1255.11 – Scales

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 cc be the weight of a cylinder, bb be the weight of a cube, and ss 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:

2c+s=3b+26s=1+c+bc+1=b+2s\begin{align*}2c + s &= 3b + 2 \\6s &= 1 + c + b \\c + 1 &= b + 2s\end{align*}

Moving all the terms to one side we get the following:

0=3b−2c−s+2=b+c−6s+1=b−c+2s−1\begin{align*}0 &= 3b - 2c - s + 2 \\&= b + c - 6s + 1 \\&= b - c + 2s - 1\end{align*}

We can add linear combinations of these expressions to each other and know that the resulting sum will be 0.

0=3b−2c−s+2+2(b+c−6s+1)=3b−2c−s+2+2b+2c−12s+2=5b−13s+4\begin{align*}0 &= 3b - 2c - s + 2 + 2(b + c - 6s + 1) \\&= 3b - 2c - s + 2 + 2b + 2c - 12s + 2 \\&= 5b - 13s + 4\end{align*}

0=3b−2c−s+2+2(b−c+2s−1)=3b−2c−s+2+2b−2c+4s−2=b−5s+4\begin{align*}0 &= 3b - 2c - s + 2 + 2(b - c + 2s - 1) \\&= 3b - 2c - s + 2 + 2b - 2c + 4s - 2 \\&= b - 5s + 4\end{align*}

So b=5s−4b = 5s - 4. We can substitute this into the earlier equation:

0=5(5s−4)−13s+4=25s−20−13s+4=12s−16\begin{align*}0 &= 5(5s - 4) - 13s + 4 \\&= 25s - 20 - 13s + 4 \\&= 12s - 16\end{align*}

So s=1612=43s = \dfrac{16}{12} = \dfrac{4}{3}. Then b=5(43)−4=203−123=83b = 5\left(\dfrac{4}{3}\right) - 4 = \dfrac{20}{3} - \dfrac{12}{3} = \dfrac{8}{3}.

Finally, we can use one of our original equations to solve for cc.

c+1=83+2(43)=163c=133\begin{align*}c + 1 &= \dfrac{8}{3} + 2 \left(\dfrac{4}{3}\right) \\[1em]&= \dfrac{16}{3} \\[1em]c &= \dfrac{13}{3}\end{align*}

Thus the relative weights are:

  • cylinder = 133\dfrac{13}{3}

  • cube = 83\dfrac{8}{3}

  • sphere = 43\dfrac{4}{3}

  • cone = 11

We need to balance 4 cones. That is, we need to balance a weight of 4. So it's easy:

cube + sphere = 83+43=123=4\dfrac{8}{3} + \dfrac{4}{3} = \dfrac{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.