I have a box.
Ratio of length, width and height of the box is $a:b:c$.
Ratio of sum of all 12 edges' length, surface area (all 6 sides) and volume is also $a:b:c$
Find the length, width and height of the box.


$a$, $b$ and $c$ are integers and can be the same number.

  • $\begingroup$ Downvotes, and votes to close this? Why? Tell me the reason, so I will not post the puzzle like this again. $\endgroup$ – Jamal Senjaya Aug 22 '16 at 9:41
  • $\begingroup$ Is there any way to find the answer ?? or just trial and error method ?? $\endgroup$ – Amruth A Aug 22 '16 at 9:45
  • $\begingroup$ @Amaruth : Math to find some candidate, and a bit of calculation between the candicate, will come to the right answer. $\endgroup$ – Jamal Senjaya Aug 22 '16 at 9:49
  • $\begingroup$ I didn't downvote, but voted to close as this is a typical math textbook problem, thus offtopic. $\endgroup$ – elias Aug 22 '16 at 9:54

I think

$a=3$, $b=12$, $c=12$ works well.


Sum of all the edges' length: $4\times(a+b+c)=4\times(3+12+12)=4\times27=108$
Surface: $2\times(a\times b+a\times c+b\times c)=2\times(3\times12+3\times12+12\times12)=$ $=2\times(36+36+144)=2\times216=432$
Volume: $a\times b\times c=3\times12\times12=432$


How to find this:

The four equations can be written as:
$4\times(a+b+c)=k\times a$
$2\times(a\times b+a\times c+b\times c)=k\times b$
$a\times b\times c=k\times c$

From the last one, we immediately get $k=ab$.
Using this on the second one, we get $c=\frac{a^2 b}4-a-b$.
All this turns the first equation into $b^2(-\frac{a^2}4+\frac{a}2+1)+b(-\frac{a^3}4+a)+a^2=0$

The square root of the discriminant has to be a rational number, so $a^4+8a^2-32a-48$ has to be a perfect square.

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    $\begingroup$ 2 same answer, but I will choose @elias answer as accepted, because elias is the first person answering rightly $\endgroup$ – Jamal Senjaya Aug 22 '16 at 9:26


a = 3, b = 12, c = 12


a = 3, b = 12, c = 12
P=108, A = 432, V = 432
P:A:V = a:b:c

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Solve the below eqn -

4a + 4b +4c =ak ; 2ab +2bc+2ca =bk ;abc=ck

So , ab =k

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  • $\begingroup$ you forgot a factor in all the equations $\endgroup$ – elias Aug 22 '16 at 9:27
  • $\begingroup$ @elias the factor gets cancelled on both sides $\endgroup$ – Amruth A Aug 22 '16 at 9:28
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    $\begingroup$ it does not! the right sides of your three equations all need the same factor added $\endgroup$ – elias Aug 22 '16 at 9:29
  • $\begingroup$ @elias they how did you find answer $\endgroup$ – Amruth A Aug 22 '16 at 9:43
  • $\begingroup$ After introducing the factor, you can start eliminating the variables. First $k$, then $c$. After that you get a quadratic equation for $b$, with parameter $a$. $\endgroup$ – elias Aug 22 '16 at 9:53

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