NO-ABEL 2000 Problem 4

For some values of c, the equation $x^c + y^c = z^c$ can be illustrated geometrically. For example, the case $c = 2$ can be illustrated by a right-angled triangle. By this we mean that, x, y, z is a solution of the equation $x^2 + y^2 = z^2$ if and only if there exists a right-angled triangle with catheters $x$ and $y$ and hypotenuse $z$ . In this problem we will look at the cases $c = -\frac{1}{2}$ and $c = - 1$ . a) Let $x, y$ and $z$ be the radii of three circles intersecting each other and a line, as shown, in the figure. Show that, $x^{-\frac{1}{2}}+ y^{-\frac{1}{2}} = z^{-\frac{1}{2}}$ b) Draw a geometric figure that illustrates the case in a similar way, $c = - 1$ . The figure must be able to be constructed with a compass and a ruler. Describe such a construction and prove that, in the figure, lines $x, y$ and $z$ satisfy $x^{-1}+ y^{-1} = z^{-1}$ . (All positive solutions of this equation should be possible values for $x, y$ , and $z$ on such a figure, but you don't have to prove that.)