The picture shows some equalities of weight among objects of four kinds: cylinders, spheres, cones, and cubes. At the bottom, four cones are placed in the left pan. What is the smallest number of objects we can put on the right pan to balance the cones?
One way that will work is to make equations. Let c be the weight of a cylinder, b be the weight of a cube, and s be the weight of a sphere. We can arbitrarily let the weight of a cone be 1, since we can find all of the relative weights in terms of a cone. Now we have three equations in three variables:
| (a) | 2c + s = 3b + 2 | → | 3b – 2c – s + 2 = 0 |
| (b) | 6s = 1 + c + b | → | b + c – 6s + 1 = 0 |
| (c) | c + 1 = b + 2s | → | b – c + 2s – 1 = 0 |
| (a) | 3b – 2c – s + 2 = 0 |
| 2(b) | 2b + 2c – 12s + 2 = 0 |
| (d) | 5b – 13s + 4 = 0 |
| (a) | 3b – 2c – s + 2 = 0 |
| 2(c) | 2b – 2c + 4s – 2 = 0 |
| (e) | b – 5s + 4 = 0 → b = 5s – 4 |
Substitute (e) into (d):
5(5s – 4) – 13s + 4 = 0
25s – 20 -13s + 4 = 0
12s – 16 = 0
s = 16/12 = 4/3.
25s – 20 -13s + 4 = 0
12s – 16 = 0
s = 16/12 = 4/3.
Then b = 5s – 4 = 5 · 4/3 – 4 = 20/3 – 12/3 = 8/3.
Then, using (c), c + 1 = 8/3 + 2 · 4/3 = 16/3 → c = 13/3.
Thus the relative weights are:
cylinder = 13/3, cube = 8/3, sphere = 4/3, and cone = 1.
We need to balance 4 cones. That is, we need to balance a weight of 4. So it’s easy:
cube + sphere = 8/3 + 4/3 = 12/3 = 4.
We can do it with just two blocks (a cube and a sphere).
Query: Is there a snazzy way to do this visually?