= Solution
The sharp bound form of the <Erdős-Ko-Rado theorem> is as follows. If $1\leq r\leq n/2$ and $\mathcal F\subseteq\binom{[n]}r$ is an <intersecting family>, then
$$
\boxed{|\mathcal F|\leq\binom{n-1}{r-1}.}
$$
The <uniform set family> of all $r$-element <subsets> containing one fixed point attains the bound. We prove it by the <Katona circle method>.
First establish the <cyclic interval intersection bound>. In any fixed cyclic ordering of $[n]$, at most $r$ cyclic intervals of length $r$ can form an <intersecting family>. If there is a selected interval, rotate the ordering so that it ends at position $n$ and consists of positions $n-r+1,\ldots,n$. Intervals ending at positions $r,\ldots,n-r$ are disjoint from it and cannot be selected. Of the remaining intervals, one ends at $n$; the others have endpoints paired as
$$
(j,\ j+n-r),\qquad 1\leq j\leq r-1.
$$
The two intervals in each pair are disjoint: one wraps around position $n$ and the other lies in the intervening gap, since $n\geq2r$. At most one of each pair belongs to the <intersecting family>, giving at most $1+(r-1)=r$ intervals.
Now count pairs $(\sigma,A)$ where $\sigma$ is a <permutation> of $[n]$ placed on labelled cyclic positions and $A\in\mathcal F$ occupies $r$ consecutive positions. Each <permutation> contributes at most $r$ pairs by the <cyclic interval intersection bound>. Each fixed $A$ contributes $n\,r!\,(n-r)!$ pairs: choose its starting position and order the elements inside and outside its interval. Since $r<n$, the starting position is unique for each such placement. Consequently
$$
|\mathcal F|\,n\,r!\,(n-r)!\leq r\,n!,\qquad |\mathcal F|\leq\frac{r}{n}\binom nr=\binom{n-1}{r-1}.
$$
This completes the <Erdős-Ko-Rado theorem> proof. At $n=2r$, complementary $r$-element <subsets> are the only disjoint pairs, so choosing one member of every complementary pair gives an extremiser. This describes all equality cases at that boundary. The range $n\geq2r$ is essential: for $n<2r$, the entire <uniform set family> is an <intersecting family>.
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