Hello! :)

What you think about Kosovo?

Thank you!

I think it's a little late to start talking about Kosovo, because they declared independence on February 17.

I do not believe that the independence of Kosovo should have been backed up by the US and many EU countries. First of all, Kosovo will not be able to survive well economically. The only industry Kosovo has is a small coal industry. As a result, it will rely on EU economic support. The US and EU do not need another impoverished nation to support.

Also, the issue of Kosovo independence has divided the EU and all western nations. Spain, for example, refuses to recognize Kosovo and Canada considered the issue for at least a month or two before recognizing Kosovo. Most notibly, however, Russia has stood firmly against Kosovo's independence. Russia under Putin has vetoed Kosovo's independence in the UN Security Council. Kosovo, as a result, has not been admitted into the UN, meaning that it isn't really a part of the international community (at least according to Putin). Putin, as a result, has accused the US and EU of recognizing Kosovo illegitimately. Putin therefore decided that Russia should take a more belligerent attitude towards breakaway areas in Georgia (Abkhazia and Southern Ossetia). Putin has given Russian passports and citizenship to most citizens in these breakaway regions. Putin has acted against the will of the UN, for the UN has not recognized the independence of these breakaway regions of Georgia, just like the US and many EU countries have done according to Putin.

The last thing the EU and US needs is a loss of unity. Recognizing Kosovo led to problems with unity amongst EU members. At least Kosovo has not yet been pillaged by violence like it was under Milosevic in 1998 and 1999.

EDIT: Thank you very much for bringing up this topic. I have been anxiously waiting to discuss this, as I am very interested in this issue.

Thank You!  :)

Abkhazia and Southern Ossetia have nothing similar situation with Kosovo. Abkhazia and S. Ossetia were mostly with georgians, after war in year 1990 in this place mostly georgians are displaced in Georgia. In Kosovo now has 90% albanians and before war were also mostly albanians ( up to 80%).
In another way, is so stupid when you think why S. Ossetia (under Georgia) like independence, while North Ossetia (under Russia) does'nt like.  :roll:

[quote="Fitim"]

Abkhazia and S. Ossetia were mostly with georgians  :roll:[/quote]

Actually, I have heard that Abkhazia especially has a large Russian minority, like how Kosovo has a large Albanian population. Although this is a completely different situation from Kosovo, Putin has excited tensions in the area for the first time since 1990 but shooting down a Georgian drone with a MiG fighter plane (which has now been confirmed by a UN investigation). In fact, Russia and Georgia have gone to the brink of war over Abkhazia and Southern Ossetia in the past few months.

[quote="knexpert"][quote="Fitim"]

Abkhazia and S. Ossetia were mostly with georgians  :roll:[/quote]

Actually, I have heard that Abkhazia especially has a large Russian minority, like how Kosovo has a large Albanian population.[/quote]

This is not true. In Abkhazia live mostly abkhaz and in Ossetia mostly are ossetians (60% osetians, other are georgians and russian). But before war in year 1998 in Abkazia over 60% were georgians. In Kosovo all time majority were albanian and genocide was over albanians.

[quote="knexpert"]In fact, Russia and Georgia have gone to the brink of war over Abkhazia and Southern Ossetia in the past few months.[/quote]

I hope this is does'nt true!  :(

Well this could become an interesting topic, but also a sad one, since Mathlinkers will probably have fiercely opposed views about this. :maybe:

All wish to cast a stone in matters like these.  The problem is that we (the West) have been fed one-sided propaganda ever since the start of the desintegration of Yugoslavia.  Serbs were genocidal, fascist, almost genetically communist, imperialist,....
I don't remember many attempts to understand all sides in our media.

Kosovo has now claimed independence...but what exactly is independence?Is it this ?

[img]http://newsimg.bbc.co.uk/media/images/44432000/jpg/_44432235_kosovoflagboys_getty416.jpg[/img]

What I'm trying to say is : independence is not a boolean.  It's not just true or false.  Kosovo is now a landlocked, poor tiny country, and apart from Albania, its neighbours are not so happy about that.  That means that there is a huge gap to be filled by eager superpowers, not in the least the United States.  In any case, I find it hard to believe that the United States sponsored this secession because of deep religious, historic and linguistic ties with the ethnic Albanians. :huh:


modular arithmetic
number theory unsolved
number theory

Determine all pairs (a,b) of natural numbers which are not divisible by 3 and whose square sum cannot be represented as square sum of three positive integers.

I hope it's correct, but I'm not sure...

We have to find those $a, b$ such that $a^{2}+b^{2}\neq x^{2}+y^{2}+z^{2}$ for all integers $x, y, z$.
Let $n = 4^{k}u$ with $4 \nmid u$. Then $n$ cannot be written as sum of three squares iff $u \equiv 7 \pmod 8$.
Also $a^{2}+b^{2}= n = 4^{k}u \equiv 2 \pmod 3$, from which $u \equiv 2 \pmod 3$. Hence by CRT $u \equiv 23 \pmod{24}$.
But $n$ has to be written as a sum of two squares. So all prime factors $p \equiv 3 \pmod 4$ occur an even number of times in the factorisation of $n$. So $u = p^{\alpha_{1}}_{1}p^{\alpha_{2}}_{2}\cdot\cdot\cdot p^{\alpha_{n}}_{n}\equiv-1 \pmod 4$. Hence there is at least one prime $p_{i}$ such that $\alpha_{i}$ is odd and so there are no such $a, b$.


what makes a good mathematicians ?

i know this question has been asked many many times in this forum.
nevertheless one more time
well i don't want to be a mathematicians.
as i not good in maths.this question sparked in my mind :lol:
but i have  to learn this   brain killing subject for my higher educations.
i hope you can give some good tips.

Well, I guess you have to be like naturally smart and curious, like Albert Einstein. His theories came out of nowhere but his brain, and he was born curious and math loving, but hated every other subject except physics

You have to really enjoy and be good at math.  There is no point being a mathmatician if you don't like it.

[quote="thatin"]genius is 30% inspiration and 70% perspiration... so I think we shall include hard work in the recipe[/quote]

Does the quote really go like that... I though Thomas Edison said something 1% inspiration and 99% perperation

[quote="tspa"]Albert Einstein.  was born...math loving, but hated every other subject except physics[/quote]
I thought Einstein was horrible at math and hated it.  He did have to learn Riemannian geometry for his general theory of relativity, but I don't remember reading/hearing that he [i]liked[/i] it.  There was a quote I had on my signature before, "Since the mathematicians have invaded the theory of relativity, I do not understand it myself any more."  [url]http://www.hal-pc.org/~wtb/einstein&mathematics.html[/url] is the wesite I found looking up that quote, and there are more useful quotables about Einstein and math.

No, I read his biography. He hated school. The only subject he liked was math, and looked forward to geometry

Ansering the original question, you need CURIOUSITY, and, of corse, intelligence (you don't need to be a genius however). Another important thing is that you have to love it, and with these things you can start

genius ??? for me that is even hard to pronounce.
i am hard working as i haven't  studied maths for 2 years, i rear remember some of the basics.
i am just searching for the website or books  to find out something.

one more tip if you can give me , is better to stick with websites rather than mathematic's books?

I'd go with both. As some of you may have read in several of my previous posts, i have been teaching myself some basic calculus before i take the class. I have done this ever since post-geometry, meaning that the summer after i took geometry, i started teaching myself Algebra 2, and then the next summer i started teaching myself precalc etc. Actually, i have been easing my way slowly into calculus ever since i was in geometry. Yeah i am a nerd like that. I have found that for many concepts, websites and flash videos explain things WAY better than any textbook. For example, back in my early days of introducing myself to calculus, i came across this site:

[url]http://www.calculus-help.com/funstuff/phobe.html[/url]

It explains the concept of a limit and the limit definition of the derivative WAY better than ANY other website or textbook i have seen. However, that may be due to the fact this site provides audio, visual, and verbal methods of teaching simultaneously. I learn math very well from flash videos


hi folks.
just want to hear out your opinions / sugestions about this particular issue. the beginners. those students / people that are in early stages of math olympiad problem solving ... do they need a special category ? are there any of you on the forum ? (you can vote, nobody will know what you voted :D) ...

you also don't know what one votes???  :D

yes I also don't know :) but it is possible to find out, but I would be wasting a loooot of time to do that :)


At a party, each man danced with exactly three women and each women danced with exactly two men.  Twelve men attended the party.  How many women attended the party?

(A) 8        (B) 12          (C) 16         (D) 18        (E) 24

[hide]
Clearly there were $12 \cdot 3 =36$ dances.
So if each woman danced with 2 men, then there are $36/2=18$ women.
$\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{\boxed{D}}}}}}}}}$
[/hide]

[hide]Since each women danced with 2 men, 6 would be the number of groups of 3 women (12/2)
Thus the number of women would be 18 (3*6),
(D)[/hide]


algebra
polynomial
quadratics
complex numbers
algebra proposed

Scientist have succeeded to find new numbers between real numbers with strong microscopes. Now real numbers are extended in a new larger system we have an order on it (which if induces normal order on $ \mathbb R$), and also 4 operations addition, multiplication,... and these operation have all properties the same as $ \mathbb R$.
[img]http://i14.tinypic.com/4tk6mnr.png[/img]
a) Prove that in this larger system there is a number which is smaller than each positive integer and is larger than zero.
b) Prove that none of these numbers are root of a polynomial in $ \mathbb R[x]$.

Let $ \alpha$ be a new number. If $ \alpha<0$ one can consider $ \minus{}\alpha$ which is greater than $ 0$. So assume that $ \alpha>0$.
If for each positive real number $ r$, $ \alpha>r$ then, $ \frac 1{\alpha}$ is the number we are after, because for each positive number $ r$, $ \alpha>\frac 1{r}$ which gives $ \frac 1{\alpha}<r$. But $ \alpha>0$ which shows that $ \frac 1{\alpha}>0$.
So assume that there is a positive real number $ R$ such that $ \alpha<R$. Let $ r$ be the superimum of all the real numbers less than $ \alpha$. (This exists because $ \alpha<R$ and $ 0$ is less than $ \alpha$)
Let $ \epsilon\equal{}\alpha\minus{}r$. Then $ \epsilon\neq0$ because $ \alpha$ is not a real number.
If $ \epsilon>0$, there is no positive real number $ a$, less than $ \epsilon$. Because if $ a$ is such a number, $ r\plus{}a<\alpha$ which contradicts $ r$ being the superimum.
But if $ \epsilon<0$ then there is no negative real number $ a$ greater than $ \epsilon$. Because if $ a$ is such a number, then $ \alpha<r\plus{}a$ which shows that the superimum should have been less than or equal to $ r\plus{}a$. So in this case, $ \minus{}\epsilon$ is the required number.
For the second part of the problem, assume $ P$ is a polynomial with real coefficients. $ P$ can be factored into linear and quadratic terms, each with real coefficients. So for $ \alpha$ to be a root of this polynomial, it has to be a root of one of the terms. (because the new system of numbers is a field) If $ ax\plus{}b$ is a linear term ($ a\neq 0$),	then $ a\alpha\plus{}b\equal{}0$ gives $ \alpha\equal{}\frac{\minus{}b}{a}$ which shows that $ \alpha$ is a real number.
If $ ax^2\plus{}bx\plus{}c$ is a quadratic term, ($ a\neq 0$ and $ b^2<4ac$ so that this term cannot be factored into linear ones) then $ a\alpha^2\plus{}b\alpha\plus{}c\equal{}a\left((\alpha\plus{}\frac{b}{2a})^2\plus{}\frac{4ac\minus{}b^2}{4a^2}\right)$
But each term of the last expression is positive, because if $ c$ is a number in the new system, either $ c\leq 0$ which shows that $ c\times c\geq0\times 0\equal{}0$, or $ c\geq 0$ which shows that $ c\times c\geq0\times0\equal{}0$. So $ c^2\geq 0$. So $ (\alpha\plus{}\frac{b}{2a})^2\geq0$. And $ 4ac\minus{}b^2>0$ which gives $ \frac{4ac\minus{}b^2}{4a^2}>0$.
Hence $ \alpha$ can not be a root of this term, which shows that $ \alpha$ is not a root of the original polynomial.

[quote="Omid Hatami"]$ P$ can be factored into linear and quadratic terms, each with real coefficients. So for $ \alpha$ to be a root of this polynomial, it has to be a root of one of the terms. (because the new system of numbers is a field) [/quote]

Sorry, but how did deduce that $ P$ can be factored into such terms?

It is a classical result of Fundamental theorem of algebra. $ P$ can be factored to linear terms in $ \mathbb C[x]$ and for each non-real complex root $ \alpha$ of $ P$, $ \bar\alpha$ is also another root. So it can be easily deduced that $ P$ can be factored into linear and quadratic terms, each with real coefficients.

Additional problem:
c) Construct such numbers.

A more abstract proof for b):

For a field to be able to carry an ordering (coherent with the field's operations) is equivalent for $ \minus{}1$ not being a sum of squares. Thus in total, an ordered field cannot have a square root of $ \minus{}1$.
As $ \mathbb C$ is an algebraic extension of degree $ 2$ of $ \mathbb R$, we just need to show that $ \minus{}1$ doesn't have a square root in this new numbers, so we are done.

part b) is just the fundamental theorem of Algebra, since the polynomial only has roots in C[x], and the only non-complex numbers in this set are the real numbers, so you're done...

You have to show that there are no complex numbers there  :wink:
(Not that this is really difficult...)

Like since no complex numbers are between any 2 real numbers, they cannot be complex?

A more mathematical approach (reasoning from given properties) would be usefull.
There is a square root of $ \minus{}1$ in appropriate completions of $ \mathbb Q$, thus there are complex numbers even between rational ones ;)

OK, but proving that $ F$ can be ordered if $ \minus{}1$ is not square is not trivial.

If there is a non-real root of a polynomial, it must have the form a+bi, with $ b\ne 0$. Let this number be right after the real number c.

Then 2(a+bi) will be right after 2c, since otherwise if there was another number in between, then there would be a number between a+bi and c. Similarly, a+c+bi will be right after c, since otherwise if there was a number in between, then there would have been a number between a+bi and c.

Therefore, 2a+bi=a+c+bi so b=0, contradiction!

[quote="Omid Hatami"]OK, but proving that $ F$ can be ordered if $ \minus{} 1$ is not square is not trivial.[/quote]
You mean: not sum of squares.
But here you just need the the trivial direction (ordered => $ \minus{} 1 < 0$ => not a square).

[quote="rofler"]Let this number be right after the real number c.[/quote]
There is no reason why there should be a "next" number. You didn't even use any property of complex numbers.


Let non emptysets $ A, B \subset M$, with $ M$ compact, and $ A\cap B\equal{}\emptyset$

prove that if  $ d(A,B)\equal{}0$, then $ \exists \, p\in (\partial A\cap \partial B)$

Lemma :If $ A$ is compact and $ B$ is closed and $ d(A,B)\equal{}0$ then $ A\cap B \neq\emptyset$

the proof is very simple.

solution of the problem :

$ d(\bar A,\bar B)\leq d(A,B)$ and  $ d(A,B)\equal{}0$ $ \Rightarrow$  $ d(\bar A,\bar B)\equal{}0$

since $ d(\bar A,\bar B)\equal{}0$ and $ \bar A$ , $ \bar B$ are compact $ \Rightarrow$  $ \bar A\cap\bar B \neq\emptyset$

  $ A\cap B\equal{}\emptyset$  $ \Rightarrow$ $ \bar A\cap\bar B  \subset  (\partial A\cap \partial B)$ $ \Rightarrow$ $ (\partial A\cap \partial B)\neq\emptyset$

 :)

------------------------------------------------------------------------------------------------------------
The incidents and difficulties, which come into human life can, measure and find out the extent of his energy and faith.

[quote="jorgeston"][quote="hamidreza"]

  $ A\cap B \equal{} \emptyset$  $ \Rightarrow$ $ \bar A\cap\bar B \subset (\partial A\cap \partial B)$ $ \Rightarrow$ $ (\partial A\cap \partial B)\neq\emptyset$

[/quote]

i don' understand this...[/quote]

$ A\cap B \equal{} \emptyset$ $ \Rightarrow$ $ A \subset B^c$ and $ B\subset A^c$ $ \Rightarrow$

$ (\partial A\cap \partial B) \equal{} (\bar A\cap\bar {A^c})\cap (\bar B\cap\bar {B^c}) \equal{} \bar A\cap\bar B \neq\emptyset$


Fill the squares of an $n\times n$ grid with the integers $\{1,2,\ldots,n^{2}\}$. Prove there exists two adjacent squares that differ by at least n/2. Also prove there exists two adjacent squares that differ by at least n (a stronger result).

BTW - Adjacent means two squares share an edge (not just a corner).

I havent really thought this through, but to reduce the sum of the differs you list the numbers up like; 1,2,3,4,5.. that gives the smallest sum of differs.
Youll find the biggest diff to be n and that occurs between all rows and between the columns you get 1. This should be the most even dist. of diffs, so you get some n:s and that settles it.

[quote="srulikbd"]1,2,3,4
5,6,7,8
9,10,11,12
13,14,15,16
where are the squares?[/quote]
The adjacent squares are not refering to perfect square numbers.


geometry
circumcircle
incenter
geometry proposed

Let ABC be a triangle that has midpoints L,M,N on BC,CA,AB respectively. Let O be a circumcircle of the ABC and D,E be a midpoint of the minor arc CA,AB respectively. O' is a circle that touches CA at M and O at D. Circle W is a external circle of the LMN for L. W touches MN, LM at X,Y respectively. Show that XY, DE, O' is meet at single point.

There may be a mistake on gramma. hh.. :)

$ BD, CE$ bisect $ \angle B, \angle C,$ they intersect at the incenter $ I.$ $ AI$ cuts $ (O)$ again at $ F$ and $ I$ is orthocenter of $ \triangle DEF.$ Therefore, $ DE$ cuts $ AI$ at its midpoint $ J,$ the incenter of $ \triangle AMN.$ Incircle $ (J)$ of this triangle is tangent to $ MN$ at $ X.$ Let $ (P)$ be circumcircle of $ \triangle AMN.$ $ AJ$ cuts $ (P)$ again at $ K,$ the circumcenter of $ \triangle JMN.$ $ DE$ is tangent to the circumcircle $ (K)$ of $ \triangle JMN$ at $ J.$ $ XY \parallel NJ,$ both perpendicular to the external bisector of the $ \angle LMN.$ Let $ XY$ cut $ DE$ at $ Q.$ $ \angle QXM \equal{} \angle YXM \equal{} \angle JNM \equal{} \angle QJM$ $ \Longrightarrow$ $ JXMQ$ is cyclic. Since the angle $ \angle JXM$ is right, the angle $ \angle JQM$ is also right and $ MQ \perp (QD \equiv JQ).$ As a result, $ Q$ is on a circle with diameter $ MD.$


You need to use text mode for the text :)
So the last 3 lines would be
[code]\put(10.7833,1.426){\circle*{0,1}} \put(0.5,-0.5){\color{green}1900}
\put(5.5,-0.5){\color{green}1950} \put(10.5,-0.5){\color{green}2000} \put(-1,2){\color{red}10\%}
\put(-1,4){\color{red}20\%} \put(-1,6){\color{red}30\%} [/code]
which also colours red the numbers on the y-axis and green on the x-axis (you will need \usepackage{color} in the preamble}.

Thanks Steve.

So you would suggest :

\put(x,y){\text{ Picture 1 }}  ?

In terms of coloring, I'm more interested how to color a whole region. Say I want to color a rectangular with a color yellow, how to do that?

Why not just use \put(x,y){Picture 1} or, if you want more than one line, use \parbox.
If you want a rectangular box with a coloured backgroud use \colorbox or \fcolorbox (this has a frame). If you want to colour an irregular shape then you'll need to use something like [url=http://www.tug.org/PSTricks/main.cgi?file=Examples/Colors/colors]PSTricks[/url].


MATHCOUNTS AMC AIME email ratio FTW blogs

These are the AoPSers I know from Alabama going to State on the 28th:

dynamo729
chestnut
weird888
ash-Pokemon
amath!
CubeX
mc??
yaofan

And possibly some high schoolers scoring?

That could possibly be a record high, go Alabama! If you guys know of any other Alabamians here on AoPS going to state, post here!

I think the scores will be a lot higher this year, what about you guys? i mean, I know of 12 chapter scores above 35!

farhan (last member on pizitz team) has one but idk on what his username is...

dynamo, which AIME are you taking?

Yah, when I fail I'll blame it on getting up so early.

The II, probably. Mostly because I havent been told to take it tomorrow  :)

does challenger come the day before or early in the morning

Ehhh people come when they want to. Last year it was Mobile so 6 hours long drive! We left Friday night but this year its the Ham so is only 1.5-2 hours so we're going early in the morning. One of our team members is riding with me and I'm pretty sure ash-Pokemon and our other member is riding that morning...

botong is luckyducky
farhan is furry____ but he is not furrykid

[quote="dynamo729"]

And possibly some high schoolers scoring?

[/quote]

dude i wish i could score like i did at states so that i could meet yall but

i doubt any (well, very few if any...not me in any case :( hm maybe i can get my parents to drop by hmmmm ) high schooler's gonna come since we have state written the same day (its like a math tournament, but its just called state written for some reason...nothing to do with mathcounts..

ash-Pokemon. I read too fast and buzzed in too fast and got it wrong like three times I've kinda given up practicing for cd, I'm focusing only on written bcuz that's what'll get me to nats (hopefully) and that will help challenger win 1st place! W00t!

berry will win first

(even though we suck)

ummm, guys guys guys
dont you thihnk the team if the kid who won first last year will win?
bcuz it will intimidate the teams that are good.

Nonsense, Gene. I have confidence in my own abilities as well as those of my team mates. I tied Tony at Nationals, too, and I did not have MIST nor AoPS books to help me. But now I do. Mwahahaha.

wow good job yaofan

THE FOLLOWING IS EDITED cuz i accidentally mentioned scores, which I doubt I can do right now:

however, i think putting a rank is ok

since my dad doesn't really know who is who, don't trust the following (so why am i wasting my time.. idk)
[hide="ranks"]
written:
1. tony (good job!) from Pizitz
2. Asutosh (?) (good job) from challenger
3.idk
4.idk (hmmm top four from four different schools)
5.Botong (good job) from Pizitz
6. idk
7. yaofan from Berry

others in top 12 written: (aww poor dynamo), nick, gene, Abdullah (hmmm i cant spell), other ppl my dad doesnt know

cd:
1. yaofan (nice! good job)
2. Nick (unless it is some other tall white person that my dad is referring to, not that I ever knew Nick was tall, cuz im just guessing who's who) from Challenger
3.whoever got 4th written
4. botong (good job.... too bad cd doesnt count :( )
5. idk
6.idk
7.idk
8.tony (well it doesnt matter so yay? wow im bipolar too)
other ppl

school
1. Pizitz
2. Challenger
3. "Auburn" (??) (hmmm maybe my dad meant Baldwin but idk)

again THE ABOVE IS UNOFFICIAL AND PROBABLY WRONG ... of course yaofan and weird and dynamo and chesnut can fix it[/hide]

i have no clue if cubeX did well cuz i dont know him and my dad obviously doesnt either, so no comments here

GOOD JOB AND GOOD LUCK to the team and team members.

@yaofan, yea my dad said he was the only chinese adult male helping out

since i cannot edit my last post so many times:

my dad was in the tabroom the whole time (except cd)...he didn't help distribute and collect the tests as I had thought he would do....

this also meant he got to grade all the papers from Challenger and Pizitz and Randolph

hmm i have to be careful to not discuss scores?

chestnut got 1st
some1 from auburn got 2nd
ash got 3rd
some1 from baldwin got 4th
botong got 5th
zaki got 6th
i got 7th
dynamo got 8th
some1 from auburn got 9th
weird888 got 10th
some1 from simmons got 11th
some1 from challenger got 12th

and hurdler

you're probably in pre-ap right?
so is it really hard as teachers say?

NO

Ignore what teachers say. Do Pre-AP, it's pretty much ridiculous if you don't do Pre-AP everything.

High school feels hard (at least in Hoover, idk elsewhere) for like the first 9 weeks of 9th grade, only since middle school is like ridiculously easy. Then you get used to it, and it feels just as easy as middle school.

Pre-AP is easy. AP is also pretty easy (once you get used to it).

i'm doing all pre-ap and all ap through high school except for math team

are u saying that just because ur a genius and it seems easy to u?

and does hoover have to read 2 assigned books over the summer?

hmmm i am not that smart, but I do work pretty hard. If you aren't that smart but you work hard, then high school is not hard, but you still hard to work hard (and this will vary from person to person)

hmmm something like that...you will probably need to read one or two books over the summer.

spain park is probably better than hoover and vestavia for that matter

who's ur piano teacher?

[quote="weird888"]connecticut had socres like 45 on average...
whos IHATEMATHALOT????[/quote] Us? 45? States?

[quote="lifesapain"]What are the complete top ten results? And top 4 team?[/quote]

Well, since we can discuss scores,

I think that chestnut got a 43 in first
Ash got a 35 (? maybe it was 34) in 2nd
someone got a 35 and tied ash
CubeX got...idk...34?

No clue what went on from there.

i got a 33...

1 point away from nats?!?!?!?

yea my dad was in the score room and scored the pizitz papers, so he told me about how 2nd through 12th were really really close.

apparently tiebreakers are whoever is the first to miss one going backwards (if both miss one then they look at next one backwards)...I never knew this during my mathcounts years.

i [b]REALLY, REALLY, REALLY, REALLY, REALLY, REALLY, REALLY, REALLY, REALLY ,REALLY, REALLY, REALLY[/b] hate myself

i could've made nats but the 1st question on the target round was unclear


If $ \overline{AD} \parallel \overline{FG}$, how many degrees are in angle $ EFG$?

[asy]import olympiad;
pair A,B,C,D,E,F,G;
E = (0,0);
A = dir(180); B = dir(150); C = dir(60); D = (.5,0);
F = dir(240); G = F + (1,0);
draw(A--D); draw(C--F--G); draw(B--E);
label("$A$",A,W); label("$B$",B,NW); label("$C$",C,N);
label("$D$",D,E); label("$E$",E,SE); label("$F$",F,W);
label("$G$",G,E);
draw(scale(2)*anglemark(C,E,B));
draw(scale(3)*anglemark(B,E,A));
draw(shift(F)*scale(2.5)*shift(-F)*anglemark(G,F,C));
label("$x$",(A+2E)/3,N);
label("$2x$", (C+3E)/4,W);
label("$1.5x$",(F+G)/2,NW);[/asy]

By the Parallel Postulate thingy (whatever it's called), we have $ 1.5x \equal{} m\angle EFG \equal{} m\angle CED$, and since $ m\angle ABE \plus{} m\angle BEC \plus{} m\angle CED \equal{} 180^\circ$, we have $ x \plus{} 2x \plus{} 1.5x \equal{} 4.5x \equal{} 3(1.5x) \equal{} 180^\circ\implies m\angle EFG \equal{} 1.5x \equal{} \boxed{60^\circ}$.


inequalities
trigonometry
inequalities unsolved

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=107262]IMO 2007 HKTST 1[/url]
Problem 3

Let $A$, $B$ and $C$ be real numbers such that
(i) $\sin A \cos B+|\cos A \sin B|=\sin A |\cos A|+|\sin B|\cos B$,
(ii) $\tan C$ and $\cot C$ are defined.
Find the minimum value of $(\tan C-\sin A)^{2}+(\cot C-\cos B)^{2}$.

$\sin(a+b)=\cos(a-b)=\sin(90-a+b)\implies 90-a+b=a+b \implies a=45$
anyway AM-GM: $(\tan C-\sin A)^{2}+(\cot C-\cos B)^{2}\geq 2(\tan C-\sin A)(\cot C-\cos B)=1-\cos B \tan C-\cot C \sin A+\sin A \cos B$
welll tired now but you can keep going

[quote="mathpk"]Isn't it problem 2? :P[/quote]
Yes, I think Problem 3 should be this one:
Let$ABC$ be a triangle with $AB=c$,$BC=a$ and $CA=b$. Show that
$\frac{\cos A}{a^{3}}+\frac{\cos B}{b^{3}}+\frac{\cos C}{c^{3}}\geq \frac{3}{2abc}$.
Using cosine law, the origin inequality become
$\sum_{cyc}\frac{b^{2}+c^{2}-a^{2}}{2a^{3}bc}\geq \frac{3}{2abc}$
$\sum_{cyc}\frac{b^{2}+c^{2}-a^{2}}{a^{2}}\geq 3$
$\sum_{sym}a^{4}b^{2}\geq 6a^{2}b^{2}c^{2}$
which is trivial by A.M-G.M.


function
calculus
integration
algebra
domain
complex analysis
complex analysis unsolved

Suppose I have a function of two complex variables $f(w,z)$.  It's beginning to occur to me that such a function could be singular along a contour $C_{1}$ in the w plane and/or singular along a contour $C_{2}$ in the z-plane.  So, is there a multiple-variable analog of Cauchy's Residue Theorem that would apply.  For example say:

$\oint f(w,z)dwdz=I(R)$

where the $I(R)$ is some operator akin to a multiple-variable residue of some sort along these contours?  Incidentally, I'd really like to be able to post that double integral with the circle around it even though I don't know exactly what that would mean.  It's the "oiint" command that apparently does not work.

So, that's the proposed unsolved problem then:  If there isn't one, which I bet a dollar there is, then come up with one. :)

I'm not sure what you're visualizing as a "contour" in $\mathbb{C}^{2}.$ If you're going to do an integral $dw\,dz$ then you need an object to integrate on that is 2-dimensional (two real dimensions, that is.) At the same time, a "typical" singularity set (or zero set - it's essentially the same thing) for an analytic function of two complex variables will be locally on-dimensional most of the time. But that's one complex dimension, which is the same as two real dimensions. Furthermore, such a function cannot have an isolated singularity, or even a singularity set confined to some compact set. Singularity sets must extend outward to the boundaries of the domain of the function. You can't "enclose" a singularity.

You can take slices of various sorts along one-complex-dimensional sets and use the ordinary residue theorem on those. In that sense, there's some sort of answer to your question, but I don't think it's a particularly enlightening answer.

A word of warning about several complex variables: the subject is different. Not all of the theorems you're used to in one variable carry over, and some new and different properties appear.

Thanks Kent.  Been looking into it on the web and found  the residue formula corresponding to residue forms as a generalization of the concept of a residue of an analytic function of one complex variable to several complex variables.

But again it involved concepts way beyond my level of familiarity with the subject; the only part of the expression I understand is the $2\pi i$ part.   :oops:  The idea looks interesting though.


Let ABC be a triangle the lengths of whose sides BC, CA, AB respectively, are denoted by a, b, c, respectively. Let the internal bisectors of the angles <BAC, <ABC, <BCA, respectively meet the sides BC, CA, AB respectively at D, E, F, respectively. Denote the lengths of the line segments AD, BE, CF, respectively, by d, e, f, respectively. Prove that:

def= [4abc(a+b+c)Δ]/[(a+b)(b+c)(c+a)]

where Δ stands for the area of the triangle ABC.

$AE=\frac{bc}{a+c}$ and $EC=\frac{ab}{a+c}$
so Stewart's Theorem gives:
$\begin{eqnarray*}\frac{abc^{2}}{a+c}+\frac{a^{2}bc}{a+c}&=&\left(e^{2}+\frac{ab^{2}c}{(a+c)^{2}}\right)b\\ \frac{ac^{2}+a^{2}c}{a+c}&=&e^{2}+\frac{ab^{2}c}{(a+c)^{2}}\\ ac&=&e^{2}+\frac{ab^{2}c}{(a+c)^{2}}\\ e^{2}&=&ac\left(\frac{(a+c-b)(a+c+b)}{(a+c)^{2}}\right)$
so similarly $d^{2}=bc\left(\frac{(b+c-a)(b+c+a)}{(b+c)^{2}}\right)$ and $f^{2}=ab\left(\frac{(a+b-c)(a+b+c)}{(a+b)^{2}}\right)$
multiply these together and you are done :huh:

$1\blacktriangleright$ $\frac{DB}{c}=\frac{DC}{b}=\frac{a}{b+c}$ $\Longrightarrow$ $\boxed{DB=\frac{ac}{b+c}}\end{array}$.

$2\blacktriangleright$ $\frac{d}{\sin B}=\frac{BD}{\sin\frac{A}{2}}$ $\Longrightarrow$ $\left\{\begin{array}{c}d=\frac{c\cdot a\sin B}{(b+c)\sin\frac{A}{2}}\\\\ a\sin B=b\sin A\end{array}\right\|$ $\Longrightarrow$ $\left\{\begin{array}{c}d=\frac{c\cdot b\sin A}{(b+c)\sin\frac{A}{2}}\\\\ \sin A=2\sin\frac{A}{2}\cos\frac{A}{2}\end{array}\right\|$ $\Longrightarrow$ ${\boxed{d=\frac{2bc}{b+c}\cdot\cos\frac{A}{2}}}$.

$3\blacktriangleright \left\{\begin{array}{c}\cos \frac{A}{2}=\sqrt{\frac{p(p-a)}{bc}}\mathrm{\ a.s.o.}\\\\ S=\sqrt{p(p-a)(p-b)(p-c)}\end{array}\right\|$ $\Longrightarrow$ $\boxed{\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}=\frac{pS}{abc}}$.

$4\blacktriangleright def=\frac{8(abc)^{2}\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}}{(b+c)(c+a)(a+b)}=\frac{8(abc)^{2}pS}{abc(b+c)(c+a)(a+b)}$ $\Longrightarrow$ $\boxed{\boxed{def=\frac{4abc(a+b+c)S}{(a+b)(b+c)(c+a)}}}$.

[b][u]Exercises :[/u][/b]

$1\blacktriangleright$ $\frac{2r}{R}\le\prod\sin\left(A+\frac{B}{2}\right)\le 1$.
$2\blacktriangleright$ $\left(\frac{2r}{R}\right)^{2}\le\prod (\cos A+\cos B)$.


[quote="chuyue"]If $a,b>0$  then  $(a^a b^b)^{\frac{2}{a+b}}\geq \frac{a^2+b^2}{2}$[/quote]

 ;)

[quote="chuyue"]If $a,b>0$  then  $(a^a b^b)^{\frac{2}{a+b}}\geq \frac{a^2+b^2}{2}$[/quote]
Beautiful inequality!
I look only $(a^ab^b)^{\frac{2}{a+b}}\geq(\frac{a+b}{2})^2$

Equality part:
[hide]Let's rearrange this equation.
$\frac{(a^2+b^2)^{a+b}}{2^{a+b}}\leq a^{2a}+b^{2b}$.
Next, set the powers of two on one side and the powers of a and b on the other.  Binomially expand $(a^2+b^2)^a and (a^2+b^2)^b$ and multiply the two together.  Then divide by $a^{2a}+b^{2b}$.
$\frac{({a\choose0}a^{2a}+{a\choose1}a^{2a-2}b^{2}+...+{a\choose a}b^{2a})({b\choose0}a^{2b}+{b\choose1}a^{2b-2}b^{2}+...+{b\choose b}b^{2b})}{a^{2a}*b^{2b}}\leq 2^a*2^b$.  Now we multiply the two terms in the numerator together and divide to get:
${(\frac{a^{2b}}{b^{2b}}+{b\choose 1}\frac{a^{2b-2}}{b^{2b-2}}+..+1)+({a\choose1}\frac{a^{2b-2}}{b^{2b-2}}+{a\choose1}{b\choose 1}\frac{a^{2b-4}}{b^{2b-4}}+..+{a\choose1}\frac{b^2}{a^2})+...+(a^{2b-2a}b^{2a-2b}+{b\choose 1}a^{2b-2a-2}b^{2a-2b+2}}+..+\frac{b^{2a}}{a^{2a}})\leq 2^a*2^b$.
Notice that if a=b, the first term is ${b\choose 0}+{b\choose 1}+...+{b\choose b}$.  That's $2^b$.  The second term is that multiplied by ${a\choose 1}$.  The next is that multiplied by ${a\choose 2}$.  The last obviously is multiplied by ${a\choose a}$ and the first is then multiplied by ${a\choose 0}$.  What we have is $2^b({a\choose 0}+{a\choose 1}+...{a\choose a})=2^b*2^a.$  Dividing out 2^b, we have ${a\choose 0}+{a\choose 1}+...{a\choose a}=2^a$, which is true.[/hide]

Actually, the equality statement could have been shown without that.
Setting a=b, we have $(a^a*a^a)^{\frac{2}{2a}}=a^2$.  The left term is $(a^{4a})^{\frac{1}{2a}}$ which of course equals $a^2$

[hide="Incomplete proof of inequality"]If $a=b$, equality holds as proven. If $a\neq b$, we can assume $a<b$ because the inequality is symmetrical in $a$ and $b$. Let's substitute $b=az, z>1$:

${({a^a}(az)^{az})}^{2 \over {a+az}}\geq{{a^2(1+z^2)} \over 2}$
${(a\cdot{a^z}{z^z})}^{2 \over {1+z}}\geq{{a^2(1+z^2)} \over 2}$
${(a^{1+z}{z^z})}^{2 \over {1+z}}\geq{{a^2(1+z^2)} \over 2}$
${a^2}{z^{2z \over {1+z}}\geq{{a^2(1+z^2)} \over 2}}$
${z^{2z \over {1+z}}\geq{{1+z^2} \over 2}}$

For $f(z)=z^{2z \over {1+z}}$ we can calculate $f(1)=1, f'(1)=1, f''(1)=1$, so taking first three terms of Taylor series in $z_0=1$, with the assumption that Taylor remainder is positive for $z>1$, we have:

${z^{2z \over {1+z}}>1+(z-1)+{1 \over 2}(z-1)^2={{1+z^2} \over 2}}$. All that remains is to prove that the Taylor remainder is positive for $z>1$, possibly by proving that $f^{(n)}(1)>0$ (it's easily proven that functions $z^{2z \over {1+z}}$ and $e^{z-1}$ are asymptotically close when $z\rightarrow 1$).[/hide]


[hide]
$(d-a)(\frac{a^{2}}{b-a}+\frac{b^{2}}{c-b}+\frac{c^{2}}{d-c})=(a+b+c)^{2}$
$((b-a)+(c-b)+(d-c))(\frac{a^{2}}{b-a}+\frac{b^{2}}{c-b}+\frac{c^{2}}{d-c})=(a+b+c)^{2}$
Cauchy's inequality says:
$((b-a)+(c-b)+(d-c))(\frac{a^{2}}{b-a}+\frac{b^{2}}{c-b}+\frac{c^{2}}{d-c})\ge (a+b+c)^{2}$
The equality condition of Cauchy is that
$\frac{\frac{a^{2}}{b-a}}{b-a}=\frac{\frac{b^{2}}{c-b}}{c-b}=\frac{\frac{c^{2}}{d-c}}{d-c}$
$\frac{a^{2}}{(b-a)^{2}}=\frac{b^{2}}{(c-b)^{2}}=\frac{c^{2}}{(d-c)^{2}}$
$\frac{a}{b-a}=\frac{b}{c-b}=\frac{c}{d-c}$

The first equality gives
$ac-ab=b^{2}-ab$
$ac=b^{2}$
$\frac{a}{b}=\frac{b}{c}$

Likewise the second equality gives
$\frac{b}{c}=\frac{c}{d}$

This shows that a,b,c,d are in geometric progression. Since $d=2a$, if $r$ is the common ratio then $r^{3}=2\implies r=\sqrt[3]{2}$

$\frac{bcd}{a^{3}}=\frac{(a\sqrt[3]{2})(a\sqrt[3]{4})(2a)}{a^{3}}=\frac{4a^{3}}{a^{3}}=\boxed{4}$
[/hide]

Good observation on the first step: $(d-a) = (d-c)+(c-b)+(b-a)$  :P  Nice one! I didn't "see" this during the actual contest.


graph theory
combinatorics unsolved
combinatorics

Twenty football teams take part in a tournament. On the first day all teams play one match. On the second day all teams play a further match. Prove that after the second day it is possible to select 10 teams, so that no two teams have yet played each other.

We have graph with 20 vertexes.The first vertex corresponds to the first  team,
  the second vertex corresponds to the second team, the third vertex corresponds
  to the third team,.... the 20th vertex corresponds to the 20th team.Between two
  vertexes there is an edge, if the two teams  that are presented by these vertexes
  have played a match.It is easy to see that this graph does not contain triangles.
  Also is easy to see that the degree of every vertex is 2.
  Now we have to prove that in this graph there is  SET of 10 vertexes in which there
  isn't edge, e.g. there isn't two vertexes in the set that are connected with edge.
  We divide the graph on its components of connectivity(i don know what is the
  english word for this... for example the graph v1-v2-v3-v1    v4-v5  there is
   2 components of connectivity - the first is the v1-v2-v3-v1 and the second
  v4-v5 ,e.g. a component of connectivity is subgraph that is connected and that
  is maximised).Let assume that the number of components of connectivity of our
  graph is k  and the number of vertexes of the components of connectivity
  are a1,a2,... ak  , 1≤k≤5.
  a1+...+ak=20.( the number of vertexes of the whole graph).If we look at one
  component of connectivity, we will see that it looks similarly like this:
   v1-v2-v3-...-vk-v1. From every component of connectivity we can choose [ak/2]
   vertexes for our SET - for example , from the component of connectivity v1-v2-v3-v4-v5-v1 we can choose the vertexes v1,v3,v5 . Here  [m] is the smallest interger
   that is equal or exceeds m - [3]=3, [2,5]=3.
   From every component of connectivity we choose [ak/2] vertexes.
   The number of vertexes that we choose at all is:
   [a1/2] +...+[ak/2] = [a1/2]+...+[a(k-1)/2] + [(20-a1-a2-...-a(k-1))/2] ≥10.

[quote]Problem is clearly wrong. Consider a graph which has 5 triangles and one pentagon (every -gon disconnected) - you can find at most 7 teams which haven't played each other yet (as you can take no more than 1 team from each triangle and no more than 2 teams from the pentagon).

Number of team has to be at least 31 I suppose. [/quote]

 Explain me pls why do u think that this graph can contain triangles? I  think this is not possible...

Of course you are right - I just forgot about it somehow while translating problem into graph theory language  :blush:. And you proof can be much shortened.

We can divide our graph (vertices - teams, edges - matches) - into $k$ disconnected cycles. It is enough to prove that there's no odd cycle, but that's clear because we can paint each cycle red-blue (red if match is on the first day, blue when on the second). Then from each vertex emanate one red and one blue edge - and it is not possible for odd number of vertices (as we have odd number of edges).  Hence we have only even cycles and from each we can choose half of vertices - hence the result.


Triangle ADE and Triangle BDC are triangles.  :ang: EAB and :ang: ABC are right angles, AB = 4, BC = 6, AE = 8, and segment AC and segment BE intersect at D. What is the difference between the areas of triangle ADE and triangle BDC?

A little help please... could someone suggest a theorem maybe?

Draw the picture and then look for parallel lines and similar triangles.

I have spent a while looking at it... I am well past that stage, I got nowhere... Geometry is a major weak point of mine.

Do you see the similar triangles?  Similar parts (like altitudes) of similar triangles are in ratio.  That's all you need.

This problem is all about efficient labeling.

Call the area of triangle ADE I, triangle ABD II, and triangle BCD III. Then ask yourself, what is the area of triangle ABE (both in numbers and in terms of the letters we just gave). Then do the same thing for triangle ABC.

Now you're looking for I - III. How can you get that?

holy crap that was easy... wow I feel stupid now...


limit
trigonometry
calculus
calculus computations

Calculate $ \lim_{n \to \infty}{sin^{2008}(\pi\sqrt{n^2\minus{}n})}$

hello, i think it doesn't exist a limit.
Sonnhard.

Oh, but the limit does exist after all.

The key: how close is $ \sqrt {n^2 - n}$ to an integer?

$ \sqrt {n^2 - n} = n\sqrt {1 - \frac1n} = n(1 - \frac1{2n} + O(n^{ - 2})) = n - \frac12 + O(n^{ - 1})$

So $ \sin^{2008}(\pi\sqrt {n^2 - n}) = \sin^{2008}(\pi n - \frac {\pi}2 + O(n^{ - 1})) = \sin^{2008}(\frac {\pi}2 + O(n^{ - 1})).$

The limit of this is ${ \sin^{2008}(\frac {\pi}2}) = 1.$

I did use both that $ \sin^{2k}(x)$ is periodic of period $ \pi$ (rather than $ 2\pi$) and that $ \sin^{2k}(x)$ is an even function. The same result would hold if you replaced $ 2008$ by any even integer but would fail if you replaced it by an odd integer.


Okay, I'm going to try to do the impossible!

Make a *whole* new game for this category.  Since mafia seems to have been here for a while, let's see if anyone likes bs?

Okay, seeing as I'm the most online person in this forum all the time..I'll be the first mod to see how it goes.  Anyone else that is extremely addicted to this forum may also apply to try game 2 or 3 with my prior permission.

I would like 4 people to sign up...I will take my mafia cards (by the way, kheg, murton, nomar, slam, check your laundry, i can't find the 2 -.-), and 1 two of a different type of card.  Shouldn't matter anyway...

Okay I pm the people the cards that they each got, and then I will officially declare the game open.

Here's how to play:
-There are an equal amount of cards dealt to each of the 4 players.
-I do eenky peenky ponky, then the eeny meeny miny moe, then the my mother said you are the one that's it...play continues alphabetically until the last alphabetical person has taken their turn.  Then play resumes to the first alphabetical person.
-The chosen person will place how many aces they have.  Then next person does 2's, then 3's, then 4's, all the way to j's, q's, and k's, then back down to aces.
-Jokers are wilds and can be used as ANY type of card.
-Before you post, private message me the cards you are putting down.  Then I'll respond and allow that move.
-Once you move, someone can claim "BS!"  If no one claims it 2 hours after the card(s) have been placed, play continues as if there was no call on it.
-You may lie if you don't have the card, and place cards that don't match your number.
-If your BSed and the claim was true, you take the WHOLE pile.  If your BSed and it was false, the bser then must take the pile.
-Try to get rid of all your cards.
-Keep going until one person has all the cards.  That person is out and the round is over.  I will begin the new round (NOT a new game).  Then same thing.
-Once someone wins, I'll give them a) A notification on following bs games that they won, and b) permission to use a unique symbol string in their sigs.

Current Round: 1
Current players: biffandoc, slamchess, klebian, h_s_potter_2002
Cards in Pile: 1
Cards that have been placed recently: 8's
Last card placer: Slamchess
Current card placer: Biffandoc
Current card type: 9's
---

First 4 to sign up will be allowed in.  I hope this becomes a successful game.  This is great if you've been lynched in mafia and are bored, want that cool sig, or just want to kick some ***.

First 4 to respond...You'll understand form there...make sure you read the WHOLE first post (if you haven't already).

And have fun!  :w00t:

k your in...i gotta do a project now it'll start tomorrow

[quote="la mafioso"]
-I do eenky peenky ponky, then the eeny meeny miny moe, then the my mother said you are the one that's it...play continues alphabetically until the last alphabetical person has taken their turn.  Then play resumes to the first alphabetical person.
[/quote]

Huh?
Can you explain that?

Anyways, I'm in, and I call mod for game 2

i will play
and i thought we established the fact that we emptied our pockets...

This:

Eenky, peenky, ponky, daddy bought a donkey, donkey die, daddy cry, eenky peenky ponky, eenie meenie minie moe catch a tiger by the toe if he hollers let him go eeny meeny miney moe my mother said to choose the very best one and you are it.

Like that!

I will begin shuffling the cards and you should have a pm a bit after supa bowl!

ALL RIGHT...NO JOKERS IN THIS ROUND!!!

I think I have the pmed...next I will begin the choosing.

I DID NOT GET ONE...

SINCE WHEN DO U LIKE FOOTBALL?

i think there's already 4 people this round   :roll:

Sorry...it'll start tomorrow..i'm bogged down with homeowrk at the moment.

[quote="h_s_potter2002"]2 Ace's.

And BS on Biffandoc.[/quote]

so how many did he get???

and what are the standings?

what happens if you run out of cards, too? As well as it being your turn.

You know what? I think we should start a new game altogther....

A bunch of the people here clearly do not care about this game as much as others.

I am going to make a new game for these reasons

a) the cards were messed up and no one has even *bothered* to tell me what they were

b) many people here are inactive

and there will only be 2 in the next game....

A vote...and if you want to be in the next game...care about it...for the other people and me.

[quote]and there will only be 2 in the next game....[/quote]

two people????
that would be stupid.

1. im in
2. well look in ur inbox
3. same as kheg

i m in :)

typing sentence to get post in

well i'll put some cards in the pile to begin with, so you don't EXACTLY know the other's cards...

besides, I can tell that there is almost no one here that is serious about this game...

I'd like to play, and I was pretty serious..

I was serious too :?
It just wasn't my turn to go


Find all integer solutions to
\[x^2+y^2+z^2=2xyz\]

what I have tried so far is taking the mod 16 on both sides

so, on the LHS, we can have 0,1,2,3 in mod 16

on the RHS, we can only have even numbers in mod 16

Since they are equal,

So, we can only have 0,2 in mod 16 on both sides.

What should I do for the rest?

Then, I tried this
\[x^2+y^2+2xy+z^2 = 2xyz+2xy\]
\[(x+y)^2+z^2 = 2xy(z+1)\]

And look at this equation in $Z_{16}$

So, we can only have 0,2 in $Z_{16}$ because on the RHS is even.

Thus, we get the same result as my previous post.
 :?:

Hello? Can anybody at least give me a hint?

I know this question is not that hard, but I can not see the answer for some reason  :(

Hint:

[hide]Infinite Descent[/hide]

Full Solution:

[hide]At least one number on the left hand side is even; wlog let it be x = 2a.

4a^2 + y^2 + z^2 = 4ayz
Mod 4:
y^2 + z^2 = 0 mod 4

So y and z are even. y=2b, z=2c.  So we have (after divide 4 from both sides)

a^2 + b^2 + c^2 = 4abc.

Continuing in this manner, we find that x,y, and z are divisible by 2^n for all n.  This is obviously impossible unless x=y=z=0.[/hide]

very interesting solution!
I have never seen something like this for a number theory question. :blush:


[color=darkblue]Let $ x,y,z\in R$. Prove that: $ \frac {9x^2}{9 \plus{} x^2} \plus{} \frac {16y^2}{16 \plus{} y^2} \plus{} \frac {25z^2}{25 \plus{} z^2}\ge \frac {(3x \plus{} 4y \plus{} 5z)^2}{50 \plus{} x^2 \plus{} y^2 \plus{} z^2}$[/color]

It's obviously true by Cauchy-Schwarz Engle form.  :P
"\[ \frac {a^2}{x} \plus{} \frac {b^2}{y} \plus{} \frac {c^2}{z} \geq \frac {(a \plus{} b \plus{} c)^2}{x \plus{} y \plus{} z}
\]"


$L_1=1,L_2=3$
$L_n=L_{n-1}+L_{n-2}$
Prove that if $p$ is a prime $L_p-1$ divisible by $p$

[quote="ZetaX"]I think it was posted before...
It works for $p=2$ so let $p$ be odd.
We have $L_n= \left( \frac{1+\sqrt 5}{2}\right) ^n+\left( \frac{1-\sqrt 5}{2}\right) ^n$.
Now , depending on if $\sqrt 5 \in \mathbb{F}_p$ or not, we work in $\mathbb{F}_p$ or $\mathbb{F}_{p^2}$.
Also $\left( \frac{1+\sqrt 5}{2}\right)^p \equiv \frac{1+\sqrt 5^p}{2} \mod p$, so we get $\left( \frac{1+\sqrt 5}{2}\right) ^p+\left( \frac{1-\sqrt 5}{2}\right) ^p \equiv \frac{1+\sqrt 5^p}{2}+ \frac{1-\sqrt 5^p}{2} \equiv 1 \mod p$ and are done.[/quote]
but, $L_n=(2-\sqrt 5)\left( \frac{1+\sqrt 5}{2}\right) ^n+\sqrt 5\left( \frac{1-\sqrt 5}{2}\right) ^n$
and can you explain why $\left( \frac{1+\sqrt 5}{2}\right)^p \equiv \frac{1+\sqrt 5^p}{2} \mod p$ ?

Pay attention that my formula gives $L_2=3$ and not $1$ !

sorry, my mistake  :?
and can you explain to me about the mod p thing please ?

[quote="pavel25"]and can you explain why $\left( \frac{1+\sqrt 5}{2}\right)^p \equiv \frac{1+\sqrt 5^p}{2} \mod p$ ?[/quote]
Expand it as binom: $(x+y)^p = \sum_{k=0}^p \binom{p}{k} x^k y^{p-k} \equiv x^p+y^p \mod p$ since $p|\binom{p}{k}$ for $1\leq k \leq p-1$.


trigonometry probability calculus function conics

I didnt find it suitable anywhere else, so I posted here.

I am taking a Pre-Calc class right now, that is also called Trig. Since I am in Math Team, I tend to not take much of class notes and hmwk assignment. (In fact, I do a lot of Math team tests---- since I know a lot of things learned in class and other things I dunno I can self-study). Recently, for the final unit of the year, our math department head created a new section called "Statistics and Probability" Unit, where we learn the Binomial Expansion, Standard Deviation, etc. Since I didnt take much note on it, I am bound to fail the test, which does not really disappoint me, because I dont think Stats is appropriate for Trig class.

So My main question is, is Statistics Suitable for a Pre-Calculus class?? Is it even necessary?

I don't even consider stat as math, as you can see it from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38784&postorder=asc&start=8]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30483&start=146]here[/url]
I never took calculus class(I studied it with my books. Maybe that's why I'm so bad at calculus), but I really don't think statistics is necessary. :D

Statistics is really tedious even with a computer, but any job you take will probably require at least a vague understanding of statistics.
I was taught a statistics unit when I was in Trig, but that was different from my Pre-Calc. In Pre-Calc, we worked mostly with limits, function transformations, and conic sections.

Statistics is becomes a bit more interesting when you go up a dimension or two, and when you begin proving all those formulas the teachers make you memorize.

[quote="blahblahblah"]The word Trigonometry is absurd for a (non-AoPS  :) ) course.  Since there is no way you need to spend a whole year on trigonometry, what's the big deal on having some probability/statistics/combinatorics in it?  The statistics is a bit fishy, since it's inevitably done mostly on calculator, but at least it has some real life applications.[/quote]
I guess learning statistics won't harm you.  :D

Stuff like AP level statistics is pretty boring.  The main thing is that it's quite difficult to derive most of the formulas, so it concentrates more on memorization and application.  However, one unit can't hurt--it's good to know what tools are available.


Find all solutions if any of $a,b,c \in \mathbb{R}$ to the equations:

$a^{3(b + c)} = 36$
$b^{3(c + a)} = 81$
$c^{3(a + b)} = 169$


function
inequalities
quadratics
algebra
domain
rearrangement inequality
Olimpiada de matematicas

Quisiera que este topic sea un lugar
donde los participantes de esta comunidad en español
tengan un grupo de preguntas  de manera periodica ...
ire colgando cada semana (un grupo de 03 preguntas): :lol:

Espero sus soluciones y/o sugerencias para resolverlas.....

[color=blue][size=184]WEEK 1 (Algebra)[/size] [/color]

1.- Probar que, para todo real $a, b, c$
\[ \frac{a^3}{bc} + \frac{b^3}{ca} + \frac{c^3}{ab} \geq a + b +c ,  \]
y cuando sucede la igualdad.

2.- Sean $a, b$ numeros reales para lo cual la ecuacion:
\[ x^4 + ax^3 + bx^2 +ax + 1 = 0,  \]
tiene al menos una solucion real. Para todo par $(a,b)$, encontrar el minimo valor de $a^2 + b^2$.

3.- Sea $f$ una funcion definida en $\mathbb{Z}^+$ por:
   $f(1) = 1,$
   $f(2n+1) = f(2n) + 1,$
   $f(2n) = 3f(n).$

 Encontrar el conjunto de valores que toma la funcion $f$.

Suerte!!  ;)

Carlos Bravo
Lima Peru
-----------
 PD:haber si alguien mas me da una manita para hacer caminar este topic!!!  :idea:

Hi!
I don't know spanish.
So i dont understand the problems completely! :(

However I think the ineq (problem 1) is reversed, that is
$\dfrac{a^3}{bc} + \dfrac{b^3}{ca} + \dfrac{c^3}{ab} \geq a + b +c ,$

The numbers (a, b, c) maybe shoud be positive, too!

Thanks

The problems read as follows (I corrected a sign error on the first problem):

1. Prove that for all real $a,b,c > 0$

$\dfrac{a^3}{bc} + \dfrac{b^3}{ca} + \dfrac{c^3}{ab} \geq a + b +c ,$

and show when equality holds.

2. There exist real numbers $a,b$ such that the equation

$x^4 + ax^3 + bx^2 +ax + 1 = 0$

 has at least one real solution.  For all $(a,b)$, find the minimum value of $a^2 + b^2$.

3. There exists a function $f$ defined for positive integers such that

$f(1) = 1,$
$f(2n + 1) = f(2n) + 1,$
$f(2n) = 3f(n).$

Find the set of values that $f$ assumes.

Here I go for the first one (I don't understand Spanish by the way, even though French allows me to decypher more or less). As noted, the inequality is the other way and the numbers must then be positive (otherwise consider the same inequality for $-a,-b,-c$).

Now, assuming $a \geq b \geq c$ and using the rearrangement inequality gives:

$\dfrac{a^3}{bc} + \dfrac{b^3}{ca} + \dfrac{c^3}{ab} \geq$

${\dfrac{a^{3}}{ab} +\dfrac{b^{3}}{bc}} +\dfrac{c^{3}}{ca} =$

${\dfrac{a^{2}}{b} +\dfrac{b^{2}}{c}} +\dfrac{c^{2}}{a} \geq$

${\dfrac{a^{2}}{c} +\dfrac{b^{2}}{b}} +\dfrac{c^{2}}{a} =$

$\dfrac{a^{2}}{c} + b +\dfrac{c^{2}}{a} \geq$

$a+b+c$.

x^4 + 1 + a(x^3 + x) + bx^2 = 0

If x is a root, so is 1/x.  Obviously x = 0 is not a root; so divide both sides of the eqn by x^2.
Put z = x + 1/x.  Then the equation is $z^2 - 2 + az + b = 0$, a quadratic with roots $z = \dfrac{-a \pm \sqrt{a^2 - 4(b-2)}}{2}$.  Then x satisfies $x^2 - zx + 1 = 0$  and so because it is real, its discriminant $z^2 - 4 \ge 0$.  Thus $[-a \pm \sqrt{a^2 - 4(b-2)}]^2 \ge 16$, so that $2a^2 + 8 - 4b \pm 2a\sqrt{a^2-4(b-2)} \ge 16$

After squaring again, we get $4a^2(a^2 - 4b + 8) \le (2a^2 - 4b - 8)^2$.  Some expanding yields $-16b^2 + 64 +64b -64a^2 \ge 0$

Then $4+4b \ge 4a^2 + b^2$.  It implies $4(2-a^2) \ge (b-2)^2$  So $a^2 \le 2$.  From our original discriminant eqn in z, which was $a^2 + 8 \ge b$, we get that $b \le a^2 + 8 \le 10$.  So in total $102$ realized when $a^2 = 2$ and $b = 10$.

[quote="zanttrang"]
3. There exists a function $f$ defined for positive integers such that

$f(1) = 1,$
$f(2n + 1) = f(2n) + 1,$
$f(2n) = 3f(n).$

Find the set of values that $f$ assumes.[/quote]

From the third rule, $f(2^k) = 3^k$ and this is obvious.

Suppose we have determined the set of values that f assumes in the domain $[0, 2^k] \cap \mathbb{N}$.  To determine the value of the function at a number n between $[2^k ,2^{k+1}] \cap \mathbb{N}$, if n is even we have f(n) = 3f(n/2), so that f(n) is 3 times one of the numbers in the set.  Infact this happens to be all numbers, because every even number in  $[2^k ,2^{k+1}] \cap \mathbb{N}$ is double some number in $[0, 2^k] \cap \mathbb{N}$.  If n is odd, then it is one more than some even number.

Thus we may say that if $S$ is the set of values f assumes for $n \in [0, 2^k] \cap \mathbb{N}$, then the set of values $S'$ that f assumes for $n \in [0, 2^{k+1}] \cap \mathbb{N}$ is $\bigcup_{x \in S} \{x, 3x, 3x+1\}$.

Express numbers in S in [b]base 3.[/b]  For k = 0 we find "1" is the only number.  For k = 1 we find the set {1,10}.  For k = 2 we find {1, 10, 11, 100}.  Put k sufficently large.  Then we can always find numbers from our transformation in S. We deduce that we get all new numbers from old ones, old ones with a zero appended at the end, or old ones with a 1 appended at the end.  So essentially, S is the set of all base 3 numbers with no "2" written its expansion.

Carlos, muy buena iniciativa pero creo que seria mejor crear un Mailing list, porque o si no, alguien no va a entender la diea y postear soluciones como paso aca. :)

To other forum participants. the idea with this topics is that we should mail solutions, or post them but after a week or something like that so everyone can have some thoughts on it.

Me parece tambien una buena idea.. pero la verdad  :?  no entendi la idea de este topic. Se pueden mandar las soluciones o tenemos que esperar una semana? (dicho sea de paso ya pasó más de una semana)

Si esta buena la idea, haz un mailing list.
mi correo: jibarotimbalero@hotmail.com
posteen sus correos los de mas.

Hola amigos, en cuanto al problema 3 hay vatias variantes de este estilo en las olimpiadas a todo aquel que desee resolver este tipo de problemas les suguiero que tomen bloque en formas binarias es mucho mas rapido y elegante, claro esta las que sean con las del sistema de base 2, ya que hay otras de otro estilo de otros sistemas de numeracion, miren este tambien es de sistema binario es muy interesante encontrar la funcion
   $f(1) = 3,$
   $f(2n+1) = f(n) + 3,$
   $f(2n) = f(n)-2.$

Suguerencia trabajar con bloques de base 2 y una simple induccion y el problema se vuelve trivial, bueno amigos  saludos a todos

Que Dios les Bendiga


2x^2 - ax + 4bx - 2ab=0

how can i solve for x?

[quote="mathamatix"]i dont believe that you can unless you know a&b however if you want an awnser in terms of a & b then use the quadratic formula[/quote]

how could i get that started..?

unless "solving for x" means have it in an $x=$ format?

how do you solve for x in terms of a & b using the quadratic formula?

[hide="Using Quadratic Formula"]first rewrite it so you can clearly see each coefficient;
$(2)x^{2}+(-a+4b)x+(-2ab).$ The numbers in parentheses are the a, b, and c terms, respectively. Now plug it into quadratic formula:$\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
$\frac{a-4b\pm\sqrt{16-8ab+a^{2}+16ab}}{4}.$
$\boxed{x=\frac{a-4b\pm\sqrt{16+8ab+a^{2}}}{4}}$ I don't think that can be simplified. I probably messed up somewhere in there.[/hide]

Or you could just factor:

$2x^{2}-ax+4bx-2ab=(2x-a)(x+2b)$
so $x=\frac{a}{2}$ or $x=-2b$.

In your quadratic formula, it should be $\frac{a-4b\pm \sqrt{a^{2}-8ab+16b^{2}+16ab}}{2\times 2}=\frac{a-4b\pm (a+4b)}{2}$ which gives you the same answer.


Let $ G$ be a compact connected Lie-Group and let $ \psi_k: G\to G,\ g\mapsto g^k$ be the $ k$-th power map. Find the mapping degree of $ \psi_k$.

[i]Hint:[/i] Choose a maximal torus $ T$ and consider the map $ G/T\times T\to G, ([g],t)\mapsto gtg^{\minus{}1}$.


The AMC-12 A 2003 #25:

Let $f(x)=\sqrt{ax^2+bx}$. For how many real values of $a$ is there at least one positive value of $b$ for which the domain of $f$ and the range of $f$ are the same set?

A) 0
B) 1
C) 2
D) 3
E) Infinitely many

[hide="I found two values for a"]

i) $a=0$

$f(x)=\sqrt{bx}$

for every $b>0$ we have:

domain: $[0,+\infty)$
range: $[0,+\infty)$

ii) $a>0$

$ax^2+bx=x(ax+b)\geq 0$

The domain consist of two parts: $(-\infty,-\frac{b}{a}) \cup [0,\infty)$ such that $x$ and $ax+b$ have equal signs.
But the value of $f(x)$ is always $\geq 0$, so the range can't be equal to the domain.

iii) $a<0$

$x(ax+b)\geq 0 \Rightarrow$

domain : $[0,-\frac{b}{a}]$

$x(ax+b)$ gets a maximum value for $x=-\frac{b}{2a}$

This maximum is $-\frac{b}{2a}(-\frac{b}{2a}a+b)=-\frac{b^2}{4a}$

So the range is $[0,\sqrt{-\frac{b^2}{4a}}]$

$\sqrt{-\frac{b^2}{4a}} = -\frac{b}{a} \Rightarrow$

$-\frac{b^2}{4a} = (-\frac{b}{a})^2 \Rightarrow$

$-\frac{b^2}{4a} = \frac{b^2}{a^2} \Rightarrow$ (we assumed $b \neq 0$)

$-4a=a^2\Rightarrow$

$a=-4$

So, if $a=-4$, then for every $b>0$ the domain and the range are equal sets[/hide]


Im in uni now, no more olympiad for me :-(

Given a bag with two black balls and three with balls, you randomely select 2 of them, in order. If the second ball you take out was white, whats the probability the first one you took out was white??

Yea, it's like "If you draw 20 cards out of a regular deck, what's the probability the next card is red?" The previous draws don't change your chances. Likewise, the next ball you get is like the first ball.
This is obviously based on the assumption that you don't know what has been drawn previously. Otherwise, it would a conditional probability.

[quote="deej21"]I assume you mean "three white balls" there.  And... this seems like a trick question.  The second one doesn't have anything to do with the first one, so wouldn't it just be $\frac 3 5$ still?[/quote]
if you had one black ball and one white ball, and the second ball taken out was white, the probability that the first one was black is 1, not 1/2

since the order doesnt inherently matter, (and thus we switch the order taken out) whether we picked then if you picked the first ball white, the second ball would have a 1/2 probability of being white

taking a more formal approach,we can attack the problem by applying Baye's theorem

 p(W1/W2) =  p(W1).P(W2/W1) / [p(W1).P(W2/W1)   +  p(B1).P(W2/B1) ]

               Where p(W1) = prob. that the first ball drawn is white and similarly the other terms are defined.

 The values are p(W1) = 3/5 , p(W2/W1) = 2/4 , p(B1) = 2/5 , P(W2/B1) = 3/4

Substituing these, we get   p(W1/W2)  = 1/2

$\frac12$ is indeed correct. Good one guys.


Six points on a circle are given. Four of the chords joining pairs of the six points are selected at random. What is the probability that the four chords are the sides of a convex quadrilateral?

$ \textbf{(A)}\ \frac{1}{15}\qquad
\textbf{(B)}\ \frac{1}{91}\qquad
\textbf{(C)}\ \frac{1}{273}\qquad
\textbf{(D)}\ \frac{1}{455}\qquad
\textbf{(E)}\ \frac{1}{1365}$

[hide]
There are $ \binom{6}{2} \equal{} 15$ pair of points and therefore $ 15$ different chords.

Of those $ 15$ chords, there are $ \binom{15}{4} \equal{} \frac{15\cdot 14 \cdot 13 \cdot 12}{4\cdot 3\cdot 2}$ ways to pick four of them

There are $ \binom{6}{4}\equal{}15$ ways to select four points and make a convex quadrilateral,

$ \Rightarrow \dfrac{15}{\frac{15\cdot 14 \cdot 13 \cdot 12}{4\cdot 3\cdot 2}} \equal{} \frac{1}{7\cdot 13}\equal{}\frac{1}{91} \longrightarrow \boxed{B}$[/hide]


inequalities
real analysis
real analysis unsolved

Let $ d$ be an element of $ \mathbb R$ satisfy $ d > 1$. Use Bernoulli's Inequality to show that the sequence $ (d^{n})$ is not bounded in $ \mathbb R$.

My question is should I assume it is bounded and try to derive a contradiction from using Bernoulli's Inequality? I clearly don't have a good idea of where to start, so any hints or help would be appreciated.

since $ d >1\rightarrow d\equal{}x\plus{}1,x>0$
thus $ d^{n}\equal{}(x\plus{}1)^{n}\geq 1\plus{}nx$


geometry
circumcircle
analytic geometry
perpendicular bisector
RMO

If the circumcenter and centroid of a triangle coincide, prove that it must be equilateral.

One way is to use coordinate geometry...
[hide]
Consider $\triangle OAB$ where the coordinates are $O(0,0)$, $A(a,0)$ and $B(b,c)$.

The coodinates of the centroid is $\left(\frac{0+a+b}{3},\frac{0+0+c}{3}\right)=\left(\frac{a+b}{3},\frac{c}{3}\right)$

In order to find the coodinates of the circumcenter, we need to find the equations of perpendicular lines of $OA$ and $OB$ and then find their intersections.

The equation of the perp. bisector of $OA$ is simply $x=\frac{a}{2}$.

The equation of the perp. bisector of $OB$ is $y-\frac{c}{2}=-\frac{b}{c}(x-\frac{b}{2})$.

So we substitute $x=\frac{a}{2}$ to this equation.
$y=-\frac{b}{c}\left(\frac{a}{2}-\frac{b}{2}\right)+\frac{c}{2}=\frac{b}{c}\left(\frac{b-a}{2}\right)+\frac{c^2}{2c}=\frac{b(b-a)+c^2}{2c}$

So the coordinates of the circumcenter is $\left(\frac{a}{2},\frac{b(b-a)+c^2}{2c}\right)$.

Finally, we compare the coordinates of the centroid and the circumcenter.

$\frac{a+b}{3}=\frac{a}{2}\Longleftrightarrow 2(a+b)=3a\Longleftrightarrow 2b=a\Longleftrightarrow b=\frac{1}{2}a$

$\frac{b(b-a)+c^2}{2c}=\frac{c}{3}$
(Substituting $b=\frac{1}{2}a$,)
$\frac{-\frac{1}{4}a^2+c^2}{2c}=\frac{c}{3}$
$\Longleftrightarrow -\frac{3}{4}a^2+3c^2=2c^2$
$\Longleftrightarrow c^2=\frac{3}{4}a^2$
$\therefore c=\frac{\sqrt{3}}{2}a$

From here, it is easy to see that $\triangle OAB$ is an equilateral triangle.

[hide="To complete the proof"]
Now we know that the coordinates of the vertices are $O(0,0)$, $A(a,0)$, and $B\left(\frac{1}{2}a,\frac{\sqrt{3}}{2}a\right)$. You can use distance formula to show that the side lengths are all equal, or you can find $OA$ and $OB$ and then show that $\angle AOB=60^\circ$. Of course, there are many other simple ways to show it :) ) [/hide]

[/hide]
P.S.- Yeah, mine is really long :D

Note that it is a well-known fact that the circumcenter is the point of intersection of a triangle's perpendicular bisectors and that the centroid is the point of intersection of the triangle's medians. If the centroid and circumcenter are concurrent, the medians must also be perpendicular bisectors. This implies that the triangle is isosceles when viewed from any base, as the perpendicular bisector bisects the base, so since the triangle is isosceles no matter from which side we look at it as isosceles, the triangle must be equilateral due to symmetry.

Is this rigorous enough or would this have to be tightened? I think it might be the latter...

I think the simplest approach is $complex$ $ numbers$. Set $A,B,C$ as$ a,b,c$ resp. also set $circumcentre$ as the $origin$ of the complex plane.
clearly $H=a+b+c$and $G=a+b+c/3$.(Use the fact that $HG:GC=2:1$)... Now as $G=C$ is given we get (a+b+c=0)!!!
this implies centroid and orthocentre coincide..
the result follows by congruences.
QED

What I think is the simplest approach:
Let $G$ be the centroid and circumcenter, and $X ,Y, Z$ the midpoints of $BC , AC, AB$ respectively. Using pythagoras in $\Delta GXB$ and $\Delta GXC$ we get $GB = GC$. Analogously we have $GC = GA = GB$. Since medians trisect each other we also have $GZ = GX = GY$. Thus we have all medians equal, and the triangle is equilateral.

if $O$ is the circumcenter, then:
$OA = OB = OC \Leftrightarrow \frac{2}{3}{m_a} = \frac{2}{3}{m_b} = \frac{2}{3}{m_c} \Leftrightarrow $ $\boxed{a=b=c}$

Recall that the area of the triangle formed by the centroid and any other two vertices of a triangle is one third of the area of the original triangle. This gives
$$ \frac{1}{2}R^2 \sin (2A)= \frac{1}{2}R^2 \sin(2B)=\frac{1}{2}R^2\sin(2C), $$
which implies
$$A=B=C$$.

How do people find these posts that were posted 12 years ago.....

Scroll down for a long time, just like if you scroll down long enough in $\pi$, you will eventually find 6 9s in a row.

No they are preparing for RMO
So they bump in past year questions

We know that, $G\equiv (1:1:1)$ and $O\equiv (\sin 2A :\sin 2B :\sin 2C)$
$$ \implies \sin 2A= \sin 2B = \sin 2C \implies \angle A = \angle B = \angle C \text{ (Since, A+B+C = 180)} \implies \boxed{\angle A= \angle B = \angle C =60^{\circ}}$$
Hence, Equilateral

Let $l$ be the perpendicular bisector of $BC$ and $A'$ the midpoint of $BC$. Then $Z=M\in l, A'\in l$ hence $l=ZA'$ which is the median from $A$. Hence $A\in l$ and $AB=AC$. In the same way $BA=BC$.

Yes, but then they wouldn't be the medians and altitudes of the same triangle

[quote=dchenmathcounts][quote=Valentin Vornicu][hide="A much simpler way ..."]
Let $O$ be the circumcenter and the centroid. Let $M,N,P$ be the midpoints of $BC,CA$ and $AB$ respectively. Then as $O$ is the circumcenter $OA=OB=OC$ (as rays). This means that the triangles $OBC$, $OCA$ and $OAB$ are isosceles. Thus $OM$, $ON$ and $OP$ are also altitudes in these triangles. However they are segments lying on the medians of the triangle $ABC$, which means that all the medians in the triangle $ABC$ are also altitudes, thus the triangle is equilateral.
[/hide][/quote]

Is this necessarily true? This argument can be applied for altitudes, angle bisectors, and medians, but theoretically (I know this doesn't actually happen), you could have the perpendicular bisectors and the medians concur at one point while being six separate line.[/quote]

The reason this argument works in this case is because the midpoint of a side is on both its perpendicular bisector and the median to it.

the idea or the motivation behind the problem any how is because of incircle it is the point of concurrence of angular bisectors which make angles equal and due to median which bisect sides makes sides equal and due to sas criterion we get congruent triangles on aplying cpct we get it to be equilateral

[quote=dchenmathcounts][quote=Valentin Vornicu][hide="A much simpler way ..."]
Let $O$ be the circumcenter and the centroid. Let $M,N,P$ be the midpoints of $BC,CA$ and $AB$ respectively. Then as $O$ is the circumcenter $OA=OB=OC$ (as rays). This means that the triangles $OBC$, $OCA$ and $OAB$ are isosceles. Thus $OM$, $ON$ and $OP$ are also altitudes in these triangles. However they are segments lying on the medians of the triangle $ABC$, which means that all the medians in the triangle $ABC$ are also altitudes, thus the triangle is equilateral.
[/hide][/quote]

Is this necessarily true? This argument can be applied for altitudes, angle bisectors, and medians, but theoretically (I know this doesn't actually happen), you could have the perpendicular bisectors and the medians concur at one point while being six separate line.[/quote]

Not in this case, since a line is defined by two points, and each of the perpendicular bisectors/medians pass through two defined points (the common point where all 6 intersect, as well as one vertex of the triangle)

The main idea is to note the fact that the Altitudes bisect the sides of a triangle $\iff$ it is an equilateral triangle

[hide=Solution]Let the circumcenter be $O$. Let the midpoints of $AB,AC,BC$ be $P,N,M$ respectively. Let the intersection of the median from $A$ with $BC$ be $M'$ and define $P', N'$ similarly. Then $CM'=\frac{1}{2}BC=CM$ and $M, M'$ are both on segment $BC$, so $M'=M$. This means all the altitudes are medians, so the triangle is equilateral as desired.[/hide]

simple just use coordinate geometry
its  simple


See attached. It's not B. We have to start with the delta G values and use delta G=-nFE apparently because we have Co in both half given reductions.

I read that we must use $ \bigtriangleup G\equal{}\minus{}nFE$ in this case. Why do we need to use $ \bigtriangleup G\equal{}\minus{}nFE$ in this case but not other cases?

This question was asked in this thread:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273512


Dr. Dominguez charges each patient according to the length of
the patient's visit, as indicated in the chart below.  If she earned
exactly $ \$500$ today, what is the positive difference between the
fewest number and the greatest number of patients that she
could have seen today?
\begin{tabular}{||c c||}
\hline
Length of Visit & Cost\\
\hline
$ \leq$ 10 minutes & $ \$$15\\
$ 10 \minus{} 30$ minutes & $ \$$25\\
$ 30 \minus{} 60$ minutes &$ \$$40\\
$ 60 \minus{} 120$ minutes & $ \$$75\\
\hline
\end{tabular}

The greatest amount of patients she could have seen:
15*30+25*2 (32)
The least amount of patients she could have seen:
75*6+25*2 (8)

So, 32-8=24

\begin{tabular}{\parallel{}c c\parallel{}}
\hline
Length of Visit & Cost\\
\hline
\leq 10 minutes & \$15\\
10 - 30 minutes & \$25\\
30 - 60 minutes & \$40\\
60 - 120 minutes & \$75\\
\hline
\end{tabular}

[quote="james4l"][quote="ernie"]\begin{tabular}{\parallel{}c c\parallel{}} \hline Length of Visit & Cost \\
\hline \leq 10 \text{ minutes} & \$15 \\
10 - 30 minutes & \$25 \\
30 - 60 minutes & \$40 \\
60 - 120 minutes & \$75 \\
\hline \end{tabular}[/quote]

+text[/quote]
har har


A game is played using two special cubical dice: one die has faces labeled 1,1,1,2,2,3; the other die has faces labeled 1,2,3,5,5,6. When these dice are rolled, what is the expected value of thesum of the numbers shown on their upper faces?

The expected value of each individual die is the average of the numbers on the die, so the expected value of the sum is $ \frac {1 \plus{} 1 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 3 \plus{} 1 \plus{} 2 \plus{} 3 \plus{} 5 \plus{} 5 \plus{} 6}6 \equal{} \boxed{\frac{16}3}$.

Edit: Thanks for the correction, ernie.

What math154 means is $ \boxed{\frac{16}{3}}$.

It appears thats what he has....

EDIT: Oh he edited  :blush:


ratio
geometry
trigonometry
trig identities
Law of Sines

This is intended for those who are truly getting started. Please don't answer if you spend most of your time posting, for example, in the advanced or intermediate forums.

Triangle $ABC$ has $\angle ABC = 75^\circ$ and $\angle BAC = 45^\circ$. $\overline {AB} = 10$. Point $D$ is on $AB$, and point $E$ is on $BC$ such that $DE \parallel AC$. If $\overline {BD} = 6$, find the ratio of the area of triangle $DBE$ to the area of triangle $ABC$.

I really have no idea and I did it the way I imagine how it should be done:

[hide]area of ABC = 6*10/2 = 30
if the DE line is parrallel to AC and BD is equal to 6 then the equation would be(I think, total guess) : 6/10 = x/6 x = 3.6

BDE  area = 3.6*6/2 = 10.8

so it would be 10.8: 30 dunno if u can simplify that[/hide]

A diagram on this one would really help.

I know I'm a moderator and shouldn't interfer in this but just my curiosity:

What makes you said that 6*10/2 is the area of triangle ABC? In any part of this problem, did it tell you that this is a right triangle?

Maybe I'm not seeing this right but according to my way, it's not like that... ;)

Oh, Yeah. My fault. Express your answer as a reduced common fraction.

only the result, I'm way too lazy to write down my steps.
the picture proves that my answer is correct.

[hide]
S :tri: BDE/S :tri: ABC=9/25 :approx: 0.36
[/hide]

[hide]it is easy to deduce that angle C is 60.  Since DE || AC, [tex]\angle BAC = \angle BDE [/tex] and [tex] \angle BCA = \angle BED [/tex].  Draw an altitude from B to DE.  Since [tex]\angle BED = 45[/tex] the height of the small triangle is 6/ :sqrt: 2.  [/hide]

but now what do i do?  law of sines?

[hide]We know that $\overline {BD} = 6$, $\overline {BA} = 10$ and $\overline {DE} \| \overline {AC}$.  Since $\overline {DE} \| \overline {AC}$, We know that $\overline {BD}$ is proportional to $\overline {BA}$, and $\overline {BE}$ is proportional to $\overline {BC}$, making the ratio of the sides of the two triangles $\frac{6}{10}$.  You need to square that to get the area, getting $\frac{36}{100}$, which simplifys to $\frac{9}{25}$.[/hide]

[quote="Jeremy"]it is easy to deduce that angle C is 60.  Since DE || AC, [tex]\angle BAC = \angle BDE [/tex] and [tex] \angle BCA = \angle BED [/tex].  Draw an altitude from B to DE.  Since [tex]\angle BED = 45[/tex] the height of the small triangle is 6/ :sqrt: 2.
but now what do i do?  law of sines?[/quote]
Jeremy, you have got very close to the final result, you needn't use law of sines at all(and I don't see anything you can do with law of sines here ;) )
[hide]
Think about it again!
[/hide]

[b]
9/25
That is the exactly problem we solved when we were in Junior 2(7th year)[/b]

h_s_potter2002 your solution is really nice, my solution is the same :D

I have the answer: (click below to view)
[hide="My Answer"]
Since the ratio of areas of two similar figures is just the square of the ratio of the lengths of two similar figures. The ratio of the lengths of line BD to line AB is 6 to 10, and the two triangles are similar. Therefore the ratio of their area is just the square of 6:10 which is 36:100. This problem was two easy, maybe I should try intermediate forums?
[/hide]
Btw, I love your Dark Templar avatar.


Solve this system
$ x^2(y\plus{}1)\plus{}2xy\plus{}4x\plus{}y\plus{}2\equal{}0$
$ y^2(z\plus{}1)\plus{}2yz\plus{}4y\plus{}z\plus{}2\equal{}0$
$ z^2(x\plus{}1)\plus{}2zx\plus{}4z\plus{}x\plus{}2\equal{}0$

hello, from your first equation we get $ y\equal{}\minus{}\frac{x^2\plus{}4x\plus{}2}{(x\plus{}1)^2}$,substituting this in the second equation we get $ \minus{}2\plus{}8x\minus{}z\plus{}12x^2\plus{}x^4\plus{}8x^3\minus{}4zx\minus{}4x^2z\equal{}0$, from here we have $ z\equal{}\frac{2\plus{}8x\plus{}12x^2\plus{}x^4\plus{}8x^3}{(2x\plus{}1)^2}$,setting this in your third equation and factoring we have
$ (x^6\plus{}14x^5\plus{}65x^4\plus{}117x^3\plus{}102x^2\plus{}43x\plus{}7)(x^3\plus{}3x^2\plus{}5x\plus{}2)\equal{}0$, from here you will get the possible solutions in $ x$.
$ x\equal{}\minus{}1$ is impossible, because it gives in the first equation a contradiction, $ x\equal{}\minus{}\frac{1}{2}$ is also impossible.
Sonnhard.

Oh thank you Dr Sonnhard Graubner,but I suppose we get it in a contest,I dont think your way is useful( I think also the computer helped you on some steps :) ).so can you show me another solution?
I really need all your help :lol:

I think that introducing $ a \equal{} 1 \plus{} x, b \equal{} 1 \plus{} y, c \equal{} 1 \plus{} z$ would be helpful. The system becomes:
$ \begin{cases}
a^2b \equal{} 1 \minus{} 2a \\
b^2c \equal{} 1 \minus{} 2b \\
c^2a \equal{} 1 \minus{} 2c
\end{cases}$
But I cannot realize how could we proceed (without brute force) from here. :huh:

Hi allnames,
"I suppose we get it in a contest"
I don't think so because sonnhard is right:
    $ p_6q_3\equal{}(x^6\plus{}14x^5\plus{}65x^4\plus{}117x^3\plus{}102x^2\plus{}43x\plus{}7)(x^3\plus{}3x^2\plus{}5x\plus{}2)\equal{}0$ is the correct solution. $ p_6$ has no real root. $ q_3$ has one real root $ u\approx{\minus{}0.5466}$.
In fact the unique solution for $ x$ is $ u\equal{}a/6\minus{}4/a\minus{}1$ where $ a\equal{}(108\plus{}12\sqrt{177})^{1/3}$.
Conclusion: the unique solution is $ x\equal{}y\equal{}z\equal{}u$.

[quote="loup blanc"]Hi allnames,
"I suppose we get it in a contest"
I don't think so because sonnhard is right:
    $ p_6q_3 \equal{} (x^6 \plus{} 14x^5 \plus{} 65x^4 \plus{} 117x^3 \plus{} 102x^2 \plus{} 43x \plus{} 7)(x^3 \plus{} 3x^2 \plus{} 5x \plus{} 2) \equal{} 0$ is the correct solution. $ p_6$ has no real root. $ q_3$ has one real root $ u\approx{ \minus{} 0.5466}$.
In fact the unique solution for $ x$ is $ u \equal{} a/6 \minus{} 4/a \minus{} 1$ where $ a \equal{} (108 \plus{} 12\sqrt {177})^{1/3}$.
Conclusion: the unique solution is $ x \equal{} y \equal{} z \equal{} u$.[/quote]
Yes,I know,but in a contest this way is time-consuming and the calculus  is very hard
I hope there is a easy proof

Another solution using mathangel's idea.
$ a\equal{}1/2$ and $ a\equal{}0$ are impossible (easy). If $ a\equal{}b$ then we are done.
Thus we may assume $ a>b>c$. If $ a<0$ then $ b>0$, a contradiction. If $ b<0$ then $ c>0$, a contradiction. $ a<1/2$ therefore $ 1/2>a>b>0$. $ b<1/2$ then $ c>0$ and $ 1/2>a>b>c>0$.
$ b(a^2\minus{}bc)\equal{}2b\minus{}2a<0$;  then $ a^2<bc$, a contradiction.
Conclusion: $ a\equal{}b\equal{}c$.
Now we must solve: $ x^3\plus{}2x\minus{}1\equal{}0$; let $ x\equal{}y/6\minus{}4/y$ ("obvious" change of unknown); then $ y^6\minus{}216y^3\minus{}13824\equal{}0$; after it's easy.

Do you want to solve (over $ \mathbb{C}$) an equation of degree 3: $ p(x)\equal{}x^3\plus{}ux^2\plus{}vx\plus{}w\equal{}0$ ?
1) $ x\equal{}y\minus{}u/3$ gives $ y^3\plus{}px\plus{}q\equal{}0$.
2) If $ p\equal{}0$ it's easy, else:
$ y\equal{}z\plus{}b/z$ gives (1) $ z^6\plus{}a_4z^4\plus{}a_3z^3\plus{}a_2z^2\plus{}a_0\equal{}0$.
The system $ a_4\equal{}a_2\equal{}0$ gives $ b\equal{}\minus{}p/3$.
With this value of $ b$, (1) is easily solved: $ 2$ values for $ z^3$, therefore $ 6$ values for $ z$. With the formula $ y\equal{}z\plus{}b/z$ we obtain $ 2$ times each root of $ y^3\plus{}px\plus{}q$.
Easy, isn't it ?

"Thus we may assume $ a > b > c$."

I am not sure about it.

If you can't write $ a>b>c$ then 2 among $ \{a,b,c\}$ are equal. For example $ b\equal{}c$.
Then $ b^3\equal{}1\minus{}2b,b^2a\equal{}1\minus{}2b$ and $ a\equal{}b$ because $ b\not\equal{}0$ (else $ 1\equal{}0$).

But $ a > b > c$ is not analogous to $ a > c > b$ because the system is not symmetric.

Hi Feliz;
you are right; we must consider also the case $ a>c>b$. But I worked enough about this subject. Now it's your problem  :rotfl:


function
calculus
derivative
real analysis
real analysis unsolved

Following Moubinool, I will post here some (actually, all) preparatory questions to the main result:
    1) Prove the existence of a sequence of real numbers $a_n$ such that for all $p$ the series $\sum_{n\geq 0}{(-1)^{np}a_n}$ converges absolutely and has sum $(-1)^{p}$.
    2) Suppose that $G$ is a $C^{\infty}$ function that vanishes for $x\geq 2$ and is equal to 1 if $x\leq 1$. For a function $f$ defined on $P=[0,\infty)*R^n$ ( cartesian product)  define a function $F(x,y)$ which is equal to $f(x,y)$ if $x\geq 0$ and to $\sum_{n\geq 0}{G(-2^n x)\cdot f(-2^n x,y)}$ otherwise. Prove that for a nonnegative integer $p$, $F$ is of $C^p$ class if and only if $f$ is $C^p$ on $(0,\infty)*R^n$ and all its partial derivatives of order at most $p$ have continuous extensions to $P$. ( could anyone show an elegant proof for this thing???)
  3) Suppose that $f(x,y)=f(-x,y)$ and $f$ is of $C^p$ class. Then what is the best $k$ for which one can always find a function $g$ of $C^k$ class such that $f(x,y)=g(x^2,y)$?


Hi everybody,,
I need some help.. :( .. I shows me that i have no a viable lessons and no budges., My report is empty..
yesterday I had a lot of budges and lessons.. but now I have nothing..
I can't even get a question!!!
I have 250 points and 2400 karma.. I don't know what is going on wrong with me.. :o
I hope that you can help me  :oops:

me too

i have got nothing at all

helppppp!!!!
 :mad:  :mad:  :mad:

Hey Alya,
It's the same case for me too! I don't know what's going on. Maybe there's a glitch, the Art of Problem Solving People did it on purpose, or there's some requirement we have to meet.

i cant retrieve my info either, dont worry its probably just down for some maintenance or something.
it will probably be up soon..... i hope

It should be up and running now.  I think we have everything running now except FTW.

Thank you very much rrusczyk, It all works now  :lol:


linear algebra
matrix
geometry
geometric transformation
rotation

a magical $3\times3$ square is a $3\times3$ matrix containing all number from 1 to 9, and of which the sum of every row, every column, every diagonal, are all equal.

Determine all magical $3\times3$ square

[tex] \sum_{i=1}^{9}=45 [/tex] so the sum we are looking for in each row, column and diagonal is 15.

in the square
[tex]a_{1}   a_{2}   a_{3}[/tex]
[tex]a_{4}   a_{5}   a_{6}[/tex]
[tex]a_{7}   a_{8}   a_{9}[/tex]

If we put a nine, the ways to sum 15 are 9+1+5 or 9+2+4 an if we put a one the ways are 1+5+9 or 1+6+8, so the only positions to the one and nine are [tex] a_{4}[/tex]  and [tex] a_{6}[/tex] or [tex]a_{2} [/tex] and [tex] a_{8} [/tex]

now it's easy. we see that [tex]a_{5}[/tex] needs to be 5 and so on.

[tex]276[/tex]
[tex]951[/tex]
[tex]438[/tex]

and the other 3 squares obtained by rotation of this first.

On a minor note, you will get double the number you have (8 instead of 4) from reflections.

Yes, $2 \times 4 = 8$. :)

Still, it doesn't break his sol. Won't cost him more than one mark...

hi.
I believe it could be a good problem try to solve this to a square [tex]m \times m[/tex]
First fund to wich m we can make a magical square and then find how many diferent squares we can make.

there is a general technique for this. only works for odd ones though. Besides it's easy to prove that it doesn't work for even values that are no multiple of 4.

Anyway for odd ones: you start in the top middle square. Each 'n+1' is placed to the top right of 'n'. (if you're on the border, just start the other side as if it were on a ball). If there is already a number in that case, instead write 'n+1' in the case right below 'n'. If you do everything well, you'll always be able to go on, and you'll end up in the bottom middle square.

Greetz.


geometry
linear algebra
matrix
LaTeX
calculus
integration
trigonometry

Source: Georgia Tech Math Tournament, 2004

9. How many distinct solutions x are there for the equation det(A-xI)=0 where det stands for determinant, I is the identity matrix, and A is the matrix

1  0 -3  0
0  2  0  1
1  0  3  0
1 -2  0  1

a) no solution   b) 1   c) 2   d) 3   e) more than 3

16. If 11^100  -1= m(10 ^n)  for intagers m and n with m not divisible by 10, then what is n?

a)0   b)1   c)2   d)3   e)4

19. (modified) Please explain why the equation sqrt(x+5) - sqrt(x-2) +1 =0  has no root.

20. A square of side length 5 begins nestled in a corner of a square of side length 8. The small and large square share corner P, and Q is the opposite corner of the small square. The small square begins to roll around inside the large square, pivoting on its corners as it hits the sides of the large square. Eventually the corners P and Q will return to their original positions. If the first time this happens P has traveled a distance d, how far has Q traveled? Express in terms of d and or pi, if necessary. (there are answer choices but I don't know enough latex to type them all out, sorry)[/code]

#16
[hide]$11^{100}-1=(11+1)^{100}-1$. Because the second to last term in the binomial expansion of $(11+1)^{100}$ is ${100\choose99}(10)^1(1)^{99}=1000$, and the last digit of a positive integral power of $11$ is always $1$, we know $n$ is at most $3$. Also, any term in the binomial expansion before the third to last term will be multiplied by a power of $10$ greater than or equal to $3$, we need only check the third to last term. The third to last term equals ${100\choose98}(10)^2(1)^{98}=495000$, which makes the sum of the last $3$ terms $496001 \Rightarrow 496001-1=496000 \Rightarrow n=3 \Rightarrow \framebox{D}$.[/hide]

#19
[hide]Let $y=x+5$, then the equation becomes $\sqrt{y}-\sqrt{y-7}+1=0$. Simplifying this, we get $\sqrt{y}=-4$. Since the square root of a real number cannot be negative, there are no real roots to the equation.[/hide]

Not a great  method but,

[hide="9"]
$\mathbf{A}-\mathbf{I}x = \left[\begin{array}{lccr} 1&0& -3& 0\\ 0 &2& 0& 1\\ 1 &0& 3& 0\\ 1&-2& 0& 1 \end{array}\right] - x\left[\begin{array}{lccr}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{array}\right]= \left[\begin{array}{lccr} 1-x&0& -3& 0\\ 0 &2-x& 0& 1\\ 1 &0& 3-x& 0\\ 1&-2& 0& 1-x\end{array}\right]$

Now, we can compute the determinant using expantion by minors.  Since the last column has a pretty good number of zeros, I will use that as the piviot.

$\left|\begin{array}{lccr} 1-x&0& -3& 0\\ 0 &2-x& 0& 1\\ 1 &0& 3-x& 0\\ 1&-2& 0& 1-x\end{array}\right| = \left|\begin{array}{lcr}1-x&0&-3\\1&0&3-x\\1&-2&0\end{array}\right| + (1-x)\left|\begin{array}{lcr}1-x&0&-3\\0&2-x&0\\1&0&3-x\end{array}\right| = (1-x)(6-2x) + 6 + (1-x)((1-x)(2-x)(3-x) + 3(2-x))=0$

If you expand this (which I don't feel like doing right now), then you'll get a 4th degree polynomail.  So there are at most 4 solutions for $x$, but you could expand it and use Descarte's Law of Sines or some other stuff to see how many real-valued roots it has.

I actually just plugged it into mathematica and it has 4 imaginary solutions.  So the answer is A (assuming I expanded the determinant right). I'm sure there's a nice way to do this using some theorems that I don't know from Linear Algebra.
[/hide]


in texas hold'em, there are 1326 two-card hands, 78 of which are pairs.  If you are playing a 5-person poker game, what is the probability that somebody has a pair?[/tex]

Well from what you give, the probability of one person having a pair is (78/1326), so the probability that someone has a pair is 5(78/1326), or (390/1326)?

Having said that I haven't done probability since GCSE which was a while ago..

I don't know why this way works, but I know it does.

Could some explain why using the other way [hide](78/1326)^5[/hide] doesn't work?

[hide]First we have to find out the probability of one person not having a pair.  That is (1326-78)/1326 or 1248/1326.  Since there are five people, we would then raise it to the power of 5, which is 3027421750099968/4099374736625376 or ~0.7385 (to tired to simplify it all the way.).

SInce this is the probability that someone does NOT have a pair, we need to subtract the numerator (~7385) from the denominator (~10000) and put it over the denominator (~10000) getting us ~2615/10000.

The probability that a person has a pair is ~2615/10000 or ~1/4.[/hide]

Both of the previous postings are only approximations.  Neither one provides the correct probability.

mikewd's posting relies on the approximation:  [tex](1-\epsilon)^p\approx(1-p\epsilon)$, where $p$ is small and $p\epsilon$ is small$[/tex].

[Yuk!  I don't know why the above looks so awful.]  Try again...

mikewd's posting is an approximation of hspotter's method and relies on the approximation:  [tex](1-\epsilon)^p\approx(1-p\epsilon)[/tex], where [tex]\epsilon[/tex] is "small" and [tex]p\epsilon[/tex] is "small".  Since [tex]p\epsilon=5\frac{78}{1326}\approx 0.294[/tex], the approximation gets you in the ballpark, but it isn't very accurate.

hspotter's posting only works if each of the 5 players is dealt 2 cards from his own fresh deck.  His answer is much closer, but it isn't exact.

Are you seeking an exact answer or a description of the difficulties or ...?

Use dollar signs rather than the $\text{\TeX}$ tag.

rcv wrote:hspotter's posting only works if each of the 5 players is dealt 2 cards from his own fresh deck.  His answer is much closer, but it isn't exact.

Okay, I get what I got wrong.

[hide]instead of 78/1326 ^ 5, you do 1326-78/1326 * 1326-77/1325 * 1326-76/1324 * 1326-75/1323 * 1326-74/1322

Way too tired to use a calculator, but once you multiply everything out, the latter part of my post should work fine, substituting the right number for my number.[/hide]

[quote="h_s_potter2002"]Okay, I get what I got wrong.[/quote]

This is still incorrect.  To follow your general method in finding the probability that no player will have a pair, the correct formula would be of the form:
[tex]\displaystyle \frac{1326-a}{1326}\times\frac{1225-b}{1225}\times\frac{1128-c}{1128}\times\frac{1035-d}{1035}\times\frac{946-e}{946}[/tex],

where the denominators are:
[tex]{52 \choose 2}=1326, {50 \choose 2}=1225, {48 \choose 2}=1128, {46\choose 2}=1035, {44\choose 2}=946[/tex].

"a" is the number of ways of getting a pair in a fresh 52-card deck.  [tex]a=13\times {4 \choose 2}=13\times 6=78[/tex].

"b" is the number of ways of getting a pair in the remaining 50 cards, assuming the first player did not get a pair.  [tex]b=11\times {4\choose 2}+2\times {3 \choose 2}=11\times 6+2\times 3=72[/tex].

Now the situation gets complicated, and this simple formula won't work.  Neither of the first two players have a pair.  Between them, they may have 2, 3, or 4 different ranks of cards.  If they have 4 distinct ranks, then [tex]c=9\times 6+4\times 3=66[/tex].  If they have 3 distinct ranks, then [tex]c=10\times 6+2\times 3+1\times 1=67[/tex].  If they have 2 distinct ranks, then [tex]c=11\times 6+2\times 1=68[/tex].

To follow your method, a probability tree is required.

Your previous method was "close" because these fractions remain "close" to 78/1326:

[tex]\frac{78}{1326}\approx 0.058824[/tex]

[tex]\frac{72}{1225}\approx 0.058776[/tex]

[tex]\frac{66}{1128}\approx 0.058511[/tex]

[tex]\frac{67}{1128}\approx 0.059397[/tex]

[tex]\frac{68}{1128}\approx 0.060284[/tex]


integration
calculus
trigonometry
complex analysis

Prove that $ \int_{0}^{\pi} e^{cos(\theta)} cos(sin(\theta)) d\theta \equal{} \pi$ by considering $ \int_\gamma \frac{e^z}{z} dz$ where $ \gamma$ is the unit circle.

okay, so just use some parity-type arguments to show the original integral is half of the integral from $ 0$ to $ 2\pi$ and then use residue calculus on the latter integral you suggested

$ e^{\cos \theta} \cos(\sin \theta) \equal{} \sum_{n\equal{}0}^\infty \frac{\cos n\theta}{n!}$


AMC
AIME
USA(J)MO
USAMO
invariant
inequalities
pigeonhole principle

Many here on this community say that AoPs Volume 1 will prepare yo for the AMC's. But does this book have actual AMC-type problems, or does it only have AMC-level problems?
For AOPS Volume II, does it have AIME-type problems, or just AIME level problems? Are the problems good enough so that you can see problems like what yo've seen on the book in the 2005 AIME? Are there any USAMO, or other national level problems there, as well?
Similarly, in the Art and Craft of Problem Solving, are there any USAMO/IMO-type problems or are there only advanced level problems not related to USAMO,etc. How much does this book really prepare you for the USAMO?

Sorry if I'm unclear. :(

In my opinion Paul Zeitz' book prepares you to get to USAMO, it doesn't help you too much for USAMO.

I will leave Richard to answer the questions about the AoPS 1 & 2 volumes ;)

The AoPS books have many actual AMC problems, but no AIME problems.  However, there are many AIME-level problems.

I do disagree with Valentin a bit on Art and Craft - I do think it is helpful for the USAMO, particularly if you haven't been elsewhere exposed to ideas like extremal principle, invariants, various inequalities, etc.  However, it won't get you anywhere near mastering the USAMO.  Only practice will do that.  It will get you started on it, and it does have lots of olympiad problems in it.  It's possible that my disagreement with Valentin is largely cultural - I think students in Romania are more regularly exposed to the basic problem solving approaches Zeitz 'introduces' than Americans are.  I, for example, had never even heard of Pigeonhole Principle before MOP, and I never knew I was pronouncing Euler wrong until I got there (none of my teachers or peers ever corrected me in high school, so I assume they didn't know it either).

Indeed, Richard is right. I did heard about the Pigeonhole principle and Euler since 5th grade :), so maybe that is why I think that it does not help you too much. Also this is because Zeitz approaches many topics that are suited more for Putnam (like integral calculus). Nevertheless, you should definitely read the book if your goal is making top 30 in USAMO.

Thanks! What books actually help you master USAMO-level problems (though I'm nowhere near that level now :) ?

I think that you must have both natural talent (25%), and a lot of practice (70%) and a little bit of luck (5%) to make a good usamo (or for that matter any national math olympiad).
Since the first and the third you cannot possibly influence, you must work on your no. 2 ;)
Solve a lot of problems. For books: geometry revisited, engel, Ttitu's books, but I think that solving all past usamo's, taking Olympiad Independent Study course at AoPS and being active in the Advanced forums on the site should basically do it.