The problem is as follows:

The figure from below represents two scales, one is not in equilibrium while the other is at equilibrium. Assume both identical objects have the same weight and those with different shapes have different weights and each object weights an integer amount of pounds. With all this information, what would it be the least amount of the orange sphere?.

Sketch of the problem

The choices given are:

$\begin{array}{ll} 1.&\textrm{3 pounds}\\ 2.&\textrm{2 pounds}\\ 3.&\textrm{1 pounds}\\ 4.&\textrm{4 pounds}\\ \end{array}$

From looking at the figure what I've attempted to do was as follows:

I'm using the notation:

$\textrm{Circle = C}$

$\textrm{Triangle = T}$

$\textrm{Square = S}$

The first figure shows:


The second figure shows:


Hence it can be inferred that:







Thus the triangle weighs more than the square.

But from only these it is not possible to know an exact weight for each of these figures. How can I assume which weight goes to which?

If I assume the square weighs $1$ pound, then triangle can be $2$, $3,$4$, etc.

Because the weights of the square and circle must account for two triangles and the weight of all objects is an integer this means that either the square and circle are both odds or both evens.

We don't know which is more heavier than the other so I'm setting that the sphere might be heavier.

Since I set the square to be $1$ pound, then the circle must be $3$, $5$, $7$, etc.

In order to reassure the least possible weight for that sphere, I shall assume that it is ligther in weight than the square.

Thus I rearrange my initial analysis and get to:

The square is $3$ pounds, the circle is $1$ pound, but triangle would be $2$ pounds, this is a contradiction as the triangle weight is more than the square.

$s=2$, $c=2$, $t=2$, also yields a contradiction.

I can't go up for the square as it will remain in contradiction. Thus the only logical choice would be what I originally set. $s=1$, $c=3$, $t=2$.

Therefore the minimum weight that it can have the sphere is three pounds or choice 1.

Is there any way to confirm this?. Is my logic correct?.


2 Answers 2


Here is a concrete way to think. $T$ can be identified as arithmetic mean of $C,S$. As they are distinct, we have two possibilities : $C<T<S$ or $C>T>S$.

From first diagram, $2C > T+S$.

First case : $C<T, C<S \Rightarrow 2C < T+S$. Second case : $C>T, C>S \Rightarrow 2C > T+S$. Thus $C>T>S$ is true.

For minimum, let $S=1, C=3$ so that $T=(S+C)/2=2$ is an integer.

  • $\begingroup$ It is not yet very clear to my why have you discarded the possibility of $C>T>S$? From where did you concluded that?. Does this stems from that the definition of arithmetic mean is that $T$ must lie in between $C$ and $S$, thus this is the answer?. I'm sorry to be slow at this but were you thinking this?. I don't know how to justify that because $2C>T+S$ makes $C>T>S$ to be true, can you state why?. This part is where I'm stuck. Why it cannot be otherwise, I mean the other inequation?. $\endgroup$ Mar 12, 2021 at 1:53
  • $\begingroup$ @ChrisSteinbeckBell See the edit. $\endgroup$
    – cosmo5
    Mar 12, 2021 at 6:51

You have already found that $$2c > s+t\quad (1)$$

Now, since you are interested in the weight of $c$, it might be better to search for a relation containing $c$ instead of removing it by using the equation

$$2t = c+s \iff t =\frac{c+s}{2}\quad (2)$$

Plugging $(2)$ into $(1)$ and rearranging gives

$$c >s$$

So, now you are almost done. Since $s$ is natural: $c > s \geq 1\Rightarrow c\geq 2$. But $t=\frac{c+s}{2}$ is also a natural number, hence $\boxed{c\geq 3}$ while choosing $s=1$ also to be minimal.


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