As I said, we shouldn't talk about organizing a tournament, we should organize a tournament.

So the first matter of business is teams. Teams or no? If yes, how many on a team?

By choosing 2,3,4, you mean EXACTLY that number of people, no more, no less. Up to 4 means 1, 2, 3, or 4.

I like the idea of teams by region, but don't think it's necessary to have a team specifically based on region, if any N people want to be on a team, they should be able to be on a team.

And we should have team names, like in the team chess tournament.

I will organize, but not be the head of, (I don't think that there will be "heads" of teams) team $ \alpha$.(alpha). I will periodically post all of the teams so far.

[hide="alpha"]1=2[/hide]

[hide="Team beta"]
#H34N1
knusperfukels123/George S Bashour
[/hide]

[hide="Team Moldova"]
Klebian
stupidityismygam
[/hide]

Team Delta[hide]

korammath
hunter34
or if im lucky, fightthetide
[/hide]

Team Zeta
[hide]lotsofmath
fireemblem13[/hide]

Team FURBIES ON FIRE  :diablo:
[hide]
infinity4ever
[/hide]

ok i think 6 teams is enough... that would make 24 people.

Dude, it seems like 2 is the majority. Sweet.

[hide="Team Carmlsaml"]
CatalystOfNOstalgia
perfect628[/hide]

i'll join infinity's team.
if she thinks i suck too much, then unimpossible can play for me

nah, u don't suck  :)

wait, so i'm assuming these are 2-person teams? or can we have more ppl?

[hide=" FURBIES ON FIRE"]
infinity4ever
mz94[/hide]

random person from tx want to be on my team?

[quote="alkjash"]I'll join carl's team. MA pride...[/quote]

Two person teams...and what clever play on words from a math comp could we turn your name into anyway?

If anyone here wants to team up with me, that would be great. :lol: And if we're doing it by states, I'm in PA.

Ok, fine then.
Be that way...

I'll join anyone's team.
But I'm not wanted...
 :(

[quote="ra5249"][hide="Team Plastic"]ra5249
stevenmeow (if he wants to...)
budi713 (also if he wants to...)[/hide][/quote]
once again
2 person teams

oops fine
i guess i might have to kick off budi for inactivity...
we'll see who replies first

ooh ooh me!!!!!!!!!! pick me!!!!!

btw i raised policecap's rating 60 pts for free  :D  :lol:

[u]Team Hunchback[/u]
me
me

We shall pwn.

Team annoying hobbits that kill all pandas

team member:

dingzhou
dingzhou
dingzhou
dingzhou
dingzhou
dingzhou
dingzhou
dingzhou
dingzhou
dingzhou

be scared....be very scared

*evil laugh.

dingzhou, i'll be on your team if you want.

I want a team... waaah. If I do get a team, I want it to be called ...

Team Omega:
ME!!!

(if you want to join, pm me)

[hide="Team Beast"]mathcrazed
unimpossible (if he wants to =)) [/hide]

Lol team beast xD... so is this for sure 2 person teams?

If it isn't, I'll go with klebian and stupidity lol xD

Oh my. Looks like I am with koram. When are we having the games?

absolutely, so really, when is this going to happen, this is dumb, we have been planning this since like 2 months ago.

Maybe it would just be easier if someone just kind of randomly assigned teams or if you guys went 1 on 1 in a countdown or something and did a tournament bracket.

Yeah, I agree. Although I'm not joining, when I first thought of tournament what I had in mind was a lot of 1 vs 1 matches.

Can I join a team???? PLEASE????? like 1=2's???

is this about the march 08 tournament or wat???

is this another tourny cuz  if it is, my team would like to compete

You are registered with "Boyz of Darkness".
Take a look at your team here:
http://aopsftw.googlepages.com/boyzofdarkness
Or go [url=http://aopsftw.googlepages.com/march08]Here[/url] for more info on the march 2008 tournament.  Google is so cool...
You can email me at aopsftw@gmail.com
if you have questions.
:)

i am registered for that tourny,

but is this a different one?

or is this sum random topic now??

Well, we have the February Tournament, which you are registered in, but the March tournament will probably be mostly newer teams.  Possibly in April we will have a sixteen team tournament.
Yes, this is basically a random topic.  Also the topic "AoPS March 2008 Tournament" should be February now.  I'll try to get avatars then we can start the games!
:lol:
P.S. I am still laughing at the "wires" part of verizon.   :rotfl:

thank you man

P.S. i no, its a great username

ill join your team lawrence...but my IP is banned until march 29  :(

hey dude I'll make a team

if anybody really wants to join it for some bizarre reason they can

it'll be like iTest

I will also join anyone's team. >.> if i make my own probably no one will want to join.
EDIT: Preferrably I should have met them in AoPS.

FYI, this tournament was cancelled.
All team tournaments are cancelled for now.
If you want, I can reserve you a spot whenever someone can't play a game.
:)

I want to join a team, if there's one open...

Yeah, I didn't read any posts besides the first, so if this started or

something, sorry.


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Given $u(r,\theta)=\frac{ (a^{2}-r^{2})}{2\pi}\int_{0}^{2\pi}\frac{h(\phi)}{(a^{2}+r^{2}-2ar\cos({\theta-\phi})}d\phi$.  How do we solve for u, with $h(\phi)=1+3\cos(\phi)$.   This seems like a relatively elementary integral, but I can't think of an applicable technique to solve it (the solution is $1+\frac{a}{r}\cos(\theta)$)

Errr... this should actually be in computations in calculus, if a moderator could put it there, please.  Sorry!

I would look for a harmonic function with given boundary values. The function $u(z)=1+\frac{3}{a}\mathrm{Re}(z)$ seems to work. (Here $z$ is complex, $z=re^{i\theta}$.) Hence $u(r,\theta)=1+\frac{3 r}{a}\cos\theta$.
In general you can expand $h(\phi)=\sum_{n\in\mathbb Z}c_{n}e^{i n\phi}$ and get $u(r,\theta)=\sum_{n\in\mathbb Z}c_{n}(r/a)^{n}e^{i n\phi}$. The sequence $((r/a)^{n})_{n\in\mathbb{Z}}$ is the Fourier multiplier of the Poisson kernel.


MATHCOUNTS
complex numbers
imaginary numbers

there is an algebraic way to proove that 2=1. can you find it.

i got it off a website.

impossible.  you would have to make a mathematical mistake. :wink:

1337 is right...iv seen that proof numerous times..
at some point in the proof you divide both sides of the equation by zero therefore it's an invalid proof...

[quote="13375P34K43V312"]It cannot be done without division by zero.[/quote]
How about this:

$ \infty+1=\infty+2$

$ 1=2$

But seriously, you can't algebraically prove 1=2.

i_like_pie, you must realize that $ \infty\neq\infty$

Yes... he was just posting a classic bogus proof that didn't make use of division by zero. Why was this thread started anyway? It should be in G&FF.

yeah you can also make an imaginary number 'mistake'

basically \[ \sqrt{36}=\sqrt{36}\implies \sqrt{6}\sqrt{6}=\sqrt{-6}\sqrt{-6}\implies 6=-6 \implies 12=0 \implies 1=2\]

[quote="indianamath"]how does 12=0 imply that 1=2?[/quote]
This is so silly...
$ 12=0$ divide by 12
$ 1=0$ add one
$ 2=1$

This is a really simple thing we learned in our Algebra textbooks.

Let us say a = b
a squared = ab
a squared plus a squared = a Squared + ab
2a squared = a squared plus ab
2a squared -2ab = a squared plus ab - 2ab
2a squared -2ab = a squared - ab
2(a squared minus ab) = 1(a squared minus ab)

Therefore, 2 = 1

 :harhar:

why are we discussing bogus proofs on a statement which it is impossible to prove? :maybe:
there's no point...it can't be done and that's final!!!

[quote="ordinary boy"]guys guys, if you want to think 2=1 go ahead.  No ones stopping you.[/quote]

I would just like to say
ordinary boy has made the 64000th post in the Mathcounts forum.

Is that sarcasm?  because I dont see where you get that number

it's definitely possible...on the forum index look at the mathcounts topic, there is the number of posts and topics in that forum...right now its like 64009 or something, so you getting the 64000th post is definitely possible..

sorry this is problaby spam but I dont see how many posts there are.

I only see the how many topics htere are.

there's supposed to be a column right next to that one with the number of posts..

a=b
Multiply both sides by a : a^2=ab
Subtract b^2 from both sides: a^2-b^2=ab-b^2
Factor each side: (a+b)(a-b)=b(a-b)
Divide each side by (a-b): a+b=b
Since a=b: 2b=b
Divide each side by b: 2=1

[quote="13375P34K43V312"]It cannot be done without division by zero.[/quote]
it is true. you do need to divide by zero because since a and b are equal, a-b=0, therefore, it is division by zero.


In Rack 'n' Roll, one must hang inner tubes on a structure known as the rack.  The rack consists of three tiers, and each tier consists of 8 pegs placed in a circle.  Each alliance has access to 21 inner tubes that they can score with.

The scoring is as follows: a row (which can be either horizontal or vertical) of $n$ tubes that belong to the same alliance is worth $2^{n}$ points for all $n$ such that $2\ge n\ge 8$.  A tube that is not part of a row is worth 2 points.

If the blue alliance places all 21 of its tubes randomly on the rack such that no peg contains more than 1 tube for each peg, compute the probability that the blue alliance has a score of 596 points.

[hide]

There are 3 pegs without a tube; if they are all i a different row, the max points you can get is $3(2^{7})$ for the rows and $7(2^{3})$ for the columns. $3(2^{7})+7(2^{3})=440$, so this wouldn't be enough.

If one row has all the tubes, the max points you can get is $2^{8}+2^{7}+2^{6}=448$ for the rows, and $5(2^{3})+2(2^{2})=48$ and $448+48=496$, still not enough.

Now if all the missing tubes are in one row, The other two rows would get a total of $2(2^{8})=512$ points. There would be 5 colums of 3, and $5(2^{3})=40$ The max points to get in the row with the 3 missing tubes would be if the missing tubes were all together, $2^{5}=32$. If this row is the top or bottom row, we would attain three additional columns of 2 above or below the missing tubes. $3(2^{2})=12$. $512+40+32+12=596$

This score was only obtained by maximizing the points as much as possible; there is no other way it would work. Now to find the probability it will be in such a position.

There are $\binom{24}{3}$ ways for the missing tubes to be positioned. We see that there are 16 ways to place the three missing tubes to obtain the score of 596. (8 ways to place them within one row, and both the top or bottom row work). $\frac{16}{\binom{24}{3}}=\frac{16}{2024}=\boxed{\frac{2}{253}}$

[/hide]


a,b,c, x,y,z are positive reals such that: $cy+bz=a$, $az+cx=b$, $bx+ay=c$. Find the minimum of:
\[ f(x,y,z)=\frac{x^2}{1+x}+\frac{y^2}{1+y}+\frac{z^2}{1+z} \]


[quote="shadysaysurspammed"][hide]$p+q: 2$?[/hide][/quote]

That's what I thought but no  :(

Ratio is p:1 and q:1.....

So then total volume mixed will be px+x+qy+y..not px+x+qx+x...that's where i went wrong.

[hide]
Assuming both container have the same amount of water.
the amount of alcohol can be expressed as p/(p+1) and water is 1/(p+1)
similarly q/(q+1) and 1/(q+1) for the other container
q/(q+1)+p/(p+1):1/(p+1)+1/(q+1)
simplfy that and u'll get 2qp+q+p : P+q+2
is it right? :roll: [/hide]

[quote="236factorial"]In one container, the ratio of alcohol to water was $p: 1$. In another container the ratio was $q: 1$. If these two containers were mixed, what would be the ratio of alcohol to water?[/quote]

This is  sloppy math problem, and almost meaningless unless one  gives the size (Or ratio  of sizes) of each container.

For example if the first container is HUGE  (compared to the second one) the answer would be close to   $p: 1$  .. Thus depending on the size, the answer could be any value between $p: 1$ and $q: 1$  If the tow conainers are of equal size the answer would be something like  $(2pq+p+q): (p+q+2)$  ..

Hope this helps.

Hope this helps.

Sorry, I meant that they were equal in size  :blush:


function
algebra
polynomial
algebra unsolved

Hint: Use following lemma (romanial selection test 2003)

If P(x) from Z[x] is irreducible and f(0) is not square of an integer then p(x^2) is irreducible.

This lemma is interesting itself.

Namdung

I have near problem :
   let f  \in Z[x] .Known that for all g  \in Z[x] , deg=2 then f(g(x)) reducible over Z[x] .Prove that f(x) reducible over Z[x]

How are they proven, Namdung and A1lqdSchool?

You want to have solution of Problem Japan 99 &  my near problem? :)  :D

Here is my approach by using complex number
suppose f(x)=g(x)h(x)->1=g(mi)h(mi) for m=+/-1,..+/-n.And we have g(mi)=h(mi)=+/-1 for m not equal =/-1.-->2\leq |g(i)-h(i)|=|(-1+2^2)..(-1+n^2)k(x)| for any k(x) has degree <=1->k(x)=0 for all x.
Hence g(x)=h(x),But f(0) is not square.Contradiction...
  EDIT:Sorry i don't read A1lqdschool's notation. :D

[quote=Japan MO 1999 Problem 4]Prove that the polynomial $f(x)=(x^2+1)(x^2+2^2)\cdots (x^2+n^2)+1$ cannot be expressed as a product of two polynomials with integer coefficients with degree greater than $1$.[/quote]

[hide = Japan MO 1999 Problem 4 Solution]
[color=#134F5C]Lemma 1 : [/color] Let $f \in \mathbb{Z}[X]$ be a monic irreducible polynomial over the integers such that $\lvert f(0) \rvert$ is not a perfect square, then $f(x^2)$ is also an irreducible polynomial.

[color=#45818E]Proof : [/color] Let us say for the sake of contradiction that $g(x) \mid F(x^2)$ and $f(x^2)$ in short is reducible and $g(x)$ is irreducible. We see that $\mathrm{deg} g(x) < \mathrm{deg} f(x^2)$. Now, $g(x) \mid f(x^2)$ and $g(-x) \mid f(x^2)$. If $g(x) = g(-x)$, then we get a contradiction since then $g(x) \mid f(x)$ but $f(x)$ is irreducible (unless all terms of $g(x)$ are multiplied by a constant $c$ which is also not possible since $f$ is monic). So, $g(x) \neq g(-x)$ and moreover both of these terms are irreducible, hence are also relatively prime. Now, we see that $\mathrm{deg} g(x) < \mathrm{deg}f(x^2)$ or else $2\mathrm{deg} f(x) = \mathrm{deg} f(x^2) \geq \mathrm{deg} g(x)g(-x) > 2 \mathrm{deg} f(x)$. Also, if $g(x)g(-x) = t(x^2)$ for a polynomial $t$, if $\mathrm{deg} g < \mathrm{deg} f$, then $t(x^2) \mid f(x^2)$ is equivalent to $t(y) \mid f(y)$, a contradiction since $f$ is irreducible. Therefore, $\mathrm{deg} g = \mathrm{deg} f$. So $g(x)g(-x) = f(x^2)$. Now putting $\boxed{x = 0}$, we get that $f(0) = g(0)^2$, a contradiction to the fact that $\lvert f(0) \rvert$ is not a perfect square. Therefore, our assumption that $f(x^2)$ is reducible is incorrect and indeed $f(x^2)$ is irreducible as desired.
-----------------------------------------------------------------------
We use the lemma to crush this problem. We will show that $f_1(x) = (x+1)(x+4)(x+9)(x+n^2) + 1$ is irreducible. So $f(x) = f_1(x^2)$ and we need to continue like previous problem. Let us say for the sake of contradiction that $f_1(x) = g(x)h(x)$. We see that if $k(x) = g(x) - h(x)$, for the values $x = -1^2, -2^2, -3^2, -4^2 \dots -n^2$, the value of $k(x)$ would be $0$. We see that because of this, $\mathrm{deg} k(x) \geq n$. But $\mathrm{deg} k(x) \leq \max \{ \mathrm{deg}$ $ g(x), \mathrm{deg}$ $ h(x) \} \leq n$, and so it is only possible for both inequalities to hold when either $f$ or $g$ is constant, implying that $f_1 (x)$ is irreducible over integers. Now, $f_1(x)$ is monic and irreducible over integers and moreover $f(0) = (n!)^2 + 1$ which is never a perfect square for all naturals $n$, hence we have using lemma proved that $f_1(x^2) = f(x)$ is irreducible as desired.

[quote=HamstPan38825]Suppose that the polynomial factors as $fg$. Then $f(x)g(x) = 1$ for $2n$ values of $x$ of the form $\pm ki$ for $1 \leq k \leq n$, and in particular $f(x)-g(x)= 0$ for all these $x$.

This implies that $f=g$. Hence $(f(X)-1)(f(X)+1) = \prod_{k=1}^n (X^2+k^2)$. By setting $x=0$ and noting that two perfect square never differ by $2$, we get a contradiction.[/quote]

I'd like to ask the OP how does $f(x)g(x) = 1$ for $x = -ki$(with $1 \le k \le n$ gives that $f(x)-g(x)$ vanishes at $2n$ points? Say $f(ki) = a+bi$ and $g(ki) = c+di$ for some $a,b,c,d \in \mathbb{Z}$, then there is a possibility that $f(ki) = \pm i$ and $g(x) = \mp i$. For such values, I don't think $f(x)-g(x)$ vanishes at all...


Suppose A is a complex number and n is a positive integer such that An = 1 and (A+1)n = 1. Prove that n is divisible by 6 and that A3 = 1.


Find all positive integers pairs $ (a,b)$ such that $ \frac{\sqrt{2}\plus{}\sqrt{a}}{\sqrt{3}\plus{}\sqrt{b}} \in\mathbb{Q}.$

[quote="rachid"]Find all positive integers pairs $ (a,b)$ such that $ \frac {\sqrt {2} \plus{} \sqrt {a}}{\sqrt {3} \plus{} \sqrt {b}} \in\mathbb{Q}.$[/quote]

$ \frac {\sqrt {2} \plus{} \sqrt {a}}{\sqrt {3} \plus{} \sqrt {b}}\equal{}\frac {\sqrt {2} \plus{} \sqrt {a}}{\sqrt {3} \plus{} \sqrt {b}}*\frac{\sqrt {3} \minus{} \sqrt {b}}{\sqrt {3} \minus{} \sqrt {b}}\equal{}\frac{\sqrt{6}\plus{}\sqrt{3a}\minus{}\sqrt{2b}\minus{}\sqrt{ab}}{3\minus{}b}\in\mathbb{Q}$------->

$ \sqrt{6}\equal{}\sqrt{2b}$-----> $ b\equal{}3$----> $ \sqrt{6}\plus{}\sqrt{3a}\minus{}\sqrt{2b}\minus{}\sqrt{ab}\equal{}\sqrt{6}\plus{}\sqrt{3a}\minus{}\sqrt{2*3}\minus{}\sqrt{3a}\equal{}0$------> $ b\equal{}3,a\equal{}n\in\mathbb{Z}$

$ \sqrt{6}\equal{}\sqrt{ab}$------>$ ab\equal{}6$----> $ \sqrt{6}\plus{}\sqrt{3a}\minus{}\sqrt{2b}\minus{}\sqrt{ab}\equal{}\sqrt{3a}\minus{}\sqrt{2b}\in\mathbb{Q}$---->$ a\equal{}3;b\equal{}2$ or $ a\equal{}2;b\equal{}3$

$ \sqrt{6}\equal{}\sqrt{2b}\plus{}\sqrt{ab}$---> $ 6\equal{}2b\plus{}2b\sqrt{2a}\plus{}ab$------->$ 2\sqrt{2a}\equal{}\frac{6}{b}\minus{}2\minus{}a\in\mathbb{Z}$----->$ b\equal{}1;2;3;6$ ------> $ 2\sqrt{2a}\equal{}4\minus{}a;1\minus{}a;\minus{}a;\minus{}1\minus{}a$-----> $ 2\sqrt{2a}\equal{}4\minus{}a$----->$ \phi$

$ b\equal{}3,a\equal{}n\in\mathbb{Z}$ and $ b\equal{}2;a\equal{}3$

Hmmm. If $ b\equal{}3$ then the first manipulation you perform is meaningless. As you can easily check directly, taking $ b\equal{}3$ and an arbitrary $ a \in \mathbb{Z}$ does [b]not [/b] yield a rational number.

In addition, it's not clear why having a rational denominator forces $ \sqrt{6} \equal{} \sqrt{2b}$ or $ \sqrt{6} \equal{} \sqrt{ab}$ etc. It is true that $ b\equal{}2$, $ a\equal{}3$ is a solution, but you didn't quite prove it's the only one.

Yes, $ b = 3, a = 2$ will definitely make the first step impossible.

Once you rationalize the denominator, we just need to mention that $ a, b$ are positive integers.
Thus, $ \sqrt {3a} > 0$.

Now, suppose that $ \sqrt {6}\neq \sqrt {3a}$ and $ \sqrt {6}\neq\sqrt {ab}$
Then after simplifying the terms with $ a$ and $ b$, we will be left with $ \frac {\sqrt {6} + ...}{3 - b}$
This expression will never be $ \in%Error. "bbmath" is a bad command.
{Q}$

Therefore, $ \sqrt{6} = \sqrt {3a}$ or $ \sqrt {6} = \sqrt {ab}$ and after calculation, we get:

$ a = 3, b = 2$ as our only solution.


just a typing mistake.
in french, it's "nous sommes les meilleures" for girls, and "nous sommes les meilleurs" for men. french grammar is something !

We are the best - English

:D

These are from Babelfish:

Wir sind das beste - German
Мы самые лучшие - Russian (this might not show up right)
Vi er det beste - Norwegian

Wo(3) Men(2) Zui(4) Hao(3) - Chinese

Nosotros somos los ms grandes - Spanish (I think)... Only on Spanish 2...

I think in Spanish it should be
Nosotros somos mejores.

[quote="darij grinberg"][quote="nr1337"]Wir sind das beste - German[/quote]

That should be "Wir sind die besten".

But why on the earth do you need this, Iris? Is this prepared to be the slogan of the next Albanian IMO team?

  Darij[/quote]

Well, you can't blame me - blame altavista!

Humlog sabse accha hai. (Hindi- wish I knew how to get a Hindi font)

[quote="white_horse_king88"]Wo(3) Men(2) Zui(4) Hao(3) - Chinese
[/quote]
我們最好
wo3 men5 zui4 hao3

hm...i can't read simplified, i can only read traditional

[quote="white_horse_king88"]Nosotros somos los ms grandes - Spanish (I think)... Only on Spanish 2...[/quote]

mas grandes? Isn't that the biggest?

Nosotros somos los mejores. That would probably be my guess. No wonder I'm in such a low spanish class...

hey i think in spanish is nosotros somos los mejores just check in a spanish dictionary :lol:

Yeah everyone is right. Like I said, I'm not the greatest at Spanish... I was under the impression that grande meant "great" also...

[quote="darij grinberg"]

But why on the earth do you need this, Iris? Is this prepared to be the slogan of the next Albanian IMO team?

  Darij[/quote]

Oh no theres not going to be our slogan. Wish it could be but we are far from it.
Its just an idea of mine for the class table which is something where every class is free to put whatever it wants: pictures, poems, curiosities, quotes, comics whatever nice.

Iris

[quote="white_horse_king88"]Wow. Oldish chinese. Simplified Chinese is easier to understand for me.[/quote]
[size=150]
我们最好[/size] Better?

[size=200]हम सबसे अच्छे हैं[/size]

Hindi above  Approx.   Hum sabse achche hai

 Urdu  would be  " Hum sabse badhiya hai "

Sanskrit:
[size=150]वयम् सर्वोत्वम्[/size]

[quote=Iris Aliaj]I'd like to have the phrase "we are the best" in as many languages as possible.
Come on people!
The list till now is short. Hope it will get longer with your help.

Ne jemi me te miret-  Albanian
Noi siamo i migliori-    Italian
Nous sommes les meigleures.-  French

I guess i have some spelling mistake.[/quote]

first, how was this post made in 2004 when you joined in 2009?
second, how does your username have a space
[quote=TTZ]lemme casually post on a thing that is 19 years old[/quote]

how did you even find this thread

found the oldest unlocked thread in ff but it has bad words in it so i won't post or link but it was from 2004
(did ff even exist back then? cuz i dont think aops was aops yet)

우리가짱이야 my korean's bad despite my parents efforts to teach me, but I think this should at least be pretty close.

[quote=MineCuber][quote=Iris Aliaj]I'd like to have the phrase "we are the best" in as many languages as possible.
Come on people!
The list till now is short. Hope it will get longer with your help.

Ne jemi me te miret-  Albanian
Noi siamo i migliori-    Italian
Nous sommes les meigleures.-  French

I guess i have some spelling mistake.[/quote]

first, how was this post made in 2004 when you joined in 2009?[/quote]
umm he joined in 2004, last visited in 2009

In my native language:
আমরা সবচে ভালো
(in Bengali)


AMC
AIME
AMC 12
AMC 12 A
AMC 12 B
USA(J)MO
USAMO

I took AMC12A at my school but AMC12B at another school because my school didn't host the B test.
Base on the answers I have, I got a higher score for the AMC12B test. Thus I hope my AMC12B score will be used to calculate my USAMO index.
However, both scores are above the AIME-qualification line. Therefore both school will ask me to write AIME I at their place.
I prefer to write it at my school for convenience. [i]If I take AIME at my school, will my score of AMCB test, which was taken at another school, be used to calculate my USAMO index?[/i]I am quite confused by this school dilemma. DO the MAA people count everything by individual no matter where one might take the AMC tests? PLEASE HELP

I believe your higher score will be used regardless.

Go ahead and take the AIME at your own school. And do the courtesy of informing the contest manager at the other school of your plans.

In principle, both of your scores count, and whichever one was more advantageous to you should determine your USAMO qualification index.

In practice, this is a clerical nightmare for the AMC office, since you are identified only by your name - the AMC does not issue unique identification codes for each participant. That presents a problem with joining the disconnected pieces of a record. It's probably especially different for them if you have an extremely common name like "Wang" or "Kim" or "Nguyen", and it would be especially difficult for them if there are any spelling variations in the name between the two sources.

There was one instance last year of a student taking an AMC B exam at my location (a university campus) and earning a higher score than he had on the AMC A at his own high school, but he qualified for the AIME in both places. He took the AIME at his own high school. When the USAMO qualifiers list came out, since I knew his AIME score, I knew that he had been omitted. I immediately contacted the AMC office, and they fixed their listing and added him to the USAMO qualifiers. The AMC office does fix errors of this type.

On the other hand, if it becomes necessary to contact the AMC office about your case, it would probably be more effective to have the contest manager of one of the schools do the contacting.

Thank you very much.
It really doesn't matter which score they are going to use. AIME+AMC only for USAMO qualification right? If they have some sorts of awards to the top index people. HAHA then I will seriously think about this issue.
GO AHEAD!

[quote]If I take AIME at my school, will my score of AMCB test, which was taken at another school, be used to calculate my USAMO index?[/quote]

Yes.  However, you must be careful to use the same name each time you take the contest.  Do not use nicknames, shorten your name or add or delete initials.  Also, include the same home street address each time you take the contest and fill out that information on the answer forms carefully, completely and identically each time.  These help us to isolate your scores and report you correctly.

In addition, when your teacher sends  the AIME answer from your school, include a note of explanation with the locations of your A-date testing and your B-date testing, along with your name, address, and A-date and B-date scores.

Steve Dunbar
MAA Director, American Mathematics Competitions


induction
geometry
3D geometry
algebra
factorization
sum of cubes

Solve the equation of differences  $y_{n+2}-2y_{n+1}+y_n=n^3$  when  $y_0=0$ and $y_1=0$

[quote="dengmi"][hide]$y_n$=$\displaystyle\sum_{0}^{n}(\frac{(n-2)(n-1)}{2})^2$ where n is a natural number.  When n=0, $y_n$ is defined as 0.  The proof of this is inductive.  First, establish the base case triple.  The first non-trivial base case would be n=3,4,5 where $y_3=1$, $y_4=10$, $y_5=46$.  A check from the equation shows that $3^3=46(y_5)-2*10(y_4)+1(y_3)$.
Then, assume it true for n, n+1, n+2 and we'll prove it for n+1, n+2, n+3.  Using the knowns from the n and applying it into the equation that is known, one gets (after simplification), $(n+1)^3+y_{n+2}+(\frac{n(n+1)}{2})^2=y_{n+3}$.  Rearranging the terms with n, one sees it's $(\frac{(n+1)(n+2)}{2})^2$.  $(\frac{(n+1)(n+2)}{2})^2+y_{n+2}=y_{n+3}$, which does satisfy the summation.[/hide]
Does this help?[/quote]
I am curious about how you obtained the expression. Is there a more algebraic way then guessing and proving by induction?

first you can notice this proprieties by samples and proves...

or else you can also demostrate it by $y_n=2y_{n-1}-y_{n-2}+{(n-2)}^3$ and recursivity : by transformating and passing from $n\to n-1$ and substituing in the formula (call this point (i.)); remembering that $S_1=\frac{n(n+1)}{2}=\sum_{i=1}^n i$ and $S_3=\sum_{i=1}^n i^3=S_1^2$, you have done (simply use the formula to find a close formula to sum of cubes in point (i.)


trigonometry
algebra
difference of squares
special factorizations

Prove

(tan x + sec x - 1)/(tan x - sec x + 1) = tan x + sec x

It's an interesting problem...

Hint in the spoiler:

[hide]

Multiply the numerator and the demonimator by a certain expression...

[/hide]

Another hint:

[hide]

Get a difference of squares

[/hide]

Spoiler:

[hide]equivalent to showing

tanx + secx - 1 = (tanx - secx  + 1)(tanx +secx)

After expanding, we get on the left hand side

tan^2x - sec^2x + tanx + secx

=tanx + secx -1,

as desired.

[/hide]

This is my long, scary, and inelegant proof:

[hide]

(continued from the beginning)

(tan x + sec x - 1)(tan x + sec x - 1)/(tan^2 x - (sec x - 1)^2)

(tan^2 x + sec x tan x - tan x + sec x tan x + sec^2 x - sec x - tan x - sec x + 1)/(tan^2x - sec^2 x + 2sec x - 1)

(tan^2 x + 2tan x(sec x - 1) + (sec x - 1)^2)/(2 sec x - 2)

(sec x - 1)(2sec x + 2 tanx)/(2)(sec x - 1)

sec x + tan x

[/hide]


inequalities AMC USA(J)MO USAMO geometry trigonometry

[hide="A rewrite"]
Rewrite as

$\frac{ab}{ \sqrt{(b^2 + c^2)(c^2 + a^2)}} + \frac{bc}{ \sqrt{(a^2 + b^2)(c^2 + a^2) }} + \frac{ca}{ \sqrt{(a^2 + b^2)(b^2 + c^2) }} \ge 1$

Substitute $u = \frac{a}{ \sqrt{b^2 + c^2} }, v = \frac{b}{ \sqrt{a^2 + c^2} }, w = \frac{c}{ \sqrt{a^2 + b^2} }$.  The problem becomes

$uv + vw + wu \ge 1$

And it's just begging for a geometric interpretation... but I can't think right now.  Post-USAMO brain freeze.   :(
[/hide]

Let $u=a^2, v =b^2, w = c^2$, the given inequality is equivalent to
$\sqrt{(u+v)(v+w)(w+u)}\leq \sqrt{uv(u+v)}+\sqrt{vw(v+w)}+\sqrt{wu(w+u)}$.
It suffices to show that
$uv(u+v)+vw(v+w)+wu(w+u)+2\sqrt{uv^2w(u+v)(v+w)}+2\sqrt{vu^2w(v+w)(w+u)}+2\sqrt{wu^2v(w+u)(u+v)}\geq (u+v)(v+w)(w+u)$
In other words, $v\sqrt{uw(u+v)(v+w)}+w\sqrt{uv(u+w)(v+w)}+u\sqrt{vw(u+v)(u+w)}\geq uvw$
Apply AM-GM inequality to LHS:
$LHS \geq 3 (u^2 v^2 w^2 (u+v)(v+w)(w+u))^{2/3}\geq 3uvw\geq uvw$.
Q.E.D.

[quote="kar"]Prove that if $a \geq 0, b \geq 0, c \geq 0$ then

$\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \leq ab\sqrt{a^2+b^2} + bc\sqrt{b^2+c^2} + ca\sqrt{c^2+a^2}$[/quote]

i believe that this is homogenous, or at least in the sense where you can scale the numbers by a positive constant, so WLOG $a^2+b^2+c^2=1$, also divide out by the LHS to get

$\sum\frac{a}{\sqrt{1-a^2}}*\frac{b}{\sqrt{1-b^2}}>1$ or

$\sum\frac{1}{\sqrt{\frac{1}{a^2}-1}}*\frac{1}{\sqrt{\frac{1}{b^2}-1}}>1$

i don't know what to do now...i know little about inequalities, but this looks like it should be easy...lol

[i]luimichael[/i] - very nice!
(My proof is about the same, but I didn't replaced anything, just squaring and AM-GM)

[i]t0rajir0u[/i], can you continue your proof? (it indeed come from geometry.)

[quote]Nice inequality. My congratulations, Karina![/quote]
Thank you, [i]Michael[/i]!  :lol:
By the way, it's from an inequality from you. Do you know what's the inequality?

[quote="kar"]
By the way, it's from an inequality from you. Do you know what's the inequality?[/quote]
Maybe you mean to this inequality:
$x,y,z > 0.$ Prove that :

$(x+y)(y+z)(z+x)+4xyz \le$
$\leq \sqrt{(x+y)(y+z)(z+x)}\left(\sqrt{xy(x+y)}+\sqrt{yz(y+z)}+\sqrt{xz(x+z)}\right)$ $?$
But it is not mine.

Hi! Karina,
Can you explain about how the problem is related to Geometry?
It may help us to understand the whole picture.  :)

Hello everyone,
All right, the geometry problem is the following.
[hide]Let $ABCDA'B'C'D'$ be a rectangular prism. Prove that
$\angle C'AC + \angle C'AB' + \angle C'AD' > 90^\circ$.
[/hide]
[i]Michael[/i], I did not mean it's a problem you derived (not a problem of you), but it's a beautifull problem I know from you. I'm not sure that the inequality you writted is related to the inequality I have posted and I cannot check it now, but I'll try.

[i]luimichael[/i] or anybody else, I think you can connect it now?
[hide="A hint"] Take $AB=a, AA'=b, AD=c$. The next step is to try to use trigonometry to get the inequality (or maybe a geometric way is possible. Then it would be very-very nice).
[/hide]

[hide="Aaah..."]
I will have to rename a few points, but the geometric interpretation is mostly clear to me now.

Consider parallelpiped $OABCO'A'B'C$ with $OA = a, OB = b, OC = c, AA' = BB' = CC' = OO' = \sqrt{a^2 + b^2 + c^2}$, etc.

Consider $\triangle AOA'$.  $OA' = \sqrt{b^2 + c^2}$ and $OA = a$, so $\tan \alpha = \tan AA'O = \frac{a}{ \sqrt{b^2 + c^2} }$.  Similarly for the other two.

The problem we are given then becomes, given $\alpha = \angle AA'O, \beta = \angle BB'O, \gamma = \angle CC'O$,

$\tan \alpha \tan \beta + \tan \beta \tan \gamma + \tan \gamma \tan \alpha \ge 1$

I am still working on the solution.  :P
[/hide]

Here's a similar FUN problem:

$3\sqrt{(a^2+b^2)(b^2+c^2)(c^2+a^2)} \geq 2ab\sqrt{a^2+b^2} + 2bc\sqrt{b^2+c^2} + 2ca\sqrt{c^2+a^2}$

Каrinа, I have recollected. :) Thank you!


number theory
relatively prime
number theory proposed

It is easy to prove that if $b>3$ is an odd prime then $2^b + b^2 = 3m_b$ where $m_b$ is a positive integer.

Prove that, for each prime $b>3$ then there exist an infinite number of primes $c>b$ such that $\gcd(m_b,m_c) > 1.$

Pierre.

Note that $m_{b}\neq 0(\mod 3)$
 Assume that $p>3$ is a prime divisor of $2^{b}+b^2$.
Note that $-2$ is square residue $\mod p$ so we can choose infinite many  prime $c$ s.t
 $c\equiv 1(\mod p-1),c^2\equiv -2(\mod p)$.
Those $c$s have required property.

[quote="nalpaction"]Note that $m_{b}\neq 0(\mod 3)$[/quote]

I don't think it is true. Just choose $b=11$... ;)

Pierre.

[quote="nalpaction"]:blush: so for $b\equiv 11(\mod 18) \iff 3|m_{b}$,a minor fixing could be easily done.[/quote]

And for $b = 13 \mod [18]$ too  :P

Pierre.

[quote="Rafal"]I assume you use the heavy-weight Dirichlet's theorem (plus the Chinese residue theorem) ?[/quote]

I think you are asking to nalpaction?
Anyway, my solution only used Dirichlet's theorem and Fermat's little one, but the ideas seem to be equivalent to nalpaction's ones.

Pierre.

Nice problem, Pierre. As usual... I tried to avoid the use of Dirichlet theorem, but it seems difficult to find elementary solution. Here is what I thought. First, let's prove we can find a prime factor $q$ of $2^b+b^2$ different from 3 and such that $b$ does not divide $q-1$.  First of all, if any prime factor $q$ of $ (2^b+b^2)/3$ is $1$ mod $b$, then the whole number is $1$ mod $b$, thus $2^b$ is congruent to $3$ mod $b$, which contradicts Fermat's theorem. Thus there is a prime factor $q$ of $(2^b+b^2)/3$ which is not congruent to $1$ mod $b$. But hopefully, $ (2^b+b^2)/3$ is never a multiple of $9$, thus our $q$ is never equal to 3. For this $q$ we have obviously that $b$ is relatively prime with $ q(q-1)$. Next, from Dirichlet theorem we can find infinitely many primes $c$ congruent to$b$ mod $ q(q-1)$. But for these numbers we will have $ (2^c+c^2)/3$ is a multiple of $q$. I hope I didn't do stupid mistakes.

That solution is good (I don't believe there is any other)

I think it's impossible to avoid the Dirichlet's theorem, however in this particular example...I'm not sure.

Yes Harazi, that's exactly how I did it  :)

Pierre.


geometry
inequalities
trigonometry
inequalities unsolved

H is the orthocenter of an ancute triangle ABC, note $ BC\equal{}a, CA\equal{}b, AB\equal{}c$.
Prove that $ \frac{HA^2\plus{}HB^2\plus{}HC^2}{a^2\plus{}b^2\plus{}c^2} \leq (cotA.cotB)^2\plus{}(cotB.cotC)^2\plus{}(cotC.cotA)^2$

Must we compare the expression $ HA^2\plus{}HB^2\plus{}HC^2$ with what?

Help me to solve it, please!
Ricardo de Leite!

[quote="ricardokaka"]H is the orthocenter of an ancute triangle ABC, note $ BC = a, CA = b, AB = c$.
Prove that $ \frac {HA^2 + HB^2 + HC^2}{a^2 + b^2 + c^2} \leq (cotA.cotB)^2 + (cotB.cotC)^2 + (cotC.cotA)^2$[/quote]

İ have a solution but it is too ugly.Just notice that if we take $ D$ the midpoint of $ BC$
then $ HA=2OD=2RcosA$ and $ a=2RsinA$

$ \Rightarrow \frac {cos^2A + cos^2B + cos^2C}{sin^2A + sin^2B + sin^2C} \leq (cotA.cotB)^2 + (cotB.cotC)^2 + (cotC.cotA)^2$

$ cosA=\frac{b^2+c^2-a^2}{2bc}$ and $ sinA=\frac{2S}{bc}$ so we must prove that

$ \frac{\sum{(\frac{a^2+b^2-c^2}{2ab})^2}}{\sum{(\frac{2S}{bc})^2}} \leq \sum{(\frac{\frac{a^2+c^2-b^2}{2ac}\cdot \frac{b^2+c^2-a^2}{2bc}}{\frac{2S}{bc}\cdot\frac{2S}{ac} })^2}$

$ \frac{\sum{c^2(a^2+b^2-c^2)^2}}{16S^2\sum{c^2}} \leq \sum{(\frac{(a^2+c^2-b^2)\cdot(b^2+c^2-a^2)}{16S^2})^2}$

$ 16S^2\cdot\sum{c^2(a^2+b^2-c^2)^2} \leq \sum{c^2}\cdot \sum{((a^2+c^2-b^2)\cdot(b^2+c^2-a^2))^2}$

I opened the brackets and got solution.İ have no time now.İ will write later if anyone wants...

Oh, thanks! I don't think it's easy like that! I understood all!

ricardokaka, I think your inequality is true for all $ \Delta ABC.$ :wink:

@cefer: Please continue! I think that you'll use the Heron's formule. I tried but it's hard for me! I'm not good at teh ineqs!
@arqady: That's right? Can you post your solution? I don't believe that my inequality is true for all triangles!

[quote="cefer"][quote="ricardokaka"]H is the orthocenter of an ancute triangle ABC, note $ BC = a, CA = b, AB = c$.
Prove that $ \frac {HA^2 + HB^2 + HC^2}{a^2 + b^2 + c^2} \leq (cotA.cotB)^2 + (cotB.cotC)^2 + (cotC.cotA)^2$[/quote]

İ have a solution but it is too ugly.Just notice that if we take $ D$ the midpoint of $ BC$
then $ HA = 2OD = 2RcosA$ and $ a = 2RsinA$

$ \Rightarrow \frac {cos^2A + cos^2B + cos^2C}{sin^2A + sin^2B + sin^2C} \leq (cotA.cotB)^2 + (cotB.cotC)^2 + (cotC.cotA)^2$

$ cosA = \frac {b^2 + c^2 - a^2}{2bc}$ and $ sinA = \frac {2S}{bc}$ so we must prove that

$ \frac {\sum{(\frac {a^2 + b^2 - c^2}{2ab})^2}}{\sum{(\frac {2S}{bc})^2}} \leq \sum{(\frac {\frac {a^2 + c^2 - b^2}{2ac}\cdot \frac {b^2 + c^2 - a^2}{2bc}}{\frac {2S}{bc}\cdot\frac {2S}{ac} })^2}$

$ \frac {\sum{c^2(a^2 + b^2 - c^2)^2}}{16S^2\sum{c^2}} \leq \sum{(\frac {(a^2 + c^2 - b^2)\cdot(b^2 + c^2 - a^2)}{16S^2})^2}$

$ 16S^2\cdot\sum{c^2(a^2 + b^2 - c^2)^2} \leq \sum{c^2}\cdot \sum{((a^2 + c^2 - b^2)\cdot(b^2 + c^2 - a^2))^2}$

I opened the brackets and got solution.İ have no time now.İ will write later if anyone wants...[/quote]

Replacing $ 16S^2$ with $ 2a^2b^2 + b^2c^2 + c^2a^2 - a^4 - b^4 - c^4$ and after opening the brackets I got
$ LHS = - \sum{a^{10}} + 3\sum{(a^8b^2 + b^8a^2)} - 2\sum{(a^{6}b^4 + b^6a^4)} - 8a^2b^2c^2\sum{a^{4}} + 10a^2b^2c^2\sum{a^{4}}$
$ RHS = 3\sum{a^{10}} - \sum{(a^8b^2 + b^8a^2)} - 2\sum{(a^{6}b^4 + b^6a^4)} - 4a^2b^2c^2\sum{a^{4}} + 10a^2b^2c^2\sum{a^{4}}$

$ \Rightarrow RHS - LHS = 4(\sum{a^{10}} - \sum{(a^8b^2 + b^8a^2)} + a^2b^2c^2\sum{a^{4}})$
$ \Rightarrow RHS - LHS = 4\sum{a^{6}(a^2 - b^2)(a^2 - c^2)} \geq 0$
İt is obvious by Schur

[quote="ricardokaka"]
@arqady: That's right? Can you post your solution? I don't believe that my inequality is true for all triangles![/quote]
For all triangle holds: $ AH\equal{}a|\cot\alpha|\equal{}2R|\cos\alpha|.$ :wink:

@cefer: Please check your solution? The formule "$ a^{14}$" isn't true, i think that it's "$ a^4$" . Your solution is true if we replace $ a^{14}$ by $ a^4$

[quote="ricardokaka"]Let $ H$ be the orthocenter of an acute triangle $ ABC$. Prove that $ \frac {\sum HA^2}{\sum a^2}\le\sum\cot^2B\cot^2C\ .$[/quote]
[b][u]Proof.[/u][/b] Denote $ \boxed {\begin{array}{c} x = \cot A > 0 \\
 \\
y = \cot B > 0 \\
 \\
z = \cot C > 0\end{array}}$ . Observe that $ \|\begin{array}{c} xy + yz + zx = 1 \\
 \\
x^2 + 1 = (x + y)(x + z) \\
 \\
x^2y^2 + y^2z^2 + z^2x^2 = 1 - 2xyz(x + y + z)\end{array}$

and $ \|\begin{array}{c} \frac {HA}{|\cos A|} = \frac {HB}{|\cos B|} = \frac {HC}{|\cos C|} = 2R \\
 \\
\frac {a}{\sin A} = \frac {b}{\sin B} = \frac {c}{\sin C} = 2R \\
 \\
\frac {\sin^2 A}{1} = \frac {\cos^2 A}{x^2} = \frac {1}{x^2 + 1}\mathrm {\ \ a.s.o.}\end{array}$ . Therefore, $ \frac {\sum HA^2}{\sum a^2}\le\sum\cot^2B\cot^2C$ $ \Longleftrightarrow$

$ \sum\frac {x^2}{x^2 + 1}\le\sum\frac {1}{x^2 + 1}\cdot\sum (yz)^2$ $ \Longleftrightarrow$ $ \sum\frac {1}{x^2 + 1}\ge\frac {3}{1 + \sum (yz)^2}$ $ \Longleftrightarrow$

$ \sum[\frac 12\cdot(\frac {1}{y^2 + 1} + \frac {1}{z^2 + 1}) - \frac {1}{1 + \sum (yz)^2}]\ge 0$ $ \Longleftrightarrow$

$ \boxed {\sum(x^2 + 1)\{[2 + (y^2 + z^2)][1 + (yz)^2 + x^2(y^2 + z^2)] - 2[1 + (yz)^2 + (y^2 + z^2)]\}\ge 0}$  $ \Longleftrightarrow$

$ \sum(x^2 + 1)(y^2 + z^2)[(x^2 + y^2)(x^2 + z^2) - (x^2 - 1)^2]\ge 0$ $ \Longleftrightarrow$

$ \sum(x^2 + 1)(x^2 - 1)^2(y^2 + z^2)\le(x^2 + y^2)(y^2 + z^2)(z^2 + x^2)\sum(x^2 + 1)$ $ \Longleftrightarrow$

$ \sum\frac {(x^2 + 1)(x^2 - 1)^2}{(x^2 + y^2)(x^2 + z^2)}\le\sum (x^2 + 1)$ $ \Longleftrightarrow$ $ \sum (x^2 + 1)[1 - \frac {(x^2 - 1)^2}{(x^2 + y^2)(x^2 + z^2)}]\ge 0$ $ \Longleftrightarrow$

$ \sum (x^2 + 1)(y^2 + z^2)[y^2z^2 + x^2y^2 + x^2z^2 + 2x^2 - 1]\ge 0$  $ \Longleftrightarrow$

$ \sum x(x + y)(x + z)(y + z)(y^2 + z^2)(x^2 - yz)\ge 0$ $ \Longleftrightarrow$ $ \sum x(y^2 + z^2)(x^2 - yz)\ge 0$ a.s.o.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=176688


Let  $a$ a natural number  such that  $3^a-2^a=p^b$ with  $b>1$ and  $p$ a prime number. Prove that the number  $a$ is also prime

just a hint: make an use of following well-known lemma  $(a^n-b^n,a^m-b^m)=a^{(n,m)}-b^{(n,m)}$.

Megus can you please prove this lemma
My teacher said that it is not so easy to prove.

Dont you think $a,b$ should be relatively prime.for example according to your formula we have:
$(4^3-2^3,4^2-2^2)=4-2=2$ that is obviously wrong.

mahmood: You are right. The condition for $(a^n-b^n,a^m-b^m)=a^{(n,m)}-b^{(n,m)}$ to hold is that $(a,b)=1$.

of course $a,b$ are supposed to be relatively prime but in this problem it is clearly met.

Beat: you can find proof of this somewhere on the Mathlinks. I remember Peter Scholze posting a solution for a generalization of this (namely considering this relation as ideals relation in some ring with identity), also I believe "normal" version has been proved somewhere - if I remember correctly you can prove it using Euclidean algorithm.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=pgcd&t=36731]here[/url] a interesting link about that  :D .

Could anyone give a complete  solution  to the problem with the lemma?

Suppose $a$ be composite then $a=mn$ ,where $m,n>1$ and

$3^a -2^a = p^b \Longleftrightarrow (3^m)^n - (2^m)^n =p^b \Longleftrightarrow \\ (3^m-2^m)(3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)})=p^b \ \ (1) ,$

hence
$3^m \equiv 2^m (mod p) \Longrightarrow 3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)} \equiv n3^{m(n-1)} (mod p)$,

but  $3^{m(n-1)} +3^{m(n-2)}2^m+....+2^{m(n-1)} \equiv 0 (mod p)$  (from (1))

hence $n3^{m(n-1)} \equiv 0 (mod p) \Longrightarrow p=3$ (contradiction) or $p|n \Longrightarrow p|a$.

So suppose $a=rp$ with $r>1$ and by a similar way we prove that $p|r \Longrightarrow p^2|a$.If we continue as above we have $p^x|a$ for any $x\geq 1$ (contradiction) or $a=p^k$ with $k\geq 2$ {it is also impossible because by  (1) for $m=p$ we have that $3^p\equiv2^p(mod p) \Longrightarrow 3\equiv2(mod p)$ (contradiction) }.

So $a$ must be prime.

Στέλιος... :!:


Suppose a very poorly designed bridge 10,000 feet long expands in length by one foot on a warm summer day. Because the side supports do not move, the middle of the bridge sags downward, making two straight, congruent line segments as shown. How many feet below the original bridge's height is the middle (A) of the lengthened bridge? Express your answer to the nearest whole number.

[asy]draw((0,0)--(5,7)--(20,7)--(25,0));
draw((5,8)--(5,10));
draw((20,8)--(20,10));
draw((5,9)--(20,9));
label("10000 feet",(12.5,10),N);
draw((23,5)--(27,5),Arrow);
draw((27,0)--(32,7)--(39.5,5)--(47,7)--(52,0));
draw((32,8)--(32,10));
draw((47,8)--(47,10));
draw((32,9)--(47,9));
draw((32,7)--(47,7),dashed);
label("$A$",(39.5,5),S);
dot((39.5,5));
label("10000 feet",(39.5,9),N);[/asy]

[quote="Jongy"]5000.5^2-5000^2 and take sqrt.
70.71 rounds to 71[/quote]

Could you please elaborate more on that?

[hide="Solution"]
The distance from one of the supports to point $A$ is $\frac{10001}{2}=5000.5$. Drop the altitude to the dotted line. The longer length is $\frac{10000}{2}=5000$. Using the Pythagorean Theorem, we get the height of the triangle (how many feet below the original bridge's height) to be $\sqrt{5000.5^2-5000^2}\approx 70.71$ which gives an answer of $\boxed{71}$.

[/hide]

I don't know where this question showed up, but I think it would help if we could do this question without using a calculator.  Knowing the approximation of the square root of 2 (about 1.41) would help.

We take note that $5000.5^2 - 5000^2 = (5000.5-5000) * (5000.5+5000) = 0.5 * 10000.5$

Now we can approximate $0.5 * 10000.5$ to be very close to $0.5 * 10000$.

So now we want $\sqrt{0.5 * 10000}$, which is equal to $100 * \sqrt{0.5} = 50 * \sqrt{2}$

Finally, we know that $\sqrt{2} \approx 1.41$.  $50*1.41 = 70.5$, which rounds to $\boxed{71}$.

We also need not worry that it would round up to 72 since: $50*1.42 = 71$, and that $1.42$ is greater than the actual value of $\sqrt{2}$.

This was a target question, so you would've been able to use a calculator.


What did you do for Arbor Day?

[hide="Here is what I did"]i cleaned my room again[/hide]

I cut down several trees, burned them, created a bonfire out of old tires, and then decided that I would burn down some more trees.

Happy Arbor Day- one of the nation's least recognized holidays!

Anywho, I did nothing out of the ordinary.

i am sorry to ask this, but when IS Arbor's Day???


I need some help.

Jim runs around the track in forty seconds.  Mark running in the opposite direction, meets Jim every 15 seconds.  How many seconds does it take Mark to run once around the track?

Thx.  :)

[hide]It takes Mark 24 seconds to run around the track. When Jim meets Mark after the first 15 seconds, Jim is 15/40 or 3/8 of the way around the track. That means that Mark has run 5/8 of the way around the track. In turn, for Mark to go 1/8 of the track, it takes him 3 seconds, and therefore a full loop takes 24 seconds. [/hide]

Thanks!
Also thanks for telling me how you did it!
 :D


A_1= 1
A_(n+1)= sqrt(A_n)+A_n+1/2
A_n=?

it equal to solve T_(n+1)*(T_(n+1)-1)=(T_n)^2
but i don't know how to do next...

[quote]
\[ A_{1}\equal{}1\]
\[ A_{n\plus{}1}\equal{}\sqrt{A_{n}}\plus{}A_{n}\plus{}\frac{1}{2}\]
\[ A_{n}\equal{}?\]
[/quote]

So, do you want to find an explicit formula for $ A_{n}$?

yes,I know ignore some items can work out the approximate solution,
but i want to know the exact solution.
I doubt if it have elementary functions solution.


show that
(a-b)(b-c)(c-a) = a^2 + b^2 + c^2 has infinite number of solutions in integers.

another that I found not all that correct: or may be there was some printing error.

show that 1/5(x^5 - (x-1)^5) + k = 0 has atleast one real solution for every k. Also find the sum of non-real roots.

i think the second problem is wrong because for k=1 there is no solution! But for every (1+5k)/5 < 0 there is a real solution!

[quote="mahbub"]thew first one is quite easy. for fixed b and c, write the equation in quadritic form of a. Now look for b,c > 0 and b > 1/6 (probably) we get real; value of a. Thus the conclusion follows :D[/quote]

Mae culpa, we are looking for integer solutions.
Now the proof?

[quote="mahbub"]Well, so i need some time. May be I can't answer this time :(  Because i am not so advanced :blush:[/quote]

Take your time :-)

The solution is not in any way related to being your advanced or not. It just strikes you or it just doesn't.

One more from INMO. Actually, this should go in number theory section, but me reluctant to open a new thread.

S is the set of all(a,b,c,d,e,f) where a,b,c,d,e,f are integers such that a^2 + b^2 + c^2 + d^2 + e^2 = f^2. Find the largest k which divides abcdef for all members of S.

Guys, please solve the one's asked here :-)

Hm, what about $(a - b)(b - c)(c - a) = a^2 + b^2 + c^2$?

I don't manage to solve that question :blush:

can u finf any (a,b,c) which satisfy that equation?

Well, mahbub. There are at least 3 solutions that you can easily see by setting $b = -c$, which are $(0,0,0)$, $(0,1,-1)$, $(2,-1,1)$. I can't find the solution to the real question yet, though.

does this help:
$(a-b)(b-c)(c-a)=a^2(c-b)+b^2(a-c)+c^2(b-a)=a^2+b^2+c^2$(just a simple change)

[quote="armanf"]I think the last question is so easy because we have 16=9+4+1+1+1
and by checking he equality mod 8 and 3 we have the solution easily.!!! :!:
_____[/quote]

LOL!

What is so special about 16 that covers every "a"?

I liked your style when you said it's very easy :D  :D

Can anybody solve the very first question here?

as ak007 said, one of the solutions are $(2,1,1)$.

Let's multiply every number from this triple by $t$ - a positive integer, so
the left side of the given equality is $L=6t^3$, the right is $R=6t^2$, so $L-R=6t^2(t-1)$.
Let's transform our triple a bit from $(2t,t,-t)$ to $(2t+y,t+y,-t+y)$ where $y$ is an integer.
Now let's note that $L$ remains unchanged, but the right side changes from $6t^2$ to
$6t^2+4yt+3y^2$.
So after the transformation the equality holds if and only if $3y^2+4yt=L-R=6t^2(t-1)$.
Let's proove that there are infinitely many such integer $y$ and positive integer $t$ for which the above equality is true.
Putting $y$ as a variable we have a solution (one of two solutions) $y=\frac{-4t-\sqrt{16t^2+72t^2(t-1)}}{6}=
\frac{-4t-2t\sqrt{4+18(t-1)}}{6}=\frac{t(-2-\sqrt{4+18(t-1)}}{3}$, so $18(t-1)+4=18t-14$ must be a perfect square and we wan't in addition $t$ divisible by $3$ - then $y$ is an integer and the pair $(t,y)$ satisfies $3y^2+4yt=L-R=6t^2(t-1)$, but there are infinitely many of such $t$ for which $3|t$ and $18t+14$ is a square (I skip the proof of this easy fact), so for significant $t$ we can obtain an integer $y$ - therefore we have infinitely many solutions $(t,y)$, so there are infinitely many solutions of the form $(2z+y,z+y,-z+y)$.

ps. As we can see, $y$ is always negative - if it's an integer, so if there is a satyfying triple $(2z+y_z,z+y_z,-z+y_z)$ then we can choose sufficiently big $z'$ for which there exists a solution and $-z'+y_z'<-z+y_z$, so it is not possible that we will obtain infinitely many triples which are simply the same.

[quote="ak007"]Well, mahbub. There are at least 3 solutions that you can easily see by setting $ b \equal{} \minus{} c$, which are $ (0,0,0)$, $ (0,1, \minus{} 1)$, $ (2, \minus{} 1,1)$. I can't find the solution to the real question yet, though.[/quote]

(2, - 1,1) is obviously not a solution!!

can anybody please provide a  good solution for this??


Prove that, in any 16-digit positive integer, there is some consecutive string of digits whose product is a perfect square.

n = d1d2...d16

If one of the di is 0,1,4 or 9 : done, just take that digit so let's suppose di != 0,1,4,9.

Let's consider the 16 numbers : D1=d1, D2=d1*d2, ..., D16=d1*d2*..*d16.

These 16 numbers are of the form 2^a * 3^b * 5^c * 7^d. There are 2^4 parity patterns for (a,b,c,d).
If one is (even,even,even,even) then finished : take the corresponding number.
Else by Pigeon-Holing there are to numbers Di and Dj with the same parity pattern. Then Dj/Di = d(i+1)*...*d(j) is a perfect square.


if n is a positive integer such that 3n+1 is a perfect square, then show that, n+1 is the sum of three perfect squares.

thank you...

$ 3n\plus{}1\equal{}0,1,4\mod 8 \to n\plus{}1\equal{}6,1,2\mod 8\not \equal{}0,4,7\mod 8$.
$ N\equal{}x^2\plus{}y^2\plus{}z^2$ for any $ N\not \equal{}4^a(8k\plus{}7)$.

Suppose $ 3n\plus{}1 \equal{} x^2$, x is positive integer.
If $ x\equiv 1$ mod 3, let $ x\equal{}3k\plus{}1$, then $ n \equal{} \frac{(3k\plus{}1)^2\minus{}1}{3} \equal{}3k^2\plus{}2k$
Thus $ n\plus{}1\equal{}3k^2\plus{}2k\plus{}1\equal{}k^2\plus{}k^2\plus{}(k\plus{}1)^2$ is a sum of three perfect squares.
If $ x\equiv 2$ mod 3, let $ x\equal{}3k\plus{}2$, then $ n \equal{} \frac{(3k\plus{}2)^2\minus{}1}{3} \equal{}3k^2\plus{}4k\plus{}1$
Thus $ n\plus{}1\equal{}3k^2\plus{}4k\plus{}2\equal{}k^2\plus{}(k\plus{}1)^2\plus{}(k\plus{}1)^2$ is a sum of three perfect squares.
If x is divisible by 3, then its square will not be in form of $ 3n\plus{}1$, so the proof is completed.


function
trigonometry
calculus
integration
Miscellaneous Problems

The function $\mu: \mathbb{N}\to \mathbb{C}$ is defined by \[\mu(n) = \sum^{}_{k \in R_{n}}\left( \cos \frac{2k\pi}{n}+i \sin \frac{2k\pi}{n}\right),\] where $R_{n}=\{ k \in \mathbb{N}\vert 1 \le k \le n, \gcd(k, n)=1 \}$. Show that $\mu(n)$ is an integer for all positive integer $n$.

In other words: find the sum of the primitive $n$-th roots of unity. By what
I wrote in http://www.problem-solving.be/pen/viewtopic.php?t=125 , this just runs out to finding the coefficient of $x$ in $\Phi_n(x)$.
As a factor of $x^n-1$, the coefficients of $\Phi_n(x)$ are always integral, and as shown there the one under observation is equal to $\mu(n)$.


calculus integration function trigonometry logarithms algebra domain

Hello,
I would like to know how to find
$ \int _0 ^{\infty} e^{\minus{}x} \frac{sin^2 x}{x} dx$.

There was a hint given: use the function $ e^{\minus{}x} \sin {2xy}$, but somehow I don't know how it is useful.
Any ideas are welcomed (maybe there are several ways to find this integral..).

Thanks in advance.

I'd start with the identity $ \sin^2 x\equal{}\frac12(1\minus{}\cos 2x)$. Now the hint makes sense: $ \int_0^1 \sin(2xy)\,dy\equal{}\left[\minus{}\frac1{2x}\cos(2xy)\right]_0^1\equal{}\frac{1\minus{}\cos(2x)}{2x}$.

The original integral is $ \int_0^{\infty}\int_0^1 e^{\minus{}x}\sin(2xy)\,dy\,dx$. Can you see what to do next?

Thank you!
.. next (by Fubini's theorem) we can change the order of the integrals, right?
And then (some work needed), if I'm not mistaken, the needed integral equals $ \frac{\ln 5}{4}$.
Don't have time to write it all now, so if it's true I'll write it tomorrow (after the exam  :ninja: )

I haven't checked the answer, but that's the method. Switching the order is the reason to write down an iterated integral at all.

With the infinite domain, Fubini's theorem needs justification: in this case, absolute convergence. Since $ |\sin(2xy)|\le 1$ everywhere, $ \int_{[0,\infty)\times [0,1]} |e^{ \minus{} x}\sin(2xy)|\,dA\le \int_{[0,\infty)\times [0,1]} |e^{ \minus{} x}|\,dA \equal{} 1$.

[hide="Another solution"]$ \int \frac{\sin^2x}{x} \cdot e^{\minus{}x}\,dx \equal{} \frac12 \int \frac{1\minus{}\cos(2x)}{x} \cdot e^{\minus{}x}\,dx \equal{} \lim_{s \to 1} \frac12 \int \frac{1\minus{}\cos(2x)}{x} \cdot e^{\minus{}sx}\,dx$

$ \equal{} \frac12 \, \lim_{s \to 1} \,\, \mathcal{L} \left[\frac{1\minus{}\cos(2x)}{x}\right] \equal{} \frac12 \, \lim_{s \to 1}\,\, \frac12 \ln\left(1 \plus{} \frac{4}{s^2}\right) \equal{} \frac{\ln 5}{4}$.[/hide]

jmerry, thanks for pointing out the difference when it's an infinite domain.

Carcul, Can you please explain more your solution? What the $ \mathcal{L}$ sign denotes (well, maybe I understand what is denoted by it, but don't understand the meaning) ? Where the next step comes from?
Maybe you can give some link.

Kar, $ \mathcal{L}$ refers to the laplace transform.


Let $ a,b$ be two nonzero real numbers, and the function $ f: \mathbb{R}\to (0;\infty)$ with $ f(x \plus{} a \plus{} b) \plus{} f(x) \equal{} f(x \plus{} a) \plus{} f(x \plus{} b), \forall x \in \mathbb{R}$ . Prove that if  $ \frac {a}{b}$  is rational, then the function $ f$ is periodic.

First a couple of reduction steps: Wlog we can assume $ a \equal{} 1$ (consider $ f(ax)$). Then $ b \equal{} \frac {m}{n}$ with integers $ m,n$. Wlog we can assume $ m,n > 0$. Define $ g_c(x) \equal{} f(x \plus{} c) \minus{} f(x)$ for $ c\in \mathbb R$. Then $ g_1$ is $ \frac {m}{n}$-periodic $ \Rightarrow$ $ g_1$ is  $ m$-periodic $ \Leftrightarrow$ $ g_m$ is $ 1$-periodic $ \Rightarrow$ $ g_m$ is $ m$-periodic. This shows that wlog we can assume  $ a \equal{} b \equal{} 1$.

$ a \equal{} b \equal{} 1$: The condition on $ f$ is equivalent to
\[ f(x) \equal{} \frac {1}{2}(f(x \minus{} 1) \plus{} f(x \plus{} 1))
\]
for all $ x\in\mathbb R$. Then $ f$ is 2-periodic: If there is some $ x\in\mathbb R$ with $ f(x \minus{} 1) > f(x \plus{} 1)$, then the sequence $ (f(x \plus{} n))_{n\ge 0}$ is decreasing with constant subsequent differences. This implies that $ f(x \plus{} n) < 0$ for some $ n > 0$ which contradicts to $ f > 0$. The other case  $ f(x \minus{} 1) < f(x \plus{} 1)$ analogously yields to a contradiction. Hence $ f(x \minus{} 1) \equal{} f(x \plus{} 1)$  for all $ x\in\mathbb R$.

[quote="Third Edition"] The other case  $ f(x \minus{} 1) < f(x \plus{} 1)$ analogously yields to a contradiction. [/quote]

Please explain, what contradiction is in this case? :-??

Thanks a lot.

Apply the trick to $ \tilde{f}(x)\equal{}f(\minus{}x)$ or, alternatively, consider the sequence $ (f(x\minus{}n))_{n\ge 0}$.


topology
geometry
advanced fields
advanced fields unsolved

How to prove that a covering map is an open immersion?

A covering map $p\colon C\to X$ is a local homeomorphism by its definition, since for any $c\in C$ there is an evenly covered neighborhood of $p(C)$. As for immersion... isn't it something from differential geometry, with $C^{1}$-smooth maps and such?


Let $ABC$ be a non-isosceles triangle such that $\displaystyle \measuredangle B < \frac{\pi}{4}$. Let $AD$ be the $A$-altitude, $D \in \left( BC \right.$, and $E$ be point on $\left( AC \right)$. Let $P \in AD$ and $Q \in BE$ such that $BD=PD=QD$ and $R \in BE$, $S \in AD$ s.t. $AE=RE=SE$.

Prove that $P,Q,R,S$ are concyclic iff $BE \perp AC$.

Sorry if the problem was discussed before, but I can't find it. (I'm not really good with the "search" button)

$\angle RQP=180^{\circ}-(\angle PQD +\angle BQD)$$=180^{\circ}-(90^{\circ}-\frac{\angle PDQ}{2}+90^{\circ}-\frac{\angle QDB}{2})=180^{\circ}-(180^{\circ}-45^{\circ})=45^{\circ}$

$\angle RSP=180^{\circ}-(90^{\circ}-\frac{\angle SER}{2}+90^{\circ}-\frac{\angle AES}{2})=\frac{\angle AER}{2}$

so quadrilateral $PQRS$ is cyclic iff $45^{\circ}=\angle RQP=\angle RSP=\frac{\angle AER}{2}\longrightarrow \angle AEB=90^{\circ}$


Yes, thanks Enescu, and sorry, I didn't know it was IMO's question

[quote="Lovasz"][quote="Diogene"][quote="Lovasz"]Find all the function $f: (-1,+\infty)\to(-1,+\infty)$ satisfying two conditions:
$a)$  $f(x+f(y)+xf(x))=y+f(x)+yf(x)$
$b)$  $\frac{f(x)}{x}$ increases $\forall x\in(-1,+\infty)$, $x\neq0$.[/quote]

A such function does'nt exist ...

 :cool:[/quote]
Why did you say that, Diogene? (It was IMO's problem)?!?!?![/quote]

Are you sure Lovasz ?? :P The correct relation ship of the IMO's problem is $f(x + f(y) + xf(y)) = y + f(x) + yf(x)$
 :cool:


geometry
function
perimeter
geometry proposed

$f:\mathbb{R}^2 \to \mathbb{R}^2$ is injective and surjective. Distance of $X$ and $Y$ is not less than distance of $f(X)$ and $f(Y)$. Prove for $A$ in plane:
\[ S(A) \geq S(f(A))\]
where $S(A)$ is area of $A$

Maybe it's time to post your solution, since you say it's easy...

Hint : u should solve it with circle packing :D

Maybe the best idea is to break it in small squares, or any regular figure..., but I think the problem is the function is not continuous.......
Also, here arises a question, is A a contour of the figure, or is A the entire figure, because some continuity problems appear....
What do you think, sam-n ????

Well for any figure $B$ you must first put some cirlces in $B$ that they are distinct and sum of their areas is $S(B)- \epsilon$.

Well consider circles with centers $f(A_1),f(A_2),...,f(A_n)(A_i \in A)$ with radius $r_1,...,r_n$ that are distinct and sum of ther areas is $S(f(A))- \epsilon$
Now draw circles with centers $A_1,...,A_n$ and radius $r_1,...,r_n$ .
Prove they are distinct and are in $A$ so for every $\epsilon$:
\[ S(A) \geq S(f(A))-\epsilon\]
And this proves the problem

For circle packing you must use this that for every $\epsilon$ there are finite squares in $A$ sum of their area is at least $S(A) - \epsilon$

[quote=Simo_the_Wolf]Take a point P inside A.
Take all the distances from P to all points of the perimeter of A.

we take $k=max[d(P,f(B))/d(P,B)]$ where B is a point of the perimeter.

We know that k<1 and all points of f(A) are in A' where A' is A dilatate about P of the coefficient k. So S[f(A)]<=S(A') but S(A')=kS(A)<S(A) and so
S[f(A)]<S(A)

c.v.d.[/quote]

But the figure may not be convex so why f(A) is in A'?


[quote="sinajackson"]John has gone to a university far away from his family to continue his studies. One of these days he realizes that he needs more money so he telegraphes to his family the following Codes:      $SEND$ $+$ $MORE$ $=$ $MONEY$
Each word has a value between numbers $0,1,2,\cdots,9$ and there are no two words that have the same value. How much money did John need?[/quote]
Don't you mean each letter has a different value?

[hide]

Since you can't carry more than 1 in an addition, $M=1$.  $S+1$ can be at most 11, and $O\neq 1$, so $O=0$.  In the next column we can't have $N=0$, so $E+0<10$, so nothing carries and $S=9$.  $N\neq E$, so $N=E+1$.  From the next column we have $E+1+R=E$.  $R\neq 9$, so $R=8$.  We need both of the last two columns to carry.  We have $E<7$, since $N=E+1<8$, but also $E>4$, since in the units column $D<8$, $D+E>10$, and $Y>1$.  For $E=6$ we have $N=D=7$, so $E=5$.  So $N=6$, $D=7$, $Y=2$.  Our addition is $9567+1085=10652$

[/hide]

Oops I made a mistake, each $letter$ has a different value.
nice solution E^(pi*i)=-1

I think this would fit better into "other problem solving strategies"

[quote="sinajackson"]dont listen to him; He doesn't know what he's talking about, he just pretends to  :D[/quote]
 :rotfl:  :rotfl:


abstract algebra
group theory
superior algebra
superior algebra unsolved

Determine for what values of n the dihedral group D_n is solvable and for what values of n A_n is solvable and why in both cases.

I'm not sure how to do this problem.  Just providing examples can't be sufficient.

this is really a nice and deep problem :)

so, the case of $D_{n}$ is easy to treat, since there's the easy chain $(e) \lhd (r) \lhd D_{n}$ which has abelian quotients.

the latter case is pretty more challenging...
let me recall that, given a group $G$ we call the derived group $DG : = ([g: h] | g,h \in G)$ the group generated by commutators of couples in $G$.
you should just prove two lemmas
1. $A_{n}$ is generated by $3$-cycles.
2. (which i seem to remember valid for all groups, but that surely works for finite ones): a group $G$ is solvable iff the chain $(D^{k}G)_{k\in\mathbb{N}}$ eventually reaches the trivial group $(e)$.
once you have that, just prove that $DA_{n}= A_{n}$ for $n\ge 5$, using first lemma...
so we've settled $A_{n}$ for $n\ge 5$.
$A_{2}$ and $A_{3}$ are trivial, and $A_{4}$ admits the normal subgroup called klein group, which is $\{e, (1\,2)(3\,4), (1\, 3)(2\, 4), (1\, 4)(2\, 3)\}$ (it has order $4$, and hence is abelian, while the quotient has order $3$, so it's cyclic).

$A_{n}$ is simple if and only if $n\neq 4.$ Then $A_{n}$ is not solvable if and only if $n\geq 5.$


What does it mean when it says directed angles modulo 180 degrees?

For example in a cyclic quadrilateral it says that the opposite angles I believe are equal in directed angles modulo 180.

I have tried to give an introduction at
http://www.mathlinks.ro/Forum/viewtopic.php?t=47944 .

See also chapter 1.7 of http://www-math.mit.edu/~kedlaya/geometryunbound/ .

  darij

Thanks

Sorry  I didn't see these earlier.


topology
inequalities
triangle inequality
calculus
calculus computations

$ (M,d)$ is a metric space. $ S \subset M$, define $ x\in M$ and $ d(x,S) \equal{} inf\{d(x,y): y \in S\}$.

a) Show $ |d(x,S) \minus{}d(y,S)| \le d(x,y)$

b) Show that $ \{x| d(x,S) \equal{}0\}$ = closure of S

I have a hunch about a), since the problem has what $ d(x,S)$ is equal to, but I have no idea about how to answer b). Any ideas?

One may start as follows:

a) $ |d(x,S) \minus{} d(y, S)| \leq d(x,y) \iff d(y, S) \minus{} d(x, y) \leq d(x, S) \leq d(x, y) \plus{} d(y, S)$

Now use the triangle inequality to justify the right-hand inequality and for the left-hand inequality you know by the definition of infimum that for each y in S $ d(x, S) \leq d(x, y)$

b) Choose $ a \in cl(S)$. Suppose U is an open set containing a, then U must intersect S at some point, say z.
...

Choose $ b \in B\equal{}\{x: d(x,S) \equal{}0\}$. Then $ b \notin cl(S) \iff$ there exists an open set containing b that does not intersect S.
...


Let S(n) be the sum of all digits of n $(n\in N)$
Prove that there are unlimited $n\in N$ such that: $S(3^n)\ge S(3^{n+1})$

I have a solution about this problem.hope that it is right.
if there only finite number $n$ such that $S(3^n)\ge S(3^{n+1})$
we can know there exist a $N$ when $n>N$    $S(3^n)<S(3^{n+1})$
first we prove $9\mid S(3^n),n\geq2$,then when get $S(3^{n+1})\geq S(3^n)+9,n>N$
and $3^n=9^{\frac{n}{2}}\leq 10^{\frac{n}{2}}-1$ when n is even.
$3^n<9^{\frac{n+1}{2}}\leq 10^{\frac{n+1}{2}}-1$ when n is odd.
so we get $S(3^n)\leq \frac{9(n+1)}{2}$
but  $S(3^{n+1})\geq S(3^n)+9,n>N$ tell us  $S(3^{n})>9(n-N)$
so  $9(n-N)<\frac{9(n+1)}{2}$     when  $n\rightarrow\infty $     contradiction. :)

Not so hard...  :)
And if we replace 3 by 2? :D

Well in this case, its NOT a hard problem only need to make some computations and you get that      36=S(3^15)<S(3^14)=45  .

Best regards.

In fact, 7=S(2^4)<S(2^3)=8   , and 11=S(5^5)<S(5^4)=13 .

[quote="ikap"]Robertux looks a little bit off the track.
For $n=2$, one should either make some computations to show that there exists $a>1$ such that $2^n<10^{\frac {n}{3a}}$, for $n$ bigger than some $N$, or to prove that we can't find $n>M$, where $M$ is the point from which $s(2^n)$ is incresing such that $s(2^{6n})=x, s(2^{6n+1})=x+2,....s(2^{6n+6})=x+18$. I don't know which one is simpler.
But what about putting $>$ instead of $\geq$ in the initial problem. It looks a little bit tougher.[/quote]
yes.if put $>$ instead $\geq$,I think it is so hard.and I think it is also a hard problem in general.3 is too sepical. :)


conics
parabola
quadratics
algebra
calculus
calculus computations

A car is traveling at night along a curve in the road that is shaped like a parabola opening northward.  A deer is munching geraniums off to the side of the road 100 feet east and 50 feet north of the vertex.  The car starts out 100 feet west and 100 feet north of the vertex.  How far away is the car from the deer when the car's headlights illuminate the deer?

Just a side note: at some point in this problem, you'll have a quadratic equation to solve which has two real roots. Both roots are meaningful, although only one is the answer to this particular problem. One root corresponds to this problem, in which the car is traveling in a net east-to-west direction; the other would correspond to the situation in which the car is traveling in the opposite direction on the same road.

To other solvers: you don't have to download this .doc file (and I haven't) because the verbal description, read carefully enough, is unambiguous.


There are a lot of math majors on this website. But I am curious if anyone is interested in studying engineering. I want to study electrical engineering, and research high speed magnetic levitation systems.

If you are not in college yet, consider [url=http://www.ece.tamu.edu/Undergrad/UndProspective.htm]Texas A&M[/url], which has a highly regarded double E program with about $ 1000$ undergraduate majors. But in that case it is probably too early to lock yourself into a narrow area such as magnetic levitation; your interests may well change when you actually start working on projects.

I'll probably do engineering in college too.  Not sure on which type yet though.

Right now I'll either do EECS or Math for my major.  Engineering is fun, thought not usually as beautiful as math.  And I'd like to note that you shouldn't feel like you have to go into engineering to research that topic.

I'm going to go into engineering!  :)

I think engineering is a great undergraduate option for anyone who is more interested in real world problem solving than in touching up on the finer details of analysis or trying to prove the Riemann Hypothesis.

High speed magnetic levitations systems are fascinating! Where are you planning on applying to?

Engineering is the way to go!  Decent money, but more importantly, I like using the math and science I learn to solve real world problems.  I plan to go into ChemE right now, as it seems to have a lot of broad applications anywhere from materials to biology research to electronics and nanotechnology.


I am confused with "combining proportions", which I read about in Volume 1. For example, in the problem:

Given that $ x$ is proportional to $ y$ and to $ z$ and is inversely proportional to $ w$, and that $ x\equal{}4$ when $ (w, y, z)\equal{}(6, 8, 5)$, what is $ x$ when $ (w, y, z)\equal{}(4, 10, 9)$?

In the solution, it combines them to get $ \frac{xw}{yz}\equal{}\frac{4 \cdot 6}{8 \cdot 5}\equal{}\frac{3}{5}$, so $ x\equal{}\frac{3yz}{5w}\equal{}\frac{27}{2}$. But if you compare them individually, $ xw$ should equal 24, but it doesn't. Why?

I don't really know how to explain this well...

First of all you have to simplify the fraction.
Also, you have to look at the fraction as ONE, not two seperate results...if you know what I mean...

If $ x$ is proportional to $ y,$ then their quotient must be constant, and if $ x$ is inversely proportional to $ y,$ their product is constant. So \[ \frac{xw}{yz}\equal{}\frac{4\cdot6}{8\cdot5}\equal{}\frac{3}{5}.\] You simplified your fraction wrong at this point. \[ x\equal{}\frac{3\cdot10\cdot9}{5\cdot9}\equal{}\boxed{6}\neq\frac{27}{2}.\] As a check, \[ xw\equal{}6\cdot4\equal{}24,\] which is still the same.

I think you put in the wrong numbers. $ (w, y, z) \equal{} (4, 10, 9)$, so:

$ x \equal{} \frac {3}{5} \cdot \frac {yz}{w} \equal{} \frac {3 \cdot 10 \cdot 9}{5 \cdot 4} \equal{} \frac {27}{2}$

But still, 6 doesn't work with anything other than $ w$ either.

So the possible answers we have:

$ \boxed{\frac {27}{2}}$ which is the correct answer from the book.

Since $ xw \equal{} 4 \cdot 6 \equal{} 24$, when $ w \equal{} 4$, then $ x \equal{} \boxed{6}$.

Since $ \frac {x}{y} \equal{} \frac {4}{8} \equal{} \frac 12$, when $ y \equal{} 10$, then $ x \equal{} \boxed{5}$.

Since $ \frac {x}{z} \equal{} \frac {4}{5}$, when $ z \equal{} 9$, then $ x \equal{} \boxed{\frac {36}{5}}$.

You get a different answer if you do each individually. [b]Why is the first one correct?[/b] Why are the rest incorrect?

[quote="PowerOfPi"]You get a different answer if you do each individually. [b]Why is the first one correct?[/b] Why are the rest incorrect?[/quote]
The answer to your questions is quite subtle.  The thing is, the problem statement implies that $ x$ depends on $ y,z,$ and $ w$ collectively since all of the variables are mentioned in the same sentence.  If the problem was "$ x$ is proportional to $ y$.  $ x$ is also proportional to $ z$.  Furthermore, $ x$ is inversely proportional to $ w$.  Finally, $ x\equal{}4$ when $ (w,y,z)\equal{}(6,8,5)$.  What is $ x$ when $ (w,y,z)\equal{}(4,10,9)$?", then your lines of reasoning would be correct. (I know that my example is a strange way to write a math problem, which is why that type of problem never occurs; it provides too much information and the wording is strange.)

I know my explanation is strange, and I am no expert or anything, but I'm pretty sure that what I'm saying is correct.  You can't isolate "$ x$ is proportional to $ y$" just because of the fact that everything is in one sentence.

Granted, I understand what you're saying.  If this were an AMC problem (which it very well may be since I dont know the source), then I would expect there to be complaints about it concerning the wording or the clarity of the problem's meaning (you might have seen the fiasco this year on the forums about an unrelated problem on the AMC but with the similar situation of ambiguous wording).
You just have to assume the most likely solution method will work (and, again, the "most likely solution method" is the book's solution because everything is one sentence).  If it were a contest and you found 4 different answers as you did, then which one would you go with?  :P

I am starting to understand. The source of the problem was AoPS Volume 1 (I couldn't make one up or find one from somewhere else).

Thanks for the help.


inequalities
induction
combinatorics theorems
combinatorics

[b]"Sperner's theory"[/b] Maybe everyone study combinatorics have known it. That is a beautiful theorem. Here is problems about it:
Suppose $X$ is a set such that $\left|X\right|=n$. A collection $A_{1},...,A_{k}$ of non-empty sets is a [i]Sperner System[/i] in $X$ if every $A_{i}$ is a subset of $X$ and there don't exist two distinct integers $i,j$ with $1\leq i, j\leq k$ such that $A_{i}$ is a subset of $A_{j}$. Prove that:
a) If $A_{1},...,A_{k}$ is a Sperner system in $X$, then $k \leq \binom{n}{[n/2]}$ (Sperner's theorem).
b) Let $A_{n}$ be the number of all Sperner Systems in $X$. Then we have:
  $2^T < A_{n} < \binom{2^T}{T}$ with $T=\binom{n}{[n/2]}$

Here $\binom{x}{y}$ means binomial coeficients.

 I have just done a) but not b). :maybe:

Well, with your notation $T = C_n^{[\frac n 2 ]}$, I can prove that
\[2^T \leq A_n \leq 3^T\]
which gives a better upper bound than yours, at least for $n \geq 3$ (check the cases $n=1,2$ by hand to get yours).

The proof is a bit technical (but quite elementary) and its simplicity may depend on what you are knowing about filters and the proof you have to a) (since there are many proofs of Sperner's theorem, The one which is the best here is to use De bruijn et al. proof using partition into symmetric chains).

It is an excelent result,Pbornsztein.I don't know how to find out and show it.It is hard for me. :blush: Is it necessary to prove by more elementary knowledge?Can you give me a way?I want to know more problems related antichain or symetric chain because I am going to propose a combinatoric topic for my class's articles.(articles about only maths!).Can you help me? ;)
With the lower bound of $|A_{n}|$ I have just found out an expression (ugly) stronger than $2^T$.(not very strong!)

For convenience, let $X = \{1,2,...,n \}$.
Let $P$ be the set of the subsets of $X$ and $P_k$ be the set of all the subsets of $X$ which contains exactly $k$ elements.

The lower bound $2^T$ is trivial since any subset of $P_{[n/2]}$ is a Sperner family. You can improve it easily since, for each $k$, any subset of $P_k$ is a Sperner family.
It leads to :
\[A_n \geq ( \sum_{k=0}^n 2^{C_n^k} ) - n\]
(since you have to count the emptyset only once) but it doesn't give an asymptotical improvement.

I will try to give you a readable proof of the upper bound soon.

Pierre.

I still use the notations of the previous post.
In this post, I give a useful proof of Sperner's theorem (To be rigourous, only the inequality you stated in the first post. The whole theorem also give the cases of equality...).
Since it's late now, I will send the rest of the promised result (probably) tomorrow.

The relation $ \subset $ (I use the symbol to denote a strict inclusion) leads to a partially ordering of the elements of $P$ (thus, we say that $P$ is a partially ordered set, also known as 'poset')
If $p,q$ are two elements of $P$ we will say that $q \succ p$ when $p \subset q$ and if $ p \subset q' \subseteq q$ then $q'=q$. This is a standard terminology, but here it just corresponds to $p \subset q$ and $|q| = |p| + 1$.

A chain of the form $C = ( A_0 \prec A_1 \prec \cdots \prec A_h )$ such that $|A_0| + |A_h| = n$ is called a symmetric chain.

Now, I give a proof of Sperner's theorem, due to De Bruijn, Tengbergen and Kruyswijk, which is based upon the following lemma :

Lemma.
There exists a partition of $P$ into symmetric chains.

Proof.
By induction on $n \geq 3$. This is obvious for $n=3$ so let's assume that the result holds for some given $n$.
Thus, let's consider a partition of $P$ into symmetric chains. For each chain $C = ( A_0 \prec A_1 \prec \cdots \prec A_h ) $, we construct the new chains :
$C' = (A_0 \prec A_1 \prec \cdots \prec A_h \prec A_h \cup \{n+1 \} )$ and
$C'' = \{ A_0 \cup \{n+1 \} \prec A_1 \cup \{n+1 \} \prec \cdots \prec A_{h-1} \cup \{n+1 \} )$,
where $C''$ does not exist if $h=0$.
It is easy to verify that all these chains form a partition of the set of the subsets of $\{1,2, \cdots ,n,n+1 \}$ into symmetric chains, which proves the lemma.

Now, from the remark above, each symmetric chain in $P$ contains exactly one member $A$ such that $|A| = [ \frac n 2 ]$. Thus, each partition of $P$ into symmetric chains contains exactly $T$ chains. Moreover, if $S$ is such a partition and $F$ is a Sperner family in $P$ then $|F \cap C| \leq 1$ for each chain $C$ which belongs to $S$.
It follows that :
$|F| \leq \sum_{C \in S} |F \cap C| \leq \sum_{C \in S} 1 = \left|S\right| = \left|T\right|$.

Pierre.

Ok, let's go for chapter two. The claimed upper bound is due to Hansel.
Following his reasoning, we will first prove the following lemma :

Lemma :
Let $S$ be the symmetric chains partition of $P$ as given in the previous post (and constructed inductively).
For each chain $C \in S$ such that $|C| \geq 3$, and three consecutive members $X_{a-1} \prec X_a \prec X_{a+1}$ of $C$, there is some $X'_a$ such that $X_{a-1} \prec X'_a \prec X_{a+1}$ and $X'_a$ belongs to a chain $D \in S$ such that $|D| = |C| - 2$.

Proof :
We use induction on $n \geq 3$.
The result is trivial for $n=3$, so let's assume that it holds for a given $n \geq 3$.

- First case : If the three members $X_{a-1}, X_a, X_{a+1}$ belong to a chain of type $C'$ :
If $n+1 \notin X_{a+1}$, then the three members belong to the chain $C \in S$ (for the step $n$), and from the induction hypothesis, we can find a chain $D \in S$ with the corresponding member $X'_a$ which also belongs to the chain $D'$ constructed from $D$ for the step $n+1$.
And, since $|D| = |C|- 2$ we have $|D'| = |C'| - 2$.
If $n+1 \in X_{a+1}$ then $X_a = X_h$ and we can take $X'_a = X_{h-1} \cup \{n+1 \}$. The corresponding chain is $C''$.

Second case : The three members $X_{a-1}, X_a, X_{a+1}$ belong to a chain of type $C''$ :
Then $Y_j = X_j \setminus \{n+1 \}$ (where $j=a-1,a,a+1$) belongs to $C \in S$ (for the step $n$) and, from the induction hypothesis, we find a corresponding chain $D \in S$ with the corresponding member $Y'_a$. Because $Y_{a+1}$ cannot be maximal in $C$ (from the construction of $C''$), the member $Y'_a$ is not maximal in $D$ (otherwise, we would have $|D| < |C|-2$). Thus, $X'_a = Y'_a \cup \{ n+1 \}$ belongs to $D''$, and $|D''| = |D| -1 = |C| -3 = |C''| -2$, so we are done again.
This ends the proof of the lemma.

Pierre.

Now, the proof (at last!).

First, a subset $F$ of a poset $P$ is called a filter if $p \in F$ and $q \geq p$ imply $q \in F$.

Now, remark that there is a bijection between filters and antichains (by associating with a filter the set of its minimal elements).
Thus, we will count filters of the set $P$ of all the subsets of $ \{1,2,...,n \}$, which is $A_n$ and prove that $A_n \leq 3^T$.

Let's consider $S$ be a partition of $P$ with symmetric chains as given above.
Order the members of $S$ by increasing size : $C_1,C_2, \cdots , C_T$, with $|C_1| \leq |C_2| \leq \cdots \leq |C_T|$ (since, as we seen before $S$contains exactly $T$ chains).
Now we construct step by step all filters $F$ in $P$ by deciding whether the members of $C_j,j=1, \cdots , T$ should belong to $F$ or not.
For the chains of size at most 2, we have at most three choices, namely taking no, or the maximal, or both members into $F$.
Suppose that we are done with all chains before $C = X_0 \prec \cdots \prec X_h$, and we have fixed some part of a filter $F$.
Let $X'_1, \cdots , X'_{h-1}$ be the corresponding members described in the lemma proved in my previous post.
Of them we already know whether they belong to $F$ or not.
Let $p$ (resp. $q$ ) be the largest (resp. smallest) index such that $X'_p \notin F$ (resp. $X'_q \in F$). If there is no such index put $p=0$ (resp. $q = h$).
Clearly, $q-p \leq 1$.
In order to obtain a filter, we need that $X_1, \cdots ,X_{p-1} \notin F$ and $X_{q+1}, \cdots , X_h \in F$, since $X_1 \prec \cdots \prec X_{p-1} \prec X'_p$  and $X'_q \prec X_{q+1} \prec \cdots \prec X_h$. Thus we have at most three choices to extend the fixed part of $F$ : This is clear if $p=q$ and if $q=p+1$ then at most $X_p$ and $X_q$ are free but if $X_p \in F$ it forces $X_q \in F$. Thus we have at most $3^T$ possibilities to construct all filters, and we are done. :P

A small remark :
This leads to $T \cdot ln(2) \leq ln(A_n) \leq T \cdot ln(3)$
In fact, Kleitman has proved that $ln(A_n) \sim T \cdot ln(2)$, but the proof is complicated so I will not give it here.

Pierre.

Thank you a lot!
All things which you posted are great for me!  :D

Well, in fact Korshunov has found an asymptotic for $A_n$ which is an improvement of Kleitman's result :

- If $n$ is even :
\[A_n \sim 2^Te^{C_{ \frac n 2 - 1}^n( 2^ {- \frac n 2} + \frac {n^2} {2^{n+5}} - \frac n {2^{n+4}} )}.\]

- If $n$ is odd :
\[A_n \sim 2^Te^{C_{ \frac {n-3} 2}^n( 2^ {- \frac {n+3} 2} - \frac {n^2} {2^{n+6}} - \frac n {2^{n+3} }) + T(2^{- \frac {n+1} 2} + \frac {n^2} {2^{n+4}})}.\]

Do never ask me to prove that  :D

Pierre.


The 2006-2007 Johns Hopkins Math Tournament (JHMT) will be held on Saturday November 11, 2006 at the JHU campus in Baltimore, Maryland (going from about 9:30 AM to 5:00 PM).

Our website is http://www.tripleintegral.org/jhmt but is currently down right now -- sorry.

Teams from anywhere are invited but most come from Maryland schools.  Information and registration forms will be mailed to most schools in Maryland, and the same information and an online registration form will be on the website when it is back up.  The website will also have old exams and solutions, results, photos, and other information.  In the meantime, please email mathclub@jhu.edu with any questions.

Thanks and we hope to see you there.

Regards,
JHU Math Club

Why are you having it in November instead of April this year?  Less chance of the delightful weather conditions we had last year? :lol:

Yes, everyone should do JHMT! My team didn't have enough competition last year. :P

JHMT has been going on for a number of years but the group of students running it changes every few years (as students graduate) and so the style/publicity of the tournament varies a lot.  I think this is the 7th or 8th year overall though.

The Spring semester seems to have a lot more time constraints; last year's JHMT was scheduled right in Baltimore County's spring break I think so not many schools came from there.  Plus students on campus are very busy during that time of year.

oh man, i wish i could get a team together especially since I'm a fan of JHU (I'll be applying to JHU)  :(

If you live in states around Maryland, you should definitely try to come. If you don't have a team, contact the director and see if you can get paired up or something. I have a feeling that this year will be tougher than last year though.

As a follow up on Silverfalcon's comment.

We want both to challenge experienced students and encourage students who have never done contest math before.  Baltimore County is only now getting serious about ARML and many of Baltimore's students have never been exposed to contest math.  JHMT was originally started to spark an interest in contest math in Baltimore County, but is clearly drawing contestants from other areas as well (in fact, a mistake in scheduling ensured minimal Baltimore County participation last time).

Given the large range of skill we come across each year, it is difficult to make one round that does justice to all our aims; a round made less challenging to encourage new students will be pointlessly easy for experienced solvers, and of course a round too difficult will fluster newer students and may even discourage their future participation.

If the tournament reaches a certain size, we may create two divisions.  For now, we'll keep doing our best :).


function
calculus
derivative
real analysis
real analysis unsolved

[quote="countdownmath"]If $f$ is differentiable on $(a, b)$ and $f'(x)\neq 0$ for all $x$ in the interval, prove that either $f'(x)>0$ or $f'(x)<0$ on the entire interval.[/quote]

Function $f'$ has the Darboux propetry. That means that the range of $f'$ is also an interval. If $f'$ changes semn, then it must also take the value $0$ , which contradicts the hypothesis. So $f'$ is always positive on this interval , or always negative.

 [u]Babis[/u]

we proove that

there exist x such that f(x)>=0 and there exist x such that f(x)<=0
 ======> there exist x such that f'(x)=0

we use simply the Intermediate value theorem and eureka we proove the ahead result

then
there isn't x such that f'(x)=0
====> for all x  f(x)>=0 or f(x)<=0

and we take f'(x) in the place of f(x) (f' is also continue)

regards

@goldbach: Nope, it's not THAT EASY. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=derivative+value+intermediate&t=85658]this topic[/url] for a proof.

[quote="stergiu"]Function $f'$ has the Darboux propetry. That means that the range of $f'$ is also an interval. If $f'$ changes [b][color=red]semn[/color][/b], then it must also take the value $0$ , which contradicts the hypothesis. So $f'$ is always positive on this interval , or always negative.[/quote]
$f$ changes "semn"? What the? Are you Romanian? I thought you were Greek.


ARML function algebra polynomial geometry trigonometry probability

Since ARML is coming up June 3rd and 4th this year it would probably be a good idea to start practicing.  ARML is predominantly a team event.  So if you really want to earn prestige at ARML you will not only have to practice yourself but get your teammates into the act.

towersfreak2006 and I will be maintaining this thread by trying to post a problem a day.  We will probably post various types of problems, i.e. individual, team, relays, etc.  As for Power Rounds I'm not sure how that would go.  We would definitely have to break that up into at least a few days if we did do it.

Also, I will have a poll running for a week.  Vote for which format you want it to be.  We can either have it where you PM us with an answer or where the problems are just freely discussed.  The first couple you will PM us with the answers until we know what the general consesus of how you want this to be conducted is.  [i]Finally, please PM the person who posts the problem with the answer.[/i]  So, for example, today I am posting a problem so send me the answer, not towersfreak2006.

[size=134][b]Day 1[/b][/size]

Let $Q(x)$ be the quotient when $37x^{73}-73x^{37}+36$ is divided by $x-1$.  Compute the sum of the coefficients of $Q(x)$.

Joseph, stop the thread until someone (admin or ARML authority) gives you permission to use ARML problems.

Ok, I forgot to make the poll anyways, lol.  But I wont post any more problems until I hear some word.

I solved this first problem should I send it to you.  Also, do you want explanations or only answers.

Explanations would make it easier to show a number of methods for a particular problem. Also, Joseph, if you wouldn't mind, could you post the source of the problem? Thanks.

[quote="joml88"]Let $Q(x)$ be the quotient when $37x^{73}-73x^{37}+36$ is divided by $x-1$.  Compute the sum of the coefficients of $Q(x)$.[/quote]

With all the zeros, you multiply $37$ by $(73-37+1)$ then you add the product of $(37-73)$ with $(37-1+1)$, the sum of the two additive inverses cancel out.

Maybe I'm not reading it correctly...

I kind of brute forced it...
[hide]
I set up a synthetic division table, although just counted how many zeroes there would be instead of writing all them out (that would be rough).  I counted (although I wouldn't doubt that I miscounted...) 35 zeroes between the $x^{73}$ term and the $x^{37}$ term, thus there are 36 terms with coefficent 37.  Between the $x^{37}$ term and the $x^0$ there are 36 terms, so there are 37 terms with coefficient 110.  the remainder is 146, but that doesn't matter.  I multiplied and added to find the solution: $37 * 36 + 110 * 37 = 5402$
Of course, I haven't been doing so well the past few days, and functions certainly aren't my strong suit, so it is very possibly wrong.
[/hide]

Ok, I think it's ok to do this just as long as we don't post a whole test at a time (which we won't).  So people who got points for the first question are Rep123Max and miraculouspostmaster.  I think it would be more fun if the problems were worth a different number of points based on the difficulty on a scale of one to ten.  I will let the last problem be a 4.  Also, I think that for the top scorer (I'll keep a tally for every day) who goes to Penn State I'll offer to buy them something around the town.  So there is some motivation for ya!  It will be something up to $\$ 10 like an icecream or whatever you want and can find aroudn town.
Ok, so next problem:
[size=150][color=blue][b]Day 2[/b][/color][/size]
[u]Point Value: 6[/u]
[i]Calculator use: YES[/i]
Let $P(x)$ be a polynomial whose degree is 1996.  If $P(n)=\frac 1n$ for $n=1,2,3,\ldots,1997$, compute the value of $P(1998)$.

The format of the contest is you PM me or towersfreak2006 (but today PM me)  the answer and then, after the answer is revealed (by towersfeak or I), you can discuss the problem openly.  So please do not discuss it yet.

do you have the older problem books (before 1989).  if you do, can you post some of those?  thanks

I know towersfreak has some older than 1995 but I'm not sure how old...

Sorry. I only have the ARML-NYSML contests book from 1989 to 1994.

[size=150][color=cyan]Day 3[/color][/size]
And good for 2 (whole) points. :D

No calculators for this one.

If $P(x)$ is a polynomial in $x$, and $x^{23}+23x^{17}-18x^{16}-24x^{15}+108x^{14}=(x^4-3x^2-2x+9)\cdot P(x)$ for all values of $x$, compute the sum of the coefficients of $P(x)$.

[size=150][color=black]Day 4[/color][/size]
Point value: 6

No calculators!  :P

The number $85^9-21^9+6^9$ is divisible by an integer between $2,000$ and $3,000$. Compute that integer.

[size=150][color=darkblue]Day 5[/color][/size]

3 points this time, and still no calculator...but at least we get a break from too much more algebra. :)

You have 121 marbles, some of which are red, some white, and the rest blue. You also have 10 jars. If all the marbles are distributed into the jars, there [i]must[/i] be a jar with at least $n$ marbles of the same color. Compute the maximum possible value for $n$.

:lol: it's nice and it has 7 points also. For me is kind of very hard :)

*blink blink.* huh??? i'm kind of going in a big loop, and my result is proposterous. . . or is it? i can't tell. . .

:what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :what?:  :ewpu:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:  :fool:

i wish it were (8/5)*f(3/8) inside the f inverse.  that would make it so much easier. . .

[size=150][color=darkred]Day 45[/color][/size]

No calculators, 5 points

Silas once had between 30 and 35 special coins, each marked with a different "value". These values were successive powers of 2, namely: 1, 2, 4, 8,... . Some of these coins were then lost. The total value of the lost coins was exactly 1/5 of the total value of all the original coins. Compute the number of lost coins.

We can keep working on Day 44 until the answers are posted, right? I probably still won't get it, but just in case I see something...

Yes, you can always work on a problem until the solutions are posted. And the day 44 problem was really tough, as only three people got the right answer. I also stole it from joml88's book :) and the recent years seem to be more tricky.

[size=150][color=indigo]Day 46[/color][/size]

No calculators, 6 points

Point $P$ is inside a rectangular box. The distances from point $P$ to four of the vertices of the box are $1, 2, 3,$ and $4$. If the distance from $P$ to another vertex is greater than $5$, find that distance.

Well... I found an answer that I'm 99% sure works, but I don't know how to explain it, since I kind of guessed-and-tested. Should I still try to show my work, or can I just send you my answer?

[size=150][color=violet]Day 47[/color][/size]

No calculators, 6 points

There are several values for a prime, $p$, with the property that any 5-digit multiple of $p$ remains a multiple of $p$ under "cyclic permutation". One such value is $41$ (for example, since $50635$ is a multiple of $41$, so are $55063$, $35506$, $63550$, and $[0]6355$); another such value of $p$ is $3$. Compute the value of $p$ that is greater than $41$.

There will be two more problems posted before I leave for ARML.  Try to get the answers in before Thursday night since I leave for ARML very, very early Friday morning.  I'm posting Day 48 now and I will post Day 49 a little bit later today.

[size=150][color=green]Day 48[/color][/size]

[i]Calculator Allowed[/i]
[i]5 points[/i]

Let $2^x3^y=\left(24^{\frac 12+\frac 13 +\cdots+ \frac{1}{60}}\right)^1\cdot\left(24^{\frac 13+\frac14+\cdots+\frac{1}{60}}\right)^2\cdot\left(24^{\frac 14+\frac 15+\cdots+\frac{1}{60}}\right)^3\cdot \ldots \cdot\left(24^{\frac{1}{60}}\right)^{59}.$  Compute the value of $x+y$.

I made a typo for Day 48...you [color=red]can[/color] use a calculator.

[size=150][color=blue]Day 49[/color][/size]

[i]No Calculuator[/i]
[i]4 points[/i]

Compute the largest factor of $1001001001$ that is less than $10,000.$

[quote="mkkool64"]Hey thanks for posting all of these problems. I think they helped me out alot.  I only hope doing them will help me at ARML.  :P[/quote]

You're welcome!  It was probably just as big a help for Samson (towerfreak2006) and I as it was for you.

Also, I just thought I would alert anyone of the threads going on in the Other US Contests & Programs forums (if you haven't seen them already).  Basically you just post and say if you are going to attend a certain site for ARML and you can coordinate meeting other AoPSers in the threads.  Here are the links:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34540]Penn State ARML[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=34957]Iowa ARML[/url]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35105]San Jose ARML[/url]

Well, good luck to everyone!  :thumbup:

Hey can we do something like this again? ARML is coming up :!:


How many factorials are palindromes smaller than 10^100?

Have fun! :D  :rotfl:

[quote="cgyao15"]I believe that all factorials are not palindromes, so the answer to your problem is none.[/quote]
EXACTLY! :D

$ 0! \equal{} 1$
$ 1! \equal{} 1$
$ 2! \equal{} 2$
$ 3! \equal{} 6$

Those are all palindromes
so there are 4?

[quote="funtwo"]trivial. factorials greater than or equal to 5! end in 0. the rest can easily be checked[/quote]

Ah... blonde moment.  I was ignoring all the zeroes haha.

rubik is right!!! one digit numbers ARE palindromes, so therefore there are four!  :o you even had me convinced there were none!! :o

err, yeah. i didnt realize that until a week ago
:noo:

Well I kind of guessed zero, so I think the answer is zero.

hmm...
i think we already kinda proved that there are 4?? :o

Yeah. I think there was another thread about this.  :maybe:

[quote]I don't even know what Moems is....
[/quote]
MOEMS stands for math olympiad for elementry and middle school

WTH don't post replies when it is not pertaining to the subject. @ the guy who created this thread this, how is this impossible? :huh:

You realize that you are reviving a month-old thread to make a useless comment.  Thus, your post was more spam than the other posts here.  Moderator(s), please lock.


Since $ \deg f \equal{} 5$, $ \Delta^5f$ must be constant.

Constructing a chart, we see that $ f(6) \equal{}44\implies\boxed D$.

[geogebra]da02a63aae30ac8a5c949e8bd24949779cd678fc[/geogebra]

[hide="Brute force"]Let $ f(x)\equal{}a_5x^5\plus{}a_4x^4\plus{}a_3x^3\plus{}a_2x^2\plus{}a_1x\plus{}a_0$.
Immediately, $ a_0\equal{}3$ because $ f(0)\equal{}3$.

Using that information and the rest of the given values, we get this system:

$ a_5\plus{}a_4\plus{}a_3\plus{}a_2\plus{}a_1\equal{}2$
$ 32a_5\plus{}16a_4\plus{}8a_3\plus{}4a_2\plus{}2a_1\equal{}3$
$ 243a_5\plus{}81a_4\plus{}27a_3\plus{}9a_2\plus{}3a_1\equal{}7$
$ 1024a_5\plus{}256a_4\plus{}64a_3\plus{}16a_2\plus{}4a_1\equal{}8$
$ 3125a_5\plus{}625a_4\plus{}125a_3\plus{}25a_2\plus{}5a_1\equal{}9$

And then doing ridiculous amounts of annoying system solving yields:
$ a_5 \equal{} 19/120$
$ a_4 \equal{} \minus{}2$
$ a_3 \equal{} 209/24$
$ a_2 \equal{} \minus{}15$
$ a_1 \equal{} 152/15$
$ a_0 \equal{} 3$

So that $ f(6)\equal{}44\implies\boxed{\text{D}}$[/hide]

[quote="math154"]Since $ \deg f \equal{} 5$, $ \Delta^kf$ must be constant.[/quote]

What does $ \Delta^5f$ mean?    :)


Find the number of integer solutions of x^4 = y^2+z^2+4.

[hide="Solution"]If $ x$ is even, then we let $ x\equal{}2k$ and we have $ 16k^4\minus{}4\equal{}y^2\plus{}z^2$ so $ y$ and $ z$ are both even. Then we have $ 4k^4\minus{}1\equal{}y_1^2\plus{}z_1^2$ where $ y\equal{}2y_1$ and $ z\equal{}2z_1$. But the sum of two squares cannot be $ \equiv{\minus{}1} \pmod{4}$ so there are no solutions if $ x$ is even.

If $ x$ is odd, we write $ y^2\plus{}z^2\equal{}(x^2\plus{}2)(x^2\minus{}2)$. Because $ x^2\plus{}2\equiv{3} \pmod{4}$, there must exist at least one prime factor of $ x^2\plus{}2$ that is in the form $ 4k\plus{}3$ AND appears an odd number of times in its prime factorization. But it is well-known that all numbers representable as the sum of two squares cannot have a prime  of the form $ 4k\plus{}3$ raised to an odd power in their prime factorizations. Thus, there are no solutions in this case either.   [/hide]

[quote="RunpengFAILS"]But it is well-known that all numbers representable as the sum of two squares cannot have a prime  of the form $ 4k \plus{} 3$ raised to an odd power in their prime factorizations. [/quote]

Can someone provide a proof of this?

This is a good book for learning Number Theory: [url=http://books.google.com/books?id=uk2U2_HDkkoC&pg=PA115&lpg=PA115&dq=Numbers+representable+as+the+sum+of+two+squares&source=bl&ots=YrrsgHPYOH&sig=DPaToSRheCEqyaRA2jBcNd4kJGs&hl=en&ei=FAJ-SsXJJo7mM-HNgOMC&sa=X&oi=book_result&ct=result&resnum=4#v=onepage&q=&f=false]You can read the entire book online[/url].

Go to page 116 for the proof. (It is beyond this forum though)


The problem, due to computer delay, is posted on 7:10. So, the test ends in 8:25, not 8:15.

Sorry for the delay!

In number 8, I assume you meant cube-free?

PS: Ugh, got home late!

Yes. That should be cube-free (old habit of saying "square free...").