2000 Pan African MO Problems/Problem 4

Problem

Let $a$ , $b$ and $c$ be real numbers such that $a^2+b^2=c^2$ , solve the system: $z^2=x^2+y^2$ $(z+c)^2=(x+a)^2+(y+b)^2$ in real numbers $x, y$ and $z$ .

Solution

Expanding the last equation and simplifying results in $\begin{align*} z^2 + 2zc + c^2 &= x^2 + 2xa + a^2 + y^2 + 2yb + b^2 \\ 2zc &= 2xa + 2yb \\ zc &= xa + yb. \end{align*}$ Isolating $z$ means that $z = \frac{xa+yb}{c}$ . Substituting $z$ in the second equation results in $\begin{align*} \frac{(xa)^2 + 2xayb + (yb)^2}{c^2} &= x^2 + y^2 \\ (xa)^2 + 2xayb + (yb)^2 &= (xc)^2 + (yc)^2 \\ 2xayb &= (xb)^2 + (ya)^2 \\ 0 &= (xb-ya)^2 \end{align*}$ By the Zero Product Property, $xb = ya$ . If $a = 0$ , then either $b = 0$ or $x = 0$ . Thus, ordered pair $(a,b,c,x,y,z)$ can be $(0,0,0,x,y,z)$ or $(0,b,b,0,y,y)$ . Otherwise, $x = \frac{ya}{b}$ , so another ordered pair can be $(a,b,c,\frac{ya}{b},y,\frac{yc}{b})$ . Setting $\frac{y}{b} = n$ means that the ordered pair can be rewritten as $(a,b,c,na,nb,nc)$ .

Because the case $(0,b,b,0,y,y)$ is part of $(a,b,c,na,nb,nc)$ , the ordered pairs that are solutions are $\boxed{(0,0,0,x,y,z), (a,b,c,na,nb,nc)}$ .

2000 Pan African MO (Problems)
Preceded by Problem 3	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 5
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2000 Pan African MO Problems/Problem 4

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