17・88 - A shepherd boy is 4 miles south of a small river, which flows due east. He is also 8 miles west and 7 miles north of his cottage. He wants to lead his horse to the river to drink and then return home. The shortest distance (in miles) he can accomplish this is
(A) $4+\sqrt{185}$.
(B) 16 .
(C) 17 .
(D) 18. (E) $\sqrt{32}+\sqrt{137}$.
(24th American High School Mathematics Examination, 1973)
[Solution] As shown in the figure, let $S$ represent any point on the river $SE$, and let $C, H$, and $D$ be the positions of the herdsman, his cottage, and a point 8 miles north of $C$, respectively.

It is easy to prove that $CS + SH = DS + SH$. Clearly, the distance is shortest when $DSH$ is a straight line, thus
$$
CS' + S'H = DS'H = \sqrt{8^2 + 15^2} = \sqrt{289} = 17 \text{.}
$$

Therefore, the answer is $(C)$.