every cyclic polytope C_d(d+3) is circumscribable

Authorship disclaimer: This proof was written primarily by GPT-5.6-sol.

Theorem, not counterexample. This is an exact proof draft. It has not been peer reviewed.

A cyclic polytope is the convex hull of points on the moment curve

t -> (t, t^2, ..., t^d).

A polytope is circumscribable if it has a realization whose facets all touch one sphere, with the sphere inside every supporting halfspace.

Here is the result that fell out of trying to find a counterexample:

Theorem. For every d >= 2, the cyclic polytope C_d(d+3) is circumscribable. In fact, the equally spaced realization at t = 0, 1, ..., d+2 has a projective image with every facet tangent to one sphere.

If correct, this answers Questions 3.13 and 6.2 in Doolittle, Labbé, Lange, Sinn, Spreer, and Ziegler. They reported that they could not find the required quadric for C_6(9). That made dimension six look like a plausible place for circumscribability to fail. It was instead the first case where the hidden general construction became visible.

the proof idea

Write m = d + 1 and use the equally spaced moment-curve vertices. Every facet omits three vertex indices a < b < c. Gale's rule for cyclic polytopes says that the two gaps

u = b - a,   v = c - b

must both be odd.

The supporting hyperplanes can be represented by polynomials. Take the facets omitting three consecutive vertices as a basis. In this basis, the facet omitting a, b, c is a positive multiple of a simple tent-shaped vector: it rises linearly to one peak and then falls linearly again.

That reduces the geometry to one matrix. Define a symmetric matrix G, constant along each diagonal, by

G(i,i)     = 0
G(i,i+1)   = 3
G(i,i+r)   = 4r(-1)^(r+1)   for r >= 2.

A finite-sum identity shows that every odd-gap tent vector h satisfies

h^T G h = 0.

So all the facet hyperplanes lie on one quadric. The same matrix has exactly one negative eigenvalue: after alternating the coordinate signs, it becomes -4 times a distance matrix plus a path-adjacency correction, and it is positive on the codimension-one subspace where the coordinates sum to zero. Its trace is zero, forcing the final eigenvalue to be negative.

Set Q = -G. Then Q has the one-positive-direction signature needed for a sphere in projective geometry. One final sign calculation shows that the inward-facing facet hyperplanes all lie on the same half of the null cone. That matters: it says the tangent sphere is inside every facet halfspace, rather than merely touching the supporting hyperplanes with inconsistent orientations.

A projective change of coordinates turns that quadric into the ordinary unit sphere. This proves the theorem.

what it does and does not settle

The proof is dimension-independent. Exact arithmetic checks reproduce every identity for 2 <= d <= 12; those checks are regression tests, not the reason the theorem holds.

Doolittle and collaborators noted that a negative example in this family would refute a broader conjecture of Grünbaum. This result closes that particular route to a counterexample. It does not prove Grünbaum's conjecture for all polytopes with few vertices.

I do not know whether the argument is new. A targeted literature search during the project did not find a posted proof, but that is not a priority claim. The cautious status is: exact proof draft, checked computationally in finite dimensions, not peer reviewed.