[math-fun] Cylinder puzzle
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin. Let X denote the intersection of all the C_k's. Prove that the surface area of X is exactly three times its volume. --Dan Those who sleep faster get more rest.
Presumably "unit" cylinders have diameter 2 and height 2 x oo ? I don't believe this. It's obviously true in the limit of infinitely many cylinders with axes distributed uniformly, around a circle or sphere, say. On the other hand, if there's just one cylinder, the ratio is only two! Fred Lunnon On 1/3/11, Dan Asimov <dasimov@earthlink.net> wrote:
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan
Those who sleep faster get more rest.
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If there is just one cylinder, X has infinite volume and infinite surface area, so the ratio can be anything you want it to be! Sincerely, Adam P. Goucher
On the other hand, if there's just one cylinder, the ratio is only two!
Fred Lunnon
On 1/3/11, Dan Asimov <dasimov@earthlink.net> wrote:
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan
Those who sleep faster get more rest.
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I'm guessing a unit cylinder has diameter = height = 1. --ms On Monday 03 January 2011 10:36:37 Adam P. Goucher wrote:
If there is just one cylinder, X has infinite volume and infinite surface area, so the ratio can be anything you want it to be!
Sincerely,
Adam P. Goucher
On the other hand, if there's just one cylinder, the ratio is only two!
Fred Lunnon
On 1/3/11, Dan Asimov <dasimov@earthlink.net> wrote:
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan
Those who sleep faster get more rest.
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From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Sun, January 2, 2011 11:54:06 PM Subject: [math-fun] Cylinder puzzle Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin. Let X denote the intersection of all the C_k's. Prove that the surface area of X is exactly three times its volume. --Dan _______________________________________________ Here is a proof for finitely many cylinders. Consider a cone of solid angle dΩ with vertex at the origin. Except for a set of measure zero, the cone intersects the boundary of X on a single cylinder. If r is the distance from the origin to the boundary point on the cone axis, then the cone has volume dV = r^3 dΩ/3, and intercepts surface area dS = r^2 dΩ sec i, where the inclination angle i is that between the cone axis and the normal to the surface element. For cylinders of unit radius r = sec i, so dS = 3 dV. -- Gene
Minor simplification: Each element of the surface dS is tangent to a plane whose distance from the origin is 1, so dV = (1/3)1 dS. (We need at least two cylinders with non-colinear axes for X to be compact.) Veit On Jan 3, 2011, at 12:22 PM, Eugene Salamin wrote:
From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Sun, January 2, 2011 11:54:06 PM Subject: [math-fun] Cylinder puzzle
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan _______________________________________________ Here is a proof for finitely many cylinders. Consider a cone of solid angle dΩ with vertex at the origin. Except for a set of measure zero, the cone intersects the boundary of X on a single cylinder. If r is the distance from the origin to the boundary point on the cone axis, then the cone has volume dV = r^3 dΩ/3, and intercepts surface area dS = r^2 dΩ sec i, where the inclination angle i is that between the cone axis and the normal to the surface element. For cylinders of unit radius r = sec i, so dS = 3 dV.
-- Gene
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Very neat argument --- but I still don't believe it. Why should it not work for a single cylinder? Just to placate those who haven't heard of principal values, or get distracted by minor matters like compactness (?), I'll give this large finite length and assert that the ratio S/V approaches 2 rather than 3, cones notwithstanding. On 1/3/11, Veit Elser <ve10@cornell.edu> wrote:
Minor simplification:
Each element of the surface dS is tangent to a plane whose distance from the origin is 1, so dV = (1/3)1 dS.
(We need at least two cylinders with non-colinear axes for X to be compact.)
Veit
On Jan 3, 2011, at 12:22 PM, Eugene Salamin wrote:
From: Dan Asimov <dasimov@earthlink.net> To: math-fun <math-fun@mailman.xmission.com> Sent: Sun, January 2, 2011 11:54:06 PM Subject: [math-fun] Cylinder puzzle
Let C_k, k = 1,2,3, . . . , n, . . . be solid unit cylinders in 3-space whose axes all contain the origin.
Let X denote the intersection of all the C_k's.
Prove that the surface area of X is exactly three times its volume.
--Dan _______________________________________________ Here is a proof for finitely many cylinders. Consider a cone of solid angle dΩ with vertex at the origin. Except for a set of measure zero, the cone intersects the boundary of X on a single cylinder. If r is the distance from the origin to the boundary point on the cone axis, then the cone has volume dV = r^3 dΩ/3, and intercepts surface area dS = r^2 dΩ sec i, where the inclination angle i is that between the cone axis and the normal to the surface element. For cylinders of unit radius r = sec i, so dS = 3 dV.
-- Gene
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From: Fred lunnon <fred.lunnon@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Mon, January 3, 2011 10:06:44 AM Subject: Re: [math-fun] Cylinder puzzle Very neat argument --- but I still don't believe it. Why should it not work for a single cylinder? Just to placate those who haven't heard of principal values, or get distracted by minor matters like compactness (?), I'll give this large finite length and assert that the ratio S/V approaches 2 rather than 3, cones notwithstanding. ________________________________ For a single cylinder of finite length, my proof applies to the volume of the cylinder contained within the solid angle subtended at the origin by the lateral surface. This is 2/3 of the total volume of the cylinder. The remaining 1/3 of the volume lies within the solid angle subtended by the end caps. Requiring the intersection of more than one cylinder not only renders the object compact, but also eliminates the end caps. -- Gene
Sorted! WFL On 1/3/11, Eugene Salamin <gene_salamin@yahoo.com> wrote:
From: Fred lunnon <fred.lunnon@gmail.com>
To: math-fun <math-fun@mailman.xmission.com> Sent: Mon, January 3, 2011 10:06:44 AM Subject: Re: [math-fun] Cylinder puzzle
Very neat argument --- but I still don't believe it.
Why should it not work for a single cylinder?
Just to placate those who haven't heard of principal values, or get distracted by minor matters like compactness (?), I'll give this large finite length and assert that the ratio S/V approaches 2 rather than 3, cones notwithstanding. ________________________________ For a single cylinder of finite length, my proof applies to the volume of the cylinder contained within the solid angle subtended at the origin by the lateral surface. This is 2/3 of the total volume of the cylinder. The remaining 1/3 of the volume lies within the solid angle subtended by the end caps. Requiring the intersection of more than one cylinder not only renders the object compact, but also eliminates the end caps.
-- Gene
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participants (6)
-
Adam P. Goucher -
Dan Asimov -
Eugene Salamin -
Fred lunnon -
Mike Speciner -
Veit Elser