solve equations - find a short piece of a wire 
Problem: A piece of wire 20 feet long is cut into two pieces so that the sum of the squares of the lengths of the two pieces is 202 square feet. What is the length, in feet, of the shorter piece of wire?

This problem looks pretty simple, but while I am working on it, I got confused. Here is what I did：I assumed the lengths of the two pieces are $x$ and $y$ and got $$x+y=20 ----(1)$$ and $$x^2+y^2=202 ----(2)$$ Then I am confused: Since $x$ and $y$ are symmetrical in both equations (1) and (2), the value of $x$ should equal to that of $y$, right? I don't see how one is shorter than the other.
 A: If we look at this problem geometrically, we are trying to find the intersection between a line and a circle. In this particular scenario, there are two intersection points. if you reflect one intersection point over the line $y=x$, you get the other intersection point, and vice versa. Another way to think about this is to see that, if $(x,y)$ is a solution, then $(y,x)$ is also a solution.

Now, back to the problem. Let $a=x+y$ and $b=xy$. Then $a=20$ and $a^2-2b=202$. Using substitution, we find that $b=99$. So $x+y=20$ and $xy=99$. In other words, $x$ and $y$ are the solutions to the equation $k^2-20k+99=0$. Using Vieta's, we can see that $k=11$ or $k=9$, so the shorter string has length 9. 
A: $400=20^2=(x+y)^2=x^2+y^2+2 x y=202 +2 x y.$ Therefore $x y=99.$
For all $x, y, z$ we have $$0=z^2-(x+y)z +x y\iff 0=(z-y)(z-y)\iff (z=x \text { or } z=y).$$  So with $x+y=20$ and $x y=99$ we have $$(z=x \text { or } z=y)\iff 0=z^2-20 z+99 \iff 0=(z-10)^2-1 \iff$$ $$ \iff z-10=\pm 1\iff (z=11 \text  { or  } z=9).$$ So the lengths are $9$ and $11. $  They cannot be $9$ & $9$, nor $11$ & $11$ as they must add to $20.$
Note that $x\ne y.$ The symmetry is in the ability to interchange the symbols $x , y$ (unless we specify  something extra, like $x>y$),not in the equality of $x$ and $y.$
