Max(Min(Sum(Distance)))=...

Geometry Level 5

There are four points on the plane, $A(1,0),B(-1,0),C(0,1),D(0,-1)$ .Choose a point $P$ and so the sum of the distances $\overline{PA}+\overline{PB}+\overline{PC}+\overline{PD}$ is minimized. Of course,now point $P$ is at $(0,0)$ .

If we add one more point $E(a,b)$ ,and choose a point $P$ so the sum of the distances $\overline{PA}+\overline{PB}+\overline{PC}+\overline{PD}+\overline{PE}$ is minimized. When $E$ is at different places,point $P$ will change its position,too.

So,what is the maximum distance from $P$ to the origin?

The answer is 0.5774.

1 solution

Jeremy Galvagni
Jul 9, 2018

I expect a better solution will come but here's what I did. I was surprised that $P$ had a maximum, but it does. My solution also uses calculus, so unless this can be solved without calculus, the category should be changed.

Importantly, I was able to convince myself using Geometer's Sketchpad, but I do not have a proof, $P$ is maximized when $E$ is on the X-axis and $P$ will also be on the X-axis. So this solution should not be considered complete.

$E=(a,0)$ where $a>1$ and $P=(x,0)$

$PA=(x-1)$ , $PB=(x+1)$ , $PC=\sqrt{x^{2}+1^{2}}$ , $PD=\sqrt{x^{2}+(-1)^{2}}$ , $PE=(a-x)$

Sum of distances as a function of $x$ is $T(x)=a+2-x+2\sqrt{x^{2}+1}$ . Check graphically there is a mimimum around $x=0.58$

$T'(x)=-1+\frac{2x}{\sqrt{x^{2}+1}}$

Surprisingly the $a$ is gone, implying this minimum doesn't depend on $a$ (I had expected a limiting value for $P$ )

Solving $T'(x)=0$ gives $x^{2}=\frac{1}{3}$ and since $x$ is the distance from the origin $P=\sqrt{\frac{1}{3}}\approx\boxed{0.57735}$

Max(Min(Sum(Distance)))=...

The answer is 0.5774.

1 solution

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