Wikipedia:Reference desk/Archives/Mathematics/2010 November 16

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November 16

Maximize volume of a cone

Hi all,

How do you find out the angle that maximizes the volume of a cone using calculus?

More specifically, http://jwilson.coe.uga.edu/emt725/Class/Lanier/Cone/image6.gif

Given that R is constant, what angle (as shown in the diagram) maximizes the volume of the cone? Here is what I got so far:

• Arc length = (1-θ/2π)2πR
• Circumference of base = 2πr

Arc length = Circumference of base

• (1-θ/2π)2πR = 2πr
• r = (1-θ/2π)R

Height of cone

• h = sqrt(R^2 - r^2)
• h = R sqrt(1-(1-θ/2π)^2)

Volume of cone

• V = (1/3) π r^2 h
• V = (1/3) π (1-θ/2π)^2 (R^2) (R sqrt(1-(1-θ/2π)^2))
• V = (1/3) π R^3 (1-θ/2π)^2 sqrt(1 - (1-θ/2π)^2)

What do I do from here? I tried differentiating the function with respect to θ and let it equal to 0:

• 0 = (1-θ/2π)sqrt(1-(1-θ/2π)^2) + (1/2)(1-θ/2π)^3(1-(1-θ/2π)^2)^(-1/2)

How do I solve for θ and how do I justify that the θ I find gives the maximum volume?

Thanks in advance! —Preceding unsigned comment added by 169.232.246.52 (talk) 05:01, 16 November 2010 (UTC)

The expressions will be simpler if you use the other angle (α=2π-θ) and convert afterward; and the derivative will be simpler if you square the volume. I'm just sayin'. —Tamfang (talk) 05:32, 16 November 2010 (UTC)

I didn't check your work, but from your final equation you can substitute $\cos(\beta )=1-\theta /2\pi$ to get

$0=\cos(\beta )\sin(\beta )+{\frac {1}{2}}\cos ^{3}(\beta )/\sin(\beta )$

which is easy to solve--multiply by sin, divide by cos, and replace the sin^2 with 1-cos^2. Any extrema of a differentiable function on a closed interval occurs either at the endpoints or at a point where the first derivative is zero. So, check the volume at the extreme values of theta and at the value induced by the above equation. If the latter is the largest, it's the global maximum. 67.158.43.41 (talk) 08:29, 16 November 2010 (UTC)

Using $\cos(\beta )=1-\theta /2\pi$

0 = cosβ sinβ + (1/2) (cosβ)^3 / sinβ

0 = cosβ (sinβ)^2 + (1/2) (cosβ)^3

0 = cosβ (1 - (cosβ)^2) + (1/2) (cosβ)^3

0 = cosβ - (cosβ)^3 + (1/2) (cosβ)^3

0 = cosβ - (1/2) (cosβ)^3

0 = cosβ( 1 - (1/2) (cosβ)^2)

First case

cosβ = 0

1-θ/2π = 0

θ = 2π

Second case

( 1 - (1/2) (cosβ)^2)= 0

cosβ = sqrt(2) / cosβ = -sqrt(2)

1-θ/2π = sqrt(2) / 1-θ/2π = -sqrt(2)

θ = (1-sqrt(2)) (2π) / θ = (1 + sqrt(2)) (2π)

Is this how I do it? —Preceding unsigned comment added by 169.232.246.199 (talk) 20:28, 16 November 2010 (UTC)

Your second case is off; it should be |cos beta| = sqrt(2). I get theta = (1 +/- sqrt(2)) * 2pi, as does Wolfram Alpha (which I'd say you should only use to double-check your work). 67.158.43.41 (talk) 21:21, 16 November 2010 (UTC)

Homeomorphisms of the closed disk

Perhaps this is a dense question, but how do I go about seeing that all orientation-preserving homeomorphisms of the closed disk are isotopic to the identity map? It's obvious in the hand-wavey case of homeomorphisms which can be visualised by stretching/contorting the space in a continuous manner, but how do I know that all (orientation-preserving) homeomorphisms are of this type, and that there aren't some exotic ones which can't be isotop'd to the identity map? Thanks, Icthyos (talk) 16:16, 16 November 2010 (UTC)

(I did of course mean to say orientation-preserving homeomorphisms from the closed disk to itself...) Icthyos (talk) 10:54, 17 November 2010 (UTC)

See Alexander's trick (is having a 'trick' named after you ven better than a lemms?). I'll put a reference to this under isotopy. Dmcq (talk) 11:58, 17 November 2010 (UTC)

Ah-hah, I see. Thanks! (I'd argue that calling something a 'trick' makes it seem less grand (and a bit more trivial) than a lemma - even if it's not...) Icthyos (talk) 12:14, 17 November 2010 (UTC)

There was a discussion here about whether it was better to have a lemma named after you or a theorem ;-) Dmcq (talk) 12:59, 17 November 2010 (UTC)

I'm not altogether sure about the homeomorphisms where the boundary is not fixed and is a 4 sphere. Is this a part of the Generalized Poincaré conjecture? Dmcq (talk) 14:23, 17 November 2010 (UTC)

I feel that question is a little over my head! I'm working with mapping class groups though, so I always assume that the boundary components are fixed, pointwise. On a related note, how does one go about showing that every orientation-preserving self-homeomorphism of the 2-sphere is isotopic to the identity? I thought about decomposing the sphere as two closed disks, joined along their boundaries, and using Alexander's trick simultaneously on both, but then I realised that would require the homeomorphism of the sphere to leave a great circle fixed, point-wise, which I can't see always being true. Is it possible to adapt the Alexander argument to see that the mapping class group of the 2-sphere is trivial, or is another insight needed? Thanks, Icthyos (talk) 19:55, 17 November 2010 (UTC)

Pi = 4

I'm so confused, was everything I was told about math a lie?

http://tinypic.com/r/27wufn/7 76.68.247.201 (talk) 16:50, 16 November 2010 (UTC)

The problem is with the last step. How do you know that the limit of the cutting process gives a circle? Something is clearly special close to the points where the original square was tangent to the circle (at 12 o'clock, 3 o'clock, 6 o'clock and 9 o'clock). My imagination tells me that the limit will be a hexagon octagon and not a circle. — Fly by Night (talk) 17:02, 16 November 2010 (UTC)

No, the limit is a circle. The problem is with the implicit assumption that the length of the limiting curve is the limit of the lengths of the curves. No argument is given for this assumption and it is in fact false. Algebraist 17:06, 16 November 2010 (UTC)

(edit conflict) Whoops, I meant octagon. — Fly by Night (talk) 17:07, 16 November 2010 (UTC)

Interesting... Is there a link to show that the assumption is wrong? — Fly by Night (talk) 17:09, 16 November 2010 (UTC)

The OP's original link shows this, since it proves a false conclusion from this assumption. You can do the same thing without the complication of the circle by considering instead the diagonal of the unit square. If we approximate this in the same way with curves that are everywhere either horizontal or vertical, then all the approximating curves will have length 2, but will converge uniformly to the diagonal. Algebraist 17:13, 16 November 2010 (UTC)

Does that show that the lengths of the curves don't converge to length of the diagonal, or that the curves themselves don't converge to the diagonal? In the circle case you have a sequence of lengths (s_n) with the property that s_n = 4 for all n. It stands to reason that the limit of s_n as n tends to infinity is 4; while the circumference of the circle is 2π ≠ 4. This would suggest to me that the limit of the zig-zags is not actually the circle. — Fly by Night (talk) 17:22, 16 November 2010 (UTC)

The limit of the zig-zags is most definitely the circle. The original construction may be a bit messy to rigorously analyze, so here's a different construction: take a regular square grid with distance ε between the grid lines, and consider a zig-zag line on the grid which goes as close to the circle as possible. Its length will be still 4, but it will be within distance ε√2 of the circle. By taking a sequence of such zig-zags for ε_n → 0, you will get a sequence of curves of length 4 whose limit is the circle.—Emil J. 17:44, 16 November 2010 (UTC)

Another example: let c_k be the piecewise linear function going through the points (0,0), (1/2k²,1/k),(2/2k²,0),(3/2k²,1/k),(4/2k²,0),...,((2k² − 1)/2k²,1/k),(1,0). Then c_k converge uniformly to the line from (0,0) to (1,0) as k → ∞, but their lengths, bounded below by 2k, go to infinity.—Emil J. 17:53, 16 November 2010 (UTC)

It's some very funny stuff this; if you ask me. It just doesn't feel right. Each successive zig-zag curve touches the circle in a finite number of points. But in the limit, it touches at all points. So a sequence of discreet points has a limit of the whole space. But the cardinality of a finite, discreet set is different to the line. I guess the fact that the circle is compact plays a role here. In a compact space, every sequence has a convergent subsequence; but we're asking that every point be an adherent point of the limit of the sequence. Consider the map g : Z → S¹ given by g(n) = (cos(n), sin(n)), is g(Z) = S¹? It's very much like space filling curves. — Fly by Night (talk) 18:05, 16 November 2010 (UTC)

I don't understand why it doesn't feel right to you. The curves $y=1/n$ obviously approach the limiting curve $y=0$ as $n\to \infty$, but of course each successive curve doesn't intersect the limiting curve at all. —Bkell (talk) 19:58, 16 November 2010 (UTC)

Yeah, after thinking about it again, your topological arguments don't make sense. It isn't the sets of points of intersection whose limit is the circle; the circle is the limit of the zigzag curves. As another example, consider the curves $y=\sin(x)/n$ for $n\in \mathbb {N}$. The limiting curve here is plainly $y=0$, but the set of intersection points at every stage is the same set $\{\,(k\pi ,0)\mid k\in \mathbb {Z} \,\}$. To answer your question ("is g(Z) = S¹?"), the answer is no: the set g(Z) is countable, while the set S¹ is uncountable. For a specific example of a point in S¹ which is not in g(Z), consider the point (−1, 0). —Bkell (talk) 20:59, 16 November 2010 (UTC)

What is true is that g(Z) is dense in S¹. —Bkell (talk) 21:01, 16 November 2010 (UTC)

(edit conflict) By giving the height and depth of each "step" appropriate values, you can fit the staircase as closely as you like to the graph of any montonic function between the points (-1/2,0) and (0,1/2), thus "proving" that the length of any such curve is 1. Gandalf61 (talk) 17:16, 16 November 2010 (UTC)

Don't you mean 1/√2? 76.68.247.201 (talk) 18:14, 16 November 2010 (UTC)

No, he doesn't. Algebraist 18:26, 16 November 2010 (UTC)

How silly of me, for some reason I was using Pythagoras...76.68.247.201 (talk) 19:39, 16 November 2010 (UTC)

I couldn't find an article about this fallacy, Mathematical fallacies has nothing like it. I'm pretty certain I've seen things like this in reliable sources and it is notable. Any idea if it is covered somewhere else? Dmcq (talk) 17:47, 16 November 2010 (UTC)

It reminds me of something I read about a long time ago. All of these curves lie outside the circle and they all have greater lengths and internal areas than the circle. This gives upper bounds. Then you start with a square inside the circle, with the four vertices lying on the circle. You add little squares, and then some even smaller squares, etc. You get a zig-zag inside the circle whose length and internal area are lower bounds for the circle. It's quite an old idea. — Fly by Night (talk) 17:55, 16 November 2010 (UTC)

You can do the same zigzag on the inside of the circle, again with length 4. Thus this doesn't give a lower bound. If you want a lower bound you need to take a piecewise linear curve all of whose corners lie on the circle, as described in arc length#Definition. Algebraist 18:26, 16 November 2010 (UTC)

I'm sure you know exactly what I was trying to get at. You inscribe a 4-gon, 5-gon, 6-gon, … to give a lower bound and you circumscribe a 4-gon, 5-gon, 6-gon, … to give an upper bound. — Fly by Night (talk) 18:39, 16 November 2010 (UTC)

Algebraist, I don't think a similar rectilinear zigzag on the inside of the circle would have length 4. In the simplest case, a square inscribed in a circle of diameter 1, we have a perimeter of $2{\sqrt {2}}$. In general, such a zigzag would have length $2(x_{\mathrm {max} }-x_{\mathrm {min} })+2(y_{\mathrm {max} }-y_{\mathrm {min} })$, where $x_{\mathrm {max} },x_{\mathrm {min} },y_{\mathrm {max} },y_{\mathrm {min} }$ are the minimum and maximum x- and y-values attained by the curve, which must lie strictly within the interval (−1/2, 1/2), assuming the circle is centered at the origin. So, as the inside zigzag approaches the circle more and more closely, its length will increase to a limit of 4. (So of course you're right that it doesn't provide a lower bound.) —Bkell (talk) 19:48, 16 November 2010 (UTC)

Oh, true. I was thinking of a different and rather silly zigzag. Algebraist 19:50, 16 November 2010 (UTC)

Alright, I think I'm starting to understand it, but I don't yet completely...is this anything like the coastline of england? 76.68.247.201 (talk) 18:14, 16 November 2010 (UTC)

Similar; except without the rain and the wind. Another article that you might like is Koch snowflake. It's a curve of infinite length that fits into a finite area. Check it out. There are some really nice pictures. — Fly by Night (talk) 18:23, 16 November 2010 (UTC)

How is it similar? They're both about approximating arc lengths using piecewise linear curves, but I can't see any more connection than that. Algebraist 18:26, 16 November 2010 (UTC)

I didn't say they (the snow flake and the zig-zag) were similar, nor connected. Simply that the OP might like the article. There's no need to be so cantankerous. Posts should add to the discussion, and not just take away from what others write. This post was totally unrelated to anything I wrote; you just wanted to have a go. Please stop. — Fly by Night (talk) 18:39, 16 November 2010 (UTC)

I know you didn't. You suggested the case at hand is similar to the coastline of England, by which the OP presumably meant the coastline paradox as described in the famous article How Long Is the Coast of Britain? Statistical Self-Similarity and Fractional Dimension. I'd like to know what similarity you see there. Please do not invent motivations for my actions. Algebraist 18:44, 16 November 2010 (UTC)

I don't invent; I have a body of knowledge to demonstrate. — Fly by Night (talk) 19:02, 16 November 2010 (UTC)

Wow, I've seen my fair share of immaturity on the internet, but I guess I was naive to think that a question on a wikipedia reference desk would be immune to it. I'm not a Wikipediean myself, so perhaps I'm not in a position to comment, but I think you two should resolve your bickering elsewhere. It's juvenile.

The reason I brought up the coastline paradox is that I saw an analogy between the pi = 4 "proof" and fractals (although, my knowledge of fractals is very poor, so I expect I'm in the dark about this). Working off what Emil J said, as the length ε of each step decreases, so does the distance between each step and the circle (ε/√2). As ε --> 0, this distance tends to zero as well. But because the total number of steps increases as well, the two end up canceling each other out, so that there still remains a net difference, in a sense, between the perimeter of the circle and the perimeter of the staircase.

My understanding of the coastline paradox is something as follows: If there existed a true length for the coastline of Britain (with a corresponding true coastline of Britain), adding a single grove to it, like a rock, would have a negligible effect on the total length of the coastline. But if small rocks were placed all over the coastline, there would be a finite addition to the length of the coastline, and hence it is really impossible to measure the coastline of Britain, because every speck of sand matters.

There appears to be an analogy between the two, although I might be way over my head. 76.68.247.201 (talk) 19:39, 16 November 2010 (UTC)

The perimeter at any point in the operation is constant at 4; I don't understand what you mean is canceling what. Emil's construction just shows more clearly than the original that the circle is approximated arbitrarily well by this type of operation, but that its perimeter is not. 67.158.43.41 (talk) 21:34, 16 November 2010 (UTC)

Silly off-topic humour

I'm going to tell my mum what you said! — Fly by Night (talk) 20:13, 16 November 2010 (UTC) I'd be delighted to hear her thoughts on approximating curves with piecewise linear functions! I could have easily made a 'your mom' joke, but I thought it best to spare you 76.68.247.201 (talk) 20:20, 16 November 2010 (UTC) Share me? Don't you mean spare yourself from becoming a hypocrite? I'll let you know what she says. — Fly by Night (talk) 20:29, 16 November 2010 (UTC) You're implying yo mama jokes are in any way childish. They are, in fact, the highest form of humor. Only an urbane gentleman could appreciate this. 76.68.247.201 (talk) 20:37, 16 November 2010 (UTC) ;-) — Fly by Night (talk) 21:05, 16 November 2010 (UTC)

By the way something nasty related to this happens in the Calculus of variations called the Lavrentiev phenomenon where one might find that a route that zig zags back and forth gives a better result than a smooth route along the same path. Dmcq (talk) 10:39, 17 November 2010 (UTC)

This paradox (mostly the straight-line case) is discussed in Riddles in Mathematics by Eugene P. Northrop, where he incorrectly asserts that the limiting shape is not the same as the line being estimated. (This is by no means the only error in the book.) AndrewWTaylor (talk) 14:44, 17 November 2010 (UTC)

By the way, it is possible to turn the paradox into a working definition. Namely, consider the definition of the length of a curve by rectification as in Arc length#Definition, but use the L₁ metric instead of the Euclidean metric. Then the length of the line segment from (x,y) to (u,v) is |x − u| + |y − v|, the length of the unit circle is 4, and in general, the length of any curve can be found by approximating it with zig-zags (using horizontal and vertical lines only).—Emil J. 15:18, 17 November 2010 (UTC)

Interesting, are there any areas of math where this is useful? 76.68.247.201 (talk) 01:02, 18 November 2010 (UTC)

Question about a Clock

Hi Wikipedia,

I have a question that I couldn't solve:

'The big hand and the small hand on a clock meet at noon and then again a little over an hour later. What’s the exact time difference between these two events?'

Analytically this wouldn't be a hard problem at all, but when I was asked to use differential calculus to solve this I simply had no idea.

Does anyone have any idea how to do this?

Thanks! —Preceding unsigned comment added by 169.232.246.199 (talk) 18:36, 16 November 2010 (UTC)

The problem doesn't have anything to do with differential calculus as far as I can see. The equations are already linear so using a linear approximation doesn't simplify them.--RDBury (talk) 18:57, 16 November 2010 (UTC)

You don't need any sort of calculus. An easy observation is that the big and small hands meet at regular intervals. Just count how many times they meet between noon and midnight (excluding one of the endpoints), and divide 12 hours by that.—Emil J. 18:59, 16 November 2010 (UTC)

You should just use vectors. Imagine the hour hand and the minute hand as vectors of equal length that move around the unit circle. The hour hand makes one revolution per hour, and so assuming that we start at midnight the hour hand can be written as

${\mathbf {h} }(t)=(\sin(2\pi t),\cos(2\pi t)),$

where t is measured in hours. The minute hand rotates sixty times per hour and so

${\mathbf {m} }(t)=(\sin(120\pi t),\cos(120\pi t)).$

If they point in the same direction then h and m will be linearly dependent vectors. If they point in opposite directions they will be too, but we will sort that out. The vectors h and m are linearly dependent if and only if

$\sin(2\pi t)\cos(120\pi t)-\sin(120\pi t)\cos(2\pi t)=\sin(-118\pi t)=-\sin(118\pi t)=0\,.$

We can solve this using the periodicity of the sine function:

$\sin(118\pi t)=0\iff 118\pi t\in \left\{\pi n:n\in {\mathbf {Z} }\right\}.$

If n is even then the hour and minute hands point in the same direction; if n is odd then they point in opposite directions. It follows that we want

$118\pi t\in \left\{2\pi n:n\in {\mathbf {Z} }\right\}\iff t\in \left\{{\frac {n}{59}}:n\in {\mathbf {Z} }\right\}.$

This gives you the time t, in hours, when the minute and hour hands point in the same direction. The times after midnight are given by

$0\,{\text{h}},{\frac {1}{59}}\,{\text{h}},{\frac {2}{59}}\,{\text{h}},{\frac {3}{59}}\,{\text{h}},{\frac {4}{59}}\,{\text{h}},\ldots$

I hope this helps. — Fly by Night (talk) 20:09, 16 November 2010 (UTC)

Fly, you might want to watch a clock carefully sometime. You write "The hour hand makes one revolution per hour" and "The minute hand rotates sixty times per hour", but actually an hour hand makes one revolution every 12 hours and the minute hand makes one revolution every hour. You seem to have "proved" that the hour hand and the minute hand point in the same direction 59 times every hour, which of course is nonsense. —Bkell (talk) 20:44, 16 November 2010 (UTC)

…Of course, what you write is true if the problem is changed to consider the minute hand and the second hand. —Bkell (talk) 20:48, 16 November 2010 (UTC)

(edit conflict) Any need for the comment "you might want to watch a clock carefully sometime"? How does that help anyone?! The method is perfect, some of the constants are wrong. It seems that I proved when the second hand and the minute hand point in the same direction. Why not change them yourself and give the OP a nice answer? Why is the maths reference desk full of people that tell people their answers are wrong without offering their own correct solutions? It's just a drain on people's energy. I'm done with this page. Have fun everyone. — Fly by Night (talk) 20:50, 16 November 2010 (UTC)

Wow, I'm sorry I offended you. I meant it as a friendly ribbing. —Bkell (talk) 21:10, 16 November 2010 (UTC)

And the reason I didn't "offer my own correct solution" is because EmilJ had basically given a correct solution above, which I didn't feel any need to reiterate or improve upon. —Bkell (talk) 21:14, 16 November 2010 (UTC)

Because the information that a certain answer is wrong is useful to all parties involved. If we didn't correct wrong statements, we'd turn into [enter name of an internet forum full of crap here]. Same goes if we implemented some stupid code of honor which says you are only allowed to comment if you also give a full solution to the OP's problem. -- Meni Rosenfeld (talk) 10:15, 17 November 2010 (UTC)

A contrived solution with differential calculus is to consider the hands as vectors parametrized by time. The local minima of the distance between the vectors are precisely the points at which they overlap. 67.158.43.41 (talk) 21:39, 16 November 2010 (UTC)

How would I model it using vectors parametrized by time then? —Preceding unsigned comment added by 169.232.246.107 (talk) 04:07, 17 November 2010 (UTC)

The vector for each hand is just where the end of the hand is at that time, so for the hour hand you could use (sin(2πt/12h), cos(2πt/12h)) and for the minute hand (sin(2πt/1h), cos(2πt/1h)). I would suggest maximizing the dot product of the two vectors, rather than minimizing the distance since things will work out much nicer that way. Still it's much easier to do what Emil J. suggested than to consider calculus at all. (In fact even this method can get you the answer without calculus if you recognize the right trig identity.) Rckrone (talk) 04:30, 17 November 2010 (UTC)

This method is quite similar to Fly by Night's above (with errors corrected), but motivated by calculus instead of linear algebra. I agree Emil's solution is the most elegant on the page. However, the OP wanted a method using differential calculus, saying "Analytically this wouldn't be a hard problem at all", to me indicating they were capable of creating their own solution, just not with calculus. 67.158.43.41 (talk) 21:21, 17 November 2010 (UTC)

If t is time in minutes after noon, this problem can be reduced to solving t = t/12 + 60. This equation is derived by noticing that the hour hand moves one-twelfth as fast as the minute hand and by recalling the condition that the minute hand is one revolution ahead of the hour hand. 720/11 minutes after noon on a clock face. —Anonymous Dissident^Talk 12:13, 17 November 2010 (UTC)

The hour hand and minute hand first coincide at (5 and 5/11) minutes after 1 o'clock, at which time the second hand is at (27 and 3/11) seconds. Am I right in my belief that at no time in the 12 hours do all three hands coincide, except at 12 o'clock?→86.132.164.178 (talk) 10:54, 18 November 2010 (UTC)

Sums of squares.

(Sorry for the vague section title: can't think of anything appropriate.) If I have N things, and I can group them any way I like, eg. (XXX)(XX)(XXXXX), N=10. Once I've done this, S is the sum of the squares of the items in each group, in this case 9+4+25=38. [You'll notice the number of groups, and their order, is not important.] I supposed that for a given N and S, such a make-up might be unique. However, through mostly guesswork I found the case S=14, N=8:

• Three lots of (NN), and two lots of of (N)
• One lot of (NNN), and five lots of (N)

How common a property is it that a given number can be expressed in two ways? Since I imagine this property becomes more common as N increases, are there any examples of three ways, or more, for N≤100? As an additional point, is there any easy way of finding a solution for given S,N? Thanks, - Jarry1250 ^{[Who? Discuss.]} 21:25, 16 November 2010 (UTC)

A related conjecture is Euler's sum of powers conjecture which you may be interested in, and Euler net has computed a fair amount in relation to that conjecture. For the k=2 case which the additional constraint that the left and right bases sum to the same number, I don't know anything more than you, sorry. 67.158.43.41 (talk) 21:49, 16 November 2010 (UTC)

(edit conflict) For a given value of N, the smallest possible S is N and the largest possible S is N². Moreover, N and S must have the same parity (that is, they are both even or both odd)—to see this, consider any partition of your N things; if N is even, then the number of lots that contain an odd number of things must be even, and so an even number of the squares in your sum are odd, which means S is even (a similar argument applies when N is odd). So the number of possible values of S for a given value of N is approximately $(N^{2}-N)/2=\textstyle {N \choose 2}$. (This count is actually off by 1, because I committed a fencepost error in my counting.) Now, the number of possible ways to partition your N things into lots is given by the partition function p(N), which is approximately $p(N)\sim {\frac {\exp \left(\pi {\sqrt {2N/3}}\right)}{4N{\sqrt {3}}}}$. As N gets larger, this exponential function grows much more quickly than the quadratic function $\textstyle {N \choose 2}$, so your intuition is correct that the property you refer to becomes more common as N increases. For N = 100, the number of possible values of S is $\textstyle {100 \choose 2}+1=4951$, while the number of ways to partition 100 items into lots is p(100) = 190,569,292. So, by the pigeonhole principle, there is some value of S for which there are at least 38,492 different solutions! —Bkell (talk) 21:50, 16 November 2010 (UTC)

For any integer n > 1 there are two distinct partitions of n² + n with the same sum of squares:

$n^{2}+n=(n-1)(n+1)+(n+1)(1)\,;\,(n-1)(n+1)^{2}+(n+1)(1^{2})=n^{3}+n^{2}$

$n^{2}+n=(n+1)(n)\,;\,(n+1)(n^{2})=n^{3}+n^{2}$

Then you can add m lots of 1 to each partition to create two partitions of n² + n + m with the same sum of squares. Your example is the case n=2, m=2. By expressing any integer k ≥ 6 in the form n² + n + m (which can be done in roughly ${\sqrt {k}}$ different ways) this allows us to find pairs of partitions of k with the same sum of squares. Gandalf61 (talk) 17:23, 17 November 2010 (UTC)