Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 163 / 3

One form of the trilinear Kakeya inequality in three dimensions is the following. If ${\mathcal{T}}_j$ are finite families of $R^{1/2}\times R^{1/2}\times R$ tubes whose directions lie within $1/30$ of $e_j$, then for every $ϵ>0$,The same statement holds for any three uniformly transverse direction caps, with a constant depending on their transversality.

Solved by gpt-5.6-sol high.

For $t_j=j/3$, the three central radial directions are proportional to $(1,t_j,t_j^2)$. Their determinant is the nonzero Vandermonde productChoose $c>0$ small enough that every triple ${\omega }_j\in {\Sigma }_j$ still has determinant bounded away from zero. An invertible linear transformation sends the three central directions to the standard basis; it sends the given tubes to comparable tubes with directions in fixed small neighborhoods of $e_1,e_2,e_3$ and distorts volume and dimensions by fixed factors.

Applying part a after this transformation gives, uniformly in the three families,This is the required moment-curve version; its uniformity is precisely the transversality supplied by the moment curve and the Vandermonde determinant.

Solved by gpt-5.6-sol high.

The three unit normals of any selected blocks remain uniformly linearly independent. Hence the intersection of three thickness-one slabs with these normals has bounded volume, and thereforeExpanding the cube of the requested $L^3$ norm and using nonnegativity givesTaking cube roots proves the estimate with a constant independent of $r$, which is stronger than the allowed factor $C_{ϵ}{r}^{ϵ}$.

Solved by gpt-5.6-sol high.