= John contact decomposition
{c}
{title2=$\sum_j c_j u_j\otimes u_j=I$}
For a centrally symmetric convex body in <John position>, there are contact points $u_j\in\partial K\cap S^{d-1}$ and positive weights $c_j$ such that $\sum_jc_j u_j\otimes u_j=I$. Taking the <trace> gives $\sum_jc_j=d$. A supporting hyperplane at $u_j$ shows $|\langle u_j,x\rangle|\leq\|x\|_K$.
Here is the maximum-volume argument for the decomposition. If $I$ were outside the <convex cone> generated by the contact matrices, a <separating hyperplane> would give a <symmetric matrix> $A$ with $\operatorname{tr}A>0$ and $\langle Au,u\rangle\leq0$ at every contact. Subtract a small positive multiple of $I$ so these inequalities become strict while the <trace> remains positive. The <support function> of $(I+tA)B_2^d$ is $|(I+tA)v|$. It decreases below $1$ near all contact directions for small positive $t$; away from them, the strict gap $h_K(v)>1$ preserves containment. Its volume increases, since $\det(I+tA)=1+t\operatorname{tr}A+O(t^2)$, a contradiction. The cone is closed because its generators have <trace> $1$ and form a <compact set>; finite-dimensional <convexity> gives a finite decomposition.
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