calculus
derivative
induction
limit
real analysis
real analysis unsolved

Let r be defined on V (x0) = {x : |x − x0| < a}. If r is n-times differentiable at x0 and r(x0) = r'(x0) = r''(x0) = · · · = r(n)(x0) = 0 (nth derivative),
prove that the limit as x approaches x0 of r(x)/[(x-x0)^n] = 0.
I use Taylor's Theorem, but all of the terms go away since the r(x0)'s all equal zero. It seems I'm left only with the remainder:
r(n+1)(c) * (x-x0) / (n+1)! . Somehow I need to show that the limit of the remainder over (x-x0)^n+1 is zero.

I'm trying to prove it by induction. I can show the base case easily. If I assume the claim is true for all n, then I'm left showing that the limit of the Lagrange Remainder as x goes to x0
r(n+1) (c) *(x-x0)^(n+1)
----------------------------
(n+1)!
-------------------------
(x-x0)^(n+1)
equals zero right? The r(n+1) (c) is the n+1th derivative of r at a point c between x0 and x. How do I do this? I'm pretty sure this problem needs to incorporate the remainder somehow.
If I assume the claim for (n-1) and try to show that the claim holds for n, it seems trivial since r(n) (x0) = 0. I'm seriously missing something.

You don't know if the derivative of order $ (n \plus{} 1)$ exists. You don't even know if the $ n$th derivative exists at points other than $ x_0$. This makes it hard to use Lagrange form of the remainder. But the much-maligned L'Hopital's rule gives an easy inductive proof.

Yes, I've been thinking about using L'Hopital's Rule.  Do you mind just quickly showing me how I would use it here?  I've been trying, but I get lost after a while.  BTW, thanks for responding.

$ r'$ satisfies all the assumptions with $ n\minus{}1$ in place of $ n$. By the induction hypothesis, $ \lim_{x\to x_0}\frac{r'(x)}{(x\minus{}x_0)^{n\minus{}1}}\equal{}0$. By the L'H Rule, $ \lim_{x\to x_0}\frac{r(x)}{(x\minus{}x_0)^{n}}\equal{}\lim_{x\to x_0}\frac{r'(x)}{n(x\minus{}x_0)^{n\minus{}1}}\equal{}0$.


[quote="reza1370"]let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} 1$. prove that
$ \frac {a}{\sqrt {b \plus{} c}} \plus{} \frac {b}{\sqrt {c \plus{} a}} \plus{} \frac {c}{\sqrt {a \plus{} b}} \geq \sqrt {\frac {3}{2}}$.[/quote]
By Holder we have:

$ ( \sum \frac {a}{\sqrt {b \plus{} c}})^2.(2ab\plus{}2ac\plus{}2bc) \geq (a\plus{}b\plus{}c)^3\equal{}1,$

$ \Leftrightarrow \sum \frac {a}{\sqrt {b \plus{} c}} \geq \frac{1}{\sqrt{2ab\plus{}2ac\plus{}2bc}}\geq \sqrt {\frac {3}{2}}$

[quote=reza1370]let $ a,b,c >0$ such that $ a\plus{}b\plus{}c\equal{}1$. prove that
$ \frac {a}{\sqrt {b\plus{}c}}\plus{}\frac {b}{\sqrt {c\plus{}a}}\plus{}\frac {c}{\sqrt {a\plus{}b}} \geq \sqrt {\frac {3}{2}}$.[/quote]
[url=https://artofproblemsolving.com/community/c6h1184978p5754753]Day #3 inequality 3[/url]
Let $ a,b,c $ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h348202p1867848]Prove that :[/url]
$$ \frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}}\geq\sqrt{\frac{3}{2}(a+b+c)} $$

[quote=reza1370]let $ a,b,c >0$ such that $ a\plus{}b\plus{}c\equal{}1$. prove that
$ \frac {a}{\sqrt {b\plus{}c}}\plus{}\frac {b}{\sqrt {c\plus{}a}}\plus{}\frac {c}{\sqrt {a\plus{}b}} \geq \sqrt {\frac {3}{2}}$.[/quote]

Simple Jensen, for the function $f(x)=\frac{x}{\sqrt{1-x}}$.


AMC
USA(J)MO
USAJMO
USAMO
AIME
AMC 12
AMC 10

Your reading load is a ton for 2 years, but if you really like it and commmit to finishing all of it in that time frame, you should be good to go. However speed isn't the only thing. Make sure you don't go so quickly that you're missing a lot of important details. Personally, after each (or every few) chapters I do, I go back and work through most of the starred exercises, a few review problems, all of the challenge problems to make sure I know everything.

And sorry, I'm not sure about your state winner question.

"...Among these contestants there were 12 states and U.S. territories not already represented, so the contestant with the highest index in each of those states and territories was added to the list. That makes a total of 498 students invited to the USAMO..."

so yeah, pretty much if your state doesn't have anyone make it to the usamo like.. normally(?), top index gets to go.

[quote="PowerOfPi"][quote="themorninglighttt"]so yeah, pretty much if your state doesn't have anyone make it to the usamo like.. normally(?), top index gets to go.[/quote]
Even if you are in eight grade? This is my hopeful schedule:

Finish the rest of Volume 1 (I have 13 chapters left) by the end of the first/second week of September.
Reveiw Intro to C&P and start Intro to Geo. by November
Take AMC8, get a perfect
Finish Intro to Geo. by January
Review Volume 1 in February
Take MC School, Chapter, State, get first in each
Take AMC10, get at least 140, stretch goal: perfect
Might take AMC12 or take a second AMC10
Skim through Volume 2
Take AIME, get at least 5-6 and make USAMO (through first in state)
Patch up my weak spots
Do good on the USAMO compared to my age (13)
Read through the intermediate series/volume 2/acops/pss the next summer and winter
Perfect on AMC10
110+ on AMC12
8+ on AIME
10+ on USAMO
Make Red MOP (and take W00t for free)

I don't know if I can read that fast, but that is my goal. I want to go medium speed on Volume 1, fast on C&P (mastered through Alcumus), slow on Geo., one quick read through Volume 2, but come back and read more detailed later.[/quote]

I made that kind of a schedule last year and i guarantee that you will be far from that (doesn't matter what direction). You cannot make a longterm schedule like that. It just doesn't work. Make a schedule to at most 2 months at a time and then make a new schedule after observing your current situation.

Well the thing is, his main schedule just consists of books, which is good, but you should also do a lot of problems.

@Poincare:Most of that post is goals and if you think about it, the intro series and volume 1 shouldn't be too hard. Skimming through volume 2 also shouldn't be too hard. Intermediate series, ACoPS and PSS are the most difficult part of his schedule. Some people DO work that way. They can set a long-term schedule and stick to it. My friend went from 52nd in state to 7th in state MC from 6th to 7th grade by setting a long-term schedule.

I will be doing problems constantly as well. I am going to take an AMC8 once every week, because I only have 12 left, and there are 12 weeks to the competition (I left a couple more for the days before the competitions, so I actually have 14). I will take AMC10s every two weeks, AMC12s every month when I feel comfortable taking it (after Intro to C&P maybe), AIMEs after Volume 1, and MathCounts the rest of my problem-doing time.

Also, my short term schedule is just to finish Volume and Intro to C&P in a month or so. I just finished chapter 16 today.

Also, I checked the USAMO participants this year, the only person from my state was the person I know who got a 175 index (135+4*10). Does this change in 2010?

[quote="PowerOfPi"]Also, I checked the USAMO participants this year, the only person from my state was the person I know who got a 175 index (135+4*10). Does this change in 2010?[/quote] How can you find the index of people based on their state?

In a previous post, he said he knew the person so he probably just asked.

you can make educated guesses using the amc10/12/aime state statistics listed online.

On the AMC website, they have the state statistics for AMC8/10/12/AIME, so I just looked at the AMC10 and AMC12s, and found those three people's scores. Then I looked at the USAMO participants (again on the website) and I saw the only person in my state that made USAMO was the person I knew. I didn't ask him, it was all online.

agh... PoP... Believe me, when you get to reading all those books you'll barely have time for an AMC10 or AMC12...

practice what's coming immediately ahead. don't read PSS and practice for USAMO before the amc's. what if you don't make it to the level you're practicing for? what a huge disappointment that would be. just concentrate on moving on from the current stage

[quote="not_trig"]About the state thing... I think they'll take the top AMC12 index for USAMO qualification. You should take 10A/12B or something, if possible.[/quote]

would they possibly do the same thing when trying to qualify for usajmo?

[quote="batteredbutnotdefeated"][quote="not_trig"]When you say "finish x book," do you mean... read through it, or do all the problems and 100% understand? There's no way you'd have time to do that latter for THAT MANY BOOKS. Especially Engel... you gotta put in a solid 2-300 hours at least to do all the problems. I haven't finished Engel, and I've been doing it for like 4 years (then again, I don't have a schedule or anything.)

It's nice to have the ambition, but try to set realistic goals.

About the state thing... I think they'll take the top AMC12 index for USAMO qualification. You should take 10A/12B or something, if possible.[/quote]

incorrect. I can, and know many others who can, finish engel in about a month or 2 and completely absorb the info. It's not as difficult as it seems.

EDIT: I HATE being braggy... sorry, if that's what some of you thought that, but I was just trying to show not_trig that it's possible.[/quote]

Then you should be at least blue mop level if you've completely absorbed it... probably higher. Are you?

[quote="batteredbutnotdefeated"]That's not for me to decide. :)[/quote]

Ehh... I'm not sure if you actually completely absorbed Engel....

[quote="batteredbutnotdefeated"][quote="not_trig"]When you say "finish x book," do you mean... read through it, or do all the problems and 100% understand? There's no way you'd have time to do that latter for THAT MANY BOOKS. Especially Engel... you gotta put in a solid 2-300 hours at least to do all the problems. I haven't finished Engel, and I've been doing it for like 4 years (then again, I don't have a schedule or anything.)

It's nice to have the ambition, but try to set realistic goals.

About the state thing... I think they'll take the top AMC12 index for USAMO qualification. You should take 10A/12B or something, if possible.[/quote]

incorrect. I can, and know many others who can, finish engel in about a month or 2 and completely absorb the info. It's not as difficult as it seems.

EDIT: I HATE being braggy... sorry, if that's what some of you thought that, but I was just trying to show not_trig that it's possible.[/quote]

You obviously misinterpreted the meaning of "absorbing the info"(i.e. you obviously did not read not_trig's entire post). There is a difference between understanding what they are saying and seeing it used in a problem, and being able to apply it to problems(not trivial problems).

oh... Either way, I was being hypothetical. I meant 1 to 2 months to actually read the book and get all the info, and a couple more months of constantly applying the info to various problems.

anyways... I'm still a [b]noobcheese[/b] by pythag's definition... :(

uh...how does it take as much time reading the book and understanding everything as solving all the problems in the book. Like for a subject oriented it takes like a month to read and understand everything and several months to do the problems(this is probably an understatement). But Engel is like a subject oriented book*10...

EDIT: this arguement is going nowhere, i should probably stop...

EDIT2: getting back on topic, im not sure if this has already been said, but in my opinion, its better NOT to have a fixed schedule just because a) you get bored very very quickly(especially with a schedule including lots of books) and b) if you are failing to keep up with the schedule, and try to go faster, you most likely will not learn all the information you need from the books.
also, im not sure if that one question was answered, but i believe you can still make usamo through taht process if you are in 8th grade...

[quote="PowerOfPi"]I am going to take the AMC 10 for the first time this year. I looked at my state's statistics (ND), and only three people qualified for the AIME, two from the AMC12, who both got a 2 on the AIME, and one (someone I know in ninth grade) from the AMC10, who got a 4 on the AIME. I think (correct me if I am wrong) the person who got a 4 on the AIME qualified for the USAMO because he was the best in the state.

Now that the new USAJMO is going to start in 2010, would the first place in the state (higher than 11-12 graders) go to the USAMO even though he is in eighth grade (me) or tenth grade (the person I know)? Or the USAJMO?

Also, right now I have only taken two AMC10s. Last month I got a 103.5. Yesterday, I got a 133.5 and I after reading the solutions I realized that my problem was time pressure for an incorrect and a blank and I marked the wrong answer in the bubble sheet on another incorrect. I haven't taken any full AIME's yet, but I can get the first few questions. I have read half of Volume 1. If I read the rest of Volume 1, review Intro to C&P, read Intro to Geo. and the intermediate series/Volume 2/PSS/ACoPS (don't know which order yet) and continue doing past tests, will I be able to get to the Red MOP level in two years?

Thanks in advance![/quote]

If you understand it all, and can do that much, probably yes. But don't exhaust yourself.

[quote="mathemonster"][quote="not_trig"]When you say "finish x book," do you mean... read through it, or do all the problems and 100% understand? There's no way you'd have time to do that latter for THAT MANY BOOKS. Especially Engel... you gotta put in a solid 2-300 hours at least to do all the problems. I haven't finished Engel, and I've been doing it for like 4 years (then again, I don't have a schedule or anything.)

It's nice to have the ambition, but try to set realistic goals.

About the state thing... I think they'll take the top AMC12 index for USAMO qualification. You should take 10A/12B or something, if possible.[/quote]
off topic but that was an epic frisbee game vs. RSI. did  u play?[/quote]

Way to almost kill the topic...

Back to the topic, I have to agree with Damien... it takes a long time... but it's worth it...

For example, I saw a TST (2000/4) problem that basically copied one of Engel's problems (or vice versa)... But for the meantime, judging at what level you (BBND) are at I would say USAJMO would be the route to take FOR NOW...

[quote="batteredbutnotdefeated"][quote="not_trig"]When you say "finish x book," do you mean... read through it, or do all the problems and 100% understand? There's no way you'd have time to do that latter for THAT MANY BOOKS. Especially Engel... you gotta put in a solid 2-300 hours at least to do all the problems. I haven't finished Engel, and I've been doing it for like 4 years (then again, I don't have a schedule or anything.)

It's nice to have the ambition, but try to set realistic goals.

About the state thing... I think they'll take the top AMC12 index for USAMO qualification. You should take 10A/12B or something, if possible.[/quote]

incorrect. I can, and know many others who can, finish engel in about a month or 2 and completely absorb the info. It's not as difficult as it seems.

EDIT: I HATE being braggy... sorry, if that's what some of you thought that, but I was just trying to show not_trig that it's possible.[/quote]

lol what? maybe if you spend like 16 hours a day

but then you'd burn out and be screwed...

so pretty much it is highly unlikely. unless perhaps if you are Alex Zhai. maybe.

also @mathemonster: yes I did (I scored once yay). also I yelled "defense" very loudly.


I'd consider my family to be upper middle class, but the camp fee of nearly $\$$3000 is rather big; I'm already taxing up about $\$$5000 for chess trips & camps and math textbooks & trips. I don't want to add another $\$$3000 onto that load, so are there any scholarships available (I don't think my family's income is allegable for financial aid)?

Yes, I also would like to know if there are any scholarship opportunities available to Canadian students, and if canadians also have to pay the $\$3000$ in U.S. Funds...  :maybe:
Right now, the cost is the biggest factor kepping me from applying..  :(

- Nawahd Werdna

Mathcounts offers scholarships (or did last year)

The finances section of the FAQ on our website has a short list of organizations that offer some form of financial aid for AwesomeMath.

The Camp FAQ:
[url]http://www.awesomemath.org/faq.shtml[/url]


inequalities
function
number theory unsolved
number theory

In [url=http://en.wikipedia.org/wiki/Brun_sieve]the Brun sieve[/url], we estabilish an inequality of the type
For all  positive integers $ N$ and $ j \geq 1$

$ \sum_{d|N;\nu(d) < 2j}\mu(d) \leq \sum_{d|N}\mu(d) \leq \sum_{d|N;\nu(d) < 2j \minus{} 1}\mu(d)$

Where $ \mu$ is the möbius function and $ \nu(d)$ is the number of prime divisors of $ d$.

Do you have a proof of this without using the methods of the sieve. I mean can we show these inequalities directly with less effort ?

I finally guessed that it is the same as showing that (if i am not mistaken)

$ \sum_{k \equal{} 0}^{j \minus{} 1}\Big{[} \dbinom{r}{2k} \minus{} \dbinom{r}{2k \plus{} 1} \Big{]}\leq 0 \leq \dbinom{r}{2j \minus{} 2} \plus{} \sum_{k \equal{} 0}^{j \minus{} 2} \Big{[}\dbinom{r}{2k} \minus{} \dbinom{r}{2k \plus{} 1}\Big{]}$
where $ r\equal{}\nu(N)$.
which is easy.


algebra
polynomial
quadratics
search
complex numbers

This solution involves complex numbers.  Are you familiar with those?

[hide]You really don't need complex numbers to do this problem. Instead, use long division and divide the 4th degree polynomial by the quadratic, which gives you x^2-5x-8 with a remainder of 55. Since the original given quadratic has a value of 1, the quotient quadratic has a value of -15, so the positive value of the 4th degree polynomial is 55-15=40.[/hide]

Well, [hide="we proceed in this way"]$ 0 \equal{} (x^2 \minus{} 5x \plus{} 7)^2 \equal{} x^4 \minus{} 10x^3 \plus{} 25x^2 \minus{} 9 \plus{} 58 \minus{} 70x \plus{} 14x^2$. So $ x^4 \minus{} 10x^3 \plus{} 25x^2 \minus{} 9 \equal{} \minus{} 14(x^2 \minus{} 5x \plus{} 7) \plus{} 40 \equal{} 40$[/hide] and that's all.

Good night
Gio


Find all real numbers $c$ such that the equation $|x|+c=|x-2|+|x+3|$ has exactly 3 real roots.

[quote="10000th User"]Find the real number $c$ such that the equation $|x|+c=|x-2|+|x+3|$ has exactly 3 real roots.[/quote]
[hide] I just plugged some values in for x on the RHS. The function $f(x)=|x-2|+|x+3|$ consists of the system: $f(x)=-2x-1$ for $x<-3$, $f(x)=5$ for $-3 \le x \le 2$, and $f(x)=2x+1$ for $x>2$. I think $c$ could either be $5$, with one root in the middle section and one root each on the outer 2 sections. Or $c=3$, with a root at the intersection of the middle and right sections and one root each on the middle and left sections. So $c=\boxed{3, 5}$.  :maybe: [/hide]

Yea, my bad, I did have a mistake too in my calculations. :oops: I'll correct the wording of the problem.


I said attention all AOPSers only so that you would vote on the poll. You do not need to post here.

What's the whole point of this if it's just basically spam...?

[size=9]I am trying to see how many people have common sense. :rotfl:  :rotfl:  :rotfl:   [/size]

[quote="mathelete"]I am trying to see how many people have common sense. :rotfl:  [/quote]

A better idea would be to stop spamming.

inb4lock.

This post is spam. As is this poll.

And now I shouldn't think that you are spamming and ruining (e.g. Never-Ending Story) :wink:
And, djimenez isn't active for over 2 years.


im confused about how to post polls on this site. i understand the poll question and run poll for part, but the poll option. do we just list it out ie: China, Russia, U.S.A. or do we have it like China or Russia or U.S.A. i may be totally off but please help. thanx. nuff said

To put up poll options, you enter one option where it says Poll option:, then you click 'Add option'.  The page reloads and you'll see everything you had before except you'll get another Poll option line that you can enter another option in, and so on.  Not sure why this isn't designed (this is an open source board produced by phpbb & custom modified by us where we could/wanted to) to take 'em all at one, but that's the way it works.


I might have made a careless mistake somewhere...
[hide="Probably not the most elegant way"] So first, what I did, was expand and combine all three equations. This gave me $ ab+ac+bc+ba+ca+cb=62$. Next, I combined like terms and got $ 2ba+2ac+2bc=62$. Then, I factored out the 2, and got $ 2(ba+ac+bc)=62$. Then, I divided both sides by 2 and got $ ba+ac+bc=31$. Then, I factored out the $ a$ from $ ba+ac$. This transformed the equation into $ a(b+c)+bc=31$. Since we know that $ a(b+c) = 17$, we know that $ bc=14$. Next, since we know that $ ca+cb=23$, we know that $ ca = 9$. Since we know that $ ab+ac=17$, we know that $ ab=8$. I found that $ abc=8c$ and that $ c=\frac{3\sqrt{7}}{2}$. Thus, $ abc=12\sqrt{7}$.[/hide]

[hide]We expand and add all of the given equations to get $ 2ab+2ac+2bc=62$. Factoring out 2 from each side gives us $ ab+ac+bc=31$. Subtracting each of the individual equations that we were given at the start gives shows us that $ bc=14$, $ ac=9$, and $ ab=8$. Multiplying these together gives us $ a^{2}b^{2}c^{2}=14\cdot8\cdot9$. Find the square root of each side, and we see that $ abc=\boxed{\pm12\sqrt{7}}$, as desired.[/hide]

Thanks, everyone who gave an answer... indianamath, you answered, but without a solution.

SImplest way :
Try ab=x,bc=y and ca=z. Then abc=sqrt(abc)

[quote]$ a(b+c)=17,\newline b(c+a)=22,\newline c(a+b)=23,$[/quote]

Expand each of the equations:
$ a(b+c)=ab+ac=17\newline b(c+a)=bc+ab=22 \newline c(a+b)=ac+bc=23$

Add the 3 equations: $ ab+ac+bc+ab+ac+bc=62%Error. "arrow" is a bad command.
ab+ac+bc=31$
$ ab=8\newline ac=9\newline bc=14$

Just make sure you solve the equation carefully and you should get:

$ {a^{2}b^{2}c^{2}}=14\times9\times8$
$ \sqrt{(a^{2}b^{2}c^{2})}=\sqrt{(14\times9\times8)}$
$ abc=\pm 12\sqrt7$

[hide]Let them be equations I,II,III respectively.

II-I gives $ bc-ab=5$.
III is $ bc+ca=23$. Solve the system, bc=14, ca=9.

III-II gives $ ca-ab=1$.
I is $ ab+ca=17$. Solve, ab=8.

Multiply, and take square root, $ abc=\sqrt{14\times9\times8}=12\sqrt{7}$.[/hide]

[quote="AnonAmoz"]Here was a question on the marathon that a solution was never posted for:

$ a(b+c)=17$ ,
$ b(c+a)=22$  ,
$ c(a+b)=23$, where $ a,b,c$ are real numbers.  Find $ abc$[/quote]

[hide="my solution which has no end and is wrong"]add all of them together and get $ 2ab+2ac+2bc=62$ so $ ab+ac+bc=31$ do factoring by grouping $ a(b+c)+bc=31$ we know $ a(b+c)=17$ so $ bc=14$. we can do the same for $ ac$ and $ ab$ by doing $ b(a+c)+.....$ and get $ ac=9$ and $ ab=8$ multiply them together and get $ a^{2}b^{2}c^{2}=72*14$(since i cant do this in my head). take them square root of both sides and get $ abc=12\sqrt7$[/hide]

[quote="AnonAmoz"]Here was a question on the marathon that a solution was never posted for:

$ a(b+c)=17$ ,
$ b(c+a)=22$  ,
$ c(a+b)=23$, where $ a,b,c$ are real numbers.  Find $ abc$[/quote]

add up the three equations and divide by two and you get ab+ac+bc=31.

so bc=14, ac=9, ab=8.

so the answer is the $ \sqrt{14*9*8}$

[quote="nshadow12"]its just $ \sqrt{2^{10}-2^{4}}=12\sqrt{7}$

2^4 because of 22-17=5 and 23-22=1 so 5-1=4
2^10 because 23-17=6, 22-17=5 and 23-22=1 so 6+5-1=10

then square root because we have counted everything to the power of 2.

Its 2 because there are 3 varables and 3 equations 3/3=1, again 3-1=2

This is the more general solution for any case with 3 variables and 3 equations.[/quote]

I don't understand what you mean by this

Some of you guys think the answer is

$ \pm12\sqrt{7}$, however, thats not right.  It is ONLY the positive root because it said a,b,c were real numbers.  If it were negative then either a,b,or c (one of the three) is negative. and the other two are positive, but we know thats false becuase none of the three original equaitons are negative.  The only way its negative otherwise if they were nonreal which is false.

Actually, if the answer were negative, then a, b, and c could all be negative. I actually think $ -12\sqrt{7}$ is a valid answer.


Are there $ n\in \mathbb{N}$ such that the sum of the digits of $ 2^n$ is equal to $ 2010$ ?

[hide="solution"] We use the fact that a number is congruent to the sum of its digits in (mod 9).

[hide="proof"] Consider some number $ x_1x_2x_3...x_n$. We can write it as $ 10^{n\minus{}1}x_1\plus{}10^{n\minus{}2}x_2\plus{}...\plus{}10x_2\plus{}x_1$. Considering this (mod 9), we find it is congruent to $ x_1\plus{}x_2\plus{}...\plus{}x_n$, which is the sum of its digits, as desired.[/hide]

Now, 2010 is congruent to 3 (mod 9). We know that $ 2^6$ is 1 (mod 9), so $ 2^n$ can either be 1, 2, 4, 8, 7, or 5 (mod 9). Thus, it cannot be congruent to 3 (mod 9), so there is no such value of n.[/hide]

[hide="Quicker solution"]Examine it mod 3:
$ 2^{n}\equiv (\minus{}1)^{n}\pmod{3}$. This is equivalent to the sum of the digits of $ 2^{n}\pmod{3}$.
(Here's a proof of why): [hide="Proof"]
Simply note that $ 10\equiv 1\pmod 3$. Any number $ n\equal{}\overline{a_0a_1\cdots a_n}$ can be expressed as $ a_0\plus{}10a_1\plus{}\cdots\plus{}10^na_n\equiv a_0\plus{}a_1\plus{}\cdots\plus{}a_n\pmod{3}$, abusing the fact that multiplying by $ 10$ changes nothing $ \mod 3$.
[/hide]
$ 2010\equiv 0\pmod 3$.
So we must have $ (\minus{}1)^{n}\equiv 0\pmod 3$, but this is impossible, so there are no solutions.[/hide]

If sum of digits would be 2010 then number $ 2^{n}$ is dividiable by $ 3$.But there's no number $ 2^{n}$ who is dividiable by $ 3$.

In an olympiad situation, would you have to prove the "sum of digits congruent to number mod 9" fact, or just state it as obvious?  Every problem having to do with it, part of the solution has to do with proving this fact, but I see it as obvious.

It is, I guess. I just feel like proving it :wink: . Also, this is HSB, so it might be something new for people who didn't realize that yet.


A binary operation $a*b$ has the property $(a*b)*c=a+b+c, \forall a,b,c$. Prove that $a*b=a+b$.

We have $((a*b)*c)*d=(a+b+c)*d$ on the one hand, and $(a*b)+c+d$ on the other hand, so these two must be equal for all $a,b,c,d$: $(a+b+c)*d=(a*b)+c+d,\ \forall a,b,c,d\ (*)$. If we take $b=c=0$, we find $a*d=(a*0)+d,\ \forall a,d\ (**)$, which is almost what we want :). We have $(a+b)*d=(a*b)+d$, which is obtained from $(*)$, by making only $c=0$ this time. From $(**)$ we find that $((a+b)*0)+d=(a*0)+b+d$, and, after deleting $d$ from both sides, we get $(a+b)*0=(a*0)+b$. $a=0$ gets us $b*0=(0*0)+b$, so $(**)$ now turns into $x*y=(0*0)+x+y,\ \forall x,y\ (***)$.

From $(***)$ and the relation we are given in the beginning, it's easy to conclude what we want ($0*0=0$, and we're done).

I'm sure some shortcuts can be made :).

[quote="grobber"]I'm sure some shortcuts can be made :).[/quote]
The one I've noticed is to put $a=b=0$ into $(*)$ to obtain $(***)$.

Nice solution, grobber!

This one is from Russia, 1998: [url]http://www.kalva.demon.co.uk/russian/rus98.html[/url] (problem 14)


geometry
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MATHCOUNTS
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What would be the best way to introduce a school to extracurricualr math?
I think that some kind of quick panphlet telling all about how the extra math is needed and how it its easy to implement would be nice. A lot of kids at my school are midly gifted(though not as much as me:P (/ego)) and it would be nice to do something with them, seeing as now many of them ace the courses and are bored in school (like me). Also, are there some books similar to AoPS for other subject matters, like physics or informatics? Those would be extremely helpful in my gifted program! That could be a gifted program(!) if you added things like rubiks cubes and debate club into the mix...

A big problem I see at my school is that even some more moderately smart kids go unchallenged. School today isn't learning, its just rote memoriztion. Kids need to learn and be creative, not memorize. You should understand  the metric system. Learning that milli means "one thousandth" and how it applies to other SI units, not just length(or whatever), is essiansial. Making connections and critical thinking by kids would be great. I see too many times a question just rephrased, and kids can't get it, even though they did something similar a few seconds ago. Same with progressivly using the answers from questions to form an overlaying thesis. They can't do it. So much is stressed at my school about honor and respect, but no creativity or critical thinking.

Some of my teachers are dumb. Really dumb. They seem to be products of this education systems too. They have memorized the curiculum, but can't answer my questions about the details.

Also, I would like to touch upon the issue of "hot topics" being taught in school. Many of the currently debated things like evolution and the iraq war are not being taught. I am not advocating teaching one side, but rather presenting the facts and letting the kids decide. Or debate.

My teacher is on the verge of registering for MC, and I need some kind of flyer or quick info sheet for her.

[quote="h_s_potter2002"]Well, you could show your teachers the websites for MathCounts and all the competitions.

And if you want materials for informatic-type contests, there a website, [url=http://www.usaco.org]www.usaco.org[/url], that has links to some training materials and stuff.  THat website also has information about the USACO contests, which are held one time each month from November to April.  And these are the contests that will help decide the US team for the IOI, the International Olympiad in Informatics.

And I have a question kind of similar to PenguinIntegral's - I go to a school, where there are people who are smart, and there are people that are good at math, but how would I get them interested, since a lot of compeitions need teams and stuff.[/quote]

Well, for the infomatics contest, I was thinking more of a book type thing. It alows me to read faster somehow, and its better for reference.

But the only suggestion I have for getting kids interested would be to grab some puzzles, like a rubiks cube, and show them how fun that is. Then once they are receptive to your ideas, point them at AoPS.

[quote="AntonioMainenti"]Accessing advanced material in areas that certain students have an interested in sadly isn't a first priority of a public school system that was designed not to encourage individuality. Sorry in advance to any fans of our school system.[/quote]
They just care about funding....

Id rather live in the old soviet union, they had a great education system. If you would good at something in childhood, you recieved training for that and usually took that job.
Everything else sucked, but I really don't care as much about everything else as I do math and the sciences.

[quote="-Job-"]I think what you're looking for is not so much the challenges or entertainment that extracurricular math could provide but the society of people who are interested in that which you do best; you're looking for people who are interested in you.
The psychology of the day...  ;)[/quote]
No, the first one.

Everyone else seems ignorant.

It just seems odd.

I see, you probably want to establish a math group whilst avoiding looking like a geek. I'm not going lie to you, it's not going to be easy. I have a plan though, how about you give us the email address of one or more key people (teachers, directors, etc) who can make this a reality. Then, we'll have some of the people here at AoPS email them under the pretext that several students in your school who frequent this forum would be interested in extra-curricular math (maybe there's only one, but they don't have to know that). This way we get the ball rolling for you (coming from an accredited institution such as the AoPS) you don't look like a geek, and you'll be all set to rock n'roll in math pretty soon.  :ninja:
Of course, we should probably delete this thread then, to completely erase all traces, in case they came around, which they always do.  :ninja:
You're welcome...  :ninja:

[quote="-Job-"]I see, you probably want to establish a math group [b]WHILE[/b] avoiding looking like a geek. I'm not going lie to you, it's not going to be easy.[/quote]

Fixed.
[quote] I have a plan though, how about you give us the email address of one or more key people (teachers, directors, etc) who can make this a reality. Then, we'll have some of the people here at AoPS email them under the pretext that several students in your school who frequent this forum would be interested in extra-curricular math (maybe there's only one, but they don't have to know that). This way we get the ball rolling for you (coming from an accredited institution such as the AoPS) you don't look like a geek, and you'll be all set to rock n'roll in math pretty soon.  :ninja:
Of course, we should probably delete this thread then, to completely erase all traces, in case they came around, which they always do.  :ninja:
You're welcome...  :ninja:[/quote]
I will not commision spam.....yet.

[quote="-Job-"]Whilst is a perfectly valid substitute for while:
http://education.yahoo.com/reference/dictionary/entry/whilst

If you do opt for spam, you know where to find me. I have my own server and could spam those suckers with little effort. Of course that probably wouldn't help your cause, but it doesn't matter because i don't go to your school.
Good luck...[/quote]

I meant to remove the avoiding part too.......

Maybe a "You have been selected for you aptness in math" email? Maybe AoPS could officialy do these notifications.

Fine you can look like a geek if you want to, that's up to you, but let's not make it the group's visual. :ninja:
To be a little more serious, maybe one of the AoPS people wouldn't mind taking up the cause, or maybe set up a program geared towards the development of math activities in schools such as yours. If not we can spam them then, they're a non-profit organization so they don't have alot of political muscle; we wouldn't have to worry about it.  :ninja:

[quote="236factorial"][quote="PenguinIntegral"][quote="-Job-"]All right then :what?: . Isn't there a Math Club or something in there?[/quote]
No. Thats the point of this thread. Many schools have absolutly none/little gifted education.[/quote]

In my school, at least there's mathcounts. However, if you read my blog (entry on 8/29), I complain about the easiness of the curriculum. Our curiculum is pathetic. It is not fast or challenging.[/quote]

I wonder is Richard Rusczyk will adopt us :rotfl:

Seriously, that would rock :ninja:

A few thoughts:

Write real letters, not emails.  Real letters are 10 times more effective.  If you don't already know how to write a nice formal letter, learn how -- it's an invaluable skill.

If your school's administration is unresponsive, contact your representative(s) on your local school board.  School board members are elected officials, and therefore (theoretically) more responsive to their constituents.  Get every parent/voter/taxpayer you know to write a letter to your school board representative(s).  Also, most school boards have public meetings, and often have an "open-mic" period where anybody can speak about anything -- consider preparing a well-thought proposal and presenting it at a school board meeting.

Schools are most inclined to support activities that will (a) not cost them any money and (b) bring them positive publicity.  You can try to raise money to help offset (a), and for (b) you can emphasize how cool it will look in the local paper if you and/or your school does well in Mathcounts.

Unfortunately, the situation you describe is all too common -- it's a large part of the reason why Richard Rusczyk founded AoPS.  It is also an unfortunate consequence of the No Child Left Behind act: as more and more schools spend more and more money to get into compliance with NCLB, less and less money is available for "gifted" programs and other perceived non-necessities.

I also suggest [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_MATHCOUNTS.php]this article[/url] by Jeff Boyd, coach of the national champion MATHCOUNTS team from Texas, entitled "Building a Successful MATHCOUNTS Program".

For MATHCOUNTS info, point your teacher to [url=http://www.mathcounts.org/webarticles/articlefiles/714-05-06kickoff.pdf]this press release[/url].  Emphasize the word "free".  The handbook itself is available for free(!) download at [url]http://www.mathcounts.org/webarticles/anmviewer.asp?a=132&z=29[/url].

Finally, a quick correction:

[quote="-Job"] (coming from an accredited institution such as the AoPS)[/quote]
We are not accredited, meaning we have very little influence over schools.

Or we could spam him if that doesn't work.  :ninja:

[quote="DPatrick"][quote="PenguinIntegral"]I wonder is Richard Rusczyk will adopt us :rotfl: [/quote]I doubt it, but I'll ask him. :rotfl:  You could try bribing him with Thai food and chocolate chip cookies.[/quote]

Thai food, chocolate chip cookies......

.....hehehe

*Evil Grin*

Are you sure he doesn't want pie?
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[img]http://www.brian1.net/gallery/random/pipie.jpg[/img]

[quote="DPatrick"][quote="PenguinIntegral"]I wonder is Richard Rusczyk will adopt us :rotfl: [/quote]I doubt it, but I'll ask him. :rotfl:  You could try bribing him with Thai food and chocolate chip cookies.[/quote]

Haha  :rotfl:  I don't even know where he lives.

I read an article [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_Calculus.php]here[/url] about something called the "calculus trap" written by Mr. Rusczyk and is similar to why this site was started.


A uniform block of ice of mass m has the shape of a cylinder of radius a.It is contained in a close fitting,  thin-walled  cylindrical  drum of radius a and mass M.

The atmosphere is sufficiently warm for a very thin layer of water to form between the ice and the drum,so that slipping can occur between them resisted by a couple equal to K times the relative angular velocity.

Starting from a position where both the  drum and ice are at rest, the drum rolls without slipping down the line of greatest slope of a plane which makes an angle $\alpha$ with the horizontal.

Show that the difference between the  angular velocities of the ice and the drum tends to the constant value

                          $\frac{mgasin\alpha(m+M)}_{k(3m+4M)$


function
advanced fields
advanced fields unsolved

Let $ A$ be a C*-algebra and $ X$ a locally compact space. Show that $ C_{0}(X,A)$, the set of all continuous functions $ f: X\rightarrow A$ such that $ \forall \epsilon > 0$, $ \{x : \|f(x)\| \geq \epsilon\}$ is compact, is a C*-algebra (where all the operations are defined pointwise and the norm is the sup norm).

Why is the condition that $ \forall \epsilon > 0$, $ \{x : \|f(x)\| \geq \epsilon\}$ is compact necessary?

this guarantees that the sup norm is finite. do you have other difficulties? the proof for the claim is like in the well-known case $ A=\mathbb{C}$. actually, one can show $ C_{0}(X,A) \cong C_{0}(X) ~ \hat{\otimes}~A$.

It may be that [b]dDI[/b] does not like the compactness condition because it deprives the new C*-algebra of a unit element. That can be fixed by adding all constant functions to $ C_{0}(X,A)$ and taking the linear span of the union. What if one goes further and replaces $ C_{0}(X,A)$ with $ C_{b}(X,A)$, the algebra of all bounded continuous functions?

I don't think this has to do something with units ... and what is your unit if $ A$ is not unital?

in general, the unitarization $ \hat{A}$ of a $ \mathbb{C}$-algebra $ A$ is the direct sum $ A \oplus \mathbb{C}$ with an obvious product, and this turns out to be a $ C^{*}$-algebra if A is one. in the special case, $ \hat{C_{0}}(X,A) \cong C(\hat{X},\hat{A})$ where $ \hat{X}$ is the one-point-compactification of $ X$.

what do you want to know about $ C_{b}(X,A)$? this is already unital; I just guess that it is isomorphic to $ C(\beta(X),A)$.


calculus integration trigonometry logarithms function induction algebra

find $\int\frac{dx}{1+x^{2n}}$ where $n$ is positive integer. :!:

I have an integral formula for $\int \frac{x^{m}}{1+x^{n}}\, dx$, where $0 \leq m < n$ are integers.

pu tehre $m=0$ and write here. if your formula is big then I think it's correct

[quote="sos440"]Let $m, n$ be integers with $0 \leq m < n$, and $\theta_{k}= \frac{2k-1}{n}\pi$. Then

$\int \frac{x^{m}}{1+x^{n}}\, dx = \frac{1}{n}\sum_{k=1}^{n}\left( \sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)-\frac{1}{2}\cos \left((m+1) \theta_{k}\right) \log \left(x^{2}-2 \cos \theta_{k}x+1\right)\right)$

For odd $n$, $\sin \theta_{k}$ becomes zero for $k = (n+1)/2$. In this case, we consider the term

$\sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)$

to be also zero.[/quote]
$sos440$ can you poste your sulution?
/////////////////////////////////
$pardesi$ Can you poste full sulution?

My proof is rather long and complicated, so I can't write down the whole process. But I can show you some key-facts.

(1) For an integer $m \geq 0$,

$\frac{x^{m}(x \cos \theta-1)}{x^{2}-2x \cos \theta+1}$
$=\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)+\frac{x \cos (m+1) \theta-\cos m \theta}{x^{2}-2x \cos \theta+1}$
$=\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)+\frac{\cos (m+1) \theta}{2}\frac{2x-2 \cos \theta}{x^{2}-2x \cos \theta+1}-\frac{\sin \theta \sin (m+1) \theta}{(x-\cos \theta)^{2}+\sin^{2}\theta}$

Try this with the mathematical induction!

(2) For integers $m, n$ with $m \geq 0$ and $n \geq 1$,

$\frac{x^{m}}{x^{n}+1}=-\frac{1}{n}\sum_{k=1}^{n}\frac{x^{m}(x \cos \theta_{k}-1)}{x^{2}-2x \cos \theta_{k}+1}$

where $\theta_{k}= \frac{2k-1}{n}\pi$. You can verify this identity even it is not so easy.

(3) For natural numbers $m, n$ with $n \nmid m$,

$\sum_{k=1}^{n}\cos \left( \frac{m(2k-1)}{n}\pi \right)$ = 0

(Hint: consider each summand to be the real part of $e^{i \pi m(2k-1)/n}$)

[quote="Extremal$pardesi$ Can you poste full sulution?[/quote]

now we r left with getting the following integral

$\int tan^{n-1}x$ $dx$
this can be easily evaluated by the following recurrence relation

$\int tan^{n}x$ $dx$ $=$ $\int tan^{n-2}x(sec^{2}x-1)$ $dx$ $=$ $\int tan^{n-2}x$ $dtanx$ $-$ $\int tan^{n-2}x$ $dx$

this is equal to $\frac{tan^{n-1}x}{n-1}$ $-$ $\int tan^{n-2}x$ $dx$

this will give u
$\int tan^{2n}x$ $dx$ $=$ $\sum_{k = 1}^{n}\frac{(-1)^{k-1}tan^{2n-2k+1}}{2n-2k+1}$ $+$ $(-1)^{n}x$

$\int tan^{2n+1}x$ $dx$ $=$ $\sum_{k =0}^{n-1}\frac{(-1)^{k}tan^{2n-2k }}{2n-2k }$ $+$ $(-1)^{n}log(sec x)$

[quote="pardesi"]let's have a direct approach .

we need to get
$\int\frac{1}{1+x^{2n}}$ $dx$ $=$ $I$ (say)
put $x^{\frac{1}{n}}$ $=$ $tan\alpha$

hence we get ,
$I$ $=$ $\int n sin^{n-1}\alpha cos^{1-n}\alpha$ $d\alpha$

the last part can easily be derived from the $gamma$ $functions$[/quote]

$x^{\frac{1}{n}}= \tan\alpha$ does not give $1+x^{2n}= 1+\tan^{2}\alpha$, but $1+x^{2n}= 1+\tan^{2n^{2}}\alpha$,

[quote="sos440"]Let $m, n$ be integers with $0 \leq m < n$, and $\theta_{k}= \frac{2k-1}{n}\pi$. Then

$\int \frac{x^{m}}{1+x^{n}}\, dx = \frac{1}{n}\sum_{k=1}^{n}\left( \sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)-\frac{1}{2}\cos \left((m+1) \theta_{k}\right) \log \left(x^{2}-2 \cos \theta_{k}x+1\right)\right)$

For odd $n$, $\sin \theta_{k}$ becomes zero for $k = (n+1)/2$. In this case, we consider the term

$\sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)$

to be also zero.[/quote]

I think you have somewhere mistake because I think that I take other answer for m=0 and n=2n. OK my answer is
$\int\frac{dx}{1+x^{2n}}$=$-\frac{1}{2n}\sum^{n}_{k=1}\cos\frac{{\pi}(2k-1)}{2n}\ln(1-2x\cos\frac{{\pi}(2k-1)}{2n}+x^{2})+\frac{1}{n}\sum^{n}_{k=1}\sin\frac{{\pi}(2k-1)}{2n}\arctan\frac{x-\cos\frac{{\pi}(2k-1)}{2n}}{\sin\frac{{\pi}(2k-1)}{2n}}$
I think am assured my answer is corect
Or I am WRONG?  :maybe:

SORRY sos440
everything is ok. you are write :)

[quote="sos440"]My proof is rather long and complicated, so I can't write down the whole process. But I can show you some key-facts.

(1) For an integer $m \geq 0$,

$\frac{x^{m}(x \cos \theta-1)}{x^{2}-2x \cos \theta+1}$
$=\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)+\frac{x \cos (m+1) \theta-\cos m \theta}{x^{2}-2x \cos \theta+1}$
$=\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)+\frac{\cos (m+1) \theta}{2}\frac{2x-2 \cos \theta}{x^{2}-2x \cos \theta+1}-\frac{\sin \theta \sin (m+1) \theta}{(x-\cos \theta)^{2}+\sin^{2}\theta}$
[/quote]
Its wrong. It's not true when $m=1$  therefore can't solve by induction.

When $m=1$,

$\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)+\frac{x \cos (m+1) \theta-\cos m \theta}{x^{2}-2x \cos \theta+1}$

$= \cos \theta+\frac{x \cos 2 \theta-\cos \theta}{x^{2}-2x \cos \theta+1}$

$= \frac{\cos \theta (x^{2}-2x \cos \theta+1)+(x \cos 2 \theta-\cos \theta)}{x^{2}-2x \cos \theta+1}$

$= \frac{ x^{2}\cos \theta-2x \cos^{2}\theta+x \cos 2 \theta}{x^{2}-2x \cos \theta+1}$

$= \frac{ x^{2}\cos \theta-x (\cos 2 \theta+1)+x \cos 2 \theta}{x^{2}-2x \cos \theta+1}$

$= \frac{ x (x \cos \theta-1)}{x^{2}-2x \cos \theta+1}$

a few of simple calculation gives the desired result... -0-

[quote="sos440"]

(1) For an integer $m \geq 0$,

$\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)$

[/quote]
1) How do you define this when m=0?
[quote="sos440"]

(3) For natural numbers $m, n$ with $n \nmid m$,

$\sum_{k=1}^{n}\cos \left( \frac{m(2k-1)}{n}\pi \right)$ = 0

(Hint: consider each summand to be the real part of $e^{i \pi m(2k-1)/n}$)[/quote]
How do you use $e^{i \pi m(2k-1)/n}$ this to prove this $\sum_{k=1}^{n}\cos \left( \frac{m(2k-1)}{n}\pi \right)$ = 0?

[quote="sos440"]
(2) For integers $m, n$ with $m \geq 0$ and $n \geq 1$,

$\frac{x^{m}}{x^{n}+1}=-\frac{1}{n}\sum_{k=1}^{n}\frac{x^{m}(x \cos \theta_{k}-1)}{x^{2}-2x \cos \theta_{k}+1}$

where $\theta_{k}= \frac{2k-1}{n}\pi$. You can verify this identity even it is not so easy
[/quote]
3) can you prove this?

1) when $m=0$, then we consider $\left( \sum_{k=0}^{m-1}x^{k}\cos (m-k) \theta \right)$ to be zero.

2) when $n$ does not divides $m$, then $e^{i \pi 2m/n}\neq 1$ and we have

$\sum_{k=1}^{n}\cos \left( \frac{m(2k-1)}{n}\pi \right)$

$= \Re \left[ \sum_{k=1}^{n}e^{i \pi m(2k-1)/n}\right]$

$= \Re \left[ e^{i \pi m/n}\frac{1-e^{i \pi 2m}}{1-e^{i \pi 2m/n}}\right]$

but since $e^{i \pi 2m}= 1$,

$= \Re [0] = 0$

3) suppose the polynomial $p(x)$ of degree $n>0$ has $n$ distinct roots, namely $x_{1}, x_{2}, \cdots, x_{n}$. then

$\frac{1}{p(x)}= \sum_{k=1}^{n}\frac{1}{p'(x_{k})(x-x_{k})}$.

you can use it to prove that.

now everything is ok.  :wink:  you put here $m$

$\frac{x \cos (m+1) \theta-\cos\theta}{x^{2}-2x \cos \theta+1}$

here

$\frac{x \cos (m+1) \theta-\cos m \theta}{x^{2}-2x \cos \theta+1}$

Let consider $I_{1}=\int \frac{dx}{1+x^{2n}}$ , $I_{2}=\int \frac{x^{2n}dx}{1+x^{2n}}$. We know $I_{1}+I_{2}=x+c$.
Now we try to solve $I_{2}$.
Suppose that
\[x^{n}= \tan t \Longrightarrow dx= \frac{(1+\tan^{2}t)dt}{n \tan^{ \frac{n-1}{n}}t}\Longrightarrow I_{2}= \int \frac{\tan^{2}t(1+\tan^{2}t)dt}{(1+\tan^{2}t)n\tan^{ \frac{n-1}{n}}t}= \frac{1}{n}\int \tan^{ \frac{n+1}{n}}tdt \Longrightarrow \tan t=u \Longrightarrow I_{2}=\int \frac{u^{ \frac{n+1}{n}}du}{1+u^{2}}=... \]
I don't know :P

[quote="sos440"]Let $m, n$ be integers with $0 \leq m < n$, and $\theta_{k}= \frac{2k-1}{n}\pi$. Then

$\int \frac{x^{m}}{1+x^{n}}\, dx = \frac{1}{n}\sum_{k=1}^{n}\left( \sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)-\frac{1}{2}\cos \left((m+1) \theta_{k}\right) \log \left(x^{2}-2 \cos \theta_{k}x+1\right)\right)$
[/quote]
You are wrong.

it's true
$\int \frac{x^{m}}{1+x^{2n}}\, dx = \frac{1}{n}\sum_{k=1}^{n}\left( \sin \left((m+1) \theta_{k}\right) \tan^{-1}\left(\frac{x-\cos \theta_{k}}{\sin \theta_{k}}\right)-\frac{1}{2}\cos \left((m+1) \theta_{k}\right) \log \left(x^{2}-2 \cos \theta_{k}x+1\right)\right)$
 where

$\theta_{k}= \frac{2k-1}{2n}\pi$
you are wrong. when you prove this

[quote="sos440"](2) For integers $m, n$ with $m \geq 0$ and $n \geq 1$,

$\frac{x^{m}}{x^{n}+1}=-\frac{1}{n}\sum_{k=1}^{n}\frac{x^{m}(x \cos \theta_{k}-1)}{x^{2}-2x \cos \theta_{k}+1}$

where $\theta_{k}= \frac{2k-1}{n}\pi$.[/quote]
because it's true

For integers $m, n$ with $m \geq 0$ and $n \geq 1$,

$\frac{x^{m}}{x^{2n}+1}=-\frac{1}{n}\sum_{k=1}^{n}\frac{x^{m}(x \cos \theta_{k}-1)}{x^{2}-2x \cos \theta_{k}+1}$

where $\theta_{k}= \frac{2k-1}{2n}\pi$ by the interpolation formula Nutona


The cost of replacement windows at Wanda's Window World is $ \$0.75$ per inch of perimeter. How many dollars will it cost William to replace a rectangular window that measures $ 2$'$ 6$" by $ 5$'$ 10$"?

Um, i'm not sure about this, but i think that the answer is supposed to be 75 dollars, not 150... because 2'6" is 30", and 5'10" is 70". So, we add these together to get $ 100"*.75\equal{}75$. ;) i hope that this helps those wondering what's up with this weird answer...

$ The$ $ Origami$ $ Guy$

But, if $ a$ and $ b$ are the side lengths of a rectangle, the perimeter is $ 2(a\plus{}b)$, not $ a\plus{}b$.


MATHCOUNTS
geometry
3D geometry
prism
AMC
AIME
rectangle

Problems 4 and 5 of the ones we went over in class, as some people who had to leave early requested:

4. A spiral staircase turns 270 deg as it rises 10 feet. The radius of the staircase is 3 feet. What is the number of feet in the length of the handrail? (2001 State Team Round #8)
5. Consider the set of points that are inside or within one unit of a rectangular prism that measures 3 by 4 by 5 units. Find the volume of this set. (AoPS Intro to Geometry)

I'm currently under the possibly mistaken impression that I don't have much homework tonight. If true, I'll post the answers for the previous weeks that I still owe.

P.S. Why is everyone not discussing problems in the previous threads? Is it that you can solve all of them or do you just want JLS to lose at the chapter contest? (Yeah, false dichotomy and all that, but still.)

\begin{offtopic}
your w00t AIME score?
\end{offtopic}

10-1=9. Still beat you, thank goodness, but you seem to have solved the same number of problems...darn it. I really really need to work harder.

Which #5? The one in the first post? If so,
[hide]You have to break it into lots of little pieces. Try drawing the picture, even though it's kind of hard to draw well. First draw the 1-unit-thick boxes on the sides. Then think about what goes in the gaps.[/hide]

[quote="bookaholic"]What in the world...I should ask for mod status. Hmm. Or not. Should I ask for mod status?[/quote]

Yes you should.  I need to study now that i got super-pwned on the aime. :(

Yep, bookaholic THAT #5.

I don't get what the question even MEANS. I think I'm inhaling too much dry erase marker fumes....as I'm getting stupidder by the second.

Oh. So the 2d equivalent would be a question like, "you have a 3x4 rectangle. You want to put a border around it so that the border is 1 unit wide at all points. What's the area of the whole thing?" And you would do it by finding the area of the border that is straight and adding the quarter-circles at the corners (and the original rectangle, of course). It's the same thing here, except you want to wrap the box in something so the something is 1 unit thick at all points. It's a slightly different statement than the original, but it means the same thing.

@bookaholic: which w00t pset probs did you do?
I did {1,2,3,4,7,8,9a}

[quote="junggi"][quote="bookaholic"]I STILL HAVEN'T GOTTEN AROUND TO LOOKING AT 5-12! DON'T REMIND ME! I'M GOING TO FAIL WOOT![/quote]
whoa did ccy hack your account?
GD I just can't seem to get 9b. Maybe it's the rock im listening to right now[/quote]
No, that was me. Sigh. What I need is someone to kidnap me and lock me somewhere with no computer and nothing but the WOOT problems for a day. Then I would get something done.

SOMEBODY tell me how to find areas with diagonals. EHHHH. I'm stupid.

[hide]Draw a big rectangle around the kite with sides parallel to the diagonals. Celebrate.[/hide]


If anyone has past tests from MOEMS, I would greatly appreciate them.

Thanks,
Joe

I would too, greatly appreciate it.  I have a good amount of Mathcounts material in exchange for tests.  PM me if interested.

if u have mathcounts material then you shouldn't need moems material...

Theres a book by George Lenchner called Math Olympiad Contest Problems for Elementary and Middle Schools. Although the questions arent real MOEMS tests, they simulate actual tests pretty closely


anyone going to make some before this next year's amcs?

No.  Thats absolutely out of the question.

(yes, of course, if not... there are plenty mocks lying around the forum)

[quote="nutz_for2.718281828"]@pakman: Where can you find some good mock AMC 12's?

@NeverOddOrEven: No,I can't make a mock AMC 12. But I sure can make a mock AMC 8.[/quote]
AoPSWiki.

[url=http://www.artofproblemsolving.com/Wiki/index.php/Mock_amc]Mocks[/url]

i bet the people will... so many people are bored

[quote="nutz_for2.718281828"]@pakman: Where can you find some good mock AMC 12's?

@NeverOddOrEven: No,I can't make a mock AMC 12. But I sure can make a mock AMC 8.[/quote]

take a look? its a sticky on this forum

i might make another mock aime soon...it should be easier

Actually, if you make it harder, that might be more helpful (I'm a convert to Mildorf's theory  :wink: )

I'd be glad to write more mock tests after college apps are done...


Hello friends, please help me with this
Let $ X: I\rightarrow{\mathbb{R}^{n^2}}$ be a differential path with matrices $ n\times{n}$ as values. Fixed  $ k\in{\mathbb{N}}$, let $ f_I\rightarrow{\mathbb{R}^{n^2}}$  be a path given by $ f(t)\equal{}X(t)^k$. Show that $ f$ is differentiable and find $ f'(t)$, for all $ t\in{I}$.

Take care
Hugs

What are you having trouble with? Have you tried $ k\equal{}2$?


how do you get a hyperbola in asymptote?

You may be interested by [url=http://piprim.tuxfamily.org/asymptote/travaux/index.html#fig0034]this example[/url] (figure 0034)


induction
search
number theory solved
number theory

The sequence L_n is defined as:
[tex]L_{0}=2 [/tex], [tex]L_{1}=1[/tex] and [tex]L_{n+1} = L_{n} + L_{n-1}[/tex]
Show that if [tex]p|L_{2k} -2[/tex], so [tex]p|L_{2k+1} - 1[/tex]

Does someone know the original source of this problem??

This is a Lucas-type sequence so we may look for a classical relation of the form $L_{n-1}L_{n+1} = L_n^2 + c(-1)^{n-1}$.
And, indeed, after looking of the first terms, we find that $c=5$ works (the general case follows by induction, from the initial recurrence relation).

Let $U_n = L_n - 2$.
Suppose that a prime $p$ divides $U_{n-1}$ for some odd integer $n$ (so that $n-1 = 2k$ is even).
Then, from above :
${(U_{n-1} + 2)(U_{n+1} + 2) = (U_n + 2)^2 + 5(-1)^[n-1}$
that leads to
$2(U_{n+1} + 2) = U_n^2 + 4U_n + 5$ mod[$p$]
But $U_{n+1} = U_n + U_{n-1} + 2 = U_n + 2$ mod[$p$]
Thus
$2U_n + 4 = U_n^2 + 4U_n + 5$ mod[$p$]
That is
$(U_n + 1)^2 = 0$ mod[$p$]
Since $p$ is prime we deduce that $p$ divides $U_n + 1 = L_{2k+1} - 1$, as desired.

I do not know the origin but, according to the fact that Lucas sequences are very well-known, this kind of exercise can be found almost everywhere...

Pierre.

Here's another approach:

Let $u=\frac{1+\sqrt 5}2,\ v=\frac{1-\sqrt 5}2$. Then $u+v=1, uv=-1$. We work in the field a)$\mathbb Z_p[u]$ if $(\frac 5p)=-1$ or b)$\mathbb Z_p$ if $(\frac 5p)=1$.

Our initial condition translates to $u^{2k}+v^{2k}-2=0\iff u^{2k}+\frac 1{u^{2k}}-2=0\iff (u^{2k}-1)^2=0\iff u^{2k}-1=0\ (*)$. On the other hand, what we must prove is that ${\small u^{2k+1}-\frac 1{u^{2k+1}}-1=0\iff (u^{2k+1})^2-u^{2k+1}-1=0\iff (u^{2k+1}-u)(u^{2k+1}-v)=0}$, which does indeed follow from $(*)$.

Well it is Iran 2002.
I had posted this problem about 2 months ago you can search it.


Presto Printing charges $ \$1.25$ per book for shipping and $ \$3$ per
order for handling. What is the total cost in dollars of shipping
and handling for an order of 10 books? Express your answer as
a decimal to the nearest hundredth.

There are 10 books, so the shipping is $ 10 \times 1.25 \equal{} 12.5$. Then the 10 dollars for handling makes a total of $ 22.5$ dollars.

[quote="GameBot"]Presto Printing charges $ \$1.25$ per book for shipping and [b]$ \$3$ per
order for handling[/b]. What is the total cost in dollars of shipping
and handling for an order of 10 books? Express your answer as
a decimal to the nearest hundredth.[/quote]
I'm getting a slightly differerent answer, but using a similar strategy.

As you said, it costs 1.25 dollars per book and we are purchasing 10. Thus we would have 1.25*10 for 12.5.
However, the problem says there is a 3 dollar charge for shipping and handling, and so we add 3 dollars to the current purchase.

Thus: 12.5+3= [b]15.50 dollars total cost[/b]


algebra
polynomial
function
trigonometry
algebra unsolved

Find the formula of
$ 1,0,\minus{}1,1,0,\minus{}1,\cdots$
I want to know how to work it out. :?:

[hide]Let $ \{x\}$ denote the fractional part of $ x$. The $ n$-th term is $ \displaystyle3\left(\minus{}\left\{\frac{n}{3}\right\}\right)\plus{}2$.[/hide]

This sequence satisfies the recurrence $ x_{n \plus{} 3} \equal{} x_n$, which has characteristic polynomial $ t^3 \minus{} 1 \equal{} 0$, so we can write $ x_n \equal{} A \plus{} B \omega^n \plus{} C \omega^{2n}$ for appropriate constants $ A, B, C$ where $ \omega^3 \equal{} 1$ is a primitive third root of unity.  (This should be intuitive:  the space of sequences that are periodic with period dividing $ 3$ is clearly of dimension $ 3$.)  Solving, we find that

$ (1) A \plus{} B \plus{} C \equal{} 1$
$ (2) A \plus{} B \omega \plus{} C \omega^2 \equal{} 0$
$ (3) A \plus{} B \omega^2 \plus{} C \omega \equal{} \minus{} 1$.

$ (1) \plus{} (2) \plus{} (3)$ gives $ 3A \equal{} 0$.  $ (1) \plus{} (2) \omega \plus{} (3) \omega^2$ gives $ 3C \equal{} 1 \minus{} \omega^2$.  And $ (1) \plus{} (2) \omega^2 \plus{} (3) \omega$ gives $ 3B \equal{} 1 \minus{} \omega$.  Finally, we have

$ x_n \equal{} (1 \minus{} \omega) \omega^n \plus{} (1 \minus{} \omega^2 )\omega^{2n}$.

The generalization is given by the [url=http://en.wikipedia.org/wiki/Discrete_Fourier_transform]discrete Fourier transform[/url].  The continuous analogue of this question is a sawtooth wave, the piecewise equation of which can be given by floor functions and fractional parts; this function can be given a Fourier series.

...I'm an idiot... carry on reading...

[hide]Ma netreba..[/hide]

Edit: Milin's trig idea-revisited: $ \cos\;3\{\frac {n \minus{} 1}{3} \} \frac {\pi}{2}$.

[hide](ako hoces da obrisem post? pa da i ti obrises svoj)[/hide]

Well, $ round \left( \sin \left( \frac{\pi x}{2} \right) \right)$ (where round(x) denotes the integer nearest to x) also works...  :D

The answer I gave can be written as a Fourier series (starting the series indexed at $ \minus{}1$ to make it an odd function)

$ x_n \equal{} \minus{} \sqrt{3} \sin \frac{2 \pi n}{3} \plus{} \sqrt{3} \sin \frac{4 \pi n}{3}$.

I think the Fourier Theory is better.
Because the $ \{ x\}$ and $ [x]$is hard to analyse.
But the trigonometric function is easier to deal with I think.

Correction:  divide both of my answers by $ 3$.


Let  x,y,z>0 such that x + y + z = xyz
Prove that (x - 1)(y - 1)(z - 1) <= 6[code]3-10

[quote="rem"]What does 6[code]3-10 mean?[/quote]

I guess it's $6\sqrt{3}-10$, because we'll have equality when $x=y=z$, i.e. $x=\sqrt{3}$ and $LHS = (\sqrt{3}-1)^{3}= 6\sqrt{3}-10$.

As was noted, $6\sqrt{3}-10=(\sqrt{3}-1)^{3}$.
Consider $AM-GM$ for the quantities $x-1, y-1, z-1$: \[\frac{(x-1)+(y-1)+(z-1)}{3}\geq \sqrt[3]{(x-1)(y-1)(z-1)}\] Where the $LHS=\frac{x+y+z}{3}-1$. In fact with a little manipulation we are very close to done now. If we could prove that $x+y+z\geq 3\sqrt{3}$ then we would have: $LHS\geq \sqrt{3}-1\geq \sqrt[3]{(x-1)(y-1)(z-1)}$ whereupon cubing gives the solution. So now we set about to prove that $x+y+z\geq 3\sqrt{3}$. Note that $x+y+z=xyz$... \[\frac{x+y+z}{3}\geq (xyz)^{1/3}\implies \frac{xyz}{3}\geq (xyz)^{1/3}\implies (xyz)^{2/3}\geq 3 \implies xyz \ \ (=x+y+z) \geq \sqrt{27}=3\sqrt{3}\] Therefore, $\frac{(x-1)+(y-1)+(z-1)}{3}=\frac{xyz}{3}-1\geq \sqrt{3}-1\geq \sqrt[3]{(x-1)(y-1)(z-1)}$
Cubing gives: \[6\sqrt{3}-10=(\sqrt{3}-1)^{3}\geq (x-1)(y-1)(z-1)\] Now we will show that there is another similar quantity which is greater than or equal to $6\sqrt{3}-10$: [hide="why?..."]Because I switched a sign and proved the wrong thing when I started working on this problem[/hide]

First notice that $x+y+z\geq \sqrt{x^{2}+y^{2}+z^{2}}$ (consider three dimensional distance, or just square... whatever)
Now RMS-AM: $\sqrt{x^{2}+y^{2}+z^{2}}\geq \frac{\sqrt{3}(x+y+z)}{3}$.
Combining these and noting that $xyz=x+y+z$ we have: \[xyz\geq \frac{\sqrt{3}(x+y+z)}{3}\implies 18xyz\geq 6\sqrt{3}(x+y+z)\] Also notice that $\left(\sum_{sym}x^{2}y\right)\geq 6xyz\implies 12xyz+\left(\sum_{sym}x^{2}y\right)\geq 18xyz$ (by rearrangement, AM-GM, whatever). We combine again: \[12xyz+\left(\sum_{sym}x^{2}y\right)\geq 6\sqrt{3}(x+y+z)\] But the $LHS$ is just $9xyz+(x+y+z)\left(\sum_{cyc}xy\right)$. Divide thru the inequality by $(x+y+z)$: \[9xyz+(x+y+z)\left(\sum_{cyc}xy\right)\geq 6\sqrt{3}(x+y+z)\implies 9\left(\frac{xyz}{x+y+z}\right)+\left(\sum_{cyc}xy\right)\geq 6\sqrt{3}\] But yet again, $xyz=x+y+z\implies \frac{xyz}{x+y+z}=1$. So, \[9+\left(\sum_{cyc}xy\right)\geq 6\sqrt{3}\implies \left(\sum_{cyc}xy\right)\geq 6\sqrt{3}-9\] In fact, $\left( \sum_{cyc}xy\right)\geq 6\sqrt{3}-9\implies (x-1)(y-1)(z-1)+2((xy)_{cyc}-x_{cyc})\geq 6\sqrt{3}-10$

So our interesting inequality is: \[(x-1)(y-1)(z-1)\leq 6\sqrt{3}-10\leq (x-1)(y-1)(z-1)+2\left(\sum_{cyc}x-xy\right)\] Of course everything written in this post makes the assumption that $x+y+z=xyz$

[quote="me@home"]As was noted, $6\sqrt{3}-10=(\sqrt{3}-1)^{3}$.
Consider $AM-GM$ for the quantities $x-1, y-1, z-1$: \[\frac{(x-1)+(y-1)+(z-1)}{3}\geq \sqrt[3]{(x-1)(y-1)(z-1)}\] Where the $LHS=\frac{x+y+z}{3}-1$. In fact with a little manipulation we are very close to done now. If we could prove that $x+y+z\geq 3\sqrt{3}$ then we would have: $LHS\geq \sqrt{3}-1\geq \sqrt[3]{(x-1)(y-1)(z-1)}$ whereupon cubing gives the solution. ...[/quote] \[A < B, B > C \not\Rightarrow A < C !!!\]

I was thinking to let

$x=\tan(\alpha)$

$y=\tan(\beta)$

$z=\tan(\gamma)$

and since $x+y+z=xyz$

$\alpha+\beta+\gamma=\pi$

but am not sure if this helps at all  :blush:

ohh you're right... but I think I could still patch up the proof because the idea should work..

Well, since noone seems to have a better alternative, I will post my long and boring solution  :)

Use the substitution $x= \tan{\alpha}, ...$ to get rid of the constraint. Notice that $\alpha, \beta, \gamma$ are angles of an acute angled triangle.

Now, due to Jensen, we have that \[\tan{(X-Y)}+\tan{Y}\geq 2 \tan{\frac{X}{2}}\] and, using the sum formula for tangents and the above result, we get that \[\tan{(X-Y)}.\tan{Y}\leq \tan^{2}{\frac{X}{2}}\] So \[\tan{(X-Y)}.\tan{Y}-\tan{(X-Y)}-\tan{Y}+1 \leq 1+2 \tan{\frac{X}{2}}+\tan^{2}{\frac{X}{2}}\] Which is equal to saying that \[(\tan{\frac{X}{2}-1}-1)^{2}\geq (\tan{(X-Y)}-1)(\tan{Y}-1)\] So we get back to the question: \[(\tan{\alpha}-1)(\tan{\beta}-1)(\tan{\gamma}-1) \leq (\tan{\alpha}-1)(\tan{(90^{\circ}-\frac{\alpha}{2})}-1)^{2}\] We can rewrite this expression with the t-formula's: \[\frac{(t^{2}+2t-1)(1-t)}{t^{2}(1+t)}\leq (\sqrt{3}-1)^{3}\] And because $\left(\frac{(t^{2}+2t-1)(1-t)}{t^{2}(1+t)}\right)' = \frac{-2(3t^{2}-1)}{t^{3}(t+1)^{2}}$ we see that the inequality holds.

Equality when $x=y=z=\frac{1}{\sqrt{3}}$.  :lol:

A nice equivalent would be obtained by subbing $x,y,z\mapsto 1/a,1/b,1/c$ so that $ab+bc+ca=1$.

Then the ineq becomes after expanding the LHS and simplifying: \[\frac{2-(a+b+c)}{2-\sqrt{3}}\le 3\sqrt{3}abc\] with equality iff $a=b=c=\frac{1}{\sqrt{3}}$.

[Don't be fooled, this problem is not trivial by AM-GM or something... both sides are smaller than 1. It is really just an equivalent, maybe a good hint for a nicer solution.]

\[x+y+z=xyz \implies \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\]  \[xy+yz+zx=(xy+yz+zx)(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}) \ge 9 \implies xy+yz+zx \ge 9\]  \[\implies (x-1)(y-1)(z-1)=xyz-(xy+yz+zx)+(x+y+z)-1\]  \[\le xyz-9+xyz-1\]  \[(x-1)(y-1)(z-1) \le 6\sqrt{3}-10 \leftarrow xyz \le 3\sqrt{3}\] But by AM-GM, $xyz \ge 3\sqrt{3}$...yeah i screwed up

[quote="K81o7"]But by AM-GM, $xyz \ge 3\sqrt{3}$...[/quote]That is exactly the point in this whole problem. :)


MATHCOUNTS
LaTeX
geometry
function
rectangle
blogs
number theory

I'm hoping that somebody who knows Latex will rewrite this list and make it into a pdf for everybody to save and print out for use in preparing for MATHCOUNTS.

Area for a circle:
r^2*pi
r = radius

DO WORD PROBLEMS or do Elias Saab's MATHCOUNTS drills (google it)

Don't assume too much from a diagram

Don't forget order of operations. You may have mental lapses in remembering this. It's unusual for MATHCOUNT to give such problems but these problems pop out once in a while

ESTIMATE graphs accurately

Exterior angles of polygons:
180-x
x = angle of the polygon

Form algebraic or operational equations from word problems

Have a mental venn diagram

Input your answer back into to the question to check if it's right

It's OK to write on the competition. Just don't get too sloppy. This usually helps you on geometry problems and picture problems

Know basic algebraic operations

Know basic geometry formulas

Know divisiblity rules well:
Divisible by 1: all numbers
Divisible by 2: all even numbers
Divisible by 3: digits of the number add up to a number divisible by 3 (you can cancel if you see any digits divisble by 3)
Divisible by 4: get rid of all the digits of the numbers except the last two; if those digits are divisible by 4 then the number is divisble by 4
Divisible by 5: last digit is 5 or 0
Divisible by 6: divisible by 2 and 3
Divisible by 7: take the last digit and double it then subtract it from the remainder of the number; repeat until the number is clearly is divisble by 7
Divisible by 8: take the last three digits THEN take the last digit add it to the combination of the two digits to it's left, if that is divisible by 8 then it is divisible by 8
Divisibility by 9: digits of the number add up to a number divisble by 3 (you can cancel 9's)
Divisibility by 10: last digit is 0
Divisibility by 11: google it

Know combinations

Know exponents

Know functions

Know how to do proportions. Such as "Auston is 60 inches tall. Using the conversion 1 inch = 2.54 cm, how tall is Auston in centimeters? Express your answer as a decimal to the nearest tenth."

Know how to factor whole numbers. Such as "Find the factors of 24."

Know how to find the prime factorization of a whole number.

Know percentages

Know permutations

Look for multiplication by 0. This will make any "difficult" operation become VERY simple

Look out for units. Always look and see if your answers are in the correct units

Memorize fraction to decimal and decimal to fraction conversions

Memorize the first 30 perfect squares

Number of degrees in a n-sided polygon:
180*(n-2)

Number of diagonals in a n-sided polygon:
n(n-3)/2

Number of factors in a number:
Find the prime factorization
Take all the exponents found in the prime factorization and get rid of the bases
Add one to each of the exponents
Multiply the exponents

Number of rectangles formed by a n by n dot grid:
(n nCr 2)^2

Number of sides in a regular polygon with an angle having x degrees:
360/(180-x)

Number of squares in a n by n dot grid:
(n-1)^2 + (n-2)^2 + ...

Number of squares in a n by n line grid:
n^2 + (n-1)^2 + (n-2)^2 + ...

Pythagoren Theorem:
a^2 + b^2 = c^2

Rate * Time = Distance (depending on the question)

Read charts correctly

Read FAST. Don't ASSUME what problems want you to do. This usually leads to a reading error. The only exception is when you're almost out of time

Sum of all the exterior angles of a polygon is 360 degrees

Sum of the first x even positive numbers squared:
x^2+x
or
x(x+1)

Sum of the first x odd positive numbers squared:
x^2

Supplement of an angle minus complement of angle (less than 90 degrees) equals 90 degrees

Understand clocks

Use ALL YOUR TIME

When midpoints of a square are connected to form a square, the resulting square is half the area of the original square.

If you want to extend on any of my ideas/formulas please do so
You can post your own ideas/formulas
This is ONLY THE SCHOOL version
I will post a CHAPTER, STATE, and NATIONAL version later
Please translate this to Latex and make it into a pdf version if you know how.
PLEASE!!!!!

[quote="Aznwithabrain"]can you make it into a pdf?
please
oh and happy thanksgiving[/quote]

Dude, almost all that stuff is posted in the AoPS MathCounts Bible. :wink:

and in my website, http://themathblog.wordpress.com

There's a formula list, way more extensive then goldnedomer's.

i know
but
can you make it into a pdf for everyone to save it to their computer for easy acces
(i'm one of those people who lose things :oops: )

[quote="13375P34K43V312"]Why don't you make a pdf yourself?

And there have been better lists, like the Mathcounts Bible, and Mr. Frost's Tips and Tricks, etc.[/quote]

Speaking of Mr. Frost's Tips and Tricks, is the site back up yet?

[hide="Link"][url]http://www.sharebigfile.com/getfile.php?id=C03EE5E7&access_key=YTMzZDEwNjJlODY0YWY0OTVlOGMxOTE5MGFiZGFhZDhiMTIwMzcyMjEzMzA4OGY3NWM2NzM4YTYzOTQ1OGNkMDMw&captcha=363337[/url][/hide]

The PDF

[size=75][color=darkred]Please try to hide large URLs. Thanks!
-nebula42[/color][/size]

i posted it on the flower problem thread, but here it is again.

I never got to really update it, so this is what I made up 6 or 7 years ago.

For those of you who are here all the time and are well prepared for MC, you probably won't find it useful as you probably know most of it.  I made that up with the things my students new to math competitions wouldn't know as they aren't part of most regular math classes(especially younger students).  Every once it awhile I would add something to it and intended to add more, but I switched to a new format for that kind of thing and it isn't digital so I can't share.  13375P34K43V312, you can check with me Monday and I'll show you the "new" format I used.  I have it at school.  Not sure how much it will help you, but maybe.  For those of you confused now.....well you can only get so much out of me if you aren't in my school :lol:

Hopefully there are no viruses, but if there are...bummer.

[quote="frost13"]

Hopefully there are no viruses, but if there are...bummer.[/quote]

Whose fault would that be if there were in fact viruses?

[quote="13375P34K43V312"]

Whose fault would that be if there were in fact viruses?[/quote]

Why the fault of the person who downloaded it without proper virus protection of course!!


trigonometry
logarithms
calculus
calculus computations

The curve C has the equation $ y\equal{}x^{\sin x}$, $ x>0$.

(i) Find the equation of the tangent to C at the point where $ x\equal{}\frac{\pi}{2}$.

(ii) Prove that this tangent touches C at infinitely many points.

$ y\equal{}e^{\sin x\ln x}$
$ y'\equal{}e^{\sin x\ln x}(\cos x\ln x\plus{}\frac{\sin x}{x})$
Tangent line to C at the $ \frac{\pi}{2}$ is $ y\equal{}x$, and it touches C at points $ x\equal{}(4k\plus{}1)\frac{\pi}{2}, k\in \mathbb Z$.


geometry
rectangle
combinatorics proposed
combinatorics

A $2003\times 2004$ rectangle consists of unit squares. We consider rhombi formed by four diagonals of unit squares.
What maximum number of such rhombi can be arranged in this rectangle so that no two of them have any common points except vertices?

[i]Proposed by A. Golovanov[/i]


topology
real analysis
real analysis unsolved

Don't know if this is the correct place, but i think this is real analysis:
On Munkres chapter of compact spaces i haven't understand the example 2:
The subspace $ X=\{0\}\cup\left\{\frac{1}{n}\ | \ n \in \mathbf{Z}_{+}\right\}$ of the real line is compact.

Give an open covering $ A$ of $ X$, there is an element $ U$ of $ A$ containing 0. The set $ U$ contains all but finitely many of the points $ \frac{1}{n}$; choose, for each point of $ X$ not in $ U$, an element of $ A$ containing it. The collection consisting of these elements of $ A$, along with the element $ U$, is a finite subcollection of $ A$ that covers $ X$

"The set $ U$ contains all but finitely many of the points $ \frac{1}{n}$" that means the complement of $ U$ in $ X$ is finite? but why i can say that? Thanks for any help

The topology on $ X$ is induced from $ \mathbb R$. That is, an $ X$-open set $ U\subset X$ is nothing but the intersection of some $ \mathbb R$-open set $ V\subset \mathbb R$ with $ X$.
So, if $ 0\in U$ (as in your post), then $ V$ contains an open interval $ (-\epsilon,\epsilon)$ around zero. Hence the points $ 1/n$ with $ n>1/\epsilon$ are contained in $ V\cap X$, which is $ U$.

I think i understand now! $ U$ contains infinite  points 1/n, so there are left finite points in $ X$
Much Thanks mlok!


inequalities
geometry
incenter
circumcircle
geometry proposed

Given are $\triangle ABC$ with incenter $I$ and circumcenter $O$ and area $S$ prove that
$3OI\ge\sqrt{S}(|\frac{b-c}{a}|+|\frac{c-a}{b}|+|\frac{a-b}{c}|)$

Right now, I just have the solution to a weaker inequality   :( .

$3\sqrt{2}OI \geq \sqrt{S}(\sum |\frac{b-c}{a}|)$.

[i] Proof [/i]

WLG, we may assume that $a \geq b \geq c$. Then it is easy to see that $|\frac{c-a}{b}| \geq max \{|\frac{a-b}{c}|, |\frac{b-c}{a}|\}$. Indeed, it is equivalent to $(b-c)(b+c-a) \geq 0$ and $(a-b)(a+b-c) \geq 0$. Thus it is sufficient to show that $\sqrt{2}OI \geq \sqrt{S}|\frac{c-a}{b}|$.

Let $B', B_{1}$ be respectively the orthogonal projections of $I$ and $O$ on $BC$. Then $OI \geq B'B_{1}= \frac{|c-a|}{2}$. On the other hand, since $sinC \leq sinB$ and $sinA \leq 1$, we obtain that $sinAsinC \leq sinB$. This implies that $b^{2}\geq 2S$. Therefore

$\sqrt{2}OI \geq \frac{|c-a|}{\sqrt{2}}\geq \sqrt{S}|\frac{c-a}{b}|$.

-------------------------------------------------------------------------------------

I have just figured out another approach.
WLG, we may assume that $a \geq b \geq c$. Then $\sum |\frac{b-c}{a}| = \frac{(a-c)(ab+bc+ca-b^{2})}{abc}$. Since $S = \frac{abc}{4R}$ and $a-c \leq 2OI$, it follows that we need to prove $9Rabc \geq (ab+bc+ca-b^{2})^{2}$. Maybe it is true.  :|


calculus
integration
trigonometry
complex analysis
calculus computations

[b][u]without[/u][/b] using complex analysis, show :D

$ \int_0^\infty \left(\frac {\sin\,x}{x}\right)^4 \;dx\; \equal{} \;\boxed{\frac {\pi}{3}}$

as I remember right it was already posted by .. you .  :lol:
idea was next: at first we show that  $ \int_{0}^{\infty}\frac {\sin x}{x} dx \equal{} \frac {\pi}{2}$
by differentiating $ I(a) \equal{} \int_{0}^{\infty} \frac {\sin x}{x}e^{ \minus{} a x}$
first hint
[hide="here"]
consider
\[ I(a) \equal{} \int_{0}^{\infty} \left( \frac {\sin x}{x} \right)^{4}e^{ \minus{} a x}
\]
and consider $ I''''(a)$
[/hide]
another way
[hide="here"]
try to integrate by parts this
\[ \int_{0}^{\infty} \left( \frac {\sin x}{x} \right)^{4}
\]
and express it by $ \int_{0}^{\infty}\frac {\sin x}{x} dx$
[/hide]
there is also another way by complex analysis   :blush:


there is a triangle

1st angle is 3 X

2nd angle is 4x

the last angle is 2X

find each angles

[hide="Solution"]The sum of the angles is 3x+4x+2x=9x. The sum of the angles also equals 180, so x=20. Then the angles are 40,60,80.[/hide]

[quote="shinwoo"]Yup..

I'm actually surprised by the fact that this kind of problem comes out in standarized tests[/quote]

Well, there's an important principle behind it, that the angles in a triangle add to 180. It's used quite a bit in geometry problems and should be memorized.

[quote="shinwoo"]Well, in PSSA, they actually gives you list of formulas you have to know

such as... area of circle, area of a polygon , degrees

that's why I'm even more surprised[/quote]

can you give me a list of those formulas that you should memorize? I'm trying to study for mathcount and would really appreciate a list a formulas.

[quote="RC-7th"]Everything here:

http://www.mathcounts.org/webarticles/articlefiles/105-toolbox.pdf

and perhaps more.

But this the kind of conversation that makes some people uncomfortable because it's about a national competition. Maybe it belongs in the Mathcounts forum.[/quote]

sorry


analytic geometry
IMO Shortlist
number theory
number theory unsolved

Let $ p$ be an odd prime and $ n$ a positive integer. In the coordinate plane, eight distinct points with integer coordinates lie on a circle with radius of length $ p^{n}$. Prove or disprove that
[b](a)[/b] there exists a triangle with vertices at three of the given points such that the squares of its side lengths are integers divisible by $ p^{n \plus{} 1}$.
[b](b)[/b] there exists a triangle with vertices at three of the given points such that the sum of the squares of its side lengths are integers divisible by $ p^{n \plus{} 1}$.

I have just restored this post from my archive. It was deleted by Orl a loooong time ago because it looked like a repeated posting of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41068]IMO Shortlist 2004 number theory 8[/url], but actually it is somewhat different from that shortlist problem (it says radius instead of diameter). So I would like to ask all of you whether this problem is correct (at the very first sight, it looks like the solutions of the shortlist problem would work here as well, but I have not checked more carefully).

  darij


What is the probability that a randomly selected positive
integer less than or equal to 100 is divisible by 7? Express
your answer as a common fraction.


hi guys,

the user 1000, which is called 1000 :P has just registed on MathLinks :)
let's bet on user 2000, shall we? :P

Ladies and gentlemen, I'm the 30000^th visitor.   :cool:

hi vali,

for a mathematician I had expected you could write the number 1000 correctly in your topic title ;)

A little update to this: we now got 3500+ users and 80000+ posts in our database :D


A sphere is inscribed in a cube that has a surface area of $ 24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

$ \textbf{(A)}\ 3 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 12$

[hide]
The area of each face of the outer cube is $\frac{24}{6}=4$, so the side length of the outer cube is $2$.  This is also the diameter of the sphere, and thus the length of a long diagonal of the inner cube.

A long diagonal of a cube is the hypotenuse of a right triangle with  a side of the cube and a face diagonal of the cube as legs.  If a side of the cube is $x$, we see that

$2=\sqrt{x^{2}+(\sqrt{2}x)^{2}}\Rightarrow x=\frac{2}{\sqrt{3}}$.

Thus the surface area of the inner cube is $6x^{2}=6\left(\frac{2}{\sqrt{3}}\right )^{2}=8$.  The answer is C.
[/hide]


[quote="SplashD"][quote="Inspired By Nature"]How many ways are there to seat $n$ married couples $(n\geqslant 3)$ around a table such that men and women alternate and each women is not adjacent to her husband? :lol:[/quote]
[hide]Is the answer $(n-2)(n-1)!$? If it is, I'll post the solution.[/hide][/quote]
I don't think so, but please post your solution. :)

[quote="arsemesomething"]i'm guessing: $n(n-1)(n-1)(n-2)(n-2)\cdots2\cdot2\cdot1\cdot1=\boxed{n!(n-1)!}$[/quote]
No, but will someone please post their solution? :rotfl:
btw, please write your answer in sum notation form. A few hints...
[hide="hint 1"]discordant permutations[/hide]
[hide="hint 2"]The women don't have to be seated first (there is no "Ladies first" restriction in this question)[/hide]
[hide="hint 3"]Laisant's recurrence formula[/hide]
Hope they help. :)

[hide="just guessing, is it"]n!(n-2)! since the husband can't sit on either side of the wife no matter guy or lady so first, they can't sit next to their spouse, that works at least for 3[/hide]

[hide="Is it..."]$2n*(2n-2)*(2n-3)*(2n-4)*(2n-5).......(1)$[/hide]

[hide="solution"]At the head of the table, there can be 2n people, because there are 2n people total. Then, to the left side of the head person, there can be $2n-2$ people, because there are $2n-1$ people left, and the person can't be next to his/her spouse. Then, next to that person, there can be $2n-3$ people, and so on.
[/hide]

These are all good guesses, but they aren't correct. Could a mod please move this to pre-olympiad.

[quote="computer_nuke"]sum from k=0 to k=n of $({(2n)}/{(2n-k)}){({(2n-k)}C{(k)})}{(n-k)}!{(-1)}^{k}$[/quote]
Could you please post your solution?

haha was studying for my combinations and permutations math test and came across this...

[hide=" my guess"]

(n-2) ( n-1)! (n-1)!

[/hide]

[quote="computer_nuke"][hide]
sum from k=0 to k=n of $({(2n)}/{(2n-k)}){({(2n-k)}C{(k)})}{(n-k)}!{(-1)}^{k}$
Lol I got this from a website as you hinted, someone provide solution please.
[/hide][/quote]
lol, thats cool.
My solution is long and takes up to much latex. Anyone have a short ingenious solution? :D

[hide]
Total number of ways of seating n couples in a circular(i presume so) table is (2n-1)!
No of ways couples can be seated together always is (n-1)!x2

Required solution = (2n-1)! - 2(n-1)!

Is this right???
[/hide]

[quote="fizsle"]sry there was a mistake in my previous solution..here is the corrected one [hide]
No of ways of seating n couples in an alternating manner.. i.e one woman one man and so on.. = (n-1)!n!
No of ways couples can be seated together always is (n-1)!x2
solution = n!(n-1)! - 2(n-1)!
[/hide][/quote]
The solution to the problem is...
[hide="answer"][img]http://www.math.dartmouth.edu/~doyle/docs/menage/menage/_2448_displaymath363.gif[/img][/hide]
I will post more hints on how to prove this soon.

I think there is something wrong with the question or the solution(both mine and yours!).... if n = 3 then there is only 1 way! if n =4 then there are only 2 ways..both the solutions do not give this!

btw, anyone know for certain the the number of combinations possible for n = 3, 4, 5 and 6.

thanks.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a group with $4n+2$ elements. Prove that it has no more than one subgroup with $2n+1$ elements.

[note... the problem from Arhimede Contest 2005:
Let $G$ be a group with $2n$ elements,  $H_i\le G$ and $|H_i|=n$, $i=\overline{1,m}$, $m\ge 2$
Prove that if $H_1\cap H_2\ldots\cap H_m=\{e\}$ then $G$ is abelian.
] if $n=2k+1$ we get the hypothesis from the previous pb. but..  something is not ok... because if there exist such subgroups $H_i$ then it is only ONE!! ... Am I missing smth?

I don't think that's a problem. The problem says that there are at least two subgroups $H_1,H_2$, so you shouldn't worry about that. It means that the condition that $n$ is even must be satisfied, so you need not wotty about what would have hapened, had this condition not been satisfied :). Or maybe I didn't understand what you were saying? :?

Anyway, for this problem, assume there are at least two distinct subgroups of index $2$, $H_1,H_2$. They are normal, so we may write $H_1H_2/H_1\cong H_2/(H_1\cap H_2)$. Since $H_1$ is a proper subgroup of $H_1H_2$, it means that $H_1H_2=G$, so $|H_1H_2/H_1|=2=|H_2/(H_1\cap H_2)|$, which means that $H_1\cap H_2$ has $\frac{|H_2|}2=\frac{2n+1}2$ elements, which is absurd.

oups  :blush:  ... I was really missing smth...
thx grobber for the sollution :D it's nicer than mine :)


trigonometry
calculus
integration
geometry
rectangle
calculus computations

how is that supposed to help? As I understand it, you can't reduce apery's constant into elementary terms

[quote="0714446459923"]how is that supposed to help? [/quote]
 :lol:  yes, it helps.
now we can consider :
$ f(z) = \frac {2(48z^{2} + 1)}{\sin \pi z (16z^{2} - 1)^{2}}$  we can use contour integration on big rectangle and calculate $ \sum_{n\in \mathbb Z}\frac {2( - 1)^{n}(48n^{2} + 1)}{ (16n^{2} - 1)^{2}}$ exact.

[hide="Another hint"]Use a suitable Fourier series.[/hide]

oh sorry, I didn't notice that it was alternating

Well here's the really cheap solution

$ \sum_{k\equal{}1}^{\infty}\frac{(\minus{}1)^{k\plus{}1}}{k^3}$

$ \equal{}\frac{1}{2}\int_{0}^{1}\frac{ln(x)^2dx}{1\plus{}x^2}$

$ \equal{}\frac{1}{4}\int_{0}^{\infty}\frac{ln(x)^2dx}{1\plus{}x^2}$

$ \equal{}\frac{\pi^3}{32}$

Which would be a really strange method if I hadn't recently solved that last integral, just by coincidence

[quote="0714446459923"]Well here's the really cheap solution

$ \sum_{k \equal{} 1}^{\infty}\frac {( \minus{} 1)^{k \plus{} 1}}{k^3}$

...

$ \equal{} \frac {\pi^3}{32}$

[/quote]

that is incorrect.

$ \eta(3) \equal{} \sum_{k \equal{} 1}^{\infty}\frac {( \minus{} 1)^{k \plus{} 1}}{k^3} \equal{} \dfrac{3}{4}\zeta(3)$

what you have as  $ \frac {\pi^3}{32}$ is $ \beta(3) \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k }}{(2k\plus{}1)^3}$


AMC USA(J)MO USAMO calculus AIME MIT college

Hello,
I am a junior in high school, and have to select my courses for my senior year. My problem is this: due to demands by my parents and my school, I have to make the choice between taking a) AP Chemistry and b) AoPS Independent Studies (2 semesters) and Independent Study in Physics (I've already done AP C, so this would be explorations in college physics). I would much prefer (b), because I enjoy physics and mathematics more than chemistry, and because I would really like to do well on ARML/make USAMO (Hey, I can dream, can't I?).

My worry is that by not taking (a), I would only have one AP science (AP Physics C). I think that this is bad, given that I want to major in planetary science/astronomy/physics (Minor or double-major in math, of course). I'm thinking this might negatively impact my ability to get into top schools.

My questions, then, are these:
1. Should I do option (a) or option (b)?
2. If I do option (b), which subjects should I study in my physics independent study?

It might be useful for whoever tries to help me to have information about classes I've already taken/am taking/will take. Below is such a list. All grades were A's, all classes honors unless otherwise noted to be AP.

Math: Advanced Algebra/Trig, Precalculus, BC Calculus, MV Calc/Linear Algebra

Science: Biology, Chemistry, Physics, Astronomy and Space Sciences, Physics AP, Science Projects*.

*This class is basically a research class. You pick some topics, devise research questions, and go do some original research. You keep research journals, write proposals, grant applications, etc., and often get published. It's a stepping-stone in my school for research competitions.

AoPS**: AMC Problem Series, Intro NT, AIME Problem Seminar

**I have not really taken these yet. I don't have time to do the problem sets in from these classes right now due to taking classes at my "normal" school that mete out loads of busy work that takes an incredible amount of time to do. As Mr. Crawford notes, unless you do the problem sets, you don't learn a thing. So I am just saving up the problems until the summer, where I can go through, do them, and actually learn.

Standardized test scores might also help (I know, I'm dumping a buch of info, but please just skip it if it is useless, I figure better more than less):

PSAT: 71/80/80 (231)
AMC: 132.5

That's all. Any information or advise you give me will be appreciated.
Thank you for your time.
sran

I'd suggest you do AP Chem. Even if your'e not really interested in chemistry, basic chemistry will be useful in physics as well as a general thing. While it would not really hinder your abilility to get into top colleges, an AP class does look impressive. Can you trim down on other subjects (example: drop school merchandising to take physics independent studies in Physics, Aops, etc.)to take AoPS classes and prepare for math competitions? Why don't you propose next year's schedule, and we can edit.
But you are obviously smart, so any choice you'll make will be fine. :lol:

Thank you for your help and kind words. Here is my proposed schedule. My school operates on the block system, and there are 8 blocks total availible. AP sciences take 1.5 blocks, all other classes take 1 block.

Class name--------------------------------Blocks
AP English*----------------------------------1
AP Political Science-------------------------1
MV Calc/Linear Algebra--------------------1
Symphonic Orchestra*--------------------1
Science Projects----------------------------1
Gym*-----------------------------------------1

AP Chemistry--------------------------------1.5
                   OR
Independent Study-Mathematics-------1
Independent Study-Physics--------------1

*Required Class

The problem is, I'm really not sure what more I can cut. I need Science Projects because I'm pretty sure I can use it to compete well in Westinghouse, etc. I'll go stark raving mad before I drop MV calc/Linear Algebra. And I'm pretty sure I need a social studies class to be competetive.

Wait, actually, there's an idea. Do you happen to know if colleges will look negatively if I only have take a social studies second semester? If so, there are certain manipulations I could pull to move gym before school second semester, and do AP Macroecon second semester. Then I could go to AoPS land :) .

Thanks for this. It is helpful.
P.S. What is "school merchandising"?

Here's what I would do if I were you:
-Since most colleges/universities are pretty inflexible in accepting AP credit, I'd suggest you drop AP Political Science, leaving you with an extra 2.5 blocks if you move gym to before school, with AP Chem. Your PSAT scores and taking AP English already suggest you're not an introverted stereotypical math/science geek(great scores, by the way!). With 1 block, you could do independent studies in Physics(I'd suggest EPGY for Light/Heat- usually a full-year 2nd year course for a physics major, or half, depending on the quality of the college). You have 1.5 blocks available, so why not do 2 AOPS class with that. I'd also suggest taking 2 or 3 AOPS classes over the summer, and beginning your science research so that class is not too stressful. Taking AOPS classes over the summer will probably let you finish half the intermediate classes (I'd suggest counting and algebra) and will prepare you to take take the AIME Problem Series and an Olympaid classe, which will hopefully get you into the USAMO. Multitasking is fun...... :)

Edit: by the way, school merchandising in our school is a class in which people mostly try to sell and design dumb school sweatshirts (Avondale)

Interesting. So you're definitely thinking that lack of social studies class will not hurt? Had not even considered that. But that is why I came here: to take advantage of the insights people like you can provide.

I really like your suggestion regarding taking AoPS over the summer. If I get rejected from RSI, thats probably what I will go with. I had not really heard much of EPGY before in the context of classes in school--I think I will take a look at it.

The only concern I still have is a lack of social studies. The MIT guy who talked at my school was pretty intense regarding MIT wanting multitalents, which is why I got into speech team, music, etc. So that would be an issue I'm likely going to keep in the corner of my mind. Do you think doing a summer program like CTD or something to take care of some social studies is a decent idea?

Gratitude,
sran

[quote="sky9073"]Unless your AP Chem teacher is excellent, I would suggest that you NOT take it.
You've already taken chemistry, so it would be easy to self-study for the AP test if you really wanted to take it. (I self-studied Chemistry last summer, and am taking AP Chem this year, finding it COMPLETELY USELESS. It's basically the same stuff.)
Besides, AP Chem isn't that impressive (for example, MIT doesn't give credit).
With the classes you've taken (very impressive), not taking AP Chem won't matter much.

Cheers![/quote]

But sran said he was pressed of time. Self studying AP Chem takes me 1.5 hrs. daily this school year, with regular chem.

And yes, sran I believe your lack of social studies will not hurt you.

Just read the EPGY webpage. How many weeks should the Light and Heat course take? From the descriptions, it is a 4-unit course, and other 4-unit courses take abou 16 weeks. This course seems to be more intense, though. My best guess, then, wouldbe 16-32 weeks.

Light/Heat took me 30 weeks this year (just finished), but I am doing lots of math comp. prep.

Well, I'm pressed for time this year. But that's because of classes like Spanish (I have to study 2 hours daily independent of any assigned homework in order to keep an A) and US history (We have to outline the textbook. I've filled up 3 notebooks already. Talk about busy work). Next year, I am focusing on classes that play to my strengths. My main time-eater will be college application essays. That will kill 1st semester, but will be a breeze second semester.

One other random question: I recognize EPGY as a pretty advanced program. So do colleges, right?

Wow. I'm getting so much stuff I never would have though of. Thanks, guys!

Its recognized. Michigan gives credit for them, for example. (I'll take classical mechanics and statistical physics -3rd/4th year physics courses, when I go there).
- If you have 1.5-2 hours of spare time every day, EPGY can be finished in 30 weeks.
Maybe you should just take Light/Heat next year and start out with Modern Physics in college. By the way, do you plan to sleep 1st semester?  :lol: I've heard apps drain your time.

It's hard for me to believe that people would suggest doing the normal coursework instead of something you would enjoy much more. Take the AoPS class and the physics independent study. I wouldn't even think twice about it.

Wow. I never even realized there were so many available options. Yes, I do plan to sleep. AP English is a breeze, I've heard; my school prepares people well. I don't go to a strong magnet school, so MV/LA is pretty much a naptime class for anyone who can qualify to AIME. And again, I'm eliminating most busy-work intense classes next year. Plus, I'm planning to knock off some college/schol apps over the summer.

BTW, congrats on getting into Michigan and getting all that advanced credit. That's pretty awesome. I hope I can do stuff like that when I graduate :).

Actually, I'm still a junior; (I [i]will [/i]take those classes)

Even more impressive, them. So its people like you who send the rest of us to community college :lol: !

On a more serious note, I think I've got it figured out now, thanks to the help from all of you. I guess I'm doing Light and Heat (And maybe modern physics)/ AoPS. After seeing all the EPGY options, I'm sort of inclined to drop AP Poly Sci and do Diff Eqs or something, too.

I keep saying this, but it is necesary for the help I've recieved. Thanks!

sran

regarding Political Science:
if you're not interested in it at all, and you're just taking it to have another AP/look good for colleges, then I agree that dropping it is probably for the best.

However, I would actually recommend keeping it.  Having your brain do non-mathy things is healthy, and depending on the quality of the class (which is definitely an important factor) you might get a lot out of it.


hola estoy  intentando  aprender  geo  proyectiva  pero no  tengo  mucho  material  solo  tengo  una pequeña introduccion de lo ques  geo  proyectiva  de geometry  revisited . yo  se que  por ejemplo nuestro  administradores  deben  saber  un poco de esto.  asi  que  les pido  que  me ayuden  por ejemplo  a contestar  las preguntas posteadas en  el  area de geometry en  el forum normal .  pascual? carlos? y  bueno  si se puede que alguien  ponga problemas aca que salga con estas cosas
                                  gracias  marko avila

Un libro: Geometric Transformations 3 de I. Yaglom.

Tu podrias decierme donbde puedo bajar ese libro? (el de I. Yaglom) a mi tambien me gustaria tenerlo.
Ademas tengo una pregunta  :P , asumiendo que dadas dos rectas paralelas, ellas se intersecan en un punto al infinito, como se establece que una recta y todas sus rectas paralelas se intersecan en el mismo punto.

Gracias de Antemano,

Atte.

José Carlos

Hola, aki unos links de los articulos de la revista eureka:

aki encontraras buenas referencias .

[url=http://www.obm.org.br/frameset-eureka.htm]OBM -Eureka[/url]


Prove that there exist  many [b] infinitive [/b]positive integers $n$ such that the number $3n^2+3n+2$ only has odd prime divisors in the form $3k+1$

[quote="jin"]Let $3n^2+3n+2=2(m^2+m+1)$
So $3(2n+1)^2-2(2m+1)^2=1$
Consider the equation $3x^2-2y^2=1$
Easy to see it has only odd solution $(x,y)$ and infinite solution.
And we can prove $m^2+m+1$ has only prime factor like $3k+1$.[/quote]

what i did was $2(3n^2+3n+2)=3(n^2+(n+1)^2)+1$ Notice that the equation $n^2+(n+1)^2=m^2$ have infinitive solutions so we are done. However it exactly has the same idea with $jin$ 's solution. thanks


function
LaTeX
calculus
calculus computations

Hi

let  $ f$ and $ g$  two functions definie  in $ \mathbb{R}_\plus{}$

and  let $ x \in \mathbb{R}_\plus{}$

prove that  [img]http://img148.imageshack.us/img148/9561/sanstitreel5.jpg[/img]

Obviously false: $ fg$ can easily be identically $ 0$ with pretty large positive $ f$ and $ g$ somewhere on $ [0,x]$ (say, $ f$ on the first half of the interval, and $ g$ on the second). By the way, learning basic $ \text{\LaTeX}$ takes about half-an hour ;).

It will be good , if you present some contre exemple

Thank you for your suggestion about latex  :wink:

I have presented it, haven't I?. If you want a formal example, take $ f(t)\equal{}1$ on $ [0,x/2]$ and $ 0$ on $ (x/2,x]$, and take $ g\equal{}1\minus{}f$.