Problem: On a Louvre tablet of about 300 B.C. are four problems concerning rectangles of unit area and given semiperimeter. Let the sides and semiperimeter be $x,y$ and $a$. Then we have \begin{equation} xy=1, \qquad x+y=a. \tag{1} \end{equation} Solve this system by using the identity $$ \biggl(\frac{x-y}{2}\biggr)^2 = \biggl(\frac{x+y}{2}\biggr)^2 - xy \tag{2} $$ Attempted solution:
By elimination: To eliminate $y$, we note that $y = \frac{1}{x}$, and we substitute this $y$-value into the equation $x+y=a$ to obtain $$ x = \frac{a \pm \sqrt{a^2-4}}{2}, $$ where this solution was obtained by using the quadratic formula.
By using the given identity: Start by letting $y=a-x$. Using this and the fact that $x+y=a$, we have the following: \begin{align*} \biggl(\frac{2x-a}{2}\biggr)^2 = \biggl(\frac{a}{2}\biggr)^2 - 1 &\longleftrightarrow 4x^2-4xa+a^2 = a^2-4\\ &\longleftrightarrow x^2-ax+1=0, \end{align*} whereby we get that $x = \frac{a \pm \sqrt{a^2-4}}{2}$, as we did above by using elimination.
Main question: My solution using (2) seems to work, but is there a "slicker" approach? That is, it seems like the originally posed problem wants me to do something nifty with (2) rather than using it in such a brutish fashion. Anyone have a cleaner approach in mind in regards to solving (1) by using (2)?
I think the intended solution was as follows: $$ \left(\frac{x-y}{2}\right)^2=\left(\frac{a}{2}\right)^2-1. $$ Thus $x-y$ can be solved for: $$ x-y=\pm2\sqrt{\left(\frac{a}{2}\right)^2-1}=\pm\sqrt{a^2-4} $$ Adding and subtracting with $x+y$ gives us: $$ x=\frac{(x+y)+(x-y)}{2}=\frac{a\pm \sqrt{a^2-4}}{2}, $$ and likewise for $y$.
Note that this is a re-derivation of the quadratic formula; in particular, it defeats the purpose of this approach to use it!