I failed the finals but still am uploading my complex bash solution for this problem since I am shameless. Took me forever for me to make the asy diag cuz of me being so useless in life. :wheelchair: [asy][asy] pair M = (0,0); pair B = (-1/sqrt(3),0); pair C = (1/sqrt(3),0); pair A = dir(105); pair P = (2B+C)/3; pair Q = (B+2C)/3; pair R = (2C+A)/3; pair S = (C+2A)/3; pair U = (2A+B)/3; pair T = (A+2B)/3; import graph; size(11cm); pen dps = linewidth(0.5) + fontsize(13); defaultpen(dps); draw(circle(M, 1), linewidth(0.5) + red); draw(C--A, linewidth(0.5)); draw(A--B, linewidth(0.5)); draw(B--C, linewidth(0.5)); draw(circumcircle(R,S,Q), linewidth(0.5) + dashed + blue); dot("$M$", M, NW); dot("$B$", B, NW); dot("$C$", C, NE); dot("$A$", A, NW); dot("$P$", P, NW); dot("$Q$", Q, NW); dot("$R$", R, NE); dot("$T$", U, NW); dot("$S$", S, NE); dot("$U$", T, NW); clip((-1,-0.3)--(1,-0.3)--(1,1.2)--(-1,1.2)--cycle); [/asy][/asy] Let $P$ and $Q$ be the points of trisection of $BC$; $R$ and $S$ that of $CA$; $T$ and $U$ that of $AB$. Also let $M$ be the midpoint of $BC$. I claim that $RSTQ$ is cyclic. Firstly, from the condition we get that $\dfrac{AM}{MB}=\sqrt{3}\implies AM=MB\sqrt{3}$. Now we proceed by using complex bash. Firstly we use the following lemma where if we have two complex numbers $a$ and $b$ and a point $p$ on $AB$ such that $AP:PB=t:1-t$, then we have that,\[p=ta+(1-t)b.\] Fix $B$ and $C$ on the complex plane with $B=-\dfrac{1}{\sqrt{3}}$ and $C=\dfrac{1}{\sqrt{3}}$. Now note that this gives $MA=MB\sqrt{3}=1$. So $A$ lies on the unitcircle. Now using our lemma, we get that the values of $Q=\dfrac{\dfrac{1}{\sqrt{3}}}{3}$, $R=\dfrac{a+\dfrac{2}{\sqrt{3}}}{3}$, $S=\dfrac{2a+\dfrac{1}{\sqrt{3}}}{3}$ and that $T=\dfrac{a-\dfrac{2}{\sqrt{3}}}{3}$. Thus it suffices to show that, \begin{align*} &\dfrac{s-t}{r-t}\div\dfrac{s-q}{r-q}\in\mathbb{R}\\ =&\dfrac{ \left( \dfrac{2a+\dfrac{1}{\sqrt{3}}}{3} \right) - \left( \dfrac{a-\dfrac{2}{\sqrt{3}}}{3} \right) }{ \left( \dfrac{a+\dfrac{2}{\sqrt{3}}}{3} \right)- \left( \dfrac{a-\dfrac{2}{\sqrt{3}}}{3}\right)}\div\dfrac{\left(\dfrac{2a+\dfrac{1}{\sqrt{3}}}{3}\right)-\left(\dfrac{\dfrac{1}{\sqrt{3}}}{3}\right)}{\left(\dfrac{a+\dfrac{2}{\sqrt{3}}}{3}\right)-\left(\dfrac{\dfrac{1}{\sqrt{3}}}{3}\right)}\\ =&\dfrac{(a+\sqrt{3})(a\sqrt{3}+1)}{8a} .\end{align*} Taking the conjugate of this term and using $\overline{a}=\dfrac{1}{a}$ shows that it equals its conjugate and so its a real value and we are done.

Another quick complex solution could be done by setting $M=0,B=1,C=-1,A=a$ s.t $a.\overline{a}=3$. By section formula the trisection points are $\frac{2a+1}{3},\frac{a+2}{3},\frac{1}{3},\frac{-1}{3},\frac{2a-1}{3},\frac{a-2}{3}$. Now it could be easily proved that, $$\frac{\frac{a+2}{3}-\frac{-1}{3}}{\frac{a+2}{3}-\frac{a-2}{3}}\div \frac{\frac{2a-1}{3}-\frac{-1}{3}}{\frac{2a-1}{3}-\frac{a-2}{3}}\in \mathbb{R} \text{ }\blacksquare$$