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August 21[edit]

Building Area[edit]

If a giant building is shaped like an octagon, and each side (8) is 4000 ft. long, how much ground would the building cover? And that answer times 11 floors?Jk31213 03:51, 21 August 2007 (UTC)[reply]

Have you attempted the problem? Asking how much ground it covers means you need the area. If it is a regular octagon, there are formulae for that. Alternatively, you can decompose the area into several squares and triangles. Strad 01:23, 21 August 2007 (UTC)[reply]

But how on Earth will you tackle part two, when you have to multiply it by 11? deeptrivia (talk) 01:42, 21 August 2007 (UTC)[reply]

I'm srry. I already knew the answer. It's 64,000,000 square feet. 64,000,000 times 11 floors would equal a total floor space of 704,000,000 square feet. This, obviously would be the largest building man has ever built. Just an idea I had for a shopping mall.Jk31213 03:51, 21 August 2007 (UTC)[reply]

I got a different answer, but on the same order of magnitude. Is this a regular octagon? nadav (talk) 04:37, 21 August 2007 (UTC)[reply]

I think I get the same answer as Nadav1. It's more than 64 million, and it's not a rational number. What method did you use to get 64 million per floor, Jk31213? -- JackofOz 04:53, 21 August 2007 (UTC)[reply]

Here's a diagram:

               |   |
               |<->|<- 4000sin(45)
               |   |
         4000 
      +--------+4    ----
     /|        |\0     |
    / |        | \0  4000sin(45)
   /  |        |  \0   |
  +---+--------+---+ ----
  |   |        |   |
  |   |        |   |
  |   |        |   |4000
  |   |        |   |
  |   |        |   |
  +---+--------+---+
   \  |        |  /
    \ |        | /
     \|        |/
      +--------+
         4000

So, I get:

Central square = 1×(4000×4000)
4 rectangles   = 4×(4000×4000sin(45))
4 triangles    = 2×(4000sin(45)×4000sin(45))

You do the math (it doesn't add up to 64,000,000). StuRat 05:04, 21 August 2007 (UTC)[reply]

Strad pointed us to a simpler formula that gives the area of a regular octagon of side length x as, roughly, 4.828427 x*2. This works out to be larger than 64,000,000. -- JackofOz 05:25, 21 August 2007 (UTC)[reply]

Right, and that's what my math gives you, but I think it's helpful to show how this can be derived, rather than just looking it up, so hopefully the original poster and any other readers can learn to solve such problems themselves in the future. StuRat 13:14, 21 August 2007 (UTC)[reply]

An alternative derivation is to divide the octagon into eight congruent radial triangles by conmecting each vertex to the centre of the octagon. If the length of one side of the octagon is x then each triangle has base x and height x(1/2 + sin(45)) so the area of the octagon is

A=8\left({\frac {1}{2}}bh\right)=4x^{2}\left({\frac {1}{2}}+\sin(45)\right)=x^{2}(2+2{\sqrt {2}})

Gandalf61 14:58, 21 August 2007 (UTC)[reply]

The method I used was none other than simply conceiving the number in my head right out of the blue. Then I once did the math a long time ago to find out the dimensions to this octagonic mall. I remember that each of the 8 sides would be 4,000 ft. in length. I also remember thinking that a normal shopping mall entrance in a total guess was 20 ft. wide for the doors. Each side could fir well over 40 entrances.Jk31213 22:09, 21 August 2007 (UTC)[reply]

Maths Joke[edit]

Hi everyone,
This is a bit of a fun question - but any ideas how this works? A Missouri farmer passed away and left 17 mules to his three sons. The instructions left in the will said that the oldest boy was to get one-half, the second oldest one-third, and the youngest one-ninth. The three sons, recognizing the difficulty of dividing 17 mules into these fractions, began to argue. Their uncle heard about the argument, hitched up his mule and drove out to settle the matter. He added his mule to the 17, making 18. The oldest therefore got one-half, or nine, the second oldest got one-third, or six, and the youngest son got one-ninth, or two. Adding up 9, 6 and 2 equals 17. The uncle, having settled the argument, hitched up his mule and drove home. --124.176.151.245 07:11, 21 August 2007 (UTC)[reply]

Note that ¹/₂ + ¹/₃ + ¹/₉ does not add up to 1; in fact, the sum is ⁹/₁₈ + ⁶/₁₈ + ²/₁₈ = ¹⁷/₁₈. The final fraction gives away "why" the uncle's trick works. For any number N (here 17) and any collection of fractions ¹/_Q adding up to ^N/_N+1 such that each divisor Q divides evenly into N+1, the same trick works; for example, 7 mules with shares of ¹/₂, ¹/₄, and ¹/₈; or 11 mules with shares of ¹/₂, ¹/₄, and ¹/₆. --Lambiam 07:56, 21 August 2007 (UTC)[reply]

Yeah, the problem is that the will is impossible (even ignoring the dicing of mules) to fulfill as stated. He tells them to distribute his entire estate by distributing only part of his estate, which is silly. The goal, then, is to find a reasonable approximation of the senile old coot's instructions, one that satisfies all parties. In this case, that amounts to rounding up - the first person gets 17/2 = 8.5, rounded up to 9; the second gets 17/3 = 5.666, rounded up to 6; the third gets 17/9 = 1.888, rounded up to 2. Ignoring the uncle's add-then-subtract-a-mule-ploy, doubtless employed to keep his math-inept nephews from whining, this is a very good solution for three reasons. First, it's simple and obvious, which is a natural tie-breaker in decisions like this. Second, the result is close both numerically and in spirit to that intended by the deceased, which is the best you can do when divvying up most people's estates. Third, it gives each person more than they were technically entitled to, a rare win-win. Black Carrot 09:52, 21 August 2007 (UTC)[reply]

The truly remarkable fact about this puzzle is not the mathematics, but the history. It has been around in one form or another for millennia. --KSmrq^T 10:36, 21 August 2007 (UTC)[reply]

I have seen it several times and always with camels. [1] says "this problem is present in an ancient Egyptian document, the Ahmes Papyrus -- a.k.a. Rhind Mathematical Papyrus." PrimeHunter 14:10, 21 August 2007 (UTC)[reply]

In my favorite version of this problem, which I think was given by Dudeney, the will says that the animals are to be divided in the proportions of one-half, one-third, and one-ninth to the three heirs. This version involves a trick of wording: the non-obvious meaning is that the word "proportions" implies a ratio, and therefore a proportion of 1/2:1/3:1/9 means exactly the same as 9/18:6/18:2/18 or 9:6:2, or for that matter 900:600:200. Thus in this version the will is carried out exactly, and borrowing an animal is just a trick to help the heirs visualize the 9/18:6/18:2/18 interpretation (while confusing the solver). --Anonymous, August 21, 2007, 23:33 (UTC).

I know this with 35 camels, from the tales book The Man Who Counted; in the end, two camels remain: the one added to the amount (for 36 being divisible by 2, 3 and 9), and another one because the sum of fractions don't reach 1. This last one is used as "payment" for the wise man who helped the resolution. wildie · wilđ di¢e · wilł die 13:34, 23 August 2007 (UTC)[reply]

Somewhere or other I saw that puzzle given with a sequel: what's the smallest number of camels that can be divided among four heirs with an analogous trick required? —Tamfang 04:01, 27 August 2007 (UTC)[reply]

Multiplicative integration[edit]

An integral is essentially a sum of a lot of terms, which each are almost zero. Is there an equivalent product of factors, which are each almost one? Inspired by a recent question, I figured that since $\int _{x=a}^{b}f(x)\mathrm {d} x$ is the sum of lots of $f(x)\mathrm {d} x$ , perhaps $\mathrm {e} ^{\int _{x=a}^{b}f(x)\mathrm {d} x}$ can be seen as the product of lots of $\mathrm {e} ^{f(x)\mathrm {d} x}$ . —Bromskloss 14:04, 21 August 2007 (UTC)[reply]

So use

\exp {\int _{x=a}^{b}\log f(x)\mathrm {d} x}

for the "product" of lots of f(x). --Lambiam 17:35, 21 August 2007 (UTC)[reply]

The logarithmic derivative would seem to be the concept corresponding to the derivative for a function that is viewed as an infinite product. Seeing as complex analysis deals a lot with logarithmic derivatives, I wonder if there is a way to reinterpret many of those results from a much more algebraic or category theory perspective, i.e. as some sort of correspondence between a "multiplicative" viewpoint and an "additive" viewpoint of the functions. Sorry if this sounds hazy...any one know what I mean? nadav (talk) 18:32, 21 August 2007 (UTC)[reply]

Does the last section of Logarithmic derivative help you here? --Lambiam 19:15, 21 August 2007 (UTC)[reply]

I saw that section. Unfortunately, I don't know enough math to understand whether it fully addresses what I meant. My experience with complex analysis has been a very 19th century approach, and it's difficult for me to try to take a step back and see the forest for the trees. I just notice that certain results, such as for the gamma and zeta functions, can be derived from both a multiplicative approach (via Weierstrass products) and through an additive approach. I guess I'm wondering if anyone is familiar with some general result that connects the two approaches (Functional determinants maybe?). But then again, I might not be making any sense, so never mind. nadav (talk) 04:49, 22 August 2007 (UTC)[reply]

binomial(?) expansion[edit]

is it possible, binomially or otherwise, to expand the following.

$(a^{x}+b^{x})^{\frac {1}{x}}$

possibly by substituting for $x={\frac {1}{k}}$ giving

$(a^{\frac {1}{k}}+b^{\frac {1}{k}})^{k}$

thank you. ΦΙΛ Κ 16:25, 21 August 2007 (UTC)[reply]

The generalized binomial theorem should give you what you want. (Of course, it's an infinite series, and you need to be careful about when it converges, but unless 1/x is an integer that's about as good as you can hope for.) Tesseran 17:09, 21 August 2007 (UTC)[reply]

(Kind of repeating the above, due to edit conflict.) If

k

is an integer, you could of course use the ordinary binomial expansion,

\left(a^{1/k}+b^{1/k}\right)^{k}=\sum _{i=0}^{k}{k \choose i}a^{(k-i)/k}b^{i/k}=a\sum _{i=0}^{k}{k \choose i}a^{-i/k}b^{i/k}=a\sum _{i=0}^{k}{k \choose i}\left({\frac {b}{a}}\right)^{i/k}

.

If

k

is not an integer, there is apparently a variant for this too. (Did I even get it right, btw?) —Bromskloss 17:35, 21 August 2007 (UTC)[reply]

boat crossing in constant current[edit]

boat in current

A boat is to cross a river of width L to a dock directly opposite. The speed of the boat at full throttle in still water is constant = V. When crossing in constant current C, the boat is continuously directed at full throttle toward the dock on the opposite shore. At what angle A will the boat dock?

Please do not post the same question in multiple places, and please respect the information at the top that indicates we will not do your homework for you. --LarryMac | Talk 20:23, 21 August 2007 (UTC)[reply]

Since motion is relative and you only have information about the boat's behaviour in still water, imagine that the river is standing still while its banks, with the dock, run by at a speed of C. So the boat has to reach a moving target. Does that help? --Lambiam 20:56, 21 August 2007 (UTC)[reply]

Just to make sure we are all on the same page here. The boat will not follow a straight line, but rather a chase curve. That's at least what I thought it was called, but I can't find it in Wikipedia. —Bromskloss 22:08, 21 August 2007 (UTC)[reply]

Ah, pursuit curve. —Bromskloss 22:14, 21 August 2007 (UTC)[reply]

Perhaps that's not the case here, though, I see now. Perhaps the boat's velocity is directed toward the dock before we add to it the velocity of the water. I'm not sure how to read the question. —Bromskloss 22:20, 21 August 2007 (UTC)[reply]

Oh, man! I'm talking nonsense. Forget everything I said. —Bromskloss 23:03, 21 August 2007 (UTC)[reply]

thanks for the responses. The derivation of the pursuit curve via parametric equations produces a solution which indicates that the docking angle for 0<C<V will be uniformly = 0. (Ie, the derivative of the function describing the curve, when evaluated at the opposite dock, = 0. Thus, tanA = 0, and therefore A=0.) Since the docking angle for C=0 is pi/2, this seems highly counterintuitive; ie, that the addition of the slightest current is sufficient to cause the docking angle to discontinuously jump from a value of pi/2 to 0.). Of course, for C>V, the boat will never dock, but will be swept downstream. (Note: this is not "homework", but would clearly seem to be a 'textbook' problem, for which I am unable to Google or Wiki an intuitively satisfactory solution.) Thanks again for all suggestions... Several physicists have told me that the solution is as the equations dictate, and that is that. But something seems wrong here to me, and several of my colleagues.)

I think that solution sounds reasonable. It's what I guessed it would be, actually. —Bromskloss 23:44, 21 August 2007 (UTC)[reply]

For motivation, think of it like this. The velocity of the boat is never directed toward the dock unless the angle is

0

(or

\pi

). Therefore, it will always drift a little downstream than what would be directly toward the dock. When it does that, the angle decreases, and this goes on until it's zero. —Bromskloss 23:53, 21 August 2007 (UTC)[reply]

If C << V, the angle of approach is small till the boat is close to the other side. As C approaches zero, if C is chosen sufficiently small the boat will appear to remain at an arbitrarily small angle until the last moment, when it makes an almost-quarter turn. At least, in the mathematical analysis, in which the boat is reduced to a point. A real boat will bump into the dock before its centre is there. --Lambiam 02:38, 22 August 2007 (UTC)[reply]

Again, thanks, both. First (Bromskloss), by assumption, the boat is "continuously" pointing to the opposite dock. "For motivation", think of it as being "reeled", via a connecting line, directly into the dock. Secondly (Lambian), I will at this point merely note Zeno's paradoxes of motion, particularly 'Achilles and the Tortoise', as caveat against too readily approaching infinitesimal motion.

World Mall[edit]

Hi it's me again on my mathematical question on this enormous building. If it was shaped like a octagon and each of the 8 sides were 4000 ft. long, how much area would it cover? I remember doing the math a long time ago and it turned out to be the right answers. Somebody else try this method. Draw this octagon out. Then draw a straight line right to the center from each of the 8 points. You now have 8 triangles. Find out the area of each triangle, times that by 8. There you have the area of just one of the 11 floors I plan to have in the building (including the roof).Jk31213 22:17, 21 August 2007 (UTC)[reply]

That seem like a good way. Is there something you would like to get help with? —Bromskloss 22:22, 21 August 2007 (UTC)[reply]

If we wanted to do it that way we would need to find the length of each triangle. I get (0.5 + sin45)×4000 or 4828.427. We can then get the area of each triangle by multiplying this length by the width of 4000 and dividing by 2. I get 9,656,854. When we multiply that by 8 triangles we get an area of 77,254,834. StuRat 02:07, 23 August 2007 (UTC)[reply]

There are several ways to get to the answer, which is $32,000,000\times (1+{\sqrt {2}})$ (or numerically 77,254,833.996) square feet. The above is one of them; another was given in the previous answer.

Subsidiary question, just for fun: Instead of taking the Earth's surface as level, take it to be a sphere of radius 20,900,000 feet. So the floors are not parallel planes but concentric spherical surfaces, say 20 feet apart (hey, it's a monumental building). Now how does the area of each floor compare to 77,254,833.996 square feet, (1) if the given measurement of 4,000 feet for the walls is a great-circle measure along their bases, or (2) if it is a straight-line measure from vertex to vertex of the ground floor through the ground?

I repeat, this is just for fun; I don't have the answers, but I thought someone with better spherical-geometry skills than mine might enjoy working them out.

--Anonymous, August 21, 2007, 23:52 (UTC).

You could also solve it by dividing up the octagon, like I did in this (rather crude) picture. The octagon is divided into a square with a side length of 4000, four isosceles right triangles with a hypotenuse length of 4000 and a leg length of n, and four rectangles with side lengths of 4000 and n. First use the Pythagorean theorem to find n, which turns out to be

2000{\sqrt {2}}

. Then use basic area formulas to find the area of one of the triangles (times 4), the area of one of the rectangles (times 4), and the area of the square, then add them all up to find the total area of the octagon, which is the answer given above.

And here's a somewhat simpler method I noticed while I was doing the above. The octagon can be seen as a big square with its corners cut off. The side length of the big square would be 4000 + 2n (n being the same value given before), and the missing corners could be put together (at their right angles) to make a smaller square with a side length of 4000, so find the area of those two squares, subtract the smaller from the larger, and you'd have the area of the octagon. --CrazyLegsKC 02:11, 22 August 2007 (UTC)[reply]

Hasn't anyone noticed that the same person asked this same question earlier on this same day, as Wikipedia:Reference_desk/Mathematics#Building Area ? I also gave an answer virtually identical to CrazyLegsKC's first answer, but with diagrams. StuRat 01:53, 23 August 2007 (UTC)[reply]