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# geometry problem

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Five friends live on the same street. Their houses are at points A, B, C, D, and E, with the distances shown. The five friends decide to meet at point P so that the total walking distance for all five friends is minimized. What is AP?

Jun 13, 2022

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Here's my attempt at this one .....

Note

P = C  ......    total distances walked  = 9 + 4 + 2 + 5  =  20

P =  midpoint of    BC    .....total distances  walked  = 7 + 2 + 2 +  4 +  7 =  22

P = one unit to the right of B.....total distances walked  =  6 + 1 + 3 + 5 + 8  =  23

P = one unit to the left of C  ....total distances walked  = 8 + 3 + 1 + 3 + 6  = 21

P =  one unit to the right of C  .....total distances walked  = 10 + 5 + 1 + 1 + 4  = 21

Let P   be somewhere to the left of C

With a little help from WolframAlpha

Min

sqrt(9-x)2 +  sqrt (4-x)^2  +  sqrt (x^2) + sqrt (2 + x)^2 + sqrt (5 + x)^2   = 20 and   x = 0 ....P =  (9,0) ,  AP  = 9

Now....let P  be somewhere to the right of C

Min

sqrt (9 + x)^2  + sqrt  (4 + x)^2  + sqrt (x^2)  + sqrt (2 -x)^2 + sqrt (5 - x)^2   = 20  and  x = 0....P =(9,0), AP  = 9

So...it appears that the walking distances are minimized if all meet at C  =  P   Jun 13, 2022