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
would 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
.