IMO 1998 Q2 #
In a competition, there are a
contestants and b
judges, where b ≥ 3
is an odd integer. Each
judge rates each contestant as either "pass" or "fail". Suppose k
is a number such that, for any
two judges, their ratings coincide for at most k
contestants. Prove that k / a ≥ (b - 1) / (2b)
.
Solution #
The problem asks us to think about triples consisting of a contestant and two judges whose ratings
agree for that contestant. We thus consider the subset A ⊆ C × JJ
of all such incidences of
agreement, where C
and J
are the sets of contestants and judges, and JJ = J × J - {(j, j)}
. We
have natural maps: left : A → C
and right: A → JJ
. We count the elements of A
in two ways: as
the sum of the cardinalities of the fibres of left
and as the sum of the cardinalities of the
fibres of right
. We obtain an upper bound on the cardinality of A
from the count for right
,
and a lower bound from the count for left
. These two bounds combine to the required result.
First consider the map right : A → JJ
. Since the size of any fibre over a point in JJ is bounded
by k
and since |JJ| = b^2 - b
, we obtain the upper bound: |A| ≤ k(b^2-b)
.
Now consider the map left : A → C
. The fibre over a given contestant c ∈ C
is the set of
ordered pairs of (distinct) judges who agree about c
. We seek to bound the cardinality of this
fibre from below. Minimum agreement for a contestant occurs when the judges' ratings are split as
evenly as possible. Since b
is odd, this occurs when they are divided into groups of size
(b-1)/2
and (b+1)/2
. This corresponds to a fibre of cardinality (b-1)^2/2
and so we obtain
the lower bound: a(b-1)^2/2 ≤ |A|
.
Rearranging gives the result.
An ordered pair of judges.
Equations
- Imo1998Q2.JudgePair J = (J × J)
Instances For
A triple consisting of contestant together with an ordered pair of judges.
Equations
- Imo1998Q2.AgreedTriple C J = (C × Imo1998Q2.JudgePair J)
Instances For
The first judge from an ordered pair of judges.
Equations
- Imo1998Q2.JudgePair.judge₁ = Prod.fst
Instances For
The second judge from an ordered pair of judges.
Equations
- Imo1998Q2.JudgePair.judge₂ = Prod.snd
Instances For
The proposition that the judges in an ordered pair are distinct.
Instances For
The proposition that the judges in an ordered pair agree about a contestant's rating.
Equations
- Imo1998Q2.JudgePair.Agree r p c = (r c p.judge₁ ↔ r c p.judge₂)
Instances For
The contestant from the triple consisting of a contestant and an ordered pair of judges.
Equations
- Imo1998Q2.AgreedTriple.contestant = Prod.fst
Instances For
The ordered pair of judges from the triple consisting of a contestant and an ordered pair of judges.
Equations
- Imo1998Q2.AgreedTriple.judgePair = Prod.snd
Instances For
The set of contestants on which two judges agree.
Equations
- Imo1998Q2.agreedContestants r p = Finset.filter (fun (c : C) => Imo1998Q2.JudgePair.Agree r p c) Finset.univ
Instances For
All incidences of agreement.
Equations
- Imo1998Q2.A r = Finset.filter (fun (a : Imo1998Q2.AgreedTriple C J) => Imo1998Q2.JudgePair.Agree r a.judgePair a.contestant ∧ a.judgePair.Distinct) Finset.univ