See Regular Tetrahedron 1234.

Construct an equilateral triangle such that its vertices and circumcentre are (in any order) 1,2,3, and 4 units from the origin.

What is the length of a side of the triangle?

broll, I corrected my post. An error in calculation provided for an incorrect assumption.

Following is my finding for given the Origin is 1 unit from the circumcenter instead of 2.

Given the lengths of OA, OB, OC and OD as 1,2,3, and 4,
let them be assigned as follows:
OA = 2
OB = 3
OC = 4
OD = 1

Using the Heron's formula for the calculation of area and
iterating through the possible range that x might be if all
points are on a single plane, I find there is no solution
for the assumed assigned lengths.

Assuming a solution existed, then the sum of the areas of
triangles AOB, BOD, COD, BCD and AOC would equal the area of
the equilateral triangle ABC. Yet iterating between 2 and 5
(the possible lengths given the max differences OC-OA = 4-2 = 2 and the minimum sum OC+OD = 5) it is found that range of
real values for area would lie between (approximately) 3.4642 and 3.5145, yet, the total sum of these calculated areas exceed the calculated area for triangle ABC by nearly 2 units square to almost 2 1/2 units square.

=================

Given the lengths of OA, OB, OC and OD as 1,2,3, and 4,
let them be assigned as follows:
OA = 4
OB = 1
OC = 2
OD = 3

Using the Heron's formula for the calculation of area and
iterating through the possible range that x might be if all
points are on a single plane, I find there is no solution
for the assumed assigned lengths.

From these ventures into different length assignments, I do not believe there is a mono-planar solution.

Edited on March 30, 2012, 1:57 pm

Posted by Dej Mar on 2012-03-30 12:57:26