Is it a Coincidence?

Geometry Level 3

If $A$ is the number of different ways that three regular polygons can be chosen such that their three interior angles add up to $360°$ (for example, the interior angles of one square and two regular octagons add up to $360°$ ), and if $B$ is the number of different cuboids with integer sides that exist with the same numerical volume and surface area (for example, $V = S = 250$ for a $5 \times 5 \times 10$ cuboid), then find $A - B$ .

The answer is 0.

2 solutions

Chris Lewis
Jun 12, 2019

The interior angle of an $n$ -gon is $\frac{180(n-2)}{n}$ . For an $a$ -gon, $b$ -gon and $c$ -gon to meet as needed,

$\frac{180(a-2)}{a}+\frac{180(b-2)}{b}+\frac{180(c-2)}{c}=360$

Rearranging,

$\frac{a-2}{a}+\frac{b-2}{b}+\frac{c-2}{c}=2\\ (a-2)bc+(b-2)ca+(c-2)ab=2abc\\ abc-2bc+abc-2ca+abc-2ab=2abc\\ abc=2(ab+bc+ca)$

But the left-hand side of this is the volume of a cuboid with dimensions $a,b,c$ , and the right-hand side is its surface area! So there is a one-to-one correspondence between the set of polygon triples and the set of cuboids; so $A-B=\boxed0$ .

EDIT: At least, nearly! See the comment below from @Russell Turner for an explanation of the special cases with $a=1$ or $a=2$ .

Incidentally, it's not hard to work out the value of $A$ and $B$ , with a bit of bounding on the smallest of $a,b,c$ and then some case analysis - but this isn't what the question asked for.

Chris Lewis - 1 year, 12 months ago

Bonus question: Prove that $A=B = 10$ .

Pi Han Goh - 1 year, 11 months ago

Rearranging gives c = 2ab / (ab - 2(a + b))

Assuming 3 <= a <= b <= c and doing the case analysis gives the 10 possible triples:

(3,7,42), (3,8,24), (3,9,18), (3,10,15), (3,12,12), (4,5,20), (4,6,12), (4,8,8), (5,5,10), (6,6,6)

Russell Turner - 1 year, 11 months ago

Excellent analysis, but it doesn't quite fully answer the question. Since a,b,c are all at least 3 for polygons, we have to eliminate the possibility that a relevant cuboid might exist with at least one side of length 1 or 2:

For a = 1, we have bc = 2(b + bc + c) => bc + 2(b + c) = 0, impossible for positive b,c

For a = 2, we have 2bc = 2(2b + bc + 2c) = 0 => 4(b + c) = 0, again impossible for positive b,c

Russell Turner - 1 year, 11 months ago

Oh whoops, you're right.

Pi Han Goh - 1 year, 11 months ago

This is a good point - some hand-waving arguments regarding the behaviour of $1$ - and $2$ -gons might be possible, but the clearest explanation is the one you've given. Thanks!

Chris Lewis - 1 year, 11 months ago

Russell Turner
Jun 21, 2019

I just assumed it was zero because of the title of the question ;)

Is it a Coincidence?

The answer is 0.

2 solutions

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