number theory
relatively prime
number theory proposed

Let $ S$ be a set of numbers containing 2 and the the primes 1 mod 4 and $ T \equal{} \{ x | x \equal{} p_1 p_2 \cdots p_k \text{ for  distinct } p_i \in S\}$. Prove or disprove: there exists infinitely many $ a \in T$ such that $ a \plus{} 1\in T$

[quote="t0rajir0u"]Let $ m, n$ be relatively prime and take $ a \equal{} (m^2 \plus{} n^2)^2$.  (Why did you ask this question in such a complicated way?)[/quote]

There's also a squarefree condition for the elements of $ T$.


I came across another question which has just stumped me because I worked it over about 5 times and I couldn't figure out how the book got the answer. I've obtained messed up answers and none have them have been...

Answer: 6

Problem:

2x - 3(x-2)/2 = 7 - x-3/3

The thing that probably messed me up the most was the fact that I didn't know how to get the variable fractions out of the way.

I really don't remember the book covering that, therefore I sucked at it.

I did somehow try to solve it: :?

2(2x - 3(x-2)/2) = (7 - x-3/3)2
4x - 3x - 6 = 14 - 2x-6/3

x - 6 = 14... +6+14...

x = 20 - 2x-6/6
3(x) = (20 - 2x-6/3)3
3x = 60 - 2x - 6
+2x
8x = 54
8/8 = 54/8
= 6.75
 :?

So basically i'm screwing up somewhere and I don't know where but I'm not getting the right answer.   :ninja:

BTW, i've never liked the way the forum was layed out.

I kinda like the idea of Pre-algebra, Intermedite Algebra, College Algebra,... stuff like that being sectioned off.
Should I start moving this stuff away from the middle school section? When I was in middle school I learned pre-algebra. So to me this is highschool math.

[quote="236factorial"]You screwed up on the second line, where you expanded -3(x-2) to be -3x-6, where the correct way should be -3x[b]+[/b]6

Now try it.[/quote]

[quote="math92"]I looked at your work and i think there was a mistake on the second line...[/quote]

ok math92, we know the mistake is at second line, you dont have to repeat it. stop spamming and i mean zero spams from now on PLEASE OK?

[quote="ritchjp"]I would have to disagree with math92 and 236factorial.
mathmanman's solution and answer look correct. As a check, substitute 6 into the equation for x and see if it agrees.[/quote]You are wrong. math92 and 236factorial say Cyberman's answer is incorrect, not mathmanman's answer. Cyberman ask where he do wrong, so math92 and 236factorial answer his question.

you're right 10000th user. my apologies to math92 and 236factorial.


[color=purple] I think it look more beautiful if you write it in the following form
        ${\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\ge \sqrt{\frac{3}{2}(a^{2}+b^{2}+c^{2}+4\sqrt{3}S})}$
     with a,b,c are three side length of a triangle which has are S [/color]

What is it? I don't understand? can you explain?

It's just from a Contest in Danang city, Viet Nam, 2004-2005

[color=purple]I can show the geometric sense.  Apply the property of  Torricenlli point we can change your algebraic inequality to the form
       $ab+bc+ca \ge \frac{a^{2}+b^{2}+c^{2}+4\sqrt{3}S}{2}$
  which is clearly true because it's equipvalent to $4R+r\ge \sqrt{3}p$  [/color]


Find all functions $ h : \mathbb Z \longrightarrow \mathbb Z$ that for every $ x,y \in \mathbb Z$,
\[h(x+y) + h(xy)= h(x)h(y) = 1.\]

[quote="behdad.math.math"]Find all functions $ h : Z \longrightarrow Z$ that for every $ x,y \in R$ :

$ h(x \plus{} y) \plus{} h(xy)$ $ \equal{}$ $ h(x)h(y) \plus{} 1$[/quote]

Try to be careful on your statements : If $ h(x)$ is only defined on $ \mathbb{Z}$, the equation cannot be defined "for every $ x,y \in R$ "

[quote="behdad.math.math"]Find all functions $ h : Z \longrightarrow Z$ that for every $ x,y \in R$ :

$ h(x \plus{} y) \plus{} h(xy)$ $ \equal{}$ $ h(x)h(y) \plus{} 1$[/quote]

I will modify the problem as :

Find all functions $ h(x)$ : $ \mathbb{Z}\longrightarrow\mathbb{Z}$ such that $ E(x,y)$ : $ h(x \plus{} y) \plus{} h(xy) \equal{} h(x)h(y) \plus{} 1$ is true $ \forall x,y\in\mathbb{Z}$

$ E(0,0)$ $ \implies$ $ h(0) \equal{} 1$

$ E(1, \minus{} 1)$ $ \implies$ $ h( \minus{} 1)(h(1) \minus{} 1) \equal{} 0$ and so either $ h(1) \equal{} 1$, either $ h( \minus{} 1) \equal{} 0$

1) $ h(1) \equal{} 1$
$ E(x \minus{} 1,1)$ $ \implies$ $ h(x) \equal{} 1$ and we can verify that this is a solution

2) $ h(1)\neq 1$ and $ h( \minus{} 1) \equal{} 0$
$ E( \minus{} 2,1)$ $ \implies$ $ h( \minus{} 2)(1 \minus{} h(1)) \equal{} 1$ and so $ 1 \minus{} h(1)$ divides 1 and so $ h(1) \equal{} 0$ or $ h(1) \equal{} 2$

2.1) $ h(1) \equal{} 2$
$ E(x,1)$ $ \implies$ $ h(x \plus{} 1) \equal{} h(x) \plus{} 1$ and so $ h(x) \equal{} x \plus{} 1$ and we can verify that this is a solution.

2.2) $ h(1) \equal{} 0$
$ E(x,1)$ $ \implies$ $ h(x \plus{} 1) \equal{} 1 \minus{} h(x)$ and so $ h(x \plus{} 2) \equal{} h(x)$ and so $ h(x) \equal{} 0$ $ \forall x$ odd and $ h(x) \equal{} 1$ $ \forall x$ even and we can verify that this is a solution.

3) synthesis:
We have 3 solutions :
$ h(x) \equal{} 1$
$ h(x) \equal{} x \plus{} 1$
$ h(x) \equal{} 0$ $ \forall x$ odd and $ h(x) \equal{} 1$ $ \forall x$ even


is there free version of the book "contest around the world" from
various year. I download year97 and 98 on IMO-USA webpage but i did't find any other years?

No , the other books aren't for free . There are quite a few : 05-96,96-97,97-98,98-99,99-00,00-01.

You can buy the rest from American Mathematical Society :D


There is a factory producing 2 very similar product [b]A[/b] and [b]B[/b] which are then packed into boxes . Suppose the amount of every boxes containing product [b]A[/b] and [b]B[/b] are the same and the probability of combination in each boxes(i.e. 1 product A and the rest are product B ,  or maybe all product A ) are also the same . One worker is opening a sealed box and obtain a product [b]A[/b] .Whats the probability for him to get another product [b]A[/b] in the same box ? ( He cant see what is inside the box)

[quote="K81o7"]So if there are n objects inside one box, then,
The probability for any given combination is $\frac{1}{n+1}$
After the worker picks the first object, there are n-1 objects left.  The probability that there are 2 or more object A is $\frac{n-1}{n+1}$.  Thus, that is the probability, because the worker must have 2 or more of object A.[/quote]

Dear K81o7.

The answer you given can only explain that if there are [b]n+1[/b] boxes with different combination in each and the probability for the worker to get 2 [b]A[/b] in a box is (n-1)/(n+1) ..... Well , that doesnt actually answer the question . Ok i will make it more clear that the sample of the boxes are extremely large ( say infinity) , and the worker just picks one box at random and opens that box . He gets a [b]A[/b] in first attempt , and what is the probabilty that he will get another [b]A[/b] in same box at second attempt (without replacement of course )

By the way , the answer is a fraction consist of only numbers ....  :oops:  :oops:

[quote="K81o7"]Well, given one box, any combination is possible (any combination with one A at least...of which there are n).  Therefore...
Ah!  Come to think of it it's probably $\frac{1}{2}$, right?  For the next object, A and B are equally likely.  I can't believe I didn't see that before.  :D[/quote]

 Nope , it is not 1/2 :P  :D

But something cute about this question is no matter what value of [b]n[/b] you plug in , you get the same probability  :roll:  :roll:  :roll:

[quote="hmoonster"]please clarify:
[quote="shyong"]Suppose the amount of every boxes containing product [b]A[/b] and [b]B[/b] are the same
[/quote]
I don't get what this is saying
[quote="shyong"]
Whats the probability for him to get another product [b]A[/b] in the same box ? ( He cant see what is inside the box)[/quote]
does that mean that after he takes out the first A, to find the probability that another A comes out right after it (i. e. does order matter) or to find the probability that he will find an A after searching extensively through the box?[/quote]

Every box has same amount of product (i.e every box has 10 product inside )

Yes , it does mean that another A comes out right after the first one .Note here that he only takes two products out from the box .

[quote="antimorph"]How does 1/4 sound

P.S. how do u use the special text? (I am new)[/quote]

Not $\frac 14$ .The probability is higher . By the way , what special text you mean ? $ \text {You mean this}$ ?


logarithms
calculus
integration
calculus computations

1. $ \sum_{n\equal{}3}^{\infty} \frac{1}{(3n\minus{}5) \ln^2 (4n\minus{}7) }$
2 $ \sum_{n\equal{}3}^{\infty} \frac{ (\minus{}1)^n}{n (\ln \ln n ) \ln n}$

Caculate sum of series:
$ \sum_{n\equal{}0}^{\infty} (2n^2\plus{}4n\plus{}3) x^{n\plus{}2}$

Or  caculate integral:
$ \int_{2}^{\infty} \frac{ dx}{(3x\minus{}5) \ln^2 (4x\minus{}7)}$

For #1 (or for that integral, which is the same thing), we have limit comparability to either the sum $ \sum_{n \equal{} ?}^{\infty}\frac1{n\ln^2n}$ or the integral $ \int_?^{\infty}\frac {dx}{x\ln^2 x}.$

(That question mark is for a starting point that doesn't itself cause any trouble.)

$ \int_2^{\infty}\frac {dx}{x\ln^2x} \equal{} \left. \minus{} \frac1{\ln x}\right|_2^{\infty}$ converges.

Hence your original series converges by limit comparison.

(The condensation test would also be effective at showing the convergence.)

For #2, if you left the alternating sign out, you'd have the sum $ \sum\frac {1}{n\ln n\ln\ln n}.$ By either the integral test or the condensation test, that diverges. But then, the alternating series test does apply and the original series does converge conditionally.

For #3: you're going to have to assume that $ |x| < 1.$ This is a power series with radius of convergence 1. For computing its exact value, let me give you some ideas:

$ \sum_{n \equal{} 0}^{\infty}x^n \equal{} \frac1{1 \minus{} x}.$

Differentiate that to get

$ \sum_{n \equal{} 1}^{\infty}nx^{n \minus{} 1} \equal{} \frac {1}{(1 \minus{} x)^2}$

$ \sum_{n \equal{} 1}^{\infty}nx^{n} \equal{} \frac {x}{(1 \minus{} x)^2}$

Differentiate that again to get $ \sum_{n \equal{} 1}^{\infty}n^2x^n$ or something related to that.

I understood your ideas, but  i can't find relations between expression $ (2n^2\plus{}4n\plus{}3)x^{n\plus{}2}$  and $ n^2x^{n}$

i have 1 question, this is right $ \sum (2n^2\plus{}4n\plus{}3)x^{n\plus{}2} \equal{}\sum 2(n\plus{}1)^2x^{n\plus{}2} \plus{} \sum x^{n\plus{}2}$


You know that one square root of a number X is -9. What is the other square root?What is the value of X?

Basically you are saying $ \sqrt{x}\equal{}\minus{}9$.  Remember that a square root of a number will always have 2 square roots, a negitive and positive.  $ \pm\sqrt{81}\equal{}9,\minus{}9$, Therefore, X is 81 and the other square root is 9.  The two square roots are always teh same, just the opposite negitive/postive sign.

Squares and Squareroots answers will always end up Positive. Thats y the sign+- is given.

by definition, the square root of a positive number $ a$ is the [b]positive[/b] solution to the equation $ x^2\equal{}a$.  Thus, the equation $ \sqrt{x}\equal{}\minus{}9$ has no solutions.

Technically, using the square root [b]symbol[/b] imples that the answer is positive. By saying the word "square root" as opposed to the symbol $ \sqrt{}$, you imply both answers. Especially when it says "[b]one[/b] square root."

just to add a little extra info here. If u have a function $ f(x) \equal{} x^2$, then the answer is always positive or 0 (I think it was the "trivial inequality" that state that). Oh, and a function f(x) is basically saying that I'm replacing a variable (in this case, x), such as -9, into the function, which means $ f( \minus{} 9) \equal{} ( \minus{} 9)^2$, or 81. Oh, and a function always only has one solution. On the other hand, $ g(x) \equal{} \sqrt x$ is only a function when stated that only positive values are the output (solution), at least most people would suppose that u are wanting the positive solution. If u don't state that, then there would be a positive and negative output. Oh, BTW, u can put any letter in front of the (x), just to represent that it's a function. So, if u put 81 into g(x), then if it's a function, u would only get 9.

Most people would suppose that when u are saying g(x), then u get the positive solution. But in this case, in ur question, we have $ \sqrt x \equal{} \minus{} 9$, we square both sides of the equation, which gives x=81. So X=81. The other solution would be $ \sqrt 81 \equal{} 9$.   :lol:


inequalities
logarithms
probability and stats

Can someone motivate the following definition?
A distribution $ P$ on the nonnegative integers with mean $ \lambda$satisfying the inequality $ \log \frac{P(j)}{Po(\lambda,j)}\ge \frac{1}{2} \log \frac{P(j\minus{}1)}{Po(\lambda,j\minus{}1)}\plus{}\frac{1}{2} \log \frac{P(j\plus{}1)}{Po(\lambda,j\plus{}1)}$ for $ j\ge1$, is said to be ultra log-concave(ULC), where $ Po$ stands for Poisson.

Actually the above is also equivalent to $ (j\plus{}1)P(j\plus{}1)P(j\minus{}1)\le jP(j)^2$.

The Poisson distribution is log-concave and the definition of ULC is "more log-concave" than the Poisson distribution, which is already log-concave  :wink:

Thanks Albanian Eagle. Can you also give some motivation for the inequality? Thanks.


geometry
geometric transformation
rotation
counting
distinguishability

A Senate committee has 5 Democrats and 5 Republicans.  In how many ways can they sit around a circular table if all the members of each party all sit next to each other?

Hmm I may be wrong, but it might only be $ 120\cdot 120\equal{}14400$

There's only one way to seat the republicans and democrats, as there are rotations or flips or something like that.

Dojo is correct. Counting rotations, there is only one way to order RRRRRDDDDD, and then $ 5!^2$ ways to assign the Senators to their seats.

I guess I have difficulty with what the word "distinguishable" means.  If the seats are numbered 1 through 10, then I get 10 choices on which 5 adjacent seats we pick for the Republicans.  Namely {1,...5}, {2,..,6}  through to {10, 1, 2, 3, 4}.

Once we pick the 5 seats for the Republicans we then have (5!)(5!) ways to seat everybody.  This gives 144000 distinguishable ways to seat the politicians.

Why does the solution not account for the 10 possible rotations of the seats?

I feel dumb haha, but I'm reviving this thread because I have the same question about this problem:

[quote="66phoenix"]
Why does the solution not account for the 10 possible rotations of the seats?[/quote]

I mean, it's circular seating, so it should...right?

This problem utilizes a strategy called grouping.
Denote D as the pack of Democrats and R as the pack of Republicans.
We need to arrange D and R around a circle. There is only one way.
Now, we need to arrange the D and R separately.
There are $5!=120$ ways to arrange the Democrats and $5!=120$ ways to arrange the Republicans.
Hence, the answer should be $5!\cdot5!=120\cdot120=120^2=\boxed{14400}$

[quote=vallon22]well, consider the republicans as one group and the democrats as another
now pretend it wasn't a round table, think of it as a straight row
using these two things, there are $ 2$ ways of arranging the democrats and republicans

however, in each of the groups, there are $ 5!$ ways to seat the members of each party
that means that within those $ 2$ ways, there are $ 120$ ways to seat the democrats, and $ 120$ ways to seat the republicans

so now we have $ 2\times120\times120\equal{}28800$ ways

however, they are sitting at a round table, not a straight row, so we have to divide by $ 10$ for rotations

therefore,  our answer is $ \frac{28800}{10}\equal{}\boxed{2880}$

i feel like i made a mistake in there somewhere, plz correct me if i'm wrong[/quote]

Not quite. Since there are 120 ways (5!) for the Democrats to sit next to each other, and 120 ways (5!) for the Republicans to sit together. 120 times 120 equals 14400 arrangements total.

There would be precisely 0 possible arrangements since the Democrats don't have a Senate majority(so there is a Republican one), and their arguments will be heated enough so that all people on one party leaves, and the President would convince everyone else to resign their position, and thus leaving the United States without a Congress.

[quote=awesomeguy856]I got this problem right after i got [url=https://artofproblemsolving.com/community/c63h286667]this[/url] problem ._.[/quote]

omggggg same

i still don't get why this is level 21 tho

and i was expecting the answer of [hide=posted already in the discussion but whatever]5!*5!=14400[/hide] to be wrong bc there's a big chance of a problem being a trick question

[quote=JacobJB][hide=sol]We have democrats on one side and reps on the other.There are $120$ ways to find each[hide=explanation]5![/hide]. So the answer is $120\cdot120=\boxed{14400}$[/hide][/quote]

Wouldn't you have to divide by 10 because it is a round table?

I had $5! \cdot 5! = 5!^2 = 120 \cdot 120 = 14400$ as my final answer.

[quote=Mathsweat_notreally]can someone plz answer my question lol[/quote]

no bc we arrange the groups relatively to each other

[quote=Hypatia1728]I don't get how the solution accounts for rotations?
[hide=What I did]If this was a row arrangement we would have $5! \cdot 5!$ ways. But since this is a circle our two blocks can also be interchanged in $2$ ways, and to account for rotation, we divide by $10$, giving us $\frac{5! \cdot 5! \cdot 2}{10}$ ways.[/hide]
Confused pls help[/quote]

i believe you have it backwards
if it were a row arrangement, you would have to multiply by 2 to account for either the democrats or republicans being on the left

however, if you have a circle arrangement, by rotating the table $180^\circ$, you'd have the opposite, and thus you dont need to multiple by 2

Rotations and reflections are considered the same, so there's only one possible seating arrangement, making the answer $(5!)^2 = \boxed{14,400}$.

why would it not be zero, because the people on the end cant sit next to another person from their party


calculus
integration
real analysis
real analysis unsolved

1. integrate sin^(5)zcos3zdz

2. Use half angle identities to evaluate the integral of cos^(2)xdx

I deleted my answers when I noticed that you double-posted. Don't do that.


In general, transcendental equations like this don't have closed form solutions.  Why would you call this "easy"?

Maybe he meant $ \frac{a}{\sin{a}}$? I'm not seeing how "a-sina=1" would be easy to solve.

Well i dont think that we can actually get a value for this.
The max. which can be done is Taking series expansion of sina and then considering only the first two possibilities or using Descartes rule of signs

[quote="21 7 15"][quote="t0rajir0u"]Descartes' rule of signs doesn't apply to Taylor series (as far as I know).  The best we can do is to find a numerical answer using one of several methods (say, [url=http://mathworld.wolfram.com/NewtonsMethod.html]Newton's method[/url]).[/quote]it is called an iterative equation.Value between 4 and 5[/quote]Oh yeah? Find the exact value without the use of a calculator or computing device, my friend. Tell me again if it is 'easy'. If you find a decimal answer or an interval of it, that's not finding a. My answer can be: between $ \minus{}\infty$ and $ \plus{}\infty$; and I'm still correct and I can still call it EASY. Go ahead, tell me a [b]closed-form[/b] answer if you can.


[quote="jsc"]3 lights in concentric circular tracks are going around a pole in the center. the first light (A) makes one revolution per minute. The second (B) makes one revolution in 5 minutes. The third (C), in 7 minutes. How frequently will the three lights and the pole be collinear (They don't have to be in the same side of the pole; A-B--Pole---C is acceptable)?[/quote]

I can't edit this post anymore, so I'l ljust put the diagram here...Fig1 is the inital positions, and Fig.2 is counted in the question.

I would guess 17.5. Find the LCM, divide it in two.


oi rizes ths e3iswshs $ x^3\plus{}ax^2\plus{}bx\plus{}c\equal{}0$ einai k,m,n Na bre8ei h e3iswsh tritou ba8mou pou exei rizes to k^3,m^3,n^3

Αρκετες πραξεις οποτε δεν ξερω αν θα αρεσε κατι τετοιο.
Αnyway σε hidden μια ταυτοτητα για βοηθεια.

[hide] $ a^3\plus{}b^3\plus{}c^3\equal{}(a\plus{}b\plus{}c)^3\minus{}3(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\plus{}3abc$ [/hide]

Καλη επιτυχια σε ολα τα μελη το Σαββατο.

[quote="Athinaios"]Αρκετες πραξεις οποτε δεν ξερω αν θα αρεσε κατι τετοιο.
Αnyway σε hidden μια ταυτοτητα για βοηθεια.

[hide] $ a^3 \plus{} b^3 \plus{} c^3 \equal{} (a \plus{} b \plus{} c)^3 \minus{} 3(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \plus{} 3abc$ [/hide]

Καλη επιτυχια σε ολα τα μελη το Σαββατο.[/quote] ne odos exei pra3eis arketes alla sundiazei arketa kai dn einai polu diskolh


let $f: \mathbb{N}\rightarrow \mathbb{N},$ s.t. $f(f(n))+2f(n)=3n+4$ for all $n\in\mathbb{N}$.  Find $f(2006)$


Θεωρούμε το τετράεδρο $ ABCD$.α)Να βρεθεί ο γεωμετρικός τόπος σημείου $ M$ όταν $ MA^2\plus{}MB^2\plus{}3MC^2\minus{}4MD^2\equal{}k^2$. β) Να βρεθεί το σημείο $ M$ ώστε $ MA^2\plus{}MB^2\plus{}3MC^2\minus{}4MD^2\equal{}min$.

[quote="r_boris"]Αν Ε το μέσον του ΑΒ ,Ζ στην EC :ZE/ZC=3/2 και Η στην DZ προς το μέρος του Ζ : ΖΗ=4DZ τοπος του Μ είναι  σφαίρα κέντρου Η και σταθερής ακτίνας(2 Stewart+1 διαμέσων) Χρειάζεται διερεύνηση[/quote]

nice :D


calculus
integration
geometry
incenter
3D geometry
tetrahedron
induction

- anyone did it? If so, how was it?

Personally, the third question confused me [b]a lot[/b].

I'm not sure it's legal to... yet. I'll post it when I get the go-ahead.

Well, they did tell us to take the contest papers home, so I suppose it's ok to share them.

To avoid copyright problems, I changed the wording around.

1. [3 marks] Let D (n) = the largest odd divisor of a natural number n. Let S (n) = :Sigma: k from n + 1 to 2n, n natural D (k). Prove that S (n) = n2.

2. [4 marks] You have a blank piece of paper, and you place unit squares onto it. At least how many do you have to place to make a grid of 25 squares by 25 squares with all the lines present?

3. [5 marks] You and a merchant have $1999 in total, consisting of $1, $5, $10, $50, $100, $500 and $1000 coins and bills. The merchant has an object of some integral value. You have enough money to buy it. Prove that you can get correct change.

4. A square of area 1 has 4 external right triangles with the sides of the square as their hypoteni. Let A, B, C and D be their right vertices, and O1, O2, O3 and O4 be their incenters respectively.

a) [3 marks] Prove that the area of quadrilateral ABCD does not exceed 2.
b) [3 marks] Prove that the area of quadrilateral O1O2O3O4 does not exceed 1.

5. [6 marks] A paper tetrahedron is cut along some of its edges so that all of its faces are parallel to the plane (i.e. flat on the plane). Can the tetrahedron be shaped, and cut, so that some faces have to overlay each other?

Okay, so number 1 isn't bad.  This is not the best write-up, but I'm fairly certain it's correct:

[hide]What we need to show is that

D(n + 1) + D(n + 2) + ... + D(2n) = 1 + 3 + 5 + ... + (2n - 1).  There are exactly the same number of terms on both sides.  I claim that there is a 1-to-1 correspondance between the terms on both sides -- each term on the right matches exactly with one term on the left.  For the odd numbers above n, this is obvious.  For the odd numbers below n, we should be able to simply say that for any even number in the interval, twice the same number cannot also be in the interval.  Thus, each odd part D(i) can appear at most once, since otherwise it would be required to appear multiplied by 2 distinct powers of 2, and one of the numbers would be at least twice the other.  Thus, 1-to-1 correspondance.  Done.[/hide]

I feel the need to explain the marking scheme. They'll mark all your questions, but your final mark is the combined mark of the three questions for which you got the highest marks. This mark is then multiplied by a coefficient to compensate for different grades. For a grade 11 / junior student, this coefficient is 1.25. (I think it's 1.1 or 1.0 for grade 12, not sure.)

Last year, I got 11 for the competition, so my final mark was 13.75.

(You got both #1 and #2. #4 may look intimidating, but it's [b]really[/b] easy.)

Okay, I've got number 4a.  Actually got it in bed last night, but whatever.

The area of a quadrilateral is d[size=75]1[/size]d[size=75]2[/size]sin(:theta:)/2, where d[size=75]1[/size] and d[size=75]2[/size] are the diagonals and :theta: is the angle between them.  If we draw two semicircles on opposite sides of the square, they represent all possible positions for A and C.  If we draw a circle of radius 1, centered at the center of the square, it contains the previous entire diagram.  Thus, any two points on the square-and-semicircle-shape are at most 2 units distant from each other.  Thus AC :le: 2.  Similarly BD :le: 2.  sin:theta: :le: 1.  So the area of ABCD :le: 2*2*1/2 = 2.

For part b, I was thinking of trying a similar argument, but I'll have to figure out how to track that incenter.

Okay, for number 3, I think I can do it like this:

I would like to do this inductively.  If it costs 0 dollars, I can pay and get exact change.  Yay.  I win.

Assume I can do it for n dollars.  Let me try topurchase an object that costs (n + 1) dollars.

Assume my smallest bill is $1000.  Well, in that case, the merchant has the other $999, so we trade.  Now, I've given him a dollar, and I had more than (n + 1) dollars to start, so I now have more than n.  I now must transfer $n to him to complete the purchase.  I can do this, by inductive hypothesis.  I win.
For $500, it works exactly the same -- I know that the merchant must have at least $499.  For $100, he must have at least $1999, and so on.  This works because for every single bill amount b that is available, we have 1999 :equiv: -1 (mod b).  So I can do it no matter what my smallest bill is, by equivalent logic.  Thus, I win.

By induction, I win for all n. :)

Okay, I think I have the fifth one as well:

Let the triangles be denoted as follows:
Triangle 1: ABC, with angles a1, b1, c1
Triangle 2: ABD, with angles a2, b2, d2
Triangle 3: ACD, with angles a3, c3, d3
Triangle 4: BCD, with angles b4, c4, d4.

Now, without loss of generality, assume triangles 1, 2, and 3 will overlap at vertex A.  In that case, a1 + a2 + a3 > 360.  Then since the angles in a triangle add to 180,
a1 + b1 + c1 + a2 + b2 + d2 + a3 + c3 + d3 = 540  and
b1 + c1 + b2 + d2 + c3 + d3 < 180.

We also know that at any vertex, the sum of two angles must be larger than the third angle.  Otherwise, they will not be able to form a corner.  (This is essentially the triangle inequality for tetrahedra -- the sum of A1 + A2 > A3 where A1, A2, and A3 are the three angles at any vertex of a tetrahedron.)  Thus we have
b1 + b2 > b4,  c1 + c3 > c4,  and d2 + d3 > d4, so
b1 + b2 + c1 + c3 + d2 + d3 > b4 + c4 + d4 = 180

This is a clear contradiction, thus our original assumption is false and the tetrahedron cannot be disected so that it must overlap itself.

My only fear in this solution is that the initial description of what must happen for it to overlap is not really verified -- but there clearly isn't any other way.


In each field of 2009*2009 table you can write either 1 or -1.
Denote Ak multiple of all numbers in k-th row and Bj the multiple of all numbers in j-th column.
Is it possible to write the numbers in such a way that
$ \sum_{i\equal{}1}^{2009}{Ai}\plus{}	\sum_{i\equal{}1}^{2009}{Bi}\equal{}0$?

[hide="Is it really this easy?"]Consider a board with all the numbers already filled in.  Any other possible configuration of numbers can be formed by changing numbers one at a time.  When a number is changed, it will change the parity of the number of -1s in a row and in a column and will therefore change two of the products.  Let's examine the possible effects of changing a number:

1) Two products change from -1 to 1.
Net result:  +4

2) Two prodcuts change from 1 to -1.
Net result:  -4

3)  One product changes from 1 to -1 and the other product changes from -1 to 1.
Net result:  no change

Clearly, all possible sums must be congruent mod 4.  Since a field completely filled with 1s would yield a sum of 4018, which is not congruent to 0 mod 4, a sum of 0 is impossible.[/hide]

Please, teach me to solve this type of problems.

Consider what happens when we change one entry. Each sum $\sum A_k,\sum B_k$ increases or decreases by $2$. Therefore, the total sum $\sum A_k+\sum B_k$ changes by a multiple of $4$. In other words, $\sum A_k+\sum B_k$ is invariant mod $4$. However, when all the entries are $1$, this sum is $2\cdot 2009\equiv 2\pmod 4$, so it is impossible. $\blacksquare$


conics
ellipse
analytic geometry
trigonometry

The equation of an ellipse in polar coordinates is given by $ r\equal{}\displaystyle{\frac{10}{3\minus{}2\cos (\theta)}}$.  Which one of the following is the polar coordinates of one of the foci of the ellipse?

(a) (0,0)   (b) (10/3, pi)   (c) (4,0)   (d)   (2,pi)   (e) None of these

The competition is tomorrow, help please!

[hide="Magical hint"] The polar equation of an ellipse is constructed so that the origin is a focus. [/hide]

[quote="kyyuanmathcount"]The equation of an ellipse in polar coordinates is given by $ r \equal{} \displaystyle{\frac {10}{3 \minus{} 2\cos (\theta)}}$.  Which one of the following is the polar coordinates of one of the foci of the ellipse?

(a) (0,0)   (b) (10/3, pi)   (c) (4,0)   (d)   (2,pi)   (e) None of these

The competition is tomorrow, help please![/quote]

[hide] Note that $ x\equal{}r\cos \theta$ and $ y\equal{}4\sin \theta$ (thus $ x^2\plus{}y^2\equal{}r^2$).
Okay, so we have that \[ 3r\minus{}2r\cos \theta\equal{}10\] \[ 3r\minus{}2x\equal{}10\] \[ 2x\plus{}10\equal{}3r\] \[ 4x^2\plus{}100\plus{}40x\equal{}9r^2\] \[ 4x^2\plus{}100\plus{}40x\equal{}9x^2\plus{}9y^2\] \[ 5x^2\minus{}40x\plus{}80\plus{}9y^2\equal{}180\] \[ 5(x\minus{}4)^2\plus{}9(y\minus{}0)^2\equal{}180\] \[ \frac{(x\minus{}4)^2}{6^2}\plus{}\frac{y^2}{(2\sqrt{5})^2}\equal{}1\] So, we can say that the center of the ellipse is (4,0) and the foci have length $ \sqrt{36\minus{}20}\equal{}\sqrt{16}\equal{}4$.  Now, the minor axis is the X-Axis, so we have that the two foci are $ (4\minus{}4,0)\equal{}(0,0)$ and $ (4\plus{}4,0)\equal{}(8,0)$.  Only the origin shows up as a choice, so the answer is $ \boxed{A}$.  [/hide]


Prove that if $2^{n+1}-1$ is prime then $2^{n}(2^{n+1}-1)$ is a perfect number. $x$ is a perfect number iff $S(x)-x=x$, i.e. $S(x)=2x$ where $S(x)$ denotes the sum of the divisors of $x$ (including $1$ and $x$.


calculus
integration
logarithms
calculus computations

Find

$\int{\frac{\ln{ax}}{x^{b}}}\,dx$ where $a,b\in\Re$

[b][u]Extension:[/u][/b]

Assume the constant of integration is $0$.  Find a set of necessary conditions for

$\int^{d}_{c}{\frac{\ln{ax}}{x^{b}}}\,dx<0$ where $a,b,c,d\in\Re$ and $d>c$

Let $f(a,b) = \int\frac{\ln(ax)}{x^{b}}dx$.

Case 1: $a$ is such that $\ln(ax)$ is defined and $b = 1$.

$f(a,1) = \ln(ax)\ln(x)-\int\frac{\ln x}{x}dx = \ln(ax)\ln(x)-\frac{(\ln x)^{2}}{2}+C$.

Case 2: $a$ is such that $\ln(ax)$ is defined and $b \ne 1$.

$f(a,b) = \frac{x^{1-b}\ln(ax)}{1-b}-\frac{x^{1-b}}{(1-b)^{2}}+C$.

$ab>0\Longrightarrow \ln (ab)=\ln |a|+\ln |b|\ .$ Particularly, $n\in N^{*}\ ,\ a\ne 0\Longrightarrow \ln a^{2n}=2n\cdot\ln |a|\ .$


3.$ a,b,c$ are complex numbers.

$ \forall z \in \mathbb{C},|z| \le 1 \Rightarrow |az^2 \plus{} bz \plus{} c| \le 1$,

Try to find out the maxinum of $ |bc|$.

[url=http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=66&Id=18787]Click here to view chinese version.[/url]

[quote="frankvista"]3.$ a,b,c$ are complex numbers.

$ \forall z \in \mathbb{C},|z| \le 1 \Rightarrow |az^2 \plus{} bz \plus{} c| \le 1$,

Try to find out the mininum of $ |bc|$.

[url=http://www.aoshoo.com/bbs1/dispbbs.asp?boardid=66&Id=18787]Click here to view chinese version.[/url][/quote]
i think we should find the maximum instead of minimum  :wink:
BTW, can you type the rest ? because i cant see the symbol in the document. :oops:

[hide]For the minimum case, $ 0$ should hold.....let's basically try to figure out a case, when both $ b$, $ c$ equal $ 0$.

What we get is that if $ a$ be of the form $ e^{i\theta}$, we have $ |z|\le 1 \iff |az^2\plus{}bz\plus{}c|\le 1$.[/hide]

[quote="soumik"][hide]For the minimum case, $ 0$ should hold.....let's basically try to figure out a case, when both $ b$, $ c$ equal $ 0$.

What we get is that if $ a$ be of the form $ e^{i\theta}$, we have $ |z|\le 1 \iff |az^2 \plus{} bz \plus{} c|\le 1$.[/hide][/quote]

I've edited.It's maxinum case.


Gauss
number theory unsolved
number theory

I study long prime and I would find a proof to prove that:

$m=\frac{10^{p-1}-1}{p}$, $10$ is a primitive root modulo $p$ (i.e. $p$ is a long prime). For all divisor $d$ of $p-1$ show that if we separate $m$ in $\frac{p-1}{d}$ parts and we sum them, then this sum is divisible by 10^d-1.

For example with p=7 and d=2:
m=142857, 14+28+57=99

Thanks a lot if you can help me !

$ Question $

We want to seperate $m$ in $\frac{p-1}{d}$ parts, but how can be sure that the digits of number $m$ are exactly $p-1 ?$

For $p=17$ why does $m=588235294117647$ has 15 digits though 10 is a primitive root modulo 17?

we need to take $m=0588235294117647$
For every $m$ you have $E(log(p))$ zero at start

I've solved the problem, I will post a solution if it interests someone

[quote="Rodolphe2005"]we need to take $m=0588235294117647$[/quote]
oh ok!

I think I got the idea!

Since you have solved it, I won't give a proof, but I'll just post my thoughts through an example

For the example you gave, p=7 and d=2

First I notice that
$-1st \ $ 10 is a primitive root modulo 7, so the smaller positive integer k, with $10^k=1mod\ 7$ is k=6.

That means, $7|999999=10^6-1,$ but $7\ doesn't\ divide\ 10^k-1\ \forall\ 1<=k<6$
For instance, $7\ doesn't\ divide\ 99$

$-2nd \ \ \ $ $99|999999$ and $99|9999$
because
$999999=99*10^4+99*10^2+99$
$9999=99*10^2+99$

$m=\frac{999999}{7}=142857$

I'll proof that m and the sum are equivalent modulo 99

$m = 14*10^4+28*10^2+57$
$n = 14+28+57$

$m-n = 14(10^4-1) +28*(10^2-1)$
$m-n = 14*9999 +28*99$

Since $99|9999$ we have $99|m-n=>m=nmod99$
So we can prove that $99|n$ if we prove that $99|m$

But $m=\frac{999999}{7}=\frac{99*10101}{7}=99*\frac{10101}{7}$

And since $7\ doesn't\ divide\ 99$ we find out that $99|m=>99|n$

If I'm not wrong, I think that the general case (for every p) must look like this  :huh:

general case is not like this, it use the fact that for all divisor $d$ of $p-1$, $p$ divide $1+10^d+...+10^{p-1-d}$

you used that but you used directly that $n=14+28+57$ but in the general case you need an expression depending on $p$ for $n$


Chao cac ban .O dd manh co bao nhieu nguoi minh nhi .Co bac nao o hd khong ?

[quote="The soul of rock"]Đã lâu lắm rồi analysis không lên forum này, [b]có del bài cũng chẳng ảnh hưởng tới ai[/b].[/quote]
Không đúng! Tất cả các thành viên đều được anh em tôn trọng như nhau. Làm Mod cũng chỉ là ong ve của nhân dân mà thôi :P


function
inequalities unsolved
inequalities

[quote="tkhtn"]Let $a,b,c\ge 0$.
Prove that: $\sum \sqrt{a+b}\le \frac{5}{2}\sqrt{a+b+c}$.[/quote]
$\sum_{cyc}\sqrt{a+b}\leq3\sqrt{\frac{\sum(a+b)}{3}}=\sqrt{6(a+b+c)}\leq\frac{5}{2}\sqrt{a+b+c}.$

[quote="justdudxit"]Is that AMGM, for clarity?[/quote]
$f(x)=\sqrt x$ is concave function above $[0,+\infty).$
Hence, $\frac{f(a+b)+f(a+c)+f(b+c)}{3}\leq f\left({\frac{a+b+a+c+b+c}{3}}\right).$
Or $(x+y+z)^2\leq3(x^2+y^2+z^2).$
Id est, $\left(\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c}\right)^2\leq3(a+b+a+c+b+c).$


Suppose $ABC$ is a triangle with $M$ the midpoint of $BC$.
Suppose that $AM$ intersects the incircle at $K,L$.
We draw parallel line from $K$ and $L$ to $BC$ and name their second intersection point with incircle $X$ and $Y$. Suppose that $AX$ and $AY$ intersect $BC$ at $P$ and $Q$.
Prove that $BP=CQ$.

Easy problem for [b]IRAN[/b] TST.

[b][color=blue]Lemma[/color][/b][color=darkblue].let $D$ , $E$ and $F$ be intesection points of incircle with $BC,AC,AB$ respectively.
prove that median $AM,DI,EF$ are concurrent.[/color]
[i]Proof.[/i] with sinus law.

back to the problem from the lemma we can get that $(A,K,X,L)=-1$ hence the the polur of $X$ is a line($\ell$)

parallel to $BC$ and pass from $A$.it's simple from here

Yes. Almost all of the participants will get full mark for this day! (I think about 14 out of 16)

Hi,amir2,  please help me, I can't understand your solution.
The lemma is OK, actually it helped me solve the problem in my way but I don't know what you mean by  (A,K,X,L)=-1; and the polur of X through which inversion, (or maybe this means something else ;? )
Thank you.

also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=80872 .

  darij

This problem apeared in a romanian competition this year, so I can guess where it comes from :P


this is just $ 5(5^5)$ which is $ 5^6$
so $ n\equal{}6$

[hide="answer"]1.we're looking for 5^n

2.we know that there are 5 5^n

3.so when we simplify we get what AD got

5(5^5)=5^6

4.so the answer is $\boxed{5^6}$ [/hide]

[b][i][u]please rate my post thank you so much!![/u][/i][/b]

please be fair though [=

:trampoline: :trampoline: :trampoline: :trampoline:


[hide="Clickie"]

Let's take a look at the set of perfect squares: $\{1,4,9,16,25,36,49,64,...\}$
As you can see the difference between every two consecutive numbers is $3,5,7,9,11,13,15,...$

So: $1-2^2=-3$
$1-2^2+3^2=6$
$1-2^2+3^2-4^2=-10$
$1-2^2+3^2-4^2+5^2=15$
etc.

You can easily see that the absolute value is given by the formula $\frac{n(n+1)}2$ (which is the formula for n consecutive numbers).
Seeing that we end with a positive number, we can let away the absolute value part and fill in $n=1999$ to obtain $1999\cdot 10^3$[/hide]

[hide="yet another cool solution"]
$1 + 3 + 5 + \dotsm + (2n-1) = n^2$

$\therefore 1^2 - 2^2 + 3^2 + \dotsm + 1999^2 = (1) - (1 + 3) + (1 + 3 + 5) - (1 + 3 + 5 + 7) + (1 + 3 + 5 + 7 + 9) + \dotsm + (1 + 3 + \dotsm + 3997)$

When all the opposite terms are canceled out the remaining sequence is:

$1 + 5 + 9 + \dotsm + 3997$      (there are 1000 terms in this sequence)

Using the formula tor the sum of an arithmetic sequence:

(500)(3997 + 1) = 1999000

[/hide]

[quote="Palytoxin"][quote="Iversonfan2005"]$1^2-2^2+3^2-4^2+5^2-6^2........-1998^2+1999^2$ is equal to....?[/quote]
[hide="simple generalization---view at your own discretion hehe"]find a formula for
\[1^2-2^2+3^2-4^2+\cdots+(-1)^{n+1}n^2\]
[hide="please tell me the formula"]you should do the problem, you know, but if you are still desperate...
[hide="yes now show me that formula"]\[1^2-2^2+3^2-4^2+\cdots+(-1)^{n+1}n^2=\frac{n(n+1)}2\]
when $n$ is odd. you go find the formula when $n$ is even. then go prove it!!![/hide][/hide][/hide][/quote]
[hide]
when you look at my solution, isn't it just $\frac{-n(n+1)}{2}$?[/hide]

Can we give some combinatorail argument for this. I thought I read one somehwere!!!!

Doesn't this give $ 1+ 5 + 9 +..... $??? gunes.

yeh i know i didnt say that one is wrong,i just write my solution.


Many handshakes are exchanged at a big international congress. We call a person an odd person if he has exchanged an odd number of handshakes. Otherwise he will be called an even person. Show that at any time, there is an even number of odd persons.

[hide="hint"]Consider the parity of the sum over all people of the number of handshakes of the number of handshakes these people have made.[/hide]

[hide="In simpler terms"] If I shook hands with you, you shook hands with me. [/hide]

Let me try this one...
[hide]If there is an odd number of odd people shaking hands, then there is an odd number of total handshakes because odd x odd = odd, and if we add the number of even people which will be an even number, the total number of handshake is odd + even = odd. This case is impossible because shaking hands is symmetric and should therefore always be even (for example, if a shook hands with b, then b also shook hands with a, so there is an even number of handshakes: a->b and b->a).

So we need an even number of odd people to produce an even number of handshakes.[/hide]


Fie ABC un triunghi oarecare si M mijloc pentru [AB] ,iar N mijloc pentru [AC] . Fie AE perpendiculara pe CM si AF perpendiculara pe BN . Demonstrati ca NE=MF daca si numai daca triunghiul ABC este isoscel de baza BC.

NE mediana in triunghiul dreptunghic AEC deci NE=AC/2
Analog MF mediana in triunghiul dreptunghic AFB deci MF=AB/2
De aici rezulta AB=AC

In al doilea sens e la fel

eu nu am facut-o asa.Oricum e mai scurt cum ai zis tu :blush:


inequalities
inequalities solved
3-variable inequality

Prove that $\forall x,y,z \in R$ we have:

$xy(y-z)(z-x)+yz(z-x)(x-y)+xz(x-y)(y-z) \leq 0$

The inequality is equivalent to $\sum x^2y^2 \ge xyz(x+y+z)$ or $\sum (xy-yz)^2 \ge 0$.


real analysis
function
integration
real analysis unsolved

I am having trouble with another Real and complex Rudin problem:
chapter 4, problem 7 (a):

Assume that both f and its maximal function Mf are in L^1(R). Prove that then f(x) = 0 a.e. [m].

Hint: To every other f in L^1(R) corresponds a constant с = c(f) > 0 such that
                Mf(x) >=  c|x|^{-k}
whenever x is sufficiently large.

Thanks a lot in advance.

Idea: If $ f$ is not $ 0$ a.e., then for a sufficiently large $ R$ you have $ \int_{|x|<R}|f|>0$. This implies a lower bound for $ Mf$ of the kind that you need.

Sorry; I am still not sure I understand exactly how to do it. Could you please maybe give me more details? Thanks a lot for your patience.

Your $ k \equal{} 1$; that's important. Try it first for $ f$ equal to the characteristic function of $ [0, 1]$. Then you should be able to follow mlok's hint.


probability
geometry
geometric transformation
invariant
superior algebra
superior algebra unsolved

this is not hard but i suspect it might require some knowledge in probability of things in infinite countable sets in order to form correct arguments
find the probability of a random rational number in REDUCED form has an even denominator
please write a complete proof as i would like to see how some details are treated

First explain what you mean by a random rational number in reduced form. Do you pick two positive integers and then reduce it, or do you throw them out if they aren't coprime, or do you pick a numerator and then pick a denominator from the set of coprime integers? I would assume that the answer varies depending on what you mean.

The question said to determine the probability that a random rational number has an even denominator.

so i would interpret as take a random rational numbers and determine the probability of it having an even denominator, ie all random numbers of equal chance of being picked.  personally i interpeted as picking two integers a and b (not necessarily positive though), then take probability of b/gcd(a,b) is even

Well, here's one way to interpret the question.

Pick two random positive integers $a$ and $b$. We consider $\frac{a}{b}$ and reduce it if necessary. There are four cases we must consider.

Case 1: $a$ is odd, $b$ is odd. Then the reduced form of $\frac{a}{b}$ is odd/odd, so in this case the denominator is odd.

Case 2: $a$ is even and $b$ is odd. Then the denominator of the reduced form will still be odd.

Case 3: $a$ is odd and $b$ is even. Then the denominator of the reduced form will be even.

Case 4 (the tough one): $a$ is even and $b$ is even. What is the probability that $b$ divides more powers of two than does $a$? We'll define $f(a)$ to be the number of factors of 2 in $a$. The probability that $f(a)=1$ is 1/4. Then we need $f(b)\ge 2$. The probability of that is 1/4. Similarly, the probability that $f(a)=k$ is $1/2^{k+1}$, and so is the probability that $f(b)\ge k+1$. So we sum all these up, we get 1/12 probability that $f(b)>f(a)$.

The answer is the sum of the answers of cases 3 and 4. The probabilities for cases 1-3 are each 1/4, so p(case 3) + p(case 4)=1/4+1/12=1/3.

Well, the integers is a countably infinite set... so I was told that you need to establish some kind of distribution before applying the probabilities.  ie you cannot just say the probability of picking an odd integer from the set of [1,inf) is 1/2

but uniform distribution wont work here... thats why i asked for a rigerous proof in the beginning

You can define it rigorously using Cesaro means, it appears. It will come out the same as uniform distribution.

... and ComplexZeta's nice answer will apply for any probability distribution for which the probability that the exact power of 2 divisding $n$ is $k$ turns out to be $2^{-k}$.

Another probability distribution that does it: start with the 2-adic integers, and use the canonical (translation-invariant) measure, scaled to total measure 1. This has the slight disadvantage that the value is not in $\mathbb{Z}$ with probability 1, but the probability of being divisible by $2^n$ is exactly $2^{-n}$., and the question still makes sense.

How does this work? Take an infinite sequence of independent identically distrubuted variables taking values of $1$ and $0$ each with probability $\frac12$. The $n$-th such variable determines the $n$-th digit left of the decimal point.


function
algebra
polynomial
algebra proposed

Let $a, b > 0$ be reals such that
\[ a^3=a+1\\ b^6=b+3a  \]

Show that $a>b$

Yes it is easy. From $a^3=a+1$ we have a>1. From $b^6=b+3a$ we have b>1. Function $b^6-b$ increase, when b>1. Therefore statement equavalent to $a^6-a>3a$ or $(a+1)^2>4a$ or $(a-1)^2>0$.

I see that (a - 1)^2 > 0 should give a > 1, but you got that earlier along with b > 1.  I don't see any steps here where you conclude that, through transitivity or else, that a > b.

There's no guarantee that $b>1$... but if $b<1$, $b<a$ automatically.

[quote="José"][quote="Johann Peter Dirichlet"]$a^3=a+1$[/quote]

Just find $a$ !!!!!!!.[/quote]
Well, $a=\frac{1}{6} \cdot \sqrt[3]{108+12\cdot \sqrt{69}}+\frac{2}{\sqrt[3]{108+12\cdot \sqrt{69}}}$, but I wanna see anyone solving this manuelly (I used Maple).
And then the next problem is to find $b$, or at least good bounds using this value (the greatest $b$ is very close to $a$).

Well, it is not so hard...
If $a^3=a+1$ then $a>1$ and $b^6=b+3a>3$, and so $b>1$.
Well, if a<b, we have

\[ (b-a) > (b-a)-(a-1)^2=(b+3a)-(a+1)^2=b^6-a^6=(b-a)(a^5+a^4b+a^3b^2+a^2b^3+ab^4+b^5) \geq b-a  \]

absurd!
Well, someone have an estimate of a-b? I think it is really small...

Equation $a^3=a+1$ give a>1. If $b^6-b=3a$ and $b\ge a>1$ we have
 $a^6-a \le b^6-b=3a \Longrightarrow a^6\le 4a \Longrightarrow (a-1)^2 \le 0$ give contradition with a>1.

Suppose that a=b. Then a^3=a+1 a6=4a. (a^3)^2=(a+1)^2 = a^2+2a+1 =4a
a^2-2a+1=0 a=1 . But this doesn't satisfy "a^3=a+1 ". Contradiction .

Suppose that a＜b. Then (a^3)^2＜b^6. a^2+2a+1＜b+3a , a^2-a+1＜b .
a＜a^2-a+1.(if a=1 , this doesn't satisfy "a^3=a+1). Hence a＜a^2-a+1＜b. For any a^2-a+1, this is true. so
3/4≤a^2-a+1, a≤3/4 . but a^3-a=1 a(a^2-1)=1 a＞0 , a^2-1＞0 and a ＞1. but a≤3/4. Contradiction.

Therefore a＞b

[quote="kokosai"][quote="Rust"]...$a^6\le 4a \Longrightarrow (a-1)^2 \le 0$...[/quote]
What's the logic between the two steps?[/quote]
We have $a^3=a+1$ therefore $a^6=(a^3)^2=(a+1)^2$

[quote="Kim Tae Soo"]Hence 4/5＜b.    a^3＜b^3 . a+1＜125/64.[/quote]
It should be 5/4<b, but why did you assume a^3<(5/4)^3?
a is less than b, and 5/4 is less than b, but should a be less than 5/4?

Sorry, if this is a stupid mistake, but i don't get it...

It should be clear from graphing $x^{3}$ and $x+1$ that $x^{3}=x+1$ has a positive root (our $a$).  Also if $x^{3}<x+1$ then $x<a$ and if $x^{3}>x+1$ then $x>a$.

Assume the contrary that $b>a$.  Then the second equation is equivilent to $b^{5}=1+3\frac{a}{b}<4$ or equivilently $b<4^{1/5}$.  Now recall that if $x^{3}<x+1$ then $x<a$.  Thus if $4^{3/5}<4^{1/5}+1$ then $b<4^{1/5}<a$ which is a desired contridiction to our asumption.

$4^{3/5}<4^{1/5}+1$ is equivilent to $4<4^{3/5}+4^{2/5}$ which is true by AM-GM.

[quote="Johann Peter Dirichlet"][color=darkred]Let $ a > 0$, $ b > 0$ be two real numbers such that $ \{\begin{array}{c}a^{3}= a+1\\ \ b^{6}= b+3a\end{array}$. Show that $ a > b\ .$[/color][/quote]
[color=darkblue][b]Proof.[/b] Observe that $ a > 1$ and $ b > 1$. Thus, $ \{\begin{array}{ccc}a^{3}= a+1 &\implies & a^{6}= (a+1)^{2}\\ \\ b^{6}= b+3a &\implies & b^{6}+a-b = 4a\end{array}\|$ $ \implies$ $ (a^{6}-b^{6})-(a-b) = (a-1)^{2}> 0$ $ \implies$
$ (a-b)(-1+\sum_{k = 0}^{5}a^{5-k}b^{k}) > 0$ $ \implies$ $ a > b$ because $ -1+\sum_{k = 0}^{5}a^{5-k}b^{k}> 5 > 0\ .$[/color]

[quote="Virgil Nicula"][color=darkred][b]An easy extension.[/b] Let $ \{m,n,p,r,s\}\subset [1,\infty )$ for which $ s\in N^{*}$, $ p\le 2s$ and $ 4mn = p+r$.
Let $ x > 0$, $ y > 0$ be two real numbers such that $ \{\begin{array}{c}x^{s}= mx+n\\ \ y^{2s}= py+rx\end{array}$. Prove that $ x\ge y$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Observe that $ x > 1$ and $ y > 1$. Thus, $ \{\begin{array}{ccc}x^{s}= mx+n &\implies & x^{2s}= (mx+n)^{2}\\ \\ y^{2s}= py+rx &\implies & y^{2s}+p(x-y) = 4mnx\end{array}\|$ $ \implies$
$ (x^{2s}-y^{2s})-p(x-y) = (mx-n)^{2}\ge 0$ $ \implies$ $ (x-y)(-p+\sum_{k = 0}^{2s-1}x^{2s-1-k}y^{k})\ge 0$ $ \implies$ $ x\ge y$ because $ -p+\sum_{k = 0}^{2s-1}x^{2s-1-k}y^{k}> 0\ .$
[b]Remark.[/b] For $ m = n = p = 1$ and $ r =s= 3$ obtain the nice proposed problem. [/color]

[quote="Virgil Nicula"][color=darkred][b]A similar exercise.[/b] Let $ a > 0$, $ b > 0$ be two real numbers such that $ \{\begin{array}{c}a^{3}= 2a+1\\ \ b^{6}= 6b+2a\end{array}$. Show that $ a > b\ .$[/color][/quote]

I think you make the problem more complicated than it is. It is particular case from USA Math Olympiad 1993 - See the problem solutions in the resources section. Or the solution in kalva web site( http://www.kalva.demon.co.uk/ ).

More general (n>2)
$ a^{n}\equal{}a\plus{}1$
$ b^{2n}\equal{}b\plus{}3a$
Then $ a>b$

[quote="TTsphn"]More general (n>2)
$ a^{n}\equal{} a\plus{}1$
$ b^{2n}\equal{} b\plus{}3a$
Then $ a > b$[/quote]
It is true for positive a,b. If n is even $ a^{n}\equal{}a\plus{}1$ had negative solution.
$ (a\minus{}b)(\frac{a^{2n}\minus{}b^{2n}}{a\minus{}b}\minus{}1)\equal{}(a\minus{}1)^{2}$ give solution, because $ a>1\to b>1$ for positive solution.

Here is my solution.

First we got a^3-a-1=0
We use formula for polynomials on power 3 to determine that polynomial a^3-a-1=0 has only one real root. (a=1 b=0 c=-1 d=-1; q=1/2 p=1/3 D=q^2+p^3 ---> D>0 ----> onlz 1 real root)
We can easily find that root is between 1 and 2 (f(1)<0, f(2)>0; a>1 a<2)

a=(b^6-6)/3
3<b^6-b<6
Therefore b>1 and b<2

Now we define function a=f(b)=(b^6-b)/3
Everything we have to do is to show this function is strictly increasing on interval (1,2) (b1>b2 ---> a1>a2)

f(b)=(b^6-b)/3
f(b+1)=((b+1)^6-(b+1))/3
f(b+1)-f(b)=(b^6+6b^5+......6b+1-b-1-b^6-b)/3 which is clerly greater then zero. And because b+1>b in interval (1,2) we proved that f(b) is strictly increasing function.

Now f(3/2)=(3^6)/(2^6)-3/2 we can clealy see that f(3/2)>3/2
So we for every b we chose there is f(b) which is greater then b.
So we proved that whichever value does a take between (1,2); b is smaller then a.

Im sorry that this is so hard to read, but I dont have latex.


Circular motion-Multiple choise questions
Circular motion
1. If a certain car, going with speed v1, rounds a level curve with a radius R1, it is just on the verge of skidding. If its speed is now doubled, the radius of the tightest curve on the same road that it can round without skidding is:
A) 2R1 B) 4R1 C) R1/2 D) R1/4 E) R1

2. An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of friction between tires and road is 0.50. The maximum speed with which this car can round this curve is:
A) 3.0 m/s B) 4.9 m/s C) 9.8 m/s D) 12 m/s E) 13 m/s

3. The driver of a 1000-kg car tries to turn through a circle of radius 100 m on an unbanked curve at a speed of 10 m/s. The actual frictional force between the tires and slippery road is 900 N. The car will:
A) slide into the inside of the curve D) make the turn only if it went faster
B) make the turn E) slide off to the outside of the curve
C) slow down due to the frictional force

4. A giant wheel, 40 m in diameter, is fitted with a cage and platform on which a man can stand. The wheel rotates at such a speed that when the cage is at X (as shown) the force exerted by the man on the platform is equal to his weight. The speed of the man (in m/s) is:
A) 14 B) 20 C) 28 D) 80 E) 120

5. A coin is placed on a horizontal phonograph turntable. Let N be the normal force exerted by the turntable on the coin, f be the frictional force exerted by the turntable on the coin, and fs, max be the maximum force of the static friction. The speed of the turntable is increased in small steps. If the coin does not slide, then
A) N increases, f increases, and fs, max stays the same
B) N increases, f increases, and fs, max increases
C) f increases and both N and fs, max stay the same
D) N, f, and fs, max all stay the same
E) N, f, and fs, max all increase

6. The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. How slowly, in ft/s, can the ball go through its top position without having the string go slack?
A) 1.3 m/s B) 2.6 m/s C) 3.9 m/s D) 6.9 m/s E) 9.8 m/s

7. An 800-N passenger in a car, presses against the car door with a 200 N force when the car makes a left turn at 13 m/s. The (faulty) door will pop open under a force of 800 N. Of the following, the least speed for which the man is thrown out of the car is:
A) 14 m/s B) 19 m/s C) 20 m/s D) 26 m/s E) 54 m/s

It looks like you are trying to get us to do your homework for you.  What, in particular, are you struggling with?  Show that you've made some attempt at a solution.  What are the fundamental equations at play?  All of these problems rely on this equation.

So show me please that equation. And please give me just one example of the solution for any of these problems.

In general, in circular motion, there is a force on the object that always points towards the center of the circle. This force is called centripetal force, and the equation is $ F_{c} \equal{} ma_{c}$ and $ a_{c} \equal{} \frac {v^{2}}{r}$.


Solve each equation. Express irrational solutions in EXACT form and as a decimal rounded to 3 decimal places.

(1) (0.3)^(1 + x) = 1.7^(2x - 1)....Notice how the right side of this equation does not have parentheses around the decimal number 1.7. What does that mean as you explain the question?

(2) e^(x + 3) = pi^(x)

1.) There really is no difference because you are exponentially increasing $ 1.7$ by $ 2x\minus{}1$.

[hide="solution 1"]
$ (0.3)^{1\plus{}x}\equal{}1.7^{2x\minus{}1}$ is equivalent to $ (3/10)^{x\plus{}1}\equal{}(17/10)^{2x\minus{}1}$
Taking the natural logarithm of both sides gives $ (x\plus{}1)(\ln{(3/10)})\equal{}(2x\minus{}1)(\ln{(17/10)})$
Expand $ x\ln{(3/10)}\plus{}\ln{(3/10)}\equal{}2x\ln{(17/10)}\minus{}\ln{(17/10)}$
Isolate the variable $ 2x\ln{(17/10)}\minus{}x\ln{(3/10)}\equal{}\ln{(17/10)}\plus{}\ln{(3/10)}$
Extract the variable $ x(2\ln{(17/10)}\minus{}\ln{(3/10)})\equal{}\ln{(17/10)}\plus{}\ln{(3/10)}$
Divide $ \boxed{x\equal{}\frac{\ln{(17/10)}\plus{}\ln{(3/10)}}{2\ln{(17/10)}\minus{}\ln{(3/10)}} \equal{}\minus{}0.297}$
My calculator simplifies that ugly fraction to $ \frac{\ln{(51/100)}}{\ln{(289/30)}}$
[/hide]
[hide="solution 2"]
$ e^{x\plus{}3}\equal{}\pi^x$
Take the natural logarithm $ (x\plus{}3)\ln{e}\equal{}x\ln{\pi}$
$ \ln{e}\equal{}1$ (by definition) Isolate $ x\minus{}x\ln{\pi}\equal{}\minus{}3$
Extract $ x(1\minus{}\ln{\pi})\equal{}\minus{}3$
Divide $ \boxed{x\equal{}\frac{3}{\ln{\pi}\minus{}1}\equal{}20.728}$
[/hide]


When you add 12 to a number and then divide the sum by 13, you get the same result as when you subtract 13 from the number and then divide the difference by 12. What is the number?


geometry
3D geometry
real analysis
geometric transformation
invariant
advanced fields
advanced fields unsolved

How would we extend the ideas of finite-dimensional Euclidean space to an "infinite-dimensional Euclidean space"? I would assume this would be unitarily equivalent to the other infinite-dimensional Hilbert spaces, but I don't really have any intuition for how it would look compared to finite-dimensional Euclidean space. What kind of sets (if any) would have finite nonzero measure? How would we extend the idea of measure anyway? It seems like the measure of the unit ball would be 0, even though it isn't compact.... Would the unit cube still have measure 1? Would compact sets have to have measure 0?

If you don't want to explain all these things, you could also point me to a good text (online or offline; the math library here is extensive).

Thanks!

what is THE measure you are talking about ?

I don't know, that's why I was asking. I mean, it is not obvious to me how you would extend the standard Lebesgue measure on finite-dimensional Euclidean space... However, my conjectures were based on the fact that the measure of the unit ball in $R^d$ is maximal for rather small dimension and then decreases fairly quickly. But I don't think that an $R^\infty$ measure is going to be very precisely defined by "the limit of a similar-looking object in $R^d$ as $d\to \infty$ ...  :P

[quote]How would we extend the ideas of finite-dimensional Euclidean space to an "infinite-dimensional Euclidean space"?[/quote]
You have to give up the invariance of measure under translation (in an infinite-dimensional space, a ball of radius 2 contains infinitely many disjoint balls of radius 1). Thus it is unnatural to use just one fixed measure -- this would not fit the translation-invariant structure of the space. A translation-invariant [i]family[/i] of measures can be used instead, such as the Gaussian measures or cube measures. A null set is defined as a set that has measure 0 for every member of the family. Marianna Csornyei [url=http://www.homepages.ucl.ac.uk/~ucahmcs/publ/aro.ps]proved[/url] that null sets are the same whether you use Gaussians or cubes. For example, a hyperplane is a null set, but a ball is not. And not every compact set is Gaussian null.


In $ \triangle ABC$, $ D(7,3)$ is the midpoint of side $ \overline{AB}$. $ E(10,9)$ is the midpoint of side $ \overline{BC}$, and $ F(5,5)$ is the midpoint of side $ \overline{AC}$. Find the coordinates of $ A$.

The centroid of the triangle with vertices $ (x_1,y_1),(x_2,y_2),(x_3,y_3)$ is $ \left(\frac{x_1\plus{}x_2\plus{}x_3}3,\frac{y_1\plus{}y_2\plus{}y_3}3\right)$,

and since the midpoints are the coordinates

$ \left(\frac{x_1\plus{}x_2}2,\frac{y_1\plus{}y_2}2\right),\left(\frac{x_2\plus{}x_3}2,\frac{y_2\plus{}y_3}2\right),\left(\frac{x_3\plus{}x_1}2,\frac{y_3\plus{}y_1}2\right)$,

we can find the average of the coordinates of the midpoints to find the centroid of

$ \triangle ABC$, $ G\equal{}\left(\frac{7\plus{}10\plus{}5}3,\frac{3\plus{}9\plus{}5}3\right)\equal{}\left(\frac{22}3,\frac{17}3\right)$.

Since $ AG\equal{}2GE$ and $ A,G,E$ are collinear, if we let $ A\equal{}(x,y)$, then

$ G\equal{}\left(10\plus{}\frac{x\minus{}10}3,9\plus{}\frac{y\minus{}9}3\right)\equal{}\left(\frac{22}3,\frac{17}3\right)\implies A\equal{}(x,y)\equal{}\boxed{(2,\minus{}1)}$.


AMC
AIME
number theory
least common multiple
function
prime factorization

[quote="PenguinIntegral"][quote="cincodemayo5590"]Find the remainder when
\[2004^{ 2003^{ 2002^{ 2001}}}\]
is divided by $1000$.[/quote]
You see a number theory question with powers. Immediately think Euler's Theorem.[/quote]

I know how to do it. I just think it's a neat problem, so I'm putting it up here so others can see it as well. :)

[quote="Kamil Witek"]How could this be on Mildorf's AMC if the sdame thing was on the canadian math olympiad?[/quote]
Because even Mildorf steals problems sometimes?

Don't you have to know the totient of 1000? How would you find that?

Euler's Totient Theorem is not actually directly applicable here because $gcd(2004, 1000) = 4$.

What [i]is[/i] applicable here is [hide="magic"] the Chinese Remainder Theorem. [/hide]
[b]worthawholebean:[/b]  $\varphi(2^{3}\cdot 5^{3}) = (2-1)2^{2}\cdot (5-1) 5^{2}= 400$

[quote="Zellex"]Woah, I got totally lost. I thought
[hide]
Chinese Remainder Theorem only guarantees a solution when you have multiple mods...something along those lines.. how did you get to the next line?[/hide][/quote]
You need to do [hide="feats of strength"] $\bmod 125$ and $\bmod 8$. [/hide]to use CRT.

Few comments:

[quote="t0rajir0u"]Euler's Totient Theorem is not actually directly applicable here because $gcd(2004, 1000) = 4$. [/quote]
Actually it may still be applicable here,  For example,
$4^{n}\equiv 4^{(2000+n)}\bmod 1000$ as long as $n \ge 3$  (IOW  It's still cyclic ( even for bases  which are not co-prime to 10) except for the first few terms)

Of course,  this is can be easily explained by:
[quote]
What [i]is[/i] applicable here is [hide="magic"] the Chinese Remainder Theorem. [/hide][/quote]
Also
[quote][b]worthawholebean:[/b]  $\varphi(2^{3}\cdot 5^{3}) = (2-1)2^{2}\cdot (5-1) 5^{2}= 400$[/quote]

One can do a little better here and get $100$ instead of $400$ (You do not have to multiply just take LCM of 4 and 100)

The "generalized" phi function for $10^{n}$ for  (n=1  is 4), (n=2 is   20), (n=3 is 100) and for n>3 is ($10^{n-1}/2)$

Also it may be fun to actually dp this problem, and check the answer
[hide][hide="peek only if you want the answer"]The last 5 digits are 12704[/hide][/hide]

[quote="Gyan"][quote="t0rajir0u"]Euler's Totient Theorem is not actually directly applicable here because $gcd(2004, 1000) = 4$. [/quote]
Actually it may still be applicable here,  For example,
$4^{n}\equiv 4^{(2000+n)}\bmod 1000$ as long as $n \ge 3$  [/quote]

Yes, hence my use of the word "directly" :)

I myself brought up this point in a topic awhile ago; generally we have $a^{\varphi(n)+k}\equiv a^{k}\bmod \varphi(n)$ iff $k$ is at least the largest exponent appearing in the prime factorization of $n$.

[quote="Gyan"]One can do a little better here and get $100$ instead of $400$ (You do not have to multiply just take LCM of 4 and 100)[/quote]

Yes, this is called the [url=http://mathworld.wolfram.com/CarmichaelFunction.html]Carmichael function[/url] $\lambda(n)$.


induction
combinatorics unsolved
combinatorics

I started reading the proof of "if G is a connected graph (with n vertices), then it has atleast n-1 edges.", and reached a problem.

The proof is done by induction, and after assuming the statement is true for a graph with n vertices, we look at a graph with n+1 vertices. Then it says: let's take off one vertex (and all of the edges connected to it) so the remaining graph stays connected, ...

What I couldn't understand is how can you know for sure such a vertex exists?

The existence of such a vertex is probably just as hard to verify as the theorem we're trying to prove. How about a workaround:

If some vertex has valence 1, we can remove it and its edge while leaving everything else connected, and the induction applies. If every vertex has valence at least 2, there are at least $\frac{2(n+1)}2=n+1$ edges, and we are done.


vector
combinatorics proposed
combinatorics

a country contains two province ,and each province contains $2^n$ cities.we assign to each city n+1 binary code st each city's code in first province have odd number of 1 and each city's code in other city have even number of 1,and also all codes are diffrent.between any to cities, which their code differ just in one place,there is telephone connection.assume that A is set of cities we define F(A) ,set of cities that there is connection between each of them and odd number of cities in A ,prove that if n be even and all members of A is from the first province  then $F(F(A))=A$

Here's an observation that helps. I think I can solve it, but the actual solution is a bit hard to explain.

We should show two things: $F(F(A))\supseteq A,F(F(A))\subseteq A$. In other words, picking a city $x$ ($x$ is the binary vector representing it) in the first province, we should show that $x$ is connected to an odd number of cities which are connected to an odd number of cities in a set included in the first province and which contains $x$ $(*)$, and it should be connected to an even number of cities which are connected to an odd number of cities in a set included in the first province which does not contain $x$ $(**)$.

The observation is that we may assume $x=(00\ldots 0)$, because there is an automorphism of the graph defined on the cities which maps $x$ to $(00\ldots 0)$, namely $a\to a\mbox{ XOR }x$. It's much easier to work with these binary codes: $x$ is assumed to be the $0$ vector, and the cities to which $x$ is connected become those vectors which have $1$ digit equal to $1$, and the neighbours of those are vectors which have two digits of $1$. It's not hard to prove $(*),(**)$ for these vectors.


calculus
integration
trigonometry
function
symmetry
derivative
calculus computations

You guys already think I ask weird questions on my tests. So here's another one, only this time you'll have to guess what the question was.

The subject is Calculus III, multivariable calculus. The particular test covered certain limited material: the calculus of the $\nabla$ operator, including curl, divergence, and gradient. Line integrals in the plane and space. Green's Theorem. Surface integrals. (No Stokes' Theorem or divergence theorem - yet.) This was question 1. It started out:

Let $u(x,y,z)=\frac{\sin\left(\frac{x}{y^{2}+z^{2}+1}\right)+e^{2xz^{2}}}{x^{2}+y^{2}+z^{2}+12}.$ Compute ...

So now, put yourself into my shoes. If you were the instructor of the course, and you had that exact setup, how would you continue the problem? What would you ask to see?

If I had an intention to be easy on my students, I would ask to compute the gradient at $0$ in 1 minute :)

How is that possible? Please comment.
How much work is needed?

Let's see: the function is a specific formula, but it's messy. Messy without much in the way of pattern or symmetry - it seems to simply represent complication for the sake of complication. But it is smooth, and smooth everywhere. I did take care of that. Now why would I use such a (scalar-valued) function?

[hide="What I really asked."]Compute $\nabla\times\nabla u=\text{curl}(\text{grad}(u)).$[/hide]

And if you actually learned the material, that's a lot less than one minute.

Well, what caught my eye was an obvious abundance of higher than first order (at 0) expressions in the formula. So, I just decided that Kent might want to exploit this somehow and what I proposed was the easiest way to do it (1 minite is not the time needed to answer my question but just the maximal time about which I was more or less sure that it would be insufficient for an average student to write down all three expressions for the partial derivatives and plug (0,0,0) into them).

that is an "nabla-calculus" identity and it is $0$.

I would ask to determine the range of $u$, with some justification. In fact, I put a question of this kind on a Calculus III exam for the past 4 semesters. My reason for doing this: although many of my students have mastered the ancient art of decreasing an exponent by one and putting the old exponent in front of a variable, they still don't know what a function is.

So, what is the range of $u$?  :wink:

Not a question I was anticipating when I wrote this thing down. The range of this function is $[c,\infty)$ for some $c\in(-1,0),$ but I don't particularly want to figure out exactly what $c$ is.


Prove for a+b+c=1 that

$ ab\plus{}bc\plus{}ac \sqrt{3abc}$

[quote="NikolayKaz"]Prove for a+b+c=1 that

$ ab \plus{} bc \plus{} ac \sqrt {3abc}$[/quote]

is it

$ ab \plus{} bc \plus{} ac \geq \sqrt {3abc}$   :?:

Well something missing indeed.
they are positive or not ???
\[ \left( {\sum {ab} } \right)^2  \ge \left( {\sqrt {3abc} } \right)^2 \,\,\, \Rightarrow \,\,\,\sum {a^2 b^2 }  \ge abc\,\,\, \Rightarrow \,\,\,\sum {{{ab} \over c} \ge 1}\]
which is easy to prove if they are positive.
I think....
 :P

Sorry, apparently forgot to write this. Yes, they are positive real numbers and it should look

$ ab\plus{}bc\plus{}ac \geq \sqrt{3abc}$

a+b+c=1
\begin{align*}
\sum ab\geq\sqrt{3abc}\\
\sum ab\geq\sqrt{3abc(a+b+c)}\\
\left(\sum ab\right)^2\geq 3\sum a^2bc\\
\sum a^2b^2\geq \sum a^2bc
\end{align*}
Multplying by 2, it`s true by Muirhead

Newton:
\begin{align*}
d_2^2\geq d_1\cdot d_3\\
\left(\frac{\sum ab}{3}\right)^2\geq \frac{a+b+c}{3}\cdot abc\\
\left(\sum ab\right)^2\geq 3abc
\end{align*}


inequalities
advanced fields
advanced fields unsolved

I hope this in the right section.

Subbose $x_{n}\to 0$ weakly in a commutative Banach algebra $A$ and $\phi_{n}\to \phi$ weakly* in the Gelfand spectrum of  $A$.  I.e. each $\phi_{n}$ is a multiplicative linear functional on $A$.

What are some sufficient conditions on $A$ to ensure that $\langle \phi_{n}, x_{n}\rangle \to 0$?

Please note that these are sequences.

Remark: This holds when $A$ has the Schur Property.

Any ideas or references would be appreciated.

What is $A^{*}$ here - the Banach space dual (the bounded linear functionals) or is it the muliplicative linear functionals?

Sorry about that.

$A^{*}$ is the dual space, but I am only interested in the characters.


I need to figure out whether or not the rationals with the 2-adic topology are connected or not.

I know that to prove a set is connected, the set cannot be partitioned into 2 disjoint, non-empty, open subsets. Therefore I think the right way to go about it would be to investigate whether it is true any 2 open sets intersect because then clearly the space would be connected.

Please can someone show me how to do this(or suggest an alternative way to prove connectedness) if my approach is extra-cumbersome/totally heading down in the wrong direction.

Thanks

You should probably investigate the possibility that they're actually disconnected first.

In particular, note my answer to you [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306255]here[/url].

In fact, any $ p$-adic topology is totally disconnected because it is induced by a non-archimedean absolute value.

But, this is how you show disconnectedness: Let $ F$ be a field and $ |.|$ a non-archimedean absolute value (i.e. the $ q$-adic absolute value). Then closed balls are open.

Proof: let $ a\in F$ and $ r \in \mathbb{R}_ > 0$ Consider the closed ball $ \bar{B}_r(a)$ (defined as usual). Let $ b\in \bar{B}_r(a)$ Consider the open ball $ B_{r/2}(b) \equal{} \{x\in F| |x \minus{} b| < r/2\}$ Now if $ c\in B_{r/2}(a)$ then $ |c \minus{} a| \equal{} |(c \minus{} b) \plus{} (b \minus{} a)| \leq Max(|c \minus{} b|,|b \minus{} a|) < r$ (by hypermetric property of non-archimedean absolute values). Therefore $ B_{r/2}(b)\in \bar{B}_r(a)$ hence the closed ball is open!

Dear LKNsngth

Please check your inbox - I have just sent you a private message regarding your last post.
Thank you for replying to my message.
Best Wishes
crackersncheese


In which states have the full score lists been released? And the sheets where each team can see how the members did?

I heard that some states have already received them; however, Texas has yet to give anything... :?

[quote="xxm0rg07hxx"]Arkansas has a pdf of scores and everything.  But that's our info.[/quote]

So did CT, each school got their sheets with scores right after the competition and a PDF was sent a week or two after. If anyone wants the PDF, PM me

I don't think you should let people have it, because some people on the pdf may not want their score known to other people.

The PDF would show how many students got a certain score on each round and total, but not a student's name.

Then again, It is possible to match students to scores using the individuals ranking list. I did this in 7th grade and found out that we lost to the #1 team by .25 points.  :(

It also says the school by the name and the score.

So does GA's list except no score. You need to look on the graph for that.

The .pdf looks like

Individuals
1  46
2  46
3  46
4  2

Team
1 46

It only shows scores like Xantos said ;) However, they did release a list of the top 54 individuals and top 9 teams so you could match those up and no one else

was that a serious score list (wondering if there were any perfect scorers)

So you think the difference between 3rd and 4th is 44?

Fl didn't give any sheets and never will :(

[quote="Ignite168"]The .pdf looks like[/quote]
You mean VA's and SOME other states, because ours doesn't.  Using "the" implies that everyone has the same pdf format.

Ok clarifying: That pdf is for CT scores only, I know some states don't even have scores for anyone. Also, the scores were fake, they were just to give an example.


probability
combinatorics unsolved
combinatorics

Let $ |A| \equal{} 4n$ and $ A_1,A_2,..,A_k$ are subsets of $ A$ with $ 2n$ elements. Prove that :there exist $ i,j$ such that $ |A_i \cap A_j| \geq n(1 \minus{} \frac {1}{k})$

Very similar, maybe equivalent topic [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=297548[/url]

[quote="sobe86"]Should it be prove that there exist $ i,j$ such that $ |A_i \cap A_j| \geq n(1 \minus{} \frac {1}{k \minus{} 1})$ ?

Otherwise it definitely isn't true for k=2 or k=3.[/quote]
I had mistake:$ A_0,A_1,...,A_k$ no $ A_1,..,A_k$

[color=red][[b]Moderator edit:[/b] The corrected problem is a particular case of http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=83740 .][/color]

this problem can be done easily by probability method!


I know this probably isn't very advanced, but I really need help on this!

Zinc Sulphide undergoes pyrometallurgy, oxidizing to become Zinc Oxide. How would I express this in a chemical formula?
Like this:
$ ZnS \rightarrow ZnO \plus{} S$       ?

Also, when Zinc Oxide is leached to Zinc Sulfate using Sulfuric Acid, is that simply

$ ZnO \plus{} H_2SO_4 \rightarrow ZnSO_4$

And finally, can someone explain in simple terms what Electrowinning is?

This is for a project, so that's why we haven't learnt it in class yet. Thanks for help

[quote="Identity"]I know this probably isn't very advanced, but I really need help on this!

Zinc Sulphide undergoes pyrometallurgy, oxidizing to become Zinc Oxide. How would I express this in a chemical formula?
Like this:
$ ZnS \rightarrow ZnO \plus{} S$       ?

Also, when Zinc Oxide is leached to Zinc Sulfate using Sulfuric Acid, is that simply

$ ZnO \plus{} H_2SO_4 \rightarrow ZnSO_4$ [/quote]

I think they should be:

$ 2ZnS \plus{} 3O_2 \rightarrow 2ZnO \plus{} 2SO_2$

$ ZnO \plus{} H_2SO_4 \rightarrow ZnSO_4 \plus{} H_2O$

oops forgot the $ O$.

In chemistry (with the exception of Organic Chemistry) we are supposed to always write balanced chemical equations.

For your first reaction, BanishedTraitor has already given a reasonable and balanced equation. However for your second reaction (which is an acid-base reaction) we should have

$ ZnO(s) \plus{} H_2SO_4(aq) \longrightarrow ZnSO_4(aq) \plus{} H_2O(l)$.

BanishedTraitor's second reaction would be reasonable if we were considering the direct reaction between metalic zinc and sulphuric acid.


I won't say that the post count should be removed. I have only one demand: there will be no discrimination between posts in serious maths and posts made in games and chain threads. I have always believed that maths is incomplete without games and fun and so is a mathematician. I have already suggested a system: there should be a record of the posts of each person in each forum underhis post count thet everyone can, given the inclination, see.

[quote="jhollenbeck"]This is a topic to see what other AoPSers think about the post count system.  I personally think that AoPS puts way too much focus on the post count factor, which, in some cases, causes spam, and that the post counts should either be kept personal (not revealed every post to others) or be completely removed from AoPS.  What do you think?   :)[/quote]

I think WE are the ones who put too much focus on it.
It's alot like a website counter, it really doesn't matter much...

I think that we should ingore all discussion about post count, leave it as it is, and let the people who want to spam spam.  They can face the consequences :?

All forums have post count btw.  It's kinda cool actually. :?

[quote="mathnerd314"]Then again, I've also heard the argument somewhere that the people who are really determined to get a high post count (for some strange reason) would spam in the real math forums to get posts instead of in G&FF, which would be even worse. Wow, that was a long run-on sentence.[/quote]
Well, really spamming should be bounded since total spamming there will have consequences.

[quote="ragnarok23"]I think there should be a counter for how many solutions to different problems you have given or how manyu people you have helped. After all, that is the main point of the site :D.[/quote]
nice idea, but that would be hard to implement because the admin would have to look for every single solution by the poster.

I think the post count function is just for fun and should be kept.

i think the post counts are a big problem unless we make it one...what is the point it having a high post count anyway? its just there

Ok so how is somebody harmed if somebody else notches up 10000 posts through sheer spamming?

[quote="bubka"][quote="lightrhee"]or we could weigh it differently(eg. post count increases by 1 if you posted in Algebra Unsolved Problems, by 0.1 if you posted in Games & Fun Factory) :D
probably not going to happen though[/quote]

Or the other way round  :D  :rotfl:[/quote]
That would make my count go down to around 200...not that it matters at all...

Instead of post count, there should be a non-quoted word count.

I still like the idea of double post count the best!

The word count is TOO pressuring on the mods.
"Ok. That one has 1,2,3,4,5,6,7,8,9,10,11......"
The post count should be left the way it is, but the old ToWs did increase post count. Too bad someone deleted them all.

Word count would keep the TOW's from getting too absurd though.

I think that the post count should stay. If we make a word count, I would be a Birch and Swane Dyer.


The $ 5$ will not be adjacent to the $ 1$ and so $ 2(3)(4)(6)\equal{}\boxed{144}$.


Problem: The complex numbers $z_{1}$, $z_{2}$,..., $z_{n}$ all belong to a disk, $D$. Prove that there exists a complex number, $z$, belonging to the same disk such that $z_{1}z_{2}...z_{n}=z^{n}$.

Confusion: I guess I might be misunderstanding what a disk is, but as far as I know, it should be a circle on the complex plane. However the theorem becomes trivially false; consider $n=2$, $(\Re[z]-3)^{2}+\Im[z]^{2}=1$, $z_{1}=2$, $z_{2}=4$. The question then says that $z=\sqrt{2*4}$ lies on that circle...What am I doing incorrectly?

Perhaps a disk looks like a CD rather than a circle?

Edit: Never mind. http://en.wikipedia.org/wiki/Disk_%28mathematics%29

I believe a disk is the region bounded by a circle, as stated in that wikipedia article. So all the $z$'s lie within a circle, not necessarily on the circle.

Paladin8 is right -a disk is the region bounded by a circle. Thus $z$ should be inside the disk or on its boundary.


In a rectangle $ABCD$, $P$ is on $BD$ such that $AP \perp BD$. Conctruct $PE \perp BC$, $PF \perp DC$ with $E$, $F$ $\in$ $BC$, $DC$. Prove that:
\[ \left( \frac{PE}{BD} \right)^{\frac{2}{3}}+ \left( \frac{DF}{BD} \right)^{\frac{2}{3}}=1 \]

(bump)

This is a nice question, and approachable -- someone should give it a try.

Yeah
I think it was given in some canadian contest (probably the open one) :D

Cool problem.  Here's a sketch: using a whole bunch of similar triangles,

$\frac{PE}{BD} = \frac{PE}{BP}\cdot \frac{BP}{BD} = \frac{AB}{BD}\cdot \frac{AB\cdot BF}{BD\cdot AB} = \left(\frac{AB}{BD}\right)^3$

Similarly, $\frac{FD}{BD} = \left(\frac{AD}{BD}\right)^3$ so the result follows, as $AB^2 + AD^2 = BD^2$.

[quote="me@home"]it would help if someone posted a diagram...[/quote]
Do it yourself.


Prove that zeta(3) is equal

$ \frac{5}{2}\sum_{n\equal{}1}^{\plus{}\infty}\frac{(\minus{}1)^{n\minus{}1}}{n^3C_{2n}^{n}}$

$ \binom{n}{k}$ is sometimes written as $ C_n^k$ (note the reversal of integers). This was the standard notation in my (Russian) textbooks.


calculus
integration
trigonometry
logarithms
complex analysis

using complex analysis techniques, show :D

$ \int_{0}^{\infty}\;\;\frac{\tan^{\minus{}1}\,x}{\sqrt{\;x\;}\cdot (1\plus{}x)}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{\pi^{2}}{4}}$

now, how about $ \int_{0}^{\infty}\;\;\frac{\ln\,x}{\sqrt{\;x\;}\cdot (1\plus{}x)}\;\;\textbf dx$

why were the choices of your contours [i]different [/i]? :|

Complex analysis techniques are here not necessary.

In both cases substitute $ x\to x^2$ then $ I_1\equal{}2\int_0^\infty \frac{\arctan x^2}{1\plus{}x^2}\, dx$ and $ I_2\equal{}2 \int_0^\infty \frac{\ln x^2}{1\plus{}x^2} \, dx$ .

After that substitute $ x\to\frac{1}{x}$ then $ I_1\equal{}\tilde{I_1}: \equal{}2\,\int_0^\infty \frac{\arctan \frac{1}{x^2}}{1\plus{}x^2}\, dx$ and $ I_2\equal{}\tilde{I_2}: \equal{}2 \int_0^\infty \frac{\ln \frac{1}{x^2}}{1\plus{}x^2} \, dx$ .

Because of $ \arctan x^2\plus{}\arctan \frac{1}{x^2}\equal{}\frac{\pi}{2}$ and $ \ln x^2\plus{}\ln \frac{1}{x^2}\equal{}0$

$ 2I_1\equal{}I_1\plus{}\tilde{I_1}\equal{}2\int_0^\infty \frac{\frac{\pi}{2}}{1\plus{}x^2}\, dx\equal{}\frac{\pi^2}{2}$ and $ 2I_2\equal{}I_2\plus{}\tilde{I_2}\equal{}0$ .

For the first, I'd start by considering the possibility of flipping the square root branch cut over and integrating over the following contour:

[img]http://img164.imageshack.us/img164/6827/arctancontourap0.jpg[/img]

Assuming I could show it's valid and all the integrals except the two on the positive real axis cancel, I'd end up with something like:

$ - 2\int_0^{\infty}\frac {ArcTan(r)}{\sqrt {r}(1 + r)} = \mathop\text{Res}\limits_{z = - 1}{\frac {ArcTan(z)}{\sqrt {z}(1 + z)}}$

which equates to the desired answer . . . although this answer still needs a lot of work.


A cuboid and a cube have the same volume and height.The lenght of the cuboid is thrice greater than that of the cube and the width is twice smaller cal. the longest diagonal of both the cube
(question2)find the sum of a series if the nth term is 1/((n+1)(square root of n)+n(square root of n+1)).
(question3)find the number of all positive intergers abcde such that:b=a+c,d=e+c.(i got 729 digits)
(question4)if f(n)=n^2f(n) and f(1)=2009.find f(2008).
(question5)find the value of cos1+cos2+...+cosn[quote][i]each question scores you 5 marks.thank you[/i][/quote]

[color=darkred][b]ans 1[/b]\
since volume same and hieght, h is same so base area is also same.
let cube's side = x
  so length of the cuboid = 3x
and,breadth  " " "  " " " " "= x/2
   diagonal of a cuboid = sqrt(l^2+b^2+h^2)
                               =sqrt(9x^2+x^2/4+x^2)
                               =x*sqrt(9+0.25+1)
                               =x*sqrt(10.25)

        diagonal of cube = x*sqrt(3)[/color] :D  :D  :D  :D

[color=red][b]ans 3)[/b]
number=abcde
assuming that all letters stand for different digits,let 'c' be constant.
 NOTE- wherever mentioned, for any number n its compliment is here defined as (10-n)
if any digit = 0 , there will be repetition of digits.
  for 'c '=1,the follwing choices are crossed out for 'a' ,
  1.)0
  2.)compliment of 'c'' and numbers greater than compliment.
  3.)c itself or there will be repetition.
so there will be 7 choices.
  and for 'e' similarly there will be 6 choices after excluding value of digit of 'a '
 so , combination for value of c be c(n)

 c(1)=7*6 =42
similarly,
 c(2) = 6*5=30
 c(3) = 5*4 = 20
 c(4) = 4*3=12
  but from 5 onwards, point 3.) above will not  followed as it will be covered by point 2.)
 so,
 c(5)=4*3=12
 c(6)=3*2=6
 c(7)=2*1=2
and 8 and 9 cannot form combinations.
 so, adding all
  42+30+20+12+12+6+2=124
                                   if my calculations are right. :D  :D  :D  :D  :D  :D [/color]

you got it all wrong so,0mark to you.hard luck try better next time

why do you think so? :mad:  :mad:  :mad:

[hide="Problem 2"]

$ \boxed{\textbf{2}} \qquad f(n) \equal{} \frac {1}{(n \plus{} 1) \sqrt {n} \plus{} n \sqrt {n \plus{} 1}} \equal{} \frac {1}{\sqrt {n}} \minus{} \frac {1}{\sqrt {n \plus{} 1}}$

$ \therefore \sum_{n \equal{} 1}^k f(n) \equal{} 1 \minus{} \frac {1}{\sqrt {k \plus{} 1}}$

$ \therefore \sum_{n \equal{} 1}^{\infty} f(n) \equal{} 1$[/hide]

 :)

1) what's a cuboid?
3)Are $ a$, $ b$, $ c$, $ d$, and $ e$ the digits of a number?
4) Am I interpreting this correctly? $ f(n) \equal{} n^2f(n) \implies 1 \equal{} n^2 \implies n \equal{} \pm 1$
5) degrees or radians?

yes. i agreee with facis
 please reformat the questions to make them more understandable


In a triangle ABC

CosACosB +SinASinBCosC=1

Prove A=B= 45 degrees

Are you sure about this? Spoiler below

[hide]To check the validity, let A=B=45 and C=90.

cos A cos B + sin A sin B cos C = 1/2 + 1/2 * 0 = 1/2 !!!!!

It is not equal to 1, maybe you meant     instead of .[/hide]

Yes, he means $\cos A\cos B+\sin A\sin B\sin C=1$.
At least, his statement is true in this case.

$\cos{A} \cos{B} + \sin{A} \sin{B} \sin{C} \leq \cos{A} \cos{B} + \sin{A} \sin{B} \cdot 1 = \cos{(A-B)}$, so $\angle A = \angle B$.

This implies that we must have equality, so $\sin{C}=1$, and $\angle C = 90^{\circ}$, $\angle A=\angle B=45^{\circ}$.


IN an analysis of the nucleotide composition of DNA, which of the following is true?

a. A=C
b. A=G and C=T
c. A+C = G+T
d. G+A=T+C
e. both C and D

I think that B and C are right because A and G are purines and C and T are pyrimidines.  All I know for sure is that  B is not correct.  Can anyone help me?  Thanks.

Well in DNA, the complementary base pairing is: A-T and C-G.

So we have A=T and C=G. That should help with the question.

I'm thinking that it might be E, can someone confirm that?

Yes, the answer is e because both c and d are true.

As TaiPan~SP stated, A=T and C=G, so you can start out with A=T, add C to both sides to get A+C=T+C, and substitute in G to get A+C=G+T.


Show that

\[i\left(p-1\right)\left(p-2\right)\cdots\left(p-i+1\right)\equiv\left(-1\right)^{i-1}i!\bmod p,\]
and use it to show that

\[\sum_{i=1}^{p-1}2^{i}i^{p-2}\equiv\frac{2-2^{p}}{p}\bmod p.\]

I heard that the proof of the second equivalence requires the Binomial Theorem.  Do you know how it applies in this case?


LaTeX
number theory unsolved
number theory

Find all sequences a_1; a_2; ... ; a_2000 of real numbers such that sum  a_n = 1999 ( n=1-->2000 )and for
any n >=1 we have 1/2<a_n<1 and a_(n+1)=a=n(2-a_n ) :D

quangpbc, Why you not use LaTex? Please learn here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php  :wink:

Sorry N.T.TUAN , because i have not learned about LATEX .I will learn to use LATEX . :D


The Countdown is not official. Good for me, otherwise I'd probably not have gone to Nationals last year.

It's March 14th. I assume you'll be going?

Yes, I am, you will probably beat me though.

Out of curiosity, did you go last year? How'd you do?
I'll see you in 2 weeks guys (xp, if you don't get first, I don't know what I'll do with you haha)

This was superwatchdog's first time in Chapter, according to him. He also got 2nd in his chapter (which is around Columbus), which is pretty dang good for a first timer.

Heh if I don't make nats, I will be pretty sad. Can't wait until the competition.

Same here. I just feel guilty that I haven't been studying hard enough... last minute regrets, I guess  :(


inequalities
trigonometry
Cauchy Inequality
inequalities open

Prove that \displaystyle \frac{1}{\tan x \sin x (\cos x -\sin x)}>8 \ \ \text{for} \ \ 0<x<\frac{\pi}{4}.

[quote="Stas"]Prove that $\displaystyle \frac{1}{\tan x \sin x (\cos x -\sin x)}>8 \ \ \text{for} \ \ 0<x<\frac{\pi}{4}$.[/quote]

\[
\begin{gathered}
  \frac{1}
{8} > \tan x\sin x(\cos x - \sin x) \hfill \\
   \Leftrightarrow \frac{{\cos x}}
{8} > \sin ^2 x(\cos x - \sin x) \hfill \\
\end{gathered}
\]
Using AMGM,
\[
\frac{{\cos ^2 x}}
{4} > \sin x(\cos x - \sin x)
\]
it remains to prove,
\[
\frac{1}
{2} > \sin x\cos x
\]
which follows from
\[
\frac{1}
{2} = \frac{{\sin ^2 x + \cos ^2 x}}
{2} > \sin x\cos x
\]

[quote="siuhochung"][quote="Viet Math"][quote]$\displaystyle \frac{1}{2} = \frac{{\sin ^2 x + \cos ^2 x}}{2} > \sin x\cos x$[/quote]

 I think you like Cauchy's Inequality very much ???  :D  :D

 Beacause

$\displaystyle \sin x\cos x=\frac{sin 2x}{2}$

 Then it equavelent

$1 > \sin 2x$

:D  :D  :D[/quote]

Actually, may i know what is the cauchy inequality you are talking about ?
To me,
AMGM means $\frac{\sum a_i}{n} \ge (\prod a_i)^{\frac{1}{n}}$
Cauchy means $(\sum a_1^2)(\sum b_1^2) \ge (\sum a_1b_1)^2$[/quote]

 But in Viet Nam somethings is different :

$\frac{\sum a_i}{n} \ge (\prod a_i)^{\frac{1}{n}}$ called Cauchy

$(\sum a_1^2)(\sum b_1^2) \ge (\sum a_1b_1)^2$ called Bunhiacopxki

 :)  :)  :)