Hello, should $T$ be the foot of the altitude from $A$? If so, it suffices to prove that $A',N,M$ are collinear, where $A'$ is the reflection of $A$ over $BC$. We have $AMA'$ isoceles, so it suffices to prove $\angle AMB=\angle NMB$. There is a spiral similarity centered at $N$ taking $AB$ to $MC$. To prove this, first note $\angle MCN=\angle BAN$. We can then find $\angle MNC=\angle BNA$ by using trig/ratio lemma on $BNC$ (using $\frac{KB}{KC}=\frac{c^2}{b^2}$ where $K=AC\cap AN$). Then we have $\triangle NBM\sim NAC$ and clearly $\triangle NAC\sim \triangle BAM\implies \triangle NBM\sim BAM$ and $\angle AMB=\angle NMB$ as desired.

droid347 wrote: Hello, should $T$ be the foot of the altitude from $A$? Agreed, my current diagram is definitely not showing $R,S,T$ are collinear, $T$ being the foot of the altitude however makes sense.

I consider $T$ the foot of the A-altitude: Let $D$ be a point such that $DB$ and $DC$ are tangent to $\odot (ABC)$ $\Longrightarrow$ since $AN$ is A-symmedian we get $A$, $N$, $D$ are collinear. Let $E$ $=$ $AN$ $\cap$ $BC$ $\Longrightarrow$ $(A,N,E,D)=-1$ and $\measuredangle EMD$ $=$ $90^{\circ}$ combining these two results we get $\measuredangle AME$ $=$ $\measuredangle EMN$, but since $TR$ $=$ $RM$ we get $\measuredangle RTM$ $=$ $\measuredangle RMT$ $\Longrightarrow$ $\measuredangle RTM$ $=$ $\measuredangle EMN$ $\Longrightarrow$ $TR\parallel MN...(1)$, but since $S$ and $R$ are the midpoints of $AN$ and $AM$ we get $SR\parallel MN...(2)$ $\Longrightarrow$ by $(1)$ and $(2)$ we get $R,S,T$ are collinear.

Jutaro wrote: Let $ABC$ be an acute-angled triangle, $\Gamma$ its circumcircle and $M$ the midpoint of $BC$. Let $N$ be a point in the arc $BC$ of $\Gamma$ not containing $A$ such that $\angle NAC= \angle BAM$. Let $R$ be the midpoint of $AM$, $S$ the midpoint of $AN$ and $T$ the foot of the altitude through $A$. Prove that $R$, $S$ and $T$ are collinear. if so then: $AN$ is the symmedian,it s easy to see that $NM$ is the symmetric of the median wrt the $BC$-bisector then also wrt $BC$ ,let $A'$ be the symmetric of $A$ wrt $BC$ thus $N,M,A'$ are collinear therefore $S,T,R$ are collinear. R HAS

Take a homothety with factor 2 at $A$ and the problem becomes $M$, $N$, and $A'$, the reflection of $A$ over $BC$, collinear. A $\sqrt{bc}$ inversion on $ABC$ swaps $M$ and $N$ and takes $A'$ to $O$, the circumcenter of $ABC$. An inversion about the circumcircle of $ABC$ fixes $A$ and $N$ and sends $M$ to the intersection of the tangents to the circumcircle at $B$ and $C$, $D$. Since $AND$ is a symmedian, $O$ lies on the circumcircle of $AMN$, so $A'$ lies on $MN$.

$\angle RTM= \angle AMB= \angle MAC+ \angle ACB= \angle BAN+ \angle ANB= \angle BCN+ \angle MNC= \angle BMN$ $\Longrightarrow TR \parallel NM$ $\Longrightarrow TR$ passes through the midpoint of $AN$ which is $S$. Done!

I am sorry for the mistake. Indeed $T$ is meant to be the foot of the altitude through $A$.

Hello. A different approach. [asy][asy]import graph; size(10cm); 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 = -6.462330210644301, xmax = 12.702791963393949, ymin = -4.825510780754414, ymax = 6.017913607188257; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((0.9045997773425172,3.386664773003271)--(0.,0.)--(5.,0.)--cycle, aqaqaq); /* draw figures */ draw((0.9045997773425172,3.386664773003271)--(0.,0.), uququq); draw((0.,0.)--(5.,0.), uququq); draw((5.,0.)--(0.9045997773425172,3.386664773003271), uququq); draw(circle((2.5,1.1463786166634775), 2.7503061525479793)); draw((0.9045997773425172,3.386664773003271)--(0.9045997773425171,0.)); draw((-3.8510083135758397,0.)--(0.,0.)); draw((-3.8510083135758397,0.)--(0.9045997773425172,3.386664773003271)); draw((0.9045997773425172,3.386664773003271)--(2.5,0.)); draw((0.9045997773425172,3.386664773003271)--(1.7885199017311508,-1.5103072892182503)); draw((-3.8510083135758397,0.)--(1.346559839536834,0.9381787418925103)); draw((1.346559839536834,0.9381787418925103)--(0.9045997773425171,0.), linetype("2 2")); draw((2.5,0.)--(1.7885199017311508,-1.5103072892182503), linetype("2 2")); /* dots and labels */ dot((0.9045997773425172,3.386664773003271),linewidth(3.pt) + dotstyle); label("$A$", (0.8326646018420635,3.550224003952134), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.24807537037813854,-0.40), NE * labelscalefactor); dot((5.,0.),linewidth(3.pt) + dotstyle); label("$C$", (5.065562826371188,-0.41248922752193157), NE * labelscalefactor); dot((0.9045997773425171,0.),linewidth(3.pt) + dotstyle); label("$T$", (0.75,-0.4665262261329416), NE * labelscalefactor); dot((-3.8510083135758397,0.),linewidth(3.pt) + dotstyle); label("$D$", (-4.012652940278509,0.18191775719917827), NE * labelscalefactor); dot((2.5,0.),linewidth(3.pt) + dotstyle); label("$M$", (2.5078115587833767,-0.41248922752193157), NE * labelscalefactor); dot((1.7022998886712586,1.6933323865016354),linewidth(3.pt) + dotstyle); label("$R$", (1.9674415726732757,1.6229043868261113), NE * labelscalefactor); dot((1.515904295599791,0.),linewidth(3.pt) + dotstyle); label("$Y$", (1.65,-0.37646456178125826), NE * labelscalefactor); dot((1.7885199017311508,-1.5103072892182503),linewidth(3.pt) + dotstyle); label("$N$", (1.6792442467478885,-1.943537521500548), NE * labelscalefactor); dot((1.346559839536834,0.9381787418925103),linewidth(3.pt) + dotstyle); label("$S$", (1.5,0.7763247419202881), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Let $D$ be the intersection of $BC$ with the tangent to $\Gamma $ at $A$ and suppose that $Y\equiv AN\cap BC$. The line $AN$ is the $A-$symmedian of $\triangle{ABC}$.Hence,we know that the quadruple $D,Y/B,C$ is harmonic and that $AN$ is the polar of $D$ with respect to $\Gamma $ (the above are well known but whoever needs more explanation may tell me). Thus,$DS\perp AN$ which means that $ASTD$ is cyclic.It suffices to show that $AMND$ is also cyclic,because,then, $\angle{MNA}=\angle{MDA}=\angle{MTS}\Rightarrow MN\parallel ST$,which is the desired result.Equivalently,it suffices to show that $DY\cdot MY=AY\cdot NY\Leftrightarrow DY\cdot MY=BY\cdot CY\Leftrightarrow (DM-MY)MY=(BM-MY)(BM+MY)\Leftrightarrow $ $\Leftrightarrow MD\cdot MY=MB^2$ which is Newton's relation for the harmonic quadruple $D,Y/B,C$.

Consider $AM \cap \Gamma =D$ Since $RS||MN$ it's enough to prove that $TR||MN$ 1) $R$ is the midpoint of $AM$ $ \Rightarrow TR=AR=RM \Rightarrow \angle RTM = \angle RMT$ 2)$\angle NAC = \angle BAM \Rightarrow \angle BAN = \angle MAC = \angle DAC$ then $BN=CD$ 3) $\angle MBN = \angle MCD$ and $BM=MC \Rightarrow \triangle BMN \cong \triangle CMD$ ($SAS$) $\longrightarrow \angle BMN = \angle CMD = \angle AMB = \angle RMT$ 4)Then, $\angle BMN = \angle RMT = \angle RTM \Rightarrow TR||MN$ $Q.E.D$

I found a tricky solution using spiral similarity twice and I thought this problem was too much for a 2nd problem in a omcc, but maybe the solution of Packito gives sense to the selection,...

¿Existe forma de resolverlo partiendo de que, por menelao en el triangulo AMX, con x el corte de AN con BC, y los puntos R,T y S, se tiene que para que sean colineales es porque AR/RM * MT/TX * XS/SA = 1, y como AR/RM=1, entonces MT/TX=SA/SX

Using Menelaus? I think you outlined it. I think spiral similarity is more straight-forward.

Let $A' \in \odot(ABC)$ such that $AA'\parallel BC$ and let $T'$ be the reflection of $A$ over $BC$; it's easy to see that $T', M, A'$ are collinear, so $$-1=(B,C;M,P_{\infty,BC})\stackrel{A'}{=}(B,C;N,A)$$and a homothety at $A$ with ratio $\tfrac{1}{2}$ finishes the problem.

Hello. I'm the happy author of this problem. I hope everyone who participated in OMCC 2016 had overjoyed it. I used it as a lemma to solve a problem from Sharygin Geometry Olympiad, correspondence round.

I am giving a solution; assume AM intersects the circumcircle at D here is RS || MN if R,S,T is collinear, it should be RT || MN in triangle RTM , R is the midpoint of hypontenous AM so ,easily AR = RM = RT; <RTM = <RMT (1) <RTM = <CMN = 180 - <BMN; <RMT = 180 - <AMB = 180 - <CMD; from (1) ,we need to prove that, <BMN = <CMD; <NAC = <BAM; or, <CAD = <BAN; from sine law, BN = CD; also, <CAN = <BAM; or, <CAN = <BAD; or, <CBN = <BCD; or, <MBN = <MCD; in triangle BMN and CMD, BM = CM; BN = CD; and <MBN = <MCD; so, triangle BMN and triangle CMD is simmilar, now, <BMN = <CMD this proofs the concurrency of R,S,T btw my first solution of higher olympiad level geometry as centroamerican

brothers can you give me any idea about homothaty and reflection in mathmetics

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ 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 = -15.653987447520949, xmax = 21.54578172460581, ymin = -17.457492228471796, ymax = 4.6489279762757745; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); draw((-3.7012747299195645,3.4292634132552204)--(-6.08,-3.99)--(5.04,-4.15)--cycle, linewidth(4) + rvwvcq); /* draw figures */ draw((-3.7012747299195645,3.4292634132552204)--(-6.08,-3.99), linewidth(4) + rvwvcq); draw((-6.08,-3.99)--(5.04,-4.15), linewidth(4) + rvwvcq); draw((5.04,-4.15)--(-3.7012747299195645,3.4292634132552204), linewidth(4) + rvwvcq); draw(circle((-0.48579318418222933,-1.692626300664937), 6.047567760061765), linewidth(2.8)); draw((-3.7012747299195645,3.4292634132552204)--(0.9623058187584692,-7.564260237672644), linewidth(2.8)); draw((-2.1022256244814166,-7.520166260072069)--(-3.7012747299195645,3.4292634132552204), linewidth(2) + wrwrwr); draw((-3.7012747299195645,3.429263413255214)--(xmin, 69.5000000000002*xmin + 260.6678571426657), linewidth(2.8)); /* ray */ draw((-6.08,-3.99)--(-0.7070972157740317,-17.073256496295194), linewidth(2.8)); draw((-0.7070972157740317,-17.073256496295194)--(5.04,-4.15), linewidth(2.8)); draw((-0.7070972157740317,-17.073256496295194)--(-3.7012747299195645,3.4292634132552204), linewidth(3.2)); draw((-0.52,-4.07)--(-0.7070972157740317,-17.073256496295194), linewidth(2.8) + dtsfsf); draw((-0.52,-4.07)--(xmin, 2.1805779192856143*xmin-2.93609948197148), linewidth(2.8) + dtsfsf); /* ray */ /* dots and labels */ dot((-3.7012747299195645,3.4292634132552204),linewidth(6pt) + dotstyle); label("$A$", (-4.463565081807408,3.886637624387928), NE * labelscalefactor); dot((-6.08,-3.99),linewidth(6pt) + dotstyle); label("$B$", (-7.055352278226076,-4.4070814041518505), NE * labelscalefactor); dot((5.04,-4.15),linewidth(6pt) + dotstyle); label("$C$", (5.38522626458353,-4.6815059308314755), NE * labelscalefactor); dot((-0.52,-4.07),dotstyle); label("$M$", (-0.19473911123548462,-3.705774280415031), NE * labelscalefactor); dot((0.9623058187584692,-7.564260237672644),dotstyle); label("$M'$", (0.537059626576845,-8.49295769027071), NE * labelscalefactor); dot((-2.110637364959782,-0.3203682933723899),dotstyle); label("$R$", (-1.7498114290866853,-0.016288977277850253), NE * labelscalefactor); dot((-3.7012747299195623,3.429263413255221),dotstyle); dot((-2.1022256244814166,-7.520166260072069),dotstyle); label("$N$", (-3.1219340624848035,-8.035583479138003), NE * labelscalefactor); dot((-2.9017501772004906,-2.0454514234084247),dotstyle); label("$S$", (-2.6035766232010698,-1.784802593657656), NE * labelscalefactor); dot((-3.8084969876150505,-4.022683496581078),dotstyle); label("$T$", (-4.616023152184977,-4.803472387133531), NE * labelscalefactor); dot((-3.915719245310536,-11.474630406417367),dotstyle); label("$D$", (-4.920939292940114,-11.969001694879294), NE * labelscalefactor); dot((-0.7070972157740317,-17.073256496295194),dotstyle); label("$K_A$", (-0.5911300942171632,-16.756185104734975), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $MN \cap AT=D$, and let the tangents at $B$ and $C$ intersect at $K_A$, $$-1=(A,N; AN \cap BC,K_A) \overset{M}{=} (A,D;T, \infty_{MK_A}) \Longrightarrow AT=TD$$and then finish off using mid-point theorem

Let $A'$ be the reflection of $A$ over $BC$. It suffices to show that $M,N,A'$ are collinear. We know that the reflection of the $A-$ HM point over $BC$ is $N$. Thus it follows that if $N'$ is the reflection of $N$ over $BC$ then $N'\in \odot(A'BC)$. Also, $N'\in AM$ from definition. Reflecting $A,N',M$ over $BC$ yields that $A',M,N$ are collinear. $\blacksquare$.

Here is my solution for this problem Solution Let $P$ $\in$ ($O$) such as $AT$ $\parallel$ $BC$; $A'$ be reflection of $A$ through $BC$ We have: 1 = $\dfrac{MB}{MC}$ = $\dfrac{AC}{AB}$ . $\dfrac{NB}{NC}$ = $\dfrac{PB}{PC}$ . $\dfrac{NB}{NC}$ or $N$, $M$, $P$ are collinear But: $\dfrac{A'T}{A'A}$ = $\dfrac{TM}{AP}$ = $\dfrac{1}{2}$ so: $A'$, $M$, $P$ are collinear Then: $P$, $M$, $N$, $A'$ are collinear or $R$, $S$, $T$ are collinear

I've got synthetic one. I used notations as in diagram in post #10 $AB=c\wedge BC=a\wedge CA=b$ We want to prove $MN||ST$ which is $\frac{MY}{NY}=\frac{TY}{SY}$ Line $AN$ is $A-$symedian so $BY=\frac{ac^2}{b^2+c^2}\wedge CY=\frac{ab^2}{b^2+c^2}$ $MY=|\frac{a}{2}-\frac{ab^2}{b^2+c^2}|=\frac{a|b^2-c^2|}{2(b^2+c^2)}$ In $BT$ exceptionally the length is signed, so we can work even if $ABC$ isn't acute $BT=c\cos\angle ABC=\frac{a^2+c^2-b^2}{2a}$ $TY=|BY-BT|=|\frac{ac^2}{b^2+c^2}-\frac{a^2+c^2-b^2}{2a}|=\frac{|b^2-c^2||a^2-b^2-c^2|}{2a(b^2+c^2)}$ Concyclicity yields $AY\cdot NY=BY\cdot CY=\frac{a^2b^2c^2}{(b^2+c^2)^2}$ $SY=|NS-NY|=\frac{|AY-NY|}{2}$ which is true By Stewart's theorem $AY^2=\frac{BY\cdot AC^2+CY\cdot AB^2}{BC}-BY\cdot CY=\frac{2b^2c^2}{b^2+c^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}$ $\frac{MY}{NY}=\frac{TY}{SY}\iff \frac{\frac{a|b^2-c^2|}{2(b^2+c^2)}}{NY}=\frac{\frac{|b^2-c^2||a^2-b^2-c^2|}{2a(b^2+c^2)}}{\frac{|AY-NY|}{2}}\iff \frac{a^2}{ AY\cdot NY}=\frac{2|a^2-b^2-c^2|}{|AY^2-NY\cdot AY|}\iff$ $\frac{a^2}{\frac{a^2b^2c^2}{(b^2+c^2)^2}}=\frac{2|a^2-b^2-c^2|}{|\frac{2b^2c^2}{b^2+c^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}-\frac{a^2b^2c^2}{(b^2+c^2)^2}|}\iff \frac{1}{b^2c^2}=\frac{2|a^2-b^2-c^2|}{2b^2c^2|b^2+c^2-a^2|}$

It is clear that $AN$ is the $A$-symmedian of $\triangle ABC$. Let $A'=AT\cap MN$. Also let $R$ and $K$ be the intersections of the $\angle A$ bisector with $(ABC)$ and $BC$ respectively. By Russia 2009 $NMKR$ is cyclic. So by $\angle MA'T=90-\angle NMK=90-\angle NCA=90-\angle C-\angle CAM=\angle AMC-90=\angle MAT$. we get that $T$ is the midpoint of $AA'$ and by homothety we are done.

First, $\angle MAC + \angle NAM = \angle NAC = \angle BAM = \angle BAN + \angle NAM$ $\Rightarrow$ $\angle BAN = \angle MAC$. In the $\triangle AMN$ we have that $\frac{AS}{AN} = \frac{AR}{AM} = \frac{1}{2}$, so by Thales's theorem we get that $SR \parallel MN$. As $\angle BAM = \angle NAC$ and $\angle ANC = \angle ABC$ by case $AA$, $\triangle BAM$ $\sim$ $NAC$ . Then, $\frac{MA}{AC} = \frac{BM}{NC} = \frac{MC}{CN}$. Notice that $\angle BAN = \angle BCN = \angle MAC$ so by case $LAL$ we get that $\triangle MAC \sim \triangle MCN$. Hence, $\angle AMC = \angle NMC = 180^{\circ} - \angle BMA = 180^{\circ} - \angle BMN \Rightarrow \angle BMA = \angle BMN$ We see that $R$ Is circumcenter of $\triangle ATM$ $\Rightarrow$ $RA = RT = RM \Rightarrow \angle TMR = \angle MTR = \angle BMA = \angle BMN$. Hence, $\angle ART = 2 \angle AMT = 2 \angle BMA = \angle BMA + \angle BMN = \angle AMN$ It follows that $TR \parallel MN $ and since $SR \parallel MN$, then $R$, $S$ and $T$ lies on the same line parallel to $MN$ $\blacksquare$

Attachments:

Let $P_A$ be the $A$ Humpty point of triangle $ABC$, $A'$ is the reflection of $A$ over line $BC$. We have $P_A$ is the reflection of $N$ over line $BC$. Thus, $M,N,A'$ are collinear. Then, $TS//A'N, TR//A'M \Rightarrow R,S,T$ are collinear.