- #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?