articles search calculus integration Support AMC AIME

So I'm hearing a lot of good things about how Firefox 3 p00ns everything else, but I'd like to hear it from you people (since most of you are actually intelligent).  Even Time magazine endorses it, and you know how they are.

Any comments?

And no, I don't want a biased opinion (unless you can control yourself) from someone who loves Firefox  :love: .

Yeah i like it. I think its quite a bit faster.

It saves sessions. Sure, Opera's done that for a long time, but it's a good improvement from FF2. Opera's session saves are still better though, cause you can save multiple session and load each separately. FF automatically saves just the current one. =P

The address bar matches up page titles, not just the URL.

why has 1337 suddenly become a synonym for cool/awesome/pwn/beast/fantastic/incredible W.T.H.

???

An article on TIME is meaningless comppared to what certain individuals tell you.  What are you trying to say?  Also, why on earth do you care?  And why are you posting this here?

I'm not obsessed with Firefox.  I use it pretty much only for testing.  I use Opera here and IE7 most other places.  Opera is sooo much more useful and productive once you get used to it.  My take on FF3?...

Customization: I like how much more you can customize Firefox in the new version.  about:config, though it's much more complicated, offers some more than opera:config, but Opera has always had things that I just can't let go of.  I don't think Firefox's extensions are REAL customization...

Security & Privacy: This is really the only one where I'll give it to Firefox.  That too, mainly because of resource hogging extensions.

Performance:  Opera, without doubt.  It loads much faster than any browser and runs tons of features with no cost to performance.

Is that what you wanted? *sigh*, this is spam.

EDIT: undefined: Opera's address bar always searched titles and now search page content from cache :)  One step ahead...always....

Things about firefox: I need it to delete EVERYTHING when I exit.

Just installed Opera.. I'm having integration issues. :( like text boxes won't let me type anything in the preferences menu.. and the skin isn't integrated with the system's.. Just from the first few minutes of usage.

But font rendering looks better than in FF. idk it's smoother? Though I don't really mind the "raw" appearance of FF's font rendering. At startup, Opera took up like 13 MB RAM, but after those few minutes, it increased to 50 MB wth.. I only had one tab open to AoPS. In comparison, I have three windows open in FF, with ten tabs in this one (two with flash, one with java) and five in another (one with flash), all taking up 160 MB.

@MustIt: But yeah you'd also mentioned CSS3 support in Opera. :) that's a good plus.

Of course, this is in Ubuntu 8.04

undefined, try:

[code]sudo apt-get install msttcorefonts[/code]

It makes font rendering a miracle in FF since you have the windows fonts

That and installing ttf-dejavu are one of the first things I did when I installed Ubuntu.

You should also use aptitude instead of apt-get because the latter doesn't purge properly, and aptitude manages dependencies better.

[quote="#H34N1"]FIREFOX IS BEAST. THIS IS SPAM[/quote]
Read this:
[quote="MathAndKnowlege"]And no, I don't want a biased opinion (unless you can control yourself) from someone who loves Firefox  :love: .[/quote]

[quote="MustItAlwaysBeMe"]An article on TIME is meaningless comppared to what certain individuals tell you.  What are you trying to say?  Also, why on earth do you care?  And why are you posting this here?[/quote]
Answer to first question: Should I use Firefox 3 or not?
Answer to second question: Would you like to be missing out on the best possible Internet browser?
Answer to third question:
[quote="MathAndKnowledge"]So I'm hearing a lot of good things about how Firefox 3 p00ns everything else, but I'd like to hear it from you people (since most of you are actually intelligent).[/quote]

But, yeah, I've now officially decided to use Firefox 3 (not that it affects any of your lives...).

It affects sooey's life whoever she may be :yup:

So MathAndKnowledge opened this topic just so he could ask, "Should I download Firefox 3?" and answered it himself without a reason.  Undefined, lock this up.

But first, my less than two cents on fonts.  Undefined, I thought you said that you weren't able to view Tahoma on Linux, but you say that dling the mscorefonts was one of the first things you did.  Anyway, if you download Powerpoint Viewer 2007 on XP, you get all the new Vista fonts, Segoi, Calibri, etc. (even though they'll look ugly without Cleartype :) )

Also check out sIFR; it replaces text on web pages with flash if it can be found.  Total beastus.

EDIT: Oh yeah, btw, Opera doesn't use up nearly the amt of memory as FF per tab; I have 16 tabs open and Opera is using about 45MB

oh crap dangggg that's good.

lol FF3 still has memory leak issues I think.. I left my laptop on for three days and it got fatter and fatter until it killed compiz fusion (idk why) taking up 260 MB

um okay wth. Tahoma works in Opera.. apparently FF can't use msttcorefonts?

meh I'm switching :)

dude
http://www.ubuntu1501.com/2007/07/how-to-install-ttf-font-in-ubuntu_11.html

easy

use virtualbox to get a copy of xp virtualized on ubuntu
then copy the fonts from XP onto linux

and why is everyone so worried about memory?
get more ram
its not that expensive now
its like what 50 dollars for 4 gigs of ddr2 800?

um virtualbox runs at half processor speed and when you have a core 2 duo @ 2.6 a single core 2.6 isn't really that bad

and photoshop, premiere, avid?

come on avid is like a necessity. what about office 2007 if you're doing like business and people keep sending you docx's

I really don't expect much out of a browser other than decent speed and adblock. That being said, I like Firefox 3's features and add-ons. Plus you know Firefox is constantly getting better due to add-ons and open source code.

[quote="MustItAlwaysBeMe"][quote="#H34N1"]and why is everyone so worried about memory?
get more ram
its not that expensive now
its like what 50 dollars for 4 gigs of ddr2 800?[/quote]It's not particularly healthy to keep adding RAM. :)  Also, I've been looking for a way to compile FF on windows with OpenType instead of TrueType.  No luck whatsoever :( .  Someone wanna do that for me? (I don't have cygwin :D )

Why dl virtualbox; there's no reason to.  If you can't run an app with WINE, why do you even want the app?[/quote]

before the subject diverts too far from RAM, i installed adobe cs3 on my laptop (which comes with 512mb RAM) even though the install program gave me a warning message saying that cs3 needs at least 1gb RAM to function. i haven't gotten any more RAM, but the only programs that i've used extensively are photoshop and illustrator (i'm kind of avoiding the flash programs for fear my computer will get fried).

am i in way over my head trying to keep this going with insufficient RAM (i have a feeling it might have had something to do with my operating system getting wiped earlier, but maybe that's just paranoia)? if i don't absolutely need to buy RAM, i won't.

@#H34N1: your link is not the kind of solution I want :P and I'm (my parents are, anyway) too lazy to buy any RAM. I currently have 1 GB, and for some dumb reason, this particular generation of macbook is only expandable (officially) to 2 GB. There are reports of 4 GB working on the second gen. macbook, but most say that the system only sees 3.3 GB. But the thing is, that would be on OS X. I wonder if it'll work in Linux.. :) crucial offers a 2 GB kit for 48 dollars, and a [u]4 GB kit for 104 dollars.[/u] That, and probably 25 - 50 dollar S&H fees.

and dude, photoshop? I switched to Linux for the open-source/freeware. I don't want photoshop ;)

hehe I agree with MustIt that if you can't run it with WINE, why want it? Actually, if I can't run it natively, I don't want it. Picasa looked good for a while, but I didn't like the windows-y interface and its import methods so I abandoned it.

You won't fry your computer just because you have insufficient RAM. You don't absolutely need it, but in the long run, more RAM will help save time and stuff.

meh you have a a mac

http://www.newegg.com/Product/Product.aspx?Item=N82E16820141366

[quote="undefined117"]hehe I agree with MustIt that if you can't run it with WINE, why want it? Actually, if I can't run it natively, I don't want it. Picasa looked good for a while, but I didn't like the windows-y interface and its import methods so I abandoned it.[/quote]Talk about jumping to conclusions... :)

You're wrong, Linux CAN use a native GUI in WINE.  http://lifehacker.com/341498/improve-windows-apps-looks-in-wine  Be glad I remember reading this.

[quote="MustItAlwaysBeMe"]So MathAndKnowledge opened this topic just so he could ask, "Should I download Firefox 3?" and answered it himself without a reason.  Undefined, lock this up.[/quote]
Umm... I answered my question after, you know, about three gazillion people (if that's a number) supported Firefox 3.  I don't think that qualifies as "answering it myself without a reason."

And, yes, I opened this topic so I could ask a question about internet browsers.  Have you checked other parts of the forum that aren't strictly for math?  They are flooded with topics that ask non-math related questions.  I don't remember an announcement saying that the Indiana forum is only for math discussion; and if you truly hate topics like this so much, then don't post in it.

But, yes, I agree that further discussion is not needed (but I'm not going to try to boss undefined to lock this up, which apparently you have the authority to do) since my question has been answered.

@MustIt: huh, it's not native.. I have to configure ~/.wine/user.reg each time I change the skin and stuff.. full theming is also notoriously slow. Just tried it. Thanks anyway. looks a bit better :)

[quote="undefined117"]er do you even know what you're talking about anyway? The things actually need to [i]fit[/i] in my laptop: [url]http://crucial.com/store/mpartspecs.aspx?mtbpoid=C599448AA5CA7304[/url]

Like I said, read things through before jumping to conclusions :) your link was for a desktop computer, not a laptop.[/quote]

Um that was what RAM i was mentioning (4 gigs for 50). It was 49.99 a couple weeks ago

@[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1183551#1183551]MathAndKnowledge[/url]: I'm not saying that posting non-math-related stuff here is bad. (In fact, that's probably the best part of this forum :) ) But the fact that you created this particular topic, I just don't get...if you're looking for specific reasons to switch, all you'll get here is stuff like "i like it", and if you just wanted to download Firefox just because it's out there, you didn't need to start the topic either.  Meh, sorry about the ranting.

@[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1183556#1183556]undefined117[/url]: Dang, I thought it was actually native.  At least you can do that much, I guess.

@#H34N1: Is adding RAM going to make that much of an impact on productivity?  Do you think I could get something at RadioShack rather than looking online?  Also, why use Linux if you'd still rather use proprietary software.

um because linux is a) more efficient, b) better, c) more stable than windows

and but for a couple programs that won't WINE I don't need it

so I just virutalize an XP on top like what solafidefarms does

If you don't mind, #H34N1, check my solution to your problem in the AIME thread? :) thanks.

Get firefox. =D. And wtf most new computers come with big monitors, wireless keyboard + mouse, 2 gigs of ram, and a decent video card. So just buy a new comp.

RAM = what you need for pretty much anything... except word processing lol. But more RAM = faster gameplay (!!!!!) and being more pro at games =D... as well as being able to run photoshop and big memory taking processes.

omg mathcrazed, do you really like firefox. looking at your avatar....

most new computers? you can't find a computer with less than 2 gigs of ram

and as i said....ram isn't that expensive
if you are cool, just buy a couple sticks get a screwdriver and stick em in

okay it's been like five hours using Opera.. it went up from like 60 MB to this 210 MB wth.

[url]http://avencius.nl/content/firefox-3-vs-opera-950-memory-usage[/url]

and [url]http://dotnetperls.com/Content/Browser-Memory.aspx[/url]

FF3 FTW. w00t sorry MustIt. :P

wtf that computer is like from 2000 then

wtf most computers DO NOT have < 1 GB of RAM

and where'd you get ALL?

http://myfiles.zapto.org/worstofworst.gif

take dell's worst model and it STARTS with a gig of ram.

that's for 300 dollars

yeeesh

[quote="MustItAlwaysBeMe"]@[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1184063#1184063]mathcrazed[/url]: Of course everyone trashes their computer and gets a new one for over 1000 dollars just to add half a GB of RAM... How is that an acceptable solution?

@[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1184248#1184248]#H34N1[/url]:  I have 256MB RAM now and my computer's max is 512MB.  Worth it? :D (I know, it hurts)

@[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1184886#1184886]undefined117[/url]:  That's pretty interesting.  I dunno why it leaked crazy for you. :) Meh, I still cant go back to Firefox because Opera has:

better load speed
"." and "," text and link searching
mouse gestures
being able to put bookmarklets anywhere
panel
ridiculously simple to enable/disable js,java,plugins,cookies (F12)

Sorry, you need to do MUCH more to get me to switch back. :( Honestly on "download day", I dld ff and then closed it in 20 seconds because Opera already had a lot of the features that were added anyway. (and 75% of my extensions got outdated :mad: )[/quote]
errr... half a gig? More like 1.5 gigs plus a better video card and office already installed... And anyways... I didn't say trash your old computer did I? And if your comp only has .5 gigs of RAM, it's probably from before like 2003. I think 5 years is a long enough time for a comp.

[quote="logic812"]Yeah, most (if not all) computers have <1 GB of RAM, take for example any Win 98 (or 95), it has up to 256 or 512. (This is so small;not)
But my XP computer has 512 MB of RAM (yes, it's XP)(bought in 2003?) but tiny RAM doesn't neccessarily mean the computer is like crap. It's just less RAM. (get an extension)[/quote]WHAT?!  That makes no sense.  Ok so most [i]32 bit[/i] operating systems have somewhere around a 3.5GB limit of RAM, not 512MB.  The motherboard on this computer, on the other hand, only supports 512MB RAM and only has 2 RAM slots.  So if you want more than this much RAM, you would need to get a 64 bit OS and run 64 bit apps, so unless you need to run those applications, you have no advantage of doing this.  This post is really uninformed.

Also, most new computers have at least 1GB of RAM.  Whether or not your statement was true, I have no idea.  And what fraction of computers do you think run 98 or 95?  How about this: http://marketshare.hitslink.com/report.aspx?qprid=10&qpmr=24&qpdt=1&qpct=3&qptimeframe=M&qpsp=113 But once again, it can't be said that has anything to do with RAM.

And yeah, this comp is from about 2000ish. :)

Well when it comes to Internet browsers it seems better to have a browser thats efficient memory wise rather than upgrading the whole systems memory.


The first term of a particular sequence is 5, and each successive term is one less than twice the previous term. What is the value of the sixth term?

So we are told that $ a_{1}\equal{}5$ and $ a_{n}\equal{}2a_{n\minus{}1}\minus{}1$.  We see that $ a_{2}\equal{}9$, so $ a_{3}\equal{}17$, $ a_{4}\equal{}33$, $ a_{5}\equal{}65$, and so $ a_{6}\equal{}129$.  Or, you can use the formula to get that $ a_{6}\equal{}32a_{1}\minus{}31\equal{}32(5)\minus{}31\equal{}129$.

I've had this problem before. The fastest way that I did it was to just bash it out.

[quote=mathisdecent]Isnt there a faster way to calculate terms in a xn-y sequence? isnt it called infinite adding? im not 100% sure[/quote]

There is. $5$ is $2^2+1$, and because the pattern is $2(2^2+1)-1 \implies 2^3+2-1 \implies 2^3 + 1$. If you look at the pattern, if $a_1 = 2^2+1, a_2 = 2^3+1,$ then $a_n=2^{n+1}+1$. We are looking for $a_6, = 2^{6+1}+1 \implies 128+1 = \boxed{129}$

Yeah, the fastest way I'd do it calculates all the terms to a_6 and formulate equations to solve or wishful thinking if bash needed.


inequalities
Gauss
inequalities unsolved

This appeared as problem 2 of the 1st round of the 3rd MathLinks contest in the following form:

[color=blue]Prove that for all positive reals a, b, c, the following double inequality holds:

$\frac{a+b+c}{3}\geq\sqrt[3]{\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{8}}\geq\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}$.[/color]

Some solutions were discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=4526 .

  darij

[quote="Gauss_87"]Is this Carlson Inequality?[/quote]

Not exactly; see http://www.mathlinks.ro/Forum/viewtopic.php?t=27766 for the Carlson inequality.

  Darij

[quote="Gauss_87"]Can someone post a solution?[/quote]
Gauss_87, darij   has proved it here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27766
since, $\sqrt{\frac{bc+ca+ab}{3}}\geq\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{3}$ ;)
Interesting, that $\sqrt[4]{\frac{(a+b+c)(a+c+d)(a+b+d)(b+c+d)}{81}}\geq\sqrt\frac{ab+ac+ad+bc+bd+cd}{6},$
where $a>0,b>0,c>0,d>0,$  is true too!

[quote="Gauss_87"]Let $a,b,c \in R^+$. Prove that:
$\frac{\sqrt{ab} + \sqrt{bc}+ \sqrt{ac}}{3} \leq \sqrt[3]{\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{a+c}{2} }$[/quote]
hi gauss87..is this from winter camp? :roll:

[quote="Gauss_87"]Hi Eucla (ciao, siamo italiani, no?!), non è del winter camp  :blush:[/quote]
yes italiani :D


This is easy problem but the problem that developed from this occurs in a lot.

How many 5-digit numbers are there?

How many 5-digit numbers with no repeated numbers are there?

How many 5-digit palindrome (read same in any direction; both front and back like 12321) are there?

If the first digit of the 5-digit palindrome must be 1 and the middle one should be either 2 or 3, how many such 5-digit numbers are there?

Try this. They're very important problems.

There are [hide]9*10^4[/hide] 5 digit numbers
There are [hide]9*9*8*7*6[/hide] 5 digit numbers with no repeats
There are [hide]9 * 10^2 [/hide] palindromes

With those conditions on the palindromes there are [hide]10*2[/hide]

How many 5-digit numbers are there?
[hide]_ _ _ _ _
9*10*10*10*10
90000[/hide]

How many 5-digit numbers with no repeated numbers are there?
[hide]_ _ _ _ _
9*9*8*7*6
27216[/hide]

How many 5-digit palindrome (read same in any direction; both front and back like 12321) are there?
[hide]_ _ _ _ _
9*10*9*1*1 since it's a palindrome, the last 2 numbers only have one option.
810[/hide]

If the first digit of the 5-digit palindrome must be 1 and the middle one should be either 2 or 3, how many such 5-digit numbers are there?
[hide]1 _ 2,3 _ 1
1*10*2*1*1
20[/hide]

I think those are right.

To mathfiend [hide]There are 10 choices for the middle number, since there weren't any restrictions on it, not 9.[/hide]

How many 5-digit numbers are there?
[hide] 9x10x10x10x10 .[/hide]
How many 5-digit numbers with no repeated number?
[hide] 9x9x8x7x6[/hide]
How many 5-digit palindrome (read same in any direction; both front and back like 12321) are there?
[hide]10x10x9[/hide]
If the first digit of the 5-digit palindrome must be 1 and the middle one should be either 2 or 3, how many such 5-digit numbers are there?
[hide]1x2x10[/hide]


modular arithmetic
quadratics
number theory unsolved
number theory

Let $d$ be a positive divisor of $5 + 1998^{1998}$. Prove that $d = 2 \cdot x^2 + 2 \cdot x  \cdot y + 3 \cdot y^2$, where $x, y$ are integers if and only if $d$ is congruent to 3 or 7 $\pmod{20}$.

Since $d$ divides $x^2 + 5y^2$, where $y = 1998^{999}$, $-5$ is a quadratic residue of $d$ and so every prime factor of $d$ is $\equiv 1,4 \bmod 5$.  Hence $d$ is representable by a quadratic form of discriminant $-20$.  There are just two classes, with reduced forms $X^2 + 5Y^2$ and $2X^2 + 2XY + 3Y^2$ respectively.  If $p$ is represented by one form then $2p$ is represented by the other.  The primes $\mathcal I$ represented by $X^2 + 5Y^2$ are $\equiv 1,4 \bmod 4$ and are $1 \bmod 4$; the primes $\mathcal J$ represented by $2X^2+2XY+3Y^2$ are $\equiv 2,3 \bmod 5$ and $3 \bmod 4$.  Since ${\mathcal I}{\mathcal I} = {\mathcal I} = {\mathcal J}{\mathcal J}$ and ${\mathcal I}{\mathcal J} = {\mathcal J}$, we see that $d$ is in $\mathcal J$ iff it is divisible by an odd number of primes in $\mathcal J$ iff it is represented by $2X^2 + 2XY + 3Y^2$.

[quote="rgep"]Since $d$ divides $x^{2}+5y^{2}$, where $y = 1998^{999}$, $-5$ is a quadratic residue of $d$ and so every prime factor of $d$ is $\equiv 1,4 \bmod 5$.  Hence $d$ is representable by a quadratic form of discriminant $-20$.  There are just two classes, with reduced forms $X^{2}+5Y^{2}$ and $2X^{2}+2XY+3Y^{2}$ respectively.  If $p$ is represented by one form then $2p$ is represented by the other.  The primes $\mathcal I$ represented by $X^{2}+5Y^{2}$ are $\equiv 1,4 \bmod 4$ and are $1 \bmod 4$; the primes $\mathcal J$ represented by $2X^{2}+2XY+3Y^{2}$ are $\equiv 2,3 \bmod 5$ and $3 \bmod 4$.  Since ${\mathcal I}{\mathcal I}={\mathcal I}={\mathcal J}{\mathcal J}$ and ${\mathcal I}{\mathcal J}={\mathcal J}$, we see that $d$ is in $\mathcal J$ iff it is divisible by an odd number of primes in $\mathcal J$ iff it is represented by $2X^{2}+2XY+3Y^{2}$.[/quote]

How do you know "d is representable by quadratic forms of discriminant -20" ? can you give a proof and generalisation of this usage ? I mean, I know that if d is prime it is representable, however it it not prime? Actually, I would like to learn that when a number can be represented by a quadratic form?

First note that $gcd(3,d)=gcd(2,d)=1$
Let $1998^{999}=a$. Then by Thue's lemma we have $a^2x^2 \equiv y^2(mod$ $d)$ for some positive integers $x,y\le \sqrt{d}$

And since $a^2\equiv -5(mod$ $d)$ then $a^2x^2\equiv -5x^2\equiv y^2(mod$ $d)$
So $5x^2+y^2$ is divisible by $d$ and since $x,y\le \sqrt{d}$ then all posible cases are $k=5x^2+y^2:=d,2d,3d,4d,5d,6d$

$A)$ If we have $d\equiv 3,7(mod$ $20)$ then $k$ cannot be $d,4d,5d,6d$
So we must have $5x^2+y^2=2d$ or $3d$.
$1)$ When $5x^2+y^2=2d$ and since $d$ is odd then $x,y$ must have same parity.
So we have $y=x+2s$ for some integer $s$ and $d=2xs+2s^2+3x^2$ q.e.d
$2)$ When $5x^2+y^2=3d$ and since $gcd(3,d)=1$ then we must have $y-x=3t$ or $y+x=3t$ for some integer $t$.
If $y-x=3t$ then $d=2x^2+2xt+3t^2$
If $y+x=3t$ then $d=2x^2+2xr+3r^2$ where $r=-t$
q.e.d
$B)$ If  $d=2x^2+2xy+3y^2$ then we must have $d\equiv 3,7(mod$ $20)$ which is easy to prove.


AMC
USA(J)MO
USAMO
AIME
AMC 10
trigonometry
AIME II

Now that the AIME II is out into the open, what are your predictions for a USAMO Index? In particular, do you believe a 224 will qualify?

I would guess the index winds up as a 217, with a floor of 8.

I would guess that your guess is slightly low. :-) (note the key word: slightly)

i think the cutoff will be about 235 and 9 for the floor.  Comparing with last year, there are a lot more people who got high scores this year.

I'll say 226.5 and 8. It's a little low, and probably won't happen (I believe the floor will probably be 9, with a small chance of 8) but I think it's feasible.

Let's think about this analytically. The 2003 Index was a 226.5 and the AMC 12 was obscenely easy that year- >30 perfect scores. This year, only a few select individuals scored more than a 143 or even a 138 on the AMC 12. That said, the AMC 12 this year was also easier than last year's AMC 12. Last year had an index of 209.5. Now I can't compare the AIME's very well, since I took 2003 as a freshman who didn't know much math, and this one with a lot more preparation, but at the least I'd say that this year's AIME has been somewhat easier than average, not by any means the "easiest in history". I would put it between last year's AIME and the 2003 AIME. Based on these conclusions, the index should be between 226.5 and 209.5, which was how I calculated my prediction for the index. How could the index ever conceivably be 235? Anyone outside of the ~20 people with a score above a 145 would need at least nine, and people with scores in the low 130's would need 11. I really don't believe the AIME was easy enough for that many people to score that well.

I like Fountainhead's analysis. I predict the index cutoff to be somwhere between 220.5 and 225.0 and the floor to be 8.

I agree with JP's prediction, around 235 and a floor value of 9. I think to compare this year's AIME scores with last year's AIME scores, one should subtract 2.5 points from this year's. I would say these AMCs are a tad easier, but not by much. Many many sophomores and below scored at least a 9; there exist many more people than just those on AOPS.

Yes, good point Fountainhead. 235 is ridiculously high. However, the index you predicted seems quite low for this year. If you look at the USAMO index polls, over 50 people on this site scored above a 220. For the index to be below 220, we would need to have more than a fifth of the USAMO qualifiers on this site (assuming 250 qualifiers). This seems like a high fraction, so I am not sure how feasible an index lower than 220 is.

[quote="JGeneson"]I think the index is 227, with an AIME floor of 9. I had a dream about this, so I am pretty sure of it. I do not think there is any way the index will go under 220 (unless they decide to make different cut-offs for AIME 1 and AIME 2).[/quote]

You are pretty sure because of a dream?  Well, I hope you're right, 'cause I've got 227.

interesting_move

[quote="filletwho"]I'm not sure about the index, but I'm willing to assume that this site holds atleast a fifth of the top ppl in the nation, if not more.[/quote]

I'm willing to bet at least a third of this year's qualifiers will have accounts here. They may not all necessarily be active though.

In any case, I'm up for a 222 - 224 and 9.

And I would debate the argument that this is not the easiest set of AIMEs ever. The 2003 AMC was eaiser, but its AIME wasn't. The cutoff will probably be slightly lower than 226.5, but not by much at all. And as for AIME floor, I say 9. Of course, 8 is also a possibility. 10 if we're extremely unlucky. 11 and above are just about impossible b/c that would be asking for a qualifying index of at least 245...

[quote="JP"]almost all my classmates improved a lot.[/quote]

Er... all your classmates are in high school.  Isn't improving a lot year-to-year to be expected? (not to demean their accomplishments, but 16-year-olds do learn fast...)  They might have made almost the same improvement had they taken last year's test again (assuming that they could forget it completely).

[quote="JP"]Fred, honestly yo know the cutoff is going to be higher than 220.  230 is certainly reasonable, but i don't see anyway it is going to be less than 230 though.  a guy who god 1 on AIME last year got 12 this year. almost all my classmates improved a lot.[/quote]

Well people improve. There are people who score like 1-2 on their first AIME, and like 7-8 on their second AIME, and 10+ on their third AIME.

I predict the qualifying index will be 230 with an AIME floor value of 9.  This year's contests were slightly easier than usual, especially the AIME1, so I think that the index and floor value will be higher than normal.

hoping for a floor value of 8 or else i will be  :(  index maybe of about...225ish

I haven't looked very intently at past stats and stuff so I have very little to base my predictions off of but I ll give it a guess anyways.

Index: 223
Cutoff: 8

I'd be willing to say Fountainhead's idea of it is fair at least to what he knows.  I flopped the AIME 1 enough because it was my first sitting that I have trouble believing that index will shoot much above 220.  Has it ever broken say a 226 before?   Still, that AIME was Easy--they were dumb mistakes that got me.  And the AIME 2 wasn't too bad at all.  I only half-way took it because it wasn't official and still did as good or better than on the first test, which tells me I could've hoped for more like an 8 or nine the first time...

Is there any chance of that index getting split?--It's true, there was a pretty huge difference between those tests.

[quote="MithsApprentice"]I'm willing to bet at least a third of this year's qualifiers will have accounts here. They may not all necessarily be active though.

In any case, I'm up for a 222 - 224 and 9.

And I would debate the argument that this is not the easiest set of AIMEs ever. The 2003 AMC was eaiser, but its AIME wasn't. The cutoff will probably be slightly lower than 226.5, but not by much at all. And as for AIME floor, I say 9. Of course, 8 is also a possibility...[/quote]Dido

one year the cutoff index was a 228 i think.

[quote="spt10v"] How would I calculate my index and floor?[/quote]

204.

You don't have a personal floor. The "floor value" we're referring to is calculated by the top (approx.) 160 indices.

HOW DO U CALCULATE THE INDEX? AND IS IT ONLY FOR AMC 12 TAKERS?

COULD YOU PLEASE STOP TYPING WITH CAPS LOCK ON?

Treething posted the formula for the index just a couple posts above.

[quote="fedwinri"]

Is there any chance of that index getting split?--It's true, there was a pretty huge difference between those tests.
[/quote]

Nope, there wasn't a huge difference between the two tests
#1-5s on AIME2 are a bit harder.

#6, 7, 8's are about the same,
#9 on AIME 2 is easier,
#10 on AIME 2 is easier,
#11 on AIME2 is harder,
#12 on AIME2 is harder
#13 on AIME2 is easier
#14 on AIME2 is easier
#15 on AIME2 is harder

So overall I think they're about the same, actually AIME2's slightly harder, but not by very much.

Ah. Difference of opinion. I think of the first 13, a good half of the AIME Alt problems are easier (not harder, mind you) than AIME. 14 has a nice synthetic solution (which I, unfortunately, did not find) but the trig one was not bad. 15 is not too bad either. So in my opinion, the two tests are about the same: CAMC did a good job balancing the difficulty level.

[quote="MithsApprentice"][quote="yif man12"]I don't think there is an index for AMC10 takers[/quote]

There is. It's the same thing. They're just not considered in the 160 scores. If there too many underclassmen who make the AIME floor, I believe it goes into index selection.[/quote]

But would that be better for the people who took the 12 or the 10? Or is the index for the 10 adjusted?

Actually, I think it's the AMC 12 takers that are adjusted. ^-^ *hop* (adjusted to 150)

[quote="MithsApprentice"][quote="yif man12"]I don't think there is an index for AMC10 takers[/quote]

There is. It's the same thing. They're just not considered in the 160 scores. If there too many underclassmen who make the AIME floor, I believe it goes into index selection.[/quote]Yes, of course.  I think it only works in the event of a tie--in other words, the AMC (ppl) use only a 10*AIME + 1% AMC in a sence, so that only in the event of a tie does it matter for thenth-graders$-$.
[quote="spt10v"]I got 124 (incredibly dumb mistakes) on AMC 10 and 8 right on the Alternate AIME.  How would I calculate my index and floor?[/quote]And that means the only thing you're worried about is that AIME of an 8 and the floor.  If it so happens that this is the cut (which it could well be), you'll need to watch your AMC score in comparison to the others because there's a chance that they'll stop it somewhere in the between and say: You need an 8 (iff floor 8) and a 130" or something.


Two cards are chosen at random, with replacement, from a standard deck of 52 cards. If both cards are face cards, what is the probability that both cards are jacks?

Many people that I know, including myself, consider aces to be face cards, so that the order goes 2-3-4-5-6-7-8-9-J-Q-K-A. In the solution, it says that only jacks, queens, and kings are face cards. Could you make this clear in the problem?

The only face cards are JQK. You are wrong. They are called this because they (usually) have a person's FACE on them.

Of course, the makers of Alcumus could make it a bit clearer :wink:

Well its just 1/3 ^2=1/9. If you are weird and think aces have faces (rhyme time!) then its 1/4^2=1/16

Hey, I didn't make up the idea that aces were face cards! I learned it from someone else. And I've never been contradicted on that point.

I did get 1/16, by the way.

EDIT: With replacement! D'oh! :-/

Wait, why is it (4/12)^2? After you pick the first jack, only 11 face cards remain, and only 3 jacks remain.

Why wouldn't the answer be (4/12)(3/11) = 1/11?

There are 4 jacks out of 12 cards. It says with replacement, so we multiply this by 4/12 again to get $\left( \dfrac{4}{12} \right)^2 = \left( \dfrac{1}{3} \right)^2 = \boxed{\dfrac{1}{9}}$.


Hello,

I'm trying to prove that if a simple graph G with $ e$ edges and $ v$ vertices satisfies $ e > \frac{(v\minus{}1)(v\minus{}2)}{2}$ then G is connected.

Here is how I proved it :

Suppose G is not connected, and without loss of generality suppose that $ u,v \in V(G)$ are two non-connected points in G. It follows that G has more than one component, suppose two, one with $ l$ $ (0 < l < v)$ vertices and the other with $ v \minus{} l$ vertices. In the "worst" case these two graphs are complete giving us : $ \frac{(l(l\minus{}1) \plus{} (v \minus{} l)(v \minus{} l \minus{} 1)}{2}$ edges in total. It remains to show that $ \frac{l(l\minus{}1) \plus{} (v \minus{} l)(v \minus{} l \minus{} 1)}{2} < \frac{(v\minus{}1)(v\minus{}2)}{2}  \forall{0 < l < v}$ giving us a contradiciton.

The proof is ugly and it may even be incorrect, so I'm wondering if someone is aware of a more elegant proof. Thanks

A possibly more refined argument is to say WLOG that $ l < v \minus{} l$. Then show that decreasing $ l$ by $ 1$ gives more edges (increases that nasty thing you showed), allowing you to consider only the case $ l \equal{} 0$.

That's not ugly at all, although I would dispense with the words "worst case" and say instead, "There are no edges joining the two components, at most $ blah$ edges on one side and at most $ blah$ edges on the other, so at most $ blah \plus{} blah$ edges total."  The final computation is just to maximize or minimize a quadratic, which on the scale of ugliness is pretty minor.

Another way to look at this is that your graph $ G$ arises by removing at most $ v \minus{} 2$ edges from a complete graph -- that shows intuitively why the graph can't be disconnected this way, but the final proof will still involve a quadratic polynomial.


A whole number is said to be ''9-heavy'' if the remainder when the number is divided by 9 is greater than 5. What is the least three-digit 9-heavy whole number?

greater than 5 mod 9 is minimized when its 6 mod 9, so we try 9*10+6=96. too small. 9*11+6=105. that'l work. $ \boxed{105}$


Given $ a, b, c > 0$ and $ abc \equal{} 1.$
Prove that: $ \frac{(a^3\plus{}b^2c)^3}{a^5\plus{}b^4c} \plus{} \frac{(b^3\plus{}c^2a)^3}{b^5\plus{}c^4a} \plus{} \frac{(c^3\plus{}a^2b)^3}{c^5\plus{}a^4b} \geq\ 12$
It's easy but I hope you'll like it, my friend.  :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ abc \equal{} 1.$
Prove that: $ \frac {(a^3 \plus{} b^2c)^3}{a^5 \plus{} b^4c} \plus{} \frac {(b^3 \plus{} c^2a)^3}{b^5 \plus{} c^4a} \plus{} \frac {(c^3 \plus{} a^2b)^3}{c^5 \plus{} a^4b} \geq\ 12$
It's easy but I hope you'll like it, my friend.  :)[/quote]
Sorry,I'm very busy,so I don't many times.
With this inequality:
use AM-GM,I will prove that:
$ ((a^3\plus{}b^2c)(b^3\plus{}c^2a)(c^3\plus{}a^2b))^3 \ge\ 64a^4b^4c^4(a^5\plus{}b^4c)(b^5\plus{}c^4a)(c^5\plus{}a^4b)$
we have:
$ (a^3\plus{}b^2c)(b^3\plus{}c^2a) \ge\ 2\sqrt{ab^2c(b^5\plus{}a^4c)(c^3\plus{}a^2b)}$
similar with two inequalies,we have:
$ ((a^3\plus{}b^2c)(b^3\plus{}c^2a)(c^3\plus{}a^2b))^3 \ge\ 64a^4b^4c^4(a^5\plus{}bc^4)(b^5\plus{}ca^4)(c^5\plus{}ab^4)$
Suppose $ b\equal{}min(a,b,c)$
-if $ a\ge\ b\ge\ c$,then
 $ (a^3\plus{}b^2c)(b^3\plus{}c^2a)(c^3\plus{}a^2b) \ge\ (a^3\plus{}bc^2)(b^3\plus{}ca^2)(c^3\plus{}ab^2)  \ge\ RHS$.
-if $ a\le\ b \le\ c$,then
$ LHS\ge\ 64a^4b^4c^4 (a^5\plus{}bc^4)(b^5\plus{}ca^4)(c^5\plus{}ab^4) \ge\ 64a^4b^4c^4(a^5\plus{}b^4c)(b^5\plus{}c^4a)(c^5\plus{}a^4b)$ :)

[quote="nguoivn"]Yes, I'm happy to see your proof for it. It's also my way to creat it. I hope that you'll finish fastly 2 old gifts of me.  :)
PS: [hide]ANh nho' bai` toan nay` anh post kha lau roi`, vi` the van' de` "ban." cua? em chac cung~ ko anh? huong? nhieu` lam'. Anh doan' la` em tim` ra loi` giai? nay` sau khi anh post n~ bai` nay`:
http://www.mathlinks.ro/Forum/viewtopic.php?t=262448
http://www.mathlinks.ro/Forum/viewtopic.php?t=262971
Tuy nhien, voi 2 bai` toan con` lai anh tang. em thi` anh se~ ko co bat ki` goi y' nao`, hi vong. em se~ thuc. su. "thu? suc" voi chung vi` chung ko fai? n~ bai` toan danh' do' kieu? $ abc \equal{} 1$, thuc ra anh ko muon em fai mat tg cho dang toan nay` nen anh da~ post cai topic dau` tien de? "goi y", ko fai? ngau~ nhien anh post dau :) [/hide][/quote]
Thank you very much :)
@ nguoivn[hide] ngai qua. dung la bai nay em tim ra loi giai sau khi anh post nhung bai do :blush:  2 bai con lai em se nghi them. Dao nay em ban hoc doi tuyen math qua  :)  [/hide]

[quote="quykhtn-qa1"]
-if $ a\ge\ b\ge\ c$,then
 $ (a^3 \plus{} b^2c)(b^3 \plus{} c^2a)(c^3 \plus{} a^2b) \ge\ (a^3 \plus{} bc^2)(b^3 \plus{} ca^2)(c^3 \plus{} ab^2)$.
[/quote]
It's wrong. Try $ a\equal{}5,$ $ b\equal{}2$ and $ c\equal{}1.$

[quote="nguoivn"]Yes, arqady. But indeed:
[quote="quykhtn-qa1"]With this inequality:
use AM-GM,I will prove that:
$ ((a^3 \plus{} b^2c)(b^3 \plus{} c^2a)(c^3 \plus{} a^2b))^3 \ge\ 64a^4b^4c^4(a^5 \plus{} b^4c)(b^5 \plus{} c^4a)(c^5 \plus{} a^4b)$
we have:
$ (a^3 \plus{} b^2c)(b^3 \plus{} c^2a) \ge\ 2\sqrt {ab^2c(b^5 \plus{} a^4c)(c^3 \plus{} a^2b)}$
similar with two inequalies,we have:
$ ((a^3 \plus{} b^2c)(b^3 \plus{} c^2a)(c^3 \plus{} a^2b))^3 \ge\ 64a^4b^4c^4(a^5 \plus{} bc^4)(b^5 \plus{} ca^4)(c^5 \plus{} ab^4)$
[/quote]
It's enough. I also don't understand the final part of quykhtn-qa1. It's not necessary  :oops:[/quote]
Because $ (a^5 \plus{} b^4c)(b^5 \plus{} c^4a)(c^5 \plus{} a^4b)\neq(a^5 \plus{} bc^4)(b^5 \plus{} ca^4)(c^5 \plus{} ab^4).$

Very, very sorry, quykhtn-qa1 and arqady. I had a big mistake in type my ineq  :(  :(
It's exactly:
Given $ a, b, c > 0$ and $ abc \equal{} 1.$
Prove that: $ \frac {(a^3 \plus{} b^2c)^3}{a^5 \plus{} bc^4} \plus{} \frac {(b^3 \plus{} c^2a)^3}{b^5 \plus{} ca^4} \plus{} \frac {(c^3 \plus{} a^2b)^3}{c^5 \plus{} ab^4} \geq\ 12$

nguoivn, maybe your first inequality is true too.


if you were to write a string of continuous integers starting with 1, what would the 2000th digit be and how did you get it?

example: 123456789101112131415...etc

I need more repetitions to determine the pattern.

Is the next cycle

110111112 etc.
or
120121122123 etc.
or tons more possibilities

The patterns the concatenation of the natural numbers.

[hide]
0[/hide]

Do you just have to figure the number of 2-digit numbers then 3-digit and so on until you get to 2000? Or am I missing something? It just seems like a very brute force method. Thanks.

Outline Solution:

[hide]
9 1 digit numbers = 9
90 2 digit numbers = 180
Sum = 189
$floor(\frac{2000-189}{3})= 603$
$603*3+189=1998$ (to see how many digits we lost by taking floor (in this case 2))
So after 1998 digits, it looks like this: ...601602603
So digit 1999 is $6$
digit 2000 is $\boxed{0}$
(and digit 2001 is $4$)
[/hide]

EDIT:
THIS SOLUTION IS A BIT FLAWED; SEE MY FULL SOLUTION BELOW

Thank you smartmouth. Much easier than the method I had in mind.

[quote="maedko"][quote="mysmartmouth"]Outline Solution:

[hide]
9 1 digit numbers = 9
90 2 digit numbers = 180
Sum = 189
$floor(\frac{2000-189}{3})= 603$
$603*3+189=1998$ (to see how many digits we lost by taking floor (in this case 2))
So after 1998 digits, it looks like this: ...601602603
So digit 1999 is $6$
digit 2000 is $\boxed{0}$
(and digit 2001 is $4$)
[/hide][/quote]

i dont follow that.  can you explain why you did what you did?  and how do you know that it started back up again with 601602...?[/quote]

OK, let me give a full solution.  Start by realizing a few facts:
There are $9\cdot 10^{(x-1)}$ x-digit numbers.  So there are 90 2 digit numbers, etc.  Don't memorize that, it's easy to see where it comes from.

So now we know that the first 9 numbers will be just 1 thru 9.  The 10th and 11th digits will be 10, etc.  However, we don't want to list ALL these numbers.  So since we know we have 90 2-digit numbers, with 2 digits each, they take up $90\cdot2=180$ digits.  Add that to the 9 digits from above, for a total of 189.  This means that 1-digit and 2-digit numbers take up a total of 189 digits.

So how many digits are left to use up before we get to 2000?  Well, 2000-189=1811.  But that is 1811 digts, and we are looking at [b]3-digit[/b] numbers.  So really only 1811/3 numbers, which comes out to be 603.66.  We disregard the .66, because rounding up will take us past the 2000th digit.  So we want to add 603 3-digit numbers.  Once we add these, we will have listed how many digits?

9 digits: 9 1-digit numbers
180 digits:90 2-digit numbers
1809 digits: 603 3-digit numbers

So 1998 digits.

Now we need to find the last number we wrote.  We wrote 603 3-digit numbers, and the 603rd 3-digit number is 702.  Thus the last number we wrote was 702.  Thus, the next number we will write is 703.  The 7 is the 1999th digit, and the $\boxed{0}$ is the 2000th digit. (The 3 is the 2001st digit).

$\boxed{QED}$

(P.S. - Sorry, in my outline solution, I forgot the fact that we add 603 3-digit numbers, not have 603 numbers all together.  Thus I made an error that threw me off by 99 digits (which is how many 1 digit and 2 digit numbers I forgot to include).  SORRY!!!)

P.P.S. - You can do it the way I did above in the outline, but you must add 99 numbers in the final step to account for the 99 I skipped.

alright, thanks a lot.  that was very thorough and i get it now ;-)


floor function
number theory proposed
number theory

Let $n\geq 2$ be integers  and $r$ be positive integers such that $r$ is not the multiple of $n$, and let $g$ be the greatest common measure of $n$ and $r$. Prove that \[ \sum_{i=1}^{n-1} \left\{\frac{ri}{n} \right\}\ =\frac{1}{2}(n-g).  \] where $\left\{x\right\}$ is the fractional part, that is to say, which means the value by subtracting $x$ from the maximum integer value which is equal or less than $x$.

First we evaluate $\sum_{j = 1}^{n - 1} ( rj \mbox{ mod } n )$, then having note that $rj - n \lfloor \frac{rj}{n} \rfloor = rj \mbox{ mod } n$, we obtain an
estimate for $\sum_{j = 1}^{n - 1} \lfloor \frac{rj}{n} \rfloor$, and then $\sum_{j = 1}^{n - 1} \left\{ \frac{rj}{n} \right\}$ is straightforward.

Begin by the remark that $rj = a ( \mbox{mod } n ) \Leftrightarrow ( r, n ) =g|a$, and we have exactly $( r, n ) = g$ integers $j$ (in Z/nZ) veryfying $rj = a ( \mbox{mod } n )$ for a given $a.$ So
\[ \sum_{j = 1}^{n - 1} ( rj \mbox{ mod } n ) = \sum_{g|a} g \cdot a = g \sum_{dg \leq n - 1} dg = g^2 \sum_{d \leq ( n - 1 ) / g} d = g^2 \cdot \frac{\left\lfloor \frac{n - 1}{g} \right\rfloor \cdot \left( \left\lfloor \frac{n - 1}{g} \right\rfloor + 1 \right)}{2}  \]
Homever $g = ( r, n ) |n$, so $g \nmid n - 1$ and thus $\left\lfloor \frac{n -1}{g} \right\rfloor = \frac{n}{g} - 1$, so
\[ g^2 \cdot \frac{\left\lfloor \frac{n - 1}{g} \right\rfloor \cdot \left( \left\lfloor \frac{n - 1}{g} \right\rfloor + 1 \right)}{2} = g^2 \cdot \frac{\frac{n}{g} \cdot ( \frac{n}{g} - 1 )}{2} = g^2 \cdot \frac{n \cdot ( n - g )}{2 g^2} = \frac{n \cdot ( n - g )}{2}  \]
Now remark that $rj \mbox{ mod } n = rj - n \lfloor \frac{rj}{n} \rfloor$, this leads us to
\[ \sum_{j = 1}^{n - 1} rj - n \sum_{j = 1}^{n - 1} \lfloor \frac{rj}{n} \rfloor = \sum_{j = 1}^{n - 1} ( rj \mbox{ mod } n ) = \frac{n \cdot ( n - g )}{2}  \]
And so
\[ \sum_{j = 1}^{n - 1} \lfloor \frac{rj}{n} \rfloor = \frac{r}{n} \sum_{j = 1}^{n - 1} j^{} - \frac{n - g}{2}  \]
Finally
\[ \sum^{n - 1}_{j = 1} \left\{ \frac{rj}{n} \right\} = \frac{r}{n} \sum_{j = 1}^{n - 1} j - \sum_{j = 1}^{n - 1} \lfloor \frac{rj}{n} \rfloor = \frac{r}{n} \sum_{j = 1}^{n - 1} j - \frac{r}{n} \sum_{j = 1}^{n - 1} j^{} + \frac{n - g}{2} = \frac{1}{2} ( n - g )  \]

A much shorter solution: Let $n=gn_1,r=gr_1$ with $(n_1,r_1)=1$, and note that, $\frac{r_1i}{n_1}=\frac{q_i}{n_1}$, where $r_1i \equiv q_i \pmod{n_1}$. Observe also that, $r_1(n-i)\equiv n_1-q_i\pmod{n_1}$ (if $q_i\neq 0\pmod{n_1}$), and $r_1(n-i)\equiv 0\pmod{n_1}$ otherwise. Now, $r_1i \equiv 0\pmod{n_1}\iff n_1 \mid r_1i \iff n_1\mid i$ since $(n_1,r_1)=1$, and this holds only for $i =g'n_1$, where $g'\in\mathbb{Z}$ such that $1\leq g'\leq g-1$. In particular, except $(n-1)-(g-1)=n-g$ terms in the sum, the remaining terms are all zero, modulo $n_1$. With this, $2 \sum_{i=1}^{n-1} \left\{\frac{ri}{n} \right\}\ = \frac{n_1(n-g)}{n_1}$, from here the conclusion follows.


If x and y are integers, compute the number of ordered pairs (x,y) that satisfies the equation 24 + x^2 = y^2

[hide="Solution"]
Since the difference of consecutive squares is odd, $|x-y| \ne 1$.  We need to find an even number of consecutive odds that add to $24$.
- $11$ and $13$ works.  $(5, 7)$ is an answer.
- $3, 5, 7, 9$ works.  $(1, 5)$ is an answer.
It is simple to prove that there are no more.
Therefore our answer is $\boxed{(5, 7), (1, 5)}$.
[/hide]

They do.  I guess miyomiyo forgot those.  In that case, there would be four times as many solutions as miyomiyo said.

Whoops - I though the question asked for positive integers.


induction
IMO Shortlist
Sequence
inequality system

The sequence $ \{a_n\}$ of integers is defined by
\[ a_1 \equal{} 2, a_2 \equal{} 7
\]
and
\[ \minus{} \frac {1}{2} < a_{n \plus{} 1} \minus{} \frac {a^2_n}{a_{n \minus{} 1}} \leq \frac {}{}, n \geq 2.
\]
Prove that $ a_n$ is odd for all $ n > 1.$

Let $x_{n}=3x_{n-1}+2x_{n-2}$ and $x_1 =2, x_2 =7$
now $x_{n+1}.x_{n-1} -x^2_n=(-2)^{n-2}$ it also satisfy that inequality of $a_n$
and also $x_n>2^n$ so  $x_n=a_n$ ,so done.


I don't understand how to solve this question.

For how many non-negative real values of x is $ \sqrt {144 \minus{} \sqrt[3]{x}}$ an integer?

[hide="solution"] To make $ \sqrt{144\minus{}\sqrt[3]{x}}$ be an integer, $ 144\minus{}\sqrt[3]{x}$ must be a perfect square. There are 13 squares from 0 to 144 inclusive (0 squared to 12 squared) and we have an x for each one, so 13.[/hide]


group theory
abstract algebra
algebra
function
domain
Ring Theory
modular arithmetic

Prove that every subgroup of a cyclic group is also cyclic. Thanks in advance.

A subgroup of the cyclic group $ \mathbb{Z}/n\mathbb{Z}$ has the form $ H/n\mathbb{Z}$ for some subgroup $ H\le \mathbb{Z}$.
But $ H$ is also an ideal of $ \mathbb{Z}$, because $ mh\equal{}\underbrace{h\plus{}\cdots \plus{}h}_{m\times}\in H$ for all $ m\in\mathbb{N},h\in H$.
But $ \mathbb{Z}$ is [url=http://en.wikipedia.org/wiki/Principal_ideal_domain]PID[/url], so $ H\equal{}t\mathbb{Z}$ for some $ t\in\mathbb{Z}$.
This gives $ H$ and $ H/n\mathbb{Z}$ cyclic.

In fact, the inverse is right, i.e., every proper subgroup of finite group G is cyclic, then G is cyclic.

first, G is soluble since Sylow 2-subgroup is cyclic. second, since every Sylow subgroup is cylic, then G is metacyclic, then G is cyclic or <a,b| a^m=b^n=1, a^b=a^r> with ((r-1)n,m)=1, r^n=1(mod m) and |G|=nm, of course,  it is cyclic since every proper subgroup is cyclic.

[quote="rulin"]G is soluble since Sylow 2-subgroup is cyclic.[/quote]
You want to use [url=http://en.wikipedia.org/wiki/Odd_order_theorem#CITEREFPeterfalvi2000]Feit–Thompson[/url]?
And you missed the exceptions $ C_p\times C_p$, $ p$ prime and $ Q_8$ and
$ \langle x,y: x^p \equal{} y^{q^n} \equal{} 1, yxy^{ \minus{} 1} \equal{} x^r\rangle$ with $ p,q$ primes, $ q|p \minus{} 1$ and $ r,n\in \mathbb{N}$ with $ r\not\equiv 1\equiv r^q\pmod{p}$.
($ r$ always exists for $ q|p \minus{} 1$)
The last example include the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral[/url] and [url=http://en.wikipedia.org/wiki/Generalized_quaternion_group]quaternion[/url] groups $ D_{2p}$ and $ Q_{4p}$ for any odd prime $ p$. (like $ D_6 \equal{} S_3$)

More elementary solution of original problem:
Let $ G \equal{} \langle a\rangle$ and $ H < G$. Assume that $ k$ is the minimal natural number s.t. $ a^{k}\in H$. For $ h\in H$ there exists $ l \in \mathbb{N}$ s.t. $ h \equal{}  a^{l}$. If $ k\not | l$, then $ \gcd(k,l) < k$ and form Bezout Identity we have $ m,n \in \mathbb{Z}$ s.t. $ \gcd(k,l) \equal{} mk \plus{} nl$. But then $ a^{\gcd(k, l)}\in H$ - contradiction, which implies $ \forall h \in H\ \exists s\in \mathbb{N} \quad h \equal{} a^{ks} \equal{} (a^{k})^{s}$ and $ H \equal{} \langle a^{k}\rangle$.


calculus
integration
quadratics
function
derivative
algebra
calculus computations

I'm working on these 2 following problems for like 30 minutes and still get stuck. Hopefully someone can help me solve them:
1.A particle moves on a straight line so that its velocity at time t is given by $ v\equal{}12\sqrt{s}$, where s is its distance from the origin. If $ s\equal{}1$ when $ t\equal{}0$, then find s when $ t\equal{}1$.
I keep getting the quadratic equation after integrating both sides of the given equation like this one: $ s\equal{}12t\sqrt{s}\plus{}1$. But solving this equation give me no ans suited like the multiple choice's answers.
2. After t years, $ 50e^{\minus{}0.015t}$ pounds of a deposit of a radioactive substance remain.What is the average amout per year not lost by radioactive decay during the second hundred years?
I don't really get this problem, so I just think taking the definite integral from t=0 to t=200 of the given function and dividing by 200. Again, no suitable answer like those given.
Sorry if those are really easy for you. I'm disappointed by the fact that these are AP questions.

[quote="ghjk"]I'm working on these 2 following problems for like 30 minutes and still get stuck. Hopefully someone can help me solve them:
1.A particle moves on a straight line so that its velocity at time t is given by $ v \equal{} 12\sqrt {s}$, where s is its distance from the origin. If $ s \equal{} 1$ when $ t \equal{} 0$, then find s when $ t \equal{} 1$.
I keep getting the quadratic equation after integrating both sides of the given equation like this one: $ s \equal{} 12t\sqrt {s} \plus{} 1$. But solving this equation give me no ans suited like the multiple choice's answers.
2. After t years, $ 50e^{ \minus{} 0.015t}$ pounds of a deposit of a radioactive substance remain.What is the average amout per year not lost by radioactive decay during the second hundred years?
I don't really get this problem, so I just think taking the definite integral from t=0 to t=200 of the given function and dividing by 200. Again, no suitable answer like those given.
Sorry if those are really easy for you. I'm disappointed by the fact that these are AP questions.[/quote]

1. $ v \equal{} 12\sqrt {s}$

 $ \int {v} \equal{} s(t)$

$ \int{ v} \equal{} 8s^{3/2} \plus{} C$

Then, $ s(0) \equal{} 8 * (0)^ {3/2} \plus{} C \equal{} 1.$ Therefore, $ C \equal{} 1$

Next:

$ s(1) \equal{} 8 * (1) ^{3/2} \plus{} 1 \equal{} 8 \plus{} 1 \equal{} 9$

2.

You're on the right track, but notice it asks for the [b]second[/b] hundred years.

$ 1/ (200 \minus{} 100) * \int_{100}^{200} 50e^{ \minus{} 0.015t} dt$

[quote="An0maly"]

1. $ v \equal{} 12\sqrt {s}$

 $ \int {v} \equal{} s(t)$

$ \int{ v} \equal{} 8s^{3/2} \plus{} C$

Then, $ s(0) \equal{} 8 * (0)^ {3/2} \plus{} C \equal{} 1.$ Therefore, $ C \equal{} 1$

Next:

$ s(1) \equal{} 8 * (1) ^{3/2} \plus{} 1 \equal{} 8 \plus{} 1 \equal{} 9$[/quote]
Notice that if the equation after integrating both sides are: $ 8s^{3/2} \plus{} C\equal{}s(t)$, you won't have the same equation if you differentiate this whole equation again, with [b]respect to t[/b]. This is why my equation has a form: $ s(t)\equal{}12.t\sqrt{s}\plus{}C$, which I assume $ 12\sqrt{s}$ is no more than a constant. Perhaps I'm wrong with this assumption, but that's the only way I can think of to make sure that going backward-taking derivative-won't be wrong. Plus, your answer is not on the given answers. :(

$ v(t) \equal{} \frac {ds}{dt} \equal{} 12\sqrt {s}$
This is a separable differential equation. It separates as $ \frac {ds}{\sqrt {s}} \equal{} 12 \,dt$
Taking the anti-derivative of both sides gives, $ 2\sqrt {s} \equal{} 12t \plus{} C$
Since $ s \equal{} 1$ when $ t \equal{} 0$, we have $ C \equal{} 2$
Therefore when $ t \equal{} 1$, $ 2\sqrt {s} \equal{} 12 \plus{} 2$. Solving for $ s$ gives $ s(1) \equal{} 49$

[quote="gauss202"]$ v(t) \equal{} \frac {ds}{dt} \equal{} 12\sqrt {s}$
This is a separable differential equation. It separates as $ \sqrt {s} \,ds \equal{} 12 \,dt$
Taking the anti-derivative of both sides gives, $ \frac {2}{3}s^{3/2} \equal{} 12t \plus{} C$
Since $ s \equal{} 1$ when $ t \equal{} 0$, we have $ C \equal{} \frac {2}{3}$
Therefore when $ t \equal{} 1$, $ \frac {2}{3}s^{3/2} \equal{} 12 \plus{} \frac {2}{3}$. Solving for $ s$ gives $ s(1) \equal{} 19^{2/3}$[/quote]

Do you mean

$ \,ds / \sqrt {s}\equal{} 12 \,dt$ ?

[quote="gauss202"]
Taking the anti-derivative of both sides gives, $ 2\sqrt {s} \equal{} 12t \plus{} C$[/quote]
Just one minor mistake: $ 2\sqrt {s} \equal{} 12t \plus{} C$ should be $ 2\sqrt {s} \equal{} 12t \plus{} C$
Therefore, the final ans is: $ s \equal{} 49$.
Thank you very much for your help, Mr. Gauss202.I'm too embarrassed. :oops:
[Edit: He did notice that. So ignore this post except my thank for you, Mr. Gauss202]


[quote="marshell"]Let $ a,b,c \ge 0: a \plus{} b \plus{} c \equal{} 3$. Find the maximum of the following expressions:

$ A \equal{} \frac {(a \minus{} b)^2}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}} \plus{} \frac {(b \minus{} c)^2}{\sqrt {b^2 \plus{} c^2 \plus{} 2a^2}} \plus{} \frac {(c \minus{} a)^2}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}$

[/quote]

Now let's start with some ideas  :)
Assume that $ a\ge b\ge c\ge 0$. Analysing this expression(using $ (a\minus{}c)^2\equal{}(a\minus{}b\plus{}b\minus{}c)^2\equal{}(a\minus{}b)^2\plus{}2(a\minus{}b)(b\minus{}c)\plus{}(b\minus{}c)^2$)

$ A\equal{}(a\minus{}b)^2(\frac{1}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}}\plus{}\frac{1}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}})\plus{}(b\minus{}c)^2(\frac{1}{\sqrt {b^2 \plus{} c^2 \plus{} 2a^2}}\plus{}\frac{1}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}})\plus{}\frac{2(a\minus{}b)(b\minus{}c)}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}$

Note that $ a\ge b\ge c\rightarrow \sqrt {b^2 \plus{} c^2 \plus{} 2a^2}\ge \sqrt {c^2 \plus{} a^2 \plus{} 2b^2}$ so

$ A\le (a\minus{}b)^2(\frac{1}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}}\plus{}\frac{1}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}})\plus{}\frac{2(b\minus{}c)^2}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}\plus{}\frac{2(a\minus{}b)(b\minus{}c)}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}\equal{}(a\minus{}b)^2(\frac{1}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}}\plus{}\frac{1}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}})\plus{}\frac{2(a\minus{}c)(b\minus{}c)}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}$

Note that $ \frac{1}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}}\plus{}\frac{1}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}\le \frac{2}{a}$ and $ 0\le a\minus{}c\le\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}$ so

$ A\le \frac{2(a\minus{}b)^2}{a}\plus{}2(b\minus{}c)\le2(a\minus{}b)\plus{}2(b\minus{}c)\equal{}2(a\minus{}c)\le6$

Finally,$ A\le 6$. Equality occur if and only if $ (a,b,c)\equal{}(3,0,0);(0,3,0)$ or $ (0,0,3)$

[quote="marshell"]Let $ a,b,c \ge 0: a \plus{} b \plus{} c \equal{} 3$. Find the maximum of the following expressions:

$ A \equal{} \frac {(a \minus{} b)^2}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}} \plus{} \frac {(b \minus{} c)^2}{\sqrt {b^2 \plus{} c^2 \plus{} 2a^2}} \plus{} \frac {(c \minus{} a)^2}{\sqrt {c^2 \plus{} a^2 \plus{} 2b^2}}$

$ B \equal{} \sum\frac {a^2 \plus{} b^2 \minus{} ab}{\sqrt {a^2 \plus{} b^2 \plus{} 2c^2}}$[/quote]
We will show $ A \le B \le 6$. The first is trivial, for the second, we have
\[ B \le \sum \frac{a^2\minus{}ab\plus{}b^2}{\sqrt{a^2\plus{}b^2\plus{}c^2}} \equal{}\frac{2\sum a^2 \minus{}\sum ab}{\sqrt{\sum a^2}}
\]
It suffices to prove that
\[ 2\sum a^2 \minus{}\sum ab \le 6\sqrt{\sum a^2}
\]
Denote $ x\equal{}\sqrt{\sum a^2}$ then $ x \le 3$, then inequality becomes
\[ 2x^2 \minus{}(9\minus{}x^2)/2 \le 6x
\]
\[ \Leftrightarrow 5x^2 \minus{}12x\minus{}9 \le 0
\]
or
\[ x \le 3
\]
:)


trigonometry
integration
real analysis
real analysis unsolved

Hey all,

doing some exam revision....   i can't work out how the book got to the answer for this question.....

[img]http://img62.imageshack.us/img62/4913/picture25dn5.png[/img]

[img]http://img62.imageshack.us/img62/4871/picture26gd8.png[/img]

could some please walk me through it?

Cheers

Sarah :)

[quote="mlok"]Since $f$ is even, its Fourier series has only cosines. Find their coefficients by integrating $f(x)\cos nx$ - this reduces to $\int_{0}^{\pi}\sin x\cos nx dx$, which is handled with that identity.[/quote]

ok, yep got it. thanks :)


[color=blue][size=150]چیه Isogonal Conjugateسلام کسی می دونه معنی[/size] [/color]

dar $\triangle{ABC}$, agar $D$ noghtei dar mosalas bashad, "ISOGONAL CONJUGATE" noghteye $D'$ mibashad. ke agar khotoot e AD, BD, CD dor mehvar e nimsaz e zaviye haye A, B, C baztabideh shavad mahal e barkhord e in khotoot e baztabideh shode $D'$ khahad bood.

[quote="amir2"]ThanX a lot $\text{amirhtlusa}$

Yani hamoon mozdavaj ham zaviye khodemoon. :)[/quote]

daghighan. :)

[quote="fatehi"]Karbord in noghte chiyeh?[/quote]

[url=http://mathworld.wolfram.com/IsogonalConjugate.html]Isogonal_conjucate: Math_World[/url]

akh joon... :D

amir joon to az che ketabaiy dars mikhooni?

yani mikham bedoonam,ketabaye maro dari?(mesle court,mariam mirza khani,...)

divooneh manzooram amirhtlusa ast :rotfl:

:blush:

man, vaghti ke dashtam miyoomadam inja, chata ketab e olampiad avordam....vali na, man hameye ketabayi ke shoma migin ro nadaram... :(

vali inja Ketab haye ingilisi ham ziyad hast ke man oon a ro ham mikhoonam.


geometry
3D geometry
tetrahedron
sphere
rhombus

Prove that, in every tetrahedron, you can circumscribe a sphere around it and inscribe a sphere inside it.

[quote="Treething"][hide="Hint"]
Prove the vertices are equidistant from the center of the tetrahedron and the centroids of the faces are also equidistant from the center.
[/hide][/quote]
well thats not true, from what the problem is asking, you can have an irregularly shaped tetrahedron and still be circumscribable and yet that center is not equidistant to the vertices!
center does not mean equidistant from vertices, like the center of a rhombus, it is not equidistant from all vertices of the rhombus in general

The center, as in the centroid. The intersection of the four lines connecting a vertex to the centroid of the opposite face. At least that's what I think Treething meant by center.

if this is what he meant, then the second part of his hint isn't true

My thought-- I don't know if it's good enough,. or right for that matter...

A tetrahedron is made of 4 points that are not coplanar. We take 3 of those points, and it can be shown that there are infinitely many spheres that have those 3 points on the surface. Therefore any 4th point lies on one of those spheres, and any tetrahedron can be circumscribed. :)

And I don't know how to do the second part, maybe it can be done in the same fashion.


linear algebra
matrix
linear algebra unsolved

Let  A be the matrix   $\left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array}\right)$ , How many orthogonal matrix P exist so that $P^{-1} A P$ is a diagonal matrix.  ( X is a  orthogonal matrix if $X^{-1}=X^t$)

Looks like that there are infinity matrix..
Hmmm.. can someone proof???
[]´s


geometry
trapezoid
algebra
polynomial
geometric transformation
LaTeX
analytic geometry

I am interested in mediterranean mathematical olympiads.
It started in 1998 and I have found all of them except 1999.

Can anyone provide me with the missing one?

Also did the 2005 MMC(Medit. Math Contest as its official name is) take place?
If yes can anyone post the problems? If not when?...

could you please post the previous mediteranian olympiads that you have ?  :lol:

Here are the problems for year 2001:

1 Let C (O, R) be a circle and PQ a chord of this
circle. Let M be the midpoint of PQ and A, C points on the circle C
such that the chord AC intersects PQ at M. The quadrilateral ABCD is
a trapezium inscribed in the circle C and AB, CD are parallel to PQ.
Prove that the intersection point X of the lines AD, BC does not
depend on the choice of A on the circle C.

2 Find all integers n such that the polynomial
P(x) = x^5 - nx - n - 2
can be written as the product of two non-constant polynomials with
integer coefficients.

3.Prove that there exist positive integer N such that the decimal
representation of the number 2000^N starts with the sequence of the
digits 200120012001.

2.4 Let ABC be an equilateral triangle of side 1. Let S be
the set of points that belong on the sides of ABC or inside the
triangle ABC. For any point M of the set S we call a(M),b(M) and c(M)
the distances of M from the sides BC, CA and AB respectively.
We define the expression f(M) (for M in S) as follows
f(M) = a(M)^3(b(M) - c(M)) + b(M)^3( c(M) - a(M)) + c(M)^3(a(M) - b
(M))
a) Describe geometrically the set of points {M in S : f(M)>=0 }
b) Find the maximal and minimal value of the expression f(M), where M
is in S, and the points M for which we take these values.

But I have been informed that the 2005 MMC has already taken place.
Do you know the problems?
If you have them post them, and I have more to post too :lol:

As far as i know 2005MMC is going to take place the next sutarday.
In Greece, we use the problems of the MMC as a TST examination for BMO and IMO.
But i have been informed that the TST will be held the next sutarday, so...

I have the problems from MMC2003 and MMC2004, I can post them if you miss any of those.

No 2003 and 2004 I have.

Anto do you have 1999MMC?

I'd like to see the problems from 1998 to 2005, Year 2001 problems had been posted (see above), can you post other years?
By the way, does any website have the problems or links to this competition?

So did the 2005 MMC take place?

Anto can you post here the problems?

[quote="Anto"]I was mistaken.  :blush:
This year, for some reason (i don't know why...), we didn't use the problems of MMC as TST.
I don't even know whether or not the MMC 2005 took place.
If you can shed some light on it, i would be gratefull.[/quote]

in Croatia it will be held in saturday, 16th april

Sorry about bad translation and writing without latex:

1999 MMC

1. Is there a circle and infinite set of points on it such that the distance of each two points (from that set) is rational number?

2. There is a figure (in plane) with area A. If A>n for some positive integer n, prove that this figure can be placed in Cartesius system so that it covers at least n+1 points with integer coordinates ((a, b) e Z^2).

3. a, b, c e R/{0}. x, y, z e R+,  x+y+z+3. Prove that

3/2 (sqrt(1/a^2 + 1/b^2 + 1/c^2)) >= x/(1+a^2) + y/(1+b^2) + z/(1+c^2)

4. Let ABC be a triangle with lenghts |BC|=a, |CA|=b, |AB|=c and angleABC=4*angleBAC. Prove that
a*b^2*c^3=(b^2-a^2-ac)[(a^2-b^2)^2-a^2*c^2]

Thanks a lot idioteque!

Hope you will be able to provide me with the 2005MMC as well(after Aplril 16).

I will try to make an effort and post the rest of the years that I have
in the following days: ie 1998,2000,2002,2003,2004(2001 I have already posted)


search
combinatorics unsolved
combinatorics

Hi everybody, i have a question for you, i m a bit in trouble with this problem,

the questions is: Suppose $ G$ is regular of degree r and has a girth g or greater. Find the lower bound for $ |V(G)|$. (consider the case g even and odd separately)

Thanks a lot to everybody

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=496069271&t=3503

Pierre.


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

let C be a 4x4 matrix on $ \mathbb{C}$; let $ P\equal{}(x\minus{}3)^2(x\minus{}2)^2$ be its characteristic polynomial, equal to its minimal polinomial; why (if correct) from this does it follow that $ rank(C\minus{}2)^2\equal{}2$
?

In fact, that's wrong. If this is its minimal polynomial, then $ \text{rank}(C\minus{}2I)\equal{}3,$ not $ 2.$

A matrix in Jordan form that has this characteristic and minimal polynomial would be:

$ C\equal{}\begin{bmatrix}3&1&0&0\\ 0&3&0&0\\ 0&0&2&1\\ 0&0&0&2\\ \end{bmatrix}.$

Then $ C\minus{}2I\equal{}\begin{bmatrix}1&1&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&0&0\\ \end{bmatrix},$ which clearly has rank $ 3.$

If the matrix $ A$ has this characteristic polynomial but minimal polynomial $ (x\minus{}3)^2(x\minus{}2),$ then $ A\minus{}2I$ would have rank $ 2.$

ok, but I wrote[quote="Clausius"]$ rank(C \minus{} 2)^2 \equal{} 2$[/quote], not "$ rank(C\minus{}2)\equal{}2$"

Oh - I did misread that. You should be able to use the fact that this matrix can be put into Jordan form in your proof.


geometry
incenter
circumcircle
geometric transformation
homothety
angle bisector
geometry proposed

Let ABC a triangle and I the incenter. Let d the tangent in A to the circumcircle and p a line parallel to BC through I. Let C(M) and F(N) the circles tangent to AB,d,p and AC,d,p respectively (M and N are the centers). Prove that AI $\perp$ MN.

[color=indigo][i]Here is the Solution[/i]

Let $T$ be the intersection of lines $d,p$ and $M'$ the midpoint of the arc $AB$.
First of all, we prove that $AM'$ is the angle bisector of $\angle{BAT}$ (angle chasing)
By angle chasing, we can also prove that $TA=TI$. It is well-known the fact that $M'A=M'I$.
Thus, $TM'$ is the angle bisector of $\angle{ATI}$ and therefore we get that $M'=M$
Similarly, we prove that $N'=N$ where $N'$ is the midpoint of the arc $AC$.
Now it is well-known and easy to prove that $M'N'\perp AI$

[i]Nick Rapanos[/i][/color]

Well I think this was just a copy (made by me :oops: ) of a problem given in Russia in 1999. I am curious if the problem is still true if we take the circles in the 'other direction' - > observe that are 2 circles tangent to AB, d, p for examples. So if we take the other 2 circles what we can say about the configuration? ( I have not spend time on it so it may be a stupid question  :blush:  )

EDIT: Ok I think Virgil Nicula's post solves it :)

[quote="Virgil Nicula"][quote="ciprian"][color=darkred][b]Let ABC a triangle and I the incenter. Let d the tangent in A to the circumcircle and p a line parallel to BC through I.
Let C(M,r_1) and F(N,r_2) the circles tangent to AB,d,p and AC,d,p respectively (M and N are the centers). Prove that AI $\perp$ MN.[/b][/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the intersection $T\in d\cap p$. Thus, $\{\begin{array}{c}TA=TI\\\ T\in MN\end{array}$ $\implies$ $\overline{TMN}\perp AI$ $\implies MN\perp AI$.
[b]Remark.[/b] Denote the intersections $\{\begin{array}{c}U\in AB\cap p\\\ V\in AC\cap p\end{array}$ and the second intersection $A'$ of the circumcircle with the line $AI$. Then $A'\in MU\cap NV$ and $\frac{c}{r_{1}}+\frac{b}{r_{2}}=\frac{2a}{r}$.[/color][/quote]

[color=blue][b]Another proof.[/b] Denote the intersections $D\in AI\cap BC$ and $V\in AA\cap BC$.
Prove easily that the triangle $AVD$ is $V$- isosceles.
Thus, the bisector of the angle $\widehat{AVD}$ is perpendicularly to the line $\overline{AID}$.
Therefore, the our problem is a consequence from homothety with the vertex $A$ a.s.o.
[b]Remark.[/b] Study and the intersection $U\in AA\cap A'A'$ !?[/color]


probability
function
logarithms
number theory
relatively prime
totient function
number theory open

Let $k$ be a (given) positive integer and choose two positive integers
$m,n$ at random. What are the probabilties
$w_1(k)$ that $k|\phi (n)$
$w_2$ that $m$ and $\phi (n)$ are relatively prime
( $\phi (n)$ is the totient function )

I remember this problem. It was based on the fact that the probability that two randomly chosen integers are relatively prime is [tex]\frac{6}{\pi^2}[/tex]. I still don't know more than [tex]w_2<\frac{1}{2}[/tex]  :P

ok, you know i solved it but you wanted a clear write-up, so here it goes:

we first show that $\omega_1(k)=1$ for all $k$. indeed, it suffices for $n$ to have a prime factor $\equiv 1\mod k$. let $p_1,...,p_l$ be the first $l$ primes $\equiv 1\mod k$.
take $n=p_1\cdots p_l N$ and only consider the probability for numbers $\leq n$. one easily sees that exactly $(p_1-1)\cdots (p_l-1) N$ numbers $\leq n$ are not divisible by any of the $p_i$. therefore the probability for a number not to be divisible by any of the $p_i$ is $(1-\frac 1{p_1})\cdots (1-\frac 1{p_l})$. by simple arguments one sees that actually for any $n\geq N_{l,\epsilon}$ we get the probability for a random number $m\leq n$ that $k\not|\phi(m)$ smaller than $(1-\frac 1{p_1})\cdots (1-\frac 1{p_l})+\epsilon$ for any $l$, $\epsilon>0$(approximate $n$ by some nearby number which is divisible by $p_1\cdots p_l$, the probability doesn't change much). but it is well-known that $\prod_{p\equiv 1\mod k} (1-\frac 1p)=0$ (the series $\sum_{p\equiv 1\mod k,p\leq n} \frac 1p$ diverges like $\frac 1{\phi(k)} \log\log n$ by some refined versions of the dirichlet prime number theorem). therefore the probability that $k\not|\phi(m)$ gets zero, as claimed.

now, knowing this, it is easy to compute $\omega_2$. since $\omega_1(p)=1$ for all primes $p$, we get the probability that the first $l$ primes $p_1,...,p_l$ divide $\phi(m)$ larger than $1-\epsilon$ for all $\epsilon$ and all $n\geq N_{\epsilon}$ where we only consider $m\leq n$. now, in order to have $\gcd(k,\phi(m))=1$, we either need to have the probability $<\epsilon$ that none of the primes $p_1,...,p_l$ divides $\phi(m)$ or the probability $<\frac{C}{p_1\cdots p_l}$ that $k$ is not divisible by any of the $p_i$ for some $C$ independent of $n$. by restricting to certain $n\geq N_{l,\epsilon}$ we may even get $C$ independent of $l$. but then, the probability is at most $\epsilon+\frac{C}{p_1\cdots p_l}$ for all $\epsilon, l$ and is therefore 0.

i hope this is clear enough.

Peter


In right angled triangle $ ABC$, $ \angle A\equal{}30^\circ$,  $ \angle C\equal{}90^\circ$, $ BC\equal{}1$. Among all the equilateral triangle with all the vertices on the sides of the triangle, which one has maximum area?

[quote="borislav_mirchev"]It is Vietnamese Math Olympiad 1996 problem.[/quote]

where I can find the discussion?

isn't it solved in All Over The world book?
http://www.mathlinks.ro/Forum/viewtopic.php?p=1247512#1247512 - it is the topic dedicated to that VMO problem ... but it is still empty.

i am still clueless. Anyone can help me? So far, I only can conjecture that the max area is $ \frac{1}{4}\sqrt{3}$.


Find the values of a and b if

lim x tends to infinity for
  (x^2+1)/(x+1) - ax+b = 2

$ \frac{x^2\plus{}1}{x\plus{}1}\minus{}ax\plus{}b\equal{}\frac{x^2\plus{}1\minus{}ax^2\minus{}ax\plus{}bx\plus{}b}{x\plus{}1}\equal{}\frac{x^2(1\minus{}a)\plus{}x(b\minus{}a)\plus{}b\plus{}1}{x\plus{}1}$.

Now, $ \lim_{x\to \infty}\frac{x^2(1\minus{}a)\plus{}x(b\minus{}a)\plus{}b\plus{}1}{x\plus{}1}\equal{}2$ implies $ a\equal{}1$ and $ b\equal{}3$.

Apply $ \frac{x^2\plus{}1}{x\plus{}1} \equal{} x\minus{}1 \plus{} \frac{2}{x\plus{}1}$


ok... im not getting this problem... someone plz help me

John has an ample supply of 10 cent and 15 cent stamps. He will buy an ample supply of stamps of one more denomination, which is a whole number. He needs to be able to combine the stamps to make any value of 30 cents or more. In cents what is the greates denomination of stamp he can buy which will enable him to do this???   :huh:

[hide="answer :censored: "] the answer is 6, but i have no idea how to get it [/hide]

So, using only 10 and 15 cent stamps we can only make multiples of 5.  Note that 31, 32, 33, and 34 all must be made, and they are 1, 2, 3, and 4 modulo 5.  Let the third denomination be $ k$.  If $ 4k\geq 35$, then we will be unable to make all four of these - we need four different residues mod 5 from multiples of $ k$.  So an upper bound is $ \frac{35}{4}$, thus $ k\leq 8$.

For 8, we cannot make 37.

For 7, we cannot make 33.

But for 6, we check that we can make 30, 31, 32, 33, 34, and 35.  We can then add 6 repeatedly to any of these to obtain any number greater than 30.  So 6 is the answer.

If you want some more information on these kinds of problems, check this out:

[url=http://en.wikipedia.org/wiki/Coin_problem]Chicken McNuggets and Coins![/url]

Not completely. I mean, is there a way to do it with out leaving mod 8 to test 31-39?

Well, his way is the way I would approach the problem. I can't think of a faster way right now...

kk, thnx for the solution everyone... especially E^(pi*i)=-1 and buzzer11 for the full soultion and the link... yay! 25 posts! :P

Well, if we take things mod 8, then 15, 10, and 8 reduce to -1, 2, and 0.  We see that getting something 5 mod 8 like 37 would require 3(2)+(-1) = 30+15 = 45 which is too big, or 3(-1) = 3(15) = 45 which again is too big. So you can do this part with mods.


superior algebra
superior algebra theorems

Every group $ G$ of order $ p^{2}$ $ \right(p\ prime)$ is abelian.

If $ G$ is cyclic proof complete. If not let $ a\in G$ with $ a\not \equal{} 1$ and construct $ \left< a\right>$ then it means there exists $ b\in G$ so that $ b\not\in\left< a\right>$ because we are assuming $ G$ isn't cyclic. Then means $ \left< a\right>\left< b\right> \equal{} G$ while $ \left< a\right>\cap\left< b\right> \equal{}\{ e\}$. So it means $ G\simeq\left< a\right>\times\left< b\right>$. So $ G$ is abelian. Q.E.D.

your condition for a group to be a direct product is wrong
for a correct proof, consult the link given by zetax


Let ABC be a  triangle inscribed in a circle of center O. A line d intersects BC, CA, AB in D,E,F respectively. Let X be a point on the circle of center O. The lines XD, XE, XF intersect again the circle in M,N,P. Prove that AM, BN, CP are concurrent in a point T on line d.

I will post a solution to this problem just for not having anymore the satanic number of posts you mentioned  :D . [I will use $K$ instead of $T$]

So here it goes:
Denote $I \in BC \cap PN$, $J \in AN \cap MB$, $H \in PM \cap AC$.

Now from applying the Pascal theorem in the cyclic hexagon $ANPMBC$ we obtain that the points $I,J,H$ are collinear. $(1)$.

But $AM, BN, CP$ are concurrent in $K$ if and only if $I,J,H$ are collinear, because $IJ, JH, IH$ are the polars of each pair of any two intersections from the trisegment set $\left\{AM, BN, CP\right\}$.

Thus by $(1)$ $AM, BN, CP$ are concurrent.

Now from proving that $K \in AM \cap BN \cap CP$ is on $D-E-F$, I went again for a projective method, I will return.


ahora si espero que lo resuelvan  :D

1) sean $AD , BE, CF$las bisectrices interiores del triangulo ABC , demostrar que si uno de los angulos $<ADF$, $<ADE, <BED, <CFE, <CFD$ mide 30° entonces al menos uno o mas de estos angulos mide 30°

bueno tengo problemas con el ingles , no se si alguien podria traducir este problema a ingles para ponerlo en el foro de geometria por favor  :wink:

TE DOY UNA AYUDA USA TRIGO. BUENO ASI LA ESTOY VIENDO YO, Y ME FALTA UN POCO, SI LA TERMINO LA POSTEO.CHAU :roll:

el <BEF no esta en esos angulos,te confundiste o de porsi a si es el problema????

si tienes razon me olvide de ese angulo  :blush:

yo tengo la solucion de ese problema ya que estuvo en el banco de problemas de la rioplatense 2001 y por cierto fue usado por peru en el selectivo conosur 2001 .

Pero no se si tenag tiempo de escribirla peor si la queiren escaneada pidanmela nomas


conics
ellipse
geometry
geometric transformation
reflection
AMC
AIME

[hide]

There are two key observations we state without proof.  First, if A and B are the foci of an ellipse, and T is the point of tangency of a line $ PQ$ tangent to the ellipse, then angle $ ATP$ is congruent to angle $ BTQ$.  That is to say, the tangent line forms equal angles with the segments joining the foci to the point of tangency.  The second observation is that the length of the major axis of an ellipse is equal to the sum of the distances from the foci to any point on the ellipse.

Hence, let A = (9,20) and B = (49,55).  Then let T = (x,0).  If we reflect A about the x-axis to A' = (9,-20), then by the first observation, it follows that A', T, and B are collinear.  By the second observation, the length of the major axis is equal to AT + BT = A'T + BT = A'B, from which we obtain

$ A'B \equal{} \sqrt{(49\minus{}9)^2 \plus{} (55\plus{}20)^2} \equal{} 5(17) \equal{} 85$.
[/hide]

[quote="CatalystOfNostalgia"][quote="sonny"]An ellipse has foci at $ (9,20)$ and $ (49,55)$. Find the length of it's major axis if the ellipse is tanget to the x-axis.[/quote]

Please use the contests section. I have no idea what year this is problem, so cannot point you to its thread.[/quote]

It is [url=http://www.artofproblemsolving.com/Wiki/index.php/1985_AIME_Problems/Problem_11]1985 AIME, #11[/url].

[quote="sonny"]CatalystOfNostalgia, thanks for the heads up!

Hmm, what I can't seem to figure out is WHY $ \angle{ATP}\cong\angle{BTQ}$[/quote]
The reason for this because of a fairly-well-known property of ellipse, which states that if we shoot a pool ball from one focus of an elliptical pool table, the ball will bounce off of the table's wall and travel into the second focus.


The sum of three positive integers is 20. If the smaller two integers differ by five, what is the smallest possible value of the largest integer?

Let the two smaller integers be $ x,x \plus{} 5$ such that $ x\geq1$ and the largest integer is equal to

$ 20 \minus{} x \minus{} (x \plus{} 5) \equal{} 15 \minus{} 2x\leq15 \minus{} 2\cdot1 \equal{} \boxed{13}$.

Edit: Sorry; you're correct.

Another way to notice this is to see that we must have $ 15\minus{}2x>x\plus{}5\implies 10>3x\implies x\leq3\implies15\minus{}2x\geq\boxed9$.

[quote=5849206328x]However, they want the smallest possible value of the largest integer
There are other ways, but we'll use guess and check for this:
1,6,13
2,7,11
3,8,9
Anything past this won't satisfy the given requirements, so the answer is $ \boxed{9}$[/quote]

an easy way to notice this is simply think of different sets of numbers under 20 that differ by 5.  there are only three sets that work when you try to add a third, bigger number in. Then it's a simple matter of finding the $ \boxed{3,8,9}$.


A public garden measuring 56 by 105 feet is to be surrounded by shrubbery. It is also to be divided in half by a wall of shrubbery running between the 56-ft sides and parallel to the other sides. If shrubbery costs $ \$30$/ft, what will be the total cost in dollars?

We know that the perimeter of the garden is $ 56\times2\plus{}105\times2\equal{}322.$
We also know that the garden is bisected by a shrub wall connecting the two 56-foot walls, or 105 feet. Add this to our total, and you get $ 427.$ Shrubbery is $ 30$ dollars a foot, so our answer is $ 30\times427\equal{}12810.$


algebra
polynomial
AMC
AIME
number theory
relatively prime
complex numbers

The equality:  $\left(\frac{1}{2}x^{2}-3x+5\right)^{2}-6\left(\frac{1}{2}x^{2}-3x+5\right)+10-2x=0$ has 4 solutions $a<b<c<d$. If $\frac{a}{b}+\frac{c}{d}=\frac{p-\sqrt{q}}{r}$ for positive integers $p$, $q$, $r$ where $p$ and $r$ are relatively prime, find $p+q+r$.

[hide="hint:"] $[f(x)-x]$ divides $[f(f(x))-x]$[/hide]

Elemennop wrote:[hide]It's just a root recursion thingy f(f(x))=x for f(x)=... so two of your roots are the roots of  and the other two are the roots of the remaining polynomial of that equation..

I'm too lazy to do the computation though.

edit> even the roots of f(x) are nonreal so how do you define greater than with complex numbers?[/hide]

Hidden Textuhhhh...you should be solving 1/2x^2-4x+5=0, because you have to subtract that x

Oh yea oops

Still kinda tedious computation, don't you think?

[hide]divide the equation by 2, let $f(x)=\frac{1}{2}-3x+5$, and $u=f(x)-x$

the equation is $f(f(x))=x$, or

$\frac{1}{2}(u+x)^{2}-3(u+x)+5=x$

$\frac{1}{2}u^{2}-3u+2ux+[\frac{1}{2}x^{2}-3x+5-x]=0$

$\frac{1}{2}u^{2}-3u+2ux+[u]=0$
$(u)(\frac{1}{2}\cdot u+2x-2)=0$

then solve the quadratics... $2-\sqrt{2}<4-\sqrt{6}<2+\sqrt{2}<4+\sqrt{6}$[/hide]

i think the computation is quite reasonable...an aime problem has a tricky step (that i gave in the hint more or less), then some computation to finish, i cannot see it taking more than 5 mins to churn out the algebra, but then again, i removed this problem because it was a bit too contrived, it is sort of like you see the solution or you don't...


What is the area of the plane of the inequality $3\le|x|+|y|\ge4$?

[hide]
Splitting up each inequality and solving for $|y|$ gives
$|y|\geq 3-|x|$
$|y|\geq 4-|x|$
To solve for $y$ in both equations, and graph them, we must set the other side to both positive and negative of itself.
So now we have:
$y\geq 3-|x|$
$y\geq |x|-3$
$y\geq 4-|x|$
$y\geq |x|-4$
But since we multiplied the right side by -1 in two of the cases, we must flip the $\geq$ sign to $\leq$.
So now our inequalities are:
$y\geq 3-|x|$
$y\leq |x|-3$
$y\geq 4-|x|$
$y\leq |x|-4$

Graphing these in the plane gives an squares inside of each other, with the inside one unshaded but the rest of the outside one unshaded.
So we must find the area of the larger and subtract from the smaller.
Since the squares are also rhombuses, we can use the rhombus are formula
$A=\frac 12 d_1d_2$
Since it is easier when you know the diagonals and not sides.
The larger one has diagonals of 8. So the area is 32. The smaller has diagonals 6 and the area is 18.
Subtracting gives $32-18=\boxed{14}$[/hide]


function
logarithms
real analysis
real analysis unsolved

Give a proof of

$exp(x)=\prod_{n \geq 0}(1-x^n)^{-\frac{u(n)}{n}}$ where u(n) is Moebius function

nice one! really cool!
let me denote $u$ with $\mu$, which i think is more standard :p
then.. i guess it's a formal identity, so..
take logarithms of both sides, to get: \[ x = \sum_{k=1}^{\infty}-\frac{\mu(k)}{k}\ln(1-x^k) = \sum_{k=1}^{\infty}\sum_{h=1}^{\infty} \frac{\mu(k)}{hk}x^{hk} \] now fix $hk = n$ and you get \[ x=\sum_{n=1}^{\infty} \frac{x^n}{n}\sum_{d|n} \mu(d) = \sum_{n=1}^{\infty} \frac{(\mu*1)(n)}{n}x^n \] but $\mu * 1 = e$ (where $e(k) = \delta_{1k}$, which is $1$ if $k=1$ and $0$ otherwise).
so you have the identity you're seeking for :)


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Let $G$ be a group of order $2n$, where $n$ is odd, with a subgroup $H$ of order $n$ satisfying $xhx^{-1}=h^{-1}$ for all $h$ in $H$ and all $x$ in $G\setminus H$. Prove that $H$ is commutative and that every element of $G\setminus H$ is of order 2.

I proved that $H$ is comutative, but i'm really not sure to say that $ord(G\setminus H)=2$ .
Anyway look at my solution (i hope is not wrong    )

From $xhx^{-1}=h^{-1}$ we get $xh=h^{-1}x (*)$ for any $x \in G\setminus H$ and $h \in H$. Now let $h \rightarrow h^{-1}$ so we have $xh^{-1}=hx$. Multiplying by $x$ in the left part and using $(*)$ we get $x^2h=hx^2$ for anyl $x \in G\setminus H$ and $h \in H$.
Using induction we can prove easily that $x^kh=hx^k$ for $k$ even and $x^kh=h^{-1}x^k$ for $k$ odd. This is true so for $k=n$(odd) we have $x^nh=h^{-1}x^n$, but $x^n=e$ for any $x \in G\setminus H \subset G$ so we get that $h=h^{-1}$ for any $h \in H$ from where $h^2=e$ for any $h \in H$. Well this is trivial. From this we get $hg=gh$ for any $g,h \in H$ which means $H$ is abelian (comutative).
Since $n$ is odd, let $n=2p+1$ where $p \in {\mathbb N}$. We have that $ord(H)=n=2p+1$ this means $h^{2p+1}=e$ for any $h \in H$. But we proved that $h^2=e$ for any $h \in H$ so from these results $h=e$ for any $h \in H$. So this means that $n=1$ so,
the group $G$ has $2$ elements and is of the form $G=\{x,e\}$. Evidently $H=\{e\},\ G\setminus H=\{x\}$. It is clearly that $x^2=e$.

q.e.d.

Let $x\in G\setminus H$. We have $[G:H]=2$ hence (it is well-known) $x^2\in H$. Then we have $xx^2x^{-1}=x^{-2}$, or $x^4=e$. The order of $x$ is 1, 2 or 4. But $n$ is odd, hence 4 is not the order from Lagrange's Theorem. And it is clear $x\neq e$ since $x\notin H$. It must have the order of 2.

p.s. The well-known fact is: If $H$ is a normal subgroup of $G$, and $[G:H]=m$ finite, then $x^m\in H$ for all $x\in G$.

Hi Liyi,

How do you know that $[G:H]=2$?

Hi, Liyi

Soory I didn't read the problem carefully before.  NIce solution!


Bisectors of angles $ \angle A$, $ \angle B$ and $ \angle C$ intercept opposite sides of triangle in $ A_1$, $ B_1$ and $ C_1$ respectively. Let M be an arbitrary point on one of the segments $ \bar{A_1B_1}$, $ \bar{B_1C_1}$ and $ \bar{C_1A_1}$, and let $ M_1$, $ M_2$ and $ M_3$ be its orthogonal projections to the lines $ BC$, $ CA$ and $ AB$ respectively. Then one of segments $ MM_1$, $ MM_2$ and $ MM_3$ is equal to the sum of the remaining two. Prove!


Ελπίζω να 'ναι σωστή, γιατί κοντεύει 5 τα χαράματα και τώρα βρήκα χρόνο να ασχοληθώ λίγο ... Τέλος πάντων!

Για $ a,b,c,d\in\mathbf{R}$ , τα οποία ικανοποιούν $ a\cdot b \equal{} \sin t$,   $ c\cdot d \equal{} \sin z$, με $ a,b$ και $ c,d$ συζυγείς ανά ζεύγη, να δειχθεί ότι:

$ (ab \minus{} cd)^2 \plus{} [(a \minus{} c)(d \minus{} b)]^2\leq 4$

Στέλιος

Αφού $ a,b,c,d \in \mathbb{R}$ τι εννοείς "συζηγείς"?

Σε παραπέμπω εδώ:

http://heron.hms.gr/gr/activities/competitions/0506/bllthal.gif

Look at the 1st one...

Ούτε εδώ "παίζουν" complex nums, right ?

Συζυγή εννοώ κάτι της μορφής:

Αν x+5y είναι ο ένας όρος, ο συζυγής του θα είναι ο x-5y.

Είναι όπως στα όρια π.χ. που για να εξαλείψουμε απροσδιοριστία λέμε, "πολ/ζουμε με τη συζυγή παράσταση, άρα ..."

Στέλιος

Παιδιά , σε μισή ώρα φεύγω 7ήμερη , αλλά ακόμη είμαι πάνω στο pc.. :P

Anyway.. Βάζω τη λύση (in english - no time for shift+alt all the time... ):

Let $ \cos t \equal{} x$. It clearly follows that $ \sin t \equal{} \sqrt {1 \minus{} x^2} \equal{} \sqrt {1 \minus{} x}\cdot\sqrt {1 \minus{} x}$. That is, $ a \equal{} \sqrt {1 \minus{} x}$, $ b \equal{} \sqrt {1 \plus{} x}$, according to the hypothesis of the problem. The same goes for $ c,d$, as well. So, the ineq becomes:

$ (\sqrt {1 \minus{} x^2} \minus{} \sqrt {1 \minus{} y^2})^2 \plus{} (x \minus{} y)^2\leq4$

or

$ \sqrt {(\sqrt {1 \minus{} x^2} \minus{} \sqrt {1 \minus{} y^2})^2 \plus{} (x \minus{} y)^2}\leq 2$

Let $ f(z) \equal{} \sqrt {1 \minus{} z^2}$. It's easy to check that this function is a semicircle. The diametre has length $ 2$, so the distance of every two arbitary points on the semicircle is less than $ 2$ (the diametre). Take the type of the distance for $ 2$ arbitary points on that circle and the solution is obvious!

See you in a week!

Cheers,

Stelios


inequalities
geometry
rectangle
geometry proposed

Let $ A_1,A_2,...,A_{2002}$ be arbitrary points in the plane. Prove that for every circle of radius $ 1$ and for every rectangle inscribed in this circle, there exist $3$ vertices $ M,N,P$ of the rectangle such that
$ MA_1 + MA_2 + \cdots + MA_{2002} + $ $NA_1 + NA_2 + \cdots + NA_{2002} + $ $PA_1 + PA_2 + \cdots + PA_{2002}\ge 6006$.

Let $W,X,Y,Z$ be the vertices of the rectangle, and $A$ a point in the plane. Then $WA+XA+YA+ZA = (WA+YA)+(XA+ZA) \geq WY + XZ = 4$. This means $\sum_{k=1}^n WA_k + \sum_{k=1}^n XA_k + \sum_{k=1}^n YA_k + \sum_{k=1}^n ZA_k \geq 4n$, so the larger three terms of the LHS will sum up to at least $3n$.


En la olimpiada rusa se proponen muchos [b]buenos[/b] problemas, algunos fáciles otros díficiles y otros muy difíciles, pero todos son interesantes. Voy a ir posteando problemas de esta Olimpiada, cuando alguien lo haya resuelto completamente, posteo otro. Que disfruten los problemas!!

 [color=darkblue]Problema 1[/color]
Los enteros $1,2,3, \ldots, 200$ son particionados en 50 conjuntos. Pruebe que alguno de estos conjuntos contiene tres números que son las longitudes de un triángulo.

$Tipe$

fijemonos en el conjunto :
${100;101;102;.....;200}$ el cual posee 101 elementos , vemos que al l tomar 3 elementos de este conjunto, estos forman loslados de un triangulo ya q la maxima diferencia entre los lados es 100, y la minima suma es 201 , por ello en cada triangulo se formara la ley de existencia de los triangulos , por ello por casillas tenemos que al menos en uno de los 50 subconjuntos se tendra al menos 3 de estos elementos y por ello seran parte de llos lados de un triangulo

No me acordé que alguna vez te dejé ese problema, bueno, como prometí, acá el segundo:

[color=darkblue]Problema 2[/color]
Sean $a_{1}, a_{2}, \ldots, a_{10}$ enteros positivos, $a_{1}<a_{2}<\ldots <a_{10}$. Sea $b_{k}$ el mayor divisor de $a_{k}$ tal que $b_{k}<a_{k}$. Asuma  que $b_{1}>b_{2}>\ldots >b_{10}$, pruebe que $a_{10}>500$.

$Tipe$

¿está muy difícil el problema o es que aun no lo han intentado?  :maybe:
o quizás alguno tiene una idea para resolver este problema?  :P

[hide="Sugerencia"]$\frac{a_{k}}{b_{k}}$ también es un divisor de $a_{k}$, analicen cómo debe ser este divisor[/hide]

$Tipe$

[color=blue]PROBLEMA:

Existen 3 numeros naturales mayores que 1, tales que el cuadrado de uno, menos 1, sea divisible por cada uno de los otros numeros??

Carlos Bravo
Lima - PERU[/color]

Carlos, mi idea era que alguien postee la solución antes que se postee otro problema  :maybe:, algo así como una maratón que he visto en otro foros  :lol: ... aún nadie ha resuelto el problema 2?

$Tipe$

[quote="Zaratustra"]
PD: El $a_{9}$ tambien es mayor que $500$ si mi solu esta bien[/quote]

Tu solución está bien, y tienes razón al decir que $a_{9}$ también es mayor que 500.

Como el problema 3 no lo puse yo  :lol: , voy a escribir la solución:

[b]Solución 3[/b]
No existen, suponiendo lo contrario, sean los números $a,b,c$ mayores que 1 que cumplen la condición del problema.
Vamos a demostrar que deben ser coprimos 2 a 2. Sea $d$ un factor común de $a$ y $b$, como $b \mid a^{2}-1$ entonces $d \mid a^{2}-1$ y como $d\mid a^{2}$, concluimos que $d=1$, es decir, $a$ y $b$ son coprimos (Análogo para los otros pares).
Supongamos que $a\leq b \leq c$ , como $b$ y $c$ son coprimos  y ambos dividen a $a^{2}-1$ entonces $bc \mid a^{2}-1$, por otro lado: $a^{2}\leq bc$, de donde $a^{2}-1 <bc$   y como $bc \mid a^{2}-1$ debería ser $a^{2}-1=0$, que no es posible.

$Tipe$

[color=darkblue]Problema 4[/color]
Sea $S(x)$ la suma de dígitos (en representación decimal) del número $x$. ¿Existen tres números naturales $a,b,c$ tales que $S(a+b)<5$, $S(c+a)<5$, $S(b+c)<5$, pero $S(a+b+c)>50$ ?

$Tipe$

[quote="johncmasb"][quote="Zaratustra"]
$\text{Luego como }b_{k}\text{ es el mayor divisor de }a_{k}:$

$b_{10}\ge \frac{a_{10}}{b_{10}}\ge 29$

$b_{10}\cdot\frac{a_{10}}{b_{10}}\ge 29^{2}= 841 >500$

$\boxed{a_{10}> 500}$

PD: El $a_{9}$ tambien es mayor que $500$ si mi solu esta bien[/quote]

pero $a_{10}$ puede ser primo, o sea, $b_{10}=1$  no necesariamente es primo, en este caso como $b_{9}>b_{10}$, sí sería primo, luego
${b_{9}}\ge \frac{a_{9}}{b_{9}}\ge 23$ y $a_{10}>a_{9}\ge 23^{2}>500$.[/quote]

Tienes razón, yo tampoco considere el caso en que $a_{10}$ es primo.

[quote="JOTAJOTARIM"]Solucion dudosa del problema 2

Si $b_{k}$ es el mayor divisor propio de $a_{k}$ entonces no necesariamente $\frac{a_{k}}{b_{k}}\leq b_{k}$. Verificar cuando $a_{k}$ es primo.

Juan Neyra[/quote]
La desigualdad $\frac{a_{k}}{b_{k}}\leq b_{k}$ es cierta siempre que $b_{k}$ es distinto de 1, es decir, falla únicamente cuando $b_{k}=1$ que equivale a decir que $a_{k}$ es primo. Como $b_{2}>b_{1}$ necesariamente $b_{2}$, $b_{3}$, ..., $b_{10}$ son > 1, luego sí se puede aplicar dicha desigualdad para $b_{2}$, $b_{3}$,...$b_{10}$.

Salvo la desigualdad aplicada a $a_{10}$ y $b_{10}$, la solución está bien.

[b] Alguien se anima con la solución del problema 4???[/b]

$Tipe$

[quote="johncmasb"]$a=5545545545, b=5455455455, c=4554554555$[/quote] ¿Podrías explicar cómo llegaste a ese ejemplo?  :lol:

$Tipe$

[quote="tipe"][quote="johncmasb"]$a=5545545545, b=5455455455, c=4554554555$[/quote] ¿Podrías explicar cómo llegaste a ese ejemplo?  :lol:
[/quote]

Mientras espero la explicación de John.....

[color=darkblue]Problema 5[/color]
Las medianas de un triángulo ABC determinan  6 triángulos. Demuestre que los centros de las circunferencias circunscritas a estos triángulos están sobre una misma circunferencia.

$Tipe$

Este resultado es el teorema de F. van Lamoen.
Una pregunta: como puedo adjuntar archivos.


is there a way to:
1. reduce font in a latex document to under 10 point?
2. implement columns?
3. change from portrait to landscape?

thanks

Hi,
1. \small{}, \footnotesize{}, \scriptsize{} and \tiny{} will make the font smaller than 10 pt, provided that the document is formatted in 10 pt.
2. I think this is what you mean by implement columns:
[code]
\twocolumn
First column
\newpage
Second column
\onecolumn
[/code]
3. The landscape page will preview correctly only in PDF format:
[code]
\begin{landscape}
\end{landscape]
[/code]

SSN.

1. so there's no real way to set a font at 9, for example?
2. that was exactly what i was looking for.  but can you explain to me how it works?
3. it doesn't seem to work for me >_<

thanks!

Hi,
1. According to "The Not So Short.....", for a default document font size of 10 pt, the four font sizes in my previous post represent 9, 8, 7 and 5 pt respectively.
2. Not really sure what more I can explain here - \twocolumn makes the document two column, then you enter the left column matter, then \newpage and then the right column matter. \onecolumn makes the document single column again if you so desire.
3. You need \usepackage{landscape} in your preamble.
SSN.

thanks a lot!  btw, its usepackage{lscape}, as i found out by checking the index, and trying stuff out when usepackage{landscape} didn't work ;)

EDIT: final question, i promise!  when i try to combine both two-column and landscape, something weird happens.  it prints text exclusively on the right side of the page (at least according to the pdf previewer).  some text will even overlap.  then a new page will be created, and it will print again on the right side.  help?  >_<

EDIT 2: i found out about \usepackage{multicols} as an alternative to \twocolumn.  (<3 google)  now the problem is fixed.  thanks for all your help!


Let $f(x)=2^{x+1} +3^{x+1} -6^x $.Prove that the equation $f(x)=f(\frac{1}{x})$ has solution. ;)

  I forgot to say that $x>0$ and $x\neq1$

Can you repair your problem, I ca not read it.

nntrkien,I think the problem looks clear now. ;)


function
topology
calculus
derivative
calculus computations

If a function is continuous and injective, is it differentiable?

I can think of functions that are continuous but not differentiable ($ |x|$, eg) and differentiable functions that are not injective ($ x^{2}$) but is there a function that is continuous and injective but not differentiable?

[hide="Counterexample."]$ \mathrm{f}(x)\ \equal{} \ \left\{\begin{array}{ll} x & x\leq0 \\\\
2x & x > 0\end{array}\right.$

is continuous and injective, but not differentiable at $ x \equal{} 0$.[/hide]

We can extend such counterexamples to functions that are not differentiable countably often.  In the spirit of Weierstrass, can we find a function that is continuous and injective but non-differentiable on a set of nonzero measure (or, well, everywhere)?

Theorem: A monotone function is differentiable almost everywhere.

However, there is a homeomorphism $ F\colon\mathbb R^2\to\mathbb R^2$ that is nowhere differentiable. For example, $ F(x,y) \equal{} (x,y \plus{} h(x))$, where $ h\colon\mathbb R\to\mathbb R$ is a continuous nowhere differentiable function.

Can we construct a homeomorphism $ F\colon\mathbb R^2\to\mathbb R^2$ that has no directional derivatives anywhere?