Leap Frog Solutions 2015
... If b = 1, then we can take a = 1 (a = 2 is not possible since m < 45), giving m = 15. But if b = 0, we can satisfy both inequalities with a = 3, giving m = 27. This is thus the largest possible such m. ...
... If b = 1, then we can take a = 1 (a = 2 is not possible since m < 45), giving m = 15. But if b = 0, we can satisfy both inequalities with a = 3, giving m = 27. This is thus the largest possible such m. ...
Answers
... 2. Since the three angles add to 180, 2b + b + 3b = 180 and b = 30. So 3b = 90 and we have a right triangle. 3. Trial and error shows the numbers are 45, 47, 49. 4. Since most of these cancel in pairs we just add 24 + 25 + 26 . . . + 31. These 8 integers average (24 + 31)/2 and so the total is 4 × 5 ...
... 2. Since the three angles add to 180, 2b + b + 3b = 180 and b = 30. So 3b = 90 and we have a right triangle. 3. Trial and error shows the numbers are 45, 47, 49. 4. Since most of these cancel in pairs we just add 24 + 25 + 26 . . . + 31. These 8 integers average (24 + 31)/2 and so the total is 4 × 5 ...
Weber problem
In geometry, the Weber problem, named after Alfred Weber, is one of the most famous problems in location theory. It requires finding a point in the plane that minimizes the sum of the transportation costs from this point to n destination points, where different destination points are associated with different costs per unit distance.The Weber problem generalizes the geometric median, which assumes transportation costs per unit distance are the same for all destination points, and the problem of computing the Fermat point, the geometric median of three points. For this reason it is sometimes called the Fermat–Weber problem, although the same name has also been used for the unweighted geometric median problem. The Weber problem is in turn generalized by the attraction–repulsion problem, which allows some of the costs to be negative, so that greater distance from some points is better.