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# Math Problem -- Tough

 Posts: 998 Add as FriendChallenge to a DebateSend a Message 5/27/2016 6:49:04 AMPosted: 1 year agoI have a fantastic problems for all lovers of math and puzzles out there. I call this a "math" problem because it does require some trigonometry and geometric constructions, but definitely nothing a first year math student wouldn't be able to do.Before I state it, you should know that I read this problem *ages* ago and that you will probably not find the answer online. I do remember the answer provided, but the proof has been lost to my mind. So I'm here to get your input and test your abilities!Problem: Consider two types of objects called "black stars" and "white stars". Both are spheres of definite size, but each has a unique property. Black stars absorb every light particle that hits them. They do not produce reflections. White stars, on the other hand, emit light from every point in all directions.Your job is to take one white star and surround as many black stars as necessary around it such that all light from the white star is being blocked by black stars. In other words, if I get in my space ship and drive around your construction, I should not be able to see even a fleck of light. Black stars should not go through each other or the white star. They may touch, however.Mathematically stated: Construct a three dimensional model that contains a central sphere of definite size surrounded by other spheres of any size and in any number, such that one can construct a "boundary sphere" that encloses the construction where there does not exist a ray that can be drawn from a point on the central sphere and extended to the boundary sphere without passing through any internal sphere.Clarifications: This is not a physics problem. You cannot make the white star so big that it turns into a black hole, absorbing all light. You cannot consider large dark stars as having gravity. Light waves do not bend at all.Additionally, you can arbitrarily change the size of the stars as you'd like, black or white. You do not need to have all your dark stars be the same size. However, all stars must have a *definite size*. The white star cannot be infinitely small (a point) and the black stars cannot be infinitely large.Think on this for a while, do some maths, then keep reading. Below here I will discuss what I believe the answer to be, though I have yet to prove it.So really, go away if you don't want hints.Leave now.Ok, good.The purported answer was that it would require an infinite number of black stars, so long as you remember that the white star still must have a physical size. This is surprising, as intuition tells us that *surely* only a finite number of black stars would be needed.If the answer is indeed infinity, the implications of this are surprising. It means you could fill up a glass container of any size with as many marbles as you wanted and, if you look from the right place, you'd still be able to see every single marble in the jar.If any other math lovers express interest in this, I'll share where I'm at with the proof. I believe I'm nearly at the end and that the answer may well be infinity.
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 9:20:22 AMPosted: 1 year agoInteresting! My reaction was also infinite, but I haven't proven it yet.BTW I haven't taken multi calc - is a multi calc course necessary to do this? (I've only covered up to single-variable calc, and I'm preparing lin alg I, which I'll take next year. In all likelihood, multi calc will wait till my third year...)I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 9:24:06 AMPosted: 1 year agoSome thoughts (please tell me if this is right) - if the required condition is achieved, we would need to be able to draw a ray from the centre in any direction and come into contact with a black star on the way.I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 5/27/2016 9:31:59 AMPosted: 1 year agoAt 5/27/2016 9:24:06 AM, Diqiucun_Cunmin wrote:Some thoughts (please tell me if this is right) - if the required condition is achieved, we would need to be able to draw a ray from the centre in any direction and come into contact with a black star on the way.Yes, that's right. More specifically, there *should not* exist a ray that does not hit any black star on the way out. It's good to hear that your intuition also went toward infinity right off the bat.It's frustrating because I a) remember the answer being infinity and b) I remember that the proof provided was complete. I am just having serious difficulties reconstructing the proof.Currently, I'm stuck on a geometric construction. I'm some 99% confident that maths above geometry and trigonometry are not needed. That being said, I have a suspicion that the geometric theorem needed here wasn't one taught in school. If you like geometry, I can explain to you where I'm at right now with the problem.
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 1:15:06 PMPosted: 1 year agoAt 5/27/2016 9:31:59 AM, Cobalt wrote:At 5/27/2016 9:24:06 AM, Diqiucun_Cunmin wrote:Some thoughts (please tell me if this is right) - if the required condition is achieved, we would need to be able to draw a ray from the centre in any direction and come into contact with a black star on the way.Yes, that's right. More specifically, there *should not* exist a ray that does not hit any black star on the way out. It's good to hear that your intuition also went toward infinity right off the bat.It's frustrating because I a) remember the answer being infinity and b) I remember that the proof provided was complete. I am just having serious difficulties reconstructing the proof.Currently, I'm stuck on a geometric construction. I'm some 99% confident that maths above geometry and trigonometry are not needed. That being said, I have a suspicion that the geometric theorem needed here wasn't one taught in school. If you like geometry, I can explain to you where I'm at right now with the problem.Haha ok, I'll try it then. Geometry is my weakest area tbh, but I'll see if I can do it...I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 1:45:48 PMPosted: 1 year agoOkay, geometric model. Let the centre of the sphere be the origin. Then the sphere is x^2 + y^2 + z^2 = r^2 where r is the radius.Assume there is a finite number of black stars. Each black star is defined by the set of points B_n = {(x, y, z) in R^3: (x-a)^2 + (y-b)^2 + (z-c)^2 = r_n^2} where r_n is the radius of the black star. For every B_n, let the set of points inside the black star be C_n {(x, y, z) in R^3: (x-a)^2 + (y-b)^2 + (z-c)^2 < r_n^2}.Let there be k black stars. Let C = U_(1 to k) C_n. Then C U {(x, y, z) in R^3: x^2 + y^2 + z^2 < r^2} = empty set.Now just to figure out how to express a ray...I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 1:52:23 PMPosted: 1 year agoAt 5/27/2016 1:45:48 PM, Diqiucun_Cunmin wrote:Okay, geometric model. Let the centre of the sphere be the origin. Then the sphere is x^2 + y^2 + z^2 = r^2 where r is the radius.Assume there is a finite number of black stars. Each black star is defined by the set of points B_n = {(x, y, z) in R^3: (x-a)^2 + (y-b)^2 + (z-c)^2 = r_n^2} where r_n is the radius of the black star. For every B_n, let the set of points inside the black star be C_n {(x, y, z) in R^3: (x-a)^2 + (y-b)^2 + (z-c)^2 < r_n^2}.: Let there be k black stars. Let C = U_(1 to k) C_n. Then C U {(x, y, z) in R^3: x^2 + y^2 + z^2 < r^2} = empty set.:Now just to figure out how to express a ray...Lol ignore the bold part...I meant this:Let C = {C_1, C_2... C_k, {(x, y, z) in R^3: x^2 + y^2 + z^2 < r^2}}. Then a cap b = empty set for any two a, b belonging to C.I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 2:03:19 PMPosted: 1 year agoOh wait. The rays aren't that hard...It's a collection of points satisfyingx = tx_ay=tx_bz =tx_cx >= 0cup the collection of point satisfyingx = tx_ay=tx_bz =tx_cx <= 0Let's call the union of these two monsters S, and we need to show that S cap C_n = empty set for all n from 1 to k.Phew, that's the model... Now how do we prove this beast? :OI think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 2:13:33 PMPosted: 1 year agoNononono ignore previous post.Let F(a,b,c) be the set of points satisfyingx = tx_ay=tx_bz =tx_cx >= 0and let G(a,b,c) be the set of points satisfyingx = tx_ay=tx_bz =tx_cx <= 0where (a,b,c) belongs to x^2 + y^2 + z^2 = r^2.Aha! That works :DAnd what we need to show is that F(a,b,c) cup C_n = empty set for all C_0 to C_k OR G(a,b,c) cup C_n = empty set for all C_0 to C_k for some values of A, B, C in R.Phew! Like I know how to proceed from here, lol...I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 2,762 Add as FriendChallenge to a DebateSend a Message 5/27/2016 2:18:54 PMPosted: 1 year agoOh wait. No, what we need to show is that there is some way for which as the number of spheres -> infinity, the the amount of ground left uncovered -> 0. Oh darn... I give up for today lol, perhaps I shall return tomorrow 8)I think it is well established that the only reason aliens come to earth is to slice up cows and examine inside peoples' bottoms. Unless you are a cow or suffer haemerrhoids I don't think there is anything to worry about from aliens. - keithprosser Don't be a stat cynic: http://www.debate.org... Response to conservative views on deforestation: http://www.debate.org... Topics I'd like to debate (not debating ATM): http://tinyurl.com...
 Posts: 304 Add as FriendChallenge to a DebateSend a Message 5/28/2016 1:52:19 AMPosted: 1 year agoI pretty sure this can be done with a finite number of black stars. One simple solution involving 6 black stars would be put 2 black stars 1000 times bigger than the white star right up next to the white star on opposite sides, leaving only a very thin ring of light showing. then place the next two black stars a few times bigger than the previous black stars on opposite sides of the ring of light, leaving only two spots of light on opposite sides of the white star. then place the last two black stars to cover up those last two spots. There is probably a more optimal way to use fewer black stars, but I think this shows the answer is definitely finite.
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 5/28/2016 2:19:53 AMPosted: 1 year agoAt 5/27/2016 6:49:04 AM, Cobalt wrote:I have a fantastic problems for all lovers of math and puzzles out there. I call this a "math" problem because it does require some trigonometry and geometric constructions, but definitely nothing a first year math student wouldn't be able to do.Before I state it, you should know that I read this problem *ages* ago and that you will probably not find the answer online. I do remember the answer provided, but the proof has been lost to my mind. So I'm here to get your input and test your abilities!Problem: Consider two types of objects called "black stars" and "white stars". Both are spheres of definite size, but each has a unique property. Black stars absorb every light particle that hits them. They do not produce reflections. White stars, on the other hand, emit light from every point in all directions.Your job is to take one white star and surround as many black stars as necessary around it such that all light from the white star is being blocked by black stars. In other words, if I get in my space ship and drive around your construction, I should not be able to see even a fleck of light. Black stars should not go through each other or the white star. They may touch, however.Mathematically stated: Construct a three dimensional model that contains a central sphere of definite size surrounded by other spheres of any size and in any number, such that one can construct a "boundary sphere" that encloses the construction where there does not exist a ray that can be drawn from a point on the central sphere and extended to the boundary sphere without passing through any internal sphere.Clarifications: This is not a physics problem. You cannot make the white star so big that it turns into a black hole, absorbing all light. You cannot consider large dark stars as having gravity. Light waves do not bend at all.Additionally, you can arbitrarily change the size of the stars as you'd like, black or white. You do not need to have all your dark stars be the same size. However, all stars must have a *definite size*. The white star cannot be infinitely small (a point) and the black stars cannot be infinitely large.Think on this for a while, do some maths, then keep reading. Below here I will discuss what I believe the answer to be, though I have yet to prove it.So really, go away if you don't want hints.Leave now.Ok, good.The purported answer was that it would require an infinite number of black stars, so long as you remember that the white star still must have a physical size. This is surprising, as intuition tells us that *surely* only a finite number of black stars would be needed.If the answer is indeed infinity, the implications of this are surprising. It means you could fill up a glass container of any size with as many marbles as you wanted and, if you look from the right place, you'd still be able to see every single marble in the jar.If any other math lovers express interest in this, I'll share where I'm at with the proof. I believe I'm nearly at the end and that the answer may well be infinity.I don't see the problem. Just put black stars in a tight geodesic dome (spherical shell) arrangement around the white star. Then plug in black stars of appropriate size to fill the gaps. I think only one extra layer would be needed.Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 5/28/2016 4:33:53 AMPosted: 1 year agoAt 5/28/2016 2:19:53 AM, dee-em wrote:I don't see the problem. Just put black stars in a tight geodesic dome (spherical shell) arrangement around the white star. Then plug in black stars of appropriate size to fill the gaps. I think only one extra layer would be needed.Intuitively, that seems like a valid solution. But when you start doing the math on finding what this "spherical shell" would look like, you start to see that there may very well not exist any configuration that works.
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 5/28/2016 7:57:45 AMPosted: 1 year agoAt 5/28/2016 4:33:53 AM, Cobalt wrote:At 5/28/2016 2:19:53 AM, dee-em wrote:I don't see the problem. Just put black stars in a tight geodesic dome (spherical shell) arrangement around the white star. Then plug in black stars of appropriate size to fill the gaps. I think only one extra layer would be needed.Intuitively, that seems like a valid solution. But when you start doing the math on finding what this "spherical shell" would look like, you start to see that there may very well not exist any configuration that works.I don't buy it. If lead shielding can block x-rays then it is possible. After all an atom can be considered a sphere and even then it is mostly space. If such lead atoms in a dense arrangement can block radiation, think how more effective your "black stars" would be in their stead. I still fail to see the problem.Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 5/28/2016 10:40:57 AMPosted: 1 year agoAt 5/28/2016 7:57:45 AM, dee-em wrote:I don't buy it. If lead shielding can block x-rays then it is possible. After all an atom can be considered a sphere and even then it is mostly space. If such lead atoms in a dense arrangement can block radiation, think how more effective your "black stars" would be in their stead. I still fail to see the problem.When I've finished the proof, I'll let you know.This problem takes place entirely in the mathematical realm, however. It's not a physics problem. I don't think the x-ray comparison is apt, since x-rays are blocked once a particular density in the "shield" is met, not once there is absolute mathematical blockage.But until I churn out the proof, I can't be sure that I'm right.
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 5/28/2016 11:22:15 AMPosted: 1 year agoAt 5/28/2016 10:40:57 AM, Cobalt wrote:At 5/28/2016 7:57:45 AM, dee-em wrote:I don't buy it. If lead shielding can block x-rays then it is possible. After all an atom can be considered a sphere and even then it is mostly space. If such lead atoms in a dense arrangement can block radiation, think how more effective your "black stars" would be in their stead. I still fail to see the problem.When I've finished the proof, I'll let you know.This problem takes place entirely in the mathematical realm, however. It's not a physics problem. I don't think the x-ray comparison is apt, since x-rays are blocked once a particular density in the "shield" is met, not once there is absolute mathematical blockage.You will need to elaborate because I don't understand the distinction.But until I churn out the proof, I can't be sure that I'm right.This link might help to illustrate what I mean:http://www.seas.upenn.edu...This lattice arrangement (cubic close packed) of equal spheres blocks light. I suspect it might be even easier with unequal spheres. I don't think the integrity would be effected by wrapping this lattice structure on a plane around a central "white star" at a sufficient distance. I could be wrong though, so I'll await your proof.Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 5/28/2016 11:42:08 AMPosted: 1 year agoIt occurs to me that we could consider the Sun as the white star in your puzzle. That being the case, what you are essentially asking is whether it is possible to find a place of perfect darkness here on Earth (treating all atoms of matter as black stars), ie. Is there a place where not a single photon can penetrate even in principle?Does anyone have an answer to this? Is it possible to create an environment with some material which is totally 100% impervious to light? I mean in a mathematically pure sense not in practical terms.Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 6/4/2016 12:59:29 AMPosted: 1 year agoAt 5/28/2016 10:40:57 AM, Cobalt wrote:At 5/28/2016 7:57:45 AM, dee-em wrote:I don't buy it. If lead shielding can block x-rays then it is possible. After all an atom can be considered a sphere and even then it is mostly space. If such lead atoms in a dense arrangement can block radiation, think how more effective your "black stars" would be in their stead. I still fail to see the problem.When I've finished the proof, I'll let you know.This problem takes place entirely in the mathematical realm, however. It's not a physics problem. I don't think the x-ray comparison is apt, since x-rays are blocked once a particular density in the "shield" is met, not once there is absolute mathematical blockage.But until I churn out the proof, I can't be sure that I'm right.Hey Cobalt, still awaiting your mathematical proof.You wouldn't be a tease, would you?Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 6/4/2016 1:32:39 AMPosted: 1 year agoAt 6/4/2016 12:59:29 AM, dee-em wrote:Hey Cobalt, still awaiting your mathematical proof.You wouldn't be a tease, would you?Haha, I was *literally* just working on it. It's not easy to prove. I conjecture that any purely convex shape with zero surface-area contact results in the inability to occlude the white star. (A bundle of spheres only touch each other at single point, so the surface area is mathematically zero, if we only considering where they touch.)
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 6/4/2016 1:37:00 AMPosted: 1 year agoAt 5/28/2016 11:42:08 AM, dee-em wrote:It occurs to me that we could consider the Sun as the white star in your puzzle. That being the case, what you are essentially asking is whether it is possible to find a place of perfect darkness here on Earth (treating all atoms of matter as black stars), ie. Is there a place where not a single photon can penetrate even in principle?To this, no. In the mathematical problem I am suggesting, there would be places of absolute darkness on Earth. (More than 50% of the Earth, in fact.)Imagine a "boundary sphere" that is hollow and contains the white star and all needed black stars. You have succeeded if no light touches any point on the boundary sphere. Some areas of darkness are easy to achieve. In fact, as the size of the white star approaches zero, the optimal configuration has the occlusion tending toward 100%. But since the white star must have a definite size, this complete occlusion can never take place.Does anyone have an answer to this? Is it possible to create an environment with some material which is totally 100% impervious to light? I mean in a mathematically pure sense not in practical terms.This is easily achievable in a mathematical sense. For instance, one could simply consider a 3x3x3 grid of cubes in which the middle cube is missing and replaced with the white star. The cubes would form a complete barrier, occluding all light. (Mathematically. The physics of this may not be true, but that is irrelevant to the problem.)
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 6/4/2016 4:43:26 AMPosted: 1 year agoAt 6/4/2016 1:32:39 AM, Cobalt wrote:At 6/4/2016 12:59:29 AM, dee-em wrote:Hey Cobalt, still awaiting your mathematical proof.You wouldn't be a tease, would you?Haha, I was *literally* just working on it. It's not easy to prove. I conjecture that any purely convex shape with zero surface-area contact results in the inability to occlude the white star. (A bundle of spheres only touch each other at single point, so the surface area is mathematically zero, if we only considering where they touch.)Yes, but you seem to be only considering a single layer of black spheres. Was that one of the constraints? If not, it is easy to conceive of a square lattice of spheres in one layer and then a second layer in which new spheres plug the gaps. That would definitely not let any light through when the layers are flat. I see no problem in forming such an arrangement as a shell around the white star at a sufficient distance.I don't agree that this is a purely mathematical problem. The physical arrangement of spheres in the lattice which is used comes into play.Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 9,792 Add as FriendChallenge to a DebateSend a Message 6/4/2016 4:54:14 AMPosted: 1 year agoAt 6/4/2016 1:37:00 AM, Cobalt wrote:At 5/28/2016 11:42:08 AM, dee-em wrote:It occurs to me that we could consider the Sun as the white star in your puzzle. That being the case, what you are essentially asking is whether it is possible to find a place of perfect darkness here on Earth (treating all atoms of matter as black stars), ie. Is there a place where not a single photon can penetrate even in principle?To this, no. In the mathematical problem I am suggesting, there would be places of absolute darkness on Earth. (More than 50% of the Earth, in fact.)How so if the spherical atoms of the Earth cannot perfectly block light as you claim?Imagine a "boundary sphere" that is hollow and contains the white star and all needed black stars. You have succeeded if no light touches any point on the boundary sphere. Some areas of darkness are easy to achieve. In fact, as the size of the white star approaches zero, the optimal configuration has the occlusion tending toward 100%. But since the white star must have a definite size, this complete occlusion can never take place.Yes, I know you are asserting this, but you don't give a reason.Does anyone have an answer to this? Is it possible to create an environment with some material which is totally 100% impervious to light? I mean in a mathematically pure sense not in practical terms.This is easily achievable in a mathematical sense. For instance, one could simply consider a 3x3x3 grid of cubes in which the middle cube is missing and replaced with the white star. The cubes would form a complete barrier, occluding all light. (Mathematically. The physics of this may not be true, but that is irrelevant to the problem.)I wasn't discussing cubes. I was referring to atoms of matter which can be considered spherical (assuming the electron shells were impenetrable which I know they aren't).Fu-Ming: "So explain to us how wasps look like wasps 100 million years later. I'll wait". Me: What part of the theory of evolution states that a species must change? I'll wait. Fu-Ming: *crickets*
 Posts: 393 Add as FriendChallenge to a DebateSend a Message 6/4/2016 1:56:31 PMPosted: 1 year agoYour problem breaks down when applied to reality. Light has wave-particle duality, thus is cannot propagate through a theoretical point in space smaller than the plank length. It would be blocked instead. Therefore the solution is to surround the white star with black stars each 1 plank metre in diameter. This is a finite number, not an infinite number. And yes, this finite number of black stars would absorb every single emitted photon.For Mother Russia.
 Posts: 393 Add as FriendChallenge to a DebateSend a Message 6/4/2016 2:04:55 PMPosted: 1 year agoAlso, another solution to your problem using a finite number of stars is to solve it using fractal geometry. Let's say you take one hemisphere of a white star, you can block half of the hemisphere's light with a star the same size as the original star and then surround that black star with more black stars which are half the size of the first black star. Then surround those smaller black stars with stars half the size again, and continue until you reach the plank length. Congrats, you've just blocked out an entire region of the white star's accessible light. Restart that process on the opposite hemisphere and all light will be blocked.Normally we'd use this fractal theorem to satisfy the problem 1/2 + 1/4 + 1/8 + 1/16 .... =1 however that requires an infinite number of black stars, which we aren't allowed to use. So stopping at the plank length is sufficient.For Mother Russia.
 Posts: 4,509 Add as FriendChallenge to a DebateSend a Message 6/4/2016 7:23:15 PMPosted: 1 year agoSee if you buy this. We'll start with occluding plane problem. Start with an infinite array of hexagonally packed spheres. If the array is projected on a plane it looks like circles touching each other with six circle around each circle and the centers on a hexagonal lattice. Another layer of spheres floating above the first produces the same projected pattern with some offset. the second layer is moved so that the centers of the spheres are moved to the centers of the gaps in the first array, is the whole plane covered?It is. Think of each projected circle as having a center circle that is completely covered, plus another area that is not completely covered. From this viewpoint, we have an array of touching triangles that we know completely covers half the plane. The array of not-completely covered triangles has triangles of exactly the same size as the triangles of complete coverage, and the lattice arrangement is exactly the same. Therefore a translation of the second layer on top of the first will completely cover the plane. Therefore atoms in two layers can completely block light.Moving to 3D, we surround the white star with a tetrahedral pyramid of four black stars. Each sphere is touched by three other spheres at three points. The touch points are not on the equator of the sphere. For any one of the spheres, the three touch points defines a triangle that lies entirely inside that sphere. Considering the four triangles defined inside the spheres, they are four of the eight faces of a regular octahedron than surrounds the white star. The other four faces of the octahedron are spaces left to be filled. Now construct a second pyramid of four black stars so large that it encloses the first pyramid. The same technique yields an octahedron with four covered triangular faces and four triangular spaces to be filled. The scale of the second open octagon does not affect its occlusion coverage, so we can rotate and shrink it to fill the spaces left open in by the first open octagon. All the triangles in both constructions are equilateral so the fit can be accomplished.The white star can therefore be completely occluded by eight black stars.I have never seen this problem before either, and it's really a great head-scratcher.
 Posts: 4,509 Add as FriendChallenge to a DebateSend a Message 6/4/2016 7:27:59 PMPosted: 1 year agoIn paragraph two above, I should have said " a center triangle that is completely covered."
 Posts: 2,626 Add as FriendChallenge to a DebateSend a Message 6/6/2016 8:10:23 AMPosted: 1 year agoAt 5/27/2016 6:49:04 AM, Cobalt wrote:I have a fantastic problems for all lovers of math and puzzles out there. I call this a "math" problem because it does require some trigonometry and geometric constructions, but definitely nothing a first year math student wouldn't be able to do.Before I state it, you should know that I read this problem *ages* ago and that you will probably not find the answer online. I do remember the answer provided, but the proof has been lost to my mind. So I'm here to get your input and test your abilities!Problem: Consider two types of objects called "black stars" and "white stars". Both are spheres of definite size, but each has a unique property. Black stars absorb every light particle that hits them. They do not produce reflections. White stars, on the other hand, emit light from every point in all directions.Your job is to take one white star and surround as many black stars as necessary around it such that all light from the white star is being blocked by black stars. In other words, if I get in my space ship and drive around your construction, I should not be able to see even a fleck of light. Black stars should not go through each other or the white star. They may touch, however.Mathematically stated: Construct a three dimensional model that contains a central sphere of definite size surrounded by other spheres of any size and in any number, such that one can construct a "boundary sphere" that encloses the construction where there does not exist a ray that can be drawn from a point on the central sphere and extended to the boundary sphere without passing through any internal sphere.Clarifications: This is not a physics problem. You cannot make the white star so big that it turns into a black hole, absorbing all light. You cannot consider large dark stars as having gravity. Light waves do not bend at all.Additionally, you can arbitrarily change the size of the stars as you'd like, black or white. You do not need to have all your dark stars be the same size. However, all stars must have a *definite size*. The white star cannot be infinitely small (a point) and the black stars cannot be infinitely large.Think on this for a while, do some maths, then keep reading. Below here I will discuss what I believe the answer to be, though I have yet to prove it.So really, go away if you don't want hints.Leave now.Ok, good.The purported answer was that it would require an infinite number of black stars, so long as you remember that the white star still must have a physical size. This is surprising, as intuition tells us that *surely* only a finite number of black stars would be needed.If the answer is indeed infinity, the implications of this are surprising. It means you could fill up a glass container of any size with as many marbles as you wanted and, if you look from the right place, you'd still be able to see every single marble in the jar.If any other math lovers express interest in this, I'll share where I'm at with the proof. I believe I'm nearly at the end and that the answer may well be infinity.Tough indeed."Hate begets hate"
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 6/6/2016 3:35:28 PMPosted: 1 year agoAt 6/4/2016 4:54:14 AM, dee-em wrote:Ah, I see what you meant. (If what I'm saying is correct) it would be possible to have localized areas of darkness on one side of the Earth. I don't know if this occlusion would be complete, since it is an entirely different problem. (White star located externally as opposed to internally)
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 6/6/2016 3:37:31 PMPosted: 1 year agoAt 6/4/2016 2:04:55 PM, Syko wrote:Also, another solution to your problem using a finite number of stars is to solve it using fractal geometry. Let's say you take one hemisphere of a white star, you can block half of the hemisphere's light with a star the same size as the original star and then surround that black star with more black stars which are half the size of the first black star. Then surround those smaller black stars with stars half the size again, and continue until you reach the plank length. Congrats, you've just blocked out an entire region of the white star's accessible light. Restart that process on the opposite hemisphere and all light will be blocked.Normally we'd use this fractal theorem to satisfy the problem 1/2 + 1/4 + 1/8 + 1/16 .... =1 however that requires an infinite number of black stars, which we aren't allowed to use. So stopping at the plank length is sufficient.That's why I clarified that this is a pure maths problem, not a physics problem. There are (many) physics related reasons as to why this is possible. When looked at as a math problem, we assume there is no Planck length and that light can "squeeze in" through any size gap, provided the passage is a straight line.
 Posts: 998 Add as FriendChallenge to a DebateSend a Message 6/6/2016 3:54:48 PMPosted: 1 year agoAt 6/4/2016 7:23:15 PM, RoyLatham wrote:See if you buy this. We'll start with occluding plane problem. Start with an infinite array of hexagonally packed spheres. If the array is projected on a plane it looks like circles touching each other with six circle around each circle and the centers on a hexagonal lattice. Another layer of spheres floating above the first produces the same projected pattern with some offset. the second layer is moved so that the centers of the spheres are moved to the centers of the gaps in the first array, is the whole plane covered?It is. Think of each projected circle as having a center circle that is completely covered, plus another area that is not completely covered. From this viewpoint, we have an array of touching triangles that we know completely covers half the plane. The array of not-completely covered triangles has triangles of exactly the same size as the triangles of complete coverage, and the lattice arrangement is exactly the same. Therefore a translation of the second layer on top of the first will completely cover the plane. Therefore atoms in two layers can completely block light.Moving to 3D, we surround the white star with a tetrahedral pyramid of four black stars. Each sphere is touched by three other spheres at three points. The touch points are not on the equator of the sphere. For any one of the spheres, the three touch points defines a triangle that lies entirely inside that sphere. Considering the four triangles defined inside the spheres, they are four of the eight faces of a regular octahedron than surrounds the white star. The other four faces of the octahedron are spaces left to be filled. Now construct a second pyramid of four black stars so large that it encloses the first pyramid. The same technique yields an octahedron with four covered triangular faces and four triangular spaces to be filled. The scale of the second open octagon does not affect its occlusion coverage, so we can rotate and shrink it to fill the spaces left open in by the first open octagon. All the triangles in both constructions are equilateral so the fit can be accomplished.The white star can therefore be completely occluded by eight black stars.I have never seen this problem before either, and it's really a great head-scratcher.I like where your head is at.This is exactly the geometric pattern that my mind went to initially. We suppose that we have this four sphere "first shell" around the white star. The "holes" that are formed don't quite produce equilateral triangles, since the white star has definite size and the holes are three dimensional. The would form a sort of exaggerated triangle with each side concave. All of these triangles would meet one another at their vertexes (since the spheres touch with a total surface area of zero, the total light "shape" would effectively be one large piece, only technically separated at single points where two spheres met.)This is why I'm having such difficulties. Imagine a white star of size r, a black star of size r' and a distance between them as d.When d = 0, the occlusion circle cast on the boundary sphere has some arclength A. (This arclength approaches 180 degrees as r tends toward 0, interestingly.) If we increase d by some amount, the arclength of occlusion A diminishes. Basically, objects closer to the star produce bigger shadows than objects further from the star.At the point where you start making shells, you begin limiting the "closest" place you can put another black star. You either have to place dark stars further away (increase d) or you have to make them smaller (reduce r'). Both of these actions decrease the occlusion arclength very quickly. (The falloff is initially very fast, followed by a slower falloff).My current struggle is in determining whether occlusion happens more quickly than "occlusion falloff" or less quickly. If it's the latter, then the problem can be solved finitely. If it isn't, then the problem is indeed an issue of infinity.