The previous topic on Wikipedia died 10 months ago, so I want to start a new one and hear new opinions.

[list]Do you consider Wikipedia to be a reliable source of information?

Do your teachers let you use it for research?[/list]
I think it's pretty reliable, with so many people constantly reading articles. If somebody vandalizes an article, the change is usually reverted very quickly, and false information that isn't vandalism is also corrected in a decent amount of time.

My previous English/Hisory teacher hated Wikipedia for some reason, but she never actually stopped us from using it as a source of information. Her argument was "So many people all contributing into one place can never be a good thing," which I find to be quite untrue.

What are your opinions?

Wikipedia is usually pretty, good, there's some really good information out there on the website. Teachers let me use it, but they generally don't recommend it as a main source.

Wikipedia is great for an overview about any topic. For specifics, it'd be better to go read another source, since Wikipedia almost never goes into much depth and other sources are much more thoroug. I never cite Wikipedia, and rather cite the source from which Wikipedia got the information. Of course, I do use Wikipedia before Mathworld for math stuff (I go to Mathworld afterwards to get the details) but Wikipedia is generally more clear.

1. Wikipedia is one of my top resources for information. In my couple years of experiance with Wiki, I have never found any of the information on there unreliable. Plus, there are also lots of references listed on that site, so Wikipedia is also a good reference source.

2. My AP bio teacher doesn't let us cite Wikipedia. He'll take 10 points off our lab if we cite it  :( . I'm not sure about the others, we didn't have to do any citations in the other classes.

The first place I usually go to when I want to look at something is Wikipedia, mainly because it's very clear and easy to understand.  However, my Math Research teacher doesn't want us using Wikipedia as a source because it might be inaccurate.  I think that Wikipedia is a good place to get started and to find links to other sites that might be more reliable, though.

[quote="lingomaniac88"]...very clear and easy to understand.  [/quote]

Some math Wiki articles aren't as clear and easy. If you were an 8th grader studying natural logs, you could easily find limit references on that Wiki article.

My old librarian LOATHED Wiki; she had hate ads on nearly every wall of my MS library.

I personally think Wiki is cool and ususally reliable source, but the thing is you can make your own article and stuff.

My teachers are the same way. They tell us not to use it as an first-hand source to go.

[quote="DonkeyKong"]They tell us not to use it as an first-hand source to go.[/quote]

Yup, instead for me they recommend a county-paid website that has links to a lot of seach engines (of course all these suck and have little to no information).

[quote="7h3.D3m0n.117"]In my couple years of experiance with Wiki, I have never found any of the information on there unreliable. Plus, there are also lots of references listed on that site, so Wikipedia is also a good reference source.
[/quote]

I've found misinformation on Wikipedia reasonably frequently---the most recent is misattribution of mathematical theorems.  On the other hand, I do think it's a good source of reliable references.  As a faculty member, I wouldn't allow citation of Wikipedia precisely because I know there's misinformation there.

Most of the articles I search on Wiki are usually chemistry related so if there was an error then it should be at least obvious (but when talking last night with i_like_pie we found a botch in the $ \text{HCl}$ article, saying how its pH was negative.

Actually the $ \mathrm{HCl}$ article doesn't have any faults, but [url=http://en.wikipedia.org/wiki/PH#Measurement]this table[/url] shows $ \mathrm{pH}\left(\mathrm{HCl}\right) \equal{} \minus{} 1.0$.

EDIT: The [url=http://en.wikipedia.org/wiki/Hydrochloric_Acid]$ \mathrm{HCl}$ article[/url] states that $ \mathrm{pH}\left(\mathrm{HCl}\right) <1.0$ also.

Meh, I'm actually a former sysop on Wikipedia; until a few months ago, when time constraints forced me to stop spending so much time on it; so I resigned.

All my teachers actually like Wikipedia, but they don't let us use it at all. If Wikipedia is in the bibliography, the project receives a 0. Go figure that out.

Personally, I agree with i_like_pie on the fact that Wikipedia's sheer popularity keeps it fairly accurate, though articles on very advanced/obscure subjects are often very flawed

I think Wikipedia is a quite a bit of fun to read and a great idea, especially for pop-culture types things, but it's certainly out of the question as an academic citation source.  A publicly-edited source simply can't be considered academically reputable, and this is obvious from the great deal of misinformation in Wikipedia articles.  It is useful however, since it often provides links to other, more reputable sources, and is typically correct for general information about topics.

Usually, most teachers in my district don't allow wikipedia to be cited as a source.
However, one teacher from a different place had said that wikipedia is a pretty reliable source of information. In most topics that you'd look at (except for maybe footnotes in history), vandalism is reverted almost immediately.
I sometimes find wikipedia hard to understand for some concepts that are new to me. I don't think wikipedia is a good source for reading up on something that i've never heard of before.

Also, here's what my biology student-teacher basically said:
Wikipedia should not be accepted as a source, but a good idea is to look at the citations that the wikipedia articles give, because they're usually good sources.

IRT The Zuton Force:
Even though it changes so much, each change is not an actual [b]edit[/b], but actually a duplication of the article. There is a unique link in the database for every single edit. You can find the article for a certain subject from any day.

Wikipedia is great to use for research, instead of those websites that you find on Google that are hardly accurate. I used Wikipedia for my report on Mount Everest. There was loads of information, and nearly all of it was correct. Except for the fact that Mount Everest is actually like 30,000 ft now, but who knows.

Wikipedia is edited, so sometimes people edit for the fun of it and get the information all messed up.

I did a report on Limestone, before, and there were so many words I couldn't even begin to say that I had to look up at dictionary.com  :D

Wiki is pretty great, overall.

[quote]I never cite Wikipedia, and rather cite the source from which Wikipedia got the information.[/quote]

Thats exactly what I do. It works brilliantly when you have to scourge up resources to list in your Bibliography and you forgot the sources you actually used.

I actually dont know my teacher's preferences on Wikipedia. My Biology teacher last year said that if we cited Wikipedia, we needed to cite 3 other sites or something like that.

[quote="jimhu"]I think its one of the best sources out there, covering pretty in depth info about almost anything. Though for research papers and stuff i probably wouldnt use it, my school probably doesnt allow it anyways[/quote]

Something related to wikipedia is answers.com or sites similar to that. They sometimes pop up higher than wikipedia when googling some topics.
I guess they're similar because 1)they use each other and 2)they both cover alot of topics in general.
There's also howstuffworks.com

There are many sites that mirror Wikipedia; answers.com is one of them. Also, see http://en.wikipedia.org/wiki/User:AxelBoldt/References_don%27t_increase_Wikipedia%27s_reliability

Some of my teachers hate it, others like it. I think it has everything you need to know, most of the time the teachers are wrong, not the website. My teacher taught us about the fall of Rome and didn't include an important point, wiki had it, and she didn't believe it until CNN published it... All the basic facts are correct, though some minor details are wrong, and what teacher wants to know what color shoes Nixon wore when he resigned?(just an example, they might have it right) When so many people contribute, how much can be wrong? When it's a single person's site(most people's sites are true), it can be wrong, and than other people can't correct, so it feeds the world false info.
Wiki is especially good for fictional books, it has a list of every demon in the Bartimaeus Trilogy(a hundred or so names, most of them mentioned once in an irrelevant footnote)(incase you don't know what that series is, look it up on, you guessed it, wikipedia!).
All in all, most of the info is correct, and it's a premium site.

Wikipedia doesn't have [i]everything[/i] you need; there are specialist resources that have more on most particular topics. It is a great first source, but not very good as an only source.
Controversial issues are always risky there- the article often becomes a battle between two opposing groups of ideologues or is dominated by one side. Either way, it's a mess.

On math topics- I find Wikipedia to be more useful than Mathworld. I have both bookmarked, but I think the wiki articles are just better in general.
Also, I'm usually just looking for a definition. If I need much more, textbooks and papers are the appropriate references.

I also prefer Wikipedia to Mathworld - besides other things, equations in Wikipedia are easier to read. On two recent occasions I was chided by undergraduates for looking up math topics on Wikipedia. One said that I should use Mathworld instead and another informed me that Wikipedia articles may be inaccurate. Kids these days. :roll:

[quote="Fraenkel"]Wikipedia is so successful because it is the supermarket for information.

As far as inaccuracies, they are usually rare.
If you check the edit history it will usually show whether anyone was trying to mess up the info on purpose, and luckily you can change it back.[/quote]

actually at least 1/4 of the edits are false. I know a person who changed this one wiki page to say SO I HERD SUM THINGS ABOUT YOU LIEKING MUDKIPZ. then I also looked at some other pages... weird stuff too

Wow... you came up with $ \frac{1}{4}$ out of nowhere because of one observed deconstructive edit?

the accuracy part... it's very difficult to find which information is accurate, especially when related to people related topics. Something like math isn't too easily made inaccurate in wikipedia, but a subject like human geography is something where it's difficult to maintain what wikipedia calls as "world neutrality".

try [url=http://en.citizendium.org/wiki/Main_Page]this[/url] this was started by Larry Sagner a founder of wikipedia who had differences with the policies and credibility of wikipedia and left it

Citizendium is a rather failing experiment. It's completely fallen short of it's 6-month goals in terms of members and articles, and much key info is missing, such as articles about major countries.

yes but it's just the start.i use wiki and like the open source thing of wiki but credibility issuse are a must

[quote="Fraenkel"][quote="7h3.D3m0n.117"]About 43.5% of all statistics are made up on the spot  :D .

Wikipedia limits destructive editting very effectively I believe. Actually, our school county's IP address is banned so we aren't allowed to modify the Wiki at all  :oops: .[/quote]

No it's 57.6% of all statistics are made up on the spot.[/quote]

ok.
I just went through the newest 50 edits/creations
[b]Most were useless. They only change 1 thing, that one thing is usually a link added or rewording of a paragraph.[/b]

[quote="moogra"][b]Most were useless. They only change 1 thing, that one thing is usually a link added or rewording of a paragraph.[/b][/quote]
And you call that useless?  :|

I guess moogra would like Wikipedia to be full of spelling errors, confusing terminology without links to other articles, badly worded paragraphs, etc.

In all seriousness, please try to have a concept of what you're talking about before you make a statement like the edits are "useless". I frequently go around making "useless" edits, and I'm fairly sure nearly all of the "useless" edits are very helpful.

I think most edits I see are useless (at least on common links)

In the edit history I constantly see a back and forth tug of war between good and evil.
Pranksters vs. Websters

Of course. There are many vandals on Wikipedia, so concordantly there have to be many vandal fighters.

Calling minor edits useless is calling the job of editors, revisors, critics, etc. every useless.

Uh you shouldn't be using any encyclopedia much less Wikipedia for a research paper. It's a good starting point for background info but you should never be citing Wikipedia. The only nice thing about Wikipedia is that they often have a list of references and you can go to those to do your research.

[quote="The Zuton Force"]Wouldn't the pH of HCl depend on it's molarity? I mean, it cites 10M as -1 on that chart and 160mM as 0.8 in the stomach acid article, which check perfectly ($ log_{10} (10) \minus{} log_{10} (0.16) \approx 1.8$, which is the difference).

Anyway, [i]using [/i]Wikipedia is absolutely fine. It's like asking a friend who's an expert. It has a lot of information and the vast majority of it is accurate and well presented.

[i]Citing [/i]it is what's the problem. [u]You really can't cite it[/u] because its changing all the time. If you were to cite it in a paper, someone else who reads your paper could go there and find a very different article, and never be able to find the article you cited because it's been... well, not revised into a new edition, but actually edited.[/quote]

This is incorrect, if you're citing it (which you might do for a non-formal non-research paper) you should be putting the date accessed as well as the revisionid so someone can check what version you used. There's nothing wrong with citing it, it's just that you shouldn't be using encyclopedias in the first place for research.


search
AMC
AIME
MATHCOUNTS
probability
expected value

please sign up:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=475480#p475480

we are planning to have approximately 40-50 people taking the test

Lol. Everyone has this in their signature. :) I stickied for you too, but I'm not sure that's even necessary :P

wow, 61 people already.. crazy. oh and btw how do you get the link to show as "Math contest" in your sig?

can someone at least repost rules, dates, format, and where we should expect the questions to be HERE??? I don't want to search through 6 pages...

Eh?  All the information about the contest is in the very first post of that topic...

Anyway:

[quote]Hello everyone, here are the details of the newest (unofficial) AoPS contest:

[u]Contest Format:[/u]
Iversonfan2005 and myself are writing/sponsoring a [b]multiple choice[/b] math contest called the Masters Invitational Math Contest.  It is on a middle to high school level, but anyone can participate.  There will be awards on for a high-scorer on every grade level, and an award for the top three scorers overall.

[u]Difficulty:[/u]
The contest will be [b]50 questions[/b], multiple choice, ranging from National MathCounts to upper level AIME difficulty.  We would like every AoPSer to participate, but do not be discouraged by harder problems towards the end.  Although we encourage middle-schoolers to participate, middle schoolers are not expected to be able to anwer the higher numbered questions.

[u]Scoring:[/u]
Each correct answer will earn 5 points, a question left blank will earn 1 point, and an incorrect answer will earn 0 points.  Scores will range from 0 - 250.

[u]Time Limit:[/u]
Due to our inability to accurately track time, all participants will have [b]3 weeks [/b]from the time of contest posting to turn in their answers.

[u]Participation:[/u]
If you want to participate, just reply to this thread indicating so.  The contest will be posted in a week or two.  This was posted in the SC forum because I moderate there, but it is for EVERYONE to participate in!

[b][u]IMPORTANT EDIT[/u][/b]
[u][b]The PDF will be posted on April 16th.  All answers will be due exactly 3 week later, on May 7th. [/b][/u] We added two weeks due to USAMO and AP exams.  Also, due to the added time, the exam has gone from 40 questions to 50 questions.  Please note these changes![/quote]

Finally...a nice, quick, AIME level contest where careless mistakes won't kill me!

Date: April 16th - May 7th
Duration: 3 weeks, in case you don't have a calendar
Difficulty: National MathCounts thru hard AIME
Format: Multiple Choice
Calculator Use: Yes
Turn in Answers: via PM to [b]MIMC Director[/b]
Contest Posted: In the original MIMC thread in the SC forum.  Also, we'll announce it in other forums as well.

[quote="Iversonfan2005"][quote="K81o7"]So showing work doesn't matter?  So careless mistakes still could do some damage...[/quote]

showing work!? :rotfl: , who has the time and the interest to look at people's work when grading this? :rotfl:  :rotfl: , besides, you don't get partial credit (it's not like if you say $1+2=4$, we'll give you $4.9$ points instead of $5$ but if you say $1+2+3=8$, we'll give you $4.8$, besides, how do you measure careless mistakes?), think of it like the AMC.[/quote]

It's not really like the AMC, because the expected value of a guess is the same as a blank.


number theory
greatest common divisor
Diophantine equation
number theory proposed

I found it!!!!

Algo to compute any semi-prime n=p*q.

1. Compute a= 2^n mod n
2. Compute b= ((2^n)-a)/n
3. Compute GCD (b,n)

GCD(b,n) is equal to one factor of n.

I checked it for some numbers it works!!!!!
I hope it will work for big numbers!

Example n=77

1. a= 2^77 mod 77 =18
2. b= ((2^77)-18)/77 = 1962541914958813595302
3. GCD(77,1962541914958813595302)=7

I hope I did not make a mistake

Great, so now factoring a large number n is only as hard as computing $ GCD(k,n)$, where $ k\equal{}O(2^n)$.   :roll: Real useful, yeah.

Does it work that is the first question.
Then why it works or not second question.
If it works how to improve it and to implement it.

Try n = 35, and you'll run into trouble.

2^35 = 34359738368 = 18 (mod 35),

(2^35 - 18)/35 = 981706810, which
is divisible by 35 once again.

[quote="CaptainCanada"]It doesn't even matter if it works. It would be way way WAAAAY too slow to ever be a feasible method of factoring large numbers. It's really amazing that you've managed to come up with an approach that is dramatically less efficient than a trivial brute force factorization.  :rotfl:[/quote]

In fact, for example using Wilson's theorem is much slower than this. But Wilson needs much less memory.

Why not make it better CaptainCanada? Create one better and show him how he can improve his idea. And by create I don't mean post one someone else figured out. That would be more helpful no?

That might work if Matharith listened to any ideas besides his own.

Maybe there is another better solution.
I will post it in 2 or 3 days.
Another new idea to factorize more easily.

Goog night!
It is time to go to bed.
Thank you for all your comments.

Good evening,

I'm looking for some documentation to solve this diophantine equation

(a*(x+y) - xy) mod k = b

where a,b,k known integers
x,y unknown

What I need is just documentation about this kind of equation. It is maybe elementary.
Thnk you for any help.

If we can solve this equation

(a*(x+y) - xy) mod k = b

it will be VERY VERY helpful.
I do not know if it has one solution or more.
I have found this on Internet

http://www.alpertron.com.ar/METHODS.HTM#SHyperb

I have to read it carefully.

Thank you for any help or comment.

Hi,

Here is the new algo rectified.

Factorisation of n

1. Compute a=2^n mod n
2. Compute b=a^2 mod n
3. Solve the equation
(a*(x+y) - xy) mod n = b
4. Compute GCD(a-x,n) or GCD(a-y,n)

Example n=77

1. a=2^77 mod 77 = 18
2. b=18^2 mod 77 = 16
3. 18*(x+y) - xy mod n = 16
Solutions : x=4 y=7
18*(4+7) - (4*7) mod 77 = 170 mod 77 = 16
4. GCD(18-4,77)=7 or GCD(18-7,11)=11

Done! 77=7*11

Have a good day.
Thank you for any comment.

[u]Very important [/u]:

I forgotten to precise that n is [u]odd semi-prime.[/u]


algebra
polynomial
calculus
integration
function
analytic geometry
vector

Is given a space of polynomials of degree <= n. The scalar product between two polynomial p_1 and p_2 is defined by the definite integral of function p_1· p_2. between the limits(0,1). Find the distance between the origin of coordinates and the space of polynomials.

What do you mean by "the origin of coordinates"? The only meaning I can think of is the zero function - which is a polynomial, and would make the distance in question zero. This is a problem about orthogonal projection in an inner product space, but it looks like some piece of information is missing.

I think the question means that:
the set U={p: p(x) = x^n +a_1x^(n-1)*........a_n} ,whose distance from origin is asked, is not a vector space because does not contain the origin. Then the distance is not zero.

The answer of the book is:
\[
\frac{{n!}}{{\left( {n + 1} \right)\left( {n + 2} \right)...2n\sqrt {2n + 1} }}
\]

Solution of the problem:

The set $ U = \{p: p(x)=x^n+a_1x^{n-1}+....+a_n\} $ has the form
$ U=u+V_{n-1} $ where u is a monic polynomial of degree n and $ V_{n-1} $
is the vector space of polynomials of degree n-1. If you choose as u the
Legendre polynomial $ P_n $, divided for is first coefficent
$ c_n= \frac {(2n-1)!!} {n!} $, the asked distance is the norm of u (||u||). Infact
u is orthogonal to $ V_{n-1} $ which we can think to be made of Legendre
polynomial of degree less than n. The norm of $  P_n $   is $ \frac {\sqrt {2}} {\sqrt {2n+1}}$,
and then the distance of U from the origin is
\[ \frac {n! \sqrt {2}} {(2n-1)!! \sqrt {2n+1}} = \frac {(n!)^2 2^{n+ \frac {1} {2}}}      {(2n)! \sqrt {2n+1}}
\]
Now we must change the interval (-1,1) in the interval (0,1) with
the relation x = 2t - 1 and we get the expected result:
$ ||P_n||=\displaystyle \int^1_{-1} (P_n(x))^2 \,dx=2 \displaystyle \int^1_0 (P_n(2t-1))^2 \,dt $


Let $ x,y,z$ be positive real numbers, then prove:
$ 10x^2 \cdot z^2\plus{}10y^2 \cdot z^2 \plus{}1 \ge x\cdot y \cdot z^2\plus{}x\cdot z \plus{}y \cdot z$.

:huh:
using AM-GM we can prove that:
$ LHS\geq xyz^2\plus{}\sqrt{19}xz\plus{}\sqrt{19}yz$
so your inequality doesnt make any sense... :maybe:


How many roots are there in the equation??

[hide]$\sin{\theta} - \sqrt{3}\cos{\theta} = \sqrt{2}$
$\frac12 \sin{\theta} -\frac{ \sqrt{3}}{2}\cos{\theta} = \frac{\sqrt{2}}{2}$
$\cos{\frac{\pi}{3}} \sin {\theta} - \sin{\frac{\pi}{3}} \cos{\theta} = \frac{\sqrt{2}}{2}$
$\sin{\left(\theta - \frac{\pi}{3}\right)} = \frac{\sqrt{2}}{2}$
Hence...
$\theta - \frac{\pi}{3} = \frac{\pi}{4} \quad \Rightarrow \quad \theta = \frac{7\pi}{12}$
Or
$\theta - \frac{\pi}{3} = \frac{3\pi}{4} \quad \Rightarrow \quad \theta = \frac{13\pi}{12}$

Two solutions[/hide]

[quote="Farenhajt"]Actually, there should be infinitely many solutions, since you can always add $2n\pi$ to any solution.[/quote]The problem asked the solutions in $[0,2\pi]$

[hide="Another solution"]
$sin \theta - \sqrt{3} cos \theta = \sqrt{2}$

$(sin \theta - \sqrt{3} cos \theta)^2 = 2$

$sin^2 \theta - 2\sqrt{3} sin \theta cos \theta + 3 cos^2 \theta = 2$

$(sin^2 \theta+cos^2 \theta) - 2\sqrt{3} sin \theta cos \theta + 2 cos^2 \theta = 2$

$1 - \sqrt{3} sin 2 \theta + (1+cos 2\theta) = 2$

$- \sqrt{3} sin 2 \theta + cos 2\theta = 0$

$\sqrt{3} sin 2 \theta = cos 2\theta$

$\sqrt{3} tan 2 \theta = 1 \ \ \ (cos 2 \theta\neq 0 )$

$tan 2 \theta = \frac{\sqrt{3}}{3} = tan (\frac{\pi}{6})$

$\boxed{2 \theta = k \pi + \frac{\pi}{6}} \mbox { or } \boxed{2 \theta = k \pi + \pi+\frac{\pi}{6}} \Leftrightarrow$

$\boxed{\theta = k \pi + \frac{\pi}{12}} \mbox { or } \boxed{\theta = k \pi + \frac{7\pi}{12}} \Leftrightarrow$

The first set of solutions gives in $[0, 2\pi]$: $\theta_1 = \frac{\pi}{12}$ and $\theta_2 = \frac{7\pi}{12}$

The second set gives $\theta_3 = \frac{13\pi}{12}$ and $\theta_4 = \frac{19\pi}{12}$

So we have four roots for the equation $(sin \theta - \sqrt{3} cos \theta)^2 = 2$

But $\theta_1$ and $\theta_4$ satisfy the equation $sin \theta - \sqrt{3} cos \theta = -\sqrt{2}$

So we only accept $\theta_2$ and $\theta_3$[/hide]

You don't need to find the roots in detail.Unit circle gives the answer. :D


AMC AIME USA(J)MO USAMO email calculus geometry

I am thinking of taking the Chemistry Olympiad next year.
Right now I am in physical science and am scheduled for AP Chemistry next year.
I am planning to start preparing this summer.

How is the Chemistry Olympiad series of tests run (i.e. AMC10/12 -> AIME -> USAMO -> MOP etc...)?

What books should I read and what should I do to prepare?
Will half a year of AP Chemistry alone be enough?
Personally, I have found that math classes at my school (including BC Calc) have not helped me significantly in math competitions as all my knowledge comes from self and independent efforts.

What caliber is expected to make the chemistry Olympiad summer camp?

Chemistry Olympiad works like this:
Local -> National -> Camp -> IChO -> Top Gold ( :D )

If you are solely preparing for the camp, i recommend Peter Atkin's gen. chem book. It's been recommended before, and I used it to get to the camp. I would say normally, a full year of AP chemistry is not enough. I took two years of chemistry (honors and AP) and took the local in the middle of my AP year. I did not know a lot of stuff on the local, and the national exam brings topics that go deeper than the AP level.

*Most* people at the camp really know their chemistry. Basically, get approx. a 90% or higher on all the sections of the national test. I would refer a post on this, but I don't remember the poster's nick nor the content.

So first of all, for a lot of random (but hopefully good!) information/advice on the Olympiad, check out http://www.scientificendeavors.com/chemistry/

Regarding preparation, I also vote Atkins.  My first year of camp qualification, I had taken the 1 year of high school chemistry, 1 semester of college general chemistry (using Atkins), and studied out of Atkins a fair amount.  My second year, I just read Atkins cover to cover (and took orgo, but that's another story).

If you have any questions, feel free to PM me or email me at greg [at] scientificendeavors [dot] com

How do the local competitions work?
I might have had the opportunity to take it this year but I did not hear about it or frankly, know about it, until one of my mom's friends told my mom that her daughter took it; so Ill try to prepare for next year.  I also heard that only 2 people from each school can make it (is this true?  because it seems kinda unfair considering my school)
I went to the website link on wikipedia and it seems to be that there are about 10,000 people that take the local competition and there are about 1000 that qualify for the USNCO (is that right?)

For this competition, do you think that just by taking two years of chemistry (honros chem in 10th grade, and next year, AP chem), I will have the ability to pass the local test (I would think this is different than BC calc-AIME argument)? I am probalby just preparing to qualify and take the USNCO...i dont think i will have a chance at the camp

Also, what is the prestige of this test?  Would qualifyin for USNCO be just as prestigious as qualifying for USAMO?  And for those that have, which ones harder to make?

ANd is there a limit on how many poeple from one school can take the local exam (it seems like a couple of people-like 3-from the 150+ students that take honros chem with my teacher this year got invited to take it because they had the three highest percentages in the class).  I hope this isnt the case because although I get like 93-94s in chemistry, I probalby dont have the ability to get top 3 percnetages (like 99.5) in the class

About 800-900 qualify. If you LIKE chemistry, go ahead and take it. Talk to your teacher, and the sign-up for you is free. (The school has to pay though.)

No, it's definitely not as prestigious as USAMO, but qualifying for the USNCO national exam is definitely a prestige and a honor (I would say slightly above qualifying for AIME), especially if you qualify in a competitive region such as New Jersey, New York, or California. However, if you make the international study camp (Top 20), then it's more prestigious than USAMO. Lastly, there's no limit from a school on who can take the local exam.

To the original questions:
What books should I read and what should I do to prepare?
Your AP Chemistry Textbook shoudl suffice.
Also, Chemical Principles (Atkins) is a nice book too.

Will half a year of AP Chemistry alone be enough?
Enough to qualify for the national Exam? Possibly...You'd have to learn some material you haven't covered though.
Enough to qualify for the camp? No.The camp is very hard to qualify for. Take a look at the old national exams to get an idea.

i mite seem nosy, but im just curious.
dont you ppl study all 3 subjects phy, math, chem along wid some others every year or is it basically one core subject per year.

it depends on the school (U.S.)

generally a typical schedule is:

Math (average senior level course would be a calculus course)
Science (usually one of Physics, Chemistry, Biology, or others)
Language (depends on the school)
English (writing)
History (depends on the school, but with one mandatory albeit useless U.S. History course)
Others, as in required extracurricular classes, or additional sciences or languages.

oh, ok.
in india,
 physics, chem and math are mandatory.
bio is optional.
languages(depends on school) english is compulsory, and another language is optional.
luckily we dont have to study history.
but we have to study everything in math - calculus, algebra, geometry, but i dont suppose they are very advanced.

@ varunarasu: Hey, Im just wondering, do you go/have plans to go to IIT? How competitive is it to make it there? Im an NRI and am just curious about what education is like in india.(My grandfather went to IIT back in the day and then got a job in USA. Thats why I live here now)

i gave jee but im not really interested in joining iit, coz i wanna do research. but ur rite abt one thing tho, dere's one hell of a competition to get into the iits. you check out the india forum and u'll realize how many ppl aspire to become iitians.


limit
Euler
trigonometry
LaTeX
calculus
calculus computations

[1] $ lim\frac{(1\plus{}x)^{1/x}\minus{}e}{x}$

[2] $ lim(\frac{sinx}{x})^{1/x^2}$

[note] for both case x is approaching to 0.

hello, it is
$ \lim_{x\rightarrow 0}\frac{(1\plus{}x)^{1/x}\minus{}e}{x}\equal{}\minus{}\frac{1}{2}e$, where $ e$ is the Euler number.
Sonnhard.

[quote="Davron"][1] $ lim\frac {(1 + x)^{1/x} - e}{x}$

[2] $ lim(\frac {sinx}{x})^{1/x^2}$

[note] for both case x is approaching to 0.[/quote]

$ \text{\LaTeX}$ tip:
 use [code]$\sin x$[/code] instead of [code]$sin x$ [/code] to get $ \sin x$ instead of $ sin x$

and [code]$\lim$[/code] instead of [code]$lim$[/code] to get $ \lim$ instead $ lim$


[quote="hainam-sp"]Find the largest real number $\alpha$ and the smallest real number $\beta$ such that:
$\alpha < \frac{a}{a+b}+ \frac{b}{b+c}+ \frac{c}{c+d}+ \frac{d}{d+e}+ \frac{e}{e+a} < \beta \forall a,b,c,d,e>0$[/quote]
the largest $\alpha$ is 1
and the smallest $\beta$ is four
because
 $\frac{a}{a+b}+ \frac{b}{b+c}+ \frac{c}{c+d}+ \frac{d}{d+e}+ \frac{e}{e+a}\leq  \frac{a+c+d}{a+b+c+d+e}+ \frac{b+d+e+a}{b+c+d+e+a}+ \frac{c+e+a+b}{c+d+e+a+b}+ \frac{d+a+b+c}{d+e+a+b+c}+ \frac{e+b+c+d}{e+a+b+d+c}=4$
and$\frac{a}{a+b}+ \frac{b}{b+c}+ \frac{c}{c+d}+ \frac{d}{d+e}+ \frac{e}{e+a}\geq  \frac{a}{a+b+c+d+e}+ \frac{b}{a+b+c+d+e}+ \frac{c}{a+b+c+d+e}+ \frac{d}{a+b+c+d+e}+ \frac{e}{a+b+c+d+e}=1$
and when $a=1,b=t,c=t^2,d=t^3,e=t^4$
and $a=t^4,b=t^3,c=t^2,d=t,e=1$
$t\rightarrow0$
$RHS\rightarrow1,4$respectively
So $\alpha=1,\beta=4$ is the proper constraints.


partition
combinatorics
Additive combinatorics
Additive Number Theory
IMO Shortlist

Does there exist an integer $ n > 1$ which satisfies the following condition? The set of positive integers can be partitioned into $ n$ nonempty subsets, such that an arbitrary sum of $ n \minus{} 1$ integers, one taken from each of any $ n \minus{} 1$ of the subsets, lies in the remaining subset.

[hide]

I claim that no such partition exists.

Proof:

Case 1: At least one set in the partition contains at least one odd element and at least one even element.

Lemma 1. All the sets have at least one odd element and at least one even element. To see this: (Letting the sets be $ A_1,A_2,...,A_n$.) suppose $ A_1$ has an even and an odd element. Then for $ i\ge 1$, we take $ n \minus{} 1$ integers from each set besides $ A_i$, then their sum plus $ t$ must lie in $ A_i$. By changing the one integer chosen from $ A_1$, we can make this integer in $ A_i$ take on either parity, so $ A_i$ has at least one odd and at least one even integer.

Lemma 2. There exists $ a$ such that $ a$ and $ a \plus{} 1$ are in different sets. This is clearly true, for otherwise one set would contain all the positive integers.

Lemma 3. Let $ d \equal{} 2(a_2 \plus{} a_3 \plus{} \cdots \plus{} a_{n \minus{} 1})$, where $ a_i\in A_i$. Then if $ a_1\in A_1$, $ a_1 \plus{} d\in A_1$.

To prove this, we simply have:
$ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{n \minus{} 1}\in A_{n}$.
$ a_2 \plus{} \cdots \plus{} a_{n \minus{} 1} \plus{} (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{n \minus{} 1}) \equal{} a_1 \plus{} d\in A_1$

Now we arrange the subsets in order $ A_1,A_2,...,A_n$ with $ a_i\in A_i$ and $ a_1'\in A_1$ satisfying:

$ a_1$ is even and $ a_1'$ is odd.
$ a_n \equal{} a_{n \minus{} 1} \plus{} 1$.

Let $ d_1 \equal{} 2(a_2 \plus{} \cdots \plus{} a_{n \minus{} 2} \plus{} a_{n \minus{} 1})$ and $ d_2 \equal{} 2(a_2 \plus{} \cdots \plus{} a_{n \minus{} 2} \plus{} a_{n})$. Then $ d_2 \equal{} d_1 \plus{} 2$ and $ \gcd(d_1,d_2) \equal{} 1$. It is well known then that there is some $ N$ such that any even integer greater than $ N$ can be expressed as $ Xd_1 \plus{} Yd_2$ for some nonnegative integers $ X,Y$. But since $ a_1 \plus{} Xd_1 \plus{} Yd_2\in A_1$ and $ a_1' \plus{} Xd_1 \plus{} Yd_2\in A_1$ by Lemma 3, so eventually all integers are in $ A_1$. But clearly this cannot be, since if we take some high enough integer $ z\in A_1$, there will be a higher integer $ z \plus{} a_2 \plus{} a_3 \plus{} \cdots a_{n \minus{} 1}\in A_n$. Hence this is impossible, so no such partition exists.

Case 2: All sets in the partition contains either only odd integers or only even integers.

We can do the same thing; we just have to say that for all even (or all odd) numbers only, all even (or odd) integers starting at some point will be in $ A_1$. This is also clearly impossible. If we take $ z\in A_1$ where $ z$ is high enough, then $ z \plus{} a_2 \plus{} a_3 \plus{} \cdots a_{n \minus{} 1}\in A_n$. If $ z \plus{} a_2 \plus{} a_3 \plus{} \cdots a_{n \minus{} 1}$ is the same parity as $ z$ then this is a contradiction. Otherwise, we can show that $ A_n$ contains all odd numbers greater that a certain number. Then all integers after some point are in either $ A_1$ or $ A_n$, so we can apply the same argument and reach a contradiction.
[/hide]

[hide]
Assume that there is such an $n$ that works.
Let the sets be $A_1,\hdots, A_n$.  Let $x_i$ be the least element of $A_i$.  Assume with loss of generality that
$1=x_1<x_2<\cdots <x_n$.
The case $n=2$ is obviously impossible.
If $n=3$, then $1+x_2\in A_3$.  We can prove inductively that $x_2+2k\in A_2$ and $x_2+2k+1\in A_3$ for all $k\in \mathbb N$.  Thus, all elements of $A_1$ are less than $x_2$.  This is impossible since $A\ni 2x_2+1>x_2$.
If $n\ge 4$, then we have two cases to consider $x_2=2$ and $x_2\neq 2$.
In the first case, let $S=x_4+\cdots, x_n$.  Therefore, $S+3\in A_3$.  This implies that $2S+4\in A_2$ and $2S+5\in A_1$.  We can now prove inductively that for all $k\in \mathbb N$,
\begin{align*}(2k-1)S+2a+1\in A_3&\text{ if }a\in \{k,\hdots, 2k-1\}\\
     2kS+2a+1\in A_1&\text{ if }a\in \{k+1,\hdots, 2k\}\\
2kS+2a\in A_2&\text{ if }a\in \{k+1,\hdots, 2k\}\end{align*}
If $S$ is even, then \[A_3\ni (2S-1)S+2S+1=2S^2+S+1=2(S-1)S+2\cdot \frac{3S}{2}+1\in A_1\] contradiction.
If $S$ is odd, then \[A_3\ni (2S-1)S+2S+1=2S^2+S+1=2(S-1)S+2\cdot \frac{3S+1}{2}\in A_2\] contradiction.
Hence, this case is impossible.
In the second case, again let $S=x_4+\cdots, x_n$.  We have that $\{1,\hdots, x_2-1\}\subset A_1$.  Therefore, $\{S+x_2+1,\hdots, S+2x_2-1\}\subset A_3$.  Thus, \[A_2\ni x_2-1+S+(S+2x_2-1)=2S+3x_2-2=S+x_2+S+2x_2-2\in A_1\]contradiction.
All cases lead to a contradiction so no such $n$ exists.
[/hide]

Am I missing something? Proof sketch:
[hide]
Suppose $A_1, A_2, \hdots, A_n$ satisfies the condition.
Claim: All the subsets are eventually periodic.
Proof: The condition states that for each $k$, $\sum_{i \neq k} A_i \subset A_k$. Thus, each subset contains a translate of each other subset. It follows that each subset contains a translate of itself, and is therefore eventually periodic.

Since each subset contains a translate of each other subset, all subsets eventually have the same period $p$. A simple argument shows that the subsets must eventually be congruence classes modulo $p=n$, which doesn't work. Contradiction.
[/hide]

First, assuming there exists such a n with sets $A_1,A_2,A_3,...,A_n$ with set $A_1$ containing elements $a_{11},a_{12},a_{13},...$, $a_{11}<a_{12}<a_{13}<...$, $A_2$ with elements $a_{21}<a_{22}<a_{23}<...$, and so on with $A_n$ having elements $a_{n1}<a_{n2}<a_{n3}<...$.

First observe, that if such a partition exists, then each set must contain infinitely many numbers- if, for example set $A_i$ contains finitely numbers with largest number $t$, then just choose the number $t+1$ from the remaining $n-1$ sets and then arbitrarily choose elements from the rest of the $n-2$ sets to get a contradiction.

Now consider the sum $S_1=a_{11}+a_{21}+a_{31}+...+a_{n1}$. Notice that $S-a_{11}$ must belong to set $A_1$. Moreover, now consider the set $A_i$ such that $a_{i2}-a_{i1}$ has the least value. Also notice that $a_{i2}-a_{i1} \leq n$. Clearly $S+(a_{i2}-a_{i1})-a_{11}$ must belong to $A_1$. Furthermore, as there was nothing special about choosing $A_1$, we repeat the process with all $A_j$, $1 \leq j \leq n$, getting $S+(a_{i2}-a_{i1})-a_{j1}$ must belong to $A_j$.

Now we replace the sum $S_1$ by $S_2=S_1+a_{i2}-a_{i1}$ to generate new numbers in each set- we now again consider $S_2+a_{i3}-a_{i2}$ for a new $i$ such that $a_{i3}-a_{i2}$ is minimum, thus $S_2+a_{i3}-a_{i2}+a_{j1}$ belongs to $A_j$. Now we keep going down "rows" of $A_j$, tweaking $S$ wherever $a_{jt}-a_{j(t+1)}$ is minimum for the $t$ which is not covered- clearly, as $a_{jt}-a_{j(t+1)}$ will always be $\leq n$, thus the elements must become eventually periodic with period $n$- if not, then elements will start to overlap $=>$ Contradiction.

Thus, our sets must become divided into modular classes eventually- clearly this can't work as this will require $0+1+2+...+(n-1) -k \equiv k \pmod{n} => n(n-1) \equiv 4k \pmod{n}$ for all $k$. As both $n=2$ (obviously) and $n=4$ ($1+2+3 \not \equiv 0 \pmod{4}$) don't work, and we're done.


inequalities
geometry
geometric transformation
number theory
graph theory
algebra
functional equation

Given $n$ points lying on a circle with radius $1$. If the distance between some two points (among these $n$ points) is $> \sqrt{2}$, then we call these two points a good pair. (Pairs mean unordered pairs here.) Let $q$ be the number of good pairs. Prove that $3q\le n^2$.

Hmmm... be careful while posting. This one is not number theory, it is combinatorics.
Moreover, it is a direct consequence of Turan's theorem.

I move it into the combinatorics section...

Pierre.

Just writing the proof I think Pierre was referring to:

Turan's Theorem states that a graph with $n$ vertices and no complete graph with $t$ vertices as a subgraph has at most $(1-\frac 1{t-1})\cdot \frac{n^2}2$ edges.

In our case, consider the graph having the $n$ points as vertices and in which two points are connected iff the distance between them is $>\sqrt 2$. This graph contains no complete graph with $t=4$ vertices, so it has at most $(1-\frac 13)\cdot \frac{n^2}2=\frac{n^2}3$ edges.

more over it appears as an exercise in Pierre's Graph paper.

Exactly again ! (even if I didn't remember of it)  :D

Pierre.

Graph paper, but also functional equation , inequalities, number theory part I...are available here :
http://www.animath.fr/tutorat.html#Les%20cours%20de%20l'Olympiade%20française%20de%20mathématiques

Hope it works...

Pierre.

I can't read French at all..  :(
Any English translation?


In the three-dimensional plane, it is often hard to discern which points are in which octant (on paper). Is there an official notation for this? When I draw out points, I just name the point $ \text{Letter}_x$ where $ x$ is the octant in which the point resides in.

I just label the points $ (x,y,z)$ based on their coordinates.


[b]A positive integer belongs to the set S if the sum of all its integer factors is 256.
Find the sum of all distinct elements of S.[/b]

It is given that this sum is a non-negative integer under 1000.

I may or may not be using this for math team at our school, so I would appreciate it if anyone would, apart from posting the solution, please post how long it took and/or difficulty rating.

Thanks,

~ tZf

Hmm... rather challenging...

Well, for $\sigma(n)=256$, we've got that $\frac{(p^{a+1}-1)}{p-1}=256$ for n having only one prime component.  Therefore, $p(p^{a}-256)=-255$, yet the only paired factors of -255 are $(1)(255)$, $(3)(85)$, $(5)(51)$, and $(15)(17)$, with the negatives in place.  Testing for $p^{a}-256$ yielding one of these factors returns false on each account, so we know that there is no number consisting of one prime component satisfying this condition.

As for numbers with more than one prime component, it gets pretty complicated, as you have to solve for integer solutions of $\frac{(p_{1}^{a_{1}+1}-1)(p_{2}^{a_{2}+1}-1)\cdots(p_{n}^{a_{n}+1}-1)}{(p_{1}-1)(p_{2}-1)\cdots(p_{n}-1)}=256$.  Maybe someone else wants to tackle this...

Just checking Sloane's I see that the answer is 217.

The sigma function, used to find the sum of the divisors of n: $\sigma(n)$.
Basically, you just find the prime components of n and its powers, and solve.  For example, the sum of the divisors of 208 is:

$208=2^{4}13^{1}$

$\frac{(2^{4+1}-1)(13^{1+1}-1)}{(2-1)(13-1)}=434$

[quote="Potato Theory"]As for numbers with more than one prime component, it gets pretty complicated, as you have to solve for integer solutions of
$\frac{(p_{1}^{a_{1}+1}-1)(p_{2}^{a_{2}+1}-1)\cdots(p_{n}^{a_{n}+1}-1)}{(p_{1}-1)(p_{2}-1)\cdots(p_{n}-1)}=256$.[/quote]

If this relation holds, then $\frac{p_{i}^{a_{i}+1}-1}{p-1}= 2^{k_{i}}, \ \forall i \in \{ 1,2,...,n\}$, such that $\prod{2^{k}_{i}}=2^{8}\Rightarrow \sum{k_{i}}=8$

Let's find prime numbers $p$ such that $\frac{p^{a+1}-1}{p-1}= 2^{k}$

We have $1+p+...+p^{a}=2^{k}\Rightarrow $
$p+...+p^{a}=2^{k}-1\Rightarrow $
$p(1+p+...+p^{a-1})=2^{k}-1 \ \ (1)$

Of course $p\neq 2$. Also, $k$ can't be $0$ or $1$, because then $2^{k}-1$ becomes $0$ and $1$ respectively. So $k\geq 2$
It follows $8\sum_{k=1}^{n}{k_{i}}\geq 2n\Rightarrow n\leq 4$

$\blacksquare$ If $n=4\Rightarrow k_{i}=2$
$\blacksquare$ If $n=3$ then $k_{1}+k_{2}\geq 4 \Rightarrow k_{3}\leq 8-4=4$. So $2\leq k_{i}\leq 4$
$\blacksquare$ If $n=2$ then $k_{1}\geq 2 \Rightarrow k_{2}\leq 8-2=6$. So $2\leq k_{i}\leq 6$
$\blacksquare$ [b][color=#006699]Potato Theory[/color][/b] showed that $n>1$

For $k=2,3,4,5,6$ we get $2^{k}-1=3,7,15,31,63$

$\blacksquare$ If $2^{k}-1=q$ is prime (mersenne) number, then $p \mid q \Rightarrow p=q$.
Then $(1)$ gives $1+p+...+p^{a-1}=1 \Rightarrow  a=1$

$\blacksquare$ If $2^{k}-1=q$ is not a prime. Then $2^{k}-1=15$ or $63$.
By checking the possibe cases, $2^{k}(p-1)+1$ is not a power of $p$, so we don't have solution

So the possible prime divisors are $3,7,31$ with $k=2,3,5$ respectivley

We want to choose each of the numbers $2,3,5$ at most once, so that we make a sum $8$.

Only possible $3+5=8 \Rightarrow p_{1}=7,\ p_{2}=31 \Rightarrow x=p_{1}\cdot p_{2}=217$


number theory
greatest common divisor
algorithm
Euclidean algorithm
number theory solved

Let $m,n$ be natural numbers and let $d = gcd(m,n)$. Let $x = 2^{m} -1$ and $y= 2^n +1$
(a) If $\frac{m}{d}$ is odd, prove that $gcd(x,y) = 1$
(b) If $\frac{m}{d}$ is even, Find $gcd(x,y)$

Putting $a=2^d$ and $k=\frac{m}{d}, l=\frac{n}{d}$ we have
- in (a)
$gcd(2^m-1,2^{2n}-1)=gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1=a-1$ but $gcd(a-1,a^l+1)=1$ so indeed $gcd(2^m-1,2^n+1)=gcd(a^k-1,a^l+1)=1$ so we are done

- in (b)
$gcd(2^m-1,2^{2n}-1)=gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1=a^2-1$ (cause $l$ is odd and $k$ is even),
so for sure $gcd(a^k-1,a^l+1)|a^2-1$, and it's easy to see that $a+1$ divides both $a^k-1$ and $a^l+1$
and also that $a-1$ does not divide $a^l+1$ - if not $a-1=2^d-1|2$ (but for $2^d-1|2$ we have $d=1$ and $a-1=1$ so that doesn't change our gcd) thus
$gcd(2^m-1,2^n+1)=a+1=2^d+1$ .

[quote="Rushil"]Can you plz explain $gcd(a^k-1,a^{2l}-1)=a^{gcd(k,2l)}-1$???[/quote]
Because $gcd(r,s)=gcd(r-s,s)$.
$gcd(a^x-1,a^y-1)=gcd(a^x-a^y,a^y-1)$. Because $gcd(a,a^y-1)=1$,
$gcd(a^x-a^y,a^y-1)=gcd(a^{x-y}-1,a^y-1)=gcd(a^{x-2y}-1,a^y-1)=...=gcd(a^{x mod y}-1,a^y-1)$.
Repeating this the exponents will be the residues in Euclidean algorithm, so $gcd(a^x-1,a^y-1)=a^{gcd(x,y)}-1$. Of course only for a>1.


[hide]
Using factorization of $a+b=(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2})$, We get
$\frac{(8-x)+(27-x)}{(\sqrt[3]{8-x})(\sqrt[3]{27-x})}=7 \iff 5-\frac{2}{7}x=\sqrt[3]{8-x}+\sqrt[3]{27-x}$. Well, from there we can see that $5=\sqrt[3]{8}+\sqrt[3]{27} \therefore \boxed{x=0}$. I'm not sure what to do next.. dont really feel like turning it into a 9th degree polynomial... any help?[/hide]

Hint
[hide]
1. [i]There are two real roots only and there is no need to deal with any degree nine polynomial.
[/i]
2. [i]Try to apply a suitable transformation instead of expanding the equation by brute force.[/i]
[/hide]

Please check your answer again.

Sorry, I have checked your answers. Both are correct.
My last post is misleading as there are more than 2 roots indeed.  :blush:

M4RIO's method is correct.
The full set of solutions should be
[hide]
$x = 0$,   $35$ or $27-\frac 1 {64}\displaystyle \left(-1+\sqrt{39}+\sqrt{-8+6 \sqrt{39}} \right)^3$.
Somehow I missed the root $x=35$ in my first calculation.   I apologize for that. :blush:
[/hide]


Apurv Nakade - 18 - Bronze
Riddhipratim Basu - 16 - Bronze
Ashay Arvind Burungale - 15 - Bronze
Varun Suhas Jog - 15 - Bronze
Manas Rajendra Joglekar - 15 - Bronze
Abhishek Hemantkumar Dang - 13 - Honourable Mention

I expected at lest one silver. Dang's result is a shock anyway. Check out their pics at [url]http://imo2006.dmfa.si/participants/IND.html[/url]

Here's the breakdown-
Apurv Nakade - 7 1 0 7 3 0
Riddhipratim Basu - 7 1 1 7 0 0
Ashay Arvind Burungale - 7 0 1 7 0 0
Varun Suhas Jog - 7 0 1 6 1 0
Manas Rajendra Joglekar - 7 0 1 7 0 0
Abhishek Hemantkumar Dang - 7 1 0 2 0 3

Hello,
    The results are sorry.We are 35th, an improvement of one place over last year.Apurv missed the silver by just one point. The 6th problem, as some AOPSers put it seems 'insanely difficult.'All Indians got 7 points in the first problem which was trivial.Surprisingly, the second problem has proved to be difficult for Indians when all others have found it relatively easy.And Dang, yes, shocking.A bronze and a silver in the last two IMOs, he was in the running for a gold here.Anyway, it happens for everyone.Maybe, (just maybe), Dang (and some of the others, too) has switched gears for IIT just as many in India do (rather sad).And the news is,(for those who do not know), we Indians now have a truly world class institute for Science too like the IITs for engineering.The IISERs are now functional at Kolkata and Pune. For more, you can visit http://www.iiserpune.ac.in. It is really great for students aspiring to study the Pure Sciences and Mathematics.
Sathej

Yes,what a beautiful prediction,No silver.Yep,Dang..don't respect him no more i guess :D   this was pretty bad though.

dang's result is really a shock!
i expected him to get a gold medal

what went wrong with him?


Monica reduced $ \frac {154} {253}$ to a common fraction by canceling the middle 5s to get $ \frac {14} {23}$ . This answer is correct, but the procedure is incorrect. Kia’s example worked with Monica’s incorrect procedure,

too; the fraction $ \frac {242} {x}$ also reduced to its equivalent common fraction when the identical middle digits of the two three-digit integers were canceled out. What is the greatest possible value of x?

[hide="my solution"]i get $ \frac{242} {100x\plus{}40\plus{}y}\equal{} \frac{22} {10x\plus{}y}$ to get $ x\plus{}y\equal{}4$ so x=4 and y=0....but the answer is 341[/hide]

there has been discussion about this before, but when you reduce, you get $ \frac{22}{40}$, which by mathcounts definition is not a common fraction.

[hide="My kinda brute-forcey solution"] So note that x must be divisible by 11 and odd, since it's then in reduced form. The ones you check are 341, 847, 649. Too bad for us that 121 goes into 847 and 649, so it won't be in lowest form. That leaves us with 341.[/hide]

I might have missed a few cases.


Aca os muestor dos problemas de goemtria q tuvieron cabezones a unos amigos, este problema es de un amigo y em dijo loposteara para q les dieran una mano.

Gracias

Esos óvalos celestes significan que los dos segmentos son iguales????

Llego a la ecuación: $2cos8x=cos6x+cos12x$, alguna otra idea :?:

En el problema 2,  la circunferencia esta inscrita en un triangulo equilatero?  :maybe:


The operation $ \otimes$ is defined for all nonzero numbers by
$ a\otimes b\equal{}\dfrac{a^2}{b}$. Determine $ [(1\otimes 2)\otimes
3]\minus{}[1\otimes (2\otimes 3)]$.

I shall officially be Ernie's much-needed (no offense) $ \text{\LaTeX}$ instructor!  :D

[b]Lesson 1[/b]

To make the multiplication symbol, there are two things you can do.

"\times" makes a $ \times$ sign. For example, $ 4 \times 5 = 5 \times 4$ is written "4 \times 5 = 5 \times 4".

"\cdot" makes a $ \cdot$ sign. For example, $ 4 \cdot 5 = 5 \cdot 4$ is written "4 \cdot 5 = 5 \cdot 4".

[b]Lesson 2[/b]

To make fractions, you can write "\frac{a}{b}", which will turn out as $ \frac{a}{b}$.

[quote="isabella2296"]I shall officially be Ernie's much-needed (no offense) $ \text{\LaTeX}$ instructor!  :D

[b]Lesson 1[/b]

To make the multiplication symbol, there are two things you can do.

"\times" makes a $ \times$ sign. For example, $ 4 \times 5 = 5 \times 4$ is written "4 \times 5 = 5 \times 4".

"\cdot" makes a $ \cdot$ sign. For example, $ 4 \cdot 5 = 5 \cdot 4$ is written "4 \cdot 5 = 5 \cdot 4".

[b]Lesson 2[/b]

To make fractions, you can write "\frac{a}{b}", which will turn out as $ \frac {a}{b}$.[/quote]

Np, izzy.  Oh, and I do know already know Lesson 1.

PM me Lessons 3 and beyond.

And thank you for enlightening me.

[quote="dragon96"]...

[b]Lesson 3[/b]

To do the the x in the circle, do \otimes

Example: $ a\otimes b$[/quote]

k thx, but mine was pretty good, dontcha think?

Close enough

Anyways, when do you use $ \otimes$ anyways?

Oh, it's just a symbol they made up, so you can make problems like this one. It gets confusing if you use everyday symbols like $ \div$ instead of something like $ \zeta$.

try puting curly braces around b or a space before b

Ok, this is getting kinda out-of-hand.  I suggest letting this thread die.

[quote="dragon96"]...

[b]Lesson 3[/b]

To do the the x in the circle, do \otimes

Example: $ a\otimes b$[/quote]

back to the problem real quick: so you just plug in whatever $ a\otimes{b}$ is into the problem?


Do solutions have to be emailed by midnight tonight, midnight tomorrow, or some other time tomorrow?  Thanks.

Midnight Tuesday night, Pacific time.  (Meaning that as I post this, you have about 38 hours.)

Will Round 2 be out as soon as Round 1 is over?

I don't believe so.  Round 2 will probably be released after the Round 1 Math Jam on Thursday evening.

The Round 2 deadline is November 20, so that gives you 5+ weeks to work on the problems.


Prove that f(n) = the largest integer less than or equal to pi x n has infinitely many values which are not prime.

[hide="Hint"]
$ f(n\plus{}1)\minus{}f(n)$ is bounded above.  When would an integer being only a fixed amount greater than another force it to be composite?
[/hide]

Thanks. I considered the base two expansion of pi and it seems to me that, since pi is irrational, this must contain infinitely many 0s and infinitely many 1s. If position 1/2^n contains a zero, then f(2^n) is even and therefore composite. Is this correct?

My first thought was similar, except with base 10 and powers of 10.

[quote="AndrewTom"]it seems to me that, since pi is irrational, this must contain infinitely many 0s and infinitely many 1s. [/quote]
This is much harder to prove than it looks.  I wouldn't go down this path.

We know a lot more about density than we do about digit distributions.  If the problem statement isn't true, then the primes have asymptotic density at least $ \frac{1}{\pi}$.

It seemed to me that if it contained finitely many 1s, then it would be rational; if it contains finitely many 0s then truncating after the last zero gives a rational number followed by 1s forming a convergent geometric series which is also rational and so pi would be rational, giving a contradiction.

Point. :oops: My concern was about two related stronger results, one about the relative frequencies of the digits $ 0$ and $ 1$ and one about a corresponding result in base $ 10$ mentioned by Darmani.  Here there is no simple argument (as far as I know) to show that any particular digit occurs infinitely often.

Thanks. (Have to go out and so I won't be able to respond to any more posts for a while.)

Hm, that might be easier than my method.  I was thinking that since $ f(n\plus{}1)\minus{}f(n)$ is at most $ 4$, and $ n!\plus{}2, n!\plus{}3, \ldots , n!\plus{}n$ are $ n\minus{}1$ consecutive composite numbers, for $ n\geq 4$ we cannot skip all these composites.  Thus we pick up at least one more composite for each factorial.


Imagine a 6-sided fair die, with labels A, B, C, D, E and F on its faces.  What is the probability of rolling the die 5 times, and getting exactly 3 A's?

[hide]
This is just a binomial probability I think so
${5\choose3}(1/6)^{3}(5/6)^{2}= 125/3888$
[/hide]

I got something else...can someone check to see if this is right?

[hide]Isn't binomial only used when there are 2 possible outcomes, for example heads or tails in a coin problem?

This is how I did it

Consider rolling the die and getting AAABC, exactly in that order.  Now treat this as a word and count the number of ways we can rearrange the letters in this word.  Since the A's repeat, we would have $\frac{5!}{3!}$.  The probability of getting AAABC in that exact order is $\frac{1}{6}^{5}$.  Multiply this probability by the amount of ways we can change the order but still have 3 A's, 1 B and 1 C.  \[\frac{1}{6}^{5}\cdot \frac{5!}{3!}\]

Now consider AAACD, AAADE etc etc.  Then consider AAABB where the B's repeat twice.  We can rearrange this order in $\frac{5!}{3!2!}$ different ways.  The same applies to AAACC, AAADD, etc.

After that, we just add all the probabilities.[/hide]

[quote="Phoenix257"]What is the probability of rolling the die 5 times, and getting exactly 3 A's?[/quote]

This depends rather much on the disposition of the person rolling the die, don't you think?

SnowStorm's answer looks correct to me: there need to be 3 places for the letters $A$, and then there are 5 choices for each of the other two spaces.

Of course, your solution should work as well.  Since you didn't actually show any calculations, I have to guess at what error you made: it looks like you might not have counted all possibilities of the form AAAXY; there are 10 such forms.  Both methods should give a total of 250 "good" possibilities out of $6^{5}$ total options.

Another method: imagine multiplying out $(A+B+C+D+E+F)^{5}$.  Each term in the resulting product (before we combine like terms) corresponds to a single roll; we're interested in the number of such rolls with 3 As.  This is equivalent to asking for the coefficient of $A^{3}$ in the product $(A+5)^{5}$.

Wow, I did all the calculations and got exactly the same thing.  3 completely different ways to think of the problem and still get the correct answer.  That's pretty cool!  and that's a neat trick with the expansion of that polynomial, thanks JBL

[quote="Phoenix257"]3 completely different ways to think of the problem and still get the correct answer.[/quote]
That's math for you.  Many roads, they all lead to the same place.


If $ a$ and $ b$ are positive integers, find the probability that $ \frac{a^2\plus{}b^2}{5}$ is a positive integer.

I think it easy to see that the answer is $ \frac{9}{25}$

All :
a=0(mod 5), b=0(mod 5)
a=0(mod 5) , b=1(mod 5)
...
a=4(mod 5) , b=4(mod 5)

25 variants

And variants , in which $ \frac{a^2\plus{}b^2}{5}$ is an integer are :
a=0(mod 5) , b=0(mod 5)
a=1(mod 5) , b=2(mod 5)
a=1(mod 5) , b=3(mod 5)
a=2(mod 5) , b=1(mod 5)
a=2(mod 5) , b=4(mod 5)
a=3(mod 5) , b=1(mod 5)
a=3(mod 5) , b=4(mod 5)
a=4(mod 5) , b=2(mod 5)
a=4(mod 5) , b=3(mod 5)

9 variants , so probability is 9/25 is 36%


...why are you posting trivial stuff on the ohio forum?
please dont insult ohio..... :mad:
$\boxed D$

[quote="anirudh"]For $x \neq 0$, $\frac{1}{x}+\frac{1}{2x}+\frac{1}{3x}$ equals

$\text{(A)}\ \frac{1}{2x}\qquad \text{(B)}\ \frac{1}{6}\qquad \text{(C)}\ \frac{5}{6x}\qquad \text{(D)}\ \frac{11}{6x}\qquad \text{(E)}\ \frac{1}{6x^{3}}$[/quote]
I agree, that was very easy. I have a harder question (AMC-12):

A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. The radius of the third circle is

$ \text{(A)}\ \frac{1}{3}\qquad \text{(B)}\ \frac{2}{5}\qquad \text{(C)}\ \frac{5}{12}\qquad \text{(D)}\ \frac{4}{9}\qquad \text{(E)}\ \frac{1}{2}$
 :D


Hello forumers. Before I ask my questions, let me explain my math hardships so as to better explain my situation. I was once good at math. Even though it was very early grades until grade 6, I was a very good math student. Then I moved and the school board and teachers changed. I was exposed to teachers teaching 4 or more subjects spread out across different grades. It was terrible. And I suffered for 2 and a half years with that crap. So basically, grades 7, 8 and part of 6 were lost to absolutely terrible teaching. It was like I didnt goto school. Well, in grade 9 I recovered from everyhting quite well except math. They never explained anything properly in 7 or 8 so many "building blocks" were forgotten. I passed 9 applied with a 51 and now Im in 10 applied. I seemed to be ok on most things that didnt depend on the things I missed in 7 and 8. But now in 10 I refuse to be bad at math. My teacher is willing to give up lunch to help me out and so am I. Only thing is, theres a lot to fill in that I missed. So hes been giving me practice sheets to get my, as he says, "Foundations poured, smoothed and set." He told me that the lunchtime help is good, but its still not enough. I told him I find a way to get some more help and here I am. Im getting more and more mathliterate (dont ask :P) the more I do but its slow going. But my idea is this: I get some worksheets or textbook work and I post what I dont get here to get it cleared up. And now after all that crap, the work. Im working on Standard Form equation of a Line. I have the following questions:

(1) For the equation: y=5(x-3)+20

     (a) Write the equation on the form of y=mx+b.
     (b) Write the equation in the form of Ax+By+C=0 (standard form isnt it?)

(2) Find two coordinates of the following line: 5x+2y+10=0

(3) Write the following equation in the form of Ax+By+C=0

     y= -3/4x+2  (thats negative 3 over 4)

Thats all. Theres more, but its textbook work that Ill post a bit later tonight. I know theres a lot of reading at the start, but you can skip most of it. but if someone could show how to do it thatd be cool. So, thanks a ton in advance :).

PS - How do you do that fancy looking way of typing out the math formulas? How you have to click the button to show the equations? Its bolded and its got the proper way to show integers... Well, whatever that is, I think its better to use that than regualr typing so if you could tell me how todo that aswell, thanks again. So yea. Ill be back at around 10PM EST.

EDIT: I know I said Id be back at 10, but I meant to type 9. My bad. But I just realized that my textbook work is basically the same as the questions above from my sheet. So no textbook just the sheet ones. If I get more work tomorrow that I dont get, Ill post it here. Thanks again.

[hide="1"]that's gotta suck. if u want to improve, u might want to try some online tutorials that u could search from any search engine. That's what i used to do. anyways,
1. You use distributive property and simplify.
y = 5x - 15 + 20
y = 5x + 5
standard form as i know is
Ax + By = C
5x - y = -5[/hide]

[hide="2"]2. i'll assume u mean intercepts. For x intercept, y = 0
x = -2
For y intercept, x = 0
y = -5[/hide]

[hide="3"]3. y= -3/4x+2
-4y = 3x - 8
3x + 4y - 8 = 0

If u need me to show more process, just tell me.

PS - I think its called latex. i downloaded it on my laptop but im too lazy to learn how to use it.[/hide]

Syrin the fancy thing is done by simply placiong the subject  between $\$$ signs.Eg....

y=\frac{-3}{4x}+2

place this between the dollar sign and you get

$y=\frac{-3}{4x}+2$

Open the attachment i've added for more help.

$*edit*--My 800th post.$ :D

Ok, I downloaded the latex help. Thanks for that shady. And thank you fireemblem for solving the questions :).


Euler
advanced fields
advanced fields unsolved

The euler totient of some number n is 12. Prove there are finitely many n such that this is true.

Forgive me if I'm missing something, but since $n$ cannot have any prime factors larger than $13$ it follows that we must have

$n = 2^{a}3^{b}5^{c}7^{d}13^{e}$

For some $a, b, c, d, e$; if any of $a, b, c, d, e$ get "too large" then the term that they contribute to $\varphi(n)$ will be "too large" to divide $12$, so the conclusion follows.

I posted a proof of something stronger yesterday, namely $\varphi(n)^{x}> n$ for all but finitely many $n$ ($x>1$ fixed, take $x=2$ here). Especially $\varphi(n)=a$ happens only a finite number of times ($a$ fixed).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=137813 .


Un multipack de 20 de teste de capacitate pentru elevii de clasa a opta pentru Testarile Nationale :D Succes!

[quote="matrix1442001"]un site ce contine simulari si teste de evaluare si pentru elevii de clasa a VIII-a.
                                http://www.stanica23.as.ro[/quote]
  Teste pentru capacitate se gasesc si pe http://www.epsilon.ro.


trigonometry algebra polynomial quadratics

Let be given a polymial with rational coefficients $f(x)$ satisfying the condition: $f(\cos\frac{\pi}{5})=0.$ Prove that:$f(\cos\frac{3\pi}{5})=0.$

[hide]

Notice that $\cos{\frac{3\pi}{5}}= 4\cos^{3}{\frac{\pi}{5}}-3\cos{\frac{\pi}{5}}$ and that $\cos{\frac{2\pi}{5}}= 2\cos^{2}{\frac{\pi}{5}}-1$. Since $\cos{\frac{3\pi}{5}}=-\cos{\frac{2\pi}{5}}\Rightarrow \cos{\frac{3\pi}{5}}+\cos{\frac{2\pi}{5}}= 0$, we substitute and have the equation:

$4\cos^{3}{\frac{\pi}{5}}+2\cos^{2}{\frac{\pi}{5}}-3\cos{\frac{\pi}{5}}-1 = 0$

Let the polynomial $P(x) = 4x^{3}+2x^{2}-3x-1$. We know already that $P\left(\cos{\frac{\pi}{5}}\right) = 0$. We can also see that $P(-1) = 0$. Since $P(x)$ is a third degree polynomial, it has 3 roots; hence 2 of these roots are $-1$ and $\cos{\frac{\pi}{5}}$. The product of these roots and a third root $r$ is $\frac{1}{4}= \left(-1\right)\left(\cos{\frac{\pi}{5}}\right)\left(r\right) \Rightarrow r\cos{\frac{\pi}{5}}=-\frac{1}{4}$ and the sum of the roots is $-\frac{1}{2}= (-1)+\left(\cos{\frac{\pi}{5}}\right)+r \Rightarrow r+\cos{\frac{\pi}{5}}= \frac{1}{2}$.

$\left(r+\cos{\frac{\pi}{5}}\right)^{2}= r^{2}+\cos^{2}{\frac{\pi}{5}}+2r\cos{\frac{\pi}{5}}= r^{2}+\cos^{2}{\frac{\pi}{5}}-\frac{1}{2}= \frac{1}{4}$
$\Rightarrow r^{2}= \frac{3}{4}-\cos^{2}{\frac{\pi}{5}}$.
Also, $r^{2}= \left(\frac{1}{2}-\cos{\frac{\pi}{5}}\right)^{2}\Rightarrow r^{2}= \frac{1}{4}-\cos{\frac{\pi}{5}}+\cos^{2}{\frac{\pi}{5}}$.
Then, $\cos^{2}{\frac{\pi}{5}}-\cos{\frac{\pi}{5}}+\frac{1}{4}= \frac{3}{4}-\cos^{2}{\frac{\pi}{5}}\Rightarrow 2\cos^{2}{\frac{\pi}{5}}-\cos{\frac{\pi}{5}}-\frac{1}{2}= 0$.
Now we can determine $\cos{\frac{\pi}{5}}= \frac{1+\sqrt{5}}{4}\Rightarrow r =-\frac{1}{1+\sqrt{5}}$, and $\cos{\frac{3\pi}{5}}= 4\left(\frac{1+\sqrt{5}}{4}\right)^{3}-3\left(\frac{1+\sqrt{5}}{4}\right) = \frac{1-\sqrt{5}}{4}=-\frac{1}{1+\sqrt{5}}$.

Thus $\cos{\frac{3\pi}{5}}= r$ and is a root of $P(x)$, so $P\left(\cos{\frac{3\pi}{5}}\right) = 0$, and since $P(x)$ satisfies all the conditions for f(x), $f\left(\cos{\frac{3\pi}{5}}\right) = P\left(\cos{\frac{3\pi}{5}}\right) = 0$.

[/hide]

We have: $\begin{cases}\cos\frac{\pi}{5}+\cos\frac{3\pi}{5}=\frac{1}{2}\\ \cos\frac{\pi}{5}\cdot\cos\frac{3\pi}{5}=-\frac{1}{4}\end{cases}$
$\Longrightarrow\cos\frac{\pi}{5},\,\cos\frac{3\pi}{5}$ are roots of the equation: $g(x)=x^{2}-\frac{1}{2}x-\frac{1}{4}=0$ e.t $g\left(\cos\frac{\pi}{5}\right)=g\left(\cos\frac{3\pi}{5}\right)=0$
Assume that: $f(x)=g(x).h(x)+ax+b$ with $h(x)$ is a polymial with rational coefficients, and $a,\,b\in\mathbb{Q}$
We have: $f\left(\cos\frac{\pi}{5}\right)=g\left(\cos\frac{\pi}{5}\right).h\left(\cos\frac{\pi}{5}\right)+a\cos\frac{\pi}{5}+b=0\Longrightarrow a\cos\frac{\pi}{5}+b=0$
If $a\neq0$ then $a\cos\frac{\pi}{5}+b=0\Longrightarrow\cos\frac{\pi}{5}=-\frac{b}{a}\in\mathbb{Q}$ Contradiction.
Then $a=0\Rightarrow b=0\Longrightarrow f(x)=g(x).h(x)$
Hence $f\left(\cos\frac{3\pi}{5}\right)=g\left(\cos\frac{3\pi}{5}\right)\cdot h\left(\cos\frac{3\pi}{5}\right)=0\quad\mathbb{QED}$

uhh...$\cos \frac{\pi}{5}=\frac{1+\sqrt{5}}{4}$, by the conjugate root theorem, $\frac{1-\sqrt{5}}{4}=\cos \frac{3\pi}{5}$ is a root also

[quote="seraph283"] write[/quote]

[quote]
Thus $\cos{\frac{3\pi}{5}}= r$ and is a root of $P(x)$, so $P\left(\cos{\frac{3\pi}{5}}\right) = 0$, and since $P(x)$ satisfies all the conditions for f(x), $f\left(\cos{\frac{3\pi}{5}}\right) = P\left(\cos{\frac{3\pi}{5}}\right) = 0$.

[/quote]
It's wrong[/quote]

I personally like Altheman's solution
no real room for careless error :D

April's solution is more complete.  It suggests the general theorem of which the conjugate root theorem is a special case:

[b]Theorem:[/b]  Given an algebraic number $r$, let $m(x)$ be the (monic, rational) polynomial of lowest degree such that $m(r) = 0$.  Then $m(x)$ must divide any (rational) polynomial $P(x)$ such that $P(r) = 0$.

[b]Case:[/b]  When $r = p+\sqrt{q}, p, q \in \mathbb{Q}$ its minimal polynomial is the quadratic $m(x) = (x-p)^{2}-q$.  Hence $\bar{r}= p-\sqrt{q}$ is also a root.


find x and y

x + 5i = y - yi

I'm looking at this problem, and I can't understand how to solve it. I think the y - yi part is the conjugate, but I'm not sure.

[i]I decuded[/i]...
y = -5

and if x = y, then x = -5

(-5, -5)

Right?

The real parts are equal, so $x=y$ and $xi=-yi$.  However, this is only true because y is real, not imaginary.  Since $y=-5$, $x=-5$.  And $-yi=-(-5)i=5i$.

Your solution seems fine.  However, isn't there another solution? You assumed that the $y$ pairs up with the $x$.  However, what happens when it is paired up with the $5i$? Say that $y=5i$.  Then $-(5i)i=5$.  This would mean that $x=5$.  So the other solution is $(5,5i)$

So there is no solution for this? Or was my solution correct?

For the record, I assumed y.

[quote="Aegor"]y can be either real or imaginary? [/quote]

It is not explicitly stated, but it is safe to assume that $x, y$ must both be real.

[quote="Aegor"][quote="t0rajir0u"]Aegor, there isn't a unique solution unless you assume $x, y$ are real because any complex $y$ has a corresponding $x = y+(5-y)i$.[/quote]

Exactly.  So there are two solutions to the problem, correct?  The one I found and the other one, since y can be either real or imaginary?  Or I missing something entirely?  Are there any solutions beyond $(-5,-5)$ and $(5,5i)$?  I am very confused.[/quote]
Well, then we could go into complex numbers, such as the possibility of $(2-5i,1+i)$, so there are infinite solutions if comlex nubmers are allowed.

Oh.   :blush: Okay, they are real.  I assumed that they were a single term, but could be real or imaginary.  Ignore my first post then.

P.S.:  Your solution is correct. :P


geometry
incenter
trigonometry
angle bisector

:D
(Let the area of $\triangle PQR$ be denoted as $[\triangle PQR]$)
1) Given $\triangle ABC$ with $AB=10,BC=12,AC=14$. Draw the altitude to AB (meeting AB at $D$), the angle bisector of $\angle A$ (meeting BC at $E$), and the [i]median[/i] of AC (meeting AC at $F$). Call the triangle enclosed by the three drawn lines $\triangle XYZ$. Find $\frac{[\triangle ABC]}{[\triangle XYZ]}$.

2) Find the generalization of $\frac{[\triangle ABC]}{[\triangle XYZ]}$ with $BC=a,AC=b, AB=c$, and find when $[\triangle XYZ]=0$ (when the lines are concurrent).

EDIT: I changed $AB$ to [u]10[/u] for convenience's sake.

[hide="part of it"]Well, I can tell you the last part.  It's easy to see that $[XYZ]=0$ when $\triangle ABC$ is equilateral...[/hide]

[quote="Altheman"]the lines concurr iff the triangle is isosceles, to prove:
[hide="hint"]the incenter lies on an altitude[/hide]

when you say bisector for the 3rd one, do you mean angle bisector, or median...[/quote]

oops  :P  i mean the median... i'll change it

[hide="hint"]How i would do it: line equations $\rightarrow$ shoelace[/hide]

[hide="Here is the a well-known lemma (a general case)."][color=darkblue][b]Lemma.[/b] Given are the triangle $ABC$ and the points: $D\in (BC)$, $\frac{DB}{DC}=x$; $E\in (CA)$, $\frac{EC}{EA}=y$; $F\in (AB)$, $\frac{FA}{FB}=z$. Denote the intersections $X\in BE\cap CF$, $Y\in CF\cap AD$, $Z\in AD\cap BE$. Then the area $[XYZ]$ of the triangle $XYZ$ is given by the following relation:[/color]\[ \boxed {\ [XYZ]=\frac{(1-xyz)^2}{(1+x+xy)(1+y+yz)(1+z+zx)}\cdot [ABC]\ }\ .  \] From the above well-known lemma, for $x=\frac cb$, $y=1$, $z=\frac{b\cdot \cos A}{a\cdot \cos B}$, you obtain what you wish.[/hide]


$ \sum_{k \equal{} 1}^{n}k^2\ge \frac {n\sqrt [6]{48}}{2}$

I bet you can guess the restrictions.

No, I can't guess the restrictions.  :(

$ n \geq 1 \implies \sum_{k \equal{} 1}^{n}k^2 \geq n^2 \geq n > \frac {n\sqrt [6]{48}}{2}$


Hi, I need some help with Verdana Font.
I searched on CTAN and  I found a .zip that contains all the files that I need (winfonts package http://www.ctan.org/tex-archive/fonts/winfonts/ ). But Looking the installation guide, I have doubts on where to paste the entire folder within the miktex 2.7 folder.
There are others instructions that I don't understand well.

0) I have installed the winfonts package.

1) Here, I'm not sure where to unzip. Must it necessary be unzipped in c:localtexmf ??

2) I found that file 'updmap.cfg' here: C:\Archivos de programa\MiKTeX 2.7\miktex\config and then, I added the line 'Map winfonts.map'.

3) I didn't have the 'file ttf2pk.cfg'. So I have to create it and put map 'ttfonts.map' then 'map +winfonts_ttf2tfm.map' on it. I created it on C:\localtexmf\ttf2tfm. Am I right?

4) I refreshed the database of miktex.

5) How can I run this file?

6) I couldn't enjoy .. :( I have 1 error Font T1/Verdana/m/n/10=jvn8t at 10 pt not loadable: Metric  (TFM) file not found.

I hope to solve my problem.
Thanks

You don't need to do any of this as you have MiKTeX 2.7 which includes XeTeX which does it all for you :)
All you do is use the fontspec package:
[code]\documentclass{article}
\usepackage{fontspec}
\setmainfont{Verdana}
\begin{document}
The quick brown fox jumps over the lazy dog.
\textbf{The quick brown fox jumps over the lazy dog.}
\emph{The quick brown fox jumps over the lazy dog.}
\end{document}[/code]
and then compile using xelatex instead of pdflatex or whatever. Simple!

Ohh thank you. I have one question: How can I compile with xelatex?
The problem is that I use Windows Xp and I see that Xetex is better for Mac. Isn't it?

You misunderstand. XeTeX is a version of TeX suitable for [b]all[/b] operating systems including Windows, which is why it is bundled with MiKTeX 2.7.
In TexnicCenter for example, Build, Define Output Profiles, make a copy of LaTeX=>PDF and rename it XeLaTeX=>PDF and then in the Path to the (La)TeX compiler change pdflatex.exe to xelatex.exe.
You can do similar things in other editors, changing pdflatex.exe to xelatex.exe.

It works.
Now I have another trouble. I need to use pstricks and some color (pstcol) and thi works with ps as output. So How can I to configure an output using verdana (xelatex) and ps format.

I added my .tex

Thanks again

I haven't tried it but it should be the same as you would normally do PS but instead of using latex.exe you use  xelatex.exe.

I've looked elsewhere and it seems that XeTeX can't do PS though that may come in the future. A workaround suggested is to compile the PSTricks parts to PDF separately then include them in the main document (compiled with xelatex) as graphics. The [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/pdfpages.html]pdfpages package[/url] may help here as it allows you to include bits of a pdf document.

There's also a suggestion to try [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/pdftricks.html]pdftricks[/url] instead of pstricks.


geometry
circumcircle
vector
geometric transformation
homothety
ratio
trapezoid

Let $O$ be the circumcenter of the triangle $ABC$. The segment $XY$ is the diameter of the circumcircle perpendicular to $BC$ and it meets $BC$ at $M$. The point $X$ is closer to $M$ than $Y$ and $Z$ is the point on $MY$ such that $MZ = MX$. The point $W$ is the midpoint of $AZ$.

a) Show that $W$ lies on the circle through the midpoints of the sides of $ABC$;

b) Show that $MW$ is perpendicular to $AY$.

Since M is the midpoint of XZ and W is the midpoint of AZ, MW is a midline of the triangle $\triangle AXZ$, $MW \parallel AX$ and $MW = \frac{AX}{2}.$ Since XY is a diameter of (O), $\angle XAY = 90^\circ$ is right, $AX \perp AY$ and $MW \perp AY.$ X is the midpoint of the arc BC of the circumcircle (O) opposite to the vertex A, hence AX bisects the angle $\angle CAB.$ Let K, L be midpoints of the sides AB, AC, respectively. Since $MW \parallel AX,$ MW bisects the angle $\angle LMK$ of the medial triangle $\triangle MKL.$ Since $MW = \frac{AX}{2},$ W is the midpoint of the arc KL of the circumcircle (N) of the medial triangle $\triangle MKL$ opposite to the vertex M.

[hide="a)"]We'll use vectors:
$ M \equal{} \frac {B \plus{} C}{2}$, $ W \equal{} \frac {A \plus{} Z}{2}$. Also $ Z \plus{} X \equal{} 2M \equal{} B \plus{} C$.

$ \frac {2}{3}W \plus{} \frac {1}{3}X \equal{} \frac {2}{3}\frac {A \plus{} Z}{2} \plus{} \frac {1}{3}X \equal{} \frac {1}{3}A \plus{} \frac {Z \plus{} X}{3} \equal{} \frac {1}{3}A \plus{} \frac {1}{3}B \plus{} \frac {1}{3}C \equal{} G$, where $ G$ is the centroid of $ ABC$.

So a homothety from $ G$ and ratio $ \minus{} \frac {1}{2}$ sends the circumcircle of $ ABC$ to its nine point circle and $ X$ to $ W$, so $ W$ is on the nine point circle, as desired.
[/hide]
[hide="b)"]We have that $ MW\parallel AX$ because $ M$ is the midpoint of $ AZ$ and $ M$ is the midpoint of $ XZ$. But since $ XY$ is a diameter, $ AY\perp AX$, so we conclude that $ MW \perp AY$, as desired.
[/hide]

My solution is a bit different.
a) Let $ O$ and $ H$ be the circumcenter and orthocenter. Let $ AH$ meet $ BC$ and the circumcircle at $ P$ and $ K$ respectively, let $ Q$ be the midpoint of $ AH$. Let $ R$ be the midpoint of $ OH$, which is the nine-point center, let $ WR$ meet $ HZ$ at $ S$. Then $ WS \equal{} \frac {1}{2}AH \equal{} OM$ and $ RS \equal{} \frac {1}{2}OZ$, so $ WR \equal{} WS \plus{} RS \equal{} OM \plus{} \frac {OZ}{2} \equal{} \frac {R}{2}$, so $ W$ lies on the nine-point circle
b) $ WQPM$ is cyclic, since all four points lie on the nine-point circle. Since $ \angle QPM \equal{} 90, \angle QWM \equal{} 90$. But $ HP \equal{} PK$ (well-known), so both $ HKXZ$ abd $ AKXY$ are isoceles trapezoids, so $ AY \parallel HZ \parallel QW$. Thus, $ MW \perp AY$.
 [geogebra]d8d2ea0b896ef2d63efd426b3be104c594edf324[/geogebra]

a)we note N and Q the middle points of AB and AC
BXCZ-paralelogram BXC=BZC=B+C
BZ and ZC intersect NM and MQ in K1 and K2
NAQM paralelogram so NMQ=A so K1ZK2M is cyclic but BZ,ZC parallel to NW and QW so the quadrilateral NWQM is cyclic
b)AX paralel to WM but AX perpendicullar to AY (YX diameter) so MW perp.to AY
so we are done

frenchy you have a nice solution
and i tried to draw a figure :)
 [geogebra]cd5efd9290267fd0d85873a0bd2e1bd6c8c16dee[/geogebra]

and looks like homothety centered at A with ratio $\dfrac{1}{2}$
then $BZC \Rightarrow NWQ$

As constructed, the triangles $\triangle ABC$ and $\triangle AXZ$ share the same $G$, having $AM$ common median, hence the homothety centered $G$ with ratio $-2$ transforms $ABXC$ into $MLWK$.

Best regards,
sunken rock

Let $ N $ and $ P $ be the midpoints of $ AB $ and $ AC $. Since $\angle NMB = \angle C $ and $ \angle PMC =\angle B, \angle NMP= \angle A $ so we have to prove that $ \angle PWN= \angle B + \angle C $. Note that $ BXCZ $ is a parallelogram and $ ABXC $ is cyclic so $ \angle BZC= \angle BXC= \angle B + \angle C $ Now $ WN $ is the midline of $ ABZ $ so $ \angle NWZ=180-\angle AWN=180- \angle BZA $ and similar $ \angle ZWP=180- \angle AZC $ So $ \angle NWP=\angle NWZ + \angle ZWP =180- \angle BZA + 180- \angle AZC = 360- \angle BZA - \angle AZC = \angle BZC = \angle B + \angle C $ so  $ MPWN $ is cyclic.

Here is a solution using Complex bashing...

As it is said to prove that $W\in$ to the circumcircle of the medial triangle of $\triangle ABC$, which is the Nine-Point Circle, so we will introduce the Nine- Point Center. Let the Nine-Point Center be $N$. So, $n=\frac{a+b+c}{2}$. So, $NM=|(\frac{a+b+c}{2})-(\frac{b+c}{2})|=\frac{1}{2}$.

Now, $z=2m-x=b+c-\sqrt{bc}\implies w=\frac{a+b+c-\sqrt{bc}}{2}$. So, $NW=|\frac{\sqrt{bc}}{2}|=\frac{1}{2}$. Hence, $W$ lies on the circumcircle of the Medial triangle.

Now for the perpendicularity part we will use the perpendicularity lemma.

$(\frac{m-x}{a-y})=-(\overline{\frac{m-x}{a-y}})$ which can be easily proved. $\blacksquare$.

Let $H$ - orthocenter, $A'$ - antipode of $A$.
Note that $YZ=AH$ , $AHZY$ - paralellogram and $W$ - its center. Dilate from $H$  with ratio $2$.
a) euler circle goes to circumcircle.
b) $MW$ goes to $A'Y$.


geometry
geometric transformation
reflection

Are lenses or mirrors affected by chromatic aberration?

Lenses are affected by chromatic aberration.The gaussian formula for the focal distance of a lense is 1/f=(n2/n1-1)*(1/R1-1/R2) where:
f-focal distance
n2-refraction index of the lens
n1-refraction index of the exterior of the lens(from which the ray comes into the lens)
R1-first radius
R2-second radius

And what it is known is that refraction index is different for different monochromatic light.if the wavelength is big then the index of refraction is low so a ''white light''(no colour at all) is decomposed by a lenses in other colours because you can write different ecuations for the different colours(you should know the relation between the object and the image)

A perfect mirror (which reflects all the electromagnetic energy will not be affected by this abberation)


algebra
polynomial
algorithm
number theory open
number theory

Has the Frobenius problem (also called Postage Stamp Problem) been solved for the case $n=3$ ?
Can you give a link to the solution if it exist ? Thanks in advance.

It was solved in the general case already (for any $n$), if a polynomial time algorithm can be called a solution. I came across a paper in which such a algorithm was given but couldn't find it now  :(

Is this a recent discovery (something between 2000-2005) ?

Not really  :) I eventually found the paper so I'm attaching it below. It's from 1989...

Googling "frobenius postage stamp problem" gives this [url=http://mathworld.wolfram.com/PostageStampProblem.html]article from Mathworld[/url] as the first item.


Given a finite number of parallel segments in the plane s.t. for each three there is a line intersecting them, prove that there is a line intersecting all the segments.

Suppose the segments are $M_iN_i$ and they are parallel to the $Oy$ axis.
Let $L_i=\{(a,b)\in \mathbb {R}\times \mathbb{R}|\quad \mbox{the line of equation} \quad y=ax+b \quad \mbox{intersects} \quad M_iN_i\}.$
It is not difficult to check that all $L_i$'s are convex sets, hence the result follows from Helly's theorem.


calculus
integration
limit
trigonometry
ratio
logarithms
real analysis

Calculate
\[ \lim_{n \rightarrow \infty} \int_{- \pi}^{\pi} \frac{n! \, 2^{2n \cos \, \theta} }{|(2ne^{i \theta} - 1)(2ne^{i \theta} - 2)(2ne^{i \theta} - 3) \cdots (2ne^{i \theta} - n) |} \, d \theta  \]

If I haven't made a stupid mistake in my computations, the answer is $2\pi$. Such a simple answer suggests that there might be some nice trick leading to the result immediately. Unfortunately, I found none, so below is a "brute force" approach. I'll just sketch the main steps leaving the technical details to you.

The first thing to note is that the integrand has a sharp peak at $0$ for large $n$, so sharp that almost the whole integral comes from an arbitrarily small interval around $0$.

The second thing to observe is that the value of the integrand $I(\theta)$ at $0$ is $\frac{2^{2n}n!(n-1)!}{(2n-1)!}\approx 2\sqrt{\pi n}$ (just use the Stirling formula).

Now comes the main part: we have to compare $I(\theta)$ to $I(0)$ for small $\theta$. To this end, write  $I(\theta)=I(0)2^{2n(cos\theta-1)} \prod_{k=1}^{n}\frac{2n-k}{|2ne^{i\theta}-k|}$. Note that $\cos\theta-1\approx -\frac{\theta^2}{2}$. Also, if $a,b>0$, $a>2b$  and $\theta$ is small, we can write
$|ae^{i\theta}-b|=\sqrt{(a-b)^2+2ab(1-cos\theta)} \approx (a-b)e^{\frac{ab\theta^2}{2(a-b)^2}}$, so the ratio $I(\theta)/I(0)$ can be well approximated by $\exp\left\{-n\theta^2\left(\ln 2+\frac 1n\sum_{k=1}^n \frac{nk}{(2n-k)^2}\right)\right\}$. The sum in the exponent is just the Riemann sum for the integral $\int_0^1\frac{t}{(2-t)^2}\,dt=1-\ln 2$, so, we finally get $I(\theta)\approx I(0)e^{-n\theta^2}$. Integrating this, we see that the integral in question is approximately $2\sqrt{\pi n}\int_{\mathbb{R}}e^{-n\theta^2}\,d\theta=2\pi$.

:thumbup: You are really great...
Yes the answer is $2 \pi$. :lol:


In $ \triangle ABC, \angle A \equal{} 75^\circ,\angle B \equal{} 35^\circ, D$ is a point on $ BC, BD \equal{} 2CD.$
Prove: $ AD\cdot AD \equal{} (AC \plus{} BD)(AC \minus{} CD).$

$ \angle C \equal{} 180^\circ \minus{} (\angle A \plus{} \angle B) \equal{} 70^\circ \equal{} 2 \angle B.$ Circle with center $ A$ and radius $ AC$ cuts $ BC$ again at $ E.$ $ \triangle AEC$ is isosceles with $ AE \equal{} AC$ $ \Longrightarrow$ external angle of $ \triangle ABE$ at $ E$ is $ \angle AEC \equal{} \angle C \equal{} 2\angle B$ $ \Longrightarrow$ $ \triangle ABE$ is also isosceles with $ AE \equal{} BE$ $ \Longrightarrow$ $ BC \equal{} BE \plus{} EC \equal{} AC(1 \plus{} 2 \cos C)$ $ \Longrightarrow$ $ \cos C \equal{} \frac {BC \minus{} AC}{2 AC}.$ By cosine theorem for $ \triangle ADC,$

$ AD^2 \equal{} AC^2 \plus{} \frac {_1}{^9} BC^2 \minus{} \frac {_2}{^3} AC \cdot BC \cdot \cos C \equal{} AC^2 \plus{} \frac {_1}{^9} BC^2 \minus{} \frac {_1}{^3}BC (BC \minus{} AC) \equal{}$

$ \equal{} AC^2 \plus{} \frac {_1}{^3} AC \cdot BC \minus{} \frac {_2}{^9} BC^2 \equal{} AC^2 \plus{} AC (BD \minus{} CD) \minus{} BD \cdot CD \equal{}$

$ \equal{} (AC \plus{} BD)(AC \minus{} CD)$


geometry
analytic geometry
integration
trigonometry
calculus
calculus computations

OK, Armo, you asked for more:

A cow is tethered to the outside wall of a circular silo of radius $r$. The length of the tether is equal to half the circumference of the silo. What is the area of the land that the cow can graze?

Express your answer as the fraction that area is of the area the cow could graze if the silo weren't in the way.

[Uses: parametric representations of curves.]

[Sanity check: expressed as a fraction, the answer is obviously greater than 1/2 and obviously less than 1.]

Hey - it's a cute problem. Sombody give it a try.

:rotfl: really , very hard  ??!! :rotfl:
 :cool:

Put the origin $O$ of polar coordinates $(r, \phi)$ at the center of the silo with radius $R$ and the ray $\phi = 0$ through the point $A$ the cow is attached Let $P$ be an arbitrary point on the silo circumference, such that $0 < \phi(P) < \pi$. The total length of the leash is $\pi R$ and remaining length of the leash at the point $P$ is is $(\pi - \phi)R$. As the leash is tangent to the silo at the point $P$ and perpendicular to its radius $OP$, the area $dA$ gained by the cow when increasing the leash angle by $d\phi$ is

$dA = \frac 1 2 [(\pi - \phi)R]^2 d\phi$

$A_1 = \int_{A_1} dA = \frac {R^2}{2} \int_0^\pi (\pi - \phi)^2 d\phi = \frac {R^2}{2} \int_0^\pi (\pi^2 - 2\pi \phi + \phi^2)\ d\phi =$

$= \frac {R^2}{2} \left [\pi^2 \phi - \pi \phi^2 + \frac{\phi^3}{3} \right ]_0^\pi = \frac 1 6 \pi^3 R^2$

The area of the semicircle with radius $\pi R$ that can be grazed without the silo being in the way is equal to

$A_0 = \frac 1 2 \pi (\pi R)^2 = \frac 1 2 \pi^3 R^2$

The total area that can be grazed is

$A_0 + 2A_1 =  \frac 1 2 \pi^3 R^2 + \frac 2 6 \pi^3 R^2 =  \frac 5 6 \pi^3 R^2$

Without the silo, the total area would be $2A_0 = \pi^3 R^2$, so that

$\frac {A_0 + 2A_1}{2 A_0} = \frac 5 6$

Thanks, yetti. That (or some slight variant of it) is clearly the best argument.

Here's a more painful approach:

As you did, I'll take the origin to be the center of the silo. I'll flip the picture around from yours, and let $\theta=\pi$ be the point at which the leash is attached. We can descibe one portion of the outer boundary by the parametric curve, for $\theta\in[0,\pi]$:

$x=R\cos\theta+R\theta\sin\theta$
$y=R\sin\theta-R\theta\cos\theta$

We can differentiate the latter to get $dy = R\theta\sin\theta\,d\theta.$

When $\theta=\pi$, the rope is pulled out straight and the cow is on the line $x=-R.$ We calculate the following integral:

$\int(x+R)dy=\int_0^{\pi}(R+R\cos\theta+R\theta\sin\theta)(R\theta\sin\theta) d\theta.$

Once you have computed this (pretty ugly, but possible), you double it, subtract $\pi R^2$ (the inaccessible area of the silo), then add $\frac12\pi^3R^2$ (the unobstructed semicircle). The answer is the same.

[quote="yetti"]As the leash is tangent to the silo at the point $P$ and perpendicular to its radius $OP$, the area $dA$ gained by the cow when increasing the leash angle by $d\phi$ is

$dA = \frac 1 2 [(\pi - \phi)R]^2 d\phi$[/quote]

How does one derive this?

EDIT: Thanks for the explanation, yetti!

This is the area of an isosceles triangle with the shoulder sides $(\pi - \phi) R,$ which form an angle $\text d \phi.$ Since $\text d\phi$ is so small, $\sin (\text d \phi) = \text d \phi.$ Or you could say that the triangle base is $(\pi - \phi)R\ \text d \phi,$ the shoulder side $(\pi - \phi) R$ and the base angles both right. In either case, $\text dA = \frac 1 2 [(\pi - \phi)R]^2\ \text d \phi.$


ΑΣΚΗΣΗ 1- ΕΚΦΩΝΗΣΗ
Έστω $n$ ένας θετικός ακέραιος . Για $1\leq k\leq n$ έστω $a_{k}$ ο αριθμός των ζευγών  των μη αρνητικών $(x,y)$ που ικανοποιούν την εξίσωση
$kx+(k+1)y=n-k+1$ .Δείξτε ότι $a_{1}+a_{2}+...+a_{n}=n$

[color=darkblue][b]ΑΣΚΗΣΗ 1-ΛΥΣΗ[/b][/color]

Όλες οι λύσεις της $kx+(k+1)y=n-k+1$ δίνονται από τους τύπους $x=n-k+1+kt$ , $y=-n+k-1-(k+1)t$ ,όπου $t \in \mathbb Z$.

Επειδή ψάχνουμε το πλήθος των μη αρνητικών  ζευγών $(x,y)$ θέλουμε να ισχύει $x=n-k+1+kt\geq 0$ και $y=-n+k-1-(k+1)t\geq 0$ από όπου προκύπτει ότι $-\frac{n-k+1}{k}\leq t\leq-\frac{n-k+1}{k+1}$.
Άρα το $a_{k}$ είναι το πλήθος των ακεραίων στο διάστημα $\left[-\frac{n-k+1}{k},-\frac{n-k+1}{k+1}\right]$.

Συνεπώς το άθροισμα $a_{1}+a_{2}+...+a_{n}$ ισούται με το πλήθος των ακεραίων στο σύνολο $\left[-n,-\frac{n}{2}\right] \cup \left[-\frac{n-1}{2},-\frac{n-1}{3}\right] \cup ... \cup \left[-\frac{1}{n},-\frac{1}{n+1}\right]$ ,που ισούται με το πλήθος των ακεραίων στο σύνολο $\left[-n,-\frac{1}{n+1}\right]$ ,που ισούται με $n$.

Στέλιος...

Πάρα πολύ ωραία Στέλιο . :lol:  Μπράβο.......

[b]Να δείξετε ότι για κάθε πρώτο αριθμό p υπάρχουν ακέραιοι
χ , ψ , z , w έτσι ώστε χ^2 + ψ^2 + z^2 = wp και 0 < w < p .[/b]

Η λύση βρίσκεται αναλυτικά στο τεύχος 59 του Ευκλείδη Β στις σελίδες 16-17-18-19  :wink:

Εστω $k,m,n$ ειναι θετικοι ακεραιοι αριθμοι τετοιοι ωστε ο $m+k+1$ να ειναι πρωτος αριθμος μεγαλυτερος του$n-1$.Θετουμε $c_{s}=s(s+1)$

N.δ.o το γινομενο $(c_{m+1}-c_{k})(c_{m+2}-c_{k})...(c_{m+n}-c_{k})$ διαιρειται απο το γινομενο $c_{1}c_{2}...c_{n}$

 :D  :D

υπεροχες ασκησεις αλλα μπορει κανεις να δωσει μια λυση για τη δευτερη ασκηση? δεν εχω Ευκλειδη Β  :D

Βάζω την απόδειξη (αλλά όχι αυτή που έχει ο Ευκλείδης Β)

[color=red]ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ-2-ΛΥΣΗ[/color]
Θα δείξουμε ότι υπάρχουν :
 $x,y,z$, όχι όλοι $\equiv 0 \mod p$ που ικανοποιούν
 $x^{2}+y^{2}+z^{2}\equiv 0(mod\ p)$. Πράγματι
Για  $p=2$ είναι εύκολο άρα έστω $p>2$ .
Έστω $R: =\{ 0 , 1 , 2^{2}, ... , \left( \frac{p-1}{2}\right)^{2}\}$ και $S: =\{-1-0 ,-1-1 ,-1-2^{2}, ... ,-1-\left( \frac{p-1}{2}\right)^{2}\}$.
Από την αρχή της περιστεροφωλιάς υπάρχουν κάποια  $r \in R, s \in S$ με $r \equiv s \mod p$, άρα υπάρχουν  $x,y$ με  $x^{2}\equiv-1-y^{2}\mod p$ or $x^{2}+y^{2}+1^{2}\equiv 0 \mod p$.

QED

Πολύ ωραία Σιλουανέ !
 Περίπου ίδια με τη δική μου λύση .

η καταλληλη μεθοδος :!:

Ράλλη

 μήπως το p είναι μεγαλύτερο του (n + 1) και όχι του (n- 1) ??

:blush::blush: ..............
Sorry...εχεις δικιο Νικο.....
Ευτυχως αυτη η λεπτομερεια δεν επιρεαζει τοσο πολυ την κεντρικη ιδεα του προβληματος....
Το παραθετω πλεον ολοκληρωμενο..
[quote="RAKAR"]Εστω $k,m,n$ ειναι θετικοι ακεραιοι αριθμοι τετοιοι ωστε ο $m+k+1$ να ειναι πρωτος αριθμος μεγαλυτερος του$n+1$.Θετουμε $c_{s}=s(s+1)$

N.δ.o το γινομενο $(c_{m+1}-c_{k})(c_{m+2}-c_{k})...(c_{m+n}-c_{k})$ διαιρειται απο το γινομενο $c_{1}c_{2}...c_{n}$

 :D  :D[/quote].

 :D

[color=red]ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ-3-ΛΥΣΗ [/color]

 Τώρα εντάξει .
 Η λύση μου στο συννημένο αρχείο .

 Για ποιο λήμμα λες Ράλλη ?

[b]ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ - 4 - ΑΣΚΗΣΗ [/b]

[b]Να εξετάσετε αν υπάρχει λύση {χ ,y ,z ,w ,v} στους ακέραιους μεγαλύτερους του 1998 , της εξίσωσης
     χ^2 + y^2 + z^2 + w^2 + v^2 = xyzwv - 65[/b]

Πολυ ωραια Νικο!!Σαν λημμα εννοουσατο<το γινομενο ν διαδοχικων αριθμων διαιρειται απο το ν!>.Ακριβως ετσι ειναι +η λυση μου...
Να υπεν8ιμισω σε οποιον δεν το κσερει οτι ειναι 8εμα μιασ ιμο...(παλια)

Για την θεωρια αριθμων 4

[hide]Γνωστο...ειναι το δευτερο θεμα απο την ολυμπιαδα της Ταιβαν το 1998...
Θα εδινα το link,αλλα τωρα δεν ανοιγει.....[/hide]

 :D

μπορει να δωσει κανεις μια ελαχιστη βοηθεια για την ασκηση 4 ? :)

[u][b]ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ 5[/b][/u]

Για κάθε πρώτο $p$ και για όλους τους φυσικούς $a$ και $n$ να αποδείξετε ότι:

$p^{n}$ $/$ $a^{p^{n}}-a^{p^{n-1}}$

[b] ΘΕΩΡΙΑ ΑΡΙΘΜΩΝ - 5 - ΛΥΣΗ[/b]

 Νίκο
 η άσκηση είναι εφαρμογή του θεωρήματος Fermat .
 Εκτός αν θέλεις και την απόδειξη του θεωρήματος .

 Kosta
 Για την άσκηση 4 θα βοηθήσει μία ακολουθία λύσεων .

Akrivos...auto eixa sto myalo mou otan tin kataskeuasa!

[color=red]Θεωρία αριθμών 6[/color]
Οι αριθμοί $a_{1},...,a_{100}$ είναι μια αναδιάταξη των αριθμών 1,2,3,...,100 . Για $1\leq i\leq 100$ έστω $b_{i}=a_{1}+...+a_{i}$ και έστω $r_{i}$ το υπόλοιπο της δίαίρεσης του $b_{i}$ με το 100 . Δείξτε ότι οι αριθμοί $r_{1},r_{2},...,r_{100}$ επιτυγχάνουν 11 διαφορετικές τιμές .

siluan na se rotiso kati..an ine a1=1,a2=2 klp..klp tote
b1=1,b2=3,b3=6,b4=10,b5=15,b6=21,b7=28,b8=36,b9=45,b10=55,b11=66,b12=78...   (i trigoniki ari8mi) kai afinun diaforetika ypolipa dierumeni me to 100...
mipos den katalava kati kala??..

[b]askisi 7[/b]
esto akolu8ia $a_{n}$ me $a_{1}=2$ kai $a_{n+1}=2^{a_{n}}$
n.d.o $a_{n+1}$isoipolipo $a_{n}$ (mod n)

 :D

den tin exo lisi akoma,aplos tin exo epali8eusi gia arketes times....

Ipothetw tha thelei na pei "epitigxanoun [b]toulaxiston[/b] 11 diaforetikes times". Etsi einai silouan?

Alexandros

Y.G. Xronia polla stous Nikous kai kiriws ston nickolas kai ston nikos pou gnwrizw! An einai kai kapoios pou lismonw xronia polla+se ekeinon!!

Ναι αυτό εννοώ,  :)

Επίσης μια παρόμοια της 7 (και λίγο πιο δύσκολη...) είναι η τρίτη από το διαγωνισμό USAMO του 1991 ([url=http://www.kalva.demon.co.uk/usa/usoln/usol913.html]link[/url]).

[color=red]ΑΣΚΗΣΗ  -  8[/color]
[i]
 Δεν είναι άσκηση ΙΜΟ , αλλά την βάζω εδώ για να τις έχουμε συγκεντρωμένες. Τη δίνω στα αγγλικά , όπως τη βρήκα σε ένα διαγωνισμό από πανεπιστήμιο.Ας κάνει κάποιος την απόδοση στα ελληνικά και τη λύση σε LAtex.Φαίνεται να έχει ενδιαφέρον ![/i]

[b]Question 5[/b]
Professor Kato has forgotten her $8$ digit phone number. But she remembers that the first digit is the number of 0’s in the number, the second digit is the number of 1’s, the third digit is the number of 2’s and so on. What is her phone number?
[hide][b]Solution[/b]
 42100100
This is found by a combination of trial and error and reasoning.
Let the entry in the ith position be $a_{i}, i = 0, 1, . . . , 7$. It’s clear that most of the entries will
have to be $0$. For example, if $a_{4}$ and $a_{5}$ are both $1$ the list must contain a $4$ and a $5$, so four
of something and five of something else, which already means at least 9 entries are needed.
Indeed, the same argument shows $a_{4}, . . . , a_{7}$ contains at most one non-zero entry, which can
only be a $1$. (The non-obvious case is if $a_{4}= 2$. Then we have exactly two $4'$s in our list and a $2$,
but the two fours mean the list has four entries of something and four of something else; theseentries can only be 2’ s and 4’ s, so the entire list must consist of 4’s and 2’s but then $a_{4}= 4$  not 2.)
2
On the other hand, the four entries$a_{4}, . . . , a_{7}$ cannot all be zero, since then we would have at
least $4$ zeros, so $a_{0}> 4$ and hence one of $a_{4}, . . . , a_{7}$ would have to be at least 1 to indicate this.
So $a_{4}, . . . , a_{7}$  contains a $1$ and three zeros. Thus $a_{0}> 3$, and $a_{1}> 1$. The list has at least $3$
non-zero entries, so $a_{0}= 3, 4$ or  $5$. And so on. Eventually we get the solution $42100100$.[/hide]

[color=red][b]ΑΣΚΗΣΗ 9[/b][/color]

Σας παραθέτω μία ωραία άσκηση από το CRUX.

Είναι εύκολο να δείξουμε ότι το γινόμενο τεσσάρων διαδοχικών θετικών ακεραίων αυξημένο κατά 1 είναι πάντα τέλειο τετράγωνο. Επίσης, είναι εύκολο να δείξουμε ότι το γινόμενο οποιωνδήποτε δύο διαδοχικών θετικών ακεραίων αυξημένο κατά 1 δεν είναι ποτέ τέλειο τετράγωνο. Επίσης, παρατηρήστε ότι:

$2\cdot3\cdot4+1=5^{2}$ και $4\cdot5\cdot6+1=11^{2}$

α) Βρείτε έναν άλλον φυσικό $n$, τέτοιον ώστε ο $n(n+1)(n+2)+1$ να είναι τέλειο τετράγωνο.

β) Υπάρχουν περισσότερα παραδείγματα;

[quote="Αρχιμήδης 6"]Οι μονες λυσεις της εξισωσης  ειναι n=0,n=2 &n=4. :wink:[/quote]

Όχι, δεν είναι μόνο αυτές οι λύσεις της εξίσωσης.

PS: Αν θες εξήγησε και το σκεπτικό σου. Αυτό είναι καλό γιατί οι ξερές απαντήσεις δε χρησιμεύουν σε τίποτα, ενώ από τις λύσεις των εκάστοτε ασκήσεων όλο και μαθαίνουμε νέα πράγματα. Ευχαριστώ!

Ναι mostel πραγματι εκανα παιδαριωδες λαθος καθαρα απο λογους ταχυτητας....οσο για την λαθος λυση μου δεν υπαρχει λογος να την γραψω αφενος γιατι καταλαβα το λαθος μου και αφετερου επειδη δεν ξερω latex. :)

Δεν υπάρχει πρόβλημα πάνω στους δύο τομείς που αναφέρεις!

Πρώτα από όλα, πολλές φορές από τα λάθη μας μαθαίνουμε κιόλας, αφού το κάθε σκεπτικό μπορεί να κρύβει κάτι το ξεχωριστό, που έστω και αν είναι λάθος ...δεν παύει να είναι ξεχωριστό! Όσον αφορά τον τομέα του Latex, αν θες κάνε ένα πρόχειρο post τη λύση σου εδώ, και θα την περάσω εγώ σε latex. Αυτά!

PS: Πλησιάζει 2 η ώρα, άρα λοιπόν ΧΡΟΝΙΑ ΠΟΛΛΑ ΣΕ ΟΛΟΥΣ ΚΑΙ ΠΑΝΩ ΑΠΟ ΟΛΑ ΥΓΕΙΑ! Καλές γιορτές παιδιά  :)

(Mιας και δεν με παιρνει υπνος)Να ευχηθω και γω ΧΡΟΝΙΑ ΠΟΛΛΑ σε ολους και κυριως στους εορταζοντες και να ευχηθω σε ολα τα μελοι του φορουμ να προσπαθουν για το καλυτερο.Οσο για την λυση μου αφου επιμενεις το λαθος  σε γενικες γραμμες ηταν οτι  αν για τους a.b.c.d ισχυει οτι ab=cd και (a.b)=(c,d)=1 τοτε a=c  και b=d η b=c και a=d.Κατι τετοιο θα ισχυε αν (a,c)=(b,d)=1 και οχι (a,b)=(c,d)=1  :) (μεγαλο λαθος εεε?)

[color=red][b]ΑΣΚΗΣΗ 10[/b][/color]

[i]Δεν είναι δύσκολη, αλλά ας κλείσει ο κύκλος των ασκήσεων με μία άσκηση κλασική και συνάμα όμορφη. Η συγκεκριμένη άσκηση βρίσκεται στο βιβλίο [b]Ολυμπιάδες Μαθηματικών [/b]του κ. Στεργίου. Την άλλαξα λίγο στα νούμερα για να ταιριάζει στην χρονιά που έρχεται![/i]

Να αποδειχθεί ότι το κλάσμα:

$A=\frac{2007n+1}{2007n^{2}+n+1}$

είναι ανάγωγο για κάθε τιμή του φυσικού $n$.

ΧΡΟΝΙΑ ΠΟΛΛΑ σε όλους κατ'αρχήν .Δεύτερον έχω μια λύση για την άσκηση 9

Λοιπόν αν δεν κάνω λάθος στην απόδειξή μου η εξίσωση
$n(n+1)(n+2)+1=y^{2}$  έχει άπειρες λύσεις (το απέδειξα με πολυωνυμικές ισοτιμίες )
Λοιπόν μια λύση είναι $n=55$   :wink:
Τώρα άλλες λύσεις είναι οι εξής :
[color=red]782 ,784[/color] ,879 ,[color=red]898 ,900[/color] ,[color=red]1022 , 1024 [/color],1117 , [color=red]1154 , 1156 [/color], 1271 ,1278, [color=red]1294,1296 [/color],......

Παρατηρήσεις :όπως μπορείτε να δείτε πολλοί από τους παραπάνω διαφέρουν κατά δύο (αυτοί με κόκκινο) . Ψάψνω να βρω μια περιοδικότητα αυτών που διαφέρουν κατά δύο (μέχρι τώρα έχω καταφέρει λίγα πράγματα )

Επίσης οι περισσότεροι από τους αριθμούς είναι <<square free>> .

Επίσης είμαι πολυ κοντά στο ναδείξω ότι αν σε μια εκατοντάδα υπάρχει ένας τέτοιος αριθμός τότε θα υπάρχει ακόμη ένας

Πώς σας φαίνεται ?

Ωραία Σιλουανέ. To 55 ήταν η άλλη λύση.  :lol:

[b]Λύση - Spoiler[/b]

[hide]Η λύση της άσκησης όπως την έστειλε ο κος Λάμπρου στο CRUX:

http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv26n1/body/HTML/59?template=CRUX

http://journals.cms.math.ca/cgi-bin/vault/public/view/CRUXv26n1/body/HTML/60?template=CRUX[/hide]

Λυση της ασκησης 10:Εστω d=(2007n+1,2007n^2+n+1) τοτε d/2007n^2+n & αφου d/2007n^2+n+1 τοτε d/1 αρα d=1.Επομενως το κλασμα ειναι αναγωγο.Χωρις παρεξηγηση αρκετα ευκολη νομιζω....... :wink:

Θα ηθελα να δω την λυση σου Σιλουαν για την ασκηση 9. :wink: