AMC
USA(J)MO
USAMO
AIME
AMC 10
AMC 12
AIME I

so how eactly does selection for USAMO work?

Also, do they set different floors and indexs for AIMEI and AIMEII (cuz if so, I'm screwed)?

From the AMC website:
"Selection for the USAMO will be made according to the follwing rules:

1. The goal is to select about 500 of the top scorers from the prior AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
2. Selection will be based on the USAMO index which is defined as 10 times the student's AIME score plus the student's score on the AMC 12 or the AMC 10.
3. The first selection will be the approximately 300 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
4. The lowest AIME score among those 300 first selected will determine a floor value. The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value.
5. The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.
7. The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests."

In the past the index/floor value for the AIME I and AIME II have occasionally differed. I believe all of this info can be found in the FAQ's at the top of the Forum, so read those next time.

wat oh crap. i thot since i was grade 10 i just had to meet a floor value on aime so i got like 109.5 on amc and then 8 on AIME

Actually, I think that there are a couple of mistakes in this version of the USAMO selection process. You should probably reference the 2007 USAMO Teacher's Manual Online instead.

Basically, if you're in 12th/11th grade, you have to meet a certain cutoff for your USAMO index (AMC + 10*AIME).

If you're in 10th grade, you have to meet a certain cutoff for your AIME score.

There are different cutoffs for the AIME I and the AIME II.

Unless there are too many people with the floor score, in which case your index does count, even if you're in 9/10th grade.

Yeah, this whole cutoff thing is completely vague at this point, so my strategy was to get way above the cutoff so no matter what system they use I don't get screwed. :P

haha yeah i was originally cut out of the USAMO until the whole scrap-the-tie-breaker last year.  I had like a 210.5 (120.5 AMC, 9 AIME), and the tiebreaker was a 212.  It was rather unfair....

what they should really do is say, everyone below grade 11 takes the 10 or just scrap the AMC10 altogether.

it's all been figured out before though, it's much less stressful to sit back and wait for them to make the decision than argue about what they should and shouldn't do :)

[quote="Mathlete9366"]well, they (the AMC coordinators) haven't announced the actual floor-value-breaks-ties-for-students-below-11th grade policy... last year, the original floor cutoff was an 8, with a certain Index, being 10 * Aime + AMC 10 or 12. Yet some people didn't even take the 10, so these people were extremely at a disadvantage - that is, a 120 on the AMC10 was "better" than a 115 on the AMC 12. As this aroused quite a bit of uproar, (especially from calc rulz, bless his efforts) the AMC completely did away with their whole index-breaks-tie crud and let ALL people who had an 8 on AIME (in 10th grade or below, of course) to take the USAMO.

Bottom line : nothing's for sure yet, I don't think. I could be wrong, but I haven't heard anybody explicitly say that anything has been decided.

If I don't make any sense, ignore me, as I'm also trying to enjoy my WOOT class atm ^_^[/quote]

The teacher's handbook says something to the effect of "the amc10/12 will be used without discrimination for the index in case of tiebreakers"

Okay, I'm not really getting this, so I'll talk about my specific case and wait for some people to explain.

10th grade - decided to take the 12 instead of the 10 and scored a 123.  AIME II - let's just say an 11.

My index is a 233, which is actually lower than last year (146.5 amc10, 9 AIME = 236.5)

This will probably put me on the borderline of top 300 indices, so let's play devil's advocate and assume I am not in the top 300 AMC12 indices.

Based on the rules according to vjneb's post, the fact that I got an 11 on the AIME, which could have easily given me a 255+ index if I had taken the AMC10, is now unimportant because nowhere are people who exceeded the floor value but not the 300th AMC12 index score are not included.

Am I missing something here?  Based on all the complaining about this year's AIME, my usamo index prediction may be a little high, but I still think that the process may be missing something.  Getting an 11 on the AIME was not easy, and if I don't make it I will have to pull a bubala.

EDIT---------------------------------------------------

Never mind.  The new selection process (copied below) takes the AIME score for grades 10 and below.  An 11 should put me in.

Here is the selection process from the 2007 USAMO teacher's manual.

Selection for the USAMO will be made according to the following rules:
1. The goal is to select about 500 of the top scorers from this years’s AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
2. Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.
3. The first selection will be the approximately 330 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
4. The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value. If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.
5 The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
6. To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the A & B Contests.
7. In advising young students (in grade 10 or below) who desire to be selected for the USAMO whether to take the AMC 12 contest or the AMC 10 contest, please be aware of the following facts:
a. In 2006, among 427 students invited to take the USAMO, 218 were in 10th grade and below. Those students had scored 8 or greater on the AIME.
b. Among those 218 students, 106 had their AIME qualifying high score based on the AMC 12 and 112 had their AIME qualifying high score based on the AMC 10.
c. In 2006, among 24,445 students invited to take the AIME, 3,994 were in grades 10 and below. Of those, 2,496 qualified for the AIME from the AMC 12 and 1,498 qualified from the AMC 10.

so if i took both 10 and 12 and reported my 12 score on AIME... i would have a hope of making it based on floor... despite my pathethic scores (12A 115.5, 10B 145.5, AIMEII 7).

It doesn't really matter which score you put on the AIME sheet, it's just to make it easier for them to find your name or something. Both your scores + AIME will count.

i don't really care what the cutoff is because i know its not 2.

[quote="Teki-Teki"]My index is a 233, which is actually lower than last year (146.5 amc10, 9 AIME = 236.5)

This will probably put me on the borderline of top 300 indices, so let's play devil's advocate and assume I am not in the top 300 AMC12 indices.[/quote]

Umm, right. I have a hard time believing that, no matter how many times you say it. I don't believe the index cutoff has ever exceeded 230, at least not recently, and the floor has not exceeded 9.

I firmly believe that they design the selection process to be confusing enough that only people capable of doing well on the USAMO understand it. Scare away the others....
I don't really understand it, in case you were wondering.


This inequality is not difficult (I found the solution) :

Prove that if a, b, c > 0 such that abc = 1, then

a + b + c >= a + b + c.

Can't you just do this with [hide]the rearrangement inequality?  Let

x^3 = a, y^3 = b and z^3 = c.  Then xyz = 1 and we want to show

x^6 + y^6 + z^6 >= yzx^4 + zxy^4 + xyz^4.  However, since we can assume WLOG x >= y >= z, we can consider this the case of the rearrangement inequality with 18 numbers, 6 x's, 6 y's, and 6 z's, which we need to multiply in three groups of 6.  We know the maximum occurs when we group the three largest together and so on, so we're done.[/hide]

Indeed Joel. I found three proofs :

1) Let a = x/y, b = y/z and c = z/x. Then we have to prove that
xyz(xy + yz + zx) <= (xy) + (yz) + (zx)
after expanding and simplifying. Now we have
(xy) + (yz) >= 2xyz
(yz) + (zx) >= 2yzx
(zx) + (xy) >= 2zxy
by AM GM. Add these inequalities and divide by 2 to get the result.

2) Cauchy's inequality gives 3(x + y + z) >= (x + y + z) so
x+y+z >= (x+y+z)(x+y+z)/3 >= (x+y+z)(xyz)^(1/3) = x+y+z
where (x + y + z)/3 >= (xyz)^(1/3) follows from AM - GM.

3) Let a = x, b = y, c = z. We want to prove that
a^6 + b^6 + c^6 >= a + b + c
so homogenising we want to prove that
a^6 + b^6 + c^6 >= (a^4)bc + a(b^4)c + ab(c^4).
Now we have by AM - GM
(a^4)bc <= 4(a^6)/6 + (b^6)/6 + (c^6)/6
a(b^4)c <= (a^6)/6 + 4(b^6)/6 + (c^6)/6
ab(c^4) <= (a^6)/6 + (b^6)/6 + 4(c^6)/6.
Adding these gives the result.
(Alternatively this can be done by rearrangement or Muirhead.)

[color=cyan]I love Cauchy for problems like this.  Good stuff.[/color]

Now try the other inequalities Joel !
This is a good opportunity to become better in proving inequalities ...

Indeed, Cauchy is really great :D

[color=cyan]I'm trying, don't worry . . . that doesn't mean too much, though.  I never got enough practice doing more complicated inequality things, although I really love them.[/color]

I might have a slightly shorter soln for this one  :D :
We now by AM-GM (a+b+c)/3 >= (abc)^1/3=1 and we also know (this is well known) a^2+b^2+c^2 >= 1/3*(a+b+c)^2 = (a+b+c)*(a+b+c)/3 >= a+b+c by using the previous ineq.

You know Grobber, I found three solutions (and I posted them).
The second one is EXACTLY the same as yours :D

Oh, so sorry abt that! I thought it was just one long proof!  :D  :D   :D My mistake!


logarithms
integration
trigonometry
calculus
calculus computations

Find the general solution of each of the following differential equations for $ y \equal{} y(x)$:

(i) $ y^y \equal{} e^{y'}$
(ii) $ y' \equal{} y \ln y \plus{} y\, e^x$
(iii) $ y'' \equal{} y \, y'$

hello, for i) rewriting your equation in the form
$ y\ln(y)\equal{}y'$
$ \int\frac{dy}{y\ln(y)}\equal{}\int\,dx$
$ \ln(\ln(y))\equal{}x\plus{}C$
$ y(x)\equal{}e^{Cx}$
Sonnhard.

Yes, except in the final answer I suppose you mean $ e^{e^{Cx}}$?

hello, yes i mean $ y(x)\equal{}e^{e^{Cx}}$
Sorry,Sonnhard.

hello, for ii) rewriting your equation in the form
$ \frac{y'}{y}\equal{}\ln(y)\plus{}e^x$
$ (\ln(y))'\equal{}\ln(y)\plus{}e^x$
setting
$ \ln(y)\equal{}p$
we get
$ p'\minus{}p\equal{}e^x$
this equation has the solution
$ p(x)\equal{}(x\plus{}C)\cdot e^x$
from here you will get
$ \ln(y)\equal{}(x\plus{}C)e^x$
$ y(x)\equal{}e^{xe^{x}}e^{Cx}$
Sonnhard.

hello, for iii) setting $ y'\equal{}p$ we get $ y''\equal{}p'$ then we have
$ p'\equal{}y\cdot p$
$ \frac{dp}{p}\equal{}ydy$
$ \ln(p)\equal{}\frac{y^2}{2}\plus{}C$
$ y(x)\equal{}\frac{\sqrt{2}}{C_1}\tan\left(\frac{1}{2}\frac{(x\plus{}C_2)\sqrt{2}}{C_1}\right)$
Sonnhard.

Thanks for (i) and (ii), in the case of (iii), how do you get from $ \ln \, (p) \equal{} \frac{y^2}{2} \plus{} C$ to $ y(x) \equal{} \frac{\sqrt{2}}{C_1} \tan \left(\frac{1}{2} \frac{(x \plus{} C_2)\sqrt{2}}{C_1} \right)$?

iii) Solve $ y'' \minus{} yy' \equal{} 0$.

Setting $ y' \equal{} p \implies y'' \equal{} \frac{dp}{dx} \equal{} \frac{dp}{dy}\frac{dy}{dx} \equal{} p \frac{dp}{dy}$.

We get:
 $ p\frac{dp}{dy} \minus{} py \equal{} 0 \implies \frac{dp}{dy} \equal{} y \implies 2p \equal{} y^2 \plus{} k^2 \implies \frac{2}{y^2 \plus{} k^2}dy \equal{} dx$.

Now integrate both sides of the last equality and you are done.


Two circles whose centers are A and B touch each other externally at C.A common tangent touches the circles at P and Q receptively.AP=8m and BQ=2m,denoting the angle PAB by x radians,show that the area bounded by PQ and minor arcs PC and QC is (40-30x-2pi)m^2.

[quote="rt_08"]
Two circles, whose centers are A and B, touch each other externally at C.
A common tangent touches the circles at P and Q, respectively.
The length of AP = 8m, and the length of BQ = 2m.
Denoting the angle PAB by x radians, show that the area bounded
by PQ and minor arcs PC and QC is (40 - 30x - 2pi)m^2.[/quote]
I mostly spaced out characters and added commas for breaks.

I added a couple of phrases for more accuracy.

Hello
[hide="solution"]
Well its easy.Just draw a line perpendicular on PA from B parallel to PQ let the point be M.
U will get a rectangle PQBM of area $ 16 m^2$ and a right trianle MAB of area $ 24 m^2$
as u said angle PAB=x radian and angle ABQ =$ pi \minus{} x$
Thus area of sector PAC=$ 32x$ and area of sector CBQ =$ 2pi\minus{}2x$
thus the area u want = area of PQBM + area of MBA-area of sector PAC -area of sector CBQ
                              =$ (40\minus{}30x\minus{}2pi)m^2$

 [/hide]
Thank u.

thanks kabi
and thanks a lot arrange your tan,u did a [u]herculean[/u] task for me


MIT
college
LaTeX
Harvard
complex analysis

Anyone know of a simple package I can download and install to be able to read and write TeX files? I have looked on ctan, but everything there seems so complicated and confusing.  :(

Fierytycoon

I use emacs as my editor and MiKTeK for TeX.  I hear there are better ones as far as error messages and such, but this is sufficient for the short documents I produce (also comes with pdflatex, which produces pdf docs directly).

MiKTeK is at miktek.org, I think.  With a little rooting around, you can get an executable version of emacs from the gnu site (just google emacs and you'll hit it).  Make sure you get the executable or you'll have to compile it yourself, which is no fun.  May just want to google 'emacs executable' to find it.

[quote="Mindspa"]I run miktex along with tecnixcenter on Windows XP.[/quote]

Er, that would be texniccenter?  Never heard of tecnixcenter.

[quote="rrusczyk"]MiKTeK is at miktek.org, I think.  [/quote]

Er, miktek.org comes up with nothing.  There [i]is[/i] a miktex.org, though, and I think this is what you want.  I found something for Miktek on google, but it may have been another person's typo.

Thanks guys, MiKTeX works great.

Fierytycoon

I'm sure it would work great...only my DSL is getting finicky again.  Can't download...waaah!  :cry:

[color=cyan]Anyone with DSL has absolutely no right to complain about their internet connection.

Joel, at 56K[/color]

Specific to JBL:

Gah!  You did CMRA last year, right?  You should see if you can get a claymath account from them.  (Just don't tell Vida I sent you!)  You can do it all through their web interface.  I usually use my MIT account for LaTeX stuff.  Hopefully our Harvard accounts will make it similarly easy for us.  I've tried miktex along with five or six other things, and they disagreed with my computer.  Then again, my computer is pretty sensitive to just about anything.

For anyone:

I use emacs and WinEdt (http://www.winedt.com/) as my editors.  I definitely prefer WinEdt.  It has lots of clickable LaTeX commands (you can do things like click  :pi: and it will insert \pi into your document) and pretty colors that make it fairly obvious if you forget a $.

Anyway, good luck.  Getting everything to run smoothly on Windows can be a real pain.

[quote="JBL"][color=cyan]Anyone with DSL has absolutely no right to complain about their internet connection.[/color][/quote]

True, but at least you can go to math jams.  I can't because my server times out every minute or so (I can be on math jam for at most a minute!).  And I can't download anything larger than about a meg, since once again my server times out...yours may be lower, but it's steadier.

Okay, enough complaining.  Wah!

I use PCTeX for my TeX stuff, and I will see if I can get Y&Y cheap from my college next year. (I hope...)

Maybe I missed something, but what is Y&Y?

It's supposedly the best TeX editor. I believe Tristan Needham in [i]Visual Complex Analysis[/i] said it's the ``Cadillac of LaTeX systems.'' By the way, VCA is a great book if you want to learn complex analysis.

I would say it's not that good. Try MiKTeX with TeXnicCenter.

[quote="ComplexZeta"]... it's the ``Cadillac of LaTeX systems.'' [/quote]

Of course, not everyone wants a Caddy.  Some people would prefer to have a Hummer or something.  *claps hands*

Well, no, just using analogies to launch meself off onto a tangent.  But having been unable to download any TeX editor, I can't say which is better (or worse) than all the others.

I couldn't figure out how to download one either. I downloaded a zip file, and when I extracted it, I would get 500 files, and I didn't know what to do with them. I suggest buying one, but they aren't cheap.

I can't even download the files due to stupid internet connection (yes, DSL, so?) that times out every minute... but even if I did get the files downloaded, there's the matter of what to do with all the .cabs, .dlls, etc.

Unless you are a serious computer jock, you want to download executables, not source code (which give you all those files you don't know what to do with).  Sometimes these can be hard to find, but googling usually gets you there ('download emacs executable', for example, though there are certainly better TeX-specific editors out there from what it sounds like on this thread).

[quote="rrusczyk"]...a serious computer jock...[/quote]

I like that quote.  It reminds me of being in 4th grade and laughing at the 'pretty ugly' joke.

Unless you are planning on writing some long, long TeX files, I think that using something simple like notepad would be fine. I personally use Textpad.

Fierytycoon


Will FLT (Fermat's Last Theorem) have any implications in math or physics? The Riemann hypothesis is applicable in proving the random distribution of primes, and a lot more ideas in math and physics. I am only wondering if this is also true of FLT.

Wiles' proof of FLT is not celebrated because the theorem itself has important implications.  Rather, it is the [i]tools that were developed as part of the proof[/i] that are the real source of exciting new mathematics.

His question is not WHY Fermat's Last Theorem is celebrated, but WILL the theorem itself one day have any important ramifications.

In my opinion, it's a bad question though. Who can predict the future well enough to tell what implications a theorem [i]will[/i] one day have?

If he had asked: Are there any [i]known[/i] implications of Fermat's Last Theorem? - that would be a more tractable question.

Well sure, but I imagine the implied question behind the question was "why did we put so much effort into proving FLT in the first place?"


we can write x=(x-1)+1, then use the AM>GM. The problen is solve after we will resolve the following ineq.: 4x*x*x>(x+1)*(x+1)*(x+1), which is very simple after we divide all at x*x*x and using the condition x>1.

[quote="mychrom"]4x*x*x>(x+1)*(x+1)*(x+1), which is very simple after we divide all at x*x*x and using the condition x>1.[/quote]

But it is not true for $x = 1+\epsilon$.  :huh:


"Equation" means there's an equal sign somewhere, yet I don't see one. Typo?

It's set to = 0

Sorry for not typing that in the first one.

I can't believe I made that mistake, thanks for pointing that out.

[quote="nkouevda"][quote="JZambrano201"]$ 2(3^x) + 3 - 9 = 0$

$ 6^x - 6 = 0$[/quote]
The fact that your solution gives the right answer is merely a coincidence, but $ 2\left(3^x\right)\ne6^x$. :wink:

[hide="Solution"]\begin{align*}2\left(3^x\right) + 3 - 9 & = 0 \\
 \\
2\left(3^x\right) & = 6 \\
 \\
3^x & = 3 \\
 \\
x & = 1\end{align*}[/hide][/quote]

2*3^x=6=2*3
we cancel  2 from both sides and get 3^x=3.   from here  x=1.

[quote="Ansimax"]2(3^x)+3-9

I am pretty sure that because the exponent is a variable, I will need to use logarithmic function to solve for it. If I am wrong please help me if I am write then I have no idea how to solve for x in this situation. Any help would be greatly appreciated[/quote]

[hide="SOLUTION"]

$2(3^x)+3-9=0$

$\implies 2(3^x)=6$

$\implies 3^x = 3$

$\implies \log_3 3^x = \log_3 3$

$\implies x \cdot \log_3 3 = 1$

$\implies x = \boxed{1}$

[/hide]

[quote="Ansimax"]2(3^x)+3-9

I am pretty sure that because the exponent is a variable, I will need to use logarithmic function to solve for it. If I am wrong please help me if I am write then I have no idea how to solve for x in this situation. Any help would be greatly appreciated[/quote]
alternatively, one side is [b]constant[/b] and the other is exponential, thus there can be at most 1 solution. trying the trivial values we see that x=1 gives us the answer.

edit: correction in bold.

^There can be more than one solution with linear and exponential. Try $2^x = x + 1$. $x = 0, 1$ work. However, I think you mean there can be at most $2$ solutions. But then your solution is incomplete.

[quote="talkinaway"]Restate:  One side is exponential and the other side is a CONSTANT.  Since exponentials have an inverse (aka the logarithm), exponential functions are one-to-one.  Therefore, there can be only one solution.

Right idea, slightly off terminology.  All constant functions are linear, but not all linear functions are constant....[/quote]
yes i meant 'constant'.

Another way to say it for other, more involved problems, is that one side or funciton is monotonically increasing/decreasing.

For example, find all $x$ such that:

$3^x+4^x = 5^x$

$1,1,1$ is an obvious solution to get.

Now divide by $5^x$

$(3/5)^x+(4/5)^x = 1$

Since both $(3/5)^x$ and $(4/5)^x$ are monotonically deceasing, there can be only one solution.

[quote="ssilwa"]Another way to say it for other, more involved problems, is that one side or funciton is monotonically increasing/decreasing.

For example, find all $x$ such that:

$3^x+4^x = 5^x$

$1,1,1$ is an obvious solution to get.[/quote]
1,1,1 doesnt work, i think you meant 2,2,2 though.

oh oops, maybe its not so obvious after all... jkjk

[quote="mathdebam"]Basically if the eqation is 2.3^x-6=0Then taking log both sides
log(2.3^x)=log 6
log 2+log 3^x=log 2+log 3
xlog 3=log3
x=1[/quote]

Your solution is incorrect because $\log 2.3^x \neq \log 2 + \log 3^x$, unless if you mean that 2.3 is $2 \cdot 3$

It would be better to solve it like this:  Converting $2.3^x = 6$ to log form we get \[\log_{2.3}6 = x\] So then $x = \frac{\log 6}{\log 2.3}$ by change-of-base.
Of course, you need a calculator to get an approximate number.

What!!!!!!!!!!!!!!.No!No!No! I never meant 2.3 as 2+0.3.Rather I meant 2 multiplied by 3.

Then your equation would be $\log 6^x = \log 6$, which has an obvious solution of 1.

EDIT:  disregard. Thanks bogtro


search
number theory unsolved
number theory

$ x,y,z$  are positive integers.  then  $ 4xyz \minus{} x\minus{} y$  is  not    perfect square

how  to  prove  it  ?

thanks!

I think this camew up a few times before e.g.

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2123787062&t=291225


http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2006-aa/06aimestats.html

Seems like a new thing this year.

lol, the numbers are kind of shifted to the left side (they are not directly below the grade levels)
anyways this is very nice!!

Tied for 2nd in 9th grade in GA.
Tied for 14th overall

#1 in WY got a 2.

Arrrrrgh, Im not even 2nd  :( .  Yet our state only have 1 qualifier...

Grr...I should be higher up for NY...had it not been for careless mistakes...gah

Tied for 2nd in MA  :)

And I [i]possibly[/i] know who got the 14.

[quote="biffanddoc"]:huuh:
WOW
thats sad about Wyoming.[/quote]

Not really.  With only 4 total high school students in the whole state, that probably beats the national average.  :rotfl:

I hear that some years, they don't have high school students there.

No, WY had a 1 and a 2.
The other 2 didn't take the AIME

VI on the other hand had 1 AIME qualifier that got a 5.
Average:5

I wish I was in one of these kind of states.

Thx to pieterminate for the correction. I'm being dyslexic again.

Maybe a hair off topic, but...

I can report that all of the people in my city that took the AMC's made it to the AIME.

Yes, that's right.  [i]Both[/i] of them.

[quote="mkkool64"][quote="Xantos C. Guin"]No, WY had a 1 and a 2.
The other 2 didn't take the AIME

VT on the other hand had 1 AIME qualifier that got a 5.
Average:5

I wish I was in one of these kind of states.[/quote]

Hey don't be dissin' Vermont, that's my state.  what do you mean about one person getting a 5?  our average is 2.41 and we had 3 people get a 5. (the 9 is me :) )[/quote]

I believe that's Virgin Islands with 1 5... :)

I tied for second in New York.

I was the top junior in the state as well.

[quote="Phelpedo"][quote="kevin gong"]CA
IL
IN
AB

^where those 4 perfect scores are from

congrats to them and everyone else :lol:[/quote]

I know the IN is David Benjamin...Heard Mr. Ruscyk say that somefrom from SD math circle got a 15...Who got it from IL? Greg Gauthier maybe?[/quote]

Tony Liu (aka thAzn1)

Tied in 6th overall in PA, 1st 9th grade in PA.

Wow, there are 2 people with 14's from my state.  I bet I know exactly who they are... :)

[quote="tongchen1226"]i am 3rd (last out of the qualifiers) in TN.
I would be tied for first in LA if there were no Katrina.  :lol:[/quote]

Well, Weining Liu made the first with AMC 143 AIME 12, and I made the second with AMC 126.5 AIME 10... ;) Still much difference between Liu and me. But he is senior now and graduate very soon, so next year it will be Chen and Me competing.
Well, still nice job, Tong. I am junior now, and next year or after I graduate you will be sure the first!!! :P  Come on!!! (the secret is I do not plan to study math contest this year, because I am preparing college, but you can study!!! Next year, the first will be yours!!!)

I think I scored like a 3...that's .25 above the average....man I am pitiful   :(

[quote="scorpius119"]hmm... that's odd, i see no 11's in PA 11th grade. Did I end up with a 10 due to bad erasing on my answer sheet?

That's not good. At least it's better than having a 15 bumped to a 14. Or a 9 to an 8 for me (that would have meant no USAMO).[/quote]

You still haven't gotten your official scores back at your school?


induction
combinatorics unsolved
combinatorics

[quote="thonglao"]There is a large pile of cards. On each card one of the number $1,2,3,.....n$ is written. It is known that the sum of all numbers of all the cards is equal to $k n!$ for some integer $k$. Prove that it is possible to arrange cards into $k$stacks so that the sum of the numbers written on the cards in each stack is equal to $n!$[/quote]

Let's prove via induction (on n)
THE BASE: n=1 (The statement is evident)
THE ASSUMPTION: $(n-1)\Rightarrow n , n\geq 2$
Then:
The well known statement: if we have a set of $\geq n$ integers then we could choose its subset of $\leq n$ integers such as their sum is devided by n.

So let's make up the substecks this way:
1)Every card with number n on it is a substeck
2)For cards with another numbers we'll do such process: if there remains $\geq n$ cards of them then we choose $\leq n$ of them with sum devided by n

At first let's note that this process will finish.
At second: if there remains some cards at the end of process (<n cards) then their sum is devided by n $(k*n!-(...)*n)$ and it will be the last substeck
At third: the sum of numbers in every substeck is devided by n and is $\leq n(n-1)$  and $\geq n$

Then lets devide the sum in every substeck and the common sum by n.
Then we get such situation:
There are some substecks marked with integer from {1, 2, ..., (n-1)}. The sum of this integers is equal to $k*(n-1)!$.
We have to arrange them into k stecks with equal sums.

But it's the assumption of induction. Then let's multipluy everything by n  ;)
And it's the end of our proof.
V:)


geometry
perimeter
graph theory
combinatorics unsolved
combinatorics

[quote="orl"]A white plane is partitioned onto cells (in a usual way). A finite number of cells are coloured black. Each black cell has an even (0, 2 or 4) adjacent (by the side) white cells. Prove that one may colour each white cell in green or red such that every black cell will have equal number of red and green adjacent cells.[/quote]

I think it works for an infinite number of black cells too.

[quote="Summerburn"][quote="weiquan"]

I think it works for an infinite number of black cells too.[/quote]

Can you post your solution, then?[/quote]

Not yet :P haven't got the details filled in...

I'm wondering why nobody posted the solution for this beautiful problem. I guess it's hard to write it down.
We will prove the following: if we color any cell, this will never contradict itself, which means by a number of "forcings" we never get that if we color it green for example, it has to be red. This will be sufficient to conclude. $ 3$ cases:
If a black cell has $ 4$ black adjacent cells there is nothing to color.
If a black cell has $ 0$ black adjacent cells, and if a white cell will be "forced" to be red for example, we will always move by "corner moves"(I'll explain it later), which means we will have smth like: (r-red,g-green,b-black)

_[color=red]r[/color]_
[color=green]g[/color][b]b[/b][color=green]g[/color]
_[color=red]r[/color]_

If a black cell has $ 2$ black adjacent cells, there can be $ 2$ cases:

_[color=red]r[/color]_
b[b]b[/b]b , where if we are forced to color one cell (red for example), we are also forced to color the other green. Call this move a jump-move. Or
_[color=green]g[/color]_

_[color=red]r[/color]_
[color=green]g[/color][b]b[/b]b   ,this will be a corner-move.
 _b_

If we color a cell, it can force the color of other cells by jump-moves or corner-moves, and these can too, and so on, and we want to prove that it can't force itself to change it's color. This means that all "cycles" constructed with corner-moves and jump-moves (which begin and end in the same cell) have an even number of moves(with each move the color changes). Two observations: with a corner-move, the parity of the row and column changes(if we number the columns and rows), while with a jump-move it doesn't. So if we want to end in the same cell we must have an even number of corner-moves. The second observation (it's really beautiful): The black cells form territories, so with a jump-move we enter or exit a territory, while with a corner-move we stay in the same territory(this is why when a black cell has no black adjacent cells we always move by corner-moves so there will be no problem). We must come back to the same territory, so each "entering" in a territory will have it's "exit-pair" so we can group the jump-moves in pairs, so we have an even number of jump-moves too. Therefore each cycle will have an even number of moves, so we can color the plane the desired way.
If smth is not clear, just ask.

a bueautiful problem.my solution:
we define a graph $G$ for center of white cells that have at least one black cell as a neighbour.connect two vertices of graph if they have a black cell as common neighbour and they have common vertices (it means the two representive square with the two vertices of graph have common vertice.) or they have a common black neighbour cell that the black cell has exactly two black neighbour.then it is easy to see that if graph $G$ be a Bipartite graph then we are done.now suppose the contrary and that we have an odd cycle in graph $G$.if we see the odd cycle in the plane we see a polygon with sides are parallel to the axises with lenght $1$ or sides that construct a $45$ degree angle with the axises and have lenght $\sqrt{2}$ now we see that the number of sides with lenght $\sqrt{2}$ is even and so the number of sides is parallel to axises are odd and the number is equal to the number of black cells in the perimeter of the polygon and so it is odd.now for every black cell on perimeter the cell has exactly two black neighbour that exactly one of them is in the interior of polygon,now we cancel the black neighbours that are not in the interior of the polygon and now we concentrate on black cells are in the interior or perimeter of the polygon.each of them has even number of black neighbours except those are on the perimeter that have odd number,so it is a contradiction because the induced graph of black cells into the polygon has an odd number of vertices with odd degree.so the graph $G$ has no odd cycle and we are done.

Here's a method that explicitly constructs a coloring.

Consider the black cells as a planar graph $G$, with an edge between two cells if they share a side in the plane. Partition the edges into cycles that don't reuse vertices (possible since all degrees are even). Then consider the faces, and color a face red if it is contained inside an odd number of cycles and blue otherwise; this obtains a 2-coloring of the faces of $G$.

Now we color the white cells of the original plane. Number the columns sequentially and color all cells in an odd column green and all cells in an even column red. With one exception: if the white cell was part of a face we colored red in the graph $G$, swap the color (green<->red).

Checking this works:
- If we have a black cell surrounded by four others, there's nothing to check.
- If we have a black cell surrounded by two other black cells which are adjacent diagonally, then the two white cells are part of the same face in $G$ and also are one column away from each other. So they have opposite colors.
- If we have a black cell surrounded by two black cells horizontally or vertically, then the two white cells below are in oppositely colored faces in $G$ and are an even number of columns away from each other. So they also have opposite colors.
- If we have a black cell surrounded by four white cells, all four are in the same face in $G$. So the two vertically touching white cells have one color and the two horizontally touching ones have the opposite.


induction
number theory
prime factorization
combinatorics unsolved
combinatorics

Theorem : All marbles have the same color

Proof : Let A be any set of n marbles, where n >= 1. The proof by induction on n.

Step : Assume that if A is any set of k marbles , then all marbles in A have the same color. Let A' be a set of k+1 marbles, where K>= 1. Remove one marble m from A'. We are then left with a set A'' of k marbles. By the induction hypothesis, A" has all the marbles of the same color. Remove from A" a scond marble m', and then add back to A" the marble m originally removed. We again have a set of k marbles, which by the induction hypothesis has marbles all the same color. Thus the two marbles m, m' removed must have been the same color, and so the set A' must contain marbles all the same color.

We conclude that in any set of n >=1 marbles, all marbles are the same color.

[quote="t0rajir0u"][quote="EquinoX"]Thus the two marbles m, m' removed must have been the same color.[/quote]
This is the step that's absurd (you can verify that it fails for $ k \equal{} 1$).  "All marbles the same color" does not mean "all marbles the same [b]fixed[/b] color."[/quote]

sorry I am not getting it, can you please clarify the same color is not the same as fixed color, an example would help maybe

[quote="coquitao"]As I said before, the real problem here is the [b]induction basis[/b]. There is no way to prove that every two marbles have the same color. Forget the case n=1, the notion "have the same color" requires at least two marbles to stare at.

Hope it helps.  :)[/quote]

are you saying the base case?

[quote="EquinoX"]
... clarify the same color is not the same as fixed color...[/quote]

That statement is redundant. That's exactly what we need to disprove.

[quote="coquitao"][quote="EquinoX"]
... clarify the same color is not the same as fixed color...[/quote]

That statement is redundant. That's exactly what we needed to disprove.[/quote]

then can you help me to disprove this statement so that it makes more sense to me

[quote="EquinoX"]... are you saying the base case?[/quote]

Yes, I'm referring to it.

---

In a typical inductive argument you have to show that: a given proposition p(n) holds for a starting value of n and also that p(n+1) holds provided that p(n) holds. So, there are [b]two[/b] parts to it.

In your first post it is shown that everything is alright with the second requirement for the validity of the argument. The first part or case base is where the flaw resides. It is not [i]true[/i] even for n=2.

[quote="coquitao"][quote="EquinoX"][quote="coquitao"] ...are you saying the base case?[/quote]

Yes, I'm referring to it.

---

In a typical inductive argument you have to show that: a given proposition p(n) holds for a starting value of n and also that p(n+1) holds provided that p(n) holds. So, there are [b]two[/b] parts to it.

In your first post it is shown that everything is alright with the second requirement for the validity of the argument. The first part or case base is where the flaw resides. It is not [i]true[/i] even for n=2.[/quote][/quote]

ok, so the statement above flaws at the base case where n = 2 or when k = 1 am I right?

Right.

If you have two marbles you can try to emulate the argument of the first post: from a set of 2 marbles you can create two marble-sets, A and B, by placing only one marble in each. All the [color=red]marbles[/color] in A have the same color and the same applies to the [color=blue]marbles[/color] in B.  The colors between different sets might well be different, though.

Salutations.

[quote="coquitao"]Right.

If you have two marbles you can try to emulate the argument of the first post: from a set of 2 marbles you can create two marble-sets, A and B, by placing only one marble in each. All the marbles in A have the same color and the same applies to the marbles in B.  The colors in between different sets might well be different, though.[/quote]

Yes that is correct, to say that all the set in A has the same color there has to be at least 2 marbles right? So is there by any chance to fix this wrong proof? Maybe by changing the base case?

[quote="Prime factorization"]You can't prove something that isn't true.[/quote]

ok thanks for the clarification, is there any way to disproof this in a more mathematical way?

I'm just trying to clarify what coquitao and t0rajir0u already correctly pointed out.

It doesn't work when getting from 1-element-sets to 2-element-sets.
Because when you remove one marble from a set of 2 marbles, there is one marble, which has some color. And when you are removing that and adding the previous one, it could have some [b]other[/b] color because [b]there is no intersection between the 2 1-element-sets[/b]. Thus you can't establish the statement for 2-element-sets.

I might have misunderstood you. But please forget k and clearly write down the proof from 1-element-sets to 2-element-sets only, which I think won't work.

If the statement is true for any set of 2 marbles there are much easier ways to establish the result for longer sets.

(Take any marble as the base marble. Then take an arbitrary other marble with it into a 2-element-set. the other marble will have the same color as the base marble. But the 2nd marble was arbitrary, meaning that we could replace it by any other marble. and thus all marbles have the same color as the base marble.)

Prime factorization was talking about this:
[quote="EquinoX"]So is there by any chance to fix this wrong proof? Maybe by changing the base case?[/quote]
No. None can fix it. In first sight base of induction looks minor in the proof. But it's actually almost everything in the proof.


geometry
3D geometry
logarithms
LaTeX
sphere

help me solve these 4 problems
please include explanations...

1.)  How many digits are in $2^{100}$

2.)  If $(4^x+4^{-x})=23$, find $(2^x+2^x)$

3.)  Find the number of zeros at the end of 500!

4.)  The volume of a cone with radius R is $\pi*R^3/3$ Find its surface area, excluding the base. Express your answer in simplest possible radical form.

thnx a bunch :)

100log2 ~ 30.1
2^100 has 31 digits

however, we need to find a soution without logarithms and without using a calculator
 :mad: i'm so used to using log to solve exponential equations

[hide="3"]
500! has (500/5 + 100/5 + 20/5) zeroes
or 124 zeroes
[/hide]

 :sleeping: i'll edit this post tommorow with latex and add the rest of my solutions

No need to use logarithms in question 1

$2^{10}$ have 4 digits

$2^{20}$ have 7 digits

$2^{30}$ have 10 digits

$2^{100}$ have 3*10+1=31 digits

Warning: you cannot use this method if exponent is larger than 971

I assume on 2 you mean find $2^x+2^{-x}$. If so, [hide]call $2^x=y$, so the equation gives us $y^2+\frac1{y^2}=23$.  If we square what we are tring to find we get $\left(y+\frac1y\right)^2=y^2+\frac1{y^2}+2$ So $\left(y+\frac1y\right)^2-2=y^2+\frac1{y^2}$. Plugging in:
$\left(y+\frac1y\right)^2-2=23$
$\left(y+\frac1y\right)^25$
$y+\frac1y=5$ (since we can't have exponentials be negative if the base is positive). So
$2^x+2^{-x}=5$[/hide]

For number 4, I doubt that's the volume of a cone. That looks like a sphere divided by 4.

[quote="JesusFreak197"]No, it's a cone with height $R$.

I [b]think[/b] I did this correctly, but I'm not quite sure...

[hide="Solution"]The surface area of a cone, excluding the base, is some fraction of a circle. In this case, the circle has radius $R\sqrt{2}$, because the cone has height $R$ and radius $R$, so we use Pythagorean Theorem. The circumference of the fraction of the circle, though, is only $2\pi R$, while the circumference of the entire circle of radius $R\sqrt{2}$ would be $2\pi R\sqrt{2}$. The area of the entire circle would be $2\pi R^2$, but we have only $\frac{1}{\sqrt{2}}$ of the circle, so our answer is $\frac{1}{\sqrt{2}}\cdot 2\pi R^2=\boxed{\pi R^2\sqrt{2}}$.[/hide][/quote]
yeah I think thats right because the area of the lateral surface or a cone is $\pi{r}l$ with $l$ being the slant height, which in this case is $r\sqrt2$, so the answer is $\pi{r^2}\sqrt2$

[quote="GoBraves"][quote="G-UNIT"]Hmmmm....is "lighting round" some sort of contest, in running, these look afwul lot like contest problems.[/quote]

The lightning round is a round in the North South Math Bee competition. The competition is only for indian people, made by indian people. I participated in chapter this year and made nationals (my score was 18/20) but i wont go to nationals because I'm busy. The chapter competition dosnt have lightning round, just 20 questions (about easy state/hard chapter target level questions), and you have 60 min. to do those (3 min. per problem). In nationals there is lightning round, which is like 50 secs. per problem, total of 20 problems.

My opinion on this competition is-

Its was a terrible attempt to copy mathcounts. A lot of the problems are very unclear, and on a lot of the problems they provide graphics, and a lot of the time the picture was screwed up. They dont even pay for you trip to nationals if you make it.[/quote]

  Thanks for briefing us on that.  However, my original intention for the question was to alert other AoPs'ers that this could potentially be a contest in-running, which would be against AoPs policy to help him/her.  Obviously, people dont listent me. :D


function trigonometry limit calculus algebra domain quadratics

Suppose that the function $ f(x)$ is defined to be $ \frac {ax}{\tan{x}}$ for $ x \ge 0$ and is defined to be $ a^2 \minus{} 2$ for $ x < 0$. Find all values of $ a$ such that $ f(x)$ is continuous.

[hide="My Solution"]
(This was an extra credit problem on a test I had recently.)

We must have

$ \frac {a\cdot 0}{\tan{0}} \equal{} a^2 \minus{} 2 \implies \frac {0}{0}\cdot a \equal{} a^2 \minus{} 2$.

Originally, I thought the left side was 0, so it implies that $ a \equal{} \pm \sqrt {2}$ (which doesn't work by the way). But then I wondered what would happen if I were to have broken the rules and defined $ \frac {0}{0} \equal{} 1$. Applying that, we have

$ a^2 \minus{} a \minus{} 2 \equal{} 0\implies (a \minus{} 2)(a \plus{} 1) \equal{} 0\implies \boxed{a \equal{} 2, \minus{} 1}.$

Ironically, it turns out that these values of $ a$ [b]do[/b] create a continuous function. I numerically evaluated
the following limits:

$ \lim_{x \to 0^ \plus{} }{\frac {2x}{\tan{x}}} \equal{} 2 \\
\lim_{x \to 0^ \plus{} }{\frac { \minus{} x}{\tan{x}}} \equal{} \minus{} 1.$

Was it pure coincidence that my definition worked out?
[/hide]

[quote="sunehra"]...if I were to have broken the rules and defined $ \frac{0}{0} \equal{} 1.$ [/quote]This is a calculus context, in which that is never the right definition. What do we do instead? That's really $ \lim\frac{\text{near }0}{\text{near }0}$, and we need to evaluate that limit involving the functions we have.

With this piecewise defined function, the function is naturally continuous except possibly where we divide by zero or switch formulas. The first place to check is the formula switch at zero, where we compare $ a^2\minus{}2$ from the left and $ \frac{ax}{\tan x}$ from the right. The limit of the former is $ a^2\minus{}2$ (it's just a constant) and the limit of the latter is $ a$ since $ \lim_{x\to 0}\frac{\tan x}{x}\equal{}\lim_{x\to 0}\frac{\sin x}{x}\cdot\lim_{x\to 0}\frac1{\cos x}\equal{}1\cdot 1\equal{}1.$ These limits agree when $ a^2\minus{}2\equal{}a,$ which is when $ a\equal{}2$ or $ a\equal{}\minus{}1.$
But wait- what about the value at zero? We're evaluating that on the $ \frac{ax}{\tan x}$ side, where it is [i]literally[/i] $ \frac{0}{0}$ and thus never defined. That's a discontinuity no matter what. For that matter, what about all those places with $ x>0$ where $ \tan x\equal{}0$ or $ \tan x$ goes to $ \infty$? Those are all discontinuities as well.

The correct answer is that the function is never continuous on $ \mathbb{R}$. It has removable discontinuities at $ x\equal{}(n\plus{}\frac12)\pi$ for $ n\ge 0$, at $ x\equal{}0$ when $ a\in \{\minus{}1,2\}$, and at $ x\equal{}n\pi$ for $ n>0$ when $ a\equal{}0$. It has a jump discontinuity at zero for other values of $ a$. It has infinite discontinuities at $ x\equal{}n\pi$ for $ n>0$ when $ a\neq 0$.
We could also say that the function is always continuous on its domain, which is $ \mathbb{R}$ minus all those points where we're dividing by zero or $ \infty$.

Yeah, it was coincidence and you got lucky. Had the definition been $ \frac{ax}{\tan 2x}$ for $ x>0,$ you would have been out of luck.

Incidentally, the wording of the problem should have been "Find all values of $ a$ such that $ f(x)$ is continuous at $ 0.$" (since there are some other discontinuities for $ x>0.$) And the definition should have been stated as $ \frac{ax}{\tan x}$ for $ x>0$ and $ a^2\minus{}2$ for $ x\le 0.$ In other words, $ >$ and $ \le,$ not $ \ge$ and $ <.$

But you've got to learn to be quick about these things. Here's what my mental processess were like upon seeing that problem:

"As $ x\to 0,$ $ \tan x$ might as well be $ x,$ so $ \frac{x}{\tan x}$ might as well be $ 1,$ so that's $ a$ from the right and $ a^2\minus{}2$ from the left - to match them, set $ a^2\minus{}2\equal{}a,$ which is a quadratic equation, which I know how to solve."

And if you're wondering where I got "$ \tan x$ might as well be $ x,$" the original fact is really that $ \sin x$ might as well be $ x$ for $ x$ near $ 0.$ Divide that by $ \cos x,$ but that might as well be $ 1.$


Five socks, colored blue, brown, black, red, and purple are in a drawer. How many combinations of three socks can be randomly selected from the drawer?

This is simply $ \dbinom53\equal{}\boxed{10}$.

Anyways if you don’t like combinations
[hide]the ways to get the first sock is 5, second 4, third 3 so we do 5x4x3=60 now we divide it by 3! to get 10[/hide]

My solution:
[hide]The first part is 5, and the second is 3. Therefore, we have:
5*4*3/3*2*1. This gives us 10.[/hide]


We have $ (x\plus{}y^2)(x^3\plus{}y^4)\equal{}z^8\equal{}(x^2\plus{}y^3)^2\Leftrightarrow x^3y^2\plus{}xy^4\equal{}2x^2y^3\Leftrightarrow xy^2(x\minus{}y)^2\equal{}0$
-) $ x\equal{}0$ or $ y\equal{}0$: Easy!
-) $ x\equal{}y$ then $ x\plus{}x^2\equal{}z^3$ and $ x^2\plus{}x^3\equal{}z^4$, it mean $ (x\plus{}x^2)^4\equal{}(x^2\plus{}x^3)^3$. OK!


factorial
inequalities
vector
induction
number theory solved
number theory

An=product(2^n-2^k, k=0..(n-1))

how do you prove that n! divides An
for every n ??

[color=red][Admin Edit][/color]

the sum is like so:

2^n-1+2^n-2+2^n-2^2+...+2^n-2^(n-1)?

because I think it is wrong!  try n=2 and 2 does not divide 5!

I hope I am not mistaking!

cheers!

i was a little tired ..

An=PRODUCT(2^n-2^k, k=0..(n-1))

how do you prove that n! divides An
for every n ??

OK, I'm really not so sure about this one, but here it goes anyway :D

First of all, we know that the power of a prime p in n! is [n/p]+[n/p^2]+...
We denote by Pow(p) the power of p in n!. The power of 2 in An is 1+2+..+n-1=n(n-1)/2, which is obviously greater than [n/2]+[n/2^2]+.., so 2^Pow(2)/An. Now we have to prove that n!/2^Pow(2) divides Bn= product from k=1 to n of (2^k-1), which is actually An divided by the greatest possible power of 2.

We know that for an odd prime p, p/2^(k*(p-1))-1. Now let's take an odd prime p/n!. If we prove that p^Pow(p)/Bn for any p, then it's finished.
IF we could prove that Pow(p) <= [n/(p-1)] then it would be finished, because there are [n/(p-1)] numbers  smaller or equal than n which are divisible with p-1, so we would have more factors in Bn which are divisible with p (factors of the form 2^(k*(p-1))-1, with k*(p-1)<=n) than Pow(p), so we would have what we want. Any idea about how to prove that inequality? Notice that Pow(p)= [n/p]+[n/p^2]+.. and n/p+ n/p^2+.. is, in fact, equal to n/p-1 (just compute the infinite sum of the terms in geometric progression). Maybe this helps. What do you think? Is this right?

Thanks! And here's another way:

As I said, n/(p-1)=n/p+n/p^2+..., so [n/(p-1)]=[n/p+n/p^2+...] so [n/(p-1)] >= [[n/p]+[n/p^2]+..] = [n/p]+[n/p^2]+... I should have seen this before... :D. Besides, it's obvious that [x+y] >= [x]+[y].

This is a funny solution:
Consider the (Z_2)^n space and count the number of ways you can choose a system {x1,x2,...,xn} of n nonzero points in that space so that we cannot find two distinct subsets with equal sum (e.g. a base for the vector space Z_2^n).
One can chose x1 in 2^n-1 ways (all but 0)
x2 in 2^n-2^1 (all but 0 and x1)
.
.
.
xk in 2^n-2^(k-1) ways (all but 0,x1,x2,x1+x2,x3,x1+x3,x2+x3,x1+x2+x3,...).
In the end this numbers must be multiplied and divided by n! (the numbers of apparences of each system).
The result must be an integer since you cannot pick 1.5 whole objects out of a box.

You can prove a simmilar result by replacing 2 by any prime p. (Of course using my previous wonderful proof)

finally you put that article to good use Mihai :D nice soln, keep up the good job!

I didn't all unterstand in Mihai proof. Can someone
explain me ?

On popular demand here is a more explicit form of my previous proof.

Let Z_2={0,1}. On Z_2 define + by the following addition rules:
0+1=1+0=1 and 1+1=0+0=0.

Let A=(Z_2)^n=(Z_2)*(Z_2)*...*(Z_2)= {(x1,x2,...,xn)| xi are 0 or 1 for all i}. On A define + by (x1, x2, ..., xn) + (y1, y2, ..., yn) = (x1+y1, x2+y2, ..., xn+yn).

For all x in Z_2 or A it is obvious that x+x=0 (or (0,0,...,0)) respectively). So x=-x, where - is the inverse operation of +. It is also obvious that x+0=x.

Define S={{s1,s2,..,sn} |so that si are all in A and for each non-void subset T of {s1,s2,..,sn}, the sum of all elements of T is not (0,0,...,0)}. (Because x=-x, this condition is equivalent to the one I used in my previous post reply). Notice that the elements of S are sets.
 Lets count the elements of S. It is the same as counting the number of possibilities in which a set r={s1, s2, ..., sn} can be chosen to respect the definition of S.
I will use 0 instead of (0, 0, ..., 0).
First notice r has no 0 element or the subset {0} can be chosen with sum 0.
I will prove by induction that sk can be chosen in (2^n)-(2^(k-1)) ways.
For k=1, we can choose for s1 all the elements of A but 0, so (2^n)-(2^0) possibilities result.
For the step from k to k+1 I prove that the already picked elements s1, s2, ..., sk, cancel 2^k possibilities for s(k+1).
The set {s1, s2, ..., sk} has 2^k possible subsets. To respect the definition of S, the sums of distinct subsets are distinct, or else their symmetrical difference (their reunion except for their intersection) would have 0 sum. Each sum of the elements of a subset of A is an element of A. Consider that the void set has sum 0. If s(k+1) would be one of the 2^k sums defined earlier, then s(k+1) would be si for some i<=k (absurd because r would not be a set  it has 2 identical elements} or the set defined as the reunion of the set whose sum is s(k+1) and s(k+1) would have sum s(k+1)+s(k+1)=0 (because x+x=0 for all x).
So s(k+1) is none of the 2^k sums. It is easy to see that if s(k+1) cannot be picked, then s(k+1) must be one of that sums. The conclusion is that s(k+1) can be chosen in (2^n)-(2^k) ways.
It is now clear that the number of ordered elements of S is ((2^n)-1)((2^n)-(2^1))...((2^n)-(2^(n-1))). The number of elements of S is the number of unordered elements of S. Each unordered element r of S appears counted n! times as an ordered set (for each permutation p of n elements, (s1, s2, ..., sn) and (sp(1), sp(2), ..., sp(n)) are counted).
So the number of elements of S is ((2^n)-1)((2^n)-(2^1))...((2^n)-(2^(n-1)))/n!. This must be an integer, so n! | ((2^n)-1)((2^n)-(2^1))...((2^n)-(2^(n-1)))

 When going from 2 to a prime p some of the elements of the proof must be changed. Z_p={0, 1, ..., p-1}. Define i+j=i+j (mod p).
Define A=(Z_p)^n in a similar manner (xi are elements of Z_p).
Define a scalar multiplication: kx=x+x+...+x (k times) where k is in Z_p and x in A.
2x=x+x is no longer 0, but px is.
The condition in the definition of S changes to: for each non-void subset T={t1, t2, ..., tl} of {s1, s2, ..., sn} there exists no system e1, e2, ..., el of nonzero elements of Z_p such that t1s1+t2s2+...+tlsl=0.
The induction hypothesis changes to: sk can be chosen in p^n-p^(k-1) ways.
To prove the step from k to k+1 I effectively construct the elements that s(k+1) cannot be. They are all the elements of the form e1s1+e2s2+...+eksk with ei in Z_p for all i in {1,...,k}. By the definition of S, similar to the proof for p=2, those sums are all distinct and it is easy to see that there are p^k elements of the form e1s1+e2s2+...+eksk when ei are in Z_p.
From this point to the end, the proofs are identical.

These proofs would have been easier to explain if vector spaces would have been elementary notions to all of you.
(Z_p)^n is a vector space over Z_p, and S is the set of all the bases of that vector space.

Hello mihai, thanx a lot for your explanation ;-)
It seems it also works with composite number, doesn't it?


trigonometry
geometry
circumcircle
symmetry
rotation
geometric transformation
reflection

A symmetrical figure composed of 3 squares with side 1 is enclosed (not necessarily inscribed) in a circle.  What is the smallest possible radius of the circle?

I've no clue how to begin.

What symmetrical figure? Is it like 3 out of four squares in a 2x2 chessboard? or what?

That's why it's hard for me.  You have to figure out what the figure is first.  All you know is it's composed of 3 congruent squares of side length 1.

all the squares have to be conected  to make the shape so its kind of a honeycomb shape then maybe you can use trig let me just sketch the figure..
here it is  : < asy > draw((0,0) - - (0,1)); draw((0,1) - - (1,1)); draw((1,1) - - (1,0)); draw((1,0) - - (0,0); dot((0,0)); dot((0,1));dot((1,1)); dot((1,0)) < /asy >

EDIT:
why won't the asymtote work?!

[quote="facis"]Are you trying to draw an equilateral triangle with a square attached to each side?[/quote]
That couldn't be right because you could draw a figure with the 3 squares all sharing 1 vertex and then the radius would be the square root of 2.

[quote="Poincare"]why won't the asymtote work?![/quote]
Because you need square brackets.  You should enclose them in [asy'] and [/asy'] (ignore the apostrophes).

[quote="kysuke"]That couldn't be right because you could draw a figure with the 3 squares all sharing 1 vertex and then the radius would be the square root of 2.[/quote]
Actually:
[asy]pair A=dir(0)/sqrt(3); pair B=A*dir(120); pair C=B*dir(120);
pair A1=A+dir(60); pair A2=A1+dir(150);
pair B1=B+dir(180); pair B2=B1+dir(270);
pair C1=C+dir(-60); pair C2=C1-dir(-150);
draw(A--B--C--A--A1--A2--B--B1--B2--C--C1--C2--cycle);
draw(Circle((0,0),sqrt((1+sqrt(3)/6)^2+0.25)));
dot((0,0));
draw((-1-sqrt(3)/6,0)--(0,0)--B1);[/asy]
You get a radius of $ \sqrt{\left(1\plus{}\frac{\sqrt{3}}{6}\right)^2\plus{}\left(\frac{1}{2}\right)^2}\approx 1.38 < \sqrt{2}\approx 1.41$.

But you can do better:

[asy]pair O=(0,0);
pair A=(-.5,-1), B=(.5,-1), C=(1,1), D=(-1,1);
pair X=(0,3/16);
pair L1=(-1,0), L2=(-.5,0), L3=-L2, L4=-L1;
pair T=(0,1);
draw(O--T--C--L4--L1--D--T);
draw(L2--A--B--L3);
draw(A--C--B);
dot(X);
draw(Circle(X,5*sqrt(17)/16));
draw(L2--(D+T)/2,dashed);
draw(L3--(C+T)/2,dashed);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);[/asy]
Note that $ AB\equal{}1$, $ AC\equal{}\frac{5}{2}$, and $ BC\equal{}\frac{\sqrt{17}}{2}$.  The area of $ \triangle ABC$ is 1 (the length of the altitude dropped to $ \overline{AB}$ is 2).  Then $ 1\equal{}\frac{abc}{4R}\Rightarrow abc\equal{}4R$, where $ R$ is the circumradius of $ \triangle ABC$, and therefore the radius of the circle.  This becomes $ \frac{5\sqrt{17}}{4}\equal{}4R\Rightarrow R\equal{}\frac{5\sqrt{17}}{16}\approx 1.29$.

I screwed up typing into my calculator, so I thought I could beat the above bound, but I figured that I might as well post this anyway.

[asy]real m = 0.5, t = 1/(2*3^.5); pointpen = black; pathpen = black+linewidth(0.6); D(shift( (-m/2,-1-m*t) ) *unitsquare); D(shift( (m/2,-m*t) )*rotate(30)*unitsquare); D(shift( (0,2m*t) )*rotate(150)*unitsquare); D(CR((0,0), ((1+m*t)^2+(1-m/2)^2)^.5 ));

D((0,0)--(0,-1-m*t)--(1-m/2,-1-m*t)--cycle,linetype("4 4"));[/asy]

Let the squares be positioned as shown above to form an equilateral triangle center (so it is symmetrical), with each side of the triangle having length $ m$. Then by the Pythagorean Theorem on the dashed triangle, we can find the radius of the circle to be $ \sqrt{\left(1\plus{}\frac{m}{2\sqrt{3}}\right)^2 \plus{} \left(1\minus{}\frac m2\right)^2} \equal{} \sqrt{\frac {m^2}3 \plus{} \left(\frac 1{\sqrt{3}} \minus{} 1 \right)m \plus{} 2}$. This minimizes at $ m \equal{} \frac{3\minus{}\sqrt{3}}{2} \Longrightarrow R \approx 1.36$.

[quote="lingomaniac88"]Well, you did beat my radius for an arrangement with an equilateral triangle in the center.  In fact, you minimized it. :)[/quote]

actually when minimizing the equilateral triangle as small as possible, it results the second image.

AZJPS's solution is incorrect since it's not symmetrical.

Lingomaniac's second solution is correct.  I only knew the multiple choice answers and had no solution and this question was driving me crazy.  Thanks!

what? azpjs's solution is definitely [i]more[/i] symmetrical than the other one :wink:

unless of course you mean reflective symmetry

not_trig, I think you are confusing symmetry with the ability to rotate a figure onto itself, which is not necessarily symmetry.

Typically, a symmetry of a set refers to an action (esp. a rigid motion on a subset of the plane) such that the image of the set under the action is the same as the set we started with.  Following the parenthetical in the previous sentence, common examples of symmetries are the $ 2n$ symmetries of a regular $ n$-gon, $ n$ of which are rotations and $ n$ of which are reflections, or symmetries of reflection through some point (which in the plane correspond to rotations by 180 degrees).  To say that something is symmetric is to say that it possesses symmetries.  I have never heard the word symmetric used to mean "possessing mirror, but not rotational, symmetries," and if that is what the author intended then it's a pretty awful question, in my opinion.

Ok so I guess I was wrong - I was thinking that the symmetry referred to is reflections, not rotations.  I'm not sure why I was saying that rotations are not a form of symmetry... In any case, the question doesn't say how symmetrical the shape is, just that it is symmetrical, and asks us to find the minimum circumradius.  Of course, there is only one minimum, but the question is not especially clear about how the squares are arranged.

This was a Mu Alpha Theta question from the recent national convention.  The question only said symmetrical which I guess everyone was to assume meant reflectional symmetry.  I wish I had caught that so I could dispute the question so everyone else would get it wrong.  That is after all the best part of Mu Alpha Theta.

[quote="facis"]I don't think you were supposed to assume that it means reflectional and not rotational symmetry; I think you were expected to figure out that the specific arrangement that minimized the circumradius had reflective instead of rotational symmetry.[/quote]

Oh wait.  I misread 1.36 as 1.26 in the rotational symmetry solution.  Carry on.

Of course, the asymmetric solutions are much more interesting than the symmetric ones.  Just look at all the pretty pictures [url=http://mathworld.wolfram.com/SquarePacking.html]here[/url], for example.


[quote="b-flat"]I noticed that there was a 404 error earlier when I was posting...

Does anyone know why?[/quote]Yes, I was updating the site. During the update some of the files were incomplete thus the 404 error.


inequalities
geometry
geometric transformation
reflection
inequalities proposed

Let $x, y, z$ three nonnegative numbers. Prove that:

                        \[ x^{4}(y+z)+y^{4}(z+x)+z^{4}(x+y)\leq\frac{1}{12} (x+y+z)^{5}. \]

Can you give the solution for this problem ??? :|

[b]Please don't post anything that has recently appeared in journals.[/b]

This inequality is found in [url=http://reflections.awesomemath.org/]Mathematical Reflections[/url], and we should wait until at least March 1, 2006 before discussing it.

Don't worry, bilzhao! I had no intention in posting a solution, although the solution I have found is not a nice one. :)  ;)

That's not the point.  People shouldn't be able to post answers to journal problems here.

Topic Locked.


I really want to see proof ineq Nesbit case n = 7,8,9
Can anyone post solution?

Maybe he meant 7,8,9 variables? But he sure didn't mean that above...

Well, apply Cauchy Schwarz

$ A\equal{}\sum_{i\equal{}1}^n \frac{a_i}{a_{i\plus{}1}\plus{}a_{i\plus{}2}}\ge \frac{n}{2}$ - the inequality

$ A(\sum_{i\equal{}1}^n a_ia_{i\plus{}1}\plus{}a_ia_{i\plus{}2})\ge (\sum_{i\equal{}1}^n a_i)^2$, and now use $ \sum_{i\equal{}1}^n (a_i\minus{}a_{i\plus{}1})^2\ge 0$, in many forms... :D

That should be it, if I have done it correctly


A cryptographic code is designed as follows. The first time a letter appears in a given message it is replaced by the letter that is $ 1$ place to its right in the alphabet (assuming that the letter $ A$ is one place to the right of the letter $ Z$). The second time this same letter appears in the given message, it is replaced by the letter that is $ 1\plus{}2$ places to the right, the third time it is replaced by the letter that is $ 1 \plus{} 2 \plus{} 3$ places to the right, and so on. For example, with this code the word "banana" becomes "cbodqg". What letter will replace the last letter $ \text{s}$ in the message "Lee's sis is a Mississippi miss, Chriss!"?

$ \textbf{(A)}\ \text{g}\qquad
\textbf{(B)}\ \text{h}\qquad
\textbf{(C)}\ \text{o}\qquad
\textbf{(D)}\ \text{s}\qquad
\textbf{(E)}\ \text{t}$

[hide="Click for solution"]
There are $ 1 \plus{} 2 \plus{} 1 \plus{} 4 \plus{} 2 \plus{} 2 \equal{} 12$ s's in total, so we shift it over $ 1 \plus{} 2 \plus{} 3 \plus{} \ldots \plus{} 12 \equal{} 78$ over.  $ 78 \equiv 0 \pmod{26}$, so it is the same letter, or $ \boxed{\textbf{(D)}}$.
[/hide]

[hide="Solution"]s is the 19th letter of the alphabet. With 12 s's in the sentence, we have $ \frac{12(13)}{2}\equal{}78\plus{}19\equiv 19\pmod{26}$.
So the answer ends up being $ \boxed{\textbf{(D)}\ \text{s}}$.[/hide]


[quote="Inequalities Master"]Let $ a,b\in \mathbb N^*$. Prove that:

$  (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^b + (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^a\ge (a^a + b^b )^{a + b + 1}$[/quote]

It's wrong; try $ a=1$, $ b=2$.

[quote="Mij"][quote="Inequalities Master"]Let $ a,b\in \mathbb N^*$. Prove that:

$  (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^b + (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^a\ge (a^a + b^b )^{a + b + 1}$[/quote]

It's wrong; try $ a = 1$, $ b = 2$.[/quote]

I'm very sorry. The correct enunciation is:

Let $ a,b\in \mathbb N^*$. Prove that:

$  (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^b + (a^{a + 1} + b^{b + 1})^b(a^a + b^b)^a\ge (a^a + b^b )^{a + b + 1}$

Please, everyone, try to prove it.

[quote="Inequalities Master"]Let $ a,b\in \mathbb N^*$. Prove that:

$  (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^b + (a^{a + 1} + b^{b + 1})^a(a^a + b^b)^a\ge (a^a + b^b )^{a + b + 1}$[/quote]

lemma)$ x\ge a\ge b\ge y>0$, $ a,b$are nature numbers, $ x+y=a+b\Rightarrow x^a+y^b\ge a^a+ b^b$
proof)
$ x^a-a^a\ge b^b-y^b\Leftrightarrow (x-a)(x^{a-1}+\cdots+a^{a-1})\ge(b-y)(b^{b-1}+\cdots+y^{b-1})$,
It is obvious because $ x+y=a+b$ and $ a,b\ge1$

Proof of problem)
Assume $ a\ge b$.
Probelm$ \Leftrightarrow (\frac{a^{a+1}+b^{b+1}}{a^a+b^b})^a+ (\frac{a^{a+1}+b^{b+1}}{a^a+b^b})^b\ge a^a+b^b$
$ (\frac{a^{a+1}+b^{b+1}}{a^a+b^b})^a+ (\frac{a^{a+1}+b^{b+1}}{a^a+b^b})^b$
$ \ge(\frac{a^{a+1}+b^{b+1}}{a^a+b^b})^a+ (\frac{ab(a^{a-1}+b^{b-1})}{a^a+b^b})^b\ge a^a+b^b$
Last inequality is true by lemma.

The solution is wrong, since $ \frac{a^{a+1}+b^{b+1}}{a^{a}+b^{b}}\le a $


Find the integer solutions $ (x,y)$ of the equation $ (x+y)k=xy$ if $ k$ is a prime number, and $ x$,$ y \in \mathbb{Z}$.

[quote="N.T.TUAN"][hide="hint"]It is equivalent to $ (x-k)(y-k)=k^{2}$.  :wink: [/hide][/quote]

Since $ (x-k)\in\mathbb{Z}$ and $ (y-k)\in\mathbb{Z}$, hence the solutions are:

$ (x-k) = k^{2}$ and $ (y-k) = 1$
$ (x-k) = k$ and $ (y-k) = k$
$ (x-k) = 1$ and $ (y-k) = k^{2}$

Therefore $ (x,y)=(k^{2}+k,k+1),(2k,2k),(k+1,k^{2}+k)$

Hmm, if $ x-k<0$ and $ y-k<0$ then...?  :maybe:


inequalities
trigonometry
geometric inequality
inequalities proposed

[color=purple]Let triangle ABC and denote its element practice. Prove that
   $\sqrt{4+\frac{r}{R}}\le \sqrt{\frac{\cos\frac{A}{2}}{\cos\frac{B}{2}\cos\frac{C}{2}}}+\sqrt{\frac{\cos\frac{B}{2}}{\cos\frac{C}{2}\cos\frac{A}{2}}}+\sqrt{\frac{\cos\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}}}\le \sqrt{\frac{5R}{2r}-\frac{1}{2}}$
   Hungkhtn i know you hate trigonometric inequality and geometric inequality but i hope the following inequality can change your conception and if you want try with GI 18 :D [/color]

Thank you Evarist. I will try but not promise when I could complete it.  :lol:

few comment: Do mean that this is hard inequality? I not think so. it's only combinations some simpl inequalitys in trigonometry. therefore it's not interesting and there are not so original idea. if some one have much free time then they can solve it :wink:

[color=purple]Ok , post your solution please and post the stronger if you can :) [/color]


Which do you use, AOPS or Mathlinks?
I use AOPS.

I like MathLinks. The colors are better.  :)

I use AOPS. :)

I am adding this line because this post is too short ;)

Agreed, too.
Mathlinks looks better (and Aops caused some problems with my PC one year ago...)!

AoPS!  I've used it ever since I became a member before the merge...never felt the urge to switch  :P  ;)

I like AoPS. I don't really know why :roll:

i use aops. i switched to mathlinks once to see what it was like and i didn't really like it, so aops it is.

aops i dont know i guess its just more popular around here.

$\mathcal{M}ath\mathcal{L}inks$ R00Lz

AoPS skin seems somehow overloaded.

[quote="math92"]i made a topic like this once...
i  use mathlinks[/quote]

Chances are, it was quite some time ago, and we're not supposed to revive month+ topics.


x + y = 11
x - y = -7
x = y - 7

y - 7 + y = 11
2y = 18
y=9
x=2

The order of the matrix is 2 x 2

[hide]$x+y=11\Rightarrow y=11-x$.

Now, substitute this into the other equation to get

$x-(11-x)=-7\Rightarrow2x=4\Rightarrow x=2$

Now that we have $x$, substitute back in to solve for $y$.

$2+y=11\Rightarrow y=9$

$(2,9)$

The other one, it is $2\times2$[/hide]

[quote="jenny221"]1.) Solve using the substitution method x + y = 11, and x - y = -7

2.) 12.	Determine the order of the matrix [ 4   6 ]
                   		                     -5 –3
                             (The -5 and –3 are included in the matrix)[/quote]

[hide]$x+y=11$
$x-y=-7$

$y=x+7$
$2x+7=11$
$x=2$
$y=9$
[b](how is this wrong?)[/b][/hide]


calculus
integration
number theory unsolved
number theory

it seems like an open problem to me, at least i am sure it uses advanced algebraic number theory.

[quote="kiemkhach"]find $n \in Z$ such that $x^2-y^3=n$ has a integral solution.[/quote]

Well, I think you need to uses a stronger language to define the problem, because takin $x=0$ and $y=-1$ you get that one $n$ is $n=1$, plese explain better the problem, if not this problem ... is for newbies. ;)

RX TCM

[quote="kiemkhach"]find $n \in Z$ such that $x^2-y^3=n$ has a integral solution.[/quote]

Good luck.  See    http://arXiv.org/abs/math/0005139

Also relevant is Tunnell's solution (conditional on Birch-Swinnerton-Dyer conjecture for elliptic curves) of the Congruent Number Problem, i.e. which $n$ are values of $x^3 + y^3$.


Let $P$ and $Q$ be two isogonal conjugates WRT $\triangle ABC$ and let $\triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c$ be the pedal triangles of $P,Q$ WRT $\triangle ABC.$ Show that the side triangle of $\triangle P_aP_bP_c$ and $\triangle Q_aQ_bQ_c$ is perspective with $\triangle ABC$ with perspector the infinite point of $\perp PQ.$

See [url=http://www.xtec.cat/~qcastell/ttw/ttweng/definicions/d_lateral_tr.html]this link[/url] for a definition of the side triangle.

Dear Luis and Mathlinkers,
thanks for evacating so nice situation.
The case with two pedal triangle coming from two isogonal points has been evocated in my article

http://perso.orange.fr/jl.ayme   vol. 2  Les deux points de Schroeter, see my "commentaire p. 5". (the proof is the same)

Sincerely
Jean-Louis

Thanks dear Jean-Louis, your article is very nice. I note that this triangle has lots of  projective properties. I have a different approach to this one. Define the points

$ M \equiv{} AQ \cap P_bP_c ,\ N \equiv{} AP \cap Q_bQ_c , \ K \equiv{} AQ \cap Q_bQ_c , \ L \equiv{} AP \cap P_bP_c.$

Let $ \triangle A'B'C'$ be the side triangle of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c.$ It is clear that $ AM$ and  $ AN$ are perpendicular to  $ P_bP_c$ and  $ Q_cQ_b.$ Thus, $ A'$ is the orthocenter of $ \triangle AKL.$ Then from $ \triangle AQQ_c \sim \triangle APP_b$ and $ \triangle AKQ_c \sim \triangle ALP_b,$  it follows that $ KL \parallel PQ$  as a consequence of  Thales theorem. Therefore, $ AA' \perp PQ$ and cyclically $ BB',CC'$ are perpendicular to $ PQ$ $\Longrightarrow$ $ \triangle ABC$ and $ \triangle A'B'C'$ are perspective with perspector at the infinite point of the direction $\perp PQ.$

Let $ \triangle A'B'C'$ be the side triangle of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c.$. Let $O$ be the center of the pedal circle, if $J \equiv OP_b \cap QQ_b, K \equiv OP_c \cap QQ_c, X \equiv Q_bP_c \cap Q_cP_b$ Now we apply Pascal's Theorem on $P_c,P_b,Q_c,Q_b,J,K \Rightarrow X \in OQ$ now by Brocard's Theorem $OX \perp AA' \Rightarrow PQ \perp AA'$ analogously $PQ \perp BB'$ and $PQ \perp CC'$ this is sufficient.

[b]Another proof[/b]

Let $ \triangle A'B'C'$ be the side triangle of $ \triangle P_aP_bP_c$ and $ \triangle Q_aQ_bQ_c.$, denote $\mathcal{P} (X)$ the powe of $X$ WRT the pedal circle of $\{Q,P\}$.Let $W$ be the center of $\odot (QQ_bQ_cA)$
Since $AA'^2+QA'^2=2QW^2+2WA'^2=2\mathcal{P}(A')+QA^2 \Rightarrow QA'^2-QA^2=2\mathcal{P}(A')-AA'^2=PA'^2-PA^2 \Rightarrow PQ \perp AA'$ analogously $PQ \perp BB'$ and $PQ \perp CC'$ this is sufficient.


Given $f(x)=x^{3}-3x^{2}+1$ where $\alpha, \beta, \gamma$ are its roots. Find a polynomial with $\frac{\alpha}{\alpha-2}, \frac{\beta}{\beta-2}, \frac{\gamma}{\gamma-2}$ as its roots.

Hence, or otherwise, find:

1. $(\alpha-2)(\beta-2)(\gamma-2)$

2. $\alpha(\beta-2)(\gamma-2)+\beta(\gamma-2)(\alpha-2)+\gamma(\alpha-2)(\beta-2)$


function
analytic geometry
graphing lines
slope
geometry
perpendicular bisector
real analysis

if you take a function f(x) = x^n, chose an arbitrary point P on the curve and find the perpendicular bisector of the line OP, then find the intercept of this bisector with the y-axis (0,k), what happens to k as P approaches the origin ? How does this change as n increases?

For n=2 I found the limit was 1/2, but couldn't find a limit for n=3 or 4.

I am trying to find (and prove) something about the existence of limits for n>2, but have only recently learnt about limits and I'm not very confident with them, and especially with rigorous proofs involving them, but I think that no limit exists..

Can anyone point me in the right direction if they understand the question

thanks

For $n>2$ and $n$ even, the intercept tends to $\infty$. For $n$ odd, the intercept tends to $\infty$ or $-\infty$, depending on which side of the origin $P$ is on. Whether you want to call this existence of a limit or not is up to you.

Let $P=(a,a^n)$. The slope of $OP$ is $a^{n-1}$, so the perpendicular bisector has slope $-\frac{a}{a^n}$ and passes through $(\frac a2,\frac{a^n}2)$.
That line has equation $a^ny+ax=\frac12(a^2+a^{2n})$, so its $y$-intercept is at $\frac{a^2+a^{2n}}{2a^n}=\frac1{2a^{n-2}}+\frac{a^n}2$.
The second term clearly goes to zero as $a\to0$. The first term is $\frac12$ when $n=2$ and blows up when $n>2$.

ah, that's exactly what I was searching for! thanks

so I guess no "limit" exist since infinity cannot be a limit.. just struck me as extremely odd that the limit for x^2 was 1/2. huh!


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a+b+c=1.$ Prove that
 $ \sqrt{a+2b}+\sqrt{b+2c}+\sqrt{c+2a}\geq\sqrt{5+12(ab+ac+bc)}.$

[quote="kunny"]That's the proof for your inequality, arqady.[/quote]
Nice, kunny. :coolspeak:


function
group theory
abstract algebra
number theory
number theory proposed

Find all real-valued functions on the positive integers such that $f(x + 1019) = f(x)$ for all $x$, and $f(xy) = f(x) f(y)$ for all $x,y$.

Assuming that 1019 is prime:
the value depends only on the residue $\mod 1019$.
take some primitive root $\zeta \mod 1019$ and define $f(\zeta)$ arbitrary; then all residue classes coprime to $1019$ are already defined. Also $f(\zeta^{1019})=f(\zeta)$, so $f(\zeta) \in \{ 0 , \pm 1\}$ since it's real and all other solutions are complex.
Now for $x$ being a multiple of $1019$, $f(x)$ can be defined as $0$ or $1$ (but everytime the same) when $f(\zeta)=1$, but must equal $0$ otherwise.

i am pretty sure it's f(x) = 0 or f(x) = 1 except for 1019, f(1019) = 1 or 0 regardless of the other function..
if x = y = 1, subst in 2nd equation, we get f(1) = 0, or f(1) = 1.
If f(1)=0,then f(a) = 0 from 2nd equation, that's the first answer.
Now look at x = y = 1019. Get f(1019^2)=f(1019).
So, f(1019)= 1 or 0.
If it is 1, then f(a) * f(1019) = f(1019 * a), or
f(1019) * f(a) = f(1019), so f(a) = 1.
if f(1019) = 0, then this case is bad because 1019 is prime. :(
[hide="But now this part is ownage."]
]Use Fermat's little theorem, so for every k not divisible by 1019, have:
(f(k))^1018 = f(number with 1 mod1019) = f(1). So f(k)^1018 = 1 looking back at the beginning of the solution.
Can f(k) be negative?No, because for some l, look at the first equation and get f(l*1019 + k) = f(k), but for some l, l*1019 + k is a perfect square and so:
f(l*1019 + k) = (f(i))^2 , which is positive.
So f(k) is positive, (f(k))^1018 = 1, so f(k) = 1. Thus get the answer.[/hide]

[quote="rem"]but for some l, l*1019 + k is a perfect square[/quote]
no. E.g. $k=7$ is no square $\mod 1019$.

All the solutions I gave above should work, and they should be all.

ZetaX: Recapitulating we have
\[ f ( x ) = \left\{\begin{array}{l} 0, \mbox{ when } f ( \zeta ) = 0\\ 1, \mbox{ when } f ( \zeta ) = 1\\ \left( \frac{x}{1019} \right), \mbox{ when } f ( \zeta ) = - 1 \end{array}\right.  \]
where $\left( \frac{x}{p} \right)$ is the legendre symbol. Right ?

Didn't you forgot the $f ( \zeta ) = - 1$ case ?

Right, I think there are four solutions: the solutions are two constant ones ($0,1$) and the Legendre symbol with $f (0 \mod 1019)=0$. Then there is one more solution with $f(x)=1$ for all $x$ coprime to $1019$ and $0$ otherwise.
But which one is missing¿

[i]Par ailleurs[/i], isn't the number of solutions, equal to the number of multiplicative subgroups of $\mathbb{Z}_{1019}$ ? :)
That would be quite nice :)

Yes there are four multiplicative subgroups here since it's a Sophie-Germain-prime, but that should be just random:

Well, for a general prime $p$, the number of solutions will be $4$ again, having the structure (and same proof as) above.
It's interesting to look at the number of solutions when the values are also allowed to be complex, then there are $p+1$ solutions (the equation $x^p-x$ has $p$ solutions, and $f(0 \mod p)=0$ is necessary when $f(\zeta) \neq 1$; when $f(\zeta)=1$, then there are again the two choices for $f(0 \mod p)$ as above).

When $p$ isn't prime, it gets interesting, but the same methode (that of describing the multiplicative structure $\mod p$) should solve it again.

I am so confused. What about FLT which leads to f(k) = 1 or -1 if f(1019) = 0 for every natural k not divisible by 1019. :?
then, for every number k, for which for some m , 1019m + k is a perfect square, f(k) = 1.Let such numbers be in set A. For the rest, if for some i,f(i) -1, multiply f(i) by f(h) for h belonging to set A, and i am pretty sure the result mod 1019 will for some h be a number k belonging to A. So , get f(i) = 1, which is a contradiction. So, if f(1019) = 0, then f(x)=1 if x isn't divisible by 1019.
This is such a cool problem!

[quote="rem"] For the rest, if for some i,f(i) -1, multiply f(i) by f(h) for h belonging to set A, and i am pretty sure the result mod 1019 will for some h be a number k belonging to A[/quote]
No, it will never belong to that set A; see the part about Legendre/Jacobi symbols in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=76610]Formulary[/url]  :)


superior algebra
superior algebra unsolved

Let $ (A,\mathfrak{m})$ be a local CM ring. If $ ht \ I \equal{} r$ then we can take $ a_1,...,a_r \in I$ such that $ ht(x_1,...,x_i) \equal{} i$ for $ 1 \leq i \leq r$.

The statement is written before the 'unmixedness theorem" or whatsoever was introduced...


modular arithmetic
function
Euler
number theory
totient function
greatest common divisor

hi
     i want to find the last 10 digits in
[code]
 a^(a^(a^(a^(a .... z times where z < 1000 and a<1000 , both z and a are postive integers.
[/code]
 i couldn't think of any approach . Can please tell me how to go about doing this. A little explanation would help me a long way.

An example is 2 and z =4 , last 4 digits are

2^2^2^2= 2^16=5536 , when the z increases the problem arises.

We want to find $ \overbrace{a^{a^{a^{a^{\cdots ^a}}}}}^{z \text{ times}} \pmod{10^{10} \equal{} 2^{10} \cdot 5^{10}}$, or, at least, a form with smaller powers. To do this, we will repeatedly apply Euler's totient function, noting that $ \varphi(2^n) \equal{} 2^{n \minus{} 1}$ and $ \varphi(2^j \cdot 5^k) \equal{} 2^{j \plus{} 1} \cdot 5^{k \minus{} 1}$. This will allow us to find the last digits for all $ a$ such that $ (a,5) \equal{} (a, 2) \equal{} 1$.

$ a^{a^{a^{a^{\cdots ^{\text{continued below...}}}} \pmod{2^{12}\cdot 5^{8}}} \pmod{2^{11}\cdot 5^{9}}} \pmod{2^{10}\cdot 5^{10}}$
The pattern that you see in the mods continues, until the multiples of 5 vanish and we are left with $ \bmod{2^{20}}$. Now the powers of 2 in the mods start dropping...
$ \dots^{a^{a^{a^{\text{continued below...}}} \pmod{2^{20}}} \pmod{2^{19}\cdot 5}}$
... until they reach $ 2$.
$ \dots^{a^{a^{a^{\dots} \pmod{2}} \pmod{2^2}}\pmod{2^3}}$
However, we can only find the last digits for $ a$ such that $ (a,2) \equal{} 1$ -- in other words, only for odd $ a$, forcing this last term to be congruent to $ 1$. This means that the last 10 digits are completely determined by the first 29 hyperexponentiations. This is true more generally -- the last $ n$ digits of a power tower are completely determined by the first $ 3n \minus{} 1$ hyperexponentiations (of course, only with $ (a,2) \equal{} (a,5) \equal{} 1$).

This is, of course, of somewhat limited use, since $ z \equal{} 29$ usually results in a very large number.

thanks akessler,

what i'm unable to understand is how do we get a^a^a... mod 2^2*5^2
= a^a.... mod ( 2^3*5), we can only apply a^phi(n) = 1 mod n if gcd(a,n)=1 ? can you please explain that a bit how we got that even though a=4,2 or 6.
Can you give me some links to read upon for the above.

Also where can i find more information about

[quote] "this is true more generally -- the last  digits of a power tower are completely determined by the first n  hyperexponentiations" [/quote]

wanted to read more about this quote .

[quote="tryitn1"]thanks akessler,

what i'm unable to understand is how do we get a^a^a... mod 2^2*5^2
= a^a.... mod ( 2^3*5), we can only apply a^phi(n) = 1 mod n if gcd(a,n)=1 ? can you please explain that a bit how we got that even though a=4,2 or 6.
Can you give me some links to read upon for the above.
[/quote]
I'm sorry, I can't understand what you're asking.
We can only use the theorem that the last $ n$ digits of a power tower are determined by the first $ 3n \minus{} 1$ hyperexponentiations in the case that $ (a,2) \equal{} (a,5) \equal{} 1$, since we use Euler's totient to prove it. The theorem might be true in the more general case, but I'm not totally sure how one would go about proving it then.

I can't give you any specific links, because I've never seen this somewhere else. I came up with that theorem after reading your question. The Wikipedia article on [url=http://en.wikipedia.org/wiki/Tetration]Tetration[/url] may help though.

1) What is (a,2)= (a,5) =1 , i think this is what is confusing me.

2)

 We got phi(100) = 40
  (7^(7^7) ) mod (100) = 7^(7^7 mod 40) mod 100 , This is fine because a^phi(n) =1 mod (n) if gcd(a,n) is 1,
as per euler totient property.But what if instead of 7 we have 6 ? 6^phi(n) =1 mod (n) is never true because gcd(6,n)=2 , n is like 10,100 etc
so how do we get 2^j+1 . 5^k-1 if  gcd(a,10^k) > 1

[quote="tryitn1"]1) What is (a,2)= (a,5) =1 , i think this is what is confusing me.[/quote]
$ (a,b)$ denotes the greatest common factor of $ a$ and $ b$. So, $ (a, b) \equal{} 1$ means that $ a$ and $ b$ are relatively prime.

[quote="tryitn1"]
2) We got phi(100) = 40
  (7^(7^7) ) mod (100) = 7^(7^7 mod 40) mod 100 , This is fine because a^phi(n) =1 mod (n) if gcd(a,n) is 1,
as per euler totient property.But what if instead of 7 we have 6 ? 6^phi(n) =1 mod (n) is never true because gcd(6,n)=2 , n is like 10,100 etc
so how do we get 2^j+1 . 5^k-1 if  gcd(a,10^k) > 1[/quote]
Well, we can't use Euler's totient if $ a$ and $ 10^{10}$ aren't relatively prime. I don't know how you would "reduce" the power tower in this case.


limit
algebra
polynomial
calculus
real analysis
real analysis unsolved

$ \lim_{x\rightarrow a^2} \frac{\sqrt{x}\plus{}x\minus{}a\minus{}a^2}{x\sqrt{x}\plus{}x\sqrt{a}\minus{}a^2\sqrt{x}\minus{}a^2\sqrt{a}}$ with $ a>0$

First suggestion: make the substitution $ u\equal{}\sqrt{x}$ so that $ x\equal{}u^2$ and this becomes a limit as $ u\to a.$

Second suggestion: that would make it a quotient of polynomials, and a $ \frac00$ case. When that happens, the numerator and denominator must have a common factor of $ (u\minus{}a).$ Factor that out.

Third suggestion: give your posts titles that actually indicate some of the content- and no, "limit" doesn't say much of anything in the calculus forum.


AMC
USA(J)MO
USAMO
quadratics
algebra
quadratic formula

Find all solutions to [tex](m^2 + n)(m + n^2) = (m - n)^3[/tex], where m and n are non-zero integers.

Since no one is going to try this . Here is my solution. But I have never taken a proof contest before, can someone tell me how many points i would've got(out of 7)?

[hide]Expand both side to get:

divide by n to get

Applying Quadratic Formula

[tex]n=\frac{3m-m^2 \pm \sqrt{m^4-6m^3+9m^2-8m-24m^2}}{4}[/tex]

If n is an integer,

 must be a perfect square.

If m=-1, then we have n=-1.

Else , where p is an integer.

Complete the square to get

(m-4)^2=p^2+16

(m-4+p)(m-4-p)=16.

Both m-4+p and m-4-p both must be even divisors of 16.

If m-4+p=a, m-4-p=b, then

If m-4+p=4, then m-4-p=4. Thus m=8. Substitute into quadratic formula to get n=-10.

If m-4+p=-4, then m-4-p=-4. So m=0, but m can't be 0 by definition.

If m-4+p=2, m-4-p=8. m=9. Subsititute into quadratic formula to get n=-6, -21.

If m-4+p=-2, m-4-p=-8, then m=-1, which gives us n=-1

Thus all solutions for (m, n) are (-1, -1), (8, -10), (9, -6), (9, -21).[/hide]

[quote="beta"]Since no one is going to try this :(.[/quote]
Well, I was going to try to finish it - I have a partial solution typed, but I had to go last night before I could finish. Oh well.

When the post is older than you and it gets bumped


geometry
circumcircle
inequalities proposed
inequalities

Let:$ a^2\plus{}b^2\equal{}c^2\plus{}d^2\equal{}5$.Prove that:
$ (5\minus{}a\minus{}2b)(5\minus{}c\minus{}2d)(5\minus{}ac\minus{}bd) \le 422$

[color=blue]Let $ a^2 \plus{} b^2 \equal{} c^2 \plus{} d^2 \equal{} 5$. Find the maximum of the expression $ E \equal{} (5 \minus{} a \minus{} 2b)(5 \minus{} c \minus{} 2d)(5 \minus{} ac \minus{} bd)$.[/color]

Consider the circle $ \Gamma$ of center $ O(0,0)$ and radius $ R \equal{} \sqrt {5}$, and the points $ P(a,b)$, $ Q(c,d)$ and $ X(1,2)$, clearly situated on $ \Gamma$. Under the given conditions we have
\[ 5 \minus{} a \minus{} 2b \equal{} \frac {1} {2} \left ( (a \minus{} 1)^2 \plus{} (b \minus{} 2)^2 \right ) \equal{} \frac {1} {2} XP^2,\]

\[ 5 \minus{} c \minus{} 2d \equal{} \frac {1} {2} \left ( (c \minus{} 1)^2 \plus{} (d \minus{} 2)^2 \right ) \equal{} \frac {1} {2} XQ^2,\]

\[ 5 \minus{} ac \minus{} bd \equal{} \frac {1} {2} \left ( (a \minus{} c)^2 \plus{} (b \minus{} d)^2 \right ) \equal{} \frac {1} {2} PQ^2.\]
Then $ E \equal{} \frac {1} {8} (XP\cdot XQ\cdot PQ)^2 \equal{} \frac {1} {8} (4R\cdot [XPQ])^2 \leq \frac {1} {8} \left ( 4R\cdot \frac {3\sqrt {3}} {4} R^2 \right )^2$. Substituting $ R$ yields $ E \leq \left ( \frac {3\cdot 5} {2} \right )^3 \equal{} 422 \minus{} \frac {1} {8} < 422$.
We have used the known formula that links the area $ [XPQ]$ of the triangle $ \Delta XPQ$ to its circumradius and the product of the lengths of its sides, and the as-well-known fact that the maximum area of an inscribed triangle is that of the equilateral one.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57609


How many positive perfect squares less than $10^{6}$ are multiples of $24$?

I'm posting all of these for the resource section.  If someone wants to start from the back of the test, I wouldn't mind the help.

i dunno how to do latex, but here is my answer:

[hide]first, in order to find the no. of perfect squares that are factors of 24, u need to know the factors of 24. that is, 2^3*3. From this, u know that the number being square should have at least 2^2 *3 in it, b/c when u square that, u get 2^4*3^2, which is definitely a multiple of 2^3*3. thus, now that we are only looking at the square root of the numbers we need, we should take the square root of 10^6. Thus, we need to look for all the numbers between 0 and 10^3 which are also multiples of 2^2*3 , or multiples of 12. Thus, u take 1000/12, which gives u 83 with some change, and that is the answer. [/hide]
Sorry if this is unclear, i am not very good at explaining my answers.. :oops:

[hide]
$24=2^{3}\cdot 3$
All the exponents must be even to have it be a perfect square. So,
it must be a multiple of 144.
$144x^{2}\leq 10^{6}$
$12x \leq 1000$
$x<84$
x can be any integer from 1 to 83, so our answer is $\boxed{83}$[/hide]

Wow! Why didn't I think of it like that??

[hide=Solution]
We know that $10^6=1000^2$ and $24=2^3\cdot3$.

We know that a perfect square must have an even amount of every prime factor, so if a perfect square is a multiple of $24$, it must be a multiple of $144$.

We know that $\sqrt{144}=12$ so a square is a multiple of $24$ if and only if its square root is a multiple of $12$.

We know that there are $\lfloor\frac{999}{12}\rfloor=\boxed{083}$ numbers less than $1000$ that is a multiple of $12$.
[/hide]

[quote=sugar_rush]lol this is probably the easiest AIME problem ever[/quote]

Well this is the first problem in the 2007 AIME...

Nice bump @RedFireTruck :D

BUMPY :)

[hide=Solution]
Notice that $$24=2^3 \cdot 3$$
Therefore, we must multiply $24$ by $6n$ where $n$ is a perfect square and $144n$ is less than $10^6$. Therefore, notice that $$\left\lfloor \frac{10^6}{144} \right \rfloor=6944$$
Then taking the floor of the square root of $6944$ we get $\boxed{83}$.

[quote=sugar_rush]lol this is probably the easiest AIME problem ever[/quote]

no, there were a lot of easier problems

First AIME problem I've solved! (?) [hide = Sol]
The prime factorization of $2$ is $2^3 \cdot 3.$
This means that all positive perfect squares of $24$ are also multiples of $2^4 \cdot 3^2 = 144.$

Every perfect square is just $144$ times another perfect square. This means that $144x^2 < 10^6$ and we can take the square root of both sides, getting us $12x < 10^3.$

$\lfloor 1000/12\rfloor = 83$

That means that all of the perfect squares under $10^6$ which are divisible by $24$ are products of $144$ and all of the squares up to (and including) $83^2,~\therefore$ the answer is $\boxed{83}.~\blacksquare$
[/hide]

[quote=sugar_rush]lol this is probably the easiest AIME problem ever[/quote]

yes I agree
my solution is very similar to ones already posted so i will not post my solution

[hide=Solution]Call the squares $k^2$, where $k$ is a positive integer. Obviously $k<10^3=1000$.

For $k^2$ to be a multiple of $2^3\cdot3$, $k$ must be a multiple of $2^2\cdot3=12$ and all multiples of $12$ work.

So the answer is the number of multiples of $12$ less than $1000$, which is $\frac{996}{12}=\boxed{083}$.  [/hide]

Oh right isn't the answer to AIME problems always a 3 digit integer?

So my answer is not accepted??

it is accepted but the suggested format is $\boxed{083}$

It, must be in the form of $(12n)^2$, note that $12n<1000$, using the first condition, hence, the number of multiples of $12,$ less than $1000,$ is $\frac{996}{12}=\boxed{83}.$

If a perfect square is divisible by $2^3 \cdot 3=24,$ it must also be divisible by $2^4 \cdot 3^2=144.$ Thus, we are looking for the number of squares less than $10^6$ which are multiples of $144.$ This is the same as the number of squares less than $10^3$ which are multiples of $12,$ which is $\frac{996}{12}=\boxed{83}.$


linear algebra
matrix
algebra
polynomial

[b]Let [/b] $ A \in M_{n} \mathbb{(R)}$ [b]Such that [/b]:  $ 16A^{3}=4A^{2}+4A-I$
[b]Put [/b]: $ S_{k}=I+A+A^{2}+A^{3}+....+A^{k}$ , $ I_{n}$ [b]is identity nxn matrix[/b].
[b]Find[/b]  $ Lim_{k \to \infty} S_{k}$ ?
[b]"...I will try to become good student of Jean..."[/b]

Give Answer : $ Lim_{k \to \infty} S_{k} \equal{} (I \minus{} A)^{ \minus{} 1}$

It's sufficient to show that $ \rho(A)\equal{}Max_{\lambda\in{spectrum(A)}}|\lambda|<1$; then it remains to prove that the $ 3$ complex roots of the polynomial $ 16x^3\minus{}4x^2\minus{}4x\plus{}1$ have a modulus $ <1$.


Investigate the asymptotic behaviour of the solutions y of the equation
$x^5+x^2y^2=y^6$ that tend to zero as x tend to zero


let S1= \sum X_i and S2= \sum (X_i)^2
prove that
S1*( \sum  [(X_i)^3)/(S1-X_i)] ) \geq  S2*( \sum [(X_i) 2 /(S1-X_i)])
where X_i are positive reals

nice one harazi (dude,i'm amazed by ur proficiency at inequalities).my solution is the same
actually i had created that problem(easy but that's an achievement for guys like me :D  :D )


Point $O$ is equidistant from all the points in regular tetrahedron $ABC-D$.  Compute $m\angle AOB$.

[hide="Quickie"]If we imagine a unit cube and the tetrahedron having $6$ cube's side diagonals for edges, then the cube and the tetrahedron share both the center and the circumscribed sphere. From there it's obvious that the sides of $\triangle AOB$ are $\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},\sqrt{2}$, giving $\angle AOB=2\arctan{\sqrt{2}}=\pi-\arccos{1\over 3}$[/hide]


trigonometry
limit
real analysis
real analysis unsolved

Determine wether $\sum_{i=1}^{+\infty}{\sin{\frac{1}{i}}}$ converges.

The sum diverges. To see this, use tle limit test.

$\lim_{i\rightarrow \infty}\frac{\sin \frac{1}{i}}{\frac{1}{i}}=1$, hence $\sum\sin \frac{1}{i}$ and $\sum\frac{1}{i}$ have the same nature.

I have a solution,but I think it has a mistake!

[hide][u][b]Problem:[/b][/u]Find all $\alpha \in R$ such that :$\sum_{n=1}^{\infty}(\sin\frac{1}{n})^{\alpha}$ is convergent.
[u][b]Solution:[/b][/u]
Because $\sin x=x-\frac{x^{3}}{6}+o(x^{3})$ when $x\to 0$so if $\alpha \le-1$ then $\sum_{n=1}^{\infty}(\sin\frac{1}{n})^{\alpha}=\infty$
If $\alpha >1$ then $(\sin x)^{\alpha}=x^{\alpha}(1-\frac{x^{2}}{6}+\frac{o(x^{3})}{x})^{\alpha}\approx x^{\alpha}(1-\alpha .\frac{x^{2}}{6}+\binom{\alpha}{2}.o(x^{3}))$
So $\sum_{n=1}^{\infty}(\sin\frac{1}{n})^{\alpha}$=$\sum_{n=1}^{\infty}(\frac{1}{n})^{\alpha}-\frac{\alpha}{6}.\sum_{n=1}^{\infty}\frac{1}{n^{\alpha+2}}+\binom{\alpha}{2}.\sum_{n=1}^{\infty}\frac{1}{n^{\alpha}}.o(\frac{1}{n^{3}})<\infty$[/hide]


function
real analysis
real analysis unsolved

Prove that $ L^{\infty}(E)$ isn't separable for $ E,|E|>0$.

To do that, you must construct a sequence of decreasing subsets of $ E$ whose measures strickly decrease. Then consider the characteristic function of the class of sets obtained by talking all possible unions of the differences of these subsets. Have fun.

Hint: find an uncountable subset $ S$ of $ L_{\infty}(E)$, s.t. the distance between every two distinct points in $ S$ is at least $ 1$.


Una casa tiene un numero par de lamparas distribuidas es sus cuartos de tal manera que existen al menos 3 lamparas en cada cuarto.
Cada lampara comparte un  switch con exactamente una lampara, no necesariamante del mismo cuarto.
Cada cambio en el switch compartido por dos lamparas, éste cambia sus estados simultaneamente.
Probar que para cada estado inicial de las lamparas existe una secuencia de cambios en los switches que al final del cual cada cuarto tiene tantas lamparas encendidas como apagadas.

Carlos Bravo ;)
Lima -PERU

esta bonito este problemita., pero facil para la imo... creo yo...

Asumamos que existe un cuarto con todas las lamparas en el mismo color ( de lo contrario tendriamo la configuracion que buscamos) ahora mostremos que el numero de estos cuartos se puede reducir por uno.

Sea $C_{1}$ un cuarto cn todas las lamparas del mismo color. Escojamos una lampara y apretamos el switch. Si el switch esta conectado a otra lamapra en $C_{1}$ pues ya acabariamos pq $C_{1}$ tendria lamparas de los dos colores. Asumamos entonces que el switch esta conecado a $C_{2}$. Si despues de apretar el switch $C_{2}$ tiene lamparas de los dos colores pues se acabaria el problema, asi que asumamos entonces que en $C_{2}$ despues del switch todas las lamparas son del mismo color.  Seguimos este proceso hasta que llegaremos a un cuarto que habiamos estado antes, sea $C_{k}$. Antes de que apretemos el switch que afecta a $C_{k}$ por segunda vez, hay dos lamparas de $C_{k}$ de negro (sin perdida de generalidad) y una blanca. Este ultimo switch que afectara a $C_{k}$ afectara una lampara negra (por que la blanca era la que habiamos afectado la primera vez para cambiar a $C_{k}$ su estado con todas del mismo color). Como el ultimo switch afecta una negra tendremos en $C_{k}$ dos blancas y una negra.
Asi que hemos reducido el numero de cuartos del mismo color por uno.


I ran across this website the other day, and I've never heard about it on AoPS before.  Since it seems like something that people here might find interesting, I thought I'd share the link:

http://www.girlsangle.org/

[quote="pianoforte"]More gender discrimination in math.[/quote]
Darn those massive cash prizes for the "Math Prize for Girls" thing still make me feel like I'm being robbed.


The sequence $\{a_{n}\}$is defined as:$a_{0}=2,a_{1}=1,a_{n+1}=a_{n}+a_{n-1}$.Prove that for any prime $p$ greater than 3,we have $a_{p}|a_{\frac{p+1}{2}}^{p}-1$.I have worked on it for days, but in vain.

Sorry that I've made an awful mistake.Now I've corrected it. :blush:

This problem has something to do with the Lucas Sequence. :)


geometric series
calculus
calculus computations

The Number s=4,513513513....Find the serie that produces the number s.....

The decimal expansion that you have is actually an infinite series written in a special notation: $4+\frac{5}{10}+\frac{1}{100}+\frac{3}{1000}+\frac{5}{10000}+\dots$ You can also write it as a geometric series: $4+\frac{513}{1000}+\frac{513}{1000000}+\dots$