By Murray Bourne, 29 Aug 2007
I came across this interesting story at North Carolina Association of Advanced Placement Mathematics Teachers. The problem statement:
If a triangle is chosen at random, what is the probability that it is acute?
The great thing about this story (from Stu Schwartz, of Wissahickon High) is the grade 10 boy:
Kurt is a student of mine, the smartest math student I have ever encountered in 33 years of teaching. He is in 10th grade, having completed Calculus AB in 8th grade, BC in 9th. He is now taking multivariable calculus online given by Stanford. This morning he came into my stat class. I casually said to him, "Hey Kurt, what is the probability that a triangle chosen at random is acute?"
He appeared disinterested.
So when he stopped into my room for a study hall 2 hours later, I asked him if he had a solution. He said that he didn't work long on it but he believed the answer was 19.3%.
I was stunned because when I simulated the problem using Fathom (a terrific statistical program put out by Key Curriculum Press), I suspected the answer was 20%. How in the world did this kid do the problem?
First---my definition of a random triangle:
A = a random number between 0 and 180;
B = a random number between 0 and 180-A;
C = 180 - A - B.
Here is Kurt's logic:
First, using the definition above, the chance of choosing angle A to be obtuse is 0.5. So the chance of all three angles being acute must be less than 0.5.
If A=30, then the other two angles must add to 150. That means that one of the other two angles must be between 60 and 90 in order for the triangle to be acute (neglecting right triangles whose probability is zero). So the probability of the triangle being acute is (90-60)/150 = 30/150.
If A=70, the other two angles must add to 110. That means that one of the other two angles must be between 20 and 90 in order for the triangle to be acute. So the probability of the triangle being acute is (90-20)/150 = 70/150.
Suppose that A is chosen and is acute. That will leave (180-A)-90 and 90. 90-[(180-A)-90] = A. So the probability of the triangle with acute angle A being acute is A/(180-A).
Obviously there are infinite possibilities of angle A giving infinite probabilities of the angle being acute, each infinitely small.
So Kurt decides to sum these probabilities up and takes the average probability. So he integrates the expression x/(180-x) from 0 to 90 and divides by 90. This, of course, is the average value of a function. He ends up with 0.386.
He then multiplies by 0.5 because he must multiply the result by the probability that A was acute to begin with. He gets 0.193.
I have simulated this in Fathom 10,000 times and my results are just about the same. A terrific use of probability theory and calculus. This from a 10th grader.
Still has me in shock.
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