Box Mystery

According the previous question Box Constant , we learnt that the box's volume $V$ can be calculated from the arithmetic mean of face area $A$ and the harmonic mean of the dimension $F$ :

$V = F\cdot A$

In this case, $A = a^2$ , and $V = 9a^2$ . Thus, $F = 9$ , so we can setup a series of fractions for this equation:

$9 = \dfrac{3}{\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}}$

$\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{3}$

Also, we know that $V = abc = 9a^2$ , so $bc = 9a$ . That means there are $2$ possible scenarios:

$1.$ $b$ or $c$ is a multiple of $9$ .

$2.$ $b$ & $c$ are both multiples of $3$ .

Considering the first one, suppose $b = 9n$ for some integer $n$ . Then $nc = a$ .

Then $\dfrac{1}{nc} + \dfrac{1}{9n} + \dfrac{1}{c} = \dfrac{1}{3}$

$\dfrac{1}{nc} + \dfrac{1}{c} = \dfrac{1}{3} - \dfrac{1}{9n}$

$\dfrac{1+n}{nc} = \dfrac{3n-1}{9n}$

$9n(n+1) = nc(3n-1)$

$9(n+1) = c(3n-1)$

Since $3n - 1$ is not a multiple of $9$ , $3n - 1 | n+1$ . The only possible value is $n = 1$ , making $a = b = c = 9$ .

However, we know that this box is not a cube, so the latter scenario must apply, and considering the multiple of $3$ between integers $3$ and $9$ , only $6$ exists. Therefore, $6$ will be one of its dimension. For generality, suppose $b=6$ .

$\dfrac{1}{a} + \dfrac{1}{6} + \dfrac{1}{c} = \dfrac{1}{3}$

$\dfrac{1}{a} + \dfrac{1}{c} = \dfrac{1}{6}$

Now plugging the value into another equation, $6c = 9a$ . Thus, $c = \dfrac{3a}{2}$ .

$\dfrac{1}{a} + \dfrac{2}{3a} = \dfrac{1}{6}$

$\dfrac{5}{3a} = \dfrac{1}{6}$

$30 = 3a$

$a = 10$ .

Then $c = 15$ .

Checking the solutions, $V = abc = 6\cdot 10 \cdot 15 = 900$ . Then the original $a^2 = 100$ , which correlates with $a = 10$ .

As a result, $a + b + c = 6 + 10 + 15 = \boxed{31}$ .