Board 1-- can't be completed
T2 triangle has lengths 1, 1 and
T1 triangle has sides of length 1.
Both triangles have areas of
This means that the total area of the equilateral triangle
must be a rational multiple of
Now, if both of these triangles are found at one of the
vertices then one side of the triangle must contain a piece of
length
and also a piece of length 1. Thus the length of the sides of
the equilateral triangle must be
where
neither of
and
which can be zero.
and
Board 2 -- can't be completed
No. If you could, its area would be a natural number because it
is a sum of 1s and 2s. However, is would also be L*L*sqrt{3}/4
using the formula for the area of a triangle. You could only
have one type of triangle on the edges for this to work.
Imagine that the triangle was solely made up from one sort of
triangle now. You'd have to remove a bit from the middle to get
it to work. But you can't because youd need to replace one of
the lengths with other lengths which wouldn't fit. Something
like that.
Board 3 --Not sure....