Erdős Problem #217

Erdős Problem #217 (Round 2)

Question recalled. For which $n$ do there exist $n$ points in $\mathbb{R}^2$, no three on a line and no four on a circle, which determine $n-1$ distinct distances and such that, after some ordering of those distances, the $i$th distance occurs exactly $i$ times?

1. ROUND-2 OBJECTIVE

Path C: exact construction/verification up to the current frontier. Round 1 had only fully verified the cases $n=3$ and $n=4$. The most promising strict advance is therefore to recover and verify the known planar constructions at the current boundary. In this round I derive explicit coordinates for Palásti's $7$-point construction from her geometric description, exactly verify Palásti's $8$-point construction, and extract exact $5$-point and $6$-point crescent configurations from the $8$-point example. The net result is a fully checked existence theorem for every $n=4,5,6,7,8$.

2. Round-1 FOUNDATION USED

I use the following Round-1 facts.

1. Proposition 4.1: the trivial $n=3$ sanity check.
2. Proposition 4.2: an explicit verified $n=4$ configuration.
3. The Round-1 gap statement: no exact planar construction beyond $n=4$ had been verified inside the write-up.

3. NEW INSIGHT / TOOL (ROUND 2)

This round adds three new ingredients.

1. An explicit coordinate derivation for Palásti's $7$-point construction from the original geometric description.
2. An exact symbolic verification of the distance multiplicities and general-position conditions for the $7$-point and $8$-point configurations.
3. A new observation from exact computation: suitable subsets of the $8$-point configuration already realize the cases $n=5$ and $n=6$, whereas no $7$-point subset of that $8$-point configuration is itself a $7$-point crescent configuration.

4. ATTACK PLAN (ROUND 2)

Round 1 had no verified examples past $n=4$. The exact tasks are therefore:

1. derive coordinates for a genuine $7$-point configuration;
2. verify the $8$-point Palásti coordinates exactly;
3. use subsets of the $8$-point configuration to fill the cases $n=5$ and $n=6$;
4. check whether one can obtain $n=7$ simply by deleting one point from the $8$-point construction.

This closes the entire verified small-$n$ range up to the known planar frontier $n=8$.

5. WORK (ROUND 2)

Proposition 5.1 (Palásti's $7$-point crescent configuration in coordinates). Let \[ \begin{array}{rcl} P_1&=&(0,0),\\[1mm] P_2&=&(1,0),\\[1mm] P_3&=&\left(1+\dfrac{\sqrt6}{2},\dfrac{\sqrt2}{2}\right),\\[3mm] P_4&=&\left(\dfrac{1+\sqrt6}{2},\dfrac{\sqrt2+\sqrt3}{2}\right),\\[3mm] P_5&=&\left(\dfrac12,\sqrt2+\dfrac{\sqrt3}{2}\right),\\[3mm] P_6&=&(0,\sqrt2),\\[1mm] Q&=&\left(\dfrac12,\sqrt2-\dfrac{\sqrt3}{2}\right). \end{array} \] Then the set \[ S_7:=\{P_1,P_2,P_3,P_4,P_5,P_6,Q\} \] is a crescent configuration of size $7$ in the plane.

Derivation of the coordinates. Palásti's geometric description starts from a convex hexagon with alternating side lengths \[ 1,\sqrt2,1,\sqrt2,1,\sqrt2 \] and alternating interior angles \[ 150^\circ,90^\circ,150^\circ,90^\circ,150^\circ,90^\circ. \] Fix $P_1=(0,0)$ and $P_2=(1,0)$. Because the exterior turning angles are \[ 30^\circ,90^\circ,30^\circ,90^\circ,30^\circ,90^\circ, \] the six edge directions are \[ 0^\circ,\ 30^\circ,\ 120^\circ,\ 150^\circ,\ 240^\circ,\ 270^\circ. \] Adding the corresponding side vectors gives exactly the six coordinates above. The perpendicular bisector of $P_1P_2$ is $x=\frac12$, and the displayed point $P_5$ lies on it. The condition $P_5Q=\sqrt3$ with $Q$ inside the hexagon forces \[ Q=\left(\frac12,\sqrt2-\frac{\sqrt3}{2}\right). \] Then \[ |P_6Q|=1,\qquad P_6Q\parallel P_4P_3, \] exactly as in Palásti's construction.

Proof that $S_7$ is crescent. A direct exact computation gives the following multiplicity pattern for the six distinct distances: \[ \sqrt{5-\sqrt6}\ \mbox{occurs once}, \] \[ \sqrt{3-\sqrt6}\ \mbox{occurs twice}, \] \[ \sqrt2\ \mbox{occurs three times}, \] \[ 1\ \mbox{occurs four times}, \] \[ \sqrt3\ \mbox{occurs five times}, \] \[ \sqrt{3+\sqrt6}\ \mbox{occurs six times}. \] Thus the distance multiplicities are exactly $1,2,3,4,5,6$.

To verify general position, use the determinant criteria \[ \Delta_{ijk}:=\det\begin{pmatrix} x_i&y_i&1\\ x_j&y_j&1\\ x_k&y_k&1 \end{pmatrix} \] for collinearity, and \[ \Gamma_{ijkl}:=\det\begin{pmatrix} x_i&y_i&x_i^2+y_i^2&1\\ x_j&y_j&x_j^2+y_j^2&1\\ x_k&y_k&x_k^2+y_k^2&1\\ x_l&y_l&x_l^2+y_l^2&1 \end{pmatrix} \] for concyclicity. All these determinants lie in the field $\mathbb{Q}(\sqrt2,\sqrt3)=\mathbb{Q}\oplus \mathbb{Q}\sqrt2\oplus \mathbb{Q}\sqrt3\oplus \mathbb{Q}\sqrt6$. Expanding every determinant in the basis $\{1,\sqrt2,\sqrt3,\sqrt6\}$, an exact symbolic check shows that all \[ \binom73=35 \] collinearity determinants and all \[ \binom74=35 \] concyclicity determinants are nonzero. Therefore no three points are collinear and no four are concyclic. $\square$

Proposition 5.2 (Palásti's $8$-point crescent configuration). Let \[ \begin{array}{rcl} A_1&=&(0,1),\\ A_2&=&(\sqrt3,0),\\ A_3&=&(2\sqrt3,0),\\ A_4&=&\left(\dfrac{5\sqrt3}{2},\dfrac52\right),\\ A_5&=&\left(\dfrac{3\sqrt3}{2},\dfrac92\right),\\ A_6&=&\left(\dfrac{\sqrt3}{2},\dfrac72\right),\\ A_7&=&\left(\dfrac{3\sqrt3}{2},\dfrac72\right),\\ A_8&=&(\sqrt3,2). \end{array} \] Then \[ S_8:=\{A_1,\ldots,A_8\} \] is a crescent configuration of size $8$ in the plane.

Proof. A direct exact computation of the $28$ squared distances gives the multiplicities \[ 1 \ \mbox{once},\qquad 19 \ \mbox{twice},\qquad 21 \ \mbox{three times}, \] \[ 3 \ \mbox{four times},\qquad 4 \ \mbox{five times},\qquad 7 \ \mbox{six times},\qquad 13 \ \mbox{seven times}. \] Thus, after ordering the distinct distances as \[ 1,\ \sqrt{19},\ \sqrt{21},\ \sqrt3,\ 2,\ \sqrt7,\ \sqrt{13}, \] their multiplicities are exactly \[ 1,2,3,4,5,6,7. \]

For general position, the same determinant criteria as in Proposition 5.1 apply. Now all determinants lie in the quadratic field $\mathbb{Q}(\sqrt3)$, so vanishing is equivalent to the vanishing of both coefficients in the basis $\{1,\sqrt3\}$. An exact symbolic check shows that all \[ \binom83=56 \] collinearity determinants and all \[ \binom84=70 \] concyclicity determinants are nonzero. Hence $S_8$ is in general position. $\square$

Proposition 5.3 (exact $5$-point and $6$-point subsets). With the notation of Proposition 5.2, set \[ T_5:=\{A_1,A_2,A_3,A_7,A_8\}, \qquad T_6:=\{A_1,A_2,A_3,A_4,A_7,A_8\}. \] Then $T_5$ is a planar crescent configuration of size $5$, and $T_6$ is a planar crescent configuration of size $6$.

Proof. General position is inherited from $S_8$.

For $T_5$, the exact squared-distance multiplicities are \[ 7 \ \mbox{once},\qquad 3 \ \mbox{twice},\qquad 4 \ \mbox{three times},\qquad 13 \ \mbox{four times}. \] Hence the four distinct distances have multiplicities $1,2,3,4$.

For $T_6$, the exact squared-distance multiplicities are \[ 21 \ \mbox{once},\qquad 3 \ \mbox{twice},\qquad 7 \ \mbox{three times},\qquad 4 \ \mbox{four times},\qquad 13 \ \mbox{five times}. \] Hence the five distinct distances have multiplicities $1,2,3,4,5$.

Therefore $T_5$ and $T_6$ are crescent configurations. $\square$

Corollary 5.4. Planar crescent configurations exist for every \[ n=4,5,6,7,8. \]

Proof. Use the Round-1 verified example for $n=4$, Proposition 5.3 for $n=5,6$, Proposition 5.1 for $n=7$, and Proposition 5.2 for $n=8$. $\square$

Proposition 5.5 (no $7$-point subset of $S_8$ is crescent). No $7$-point subset of the $8$-point configuration $S_8$ from Proposition 5.2 is itself a $7$-point crescent configuration.

Proof. There are only $8$ such subsets. An exact check of the distance multiplicities of each subset shows that none has multiplicity pattern $1,2,3,4,5,6$. Hence a separate $7$-point construction really is needed. $\square$

6. ADVERSARIAL VERIFICATION

The new work was checked against the main possible vulnerabilities.

1. Derivation of the $7$-point coordinates. I verified that the alternating side-length and angle data lead to the displayed edge directions and hence to the displayed vertex coordinates. I also checked directly that $P_5$ lies on the perpendicular bisector of $P_1P_2$, that $P_5Q=\sqrt3$, that $P_6Q=1$, and that $P_6Q\parallel P_4P_3$.
2. Exactness of the checks. For $S_7$, all equalities were checked in the basis $\{1,\sqrt2,\sqrt3,\sqrt6\}$. For $S_8$, all equalities were checked in the basis $\{1,\sqrt3\}$. No floating-point arithmetic was used.
3. Subset inheritance. Because general position is hereditary under taking subsets, only the distance multiplicities needed to be checked for $T_5$ and $T_6$.
4. Need for a separate $7$-point construction. The exact failure of all eight $7$-point subsets of $S_8$ confirms that Proposition 5.1 is not redundant.

7. FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The Round-2 advance is substantial: I have now given fully checked planar crescent configurations for every \[ n=4,5,6,7,8. \] This matches the known planar frontier. The remaining open range is therefore \[ n\geq 9. \]

8. COMPLETION ESTIMATE

COMPLETION: 60%

9. REFERENCES

1. I. Palásti, On the seven points problem of P. Erdős, Studia Sci. Math. Hungar. 22 $1987$, 447--448.
2. I. Palásti, Lattice point examples for a question of Erdős, Period. Math. Hungar. 20 $1989$, 231--235.
3. D. Burt, E. Goldstein, S. Manski, S. J. Miller, E. A. Palsson, and H. Suh, Crescent configurations, preprint $2015$, used here for the explicit $8$-point coordinates.
4. T. F. Bloom, Erdős Problem #217, problem page accessed 2026-03-07.

Problem 217: a bounded exact search for a ninth point

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. The target is to determine the sizes of planar crescent configurations: no three collinear, no four concyclic, and distance multiplicities $1,2,\ldots,n-1$. I read gpt_pro_5.4/217.tex in full, including the seven-point construction, eight-point construction, and subset checks. I tested a concrete continuation of its eight-point configuration, instead of treating those already supplied constructions as new progress.

2) WORK

Use coordinates $(a\sqrt3/2,b/2)$ and write the supplied eight points as \[ (a,b)\in\{(0,2),(2,0),(4,0),(5,5),(3,9),(1,7),(3,7),(2,4)\}. \] Four times the squared distance between two such points is the integer $3(a-a')^2+(b-b')^2$. Exact integer computation verifies the eight-point squared-distance multiplicities \[ 1:1,\quad19:2,\quad21:3,\quad3:4,\quad4:5,\quad7:6,\quad13:7. \] All 56 triple determinants and all 70 circle determinants are nonzero; for the latter one can use rows $(3a^2+b^2,a,b,1)$ because the omitted scale factors are nonzero.

For a ninth point, the seven existing distance classes survive. Exactly one new class is therefore permitted. Its eight new edges must raise the old multiplicities and introduce the new one in such a way that the resulting multiset is exactly $\{1,2,\ldots,8\}$. This gives an exact necessary filter without geometry tolerances.

I executed this filter for every integer coordinate pair $-20\le a\le20$, $-20\le b\le30$, excluding the eight existing points: 2083 candidates. None even passed the multiplicity filter. Consequently there is no one-point extension on this finite grid box. The standard-library code and result are in verification/batch_1_checks.py and verification/batch_1_checks.json.

3) ADVERSARIAL VERIFICATION

The integer representation is exact and distinguishes squared distances without floating-point rounding. The result does not exclude points outside the box, points outside this lattice, moving the original eight points, or nine-point configurations having no such eight-point subset. General position alone is hereditary; the crescent multiplicity property is not, as the supplied seven-point deletion check already illustrates.

4) FINAL

UNRESOLVED. The attempted lattice extension failed in the precisely stated finite search region. No ninth-point construction or universal nonexistence proof was obtained, and this negative finite result is not evidence of a full solution.

5) SOURCES CHECKED

Complete supplied file and exact local computation dated 2026-09-23. Burt et al., Crescent configurations, https://arxiv.org/abs/1509.07220, checked for the problem and existing construction context. The grid search is the only additional computational claim.

6) COMPLETION ESTIMATE

Subjective progress toward the unrestricted existence problem. COMPLETION: 8%

Erdős Problem #217

1. Formal Restatement

Let $n\in\mathbb{N}$ and let $S\subset\mathbb{R}^2$ be a set of $n$ points in general position, meaning:

1. no three points of $S$ are collinear;
2. no four points of $S$ are concyclic.

Let \[D(S):=\{d(p,q): p,q\in S,\ p\neq q\}\] be the set of distinct pairwise distances determined by $S$.

Problem #217. For which $n$ does there exist $S\subset\mathbb{R}^2$ in general position with

1. $|D(S)|=n-1$ (exactly $n-1$ distinct distances), and
2. there is an ordering $\delta_1,\dots,\delta_{n-1}$ of the elements of $D(S)$ such that the distance $\delta_i$ occurs exactly $i$ times among the $\frac{n(n-1)}{2}$ unordered pairs of points in $S$.

(Note that the total number of pairs is $\frac{n(n-1)}{2}=\sum_{i=1}^{n-1} i$, so these multiplicities exhaust all pairs.)

2. Quick Literature/Context Check

The prompt states:

• There is an explicit example for $n=4$ ("an isosceles triangle with the point in the centre").
• For $n=5$ there is a construction due to Pomerance, and Palásti proved existence for all $n\le 8$.
• Erdős believed nonexistence for all sufficiently large $n$; one route to prove this would be via strong lower bounds on the number of distinct distances determined by $n$ points in general position.

This strongly suggests the classification of all $n$ is open.

3. Attack Plan

Proof approach (construct for many $n$). Try to build crescent configurations inductively:

• Start from a known small configuration.
• Add a point so that it introduces exactly one new distance and increments multiplicities in a controlled way.

The challenge is simultaneously preserving general position and achieving the exact multiplicity pattern $1,2,\dots,n-1$.

Disproof approach (show impossible for large $n$). Use lower bounds on the number of distinct distances in general position. If one could prove that any $n$ points in general position determine at least $n$ distinct distances for all sufficiently large $n$, then crescent configurations (which have only $n-1$ distinct distances) would be impossible for large $n$.

4. Work

I can fully verify the claimed example for $n=4$ (and note the trivial $n=3$ case), but I cannot determine existence/nonexistence for all $n$.

Proposition 4.1 ($n=3$ sanity check). For $n=3$, an isosceles triangle with distinct side lengths gives $|D(S)|=2=n-1$ and multiplicities $1$ and $2$.

Proof. Let $S=\{A,B,C\}$ with $AB=AC\neq BC$. Then the distinct distances are $\{AB,BC\}$; the distance $AB$ occurs twice (pairs $\{A,B\},\{A,C\}$) and $BC$ occurs once. General-position constraints are vacuous for $n=3$ beyond "not collinear", which is satisfied by any nondegenerate triangle. $\square$

Proposition 4.2 (Explicit crescent configuration for $n=4$). Let \[B=(-1,0),\qquad C=(1,0),\qquad A=(0,2). \] Let $O$ be the circumcenter of $\triangle ABC$; for these coordinates one computes \[O=\left(0,\frac{3}{4}\right). \] Set $S:=\{A,B,C,O\}$. Then $S$ is in general position and determines $3$ distinct distances whose multiplicities are $1,2,3$.

Proof. Step 1: compute the three distance values and multiplicities. First, \[BC=2. \] Next, \[AB^2=(0+1)^2+(2-0)^2=1+4=5,\quad AC^2=(0-1)^2+(2-0)^2=5, \] so $AB=AC=\sqrt{5}$. Finally, since $O$ is the circumcenter, the three distances $OA,OB,OC$ are equal to the circumradius $R$. Because $A$ and $O$ lie on the $y$-axis, \[R=OA=\left|2-\frac{3}{4}\right|=\frac{5}{4}, \] and one may check similarly that $OB=OC=5/4$. Thus the multiset of distances among the $6$ pairs consists of:

• $2$ occurring once (pair $\{B,C\}$),
• $\sqrt{5}$ occurring twice (pairs $\{A,B\},\{A,C\}$),
• $5/4$ occurring three times (pairs $\{O,A\},\{O,B\},\{O,C\}$).

Hence there are $3=n-1$ distinct distances with multiplicities $1,2,3$.

Step 2: general position. No three of $A,B,C$ are collinear since they form a nondegenerate triangle. The point $O=(0,3/4)$ is not on the line $BC$ (which has equation $y=0$), so $B,C,O$ are not collinear. Also $A$ is not on the line through $B$ and $O$ (slope $3/4$) nor on the line through $C$ and $O$ (slope $-3/4$), so no triple involving $O$ is collinear.

For concyclicity: the points $A,B,C$ determine a unique circle, namely their circumcircle, whose center is $O$ and whose radius is $R=5/4$. Since $d(O,O)=0\neq R$, the point $O$ is not on this circle. Therefore $A,B,C,O$ are not concyclic. $\square$

Sticking point for #217. Beyond small $n$, achieving the exact multiplicity pattern appears to require delicate global control of the distance multiset; no general inductive step is provided here.

5. Verification

• The $n=4$ configuration was checked explicitly: all pairwise distances computed, multiplicities matched, and the general-position conditions verified.
• Edge cases: for $n<4$, the "no four concyclic" condition is vacuous; for $n=3$ the required multiplicities $1,2$ are achievable as shown.

6. FINAL

UNRESOLVED.

1. (i) Strongest fully proved partial result obtained here: An explicit crescent configuration exists for $n=4$ (Proposition 4.2), and the multiplicity condition is trivially achievable for $n=3$ (Proposition 4.1).
2. (ii) First gap blocking completion: No method is given to either construct crescent configurations for all $n$ (or all $n\le N$) or to prove a lower bound on distinct distances in general position that would rule them out for large $n$.
3. (iii) Most promising next moves:
   • Develop or use stronger lower bounds for the number of distinct distances determined by $n$ points under the "no three collinear/no four concyclic" constraints, aiming to prove $|D(S)|\ge n$ for large $n$.
   • Search for an inductive "add one point" mechanism that adjusts distance multiplicities in the required way while maintaining general position.
   • Translate the multiplicity constraints into constraints on distance graphs (edge-colored complete graphs) and use extremal graph theory to force forbidden geometric patterns.
4. (iv) Conjectured final answer and minimal counterexample shape: I conjecture the answer is: such configurations exist only for bounded $n$ (in line with Erdős's belief). A minimal counterexample to this conjecture would be an explicit crescent configuration with $n=9$ in general position (since existence is known up to $n=8$ per the prompt).

7. Completion Estimate

20%. Small examples are verified, but there is no progress toward the main classification "for which $n$" beyond what is already indicated in the prompt.