$17 \cdot 62$ Let $\triangle A B C$ satisfy the following conditions: $A B=A C$, and $D$ is a point on $A C$ such that $B D \perp A C$, and the lengths of $A D$ and $C D$ are integers and $B D^{2}=57$. The smallest value of $A C$ among all such triangles is
(A) 9 .
(B) 10 .
(C) 11 .
(D) 12 .
(E) 13 .
(50th American High School Mathematics Examination, 1999)
[Solution] Since $A D$ and $C D$ are integers, let $A D=m, D C=n$. Then $A B=A C=$ $m+n$.

In the right triangle $\triangle A B D$, $A B^{2}-A D^{2}=B D^{2}$, which means $(m+n)^{2}-m^{2}=57$, so $n(2 m+n)=57$.
Since $57=3 \times 19=1 \times 57$, it is clear that $n=3, 2 m+n=19$,
yielding $m=8$; or $n=1, 2 m+n=57$, yielding $m=28$.
Thus, $m+n=11$, or $m+n=29$.
Therefore, the smallest value of $A C$ is 11.
Hence, the answer is $(C)$.