Problem: In an acute triangle $ABC$, let $M$ be the midpoint of $\ol{BC}$.
Let $P$ be the foot of the perpendicular from $C$ to $AM$.
Suppose that the circumcircle of triangle $ABP$
intersects line $BC$ at two distinct points $B$ and $Q$.
Let $N$ be the midpoint of $\ol{AQ}$.
Prove that $NB = NC$.

Solution: We show several different approaches.
In all solutions, let $D$ denote the foot of the altitude from $A$.

\begin{center}
\begin{asy}
size(10cm);

pair A = dir(115);
pair B = dir(210);
pair C = dir(330);
pair D = foot(A, B, C);
pair M = midpoint(B--C);
pair P = foot(C, A, M);
pair Q = 2*M-D;

filldraw(A--B--C--cycle, opacity(0.1)+cyan, blue);
filldraw(circumcircle(A, B, P), opacity(0.1)+yellow, orange+dashed);
draw(A--D, blue);
draw(A--P--C, deepgreen);

draw(A--Q, red+dashed);
pair N = midpoint(A--Q);
draw(B--N--C, red+dotted);

pair R = foot(B, A, M);
draw(B--R, dashed);

dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$D$", D, dir(D));
dot("$M$", M, dir(45));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(315));
dot("$N$", N, dir(45));
dot("$R$", R, dir(30));

/* --------------------------------+
| TSQX: by CJ Quines and Evan Chen |
| https://github.com/vEnhance/tsqx |
+----------------------------------+
!size(10cm);
A = dir 115
B = dir 210
C = dir 330
D = foot A B C
M 45 = midpoint B--C
P = foot C A M
Q 315 = 2*M-D
A--B--C--cycle / 0.1 cyan / blue
circumcircle A B P / 0.1 yellow / orange dashed
A--D blue
A--P--C deepgreen
A--Q / red dashed
N 45 = midpoint A--Q
B--N--C / red dotted
R 30 = foot B A M
B--R / dashed
*/
\end{asy}
\end{center}

\paragraph{Most common synthetic approach.}
The solution hinges on the following claim:
\begin{claim*}
  $Q$ coincides with the reflection of $D$ across $M$.
\end{claim*}
\begin{proof}
  Note that $\dang ADC = \dang APC = 90\dg$, so $ADPC$ is cyclic.
  Then by power of a point (with the lengths directed),
  \[ MB \cdot MQ = MA \cdot MP = MC \cdot MD. \]
  Since $MB = MC$, the claim follows.
\end{proof}
It follows that $\ol{MN} \parallel \ol{AD}$,
as $M$ and $N$ are respectively the midpoints of $\ol{AQ}$ and $\ol{DQ}$.
Thus $\ol{MN} \perp \ol{BC}$,
and so $N$ lies on the perpendicular bisector of $\ol{BC}$, as needed.
\begin{remark*}
  [David Lin]
  One can prove the main claim without power of a point as well, as follows:
  Let $R$ be the foot from $B$ to $\ol{AM}$, so $BRCP$ is a parallelogram.
  Note that $ABDR$ is cyclic, and hence
  \[ \dang DRM = \dang DBA = QBA = \dang QPA = \dang QPM. \]
  Thus, $\ol{DR} \parallel \ol{PQ}$, so $DRQP$ is also a parallelogram.
\end{remark*}

\paragraph{Synthetic approach with no additional points at all.}
\begin{claim*}
  $\triangle BPC \sim \triangle ANM$ (oppositely oriented).
\end{claim*}
\begin{proof}
  We have $\triangle BMP \sim \triangle AMQ$ from the given concyclicity of $ABPQ$.
  Then
  \[ \frac{BM}{BP} = \frac{AM}{AQ} \implies
    \frac{2BM}{BP} = \frac{AM}{AQ/2} \implies
    \frac{BC}{BP} = \frac{AM}{AN} \]
  implying the similarity (since $\dang MAQ = \dang BPM$).
\end{proof}
This similarity gives us the equality of directed angles
\[ \dang \left( BC, MN \right) = -\dang \left( PC, AM  \right) = 90\dg \]
as desired.

\paragraph{Synthetic approach using only the point $R$.}
Again let $R$ be the foot from $B$ to $\ol{AM}$, so $BRCP$ is a parallelogram.
\begin{claim*}
  $ARQC$ is cyclic; equivalently, $\triangle MAQ \sim \triangle MCR$.
\end{claim*}
\begin{proof}
  $MR \cdot MA = MP \cdot MA = MB \cdot MQ = MC \cdot MQ$.
\end{proof}

Note that in $\triangle MCR$, the $M$-median is parallel to $\ol{CP}$
and hence perpendicular to $\ol{RM}$.
The same should be true in $\triangle MAQ$ by the similarity,
so $\ol{MN} \perp \ol{MQ}$ as needed.

\paragraph{Cartesian coordinates approach with power of a point.}
Suppose we set $B = (-1,0)$, $M = (0,0)$, $C = (1,0)$, and $A = (a,b)$.
One may compute:
\begin{align*}
  \overleftrightarrow{AM} : 0 &= bx - ay \iff y = \frac ba x \\
  \overleftrightarrow{CP} : 0 &= a(x-1) + by
    \iff y = -\frac ab (x-1) = -\frac ab x + \frac ab. \\
  P &= \left( \frac{a^2}{a^2+b^2}, \frac{ab}{a^2+b^2} \right) \\
\end{align*}
Now note that
\[ AM = \sqrt{a^2+b^2}, \qquad PM = \frac{a}{\sqrt{a^2+b^2}} \]
together with power of a point
\[ AM \cdot PM = BM \cdot QM \]
to immediately deduce that $Q = (a,0)$.
Hence $N = (0, b/2)$ and we're done.

\paragraph{Cartesian coordinates approach without power of a point (outline).}
After computing $A$ and $P$ as above, one could also directly calculate
\begin{align*}
  \text{Perpendicular bisector of $\ol{AB}$}:
    y &= -\frac{a+1}{b} x + \frac{a^2+b^2-1}{2b} \\
  \text{Perpendicular bisector of $\ol{PB}$}:
    y &= - \left( \frac{2a}{b} + \frac{b}{a} \right) x - \frac{b}{2a} \\
  \text{Perpendicular bisector of $\ol{PA}$}:
    y &= - \frac{a}{b} x + \frac{a+a^2+b^2}{2b}. \\
  \text{Circumcenter of $\triangle PAB$}
    &= \left( -\frac{a+1}{2}, \frac{2a^2+2a+b^2}{2b} \right).
\end{align*}
This is enough to extract the coordinates of $Q = (\bullet, 0)$,
because $B = (-1,0)$ is given, and the $x$-coordinate
of the circumcenter should be the average of the $x$-coordinates of $B$ and $Q$.
In other words, $Q = (-a,0)$.
Hence, $N = \left( 0, \frac b2 \right)$, as needed.

\paragraph{Ill-advised barycentric approach (outline).}
Use reference triangle $ABC$.
The $A$-median is parametrized by $(t:1:1)$ for $t \in \RR$.
So because of $\ol{CP} \perp \ol{AM}$, we are looking for $t$ such that
\[
  \left( \frac{t \vec A + \vec B + \vec C}{t+2} - \vec C \right)
  \perp \left( A - \frac{\vec B + \vec C}{2} \right).
\]
This is equivalent to
\[
  \left( t \vec A + \vec B - (t+1) \vec C \right)
  \perp \left( 2 \vec A - \vec B - \vec C \right).
\]
By the perpendicularity formula for barycentric coordinates (EGMO 7.16),
this is equivalent to
\begin{align*}
  0 &= a^2t - b^2 \cdot (3t+2) + c^2 \cdot (2-t) \\
  &= \left( a^2-3b^2-c^2 \right) t - 2(b^2-c^2) \\
  \implies t &= \frac{2(b^2-c^2)}{a^2-3b^2-c^2}.
\end{align*}
In other words,
\[ P = \left( 2(b^2-c^2) : a^2-3b^2-c^2 : a^2-3b^2-c^2 \right). \]
A long calculation gives
$a^2 y_P z_P + b^2 z_P x_P + c^2 x_P y_P
= (a^2-3b^2-c^2)(a^2-b^2+c^2)(a^2-2b^2-2c^2)$.
Together with $x_P+y_P+z_P=2a^2-4b^2-4c^2$,
this makes the equation of $(ABP)$ as
\[
  0=-a^2yz-b^2zx-c^2xy
  + \frac{a^2-b^2+c^2}{2} z(x+y+z).
\]
To solve for $Q$, set $x=0$ to get to get
\[ a^2yz = \frac{a^2-b^2+c^2}{2} z(y+z)
  \implies \frac yz = \frac{a^2-b^2+c^2}{a^2+b^2-c^2}. \]
In other words,
\[ Q = \left(  0 : a^2-b^2+c^2 : a^2+b^2-c^2 \right). \]
Taking the average with $A = (1,0,0)$ then gives
\[ N = \left(  2a^2 : a^2-b^2+c^2 : a^2+b^2-c^2 \right). \]
The equation for the perpendicular bisector of $\ol{BC}$
is given by (see EGMO 7.19)
\[ 0 = a^2(z-y)+x(c^2-b^2) \]
which contains $N$, as needed.

\paragraph{Extremely ill-advised complex numbers approaches (outline).}
Suppose we pick $a$, $b$, $c$ as the unit circle, and let $m = (b+c)/2$.
Using the fully general ``foot'' formula, one can get
\[
  p = \frac{(a-m) \ol c + (\ol a - \ol m)c
    + \ol a m - a \ol m}{2(\ol a - \ol m)}
    = \frac{a^2 b - a^2 c - a b^2 - 2 a b c - a c^2 + b^2 c + 3 b c^2}
    {4bc-2a(b+c)}
\]
Meanwhile, an extremely ugly calculation will eventually yield
\[ q = \frac{\frac{bc}{a}+b+c-a}{2} \]
so
\[ n = \frac{a+q}{2} = \frac{a+b+c+\frac{bc}{a}}{4}
  = \frac{(a+b)(a+c)}{2a}. \]
There are a few ways to then verify $NB = NC$.
The simplest seems to be to verify that
\[ \frac{n - \frac{b+c}{2}}{b-c} = \frac{a-b-c+\frac{bc}{a}}{4(b-c)}
  = \frac{(a-b)(a-c)}{2a(b-c)} \]
is pure imaginary, which is clear.