Jim Boyce of Mountain View CA solved the problem. Here is his solution:
In each case, I have a countable set of points. The points lie on a
circle of radius sqrt(2).
{ ( root2*cos (2*i*A), root2*sin(2*i*A)) } for carefully chose angle A.
For irrational length and rational area, I use A = arccos(4/5).
For rational length and irrational area, I use A = arccos(1/3).
In each case, all the points are on a particular circle, so no three
points are collinear.
Look at the values of cos(i*A) and sin(i*A).
For A = arccos(4/5), sin(A) is 3/5. It is easy to see that cos(i*A) and
sin(i*A) are all rational numbers.
For A = arccos(1/3), sin(A) is (2/3)*root2. It is easy to see that
cos(i*A) is always rational, and sin(i*A) is always a rational multiple
of root2.
If we look more carefully at these sine and cosine values, we can see
the sin(i*A) is zero only when i is zero. (This means we have countably
infinite points rather than finitely many points.)
For A=arccos(4/5), we can write cos(i*A) as p/q and sin(i*A) as r/q,
where q is 5**i and p and r are not divisible by 5. The residues (mod
5) of p and r are periodic.
i cos(i*A) sin(i*A) p(mod 5) r(mod 5)
1 4/5 3/5 4 3
2 7/25 24/25 2 4
3 -44/125 17/125 1 2
4 -527/625 336/625 3 1
5 -3116/3125 -237/3125 4 3
For A=arccos(1/3) we can write cos(i*A) as p/q and sin(i*A) as
(r/q)root2, where q is 3^i and p and r are not divisible by 3. The
residues (mod 3) of p and 4 are periodic.
i cos(i*A) sin(i*A) p(mod 3) r(mod 3)
1 1/3 (2/3)root2 1 2
2 -7/9 (4/9)root2 2 1
3 -23/27 (-10/27)root2 1 2
The length of a chord that subtends an angle of (2*i*A) in the unit
circle is 2 sin(i*A). So, for A=arccos(4/5), the (unit-circle) chords
all have rational length and for A=arccos(4/5) the (unit-circle) chords
are all rational multiples of root2. When we scale up to the circle
with radius root2, the arccos(4/5) chords are all rational multiples of
root2 and the arccos(1/3) chords are all rational.
The area of a triangle whose vertex coordinates are all rational is
rational. That is true for the unit-circle using A=arccos(4/5). When
we scale the radius up to root2, the areas of all the triangles doubles,
so those areas are still rational.
When we look at the arccos(1/3) case, the vertices (in the unit-circle)
all have rational x-coordinate and the y-coordinates are all rational
multiples of root2. The triangles all have areas that are rational
multiples of root2. When we scale the radius up to root2, the areas
double, so they are still rational multiples of root2.