Lost in a Square

[Anup183]

Mahesh is lost (at point M) in a square forest ABCD.
His friends Abhay, Brijesh and Elena are respectively located at A, B and E, point E being the midpoint of side DC.
If the side of the forest is 2 km, what is the minimum value of the
sum of the distances between
Mahesh and his three friends?
Note: the figure shows approximate position of Mahesh.

 

[Deleted member 4993]

Mahesh is lost (at point M) in a square forest ABCD.
His friends Abhay, Brijesh and Elena are respectively located at A, B and E, point E being the midpoint of side DC.
If the side of the forest is 2 km, what is the minimum value of the
sum of the distances between
Mahesh and his three friends?
Note: the figure shows approximate position of Mahesh.
View attachment 30538

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[lev888]

Mahesh is lost (at point M) in a square forest ABCD.
His friends Abhay, Brijesh and Elena are respectively located at A, B and E, point E being the midpoint of side DC.
If the side of the forest is 2 km, what is the minimum value of the
sum of the distances between
Mahesh and his three friends?
Note: the figure shows approximate position of Mahesh.
View attachment 30538

What topic are you studying? Did you go through any example problems?

 

[Steven G]

Put the square on an x-y coordinate system.
Use A= (0,0), B=(2,0), C = (2,2), E= (1,2) and M = (x,y).
Now find the required distances using the distance formula and see if you can figure out what the minimum distance would be.

 

[Dr.Peterson]

The direct approach seems likely to be quite difficult; at the least, I would probably try a symmetrical coordinate system like this:

​

I'd also take symmetry into account to simplify the work.

But we don't know whether you are expected to use multivariable calculus, or something else. (You called this algebra, but it could also be geometry.) I happen to know a theorem about the minimal sum of distances to vertices of any triangle that makes this easy, if you know it. I don't recall offhand how to prove it geometrically, but it may come to me.

 

[Steven G]

Dr P,
What are the numbers 1.38, 1.75 and 0.87 supposed to represent?
If they represent the length of the line segments then you can only get these values if you know where M is located.

 

[Dr.Peterson]

Dr P,
What are the numbers 1.38, 1.75 and 0.87 supposed to represent?
If they represent the length of the line segments then you can only get these values if you know where M is located.

I just picked a point randomly to look like the given figure. Those are the actual measurements for that point.

Don't worry, I know the correct answer, and that is not it. It's nowhere near that point.

What I'm struggling with is how to efficiently get the answer without either using GeoGebra and moving the point around, or using the theorem I know but have forgotten how to prove. (I probably first learned it in some discussion of soap bubbles, if that's a useful hint.)

 

[Dr.Peterson]

I'll add some hints.

First, if you want to know what theorem I used, search for "Fermat point". The proof in general is complicated; but we are applying it to a special case, the isosceles triangle ABE in my image. The proof in that case is relatively simple.

On the other hand, symmetry, as I mentioned, reduces the complexity tremendously, no matter what method you use. Can you see why, in my drawing, M must lie on the y-axis? One approach focuses on the locus of M for a given sum AM + BM.