Problem: Let $ABC$ be an acute, scalene triangle,
and let $M$, $N$, and $P$ be the midpoints of
$\ol{BC}$, $\ol{CA}$, and $\ol{AB}$, respectively.
Let the perpendicular bisectors of $\ol{AB}$ and $\ol{AC}$
intersect ray $AM$ in points $D$ and $E$ respectively,
and let lines $BD$ and $CE$ intersect in point $F$, inside triangle $ABC$.
Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.

Solution: We present four solutions.

\paragraph{Barycentric solution.}
First, we find the coordinates of $D$.
As $D$ lies on $\ol{AM}$, we know $D=(t:1:1)$ for some $t$.
Now by perpendicular bisector formula, we find
\[ 0 = b^2(t-1) + (a^2-c^2) \implies t = \frac{c^2+b^2-a^2}{b^2}. \]
Thus we obtain \[ D = \left( 2S_A : c^2 : c^2 \right). \]
Analogously $E = (2S_A : b^2 : b^2)$,
and it follows that \[ F = \left( 2S_A : b^2 : c^2 \right). \]

The sum of the coordinates of $F$ is
\[ (b^2+c^2-a^2)+b^2+c^2=  2b^2+2c^2-a^2. \]
Hence the reflection of $A$ over $F$ is simply
\[ 2F-A = \left( 2(b^2+c^2-a^2)-(2b^2+2c^2-a^2) : 2b^2 : 2c^2 \right)
  = \left(-a^2 : 2b^2 : 2c^2 \right). \]
It is evident that $F'$ lies on $(ABC) : -a^2yz - b^2zx - c^2xy = 0$,
and we are done.

\paragraph{Synthetic solution (harmonic).}
Here is a synthetic solution.
Let $X$ be the point so that $APXN$ is a cyclic harmonic quadrilateral.
We contend that $X = F$.
To see this it suffices to prove $B$, $X$, $D$ collinear
(and hence $C$, $X$, $E$ collinear by symmetry).

\begin{center}
\begin{asy}
pair A = dir(130);
pair P = dir(210);
pair N = dir(330);
pair B = 2*P-A;
pair C = 2*N-A;
pair O = -A;
pair T = midpoint(P--N);
pair D = extension(A, T, P, O);
pair E = extension(A, T, N, O);
pair X = extension(B, D, E, C);
pair M = extension(A, T, B, C);
draw(N--T, heavygreen);
draw(T--P, lightblue);
draw(X--P, heavygreen);

draw(B--C--N, lightblue);
draw(P--O--N, lightblue);
draw(N--E, lightblue);
filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

draw(D--X--C, red+dashed);

filldraw(A--B--X--cycle, opacity(0.1)+lightgreen, heavygreen);
filldraw(A--M--N--cycle, opacity(0.1)+lightgreen, heavygreen);

dot("$A$", A, dir(A));
dot("$P$", P, dir(P));
dot("$N$", N, dir(10));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$O$", O, dir(O));
dot("$T$", T, dir(45));
dot("$D$", D, dir(60));
dot("$E$", E, dir(240));
dot("$X$", X, dir(X));
dot("$M$", M, dir(M));

/* TSQ Source:

A = dir 130
P = dir 210
N = dir 330 R10
B = 2*P-A
C = 2*N-A
O = -A
T = midpoint P--N R45
D = extension A T P O R60
E = extension A T N O R240
X = extension B D E C
M = extension A T B C
N--T heavygreen
T--P lightblue
X--P heavygreen

B--C--N lightblue
P--O--N lightblue
N--E lightblue
unitcircle 0.1 lightcyan / blue

D--X--C red dashed

A--B--X--cycle 0.1 lightgreen / heavygreen
A--M--N--cycle 0.1 lightgreen / heavygreen

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

Let $T$ be the midpoint of $\ol{PN}$,
so $\triangle APX \sim \triangle ATN$.
So $\triangle ABX \sim \triangle AMN$, ergo
\[ \dang XBA = \dang NMA = \dang BAM = \dang BAD = \dang DBA \]
as desired.
\paragraph{Angle chasing solution (Mason Fang).}
Obviously $ANOP$ is concyclic.
\begin{claim*}
  Quadrilateral $BFOC$ is cyclic.
\end{claim*}
\begin{proof}
  Write
  \begin{align*}
  \dang BFC = \dang FBC+\dang BCF
  &=\dang FBA +\dang ABC+\dang BCA+\dang ACF\\
  &=\dang DBA +\dang ABC+\dang BCA+\dang ACE\\
  &=\dang BAD +\dang ABC+\dang BCA+\dang EAC\\
  &=2\angle BAC=\angle BOC. \qedhere
  \end{align*}
\end{proof}
Define $Q = \ol{AA} \cap \ol{BC}$.
\begin{claim*}
  Point $Q$ lies on $\ol{FO}$.
\end{claim*}
\begin{proof}
  Write
  \begin{align*}
     \dang BOQ = \dang BOA+\dang AOQ
     &=2\dang BCA+90^{\dg}+\dang AQO\\
     &=2\dang BCA+90^{\dg}+\dang AMO\\
     &=2\dang BCA+90^{\dg}+\dang AMC+90^{\dg}\\
     &=\dang BCA+\dang MAC =\dang BCA+\dang ACE \\
     &=\dang BCE =\dang BOF. \qedhere
  \end{align*}
\end{proof}
As $Q$ is the radical center of $(ANOP)$, $(ABC)$ and $(BFOC)$,
this implies the result.
\paragraph{Inversive solution (Kelin Zhu).}
Invert about $A$ with radius $\sqrt{bc}$
followed by a reflection over the angle bisector of $\angle A$,
and denote the image of a point $X$ by $X'$.
The inverted problem now states the following:

\begin{quote}
  In triangle $AP^{\ast}N^{\ast}$,
  let $B^{\ast}$, $C^{\ast}$ be the midpoints of $AP^{\ast}$, $AN^{\ast}$
  and $D^{\ast}$, $E^{\ast}$ be the intersection of the $A$ symmedian
  with $(AP^{\ast})$, $(AN^{\ast})$, respectively.
  $(AB^{\ast}D^{\ast})$, $(AC^{\ast}E^{\ast})$
  intersect at a point $F^{\ast}$; prove that it lies on $P^{\ast}N^{\ast}$.
\end{quote}

I claim that, in fact, the midpoint of $P^{\ast}N^{\ast}$ is the desired intersection.
Redefine that point as $F^{\ast}$ and I will prove that
$(AB^{\ast}D^{\ast})$, $(AC^{\ast}E^{\ast})$ pass through it.

Note that
\[ \angle AD^{\ast}B^{\ast}=\angle D^{\ast}AB^{\ast}=\angle F^{\ast}AN^{\ast}=\angle AF^{\ast}B^{\ast}, \]
where the first equality is due to $B^{\ast}$ being the circumcircle of $AD^{\ast}P^{\ast}$,
the second equality is due to the definition of the symmedian,
and the third equality is due to the parallelogram $AB^{\ast}F^{\ast}C^{\ast}$.
A symmetric argument for $C$ finishes.