# Find All Dimensions such that Volume of Box = Surface Area

A rectangular prism has integer edge lengths. Find all dimensions such that its surface area equals its volume.

My Attempt at a Solution:

Let the edge lengths be represented by the variables $l, w, h$.

Then $$lwh = 2\,(lw +lh + wh) \implies lwh = 2lwh\left(\frac{1}{l} + \frac{1}{w} + \frac{1}{h}\right).$$

Dividing both sides of the equation by $lwh$ yields $$1 = 2\left(\frac{1}{l} + \frac{1}{w} + \frac{1}{h}\right)$$

Or, $$\frac{1}{l} + \frac{1}{w} + \frac{1}{h} = \frac{1}{2}$$

Though perhaps a bit unnecessary, I used some algebraic deduction and number theory to find all the possible ordered triple pairs for the dimensions of the rectangular prism in the cases where all the dimensions are the same and two of the dimensions are the same.

My answers were: $(6,6,6),(5,5,10),(8,8,4),(12,12,3)$

I have a hunch that no ordered pair exists where all three values are distinct, but is there a way to rigorously prove this?

Note: By the AM-GM Inequality, $lwh \geq 216$. (I haven't been able to make use of this fact, but I just noted it here in case)

Edit: After researching a little more about Egyptian Fraction Analysis, inspired by marty cohen's answer, I found from a Mathworld Wolfram page about Egyptian Fractions that a unit fraction can be (infinitely) split into two more unit fractions: $$\frac{1}{a} = \frac{1}{a+1} + \frac{1}{a(a+1)}.$$ I then used this idea to successfully generate a few ordered triplet solutions ($l,w,h$).

WLoG, assume $l \leq w \leq h$. Then the following ordered triplets ($l,w,h$) are solutions of the equation $\frac{1}{l} + \frac{1}{w} + \frac{1}{h} = \frac{1}{2}$.

$$(4,6,12), (3,7,42), (3,8,24)$$

But, I suspect there are many more solutions since you can mix and match the fractions that you decompose. This results in countless more combinations of ($l,w,h$), and I don't know how to use casework or an otherwise more organized approach to solve this problem. But, I do believe it can still be solved using this approach.

If anybody has any idea on how to solve the problem using the "splitting unit fraction" method, or has any other working solution, I would greatly appreciate it if you shared it with me.

Apologies for the long post and thanks for reading.

Since you have already determined all solutions where two or three of the variables are equal it remains to find all integer triples $(a,b,c)$ with $$a<b<c\quad{\rm and} \qquad{1\over a}+{1\over b}+{1\over c}={1\over2}\ .\tag{1}$$ According to your edit you suspect that there could be an infinity of solutions. But this is not the case.

From $(1)$ one immediately infers that ${1\over6}<{1\over a}<{1\over 2}$, which implies $3\leq a\leq 5$. We now treat these cases separately.

(i) When $a=3$ we have to solve $${1\over b}+{1\over c}={1\over 6}$$ with $b<c$, and this implies $7\leq b\leq11$. It follows that $$(b,c)\in\bigl\{(7,42), (8,24), (9,18), (10,15), (11,{\textstyle{5\over66}})\bigr\}\ .$$ This leads to the admissible triples $$(3,7,42), (3,8,24), (3,9,18), (3,10,15)\ .\tag{2}$$ (ii) When $a=4$ we have to solve $${1\over b}+{1\over c}={1\over 4}$$ with $b<c$, and this implies $5\leq b\leq7$. It follows that $$(b,c)\in\bigl\{(5,20), (6,12), (7,{\textstyle{3\over28}})\bigr\}\ .$$ This leads to the admissible triples $$(4,5,20), (4,6,12)\ .\tag{3}$$ (iii) When $a=5$ we have to solve $${1\over b}+{1\over c}={3\over 10}$$ with $b<c$. This implies ${1\over b}>{3\over20}$, or $b\leq6$, and the condition $a<b$ then enforces $b=6$. This would imply $c={15\over2}$, which is not admissble.

All in all there are the $6$ triples listed in $(2)$ and $(3)$.

• It's not required that a<b<c, it should be a <= b <= c. – RoadieRich Apr 2 '15 at 18:46
• @RoadieRich: See my edit (first two lines). – Christian Blatter Apr 2 '15 at 19:23
• Hello Christian :) Thanks for presenting such an excellent solution. I just wanted to clarify, though, I didn't mean to write that I suspected there were an infinite number of solutions. I just meant that any given unit fraction could be infinitely decomposed into smaller unit fractions. – A is for Ambition Apr 2 '15 at 21:20

Starting with your equation $\frac{1}{h} + \frac{1}{w} + \frac{1}{l} = \frac{1}{2}$, it looks to me like straightforward Egyption fraction analysis.

Assume that $h \le w \le l$.

First, $h \ge 3$ (or else the sum exceeds $\frac12$) and $h \le 6$ (or else the sum is less than $\frac12$).

For any of these values, $\frac1{w}+\frac1{l} =\frac12-\frac1{h}$.

To get a lower bound on $w$, $\frac1{w} <\frac12-\frac1{h}$ or $w >\frac{1}{\frac12-\frac1{h}} =\frac{2h}{h-2} =\frac{2h-4+4}{h-2} =2+\frac{4}{h-2}$ or $w \ge 3+\lfloor \frac{4}{h-2} \rfloor$.

To get an upper bound on $w$, $2\frac1{w} \ge\frac12-\frac1{h}$ or $w \le\frac{2}{\frac12-\frac1{h}} =\frac{4h}{h-2} =\frac{4h-8+8}{h-2} =4+\frac{8}{h-2}$ or $w \le 4+\lfloor \frac{8}{h-2} \rfloor$.

Here are these bounds (remembering that $w \ge h$):

$\begin{array}{lll} h & low & high\\ 3 & 6 & 12\\ 4 & 5 & 10\\ 5 & 5 & 6 \\ 6 & 6 & 6 \\ \end{array}$

For each possible $(h, w)$ pair, check if $l =\frac{1}{\frac12-\frac1{h}-\frac1{w}} =\frac{2hw}{hw-2w-2h} =\frac{2hw-4w-4h+4w+4h}{hw-2w-2h} =2+\frac{4w+4h}{hw-2w-2h}$ is an integer.

• hi marty, great solution and thanks for introducing me to Egyptian fraction analysis :) i just have one question, though, which is why you used the floor function to define both the lower and upper bounds. thanks! – A is for Ambition Mar 20 '15 at 21:53
• For the lower bound, if the expression ($4/(h-2)$) is an integer, then we want the next integer, since the inequality is strict. That's why I used floor + 1. – marty cohen Mar 20 '15 at 23:16

There is $(4,5,20)$.

I can't think of a general approach that doesn't involve brute force, however.

Here's a somewhat ad hoc method for generating solutions to $\frac1h+\frac1w+\frac1l=\frac12$. Find a number $n$ for which there are three factors, $a,b,c$ of $n$ that add to $\frac{n}2$. We then divide the equation $a+b+c=\frac{n}2$ by $n$ to get the desired form.

For example if $n=30$, we can take factors $2, 3, 10$. Then dividing $2+3+10=15$ by $30$ gives $\frac{1}{15}+\frac{1}{10}+\frac{1}{3}=\frac{1}{2}$. Thus a box of dimensions $15$ by $10$ by $3$ should work.