Mock AIME 3 2006-2007 Problems/Problem 11
If and are real numbers such that find the minimum value of .
Factoring the LHS gives .
Now converting to polar:
Since we want to find ,
Since we want the minimum of this expression, we need to maximize the denominator. The maximum of the sine function is 1
(one value of which produces this maximum is )
So the desired minimum is
Since , finding the minimum value of is similar to finding that of . Let , where is the minimum value. We can rewrite this as and . . . . We want this polynomial to factor in the form , where at least one of . ( If , the equations and have no real solutions). Since , both and , so .
We can now use the “discriminant” to determine acceptable values of . simplifies to . Since , the minimum value of .