interesting

Suppose the circumcenter of $\triangle ABC$ is $O$,Let $AS\cap BC=A',SP\cap BC=P'$ Let $OB\cap \odot(A'SB) = T$ We say that the circle that goes through $P$ and $S$ and is tangent to lines $AB$, $AC$ is $\omega$ $\mathbf{Lemma.1} O,H,T,S$ are cyclic $\mathbf{Proof.}$ Note that $\angle OBC=\angle A'BS=30^{\circ}$ Hence $TA'=SA'$ Familiar with $H,O$are symmetric about $AS$ Hence $A'H=A'O=A'T=A'S$ Which means that $O,H,T,S$ are cyclic $\mathbf{Lemma.2} H,P',B,T$ are cyclic $\mathbf{Proof.}$ Note that $SP\cdot SP'=SB^{2}=SO^{2}=SH^{2}$ Hence $\angle P'HS=\angle HPS=90^{\circ}$ So $\angle HP'A'=\angle OSH=\angle HTO$ Which means that $H,P',B,T$ are cyclic $\mathbf{Lemma.3} P'T$ is the common tangent of circle $\odot(A'BS)$ and circle $\odot(A'SC)$ $\mathbf{Proof.}$ Suppose the radius of $\odot(A'BS)$ and $\odot(A'SC)$ is $r_{1}$ and $r_{2}$ respectively Note that $r_{1}=A'S=r_{2}$ So we need to prove that $PT\bot A'S$ Consider that $\angle P'TA'=\angle P'HB+180^{\circ}-\angle ACB=90^{\circ}-\angle ABC+180^{\circ}-\angle ACB=150^{\circ}=\angle TBP'$ Hence $P'T$ is tangent to $\odot(A'BS)$ Since $\angle HP'T=\angle HBT=\angle BHT=\dfrac{1}{2}\angle OSH=\angle A'SH$ Hence $\measuredangle (P'T,AS)=\angle P'HS=90^{\circ}$ Which means that $P'T$ is the common tangent of circle $\odot(A'BS)$ and circle $\odot(A'SC)$ $\mathbf{Finally}$ Consider an inversion $\mathcal{I}_{S}^{SC^{2}}$ So $A\mapsto A',B\mapsto B,C\mapsto C,H\mapsto H,P\mapsto P',\omega \mapsto PT,AB\mapsto \odot(A'BS),AC\mapsto \odot(A'CS)$ Note that $P'T$ is the common tangent of circle $\odot(A'BS)$ and circle $\odot(A'SC)$ Hence $\omega$ is tangent to $AB,AC$

Let $M$ be a point on $\Omega$ such that $SM$ is a diameter of $\Omega$. Let $O$ be the center of $\Omega$ and $Y$ be the midpoint of $BC$, hence $S,Y,O,M$ are collinear. Since $\angle HPS=90^\circ$, then $P,H,M$ are collinear. Let $PM$ meet $AS$ at $X$. Since $\angle COY=60^\circ$, then we have $AH=2OY=OC=OM$. Consider $\triangle AXH\sim\triangle SXM$ since $AH\parallel MS$. But $MS=2OM=2AH$, so $XS=2AX$. Let $I$ be midpoint of $XS$ and $N$ be a projection of $I$ on $AC$. Since $\angle HPS=90^\circ$, we have $AX=XI=IS=IP$. But $\angle IAN=30^\circ$, then $AI=2IN$ and hence $IP=IS=IN$. Since $I$ lies on the internal angle bisector of $\angle BAC$. then the circle centered at $I$ and radius $IP=IS=IN$ passes though $P,S$ and touches $AC$ at $N$ and also touches $AB$.

Solved this with only geogebra on a bus. ( I am typing from a phone because I have nothing better to do ) Let the line passing trough $H$ perpendicular to the angle bisector intersect the circle at $L$ and $M$ and the sides of the triangle at $G$ and $F$. Also define the projection of $A$ on the line as $N$ note that $NHSP$ is cyclic We claim that $(GFSP)$ is the desired circle. Note that clearly $(GFS)$ is tangent to both sides of the triangle so it is only left to prove that $P$ indeed lies on the circle. Now define the function $f(P)= \pm \frac{PL}{PM}$ to be negative for every point on segment $LM$ and below and positive otherwise. For the sake of simplicity Let $a=f(A)$ and similarly for $b,c$. Therefore we directly compute: $f(S) = - a$, $f(G) = ac$ , $f(F) =ab$, $f(N)= -a^2$ Claim: $f(H) = -bc$ Proof: (ratio lemma spam): Let: $\angle{LAB} = x$. The following follow from easy angle chase: $\angle{LAH}=3x$, $\angle{MAH}= 90-3x$, $\angle{CLM} = 90-x$, $\angle{CML}=x$, $\angle{BLM}=30-x$, $\angle{LMB} = 60+x$ and finally. $\angle{ALM} = 60-x$, $\angle{AML} = 30+x$. We now just have yo verify an easy trig identity: $$\frac{\sin(3x)}{\sin(90-3x)} = \frac{\sin(x)}{\sin(90-x)} \frac{\sin(60+x)}{\sin(30-x)} \frac{\sin(60-x)}{\sin(30+x)}$$With this lemma we can compute $f(P) = \frac{f(N)f(H)}{f(S)}= -abc$. Checking if it is cyclic is easy now: $f(P)f(S) = f(G)f(F) \Leftrightarrow a^2bc=a^2bc$ so we are done!

Let $O$ be the center of $(ABC)$. Lemma: $AHSO$ is a rhombus. Proof. If $M$ is the midpoint of $BC$, then $AH=2OM=OC=OA$ (since $BHOC$ is cyclic as $A=60^{\circ}$). Let $OH$ intersect $AB$ and $AC$ at $E$ and $F$ respectively. Since $S$ is a reflection of $A$ about $EF$, $\triangle AEF$ and $\triangle SEF$ are equilateral, and it's easy to see that $AE$ and $AF$ are tangents to $(SEF)$. It remains to show that $P$ lies on this circle. Let the circle with diameter $AH$ intersect $ABC$ at $P$. Note that the nine-point center $N$, which is the midpoint of $OH$, lies on both $(PSH)$ and $(AGH)$. Then, by the Radical Axis Theorem, $AG$, $NH$, and $SP$ are concurrent at some point $X$. It is also well-known that $(AG,AH;AE,AF)=-1$. It follows that $XE*XF=XH*XN=XP*XS$, which solves the problem.

Let $N$ be the midpoint of the arc $\widehat{BAC}$;${X}=HN\cap AS$; $O_1$ be the midpoint of $XS$ and $F,G$ be the feet from $O_1$ to $AB,AC$. Obviously $P,H,X,N$ are collinear and we will prove that the circle $\omega$ with center $O_1$ and with radius $O_1S$ is the desired circle. Since $\angle XPS=90^\circ$ we have $X,P,S\in \omega$. Also we have $O_1\in AS$ so $O_1F=O_1G$. If we prove $O_1F=O_1S$ we are done because then $F,X,P,G,S\in \omega$ and since $O_1F\perp AB$ and $O_1G\perp AC$ we have that $\omega$ is tangent to $AB$ and $AC$ and passes through $P,S$ as wanted. For this just notice that $\angle O_1AF=30^\circ$ so $O_1F=\frac{AO_1}{2}$. But since $AH\parallel NS$ we have $\frac{AH}{NS}=\frac{AX}{XS}=\frac{1}{2}$(easy trig) so $AX=O_1S$ so $AO_1=XS$,but then $\frac{AO_1}{2}=\frac{XS}{2}=O_1S=O_1F$ and we are done.

We complex bash with $\Omega$ as the unit circle. Let $\omega = e^{\pi i/3}$, so that $$|a|=|s|=1$$$$b = \omega s$$$$c = \omega^{-1}s$$$$h = a+b+c = a+s$$$$p = \frac{s+h}{s\overline{h}+1} = \frac{a(a+2s)}{2a+s}$$Now let $O$ be the intersection of the perpendicular bisector of $\overline{PS}$ and the internal bisector of $\angle BAC$. (If such a circle as requested in the problem does exist, its center necessarily lies on both lines.) Then $$\overline{o} = \frac{o}{ps} = \frac{a+s-o}{as}$$and so $$o = \frac{p(a+s)}{a+p} = \frac{a+2s}3$$Then $$|o-s| = |o-p| = \frac{|a-s|}3$$and the distance from $O$ to line $AB$ is $$\frac{|o-a-b+ab\overline{o}|}2 = \frac12\left|\frac{a+2s}3-a-\omega s+\frac{\omega(2a+s)}3\right| = \frac12\left|\frac{2a(\omega-1)}3 + \frac{2s(1-\omega)}3\right| = \frac{|a-s||\omega-1|}3 = \frac{|a-s|}3$$since $|\omega-1| = 1$. Similarly, this also equals the distance to line $AC$ by replacing $\omega$ with $\omega^{-1}$. So the desired circle is the one centered at this point $O$ with radius $\frac13AS$. $\blacksquare$

Nice Let $A'$ be the reflection point of $A$ over $H$, and $M$ be the antipode of $S$ in $\Omega$. $\angle MPS=\angle HPS=90^{\circ}$, so $(M,H,P)$ collinear. Let $O$ be the center of $\Omega$. Then by Servoi's theorem $MO=OS=AH=A'H$ because $\angle BAC=60^{\circ}$. Since $MS//AA'$, $\square MOA'H$ is a parallelogram. Then $MP//OA'$ and $MO=OS\Rightarrow OA'$ is the perpendicular bisector of $PS$.$\cdots (1)$ If we let $OA'\cap AS=I$, $AI:IS=AA':OS=2:1$ and $\angle IAB=\angle IAC=30^{\circ}$. Therefore the circle with center $I$ containing $S$ is tangnet to $AB$, $AC$. Name this circle $\Gamma$. By $(1)$, $P\in \Gamma$, so $\Gamma$ is a circle that goes through $P$ and $S$ and is tangent to lines $AB$, $AC$. $\blacksquare$

Let $O$ the center of $\Omega$ and let $SO \cap \Omega=S'$, now $S',H,P$ colinear is trivial, also let $M$ the midpoint of $BC$, clearly since $\triangle S'BC$ is equilateral we have $AH=2MO=OS$ therefore $2AH=SS'$, let $HS' \cap AS=D$ then by thales we have $DS=2AD$, so now let $O'$ the midpoint of $DS$ and the projections from $O'$ to $AB, AC$ be $E,F$ respectively. Now because $\angle DPS=90$ we have $O'$ center of $(DPS)$, but also since $AD=DO'$ we must have $D$ center of $AFO'E$ and in fact because $\angle FO'A=60=\angle EO'A$ we must have that triangles $\triangle FDO', \triangle EDO'$ are equilateral and therefore we have: $O'F=O'E=O'D=O'P=O'S$ so $O'$ is center of $(FDESP)$ and clearly this is tangent to lines $AB, AC$ so we are done .

Let $O$ be the center of $(ABC)$, and $Q$ be on $(ABC)$ such that $\angle AQH = 90^\circ$. Since $AHSO$ is a rhombus, we get that $PQ \parallel AS$. Let $M$ and $N$ be the midpoints of $BC$ and $PS$ respectively (it is well-known that $Q,H,M$ are collinear), and let $E = AS \cap HM$. Then \[ \frac{AE}{ES} = \frac{AH}{MS} = \frac{AH}{OM} = 2\]so $E$ is the incenter of the equilateral triangle whose sides are $AB$, $AC$ (both extended) and the line through $S$ perpendicular to $AS$. It suffices to show that $EP = ES$, or $\angle ENS = 90^\circ$. But now note that $\angle QAE = \angle NSE$ and \[ \frac{QA}{NS} = \frac{QA}{\frac 12 PS} = 2 = \frac{AE}{SE} \]so $\Delta QAE \sim \Delta NSE$, so $\angle ENS = \angle EQA = 90^\circ$, hence proved.