Before you begin this section, it is imperative to review the idea of the wedge product from multivariable calculus and to understand its application to geometric probability. To that end, please study our page on differential forms.

To study triple points, let us pick three points at random in the plane. Denote these three points by and let denote the center of the circumcircle about triangle P₁P₂P₃ . Also, let denote the polar coordinates of P_i with respect to Q. The coordinates of P_i are thus

			(1)

For sets of n-many independent points , the density is . Also, density is always considered to be a positive meaure.

Note that a triple of points is not the same thing as a labeled triangle, as the triangle P₁P₂P₃ is not identical to the triangle P₂P₁P₃ even if the two triangles have the same sets of vertices. Letting one can show by a very long calculation (please don't try it) that the density of triples of points in the plane is

		(2)

where

		(3)

Problem 9: (a) Show that the area of the triangle P1P2P3 above is equal to . Hint: use equation 12 from this link on triangle area.

Let T = denote the area of triangle P₁P₂P₃. So, - T = 0. If we differentiate this equation, we get:

Using the properties of exterior multiplication, it follows that

		(4)

Now, consider the set of all triples P₁, P₂, P₃ such that the circumdisk about the triangle P₁P₂P₃, denoted C in green, lies within the red circle of given radius, denoted . In other words, consider the set of triples of points such that when you throw them down randomly on the plane inside the circle , the circumcircle drawn through the three points is entirely contained in . You can move the vertices of the triangles around to see when the green circumcircle falls entirely in the red circle . Can you arrange the points so that the green circle does not lie entirely within the red circle yet all three points do? By doing this, one can see that this exercise differs from randomly choosing a triangle that lies within a circle.

Please enable Java for an interactive construction (with Cinderella).

Thus, the measure of this set of triples, letting r and be the polar coordinates of Q with respect the center of the circle , is given by equation (5).

From equation (2), we have

		(a)
				(5)
		(b)

In step (a), dQ has been converted to polar coordinates (r,).

In step (b), , where is the restricion on the position of Q.

If you also consider the restriction that the triangle P₁P₂P₃ be an acute angled triangle, then the measure of this set is

		(6)

Note that these are very complicated integrals; so complicated, in fact, that Maple would spend many hours evaluating each one. It should also be noted that the probability that a triple (from the set of triples where ) forms an acute triangle is 1/2. Then as gets larger and approaches infinity, approaches the entire plane. So then the probability that any random triangle on the plane is acure is 1/2.

Problem 9: (b) Choose random equilateral triangles P1P2P3 inside . Find the measure of such equilateral triangles given by . Solutions

Consider the animation below. It is called disk triangle picking and it involves picking three random points on a disk, and then forming a random triangle with these points. Thus, let the three random points be denoted . Then the average area of the triangles is given by the following integral:

		(7)

Once again, this integral is very complicated.

There are several other ways of picking randome triangles. We can pick triangles inside another triangle, inside a disk, or inside a square.

References: Santalo; Eric Weinnstein's Mathworld