Here it is: (the underfull box is at Lorenz line, shows Texnic Center.)

[quote]

\documentclass[a4paper,10pt,oneside]{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}

\begin{document}

\begin{thebibliography}{13}
\bibitem{Agrawal}Agrawal, O. P., Formulation of
Euler-Lagrange equation for fractional variational
problems, J. Math. Anal. Appl. 272(2002), 368-379
\bibitem{Albu}Albu, I.D., Neamtu, M., Opris, D., The
geometry of higher order fractional osculator bundle
and applications, preprint, West University of
Timisoara, 2007
\bibitem{Bolean}Boleantu, M., The Lagrange formalisme of
classical Mechanics and conservation laws deduced
from it, Semin. Mec. (58), West Univ. of Timisoara,
1998
\bibitem{eu}Boleantu, M., Opris, D., Fractional dynamical
systems and applications in mechanics and economics,
arXiv: 0709.1681v1 [math.DS], (2007)
\bibitem{Caputo}Caputo, M., Kolari, J., An analytical
model of the Fisher Equation with Memory Functions,  Alternative Perspectives in Finance and Accounting, 1,
http://www.departments.bucknell.edu (electronic
journal), 2001
\bibitem{Cottrill}Cottrill-Shepherd, K., Naber, M.,
Fractional differential forms, J. Math. Phys. 42(2001),
2203-2212
\bibitem{Cresson}Cresson, J., Fractional embedding of
differential operators and Lagrangian systems,
Journal of Mathematical Physics, vol. 38, Issue 3,
(2007)
\bibitem{Nabulsi}El-Nabulsi, R. A., A fractional approach
of  nonconservative Lagrangian dynamics, Fizika A14,
4(2005), 289-298
\bibitem{Lorenz} Lorenz, W.H., Nonlinear dynamical economics and chaotic motion, Springer-Verlag, 1993
\bibitem{Miron}Miron, R., The geometry of higher order
Lagrange spaces. Applications to Mechanics and Physics.
Kluwer Academic Publishers, FTPH no. 82, 1997
\bibitem{Podlubny}Podlubny, J., Fractional differential
equation, Acad. Press, San Diego, 1999
\bibitem{broasca}Sommacal, L. \textit{and all},
Fractional model of a gastrocnemius muscle for tetanus
pattern, Proceedings of IDETC/CIE 2005, ASME 2005
International Design Engineering Technical
Conferences, September 24-28, 2005,
Long Beach, California, USA
\bibitem{Vedham}Vedham, K. B., Spanier, J.,
The fractional  calculus, Acad. Press, New York, 1974
\end{thebibliography}
\end{document}

[/quote]

Thanks

Jean

P.S. I also tried to attach the file "trouble"

The problem is the line
[code]\begin{thebibliography}{13}[/code]which says that the widest label (the text inside {} after \bibtem) is 2 characters wide. But that's too small and needs at least 3 characters. The warning is telling you that there isn't room for the label making spacing go wrong for that item - as the badness is less than 2000 it's practically undetectable in the printout so can be ignored.

You should only use numbers in \begin{thebibliography} if the labels are numbers 1 to n, but you are using text so you should use the widest one you have which I think is Cottrill. So you should use
[code]\begin{thebibliography}{Cottrill}[/code]instead.


Here it is! Make sure you are being good about time and stuff. Don't go on to the 2nd page of the file until setting aside 40 minutes.

Deadline: Saturday, September 2 at 11:59:59 eastern

questions you might want to check

#8,14,15

[quote="alex__ib"]at #14 you might want to add that x isn't 0[/quote]

not only that

I think he means x in #14.

P.S. There is no answer, because x can be greater than 15.

EDIT: I can't compete, it's too frustrating.

[quote="lotrgreengrapes7926"]There's a typo on #14, but you smart people can figure out what he meant. :P[/quote]

I just want jclarke to say it himself

Hey jclarkemathL314159 can you tell us when is the latest we can pass in the answers?

Um sorry to double post but like SplashD said, how do we know what we got wrong? Can you pm us the answers or something? :)

[quote="SplashD"][quote="Ignite168"]ugh, I think I got the 15. Oh well, at least I didn't get last but I agree, these problems were a bit hard for AMC8 difficulty, that's why my score is so low.[/quote]

I agree it's harder than average but all mock tests are.

no offense, but why are you complaning? you're going into 9th grade[/quote]
Wait. If this is "hard" for an AMC 8 then what is easy for one?

[quote="i_like_pie"][quote="daermon"]but are you going into ninth?[/quote]
Vishalarul is in 8th grade right now. He goes to my school.[/quote]

I apologize, but then you should just submit your answers to jclarke and tell him you're an 8th grader.

[quote="goldendomer"]No offense but that test was not the same difficulty of the AMC 8. It was like an easy version of the AMC 10. By the way, someone should transfer this to the AMC forum.[/quote]
A mod should move this to the AMC forum...

This is a hard version of the AMC-8 :rotfl:

jclarkemathL314159, you said that you would be able to post more AMC's. Not to be impatient, but do you know when you would be able to? Thanks. Also, like a few people have said earlier, a moderator should move this to the AMC forum. :D

jclarkemath, could you tell us who got the 8, 15, and 19?  i really want to know my score  :D

daermon, I just pmed you.

Anyway, since no one else is responding, I'm posting the ANSWERS now. Use this thread to post solutions.

Mock AMC 8 A answers

1. D
2. B
3. A
4. B
5. E
6. C
7. D
8. B
9. D
10. C
11. E
12. D
13. C
14. A
15. C
16. E
17. B
18. C
19. D
20. D
21. E
22. C
23. D
24. C
25. B

[quote="now a ranger"]is this a very hard version of the amc 8?[/quote] It is a mock amc 8. Generally, anything mocks that are on AoPS are much harder than the real deal.

Is this going to be moved to the AMC forum?


[quote="Linh.NDD"]Let $a,b,c$ be positive real numbers such that $a^{3}+b^{3}+c^{3}=3$. Prove that:
$\frac{a^{2}}{\sqrt{3-a^{2}}}+\frac{b^{2}}{\sqrt{3-b^{2}}}+\frac{c^{2}}{\sqrt{3-c^{2}}}\geq \frac{3\sqrt{2}}{2}$[/quote]
I think, your inequality is wrong.
Try $a=b\rightarrow\sqrt[3]{1.5}^{-}$ and $c\rightarrow0^{+}.$ :wink:

[quote="arqady"][quote="Linh.NDD"]But $a,b,c\neq 0$  :D[/quote]
I don't say that $c=0.$ I say that if $c=0$ then ... :wink:
If you want try $a=b=\sqrt[3]{1.4}$ and $c=\sqrt[3]{0.2}.$[/quote]

Hello Linh, a advice for you is that, if Arqady checked and said that you are wrong, you can be sure that this is a fact. As I see, you are actually wrong, too.  :lol:


(1)

The sequence R = {r_0,r_1,r_2,...} = {3,7,47,2207,...} begins with r_0 = 3 and thereafter r_n = (r_(n-1))^2-2. From R, the kth term of a second sequence S is determined from the product of the first k terms of R by taking the square root once for each term in the product:

S = {3^0.5, ((3*7)^0.5)^0.5, (((3*7*47)^0.5)^0.5)^0.5,...}

To what limit does S converge ?

(2) Determine the sum S = \sum^{infinity}_{n=1} arctan (2/n^2).

(3) If f_n the denotes the nth term of the Fibonacci sequence, determine the sum of the series: S = \sum^{infinity}_{n=1} (-1)^n/(f_n*f_(n+2)) .

(2)

In the correct domain :
arctan (a) - arctan(b) = arctan((a-b)/(1+a*b))

So

arctan(2/n^2) = arctan(1/(n-1)) - arctan(1/(n+1))

So we have a telescoping sum, if I didn't make error, it is : arctan(1/2) + arctan(2) + Pi/4 = 3Pi/4

(1):
a_n=r_n (I like a better :D )

(*) a_n-1=(a_(n+1)+1)/(a_n+1)
(**) a_n+2=(a_(n+1)-2)/(a_n-2)

We multiply all relationships of the type (*) from 0 to n to get P_n=(a_0-1)*..*(a_n-1)=(a_(n+1)+1)/4.

We multiply all relationships of the type (**) from 0 to n to get R_n=(a_0+2)*..*(a_n+2)=a_(n+1)-2.

It's obvious that P_n<S_n<R_n and lim (P_n^(1/2^(n+1)) = lim (R_n^(1/2^(n+1))) = lim (a_(n+1)^(1/2^(n+1))) when n->00. This means that the limit we're looking for is L=lim {n->00} (a_n^(1/2^n)).

There exists an a s.t. e^a+e^(-a)=3=a_0. Then a_1=a_0^2-2=e^(2a)+e^(-2a), .., a_n=e^(2^n*a)+e^(-2^n*a), which means that L=e^a=e^(arccosh(3/2)). Is it OK orl?

Hi grobber,

I don't know whether your solution is correct as I wonder how arccosh is defined. I know cosh(x) is defined as 1/2*(e^x+e^(-x)). So what is the inverse function ? Anyway the limit is the square of the golden mean: ((1+5^0.5)/2)^2.

Well, I don't know what the inverse function is either, but could you check my solution and see whether there's something wrong with it or not? And could you please post you solution or send it to me by PM?

arccosh(x) = ln(x \pm  \sqrt(x^2 - 1)) for real x

mathworld.wolfram.com is a good mathematical resource for that sort of stuff.

Well, we now know what arccosh(x) is, but there's still a problem: I got e^arccosh(3/2), and orl says it's ((1+5^0.5)/2)^2, so are these 2 equal? Or maybe I made some mistake, but in this case I'd like to know what it is. My soln looks Ok to me... Could you people check this out?

let us denote arccosh 3/2 = x. Then cosh x = 3/2 :) which means that

ex+1/ex = 3 => y2 - 3y + 1 = 0 => y = ( 3 +\sqrt 5) / 2 = ( 1+\sqrt 5 )2 / 4, which is exactly the number orl indicated. :)

I have denoted by y grobber's result.

Oh... So it WAS correct! :D Thanks Valentin!

Here's another way to do it:

e^(arccosh(3/2)) = e^(ln(3/2 \pm \sqrt((3/2)^2 - 1))) = 3/2 \pm \sqrt (9/4 - 1) = 3/2 \pm \sqrt(5/4) = (3 \pm \sqrt(5))/2

Take the addition sign, and you get your answer.


Let's say i qualified for USAMO. (although im not so good)
it says in the qualification that i have to be either

1. a citizen of U.S.

or

2. possess a green card.

What happens if I am in the process of green card and INS
has issued an alien number so that i am legally in the United states?

The person to ask about things like that is someone from the AMC.  Go e-mail them.  Steve Dunbar's e-mail address is available here: http://www.unl.edu/amc/whoswho.html


quadratics
arithmetic sequence
algebra
geometric sequence
geometric series

[hide="Solution"]

  Albatross drops apples in the following arithmetic progression $ 1, 2 ,3 ,4 ,5\cdots$  so when they ask you how many times he has repeated this process there are simply asking you to find $n$ in the equation:

$ 1 + 2 + 3 + 4 + 5 + 6 \cdots + n = 5253 $

  Applying Gauss's formula we see that $\frac{n(n+1)}{2}=5253$

  Thus,$ n(n+1)=10506 $

  Now we can solve this equation two ways.

  The first solving as a quadratic equation: $ n^2 + n -10560=0$, but this becomes very messy, and is unesessary for such a simple problem.

  The other solution involves noticing that $ n$ and $ n+1$ are so close to together, we could just as easily take the square root of 10506.

$\sqrt{10506}=102.4987805\cdots$

  So lets check:

$ \frac{102(103)}{2}=  \frac{10506}{2} = 5253$.

  And it works!  Thus, the number of times the process was repeated is $\boxed{102}$ times.[/hide]

good work G-UNIT.  Your posts have been good.

Hi everyone:

[hide]
 I used the formula for an arithmetic progression for this problem.

 [tex]s = \frac {n}{2}[2a + (n - 1)d][/tex]

 The sum [tex]s[/tex] is 5253
 The value of [tex]a[/tex] is 1
 The difference [tex]d[/tex] is 1

 We are trying to find the value of [tex]n[/tex]

 Substituting the known values into the formula we have:

 [tex]5253 = \frac {n}{2}[2(1) + (n - 1)(1)] = \frac {n}{2}(n + 1)[/tex]

 We end up with the quadratic equation:

 [tex]n^2 + n - 10506 = 0[/tex]

 solving for [tex] n [/tex] we have [tex] n = 102 \ or \ -103[/tex]

 since [tex] n [/tex] can't be negative the answer must be [tex] 102 [/tex]

 So I agreed with G-unit up to the quadratic equation, but obtained it a different way.

 Also, I didn't see solving the quadratic as "messy".[/hide]

[quote="10000th User"][quote="rcv"]I'm afraid I must disagree with the previous answers and solutions.[/quote]
I agree with rcv.
[hide]He repeated 5253 times!!![/hide][/quote]

  I guess the solution to this problem depends on how we define "drops 5253 apples".  Whether we interpret this in "one go" meaning that was the last drop he made was the 5253rd, then yes the answer is 5253.  However, if you define "drops 5253 apples" as the sum of all the apples he has already dropped on the floor, then I would disagree and say the answer is 102.  Also note that the word "total" in "dropped a total of 5253 apples"  also points in the direction that we are looking for the sum of all the apples rather than the last one made.  Well, thats just how I see it, mabye theres another angle im not seeing..............

I see the word total, which all the previous too, I guess.
I still think the answer is 102 times.

[quote="A+MATH"][quote="Dr. No"][quote="IntrepidMath"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]

102 times ... if the Albatross could pick 102 apples at once though, why didn't he do that first? [i]Sheesh[/i] ... talk about a bird brain.[/quote]

Its better to actually post a solution than pointlessly spamming like this.[/quote]

I resent the accusation that anything I post is [i]spam[/i].

This problem is easily solved but given that I was the first to post a reply I didn't wish to undermine anyone else's learning by posting a full solution. In other words, the joy of doing this stuff is found in doing it for one's self.

I have a feeling it's 5253/3 or 1751 times. he repeated it 1750 times. If the process mentioned is drop one, then drop two.

With regards to rcv's posts, perhaps the pattern is not clear?  We immediately assumed Albatross would then drop 3, then 4 etc. Many sequences begin with 1, 2, ...

[quote="xenon_nightmare152"][quote="Palytoxin"][quote="rcv"]I still disagree with all previous answers and solutions.[/quote]
hmm maybe [hide="this?"]5253 divided by 3apples per repeat of dropping 1,2,1,2,1,2...?[/hide][/quote]

I agree...[/quote]

  So if you assume the amount of apples dropped ends at 2, then I suppose 1751 is right.  Unless you are counting the individual steps in which case it would be twice the amount mentioned above, 3502.

[quote="IntrepidMath"]Suppose Albatross Picks an apple. he drops it. Suppose he picks 2 apples now. He drops the apples. He repeats this process until he has dropped a total of 5253 apples. How many times has he repeated this?[/quote]
I think 236factorial has found the chief fault in the previous answers.  The question states that he "repeats this process".

I believe the question defines "the process" as "pick an apple; drop it; pick 2 apples; drop them".  (I see nothing in the problem statement to suggest that 1 and 2 are the first two terms of a sequence.  Also, why do we want to assume this is an arithmetic sequence, rather than a geometric sequence?  If we change the problem from "5253 apples" to "4095 apples", do we now assume that a geometric sequence is involved?  Or should common sense lead us to ask if he can hold 2048 apples at once.  Or 102 apples at once?  If Albatross is a human, he can't hold that many apples.  If Albatross is a mythical beast, perhaps he can hold that many apples.)

We're not quite done.  I'm not sure which answer 236factorial is claiming -- 1751 or 1750?  What is the problem asking for?

IntrepidMath:  If you have the source for this problem, this would be a good time to share the souce and the solution with this thread.  If you created this problem on your own, did you intentionally choose the wording so this would be a tricky question?  Or was the wording unintentionally ambiguous?

If he repeated the process 5253 times, then the albatross would have had to pick up only one apple each time. But the problem states that after he picked up one apple and dropped it, he then picked up [tex] 2 [/tex] apples and dropped those.
It was only an assumption on my part that each time he picked up more apples, it was always one more from the time before. But if that wasn't the case, then what did he do? But just due to that one statement I can't say that it was 5253 times.

[quote="rcv"]We're not quite done.  I'm not sure which answer 236factorial is claiming -- 1751 or 1750?  What is the problem asking for?[/quote]

I see that he [i]repeated[/i] it 1750 times.

The way it works is an arithmetic sequence, 1, 2,3, 4,5 ,6, 7,8,9,10..., sorry for any confusion. [hide="the answer"] is 102[/hide]

As I see it, because the problem was rather vague and did not specify on what exactly the process was; to solve this problem one had to use his common sense and make assumption concerning the "proccess" given.  We are given that the first two terms are 1 and 2.  From this I believe we can make 3 common conclusions.  The first, it repeats in a geometric progression, 1, 2 ,4 , 8......., and so on.  The second, it repeats in a arithmetic progression, 1, 2, 3, 4, 5, 6......, and so on.  Lastly, the only two terms in the sequence are 1 and 2, thus' repeating as 1, 2, 1, 2, 1, 2.....,and so on.  In my case, I felt the arithemetic progression was most suitable.  Why?  I will explain.  I immediately ruled out the geometric series in my mind.  Why?  Well, assumptions, and conclusions can only come from you experience or rather background, with what you are dealing with.   In this case it problem solving; and in the bulk of geometric series, concerning 2, I have seen, almost 99% of them give at least 3 or 4 terms in the series before ending with continued line of dots.  For example, in a problem concerning geometric serieses you might see:

1, 2, 4,........               OR        1, 2, 4, 8.............   But in my personal experience I have never seen a problem give only two terms, 1,2....... ,and expect that you can tell they are reffering to an geometric progression.  The other two I found equaly likely to occur.  However, I chose the arithmetic progression over the re-occuring 1,2 progression. Why?  Simply because I have seen more problem, where the key principle is arithmetic progression, and the number of term that are given are generally limited.  For example:

1, 2, 3.............   OR   1, 2, 3 ,4......................  I do understand that only the first two terms are given, and if INTREPIDMATH had only written a few more lines concerning how many apples Albatross was dropping, it would have become obvious what the problem was asking for and we could have avoided this delimma.  Nontheless, in problems we have to work with whats given and in this case I made an assumption based on my personal experience and backround with problems, since my backround is likely to differ from yours, it is most likely that different answers will occur (which they have).  However, let this be a learning lesson to users posting problems, to     [b] specify[/b] things that are crucial in the problem,(in this case it was the repreating pattern) and not force the problem solver to make controversial assumptions, which can turn out wrong.

I don't think we received enough information to know exactly what the "process" is.

[hide]

I started out by writing the equation: $1+2+...n = 5253$ Then, I converted the equation into $\frac{n(n+1)}{2} = 5253$
Multiply by 2 on each side: $n(n+1) = 10506$
So:
$n = \framebox{102}$[/hide]

[quote="math92"][quote="pascal_1623"][hide]

I started out by writing the equation: $1+2+...n = 5253$ [/hide][/quote]
why does it have to equal 5253?[/quote]

That's how many apples he dropped total.


geometry
3D geometry
sphere
tetrahedron
calculus
integration

You can find all of the 2007 PUMaC test documents [url=http://www.princeton.edu/~nsavir/PUMaC/]here[/url].

Tests labeled "Subject A" or "Subject B" are the documents that we gave out at the contest. If I remember correctly, we issued the following errata before the respective tests began:
Geometry A1/B6: In the diagram, A and B are switched.
Geometry A6: Change 80-whatever to 37.
Team 7: CP = PQ should be CQ = PQ
Team 9: r should be positive.

Tests labeled "Subject 3" contain all 15 questions for the subject and have solutions.

Feel free to discuss the problems and solutions here!

Here is what I tried to do for combo #8, though I ended up getting the wrong answer somehow. This time, I got the right answer.
[hide="Combo 8"]Let f(n) be the number of pairs of n-digit numbers with even dot-product of digits (this is what I am calling the operation in the problem). We may verify that f(1)=56 and there are 75 ways for the product of the ith digits to be even, and 25 ways for the product to be odd, i>1. Then $ f(n \plus{} 1) \equal{} 75f(n) \plus{} 25((9*10^{n \minus{} 1})^2 \minus{} f(n)) \equal{} 50f(n) \plus{} 2025*10^{2n \minus{} 2}$. We may see that this recurrence is satisfied by $ f(n) \equal{} 775*50^{n \minus{} 2} \plus{} 4050*100^{n \minus{} 2}$.[/hide]
And here is our awesome solution to #7 on the team round:
[hide="Team 7"]Take Menelaus on triangle ABQ and colinear points R, P, C. We find AQ=BP. Now choose a point R' on ray CR so that BR'=BR. Note BR'=AR, BP=AC, R'PB=RCA. We have a SSA congruence, which is satisfied by triangles RBP and R'BP. One of these is congruent to ARC. Obviously, RBP is not, so R'BP is, and in particular R'P=RC, so R'P=PC=RB. Hence R'RB is equilateral, so R'BR=60 and CRB=120.[/hide]

Yeah WindingFunction will post them some time soon I think.

[quote="diophantient"]In number four on the team round, it says that x can not be larger than 3, but how does that tell us that there is not a nonintegral value of x?[/quote]
We never showed that $ x$ must be an integer. When I was solving the problem, I first evaluated at $ x\equal{}0, 1, 2, 3$ to check what it looked like. I got $ x\equal{}1, 2$ as solutions. Then I showed that solutions are at most 3. Then I realized that in fact there can only be 2, so I was already done.

An alternative sequence of thoughts might be to first check that $ x$ can't be too small or big and show that $ 0 < x < 3.$ Then check how many possible solutions there are (2). Then you might want to get a better idea of where exactly they are, so you can check $ x\equal{}1, 2,$ and you get lucky.

To be honest I don't know how I would approach the problem if there was a non-integer solution.

[quote="yjneb"]Are you going to put up official solutions to the Power Test?[/quote]

Sorry it's been slow...I have to fix a couple of things in the document, but I also have to catch up on the work I neglected last week due to PUMaC preparations, and that's unfortunately taking priority.  Please be patient.  :(

What was the breakdown of points for the problems?

Was it 3,3,3,4,4,4,4,5,5,5

[quote="CatalystOfNostalgia"]Can someone post the Power Test problems for those of us that did not participate?[/quote]

That I can do.

[hide="win solution for geometry 9"]Put stuff in the 4-D plane. $ A\equal{}(\sqrt{2},0,0,0),B\equal{}(0,\sqrt{2},0,0),C\equal{}(0,0,\sqrt{2},0),D\equal{}(0,0,0,\sqrt{2})$.The center of the sphere in the middle is the center of this tetrahedron, $ O\equal{}(\sqrt{2}/4,\sqrt{2}/4,\sqrt{2}/4,\sqrt{2}/4)$. The answer is just $ OA\minus{}1$.[/hide]

For the 2009 pumac, are you still going to have 15 questions for each subject, and give #1-10 to Division B, and # 6-15 to division A?

[quote="mathemonster"][hide="Win Solution for NT #7 better than functionology"]$ 8(x^3 \plus{} x^2y \plus{} xy^2 \plus{} y^3) \equal{} 15(x^2 \plus{} y^2 \plus{} xy \plus{} 1) \implies gcd(15,8) \equal{} 1 \implies (*)x^3 \plus{} x^2y \plus{} xy^2 \plus{} y^3 \equal{} 15 , (**)x^2 \plus{} y^2 \plus{} xy \plus{} 1 \equal{} 8$Multiplying (**) by y and subtracting from (*) we have $ x^3 \equal{} 15 \minus{} 7y, x > 0 \implies y \equal{} 1,2$ both of which yield integrals solutions for x and give our only solutions $ (1,2) (2,1)$[/hide][/quote]

Your solution is false; from $ 8(x^3 \plus{} x^2y \plus{} xy^2 \plus{} y^3) \equal{} 15(x^2 \plus{} y^2 \plus{} xy \plus{} 1)$ we can only conclude that there is some integer k such that $ x^3\plus{}x^2y\plus{}xy^2\plus{}y^3 \equal{} 15k$ and $ x^2\plus{}y^2\plus{}xy\plus{}1 \equal{} 8k.$

yes i realize. You just have to test a few cases though :D

Can we get the problem from the 2009 competition(it was yesterday)?

Be patient. The results aren't even up yet, and I'm sure the organizers are pretty tired from all the difficulties they had yesterday


So I read the FAQ on the AMC site and it said that you could take the AMC 12A and 12B if your school registered for both.  However, I know my school doesn't do both (we only do B) - so could I take the "A" at another school?  How would I could about finding a local school which offers it?

Usually, if schools want to offer only one of the dates, they do the A. For that reason, there is a list of places [url=http://www.unl.edu/amc/b-registration/b1-archive/2009-2010/2010-HS-CU-list.shtml]here[/url] that offer the B test. I am not sure about where you can take the A test though.

You are at an advantage, as PowerOf$ \pi$ said, most schools do A. So do the A at one of those schools, and take B at your regular school


Let $ A,B,C,D \in S_n{(\mathbb{R})},n,m \in \mathbb{N^{*}}$
Prove inequality determinant  of matrix :
  $ \sqrt [n]{\det(A^{2m} \plus{} B^{2m} \plus{} C^{2m} \plus{} D^{2m})}\geq\sqrt [n]{\det(A^{2m} \plus{} B^{2m})} \plus{} \sqrt [n]{\det(C^{2m} \plus{} D^{2m})}$
$ S_n(\mathbb{R})$ are set  $ nxn$ real symmetric matrices.
[hide=Answer][/hide]

It is a kind of Minkowski's determinantal inequality.


geometry
geometric transformation
homothety
parallelogram
IMO Shortlist
geometry proposed

Let $ ABC$ be a triangle with incircle $ (I)$.$ (I)$ touch $ BC;CA;AB$ at $ D;E;F$,respectively.$ DF$ cuts $ BE$ at $ M$;$ DE$ cuts $ CF$ at $ N$
$ K;L$ are midpoints of $ ME;NF$,respectively.
Prove that $ AI;FK;EL$ are concurrent.

It is suffice to show $ \angle FEL\equal{}\angle KFE$ which is imo shortlist problem.

[color=blue][b]Lemma[/b]: Let $ ABC$ be a triangle and $ D$ be a point on $ CA$. $ X,Y$ are lie on $ BD,BC$ such $ AX,AY$ are isogonal line wrt $ \angle A$ and $ E,F$ be on $ BD,BC$ such $ XF\parallel YE$. Then $ AF,AE$ are isogonal line wrt $ \angle A$.[/color]
proof
Let $ P$ be a point of $ AB$ such triangle $ PXF$ and $ AEY$ are homothety with homothety center $ B$. It is obvious that $ P$ is well defined. From $ \angle PAX\equal{}\angle BAX\equal{}\angle YAC\equal{}\angle AYE\equal{}\angle PFX$,  $ A,P,X,F$ are cyclic. So $ \angle BAF\equal{}\angle PAF\equal{}\angle PXF\equal{}\angle AEY\equal{}\angle EAC$. So $ AF,AE$ are isogonal line wrt $ \angle A$.
I remember that this lemma is in crux.

Let return to above problem.
Let $ G$ be intersection of $ BE,CF$(or equivalently, $ G$ is symedian point of triangle $ DEF$),$ P$ be a midpoint of $ FD$ and $ X,Y,$ be a point on $ DF,DE$ such $ GX\parallel DE,GY\parallel DF$. By well known theorem,$ GE$ and $ PE$ are isogonal line wrt $ \angle DEF$. Hence by lemma, we have $ \angle LEF\equal{}\angle DEX$ and similarly, $ \angle KFE\equal{}\angle YFD$. So it is suffice to show $ E,F,X,Y$ are cyclic. Let $ Z$ be a point on $ DF$ such $ E,F,Z,Y$ are cyclic. I will show $ Z\equal{}X$. Since $ GXDY$ is parallelogram, $ GD$ pass midpoint of $ XY$ and $ GD$ pass midpoint of $ YZ$ because triangle $ DYZ$ and $ DFE$ are inverserly similarl and $ GD$ is isogonal line of $ D\minus{}$median. Hence midpoint of $ YZ,YX$ and $ D$ are collinear which imply $ Z\equal{}X$. So we are done.

[quote="kritikos"]The Lemma is correct for $ XF//EY//$[color=darkred]$ AC$[/color] and not only for $ XF//YE$. Besides you take that to your proof.[/quote]
 You are right. Sorry for my mistake. I forgot writing $ \parallel AC$ in lemma.


let $n\geq 2$ an integer and $x_{1},x_{2},...,x_{n},y_{1},y_{2},...,y_{n}$ real numbers such that : $\underset{i=1}{\overset{n}{\sum }}x_{i}^{2}=\underset{i=1}{\overset{n}{\sum }}y_{i}^{2}=1$
Prove that : $\left( x_{1}y_{2}-x_{2}y_{1}\right) ^{2}\leq 2\left| 1-\underset{i=1}{\overset{n}{\sum }}x_{i}y_{i}\right| $

really ? do you have the solution ? :lol:

[quote="siuhochung"][quote="hungkhtn"]Easy,I think

$\sum{x_i^2}\sum{y_i^2}-(\sum{x_iy_i})^2=\sum{x_i-y_j)^2 \ge (x_1y_2-x_2y_1)^2}$

Then  $(1+\sum{x_iy_i})(1-\sum{x_iy_i})  \ge (x_1y_2-x_2y_1)^2$

Because $\sum{x_iy_i} \le 1$ then we have the result![/quote]

hungkhtn, do you mean

$\sum{x_i^2}\sum{y_i^2}-(\sum{x_iy_i})^2=\sum(x_iy_j-x_jy_i)^2 \ge (x_1y_2-x_2y_1)^2$[/quote]

A bit mismath is typing:

$\sum{x_i^2}\sum{y_i^2}-(\sum{x_iy_i})^2=\sum_{i<>j}{(x_iy_j-x_jy_i)}^2 \ge (x_1y_2-x_2y_1)^2$

It's very simple!

Indeed, this one is the solution that I have posted some time ago on the forum and that appears in Old and New Inequalities. It is really simple problem.


trigonometry
inequalities unsolved
inequalities

In triangle $ABC$, prove that
$\frac{(b+c)cosA}{a^2} + \frac{(c+a)cosB}{b^2} + \frac{(a+b)cosC}{c^2} \geq \frac{9}{a+b+c}$

write $cosA = \frac{b^2+c^2-a^2}{2bc}$ and so on,
\[
\begin{array}{l}
 (a + b + c)\sum {\frac{{\left( {b + c} \right)\cos A}}{{a^2 }}}  \\
  = (a + b + c)\left( {\frac{1}{{2abc}}\sum {\frac{{b + c}}{a}} \left( {b^2  + c^2  - a^2 } \right)} \right) \\
  = (a + b + c)\left( {\frac{1}{{2abc}}\sum {\frac{{b + c}}{a}} \left( {b^2  + c^2  - a^2 } \right)} \right) \\
  = (a + b + c)\left( {\frac{1}{{2abc}}\left( {\left( {a + b + c} \right)\left( {\sum {\frac{{b^2  + c^2  - a^2 }}{a}} } \right) - \sum {a^2 } } \right)} \right) \\
  \ge \frac{{a + b + c}}{{2abc}}\left( {\left( {a + b + c} \right)\left( {a + b + c} \right) - \sum {a^2 } } \right) \\
  = \frac{{a + b + c}}{{abc}}\left( {ab + bc + ca} \right) \\
  \ge 9 \\
 \end{array}
\]


[quote="yif man 12 (in the College forum)"]The SAT problems are probably even easier than that. I'd say they're mainly AMC 8 level, with a couple that might be on the AMC 10[b]B[/b].[/quote]

Are the problems in AMC 10B generally easier than those in AMC 10A?

The AMC 10 A and B are supposed to be the same difficulty.  Of course, the exams are different so you would probably find one to be easier than the other depending on what questions appear on each one.  The same goes for the AMC 12 A and B.

Yes, I know, but I was wondering why he mentioned 10[b]B[/b]...


Alright, so it seems that the Open response (Part II) of the USNCO is worth 100 points.
How many points is the MC worth? I am guessing 60, but it could also be 120. And I have no idea about how much the lab worth but some of you say it is worth very little.
Does anyone have a clearer idea on how it all works? There's like a few days left before the exam, so I want to get as much information as possible. Thanks!  :)

I maintain my stated belief that the lab doesn't count for beans, based on the fact that I completely blew it last year and it ended up not mattering.

Yes.  I screwed it up too.  Actually, I've heard that you can get a good chunk of the points for the lab just by having organized work (i.e. a chart) even if your method/results suck.  Sorry though, I have no idea how its all weighted.  I would just say, try your best on all of it:).  Yes, I have a feeling that won't help at all.


calculus
integration
probability and stats

Hello,

I've come across this integral when trying to find an expression for $ \displaystyle\int_t^Tr(u) \, du$ with $ dr$ following the Vasicek model and $ W_s$ being a standard Brownian motion:

$ \sigma\displaystyle\int_t^Te^{-au} \displaystyle\int_t^ue^{as} dW_s \, du
= \dfrac{\sigma}{a}\displaystyle\int_t^T(1 - e^{-a(T-s)}) \, dW_s$

The expression given above is supposed to be correct. I tried to get this answer using standard integration by parts as follows:
\begin{align*}
\sigma\displaystyle\int_t^Te^{-au} \displaystyle\int_t^ue^{as} dW_s \, du
&=&
\left[-\dfrac{\sigma}{a}e^{-au}\displaystyle\int_t^ue^{as}\, dW_s\right]_t^T +
\dfrac{\sigma}{a}\displaystyle\int_t^Te^{-au} \dfrac{d}{du}\displaystyle\int_t^u e^{as} \, dW_s\, du
\\&=&
\dfrac{\sigma}{a}\displaystyle\int_t^Te^{-au} \dfrac{d}{du}\displaystyle\int_t^u e^{as} \, dW_s\, du
-\dfrac{\sigma}{a}\displaystyle\int_t^Te^{-a(T-s)}\, dW_s
\end{align*}

Now clearly I would like $ e^{-au}\dfrac{d}{du}\displaystyle\int_t^u e^{as} \, dW_s\, du = 1dW_s$ but I have no idea how to justify this. Any help?

$ \equal{} \sigma \int_t^T e^{ \minus{} au} \displaystyle \int_t^T 1_{s < u} e^{as} dW_s \, du$ and then integrate first along $ du$ then $ dW_s$

So the interchange of $ dW_s$ and $ du$ can be justified?

Doing this, I get:

$ \equal{}\sigma\int_{t}^{T}e^{\minus{}au}\int_{t}^{T}1_{s < u}e^{as}dW_{s}\, du
\\\equal{} \sigma\int_{t}^{T}e^{as}\int_{s}^{T}1_{s < u}e^{\minus{}au}\, du\, dW_{s}
\\\equal{} \dfrac{\sigma}{a}\int_{t}^{T}e^{as}(e^{\minus{}as} \minus{} e^{\minus{}aT})\, dW_{s}
\\\equal{} \dfrac{\sigma}{a}\int_{t}^{T}(1 \minus{} e^{\minus{}a(T\minus{}s)})\, dW_{s}$

Thanks.

[quote="jshen"]So the interchange of $ dW_s$ and $ du$ can be justified?
[/quote]

This is the "Stochastic Fubini" theorem. I don't remember the proof though, and I think that it could be interesting to try to prove it again. Let's do it  :)

I think that it can be done via Monotone Class Theorem


$ a$ and $ b$ are positive real numbers with $ a\le b\le1$. Show that
\[ 1\minus{}a^6b^6\ \ge\ a^2b^3(1\minus{}a^2)\plus{}a^3b(1\minus{}b^4)
\]
:?:

Let

 $ a \equal{} \frac {1}{1 \plus{} x \plus{} xy}, b \equal{} \frac {1 \plus{} x}{ 1 \plus{} x \plus{} xy} (x,y\geq0)$

then

$ 1 \minus{} a^6b^6 \minus{} a^2b^3(1 \minus{} a^2) \minus{} a^3b(1 \minus{} b^4)$

$ \equal{} \frac {x(4 \plus{} 6y \plus{} 34x \plus{} 72xy \plus{} \ldots \plus{} 16611x^9y^5)}{(1 \plus{} x \plus{} xy)^{12}}\geq0$.

$ 1 \minus{} a^6 b^6 \; \ge \;a^2 b^3 (1 \minus{} a^2 ) \plus{} a^3 b(1 \minus{} b^4 ) \Leftrightarrow \frac{1}{a} \plus{} \frac{1}{{b^2 }} \plus{} a^3 b^3  \le a \plus{} b^2  \plus{} \frac{1}{{a^3 b^3 }}$.
That seems easier.

[quote="Mary Barton"]$ a$ and $ b$ are positive real numbers with $ a\le b\le1$. Show that
\[ 1 \minus{} a^6b^6\ \ge\ a^2b^3(1 \minus{} a^2) \plus{} a^3b(1 \minus{} b^4)
\]
:?:[/quote]
Nice inequality. Here is my solution :)

I think I used to learn Vietnamese at past......but forgot due to long gap!!! :P
So, can anyone translate them for me, PLEASE!

[P/s: Ami ekjon veto bangali, vietnami vasha jani na, doya kore ektu keu onubaad kore deben!!] :P

[quote="Mary Barton"][quote="ZHANGWENZHONGKK"]$ 1 \minus{} a^6 b^6 \; \ge \;a^2 b^3 (1 \minus{} a^2 ) \plus{} a^3 b(1 \minus{} b^4 ) \Leftrightarrow \frac {1}{a} \plus{} \frac {1}{{b^2 }} \plus{} a^3 b^3 \le a \plus{} b^2 \plus{} \frac {1}{{a^3 b^3 }}$.
That seems easier.[/quote]
But that was where I took the problem from!

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=219432

I narrowed it down to the case $ a\le b\le1\le c$, which is equivalent to this problem.

So if this problem is solved, so is the other inequality! [img]http://i41.photobucket.com/albums/e271/Carduelis_carduelis/Tongue.gif[/img]

Thanks fjwxcsl and can_hang2007 for your solutions! :omighty:[/quote]
haha, it's equivalent to my inequality but I dont see it.  :blush:
By the way, the original inequality on that link, there still has another simpler solution for it. :)


Let $ 0 < a_1 < a_2 < ... < a_n$ be integer.A positive integer $ m$ such that: we have find the integer $ 0 < b_1 < b_2 < ... < b_m$ such that:
         $ \sum_{k \equal{} 1}^{n}2^{a_k} \equal{} \sum_{k \equal{} 1}^{m}b_k$ and $ \prod_{k \equal{} 2}^{n}(2^{a_k})! \equal{} \prod_{k \equal{} 1}^{m}b_k!$
Prove that: $ n \equal{} m$ and $ b_i \equal{} 2^{a_i}$  $ \forall i \equal{} 1,2,...,n$

It's quite easy.
From the second indenty, we have:
$ \begin{array}{l}
 \sum\limits_{k \equal{} 1}^n {v_2 (2^{a_k } !) \equal{} \sum\limits_{k \equal{} 1}^m {v_2 (b_k !)} }  \\
  \Leftrightarrow n \equal{} \sum\limits_{k \equal{} 1}^m {s_2 (b_k )}  \\
 \end{array}$.
Thus $ n\equal{}m.$ Because $ 0<b_1<...<b_n,$ so $ b_k  \equal{} 2^{x_k } (x_k\in N; x_1<x_2<...x_n).$
From that, we can check easy that$ x_k\equal{}a_k$( k=1,2,...,n)

[quote="tronghieu"]It's quite easy.
From the second indenty, we have:
$ \begin{array}{l} \sum\limits_{k \equal{} 1}^n {v_2 (2^{a_k } !) \equal{} \sum\limits_{k \equal{} 1}^m {v_2 (b_k !)} } \\
\Leftrightarrow n \equal{} \sum\limits_{k \equal{} 1}^m {s_2 (b_k )} \\
\end{array}$.
Thus $ n \equal{} m.$ Because $ 0 < b_1 < ... < b_n,$ so $ b_k \equal{} 2^{x_k } (x_k\in N; x_1 < x_2 < ...x_n).$
From that, we can check easy that$ x_k \equal{} a_k$( k=1,2,...,n)[/quote]

What are $ v_2$ and $ s_2$?

Thanks,

denote $ v_p(a)$ the exponent of the prime number p in the decomposition of a.
$ s_b(n)$ is the sum of digits of n when written in base b.
this problem is not difficult,and I have seen a simple results in  book by Titu :D


An ancient human tribe had a hierarchical system where there existed one chief with 2 supporting chiefs, each of whom had 2 equal, inferior officers. If the tribe at one point had 10 members, what is the number of different ways to choose the leadership of the tribe?  That is, in how many ways can we choose a chief, 2 supporting chiefs, and two inferior officers reporting to each supporting chief?

There is one chief, two under him, and two more under each supporting chiefs for a total of $ 1\plus{}2\plus{}4\equal{}7$ members in the leadership. If 10 members are in the tribe, there are 10 ways to choose the chief, 9 to choose the first supporting chief, 8 to choose the second and so on. $ 10\times9\times8...\times4\equal{}604,800.$

Your problem is that each pair of leaders are not distinct. Therefore, you must use some combinations also.
10 for chief
9C2 for supporters
7C2 for supporters of supporter1
5C2 for supporter2
Multiply them all for $ 75600$.

You probably don't need the answer anymore, but to save future questions too.

There are two different supporting chiefs who need two inferior officer. It would be $\dbinom{7}{4}$, it it were one supporting chief who needed four inferior officers.

[quote=n1000]Wait, so if the supporting chiefs are not distinct, why are their respective pairs of inferior chiefs distinct?[/quote]

Because two of them report to one chief, two to the other.
The two supporting chiefs are at the same level of power, so they are chosen in the same time. Does this make sense?

[quote=TheMaskedMagician]Can someone explain why we don't have $\binom{7}{4}$ choices for the inferior officers instead of $\binom{7}{2}\binom{5}{2}$?[/quote]

Let's start at the very beginning. There is 1 chief, 2 supporting chiefs, and 4 servants to the supporting chiefs. So you have $10\cdot\dbinom92$ (for the supporting chiefs) $\cdot \dbinom74$ (for the two inferior officers) $\cdot \dbinom42$ (how many ways to assign the officers to a sup. chief. Does this make sense?

Hello,
It seems like the [b]supporting chiefs are distinct[/b] since the problem doesn't state "equal supporting chiefs" (unlike the inferior officers) AND it specifies in parentheses, "supporting chief A and supporting chief B".
Because of this, I think that instead of combinations (7C2) you have to use permutations for the distinct supporting chiefs (A and B). So, my answer was double of 75600.

Yeah I would say that this problem is a little misleading. They do specify that there is a supporting chief A AND B, so I first assumed it they were distinct.

We don't multiply by 9 choose 2 instead of 9 * 8 because the support officers are DISTINCT and not equal, right?

@above that's correct.

[hide=Solution]
There are $10\cdot9\cdot8$ ways to choose the two supporting chiefs, since all three of them are distinct. Then, there are $\binom{7}{2}$ ways to choose two (indistinguishable) officers for chief A, and $\binom{5}{2}$ ways to choose two (indistinguishable) officers for chief B. So, in total, there are
$$10\cdot9\cdot8\cdot\binom{7}{2}\cdot\binom{5}{2} = \boxed{151200}$$
ways to choose the leadership of the tribe.
[/hide]

[hide=Think About It Solution]Label the ten people in the tribe with the numbers $1$ through $10$. We arrange the characters 1ABaabbXXX in a permutation. There are $\frac{10!}{3!2!2!} = 151200$ ways to do this. Then, whichever person is assigned the $1$ is the chief, the people who get A and B are supporting chief A and supporting chief B, the two people with a are inferior officers to supporting chief A, the people with b are inferior officers to supporting chief B, and the people who get X do not get a role. This is a bijection so the number of ways to assume roles of leadership is $151200$.[/hide]

I made a mistake here for the first time. It's not that this is a tough one but there are some confusions. At first I thought there were just five members in the committee because I didn't read the last five words of the question. Then I failed to take into account that supporting chiefs A and B are distinct. I'm so careless.

[quote=ilovepizza2020]This question bullied me :([/quote]

I remember…I do remember…

[hide=First Try]There is 1 chief and 2 of each of the others, so it's $10\binom92\binom72\binom52=\boxed{75600}.$[/hide]
[hide=Second Try]Oh, there's actually 4 inferior officers. So it's $10\binom92\binom74=\boxed{12600}.$[/hide]

:(

Actually, this sounds more like middle school

[hide=First Answer]That's just 10x9x8x7x6x5x4 = 604800[/hide]
[hide=Second Answer]I just realised: the pairs must be distinct! It's going to be 75600[/hide]
 :|

Haha rip. I answered wrong the first time by assuming they were indistinguishable, then wrong a second time by assuming all the chiefs in each rank were distinguishable


limit
function
calculus
derivative
real analysis
real analysis unsolved

Let $ (x_{n})_{n\geq 1}$ ,$ x_{1}\equal{}3$ and $ x_{n}\equal{}\frac{x_{n}^2\plus{}3}{3x_{n}},\forall n\geq 1$
Prove that $ x_{n}$ converges and find $ \lim_{n\rightarrow \infty} x_{n}$

[quote="Svejk"]Let $ (x_{n})_{n\geq 1}$ ,$ x_{1} \equal{} 3$ and $ x_{n} \equal{} \frac {x_{n}^2 \plus{} 3}{3x_{n}},\forall n\geq 1$
Prove that $ x_{n}$ converges and find $ \lim_{n\rightarrow \infty} x_{n}$[/quote]
first: I don't think that it's hard problem. good place will be Computations and tutorials.
Seccond:
you have mmistake in your problem please correct your problem
Third:
After correction and direct calculation answer will be $ \lim\equal{} \sqrt{\frac{3}{2}}$ prove it yourself

[quote="Svejk"] $ x_{n} \equal{} \frac {x_{n}^2 \plus{} 3}{3x_{n}},\forall n\geq 1$

[/quote]

LHS should be $ x_{n\plus{}1}$

We could use the Inverse Newton Approximation (I'm not sure there is a thing called like that, it just something I tend to do with problems like this). So, the NA says:

If:

$ {x_{n + 1} = x_{n} - \frac {f(x_{n})}{f'(x_{n})}}$

then

$ {f(x_{\infty}) = 0}$

Now, we have:

${ {x_{n + 1} = \frac {x^{2}_{n} + 3}{3x_{n}}} = x_{n} - (x_{n} - \frac {x^{2}_{n} + 3}{3x_{n}}) = x_{n} - \frac {2x^{2}_{n} - 3}{3x_{n}}}$

So, we can say that:

$ {\frac {f(x)}{f'(x)} = \frac {2x^{2} - 3}{3x}}$
$ {\frac {f'(x)}{f(x)} = \frac {3x}{2x^{2} - 3}}$
$ {ln(f(x)) = \frac {3}{4}ln(2x^2 - 3) + C}$
$ {f(x) = C(2x^2 - 3)^{\frac {3}{4}}}$

And finaly, for the solution, we get:

$ {f(x_{\infty}) = C(2x_{\infty}^2 - 3)^{\frac {3}{4}} = 0}$

$ {x_{\infty,1} = - \sqrt {\frac {3}{2}}}$

and

$ {x_{\infty,2} = \sqrt {\frac {3}{2}}}$

Since $ {\forall x>0 \rightarrow f(x)>0}$ and $ x_{0}=3$ we take $ x_{\infty}=x_{\infty,2}=\sqrt {\frac {3}{2}}$

milin, that's a clever technique!  I'll keep it in mind.  :)

[quote="milin"]Tnx.. :) The theory around it is something I didn't have time to do (and probably wouldn't know how to do), but for nice, continues functions it work's OK![/quote]
It's called the Newton-Raphson method, and will work for almost any differentiable function in a suitable region around the zero you are approximating to.

Another way of solving this question would be to use the Contraction Mapping theorem on a suitable interval.

Can anyone give some additional information (or post a link) on how the Newton Raphson method is rigorously justified in application to problems with sequences, such as this one. What I'm seeing doesn't have to do with sequences so I'm a little confused about motivation and the proof that it works.

But it is beautiful. Such a handy thing! Where did you learn about it, Milin?

It's a standard method for finding solutions for equations, and it's very hard to use for anything in proofs. You always get convergence on some interval when trying to solve $ f(x)\equal{}0$ for a nice enough function $ f$- but the size of that interval depends on $ f'$ and $ f''$. The Banach contraction principle gives convergence on the largest symmetrical interval around a root where $ \left|\frac{f(x)f''(x)}{(f'(x))^2}\right|\le 1$.

The reason it's a useful method is that the convergence is [i]fast[/i] once you're close enough.

This particular application of the method (finding square roots) is also very old; it goes back to the Babylonians. In this case, the convergence is much more reliable and works from any starting point, since you're above the root after one step, and it's a contraction out to $ \infty$ on that side.

milin can you give more information about your method pls ?

Ok, I don't want to sound stupid, but I didn't found out about it.. I made it up by my self.. So, sorry.. I can't give any clues in where to find some informations about is.

The general Babylonian square root iteration for $ \sqrt{a}$: Start with any positive $ x_0$ (preferably close to $ \sqrt{a}$), and let $ x_{n\plus{}1}\equal{}\frac{x_n\plus{}\frac{a}{x_n}}{2}$. This sequence always converges to $ \sqrt{a}$, and the convergence is rapid once we get close; $ |x_{n\plus{}1}\minus{}\sqrt{a}|\le C|x_n\minus{}\sqrt{a}|^2$ for some constant $ C$.

This is a specialization of Newtons method; this problem behaves better than the general case, since it doesn't blow up for large $ x$.

For $ a\equal{}\sqrt{\frac32}$, this iteration would be $ x_{n\plus{}1}\equal{}\frac{x_n\plus{}\frac{3}{2x_n}}{2}\equal{}\frac{2x_n^2\plus{}3}{4x_n}$. The problem's iteration is an extrapolation; $ \frac43$ of this and $ \minus{}\frac13$ of staying put. That drastically slows convergence; the new error is about $ \minus{}\frac13$ of the old, rather than its square.
It's easy to prove local convergence; that's all about the derivative of the itaration function at the fixed point, which is less than $ 1$ in absolute value. To make it work globally, we need more about the derivative so we can apply the contraction principle on a larger interval.

Let $ f(x)\equal{}\frac{x^2\plus{}3}{3x}\equal{}\frac{x}{3}\plus{}\frac1x$; $ f(x)\ge\frac{2}{\sqrt{3}}$ for all $ x>0$, with equality iff $ x\equal{}\sqrt{3}$. We have $ f'(x)\equal{}\frac13\minus{}\frac1{x^2}$, so $ \minus{}1\le f'(x)\le \frac13$ for all $ x$ in the range of $ f$. That's almost enough; if that $ \minus{}1$ were $ c$ for any $ c<1$, we could apply the Mean Value Theorem and the contraction principle directly.

In this case, the simplest fix is probably to look at $ f(f(x))\equal{}\frac{x}{9}\plus{}\frac1{3x}\plus{}\frac{3x}{x^2\plus{}3}$ instead; its derivative is less than $ \frac19\plus{}\frac34$ for all $ x>1$, which includes all $ x$ in the range of $ x$.


A rapidly growing tree grows 24 inches in one week. On average, how many inches does the tree grow in one hour? Express your answer as a common fraction.

There are 7 days in a week and 24 hours in a day, so the answer is $ 24/(24*7)\equal{}\boxed{1/7}$


geometry
integration
calculus
calculus computations

Use Green's Theorem to find the area of the region between the graphs of $ y \equal{} x^2$ and $ y \equal{} x^3$.

So $ A \equal{} \oint_{C} x \ dy \equal{} \minus{} \oint_{C} y \ dx \equal{} \frac{1}{2} \oint_{C} x \ dy \minus{} y \ dx$

I used $ A \equal{} \minus{} \oint_{C} y \ dx$

So $ A \equal{} \minus{} \int_{1}^{0} x^2 \minus{} x^3 \ dx \equal{} \frac{1}{12}$

Is this the correct way of doing the problem?

How about this way:
$ A\equal{}\frac{1}{2}\oint_{\partial D}\minus{}y\mbox{ d}x\plus{}x\mbox{ d}y\equal{}\frac{1}{2}\iint_{D}2\mbox{ d}x\mbox{ d}y$
So we need to find the intersection points, which are $ (0,0)$ and $ (1,1)$.  So then we have
$ \frac{1}{2}\int_{0}^{1}\int_{x^3}^{x^{2}}2\mbox{ d}x\mbox{ d}y\equal{}\int_{0}^{1}y^{2}\minus{}y^{3}\mbox{ d}y\equal{}\frac{1}{12}$
Two different methods, same result.

[b]ragingbullrocky[/b]: your solution is correct, although on a test I'd like to see a little explanation of the limits and signs.

[b]JRav[/b]: you did not really use Green's theorem - you just calculated the double integral that represents the area.

My solution is an application of Green's Theorem (at least according to my textbook, in which this formula is listed in the section devoted to Green's Theorem).

This is actually an interesting point. If an exam problem asks to compute a double integral (such as $ \iint_D 1\,dA$) using Green's theorem, should a solution with $ \iint_D 1\,dA\equal{}\minus{}\oint_C y\,dx\equal{}\iint_D 1\,dA$ count? After all, Green's theorem was used - even twice!

Probably not, because its probably means using only Green's Theorem once and using a line integral.

I guess its at the discretion of the professor then.


[quote="Vasc"]Let $a,b,c$ be distinct real numbers such that
$a^4+b^4+c^4+ab^3+bc^3+ca^3= 2(a^3b+b^3c+c^3a)$ .
Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$.[/quote]

The equation $a^4+b^4+c^4+ab^3+bc^3+ca^3=2(a^3b+b^3c+c^3a)$ rewrites as $(a^4+b^4+c^4)+(b^3a+c^3b+a^3c)-2(a^3b+b^3c+c^3a)=0$. Because of the identity

$(a^4+b^4+c^4)+(b^3a+c^3b+a^3c)-2(a^3b+b^3c+c^3a)$
$=\frac12((a^2-b^2+bc-ab)^2+(b^2-c^2+ca-bc)^2+(c^2-a^2+ab-ca)^2)$,

this equation is equivalent to $(a^2-b^2+bc-ab)^2+(b^2-c^2+ca-bc)^2+(c^2-a^2+ab-ca)^2)=0$. But since the numbers a, b, c are real and squares of reals are always $\geq 0$, this yields $a^2-b^2+bc-ab=0$, $b^2-c^2+ca-bc=0$ and $c^2-a^2+ab-ca=0$. Now, $a^2-b^2+bc-ab=0$ rewrites as $a^2-ab=b^2-bc$, or, equivalently, a (a - b) = b (b - c), and similarly we get b (b - c) = c (c - a). Thus, a (a - b) = b (b - c) = c (c - a). If we denote a (a - b) = b (b - c) = c (c - a) = k, we thus have $a-b=\frac{k}{a}$, $b-c=\frac{k}{b}$, $c-a=\frac{k}{c}$, so the trivial equality (a - b) + (b - c) + (c - a) = 0 becomes $\frac{k}{a}+\frac{k}{b}+\frac{k}{c}=0$, so that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$, and we are done.

Actually, the above argument is not 100% complete, since difficulties might occur in the case when k = 0. However, this case is easy to filter out: Since a (a - b) = b (b - c) = c (c - a) = k, when k = 0, we must have a (a - b) = b (b - c) = c (c - a) = 0. Thus, either the numbers a - b, b - c and c - a are all 0, what yields a = b = c and thus contradicts the condition that the numbers a, b, c are distinct, or (at least) one of the numbers a, b, c is 0; in let's WLOG assume that a = 0. In this case, the equation a (a - b) = b (b - c) = c (c - a) = 0 becomes 0 (0 - b) = b (b - c) = c (c - 0) = 0; hence, $c^2=c(c-0)=0$, so that c = 0, and thus b (b - c) = 0 becomes $b^2=0$, so that b = 0, what leads to a = b = c = 0, what is impossible again since a, b, c must be distinct. Thus, the proof is completed.

[quote="siuhochung"]what if a=b=-2 and $c=\sqrt{5}+1$?[/quote]

This triple doesn't satisfy $a^4+b^4+c^4+ab^3+bc^3+ca^3=2(a^3b+b^3c+c^3a)$, does it?

  Darij

[quote="Vasc"]My solution is just same.
Congratulations Darij!  :)[/quote]
But if i cannot find mr. stefan's identity, what can i do? :?


Hello everyone.

For years, geometry has been most challenging me of all the subjects in olympiad-style questions. What good books are out there that go over Olympiad-level geometry concepts and have good explanations and problems?

Thank you so much for your help.

JB

Theres a good book on how to...

nevermind.

I can't even try to make this look like a real post. See you tomorrow.

Seahawks v. Steelers!
SUPER BOWL EXTRA LARGE! (XL)

For online stuff, try this:

http://math.mit.edu/~kedlaya/geometryunbound/

For TONS AND TONS of problems, you can dig through the files here:

http://www.math.su.se/~mleites/Prasolov/

A compiled version of the plane geo:

http://students.imsa.edu/~tliu/Math/planegeo.pdf

You've probably heard of books like Geometry Revisited, etc -- those are pretty good too.  The AoPS class is also pretty amazing.  Hmmm.. yeah hope that helps.  :)


ratio
geometry
similar triangles
power of a point

circle C has a radius of 4 and circle A has a radius of 6. if the length of segment AC is 15, and segment BE is tangent to both circles, what is the length of segment BE?

can i check my answer?

[hide="answer?"]$5\sqrt{5}$[/hide]

could someone post a correct diagram? im not sure im drawing this out correctly...

AC is a line segment connecting the centers of circles A and C correct? there is more than one way to have a line tangent to both circles it seems, are we looking for the shortest?

There are both internal and external tangents  :( , but we can find both :)

What iamagenius gives as an answer is for the internal tangent:

[hide="internal"]Sketch the diagram and create two right triangles that are similar by Vertical Angle Theorem. Two triangles have side ratio of $4/6=2/3$. Hypotenuse has length $15 \cdot \frac{2}{5}=6$ and $15 \cdot \frac35=9$ each. So the other legs have length $\sqrt{36-16}$ and $\sqrt{81-36}$. Add these together to get $2\sqrt{5}+3 \sqrt{5}=\boxed{5 \sqrt{5}}$.[/hide]
but there is also external one
[hide="external"]Again, sketch the diagram and draw radii segments perpendicular to the external tangent. Draw altitude from $C$ to the radius drawn from $A$ perpendicular to the tangent. If we let that point $H$, we can find the length of $CH$ from $\triangle ACH$, which is equivalent to $\sqrt{15^{2}-2^{2}}=\sqrt{221}$. $221=13 \cdot 17$ has no square factors , so the answer is just $\boxed{\sqrt{221}}$. [/hide]

[quote="Glen Mann"]okay I did this out and got the same answer as the original poster. I used the fact that there were similar triangles and then I used power of a point to find the segment tangent length..

kohjhsd your solution seems much simpler though, and Id like to know your reasoning behind how you found the hypotenuse length so easily, where did $\frac{3}{5}$ and $\frac{2}{5}$ come from? and then you find the other lengths so simple.. is it a case of not showing your work or is there another theorem you used that I am unaware of?

thanks :D[/quote]

Well, $\frac35$ and $\frac25$ is from the similarity. Because the triangles have ratio of sides 2 to 3, their hypotenuses will also be in ratio 2/3. Because the sum is 15, one is 2/5 and the other is 3/5. There's no need to apply these fancy power of a point methods, I think.

The hypotenuse lengths are just from Pythagorean Theorems. Nothing unusual.

However, this is the faster way for the internal one:

[hide]There are 2 similar triangles. Extend the radius 4 to have radius 10, and move the 15 segment down and you have a right triangle whereas you can find the hypotenuse with simple $5 \sqrt{3^{2}-2^{2}}$.

See picture below.
[/hide]

oh right, right angle triangles are formed.. DOH!

:blush:

your second method is much quicker

great! sorry, i was referring to the external tangent. the problem has a diagram with it, but i couldn't provide it :(


find all the function $ f: R\minus{}\minus{}\minus{}>R$ such that:
$ f(f(x)\plus{}y)\equal{}f(x^2\minus{}y)\plus{}4f(x)y$

Setting $ y\equal{}\frac{x^2\minus{}f(x)}{2}$ yields (*) $ 0\equal{}f(x)[x^2\minus{}f(x)]$. Assume for a moment the existence of some $ z,w\not\equal{}0$ with $ f(z)\equal{}0,f(w)\not\equal{}0$. (*) and the original equation imply $ 0\not\equal{}w^2\equal{}f(w)\equal{}f(f(z)\plus{}w)\equal{}f(z^2\minus{}w)\Longrightarrow (z^2\minus{}w)^2\equal{}w^2\Longrightarrow 2w\equal{}z^2$.

Hence, either the set $ \Omega\equal{}\{x\not\equal{}0|f(x)\equal{}0\}$ or the set $ \Omega'\equal{}\{x\not\equal{}0|f(x)\not\equal{}0\}\equal{}\{x\not\equal{}0|f(x)\equal{}x^2\}$ is empty. This corresponds to the solutions $ f(x)\equal{}x^2$ and $ f(x)\equal{}0$.

[quote="hjbrasch"]Setting $ y \equal{} \frac {x^2 \minus{} f(x)}{2}$ yields (*) $ 0 \equal{} f(x)[x^2 \minus{} f(x)]$. Assume for a moment the existence of some $ z,w\not \equal{} 0$ with $ f(z) \equal{} 0,f(w)\not \equal{} 0$. (*) and the original equation imply $ 0\not \equal{} w^2 \equal{} f(w) \equal{} f(f(z) \plus{} w) \equal{} f(z^2 \minus{} w)\Longrightarrow (z^2 \minus{} w)^2 \equal{} w^2\Longrightarrow 2w \equal{} z^2$.

Hence, either the set $ \Omega \equal{} \{x\not \equal{} 0|f(x) \equal{} 0\}$ or the set $ \Omega' \equal{} \{x\not \equal{} 0|f(x)\not \equal{} 0\} \equal{} \{x\not \equal{} 0|f(x) \equal{} x^2\}$ is empty. This corresponds to the solutions $ f(x) \equal{} x^2$ and $ f(x) \equal{} 0$.[/quote]
A nice solution gjbrasch.But i want to ask you a question:
$ 2w\equal{}z^2$ it's used for ???

It is used to show that exactly one of the sets $ \Omega$ or $ \Omega'$ is empty, since $ \forall z\in\Omega\,,\,w\in\Omega': 2w\equal{}z^2$ holds (and note, that $ \Omega\cup\Omega'\equal{}\mathbb{R}^*$)


Consider all functions $f$ from the positive integers to the positive integers such that

i) For each positive integer $m$, there is a unique positive integer $n$ such that $f(n)=m$;
ii) For each positive integer $n$, we have
$f(n+1)$ is [b]either[/b] $4f(n)-1$ [b]or[/b] $f(n)-1$.

Find the set of positive integers $p$ such that $f(1999)=p$ for some function $f$ with properties i) and ii).

Should be pretty easy to calculate... there are only two possible functions.

I dont think so. It can be different for each $n$ surely?

I don't have any experience about functional equations, but I'll try making some considerations...

 [hide] From i) we can deduce that $f$ is injective.
From ii) we can say that either the function is increasing or it's decreasing. In fact, since the function is defined on the positive integers, we see that $f(n+1)=4f(n)-1$ is monotonically increasing, while $f(n+1)=f(n)-1$ is monotonically decreasing. Obviously, $f$ can't be increasing and decreasing at the same time, so we have to distinguish between two cases...
I) $f$ is increasing
II) $f$ is decreasing
[/hide]

After this, I don't know how to proceed...  :blush:

EDIT: Ok, after what tim wrote next I must cancel my (confused!) ideas...  :wink:

To clarify, for each different $n$ we can choose one of the two possibilities independently, as seamusoboyle suggested.

Thus for any particular function, we might have

$f(3)=4f(2)-1$ but $f(7)=f(6)-1$


function
real analysis
real analysis unsolved

I believe that this is more a real analysis problem than complex and that's why i post it here..
I need some help on proving that the complex function f(z)=|z|/(i+Imz) is continuous.
I multiplied the numerator and denom. with the conjugate, but i should know that Imz is continuous..Can i be sure about that or is there any other way?
Thank you very much.


integration
conics
function
complex analysis
complex analysis unsolved

let $ \gamma$ be a piecewise smooth curve in disk $ |z| < 1$ then $ \int_{\gamma} \frac {|dz|}{1 \minus{} |z|^2}$ is called [i]conic lenght(?)[/i] of $ \gamma$ prove that an analytic function from a disk to it self doesn't increase conic length of curves.

I have not heard the term conic length, I had always heard hyperbolic length.  (Hopefully they are the same  :) ).

Ahlfors' book "Complex Analysis" (3rd edition) takes you through the proof on pages 135 and 136.  He then concludes by giving a first exercise which is precisely what your question is asking for.  It results from using formula (36) on page 136 to compute expression.


topology
geometry
3D geometry
sphere
function
calculus
integration

I am trying to write up the solutions to the problems below, but I often find that I have problems explicating clearly to the grader, either analytically or verbally my reasoning, so I was hoping that any comments would serve to refine my approach at reasoning.

1. Consider a sequence $ (x_{n})$ in the metric space $ \mathbb{R}$. If $ (x_{n})$ converges in $ \mathbb{R}$ prove that the sequence of absolute values $ (\mid x_{n}\mid)$ converges in $ \mathbb{R}$.

[hide]
my attempt:
I can argue this pretty well for positive sequences that converge to a positive limit or a sequence of negative numbers that converges to a negative limit. But for the case where the limit $ x < 0$, how would I show analytically that $ d(x_{n}, x) < d(\mid x_{n} \mid , \mid x \mid) < \epsilon$? Should I just write that and then give an example of how this would be true?
[/hide]

2. A map $ f: M \rightarrow N$ is open if for each open set $ U \subset M$, the image set $ f(U)$ is open in $ N$. If $ f$ is open, is it continuous? If $ f$ is an opn , continuous bijection, is it a homeomorphism?

[hide]
my attempt:
What would be solid counterexamples for both these questions?
[/hide]

3.Prove that the 2-sphere is not homeomorphic to the plane.

[hide]
my attempt:
I can see that for some radius r, I claim that the 2-sphere is not homeomorphic because then our 2-sphere is bounded and the plane is not. However, couldn't someone say if my r is infinity then my 2-sphere is not bounded anymore? How can I counter this last case?
[/hide]

4. Consider the identity map $ id: C_{max} \rightarrow C_{int}$ where $ C_{max}$ is the metric space $ C([a,b], \mathbb{R})$ of continuous real valued functions defined on [a,b], equipped with the max metric $ d_{max}(f,g) \equal{} max \mid f(x) \minus{} g(x) \mid$, and $ C_{int}$ is $ C([a,b],\mathbb{R})$ equipped with the integral metric,

$ d_{int}(f,g) \equal{} \int_{a}^{b} \mid f(x) \minus{} g(x) \mid dx$.

Show that id is a continuous linear bijection.

[hide]
my attempt:
The linear bijection is easy to prove, but I am having trouble explaining clearly the continuity part. My rough idea was to show that $ \forall \epsilon > 0$ and $ \forall f,g \in C_{max} \exists \delta > 0$ such that $ d_{max}(f,g) \equal{} max \mid f(x) \minus{} g(x) \mid < \delta$. I started with showing:

$ d_{int}(id(f),id(g)) \equal{} \int_{a}^{b} id(f) \minus{} id(g) dx \equal{} \int_{a}^{b} f \minus{} g dx \equal{} [b \minus{} a] (f_{1} \minus{} g_{1}) \cdots (f_{i} \minus{} g_{i})$ somehow $ < d_{max}(f,g) < \delta \equal{} \epsilon \cdot [b \minus{} a]$

I 'm not sure if we can use the Riemannian definition of integrals yet, but this was the only way I could think of showing it.
[/hide]

1. Just use the reverse triangle inequality $ d(|x_n|, |x|) \leq d(x_n, x)$. You should be able to fill in the details from there.

2. Make up a simple map $ f$ that maps at least one closed set to an open set to show that openness of a map does not imply continuity.

3. The sphere is compact while the plane is not. However, I do not know if you have studied compactness in your course.

4. Use the fact that for a continuous function $ h$ on a compact interval $ [a,b]$, $ \int_a^b h(x)\mbox{ }dx \leq (b \minus{} a) \mbox{max }_{t \in [a,b]} h(t)$ and the result will follow almost immediately.

5. Use the hide tags:
[code][hide][/hide][/code]

Post your solutions after considering my hints and I can tell you how they are.

1. I was able to figure out, I used triangle inequality as well.

2. Can I just make up some [a,b] - in $ M$ and call that a closed set within the open set $ M$? Or can I use $ \mathbb{Q}$ and say that it's closed in $ \mathbb{R}$- which is open and if $ f$ maps from $ \mathbb{R} \rightarrow \mathbb{R}$, then a mapping from the rationals to the reals would show that it's not continuous? Can I use this same example to disprove that $ f$ being an open, continuous bijection is not a homeomorphism?

3. Yes, we studied compactness. But how do I show that it's closed and bounded which I think would really help in showing that it's compact. From above, I have a hard time trying to argue the 2-sphere is closed especially if we have a 2 sphere that was formed from a sphere of radius infinity, because then I feel that I can run into the problem, with a 2-sphere of that "size" , by being able to map every point on my 2-sphere onto $ \mathbb{R}^{2}$

3. For simplicity, consider the unit sphere. It is closed because it is the inverse image of the closed set $ \{1\}$ under the norm induced by the metric, which is a continuous function. It is bounded because it can be contained within the ball of radius $ 2$.

The sphere is the inverse image of a one-point set under the continuous function $ \parallel{}x\parallel{}$. Therefore the sphere is closed. $ S_{n}$ is also bounded. Therefore it is compact.

Edit: sorry Zhylliolom, I was writing a reply at the same time you submited your post and it was exactly the same....  :P

I see what you're saying with the unit ball, but in general would I have to argue ad infinitum that there will always be a 2-sphere formed from radius r that will always be bounded by a 2-sphere formed from r-1?

Edit: ok, I will take some time to look into the inverse image.

[quote]I can see that for some radius r, I claim that the 2-sphere is not homeomorphic because then our 2-sphere is bounded and the plane is not.[/quote]

This is wrong, I mean, if $ A$ is bounded and $ B$ is not, this is not enough to claim that are not homeomorphic, in fact, consider for example $ \mathbb{I} \equal{} (0,1)$ and $ \mathbb{R}$, you can see that $ \mathbb{I}$ is bounded and is! homeomorphic to $ \mathbb{R}$, can you find the homeomorphism?

But, technically isn't $ I$ an open interval in $ \mathbb{R}$ ? I thought that means it is open. However if $ I$ were in the $ \mathbb{Z}$ then it is closed. Perhaps I'm wrong, any thoughts or correction on this would be helpful.

4. Do I take $ \delta \equal{} \frac {\epsilon}{[b \minus{} a]}$? And what do I use to justify this inequality (name of the rule/thm) $ \int_{a}^{b}h(x)\mbox{ }dx\leq (b\minus{}a)\mbox{max }_{t\in [a,b]}h(t)$?

yes, $ \mathbb{I}$ is open in $ \mathbb{R}$, why this question?
What I wanted to mean is that this is not the way of proving that they're not homeomorphic, I think the properly way is by compactness.

I guess I associated openness with boundedness, which I now see is not necessarily true.

In fact, in a metric space with norm $ \parallel{}x\parallel{}$ to be bounded means that you can put it into a sphere (as a posible definition). but for example in $ \mathbb{R}$, the interval $ [0,1]$ is a subset of the sphere of center $ 0$ and radius $ 10$, so it's bounded, do you get the point? ...

Yes, I think I just need some more time to mull over that. I think it's starting to make sense as I pore over the definition of boundedness.

Are there any thoughts on what I posted above for 2. and 4.?


[hide="One line solution"]
By Holder's inequality,
$ (ab\plus{}a^2\plus{}b^2)(b^2\plus{}c^2\plus{}bc)(a^2\plus{}ac\plus{}c^2) \ge (ab\plus{}bc\plus{}ca)^3$[/hide]

@akashram if you wanted the proof for this inequality then you can find it at this link: http://www.artofproblemsolving.com/Wiki/index.php/Holder%27s_Inequality

Ive never used Holders. Please provide details on the transformation a what you set the variables to be  :blush:

To avoid using confusing notation,
Holder's says that if you have n parentheses of sums you can take the nth root of the product of all the first terms in each parentheses, plus the nth root of the product of the second terms, etc, add it up and take it to the nth power and it will be less than the original.

I see... Now I just have to put it into good use.


Hey here are some problems that i am enjoying from a book our teacher gave us to do out of class like in spare time. see if you find them difficult:

[b]1. [/b]Jack, John and Jill go up the hill with five empty pails to a place where three springs are. Each of the five pails has a $8$ litre capacity. One of the springs gives $2$ litres of water in one minute while each of the other two give $1$ litre of water in one minute. It is possible to use one spring to fill two pails at the same time, and it takes less than $2$ minutes but more than $1$ minute to take a pail from one spring to another one. What is the shortest time Jack, John, and Jill need to fill all five pails to their capacity?

[b]2. [/b]When Adam saw the diagram below, he wrote a number in each of the empty circles so that the sum of all the numbers in the circles equals 25. Then he noticed that whenever three circles are such that every two of them are connected with a single line, the numbers in these three circles add up to 12. What number did he put in the central circle? Find all possible answers. [b](refer to the picture i have attatched).[/b]

[b]3. [/b]When a positive integer $N$ is written in base $9$, it is a two-digit number. When $6N$ is written in base $7$, it is a three-digit number obtained from the two-digit number by writing digit $4$ to its right. Find the decimal representations of all such numbers $N$.

[b]4.[/b] Find all pairs of positive integers $a$ and $b$ such that
                       $a^2 + b^2 - 7 = ab$

[b]5.[/b] A bus route from town $A$ to town $D$ goes along a straight line and passes first through town $B$ and then through town $C$. Ann has a positioning device that can tell her the distance from her current location to any of the four towns on the route. When she is somewhere between $A$ and $B$, from the information supplied by the device Ann learns that the sum of the distances to $A$, to $B$ and to $D$ is greater than the distance to $C$ by $5.9$ km. When Ann is between $B$ and $C$, she finds out that the sum of the distances to all four towns in $16.7$ km. When she is between $C$ and $D$, Ann discovers that the sum of the distances to $B$ and to $D$ is $8.1$ km. What is the distance between $C$ and $D$?

These are meant to be cryptic and challenging, lets see how you all go  :)

Enjoy,
Kesh[img]My%20Documents\question%202%20pic[/img]

Yor diagram hasn't come. Can you please post it!!!

[quote="keshin"]
[b]4.[/b] Find all pairs of positive integers $a$ and $b$ such that
                       $a^2 + b^2 - 7 = ab$
[/quote]

I 'll try this problem... :)
rewrite the given equation as $(a-b)^2+a^2+b^2=14$
so $a^2=1$ or $4$ or $9$ and then you find the pairs easily.. :)

And how did you conclude the last step????

[quote="socrates"]because  we must have $a^2\leq14$ and $a \in N$..... :)
ok?[/quote]

$\displaystyle (a-b)^2+a^2+b^2=14$
add that all out :P


Determine all pairs of positive integers $ a$ and $ n$ for which each prime divisor of $ a^{\phi (n)} \minus{} 1$ divides $ n$

[quote="jbmorgan"]Determine all pairs of positive integers $ a$ and $ n$ for which each prime divisor of $ a^{\phi (n)} - 1$ divides $ n$[/quote]
Hmm may be I have some mistakes ...
Only need to consider $ a > 1$
The first we have an old result : Let $ p$ is an odd  prime divisor of $ a^{2^k} + 1$ then $ 2^{k + 1}|p - 1$
It guesses us to consider prime divisors of number ${ a^{\frac {\varphi(n)}}{2}} + 1$ .
Step 1 : $ n$ has at most one odd prime divisor . If not we consider a odd prime divisor q  of $ a^{\frac {\varphi(n)}{2}} + 1$ (Notice that if $ a^2 + 1$ can't be a power of 2 for $ a > 1$  .From the lemma we have : ${ v_2(q - 1)\geq v_2(\varphi(n))\geq \sum_{p|n} v_2(p - 1}$ it gives contradiction .
Step 2 : Similarity to step 1 ,it is easy to show that $ n$ must be one of following forms $ {p,2p}$
Now consider two case
i) $ n = p$ then we must have :
\[ a^{p - 1} - 1 = p^k
\]
It is not hard to shows that $ p$ must be equal 2 or 3 .
For the case $ p = 2$ we have $ a = 2^{t} + 1$
For the case $ p = 3$ then we have $ a = 2$
ii ) $ n = 2p$ then we have :
\[ a^{p - 1} - 1 = 2^k.p^t
\]
\[ \equiv alent (a^{\frac {p - 1}{2} + 1)(a^{\frac {p - 1}{2}} - 1) = 2^k.p^t
}\]
This equation can be solved by some calculations ...


topology
function
advanced fields
advanced fields unsolved

Let g:X->Y be a continuous mapping of a separable metric space X onto a compact Hausdorff space Y.  Show that Y is metrizable.    Also what is an example of a continuous mapping of a separable metric space onto a non-metrizable normal space?

Al

First the counterexample:

Any countable topological space is a continuous image of a separable metric space, namely a discrete countable space. This means that in order to find a counterexample, it suffices to find a countable $T_{4}$ (normal and Hausdorff) space $X$ which is not metrizable.

Let $X$ be the set of non-negative integers. All points are open except for $0$, and the neighborhoods of $0$ are $\{0\}\cup A$, as $A$ ranges through the elements of a [url=http://en.wikipedia.org/wiki/Ultrafilter#Types_and_existence_of_ultrafilters]free ultrafilter[/url] on $X\setminus\{0\}$. $X$ is clearly Hausdorff, since all points different from $0$ are open. It's also normal, because among any two disjoint closed sets, one will also be open, being contained in $X\setminus\{0\}$. Finally, $X$ is not metrizable because it's not first-countable at $0$, as a simple argument shows.

Now the first part (the proof of the fact that compact Hausdorff continuous images of separable metric spaces are metrizable):

The conclusion follows from these two results:

[i]Proposition 1[/i]

A compact Hausdorff space $X$ is metrizable if the diagonal $\Delta=\{(x,x)\ |\ x\in X\}$ is a $G_\delta$ subset of $X\times X$.

(the converse is also true, but that's obvious)

Even though this is not difficult, I'll just outline the proof (just don't feel like typing that much :)). As is well-known, a compact Hausdorff space is metrizable iff it has a countable basis of open sets. If $\Delta=\bigcap_{n\ge 1}G_{n}$ where $G_{n}$ are open in $X\times X$, take, for each $n$ and each $x\in X$, and open neighborhood $U_{n,x}$ of $x$ such that $x\times U_{n,x}\subset G_{n}$. For each $n$ the sets $U_{n,x},\ x\in X$ cover $x$, so we can extract a finite subcovering. The union of all these finite coverings as $n$ ranges over the positive integers is a countable family of open sets, and the idea is to prove that it's actually a basis.

[i]Proposition 2[/i]

If $f: T\to Y$ is a continuous function of the separable metric space $T$ onto the T_4 topological space $Y$, then every closed subset of $Y$ is a $G_\delta$.

[i]Proof[/i]

Let $F$ be closed in $Y$, and let $(F_\alpha)$ be a family of closed neighborhoods of $F$ in $Y$ with intersection $Y$ (such a family does exist, because $Y$ is T_4; less than the T_4 condition may be enough, but I'm not interested in that :)). Now $f^{-1}(F)$ is a closed subset of $T$, and $(f^{-1}(F_\alpha))$ is a family of closed neighborhoods of $f^{-1}(F)$ with intersection $f^{-1}(F)$. The open complements $D_\alpha=T\setminus f^{-1}(F_\alpha)$ cover the separable and hence Lindelof metric space $T\setminus f^{-1}(F)$, so countably many cover it. If these are, say, $D_{n},\ n\in\mathbb N$, then $F$ is the intersection of the closed neighborhoods $f(T\setminus D_{n})$ (these are some of the $F_\alpha$ we considered initially) and hence of the interiors of $f(T\setminus D_{n})$. This makes $F$ a $G_\delta$, as desired.

In order to complete the proof, consider the continuous map $g: S\to X$ of the separable metric space $S$ onto the compact Hausdorff space $X$. It induces, in an obvious way, a continuous onto map $f: S\times S\to X\times X$. Apply Proposition 2 to $T=S\times S$ (which is clearly separable metrizable) and $Y=X\times X$ to conclude that every closed subset of $X\times X$ is a $G_\delta$. Since $X$ is Hausdorff, $\Delta$ is closed in $X\times X$ and hence a $G_\delta$. Now an application of Proposition 1 finishes the whole thing.


[quote="Farazy"]What are the odds of you flipping two heads and a tails with a coin?[/quote]

[hide="the answer is"] 1/8?[/hide]

[quote="Farazy"]Crystal has 3 nickels, 2 dimes, and 1 quarter. How many different sums of money can Crystal obtain by using exactly four of these coins?

Bob has 3 pennies 2 nickels and 1 quarter.  How many different sums of money can he make using these coins?

What are the odds of you flipping two heads and a tails with a coin?[/quote]

In exactly 3 flips, right?

Yeah, three filps.  I know the answer...

[hide="Here it is"]3/8[/hide]

BUT

There should be a really quick method of solving this.


Let $A,B$ be two matrices with eigenvalues being in ${\mathbb R}_+$ and $det(AB)=1$. Show that $Tr(A) + Tr(B) \geq 2n$.
 :ninja:

[quote="Moubinool"]$a_1...a_nb_1...b_n=1$
[/quote]

I can't understand that ?

He just used that $det(AB)=detA detB$, alekk.


trigonometry
circumcircle
geometry
trig identities
Law of Sines

If one side of a triangle is $ 12$ inches and the opposite angle is $ 30$ degrees, then the diameter of the circumscribed circle is:
$ \textbf{(A)}\ 18\text{ inches} \qquad\textbf{(B)}\ 30\text{ inches} \qquad\textbf{(C)}\ 24\text{ inches} \qquad\textbf{(D)}\ 20\text{ inches}\\
\textbf{(E)}\ \text{none of these}$

[hide]By the extended Law of Sines, we have $ \frac{a}{\sin A}\equal{}2R$, so, $ \frac{12}{\frac{1}{2}}\equal{}24\equal{}2R$. So, the diameter is $ 24$, or $ \textbf{(C)}$[/hide]

[hide="Without trig"]
Draw a diagram and label the triangle $ ABC$, where $ B$ is the vertex of the $ 30 ^\circ$ angle and $ AC\equal{}12$.  We know that arc $ AC\equal{}60^\circ$, so $ AC\equal{}r \implies 2r\equal{}d\equal{}24$, or $ \boxed{\textbf{(C)}}$.
[/hide]


$\frac{1}{n(n+1)(n+2)(n+3)}={1\over 6n}-{1\over 2(n+1)}+{1\over 2(n+2)}-{1\over 6(n+3)}$ using [url=http://en.wikipedia.org/wiki/Partial_fraction]partial fraction decomposition[/url].
The rest is easy.
The answer is ${1\over 18}-{1\over 3\cdot 2005\cdot 2006\cdot 2007}$.

[quote="me@home"]olorin's is textbook method, while n.t.tuan's is simpler solution using "insight method"... nice solution n.t.tuan (it generalizes, right?)[/quote]
$\frac{1}{n(n+1)(n+2)...(n+k)}=\frac{1}{k}(\frac{1}{n(n+1)...(n+k-1)}-\frac{1}{(n+1)(n+2)...(n+k)})$ :D

How about $\frac{1}{[n(n+1)(n+2)...(n+k)}]^{2})$

(note. Can you always find the sum of squares of a sequence given its sum of linears? Third powers? etc? (I think you can't))

[quote="smart of math"]I have question :

$ \sum_{i=0}^{n}\frac{1}{(4i+1)(4i+2)(4i+3)(4i+4)} $[/quote]

For $n=\infty$ my CAS yields $\frac{1}{24}\left(6\ln 2-\pi\right)$, so I don't expect that there would be a closed form for $n$ finite (unless you can find a closed form whose $n\to\infty$ limit is $\frac{1}{24}\left(6\ln 2-\pi\right)$).

here is his original post : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=427525&p=2422336#p2422336
he asks for the limit too! so, i think you have the solution he is asking for.


search
inequalities unsolved
inequalities

It's here

[url]http://www.artofproblemsolving.com/Forum/topic-55164.html[/url]

Please search before posting


linear algebra
matrix
linear algebra unsolved

let A and B be two n x n complex matrices. let C=AB-BA. if CA=AC , then show that C is nilpotent

[quote="prime"]It follows that each C^k is a commutator, where k>0, and hence has trace zero. Thus C is nilpotent.[/quote]

Could you explain more please ???  :?

sorry, i dint quite follow your proof because i dont no what commutator means. when u say that $C^k$ is a commutator, what does that mean?

For every k>0, one easily checks that C^k=AC^(k-1)B - C^(k-1)BA. Thus TrC^k=0 for all k>0. It is known (and a nice exercise) that this last condition is equivalent to A being nilpotent.

[quote="prime"]It follows that each C^k is a commutator, where k>0, and hence has trace zero. Thus C is nilpotent.[/quote]

I'll try to explain what he probably meant: for every to matrix $A,B$ you can define the matrix $[A,B]=AB-BA$ which is called a commutator of $A,B$. Notice that every matrix which is commutator of some $A,B$ has zero trace since $tr(AB)=tr(BA)$. Now, according to our problem, since $C=[A,B]$ multiply both sides by $C$ from left and use $[A,C]=0$ you'll get $C^2=CAB-CBA=ACB-CBA=[A,CB]$. Now multiply this by $C$ from right and you'll get $C^3=[A,CBC]$ and so on... So you see that $tr(C^k)=0$ for $k=1,2,...,n,...$. Now try to prove that this follows that all
the eigenvalues of $C$ are zero and thus $C$ is nilpotent.

Edit. I didn't see that prime replied,


$ x\equal{}0$ doesn't work...your solution set is wrong

it should be

$ 5<x\le 8$.

[quote="i_like_pie"]This is too easy for Intermediate and should probably go in Classroom Math.

[hide="Solution"]$ \frac{2x\minus{}7}{x\minus{}5}\ge3$

$ 2x\minus{}7\ge3x\minus{}15$

$ \boxed{x\le8\text{ and }x\ne5}$[/hide][/quote]

typical mistake. never multiply through by denominator.  move everything to one side instead.

[quote="grizzland"][img]http://webwork2.math.utah.edu/math1050fall2007-4/tmp/l2h/set2/42-19754108/19754108img1.gif[/img][/quote]
Solution(by galois01)
$ \frac{2x\minus{}7}{x\minus{}5}\geq3$ where $ x\ne5$
$ \frac{2x\minus{}7}{x\minus{}5}\minus{}\frac{3(x\minus{}5)}{x\minus{}5}\geq0$
$ \frac{8\minus{}x}{x\minus{}5}\geq0$
Now we have two cases:
1st case
$ 8\minus{}x\geq0$
 $ x\minus{}5>0$
2 case
$ 8\minus{}x\leq0$
$ x\minus{}5<0$
From the first case we take $ 5 < x\leq8$  and the second has no solution
So the only solution to the problem is
$ 5 < x\leq8$

[quote="Hetidemek"][quote="i_like_pie"]
$ \frac{2x\minus{}7}{x\minus{}5}\ge3$

$ 2x\minus{}7\ge3x\minus{}15$

$ \boxed{x\le8\text{ and }x\ne5}$[/quote]
Actually, you can't multiply by $ x\minus{}5$ 'cause we don't know if it less or great than zero.[/quote]
That's correct  :wink:

[quote="galois01"]1st case
$ 8\minus{}x\geq0$
$ x\minus{}5\geq0$
2 case
$ 8\minus{}x\leq0$
$ x\minus{}5\leq0$[/quote]

You mean $ x\minus{}5>0$ and $ x\minus{}5<0$  :wink:  :P

[quote="Anaxerzia"][quote="galois01"]1st case
$ 8\minus{}x\geq0$
$ x\minus{}5\geq0$
2 case
$ 8\minus{}x\leq0$
$ x\minus{}5\leq0$[/quote]

You mean $ x\minus{}5 > 0$ and $ x\minus{}5 < 0$  :wink:  :P[/quote]
Thank's for correcting me you are right I forget it I edit immediately  :wink:  :P

[quote="i_like_pie"]This is too easy for Intermediate and should probably go in Classroom Math.

[hide="Solution"]$ \frac{2x\minus{}7}{x\minus{}5}\ge3$

$ 2x\minus{}7\ge3x\minus{}15$

$ \boxed{x\le8\text{ and }x\ne5}$[/hide][/quote]

maybe it is not so easy :P ?

$ \frac{2x\minus{}7}{x\minus{}5}\geq 3$

$ 2\plus{}\frac{3}{x\minus{}5}\geq3$

$ \frac{3}{x\minus{}5}\geq 1$

Clearly $ 0<x\minus{}5\leq3$, so $ \boxed{5<x\leq8}$.


This is the same as this:

$ \sqrt [3]{(7!)(7!)(7!)(8)}$

So, this is $ 7! \sqrt [3]{8}$, which becomes $ 7! \cdot 2$, so the answer is $ \boxed{10080}$.

[color=blue]Fixed $ \text{\LaTeX}$. -Izzy[/color]

[quote="ernie"]This is the same as this:

$ \sqrt [3]{(7!)(7!)(7!)(8)}$

So, this is $ 7! \sqrt [3]{8}$, which becomes $ 7! \cdot 2$, so the answer is $ \boxed{10080}$.[/quote]

fixed your "unparseable or potentially dangerous latex formula"


1.Prove that with a prime p=7 or 11(mod16) the equation
       (1) x^4+p*y^4=z 2  has no solution in Z+.
2.With p=3,(1) has a solution x=1,y=1,z=2
3.With p=5,(1) has a solution x=1,y=2,z=9

yes , and I think this problem is not difficult.


1.        IN TRIANGLE ABC TWO STRAIGHT LINES ARE DRAWN FROM AN ARBITARY POINT  D ON SIDE AB WHICH ARE PARRALLEL TO AC AND BC  .THESE LINES MEET BC AND AC AT F AND G RESPECTIVELY .PROVE THAT SUM OF CIRCUMFERENCES OF CIRCUMSCRIBED CIRCLES AROUND TRIANGLE ADG AND TRIANGLE BDF IS EUQAL TO THE CIRCUMFERENCES OF THE CIRCUMCIRCLE AROUND THE TRIANGLE ABC.

$ \triangle ACB \sim \triangle AGD \sim \triangle DFB$

Denote by $ l(C)$ the length of the circumference of the circle $ C$

$ \frac {l(C_{ADG})} {l(C_{ABC})}\equal{}\frac {AD} {AB}$

$ \frac {l(C_{DFB})} {l(C_{ACB})}\equal{}\frac {DB} {AB}$

$ \Rightarrow l(C_{ADG})\plus{}l(C_{DFB})\equal{}l(C_{ACB})$


number theory
greatest common divisor
modular arithmetic
quadratics
number theory solved

Let a_1, b_1 be positive integers satisfying the required condition.
Then a_1=4a_2 and b_1=4b_2 for some a_2 and b_2 as a_1 and b_1 divisible by 4.
So smaller a_2 and b_2 also satisfies the condition.
But there are not infintely many positive integers less than a_1 and b_1, contradiction.

We see that d=gcd (a,b) is a power of 2. Let a=dA, b=dB. Cancelling both sides by d and observing, we get A and B are both even, contradiction.

Another approach - the system $ 36a \plus{} b \equal{} 2^s, 36b \plus{} a \equal{} 2^t$ has integer solutions iff $ 36^2 \minus{} 1 | 36 \cdot 2^t \minus{} 2^s, 362^s \minus{} 2^t$ (just solve the system explicitly).
$ 36^2 \minus{} 1 \equal{} 35 \cdot 37$. $ 35, 37$ are coprime.
$ 35 | 36 \cdot 2^t \minus{} 2^s, 36 \cdot 2^s \minus{} 2^t \leftrightarrow 2^{s \minus{} t} \equiv 1 \pmod {35}$
$ 37 | 36 \cdot 2^t \minus{} 2^s, 36 \cdot 2^s \minus{} 2^t \leftrightarrow 2^{s \minus{} t} \equiv \minus{} 1 \pmod {37}$

$ 2^{s \minus{} t} \equiv 1 \pmod {35} \implies s \equiv t \pmod {12}$ (because the order of $ 2$ is $ 12$)
$ 2^{s \minus{} t} \equiv \minus{} 1 \pmod {37} \implies s \equiv t \pmod {18}$ but  $ s \neq t \pmod {36}$ (because the order of $ 2$ mod $ 37$ is $ 36$, and $ 2$ is not a quadratic residue)
Yet, $ 12,18 |a \implies lcm(12,18) \equal{} 36 |a$, a contradiction.

[EDIT: Now when I think of it, it reminds me of some IMO problem. See this - [url]http://projectpen.files.wordpress.com/2008/12/pen-16sol.pdf[/url] ]

[quote="shobber"]Show that for any positive integers $ a$ and $ b$, $ (36a \plus{} b)(a \plus{} 36b)$ cannot be a power of $ 2$.[/quote]
Assume that there exists a pair. Then there must a exist a pair such that $ (a,b) \equal{} 1$. Then $ 36a \plus{} b \equal{} 2^n, 36b \plus{} a \equal{} 2^m$. So $ n,m > 5$. Thus $ 2^5 \mid (36a\plus{}b)\plus{}(36b\plus{}a) \equal{} 37(a\plus{}b) \iff 2^5 \mid a\plus{}b$. Thus $ a \equiv b \equiv 1 \bmod 2$. And thus $ 36a \plus{} b \equiv 1 \bmod 2$. Contradiction..

let $ (a\plus{}36b)(36a\plus{}b)\equal{}2^z$. so $ 36a\plus{}b\equal{}2^x$ and $ a\plus{}36b\equal{}2^{x\plus{}y}$. (assume that a+36b is not less thsn 36a+b). So  $ 36a.2^y\plus{}b.2^y\equal{}2^{x\plus{}y}$ . Hence $ y<6$. but $ 37|2^x\plus{}2^{x\plus{}y}$.absurd.

Assume there is a solution. Take $a < b$ and the smallest possible $a$. Now $(36a + b)$ and $(a + 36b)$ must each be powers of $2$. Hence $4|b$ and $4|a$. So $a/2$ and $b/2$ is a smaller solution with $a/2 < b/2$. Contradiction.

WLOG assume that $v_2(b)\ge v_2(a)$. Then we can divide by by $2^{2v_2(a)}$ to get $(36m+2^km)(2^k\cdot 36n+m)$ where $m$ and $n$ are odd positive integers. Since $2^k\cdot 36n+m$ is odd and not equal to $1$, then $(36a+b)(a+36b)$ cannot be a power of $2$.

Would this solution be acceptable?

[hide=Solution]Suppose for the sake of contradiction that there is a solution. Consider the solution such that $a+b$ takes on the smallest value, we will use infinite descent to show that there is always a smaller ordered pair that works. If $(36a+b)(36b+a)=2^k$ where $k>10$, then we must have that both $36a+b$ and $36b+a$ are even. This means both $a$ and $b$ must be even. Let $a=2x$ and $b=2y$. Our equation is now $(72x+2y)(72y+2x)=2^k$ or when divided by 4, $(36x+y)(36y+x)=2^{k-2}$. Clearly this is another solution in which $x+y=\frac{a+b}{2}$ contradicting the minimality of $a+b$. Thus, we are done $\blacksquare$. [/hide]

It looks fine, just make sure the grader knows that you know that $2^k\geq 2^2$ :)

[quote=mathlete3141592][quote=subham1729]First $a$ and $b$ must be distint
so WLOG $a>b$ so $36b+a$ divides $36a+b$
hence $36b+a$ divides $a-b$ (impossible)[/quote]

How do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?[/quote]

Does anybody see why this is true?

[quote=rafayaashary1]i think it should be $35a-35b$ by Euclid's Algorithm, I can't get any farther :([/quote]

Then why can't 36b+a divide 35a-35b? I can't understand subham's solution.

[quote=mathlete3141592][quote=subham1729]First $a$ and $b$ must be distint
so WLOG $a>b$ so $36b+a$ divides $36a+b$
hence $36b+a$ divides $a-b$ (impossible)[/quote]

How do you conclude that 36b+a must divide a-b from that fact that 36b+a divides 36a+b?[/quote]

Remember we are assuming that $36b+a$ is a power of two. Then $(36b+a,35)=1$ so $36b+a|35a-35b$ implies $36b+a|a-b$...

Just write $a=2^{\alpha}a',b=2^{\beta}b'$ to get $2^{\alpha +2} \cdot 9a' +2^{\beta}b'=2^m$, and $2^{\beta+2} \cdot 9b' +2^{\alpha}a'=2^n$- thus, as $a,b$ are positive integers, $\alpha+2=\beta, \beta+2=\alpha$ which is absurd.

I'm not quite sure about this, can someone :help: or Point any flaws?
[quote=APMO 1998 P2]Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.[/quote]
[color=#000][b]Solution:[/b][/color] Since, $(36a+b)(a+36b)$ is symmetric. Hence, Suppose, $a \geq b$. Now suppose, there exists a $k$
$$(36a+ ~b)(a+36b)=2^k=2^{x+y}$$
Suppose, $$\begin{cases} 36a+ ~b=2^x \\ 36b+ ~a=2^y \end{cases}$$

Hence, $a,b$ are even $\implies$ $a=2a_1$ and $b=2b_1$ for $a_1,b_1 \in \mathbb{Z}^+$ [size=50](#27) Thanks[/size]
$$\begin{cases} 36a+b=2(36a_1+b_1)=2^x \implies 36a_1+b_1=\frac{2^x}{2} =2^{x-1}=2^{x_1} \in \mathbb{Z}^+ \\ 36b+a=2(36b_1+a_1)=2^y \implies 36b_1+a_1=\frac{2^y}{2} =2^{y-1}=2^{y_1} \in \mathbb{Z}^+ \end{cases}$$
Hence, now,
$$(36a_1+b_1)(36b_1+a_1)=2^{x_1+y_1}$$
Repeating the Same Procedure, we find, $a_1=2a_2$ and $b_1=2b_2$ for $a_2,b_2 \in \mathbb{Z}^+$. We can continue This till $x_n+y_n$ doesn't become negative and Since, $a,b$ are positive integers, we get a Contradiction! $\qquad \blacksquare$

---------------------------------------------
@Jupiter and agastya,  Thanks

[hide=possible fault][quote=AlastorMoody]I'm not quite sure about this, can someone :help: or Point any flaws?
[quote=APMO 1998 P2]Show that for any positive integers $a$ and $b$, $(36a+b)(a+36b)$ cannot be a power of $2$.[/quote]
[color=#000][b]Solution:[/b][/color] Since, $(36a+b)(a+36b)$ is symmetric. Hence, Suppose, $a \geq b$. Now suppose, there exists a $k$
$$(36a+ ~b)(a+36b)=2^k=2^{x+y}$$
Suppose, $$\begin{cases} 36a+ ~b=2^x \\ 36b+ ~a=2^y \end{cases}$$
Now, $a \geq b \implies x \geq y$
$$35(a ~ - ~ b)=36a ~ + ~ b -(36b ~ + ~a)=2^x ~ - ~2^y \implies \qquad a ~ - ~b=\frac{2^x-2^y}{35} \qquad - - - \text{[1]}$$
$$37(a ~ + ~b)=36a ~ + ~b + ~36b ~+~ a=2^x~ +~ 2^y \implies \qquad a ~ + ~b=\frac{2^x+2^y}{37} \qquad - - - \text{[2]}$$
Hence,
$$\text{[1]} + \text{[2]} \implies a = \frac{36 \cdot 2^x ~ - ~ 2^y}{35 \cdot 37} \in \mathbb{Z}^+$$
Hence, $a,b$ are even $\implies$ $a=2a_1$ and $b=2b_1$ for $a_1,b_1 \in \mathbb{Z}^+$
$$\begin{cases} 36a+b=2(36a_1+b_1)=2^x \implies 36a_1+b_1=\frac{2^x}{2} =2^{x-1}=2^{x_1} \in \mathbb{Z}^+ \\ 36b+a=2(36b_1+a_1)=2^y \implies 36b_1+a_1=\frac{2^y}{2} =2^{y-1}=2^{y_1} \in \mathbb{Z}^+ \end{cases}$$
Hence, now,
$$(36a_1+b_1)(36b_1+a_1)=2^{x_1+y_1}$$
Repeating the Same Procedure, we find, $a_1=2a_2$ and $b_1=2b_2$ for $a_2,b_2 \in \mathbb{Z}^+$. We can continue This Infinitely, But Since, $a,b$ are positive integers, Hence by FMID we get a Contradiction! $\qquad \blacksquare$

---------------------------------------------

@below Am I making any mistake :maybe: ? Also, where shall I fix my solution? :maybe: Pls Helpp :help: I thought FMID works in this way? (Pls correct, me a total noob)[/quote]
[/hide]
Note that $x_1+y_1<x+y$ and thus going towards $-\infty$. Therefore, there will be no infinite descent because it will be stop when $x_n+y_n<0$. But you may lead the proof :
(i) When $x+y$ is even : In this case once you will get $x_n+y_n=0$ which is absurd since $m,n $ are positive integers.
(ii) When $x+y$ is odd : In this case once you will get $x_n+y_n=1$ which is again not possible since $m,n$ are positive integers

@pluto1728 I don't understand what are you telling. I think his solution is completely  fine.
Btw @alastor
From the condition
$$\begin{cases} 36a+ ~b=2^x \\ 36b+ ~a=2^y \end{cases}$$
Isn't it obvious that both of $a,b$ are even.
Thanks Jupiter.

@above Lol stupid me...
@Jupiter...I meant there is decreasing sequence of positive integers right ($a,b $)? So FMID should work?

@below Oh! I kind of get it... Thanks! :D

[quote=Mr.Chagol]^^ try solving this with geo[/quote]
Whut? How :ewpu: ?

[quote=AlastorMoody]@above Lol stupid me...
@Jupiter...I meant there is decreasing sequence of positive integers right ($a,b $)? So FMID should work?[/quote]

The argument that you used to proof that $a$ and $b$ are even used the fact that $x+y>0$. However, the sequence $\{x_i+y_i\}_{i=1}^{\infty}$ is strictly decreasing by a gap of $2$. Thus, once $x_i+y_i$ becomes negative (for some $i$), you lose the power to prove that $a_i\equiv b_i\equiv 0\pmod{2}$

^^ try solving this with geo

@below thanks for reminding

$\spadesuit \ \textbf{\textrm{A different solution without infinte decent:}}$
Let $a=2^c \cdot p, b=2^d \cdot q$ where $p,q$ are odd numbers and $c,d$ are non-negative integers, WLOG, assume that $c\ge d$, \[(36a+b)(a+36b)=2^d(36\cdot 2^{c-d} \cdot p+q)(36b+a)\] since $36\cdot 2^{c-d} \cdot p+q$ is a non-trivial odd positive integer, $(36a+b)(a+36b)$ cannot be a power of $2$. $\quad \blacksquare$

36a+b and 36b+a are both powers of 2 we can assume WLOG that : a+36b|b+36a
That gives  a+36b|35(a-b)  but a+36b is a power of 2 so  a+36b|a-b  that gives a=b and that also is a contradiction


Help required :- :?

1. A cork floating in a calm water executes SHM of frequency $\nu$ when a wave generated by a boat passes by it. The frequency of the wave is
(a) $\nu$ (b) $\frac{\nu}{2}$ (c) $2\nu$ (d) $\sqrt{2} \nu$

2. A sonometer wire of length $l$ vibrates in fundamental mode when excited by a tuning fork of frequency 416 Hz. If the length
is doubled keeping other things same, the string will vibrate with a frequency of
(a) 416 Hz (b) 208 Hz (c) 832 Hz (d) stop vibrating

3. A wave moving in a gas
(a) must be longitudinal                    (b) may be longitudinal
(c)  must be transverse                     (d) may be transverse

4. Two particles $A$ and $B$ have same phase difference of $\pi$ when a sine wave passes through a region

(a) $A$ oscillates at half the frequency of $B$.
(b) $A$ and $B$ move in opposite directions.
(c) $A$ and $B$ must be separated by half of a wavelength
(d) The displacements at $A$ and $B$ have equal magnitudes.

5. In a stationary wave,
(a) all the particles of the medium vibrate in phase
(b) all the antinodes vibrate in phase
(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase.

Try to support your answers with reasons.

[quote="bus"]I'd say that number 4 is obviously (c), a little less obviosly (d) and could be (b) if we assume that they [i]are[/i] moving at all.[/quote]

Try to provide reasons. The anwer given is (b) & (d).
Shocked at not having (c), I too was.
Try for other problems also.
Thanks for replying.

[quote="bus"]OK, I get it... they don't have to be separated by 1/2 wavelength, it can be k + 1/2 wavelength for any integer k. (b) and (d) are kinda obvious, the displacements are given by equations:

$x_1=A\sin\phi_1$
$x_2=A\sin\phi_2=A\sin(\phi_1+\pi)=-A\sin\phi_1=-x_1$

and the velocities are:

$v_1=v_\text{max}\cos\phi_1$
$v_2=v_\text{max}\cos\phi_2=v_\text{max}\cos(\phi_1+\pi)=-v_\text{max}\cos\phi_1=-v_1$.[/quote]

The equation of a travelling wave is $y = A\sin (kx - wt)$.
So, we cannot stick only to path difference $\frac{\lambda}{2}$
we can also have some condition for $t$.
I have got the solutions of all the problems which were submitted by me. If you stick to some of them then you can consult me.
Bye... :lol:

[quote="Bacco"]I'd say: 1a, actually the cork is always following the wave and both of them have the same frequency
2b, actually if the lenght of the wire is doubled the lenghtwave will be doubled too (in foundamental mode) so the frequency'll be the half.
3a(i think)
5: i don't know what an antinode is! I think a,d are wrong, if i understand right what an antinode is the answer should be b (if an antinode is a point that moves with the maximum allowed amplitude)[/quote]

You are true about Q1, but for Q2 the frequency will be the same as was before because the wire is in resonance with a tuning fork.
What you are thinking about an antinode is correct, but the answer is (c) & (d).
Thanks for your reply.
Bye...  :lol:

Sorry...
the fact is that i disn't exactly understand what's a tuning fork...
Sorry for 5 too, actually d is true because all particles have the same phase but different amplitudes


Walker Middle School sells graphing calculators to raise funds. The school pays $ \$90$ for each calculator and sells them for $ \$100$ apiece. They hope to earn enough money to purchase an additional set of 30 calculators. How many calculators must they sell to reach their goal?

[quote="JRav"]If they want to buy an additional 30 calculators, they need another $ \$2700$.  Since they make $ \$100$ on each calculator sold, they need to sell 27.[/quote]
Actually, they make $\$10$ on each calculator sold since $\$100-\$90=\$10$ so they need to sell $\frac{2700}{10}=\boxed{270}$ calculators.


$e \leq x \leq y$ . Prove that: $x^{y}\geq y^{x}$

[quote="kunny"]You are to examine the behaviour of the graph of $y=\frac{\ln x}{x}.$[/quote]
[u][b]Similar[/b][/u]
Find all  $x,y \in$[b]Z[/b] : $m^{n}=n^{m}$


[quote="ifai"]Calculate $\left\{\left[i^{100}-\left(\frac{1-i}{1+i}\right)^5\right]^2+\left(\frac{1+i}{\sqrt 2}\right)^{20}\right\}(1+2i)$.[/quote]
[hide]First, $i^{100}=1$. Now $\frac{i-1}{i+1}$ must have a length of one as $i+1$ and $i-1$ have the same length. And it must have an angle of $\frac{- \pi}{2}$ since $i-1$ has an angle of $\frac{-\pi}{4}$ and you're subtracting another $\frac{\pi}{4}$, so it equals $-i$. $(-i)^5 = -i$, so we have ${\left[i^{100}-\left(\frac{1-i}{1+i}\right)^5\right]^2=(1+i)^2 = 2i}$. Now $\displaystyle (\frac{i+1}{\sqrt 2}) ^2 = \frac{2i}{2} = i$. $i^{10} = -1$. We end up with $(2i-1)(2i+1) = -4-1 = -5$. So $-5$ is our answer.[/hide]


Hey folks!!!
I had trouble understanding the explanation at the back of the mathcounts handbook for this problem.
If you have the handbook, can you plz explain to me, in layman's terms, how to solve the problem?
I would VERY much appreciate it.

Can some1 plz help me???
Only 8 views, come on!

I think most people here don't have the handbook with them right now.

If you can post the problem, that would be great. :)

I have the handbook. The problem is

A bird collection has exactly four types of birds (eagles, doves, crows and sparrows). The eagles and doves make up 60% of the collection, and the doves and crows make up 20% of the collection. If the 18 crows in the collection represent 5% of the total number of birds, how many of the birds are sparrows?

*checks watch* *counts words* YAY typing record!  :lol:

If 18 crows is 5% of the total number of birds, then there are 360 birds total. Since doves and crows are 20% of the total birds together, the doves make up 15% of the total (20-5). And eagles and doves make up 60% together, so the eagles make up 45% of the total birds. So crows (5%), doves (15%), and eagles (45%) make up 65% of the total birds. What's left for the sparrows is 35%.

35% of 360 is 126, so there are 126 sparrows.

Hope this helps.