calculus
geometry
3D geometry
derivative
MATHCOUNTS

Who's coming to the Cobb County tournament next week? 50 problems in 60 minutes scares me.

My team, namely me, Bethany, Val, and 7th grade bro Danny. We're doing varsity, even though only half of us are V age... :| :D :huh:

Northview'll be there... make sure you know calculus OR ELSE  :evilgrin:

I really hope I can remember all the calculus I taught myself (with the aid of a book)... :)

I think / hope any calculus on the written test will be very basic.

I would guess limits and maybe basic derivatives. I would think that it's too early for a related rates problem. But then, I haven't seen the test. (Well, all of it! :-)

Pretty much all of the Cobb County schools will be there. But it's also band and orchestra (I think) all-state auditions so they will be some folks missing.

[quote="gtf"]I would think that it's too early for a related rates problem.[/quote]
I hope there are related rates problems as I just learned about those today (and have a test on them on Thursday).

Could you give an example of a related-rates problem? I dont' think I know what you mean.

I'm not going to be there because of all-state.  :(

Classic related rates problem:

A cone of some dimension (it should be specified, but I'm too lazy) is being filled at a rate of 1 ft^3/min. When the height of the water is 4 inches, how fast is the height rising?

All-state auditions are claiming me this weekend.  They claimed me two weekends ago, also, but that was for chorus.  I think, in fact, that all-state auditions have thus far prevented me from ever attending the Cobb Invitational.

Yes, "related rates problems" essentially means "applications of the derivative."

And why, Karthik, should we know calculus?  Calculus in high-school math competitions is bad manners.

Heh, the funny thing is, you don't even need to take a derivative to solve a problem such as the one Mr. Fulton has shown.  In fact, it might be possible to solve it without any concious knowledge of calculus whatsoever.

[quote="Elemennop"][quote="Twiz"][quote="Boy Soprano II"]Calculus in high-school math competitions is bad manners.[/quote]
As is statistics - *cough cough* north springs tournament *cough cough*[/quote]

As is geometry[/quote]

As is anything above the MathCounts level. [/sarcasm] *coughs, gets cough drop and finally stops coughing*

[quote="funcia"]
As is anything above the MathCounts level. [/sarcasm] *coughs, gets cough drop and finally stops coughing*[/quote]

Statistics is not necessarily "above the MathCounts level," but it's something that most people learn the basics of later in high school, so like calculus, it is not usually an indicator of skill but of what math courses one has taken.

[quote="Boy Soprano II"]
And why, Karthik, should we know calculus?  Calculus in high-school math competitions is bad manners.[quote]

No dude seriously. Some of the questions present are brutally calculus. They ask you to find tangent lines to implicit functions. It's practically impossible to do them without calc unless you really want to plug in $ y \equal{} mx \plus{} b$ and set the resulting discriminant (if you're lucky enough for the degree to be $ 2$) to $ 0$ or something. Sometimes, they just straight out ask you if $ f(x) \equal{} x^2$ what is $ f'(x)$ and then you have to know calc.

[quote="funcia"][/quote][quote="Elemennop"][quote="Twiz"][quote="Boy Soprano II"]Calculus in high-school math competitions is bad manners.[/quote]
As is statistics - *cough cough* north springs tournament *cough cough*[/quote]

As is geometry[/quote][quote="funcia"]

As is anything above the MathCounts level. [/sarcasm] *coughs, gets cough drop and finally stops coughing*[/quote]

The thing i,s i'm not sarcastic

congrats to heng, harrison, and tomas! i got sixth place, so I'm pretty happy. my team (of me) got 19th place, but I pointed out to the people that they forgot to give me credit for my power round, so I got bumped up 75 points. i looked at the scores and thought I had gotten 8th place, but I was looking at the jv scores, so my team's final place was actually 14th place. yay.

I understand lots of scores were messed up, but oh well, it was approximate ranking :).

Congratulations to all the winners  :) I thought I got 41, but I guess that at the end of the confusion, my score dropped a bit  :oops: I'm happy about my 4th though  :) But I liked the test much better than that of last year or the year before; the problems were much more of a "Lassiter-like" competition level, which was good, because you had to work fast to finish.

Yes, i did not finish. I missed out on the last page, which was perhaps the easiest page of the test looking back at it. :( I also think they need to make the group ciphering harder next year. I don't really like the idea of item analysis either, but maybe that's just because it didn't work out for me at this tournament :-D (I tied with fourth and fifth place and got sixth).

Edit: who keeps rating seva's posts as spam? :rotfl:

Hmm... how come neither Miles nor Rohan came for Lassiter? Could it have been that they wrote a few of the questions?

Not likely. Miles had an all-state audition. Not sure where Rohan was. A couple of the last page where mine. I especially liked:

If $ \displaystyle\sum_{n \equal{} 0}^\infty sin^{2n}\theta \equal{} 3$, then $ cos 2\theta \equal{}$

Really easy if you see the geometric series. Maybe really hard otherwise.  :D It's actually a take off from an old AMC problem in the past few years. (Well, maybe more of a ripoff than a take off.)

BTW, I caught a scoring error. Dodgen Middle School actually came in 4th! Not 9th. Scary how smart those young kids are!

Fulton

I think Rohan had all-state auditions, or honor orchestra auditions, which were done at the same time as all-state auditions.

Anyhow, item analysis is really the only way to break ties without tiebreakers.  So unless you are [i]planning[/i] to have ties and have a bunch of tiebreakers ready, it's really the only way to deal with ties.  I think we might have had to do that at Lassiter last year.

(Edit :  I'm Miles.)

My problem with item analysis is that in a race-like test such as Cobb, working the easy problems first and then going back to the harder ones is a good way to be able to finish as many problems as possible, so if you get penalized for not solving the harder problems, you are essentially being penalized for having a good strategy. That being said, I understand that tiebreakers can take a lot of time and may not always be a possibility. I think the best thing to do is to have a scoring system complicated enough to avoid many ties (like Billy's differently weighted questions and Lassiter's more valuable free response questions).

Really???

I thought I saw Rohan! I must be hallucinating, or maybe it was some other guy who looked like Rohan...but I remember he was sitting by Lassiter.

I didn't do that well, I got 15th :blush: . At least Northview Varsity won.

And I had a question: I heard that there was a mistake with two of the problems (in Varsity) and that Mr. Slater counted those, but there was another that was a mistake, but he acknowledged that but did not count it in scoring. What exactly happened there?

Anyway good job every1!!!

Wow, I'm pretty slow.  Who's Boy Soprano II and who's funcia?  No, this is not spam, those who will mark this post as spam!

Boy Soprano = Miles
funcia = Val, kinda shorter girl who was on my team yesterday and was practically on my team at Alpharetta. Has not placed [i]yet[/i], but is working on her pwnage skills. :)

I think that I completely agree with Miles. (As much as that hurts! :-) ) Item analysis is the only way to go about it. Eric, I agree that your strategy is the best, but the person that beats you solved the same number of questions, but somewhere did a "harder" question. With the "superior" strategy, you would have answered more correctly.

BTW, after winning, what, 2 tournaments in a row? It's ok to have an "off" week!

The mistake was in the JV test. There was an error on #23. The wrong answer was counted as correct. Unfortunately, no one brought it to anyones attention until the awards were being announced.

[quote="Twiz"][quote="gtf"]I think that I completely agree with Miles. (As much as that hurts! :-) ) Item analysis is the only way to go about it. Eric, I agree that your strategy is the best, but the person that beats you solved the same number of questions, but somewhere did a "harder" question. With the "superior" strategy, you would have answered more correctly.

BTW, after winning, what, 2 tournaments in a row? It's ok to have an "off" week![/quote]

I wasn't really complaining about item analysis in reference to my place at the last tournament. I really feel like I placed how I deserved to place, and I'm completely happy with sixth place :) . I think that I was really fortunate at North Fulton and UGA and don't expect to be among the top three too often. I didn't even use the strategy I was talking about, but I probably should have because I didn't finish. I just think that tie breakers and complicated scoring schemes are a more definitive way of breaking ties, but I guess I have to admit that item analysis isn't too bad.[/quote]

Though item analysis isn't too bad ( :D ) tiebrakers like a speed round or something would be pretty interesting.

Heh Karth would you have liked a speed round for my tiebreakers? ;)

I feel pretty bad about getting a lower place than Arvind did...20th is not the best place I could have gotten.

Well actually everyone who got a place lower than 10th place may bump their place up 1 because there was no one who got 11th place. It was completely missing from the list. :wink:

OK, so i got 14th...anyway, Karthik, that would be exiting but since there were a lot of ties it would be frustrating...just imagine, like 7 tiebrakers...each with like 8 people. ouch. :P

Isn't that how they broke ties at UGA if there were ties? I know there is at least one other tournament where they use this kind of problem analysis to break ties, but i can't think of it right now off the top of my head...

No, actually, ties would be broken probably only where it mattered in the top 10. The rest could be broken simply by item analysis, I guess. Besides, it provides entertainment too.

gr. cobb was not pretty. at all. didn't really wake up until the individual test was over. i'm still happy with the team results though. i liked the competition overall, although they could use some work on the mistakes.

OK, major brain fade. How do you work the only where you need the remainder of the number that you get by placing all the numbers from 1 to 100 end to end and divide by 11?

Poorly stated because I'm doing it quickly and class is starting.

Thanks,

Okay, unfortunately, I didn't think of this last saturday, but here's a way to do that problem:
[hide]
Let $ x\equal{}1234567891011...979899$
For any integer, it is congruent mod 11 to the number obtained by adding the numbers in the ones place, the hundreds place, the ten thousands place,... and subtracting the numbers in the tens place, the thousands place, the hundred thousands place,... (this can be proved using mods). The first digit in $ x$ is $ 1$ and it is the 189th digit to the left. It is one of the numbers that must be subtracted. So we start, $ \minus{}1\plus{}2\minus{}3...\plus{}8\minus{}9\equal{}\minus{}5$ Now the rest of the digits come in pairs that form two digit numbers, so the remaining sum we are looking for should be $ (1\minus{}0)\plus{}(1\minus{}1)\plus{}(1\minus{}2)\plus{}...\plus{}(9\minus{}9)$ Each two digit number except the ones that end in 0 and the ones that have two of the same digits has a reverse number that it cancels with (for example $ (4\minus{}6)$ cancels with $ (6\minus{}4)$). The numbers that have two of the same digit cancel themselves out because their sum is zero, so we need only to add $ (1\minus{}0)\plus{}(2\minus{}0)\plus{}(3\minus{}0)\plus{}...\plus{}(9\minus{}0)\equal{}45$. $ \minus{}5\plus{}45\equal{}40$ which is congruent to $ 7$ (mod 11) I think the answer is $ 7$, but I have to go to class. I'll come back and check this later. I realize this solution would probably take longer than one minute, so it might not be worthwhile in a race situation like Cobb.[/hide]

So can't we just consider the 2-digit numbers: we have 9 of each digit 0...9 in the even places, and 10 of each digit 1..9 in the odd places, so we have -(1...9) in the sum, which is 45, and then we have 1-2+3-4..9=-5, so we end up with 40==7 mod 11.


Let a,b,c positive real numbers such that abc=1.  Prove that

  $ 1\plus{}\frac{3}{a\plus{}b\plus{}c\plus{}}\geq2\sqrt{\frac{3}{ab\plus{}bc\plus{}ca}}$.

It's an easy one. Here's the solution:

Given that abc=1
since, a+b+c <= ab+bc+ca
so, 3/(a+b+c) => 3/(ab+bc+ca) [Equality holds iff a=b=c=1]
Thus, 2[3/(a+b+c)]^1/2 => 2[3/(ab+bc+ca)]^1/2
again by AM-GM,
1+3/(a+b+c) => 2[3/(a+b+c)]^1/2 => 2[3/(ab+bc+ca)]^1/2.......Done! :lol:

[quote="Sunkern_sunflora"]It's an easy one. Here's the solution:

Given that $ abc\equal{}1$
since, $ a\plus{}b\plus{}c \leq ab\plus{}bc\plus{}ca$
so, $ \frac{3}{a\plus{}b\plus{}c}\geq\frac{3}{ab\plus{}bc\plus{}ca}$ [Equality holds iff a=b=c=1]
Thus, $ 2\sqrt{\frac{3}{a\plus{}b\plus{}c}}\geq 2\sqrt{\frac{3}{ab\plus{}bc\plus{}ca}}$
again by AM-GM,
$ 1\plus{}\frac{3}{a\plus{}b\plus{}c}\geq 2\sqrt{\frac{3}{a\plus{}b\plus{}c}}\geq 2\sqrt{\frac{3}{ab\plus{}bc\plus{}ca}}$.......Done! :lol:[/quote]

[quote="Marius Mainea"]
Given that $ abc \equal{} 1$
since, $ a \plus{} b \plus{} c \leq ab \plus{} bc \plus{} ca$
[/quote]

 It is wrong.  Try $ a\equal{}b\equal{}\frac{1}{2}$,  $ c\equal{}4$

[quote="Marius Mainea"]Let a,b,c positive real numbers such that abc=1.  Prove that

  $ 1 \plus{} \frac {3}{a \plus{} b \plus{} c}\geq2\sqrt {\frac {3}{ab \plus{} bc \plus{} ca}}$.[/quote]
Try $ a \equal{} 4$ and $ b \equal{} c \equal{} \frac {1}{2}.$ :wink:
But the following inequality is true already:
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that:
\[ 1 \plus{} \frac {1}{a \plus{} b \plus{} c}\geq\frac {4}{\sqrt {3(ab \plus{} ac \plus{} bc)}}
\]

[quote="arqady"]
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ abc \equal{} 1.$ Prove that:
\[ 1 \plus{} \frac {1}{a \plus{} b \plus{} c}\geq\frac {4}{\sqrt {3(ab \plus{} ac \plus{} bc)}}
\]
[/quote]

  Denote $ a\equal{}\frac{1}{x}, b\equal{}\frac{1}{y}, c\equal{}\frac{1}{z}$  we must prove that

 $ 1\plus{}\frac{1}{xy\plus{}yz\plus{}zx}\geq\frac{4}{\sqrt{3(x\plus{}y\plus{}z)}}$

 But

$ LHS\geq1\plus{}\frac{3}{(x\plus{}y\plus{}z)^2}\equal{}\frac{1}{3}\plus{}\frac{1}{3}\plus{}\frac{1}{3}\plus{}\frac{3}{(x\plus{}y\plus{}z)^2}\geq4\sqrt[4]{\frac{1}{3}\frac{1}{3}\frac{1}{3}\frac{3}{(x\plus{}y\plus{}z)^2}}\equal{} RHS$


inequalities
geometry
geometric transformation
reflection

$ x,y,z,t >0$
prove that  $ x^2 \plus{}y^2\plus{}z^2 \plus{}t^2 \plus{}xyzt\plus{}1 \geq xy\plus{}yz\plus{}zt\plus{}tx \plus{} xz\plus{}yt$

It guess the follow Lemma can be useful.
[b]Lemma(Turkevici, Kvant):[/b]
[i]For any positive real numbers $ x,y,z,t$ we get that $ x^4\plus{}y^4\plus{}z^4\plus{}t^4\plus{}2xyzt \geq x^2y^2\plus{}y^2z^2\plus{}z^2t^2\plus{}t^2x^2\plus{}x^2z^2\plus{}y^2t^2$[/i]

it was posted in the iranian forum too [url]http://www.mathlinks.ro/viewtopic.php?t=206550[/url] but it's in persian,anyway here's the solution:

first of all WLOG assume that $ x\geq y\geq z\geq t\geq 0$,now let:

$ f(x,y,z,t) \equal{} x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2 \plus{} xyzt \plus{} 1 \minus{} xy \minus{} yz \minus{} zt \minus{} tx \minus{} xz \minus{} yt$

we will prove that $ f(x,y,z,t)\geq f(\sqrt {xz},y,\sqrt {xz},t)$,note that:

$ f(x,y,z,t)\geq f(\sqrt {xz},y,\sqrt {xz},t)$

$ \iff x \plus{} z \plus{} 2\sqrt {xz}\geq y \plus{} t$

which is true according to the assumption $ x\geq y\geq z\geq t$

hence according to SMV-theorem we get that $ f(x,y,z,t)\geq f(a,a,a,t)$,so it's sufficient to show that:

$ f(a,a,a,t)\geq 0$

$ \iff t^2 \plus{} a^3t \plus{} 1\geq 3at$

which is true by AM-GM:

$ t^2 \plus{} a^3t \plus{} 1\geq 3\sqrt [3]{a^3t^3} \equal{} 3at$

______________________________________________________________________

P.S:also note that if we set $ t \equal{} 1$ then we would end up with the inequality:

$ x^2 \plus{} y^2 \plus{} z^2 \plus{} xyz \plus{} 2\geq xy \plus{} yz \plus{} zx \plus{} x \plus{} y \plus{} z$

which is another interesting inequality (with a more "elementary" solution).

Of course this problem is pretty much a consequence of the turkevici inequality but I feel like that just trivializes the problem and makes it uninteresting.

I proposed this problem in mathematical reflections with that knowledge. My proof also used mixing variables.

Problem:
Let $ a,b,c,d$ be non-negative real numbers. Prove that the following inequality holds:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} 1 \plus{} abcd\ge ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd
\]

Solution:
Observe that the inequality is symmetric. Hence, we can say without loss of generality that $ ab\minus{}1\ge cd\minus{}1$.

Case 1: $ (ab\minus{}1)(cd\minus{}1)\ge 0$
Rewrite the inequality as follows: \[ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \minus{} (a \plus{} b)(c \plus{} d) \plus{} (ab \minus{} 1)(cd \minus{} 1)\ge 0
\]
Then $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\ge \frac{1}{2}\cdot (a\plus{}b)^2\plus{}\frac{1}{2}\cdot (c\plus{}d)^2\ge (a\plus{}b)(c\plus{}d)$ by the AM-GM inequality twice.

Case 2: $ (ab \minus{} 1)(cd \minus{} 1) < 0$

Since $ ab \minus{} 1 \ge cd \minus{} 1$. Then $ ab \minus{} 1 > 0$, and $ cd \minus{} 1 < 0\implies 3\minus{}cd>0$.

Let $ f(a,b,c,d)$ denote the LHS minus the RHS. Observe that
\[ f(a,b,c,d) \minus{} f\left(\frac {a \plus{} b}{2},\frac {a \plus{} b}{2},c,d\right) \equal{} \frac {1}{4}\cdot (a \minus{} b)^2(3 \minus{} cd)\ge 0
\]
It suffices to prove the inequality
\[ (cd \plus{} 1)\cdot f(a,a,c,d)\ge 0\iff
\]

\[ ((a)(cd \plus{} 1) \minus{} (c \plus{} d))^2 \plus{} \frac {1}{2}\cdot (c^3d \plus{} d^3c \minus{} 2c^2d^2) \plus{} \frac {1}{2}\cdot (c^3d \plus{} d^3c \plus{} 1 \plus{} 1 \minus{} 4cd)
\]
which is true by AM-GM.


Let $ a,b,c \geq 0$ such that $ 3a^2 \plus{} 2b^2 \plus{} c^2 \equal{} 6$
Find the maximum value of  $ C \equal{} a \sqrt {1 \plus{} b^3} \plus{} b \sqrt {1 \plus{} c^3} \plus{} c \sqrt {1 \plus{} a^3}$

It's weaker than hungkhtn's old result  :wink:
Given $ a, b, c \geq\0$ and $ a + b + c \leq\ 3$.Prove that:
$ a\sqrt {1 + b^{3}} + b\sqrt {1 + c^{3}} + c\sqrt {1 + a^{3}} \leq\ 5$
Equaity occurs when $ (a, b, c) = (0, 1, 2)$


[hide]checking with a calculator, this is as simple as I got it

I can't latex
factor sqrt(11-6sqrt(2))(11+6sqrt(2)))(sqrt(11+6sqrt(2))-sqrt(11-6sqrt(2)))
by difference of two squares the right term is 7

now we set up an equation

x=sqrt(11+6sqrt(2))-sqrt(11-6sqrt(2))
x=22+2sqrt(11+6sqrt(2))qrt(11-6sqrt(2))

by difference of two squares the left term is 14 and x^2=8 and x=2sqrt(2)

the top is 14sqrt(2)

set up another equation

x=sqrt(sqrt(5)+2)+sqrt(sqrt(5)-2)
x^2=2sqrt((sqrt(5)+2)(sqrt(5)-2))+2sqrt(5)

by difference of two squares the left term is 2

and the denominator is sqrt(2sqrt(5)+2)-sqrt(sqrt(5)+1))

and factoring

sqrt(sqrt(5)+1)(sqrt(2)-1)

multiplying the numerator and the denominator by sqrt(2)+1

we have (28+14sqrt(2))/(sqrt(sqrt(5)+1))

and multiplying both the numberator and denominator by sqrt(sqrt(5-1))

we have (14+7sqrt(2))(sqrt(sqrt(5)-1)[/hide]

I got the numerator but have no idea how to work out the denominator.
numerator first half:
sqrt(11+6sqrt2)*sqrt{(11+6sqrt2)(11-6sqrt2)}=7*sqrt(11+6sqrt2)

similarly, the second half is 7*sqrt(11-6sqrt2). so substract them get 49.

for the denominator I'm clueless, but I think the answer should be simpler... :maybe:


To attend a wedding, Barbara starts driving west from Woodstock at 80km/h. One hour later, Barbara's parents leave Woodstock and drive along the same road at 100km/h. At what distance from Woodstock will Barbara's parents pass her?

I think the answer is 400 km but could someone show me how to figure it out using linear systems?  :)

I believe to find this, you would want to do:
100X=80X+80
20X=80
x=4 hours, which would mean that the parents traveled 400 km.

....but would that qualify as a linear system?


1. A spacecraft receding from the Earth at $ 0.97c$ transmits data at the rateof $ 1 \times 10^4$ pulses per second. At what rate are they received?

2. A spectral line whose wavelength in the laboratory is 400 nm is found to be shifted to 600nm in the spectrum of a distant galaxy. What is the receding speed of the galaxy?

Use the Relativistic doppler effect: $ \nu\equal{}\nu_0 \sqrt{\frac{1\minus{}\beta}{1\plus{}\beta}}$ where $ \beta\equal{}\frac{v}{c}$

Thanks, what is  $ v_0$ in this case? The speed from whose inertial reference frame (wrt question 1). Also, does question 2 imply that light changes colour when it passes from one galaxy to the next?

[quote="aidan"]Thanks, what is  $ v_0$ in this case? [/quote]
Sorry about the ambiguity. $ \nu_0 \equal{} 1\times 10^4 \text{Hz}$ is actually the source frequency. Velocity $ v$ is accounted in the $ \beta \equal{} \frac {v}{c}$ term.

[quote="aidan"]Also, does question 2 imply that light changes colour when it passes from one galaxy to the next?[/quote]
I wouldn't put it like that, as it carries the notion that light has some 'absolute' wavelength, which is not true. Wavelength, like any length, depends on frame of referance. Say rather, that seen from one galaxy, another galaxy appears to emit longer wavelenghts.

Ok.. Thanks. Two more challenging problems are here:

3. Some particle accelerators allow protons $ (p^\plus{})$ and antiprotons $ (p^\minus{})$ to circulate at equal speeds in opposite directions in a device called a storage ring. The particle beams cross each other at various points to produce $ p^\plus{}$ and $ p^\minus{}$ collisions. In one collision, the outcome is $ p^\plus{} \plus{} p^\minus{} \rightarrow e^\plus{} \plus{} e^\minus{} \plus{} \gamma \plus{} \gamma$, where $ \gamma$ represents a high energy gamma ray photon. The electron and positron are ejected from the collision at $ 0.9999995c$, and the gamma photon wavelength was found to be $ 1.0 \times 10^{\minus{}6} \text {nm}$. What were the proton's and anti-proton's speeds prior to the collision?

4. The star Alpha goes supernova. Ten years later and 100 light years away, as measured by astronomers in the galaxy, star Beta explodes. (Astronomer's frame need not be at rest relative to the stars). An alien spacecraft passing through the galaxy finds that the distance between the two explosions is 120 light years. According to the aliens, what is the time between the explosions?


How many digits are used to number the first 228 pages in a phone book?
[hide="answer"]576 if I'm not mistaken[/hide]

I dunno if there is a shorter way for this but...
[hide]9 numbers from 1-9, 9 digits
90 numbers from 10-99 so $90 \times2=180$ digits
129 numbers from 100-228 so $129 \times3=387$ digits
$9+180+387=\boxed{576}$ Digits. [/hide]
Why does LaTeX look different?

[hide]9 one-digit numbers,
90 two-digit numbers,
129 three-digit numbers,
for a total of $9+90\times 2+129\times 3=576$.[/hide]

[quote="solafidefarms"]To answerers of problems: Please don't just post the answer; explain the solution...If you get it wrong, that way we can point out where you went wrong. Also, unless otherwise specified, put your solution in [hide ] tags.

To posters of problems: Wait a few days before posting the right answer or giving any indication of who is correct. People then post more solutions. Also, unless it's a problem you've been having trouble with, include in your first post "Hide your answers!".

I don't always follow my own rules either, but try to follow them. It makes for a better forum.[/quote]
 :roll:

[quote="Ignite168"][quote="pgpatel"][quote="solafidefarms"]To answerers of problems: Please don't just post the answer; explain the solution...If you get it wrong, that way we can point out where you went wrong. Also, unless otherwise specified, put your solution in [hide ] tags.

[b]To posters of problems: Wait a few days before posting the right answer[/b] or giving any indication of who is correct. People then post more solutions. Also, unless it's a problem you've been having trouble with, include in your first post "Hide your answers!".

I don't always follow my own rules either, but try to follow them. It makes for a better forum.[/quote]
 :roll:[/quote][/quote]
I'm known to be a hypocrite :D. I was following treething's format.

Digits(for a 3 digit page number book) = 3p-108, where p is the number of pages.

[quote="pgpatel"][/quote][quote="Ignite168"][quote="pgpatel"][quote="solafidefarms"]To answerers of problems: Please don't just post the answer; explain the solution...If you get it wrong, that way we can point out where you went wrong. Also, unless otherwise specified, put your solution in [hide ] tags.

[b]To posters of problems: Wait a few days before posting the right answer[/b] or giving any indication of who is correct. People then post more solutions. Also, unless it's a problem you've been having trouble with, include in your first post "Hide your answers!".

I don't always follow my own rules either, but try to follow them. It makes for a better forum.[/quote]
 :roll:[/quote][/quote][quote="pgpatel"]
I'm known to be a hypocrite :D. I was following treething's format.[/quote]

 :spam:  :spam:  :spam:  :spam:


[quote="nonie"]The length in feet of 3 pieces of timber are 48, 72, and 40. The sawmill operator needs to cut the timber into logs of equal length with no waste. how many feet long is the greatest possible lenth she can cut.[/quote]
[hide]
What we're trying to find is the greatest common denomitor, or GCD.

Which is 8.[/hide]

[hide]
the gcf of 48, 72, and 40 is
8 so $\boxed{8}$ is the answer[/hide]

Yeah you're right... :) . That was an easy one....

[quote="berry"]i dont get how you guys got that method and answer.could you explain it?? :blush:[/quote]

[hide=" explanation"]Sure...

I assume you know gcf is greatest common factor and if you don't well it is.

Th problem asks for the greatest possible length the sawmill operator can cut the logs into to equal pieces  with out a remainder The large logs are of lengths 48, 72, and 40. So you need to find the GCF of 48, 72, and 40. To find the GCf you can prime factorize the numbers 48 is $2^4* 3$ 72 is $3^2* 2^3$ and 40 is $5* 2^3$ take the numbers that are common and have the least value for the  exponent so it is just $2^3$ = 8 because the 5's and 3's arent common throughout. You can also find GCF's by euclids method( I think that's what its called) You use dividing for that one.. [/hide]

[hide] All you need to do is find the GCD of all of the numbers, which is 8.[/hide]

GCF is 8, so that is our answer :D nice problem


inequalities
function
inequalities proposed

Let $a,b,c$ is possitive number. Prove that:
$\frac 1a+\frac 1b+\frac 1c + \frac{27}{a+b+c} \geq 16 \sum \frac{1}{a+2b+c}$
Personal I think it isn't difficult. Hope that it will receive nice sols.

[quote="e.lopes"]by cauchy, $(a + (a+b+c))(\frac{1}{a}+ \frac{9}{a+b+c}) \geq 16$[/quote]

  Wow, I admit it is a very nice sols. Indeed, I generate it through this inequality:
$\frac1a+\frac1b+\frac1c +\frac{9}{a+b+c} \geq 4 \sum{\frac{1}{a+b}}$ and apply it twice. :lol:

[quote="Ezbakhe Yassin"][quote="Cezar Lupu"]Yes, The inequality posted by Gibbenergy can be easily proven by Popoviciu' inequality.
 Just take the convex function $f(x)=\frac{1}{x}$ and then you will get the result. :)  ;)[/quote]

Could you post your solution? :)[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?t=68467 and look at post #5 . You will see Popoviciu. Acording to Cezar Lupu, you will see that my inequality is very easy since it follows immediately Popoviciu :)

[quote="konlsy"][quote="e.lopes"]by cauchy, $(a + (a+b+c))(\frac{1}{a}+ \frac{9}{a+b+c}) \geq 16$[/quote]
could u write it more details.thanks :ninja:[/quote]

  It is very easy.
Applying AM-GM:
$a+(a+b+c)=a+\frac{a+b+c}{3}+\frac{a+b+c}{3}+\frac{a+b+c}{3} \geq 4 \sqrt[4]{\frac{a(a+b+c)^3}{3^3}}$
$\frac1a+\frac{9}{a+b+c}=\frac1a+\frac{3}{a+b+c}+\frac{3}{a+b+c}+\frac{3}{a+b+c} \geq 4 \sqrt[4]{\frac{3^3}{a(a+b+c)^3}}$
Then time them and get 16. :)

[quote="konlsy"][quote="e.lopes"]by cauchy, $(a + (a+b+c))(\frac{1}{a}+ \frac{9}{a+b+c}) \geq 16$[/quote]
could u write it more details.thanks :ninja:[/quote]

Let's remind what Cauchy's inequality says: $(a^2+b^2)(x^2+y^2)\geq (ax+by)^2$.

So, Cauchy's inequality yields that $[a + (a+b+c)]\cdot\left(\frac{1}{a}+ \frac{9}{a+b+c}\right) \geq \left(\frac{\sqrt{a}}{\sqrt{a}}+3\cdot \frac{\sqrt{a+b+c}}{\sqrt{a+b+c}}\right)^2=(1+3)^2=16$

So:

$\frac1{a}+\frac9{a+b+c}\geq \frac{16}{2a+b+c}$, $\frac1{b}+\frac9{a+b+c}\geq \frac{16}{a+2b+c}$ and $\frac1{c}+\frac9{a+b+c}\geq \frac{16}{a+b+2c}$

Summing those three inequalities, we have that:

$\frac1{a}+\frac1{b}+\frac1{c}+\frac{27}{a+b+c}\geq \frac{16}{2a+b+c}+\frac{16}{a+2b+c}+\frac{16}{a+b+2c}= 16 \sum \frac{1}{2a+b+c}$

Hope it's a bit more clear now :) ;)


limit
inequalities
calculus
calculus computations

Is there a way to find this limit without using Stirling's approximation?

Due to the inequality $ (2k \plus{} 1)^2 > 2k(2k\plus{}2)$ for all $ k$, we have that
$ ((2n \plus{} 1)!!)^2 \equal{} \prod_{i \equal{} 1}^{n} (2i \plus{} 1)^2 > \prod_{i\equal{}1}^{n}(2i)(2i\plus{}2) \equal{} ((2n)!!)^2 * \frac{2n\plus{}2}{2}$
Therefore, $ \frac {(2n)!!}{(2n \plus{} 1)!!} < \frac {1}{\sqrt {n \plus{} 1}}$ holds.
Taking limits, we have the result immediately.


Euler
function
limit
number theory open
number theory

[quote="dule_00"]For every $\delta >1$ and any positive $\epsilon$ there exist $n$ such that $|\frac{\phi(n)}n-\delta|<\epsilon$.

True or false???[/quote]

Obviously, if $\delta > 1$ this is false because $\frac{\phi(n)}{n} \le 1$ for all integers $n$.

However it is true for [u]infinitely many[/u] $n$ if $0\le \delta \le 1$ and $\epsilon > 0$.

[b]Hint:[/b] Consider integers $n$ of the form $n = p_{l} p_{l+1} p_{l+2} \dots p_{l+r}$, where $p_k$ is the $k$-th prime. In this case $\frac{\phi(n)}{n} = \prod_{ i= l }^{l+r} \left( 1 - \frac{1}{p_i} \right)$.
Use that $\; \; \lim_{x \to \infty} \prod_{ p \le x } \left( 1 - \frac{1}{p} \right) = 0$ and show that there are infinitely many pairs $l, r$ such that $\delta - \epsilon < \frac{\phi(n)}{n} < \delta + \epsilon$.

[b]Questions:[/b]

Let $0 \le \delta \le 1$ and $\epsilon > 0$.

Are there iffinitely many integers $n$ such that $\delta - \epsilon < \frac{\phi(n^2+1)}{n^2+1} < \delta + \epsilon$?

Are there iffinitely many primes $p$ such that $\delta - \epsilon < \frac{\phi(p-1)}{p-1} < \delta + \epsilon$?

Doychin: Maybe dule_00 meant $\frac{n}{\varphi(n)}$ ?


geometry
circumcircle
trigonometry
calculus
derivative
trig identities
Law of Sines

A triangle has sides of length a, b, c and circumradius R.

If a2 + b2 = c2 - R2,

find all possibilities for the angles of the triangle.

This is not very elegant.
Law of sines:

a/Sin[A]=b/Sin[B]=c/Sin[C]=2R.

Substituting, we have (c*Sin[A]/Sin[C])^2+(c*Sin[B]/Sin[C])^2=c^2-(c/(2Sin[C]))^2, (c^2*Sin[A]^2/Sin[C]^2)+(c^2*Sin[B]^2/Sin[C]^2)=c^2-(c^2/(4*Sin[C]^2), 4c^2*Sin[A]^2+4c^2*Sin[B]^2=4c^2*Sin[C]^2-c^2. Divide by c^2 since c != 0. So, 4*Sin[A]^2+4*Sin[B]^2-4*Sin[C]^2=-1, Sin[A]^2+Sin[B]^2-Sin[C]^2=-1/4, Note that Sin[Pi-x]=Sin[x] and since A+B+C=Pi, Sin[A]^2+Sin[B]^2-Sin[A+B]^2=-1/4, (after simplifying), Cos[A+B]*Sin[A]*Sin[B]=1/8.

Lemma. f[x,y]=Cos[x+y]*Sin[x]*Sin[y] attains a maximum at x=y=Pi/6 of 1/8 bounded by x+y<=Pi/2, x>=0, y>=0.

f_x is partial derivative with respect to x.

Proof. Checking boundary points, we have Cos[Pi/2]*Sin[x]*Sin[y]=0. Cos[x+y]*Sin[0]*Sin[y]=0. Cos[x+y]*Sin[x]*Sin[y]=0. Now checking interior points. f_x=Cos[x] Cos[x + y] Sin[y] - Sin[x] Sin[y] Sin[x + y], f_y=Cos[y] Cos[x + y] Sin[x] - Sin[x] Sin[y] Sin[x + y], Setting f_x, f_y=0, yields, Cos[x] Cos[x + y] Sin[y]=Cos[y] Cos[x + y] Sin[x]. Divide through by Cos[x+y] since x+y<Pi/2,x>0,y>0. Cos[x]Sin[y]=Cos[y]Sin[x], Divide through by Cos[x]Cos[y] since x,y<Pi/2. So Tan[y]=Tan[x], which holds true obviously when y=x+(Pi/2)k, but x+y<Pi/2, so y=x. Substituting back into f_x or f_y, we have Sin[x]*Cos[x]*Cos[2x]-Sin[x]*Sin[x]*Sin[2x]=0, divide through by Sin[x], Cos[x]*Cos[2x]-Sin[x]*Sin[2x]=0, Cos[3x]=0, 3x=Pi/2+Pi*k. Obviously the only x such that 0<x<Pi/2 is x=Pi/6. And y=x, critical point at (Pi/6,Pi/6). f[Pi/6,Pi/6]=1/8, the maximum on x>0, y>0, x+y<Pi/2. You can prove it is a local maximum by observing that D=f_{xx}(x,y)*f_{yy}(x,y)-(f_{xy}(x,y))^2 evaluated at x=y=Pi/6, is D=3/4. and f_{xx}(x,y) evaluated at x=y=Pi/6 is -1 (Hessian matrix).

So the only possibility is: 30-30-120 triangle.

Good job !

A somewhat more elegant solution can be found at

http://www.kalva.demon.co.uk/bmo/bsoln/bsol923.html


I identify as a mathematician and as genderqueer and pansexual.

Iowa Pride Network has connected me to a wonderful community of intelligent queer and allied youth, working together for social change.

My high school has a phenomenal math scene, where I've made my best friends and spent hours doing math with them.

But, like the rest of America, most of the queer youth community doesn't speak math.  And I don't know about any other GLBT students on my math team.  So sometimes I get a little lonely in both communities.

I'd like to find the intersection.  Are you here?

Yes, the intersection is here. Then again, I don't know about any others either; nearly everyone I know in math is indefinite to me, since they just don't talk about sex or relationships.

My approach to the whole thing: I just focus on different aspects of myself in different spaces. Here, I'm a combinatorist/analyst/teacher/dabbler in everything, and my non-mathematical identity mostly doesn't matter.

Although this is slightly different than what you seem to be looking for, it is good to keep in mind that there are also supportive professional organizations. NOGLSTP is National Organization of Gay and Lesbian Scientists and Technical Professionals and the math portion of this group meets every year at the joint math meetings. I'm on their mailing list.

I also probably shouldn't mention this since some people on here know my actual identity and the news isn't really public yet, but I'll keep it as anonymous as I can. It isn't impossible to find fellow gay mathematicians (although it is usually awkward). I recently started dating another grad student. So don't lose hope at finding others.

[quote="HilbertThm90"](although it is usually awkward)[/quote]  Isn't this true of every social interaction involving mathematicians?  :wink:

This isn't really advice for the original poster, but it's still something to consider. If you don't have much support or a big community where you are, move. Go to college in a big city (like Seattle).


geometry
geometric transformation
rotation
3D geometry
sphere
invariant

A rotating sphere contract slowly due to internal forces to (1/n)th of its original radius.What happens to its angular velocity.Show that increase in its energy equals the work done during its contraction.

(2/5)MR^2*w_1=(2/5)M(R-R/n)^2*w_2

From this we should find the change in w.Am I correct?

Regarding the workdone: Please help me to start with.

[hide="hint"]use adiabatic invariants :D [/hide]

I do not know about adiabatic invariant.
However,I should have written R/n in place of (R-R/n) in RHS

Your formula expresses the conservation of KE,well it isn't quite correct because you must have w1 and w2 squared.But KE is not conserving,the entire energy is conserving because the system is isolated.Now,if the entire energy is conserving shouldn't the question sound like this?:show that increase in it's KE equals the work done in it's contraction? I'm saying this because only conservative forces are acting so L=d KE=d PE(we've discussed this in another topic)
     Now about the entire energy that is conserved:there must be a potential energy of each dm situated at r from the center but only a sphere of radius r acts upon dm.So you integrate to fiind the potential energy of the system.The conservation clearly shows that angular velocity gets bigger but i'm not sure about the method.Don't you have answers to these beautiful problems? :)

i don't know how far this will be relevant to the discussion here but still if u want you can go through an almostr similar question like this
http://www.mathlinks.ro/Forum/viewtopic.php?t=135896

radac_mail,what I wrote in my 1st thread was conservation of angular momentum,not KE.That gives you the relation w'=n^2*w

change in KE:
DeltaT= T_aft - T_init =[1/2 * 2/5 * M * (R/n)^2 * (w*n^2)^2] -[1/2 * 2/5 * M * R^2 * w^2] = 1/5 * M * R^2 * w^2 * (n^2 - 1)

Change in pE:
I used the self energy concept.For a sphere it is (-3/5)[(GM^2)/R]
Now I did a trick.[GmM/(R^2)]=m(w^2)*R......force equation....
This gives GM=w^2*R^3
Now I compared final self energy and initial self energy.

Should have been equal,but came 3 times d(KE)
Would you please check if I went wrong?

sorry for the wrong observation.i think you should have dPE/dKE=-3/(n+1) but this doesn't answer the question...

As the contraction is quasistatic (or adiabatic) the resultant force in each element is always the centripetal force, and a careful calculation of its work leads to the same expression of the KE:
$\,\,{1\over5}M{\omega}^{2}R^{2}(n^{2}-1).\,\,$

It was a long time ago...as far as I remember I intrgrated mw^2/R w.r.t. dr and used a trick...GM=w^2*r^3...or somethink like that.I am to recheck again.


limit
logarithms
calculus
calculus computations

How to calculate
1) $ lim_{x\to \infty}$ $ \left(\frac{2^x\minus{}xln2}{3^x\minus{}xln3} \right)  ^\frac{1}{x^2}$
2) $ lim_{x\to \infty}$ $ \left( \sqrt[3]{x^3\plus{}x^2\plus{}x\plus{}1}\minus{}\sqrt{x^2\plus{}x\plus{}1}\frac{ln(e^x\plus{}x)}{x}\right)$
3) $ lim_{x\to \infty}$ $ \left( (x\plus{}3)^{1\plus{}\frac{1}{x}}\minus{}x^{1\plus{}\frac{1}{x\plus{}3}}\right)$

hello for 1) writing your searched limit in the form
$ e^{\lim_{x \to \infty}\frac{\ln\left(\frac{2^x\minus{}x\ln(2)}{3^x\minus{}x\ln(3)}\right)}{x^2}}$
and applying L'Hospital's rule we get $ 1$.
Sonnhard.

hello, for 2) we get
$ \lim_{x\to \infty}\left(\sqrt[3]{x^3\plus{}x^2\plus{}x\plus{}1}\minus{}\sqrt{x^2\plus{}x\plus{}1}\frac{\ln(e^x\plus{}x)}{x}\right)\equal{}\minus{}1/6$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, for 2) we get
$ \lim_{x\to \infty}\left(\sqrt [3]{x^3 \plus{} x^2 \plus{} x \plus{} 1} \minus{} \sqrt {x^2 \plus{} x \plus{} 1}\frac {\ln(e^x \plus{} x)}{x}\right) \equal{} \minus{} 1/6$
Sonnhard.[/quote]
Is this L'Hospital's rule applied again or  some different method?

hello, for 3) we get
$ \lim_{x \to \infty}\left((x\plus{}3)^{1\plus{}\frac{1}{x}}\minus{}x^{1\plus{}\frac{1}{x\plus{}3}}\right)\equal{}3$
Sonnhard.

Excuse me but could you please explain how you obtain 2) and 3)


calculus
real analysis
real analysis theorems

hi folks,
I'm an audio engineering student and am quite interested in DSP (Digital Signal Processing). Unfortunately, this topic is not fully covered in my school, so I picked up some books. Upon reading, I found that I apparently need to refresh my math knowledge. I had Calculus I at school, liked it and was quite good at it, so I should be able to start right there.
Basically, I need a good book that explains higher mathematics, preferably in a technical context. It doesn't have to go into detail, since I'll be only needing the basics of higher mathematics (is that a contradiction?), like Fourier series.

Anybody got any ideas?

Schaum's Outline of Advanced Calculus by Murray R. Spiegel


AMC
AIME
geometry
geometric transformation
reflection
USA(J)MO
USAMO

[quote="jnApp66"]I took the AMC 10a and I'm going to take the AMC 12b.  When they combine my AMC and AIME scores, will they use both the 10a score and 12b score or will they just use the one with the higher score?[/quote] Since you can qualify with a floor AIME score, a 10th grader or below will almost never qualify using the index. If you do, it will be your AMC 12 score.

[quote="lotrgreengrapes7926"] Since you can qualify with a floor AIME score, a 10th grader or below will almost never qualify using the index.[/quote]
This is false.  However, it is true that since you can qualify using floor score, all you have to worry about is your AIME score, and it is also true that you cannot qualify via index using the AMC 10.

[quote="jnApp66"][quote="Boy Soprano II"]
This is false.  However, it is true that since you can qualify using floor score, all you have to worry about is your AIME score, and it is also true that you cannot qualify via index using the AMC 10.[/quote]
What does "floor score" mean and what's the "index"?[/quote]This information is all on the [url=http://www.unl.edu/amc]AMC web site[/url] (click on USAMO).  (The numbers are out of date though; they don't yet reflect the USAMO expansion to 500 students.)

That's not true. Regardless of whether you take the 10 or the 12, if you get above the floor score, you will qualify for the USAMO.

[quote="13375P34K43V312"]In a nutshell, if you take the 10, they only look at your AIME score, and if you get above a certain number (it was 8 last year), you make USAMO. If you take the 12 only, they combine your AMC 12/AIME score, they take the higher of the A and B, and you get above a certain number (it was around 237.5? last year).[/quote]

Actually, with the expansion, it hasn't been as high.  It was only 217.5 last year.

Also, I thought you could get in by the floor just by being 10th grade or under even if you took the 12?

Yes. 10th graders and below taking the 12 can qualify via the floor. However, the term "index" is often used to describe AMC10 + AIME score as well as AMC12+AIME score. However, the AMC10 version of index doesn't really matter because its only effect (which was avoided last year, and will likely be avoided again) is to limit the number of floor qualifiers.

EDIT:
OOOPS I must be really confused.

Sorry deleted my post... :roll:


geometry
circumcircle
power of a point
radical axis
geometry unsolved

In a convex pentagon $ABCDE$, the sides $AE,BC$ are parallel and $\angle ADE=\angle BDC$. The diagonals $AC,BE$ meet at $P$. Show that $\angle EAD=\angle BDP$

[quote="cancer"][color=darkred]In a convex pentagon $ABCDE,$ the sides $AE,\,BC$ are parallel and $\angle ADE=\angle BDC,$ The diagonals $AC,\,BE$ meet at $P.$
Show that $\angle EAD=\angle BDP$ [size=150]Very nice problem![/size][/color][/quote]
[color=darkblue][b]Proof.[/b] Denote: $\begin{cases}G=AD\cap BC,\,H=BD\cap AE\\ I=CD\cap (BDG),\,J=AI\cap (BDG)\\ K=DE\cap (ADH),\,L=BK\cap (ADH)\end{cases}$

We have:
$\widehat{AED}=180^\circ-\widehat{EAD}-\widehat{EDA}=180^\circ-\widehat{DGB}-\widehat{BDC}\\ =180^\circ-\widehat{DIB}-\widehat{BDI}=\widehat{IBD}=\widehat{IJD}$
So $A,\,E,\,D,\,J$ are concyclic, thus $\widehat{AJE}=\widehat{ADE}=\widehat{BDI}=\widehat{BJI}\implies J\in BE$
Similarly, we also have: $L\in AC$

We have: $\widehat{ALB}=\widehat{ADK}=\widehat{AJE}\implies A,\,B,\,L,\,J$ are concyclic.
$\implies PA\cdot PL=PB\cdot PJ\implies P$ lies on the radical axis of two circles $(ADH)$ and $(BDG)$

On the otherhand, $(ADH),\,(BDG)$ are ex-tangent at $D,$ so $PD$ is a common tangent of $(ADH)$ and $(BDG)$
Hence: $\widehat{PDB}=\widehat{DGB}=\widehat{DAE}$
[/color]

[color=blue][b]P.S.[/b] Thank you, Huyen Vu - Who helped me very much![/color]

[quote="April"] so $PD$ is a common tangent of $(ADH)$ and $(BDG)$

[/quote]

Why is this?  :maybe:

Let $F$ be a point such that $FE//BD,FA//CD$, $BE$ meets $DF$ at $P'$
we have triangles $AEF$ and $CBD$ are similar
So $AE/BC=EF/BD=EP'/P'B$
Then $A,P',C$ are collinear.Thus $P' \equiv P$
otherwise $A,E,D,F$ are cyclic (because $\angle AFE=\angle BDC=\angle ADE$) so $\angle EAD=\angle EFD=\angle PDB$ $q.e.d$

Thank you dear [b][size=100]Huyen Vu[/size][/b], for a nice ( I think the simplest ) solution.

Let us to present another approach.

$\bullet$ Through vertex $D,$ we draw a line parallel to $AE\parallel BC,$ which intersects the sideline $AB$ of the given pentagon $ABCDE,$ at a point so be it $M.$

So, it is enough to prove that the segment lines $DP,$ $DM,$ are isogonal conjugates with respect to the angle $\angle D.$

Through vertices $A,$ $B,$ we draw two lines as the isogonal conjugates of $AE,$ $BC,$ with respect to the angles $\angle A,$ $\angle B,$ respectively and we denote as $Q,$ their intersection point.

Based on the [b][size=100]isogonic theorem,[/size][/b] we conclude that the points $D,$ $P,$ $Q,$ are collinear $($ because of $P\equiv AC\cap BE\cap DQ$ $).$

We draw the circumcircles $(O_{1},)$ $(O_{2}),$ of the triangles $\bigtriangleup AMD,$ $\bigtriangleup BMD,$ respectively.

Because of $\angle MDB = \angle CBD = \angle QBM,$ we conclude that the segment line $QB,$ is the tangent line of $(O_{2})$ at point $B$ and similarly the segment line $QA,$ is the tangent line of $(O_{1}),$ at point $A.$

Hence, based on the below [b][size=100]Lemma[/size][/b], we conclude that the segment lines $DP,$ $DM,$ are isogonal conjugates with respect to the angle $\angle D$ and the proof is completed.

$\bullet$ This proof is dedicated to [b][size=100]Stelios Moshoglou.[/size][/b]

[b][size=100]LEMMA.[/size][/b] – A triangle $\bigtriangleup ABC$ is given and let $M$ be, an arbitrary point on the side segment $BC.$

We draw the circumcircles $(O_{1}),$ $(O_{2})$ of $\bigtriangleup AMB,$ $AMC$ respectively and we denote as $P,$ the intersection point of their tangent lines, at points $B,$ $C.$

Prove that the segment lines $AP,$ $AM$ are isogonal conjugates, with respect to the angle $\angle A.$

[b][size=100]PROOF OF THE LEMMA.[/size][/b] – We denote as $D,$ the intersection point of $(O_{1}),$ from $AC$ and so, we have $\angle MBD = \angle MAC = \angle MCP$ $\Longrightarrow$ $BD\parallel PC.$

Let $X$ be an arbitrary point on the sideline $AB$ of $\bigtriangleup ABC,$ $($ $B,$ between $A,$ $X$ $).$

It is easy to show that the quadrilateral  $ABPC$ is cyclic, because of $\angle XBP = 180^{o}-(\angle PBM+\angle MBD+\angle DBA) =$

$= 180^{o}-(\angle BAM+\angle MAD+\angle DBA) = 180^{o}-\angle ADB = 180^{o}-\angle ACP.$

Hence, we have that $\angle BAP = \angle BCP = \angle MAC$ and the proof is completed.

Kostas Vittas.

Can anyone answer my question above?  :|


How can I split up this sum into even and odd "n?"

$ \sum^{\lfloor \frac{n}{2} \rfloor }_{k\equal{}0} f(n, k)$

This is what I got, but I don't think it's right.

$ \sum^{n}_{k\equal{}0} f(2n, k)\plus{}\sum^{n}_{k\equal{}0} f(2n\plus{}1, k)$

Would I have to put this into a piecewise? What if there is an [b]infinite[/b] outside sum of n?

The summation you gave returns a function in terms of $ n$, so I think whatever advice you were given meant to split up the statement of the sum depending on whether $ n$ is even or odd.  That is,

- $ n$ even:  the sum is $ \sum_{k\equal{}0}^{\frac{n}{2}} f(n, k)$.
- $ n$ odd:  the sum is $ \sum_{k\equal{}0}^{ \frac{n\minus{}1}{2}} f(n, k)$.

[quote="t0rajir0u"]The summation you gave returns a function in terms of $ n$, so I think whatever advice you were given meant to split up the statement of the sum depending on whether $ n$ is even or odd.  That is,

- $ n$ even:  the sum is $ \sum_{k \equal{} 0}^{\frac {n}{2}} f(n, k)$.
- $ n$ odd:  the sum is $ \sum_{k \equal{} 0}^{\frac {n \minus{} 1}{2}} f(n, k)$.[/quote]

Ahh I see, I overlooked that.

My next question is can I rewrite it as an equality with my first statement?

$ \sum_{k \equal{} 0}^{\frac {n}{2}} f(n, k) \plus{} \sum_{k \equal{} 0}^{\frac {n \minus{} 1}{2}} f(n, k) \equal{} \sum^{\lfloor \frac {n}{2} \rfloor }_{k \equal{} 0} f(n, k)$

Obviously this is not true, [b]BUT[/b], what if there was an infinite sum of n before my initial sum? Is the following statement true?

$ \sum^{\infty}_{n \equal{} 0} \sum_{k \equal{} 0}^{\frac {n}{2}} f(n, k) \plus{} \sum^{\infty}_{n \equal{} 0} \sum_{k \equal{} 0}^{\frac {n \minus{} 1}{2}} f(n, k) \equal{} \sum^{\infty}_{n \equal{} 0} \sum^{\lfloor \frac {n}{2} \rfloor }_{k \equal{} 0} f(n, k)$

Would it be necessary to change the index of the infinite sums to the appropriate odd or even n, or doesn't that matter because it goes to infinity?

And are you sure f(n,k) stays f(n,k) and not f(2n,k) and f(2n+1,k)?

[i]Assuming absolute convergence[/i], it would be legitimate to write

$ \sum_{\text{n is even}}^{\infty} \sum_{k\equal{}0}^{ \frac{n}{2} } f(n, k) \plus{} \sum_{\text{n is odd}}^{\infty} \sum_{k\equal{}0}^{ \frac{n\minus{}1}{2} } f(n, k) \equal{} \sum_{n\equal{}0}^{\infty} \sum_{k\equal{}0}^{ \lfloor \frac{n}{2} \rfloor } f(n, k)$.

It would be equally legitimate to write the left and right sums in terms of $ m \equal{} \frac{n}{2}$ and $ m \equal{} \frac{n\minus{}1}{2}$, respectively, in which case we would have $ f(2m, k)$ and $ f(2m\plus{}1, k)$ terms.  Just be aware that you're relabeling indices when you do this; don't use the same one.

[quote="t0rajir0u"][i]Assuming absolute convergence[/i], it would be legitimate to write

$ \sum_{\text{n is even}}^{\infty} \sum_{k = 0}^{\frac {n}{2} } f(n, k) + \sum_{\text{n is odd}}^{\infty} \sum_{k = 0}^{\frac {n - 1}{2} } f(n, k) = \sum_{n = 0}^{\infty} \sum_{k = 0}^{\lfloor \frac {n}{2} \rfloor } f(n, k)$.

It would be equally legitimate to write the left and right sums in terms of $ m = \frac {n}{2}$ and $ m = \frac {n - 1}{2}$, respectively, in which case we would have $ f(2m, k)$ and $ f(2m + 1, k)$ terms.  Just be aware that you're relabeling indices when you do this; don't use the same one.[/quote]

Can I rewrite both sums in terms of $ \sum^{\infty}_{n=0} \sum^n_{k=0}$ like so?

${ \sum_{2n}^{\infty} \sum_{k = 0}^{\frac {n}{2} } f(n, k) + \sum_{2n+1}}^{\infty} \sum_{k = 0}^{\frac {n - 1}{2} } f(n, k) = \sum_{n = 0}^{\infty} \sum_{k = 0}^{\lfloor \frac {n}{2} \rfloor } f(n, k)$

Now take the starting point of the infinite series and plug it into all n's.

$ \sum_{n=0}^{\infty} \sum_{k = 0}^{\frac {(2n)}{2} } f((2n), k) + \sum_{n=0}^{\infty} \sum_{k = 0}^{\frac {(2n+1) - 1}{2} } f((2n+1), k) = \sum_{n = 0}^{\infty} \sum_{k = 0}^{\lfloor \frac {n}{2} \rfloor } f(n, k)$

Which then simplifies to...

$ \sum_{n=0}^{\infty} \sum_{k = 0}^{n} f(2n, k) + \sum_{n=0}^{\infty} \sum_{k = 0}^{n} f(2n+1, k) = \sum_{n = 0}^{\infty} \sum_{k = 0}^{\lfloor \frac {n}{2} \rfloor } f(n, k)$

Yes.  That's what happens if you write everything in terms of $ m$.


Just wondering. When are we allowed to post questions for the Euclid. O_O.

no need for 2 posts but if u got them feel free to post them now we dont mind LOL

Ugh, i couldn't find the delete button >.< . LOL, i meant after the Euclid ; )

i think we can start talking about it now.
Is it just me or is it harder than last year...so much geometry.

I did last year as practice, and this was the same difficulty except for 9, it wasn't as straight forward. I think ill be getting anywhere between 86-91.. >.< Meh, i got 2 more years. Again anyone have answers for #9

Honestly, I think #9 and #10 were much easier than last year's.. although i didn't completely get themm :(..

Hypatia.. ?? it was jokes :D If only I had enough time to write one more sentence and conclude number #4 c).. :P

ok 10 is not easier than last year's 10. and 8 is so much harder than last year. last year was a geometry problem that asked you to solve for something, this year is prove.

o. O

well, yea, prolly...

but, 10 was pretty much approachable using simple compinatorics and possibilities.

in my solution, I fixed the ten gifted plates, and inserted 1 no-gift plates in between, and found the possibility of inserting the remaining plates..

apparently, it's not correct :P


do problem series classes only go over problems from past exams? if so, can't you just buy the past exams and do the problems yourself and check the answers?

i'm thinking it is more beneficial to take a subject class.

yes, past exams. most of them though, you don't even know that they are from past exams. If you're improving a new subject, take the subject class. If you're just reviewing for a competition, take the series.

A lot of the class is working through the problems and learning how to approach problems; most people just starting into AIME level problems tend to fall back on brute force techniques, and the goal of the series is to step back from this and apply more elegant methods.


number theory
combinatorics unsolved
combinatorics

Does there exist $n$ ($n$ is an integer) such that $n^2 =222222.....25$ (the number of $2\ \geq 3$).

[i]Edited by Myth[/i]

Hi!

Please, read http://www.mathlinks.ro/Forum/viewtopic.php?t=21455
The title "a nice question" is out of law. Title should describe the essence of problems!

Also it is a Number Theory question, not Combinatorics.

And I deleted your double post on this problem.


invariant
group theory
abstract algebra
superior algebra
superior algebra unsolved

Let A be a group of automorphisms acting
on a group G leaving invariant some coset Hx of a subgroup H. Then the
following hold:
1. H is itself A-invariant.
2. If A and H have coprime orders (so that at least one is solvable), then
A fixes an element of the coset Hx.

This result belongs to Glauberman, but I don't have a proof.

I think $1.$ isn't too hard: if $f\in A$, then $f(h_1x)=h_2x,\ h_2\in H,\ \forall h_1\in H$. This means that $f(h_1)f(x)=h_2x\iff f(h_1)=h_2xf(x^{-1})=h_2t$. In turn, this means that $e=f(e)=ht,\ h\in H\Rightarrow t\in H$.

[Edit: Sorry; this only shows that $f(H)\subseteq H$, not that $f(H)=H$. If $H$ is infinite, the equality does not follow immediately; I'll think about it some more]

[Second edit :): The surjectivity of $f$ on $H$ follows easily from the fact that $t\in H$ and from the surjectivity on $Hx$]

[quote="grobber"]
[Second edit :): The surjectivity of $f$ on $H$ folows easily from the fact that $t\in H$ and from the surjectivity on $Hx$][/quote]

Or from $f^{-1}(H)\subset H$.

Thanks for your solutions but I'd like a proof for 2. 1 is just for fun :lol:


se yali ke az yek raase motevaziosotoohe ghaem ekhraj mishavand,ba ghotre an zavayaye $a,b,c$ misazand sabet konid :
$f=(1+\frac{1}{cos^{2}a})(1+\frac{1}{cos^{2}b})(1+\frac{1}{cos^{2}c})\geq 64$

[quote="amirhtlusa"][quote] motevaziosotoohe ghaem[/quote]

 :blush:
kalameye Ingilisyi in chiye?[/quote]
rectangular parallelepiped


Hey, I've been having a hard time with this problem (it's supposed to be pretty easy) so any help would be appreciated.

Let $V$ be a finite-dimensional vector space, and let $T: V\to V$ be linear. Prove that if $\text{rank}(T) = \text{rank}(T^{2})$ then $R(T)\cap N(T) = \{0\}$. Deduce that $V=R(T) \oplus N(T)$.  Where $\oplus$ denotes the direct sum.

Thanks.

[quote="blahblahblah"]Say $y\in R(T)\cap N(T)$.  Take a basis $B$ for $N(T)$, since $\text{rank}(T)=\text{rank}(T^{2})$ we see that it is also a basis for $N(T^{2})$.

If there is some nonzero $y\in R(T)\cap N(T)$, we have $v\in V$ such that $Tv=y$, and $TTv=Ty=0$.  Then $v$ is a linear combination of elements from the basis for $N(T)$, so $Tv=0=y$, a contradiction.

This proof seems wrong (it's 2am over here) but I don't see any obvious mistakes.

For the latter part we trivially have $V=R(T)+N(T)$, and since $R(T)\cap N(T)=\{0\}$ we see that $V=R(T)\oplus N(T)$ must actually hold.[/quote]

Hey, thanks a lot (sorry I took so long to get back).  Anyway this is the same way I ended up doing it.

[quote="perfect_radio"]The general formula is $\textrm{rank}(AB)+\textrm{dim}\left( N(A) \cap R(B) \right) = \textrm{rank}(B)$. See [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=60203]this thread[/url] for example.

By the way, from this you can easily deduce that if $\textrm{rank}\left( A^{2}\right) = \textrm{rank}(A)$ then $\textrm{rank}\left( A^{n}\right) = \textrm{rank}(A), \, \forall n \in \mathbb N$ (this result has also been posted a couple of times).[/quote]

Thanks for these results, they are pretty interesting.


Prove that in every traingle ABC we have:

[tex]\prod \cos(\frac{A}{2}) >\frac{3}{8}[/tex]

This is not true: take a triangle with angles 150o, 15o, and 15o.

cos 75o :approx: 0.2588
cos 7.5o :approx: 0.9914

cos 75o * (cos 7.5o)2 :approx: 0.2544 < 0.375.

Sure ?

if A=B, C=Pi-2A we would have cos(A)sin(A) > 3/8 which can't be true.

Edited : sorry I didn't see Osiris post.

An extra :

Since log(cos(x/2)) is concave we have Prod cos (A/2) <=  cos(Pi/6)^3  :P

I am sorry it should have been sum instead of the product.

its still a weird inequality, but w/e:

[hide]

for cos(x/2) < 3/8, we know x/2 > 60 at least because cos60 = 1/2 and cos decreases from 0 to 90.

so x > 120. since its a triangle, not all the angles can be greater than 120 so there is always at least one angle for which cos(x/2) > 1/2 > 3/8.[/hide]

Er, I think what georg means is that

In a triangle ABC, cos (ABC/2) + cos (BCA/2) + cos (CAB/2) > 3/8.

yeah, i proved at least one of them is greater than 3/8 so it must be true.

Hmm, yeah, I see what you mean.

Maybe a better question would be to find the minimal value of this expression.

Edit: it's 0?
Edit: gah,I was thinking of the products.

Prove that in a triangle we have:

[tex]\frac{3}{8} < \sum \cos \frac{A}{2} < \frac{3}{4}$.

[/tex]

Ignore this -- mark me down for not reading carefully.

Yes, but I think the problem is asking about the cosine of half-angles.

i'm not too sure i understand the problem.  we are attempting to prove that

3/8<cos(A/2)+cos(B/2)+cos(C/2)<3/4
given A+B+C=180 degrees

right?

And given that A, B and C are positive.  Correct.

hmm... the maximum value of that expression seems to be 3 :rt3: /2
when A=B=C=60 degrees

Can't we use sum-product identities somehow?

if my interpretation of the problem is indeed correct, then the statement we're trying to prove seems to be false.  the minimum value is two, and the maximum is 3rad3/2

max value can be proven with jensen, and minimum with smoothing (or roughing, if you prefer to think of it that way, because we're moving the numbers further away from each other)


if i do not have any knowledge in number theory and probability(for the AMC and AIME), does anyone know any resources or any classes i could take here at AoPS. if i take a class here, how long do the classes last? how many times a week are they? and how long are the classes?

The answer to this largely depends on where you are now with these subjects.  You can find a list of our classes here:

http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php

Each class has a pre-test and a post-test, which should give you some idea which class suits you best.  Most classes are 12-week classes that meet once a week for 90 minutes (the intermediate number theory class is only 8 weeks currently).

If you have very little experience with counting concepts, we recommend our new Introduction to Counting & Probability textbook, which you can read about here:

http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?page_id=3


Let$ a,b$ and $ c$   are   positive   real  numbers and a^2+b^+c^2=3. Prove that
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})\ge9(\frac {1}{a \plus{} b \plus{} c})$

[quote="ABRORBEK"]Let$ a,b$ and $ c$   are   positive   real  numbers and a^2+b^2+c^2=3. Prove that
$ (\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a})\ge9(\frac {1}{a \plus{} b \plus{} c})$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33612


is there a good text which has some commentary of Lamé's proof of fermat's last th.? I know the general idea here is to factor $ x^n\equal{}y^n\plus{}z^n$ as $ (z\minus{}y)(z\minus{}\omega y)(z\minus{} \omega^2 y)\cdots(z\minus{}\omega^{n\minus{}1}y)$ where $ \omega\equal{}e^{\frac{2\pi i}{n}}$ is the $ n^{th}$ root of unity of 1. I suppose it would be a good algebra text which would explain why there is no unique factorization in $ \mathbb{Z}[\omega]$

[quote="Los"]is there a good text which has some commentary of Lamé's proof of fermat's last th.? I know the general idea here is to factor $ x^n \equal{} y^n \plus{} z^n$ as $ (z \minus{} y)(z \minus{} \omega y)(z \minus{} \omega^2 y)\cdots(z \minus{} \omega^{n \minus{} 1}y)$ where $ \omega \equal{} e^{\frac {2\pi i}{n}}$ is the $ n^{th}$ root of unity of 1. I suppose it would be a good algebra text which would explain why there is no unique factorization in $ \mathbb{Z}[\omega]$[/quote]
I've [b]heard[/b] (haven't tested) from someone that $ \mathbb{Z}[\omega]$ is a UFD for n=3, 5, 7, 11, 13, 17, but that it isn't for 23, so in order to understand why there isn't a UFD in all $ \mathbb{Z}[\omega]$, perhaps try to find a counterexample for n=23?

In fact, the class number $ h_n$ of $ \mathbb{Q}[\zeta_n]$ goes to infinity if $ n$ does (e.g. the second factor $ h_n^ \minus{}$ behaves like $ n^{\frac 14 \varphi(n)}$. Especially, only finitelly many such fields have unique factorisation.

But I strongly recommend to read some modern book on general algebraic number theory. Only from them, you may understand everything, and only then you may really understand why this "old stuff" is wrong.


So...the ESPN broadcast is June 3 at 12 ET on ESPN2......I started this thread just for any discussion on it; do you think you might have your 5 seconds of fame on it....what do you think they'll show......how do you think it'll turn out......will they do a good job on it........how much of what different activities will they show......if they'll show the parade....etc.......

i know there's already a  thread on this topic but its gotten lost in all the new topics..so i figure i'd start a new one

You never know how it'll be till it happens. I'll probably only have 1 more second of TV time, if that much. But I've had lots of TV time already...no need to hog it! ;)  AoPSers should be visible most of the time...too bad none of the finalists were able to give a plug for AoPS. :(

[quote="rrusczyk"]I think the commentary will be insightful  :lol: .[/quote]
What will you talk about?  Will you talk about I_Would_Say, my twin, and my other twin, nat mc.  Or will you talk about how awesome the 7th graders were this year

I think he'll talk about how amazing I was.

But then how they were destroyed by Indiana at laser tag ^_^.

And maybe he will talk about how I asked him for his autograph....  :D

Well Indiana was pwned at dinner :? Ooh, maybe Mr. Rusczyk will talk about the aimlessly milling about/standing in line/I wear a sign on my head for an hour and a half round while the press sets up for the countdown round round. It was truly the most grueling competition

I think the whole broadcast will be about me scoring an 18 on the mini-gold course and winning the nearby National ModestyCounts contest.

I'm not in the countdown, but I hope there's a shot of me...

They should just do a shot of me, with the title: Most Jipped individual(14th place, 4th place team)

I think it had a few too many commercials. One of the broadcasters seemed to know his math very well :D .

[quote="rrusczyk"]I think the commentary will be insightful  :lol: .[/quote]
 :lol:

[quote="sirorange"]I think he'll talk about how awesome the Mass. team was with their ties and bandannas. And X-treme math shirts![/quote]

Speaking of ties: I think Sergei had the best tie ever.

they should have showed more of the other activities besides just countdown

nope, no more. Thats why you should tape it

i taped it someone wanna buy it off of me?

lol

lol, Well I haven't yet so I just might *reaches for penny* nah, I'll watch it Tues

[quote="Phelpedo"]I'll buy it if you edit it and put my picture in place of Neal's and Sergei's... you know what just replace all the Countdown and team winners with me.[/quote]

email me ur picture

So, tell me about the countdown on ESPN. Was it good, bad compared to last years? Was there anything different, new or exciting? I want to hear details people, details! :P

Well, I was talking about the overall showing, I was actually at the countdown, and I saw it on Monday, so forget my question :P

Neither did I. Maybe if the participants gave actual good answers to when the moderator asked them a question, or just use footage from the interviews, they could give the contestants actual intros to themselves.

You never really got a feel for how great of an experience it was

i didn't get to watch it because I don't have ESPN, but my friend taped it, so hopefully I'll get to watch it...

they did ok...

they definitely could have explained harder problems


For how many ordered pairs of positive integers is $3x+2y<24$?

Rearranging this equation to make it easier to solve for ordered pairs, we get:

\[ 3x+2y<24 \]
\[2y<-3x+24\]
\[y<-\frac{3}{2}x+12\]

Testing until $y$ is non-positive, we see that the ordered pairs are:

\[(2,9)\]
\[(4,6)\]
\[(6,3)\]

That makes a total of $\boxed{3}$ ordered pairs of positive integers.

[quote="gaussintraining"]Note that the question is $3x+2y<24$, not $3x+2y=24$! Testing values based on the values of $x$, since $1\le x\le 7$, there should be $1+2+4+5+7+8+10=\boxed{37}$ ordered pairs.[/quote]
Oh, woops.  I completely disregarded that.  :oops:

Sorry about that.  And thank you gaussintraining!


Find the structures of compounds A - E, with justification:

1) Nicotinic acid + $ \ce{SOCl2}$/heat -----> A

2) A + $ \ce{CH3CH2OCH2CH2CH2CdCl}$ -----> B

3) B + ${ \ce{NH3}/\ce{H2}/\ce{Pt}}$ ----- > C

4) C + $ \ce{HBr}$/heat -----> D

5) D + $ \ce{CH3I}/\ce{NaHO}$ -----> E

is the final product this--> http://en.wikipedia.org/wiki/Nicotine ??

Propose structures for compounds A to E, with mechanistic explanation.

in the first step- OH is replaced by Cl
second- nucleophilic substitution again with the organometallic replacing Cl
third-nucleophilic addition with nh3 attacking the carbonyl group
fourth-cleavage of ether to give bromide which on heating cyclizes by nucleophilic subs as nh3 attacks Br.
fifth-methylation of amine.


group theory
abstract algebra
superior algebra
superior algebra solved

Can anyone find all subgroups of $\mathbb{Z}\times\mathbb{Z}$? :maybe:

Every nontrivial subgroup of $\mathbb{Z}\times\mathbb{Z}$ is a free abelian group with a basis of one or two elements. If there are two elements in the basis, they are linearly independent over $\mathbb{Q}$.

They are just (m) x (n).

If A and B are abelian groups the subgroups of AxB are A'xB' where A' and B' are subgroups of A and B respectively.

Am I wrong? Why can't I use the proof given here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102685

for ideals in rings?

Ideals $\neq$ subgroups.

Amfulger explicitely used the property $a \in I \implies ar \in I$ for all $r$ from the ring.

You use that $f(x)=g(x)+h(x)$ for all elements $x$ implies $f(X)=g(X)+h(X)$ for all sets $X$. This is wrong.

Ah, now I see my mistake. How silly!   :rotfl:

Since Z is noetherian, every finitely generated abelian group is noetherian, so every subgroup is finitely generated.

We have the structure theorem for finitely generated abelian groups: If A is such a group then A = A1 + ... + An (direct sum) where each Ai is either Z or Z/(m) (m depending on i).

It follows that A is the direct sum of a free group and a torsion group.  Therefore A is free if and only if A is torsion free.

If A is torsion free and B < A a subgroup, B is obviously torsion free. Therefore all subgroups of free finitely generated abelian groups are free.

The rank of an abelian group is defined as the maximal number of linearly independent elements. With Zn = Z x ... x Z (n times) rank Zn = n. If A is free, rank A = number of elements in any basis.

If B < A is a subgroup we obviously have that rank B <= rank A. If A = F + T is the decomposition of A into a free group and a torsion group, then rank A = rank F. In particular rank T = 0.

For any groups A and B (finitely generated abelian) we have the general formula:  rank(A+B) = rank A + rank B.

Let's now return to the original question: The subgroups of ZxZ. jmerry is right. I only wanted to work it out. Let A be a subgroup. Then rank A <= rank (ZxZ) = 2 so is 0,1 or 2. A is free so has a basis of 0,1 or 2 elements. If 1 then it has the form Z(m,n) , a discrete line. Think of it as walking in the plane, taking steps of size and direction (m,n) starting at the origin. If A has a basis of 2 elements then it is Z(m,n) + Z(m',n'). You walk as before but can stop anywhere and change direction and step size to (m',n'). A is the set of points you can reach walking like that. It is a lattice or net in the plane with masks that are parallellograms.


show that there is no biggest number a such that $ a^2$ $ <$ $ 2$

we get $ a < 1.414$.now it does not matter if we increase the decimal places.so $ (1.414367890.......)^2  <2$.these decimal places can keep stretching on and on so we cant find the largest no.$ <2$ .

@aadil your proof is not that rigorous ,
There is a very simple following proof given by a 9th standard guy and was asked in a Indian exam conducted by association of mathematics teachers of India.
Let $ a$ be the maximum such number satisfying the condition .
But we know that  $ a < \frac{a \plus{} \sqrt{2}}{2} < \sqrt{2}$
$ a^{2} < (\frac{ a \plus{} \sqrt{2}}{2})^{2} < 2$....so we cannot have a maximum number.. :wink:

One can give a stronger result, though: there is no largest [b]rational[/b] number with the desired property.

[quote]@aadil your proof is not that rigorous [/quote]
it was only an idea not a proof

i dont get it man..of course aadil's argument is sufficient to prove it
it actually goes to hyperreal arguments, which provide a quality satisfactory answer.
[url]http://en.wikipedia.org/wiki/Hyperreal_number[/url]

No, it doesn't.  aadil's argument only works if he knows that the limit of the sequence he's trying to define is $ \sqrt {2}$, and that brings in a stronger claim (both the irrationality of $ \sqrt {2}$ and the existence of its decimal expansion) than is necessary for this problem.  This problem is much simpler than that.

[b]Edit:[/b]  I guess I should put my money where my mouth is.  If $ a^2 < 2$ then I claim that $ a^2 < \left( \frac{2a \plus{} 2}{a \plus{} 2} \right)^2 < 2$.


i wuold like to have an information i'm gonna aplly for university next year, if i'll do SAT in october next eyar will i be able to have the results before university deadline?

[quote="JRav"]It depends if you are going to apply early or not.  I think most schools will accept October scores because they usually get sent in time but check with where you are applying.  If you are applying RD, then your scores will definitely be sent in time.  But check first.[/quote]
what is RD ? :blush:

[quote="JRav"]No, I meant RD.  Regular Decision.  If you are applying Early Decision, contact the school.  Regular Decision is if you are applying at the normal time (usually January).[/quote]
is there any advantage by applying with ED?

You really need to do your research.  You are advised not to apply ED unless you are 100% sure that is where you want to go because if you are admitted, you must withdraw all other applicants and you can't apply anywhere else, meaning, you're stuck with your ED school.  You also will not have the opportunity to negotiate your financial aid package should you apply for one.


calculus
integration
trigonometry
vector
derivative
calculus computations

I've been having a lot of difficulty grasping line integrals in general, but I'm particularly stuck on this question.

Given the equation:
[img]http://img218.imageshack.us/img218/3145/eq1ta1.gif[/img]

and the counter Clockwise circle C, denoted by the polar equation:
[img]http://img262.imageshack.us/img262/2724/eq2qd7.gif[/img]

use Green's theorem to evaluate:
[img]http://img230.imageshack.us/img230/9721/eq3om6.gif[/img]

It doesn't seem too hard but I can't seem to get it. Any solutions would be appreciated.

One way to do this problem is with a direct calculation. The key to directly calculating a line integral is to parameterize everything using one variable. Since this curve is defined in polar terms it is naturally parameterized using $ \theta$ (this curve by the way is not a circle. It is "roundish", but not exactly a circle). So let $ r(\theta) \equal{} (r \cos \theta, r \sin \theta)$.

Therefore $ F(x,y) \equal{} \Big(\frac { \minus{} r \sin \theta}{r^2}, \frac {r \cos \theta}{r^2}\Big) \equal{} \Big(\frac { \minus{} \sin \theta}{r}, \frac {\cos \theta}{r}\Big)$

and by the product rule, $ dr \equal{} \Big(\frac {dr}{d\theta} \cos \theta \minus{} r \sin \theta, \frac {dr}{d\theta} \sin \theta \plus{} r \cos \theta\Big)$

Taking the dot product, $ F \cdot dr \equal{} \Big( \minus{} \frac {1}{r} \frac {dr}{d\theta} \sin \theta \cos \theta \plus{} \sin^2 \theta\Big) \plus{} \Big(\frac {1}{r} \frac {dr}{d\theta} \sin \theta \cos \theta \plus{} \cos^2 \theta\Big) \equal{} 1$

Therefore [img]http://img230.imageshack.us/img230/9721/eq3om6.gif[/img] = $ \int_0^{2\pi} d\theta \equal{} 2\pi$.

I don't have time to demonstrate why, but Green's Theorem does not apply here because the curl of F = 0 everywhere except at (0,0) where it has a singularity. For Green's Theorem to work, the vector field F must be defined and have continuous partial derivatives everywhere inside the region of integration.


function
algorithm
combinatorics unsolved
combinatorics

two persons were last night together
they talked and planed for todays game the game is this:
we give the first person 5 cards
the first one gives 4 of them to the person number 2
and person number 2 tells us the 5th card
how is this possible??
P.S:the cards are in 4 colours and are from 1 to 13

The first one (call him Peter) should give the cards to the second(John) in some order.

We will built a function $f$ from the set of 5-combinations(set A)(combinations are unordered) to the set of 4-variations(set B)(variations are ordered) such that
1)if $f(a_1,a_2,a_3,a_4,a_5) = (a{}_i{}_1,a{}_i{}_2,a{}_i{}_3,a{}_i{}_4)$ then ${i_1,...,i_4}\subset {1,2,3,4,5}$
2)$f(x)\neq f(y)$ for $x\neq y$ and x,y from A;

By this function if Peter has the cards a_1,a_2,a_3,a_4,a_5 then he gives to John
$f(a_1,a_2,a_3,a_4,a_5)$ and because of 2) John can find the original 5-combination :)

So how will we build this function?
By M.Holl's theorem for SDM(system of different members) (I apologize if I have mistaken the name, I had no idea how is this in English :)
- it says that if we have p sets $S_i={a{}_i{}_1,....,a{}_i{}_t} i=1...n$ and if $|\bigcup S_i|\geq q$, where the union is for any arbitrary q sets, then there exists
$B={b_1,...,b_n}$ such that $b_i \in S_i$ and all $b_i$ are different!!!!

So for every 5-combination consider the set all 4-variations posible due to 1)
If we have q 5-combinatios then all possible from them 4-variations are more than q because every 5-combination leads to 120 4-variations and every 4-variation can be obtained from only 48 5-combinations!!!!

And by the Holl's theorem we see that there is SDM and by this we solve the problem!!

Well, it's a very good solution :)
I just want to say that you probably meant [url=http://mathworld.wolfram.com/HallsTheorem.html]this[/url] theorem. It's known as Hall's Theorem for system of distinct representatives and it's very useful as we can see

right, toshko :)
system of distinct representatives
sounds better :D

Well, Mitko, your solution is very good.
I must confess I wouldn't probably thought of it :).

P.S. I'm very much used to reading Maths in English so sometimes I can't translate theorems into our native language ;)

P.P.S I'm sorry if this message was out of topic  :oops:

sorry but can you find some algorithm for it??
I mean how should peter give the cards to john??
 :D

I have said it:

[quote]By this function if Peter has the cards a_1,a_2,a_3,a_4,a_5 then he gives to John
 and because of 2) John can find the original 5-combination [/quote]

Next time, please read more carefully :)


limit
real analysis
real analysis unsolved

calculate

\[ \mathop {\lim }\limits_{n \to \infty } \left( {\frac{{9n^2 }}
{{21n^3  \plus{} 1}}} \right)^{\frac{{ \minus{} 5n^2  \plus{} 1}}
{{10n \minus{} 3}}}
\]

It is $ \mathop {\lim }\limits_{n \to \infty } \left( {\frac
{{21n^3 \plus{} 1}}{{9n^2 }}} \right)^{\frac{{  5n^2 \minus{} 1}}
{{10n \minus{} 3}}}$. The form is $ \infty^{\infty}$, hence the limit is $ \infty$.

Once again, $ \lim _{n \to \infty } \left( {\frac{{9n^2 }}
{{21n^3 \plus{} 1}}} \right)^{\frac{{ \minus{} 5n^2 \plus{} 1}}
{{10n \minus{} 3}}} \equal{} \plus{}\infty$


Incircle $ (I)$ of triangle $ ABC$ touch $ BC;CA;AB$ at $ A_1;B_1;C_1$
$ AA_1$ cut $ (I)$ at $ Q$($ Q \neq A_1$)
The line $ l // BC$ and through $ A$.
The lines $ A_1C_1$ and $ A_1B_1$ cut $ l$ at $ P;R$
Prove that
$ \angle PQR\equal{}\angle B_1QC_1$

You can easily prove that PC1B1R lie on a circle with center A. Next you prove that PC1QA and AQB1R are concyclic (in both proofs you use that fact BC||l and properties of angle between tangents and chords) => $ \angle C_1QB_1\equal{}180^{\circ}\minus{}(\angle C_1A_1Q\plus{}\angle QA_1B_1)\equal{}180^{\circ}\minus{}(\angle QPA\plus{}\angle ARQ)\equal{}\angle PQR$ :)


$71^{43}$ has 80 digits. Prove some digit occurs in the number at least nine times.

Well obviously pigeonhole wouldn't work here, so...

Well, if there were eight of each digit, it would have to be divisble by 9, which it obviously isn't.

!!! Treething with a good solution FTW.

Suppose no digit occurs more than 8 times. Then every digit occurs exactly 8 times, making the sum $8(45) = 9*40$. But $71$ is not divisible by 9, so $71^{4}3$ can't be.

[quote="DiscreetFourierTransform"]I'm new at this whole number theory thing (not my strong suit in math class), but could you count the number of zeroes that it ends in? I'm not sure if there's an easy way of doing this.[/quote]

Try to remember what the prime factors of 10 are, and then try to find those in $71^{43}$, and you'll conclude in how many zeroes the number ends :)


The sum of two numbers is 25 and their difference is 11. What is
the smaller of the two numbers?

We have a system of equations:
                                   $ x \plus{} y \equal{} 25$
                                   $ x \minus{} y \equal{} 11$
Adding,
                    $ 2x \equal{} 36 \implies x \equal{} 18$

Plugging this in, we find that $ y \equal{} \boxed{7}$.

1 ST of all we know that the sum of 2 is 25 and differ is 11 so lets call these # x and y

so x+y=25
    x-y=11
you could substitute but we just add the # and get 2x=36 x=18 y=7 so the answer is

[hide="the substitution answer is hear"]7 IS THE ANSWER YOU SEEK [/hide][/quote]

[size=200][color=#8080FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!!! IT IS GREATLY APPRECIATED !!![/u][/i][/b][/color][/size]

[size=50][color=#808000]I HAVE 99.1546514654% THAT IM CORRECT[/color][/size]

 :bruce:  :bruce:  :bruce:  :bruce:


If $ b$ and $ c$ are constants and:

$ (x\plus{}2)(x\plus{}b) \equal{} x^2 \plus{} cx \plus{} 6$

then $ c$ is:

$ \textbf{(A)} \minus{}5 \qquad \textbf{(B)} \minus{}3 \qquad \textbf{(C)} \minus{}1 \qquad \textbf{(D)} 3 \qquad \textbf{(E)} 5$

hello, we have $ x^2\plus{}x(2\plus{}b)\plus{}2b\equal{}x^2\plus{}cx\plus{}6$, from here we get
$ 2b\equal{}6 \Leftrightarrow b\equal{}3$ and $ 2\plus{}b\equal{}c \Leftrightarrow c\equal{}5$.
So the answer ist $ \text{E}$.
Sonnhard.


geometry
3D geometry
sphere
analytic geometry
trigonometry
linear algebra
matrix

[b]Compute the expression for $ \Omega$ (canonical volume form) on $ S^2$ (with induced metric of $ R^3$) in terms of the coordinates given by spherical coordinates $ (\rho,\theta,\phi)$. The sphere is of unit radius[/b]

This is what I did. This is a very simple problem, but I am trying to follow the things taught in the differential class.  I dont think I am getting the right answer. I would appreciate if anyone can point out the mistake I am making.

Consider the open set $ U \equal{} (0,2\pi)\times (0,\pi)\subset R^2$ and a mapping $ \chi: (0,2\pi)\times (0,\pi)\rightarrow S^2\triangleq \left[\begin{array}{c} \cos\theta\sin\phi \\
\sin\theta\sin\phi \\
\cos\phi \end{array}\right]$
We plan to treat $ (U,\chi)$ as the coordinate chart which takes the sphere to $ R^2$ (except three points, of course).

Now, we will try to find out the local representation of $ \Omega$ and we can write:

$ \chi^\star \Omega \equal{} \sqrt {g}~dx^1\wedge dx^2\wedge\cdots dx^n$ where $ ^\star$ is the pullback and $ g$ is the determinant of the metric tensor.

$ g \equal{} \left[\frac {\partial(x,y,z)}{\partial(\theta,\phi)}\right]^T \left[\frac {\partial(x,y,z)}{\partial(\theta,\phi)}\right] \equal{} \left[\begin{array}{cc} \sin^2\phi & 0 \\
0 & \cos^2\phi \end{array}\right]$

Hence, we have

$ \chi^\star \Omega \equal{} \sin\phi\cos\phi~d\phi\wedge d\theta$

[color=red]But, shouldn't it be $ \sin\phi~d\phi\wedge d\theta$[/color]

The lower left entry of that matrix should be $ 1$; there are three nonzero entries in the sum, not two.


I do not know whether this has already been posted here...

$Q)$If you start from your house,walk south then east and then north,you reach back at your house.How is this possible...There are more than 1 solution(s).

you could be walking on the surface of a sphere and not on a plane. you start at a pole, go south to the equator, go east so that you cover a quarter of the equator, then go north, you end up at the same pole.

The distance that you walked east was $0$, maybe? Or you walked a small enough distance that you still arrive back at your house, but not at the exact point that you left from. :lol:

You could also be walking east all the way around a line of latitude.

Maybe your house is really big.

Maybe the directions are defined differently.

Maybe, although you walked south, east, and north, you conveniently forgot to mention that you also walked west. Because you don't reach your house immediately...

Here are those more than 1 solution(s).

(i) Your house is on the North Pole from which you could walk a mile (or any given distance) south, a mile east, and a mile north and find yourself back at the starting point.

(ii) (a) Your house could be at any point on a circle drawn starting from below the North Pole to a distance away from the South Pole to consider and take into account the curvature of the earth. After walking 'x' mile south, your next walk of 'x' mile east will take you on a complete circle around the Pole, and the walk 'x' mile north from there will then return you to the starting point.

     (b) You could also start at points closer to the Pole, so that the walk east would carry you just twice around the Pole, or three times, and so on.

Hmm...Maybe walk south to the South Pole, go east as much as you like there, then go back north (towards where you used to be).

How about you only walk one km each distance and reach back?

Maybe you walk the first three directions, and then DRIVE north?

mabe u dont have a home.. and u live on the streets

Guys please give proper explainations..this problem is in ACPOS,so it's not a stupid one :P


geometry
rectangle
probability
inequalities unsolved
inequalities

[quote="karis"]Where's y?
And m,n? :maybe:[/quote]
sorry,where m,n are positive integer number.

[quote="math10"]Let $ x,y$ are nonegative real  number such that $ x \plus{} y \equal{} 1$.Prove that: $ (1 \minus{} x^n)^m \plus{} (1 \minus{} x^m)^n\geq 1$[/quote]

I think must be so : $ m$ and $ n$ positive integer , $ p \ge 0$ and $ q \ge 0$ $ p \plus{} q \equal{} 1$ . Prove that : $ (1 \minus{} p^n)^m \plus{} (1 \minus{} q^m)^n\geq 1$.

Let we have rectangle $ (m,n)$ , n -strings and m-columns of black or white points.

Probability of white color point - $ p$ . Probability of black color point - $ 1 \minus{} p \equal{} q$ .

$ P(A) \equal{} (1 \minus{} p^n)^m$ - probability have not "white" column.

$ P(B) \equal{} (1 \minus{} q^m)^n$ - probability have not "black" string.

Obviously $ P(A) \plus{} P(B) \ge 1$ .

Equality holdes for $ p \equal{} 1$ or $ q \equal{} 1$.


probability
arithmetic sequence
combinatorics unsolved
combinatorics

show that  the set $\{1,2,...,2^{n}\}$ can be partitioned in two classes,none of which contains an arithmetic progression of lenght $2n$.

wow i spent a few minutes on it thinking the set was $2^0, 2^1, ... , 2^n$ until I realized that would make the problem trivial.  good problem though.

Barnacle, i dont understand ur solution, would u plz explain it in detail? thanx in advance.

Probabalistic method, with any luck:

There are certainly no more than $2^n$ possible first terms for an arithmetic progression contained in $\{1, 2, 3, \ldots, 2^n\}$ and no more than $\frac{2^n} {2n} = \frac{2^{n-1}} n$ possible common differences.  Thus, we have as a very rough upper estimate that there are no more than $\frac{2^{2n - 1}} n$ arithmetic progressions of length $2n$ in $\{1, 2, 3, \ldots, 2^n\}$.  Suppose we generate the partition randomly, all partitions being equally likely.  Given any subset of $2n$ members, the probability they end in the same half of the partition is $\frac 1 {2^{2n-1}}$.  Thus, the probability that a given $2n$-term arithmetic progression falls is in one of the partitioning sets is $\frac 1 {2^{2n-1}}$ and so the probability that [i]some[/i] $2n$-term a.p. is in one of the partitioning sets is at most $\frac{2^{2n - 1}} n \cdot \frac 1 {2^{2n-1}} = \frac 1 n < 1$ for $n > 1$.  Thus, for $n > 1$, we have with probability greater than 0 that there is a partition in which no a.p. is contained in one of the partition sets, so there exists at least one partition in which no a.p. may be found in either partition set, and we're done.  (The case $n = 1$ is trivial.)


calculus
integration
LaTeX
logarithms
real analysis
real analysis unsolved

f(x) is continuous and integral(0 to 1)f(x)dx=ln(1+sqrt(2)). prove there exist x_0 in(0,1) such that 1/sqrt(2)<f(x_0)<sqrt(2)/(1+x_0).

notes:let g'(x)=f=>g(1)-g(0)=ln(1+sqrt(2)) and now i put t(x)=g(x)-sqrt(2)*ln(1+x) and s(x)=g(x)-x/sqrt(2). using MVT i can prove that there exist a,b in(0,1) such that t'(a)<0 and s'(b)>0 but i can't say a=b.

please read somwhere how to wrote it in $LaTeX$.  :(

[url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]Like here for example[/url]  :)

[b]martin1985 [/b]: Is this what you meant?
$\hline$

$f(x)$ is continuous and $\int_{0}^{1}f(x)dx = \ln{(1+\sqrt{2})}$. prove there exist $x_{0}\in (0,1)$ such that $\frac{1}{\sqrt{2}}< f(x_{0}) < \frac{\sqrt{2}}{(1+x_{0})}.$

notes:let $g'(x)=f \geq g(1)-g(0) = \ln{(1+\sqrt{2})}$ and now i put $t(x)=g(x)-\sqrt{2}*\ln(1+x)$ and $s(x)=g(x)-\frac{x}{\sqrt{2}}$. using MVT i can prove that there exist $a,b \in (0,1)$ such that $t'(a)<0$ and $s'(b)>0$ but i can't say $a=b$.

Thanks for latexing, color.

The first step of a solution is to prove by continuity that either $f(x) \leq \frac{1}{\sqrt{2}}$ or $f(x) \geq \frac{\sqrt{2}}{1+x}$ for all $x \in [0, 1]$.