Finding the Fermat Point Given Three Arbitrary Points

  At the end of a letter to Mersenne in about 1640 concerned with finding maxima and minima Fermat proposed this problem:

  "Datis tribus punctis, quartum reperire, a quo si ducantur tres rectæ ad data puncta, summa trium harum rectarum sit minima quantitas."

  "Given three points, the forth to be found, from which you draw three lines to the given points, the sum of these three lines is to be a minimum quantity."

So, given three arbitrary points, and using the method in the previous blogs, we can find the Fermat point as follows. The distances are the ℓ_i whose sum is to be minimized. Taking the derivative we find for an assumed value of x that dL=f^Tdx where f is the sum of three unit vectors pointing to x. For the position of x for the minimum value of L the change dL has to be zero for arbitrary changes in position, dx, and the only way that this can happen is if f is equal to zero too. But the value of f at the assumed point is not necessarily zero so we look at changes in f with position and see find the value of dx for which f+df=f+Mdx=0. These are the correction equations for f. The matrix M is found by extracting the derivative of the vector function f(x). Using the method of least squares one can show that dx is a solution of the normal equations.

Using the above equations in Excel and repeatedly correcting the value for x we arrive at the Fermat point after just a few iterations.

Checking the angles between the lines from x to the given points we find they are all 120° which was deduced from the minimum condition.

The correction equations for Fermat's problem are simpler than the reflection problem since we do not have a constraint on the change for dx.