# Find the length, width and height of the box [closed]

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.

Note:

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

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

I think

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

Check:

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: $a\times b\times c=3\times12\times12=432$

$3:12:12=108:432:432$

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.

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

Possible

a = 3, b = 12, c = 12

so,

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

Solve the below eqn -

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

So , ab =k

• you forgot a factor in all the equations – elias Aug 22 '16 at 9:27
• @elias the factor gets cancelled on both sides – Amruth A Aug 22 '16 at 9:28
• it does not! the right sides of your three equations all need the same factor added – elias Aug 22 '16 at 9:29
• @elias they how did you find answer – Amruth A Aug 22 '16 at 9:43
• 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 $a$. – elias Aug 22 '16 at 9:53