Given an equilateral triangle with its side unknown, there are three unequal lines originating from each vertex and converging at a certain point inside the triangle. The lengths of the segments are 3,4,5.
What is the length of any side of the equilateral triangle?
I remembered seeing a tidy relationship that allows you to determine the length of the side of an equilateral triangle given the distances from a point inside the triangle to each of the three vertices. It appears in Martin Gardner's book referenced below. It goes like this:
. . .3(a^4 + b^4 + c^4 + d^4) = (a^2 + b^2 + c^2 + d^2)^2
Taking the distances as a, b, and c, the side of the triangle is d.
One solution was offered by Charles W. Trigg in Mathematical Quickies and appears in Mathematical Circus by Martin Gardner, Mathematical Association of America, 1992, pp 64.
1--Draw equilateral triangle ABC, A and C on the bottom, A on the left, B at the top.
2--Locate the point 3 units from C, 4 units from A, and 5 units from B.
3--Construct equilateral triangle PCF below PC with PC = PF = CF.
4--Extend PC out to the left to E making AE perpendicular to PC extended.
5--/_PCB = 60º minus /_PCA = /_ACF.
6--Triangles PCB and FCA are congruent and AF = BP = 5.
7--Since APF is a right triangle, /_APE = 180º - 60º - 90º = 30º.
8--Thus, AE is 2 ans EP is twice the sqrt of 3.
9-- This allows the equation AC = sqrt[2^2 + (3 + 2sqrt(3)^2] = sqrt[25 + 12sqrt(3)]
10--This results in a side of the triangle, AC = 6.766....
Another solution exists in L.A. Graham's Ingenious Mathematical Problems and Methods, Dover Publications, Inc., 1959, pp 189.
There are several solutions.
