What is the probability that a triangle picked at random will be acute? An average value problem.

The thing here is to define what you mean by picking a triangle “at random.” You could open a Geometry book and take the first triangle you come to, but are the triangles in a Geometry book really a good sample space? I doubt it.

Let’s try this: let A, B, and C be the measures, in degrees, of the angles of . Let A be a random number between 0 and 180, and let B be a random number between 0 and 180 – A. Then, let C = 180 – A – B.

Then

For any A < 90, B and C are chosen from the interval . In order for the triangle to be acute B and C must be within 90 of both ends of this interval. That is, B and C must both be in the interval .

This is an interval of length A and the probability of picking numbers, B and C, at random in this interval is . At this point you may want to stop and calculate a typical probability. For instance, if A= 30 then the probability of both B and C being acute is .

In general . What we need is the average of (all) these values.

This average is

So about 19.3% of triangles are acute, the rest are obtuse (or right). Leaving one to believe that most triangles are obtuse.

Challenge: Try writing a calculator or computer program that picks the measures of the angles of a triangle as described above. Repeat this many times while the program counts the number of acute triangles and the total number of triangles and finds their ratio. Does it come close to 19.3%?
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This problem was posted on the AP Calculus Electronic Discussion Group (11/22/03) by Stu Schwartz an AP Calculus teacher at Wissahickon High School in Ambler, Pennsylvania. The solution is by one of Mr. Schwartz’s students Kurt Schneider, a tenth grader at the time who completed AB calculus in eighth grade and BC calculus in ninth grade!