Random triangles in a bounded measurable set

An application of Kendall’s “spherical blackboard”

These notes were written nearly 30 years ago. They were not meant for publication, but are put on this website because David Kendall mentions them in [Ref. 2]. They provide an alternative derivation of Kendall’s formula for the shape density of a random triangle formed by 3 points with independent uniform distributions inside a circle. MB, May 2013

Let K be a bounded measurable set in R2 with |K|>0. Let three distinguishable points A,B,C have independent uniform distributions in K. We describe a method for finding the shape distribution of triangle ABC, where following Kendall [Ref. 1] the shape is identified with a point σ on S2(½) (surface of the 3-dimensional sphere of radius ½). We use spherical coordinates θ (co-latitude) and φ (longitude) on S2(½) identical with those on page 97 of [Ref. 1]. The total measure of S2(½) is taken to be 1, so that

dσ 1 sinθ dθ dφ .

Suppose first that A,B,C have independent identical circular gaussian distributions in R2, say

prob(x,y)=λ exp(−λπ(x2+y2)) dxdy,

where the origin 0 has any convenient location w.r.t. K. Since K is bounded, the induced distribution over K will tend uniformly to the uniform distribution as λ0. The required shape distribution over K can therefore be obtained as the limit of the shape distribution induced by (2). The positions of A,B,C are determined by the six coordinates xA,yA, etc, or alternatively by the following:

(a) the shape σ of triangle ABC;
(b) the coordinates (xG,yG) of the centroid G;
(c) the size g defined by

g=(|GA|2+|GB|2+|GC|2) ;

(d) the orientation ω measured w.r.t. any convenient origin.

Fig. 1 Notation for defining the position of a triangle in the plane

We use the notation of Fig. 1. As shown in [Ref. 1], or on this website, the position of the shape ABC on S2(½) (the “spherical blackboard”) is given by

|GC|:|AB|=sin(θ/2):3cos(θ/2) .

We next find the joint p.d.f. of σ,G,g,ω. It follows from (3) and (4) that

xA=xG  1 gsin(θ/2) cos(φ+ω)    1 gcos(θ/2) co
√6 √2
yA=yG  1 gsin(θ/2) sin(φ+ω)    1 gcos(θ/2) si
√6 √2
xB=xG  1 gsin(θ/2) cos(φ+ω)  +  1 gcos(θ/2) co
√6 √2
yB=yG  1 gsin(θ/2) sin(φ+ω)  +  1 gcos(θ/2) si
√6 √2
xC=xG  2 gsin(θ/2) cos(φ+ω)
yC=yG  2 gsin(θ/2) sin(φ+ω)

and from (3) that


The Jacobian of (5) turns out to have absolute value

| (xA,yA,xB,yB,xC,yC) |  =  3 g3sinθ ,
(xG,yG,g,θ,φ,ω) 4

so from (1) and (6) the required p.d.f. is

prob,G,g,ω)=3πλ3g3exp(−πλ(3xG2+3yG2+g2)) dσ dGdgdω.

Here the function on the RHS is independent of ω and σ. This shows that the orientation ω is uniformly distributed, as expected, and also that the circular gaussian induces a uniform distribution of the shape σ on S2(½), as already shown by Kendall in [Ref. 1].

Now consider some fixed shape, size and orientation σ,g,ω of ABC. Conditionally on these, the centroid G ranges over R2 with the circular gaussian distribution implied by (8). We need the probability that all three points A,B,C then fall inside K. Define points A0,B0,C0 such that OA0 = AG, etc. Then A0B0C0 is a fixed triangle congruent to ABC and rotated through π w.r.t. ABC (Fig. 2).

Fig. 2 Three positions of a shape, and their intersection

Let KA,KB,KC be the sets obtained by translating K through the vectors OA0, OB0, OC0, and define

J=J (σ,g,ω)=KAKBKC.

Then clearly A,B,C, all fall in K if and only if G falls in J. Since K is measurable, so is J. Now keep σ fixed and allow g and ω to vary. Then from (8) the probability that A,B,C all fall in K is

dσ 3πλ3g3exp(−πλg2)(1+O(λ)) |J (σ,g,ω)| dgdω.
g= ω=

Since K is bounded there is some g0>0, independent of σ and ω, such that |J (σ,g,ω)|=0 whenever g>g0 (because ABC is then too large to fit inside K in any position). Hence the exponential in (10) can be absorbed into the term 1+O(λ), and that term can then be taken outside the integral since O(λ)0 uniformly over all ω and gg0. The overall probability that A,B,C all fall in K is

λ3 |K|3 (1+O(λ)).

Dividing (10) by (11) and letting λ0 we get the desired p.d.f.

prob(σ)=3π|K|3dσ g3 |J (σ,g,ω)| dgdω.
g= ω=

Recall that in Fig. 2 the size of A0B0C0 is the variable g while KA,KB,KC are congruent to K. It may be more convenient to apply the dilatation ƒ=g1 to Fig. 2 so that A0B0C0 has size 1 while KA,KB,KC undergo equal dilatations ƒ about A0,B0,C0. Define

I (σ,ƒ,ω)=ƒKAƒKBƒKC.


|J (σ,g,ω)|=ƒ2|I (σ,ƒ,ω)|

and (12) can be rewritten as

prob(σ)=3π|K|3dσ ƒ7 |I (σ,ƒ,ω)| dƒ dω,
ƒ= ω=

where the integrand vanishes for all ƒ<g01.

If K is star-shaped a further transformation is possible. Take O to be one of the points from which the whole boundary of K is visible. For any PR2 define

ƒ*=ƒ*(P,ω)=inf{ƒ|PI (σ,ƒ,ω)} .

Then PI (σ,ƒ,ω) for all ƒ>ƒ* and the contribution to (15) for a given P and ω is

dPdω  ƒ7dƒ=  1 ƒ*6dPdω.
ƒ* 6

Hence (15) can be rewritten for star-shaped K as

prob(σ)=  π |K|3dσ   ƒ*(P,ω)6dPdω .
2 PR2 ω=

If one were interested in some statistic other than shape, e.g. the minimum width of triangle ABC as a criterion for approximate alinement, one could perhaps apply a similar method with integration over a suitable combination of σ,g,ω. This has not yet been tried.


Since triangle A0B0C0 has the same shape as ABC we can and will drop the suffix 0 from now on. As an external check on the method we calculate the shape distribution when K is a circular disk; this has already been done by Kendall [Ref. 2]. Let K have radius 1 and centre at O. The intersection J, when non-empty, is bounded by three circular arcs, and the expression for | J | is complicated. It is easier to use formula (18), where now f*(P,ω) is independent of ω and equals the distance from P to the furthest vertex of ABC.

Fig. 3 Infinite wedge over which integral is taken

Suppose first that A,B,C are not collinear. The set of P for which C (say) is the furthest vertex forms an infinite wedge W (Fig. 3, shaded) bounded by the perpendicular bisectors of CA and CB. The function f*(P,ω) in (18) is now |CP|. Fig. 3 shows that

  |CP|6dP  =  A sdsdψ .
W ψ=B s= (s2+2Rsco+R2)3

We find first that

sds  =  1 (3cosin2ψ)
s= (s2+2Rsco+R2)3 8R4sin4ψ

(where the RHS is O(1) for small ψ). Hence

  |CP|6dP  =  1 [t(A)csc4(A)+t(B)csc4(B)] ,
W 4R4

where R is the circumradius of ABC, given by

4R2(sin2A+sin2B+sin2C)=3 ,

and the function t is given by

t(A)=  3 A  1 sin2A+  1 sin4A.
8 4 32

Integration over ω contributes a factor , so that (18) yields, after summing over A,B,C,

prob(σ)=  8 (∑sin2A)2 ∑t(A)csc4Adσ .

This formula is the same as that given by equations (20) and (21) of [Ref. 2]. It is interesting to note that the function we have called t(A) is obtained by Kendall in a different way, namely as

t(A)=  A sin4ψ dψ .

When ABC degenerates into a straight line, say with C between A and B, then the contribution from C vanishes and (18) becomes the sum of two equal integrals over half-planes separated by the perpendicular bisector of AB. If following [Ref. 2] we write

|AC|:|CB|=q:p,p+q=1 ,

then we easily find

prob(σ) 1 (1+p2+q2)2dσ ,

again in agreement with [Ref. 2].

[1] D. G. Kendall, “Shape manifolds, procrustean metrics, and complex projective spaces”, Bull. London Math. Soc. 16, 81–121 (1984).
[2] D.G. Kendall, “Exact distributions for shapes of random triangles in convex sets”, Adv. Appl. Prob. 17, 308–329 (1985).