Problem: An equilateral pentagon $AMNPQ$ is inscribed in triangle $ABC$
such that $M \in \ol{AB}$, $Q \in \ol {AC}$, and $N,P \in \ol{BC}$.
Let $S$ be the intersection of $\ol{MN}$ and $\ol{PQ}$.
Denote by $\ell$ the angle bisector of $\angle MSQ$.

Prove that $\ol{OI}$ is parallel to $\ell$,
where $O$ is the circumcenter of triangle $ABC$,
and $I$ is the incenter of triangle $ABC$.

Solution: \paragraph{First solution (complex).}
In fact, we only need $AM = AQ = NP$ and $MN = QP$.

We use complex numbers with $ABC$ the unit circle,
assuming WLOG that $A$, $B$, $C$ are labeled counterclockwise.
Let $x$, $y$, $z$ be the complex numbers corresponding to the arc midpoints
of $BC$, $CA$, $AB$, respectively; thus $x+y+z$ is the incenter of $\triangle ABC$.
Finally, let $s > 0$ be the side length of $AM = AQ = NP$.

Then, since $MA = s$ and $MA \perp OZ$, it follows that
\[ m - a = i \cdot sz. \]
Similarly, $p-n = i \cdot sy$ and $a-q = i \cdot sx$, so summing these up gives
\[ i \cdot s(x+y+z) = (p-q) + (m-n) = (m-n) - (q-p). \]
Since $MN = PQ$, the argument of $(m-n) - (q-p)$ is along
the external angle bisector of the angle formed, which is perpendicular to $\ell$.
On the other hand, $x+y+z$ is oriented in the same direction as $OI$, as desired.

\paragraph{Second solution (trig, Danielle Wang).}
Let $\delta$ and $\epsilon$ denote $\angle MNB$ and $\angle CPQ$.
Also, assume $AMNPQ$ has side length $1$.

In what follows, assume $AB < AC$.
First, we note that
\begin{align*}
BN   &=   (c-1) \cos B + \cos \delta \textrm{,}    \\
CP   &=   (b-1) \cos C + \cos \epsilon \textrm{, and}   \\
a   &=   1 + BN + CP
\end{align*}
from which it follows that
\[
\cos \delta + \cos \epsilon = \cos B + \cos C - 1
\]
Also, by the Law of Sines, we have $\frac{c-1}{\sin\delta} = \frac{1}{\sin B}$
and similarly on triangle $CPQ$, and from this we deduce
\[ \sin \epsilon - \sin \delta = \sin B - \sin C. \]
The sum-to-product formulas
\begin{align*}
\sin{\epsilon} - \sin{\delta} &= 2 \cos\left( \frac{\epsilon + \delta}{2} \right) \sin\left( \frac{\epsilon - \delta}{2} \right) \\
\cos{\epsilon} - \cos{\delta} &= 2 \cos\left( \frac{\epsilon + \delta}{2} \right) \cos\left( \frac{\epsilon - \delta}{2} \right)
\end{align*}
give us
\[
\tan \left( \frac{\epsilon-\delta}{2} \right) = \frac{\sin{\epsilon} - \sin{\delta}}{\cos{\epsilon} - \cos{\delta}}
= \frac{\sin B - \sin C}{\cos B + \cos C - 1}. \]
Now note that $\ell$ makes an angle of
$\frac{1}{2}(\pi+\epsilon-\delta)$ with line $BC$.
Moreover, if line $OI$ intersects line $BC$ with angle $\varphi$ then
\[ \tan\varphi = \frac{r - R \cos A}{\frac{1}{2}(b-c)}. \]
So in order to prove the result, we only need to check that
\[ \frac{r - R \cos A}{\frac{1}{2}(b-c)}
  = \frac{\cos B + \cos C + 1}{\sin B - \sin C}. \]
Using the fact that $b = 2R\sin B$, $c = 2R\sin C$, this  reduces to the fact that
$r/R + 1 = \cos A + \cos B + \cos C$, which is the so-called Carnot theorem.