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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.

$(1) \quad 4$

$(2) \quad 5$

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

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

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Taha Akbari
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  • Since this is the sum of the distances from $(x,y)$ to $(-1,1)$, $(-2,-1)$, and $(1,-1)$ respectively, then it would seem to be minimized at the centroid of the triangle with vertices $(-1,1),(-2,-1),(1,-1)$, i.e., at the point of coincidence of the medians, which in this case is $(-2/3,-1/3)$ – Nicholas Stull Dec 06 '16 at 14:52
  • @NicholasStull I think that point is fermat point but how to find the distance? – Taha Akbari Dec 06 '16 at 15:08
  • The point $(-2/3,-1/3)$ gives approximated $7.33$ which is larger than $5.06449$ gives by the point $(-1,1)$ – Piquito Dec 06 '16 at 15:29
  • The point $(-2/3,-1/3)$, when plugged into this function, gives $$\frac{1}{3}(2\sqrt{5}+\sqrt{17}+\sqrt{29})$$ which gives the approximate decimal $4.66$ – Nicholas Stull Dec 06 '16 at 15:40
  • @Macavity Oops. Forgot about that. So while my answer gives an important starting point, it's unfortunately not going to be the minimum. (It does, however, appear to eliminate all but one answer, if I'm not mistaken). – Nicholas Stull Dec 06 '16 at 16:32

2 Answers2

2

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.

enter image description here

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.

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Ng Chung Tak
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1

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.

enter image description here

Max Payne
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