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?

Let the vertices of the equilateral triangle be denoted ABC with the circumcenter and origin labeled D and O, respectively.

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

Let x be the length of one side of the equilateral triangle.
Then x = AB = AC = AB, and
AD = BD = CD = (SQRT(3)*x)/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.

Assuming a solution existed, then the sum of the areas of triangles AOB, BOD, COD and BCD minus the area of AOC would equal the area of the equilateral triangle ABC. Yet iterating between 2 and 4 (the possible lengths given as OC-AO = 3-1 = 2 and OC+OA = 3+1 = 4) it is found that only lengths 3.5 to 4 provide real values. And again yet, all these calculated areas for triangle ABC are not equal but are less than the sum of the "4 - 1 sub-triangles" for each of these values within the range.

This analysis does not preclude the possibility that the Origin might have a different Z-factor than the other points. Nor that the assigned lengths -- 1,2,3, and 4 -- might apply to different line segments. 

Edited on March 30, 2012, 12:59 pm

Posted by Dej Mar on 2012-03-30 05:25:10