Problem: Let $\triangle ABC $ be a non-equilateral,
acute triangle with $\angle A = 60\dg$,
and let $O$ and $H$ denote the circumcenter and orthocenter
of $\triangle{ABC}$, respectively.
\begin{enumerate}[(a)]
  \ii Prove that line $OH$ intersects both segments $AB$ and $AC$
  at two points $P$ and $Q$, respectively.
  \ii Denote by $s$ and $t$ the respective areas of triangle $APQ$
  and quadrilateral $BPQC$.
  Determine the range of possible values for $s/t$.
\end{enumerate}

Solution: We begin with some synthetic work.
Let $I$ denote the incenter, and recall (``fact 5'')
that the arc midpoint $M$ is the center of $(BIC)$,
which we denote by $\gamma$.

Now we have that
\[ \angle BOC = \angle BIC = \angle BHC = 120\dg. \]
Since all three centers lie inside $ABC$ (as it was acute),
and hence on the opposite side of $\ol{BC}$ as $M$,
it follows that $O$, $I$, $H$ lie on minor arc $BC$ of $\gamma$.

We note this implies (a) already,
as line $OH$ meets line $BC$ outside of segment $BC$.

\begin{center}
\begin{asy}
pair A = dir(50);
pair B = dir(210);
pair C = dir(330);

pair M = dir(270);
pair I = incenter(A, B, C);
pair O = circumcenter(A, B, C);
pair H = orthocenter(A, B, C);
pair P = extension(O, H, A, B);
pair Q = extension(O, H, A, C);

filldraw(A--P--Q--cycle, opacity(0.2)+lightgreen, deepgreen);
filldraw(B--P--Q--C--cycle, opacity(0.2)+lightcyan, deepgreen);
filldraw(circumcircle(A, B, C), opacity(0.1)+yellow, red);
filldraw(CP(M, I), opacity(0.1)+lightblue, dotted+blue);
draw(A--M, red);

filldraw(A--O--M--H--cycle, opacity(0.1)+orange, lightred);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$M$", M, dir(M));
dot("$I$", I, dir(20));
dot("$O$", O, dir(90));
dot("$H$", H, dir(50));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(Q));

/* TSQ Source:

A = dir 50
B = dir 210
C = dir 330

M = dir 270
I = incenter A B C R20
O = circumcenter A B C R90
H = orthocenter A B C R50
P = extension O H A B
Q = extension O H A C

A--P--Q--cycle 0.2 lightgreen / deepgreen
B--P--Q--C--cycle 0.2 lightcyan / deepgreen
circumcircle A B C 0.1 yellow / red
CP M I 0.1 lightblue / dotted blue
A--M red

A--O--M--H--cycle 0.1 orange / lightred

*/
\end{asy}
\end{center}

\begin{claim*}
  Triangle $APQ$ is equilateral with side length $\frac{b+c}{3}$.
\end{claim*}
\begin{proof}
  Let $R$ be the circumradius.
  We have $R = OM = OA = MH$, and even $AH = 2R \cos A = R$,
  so $AOMH$ is a rhombus.
  Thus $\ol{OH} \perp \ol{AM}$ and in this way
  we derive that $\triangle APQ$ is isosceles, hence equilateral.

  Finally, since $\angle PBH = 30\dg$, and $\angle BPH = 120\dg$,
  it follows that $\triangle BPH$ is isosceles and $BP = PH$.
  Similarly, $CQ = QH$.
  So $b+c = AP + BP + AQ + QC = AP + AQ + PQ$ as needed.
\end{proof}

Finally, we turn to the boring task of
extracting the numerical answer.
We have
\[ \frac{s}{s+t} = \frac{[APQ]}{[ABC]}
  = \frac{\frac{\sqrt3}4 \left( \frac{b+c}{3} \right)^2}%
  {\frac{\sqrt3}4 bc}
  = \frac{b^2+2bc+c^2}{9bc}
  = \frac19 \left( 2 + \frac bc + \frac cb \right). \]

So the problem is reduced to analyzing the behavior of $b/c$.
For this, we imagine fixing $\Gamma$ the circumcircle of $ABC$,
as well as the points $B$ and $C$.
Then as we vary $A$ along the ``topmost'' arc of measure $120\dg$,
we find $b/c$ is monotonic with values $1/2$ and $2$ at endpoints,
and by continuity all values $b/c \in (1/2,2)$ can be achieved.

So \[ \half < \frac bc < 2
  \implies 4/9 < \frac{s}{s+t} < 1/2
  \implies 4/5 < \frac st < 1 \]
as needed.