Fixing a Point - Futility Closet

A problem proposed by Richard Hoshino and Sarah McCurdy for Crux Mathematicorum, September 2008:

Five points lie on a line. Here are the 10 distances between pairs of points, listed from smallest to largest:

2, 4, 5, 7, 8, k, 13, 15, 17, 19

What’s k?

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This solution is by Gustavo Krimker of Universidad CAECE, Buenos Aires. The distance between the first and fifth points is 19. That means that 19 is also the sum of each of three pairs of intermediate distances: (the distance from the first point to the second point) + (the distance from the second point to the fifth point) = 19 (the distance from the first point to the third point) + (the distance from the third point to the fifth point) = 19 (the distance from the first point to the fourth point) + (the distance from the fourth point to the fifth point) = 19 Looking over the list of distances we’ve been given, the only pairs that add to 19 are (2, 17) and (4, 15). So those pairs correspond to two of the bulleted sums above. What’s the third pair? It must be (7, k) or (8, k). (It can’t be (5, k) or (k, 13) because we know that k falls between 8 and 13 and so those pairs can’t be made to total 19.) And if the third case is (7, k) or (8, k), then k must be 11 or 12. Now, 17 is the second-greatest distance on the list; that means it’s the distance between one of the end points in the row of five and the penultimate point on the opposite side. Either way, that means that 17 is the sum of two pairs of intermediate distances, using the same reasoning as above. One of those pairs is (4, 13), and the other must be (5, k), (7, k), or (8, k). That means that k must be 12, 10, or 9. The only one of these that matches a candidate above is 12. (Ian VanderBurgh, “Mathematical Mayhem,” Crux Mathematicorum 34:5 [September 2008], 275-286.)

This solution is by Gustavo Krimker of Universidad CAECE, Buenos Aires.

The distance between the first and fifth points is 19. That means that 19 is also the sum of each of three pairs of intermediate distances:

(the distance from the first point to the second point) + (the distance from the second point to the fifth point) = 19
(the distance from the first point to the third point) + (the distance from the third point to the fifth point) = 19
(the distance from the first point to the fourth point) + (the distance from the fourth point to the fifth point) = 19

Looking over the list of distances we’ve been given, the only pairs that add to 19 are (2, 17) and (4, 15). So those pairs correspond to two of the bulleted sums above. What’s the third pair? It must be (7, k) or (8, k). (It can’t be (5, k) or (k, 13) because we know that k falls between 8 and 13 and so those pairs can’t be made to total 19.) And if the third case is (7, k) or (8, k), then k must be 11 or 12.

Now, 17 is the second-greatest distance on the list; that means it’s the distance between one of the end points in the row of five and the penultimate point on the opposite side. Either way, that means that 17 is the sum of two pairs of intermediate distances, using the same reasoning as above. One of those pairs is (4, 13), and the other must be (5, k), (7, k), or (8, k). That means that k must be 12, 10, or 9. The only one of these that matches a candidate above is 12.

(Ian VanderBurgh, “Mathematical Mayhem,” Crux Mathematicorum 34:5 [September 2008], 275-286.)