Source: USATST-2024-P4
Category: Existence and Construction

Problem:
Find all integers $n \ge 2$ for which there exists a sequence of $2n$ pairwise distinct points $(P_1, \dots, P_n, Q_1, \dots, Q_n)$ in the plane satisfying the following four conditions: 
\begin{enumerate}[label={(\roman*)}]
\item no three of the $2n$ points are collinear;
\item $P_iP_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $P_{n+1}=P_1$;
\item $Q_iQ_{i+1} \ge 1$ for all $i=1,2,\dots,n$, where $Q_{n+1}=Q_1$;
\item $P_i Q_j \le 1$ for all $i=1,2,\dots,n$ and $j=1,2,\dots,n$.
\end{enumerate}

Solution:
The answer is even integers only.

\paragraph{Proof that even $n$ work.}
If we ignore the conditions that the points are pairwise distinct and form no collinear triples, we may take 
\[ 
P_{2i-1}=(0.51, 0), \quad P_{2i}=(-0.51, 0), \quad Q_{2i-1}=(0, 0.51), \quad Q_{2i}=(0,-0.51) 
\] 
for $i = 1, 2, \dots, n/2$. The distances $P_iP_{i+1}$ and $Q_iQ_{i+1}$ (including $P_nP_1$ and $Q_nQ_1$) are $1.02 > 1$, while all distances $P_iQ_j$ are $0.51\sqrt{2} \approx 0.72 < 1$. Because these inequalities are strict, we may perturb each point by a sufficiently small amount to ensure that the strict distance inequalities still hold, while making the points pairwise distinct and ensuring no three points are collinear.

\paragraph{Proof that odd $n$ do not work.}
The main claim is the following. 
\begin{claim*} 
For $1\leq i\leq n$, points $Q_i$ and $Q_{i+1}$ must lie on opposite sides of line $P_1P_2$. 
\end{claim*} 
To isolate the geometry component of the problem, we rewrite the claim in the following contrapositive form, without referencing the points $Q_i$ and $Q_{i+1}$: 
\begin{lemma*} 
Suppose $P_1, P_2$ are points with $P_1P_2 \ge 1$, and $A, B$ are two points such that $\max(P_1A, P_1B, P_2A, P_2B) \le 1$. Moreover, assume $A$ and $B$ lie on the same side of line $P_1 P_2$, and no three of $\{P_1, P_2, A, B\}$ are collinear. Then $AB < 1$. 
\end{lemma*} 
\begin{proof}[Proof of Lemma] 
Since $A$ and $B$ lie on the same side of $P_1P_2$, the convex hull of these four points is either a quadrilateral or a triangle. 
\begin{itemize} 
    \item If the convex hull is a quadrilateral, assume without loss of generality that the vertices are $P_1, P_2, A, B$ in order. Let $X$ denote the intersection of the diagonals $P_1A$ and $P_2B$. Since $P_1P_2 \ge 1$, we have
    \[ 
    1 + AB \leq P_1P_2 + AB < (P_1X + XP_2) + (AX + XB) = P_1A + P_2B \leq 2. 
    \] 
    Thus, $AB < 1$.
    \item Otherwise, the convex hull is a triangle. Since $A$ and $B$ are on the same side of $P_1P_2$, the segment $P_1P_2$ must be an edge of this convex hull. Assume without loss of generality that $B$ is in the interior of triangle $P_1P_2A$. Since $\angle P_1BA + \angle P_2BA = 360^\circ-\angle P_1BP_2 > 180^\circ$, at least one of $\angle P_1BA$ and $\angle P_2BA$ is obtuse. Assume without loss of generality the former angle is obtuse; then in $\triangle P_1BA$, the side opposite the obtuse angle is strictly the longest, so $AB < P_1A \leq 1$.
\end{itemize} 
\end{proof} 
Applying the lemma to $A = Q_i$ and $B = Q_{i+1}$, we are given $P_1P_2 \ge 1$ and $P_1Q_i, P_1Q_{i+1}, P_2Q_i, P_2Q_{i+1} \leq 1$. Since $Q_iQ_{i+1} \geq 1$, the points $Q_i$ and $Q_{i+1}$ cannot lie on the same side of $P_1P_2$. This proves the claim. 

It follows from the claim that $Q_i$ is on the same side of line $P_1P_2$ as $Q_1$ if $i$ is odd, and on the opposite side if $i$ is even. Since $Q_1 = Q_{n+1}$, this alternation means the construction is not possible when $n$ is odd.

Answer: Even integers only

Grading guidelines:
(Partial) 1. Gave a correct construction for infinitely many even n (e.g. using two alternating locations for the Pi and two alternating locations for the Qj with strict inequalities), with a brief explanation that small perturbations can enforce “pairwise distinct” and “no three collinear” while preserving all distance inequalities. 2. Proved the key geometric lemma/claim: if P1P2≥1 and A,B satisfy P1A,P1B,P2A,P2B≤1 and A,B lie on the same side of line P1P2 (with no three collinear), then AB<1; and correctly applied it to deduce that Qi and Qi+1 must lie on opposite sides of P1P2. 3. From the “opposite sides” claim, correctly derived the parity alternation of the Qi relative to line P1P2 and concluded that odd n is impossible (even if the even-n construction is missing). (Almost) 1. Correctly proved that odd n do not work via the opposite-sides lemma/claim and the resulting alternation argument, but the even-n existence part is missing or only sketched without a perturbation/distinctness justification. 2. Gave the even-n construction and the parity obstruction for odd n, but left a localized gap in the lemma proof (e.g. the convex-hull case split/diagonal inequality or the obtuse-angle argument is not fully justified). 3. Solution is almost complete, but made minor mistakes which are not negligible.