Problem: Let $ABC$ be a non-equilateral triangle
and let $M_a$, $M_b$, $M_c$ be the midpoints
of the sides $BC$, $CA$, $AB$, respectively.
Let $S$ be a point lying on the Euler line.
Denote by $X$, $Y$, $Z$ the second intersections
of $M_a S$, $M_b S$, $M_c S$ with the nine-point circle.
Prove that $AX$, $BY$, $CZ$ are concurrent.

\end{enumerate}

Solution: We assume now and forever that $ABC$ is scalene since the problem
follows by symmetry in the isosceles case.
We present four solutions.

\paragraph{First solution by barycentric coordinates (Evan Chen).}
Let $AX$ meet $M_b M_c$ at $D$,
and let $X$ reflected over $M_b M_c$'s midpoint be $X'$.
Let $Y'$, $Z'$, $E$, $F$ be similarly defined.

\begin{center}
\begin{asy}
pair M_a = dir(110);
pair M_b = dir(210);
pair M_c = dir(330);
pair A = M_b+M_c-M_a;
pair B = M_c+M_a-M_b;
pair C = M_a+M_b-M_c;
pair O = circumcenter(A, B, C);
pair S = 0.65*O;
pair N = origin;
pair X = -M_a+2*foot(N,M_a,S);
pair Y = -M_b+2*foot(N,M_b,S);
pair Z = -M_c+2*foot(N,M_c,S);
filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(M_a--M_b--M_c--cycle, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--M_a, cyan);
draw(Y--M_b, cyan);
draw(Z--M_c, cyan);

pair D = extension(X, A, M_b, M_c);
pair Dp = M_b+M_c-D;

pair E = extension(Y, B, M_c, M_a);
pair Ep = M_c+M_a-E;
pair F = extension(Z, C, M_a, M_b);
pair Fp = M_a+M_b-F;

draw(A--D, deepgreen);
draw(B--E, deepgreen);
draw(C--F, deepgreen);

draw(M_a--Dp, lightred);
draw(M_b--Ep, lightred);
draw(M_c--Fp, lightred);

draw(M_a--D, lightred+dotted);
draw(M_b--E, lightred+dotted);
draw(M_c--F, lightred+dotted);

dot("$X$", X, dir(250));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(200));

dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$S$", S, dir(310));
dot("$D$", D, dir(D));
dot("$D'$", Dp, dir(Dp));
dot("$E$", E, dir(E));
dot("$E'$", Ep, dir(Ep));
dot("$F$", F, dir(220));
dot("$F'$", Fp, dir(Fp));

/* TSQ Source:

M_a = dir 110
M_b = dir 210
M_c = dir 330
A = M_b+M_c-M_a
B = M_c+M_a-M_b
C = M_a+M_b-M_c
O := circumcenter A B C
S = 0.65*O R310
N := origin R45
X := -M_a+2*foot(N,M_a,S)
Y := -M_b+2*foot(N,M_b,S)
Z := -M_c+2*foot(N,M_c,S)
unitcircle 0.1 lightcyan / lightblue
M_a--M_b--M_c--cycle lightblue
A--B--C--cycle lightblue
X--M_a cyan
Y--M_b cyan
Z--M_c cyan

D = extension X A M_b M_c
D' = M_b+M_c-D

E = extension Y B M_c M_a
E' = M_c+M_a-E
F = extension Z C M_a M_b R220
F' = M_a+M_b-F

A--D deepgreen
B--E deepgreen
C--F deepgreen

M_a--Dp lightred
M_b--Ep lightred
M_c--Fp lightred

M_a--D lightred dotted
M_b--E lightred dotted
M_c--F lightred dotted

!dot("$X$", X, dir(250));
!dot("$Y$", Y, dir(80));
!dot("$Z$", Z, dir(200));

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

By Cevian Nest Theorem it suffices to
prove that $M_a D$, $M_b E$, $M_c F$ are concurrent.
Taking the isotomic conjugate and recalling that
$M_a M_b A M_c$ is a parallelogram,
we see that it suffices to prove
$M_a X'$, $M_b Y'$, $M_c Z'$ are concurrent.

We now use barycentric coordinates on $\triangle M_a M_b M_c$.
Let \[ S = \left( a^2S_A + t : b^2S_B + t : c^2S_C + t \right) \]
(possibly $t=\infty$ if $S$ is the centroid).
Let $v = b^2S_B+t$, $w=c^2S_C+t$.
Hence \[ X = \left( -a^2 vw : (b^2w+c^2v)v : (b^2w+c^2v)w \right). \]
Consequently,
\[ X' = \left( a^2vw : -a^2vw + (b^2w+c^2v)w : -a^2vw + (b^2w+c^2v)v \right) \]
We can compute
\[ b^2w+c^2v = (bc)^2(S_B+S_C)+ (b^2+c^2)t = (abc)^2 + (b^2+c^2)t. \]
Thus
\[ -a^2v+b^2w+c^2v
  = (b^2+c^2)t + (abc)^2 - (ab)^2S_B - a^2t =
  S_A( (ab)^2 +t ). \]
Finally
\[ X' = \left( a^2vw :
  S_A (c^2S_C+t) \left( (ab)^2 +2t \right) :
  S_A (b^2S_B+t) \left( (ac)^2 +2t \right)
  \right) \]
and from this it's evident that $AX'$, $BY'$, $CZ'$ are concurrent.

\paragraph{Second solution by moving points (Anant Mudgal).}
Let $H_a$, $H_b$, $H_c$ be feet of altitudes,
and let $\gamma$ denote the nine-point circle.
The main claim is that:
\begin{claim*}
  Lines $XH_a$, $YH_b$, $ZH_c$ are concurrent,
\end{claim*}
\begin{proof}
  In fact, we claim that the concurrence point lies on the Euler line $\ell$.
  This gives us a way to apply the moving points method:
  fix triangle $ABC$ and animate $S \in \ell$;
  then the map
  \begin{align*}
    \ell & \to \gamma \to \ell \\
    S &\mapsto X \mapsto S_a \coloneq \ell \cap \ol{H_a X}
  \end{align*}
  is projective, because it consists of two perspectivities.
  So we want the analogous maps $S \mapsto S_b$, $S \mapsto S_c$ to coincide.
  For this it suffices to check three positions of $S$;
  since you're such a good customer here are four.
  \begin{itemize}
    \ii If $S$ is the orthocenter of $\triangle M_a M_b M_c$
    (equivalently the circumcenter of $\triangle ABC$)
    then $S_a$ coincides with the circumcenter of $M_a M_b M_c$
    (equivalently the nine-point center of $\triangle ABC$).
    By symmetry $S_b$ and $S_c$ are too.

    \ii If $S$ is the circumcenter of $\triangle M_a M_b M_c$
    (equivalently the nine-point center of $\triangle ABC$)
    then $S_a$ coincides with the de Longchamps point of $\triangle M_a M_b M_c$
    (equivalently orthocenter of $\triangle ABC$).
    By symmetry $S_b$ and $S_c$ are too.

    \ii If $S$ is either of the intersections of the Euler
    line with $\gamma$, then $S = S_a = S_b = S_c$
    (as $S = X = Y = Z$).
  \end{itemize}
  This concludes the proof.
\end{proof}

\begin{center}
\begin{asy}
pair M_a = dir(110);
pair M_b = dir(210);
pair M_c = dir(330);
pair A = M_b+M_c-M_a;
pair B = M_c+M_a-M_b;
pair C = M_a+M_b-M_c;
pair O = circumcenter(A, B, C);
pair S = 0.75*O;
pair N = origin;

pair X = -M_a+2*foot(N,M_a,S);
pair Y = -M_b+2*foot(N,M_b,S);
pair Z = -M_c+2*foot(N,M_c,S);

filldraw(unitcircle, opacity(0.1)+lightcyan, lightblue);
draw(M_a--M_b--M_c--cycle, lightblue);
draw(A--B--C--cycle, lightblue);
draw(X--M_a, cyan);
draw(Y--M_b, cyan);
draw(Z--M_c, cyan);

pair H_a = M_b*M_c/M_a;
pair H_b = M_c*M_a/M_b;
pair H_c = M_a*M_b/M_c;

draw(X--A, deepgreen);
draw(Y--B, deepgreen);
draw(Z--C, deepgreen);

filldraw(H_a--H_b--H_c--cycle, opacity(0.1)+lightred, lightred);
draw(H_a--X, lightred);
draw(H_b--Y, lightred);
draw(H_c--Z, lightred);

dot("$X$", X, dir(250));
dot("$Y$", Y, dir(80));
dot("$Z$", Z, dir(200));

dot("$M_a$", M_a, dir(M_a));
dot("$M_b$", M_b, dir(M_b));
dot("$M_c$", M_c, dir(M_c));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$C$", C, dir(C));
dot("$S$", S, dir(310));
dot("$H_a$", H_a, dir(H_a));
dot("$H_b$", H_b, dir(H_b));
dot("$H_c$", H_c, dir(H_c));

/* TSQ Source:

M_a = dir 110
M_b = dir 210
M_c = dir 330
A = M_b+M_c-M_a
B = M_c+M_a-M_b
C = M_a+M_b-M_c
O := circumcenter A B C
S = 0.75*O R310
N := origin R45

X := -M_a+2*foot(N,M_a,S)
Y := -M_b+2*foot(N,M_b,S)
Z := -M_c+2*foot(N,M_c,S)

unitcircle 0.1 lightcyan / lightblue
M_a--M_b--M_c--cycle lightblue
A--B--C--cycle lightblue
X--M_a cyan
Y--M_b cyan
Z--M_c cyan

H_a = M_b*M_c/M_a
H_b = M_c*M_a/M_b
H_c = M_a*M_b/M_c

X--A deepgreen
Y--B deepgreen
Z--C deepgreen

H_a--H_b--H_c--cycle 0.1 lightred / lightred
H_a--X lightred
H_b--Y lightred
H_c--Z lightred

!dot("$X$", X, dir(250));
!dot("$Y$", Y, dir(80));
!dot("$Z$", Z, dir(200));

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

We now use Trig Ceva to carry over the concurrence.
By sine law,
\[ \frac{\sin \angle M_{c}AX}{\sin \angle AM_{c}X}
  =\frac{M_{c}X}{AX} \]
and a similar relation for $M_b$ gives that
\[ \frac{\sin \angle M_{c}AX}{\sin \angle M_{b}AX}
  = \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X} \cdot \frac{M_{c}X}{M_{b}X}
  = \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X}
  \cdot \frac{\sin \angle XM_{a}M_{c}}{\sin \angle XM_{a}M_{b}}. \]
Thus multiplying cyclically gives
\[ \prod_{\text{cyc}} \frac{\sin \angle M_{c}AX}{\sin \angle M_{b}AX}
  = \prod_{\text{cyc}} \frac{\sin \angle AM_{c}X}{\sin \angle AM_{b}X}
  \prod_{\text{cyc}} \frac{\sin \angle XM_{a}M_{c}}{\sin \angle XM_{a}M_{b}}. \]
The latter product on the right-hand side equals $1$
by Trig Ceva on $\triangle M_a M_b M_c$ with cevians
$\ol{M_a X}$, $\ol{M_b Y}$, $\ol{M_c Z}$.
The former product also equals $1$
by Trig Ceva for the concurrence in the previous claim
(and the fact that $\angle AM_{c}X=\angle H_{c}H_{a}X$).
Hence the left-hand side equals $1$, implying the result.

\paragraph{Third solution by moving points (Gopal Goel).}
In this solution,
we will instead use barycentric coordinates with resect to $\triangle ABC$
to bound the degrees suitably,
and then verify for seven distinct choices of $S$.

We let $R$ denote the radius of $\triangle ABC$,
and $N$ the nine-point center.

First, imagine solving for $X$ in the following way.
Suppose $\vec X = (1-t_a) \vec M_a + t_a \vec S$.
Then, using the dot product
(with $\left\lvert \vec v \right\rvert^2 = \vec v \cdot \vec v$ in general)
\begin{align*}
  \frac14 R^2
  &= \left\lvert \vec X - \vec N \right\rvert^2 \\
  &= \left\lvert t_a(\vec S - \vec M_a) + \vec M_a - \vec N \right\rvert^2 \\
  &= \left\lvert t_a(\vec S - \vec M_a) \right\rvert^2
    + 2t_a \left( \vec S - \vec M_a \right) \cdot \left( \vec M_a - \vec N \right)
    + \left\lvert \vec M_a - \vec N \right\rvert^2 \\
  &= t_a^2 \left\lvert (\vec S - \vec M_a) \right\rvert^2
    + 2t_a \left( \vec S - \vec M_a \right) \cdot \left( \vec M_a - \vec N \right)
    + \frac 14 R^2
\end{align*}
Since $t_a \neq 0$ we may solve to obtain
\[ t_a = -\frac{2(\vec M_a - \vec N) \cdot (\vec S - \vec M_a)}%
  {\left\lvert \vec S - \vec M_a \right\rvert^2}.  \]
Now imagine $S$ varies along the Euler line,
meaning there should exist linear functions
$\alpha, \beta, \gamma \colon \RR \to \RR$ such that
\[ S = \left( \alpha(s) , \beta(s) , \gamma(s) \right)
  \qquad s \in \RR \]
with $\alpha(s) + \beta(s) + \gamma(s) = 1$.
Thus $t_a = \frac{f_a}{g_a} = \frac{f_a(s)}{g_a(s)}$
is the quotient of a linear function
$f_a(s)$ and a quadratic function $g_a(s)$.

So we may write:
\begin{align*}
  X &= (1-t_a) \left( 0, \half, \half \right)
    + t_a \left( \alpha, \beta, \gamma \right) \\
  &= \left( t_a \alpha, \half(1-t_a) + t_a \beta,
    \half(1-t_a) + t_a \gamma \right) \\
  &= \left( 2f_a \alpha : g_a - f_a + 2f_a \beta : g_a - f_a + 2 f_a \gamma \right).
\end{align*}
Thus the coordinates of $X$ are quadratic polynomials in $s$
when written in this way.

In a similar way, the coordinates of $Y$ and $Z$
should be quadratic polynomials in $s$.
The Ceva concurrence condition
\[ \prod_{\text{cyc}} \frac{g_a-f_a+2f_a\beta}{g_a-f_a+2f_a\gamma}
  = 1 \]
is thus a polynomial in $s$ of degree at most six.
Our goal is to verify it is identically zero,
thus it suffices to check seven positions of $S$.

\begin{itemize}
  \ii If $S$ is the circumcenter of $\triangle M_a M_b M_c$
  (equivalently the nine-point center of $\triangle ABC$)
  then $\ol{AX}$, $\ol{BY}$, $\ol{CZ}$
  are altitudes of $\triangle ABC$.

  \ii If $S$ is the centroid of $\triangle M_a M_b M_c$
  (equivalently the centroid of $\triangle ABC$),
  then $\ol{AX}$, $\ol{BY}$, $\ol{CZ}$
  are medians of $\triangle ABC$.

  \ii If $S$ is either of the intersections of the Euler
  line with $\gamma$, then $S = X = Y = Z$
  and all cevians concur at $S$.

  \ii If $S$ lies on the $\ol{M_a M_b}$,
  then $Y = M_a$, $X = M_c$, and thus $\ol{AX} \cap \ol{BY} = C$,
  which is of course concurrent with $\ol{CZ}$
  (regardless of $Z$).
  Similarly if $S$ lies on the other sides of $\triangle M_a M_b M_c$.
\end{itemize}
Thus we are also done.

\paragraph{Fourth solution using Pascal (official one).}
We give a different proof of the claim that
$\ol{XH_a}$, $\ol{YH_b}$, $\ol{ZH_c}$ are concurrent
(and then proceed as in the end of the second solution).

Let $H$ denote the orthocenter, $N$ the nine-point center,
and moreover let $N_a$, $N_b$, $N_c$ denote the midpoints
of $\ol{AH}$, $\ol{BH}$, $\ol{CH}$,
which also lie on the nine-point circle
(and are the antipodes of $M_a$, $M_b$, $M_c$).
\begin{itemize}
  \ii By Pascal's theorem on $M_b N_b H_b M_c N_c H_c$,
  the point $P = \ol{M_c H_b} \cap \ol{M_b H_c}$
  is collinear with $N = \ol{M_b N_b} \cap \ol{M_c N_c}$,
  and $H = \ol{N_b H_b} \cap \ol{N_c H_c}$.
  So $P$ lies on the Euler line.

  \ii By Pascal's theorem on $M_b Y H_b M_c Z H_c$,
  the point $\ol{YH_b} \cap \ol{ZH_c}$ is collinear
  with $S = \ol{M_b Y} \cap \ol{M_c Z}$
  and $P = \ol{M_b H_c} \cap \ol{M_c H_b}$.
  Hence $YH_b$ and $ZH_c$ meet on the Euler line, as needed.
\end{itemize}