$27 \cdot 4 P$ points are in the same plane as a given square with side length 1. Let the vertices of this square be $A, B, C, D$ (counterclockwise), and the distances from $P$ to $A, B, C$ be $u, v, w$. If $u^{2}+v^{2}=w^{2}$, then the maximum distance from $P$ to $D$ is
(A) $1+\sqrt{2}$.
(B) $2 \sqrt{2}$.
(C) $2+\sqrt{2}$.
(D) $3 \sqrt{2}$.
(E) $3+\sqrt{2}$.
(34th American High School Mathematics Examination, 1983)
[Solution] Establish a Cartesian coordinate system as shown in the right figure:
Given $u^{2}+v^{2}=w^{2}$, we have
$$
\begin{array}{l}
{\left[(x-1)^{2}+y^{2}\right]+\left[(x-1)^{2}+(y-1)^{2}\right]} \\
=x^{2}+(y-1)^{2},
\end{array}
$$

which simplifies to $x^{2}-4 x+2+y^{2}=0$,
or $(x-2)^{2}+y^{2}=2$.
Thus, the locus of point $P$ is a circle with center at $(2,0)$ and radius $\sqrt{2}$.

From the figure on the right, it is clear that the point on this circle farthest from $D$ is $E(2+\sqrt{2}, 0)$.
Therefore, the answer is (C).