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Math-5425-A3 · revision r1 · on Math-5425

gpt pro 5.2 attempt on Math-5425

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Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

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  1. Erdős Problem #97

Erdős Problem #97

  1. Formal Restatement.

    Let $P$ be a convex polygon in $\mathbb{R}^2$ with vertex set $V=\{v_1,\dots,v_n\}$. Say that a vertex $v\in V$ is $4$-pinned-repetition-free if there do not exist four distinct vertices $x_1,x_2,x_3,x_4\in V\setminus\{v\}$ with \[ \|v-x_1\|=\|v-x_2\|=\|v-x_3\|=\|v-x_4\|. \] Equivalently, in the multiset $\{\|v-x\|:x\in V\setminus\{v\}\}$, every value occurs with multiplicity at most $3$.

    Question $\#97$. Is it true that every convex polygon $P$ has at least one vertex that is $4$-pinned-repetition-free?

  2. Quick literature/context check.

    The statement with "$4$" replaced by "$3$" is false: Danzer produced a convex $9$-gon such that every vertex has three other vertices equidistant from it, and Fishburn--Reeds constructed a convex $20$-gon where the witnessing distance can be chosen the same for all vertices.(See the summary and references on the Erdős Problems website.[ErdosProblems97,FiRe92])

    Erdős discussed this type of question already in 1946 and later writings; in particular, he explicitly expressed hope that a statement of this form might hold for "$4$" even though it fails for "$3$" (and remarked he could not prove even very weak quantitative variants).(See, e.g., Erdős $1986$ and the sources listed on the Erdős Problems page.[Er86, ErdosProblems97])

    As of the last edit date shown on the Erdős Problems page (27 Oct 2025), the problem is listed as open and falsifiable by a finite counterexample.[ErdosProblems97]

  3. Attack plan.

    Proof track. Try to show that assuming the negation (every vertex $v$ admits a radius $r(v)$ with at least four vertices on the circle $\partial B(v,r(v))$) forces too many circle--circle intersections. Since two circles with distinct centers meet in at most $2$ points, this becomes an extremal set-system/incidence problem. The convexity constraint might further restrict which incidences are geometrically realizable.

    Disproof track. Try to explicitly construct a convex $n$-gon where for each vertex $v$ there is a circle centered at $v$ passing through (at least) four other vertices. One approach is to prescribe an "incidence design" $(V,\{S_v\}_{v\in V})$ with $|S_v|\ge 4$ and $|S_u\cap S_v|\le 2$, then attempt to solve the resulting distance-equality constraints in coordinates and finally check convex position.

    Phase 1 small cases. Note that for $n\le 4$ the question is trivial (a vertex cannot have four other vertices). For $n=5$ it reduces to asking whether every vertex has all four other vertices at the same distance, which is impossible in $\mathbb{R}^2$ (it would force all pairwise distances equal).

  4. Work.

    I did not obtain a full proof or a counterexample for general $n$. I did, however, obtain a clean unconditional "small-$n$" obstruction that does not use convexity.

    Theorem (No $6$-point counterexample, even without convexity): There do not exist six points $p_1,\dots,p_6\in\mathbb{R}^2$ such that for every $i\in\{1,\dots,6\}$ there is a radius $r_i>0$ for which at least four of the other points lie on the circle $\{x:\|x-p_i\|=r_i\}$. Equivalently: for every $6$-point set $A\subset\mathbb{R}^2$, there exists $p\in A$ such that no four points of $A\setminus\{p\}$ are equidistant from $p$.
    Proof: Suppose for contradiction that $p_1,\dots,p_6$ satisfy the stated property. For each $i$ choose a circle \[ C_i := \{x\in\mathbb{R}^2 : \|x-p_i\|=r_i\} \] that contains at least four points of $\{p_1,\dots,p_6\}\setminus\{p_i\}$. Define the index set \[ S_i := \{j\in\{1,\dots,6\}\setminus\{i\} : p_j\in C_i\}. \] Then $|S_i|\ge 4$ for each $i$.

    Step 1: no $S_i$ can have size $5$. If some $|S_i|=5$, then $C_i$ contains all five points $\{p_j:j\ne i\}$. For any $k\ne i$, the circle $C_k$ contains at least $4$ of those same five points, hence $C_i$ and $C_k$ would share at least $4$ common points. But two distinct circles in the plane have at most $2$ intersection points, and $C_i\ne C_k$ because they have different centers $p_i\ne p_k$. Contradiction. Therefore, for each $i$ we must have \[ |S_i|=4. \]

    Step 2: derive a forced "missing index" map. Since $S_i\subseteq\{1,\dots,6\}\setminus\{i\}$ has size $4$ and $\{1,\dots,6\}\setminus\{i\}$ has size $5$, there is a unique index $f(i)\in\{1,\dots,6\}\setminus\{i\}$ such that \[ S_i = \{1,\dots,6\}\setminus\{i,f(i)\}. \] In words: $C_i$ passes through every point except $p_i$ itself and one "missing" point $p_{f(i)}$.

    Step 3: use a special pair to contradict circle intersection. Fix any $i$, and set $j:=f(i)$ (so $j\ne i$). Then by definition $j\notin S_i$, i.e., $p_j\notin C_i$. Now consider $S_j$. We know $|S_j|=4$ and $S_j = \{1,\dots,6\}\setminus\{j,f(j)\}$. Since $f(i)=j$, the set of indices excluded from $S_i\cap S_j$ is \[ \{i,j,f(i),f(j)\} = \{i,j,j,f(j)\} = \{i,j,f(j)\}, \] which has size at most $3$. Therefore \[ |S_i\cap S_j| \;=\; 6-\left|\{i,j,f(j)\}\right| \;\ge\; 3. \] Thus the two circles $C_i$ and $C_j$ share at least three common points $p_k$ with $k\in S_i\cap S_j$. This is impossible because two distinct circles intersect in at most two points.

    We have reached a contradiction, so no such six-point configuration exists. ∎

    Consequences. In particular, for every convex hexagon (indeed, for every $6$-point set in $\mathbb{R}^2$) there exists a vertex $v$ for which no four other vertices are equidistant from $v$. So any counterexample to Problem #97 must have at least $7$ vertices.

    Where the approach gets stuck for general $n$. For $n\ge 7$, the simple "missing index" argument above no longer forces a unique complement representation for each $S_i$, and extremal set systems with $|S_i|=4$ and $|S_i\cap S_j|\le 2$ can exist combinatorially (e.g. via designs). The remaining difficulty is to determine whether such incidence patterns can be realized by circles centered at the points themselves and with the points in convex position.

  5. Verification.

    The only geometric input used in Theorem thm:n6 is: (i) distinct circles intersect in at most $2$ points, and (ii) circles $C_i$ and $C_k$ are distinct because their centers $p_i\ne p_k$. All other steps are purely combinatorial on the incidence sets $S_i$. No convexity hypothesis was used, so the conclusion is valid a fortiori for convex polygons with $n\le 6$.

  6. FINAL: UNRESOLVED.

    • (i) Strongest fully proved partial result: Theorem thm:n6 (no $6$-point configuration can have the "every point has $4$ equidistant neighbors" property). In particular, Problem #97 holds for all convex polygons with $n\le 6$.
    • (ii) Most plausible route forward: Model the negation by circles $C_v$ centered at each vertex with $|V\cap C_v|\ge 4$, convert to a constrained incidence design $\{S_v\}$ with $|S_u\cap S_v|\le 2$, then incorporate convex position constraints (cyclic order/visibility constraints) to rule out realizability for large $n$, or else explicitly solve the distance-equality system for a realizable design.
    • (iii) First place the proof attempt breaks: Beyond $n=6$, the incidence sets $S_v$ need not be complements of size-$2$ subsets, so the forced pair argument $j=f(i)$ that yields $|S_i\cap S_j|\ge 3$ no longer applies.
    • (iv) What a counterexample would have to look like: A convex $n$-gon together with a choice of radii $\{r(v)\}_{v\in V}$ so that for each $v$ at least four vertices lie on the circle centered at $v$ with radius $r(v)$, and such that for distinct $u,v$ the circles share at most two vertices (automatically), yet the system is geometrically realizable in convex position.

  7. Completion estimate.

    COMPLETION: 25%.