The figure shows some sets of equal weight among objects of four kinds: cylinders, spheres, cones, and cubes. In the last one, four cones are placed in the left pan. What is the smallest number of objects that can be put on the right pan to balance the cones?
Solution
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 \rightarrow 3b – 2c – s + 2 = 0
(b) 6s = 1 + c + b \rightarrow b + c – 6s + 1 = 0
(c) c + 1 = b + 2s \rightarrow b – c + 2s – 1 = 0
(a) + 2(b) \rightarrow 3b – 2c – s + 2 + 2b + 2c – 12s + 2 = 0 + 0 \rightarrow 5b – 13s + 4 = 0 (d)
(a) + 2(c) \rightarrow 3b – 2c – s + 2 + 2b – 2c + 4s – 2 = 0 + 0 \rightarrow b – 5s + 4 = 0 \rightarrow b = 5s – 4 (e)
Substitute (e) into (d):
5(5s – 4) – 13s + 4 = 0
25s – 20 –13s + 4 = 0
12s – 16 = 0
s = 16/12 = 4/3.
Then b = 5s – 4 = 5 \cdot 4/3 – 4 = 20/3 – 12/3 = 8/3.
Then, using (c), c + 1 = 8/3 + 2 \cdot 4/3 = 16/3 \rightarrow 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? It seems a shame to turn the pretty pictures into equations.