jmc

We color one of the triangles blue, and draw three blue segments connecting its points of intersection with the other triangle. Because of the symmetry inherent in the grid and the two triangles (which are both isosceles), these three blue segments divide the blue triangle into congruent smaller triangles. The blue triangle contains 9 of these congruent smaller triangles.

The region of overlap of the two triangles is a hexagonal region. As per the diagram above, this hexagonal region contains 6 of these congruent smaller triangles. Thus, the area of the hexagonal region is $6/9=2/3$ of the area of one of the isosceles triangles. We compute the area of one isosceles triangle as follows:

Label points $A,B,C,D,E,F$ as above. To compute the area of this triangle ($△AEF$), notice how it is equal to the area of square $ABCD$ minus the areas of triangles $△ADE$, $△ABF$, and $△ECF$. The square has side length 2 units, so the area of $△ADE$ and $△ABF$ is $\frac{1}{2}(2)(1)=1$ and the area of $△ECF$ is $\frac{1}{2}(1)(1)=\frac{1}{2}$. The area of square $ABCD$ is $2^2=4$, so the area of triangle $△AEF$ is equal to $4-2(1)-\frac{1}{2}=\frac{3}{2}$.

Finally, remember that the hexagonal region has area $2/3$ of the area of the triangle, or $\frac{2}{3}\cdot \frac{3}{2}=1$. Thus, the answer is $\boxed{1}$.