Can you solve the volume of a cube with unequal heights?

[256bits]

I was referring to this:



That's right, but the two solids are complementary, if you put (as you showed earlier) one on top of the other they fit. If you are considering two planes [not arbitrary convex/concave shape], the slice is plus or minus, but volume is the same.
does problem boil down to: calculate the volume of a triangular wedge.

find the volume of the wedge and either subtract or add the wedge volume depending upon whether or not the 4th point is above or below the planar surface,
See post 17

Assuming corners of unequal heights and the top formed by the intersection of planes:
Case 1: The top formed by the sides intersected by a 6th plane - solved by Dave by symmetry, by Vmuth analytically, and Number Nine by calculus.

Case 2: top formed by the intersection of 2 planes. The line of intersection runs from one corner to the opposite, and forms a ridge or a valley.

Case 3: top formed by the intersection of 3 planes - The line of intersection in Case 2 is cut somewhere along its length.

Case 4: top formed from the intersection of 4 planes. The line of intersection from Case 2 is cut in at 2 points.

That's it for planes having on their surface at least 2 points from the corners.

 

[256bits]

sorry it's actually \frac{1}{6}a^{2}(h1+2h2+h3+2h4) where 'a' is the side of the base which is a square and i meant that the points h2 and h4 makes the edge on the top surface , i got this formula two days ago

Your formula was correct, except you failed to mention which point h1 ... h4 were at which corner.

 

[256bits]

Particular cases of my formula:
1) if the four points in the top are in a plane , they satisfy the following
a) shortest and the tallest are always on the opposite corners

b) h1+h3 = h2+h4 ( any proof ?)

Proof? Since the top surface is a plane, it intersects 2 parrallel side planes forming equal sized triangles, with one (bottom) side of the triangle parrallel to the base. need I go on...

 

[vrmuth]

Proof? Since the top surface is a plane, it intersects 2 parrallel side planes forming equal sized triangles, with one (bottom) side of the triangle parrallel to the base. need I go on...

yes , could you elaborate ?

 

[vrmuth]

Case 3: top formed by the intersection of 3 planes - The line of intersection in Case 2 is cut somewhere along its length.

Case 4: top formed from the intersection of 4 planes. The line of intersection from Case 2 is cut in at 2 points.

That's it for planes having on their surface at least 2 points from the corners.

There are only 4 vertices on the top of this shape , intersection of 3 planes will give rise to 5th point which in turn will give 5th height so that is a different figure , i think that's the next problem to be discussed in this thread, i like to try to find that volume also

 

[logics]

* a) shortest and the tallest are always on the opposite corners... b) h1+h3 = h2+h4 ( any proof ?)

reductio ad absurdum [\rightarrow\leftarrow] and symmetry give you an instant proof by contradiction. I do not know if there is another proof, but surely it would be complex:
Axis of rotation can be: space diagonal or face diagonal.
A plane rotating on a space diagonal [h₁,h₃] divides a cuboid (V=a²h₃) into two symmetrical, complementary solids [h₂ = h₃-h₄]: V₁= a²h₃/2 (case 1, h: 0,2,8,6 + 8,6,0,2)
Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)
If we are to exclude concave, convex or goofy surfaces, other cases are determined by: additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.
The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1, brilliantly solved by Dave: V = a²h₁₊₃/ 2

Edit: V = a²h_(0-1)+3/2

 

[vrmuth]

If we are to exclude concave, convex or goofy surfaces,other cases are determined by:additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.

I don't understand what you are saying here , and i don't know what's face diaonal and how a plane rotating about it. explanation pls ...

 

[logics]

I don't understand what you are saying here , and i don't know what's face diaonal and how a plane rotating about it. explanation pls ...

you're welcome, vrmuth: the best explanation is a picture http://wikipedia.org/wiki/Space_diagonal"

 

[DaveC426913]

reductio ad absurdum [\rightarrow\leftarrow] and symmetry give you an instant proof by contradiction. I do not know if there is another proof, but surely it would be complex:
Axis of rotation can be: space diagonal or face diagonal.
A plane rotating on a space diagonal [h₁,h₃] divides a cuboid (V=a²h₃) into two symmetrical, complementary solids [h₂ = h₃-h₄]: V₁= a²h₃/2 (case 1, h: 0,2,8,6 + 8,6,0,2)
Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)
If we are to exclude concave, convex or goofy surfaces, other cases are determined by: additional plane on axis [h₂, h₄], planes rotating on face diagonal axis etc.
The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1, brilliantly solved by Dave: V = a²h₁₊₃/ 2

I do not know if your "logics" incorporates this but as you can see by my earlier analysis, the diagonal's endpoints are not actually known. They could be from h₁ to h₃ or from h₂ to h₄ - even on the same object.

i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup.

See diagram in post 22.

 

[logics]

... given the surface area of all sides, ... The sides of this object would be four trapezoids of which two, and only two, heights each side would be equal (where the two trapezoids meet to create a corner

I do not know if your "logics" incorporates this ...i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup..

if ...

OP meticolously formulated the problem and gives the impression he would give Areas in an ordered sequence. If gives him the benefit of doubt, if we get an ordered sequence A_2,3,4,1 we can solve any rotation on space diagonal. Now that you know it is so, your logics , will easily lead you to the solution. I imagine you want the satisfaction of finding it yourself.

 

[vrmuth]

Two such planes delimit a (V_w= 1/6...) wedge: V₂ = V₁ \pm V_w (case 2, h: 0,2,8,5 + 8,6,0,3, h_w =1)

Calculating the volume of the wedge separately may seem to be a good logic but it will complicate the problem because even if you deduce all h and their order , the order of the heights won't decide whether is there a wedge or a valley on the top surface

 

[DaveC426913]

OP meticolously formulated the problem and gives the impression he would give Areas in an ordered sequence. If gives him the benefit of doubt, if we get an ordered sequence A_2,3,4,1 we can solve any rotation on space diagonal. Now that you know it is so, your logics , will easily lead you to the solution. I imagine you want the satisfaction of finding it yourself.

No. You are not understanding the import of my assertion. You do not have a unique solution yet.

Even with all corners precisely defined in an ordered sequence, and making no change to them of any kind, it is still possible to construct the box two ways after the fact. you have two choices for the space diagonal and you cannot specify them. Only the OP can.

You could physically build the entire structure, floor and four walls, including precisely the heights of all four corners, made out of steel bars - yet you still do not have enough information to determine the shape of the top, and thus determine the volume.

Please see attached greatly clarified diagram. This should make it obvious.

 

[phinds]

Even with all corners precisely defined in an ordered sequence, and making no change to them of any kind, it is still possible to construct the box two ways after the fact.

Yep, With the single exception of the case where all 4 upper points are on the same plane.

EDIT: Hm ... not even sure why I posted that Dave, since I KNOW you already know that. That's one of the problems of being a speed typist with no mental bladder control.

 

[DaveC426913]

Yep, With the single exception of the case where all 4 upper points are on the same plane.

EDIT: Hm ... not even sure why I posted that Dave, since I KNOW you already know that. That's one of the problems of being a speed typist with no mental bladder control.

Right. OP specified that no two heights are the same.

 

[vrmuth]

The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4 and the shape of the top, else if * we can only solve the trivial case 1

the diagonal's endpoints are not actually known. They could be from h₁ to h₃ or from h₂ to h₄ - even on the same object.

i.e. any solution is going to have to spit out two equally possible answers, from which the OP will have to choose which matches his setup.

here is the proof
assume all the h are known , and the base be a rectangle with dimension a and b , divide the figure into two volumes by a vertical plane along the diagonal in the top (say h2h4)
lets take anyone (say with h1,h2 and h4 ) let x,y axes be along two sides that make right angle in the base and z be along h1 take a series of thin vertical slices parallel to x-axis with thickness dy , its face is a trapezium , ( see the picture for its dimensions )
this volume is given by \frac{a(b-y)}{2b} \left[ \frac{(2h4-h1-h2)y}{b}+ (h1+h2) \right] dy integrating this from 0 to b gives V_{1}=\frac{ab(h1+h2+h4)}{6} since the other volume is a similar shape with heights h2 , h3 and h4 its volume is given by V_{2}=\frac{ab(h2+h3+h4)}{6} hence the total volume V=V1+V2=\frac{ab(h1+2h2+h3+2h4)}{6} where the heights h2 and h4 are the end points of the diagonal in the top surface , the choice of heights take care of whether is there wedge or valley

 

[phinds]

Right. OP specified that no two heights are the same.

Yeah, but that doesn't keep them from being in the same plane, as you pointed out in post #10 when you solved the original problem.

 

[DaveC426913]

Yeah, but that doesn't keep them from being in the same plane

Apologies. Correct.

 

[DaveC426913]

here is the proof
assume all the h are known , and the base be a rectangle with dimension a and b , divide the figure into two volumes by a vertical plane along the diagonal in the top (say h2h4)
lets take anyone (say with h1,h2 and h4 ) let x,y axes be along two sides that make right angle in the base and z be along h1 take a series of thin vertical slices parallel to x-axis with thickness dy , its face is a trapezium , ( see the picture for its dimensions )
this volume is given by \frac{a(b-y)}{2b} \left[ \frac{(2h4-h1-h2)y}{b}+ (h1+h2) \right] dy integrating this from 0 to b gives V_{1}=\frac{ab(h1+h2+h4)}{6} since the other volume is a similar shape with heights h2 , h3 and h4 its volume is given by V_{2}=\frac{ab(h2+h3+h4)}{6} hence the total volume V=V1+V2=\frac{ab(h1+2h2+h3+2h4)}{6} where the heights h2 and h4 are the end points of the diagonal in the top surface , the choice of heights take care of whether is there wedge or valley

Yes, there is a solution, provided the OP makes one more decision - whether the roof is concave or convex (i.e. where he chooses to build the diagonal).

Any proofs not given this piece of information will have to spit out two answers.

 

[vrmuth]

[/itex]

Yes, there is a solution, provided the OP makes one more decision - whether the roof is concave or convex (i.e. where he chooses to build the diagonal).

Any proofs not given this piece of information will have to spit out two answers.

yes you are right, the orientation of the diagonal decides whether the top is concave or convex. But since its meaningless to talk about the volume and its formula when the shape of one surface is not specified , the diagonal should be known ,
then if the heights h_{2,4} correspond to the diagonal , where is the another answer ?

 

[phinds]

[/itex]

yes you are right, the orientation of the diagonal decides whether the top is concave or convex. But since its meaningless to talk about the volume and its formula when the shape of one surface is not specified , the diagonal should be known ,
then if the heights h_{2,4} correspond to the diagonal , where is the another answer ?

What are you on about? IF the diagonal is known then there is NOT "another" solution. The statement that there has to be two solutions is based on not knowing which diagonal is being used so you have to solve for both.

 

[vrmuth]

What are you on about? IF the diagonal is known then there is NOT "another" solution. The statement that there has to be two solutions is based on not knowing which diagonal is being used so you have to solve for both.

what do you mean ? two different formulas or one formula that suits both cases?

 

[DaveC426913]

what do you mean ? two different formulas or one formula that suits both cases?

It's one formula (or possibly one algorithm, which is several steps of formulae), that will leave it up to the user to plug in values. The user has a choice of which values she plugs in, based on what design she intends to solve for. The onus will be on the user to decide, at solving time, where the diagonal is.

 

[phinds]

what do you mean ? two different formulas or one formula that suits both cases?

The "general" solution is TWO solutions (or a "solution" that gives two answers) but if you SAY which diagonal you are talking about then there is only one solution (yes, one formula) needed.

 

[vrmuth]

It's one formula (or possibly one algorithm, which is several steps of formulae), that will leave it up to the user to plug in values. The user has a choice of which values she plugs in, based on what design she intends to solve for. The onus will be on the user to decide, at solving time, where the diagonal is.

The "general" solution is TWO solutions (or a "solution" that gives two answers) but if you SAY which diagonal you are talking about then there is only one solution (yes, one formula) needed.

very good , thanks dave that's what exactly i was trying to say ( because of my poor communicatin skill i was unable ) you know my formula is such a one ?, the user must choose h2 and h4 as the heights of the end points of the diagonal without worrying about whether its a wedge or a valley . if its a valley the other two heights h1 and h2 will be taller and shorter if its a wedge , the same formula will take care of everything b'cause i derived the formula for any values of h1 and h3 , the formula for area of a rectangle(with a>b) hold good for area of a square and we don't have a different formula for b>a

 

[vrmuth]

if its a valley the other two heights h1 and h2 will be taller and shorter if its a wedge...

sorry it's h1 and h3

 

[logics]

...which I plan on working on this weekend myself when I have free time from my homework...

... the order of the heights won't decide whether is there a wedge or a valley on the top surface

PF policy is to give hints, not solutions. I was waiting for OP to turn up: I want to give him a chance to solve the problem himself. In the meanwhile, I'll give him [and you] another hint.
The order, of course, does not decide but every sequence has only one possibility. General solution is very simple: just a few symbols.

 

[vrmuth]

PF policy is to give hints, not solutions.

yes, but this not a home work :), and the op also not a math student it seems , so he may not be able to solve even if you give him hints

 

[DaveC426913]

yes, but this not a home work :), and the op also not a math student it seems , so he may not be able to solve even if you give him hints

Technically he's got a point though. Regardless of whether an OP is a student or not, we are not supposed to solve homework-like problems. Students read the forum too.

But I felt this was an exception since I see it as not a homework problem (that's the debatable point) But we didn't actually know if there was a solution. Also because it was really interesting.

 

[vrmuth]

if you are given the surface area of all sides, including the base, except for its top? The sides of this object would be four trapezoids

Will the area of all the sides be given seperately in order? ,There are infinite number of possible values of h , so any one value of h must be chosen(say k) and then the resulting volume will depend upon it (k) , so for each value of fixed h we 've a different volume , all having the given suface area , in otherwords there are infinite volumes for the "one set of surface area" ( i will give the proof if allowed )

The problem says: given the "surface areas" of all sides: we can deduce the values of, but we need to know the order of h _1,2,3,4

how can we deduce ?

The order, of course, does not decide but every sequence has only one possibility. General solution is very simple: just a few symbols.

what sequance you are talking about ? i cann't logic it

But I felt this was an exception since I see it as not a homework problem (that's the debatable point) But we didn't actually know if there was a solution. Also because it was really interesting.

yes its geting more and more interesting now

 

[DaveC426913]

Will the area of all the sides be given seperately in order? ,There are infinite number of possible values of h , so any one value of h must be chosen(say k) and then the resulting volume will depend upon it (k) , so for each value of fixed h we 've a different volume , all having the given suface area , in otherwords there are infinite volumes for the "one set of surface area" ( i will give the proof if allowed )

how can we deduce ?

I thought we'd settled this. We are given the height of the four corners, in order. The OP has base and all sides already built, so he knows what they are.