# Mock AIME 3 2006-2007 Problems/Problem 11

## Problem

If and are real numbers such that find the minimum value of .

## Solution 1

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

## Solution 2

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 .

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