$37 \cdot 14$ If $(x, y)$ where $x, y$ are both integers is called a lattice point, the maximum area of a square that contains 3 lattice points is closest to
(A)4.0.
(B) 4.2.
(C) 4.5.
(D)5.0.
(E)5.6.
(49th American High School Mathematics Examination, 1998)
［Solution］The 3 lattice points contained within the square are not collinear. Let them be $A, B, C$.

The nearest lattice point on the same side of the line connecting $A$ and $C$ is $D$. Suppose $l_{1}, l_{2}$ are the lines on which the sides of the square lie. Clearly, $l_{1}$ cannot pass beyond $D$, otherwise $D$ would be between the perpendiculars from $A$ and $C$ to $l_{1}$ (as shown in the figure), which would mean there are four lattice points within the square, which is not acceptable.
If $l_{1}$ is not perpendicular to $BD$, then the distance from $B$ to $l_{1}$ is less than $BD$, so the position of $l_{1}$ is as marked in the figure.
By similar analysis, the position of $l_{2}$ should also be as shown in the figure, at which point the distance between $l_{1}$ and $l_{2}$ is $\frac{3}{2} \sqrt{2}$.
$$
\therefore \quad S_{\text {square }} \leqslant\left(\frac{3}{2} \sqrt{2}\right)^{2}=4.5 \text {. }
$$

Therefore, the answer is $(C)$.