The following problem appeared on the American High School Mathematics Examination (now called the AMC 12) in 1988:
If , what is ?
When I presented this problem to a group of students, I was pleasantly surprised by the amount of creativity shown when solving this problem.
Yesterday, I presented a solution using triangles. Here’s a second solution that I received: begin by squaring both sides and using a Pythagorean trig identity.
We use the Pythagorean identity again to find :
Therefore, we know that
so the answer is either or . However, this was a multiple-choice contest problem and was not listed as a possible answer, and so the answer must be .
For a contest problem, the above logic makes perfect sense. However, the last step definitely plays to the fact that this was a multiple-choice problem, and the concluding step would not have been possible had been given as an option.