Problem: Let $ABC$ be a triangle with $CA=CB$ and $\angle{ACB}=120^{\circ}$,
and let $M$ be the midpoint of $AB$.
Let $P$ be a variable point of the circumcircle of $ABC$,
and let $Q$ be the point on the segment $CP$ such that $QP = 2QC$.
It is given that the line through $P$ and perpendicular to $AB$
intersects the line $MQ$ at a unique point $N$.
Prove that there exists a fixed circle such that $N$
lies on this circle for all possible positions of $P$.

Solution: Since $\vec N = 3\vec Q - 2\vec M$,
it suffices to show $Q$ moves on a fixed circle.
That fixed circle is the image of $(CAB)$ under a homothety at $C$ with ratio $1/3$,
so we are done.
\begin{center}
\begin{asy}
pair C = dir(90);
pair A = dir(150);
pair B = dir(30);
pair M = midpoint(A--B);
pair P = dir(-40);
pair Q = (2*C+P)/3;
pair N = extension(M, Q, P, P+C);
draw(unitcircle);

draw(A--B--C--cycle, dotted);
draw(C--P--N--M--cycle);

dot("$C$", C, dir(C));
dot("$A$", A, dir(A));
dot("$B$", B, dir(B));
dot("$M$", M, dir(270));
dot("$P$", P, dir(P));
dot("$Q$", Q, dir(225));
dot("$N$", N, dir(N));

/* TSQ Source:

C = dir 90
A = dir 150
B = dir 30
M = midpoint A--B R270
P = dir -40
Q = (2*C+P)/3 R225
N = extension M Q P P+C
unitcircle

A--B--C--cycle dotted
C--P--N--M--cycle

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

\begin{remark*}
  Note that points $A$ and $B$ are superfluous,
  and the choice of constant $2$ or the definition of $M$
  is also arbitrary (only the fact that $\ol{CM} \parallel \ol{PN}$ matters).
  Moreover, the condition $\angle ACB = 120^{\circ}$ is never used either.

  As a result, I did not find this problem very inspiring.
\end{remark*}