17・186 $P$ is a point on the known line segment $M N$, and $\frac{M P}{P N}=\frac{P N}{M N}$. Using the lengths $|M P|^{2}$, $|N P|^{2}$, $|M N|^{2}$ as sides, then
(A) a right-angled triangle can be formed.
(B) a right triangle can be formed.
(C) an obtuse triangle can be formed.
(D) a triangle cannot be formed.
(Shenyang City, Liaoning Province, China "Yucai Cup" Junior High School Mathematics Competition, 1988)
[Solution] Let $M N=a, P N=x$, then $M P=a-x$. Substituting into the equation
$$
\frac{M P}{P N}=\frac{P N}{M N},
$$

we get $\frac{a-x}{x}=\frac{x}{a}$, which simplifies to $x^{2}+a x-a^{2}=0$.
Solving for $|P N|=x=\frac{\sqrt{5}-1}{2} a,|M P|=a-x=\frac{3-\sqrt{5}}{2} a$.
Thus, $|P N|^{2}=\frac{3-\sqrt{5}}{2} a^{2},|M P|^{2}=\frac{7-3 \sqrt{5}}{2} a^{2}$.
$$
\begin{aligned}
\because \quad & |M P|^{2}+|P N|^{2}-|M N|^{2} \\
= & \frac{7-3 \sqrt{5}}{2} a^{2}+\frac{3-\sqrt{5}}{2} a^{2}-a^{2} \\
= & (4-2 \sqrt{5}) a^{2}<0,
\end{aligned}
$$
$$
\therefore \quad|M P|^{2}+|P N|^{2}<|M N|^{2} \text {. }
$$

Therefore, the segments $|M P|^{2},|P N|^{2},|M N|^{2}$ cannot form a triangle. Hence, the answer is $(D)$.