Find the minimum of $\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x+2)^2+(y+2)^2}+\sqrt{(x-1)^2+(y+1)^2}$ For real $x,y$

Question

Find the minimum of $\sqrt{(x+1)^2+(y-1)^2}+\sqrt{(x+2)^2+(y+2)^2}+\sqrt{(x-1)^2+(y+1)^2}$ For real $x,y$.

It is the distance of a point in a triangle from its vertexes which gives the minimum when $(x,y)$ is on the Fermat's point. But now I am stuck in finding the location of Fermat's point.

$(1) \quad 4$

$(2) \quad 5$

$(3) \quad \sqrt{3}+2\sqrt{5}$

$(4) \quad 4\sqrt{2}$

Answer 1

Method I

Let $A(-1,1)$, $B(-2,-2)$ and $C(1,-1)$ be the vertices.

We have $a=\sqrt{10}$, $b=2\sqrt{2}$, $c=\sqrt{10}$ and \begin{align*} \cos B &= \frac{c^2+a^2-b^2}{2ca} \\ &=\frac{10+10-8}{2(10)} \\ &= \frac{3}{5} \\ \sin B &= \frac{4}{5} \end{align*}

According to the result here

\begin{align*} S^2 &= c^2+a^2-2ca\cos (B+60^{\circ}) \\ &= 10+10-20 \left( \frac{3}{5} \cos 60^{\circ}- \frac{4}{5} \sin 60^{\circ} \right) \\ &= 20-2(3-4\sqrt{3}) \\ &= 14+8\sqrt{3} \\ &= 8+8\sqrt{3}+6 \\ &= \left( 2\sqrt{2}+\sqrt{6} \right)^2 \\ S &= 2\sqrt{2}+\sqrt{6} \\ &= 5.2779 \ldots \end{align*}

which is none of the options given.

Method II

By symmetry, let the Fermat point be $(x,x)$, then

\begin{align*} S &= 2\sqrt{2(x^2+1)}+\sqrt{2} \, |x+2| \\ &= \sqrt{2} \left( 2\sqrt{x^2+1}+|x+2| \right) \\ &= \sqrt{2} \left( 2\sqrt{x^2+1}+|x+2| \right) \\ S' &= \sqrt{2} \left[ \frac{2x}{\sqrt{x^2+1}}+\operatorname{sgn} (x+2) \right] \\ \end{align*}

Now $$S'=0 \implies x=-\frac{1}{\sqrt{3}}$$ which gives the same result.

Answer 2

According to this wikipedia article on fermat points, we can find the fermat point by following method if all angles of triangle are less than $120^\circ$

The Fermat point of a triangle with largest angle at most 120° is simply its first isogonic center or X(13), which is constructed as follows:

1. Construct an equilateral triangle on each of two arbitrarily chosen sides of the given triangle.
2. Draw a line from each new vertex to the opposite vertex of the original triangle.
3. The two lines intersect at the Fermat point.

For this question, the point comes close to $(0.8722881509, 0.1744576301)$ and the value of function is approximately $4.625181601$

Note: If you wish to calculate the exact position, you need to first find the three points corresponding to equilateral triangles, then find equations of lines, and solve them.