James Tanton (of G’day Math fame) proposes an elementary problem in probability in his course Geometry: An Interactive Journey to Mastery: What is the probability that a randomly chosen chord of a circle will be longer than the side of an inscribed equilateral triangle? A chord is a straight line segment joining two distinct points of the circle, and the three vertices of an inscribed triangle all lie on the circle. In this essay, I take a close look at this problem
The problem statement begins with a circle and its chords. The required probability is the fraction of the set of all chords of the circle which are longer than the side of an inscribed equilateral triangle. To compute such a fraction requires a measure of the size of a set of chords. One solution for this requirement is to establish a correspondence between sets of chords and objects that have computable sizes. Then the size of a set of chords is the size of its corresponding object.
A more precise formulation of the problem is:
A = the set of all chords of the given circle, known as the population; and
B = the subset of A comprising those chords longer than the side of an inscribed equilateral triangle.
The desired probability is the size of B divided by the size of A.
Method 1
In Method 1, a set of chords corresponds to an arc of the given circle. The size of a set of chords is the length of its corresponding arc.
Choose a point P on the circle.
Every chord with end point P corresponds to the point of the circle at its other end.
Hence, the set A corresponds to the entire circle.
The size of A is the circle’s circumference.
Inscribe an equilateral triangle in the circle with one vertex at P.
The three vertices of the triangle divide the circle into three arcs of equal length.
Chords that join P to a point in the arc opposite P are longer than a side of the triangle.
Thus the set B corresponds to the arc opposite P.
The length of this arc is one-third the length of the entire circle.
Consequently, the size of B is one-third the size of A.
The probability that a random chord is longer than the side of an inscribed equilateral triangle is ⅓.
Method 2
For method 2, a set of chords corresponds with a segment of a diameter of the circle. The size of a set of chords is the length of its corresponding line segment.
Choose a diameter d of the circle.
Designate by e the diameter perpendicular to d.
Every chord parallel to d corresponds to the unique point where it intersects e.
Hence, the set A corresponds to the entire diameter e.
The size of A is the length of e.
Inscribe an equilateral triangle in the circle with one side parallel to d.
Inscribe another equilateral triangle with one side parallel to d, but oriented 60º with respect to the first one. One of its vertices will fall at an end point of the diameter e.
Chords parallel to d and lying between the two triangle sides parallel to d are longer than a triangle side.
Thus the set B corresponds to the segment of e bounded by the triangle sides parallel to d.
This segment is one-half the length of the diameter e. [See box.]
Consequently, the size of B is one-half the size of A.
The probability that a random chord is longer than the side of an inscribed equilateral triangle is ½.
============================================================== Geometry of an Inscribed equilateral triangle A line between a vertex and the midpoint of the opposite side divides the triangle into two equal right triangles. In both right triangles, the length of the hypotenuse is twice the length of the short side. These statements are easy consequences of the equal angles and equal sides of the inscribed triangle.
The small right triangle bounded by sides a, b, and c is similar to each of the large ones, so its hypotenuse has length twice its short side. The hypotenuse c is a radius of the circle, and a represents the distance from the center of the circle to a side of the triangle. Doubling the lengths of a and c justifies statement 10. ==============================================================
Oh My!
What is going on here? The answer to the original question cannot be both ⅓ and ½. I think neither method is correct, but maybe I can identify what went wrong.
A probability is a ratio of the size of the set of items satisfying a condition to the size of the population of all such items. Many probability calculations go wrong by not carefully specifying this population. So take a closer look at the populations implicit in the two methods.
In Method 1, a set of chords corresponds to an arc of the circle. The size of a set of chords is then the length of its corresponding arc. Steps 1 through 3 specify this correspondence. In step 3, the population becomes all chords having an endpoint at the chosen point P. But this is not the original set of all chords. Method 1 calculates the probability that a chord anchored at a given point on a given circle is longer than the side of an inscribed equilateral triangle.
In Method 2, a set of chords corresponds to a segment of a diameter. Steps 1 through 4 specify this correspondence. In step 4, the population becomes all chords parallel to the chosen diameter d. But this is not the original set of all chords. Method 2 calculates the probability that a chord parallel to a given diameter of a given circle is longer than the side of an inscribed equilateral triangle.
Both methods replace the population of all chords with a smaller subset, with different choices yielding different results. The only difference between the methods is which part of the population of all chords they use, so the difference in results must be a consequence of some difference between:
a) considering chords anchored to a chosen point, and
b) considering chords required to be parallel to a chosen diameter.
How that difference maps into different results is another interesting problem which I will not address here.
Is there a correspondence of sets of chords to measurable objects that avoids the problems of these two methods? Is the problem even well-posed so that it is solvable? The original problem has become clearer by making plausible attempts that failed. The next attempt will find a correspondence to the set of all chords, not just a subset. Stay tuned.
