algebra
binomial theorem
number theory proposed
number theory

Let the sequence $a_i$ be defined as
$a_1=3$ and $a_{i+1}=3^{a_i}$.

Which integers from $00$ to $99$ appear as the last two digits of infinitely many $a_i$'s?

First note that
\[3^{4k}=9^{2k}=(1-10)^{2k}\]
Using the binomial theorem to expand
\[(1-10)^{2k}=1-20k+...\]
Notice that from the third term on all are divisible by 100.  So considering $\mod{100}$
\[3^{4k}\equiv 1-20k \mod{100}\equiv 80k+1\mod{100}\]
Looking at the exponents of each $a_i \mod{4}$ we see all are congruent to $3\mod{4}$.  That is to say each $a_i$ can be expressed in the form $3^{4k+3}$.  Expanding from the congruence above we have
\[3^{4k+3}\equiv 60k+27\mod{100}\]
Implying the last two digits of each $a_i$ belongs to the set $\left\{ 07, 27, 47, 67, 87 \right\}$

[quote="amd2357"]
Implying the last two digits of each $a_i$ belongs to the set $\left\{ 07, 27, 47, 67, 87 \right\}$[/quote]

How do you know all of them work?

Indybar, you're right I forgot to finish the problem.  Actually, my solution set is incorrect.  It should be
\[\left\{03,27,87\right\}\]
I included the 03 for the first number.  Anyways, the last two digits associated with $3^{4k+3}$ cycles every 5 values of $k$.  So it makes sense to analyze the exponents $\mod{20}$.  Looking at exponents only we have $3\mod{20}, 3^3\equiv 7\mod{20}$  All terms after this we look at the exponent of the exponent $\mod{8}$ (Fermat's theorem) to see all exponents after $3^3$ are congruent to $7\mod{20}$  The three numbers given are the corresponding values of the final two digits


Refer to the first post of the second page in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259275]this thread.[/url] If you would like to join the next available tournament, post something along the lines of "I would like to join [whatever the next available tournament is.]"

Sign-ups for FTW Tourney 1, hosted by AIME15, may now begin.

I sign-up, by the way.

1. isabella2296
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.

Please also post your highest rating.  Seeding will be done using this AND also with the help of random.org.

FTW Tourney #1:
1. isabella2296(1775)
2. pythag011
3. Jongao
4. BOGTRO
5. nikeballa96(1701)
6.
7.
8.
9.
10.
11.
12.

Guys, from now on, please copy+paste the sign up list into your post and add your name on to it. Thanks!

[quote="isabella2296"]
[b]Guys, from now on, please copy+paste the sign up list into your post and add your name on to it. Thanks! [/b] [/quote]

really, jjx?

FTW Tourney #1:
1. isabella2296(1775)
2. pythag011
3. Jongao
4. BOGTRO
5. nikeballa96(1701)
6. jjx1 (1725)
7.
8.
9.
10.
11.
12.

Woah wait.  [b]My[/b] tourney gets friggin run over by this?  That rule sucks.  No offense mods.

[quote="isabella2296"]Your tourney was in no way run over. You can still continue it.[/quote]

Uhh... yeah it was.

1:  You completely killed the popularity of my tourney.
2:  In a physical manner (if necessary), you un-sticky-fied it, clearly meaning that it is not active or [b]insignificant[/b].
3:  You completely insulted my tourney.

* Note: My reference of you refers to any of the mods.

Apparently, it is time for the next tourney. (#9 i think)
Would all willing to join plz enter their names and ratings
thank you

Perhaps you should sign up to be a moderator and wait your turn :wink:

he is a tourney moderator
look to the third uncolored tourney in the list
the other two pplz aren't active

Then why was the tourney he tried to make locked and deleted?

i join.
highest 1446 current 1180 or 1150.

This tourney is not working. I suggust that we let the next mod start.

can i join?
Highest rating- (1661)
Current- (1587)

I will join.

I have a rating around 1300

IF this thread is still alive, I'll join. My rating is around 1370 at the moment, high of 1440.

If you want to join a tournament, find a tournament that is looking for players and sign up on that thread. This thread is not in use anymore.


[u][i]Problem
 [/i][/u]

[i] Solve the system  of three equations[/i]

$ \{\begin{array}{c}  x^{5} - x^{4} + 2x^{2}y = 2 \\
y^{5} - y^{4} + 2y^{2}z = 2\\
 z^{5} - z^{4} + 2z^{2}x = 2
\\\end{array}$
[i]
   Easy and nice :lol: [/i]

[i] Love An Forever :blush:
      Huỳnh Võ Phương An[/i]

[i]Noone wants to solve this problem?[/i] :(

       [i] I hope Scorpius_119 could  help me but he had offline a long time[/i]

i  have 1 solution    but  it's  complete.
        suppose $ x >1$    then $ y <1.$  (  from  first  equation )
     from  second  equation    with $ y<1$ then $ z>1.$
  from  thirth  equation $ x<1.$
contract    with  suppose .
  hence$ x \equal{}1    \to y\equal{}1  \to z\equal{}1 .$
     i  also   have not yet  solve     with$ x<0$

if   this  problem   solve with $ x,y,z  >0$ then   very  easy

i  have 1 solution    but  it's not  complete.
        suppose $ x > 1$    then $ y < 1.$  (  from  first  equation )
     from  second  equation    with $ y < 1$ then $ z > 1.$
  from  thirth  equation $ x < 1.$
contract    with  suppose .
  hence$ x \equal{} 1 \to y \equal{} 1 \to z \equal{} 1 .$
     i  also   have not yet  solve     with$ x < 0$
  2 case :    -1<x <1   then  we  also  prove contracdict.


I have some questions for the IMSA admission

First: what's a good SAT score for admission?

Second: how does the admission work. Do you have to have an interview, write an essay, do research, or what?

Third: What is the best thing to do for recommendations, etc. (The part that's 1/3 of admission)?

Please answer these questions before next year's admission  :D

See here and the links to the left:

[url]https://www3.imsa.edu/admissions/apply[/url]

Yeah, I saw that but I want it from the geniuses going to IMSA  :D

Well, basically the admissions compose of a variety of things:

SAT Scores
Recommendation letters from teachers (math, science, English)
Recommendation letter from principal
Four student-written essays (make sure that you put your best effort in these, as they certainly test your essay-writing skills as well)
Basic Application form (Make sure you include all your academic-related activities on that)
One parent-written essay

That's pretty much all...

Note that in order for your application to get full consideration, you [b]MUST have completed one year of algebra and one year of a high school level science course before admission. If you don't have both of these criteria satisfied, your application won't even be considered![/b] (I mean, seriously, I know one of my friends who didn't get in only because of missing HS level science coursework verification!)

For SAT scores, I believe the average math score of accepted applicants was a 710, and the average reading score was a 590. They only consider reading and math scores during the admissions process. Also note that you can take the SAT more than once, and your highest scores would be considered in the application process.

[quote="suma_milli"][quote="gauss1181"]

Note that in order for your application to get full consideration, you [b]MUST have completed one year of algebra and one year of a high school level science course before admission. If you don't have both of these criteria satisfied, your application won't even be considered![/b] (I mean, seriously, I know one of my friends who didn't get in only because of missing HS level science coursework verification!)[/quote]

Yeah I feel bad for that poor person although it was her fault for not having researched before and done high school science.  :roll:

Btw you mean HS math I don't think just one year of algebra qualifies.[/quote]

Actually, it does. It literally says so on the application info. Also, I know one freshman girl (who only took algebra) who applied and got accepted.

[quote="helagha"]But what's a good SAT score?

Oh, and suma_milli, don't you live in IL?[/quote]

I did,  but now I live in MO.

The SAT score isn't the most important part of your application.  As long as you get decent scores you should be fine.  It seems like you're almost searching for a "guaranteed" way to get into IMSA.  Unfortunately there is no guaranteed way to get in.  You have to have decent scores [well I guess you don't HAVE to.. but if you're asian.. you pretty much do] but I think most importantly your love of math and science has to show through your application.  You should also do lots of math and science activities and hopefully you'll get in :)

@gauss: wow, wait like algebra I math?

[quote="suma_milli"]

@gauss: wow, wait like algebra I math?[/quote]

yes, as in terms of school math. lolz they don't really have high standards for school math requirements...it's mostly stuff that's done out of the standard math classroom.

=_= And they call it [i]math[/i] and science academy.  They should be ashamed.   :ninja:

 :rotfl:  :P

Remember that IMSA has applicants from disadvantaged backgrounds, who don't get the opportunity to go beyond algebra I freshman year.

Hmm then wouldn't IMSA have disadvantaged applicants that didn't get to do biology/other science in 9th grade?

The application process is intended for high school freshman.  Presumably, freshman applicants (from any background) would be taking science at their high school, which automatically qualifies as high school science.

But you can get in as an underclassman (8th grade or below) if you're smart enough, much like a standard freshman applicant, and take classes just as advanced as a standard freshman's.

For math I'm pretty sure I'm fine (I'm taking what most people take only in college).
For science, I already took physics and chemistry (well, only about eight chapters of each) and i'm taking honors biology next year.

I think i'm ahead of the standard for freshmen....

Well that's great then! :D  I think you'll have a good chance at getting in as long as you do plenty of math/science activities.

[quote="asethi01"]Isn't getting into national MATHCOUNTS an advantage?
Do they look at that for consideration?????[/quote]
eh...of course. that adds to your math activity advantage.

Lol yeah I doubt you wouldn't get in if you made nat mathcounts.

I made state.....

And for the science activities, do they mean taking extra science stuff? (Like taking Honors Physics in ninth grade?)

Well taking an honors level course would give you better chances of getting in...

And you should probably mention that you made state countdown, which means being in top 16. It's actually a pretty good accomplishment in terms of IMSA acceptance standards.

It's actually really easy to make it into IMSA.  Get a ~2000 SAT (~1400 CR/Math), take a few good math/science classes, get good grades, and express interest for IMSA and you're pretty much in.  If you're a girl it's even easier.
IMSA accepts a ton of 9th graders with this kind of resume (although not necessarily 8th graders; they require slightly more but not much more), so it should be a breeze for anyone who's decent.


我都係喜愛數學的人~

請多多指教！

有朋自远方来，不亦乐乎！

君兮，君兮，不亦乐兮！

懂得规避，知晓礼仪，何以迁求所以——


Prove that if $ m^{4}\plus{}4^{n}$ is prime then $ m$ is odd and $ n$ is even, except $ m \equal{} n \equal{} 1$

Easy to check m is odd
Suppose  $ n$ is odd then
$ n\equal{}2k\plus{}1$
Use
$ x^{4}\plus{}4y^{4}\equal{}(x^{2}\plus{}2y^{2}\plus{}2xy)(x^{2}\plus{}2y^{2}\minus{}2xy)$


Problem that's been keeping me very annoyed, I've tried everything!

Find the value of $ x$ in the following equation:

$ 1 \plus{} \frac {x \minus{} 5}{\sqrt {x^2 \minus{} 10x \plus{} 29}} \plus{} \frac {x \minus{} 5}{\sqrt {x^2 \minus{} 10x \plus{} 34}} \equal{} 0$

[hide="My attempt"]

The problem can be factored as such :

$ 1 \plus{} \frac {x \minus{} 5}{\sqrt {(x \minus{} 5)^2 \plus{} 2^2}} \plus{} \frac {x \minus{} 5}{\sqrt {(x \minus{} 5)^2 \plus{} 3^2}} \equal{} 0$
[list][*]$ k \equal{} 5 \minus{} x$
$ \Downarrow$[/list]
$ \frac {k}{\sqrt {k^2 \plus{} 2^2}} \plus{} \frac {k}{\sqrt {k^2 \plus{} 3^2}} \equal{} 1$

Which is equivalent to adjoing two triangles in the side that measures $ k$ , having a triangle like this:

[geogebra]e32eaf583d4dfd1145a19536e1c5759e20c68ba6[/geogebra]

And that's as far as I went. Any help at all is very appreciated, thanks.[/hide]

[hide=" A hint"]We must have $ x<5$! The equation is equivalent to $ 1\equal{}\frac{5\minus{}x}{\sqrt{(5\minus{}x)^2\plus{}4}}\plus{}\frac{5\minus{}x}{\sqrt{(5\minus{}x)^2\plus{}9}}$, or $ f(t)\equal{}\frac{1}{\sqrt{1\plus{}4t}}\plus{}\frac{1}{\sqrt{1\plus{}9t}}\equal{}1(t\equal{}\frac{1}{(5\minus{}x)^2}>0)$. Now note that $ f(t)$ is monotone!.[/hide]

Hello N.T.TUAN, thanks for the hint, but I'm not familiarized with those terms. First of all what are you substituting $ t$ for in the denominator and what is monotone?
Could you be kind enough to solve it my good sir ?  :lol:

See here http://en.wikipedia.org/wiki/Monotonic_function  :D

Thanks for the link, but that didn't help me very much. Can anybody else help me?

Hi! triplebig. This hint will hepl you beter.

Equation $ \Longleftrightarrow \frac{5\minus{}x}{\sqrt{(5\minus{}x)^2\plus{}4}}\plus{}\frac{5\minus{}x}{\sqrt{(5\minus{}x)^2\plus{}9}}\equal{}1$

Let $ t\equal{}\frac{1}{(5\minus{}x)^2}$, we get equation: $ \frac{1}{\sqrt{1\plus{}4t}}\plus{}\frac{1}{\sqrt{1\plus{}9t}}\equal{}1$

$ \Longleftrightarrow \sqrt{1\plus{}4t}\plus{}\sqrt{1\plus{}9t}\equal{}\sqrt{1\plus{}4t}.\sqrt{1\plus{}9t}$

Let $ u\equal{}\sqrt{1\plus{}4t}, v\equal{}\sqrt{1\plus{}9t}$, we get system: $ \begin {cases} u\plus{}v\equal{}uv\\9u^2\minus{}4v^2\equal{}5 \end {cases}$

$ \Longleftrightarrow \begin {cases} v\equal{}\frac{u}{u\minus{}1}\\9u^2\minus{}4v^2\equal{}5 \end {cases}$

$ \Longrightarrow 9u^2\minus{}\frac{4u^2}{(u\minus{}1)^2}\equal{}5$

$ \Longleftrightarrow 9u^4\minus{}18u^3\plus{}10u\minus{}5\equal{}0$

Hence we only solve this equation: $ 9u^4\minus{}18u^3\plus{}10u\minus{}5\equal{}0$

Let $ y \equal{} 9x^4 \minus{} 18x^3 \plus{} 10x \minus{} 5$, we get graph of this funtion:

So this equation get only one root $ u$, because $ u\ge 1$.

Now I'm buzy. I will give you formula of this root when I'm free. I wish you good luck.

Thank you very much thanhnam2902

Now it comes down to solving that quartic equation, which I can't seem to find anything...

Hope to be seeing the solution

See here http://en.wikipedia.org/wiki/Quartic_equation .


geometry
perimeter
area of a triangle
Heron\u0027s formula

What is the number of square units in the area of a triangle whose
sides are 9, 12 and 15 units?

This is a 3-4-5 right triangle with similarity factor 3. So just take half the product of the two legs, 9 and 12.

$ (\frac{1}{2})(9)(12)\equal{}\boxed{54}$ square units.


Prove that the sequence an={1+1/2+...+1/n} is not convergent in a solution of at most 3 lines!

For all i ,there exists c in (i,i+1) so that 1/c =ln(i+1) -ln(i)
   -->     1/(i+1)< ln(i+1)-ln(i) <1/i
   -->    a(n) > sum( ln(i+1)-ln(i)) {i=1..n} = ln(n+1)
   -->    a(n) is disverge

No-no-no! {x} means the fractional part of x ie x-[x] !Otherwise the problem would be too simple!

A more general result holds:

If (ab)n>=1 is a sequence which tends to 00 s.t. an+1-an->0 then {an} is divergent.

It seems, we solved a similar or maybe the same problem

I also seem to remember it, but I don't know if I can solve it in at most 3 lines :D. I think the problem was posted by Kuba (I might be wrong).

It is here  :)
http://www.mathlinks.ro/viewtopic.php?t=1420

Here the solution in three lines
line1 denote by bn=1+1/2+...+1/n,suppose an is convergent
line2 compute [b2n] -[bn] =b2n-a2n-bn+an=1/n+1 +...+ 1/2n -a2n+an.
line3 first member an integer, the second converges to ln2, false!


How to play the stupid Jedi game:
1. Sign up ( Just write like I'm in )
2. When the round starts name 1 Jedi
3. whoever does not name a jedi or repeats one is out
4. WARNING: this game is so STUPID
( I learned it from my friend)

Ok.  I'm in though not a big StarWars fan.

OK anyone else? Anyway I'll start the round when someone else comes in. :starwars:

i'm in. i think we should wait for more people.

Anyone else? After like 2 more people we are strarting :starwars:

C'mon people this game may (even though i know it) be FUN!

come on people come on people come on people  come on people  come on people  come on people come on people come on people come on people come on people come on people come on people come on people come on people come on people  :furious:

Stop spamming and bumping a topic. One post is enough, three's unnecessary.

So people in:
math92
jimli
mathbomber
juicy
yifman?  Are we gonna start?  You said after 2 more people.

Anybody that used a lightsaber so yes.
 Okay, the moment you have been waiting for- ROUND 1
I randomly picked who was going to go first so math92-
 :play_ball: Play Ball!

 :starwars:

( Oh yeah, People who wanted to play but are not in this round can sign up in the next game. :D )

I "would" join in, but somehow I don't think I'm welcome (from the posts above ;) )

you are welcome because nobody else is going.

ok... so far, there's Yoda, Qui-Gon Jinn, Anakin, Luke, Lowbacca, Plo Koon... Palpatine is NOT a Jedi, and he never was.  Palpatine is Sith.  So math92, between you and me!  I'll say Mace Windu.

By the way, are you familiar with expanded universe (EU), or are you just for the six movies?  (Anything outside the movies - books, games, TV series, etc. is EU)  jimli said he is not a big Star Wars fan, but I suspect he does know EU because Lowbacca only appears in EU.  If you know EU, we have like ten times as many choices to pick from.  ;)  This kinda goes for stupid game 2, also - we need to call this something other than "stupid game"!

math92, it's my turn to ask: are you going to go, or is this game over?

Ulic Qel-Droma
she was on the dark side before darth vader

yay!  i'll take that to mean you know expanded universe.  I'll say Nomi Sunrider. (It [i]is[/i] Nomi, right?)

Zez-Kai Ell
he was on the dark and light side :o

yeah, well, so was anakin... did you get zez-kai ell from KotOR II?  I think he appears there somewhere... or so I've heard.  If you give up on my avatar by next post, I'll say it next time... anyways, I'll say Kyp Durron!

yeah, I know Ikrit ;)  Junior Jedi Knights series, brief mention in [u]Crisis at Crystal Reef[/u] in Young Jedi Knights series.  Brief mention of his passing away in some book in the NJO series...  yeah.  The numbers of Jedi and planets are staggering.

Jedi named: Yoda, Qui-Gon Jinn, Anakin, Luke, Lowbacca, Plo Koon, Mace Windu, Ulic Qel-Droma, Nomi Sunrider, Zez-Kai Ell, Kyp Durron, Ikrit, and now I say Mara Jade Skywalker.  Of course.  After the nice little discussion I had in Stupid Game 2...

EDIT: Ikrit's really a meter tall?  I thought he was smaller than that!  Well, from the cover of [u]The Golden Globe[/u], anyway...

Juhani, she is a Jedi character from Star Wars: Knights of the Old Republic.

oh no!  not video games... only one I have is Jedi Knight II: Jedi Outcast.  I've heard about KotOR, Jedi Academy, Dark Forces I and II, Battlefront, X-Wing Commander, and so on and so forth, though.  I don't know much of anything about them other than that they exist, though.  I'll stick with the classics for now.  Obi-Wan Kenobi!

I surrender...i can't think of anymore!!!!!
 :surrender:

yeah, comes with reading a lot.  I guess it could come with watching and re-watching SW movies (which I don't do), too, though.

you know so much about star wars.  If they make a Stupid Game 3 then it should be something else then SW. :)


[quote="xzlbq"]Let $ x,y,z > 0$,prove that:

$ \frac {2}{3} \geq \frac {1}{2}\frac {y \plus{} z}{x \plus{} y \plus{} z} \plus{} \frac {2\sqrt {3}}{3}\sqrt {\frac {xyz}{(x \plus{} y \plus{} z)(y \plus{} z)^2}}$

BQ

Geometric background is:
in triangle,
$ \frac {2}{3} \geq \frac {r}{h_a} \plus{} \frac {h_a}{\sqrt {3}s}.$
BQ[/quote]

$ \frac {2}{3} \geq \frac {1}{2}\frac {y \plus{} z}{x \plus{} y \plus{} z} \plus{} \frac {2\sqrt {3}}{3}\sqrt {\frac {xyz}{(x \plus{} y \plus{} z)(y \plus{} z)^2}}$

$ \Longleftrightarrow (\frac {2}{3} \minus{} \frac {1}{2}\frac {y \plus{} z}{x \plus{} y \plus{} z} )^2\geq ( \frac {2\sqrt {3}}{3}\sqrt {\frac {xyz}{(x \plus{} y \plus{} z)(y \plus{} z)^2}})^2$

$ \Longleftrightarrow (16y^2 \minus{} 16yz \plus{} 16z^2)x^2 \plus{} ( \minus{} 24y^2z \plus{} 8y^3 \plus{} 8z^3 \minus{} 24z^2y)x$ $ \plus{} z^4 \plus{} y^4 \plus{} 4y^3z \plus{} 6y^2z^2 \plus{} 4yz^3\ge 0$

and

$ 16y^2 \minus{} 16yz \plus{} 16z^2 > 0$

$ \Delta \equal{} 576yz(y \minus{} z)^2(y \plus{} z)^2\geq 0$

so.............

[quote="hedeng123"][quote="xzlbq"]Let $ x,y,z > 0$,prove that:

$ \frac {2}{3} \geq \frac {1}{2}\frac {y \plus{} z}{x \plus{} y \plus{} z} \plus{} \frac {2\sqrt {3}}{3}\sqrt {\frac {xyz}{(x \plus{} y \plus{} z)(y \plus{} z)^2}}$

BQ

[/quote][/quote]

It's obviously trues by AM-GM, hedeng123  :)

[quote="nguoivn"][quote="hedeng123"][quote="xzlbq"]Let $ x,y,z > 0$,prove that:

$ \frac {2}{3} \geq \frac {1}{2}\frac {y \plus{} z}{x \plus{} y \plus{} z} \plus{} \frac {2\sqrt {3}}{3}\sqrt {\frac {xyz}{(x \plus{} y \plus{} z)(y \plus{} z)^2}}$

BQ

[/quote]

It's obviously trues by AM-GM, hedeng123  :)[/quote][/quote]

 :D

"This ineq does not make number" ?  What do you mean?  I cannot understand. :?:

Where is wrong？
x=y=z,Is not equal to zero。
BQ

hello, sorry, i have made a mistake the given inequality is equivalent to
$ a\sqrt{3}(2b\plus{}2c\minus{}a)\geq12A$ with
$ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ squaring both sides we get
$ 12a^4\minus{}12a^3b\minus{}12a^3c\minus{}6a^2b^2\plus{}24a^2bc\minus{}6a^2c^2\plus{}9b^4\plus{}9c^4\minus{}18b^2c^2\geq0$
this is equivalent with
$ 12a^2(a\minus{}b)(a\minus{}c)\plus{}3(b\minus{}c)^2(3(b\plus{}c)^2\minus{}2a^2))\geq0$
this is true if $ a\equal{}\max\{a,b,c\}$.
Sonnhard.


calculus
integration
trigonometry
LaTeX
function
logarithms
calculus computations

^Thank you

that helps slightly, but now i end up with:

$\int \frac{1} {x\sqrt{25x^2+16}}$

$\int \frac{1} {\frac{4}{5}tan\theta 4sec\theta}$

with no $d\theta$ to help me integrate.

are u proposing that there is something wrong with the question?  If not, please can someone show me how to calculate this with some more working?

thanks :D

If $5x=4\tan\theta,$ then

$5dx = 4\sec^2\theta\,d\theta$ or $dx=\frac45\sec^2\theta\,d\theta.$

This is a substitution. You have to use the chain rule and account for everything.

Note also that you should be writing the original integral as

$\int\frac{dx}{x\sqrt{25x^2+16}}$

^^ Thanks.

I've got much further through  this question, but i'm stuck again. :(

$\int\frac{dx}{x\sqrt{25x^2+16}}dx$

$\int \frac{\frac{4}{5}sec^2\theta} {\frac{4}{5}tan\theta 4sec\theta}d\theta$

$\int \frac{sec\theta}{4tan\theta}d\theta$

$\frac{1}{4}\int \frac{d\theta}{sin\theta}$

$\frac{1}{4}ln|csc\theta - cot\theta|$

i know that $\theta = tan^-1(\frac{5}{4}x)$ but i have no idea what happens when i substitute theta into the above answer :S

Thanks guys

Draw a right triangle and give it some side lengths so that $\theta$ has the tangent you need it to. Then you can just read off what those other ones are.

Thanks dude :D

sorry guys i have just got 1 more question:

since cosec and cot can be negative between $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$
should my answer read:

$\frac{1}{4}ln|+/-csc\theta -/+cot\theta|$

+/- meaning plus or minus (couldnt find the symbol)

No, that's what the absolute value sign is there for.

Two $\LaTeX$ tips:

1. The symbol you couldn't find is \pm (short for plus-minus): $\frac{1\pm\sqrt{5}}2.$

2. Put a backslash in front of the names of standard named functions: \csc\theta, not csc\theta; \ln(1+t), not ln(1+t).

$\csc\theta$ looks better than $csc\theta.$

$\ln(1+t)$ looks better than $ln(1+t).$

$\int \frac{1}{\sin \theta } d\theta =\ln \left|\tan \frac{\theta}{2}\right| +C.$


If [x] denotes the floor function and m, n are integer,

[(x+m)/n] = [([x]+m)/n]

How can we prove this?

[hide]

Let x = kn + a + b, where k and a are integers, and 0<=b<1.

Then we want to prove [k + (a+m+b)/n] \ [k+(a+m)/n], or equivalently, [(a+m+b)/n] = [(a+m)/n].

Now its pretty easy to see why thats true - the only time these would be different if a+m was just below a multiple of n and a+m+b was just above it, but b<1 and thus this is impossible (as a,m are integers.)

[/hide]


number theory
prime numbers
number theory proposed

Prove that there are infinitely many pairs of distinct prime numbers $ (p,q)$ such that $ p\ |\ 2^{q\minus{}1}\minus{}1$ and $ q\ |\ 2^{p\minus{}1}\minus{}1.$

This looks rather interesting. Does anyone know a solution of this?

See here : http://www.mathlinks.ro/viewtopic.php?t=278140


salam doostan
too in topic har bar yek nafar yek too mayehaye soale  marhale dovom mide va bad vaghti kasi halesh kar ye soale jadid post mikone ;)

[b][size=150][color=green]first question[/b][/size][/color]
mosalase $ABC$($<C=90$)mafrooz ast.$M$ vasate $BC$ vaghe shode.
$E$ noghteye delkhahi bar rooye $AC$ ast va $F$ vasate $EC$ vaghe shode $H$ paye amode vared az $C$ mibashad va $I$ markaze dayereye mohitiye $HAE$ ast sabet konid do mosalase $IFM$ va $ABC$ motashabeand.(mexico olympiad)

[quote="mohamadhosein"]ja dare az hameye doostan be khatere esteghbale por shooreshoon az in topic tashakor konam. :yup:[/quote]
solae tarkibiat bezar porshooresh mikonim  ;)

soalro mishe dobare check koni choon fekr konam ghalate?
male koodom sale Mexice?

vaghean moteasefam
 editesh mikonam
fekr konam sale 1990 bashe

[quote="BaBaK Ghalebi"]solae tarkibiat bezar porshooresh mikonim  ;)[/quote]
bashe hatman ;)

manam tarkibiyat hastam chandta ziba daaram betoon migam

ghorboonet piroosh. ;)
behtare dige hal masale ro post konid. :D

This is a nice problem that just talented one can solve it. ;)
Therfore u solve it :D
This is really like 2 step.

salam doostan
omidvaram hamamoon dar marhale dovome olampiade riazi(computer) movafagh bashim.
khoda ro faramoosh nakonid baraye ma ham doa konid estressam bikhial shid :P
 :)

bara hamatoon arezooye moovafaghiat mikonam
faghat torokhoda maro ham doa kooooonid!!!
 ;)  :10:

man esme noghat ro avaz kardam :oops:

makan hendesiye O mishe amood monasefe BH .hal tahte tajanos marpichi be markaze M va nesbate $\frac{MN}{AM}=\Cos C$ noghteye N be A  tabdil mishe.va choon zaviye bein ON va AB barabar ast ba $\angle AMN=\angle C$ pas khate ON be AB tabdil mishavad.
farz konid taht in tajanos noghteye O be E' tabdil shavad va N be A pas $\frac{E'A}{ON}=\frac{MA}{NM}=\cos C\longrightarrow E'A=ON.\cos C$ noghte barkhord in dayere ba AB ra F dar nazar begirid.dar natige $AF=AB-BF=AB-2BG=AB-2(\frac{AB}{2}-NG)=2NG=2.ON\cos C$ pas $AF=2AE'$ dar natije $E'\equiv E$

amir jan inam az rahehale  man ;)
rstesh man noghteye $M$ ro $G$ farz kardam.
ba in yeki fekr konam kolli hal koni :lol:
$E'$ ra noghteye moghabele ghotriye $E$ migirim.
kholase $GF$ va$BE$ movaziand.
hamchenin $IF$ va $E'C$ ham movaziand.
hamchenin do mosalase$AHE'$ va $CHE$ motashabehand.pas$\frac{E'A}{EC}=\frac{HA}{HC}$ pas $\frac{E'A}{EC}=\frac{AC}{BC}$ $\longrightarrow$
do mosalase $BCE$ va$CAE'$(*) motashebahnd.
$K$ ra noghteye barkhorde$BE$ va$CE'$ migirim.
$\angle AE'C=\angle BEC =\angle KEC$ pas $\angle EKC =90$
az tarafi $KC$ va $IF$ movazaiand va $BE$ va $GF$ movaziand  $\longrightarrow$
 $\angle IFG =90$
hal chon $\angle ACE' =\angle CBE$ (*)va
$\frac{AC}{E'C}= \cos {ACE'}$ va $\frac{BC}{BE}= \cos {CBE}$
pas $\frac{AC}{E'C}=\frac{BC}{BE} \longrightarrow \frac{AC}{2IF}=\frac{BC}{2GF} \longrightarrow \frac{AC}{IF}=\frac{BC}{GF} \longrightarrow$do mosalase $IFG$ va $ABC$ motashabehand. :huh:

mohammad hossein chand ta eshtebahe Typi dari bekhatere hamin rahe haleto nafahmidam ya man daram sooti midam. :blush:

masalan:

kholase $FG$ va $BE$ movaziand ke bayad beshe $MF$ va $BE$ movaziyand.

khate sheshom: $\frac{E'A}{E'C}=\frac{HA}{HC}$ bayad beshe $\frac{E'A}{EC}=\frac{HA}{HC}$.

do mosalse $BCE$ va $CAE'$ motashabehand vali fekr nakonam bashand.
noghate $K$ va $G$ koodoomand :?:


inequalities
trigonometry
inequalities proposed

$0\leq x\leq \frac{\pi}{2}$ is a real number. $y_{i}$ ($1\leq i\leq n$) are reals and $y_{i}\geq 1$ for every $1\leq i\leq n$. $z_{j}$ ($1\leq j\leq n$) are reals also. Prove:

$(sinx)^{({\sum^{n}_{i=1}y_{i}})^{\sum^{n}_{j=1}\frac{z_{j}}{z_{j+1}}}-n^{n}}\leq 1$
 :ninja:

I am not sure the problem is correct as stated. Let all the $z_{j}= 0$. Then the exponent is $1-n^{n}$. Let $n=2$ so the exponent is $-3$. Since $0 \le \sin(x)\le 1$ on the given interval, the inequality is reversed except if $x = \frac{\pi}{2}$ (when equality holds).

On the other hand if the $z_{j}$ are all large, the nested exponent is close to $n$, and we can take the $y_{i}$ enough bigger than 1 so that their sum is way larger than $n$, so that the exponent of the sine is positive. Then, the inequality holds as stated.

It is not obvious to me how to fix it. Of course, I'm a newb, so maybe there is jsut some small constraint missing. :P

well for the inequality to hold all we need is the exponent to be positive :wink:
since $\sum y_{i}\geq n$  we need $\sum \frac{z_{i}}{z_{i+1}}\geq n$ which is just AM-GM :roll:

Ah, oops... I read that as $\frac{z_{j}}{z_{j}+1}$...  :oops:  :blush:  :oops:  :blush:

Still, in the problem he must add that $z_{i}>0$ :wink:

that's right Albanian eagle, $z_{j}>0$. :D
sorryyyyyyyyyyy :oops:


Rhombus $ ABCD$ is similar to rhombus $ BFDE$.  The area of rhombus $ ABCD$ is 24, and $ \angle BAD \equal{} 60^\circ$.  What is the area of rhombus $ BFDE$?

[asy]
size(180);
defaultpen(linewidth(0.7)+fontsize(11));
pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3));
pair point=(3/2, sqrt(3)/2);
draw(B--C--D--A--B--F--D--E--B);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));[/asy]

$ \textbf{(A) } 6 \qquad \textbf{(B) } 4\sqrt {3} \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 9 \qquad \textbf{(E) } 6\sqrt {3}$

[hide="Answer"]Drawing $AE$ and $BD$, we notice by equilateral triangles that each half of rhombus $BFDE$ has one-third the area of half of rhombus $ABCD$. $8\Rightarrow \boxed{C}$[/hide]


calculus
integration
trigonometry
function
calculus computations

[quote="reedk4"]Integral of (sin x cos x) dx

Use u substitution for both sin x and cos x, correct?
But when you do that you get different answers!
You can also do a trig identity with sin x cos x, correct?
I'm sorry, just really confused!  Haven't done calculus in about 5 years.

Please someone explain! Thanks.[/quote]

 It should be in the tutorial!!
 The results are different because you're taking indefinite integral, which is determined up to an additive constant. I'm sure the two results you have differ only a real constant ;)

Expanding out what dickchimney said ...

1. Let $u=\sin(x);\,du=\cos(x)\,dx.$

$\int\sin(x)\cos(x)\,dx=\int u\,du=\frac12u^2+C_1=\frac12\sin^2(x)+C_1.$

2. Let $u=\cos(x);\,du=-\sin(x)\,dx.$

$\int\sin(x)\cos(x)\,dx=-\int u\,du=-\frac12u^2+C_2=-\frac12\cos^2(x)+C_2.$

3. Use the trigonometric identity $\sin(2x)=2\sin(x)\cos(x).$

$\int\sin(x)\cos(x)\,dx=\frac12\int\sin(2x)\,dx=-\frac14\cos(2x)+C_3.$

I wrote $C_1,C_2,C_3$ to emphasize that these constants are different from line to line. So, reed4k, here's my exercise for you. It's a trigonometric identity exercise, and a reasonably simple one. Show that it is possible to choose constants $C_1,C_2$, and $C_3$ such that

$\frac12\sin^2(x)+C_1\equiv-\frac12\cos^2(x)+ C_2\equiv-\frac14\cos(2x)+C_3.$

(I am using "$\equiv$" to mean "is identically the same function as.")

yeah Kent is right. Actually $\int \sin(x) \cos(x) dx$ is a fairly simple integration problem, but has different ways that it can be solved.


hello , help me please  :blush:

$ a,b,c \in\mathbb{R}\quad and \quad|c|<1 \quad, |a\plus{}b\plus{}c|<1\quad , |a\minus{}b\plus{}c|<1$

prove that : $ \forall x \in [\minus{}1,1], |ax^2\plus{}bx\plus{}c|<5/4$

thanks

[hide="Hint"]Let $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$.  The given information is equivalent to $ |f(x)| < 1$ for $ x \equal{} \minus{}1, 0, 1$.[/hide]

[quote="matt276eagles"][hide="Hint"]Let $ f(x) \equal{} ax^{2} \plus{} bx \plus{} c$.  The given information is equivalent to $ |f(x)| < 1$ for $ x \equal{} \minus{} 1, 0, 1$.[/hide][/quote]

thanks matt276eagles , but we must prove this relation $ \forall x \in [\minus{}1,1]$

hello,

let $ \quad d p(x)=ax^{2}+bx+c$
 please tell me how I can use this formula $ \quad d p(x)= \frac{p(1)}{2}x(x+1)+\frac{p(-1)}{2}x(x-1)+p(0)(1-x^{2})$ to resolve the problem, thanks


ARML geometry parameterization trapezoid

ARML is soon! Very soon!

Can someone post some test-taking strategies?

I know

1. Get lots of sleep before the test
2. Don't overeat/overexcite before the test (3 days)
3. Don't Cram, it won't help.

Does anyone have any more tips? In general or specifacally ARML.

Thanks!

On the individual round, I find that its crucial not to start panicking if it seems like neither problem is doable. That just makes things worse. Usually its just that you missed something obvious, so take a breath and look at a problem a second time.

Also, if on the first round you get both questions wrong, don't get psyched out. That happened to me once but I managed to get the last 6 right, so I'd say the key is mostly managing your nerves. :)

Have fun. This works wonders towards not letting you stress out. Hang out with your friends in the downtime, and if you don't know anyone get to know them or recruit people you know and like (I've done this a lot, obviously a little late for *this* year).

Don't worry so much about scores. Honestly, the difficulty curve on the individual round in the past few years has been pretty stupid. Power is always interesting and there are usually interesting questions on the Team round.

After the individual rounds, *run* to lunch. That's the biggest one.

Most important for me: eat. I tend to stop eating when I'm having fun/competing.

[quote="BOGTRO"]
1. Get lots of sleep before the test[/quote]

Not true! i got 2 hrs of sleep before NYSML and got a 6 on Individ round.

then i screwed up...so whatever

Oh, and BOGTRO...

WIN THE SONG CONTEST!!!!

 :lol:  :D

Don't worry about your score...it doesn't matter if you get a 2 or a 6. It is better to solve one problem than to make progress on two problems, but not solving either of them. Do casework quickly.

Don't be constipated the days before the test...I'm speaking from experience here.

[quote="cyberspace"][quote="worthawholebean"]After the individual rounds, *run* to lunch. That's the biggest one.[/quote]
Thats the best part of the whole weekend!!! :P[/quote]

...or just don't eat in the cafeteria

An excellent way to perform badly on the relays is not to get enough sleep the night before, then wear yourself out playing frisbee at lunchtime.

Also, on the team event, you have 20 minutes, 15 people, and 10 questions.  This means that if each person solves just one question in that really long time, you'll ace it.  So unless you're the designated team organizer or someone who's supposed to be jumping around, stick to the question you've been assigned until either you solve it, or it's solved and checked by two other people.  Same goes for Power round.

And on the individual rounds, don't jump between questions too much either.  Pick the one that you know you can solve within ten minutes and solve it, unless you really get stuck.  Then work on the other.  Check your answers.

Be aware of ways you can cheat the question; this tends to happen in geometry problems.  If a quantity is unspecified and doesn't look like it's determined by the problem's parameters, you can probably (if you're careful!) get away with making it whatever you want.  Make sure you make the right assumption, though--case in point: On the trapezoid problem in 2005, I spent 9 minutes assuming it was isosceles only to get an answer I didn't like, then at the last second assumed it was a right-angled trapezoid and solved it.

2008 factors as $ 2^3*251$.

Have fun.  Play some frisbee.  Meet people from other teams if you have time.  Play games.  As a general rule, the only excuse for being in your room is if you're sleeping.  Spend the rest of your time being social.

[quote="BOGTRO"]
1. Get lots of sleep before the test[/quote]
huh
get lots of sleep
-> sleep early
-> less time playing games
-> less fun
-> err whats the point of going to arml then

Yeah junggi is right, lol.
Well, it depends whether you're going to ARML mainly for the experience, or to do really well.
And my tip: don't suck on the relay, especially if you're #1 or #2. If at all possible, send back all possible answers the minute you get the problem. Last year, I narrowed it down to two answers (something like $ \frac {\pi}{4},\frac {3\pi}{4}$)and sent both back when I was #2 without ever getting the answer from our #1. This, of course, left #3 to choose b/n two answers, but it was the best I could do.

Notice I didn't say "get lots of sleep", just that you will do badly on relays if you don't get enough.  (Actually, the West Coast is rather lucky on this point in that we get relays before lunch, rather than around 1:00 or so when people are naturally less alert.)  You know your own sleep requirement, or you should.  If that requirement allows you to stay up playing games late into the night, go for it.

Also, everyone should read the new relay rules posted at the bottom of this page:
[url]http://www.arml.com/arml_registration/login.php[/url]
I don't know to what extent the new stipulations on "communication" are going to be enforced, but if they're super-strict about it, a lot of teams could be at risk of disqualification.  I don't think they will be, though, since a lot of the practices that teams use to communicate (like what davidyko said) can't be distinguished in any objective way from legal answer-passing.

Georgia's old flinging-of-confetti at each other won't work this year :(.

[quote]2008 factors as 2^3*251[/quote]

This is definitely the most useful tip.

Also, not stressing is a plus. Last year I went into the individual round not expecting much and not caring a little bit. I ended up getting a seven.

Educated guessing is worth it if there is a couple seconds left. I read individual #6, realized I had no idea how to do it, and guessed 720 correctly.

If you have a way of cheating on the relays, it is not in your best interest to tell lots of other people what you are doing.

[quote="TimReyn"][quote]You may continue to pass answers, even if they are the same as before.[/quote]

[quote]Such practices include, but are not limited to: passing back multiple pieces of paper with the same answer at the same time in an effort to communicate that the participant is very certain of their computation[/quote]

They say you can send multiple copies of the same answer back but then later cite something essentially the same as cheating.  That's a bit confusing.
It's always been common practice on my team to send back a second or third (or, in one case in practice, ~200th) copy of the answer to show that we're pretty sure that it's the answer.  Now that's not OK?[/quote]

The latter version includes the phrase "at the same time"

Thankfully our team is too crappy to even have codes in the first place (a grand total of 4 points on the relay last year! woo!)

[quote="Ignite168"]
The latter version includes the phrase "at the same time"
[/quote]
True, but the spirit is still there--sending back the same answer twice gives the same message, and that's what they don't want you to do.

Yeah, Upstate NY tends to have relay issues anyway.  We generally stack one team and put three really strong people on it--last year, that team got zilch.

[quote="TimReyn"][quote]You may continue to pass answers, even if they are the same as before.[/quote]

[quote]Such practices include, but are not limited to: passing back multiple pieces of paper with the same answer at the same time in an effort to communicate that the participant is very certain of their computation[/quote]

They say you can send multiple copies of the same answer back but then later cite something essentially the same as cheating.  That's a bit confusing.
It's always been common practice on my team to send back a second or third (or, in one case in practice, ~200th) copy of the answer to show that we're pretty sure that it's the answer.  Now that's not OK?[/quote]

My understanding of that first thing is:

it's OK to do something like pass $ 2$, then later pass $ 3$ (but ONLY because you now think $ 3$ is the correct answer), and then even later pass $ 2$ (but ONLY because you now think $ 2$ is the correct answer, again).

However, you are NOT allowed to pass $ 2$, then pass $ 2$ again to indicate that you're sure of the answer.

So yes, according to that what you are describing is now considered cheating.

Essentially, what sums up the entire rule change is:

[quote]
The method of passing of answer(s) should, in no way, indicate anything besides the answer to the question.[/quote]

Your method indicates more than this, it tells the other person how sure you are of your answer. Thus it is not allowed.

Basically that means that the only strategy you can legally use is:

Work on your problem. When you solve it, pass the answer. If you later think that your answer is wrong, pass the correct answer. When you receive an answer, you should take that to mean your teammate thinks that's the correct answer to their problem. If you later receive a different answer, you should take it to mean that your teammate now thinks that [i]that[/i] is the correct answer, and you should throw away the other thing they passed you.  (I think you're still free to do things like passing a weak guess early on, if your not sure of the answer [like if you're person 2, and you haven't gotten anything from person 1] but if you then decide that guess was correct you cannot re-pass it.)

[By the way, this is what SoCal's been using for years. I've always felt that these more complicated strategies were a bit questionable, and sort of went against the spirit of the relays. I'm glad they've finally clarified the rules.]

At least, that's my interpretation of the rules, if that's wrong, someone correct me.

[quote="Ignite168"]Thankfully our team is too crappy to even have codes in the first place (a grand total of 4 points on the relay last year! woo!)[/quote]Yeah, and I believe all those points came from one relay team that included you and me...

I highly recommend [i]not[/i] getting a gushing nosebleed during the 4th pair of individuals. I can only [i]imagine[/i] what the poor person in the grading room must have though when they saw my answer slip that year.  :oops:

Practically, though: use pencils. Write LEGIBLY on your answer sheets. If you have sloppy handwriting and use ink, and [i]then[/i] decide to change your answer on the answer sheet... well, imagine how illegible that mess would be. I don't have to imagine it because I see it routinely in the grading room, where if we can't read it clearly we can't possibly count it as correct.

[quote="tjhance"][quote="generating"]I highly recommend [i]not[/i] getting a gushing nosebleed during the 4th pair of individuals. I can only [i]imagine[/i] what the poor person in the grading room must have though when they saw my answer slip that year.  :oops:[/quote]

I have always imagined elaborate scenes where I am at ARML and my pencil breaks, and I am forced to write the answer with my own blood in the last few seconds.[/quote]

As long as your bloodsmanship is legible, you'll be fine  :lol:


limit
real analysis
real analysis unsolved

The sequence $\{x_{n}\}$ defined as follow:
                $x_{0}=\frac{1}{2},x_{n+1}=x_{n}-x_{n}^{2}$ $\forall n \geq 0$
Find all $m \in R$ such that $\exists L=\lim_{n \to \infty}n^{m}(nx_{n}-1)$ and $L \in \mathbb{R}\setminus\{0\}$

[quote="nhat"]The sequence $\{x_{n}\}$ defined as follow:
                $x_{0}=1,x_{n+1}=x_{n}-x_{n}^{2}$ $\forall n \geq 0$[/quote]

Then $x_{1}= 1-1^{2}= 0$ so $x_{n}= 0$ for all $n \geq 1$.

I think that the only value of $m$ which verifies that condition is 1.

To see this, first observe that it is not hard to prove that:

$x_{n}\rightarrow 0$ and $nx_{n}\rightarrow 1$.

On the other hand, observe that $m>0$ otherwise the limit is 0.

So we assume that $m>0$.
An application of  Cesaro Stolz lemma shows that:

$L=\lim n^{m}(nx_{n}-1)=\lim \frac{(n+1)x_{n+1}-nx_{n}}{1/(n+1)^{m}-1/n^{m}}$

$=\lim \left(\left(\frac{n+1}{n}\right)^{m}nx_{n}\frac{n^{m-1}}{(n+1)^{m}-n^{m}}n^{m}(nx_{n}-1)+\left(\frac{n+1}{n}\right)^{m}\frac{n^{m-1}}{(n+1)^{m}-n^{m}}n^{m-1}n^{2}x^{2}_{n}\right)$
The first term converges to $\frac{L}{m-1}$ and the second term converges to $\infty$ if $m>1$ and $0$ if $m<1$.
thus,

$L=\frac{L}{m-1}+\infty$ if $m>1$ which implies that $L=\infty$ but this is false

or

$L=\frac{L}{m-1}$ if $m<1$ but this implies that $L=0$ and since we need to have $L\neq 0$ it follows that $m=1$.

Now the question is what is $L$ when $m=1$?

I hope that there are no mistakes in my calculations.


algebra
polynomial
inequalities
triangle inequality
algebra proposed

the degree of the polynomial$P(x)$is atmost $2n$,we know that for any $k\in[-n,n]$ we have $\mid P(k)\mid \leq1$ prove thatfor any $x\in [-n.n]$  we have $\mid P(x)\mid \leq 2^{2n}$

[hide="ideas+sketch of a solution"]
1. VanDerMonde's determinant
or
2. Lagrange interpolating polynomial (applied to the point (k;p(k))), then, we use the triangle inequality, and we're done[/hide]
:)


I need help in proving the following inequality:

 Given $ 0<t<1$, $ 0\le w_i < 1$, with $ \sum_{i\equal{}1}^n w_i \equal{} 1$.
Prove:

$ \prod_{i\equal{}1}^n \frac{1}{1\minus{}tw_i} \le \frac{1}{1\minus{}t}$

with equality when $ w_i \equal{} 0$.

[quote="kcstevonne"]I need help in proving the following inequality:

 Given $ 0 < t < 1$, $ 0\le w_i < 1$, with $ \sum_{i \equal{} 1}^n w_i \equal{} 1$.
Prove:

$ \prod_{i \equal{} 1}^n \frac {1}{1 \minus{} tw_i} \le \frac {1}{1 \minus{} t}$

with equality when $ w_i \equal{} 0$.[/quote]

Maybe the below fact can help here. :)
\[ f\left( {a_1 ,a_2 ,...,a_n } \right) \le f\left( {0,a_1  \plus{} a_2 ,a_3 ,...,a_n } \right)
\]
\[ \Leftrightarrow \frac{1}{{\left( {1 \minus{} a_1 t} \right)\left( {1 \minus{} a_2 t} \right)}} \le \frac{1}{{1 \minus{} \left( {a_1  \plus{} a_2 } \right)t}}
\]
\[ \Leftrightarrow a_1 a_2 t^2  \ge 0
\]
which is always true. :)

[quote="kcstevonne"]Honey_S,
 Is this a well-known inequality? Also, can this be extended to the case n>2?[/quote]

Did you mean your first inequality
[quote="kcstevonne"]
 Given $ 0 < t < 1$, $ 0\le w_i < 1$, with $ \sum_{i \equal{} 1}^n w_i \equal{} 1$.
Prove:

$ \prod_{i \equal{} 1}^n \frac {1}{1 \minus{} tw_i} \le \frac {1}{1 \minus{} t}$

with equality when $ w_i \equal{} 0$.[/quote]
Or this in my post
[quote="Honey_S"]
\[ f\left( {a_1 ,a_2 ,...,a_n } \right) \le f\left( {0,a_1 \plus{} a_2 ,a_3 ,...,a_n } \right)
\]
[/quote]

[quote="kcstevonne"]in your post, of course.[/quote]
It's mixing variables.


Demonstrati ca $ |a|\plus{}|b|\plus{}|c|\plus{}|a\plus{}b\plus{}c| \geq |a\plus{}b|\plus{}|b\plus{}c|\plus{}|c\plus{}a|$ unde a,b,c sunt numere complexe.

"hint". se poate folosi identitatea \[ |a|^2\plus{}|b|^2\plus{}|c|^2\plus{}|a\plus{}b\plus{}c|^2\equal{}|a\plus{}b|^2 \plus{} |b\plus{}c|^2 \plus{} |c\plus{}a|^2\]

[quote="balan razvan"]Demonstrati ca $ |a| \plus{} |b| \plus{} |c| \plus{} |a \plus{} b \plus{} c| \geq |a \plus{} b| \plus{} |b \plus{} c| \plus{} |c \plus{} a|$ unde a,b,c sunt numere complexe.[/quote]
[hide="hint"]use popoviciu inequality[/hide]

[quote="HLAWKA"] [color=darkred]$ \{a,b,c\}\subset\mathbb C\ \implies\ |a| \plus{} |b| \plus{} |c| \plus{} |a \plus{} b \plus{} c| \geq |a \plus{} b| \plus{} |b \plus{} c| \plus{} |c \plus{} a|$ .[/color] [/quote]

[color=darkblue][b][u]Demonstratie.[/u][/b] Se aduna identitatea $ |a \plus{} b|^2 \plus{} |b \plus{} c|^2 \plus{} |c \plus{} a|^2 \equal{} |a|^2 \plus{} |b|^2 \plus{} |c|^2 \plus{} |a \plus{} b \plus{} c|^2$ la urmatoarele inegalitati [b]inmultite cu doi[/b] :

$ \left\{\begin{array}{c} |a \plus{} b|\cdot |b \plus{} c| \equal{} |b(a \plus{} b \plus{} c) \plus{} ac|\ \le\ |b|\cdot |a \plus{} b \plus{} c| \plus{} |a|\cdot |c| \\
 \\
|b \plus{} c|\cdot |c \plus{} a| \equal{} |c(a \plus{} b \plus{} c) \plus{} ba|\ \le\ |c|\cdot |a \plus{} b \plus{} c| \plus{} |b|\cdot |a| \\
 \\
|c \plus{} a|\cdot |a \plus{} b| \equal{} |a(a \plus{} b \plus{} c) \plus{} cb|\ \le\ |a|\cdot |a \plus{} b \plus{} c| \plus{} |c|\cdot |b| \end{array}\right\}$ .[/color]


[quote="tonypr"]Using only the digits $ 3$ and $ a$, we form the next number of $ 90$ digits:

$ 3a33a333a3333a\cdots33\cdots3a$

If the number is a multiple of $ 18$, then what are the possible values of $ a$?

Hide solutions[/quote]

Clearly $ a$ must be even.

Then let the number of $ a$s be $ n$, then

$ 2\plus{}3\plus{}...\plus{}(n\plus{}1)\equal{}90$
$ \Leftrightarrow \frac{n(n\plus{}3)}{2}\equal{}90$
$ \Leftrightarrow n\equal{}12$

So the sum of digits of the number is divisible by $ 9$
$ \Leftrightarrow 9|3 \times (90\minus{}12)\plus{}12a$
$ \Leftrightarrow 3|a$

So $ 0,6$ are the possible values of $ a$


Show that
\[ \sum_{( a, b ) = 1} \frac{1}{( ab )^2} = \frac{5}{2}  \]
where $a, b$ runs through $\mathbb{N}$.


A beautiful problem:

Suppose we partition the unit square into polygons (not necessarily convex). Show that there is an $r>0$ s.t. if all the polygons can be covered by a circle of radius $r$, there is at least one polygon which has no less than $6$ neighbours (two polygons are said to be neighbours if they share at least a vertex).

In other words, if the polygons are "small enough", one of them will have $\ge 6$ neighbours.

This one appears for $r = \frac 1 {30}$ as ex. #118 in
A.M. Yaglom, I.M. Yaglom, Challenging mathematical problems with elementary solutions vol. II, Dover.
But their proof is quite long....

Pierre.


Each test in Chung's math class is worth 100 points. Chung scored 78 points on the first test, and her score on the second test raised her average by exactly 2 points. On the third test, her score again raised her average by exactly 2 points, and she continued to score well enough so that her mean was increased by 2 points each time. After some number of tests, she realized that no matter how well she did on the next test, it was impossible to raise her mean by exactly 2 points. What was her mean at that time?

So her test scores were $ 78,82,86,90,...$.

Note that they must always increase by $ 4$.

So, since she can't score 102, she ends her streak at 98, at which time the mean is 88.


calculus
derivative
function
calculus computations

I need help with the below problem... I tried lots of ways and I always get stumped..

If f is a differentiable function, and g(x,y,z)=f(yz,xz,xy), then what is the partial derivative of g(x,y,z) in respect to x in terms of f?

It's the multivariable chain rule- and your problem is most likely that you're not distinguishing between two different things named "$ x$".

Let $ X(x,y,z)\equal{}yz,Y(x,y,z)\equal{}xz,Z(x,y,z)\equal{}xy$.
Now $ g(x,y,z)\equal{}f\circ (X,Y,Z)(x,y,z)$, so $ \frac{\partial g}{\partial x}\equal{}\frac{\partial f}{\partial X}\cdot \frac{\partial X}{\partial x}\plus{}\frac{\partial f}{\partial Y}\cdot \frac{\partial Y}{\partial x}\plus{}\frac{\partial f}{\partial Z}\cdot \frac{\partial Z}{\partial x}\equal{}z\frac{\partial f}{\partial Y}\plus{}y\frac{\partial f}{\partial Z}$ all evaluated at $ (X(x,y,z),Y(x,y,z),Z(x,y,z))$ or $ (x,y,z)$ as appropriate.

The other partial derivatives are similar, of course.

Your solution doesn't work out when say:
g(x,y,z) = (xyz)^2 and f(yz,xz,xy) = xyz

Or am I just confused and you can't use that example.

And if you solution is correct, then does partial f in respect to Y = partial f in respect to xz?

The expression "$ \frac{\partial f}{\partial xz}$" is nonsense. $ f$ is a function of three variables, and the quantity I've called $ \frac{\partial f}{\partial Y}$ is its derivative with respect to the second variable. It doesn't matter whether I call that first variable $ y,Y,z,\gamma$, or anything else.

We evaluate this derivative at the point $ (yz,xz,xy)$.

Your example is definitely confused. What is $ f(2,3,6)$? $ 6$ or $ 36$?
Never lose sight of what the expression $ f(a,b,c)$ means- it's a function evaluated at a specific point.


advanced fields
advanced fields theorems

In Hatcher's book page 58 some covering spaces of $ S^1 \vee S^1$ are showed.  For instance, in the first example what does $ < a,b^2,bab^{ \minus{} 1} >$ mean?   Also, in the picture, are the two edges label by $ b$ identified?

Please I'll appreciate if you can clarify this to me

Also, I guess in the picture there is only one vertex, however, it looks like there is an intersection point in the right side of the picture that is not a vertex, how is that possible?   how can I attach 1-dimensional cells with intersection points ?


Now here's a more difficult inequality, And I think that it's long time till there's anyone can solve it. Any where, It still look simple and easy....

Prove that for all $ a,b,c$ nonegative reals:

$ \frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ac}}+\frac{1}{\sqrt{4c^{2}+ab}}\ge \frac{4}{a+b+c}$

Please post here, Any idea for it?

To VASC : I want to ask you if $n-1 EVQ$ can help us to solve this problem? Because I really don't know how to use it?

[quote="hungkhtn"]No, of course. THe equal take if a=b,c=0.[/quote]
Did you mean we can use jensen? or not? Sorry, but I'm afraid that I didn't get it.
$\frac{1}{\sqrt{4a^2+bc}}+\frac{1}{\sqrt{4b^2+ac}}+\frac{1}{\sqrt{4c^2+ab}} \ge \frac{4}{a+b+c}$
Substract both sides by $\frac{3}{a+b+c} :$
 $\Rightarrow (\frac{1}{\sqrt{4a^2+bc}}- \frac{1}{a+b+c})+(\frac{1}{\sqrt{4b^2+ac}}-\frac{1}{a+b+c})+(\frac{1}{\sqrt{4c^2+ab}}- \frac{1}{a+b+c}) \ge \frac{1}{a+b+c}$
$\Rightarrow \sum (\frac{1}{\sqrt{4a^2+bc}}- \frac{1}{a+b+c}) \ge \frac{1}{a+b+c}$
It's obvious that $a,b,c$ cannot be all zero, then times $a+b+c$ for both sides:
$\sum (\frac{a+b+c-\sqrt{4a^2+bc} }{\sqrt{4a^2+bc}}) \ge 1$ $...$
Frank

Is your solution? Do you complete it? Because It's really incomplete!

Hi everyone, this inequality was solved by a Vietnamese high school student two months ago. I posed some hard inequalities to him, he solved all of them. He used mixing variable method.

Here is a line in his solution:

We need to show that $ 6303t^{5}+3320t^{4}+1776\geq 5656t^{3}$, where $ 0\leq t\leq1$.


number theory
greatest common divisor
number theory unsolved

m,n are positive integers, let (m,n)=1, find the possible value of (m^3+n^3, m^2+n^2)

(a,b) refers to the gcd of a and b.

Let $ p^k$ be a prime power dividing both $ m^3 + n^3$ and $ m^2 + n^2$.  Clearly $ p$ cannot divide either $ m$ or $ n$ or else it divides both, so $ \gcd(p^k, m) = \gcd(p^k, n) = 1$.  On the other hand,

$ m^3 \equiv - n^3 \bmod p^k, m^2 \equiv - n^2 \bmod p^k \implies$
$ \frac {m^3}{m^2} \equiv \frac { - n^3}{ - n^2} \bmod p^k \implies$
$ m \equiv n \bmod p^k$

so in fact $ m^3 + n^3 \equiv 2m^3 \bmod p^k$ and $ m^2 + n^2 \equiv 2m^2 \bmod p^k$.  On the other hand, $ p$ cannot divide $ m$, so $ p^k = \boxed{2}$ is the only possible non-unital gcd.

Algebraically (and I suppose this is much faster):
\begin{eqnarray*}
(m^3 + n^3, m^2 + n^2) &=& (n^3 - mn^2, m^2 + n^2) \\
 &=& (n - m, m^2 + n^2) \\
 &=& (n - m, 2m^2) \\
 &=& (n - m, 2)
\end{eqnarray*}

where we have used the following facts:  $ (n^2, m^2 + n^2) = 1, (m, n - m) = 1$.


The lengths of the sides of a convex hexagon $ABCDEF$ satisfy : $AB=BC$ ; $CD=DE$ ; $EF=FA$.
Show that : $\frac{BC}{BE} +\frac{DE}{DA} +\frac{FA}{FC} \geq \frac{3}{2}$

http://www.mathlinks.ro/Forum/viewtopic.php?t=26

  darij


let :0<x,y,z<1
and $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$
prove that:
$ \sum \frac {\sqrt {x(1 \minus{} x)}}{\sqrt {y(1 \minus{} y)} \plus{} \sqrt {z(1 \minus{} z)}}\geq x \plus{} y \plus{} z$

all I can get is that there exists sides of a triangle $ a,b,c$ such that

$ x \equal{} \frac {a}{b \plus{} c} ,y \equal{} \frac {b}{c \plus{} a} ,z \equal{} \frac {c}{a \plus{} b}$

not a lot, actually  :(

maybe the best method to solve it is substitution:
$ \sqrt {xy}\equal{}cos \alpha$
$ \sqrt {yz}\equal{}cos \beta$
 $ \sqrt {zx}\equal{}cos \gamma$
$ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$


I don't know if this is the right forum to post this but

Solve $n^{3}+5n^{2}-49n\equiv 45\pmod{100}$

[hide="Not too sure"]Hensel's Lemma and CRT? :blush: [/hide]

Hensel's Lemma is probably too strong of a result.  Anyway, this problem is not too hard.  It might be hard $\bmod 1000$.

[hide="Solution"] We will solve $\bmod 4, \bmod 25$.

Now, stuff in $\bmod 4$ is really simple.  The equation becomes $n^{3}+n^{2}-n \equiv 1 \bmod 4$ which has solutions $n \equiv 1, 3 \bmod 4$.  In other words, $n$ is odd.

The solution is less obvious $\bmod 25$, but let's calculate a solution $\bmod 5$ first.  Now, it reduces to

$n^{3}+n \equiv 0 \bmod 5 \implies n(n^{2}+1) \equiv 0 \bmod 5$

Which has solutions $n \equiv 0, 2, 3 \bmod 5$.

Now, $\bmod 25$ we know the solution must take the form $n = 5k+(0, 2, 3)$ (where $(0, 2, 3)$ is just my shorthand notation for a calculation of each case; I will do the calculations simultaneously).  Plugging in and taking $\bmod 25$ we have

$15k \cdot (0, 2, 3)^{2}+(0, 2, 3)^{3}+5 \cdot (0, 2, 3)^{2}+5k+(0, 2, 3) \equiv 20 \bmod 25$

Now, $(0, 2, 3)^{3}+(0, 2, 3) = (0, 10, 30)$.  Dividing by $5$ (which makes our equation $\bmod 5$) we have

$3k \cdot (0, 4, 4)+(0, 2, 1)+(0, 4, 4)+k \equiv 4 \bmod 5$

Simplifying a little,

$k \cdot (1, 3, 3) \equiv (4, 3, 4) \bmod 5$
$k \equiv (4, 1, 3) \bmod 5$

Hence $n \equiv 20, 7, 18 \bmod 25$.

Using CRT we have $n \equiv \boxed{ 45, 7, 43 }\bmod 50$. [/hide]


Zephyrus has a bowl and six lines on his table. Four of the lines are 2 inches long, and the other two are of unknown length. If the six lines can be arranged into a hexagon with four angles of measure 135 degrees and two angles of measure 90 degrees with integral area, what is the shortest possible lengths of the two lines of unknown length?

[quote="Treething"]Zephyrus has a bowl and six lines on his table. Four of the lines are 2 inches long, and the other two are of unknown length. If the six lines can be arranged into a hexagon with four angles of measure 135 degrees and two angles of measure 90 degrees, how long are the two lines of unknown length?[/quote]

[hide]they are the same length[/hide]

[quote="Treething"]gah, i left out something. Should be a lot more fun now.[/quote]

[hide]oh, the answer is $\frac{\sqrt{2}}{4}$ then[/hide]


calculus logarithms integration derivative

Uuh wrong forum, and what exactly do you want? The second fundamental theorem (isn't it just the second half of the theorem?) basically states that integration and differentiation are opposite processes.

[quote="cmrockett07"]someone explain this stuff to me. how is it applied when used with Chain Rule, logarithms, etc.???????????[/quote]

You can [b]define[/b] logarithm as,

$\ln x = \int_{1}^{x}\frac{dx}{x}$

Then, by fundamental theorem,

$\frac{1}{x}= (\ln x)'$ on $(0,\infty)$


A very small cube of mass $ m$ is placed in the inside of a funnel rotating about a vertical axis at a constant rate of $ \omega$ revolutions per second.  The wall of the funnel makes an angle $ \theta$ with the horizontal.  The coefficient of static friction between the cube and the funnel is $ \mu_{s}$ and the center of the cube is at a distance $ r$ from the axis of rotation.  Find the ($ a$) largest and ($ b$) smallest values of $ \omega$ for which the cube will not move with respect to the funnel.

Sorry I can't provide any diagrams.  The cube is placed somewhere on the wall of the funnel.

For $ (a)$; we want $ \omega$ such that cube does not slide down$ \Rightarrow$friction acts upwards. For $ (b)$; we want $ \omega$ such that cube does not slide up$ \Rightarrow$friction acts downwards.
[hide="Answer"]$ \sqrt{\frac{g(\sin\theta\minus{}\mu\cos\theta)}{r(\cos\theta\plus{}\mu\sin\theta)}}\leq\omega\leq\sqrt{\frac{g(\sin\theta\plus{}\mu\cos\theta)}{r(\cos\theta\minus{}\mu\sin\theta)}}$[/hide]


[img]http://img011.photo.21cn.com/photos/user/24454028/20070415211058/o/8547912.jpg[/img]

It seem that the formula is complex

But I found that I can't make it simple

$\frac{1}{2}\left( \sqrt{\frac{\sqrt{5}}{8}}\right) \left( \sqrt{1+\sqrt{5}}\right) a^{2}$??

What exactly do you want?

Trig functions on multiples of $\pi/5$ are going to have $\sqrt{5}$'s all over... you can't help it.

Prove that formula is equal :

[img]http://mathworld.wolfram.com/images/equations/RegularPolygon/inline116.gif[/img]

Sorry , that $a$ is unnecessary

why exactly do trig functions on multiples of $\frac{\pi}{5}$, come out with $\sqrt{5}$ in them?

Most of the $\frac{n\pi}{20}$ trig formulae are provable by making a regular pentagon... you have a bunch of 1-2-$\sqrt{5}$ triangles.


Bon, donc fidèle à moi-même, je cre un nouveau fil pour l'exo d'Alain123 :

[quote="Alain123"]Aller, on se lance !!

Calculer
$\int_0^{+\infty} \frac{\cos(t)}{1+t^2}dt$

CA marche !![/quote]

Pierre, dsol pour le titre super nul... :blush:

Bonjour,

Ca ressemble bigrement  un exo colle de sp, chapitre "intgrales pas bien propres trs dpendantes d'un paramtre"  :P

f(x) = $\displaystyle \int_0^{+\infty} \frac{\cos(x*t)}{1+t^2}dt$

C'est dfini trivialement, drivable, paire, tout ce qu'on veut et mme plus, en bidouillant un peu (avec $\displaystyle \int_0^{+\infty} \frac{t \sin(x*t)}{1+t^2} dt $ et ses petites soeurs), il vient f''(x) = f(x). Sachant que f(0)=pi/2, .. f(x) = pi/2 * e^-abs(x)

Que j'ai combien  ma colle ?

Moyen .....

Manque de justifications rigoureuses tout cela ....

Enfin, pour un retour aprs des ftes de fin d'anne , disons, CORRECT.

Bah, je rserve les justifs rigoureuses pour un exo moins triv'  :D   ;)

Que de toute faon c'est du cours direct et que l'excs de rigorisme tue l'intuition  :P

Though I don't undestand whole discussion, I see belenos used quite well known trick. :D
As I remember, we can calculate $\int_0^{+\infty}\frac{\sin t}{t}dt$ in [i]almost[/i] the same way. :)

Et voil...on cre un forum franais, o il s'agit de parler franais, et on se fait envahir par des cohortes de russes qui parlent anglais  :D  :D  :D

Pierre, que pauvre France... :P

Mais que diriez-vous des armes roumaines qui parlent franais et l'anglais aussi? :)

...[hide="venir bientot"]Valentin, que pavre Roumaine, L'anglais et un peu Franais :P[/hide]

[quote="Myth"]Hmm... I only understand that Pierre said something about "russes". I hope it were kind words ;)[/quote]

probably you understood "cohortes", too. anyway my skills in understanding french are not much better :D ...

  dg

moin mi cause crole de l'le de la  Runion
Ot zoreil  qomen y l, bon an marmailles

pou $\int_{0}^{+\infty}(sinx)/xdx$ sa ou calqil aveq transforme de
Laplace

L'le de la Runion en force

Yes, something like Laplace transformation.

[quote="pbornsztein"]Et voil...on cre un forum franais, o il s'agit de parler franais, et on se fait envahir par des cohortes de russes qui parlent anglais  :D  :D  :D [/quote]
sniff... I have translated this text... sniff... Rrrrr....

[quote="Myth"]sniff... I have translated this text... sniff... Rrrrr....[/quote]allons tous aux forum russes et soumettons à une contrainte du Mikhail en russe :P

[quote="Myth"]You may go. Anyway, no russian discussions in Russian local community.[/quote]Mais je ne peux parle pas de russe :( ...

Laissons les russes venir ici!
Aprs tout, Tolsto parlait franais !!!  :D

Et Anna Karenine, c'est vraiment gnial.

Pierre, qui fait un peu dvier le topic...


algebra
polynomial
number theory
prime divisor
IMO 2008

[quote="[S]unn[Y]"]Show that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4+1$ is greater than $2n$[/quote]
Well, I don't know if is possible to use the Dirichlet's theorem to conclude that exists infinitely $n$ such that $n^4+1$ is prime, because in this case, the problem is solved, but one comment is that [can's use here]
\[
n^4+1=(n^2-\sqrt 2n+1)(n^2+\sqrt 2n+1)=(n^2-\sqrt {2}nI-1)(n^2+\sqrt 2nI-1),
\]
and of course if $n^4+1$ is a product of two primes and $n>1$ the result holds too, and with three primes is enought that $n>7$ to the result hold. But I don't have a consistent a proof. Sorry!.

RobertuX the C.M.

[quote="harazi"]Oh, my god! Could you please at least simulate that you think a little bit before posting?[/quote]
The last prime divisor for me if number is prime is itself, at least, for my information, if not, please, correct to me. Sorry if I try to give some comments of the problem, I think is NOT bad to make some comments, if not tell me WHY!?

RX

[quote="RobertuX"]
Well, I don't know if is possible to use the Dirichlet's theorem to conclude that exists infinitely $n$ such that $n^4+1$ is prime, because in this case, the problem is solved[/quote]

Splendid!  :D  :D

Now, you have solved one of the main open question about primes of the form $n^2+1.$  :D

Pierre.

I am really curious to see who will find first the 3 lines solution of this great problem.

[quote="[S]unn[Y]"]Show that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4+1$ is greater than $2n$[/quote]
$n^4+1=(n^2-1)^2+2n^2$ if is prime the result holds, if is a compositi number, then every prime has the form $q_1^2+2q_2^2$ but don't have, for the moment, any argument to prove has infinitely greater than this maximum prime is greater than $2n$.  :?

RX

Well, If we do not consider a one-line solution using a really nice extansion of Dirichlet's theorem :D , I think I will have to use more than 3 lines... ;)

First, it is clear that there infinitely many primes $p$ which divides at least one of the numbers of the form $n^4+1$ (otherwise, as Euclid did, consider $(p_1 \cdots p_k)^4+1$ ).

Now let $p$ be any of these odd primes and $n$ such that $n^4+1 = 0 \mod [p].$
Let $n = qp+r$ be the euclidean division of $n$ by $p$, so that $0<r<p$. It is also clear that  $p$ divides $r^4 + 1 = (n-pq)^4 + 1.$ Thus $p$ also divides $(p-r)^4+1.$
Let $m = \min (r,p-r).$ Then $p > 2m$ and $p$ divides $m^4+1.$ Thus the greatest prime divisor of $m^4+1$ is greater than $2m.$

Now, for each $p$ as above we may find some $m$ which has the desired property and such that $p$ divids $m^4+1$. Since, for any of these $m$, the number $m^4+1$ has only a finite number of prime divisors, so that $m$ is associated to only a finite number of primes $p$ in the above reasoning.
Since we have seen that we have infinitely many primes $p$ to start the reasoning, it follows that we will have infinitely many $m$ with the desired property, and we are done.

Pierre.

What extension of Dirichlet theorem? I used it in its original form.

[quote="RobertuX"]
I don't know if is possible to use the Dirichlet's theorem to conclude that exists infinitely $n$ such that $n^4+1$ is prime, because

[quote="harazi"]Oh, my god! Could you please at least simulate that you think a little bit before posting?[/quote]
[b]Sorry but I can't believe. Look your differnet reaction using a similar comment [/b]

[quote="pbornsztein"][quote]that exists infinitely $n$ such that $n^4+1$ is prime[/quote]

That one  :)

Pierre.[/quote]

[quote="harazi"]Ah, ok, I understand now! No, in fact, I really used Dirichlet's theorem to have a solution for this problem.[/quote]

[/quote]
Sorry I am lost with the way you talk with the comunity, sorry. I feel excluded and confused.  :what?:

But I have solved this problem using Dirichlet's theorem! Why nobody believes me?

I suppose you used it just for $8k+1$?  ;)

So it would suffice to show that there are infinitely many $p$ s.t. there's at least one solution to $x^4+1$ in $\mathbb Z_p$, right? Because if $x$ is a solution, then $-x$ also is, so we can find one smaller than $\frac p2$. $x^4+1$ does indeed have a root in $\mathbb Z_p$ for primes of the form $8k+1$, as you guys mentioned :). That's because the polynomial $x^8-1$ has $8$ roots in that case, so $x^4+1$ must have $4$.

And one does not need to employ Dirichlet's Theorem for this. We can look at divisors of numbers of the form $2^{2^t}+1$, which are all of the form $k2^{t+1}+1$.

Is this Ok? Anyway, I suppose this is pretty much what Pierre did, if you think about it.

Yup, that was the idea and it is similar to Pierre's, but shorter. I said 3 lines, don't forget. :D

I could have spared the explanations, and the stuff about $2^{2^t}+1$. Hmm.. Yes, I suppose it would've been about three lines then :D.

[quote="harazi"]Source: one of Leningrad's mathematical olympiads (about two or three years ago). I think this is the most beautiful number theory problem that has ever been given in a mathematical olympiad. Enough with sentimentalisms...[/quote]

Oh, thank you, I am embarrased.

[quote="[S]unn[Y]"]Show that there exist infinitely many positive integers $ n$ such that the largest prime divisor of $ n^4 \plus{} 1$ is greater than $ 2n$[/quote]
It is obviosly. Let $ p\equal{}1\mod 8$ and $ a$ is primitive root and $ b\equal{}a^{(p\minus{}1)/4}\mod p,c\equal{}b^3\mod p\equal{}\frac{\minus{}1}{b}\mod p$, then exist 4 roots $ x^4\plus{}1\equal{}0\mod p$, they are $ b,c,p\minus{}b\equal{}b^5\mod p,p\minus{}c\equal{}b^7\mod p$. Obviosly one of roots less than p/2. It give $ n,p>2n, p|n^4\plus{}1$ for any prime $ p\equal{}1\mod 8$.

Rust, more than 2 years passed and you still haven't learned anything at all from this forum. You could start taking a glimpse at the previous discussion, write that obviosly obviously and so on. But I guess that's a huge effort.

This is the official solution. [St. Petersburg 2001]

[b]Proof.[/b]

We claim first that there are infinitely many numbers that are prime divisors of $ m^4 \plus{} 1$ for some $ m$. Suppose to the contrary that there is only a finite number of such primes. Let $ p_1, p_2,\ldots, p_k$ be all of them. Let $ p$ be any prime divisor of $ (p_1 p_2\cdots p_k )^4 \plus{} 1$. This number cannot equal any $ p_i$ . This contradicts our assumption, and establishes the claim.

Let $ \mathcal{P}$ be the set of all numbers that are prime divisors of $ m^4 \plus{} 1$ for some $ m$. Pick any $ p$ from $ \mathcal{P}$ and any integer $ m$ such that $ p$ divides $ m^4 \plus{} 1$. Let $ r$ be the residue of $ m$ modulo $ p$. It follows that $ r < p$ and $ p$ divides both $ r^ 4 \plus{} 1$ and $ (p \minus{} r )^4 \plus{} 1$. Let $ n$ be the minimum of $ r$ and $ p \minus{} r$ . It follows that $ n <p/2$ or $ p > 2n$. If $ n$ can be obtained using the construction above, then it satisfies the desired condition. If it is constructed using the prime $ p$, then $ p$ divides $ n^4 \plus{} 1$. Thus, any such number $ n$ can be constructed with only a finite number of primes $ p$. Since the set $ \mathcal{P}$ is infinite, and for each integer $ m$ such a number $ n$ can be constructed, there is an infinite number of integers $ n$ satisfying the desired condition.

Looking back, it is quite strange to think that they gave a really similar problem at IMO 2008. :wink:

Pierre.

[quote=SCP][quote="grobber"]

And one does not need to employ Dirichlet's Theorem for this. We can look at divisors of numbers of the form $2^{2^t}+1$, which are all of the form $k2^{t+1}+1$.

Is this Ok? Anyway, I suppose this is pretty much what Pierre did, if you think about it.[/quote]

I think this is not OK,
$n=2^{2^{t-2}}$ and you say $k2^{t+1}+1>=2^{2^{t-2}+1}$ but if $t$ is big, the $RHS$ is bigger, not?[/quote]

I think this is OK. The part about $2^{2^t}+1$ is merely to assert there are infinitely many primes which are $1 \pmod 8$.

There exist infinitely many primes that are divisors of integer $n^4+1$ for some $n\in N$.
Assume that there exist only finitely many integers $n$ so that largest prime divisor of $n^4+1$ is greater than $2n$. Let $m$ be the biggest such integer. Take a prime $p$ that divides $r^4+1$ for some $r<p$ and such that $p>m^4+1$. Note that $(p-r)^4+1$ is also divisible by $p$. One of $r$ and $p-r$ is smaller than $p/2$ , let it be $k$. So $p$ divides $k^4+1$ and $2k<p$. And so $k^4\ge p-1>m^4$ thus $k>m$ -contradiction with the maximality of $m$.


[quote="Beat"]Let $m$, $n$ be positive integers with $n > 1$, and let $a$, $b$, $c$ be positive real numbers satisfying
$a^{m+1}+b^{m+1}+c^{m+1}= 1$. Prove that $\frac{a}{1-ma^{n}}+\frac{b}{1-mb^{n}}+\frac{c}{1-mc^{n}}\ge\frac{(m+n)^{1+\frac{m}{n}}}{n}$[/quote]
I'm not sure about this one.
Take $a\to 0$,$b\to 0$ and $c\to 1$ you will get that
$0>\frac{1}{1-m}\geq \frac{(m+n)^{1+\frac{m}{n}}}{n}$
which isn't true,I think :lol: .


superior algebra
superior algebra unsolved

Hello,
Can someone please help me solve these questions with full working out please ?

1. Show that 5 is irreducible in Z[sqrt(-2)] and Z[sqrt(3)] but not in Z[i].

2. Find a factorisation of 9 + 4(sqrt(3)) into irreducibles, in Z[sqrt(3)].

3. Find four greatest common divisors of 13 + 39i and −17 + 19i in Z[i].

Thank You.

Does anyone have any ideas ? Please, can anyone help me ?

1) $ 5\equal{}(1\plus{}2i)(1\minus{}2i)$ so not irreducible in $ \mathbb{Z}[i]$.


function
real analysis
real analysis unsolved

Supposed we have two metric spaces $ X$ and $ Y$, then $ f$ is a continuous function from $ X$ to $ Y$. If $ E$ dense in $ X$, prove that $ f(E)$ is dense in $ f(X)$.

I think you should come back the definition of dense. $ A$ is dense in $ B$ if every point in B is a limit point of A, or more simple $ \overline{A}\equal{}B$.
So, we need to prove that $ \overline{f(E)}\equal{}f(X)$, which means that we have to prove every point in $ f(X)$ is a limit point for $ f(E)$.
Let $ y$ be a point in $ f(X)$, then $ y\equal{}f(x)$ for some $ x\in X$. Since $ E$ is dense in $ X$ so there is a sequence $ {x_n}$ in $ E$ such that $ lim x_n \equal{} x$. Since $ f$ is continuous, $ f(x_n)\to f(x)\equal{}y$. So, $ y \in \overline{f(E)}$. Therefore $ f(E)$ is dense in $ f(X)$


#27

If each letter in the expression $ (a \plus{} \frac {b}{c})(d \plus{} e)$ is replaced by a different digit from 1 to 9, inclusive, what is the smallest possible integer value of the expression?

[hide="Answer"]8. For some reason, I kept on getting 9...[/hide]

#28

For any a and b, we will define a "move that adds (a,b)" as a move from any point (c,d) to the point (c+a, d+b). A particle moves according to the following rules. If a move added (p,q) to get to the current position, the next move will add either (p-1,q) or (p,q-1) to move to the next position. Neither p nor q can ever be negative. Particle A starts at (0,0) and moves to (3,4). Particle A will continue to move according to the preceding rules. When the particle can no longer move according to the preceding rules, it is located at point (r,r). What is the value of r for particle A?

[hide="Answer"]14[/hide]

...then you are obviously not using the optimal positioning :P

no...
[hide="27)"]when $ a \equal{} 1, b \equal{} 3, c \equal{} 9, d \equal{} 2, e \equal{} 4$, then $ \left(1 \plus{} \frac {3}{9}\right)(2 \plus{} 4) \equal{} \left(\frac {4}{3}\right)(6) \equal{} \boxed{8}$ which is indeed the answer. Prove that $ 8$ is the minimum, for an extra challenge ;)[/hide]

*GASP* OH MY GOSH I CAN'T BELIEVE I DIDN'T FIND THAT POSSIBILITY.

But thanks anyways! :D

And after a loooong while of checking I finally realized how I got #28 wrong :mad: ...you had to keep on going until you get a move that's getting you nowhere...

Also, I was originally going to post #30 but then after I had dinner today I was able to solve all of #27, 28, and 30 without hassle, realizing "hey, it was nothing but stupid mistakes" :|

#27 just required lots of trial and error or something...experimenting with equivalent values of $ \frac{4}{3}$...?

[hide="#28"]The first move added $ (3,4)\implies{p\plus{}q\equal{}7}$. The next move will add $ (p,q)$ where $ p\plus{}q\equal{}6$. Then the next move $ p\plus{}q$ will be $ 5$ and so on until $ p\plus{}q\equal{}1$. Thus, the possible final positions of the particle will be in the form $ (3\plus{}P, 4\plus{}Q)$ where $ P\plus{}Q\equal{}6\plus{}5\plus{}4\plus{}3\plus{}2\plus{}1\equal{}21$. We are given that the $ 3\plus{}P\equal{}4\plus{}Q$ so $ P\equal{}11$ and $ Q\equal{}10$ so the answer  is $ \boxed{14}$.[/hide]


calculus
derivative
function
limit
calculus computations

Let f be a function that is everywhere differentiable and that has the following properties:

(i) f(x + h) = (f(x) + f(h)) / (f(-x) + f(-h)) for all real numbers h and x.
(ii) f(x) > 0 for all real numbers x.
(iii) f'(0) = - 1
(iv) f(-x) = 1/f(x), f(x+h) = f(x)f(h)

Use he definition of the derivative to find f'(x) in terms of f(x).

 :lol:

From the first condition, we have $f(x+h) = \frac{f(x)+f(h)}{f(-x)+f(-h)}$ for all real $x$ and $h.$

We put $h = 0,$ denote $f(x)$ by $y$ and note that $f(-x) = \frac1{f(x)}.$ We thus get

$y = \frac{y+f(0)}{\frac1{y}+f(0)}$
$\Rightarrow (f(0)-1)(y-1) = 0$

But, $y = f(x) = 1$  for all real $x$ violates condition $(3).$ So, we must have $f(0)=1.$

Thus, $f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$

$\Rightarrow f'(x) = \lim_{h\to 0}\frac{f(x)f(h)-f(x)}{h}$ ... [Using the last condition]

$\Rightarrow f'(x) = f(x)\cdot\lim_{h \to 0}\frac{f(h)-1}{h}$

$\Rightarrow f'(x) = f(x)\cdot\lim_{h \to 0}\frac{f'(h)}{1}$ ... [Using L'Hospital's rule]

$\Rightarrow f'(x) = f(x)\cdot f'(0)$ ... [Since the function is differentiable everywhere.]

$\Rightarrow \boxed{f'(x) =-f(x)},$ which leads to $f(x) = e^{-x}.$

Edit: Thanks to JBL for pointing out a "small" error in my solution earlier.

[quote="JBL"][quote="FieryHydra"]However, the problem with the above solution is that we get $f(x) = e^{-x}$ which violates condition $(1).$ Am I missing something?[/quote]

You're missing that the function $e^{-x}$ satisfies condition $(1)$ :)

You could have gotten to $f(0) = 1$ simply by plugging $x = h = 0$ into condition $(1)$ and saved yourself a little time, as well.[/quote]
Indeed, JBL, you are right! :) I wasn't paying enough attention while working on the problem. And, yes, I could have saved a lot of time by putting $x=h=0.$ Thanks!


Answer: a

Lets say that each hour the cell phone is on, it consumes 1 unit of energy from the battery.  So after 9 hours of being on, there are 24-9=15 units of energy left in the battery.    The problem said that actually using the cell phone continously lasted 3 hours.  Since there are 24 units of energy in the batter, that means that 24/3 = 8 units of energy are used per hour the cell phone is actually used.  She's been using it for an hour, so 8 units of energy are additionaly used.  So now there are 15-8=7 units of energy left in the battery.  She's now going to leave it on for the rest of the time.  Since 1 unit of energy is consumed per hour, it will last 7 more hours.

Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the batter will last for 24hours. If she is using it constantly, the battery will last for only 3hours. Since the last recharge, her phone has been on 9 hours, and during that time she has used it for 60 minutes. If she doesn't talk any more but leaves the phone on, how many more hours will the battery last?

[hide]okay, by reading the problem a few times, you can tell that 1 hr of talking on the phone is equal to 8 hrs of leaving it on without using it.... so he has left it on for 9hrs - the 1 hr she has talked, so she lefted it on without talk for 8 hrs.. and since talking for 1hr is equal to not talking for 8 hrs, you do 8 + 8 = 16, and 24-16 = 8...

oops, I got a different answer than dinoman, i think it's b....  :?  :?[/hide]

but wouldn't it be 8 + 8 because the one hour that's on the phone already include the time on... like when I said 1 hr talking is = 8 hrs on, I meant that the 1 hr one already includes the hr in which the phone have to be on for her to be talking.... i"m confused in whether or not that counts....  :?  :?

Yeah, but her cell phone was already on for 9 hours.

EDIT: OH! I GET IT NOW!

Lalala, you're right.

I remember this from a practice. It's a pretty cool problem. You've got to think about multiple ideas like ratios and such.

oooooooooops silly missreading....

ya i get it now, [b]during [/b]the 9 hours it was on she talked for an hour....

thanks lucy

o, was I right? that was just a random guess... so the answer is 8? i thought it wouldn't be 8 because 24/3 = 8 and thought people wouldn't make it that simple...

oooo,,

I think I get it now..

it was really confusing..

I missed it in AMC8..

Thanks alot guys :lol:

I think I've asked this before, and I know I should know this, but what's AMC8?

*Ahem* The AMC8 stands for the American Mathemactical Compitition 8th grade. It is a very predigous err... *loses important voice* compitition... umm yea.  :lol:

[quote]*Ahem* The AMC8 stands for the American Mathemactical Compitition 8th grade. It is a very predigous err... *loses important voice* compitition... umm yea. [/quote]

HaHaHa :lol:

it makes me laugh..

he he....

yes, it's American Mathematical Competitions for 8th graders

this is a national contest that makes people's lives miserable

confession

the first time I joined AoPS, it was because I wanted to prepare for up coming AMC8,

but I lost interest in preparing AMC8..I started to like real math... not just preparations :P

[quote="shinwoo"]I need help with this one...this one is really weird

Niki usually leaves her cell phone on. If her cell phone is on but she is not actually using it, the batter will last for 24hours. If she is using it constantly, the battery will last for only 3hours. Since the last recharge, her phone has been on 9 hours, and during that time she has used it for 60 minutes. If she doesn't talk any more but leaves the phone on, how many more hours will the battery last?

a) 7
b) 8
c) 11
d) 14
e) 15

 :?  :(  :blush:[/quote]

[hide]every hour you talk on it, it takes up 8 hours of regular time

so 9+8-1=16

24-16=8

(B)[/hide]

[quote="LuCy4EvA"]is it a National kinda thing too?[/quote]

[quote="shinwoo"]this is a national contest that makes people's lives miserable[/quote]

it's b. since she uses a third of the power she has initially, 60 min is a third of 3 hr.
and the other 8 hrs is a third of the total power. so she has a third left and a thrid of 24 is 8 or b.[/hide]


Two main points
a) They are Christian missionaries
b) They were kidnapped

Are there anyone who concern this topic like me?

[quote]What's your opinion on Korean missionaries' being kidnapped[/quote]
I strongly disapprove of it.

According to my experience, Christian--every branch--are very aggressive, believers like to convert others, reduce the number of unbelievers, mentally or physically.
Are there anyone(non Christian) who never see some "Brother John" knocking your door and begin to missionize?

Yes, Christian are good at propaganda, they are famous for their charity, which you can read on news everyday and see around you. But please remember: first, the money they spend comes from donation, and it is tax-free, second, if you know the ratio of the money that really goes to its real goals, you will know that these bristletails are exactly the same as their fathers in the midage.

Most people know that their charity never come alone, the next step must be sermon. Along with their giving you medicine, they will try to convince you that it is the God who ....

Taliban has two requests
a) Afghanistan government release 23(?) Taliban members in the jail.
b) South Korea must retreat their troops from Afghanistan.

[quote="fredbel6"][url]http://news.bbc.co.uk/2/hi/south_asia/6916762.stm[/url]

[url]http://news.bbc.co.uk/2/hi/asia-pacific/6915268.stm[/url]
Let's get ourselves some food for thought first.

Some things popping up in my mind :

- what kind of work were they doing there or planning to do? :|   I read "missionaries" : does that mean that they are catholics, who see is as their duty to convert others elsewhere, or simply to help the sick, the wounded and the poor?
-I can't figure out what exactly the demands are? If know South Korea has a military presence in Afghanistan, but do they even have the authority to meet those demands (like releasing captured Taleban warriors?)
- I don't know how society thinks about this in South Korea, but I know that Asian societies can have quite a different view on things like this.  Remember : in April 2004, three Japanese relief workers were captured and released, and back in Japan, they were not received as heroes at all.  (They even made a film [url=http://www.imdb.com/title/tt0456836/]Bashing[/url] about it.)[/quote]

Chosunilbo--(South)Korea Daily
http://english.chosun.com/w21data/html/news/200708/200708020009.html

Have the Taliban Forgotten Muhammad's Teachings?

[quote]When God created the world, he found a trash heap that couldn't be used for anything useful, so he threw it down to the earth and turned into Afghanistan, a story goes. Afghanistan is such a desolate and forlorn land. Taliban, grant "salaam" to the innocent young people who arrived in the most barren land in the world to help it become a land of peace.[/quote]

[quote="Buffalo"]According to my experience, Christian--every branch--are very aggressive, believers like to convert others, reduce the number of unbelievers, mentally or physically.
Are there anyone(non Christian) who never see some "Brother John" knocking your door and begin to missionize?

Yes, Christian are good at propaganda, they are famous for their charity, which you can read on news everyday and see around you. But please remember: first, the money they spend comes from donation, and it is tax-free, second, if you know the ratio of the money that really goes to its real goals, you will know that these bristletails are exactly the same as their fathers in the midage.

Most people know that their charity never come alone, the next step must be sermon. Along with their giving you medicine, they will try to convince you that it is the God who ....
[/quote]
 :lol:
You have some twisted ideas. :huh:

I myself was raised as a catholic, although I don't consider myself one. Not all christians are fanatics, where I live, most people know next to nothing about their own religion, which is becoming less and less important each day.

[quote="fredbel6"] :lol:
You have some twisted ideas. :huh:

I myself was raised as a catholic, although I don't consider myself one. Not all christians are fanatics, where I live, most people know next to nothing about their own religion, which is becoming less and less important each day.[/quote]

What do you mean "be raised as a catholic", you were educated as a catholic(people want you to become a catholic), or simply the fact that someone in your family is a catholic?

You mix up two concepts: individual family believers and the professional core members. Ordinary believers provides financial support so that those core members can enjoy their lives and have the energy to find more preys.

I believe that "Religion is the opium of the people". People have the right to take opium if they need it to console themselves, but if someone or some organizations try to sell it to me, I would call the policeman first.

[quote="Buffalo"][quote="fredbel6"] :lol:
You have some twisted ideas. :huh:

I myself was raised as a catholic, although I don't consider myself one. Not all christians are fanatics, where I live, most people know next to nothing about their own religion, which is becoming less and less important each day.[/quote]

What do you mean "be raised as a catholic", you were educated as a catholic(people want you to become a catholic), or simply the fact that someone in your family is a catholic?[/quote]
I mean that in my region, everyone was baptised and went  through the usual christian rites, and that I went to a catholic school. However, most people don't even go to church every week.

[quote]You mix  up two concepts: individual family believers and the professional core members. Ordinary believers provides financial support so that those core members can enjoy their lives and have the energy to find more preys.[/quote]
What is an "ordinary believer"?
Usually, they only pay taxes to the state, which (without asking them) gives some to the most important religions (here the catholic church gets the biggest chunk but mosques get money too..)

[quote]I believe that "Religion is the opium of the people". People have the right to take opium if they need it to console themselves, but if someone or some organizations try to sell it to me, I would call the policeman first.[/quote]
I agree, but why all this indignation?
We don't even know for sure what kind of work they were doing, or do we?

Update: 2 South Korean women have been freed.
http://www.cbsnews.com/stories/2007/08/13/terror/main3160692.shtml

Of course I am against kidnapping and other extremisms...  But these well-meaning and naive missionaries had no business putting themselves in harm's way.  The Moslem world largely believes Christendom, the west in general and the US  specifically are out to stamp out Islam.  US troops - Infidels - being in Saudi Arabia outraged many Moslems in 92 - never mind that they had been invited.  Tancredo's recent remarks threatening Mecca http://www.rockymountainnews.com/drmn/local/article/0,1299,DRMN_15_5673179,00.html
was about the worst possible thing to say in today's climate...  Missionaries acting in the capacity of humanitarians should be under the auspices of a UN or other ngo: Medecins sans Frontieres/Red Crescent etc. - for their own protection and to make their efforts effective.  Missionaries going as missionaries shouldn't.  Moslems converting away from their faith are liable to be put to death under Sharia law (as popularly interpreted); those urging them to do so can't fare much better.  Gack.  Utter utter foolishness...

[quote]
I agree, but why all this indignation?
We don't even know for sure what kind of work they were doing, or do we?[/quote]

Please, you two, let's not turn this into a flame war over whether missionaries are aggressive, horrible beings or not. They're human, let's leave it at that.

[color=green]Moderator edit: the first part of this post is responding to a spammy post that has been deleted.[/color]

I understand him and I don't see mistakes that makes it impossible to understand.
There is a nicer way to ask for better english, as it's not very likely that he does this on purpose.
Your amount of typing mistakes is much higher, Elemennop ;)

On the other side, I dont agree to his way of presenting his information and his rather radical (and generalising) point of view.


logarithms
integration
calculus
calculus computations

Calculate the exact value of the length of the portion of the curve
$ y \equal{} \frac12x^2 \minus{} \frac14ln(x)$ from $ x \equal{} 2$ to $ x \equal{} 4$.

[quote="AndrewTom"]
$ \equal{} [\frac {x^{2}}{2} \plus{} \frac {1}{4} \ln x]_{2}^{4}$

$ \equal{} (6 \plus{} \frac {1}{4} \ln 4) \minus{} (2 \plus{} \frac {1}{4} \ln 2)$ ... (*)

$ \equal{} 6 \plus{} \frac {1}{4} \ln 2$.[/quote]

Andrew Tom,

you had an accident in this step.  It should be:

$ \equal{} (8 \plus{}\frac 14 \ln 4) \minus{} (2 \plus{} \frac 14 \ln 2)$

Thank you for a correct procedure, work, and the final result.

Thanks Arrange your tan.

I made a transcription error in typing what I'd written on a piece of paper. I was rushing as usual. I've edited it now.

Thanks again.


AMC
USA(J)MO
USAMO
geometry
geometric transformation
rotation

So what exactly are the criteria for each color MOP?
Like red MOP is the top __ freshman
and blue mop is i don't know

Red MOP - freshmen only
Blue MOP - Anyone but seniors
Black MOP - Anyone

However, the USAMO cutoff score for each color gets higher and higher.  Usually Red is the lowest, and Black is always the highest.

As far as numbers go, I think there are 50 or 60 MOPpers all together.  And usually there are around 25 Red MOPpers.

Read page 6, Section VIII, first two paragraphs of the USAMO Teachers' Manual.  It's available on our web page.  In the first paragraph the approximately 12-18 , and in the second paragraph, the approximately 30 are chosen on the basis of USAMO score.

You will notice that  there is no mention of "colors" or "color groups".  The color designations are simply an ad hoc way of dividing the MOSP participants into groups for the purpose of assigning instructors, classrooms and topics and making sure a complete rotation of subjects at an appropriate level is assigned.  Students occasionally move from one color group to another based on past expereince and instructor recommendations.  This is to say that the color designations are nothing but an informal way of organizing one part of the program.

For some other parts of MOSP, e.g. the Team Tests, Greek letters are used and teams are mixed across age and grade levels.  For some other parts of MOSP, no designation is used.

Steve Dunbar
AMC Director

I'm pretty sure this is right...

Black: Top ~12
Blue: Next ~12 but not seniors
Red: Top ~30 freshmen not in those groups

Lately it has been blue: next 18 red: 24 froshies

Yeah, the number of blue was definately more than half the number of reds (at least when I was at mop)

What are the cutoffs for the groups usually around?

Last year, they were something like 9, 19, and 26 respectively.

From Wikipedia:

the top twelve scorers from all grades form the "black" group, which is eligible for selection to the IMO team. The approximately eighteen highest scorers among the remaining non-graduating students (i.e. non-seniors) form the "blue" group. (These colorful designations were adapted from Karate.) In 2004, the program was expanded to include approximately thirty of the highest-scoring freshmen each year, the "red" group.

In 2006 the USAMO cutoff scores were 25 for black MOP, 18 for blue MOP, and 9 for red MOP.

1) so the goal of MOP is to get as few seniors as possible? only ones that have a chance to make IMO?

2) only the people in black mop have a chance of making IMO, correct?

To (2), that's the idea.  These are the people that take TST.  There is a provision in the USAMO Manual saying that

[quote]In order to assure that the strongest possible team has been
formed, the MOSP Director/IMO Team Leader reserves the right
to appoint up to one IMO team member. This appointment would
happen only in a circumstance where a student’s performance
during the first two weeks of MOSP clearly surpasses that of an
existing team member.[/quote]

I don't know if this has ever happened, but it would be very rare.

To (1), I would say that the goal is to get as many non-seniors as possible.  MOP, besides the Top 12, is partially an investment in the future of the U.S. at the IMO.