Spring 1998
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Spring 1998 |
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Four problems draw a bead on mathematical minds
The two teachers that pushed me the hardest, and were therefore the best, were Clarence Sousley at Rose Poly in 1943 and Michael Golomb ten years later at Purdue. Professor Golomb is still mathematically active and the problems below are adapted from one that he posed in a recent issue of The College Mathematics Journal. You learned about center of mass in calculus, physics and mechanics and I’m sure you were awake in at least one of these courses. The center of gravity (in one and two dimensions) is the position of a pivot point about which an object will balance. Problem 1. A line has three beads at one end and two beads at the other. If all the beads are identical, find the center of gravity. Neglect the mass of the line. We now move gently to two dimensions and recall that any knife edge on which a system balances will pass through its center of gravity. In the following problems you are to place identical beads on the verticies of regular (equal sides and angles) polygons so that the center of gravity (CG) will be at the center of the polygon (CP). For example, if you place two beads at each vertex of a hexagon, the center of gravity will be at the center of the hexagon. Note that when the number of vertices is even, then each vertex will have another vertex directly opposite (180 degrees between them). Problem 2. Find a bead placement on the vertices of a regular octagon such that CG=CP and such that all vertices do not have the same number of beads. Problem 3. Find a bead placement on a regular hexagon such that CG=CP and such that opposite vertices do not have the same number of beads. Try to do this for an octagon. What related general result might be true? Problem 4. If the number of vertices is odd, then the situation is different. For example if the polygon is an equilateral triangle and if CG=CP, then we must place the same number of beads on each vertex of the triangle. Find a bead placement on a regular nonagon (nine sides) such that CG=CP and such that all vertices do not have the same number of beads. What related general result might be true? A number of solvers did not find all 75 possible finishes in the four horse race, but that’s OK since I also missed it on my first try. Joe Moser found a solution to this problem for the general case with n horses in the race. The n compartment bubble chamber has a maximum height of sqrt(1+n)/2 Send your solutions and or requests for previous solutions to Herb Bailey, Rose-Hulman, 5500 Wabash Ave., Terre Haute, IN, 47803 or Herb.Bailey@rose-hulman.edu. Alumni solvers (listed by class year) of problems in the winter issue were:
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