$26 \cdot 38$ On the plane, there are three point sets $M, N, P$:
$$
\begin{aligned}
M= & \{(x, y) \mid | x|+| y |<1\}, \\
N= & \left\{(x, y) \left\lvert\, \sqrt{\left(x-\frac{1}{2}\right)^{2}+\left(y+\frac{1}{2}\right)^{2}}\right.\right. \\
& \left.\quad+\sqrt{\left(x+\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2}}<2 \sqrt{2}\right\}, \\
P= & \{(x, y) \mid | x+y|<1, \quad| x|<1, \quad| y \mid<1\} .
\end{aligned}
$$

Then their relationship is
(A) $M \subset P \subset N$.
(B) $M \subset N \subset P$.
(C) $P \subset N \subset M$.
(D) None of (A), (B), (C) hold.
(China High School Mathematics League, 1988)
[Solution] As shown in the figure, set $M$ is the interior of square $ABCD$, and set $P$ is the interior of hexagon $AEBCFD$. Therefore, $M \subset P$.

Set $N$ is the interior of an ellipse with foci at points $Q_{1}\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $Q_{2}\left(\frac{1}{2}, -\frac{1}{2}\right)$, and a semi-major axis of $\sqrt{2}$.

It is easy to see that points $E$ and $F$ lie on the ellipse $\left(\because E Q_{1} + E Q_{2} = 2 \sqrt{2}\right)$, and points $A, B, C, D$ lie inside the ellipse $\left(\because A Q_{1} + A Q_{2} < 2 \sqrt{2}\right)$. Therefore, $P \subset N$. Hence, $M \subset P \subset N$.
Thus, the answer is $(A)$.