Problem: Let $P$ be a point in the plane of $\triangle ABC$,
and $\gamma$ a line through $P$.
Let $A'$, $B'$, $C'$ be the points where the
reflections of lines $PA$, $PB$, $PC$ with respect to $\gamma$
intersect lines $BC$, $CA$, $AB$ respectively.
Prove that $A'$, $B'$, $C'$ are collinear.

\end{enumerate}

Solution: We present three solutions.

\paragraph{First solution (complex numbers).}
Let $p=0$ and set $\gamma$ as the real line.
Then $A'$ is the intersection of $bc$ and $p\ol a$.
So, we get
\[ a' = \frac{\ol a(\ol b c - b \ol c)}{(\ol b - \ol c)\ol a-(b-c) a}. \]

\begin{center}
  \begin{asy}
    size(4cm);
    pair A = dir(110), B = dir(210), C = dir(330);
    Drawing("A", A, A);
    Drawing("B", B, dir(-90));
    Drawing("C", C, dir(-90));
    draw(A--B--C--cycle);
    pair P = Drawing("P", 0.1*dir(40), dir(-35));
    pair X1 = P + .5 * dir(10);
    pair X2 = P - .5 * dir(10);
    Drawing(Line(X1,X2), dashed, Arrows);
    pair T = 2*foot(A,X1,X2)-A;
    pair A1 = Drawing("A'", extension(P,T,B,C), dir(-40));
    draw(A--P--A1);
  \end{asy}
\end{center}

Note that
\[ \ol a' = \frac{a (b \ol c - \ol b c)}{(b-c) a - (\ol b - \ol c)\ol a}. \]
Thus it suffices to prove
\[ 0 = \det
\begin{bmatrix}
  \frac{\ol a(\ol b c-b \ol c)}{(\ol b - \ol c)\ol a - (b-c) a} & \frac{a (b \ol c - \ol b c)}{(b-c) a - (\ol b - \ol c) \ol a} & 1 \\
  \frac{\ol b(\ol c a-c \ol a)}{(\ol c - \ol a)\ol b - (c-a) b} & \frac{b (c \ol a - \ol c a)}{(c-a) b - (\ol c - \ol a) \ol b} & 1 \\
  \frac{\ol c(\ol a b-a \ol b)}{(\ol a - \ol b)\ol c - (a-b) c} & \frac{c (a \ol b - \ol a b)}{(a-b) c - (\ol a - \ol b)\ol c} & 1
\end{bmatrix}.
\]
This is equivalent to
\[ 0 = \det
\begin{bmatrix}
  \ol a(\ol b c -b \ol c) & a (\ol b c - b \ol c) & (\ol b - \ol c) \ol a - (b-c) a \\
  \ol b(\ol c a -c \ol a) & b (\ol c a - c \ol a) & (\ol c - \ol a) \ol b - (c-a) b \\
  \ol c(\ol a b -a \ol b) & c (\ol a b - a \ol b) & (\ol a - \ol b) \ol c - (a-b) c \\
\end{bmatrix}. \]
This determinant has the property that the columns sum to zero, and we're done.
\begin{remark*}
  Alternatively, if you don't notice that you could just blindly expand:
  \begin{align*}
    &\phantom{=} \sum_{\text{cyc}} ((\ol b - \ol c) \ol a - (b-c) a) \cdot
    - \det
    \begin{bmatrix}
      b & \ol b \\
      c & \ol c
    \end{bmatrix}
    \left( \ol c a - c \ol a \right)\left( \ol a b - a \ol b \right) \\
    &= (\ol b c - c \ol b)(\ol c a - c \ol a)(\ol a b - a \ol b)
    \sum_{\text{cyc}} \left( ab - ac + \ol c \ol a - \ol b \ol a \right) = 0.
  \end{align*}
\end{remark*}

\paragraph{Second solution (Desargues involution).}
We let $C'' = \ol{A'B'} \cap \ol{AB}$.
Consider complete quadrilateral $ABCA'B'C''C$.
We see that there is an involutive pairing $\tau$ at $P$
swapping $(PA,PA')$, $(PB,PB')$, $(PC,PC'')$.
From the first two, we see $\tau$ coincides with reflection
about $\ell$, hence conclude $C'' = C$.

\paragraph{Third solution (barycentric), by Catherine Xu.}
We will perform barycentric coordinates on the triangle $PCC'$,
with $P=(1,0,0)$, $C'=(0,1,0)$, and $C=(0,0,1)$.
Set $a = CC'$, $b = CP$, $c = C'P$ as usual.
Since $A$, $B$, $C'$ are collinear,
we will define $A = (p : k : q)$ and $B = (p : \ell : q)$.

\begin{claim*}
  Line $\gamma$ is the angle bisector of
  $\angle APA' $, $\angle BPB'$, and $\angle CPC'$.
\end{claim*}
\begin{proof}
  Since $A'P$ is the reflection of $AP$ across $\gamma$, etc.
\end{proof}

Thus $B'$ is the intersection of the isogonal of $B$ with respect to $\angle P$
with the line $CA$; that is,
\[ B' = \left( \frac pk \frac{b^2}{\ell}
  : \frac{b^2}{\ell} : \frac{c^2}{q} \right). \]
Analogously, $A'$ is the intersection of the isogonal of $A$ with respect to $\angle P$
with the line $CB$; that is,
\[ A' = \left( \frac{p}{\ell} \frac{b^2}{k}
    : \frac{b^2}{k} : \frac{c^2}{q} \right). \]
The ratio of the first to third coordinate in these two points
is both $b^2pq : c^2k\ell$, so it follows $A'$, $B'$, and $C'$ are collinear.

\begin{remark*}
  [Problem reference]
  The converse of this problem appears as problem 1052
  attributed S.\ V.\ Markelov in the book
  \emph{Geometriya: 9--11 Klassy: Ot Uchebnoy Zadachi k Tvorcheskoy, 1996},
  by I.\ F.\ Sharygin.
\end{remark*}