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4. Random Triangles

Statement of the Problem

Suppose that a stick is randomly broken in two places. What is the probability that the three pieces form a triangle?

Mathematical Exercise 1. Without looking below, make a guess.

Simulation Exercise 2. Run the triangle experiment 50 times. Do not be concerned with all of the information displayed in the applet, but just note whether the pieces form a triangle. Would you like to revise you guess in Exercise 1?

Mathematical Formulation

As usual, the first step is to model the random experiment mathematically. We will take the length of the stick as our unit of length, so that we can identify the stick with the interval [0, 1]. To break the stick into three pieces, we just need to select two points in the interval. Thus, let X denote the first point chosen and Y the second point chosen. Note that X and Y are random variables and hence the sample space of our experiment is

S = [0, 1]2.

Now, to model the statement that the points are chosen at random, let us assume, as in the previous sections, that X and Y are independent and each is uniformly distributed on [0, 1].

Mathematical Exercise 3. Show that (X, Y) is uniformly distributed on S = [0, 1]2.

Hence, P(A) = area(A) / area(S) = area (A) for A S.

The Probability of a Triangle

Mathematical Exercise 4. Argue that the three pieces form a triangle if and only if the triangle inequalities hold: the sum of the lengths of any two pieces must be greater than the length of the third piece.

Mathematical Exercise 5. Show that the event that the pieces form a triangle is T = T1 union T2 where

  1. T1 = {(x, y) in S: y > 1/2, x < 1/2, y - x < 1/2}
  2. T2 = {(x, y) in S: x > 1/2, y < 1/2, x - y < 1/2}

A sketch of the event T is given below:

The event that the pieces form a triangle

Mathematical Exercise 6. Show that P(T) = 1/4.

How close did you come with your guess in Exercise 1? The relative low value of the probability in Exercise 6 is a bit surprising.

Simulation Exercise 7. Run the triangle experiment 1000 times, updating every 10 runs. Note the apparent convergence of the empirical probability of Tc to the true probability.

Triangles of Different Types

Now let us compute the probability that the pieces form a triangle of a given type. Recall that in an acute triangle all three angles are less than 90°, while an obtuse triangle has one angle (and only one) that is greater than 90°. A right triangle, of course, has one 90° angle.

Mathematical Exercise 8. Suppose that a triangle has side lengths a, b, and c, where c is the largest of these. Recall (or show) that the triangle is

  1. acute if and only if c2 < a2 + b2.
  2. obtuse if and only if c2 > a2 + b2.
  3. right if and only if c2 = a2 + b2.

Part (c), of course, is the famous Pythagorean theorem, named for the ancient Greek mathematician Pythagoras.

Mathematical Exercise 9. Show that the right triangle equations for the stick pieces are

  1. (y - x)2 = x2 + (1 - y)2 in T1.
  2. (1 - y)2 = x2 + (y - x)2 in T1.
  3. x2 = (y - x)2 + (1 - y)2 in T1.
  4. (x - y)2 = y2 + (1 - x)2 in T2.
  5. (1 - x)2 = y2 + (x - y)2 in T2
  6. y2 = (x - y)2 + (1 - x)2 in T2.

The events that the pieces form acute and obtuse triangles

Mathematical Exercise 10. Let R denote the event that the pieces form a right triangle. Show that

P(R) = 0.

Mathematical Exercise 11. Show that the event that the pieces form an acute triangle is A = A1 union A2 where

  1. A1 is the region inside curves (a), (b), and (c) of Exercise 7.
  2. A2 is the region inside curves (d), (e), and (f) of Exercise 7.

Mathematical Exercise 12. Show that the event that the pieces form an obtuse triangle is B = B1 union B2 union B3 unionB4 union B5 union B6 where

  1. B1, B2, B3 are the regions inside T1 and outside of curves (a), (b), and (c) of Exercise 7, respectively.
  2. B4, B5, B6 are the regions inside T2 and outside of curves (d), (e), and (f) of Exercise 7, respectively.

Mathematical Exercise 13. Show that

  1. P(B1) = integral[0, 1/2] [x(1 - 2x) / (2 - 2x)]dx = 3 / 8 - ln(2) / 2.
  2. P(B2) = integral[0, 1/2] [x(1 - 2x) / (2 - 2x)]dx = 3 / 8 - ln(2) / 2.
  3. P(B3) = integral[1/2, 1] [y + 1 / (2y) - 3 / 2]dy = 3 / 8 - ln(2) / 2.

Mathematical Exercise 14. Argue from symmetry that

P(B) = 9 / 4 - 3 ln(2) ~ 0.1706

You might also argue from symmetry that P(Bi) must be the same for each i, even though B1 and B2 (for example) are not congruent.

Mathematical Exercise 15. Show that

P(A) = 3 ln(2) - 2 ~ 0.07944.

Simulation Exercise 16. Run the triangle experiment 1000 times, updating every 10 runs. Note the apparent convergence of the empirical probabilities to the true probabilities.

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