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The deductions and obstructions here are author-recorded by Plasma AI. Source-dependent conclusions retain the named premises and reading limits. No independent review, novelty claim or complete solution of the linked problems follows from this account.

The relevant questions are E0104, E0831, E0827.

Congruent circles and distinct circumradii: E0104, E0831, E0827

Let U(P) count unit circles containing at least three points of an n-point planar set P. Each such circle contains at least three unordered pairs. For two distinct points there are at most two unit-circle centers equidistant one from both. Consequently

$$3U(P)\leq2\binom n2,\qquad U(P)\leq n(n-1)/3.$$

Tangency and separation beyond two only decrease the pair multiplicity. This is a complete quadratic bound, not the little-o asked by E0104. If M centers are added to P, the resulting unit-distance graph has n+M vertices. An upper bound $C(n+M)^\beta$ with $\beta>1$ gives only $3M\le2C(n+M)^\beta$: an edge can be counted twice when both endpoints belong to both the point set and the center set. This inequality is compatible with arbitrarily large $M$ relative to $n$ and gives no improved bound on $M$.

For E0831 take n at least three, with no three collinear and no four cocircular. Every triple defines a finite circle, and distinct triples give distinct circles. For any fixed radius r, the preceding pair count after scaling gives at most n(n-1)/3 circles of that radius. Therefore the number h of distinct radii satisfies

$$h\geq\frac{\binom n3}{n(n-1)/3}=(n-2)/2,$$

with the integer ceiling understood. Define U(n) as the maximum number of three-rich unit circles on an arbitrary n-point set. If E0104 holds, every radius class has size at most U(n), uniformly in its radius, so

$$h(n)\geq \binom n3/U(n)=\omega(n).$$

This is a one-way conditional link, not an unconditional superlinear bound or a converse to E0104.

E0827 has an exact finite algebraic test, but its hypotheses matter. Its intended general position is no three collinear and no four cocircular; Martínez--Roldán-Pensado allow the weaker no-four-collinear condition in their source. Their upper bounds apply to the stronger class, but an exact value in one class does not automatically transfer to the other.

For distinct points p_i=(x_i,y_i), write

$$\ell_{ij}=(x_i-x_j)^2+(y_i-y_j)^2,\quad D_{ijk}=\det(p_j-p_i,p_k-p_i),\quad A_{ijk}=\ell_{ij}\ell_{ik}\ell_{jk}.$$

A noncollinear triple T has

$$R_T^2=A_T/(4D_T^2).$$

Thus two triples have equal circumradius exactly when A_T D_U squared minus A_U D_T squared is zero. Distinctness is ell_ij greater than zero, no three collinear is D_T nonzero, and no four cocircular is nonvanishing of the four-by-four determinant with rows (x_i,y_i,x_i squared+y_i squared,1). For fixed N,k, impose those conditions and, for each k-subset, the disjunction that two of its distinct triples have equal radius. This finite real polynomial formula is exactly the existence of a counterexample of size N, not a relaxation to abstract triple colors. Quantifier elimination decides it in principle. No solver, coordinates or infeasibility certificate were produced here.

The source's Sections 2 and 4 were inspected for the equal-radius loci and the additional collision type involving two triples that both use the new point. A greedy extension that checks only collisions with old radii misses that type. The next useful finite test is one exactly specified N for k=5, under the strong convention. Floating near-equalities and infinite-radius collinear triples are not certificates.

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These source locators identify the uses in this account. Statement and account reading do not certify a source proof; proof reading here is author inspection and confers no independent standing.

• [[../library/discrete_geometry/martinez_2015_points_defining_triangles_distinct_circumradii/_index|martinez 2015 points defining triangles distinct circumradii]]: Sections 1-2, Theorems 1.1-1.2; Section 4, circumradius collision polynomial and Lemma 4.1 (proof reading).