Results are up: http://www.mathmeets.com/2008_state_meet/index.php.

Congratulations to [b]Acton-Boxborough Regional High School [/b]for claiming first place in the Large School division by 10 points, with a score of 123.
Congratulations to [b]Bedford High School [/b]for claiming first place in the Medium School division by 8 points, with a score of 71.
Congratulations to [b]Worcester Academy[/b] for claiming first place in the Small School division by 11 points, with a score of 62.

Congratulations to the following NEAML qualifiers from MA:
[b]Large[/b]: Acton-Boxborough, Lexington, Belmont, Concord Carlisle, Algonquin, and Newton South.
[b]Medium[/b]: Bedford, Winchester, Weston, Westborough, and Deerfield.
[b]Small[/b]: Worcester, Maimonides, McDuffie, Dover-Sherborn, and Joseph Case.

How did everybody do? I got an 18 :lol: But only because I did Rounds 1, 5, and 6.

Compared to last year's meet, this year's was much easier. (Note: ten 18's this year compared to a highest score of 13 last year.)

Also, this year, unlike last year, Bedford [b]got[/b] first place in the Medium Division without any ties!

[hide="Stats"]
[b]Note:[/b] This was not on the website; I came up with these stats using Excel.

Average scores:
[b]Round 1[/b] - 2.90625
[b]Round 2[/b] - 1.42519685
[b]Round 3[/b] - 1.03125
[b]Round 4[/b] - 1.307086614
[b]Round 5[/b] - 3.645669291
[b]Round 6[/b] - 2.25

So the rounds, in order of difficulty, are: 5, 1, 6, 2, 4, 3.

The average total score of all participants was: [b]6.282352941[/b].

Score distributions:

[b]18[/b]: 10
[b]17[/b]: 0
[b]16[/b]: 4
[b]15[/b]: 6
[b]14[/b]: 4
[b]13[/b]: 8
[b]12[/b]: 7
[b]11[/b]: 14
[b]10[/b]: 8
[b]9[/b]: 8
[b]8[/b]: 14
[b]7[/b]: 20
[b]6[/b]: 26
[b]5[/b]: 18
[b]4[/b]: 21
[b]3[/b]: 26
[b]2[/b]: 19
[b]1[/b]: 20
[b]0[/b]: 22
[/hide]

which additional teams qualified for NEAML?

There are definitely more NEAML qualifiers, seeing that the top 6 in large qualify.

Oh really? I never knew that. :blush: OK, I'll edit it.

I am pretty sure there are more teams from medium and small, which is what I'm wondering, but I don't know for a fact. Would you guys happen to know?

I believe the NEAML qualifiers are
[b][u]LARGE[/u][/b]
1. Acton-Boxborough
2. Lexington
3. Belmont
4. Concord-Carlisle
5. Algonquin
6. Newton South
[b][u]MEDIUM[/u][/b]
1. Bedford
2. Winchester
3. Sharon
4. Weston
5. Westborough
7. Deerfield [as WMML champion]
[b][u]SMALL[/u][/b]
1. Worcester
2. MacDuffie
3. Maimonides
4. Dover-Sherborn
7. Case [as SMCML champion]

Ooh, thanks! I thought someone told me only the top 3 schools get to NEAML... blah whatever. Editing again!

[quote="Yongyi781"]I thought someone told me only the top 3 schools get to NEAML... [/quote]  Quoting the rules: "The selection of the teams for the New England Math Team Competition will be based upon the performance at the State Tournament. At least four Large schools, three Medium schools, and two Small schools will be selected to represent Massachusetts in the New England Tournament. The overall league champion from every league is automatically invited to attend the New Englands."

Ah, so they said "at least."

And additionally, I found out that Newton North only sent 7 team members instead of 8, and the MAML data listed that team member in the Individual results. So in fact, only 255 people went to the MAML, not 256.

loool
http://www.wocomal.org/statistics/maml/alltime/team/championshipsbyyear.html

hopefully lexington won't pummel us during NEAML for breaking their 9 year championship reign at MAML...

Here is the official list of qualifiers, which I received in email today.  I had missed two (shown in boldface).

[quote]Qualified teams:

[b]Bromfield School[/b]
Joseph Case High School
Dover Sherborn High School
The MacDuffie School
Maimonides School
Worcester Academy

Bedford High School
Deerfield Academy
Sharon High School
Westborough High School
Weston High School
Winchester High

Acton Boxborough
Algonquin
Belmont High School
[b]Canton High School[/b]
Concord Carlisle High School
Lexington High School
Newton South
[/quote]

In Connecticut's state tournament, the school that won came from the Small Schools division.. (even though the Large Schools had 4 people per round, the top Large School still had less points than both the top Small and Medium school)

So yeah, don't expect much competition at NEAML from CT's Large Schools. Small Schools on the other hand though..

We got MAML (or maybe it was SAML) level 2 awards at the State Meet but are we ever going to find out our scores? I'm curious to know how many points were taken off for the oversight on the last problem, and I'm sure other people want to know their score as well.

Does anyone have a list of the top 20 on the MAML Level 2 for those that weren't at the state meet? (I actually know what I got but would like to see the whole list)

[quote="CatalystOfNostalgia"][quote="Pineappleperson"]hopefully lexington won't pummel us during NEAML for breaking their 9 year championship reign at MAML...[/quote]

hopefully[/quote]

Hope springs eternal.


Art of Problem Solving now has a [url=http://www.facebook.com/pages/Art-of-Problem-Solving/322689570276?ref=ts]page on Facebook[/url] and a [url=http://twitter.com/AoPSNews]feed on Twitter[/url].  We'll be using these to make announcements about AoPS, and to spread the word about programs of interest to AoPSers.  Those of you who host such programs should email us at aops@artofproblemsolving.com when you have news that AoPSers might want to know.

Thank you! This will be useful and exciting.

I would highly doubt that. First of all, AoPS is the main place where the main announcements are posted and the activities are held. Also, some people don't have Facebook or Twitter where they would find information.

I saw the AoPS page on Facebook! It's so cool that AoPS finally has one. :lol:

[quote="AIME15USAMO"]Will there be information posted on Facebook/Twitter that we couldn't find here, on AoPS?[/quote]

My expectation is to use it as a news feed of sorts -- people will send me information about this program or that one, and I'll post it there.

Maybe [hide="here"][url]http://www.artofproblemsolving.com/Forum/index.php?f=144[/url][/hide]?
I guess the stuff posted on Facebook/Twitter isn't really news, more just like updates..if that makes sense  :P
Also, can I add you as a friend of Facebook Mr Rusczyk?  :maybe:

[quote="PowerOfPi"]Where could we find the similar announcements on AoPS?[/quote]

Right now, nowhere.  AoPS specific announcements are made on the forum and our mailing lists.  I hope to use FB/Twitter for more general announcements.

[quote="AIME15USAMO"]
Also, can I add you as a friend of Facebook Mr Rusczyk?  :maybe:[/quote]

Sure, but include your username in the Friend request, so I know who you are ;)

Are we allowed to add our instructors on Facebook?  :D

Well I know Mr. Rusczyk friended a lot of us, but I don't know about the other instructors. But I don't see any reason why they won't either.
(Or maybe they usually don't friend much users, but Mr. Rusczyk is very sociable. :wink: )


inequalities
geometric inequality
geometry
cyclic quadrilateral

HaHaHaHa   INMO is over
Now i can rest in peace not pieces

How many didyou and others do ? What about Class XII students??? How many did they do usually - the top ones???

Hello,
    How was the INMO? How many did you do, Rushil, Vihang? I am Sanjith from Chennai. I managed only 2 and a half. In Chennai, one boy manged to do four. The rest of the boys,here, did around 2-3 like me. Let us start discussing about the problems. I got the second problem. I also did the one on combinatorics(Problem 4), I think. And the geometric inequality on cyclic quadrilateral, I got the first part.
Sanjith

Hello,
    How was the INMO?How many problems did you solve, Rushil,Vihang? Here, in Chennai, one boy managed four problems. The rest, it was 2-3. My brother Sanjith maged about 2 and a half. He got the second one, the fourth one and first part of the fifth problem. How did students do at Pune?
Sathej

hi,
i did 3 and a half problems.the 2nd the 5th and the 6th and 1st part of the first one.does anybody hear know about the usual cut off for inmo. hope u all did well :)

I got 4 and a half (1,2,3,5 and 6(a)) :D
Sahil Mhaskar (Class XII) solved 5 and a half though. :o
I have posted the problems in the respective sub forums and you may find answers there.

hi,
   i got three and a half problems right 1(a),2,5,6 and gave  some uncooked answers for the rest
how did u all do??

udbhav

[quote="Vihang"]
I have posted the problems in the respective sub forums and you may find answers there.[/quote]
Just curious, where  are the problems ? Thanks

Added later, Okay I see it. It is [url=http://www.artofproblemsolving.com/Forum/resources.php?c=78&cid=46&year=2006]Here[/url]

Well...

I got 3 and a half : 1 , 3 , 5 , 6(a)

I almost got Q 4 without the last step and I just missed the last step - dunno if they do partial marking but I'm not hopeful of getting much marks....

6(b) - couldnt do but when came home and calculated on calculator, the required $n$ is same as in part (a)

Q2 : Well, I have the following to say ---- the trick can only be found by those who have done extensive practice or those who have seen this exact question before( even for the former, they too mioght have seen this....)
I am amazed that the INMO people put a question that can only be solved if you have done it before - something which happened here... a few XII students at my centre had been told the EXACT problem in an IIT-JEE prep tutition - this goes to say that this must be a very OLD olympiad question ---- why dont they INMO people realise who much unfairness this puts into the equation --- even those who did it and are here, be frank, dont you think you could do it only coz you had done something similar before!!
Really shocking!!

Anyways,,, it went good!! How did other XII students in your centres etc. do?? Fingers crossed!!!!

Where's ashwath.rabindranath.... he;s on this forum... how did he do???? Somebody PM him!!

I solved something between three and a half and four
   I solved 1 3 5 6A and am  extremely doubtful about my solution for 2!

Hi,
I agree with Rushil on the second problem, the problem is tough if one has not seen it before, actually I had seen the series( in the solution) in a book written by Prof.G.H.Hardy 'A Course in Pure Mathematics' . If I had not seen that before, I would not have got that.
Sanjith.

Well the 1/2(a+b-1)(a+b-2) reminded me of triangular numbers.
(Thanks to the Proofs without words column here :roll: )
This cavee rise to an extremely simple solution.

See, I told ya!!!

Anyways, best of luck to all of us!!  :D

Any ideas about the general performance of Class XII students?? Please tell !

Hello,
    I think this year's INMO was just like last years,maybe a bit easier. I gather that last year the cut off was about 31-32 for IX,X and XI students and 45 for XII students. This year maybe slightly higher, say by about 5-10 marks.And in INMO, the way you solve a problem matters.They give quite an importance to writing down proofs. Proofs given have to be conclusive, taking care of all nuances of the problem and be reasonably rigorous.
Sathej

I agree with Sathej.
I know this because I missed clearing the INMO by 3 marks last year. :(
The proofs need to be up to the point and exact.(I learnt it the tough way!)
Anyways best of luck to all of you.

last year the cut off was less than 2 qns.
but it varies every year.
3 + is a sure in.(hopefully)

hey guys,
its really cool to answer inmo in the 9th.
but then, there have been guys who have cleared inmo in the 8th itself!!(they r unique)
infact 3 from india have been to the imo in the 9th, out of which 2 have been bronze medallists and 1 silver medallist.
so....

rushil,
last time the cut off for 12thers was 4 problems (approx).
its usually tough for 12thys, but then u seem to be quite good.
hope to c u at the maths camp this yr.

Hi there,

Do you when the INMO results shall be announced?? I got 3 and a half questions right, hopefully. :?

 I'm eagerly looking for the results to come up...... When do they usually arrive??
I thought this year's paper was easier than last years... what do you say :?:

signing off......
 Sudharsha (A math lunatic!) :lol: [/quote]

hi!
   the inmo results are out but i can't seem to find them on the net.do you ppl have any ideas where i can find them except on bhaskaracharya's sit because they don't have it.btw congarats to everyone who is selected for the camp.

Udbhav

I doubt they'll put them online so soon like the Physics/Chem etc.... it seems Maths is toooo underfunded in India --- we'll have to wait for the letters!! I'm not aware of any up to date site!

Well the result have arrived at Bhaskaracharya :D
I don't know the full list but will let you know about it soon

I'm in!!! I got 51 marks --- round about what I expected! In fact, the guy in XII who I told you abt that he had done one more question coz he had already done it, got just 53!! From Punjab, its just me and another guy in XII and just one in XI( he got around 35 --- so the cut off hasnt increased much!)

So, who others qualified??

Please post the full list when its out!!!

hey guys it seems that even maths is enjoying a special status in india. the whole list is online on the hbcse site.here's the link.i got in :lol:

http://www.hbcse.tifr.res.in/olympiads/inbo06_results/inmo_results06

congrats to all others who qualify.

Udbhav

That's true, venkata - the paper was easier --- but it seems the cut off is still low --- the XI guy from Punjab got in with 35 marks --- dont know abt other regions!

[color=darkblue][size=200]Please note: Kindly keep all INMO post-result discussion to the INMO 2006 result Sticky..... I shall lock this now --- reason: this was a 'pre-result' discussion!!! [/size][/color]


calculus
integration
trigonometry
induction
algebra
polynomial
real analysis

A nice integral in my textbook, but I cannot get it:
${ F(x)=\int^{2\pi}_{0}e^{xcos\alpha}cos(xsin\alpha)}\,d\alpha$, prove that $ F(x)=2\pi$ for all x in real.

Notice that $ e^{x \cos \alpha} \cos (x \sin \alpha) \equal{} \text{Re} \left[ \exp(x e^{i \alpha}) \right]$.

Then for $ x \neq 0$,

$ F(x) \ \equal{} \ \text{Re} \left[ \oint_{|z|\equal{}|x|} \frac{e^z}{iz} \, dz \right]$

with $ z \equal{} x e^{i \alpha}$. You may use residue thorem to evaluate this integral.

Seem a nice solution. I can't , however,  get it, for I am just learning multivariable calculus therefore I don't understand more than an element solution...

[hide="My second trial"]
Proposition) $ F^{(n)}(x) = \int_{0}^{2\pi} e^{x \cos\alpha}\cos(n\alpha+x\sin\alpha) \, d\alpha$.

Proof) For $ n = 0$, this equality obviously holds. Now assume this equality holds for $ n$. Then

\begin{eqnarray*}
F^{(n+1)}(x)
& = & \frac{d}{dx}\int_{0}^{2\pi} e^{x \cos\alpha}\cos(n\alpha+x\sin\alpha) \, d\alpha\\
& = & \int_{0}^{2\pi} \frac{\partial}{\partial x} e^{x \cos\alpha}\cos(n\alpha+x\sin\alpha) \, d\alpha\\
& = & \int_{0}^{2\pi} \left[ e^{x \cos\alpha}\cos\alpha \cos(n\alpha+x\sin\alpha) - e^{x \cos\alpha}\sin\alpha \sin(n\alpha+x\sin\alpha) \right] \, d\alpha\\
& = & \int_{0}^{2\pi} e^{x \cos\alpha}\cos((n+1)\alpha+x\sin\alpha) \, d\alpha
\end{eqnarray*}

and by induction, the proposition is proved.

Now we have $ F(0) = 2\pi$ and $ F^{(k)}(0) = 0$ for all positive integer $ k$.

Thus if $ F$ is analytic at $ x = 0$, then $ F(x) = 2\pi$ for all $ x$.

Hmm... well, this method also seems to need somewhat advanced arguments, since we have to prove that $ F$ is indeed analytic.
[/hide]

Slightly modified version:

$ e^{x \cos \alpha} \cos (x \sin \alpha) = \text{Re} \left[ \exp(x e^{i \alpha}) \right] = \text{Re} \left[ \sum_{n=0}^{\infty} \frac{1}{n!}x^n e^{i n\alpha} \right] = \sum_{n=0}^{\infty} \frac{1}{n!}x^n \cos (n \alpha)$

For each $ x$, this series converges uniformly for all $ \alpha$. So we can integrate this series term-by-term, and we have

$ F(x) = \sum_{n=0}^{\infty} \int_{0}^{2\pi} \frac{1}{n!}x^n \cos (n \alpha) \, d\alpha = 2\pi$

Thank you, sos440. I find that I can use the remains of Taylor polynomial to solve it with the $ F^{(n)}$. Actually it is not difflcult in that way. Thank you.

[quote="dingdongdog"]A nice integral in my textbook, but I cannot get it:
${ F(x) = \int^{2\pi}_{0}e^{x \cos\alpha}\cos(x \sin\alpha)}\,d\alpha$, prove that $ F(x) = 2\pi$ for all x in real.[/quote]Here's a simple solution, with credit to one of my colleagues: Clearly $ F(0) = 2 \pi$. Then
\[ F'(x) = \int_{0}^{2\pi} e^{x \cos \alpha} \left( \cos \alpha \cos (x \sin \alpha) - \sin \alpha \sin (x \cos \alpha) \right) d \alpha \\=
\int_0^{2\pi} \frac{d}{d \alpha} \frac{1}{x} e^{x \cos \alpha}  \sin ( x \sin \alpha) d \alpha = \frac{1}{x}
e^{x \cos \alpha}  \sin ( x \sin \alpha) \bigg|^{\alpha = 2\pi}_{\alpha = 0} = 0\]
and hence $ F(x) = 2 \pi$ for all $ x$.  :lol:

[quote="hpe"][quote="dingdongdog"]A nice integral in my textbook, but I cannot get it:
${ F(x) = \int^{2\pi}_{0}e^{x \cos\alpha}\cos(x \sin\alpha)}\,d\alpha$, prove that $ F(x) = 2\pi$ for all x in real.[/quote]Here's a simple solution, with credit to one of my colleagues: Clearly $ F(0) = 2 \pi$. Then
\[ F'(x) = \int_{0}^{2\pi} e^{x \cos \alpha} \left( \cos \alpha \cos (x \sin \alpha) - \sin \alpha \sin (x \cos \alpha) \right) d \alpha \\
= \int_0^{2\pi} \frac {d}{d \alpha} \frac {1}{x} e^{x \cos \alpha} \sin ( x \sin \alpha) d \alpha = \frac {1}{x} e^{x \cos \alpha} \sin ( x \sin \alpha) \bigg|^{\alpha = 2\pi}_{\alpha = 0} = 0
\]
and hence $ F(x) = 2 \pi$ for all $ x$.  :lol:[/quote]
Maybe what you mean is $ F'(x) = \int_{0}^{2\pi} e^{x \cos \alpha} \left( \cos \alpha \cos (x \sin \alpha) - \sin \alpha \sin (x \sin \alpha) \right) d \alpha$ ...


How many distinct prime divisors does 900 have?

To answer this question, we would first need to do a prime factorization.

There are many ways to find the prime factorization of 900. Here is one method:
$ 900$
$ \equal{}9\times100$
$ \equal{}3\times3\times10\times10$
$ \equal{}3\times3\times2\times5\times2\times5$
$ \equal{}2^2\times3^2\times5^2$

We thus see that 900 is divisible by $ \fbox{3}$ different primes: $ 2,3,$ and $ 5$.


It’s raining! Each raindrop can hit the ground anywhere within a fixed n x n region on a flat plane. The plane starts out completely dry. On impact, each raindrop leaves a wet circle of radius r, with the impact point as its center. After t raindrops have fallen, what is the area of the part of the plane that will end up wet, on average?

all of the plane, right?

or am I missing something? :?

Look at it from the point of view of a point on the plane.

[hide="Solution"]The probability that any particular raindrop makes the point wet is $\frac{\pi\cdot r^2}{n^2}$. Since each raindrop is independent, the probability that this point is not wet after $t$ drops is $\left(1-\frac{\pi\cdot r^2}{n^2}\right)^t$. This is the average fraction of the region  still dry.[/hide]
Actually, that isn't quite enough; the above only applies when the point is more than $r$ away from the boundary. It could be worked out for the other points as well, but the answer would be hopelessly ugly, requiring numerical methods.

[quote="JBL"]You're missing something -- try re-reading, perhaps?

This doesn't seem easy at all.  All that seems obvious to me is that as $r \to 0$, the solution approaches $\pi \cdot r^2 t$, but beyond that I really don't have any clue.  Any approach based on putting points on a plane seems unlikely to result in anything nice at all.[/quote]

sorry, I just thought of a rainstorm, and did not see the $t$ part. oops :blush:


[quote="Till"]Let $a,b,c$ positive reals.

Prove that
$(a+b)^{c}+(b+c)^{a}+(c+a)^{b}>2$[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133925


Anything by hannah montana

Anything by anyone like hannah montana

Umm........

There was a total failure by Beatles in White Album

Called Revolution 9, you will creep out once you hear it or just feel weird.

http://youtube.com/watch?v=_wCJ9WmX9Zw not interesting vid to watch

http://youtube.com/watch?v=PG0wksBzKSc this one gives me the creeps

NOOOOOOOOOOOOOOOOOOOOOOOOO
HANNAH MONTANA IS LIFE!!

ok i admit... shes not veryy er...professional???
 :rotfl:

ahhhh that terrible songg
i heard it on the radio last week... why would they bring that up again...

anything from disney channel, american idol, mtv, justin timberlake, soulja boys, and me.

[quote="7h3.D3m0n.117"]She's only 15 years old. How professional can a 15 year old be?

Summer Love by Justin Timberlack. Fail.[/quote]
not very professional

You're Beautiful by James Blunt.

This song is annoying.

Everything else said on this thread I agree with so far

cant believe no one said jonas brothers....yet

and yea the other ones are really annoying

In my opinion Hillary Duff's cover of The Who's My Generation is the worst.

WTF? Dream Theater is great.

Most popular songs nowadays.

anything by miley cyrus makes my ears bleed. :(

@1=2: What? that songs pretty good.....

Yea, I agree DT is an awesome band, even if LaBrie's vocals get a little annoying at times.

Anyway, to continue with the thread...This is not necessarily the worst, but I personally dislike most songs on Metallica's St. Anger album. I really hope they pick up the Master of Puppets, and Ride the Lightning feel with their next release (Death Magnetic).

[quote="MustItAlwaysBeMe"]Anything by Soulja Boy (duh)[/quote]

The only song that was ever on air by Soulja Boy was Crank Dat.

Crank Dat was OK for like 2 days. Then it got old...
An annoying song is Bleeding Love (Leona Lewis) or just Miley Cyrus (always bad)

To Mewtoo555...you created your account on my birthday.

My least favorite song: Anything by anyone who thinks they can sing but really cant ie. miley cyrus.

[quote="RussianRocket"]
An annoying song is Bleeding Love (Leona Lewis) [/quote]

**gasp** how dare you?

[quote="basted"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]

 :mad:

[quote="basted"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]
NOOOOOOO!!!!! Avril Lavigne rocks!

[quote="mewto55555"]Barbie Girl[/quote]
 :rotfl: agreed :rotfl:

[quote="cyberspace"][quote="basted"]Avril Lavigne is pretty annoying. And any song similar in style like hers...[/quote]
NOOOOOOO!!!!! Avril Lavigne rocks![/quote]

THank you cyberspace.

[quote="aanjohri"]I also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.[/quote]

Well then. I guess no one else but me likes these song...

But I also love every type of music (especially Bengali music my friend lets me listen to on her iPod...:D) except country. The one thing people here like. IN ALABAMA.

"I've got my sights set on you,
And I'm ready to aim,
I have a heart that will,
Never be tamed,
I knew you were something special
When you spoke my name,
Now, I can't wait to see you again...

I've got a way of knowing,
When something is right,
I feel like I must have known you,
In another life,
Cause I felt this deep connection,
When you looked in my eyes,
Now I can't wait to see you again,

The last time I freaked out,
I just kept looking down,
I st-st-stuttered when
You asked me what I'm thinkin' bout
Felt like I couldn't breathe
You asked what's wrong with me
My best friend Lesley said
"Oh she's just being Miley!"

The next time we hang out,
I will redeem myself,
My heart it can rest till then,
Whoa! Whoa! I,
I can't wait to see you again,

I got this crazy feeling
Deep inside
When you called and ask to see me
Tomorrow night,
I'm not a mind reader,
But I'm reading the signs
That you can't wait to see me again

The last time I freaked out,
I just kept looking down,
I st-st-stuttered when
You asked me what I'm thinkin' bout
Felt like I couldn't breath
You asked what's wrong with me
My best friend Lesley said
"Oh she's just being Miley!"

The next time we hang out,
I will redeem myself
My heart can rest till then,
Whoa! Whoa! I,
I can't wait to see you again,

I've got my sight set on you,
And I'm ready to aim,

The last time I freaked out,
I just kept looking down,
I st-st-stuttered when
You asked me what I'm thinkin' bout
Felt like I couldn't breathe
You asked what's wrong with me
My best friend Lesley said,
"Oh she's just being Miley"

The next time we hang out,
I will redeem myself
My heart can rest till then,
Whoa whoa I,
I can't wait to see you again,
Whoa whoa I,
I can't wait to see you again."

[size=200][u][i][b]HAD 2 SHARE THE PAIN... SORRY... COULDNT HELP IT[/b][/i][/u][/size] :furious:  :noo:   :cursing:

"See You Again"--miley cyrus

THe pain is too much russianrocket delete it!!

[quote="SnipedYou"][quote="aanjohri"][quote="RussianRocket"]
An annoying song is Bleeding Love (Leona Lewis) [/quote]

i agree. it keeps repeating the same line over and over.

I also don't like anything written by Natasha Bedingfield, especially Pocketful of Sunshine and Unwritten.[/quote]

Chocolate Rain is probably the most repetitive song there is, also Say by John Mayer is pretty repetitive too.[/quote]
Dude, Chocolate Rain is pretty awesome (*moves away from mic to breath*). DnB remix is pretty good, too.
Dude, it's scary, and that's an understatement; let me read my book marks bar which got screwed got by my sister:
Yahoo E-mail
AoPS
Gmail
Youtube - Hannah Montana rock star FULL + lyrics video (WTF?)
Youtube - Hannah Montana True friend FULL + lyrics
Youtube - Miley Cyrus Hannah Montana I Miss You FULL + lyrics
Youtube -Let's Dance! (Lyrics) Hannah Montana/ Miley Cyrus

AUGHAUGUAGAHG! And she sings this crap all the time.

That aside, Soulja boy is pretty dang crappy too. Actually, pretty much all the stereotype bands (Nickelback, ewww, Greenday, okayyy, Linkin Park, wtf?, etc.)

[quote="Nerd_of_the_Ages"]stereotype bands (Nickelback, ewww, Greenday, okayyy, Linkin Park, wtf?, etc.)[/quote]

OMG ARE YOU FRIKIN KIDDING ME LINKIN PARK IS AWESOME!
[color=white][size=9](maybe thats cuz i like emo music....)[/size][/color]

[quote="myyellowducky82"]i know her songs scare me[/quote]
THAT'S WHY SHE GOT OWNED BY KEVIN CHEN!!!!

OMG I WAS SO HAPPY MILEY GOT OWNED BY KEVIN CHEN!!!!

umm...

i get annoyed by miley cyrus.
a lot of high school musical songs.

now that i think about it, anything from disney channel.

but i do like the stuff thats on the radio. like Ne-o, chris brown's voice is just annoying, rihanna (except for umbrella. umbrella was the worst), and Cascada.

[quote="hunter34"]Time to bash.

Linkin fart has about the same musical talent as a rock with testosterone.

Miley Cyrus is Mainstream. Mainstream is just popular for being mainstream. Most of the musicians there have less musical talent than linkin fart.

And wtf? gauss, you're like a fan girl.

With that being said. I hate tool.[/quote]
Totally...

[quote="hunter34"]Time to bash.

Linkin fart has about the same musical talent as a rock with testosterone.

Miley Cyrus is Mainstream. Mainstream is just popular for being mainstream. Most of the musicians there have less musical talent than linkin fart.

And wtf? gauss, you're like a fan girl.

With that being said. I hate tool.[/quote]
I'm not a fan girl, I'm a tomboy. So i guess the whole AoPS community owns miley cyrus. and that's only because we're all smarter than her in math :D

GO NERDS!!!!!!!!! :D


inequalities
abstract algebra
triangle inequality
geometry proposed
geometry

Let a acute-triangle $ ABC$. Find point $ M$ inside triangle so that $ AM.BM.AB \plus{} BM.CM.BC \plus{} CM.AM.CA$ min!

We have the following inequality for every point $ M$ in the triangle $ ABC$:
\[ a\cdot MB \cdot MC \plus{} b\cdot MC \cdot MA \plus{} c\cdot MA \cdot MB \geq abc
\]
I shall prove this inequality using complex numbers.
Let $ M$ be the center of the complex plane where $ A(a)$,$ B(b)$,$ C(c)$.
We have the following identity:
\[ ab(a \minus{} b) \plus{} bc(b \minus{} c) \plus{} ca(c \minus{} a) \equal{} \minus{} (a \minus{} b)(b \minus{} c)(c \minus{} a)
\]
Aplying module and using the triangle inequality we obtain the desired inequality:
\[ \sum {\left|a\right|\left|b\right|\left|a \minus{} b\right|}\geq \left|\sum{ab(a \minus{} b)}\right| \equal{} \left|a \minus{} b\right|\left|b \minus{} c\right|\left|c \minus{} a\right|
\]
So $ min\{ a\cdot MB \cdot MC \plus{} b\cdot MC \cdot MA \plus{} c\cdot MA \cdot MB\} \equal{} abc$

[quote="Vrajitoarea Andrei"]mininum for $ M\in\{A,B,C\}$[/quote]
But $ M$ must be inside triangle!
And min when $ M$ be orthocentre of triangle!

Sorry.My mistake. Indeed , we obtain the minimum for $ a\plus{}b\plus{}c\equal{}0$ meaning that $ M\equal{}H$

[hide="Another hint:"]Let $ E,F$ are points so that: $ BCME,BCAF$ are parallelograms.
Hence $ AF\equal{}EM\equal{}BC,EF\equal{}AM,EB\equal{}CM,BF\equal{}AC$.
Apply Ptoleme inequality for $ ABEF,AEBM$ we have solution! [/hide]

[quote="Vrajitoarea Andrei"]Sorry.My mistake. Indeed , we obtain the minimum for $ a \plus{} b \plus{} c \equal{} 0$ meaning that $ M \equal{} H$[/quote]

Why for $ a\plus{}b\plus{}c\equal{}0$ ?


calculus
integration
real analysis
real analysis unsolved

Could someone help me with this problem:
If f in L^1 is differentiable a.e., and f' in L^1, what is the Fourier transform of f' ?? Is is tiF(t), where F(t) is the Fourier transform of t?
I think I can show this by using integration by parts, but I wanted to check that I am right.

Yes. (up to constant multipliers depending on which definition you're using)


Coach Frost had fifteen students attend his first MATHCOUNTS practice.  Coach Frost asked the students to shake hands with each other.  Assume each student shook hands with every other student exactly once, and no student shook hands with Coach Frost.  How many total handshakes were exchanged?

well...[hide="solution"]this is a combination problem...so just do $\frac{15*14}{2}$ to get $15*7$ which is [b]105[/b].
u also can do (tho i dont suggest it) 14+13+12+11+10+9+8+7+6+5+4+3+2+1 which comes out being the same answer, 105.[/hide]

[hide][quote="flamesofdestiny"]u also can do (tho i dont suggest it) 14+13+12+11+10+9+8+7+6+5+4+3+2+1 which comes out being the same answer, 105.[/quote]

That's because the formula for the sum of the first $n$ integers is $\frac{n(n+1)}{2}=\frac{14(15)}{2}$, which is actually where your first solution came from.[/hide]

[hide]its 15 choose 2=105 :idea: [/hide]

[quote="rcv"][quote="Treething"]this is soooo easy....................i can't believe this...
[hide="easy way"]
15 choices for person one
14 choices for person two

divide by two because you doublecount

<_<
>_>
<_<
>_>
[/hide][/quote]
Folks:  If you think a problem is very easy, why don't you leave it for those who don't think it is so easy.

While you didn't give a final answer, your solution seems to lead to an incorrect answer.  [Maybe this wasn't as easy as you thought? :) ][/quote]

It looks right to me :? $\frac{15 \cdot 14}{2}=15 \cdot 7 = 105$...

[quote="chess64"]It looks right to me :? $\frac{15 \cdot 14}{2}=15 \cdot 7 = 105$...[/quote]

I think he got lucky.... sure works for these types of problems.

[quote="236factorial"]I think he got lucky.... sure works for these types of problems.[/quote]
What do you mean? That's what I did... What's wrong with that method? It's what you're supposed to do! 15 choices for the 1st, 14 for the 2nd, divide by 2 to account for repeats. (A-B is the same as B-A)

[quote="chess64"]What do you mean? That's what I did... What's wrong with that method? It's what you're supposed to do! 15 choices for the 1st, 14 for the 2nd, divide by 2 to account for repeats. (A-B is the same as B-A)[/quote]

Oh, I was just thinking that the first person had only 14 people, second had 13, third had 12, etc.... so we wouldn't worry about repeats.  :oops:

[quote="chess64"]It looks right to me :? $\frac{15 \cdot 14}{2}=15 \cdot 7 = 105$...[/quote]

Treething's solution is correct.  My apologies to you, Treething.  I misunderstood your brief solution.


determine a,b if ax^17+bx^16-1 divides x^2-x-1

A solution using complex numbers :

p(x)=ax^17+bx^16-1 divides x^2-x-1 iff p(phi) = p(1/phi) = 0

with phi=(1+sqrt(5))/2

from phi²=phi+1 we quickly get phi^17 =1597*phi+987 and phi^16=987*phi+610

Then p(phi) = p(1/phi) = 0 gives (after some boring  :( and unchecked  :) algebra) :

a = -987, b = 1497.


${\sin (3 x)=4 \sin (a) \sin \left(a+\frac{\pi }{3}\right) \sin \left(a+\frac{2 \pi }{3}\right)}$

From Trig Rules

${\sin (\theta+\phi )=\cos (\phi ) \sin (\theta )+\cos (\theta ) \sin (\phi )}$

Simplifying

${\sin \left(a+\frac{\pi }{3}\right)=\cos (a) \sin \left(\frac{\pi }{3}\right)+\cos \left(\frac{\pi }{3}\right) \sin (a)}$
${\sin \left(a+\frac{\pi }{3}\right)=\frac{1}{2}\left(\sqrt{3}\cos (a)+\sin (a)\right)}$

${\sin \left(a+\frac{2 \pi }{3}\right)=\cos (a) \sin \left(\frac{2 \pi }{3}\right)+\cos \left(\frac{2 \pi }{3}\right) \sin (a)}$
${\sin \left(a+\frac{2 \pi }{3}\right)=\frac{1}{2}\left(\sqrt{3}\cos (a)-\sin (a)\right)}$

Substituing

${4 \sin (a) \sin \left(a+\frac{\pi }{3}\right) \sin \left(a+\frac{2 \pi }{3}\right)=\sin (a) \left(\sqrt{3}\cos (a)-\sin (a)\right) \left(\sqrt{3}\cos (a)+\sin (a)\right)}$
${4 \sin (a) \sin \left(a+\frac{\pi }{3}\right) \sin \left(a+\frac{2 \pi }{3}\right)=3 \cos^{2}(a) \sin (a)-\sin^{3}(a)}$

But we have...

${\sin (3 x)=3 \cos^{2}(x) \sin (x)-\sin^{3}(x)}$

${3 \cos^{2}(a) \sin (a)-\sin^{3}(a)=3 \cos^{2}(a) \sin (a)-\sin^{3}(a)}$

So it is correct only if

${a=x}$

And it is the proof


Our school's girls volleyball team has 14 players, including a set of 3 triplets: Alicia, Amanda, and Anna.  In how many ways can we choose 6 starters if all three triplets are in the starting lineup?

If all three triplets are in the starting line, then there are 11 players left to choose 3 people from. Thus our answer is $ \binom{11}{3} \equal{} 165$.

[quote=OlympusHero]how is this level 12

in total, the volleyball team has $14$ players. the triplets are automatically starters, so exclude them which gives $11$ remaining players.

we want to pick $6$ starters but the triplets are automatically starters, so exclude them which gives $3$ remaining starters.

therefore the answer is $\binom{11}{3}=\boxed{165}$.[/quote]

lvl 12 is supposed to be easy...


AMC
USA(J)MO
USAMO
vector
parameterization
search
linear algebra

I posted these in the Intermediate forums, but no one answered  :( ...

[quote="leoxnlin"]Let $ S$ be a set of $ n$ ordered sextuples where the $ i$th ordered sextuple is $ (i_1,i_2,i_3,i_4,i_5,i_6)$. For all $ i_u$, $ i_u \in \mathbb{Z} , 0 \le i_u \le 9$. Find the maximum possible value of $ n$ if there do not exist any $ i_u, i_v, j_u, j_v$ such that $ i_u \equal{} j_u$ and $ i_v \equal{} j_v$.[/quote]

[quote="leoxnlin"]Finitely many cards are placed in two stacks, with more cards in the left stack than the right. Each card has one [i]or more[/i] distinct names written on it, although different cards may share some names. We define a "shuffle with respect to a name" the process of moving every card that has that name written on it to the opposite stack. Prove that it is always possible have more cards in the right stack after finitely many shuffles.[/quote]

Thank you  :) .

[hide="Second problem"]Shuffle with respect to all names.[/hide]

Um, no MellowMelon, I believe there must be a misunderstanding,... counterexamples are easily constructed... perhaps you missed the "one [i]or more[/i] distinct names" part?

Some weakish bounds on the first problem:

[hide]If we have more than 100 sextuples then two of them have the same first two elements, a contradiction.  Therefore $ n\le 100$.

A construction for $ n \equal{} 52$: For 49 of our sextuples, we only use the digits 0-6.  Let the elements of a general sextuple be $ a,b,a \plus{} b,a \plus{} 2b,a \plus{} 3b,a \plus{} 4b$, reducing modulo 7.  If two pairs of corresponding elements of different sextuples are the same, then some linear combination of their formulas above gives us $ a$ or a multiple of $ b$, and from there we can show that the $ a$ and $ b$ values of the two sextuples are the same.  Therefore, making the $ a$ and $ b$ values of each sextuple different is sufficient to ensure that the 49 sextuples satisfy the requirements of the problem.

We can also throw in $ (7,7,7,7,7,7),(8,8,8,8,8,8)$, and $ (9,9,9,9,9,9)$ to give us a total of 52 sextuplets.

Therefore $ 52\le n\le 100$.[/hide]
What is the source of this problem?  It's similar to an old and fairly easy USAMO problem, but this one doesn't seem like it has a nice solution.

Thanks matt276eagles  :)  That's about the only useful response I've ever received at all on these questions...

[hide="Second problem"]
Let $ N$ be the set of all names which appear on any of the cards. for every $ A \subseteq N$ consider the effect of shuffling all the names in $ A$. If we do this for all possible $ A$'s, then each card remains on the lefts stack precisely half the time, and moves to the right half of the time. So, on average, the number of cards on the left and on the right must be equal. Since $ A \equal{} \emptyset$, which corresponds to doing nothing, yields a larger left stack, there must be at least one $ A$ yielding a larger right stack.
[/hide]

This problem can be found in Peter Winkler's "Mathematical Puzzles: A Connoisseur's Collection", p. 3, "Tipping the Scales". It's on Google Books.

Can you rephrase your first problem? I can't make sense of it... sorry.

Essentially it is asking for the maximum number of six-digit integers (where we allow leading zeroes) such that no two numbers share more than one digit.

Not quite. The precise problem is this: find a large set $ A$ of vectors $ {\bf x} \equal{} (x_1,x_2,\ldots, x_6)$ such that, for any two indices $ 1 \leq i < j \leq 6$ and any two vectors $ \bf{x}$, $ \bf{y}$ we have that $ (x_i,x_j) \neq (y_i,y_j)$. This is part of a family of such problems determined by the parameters: alphabet size (10), length (6), and length of sub-vectors we care about (2). For other parameter sets this is easily solved via linear algebra, but the choice of parameters here is a bit nastier. When I get a chance I'll see if I can imporove the bounds.

I think (2) is actually a weaker version of http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1509886817&t=5976

i think the answer is 10.
since i_u is in the interval[0,9], there are only 10 alternatives for it,
one for each set.


abstract algebra
number theory proposed
number theory

It's really obvious.
We have the only solution $(x,y)=(0,1)$.
Assume that $y>1$.
If $2x^4=(y-1)(y+1)$ then $y$ is odd (and $\gcd(y-1,y+1)=2$).
So ($y-1=2\alpha^4$ and $y+1=\beta^4$) or ($y+1=2\alpha^4$ and $y-1=\beta^4$)
The first case is impossible because $2(1+\alpha^4)=\beta^4 \Rightarrow \alpha^4\equiv-1[4]$.
Thus $\beta^4=2(\alpha^4-1)=2(\alpha-1)(\alpha^2+1)(\alpha+1)$ and $\alpha$ is odd.
We have $\gcd(\alpha-1,\alpha+1)=\gcd(\alpha-1,\alpha^2+1)=\gcd(\alpha+1,\alpha^2+1)=2$, so:
$\alpha-1=2\gamma^4$ and $\alpha+1=2\chi^4$ $\Rightarrow 1+\gamma^4=\chi^4 \Rightarrow$ impossible.

[quote="nguyenquockhanh"]solve: 2$x^4+1=y^2$[/quote]
Easy, only need to work module $5$ and then module 2, geting that $\tilde X^4=2+5D$, so if $x\ne 0$ the congruence is IMPOSSIBLE. Then the only solution is $(0,1)$ and $(0,-1)$.

Best Regards.

RX TCM

okie :D
I have three solutions for this pro, one of them similar above solutions.
And this morning, my teacher gave me below pro:
[b]PRO-2[/b]: solve: $8\cdot x^4+1=y^2$
And i have 2 sol for this pro.
I think very cleary if use THEOREM-FERMAT:
[b]THEOREM-FERMAT[/b]: if $a^2-b^2=c^2$ and $a^2+b^2=d^2$ then $b=0$ you  agree, aren't you!

[quote="romano"]THEOREM-FERMAT: if $ a^2-b^2=c^2 $ and $ a^2+b^2=d^2 $ then $ b=0 $
 Can you post the proof for this problem ?[/quote]
okie: here is solution:assume $b>0$, $gcd(a,b)=1$ and $d$ min
$2a^2=c^2+d^2$ so $2|c-d$, we obtain $a^2=(\frac{c+d}{2})+(\frac{d-c}{2})$
now, applying PITAGO-EQUATION then $2b^2=d^2-c^2=2\cdot{4mn(m^2-n^2)}$ so $b=2t$ and $t^2=mn(m^2-n^2)$ thus $m=x^2$, $n=y^2$ and $m^2-n^2=z^2$, but now, we have $x^2-y^2$ and $x^2+y^2$ are both squares, contradiction with the hypothesis $d$ min

[b]PITAGO-EQUATION[/b]:all solutions of the equation: $x^2+y^2=z^2$ are: $(k(m^2-n^2),2mn,(m^2+n^2))$ in here $m\neq {n}(mod2)$, $gcd(m,n)=1$

[quote="nguyenquockhanh"][b]PRO[/b]: assume a,b possitive integer
solve : $(a^2+b^2)\cdot x^4+1=y^2$[/quote]

NOTE: when $a=b=6$ then  equation $ 72\cdot x^4+1=y^2$ has no roots, okie?

By Lagrange:Solve the equation x 2 +y^4=2*z^4 in Z+

Here is my solution for proving x^4-2y^4=±1 has no solution and i think it can be used in the general equation       http://www.mathlinks.ro/Forum/viewtopic.php?t=21818&highlight=
ANd:[quote="nguyenquockhanh"][b]PRO[/b]: assume a,b possitive integer
solve : $(a^2+b^2)\cdot x^4+1=y^2$[/quote]
Could you post your solution,nguyenquockhanh???


I am not sure maybe the question was already asked. Is the limited version of the forum, no avatars and other stuff which is not necessarily required but causes a lot of traffic, already implemented on the new forum. I assumed the topic should be covered at:

Forum index  "user"  Options  Board layout

Thanks.

[quote="Valentin Vornicu"]I see no problem ... what is the problem? (remember this is the support forum not the question & suggestion one ;) ).[/quote]

Feel free to move the post. What I meant is what for instance was requested by Darij once. That you could choose a more text based version where some graphical components were not displayed. I supposed that it can be found here:

Forum index  orl  Options  Board layout

So please tell me how I can change the forum view to this limited edition in my profile, options or anywhere else...

[quote="Myth"][quote="orl"]stuff which is not necessarily required but causes a lot of traffic[/quote]
Oh, yes. Huge traffic is a big problem for me...  :(
And avatars and signatures are not main reason for this, but entire structure.[/quote]
Myth, the new structure is massively improved. It caches most of the stuff down into your hard disk (thus the massive speed improvement), and all the java scripts are downloaded as a temporary file.

I tried it on Romanian dial-up and it was very quick :)

But again I fail to see  what this topic is about :?

I am Opera user, so all graphical components are not important for me, because all images are cached (as stylesheets and other things). And only images I download are latex-images.

Speed were quite well all the time, even if I used dial-up. But I pay for traffic, and that is very annoying thing for me, because cost is quite high. :(

[quote="Myth"]Speed were quite well all the time, even if I used dial-up. But I pay for traffic, and that is very annoying thing for me, because cost is quite high. :([/quote]Ok you so you want another text-based (or something like that) version right?

[quote="Valentin Vornicu"][quote="Myth"]I think, it is impossible at the current stage. Just see source of this page. It is about 60KB, but useful information occupies at most 3-4KB. What are we talking about?
At the same time, I don't think we are able to compact large part of the source code without lost of site functionality.

I don't see appropriate solution for reducing page weight at the moment.[/quote]The size of the page also includes cached elements, so the download you make is actually much smaller than before.[/quote]
????
Page is what browser gets for his request. Nothing else. But page source can refer to external files, which are cached.

I looked at this and it seems that the page sizes are indeed large :( ... I can scrap off some 20kB / page if I'll get the jumpbox off.

Other stuff might affect the forum navigation ... but if you have any other suggestions I'd be glad to hear them.

Maybe the main style with jumpbox and an alternative style without. I personally use the jumpbox very often.

  Darij

[quote="Myth"]Is it possible to make new style without jumpbox? You can add link to the page with the forum site map instead of it.[/quote]:what?:  I won't make a new style for that ... I'll put an option in the board layout menu, so users can select it on/off ;)

New style means different colors and images, so I won't do that right now. We have two different styles, and that's enough worries for now  :cool:

[quote="Valentin Vornicu"][quote="Myth"]Is it possible to make new style without jumpbox? You can add link to the page with the forum site map instead of it.[/quote]:what?:  I won't make a new style for that ... I'll put an option in the board layout menu, so users can select it on/off ;)[/quote]
Indeed, I mean such solution (in other words...  :blush: )


What is the greatest prime factor of 221?

$ 221 \equal{} 13 * 17$ The greater prime factor is $ \boxed{17}$.


Κάποια (elementary) λύση? Έχω βρει λύση με χρήση (μιας ειδικής) Gronwall.Καμιά άλλη ιδέα?

Η Grönwall είναι πολύ κλασική ανισότητα στις ODEs.

Ουσιαστική η elementary λύση που ζητάς για την άσκησή σου είναι ίδια με την απόδειξη της Grönwall (απλά χρησιμοποιείς συγκεκριμένες weight/decay functions) -- και πιο elementary δε γίνεται, κανά δυο γραμμές είναι.

[quote="Fischerman"]Οκ,thx (ρωτούσα, γιατί εκεί που τη βρήκα η Gronwall θα ήταν long shot  :wink: )[/quote]

Ναι απλά όπως είπα η απόδειξή της είναι more or less straightforward, ακόμα πχ. και για μαθητές λυκείου (θέλω να ελπίζω...).

Θέτω $ g(x)\equal{}\int^x_0f(s)ds$ τότε η $ g$ είναι παραγωγίσιμη και εύκολα η $ e^{\minus{}x}g(x)$ βγαίνει φθίνουσα. Άρα $ g(x)e^{\minus{}x} \le g(0).e^0\equal{}0$ οπότε $ g(x) \le 0$ και από την δοσμένη σχέση και $ f(x) \le 0$ . Όμως $ f(x) \ge 0$ συνεπώς $ f(x)\equal{}0$ στο $ [0,1]$

[quote="r_boris"]Θέτω $ g(x) \equal{} \int^x_0f(s)ds$ τότε η $ g$ είναι παραγωγίσιμη και εύκολα η $ e^{ \minus{} x}g(x)$ βγαίνει φθίνουσα. Άρα $ g(x)e^{ \minus{} x} \le g(0).e^0 \equal{} 0$ οπότε $ g(x) \le 0$ και από την δοσμένη σχέση και $ f(x) \le 0$ . Όμως $ f(x) \ge 0$ συνεπώς $ f(x) \equal{} 0$ στο $ [0,1]$[/quote]

Με διέψευσε ο r_boris: σε μια γραμμή μπορούμε να την χωρέσουμε :D

[quote="r_boris"]Θέτω $ g(x) \equal{} \int^x_0f(s)ds$ τότε η $ g$ είναι παραγωγίσιμη και εύκολα η $ e^{ \minus{} x}g(x)$ βγαίνει φθίνουσα. Άρα $ g(x)e^{ \minus{} x} \le g(0).e^0 \equal{} 0$ οπότε $ g(x) \le 0$ και από την δοσμένη σχέση και $ f(x) \le 0$ . Όμως $ f(x) \ge 0$ συνεπώς $ f(x) \equal{} 0$ στο $ [0,1]$[/quote]

Ναι.


A free living one celled organism, lacking a cell wall is collected from a stream containing mineral rich water. The organism is placed on a microscope slide in distilled water.
1-What microscopic changes in the cell would the observer see?
2-What are the causes of the changes observed?

Did you even read your textbook?
[hide]
1. The cell will expand.
2. Osmosis.  (Water will enter the cell, because the cell is hypotonic to the distilled water.  Its natural environment is the stream, which is "mineral rich", so the concentration of water is higher outside the cell when it is placed in distilled water.)

[/hide]


A function is defined by $ f(0) \equal{} 1$ and $ f(n) \equal{} f(n \minus{} 1) \plus{} n \plus{} 1$. Find $ f(5)$.

$ f(0)\equal{}1 \implies f(1)\equal{}1\plus{}1\plus{}1\equal{}3
\\ f(1) \implies f(2)\equal{}3\plus{}2\plus{}1\equal{}6
\\ f(2) \implies f(3)\equal{}6\plus{}3\plus{}1\equal{}10
\\ f(3) \implies f(4)\equal{}10\plus{}4\plus{}1\equal{}15
\\ f(4) \implies f(5)\equal{}15\plus{}5\plus{}1\equal{}\boxed{21}$

We can use induction to show that $ f(n)$ is the $ n\plus{}1$th triangular number.


[hide]

First note that if $ a\equal{}b\equal{}c$ then it evaluates to $ \frac{1}{12}$.
We'll prove that this is indeed the minimum.

By $ \sqrt{\frac{x^2 \plus{} y^2 \plus{} z^2}{3}} \ge \frac{x\plus{}y\plus{}z}{3}$ we have $ x^2 \plus{} y^2 \plus{} z^2 \ge \frac{(x\plus{}y\plus{}z)^2}{3}$
and applying that here we get:

$ \frac{a^8}{(a^2 \plus{} b^2)^2} \plus{} \frac{b^8}{(b^2\plus{}c^2)^2} \plus{} \frac{c^8}{(c^2 \plus{} a^2)^2} \ge \frac{1}{3}\left(\frac{a^4}{a^2 \plus{} b^2} \plus{} \frac{b^4}{b^2 \plus{} c^2} \plus{} \frac{c^4}{c^2 \plus{} a^2}\right)^2$

Thus it's enough to prove that

$ \frac{1}{3}\left(\frac{a^4}{a^2 \plus{} b^2} \plus{} \frac{b^4}{b^2 \plus{} c^2} \plus{} \frac{c^4}{c^2 \plus{} a^2}\right)^2 \ge \frac{1}{12} \equal{} \frac{(ab \plus{} bc \plus{}ca)^2}{12}$

This is equivalent with

$ 2\frac{a^4}{a^2 \plus{} b^2} \plus{} 2\frac{b^4}{b^2 \plus{} c^2} \plus{} 2\frac{c^4}{c^2 \plus{} a^2} \ge ab \plus{} bc \plus{} ca$

Moreover $ \frac{a^4}{a^2 \plus{} b^2} \equal{} a^2 \minus{} \frac{a^2b^2}{a^2\plus{}b^2}$ and similarly for the other three. Thus we get the equivalent inequality

$ 2a^2 \plus{} 2b^2 \plus{} 2c^2 \ge ab \plus{} bc \plus{} ca \plus{} 2\frac{a^2b^2}{a^2 \plus{} b^2} \plus{} 2 \frac{b^2c^2}{b^2 \plus{} c^2} \plus{} 2\frac{c^2a^2}{c^2 \plus{} a^2}$

But now

$ \frac{a^2 \plus{} b^2}{2} \ge ab$ and

$ \frac{a^2 \plus{} b^2}{2} \ge 2\frac{a^2b^2}{a^2 \plus{} b^2} \Leftrightarrow (a^2 \plus{} b^2)^2 \ge 4a^2b^2 \Leftrightarrow (a^2 \plus{} b^2 \minus{} 2ab)(a^2 \plus{} b^2 \plus{} 2ab) \ge 0$

Thus $ a^2 \plus{} b^2 \ge ab \plus{} 2\frac{a^2b^2}{a^2 \plus{} b^2}$ and we get the desired inequality by doing the same for $ b,c$ and $ c,a$ and adding up.

[/hide]


calculus
integration
calculus computations

solve  :lol:

$ \int_0^1\;\sec^{\minus{}1}\,(1\plus{}\sec\,x)\;\textbf dx$

put $ {1\plus{}secx\equal{}secz}$ and solve it by integration by parts...

[OT]For misan: your avatar is very nice!!! :lol:   :rotfl: [/OT]


A man named Juan has 3 rectangular solids, each having volume 128. Two of the faces of one solid have areas 4 and 32. Two faces of another solid have areas 64 and 16. Finally, two faces of the last solid have areas 8 and 32. What is the minimum possible exposed surface area of the tallest tower Juan can construct by stacking his solids one on top of the other, face to face? (Assume that the base of the tower is not exposed.)

well if you know ab, and ac, you can find bc (xy represents a face area) by doing a^2b^2c^2/(ab*ac), so from that then you can find a,b,c for the three cubes by plugging abc/(ab) or /(bc) or /(ac) (its pretty straightforward)

anyways, then you know the side lengths are (1,4,32) for cube 1 (2,8,8) for cube 2 and (2,4,16) from cube three.

if you want tallest, then 32,8, and 16 have to be the heights.

minimum surface area means that the largest base will go on bottom, second largest in middle, thrid largest at top (so like a pyramid shape when the three cubes are stacked)

so put the 8 height cube on bottom with base area of 16, 16 height in middle with base area of 8, and 32 height on top with base area of 4.

So total surface area that is not exposed is 16+2(8)+2(4)=40, multiply by 2 for the others because its an area of 8 for first and second cube when stacked on top of eachother.

The total surface area of the 3 cubes is 728, so 728-40=688.  I probalby made a mistake through all that though


obama ftw,
but i supported clinton in the primaries

note: i don't know if this has already been set up, if yes, then just move it to there
note number 2!!!!: give reasons, not just "he's stupid" or anything like that

i feel really bad for nader,
but he supported bush for the war

and that's what made people hate him.

it must be depression knowing you wasted 20 years of your life trying to run for president.  :(

1) well he didn't JUST run for president in 20 years, cheer up

2) it's Nader

Obama has an age older than his IQ. I think we will have the Republican win by a long shot.
I wish Hillary was the democratic candidate...
Have you heard people like Rush Limbaugh or Hillary? They are like "SUPPORT OUR OWN PARTY!!!" but that is not necessarily saying "Vote for the person best suited for president." Also, people who were democrats/republicans went to the republican/democratic convention and voted for the person whom the democrats/republicans could beat easily, not who would lead this country best. Now their age is TWICE their IQ!

While I don't necessarily disagree with some of your conclusions, I don't think you have any idea what you're talking about.

actually no I don't; dad just tells me what he makes of it and forbids me to read the news. :(
I think I should start to be a very naughty boy soon...


complex analysis
complex analysis unsolved

$ K^{p,q} :$ faisceau des germes de formes différentielles de type $ (p,q)$ à coefficients distributions. Je ne sais pas ce concept? Aide-moi

Posts in the non-country-specific forums should be in English.

http://en.wikipedia.org/wiki/Dolbeault_cohomology

this might help? it looks like the general idea is simply a generalization of de Rham cohomology to the complex setting -- now one considers sheaves of holomorphic differential forms ?

actually I don't know if you know english, so this might help more:

http://fr.wikipedia.org/wiki/Cohomologie_de_Dolbeault

je suis désolé de ne pas être capable de t'aider directement mais je ne sais presque rien de la géométrie complexe; peut-être que tu pourrais poser la question encore dans << advanced topics >>. si je peux expliquer ce que je comprends, il semble que ce n'est qu'une généralisation de la cohomologie de de Rham si on considère les formes holomorphiques.

(I'm not sure where the differential forms with coefficients as distributions comes from, though)

I read this concept in artile "Un theorème de dualité" of Jean - Pierre Serre. But I  do not understand this concept. Help me, please.

Which concept is "this concept"? Sheaf? Germ? Differential form? Distribution? De-Rham cohomology? There's some pretty heavy machinery here (and in your other post on fibred spaces). Without a [b]lot[/b] more information about what you already know and what it is exactly that you don't understand, I doubt that anyone here (or elsewhere) will be able to help you.

(I trust that you are thoroughly familiar with, at least, the concepts of sheaves, differential forms, and cohomology. If you do, it would help us if you explained what it is about $ K^{p,q}$ that you're not sure about. If you don't, you should not be attempting to read this paper by Serre.)

Serre signed the $ K^{p,q}$ is the germs of the differential forms of type (p, q) with the coefficient distribution. I really do not know what $ K^{p,q}$ is ? . You help me,please.

I think I'm giving up. Just in case:

[url=http://en.wikipedia.org/wiki/Distribution_(mathematics)]Distribution[/url]

[url=http://en.wikipedia.org/wiki/Complex_differential_form]Differential forms[/url] of type $ (p,q)$

[url=http://en.wikipedia.org/wiki/Germ_(mathematics)]Germs[/url]


inequalities
induction
calculus
derivative
inequalities unsolved

Let $ a_1,a_2,...,a_n\geq0$.Prove that:
$ (a_1^t\plus{}a_2^t\plus{}...\plus{}a_n^t)^{t\plus{}1}\geq(a_1^{t\plus{}1}\plus{}a_2^{t\plus{}1}\plus{}...\plus{}a_n^{t\plus{}1})^t$,
where $ t,n$ are naturals.

Are you asking for a proof of the power mean inequality? :huh:

You could just look it up... it's Jensen on stuff that I don't quite remember.

EDIT: wait, darn there's no $ n$ and it's reversed :(

However, induction on $ n$ should work nicely... group $ a_{n}$ and $ a_{n \plus{} 1}$ together

You know $ \sqrt [t]{a_1^{t} \plus{} a_2^{t}}$ is decreasing in $ t$ by derivative testing. Then, if

$ \sqrt [t]{a_1^{t} \plus{} \ldots \plus{} a_n^{t}} \ge \sqrt [t \plus{} 1]{a_1^{t \plus{} 1} \plus{} \ldots \plus{} a_n^{t \plus{} 1}}$, we want to show

$ \sqrt [t]{a_1^{t} \plus{} \ldots \plus{} a_{n \plus{} 1}^{t}} \ge \sqrt [t \plus{} 1]{a_1^{t \plus{} 1} \plus{} \ldots \plus{} a_{n \plus{} 1}^{t \plus{} 1}} (*)$

But as $ \sqrt [t]{a_n^{t} \plus{} a_{n \plus{} 1}^{t}}\ge \sqrt [t \plus{} 1]{a_n^{t \plus{} 1} \plus{} a_{n \plus{} 1}^{t \plus{} 1}}$, we have the LHS of $ (*)$ is at least

$ \sqrt [t]{a_1^{t} \plus{} \ldots \plus{} a_{n \minus{} 1}^{t} \plus{} \left((a_n^{t \plus{} 1} \plus{} a_{n \plus{} 1}^{t \plus{} 1})^{\frac {1}{t \plus{} 1}}\right)^t}$

which by the inductive hypothesis is at least

$ \sqrt [t \plus{} 1]{a_1^{t \plus{} 1} \plus{} \ldots \plus{} a_{n \plus{} 1}^{t \plus{} 1}}$ as desired.

(note n=1 is trivial)


I realized today that I had done the entirety of number 5 correctly, except I had used the wrong formula for the area of a triangle: I used ab sin C insetad of 0.5ab sin C.  So my answer came out to exactly twice the correct answer (1004/671007), presumably without any other mistakes made.  I would say that the error was simply a memory lapse, so I'm rather disappointed.  Is there any chance that a grader might overlook that and still give full credit, if the entire solution was correct except for that one recollection of the formula?

I figured.  I just want to bash my head against the wall though!  What a dumb thing to make a mistake on!


We define a function $f$ on the positive integers as follows:\[f(1)=1,f(3)=3\]\[f(2n)=f(n)\]\[f(4n+1)=2f(2n+1)-f(n)\]\[f(4n+3)=3f(2n+1)-2f(n)\]for all positive integers n
(i)  For $x=a_k.2^k+a_{k-1}.2^{k-1}+...+a_0$, where $a_j$ is either $0$ or $1$, evaluate $f(x)$
(ii) Determine the number of positive integers $n$, less than or equal to $1994$, for which $f(n)=n$

We can easily proove that if Ks=1+2+ 2^2  + 2^3 + ....+ 2^s then f(Ks)=Ks and this can give the solution for the second question.
For the first question,I considred the fact that every natural number can  be written as sums of terms like (2^a)*Ks.
This consideration leaded me to see that f behave as a sequence of the form Un=2Un-1  + b,but i really failed to reach the closed form for f(n).I have a real feeling that this way is to be considered .

He-he.
We are to notice that $f(x)=a_02^k+a_12^{k-1}+...+a_k$ (reversed representation of $x$).

That's all.


Is there a function $f: \mathbb N \rightarrow \mathbb N$ such that for every $n \in \mathbb N$, \[ f(f(n-1)) = f(n+1) - f(n) \, ?\]

[quote="X-man"]Is there a function $ f: \mathbb N \rightarrow \mathbb N$ such that for every $ n \in \mathbb N$ is
\[ f(f(n \minus{} 1)) \equal{} f(n \plus{} 1) \minus{} f(n)
\]
[/quote]
[hide="My solution"]
$ f(n)\in\mathbb N$ $ \implies$ $ f(n)\geq 1$ $ \forall n$
Then $ f(n\plus{}1)\equal{}f(n)\plus{}f(f(n\minus{}1))$ $ \forall n>1$ $ \implies$ $ f(n\plus{}1)>f(n)$ $ \forall n>1$
So $ f(n)\geq n\minus{}1$ $ \forall n$
So $ f(f(n))\geq n\minus{}2$ $ \forall n$

So $ f(n\plus{}1)\geq f(n)\plus{}n\minus{}3\geq n\minus{}1\plus{}n\minus{}3\equal{}2n\minus{}4$ $ \forall n>1$ and so $ f(8)\geq 10$ and so $ f(f(8))\geq f(10)$
But $ f(f(8))\equal{}f(10)\minus{}f(9)$ and so $ f(f(8))<f(10)$

Hence a contradiction.
Hence no such $ f(x)$ exists.

[/hide]


Hi i'm Rafael from Venezuela and here the IMO team have been pre-selected from a test we presented on April 27th, until now the members are:
Andrés Guzmán (OMCC 04, IBERO 04)
Sofía Taylor (OMCC 05)
and Rafael Guédez (me) :D  (IBERO 05)

We don't know if there is going to be another member joining us, but if there is a chance i guess it might be on may 27th, that day we'll have a test, the problem is that i was told that the last day for registration of contestants was on may 15th so i'm guessing that if it is true then only the 3 uf us will be there, we hope to see you there!

[quote="rafaelguedez"]oops i wanted to post this on the forum about imo 2006, i made a mistake posting this in here :blush: i'm a new member so i'll post this up there in the "imo 2006 slovenia" forum[/quote]

What about posting this on the Spanish Communities?????

Hi, during IMO 2002 I had a brief encounter with the venezuelan team (there were five of them at that time) and I am quite surprised to see that there are now only four  :o

Anyway, I remember because Venezuela had the exact same score that year as our team : 58/252  :) (a shared position 52 out of 84), so that also means that the average Venezuelan is better than the average Belgian  :roll:


trigonometry
calculus
calculus computations

Please prove or disprove that this serie converges or diverges:
\[ \sum_{n\equal{}1}^{\infty}\sin\Big(\frac{1}{n}\Big)\]

The limit comparison test is the workhorse for things like this.

How does $ \sin\left(\frac{1}n\right)$ behave for large $ n?$

Hint: what does $ \sin\theta$ look like for small $ \theta?$

[quote="Kent Merryfield"]The limit comparison test is the workhorse for things like this.

How does $ \sin\left(\frac{1}{n}\right)$ behave for large $ n?$

Hint: what does $ \sin\theta$ look like for small $ \theta?$[/quote]

Thx a lot... great hint.. cya :)


Ok. Im in the olymp problemsolving class, and being that it starts in two days, Im kind of interested in why there is still no forum for it. Is this supposed to be, or am I just special like that?

Heh. Yeah I see it now. Muchos Gracias :)


AMC
AIME
geometry
3D geometry
icosahedron
rectangle
algebra

I figured that we could have a marathon where every one asks and posts a amc10/12/aime level question (real or made up).  Try to make sure that it fits the level amc10/12/aime.

Problem 1:Every card in a deck has a picture of one shape - circle, square, or triangle, which is painted in one of the three colors - red, blue, or green. Furthermore, each color is applied in one of three shades - light, medium, or dark. The deck has 27 cards, with every shape-color-shade combination represented. A set of three cards from the deck is called complementary if all of the following statements are true:

i. Either each of the three cards has a different shape or all three of the card have the same shape.

ii. Either each of the three cards has a different color or all three of the cards have the same color.

iii. Either each of the three cards has a different shade or all three of the cards have the same shade.

How many different complementary three-card sets are there?

[hide="Solution"]
Start with the case where the all the provisions of the first part of (i), (ii), and (iii) are fulfilled. Initially we may choose any of the 27 cards, but then we are forced down to 8 to pick from and then there is only 1 left that remains to be chosen. However, any particular selection of cards can be chosen 3 different ways, so $ \frac {27 \cdot 8}{3} \equal{} 72$ ways for this case.

Now, proceed to the case where only two of provisions of the first part of the given statements are fulfilled. Initially we may choose any of the 27 cards, but then we are forced down to 4 cards to pick from and then the last card gives us one option. However, any particular selection of cards can be chosen 3 different ways, so $ \frac {27 \cdot 4}{3} \equal{} 36$ ways for this case.

Next case: only one of the provisions of the first part of the given statements is fulfilled. Initially, we may choose any of the 27 cars, but then we are forced into choosing one of two possible cubes, and then the last cube is determined for us. However, any particular selection may again be chosen 3 different ways, so $ \frac {27 \cdot 2}{3} \equal{} 18$ ways for this case.

Note that the case with all cards having the same shape, color, and shade is impossible. We have shown $ 72 \plus{} 36 \plus{} 18 \equal{} 126$ ways to achieve the given conditions.
[/hide]

Problem: A cone has circular base radius 1, and vertex a height 3 directly above the center of the circle. A cube has four vertices in the base and four on the sloping sides. What length is a side of the cube?

[hide="Solution"]We can think of that problem as finding the side length of a square inscribed in a circle of radius 1. Simple arithmetic (blah pythagorean theorem) gives us $ \boxed{\sqrt {2}}$.[/hide]

[size=150]Problem 3[/size]
[size=59]forgive me---my ignorance limits me to nothing harder than:[/size]

Consider an icosahedron (regular polyhedron with 20 triangular faces). The vertices are labeled incrementally with integers, starting with $ 1$. In how many ways can we connect two vertices of this polyhedron, provided that they do not belong to the same face?

V + F - E = 2
E - V = 18 ( F = 20 )
Thats as far as I got... Is there a trick to find the # of vertices? I recall there being 1...

[quote="undefined117"][hide="Solution"]We can think of that problem as finding the side length of a square inscribed in a circle of radius 1. Simple arithmetic (blah pythagorean theorem) gives us $ \boxed{\sqrt {2}}$.[/hide]

[/quote]

Incorrect.

[quote="McDutchy"]Incorrect.[/quote]

Ah whoops sorry---I assumed the vertices touched the circle, which is impossible. I shouldn't do these problems at night. :D

[hide="hopefully CORRECT solution to Dutchy`s problem :oops: "]
[asy]size(140);
real S=(3*sqrt(2))/(3+sqrt(2));
real SD=S*sqrt(2);
D((1-SD/2,0)--(1+SD/2,0)--(1+SD/2,S)--(1-SD/2,S)--cycle,red);
D((0,0)--(2,0)--(1,3)--cycle); D((1,3)--(1,0),dashed);
MP("\frac{1}{\sqrt{2}}x",(1,S),NE); MP("x",(1+SD/2,S/2),W); MP("\frac{x}{3}",(2,0),SW);[/asy]
Side view of cone; the longer side of the red rectangle is of course, the diagonal of the cube face.

Then $ 1\minus{}\frac{x}{3}\equal{}\frac{x}{\sqrt{2}}$, and solving gives us $ \boxed{x\equal{}\frac{3\sqrt{2}}{3\plus{}\sqrt{2}}}$.
[/hide]

[quote="undefined117"]
Consider an icosahedron (regular polyhedron with 20 triangular faces). The vertices are labeled incrementally with integers, starting with $ 1$. In how many ways can we connect two vertices of this polyhedron, provided that they do not belong to the same face?[/quote]

[hide="Continuing..."]Yes there is a trick. Recall that a triangle has three vertices. So there are $ 20 \cdot 3 \equal{} 60$ vertices... but also recall that 5 triangles meet at a vertex. Therefore, divide by 5 to correct for overcounting to get $ 12$ vertices.

Now, since the faces are triangles, a line segment connecting two vertices belong to the same face iff they are on the same edge. So all we have to do in the end is just subtract the number of edges.

Now, on to the meat. There are 12 vertices, and consequently $ \binom {12}2 \equal{} 66$ ways to choose two of them to connect.

Now, subtract the number of edges. Recall that a triangle has three edges. So there are $ 20 \cdot 3 \equal{} 60$ edges... but also recall that two triangles meet at an edge. So divide by 2 to get $ 30$ edges. This is what we need to to subtract, so our final answer is...

\[ \boxed{36}\][/hide]

[b]New problem[/b]
I meant to solve the equation $ x^2\plus{}bx\plus{}c\equal{}0$, but I accidentally solved $ x^2\plus{}cx\plus{}b\equal{}0$ and got the roots $ 3\sqrt 2\plus{}1$ and $ \minus{}3\sqrt 2\plus{}1$. What were the roots of the original equation?

[hide="i think this is right..."]
By vietas formula for sum of roots, we know that c=-2, and by the formula for product of roots we have...1-36=-35?  I think...

So b=-35, and the original equation is x^2-35x-2.  Quadratic formula yields [35+-sqrt(1233)]/(2)

[/hide]
I totally think this is wrong, so I won't post another prob...

by the quadratic formula we have : $ \frac {-c + \sqrt {c^2 -4b}}{2} = 1 \+- 3\sqrt {2}$
so c = -2
and $ \sqrt{c^2 - 4b} = 2 (3\sqrt{2}) \implies c^2 - 4b = 72 \implies  b = -17$
roots are $ 17 + \sqrt { 297} / 2$ and $ 17- \sqrt { 297} / 2$

[quote="mathemonster"]by the quadratic formula we have : $ \frac { \minus{} c \plus{} \sqrt {c^2 \minus{} 4b}}{2} \equal{} 1 \ \plus{} \minus{} 3\sqrt {2}$
so c = -2
and $ \sqrt {c^2 \minus{} 4b} \equal{} 2 (3\sqrt {2}) \implies c^2 \minus{} 4b \equal{} 72 \implies b \equal{} \minus{} 17$
roots are $ 17 \plus{} (\sqrt { 297}) / 2$ and $ (17 \minus{} \sqrt { 297}) / 2$[/quote]

New problem:
consecutive  numbers beginning with 1 are written down, 1 number is removed, and the average is $ 35\frac{7}{17}$. Which number was removed?

[hide="The Uber-Slick Solution for my problem"]
Well, we already have the two roots $ r_1$ and $ r_2$ of the modified equation, so... multiply it out!
\begin{align*} (x - r_1)(x - r_2) & = (x - 3\sqrt 2 - 1)(x + 3\sqrt 2 - 1) \\
&= ((x - 1) - 3\sqrt 2)((x - 1) + 3\sqrt 2) \\
&= (x - 1)^2 - 18 \\
&= x^2 - 2x - 17
\end{align*}
Thus the original equation is obtained by swapping $ b$ and $ c$...
\[{ x^2 - 17x - 2 = 0 \implies x = \frac {17 \pm 3\sqrt {33}}2}
\]
[/hide]

[quote="maththemonster"]
New problem:
consecutive  numbers beginning with 1 are written down, 1 number is removed, and the average is $ 35\frac {7}{17}$. Which number was removed?[/quote]

Let $ n$ be the number of terms.
If the number removed was $ 1$, the minimum number of terms is $ 68$ because the average of 2-68 is 35. If the number removed was $ n$, then the minimum number of terms is $ 72$ because the average of 1-71 is 36.

Furthermore, looking at the denominator, one less than the number of terms must be divisible by 17 (since it is a prime number). The only multiple of 17 that is in the range 68-72 is 68, so there must have been 69 terms.
\[ 1 \plus{} 2 \plus{} \dots \plus{} 69 \equal{} 69\cdot 35 \equal{} 2415,
\]
but
\[ 68 \cdot (35 \plus{} 7/17) \equal{} 2408,
\]
so the number removed must have been $ 2415 \minus{} 2408 \equal{} \boxed 7$.

[b]New problem[/b]
The first 99 positive integers are written down in a line without spaces, so it is written in monospaced font like 123456789101112... Then, the sequence is reversed and placed directly below the first line, so the two lines go like this:

1234567890111213141516...
9998979695949392919089...

How many columns are there such the two digits belonging to that column are the same? For example, the first such column is the 9th column, where the two digits are both 9's.

[b]Note: It is the sequence of numbers that is reversed, not the sequence of digits. Notice that the second sequence of digits begins with 99989796... and not 99897969.[/b]

* Tentative
18?
In the icosahedron problem how did Yongyi know that 5 triangles meet at each vertice.

definition of icosahedron...use wikipedia.

--new problem--

Find the the product $ xy$ if

\[ x^2 \plus{} y^2 \equal{} 1 \]

and

\[ x^4 \plus{} y^4 \equal{} \frac{17}{18} \]

[hide="Source"]
HMMT
[/hide]

[hide]Squaring the first equation and subtracting the second gives $ 2(xy)^2 \equal{} \frac {1}{18}$. Now, divide by two and take the square root to get $ xy \equal{} \pm\frac {1}{6}$.[/hide]

Is there something that I overlooked to get both the positive and negative?

No, that's correct. In the future, please post a new problem after solving.

--new problem--

What is the 336th digit of $ \frac{5}{7}$? (Kind of easy, sorry guys.)

[hide] The digits of n/7 after the decimal point repeat in cycles of 6, for all positive numbers n (verify this). Hence, it is the sixth digit after the decimal point in 5/7, which is 5.[/hide]

New problem:

A point on a circle inscribed in a square is 1 and 2 units from the two closest sides of
the square. Find the area of the square. (HMMT)

[quote="Yongyi781"]

[b]New problem[/b]
The first 99 positive integers are written down in a line without spaces, so it is written in monospaced font like 123456789101112... Then, the sequence is reversed and placed directly below the first line, so the two lines go like this:

1234567890111213141516...
9998979695949392919089...

How many columns are there such the two digits belonging to that column are the same? For example, the first such column is the 9th column, where the two digits are both 9's.

[b]Note: It is the sequence of numbers that is reversed, not the sequence of digits. Notice that the second sequence of digits begins with 99989796... and not 99897969.[/b][/quote]

Solve Yongyi's old problem now.

Yongyi, please post the solution. I think we're all stumped.  :)

We can divide the columns into three parts:

Part 1: Row 1 is single-digit and Row 2 is double-digit. This is basically the first 10 columns.

Simple brute-forcing:
123456789...
999897969...
Yields only 1 column with a common digit.

Part 2: Row 1 is double-digit and Row 2 is single-digit. This is basically the last 10 columns.

More brute-forcing:
...596979899
...987654321
Yields NO columns with a common digit.

Part 3: Both rows are double-digit. It might be helpful to list a few numbers at the beginning and the end:
10111213141...8990919293949
59493929190...6151413121110

Looking at the even columns, we readily see that every group of 10 digits will have one column match (because the second column cycles through each of the 10 digits every 10 digits) EXCEPT the last 10 (see below). So there are 8 more columns that match.
01234567890123456789
99999888888888877777

Looking at the odd columns, we readily see that every group of 10 digits will have one column match (because the second column cycles through each of the 10 digits every 10 digits) EXCEPT the last 10 (see below). So there are 8 more columns that match.
1111111111222222222233333333334444444444....9999999999
5432109876543210987654321098765432109876....5432108642

Adding up, $ 1 \plus{} 8 \plus{} 8 \equal{} \boxed {17}$, as desired.

[b]New problem[/b]
What is the total number of non-congruent triangles that can be formed by joining three vertices of a regular dodecahdedron?


trigonometry
real analysis
real analysis unsolved

Prove that the product of the two real roots of $x^{4}+x^{3}-1 = 0$ is a root of $x^{6}+x^{4}+x^{3}-x^{2}-1 = 0$

I would do this:

Let $x_{1}$, $x_{2}$ the real roots and $a$ be the modulus of the complex root.
From Vieta`s formulae we get that
$x_{1}x_{2}a^{2}$=1, hence $x_{1}x_{2}=\frac{1}{a^{2}}$.
Thus you have to show that $\frac{1}{a^{2}}$ is a root of your equation of degree 6, i.e.
$1+a^{4}+a^{6}-a^{8}-a^{12}=0$.

Let $ae^{i\theta}$ be the complex root of the equation of degree 4.
You get that

$a^{3}(ae^{i\theta}+1)=e^{-3i\theta}$
solve the equation for real and imaginary parts and you get

$a^{4}\sin\theta=-\sin3\theta$
$a^{4}\cos\theta+a^{3}=\cos(3\theta)$
it follows that $a^{4}=-3+4\sin^{2}\theta$ or $\sin^{2}\theta=\frac{a^{4}+3}{4}$, now solve the other equation for $cos\theta$ and then use the trig, identity $sin^{2}\theta+cos^{2}\theta=1$ from which,after some calculations you should get your equation.


Ecuatia se scrie echivalent $(3^{x-1}-2^{x-1})(3^{x}+2^{x}+1)=0$ si are solutia unica $x=1$.


[quote="MoHaMeD"]I'd like to inform you that the international congress of mathematicians will be held this year in Madrid from 22 to 30 august :students are concerned too ,the official website is :http://www.icm2006.org .
I guess it might even be a chance for mathlinkers to get to know each other ,at least for those who are in Europe ,so what do u think ??[/quote]
what happen usually in similars cogresses?i have no idea!

usually there are well known mathematicians who give lectures about their recent results ,moreover the fact that students might have direct contact with those scholars is advatageous i guess,there are also some activities like a visit to Escorial .
I was informed about that event by a spaniard friend if u have specific questions check the website at first otherwise give me ur questions (smart ones of course) i'll ask her for u :)

[quote="marocleverness"]that's look very interesting,please how can i participate,apologize me coz my english is very poor and i can't undersant the whole of the sitweb contents.[/quote]

The ICM took place every four years, this is the biggest mathematical meeting. The great mathematicians are invited to give lectures on their most recent achievements. The most important event during this congress is certainly the famous "Field medal" which is decerned to (at most) four mathematician under forty who proved some extremely hard theorems or theories. In 2002, the Field medal was attributed to Lafforgue (France) and Voevodsky (Russia).
Students are young researchers can participate to this congress and meet the great mathematicians. This year, I am almost sure that the Field medal will be given to Terence Tao, if you do not know him, I will just inform you that he took part to IMO three times (1986, 1987, 1988), he won bronze, silver and gold medal, but he was at that time only 11, 12 and 13 years old. He got his PhD at age 21!!


real analysis
parameterization
logarithms
probability and stats

Let $ (X_n)_{n \ge 1}$ be independent and exponentially distribuited with parameter 1. Show
$ P(\limsup_{n\minus{}>\infty}\frac{X_n}{\ln{n}}\equal{}1)\equal{}1$.

Use Borel-Cantelli:
\[ \sum_n P(X_n\geq (1 \minus{} \epsilon)\ln(n)) \equal{} \sum_n n^{\minus{}(1 \minus{} \epsilon)} \equal{} \infty
\]

\[ \sum_n P(X_n\geq (1 \plus{} \epsilon)\ln(n))\equal{}\sum_n n^{\minus{}(1 \plus{} \epsilon)} < \infty
\]
This means exactly what you need: $ \limsup \frac {X_n}{\ln(n)} \equal{} 1 \ a.e.$


Lines L and M are given by the parametric equations

L: x = 2 + s, y = -s, z = 2 - s
M: x = -1 - 2t, y = t, z = 2 + 3t

(i) Show that L and M do not intersect

(ii) Calculate the shortest distance between L and M

[quote="archimedes1"][hide]Um, I've never worked with parametric equations before, but it doesn't look too intimidating...
We eliminate s from the first line by taking the first equation times two and adding it to the rest.
$2x+y+z=6$
We eliminate t from the second line by taking the first equation times two and adding it to the rest.
$2x+y+z=4$

So that's the answer for a).
[/hide][/quote]

wouldn't those be plains, not lines?

[hide="a"]
y=2-x
z=2+y

y=-.5(x+1)
z=3y+2

if there were a point x,y,z that was on both lines, then
2+y=z=2+3y -> y=0
y=-.5(x+1) means that x=-1
0= 2-(-1)
0=3

[/hide]

[quote="0714446459923"]
wouldn't those be plains, not lines?
[quote]
I guess so.  But does it matter for the first part?  We're still relating x y and z to some number and 4 is never equal to 6.

By the definition of intersection:

[hide]$(x_{L}, y_{L}, z_{L}) \equiv (x_{M}, y_{m}, z_{m}) \implies x_{L}=x_{m}, \; y_{L}=y_{M}, \; z_{L}=z_{M}$

(I)  $2+s=-1-2t$
(II) $-s = t$
(III) $2-s=2+3t$

By (II), (I) becomes $2+s=-1+2s \implies s=3 \implies t=-3$

However, by (II), (III) becomes $2+t=2+3t \implies t=0 \implies s=0$

This leads to a contradiction, and all three equations cannot hold true simultaneously.

Hence, the lines will never intersect.[/hide]

[hide="ii"]
Since the lines do not intersect, they are skew lines, and there must be two parallel planes such that each plane contains one of the lines. Written in vector form, the two lines are
\[<2, 0, 2>+s<1,-1, 1>, <-1, 0, 2>+t<-2, 1, 3>. \]
Let $S$ and $T$ denote the vectors parallel to the two lines. To find equations for the planes, we take their cross product
\[S\times T = \left|\begin{matrix}i&j&k\\ 1&-1&-1\\-2&1&3 \end{matrix}\right| = <-2,-1,-1> \]
Using the two points $(2, 0, 2)$ and $(1,-1,-1)$, we can write the two planes in point-norm form as
\[-2x-y-z+6 = 0,\\-2x-y-z = 0 \]
The distance between the two is then
\[d = \frac{|6-0|}{\sqrt{(-2)^{2}+(-1)^{2}+(-1)^{2}}}= \sqrt{6}\]
[/hide]

[quote="archimedes1"][hide]Um, I've never worked with parametric equations before, but it doesn't look too intimidating...
We eliminate s from the first line by taking the first equation times two and adding it to the rest.
$2x+y+z=6$
We eliminate t from the second line by taking the first equation times two and adding it to the rest.
$2x+y+z=4$

So that's the answer for a).
[/hide][/quote]

This is certainly not the way that I would solve the problem, but it is certainly correct. Let me clarify your argument:

Since those planes do not intersect, and the lines are in those planes, the lines must not intersect. This is a nice argument; it does not merit a "spam" rating.


algorithm
function
Harvard
college
ceiling function
algebra
floor function

I have posed this problem here before, but noone did post a solution. I then recall that I somehow managed to show it (I hope I managed), but now I'm stuck again when I try to solve it (I didn't write down the approach I did before), so then I suspect that I maybe didn't prove it correctly before. Anyway, here it is:

Let A = {1,2,...,2002}

Let B be a subset of A such that:
1) There are no two consecutive elements in B
2) There are no two elements in B such that one of them divides the other.

What is the size (amount of elements) of the largest subset B with those properties above?
I get the answer to be 834, by choosing the following subset: B = {2001,1999,...,669,666,...,334}, that is I choose first as many odd numbers as possible (starting with the largest one) and then switch over to even numbers and choose as many as possible. This algorithm seem to generate maximal subsets for smaller sets A (I have verified it for smaller sets A), but I find it hard to prove that the maximal subset must have that many elements.

Very thankful for a solution or some new insights in how to prove it.

Anyone has insights into this problem?

I haven't been able to solve it before, and neither has the new idea helped me:
The set I constructed has the maximal sum of all those subsets B with maximal size.

That didn't help me, but maybe I've missed out on something? Any thoughts would be welcome!

Hmm. I'm wondering if anyone has tried this problem? I see that many people have seen it...

It would be really interesting to see a solution for this one. And for the generalized version of it.

I have no good ideas at the moment :S

True :)

Do you have any idea on how to find the correct answer? :S

it would be nice to study
$f(n)=2n-[\sqrt{n}]^2$ for the subsets $A \subset N$
so that $f(A) \subseteq A$

there is some stuff that resembles to this problem

Hmm. What is so special with that function? No two values of it divide each other or what?

Hm. I don't know what [x] is denoting. Ceiling function? Floor function? A combination of the two?

If it's any of those, then look at f(4) = 2*4 - 4 = 4 and f(16) = 2*16 - 16 = 16. f(4) divides f(16)...

So it probably means something else.

What do you mean? Does [ ] denote any of those I mentioned? Then the function isn't so good for our problem (at least as far as I can see).

I've been thinking about this problem a lot lately as well (I noticed you're Swedish - did you get it from fråga lund om matematik too?), but I haven't found the answer. I too found that first taking as many odd numbers as possible from the "top", and then as many even numbers as possible, gives a good result.

I've written a program to brute force subsets of {1,...,n} for all n up to 104 (after that, it was too slow to continue). For these n, the above kind of subset gives the optimal number of elements except when n=52 and n=76. For these, the optimal subsets have one more element than the other ones. These were:

n: 52
expected: 21 (taking odd numbers, then even numbers, as above)
optimal subsets: 22 members
8 10 12 14 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 52
6 8 10 14 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 52

n: 76
expected: 31
optimal subsets: 32 members
12 14 16 18 20 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76
10 12 14 16 18 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76
8 12 14 18 20 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76
8 10 12 14 18 22 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 59 61 63 65 67 69 71 73 76

maybe if you post your solution(the algo in the language of choice) we can make some optimization and make it
print the solution for the problem faster

I first saw the problem on a homepage of a Ph.D student in Harvard, Gabriel Carroll. He had a collection of math problems and this one was rated as hard. I haven't been able to find that homepage again (it seems that it's erased). So I asked this question to 'Fråga Lund' forum, but you saw the answer :(

It's interesting that all those sets your algorithm produced are generated quite simply. By substracting a 2 or a 3 from every number. Similar to the one I got before (by taking odd numbers first). But it seems to be very hard to prove that it's the optimal way. And it seems equally hard to just prove, without constructing an algorithm, that the maximal set must have so and so many numbers. I have given up on the problem for a while now...hopefully something will happen soon.

Also, you can send your algorithm if you want as spx2 wrote.

The program I mentioned was simply a DFS on all "B"-sets. I kept a list of "candidate members" of the set, initially the whole range {2,...,n}, and then in succession chose members from it, sieving out candidates that are multiples of it. As an optimization, I wouldn't generate sets with less members than the expected value. The code was ugly, so I'd rather not post it.

[quote="Rust"]your B is not optimal (834 members), for example B={2002,2000,1998,1996, 1993,1991,...,665, 662,660,...,332} have 835 members.[/quote]
This, by the way, is wrong. 1001 divides 2002, and both are in the set.

So... the 835-member set is still a possibility! :) The algorithm for constructing it wasn't valid for all n, but maybe it is for most n - maybe for all odd ones?

Eighty there is no expected size of a set in this problem so dont expect one in the algorithm.
2nd thing please write the algorithm here dont matter if its ugly i want to see it

I used the word "expected" in the exact same sense as in my first post, ie as the size of the "B-set" generated by taking as many odd numbers "from the top" as possible, and after that, as many even numbers as possible.

I'll clean up the code a little, and post it shortly. My code can be found on: [url=http://paste.axpr.net/?show=395]http://paste.axpr.net/?show=395[/url]. The possible bottlenecks are the excessive copying of arrays (there's probably nothing that can be done about that), and the extreme recursion.


pigeonhole principle
combinatorics unsolved
combinatorics

Hello .
Given $ 2^{2n\minus{}1} \plus{}1$ odd numbers from $ {{2^{2n} \plus{}1 , 2^{2n} \plus{}2 , 2^{2n} \plus{}3 , ... ,2^{3n}}}$ . Prove that there exist two number like $ a,b$ such that $ a^2$ not divisible by $ b$ and $ b^2$ not divisible by $ a$ .

I just know that `Pigeonhole principle` helps and no more !!! Plz help me . :blush:


[quote="236factorial"]How many incongruent trapezoids ABCD with side AB parallel to side CD have sides of lengths AB=16, BC=13, CD=10, and DA=6?

a) 0
b) 1
c) 2
d) 3
d) 4[/quote]

[hide]is it a) [size=200]0[/size][/hide]

[hide]I got $0$ too. Draw $2$ altitudes from $D$ and $C$ to side $AB$. Let the intersection of the altitude from $D$ to $AB$ be $E$ and the one from $C$ to $AB$ be $F$. Let the length of the altitude be $h$. Let $AE$ be $x$, which makes $BF$ $6-x$. We have $2$ right triangles now, $AED$ and $BFC$. These give the equations $x^2+h^2=6^2$ and $(6-x)^2+h^2=13^2$. Substitute the first equation into the second and we get $x=-11\frac{1}{12}$, which cannot be possible, so the answer is $0$.[/hide]


function
integration
trigonometry
Euler
real analysis
real analysis unsolved

Prove that $\forall 0<a<1$ we have : $\int_{0}^{1}\frac{x^{a-1}}{(1-x)^{a}}dx=\frac{\pi}{\sin\pi a}$

[quote="Extremal"]it called Euler refliction formula http://en.wikipedia.org/wiki/Reflection_formula[/quote]

Yes, and here is another source: [url]http://planetmath.org/?from=objects&name=EulerReflectionFormula&id=8539&total=0&offset=0&op=getobj&watch=add[/url]


linear algebra
matrix
combinatorics proposed
combinatorics

Given a $10*10$ board, we write the numbers $1,2,3,4....,100$ in its squares, in such a way that we first put number 1, and then we put the next number (2) in an adjacent square, and continue like this. find the least sum of the numbers among a main diagonal.

The answer is $81+54 = 135$. We divide the board in two pieces across a main diagonal (without counting that main diagonal), and start putting the numbers in increasing order acording to what we can do. Now it's easy to check that to cross from one piece to another while you are putting the numbers you have to walk through the main diagonal. So if we take the biggest number from a main diagonal, then this number will be bigger than all the other numbers in one of the pieces (because when we put that number, since it's the biggest in the diagonal, it is the last number we put there, so at least one of the two pieces must be completely filled with numbers). Hence that biggest number is at least $54$ (it's bigger than all the 45 numbers from one piece, and the others nine numbers of the diagonal). The sum of the other nine numbers is at least $1+3+\dots +17=81$ (they are either all odd or all even, it's better when they are odd), and therefor the sum of the numbers of a main diagonal is at least $135$. It is very easy to find (but not to post !!) an example, starting in one corner...

Can you post the matrix with the minimum sum on the main diagonal for the whole 10*10 matrix? Or at least for a 4*4 one?

I believe that what you calculated is indeed a lower bound, but it is not acheivable.


I've been trying to find old AMC 10 contests to practice with, but I can't find the AMC 10 (B) contests for 2000 and 2001. My 21st century CD from http://www.unl.edu/amc only has one set of problems from 2001, so I came to the conclusion that there [i]is[/i] no AMC 10 (B) for those years. Can someone please confirm this? Thanks.

I think those might have been the first years of the AMC 10 since there was the AHSME before that.  So it is very well possible that there was no B edition for those years.

There was only one edition for 2001's AMC 10 and 12. At that time, they didn't split the 2 contests.

The first AMC 10 contest was in the year 2000.  Since 2002, both the AMC 10 and the AMC 12 have been offered in A-date and B-date versions, the B-date version occurring 15 days after the A-date version with a new set of questions.  There is no B-date version for the AMC 10 in the years 2000 and 2001.  Likewise, there is no B-date version of the AMC 12, or the AHSME, prior to 2002.

Steve Dunbar
MAA Director, American Mathematics Competitions


[hide="Hint for (a)"]Express $\cot A$ and $\cot B$ and prove $\cot(A+B)=-\cot\alpha$[/hide]

A)

[hide]\[ \begin{array}{l} tg(A + B) = \frac{{tgA + tgB}}{{1 - tgAtgB}} = \frac{{\frac{{\sin \alpha }}{{a - \cos \alpha }} + \frac{{a\sin \alpha }}{{1 - a\cos \alpha }}}}{{1 - \frac{{\sin \alpha }}{{a - \cos \alpha }}.\frac{{a\sin \alpha }}{{1 - a\cos \alpha }}}} \\ = \frac{{\frac{{\sin \alpha (1 - a\cos \alpha ) + a\sin \alpha (a - \cos \alpha )}}{{(a - \cos \alpha )(1 - a\cos \alpha )}}}}{{\frac{{(a - \cos \alpha )(1 - a\cos \alpha ) - a\sin ^2 \alpha }}{{(a - \cos \alpha )(1 - a\cos \alpha )}}}} \\ = \frac{{\sin \alpha (1 - a\cos \alpha + a^2 - a\cos \alpha )}}{{a - a^2 \cos \alpha - \cos \alpha + a\cos ^2 \alpha - a\sin ^2 \alpha }} \\ = \frac{{\sin \alpha (1 - 2a\cos \alpha + a^2 )}}{{a - a^2 \cos \alpha - \cos \alpha + a(1 - \sin ^2 \alpha ) - a\sin ^2 \alpha }} \\ = \frac{{\sin \alpha (1 - 2a\cos \alpha + a^2 )}}{{2a(1 - \sin ^2 \alpha ) - a^2 \cos \alpha - \cos \alpha }} \\ = \frac{{\sin \alpha (1 - 2a\cos \alpha + a^2 )}}{{2a\cos ^2 \alpha - a^2 \cos \alpha - \cos \alpha }} \\ = - \frac{{\sin \alpha (1 - a\cos \alpha + a^2 - a\cos \alpha )}}{{ - 2a\cos ^2 \alpha + a^2 \cos \alpha + \cos \alpha }} \\ = - \frac{{\sin \alpha (1 - 2a\cos \alpha + a^2 )}}{{\cos \alpha (1 - 2a\cos \alpha + a^2 )}} = - \frac{{\sin \alpha }}{{\cos \alpha }} = - tg\alpha \\ \end{array}  \]
[/hide]


A four-digit number is created by using each of the digits 4, 5, 8 and 9 exactly once. What is the probability that the number will be a multiple of 4? Express your answer as a common fraction.

[hide="Slightly more in detailed solution"]

There are 12 different combinations for the last two digits. This is because there are 4 options to choose the the third digit and 3 numbers remaining to choose the fourth digit. Of those options, only 48 and 84 are divisible by 4, so there are 2 successful outcomes out of a total of 12, so $\frac{2}{12}$ = $\frac{1}{6}$ [/hide]

Why isn't it 24 different combinations? 4x3x2?

He's only looking at the last two digits. 4x3=12.


because we are pressed for time as i won't have time to come on after break, i will start a tourny with 8 people
these people must be dedicated to finishing the tourny
anyone who does not meet the deadline, usually 1, maybe 2 days, will be eliminated
results can be pmed to me
i will not send out reminder pms due to the short amount of time
players must look at this thread for upcoming games
if your opponent does not respond to a pm, then pm me before the deadline and tell me

RULES
single-elimination
count-downs
best of 3 (you do not need to play 3 games if one person is already ahead 2 games to 0)

i hope to get this tourney started today and have it end sometime over the weekend
good luck to all that join!

participants

1. Mewto55555
2. AIME15
3. jjx1
4.
5.
6.
7.
8.

i will repeatedly edit this post
for list of participants, always look here

this is tourney is officially closed due to lack of people


KoL is a fun game. [url=http://www.kingdomofloathing.com](link to the game)[/url] I am addicted to it. Lv.6 Seal clubber, iamstupid. Create accounts!!! :D

I'll post my stats tomorrow, but I'm better than you :P

I know, Lv.7+ Sauceror. (maybe Pastamancer???)

oneequalstwo
Level 8
Vermicelli Enchanter

Pastamancer

Garyzx (#1403997)
Level 21
Pastamancer

Ascensions:	4

Clan: KOL Addicts

[quote="nick42"]hah, i'm still a noob at this game. despite being a lvl 17 turtle tamer.  still debating on whether or not to ascend. :D

TIGERFAN08
ASIAN DYNASTY[/quote]

ascend. even though I am noobiER, i think it is better to ascend. you will not lose much.

I ascended once, as a level 11 seal clubber :P
Then, it started to get boring, and my account was eventually deleted (I was like a level 12 pastamancer at that time)
The only thing I was really proud of in that game was that I got both trophies for [hide]eating 50 lucky eggs (or whatever you call them) [/hide] and for [hide]drinking 30 white canadians[/hide]
(spoiled for those who want to figure out how to get the trophies themselves)

I started this game today, and it is such a strange game...

Anyway, I'm an_adventurer_is_me, a level 2 turtle tamer.

me level 8!!! :D me=iamstupid for the countless times i have made reading errors.  :wallbash_red:

[quote="1=2"]I've almost cleared the Cyrpt, but that dragon pwns me every time, even when I use a cookie blob of Crimbo.[/quote]

Use free buffs. [url]http://www.mrkinks.com/buffme/index.pl[/url]

[quote="herefishyfishy1"]Isn't that cheating...?[/quote]

No, it's not.

blah i started this game a few days ago

this IS addicting but i cant play as much as i want because of fullness/drunkeness/etc.

how can you prevent this? (maybe a cheat code? xD)

1=2, how many days did it take for you to reach to level 8?

[quote="1=2"]Eat things that give the maximum amount of adventures.

And Once you deal with Azael in h311, you get an iron kidney, so it takes twice as much to get drunk.[/quote]

15 to 20 drunkeness.

level 9!!!

(Crimbo Cafe gives tons of adventures)

Level 11. I hate this quest for the Holy whatever! Has anyone used the Spirit of Crimbo? I have the pair of ragged claws, the mutant couture outfit, and the burrowgrub hive. I also got the rage gland from the Crimbonation.


calculus
integration
trigonometry
calculus computations

Can someone help me evaluate this integral--

Integral [tan^-1 {sinx + cosx} ]dx

i.e. tan inverse of (sinx+cosx)

$\int \tan^{-1}(\cos x + \sin x) dx$

It doesn't seem to me like the integral is expressible via normal means.

I did it on the integrator on Mathworld. It spat some polylogarithms back at me in a huge complicated expression.


[youtube]dmt2_wyDKJI[/youtube]

I just realized how good Bush's reflexes are...

this oe radio show said the followig about this incident
"The lame duck presidet, well, ducked."

tinytim also posted this on my blog

please read the last line of my sigature