Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 11"
Line 34: | Line 34: | ||
<cmath>4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)</cmath>. | <cmath>4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)</cmath>. | ||
<cmath>8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0</cmath>. | <cmath>8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0</cmath>. | ||
− | We want this polynomial to factor in the form <math>(x^2 - r)(x^2 - s)</math>, where at least one of <math>r, s \ge 0</math>. ( If <math>r, s < 0</math>, the equations <math>x^2 = r</math> and <math>x^2 = s</math> | + | We want this polynomial to factor in the form <math>(x^2 - r)(x^2 - s)</math>, where at least one of <math>r, s \ge 0</math>. ( If <math>r, s < 0</math>, the equations <math>x^2 = r</math> and <math>x^2 = s</math> have no real solutions). Since <math>a > 0</math>, both <math>-8(a + 3) > 0</math> and <math>(a + 6)^2 > 0</math>, so <math>r, s > 0</math>. |
We can now use the “discriminant” to determine acceptable values of <math>a</math>. <math>(8(a + 3))^2 - 4\cdot 8 \cdot (a + 6)^2 \ge 0</math> simplifies to <math>a^2 \ge 18</math>. | We can now use the “discriminant” to determine acceptable values of <math>a</math>. <math>(8(a + 3))^2 - 4\cdot 8 \cdot (a + 6)^2 \ge 0</math> simplifies to <math>a^2 \ge 18</math>. |
Revision as of 15:17, 8 November 2019
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
.
<baker77>