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 a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. Again, I begin by squaring both sides.

Yesterday, I used the Pythagorean identity again to find . Today, I’ll instead plug back into the original equation :

Unlike the example yesterday, the signs of and must agree. That is, if , then must also be positive. On the other hand, if , then must also be negative.

If they’re both positive, then

,

and if they’re both negative, then

.

Either way, the answer must be .

This is definitely superior to the solution provided in yesterday’s post, as there’s absolutely no doubt that the product must be positive.

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.
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One thought on “Different ways of solving a contest problem (Part 3)”

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