This week’s Riddler Classic is simple to state, but tricky to figure out:
What whole number dimensions can rectangular prisms have so that their volume (in cubic units) is the same as their surface area (in square units)?
Here is my solution:
If the side lengths of the prism are $a,b,c$, then the volume being equal to the surface area implies that
abc = 2ab+2ac+2bc.
\]Let’s assume the side lengths are ordered, so $a \ge b \ge c$. We therefore have: $ab \ge ac \ge bc$. So $abc = 2ab+2ac+2bc \leq 6ab$. Simplifying, we obtain $c \leq 6$. We can now consider each case separately:
- If $c=6$, our equation reduces to $ab=3a+3b$. Rearranging and factoring, this is $(a-3)(b-3)=9$. Since $c=6$ was assumed to be the smallest side length, each of the factors $(a-3)$ and $(b-3)$ is at least $3$. So the only solution is $(6,6,6)$.
- If $c=5$, our equation reduces to $3ab=10a+10b \le 20a$. Therefore, $3b \le 20$ and so $b \le 6$. Since $c\le b$, this means we only have to check $b=5$ and $b=6$. Checking each case, we find the only solution is $(10,5,5)$.
- If $c=4$, our equation reduces to $ab=4a+4b$. Rearranging and factoring, this is $(a-4)(b-4)=16$. Each way of factoring $16$ as a product of two factors yields a different solution. These are: $16\times 1$, $8\times 2$, and $4\times 4$. The resulting $(a,b,c)$ triples are: $(20,5,4)$, $(12,6,4)$, $(8,8,4)$.
- If $c=3$, our equation reduces to $ab=6a+6b$. Rearranging and factoring, this is $(a-6)(b-6)=36$. Each way of factoring $36$ as a product of two factors yields a different solution. These are: $36\times 1$, $18\times 2$, $12\times 3$, $9\times 4$, $6\times 6$. The resulting triples are: $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.
- If $c=2$, our equation reduces to $0=4a+4b$, which has no solutions.
- If $c=1$, our equation reduces to $0=ab+2a+2b$, which also has no solutions.
And that’s it! Overall, the problem has exactly 10 solutions. They are:
$(6,6,6)$, $(10,5,5)$, $(20,5,4)$, $(12,6,4)$, $(8,8,4)$, $(42,7,3)$, $(24,8,3)$, $(18,9,3)$, $(15,10,3)$, $(12,12,3)$.
Note: I assumed that “whole number” in this context means a positive integer. If we allow for side lengths of zero, then any degenerate prism with side lengths $(a,0,0)$ will have both area and volume equal to zero, so there are infinitely many solutions.
Another note: If there were some way to upper-bound $a$, then we could do an exhaustive numerical search to find all solutions because there would only be finitely many triples $(a,b,c)$ to search. Alas, I have not been able to find an upper bound on $a$. I suspect it might not be possible, since if we set $a\to\infty$, the problem reduces to $bc=2b+2c$, which is the 2D version of the problem (a rectangle whose area equals its perimeter)!