Balancing Problem - More Shapes!

Let the masses of circle (round), star and triangle be $r$ , $s$ and $t$ respectively. It is given:

$\begin{cases} 3s+r=3t & & \Rightarrow 3t = r + 3s &...(1) \\ 2t+s = 2r & \Rightarrow 2t = 2r - s & \Rightarrow 6t = 6r -3s &...(2) \end{cases}$

$\begin{array}{c}\Rightarrow Eq.1+Eq.2: & 9t = 4r & \Rightarrow t = \frac{7}{9}r \\ \quad \space Eq.2: & 2t=2r-s & \Rightarrow s = 2r-2t = \frac{4}{9}r \end{array}$

$\begin{cases} \text{Option 1: } 5t & = 5\times \frac{7}{9}r & = \frac{35}{9}r \\ \text{Option 2: } 3r+2s & = 3r + 2\times \frac{4}{9}r & = \frac{35}{9}r \\ \text{Option 3: } 7s & = 7\times \frac{4}{9}r & = \frac{28}{9}r \\ \text{Option 4: } 4t & = 4\times \frac{7}{9}r & = \frac{28}{9}r \\ \text{Option 5: } 3t+r & = 3\times \frac{7}{9}r + r & = \color{#D61F06} {\frac{30}{9}r} \end{cases}$

The option that will not balance any of the others is $\color{#D61F06}{\boxed{\text{3 triangles and 1 circle}}}$ .

$1T = 1$ Triangle

$1C = 1$ Circle

$1S = 1$ Star

We can write the following equations:

$3S + 1C = 3T$

$2T + 1S = 2C$

First, we can rewrite our first equation from above:

$3S + 1C = 3T \Leftrightarrow 1C = 3T - 3S$

And put it into the second equation:

$2T + 1S = 2C \Leftrightarrow 2T + 1S = 2(3T - 3S) \Leftrightarrow 2T + 1S = 6T - 6S$

Rewrite that again to:

$2T + 1S = 6T - 6S \Leftrightarrow 2T + 7S = 6T \Leftrightarrow \boxed{7S = 4T}$

Now, we can cancel the $7$ Stars option and the $4$ Triangle option from our list, since there is a solution that will balance them. Because the $7$ Stars option will be balanced by the $4$ Triangle option and the $4$ Triangle option will be balanced by the $7$ Stars option.

Now, let's see if the $5$ Triangles option can be balanced by one of the remaining options. To do so, let's rewrite our first and second equation from the beginning :

$3S + 1C = 3T \Leftrightarrow 3S + 1C + 2T = 5T$

$2T + 1S = 2C \Leftrightarrow 2T = 2C - 1S$

Now, put the second rewritten equation into the first rewritten equation to get:

$3S + 1C + 2T = 5T \Leftrightarrow 3S + 1C + (2C - 1S) = 5T$

Now, let's rewrite that again to get eventually:

$3S + 1C + (2C - 1S) = 5T \Leftrightarrow \boxed{2S + 3C = 5T}$

And we see that the $5$ Triangles option can be balanced by the $2$ Stars and $3$ Circles option and vice versa. And since we know that one option can't be balanced by any of the other, we know that our remaining option $3$ Triangles and $1$ Circles must be our desired solution, as we've showed that all other options can be balanced by at least one of the other options.

Star(S), Circle(C), Triangle(T)
Given: 3S+C=3T
2C=2T+S

replace C with (2T+S)/2 => 3S + (2T+S)/2 = 3T ----> 7S = 4T (therefore, option 4 and 5 are out)
rewrite option1: 3C + 2S in terms of T ---> (27/7)T + (8/7)T = 5T (option 1 and 2 are out)
remaining option is the 3rd option.