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 . Therefore, the minimum value of .
Solution 3 (Lagrange Multipliers)
Since , will be minimized when is at its minimum. We construct the Lagrangian by taking our value to optimize and subtracting off times our constraint, which we've set equal to zero. Now we seek points where its gradient is zero to get the three equations: All that is left to do is trudge through the algebra and solve for . Adding and subtracting the first two equations, dividing by two, and setting for convenience yields: We will use the last three equations to solve for , then square the result for our answer. These equations can be solved manually with some algebra, but there is a nicer way! Notice that this is actually the eigenvalue equation: The matrix is times the reflection matrix across the line, which means that and our should be on that line as (or on the perpendicular line with , but the answers will be the same regardless). Plugging this into our other condition, we have: . Since , our answer is thus . QED. -eiis1000