Optimizing Isosceles Triangle Problem: Find Min. Sum of Distances

[ciubba]

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 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 Their sum is my objective function, so D_t=(2-x)^2+(\sqrt{2}-y)^2 + (x-0)^2+(\sqrt{2}-y)^2 + (4-x)^2+(0-y)^2

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?

 

[Abtinnn]

Choose a coordinate system so that the perpendicular bisector becomes your y-axis. That would simplify things.

 

[ciubba]

Hmm, that's a good idea.

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

Edit: Still number crunching

 

[ciubba]

Okay, I get

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

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

D_t'=6 y-2 \sqrt{2}

Which has a root at y=\frac{\sqrt{2}}{3}

Unfortunately, that is the reciprocal of the book's answer. Where did I mess up?

 

[Abtinnn]

I think your coordinates for v2 are not correct. if the side length is 2sqrt(2), then v2 would be 2, right?

 

[Abtinnn]

Also, your values of D₁, D₂, and D₃ are the square of the distances, so you have to take the square root.

 

[ciubba]

Doing that gives me the right answer-- thanks!

 

[Abtinnn]

Any time man! :D