Ring Theory
superior algebra
superior algebra solved

Let [b]$R$[/b] and [b]$S$[/b] be rings and let $\phi$: [b]$R$[/b] to [b]$S$[/b]  be a ring
   homomorphism such that $\phi(R)$ is not equal to {$0_s$}. Show that
   if $R$ has unity $1_R$ and [b]$S$[/b]  has no 0 divisors, then $\phi (1_R)$ is
   unity for [b]$S$[/b] .
Give an example to show that if [b]$S$[/b] has zero divisors, [b]$R$[/b] need not have a unity.

Let $phi(1_R)=a$. If $a=0_S$, then for all $x\in R$, $\phi(x)=\phi(x\cdot 1_R)=\phi(x)\cdot a=0_S$ which contradicts $\phi(R)\neq0_S$.
Therefore $a\neq 0$. We have $a=\phi(1_R)=\phi(1_R)\phi(1_R)=a\cdot a$. Since $a$ is not a zero divisor in $S$, we get $a=a^2\Rightarrow a=1_S$.

For the counter-example, take $R=\mathbb Z/2\mathbb Z$, $S=\mathbb Z/4\mathbb Z$ and $\phi(\bar x)=\widehat{2x}$.


In the game of $ Mindmaster$, secret codes are created by placing pegs of any of seven different colors into four slots. Colors may be repeated, and no slot may remain empty. How many secret codes are possible?

Since there are $ 4$ slots, $ 7$ colors, and colors may be repeated, the answer is simply $ 7^4$.

If you haven't memorized $ 7^4$, here's an easy way to do it:

$ 7^4$
$ \equal{} 7^2*7^2$
$ \equal{} 49*49$
$ \equal{} (50 \minus{} 1)(50 \minus{} 1)$
$ \equal{} 50^2 \minus{} 2(50)(1) \plus{} 1^2$ (Remember: $ (a \minus{} b)^2 \equal{} a^2 \minus{} 2ab \plus{} b^2$)
$ \equal{} 2500 \minus{} 100 \plus{} 1$
$ \equal{} \fbox{2401}$

[quote="cookieboy"]easy, so easy[/quote]
Please do not post comments like that for it may offend other users.  It is spammy as well.


geometry
geometric transformation
homothety
inequalities
trigonometry
vector

To avoid everything conglumerating on a single thread, I'll stick 2001 shortlist-related discussion here.

Hmmm... where have I seen A3 before............? :mad:

I think G1 is fairly easy (easier than the ceva method in the PDF) using the quite well known fact that if we erect squares on the sides of a triangle, the cevians through their centres meet at the orthocentre of the triangle formed by their centres (see Geometry Revisited page 97). These are the same cevians as those in this question, by a simple homothety argument.

G2 is recommended, although the fun bit is the inequality, not the boring trig introduction, unless anyone did it a geometrical way..?

Edit: I've checked the PDF, and it gives 4 solutions, none of which resembles mine. Oh well.

Jack, have you done N1 and/or G3?

Any hints?  :oops:

To be fair, I should probably get N1, but it just seems a thoroughly boring question. N2 on the other hand... :)

I think N1 can be done ok with standard form (or similar). I don't believe that it's a particularly nice problem, though.

Oooh, A4 is a very cool problem. It's an awkward but fairly easy to define fun equation.

Nice, isn't it?  Shame that the most natural way to phrase the result uses post-IMO maths...

If you are still trying G3, then [hide="hint"]it dies quickly using vectors and Cauchy-Schwarz.[/hide]
G4 isn't so very nice, while G5 is beautiful, and there's a lot to be said about the configuration.

Jack

I don't understand the official solution for N1.

The paragraph between (1) and (2) doesn't seem to make sense to me. If I have $(n+1)!=2-\epsilon$ in standard form, and $(n+2)!=2.5$, surely $n+2 = 6.25$ will work just as well as 5/4?

Grrrrr.... you had to :lol: me too....

Thanks for the correction :).

I don't see why C1 is bad. It's pretty easy, I suppose, but you get to finish off using AM-GM, which is cool. :D I agree C2 is cooler however, but I think C3 was even nicer than C2.

I am trying to fight the stigma I have against combinatorics problems by doing lots and lots of them, and my mentor is Joseph Myers, so this plan should work :D.


algebra
polynomial
calculus
calculus computations

Let $P(x)$ be a polynomial with real coefficients.
Prove that if there is a point $\gamma$ such that $P(\gamma)=P'(\gamma)=0$ then $\gamma$ is a root with multiplicity greater than one.

$P(x) = (x-\gamma) Q(x)$ and $P'(x) = Q(x)+(x-\gamma ) Q'(x)$

hence $Q( \gamma ) = 0$ etc...


Let $x_{i},y_{i},1 \leq i \leq n$
be real numbers such that $x_{1}\leq x_{2}\leq x_{3}.... \leq x_{n}$ and $y_{1}\geq y_{2}\geq y_{3}.... \geq y_{n}$ and let $z_{1}, z_{2}, .... z_{n}$ be any permutation (reordering) of$y_{1}, y_{2}, .... y_{n}$

Show that $\sum_{i=1}^{\n}(x_{i}-y_{i})^{2}\geq \sum_{i=1}^{\n}(x_{i}-z_{i})^{2}$

I will post my solution up shortly.

[hide="Solution"] $\sum y_{i}^{2}= \sum z_{i}^{2}$, so square and subtract.  We then wish to show that

$\sum x_{i}z_{i}\ge \sum x_{i}y_{i}$

But this follows trivially from Rearrangement. [/hide]

nice solution...I also think that's the best way to solve thiw inequality :wink:


[quote="math92"]Is there any way we can upgrade the avatar KB from 15 to about 30?[/quote]It's fine that way. Anything larger than 15k is probably an annoying animation :)

[quote="math92"]Is there any way we can upgrade the avatar KB from 15 to about 30?[/quote]

Just decrease the quality a bit. On the size limitations that we have, 15K is plenty for a good avatar.

[quote="Valentin Vornicu"]It's fine that way. Anything larger than 15k is probably an annoying animation :)[/quote]No, it's not. A 100x80 in JPEG format has an upperlimit size of 24 kb I believe. For people's faces, a picture without too many blending easly gets to 19-20 kb, if the background has the same type of colors as the face.

I'm posting this because I'm currently in that problem: I'm tryng to upload an avatar of 19900 bytes, which is just a compressed JPEG picture of myself, but I'm disallowed, due to the 16kb limit. I would suggest raising it just a little, to 22kb or something, so everyone can at least upload their nonanimated picture...

[ I saw Valentin mentioned dial-up users in another topic on avatar size... Valentin, do you realize that with all the forums, just the jumpbox at the bottom takes 24 kb pagesize and thus 6 extra seconds just for the box only on a dial-up connection? ;) ]

[quote="Peter VDD"][ I saw Valentin mentioned dial-up users in another topic on avatar size... Valentin, do you realize that with all the forums, just the jumpbox at the bottom takes 24 kb pagesize and thus 6 extra seconds just for the box only on a dial-up connection? ;) ][/quote]That was taken into consideration "years ago" (literally).

See the 4th option in [url=http://www.mathlinks.ro/Forum/usercp.php]Options[/url] ;)

But anyway, since for now the avatar directory isn't that big, I will increase the avatar size to 22 kB :)

Ah ok... so many option one doesn't get the time to try them all. :oops:

Anyway thanks for the size!

[quote="MithsApprentice"]Lol. In the spirit of this topic, I propose 100x100 pixels on avatars. :D[/quote]Don't push it :D

Sorry but could anyone tell me where I should go to make my signature visible?

Go to http://www.mathlinks.ro/Forum/usercp.php?u=20604&mode=prefs&sub=posting

and select attach my signature. This means that each time you post the "Attach signature" checkbox is checked (which you can do manually for each post - what you set in the profile is just a default value).

Just wondering, it says on my computer the maximum size is 21KB, instead of 22KB, like you said.  This doesn't matter to me, but I thought I might point it out anyway.


rate problems
calculus
calculus computations

Hi there. I'm stuck on this one.

Find the average rate of change of f(x)= -3x^2+4:
a) From 0 to 2
b) From 4 to 6
c) From 2 to 5

Thanks!

thanks, I figured it out by myself  :blush:


I can't log in to the USAMTS website. I just reset my password. My username is chess.

When you reset your password, did you get an email with the new password?

yes. it was some password. i'll pm you with that.


search
counting
derangement
combinatorics unsolved
combinatorics

In how many different ways the numbers 1, 2, 3,.....n can be arranged so that none of these numbers take its initial place.
For example: 1 cannot be in first place. 2 cannot be in second place. etc..

[quote]Make a Google search on the keyword "derangements". Classical application of PIE.[/quote]
i dont understand how PIE is related to derangements

PIE does not mean the  mathematical symbol..

it means  PRINCIPLE OF INCLUSION AND EXCLUSION


limit
calculus
derivative
calculus computations

How would one go about solving this limit?

$ \lim_{x\to1}{\frac{x^{1000}-1}{x-1}}$

The answer is 1000, but it's the work I'm interested in.

Or, if $ f(x)\equal{}x^{1000},$ this is the definition of $ f'(1).$

This problem quite easy to use L'Hospital.

What TTsphn is talking about is that if you have a limit that is of the indeterminate form $ \frac{0}{0}$ or $ \frac{\infty}{\infty}$, then you can take the quotient of the limits of the derivative of the numerator and denominator and have this quotient be equal to the original limit.

In other words:
$ \lim_{x\to 1}\frac{x^{1000}\minus{}1}{x\minus{}1}\equal{}\lim_{x\to 1}\frac{1000x^{999}}{1}\equal{}\boxed{1000}$

Let $ S \equal{} 1\plus{}x\plus{}x^{2}\plus{}\cdots\plus{}x^{999}$

$ \Rightarrow xS \equal{} x\plus{}x^{2}\plus{}\cdots\plus{}x^{999}\plus{}x^{1000}$

$ \Rightarrow xS\minus{}S \equal{} x^{1000}\minus{}1$

$ \Rightarrow S(x\minus{}1) \equal{} x^{1000}\minus{}1$

$ \Rightarrow S \equal{}\frac{x^{1000}\minus{}1}{x\minus{}1}$ for $ x\neq 1$

$ \Rightarrow\lim_{x\to 1}\frac{x^{1000}\minus{}1}{x\minus{}1}\equal{}\lim_{x\to 1}S \equal{} 1000$


I thought I would start this thread to recap what items you should bring to Penn State.

\[ \text{Mandatory/Very Highly Recommended}  \]

$\bullet$ Change of clothes for Friday, Saturday, and Sunday; if you plan on playing frisbee, basketball, etc plan on bringing an extra outfit or two in case you get sweaty; there is a pool that can be used at certain times so if you want to swim bring a bathing suit
$\bullet$ Toiletries such as soap, shampoo,a toothbrush, toothpaste, etc.
$\bullet$ towels
$\bullet$ blanket and pillow (you do not need to bring linens though)
$\bullet$ SAT-approved calculator
$\bullet$ money for lunch and dinner Thursday, all meals Friday, dinner Saturday, and breakfast and lunch on Sunday
$\bullet$ any medicine you need

\[ \text{Additional Stuff}  \]
$\bullet$ jacket
$\bullet$ umbrella
$\bullet$ stuff for the bus ride such as books, math books, magazines, cd-players, mp3 players, etc.
$\bullet$ frisbee (Joe)
$\bullet$ basketball (...?)
$\bullet$ playing cards (Bella)
$\bullet$ DVD's (Shobhit)

If anyone else has suggestions or is bringing something that other people on the team will get use out of, please post here.

Well, I'm likely to be bringing a pack or two of cards.

um, yeah, I'm bringing all my movies (I got ALL the good ones), and my laptop to watch during the busride.

If any one else wants to bring movies, go ahead.

Hm...I think maybe I should bring my XBOX, not my XBOX 360, just in case.  I'm not very good at keeping track of stuff...I'll probably bring the XBOX.  And poker chips!

[quote="joml88"]The bus is charter and has t.v.'s.  I'm not sure whether they have a VHS or DVD player (or both).[/quote]

The bus has both. Remember that if you bring any movies make them non-R-rated. We have different age groups on the bus. Same goes with the games.

And make sure you bring a [b]Pillow[/b], an umbrella, a jacket, and any [b]medicines[/b] that you need!!

About the different age groups:  I am definitely the youngest person on the trip.  My parents and I are OK with R-rated movies, usually.  However, if we wanna keep it PG-13, that's fine.  Oh, except someone bring some Family Guy DVDs!  Mr R!

Well, it's a math competition, so we need a chessboard.  Mine is too big...any volunteers?

Guys,
We probably need to stick with PG movies.  I do not want any parent complaints (No Family Guy).  Also, if you are bringing any medicine, it better be on the release your parents sent me.  Finally, I highly suggest that you do not bring anything too expensive or bulky (i.e. an XBox).  You will have enough to carry, and I am not going to have any free hands to help you.  We can have plenty of fun without video games and tv.

[quote="MIMC Director"]Oh, and I'm bringing my 28" flatscreen TV.  Mr. R can help me carry it.[/quote]

Whoa, thanx so much MIMC Director! And mysmartmouth is bringing his xbox so that means we can play Halo/Halo 2/and other Xbox games I decide to bring on Friday.

Or wait, are we actually doing math on Friday?

Errr....I think MIMC Director was being....sarcastic?  ironic?  IDK.  Anyway, I'm not bringing the XBOX, and I'm obviously not bringing the TV.  Bring whatever you can sneak out Shobhit...lol.


Equilateral triangel $ABC$ with side length 1 is drawn. A square is drawn such that its vertex at point $A$ is opposite to its vertex at midpoint at $BC$. Find the area of the area enclosed within the intersection of the insides of the triangle and square.

[quote="rnwang2"]I'm getting an answer of
[hide]$\frac{3}{4\sqrt{3}+4}$.[/hide][/quote]

I believe thats correct. Can you post your solution?

Yeah, it might be the same...I just did some weird trig stuff.

ever since coming to this forum I try to use trig as little as possible :o

[quote="GlenthemanN"][quote="sonny"][quote="GlenthemanN"]EDITED:
$\frac{3\sqrt{3}-3}{8}$ is the answer (im almost certain)
[hide="answer"]
In our equilateral triangle ABC with side length 1, youll notice that the diagonal of the square within is the bisector of angle A whcih essentially cuts the triangle into two 30-60-90 triangles. Using this logic we find the diagonal of the square to be:
$\frac{\sqrt{3}}{2}$

We can find the area of the square by multiplying the diagonal of the square by half of the diagonal;
$\frac{\sqrt{3}}{2}* \frac{\sqrt{3}}{4}= \frac{3}{8}$

Now if you notice, we are left with two 15-75-90 triangles, one side being the side length of the square. Using our diagonal measure we can find the sidelength of the square to be
$\frac{\sqrt{6}}{4}$

Using our new side length and the tangent ratio we find the shorter side of he triangle to be:

$\tan{15}= 2-\sqrt{3}= \frac{4x}{\sqrt{6}}\Rightarrow x = \frac{2\sqrt{6}-3\sqrt{2}}{4}$

now to find the area of both 15-75-90 triangles:
$\frac{\sqrt{6}}{4}\cdot \frac{2\sqrt{6}-3\sqrt{2}}{4}= \frac{12-6\sqrt{3}}{16}= \frac{6-3\sqrt{3}}{8}$

now subtract these from the total area of the square to find the area we need:$\frac{3}{8}-\frac{6-3\sqrt{3}}{8}= \boxed{\frac{3\sqrt{3}-3}{8}}$
[/hide][/quote]

Um, thats what I did and I got the same exact answer as you. I don't know where I went wrong because their answer was the same as rnwang2[/quote]

his answer is the same, rationalize his denominator and youll be in for a surprise ;)[/quote]

wow I never thought of doing that after looking at the solutions. I just looked at the answer and went  :huh:  I guess I am wrong. Yippie I did #7 on HMMT right!!


If $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}2$, then

$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\leq\frac{3\minus{}abc}{2}$

[quote="Ligouras"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 2$, then

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\leq\frac {3 \minus{} abc}{2}$[/quote]

Happy new year!
Stronger is:

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\leq\frac {3 \minus{} abc}{\sqrt{3(a^2 \plus{} b^2 \plus{} c^2)}}.$

[quote="xzlbq"][quote="Ligouras"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 2$, then

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\leq\frac {3 \minus{} abc}{2}$[/quote]

Happy new year!
Stronger is:

$ a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2\leq\frac {3 \minus{} abc}{\sqrt {3(a^2 \plus{} b^2 \plus{} c^2)}}.$[/quote]

Thanks My Friend and Happy New Year!!!!  :)   :)


What do you guys think the Horcruxes are going to be?

CONFRIMED 1. the locket. The initials RAB point to Regulus Black, which makes sense in that he was killed because he angered Voldemort. In the 5th book they found an unopenable locket in grimmauld place and I think that is the horcrux. Kreacher probably has it.

2. Harry. Didn't the 6th book say "One cannot live while the other is alive" or something like that? Harry would have to kill himself to kill voldemort.

3. Nagini. Voldemort's snake.

CONFIRMED AND DESTROYED 4. The Diary.

CONFIRMED AND DESTROYED 5. Dumbledore's ring.

CONFRIMED 6. a piece inside voldemort's body

7. hufflepuff's cup

8. griffindor's sword?

9. something of ravenclaw?

Hmm... They all seem like good places to put them. So I agree with you.

Option 2 doesn't seem logical. Why would Voldemort try to kill harry so often after placing a part of his soul in Harry? Losing a part of one's soul can't be pleasant, right?

The other options are pretty good guesses, i think.

Voldemort has 6 more horcruxes and he can make another one.

[quote]2. Harry. Didn't the 6th book say "One cannot live while the other is alive" or something like that? Harry would have to kill himself to kill voldemort.
[/quote]

Not exactly, the thing said "Neither can live while the other survives", which means only one of Harry/Voldemort has to die. The other stays alive.

Which cup of Huffelpuff's are you talking about?

O_o  Number 1 is a great idea, I'd never thought of that.

However, I seriously doubt Harry is a horcrux.

eh i dont have many predictions for horcruxes
but i am pretty sure of one thing
percy is gonna become a death eater and kill ms. weasley
i forget the details, but there was this dude in mythology named percevius or something who turned evil and killed his mom
and rowling uses lots of mythology references

the locket isn't with kreacher!!!

in book 6, mundungus (or whatever his name was) had a brief case filled with stuff from the house of black, and in it was "something silver that ron recognized".

I forgot what happened after that, but anywho the locket at least went through Mundungus (or whatever his name was).

I agree with the RAB being regulus black.

clearly griffondor's sword is not a horcrux since it was said in book 6.  so that leaves exactly 8! 7 horcruxes plus voldemort's body

actually, voldemort's body [b]is[/b] a horcrux.
So 7 horcuxes, including voldemort's body

I dunno--there's something fishy going on.  JKR has so obviously set it up to look like RAB is Black, the locket horcrux is with Kreacher, etc.  Given the past twists, there has to be something big we don't see coming.  If those two facts are true, I think they'll be revealed very early on in the book and will just be the jumping-off point for a whole lot more.  The book won't just be a Horcrux hunt.  And for that reason, I really have a hard time believing there are even four more out there, as that seems like it would take up too much narrative space  to fit it all into the book.

That said, here's what I think:

1. Locket: Already destroyed, probably by RAB or someone working with him.  This will be revealed early on.
2. Ring: Destroyed.
3. Diary: Destroyed.
4. Hufflepuff's cup: Harry will need to destroy this.  Either this or #5 will be the tough one, where Harry needs to figure out exactly how to destroy a horcrux.
5. Unknown.  Probably something belonging to Ravenclaw.
6. Nagini: Probably destroyed shortly before or during the last battle.  Destruction might be simpler since it's hosued in a living thing.  Can you destroy the horcrux if you kill the snake?
7. The piece in Voldemort's body: Last to be destroyed, presumably.

I don't think Harry is a horcrux.  If he had a piece of someone else's soul living in him, I think we would know it.  And I think Dumbledore would certainly know it and would at least give some hint of its existence.

How about this theories? [b]Harry is horcrux.[/b] Why does Voldemort seek him and want to kill him?
Because he want to [b]move [/b] the horcrux from Harry to other thing, then after he does it, he kill Harry.
I am three quarters sure that in the ending, harry could unite Gryffindor and Slytherin to destroy the darkness.
And, maybe--- like the type of JK Rowling's story in previous book---- the good one is Voldemort, meanwhile Dumbledore is actually Evil!

[quote="boo"]

CONFRIMED 1. the locket. The initials RAB point to Regulus Black, which makes sense in that he was killed because he angered Voldemort. In the 5th book they found an unopenable locket in grimmauld place and I think that is the horcrux. Kreacher probably has it.

[/quote]

Why is this one confirmed?

The locket is a confirmed horcrux because Harry and Dumbledore found the fake one (removed by RAB) in book 6.  It's still theoretically possible that the locket was a fake planted by Voldy in the first place, but for all practical purposes, we know the locket is one of them.  What we don't know for sure is if it's the locket in Grimmauld Place.

It's a conjecture...and likely not true.  But possible.

Is Dumbledore just super smart or can he predict the future? That's weird I mean he seems to know what's going to happen before they actually do.

[quote="BanishedTraitor"]Yeah I know that but what I was asking is why is it confirmed that $ R.A.B$ is Regulus Black? Or is it still a conjecture?[/quote]

It's conjectured based on the fact that "RAB" has always been translated to match the initials of Regulus Black in other languages.  For many people, this makes it essentially a certainty.

yeah, that's basically where it becomes obvious who RAB is (unless JK rowling just wants to be evil and lead everyone astray, because that's just how horrible of a person she is lol).  still, did nobody see my evidence for why the locket is not destroyed, but rather that it went through the hands of Mundungus, the kleptomaniac?

IF I WERE VOLDY i would make DUMBLEDORE A horcrux, i know it sounds well dumb, but knowing that in order to kill voldy, harry would have to kill dumbledore something harry would never do. now if dumbledore realizes that he is a horcruz he will obvioulsy not give a crap about his life and will try to die, but only after he explains to harry about horcruxes, somehow voldy gets to know that dumby wants to die so he sends malfoy on a mission to completely destroy dumby before he can putforth the info about horcruxes to harry (thus the attempts of poisoning him etc) dumby then asks snape to kill him (thus the part where hagrid hears them overtalking about how dumby takes things for granted and crap like that) and SO SNAPE KILLS DUMBLEDORE end of story.

Wormtail owes harry a life debt so I'm sure that will come into play in the last book.
Possibly wormtail tells harry the location of a horcrux or something.
He was present on the night Voldy supposedly made harry a horcrux.
and he probably knows about nagini.

Dumbledore concluded that based on the "unusual" degree of control Voldy has over the snake.

Dumbledore's theory is:
- Voldy wanted six total horcruxes (seven-part soul), and was saving the last one for Harry's death.
- Obviously, that last horcrux didn't get made, so Voldy made it out of Nagini later on.

This is a big tangent, but I disagree with this model.  I think he was saving one for Harry's death, and one for Neville's, which would occur later, since Neville could also have been the one to whom the prophecy referred.  Then when Harry clearly became the chosen one, he forgot about Neville, made Nagini horcrux #5, and is saving #6 for when he kills Harry.  That leaves him still one short now.  The other reason I believe this is that there's not enough time in one book for a massive Horcrux hunt plus time to resolve all the other loose ends in the saga.


function
floor function
ceiling function
LaTeX

How do I do floor function and ceiling function?

To do $ \lfloor x\rfloor$:
[code]$\lfloor x\rfloor$[/code]
To do $ \lceil x\rceil$
[code]$\lceil x\rceil$[/code]

This can be found [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideSym.php#bracket]here[/url]

And, as is implied on that link, the symbol can be scaled. For instace

\left\lfloor \frac{x}{2} + 1 \right\rfloor renders as $ \left\lfloor \frac{x}{2} \plus{} 1 \right\rfloor.$


integration
limit
function
calculus
logarithms
real analysis
real analysis unsolved

If $ \int f(x) dx \equal{} 0$ where the upper limit is $ 1$ and lower limit is $ 0$, show that there exist $ c \in (0,1)$ such that $ \int xf(x) dx \equal{} 0$ where the upper limit is $ c$ and the lower limit is $ 0$. Thank you for your attention before

I am sorry but $ f$ is continuous and differentiable. Thank you before.

I'll use the hypothesis that $ f$ is continuous; I don't think I need $ f$ to be differentiable.

We are given that $ \int_0^1f(x)\,dx\equal{}0.$ Define $ g(x)\equal{}\int_0^xtf(t)\,dt.$

We are asked to show that $ g(c)\equal{}0$ for some $ c\in(0,1).$

Suppose not. Then, since $ g$ is continuous, the Intermediate Value Theorem shows that $ g(x)$ is always positive or always negative. WLOG, assume that $ g(x)>0$ for all $ x\in(0,1).$ Note that this also shows that $ g(1)\ge0.$

An application of L'Hôpital's rule shows that $ \lim_{x\to0^\plus{}}\frac{g(x)}{x^2}\equal{}\frac{f(0)}{2}$ exists. We treat $ \frac{g(x)}{x^2}$ as a continuous, hence Riemann integrable, function on $ [0,1].$ Since $ \frac{g(x)}{x^2}$ is strictly positive, its integral is strictly positive.

$ 0<\int_0^1\frac{g(x)}{x^2}\,dx\equal{} \int_0^1\int_0^x\frac{1}{x^2}\,tf(t)\,dt\,dx$

$ \equal{}\int_0^1\int_t^1\frac{1}{x^2}\,tf(t)\,dx\,dt\equal{} \int_0^1\left(\frac1t\minus{}1\right)tf(t)\,dt$

$ \equal{}\int_0^1(1\minus{}t)f(t)\,dt\equal{}\int_0^1f(t)\,dt\minus{}\int_0^1tf(t)\,dt$

$ \equal{}0\minus{}g(1)$

Thus we have $ 0<\minus{}g(1)$ or $ g(1)<0.$ But this contradicts that we have already established that $ g(1)\ge 0.$ Hence the assumption that $ g(x)$ is always positive on $ (0,1)$ must be false.

----

I'm not particularly happy with that proof. I won't be surprised if someone can do better.

[quote="perfect_radio"]That trick is very nice  :)

This also appeared [url=http://www.mathlinks.ro/viewtopic.php?p=486661#486661]here[/url]. I gave a link there to a problem ("Flett") that is equivalent to this one.
[Now I know the solution to it, at that time I didn't.][/quote]

  hey, where have u been ?   :oops:   did you hang out with bruce dickinson and gang ?  :D

An other one .. :lol:  .. Let $ F$ be such that  $ F(t) \equal{} \int _0^t f(x)dx$, with  $ f$ continuous and $ F(1) \equal{} 0 \equal{} F(0)$.
We can write $ \int _0^t xf(x)dx \equal{} t F(t) \minus{} \int _0^t F(x)dx \equal{} t(F(t) \minus{} F(u_t))$, where $ 0 < u_t < t$ .
Suppose that $ \forall _{ t\in ]0,1[}, \int _0^t xf(x)dx \neq 0$ . Because of the continuity and the Intermediate Value Theorem,   only two cases are possible :
1) $ \forall_{ 0 < t < 1}, \int _0^t xf(x)dx > 0$ , so $ \int _0^1 xf(x)dx \equal{} F(1) \minus{} F(u_1) \equal{} \minus{} F(u_1) \geq 0 \equal{} F(0) \equal{} F(1)$  with $ 0 < u_1 < 1$, so $ \exists_{0 < m < 1}, F(m) \equal{} \min_{ 0\leq x\leq 1} (F(x))$
We can write $ \int _0^m xf(x)dx \equal{} m F(m) \minus{} \int _0^m F(x)dx \leq m F(m) \minus{} \int _0^m F(m)dx \equal{} 0$ , contradiction !!
2)$ \forall_{ 0 < t < 1}, \int _0^t xf(x)dx < 0$ , so  again a contradiction by using $ f(x) \leftrightarrow \minus{} f(x)$ and the above point 1 !!
Finaly we have to have $ \exists _{0 < c < 1} , \int _0^{c} xf(x)dx \equal{} 0$
 :wink: Could you prove that $ \exists _{0 < c < 1},\int _0^{c}\ln (1\plus{}x^3) f(x)dx \equal{} 0$  ?
:cool:

Sorry for reviving, but anybody having a solution using Flett's Theorem ?

See here for Flett's Theorem: http://www.asymmetry.gr/images/various_math_documents/Theorimata_meshs_timhs.pdf?


Post your favorite falcon punch videos here. I'm too busy to find a favorite right now, but have fun searching for good FALCON PWAWNCH! videos.

[quote="randomdragoon"]Turning it into "post a video related to the previous one"

[youtube]Lziyu7gn3Q4[/youtube]

:P[/quote]
this is WAY too awesome to describe in words

not exactly a video:

[img]http://i180.photobucket.com/albums/x76/JUSTanLS/Falcon_Punch.jpg[/img]

Um... I don't know how to post a video, so here's the link:

http://www.youtube.com/watch?v=k3W2gOYzMvU


I'm getting tired of trying to find epic XKCD articles, so let's make a compendium here.

I'll start:

http://xkcd.com/311/

Mhmmm...
I love this site :)

http://xkcd.com/360/
and this;
http://xkcd.com/268/

http://xkcd.com/178/

http://xkcd.com/530/

http://xkcd.com/447/

http://xkcd.com/582/

http://xkcd.com/281/

http://xkcd.com/481/

http://xkcd.com/272/

http://xkcd.com/131/

http://xkcd.com/461/

http://xkcd.com/255/

Aren't all of them epic?   :wink:

[url]http://xkcd.com/386/[/url]

[url]http://xkcd.com/430/[/url]


In the (x,y,z) plane, what is the equation of a sphere whose center is (0,0,0)? What if the center is (h, y, k)?

gosh...

why in this forum???

and, it's not a plane...

[hide="anyways..."]1.) $ x^2 \plus{} y^2 \plus{} z^2 \equal{} r^2$, where r is the radius.
2.) $ (x \minus{} h)^2 \plus{} (y \minus{} j)^2 \plus{} (z \minus{} k)^2 \equal{} r^2$ (you mean k instead of y, right?)[/hide]

if the center is y then that would be a pretty messed up sphere (more like a cylinder)


Find the limit of $ x_n$ as $ n \rightarrow \infty$ of the following sequence:

$ x_{n\plus{}1}\equal{}a x_n\plus{}2^{\minus{}n} \cdot n^2 , n \ge 0, x_0\equal{}1, a \in \left [ 0,1 \right )$

Or, more generally, assume that $ a\in[0,1)$, $ b_n\to 0$. Find the limit of the sequence given by the recurrent relation $ x_{n\plus{}1}\equal{}ax_n\plus{}b_n$.


Let $ a,b,c$ be positive real numbers satisfying $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \frac{a(b\plus{}c)}{4\minus{}b\minus{}c}\plus{}\frac{b(c\plus{}a)}{4\minus{}c\minus{}a}\plus{}\frac{c(a\plus{}b)}{4\minus{}a\minus{}b} \ge 3abc
\]

$ \frac {a(b \plus{} c)}{4 \minus{} b \minus{} c} \equal{} \frac {a(3 \minus{} a)}{a \plus{} 1} \equal{} \minus{} a \plus{} 4 \minus{} \frac {4}{a \plus{} 1}$. We have to show
$ \frac {4}{a \plus{} 1} \plus{} \frac {4}{b \plus{} 1} \plus{} \frac {4}{c \plus{} 1} \plus{} 3abc \le 9$
$ \frac {4(ab \plus{} bc \plus{} ca \plus{} 9)}{abc \plus{} ab \plus{} bc \plus{} ca \plus{} 4} \plus{} 3abc \le 9$
Let $ p \equal{} ab \plus{} bc \plus{} ca$, $ q \equal{} abc$. Since $ a \plus{} b \plus{} c \equal{} 3$, we have $ q \le 1$ and since $ (ab \plus{} bc \plus{} ca)^2 \ge 3abc(a \plus{} b \plus{} c)$, we have $ p \ge 3 \sqrt {q}$.
We have to show $ p(5 \minus{} 3q) \ge 3q \plus{} 3q^2$. Using $ p \ge 3 \sqrt {q}$, we have to show
$ 5 \ge 3q \plus{} \sqrt {q} \plus{} \sqrt {q^3}$, which is obviously true for $ q \le 1$.

it's easy but nice,
we have, $ S\equal{}\sum\frac{a(b\plus{}c)}{4\minus{}b\minus{}c}\equal{}\sum\frac{a(b\plus{}c)}{c\plus{}1}$
By Cauchy Schwarz,
$ \left(\sum\frac{a(b\plus{}c)}{c\plus{}1}\right)(a\plus{}b\plus{}c\plus{}3)\geq \left(\sum\sqrt{a(b\plus{}c)}\right)^2$
By AM-GM,
$ \sum\sqrt{a(b\plus{}c)}\geq 3\sqrt[6]{abc(a\plus{}b)(b\plus{}c)(c\plus{}a)}\geq 3\sqrt[6]{8a^2b^2c^2}\geq 3\sqrt{2abc}$
the last one is because that, $ abc\leq1$
therefore $ \left(\sum\sqrt{a(b\plus{}c)}\right)^2\geq 18 abc$
then $ S \geq 3abc$


function
analytic geometry
Asymptote
geometry
calculus
calculus computations

(a) The function $ f(x)$ has $ f'(x) \equal{} \frac{u(x)}{v(x)}$. Given that $ f'(k) \equal{} 0$, show that $ f''(k) \equal{} \frac{u'(k)}{v(k)}$.

(b) The curve C with equation

$ y\equal{} \frac{2x^{2}\plus{}3}{x^{2}\minus{}1}$

crosses the $ y$-axis at the point A.

(i) Find the coordinates of the point A.

(ii) Find the equations of the three asymptotes of C.

The point P$ (a, b)$, $ a>0$ and $ b>0$, lies on C. The point Q also lies on C with PQ parallel to the $ x$-axis and $ AP \equal{} AQ$.

(iii) Show that the area of triangle PAQ is given by $ \frac{5a^{2}}{a^{2}\minus{}1}$.

(iv) Find, as $ a$ varies, the minimum area of triangle PAQ, giving your answer in its simplest form.

coordinates of  A  are $ (0,\minus{}3)$.

thats  the  easiest to get  :D


A box contains 3 black balls, 4 red balls and 7 green balls. Two balls are drawn at random. (a) What is the probability of drawing at least one red ball? (b) What is the proability of drawing two green balls? (c) What is the probability of balls of different colours?

For (a) I came up with  4/14, but I am not sure of the process to work out the probability of drawing multiples. :?:

Is there replacement?  I think you are right for the first one.  For the second, assuming no replacement, I think it is $ \left(\frac{7}{14}\right)\cdot\left(\frac{6}{13}\right)\equal{}\frac{6}{26}$.  Sorry, can't help you on 3 (I never liked probability).

The question doesnt mention replacement so I would assume not.

Thanks,

For sufficiently small problems like (c), casework is typically the easiest solution.
You can simply find the probability of (1) picking black, then another color, (2) picking red, then another color, and (3) picking green, then another color.

A box contains 3 black balls, 4 red balls and 7 green balls. Two balls are drawn at random. (a) What is the probability of drawing at least one red ball? (b) What is the proability of drawing two green balls? (c) What is the probability of balls of different colours?

For c) You can take total prob - prob of only one colour.
Therefore P=1-3C2/14C2 -4C2/14C2 - 7C2/17C2

hmm I'm getting a different answer.  someone please check my answer?

a.)$ 1\minus{}P($no reds$ )$.
$ P($no reds$ ):$ $ \frac{10C2}{14C2}$
$ 1\minus{}P($no reds$ )$.: $ \frac{45}{91}$
b.) $ \frac{7C2}{14C2}$ = $ 3/13$.
c.) agree with madness's answer


There's no pre/post test for independent studies, so i have no idea which one suits me the best. Is there gonna be some?   Also, I was gonna take Intro to Number Theory, When i tried the post test, i got around 50%, so should i take it?

for the intro to number theory, try the pretest. If you get a good score on that, take the class. If not, you can stil try I guess  :?

you misunderstood me, i got everything on the pretest.  i got 50% on post test, so i dont know if I'll benefit from that class.

it seems like you're just missing the knowledge from the class. If you want to learn the stuff, you should take it.

Many of the students in the Introductory Number Theory class know some of the post test material going in and still learn a lot.

We generally advise students on which Independent Study program is best for them.  I think that the Introductory B would be the best choice for you out of the Independent Study programs.

If you know 50% of the posttest, you will do very well in class. You will still learn a lot. When you don't know much of the posttest, it'll be a bit of a challenge.


In "Puzzles and Problems"
>>> In "Geometry"
>>>>>>>> Problem #8 (the diagram consisting of the 12 dots)

What is the ACTUAL answer (different from Paul Cox's)?

This is an example of a flawed question with the correct intent.
That is, his "solution" matches the intent.

But again, what is the solution to the puzzle #8, as stated?

Open hint...

He never mentions about the vertices of squares having to coincide with dots of the diagram.  It can't be assumed in there.

Could you post a link to the website or something, please?

Thanks :)

Please go to this link, or type this in if it doesn't auto link...

http://members.cox.net/mathmistakes/

This is for "The Glossary of Mathematical Mistakes."

The answer is "unlimited" or "no limit" because the phrase
"connect the dots" is sufficiently vague and does not make it rigorous that all 4 vertices of every square must coincide with
exactly 4 dots of the diagram.

I agree with box of rocks.

As it is worded, it only seems that you have to count the number squares containing at least two of the dots. You can make the side any length greater than the distance between the given dots.

Did you have some other condition in mind?


Some NYU grad students are planning a day of interesting math talks for high school students on Saturday, March 25, 2006.  You can find more details at [url=http://cims.nyu.edu/~csplash/index.php]cims.nyu.edu/~csplash/index.php[/url].  The event is free, but you must register in advance; space is limited.

I've got to pay attention to my own department a lot more--I didn't even know about this until now.  :blush:

c-splash is suppossed (sp) to be good but it's a take of of SPLASH at MIT. SPLASH is once a year and it's a weekend of fun (and sometimes crazy) math and science classes


Not uniquely defined  :P
You probably want to hear $\frac{1}{\sqrt{e^\pi}}$.

I think that's the answer. But how did you work out?

Come on. Who can tell me HOW to solve it? By formula?


geometry
incenter
geometric transformation
reflection
angle bisector
geometry proposed

Sorry if this posted again. Sorry if you will find it very easy.

Let  $ABC$ a triangle. The internal bisectors of  the angles $B,C$  intersect the sides $AC$ and  $AB$ at  $D$  and $E$  respectively. Find the angles of the triangle if
$\widehat {BDE}=24$  and  $\widehat {CED}=18$

Let $BD \cap CE = I$, so $I$ is the incentre of $\triangle{ABC}$, thus:
\[ \angle{BIC}= 90^o + \frac{\angle{A}}{2} \]
So $\angle{A}=96^o$

Now, reflect $E$ through $BD$, reflect $D$ through $CE$, call the reflections $E'$ and $D'$. Since $BD$ and $CE$ are angle bisectors, so $D'$ and $E'$ are all on side $BC$. Let $ED' \cap BD=P$.

Since $\angle{PEI}=\angle{DEI}=18^o$, so we have:
\[ \angle{EPB}=\angle{PED}+\angle{PDE}=24^o+36^o=60^o \]
Because $E'$ is the relection of $E$, so we have $\angle{EPB}=\angle{E'PB}$. Thus $\angle{E'PB}=60^o$.Also we have $\angle{E'PD'}=180^o-\angle{EPE'}=60^o$. Therefore we see that $PE'$ is the angle bisector of $\angle{BPD'}$

Let $R = DD' \cap CE$, so $\angle{CRD}=90^o$, thus:
\[ \angle{EDD'}= 90^o - \angle{RED}=90^o - 18^o = 72^o \]
Since $\angle{EDE'} = 2\angle{EDB}=48^o$, we have:
\[ \angle{E'DD'}=\angle{EDD'}-\angle{EDE'}=72^o-48^o=24^o=\angle{EDP} \]
Therefore we see that $E'D$ is the angle bisector of $\angle{PDD'}$

So now we can get a conclusion that $E'$ is the ex-centre of $\triangle{PDD'}$. Hence:
\[ \angle{PE'D'}=90^o - \frac{\angle{PDD'}}{2}=66^o \]
Thus:
\[ \angle{PBE'}=66^o - 60^o = 6^o \]
So $\angle{ABC}=12^o$ and $\angle{ACB}=72^o$.

Yes shobber nice solution. But I am wondering if we could find a trigonometric solution ?


Putnam
induction
number theory unsolved
number theory

Let $T_0 = 2, T_1 = 3, T_2 = 6$.

For $n \ge 3$, $T_n = (n + 4)T_{n-1} - 4nT_{n-2} + (4n - 8)T_{n-3}$.

The first few terms are $2, 3, 6, 14, 40, 152, 784, 5168, 40576$. Find, with proof, a formula for $T_n$ of the form $T_n = A_n + B_n$ where $\{A_n\}$ and $\{B_n\}$ are well-known sequences.

$T_n=n!+2^n$

To prove the rest, use the fact that $(a_n)_n$, where $a_n=n!$, satisfies $a_n=na_{n-1}$ and $(b_n)_n$, where $b_n=2^n$, satisfies $a_n=2a_{n-1}$.

Sorry for my ignorance, but how would you show that? I am so lost :oops:

Does anyone know how to finish the rest?

There is only one way to prove a statement like this, once you have the answer:
Induction.

Assume you have $T_k=2^k+k!$ for $k=n-1,n-2,n-3$. Show that $T_n=2^n+n!$.

But that's the problem, how would you show it by induction?

You have only one tool to prove the induction step with: the recursion. That's why you need three terms in the hypothesis.

This is a pretty easy problem- it's easy to guess the answer (40576 looks like 8! plus something...), and there is only one way to proceed once you know that answer.


superior algebra
superior algebra unsolved

Let $K$ be the conjugacy class of transpositions in $S_{n}$ and let $K'$ be the conjugacy class of any element of order two in $S_{n}$ that is not a transposition.Prove that $|K| \neq |K'|$.

Let's count all permutations that are products of $k$ disjoint transpositions, $1\le k\le n/2$. To construct such a permutation, I first choose an ordered subset of $2k$ elements -- there are  $\frac{n!}{(n-2k)!}$ ways to do this. If the chosen elements are, say, $4,2,8,1,3,9$, then I take the product of transpositions $(42)$, $(81)$, $(39)$. Since the order of pairs and the order of elements within each pair do not matter, there are \[N(k): =\frac{n!}{2^{k}k!(n-2k)!}\] permutations that are products of $k$ disjoint transposition. It remains to prove that $N(1)\ne N(k)$ for any $1<k\le n/2$. Maybe this can be done by looking at the divisors of $N(k)$.  :maybe:

Hi.

In general, this is not true. For example in $S_{6}$ there are 15 transpositions and 15 elements of type 2-2-2 (that means products of 3 disjoint transpositions). But it is true if $n\neq 6$

Proof:
Let n be greater than 6 and let $k\geq3$ be a number with $N(1)=N(k)$
$\Rightarrow \frac{n!}{2!\cdot(n-2)!}=\frac{n!}{2^{k}\cdot k!\cdot (n-2k!)}$
$\Rightarrow 2^{k-1}\cdot k!=\frac{(n-2)!}{(n-2k)!}=(n-2)(n-3)...(n-2k+1)$

Since $2k\leq n$ we have:

$2^{k-1}=\frac{(n-2)(n-3)...(n-2k+1)}{k!}$
$\Rightarrow 2^{k-1}\geq (n-2)(n-3)...(n-k+1)$

Now we have $n-k+1\geq 5$ since we had choosen $n\geq 7$. So the k-2 factors on the RHS are $\geq 5^{k-2}$. Here is a contradiction since the RHS should also be $\leq 2^{k-1}$.

The next case is $n\geq 7$ and $k=2$. This also leads to a contradiction since
$N(1)=N(2)\Rightarrow \frac{n(n-1)}{2}=\frac{n(n-1)}{2}\cdot\frac{(n-2)(n-3)}{2}$
$\Rightarrow 1=\frac{(n-2)(n-3)}{2}\geq\frac{5\cdot 4}{2}=10$

The cases n=2,n=3 are trivial. The case n=4 is well known: In $S_{4}$ there are 6 transpostions but only 3 other elements with order two. The last case is n=5 and here a little calculation shows, that there are 10 transpositions but 15 Elements of type 2-2.

The result is therefore:
N(1)=N(k) implies (n=6 and k=3) or k=1


function
modular arithmetic
algebra
polynomial
induction
IMC
college contests

Let $ n$ be a positive integer and $ a_{1}, \ldots, a_{n}$ be arbitrary integers. Suppose that a function $ f: \mathbb{Z}\to \mathbb{R}$ satisfies $ \sum_{i=1}^{n}f(k+a_{i}l) = 0$ whenever $ k$ and $ l$ are integers and $ l \ne 0$. Prove that $ f = 0$.

Assume we can find a non-zero $ f$ with the required property. Notice that the sequence $ f(n)_{n\in\mathbb Z}$ is linear recursive, and also that our non-zero $ f$ can be taken to have integer values. Moreover, the sequence becomes periodic when reduced modulo any prime. Now choose a prime $ p$ larger than $ n$ and such that the reduced sequence $ f(n)\pmod p$ is non-zero. Taking $ l$ to be a period of $ f(n)\pmod p$ and $ k$ an integer such that $ p\not|f(k)$ gives the contradiction we're after.

Another solution.
To each polynomial  $ Q(x) \in \mathbb{R}[x] = a_{n}x^{n}+\ldots+a_{1}x+a_{0}$ we assign $ g(Q) = a_{n}f(n)+\ldots+a_{1}f(1)+a_{0}f(0)$.
Let A be the set of all polynomials P such that g(P) = 0.
Obviously A is closed under addition, and the hypotesis on f says that:
- $ P(x) = x^{a_{n}-a_{1}}+\ldots+x^{a_{2}-a_{1}}+1 \in A$ (assuming $ a_{n}\le a_{n-1}\le \ldots \le a_{1}$).
- $ m,n \in \mathbb{Z}^{+}\Rightarrow x^{m}P(x^{n}) \in A$
And, using the fact that A is closed under addition, we conclude that for each polynomial Q, positive integer n:
$ Q(x)P(x^{n}) \in A$.

If we prove that, given a polynomial P(x) such that 0,1 are not roots (our P satisfies this), there exist a natural n such that $ P(x^{n}), P(x)$ are coprime, we have solved this problem, since by Bezout there are polynomials A,B with $ A(x)P(x)+B(x)P(x^{n}) = 1 \in A$, and $ x^{k}A(x)P(x)+x^{k}B(x)P(x^{n}) = x^{k}\in A$, so for each positive integer k, $ f(k) = 0$, and the result follows for the negative integers, too.
Given the finite set B of the complex roots of P, with $ 0,1 \not \in B$, we want to find an n such that $ a \in B \Rightarrow a^{n}\not \in B$. If a has modulus <1, or modulus >1, then the powers of a are infinitely many, so each n sufficiently large is good. If a is never a m-th root of unit, then the same: being B finite, each n sufficiently large is good. If a is a m-th power of unit, each multiple of m is good, since $ 1 \not \in B$.
So we take an n sufficiently large, multiple of $ m_{1}\cdot m_{2}\cdots m_{k}$ where for each i there is an $ m_{i}$th root of unit in B.
And the problem is proved. Do you like my solution?  :P

[quote="grobber"] $ f$ can be taken to have integer values.[/quote]

and why is this obvious?

Did I say it was?

First, make $ f$ rational: write each member $ f(n)$ of the sequence as $ f(n)=\sum\alpha^{n}_{i}e_{i},\ \alpha_{i}\in\mathbb Q$ for a Hamel basis $ (e_{i})$ of $ \mathbb R$ over $ \mathbb Q$, and replace each $ f(n)$ with $ \alpha^{n}_{i}$for some $ i$ chosen such that at least one $ \alpha^{n}_{i},\ n\in\mathbb Z$ is non-zero. Then notice that given $ f(n),\ |n|\le N$ for a large enough positive integer $ N$, the other values of $ f(n)$ can be found through recurrence relations with coefficients $ 1$; hence if $ f(n)$ is an integer for $ |n|\le N$, all $ f(n)$ must be integers. But right now our $ f(n)$ are rational numbers. Multiply everything by the product of the denominators of $ f(n),\ |n|\le N$, and you're done.

I think that to make it with coefficients 1 you have to multiply the $ a_{i}$'s with some integer values so that they are all different.
If some of them are zero you can "translate" them.

Looks like I did overlook the case when some of those $ a_{n}$ are equal, but that's not much of a problem. Allow them to be rational, and notice that if those first few $ f(n)$ are integers, then the only primes which can divide the denominators are those dividing the coefficients of the recurrence. Reduction modulo some prime still makes sense if you choose a prime which does not divide those coefficients.

A question: how can we avoid the use of the Axiom of Choice in grobber's solution?

I'm not sure if this works, since I haven't followed grobber's argument too closely:

Couldn't we just build a basis for the space generated by the sequence $ \left\{ f(n) \right\}_{n \geq 1}$ (via induction)?

If we start like edriv:

[quote="edriv"]
To each polynomial  $ Q(x) \in \mathbb{R}[x] = a_{n}x^{n}+\ldots+a_{1}x+a_{0}$ we assign $ g(Q) = a_{n}f(n)+\ldots+a_{1}f(1)+a_{0}f(0)$.
Let A be the set of all polynomials P such that g(P) = 0.
Obviously A is closed under addition, and the hypotesis on f says that:
- $ P(x) = x^{a_{n}-a_{1}}+\ldots+x^{a_{2}-a_{1}}+1 \in A$ (assuming $ a_{n}\le a_{n-1}\le \ldots \le a_{1}$).
- $ m,n \in \mathbb{Z}^{+}\Rightarrow x^{m}P(x^{n}) \in A$
And, using the fact that A is closed under addition, we conclude that for each polynomial Q, positive integer n:
$ Q(x)P(x^{n}) \in A$.
[/quote]

We can also finish by observing that
$ (x^{a_{1}})^{l}+(x^{a_{2}})^{l}+\dots+(x^{a_{n}})^{l}$ is in $ A$.

So we get all the power symmetric polynomials with $ l\ge 1$. From Newton's identities, it is clear that they generate $ (x^{a_{1}})(x^{a_{2}})\dots(x^{a_{n}})$ which gives
$ f(a_{1}+\dots+a_{n}+k)=0$ for all $ k$ and we are done.


trigonometry
AMC
AIME
geometry
probability
number theory
\/closed

I remember reading about this topic, but I can't seem to find the same page again.
What is the overlap for the classes, Algebra 1, 2, 3, and the books, Introduction to Algebra and Intermediate Algebra.
If I have taken Algebra, Algebra 2, and Trigonometry in school, would it be recommended that I take Alg 1, 2 or 3?
Also, does Intermediate Algebra cover all Algebra that would be on the AMC 10/12 and AIME?

Intro should cover most of the AMC 10/12 stuff, the rest of AMC 12 and AIME is just really Interm Algebra[/i]

Oh ok, ty  :lol:

What's the overlap between the classes and the books? That is, which books correspond to which classes?

Algebra I and Algebra 2 = Introduction to Algebra
Algebra 3 - Intermediate Algebra

[quote]If I have taken Algebra, Algebra 2, and Trigonometry in school, would it be recommended that I take Alg 1, 2 or 3? [/quote]Does anyone know?

Algebra 3 probably. Algebra 2 AoPS probably covers the Alg 2 school again.

You probably don't need Algebra 1 or 2.  Algebra 3 includes some basic material you probably already know, and a lot of more advanced material and harder problems unlike what you had in your regular class.  You can use the diagnostic tests on the classes pages as a guide.

Add Vol. 2 to that and see if you can get 6+ on AIME consistently, then probably.

[quote="AIME15USAMO"]One more question...If I complete Intermediate Algebra (the book), Precalculus (the class), and Intermediate Algebra (class if you are offering one this summer), do you think I'll be ready for WOOT next year?
(@ Mr Rusczyk)[/quote]

You should be (I'm assuming you have done some counting, number theory, and geometry before, or that you'll mix in a little of that over the next year).

[quote="AIME15USAMO"]One more question...If I complete Intermediate Algebra (the book), Precalculus (the class), and Intermediate Algebra (class if you are offering one this summer), do you think I'll be ready for WOOT next year?
(@ Mr Rusczyk)[/quote] Whoops, typo. I meant..
One more question...If I complete Intermediate Algebra (the book), Precalculus (the class), and Intermediate Counting + Probability (class if you are offering one this summer), do you think I'll be ready for WOOT next year?


Jake stood at the center of a circular field that had a radius of $n$. He walked [b]due north halfway [/b]to the circle. He then turned and walked [b]due east half way [/b]to the circle. He turned again and walked [b]due south half way[/b] to the circle. Finally he turned and walked [b]due west half way [/b]to the circle. When he stopped, how many feet was Jake from the center of the circle? Express your answer in terms of $n$. :wallbash_red:

put the circle on coordinate axis.

the equation for the sircle is $x^2+y^2=n^2$

[hide]1.  he first waks halfway to the north so he ends up at $(0, \frac{n}{2}$

2.  he then walks halfway to the east.  Use the equation of the circle to find the distance to the circle.
when $y=\frac{n}{2}$, $x=\frac{n\sqrt{3}}{2}$
so he goes to $(\frac{n\sqrt{3}}{4}, \frac{n}{2}$

3.  he then walked south.
when $x=\frac{n\sqrt{3}}{4}$, $y=-\frac{\sqrt{13}\cdot n}{4}$
so he goes to $(\frac{n\sqrt{3}}{4}, \frac{n(2-\sqrt{13})}{8}$

4.  Then he walked west
When $y=\frac{n(2-\sqrt{13})}{8}$, $x=-\frac{n(4\sqrt{13}+47}{64}$
so he ends at $(-\frac{4\sqrt{13}-16\sqrt{3}+47}{128}n, \frac{n(2-\sqrt{13})}{8}$

so the distance is .575421n?[/hide]

That's really close, I think there're some minor mistakes.

For one thing, you forgot to take the square root of $\frac{n^2(4\sqrt{13}+47)}{64}$, you only took the square root of $n^2$

Too lazy to use LaTeX. Actually, it doesn't even work on my computer. Anyway...
[hide] This is about setting up right triangles. Obviously, when you go north, it is n/2. From there, you extend your path going east, then draw a line from the center to the end of the extended path. Then you find that the extended east path is [(root3)/2]n, so the real east path is [(root3)/4]n. Now extend the southern path. Draw a radius to the end of the extended path, and draw a line perpindicular to the eastern path to the southern path. We now have a right triangle. The length of the long leg is [(root13)/4]n, so the length of the entire extended southern path, divided by 2, is {[2+(root13)]n}/8.
Finally, we find the length of the western path. We extend the western path, then draw a radius to the end of the extended western path. Blah blah blah blah since you know that the short leg of the new right triangle is {[(root13)-2]n}/8 (since we subtract the northern path from the southern path). I cannot believe you are still reading this we know the hypotenuse is n, so SE find that the long leg is {root[55+4root13]n}/8. We add that to the eastern path and divide by to to get {2root13+root[55+4root13]}/16. Then you subtract the lump of cheese aka eastern path, form the last right triangle, then find the answer

I am wayyy too lazy to do that.[/hide]


First, it's easy to see that a triangle can be cut into $n^2$ congruent triangles.
Then, $2005=22^2+39^2$, so we draw an Rt triangle, with its height on the longest side. The height separates the side into 2 segments in $22^2$cm and $39^2$cm. Cut the small one into $22^2$ and the big one into $39^2$.
It's easy to prove the triangles are paiewisely congruent.

can anyone give us other solution  :lol:  :lol:


A gecko is in a room that is 12 feet long, 10 feet wide, and 8 feet tall. The gecko is currently on a side wall (10' by 8'), one foot from the ceiling and one foot from the back wall (12' by 8'). The gecko spots a fly on the opposite side wall, one foot from the floor and one foot from the front wall. What is the length of the shortest path the gecko can take to reach the fly assuming that it does not jump and can only walk across the ceiling and the walls? Express your answer in simplest radical form.

I just want to know how to do this problem...I know the answer is $2\sqrt[2]{113}$

This has already been posted

Here is the link

[url=Gecko in a Room]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Gecko+++Room&t=75889[/url]

Geez I thought a wall was a wall and the ground was the ground...I thought the Gecko couldn't go across the ground, that's why I got it wrong  :wallbash_red:

edit: the ground shouldn't really matter because the "inverse" of that path will go across the ceiling

[hide="hint"]Pertend the Gecko (wat a ghetto name, lol) is in a room made of a cardboard box. Fold out the cardboard box.[/hide]


USAMTS
algebra
polynomial
linear algebra

[quote="SnowStorm"]I was wondering if someone could explain #3 from the USAMTS Round 4.  I spent forever on it, but couldn't get it.
Heres the question btw,
Find, with proof, a polynomial $f(x, y, z)$ in three variables with integer coefficients such that for all integers $a, b, c,$ the sign of $f(a, b, c)$ (that is positive, negative or zero) is the same as the sign of $a + b\sqrt[3]{2} + c\sqrt[3]{4}$.

Thanks

P.S. It's ok to discuss this right?[/quote]

Yes, I suppose that I it is totally OK to discuss this here. But if you are really unsure if you can post something or not, you can just send a message to the moderators.

Now, the problem is interesting, nevertheless let's see what can we do to it. Here you have a few ideas

[hide="Ideas"]
First than all, the expresion $a + b\sqrt[3]{2} + c\sqrt[3]{4}$ cannot be zero in the integers.
Suppose that the condition about the expression above can be extended to the reals, then, note that the triples $(\sqrt[3]{2},1,0)$, $(\sqrt[3]{4},0,1)$ and $(0,\sqrt[3]{2},1)$ are roots.
Take three polynomials that hold with such roots, and multiply them. My example, $f(x,y,z)=(x^3-2y^3)(x^3-4z^2)(y^3-2z^2))$. It is not hard to prove that $f(x,y,z)=0$ iff $x + y\sqrt[3]{2} + z\sqrt[3]{4}=0$, but I am to lazy rightnow to write it totally.
Then, everything that you have to prove is that the sign is the correct for at least two points. Unfortunatelly in my first example it is not, then, it is easy, just invert it, that is, take $f(x,y,z)=(2y^3-x^3)(x^3-4z^2)(y^3-2z^2))$. It works.
[/hide]

I hope it helps. It may be weird to understand (may be) because somehow I am trying to hide Linear Algebra ideas behind my statements, and I don't know if you know something about it.

[quote]
Take three polynomials that hold with such roots, and multiply them.
[/quote]
How can you assume that this will work?

Use the factorization $(x^3+y^3+z^3-3xyz)=(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$.

The USAMTS math jam did a good job of motivating the solution to the problem, but the final answer basically relies on a tricky factorization, as mentioned by Jimmy.  Because $x^2 + y^2 + z^2 - xy - yz - xz = \frac{1}{2}(x-y)^2+\frac{1}{2}(y-z)^2+\frac{1}{2}(z-x)^2$, it is always $\geq 0$, meaning that $\mbox{sgn}(x + y + z) = \mbox{sgn}(x^3 + y^3 + z^3 - 3xyz)$ (unless $x = y = z$, but this can't happen in the given problem except for the trivial case of $a = b = c = 0$).  From there, simply let $a = x$, $\sqrt[3]{2}b = y$, and $\sqrt[3]{4}c = y$, and you have your answer.


Evaluate $ \frac{3\plus{}x(3\plus{}x)\minus{}3^2}{x\minus{}3\plus{}x^2}$ for $ x\equal{}\minus{}2$.

brute force.

$ \frac{3\minus{}2(3\minus{}2)\minus{}3^2}{\minus{}2\minus{}3\plus{}4}\equal{}\frac{1\minus{}9}{\minus{}1}\equal{}8$

answer : 8


geometry
3D geometry
Pythagorean Theorem

What is the length of a space diagonal of a cube, given that a lateral edge of the cube is 4 cm long? Express your answer in simplest radical form.

4sqrt3 since by the pythagorean theorem, the face diagonal is $ s\sqrt{2}$and then again using pythagorean theorem you get $ \sqrt{s^2\plus{}(s\sqrt{2})^2} \equal{} s\sqrt3$ Plugging $ s\equal{}4$ into equation gives space diagonal = $ 4\sqrt{3}$


group theory
abstract algebra
superior algebra
superior algebra unsolved

For some reason I was playing around just now before I go to bed with centers of groups with examples.
So I am curious (examples work out) if $ \text{if } ab \in Z(G)$, then is $ ab\equal{}ba$?
And if $ ab \in Z(G),$ is $ a,b \in Z(G)$.

The answer to your first question is "yes" by definition.  (Otherwise neither $ a$ nor $ b$ would be in the center.)

Your second question is equivalent to the question of whether $ a \in Z(G)$ implies $ b, b^{\minus{}1} a \in Z(g)$ for arbitrary $ b \in G$, which is clearly false in a nonabelian group, again by definition.  (This is especially obvious if you know examples of groups with trivial center.)

Trivial center means stuff like the identity?

Trivial center means [i]just[/i] the identity.  For example, every nonabelian simple group has trivial center (since the center is a normal subgroup and can't be the whole group), so the smallest example here is $ A_5$.

OOh neat. Thanks. I havn't done anything with normal subgroups yet but I'll keep this in mind come that time.


induction
inequalities
analytic geometry
vector
graphing lines
slope

QUESTION:

Let $\frac{a_{1}}{b_{1}}, . . . , \frac{a_{n}}{b_{n}}$ be n fractions (with $b_{k}> 0$ for k = 1, . . . , n). Show that the fraction

$\frac{a_{1}+. . .+a_{n}}{b_{1}+. . .+b_{n}}$

lies between the smallest and the largest of the given n fractions.

My Attempted Solution:

I am hoping that somebody can check my solution, or atleast help me come up with a more straightforward solution.

$\frac{a_{1}+. . .+a_{n}}{b_{1}+. . .+b_{n}}$

=$\frac{\sum_{i=1}^{n}a_{i}}{\sum_{i=1}^{\n}b_{i}}$

=$\sum_{i=1}^{n}\frac{b_{i}}{\sum_{i=1}^{n}b_{i}}*\frac{a_{i}}{b_{i}}$

=$\sum_{i=1}^{n}\frac{a_{i}}{\sum_{i=1}^{n}b_{i}}$

I can get this far, but how does this imply that the average lies between the largest and smallest fraction? any ideas?

It's an easy induction proof. Start with two fractions.

problem_solver wrote:Elemennop wrote:Here's a hint:

[hide]

Order the fractions (ie. let  for .

[/hide]

Sorry, ordering it from smallest to biggest dosn't make it any easier...

oh boy...

You sure? Try ordering it, then assuming your inequality is true (that is,  is less than the given expression). Can you reduce that to something you already know is true?

[quote="t0rajir0u"][hide="Super hint"] Put $(b_{i}, a_{i})$ on a coordinate system.  Where is $(\sum b_{i}, \sum a_{i})$?  [/hide][/quote]

That tip helps.

Put $(b_{i}, a_{i})$ on a coordinate system, the value of the fraction is just its slope. The above sum is just the vector sum of the points.

Since we are only interested in the value of the fraction, the slope, we wil lthink of the above points as the centroid of the points, hence smaller that the biggest fraction and bigger thaan the smallest fraction?

so is $(\sum b_{i}, \sum a_{i})$ the centroid?

Has the same slope as the centroid, yeah.


geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

A plane is colored in two colors. For each triangle whose vertices are same colored, its orthocenter has the same color as its vertices. Prove that the whole plane is colored with one color.

The assertions also holds (but is much easier) for circumcenter. Is there maybe any generalization?

Hint:
[hide]There always exists a monochromatic equilateral triangle.[/hide]

I think this should work:

Take any line $\ell$ and consider it horizontal. Suppose $A,B$ belonging to this line have the same color, and let $C$ be a point whose projection on $\ell$ is neither $A$ nor $B$. The orthocenter $D$ of $ABC$ must have the same color as $C$. Now consider the reflections $A',B'$ of $AB$ in the perpendicular through $C$ to $\ell$. Since $A'B'CD$ is an orthocentroidal quadrilateral, and $C,D$ have the same color, $A'B'$ must have the same color. We have thus found the following: if $A,B\in\ell$ have the same color, then any two points $A',B'\in\ell$ with $A'B'=AB$ and s.t. $A,B,A',B'$ are all distinct must also have the same color.

We can eliminate that artificial condition that $A,B,A',B'$ are distinct as follows: suppose $A'=B$ (s.t. $B$ is the midpoint of $AB'$), and that $A,B$ have the same color. Now take a segment $UV=AB$ on $\ell$ s.t. $U,V$ do not belong to $\{A,B,B'\}$. Then $U,V$ must have the same color, because $UV=AB$ and $A,B,U,V$ are distinct, so $A'=B,B'$ must also have the same color, since $UV=A'B'$ and $U,V,A',B'$ are distinct. We thus have the following result: if $AB=UV$ are two segments on some line $\ell$, then $A,B$ have the same color iff $U,V$ have the same color.

In particular, the proposition above says that if $X,Y$ are symmetric wrt $O$, then $X$ and $Y$ must have the same color, and we're done: every two points in the plane are, of course, symmetric wrt some other point.


[quote="10000th User"]What happen? All MathLinks images gone! No Emoticons, no top images, all disappeared![/quote]They will be back soon :)

[quote="jhollenbeck"]when will they be back?[/quote]Soon ment in a couple of minutes back then.


Je habite les Etats Unis, mais je suis de Costa Rica. Oui, je sais que mon francais c'est pathetique, mais je m'efforce pour amepolier le.

Au revoir,

Bonjour Patrick,

[quote="DPatrick"]Et est-ce que vous lirez les mathématiques en français ou en anglais (ou en une langue différente) le plus souvent?[/quote]

Tant qu'à faire je préfère lire les mathématiques (ou l'informatique) dans la langue de Fermat. Mais les lire dans la langue de Wiles ne me dérange pas  :) . L'immense majorité des ressources disponibles sur le web sont en anglais de toute façon.

Amicalement

Moi aussi je lis en anglais la plupart du temps, c'est vrai qu'on trouve plus facilement ce qu'on cherche dans la langue de Thomas More  ;)

[quote="alkachi"]nous au maroc à l'école on étudie en arabe mais les recherches et les documentations en français et pour les forun en generale c'est en anglais; Bref nous utilisons plusieurs langue pour les mathématiques :lol:[/quote]
Oui j'affirme ce qu'as dit alkachi nous des "polylingue" :D

[quote="DPatrick"]Je suis curieux...

Et est-ce que vous lirez les mathématiques en français ou en anglais (ou en une langue différente) le plus souvent?[/quote]

 :D  L'avenir, c'est  le volapük   :P

Bonjour everyone, Je peux parler francais :D et J'etude francais

Bonjour!  J'etudie francais aussi.  J'habite de Cupertino, Californie.

(Pease correct me if I made a mistake.  I need to improve my French)

(Bonjour! J'étudie le français aussi. J'habite à Cupertino en Californie)
:)

Salut!
Moi aussi. J'habite en Vietnam. J'etudie francais :)

[quote="mathmanman"]:)

"J'habite [b]au[/b] Vietnam. J'étudie [b]le[/b] français."

For the accents, try the extended ascii chars ;) (http://www.lookuptables.com/ - e.g. : ALT+136 --> ê)[/quote]
 :blush:
Merci bien, mathmanman :)

Ah, et maintenant vous les deux savez parler francais!!! Vous etes des profs tres interessants!

Actually, Il y a une région française au québec, au canada.

Ouais, je sais. J'y suis allé une fois (peut-être deux). Les gens là-bas sont un peu bizarres. Ils disent "Est-ce que je peux parker mon car ici?"

En fait c'est "stationner son char" au Québec! (J'ai habité à Montréal pendant un an et en France pendant 9 mois il y a quelques années.) Mais en France, on dit "parquer sa voiture". Cool, hein? Normalement le Québec est plus à l'aise à éviter les anglicismes (par exemple, utiliser "baladodiffusion" au lieu de "podcast")

Ah, bon! Je pensais que c'était à l'envers, qu'on utilisait souvent les anglicismes au Québec.

Bonjour tout le monde! Je ne suis pas francais mais j'apprends francais a l'ecole et j'adore la langue. Mon professeur de francais est superbe. Ma mere aussi parle le francais tres bien. Je voudrais ameliorer mon francais par discuter avec vous. Tres desole pour ne pas mettre les accents car mon ordinateur ne le permit pas.

Et pour quoi les gens n'ecrit plus dans ce forum?

Est-ce que c'est correcte?

Je suis très content que tu veux participer dans ce forum! Il y a longtemps que quelqu'un a parlé ici. :(

Qu'est-ce que tu voulais dire par "Est-ce que c'est correct?"

Bonjour!!! Je veux dire que le chose que j'ai ecrit, est-ce que c'est correct?  Est-ce qu'il ya quelques fautes?

J'habite ici en etats unis et je lis en anglais au lieu de francais parce que c'est plus facile.

J'habite au Canada, mais ne pas du Québec. Je viens d'Ontario.

Bonjour tout le monde! J'habite au Wisconsin ou il fait très froid pendant l'hiver.

j'habite en l'etats-unis,

je visite france l'année dernière,

ou habites vous??


In chess, a knight moves in a L-shape manner -- two spaces in one direction and one space in a direction perpendicular to the first direction--as shown. Beginning in and including the upper left corner(marked *), what is the greastest number of squares in a 4x4 checkerboard that a knight can visit without visiting any square more than once?('95 State Sprint #30)

Billy

I saw this in the all time greatest book... the answer is 15. Is there a formula for this kind of problem?

This isn't the best of explanations, but you end up getting cornered because of the way a knight movesno matter ow much you try... sorry if this is vague but no matter how many moves you make, you always somehow run out of spaces before the 16th... some people get stuck and only get 14 because they don't move right...

[quote="solafidefarms"]But can anyone explain the [b]why??[/b] That's my problem.[/quote]
[hide="I think we can show why we cannot cover all 16 squares"]
First, let's alternately color the checkerboard black and white.

Next, note that the color a knight is on changes with each move.  In other words, every odd number will be on (say white).  Every even number will be on black.

Next, note that there are only two ways in or out of each corner.  For a white corner, the only two ways in or out are via the two black squares in the middle.  For a black corner, the only two ways in or out are via the two white squraes in the middle.

If we are to fill both white corners, we must start at a white corner (1).  Then touch one of the black squares in the center (2).  Then touch the opposite white corner (3).  Then touch the other black square in the center (4).  Then proceed with moves 5+.

If we are to fill both black corners, we must end at a black corner (16).  Prior to that, touch one of the white squares in the center (15).  Prior to that, touch the opposite black corner (14).  Prior to that, touch the other white square in the center (13).  Preceed with other moves (12-).

The remaining eight numbers (5 through 12) must fill the remaining empty squares.  But if you examine the remaining empty squares, you will find there are two separate cycles of 4 squares each, and there isn't any way to cross between the two cycles, using a knight's move.[/hide]

... and as usual rcv comes up with the perfect responses... :)
Edit:
Here's another problem, it's somewhat different from this problem, but it involves a pattern... I think it's from the 01 states.. but I'm not sure

In the counting game Blippy, one counts as such: 1,2,4,5,7,8,10,11,14...so that you all numbers except numbers that are multiples of three or contain the number three as one of its digits. What is the 41st number that in this sequence? (I wrote this from memory, so It might not exactly be the same problem)

[hide]Take a estimate. Lets say 60. There are 10 numbers with 3 in the tens place, and there are 6 with a 3 in the units place. There is 1 with a 3 in both places, and we have counted it twice, for 15 numbers that are disqualified for containing 3's.
Then we take out the multiples of 3. There are 20, but we have counted 30, 33, 36, 39, and 3 already. So there are 35-5=30 skipped numbers. So below 60 there are 30 numbers said. Then we just count up till we get the 41st:
61,62,64,65,67,68,70,71,74,75,[b]76[/b][/hide]

Billy[/b]

[quote="hockeyadi23"]There's a faster way to do this problem, but it's not too much faster than your way billy...

There's a distinct pattern in this problem... which makes it really easy. However if you don't notice this, you will have to write out 41 numbers.[/quote] What's the pattern?

This is the way I found: each group of tens has 6 numbers that are counted, for example, in the 20's, there are 20,22,25,26,28, and 29.

So...

6+6+6+6+6+7=37 (I'm just adding my way up to 40, and the 7 is there because 63 is both a multiple and contains a 3)

That's 6 groups of ten, so the number is in the 60's. But none of the 30's are counted, so the number is in the 70's. Listing out the 70's, we have 70,71,74,76,77, and 79. It's the 3rd number, 74.

Is that simliar to your way?

Actually there are six in the 60s: 61, 62, 64, 65, 67, 68. I don't know what you mean by there are seven... but the sixes was the exact same way I did it.

so there would be 6 in every group of ten, but with all the 30s excluded.

6+6+6+6+6+6=36
41-36=5
5th number in the 70's is
70,71,73,74,[b]76[/b]

Ok, I think I've got it now. Thanks

[hide]15 i do this kinda stuff when i play chess cuz i luv knights
(ever used the fried liver on a noob?)[/hide]


calculus
derivative
calculus computations

How do you calculate the shortest distance between the point (3,5) and
the circle with radius 1 on an easy way.
I used the euclidean distance and used differentiation, but i get a lot of trouble doing that.

Thanks

Try a straight line from the point to the center of the circle. (I'd do it, but I'd have to assume the center is at (0,0)).


I was wondering how USAMTS score converses into AMC score.

I think someone said AMC score = 6*USAMTS round 1 score...

what is the conversion?

There is no conversion. They are completely different contests.

However, USAMTS can just simply help you qualify for AIME or USAMO if you don't take AMC's. (I think, but I'm not sure)

[quote="gauss1181"][quote="worthawholebean"]Having a high score (usually 68) after three rounds qualifies you for the AIME.[/quote]
Wow that's cool. I'd better train for that. :D[/quote]

Well first...there isn't really anything to train for...you just like, do the problems. And also it gives you an AMC score of 100 so it isn't really [i]that[/i] helpful for qualifying for the USAMO, you'll probably need 11 or so on the AIME for it to do anything. It's significantly easier to qualify the regular way.

Catalyst is right - It does require a lot more skill to qualify via USAMTS than the normal route and gives you worse chances for USAMO. Last year I qualified for AIME by a huge margin on AMC and missed by a pretty large margin on USAMTS. That said, it's nice as a safety net in case you're sick on day of AMC/snow day/whatever.

This is true, would 4 problems on each qualify me???

[quote="AIME15"]Well 68/75 over three rounds... is what's needed.

4 problems each set, assuming you ace each, gives you 20*3/75=60/75.  Eight points less than needed.

I'd say it's easier qualifying with AMC :P[/quote]
I agree with that. So I'm doing BOTH USAMTS and AMC this year.

[quote="FantasyLover"]well i qualified for AIME last year(7th) so im just doing USAMTS for fun(go prizes)

 :rotfl:[/quote]
Same here, except I'm also doing AMC for the first time.

i dont really understand the USAMTS problems
How should i approach them?

USAMTS is designed to be much closer to real research mathematics than anything else. It's totally fine if it takes you a couple days to solve a problem - that's a lot closer to what mathematicians do than olympiads.

my school doesn't administer AMC10/12
so USAMTS is my only hope
and its pretty darn hard

Are there any local schools that do administer the tests that you could possibly test at? Or one of those colleges on the AMC website?

Yeah, like I took AMC with my local high school...

Of course, it all depends on where you live...like if you're in Alaska, and your school doesn't do AMC, drive like 300 miles to take it...

My local HS offers AMC but not many people take it...I think I should join THEM. (just to at least spread the word about AMC)

FEBRUARY
The message is too small. Please make the message longer before submitting.

[quote="gauss1181"][quote="Heero Kirashami"]OBVIOUSLY we can't tell you directly, but if you look at old solutions from previous years, you should be able to get an idea of how to approach these problems.

Remember: Sometimes, to do these problems, you need to stare at a piece of paper for hour(s) before you understand anything.[/quote]
That happens to me too. Sometimes there are even problems where I take several days. :mad:[/quote]

i just stared at 3 problems and didn't write anything down, and eventually, if you think hard enough, you'll come up with a solution in your head.

[quote="AIME15"]What's the definition of "high school", as defined by USAMTS?[/quote]
you just can't be graduated from 12th grade yet


[quote="nateharman1234"]in mod 12?[/quote]

In mod 3 also, no?

also in mod 6.... just any mod $n$ , where $n$ is a factor of twelve

[quote="swiftkodo"]11+2=1
discuss[/quote]

Please specify what you would like to know.

well its supposed to be 11:00am (or 11:00pm) add 2 hours and its 1:00pm (or 1:00am)
but the mods work fosho

sen is right.  11+2 doesn't equal one, because 13 doesn't equal 1.  No matter what modulo you're working in.

[quote="swiftkodo"]11+2=1
discuss[/quote]

this is never true...

when you say $13 \equiv 1 \mod 12$, then you are saying that they are [i]equivalent[/i], not equal

[quote="LordoftheMorons"]$\equiv\neq=$[/quote]

Nice.  :D

Of course, when you're faced with an obviously false statement like that, the natural thing to do is to assume that $= \neq =$, as it were - that $=$ is being made to stand for some other relation.

Well, if you say it as 11 p.m. (or a.m.) plus two hours is 1 a.m. (or p.m.) you should be OK.

It's more logic than math though.


currently new blogs are not working

FF 3.0 and IE 7 and 8.

i deleted my blog and started a new one but i can't acces it

the url was broken so i replaced "viewtopic" with "weblog".

then it just showed an empty page.

that is all.

There's definitely an issue here. I will investigate it soon - meanwhile don't delete your blogs :)


Prove that, if $\alpha, \beta$ and $\gamma$ are internal angles of a triangle and they satisfy the equation
\[ \sin(3\alpha)+\sin(3\beta)+\sin(3\gamma)=0  \]
then, at least one of the three angles $\alpha, \beta$ or $\gamma$ is equal $60^{\circ}$.

[quote="shyong"]$\sin 3\alpha+\sin 3\beta+\sin 3\gamma=-4\cos\frac{3\alpha}{2}\cos\frac{3\beta}{2}\cos\frac{3\gamma}{2}=0$

So one of them , say $\cos\frac{3\alpha}{2}=0$ $\implies \alpha=\frac{\pi}{3}$

 :)[/quote]

 :lol:

Ok, but there is a lot of work between $\sin 3\alpha+\sin 3\beta+\sin 3\gamma$ and $-4\cos\frac{3\alpha}{2}\cos\frac{3\beta}{2}\cos\frac{3\gamma}{2}=0$

:P

I remind you some properties from trigonometry:

[u]The reduction to the first quadrant:[/u]

$\sin x=-\sin (x\pm \pi );\ \cos x=-\cos (x\pm \pi);$

$\sin x=-\sin (-x);\ \cos x=\cos (-x).$

[u]Some trigonometrical identities[/u] $(k\in Z):$

$\blacksquare \ 1^{\circ}.\ x+y+z=2k\pi \Longrightarrow \sin x+\sin y +\sin z=4(-1)^{k+1}\sin \frac x2\sin \frac y2\sin \frac z2;$

$\blacksquare \ 2^{\circ}.\ x+y+z=(2k+1)\pi \Longrightarrow \sin x+\sin y+\sin z=4(-1)^k \cos \frac x2\cos \frac y2\cos \frac z2.$

[u]The solution of the proposed problem: I used the relation[/u] $(1)$ [u]for[/u] $k: =0$.

$\triangle ABC,\ (\pi -3A)+(\pi -3B)+(\pi -3C)=3\pi -3(A+B+C)=0\Longrightarrow$

$0=\sum \sin 3A=\sum \sin (\pi -3A)=-4\prod \sin \left (\frac{\pi}{2}-\frac{3A}{2}\right)\Longrightarrow$

$\prod \cos \frac{3A}{2}=0\Longrightarrow\prod \left(\frac{3A}{2}-\frac{\pi}{2}\right)=0\Longrightarrow \prod \left (A-\frac{\pi}{3}\right)=0\Longrightarrow$

$A=\frac{\pi}{3}\ \vee \ B=\frac{\pi}{3}\ \vee\ C=\frac{\pi}{3}.$


[quote="xxxxx"]Find the largest value of X that satisfies:
 (3x - 5) = (x + 2)
 (both sides in absolute value)

could somebody type like a basic list of LaTex to use for these signs and stuff?[/quote]

so then your problem will be

$|(3x-5)| = |(x+2)|$

[hide]$3x-5=x+2$ OR $3x-5=-x-2$
Solving each we get $x=\frac{7}{2},\frac{3}{4}$, so the higher is: $\frac{7}{2}$.[/hide]


Here's a problem that I'm having a lot of trouble making progress on.
(a) Let $ p(t) \equal{} at^2 \plus{} bt \plus{} c$ be any quadratic polynomial. Verify that $ p(1) \plus{} p(4) \plus{} p(6) \plus{} p(7) \equal{} p(2) \plus{} p(3) \plus{} p(5) \plus{} p(8)$. (This part is easy)
(b) Partition the set of numbers $ \{1,2,3,...,16\}$ into two sets such that, given any cubic polynomial $ p(t)$ with integer coefficients, the sum of the numbers $ p(k)$ where k ranges over one of the two sets is the same as the sum were $ k$ ranges over the other.
(c) Let $ m$ be a positive integer. Prove that given a set with elements being the numbers 1 to $ 2^{m \plus{} 1}$, you can divide the set into two subsets such that for any polynomial $ p(t)$ of degree not exceeding m, as $ p(t)$ ranges over the elements in one of the sets, it's equal to the sum as $ p(t)$ ranges over the elements in the other set.

Thanks for any help!

Read the sections on the generalized forward difference and characteristic polynomials in [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=227941]this[/url] and then look at $ (r\minus{}1)(r^2\minus{}1)(r^4\minus{}1)\cdots (r^{2^m}\minus{}1)$. There is a neat way to interpret the desired partition in binary.

(Also, the first practice problem in that post is basically equivalent to this.)


Let $ a,b,c > 0 $
   Prove that :
 $ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{a+b}{c+a} + \frac{b+c}{a+b} + \frac{c+a}{b+c} $

let x=a/b ,y=c/a ,z=b/c

problem is true if and only if

(x^2+y^2+z^2-x-y-z)+(x^2*z+y^2*x+z^2*y-3)>=0

first term and second term are nonegative

Because (x^2+y^2+z^2)>=(x+y+z)(x+y+z)/3 (Because Cauchy)

>=(x+y+z) (A.M.>=G.M.)

second term A.M>=G.M.
_________________


I need the following book: Local Cohomology: an algebraic introduction with geometric applications. If you have the file of this book, please send it to antrannguyen@gmail.com. Thank you very much!

1. You can buy it from Amazon
2. You can check your local university library if they have a copy of it.
3. Maybe [url=http://books.google.com/books?id=k7a5GcWkPL4C&dq=%22Local+Cohomology:%22+an+algebraic+introduction+with+geometric+application&printsec=frontcover&source=bl&ots=3e0MSTGuV0&sig=f1DoyyKIheAb6hZsNngznQ5SY2A&hl=en&ei=ceDDSdTuINWPsAawoLH-Bg&sa=X&oi=book_result&resnum=1&ct=result#PPR7,M1]this[/url] helps.

Thank you. Now I get the files of this book.


AMC
AIME
topology
videos
ARML
articles
National Assessment

Hmm. Okay I am new to AoPS, but I thought this might be the best place to post this thread... I'm sure you'll tell me if I'm wrong.

[color=blue]I'm in charge of math team at my school, which is a strong indicator of the sad state it is in (or maybe the cause of it? >.>;; ), and I was kinda wondering how other math teams are run, to figure out how to put it on a more positive trend.[/color]

My school: Mu Alpha Theta = math tutoring and that's about it, math team consists of ~30 people, and meetings generally ~10-15 people, once a week, kinda breaks up after 30 minutes. Not a lot of team spirit at all, and sadly I must admit we don't do that well at competitions (well, I was the only AIME qualifier last year...). Not my main concern if it weren't for the morale drag. I'm more interested in kinda sharing the whole “whoa-- [i]this[/i] is math” feeling I got from GHP... Uh, my officers are not that enthusiastic? er not the proactive kind. People in general are more spread out among all the different clubs... there are definitely smart people, but many aren't interested in math team, and those that are have commitments to lots of other clubs as well.... well I will stop rambling now XD

[b]So if anyone would be willing to help me out...  [/b][color=blue]I was wondering how math teams @ [i]your[/i] schools are run. [/color]
Specifically: (in no particular order) (don't feel obliged to answer all of them... it turned out to be kinda long >.>;;)
[list][*][b]Problems of the day-- [/b]where do you get them from, how do you post them, how do you collect answers, how do you award points, what awards do you give / or just bragging rights
[*][b]Organization-- [/b]how many officers / what they are in charge of, how officers are elected or chosen. How much is up to sponsor vs. students. How do you manage notifications / mailing lists. Yea...
[*][b]T-shirts-- [/b] redo every year or not, members vote on it or just officers decide, even more helpful if there was a central collection of math team t-shirts somewhere, for ideas
[*][b]Meetings-- [/b]how often, how long, requirements. How convince people to come and stay, what do you do at meetings generally? Listen to one person run through problems, or do a “problem- time to work – present solution” system? Or something else?
[*][b]Calendar-- [/b]Related to above what do you do @ mtgs, what sort of events do you do? Talk about famous mathematicians? Math beyond normal competition math, such as topology or cryptography etc. (who talks about that or just general idea)? Just run through practice problems? Do you go over problems after the competitions, vs. Do you have a crash course / any course at all (like alg > geo > or AoPS 1, 1st semester or something)? If both, how do you fit going over competitions and the crash course into your meetings? If you add in extra-math stuff, like PLUS magazine or w/e, how in the world do you fit it in? O__o
[*][b]Levels-- [/b]How do you deal with different levels and expectations? Do people in your math team actually work on problems at home, or by themselves? Do you try to make it everyone-is-welcome and then go through a crash course? Do you split up into JV/V, and how?
[*][b]Publicity-- [/b] What's it like at your school? When they say, “cheer on our X team as they play Y this saturday” do they ever include math team? I dunno everything. Oh, how do you recruit members and retain members? Free food only worked for so long XD;; I really am incredibly horrible at this stuff.
[*][b]Requirements-- [/b]again emphasizing this, what sort of requirements do you guys have. How do you (do you?) make people lean more towards quality than quantity. As in, more than just showing up at meetings, and not showing up 30 minutes in after, say, the Key club meeting is over. Do you require people attend entire meetings? etc.[/list]
Well, in general, motivation. How. >.<;; Also, more oriented towards the competition aspect, or the exploring-math / fun aspect?
Or anything in general, any comments / experience you would like to share...

Oh, what websites do you use a lot, for resources etc.? More videos like the awesome Google video on topology would be nice. Or anything, for the exploring-math/ having fun aspect. Oh, and the whole community and resources thing. I guess AoPS is as close as it gets, though.

Er, long post, kudos and thanks if you get down here. :]

[size=50]p.s. And no, normally I don't butcher grammar like this... but for some reason I'm in a fragments-mood. iunno why XD The rambling is normal 'tho... >.>[/size]

Hey qwuery, welcome to AoPS.
I think many schools share the general lack of interest in math that you described. My school (Woodward Academy) is lucky to have eight or so people show up at meetings. But I think that can be alright. The eight people who show up are usually people who are good at and interested in math. There are announcements at the beginning of the year for math team and that's about it. I don't think heavy recruitment is necessary because those who are interested will come on their own. You can only influence interest in math to an extent. Some people just aren't and will never be interested.
My school doesn't have officers, and I like this because it discourages people from doing math team simply to brag that they were an officer on their college applications.
I don't think people should be required to do anything. Those who are interested in math will attend meetings and do math on their own. Those who aren't won't come to meetings and that's fine. They can pursue different interests.
Competitions that can be taken at your school are fun, and they are good for those who are not willing to or are not able to attend Saturday competitions - here are some that I recommend:
Mandelbrot
Georgia Math League
National Assessment and Testing
Kennesaw Math Competition
(Google any of these that you have not heard of)
I know that I didn't nearly answer all of your questions, but I don't have much to say about anything else. My school is lucky enough to have a great math team sponsor, so I and other students don't really have responsibilities - he takes care of mostly everything.
Also, if you don't mind saying, who are you, and what school are you from?

[quote="Twiz"]You can only influence interest in math to an extent. Some people just aren't and will never be interested.[/quote]
I disagree with this.  However, making people become interested in math is no trivial task.  A skilled presenter can instill some interest in a group of good students with some reliability.  Instilling interest in an individual is, I imagine, harder.  You must show the individual something that somehow strikes him or her as elegant.  Because of the way math is often taught in school, this can be difficult.  However, I try from time to time, and every once in a while, I have met with success, which is immensely gratifying.  Of course, such an elegant demonstration is only a beginning.

[quote]Also, if you don't mind saying, who are you, and what school are you from?[/quote]
I second this.  Where in Georgia are you, at least?  Maybe there's an ARML coach near you who can offer you help and advice.

Our school has a similar situation officer-wise to Twiz's.  However, there is something to be said for getting people to do good things for the wrong reasons.  I imagine many people who join our math team are impressed by the piles of trophies we have on shelves—they are interested in the trophies.  But hopefully, they will become more drawn to the actual mathematics.  I think this is how I was originally drawn to math team myself in middle school—I wanted to win awards, I was already good at math compared to other students, and liked it, and math team was advertising on the announcements.   Quality of people on math team is all very fine, but quantity is nothing to scoff at.

Anyhow for advertising, our math team is barely publicized to the school at large, proportionate, say, to our football team, and each team's relative success (compare :  two-year state champions, people who lost all but one of the first seven games of the season).  But everybody takes a math class, and many of the best students pass through the hands of a few people.  When these students ask how they can do better in class, our teachers (who are also our coaches) say, "Come to math team."  Some of them do—and some of these people have become regular members of the team.  Even if they never shine, it still does them good, and they can't possibly [i]hurt[/i] the team.

Anyhow, you might want to read about some of the winners of the [url=http://www.unl.edu/amc/f-miscellaneous/f4-sliffe/sliffe.shtml]Sliffe awards[/url] for distinguished mathematics teaching.  Also, I think there are some [url=http://www.artofproblemsolving.com/Resources/AoPS_R_Articles.php]articles[/url] on this site on the subject.

[quote="Boy Soprano II"]However, I try from time to time, and every once in a while, I have met with success, which is immensely gratifying.[/quote]
I guess the reason I hold a different opinion is because I have also tried to interest people in math but have had no success whatsoever.  :maybe:

[quote="Boy Soprano II"]I imagine many people who join our math team are impressed by the piles of trophies we have on shelves—they are interested in the trophies.  But hopefully, they will become more drawn to the actual mathematics.  I think this is how I was originally drawn to math team myself in middle school—I wanted to win awards, I was already good at math compared to other students, and liked it, and math team was advertising on the announcements.[/quote]
I also was drawn to math by the awards. :-) But strangely enough, awards had the opposite effect on another kid at my school. He joined math team this year as a senior, went to Billy's tournament as his first competition ever, won first place in ciphering, and I haven't seen him at any math team practices since (I assume he quit).

And just to be clear, I am in no way trying to encourage trying to limit membership in math team. It should be open to anyone who wants to join. I just don't think it's a big deal if it's a small team (unless it's too small to regularly get four people together to attend a tournament as a team as is sometimes the case at my school :( ).

Thanks for the replies! No worries about answering everything, I blather a lot XD

[quote="I"]make people lean more towards quality than quantity. [/quote]
Er, sorry ambiguous statement, I meant more towards attending a few clubs but dedicating more time to each (quality), as opposed to joining 12 clubs but just "being there" (quantity).

[quote="Twiz"] My school doesn't have officers, and I like this because it discourages people from doing math team simply to brag that they were an officer on their college applications.[/quote]
That is something to consider, I could just dismantle the whole officer thing for next year. Yea... math team doesn't seem to be a high priority sometimes... A bit of a problem when all your officers are busy w/ APs and other clubs .____. Myself included.

We do have ~10 people who regularly come to competitions, a bit worrying sustainability-wise however is the fact that it's majority juniors and seniors. Is that normal though?

[quote="Boy Soprano II"]I imagine many people who join our math team are impressed by the piles of trophies we have on shelves—they are interested in the trophies. But hopefully, they will become more drawn to the actual mathematics. [/quote]
Uh, the trophy-attraction is less of a possibility for us I think XD;; We do alright, it's just I think it'd be more encouraging if we did a bit better. Or am I taking all this stuff to seriously, should I just relax and take the all the defeatist jokes in stride?

Compounding this is that majority people aren't willing to work on anything at home. It seems to me like they've just been good at math in middle school, and like competitions because they won some in middle school, and I'm afraid that our so-so streak loses quite a few promising people =/
(yea, I fail at inspiring people, even against 'rival' schools XD)

So, I guess it's normal for math teams to be rather incognito then. Our school is reversed-situation anyways, I'm still investigating how they turned our football team into a winning team (morale? inspirational coach? super-training? >.>;; I should pay more attention).

But... our school is pretty big-- 3000 people. They had to cap the Mu Alpha Theta memberships at 80 or something, and robotics has people staying after for 2+ hours every week (~40 people). So I don't know what I am doing wrong that it is so hard to get weekly attendance for an hour from, say, 15 people. X____x;;

[quote="Twiz"] Mandelbrot
Georgia Math League
National Assessment and Testing
Kennesaw Math Competition [/quote]
Hmm, we've only done KSU. Will look into them, thanks =)

[quote="Boy Soprano II"] A skilled presenter can instill some interest in a group of good students with some reliability. [/quote]
So, kinda like a math circle? Well... I don't qualify as a skilled presenter, judging by the amount of chatter I get when putting up a solution XD Myself, I'm kinda short on time =( and I just don't see anyone else caring enough to invest the amount of time required.
I don't know if there's a good starting point to find good presentation material? I could theoretically ask for volunteers, but would need some examples.

[quote="Boy Soprano II"] Anyhow, you might want to read about some of the winners of the Sliffe awards for distinguished mathematics teaching. Also, I think there are some articles on this site on the subject. [/quote]
Duly noted (but yet to be read), thanks!

[size=50] bleargh why are my posts always so long XD blah blah I could go on  forever... as for locality, I use this pseudonym a lot so I'll defer that to less Google searchable methods like pm-ing (yes I'm paranoid XD)[/size]