Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 109 / 3 / i

Solution

Discard ground points lying in no set. For every point

x

that lies in at least two members, form

L_{x} = {i : x \in A_{i}} \subseteq [n] .

Every pair of indices lies in exactly one

L_{x}

, and no

L_{x}

equals

[n]

because the total intersection is empty. Thus the

L_{x}

n

indices. The number of its lines through index

i

is at most

a_{i}

, with equality unless

A_{i}

contains private points.

r_{i}

is the number of lines through point

i

n

points, then

\sum_{i = 1}^{n} (2 r _{i}) \geq (2 n) .

Applying it here gives

\sum_{i = 1}^{n} (2 a _{i}) \geq \sum_{i = 1}^{n} (2 r _{i}) \geq (2 n) .

Moreover, a pair of ground points can lie in at most one

A_{i}

, since two different members meet in only one point. Hence

\sum_{i} (2 a _{i}) \leq (2 m) .

Combining the inequalities gives

(2 m) \geq (2 n)

, and therefore

m \geq n

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