$37 \cdot 13 A B C$ is a triangle, points $A, B$ have coordinates $A(0,0), B(36, 15)$, and both coordinates of $C$ are integers, then the minimum value of the area of $\triangle A B C$ is
(A) $\frac{1}{2}$.
(B) 1 .
(C) $\frac{3}{2}$.
(D) $\frac{13}{2}$.
(E) no minimum value.
(38th American High School Mathematics Examination, 1987)
[Solution] Let the coordinates of point $C$ be $(x, y)$. Then, by the triangle area formula,
$S_{\Delta}=\frac{1}{2}\left|\begin{array}{ccc}0 & 0 & 1 \\ 36 & 15 & 1 \\ x & y & 1\end{array}\right|$ in absolute value
$$
=\frac{1}{2}(36 y-15 x) \text { in absolute value, }
$$
$\because x, y$ are both integers, and $S_{\triangle} \neq 0$,
Also, $36 y-15 x=3(12 y-5 x)$, and when $x, y$ are both integers, the minimum value of $12 y-5 x$ is 1 (when $y=3, x=7$).
$\therefore$ When $x, y$ are both integers, the minimum value of $36 y-15 x$ is 3.
Therefore, $\left(S_{\triangle}\right)_{\min }=\frac{3}{2}$.
Hence, the answer is $(C)$.