Advanced book on Mathematics Olympiad

790 Combinatorics and Probability

the area we are seeking (after doubling) is

2

1 +r^2 − 2 h^2

√

r^2 −h^2

.

Dividing byπ, then integrating overr, we compute the distribution ofhto be

1

π

∫ 1

h

2

1 +r^2 − 2 h^2

√

r^2 −h^2

2 rdr=

16

3 π

( 1 −h^2 )^3 /^2.

Let us now return to the computation ofB+ 2 A. Denote byA(h)the smaller of the
two areas of the disk cut off by a chord at distanceh. The chance that the third point is
in the smaller (respectively, larger) portion isA(h)π (respectively, 1−A(h)π ), and then the
area we are trying to compute isπ−A(h)(respectively,A(h)). Using the distribution
onh, and the fact that

A(h)= 2

∫ 1

h

√

1 −h^2 dh=

π

2

−arcsin(h)−h

√

1 −h^2 ,

we obtain

B+ 2 A=

2

π

∫ 1

0

A(h)(π−A(h))

16

3 π

( 1 −h^2 )^3 /^2 dh=

35 + 24 π^2

72 π

.

Using the fact that 4A+ 3 B=π, we obtainA= 4835 πas in the first solution.

Remark.This is a particular case of the Sylvester four-point problem, which asks for
the probability that four points taken at random inside a convex domainDform a non-
convex quadrilateral. Nowadays the standard method for computing this probability
uses Crofton’s theorem on mean values. We have seen above that whenDis a disk the
probability is 1235 π 2. WhenDis a triangle, square, regular hexagon, or regular octagon,

the probability is, respectively,^13 ,^1136 ,^289972 , and^1181 +^867

√
2
4032 + 2880 √ 2 (cf. H. Solomon,Geometric
Probability, SIAM, 1978).
(first solution by D. Kane, second solution by D. Savitt)