Problem: Consider a convex pentagon circumscribed about a circle.
We name the lines that connect vertices of the pentagon with
the opposite points of tangency with the circle \emph{gergonnians}.
\begin{enumerate}
  \ii[(a)] Prove that if four gergonnians are concurrent,
    then all five of them are concurrent.
  \ii[(b)] Prove that if there is a triple of gergonnians
    that are concurrent, then there is another triple
    of gergonnians that are concurrent.
\end{enumerate}

Solution: This problem is insta-killed by taking a homography
sending the concurrency point (in either part)
to the center of the circle while fixing the incircle.
Alternatively, one may send any four of the tangency points to a rectangle.

Here are the details.
Let $ABCDE$ be a pentagon with gergonnians
$\ol{AV}$, $\ol{BW}$, $\ol{CX}$, $\ol{DY}$, $\ol{EZ}$.
We prove the following lemma, which
(up to a suitable permutation of point names)
solves both parts (a) and (b).
\begin{lemma*}
  The gergonnians $\ol{AV}$, $\ol{CX}$, $\ol{DY}$
  are concurrent if and only if the gergonnians
  $\ol{AV}$, $\ol{BW}$, $\ol{EZ}$ are concurrent.
\end{lemma*}
\begin{proof}
  We prove the first set implies the second
  (the converse direction being identical).
  Suppose $\ol{AV}$, $\ol{CX}$, $\ol{DY}$ intersect at $P$
  and take a homography fixing the circle
  and moving $P$ to its center.

  \begin{center}
  \begin{asy}
  size(7cm);
  pair Y = dir(115);
  pair X = -conj(Y);
  pair V = dir(270);
  pair Z = X*X/V;
  pair W = Y*Y/V;

  pair A = 2*X*Y/(X+Y);
  pair B = 2*Y*Z/(Y+Z);
  pair C = 2*Z*V/(Z+V);
  pair D = 2*V*W/(V+W);
  pair E = 2*W*X/(W+X);

  filldraw(A--B--C--D--E--cycle, opacity(0.1)+lightblue, blue);
  filldraw(unitcircle, opacity(0.1)+lightcyan, blue);

  pair P = origin;
  draw(C--X, red);
  draw(D--Y, red);
  draw(A--V, red);

  draw(B--W, heavygreen+dashed);
  draw(E--Z, heavygreen+dashed);

  dot("$Y$", Y, dir(Y));
  dot("$X$", X, dir(X));
  dot("$V$", V, dir(V));
  dot("$Z$", Z, dir(Z));
  dot("$W$", W, dir(W));
  dot("$A$", A, dir(A));
  dot("$B$", B, dir(B));
  dot("$C$", C, dir(C));
  dot("$D$", D, dir(D));
  dot("$E$", E, dir(E));
  dot("$P$", P, dir(0));

  /* Source generated by TSQ */
  \end{asy}
  \end{center}

  Then $X$ and $Y$ are symmetric around $\ol{APV}$
  by hypothesis.
  Since $D = \ol{VV} \cap \ol{PY}$, $C = \ol{VV} \cap \ol{PX}$,
  it follows that $C$ and $D$,
  and hence $Z$ and $W$,
  are also symmetric around $\ol{APV}$.
  Consequently $B$ and $E$ are symmetric too.
  So $\ol{BW}$ and $\ol{EZ}$ meet on $\ol{AV}$.
\end{proof}