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 Distance squared (Posted on 2018-11-24)
P is a point such that PP12 + PP22 + PP32 + PP42 + PP52 = k where Pr is a point with coordinates (r, r2).

Find the least value of k for which such a point P exists.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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Find P = (x,y) to minimize k by setting dk/dx and dk/dy to 0.

All sums are over i=1,5
(x_i,y_i) = (1,1), (2,4), (3,9), (4,16), (5,25)

k(x,y) = sum[ (x-x_i)^2 + (y-y_i)^2]
= 5 x^2 +5 y^2 + sum(x_i^2) + sum(y_i^2) - 2 x sum(x_i) - 2 y sum(y_i)

0 = dk/dx = 10x - 2 sum(x_i) --> x = sum(x_i)/5 = 3
0 = dk/dy = 10y - 2 sum(y_i) --> y = sum(y_i)/5 = 11
P=(3,11)
k = 384

(moral of the story: As formulated, x and y are completely decoupled and can be optimized independently. Also - the best x and y are the average x and y values of the points - as we have just proven.)

Edited on November 24, 2018, 8:57 pm
 Posted by Steven Lord on 2018-11-24 14:48:47

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