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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.

learning
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  • But being symmetric in $y$ and $x$ doesn't guarantee that $y = x$; you can see that in the individual equations. It just means that if $(x,y)$ is a solution, then $(y,x)$ is as well. – pjs36 May 28 '16 at 03:24

2 Answers2

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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. Not to scale

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.

Hrhm
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  • Good solution - you beat me to it...I used the two equations the OP gave. Using (2), solve for $x$, neglecting the negative root since $x$ is a length. Plug this into (1), solve for $y$, and I ended up with $y=\sqrt{101 \pm 20}$, and, again, dropping the negative root. So, $y$ being the length of the shorter piece, $y=9$ or $y=11$, measured in feet is the length(s) of the shorter piece of wire. – Procore May 28 '16 at 03:34
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$400=20^2=(x+y)^2=x^2+y^2+2 x y=202 +2 x y.$ 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 $y.$

DanielWainfleet
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