$35 \cdot 51$ Divide a square with a side length of 2 into four equal areas, such that in one of these parts, there exist three points that can form an equilateral triangle with a side length greater than 1. The number of such divisions that satisfy the above property
(A) does not exist.
(B) is exactly one.
(C) is finite but more than one.
(D) is infinitely many.
(1st "Hope Cup" National Mathematics Invitational, 1990)
[Solution] Given the side length of the square is 2, its area is 4.
When it is divided into four parts, each part has an area of 1.
It is only necessary to divide one part into an area of 1, which is a regular triangle with side length $a=2 \sqrt{\frac{\sqrt{3}}{3}}$.

And $2 \sqrt{\frac{\sqrt{3}}{3}}>1$, thus a regular triangle with side length $b$ greater than 1 can be found from it (as shown by the dotted line in the figure).

Since the position of the above regular triangle with an area of 1 is not limited, there are infinitely many ways to divide it.
Therefore, the answer is $(D)$.