A tricky question with complex numbers

[Physics lover]

Homework Statement

The question is in Attempt at a solution

Relevant Equations

Triangular inequalities in complex no.
Distance formula
Polar coordinates

All i was able to think was that i have to find a point (x,y) such that sum of its distances from points (0,0),(1,0),(0,1) and (3,4) is minimum.I tried by assuming the point to be centre of circle passing through any of the above 3 points,But it didn't helped me.

 

[mitochan]

Hi.Let (x,y) be the solution
s=\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}
should have minimum value. How about checking partial derivatives of s ?

 

[archaic]

Hi.Let (x,y) be the solution
s=\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}
should have minimum value. How about checking partial derivatives of s ?

I was thinking abou PDs too, but this is the precalc section.

 

[Physics lover]

Hi.Let (x,y) be the solution
s=\sqrt{x^2+y^2}+\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y-1)^2}+\sqrt{(x-3)^2+(y-4)^2}
should have minimum value. How about checking partial derivatives of s ?

My teacher has asked to do it without partial derivative.

 

[PeroK]

My teacher has asked to do it without partial derivative.

Can you use the symmetry of the problem? What if you only had $|z|$ and $|z-1|$? Try looking at that to start with.

 

[aheight]

My teacher has asked to do it without partial derivative.

It looks like a problem solvable by AM-GM inequality however I can't quite get it. May wish to try it.

 

[PeroK]

It looks like a problem solvable by AM-GM inequality however I can't quite get it. May wish to try it.

It's actually much simpler than that. You just have to break the problem down in the right way.

 

[Physics lover]

Can you use the symmetry of the problem? What if you only had $|z|$ and $|z-1|$? Try looking at that to start with.

i can apply triangular inequality to find that is it right?

 

[PeroK]

i can apply triangular inequality to find that is it right?

You can always use the triangle inequality. You already worked out, I think, that to minimise the sum $|z -1| + |z-i|$ is equivalent to finding a point which minimises the sum of the distances to the points $(1,0)$ and $(0, 1)$.

Can you solve that?

 

[Physics lover]

You can always use the triangle inequality. You already worked out, I think, that to minimise the sum $|z -1| + |z-i|$ is equivalent to finding a point which minimises the sum of the distances to the points $(1,0)$ and $(0, 1)$.

Can you solve that?

that may be the point (1/2,1/2)

 

[PeroK]

that may be the point (1/2,1/2)

Only that point?

 

[Physics lover]

Only that point?

I think i need to find a relation between x and y and then i can differentiate the equation to find minimum.Is there any other method?

 

[archaic]

You can always use the triangle inequality. You already worked out, I think, that to minimise the sum $|z -1| + |z-i|$ is equivalent to finding a point which minimises the sum of the distances to the points $(1,0)$ and $(0, 1)$.

Can you solve that?

It should be any point on the line passing by the two.

 

[PeroK]

It should be any point in between.

Yes, exactly, any point on the line segment between the two points in question. You can, of course, use the triangle inequality to prove that, if you wish.

Now, that leaves two more points to deal with.

 

[Mark44]

and then i can differentiate the equation to find minimum

You have posted the problem in the Precalculus section, which precludes the use of calculus techniques such as differentiation and integration. Should I move this thread to the Calculus & Beyond section?

 

[Physics lover]

You have posted the problem in the Precalculus section, which precludes the use of calculus techniques such as differentiation and integration. Should I move this thread to the Calculus & Beyond section?

Ofcourse,You can.

 

[PeroK]

You have posted the problem in the Precalculus section, which precludes the use of calculus techniques such as differentiation and integration. Should I move this thread to the Calculus & Beyond section?

I thought we were nearly there with some elementary geometry!

 

[aheight]

After Pero's suggestion, it falls out, after some algebra, without Calculus: The diagonal line $y=1-x$, minimizes the distance of one set, then minimize the distance from that set betwen $x+iy$ to $3+4i$

 

[PeroK]

After Pero's suggestion, it falls out, after some algebra, without Calculus: The diagonal line minimizes the distance of one set, then minimize the distance from that set betwen $x+iy$ to $3+4i$
View attachment 250945

That looks to me very like geometry, not algebra!

 

[aheight]

That looks to me very like geometry, not algebra!

Well, I did have to make two linear equations and then solve for their intersection (and I like making plots).

 

[aheight]

So should it be like this
I have to find a point from (0,1) and (1,0) and another loint from the other two mentioned points and then the final point will be on line joining these two points that i found.Is it correct?

Perhaps I can help with that: Find a point along the line from (0,1) to (1,0) such that the distance from that point to the origin plus the distance from that point to (3,4) is minimum. The Triangle Inequality requires the three points to be along a straight line.

Also, not wishing to cheat or anything, here's the code in Mathematica as a minimization problem. You can use it to check your work:

myVals = {x, y} /. 
  FindMinimum[{(Abs[z] + Abs[z - 1] + Abs[z - I] + 
        Abs[z - 3 - 4 I]) /. z -> x + I y, 
     Element[{x, y}, Disk[{0, 0}, 10]]}, {x, y}][[2]]

{0.4285714284248281`, 0.5714285709615166`}

 

[PeroK]

Perhaps I can help with that: Find a point along the line from (0,1) to (1,0) such that the distance from that point to the origin plus the distance from that point to (3,4) is minimum. The Triangle Inequality requires the three points to be along a straight line.

Also, not wishing to cheat or anything, here's the code in Mathematica as a minimization problem. You can use it to check your work:

myVals = {x, y} /.
  FindMinimum[{(Abs[z] + Abs[z - 1] + Abs[z - I] +
        Abs[z - 3 - 4 I]) /. z -> x + I y,
     Element[{x, y}, Disk[{0, 0}, 10]]}, {x, y}][[2]]

{0.4285714284248281`, 0.5714285709615166`}

If we go back to the original question we had to find the minimum value of the expression. We didn't actually need to find $z$.

 

[Physics lover]

If we go back to the original question we had to find the minimum value of the expression. We didn't actually need to find $z$.

i think i would have to check my calculation since i am getting the desired point to be (3/5,2/5)

 

[PeroK]

i think i would have to check my calculation since i am getting the desired point to be (3/5,2/5)

That's not correct. But, as I said, you don't actually need to find the point.

 

[WWGD]

Perhaps I can help with that: Find a point along the line from (0,1) to (1,0) such that the distance from that point to the origin plus the distance from that point to (3,4) is minimum. The Triangle Inequality requires the three points to be along a straight line.

Also, not wishing to cheat or anything, here's the code in Mathematica as a minimization problem. You can use it to check your work:

myVals = {x, y} /.
  FindMinimum[{(Abs[z] + Abs[z - 1] + Abs[z - I] +
        Abs[z - 3 - 4 I]) /. z -> x + I y,
     Element[{x, y}, Disk[{0, 0}, 10]]}, {x, y}][[2]]

{0.4285714284248281`, 0.5714285709615166`}

This seems to be the pair (3/7,4/7).

 

[Physics lover]

This seems to be the pair (3/7,4/7).

thanks i got it.

 

[Physics lover]

That's not correct. But, as I said, you don't actually need to find the point.

so how we will do it without finding the point.

 

[PeroK]

so how we will do it without finding the point.

You're only asked for the sum of the distances. You don't need to calculate the point for that.

 

[archaic]

Yes, exactly, any point on the line segment between the two points in question. You can, of course, use the triangle inequality to prove that, if you wish.

Now, that leaves two more points to deal with.

I think I got it. Any point on the line passing by $(0,1)$ and $(1,0)$ minimizes the distance between them, it is also the same for the origin and $(3,4)$, so $z$ that minimizes that equation should be the point of intersection of both lines.

 

[PeroK]

I think I got it. Any point on the line passing by $(0,1)$ and $(1,0)$ minimizes the distance between them, it is also the same for the origin and $(3,4)$, so $z$ that minimizes that equation should be the point of intersection of both lines.

Yes. And you know the sum of the distances in both cases, so you don't even need to calculate the point - as long as geometrically you can see it exists.