**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*.

Taylor Gibson of the Greenhill School in Addison, Texas wrote this Monte Carlo simulation of the situation described above. Thank you, Taylor.

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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!