2006 iTest Problems/Problem 37

Solution

To begin, let's number the equations as follows

$\begin{align} x^2 + xy + y^2 &= 1 \\ y^2 + yz +z^2 &= 2 \\ z^2 + zx +x^2 &= 3 \end{align}$

In order to solve for $y^2$ , we begin by subtracting the first equation from the second equation:

$\begin{align*} 1&=y^2 + yz +z^2 -(x^2 + xy + y^2)\\ &= z^2 + yz - xy - x^2 \\ &=(z-x)(x + y + z) \\ \end{align*}$

Similarily, subtracting the second eqaution from the third equation:

$\begin{align*} 1&= x^2 + xz +z^2 -(y^2 + zy + z^2)\\ &= x^2 + xz - zy - y^2 \\ &=(x-y)(x + y + z) \\ \end{align*}$

Subtracting the factored equations we see that

$\begin{align*} 0&=(x-y)(x + y + z)-(z-x)(x + y + z)\\ &=(x+y+z)((x-y)-(z-x))\\ &=(x+y+z)(2x-y-z)\\ \end{align*}$

Therefore, we can conclude that either $x+y+z=0 \Rightarrow x=-z-y$ , or that $2x-y-z=0 \Rightarrow x=\frac{z+y}{2}$ . As we are told that $y^2$ has a unique value, we can use either option.

Continuing with the second option, we can plug $x=\frac{z+y}{2}$ into the first equation:

$\begin{align*} 1&=x^2+xy+y^2\\ &=\left(\frac{z+y}{2}\right)^2 + \left(\frac{z+y}{2}\right)\cdot y +y^2 \\ &=\frac{z^2+2yz+y^2}{4} + \frac{yz+y^2}{2} + y^2 \end{align*}$

Multiplying each side of the resulting equation by four:

$\begin{align*} 4&=4 \cdot \left(\frac{z^2+2yz+y^2}{4} + \frac{yz+y^2}{2} + y^2 \right)\\ &=(z^2+2yz+y^2)+(2yz+2y^2)+4y^2\\ &=z^2+4yz+7y^2 \end{align*}$

We can then subtract twice our original second equation, $y^2+yz+z^2=2$ , from the equation from the previous step, yielding:

$\begin{align*} 4-2\cdot2&=(z^2+4yz+7y^2)-2(y^2+yz+z^2)\\ 0&=5y^2+2yz-z^2\\ \end{align*}$

Rearranging the new equation, we see that:

$\begin{align*} 5y^2 &=z^2-2yz\\ 6y^2 &= z^2-2yz+y^2\\ 6y^2 &=(z-y)^2\\ \sqrt{6y^2}&=\sqrt{(z-y)^2}\\ y\sqrt{6}&=z-y\\ \end{align*}$

We can ignore the negative root of the RHS because both y and z are positive, and because $z>y$ which is a result of (3) being greater than (1). Continuing to manipulate the newest equation yields:

$\begin{align*} z-y &=y\sqrt{6}\\ z&=y\sqrt{6}+y\\ z&=y(\sqrt{6}+1) \end{align*}$

Plugging this back into (2)

$\begin{align*} y^2 + yz +z^2 =& \quad 2\\ y^2+y \cdot y(\sqrt{6}+1) + (y(\sqrt{6}+1))^2 =&\\ y^2\cdot(1+(\sqrt{6}+1)+(\sqrt{6}+1)^2)=&\\ y^2\cdot(1+(\sqrt{6}+1)+(6+2\sqrt{6}+1))=&\\ y^2\cdot(9+3\sqrt{6})=& \end{align*}$

Solving for $y^2$ yields that $y^2= \frac{2}{9+3\sqrt{6}} \Rightarrow y^2=\frac{2}{9+3\sqrt{6}} \cdot \frac{9-3\sqrt{6}}{9-3\sqrt{6}} = \frac{18-6\sqrt{6}}{27} = \frac{6-2\sqrt{6}}{9}$

Therefore, the solution is $6+2+6+9 = \boxed{23}$

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2006 iTest Problems/Problem 37

Solution