# 1983 AHSME Problems/Problem 29

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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{align*}&(x^2+y^2)+((x-1)^2+y^2)=(x-1)^2+(y-1)^2\\ &2x^2+2y^2-2x+1=x^2+y^2-2x-2y+2\\ &x^2+y^2=-2y+1\\ &x^2+y^2+2y-1=0\\ &x^2+(y+1)^2=2\end{align*}

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 $\boxed{2+\sqrt{2}}.$