Problem: Alice the architect and Bob the builder play a game.
First, Alice chooses two points $P$ and $Q$ in the plane
and a subset $\mathcal{S}$ of the plane, which are announced to Bob.
Next, Bob marks infinitely many points in the plane,
designating each a city.
He may not place two cities within distance at most one unit of each other,
and no three cities he places may be collinear.
Finally, roads are constructed between the cities as follows:
each pair $A$, $B$ of cities is connected with a road along the line segment $AB$
if and only if the following condition holds:
\begin{quote}
For every city $C$ distinct from $A$ and $B$,
there exists $R\in \mathcal{S}$ such that $\triangle PQR$ is directly similar
to either $\triangle ABC$ or $\triangle BAC$.
\end{quote}
Alice wins the game if (i) the resulting roads allow for travel
between any pair of cities via a finite sequence of roads and (ii) no two roads cross.
Otherwise, Bob wins. Determine, with proof, which player has a winning strategy.

Solution: The answer is that Alice wins.
Let's define a \emph{Bob-set} $V$ to be a set of points in the plane with no three collinear
and with all distances at least $1$.
The point of the problem is to prove the following fact.
\begin{claim*}
  Given a Bob-set $V \subseteq \RR^2$, consider the \emph{Bob-graph}
  with vertex set $V$ defined as follows:
  draw edge $ab$ if and only if the disk with diameter $\ol{ab}$
  contains no other points of $V$ on or inside it.
  Then the Bob-graph is (i) connected, and (ii) planar.
\end{claim*}
Proving this claim shows that Alice wins since Alice can specify $\mathcal{S}$
to be the set of points outside the disk of diameter $PQ$.

\begin{proof}
  [Proof that every Bob-graph is connected]
  Assume for contradiction the graph is disconnected.
  Let $p$ and $q$ be two points in different connected components.
  Since $pq$ is not an edge, there exists a third point $r$
  inside the disk with diameter $\ol{pq}$.

  Hence, $r$ is in a different connected component
  from at least one of $p$ or $q$ --- let's say point $p$.
  Then we repeat the same argument on the disk with diameter $\ol{pr}$
  to find a new point $s$, non-adjacent to either $p$ or $r$.
  See the figure below, where the X'ed out dashed edges indicate
  points which are not only non-adjacent but in different connected components.

  \begin{center}
  \begin{asy}
    size(6cm);
    pair p = (-1,0);
    pair q = (1,0);
    pair r = (0.4,0.7);
    filldraw(unitcircle, opacity(0.1)+blue, blue);
    filldraw(circle((p+r)/2, abs(p-r)/2), opacity(0.1)+red, red);
    pair s = (-0.3,0.6);
    draw(p--q, dashed+blue);
    draw(p--r, dashed+red);
    draw(s--r, dashed);
    dot("$p$", p, dir(p));
    dot("$q$", q, dir(q));
    dot("$r$", r, dir(r));
    dot("$s$", s, dir(s));
    picture X;
    real eps = 0.05;
    draw(X, (-eps,-eps)--(eps,eps), black+1.4);
    draw(X, (eps,-eps)--(-eps,eps), black+1.4);
    add(X);
    add(shift(midpoint(p--r))*X);
    add(shift(midpoint(s--r))*X);
    label("$\delta_1$", (0,0), 2*dir(45), deepgreen);
    label("$\delta_2$", midpoint(p--r), 2*dir(-10), deepgreen);
    label("$\delta_3$", midpoint(s--r), 2*dir(100), deepgreen);
  \end{asy}
  \end{center}

  In this way we generate an infinite sequence of distances
  $\delta_1$, $\delta_2$, $\delta_3$, \dots\ among the non-edges in the picture above.
  By the ``Pythagorean theorem'' (or really the inequality for it), we have
  \[ \delta_i^2 \le \delta_{i-1}^2 - 1 \]
  and this eventually generates a contradiction for large $i$,
  since we get $0 \le \delta_i^2 \le \delta_1^2 - (i-1)$.
\end{proof}

\begin{proof}
  [Proof that every Bob-graph is planar]
  Assume for contradiction edges $ac$ and $bd$ meet, meaning $abcd$ is a convex quadrilateral.
  WLOG assume $\angle bad \ge 90\dg$ (each quadrilateral has an angle at least $90\dg$).
  Then the disk with diameter $\ol{bd}$ contains $a$, contradiction.
\end{proof}

\begin{remark*}
  In real life, the Bob-graph is actually called the \href{https://w.wiki/DVnd}{Gabriel graph}.
  Note that we never require the Bob-set to be infinite;
  the solution works unchanged for finite Bob-sets.

  However, there are approaches that work for finite Bob-sets that don't work for
  infinite sets, such as the \href{https://www.graphclasses.org/classes/gc_1224.html}{relative neighbor graph},
  in which one joins $a$ and $b$ iff there is no $c$ such that
  $d(a,b) \le \max \{d(a,c), d(b,c)\}$.
  In other words, edges are blocked by triangles where $ab$ is the longest edge
  (rather than by triangles where $ab$ is the longest edge of a right or obtuse triangle
  as in the Gabriel graph).

  The relative neighbor graph has fewer edges than the Gabriel graph, so it is planar too.
  When the Bob-set is finite, the relative distance graph is still connected.
  The same argument above works where the distances now satisfy
  \[ \delta_1 > \delta_2 > \dots \]
  instead, and since there are finitely many distances one arrives at a contradiction.

  However for infinite Bob-sets the descending condition is insufficient,
  and connectedness actually fails altogether.
  A counterexample (communicated to me by Carl Schildkraut) is to start by taking
  $A_n \approx (2n,0)$ and $B_n \approx (2n+1, \sqrt3)$ for all $n \ge 1$,
  then perturb all the points slightly so that
  \begin{align*}
    B_1A_1 &> A_1A_2 > A_2B_1 > B_1B_2 > B_2A_2 \\
    &> A_2A_3 > A_3B_2 > B_2B_3 > B_3A_3 \\
    &> \dotsb.
  \end{align*}
  A cartoon of the graph is shown below.
  \begin{center}
  \begin{asy}
    dotfactor *= 1.5;
    pair A1 = (0,0);
    pair A2 = (2,0);
    pair A3 = (4,0);
    pair A4 = (6,0);
    pair B1 = (1,3**0.5);
    pair B2 = (3,3**0.5);
    pair B3 = (5,3**0.5);
    pair B4 = (7,3**0.5);
    dot("$A_1$", A1, dir(-90), blue);
    dot("$A_2$", A2, dir(-90), blue);
    dot("$A_3$", A3, dir(-90), blue);
    dot("$B_1$", B1, dir(90), blue);
    dot("$B_2$", B2, dir(90), blue);
    dot("$B_3$", B3, dir(90), blue);
    label("$\dots$", A4, blue);
    label("$\dots$", B4, blue);
    draw("$2.08$", A1--A2, dir(-90), blue, Margins);
    draw("$2.04$", A2--A3, dir(-90), blue, Margins);
    draw(A3--A4, blue, Margins);
    draw("$2.06$", B1--B2, dir(90), blue, Margins);
    draw("$2.02$", B2--B3, dir(90), blue, Margins);
    draw(B3--B4, blue, Margins);
    draw(rotate(60)*"$2.09$", B1--A1, dotted);
    draw(rotate(-60)*"$2.07$", B1--A2, dotted);
    draw(rotate(60)*"$2.05$", B2--A2, dotted);
    draw(rotate(-60)*"$2.03$", B2--A3, dotted);
    draw(rotate(60)*"$2.01$", B3--A3, dotted);
  \end{asy}
  \end{center}
  In that case, $\{A_n\}$ and $\{B_n\}$ will be disconnected from each other:
  none of the edges $A_nB_n$ or $B_nA_{n+1}$ are formed.
  In this case the relative neighbor graph consists of the edges
  $A_1A_2A_3A_4 \dotsm$ and $B_1 B_2 B_3 B_4 \dotsm$.
  That's why for the present problem, the inequality
  \[ \delta_i^2 \le \delta_{i-1}^2 - 1 \]
  plays such an important role, because it causes the (squared) distances
  to decrease appreciably enough to give the final contradiction.
\end{remark*}