Optimizing Isosceles Triangle Problem: Find Min. Sum of Distances

In summary, the conversation involved finding the location of point P on the perpendicular bisector of an isosceles triangle's base that minimizes the sum of distances between P and the three vertices. After putting the triangle on a Cartesian coordinate system and using similar triangles, the objective function was derived as D_t=(\sqrt{2}-y)^2+8+y^2, with the derivative D_t'=6y-2\sqrt{2}. The correct answer was found by taking the square root of the distances and using the correct coordinates for the triangle's vertices.
  • #1
ciubba
65
2

Homework Statement


An isosceles triangle has a base of length 4 and two sides of length 2sqrt(2). Let P be a point on the perpendicular bisector of the base. Find the location P that minimizes the sum of the distances between P and the three vertices.

Homework Equations


N/A

The Attempt at a Solution



Putting this on the cartesian coordinate system leaves me with one vertex, v1, at (0,0), v2 at (2,sqrt(2)) and v3 at (4,0).

The distance between the vertices and P would then be [tex]D_1=(2-x)^2+(\sqrt{2}-y)^2 \; \; D_2=(x-0)^2+(\sqrt{2}-y)^2 \; \; D_3=(4-x)^2+(0-y)^2[/tex] Their sum is my objective function, so [tex]D_t=(2-x)^2+(\sqrt{2}-y)^2 + (x-0)^2+(\sqrt{2}-y)^2 + (4-x)^2+(0-y)^2[/tex]

I'm assuming that I can come up with a constraint by similar triangles, but this seems like an incredibly obtuse way of solving this problem. Could someone point me to a better direction?
 
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  • #2
Choose a coordinate system so that the perpendicular bisector becomes your y-axis. That would simplify things.
 
  • #3
Hmm, that's a good idea.

So, v1=(-2,0) , v2=(0,sqrt(2)) , v3=(2,0)

Edit: Still number crunching
 
Last edited:
  • #4
Okay, I get

[tex] D_1=4+y^2, \; D_2=(\sqrt{2}-y)^2, \; D_3=4+y^2[/tex]

Thus, the objective function, their sum, is [tex]D_t=(\sqrt{2}-y)^2+8+y^2[/tex]

[tex]D_t'=6 y-2 \sqrt{2}[/tex]

Which has a root at [tex]y=\frac{\sqrt{2}}{3}[/tex]

Unfortunately, that is the reciprocal of the book's answer. Where did I mess up?
 
  • #5
I think your coordinates for v2 are not correct. if the side length is 2sqrt(2), then v2 would be 2, right?
 
  • #6
Also, your values of D1, D2, and D3 are the square of the distances, so you have to take the square root.
 
  • #7
Doing that gives me the right answer-- thanks!
 
  • #8
Any time man! :D
 

Related to Optimizing Isosceles Triangle Problem: Find Min. Sum of Distances

What is the "Optimizing Isosceles Triangle Problem"?

The Optimizing Isosceles Triangle Problem is a mathematical problem that involves finding the minimum sum of distances between three points on a plane, where two of the points are fixed and the third point is free to move. The goal is to find the optimal position for the third point that minimizes the total distance between all three points.

What is the significance of this problem?

This problem has real-world applications in fields such as transportation, logistics, and network design. It can also serve as a basis for more complex optimization problems.

What are the steps involved in solving this problem?

The steps involved in solving the Optimizing Isosceles Triangle Problem include defining the problem, setting up the mathematical equations, finding the derivatives of the equations, and using optimization techniques such as setting the derivatives to zero and solving for the optimal point.

What are some common techniques used to solve this problem?

Some common techniques used to solve this problem include calculus-based methods such as Lagrange multipliers, geometric methods such as the Pythagorean theorem, and algebraic methods such as substitution and elimination.

What are some tips for finding the optimal solution to this problem?

Tips for finding the optimal solution to this problem include exploring various techniques, double-checking the calculations, and breaking down the problem into smaller, more manageable parts. It is also helpful to have a strong understanding of geometry and calculus.

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