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Difference between revisions of "Mock AIME 3 2006-2007 Problems/Problem 11"
(→‎Solution 3 (Lagrange Multipliers): Upgrade solution to use matrices for faster algebra)
 
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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 = -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>.
 
<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> would 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 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>.
−
Since <math>a^2 = (x^2 + y^2)^2</math>, the minimum value of <math>(x^2 + y^2)^2 = \boxed{18}</math>.
+
Therefore, the minimum value of <math>(x^2 + y^2)^2 = a^2 = \boxed{18}</math>.
 +
 
 +
<baker77>
 +
 
 +
== Solution 3 (Lagrange Multipliers) ==
 +
 
 +
Since <math>x^2+y^2\ge0</math>, <math>(x^2 + y^2)^2</math> will be minimized when <math>x^2 + y^2</math> is at its minimum.
 +
We construct the Lagrangian <math>\mathcal{L} = x^2 + y^2 - \lambda(x^2 - y^2 + 2xy - 6)</math> by taking our value to optimize and subtracting off <math>\lambda</math> times our constraint, which we've set equal to zero. Now we seek points where its gradient is zero to get the three equations:
 +
<cmath>2x - 2\lambda x - 2\lambda y = 0</cmath>
 +
<cmath>2y + 2\lambda y - 2\lambda x = 0</cmath>
 +
<cmath>x^2 - y^2 + 2xy - 6 = 0</cmath>
 +
All that is left to do is trudge through the algebra and solve for <math>x^2+y^2</math>. Adding and subtracting the first two equations, dividing by two, and setting <math>\mu = 2\lambda</math> for convenience yields:
 +
<cmath>x + y = \mu x</cmath>
 +
<cmath>x - y = \mu y</cmath>
 +
We will use the last three equations to solve for <math>x^2+y^2</math>, then square the result for our answer.
 +
These equations can be solved manually with some algebra, but there is a nicer way! Notice that this is actually the eigenvalue equation:
 +
<cmath>\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mu \begin{bmatrix} x \\ y \end{bmatrix} </cmath>
 +
The matrix is <math>\sqrt 2 </math> times the reflection matrix across the <math>22.5^\circ</math> line, which means that <math>\mu = \sqrt2</math> and our <math>(x,y)</math> should be on that line as <math>(r\cos22.5, r\sin22.5)</math> (or on the perpendicular line with <math>\mu = -\sqrt2</math>, but the answers will be the same regardless). Plugging this into our other condition, we have: <cmath>r^2\cos^2 22.5 - r^2\sin^2 22.5 + 2r^2\cos22.5\sin22.5 - 6 = 0</cmath> <cmath>r^2 \cos 45 + r^2\sin 45 = 6</cmath> <cmath>r^2 = 3 \sqrt 2</cmath>. Since <math>(x^2+y^2)^2 = r^4</math>, our answer is thus <math>(3\sqrt2)^2 =\boxed{18}</math>.
 +
QED. -eiis1000

Latest revision as of 00:53, 30 April 2022

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 $(x^2 + y^2)^2 = a^2 = \boxed{18}$.

<baker77>

Solution 3 (Lagrange Multipliers)

Since $x^2+y^2\ge0$, $(x^2 + y^2)^2$ will be minimized when $x^2 + y^2$ is at its minimum. We construct the Lagrangian $\mathcal{L} = x^2 + y^2 - \lambda(x^2 - y^2 + 2xy - 6)$ by taking our value to optimize and subtracting off $\lambda$ times our constraint, which we've set equal to zero. Now we seek points where its gradient is zero to get the three equations: \[2x - 2\lambda x - 2\lambda y = 0\] \[2y + 2\lambda y - 2\lambda x = 0\] \[x^2 - y^2 + 2xy - 6 = 0\] All that is left to do is trudge through the algebra and solve for $x^2+y^2$. Adding and subtracting the first two equations, dividing by two, and setting $\mu = 2\lambda$ for convenience yields: \[x + y = \mu x\] \[x - y = \mu y\] We will use the last three equations to solve for $x^2+y^2$, then square the result for our answer. These equations can be solved manually with some algebra, but there is a nicer way! Notice that this is actually the eigenvalue equation: \[\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \mu \begin{bmatrix} x \\ y \end{bmatrix}\] The matrix is $\sqrt 2$ times the reflection matrix across the $22.5^\circ$ line, which means that $\mu = \sqrt2$ and our $(x,y)$ should be on that line as $(r\cos22.5, r\sin22.5)$ (or on the perpendicular line with $\mu = -\sqrt2$, but the answers will be the same regardless). Plugging this into our other condition, we have: \[r^2\cos^2 22.5 - r^2\sin^2 22.5 + 2r^2\cos22.5\sin22.5 - 6 = 0\] \[r^2 \cos 45 + r^2\sin 45 = 6\] \[r^2 = 3 \sqrt 2\]. Since $(x^2+y^2)^2 = r^4$, our answer is thus $(3\sqrt2)^2 =\boxed{18}$. QED. -eiis1000

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