Tag Archives: Average Value

Most Triangles Are Obtuse!

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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 \Delta ABC. 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 P(A<90)=\tfrac{1}{2}

For any A < 90, B and C are chosen from the interval \left( 0,180-A \right). 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 [90-A,90].

This is an interval of length A and the probability of picking numbers, B and C, at random in this interval is \frac{A}{180-A}. 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 \frac{30}{180-30}=0.20.

In general P\left( A<90\text{ and }\left( B<90\text{ and }C<90 \right) \right)=\frac{1}{2}\cdot \frac{A}{180-A}. What we need is the average of (all) these values.

This average is \displaystyle \frac{1}{2}\frac{1}{90-0}\int_{0}^{90}{\frac{A}{180-A}dA}\approx 0.19315

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!

Average Value of a Function

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Today I want to consider a way of developing the expression for finding the average value of a function, f (x), on an interval [a, b].

Ask students how to find the average of a bunch of numbers and they will say, “add them up and divide by the number of numbers.” Then ask if they can average an infinite number of numbers. Most will say no since you cannot divide by infinity.

Well, what if the numbers were all the same?

Such as all the y-values of f (x) = 2 between x =1 and x = 5. Isn’t the average 2? So, apparently, you can sometimes average an infinite number of numbers.

Next suggest something like f (x) = x between x = 0 and x = 3. Since half the values are obviously above 1.5 and half below, can’t 1.5 be their average? Sketch this situation and draw the segment at y = 1.5 to help them see this.

Suggest another situation, say f (x) = 2 + sin(x) on the interval \left[ 0,2\pi \right] and some others. The average appears to be 2, again since half the values are above and half below 2.

When you draw the line that is the apparent average, lead the students to see that the rectangle formed by this line, the x-axis and the ends of the interval has the same area as between the function and the x-axis.

Continue with more difficult examples until someone hits on the idea of finding the “area” with an integral and then dividing the result by the width of the interval to find the height of the rectangle that is also the average value:

\displaystyle \overline{y}=\frac{\text{''area''}}{\text{length of interval}}=\frac{\int_{a}^{b}{f\left( x \right)dx}}{b-a}

Click here for an activity that you can use to develop this idea from scratch.

The next post: A fun application of average value – Most Triangles Are Obtuse.