Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 11"
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== Solution 2 == | == Solution 2 == | ||
− | Since <math>x^2 + y^2 \ge 0</math>,finding the minimum value of <math>(x^2 + y^2)^2</math> is similar to finding that of <math>x^2 + y^2</math>. Let <math>x^2 + y^2 = a</math>, where <math>a</math> is the minimum value. We can rewrite this as <math>y^2 = -x^2 + a</math> and <math>y = \sqrt{-x^2 + a}</math>. | + | Since <math>x^2 + y^2 \ge 0</math>, finding the minimum value of <math>(x^2 + y^2)^2</math> is similar to finding that of <math>x^2 + y^2</math>. Let <math>x^2 + y^2 = a</math>, where <math>a</math> is the minimum value. We can rewrite this as <math>y^2 = -x^2 + a</math> and <math>y = \sqrt{-x^2 + a}</math>. |
<cmath>2xy + 2x^2 = x^2 + y^2 + 6</cmath><cmath>2x(\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6</cmath><cmath>2x(\sqrt{-x^2 + a}) + 2x^2 = a + 6</cmath><cmath>2x^2 - (a + 6) = -2x(\sqrt{x^2 + a})</cmath>. | <cmath>2xy + 2x^2 = x^2 + y^2 + 6</cmath><cmath>2x(\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6</cmath><cmath>2x(\sqrt{-x^2 + a}) + 2x^2 = a + 6</cmath><cmath>2x^2 - (a + 6) = -2x(\sqrt{x^2 + a})</cmath>. | ||
<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>. | ||
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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>. | ||
Since <math>a^2 = (x^2 + y^2)^2</math>, the minimum value of <math>(x^2 + y^2)^2 = \boxed{18}</math>. | Since <math>a^2 = (x^2 + y^2)^2</math>, the minimum value of <math>(x^2 + y^2)^2 = \boxed{18}</math>. | ||
+ | |||
+ | <baker77> |
Revision as of 15:16, 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
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
.
<baker77>