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?
+9519 Actually, we can ignore the walking distances of the people living at A and E, because their walking distance, no matter where the point is, must add up to 14. Also, it is not optimal to place the point P between A and B or between D and E, because either the person living at B or the person living at D would have to walk a whole lot. We conclude that the optimal location of the point P must be somewhere between B and D. Using the same logic, the walking distances of the people living at B and D can actually be ignored because in that case, it would always add up to 6. We only need to care about how much the person who lives in C needs to walk. Therefore, the optimal strategy is to place the point P at C (i.e., all the friends meet at C). Then AP is just AC, which is 9.
