[Guest]

An infinite amount of spheres are placed in a cone with a half angle x (0<x<90) as in the diagram below. Find the ratio:

(total volume of spheres) / (volume of cone)

I hope this one last more than 30 minutes this time.

Edited November 8, 2009 by psychic_mind

[Guest]

This might be wrong but if you were to take half the cross-section of the top sphere and cone, you would get two shapes. a trapezium and a semicircle.

(r+q+r-q)/2*2r is the area for the trapezium (where r is the semi-circle radius) which comes to 2r^2

and for the semicircle you'd get 0.5pi*r^2

so the ratio would be 0.785 (3sf)

By relying on the fact that the volume ratio is consistent all the way down, this might work but tell me if I'm wrong!

[Guest]

On first inspection, although it does not make sense, zero!

[Guest]

On first inspection, although it does not make sense, zero!

How can that be? (total volume of spheres) / (volume of cone) = 0 means that (total volume of spheres) = 0, which is not true.

[Guest]

I would suggest that if you were to sawcut (parallel to the base of the cone) at the point of contact between the first two spheres, then calculate the ratio of volume of the first sphere and the volume of the truncated cone (or frustum). That ratio will repeat for all others so that will be constant and the answer. I'm too lazy to do the sums!

[Guest]

You are right... I said it didn't make sense! I will try to solve it.

[Guest]

An infinite amount of spheres are placed in a cone with a half angle x (0<x<90) as in the diagram below. Find the ratio:

(total volume of spheres) / (volume of cone)

I hope this one last more than 30 minutes this time.

consider only the first portion of the cone and the first sphere...

Area of the sphere=4/3*pi*r^3

if x is the half-angle... assume cone height of y

bigger radius of the frustum=ytanx

smaller radius of the frustum=(y-2r)tanx

Area of the frustum=[(1/3)*pi*(ytanx)^2*y]-[(1/3)*pi*((y-2r)tanx)^2*(y-2r)]

=(1/3)pi*y^3*(tanx)^2-(1/3)pi*(y-2r)^3*(tanx)^2

=(1/3)pi*(tanx)^2*[y^3-(y-2r)^3]

=(pi/3)*(tanx)^2*r*(6y^2-12r*y+8r^2)

i thought y would cancel out -____-

anyhow,

ratio=(2r^2)/[(tanx)^2(3y^2-6ry+4r^2)]

i haven't thought about the cone height (y) though...

[Guest]

consider only the first portion of the cone and the first sphere...

Area of the sphere=4/3*pi*r^3

if x is the half-angle... assume cone height of y

bigger radius of the frustum=ytanx

smaller radius of the frustum=(y-2r)tanx

Area of the frustum=[(1/3)*pi*(ytanx)^2*y]-[(1/3)*pi*((y-2r)tanx)^2*(y-2r)]

=(1/3)pi*y^3*(tanx)^2-(1/3)pi*(y-2r)^3*(tanx)^2

=(1/3)pi*(tanx)^2*[y^3-(y-2r)^3]

=(pi/3)*(tanx)^2*r*(6y^2-12r*y+8r^2)

i thought y would cancel out -____-

anyhow,

ratio=(2r^2)/[(tanx)^2(3y^2-6ry+4r^2)]

i haven't thought about the cone height (y) though...

It's possible that that's correct but the solution should only have the variables given in the question (i.e. x).

[plainglazed]

...the two dimensional ratio is the same: pi/{4(tan(45-x/2)+tan(x))} ???

[Guest]

Unfortunatelly I have to go. I came up with a long expression that can not be right either. It looks as if the final expression would depend on the first ratio only, but the limit for n=inf diverges so I made a mistake somewhere. Good luck. I will check the answer when I get back.

Ridelto

[plainglazed]

2/{tan(45-x/2)² + tan(45-x/2)tan(45+x/2) + tan(45+x/2)²}

[Guest]

Ok, as I see it, the answer is much less complicated than most of the other ones which have been posted:

First, note that you can cut the cone with a straight line between each sphere, giving an infinite number of sections, each with exactly the same ratio of sphere to cone. Find that ration for one section at you find the overall ratio as well.

Second, for those of you who do not already know, the volume of a sphere is (4/3)*Pi* R^3.

Next, note that looking at a 2D cross section of the sceneraio (as shown in the image included) you see cirlces within a triangle instead of spheres within a cone. Taking slices the same way as above, gives sections of a circle within a trapezoid. Now that trapezoid (which touches the circle on all 4 sides) will have the exact same area as a square with each side equal to the diameter of the circle. Similarly in our 3D case, the conic sections around each sphere will have the same volume as cylinder, with a cross section area equal to Pi*R^2 and a length equal to the diameter, 2R. This gives the cylinder, and in turn the conic section, a volume of 2*Pi*R^3.

Finaly, taking the ratio of the volume of the sphere to the volum of the conic section, we get [(4/3)*Pi* R^3]/[2*Pi*R^3] which simplifies to a nice neat 2/3.

There, now isn't 2/3 a much cleaner, simpler answer than the ones listed above?

Oh, and its actually correct

If anyone thinks they can find a flaw in this, please do so. I'ld be glad to here it.

Edited November 8, 2009 by pfox

[Guest]

Ok, as I see it, the answer is much less complicated than most of the other ones which have been posted:

First, note that you can cut the cone with a straight line between each sphere, giving an infinite number of sections, each with exactly the same ratio of sphere to cone. Find that ration for one section at you find the overall ratio as well.

Second, for those of you who do not already know, the volume of a sphere is (4/3)*Pi* R^3.

Next, note that looking at a 2D cross section of the sceneraio (as shown in the image included) you see cirlces within a triangle instead of spheres within a cone. Taking slices the same way as above, gives sections of a circle within a trapezoid. Now that trapezoid (which touches the circle on all 4 sides) will have the exact same area as a square with each side equal to the diameter of the circle. Similarly in our 3D case, the conic sections around each sphere will have the same volume as cylinder, with a cross section area equal to Pi*R^2 and a length equal to the diameter, 2R. This gives the cylinder, and in turn the conic section, a volume of 2*Pi*R^3.

Finaly, taking the ratio of the volume of the sphere to the volum of the conic section, we get [(4/3)*Pi* R^3]/[2*Pi*R^3] which simplifies to a nice neat 2/3.

There, now isn't 2/3 a much cleaner, simpler answer than the ones listed above?

Oh, and its actually correct

If anyone thinks they can find a flaw in this, please do so. I'ld be glad to here it.

I think the flaw is

"Now that trapezoid (which touches the circle on all 4 sides) will have the exact same area as a square with each side equal to the diameter of the circle." I think you're assuming that the circles touch the trapezoid at half its perpendicular height. That is not correct.

[Guest]

I believe I got the answer!!

I agree that you only need to consider the ratio of the volume of the cone up to the height of the first sphere. The ratios are the same after that and the entire ratio is the same as a result

As stated before, volume of the first sphere is 4/3*pi*r^3

Now to find the volume of the section of cone:

V = int(pi*s^2) dy where s defines an incremental radius of cone

s = yb/h where h=cone height and b=base radius

V = int(pi*(yb/h)^2 dy)

= pi*b^2/h^2 * int(y^2)dy

integration is not from 0 to h, but from 0 to 2r where r is circle radius

solving integration:

= pi*b^2/h^2 * (2r)^3/3

= pi*b^2/h^2 * 8r^3/3

Now, the half angle x can be defined as: tan(x) = h/b (height over base)

So now we can plug this into the solution above:

= pi*8*r^3/(3 * tan(x)^2)

Now for the ratio:

[pi*8*r^3/(3 * tan(x)^2)] / [4/3*pi*r^3]

= 2/tan(x)^2 <-----ans!!!

Edited November 8, 2009 by seg9585

[Guest]

This MAY be the right answer...

I agree that you only need to consider the ratio of the volume of the cone up to the height of the first sphere. The ratios are the same after that and the entire ratio is the same as a result

As stated before, volume of the first sphere is 4/3*pi*r^3

Now to find the volume of the section of cone:

V = int(pi*s^2) dy where s defines an incremental radius of cone and y defines incremental cone height

s/b = y/h where h=cone height and b=base radius

s = yb/h

V = int(pi*(yb/h)^2 dy)

= pi*b^2/h^2 * int(y^2)dy

integration is not from 0 to h, but from 0 to 2r where r is circle radius

solving integration:

= pi*b^2/h^2 * (2r)^3/3

= pi*b^2/h^2 * 8r^3/3

Now, the half angle x can be defined as: tan(x) = h/b (height over half base)

So now we can plug this into the solution above:

= pi*8*r^3/(3 * tan(x)^2)

Now for the ratio:

[pi*8*r^3/(3 * tan(x)^2)] / [4/3*pi*r^3]

= 2/tan(x)^2 <-----ans!!!

Edited November 8, 2009 by seg9585

[plainglazed]

I think this ratio reduces to cos(x)²/6

Looking at the side elevation of one sphere of radius R and the corresponding truncated cone of height 2R: the truncated cone does not have the same volume as a cylinder of radius R and height of two R but does have the volume of a cylinder of radius R/cos(x) and height of 2R. Look at the side elevation (two dimensional view of a trapezoid and enclosed circle of radius R). If we draw a horizontal line thru the center of the circle we would have the average width of the trapezoid. Volume of the cylinder equal to the volume of the truncated cone (frustum) is pi(R/cos(x))²H where H = 2R or 4R³pi/cos(x)². Volume of the sphere = 4/3piR³. Volume of the spheres to cones is cos(x)²/6 ???

[plainglazed]

2cos(x)²/3? My head hurts but this has been a good way to try to get clear out the cobwebs. Fun stuff

[Guest]

Ok, as I see it, the answer is much less complicated than most of the other ones which have been posted:

First, note that you can cut the cone with a straight line between each sphere, giving an infinite number of sections, each with exactly the same ratio of sphere to cone. Find that ration for one section at you find the overall ratio as well.

Second, for those of you who do not already know, the volume of a sphere is (4/3)*Pi* R^3.

Next, note that looking at a 2D cross section of the sceneraio (as shown in the image included) you see cirlces within a triangle instead of spheres within a cone. Taking slices the same way as above, gives sections of a circle within a trapezoid. Now that trapezoid (which touches the circle on all 4 sides) will have the exact same area as a square with each side equal to the diameter of the circle. Similarly in our 3D case, the conic sections around each sphere will have the same volume as cylinder, with a cross section area equal to Pi*R^2 and a length equal to the diameter, 2R. This gives the cylinder, and in turn the conic section, a volume of 2*Pi*R^3.

Finaly, taking the ratio of the volume of the sphere to the volum of the conic section, we get [(4/3)*Pi* R^3]/[2*Pi*R^3] which simplifies to a nice neat 2/3.

There, now isn't 2/3 a much cleaner, simpler answer than the ones listed above?

Oh, and its actually correct

If anyone thinks they can find a flaw in this, please do so. I'ld be glad to here it.

[Guest]

You mean that you'd be glad to "HEAR" it surely?

[Guest]

I'm going to take another stab at this:

The total volume of the truncate cone is the full cone minus the truncated area.

So we determine the volume of the truncated area:

if h is height of full cone, (h-2r) is height of truncated cone

now, we need to determine radius of truncated cone. That is: tan(x) = (h-2r)/s or s=(h-r2)/tan(x)

volume of truncated cone:

pi*[(h-2r)/tan(x))]^2*(h-2r)/3

= pi*(h-2r)^3 / (3*tan(x)^2)

Area we are looking for is full cone minus truncated cone:

full cone base radius and height are related as follows:

tan(x) = h/b or b = h/tan(x)

Volume of full cone = pi*b^2*h = pi*h^3/[3tan(x)^2]

Total: pi*h^3/tan(x)^2 - pi*(h-2r)^3 / (3*tan(x)^2)

Expanded:

pi/[3*tan(x)^2] * [h^3 - (h^3 - 6h^2r + 12hr^2 + 8r^3)]

= pi/[3*tan(x)^2] * [6h^2r - 12hr^2 - 8r^3]

Now let's clean this up a little:

plug in the ratio for a circle that I used before (4/3*pi*r^3):

[6h^2r - 12hr^2 - 8r^3] / [4*tan(x)^2*r^3]

Now we need to eliminate the h and r

Geometrically, the circle is bound by the cone. You can draw 2 congruent triangles from the cone's angle and the center of the circle, which provides the following relations:

tan(x/2) = r/b where b is base radius of cone

tan(x) = h/b where h is height of cone

tan(x/2) = r*tan(x) / h

Now we can sub into the above equation after simplifying it:

3/2*h^2/(tan(x)^2*r^2) - 3*h/(tan(x)^2*r) - 2/tan(x)^2

= 3/[2*tan(x/2)] - 3/[tan(x)*tan(x/2)] - 2/tan(x)^2 <----ans

SANITY CHECK: when x goes to 90, tan(x) goes to infinity. Geometically, this means the cone is now a cylinder.

The volume for a cylinder is pi*r^2*(2r) = 2*pi*r^3, so the ratio here is 2*pi*r^3 / [4/3*pi*r^3] = 3/2

plugging in 90 for my equation above gives us the following:

3/[2*tan(45)] - 3/[tan(90)*tan(45)] - 2/tan(90)^2

= 3/2 - 0 - 0 !!!! confirmed!

[Guest]

The diagram throws you off. There doesn't have to be empty spaces. The puzzle doesn't specify a mininimum size sphere - so spheres can be infinitely small

So the ratio is 1 to 1 with spheres filling 100% of the cone or really 99.999999999999999999999999999999999% rounded to 100%.

First timer.

Edited November 9, 2009 by HiddenLake

[Guest]

I'm getting (2cos²x)/(3+sin²x)

Explanation to follow.

[Guest]

Say we've got some cone with half angle x. Plop in a sphere of radius 1 and according to the method described by Forcedhand (which I agree with) cut out the portion with the sphere in it, it's volume would be (see diagram):

Volume of cone section = (pi/3)(L+1)[(L+1)tanx]² - (pi/3)(L-1)[(L-1)tanx]²

= (pi/3)tan²x[(L+1)³-(L-1)³]

= (pi/3)tan²x[(L³+3L²+3L+1)-(L³-3L²+3L-1)]

= (pi/3)tan²x(6L²+2)

= (pi/3)tan²x[(6/sin²x)+2]

= (pi/3)[(sin²x)/(cos²x)][(6+2sin²x)/(sin²x)]

= (2pi/3)[(3+sin²x)/(cos²x)]

Volume of Sphere = 4pi/3

Ratio = (4pi/3)/{(2pi/3)[(3+sin²x)/(cos²x)]}

= (2cos²x)/(3+sin²x)

Sorry, not sure how to hide the diagram...

Edited November 9, 2009 by Tuckleton

[Guest]

It doesn't matter you said infinite. That makes the shape fractal. Its just like the golden rectangle. With the golden rectangle you continue cutting the rectangle in a pattern of vertical horizontal vertical. As long as you keep zooming in you can continue cutting. The same is true with this. As long as you keep zooming into the vertex of the triangle you will always be able to add another circle. Therefore the shape is fractal and it cannot be proven. Your question is like asking how many straight lines make up the English coast line.

[plainglazed]

Edited November 9, 2009 by plainglazed

[Guest]

I think you've got your formula for the volume of a circular cone frustum wrong. By that formula you'd get pi/4*r²h for the volume of a frustum with r₁=0 (a regular cone).