geometry MIT college Harvard trigonometry algebra polynomial

Hi,
I am going to apply to RSI this year, but have encountered one significant issue. We are asked to name the field and sub-field that we want to study at the program. I am pretty sure that I want to do physics as my field, but I am unsure about the sub-field. The problem is that the research I have done has been all over the place-I have done some E-M work with regards to properties of charged particles, as well as some biology and fluid dynamics based research. As a result, I am unsure what I would like to pursue as my sub-field.
(The only unifying charectaristic of this work is that it has all been done with regard to questions centering around space- for example, studying how exposure to some Mars-type conditions (low pressure, higher radiation, high aridity) affect the growth of fungi and lichen spores, or trying to build a better model for fluid interactions in microgravity.)

From considering this, I have the following list of questions:
1. Would "space science" be acceptable as either a field or subfield? If so, what might the counterpart sub-field/field be? Would RSI look my choice of "space science" negatively?
2. Does there exist a list somewhere which gives acceptable fields/sub-fields?
3. Does anyone know of any source where I could gain information about scientific fields/sub-fields? Till now, I just studied what ever interesting topics came up.
4. Would RSI view an engineering-type field (For example, interstellar propulsion methodologies, or antiparticle studies) negatively as compared to "pure" sciences such as chaos theory or brane theory?

Any information you could give on this topic (or indeed any other pertaining to RSI) would be very helpful.

Thanks
sran

Hi Sran,

I went to RSI 2003. Needless to say, it was a totally *incredible* experience; I can assure you that it is unlike any other math/science camp. I'd be happy to help you with any questions you might have about your RSI application (although, as an International student, I went through a very different application process; I can, however, direct you to the appropriate person if I'm unable to answer your question.)

Concerning your question about field and subfield of research interest, I remind you that you are asked to give *two* such choices (so you may list two very different subfields of physics that interest you.) Example:

[b]Choice 1.  [/b]   Mathematics/Partition Theory
[b]Choice 2.  [/b]   Mathematics/Combinatorial Geometry

[These were the choices that I listed --- although I actually ended up with a research project in combinatorial group theory   :) .]

In both your choices, you should list 'Physics' as your field and a more specific subfield (Example: 1. Physics/Astrophysics;  2. Physics/Quantum Cosmology). It's great that you have a rich research experience in different fields of physics; in general, it will be to your advantage to demonstrate that you may flexibly adjust to whichever research problem your mentor assigns you, so it would probably be a good idea to list two very different areas of physics knowledge that interest you rather than unifying both of your choices under 'space science.' Whatever your choices are, *much* more important is the second part of Question 2, Part II in the application, 'In your fields of interest, please state the three to five most interesting questions/problems and why they are interesting to you.' (This is probably the most important part of the entire application.) This is your chance to demonstrate strong motivation by describing your previous research, stating specific (be specific!) research problems that interest you (in mathematics, these would most likely be concrete unproven conjectures), and showing an overall physics maturity.
Do plan your answers to this question carefully.

As for a list of 'acceptable subfields,' I don't think such exists. Again, it is your research statement that really matters, not your choice of subfield. Still, you may find it helpful to browse http://scienceworld.wolfram.com/physics/ or http://www.arxiv.org/ to best formulate your choices of subfield.

Concerning your question 4, the answer is absolutely 'no.' Every science is equally respected at RSI; come as you are, and list the fields that you feel most passionate about. And don't forget that you'll conduct your research at MIT  :) , the world's cutting-edge engineering center; you'd be provided with any facilities/resources you need to complete your project.

Hope this information helps; feel free to ask any further (or other) questions that come to your mind  :D . Just let me know if I else can be of assistance!

--Vesselin
RSI'03
Harvard'09

Thanks for your informative post vess!   :)

I plan on applying to RSI '05 with math as a field. For subfields, I am interested in combinatorics and number theory, but should I narrow down these fairly broad fields? Like, instead of combinatorics, should I say perhaps combinatorical game theory, or something else more narrow? Also, for the second part of this question, asking for problems I am interested in, would it be ok if I discussed a problem that I've researched that lies outside of the field?

Also, should we send RSI previous papers/research? Or just describe it?

Thanks in advance.

Vesselin,
Thank you for all your help. I really appreciate it; your post clarified a lot of questions I had. But three more, if I may ( :blush: ):

1. Do you happen to know if RSI admissions are rolling? I looked and looked but could not find this on their website.

2. For the section asking for SAT I and II scores: I have not taken any yet. Should I explain why or just say "I have not taken any"?

3. If both my fields were physics, and my subfields, say, stellar physics and gravitational wave theory, then would I describe 3-5 problems in physics, 3-5 problems in stellar physics and gravitational wave theory, or 3-5 problems from each stellar physics and gravitational wave theory?

(I'm guessing not the last one, because the application asks for questions in the fields, but just checking)

Thanks again,
sran

[b]@ tetrahedr0n: [/b]

Yes, you definitely should narrow down your subfields --- 'number theory' and 'combinatorics' are far, far too broad to adequately describe your areas of interest. Here are some sample subfield descriptions:

[i]Algebraic Number Theory[/i]; [i]Analytic Number Theory[/i]; [i]Additive Number Theory$^*$[/i]; $p$-[i]Adic Analysis[/i];  [i]Algebraic Combinatorics[/i]; [i]Combinatorial Game Theory[/i]; [i]Combinatorial Group Theory[/i]; [i]Ramsey Theory[/i];

[*  This is the area of combinatorial number theory that deals with zero-sum type, coverings of the integers with arithmetic progressions, and the geometry of subset sums; considering your combined interest in combinatorics and number theory, you'll probably find this interesting.]

>Also, for the second part of this question, asking for problems I am interested in, would it be ok if I discussed a problem that I've researched that >lies outside of the field?

The assignment says: '[b]In your fields of interest[/b], please state the three to five most interesting questions/problems and why they are interesting to you.' However, it is expected that your previous research is not completely independent of your current field of interest  :) . It is generally advisable to make use of the opportunity and briefly describe any previous research that you have conducted, especially if it employs a specific apparatus or theory that might be useful in your new (different) research. Perhaps if you tell me what exact problem did you research I could help you more?

>Also, should we send RSI previous papers/research? Or just describe it?

By all means, you should at least describe your previous research. Have you written a nicely structured research paper? If yes, then it *might* be a good idea to submit it as an additional material, although I can't give you the best possible advise at the moment (as I said, I am an International student, and I went through a rather different selection procedure. I also completed the CEE appl. forms, but only after I had already been selected; these forms were needed for the co-ordinators to find an appropriate mentor). In my Harvard application this fall, I submitted my RSI paper and a reprint of my publication as additional materials, and I suppose this has played quite a role in my ultimate admission; for the RSI applications, however, I'm not quite sure how good an idea it is to submit additional materials. I'll be sure to ask an appropriate person for better advice, don't worry  ;) ; personally, I *would* submit my paper. Again, what is it about?

[b]@ sran: [/b]

>If both my fields were physics, and my subfields, say, stellar physics and gravitational wave theory, then would I describe 3-5 problems in physics, >3-5 problems in stellar physics and gravitational wave theory, or 3-5 problems from each stellar physics and gravitational wave theory?

I would describe 3-5 problems in physics altogether, focusing on stellar physics and gravitational wave theory. However, if you have many concrete questions & problems and feel that you need more space to define your research interests, it is not a crime to break your answer into two parts, one for stellar physics and one for gravitational wave theory.

Good Luck, prospective Rickoids!  :) Always feel free to ask more RSI questions (not necessarily about admissions)  ;)

--Vesselin

Thanks vess. My paper (trig identities derived by use of polynomials and their graphs in the complex plane, in particular, cyclotomic polynomials) can be found here:

[url]http://www.artofproblemsolving.com/Resources/Papers/TrigIdent.pdf[/url]

It doesn't on face have a lot to do with combinatorics, but I'd like to write about the work in this paper, and some other questions I've pondered about cyclotomic polynomials.

Thanks for your advice.

[quote="sran"]
1. Do you happen to know if RSI admissions are rolling? I looked and looked but could not find this on their website.

2. For the section asking for SAT I and II scores: I have not taken any yet. Should I explain why or just say "I have not taken any"?

[/quote]

2. If you took the PSAT, I think it will be no problem at all --- the SAT's are optional.

1. What does 'rolling admissions' mean? (Forgive my ignorance  :) , but I'm not a U.S. student  :lol: ) Sorry, could you rephrase your question?

By rolling admissions, sran means that applications are evaluated as they come in, and spots are awarded before the deadline admissions date. In other words, people who apply early get a big advantage.

In preparing your research statement, perhaps a good idea will be to look at several [url=http://web.mit.edu/rsi/www/2004/help/examples/]example RSI papers[/url] in your field. This will give you a good understanding of what kinds of projects are conducted at RSI, and may help you formulate appropriate problems/questions. You may also see the [url=http://web.mit.edu/rsi/compendium/from-pton-acct/compcode/skel.pdf]2002[/url],[url=http://web.mit.edu/rsi/compendium/final/rsicomp2003.pdf]2003[/url] and [url=http://web.mit.edu/rsi/compendium/final/rsicomp2004.pdf]2004[/url] RSI Compendiums.

[b]Get Inside RSI. [/b] [i]RSI just might be the most fun your brain ever had in high school[/i]. Check out our amazing [url=http://photos.rsiaa.org/gallery/]photo server[/url] to sense only a small portion of the fun that awaits you at RSI. Also check out the [url=http://www.cee.org/news/newslr.shtml]CEE Newsletters[/url].

[b]Notable RSI Alumni. [/b] If you're a math person, you've probably heard the name of [url=http://www.math.ucla.edu/~tao/]Terence Tao[/url] (RSI'89). He is one of the two mathematicians who recently proved the long-standing conjecture that there exist arbitrarily long arithmetic progressions consisting solely of primes. Tao is also an IMO legend: he went to the IMO at the age of 11 (he's the youngest IMO contestant ever). To RSI he went at age 14.

[quote="vess"]
[b]Get Inside RSI. [/b] [i]RSI just might be the most fun your brain ever had in high school[/i]. Check out our amazing [url=http://photos.rsiaa.org/gallery/]photo server[/url] to sense only a small portion of the fun that awaits you at RSI. Also check out the [url=http://www.cee.org/news/newslr.shtml]CEE Newsletters[/url].

[/quote]

On which photos can we see you ?  :)

[quote="vess"]In preparing your research statement, perhaps a good idea will be to look at several [url=http://web.mit.edu/rsi/www/2004/help/examples/]example RSI papers[/url] in your field.
[/quote]

Some materials rather look as contents covered in a regular courses and it seems that it was re-invented like some of the material by [url=http://web.mit.edu/rsi/www/pdfs/papers/99/rkotrede.pdf]Rob Kotredes, Anti-aliasing of Images Using Line Sampling (1999).[/url] Nonetheless it is a very interesting article and a great performance. As he states that he is looking for efficient algorithms I want to add that for triangle scan conversion it is more efficient to use the "Triangle Sweep-Line Algorithm" than evaluating dot-products for every point to check whether it is a inside a triangle.  :)

[quote]tetrahedr0n, I was looking at your paper, and I was amazed by the level of mathematics in it. Did you take advanced mathematics courses before you did your research?[/quote]

Our mentor, Dorin Andrica, gave us quite a bit of reading (in Romanian :)) on the topic. Most of the ideas in the paper aren't very difficult - perhaps knowledge of calculus would be good, and we used a result from another paper on cyclotomic polynomials. Personally, I haven't taken anything more than a short course in abstract algebra.

[quote="orl"]
On which photos can we see you ?  :)
[/quote]

 :D  :D  :D  For example, on [url=http://photos.rsiaa.org/gallery/album44/RSI_016]this[/url] photo.  :lol:

[quote="orl"]
Some materials rather look as contents covered in a regular courses and it seems that it was re-invented like some of the material by Rob Kotredes, Anti-aliasing of Images Using Line Sampling (1999). Nonetheless it is a very interesting article and a great performance.
[/quote]

At the same time, sometimes unusually strong results arise from the research conducted at RSI. For example, [url=http://www.emis.de/journals/INTEGERS/papers/dg3/dg3.pdf]this[/url] paper by Steve Byrnes published in the INTEGERS journal is based on his RSI'02 project. (Steve is also the first place winner on the 2002 Siemens-Westinghouse Science competition.)

[b]Notable RSI Alumni. [/b] [url=http://www.people.fas.harvard.edu/~gcarroll/]Gabriel Carroll[/url] (RSI'00; Harvard'05) has won a silver and two gold IMO medals (including a perfect score), and is a four-times Putnam Fellow (among the Top 5 performers). His totally outstanding [url=http://www.people.fas.harvard.edu/~gcarroll/math/intel.pdf]RSI project[/url] placed 3rd on Intel.

[quote="vess"]
 :D  :D  :D  For example, on [url=http://photos.rsiaa.org/gallery/album44/RSI_016]this[/url] photo.  :lol:

[/quote]

Hmm. And where are the other images. BTW you do not seem to be very pleased that other people take a photo of you.  :lol:

[quote="orl"]
 BTW you do not seem to very pleased that other people take a photo of you.  :lol:[/quote]

 :D  :D  :D  You're right.  :D

[b]@ tetrahedr0n: [/b] [i]Do[/i] mention the work in your paper, as well as some of the questions that interest you about cyclotomic polynomials.

Let me add another entry to the never-ending list of

[b]Notable RSI Alumni. [/b] The first places on the 2002 Intel & Siemens science competitions were both occupied by Rickoids (as the RSI alumni are affectionately called): Jamie Rubin and Steve Byrnes (both are now Harvard undergraduates).

d_fok, I hope you at least applied!!!  I didn't have any prior research experience either, and I just got in.  *does a happy dance*

Yeah.  Of course, I didn't get in.  And they couldn't even spell my name correctly!  It can't be that hard to spell my name!  My name is not Frnak!

Congratulations to anyone who got in.

Hmm, I have not received anything = aargh.

w/e my odds were zero, anyways.

Wow - misspelling "Frank" takes some effort.

mcalderbank - my chances were zero, too.  Don't lose hope.  (And don't panic if the letter is skinny - all of them are, not just the rejection/waitlist letters!)  Good luck!

My condolences to people who applied and didn't get in/were waitlisted :-/

And thanks, tetrahedr0n.

Aargh, I just checked, and I did not get in. O well, I suppose I can always go to MathCamp. Rejected from Clay and RSI.

Congrats flierdeke!! Have fun :lol:

I was accepted   :D

How many people from AoPS got in? College Confidential probably has around 10....w

I'll count for both  :)

I got in. Blah, it was hideously frightening. I saw the envelope, it was thin, my heart sank...I decided to go ahead and open it anyway, and then screamed like a wounded beast (my mom's description) for a little while.

It's going to be interesting to go to a place where I will intellectually be at the bottom :? . But I anticipate learning a lot from all of you guys. Just wondering: is everyone at RSI usually a USAMO qual or something? I'd like to know exactly how behind I am :blush: .

Vesselin (Mr. Vesselin?), I'd like to thank you for all the help and advice you gave during the application process. I really appreciated it, and it was likely a deciding factor in my case.

Congrats to everyone who got in! I did as well :) I didn't have the same reaction, though. When I picked up the envelope, I had no idea who it was from, so I read it quickly before realizing what it actually said. Once again, congratulations to everyone who got in, and I'll see you there!

BTW, I (we) can't stress enough how helpful you were, vess. Thank you very much.

sran- don't worry, I'm not a USAMO qual  ;)  faaar from it (and that's why I don't post much on this board)..

Anyway, congrats fellow rickoids, and see you guys this summer!
I'm on AIM quite a bit if anyone wants to talk ^.^

I'm not a USAMO qualifier either (close this year, but only because I'm still technically a sophomore and get the benefit of the floor value system) - and math/comp sci is my research area for RSI!  Perhaps there are fewer people than we imagine who are both really into math and really interested in doing sciencey research over the summer?

"Perhaps there are fewer people than we imagine who are both really into math and really interested in doing sciencey research over the summer?"

I know two very excellent students who did not get into RSI.  Both are really into math, and interested in doing research over the summer.  One was quite close to qualifying for USAMO this year and in fact has qualified in the past.  The other knows several folks who didn't get in.

So I suspect the case is that (a) y'all who got in submitted wonderful applications (congratulations on this achievement!), and (b) USAMO qualification doesn't have a big correlation with admission to RSI.  I'm sure that contest success is one of many factors considered by the RSI decision-makers.

Yeah, I know someone who went to RSI 2003 and didn't make USAMO. I'm sure there are plenty our there. But I hope USAMO qualification counts for something... :lol:

When do they send the (Insert curses) detailed packet?

To those who didn't get in, believe me, RSI is certainly not the only great program out there.  I hope you find a way to have a fantastic summer.

To those who did, congrats!
I was RSI '03 (Vesselin was roommate, actually!).  Since I now go to MIT, and will be there over the summer, hopefully I will get the chance to meet some of you.

Andre'
RSI '03
MIT '08

P.S.  Vesslin, I hope you have a great summer and I look forward to meeting up with you in Cambridge this fall!


Detailed solutions are appreciated.

Workout3 #8 A cliff at Warloe Bay is 36 feet above the water. If an object is dropped from the
cliff, the formula h = 36  16t 2 will give the height above the waters surface, in feet,
of the object at t seconds from the time it was dropped. How many seconds does it
take for an object dropped from the cliff to reach the waters surface? Express your
answer as a decimal to the nearest tenth.

Warm-up7 #2 Kris has an endless supply of 4-cent stamps and 7-cent stamps for mailing a package.
What is the greatest amount of postage that cannot be affixed to a package using
these stamps?

Thank you for your help in advance.

[hide="#2"]#2 (Note: For #1, I have no idea what the equation is; there are a lot of squares, at least 4)==> We try to eliminate the numbers we can get.  All multiples of 4, all multiples of 7, all multiples of 11, are just some on the multiples we can get.  In addition, we can get all numbers 1 more than a multiple of 7 and 11 (because 4+4=7+[u]1[/u]). This means that if we can make a number that is composed of at least one 7, then we can get the number one more than it.  Our only problem would occur if we had all 4's.  However, if we have 5 4's, then we can replace them with 3 7's, which will give us that next amount.  And then, with those 7's, we can get the next numbers by replacing them with 4's, and then we will have 1 more 4 than before, so we can just keep on going.  So, after 20, we can get any number we want.  Since 16 can be gotten by 4 4's, we can't get 17.  We can get 18 with 2 7's and 1 4, and since this has a 7, we can get 19.  So, the answer is [u]17[/u[/hide]]

Thank you for the reply.

The equation is h = 36 - 16t^2. I think the question was wrong, but not sure.

These problems are hard if you don't see the trick to them.
First Problem
[hide]The problem can be simplified into:
Find the time it takes between t when $ h\equal{}36$ and t when $ h\equal{}0$
These are pretty simple algebra equations:
$ 36\equal{}36\minus{}t^2$
$ 0\equal{}36\minus{}t^2$
We see that the first equations simplifies to $ t\equal{}0$, and that makes sense because no time has elapsed right when you start. The second equation simplifies to $ t^2\equal{}36$, and $ t\equal{}6$ or $ t\equal{}\minus{}6$. Because we are solving for time, and time is not negative, we get $ t\equal{}6$ for the second equation. 6 seconds minus zero seconds equals [b]6 seconds[/b] [/hide]

Second Problem
[hide] This is as simple as using the [color=red]CHICKEN MCNUGGET THEOREM[/color]. It was invented by a guy who wanted to know what was the greatest number of Chicken McNuggets that he could NOT order, when buying them in sets of 6 and 9 and 20. The solution, [u]only when the numbers are relatively prime[/u], is mnp-mn-mp-np-m-n-p where m, n, and p are the values. Similarly, with two, its mn-m-n. To find out why this works, go to [url]http://en.wikipedia.org/wiki/Coin_problem[/url]. In your case, we are solving with $ m\equal{}7$ and $ n\equal{}4$. We get $ 28\minus{}7\minus{}4\equal{}17$[/hide]


induction
number theory
relatively prime
number theory proposed

Let $ a,\ b$ are relatively prime and $ a$ is odd. Define, for positive integer $ n$, $ a_n,\ b_n$ such that $ (a\plus{}b\sqrt{2})^{n}\equal{}a_n\plus{}b_n\sqrt{2}$.
Prove that $ a_n$ is odd and $ a_n,\ b_n$ are relatively prime for every $ n$.

we use induction
for $ n \equal{} 1$ it's obvious
let assume that for $ n \minus{} 1$ , ($ n \ge 2$) the proprety is also true so $ a_{n \minus{} 1}$ is odd and ($ a_{n \minus{} 1},b_{n \minus{} 1}$)$ \equal{} 1$
 we have ($ a \plus{} b\sqrt2$)($ a_{n \minus{} 1} \plus{} b_{n \minus{} 1}\sqrt2$)=$ aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1} \plus{}$($ ab_{n \minus{} 1} \plus{} ba_{n \minus{} 1}$)$ \sqrt2$) so $ a_{n} \equal{} aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1}$is odd
($ a_n,b_n$)$ \equal{}$($ aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1},ab_{n \minus{} 1} \plus{} ba_{n \minus{} 1}$)$ \equal{} \delta$,so we suppose that: $ \delta \ge 2$ and $ p$ is a prime divisor of $ \delta$ we should have ($ 2b^2 \minus{} a^2$)$ b_{n \minus{} 1} \equal{} b$($ aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1}$)$ \minus{} a$($ ab_{n \minus{} 1} \plus{} ba_{n \minus{} 1}$) is divisible by $ p$ we proof similary that ($ a^2 \minus{} 2b^2$)$ a_{n \minus{} 1}$ is divisible by $ p$ since we have ($ a_{n \minus{} 1},b_{n \minus{} 1}$)$ \equal{} 1$ then $ a^2 \minus{} 2b^2$ is divisible by $ p$, now notice that $ a$($ aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1}$)$ \minus{}$($ a^2 \minus{} 2b^2$)$ a_{n \minus{} 1} \equal{} 2b$($ b_{n \minus{} 1} \plus{} 2ba_{n \minus{} 1}$) is divisible by $ p$ too so $ b_{n \minus{} 1} \plus{} 2ba_{n \minus{} 1}$ is divisible by $ p$ and we have also $ b$($ aa_{n \minus{} 1} \plus{} 2bb_{n \minus{} 1}$)$ \plus{}$($ a^2 \minus{} 2b^2$)$ b_{n \minus{} 1} \equal{} a$($ ba_{n \minus{} 1} \plus{} ab_{n \minus{} 1}$)so $ ba_{n \minus{} 1} \plus{} ab_{n \minus{} 1}$ is divisible by $ p$
so we have both $ ba_{n \minus{} 1} \plus{} ab_{n \minus{} 1}$ and $ b_{n \minus{} 1} \plus{} 2ba_{n \minus{} 1}$ are divisible by $ p$ i think from here you can deduce an contradiction ..i just don't have time to compete this proof  :blush:

well... now i will complet this (ugly) proof :
from $ aa_{n \minus{} 1} \plus{} bb_{n \minus{} 1} \equiv ba_{n \minus{} 1} \plus{} ab_{n \minus{} 1} \equiv 0$($ mod p$) we can deduce that $ ab$($ a_{n \minus{} 1} \minus{} b_{n \minus{} 1}$) is divisible by p.
$ ab$ cannot be divisible by p because with $ a^2 \minus{} 2b^2$ we will have a contradiction ( easy !), so $ a_{n \minus{} 1} \equiv b_{n \minus{} 1}$ so we will get from : $ aa_{n \minus{} 1} \plus{} bb_{n \minus{} 1} \equiv ba_{n \minus{} 1} \plus{} ab_{n \minus{} 1} \equiv 0$($ mod p$):  $ a \plus{} b$ is divisible by p or $ a_{n \minus{} 1} \plus{} b_{n \minus{} 1}$ is divisible by p. but if $ a_{n \minus{} 1} \plus{} b_{n \minus{} 1}$ is divisible by p wi will get $ 2a_{n \minus{} 1}$ and $ 2b_{n \minus{} 1}$ are divible by p..this is a contradiction because we will have $ a_{n \minus{} 1}$ and $ b_{n \minus{} 1}$ are divible by p since p is odd
so we must have $ a \plus{} b$ and $ a^2 \minus{} 2b^2$ are devisible by p so $ 2ab \plus{} 2b^2 \equiv a^2 \plus{} 2ab \equiv 0$ and $ (a \plus{} b)^2 \minus{} a^2 \minus{} 2ab \equal{} b^2 \equiv 0$ so $ a$and $ b$are both divisible by p this is contradiction since they are coprime and this complete our proof  :)


There is a party in which there are a certain number of people and the winner of a game getas some money.The first gets 1 unit of currency,the second gets 1 more than the first,the third gets 2 more than the second and so on.The last person gets $19901$ units of currency.Find the number of people and totla money distributed.

P.s-I kknow the game is dumb cos the last person gets the most money :D

The statement is wrong, since if the last person got $10,000$ units of money, then the calculation gives that the number of people isn't integer. Please check it.

Yes farenhajit,corrected(I hope its right now  :? )

[hide]Writing out the first few terms in $(n,t)$, where $n$ is the $n^{th}$ term and $t$ is the term itself.

We get: $(1,1);(2,2);(3,4);(4,7);(5,11);(6,16);(7,22)...$

Organizing the $t's$, we get $1,2,4,7,11,16,22...$.

Subtracting consecutive terms, we have:

$1,2,4,7,11,16,22...\rightarrow 1,2,3,4,5,6...\rightarrow 1,1,1,1,1,1...$

Seeing this, we realize that the sequence is a $2^{nd}$ degree polynomial.

Thus, the sequence can be generated from an equation $ax^2+bx+c=y$.

Then, we choose $3$ ordered pairs and plug in.

Solving the equation, we get $a=\frac{1}{2},b=\frac{-1}{2},c=1$.

*EDIT* fixed $a,b,c$ because I solved them wrong.

The rest of the solution follows:

With $y= \frac{1}{2} x^2 - \frac{1}{2} x + 1$, we plug in 19901 for $y$ and factor to get $\frac{(x-200)(x+199)}{2} \rightarrow x=200 \rightarrow \boxed{200 people}$.[/hide]

[hide]

As for the total money spent, we can basically do the same way.

We list the ordered pairs $(x,y)$, where $x$ is the $x^{th}$ term and $y$ is the total sum of the first $x$ numbers in sequence.

The list goes as follows:
$(1,1);(2,3);(3,7);(4,14);(5,25)...$

Taking the $y$ terms, we list them and subtract the consecutive $y$ terms to get until we hit a sequence of 1's:
$1,3,7,14,25,41...\rightarrow 2,4,7,11,16...\rightarrow 2,3,4,5....\rightarrow 1,1,1,1..$

Thus, we know that this is a $3^{rd}$ degree polynomial which can be expressed in the form of $ax^3+bx^2+cx+d=y$.

Picking the first four ordered pairs, we plug them in to get four equations. Then we solve them to get:
$a=\frac16,b=0,c=\frac56,d=0$

Thus, we have the equation $\frac{x^3}{6}+\frac{5x}{6}=y$.

Since we know that there are $200$ people, we plug in $x=200 \rightarrow \boxed{1,333,500}$[/hide]

Zellex,could you xplain wjhy in the first solution you saisd that the thing is a second degree equation and it can be written in the form.....

I dunno if my solutions right,its much simpler....

[hide]A liitle bit of observation will give us:-
$1+1+2+3+....(x-1)=19901$
(Where x is the number of persons or the last person's number.Add x to both sides...)
${1+\frac{(x)(x+1)}{2}}=19901+x$
Solving we get $x=\boxed{200}$[/hide]

[hide]
Let's try to finish [b]shadysaysurspammed[/b]'s solution.

It's only left to calculate the total amount of money distributed.
If  $a_n$ denotes amount of  money received by $n$-th person, denote $S_n=a_1+\ldots+a_n$.

From $a_n=1+\frac{(n-1)n}{2}$ follows $S_n=S_{n-1}+1+\frac{(n-1)n}{2}$.
Thus we have $S_n=n+\frac{(n-1)n}{2}+\ldots+\frac{1\cdot 2}{2}$.
So we need to calculate the sum:

\begin{eqnarray*}\sum_{k=1}^{n-1}\frac{k(k+1)}{2}&=& \frac{1}{2}\sum_{k=1}^{n-1}((k+1)^2-(k+1))\\&=&\frac{1}{2}\sum_{k=2}^n(k^2-k)\\&=&\frac{1}{2}(\sum_{k=2}^nk^2-\sum_{k=2}^nk)\\&=&\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2})\end{eqnarray*}.

Finally $S_n=n+\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2})$
and plugging in $n=200$ we get $S_{200}=1333500$
[/hide]

Uh, I don't know the exact term for this method, but if there is a sequence of numbers, such as:

$1,4,9,16,25,36,49....$

We can subtract consecutive terms to get:

$(4-1),(9-4),(16-9),(25-16),(36-25),(49-36)....\rightarrow 3,5,7,9,11,13...$

We continue to do it:

$(5-3),(7-5),(9-7),(11-9),(13-11)....\rightarrow 2,2,2,2,2,2...$

Until we hit a repetition of a single digit, which in this case is $2$. Now, we count up how many times we had to repeat this process. We did it twice, so this is a $2^{nd}$ degree polynomial. To find this polynomial, we can write it as $y=ax^2+bx+c$. We can make $x$=the $n^{th}$ term, and we can make $y$ = what the $n^{th}$ term is equal to. So, we can make ordered pairs $(1,1),(2,4),(3,9)....$. The $3$ in $(3,9)$ represents that the $x$ is the $3^{rd}$ term of the sequence. The $9$ represents the $y$ value of the $x^{th}$ term of the sequence.

If you are not convinced or don't quite understand, you can try to solve these sequences and find the general fomula for it.

$1,5,14,30,55,91,140...$

$1,9,36,100,225,441....$

I might have some wrong numbers in those sequences...I made them up in my head :P. I'll post the answer if you need them.

*EDIT*

Find the $100^{th}$ term for the first sequence.

Find the $50^{th}$ term for the second sequence.

Thnx for your time zellex :) I understand it now.


Aratati ca exista $f: R\to R$ cu proprietatea ca $f(f (x))=x+a, \forall x\in R$ si $\lim_{x\to\infty}(f(x)-x)=0 \Leftrightarrow a=0$.

O intebare mai generala.... exista particularitati pentru o functie care satisface $f(f(x))=x+a$?  :?
EDIT:
1)$f(x+a)=f(x)+a$
2)am gasit pt problema funcia $f(x)=x+\frac{a}{2}$ ... deci exista functii... e unica?

Daca $\ a=0$ atunci functia identica verifica cerinta.
Presupunem $\ a\neq0$  $\ \lim_{x\to\infty}(f(x)-x)=0$ si $\ f(f(x))=x+a$.Atunci $\ \lim_{x\to\infty}f(x)=\infty$.
Pentru orice sir $\ {x_{n}}$  $x_{n}\rightarrow\infty \Rightarrow (f(x_{n})-x_{n})\rightarrow0 (*)$.
Deci $\ f(f(x_{n}))-f(x_{n})\rightarrow0 (**)$ pentru ca $\ f(x_{n})\rightarrow\infty$.
Adunand $\ (*)$ cu $\ (**)$ si folosind $\ f(f(x))=x+a$ obtinem $\ a\rightarrow0$ care este o contradictie, numarul $\ a$ fiind fixat si nenul.

[quote="spix"]Cred ca ai inteles gresit a doua conditie.... Caci  $f(x)=x+\frac{a}{2}$ verifica.
Daca $a\neq 0$ atunci $\lim_{x\to\infty}(f(x)-x)\neq 0$.[/quote]
$\ f(x)=x+\frac{a}{2}$ verifica $\lim_{x\to\infty}(f(x)-x)=0$ ??

:huh:  Conditia a doua e $\lim_{x\to\infty}(f(x)-x)=0 \Leftrightarrow a=0$ ... e cu daca si numai daca... de asta zic ca ai inteles gresit.

Eu am inteles problema in felul urmator: Aratati ca exista o functie cu proprietatile p1 si p2 daca si numai daca a este egal de 0. Si inca asta inteleg din cum este formulata problema.

$P_1$: $f\circ f(x)=x+a$
$P_2$: $\lim_{x\to\infty}(f(x)-x)=0 \Leftrightarrow a=0$
Poate m-am exprimat eu lacunar. :whistling:

$\ a$ este fixat ?
Pentru valori diferite a lui $\ a$ in $\ P_{2}$ obtinem functii diferite in $\ P_{1}$.

Enuntul probelemei l-am reprodus exact cum l-am gasit . Din pacate nu stiu autorul ,anul sau culegerea din care a fost luat caci e de pe o foaie cu mai multe probleme de acest fel cu notatia $***$.


Let $A_1A_2A_3 \ldots A_{21}$ be a 21-sided regular polygon inscribed in a circle with centre $O$. How many triangles $A_iA_jA_k$, $1 \leq i < j < k \leq 21$, contain the centre point $O$ in their interior?

[quote="Singular"]Choose 3 points in 21 choose 3 ways, say the 3 arbitrary points are PQR.

Fixing P,  Q and R are correct choices if they lie on opposite sides of PO.  This happens in 10 * 10 / 2 = 50 ways.  The number of ways to choose QR arbitrarily is 20 choose 2.

So in total we want (50/ 20C2) * 21C3

ps. i could be wrong here.[/quote]Yes, it's wrong, you double counted  :P

[hide="idea"]the circumcenter lies inside a triangle iff it is acute...so find all possible acute triangles (don't overcount since the verticies have to be chosen in order on the circle)[/hide]

This question is not solved yet and my previous comment is 50% correct. Singular's answer is not correct. What happen if you fix $P$ at $A_1$ and you choose triangle $Q=A_2$ and $R=A_{21}$? Clearly they are at opposite side of diameter but $\triangle PQR$ is no longer containing circumcenter $O$.

My idea is the following: Let $x,y,z\in\mathbb{N}$ be the 'arc-count' side-lengths of the triangle. So for example, if triangle is $A_3A_9A_{15}$, the 'arc-count' side-lengths of the triangle are 6,6,9 because it takes 6 arcs from $A_3$ to $A_9$, it takes 6 arcs from $A_9$ to $A_{15}$, and it takes 9 arcs from $A_{15}$ to $A_3$. Notice that $x+y+z=21$.

From what Altheman said, $O$ is inside triangle iff it is acute; therefore $x,y,z<11$.

WLOG we can assume $0<x\le y\le z<11$.

We have a few configurations:

7,7,7

--Equilateral triangle has symmetric sides so we have 21/3=7 triangles here containing $O$
___________________________

6,6,9
5,8,8
3,9,9
1,10,10

--These are non-equilateral isosceles triangles and since we can rotate the top vertex through each $A_i$, we have 21x4=84 isosceles triangles
________________________________

6,7,8
5,6,10
5,7,9
4,7,10
4,8,9
3,8,10
2,9,10

--These are scalene triangles and each configuration has 2 different rotational combinations: i.e. 6,7,8 and 6,8,7; as in the previous case we can rotate through each vertex $A_i$ and we have in total 21x2x7=294

My total number of triangles containing $O$ is 294+84+7=385

One last thing, please verify if this is correct...I don't know the official answer  :blush:

Pick a vertex and draw a line passing through $O$ and that vertex . Label that vertex as 1 . Label the vertex in clockwise as 2,3,... 11 . For anticlockwise direction , label it as -2,-3,... -11 .

Join 1 and 11 and let it be one of the side of the triangle . Now we join it with any other vertices from 2 to 10 , we will have triangle WITHOUT center in them . So there are a total of 9 ways . Now we join 1 with 10 and similarly there are 8 ways to form a triangle without center in them . So we proceed and get the number of triangle to be

1+2+3+...+9=45.

Symmetrically we can use vertex 2,3,...,11,-2,-3,...-11 also hence it is

45 x 21 = 945 triangle without the center .

Hence those with center is $21C3-945= \boxed{385}$

[hide=Answer for 2m+1 sides]$ \[\sum_{k = 1}^m k^{2}$[/hide] :P
[hide=Answer for 2m sides]$ (m-2)(m-1)m/3$[/hide] :P
[hide=Proof for odd number of sides]

Let there be 2m+1 sides.m>=1,m integer.
Fix a particular vertex A. The point diametrically opposite this one is symmetrically squeezed between the m-th and
m+1-th side of the polygon counting from A.
Now take the [b]k-th vertex from A[/b]. [b][k<=m][/b]Mark it [b]B[/b]. Let its [b]diametrically opposite point be C and that of A be D[/b].
[b]Notice that any choice of third vertex between C and D (on the arc on which A and B don't lie) is a required choice and all others are not. Now there are exactly k vertices between C and D. So we have k valid triangles.[/b]
So total number of triangles with fixed vertex A is $[2\[ \sum_{k = 1}^m k]/2$=$\[ \sum_{k = 1}^m k$.
We divided by 2 because for each triangle it was counted twice from its viewpoint of the 2 vertices other than A.
Now we do this sum of each $2m+1$ vertices(taking them as A) then divide by 3 upon observing that we counted each triangle thrice.
So final answer is $\[[\sum_{k = 1}^m k]  (2m+1)/3$ = $ \[\sum_{k = 1}^m k^{2}$.[/hide]  :trampoline:
[hide=Proof for even number of sides]
It is a tad easier. Proceed the same way as i did for odd number of sides just by noticing the minute differences about the diametrically opposite point.[/hide] :D


There are 8 adjacent squares. Your goal is to get from square 1 to square 8. You roll a fair 6 sided die to determine the number of squares to move. Here's what the squares say:

Squares 2,5,and 7: Go back to square 1 and wait 1 minute before rolling again

Squares 3 and 4: Go to square 8(game over)

Square 6: Go back to square 1 and wait 4 minutes before rolling again

Assuming that the time it takes you to roll and to move from square to square is neglegible, what is the expected number of minutes you will have to wait before getting to square 8?

Suppose having the die means you will gain on average r minutes.  We find that 1/2 the time we get 1+r mins, 1/3 the time we win, 1/6th the time we get 4+r mins.  So r = (1/2)(1+r) + (1/3)(0) + (1/6)(4+r)  implying r = 7/2.

If you are on 4 (or 3), you automatically go to square 8 and the game is over (you are successful). I added this in the problem to make it more clear.


Let n be a composite number. Consider sequence $(a_{i})$ such that $a_{1}$ is an odd prime and $a_{i+1}=a_{i}+n$ for all $i\in\mathbb{N}$. Prove that
$\overline{a_{1}a_{2}a_{3}a_{4}\ldots a_{n}1}$
is not a square.
(where $\overline{abc}$ means $100a+10b+c$ etc.)

More formal:
Prove that
\[ \sum_{i=1}^{n}({a_{i}\cdot 10^{(\sum_{j=i+1}^{n}{(\lfloor log{a_{j}}+1\rfloor)+1)}})+1  }\]
is not a square


function
limit
real analysis
real analysis unsolved

Find all continuous functions $f$:$[a,b]\rightarrow[a,b]$ satisfied:
     $|f(x)-f(y)|>|x-y|$ with every $x,y \in (a,b), x \neq y.$
And more:
  A continuous function $f: R\rightarrow R$ satisfied:
   $|f(x)-f(y)|>2|x-y|$ with every $x, y \in R, x \neq y.$
  Prove that $f$ has an unique fixed point.

Hmm, I don't think so! Can u see it again?

For the first problem,
I don't think there exist,
If there exist such f,
First prove f is strickly monotonic.
Wlog,it is strickly increasing
Then it is easy to prove $f(a)=a$,$f(b)=b$
then choose any c,d such that $a<c<d<b$
we have $f(d)-f(c)>d-c$
so there much have:
$f(c)-f(a)<c-a$ or $f(b)-f(d)<b-d$
this will lead to contradiction.
I hope it is right.

The second also seems to be easy.

We get that $f$ is injective (just check the definition). Since it's continuous, it follows that it is strictly monotonic. The limits in $\pm \infty$ must be $\pm \infty$ (otherwise it would contradict the hypothesis).

If $f$ is decreasing, then $f$ has a fixed point since $\lim_{x \to-\infty}(f(x)-x) = \infty$ and $\lim_{x \to \infty}(f(x)-x) =-\infty$.

Now let $f$ be increasing. Suppose the opposite.
[i]Case 1.[/i] $f(x) > x, \, \forall x \in \mathbb R$. Then,$f(0)-f(x) >-2x, \forall x < 0$, so$f(0)+2x > f(x)$. This yields$f(0)+2x > x \, \, \Longleftrightarrow \, \, f(0)+x > 0$, which doesn't hold as$x \to-\infty$.
[i]Case 2.[/i] $f(x) < x, \, \forall x \in \mathbb R$. Then,$f(x)-f(0) > 2x, \forall x > 0$, so$f(x) > 2x+f(0)$. This yields$x > 2x+f(0) \, \, \Longleftrightarrow \, \, 0 > x+f(0)$, which doesn't hold as$x \to \infty$.

So $f$ has at least a fixed point.

If it had two ($x_{1},x_{2}$), we would get an obvious contradiction $\left| f(x_{1})-f(x_{2}) \right| = \left| x_{1}-x_{2}\right|$, so we're done.

Hopefully, it's correct.

A differentiable function $f$  $/[a;b]$ satisfied: $0\leq f'(x)\leq \frac{1}{2}\forall x \in [a,b]$
 Prove that, if $f([a,b])$ is not a subset of $[a,b]$ then $f$ hasn't any fixed point in this segment.


That is right. Can we have that if $a,b,c\geq 1$ then?
\[ \frac1{a(b+c)}+\frac1{b(c+a)}+\frac1{c(a+b)}\leq\frac3{1+(abc)^{2/3}}.  \]

[quote="Lovasz"][quote="pvthuan"]$\text{Prove or disprove that if }\; 0<a,b,c\leq1\; \text{then}$
\[ \frac1{a(b+c)}+\frac1{b(c+a)}+\frac1{c(a+b)}\geq\frac3{1+(abc)^{2/3}}.  \][/quote]
Another:
We have a famous ineq:
      $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z} \geq \frac{3}{1+\sqrt[3]{xyz}}$
which is equivalent to:
       $\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca} \geq \frac{3}{1+\sqrt[3]{(abc)^2}}$
Now, we must prove:
       $\frac{1}{a(b+c)}+\frac{1}{b(c+a)}+\frac{1}{c(a+b)} \geq \frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}$
The ineq above is true:
consider        $...\frac{1}{a(b+c)} - \frac{1}{1+ab}=\frac{1-ac}{a(b+c)(1+ab)}\leq1$
because        $0<a,b,c\leq1$.[/quote]

That is also nice. Do NOT miss my next question.

but a,b,c=<1 so you can't use Jensen inequality as above


Prove that there are infinitely many pythagorean triples.

[hide]Well, there is the (3,4,5) triple, then we can expand on the fact that we can just say (3k,4k,5k).

These can be shown to be pythags.[/hide]

[hide]I used the fact that there are basic Pythagorean triples -- 3, 4, 5   ,    5, 12, 13     and so on.

Then, all the multiples of such are also Pythagorean triples. These go in the form of:  $(3n, 4n, 5n),\ (5n, 12n, 13n)$, where $k\le\infty$. Thus, there are infinitely many Pythagorean triples.[/hide]

[quote="bpms"][quote="t0rajir0u"][hide="Here they are"] $a = k(m^{2}-n^{2}), b = 2kmn, c = k(m^{2}+n^{2})$ [/hide]

[b]Extension:[/b]  Prove that the above expression generates all Pythagorean triples.[/quote]
[hide]$(m^{2}-n^{2})k+2kmn=(km^{2}+kn^{2})^{2}$
$k^{2}m^{4}+k^{2}n^{4}-2k^{2}m^{2}n^{2}+4k^{2}n^{2}m^{2}=k^{2}m^{4}+k^{2}n^{4}+2k^{2}n^{2}m^{2}$
Simplying and canceling terms, we get
$0=0$ which is true for all reals. So the expressions work.
[/hide][/quote]

No, you've only proved that if three numbers are of that form, then they are Pythagorean triples. The question is the converse: if three numbers form a Pythagorean triple, then they can be written in that form. ;)

[hide]Take any odd number x.  Then since x is odd, $x^{2}$ can be written as $2m+1$.  Then we have that since $(m+1)^{2}-m^{2}=2m+1$, $x, m, m+1$ is a pythagorean triplet.[/hide]


Να μια πολύ ενδιαφέρουσα   :diablo:  Αν είναι $ a,b,c$ πλευρές τριγώνου με εμβαδό $ K$ και $ u,v,w$ θετικοί πραγματικοί να δείξετε ότι:

$ \frac{ua^4}{v\plus{}w}\plus{}\frac{vb^4}{u\plus{}w}\plus{}\frac{wc^4}{v\plus{}u}\geq8K^2$ Πότε ισχύει η ισότητα????  :lol:

Ειναι μια απ τις γεωμετρικες ανισοτητες Τσιντσιφα

Έχετε καμιά λυση?


You must use English in this forum. Here is your problem:

Let  $a,b,c$ be sides of a non-obtuse triangle  . Prove that
\[\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3 \]

[quote="nguyenphihung"]Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng .
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=130812


How many numbers can be expressed as a sum of two or more distinct elements of the set {0,1,2,4,8,16}?

Since you can use a 0, you can change to problem to finding how many numbers can be expressed with [i]one[/i] or more distinct elements from $ \{1, 2, 4, 8, 16\}$.

Notice that these are all powers of two. Thus, this is basically the same as binary (or base two), where we are allowed to create numbers that use only some of the first five smallest bits (which are binary 'digits'). Thus, we can create every number from $ 1$ to $ 2^5 \minus{} 1 \equal{} 31$, so the answer is $ \boxed{31}$.

If you don't know how base two works, here's a short explanation:
We usually use numbers that are in base ten. For example, the number $ 1204$ is actually
$ 1 \cdot 10^3 \plus{} 2 \cdot 10^2 \plus{} 0 \cdot 10^1 \plus{} 4 \cdot 10^0$.

Base two is quite similar, except the only 'digits' are now $ 0$ and $ 1$, and we multiply by powers of two instead of powers of ten in the expression above. For example, $ 1011_2$ is the same as

$ 1 \cdot 2^3 \plus{}0 \cdot 2^2 \plus{} 1 \cdot 2^1 \plus{} 1 \cdot 2^0 \equal{} 11_{10}$.


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

I asked this question. A lot. I think the answer is that don't do AMC 12 problems right now. Do AIME or harder problems and keep doing those. Once its AMC 12 time, then start doing AMC 12 or you'll run out the AMC 12's before you can say "Math". And, doing AIMEs helps a lot for AMC 12 in my opinion. Also, MAKE SURE YOU UNDERSTAND GEOMETRY

I did chapter 7 in intro to geometry again my score went up 20 points. Also, if you can then get challenging problems in geometry (its a great book)

[quote="AIME15USAMO"]Okay, I'm getting ready to start preparing for AMC 10/12 and AIME.

I average around a 130 on AMC 10 and about 90-110ish on the AMC 12 and only like a 3-5 on the AIME.

I'm thinking about taking the AMC 12 problem solving series and AIME problem solving series through AoPS. Has anyone taken these, and if so, are they useful? I'm also reviewing AoPS volumes 1 and 2.

Is there anything else I can do? I'm open to anything (classes, books etc.). Don't suggest olympiad problems because those are a little too hard for me right now.[/quote]

Don't bother with AMC 10 as much, because 130 is pretty good. You might want to work on the weak parts, though. I might take AMC 12 series, then AIME also like you. Studying AoPS Volumes 1 and 2 are good, but you might have to pay really close attention to geometry and trig in Vol.2 because that can get hard. Also you might want part of Intermediate Algebra before Vol. 2

I found that trig in AoPS volume 2 quite difficult, but DEFINITELY WORTHWHILE.

Also(sorry if I'm adding another question to be answered :oops: ), I need some dire help with AMC10... I find myself scoring [b]higher[/b] on AMC12's then 10's...

I've only done a couple practice tests.... Would you say 1 AMC 10 a day and 3 AIME problems a day is good?

Do you mean 1 amc 10 test? If so, that's pretty good. If you want, you can supplement it with

these mock AMCs: http://www.artofproblemsolving.com/Wiki/index.php/Mock_AMC#Mock_AMC_.231

these PMWCs: http://www.artofproblemsolving.com/Wiki/index.php/PMWC_Problems_and_Solutions

and any math books you have around the house.

Also, when you say higher, around what scores do you get for both? It would help me decide what you should do better.

@bbnd:That's pretty much what I do except I also started reviewing AoPS volume 1. It's easy but reviewing doesn't hurt and I've realized a thing here and there that I didn't get the first time going through the book.

Would ACoPS or PSS be good reads for me at my level? If so, how long are these (it doesn't matter, I have time. Just wondering)?

I was considering reading those two books as well... Unfortunately, those books are pre-olympiad level-olympiad level and therefore don't help unless you make USAMO.

I say review AoPS volume 2. It has the material needed to score 5+ on the AIME and 120+ on the AMC12.

[quote="AIME15USAMO"]After reviewing volume 2 will I have the math skills needed to be able to understand ACoPS without struggling too much?[/quote]

No, not necessarily. It depends on the person.
And you're [i]supposed[/i] to struggle with the problems in ACoPS. That's the learning process.

Yes, struggle as in think it's difficult, not struggle as in not understand a single word it says and walk away from the book. I realize I need to challenge myself.

As for my question before, I meant to say, does AoPS volume 2 have the essential skills needed to learn from the ACoPS book?

[quote="batteredbutnotdefeated"]I was considering reading those two books as well... Unfortunately, those books are pre-olympiad level-olympiad level and therefore don't help unless you make USAMO.
[/quote]

Although ACoPS is very difficult and pre-olympiad, it will help a lot with AIME if you're one to consistently score 5+ on it.

And what about PSS? Is that Olympiad level?

Yeah. From what I've been told, it's harder than ACoPS

Would that be a good read to do good at USAMO, assuming I make USAMO? Or should I just focus on AMC12/AIME prep until the AIME is over? I don't think I really have much of a chance of making even Red MOP so maybe this would be a better approach for now?


[hide](3/14, 3/16) I put it in the form of -

5(1/x)+4(1/y)=-2

8(1/x)-4(1/y)=16

added, and solved for x then substituted x back in to solve for y.[/hide]

[hide] (3/14, 3/16)...

I set the systems up like this:

-5y+4x=-2xy

2y-x=4xy

Solving this system I got 3y=14xy.... thus getting x=3/14. Substituting for x, I found y=3/16

[/hide]


Umm... Hello everyone.
I'm new here.
I was just looking around and somehow got here..
I saw some posts about MathPath, and it sounded really cool..
Can you guys tell me information like how you can get in, the level of difficulty, costs,and bunch of other stuffs..?

they have it on the website.

[url]http://www.mathpath.org/camp.htm[/url]

everything you need to knoiw is there

Eh, it was fun, but I highly doubt I can go again

They have a bunch of classes running simultaneously, so you choose the one appropriate for your skill level. These run during the day time, after dinner is basically free time.

Should I note that no electronic games are allowed, as far as I'm concerned


[quote="AndrewTom"]Factorise $ \ 2x^2 \minus{}3xy \minus{}2y^2 \plus{}x \plus{}13y \minus{}15.$[/quote]

As $ \ 2x^2 \minus{}3xy \minus{}2y^2 \equal{}(2x\plus{}y)(x\minus{}2y)$

$ \implies \ 2x^2 \minus{}3xy \minus{}2y^2 \plus{}x \plus{}13y \minus{}15\equal{}(2x\plus{}y\plus{}\lambda)(x\minus{}2y\plus{}\mu)$

Now by equating coefficient $ \ x$ and $ \ y$ on both side, we get $ \ \lambda \plus{}2\mu\equal{}1$ and $ \ \minus{}2\lambda \plus{}\mu\equal{}13$

$ \implies \mu\equal{}3,\ \lambda\equal{}\minus{}5$

$ \implies\ 2x^2 \minus{}3xy \minus{}2y^2 \plus{}x \plus{}13y \minus{}15\equal{} (2x\plus{}y\minus{}5)(x\minus{}2y\plus{}3)$

Thanks makar. That's exactly what I did but wondered if there is a neater way of doing it.

[quote="AndrewTom"]Factorise $ \ 2x^2 \minus{} 3xy \minus{} 2y^2 \plus{} x \plus{} 13y \minus{} 15.$[/quote]

Alternate:

Consider equation: $ \ 2x^2 \minus{} 3xy \minus{} 2y^2 \plus{} x \plus{} 13y \minus{} 15 \equal{} 0$

$ \implies 2x^2\plus{}(1\minus{}3y)x\minus{}(2y^2\minus{}13y\plus{}15)\equal{}0$

$ \implies\ x \equal{} \frac {3y\minus{}1 \pm \sqrt{(3y\minus{}1)^2\plus{}8(2y^2\minus{}13y \plus{}15)}}{4}$

$ \implies\ 4x \minus{}3y\plus{}1 \equal{}\pm \sqrt{25y^2\minus{}110y\plus{}121}$

$ \implies\ 4x \minus{}3y\plus{}1 \equal{}\pm \sqrt{(5y\minus{}11)^2}$

$ \implies\ 4x \minus{}3y\plus{}1 \equal{}\pm (5y\minus{}11)$

$ \implies\ 4x \minus{}8y\plus{}12 \equal{}0$ or $ \ 4x\plus{}2y\minus{}10\equal{}0$

$ \implies\ x \minus{}2y\plus{}3 \equal{}0$ or $ \ 2x\plus{}y\minus{}5\equal{}0$

$ \implies 2x^2 \minus{} 3xy \minus{} 2y^2 \plus{} x \plus{} 13y \minus{} 15\equal{} (x \minus{}2y\plus{}3)(2x\plus{}y\minus{}5)$


calculus
integration
calculus computations

Let $ x \equal{}\frac{1}{1\minus{}t^{2}}$,

[hide="and so"]$ x\minus{}1 \equal{}\frac{t^{2}}{1\minus{}t^{2}}$, $ x' \equal{}\frac{2t}{(1\minus{}t^{2})^{2}}$ and

$ P \equal{}\int\frac{1}{x\sqrt{x}\sqrt{x\minus{}1}}\,dx \equal{}\int2\,dt \equal{} 2t\plus{}C \equal{} 2\sqrt{\frac{x\minus{}1}{x}}\plus{}C$.
[/hide]

[hide="Solution"]
$ \int\frac{dx}{x^{2}\sqrt{\frac{x\minus{}1}{x}}}$ subs. $ \frac{x\minus{}1}{x}\equal{}t^{2}$, and we'll get: $ x\equal{}\frac{1}{1\minus{}t^{2}}$, so $ dx\equal{}\frac{2t}{(1\minus{}t^{2})^{2}}dt$, hence:
$ \int\frac{1}{\frac{1}{(1\minus{}t^{2})^{2}}\cdot t}\cdot\frac{2t}{(1\minus{}t^{2})^{2}}dt\equal{}2\int dt\equal{}2t\plus{}C\equal{}2\sqrt{\frac{x\minus{}1}{x}}\plus{}C$[/hide]


limit
real analysis
real analysis unsolved

Suppose that $ (x_n)_{n\in\mathbb{N}}\subset\mathbb{C}$ and there exists a constant $ \gamma$ such that $ |x_n|\leq \frac {\gamma}{n}$ for all $ n\geq 0$. Denote by $ s_n\equal{}\sum_{i\equal{}0}^{n}x_i$. Show that if
\[ \frac {1}{n \plus{} 1}\sum_{i \equal{} 0}^{n}s_i \rightarrow s \in \mathbb{C},
\]
as $ n \rightarrow \infty$, then $ s_n\rightarrow s$, as $ n\rightarrow\infty$.

maybe im missing something but

$ |x_n| \le \frac{\gamma}{n} \implies \lim_{n \to \infty} |x_n| \equal{}0 \implies x_n \to 0$

Sorry, you are right, that was stupid. I edited with a statement that makes more sense.
EDIT : it is even correct!

The right way to say "converse-like of ... summation theorem" would be to call this a [b]Tauberian theorem.[/b]

I've seen this particular Tauberian theorem attributed to Hardy and Littlewood. Note that the "Tauberian condition" is that $ x_n\equal{}O(n^{\minus{}1}),$ which is quite a bit different in practice from giving the condition as $ x_n\equal{}o(n^{\minus{}1}).$

Ranking a few Tauberian theorems by difficulty of the proof:

(easiest)

Cesaro summation, $ x_n\equal{}o(n^{\minus{}1}).$

Abel summation, $ x_n\equal{}o(n^{\minus{}1}).$ (The original Tauber's Theorem.)

(gap)

Cesaro summation, $ x_n\equal{}O(n^{\minus{}1}).$ (Hardy-Littlewood)

Abel summation, $ x_n\equal{}O(n^{\minus{}1}).$ (Littlewood)

(hardest)


function
limit
inequalities
real analysis
triangle inequality
real analysis unsolved

If $ (f_{k})$ is a sequence of functions which are upper semicontinuous (usc) at $ \textbf{x}_{0}$ and which converge uniformly near $ \textbf{x}_{0}$, show that $ \lim_{k}f_{k}$ is usc at $ \textbf{x}_{0}$.

My attempted solution: Since we have uniform convergence near $ \textbf{x}_{0}$, for all $ \epsilon>0$ there exists $ K\geq 1$ such that for all $ \textbf{x}$ near $ \textbf{x}_{0}$ and $ k\geq K$, $ \left|f_{k}(\textbf{x})\minus{}f(\textbf{x})\right|\leq\epsilon$.  Then the triangle inequality gives $ \left|f_{k}(\textbf{x}_{0})\right|\minus{}\left|f(\textbf{x}_{0})\right|\leq\left|f_{k}(\textbf{x}_{0})\minus{}f(\textbf{x}_{0})\right|\leq\epsilon$.  Rearranging gives $ \left|f(\textbf{x}_{0})\right|\geq\left|f_{k}(\textbf{x}_{0})\right|\minus{}\left|f_{k}(\textbf{x}_{0})\minus{}f(\textbf{x}_{0})\right|\geq\left|f_{k}(\textbf{x}_{0})\right|\minus{}\epsilon$.  Since each $ f_{k}$ is usc at $ \textbf{x}_{0}$, $ f_{k}(\textbf{x}_{0})\geq\limsup_{n\rightarrow\infty}f_{k}(\textbf{x}_{n})$, where $ \textbf{x}_{n}$ is a sequence near $ \textbf{x}_{0}$ converging to $ \textbf{x}_{0}$.  Then we can replace this above to get that $ \left|f_{k}(\textbf{x}_{0})\right|\minus{}\epsilon\geq\limsup_{n\rightarrow\infty}f_{k}(\textbf{x}_{n})\minus{}\epsilon$.  Hence we get that $ \left|f(\textbf{x}_{0})\right|\plus{}\epsilon\geq\limsup_{n\rightarrow\infty}f_{k}(\textbf{x}_{n})$.  As $ \epsilon\rightarrow 0$, $ f_{k}$ approaches $ f$ so $ f(\textbf{x}_{0})\geq\limsup_{n\rightarrow\infty}f(\textbf{x}_{n})$.  Hence $ f$ is usc at $ \textbf{x}_{0}$.

My other question is if we have $ f<\infty$ and usc on $ [a,b]$, how can we find continuous $ f_{n}$ on $ [a,b]$ such that $ f_{n}$ decrease to $ f$?

I think $ 2\epsilon$ should appear somewhere in your proof, because both the value at $ x_0$ and the limsup could be off by $ \epsilon$ either way.
I would put it like this: for large $ k$
$ f(x)\minus{}\epsilon \le f_k(x) \le f(x)\plus{}\epsilon$ in some neighborhood of $ x_0$.
It follows that
$ \limsup_{x\to\ x_0}f(x)\minus{}\epsilon \le \limsup_{x\to\ x_0}f_k(x) \le \limsup_{x\to\ x_0}f(x)\plus{}\epsilon$.
Combine the above to obtain
$ f(x_0)\ge f_k(x_0)\minus{}\epsilon \ge \limsup_{x\to\ x_0}f_k(x) \minus{} \epsilon
\ge \limsup_{x\to\ x_0}f_k(x) \minus{} 2\epsilon$

For the second question, try something along the lines of $ f_n(x)\equal{}\sup_y (f(y)\minus{}n|x\minus{}y|)$.

Why does the function $ f_{n}(x)\equal{}\sup_{y}(f(y)\minus{}n\left|x\minus{}y\right|)$ necessarily converge downwards to $ f$?  And how did you come up with it?


i'm a little confused on writing electron configurations for some elements. i can see why iron is $\text{[Ar]}\, 4s^{2}\, 3d^{6}$ but why is lead $\text{[Xe]}\, 6s^{2}\, 4f^{14}\, 5d^{10}\, 6p^{2}$?

in general, how do you come up with the configuration of a higher atomic number element?

There are some rules, and there are some exceptions to those rules.

This is the rule:

http://everything2.com/index.pl?node_id=953084&lastnode_id=0

Several elements, such as copper (?) have configurations that break this rule because life is complicated.


algebra
polynomial
linear algebra
matrix
induction
topology

If $ f_1, \ldots, f_n : \mathbb C^n \to \mathbb C$ are polynomials such that $ (z_1, \ldots, z_n) \to (f_1(z_1), \ldots, f_n(z_n))$ extends to a continuous map $ \widehat f : S^{2n} \to S^{2n}$, then is the degree of $ \widehat f$ equal to the product of the degrees of $ f_1,\ldots,f_n$?

[color=red][b]EDIT:[/b][/color] [color=darkred]This is not true for all such polynomials.[/color] The question is, how can we characterize the polynomials for which this is true? Perhaps all polynomials should be of different degree?

[quote="Third Edition"]You could try the condition that every $ f$ is a product of pairwise different irreducible polynomials and that every pair of the $ f$'s does not have common irreducible factors. In this case I suspect that $ \hat{f}$ is generically a topological covering map. Then count the elements of the fibers of $ \hat{f}$.[/quote]

This is not enough, because the map $ (z_1, z_2) \to ((z_1 \minus{} 1)(z_1 \minus{} 4), (z_1 \minus{} 2)(z_1 \minus{} 3))$ is such and has degree zero. Perhaps we should add the requirement that $ f_i \plus{} c_i$ are linearly independent for any choice of $ c_i$. (Because if $ f_1, \ldots, f_n$ are nonconstant and good, the system $ f_i \plus{} c_i \equal{} 0$ would always have a solution.)

[color=gray]I was thinking along these lines; this is a special case but perhaps an interesting result.[/color]

Call a continuous map $ h = [h_1, \ldots, h_n] : \mathbb C^n \to \mathbb C^n$ [b]boxed[/b] if there are nonzero $ c_i \in \mathbb C$, called the [i]coefficient numbers[/i] of $ h$, such that for any sequence $ \{\mathbf z_k\}$ of numbers in $ \mathbb C^n$ tending to infinity, the sequence $ \{h (\mathbf z_k)\}$ tends to infinity and, if $ \{z_{ki}\}$ also tends to infinity, the sequence $ \{h_i (\mathbf z_k) z_{ki}^{ - e_i}\}$ tends to $ c_i$. Here $ \mathbf z_k = [z_{k1}, \ldots, z_{kn}]$. The [i]elevation numbers[/i] of $ h$ are the integers $ e_i$.

Every boxed map $ h : \mathbb C^n \to \mathbb C^n$ extends to a continuous map $ \widetilde h : S^{2n} \to S^{2n}$ that sends the one-point compactification of $ \mathbb C^n$ to itself. This extension induces a homomorphism of the $ 2n$th homology group of $ S^{2n}$ to itself, and since this group is $ \mathbb Z$, any such homomorphism is multiplication by an integer, which we call the [b]degree[/b] of $ h$, and denote $ \deg h$. Since homotopic maps induce the same homomorphism, if the extensions of two boxed maps are homotopic, these maps have the same degree.

As the extension of a boxed map $ h$ is homotopic to the extension of $ h [z_1, \ldots, z_n] = [z_1^{e_1}, \ldots, z_n^{e_n}]$ through the extensions of the maps
\[ [z_1, \ldots, z_n] \to (1 - t) h + t \left[c_1 z_1^{e_1}, \ldots, c_n z_n^{e_n}\right]
\]
and therefore $ \deg h = \deg \widehat h$. Here $ c_i$ are the coefficient and $ e_i$ the elevation numbers of $ h$.

How can we find the degree of $ \widehat h$? Consider the subspace $ A \simeq S^1 \times S^{2n - 2} \subset S^{2n}$ consisting of all $ (z_1, \ldots, z_n)$ such that $ |z_1| = 1$, in addition to the compactification point. It is a basic result in homology theory that there is the natural exact sequence
\[ \cdots \longrightarrow 0 \longrightarrow \mathcal H_{2n} (S^{2n}) \stackrel{j_{\ast}}{\longrightarrow} \mathcal H_{2n} (S^{2n}, A) \stackrel{\partial}{\longrightarrow} \mathcal H_{2n - 1} (A) \longrightarrow 0 \longrightarrow \cdots
\]
That this sequence is natural means the diagram obtained when $ \widehat h_{\ast}$ sends this sequence to itself is commutative. Since the groups are $ \mathbb Z$, $ \mathbb Z \oplus \mathbb Z$ and $ \mathbb Z$, respectively, the homomomorphism $ \widehat h_{\ast} : \mathcal H_{2n} (S^{2n},A) \to \mathcal H_{2n - 1} (S^{2n},A)$ is multiplication by the matrix $ \begin{pmatrix} \deg \widehat h & 0 \\
0 & \deg \widehat h\end{pmatrix}$, and then the homomorphism $ \widehat h_{\ast} : \mathcal H_{2n - 1} (A) \to \mathcal H_{2n - 1} (A)$ is multiplication by $ \deg \widehat h$. Since the map $ \widehat h$ is the composition of the extensions of the maps $ [z_1, \ldots, z_n] \to [z_1^{e_1}, z_2, \ldots, z_n]$ and $ \widehat h_0 [z_1, \ldots, z_n] = [z_1, z_2^{e_2}, \ldots, z_n^{e_n}]$, and the degree of the first map is $ e_1$, we have that $ \deg \widehat h = e_1 \cdot \deg \widehat h_0$. By induction, the degree of $ \widehat h$ is $ e_1 \cdots e_n$, and therefore the degree of $ h$ is also $ e_1 \cdots e_n$. We have proven the following theorem.

[color=darkred][b]Theorem.[/b] [i]If $ h$ is a boxed map with elevation numbers $ e_1, \ldots, e_n$, then the degree of $ h$ is $ e_1 \cdots e_n$.
[/i][/color]

[hide="Applications of the theorem."]
Notice that if $ h$ is not surjective, then $ \widetilde h$ is also nonsurjective, say, it does not contain some point $ P$ in the image. Then $ \widetilde h$ factors as $ S^{2n} \to S^{2n} - P \hookrightarrow S^{2n}$, hence the induced homomorphism $ \widetilde h_{\ast}$ factors as $ \mathcal H_{2n} (S^{2n}) \to \mathcal H_{2n} (S^{2n} - P) \to \mathcal H_{2n} (S^{2n})$. Since the group in the middle is zero, the composition, and hence the degree of $ h$, must be zero too. From this we conclude that [color=red][i]nonsurjective boxed maps have degree zero.[/i][/color] Or, in order to prove that a map is surjective, it is enough to prove that its degree is nonzero.

Consider, for example, the following problem.
[b]
Problem 1.[/b] [i]Prove that if $ f_1, \ldots, f_n$ are nonconstant polynomials (with complex coefficients) and $ g_{ij}$ are any polynomials (also with complex coefficients), and if the polynomials $ h_{ij}$ are such that $ \deg h_{ij} < \dg f_k$ for any $ i,j,k$, then the system of $ n$ equations
\[ f_k (z_k) = \sum_{i = 1}^{n} h_{ki} (z_k) g_{ki} (2^{z_i - \overline{z_i}})
\]
has a solution.[/i]
[/hide]

[b][i]Corrections or changes are welcome.[/i][/b]


What is the remainder when 2 ^ 2006 is divided by 17 ?

(A)0 (B) 1 (C) 5 (D) 11 (E) 13

[hide="Solution"]By FLT, $ 2^{16}\equiv 1\bmod 17$, and $ 2006\equiv 6\bmod 16$, so $ 2^{2006}\equiv 2^6\equiv \boxed{\textbf{(E)}\ 13}\bmod 17$.
[/hide]

[quote="AdithyaGanesh"]What is this "FLT?"   :blush:[/quote]

Fermat's Little Theorem

[quote="AdithyaGanesh"]What is this "FLT?"   :blush:[/quote]
Basically Fermat's Little Theorem states that if we have two numbers $ a$ and $ p$ are coprime then $ a^{p\minus{}1}\equiv 1\bmod{p}$

Sorry about that. By FLT I meant Fermat's Little Theorem (as multiple people have stated already).

FLT, Euler's Totient Theorem, and Carmichael's Theorem are the big three that apply to modular arithmetic problems involving numbers raised to high powers.


Please define the following terms:

1, 2, e, 4, 5, 6, 7, 8, +, =, equal

the biggest problems with these is that they do stuff and get an answer, but you could do other stuff and get a different answer. So, unless you can nicely define the terms above, this is probably a dumb puzzle.

If 1+2=e, 4+5=6, what does 7+8 equal?

does this have to  do  with the keyboard?

3 could also mean "E" as in L337Sp34k. I'm not sure if it fits into the puzzle, though. I can't imagine what 9 would be.

Along that line of thinking:

Then the answer would simply be 9.  (or perhaps, q.  It sort of looks like a 9.)

Using the keyboard explanation and the leet explanation, the answer would be p, as it is like 9 and is after 9 on the keyboard.

[quote="Arrange your tan"]Also, define what is "not a trick question."[/quote]

There is a certain pattern with each of the answers. Try to figure it out.\

BTW, the answer is a number.

Well, I think the question should say "1+2=E", or perhaps "1+2=ɛ". The answer is turned upside down. Therefore, 7+8=6.

98[size=100]㆚[/size][size=134]ɛɛ[/size]

lol, I got confused looking at 4+5=6. 51 doesn't look that much lisk 15 upside down, but I can't find anything much better. Sl, perhaps? Alternatively, ΛX.

Lingo was right. When flipped upside down on a TI-34, 1+2=e, and 4+5=6. 7+8=15, but when the calculator is flipped upside down, it is 51.

No!

lotofmath, as I thought,

you posted a FLAWED problem.

An upside-down result of entering "1 + 2 = " gives
an upside-down "3", not an "e".

Also, lingomaniac88 is not correct, because he posted
"51" and not an upside-down "15", which are NOT the
same things.

[quote="lotsofmath"]When flipped upside down on a . . .  calculator[/quote]

The puzzle might not be perfect, but the "15" "51" does work.

[quote=Arrange your tan]An upside-down result of entering "1 + 2 = " gives
an upside-down "3", not an "e".[/quote]

It may be an upside down 3, but it looks like an E. 7+8 is put in with the calculator right-side-up. I get the answer of 15, and then flip the calculator over. It now reads "51".

No, now you're partially changing your story.

You can't go with "E" when you used "e" in your problem
headline.  That's a bait-and-switch tactic.

STop Fighting!
its just a dumb problem.......

If he said E, then it would have been obvious. THIS IS A PUZZLES AND BRAINTEASERS FORUM FCOL!

Arrange your tan,

Why must you keep arguing with every single thing?????????? The point of this forum is NOT to go around with your nose in the air and doing everything possible to argue!


conics
projective geometry
geometry theorems
geometry

Please can anyone explain me something aout the Polar point, please.

Davron

[b]Theorem. [/b]Given a Conic $\zeta$ and a point $P$ in its plane, let $\{ A, B\} \subset \zeta$ such that $A, B, P$ are colinear, the locus of the harmonic conjugate of $P$ wrt $\overline{AB}$ is a line $p$.

[b]Definition.[/b] $p$ is the [i]polar[/i] of $P$ wrt $\zeta$ and $P$ is the [i]pole[/i] of $p$ wrt $\zeta$.


algebra
polynomial
articles
complex numbers
imaginary numbers

can we have imaginary number base?
a polynomialz root is also a base? right ?
if it is a base then we can have imaginary bases cant we?

can someone help me with finding a useful link for learning number bases please?

anyone ?
or imaginary base doesnt exist?

Sure, you can have a base with almost any number, although roots of unity are very limiting.

You may want to read [url=http://en.wikipedia.org/wiki/Non-standard_positional_numeral_systems]this article[/url].

A specific example of a number system with an imaginary base is [url=http://en.wikipedia.org/wiki/Quater-imaginary_base]base 2i, aka "Quarter-Imaginary Base"[/url]

[b]Related Problem:[/b] Let $ \omega \equal{} e^{i \theta}  : \frac{\theta}{\pi} \not \in \mathbb{Q}$. If we are allowed to place only 1 or 0 in each place value, can we represent the entire complex plane? What if we can also use -1? What if we can use [i]any[/i] integer?


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

Is there any efficient and effective method to check whether a polynomial in $\mathbb{Z}[x]$ has a quadratic factor?

Show that the polynom $x^5-\alpha x-1$ has a quadratic factor if and only if $\alpha= -1$

We can restrict to finitely many possibilities by restricting the first and last coefficients to divide the leading and trailing terms, then bounding the middle coefficient with simple estimates of root size. There's no way this will be as efficient as the system of checking for linear factors, but it's something. More is possible, but it would be very hard to generalize.

For your specific problem, let's test the two (families of) possibilities, $x^2+bx+1$ and $x^2+bx-1$.
$x^5+ax-1=\left(x^2+bx+1\right)\cdot\left(x^3-bx^2+(b^2-1)x-1\right)+ (-b^3+2b+1)x^2+(-b^2+b+1+a)x$
From the $x^2$ term, $b=-1$. From the $x$ term, $a=b^2-b-1=1$. We have $x^5+x-1=(x^2-x+1)(x^3+x^2-1)$, and this is the only factorization on this side.

$x^5+ax-1=\left(x^2+bx-1\right)\cdot\left(x^3-bx^2+(b^2+1)x+1\right)+ (-b^3-2b-1)x^2+(-b^2+b-1)x$
Since $b^3+2b+1=0$ has no integer roots, there are no factorizations here and we are done.


A small block of mass starts from rest at the top of a globe of radius R. At what angle theta does it slide off the surface of the globe? Assume the system is frictionless.

OK, so it seems relatively easy but I have no idea how to do this. help?

:) thanks!

Is the globe fixed??

I think this has been discussed here [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=44371]www.mathlinks.ro/Forum/viewtopic.php?t=44371[/url]

Sorry if I am wrong!!!


Δίνοντα οι πίνακες $ A \equal{} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}$ και $ B \equal{} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix}$.

Αν $ A^{n_1}B^{n_2} \ldots A^{n_k} \equal{} I$, όπου $ n_1,\ldots,n_k \in \mathbb{Z}$, να δειχθεί ότι $ n_1 \equal{} n_2 \equal{} \ldots \equal{} n_k \equal{} 0$.

Αν $ A^{n_1}B^{n_2} \ldots B^{n_k} \equal{} I$, όπου $ n_1,\ldots,n_k \in \mathbb{Z}$, να δειχθεί ότι $ n_1 \equal{} n_2 \equal{} \ldots \equal{} n_k \equal{} 0$.

(Υπάρχει μια σχετικά σύντομη απόδειξη.)

Επειδή είναι δύσκολη,

[hide]
Παίξτε ping pong...
[hide]  [geogebra]9b1959de36ac7ab6157d547c2da34c387edff1ee[/geogebra] [/hide]
[/hide]

[quote="mathsforfun!"]Φιλε μου δεν το εξηγεις λιγο γιατι δεν το καταλαβα με το πιγκ πογκ και το γραφημα...Δεν ειμαι ιδιαιτερα εξοικειωμενος στην Αλγεβρα αλλα αν μπορουσες να το εξηγησεις λιγο παραπανω θα με βοηθουσες αρκετα .Ευχαριστω πολυ[/quote]

Κοίταξε που στέλνουν οι πίνακες Α,Β τα σημεία που βρίσκονται στο μπλε και κόκκινο μέρος.

:blush: ,ναι σωστα , καταλαβα...ωραια ασκηση

Η μέθοδος απόδειξης ονομάζεται ping pong lemma. Παίρνεις ένα κόκκινο σημείο, ο Α^ν για ν/=0 το στέλνει σε μπλε, μετά το Β^μ το στέλνει πίσω σε κόκκινο κοκ.


[quote="ineq"]Let $ a,b,c>0.$ Prove:$ (a\plus{}b)^c\plus{}(b\plus{}c)^a\plus{}(c\plus{}a)^b > 2$[/quote]
It's a very known inequality. Use the following reasoning:
If $ 0<y<1$ and $ x>0$ then $ x^y>\frac{x}{x\plus{}y}.$  :wink:

Can you explain more?
a, b, c be positive real numbers.
 :|

Yes, but if at least one of them is greater than $ 1$, the inequality is trivial ;)


calculus
integration
number theory solved
number theory

Let x be real number such that  x^3 - x and x^4 - x are integral. Show that x is also integral.

Namdung

[i]unknown source[/i]

Let a = x 4  - x and b = x 3  - x.
Then, bx = a + x - x 2 .
That is x 2  = (1-b)x + a.
Thus b+x = x 3  = (1-b)x 2  + ax = (1-b) 2 x + a(1-b) + ax, from which it follows that :
(b 2  - 2b + a)x = b-a+ab.  (1)

- If b 2  - 2b + a = 0, then b-a+ab = 0, which easily leads to b = 0, and then x is in {-1;0;1}, or to (2-b)(1-b) = 1 which is impossible since b is an integer.
Thus, in that case, x is an integer.

- Now suppose that b 2  - 2b + a =/=0. Then, from (1), we deduce that x is rational. Let x = p/q with gcd(p,q) = 1, and q > 0.
Then, b = p 3 /q 3  - p/q which leads to bq3 = p 3 - pq 2 ). Thus, q divides p 3 , which forces q = 1, and x to be an integer.

Pierre.

Nice and short solution. Congratulations. In the year of 2000, when I suggest this problem in one contest (for 10 grade students) only one contestant had perfect score. Now, I'm thinking about it generalization, such as: if x^k-x, x^m-x are integral then x is integral too. For k=2, we have Ireland problem. For k=3, m=4 is my problem.

Namdung