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The construction is not possible, though it looks tantalisingly likely that it might be!
However, there is a construction that acts as a proof to the contrary.
Inscribe a circle of radius (13/3)^(1/2) based on the origin. Construct the line {x, 0.5). This intersects the circle at two points. Call either such intersection, P. Construct a further circle, Q, based on P, and whose circumference passes through {0, 1}. Mark that point, A. Construct circles of radius 3 and 4 based on the origin. Q intersects each of these circles at two points. Call the intersection that is farther from A on the radius 3 circle, e.g. at {6.75^(1/2), -1.5}, B. Call the intersection that is closer to A on the radius 4 circle, e.g. at {2*3^(1/2), 2},C. The triangle ABC is equilateral, and every triangle with vertices 1,3, and 4 from the origin is congruent to it. Unfortunately, (13/3)^(1/2) is slightly greater than 2.
Hence this solution also resolves a more tricky problem:
A regular tetrahedron lies with its four vertices at respective distances from the origin as 1, 2, 3 and 4 units. What is the length of each edge of the tetrahedron?
But if so, then 3 of those 4 vertices must lie in a plane 1, 3, and 4 from the origin, while the 4th must be 2 units from the origin, and either vertically above, or vertically below, the circumcentre of the equilateral triangle they form. Given however that such circumcentre is more two units from the origin, there is no corresponding point on the 2-shell on which to place the last vertex. In fact the last vertex of this tetrahedron is on the surface of a sphere of radius 13^(1/2) from the origin.
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