$20 \cdot 21$ In the unit circle, chords $P Q$ and $M N$ are parallel to the unit radius $O R$ through the center $O$. Chords $M P, P Q$, and $N R$ are all $s$ units long, and chord $M N$ is $d$ units long. Of the following three equations
$\mathrm{I} . d-s=1$;
II . $d s=1$;
III . $d^{2}-s^{2}=\sqrt{5}$.

which must be correct?
(A) Only I.
(B) Only II.
(C) Only III.
(D) Only I and II.
(E) I, II, and III.
(24th American High School Mathematics Examination, 1973)
[Solution 1] As shown in the figure, the length of chord $MN$ is $d$, and each chord of length $s$ subtends an arc of $\frac{180^{\circ}}{5}=36^{\circ}$. Connect the radius $OP$, intersecting $MN$ at $T$. Draw perpendiculars from $O$ and $Q$ to $MN$, denoted as $OV$ and $QU$ respectively.
From $\triangle OVN \sim \triangle QUN$, we have
$\frac{UN}{QN}=\frac{VN}{1}$, or $\frac{d}{2}-\frac{s}{2}=s \cdot \frac{d}{2}$.
Thus, $d-s=ds$.
Since $\angle MPT = \angle MTP = 72^{\circ}$, it follows that $MT = s$.
Similarly, $\angle OTN = \angle ORN$, so $ORNT$ is a parallelogram with opposite sides $TN = OR = 1$. Therefore,
$$
MN - MT = d - s = 1.
$$

From $(d-s)^2 + 4ds = 1 + 4 = 5$ or $d^2 + 2ds + s^2 = 5$,
we get $d + s = \sqrt{5}$.
Thus, $d^2 - s^2 = (d+s)(d-s) = \sqrt{5} \cdot 1 = \sqrt{5}$.
In summary, all three equations I, II, and III are correct.
Therefore, the answer is $(E)$.

[Solution 2] Given $MT = s$ and $MN = d$, and since $OTNR$ is a parallelogram, its side $TN = MN - MT = d - s = 1$.

Let $P'$ (not shown in the figure) be the other endpoint of the diameter through $P$. Then the chord $PP'$ intersects $MN$ at $T$. By the intersecting chords theorem, we have
$PT \cdot TP' = MT \cdot TN$, or $(1-s)(1+s) = s \cdot 1$.
This quadratic equation has the positive root $s = \frac{1}{2}(\sqrt{5} - 1)$. This can be used to prove that equations II and III are correct. Therefore, all three equations I, II, and III are correct.
Hence, the answer is $(E)$.