Example 1 Figure 1 is a shape composed of 4 unit squares, with each vertex of the unit squares called a grid point. By taking any three grid points as vertices, how many different isosceles right triangles can be formed?

Regarding this problem, the answer given in [1] is 24, [2] supplements it to 30, and [3] provides the correct answer as 32. This shows that although the problem seems simple, a slight oversight can lead to errors in solving it.

This article introduces an effective method for solving such counting problems—the labeling method. It can effectively avoid the aforementioned mistakes.

The basic approach is to label each grid point with a number, and the number labeled at grid point $A$ represents the number of counting objects (such as isosceles right triangles) with $A$ as the special element.

For Example 1, since each isosceles right triangle has a unique right-angle vertex, we can consider the number of isosceles right triangles with each grid point as the right-angle vertex.

Of course, once the right-angle vertex is determined, the isosceles right triangles can be divided into two major categories:
(1) The legs are parallel to the grid lines (called upright);
(2) The legs are not parallel to the grid lines (called slanted).
For each major category, they can be further classified by different directions. Once the direction is determined, they can be classified by the length of the legs.
Solution: Through the above classification, mark the number of isosceles right triangles with each grid point as the right-angle vertex (Figure 2). For example: $B_{3+1}$ indicates that there are 3 upright and 1 slanted isosceles right triangles with $B$ as the right-angle vertex; $E_{1+2}$ indicates that there are 1 upright and 2 slanted isosceles right triangles with $E$ as the right-angle vertex, among which 1 slanted one (indicated by a dashed line) is not easily noticed.

Noticing the symmetry of the figure, it is only necessary to number half of the grid points. Since each type of grid point appears twice, there are a total of $2(1+4+2+6+3)=32$ isosceles right triangles.

Interestingly, a similar problem appeared in the 2004 National Junior High School Mathematics Competition.