Re: probablility of a forth point being contained within a triangle

Fun with Triangles

Four points are placed at random on a piece of paper, connect the 3 points which create the triangle of the largest area. What is the probability that the 4th point is in the triangle?

Hello Tracy,

This does look tricky; in fact, my first inclination was to figure on multiple integrals of complex coordinates to solve the problem, but then I thought, "what do Descartes and Leibnitz and the others have over Euclid, anyway?" Since this is a highschool problem, why use college level mathematics? So I threw out the coordinates and the calculus and went with simple geometry.

The first step to solving this is to think about the area of a triangle. A triangle's area is half of its base times its height - I don't like numbers, so let's just say that the area is a function of the base and the height. But that means that for constant base and height, the area of a triangle is constant, no matter what it looks like. In other words, for a given base, we can define a line parallel to the base, which describes all points that give triangles with the base with the same areas.

"So," you're thinking, "what does this have to do with the problem?" Well, since the area is based on the height, a triangle with a higher height would have a larger area and a triangle with a shorter height would have a smaller area; or said another way, for a point to describe a triangle with the base of a smaller area, then it must lie below the height line. So, if we define the segment connecting two of your random points as the base, and find the point that gives the largest triangle, then the fourth point must lie within the height line, or it defines a larger triangle.

Now, we have a boundary for the fourth point - it must lie within a certain area, or else it would define a larger triangle. Since the fourth point is defined as not part of the largest triangle, it must be a point that would make a smaller triangle; i.e. it must have a smaller height from the base than the third point. But, which base? Triangles have three sides, so they have three potential bases, each with its own corresponding height. Since each base (side) has its own corresponding height, each one has its own parallel height line that defines a boundary for the fourth point:

There you have it: the fourth point has to be within the larger, dashed triangle, in order for the other three points to create the triangle with the largest area. Although it is obvious from looking at the figure above, you can actually prove through Euclidean geometry (opposite angles of intersected parallel lines, etc.) that each of the three outside triangles - two dashed sides and one solid side - are congruent to the center triangle. Since congruent triangles have equal areas, and since there are four total congruent triangles, the inside triangle accounts for 1/4 of the area within the boundaries; or, the probability that the fourth point is in the triangle is one in four (25%). Not so tricky after all.

Well, maybe I spoke too soon. You may notice, as I did, that sometimes your boundary triangle doesn't all fit on the page. Obviously, this reduces the outside area, increasing the probability of the fourth point lying inside. At the extreme, if the first three points are three of the corners of the paper, then half of the paper is inside and half is outside, such that the probability would be 1/2 (50%). Clearly then, the total probability for a finite sheet of paper will be higher than 25%, since a good portion of the triangles will have boundaries that extend beyond the edge of the paper. Since an accurate determination of this absolute probability would require a good deal of calculus, I have to think that your teacher was thinking in the "infinite paper" realm, in which case the original answer would hold.

Thanks for making me think,

Mike