How to fit three cones into the corresponding cylinder of equal r,h?

Question

As you know, the volume of a cylinder (radius r, height h) is 3 times the volume of a cone with equal radius r, height h

Now then. If you took three such solid cones, how could you actually pack, slice or topologically map them to fit into the corresponding solid cylinder?
Consider the cylinder (call it D to avoid confusion):
x²+y² ≤ r², 0≤z≤h
which is equivalent to:
-r≤x≤r , -√(r²-x²) ≤ y ≤ +√(r²-x²), 0≤z≤h

So say we take the first cone C1 and stand it in the middle of D with its base center on the origin. Then equation of C1 is: 0≤z≤h, x²+y² ≤ (r/h)*z

Now, how can you squeeze/map two other cones C2, C3 into the the unfilled space of the cylinder D?
Can you find me mappings f,g:R³→R³ (i.e. f(x,y,z)→(x',y',z'))
which map any point (x,y,z) in C2 and C3 respectively into the interior of D such that the three cones don't intersect with each other? Bonus points if your mappings preserve any properties (e.g. volume, angle, length), and thus involve less sawing, slicing or melting.
For common notation, let us say C2 is placed with its base-center on (-2r,0) and C3 with base-center on (+2r,0). Then their eqns are:
C2: 0≤z≤h, (x+2r)²+y² ≤ (r/h)*z
C3: 0≤z≤h, (x-2r)²+y² ≤ (r/h)*z

Here is maybe one conceptual startpoint. Slice C2 vertically into four quarters (i.e. slice it vertically along the planes x=0, y=0).
Now you have four quarter-slices of cone, each with the eqn:
0≤z≤h, x²+y² ≤ (r/h)*z and 0≤x, 0≤y
Turn those four quarter-slices upside down, pack them around C1 and try to cram them into the cylinder D. They won't quite fit since they leave gaps due to being convex - their base will occupy a square of 2r*2r. But it's a first approximation.

Another possibility:
Consider the two separate empty slices of volume D not filled by C1
x²+y² ≥ r² (GREATER THAN! i.e. the exterior of a cone), 0≤z≤h
Call them
E2: x≤0, x²+y² ≥ r², 0≤z≤h (left half-exterior of cone)
E3: 0≤x, x²+y² ≥ r², 0≤z≤h (right half-exterior of cone)

Now all you have to do think up some mappings
f:C2→E2, g:C3→E3 which fully occupy the interior of D, don't overlap with C1 or with each other.


[Perhaps you may have seen the brilliant old Laurel and Hardy carpet-cutting sketch, where they cut mismatched-shape carpet scraps to fit into holes e.g. a triangle into a rectangle, a trapezium into a square etc. Well that's the same challenge in two dimensions. See my other open question on that sketch.]

Star if you like it. I think this would make a good interview question for creativity. Topologists or algebraists, let me know which theory this relates to.

Answer 1

(Completely revised answer) The idea is to map each cone /apexes downwards/ into a cylinder with the same base and 3 times shorter altitude (hence the same volume) by a 1-1 continuous transformation, then to stack these cylinders, thus filling D (my previous suggestion did almost the same, but was not 1-1 correspondence since the solids were not closed as required)

The cone C₁: x² + y² ≤ (r/h)*z, 0 ≤ z ≤ h

can be obtained revolving the triangle with vertexes

(0, 0), (0, h) and (r, h) in xz-plane about z-axis. If this triangle is subjected to a "non-uniform stretch", pulling the hypotenuse midpoint in the direction of the point (r, 0), this deformation will map it into a rectangle with vertexes

(0, 0), (0, h), (r, h) and (r, 0).

Next a vertical contraction (3:1) will map it into a rectangle

(0, 0), (0, h/3), (r, h/3) and (r, 0)

Revolving the latter about z-axis, we'll get the desired cylinder - follow the link below t see how it is looking like:

http://farm4.static.flickr.com/3151/3061072949_483822fba2_o.gif



The first mapping can be described as follows:

(*) H → H;

(**) If P(x, z) satisfies x ≥ 0, h > z ≥ hx/r, x/r + z/h ≤ 1 then

P → P' such that |HP'| = {HP| * |HP₀| / |HP₁|

/P₀ and P₁ are intersecting points of the line HP with x-axis and the hypotenuse respectively/

or, analytically:

x' = x(r(h - z) + hx) / (r(h - z)); z' = (rz - hx) / r

(***) If Q(x, y) satisfies x > 0, z ≤ h, z ≥ hx/r, x/r + z/h ≥ 1 then

Q → Q' such that |HQ'| = {HQ| * |HQ₀| / |HQ₁|

/Q₀ and Q₁ are intersecting points of the line HQ with the line x=r and the hypotenuse respectively/

or, analytically:

x' = (hx - rz + rh) / h; z' = (2rhz - rh² + hxz - rz²) / (hx)

/(**) and (***) yield the same for points on the median HM/



The second (vertical contraction): x" = x', z" = z'/3



C₂ and C₃ should be translated horizontally along vectors (±2r, 0, 0) respectively, moving their apexes into the origin, then mapped onto same cylinders, finally the latter should be subjected to vertical shifts by h/3 and 2/3



Easily can be seen that the composition of the transformations described is continuous mapping with no singularities, hence there is 1-1 correspondence between each cone-cylinder pair.

Answer 2

I am Just copy pasting the answer given by ja.monte and Breatalo.. to the following question



https://answersrip.com/question/index?qid=20081105231913AAjh45m



(So No credit goes to me)



I am sure the following will help .

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Answer by ja.monte:

let y = r*x/h a line in the plane xy



h is the height of the cylinder



that coincide with the line y = r that r = radius of the cylinder,



then y = r*x/h is correct for x = h



so let y = r*x/h a line x = 0 and x = h its bound



if we rotate y over the x axis we obtain the volume generated by the rotation and that gives us the cone volume.

Calculate the integral from x = 0 to h

Spi*R^2dx = pi*S(r*x/h)^2dx =



=pi*S(r^2*x^2/h^2)dx



=pi*[r^2*(x^3)/3h^2] from 0 to h



=pi*[ r^2*(h)^3/3h^2 - 0]



= pi*[r^2*h / 3] = (1/3) volume of the cylinder

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Answer By Breatalo.:



The invention of calculus came about via attempting to approximate areas or volumes of objects otherwise impossible to determine.



The easy way is to take a cone and a cylinder with the same base and height as the cone, made of the same materials, and place them into a bowl of liquid. The amount of water displaced by the cone is 1/3 that displaced by the cylinder (thank you Archimedes).



That is visual proof, but a mathematical way was also desired.



Now, draw a right triangle on the coordinate system with (0,0) as the 90 degree angle, (r,0) on the y-axis and (0,h) on the x-axis.



If we rotate this triangle around the x-axis to see what shape that makes, we get a cone. Can you visualize this?



However, we still can't tell what it's volume is. The cross section in one direction is a circle. Extending to a third dimension makes that a cylinder.



We can think of a cone as a set of stacked cylinders. The circle at the base is the largest with radius r, and the one at the top has radius 0. In between are infinite cylinders with height = total height / number of cylinders. The radii from the base to the top also decrease by that same factor: base radius / number of cylinders.



But infinity doesn't help a whole lot, so we try an approxmation.



If we have only one cylinder with height h and radius r. the volume is base * height, which is pi*h*r^2. But if you drew the rectangle bounded by bottom left (0,0) to upper right (h,r), it doesn't come near the space under the triangle.



What if we had two cylinders of equal height stacked so their combined height was h (height = h/2), and the radii of the larger was r, and the other was r/2.



Stack them with the circular parts centered and you see an approximation of the cone.



Then the combined cyinder volume = pi*h/2*r^2 + pi*h/2*(r/2)^2

= pi*h*r^2 /2 + pi*h*r^2 /8

= pihr^2 ( 1/2 + 1/8 )

= pihr^2 ( 4/8 + 1/8 )

= pihr^2 ( 5/8 )



Now, what i we had 3 cylinders. This time they all have largest has height h/3 and radii r, 2r/3, r/3



pi*h/3*r^2 + pi*h/3*(2r/3)^2 + pi*h/3*(r/3)^2

= pi*h*r^2 / 3 + pi*h*r^2 * 4/27 + pi*h*r^2)^2 * 1/27

= pi*h*r^2 ( 1/3 + 4/27 + 1/27)

= pi*h*r^2 ( 9/27 + 4/27 + 1/27)

= pi*h*r^2 ( 14/27)



For 4, we'll have:

= pi*h*r^2 ( 1/4 + 9/64 + 4/64 + 1/64)

= pi*h*r^2 ( 16/64 + 9/64 + 4/64 + 1/64)

= pi*h*r^2 ( 30/64)



notice the pattern:

1: 1/1

2: 4/8 + 1/8

3: 9/27 + 4/27 + 1/27

4: 16/64 + 9/64 + 4/64 + 1/64



It is the sums of first n squares / n cubed.



If we did this up to n cylinders, the factor would be:

(n^2 + (n-1)^2 + ... + 3^2 + 2^2 + 1^2) / n^3



someone else has proved that the sum of first n squares is

(n)(n+1)(2n+1)/6





So substitute that:

((n)(n+1)(2n+1))/6

/ n^3

Now this can also be written as

=(n/n) * (n+1)/n * (2n+1)/n all over 6

=1 * (1+1/n) * (2 + 1/n) all over 6.



When n gets really high 1/n gets really close to zero, so at n = infinity

=1 * (1+0) * (2 + 0) all over 6

=1 * 1 * 2 / 6

=2/6

=1/3



So the volume of the cone = pihr^2 (1/3), which is what we wanted to prove.

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Answer 3

I think this will work, but the cones have to be plastic.



There are three phases:

I Map the cones to rectangular-based pyramids.

II Assemble the pyramids into a square-based prism.

With side, s, of the square such that, s^2 = (pi)r^2 and height, h.

III Map the square-based prism into the cylinder.



In phase I one of the of the rectangular-based pyramids has an s-by-s base and a height h. The other two have an s-by-h base and height s. These are not right pyramids. Rather the apex is above one of the corners of the base. I think the two non-square-based pyramids have to be mirror images of each other.



In phase II place the two non-square-based pyramids together so that their s-by-s right triangular faces form an s-by-s square. Then complete the square-based prism with the third pyramid.



I don't think any amount of finite cutting of solid cones can turn the cones into a cylinder. This has probably been proven, but it is beyond me.



ADDED:

Here's another plastic deformation.



Take the cone (with apex up) and squash it down into a cylinder 1/3 the height of the cone.

In polar cylindrical coordinates transform points on the cone (R, θ, z) by θ' = θ, z' = z/3, R' = Rh/(h-z).

The R transformation makes every circular cross-section expand to the radius of the base. I use R because you already defined r as the radius of the cylinder.

In cartesian: x' = xh/(h-z), y' = yh/(h-z), z' = z/3.

The other two cones undergo these transformations so that they stack

Second cone: x' = xh/(h-z), y' = yh/(h-z), z' = z/3 +h/3.

Third cone: x' = xh/(h-z), y' = yh/(h-z), z' = z/3 +2h/3.



Conceptually, I like the first transformation better, because the transformed cones, though greatly distorted, still have apexes. However, the equations are very bulky.

Answer 4

Interesting problem. It reminds me of the time during my teaching career when a quite bright class simply would not accept that a cylinder had three times the volume of the corresponding cone. They only agreed when I poured water from a hollow cylinder to fill a hollow cone three times. I wish that there had been a simple solid method of doing it. I suppose plasticine would do, but I didn't want to get into that.



As to your problem, I will be very surprised if there is a simple transformation that will work but I look forward to being proved wrong.

Answer 5

The Biker Granny

Answer 6

Take solid cylinder of equal r, h.

Immerse it in to a bucket of water having 2r, 2h.

Note the level of the water in the bucket and mark this level. Take the cylinder out of bucket. Now take three solid cones and immerse these three cones into the same water of the same bucket. You will see that this time the water level is the same as marked for the cylinder.





Thus volume of a cylinder of r, h = 3( volume of a cone of r, h).