec 26, 2020, 9:25 AM
Reason: Forgot '

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DNCT1

#6 h Dec 31, 2020, 12:20 PM • 2 Y

Y by Kanep, jhu08

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$\textit{Proof}$
Let $AE$ be the altitude of $\bigtriangleup ABC$ and $AE\cap (O)= F\neq A$
We easily to show that $E$ is the midpoint of $HF$ by angle chasing here
Now let $P$ is the midpoint of $HO$
We have $EP=\frac{OF}{2}=\frac{OA}{2}=AT$ and $TP\parallel AH\equiv AE$ so $ATPE$ is isoceles trapezoid.
So $\angle OTP=\angle OAH=\angle AEP=\angle ASP$ $\implies (AST)\quad\text{passes through}\quad P$
The end the proof. $\quad\blacksquare$

This post has been edited 1 time. Last edited by DNCT1, Dec 31, 2020, 12:23 PM

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#7 h Sep 17, 2021, 1:00 AM • 1 Y

Y by jhu08

Let $D$ be the $A$ altitude on $\triangle ABC$ , let $H_A$ be the reflection of $H$ over $D$ , let $A'$ be the antipode of $A$ on $(O)$ and $A''$ be the reflection of $H_A$ over $O$
First $180-\angle BAC=\angle BHC=\angle BH_AC$ thus $H_A$ lies on $(O)$ thus we have that $A''$ is antipode of $H_A$ on $(O)$
Now $\angle ATS=\angle ADS=90$ thus $D$ lies on $(ATS)$ and note that $AH_AA'A''$ is a rectangle thus we have that $\widehat{AA''}=\widehat{H_AA'}$ were we took arcs w.r.t. $(O)$ thus we have $\angle DAT=\angle HH_AO=\angle ADN_9$ and since by midbase $AD \parallel TN_9$ thus we have that $ATN_9D$ isosceles trapezoid thus $N_9$ lies on $(ATS)$ and we are done :blush:

:blush:

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#8 h Sep 17, 2021, 2:43 AM • 1 Y

Y by jhu08

Vlados021 wrote:

Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$ . Point $T$ is the midpoint of the segment $AO$ . The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$ .
Prove that the circumcircle of the triangle $AST$ bisects the segment $OH$ .

(M. Berindeanu, RMC 2018 book)

If someone want to avoid the synthetic solution, an overkill using $\text{Linearity of Power of Point}$ , also exist..

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#9 h Sep 17, 2021, 1:04 PM

Y by

Let $D$ be the foot of the $A$ -altitude, $A_1$ lie on $(ABC)$ such that $AA_1 \parallel BC$ , $M$ be the midpoint of $BC$ , and $N_9$ be the Nine-Point center of $ABC$ , which coincides with the midpoint of $OH$ .

Because $\angle ADS = 90^{\circ} = \angle ATS$ we know $ATDS$ is cyclic. It's easy to see $N_9T$ is a midline of $AOH$ , so $N_9T \parallel AD$ . Now, it suffices to show $ADN_9T$ is an isosceles trapezoid.

Since $DHOM$ is a trapezoid, we have $dist(N_9, BC) = \frac{OM + DH}{2}$ by midline properties. It's well-known that $2 \cdot OM = AH = 2 \cdot N_9T.$ Hence, $dist(T, AA_1) = AH - N_9T - [dist(N_9, BC) - DH]= N_9T - \left( \frac{OM + DH}{2} \right) + DH$ $= OM - \frac{OM}{2} + \frac{DH}{2} = \frac{OM + DH}{2} = dist(N_9, BC)$ which clearly finishes. $\blacksquare$

Remark: Because proving $ADN_9T$ is an isosceles trapezoid via angle chasing feels somewhat circular, we are motivated to length chase.

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primesarespecial

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primesarespecial

#10 h Sep 22, 2021, 1:51 PM

Y by

This is fairly simple by complex numbers....

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Mahdi_Mashayekhi

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Mahdi_Mashayekhi

#11 h Dec 9, 2021, 6:30 PM

Y by

Let D be midpoint of OH. We have to prove ATDS is cyclic. Let AP be altitude of ABC.
(1) ∠ATS = ∠APS = 90 ---> ATPS is cyclic.
(2) ∠PST = ∠PAT = ∠DTO ---> STDP is cyclic.
from (1) & (2) we have ATDPS is cyclic so ATDS is cyclic as well.

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#12 h Jan 4, 2022, 6:08 AM

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Let $N_9$ be the midpoint of $HO$
Let $AP$ be $A-$ altitude of $ABC$
Let $L$ such that $OL \perp BC$ i.e $L$ be the midpoint of $BC$ .
Let $U$ be the midpoint of $PL$
Note that: $ATPS$ is cyclic
$\implies \angle TSP=\angle HAO=\angle N_9TO$
Now: $TN_9LO$ is paralelogram. $\implies \angle N_9TO=\angle N_9LO=\angle UN_9L=\angle PN_9U$
$\implies SPN_9T$ is cyclic.
$\implies A,T,N_9,P,S$ is cyclic
i.e $A,S,T,N_9$ is cyclic. $\blacksquare$

This post has been edited 1 time. Last edited by Albert123, Jan 4, 2022, 6:21 PM

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#13 h Jan 4, 2022, 7:48 PM

Y by

Vlados021 wrote:

Let $O$ be the circumcenter and $H$ be the orthocenter of an acute-angled triangle $ABC$ . Point $T$ is the midpoint of the segment $AO$ . The perpendicular bisector of $AO$ intersects the line $BC$ at point $S$ .
Prove that the circumcircle of the triangle $AST$ bisects the segment $OH$ .

(M. Berindeanu, RMC 2018 book)

Let $D$ be the foot of perpendicular form $A$ upon $BC$ .

Claim: $D$ lies on circumcircle of $\Delta AST$
Proof: Follows easily from the fact that $\Delta ASD$ is a right angle triangle.
Define $M$ as the midpoint of $BC$ . Let $N$ be mid point of $OH$ .
Now $TN = \frac{AH}{2} = OM$ Also $AH \parallel TS \parallel OM$ which gives us that $TOMN$ is a parallelogram.
Extend $TN$ to meet $BC$ at $P$ . Invoking the fact that N is the nine- point center we get that $DP=PM$
Now $\angle DAO = \angle NTO = \angle PNM =\angle DNP$ Therefore $\angle DNT = \pi - \angle DNP = \pi - \angle DAT$ implying $N$ lies on circumcircle of $\Delta AST$

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BVKRB-

#14 h Jan 7, 2022, 5:24 PM • 2 Y

Y by REYNA_MAIN, nathantareep

Solved while killing irritating mosquitoes bare handed

Posting for storage
Let $N_9$ denote the nine point centre that is the midpoint of $OH$ and let $D$ be the $A-$ Altitude in $\triangle ABC$ , $M$ be the midpoint of $BC$
Observe that $\odot(ATSD)$
Now from the well known fact that the nine point radius is half of the circumradius we easily get by some angle chasing that $TN_9MO$ is a parallelogram which implies $OM \parallel TN_9 \parallel AD$ which with the fact that $AT=DN_9$ implies that $ATN_9D$ is a cyclic isosceles trapezoid (

) which implies the desired conclusion from the first observation $\blacksquare$

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1080 posts

#15 h Jan 26, 2022, 4:10 AM • 2 Y

Y by centslordm, megarnie

Let $N_9$ be the Nine-Point center of $\triangle ABC,$ noting that $N_9$ is the midpoint of $\overline{OH}.$ Also, let $D=\overline{AH}\cap\overline{BC},$ noting $D$ lies on $(AST)$ since $\angle ADS=\angle ATS=90.$ Also, $\overline{NN_9}\parallel\overline{AD}$ by similarity. Because $N_9D=\tfrac{1}{2}AO=AT,$ $ATN_9D$ is cyclic. $\square$

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#16 h Dec 6, 2022, 9:40 AM

Y by

straightforward complex bash

Let $(ABC)$ be unit circle and $N$ midpoint of $OH$ .
$t=\frac{a}{2} \wedge \bar{t}=\frac{1}{2a}$ $n=\frac{a+b+c}{2} \wedge \bar{n}=\frac{ab+bc+ca}{2abc}$ Since $T$ is the foot of perpendicular from $S$ to $AB$ we have
$t=\frac{\bar{a}s+a\bar{s}}{2\bar{a}} \iff \bar{s}=\frac{a-s}{a^2}(1)$ On the other hand since $S$ lies on $BC$ we have
$s+\bar{s}bc=b+c \iff \bar{s}=\frac{b+c-s}{bc}(2)$ İf we equalize $1$ and $2$ we easily get
$s=\frac{a(bc-ab-ac)}{bc-a^2} \wedge \bar{s}=\frac{b+c-a}{bc-a^2}$ Now it remains to show that
$\frac{(t-a)(n-s)}{(t-s)(n-a)} \in \mathbb{R} \iff \frac{-\frac{a}{2}(\frac{(a^2+bc)(b+c-a)}{2(bc-a^2)})}{\frac{a(2ab+2ac-a^2-bc)}{2(bc-a^2)}(\frac{b+c-a}{2})} \in \mathbb{R} \iff \frac{a^2+bc}{a^2+bc-2ab-2ac} \in \mathbb{R}$
so we are done

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#17 h Nov 14, 2024, 5:15 PM

Y by

My solution is a bit similar to #3 but not quite the same.
Let $D$ be the feet of the altitude from $A$ to $BC$ .

Claim: Points $A,T,D$ and $S$ are concyclic.

Proof:

Let $M$ and $N$ be the midpoints of $BC$ and $OH$ respectively.

Claim: The quadrilateral $\square TOMN$ is a parallelogram

Proof:

Claim: $ND=NM$

Proof:

Claim: The quadrilateral $\square ATND$ is an isosceles trapezoid

Proof:

Claim: Points $A,S,T$ and $N$ are concyclic.

Proof:

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460 posts

#18 h Jan 2, 2025, 3:52 AM

Y by

post for storage. Duplicate with a lot of solutions.

Let $AH$ intersects $BC$ at $H_A$ . We have $\angle AH_AS=\angle ATS=90^{\circ}$ so $ATH_AS$ is cyclic.
Then as $AT=TO$ and $ON=NH$ we know $TN$ is the $O$ -midline of $AOH$ . So $TN\parallel OH$ and $TNH_AA$ is a trapezoid.
As $AT=H_AN=\frac{R}{2}$ , we know that $TNH_AA$ is isosceles and cyclic, therefore $ATNH_AS$ is cyclic and so $STAN$ is cyclic.

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ezpotd

#19 h Jan 2, 2025, 7:19 AM

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quick complex numbers:

$t = \frac a2$ , $TS \cap (ABC) = aw, \frac aw$ for $w = e^{\frac 13 \pi i }$ , so $s = TS\cap BC = \frac{a^2(b + c) - bc(a)}{a^2 - bc}$ . Then we desire to show $a,s,t$ is cyclic with $\frac{a + b + c}{2}$ . This is equivalent to showing $\frac{t - \frac{a + b + c}{2}}{a - \frac{a + b + c}{2}}\frac{a - s}{t - s}$ is self conjugating, simplifying we get $\frac{b + c}{b + c - a}2 \frac{a^3 - a^2b-a^2c}{a^3 - 2a^2b-2a^2c+abc} = 2a \frac{b + c}{(a^2-2ab-2ac+bc)}$ , which is obviously self conjugating.

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