Answer:
see attached
Explanation:
You want two non-congruent spanning triangles of the given parallelogram, each with 1/3 the area of the parallelogram.
Spanning triangle
Each spanning triangle will have vertices such that one is on a vertex of the parallelogram, and the other two are on the sides adjacent to the opposite vertex.
Two such triangles are shown in the attachment. Each is a right triangle. The red triangle has half the base and twice the height of the blue triangle, so has the same area. (We're calling the left edge of the triangles their "base.")
Area
We can show the blue triangle is 1/3 the area of the parallelogram by considering the three triangles at its edges.
The lower left equilateral triangle has a base that is 2/3 that of the parallelogram, and the same height. Its area is (1/2)(2/3)(1) = 1/3 of the area of the parallelogram.
The triangle at lower right has a base that is 1/3 that of the parallelogram, and 1/2 the height. Its area is (1/2)(1/3)(1/2) = 1/12 of the area of the parallelogram.
The triangle at upper right has a base equal to that of the parallelogram, and half the height. Its area is 1/2(1)(1/2) = 1/4 of the area of the parallelogram.
Then the area of the blue triangle is ...
1 - (1/3 +1/12 + 1/4) = 1/3
the area of the parallelogram, as required.
See the attachment for the triangles.
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Additional comment
There may be a suitable formula for finding these triangles. We found them by trial and error, using a graphing program to compute the area as we moved the vertices between grid points.
An equation for triangle area could be written as a function of the fraction along a parallelogram edge where the triangle vertex is located. This would give an equation that would relate the two parameters. Then the trick is finding the discrete values that represent grid locations and satisfy the parametric equation. That seems like more work than what we did by trial and error.
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