[Guest]

99 athletes (consisting of three groups of 33 athletes) are lined up on a straight line. The first group of athletes are lined up so that one of them will be found in every 103 meters; the second group of athletes are lined up so that one of them will be found in every 107 meters, and the third group of athletes are lined up so that one of them will be found in every 109 meters. (If we set the starting point as 0m, the positions of the first six athletes would be as follows: 103m, 107m, 109m, 206m, 214m, 218m).

After your signal, these 99 athletes will meet on a meeting point you choose.

Where should the meeting point be, in order to minimize the total distance taken by all the athletes?

[HoustonHokie]

I'm thinking the answer should be the centroid or average of the starting points of all the athletes. The average is 1807.67 m, which means that the total distance covered by all athletes to arrive at the meeting point is 86881.33 m.

[Guest]

The average of the average positions for each group. So

((33*109 + 109)/2 + (33*107 + 107)/2 + (33*109 + 109)/2)/3 = 1807.67m

[Guest]

should be 1716 meters if first player is at 0 meters,

If first player is at 103 meter, then 1716+103 = 1819 can be the minimum. total distance to be covered = 86870 meters

Edited December 9, 2009 by deepan c

[Guest]

should be 1716 meters if first player is at 0 meters,

If first player is at 103 meter, then 1716+103 = 1819 can be the minimum. total distance to be covered = 86870 meters

As first 6 are given - nobody mentioned on distance 0, the meeting point should be on 1819 m

Edited December 9, 2009 by Hugo

[Guest]

If the closest person is at 103m, and the farthest is at 109 x 33 = 3597m, the median of these two is 1747m.

If the closest person is at 103m, and the farthest is at 109 x 33 = 3597m, the median of these two is 1747m.

[Guest]

=((3399-103)/2+(3531-107)/2+(3597-109)/2)/3=1701.33

[Guest]

The Earlier post is wrong. This should be correct

=((3399+103)/2+(3531+107)/2+(3597+109)/2)/3=1807.66

Edited December 9, 2009 by Tonybz

[Guest]

The position we want is the position where there are the same number of runners to the left as there are to the right. With 99 runners, this must be directly at the 50th runner, who will have 49 runners on either side of him. Assuming the 1st runner is on the far left and the last runner is on the far right, pick a point somewhere in the middle of the 49th and 50th runner (a stretch of 68m). If I move this point 1m to the left, 49 runners will have to run 1m less and 50 runners will have to run 1m more for an increase in the total distance of 1m. This trend continues in a linear fashion until I am between 48 and 49 where moving 1m left increases the total distance by 2m. Between 47 and 48 makes it +3m and so on. But, if I move to the right while between 49 and 50 the total distance decreases instead. But once I pass the 50th runner and am in between 50 and 51, going right means 50 people go further while only 49 have less distance to go, causing the total distance to increase again. So we reach a minimum at the position of the 50th runner (1819m)

[superprismatic]

There seems to be disagreement about the answer. So,

The point we are looking for is the point x which minimizes

F(x)=Sum{i from 1 to 33}[(x-103i)²+(x-107i)²+(x-109i)²].

Here, I summed the squares of the distances of all the

athletes to the point x. This is equivalent because the

sum of distances is minimized at the same place that the

sum of squared distances is. The advantage of using squared

distances is that we don't have to use the awkward absolute

value function. The function F is defined for all real x,

so a minimum is obtained where the slope of the tangent to F

is 0. That is, the derivative of F with respect to x must

be 0 at that point. So, taking the derivative of F with

respect to x and setting it equal to 0, we can solve for

x. Doing this we get x=1807.666....

[Guest]

This is equivalent because the sum of distances is minimized at the same place that the sum of squared distances is.

Take a simplified case where there are runners only at 0, 5 and 7.

If the meeting place is at 5: Sum of distances is 5+0+2=7. Sum of squared distances is 5²+0²+2²=31

If the meeting place is at 4: Sum of distances is 4+1+3=8. Sum of squared distances is 4²+1²+3²=26

I still stand by the answer originally given by Deepan/Hugo.

Edited December 11, 2009 by Tuckleton

[superprismatic]

Take a simplified case where there are runners only at 0, 5 and 7.

If the meeting place is at 5: Sum of distances is 5+0+2=7. Sum of squared distances is 5²+0²+2²=31

If the meeting place is at 4: Sum of distances is 4+1+3=8. Sum of squared distances is 4²+1²+3²=26

I still stand by the answer originally given by Deepan/Hugo.

Yes, I made a pretty big mistake when I tried to get around the problem of the absolute value function being non-differentiable. Thanks for pointing out my error. I didn't read the explanation in your first post carefully enough.