The South African Mathematical Olympiad Third Round

The two equations can be rewritten as $$\begin{array}{rl}(x-24{)}^2+{\left(y-\frac{29}{2}\right)}^2& =\frac{289}{4}\\ (x-24)\left(y-\frac{29}{2}\right)& =-30.\end{array}$$ By putting $X=x-24$ and $Y=y-\frac{29}{2}$, we obtain $$X^2+Y^2=\frac{289}{4}\text{\hspace{1em}(1)}$$ $$XY=-30.\text{\hspace{1em}(2)}$$ From (2) we have $Y=-30/X$. Plug this into (1) to get $X^2+\frac{900}{X^2}=\frac{289}{4}$, which simplifies to $$X^4-\frac{289}{4}{X}^2+900=0.$$ We can now solve for $X^2$, using the quadratic formula: $$\begin{array}{rl}{X}^2& =\frac{1}{2}\left(\frac{289}{4}\pm \sqrt{{\left(\frac{289}{4}\right)}^2-60^2}\right)\\ & =\frac{1}{2}\left(\frac{289}{4}\pm \sqrt{\left(\frac{289}{4}-60\right)\left(\frac{289}{4}+60\right)}\right)\\ & =\frac{1}{2}\left(\frac{289}{4}\pm \sqrt{\frac{49}{4}\cdot \frac{529}{4}}\right)\\ & =\frac{1}{2}\left(\frac{289\pm 161}{4}\right),\end{array}$$ so that $X^2=\frac{225}{4}$ or $X^2=16$. Thus, $X=\pm \frac{15}{2}$ or $X=\pm 4$, and correspondingly, $Y=\mp 4$ or $Y=\mp \frac{15}{2}$. Using $x=X+24$ and $y=Y+\frac{29}{2}$, we find that $(x,y)$ solves the original set of equations if and only if $(x,y)\in \{(28,7),(20,22),(\frac{33}{2},\frac{37}{2}),(\frac{63}{2},\frac{21}{2})\}$.

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Alternative solution.

Proceed as in Solution 1 up to the two equations $$X^2+Y^2=\frac{289}{4}\text{\hspace{1em}(1)}$$ $$XY=-30.\text{\hspace{1em}(2)}$$ By either adding $2XY=-60$ to (1), or subtracting it from (1), we get, respectively, $$\begin{array}{rl}{X}^2+2XY+Y^2& =(X+Y{)}^2=\frac{49}{4}={\left(\frac{7}{2}\right)}^2\\ X^2-2XY+Y^2& =(X-Y{)}^2=\frac{529}{4}={\left(\frac{23}{2}\right)}^2.\end{array}$$ From this, we have to solve four systems of equations: $$\begin{array}{rl}X+Y& =\pm \frac{7}{2}\\ X-Y& =\pm \frac{23}{2}\end{array}$$

X + Y	X - Y	X	Y	x	y
$\frac{7}{2}$	$\frac{23}{2}$	$\frac{15}{2}$	$-4$	$\frac{63}{2}$	$\frac{21}{2}$
$\frac{7}{2}$	$-\frac{23}{2}$	$-4$	$\frac{15}{2}$	$20$	$22$
$-\frac{7}{2}$	$\frac{23}{2}$	$4$	$-\frac{15}{2}$	$28$	$7$
$-\frac{7}{2}$	$-\frac{23}{2}$	$-\frac{15}{2}$	$4$	$\frac{33}{2}$	$\frac{37}{2}$

We conclude that there are exactly four pairs $(x,y)$ of real numbers that solve the original two equations, namely $(\frac{63}{2},\frac{21}{2})$, $(20,22)$, $(28,7)$ and $(\frac{33}{2},\frac{37}{2})$.