# Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 11"

Line 30: | Line 30: | ||

== 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 | + | 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 = \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>. |

## Revision as of 14:21, 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 . . . . 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>