|Mar12-09, 10:30 PM||#1|
Minimum distance in within a quadrilateral
1. The problem statement, all variables and given/known data
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.
2. Relevant 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.
3. 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.
|Mar12-09, 10:53 PM||#2|
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.
|distance, euclidean, geometry, quadrilateral|
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