Same Ratio

[jasen]

I have a box
Rasio of length, width and height of the box is a:b:c
Rasio of sum of sides, sum of surface area and volume is also a:b:c
find the length, width and height of the box
a, b and c are integers & can be the same number.

[bonanova]

I'm confused about the ratios.
 

Spoiler

a:b:c I understand, because they all have dimensions of length. Since they have the same dimensions, their ratios are pure numbers, without units.

But the sum of sides (which I take to mean a+b+c, but could mean 4a+4b+4c) has dimension of length, sum of surface area has dimension of area and volume has dimension of, um, volume. Normally ratios like that are meaningless without specifying the units used. The ratio of area to perimeter for a square for example is always a²/4a = a/4. But that is not a pure number, it's a measure. And it has, for a given square, say one that is 12 inches on a side, differing values: the ratio is 3 (inches), if the side is measured in inches, and 1/4 (foot), if the side is measured in feet.

I guess we can ignore all this if we say there is a box, and its linear dimensions (in some units, never mind what they are) are a, b, and c. It's just that ratios of lengths, areas and volumes will not be pure numbers. But in fairness you did not say they were.

 

[jasen]

16 hours ago, bonanova said:

I'm confused about the ratios.
 

  Reveal hidden contents

a:b:c I understand, because they all have dimensions of length. Since they have the same dimensions, their ratios are pure numbers, without units.

But the sum of sides (which I take to mean a+b+c, but could mean 4a+4b+4c) has dimension of length, sum of surface area has dimension of area and volume has dimension of, um, volume. Normally ratios like that are meaningless without specifying the units used. The ratio of area to perimeter for a square for example is always a²/4a = a/4. But that is not a pure number, it's a measure. And it has, for a given square, say one that is 12 inches on a side, differing values: the ratio is 3 (inches), if the side is measured in inches, and 1/4 (foot), if the side is measured in feet.

I guess we can ignore all this if we say there is a box, and its linear dimensions (in some units, never mind what they are) are a, b, and c. It's just that ratios of lengths, areas and volumes will not be pure numbers. But in fairness you did not say they were.

 

Sum of sides means sum of all 12 sides, sum of surface area means sum of all 6 areas.
Ignore the dimensions. Let say length in m, area in m² and volume in m³

All length are  integer.

Edited June 26, 2016 by jasen

[bonanova]

Meanderings.
 

Spoiler

Suppose the volume and area are equal: V = abc = A = 2(ab+bc+ca)

Then we have the Diophantine equation abc = 2(ab+bc+ca).

abc - 2(ac+bc) = 2ab
c(ab -2a-2b) = 2ab
c = 2ab / (ab-2a-2b) = 2 + 4(a+b)/[ab-2(a+b)

Search for integer values of (a,b) such that ab-2(a+b) divides 4(a+b) then calculate c.

If we constrain (without loss of generality) a<=b<=c, then solutions can be found,
by exhaustive search, for a=3, 4, 5, 6, as follows

a   b   c  V=A
= === === ====
3   7  42  882
3   8  24  576
3   9  18  486
3  10  15  450
3  12  12  432
4   5  20  400
4   6  12  288
4   8   8  256
5   5  10  250
6   6   6  216

Interestingly, one of the solutions is a cube, for which V = A.
Unfortunately, 4(6+6+6) is 72, not 216, a factor of 3 too small. So, (a,b,c) = (6,6,6) is not a solution.

That is, 6:6:6 is not equal to 72:216:216.

Another solution with b=c is 4 8 8.

But again, 4(4+8+8) is 80, not the required 128.

4:8:8 is not equal to 80:256:256.

We've ruled out the cube (a=b=c) and we've ruled out V=A (b=c).

The solution has a=b or all distinct a b c values.
a=c is not possible unless a=b=c (cube)

      

 

[jasen]

2 hours ago, bonanova said:

Meanderings.
 

  Reveal hidden contents

Suppose the volume and area are equal: V = abc = A = 2(ab+bc+ca)

Then we have the Diophantine equation abc = 2(ab+bc+ca).

abc - 2(ac+bc) = 2ab
c(ab -2a-2b) = 2ab
c = 2ab / (ab-2a-2b) = 2 + 4(a+b)/[ab-2(a+b)

Search for integer values of (a,b) such that ab-2(a+b) divides 4(a+b) then calculate c.

If we constrain (without loss of generality) a<=b<=c, then solutions can be found,
by exhaustive search, for a=3, 4, 5, 6, as follows

a   b   c  V=A
= === === ====
3   7  42  882
3   8  24  576
3   9  18  486
3  10  15  450
3  12  12  432
4   5  20  400
4   6  12  288
4   8   8  256
5   5  10  250
6   6   6  216

Interestingly, one of the solutions is a cube, for which V = A.
Unfortunately, 4(6+6+6) is 72, not 216, a factor of 3 too small. So, (a,b,c) = (6,6,6) is not a solution.

That is, 6:6:6 is not equal to 72:216:216.

Another solution with b=c is 4 8 8.

But again, 4(4+8+8) is 80, not the required 128.

4:8:8 is not equal to 80:256:256.

We've ruled out the cube (a=b=c) and we've ruled out V=A (b=c).

The solution has a=b or all distinct a b c values.
a=c is not possible unless a=b=c (cube)

      

 

Bonanova, you are very close, your explaination is very cool, just quite not carefull.

check again your exhaustive search list , you miss something.

[Logophobic]

12 hours ago, bonanova said:

We've ruled out the cube (a=b=c) and we've ruled out V=A (b=c).

Except that there is one result (b=c) that you did not test:

Spoiler

a = 3, b = 12, c = 12

P=108, A = 432, V = 432

P:A:V = a:b:c