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.

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.

Could there be a solution with 9-point circle?

[asy][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][/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.

$\textit{Proof}$ Let $AE$ be the altitude of $\bigtriangleup ABC$ and $AE\cap (O)= F\neq A$

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. $\quad\blacksquare$

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

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..

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.

This is fairly simple by complex numbers....

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.

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.$\blacksquare$

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 $\Delta AST$

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 $\blacksquare$

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. $\square$

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

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.

Let $M$ and $N$ be the midpoints of $BC$ and $OH$ respectively. Claim: The quadrilateral $\square TOMN$ is a parallelogram

Claim: $ND=NM$

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

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

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.

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.