Minimum distance in within a quadrilateral

  • Thread starter Thread starter bronxbombas
  • Start date Start date
  • Tags Tags
    Minimum
bronxbombas
Messages
9
Reaction score
0

Homework Statement


Let A, B, C, D be the vertices of a convex quadrilateral. Convexity means that for each lines L(ab), L(bc), L(cd), L(da) the quadrilateral lies in one of its half-planes. Find the point P for which the minimum Min(d(P,A)+d(P,B)+d(P,C)+d(P,D)) is realized.


Homework Equations


Min(d(P,A)+d(P,B)+d(P,C)+d(P,D)) is the equation we're trying to minimize.
distance=d(X,Y)=abs(X-Y)=sqrt((X-Y)x(X-Y)) where "x" is the dot product.

The Attempt at a Solution


For starters, this is for my Euclidean geometry class, so there's no coordinates or Calculus, I presume. My initial guess is that the point that would minimize those distances would be the intersection of the diagonals but I can't figure out why.
 
Physics news on Phys.org
I think you are probably right about the intersection of the diagonals. I'm not sure of a mathematical way to prove this, but I believe you can show that moving away from that point increase the sum of the lengths of those four lines. I can think of a few examples.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top