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Question 1149740: Six points are put on a segment 23 cm long (including the end points of the segment). This creates 15 different distances between the points, which are, in cm, 2, 3, 5, 6, 7, 8, 9, n, 13, 14, 15, 18, 20, 21, and 23. What is the value of n?
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Cool problem....!
I got some good mental exercise working on it.
There might well be methods for solving it that are much easier than what I ended up doing....
(1) One of the given lengths is equal to the full length of the segment; that means two of the six points are the endpoints of the segment. So there are four points on the segment between the two endpoints, dividing the segment into five pieces.
Picture the segment as a number line from 0 to 23, with two of the six points at A=0 and F=23. We need to find the locations of the other four points that will give 15 different distances between pairs of points -- the 14 given distances and the one unknown distance.
(2) The sum of the five shortest pieces has to be the length of the whole segment; so the five pieces have lengths 2, 3, 5, 6, and 7.
(3) In order to have two segments with lengths 2 and 21, one of the points has to be 2 units from one end of the segment. Without loss of generality, we can place one of the points at B=2 on our number line.
(4) In order to have two segments with lengths 3 and 20, one of the points has to be 3 units from one end of the segment. We can't put this point at 3 on our number line, because 2 and 3 are 1 unit apart, and there is no segment between two points with a length of 1. So we need to put a point at E=20 on our number line.
(5) We now have a number line from 0 to 23 with segment AB of length 2 at one end and segment EF of length 3 at the other end. We need to place two points C and D between 2 and 20 such that segments BC, CD, and DE have, in some order, lengths 5, 6, and 7, and in such a way that the distances between pairs of points are the given distances.
There are only 3!=6 different orders in which we can place the segments of lengths 5, 6, and 7 between 2 and 20, so it isn't much of a task to check to see which one is the right one.
(a) Arranging the five segments in the order 2-5-6-7-3 or 2-5-7-6-3 won't work, because there are no two points with a distance of 6+7+3=16 between them.
(b) The arrangements 2-6-5-7-3 and 2-7-5-6-3 do not work, because that would give two distances of 6+5=11 and 5+7=12 between pairs of points, but there is only one unknown distance between 9 and 13.
(c) The arrangement 2-6-7-5-3 doesn't work, because that would create distances of 2+6+7=15 and 7+5+3=15 between pairs of points, but the distances between pairs of points are all different.
(d) The only arrangement left is 2-7-6-5-3; and it produces the required distances between pairs of points, with the unknown distance being 11.
ANSWER: n = 11
The coordinates of the points on our number line are A=0, B=2, C=9, D=15, E=20, and F=23.
The distances between pairs of points are...
AB = 2
AC = 9
AD = 15
AE = 20
AF = 23
BC = 7
BD = 13
BE = 18
BF = 21
CD = 6
CE = 11
CF = 14
DE = 5
DF = 8
EF = 3
Those are the 14 given distances between pairs of points, with CE=11 being the distance we were to find.
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