duh they're gonna pwn. especially selke, as he should make masters round in '10. sorry for the slight invasion of your forum, guys.

[quote="AwesomeToad"]That's fine as long as we're not not talking about say pythag.

By the way is masters top 4 cd or written?[/quote]

written

cuz I made it...

dnkwin did u really make the nats?
what is ur real name?

yes he did and he did very well
just pm him instead of posting here
no names to usernames mentioned on the forum

=)

bleh.  In 6th grade I got 63rd(23/46. Blah.  Careless mistakes. :( )
Now I'm in 7th grade, trying again......I'm probably gonna get like 25th this time at best.

I would really love to make nats and that would be so awesome, but I live in Indiana.  CURSE THAT!  IF only I lived in say Oklahoma or North Dakota.......then I'd actually have a chance.  bleh. :(

Also, if any of you see Doom Monkey, he's in my school.

What do you think the difficulty of the test will be?  Easyish or hardish?

[quote="gauss1181"][quote="AwesomeToad"]Hey who do you think is going to pwn this year?
I practiced and did alot but I think Mark Sellke, Yushi and David are going to pwn. I don't think Vivian Gu does MC this year.

Blah got 35th-60th last year, lots of people with 25-28[/quote]\

I think Vivian Gu is still doing Mathcounts this year. In 2009, she was a 7th grader. :wink:[/quote]

Vivian is NOT doing MC this year. I go to her school and she told me. She said something about GMO.

[quote="maybach"][quote="ernie"][quote="maybach"]Easyish or hardish?[/quote]

I think it's gonna be middle-ish.[/quote]

Um.........similar to, harder than, or easier than the 2008 State?(I hope the latter)[/quote]

Maybe a bit harder?

[quote="ernie"][/quote][quote="maybach"][quote="ernie"][quote="maybach"]Easyish or hardish?[/quote]

I think it's gonna be middle-ish.[/quote]

Um.........similar to, harder than, or easier than the 2008 State?(I hope the latter)[/quote][quote="ernie"]

Maybe a bit harder?[/quote]
2008 State was kinda hard... I only got a 35 on it (compared to 46 this year)... so at the rate mathcounts problems are going I would say a LOT easier...

[quote="math154"]It's an unwritten rule that they can only repeat every problem five times so they're running out of ideas.

[quote="AwesomeToad"]Vivian is NOT doing MC this year. I go to her school and she told me. She said something about GMO.[/quote]

...Do you mean CGMO? I can't think of anything else...

Also, that would be a pretty silly reason (bleh maybe not) not to do MC?[/quote]

It is pretty silly... Also wait what unwritten rule? I thought it was two, not five? (exact wording, of course.)

[quote="AwesomeToad"]

Blah got 35th-60th last year, lots of people with 25-28[/quote]

Argh.  I got 23/46 in state.  But I took 2009 again and got 28/46.  Still total fail though.............

[quote="AwesomeToad"]I think Vivian meant CGMO. There was something else.

Someone at my school said the stupidest thing ever:
[quote]Whats the point of doing MAthcounts again if you already got a trip to Orlando?[/quote][/quote]

getting another trip to nationals..............

[quote="pinkmuskrat"]heh yeah. we all experience these ppl on a daily basis.

and, other than the whole nats thing, mc is fun! yay[/quote]

Wait so the nationals is at Disney again this year?

The guy who came 3rd at AAPotter (Purdue) chapter goes to my school, came 7th state

Does some really weird stuff at my school.

Not want to mention his name though...

[quote="AwesomeToad"]Besides, most people at my school don't do math at all (mostly jocks)
I'd like to raise my national ranking if I got another trip[/quote]

if you are supposedly in vivian's school, then you are in my school too. i have no clue who you are though.

People in 8th grade now.
But the guy who came 7th was sure he got at least a 27 on sprint and perfect target.

But he  got 24/14 so yeah

[quote="AwesomeToad"][quote="maybach"]Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read "7th place".....feel really sorry for him......poor guy :(

Also, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).[/quote]

I hate to say this but don't feel to sorry for him.

Also the 35/46 would probably just barely make state cd better not count on it.
Of course, the level of the competition and test also matter[/quote]

I agree. Only something with a 50/55 would be enough to win state mc.

Generally on an average test you would need 43/46

But this year Sellke got 45 and year before best was 41.
Also, what do you mean by 50/55?

Maybe he was referring to the team score?  But team round doesn't matter in state.  Also, it's too bad they cut the scholarship money.  I think with some practice I could get 20th place, but that wouldn't be good enough for a scholarship anymore :(

Also, my sister went to MC State in 8th grade at Rose Hulman and got 163rd place(I got 63rd :rofl:).  I thought the Rose Hulman auditorium seemed nicer.  But only 2 kids get scholarships though..........

[quote="AwesomeToad"]Team Round determines the state coach wow my school did better than last year.

2 people in CD!!/Top 10[/quote]

State Coach doesn't really matter.

Awesome Toad, you're from West Laffayette, right?

Anyways the state coach will always be Bob Fischer no matter what.

Also, I actually got a 4 on target, but I got a 5 because they misread my answer to the rectangular prism question. :)

Still did horribly though.........

Are all of you guys 8th graders and below?

Yeah.  I'm currently in 7th grade!(thank god, with selkie, I know I have know chance this year)

Sellke,
Liang, and homma are gonna pwn this year.

Vivians not doing mc, so everyone wants the 4th spot

[quote="AwesomeToad"][quote="maybach"]AwesomeToad, is that the kid who won state countdown?

I was there.[/quote]

Yes.
Examples: [quote]Oh I hit you. I get 5 points[/quote]
???
or
[quote]I was up all night playing starcraft blah pwns a korean oh i'm braindead go to bed at 5: 60[/quote][/quote]

Corrections:

Oh I hit you. I get 2 points. (Currently Retired)

I was up 1/3rd of the night playing starcraft, I pwn koreans, I'm braindead go to bed at 1:00. (Retired as of Friday)

Anyways, meh I'm doing more programming than math right now, but I'll still try for 4th  (although there are some really good 8th graders ready to pwn me) :P. Does anyone know what grade 5th/6th place are in?

[quote="Maybach"]Yeah.  I was there feeling really sad watching cause' I really wanted to be up there and I actually thought I did a good job(Man I need to practice on sprint, but I've imrpoved my average score 14 points since).  I saw that dude.  He looked like he was praying and he looked really sad when they read "7th place".....feel really sorry for him......poor guy :(

Also, could I make cd(not nats) for states with 35/46?  Because that's about what I can get on a good day assuming the test was made recently(2006-).[/quote]

Nah, I didn't really expect to make nats (although it would be awesome if it happened). My goal was to pwn people in CD, and not fail too bad in written.

Alex Barksdale came in 5th if I remember right
Correct me if I'm wrong...

Does anyone have a list of the top 16? 2009 State MC?


A circle of radius $ r$ is concentric with and outside a regular hexagon of side length $ 2$. The probability that $ 3$ entire sides of the hexagon are visible from a random point on the cirlce is $ 1/2$. What is $ r$?

just to be sure...

The area of visibiliy is enclosed by the tangent to the circle at that point and the closest part of the hexagon right??

 :)


[quote="mathVNpro"]Given $ (x_n)$ and $ (y_n)$ defined as
\[ x_1 \equal{} y_1 \equal{} \sqrt {3}\ ; \ x_{n \plus{} 1} \equal{} x_n \plus{} \sqrt {1 \plus{} \left ({x}_n \right)^2}\ ; \ y_{n \plus{} 1} \equal{} \frac {y_n}{1 \plus{} \sqrt {1 \plus{} \left (y_n \right)^2}} \ \forall n\ge 2\]
Prove that $ x_n.y_n$ $ \in$ $ (2,3)$[/quote]

hint :

for $ x_n$ : $ \cot \frac{x}{2}\equal{}\cot x\plus{} \sqrt{1\plus{}\cot^2 x}$

for $ y_n$ : $ \tan \frac{x}{2}\equal{}\frac{\tan x}{1\plus{}\sqrt{1\plus{}\tan^2x}}$

[quote="hophinhan"][quote="mathVNpro"]Given $ (x_n)$ and $ (y_n)$ defined as
\[ x_1 = y_1 = \sqrt {3}\ ; \ x_{n + 1} = x_n + \sqrt {1 +  ({x}_n )^2}\ ; \ y_{n + 1} = \frac {y_n}{1 + \sqrt {1 +  (y_n )^2}} \ \forall n\ge 2\]
Prove that $ x_n.y_n$ $ \in$ $ (2,3)$[/quote]

hint :

for $ x_n$ : $ \cot \frac {x}{2} = \cot x + \sqrt {1 + \cot^2 x}$

for $ y_n$ : $ \tan \frac {x}{2} = \frac {\tan x}{1 + \sqrt {1 + \tan^2x}}$[/quote]

We have

$ \displaystyle 2 < x_{n}\cdot y_{n} = {\frac {2 \cos^2 (\frac{\alpha}{2^n})}{2 \cos^2 (\frac{\alpha}{2^n})-1}} < 3 \qquad n\ge2, \alpha=\frac{\pi}3$


induction
number theory unsolved
number theory

[quote="marko avila"]prove that

           0 < sum g(k)/k from 1 to n - 2n/3 < 2/3

    where g(k)  is the greatest  odd  divisor of  k[/quote]

Use induction on n.

Maybe this is related:
http://www.mathlinks.ro/Forum/viewtopic.php?t=73591


Prove that if x is an integer, than x^2+6x+5/x+5 is also an integer.

note that $\frac{ x^2+6x+5}{x+5}=x+1 \in Z$ if $x \in Z$

That's it

It means the number is rational :)
N -> natural
Z -> rational
D -> decimal
Q -> irrational
R -> real
(C -> complex)


Okay, so I'm thinking of doing a Family-Feud type of thing.  Math related, of course.  But I need to do some surveys.

1. Name a word that you will often find in a problem that tells you that it will most likely require calculus to solve.

2. Besides “Q.E.D.”, name a word or phrase that is often found in proofs.

3. Name the branch of mathematics that’s the hardest to understand.

4. Name the most widely known theorem.

5. Name something that you can do with most graphing calculators, but not with scientific calculators.

6. Name a value for $x$ between 0 and 90 exclusive that you have $\sin x^{\circ}$ memorized.

7. Name a function that, when graphed, could be identified right away.

8. Name a mathematical feat that Chuck Norris is capable of.

9. Name a widely known mathematics competition.

10. Name a formula that any math student should have memorized.

11. Name a branch of mathematics that many people hate.

12. Name a famous mathematician.

13. Rational or irrational, name one of the most important real numbers.

PM me your answers, and once we get enough votes, we can play!

You may submit multiple answers.
However, these answers will be weighted so that everyone who votes gets an equal say.
So let's say one member votes for answer A, and another nominates answer B and answer C.  Answer A will count for twice as much points as B or C.

These should explain it:
http://en.wikipedia.org/wiki/Chuck_Norris
http://en.wikipedia.org/wiki/Chuck_Norris_Facts

So far, 5 people, including me, voted.  I probably will stop once I get a decent size of people, maybe 20 or 25 or something, and then just play around with the numbers.

[quote="fireemblem13"][quote="bubka"]Who is Chuck Norris?[/quote]
read my signature. :D search for chuck norris random fact generator on google.[/quote]

Ok so your signature has plenty of mathematical feats Chuck Norris is capable of. But that does not help me as regards his identity. I am too lazy for a google search now, so somebody clarify please.

as far as i know, he's a made-up person, but im not the best person to ask. my friend told me a chuck norris joke(non-mathematical) one day and it was funny... so yea...

There is a real Chuck Norris and I have seen him on tv ;)   I have signed up, too.

He played Walker in "Walker, Texas Ranger"

So far, 11 people voted.  I'm looking for more, still.

so, once there are a lot of ppl who have done this, what do we do?

let's see.... the wikipedea link above is pretty useful. basically, he's someone you associate impossible and crazy talsks with. for example, some random facts
1. Chuck Norris doesn't brush his teeth. He points his fist at his mouth and the plaque jumps out. :rotfl:  :rotfl:
2. If you are eating Alphabits and the letters somehow spell Chuck Norris, DO NOT attempt to take another bite if you value your life.
3. Contrary to popular belief, a meteor did not end the dinosaurs' existance. It was Chuck Norris' roundhouse kick; it was so powerful it left a massive crater in the earth. :rotfl:

Once we get a bunch of votes, we'll start the game.  Most likely in another topic, though.

So far, 12 votes.  If you're okay with this, we could get something going.


resolve this equation
$sin^3$x(1+cotangx)+$cos^3$x(1+tanx) = cos2x

$\sin^3x(1+\cot x)+\cos^3x(1+\tan x)=\cos 2x$
$\sin^2 x(\sin x+\cos x)+\cos^2 x(\cos x+\sin x)=\cos 2x$
$\sin x+\cos x=\cos 2x$
$\sin x\pm\sqrt{1-\sin^2 x}=1-2\sin^2x$
$\pm\sqrt{1-\sin^2 x}=1-\sin x-2\sin^2 x$
$1-\sin^2 x=(1-\sin x-2\sin^2 x)^2$
After expansion and factorization, we have
$\sin x(\sin x+1)(2\sin^2 x-1)=0$
$\sin x=0$ or $\sin x=-1$ or $\sin x=\pm\frac{\sqrt{2}}{2}$
However, from the original equation, we know that $\sin x$ and $\cos x$ must be both non-zero or otherwise $\tan x$ and $\cot x$ will be undefined. So the case $\sin x=0$ and $\sin x=-1$ (which leads to $\cos x=0$) should be rejected. So the case left is $\sin x=\pm\frac{\sqrt{2}}{2}$.
We now get $x=\frac{\pi}{4}+\frac{k\pi}{2}$ where $k\in\mathbb{Z}$
But when we put x into $\sin x+\cos x=\cos 2x$ adn check, we find that only when $x=\frac{3}{4}\pi,\frac{7}{4}\pi,etc$ the equation is satified.
Therefore, $x=\frac{3}{4}\pi+k\pi$ where $k\in\mathbb{Z}$
PS1 This is my 100th post!  :D
PS2 Is this a number theory problem?

first of all $x\neq \frac{k\pi}{2},k\in Z$ ..... [b](*)[/b]

equation is equivalent with:

$\sin x+\cos x=\cos 2x$

$\sin x+\cos x=-(\sin x+\cos x)(\sin x-\cos x)$

$(\sin x+\cos x)(\sin x-\cos x+1)=0$

$(\sin x+\cos x)\left(\cos\left(x+\frac{\pi}{4}\right)-\frac{\sqrt{2}}{2}\right)=0$

second factor of LHS is $\neq 0$ (condition [b](*)[/b]), so solution is

$\boxed{x=m\pi-\frac{\pi}{4}\ ,\ m\in Z}$


algebra
polynomial
number theory
relatively prime
AMC

The polynomial
           $ P(x) \equal{} (1\plus{}x \plus{} x^2 \plus{} . . . \plus{}x^{17})^2\minus{}x^{17}$
has $ 34$ complex zeros of the form $ z_k\equal{}r_k[cos(2\pi\alpha_k) \plus{}isin(2\pi\alpha_k)]$
$ k\equal{}1,2,3,...,34$ with $ 0<a_1<\equal{}a_2,a_3 <\equal{}... a_{34}<1$ and $ r_k>0$. given that
$ a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}a_5\equal{}m/n$, where $ m$ and $ n$ are relatively prime positive integers, find $ m\plus{}n$.


search
number theory unsolved
number theory

[quote=sankha012]one sol is $n=\left(\frac{(2+\sqrt{5})^{2k}(1+\sqrt{5})-(2-\sqrt{5})^{2k}(1-\sqrt{5})}{2\sqrt{5}}\right)^2$ for all $k\in\mathbb{N}$.Pretty bad looking,huh? :D
In this case,$n(n+1)=p^4+p^2$ and $(p^2-2)^2+(5p^2-4)$.This $5p^2-4$ is a square.[/quote]
Hey (sorry to revive this thread).. But I seriously wanna ask How the hell did you come up with that? Plz. give the motivation behind it...
 Since there is no reachable solution to this prob on this thread till now here's my idea:-
[hide]Set $n=m^2$ as then $n(n+1)$ is already a sum of two squares.
Now if we get $m^2=p^2+q^2,$ then notice that it becomes $(pm+q)^2 - (p-qm)^2$..
Now it's trivial as it's well know that there are infinite pythagorean triplets...So there are infinite $(m,n,p)$ where m,n,p are integers..
QED[/hide]

[quote=makar]Prove that there are infinitely many positive integers $ n$ such that $ n(n\plus{}1)$  can be represented as a sum of two positive squares in at least two different ways. (Here $ a^{2}\plus{}b^{2}$ and $ b^{2}\plus{}a^{2}$ are considered as the same representation.)[/quote]

If n is such a number, then so is 4n(n+1).


An open box with a square base is required to have a volume of 10 cubic feet.

NOTE: What exactly is a cubic foot?

(a) Express the amount A of material used to make such a box as a function of the length x of a side of the square base.

(b) How much material is required for a base 1 foot by 1 foot?

NOTE: Is one foot by foot the same as 144 inches?

The box have a square base.
The volume of the box is the area of the base times the heigth.
Or

${x^{2}h = 10}$

It is equal 10, because the volume was given.

The amout of material is proportional to the sufarce area. so

${S=x^{2}+4 h x}$
${h=\frac{10}{x^{2}}}$
${S=\frac{x^{3}+40}{x}}$

${A=\alpha S}$

I'm not english. so i dont know how much is one feet, and one inches. but one foot by one foot could be 144 square inches

dude, omfg.  let me introduce you to my bestest cyber friends, yahoo and his sidekick google.  you could meet them at yahoo.com and google.com.  they would love to meet you because thats how they make their money and im sure they would be extreely helpful to you.

you would be saving us a buch of time if you used them

one foot is 12 inches.  that means 1 square foot is 144 square inces and 1 cubic foot is 12^3 cibic inches.


Hello!

I have two questions.

1. I really don't get the difference between AIME A and AIME B Problem Series, diverting from Aime A is not in the summer.

2. Which one should I take AMC 12 Problem Series or AIME B?

I am going to finish Volume 1 and 2 this summer. And I really want to do well on the AIME next year. I also might take WOOT, but I am not positively sure.

Thanks for your help! :lol:  :)

[quote]Description
Our two AIME Problem Series classes prepare students for the AIME, which is the second in the series of tests used to determine the United States team at the International Math Olympiad. Many top colleges also request AIME scores as part of the college application process. Each day consists of a discussion of specific problems from past AIME competitions, or from other contests of a similar difficulty level. Important general problem solving strategies and specific math facts are taught through solving these problems.

Please note, we offer two versions of this class, AIME Problem Series A and AIME Problem Series B. These two courses are roughly the same difficulty, but the problems covered in the two courses are completely different. Many students opt to take both classes. We typically offer one in the summer and one in the fall/winter.[/quote]

that should answer question number 1.

if anyone could clear this up for me at least

#2

thanks! :)

To answer #2, we'll probably need some more background info on you.

what would you mean by some background information

[quote="crazypianist1116"]What are your goals? Have you qualified for AIME before? Do you want a higher AIME score or higher AMC score? What grade are you in? etc.

Also, I would suggest holding off on WOOT until you are positive that you will qualify for USAMO.[/quote]

I agree on more background info needed, but I don't think should avoid WOOT until you are certain you're ready for the USAMO.  We've had plenty of people take WOOT fruitfully before being sure they could pass the AIME.  Before taking on WOOT, you should be sure that you are dedicated enough to put in the time to get the most out of the course.  Mathematically, it would be best if you are confident you can get at least 5 AIME questions.  So, you should have a fighting chance on the AIME before starting WOOT, but you needn't wait until you are sure you can pass the AIME to join.


An experiment has n possible outcomes, all equally likely. Each trial is independent. We repeat until we have two trials with the same outcome. For example, we would stop after getting 1, 4, 3, 4. What is the expected number of trials f(n)? Find an asymptotic expression for f(n).

My apologies for the incorrect link.  Here is the link but a similiar (though different) problem posted eariler.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1695488873&t=158138

This is a different (although similar) problem.

[hide="Solution"]The probability we finish in exactly $ m+1$ trials, $ 0 \leq m \leq n$ is $ \frac{m}{n}\cdot \prod_{i = 0}^{m-1}\frac{n-i}{n}= \frac{m \cdot n!}{n^{m+1}(n-m)!}$.  Thus, the answer is $ \sum_{m = 0}^{n}\frac{m(m+1) \cdot n!}{n^{m+1}(n-m)!}$.  I don't, off the top of my head, see a nice form for this expression (although that shouldn't stop others from looking for one).[/hide]


AMC
AIME
ARML
USA(J)MO
USAMO
algebra
functional equation

So I've been working through AoPS vol 1&2 over the summer and have been doing random aime problems (usually around 5~6 tough ones a day). I don't know but I think I'm starting to lose my interest in math since I haven't really been doing much else. I am scared that I will burnout and abandon math altogether. Can someone please tell me how much math I should be doing. Right now I do about 7-8 hours of math per day but most of it is done in the early mornings or at night.

Thanks in Advance

There's no amount you "should" be doing. Do as much as interests you. I can say that 7-8 hours would be too much for me, at least.

I would suggest taking a break from math a little bit and start doing something else. I was burned out just after this year's AIME and didn't feel like doing math anymore. Usually, when you take breaks, you start to remember what you enjoyed about math and then you'll do math again. You just need some variety in your life once in a while so it wouldn't hurt to take an entire day or two off of math. And I also do 7-8 hrs per day (during winter, summer, and spring break) so I know how you feel. The chances that you will [b]completely[/b] lose your interest is very low if you really do have a true passion for math.

You can always diversify and study something like science one day, then math for the next few days.  Take breaks from math, but you don't necessarily have to take breaks from studying.

Personally, working through a whole problem book is too intimidating for me. I usually do problems from various sources to reduce fatigue (e.g. aops forums).

the problem is that you're doing 7-8 hours of math a day. you should barely even be awake for that long. add in meals, school, practicing instruments, sports practices, and studying/homework for other subjects and you're up until midnight
you should probably get some rest

Yeah, 7-8 is kind of overkill.

When I get bored, I do things I shouldn't be doing (cough group theory/axiomatic set theory/etc...), or I just do olympiad problems instead, which are very different from AIME problems. There's always something out there.

How quickly you burn out I think depends more on how much you force yourself to do, as opposed to how much you're naturally inclined to do. As an example of the latter, when doing a stupid assignment like writing a paper my idea of a break is finding a functional equation on these forums and busting my brain on it for a bit. So that 7-8 hours is going to read differently for different people. But if you wouldn't normally do something like 4-5 hours per day anyway, unforced, you might want to cut back.

[OFFTOPIC]
[quote="Zhero"]I do things I shouldn't be doing (cough group theory/axiomatic set theory/etc...)[/quote]
"Shouldn't be doing"? That is perfectly legitimate math, maybe more so than contest math, and for the olympiad problems motivated by advanced concepts* you may be well prepared.

* such as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159989]this IMO problem[/url] whose solution is elementary only in its presentation, though such a problem [i]should[/i] never appear on an olympiad.
[/OFFTOPIC]

EDIT to below: Of course. I suppose I implicitly meant "on average" several times in this post.

Well the amount you would do "unforced" can also vary. For example I might feel inclined to do a lot of math one day and then much less the next.

@Palmer: Is this not the case for you?

You should no when you feel tired. After you feel tired you will never remember the concepts you just studied. So if you do not feel tired doing 7-8 hours of math then I guess I should be fine.

 :surf: This is my 50th post(no celebration).

[quote]"Shouldn't be doing"? That is perfectly legitimate math, maybe more so than contest math, and for the olympiad problems motivated by advanced concepts* you may be well prepared. [/quote]
Hmm yeah, now that you mention it, I think I'll have to retract my statement. Though I didn't think much of it before, I did benefit from doing axiomatic set theory. For one, it helped me to understand mathematical rigour more (e.g., what axiomatic systems are and how to write formal proofs), and I learned much about functions.

And it certainly provides for a fine break away from the stuff like that the OP's been doing. And it does teach invaluable techniques.

I suppose this notion arose from my putting competition math above all the other math... in the past few months, in my eyes, all the other math is just a distraction from competition math.

But my eyes have opened now. It's summer now, no competitions to worry about. Time to indulge...

A great way to memorize new formulas/concepts is to do them right before you sleep. Believe it or not, your brain thinks about what you just did while you're asleep, and when you wake up again, what you just learned will stay with you for a long time.

If I do math right before I sleep, I can't sleep.

I recommend 1-3 hours of math on weekdays (depending on how much homework you have) and 4-6 hours on weekends. (each day)

That's like all I have time for. But the week after ARML (and the month before it) I barely did any math at all, because preparing for USAMO basically made me sick of it. (temporarily, of course)

Wow, do it like, few hours. 1, 1.5, 2.Something like that. I'd die if I did math stuff 7-8 hours a day. It s like, a part time job for you :D
I'd say forget about maths for like a few weeks, go do some sports, hang out with your buddies more. The reason why you lose interest is because you dedicate too much time to it, you feel a routine in it, it becomes boring. I know what I'm talking about  :roll:

For me, I deal with maths in the evening. Open up the unsolved problems section and work my magic. But I can't do it for 7 hours a day :(
You know if you do maths so long, your head is like, exhausted, usually when I feel that way I go outside jogging or riding a bike or something and when I think about the math problem I had, the answer seems to somewhat more easily reveal itself :P

So all in all, you just have to take up something else if you don't wanna lose interest.

Have fun.


Determine the number of possible values for the determinant of $ A$, given that $ A$ is a $ n*n$ matrix with real entries such that $ A^3\minus{}A^2\minus{}3A\plus{}2I\equal{}0$, where $ I$ is the identity and $ 0$ is the all-zero matrix.  [/b][/hide][/code]

All eigenvalues of this matrix are roots of $ x^3 \minus{} x^2 \minus{} 3x \plus{} 2 \equal{} 0.$ That equation has three real roots, $ 2$ and $ \frac { \minus{} 1\pm\sqrt {5}}2.$ Let's just call these numbers $ r_1,r_2,$ and $ r_3.$

The possible values of $ \det(A)$ are $ r_1^ir_2^jr_3^k,$ where $ i,j,$ and $ k$ are nonnegative integers such that $ i \plus{} j \plus{} k \equal{} n.$

We can form such a product in $ \binom{n \plus{} 2}{2}$ ways. The only remaining question is whether different values of $ (i,j,k)$ could result in the same determinant - I don't think so, but I haven't proven that yet.

Hello,

I am sure that there is an easier way to finish the last part, but this is one way:

Let $ i_1\plus{}j_1\plus{}k_1\equal{}n$ and $ i_2\plus{}j_2\plus{}k_2\equal{}n$. Then if we have two values of the determinant that are equal, we get

$ \begin{array}{lrl}
 &2^{i_{1}}\left(\frac{\minus{}1\plus{}\sqrt{5}}{2}\right)^{j_{1}}\left(\frac{\minus{}1\minus{}\sqrt{5}}{2}\right)^{k_{1}}&\equal{}2^{i_{2}}\left(\frac{\minus{}1\plus{}\sqrt{5}}{2}\right)^{j_{2}}\left(\frac{\minus{}1\minus{}\sqrt{5}}{2}\right)^{k_{2}}\\
\Leftrightarrow & 2^{i_{i}\minus{}j_{1}\minus{}k_{1}\minus{}(i_{2}\minus{}j_{2}\minus{}k_{2})}&\equal{}\left(\minus{}1\plus{}\sqrt{5}\right)^{j_{2}\minus{}j_{1}}\left(\minus{}1\minus{}\sqrt{5}\right)^{k_{2}\minus{}k_{1}}\\
\Leftrightarrow  &4^{i_{1}\minus{}i_{2}}&\equal{}\left(1\minus{}5\right)^{j_{2}\minus{}j_{1}}\left(\minus{}1\minus{}\sqrt{5}\right)^{j_{1}\minus{}k_{1}\minus{}(j_{2}\minus{}k_{2})}\\
\Leftrightarrow  &\left(\minus{}1\right)^{j_{1}\minus{}j_{2}}4^{i_{1}\plus{}j_{1}\minus{}(i_{2}\plus{}j_{2})}&\equal{}\left(\minus{}1\minus{}\sqrt{5}\right)^{j_{1}\minus{}k_{1}\minus{}(j_{2}\minus{}k_{2})}\\
\Leftrightarrow & \left(\minus{}1\right)^{j_{1}\minus{}j_{2}}4^{k_{2}\minus{}k_{1}}&\equal{}\left(\minus{}1\minus{}\sqrt{5}\right)^{j_{1}\minus{}k_{1}\minus{}(j_{2}\minus{}k_{2})}
\end{array}$

Now one can argue that the left hand side of the last equation always is rational for integer exponents while the right hand side always is irrational except when the exponent is zero (I am not giving a proof to this statement though) which forces the RHS to be equal to 1. From the LHS we therefore get $ k_1\equal{}k_2$ which, if put into the condition $ j_1\minus{}k_1\equal{}j_2\minus{}k_2$, shows that $ j_1\equal{}j_2$, but then $ i_1\equal{}i_2$ as well and we are done. The answer is thus $ \binom{n\plus{}2}{2}$, the one Kent Merryfield anticipated.


I know this problem is posted before but as I know ,noone posted solution.
Prove that for all $ n\geq 3$ and $ n\in Z_{\plus{}}$ we have $ S(9^{n})>9$ .where $ S(n)$ denote sum of all digits of $ n$.

If n is odd last digit is 9, therefore (because $ 9^{n}>9$) $ S(9^{n})>9$. If n=2k, then $ S(9^{n})\equal{}S(81^{k})$.
$ 81^{k}\equal{}1\plus{}10*k*8\plus{}100*\frac{k(k\minus{}1)}{2}64\plus{}...$
By considering many variants of k we can prove, that $ S(81^{k})>9$ for k>1.


[quote="arqady"][quote="teanleen"]Ask for help![/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=57683[/quote]Thank you!
But ,I don't think it is very nice!


limit
LaTeX
calculus
calculus computations

[color=darkblue]Let $ f(n) \equal{} (n^2 \plus{} n \plus{} 1)^2 \plus{} 1$, with $ n \in N$ and $ n\ge 1$.

Let $ x_n \equal{} \frac {f(1).f(3).f(5)...f(2n \minus{} 1)}{f(2).f(4).f(6)...f(2n)}$, with $ n \in N$ and $ n\ge 1$.

Let $ u_n \equal{} n^2.x_n$

Find $ \lim_{n \to \plus{} \infty}(u_n)$[/color]

$ \text{\LaTeX}$ note: use \cdot for the multiplication dot. $ f(1)\cdot f(2)$ looks much better than $ f(1).f(2)$. Also, \dots is a better spaced ellipsis, and \cdots is a vertically centered one.


geometry
area of a triangle
Heron\u0027s formula

In triangle $ABC$, find $\frac{\cot{\frac{A}{2}}}{s-a}$ where $s$ is the semi-perimeter. (you can answer in side lengths, etc.)

We have

\[\cot \frac{A}{2}+\cot \frac{B}{2}=\frac{c}{r}\]

\[\cot \frac{B}{2}+\cot \frac{C}{2}=\frac{a}{r}\]

\[\cot \frac{C}{2}+\cot \frac{A}{2}=\frac{b}{r}\]

so, we get $\cot \frac{A}{2}=\frac{b+c-a}{2r}$,therefore using Heron's formula

$\frac{\cot \frac{A}{2}}{s-a}=\frac{1}{r}=\frac{1}{2}\sqrt{\frac{(a+b-c)(b+c-a)(c+a-b)}{a+b+c}}$

I believe you can get it from drawing a triangle, look at $tan {\frac{A}{2}}$ and then you'll get the relation.


(*) $P(x^{2}+1)=P(x)^{2}-1$
Deriving (*) we get
(**) $xP'(x^{2}+1)=P(x)P'(x)$.

Take $g(x)=x^{2}+1$ and for all $n\geq 0$
$g_{n}(x)=\underbrace{g(g(\cdots g(x)))}_{n-{\text{times}}}$.

Then $0<g(0)<g_{2}(0)<g_{3}(0)<\cdots$, as $g(\lambda)-\lambda=(\lambda-{1\over 2})^{2}+{3\over 4}>0$ for all $\lambda\in\mathbb{R}$,
and for all $\lambda\in\mathbb{C}$ we have
$P(\lambda)=0\ \ \Rightarrow P(g(\lambda))=P(\lambda)^{2}-1=-1$ and
$P(\lambda)=-1\Rightarrow P(g(\lambda))=P(\lambda)^{2}-1=0$.

(1) $\gcd (P(x),g_{n}(x))=1$ for all $n\geq 0$:
Otherwise $P(x)$ and $g_{n}(x)$ have a common root $\lambda\in\mathbb{C}$ for some $n\geq 0$.
But then $P(\lambda)=0$ and $P(0)=P(g_{n}(\lambda))\in\{0,-1\}$.
So $P(0)=P(g_{2}(0))=P(g_{4}(0))=\cdots$ and the polynomial $P(x)-P(0)$ has an infinite number of roots.
This forces $P\equiv P(0)\in\{0,-1\}$, which contradicts (*).

(2) $g_{n}(x)\ |\ P'(x)$ for all $n\geq 0$:
We have $g_{0}(x)=x\ |\ xP'(g(x))=P(x)P'(x)$ and so $g_{0}(x)\ |\ P'(x)$ by (1)
and from $g_{n}(x)\ |\ P'(x)$ follows $g_{n+1}(x)=g_{n}(g(x))\ |\ xP'(g(x))=P(x)P'(x)$
and $g_{n+1}(x)\ |\ P'(x)$ for all $n\geq 0$ again using (1).

Now $\deg g_{n}=2^{n}\mathop{\longrightarrow}_{n\rightarrow\infty}\infty$, so $P'\equiv 0$ and $P\equiv c$ constant. Then (*) gives $c=c^{2}-1$.
So (*) has $2$ solutions $P\equiv{1+\sqrt{5}\over 2}$ and $P\equiv{1-\sqrt{5}\over 2}$.


If $ x$ is positive and $ x^2 \equal{} 729$, what is the value of $ x$?

The equation $ x^2\equal{}729$ has solutions $ x\equal{}\pm 27$.  They tell you $ x$ is positive, so $ x\equal{}\boxed{27}$

[quote="JSGandora"][quote="Kouichi Nakagawa"]$ 7+2+9=18=9 \times 2 \ \text{i.e.} \ 9 \mid 729$.
So,
$ \begin{eqnarray*}729 & = & 9\times 81 \\ & = & 9 \times 9 \times 9 \\ & = & 9^3 \\ & = & \left(3^2\right)^2 \\ & = & 3^{2 \times 3} \\ & = & 3^{3 \times 2} \\ & = & \left(3^3\right)^2 \\ & = & 27^2.$[/quote]
@Kouichi Nakagawa, for the fourth line in the equation, it should be $=\left(3^2\right)^3$, not $=\left(3^2\right)^2$.[/quote]

Yes that's right.
I've made a typo.


Solve the following, given that y(1) = 1

x(-5y^2 + 9x^2)y' = 2(y^2)(3x-5y)

Hello there. This is a first order homogeneous ODE that can be solved using the substitution, y = vx.


geometry
circumcircle
ratio
geometry unsolved

please post solutions to----

http://www.mathlinks.ro/viewtopic.php?t=238434

Mmm, It 's easy. You can figure it out for your case. In case, you need the solution. Click on the hidden message.

[hide="Solution"]
$ ND$ is the diameter of the circumcircle of triangle $ BFD$ and $ \triangle up BFD$ is the image of triangle $ ABC$ under the scale of half. (or $ BFD$ is similar to $ ABC$ with ratio $ \frac{1}{2}$).
[/hide]


number theory
least common multiple
prime factorization

This should set an example; if you want harder problems look somewhere else.

1. Two friends are playing tag. The first runs at 20 feet per minute and the other at 25 feet per minute. If the first friend has a minute head start how long will it take the second friend to catch up?

2. There are two brothers. Right now one is three times as old as the other. In three years he will only be twice as old. What are their ages now?

I hope these problems show an appropriate level of difficulty for Getting Started problems (they're easy).  ;)

Edited to change speed (but it doesn't change the answer to the first problem) and to correct spelling.

These problems look good to me.  Although 20 feet per hour is kind of slow  :D .

1.

[hide]First guy has a 20 ft head start, since he has a minute's head start.  After one minute, the first guy will have traveled 40 ft, and the 2nd guy will have traveled 25 ft.  So if we keep on going and listing numbers, we find that once 4 minutes are over, they will both have traveled 100 ft.

Edited, I found a nicer way using the least common multiple of 20 and 25.  I think there's a way to do this problem using the distance formula, but oh well.

20's prime factorization is 5*2 2 , and 25's is 5 2 .  So the LCM of these two numbers needs to have a 5 2  and a 2 2 .  Multiplying 25 and 4 out we get 100, our answer from before.  Now, we just need to figure out how long it'll take.  We can use the distance formula I think.  Distance=100, so 100=25x, x=4.  If you wanted to do it for the first guy, his distance is 80 feet, so 80=20x, and both ways we get 4.[hide][/hide]

2.

[hide] Let a be the older brother and b be the younger brother.  Thus, a = 3b.  In 3 years a will be twice as old as b.  So a + 3 = 3 (b + 3), which gives a + 3 = 3b + 6.  Using substitution, 3b + 3 = 3b + 6, so b = 3, and a = 9.[/hide][/hide]

Let's try something. If at time x the faster brother catches up to the slower brother, what's the distance that the faster brother has covered? What's the distance that the slower brother has covered? Make this into an equation and solve for x.

[quote="mikewd"]Random question but are American children taught feet and inches in school?![/quote]

Yes, as my fellow moderator explained, United States schools still mostly teach English customary units. Interestingly, the metric system is the only system of measurement ever OFFICIALLY acknowledged by Congress (back in the 1860s), but the force of custom is such that most Americans are still much more familiar with inches than with centimeters, with miles than kilometers, and so on.

[quote="joml88"]Yeah, America mostly uses feet, inches and the like.  I personally believe that we should switch to the metric system(and most other ppl who know science will agree).  We have to spend 1-2 weeks in science class just to learn how to convert from the English system to the metric...all you need to know for the metric system are the prefixes!  We Americans are too stubborn.[/quote]

English system.. how ironic, we use metric in the UK (well in school at least, my parents don't)

I tried what you suggested, Osiris, and this is what I got:

[hide]x is the time it takes for 2nd guy to catch up with 1st guy, so 25x = 20 + 20x, since the 1st guy had a 20 ft head start.  Solving for x gives x = 4.  Wow that was a lot easier.[/hide]

[quote="Duck-Billed Platypus"]Just cause I'm bored... #1 is sort of ambiguous (is that the right word for this?) because they could make turns, etc..

#2: Care to clarify?
[quote]In three years he will only be twice as old.[/quote]
Twice as old as he was, or the older brother was, or the older brother will be?[/quote]

In three years the older brother will be twice as old as the younger brother.

#2: [hide]Let first brother's age be A, second brother's age be B.

One is 3 times as old as the other, so A=3B.

In 3 years he will be twice as old as the other, so A+3=2(B+3).

Substitute.. 3B+3=2B+6

B=3

Their ages now are 9 and 3.[/hide]

1.[hide]

The second friend catches up to the first friend 5 feet / minute, since 25-20 is 5.  Since the first friend has a minute, or 20 feet head start it will take 20/5, or 4 minutes for the second friend to catch up.[/hide]

2.[hide]

a will represent the first brother's age right now, and b will represent the second brother's age right now.

3a = b

2(a+3) = b+3

2a+3 = b

6a+9 = 3b

6a+0= 2b

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

0a+9=b

So the older brother is 9 years old.  We can substitue that into 3a = b.

3a = 9

a = 3

So the younger brother is 3 years old.

[/hide]

1.[hide]20x + 20 = 25x

20 = 5x

4=x

the + 20 is the fact that 20 ft/min* 1 min = 20 ft, thus the y-interecept at 0 min mark of when the faster of the two starts to run.[/hide]

2.[hide]2 brothers x and

x = 3y

x + 3 = 2(3 + y)

x + 3 = 6 + 2y

3y + 3= 6 + 2y

y = 3

x = 3 (3) = 9

The older brother (x) is 9 years of age, while the younger brother (y) is 3 years of age.

[/hide]

1. [hide] let x = minutes they have been running for. Since the first boy (20ft per minute) has a minute head start, he is 20 feet ahead.  The first boy's distance is represented by 20x, and the second boy's distance is represented by 25x. So I set up an equation:  20 + 20x = 25x. After doing some algebra, x comes out to be 4, so therefore the answer is 4 minutes.[/hide]

2. [hide] I made the younger one's age x, and the older one's 3x (seeing that he is 3 times the age of the younger one. Three years later, the older one's age will be 3x+3. The problem says at this time, the older one will only be twice the age of the younger one. I set up this equation: 3x+3 = 2(3+x) , or 3x+3=(6+2x). x comes out to be 3, so if you plug that in for the two original ages, you'll get 3 and 9.[/hide]

Thanks for these problems. The other ones in this forum are too hard for me at this point, but these are easy (maybe even a bit too easy).


geometry
parallelogram
geometry unsolved

[quote="bbgun34"]Could someone give a solid/rigorous proof for the following:

ABCD is a convex quadrilateral. The midpoints of AB, BC, CD, DA are E, F, G, H. Segments AF, BG, CH, DE are drawn. The points at which these segments intersect are W, X, Y, Z. Then the midpoints of WX, XY, YZ, ZW form a parallelogram.

[geogebra]c2f4825bba5facac77e0148a3d467d5d29c0a8f7[/geogebra][/quote]
If $ \square ABCD$ is a convex quadrialteral then the midpoints of $ AB,BC,CD,DA$ form a parallelogram, so there is probably something wrong in the statement as one is given much more information than needed.


Let  $a_k\geq0$ for all $k=1,2,...,n$.Prove the following inequality:

               $ e^(a_1+a_2+...+a_n-n)\geq \frac^n {a_1+a_2+...+a_n}{n}$ ;)

Please, learn Latex. I learned quickly,although I know the first notions from you, for which I thank you !

Oh,levi I'm so lazy.But in the end I must learn it.Anyway,thanks for the advice. ;)


MATHCOUNTS
Pythagorean Theorem
geometry
exterior angle

anyone who has intersting geometric proofs, could ou please post them here?
for example, the pythagorean theorm.

If possible, please post a problem using that proof in your post. only if possible. Thanks.

For proofs of the Pythagorean Theorem, check [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275754]this[/url] thread.

IDK how intense you are with geometric proofs. Heres a really really basic one:

"Show that he shortest distance from a line AB to a point C not on AB is the segment CD, where line CD is perpendicular to line AB and point D lies on line AB."

Thats not very interesting, but I came up with it (yeah, obviously someone noticed it waaaaayyyyy before me and its sorta obvious, but I hadnt heard of any theorem like that before, even though im sure there is one out there)

are you talking about like, MC proofs, if higher move it there mods.

One such example:-

Prove that the sum of the 3 exterior angles of a triangle is equal to 360 degrees.

To prove arbalas exterior angle prove

let say we have a triangle with angle measures a b and c.

We know that the exterior angle is equal two the sum of the two interior angles so

exterior a=b+c
exterior b=a+c
exterior c=a+b

The sum is 2a+2b+2c and since a+b+c=180 the sum of the exterior angles ina triangle is 360 degrees

random proof that if two lines have consecutive interior angles(the ones created by the transversal) that add up to 180, then the lines are parallel: Contradiction- Draw the lines such that they intersect. This creates a triangle with a zero degree angle, which is impossible so we are done.

you guys can also post formulas also and theorms.

here's one of the most useful formulas I know:

The Pythagorean Theorem states that in a right triangle, where a and b are its legs and c is its hypotenuse, a^2+b^2=c^2

lol. So basic yet so helpful. :)


Given a circle $ k$ and a point $ P$ outside it, an arbitrary line $ s$ passing through $ P$ intersects $ k$ at the points $ A,B$. Let $ M,N$ be the midpoints of the arcs determined by the points $ A,B$. Let $ C$ be the point on $ AB$ such that $ PC^2\equal{}PA\cdot PB$. Prove that $ \angle MCN$ doesn't depend on the choice of $ s$.

[img]http://i006.radikal.ru/0804/04/4c243cc021f5.jpg[/img]
Let $ K,L \in k$, s.t. $ PK$ and $ PL$ are tangent to $ k$. Then we have $ PK^2 \equal{} PL^2 \equal{} PA\cdot PB \equal{} PC^2$, so $ PK \equal{} PC \equal{} PL$, so $ C$ always lie on the circle $ t$ with radius $ PK$. Let $ \angle{CKL} \equal{} \alpha$, then $ \angle{BPL} \equal{} 2\alpha$. But $ 2\alpha \equal{} \angle{BPL} \equal{} \dfrac{\breve{BL} \minus{} \breve{AL}}{2} \equal{} \dfrac{\breve{BM} \plus{} \breve{ML} \minus{} \breve{AL}}{2} \equal{} \dfrac{\breve{LA} \plus{} 2\breve{ML} \minus{} \breve{AL}}{2} \equal{} \breve{ML}$
so $ \angle{MKL} \equal{} \dfrac{\breve{ML}}{2} \equal{} \alpha$ and $ \angle{CKM} \equal{} \alpha \minus{} \alpha \equal{} 0$, so $ K,C,M$ are collinear.
 Let $ \angle{CLK} \equal{} \beta$, then $ \angle{BPK} \equal{} 2\beta$. But $ 2\beta \equal{} \angle{BPK} \equal{} \dfrac{\breve{BK} \minus{} \breve{AK}}{2} \equal{} \dfrac{\breve{BN} \plus{} \breve{KN} \minus{} \breve{AK}}{2} \equal{} \dfrac{\breve{AK} \plus{} 2\breve{NK} \minus{} \breve{AK}}{2} \equal{} \breve{NK}$
so $ \angle{NLK} \equal{} \dfrac{\breve{NK}}{2} \equal{} \beta$ and $ \angle{CLN} \equal{} \beta \minus{} \beta \equal{} 0$, so $ N,C,L$ are collinear.

Because $ P$ and $ k$ are constant, $ \angle{KPL} \equal{} const$. But $ \angle{MCN} \equal{} \angle{KCL} \equal{} 180^\circ \minus{} \dfrac{\angle{KPL}}{2} \equal{} const.$, q.e.d.

Thank [color=darkblue][b]Ahiles[/b][/color] for nice solution.


ARML algebra polynomial geometry circumcircle LaTeX inequalities

Post ARML problems/solutions to previous ARML problems.  To make the difficulty more varried, I suggest we post easier problems (numbers 1-3) first, then moderate difficulty (4,5), then the harder problems (6-8).  The proccess starts over every 10, so just look at the units digit to determine the difficulty of the problem.

[color=blue][size=200]PROBLEM 1[/size][/color]
Source:  2000 problem 1

For $1<x<y$, let $S=\{1,x,y,x+y\}$.  Compute the absolute value of the difference between the mean and median of $S$.

[quote="not_trig"]Is this a bit trivial? :D

[hide="Problem 1 Solution"]

Mean = $1+x+y+(x+y)=2(x+y)+1$
Median = $\frac{x+y}{2}$

|Mean-Mode|=$\frac{3}{2}(x+y)+1$

 :maybe:
Is this a trick question...  :wink:
[/hide]

[color=red]You don't need quotations for hide titles.
b-flat[/color][/quote]

Generally for mean, you divide by the number of terms  :wink:
No, I checked my book, and it is exactly the same.

the answer is $1/4$, it's self explanatory

uh, I don't really have ARML problems...but I got similar problems I guess or whatever...

so here's like a #1
[size=150]
Problem 2:[/size]

Find the sum of the prime factors of the number which is four more than the square of $625$.

[quote="NeverOddOrEven"][hide="Problem 2 Solution"]$625^{2}+4=390,629$, which i think is prime so our answer is $390,629$
not sure if i'm right though[/hide][/quote]

390,629 is not prime (I used my calculator here - it is $577\times677$, making the answer $1254$).  The problem is asking to factor the sum of squares, but I don't know an easy way to do that.

$=(n^{2}-2n+2)(n^{2}+2n+2)$. When you get simple polynomials like that that don't have real roots, look for difference of squares.

I don't know who's supposed to post, but I have to leave, so... somebody else do it.

ohh, that is cool (I can't press a key mutliple times quickly).

[size=150][color=blue]Problem 3[/color][/size]
$\triangle ABC$ is inscribed in circle $O$, the radius of $O$ is 12 and $m\angle ABC=30^\circ$.  A circle with center $B$ is drawn tangent to the line containing $\overline{AC}$.  Let $R$ be the region which is within $\triangle ABC$, but outside circle $B$.  Compute the maximum area of $R$.

This was a 1997, problem 2.

I have solved it after remembering how to divide right :wink:

[hide="solution"]

Let the radius of circle $B$, which is also the height from $AC$ of $\triangle ABC$, be $h$.

The area of the sector of circle $B$ inside $\triangle ABC$ is a $30^\circ$ sector, so it has area $(\frac{1}{12})(\pi h^{2})$.

$\triangle ABC$, on the other hand, has area $(\frac{1}{2})(12)(h)$, since $AC$ = 12.

$AC$ is 12 because $\angle AOC$ (let's call O the circumcenter of $\triangle ABC$) is 60, and $\triangle AOC$ is isoceles, which is obvious.

So, optimizing $6h-\frac{\pi}{12}h^{2}$, we get $h=\frac{36}{\pi}$, and

$R=\frac{108}{\pi}$.

[/hide]

I don't have a problem -- someone else post one. (assuming I'm right...)

[quote="not_trig"]
I don't have a problem -- someone else post one. (assuming I'm right...)[/quote]
Hey I'm new to NC. I just moved here a week ago to start the tenth grade!

If $a^{3}= b^{2}$ and $c^{3}= d^{2}$, where $a, b, c,$ and $d$ are positive integers; and if $c-a = 25$, what are $a, b, c,$ and $d$?

Congratulations on moving to the ARML national champions' state!

(thank MysticTerminator)

Your LaTeX is pretty good.

BTW, in problem solving marathons, please post the problem number.

[hide="solution to number 4"]

Since $a^{3}=b^{2}$, $\sqrt{a^{3}}$ is an integer, $\sqrt{a}$ must be an integer. Let's call $\sqrt{a}$ $j$.  Then, $b=j^{3}$.

Similarly, let $\sqrt{c}$ be $k$, so $c=k^{2}$ and $d=k^{3}$.

Since $a, c > 0$, $j, k > 0$.

Therefore, $c-a = j^{2}-k^{2}= (j+k)(j-k) = 25 \cdot 1$.

(Since j, k are positive, 5*5 wouldn't work since k would be 0)

Therefore, $j=13$, $k=12$.

So, \[a=169, b=13^{3}, c=144, d=1728\] (too lazy to calculate $13^{3}$...)  :lol:

[/hide]

Again, I have no problems to give  :(

Maybe I should check my ARML book....

[quote="Imperial Effect"]I feel like such a fool… The problem above was not an ARML problem (sorry  :blush: ).
And I have never taken any national or state math tests. Therefore, I did not make it to the ARML nationals.  :noo:

Once again I am sorry for my ignorance and any deceptiveness I may have portrayed on my part. I am just a teenage who likes math.[/quote]

er what it is all ok this is just a forum for like hanging out and stuff

For the people here who aren't familiar with Jensen's (or, for that matter, convexity), QM-HM is probably a more illustrative proof. Yes, I know it is a special case of Jensen's...

[hide=":weightlift:"]$\sqrt{\frac{x^{2}+y^{2}+z^{2}}3}\ge \frac 3{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=3$, from which the result follows immediately.[/hide]
You got to it first, Jeremy, so I grant you the burden of posting the next problem. ARML marathon, right? :roll:

Ok, while we are on the subject:

Let $a,b,c>0$ be such that $a+b+c=1$

Prove that $a^{b}\cdot b^{c}\cdot c^{a}\le ab+bc+ca$

For those who want a more ARML-style problem, compute the sum (from k=1 to infinity) of

$\frac{F_{k}}{2^{k}}$

where $F_{1}=1$, $F_{2}=1$, and the $F_{k}$ are the Fibonacci numbers

[quote="n^4+4"]Ok, while we are on the subject:

Let $a,b,c>0$ be such that $a+b+c=1$

Prove that $a^{b}\cdot b^{c}\cdot c^{a}\le ab+bc+ca$[/quote]
[hide=" :rotfl: "]It's AM-GM on $a,b,c$ with weights $b,c,a$.[/hide]
I think your second problem is harder :P

[quote="not_trig"]*ahem* any idea? :oops:

anyway I have an easy inequality...

[b]Problem 22[/b]

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1 \Rightarrow x^{2}+y^{2}+z^{2}\ge 27$[/quote]

well my solution used AM-GM and GM-HM:

$\sqrt[3]{xyz}\ge\frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}=3$; $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\Leftrightarrow xyz=xy+yz+xz \Rightarrow xy+yz+yz\ge27$

now $x^{2}+y^{2}+z^{2}\ge xy+yz+xz = 27$

EDIT: BERMAN YOU FOXY PERSON YOU KEEP USING SMILEYS AS HIDE TITLES

@boy soprano: out of curiosity, why did you (falsely) change the inequality to be strict?

Actually, I think the second problem is easier... I guess I'm just weird like that. Anyhow, it's pretty cool, so it deserves more than the workout smiley.

[hide=":starwars:"]S=1/2+1/4+2/8+3/16+5/32+...
[u]2S=1+1/2+2/4+3/8+5/16+8/32+...[/u]
S+2S=1+2/2+3/4+5/8+8/16+13/32+...
But just by quadrupling the original:
4S=3+2/2+3/4+5/8+8/16+13/32+...
Subtracting, S=2.[/hide]

That problem was really easy to cheat on (i.e. just check every one of the 5 answer choices they gave us)

As it is I see nothing better than working through about 10 cases

[quote="jb05"]@boy soprano: out of curiosity, why did you (falsely) change the inequality to be strict?[/quote]
What do you mean?

$\frac{1}{3}^{1/3}\frac{1}{3}^{1/3}\frac{1}{3}^{1/3}= \frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}= \frac{1}{3}$?

[quote="jb05"]When you quoted it, you changed <= to <.

This new problem looks disgusting. Who's up for trying all 12 cases? :wink:[/quote]
Oh, no, I think it just looks like that because the bottom of the quote box makes it hard to see the equality line.  Compare:
[quote]$a^{b}\cdot b^{c}\cdot c^{a}\le ab+bc+ca$[/quote]
[quote]
$a^{b}\cdot b^{c}\cdot c^{a}\le ab+bc+ca$

[/quote]

(And in my previous post, I forgot what "strict" meant :  I thought strict => sharp => equality included  :roll: )

The second problem is just plugging 1/2 into the generating function, is it not? So it would be $\frac{1/2}{1-1/2-1/3}= 2$

[quote="b-flat"][quote="not_trig"]erm i have another problem:

Find $a-b$ if $a^{2}-b^{2}=45$ adn $a^{3}-b^{3}=945$, $a, b \in \mathbb{Z}$.

Sure, this was #2 on our comprehensive regional, but even Vivek skipped it...[/quote]

[hide]The key part that makes it easy is the $a,b\in\mathbb{Z}$.

$a^{2}-b^{2}=45$

$(a+b)(a-b)=3^{2}\cdot5$

There are only 3 pairs of solutions $(a^{2},b^{2})$ that solve this; $(23^{2},22^{2})$, $(9^{2},6^{2})$, and $(7^{2},2^{2})$.

$(23^{2},22^{2})$ can be thrown out because $23^{3}-22^{3}$ is way greater than 945.

$(9^{2},6^{2})$ produces $(\pm9,\pm6)$, which could only possibly be true if $9^{3}+6^{3}=945$, and it does.

Thus, $a-b=3$.[/hide][/quote]

lol i think "use calculator" would be a slightly easier solution, but this is good too (kinda lucky that $9^{3}+6^{3}=945$)


ARGH, I have spent ages trying to prove this.

ab + bc + ca ≤  $a^2  + b ^2  + c^2$

for all real non-zero numbers a,b,c

I just cant do it, im missing something simple.

Your help is appreciated.

Thanks

Have you heard of the rearrangement inequality?

Well first assume WLOG that $a \ge b \ge c$.  Then by rearrangement, this is trivial.

Actually, an easier proof for you would be to multiply 2 on both sides, and then subract the $2ab+2bc+2ac$ terms over.  Factor into squares, and then you're done.

$(a-b)^2 + (b-c)^2 + (c-a)^2 \geq 0$
But you should learn "rearrangement inequality" too. It's usefull in BW  :)

WLOG, assume $a \geq b \geq c$, then: $a \cdot a + b \cdot b + c \cdot c \geq a\cdot b + b \cdot c + c \cdot a$.

BTW, what is BW?

Probably the Baltic Way competition? Am I right?

Joining in flogging  ...
if  (x,y,z)  are absolute values  of (a,b,c), and thus non-negative, by AM-GM
we get  $x^3+y^3+z^3 -  3 xyz  \ge 0$
or $(x+y+z)(x^2+y^2+z^2 -xy -yz-zx) \ge 0$
since  $x+y+z \ge 0$ we have  $a^2+b^2+c^2  \ge  xy+yz+zx$ but $xy \ge ab $ etc.. so you get what you want...


geometry inequalities AMC AIME blogs USA(J)MO USAMO \/closed

I heard that the Olympiad classes (not WOOT, of course):

- Olympiad Problem Solving
- Olympiad Geometry
- Olympiad Inequalities

are going to be cancelled. Is this true?  :noo:  :(

Of course, I may be wrong, and this might just be a whole rumour that's spreading around. If so, I completely apologize and someone can just delete this post. But please say whether the Olympiad classes are going to be continued or not, because I can't afford to pay for WOOT.

According to the classes information under "Online School," Olympiad Problem Solving will occur during Fall 2006 and Olympiad Geometry and Inequalities will occur during Winter 2006/7.

We have decided to discontinue the Olympiad classes in order to be able to spend more time on other projects.  We strongly encourage students interested in the Olympiad classes to participate in WOOT.  In order to facilitate this, we have added any early-bird discount on the WOOT price.  Students who sign up by July 15 will receive a 100 dollar discount, as now reflected on the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Enrollment page[/url].

WOOT! I think I will probably sign up for WOOT now! I just have to convince my parents, lol :)

So are the Olympiad classes the only ones being discontinued (of course along with the independent study)?  Will there be any options besides WOOT for those who were planning to take an olympiad class?  I guess those new books are coming along then.

Does this mean that the classes already planned for this year will be cancelled? (EDIT: Nvm, I see that they've been taken off the classes page.)

Also, I won't have the time or money next school year to do WOOT, but I would have enough for olympiad classes, does anyone have any recommendations as to what I should do? I'm considering taking another AIME problem series, doing an intermediate class, though for some of those classes there would be a lot of review (especially seeing as I plan to go through AoPS 2), possibly trying WOOT anyway, or even not taking an AoPS classes.

While I understand the concerns of those of you who would like to take the old Olympiad classes, I ask that you understand our position as well.  We've been heavily subsidizing the school for a very long time (and will still be subsidizing WOOT).  In order to continue to be able to produce any resources at all, we have to focus more of our time on properly priced resources, instead of spending 50-60% of our time on resources we give away either for free or for a very, very low price.  I've posted some of my thoughts about this in my blog, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=6098]here[/url].

Thankyou for your response on pricing.

However, assuming that I can afford it, I'm also worried about time, seeing as I'll have a particularly busy sophmore (next) year (I have been planning to do WOOT in junior year). How much time does one have to spend on WOOT? Could I still get as much from WOOT if I didn't spend as much time as I should on it? Or would it be recommended that I do something lighter next year(thinking of AIME Problem Series B as well as A, which I'm doing over the summer)?

[quote="calc rulz"]Thankyou for your response on pricing.

However, assuming that I can afford it, I'm also worried about time, seeing as I'll have a particularly busy sophmore year. How much time does one have to spend on WOOT? Could I still get as much from WOOT if I didn't spend as much time as I should on it? Or would it be recommended that I do something lighter next year(thinking of AIME Problem Series B as well as A, which I'm doing over the summer)?[/quote]

I think the answer to this depends on how well you can budget your time.  WOOT is somewhat more spread out than our other classes.  As I remember from your earlier postings in the AMC area, you are already pretty proficient with the AIME.  Due to the USAMO expansion, you'll also have some more breathing room.  I'd suggest you start spending more of your time with the more challenging material.  In other words, I'd recommend *not* taking the AIME PS B and focusing more of your efforts on Olympiad material.  If you dive into AIME PS A over the summer, and spend some time with books like Zeitz and such, you should put yourself in a good position with respect to the AIME in 2007.  (I'm assuming you're on your school's math team and will have plenty of opportunity to stay sharp for the AMC.)

One of our goals in expanding the USAMO was to get more students to spend some time working on olympiad-style problems.  Far too many would spend far too much time cramming away for the AIME.  I would guess that at least half the students who barely qualified for the AIME (if not more) had never tried an olympiad problem before qualifying for the USAMO.  We'd like to change that, and encourage students to start developing problem solving stamina, writing skills, and the courage to attack problems that appear impossible.

Very understandable.  Okay. Well, one final question that's been bugging me though:  Why was inversion taught in the olympiad geometry class?  To my understanding [url=http://olympiads.win.tue.nl/imo/]inversion isn't tested at the IMO[/url] but [url=http://en.wikipedia.org/wiki/Mathematical_Olympiad_Program]Wikipedia says[/url] MOP poses questions about it.  Is it just for mathematical enrichment, or can it be useful in some problems although it's not required? Thanks.

P.S. The "Naming controvesy" article on wikipedia is quite humorous

Inversion was taught for one of the reasons you surmise - it can be extremely useful, but it's not ever required.  (In fact, this is true of a lot of the advanced techniques in all areas of Olympiad problem solving.)

I also taught it because it's way cool stuff.

Mr. Rusczyk,

Although I understand the cost issue in cancelling the Olympiad classes, it is a loss nevertheless for students.  I have no doubt WOOT is a great course,  but it will not go as deep as the Olympiad subject classes therefore it is not really able to replace the Olympiad classes.  But again I understand you have no choice.

But I have an idea that may benefit both students and AoPS.  Are you willing or able to sell transcritpts of past Olympiad classes (have you nicely archived those materials?)?   Students purchasing those transcripts can study on their own pace.  It may not benefit as much as actually participating in the classes, but it should be next to the best.  I have no idea how many students will be willing to buy, but I will be willing to pay for it.

Thanks,

Kevin Yuan (I'm using my son Allen's account to post this)

Is there a chance that they will come out in early 2007 or late 2006?  :D

[quote="Knuth's_check"]I am very sorry that this is a little bit off topic and if this has been announced before, although what is the date when the Intro to Number Theory book is going to be released? :?[/quote]

Around the 27th of this month.

[quote="rrusczyk"]We will be writing Olympiad books in the coming years, and there will be beginning Olympiad material in the Intermediate texts, the first two of which will be out within a year.  Putting the transcripts in a format (and configuring the ordering system, etc) that would allow selling those is unfortunately too time-consuming to merit what they would produce.  We will be drawing on the highlights of these for WOOT, and I expect the entirety of a few years' WOOT participation will be much broader and deeper than the Olympiad classes were.[/quote]

Then would AoPS mind if we tried to get past transcripts/problems from people who took the courses when they were still offered?

No, we can't encourage that.  We may bring the olympiad classes back some day, and we'll be drawing on some of that material for WOOT.  Moreover, we cannot encourage any sort of file sharing of our online school material, as it jeopardizes the future of the online school (and has been a contributor to our discontinuing the olympiad classes for now).

That's perfectly reasonable, of course. Looks like this generation of problem solvers will have to make do without them  :(

[quote="DPatrick"][quote="horselvr101"]Speaking of books and classes and the like, what are the next books that are going to come out?[/quote]Introduction to Algebra and Intermediate Counting & Probability should come out in early 2007.  Intermediate Number Theory should come out in Spring 2007.[/quote]
What exactly will the intro to algebra include?

Basically all of algebra I, much of the algebraic portion of pre-algebra, and a healthy chunk of algebra II.

Linear equations, quadratics, functions, word problems, proportions, special factorizations, arithmetic and geometric sequences and series, radicals, exponentials, plenty of basic analytic geometry.

[quote="horselvr101"]What about Intermediate Number Theory? I just took the Intro to Number Theory course and I was wondering where I should go next to continue studying that area...[/quote]

It'll be a while before we have an Interm. Number Theory book.  The class will probably be offered again next summer.

Will Interm Algebra come out before or after the class starts?

Are the olympiad classes ever going to come back or are they gone for good?
What is the tentative schedule for books to come out this year?

-bpms

[quote="bpms"]Are the olympiad classes ever going to come back or are they gone for good?
What is the tentative schedule for books to come out this year?
[/quote]

The olympiad classes are closed for the foreseeable future.  Some of the material will be folded into WOOT.

Tentative books schedule: Intro Algebra March, Interm Counting May, Interm Algebra June.


I've come up with the following conjecture but I'm not sure how to prove it. Take a pre-solved Rubik's cube (normal 3x3x3) and perform any number of moves. Prove that for some number n, you can iterate that sequence n times such that the cube will end in a solved position. I've tried this for several sequences and so far I haven't found one that doesn't solve in less than 200 (note the sequences I've been doing are at most 4-5 moves and not very complicated). If anyone has a program simulating a cube I'd appreciate the source. I don't know too much about orbitals so if they're used to approach this problem, please tell me.

one i know off top of head; R U' repeated 64 times; 128 moves.

Let a move be any given sequence of what you normally call a move.

The order of the rubik-group is finite, so you will ever reach the solved position (=neutral element). You can even show (by Lagranges theorem) that the number of moves you did is a divisor of the total number of different positions possible.

To show it more elementary: you can't get a new constellation everytime after doing the move, since that would mean that the cube has $\infty$ many positions, which is clearly not to case. So after some time you get a position you already had before, but what is the difference between that position and a full solved cube after some recoloring¿ Nothing! So you will ever reach the position you started with (solved one) again.

Its not too hard to find a maximal period for your re-iterations. The something about it in the strategy section


find square root of:-

$11+2\sqrt{7}+\sqrt{12}+\sqrt{84}$

farenhajt , in that, how did you get to know that it would be consisiting of 3 terms , 1, sqrt3 and sqrt7 ? i mean if it was a much more complex expression then wouldnt it be hard to get to know the terms or to split them like you did?I mean isnt their some sure-shot way like using variables that you can get the answer?

Well, I think there's no general formula, since the expression must have special properties. In this example it's really obvious that we have $2\sqrt{3}, 2\sqrt{7}$ and $2\sqrt{3}\cdot\sqrt{7}$, and that points to a square-of-a-trinomial formula. Once you know where to go, the rest is easy. And of course, you must have working knowledge of polynomial factorization, in order to recognize the patterns.


not sure where to post this people told me this was one of the good books on glambling and stuff. If anybody has it le me know what you think of it.

i'm not interested in this topic. good bye.

[quote="Nick31"][quote="dreambold22"]gambling???????[/quote]

why not[/quote]

Erm.....  :what?:  :what?:  :what?:

[quote="leoxnlin"]Is it a probability book?  :|[/quote]

probabilities applied to various card games, but I don't have the book so I can't tell exactly


Given quadrilateral $ ABCD$, $ AC\cap BD=O$. Prove that $ ABCD$ is cyclic iff $ OA\sin A+OC\sin C=OB\sin B+OD\sin D$.

http://www.mathlinks.ro/Forum/viewtopic.php?t=26474
See, for the solution,
http://www.mathlinks.ro/Forum/viewtopic.php?t=15584 post #2 file IMO_1997_shortlist.pdf or
http://www.kalva.demon.co.uk/short/soln/sh9723.html

  darij


Let $ x_1,x_2,...,x_n$ be real numbers such that $ x_i > 1$ for $ i \equal{} 1,2,...,n$. Prove that
\[ \frac {x_1x_2}{x_3 \minus{} 1} \plus{} \frac {x_2x_3}{x_4 \minus{} 1} \plus{} ... \plus{} \frac {x_nx_1}{x_2 \minus{} 1} \ge 4n.
\] Determine when the equality occurs.

I am so bored today!
Now, we have
$ \sum_{i \equal{} 1}^n \frac {x_ix_{i \plus{} 1}}{x_{i \plus{} 2} \minus{} 1} \geq 4\frac {x_ix_{i \plus{} 1}}{x_{i \plus{} 2}^2}$
with the notations are $ n \plus{} 1 \equal{} 1$ and $ n \plus{} 2 \equal{} 2$
It is mere to prove that
$ \sum_{i \equal{} 1}^n \frac {x_ix_{i \plus{} 1}}{x_{i \plus{} 2}^2} \geq n$ which is true by the general form of Cauchy in my signature!! :D


inequalities ratio geometry algebra polynomial circumcircle function

Last week, the Nordic Mathematical Contest was held. Here is the problem-set.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206646#p206646]Problem 1 - link[/url]
Find all positive integers $k$ such that the product of the digits of $k$, in decimal notation, equals \[\frac{25}{8}k-211\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206634#p206634]Problem 2 - link[/url]
Let $a,b,c$ be positive real numbers. Prove that \[\frac{2a^2}{b+c} + \frac{2b^2}{c+a} + \frac{2c^2}{a+b} \geq a+b+c\]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206644#p206644]Problem 3 - link[/url]
There are $2005$ young people sitting around a large circular table. Of these, at most $668$ are boys. We say that a girl $G$ has a strong position, if, counting from $G$ in either direction, the number of girls is always strictly larger than the number of boys ($G$ is herself included in the count). Prove that there is always a girl in a strong position.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=206638#p206638]Problem 4 - link[/url]
The circle $\zeta_{1}$ is inside the circle $\zeta_{2}$, and the circles touch each other at $A$. A line through $A$ intersects $\zeta_{1}$ also at $B$, and $\zeta_{2}$ also at $C$. The tangent to $\zeta_{1}$ at $B$ intersects $\zeta_{2}$ at $D$ and $E$. The tangents of $\zeta_{1}$ passing thorugh $C$ touch $\zeta_{2}$ at $F$ and $G$. Prove that $D$, $E$, $F$ and $G$ are concyclic.

Can you say please how much time is allowed ?
P.S. the inequality seems to be very easy  :lol:

4 hours.

Yes, the inequality is very easy!

Does anyone has recent years' NMO problems?

Arccosinus, were you at Baltic Way 2004? I might know you from there  ;)

No portoseb. I wanted to go, but I gave my place in the team to an alternate, because I had to go to a wedding in the U.S.

I saw the results today, at the Finnish IMO selection in which I take part. I was the first with 18 points, so nobody got full points. As soon as the problems will be put on the internet I will let you know.

EDIT: here they are:

http://www.math.helsinki.fi/~smy/olympia/PM/NMC19.htm

19th Nordic Mathematical Contest
Results

			Problem	Sum
			1	2	3	4
1.	Dumitrescu, Sebastian	FIN	5	4	5	4	18
2.	Alanko, Samu	FIN	5	5	5	0	15
　	Pettersson, Ville	FIN	5	5	5	0	15
　	Rennemo, Jørgen Vold	NOR	5	5	5	0	15
　	Stefánsson, Örn	ICE	5	5	5	0	15
6.	Xing, Chen	SWE	4	5	5	0	14
7.	Halldórsson, Höskuldur Pétur	ICE	3	5	3	2	13
　	Troeng, Olof	SWE	5	3	5	0	13
9.	Jacobsen, Martin Wedel	DEN	5	5	0	2	12
　	Liu, Zihan	SWE	5	0	2	5	12
　	Reeh, Sune Nikolaj Precht	DEN	5	5	0	2	12
　	Vass, Marton	FIN	5	0	5	2	12
13.	Edlund, Erik	SWE	5	5	1	0	11
14.	Kolu, Iiris	FIN	5	4	1	0	10
　	Rudberg, Tore	SWE	5	5	0	0	10
　	Sigstad, Henrik	NOR	5	4	1	0	10
　	Sireni, Jukka	FIN	5	5	0	0	10
　	Terelius, Håkan	SWE	5	0	5	0	10
　	Thelin, Sam	SWE	3	5	2	0	10
20.	Holden, Nina	NOR	4	5	0	0	9
　	Johnsen, Hilde Galleberg	NOR	3	5	0	1	9
　	Johnson, Kerstin	SWE	4	5	0	0	9
　	Nors Andersen, Drees	DEN	3	4	0	2	9
24.	Edlund, Björn	SWE	3	0	5	0	8
　	Helgadóttir, Inga Steinunn	ICE	3	3	0	2	8
　	Meuller, Simon	SWE	0	3	5	0	8
　	Vesalainen, Esa	FIN	5	3	0	0	8
　	Årdal, Atle Rygg	NOR	3	5	0	0	8
29.	Abrahamsen, Mikkel	DEN	4	0	2	1	7
　	Grythe, Thomas Berge	NOR	2	5	0	0	7
　	Müller, Jeppe Krarup	DEN	5	2	0	0	7
32.	Blåsten, Eemeli	FIN	0	3	3	0	6
　	Jacobsen, Kåre	DEN	1	5	0	0	6
　	Kokkala, Janne	FIN	1	5	0	0	6
　	Peng, Gunnar	SWE	1	4	0	1	6
　	Siljander, Juhana	FIN	5	0	1	0	6
　	Wilson, Danny	DEN	5	0	0	1	6
38.	Brandt, Asbjørn	DEN	5	0	0	0	5
　	Jensen, Morten Harding	DEN	5	0	0	0	5
　	Lähteenmäki, Meri	FIN	3	0	0	2	5
　	Mäki, Sami	FIN	3	0	2	0	5
　	Poulsen, Morten Kjær	DEN	5	0	0	0	5
　	Selin, Mårten	SWE	1	4	0	0	5
　	Sjamsutdin, Aslanbek	NOR	3	0	1	1	5
　	Sparsø, Eske	DEN	5	0	0	0	5
46.	Eladhari, Matti	SWE	4	0	0	0	4
　	Guðmundsson, Jón Emil	ICE	3	0	0	1	4
　	Johannesen, David	DEN	3	0	0	1	4
　	Laaksonen, Antti	FIN	2	0	0	2	4
　	Maríusdóttir, Þorey María	ICE	3	0	0	1	4
　	Tørholm, Sebastian Paaske	DEN	3	0	0	1	4
　	Vester, Steen	DEN	4	0	0	0	4
53.	Alexanderson, Per	SWE	2	0	0	1	3
　	Egilsdóttir, Salvör	ICE	3	0	0	0	3
　	Ernvall, Toni	FIN	3	0	0	0	3
　	Guðmundsdóttir, María Helga	ICE	3	0	0	0	3
　	Gunnarsson, Einar Bjarki	ICE	2	0	0	1	3
　	Gyring, Peter	NOR	3	0	0	0	3
　	Helgason, Einar Axel	ICE	2	0	0	1	3
　	Jensen, Ole S.	DEN	3	0	0	0	3
　	Nielsen, Emil Bundgaard	DEN	3	0	0	0	3
　	Óskarsdóttir, María	ICE	3	0	0	0	3
　	Sigurðsson, Baldur	ICE	2	0	1	0	3
　	Sigurðsson, Daníel Þröstur	ICE	3	0	0	0	3
　	Sun, Lei	SWE	3	0	0	0	3
　	Yilong, Li	FIN	0	3	0	0	3
67.	Jokinen, Kosti	FIN	0	0	0	2	2
　	Klungre, Vidar	NOR	1	0	1	0	2
　	Li, Kang	DEN	0	0	0	2	2
　	Mertaniemi, Henrikki	FIN	2	0	0	0	2
　	Nielsen, Kræn Vodder	DEN	0	0	0	2	2
　	Nuñez, Walter	SWE	1	0	1	0	2
　	Parnfelt, Johan	SWE	1	0	0	1	2
　	Rydberg, Elin	SWE	2	0	0	0	2
　	Stråhlman, Christian	SWE	2	0	0	0	2
　	Sveinbjörnsson, Benjamin Ragnar	ICE	2	0	0	0	2
　	Svensson, Mats	SWE	2	0	0	0	2
　	Åstrand, Matti	FIN	0	0	0	2	2
79.	Larsson. Joel	SWE	1	0	0	0	1
　	Lode, Pål Sævik	NOR	0	0	1	0	1
　	Marcher, Jette Louise	DEN	0	0	0	1	1
　	Rantanen, Mikko	FIN	1	0	0	0	1
83.	Berg, Justus	SWE	0	0	0	0	0
　	Kettunen, Markus	FIN	0	0	0	0	0
　	Lauridsen, Nis Brix	DEN	0	0	0	0	0

[b]Nordic Mathematical Contest 2009[/b]
April 2, 2009

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028413#p3028413]Problem 1[/url]
A point $P$ is chosen in an arbitrary triangle. Three lines are drawn through $P$ which are parallel to the sides of the triangle. The lines divide the triangle into three smaller triangles and three parallelograms. Let $f$ be the ratio between the total area of the three smaller triangles and the area of the given triangle. Prove that $f\ge\frac{1}{3}$ and determine those points $P$ for which $f =\frac{1}{3}$ .
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028409#p3028409]Problem 2[/url]
On a faded piece of paper it is possible to read the following:
\[(x^2 + x + a)(x^{15}- \cdots ) = x^{17} + x^{13} + x^5 - 90x^4 + x - 90.\]
Some parts have got lost, partly the constant term of the first factor of the left side, partly the majority of the summands of the second factor. It would be possible to restore the polynomial forming the other factor, but we restrict ourselves to asking the following question: What is the value of the constant term $a$? We assume that all polynomials in the statement have only integer coefficients.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028392#p3028392]Problem 3[/url]
The integers $1$, $2$, $3$, $4$, and $5$ are written on a blackboard. It is allowed to wipe out two integers $a$ and $b$ and replace them with $a + b$ and $ab$. Is it possible, by repeating this procedure, to reach a situation where three of the five integers on the blackboard are $2009$?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028405#p3028405]Problem 4[/url]
$32$ competitors participate in a tournament. No two of them are equal and in a one against one mach the better always wins. Show that the gold, silver, and bronze medal winners can be found in $39$ matches.

[b]Nordic Mathematical Contest 2004[/b]
April 1, 2004

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028451#p3028451]Problem 1[/url]
Twenty-seven balls labelled from $1$ to $27$ are distributed in three bowls: red, blue, and yellow. What are the possible values of the number of balls in the red bowl if the average labels in the red, blue and yellow bowl are $15$, $3$, and $18$, respectively?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028439#p3028439]Problem 2[/url]
The Fibonacci sequence is defined by $f_1 = 0, f_2 = 1$, and $f_{n+2} = f_{n+1}+f_n$ for $n\ge 1$. Prove that there is a strictly increasing arithmetic progression whose no term is in the Fibonacci sequence.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028457#p3028457]Problem 3[/url]
Given a finite sequence $x_{1,1}, x_{2,1}, \dots , x_{n,1}$ of integers $(n\ge 2)$, not all equal, define the sequences $x_{1,k}, \dots , x_{n,k}$ by
\[ x_{i,k+1}=\frac{1}{2}(x_{i,k}+x_{i+1,k})\quad\text{where }x_{n+1,k}=x_{1,k}.\]
Show that if $n$ is odd, then not all $x_{j,k}$ are integers. Is this also true for even $n$?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3028454#p3028454]Problem 4[/url]
Let $a, b, c$ be the sides and $R$ be the circumradius of a triangle. Prove that
\[\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\ge\frac{1}{R^2}.\]

[b]Nordic Mathematical Contest 2010[/b]
April 13, 2010

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029723#p3029723]Problem 1[/url]
A function $f : \mathbb{Z}_+ \to \mathbb{Z}_+$, where $\mathbb{Z}_+$ is the set of positive integers, is non-decreasing and satisfies $f(mn) = f(m)f(n)$ for all relatively prime positive integers $m$ and $n$. Prove that $f(8)f(13) \ge (f(10))^2$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029733#p3029733]Problem 2[/url]
Three circles $\Gamma_A$, $\Gamma_B$ and $\Gamma_C$ share a common point of intersection $O$. The other common point of $\Gamma_A$ and $\Gamma_B$ is $C$, that of $\Gamma_A$ and $\Gamma_C$ is $B$, and that of $\Gamma_C$ and $\Gamma_B$ is $A$. The line $AO$ intersects the circle $\Gamma_A$ in the point $X \ne O$. Similarly, the line $BO$ intersects the circle $\Gamma_B$ in the point $Y \ne O$, and the line $CO$ intersects the circle $\Gamma_C$ in the point $Z \ne O$. Show that
\[\frac{|AY |\cdot|BZ|\cdot|CX|}{|AZ|\cdot|BX|\cdot|CY |}= 1.\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029748#p3029748]Problem 3[/url]
Laura has $2010$ lamps connected with $2010$ buttons in front of her. For each button, she wants to know the corresponding lamp. In order to do this, she observes which lamps are lit when Richard presses a selection of buttons. (Not pressing anything is also a possible selection.) Richard always presses the buttons simultaneously, so the lamps are lit simultaneously, too.
a) If Richard chooses the buttons to be pressed, what is the maximum number of different combinations of buttons he can press until Laura can assign the buttons to the lamps correctly?
b) Supposing that Laura will choose the combinations of buttons to be pressed, what is the minimum number of attempts she has to do until she is able to associate the buttons with the lamps in a correct way?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029757#p3029757]Problem 4[/url]
A positive integer is called simple if its ordinary decimal representation consists entirely of zeroes and ones. Find the least positive integer $k$ such that each positive integer $n$ can be written as $n = a_1 \pm a_2 \pm a_3 \pm \cdots \pm a_k$ where $a_1, \dots , a_k$ are simple.

2004, 2009, 2010 added; thanks for posting! :)

[b]Nordic Mathematical Contest 2011[/b]
April 4, 2011

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029937#p3029937]Problem 1[/url]
When $a_0, a_1, \dots , a_{1000}$ denote digits, can the sum of the $1001$-digit numbers $a_0a_1\cdots a_{1000}$ and $a_{1000}a_{999}\cdots a_0$ have odd digits only?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029944#p3029944]Problem 2[/url]
In a triangle $ABC$ assume $AB = AC$, and let $D$ and $E$ be points on the extension of segment $BA$ beyond $A$ and on the segment $BC$, respectively, such that the lines $CD$ and $AE$ are parallel. Prove $CD \ge \frac{4h}{BC}CE$, where $h$ is the height from $A$ in triangle $ABC$. When does equality hold?
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029951#p3029951]Problem 3[/url]
Find all functions $f$ such that
\[f(f(x) + y) = f(x^2-y) + 4yf(x)\]
for all real numbers $x$ and $y$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029970#p3029970]Problem 4[/url]
Show that for any integer $n \ge 2$ the sum of the fractions $\frac{1}{ab}$, where $a$ and $b$ are relatively prime positive integers such that $a < b \le n$ and $a+b > n$, equals $\frac{1}{2}$.
(Integers $a$ and $b$ are called relatively prime if the greatest common divisor of $a$ and $b$ is $1$.)

[b]Nordic Mathematical Contest 2012[/b]
27 March, 2012

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3029993#p3029993]Problem 1[/url]
The real numbers $a, b, c$ are such that $a^2 + b^2 = 2c^2$, and also such that $a \ne b, c \ne -a, c \ne -b$. Show that
\[\frac{(a+b+2c)(2a^2-b^2-c^2)}{(a-b)(a+c)(b+c)}\]
is an integer.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030023#p3030023]Problem 2[/url]
Given a triangle $ABC$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $BC$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $AB$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $AB$ (if there is no second point of intersection, choose $Q = B$). Prove that $AQ = AC$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030035#p3030035]Problem 3[/url]
Find the smallest positive integer $n$, such that there exist $n$ integers $x_1, x_2, \dots , x_n$ (not necessarily different), with $1\le x_k\le n$, $1\le k\le n$, and such that
\[x_1 + x_2 + \cdots + x_n =\frac{n(n + 1)}{2}\quad\text{, and }x_1x_2 \cdots x_n = n!,\]
but $\{x_1, x_2, \dots , x_n\} \ne \{1, 2, \dots , n\}$.
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3030045#p3030045]Problem 4[/url]
The number $1$ is written on the blackboard. After that a sequence of numbers is created as follows: at each step each number $a$ on the blackboard is replaced by the numbers $a - 1$ and $a + 1$; if the number $0$ occurs, it is erased immediately; if a number occurs more than once, all its occurrences are left on the blackboard. Thus the blackboard will show $1$ after $0$ steps; $2$ after $1$ step; $1, 3$ after $2$ steps; $2, 2, 4$ after $3$ steps, and so on. How many numbers will there be on the blackboard after $n$ steps?


calculus integration algebra function domain geometry rectangle

I'm reading about complex analysis and I have some problems translating $\epsilon$s and $\delta$s into a geometric interpretation. I have a clear geometric view on Lebesgue/Riemann-Stieltjes integration of real valued functions. I want to accomplish that with complex integration too. I'll state the theorem

[b]Theorem:[/b] Let $\gamma: [a,b] \to \mathbb{C}$ be of bounded variation and suppose that $f: [a,b] \to \mathbb{C}$ is continuous. Then there is a complex number $I$ such that for every $\epsilon > 0$ there is a $\delta > 0$ such that when $P=\{t_{0}< t_{1}< ... < t_{m}\}$ is a partition of $[a,b]$ with $\parallel P\parallel =max\{(t_{k}-t_{k-1}): 1 \leq k \leq m \}< \delta$ then
\[| I-\sum_{k=1}^{m}f(\tau_{k})[ \gamma (t_{k-1})] | < \epsilon \]
First of all, I hardly believe the author made an awfull typo and he meant to say:
\[{| I-\sum_{k=1}^{m}f(\tau_{k})[ \gamma (t_{k})-\gamma(t_{k-1})}] | < \epsilon \]
He calls $I$ the integral of $f$ with respect to $\gamma$ over $[a,b]$. I don't see the geometric meaning. First of all we are not "integrating" over the domain of $f$, which is kinda weird...

Later on, the author defines the line integral by considering the function $f \circ \gamma$. This makes already some more sense to me. But I still don't feel comfortable. If someone can tell me how to make a drawing to clear things up, that would be most helpful :)

[quote="Jacobson"]
First of all, I believe the author made an awfull typo and he meant to say:
\[{| I-\sum_{k=1}^{m}f(\tau_{k})[ \gamma (t_{k})-\gamma(t_{k-1})}] | < \epsilon \]
[/quote]
You are correct :).
[quote]
He calls $I$ [b]the integral of $f$ with respect to $\gamma$ over $[a,b]$ [/b].
[/quote]
This is the first time I see such a combination of words. Of course, formally it makes perfect sense and allows one to reduce the integration over rectifiable paths to the Stiltjes integration very easily: you just consider your curve $\gamma$ in the same way as you would consider $g$ in $\int_{a}^{b}f\,dg$: you can talk either about Riemann sums or about the (complex-valued) measure generated by $\gamma$, which is sort of convenient, but the object that arises is of hardly any use outside the particular problem of defining the integral of a function $F: \mathbb C\to\mathbb C$ over $\gamma$ (for which you just need to apply this construction to $f=F\circ\gamma$).
[quote]
I don't see the geometric meaning.
[/quote]
Do you see the geometric meaning of the Stiltjes integral?
[quote]
First of all we are not "integrating" over the domain of $f$, which is kinda weird...
[/quote]
Well, this integral should be written $\int_{a}^{b}f\,d\gamma$, so the integration actually is over the domain of $f$ (which is the interval $[a,b]$). Just don't try to think of it as of integral over $\gamma$; however strange the definition may seem, it is put together correctly: it is, indeed, "integral of $f$ over $[a,b]$ with respect to $\gamma$" in the same way as the Stiltjes integral $\int_{a}^{b}f\,dg$ is "integral of $f$ over $[a,b]$ with respect to $g$".
[quote]
Later on, the author defines the line integral by considering the function $f \circ \gamma$. This makes already some more sense to me. But I still don't feel comfortable. If someone can tell me how to make a drawing to clear things up, that would be most helpful :)[/quote]
Note that when you take $f=F\circ\gamma$, you can rephrase the definition of the Riemann sum in the following way. Take the curve $\Gamma\subset \mathbb C$ (the one parameterized by the mapping $\gamma: [a,b]\to\mathbb C$), split it into small pieces by points $z_{0},z_{1},\dots,z_{n}\in \Gamma$ ($z_{k}$ is the same as $\gamma(t_{k})$); choose a point $\zeta_{k}$ ($=\gamma(\tau_{k})$) on each piece of $\Gamma$ between $z_{k-1}$ and $z_{k}$ and consider $\sum_{k}F(\zeta_{k})(z_{k}-z_{k-1})$. In the case, when $\Gamma$ is a line segment, you'll get the usual Riemann sum. The problem with the drawing is that the product of two complex numbers doesn't have a clear "geometric meaning" unlike the product of two real numbers, which can be represented as a (signed) area of a rectangle, but, I hope, you see that the formula for the Riemann sums didn't change compared to the one for the real-valued functions defined on an interval.

Hope, this helps a bit :).


function
absolute value
calculus
calculus computations

1. using the definition of cauchy, how do I show that every cauchy sequence of  real numbers is a bounded sequence?
2. if these are true prove them otherwise provide a counterexample:
a. every unbounded sequence of positive real numbers diverges to [size=150]∞[/size]
b. every unbounded sequence of positive real numbers has a subsequence that diverges to ∞

By the Cauchy property, a Cauchy sequence is eventually bounded; there exists $N$ such that $|x_N-x_n|<1$ for $n>N$. (That specialized the property in two ways). Now, the first $N-1$ terms can't make the sequence unbounded; the first $N-1$ terms have a maximum $M$ and a minimum $C$. We have $\min(x_N-1,C)<x_n<\max(x_N+1,M)$ for all $n$, and the sequence is bounded.

This is an important method; it can be adapted to general metric spaces, and you will probably have reason to use it again.

2a: The answer is no. Try interlacing an unbounded sequence and a bounded sequence.

2b: The answer is yes. You want to construct an unbounded subsequence, so start with an element, find something bigger, find something even bigger,...

1. Show that if a sequence {an} converges to 0, and a sequence {bn} is bounded, then the sequence {an,bn} converges to 0.

2. give an example of a set S in R that has exactly three limit points.

3. Suppose {an} from n=1 to [size=200]∞[/size] is a cauchy sequence and let bn=3an+7 for each positive integer n. Using the definition of cauchy sequence to show that {bn} from n=1 to [size=150]∞[/size].

3 doesn't make any sense

$1:$ Say that the terms of $(b_n)_n$ are bounded above in absolute value by $M$.  Given $\epsilon>0$, pick $N$ large enough that $|a_n|<\frac{\epsilon}{M}$ for $n>N$.  Then $|a_nb_n|\leq \epsilon$ for $n>N$.

$2:$  I hope you know how to construct a sequence with a single limit point (  $(n^{-1})_n$ ).  Now mix this with the sequences $(n^{-1}+1)_n$ and $(n^{-1}+2)_n$.  You are going to have to fiddle around with the indices to make it work, but this isn't hard.

It looks like mikejones left out a few words in #3. I think we can figure out what he meant to say:

Suppose $(a_n)$ is a Cauchy sequence and that $b_n=3a_n+7.$ Prove that $(b_n)$ is a Cauchy sequence.

Here's a generalization: suppose $f: \mathbb{R}\mapsto\mathbb{R}$ satisfies a Lipschitz condition. That is, suppose there is a constant $K$ such that for all $x$ and $y,\,|f(x)-f(y)|\le K|x-y|.$ Suppose $(a_n)$ is a Cauchy sequence and that $b_n=f(a_n).$ Prove that $(b_n)$ is a Cauchy sequence.

(The connection between the specific case and the generalization: the function $f(x)=3x+7$ satisfies a Lipschitz condition with constant $K=3.)$

Or take the generalization one step further out: suppose $f$ is uniformly continuous. Suppose $(a_n)$ is a Cauchy sequence and that $b_n=f(a_n).$ Prove that $(b_n)$ is a Cauchy sequence.

[quote="jmerry"]By the Cauchy property, a Cauchy sequence is eventually bounded; there exists $N$ such that $|x_N-x_n|<1$ for $n>N$. (That specialized the property in two ways). Now, the first $N-1$ terms can't make the sequence unbounded;...
[/quote]
I don't quite see why you can claim the first N-1 can't be unbounded? Can you explain?

[quote="jmerry"]2a: The answer is no. Try interlacing an unbounded sequence and a bounded sequence.
[/quote]
I have my own question for this one. Do we distiguish between [b]divergent to $\infty$[/b] and [b]divergent[/b]?

[quote="Hazel"]I don't quite see why you can claim the first N-1 can't be unbounded? Can you explain?[/quote]

they are between $\min_{i \in \overline{1,N-1}} x_i$ and $\max_{i \in \overline{1,N-1}} x_i$.

For the other question, yes, divergent to $\infty$ means that the sequence has limit $\infty$, while divergent means that the sequence doesn't have finite limit.


function
integration
calculus
derivative
geometry
rectangle
real analysis

Suppose
$u(x,t)$ is continous functions
$\frac{\partial u}{\partial t}$ is continous
$u(x+1,t)=u(x,t), \; \frac{\partial u(x+1,t)}{\partial x}=\frac{\partial u(x,t)}{\partial x}$
$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$

Prove that $E(t)=\frac{1}{2}\int_{0}^{1}(\frac{\partial u(x,t)}{\partial t})^2+(\frac{\partial u(x,t)}{\partial x})^2)dx$ is constant independent of $t$

Here's a formal calculation, assuming we have as many derivatives as we need:

$\frac{dE}{dt}=\int_0^1\frac{\partial u}{\partial t}\cdot \frac{\partial^2u}{\partial t^2} + \frac{\partial u}{\partial x}\cdot\frac{\partial^2u}{\partial t\partial x}\,dx.$

Now integrate the second term by parts:

$\int_0^1\frac{\partial u}{\partial x}\cdot\frac{\partial^2u}{\partial t\partial x}\,dx = \frac{\partial u}{\partial x}\cdot\frac{\partial u}{\partial t}|_0^1 - \int_0^1\frac{\partial^2u}{\partial x^2}\cdot\frac{\partial u}{\partial t}\,dx.$

Now note that the boundary terms in this integration by parts are zero by the periodicity of $u$ and use the fact that $u$ satisfies the wave equation to replace the second derivative with respect to $x$ with the second derivative with respect to $t.$

Hence, $\frac{dE}{dt}=\int_0^1\frac{\partial u}{\partial t}\cdot \frac{\partial^2u}{\partial t^2}-\frac{\partial u}{\partial t}\cdot \frac{\partial^2u}{\partial t^2}\,dx = 0.$

Given the density of $C^{\infty}$ functions, this identity would now hold in the appropriate Sobolev space, with the derivatives interpreted as weak derivatives.

A second calculation, using a similar level of formality. We use the technique of separation of variables to find the general solution of the PDE.

$u(x,t)=\sum_{n=-\infty}^{\infty} e^{2\pi inx}\left(c_ne^{2\pi i nt}+d_ne^{-2\pi int}\right).$

Take the appropriate partial derivatives:

$\frac{\partial u}{\partial x}=\sum_{n=-\infty}^{\infty} 2\pi i ne^{2\pi inx}\left(c_ne^{2\pi i nt}+d_ne^{-2\pi int}\right).$

$\frac{\partial u}{\partial t}=\sum_{n=-\infty}^{\infty} 2\pi i ne^{2\pi inx}\left(c_ne^{2\pi i nt}-d_ne^{-2\pi int}\right).$

By Parseval's identity, if $g(x)=\sum_{n=-\infty}^{\infty}a_ne^{2\pi inx}$ then

$\int_0^1|g(x)|^2\,dx=\sum_{n=-\infty}^{\infty}|a_n|^2.$

Thus,

$\int_0^1\left|\frac{\partial u}{\partial x}\right|^2\,dx = \sum_{n=-\infty}^{\infty} 4\pi^2n^2\left|c_ne^{2\pi i nt}+d_ne^{-2\pi int}\right|^2$ and

$\int_0^1\left|\frac{\partial u}{\partial t}\right|^2\,dx = \sum_{n=-\infty}^{\infty} 4\pi^2n^2\left|c_ne^{2\pi i nt}-d_ne^{-2\pi int}\right|^2$

By direct calculation,

$\left|c_ne^{2\pi i nt}+d_ne^{-2\pi int}\right|^2=|c_n|^2 + c_n\overline{d_n} e^{4\pi int}+\overline{c_n}d_ne^{-4\pi int}+|d_n|^2$

$\left|c_ne^{2\pi i nt}-d_ne^{-2\pi int}\right|^2=|c_n|^2 - c_n\overline{d_n} e^{4\pi int}-\overline{c_n}d_ne^{-4\pi int}+|d_n|^2$

Add these together, and the result is independent of $t$. In the end,

$E(t) = \sum_{n=-\infty}^{\infty}4\pi^2n^2\left(|c_n|^2+|d_n|^2\right).$

Now that I look at it, this one has a Green's theorem solution too:
Write $E(0)-E(T)=\int_{dR}\frac12\left(\left(\frac{\partial u(x,t)}{\partial t}\right)^2+\left(\frac{\partial u(x,t)}{\partial x}\right)^2\right)dx+0\, dt$, where $R$ is the rectangle with vertices $(0,0),(1,0),(1,T),(0,T)$. By Green's theorem, this becomes
$E(0)-E(T)=\int_R\frac{\partial u(x,t)}{\partial t}\cdot\frac{\partial u(x,t)}{\partial t^2}+\frac{\partial u(x,t)}{\partial x}\cdot\frac{\partial u(x,t)}{\partial x\partial t}\, dx\, dt$
$=\int_0^T\int_0^1\frac{\partial u(x,t)}{\partial t}\cdot\frac{\partial u(x,t)}{\partial x^2}+\frac{\partial u(x,t)}{\partial x}\cdot\frac{\partial u(x,t)}{\partial x\partial t}\, dx\, dt$
$=\int_0^T\int_0^1\frac{\partial}{\partial x}\left(\frac{\partial u(x,t)}{\partial x}\cdot\frac{\partial u(x,t)}{\partial t}\right)\, dx\, dt=0$
by periodicity of $\frac{\partial u(x,t)}{\partial x}\cdot\frac{\partial u(x,t)}{\partial t}$ in $x$.

This assumes that the mixed partial derivatives are continuous, so we can set them equal to each other.


[quote="guile"]Consider the sequence below where each third number is missing.

$1, 2, 4, 5, 7, 8, 10, 11 ...$

What is the the sum of the first one hundered terms? Show as many solutions as possible.[/quote]
[hide="hint"]
First find the sum when each third number is not missing. And then subtract $3+6+9+...$.
[/hide]

[hide="Solution"] I believe it ends in 149 and there are 49 numbers missing, so $\frac{149(150)}{2}-3\frac{49(50)}{2}=149\times75-49\times25=\boxed{7500}$. :D I think. [/hide]

[hide]Well I think that we all know to add the first and last terms together and multiply that by half the number of terms to get 5050 for the sum of 1-100.  We can use the same concept for this.

$\displaystyle{(3+99)\cdot\116}$  But because there are an odd number of terms we add the middle term, the 17th.
$\displaystyle{3\cdot\117=41}$  Add these two together to find out much to subtract from 5050.  So we do $\displaystyle{5050-1673=\boxed{3377}}$  :D [/hide]

oops... I answered the sum of 1-100 not including the multiples of 3... sorry :blush:

[quote="DanK"][quote="jhollenbeck"]You can make this the sum of the first 50 terms of this sequence--3,9,15..., which ends in $3+(49*6)=297$.  Now we find the sum of such terms, $\frac{(3+297)(50)}{2}=\fbox{7500}$[/quote]

Darn it, you stole my method![/quote]

Jajajaa!

[quote="mbabbitt2003"]well.....
[hide]you add the first and second terms, you get 3. Third and fourth, 9. Fifth and sixth, 15.
There are 50 of those pairs.
You have $((2*(50*51)/2)-50)*3=50*50*3=[7500]$[/hide]

sorry, I didn't post all of the solution, because my hands are cold. Oh. I didn't steal anyone's method.[/quote]

Nice method!!!!

[hide="solution using gauss formula"]
you can make this into another sequence
$1+2+3+4+5+...$
by adding $3+6+9...$ to the original sequence
since we are adding the first 100 terms of the original sequence
we can add all the numbers from 1-150
and subtract the first 50 multiples of 3
the sum of 1-150 is
$\frac{(150)(150+1)}{2} = 11325$
the 3+6+9 sequence can be expressed as
$\frac{3\times(50)(50+1)}{2} = 3825$
our answer is then $11325-3825=\fbox{7500}$[/hide]


calculus
inequalities
calculus computations

This is a standard result proved in most good calculus textbooks, e.g. Apostol.

[quote="spider_boy"][quote="spider_boy"]We have sets $A$ and $B$ of real numbers.
The elements of the set $A+B$ are given as $a+b$ where $a\in{A}$ and $b\in{B}$.
Prove that $supA+supB=sup(A+B)$ where supA=supremum of the set A.[/quote]

I did not say that it is so hard.I am interested in people's solutions. :)[/quote]
Let $\sup A = x$ and $\sup B = y.$

We have $a \le x \ \forall a \in A$ and $b \le y\ \forall b \in A,$ which implies $a+b \le x+y.$ Hence, an upperbound for $A+B$ exists and $x+y$ is one of the upperbounds.

Claim: $\sup (A+B) = x+y.$ Suppose, for the sake of contradiction, $\sup (A+B) = x+y-\epsilon$  for some $\epsilon > 0 .$

Since $\sup A = x, \ \exists a' \in A$ such that $a' > x-\frac{\epsilon}{2}.$ Again, since $\sup B = y, \ \exists b' \in B$ such that $b' > y-\frac{\epsilon}{2}.$ Thus, we have $a'+b' > x+y-\epsilon,$ which contradicts the assumption that $x+y-\epsilon = \sup (A+B).$ And, we are done.

[I can't think of a simpler proof. May be there is one.]

1) max(A)<sup(A) and
2) max(B)<sup(B)
3) max(A)+max(B)<sup(A)+sup(B)

then max(A)+max(B)\in A+B, furthermore it is the max in A+B

suppose there is a stronger upper bound (in line 3) ), subtracting one of the inequalities implies 1) or 2) is not the min upper bound, contradiction

Most of those maximums you refer to don't actually exist.

For example, take $X=Y=(0,1)$.  Then $X+Y=(0,2)$, but $\max(X),\max(Y),\max(X+Y)$ all fail to exist.