Erdős Problem 735: positive-weight magic configurations
1. Statement and scope
Classify finite point sets $P\subset\mathbb R^2$ admitting positive weights such that every line determined by at least two points has the same total weight. For $|P|\ge2$, divide by this positive common total and normalize every determined-line sum to $1$. The known complete classification, proved by Ackerman, Buchin, Knauer, Pinchasi and Rote, consists of collinear sets, sets with no three collinear, near-pencils with all but one point collinear, and projective copies of one seven-point failed-Fano configuration. This attempt explicitly verifies all those families and their normalized weights, and proves a useful necessity reduction. It does not reconstruct the geometric classification lemma at the end of that reduction.
2. Complete verification of the listed families
If all points are collinear, any positive weights summing to $1$ work. For zero or one point there is no determined line and the condition is vacuous. If $n\ge3$ and no three points are collinear, every pair has weight sum $1$. Comparing the pairs in any triple forces each of their weights to be $1/2$, and comparison with that triple forces every remaining weight to be $1/2$. These weights work. For two points only their total is prescribed. For a near-pencil with $r\ge3$ points on a line $L$ and an off-line point $p$, every line from $p$ to $L$ contains exactly those two points. All $r$ on-line weights must therefore have one common value $u$, and $ru=1$. The unique normalized weights are \[
w(x)=1/r\quad(x\in L\cap P),\qquad w(p)=1-1/r>0.
\] The three-point near-pencil is already the no-three-collinear case. For the exceptional family take \[
A=(0,0),\ B=(2,0),\ C=(0,2),\
a=(1,1),\ b=(0,1),\ c=(1,0),\ O=(2/3,2/3).
\] The six three-point lines have point sets \[
\{A,B,c\},\ \{B,C,a\},\ \{C,A,b\},\
\{A,a,O\},\ \{B,b,O\},\ \{C,c,O\}.
\] Their pairs are all distinct and account for $18$ of the $\binom72=21$ pairs. The remaining pairs are $ab,bc,ca$, each an ordinary line, as is checked immediately from the coordinates. Assign weight $1/2$ to $a,b,c$ and weight $1/4$ to $A,B,C,O$. Every determined line has weight $1$. These weights are unique: the ordinary-line equations first force the three half-weights, the three side equations then force $w(A)=w(B)=w(C)=1/4$, and a median equation forces $w(O)=1/4$. Any projective transformation for which all seven image points are finite preserves incidence, hence preserves the line-sum condition and these weights. This coordinate configuration is projectively equivalent to the version drawn using an equilateral triangle and its angle bisectors.
3. A necessity reduction from ordinary lines
We use one explicitly external input, the Kelly--Moser ordinary-line bound: a noncollinear set of $t$ real planar points determines at least $3t/7$ lines containing exactly two of them. Suppose $P$ is not collinear and not a near-pencil. For every $p\in P$, there is an ordinary line of $P$ avoiding $p$. To see this, assume all ordinary lines pass through $p$. The set $Q=P\setminus\{p\}$ is noncollinear. Every ordinary line of $Q$ must pass through $p$, because otherwise it would be an ordinary line of $P$ avoiding $p$. There are at least $3(n-1)/7$ such lines. Their pairs of $Q$-points are disjoint, since different lines through $p$ meet only at $p$. Thus at least $6(n-1)/7$ points of $Q$ lie on these three-point lines of $P$. None can lie on an ordinary line of $P$ through $p$. There are consequently at most $(n-1)/7$ ordinary lines of $P$, contrary to the lower bound $3n/7$. Let $A$ be the set of points incident with an ordinary line of $P$. Every point in $A$ has weight $1/2$. Indeed, if an endpoint $p$ of an ordinary line had weight other than $1/2$, one endpoint of that line would have weight greater than $1/2$; call it $p$. Choose an ordinary line $qr$ avoiding $p$. Its weights sum to $1$, so one endpoint, say $q$, has weight at least $1/2$. The determined line $pq$ then has weight greater than $1$, impossible. It follows that every line through two points of $A$ is ordinary: their weights already sum to $1$, so positivity excludes a third point. Put $B=P\setminus A$. If $b\in B$ and $a\in A$, the line $ab$ cannot be ordinary, by the definition of $B$. It has a third positive-weight point, so $w(b)<1/2$. We have proved \[
w(a)=1/2\ (a\in A),\qquad 0<w(b)<1/2\ (b\in B),
\] and the ordinary lines of $P$ are exactly the lines joining pairs of points of $A$. The set $A$ is nonempty by the same ordinary-line theorem. If $B=\varnothing$, all determined lines are ordinary and the general-position family has already been obtained.
4. Exact missing step
When both $A$ and $B$ are nonempty, the remaining geometric assertion is that the positive line-sum condition and the ordinary-line pattern just proved force $|A|=3$, $|B|=4$, with precisely the exceptional incidence configuration. This is Theorem 2 in the authors' paper; their proof passes to weighted arrangements of great circles. I have verified its statement and the elementary reduction above, but have not supplied an independent reconstruction of that arrangement argument. Treating the finite weight equations alone as a classification would omit the essential real-geometric restriction.
5. Sources and assessment
Sources:
https://www.erdosproblems.com/735; E. Ackerman, K. Buchin, C. Knauer, R. Pinchasi and G. Rote,
There are not too many magic configurations, Discrete Comput. Geom. 39 $2008$, 3--16, author manuscript
https://page.mi.fu-berlin.de/rote/Papers/pdf/There+are+not+too+many+magic+configurations.pdf. The Kelly--Moser bound and its use in this reduction are explicitly recorded in that manuscript.
UNRESOLVED in this reconstruction: every listed example and its weights are verified, and necessity is reduced to the precise mixed ordinary-line classification lemma, whose proof is still missing here. The problem itself has the known complete resolution stated above. The estimate measures this attempt's proof coverage, not the likelihood of a new solution.
Completion rate estimate: 60%.