Proportion/Intermediate

Problem

$x$ is directly proportional to the sum of the squares of $y$ and $z$ and inversely proportional to $y$ and the square of $z$. If $x = 8$ when $y = 1/2$ and $z = \sqrt {3}/2$, find $y$ when $x = 1$ and $z = 6$. (Thanks to Bicameral of the AoPS forum for this one)

Solution

First, let's write these out algebraically: the first is $x=k(y^2+z^2)$, the second is $xy=k_1$, and the third is $xz^2=k_2$. Plugging the values into the first equation gives $8=k((\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2)=k(\frac{1}{4}+\frac{3}{4})=k$, so $k=8$. The second gives $8*\frac{1}{2}=4=k_1$, and the third gives $8*(\frac{\sqrt{3}}{2})^2=8*3/4=6=k_2$. We can use any of these three to solve the problem. Using the second gives us $1*y=y=4$, so the answer is $y=4$.