Find a rectangular prism where the edges are whole numbers, and the surface area is equal to the volume. What are the edge lengths for this rectangular prism?
Extra Challenge: There are only ten possible answers to this question. For big dick internet points, name all ten.
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"rectangular prism". you mean cuboid.
Both terms are acceptable, and one is not privileged over the other. Wikipedia is not the end-all be-all of knowledge.
t. not the OP
Getting philosophical: even given the absence of physical units of measure, can one /really/ equate area and volume? (Do not say: magnitudes are equal. It is still the case that distinct things are being measured.)
Here fag. It's the 7:4 ratio
Here's the seed string, 714285
Pic related
Leaving the above unit-autism aside and "just going with it", the OP clearly prompts us:
Find solutions to the diophantine equation
[math] abc = 2(ab+ac+bc) [/math]
where a, b and c are the edge lengths of the appropriate rectangular prism, or cuboid, or box, or...
Clearly a trivial solution is a=b=c=0, or (0,0,0), say-one can decide for one's self whether this "counts" or not. One may decide for one's self whether But now consider the general situation where a=b=c. Then we have the slightly non-trivial result the the cubic edge length of 6 fits the bill, producing both volume and surface area of 216, or 6x6x6, amusingly. As an aside, it also happens that 216 is a triangular number, the sum from 1 through 36.
[math] A = 6a^2 = a^3 \rightarrow a=b=c=6 [/math]
Notice that cancellation of the a^2 term in this case to complete the argument assumes nonzero a, while the case with zero a is found by simple inspection.
Implicitly the OP refers to natural number solutions, but he does not expressly specify same, so that for completeness one might speak of structures embedded in R^3 with appropriate "negative" components. Notice that such animals are precluded in the above formulation: the LHS is manifestly positive, while the RHS is manifestly negative.
The next fruitful line of attack, then, is to consider the case where two edges are equal while the remaining edge is unequal to them, just plodding through it (sage)...
6,6,6
4,8,8
8,4,8
8,8,4
5,5,10
5,10,5
10,5,5
3,10,15
10,3,15
3,15,10
10,15,3
15,3,10
15,10,3
Those are more than 10 solutions already.
I think that's only 4 solutions
1+3+3+6 = 13 =/= 4
If one distinguishes, when it's all done, between the various (a,b,c) components. Real mathematicians wouldn't bother about this bookkeeping in an effort to find "really distinct solutions", except as a formality, though you're clearly also on the right track.
Add to these the situations (12,12,3) and even (-4,-4,1), again for completeness of treatment...
well
1+3+3+6 = 13 =/= 10
either
A = a/c
B = b/c
ABc^3 = 2ABc^2+2Ac^2+2Bc^2
ABc = 2AB+2A+2B
c = 2+2/B+2/A
a = 2A+2A/B+2
b = 2B+2B/A+2
For the above terms to all be integers, A =[2,1]/[2,1]
If A = 1
c = 4+2/B
a = 4+2/B
b = 4B+2
B = [2,1]/[4,2,1]
8,8,4
6,6,6
5,5,10
12,12,3
If A = 2
c = 3+2/B
a = 6+4/B
b = 3B+2
B = [2,1]/[3,1]
6,12,4
9,18,3
c = 2+2/B+2/A
a = 2A+2A/B+2
b = 2B+2B/A+2
The rest are where 2/A is not an integer
xyz = 2(xy + yz + xz)
Obvious solutions:
if x = y = z = 0, then it's true.
If x = y = z, then you have x^3 = 6x^2 or x = 6.
so those are two solutions. for the rest, use lagrange multipliers.
Now, The next thing is to equate two edges while holding the third unequal. We may as well say that
[math] a = b \rightarrow V = a^{2}c \;\; ; \;\; A = 2(a^2 + 2ac) [/math]
so that, again assuming nonzero a=b,
[math] a(c-2) = 4c [/math]
We thus arrive at a very useful and straightforward form where a little bruteforcing makes total sense given the circumstances. Inspecting c = 0-10 with regard to this form yields the following, in order:
0 a=0
1 a=-4
2 0=8
3 a=12
4 a=8
5 a not integral
6 a=6
7 a not integral
8 a not integral
9 a not integral
10 a = 5
The above information merely reproduces the trivial zero case, the degenerate case involving negatives, a contradiction, a previously unmentioned case, a case now mentioned twice, various contradictions which require that a is not an integer, much less natural, a reproduction of the cubic case, and finally the (5,5,10) case already mentioned.
After this little jaunt, brute force stops making sense, and one should like to generally describe cases. Clearly situations leading to the case when a is not an integer are "out" yet it seems that pushing this a bit further should lead to most of what the OP has advertised as the ten "really distinct" solutions (the above poster distinguishes between cases, so for him there is some larger, equivalent combinatorial count, assuming the OP's claim is correct).
Still, playing with (varying) a would still seem to make sense, and does yield a few degenerate cases with negative components worth mentioning, in addition to contradictions and analogous reiterations of the above.
So far I count these, being trivial (zero), degenerate (having negative components) and nontrivial:
(0,0,0)
(-4,-4,1)
(2,2,-1)
(3,3,-6)
(6,6,6)
(5,5,10)
(8,8,4)
(12,12,3)
and the (3,5,10) solution (emphasis that the word solution is not plural) mentioned by another user, not to mention the other bits mentioned by a later user.
(3,10,15), excuse me.
They said Matlab wasn't useful
I'm an idiot but I got 12x1x3
2(ab+bc+ca)=abc, ab(c-2)=2c(b+a)
c=3, ab=6(b+a), b(a-6)=6a, b=2*3a/(a-6)
or letting a=a-6
b=6(a+6)/a
For a = 1-4, 6, 9, 12, 18, 36 we get natural b.
c=4 then ab=4(b+a) or b=4(a+4)/a
a = 1,2,4,8
c=5 then 3ab=10(b+a), b=10a/(3a-10)
a = 5
So, I have (7,42,3), (8,24,3), (9,18,3), (10,15,3), (12,12,3), (5,20,4), (6,12,4), (4,8,4), (5,10,5), (6,6,6)
OP here. Goddammit, I knew I would overlook some important detail. I should have specified that the rectangular prism must have positive, non-zero lengths for the sides. And I should have said that the order in which the sides are listed doesn't matter (in other words, "8,8,4" is functionally the same as "4,8,8"
But anyways, the award goes to: and The second answer is more thorough, but there is a mistake / typo regarding "4,4,8". But congratulations nonetheless! You have mentally slaughtered the other posters, and have earned your rank as the smartest posters here (at least for the next 24 hours, until the next puzzle is posted).
Here is the answer key:
>(6,6,6) = 216
>(5,5,10) = 250
>(4,4,8) = 256
>(4,6,12) = 288
>(4,5,20) = 400
>(3,12,12) = 432
>(3,10,15) = 450
>(3,9,18) = 486
>(3,8,24) = 576
>(3,7,42) = 882
FUCK I made the same mistake.
I meant to say (4,8,8).
