1983 AHSME Problems/Problem 29

Place the square on the coordinate plane with $A$ as the origin. (This means that $B=(1,0), C=(1,1),$ and $D=(0,1).$ We are given that $PA^2+PB^2=PC^2,$ so

$\begin{aligned} (x^{2} + y^{2}) + ((x - 1)^{2} + y^{2}) = (x - 1)^{2} + (y - 1)^{2} \\ 2 x^{2} + 2 y^{2} - 2 x + 1 = x^{2} + y^{2} - 2 x - 2 y + 2 \\ x^{2} + y^{2} = - 2 y + 1 \\ x^{2} + y^{2} + 2 y - 1 = 0 \\ x^{2} + (y + 1)^{2} = 2 \end{aligned}$

Thus, we see that $P$ is on a circle centered at $(0,-1)$ with radius $\sqrt{2}.$ The farthest point from $D$ on this circle is at the bottom of the circle, at $(0, -1-\sqrt{2}),$ so $PD$ is $\boxed{2+\sqrt{2}}.$

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1983 AHSME Problems/Problem 29