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

Revision as of 19:30, 15 February 2015

If $x$ and $y$ are real numbers such that $2xy+2x^2=6+x^2+y^2$ find the minimum value of $(x^2+y^2)^2$

Factoring the LHS gives $2x(x+y)=6+x^2+y^2$ .

Now converting to polar: $2r\cos(a)(r\cos(a)+r\sin(a))=6+r^2$

$2\cos(a)(\cos(a)+\sin(a))=\frac{6}{r^2}+1$

$2\cos^2(a)-1+2\cos(a)\sin(a)=\frac{6}{r^2}$

$\cos(2a)+\sin(2a)=\frac{6}{r^2}$

$r^2=\frac{6}{\cos(2a)+\sin(2a)}$

Since we want to find $(x^2+y^2)^2=r^4$ , $r^4=\frac{36}{(\cos(2a)+\sin(2a))^2}=\frac{36}{1+2\cos(2a)\sin(2a)}=\frac{36}{1+\sin(4a)}$

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 $a$ which produces this maximum is $a=\frac{\pi}{4}$ )

So the desired minimum is $\frac{36}{2}=\boxed{018}$

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+== Problem ==
+If <math>x</math> and <math>y</math> are real numbers such that <math>2xy+2x^2=6+x^2+y^2</math> find the minimum value of <math>(x^2+y^2)^2</math>
+----
+== Solution ==
+----
 Factoring the LHS gives <math>2x(x+y)=6+x^2+y^2</math>.