Advanced book on Mathematics Olympiad

226 4 Geometry and Trigonometry

Crofton’s theorem.LetDbe a bounded convex domain in the plane. Through each
pointP(x, y)outsideDthere pass two tangents toD. Lett 1 andt 2 be the lengths of the
segments determined byPand the tangency points, and letαbe the angle between the
tangents, all viewed as functions of(x, y).^1 Then
∫∫

P/∈D

sinα

t 1 t 2

dxdy= 2 π^2.

Proof.The proof becomes transparent once we examine the particular case in whichDis
the unit diskx^2 +y^2 <1. Each point outside the unit disk can be parametrized by the pair
of angles(φ 1 ,φ 2 )where the tangents meet the unit circleS^1. Since there is an ambiguity
in which tangent is considered first, the outside of the disk is in 1-to-2 correspondence
with the setS^1 ×S^1. It so happens, and we will prove it in general, that on changing
coordinates from(x, y)to(φ 1 ,φ 2 ), the integral from the statement becomes

∫∫

dφ 1 dφ 2
(divided by 2 to take the ambiguity into account). The result follows.
In the general case we mimic the same argument, boosting your intuition with Fig-
ure 31. Fix a Cartesian coordinate system with the originOinsideD. For a point(x, y)
denote by(φ 1 ,φ 2 )the angles formed by the perpendiculars fromOonto the tangents
with the positive semiaxis. This is another parametrization of the exterior ofD, again
with the ambiguity of which tangent is considered first. LetAi(i,ηi),i= 1 ,2, be the
tangency points.

O

A

P

O

A

θ

θ

1

1

1 θ +π/2^1

2

2

Figure 31

The main goal is to understand the change of coordinates(x, y)→(φ 1 ,φ 2 )and in
particular to write the Jacobian of this transformation. Writing the condition that the
slope of the lineA 1 Pis tan(φ 1 +π 2 ), we obtain

(x− 1 )cosφ 1 +(y−η 1 )sinφ 1 = 0.

Taking the differential yields

(^1) If the boundary ofDhas some edges, then there are pointsPfor whicht 1 andt 2 are not well defined,
but the area of the set of these points is zero, so they can be neglected in the integral below.