Let $ A,B, C \in \mathcal{M}_3{(\mathbb{R})}$ and $ \varepsilon \in \mathbb{C} \minus{} \mathbb{R}$ with $ \varepsilon^3 \equal{} 1$. If $ \det(A \plus{} \varepsilon B) \equal{} \det(A \plus{} \varepsilon C)$, prove that:

$ \det(A \plus{} B) \minus{} \det(A \plus{} C) \equal{} 3 \det(B \plus{} C)$.

Take $ A=0$, $ B=\left[\begin {array} {LLL} 1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \end {array} \right]$ and $ C=\left[\begin {array} {LLL} 0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end {array} \right]$,
then $ det(A+xB)=0$ and $ det(A+xC)=0$, but $ det(A+B)-det(A+C)=0$ is not equal to $ 3.det(B+C)=3$.
Sorry if I am wrong.


we know $ 3^3\equal{}27$, so let's simplify the left side of the expression

$ 27\plus{}27\plus{}27\equal{}243\cdot3^k$
$ 81\equal{}243\cdot3^k$

now we divide each side by 243

$ \frac{1}{3}\equal{}3^k$

thus we see that $ k\equal{}\boxed{\minus{}1}$

you could also solve this by making everything a power of three and proceed from there, but you can do that on your own


trigonometry
inequalities
geometry proposed
geometry

Prove that

$ \frac{1}{\cos^2 \frac{A}{2}}\plus{}\frac{1}{\cos^2 \frac{B}{2}}\plus{}\frac{1}{\cos^2 \frac{C}{2}}\geq4$

The inequality is equivalent to : $ \sum\frac{1}{\frac{1\plus{}\cos A}{2}}\ge 4\ \Longleftrightarrow\ \sum\frac{1}{1\plus{}\cos A}\ge 2$ .

By the [u][b]Cauchy-Schwarz[/b] inequality[/u] we get $ \sum\frac{1}{1\plus{}\cos A}\ge\frac{(1\plus{}1\plus{}1)^2}{3\plus{}\sum\cos A}\ge \frac{9}{3\plus{}\frac{3}{2}}\equal{}2$ , because

$ \cos A\plus{}\cos B\plus{}\cos C\le\frac{3}{2}$ .

Prove that $ \boxed {\ \frac {1}{\cos^2 \frac {A}{2}}\ \plus{} \ \frac {1}{\cos^2 \frac {B}{2}}\ \plus{} \ \frac {1}{\cos^2 \frac {C}{2}}\ \ge\ 3\ \plus{} \ \left(\frac {4R \plus{} r}{s\sqrt 3}\right)^2\ }\ \ge\ 4$ .

[hide="Proof."]
$ \left\|\begin{array}{c} \tan\frac A2 \equal{} \frac {r_a}{s}\ \mathrm{a.s.o.} \\
 \\
\sum r_a \equal{} 4R \plus{} r\end{array}\right\|\ \implies$ $ \sum\frac {1}{\cos^2\frac A2} \equal{} \sum\left(1 \plus{} \tan^2\frac A2\right) \equal{}$ $ 3 \plus{} \sum\tan^2\frac A2\ \ge$

$ 3 \plus{} \frac 13\cdot\left(\sum \tan\frac A2\right)^2 \equal{} 3 \plus{} \frac 13\cdot\left(\sum \frac {r_a}{s}\right)^2 \equal{} 3 \plus{} \left(\frac {4R \plus{} r}{s\sqrt 3}\right)^2\ge 4$ since $ s\sqrt 3\le 4R\plus{}r$ .[/hide]


Let $ k$ and $ n$ be two positive integers with $ k<n$, and let $ a_1,a_2,...,a_n$ be reals s.t. $ a_1\leq a_2\leq...\leq a_n$.Prove that:
$ (a_1\plus{}a_2\plus{}...\plus{}a_n)^2\geq n(a_1a_{k\plus{}1}\plus{}a_2a_{k\plus{}2}\plus{}...a_na_{k\plus{}n})^2$
for the following two cases:
$ (a)$ $ n\equal{}2k$;
$ (b)$ $ n\equal{}4k$.

No one can solve?   :(  :?:

[hide="to arqady"]arqady, please post your's or Vasc's solution.  [/hide]


1. Evaluate the value when x = 10.5 in:

$[x] + [x + \frac {1}{x}]+ [x + \frac {1}{\frac {1}{x}}]$

where $[x]$ denotes the greatest integer less than or equal to $x$.

2. We say that the real number is rational if it can be expressed as fraction. For example:

$0.3333333.... = \frac {1}{3}$

Find $m+n$ if $0.123123123... = \frac {m}{n}$.

3. Suppose that Jane leaves the town at 5:00 so that she can go to the mall. After 9000 seconds, she came back home to rest. If her return time looks like $\text {a : bc}$ in digital clock, where $a,b,c$ may or may not equal to zero, find $a+b-c$.

This digit clock only expresses time from 1 to 12. So, 7:00 in morning and night are expressed exactly same. Not like 7:00 and 19:00.

4. $C$ numbers of people came to the conference. Each person in the conference shook hands with other people but exactly once. If there was 55 handshakes, how many people came to the conference? In another word, what does $C$ equal?

5. What is the region enclosed by two axes and the two equations $x = 4$ and $y = 4$.

By two axes, we mean the two original x-axis and y-axis you draw on Cartesian Plane.

6. Six coins are flipped. Each coin is flipped at different time. After flipping five coins, the result was HTHT. If the probability that the next two coin flips result continues the pattern is $\frac {x}{y}$, find $x+y$.

7. $\sum^{20}_{c=1} c = \sum^{10}_{o=1} o + \sum^{5}_{p=1} p + K$.

Find the value of K.

8. $a_n = a_{n-2} \cdot a_{n-1}$. If $a_1 = 5$ and $a_2 = 2$, for what value of $z$ that $a_z$ does it have value of $2^p 5^p$ where $p$ is integer greater than 1.

Explain why there is only one valid solution for this problem thoroughly. Your point is going to be graded mostly on explanation and correct answer for this one ESPECIALLY.

9. Roy begans to number the pages from 1 through $n$. If he had added the pages he numbered right, the result would've been 7260. But he accidentally skipped to number $n$ and numbered 1 twice instead. What was his mistaken result?

10. Celebrating the COP Third, Silverfalcon made the feast to everyone who participated during the COP. If he paid 5 dollars to ones who participated in COP at least once and 8 dollars to ones who participated in both COPs and that there were 50 people on the feast, find the number of dollars he spent if number of people who received 5 dollars and number of people received 8 dollars wer equal.

[b]Answers are due 2/26/05[/b]

Rule: You must solve the problems with all the sources you can find. Using sources is optional. You may use calculators but help from other people or online tutor is perfectly disallowed. For every problem you solved, you must send the explanations telling how you solved each of them with correct answer. If your message only contains the answers but not solutions, it would be disqualified.

Qualification: If you get 7 or more with correct answer AND right explanation, you'll be qualified to the COP Round 2.

do we post our answers here?

or

give them to you privately? :?

Private messaging to me with fully-written solutions.

For 2, let's say x=m+n is the answer. Then 2x,3x,4x,5x,6x,.... are all answers. I think you want gcf(m,n)=1.

Same for number 6

Also, for number six, you say that after five coins are flipped the results are HTHT. Does that mean we don't know what the fifth coins is? Because then we don't know the pattern.

For 2, yes, m, n in lowest terms.

For 6, you don't need that expression. ;)

For number 10, can I reword as follows?

Celebratnig the Third COP, Silverfalcon invited everyone who had participated in a past COP to a feast. He gave 5 dollars to everyone who had participated in one past COP, and 8 dollars to people who had participated in both COPs. He invited 50 people. How many dollars did he give away if the number of people recieved 5 dollars was the same number of people who recieved 8 dollars.

Not that before people who had participated in both COPS may have gotten 13 or 8 dollars, depending on how it was read.

[quote]9. Roy begans to number the pages from 1 through n. If he had added the pages he numbered right, the result would've been 7260. But he accidentally skipped to number n and numbered 1 twice instead. What was his mistaken result? [/quote]

do you mean *skipped number n* or *skipped [size=150]to[/size] number n*?

that has to be stated correctly,

who solved this problem already?

I asked him about it, and it means that no page is numbered "n". However two pages are numbered "1"

[quote]
9. Roy begans to number the pages from 1 through $n$. If he had added the pages he numbered right, the result would've been 7260. But he accidentally skipped to number $n$ and numbered 1 twice instead. What was his mistaken result?[/quote]

When you asked " what was his mistaken result", do you mean what the sum was after he made the mistake?

And also, i need some more clarification on number 5. How are you sopposed to find the area between axis?

so, are you saying that everything is numbered right but the problem is, Roy forgot to count the last number and counted the page number 1 twice...

right?

 :(  :?

don't have much time left to solve all the problems .. ahh!!!

This is what you need to find:

[quote]the sum was after he made the mistake? [/quote]

For #5, it's not hard to do it. First, make your y-axis and x-axis on your paper. Then, graph those two equations. What kind of shape do you get?

That should be an easy question. :)

shinwoo, yes, you're right. $n$ is the last number. So, he didn't count $n$ but instead he counted 1 twice.

The point of that problem is to find what $n$ is equal to. Then, that should be easy. :D

[quote="Silverfalcon"]8. $a_n = a_{n-2} \cdot a_{n-1}$. If $a_1 = 5$ and $a_2 = 2$, for what value of $z$ that $a_z$ does it have value of $2^p 5^p$ where $p$ is integer greater than 1.

Explain why there is only one valid solution for this problem thoroughly. Your point is going to be graded mostly on explanation and correct answer for this one ESPECIALLY.[/quote]

Did you mean an integer greater or equal to one? Because if you didnt, then i see no way of solving this problem.

[quote="Farazy"][quote="Silverfalcon"]6. Six coins are flipped. Each coin is flipped at different time. After flipping [b][i][u]five[/u][/i] coins, the result was HTHT[/b]. If the probability that the next two coin flips result continues the pattern is $\frac {x}{y}$, find $x+y$.[/quote]

You only gave 4 coin tosses... :lol:[/quote]

LOL... That's right, HTHT is not 5...  :P

Oops. Mistake on counting. :D  :D

I guess I should go back to the classroom to count again! :D  :D

Although it's higher than my counting :P

[quote="GoBraves"][quote="Silverfalcon"]8. $a_n = a_{n-2} \cdot a_{n-1}$. If $a_1 = 5$ and $a_2 = 2$, for what value of $z$ that $a_z$ does it have value of $2^p 5^p$ where $p$ is [b]integer greater or equal to 1.[/b]

Explain why there is only one valid solution for this problem thoroughly. Your point is going to be graded mostly on explanation and correct answer for this one ESPECIALLY.[/quote]

Did you mean an integer greater or equal to one? Because if you didnt, then i see no way of solving this problem.[/quote]

Does this fixed?

Sorry, I was busy when I was making these. :(

One more thing, when will we find out our results?

[quote="Silverfalcon"]ASAP.

The thing is that I have to write second round of COP + Mock AIME + grading both rounds of COP (COP Round 1 + COSPD) + school work + other things :D .

I'll try to get them done ASAP.[/quote]

oh wow, thats going to take a long time.

Im wondering, if you do qualify, does your first round score affect your total score?

Nope.

The first round or COPSD doesn't matter your score except on the qualification to round 2.

Now, for Round 2, the score you get really matters because you will be either eliminated or survived. :)

[quote="Silverfalcon"]One COP - 8 dollar
Two COP - 13 dollar

Yes. You may translate that way. :)[/quote]

That's not how I translated it. What happened to people participating in only one COP getting 5 dollars?

Wait.

Yeah, that's wrong translation. But keep it that way.

The reason is that already several people send me the answers using that method and I made them qualified.

[quote="Silverfalcon"]Because of same reason as COP, the new deadline for the COPSD is next week.

See the COP thread for exact date.

By the way, there are three qualifiers now. :)[/quote]

Haha... And i was afraid i wouldn't get them finished in time...  :blush:

Oh well.  :P

Please note that Friday is last day to turn the answers in.

:o oooooo.....

well, I had a dream about COPSD round 2 ...

It was really hard....

I don't know how my dream made up those problems and dont really remember what they were, but I just knew that it was hard...

Was the answer to number 8 1?

Because it seems there is no other possible answer...

wait a minute.....
we still have time to solve thse problems?
great...cause i totally spaced out about COPSD....i mean my plan was to do it on Friday real quick but stuff came up  :P

[quote="shinwoo"]I don't think we are supposed to discuss problems yet...

there are still people taking this round

and I didn't even solve that one.. :D ( that's the one i missed )[/quote]

I thought round one was over on Friday.

[quote]I thought round one was over on Friday.[/quote]

I thought so, too...but Silverfalcon extended the due date for the MC students....

MC people are really busy preparing for the chapter round or something...

[quote]Um.

Next Friday Wink[/quote]

^
|
Silverfalcon..


[quote="MCrawford"]On Wednesday, April 13, AoPS will host a Math Jam that provides students with a unique opportunity to learn about the Siemens Westinghouse Competition for students of mathematics and science.  Janet DeCarlo of the College Board will host the Math Jam along with scholarship winner Po-Ling Loh, a student from Wisconsin whose project analyzed a new mathematics problem about closed embeddings of finite groups, will give interested math and science scholars their perspectives on the program and will answer questions for students, parents, and teachers.

The Math Jam will begin at 5:30 PM ET (2:30 PM PT).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31876]Learn more about this Math Jam.[/url]

You can learn more about all of the Math Jams at Art of Problem Solving by clicking on the Math Jams tab at the top of this page.

If you have any questions about this or any other Math Jam you can email us at classes@artofproblemsolving.com.

In order to attend a Math Jam you must first log on the Art of Problem Solving Forum at http://www.artofproblemsolving.com/Forum/index.php, then click the Classroom button on the left panel of the site up to 15 minutes before the Math Jam begins.  The Virtual Classroom should then open automatically.[/quote]

If you were intending to ask a question, you didn't get the text in with the quote.

[quote="xxreddevilzxx"]Would we miss a lot if we cant get in until 4PM PT?[/quote]

If you get to the Math Jam late, you can still ask your questions and read the transcript for other details.  However, your question might not get answered if it has already been answered in which case you'll have to read the transcript.

ive been looking for the "join jam" button since ten minutes before it was supposed to start but the whole thing just disappeared...

I can't get in either. The virtual classroom isn't loading.

Be patient.  I'm trying to fix the problem.  Try in a few minutes.  It hasn't started yet.

My mother really wanted to do it, and it showed up in the classroom-select page 15 minutes before, but nothing happened when we clicked on the radio button, and now it has just dissappeared! :(  :?

Mother is quite unhappy!

Billy

The classroom should be fixed now.  You can join.

Thank you!!! My mother was about to drag us all out shopping! ;)  ;)

You might need to open a fresh browser after clearing your cookies.


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

Inspired by Tare's MATHCOUNTS poll...[url]http://www.artofproblemsolving.com/wforum/viewtopic.php?t=1772[/url]

What do you think will be the highest level of competion you get to this coming year?

__

Explanation:
AMC = American Mathematics Competition.  Two levels (10th grade and 12th grade), two separate dates ("A" date and "B" date).  Multiple choice. 75 min.

AIME = American Invitational Mathematics Exam.  Students who were in the top 1% on the AMC10 or were in the top 5% on the AMC12 or earned 100 points on the AMC12 are invited to take this exam.  Answers are of the form XXX (0-999).  2 hr.

USAMO = United States of America Math Olympiad.  An olympiad-style (containing proofs).  6 problems, 9 hours, spread over two days.

MOP (or MOSP) = Math Olympiad (Summer) Program.  The top scorers on the USAMO are invited here to train before the IMO.

IMO = International Math Olympiad.  The top 6 out of MOP, plus one alternate (sometimes 2), travel to a central event and compete against teams from other countries.  IMO 2004 is in Athens, Greece.

realistically, aime, but if i worked really hard im sure usamo isnt that far from my grasps or anyone elses for that matter

i think i'll make USAMO, if not this year, then next year... but i'd like to go this year as well :P

I missed MOP by 2 points last year! Why wouldn't I get in this year?

once you qualify for aime through amc 10, it doesnt matter what that score was (basically). all you need to do is match the lowest aime score of a 10+ grader which is usually 8ish

I made AIME last year and was pretty much lost on it. 104.5 on AMC 12, 2 on AIME. I'd like to do better. So a couple questions:
1) How well do you have to do on AIME to take USAMO?
2) What are your best suggestions on things to learn for AIME?
3) What should I read other than AoPS 2 (I'm sure there are other suggestions)?
4) Other?

i said this before, but people here make it sound like it's really easy to make USAMO; i dont think it is.  You need around a 220 index score, so if you get a 104.5 on AMC, you'd need ~12 on the AIME.

this has all been said before by others, but ill say it again. cats, if you are in 11th/12th grades (like confuted), you need an index of around 220 (226 last year, in the 210s the years before i think.) but if you are in 10th or below, regardless of your amc10/12 score, you make it if you have an aime score greater than or equal to the floor aime score , that is , out of all those who qualified with the index of 220 or more, the lowest aime score. last year, the base was 8. that means 8 correct was an usamo qualifying score for students 10th grade or below. i dont think that making usamo is easy ,in fact i think its quite a challenge, but with last years aime, i thought 8 was an attainable score. i got 6, and i believe that with a little more practice, concentration, 8 was doable (consider that i missed one of the easier problem simply on the basis of misreading a parantheses.) of course, a lot of improvement takes place in the course of a year! i think many people on this site had a result similar or better than mine and think that they can get a score necessary for qualification in usamo. this is especially true for younger students. i think this is why a lot of people claim they could make usamo this year, not that they think the aime is a very easy test.

as for materials, past aimes are always good practice!

[quote="TheSouthpaw"]8ish? Really? I was more under the impression of 5 or 6.

I'm pretty sure I'll make USAMO. That's my goal, at least.[/quote]

yea, i believe the cutoff for last year's 10th and below was exactly 8 on AIME...

i got 140-sumthing on AMC 10 last year, but only a 5 on AIME...*sigh* getting a 8 is REALLY hard...

i think the base aime was 8 this year, but the previous years its been 6 ( i know for sure in 2001 and 2002 it was 6). as for getting an 8, my opinion is that its hard, but quite achievable.

So if someone else screws up on their AIME there'll be more chances of us making it to USAMO, oh great Neptune what an irony...

I'm pretty sure others have asked this before but is there any way to get past AIME problems for free?

[quote="Tare"]So if someone else screws up on their AIME there'll be more chances of us making it to USAMO, oh great Neptune what an irony...

[/quote]

if they mess up, do you really think they would make usamo? it would have to be the majority that screw up. and the base for the below 10th graders is based on the 10+ graders

I mean like have a really bad day and didn't get enough sleep or something like that.

... :cry: I just realized I won't be able to make USAMO anyways since I'm not "American citizens or permanent residents in other words, those seeking citizenship and currently possessing a U.S.A. Immigration 'green card'"...oh poo, so all this talk was like it didn't even matter...

Thus I suggest someone add "impossible" as one of the options for people living in countries other than the country they have their citizenship in.

[hide]Do you think they'll do a background check?[/hide]

same here... although I am "honoured" to unofficially write the USAMO as IMO selection in Canada... The guys in US just doesnt recognize it because I'm not "American citizens or permanent residents in other words, those seeking citizenship and currently possessing a U.S.A. Immigration 'green card'"

Is writing USAMO necessary for IMO selection in Canada? If so, I'm screwed over by my school's lame excuse for an administration.

why would you need to write the AMERICAN Math olympiad to make the CANADIAN IMO team? that would be absurd, but these days....

Tare, I have a friend who took the USAMO in 2003 (officially) even though he isn't a US citizen or permanent resident. Maybe the AMC people made a mistake, or maybe they just don't follow the stated rules.

[quote="Tare"]Is MOP considered a competition? And what should i make of that Canadian-American Math Camp thing Osiris went to?

I might make AIME if I'm lucky, but since I'll be qualifing through AMC10 my chances of making USAMO is from 0 to none :lol:[/quote]

MOP isn't a competition, but it's a level you can reach.  Just like Nirvana.

As for Mathcamp, that's completely different.  With Mathcamp, you submit an application form, try your hand at a few math questions, and pay some money (unless you're recieving financial aid), and if you're admitted you go.  With MOP, you don't have to pay anything, but you have to do well on USAMO to get in.  Thus Mathcamp isn't really a level (though it helps) because the emphasis is less on [i]earning[/i] the right to get in, and also because it isn't officially affiliated with AMC.

Oh, and with regards to quallifying for USAMO via AMC10, it can be done.  You have several options for getting in (see "USMAO [sic]" in the AMC forum), but getting above about an 8 on the AIME is only one of the ways you can get in...in fact, there are [i]more[/i] ways that you can qualify by taking the AMC10.  I wouldn't say it's as easy as it will be when you're only taking the '12 (because then you'll be older and more experienced; everyone will usually be better two years from now) - but it's possible.

mathfanatic wrote:Oh, and with regards to quallifying for USAMO via AMC10, it can be done.  You have several options for getting in (see "USMAO [sic]" in the AMC forum), but getting above about an 8 on the AIME is only one of the ways you can get in...in fact, there are more ways that you can qualify by taking the AMC10.  I wouldn't say it's as easy as it will be when you're only taking the '12 (because then you'll be older and more experienced; everyone will usually be better two years from now) - but it's possible.

If that is the case I'll be taking the AMC 10 two years from now :twisted:

...no really, I'm pretty sure if I just write what grade I'm in then I'd qualify, just like MAML Olympiad.

Simon did he state that he would like to have citizenship in the US? All he has to say is "I'd like to live here for the rest of my life" and your through, according to the rules. I don't plan on making that statement however because my Visa's going to run out permenantly within the next few years and I have no plan of staying here anymore, so I'm screwed...

I think you (ml2) missed one important word in bugzpodder's post, see if you can find it yourself

Answer: [hide]"Unofficially"[/hide]

Although you do have a point ml2. American's don't recognise Canada as a country just like most people think Singapore isn't one either. "It's a city for H*****'s sake!" copyrighted 1998 by my 5th grade teacher

Anyways Canada-US is the longest borderline between two contries that is the least guarded...Canada's practically our unofficial 51st state, at least that's what my SS teacher (current) said according to the US-UK history, Canada almost became part of the US! Wow...*views South Park* what an irony!  </jkjk>

Mmm, starting to sound like a SAT question: "MOP is to Nirvana as IMO is to..."

tare, that statement for "Permanent Citizen" states that you have "Green Card AND will apply to become a citizen." wanting to be a citizen doesnt count. consider this: lets say that i am a french citizen and im living in paris. i can take amc12 and aime, then ill claim i want to live in the US. do i get to take the usamo? i dont think so.

[quote="Tare"]Although you do have a point ml2. American's don't recognise Canada as a country just like most people think Singapore isn't one either. "It's a city for H*****'s sake!" copyrighted 1998 by my 5th grade teacher :lol:
Anyways Canada-US is the longest borderline between two contries that is the least guarded...Canada's practically our unofficial 51st state, at least that's what my SS teacher (current) said according to the US-UK history, Canada almost became part of the US! [/quote]

I think rippledance will have something to say about this...

But seriously, I think most people know that Canada is a country.  If you're interviewing 3rd or 4th graders, maybe they don't know what Canada is, but for most educated Americans?  I think at least that - if little else - is known.

[quote="mathfanatic"]But seriously,[/quote]
No buts mister :lol:

I believe rippledance already heard of this, because I said it before too. It's a famous story.
...she might even live near her, what an irony :twisted:

Anyways yes probably a statement isn't enough but still you have to take some actions to show you're willing to stay, an action I will refuse, so oh well.

[quote="MysticTerminator"]uh...tare...are you feeling ok?[/quote]
Nope, as usual :lol:


[quote="konlsy"]If a,b,c>0, prove that:
\[{ \frac{1}{a\sqrt(a+b)}+\frac{1}{b\sqrt(b+c)}+\frac{1}{c\sqrt(c+a)})} \geq \frac{3}{\sqrt(2abc)}  \]

thank u for ur help :oops:[/quote]
First solution: Let
$a=\frac{1}{x},...$
${\frac{1}{a\sqrt{a+b}}+\frac{1}{b\sqrt{b+c}}+\frac{1}{c\sqrt{c+a}})} \geq \frac{3}{\sqrt(2abc)}\Leftrightarrow\frac{x}{\sqrt{xz+zy}}+\frac{y}{\sqrt{xz+xy}}+\frac{z}{\sqrt{xz+yz}}\ge\frac{3}{\sqrt{2}}$
But we have
$\sqrt{xz+zy}\le\frac{1}{2\sqrt{2}}(2z+x+y)$
$\Rightarrow\sum\frac{x}{\sqrt{xz+zy}}\ge\sum\frac{2\sqrt{2}x}{2z+x+y}\ge\frac{2\sqrt{2}(x+y+z)^2}{\sum x^2+3\sum yz}\ge\frac{3}{2\sqrt{2}}$
Second solution:
Let $x=\sqrt{\frac{2bc}{a(a+b)}},..$
We have to prove $x+y+z\ge3$
But we have $xy+yz+zx=\frac{2a}{\sqrt{(b+c)(c+a)}}+\frac{2b}{\sqrt{(a+b)(c+a)}}+\frac{2c}{\sqrt{(a+b)(b+c)}} \ge\sum\frac{4a}{a+b+2c}\ge 4\frac{(a+b+c)^2}{\sum a^2+3\sum yz}\ge3$

u are so quickly that the typing word has some mistake. but i doesn't matter.
 :P  thank u

[quote=sqing][quote=konlsy]If a,b,c>0, prove that:
\[ \frac{1}{a\sqrt{(a+b)}}+\frac{1}{b\sqrt{(b+c)}}+\frac{1}{c\sqrt{(c+a)}} \geq \frac{3}{\sqrt{2abc}}  \]

thank u for ur help :oops:[/quote]
[url=http://artofproblemsolving.com/community/c261h59015p359641]Vietnamese Mathematics e - Olympiad (VMEO)[/url]
[b]TST VN 2000[/b][/quote]
Don't Viet Nam TST 2000.


a)Find all $k\ge 0$ such that $\forall x,y,z\ge 0$we have
$\sum(\sqrt{x^2+kyz}-\sqrt{y^2+kzx})^2\le \sum(x-y)^2$ :)  :)
b)Find all $k\ge 0$ such that $\forall x,y,z\ge 0$we have :
$\sum|\sqrt{x^2+kyz}-\sqrt{y^2+kzx}|\le \sum|x-y|$ :)  :)

[quote="hungkhtn"]I see that you are very interesting with $a^2+kbc$  :gathering:[/quote]
Because it's one of the beautiful things.  :ddr:

[quote="Theseus of Atens"][quote="nthd"]
Because it's one of the beautiful things.  :ddr:[/quote]
I am interesting with your answer ,nthd.
I think$0\leq k \le 4$ :spider:[/quote]
Oh,yes. Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80134]here[/url] to view a special case. :)


Let $ u \equal{} \left (p \plus{} \sqrt {r} \right)^{\frac {1}{3}} \plus{} \left (q \minus{} \sqrt {r} \right)^{\frac {1}{3}}$, Where $ p,q,r$ are integers and $ r \ge 0$ is not a perfect square. If $ u$ is rational and $ u \not \equal{} 0$ , prove that $ p \equal{} q$ and $ u$ is integral.

Hello Tiger100.
This is my first time typing in LaTeX. I hope it works!
I have tackled your problem this way:

Let $ x\equal{}\left(p\plus{}\sqrt{r}\right)^\frac{1}{3}$ and $ y\equal{}\left(q\minus{}\sqrt{r}\right)^\frac{1}{3}$.
Then, $ u^3\equal{}(x^3\plus{}y^3)\plus{}3xy(x\plus{}y)\equal{}(p\plus{}q)\plus{}3xyu$,
or, $ u(u^2\minus{}3xy)\equal{}p\plus{}q \Rightarrow$ $ xy$ is rational.
But $ xy\equal{}(pq\minus{}p\sqrt{r}\plus{}q\sqrt{r}\minus{}r)^\frac{1}{3}$. Hence, $ p\equal{}q$, and $ xy\equal{}(p^2\minus{}r)^\frac{1}{3} \Rightarrow xy \in \textbf{Z}$.

Let $ xy\equal{}k$ and $ u\equal{}\frac{a}{b}$ where $ k,a,b \in \textbf{Z}$, and $ gcd(a,b)\equal{}1$.

Then, $ \frac{a}{b}\left(\frac{a^2}{b^2}\minus{}k\right)\equal{}p\plus{}q$,
or, $ \frac{a(a^2\minus{}b^2k)}{b^3}\equal{}p\plus{}q \in \textbf{Z}$.
However, $ gcd(a(a^2\minus{}b^2k,b)\equal{}1 \Rightarrow b \equal{} 1$.

Q.E.D.

By the way, how do you type the set real, integer, etc. in the conventional way? I just typed them in bold.  :P


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 10 A
algorithm

Let's say somebody has pretty good grades in high school and does well in math class but had no other experience with problem solving outside of the regular high school curriculum (didn't take any AMC tests, etc.).  I know it may be hard to give an exact answer, but about how difficult do you think it would be to do well enough on the AMC 12 to qualify for the next AIME, or possibly even the USAMO, after studying and doing practice problems up until then (and when I say "somebody," although I am somewhat similar to the description I gave, I'm talking about anybody in general, so if somebody else were to ask me this question, for example)?

I'm sorry if this seems like a frequently asked question, but I didn't see any other threads quite like this that answered this exact question (I just saw threads like "how can I improve my score").

It depends on how rigorous the school's math program is. It depends a lot more on how much time you have to prepare.

The reason I say that is because I was in the same situation last year as a ninth grader. I had good grades in school and I took the AMC 10A and studied in the two weeks for the 10B. I got a 93 on the A and a 108 on the B, so that might give you some idea.

I wouldn't rely on the school curriculum for math because the classes are meant to educate the [b]average or below average[/b] student.
As I am only a 6th grader (:maybe: maybe I shouldn't be posting in this forum), I probably don't have that much experience with these things.
To do decently on the AMC 10 or 12, you have to do a [b]lot[/b] of training in higher levels of math. Even when you do that, it doesn't always work. I studied for several months (about 4), yet I could only score a 112.5 on AMC10A, and 111 on AMC10B.
This should give you the idea that AMC and AIME (especially USAMO) are [b]extremely[/b] hard.

That depends a lot on how much time you have and possibly more on natural ability.  AMC math is very different - not just harder - from most of the mathematics you're likely to have encountered in high school.  To do well in high school math is largely a matter of correctly remembering and applying a small set of algorithms, but to do well in AMC math requires learning [b]heuristics[/b] (of the problem-solving variety), which are much trickier to master and require a lot more practice and patience.  They are also, of course, more versatile, more gratifying to understand, and much more fun.

in sixth grade, i was a cocky little jerk i was, by far, the smartest kid in my school, and i thought i was just great. i got a 22/5 at MC, and got a 96 on AMC 10. this year, after realizing that i'm not god's gift to mankind, i got a 29/16 at MC chapters, a 28/18 at MC states, and a 144 AMC 10 and 121.5 AMC 12.

yay. as roflolcopt said, schools teach to the BOTTOM of the class. an average high school student in AP Calc would get a ... zero on the AIME.

Let me give an example of what I'm talking about:  proving trigonometric identities.  The sine and cosine functions are completely characterized by

- the Pythagorean theorem and
- the angle addition formulas,

in a certain precise sense.  You don't need [b]any[/b] other identities to prove, in principle, every conceivable trigonometric identity.  Of course, in high school you usually learn a larger list of identities because some can be quite convenient to apply in special cases and it is helpful to recognize certain forms.  Nevertheless, in principle there is an algorithmic way to do this, although often the neatest solution is sufficiently non-obvious that there are plenty of students who consider this difficult.

What does an AMC student think of trigonometric identities?  What an AMC student knows is that the complex exponential and the relations

$ \cos x \equal{} \frac {e^{ix} \plus{} e^{ \minus{} ix}}{2}$
$ \sin x \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$

reduce every trigonometric identity to a [b]polynomial[/b] identity in $ q \equal{} e^{ix}$, and polynomial identies are trivial to prove algorithmically.  There is a second way to reduce trigonometric identities to polynomial identities based on the tangent half-angle formulae

$ \cos 2x \equal{} \frac {1 \minus{} \tan^2 x}{1 \plus{} \tan^2 x}$
$ \sin 2x \equal{} \frac {2 \tan x}{1 \plus{} \tan^2 x}$

which reduces every trigonometric identity to a polynomial identity in $ t \equal{} \tan x$.  In other words, proving trigonometric identities reduce to a [b]calculation[/b], not a [b]problem[/b], for an AMC student.  A trigonometry problem on the AMC is either not going to rely on identities at all or is going to use them as one step in a much more interesting sequence of deductions.  This is the kind of change of viewpoint you will have to get used to.

Edit:  Perhaps an even better example here is [b]integrating[/b] trigonometric functions.  Every trigonometric integral can be rewritten as the integral of a rational function, and there is an algorithmic way to integrate rational functions.


Find the prime p if $ 7p+1=n^{2}$ where n is a positive integer.

[hide]$ 7p+1=n^{2}$

$ 7p=n^{2}-1$

$ 7p=(n+1)(n-1)$

$ p$ can be either $ 5$ or $ 9$, but since it has to be a prime, $ p=5$.[/hide]

How did you determine the possible answers as $ 5$ and $ 9$?

The thing on the right is a product of two factors. So is the expression on the left.

So we have two cases:
7 = (n + 1) OR 7 = (n - 1)
(p in this case would be the other factor.)
So from this, we can deduce what the other factor is.

N is either 6 OR 8.
if N is 6, then (n-1) (the other factor in the first case) = 5.
if N is 8, then (n+1) (the other factor in the second case) = 9.

9 isn't prime.
Therefore 5 is the solution.


trigonometry
quadratics
algebra
quadratic formula

solve for $ \theta$

$ 2cos^2\theta \plus{} 3sin\theta \minus{} 1 \equal{} 0$

[hide="Solution"]The expression is equivalent to $ 2(1 \minus{} \sin^2 \theta) \plus{} 3 \sin \theta \minus{} 1 \equal{} 2 \minus{} 2\sin^2 \theta \plus{} 3 \sin \theta \minus{} 1 \equal{} 1 \plus{} 3\sin \theta \minus{} 2\sin^2 \theta \equal{} 0$

$ 2\sin^2 \theta \minus{} 3 \sin \theta \minus{} 1 \equal{} 0$. Use the quadratic formula; $ \sin \theta \equal{} \displaystyle\frac{3 \pm \sqrt{17}}{4}$. The plus solution exceeds $ 1$, but the minus solution is valid.

So $ \theta \equal{} \sin^{\minus{}1} \displaystyle\frac{3 \minus{} \sqrt{17}}{4}$.

I don't think there's a good formula for that.

[/hide]

hello, i have got
$ \theta\equal{}2\arctan\left(\frac{3}{2}\plus{}\frac{\sqrt{17}}{2}\minus{}\sqrt{\frac{1}{2}(11\plus{}3\sqrt{17})}\right)\plus{}2\pi k_2$
$ \theta\equal{}2\arctan\left(\frac{3}{2}\plus{}\frac{\sqrt{17}}{2}\plus{}\sqrt{\frac{1}{2}(11\plus{}3\sqrt{17})}\right)\plus{}2\pi k_2$
Sonnhard.


Prove that for positive reals following inequality holds:

$ \sqrt[3]{ab}\plus{}\sqrt[3]{cd}\leq \sqrt[3]{(a\plus{}c\plus{}b)(a\plus{}c\plus{}d)}$

[quote="delegat"]Prove that for positive reals following inequality holds:

$ \sqrt [3]{ab} \plus{} \sqrt [3]{cd}\leq \sqrt [3]{(a \plus{} c \plus{} b)(a \plus{} c \plus{} d)}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3296

It can be done with Holder:
\[ \left( \sqrt[3]{ab}\plus{}\sqrt[3]{cd}\right)^3 \leq \left( \frac{a}{a\plus{}c}\plus{}\frac{c}{a\plus{}c}\right) ((a\plus{}c)\plus{}d)(b\plus{}(a\plus{}c))\]

Beautiful and very simple solution [b]scorpius119[/b]  Thanks a lot for link [b]arqady[/b]


trigonometry
LaTeX
geometry proposed
geometry

Given $\triangle{ABC}$, Prove the following equality.

\[ \frac{\sin 2A+\sin 2B-\sin 2C}{\sin 2A+\sin 2B+\sin 2C}=\frac{\tan C}{\tan A+\tan B+\tan C} \]

This problem ia a good exercise for writing in LaTeX $\sin ...,\ \cos ...,\ \tan ...$

$\frac{\sin 2A+\sin 2B-\sin 2C}{\sin 2A+\sin 2B+\sin 2C}=\frac{\sin C\cos (A-B)-\sin C\cos C}{\sin C\cos (A-B)+\sin C\cos C}=$

$\frac{\cos (A-B)+\cos (A+B)}{\cos (A-B)-\cos(A+B)}=\frac{\cos A\cos B}{\sin A\sin B}=\frac{1}{\tan A\tan B}=$

$\frac{\tan C}{\tan A\tan B\tan C}=\frac{\tan C}{\tan A +\tan B +\tan C}.$

Can anyone solve this problem geometrically? :D

I am sorry, you must begin with this !

Let $DEF$ be the orthic triangle, i.e. $D\in BC,\ AD\perp BC,\ldots$ Thus,

$\frac{EF}{\sin 2A}=\frac{FD}{\sin 2B}=\frac{DE}{\sin 2C}=R$. I note $2s'=EF+FD+DE$

and $EF=a',\ FD=b',\ DE=c'$. Therefore, the left part of the identity

is $\frac{s'-c'}{s'}$ and the right part of the identity is $\frac{a'b'}{ab}\ldots$ a.s.o.


We should have a Gunn vs Paly tournament.

does anyone know who's doing it at Paly? I want to meet him sometime.

I haven't heard anything about it, but if the club gets going, I should be able to find out on Sept. 28 (Club Day).

[quote="hunter34"]I want to watch too.[/quote]

lol Texas? We aren't going that far. You'd have to take a plane if you actually wanted to watch. Besides, we aren't that pro.

Flying is too expensive. I will just roll to the Tournament.

That's actually a good idea. I don't know who'd film though

Dude wtf... you forgot to include Saratoga high. And I'm telling you, we would crush you. (I'm only 97% sure though).

uhh I don't know Saratoga High's people... yeah...
iono who's the fastest at Saratoga?

I guess I can take Saratoga as well (since I'm not sure paly's going), but you aren't picking the site if paly is in as well.


calculus
integration
derivative
analytic geometry
real analysis
real analysis unsolved

help....im stuck with this problem???can anyone help me pls...

1. A swimming pool is to be constructed with full capacity of 400m^3. The pool is designed to have a semi circle at one side of a rectangular shape and the diameter of the circle is equal to the length of that side. Find the dimensions of the pool which will have the largest inner area.

2.the region between y=(x)^1/2 and y=x^x ; 0<=x<=1/2, is revolved about the line x=0.Find the volume of the solid which is generated.

3.evaluate lim (1+2x)^1/2-(1-2x)^1/2/x
x-0

4.compute f'(x) and f'(1) when f(x)=integration e^-t^2 dt from x^x to x

5.let f(x)=x^2+4x+5/x+2

find the coordinates of the stationary/critical points of f and use the first derivative test to determine their nature.

thanks a lot

I will give a hint.

For 3: rationalization of numerator.

For 4:$\frac{d}{dx}\int_{g(x)}^{h(x)} f(t)dt=f(h(x))h'(x)-f(g(x))g'(x)$


Problem 1. Let $ a,b,c>0$, prove that  $ \sum\frac {a^2}{b\plus{}c}\ge\frac{\sqrt[3]{9(a^3\plus{}b^3\plus{}c^3)}}2$

Problem 2. Let $ a,b,c$ be sides length of an acute triangle, prove that $ \sum\frac{a^2}b\ge \sum\frac {a^2}{b\plus{}c}$

[quote="akai"]Problem 1. Let $ a,b,c > 0$, prove that  $ \sum\frac {a^2}{b \plus{} c}\ge\frac {\sqrt [3]{9(a^3 \plus{} b^3 \plus{} c^3)}}2$

Problem 2. Let $ a,b,c$ be sides length of an acute triangle, prove that $ \sum\frac {a^2}b\ge \sum\frac {a^2}{b \plus{} c}$[/quote]

$ \sum\frac {a^2}b\ge \sum\frac {a^2}{b \plus{} c}$

$ \sum\frac {a^2}b\ge \frac {(a \plus{} b \plus{} c)^2}{a \plus{} b \plus{} c} \equal{} a \plus{} b \plus{} c$

we just need to prove

$ a \plus{} b \plus{} c\geq \sum\frac {a^2}{b \plus{} c}..............(*)$

substitution $ a \equal{} x\plus{}y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$ ($ x,y,z > 0$)

$ (*)\Longleftrightarrow 2(x \plus{} y \plus{} z)(3x^2y \plus{} 3x^2z \plus{} 14xyz \plus{} 3xy^2 \plus{} 3xz^2 \plus{} 3y^2z \plus{} 3yz^2) \geq 0$

Problem 2 is just so trivial since

\[ \frac{a^2}{b} \ge \frac{a^2}{b\plus{}c}\]

I'm sorry, it's must be
Problem 2. Let $ a,b,c$ be sides length of an acute triangle, prove that $ \sum\frac {a^2}b\ge 2\sum\frac {a^2}{b \plus{} c}$

1st problem can be easily killed with  Jensen for $ f(x)\equal{} \frac{1}{S\minus{}x}$ ($ S\equal{}a\plus{}b\plus{}c$) and weights $ \frac{a^2}{a^2\plus{}b^2\plus{}c^2}$ etc.
then it can be very easily solved with Am-GM


inequalities
number theory unsolved
number theory

[b]Find all primes p,q and even n>2 such that $ p^n\plus{}p^{n\minus{}1}\plus{}...\plus{}1\equal{}q^2\plus{}q\plus{}1$.
[/b]
My idea:

$ p(p^{n\minus{}1}\plus{}.....\plus{}1) \equal{} q(q\plus{}1)$.

Let k be the largest power of p dividing q+1, then
$ p^{n\minus{}1}\plus{}.....\plus{}1 \equal{} qcp^{k\minus{}1}$
If k>1 then the RHS is divisible by p and LHS is not(I am using n>2)
so q+1 = pc for some $ c \in \mathbb{N}$

It looks like I have deduced something, but then what do we do??
Or is this approach useless??

[quote="isomorphism"][b]Find all primes p,q and even n>2 such that $ p^n \plus{} p^{n \minus{} 1} \plus{} ... \plus{} 1 \equal{} q^2 \plus{} q \plus{} 1$.
[/b]
My idea:

$ p(p^{n \minus{} 1} \plus{} ..... \plus{} 1) \equal{} q(q \plus{} 1)$.

Let k be the largest power of p dividing q+1, then
$ p^{n \minus{} 1} \plus{} ..... \plus{} 1 \equal{} qcp^{k \minus{} 1}$
If k>1 then the RHS is divisible by p and LHS is not(I am using n>2)
so q+1 = pc for some $ c \in \mathbb{N}$

It looks like I have deduced something, but then what do we do??
Or is this approach useless??[/quote]
 :(
Anyone??

I will be terse... i think i have done it earlier . Had to use some inequality between p and q.
sorry for this (spam) post..
but on similar line

anyways i'll be back with the solution

[quote="pardesi"][url=http://www.mathlinks.ro/viewtopic.php?p=182765#p182765]here[/url] is the solution[/quote]

Thanks you  :D


If $a,b,c\in R$ prove that     $a(a+b)^5+b(b+c)^5+c(c+a)^5\geq 0$

It took a while until I realised it, but this problem is trivial. :)

Substitute $a+b \rightarrow x, b+c \rightarrow y, c+a \rightarrow z$. We get:
\[ (x+z-y)x^5 + (x+y-z)y^5 + (y+z-x)z^5 \ge 0  \]
$\frac{x^6}2+x^5z-xz^5+\frac{z^6}2 = \frac12 (x^2+z^2) (x^2-xz-z^2)^2 \ge0$
$\frac{y^6}2+yz^5-y^5z+\frac{z^6}2 = \frac12 (y^2+z^2) (z^2-yz-y^2)^2 \ge0$
$\frac{x^6}2+y^5x-x^5y+\frac{y^6}2 = \frac12 (x^2+y^2) (y^2-xy-x^2)^2 \ge0$

[quote="TomciO"]You can't suppose that $z=1$.[/quote]why not tomcio? I think it is perfectly ok to do so. Anyway I removed the part as it was not necessary, just filling in 1=z again in the proof brought back the original problem :)

[hide="(for reference, I wrote: )"]Now if it is true for $(x,y,z)$, it is also true for $(kx,ky,kz)$ with $k\in\mathbb{R}_0$,
--> If $z\not=0$ we can assume wlog that $z=1$. Then the problem transforms into:

$x^5-x^5y+x^6+y^5x-y^5+y^6+y-x+1 \ge 0$[/hide]

Because $(x, y, z)$ implies $(kx, ky, kz)$, not $(x, y, kz)$. Do you get me? I think you can only make assumption which involves all variables...

Well $(x,y,z)$ implies $(z^{-1}x,z^{-1}y,z^{-1}z) = (x/z, y/z, 1)$, right?

And by the way, I wouldn't call it trivial...

[quote="Arne"]And by the way, I wouldn't call it trivial...[/quote]Seen afterwards, it is :) but it indeed took me one hour to get this 'trivial' solution.

[quote="Peter VDD"]but it indeed took me one hour to get this 'trivial' solution.[/quote]

So, there is no reason to call it trivial at all! In fact, it's a tough problem!

$a(a+b)^7+b(b+c)^7+c(c+a)^7\geq0.$ This is not true! ;)


calculus
integration
algebra
polynomial
quadratics
number theory
advanced fields

Prove that 4^(1/3) + 22^(1/2) is irrational.

An algebraic integral number is rational iff it is integer.

As $ \bf{zhaobin}$ said,  the number $ x_0 \equal{} 4^{\frac {1}{3}} \plus{} \sqrt {22}$ is root of the polynomial $ p(x)\equal{}(x^3 \plus{} 66x \minus{} 4)^2 \minus{} 22(3x \plus{} 22)^2$.
If $ x_0$ is rational then it must be integer, but $ x_0$ is not integer, so it is irrational.

Alternately, one has $ 4^{ \frac{1}{3} } \equiv 3 \bmod 23$ but $ \minus{}1$ is not a quadratic residue $ \bmod 23$.

[quote="math beginner"]IOO:

How do you get the polynomial??[/quote]

All that really matters is that a sum of algebraic integers is algebraic (check
any introductory algebraic number theory text). Both $ 4^{1/3}$ and $ 22^{1/2}$
are clearly algebraic integers, so their sum is too.

Or you could just do it t0rajir0u's way, which requires no machinery whatsoever ;)

Both summands lie in independent field extensions of $ \mathbb Q$ (that is, their intersection is only $ \mathbb Q$) by reasons of degree.


What is the sum of the reciprocals of all whole number divisors of 6?

$ 6$ is a [url=http://en.wikipedia.org/wiki/Perfect_number]perfect number[/url], so the answer is $ \boxed{2}$.

Another way is to simply list: $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \frac16 \equal{} \boxed{2}$.

The reason the reciprocal trick works is as follows.

Let the divisors of $ p$ be $ a_1,a_2,a_3,...,a_n$ where $ a_n\equal{}p$

Because $ p$ is perfect, $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}2p$

We are looking for $ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}$

Obviously the LCM of the denominators is $ p$

$ \frac{\frac{p}{a_1}}{p}\plus{}\frac{\frac{p}{a_2}}{p}\plus{}...\plus{}\frac{\frac{p}{a_n}}{p}$

Now here's the tricky part. Let's say our number is 28. If we write out the factors in order, 1,2,4,7,14,28, we see that the numbers on the outside multiply to 28. Same with 2 and 14, and then 4 and 7. Thus, for any perfect number, $ a_1 \cdot a_n \equal{} p$ $ a_2 \cdot a_{n\minus{}1}\equal{} p$ .... $ a_m \cdot a_{n\plus{}1\minus{}m} \equal{} p$

Thus, we can simplify the fractions in the numberator. $ \frac{p}{a_1}\equal{}a_n$ and similarly, $ \frac{p}{a_n}\equal{}a_1$ YAY!!!!

Therefore,

$ \frac{\frac{p}{a_1}}{p}\plus{}\frac{\frac{p}{a_2}}{p}\plus{}...\plus{}\frac{\frac{p}{a_n}}{p}\equal{}\frac{a_n}{p}\plus{}\frac{a_{n\minus{}1}}{p}\plus{}...\plus{}\frac{a_1}{p}$

And what do we know about the sum of all factors of a perfect number? They add up to 2 times the number!

$ \frac{1}{a_1}\plus{}\frac{1}{a_2}\plus{}...\plus{}\frac{1}{a_n}\equal{}\frac{\frac{p}{a_1}}{p}\plus{}\frac{\frac{p}{a_2}}{p}\plus{}...\plus{}\frac{\frac{p}{a_n}}{p}\equal{}\frac{a_n}{p}\plus{}\frac{a_{n\minus{}1}}{p}\plus{}...\plus{}\frac{a_1}{p}\equal{}\frac{2p}{p}\equal{}\boxed{2}$


I'd really like to learn how to spin a pencil and do other stuff like that.  So far, I can just roll it over my fingers or get it to spin between two fingers for like 4 seconds.  I could use tips....  Thanks!  :)

There's some instructions
[youtube]GdVVt_jXJvM[/youtube]
 I still can't really do it, but...

Thanks for your time!  I guess I'll just keep practicing...


conics
geometry
projective geometry
linear algebra
geometry theorems

1. It is well known that every 5 point (in general position) determe some conic.

I would like to see some synthetic prof about it.

2.  Also every 5  lines (in general position) determe some conic (= eksist eksactly one conic wich is tangent to all 5 lines).

Can somebody prove this (if posible synthetic) or give me some reference to find it?

Thanks!

[quote="Number1"]1. It is well known that every 5 point (in general position) determe some conic.

I would like to see some synthetic prof about it.  [/quote]

This question really belongs to projective geometry, and there is a chance for a synthetic proof to exist if you assume enough axioms to make this equivalent to an algebraic problem about fields.  But it is unclear how to even define a projective conic synthetically.

For a synthetic Euclidean (distance) geometry proof  defining a conic synthetically is easy, but the proof must be absurdly difficult unless it is a circumlocution for the algebra.

The linear algebra proof is the simplest.


function
quadratics
LaTeX
geometry
geometric transformation
reflection
algebra

Find the inverse if

y=6x^4 - 11x^2a^4 + a^8.

Source: Something by the way of "approaching precalc." It was an old high school book I think. Have fun!

x=6y^4 - 11y^2a^4 + a^8

Then, just solve for y.

Ok. "Just solve for y is the whole problem. Haha. Try again.

[quote="paladin8"]You could apply the quadratic formula to get something of the form $x^{2}= \text{blah}$ so $x = \sqrt{\text{blah}}$ and then find the interval along which it is a function.[/quote]

You mean $x = \pm \sqrt{\text{blah}}$ ? Since you don't know if they want the positive or negative root.

Can you please type the question into latex because I am confused about the original question.

Is it:

$y= 6x^{4}-11x^{2}a^{4}+a^{8}\\ \Rightarrow x = 6y^{4}-11x^{(2a)^{4}}+a^{8}\ \text{OR}\\ \Rightarrow x = 6y^{4}-11x^{2a^{4}}+a^{8}\ \text{OR}\\ \Rightarrow x = 6y^{4}-11x^{2}a^{4}+a^{8}$

I do not know how to use latex. I will need to look at the links to the left but that is for later. It is the first equation that you have. Sorry for my shortcomings in latex but enjoy the problem!

[quote="mathgeniuse^ln(x)"]The function is an inverse iff it is one-to-one, by the way, so there could be no inverse.[/quote]

In fact, this function has no inverse, since $y(x) = y(-x) \forall x$.

[quote="spoudyal"]t0rajir0u, I do not understand your reasoning. First, because that would mean that no even function has an inverse but we know that y=x^2 has an inverse. It may just be that the domain of the inverse is not all |R.  Nice job lotrgreengrapes7926, all the math looks right but I don't think that you can actually check it with another method (eg. graphically). Before I tried the quadratic equation I had started with completing the square. It got way too messy and I gave in to the quadratic equation. Props!  :first:[/quote]
Thanks. But t0rajir0u is right. No even function has an inverse, and more generally, a function must be one-to-one (pass the horizontal line test) in order to have an inverse. This is because the inverse is required to be a function. If the function doesn't pass the horizontal line test, then its reflection over the line $y=x$ will not pass the vertical line test, and therefore, not be a function.

Ok, so now I understand the reasoning. But if I should not ask for the inverse, since one does not exist, what should I ask for. Just switch the variables first and ask to solve for y? I really did not think that even functions do not have an inverse! Thanks for clarification!  :lol:

Now that I think of it ... That is what paladin8 was talking about. So, an interval (ex: y=x^2 has the inverse sqrt(x) on [0,inf) must be specific for even/non1-1 functions? Thanks again to everyone for clearing this up for me.

[quote="spoudyal"]an interval (ex: y=x^2 has the inverse sqrt(x) on [0,inf) must be specific for even/non1-1 functions?[/quote]

Yes.  That is the essence of the inverse trigonometric functions, for example.


Hola a todos :

Estaba pensando que seria interesante crear una seccion para matematicas universitarias aqui en la comunidad hispanohablante . Y poder debatir topicos interesantes ya sea en los campos de algebra y analisis y porque no olvidar el calculo .

Este seccion seria un punto para postear topicos acerca de la olimpiada iberoamericana universitaria de matematicas .

Bueno no tengo mas nada que decir solo que Espero tener sus opiniones y sugerencias . y que es lo que se decidira en la comunidad .

Estpoy deacuerdo contigo, asi podemos tener algho mas ya que ya no participamos en las olimpiadas universitarias  :(

[quote="carlosbr"]Hola hucht ....
como ya le exprese a DarthTenebrus ...

[quote]
Buenos dias amigo mio....

he leido tu msg hace unos dias ...
me parce interesante tu idea se formar una seccion para la mate Universitaria..
lo hemos consultado los moderadores....
pero por el momento no tenemos la aprobacion de los moderadores delMathLinks para abrir una nueva seccion..
esto debido a que la comunidad en español no cuenta con un gran numero de miembros universitarios....
es por ello, que por ahora una salida seria definir bien el titulo de los mensajes.. talves de esta manera:

NIVEL U: ANALISIS REAL:P1: INTEGRALES DE LEGENDRE
NIVEL U: COMPLEJOS: Pn:XXXXXX.....

asi sabriamos d que tipo de preguntas hablamos...
bueno aunq esta recomendacion seria para todos los mensajes que iremos colgando en la comunidad
esperamos muy pronto tener una respuesta positiva de parte de los administradores...

bueno es todo lo q te puedo comentar por el momento ..
y gracias por participar en la comunidad y tus aportes ..
la comunidad y yo personalmente te agradece el ser parte de esto ... gracias!!

Carlos Bravo :)
Lima -PERU

[/quote]

Carlos Bravo :)
Lima -PERU[/quote]

Una pregunta, por que Peru no participa en la olimpiada universitaria, y alguien podria explicarme que es el PUTNAM.
De antemano gracias....

[quote="yamilimot"]alguien podria explicarme que es el PUTNAM.
[/quote]

http://www.mathlinks.ro/Forum/index.php?f=80

Si no me equivoco, PUTNAM es una olimpíada para alumnos universitarios de Canadá y Estados Unidos.

Hasta donde yo sepa Peru participo en las Olimpiadas Iberoamericanas Universitarias de matematicas los años 2000,2001,2002,2003 pero de alli no se mas . Mas bien Si tienen por alli algun problemita interesante de matematicas universitarias seria bueno discutirlo en el foro .

como hay esta secion
no se si alguien sabe sobre la interpolacion de legendre ???? :(


In acute-angled triangle $ABC$, such that $CA\neq CB$ we have that $A_1$ and $B_1$ are the tangent points with the excircles of the triangle at the sides $CA$ and $CB$ and $I$ is the centre of the incircle. The line $CI$ intersects the circumcircle of triangle $ABC$ at point $M$ and the line from $M$ perpendicular to $CM$ intersects the line $AB$ at point $N$. Prove that the lines $NI$ and $A_1B_1$ are parallel.

[quote="Beat"] we have that $A_1$ and $B_1$ are the tangent points with the excircles of the triangle at the sides $CA$ and $CB$ [/quote]
Please can you explain which excircles with which sides (respectively) do you mean ?

The excircles of the triangle ABC which are tangent to the [u][i]sides[/i] AC and BC[/u] at points $A_1$ and $B_1$, it doesn`t matter which one is $A_1$ and which one is $B_1$


A 1,000-foot long retaining wall is to be built, partly of wood and the rest of stone. The wood costs $ \$8$ per foot and the stone costs $ \$10$ per foot. The total cost can be at most $ \$$9,200. What is the greatest number of feet of stone that can be used?

[quote="N1349"]How do you solve this problem?[/quote]

Well, let number of wood be x and number of stones be y. We want 1000 ft so x+y=1000. We want the most we can have, so 8x+10y=9200 -> 4x+5y=4600.

4(1000-y)+5y=4600
4000+y=4600
y=600

You can use at most 600 pieces of wood

Dude, you got your variables mixed up, you said y = stone, not wood.

otherwise, thanks for explaining!


Λοιπόν αυτό που συμβαίνει είναι λίγο τρομερό........
Δύο άνθρωποι να έχουν την ίδια ιδέα για γενίκευση της ίδιας άσκησης  :o
Στέλιο μήπως έχεις γράψει λάθος το δεύτερο μέλος ??
Μήπως είναι $S+\sqrt{(n-1)S}$   ???

Νομίζω πως η ανισότητα ισχύει και με το $S+2\sqrt{S}$.

Αν όμως έχεις μια απόδειξη για το $S+\sqrt{(n-1)S}$ τότε αν μπορείς γράψ'την,γιατί μπορεί να βοηθήσει!

Την παραθέτω λοιπόν .
Είναι $\sqrt{a_2}<a_2+\sqrt{a_2}$
       $\sqrt[3]{a_3^2}<a_3+\sqrt{a_3}$  κτλ

Άρα το πρώτο μέλος είναι μικρότερο από
$S+\sqrt{a_2}+\sqrt{a_3}+...+\sqrt{a_n}$
Από Cauchy όμως $(n-1)S\geq (\sqrt{a_2}+....+\sqrt{a_n})^2$
άρα  $\sqrt{a_2}+\sqrt{a_3}+...+\sqrt{a_n}\leq \sqrt{(n-1)S}$
Συνεπώς το πρώτος μέλος είναι μικρότερο από
$S+\sqrt{(n-1)S}$

Ωραία απόδειξη!
Η άσκηση που έβαλα μου φαίνεται ελαφρώς πιο δύσκολη!  ;)
Μπορεί κανείς να τη λύσει?

Έχεις δίκιο είναι ελαφρώς πιο  δύσκολη .
 :)
Έστω ότι για  $k$ αριθμούς υπάρχει $m$  ώστε  $a_i^{1-\frac{1}{i}}\leq ma_i$
Τότε προφανώς το άθροισμα αυτών των αριθμών είναι μικρότερο ή ίσο από $mS$  δηλαδή
$\sum a_i^{1-\frac{1}{i}}\leq mS$ . Tώρα για τους υπόλοιπους αθροισμούς θα ισχύει $a_j^{1-\frac{1}{j}}\leq\frac{1}{m^{j(1-\frac{1}{j})}}$  .Άρα
$\sum a_j^{1-\frac{1}{j}}\leq\frac{1}{m-1}$  από την ανισότητα Choeher .
Άρα $a_2^{1-\frac{1}{2}}+a_3^{1-\frac{1}{3}}+...+a_n^{1-\frac{1}{n}}\leq mS+\frac{1}{m-1}$

Επιλέγοντας $m=1+\frac{1}{\sqrt{S}}$  παίρνουμε τη ζητούμενη

Στέλιο θα τη βάλλω και στο αγγλικό φόρουμ .Οκ ?
Πώς σας φάνηκε η λύση ??

Πως μας φαινεται η λυση? αισχος, ντροπη δεν το περιμενα αυτο απο εσενα. να παραιτηθεις αμεσως, :rotfl:  :lol: .Σοβαρα τωρα  μπραβο Σιλουανε.σπουδαια λυση. ποια ειναι η αν. του Choeher? ειναι δικο σου τεχνασμα?

Σιλουανέ ,συγχαρητήρια για την ωραία λύση !!!
Μάλλον θα πρέπει να ψάξω για πιο δύσκολες ανισότητες!  :P

Όταν λες ανισότητα Choeher εννοείς  την:

$\frac{1}{m}+\frac{1}{m^{2}}+.....+\frac{1}{m^{n}}\leq \frac1{m-1}$

για κάθε $m>1$ και $n \in \mathbb N^{*}$ ;


vector
linear algebra
matrix
projective geometry
LaTeX

I would like to produce n by 1 arrays where the entries are separated by a `colon on its side', so that they denote elements of a projective space which are acted on by matrices. I.e. the transpose of something like [a : b : c : ... : z]. I've tried doing \ddot{} for each element except the top one but it doesn't look good.

Not sure if this is the effect that you want, but on the forum you can try something like
\[ \begin{bmatrix}
a \\[\minus{}1ex]
..\\
b \\[\minus{}1ex]
..\\
c \\[\minus{}1ex]
.. \\
\vdots \\[\minus{}1ex]
..\\
z
\end{bmatrix}\] and you can adjust the vertical spacing as you wish.
In a document \renewcommand{\arraystretch}{[i]factor[/i]} can be used to change the separation of all the rows where [i]factor = 1[/i]  is the default and [i]factor = 0.5[/i] halves the separation etc.


geometry
articles
AMC
AIME
Putnam
CEMC
number theory

Please post links to resources suitable for AMC-level preparation here.

In an effort to organize this better, it would be nice if we used a classification system.  You don't have to follow this, but it'd definitely be helpful.  If you don't follow this, we can always apply it later.

Name of Resource:
URL (Web Address):
Category: (Problems - actual AMC, Problems - other, Lecture Notes, Books, Math Journals, Other - please specify)
Topics: (General, Algebra, Combinatorics, Geometry, Graphing/Functions, Number Theory, Other - please specify).
Difficulty: (1-10 scale)
Comments: May be blank.

Thanks,
mathfanatic

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

Name: [url=http://www.pims.math.ca/pi/]Pi in the Sky[/url]
URL: [url]http://www.pims.math.ca/pi/[/url]
Description: Semi-annual publication from Canada, designed to promote math, interest in math activities/contests, and math careers.  Features articles on basic math principles, math challenges, etc.
Category: Math Journals
Topics: General
Difficulty: 6
Comments: None
Problem Example: Find all distinct pairs (x,y) of integers that are solutions of the equation x:^2: - xy + y:^2: = x + y.

[url=http://www.math.uncc.edu/~hbreiter/index.html#American%20Mathematics%20Competitions]Harold Reiter's Homepage[/url]
http://www.math.uncc.edu/~hbreiter/index.html#American%20Mathematics%20Competitions
Category: AMC Problems
Topics: Just past problems from the AMC/AHSME, and links to other sites
Difficulty: 4

[url=http://www.kalva.demon.co.uk/aime.html]AIME Problems[/url]
http://www.kalva.demon.co.uk/aime.html
Category: AIME Problems
Topics: Past problems from the AIME
Difficulty: 7

[url=http://www.kalva.demon.co.uk/index.html]Kalva homepage[/url]
http://www.kalva.demon.co.uk/index.html
Category: Complete past problem sets from a variety of competitions
Topics: AIME, Dudney, IMO, Putnam, etc
Difficulty: 7-10

Name: [url=http://www.cms.math.ca/Competitions/COMC/]Canadian Open Math Challenge[/url]
URL: http://www.cms.math.ca/Competitions/COMC/
Description: Past exams from 1996 through 2002 of the COMC, which is approximately an AMC-level contest.
Category: Problems - Other
Topics: General
Difficulty: 4
Comments: Includes questions, solutions, results, and links to other competitions, including the CMO, APMO, IMO, MOCP, and IMTS.

Name: [url=http://www.cms.math.ca/Competitions/CMO/]Canadian Mathematical Olympiad[/url]
URL: http://www.cms.math.ca/Competitions/CMO/
Description: Past exams from 1969-2003, plus solutions of the problems from 1994-2003.
Category: Problems - AIME Level; Problems - Olympiad Level
Topics: General, Proofs
Difficulty: 6-8
Comments: Again, includes questions, solutions, results, and links to other competitions, including the CMO, APMO, IMO, MOCP, and IMTS.

Name: [url=http://goodlymaths.0catch.com]Goodly Maths[/url]
URL: http://goodlymaths.0catch.com
Description: Resources for students interested in math competitions, including AMC and AIME problems, tutorials, a problem of the week contest, links, a giant contest directory, and more.
Category: Problems - actual AMC/AIME, General
Topics: General
Comments: Brought to you by yours truly, and now open to the public.

[url=http://goodlymaths.0catch.com/contests/amc/amc_how_to_prepare.htm]How to Prepare for the AMC Exams[/url]
URL: http://goodlymaths.0catch.com/contests/amc/amc_how_to_prepare.htm
Description: Short article with links to resources for preparing for the AMC8/10/12.  Some comments on what to do and what not to do to prepare.
Category: FAQ

CEMC Math Competitions Previous Contests
[url]http://www.cemc.uwaterloo.ca/english/contests/index.shtml[/url]
Category: Problems - other; range from AMC to AIME-type problems
Topics: all types
Difficulty: 2-7
Comments: Many different Canadian contests that range in difficulty.


could anyone help me these questions

The nucleus is a key feature that distinguishes eukaryotic from prokaryotic cells.  as a general rule, each eukaryotic cell has a single nucleus.  discuss three exceptions to this rule (eukaryotic cells w/o a nucleus or with more than one nucleus) and how these exceptions represent adaptations fro specialized cell functions.

describe the movements of an excreted protein within the endomembrane system from the time protein synthesis begins on a free ribosome until it is released outside the cell. include an explanation of the role of the signal sequences.

thank you!

Please try to do your own homework.  This is material that should be in your textbook.  It's also freely available online.

[hide="Hint for number 2"]

1. Look up "eukaryotic codons"

2. Look up mRNA

3. How does mRNA contribute to making RNA (and how tRNA and rRNA come into play).
[/hide]


hi
when can i assume a>b>c
or abc =1
or a+b+c=1
etc :maybe: [/quote]

for 1 when the inequality is absolutely symmetric in $ (a,b,c)$
2nd and 3rd depends on the question can be used when the inequalty is homogeneous :wink:

when the inequality is cyclic????? :maybe:

when the inequality is cyclic you need to consider two cases: $ a\ge b\ge c$ and $ a\le b\le c$.

[quote="stephen38"]I  want to ask too.Do you have useful methods for solving inequality with given condition : a>=b>=c? :maybe: Thanks a lot.[/quote]

[b]Problem:[/b] Let $ a,b,c$ be positive real numbers. Prove that
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}\]
As the inequality is symmetric in $ a,b,c$, without losing generality we may assume that $ a\ge b\ge c$. Then you can easily show that $ b+c\le c+a\le a+b$. That is, $ \frac{1}{b+c}\ge \frac{1}{c+a}\ge \frac{1}{a+b}$. Combining this with $ a\ge b\ge c$, we can apply rearrangement as follows:
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\qquad(1) \]
And
\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{b}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}\qquad(2) \]
Adding $ (1)$ and $ (2)$, yields

$ 2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\ge \frac{b}{b+c}+\frac{c}{c+a}+\frac{a}{a+b}+\frac{c}{b+c}+\frac{a}{c+a}+\frac{b}{a+b}$ $ =3$, which is what we wanted to prove.

[b]Problem:[/b] Prove that for positive reals satisfying $ a\ge b\ge c$ the following inequalities hold:

\[ 1.\ a^{2}b+b^{2}c+c^{2}a\ge ab^{2}+bc^{2}+ca^{2}\]\[ 2.\ a^{5}b+b^{5}c+c^{5}a\ge ab^{5}+bc^{5}+ca^{5}\]
[b]1. Solution:[/b] Rewrite the inequality as
\[ ab(a-b)+bc(b-c)+ca(c-a)\ge 0 \]
\begin{eqnarray*}ab(a-b)+bc(b-c)+ca(c-a) &=& ab(a-b)+b^{2}c-bc^{2}+c^{2}a-ca^{2}\\ &=& ab(a-b)-c(a^{2}-b^{2})+c^{2}(a-b)\\ &=& (a-b)(ab-ca-bc+c^{2})\\ &=& (a-b)(b-c)(a-c)\\ &\ge & 0\end{eqnarray*}

As $ a\ge b,\ b\ge c$ and $ a\ge c$.

[b]2. Solution:[/b] Try yourself. [hide="hint"]The factorization should be
$ (a-b)(b-c)(a-c)(ab+bc+ca)$.[/hide]

Thanks for your help.I remembered that in the past,I've ever stucked in the IMTS problems with this given conditions. :blush: I will try it later with your method.  :D

[quote="ishfaq420haque"]thanks  :D[/quote]
[color=red]Welcome to mathlinks [/color]:D


[quote="#H34N1"]Is it possible to place 1995 different natural numbers along a circle so that for any two of these numbers, the ratio of the greater to the lesser is prime?[/quote]
i think that you wand say :

[quote="aviateurpilot"]Is it possible to find 1995 different natural numbers $ a_1,a_2,...,a_{1995}$ such that:
$ \forall k\in\{1,2,...,1994\}: \ \frac {max(a_k,a_{k \plus{} 1})}{min(a_k,a_{k \plus{} 1})}\ is\ prime$ and $ \frac {max(a_1,a_{1995})}{min(a_1,a_{1995})}\ is\ prime$[/quote]
is it that?
if yes, then there arn't such numbers. because $ 1995$ is odd.

then exists $ A,B$ partition of $ \{1,....,1995\}, and\ (p_i)_{i \equal{} 1...1995}$ such that $ \forall i\in A: \ a_{i} \equal{} p_ia_{i \minus{} 1},\forall i\in B: \ a_{i} \equal{} a_{i \minus{} 1}/p_i$ (we take here $ a_0 \equal{} 1995$)
obviouly we see that $ \prod_{i\in A}p_i \equal{} \prod_{i\in B}p_i$ then $ |A| \equal{} |B|$
so $ 1995 \equal{} |A| \plus{} |B| \equal{} 2|A|$ (impossible)

[quote="aviateurpilot"]
there exists $ A,B$ partition of $ \{1,....,1995\}, and\ (p_i)_{i \equal{} 1...1995}$ such that $ \forall i\in A: \ a_{i} \equal{} p_ia_{i \minus{} 1},\forall i\in B: \ a_{i} \equal{} a_{i \minus{} 1}/p_i$ ([color=red][b]we take here[/b][/color] $ \boxed{a_0 \equal{} 1995}$)
obviouly we see that $ \prod_{i\in A}p_i \equal{} \prod_{i\in B}p_i$ then $ |A| \equal{} |B|$
so $ 1995 \equal{} |A| \plus{} |B| \equal{} 2|A|$ (impossible)[/quote]
i have write ([color=red][b]we take here[/b][/color] $ \boxed{a_0 \equal{} 1995}$)
i want write ([color=red][b]we take here[/b][/color] $ \boxed{a_0 \equal{} a_{1995}}$)

[quote="nateharman1234"]Heres really the same argument as above but a little nicer to understand (at least in my opinion): if start at A and you go around the circle clockwise the number of primes (not always distinct) that divide the number either goes up or down by one each time you move one place around the circle.  So when you get back to A you should have gone up the same amount of times you went down, but if you moved an odd number of times thats impossible.[/quote]

This was really good :)


algebra
polynomial
real analysis
real analysis unsolved

If $x$ satisfy
\[a_{0}+a_{1}x+a_{2}x^{2}+\ldots+a_{n}x^{n}\, , n\in \mathbb{N}\]
with all $a_{i}\in\mathbb{Z}$, it is called algebraic. [color=blue]Show[/color] that the set of
all these $x$ is countable.

[color=blue]show[/color] that there are infinite number of non-algebraic real numbers.

Let $\mathbb{Z}[x]$ be the set of polynomials of integer coefficients. we define $f : \mathbb{Z}[x] \to \mathbb{Q}$ by

$f \; : \; a_{0}+a_{1}x+\cdots+a_{n}x^{n}\; \mapsto \; p_{1}^{a_{0}}p_{2}^{a_{1}}\cdots p_{n+1}^{a_{n}}$

where $p_{k}$ is the $k$-th prime number. Then you can easily check that $f$ is an injection. Since $\mathbb{Q}$ is countable, so does $\mathbb{Z}[x]$ and we can number every element in $\mathbb{Z}[x]$ as $P_{1}, P_{2}, P_{3}, \cdots$.

Suppose $P(x)$ is in $\mathbb{Z}[x]$. Then $P(x) = 0$ has only finite number of roots and $A(P) : = \{ x : P(x) = 0 \}$ is finite.

Not we are ready to prove it. Let $A = \bigcup_{n=1}^{\infty}A(P_{n})$. Then $A$ is the set of algebraic numbers. But a countable union of countable sets is again countable, $A$ is countable and hence proved.

Also, $\mathbb{R}$ is uncountable, $\mathbb{R}\backslash A$ is also uncountable.


[quote="just-abdess"]Prove the following for all n, n a positive integer:

1)

$ n^{n} > 1 \cdot 3 \cdot 5 \cdot... \cdot(2n \minus{} 1)$

2)

$ 2^{n} > 1 \plus{} n\sqrt {2^{n \minus{} 1}}$

3)

$ 1 \plus{} \frac {1}{\sqrt2} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {n}}\geq \sqrt {n}$

4)

$ \frac {1}{2}\leq \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} ... \plus{} \frac {1}{2n}\leq \frac {3}{4}$

5)

$ 1 < \frac {1}{n \plus{} 1} \plus{} \frac {1}{n \plus{} 2} \plus{} ... \plus{} \frac {1}{3n \plus{} 1} < 2$
[/quote]

Try what's in the quote box.

An integer includes {..., -3, -2, -1}, so I suggest you streamline and phrase it
closer to what I showed.  Also, don't use periods for multiplication.

[quote="mstf9696"]I think n a pozitive integer.if n is a positive integer i solved first question.already if n is a negative integer first inequality is false.
[hide]1.3.5…….(2n-1)=(1.2.3.4…….(2n))/(2.4.6…..(2n))=(2n)!/((2.1)(2.2)(2.3)….(2.n))=(2n)!/(2^n.(n)!)
So we must prove n^n>(2n)!/(2^n.(n)!). Let both sides of the inequality multiplied by  2^n.we must prove   (2n)^n>(2n)!/((n)!)=(2n)(2n-1)….(n+1) now.  Namely
(2n)(2n)(2n)……(2n)>(2n)(2n-1)……(n+1)  (there are n times 2n in the left side). 2n=2n,2n>2n-1,2n>2n-2,,,,,,,2n>n+1 so (2n)^n>(2n)(2n-1)(…..)(n+1) is true.
[/hide][/quote]

Don't use periods for multiplication, especially after I posted that to the original poster.
"2.1" is "two point one" or two and one-tenth, and not two times one.


Let $ x\equal{}a\plus{}y, \ y\equal{}\{x\}, a\equal{}[x]$ integer, then your function $ f(a,y)$ decrease by y and increase by a.
$ lim_{x\to 2, x<2} f(a,y)\equal{}\frac{16}{3}$.

[quote="Rust"] then your function $ f(a,y)$ decrease by y and increase by a.
$ lim_{x\to 2, x < 2} f(a,y) \equal{} \frac {16}{3}$.[/quote]

It's incomprehensible! Please give more details!

Could someone perhaps post the problems of Mediterranean 2008?


Solve the equation $x!+y!+z!=xyz$, where $x$, $y$ and $z$ are positive integers and $xyz$ denotes a 3-digit number.  Prove that you have found all solutions.

Thanks

Alright. For starters, we have $x!,y!,z!\leq 1000 \implies x,y,z \leq 6$.

The hundreds digit cannot be over $6$, so $720$ is not a possibility for $x!,y!,z!$, and $x,y,z \leq 5$.

Now we clearly can't create a 3-digit number without a $5!$, so one of $x,y,z$ must be $5$. However, the only digit of $5$ we can create now is $120+24+1=145$ or $120+24+6=150$.

Checking, the first gives $1!,4!,5!$, so $(x,y,z)=(1,4,5)$ is a solution.

The second requires that $z=0$, wheras $z$ should be $3$, so it doesn't work.

Our only solution is therefore $\boxed{(x,y,z)=(1,4,5)}$


Though this is a little late...

Does anyone have a video of the countdown available, just curious?

and also, to the participants from last year, did you guys get the packet with the questions and statistics, because l didn't...

I believe the coach gets 1, ours distributed that to the team

ur mario choi right?

yeah, i'm mario choi, but nothing got mailed like in 2006?

um, i think the coach gets the statistics and problems at the competition, nothing is mailed

The video of the Countdown Round is under Highlights on the MATHCOUNTS website.

Mario Choi, it actually did contain a glance of your part in the round. :) (i think it was against justin ahmann)

I remember u saying something about fermat's little theorem on the bus ride to that stupid rodeo thing

... I never went against Justin...

I played Kevin, and lost in the quarterfinals and I beat the 11th seed first round

[quote="gauss1181"][quote="tinytim"]The full video is not on the MC website though[/quote]
What do you mean by the full video?[/quote]
I mean that not all of the questions from the 2007 cd round were on the video. This year, they were.


Hello.
Here is my solution.

Solution:Since $mdc(101,100)=1$, we have that $n_{a}+n_{b}-n_{c}-n_{d}\equiv 0 (mod \space 10100)$ if, and only if, $n_{a}+n_{b}-n_{c}-n_{d}\equiv 0 (mod \space 101)$ and $n_{a}+n_{b}-n_{c}-n_{d}\equiv 0 (mod \space 100)$.SO...

1º-We can see that $n_{a}= 101a-100.2^{a}\equiv-100.2^{a}\equiv 2^{a}(mod 101)$.So $n_{a}+n_{b}\equiv n_{c}+n_{d}(mod101)$ implies $2^{a}+2^{b}\equiv 2^{c}+2^{d}(mod 101)$.And lets assume that$d=min(a,b,c,d)$ and $a \not= d$ and $b \not= d$

2º-Now, $n_{a}=101a-100.2^{a}\equiv 101a \equiv a(mod 100)$.So $n_{a}+n_{b}\equiv n_{c}+n_{d}(mod 100) \Rightarrow a+b \equiv c+d(mod100)$.And since $a+b <200$ we must have $a+b= c+d$ or $a+b= 100+c+d$(Don't forget that we are supposing that $a \not= d$ and $b \not= d$).

First case:$a+b = c+d$.
So $a-d = c-b$ , thus $2^{a-d}\equiv 2^{c-b}(mod101)$,and (by 1º) we know that $2^{a}+2^{b}\equiv 2^{c}+2^{d}(mod 101) \Rightarrow 2^{d}(2^{a-d}-1) \equiv 2^{b}(2^{c-b}-1)(mod 101)$.Hence $2^{d}(2^{c-b}-1) \equiv 2^{b}(2^{c-b}-1)(mod 101)$   $(ii)$
Now,since $0 \leq b,c \leq 99$, we have that $c-b <100 = ord_{101}(2)$.Hence $mdc(2^{c-b}-1,101)=1$, so $2^{d}\equiv 2^{b}(mod101)$.So $2^{d}(2^{b-d}-1) \equiv 0(mod101)$ but $2^{d}$ is realtively prime to $101$, so $2^{b-d}-1 \equiv 0 (mod 101)$, but that is an absurd since $b-d < 100 =ord_{101}(2)$.
So or $a=d$ or $b=d$.Suposse $b=d$, so $a+b = c+d$ implies $a=c$, and if $a=d$ , $a+b = c+d$ implies $b=c$.Hence $\{a,b\}=\{c,d\}$.

Second case:$a+b = 100+c+d$
In that case we have that $2^{a-d}\equiv 2^{100}.2^{c-b}\equiv 2^{c-b}(mod 101)$, and now, by the same way of the first case, we can conclude that $\{a,b\}=\{c,d\}$.

And the problem is finished!! :D


[quote="sagar"]patrick is trapped at the bottom of a very deep canyon with vertical walls. he has a gun with him to help signal his positon. jeff is located on the ground near the entrance to the canyon. he heard the sound of the gun shot at exactly 2.75 sec before the arrival of the bullet, which he caught when it was motionless in the air. if the air temperature was equal to 10 degree celcius in the canyon, how deep was the canyon , and how fast was the initial speed of the bullet?[/quote]
I know things like speed and velocity, but this is really advanced. :o

this is a typical question u would be happy to expect from physics 115.
intial velocity of bullet Vb = 9.8T
d is the height of the canyon
T is the time for bullet to reach motionless
4.9T^2 + T(337.5-Vb) - 928.125 = 0

solve both equations then u can sub back to the equation
d = VbT - 4.9T^2
then u'll get the answers.

hopeful i can still do arthmetic without errors...


It is a direct result of Euler's formula:

\[re^{i\theta}=r(\cos \theta+i\sin \theta).\]

You can get Euler's formula using the Taylor expansions of $e^x$, $\cos x$ and $\sin x$.  Look [url=http://mathworld.wolfram.com/EulerFormula.html]here[/url] for more info.

[quote="jli"]how does $e^{pi\cdot i}=-1$ work?[/quote]

Euler discovered that $e^{ix}$ has the same series expansion as $\cos x + i\sin x$

That means $e^{ix} = \cos x + i\sin x$

If you substitute $\pi$ in place of $x$ you get

$e^{i\pi} = \cos \pi + i\sin \pi = -1+i0 = -1$

[quote="Treething"]Erm, this is moving a little towards the calculus side :? ...[/quote]

Don't think so, Threething. The answer to the question involves infinite series, a calculus topic, but no derivation dependent on those series has been given. Thus all you have at this point is a formula: $e^{ix}= \cos x +i\sin x$, which is certainly not beyond the level of this forum ... I hope.


real analysis
limit
function
integration
calculus
real analysis unsolved

Let $(A_n)_n$ be a sequence of Lebesgue measurable subsets of $[0,1]$, such that for any interval $(a,b)$ we have that $\lim_{n \to \infty} \lambda(A_n \bigcap (a;b) )= \frac{b-a}{3}$. Prove that for any function $f: [0,1] \to \mathbb{R}$ of bounded variation $\lim_{n \to \infty} \int_{A_n}f(x)dx = \frac{1}{3}\int_0^1 f(x)dx$.
Here $\lambda$ is a Lebesgue measure on $\mathbb{R}$.

I'm sure the last integral should have a $\frac 13$ in front of it :).

Let's prove it for $L^\infty$ functions (clearly, functions with bounded variation belong to $L^\infty1([0,1])$, since they're just differences of non-decreasing functions). If for a measurable set $S,\ \chi_S$ is its characteristic function, the hypothesis says that for every interval $J$ we have $\int\chi_{A_n}\chi_J\ \mbox d\lambda\to\int\frac 13\chi_J\ \mbox d\lambda$, and the conclusion should be that for any $f\in L^\infty$, we have $\int\chi_{A_n}f\ \mbox d\lambda\to\int\frac 13f\ \mbox d\lambda$.

The dual space of $L^1$ can be identified with $L^\infty$, so we have to prove that $\chi_{A_n}\to\frac 13$ (the constant function $\frac 13$) weakly. By a theorem of John Rainwater, a sequence $x_n$ in a normed space converges to $x$ weakly iff it's bounded, and $f(x_n)\to f(x)$ for every extreme point $f$ of the unit ball of the dual space. The extreme points of the unit ball of $L^\infty$ are $1$ or $-1$ almost everywhere, so they are differences of characteristic functions of complementary measurable subsets of $[0,1]$. This means that it suffices to show that $\int\chi_{A_n}\chi_S\ \mbox d\lambda\to\int\frac 13\chi_S\ \mbox d\lambda$ for every measurable set $S$, or, in other words, that $\lambda(A_n\cap S)\to\frac 13\lambda(S)$ for all measurable $S$, and this is easier to see than to write down.


Find the number of non-negative integer solutions to the equation :
   $ x_1+x_2+x_3+x_4+ x_5 = 28 $
  in each of the following cases :

  i) $ x_i \geq 3 \forall i $
 ii) $ 0 \leq x_1 \leq 3 ; 1 \leq x_2 \leq 4 ; x_3 \geq 13 $

[quote="jarod"]Find the number of non-negative integer solutions to the equation :
   $ x_1+x_2+x_3+x_4+ x_5 = 28 $
  in each of the following cases :

  i) $ x_i \geq 3 \forall i $[/quote]
Let $y_i = x_i - 3$ then
$ y_1+y_2+y_3+y_4+ y_5 = 28 - 15 = 13$
The number of non-negative solutions is ${13+5-1\choose 5-1} = {17\choose 4} = 2380$.

[quote="jarod"]ii) $ 0 \leq x_1 \leq 3 ; 1 \leq x_2 \leq 4 ; x_3 \geq 13 $[/quote]
The number of non-negative solutions equals the coefficient of $z^{28}$ in
$(1+z+z^2+z^3)(z+z^2+z^3+z^4)(z^{13}+z^{14}+...)(1+z+...)^2 =\\ z^{14} (1-z^4)^2 / (1-z)^5$
which is equal to
$\sum_{k=0}^2 (-1)^k {2\choose k} (-1)^{14-4k} {-5\choose 14-4k} =\\
\sum_{k=0}^2 (-1)^k {2\choose k} {18-4k\choose 14-4k} =\\
{2\choose 0}{18\choose 14} - {2\choose 1}{14\choose 10} + {2\choose 2}{10\choose 6} = 1268.$


Proove that:

\[ \sum_{i=1}^{\underbrace{666 \text{\ldots 666}}_{k}}i=\underbrace{222 \text{\ldots 222}}_{k}\underbrace{111 \text{\ldots 111}}_{k}\]

Wait...are you just asking for the $ 6\left(\frac{10^{k}-1}{9}\right)$-th triangular number?

$ \sum_{i=1}^{\frac{6}{9}\left(10^{k}-1\right)}i=\frac{1}{9}\left(10^{k}-1\right) \left(2 .10^{k}+1\right)\Rightarrow \underbrace{222 \text{\ldots 222}}_{k}\underbrace{111 \text{\ldots 111}}_{k}$

What i am saying is: Is not OBviously that the second term transform in the third.

just evaluate the sum is not the solution.

on solution would be:
\[ \sum_{i=1}^{\frac{6}{9}\left(10^{k}-1\right)}i=2. 10^{k}\sum_{j=0}^{k-1}10^{j}+1 \sum_{j=0}^{k-1}10^{j}\]
i just want to know anothers ways to proof that.

:)


What is the largest positive integer that is not the sum of a positive integral multiple of 42 and a positive composite integer?

As glad as I am that you posted the original problem, this still doesn't quite solve my problem of finding a solution.

If you click on the link with the problem it will lead you to a page where several solutions are given.

Or just click here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=394508#p394508


algebra
function
domain
induction
abstract algebra
search
Functional Analysis

If $ X$, $ Y$ are Banach spaces and $ A$ is a closed linear operator from $ X$ to $ Y$ s.t. $ R(A)$ is not closed in $ Y$, show that for each $ \epsilon > 0$ there is an infinite dimensional closed subspace $ M \subset X$ s.t. the restriction of $ A$ to $ M$ is compact with norm $ {}<\epsilon$.

1. We can reduce to the situation that $ X$ has a countable, but infinite basis such that $ A$ is injective on an infinite dimensional subspace $ X'$ of the domain $ D_A$:

[hide="Proof of 1."]Since $ R(A)$ is not closed in $ Y$, there is some $ y\in Y\setminus R(A)$ together with a sequence $ \{x_n\}$ in $ X$ such that $ Ax_n\to y$.  If $ Y': = span\{Ax_n\}$ is finite dimensional, it is closed in $ Y$, such that $ y\in Y'$. But since $ Y'\subset R(A)$, we have the contradiction $ y\in R(A)$. Thus $ Y'$ has infinite dimension. Taking a subsequence of $ \{Ax_n\}$ if necessary, we can assume that the set $ \{Ax_n\}$ is linearly independent. This implies that $ X' = span\{x_n\}$ is infinite dimensional, $ A$ restricted to $ X'$ is injective and that $ A$ restricted to $ \overline{X'}$ is a closed operator whose range is not closed. Substituting $ X$ by $ X'$, we have achieved claim in 1.
[/hide]

2. Let us assume the situation as given in 1. Then $ \alpha: = \inf_{x\in X', \,\|x\| = 1} \|Ax\| = 0$.
[hide="Proof of 2."] If $ \alpha > 0$, then taking a sequence $ Ax_n\to y\notin R(A)$ as above, $ \{Ax_n\}$ is Cauchy, hence $ \{x_n\}$ is Cauchy in $ X$ because of $ \alpha\|x_n - x_m\|\le \|Ax_n - Ax_m\|$. Thus $ x_n\to x^*\in X$. Since  $ A$ is assumed to be a closed operator, the two limits $ x_n\to x_*$ and $ Ax_n\to y$ imply $ x^*\in D_A$ and $ Ax^* = y$, hence $ y\in R(A)$, a contradiction. Thus we know $ \alpha = 0$.
[/hide]

3. Given $ \varepsilon > 0$, there is a sequence $ \{z_n\}$ in $ X'$ such that $ \{Az_n\}$ is linearly independent with $ \|z_n\| = 1$ and $ \| Az_n\| < \frac {\varepsilon}{2^n}$ for all $ n$.

[hide="Proof of 3."]We define the sequence $ \{z_n\}$ by induction: For $ n = 1$ (2.) provides $ z_1\in X$ with $ \|z_1\| = 1$ and $ \| Az_1\| < \frac {\varepsilon}{2}$. By (1.) $ span\{x_n\}$ is dense in $ X$ and $ \{x_n\}$ is linearly independent. For the induction step define $ X_m: = span\{x_1,\ldots,x_m\}$ and $ X'_m: = span\{x_k|k > m\}$ such that $ X_m\oplus {X'_m}\subset X$ for all $ m$. Now let $ z_1,\ldots,z_n\in X_{m_n}$ ($ m_n\in \mathbb N$) be linearly independent with $ \|z_m\| = 1$ and $ \| Az_m\| < \frac {\varepsilon}{2^m}$ for all $ m\le n$. The premises of (2.) still hold for $ X'_{m_n}$ such that there is $ z_{n + 1}\in X'_{m_n}$ with $ \|z_{n + 1}\| = 1$ and $ \| Az_{n + 1}\| < \frac {\varepsilon}{2^{n + 1}}$. Certainly $ z_1,\ldots,z_n,z_{n + 1}$ are linear independent in $ X_{m_{n + 1}}$ for some natural $ m_{n + 1} > m_n$. Since $ A$ is injective on $ X'$ we get that $ \{Az_n\}$ is linearly independent.
[/hide]

4. Let $ Z: = span\{z_n\}$. Then $ M: = \overline{Z}$ does the job, i.e. $ A$ restricted to $ M$ is compact with norm $ \le \varepsilon$.

[hide="Proof of 4."]$ Z$ is infinite dimensional. Because $ \{Az_n\}$ is linearly independent, the relation
\[ Az = \sum_nf_n(z)Az_n\]
(where the sum is always finite) shows that $ f_n: Z\to \mathbb R$ is a linear functional with kernel $ span\{z_m|m\neq n\}$ for all $ n$. Furthermore we have $ f_n(z_n) = 1$ which implies that $ \|f_n\| = 1$, in particular $ f_n$ is continuous. Thus
\[ \|Az\|\le\sum_n\|f_n\| \|Az_n\| \|z\|\le \varepsilon \|z\|\]
for each $ z\in Z$. In other words $ \|A|Z\|\le \varepsilon$. Moreover $ A|Z$ is the uniform limit of linear maps of finite rank, hence compact. By boundedness, $ A$ can be uniquely extended to $ M = \overline{Z}$ with the same norm $ \le \varepsilon$ where the property of being compact is preserved (e.g. extend the $ f_n$'s to $ M$ first, then $ A$). By closedness of $ A$ the extension to $ M$ coincides with $ A$.[/hide]

In fact, the solution proposed by Third Edition has a gap. In step 4 the author writes:

[quote]Furthermore we have $ f_n(z_n) \equal{} 1$ which implies that $ \|f_n\| \equal{} 1$[/quote]

In reality one gets only estimate from below for $ \|f_n\|$. The estimate from above and even the continuity of $ f_n$ do not follow from the construction.

One can fix the solution if one substitutes linear independent sequences by basic sequences. To do this correctly one has to learn the Masur's technique of basic sequences selection. It is not covered by standard courses of Functional Analysis: one has to search in Banach space theory manuals. See for example Lemma 6.3.1 and Theorem 6.3.3 from

Kadets, M.I.; Kadets, V.M.  Series in Banach spaces.    Conditional and unconditional convergence / Birkhauser, 1997  (Operator Theory Advances  and Applications, vol. 94)


[quote="paradox102"]I see you guys have put pictures in the solutions, and i see that some people drew the squares for the first problem of round two.

Can you please tell me how to do that?  do you use a different system?  I heard of a thing called the geometer's sketchpad, but is that really necessary?[/quote]

For the squares in #1, just use some commands in $\text{\LaTeX}.$  For example, I used the tabular environment.


΍ ΍
      ΍ ΍
      ΍ ΍ ΍ ΍
in how many ways letters of NATION can be filled in this figure
such that no row remain empty, and each box can contain not more than one letter.

Let x=number of squares selected in 1st row; y=number of squares selected in 2nd row; z=number of squares selected in 3rd row;
$ <x,y,z>\equal{}<1,1,4>,<1,2,3>,<2,1,3>,<2,2,2>$

Required number of ways to select squares =4+8+8+6=26

Number of ways to permute the letters of the word NATION=$ \frac{6!}{2!}\equal{}360$

Therefore number of ways to put the letters of the word NATION=$ (26)(360)$


[quote="mofidy"]Let $x+y+z=1$ where $x,y,z>0.$ Prove \[\frac{{xy}}{{\sqrt{xy+yz}}}+\frac{{yz}}{{\sqrt{yz+zx}}}+\frac{{zx}}{{\sqrt{zx+xy}}}\leqslant \frac{{\sqrt 2 }}{2}\] [/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=97495
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439


There is a $ (x\minus{}x)$ term, so the whole thing equals $ 0$.

Didn't you already post this in your marathon?  :wink:

Well... Yes... am I in trouble... -gulp-. :oops:

[quote="alton66"](x-a)(x-b)(x-c)...(x-z)=?

It's really not that hard if you think about it.   :D[/quote]

Nice challenge...  :rotfl:

[quote="AIME15"]you got this in one of the books, can't remember which one...[/quote]

yeahh... page 117 of intro to alg

haha im teh only one stupid enough to look this up

in the product is an (x-x) term, which is 0.  And anything multiplyed by 0 is 0.

Please do not revive old threads. It is considered spam.


Find all solutions to the following differential equation: \[ x+tx'-4\sqrt{x'}=0. \]

notam x'=p(t) si obtinem $x+tp-4\sqrt{p}=0$, care prin derivare da $(t-\frac{2}{\sqrt{p}})\frac{dp}{dt}=-2p$, adica $t-\frac{2}{\sqrt{p}}=-2p\frac{dt}{dp}$. obtinem $t=\frac{c+lnp}{\sqrt{p}}$ si $x=\frac{-cp-plnp+4p}{\sqrt{p}}=(-c-lnp+4)\sqrt{p}$


The following facts are given:
$ 1$. $ a!$ is the smallest number divisible by the number $ 65536$
$ 2$. $ b!$ is the smallest number divisible by the number $ 1679616$
$ 3$. $ c!$ is the smallest number divisible by the number $ 95$
$ 4$. $ d!$ is the smallest number divisible by the number $ 6671808$
$ 5$. $ a, b, c, d$ are real positive integers.

List $ a, b, c, d$ in the ascending order of values.

this is really not a good way
[hide]
first , rewrite
$ 65536\equal{}2^{16}$
$ 1679616\equal{}2^{8}3^{8}$
$ 95\equal{}5.19$
$ 6671808\equal{}2^{6}3^{6}.11.13$
now do some computation we find $ a\equal{}18 ; b\equal{}18 ; c\equal{}19 ; d\equal{}15$[/hide]


Find all a,b,c,d,e positive integwers such that each one divides sum of other four.

Bomb.

The classical method can be adapted, even if I will not compute all the xases :
http://www.mathlinks.ro/Forum/viewtopic.php?t=38709

Pierre.


Prove that, For all $ n\in\mathbb{N*}$ ,always exist a prime number $ p\in (n^{2}\minus{}n,n^{2})$

Also, this is the third time you've posted it.  Stop spamming.

Four if counting a post in the Vietnamese Forum ;)

LOCKED.


Vieta
algebra
polynomial
binomial theorem

Hey, could someone please explain to me how to generalize the binomial theorem to terms that are not simple binomials raised to powers? For example : the problem

Find the $ 5th$ term in the expansion of $ (2x\plus{}7)^{10}$ is a very straightforward application of N choose K times $ x^k , y^{n\minus{}k}$
However, how would you solve the following which is a bit more complicated?

Find the $ 99th$ term in the expansion of $ (x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)$. . .$ (x\minus{}100)$. Could someone please help? Thank you in advance.

For that, you would need [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta%27s_formulas]Vieta's Formulas[/url], not the binomial theorem.

Hi,I think there's no way to solve that by Vieta's formula, and this topic does not belong to HSB !
   If you know to solve it , post it then !
Bye : endb0x.

The 99th term is $ r_1r_2 \hdots r_{99}\plus{}r_1r_3 \hdots r_{100}\plus{}r_1r_2r_4 \hdots r_{100} \plus{}\hdots$,

where the r's are the roots. A very large number.

Vieta's formulas and the binomial theorem are both consequences of the definition of polynomial multiplication, so all you really need to do is understand that thoroughly.

I think what you're really asking for is a [i]simple closed form[/i] for that term, and unless you consider $ \minus{}H_{100} 100!$ a significant simplification (I certainly don't) there's not much to say.  If you'd like to know more about these kinds of numbers, they're called [url=http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind]Stirling numbers[/url], and they're really quite interesting in their own right, but don't expect anything like a simple closed form.


calculus derivative function algebra polynomial

If f''(x) - f'(x) - 2f(x) = o,  f'(0) = -2, and f(0) = 2, then f(1) = ?

What the crap, How do you do this?

It's a second order differential equation with constant coefficients.  We want to find the roots of the characteristic polynomial --> x^2 - x - 2 = 0 --> (x-2)(x+1) = 0 --> x = 2 or x = -1.

Therefore y = ae^(2x)+be^(-x) --> 2a - b = -2 and a+b = 2 --> a = 0, b = 2

Thus y = 2e^(-x) so y(1) = 2/e.


topology
complex analysis
real analysis
Functional Analysis
\/closed

I'm wondering whether it might be a good idea to have a topology forum under "College Playground." I feel that topology is a fairly elementary subject, taught around the same time as abstract algebra and complex analysis (at least in the US), and that it shouldn't be restricted to the "Advanced Fields" forum along with algebraic and differential geometry. Thoughts?

Apart from the claim that topology is fairly elementary (algebraic topology isn't that trivial  :wink: ), I agree. I see no reason not to have an extra subforum.

Actually, it's not restricted to "Advanced fields". Some of it is in "Calculus/Analysis", since most real analysis classes also cover basic topology. "Advanced fields" is more of a pile of stuff that doesn't fit elsewhere than truly advanced material.
There would probably be about as much action in a topology subforum as in the probability/statistics subforum, combining the two current homes.

We, indeed, have enough posts related to topology to justify opening a separate subforum. Moreover, as the experience with Probability/Statistics shows, having a separate subforum for a subject increases the number of posts in that subject and attracts some new high quality posters (like, say, roah). So, I'm also in favor of getting a special space for topology and, in view of recent activity, maybe for functional analysis as well :).

I just spent 2 minutes quickly grabbing a few topics in the Advanced forum and moving them over. I'm sure there are more in other places, including Calculus/Analysis, that are also reasonable candidates to be moved. fedja should find some topics he thinks could be moved.

A little suggestion: can we please subdivide it into "General topology" and "Algebraic topology"¿


function integration limit search algebra domain calculus

Lebesgue's Dominated Convergence Theorem goes as follows:
Suppose that $\{f_n\}$ is a sequence of measurable functions, that $f_n\to f \hbox{ a.e.}$, as $n\to\infty$, and that $|f_n|\leq g \hbox{ a.e.}$ for all $n$, where $g$ is integrable. Then $f$ is integrable, and \[\int fd\mu = \lim_{n\to\infty} \int f_n d\mu.\]

Give examples to show that the condition "$g$ is integrable" cannot be removed.

They're not really two separate conditions. If you don't have (i), then you really don't have any condition at all, and you can write any example you want. The key is (ii), the integrability of the dominating function.

Consider this problem:

On [0, 2], let $f_n$ be the piecewise linear continuous function whose graph consists of the line segments from $(0,0)$ to $(\frac1n,n)$ to $(\frac2n,0)$ to $(2,0)$. Let $g(x) = \sup f_n(x)$.

Is $g$ integrable?

[quote="Kent Merryfield"]They're not really two separate conditions. If you don't have (i), then you really don't have any condition at all, and you can write any example you want. The key is (ii), the integrability of the dominating function.
[/quote]

hmm..i realized this soon after i posted it  :blush:
now i have modified the question.

A different example, this time on [$0,\infty$). Let $f_n(x) = \frac1n$ for $x\in [0,n)$ and 0 otherwise. Let $g(x) = \frac1x$ for $x>1$ and 1 otherwise. The sequence $f_n$ converges to 0 pointwise - in fact, uniformly. The function $g$ dominates the sequence, but $g$ is not integrable. We have $\lim \int f_n = 1$ while $\int \lim f_n = 0$.

This is a standard textbook example, and the problem in the previous post is a relative of a textbook example.

Let's expand the search for examples beyond the Dominated Convergence Theorem (and its partner, the Monotone Convergence Theorem).

Given a measurable function $f(x,y)$ on some nice uncomplicated product domain (like the unit square), we have that $\displaystyle \int_0^1 \int_0^1 f(x,y) dx dy = \int_0^1 \int_0^1 f(x,y) dy dx$ if either of the following conditions hold:

(i) $f \ge 0$ a.e.  (Called Tonelli's Theorem - consider it the analogue of the Monotone Convergence Theorem), or

(ii) $\displaystyle \int_0^1 \int_0^1 |f(x,y)| dx dy < \infty$ (Called Fubini's Theorem - consider it the analogue of the Dominated Convergence Theorem.)

Now: can you construct an example of a function, unquestionably measurable (it can be done as a "step function" with infinitely many steps) that is neither positive nor absolutely integrable, and for which the interchange of interated integrals fails?


As part of the solution to a problem in AoPS 2, it is necessary to solve for x in the equation

(4/3)^x = 8 :rt3: /9

To solve it, I wrote the equation in log form and slowly went about simplifying it...but the process would take too long in an actual competition. What is the quickest way to solve the equation?

Fierytycoon

I would recommend factoring it in some way to make the process quicker. Replace 8 by 4^(3/2) and sqrt(3)/9 by 1/3^(3/2), so when we take logs with base 4/3 of both sides, we immediately get x=3/2.

Thanks for the suggestion, ComplexZeta. Might take me a while to get used to these things such that I can see these kinda things as quickly as you did though.  8-)

Fierytycoon

It's all about practice. You can see stuff when you take the contest, but why would you want to rely on divine inspiration when you could have seen it all before? It's just safer to know it before.

[color=cyan]Wait, I don't understand, why is taking a logarithm difficult in that problem?  You take a log base 2, and then most of it goes away, except for a few log base 2 of 3 terms, which all go away soon afterwards.  It's like <2 minutes.[/color]

[quote]It's like <2 minutes.[/quote]

Well, if its at the end of a test such that I am getting tired and since I don't take altoids, it might well be 2 min for me. And since this is only a small part of a longer log problem, solving it may take more than the 3 min allotted for each AMC12 problem (assuming that the longer problem could potentially be on an AMC12, which looks to be the case). Gotta be as efficient as possible.

[quote="Fierytycoon"]Yeah. But I can just imagine...

Fierytycoon: I have a problem I am having trouble with. Factor
a^3x^2 + 3a^2bx^2 + 3ab^2x^2 +
b^3x^2 + 3a^2cx^2 + 6abcx^2 + 3b^2cx^2 +
3ac^2x^2 + 3bc^2x^2 + c^3x^2 + 2a^3xy +
6a^2bxy + 6ab^2xy + 2b^3xy + 6a^2cxy +
12abcxy + 6b^2cxy + 6ac^2xy +
6bc^2xy + 2c^3xy + a^3y^2 + 3a^2by^2 +
3ab^2y^2 + b^3y^2 + 3a^2cy^2 + 6abcy^2 +
3b^2cy^2 + 3ac^2y^2 + 3bc^2y^2 + c^3y^2

ComplexZeta: Oh, that's simple. Everyone knows that that expression is equal to (x + y)^2 * (a + b + c)^3. Common knowledge.

lol, just kidding with ya Simon.  :D

Fierytycoon[/quote]

Um. Right. Of course. I should remember that one. But for a problem like this (suppose some twisted testwriter put a problem like this on a test), put in 1 for each variable and then do a prime factorization. That kind of thing sometimes helps.

[color=cyan]No, on the AMC you have an average of 3 minutes.  Which is very different from having exactly 3 minutes.  Especially since the first problems are mostly very easy.  Anyhow, to suggest that you should be able to do the AMC with solving every question in three minutes, or even an average of 3 minutes, is absurd -- only very very few people do that every year, and it is almost negligeable advantage to get a 150 over a 143 or 140.5 or whatever you get for leaving a couple of them off.  At least, if you're scoring in that range anyhow, odds are very good for your making USAMO.[/color]

[quote]and it is almost negligeable advantage to get a 150 over a 143 or 140.5 or whatever you get for leaving a couple of them off. [/quote]

Yes, but this year, it would have gotten a person the advantage of an AIME problem, since the cutoff was 226. Everything counts, especially since I am not the type of guy (at least, not yet) who goes into the AIME expecting a 13 or 14.

Yeah, and about the easier problems (about #1-16), I should probably practice and find a way to do those quicker, since they are taking too long right now.

Fierytycoon

[quote="Fierytycoon"]Yeah. But I can just imagine...

Fierytycoon: I have a problem I am having trouble with. Factor
a^3x^2 + 3a^2bx^2 + 3ab^2x^2 +
b^3x^2 + 3a^2cx^2 + 6abcx^2 + 3b^2cx^2 +
3ac^2x^2 + 3bc^2x^2 + c^3x^2 + 2a^3xy +
6a^2bxy + 6ab^2xy + 2b^3xy + 6a^2cxy +
12abcxy + 6b^2cxy + 6ac^2xy +
6bc^2xy + 2c^3xy + a^3y^2 + 3a^2by^2 +
3ab^2y^2 + b^3y^2 + 3a^2cy^2 + 6abcy^2 +
3b^2cy^2 + 3ac^2y^2 + 3bc^2y^2 + c^3y^2

ComplexZeta: Oh, that's simple. Everyone knows that that expression is equal to (x + y)^2 * (a + b + c)^3. Common knowledge.[/quote]

I hope you didn't type all of that out... otherwise, you'll have sore fingers tomorrow.

No, I used mathematica. That is why (a^3)(y^2) came out to be a^3y^2.

Fierytycoon

Oh, I was wondering that a^3y^2 meant...

You lucky duck.  You have mathematica?  Oh, I guess a lot of people here do.  I'm one of the few who doesn't... :cry:


Find all positive $ a,b$ such that numbers $ \frac{a^2\plus{}b}{b^2\minus{}a}$ and $ \frac{b^2\plus{}a}{a^2\minus{}b}$ are both integers.

Look here   http://www.mathlinks.ro/Forum/viewtopic.php?t=150477
Daniel