Mock AIME 3 2006-2007 Problems/Problem 11

Problem

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

Solution 1

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 $\frac{36}{2}=\boxed{018}$

Solution 2

Since $x^2 + y^2 \ge 0$ , finding the minimum value of $(x^2 + y^2)^2$ is similar to finding that of $x^2 + y^2$ . Let $x^2 + y^2 = a$ , where $a$ is the minimum value. We can rewrite this as $y = -x^2 + a$ and $y = \sqrt{-x^2 + a}$ . $2xy + 2x^2 = x^2 + y^2 + 6$ $2x(\sqrt{-x^2 + a}) + 2x^2 = x^2 + (-x^2 + a) + 6$ $2x(\sqrt{-x^2 + a}) + 2x^2 = a + 6$ $2x^2 - (a + 6) = -2x(\sqrt{x^2 + a})$ . $4x^2 - 4(a + 6)x^2 + (a + 6)^2 = 4x^2(-x^2 + a)$ . $8x^4 - 8(a + 3)x^2 + (a + 6)^2 = 0$ . We want this polynomial to factor in the form $(x^2 - r)(x^2 - s)$ , where at least one of $r, s \ge 0$ . ( If $r, s < 0$ , the equations $x^2 = r$ and $x^2 = s$ have no real solutions). Since $a > 0$ , both $-8(a + 3) > 0$ and $(a + 6)^2 > 0$ , so $r, s > 0$ .

We can now use the “discriminant” to determine acceptable values of $a$ . $(8(a + 3))^2 - 4\cdot 8 \cdot (a + 6)^2 \ge 0$ simplifies to $a^2 \ge 18$ . Therefore, the minimum value of $(x^2 + y^2)^2 = a^2 = \boxed{18}$ .

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Mock AIME 3 2006-2007 Problems/Problem 11

Problem

Solution 1

Solution 2