Erdős Problem #97

Erdős Problem #97

1) ROUND-2 OBJECTIVE

I pursue path (A): continue the proof attempt by attacking the first unresolved cardinality after Round-1. Round-1 proved that no $6$-point counterexample exists, so the next natural gap is the $7$-point case. The goal in this round is to determine the exact combinatorial structure forced on any hypothetical $7$-point counterexample.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without re-proving them.

• Round-1 theorem: no $6$-point set in the plane can have the property that every point has four other points equidistant from it. Hence any counterexample to Problem #97 must have at least $7$ points.
• Round-1 incidence reformulation: if $A=\{p_1,\dots,p_n\}\subset \mathbf{R}^2$ is such a configuration and $C_i$ is a circle centered at $p_i$ containing at least four other points of $A$, then \[ S_i:=\{j\in\{1,\dots,n\}\setminus\{i\}: p_j\in C_i\} \] satisfies $|S_i|\ge 4$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is a double-counting argument on point--circle incidences and circle--circle intersections in the first open case $n=7$. It forces exact equalities everywhere, and these equalities imply that the complements \[ B_i:=\{1,\dots,7\}\setminus S_i \] form a Steiner triple system on $7$ points.

So Round-2 does not merely say "$n=7$ is hard"; it proves that any $7$-point counterexample must carry the rigid incidence pattern of a $2$-$(7,3,1)$ design.

4) ATTACK PLAN (ROUND-2)

The exact gap left by Round-1 is this: a counterexample, if one exists, has at least $7$ points, but nothing structural was known at $n=7$.

To advance strictly beyond Round-1, I prove the following sharp reduction:

Claim to prove. If $A=\{p_1,\dots,p_7\}\subset \mathbf{R}^2$ has the property that for every $i$ there is a circle $C_i$ centered at $p_i$ containing at least four other points of $A$, then:

• each $C_i$ contains exactly four points of $A$;
• each point of $A$ lies on exactly four of the seven circles;
• every two circles $C_i,C_j$ meet in exactly two points of $A$;
• the $7$ complements $B_i=\{1,\dots,7\}\setminus S_i$ are $3$-subsets with every pair of labels contained in exactly one $B_i$.

This overcomes the main Round-1 obstacle: instead of having many uncontrolled incidence patterns for $n\ge 7$, the case $n=7$ collapses to one highly constrained combinatorial model.

5) WORK (ROUND-2)

7-point structure theorem.

1. $|S_i|=4$ for every $i$.
2. If \[ t_j:=|\{\,i\in\{1,\dots,7\}: j\in S_i\,\}| \] is the number of circles through $p_j$, then $t_j=4$ for every $j$.
3. For every $i\neq k$, \[ |S_i\cap S_k|=2. \]
4. If $B_i:=\{1,\dots,7\}\setminus S_i$, then $|B_i|=3$, $i\in B_i$, and every pair of labels from $\{1,\dots,7\}$ belongs to exactly one block $B_i$.

Proof.

Step 2: double count pairwise circle intersections through the points of $A$. Let \[ m:=\sum_{i=1}^7 |S_i|=\sum_{j=1}^7 t_j. \] Because every $|S_i|\ge 4$, we have $m\ge 28$.

Now count triples $(j,\{i,k\})$ with $i<k$ and $j\in S_i\cap S_k$ in two ways: \[ \sum_{1\le i<k\le 7} |S_i\cap S_k| = \sum_{j=1}^7 \binom{t_j}{2}. \] Since any two distinct circles intersect in at most two points, \[ |S_i\cap S_k|\le 2 \quad\text{for all }i\neq k, \] hence \[ \sum_{j=1}^7 \binom{t_j}{2} = \sum_{1\le i<k\le 7} |S_i\cap S_k| \le 2\binom{7}{2}=42. \]

On the other hand, \[ \sum_{j=1}^7 \binom{t_j}{2} = \frac12\left(\sum_{j=1}^7 t_j^2-\sum_{j=1}^7 t_j\right) \ge \frac12\left(\frac{(\sum_{j=1}^7 t_j)^2}{7}-\sum_{j=1}^7 t_j\right) = \frac12\left(\frac{m^2}{7}-m\right), \] by Cauchy--Schwarz. Since $m\ge 28$ and the function $\frac12(\frac{m^2}{7}-m)$ is increasing for $m\ge 4$, \[ \sum_{j=1}^7 \binom{t_j}{2} \ge \frac12\left(\frac{28^2}{7}-28\right)=42. \] Therefore all inequalities above are equalities.

Equality in \[ \sum_{j=1}^7 \binom{t_j}{2}\le 42 \] forces \[ |S_i\cap S_k|=2\qquad\text{for all }i\neq k. \] Equality in Cauchy--Schwarz forces all $t_j$ to be equal, and since their sum is $m=28$, we obtain \[ t_1=t_2=\cdots=t_7=4. \] Finally, \[ \sum_{i=1}^7 |S_i|=\sum_{j=1}^7 t_j=28, \] and because each $|S_i|\ge 4$, necessarily \[ |S_i|=4\qquad\text{for all }i. \]

Step 3: pass to the complements. Define \[ B_i:=\{1,\dots,7\}\setminus S_i. \] Since $|S_i|=4$, we get $|B_i|=3$. Also $i\notin S_i$, so $i\in B_i$.

For $i\neq k$, \[ |B_i\cap B_k| = 7-|S_i\cup S_k| = 7-(|S_i|+|S_k|-|S_i\cap S_k|) = 7-(4+4-2)=1. \] Thus any two distinct blocks $B_i,B_k$ meet in exactly one label.

Now count unordered pairs of labels contained in the blocks $B_i$. Each block contributes \[ \binom{3}{2}=3 \] pairs, so altogether the $7$ blocks contribute \[ 7\binom{3}{2}=21 \] pairs, counted with multiplicity. But the whole label set $\{1,\dots,7\}$ has exactly \[ \binom{7}{2}=21 \] unordered pairs. Since two distinct blocks intersect in only one label, no pair of labels can occur in two different blocks. Hence every pair of labels occurs in exactly one block.

This is exactly the defining property of a Steiner triple system on $7$ points. End of proof.

Corollary.

Equivalently: the first open case $n=7$ is not a free geometric search problem anymore; it is reduced to realizing a very specific incidence structure with circles centered at the points themselves.

6) ADVERSARIAL VERIFICATION

I now check the new work against the main failure modes.

• Did I accidentally use convexity? No. The entire theorem is valid for arbitrary $7$-point sets in the plane. So it is genuinely stronger than the convex-polygon statement.
• Could two circles coincide, invalidating the intersection bound? No. The centers are $p_i$ and $p_j$ with $i\neq j$, so $C_i\neq C_j$.
• Did Step 1 really rule out $|S_i|=6$? Yes. If $C_i$ contains all six other points, then any other $C_j$ contains at least four points, at most one of which can be $p_i$; therefore at least three lie on both circles, impossible.
• Did the convexity inequality hide a gap? No. I wrote it explicitly as a Cauchy--Schwarz bound on $\sum t_j^2$, and the equality case is standard: all $t_j$ are equal.
• Does the pair count for the blocks really imply "exactly one"? Yes. Pairwise block intersections have size $1$, so no unordered pair of labels can lie in two different blocks. Since the total number of pair-occurrences is exactly $\binom72$, every pair occurs once.
• Interaction with Round-1. Round-1 only excluded $n\le 6$. The new theorem settles the entire combinatorial shape of the first remaining case $n=7$, so this is a strict advance.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 proves a new theorem: every hypothetical $7$-point counterexample must realize a Steiner triple system on $7$ labels through the complements $B_i=\{1,\dots,7\}\setminus S_i$. This is substantially stronger than the Round-1 statement "$n\ge 7$" and isolates a sharply formulated obstruction for the first open case.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

No new external reference is used in the Round-2 theorem; the argument above is self-contained.

Erdős Problem #97: independent continuation

1) OBJECTIVE AND SOURCE AUDIT

2) WORK: NO SEVEN-POINT CONFIGURATION

Make a graph $R$ on the seven labels, joining $i,j$ when $S_{ij}=S_{ji}=1$. Every vertex has degree at least $4+4-6=2$. On every edge of $R$, $r_i=|p_i-p_j|=r_j$, so radii are constant on each connected component. Consequently incidences between two vertices of the same component are symmetric, including pairs not originally known to be adjacent.

If $R$ is disconnected, its component sizes must be three and four. The component of size three is a triangle. Let the four-vertex component have $e$ edges. Internal directed incidences number $6+2e$. There are at most twelve directed incidences between components, because any reciprocal pair would connect them. Since the total is 28, $e\ge5$. If $e=5$, the two nonadjacent vertices of this component have its other two vertices as common out-neighbours. Each also has two out-neighbours in the three-vertex component; those two subsets intersect. Their rows then intersect in at least three places, contradicting $1$. If $e=6$, each of the four vertices has exactly one out-neighbour in the other component. Two of them choose the same vertex by pigeonhole, again giving that pair at least three common out-neighbours. Thus $R$ is connected.

All seven radii are now equal, so $S=S^t$. The all-ones vector is an eigenvector of $S$ of eigenvalue four. On its six-dimensional perpendicular space, $1$ gives $S^2=2I$, so the remaining eigenvalues are $\sqrt2$ or $-\sqrt2$. Since the diagonal is zero, the trace would give \[ 0=4+(a-b)\sqrt2,\qquad a,b\in\mathbb N_0,\quad a+b=6, \] which is impossible. No such seven-point configuration exists.

3) ADVERSARIAL VERIFICATION

7) FINAL AND REFERENCE

8) COMPLETION ESTIMATE

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%.