# The minimum value of $|z+1|+|z-1|+|z-i|$ for $z \in \mathbb C?$

I was thinking about the following problem:

How can i find the minimum value of $|z+1|+|z-1|+|z-i|$ for $z \in \mathbb C?$ There are four options which are $(a)2,(b)2\sqrt 2,(c)1+ \sqrt 3,(d)\sqrt 5.$ It is a multiple choice question and so i am looking for a shorter method so that utilizing minimum effort and time i can get the correct option. Can someone point me in the right direction? Thanks in advance for your time.

• en.wikipedia.org/wiki/Fermat_point – Rijul Saini Jan 8 '13 at 14:28

Since we don't know the context in which you confront this problem, here's an elementary geometric approach. (It might not be the most efficient way since you may have other tools to bring to bear on the problem.)

This is a classic problem (due to Fermat) for finding what is called the Fermat point (more generally the geometric median) of the triangle you gave.

As the links outline, since the triangle here involves no angle larger than $120^\circ$, the point you seek is found as follows:

1. Construct equilateral triangles exterior to each side of the given triangle.
2. Form the line segment from the new vertex to opposite old vertex.
3. The Fermat point is the intersection of these three lines.

This diagram summarizes the situation:

Computing the coordinates of the northwest black point is easy with a little trigonometry: \begin{align}x_1&=-1-\sqrt{2}\sin(15^\circ)=-{1\over 2}(1+\sqrt{3}),\\ y_1&=\sqrt{2}\sin(75^\circ)={1\over 2}(1+\sqrt{3}), \end{align} the northeast point follows from symmetry, and the south point is simple.

The $x$-coordinate of the Fermat point is zero, and a little algebra reveals the $y$-coordinate is $1/\sqrt{3}$.

Finally, to answer the question that was asked: in order to minimize $$w(z):=|z+1|+|z-1|+|z-i|, \quad z\in \mathbb{C},$$ we take $z=0+{1\over \sqrt 3}i$ which results in $w=1+\sqrt 3$.

Here's a contour plot of $w$ superimposed with items above:

• Compute the line from $(x_1,y_1)$ to $(1,0)$. It is $\ell(x)=-{1\over \sqrt 3}x+{1\over \sqrt 3}$. Since we are looking for an intersection with the line $x=0$, just evaluate $\ell(0)={1\over \sqrt 3}$ to get the $y$ coordinate of the Fermat point. – JohnD Jan 9 '13 at 17:30

Each term gives the distance from the corresponding point. So in essence you are asking to minimize the summed distance between the points $(-1,0),\ (1,0),\ (0,1)$. This is called the "Geometric median", and for the 3 point case is named the "Fermat Point". Instruction on how to find this minimum can be found on a related ME question.

Specifically in your case, All angles are smaller than $120^\circ$, so we know that this pint forms that angle with the triangle's vertices. This means that the point lies on the $y$ axis and that moreover it's $y$ coordinate is given by: $$y = 1/\tan 60^\circ =1/\sqrt{3}$$