3

Let $x,y∈R$ . Find the minimum value of this expression:

$P=\sqrt{2x^2+2y^2-2x+2y+1}+\sqrt{2x^2+2y^2+2x-2y+1}+\sqrt{2x^2+2y^2+4x+4y+4}$

We have: $P=\sqrt{(\sqrt{2}x-\frac{1}{\sqrt{2}})^2+(\sqrt{2}y+\frac{1}{\sqrt{2}})^2}+...+\sqrt{(\sqrt{2}x+\sqrt{2})^2+(\sqrt{2}y+\sqrt{2})^2}$

I think use vector or Geometric method to solve this problem. But I don't know how to choose vectors or points logically?

abcdxyz
  • 430

1 Answers1

2

Hint:

You need $P(x, y)$ to be the Fermat point of the triangle $A(\frac12, -\frac12), B(-\frac12, \frac12), C(-1, -1)$.

Note that $\triangle ABC$ is isosceles, and from properties of the Fermat Point, it should lie on the angle bisector at $C$. This gives us $x=y$. Now there are two good routes,
(i) to use calculus in one variable, or
(ii) to note an additional property of the Fermat point $\angle APB = \angle BPC = \angle CPA = 120^{\circ}$ and use coordinate geometry / vectors.

Added: To check your final answer,

I get the minimum as $P \ge 2+\sqrt3$.

Macavity
  • 46,381
  • Can you suggest to me more? Be more specific?? I don't know . Because if I use vector, i must have 4 points to apply?? And with the points: $(x;y),(\frac{1}{2};-\frac{1}{2}),(-\frac{1}{2};\frac{1}{2})$ and the last, I think this is the point $(-1;-1)$ not $(-\frac{1}{2};\frac{1}{2})$ – abcdxyz Jan 14 '14 at 04:11
  • You are of course right about the third point. Will add more to the hint. – Macavity Jan 14 '14 at 04:18
  • Thanks . I find out the result as you. :) $P\ge 2+\sqrt{3}$ – abcdxyz Jan 14 '14 at 07:53