Did anyone see the David Blaine stunt?  He was attempting to hold his breath for 9 minutes.  I think David Blaine is crazy. I would NEVER attempt that in my life.  Is there any other stunt you would like to see him do?

YEA MAN I WATCHED THAT

I actually didn't find it all that interesting to just see him in a big sphere of water - I liked the clips of card tricks better

I liked the basketball-card trick that he did, that was nice. Other bizarre things he did are ... well bizarre. But, this time he just went overboard, he has liver failure.

An amazing stunt would be:
.... spending a day in Bagdad...


integration
trigonometry
function
calculus
calculus computations

(1) Prove that $ \sum_{n\equal{}1}^{\infty} \frac{1}{x^2\plus{}n^4}$ converges and the limit value $ f(x)$ is continuous for $ \forall{x}\in{\mathbb{R}}$.

(2) Prove that $ \int_0^{\infty} f(x)\ dx$ converges and find the value.

(3) Compute $ \int_0^{\infty}f(x)\cos bx\,dx.$

(Note that this includes (2) as the $ b\equal{}0$ case.)

Ah, sorry about that. (3) doesn't work out as nicely as I thought.

As for (1): Each term of that series is bounded by $ \frac1{n^4}$ and since $ \sum\frac1{n^4}$ converges, then the series of functions converges absolutely and uniformly by the Weierstrass M test. That makes $ f(x)$ continuous on $ \mathbb{R}$ as the uniform limit of a series of continuous functions.

For (2), we will interchange the sum and the integral. (This can be justified in several different ways. The laziest justification is to note that the terms of the series are nonnegative so we may quote the Monotone Convergence Theorem.)

$ \int_0^{\infty}f(x)\,dx\equal{}\int_0^{\infty}\sum_{n\equal{}1}^{\infty}\frac1{x^2\plus{}n^4}\,dx$

$ \equal{}\sum_{n\equal{}1}^{\infty}\int_0^{\infty}\frac{1}{x^2\plus{}n^4}\,dx$

$ \equal{}\sum_{n\equal{}1}^{\infty}\frac{\pi}{2n^2}\equal{}\frac{\pi}2\cdot\frac{\pi^2}{6}\equal{}\boxed{\frac{\pi^3}{12}}.$

If we attempt (3), we note that $ \int_0^{\infty}\frac{\cos bx}{x^2\plus{}n^4}\equal{}\frac{\pi}{2n^2}\,e^{\minus{}bn^2}.$

That leaves us with the sum $ \sum_{n\equal{}1}^{\infty}\frac{\pi}{2n^2}\,e^{\minus{}bn^2}.$


category theory
abstract algebra
Support
advanced fields
advanced fields unsolved

sheafification is a free faithful functor ' : {presheaves on X} --> {sheaves on X}. is it a full functor? i.e. does every homomorphism $ F' \to G'$ come from a homomorphism $ F \to G$?

Take $ F(U)\equal{}\{*\}$ for any $ U$. Then $ F'\equal{}F$, and you are talking about the map $ G(X)\to G'(X)$ which is neither injective nor surjective, in general. owk

oh you're right. my original problem is to show that $ G \cong f_* f^{ \minus{} 1} G$ for a closed immersion $ f : X \to Y$ of topological spaces and a sheaf $ G$ on $ Y$. this is (via the adjunction between $ f_*$ and $ f^{ \minus{} 1}$) equivalent to: $ f^{ \minus{} 1}$ is a faithful and full. but I cannot manage this at all (because there arise maps [i]into[/i] associated sheaves, and [i]into[/i] direct limits...). any hints?

Somehow, you got your spaces and functors mixed up. Roughly speaking, when you consider a closed immersion $ f$, the functor $ f^{ \minus{} 1}$ discards everything outside the image of $ f$, and $ f_*$ sets all stalks outside the image of $ f$ to zero/$ \{*\}$. So, it should be $ f^{ \minus{} 1}f_* \equal{} \mathrm{id}$, not $ f_*f^{ \minus{} 1}$. owk

I found this isomorphy in the book of liu, ex. 2.2.13, but I also just found out that it's wrong. nevertheless, it seems to be used in lemma 2.23 ... I think we have $ G \cong f_* f^{ \minus{} 1} G$ in the special case $ f : supp(G) \subseteq X$. but I still don't understand why $ (f_* O_Y)_{f(y)} \equal{} O_{Y,y}$ in the context of this lemma ...

Ok, then let us consider abelian sheaves (I already had to make a minor correction above for the set case).

In my edition of Liu, it only says that $ G\to f_*f^{\minus{}1}G$ is surjective for $ f\colon X\to Y$ a closed embedding. For the stalk of $ f_*F$ in $ x\in X$, use that neighbourhoods of $ x$ in $ X$ have the form $ U\cap X$ for neighbourhoods $ U$ of $ x$ in $ Y$. You just need that $ f$ is an embedding, nothing about closedness. Stalks of $ f^{\minus{}1}G$ should be clear (if $ i_x$ is the inclusion of a point $ x$, then $ G_x\equal{}i_x^{\minus{}1}G$).

For closed embeddings, the essential image of $ f_*$ consists of sheaves on $ Y$ having support in $ X$: it should be clear that for any point $ y\notin X$, the stalk of $ f_*F$ vanishes, and as shown above, $ G\to f_*f^{\minus{}1}G$ is an isomorphism on stalks in $ x\in X$. owk

thank you!

while we're at liu ;), can someone explain the step "Hence $ Z \cong Spec A/I$." in the proof of 3.20 to me? the other steps are clear to me.

Z->X and Spec A/I->X are closed immersions with the same sheaf of ideals, hence they are isomorphic.

hi philipp   :)

thank you, now it's clear.


these kind of problems should go to [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]the calculus forum[/url]..

the top must have x^3 cancel, bottom has x^4 cancel, the x^2 terms matter, putting this together, we have

2a+b-3c=0
5a-b+4c=0

(-a+5b-c)/(2)=1

so solve that system


Hi,
I usually access the Aops page through google,And sometimes(nowadays more frequently than not), I am not able to even get to the Aops page. The message appears on the screen " You have been banned" "Please contact the web administrator".

I don't understand how do you access the page through google?  :maybe: If that message appears it means that you use a proxy that has been banned (your account is not banned simply because you were able to post).

I think he is talking about using Google Translate as a proxy to bypass a school/work filter?

Anyway, what you probably need to do is access the page directly, not through Google.  If that doesn't work, either you are IP banned from AoPS, or your employer/school administration has blocked AoPS.  (I know my school blocks AoPS because it is an open forum.)

What you need to do is determine if the error message you are getting is the AoPS IP Ban error message. (Valentin?)  If it is not, you need to contact the owner of the computer (or network) that you are using.

Clicking on the link would not cause this problem.  Going to Google Translate and typing in the AoPS address to translate it into English could possibly.

[quote="7h3.D3m0n.117"]possibly too lazy to memorize the full address of AoPS and searches it on google so he clicks on the link...?[/quote]

I type in A in teh address abr and it fills the rest up. I thought schools were not supposed to like have forum access without direct supervision :huh:

I thought you used Firefox. Make a bookmark and put it below the toolbar.

OK I will clear your doubts, what I meant was what 7h3.D3m0n.117 meant, I'm too lazy to type in the address bar and so search "aops" on google, and click on the link. And I'm accessing the forum from my home. So, I tried typing the address in the address bar getting in but that too does not work. The same message appears. But it is only temporary, say for half an hour usually in the evenings (approx.6:00 pm IST).So I don' know what to do, other websites work fine, just Aops, thats why I'm asking for help from here.And I use Internet Explorer.

hmm i access AoPS by making my own personal webpage and then using it as my homepage...it saves the clutter of too many bookmarks on Firefox...see if that helps you

[quote="Valentin Vornicu"][quote="chess64"]Could you post a screenshot of the banned error screen? It seems to me that your IP accidentally got blocked by one of the admins... *cough*[/quote]IP Banning doesn't work between 10 am - 6 pm, so if he would be banned he would not be able to post or even visit the site. There's something else here ...  :maybe:[/quote]

Which time zone is that in?

[quote="Valentin Vornicu"][quote="chess64"]Which time zone is that in?[/quote]What?  :huh: I just said it doesn't matter what time it is, if you are banned, you are [b]always[/b] banned, not only between some random hours ...[/quote]

Oh... heh, it kind of sounding like you were saying that it didn't work between those times; you were actually saying that there aren't any set times for banning.

Well, for starters the OP could post a screenshot...

Hello,
Sorry people that I didnt reply; the problem didnt occur for some time, and then yesterday I encountered the problem . I saved the screen but I unfortunately saved as .mht extension, and I cant add a .mht file as a attachment here. And even a .htm/.html file is not allowed. So here's the message that appears in a text file.

Does your IP address change while browsing?

Aaaah. When you do as you do through google, it sometimes logs you on as GoogleBot.

Maybe that's the problem?

I've encountered the problem once.
It was for an entire day in June.
I typed in the AoPS url and Mathlinks url and they all said "You been banned. Please contact the Webmaster."

I repeat do you have a different ip each session of internet you open?

I don't get you,(I am not too well versed with the computer terminology), but isn't the IP address always the same for the same internet connection? I always access the net with the same internet connection on the same computer.

Your IP address can do like 3 basic things (from what I've seen).

1) My IP address stays the same every day all day all week all year. (Static IP).

2) Some IPs change every time you power cycle (reset) your router, or change every night, but stay the same during the day.

3) Many IPs change every time you log in and out of the internet.  (Dynamic IP).

Thealchemist, you need to determine which type of IP you have.  You can find out your current IP by visiting [url=http://www.whatismyip.org/]this site[/url].


Find all functions $f$ from the integers to the integers such that:

$f(x-y+f(y))=f(x)+f(y)$ for all integers $x,y$.

Is this a good enough proof?

[hide]Consider the case $m=n=k$ where $k$ is a random integer and we have $f(f(k))=2f(k)$ and therefore $f(n)=2n$. Use this function and substitute $m$ and $n$ where they are not equal for verification.[/hide]

You assumed that $f$ is one-to-one.
[hide="For example"] $f(x)=0$ works[/hide]
I don't know if there's anything else.

You basically assumed that $f(k)$ takes on all integer values.  $f(f(k)) = 2 f(k) \forall k$ does [b]NOT[/b] immediately imply $f(n) = 2n \forall n$.


I'm looking to take some AoPS classes over the next school year and I'm looking for some advice on what they are like from those and which ones would be the best from those who have taken some before. I've covered the "typical" US school precalculus curriculum (I was planning on taking AP Calculus next year before I discovered AoPS), but I have little experience in actual problem-solving math. I've been working through the Art of Problem Solving books, and I've realized that while I may "know" most of the subjects in the first volume and several in the second, it is difficult for me to see how that knowledge fits together to solve the problems. I've been getting better, but I feel that having some personal contact would help me a lot.

I would like to take the Intermediate Independent Study, but due to money constraints I would only be able to do one semester which I would probably take in the spring. I would like to take two regular classes in the fall, however, and am leaning towards the Intermediate Trig/Complex Numbers and the AMC-12 Problem Series. For those who have taken the classes before,

1) How difficult is the Intermediate Trig/Complex Numbers class and how in-depth will it go? Is it suitable for someone who has a grasp of the basic concepts (I know almost all the trig concepts and many of the complex number concepts in AoPS Vol. 2) but has a difficult time doing "out of the box" problems?

2) What topics does the AMC-12 Problem Series cover? I've been going through some old AHSMEs and feel pretty comfortable with most of the first 10/15 problems or so but really get bogged down in the last 5. Do you think this class will be suitable for me?

Thank you very much for all your help!

The Trig class should fit you precisely.

The AMC class covers a little bit from all the main topics you'll see on the AMC->12.  The specific subject classes generally go much more deeply into each subject area.


How do you find the explicit rule for a geometric sequence?

For example, the explicit rule for the sequence: 200, 100, 50, 25, 12.5
is  [size=150]A[/size]n = 25 * 2 ^ (4-n)

Sometimes you can use the recursive rule to help derive the explicit formula but that does not work in all cases. Is there a method for finding the explicit rule for any geometric sequence?

Hi jksong.

A geometric sequence may be written $ a, ar, ar^{2}, ar^{3}, ar^{4}, ..., ar^{n\minus{}1}, ...$.

In your case $ a\equal{}200$ and $ r\equal{} \frac{1}{2}$ so that the general term is $ 200 \times (\frac{1}{2})^{n\minus{}1} \equal{} 200 \times 2^{1\minus{}n} \equal{} 25 \times 8 \times 2^{1\minus{}n} \equal{} 25 \times 2^{3} \times 2^{1\minus{}n} \equal{} 25 \times 2^{4\minus{}n}$.

I hope this is helpful; if not ask again.

Thank you very much AndrewTom. That answered my question  :)


induction
inequalities unsolved
inequalities

http://www.kalva.demon.co.uk/apmo/asoln/asol891.html

look here :)

I have another solution .
We know that $\frac{1}{k!}\geq\frac{1}{k^{k}}$  so
$\frac{S^{k}}{k!}\geq \frac{S^{k}}{k^{k}}$

Summing for k=1,...n we have
$RHS\geq 1+\frac{S}{1}+...+\frac{S^{n}}{n^{n}}$

Now $(1+a_{1})....(a+a_{n})=1+S_{1}+....+S_{n}$
where $S_{1}=a_{1}+...+a_{n}$
         $S_{2}=x_{1}x_{2}+...+x_{1}x_{n}+....+x_{n-1}x_{n}$
         .....
         $S_{n}=a_{1}....a_{n}$
An now induction to prove
$1+\frac{S}{1}+...+\frac{S^{n}}{n^{n}}\geq 1+S_{1}+...+S_{n}$

Indeed by induction it suffices to prove that
$\frac{S^{n}}{n^{n}}\geq S_{n}$  which is true since it is equivalent to AM-GM  :wink:


The arithmetic mean of 12 scores is 82. When the highest and lowest scores are  removed, the new mean becomes 84. If the highest of the 12 scores is 98, what is the lowest score.

$ 12\times 82\equal{}984$ total points.  After the two scores are removed, there are $ 84\times 10\equal{}840$ total points.  So the lowest score, $ x$ is
$ 984\minus{}98\minus{}x\equal{}840\Rightarrow x\equal{}46$.

Or, by "weighted mean"

Distance from 84 to 82: 2 (weight: 10)
So our counterbalance to 82 with weight 2 will be 72.

98-72=26, 72-26=46.

weights are a great strategy for mean/average problems. it's much quicker than the usual multiplication method :)


The Fibonacci sequence $F_{n}$ is definied by

$F_{1}=1$ , $F_{2}=1$,$F_{n+2}=F_{n+1}+F_{n}$

Show that $F_{mn-1}-(F_{n-1})^m$ is divisible by $(F_{n})^2$ for all $m \geq 1$ and $n\geq 2$
               $F_{mn}-(F_{n+1})^m+(F_{n-1})^m$ is divisible by $(F_{n})^3$ for all $m \geq 1$ and $n\geq 2$

[quote="freemind"]From what I know, the first terms of the Fibonacci sequence are: $F_{1}=1, F_{2}=1$. The sequence you wrote is Lucas.[/quote]

I think Lucas begins with $1$ and $3$, and the given sequence is just Fibonacci without the leading $1$.


A is a square matrix such that $ A^{2008}\equal{}0$. prove that for all positive integer n we have:
$ rank(A)\equal{}rank(A\plus{}A^2\plus{}...\plus{}A^n)$ :!:

[hide="hint"]Try to prove that if $ T$ is an invertable matrix then $ rank(A)\equal{}rank(AT)$.The conclusion follows by proving that $ I\plus{}A\plus{}A^2\plus{}\cdots \plus{} A^{n\minus{}1}$ is invertable [/hide]

ok, and here is my solution:
suppose A=$ [a_{ij}]_{n\times n}$, it is easy to prove that for $ X\in R^n$ we have $ AX\equal{}0$ if and only if $ (A\plus{}A^2\plus{}...\plus{}A^n)X\equal{}0$.so two equations have common root set. so $ rank(A)\equal{}rank(A\plus{}...\plus{}A^n)\equal{}n\minus{}k$. here k is the dimension of the space of root of the two above equations


Is it possible to find the foci of an ellipse if you're given the equation of the particular ellipse?

[quote="guile"]Ellipses are written in the form [(x - h)^2]/a^2 + [(y - k)^2]/b^2 = 1, where (h,k) is the center and 2a and 2b are lengths of major and minor axes (interchangeable, depending on the position of the ellipse).  If given the above equation, the distance the foci are from the center of the ellipse, but on the major axis, is...

f^2 = absolute value (a^2 - b^2).[/quote]
just to add:
the points would be (assuming $a>b$) $(h\pm\sqrt{a^2-b^2},k)$ when $ a^2$ is under $(x-h)^2$ and are $(h,k\pm\sqrt{a^2-b^2})$ when $a^2$ is under $(y-k)^2$


A picture is worth a thousand words. Reality and candidness here.

From left to right: Mrs. Cockrell (coach), me (Miriam Bailey, aka egghead91), Andrew Messinger, Karl Daningburg

Here, perfectnumber628 is waiting to give an answer at Quizzing Nationals. As I was not there, I do not kow whether or not it was correct.

Yes, it is a photoshopped Axe can. On purpose.

But unfortunately, it was not submitted on time.

Kyle Liddle with the weird snake toy they were all obsessing over, Kamillah Ramos, Mabel Hope, and Mrs. Hope. In case you're wondering why I'm posting these people, they are some of my best friends (outside of my own team, of course!).

Picture of perfectnumber628 at Nationals this past June pretending to be a ninja.


superior algebra
superior algebra theorems

Let $(B_{i}, \psi_{ij})$ be a directed family of rings, with $B = \varinjlim B_{i}$. For each $i$ suppose that $\phi_{i}: A \to B$ with $\phi_{i}= \psi_{ji}\phi_{j}$ ($j \leq i$), and let $\phi : A \to B$ be the induced map. Then
\[{^{a}\phi}(\text{Spec}B) = \bigcap_{i}{^{a}\phi_{i}}(\text{Spec}B_{i}). \]

I have seen this result quoted many time, but I would love to see a proof. Does anyone have a good reference for a proof?

At the risk of sounding "so American" may I ask for an English reference. I can go as far as reading British if I have to.  :blush:


Let T_n be the Tribonacci sequence defined by T_0 = T_1 = 0, T_2 = 1 and T_n = T_(n-1) + T_(n-2) + T_(n-3) for n \geq 3. Prove T_(2n) = n (mod 4) for all n \geq 0.

T_(2n) = n (mod 4)

Orlando, are you sure ? It seems to be false for n=2  : T(4) = 5 != 2 [4] ??

Thank you for the correction. It corrected the typo in the initial condition of the sequence.  :)

Ok, then by strong induction :

T(2*n) = n mod 4
T(2*n+1) = n mod 2

Conclusion follows.


Using only 32 cent and 20 cent stamps, Charlie put 3.36 postage on
a package he sent to his sister. He used twice as many 32 cent stamps
as 20 cent stamps. Determine how many of each type of stamp he
used.

[hide]
suppose he used x amount of 20 cent stamps
so you get the equation
$(0.2)x+(2)(0.32)x=3.36$
solve to get $x=4$
so he bought $4$ $20$ cent stamps and $8$ $32$ cent stamps[/hide]

A related (but more general) question is to work out when you can/can't pay a certain amount, given two stamps of value $m$ and $n$.

Are there values for $m$ and $n$ with which there are an infinite number of amounts you cannot pay?

What's the largest amount you can't pay (assuming there aren't an infinite number of such amounts)? (This question is a bit tough for high school basics, but it still could be fun to play around with).

[quote="Quevvy"][quote="i_like_pie"][b]CMT[/b]

$x=mn-m-n$, where $m$ and $n$ are relatively prime and $x$ is the largest amount that can't be made.[/quote]
What is CMT?[/quote]
Chicken Mgnugget Theorem


I am working with cubes (block) to build pyramids.

The first pyramid has 1 block on layer 1 and 4 blocks on layer 2.
The second pyramid has 1 block on layer 1, 4 blocks on layer 2, 9 blocks on layer 3.
The third pyramid has 1 block on layer 1, 4 blocks on layer 2, 9 blocks on layer 3, 16 blocks on layer 4.
The fourth pyramid has 1 block on layer 1, 4 blocks on layer 2, 9 blocks on layer 3, 16 blocks on layer 4, 25 blocks on layer 5.
and so on ...

If I am to build 100 pyramids of increasing layers, how many blocks ( cubes ) do I need ?   :blush:

sorry to bring back an old problem, but there were no responses. can someone please find a solution?

I think the formula only solves for the number of cubes for the last pyramid.  What about the first to the 99th pyramids?  I am stumped too ...  :surrender:

[quote="G-UNIT"][quote="xenon_nightmare152"][quote="G-UNIT"][hide] well it seems that the $nth$ pyramid is the sum of $ 1^2 + 2^2 + 3^2 \cdots n^2 + (n+1)^2$

Plugging in 100 for $n$, wee see we must find the sum of the first 101 squares which is:

$\frac{101(102)(203)}{6} = 348551$  Meaning we will need 348551 blocks.

We derived this using the sum of teh fisrt n square formula

$1^2+2^2+3^2+...+n^2=\frac {n(n+1)(2n+1)} {6}$  [/hide][/quote]

I believe you have to find the sum of (1+4) + (1+4+9) + (1+4+9+16)...[/quote]

why?[/quote]

read the problem again. the first pyramid has 1+4, the second pyramid has 1+4+9, the third has 1+4+9+16 etc.

you need to find out how many blocks are in [b]100 pyramids[/b], not the [b]100th[/b] pyramid.

this problem is kind of driving me insane. (well...more insane than i already was). is there any formula or shortcut to finding the sum of $(1+4) + (1+4+9) + (1+4+9+16)...$
 :wallbash:


Did you like the last Mock MATHCOUNTS Competition (by chess64)? For those who didn't participate, do you want another one?

Who said "No"? Why? I might be able to improve it if you tell me what's wrong.

I mostly liked the one you did. The problem was that two of the questions had no answer. Also, could they be harder (national level) as most of us are practicing for nationals?

Yes sir! Hard questions coming up... heheheheheh  :P

Make them nationals level. Not oh-my-god-this-is-the-hardest-nationals-MC-test-ever-written level.

Did either of you (chess64 or tarquin) like mine? (other than it was late :blush:  :blush: )?

I don't like it because I made so many dumb mistakes.
Other then that, pretty good test.

[quote="tarquin"]I don't like it because I made so many dumb mistakes...[/quote]lol you got a 30.

[quote="tarquin"]Yeah .. a 30 ...
but I still made a lot of dumb mistakes.[/quote]
More than I got at state :(  :(  :(  :(  :(  :(  :(  :( :(  :(  :(  :(  :(  :(  :(  :( :(  :(  :(  :(  :(  :(  :(  :( :(  :(  :(  :(  :(  :(  :(  :( . But at least im going to nationals.

Yay!
I came in 2nd on my first mock competition.

Penguin!

I forgot to compete. That'd be nice if there was a second one :) .

[quote="chess64"]Yeah, I'll make one soon, but it's hard to make up questions . :wacko:[/quote]Definetely. I made up 25, then I couldn't think of any more. I used Elias Saab's web site fro some *help*  :blush:

It takes a really long time to write a good test.


:?  Is anyone in 6th grade mathcounts?  :)

I went to MathCounts in sixth grade. Got a 34 at chapter, translating to 24th place within the chapter. Improved to 42 and third place in chapter this year, though my state ranking is probably somewhere around 50th place. Hoping to make nationals next year.

I wish that I had been in sixth grade mathcounts, but my school didn't do it.

In sixth grade I got somewhere around eighth in the chapter - I don't know the place, I just know I advanced to state - and got 42nd in the state of Ohio. I was the only person from my school to advance to state that year. It was pretty lonely. Suffice it to say I've improved over the years, now in eighth grade I'm going to nationals. (!!!)

A 6th grader in Texas is going to nationals. I thought that that was nuts!

I was in 6th grade last year and got 8th in state. (Tennessee) Improved to first in 7th grade! :)
Also there were this one guy in 6th grade that got 2nd.

I went in 6th grade, huge advantage by the time your in 8th! I'm in SC, a weak state, (53rd, yikes!) but I placed 26th in Chapter in 6th, and 12th in seventh grade.  8th grade will be fun.  But in Maryland, you could expect a 50th place or better to be very good, considering you are only in 6th grade with NO Algebra or Geometry.  Algebra I makes it so much easier....

f your in 6th grade, here is a good solid formula for an arithmetic sequence I used in 6th:

a + dn - d = z

Where
[i]a[/i] is the first term of an arithmetic sequence, [i]d[/i] is the common difference, [i]n[/i] is the number of terms, and [i]z[/i] is the last term.  Very useful if you solve for [i]n[/i].  This improved my score by at least 2 poins, translating to about 10 places in SC.

I'm in 6th grade.I got to state in vermont.

6th grade year:
37 at chapter (3rd place if my score counted)
23 at state (somewhere around 20th)

6th grade
5th at chapter
75th at state GA

7th grade
13th at chapter
13th at state GA

8th Grade
1st at chapter
1st at state GA
16th at nationals

I improved over the years.

That 8th grade bit looks good...Good luck at nationals, if you are [i]currently[/i] an eighth grader!

of course he's currently an eigth grader! He's <snip>!!!
He's going to win Nationals!
[color=darkblue]
[size=75][Moderator rcv's note:  You may disclose your own identity, but please don't disclose the identity of others.][/size][/color]

Hmmm...I never did mathcounts in 6th grade, I wouldn't have made the team.  I was only just learning material similar to Algebra I at the end of 6th grade, and learning Algebra I material was what really, really boosted my mathcounts scores.

[quote="tarquin"]You're in Texas? What school?
I'm from Flour Bluff JH we got 4th.[/quote]

um... If you're talking to me, I'm from Kleb (Hence my name).
This was our school's first year for Mathcounts at all.... We got 18th. :blush:
I got 14th as an individual though.

Hah! I knew it...
I was actually referring to eryaman.

Well, I'm in 9th grade now, so I'm not in middle school anymore, but during my Mathcounts years I went to Canyon Vista Middle School.  We never do anything good, last year was our best year.  We got 8th place team.

I made it to Nats last year. I'm hoping to make it three years...


Source - Mock Exam for Quant Ability

A box contains 100 tickets numbered 1 to 100. A person picks out 3 tickets from the box such that the product of the numbers on two of the tickets yields the number on the third ticket. How many tickets can never be picked up ?

a) 10
b) 11
c)  25
d) 26

Please help....  :?

[hide]
Consider a number, less than or equal to 100. All numbers less than or equal to 50 can be chosen, apart from 1. This is clear as x * 2 = 2x, x <= 50, 2x <= 100 which is also in the set, so it can be chosen. For numbers above 50, if it can be factored into x * y, where x and y are different numbers > 1, then it can be chosen. The numbers which cannot be chosen are the primes more than 50.

The primes more than 50 are [53,59,61,67,71,73,79,83,89,97], and 1 also cannot be chosen. So theres 11 numbers that cannot be chosen.
[/hide]

[quote="superwizard188"][hide]um, prime numbers higher than 50.

53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

So... 10.[/hide][/quote]

What about number 1?

so basically you have to do a bit of reasoning.
thankx a ton  :)

[hide]Oh, I got 10 initially, until I realized that one cannot be achieved.

So it is 11.[/hide]


I have always wondered what people used as their home page options, now with the multitude of sites out there. I realize that those of us who use Firefox to restore tabs probably haven't seen the homepage in years, but try to at least remember. It's a curious thing to see what the distribution will be like.

Google.

I know I can just Google search from my URL bar in Chrome -- I just like the look of Google's page.

Blank. Why is there no option for blank??

Erm...
I think it's either iGoogle or Google.
Something that end with a Google.

http://www.google.com/firefox?client=firefox-a&rls=org.mozilla:en-US:official

Mozilla Firefox rules!!!

There's this Firefox addon that shows nine of the most important sites you visit.

Well, it's my homepage.

cmon you aopsers you don't have aops as your home page that fails!!

PS i made up the word aopsers right now.

I actually have msn, I don't know it just sticks.  I don't even use msn

[quote="jjfun1"]Google. :)
I like the simple look when I first go on the internet. :)[/quote]

I second that.

If you want to find out, go to [url]http://www.aperturescience.com[/url].
There is a certain sequence of commands there that will take you to your home page.

Pie, your home page is NOT aops.

Mine is Wolfram|Alpha.  :P

umm i have google chrome with 4 home pages...

Firefox homepage but right after I post this I'll change it to AoPS.

First it was Lemmy's Land
Next it was Google
Then it was Art of Problem Solving
Now it is back to Google

Google, but... I rather like the Wikipedia idea.

sorry to revive thread, mine is google, bing, and aops

Mine is accuweather
Thinking about changing it to AoPS

[quote="dragon96"]I have AoPS as my home page (that shows how I'm an AoPSer in 1 of 12 ways). I've seen Google the most common though.

Besides Google, how many people here use a search engine as the home page?[/quote]
technically yahoo counts, so me

It used to be wolfram.com cause my dad works there

Please don't revive 5 1/2 year old threads.


Your friend picks two numbers in the reals, puts their hands behind their backs, and then puts one number in each hand. You pick a hand to open, and you see the number inside. Can you, with a probability of greater than 50%, now pick a hand to choose such that its number is the larger of the two?

No, if the number is in the real, then it could be any decimal number like 1.537456285639656749656749............. or anything else.
That is.... if you CAN actually put numbers in your hands

I remember that the answer is yes, but I don't remember why and I don't remember what thread discussed this problem.

Before anything starts, I'll keep a number N in my mind.  let a < b be the two numbers in your hand.  Then whatever number I see, I'll stay if the number is greater than N, and switch if the number is smaller than N.

There are only three cases:

a < b < N: Then I'd always switch, so it's 50% here.

N < a < b: Then I'd always stay, so it's 50% here.

a < N < b: Then I'd always win, so it's 100% here.

So it's definitely better by 50%, but not by much...


I constructed this before a while....
If $ p$ is a prime number and $ x^2 \plus{} y^2 \equal{} z^2 \plus{} w^2 \equal{} p^2$ find $ x,y,z,w,p$.($ x,y,z,w$ are integers)

i think there are so many answers.
it's similar to pytagoras numbers.
ex) (x,y,z,w,p)
      (3,4,0,5,5)
      (3,4,5,0,5)
      (5,12,13,0,13)
      (5,12,0,13,13)

I believe that the problem would be more intersesting if we stated:

solve the equation in integers
$ x^2 \plus{} y^2 \equal{} z^2$
for when for each z it has exactly four (with the rearangements) solutions (x,y).
However this is quite complicated for me. :blush:  (I think it would be even more complicated in case there were up to four solution.)

Guys you checked only the trivial solutions.... :lol: Ι want from us to find all the solutions of equation...What if $ x,y,z,w$ arent equal to zero?I will post my solution if want.

FOR MANIOPAS and others....
If i understanded well u said that is harder to solve the general equation $ x_1^2\plus{}y_1^2\equal{}x_2^2\plus{}y_2^2\equal{}....x_n^2\plus{}y_n^2\equal{}p^2$ for n>2 ($ p$-prime)....Can anybody solve this equation for $ n\equal{}2$ and general for $ n>2$?I give 2 days for a complete solution... :wink:

Scketch για τη λύση
[hide]Λοιπόν . Εύκολα δείχνουμε (με Ζ[i] ή εντελώς απλά ) ότι υπάρχουν  $ a,b,c,d$ me (b,d)=1 ώστε

$ 4p^2\equal{}(a^2\plus{}c^2)(b^2\plus{}d^2)$  . Μετά ακόμα μια φορά το ίδιο  :) [/hide]

(Δεν ειχα internet γιαυτο καθυστερησα...)Αυτη ηταν η σκεψη μου Σιλουανε και ετσι την κατασκευασα.... :)
GOOD LUCK ...WAGM!!! :wink:


hello, this equation has infinity solutions.
Sonnhard.

hello, I think warut_suk wants a general form for the solutions to this equation.

Is the $\cos{6}$ really that, or is it supposed to be something like $\cos{6x}$? A constant like that kind of surprises me.

Darn. It feels weird to me starting a post with "hello, "


Show that $ 121\not|\ n^2\plus{}3n\plus{}5$ for every $ n\in\mathbb{N}$.

The problem is equivalent to

$ 121 | 4n^2 \plus{} 12n \plus{} 20 \equal{} (2n \plus{} 3)^2 \plus{} 11$

but clearly this requires that the greatest power of $ 11$ that divides $ (2n \plus{} 3)^2$ is odd; contradiction.

The generalization is given by [url=http://mathworld.wolfram.com/HenselsLemma.html]Hensel's lemma[/url] with the $ p$-adic valuation.


Consider the infinitely tall system of identical massive cylinders and massless planks shown the figure below. The moment of inertia of the cylinders is $ I \equal{} MR^2/2$. There are two cylinders at each level, and the number of levels is infinite. The cylinders do not slip with respect to the planks, but the bottom plank is free to slide on a table. If you pull on the bottom plank so that it accelerates horizontally with acceleration $ a$, what is the horizontal acceleration of the bottom row of cylinders?

[img]http://img138.imageshack.us/img138/162/infinitecylinder.jpg[/img]

[hide="Hmm"]If there is no rolling resistance, shouldn't the acceleration just be $ \mathbf{a} /2$?[/hide]


This is copied from whatever the new name for the Etc. Forum is, since it was quickly pushed to the bottom of the list.

Original Question:  OK this is more of a fact than a riddle. But its real interesting.
Why are manhole covers round?

Answer:  If they were any other shape, they could be turned to fall down to the bottom of the hole. But a round manhole cover cannot.

New Question Resulting from Answer: Actually, thinking about it more, the circle is not the unique shape with this property. The thing I'm about to describe is called a Reuleaux Triangle:
Take an equilateral triangle ABC. Centered at point A, draw circular arc BC. Centered at point B, draw circular arc CA. And centered at point C, draw circular arc AB. Then the figure formed, which looks like a triangle with its sides bent outwards a little bit, also has that property.
Can you prove it?

What do you mean by that?  "width" is not something that I've ever heard defined for a curve.

This should explain it: http://mathworld.wolfram.com/CurveofConstantWidth.html

Here are some other examples of curves of constant width: http://mathworld.wolfram.com/ReuleauxPolygon.html

And here's a nice page about curves of constant width: http://www.cut-the-knot.org/do_you_know/cwidth.shtml

And there's a really great recreational math book called [u]The Enjoyment of Math[/u] by H.Rademacher and O.Toeplitz that discusses this and other cool topics. I highly recommend it.

That's right...a drill bit in that shape can make a nearly-square hole because it's a curve of constant width, if I recall correctly.  Has anyone heard of this property?

Yes, although it has to be centered somewhere other than at its center, I think.  Something like that.  I don't remember it exactly -- I think it's on mathworld, somewhere.

round manhole covers are easier to make than realux polygons.

round covers are easier to transport....


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

Let $ABCD$ be a convex quadrilateral . $ABCD$ is inscribed in $(O,R)$ . It's known that : $AB=BC=CD=l$ and $AD=l+R$ , $l<R$ . Determine $\angle A, \angle B ,\angle C , \angle D$ ?

let m<BAC=x, so on m<BCA=m<BDC=m<BDA=m<CAD=x there for because of law of sines we have l+r=2r.sin(180-3x) but l=2rsin(x)... by replacing we have:
2r.sin(x)+r=2rsin(3x) an then we have 4sen(x).cos(2x)=1 an solving this we have, let s=sinx, s=sqrt(5)-1 or s=-sqrt(5)-1 or s=1/2

If x was 30º then m<A=60º, m<B=120º, m<C=120º, m<D=60º

AS YOU SAID L<r SO THERE'S NO SOLUTION  :(

Hucht, you made a little mistake. This trigonometrical equation has two solutions !

[hide="Here is a solution"]Let $ABCD$ be a convex quadrilateral inscribed in the circle $C(O,R)$. Prove that:

$\ I.\ \ AB=BC=CD=l<R,\ DA=l+R\Longrightarrow$ $A=D=36^{\circ},\ B=C=144^{\circ}.$

$II.\ AD=l<R,\ AB=BC=CD=l+R\Longrightarrow$ $B=C=72^{\circ},\ A=D=108^{\circ}.$

[b]Solution[/b] (only for the first point). $m(\widehat {BAC})=x< \frac{\pi}{3}$, $m(\widehat {ABD})=y<180^{\circ}\Longrightarrow$

$3x+y=180^{\circ}$, $2R\sin x+R=2R\sin 3x$ $\Longrightarrow$

$1+2\sin x=2\sin 3x\Longleftrightarrow\cos x+2\sin x\cos x=2\sin 3x\cos x\Longleftrightarrow$

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

$x=\left|\frac{\pi}{2}-4x\right| \Longleftrightarrow x=\frac{\pi}{2}-4x\ \vee\ x+\left(\frac{\pi}{2}-4x\right)=0\Longleftrightarrow x\in \left\{\ \frac{\pi}{10}\ ,\ \frac{\pi}{6}\ \right\}.$

$\blacksquare\ 1^{\circ}.\ x=\frac{\pi}{10}\Longrightarrow A=D=\frac{\pi}{5},\ B=C=\frac{4\pi}{5}.$
$\blacksquare\ 2^{\circ}.\ x=\frac{\pi}{6}\Longrightarrow A=D=\frac{\pi}{3},\ B=C=\frac{2\pi}{3}.$
$l<R\Longrightarrow x<\frac{\pi}{6}$, i.e. the [u]only solution[/u] is $\blacktriangleright\boxed {\ x=\frac{\pi}{10}\ }\blacktriangleleft$ ([u]the first case[/u]}.

[b]Remark.[/b] Let $ABCDEFGHIJ$ be a regular decagon, $AB=l=2R\sin 18^{\circ}$.

$\blacktriangleright\ \ I.$ In the quadrilateral $ABCD$ we have $AB=BC=CD=l$, $DA=l+R$.

$\blacktriangleright\ II.$ In the quadrilateral $ADGH$ we have $HG=l$, $HA=AD=DG=l+R$.

A nice problem for the section [b]High School - Intermediate Topics ![/b][/hide]


What was your score on the 2008 National Sprint Round?

I didn't even make nationals... :(

BOO HOO :maybe:  :mad:

No-one got $ 30$...

Cause Bobby was the Written round winner.

i got a 17 on the sprint, placed 63rd overall :P
this year's sprint seemed kind of easy compared to other year's tests i've taken as practice though

bobby shen got i think a 29

Yes, bshen did get a 29. I agree, the sprint round was easier concept-wise but I still ran our of time like I always do.

[quote="math154"]One reason Texas happens to be so fast is because they do Number Sense and know most or all of these [url=http://www.academicmeet.com/tricks/]tricks[/url].[/quote]
Wow...

I feel week in Illinois compared to those kids.. :(

omg, it's lucien!

I got a 17, epic phail.

I got a 21 or 25, even more epic phail.

Disclaimer: Do not badger me about making excuses about blah this, blah that, etc. The reason I think I got 25 is because I meticulously checked my answers against the answer key, and was being really harsh; I got 5 wrong. However, there might be multiple factors of getting an additional 4 wrong, many of which we shouldn't discuss here (maybe I forgot to put an extra 0 at the end, or perhaps something to do with my handwriting, for example). At this point, I care too little to change it, since my target was a complete failure.

Edit: oh, and by the way, 600th post! :jump:  :D

Who the hell just said they beat bobby shen by one point, saying they got a full score?

Freaking LIARS!!!!!!!!!!!!! :censored:  :furious:

Woah, it's fine. Every single poll has a few liars in it, no need to get worked up.
I got a 19 :(

I didn't make it to nats, but i did get to try it and i got 13 out of 30.

P.S. - Didn't get to try the target round yet -  :(

[quote="isabella2296"][quote="AIME_is_hard"]i got a score of $ 30 \minus{} 9 \equal{} 21$

 :oops: epic fail.[/quote]

That's not bad! You did a lot better than me. I didn't even make it to nationals. :([/quote]
I think u would make nats now....if u could have a competition again..And I got a 28 on a practice Nats tesst.....

On ALL the practice Nationals tests I've ever taken, I've gotten in the high 30's to 46's. That means that my low score was due to a REALLY BAD DAY and/or excess of counting and probability problems :D

I almost got 8 out of 8 on a National Target, but I made an addition error, which is probably one of the worst errors you can make. If I had gotten all eight questions right, then my total would have been 37 out of 46.

Well it was one of the easiest target rounds I've ever seen. It was easier than most of the state targets that I've seen.

You talking about National Target 2008?

If you only have the sprint questions, can you post more of them here?

[quote="Wickedestjr"]I almost got 8 out of 8 on [b]a[/b] National Target, but I made an addition error, which is probably one of the worst errors you can make. If I had gotten all eight questions right, then my total would have been 37 out of 46.[/quote]

He wasn't talking about this year's target (*Whew*!!) :)
Yeah, I've gotten 8's a bunch of times on many targets. So I didn't really focus my attention so much on target rounds, and that's what killed me this year.

No it was 1993 Nats. And by the way; Seven people got full score? I feel stupid too.

I haven't even taken the 2008 National Target round yet. Can somebody please give me the questions and answers?

You must realize that Bobby Shen got a 29, which was the highest score on the sprint. That means the seven people were LYING!! Don't feel stupid!!!

Well, they weren't necessarily lying, but they are many different possibilities for what happened. These are listed below;

1. They lied, which is the the most likely reason.
2. Those seven people were all either very intelligent high schoolers, or above. (Or an unknown mathlete to this world).
3. They were thinking of a different sprint round and accidentally said 30. (Which I highly doubt.)

what was the lowest score? sprint/target/total? if anybody has all those statistics can they plz post them?

OMG!!! 2008 Nationals is soooo over...why the heck do people still care about these things???


Given a convex quadrilateral $ ABCD$, $ E$ is the point of intersection of $ AB$ and $ CD$, and $ F$ is the intersection of $ AD$ and $ BC$. The diagonals $ AC$ and $ BD$ meet at $ O$. Suppose that $ M, N, P, Q$ are the midpoints of $ AB, BC, CD$, and $ DA$. Let $ H$ be the intersection of $ OF$ and $ MP$, and $ K$ the intersection of $ OE$ and $ NQ$. Prove that $ HK \parallel{} EF$.

It's a direct application of the Gauss line theorem. In the quadrilateral $ AFBO$, we have the middle points of segments $ AB$, $ OF$ and $ CD$ are collinear, so we conclude that $ H$ is the mid-point of the segment $ OF$. Similarly, $ K$ is the mid-point of $ OE$. In conclusion, we have $ HK$ is parallel to $ EF$.


I did the Mandelbrot First round and the second round.

Are the second round results final? Because I should probably be listed somewhere there, but no one from my school is listed although I am sure their scores were high enough.

Can teachers still submit scores online?

Team scores have not come out.  However, individual rankings have come out.  You should talk to your club advisor; maybe he forgot.

so he can still submit?

I should be in one of the Tiers....

 :o

Our teacher forgot about all math leagues. Even the Alamiss.

[quote="K81o7"]My teacher DID in fact submit the second round (as far as I know) but none of us are listed in the individual scores...
I was thinking I'd wait until the thrid round and wait until they re-update.  If they didn't show our scores, I'd investigate it further...[/quote]

GOOD. I feel slightly relieved now. well not that it's THAT important.

Did you do Regional or National Level? I did regional, but then again I'm in 8th grade.

Our teacher forgot to submit our state math league scores and ALL of the other mail-in math leagues with the exception of Ciphering time trials on http://www.natassessment.com

I think the problems are pretty cool. I'm still not used to some of the logic and understanding the problems. ( especially the 3 point ones)

My name is Alan Zhao, you'll find me on the tier one western region  :lol:.  Indeed, the problems are quite cool.

Yeah, the problems are nice.  I find it fun, especially the team competition.

Oh, and I'm in national.  I don't think there's THAT big of a difference anyways.

I just pwned the regional team test with $(2rs,r^{2}-s^{2},r^{2}+s^{2})$ :lol:

bah

we pwned the team test with 2.5 people


geometry
circumcircle
number theory
greatest common divisor
cyclic quadrilateral
geometry unsolved

Let $ABCD$ be a cyclic quadrilateral such that lines $AD$ and $BC$ are not parallel. Given points $E$ and $F$ on line $CD,$ let $G$ and $H$ be the circumcenters of triangles $BCE$ and $ADF,$ respectively. Prove that lines $AB,\,CD,$ and $GH$ are concurrent or parallel if and only if $ABEF$ is cyclic.

We suppose that the quadrilateral $ABEF$ is cyclic and let $N,$ $N',$ the intersection points of the sidelines $AB,$ $CD$

of $ABCD$ respectively, from the segment line $HG$ and it is enough to prove that $N\equiv N'.$

From cyclic quadrilateral $ABEF$ $\Longrightarrow$ $\angle AFC = \angle EBN$ $,(1)$ $($ $N\equiv AB\cap HG$ $).$

From $(1)$ $\Longrightarrow$ $\angle ADC+\angle DAF = \angle EBC+\angle CBN$ $\Longrightarrow$ $\angle DAF = \angle EBC$ $,(2)$

From $(2)$ $\Longrightarrow$ $\angle DHF = \angle EGC$ $,(3)$ and so, the isosceles triangles $\bigtriangleup HDF,$ and $\bigtriangleup GEC,$ are similar.

Hence we have that $\angle DFH = \angle ECG$ $\Longrightarrow$ $HF\parallel GC$ $\Longrightarrow$ $\frac{N'G}{N'H}= \frac{DH}{EG}$ $,(4)$ $($ $N'\equiv CD\cap HG$ $).$

$\bullet$ We denote as $M,$ the intersection point of $AD,$ $BC$ and we will prove that the triangle $\bigtriangleup PAB,$

where $P\equiv AH\cap BG,$ is isosceles $($ $PA = PB$ $).$

$\angle PBA = \angle GBN = \angle GBC+\angle CBN = \angle GCB+\angle ADC =$

$= \angle BCD-\angle GCD+\angle ADC = \angle MAB-\angle HDF+\angle ADC =$

$= \angle PAB+\angle MAP-\angle HDA-\angle ADC+\angle ADC = \angle PAB$

$\Longrightarrow$ $\angle PBA = \angle PAB$ $\Longrightarrow$ $PA = PB$ $,(5)$

Apply the [b][size=109]Menelaus’s theorem[/size][/b] in the triangle $\bigtriangleup PHG,$ taken as its transversal the segment line $HG,$

we have that $\frac{NG}{NH}\cdot \frac{AH}{AP}\cdot \frac{BP}{BG}= 1$ $\Longrightarrow$ $\frac{NG}{NH}= \frac{BG}{AH}= \frac{DH}{EG}$ $,(6)$

From $(4),$ $(6)$ $\Longrightarrow$ $\frac{N'G}{N'H}= \frac{NG}{NH}$ $\Longrightarrow$ $N'\equiv N$ and the proof is completed.

Kostas Vittas.

PS. It is easy to show that $AB\parallel HG\parallel CD,$ when the cyclic quadrilateral $ABCD,$ is an isosceles trapezium.


geometry
power of a point
radical axis
IMO Shortlist
geometry proposed

For a given triangle ABC, let X be a variable point on the line BC such that the point C lies between the points B and X. Prove that the radical axis of the incircles of the triangles ABX and ACX passes through a point independent of X.

This is a slight extension of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=41033]IMO Shortlist 2004 geometry problem 7[/url] and can be found, together with the proposed solution, among the files uploaded at http://www.mathlinks.ro/Forum/viewtopic.php?t=15622 . Note that the problem was proposed by Russia. I could not find the names of the authors, but I have two particular persons under suspicion. Maybe somebody could shade some light on this...

  Darij

Let's assume first that we've shown the following:

If the incircle of $ABX$ touches $AX,BX$ at $S_b,T_b$ respectively, then $S_bT_b$ passes through a fixed point $P_b\ (*)$.

Then, since the situation is analogous, if the incircle of $ACX$ touches $AX,CX$ in $S_c,T_c$ respectively, then $S_cT_c$ passes through a fixed point $P_c$. Since the radical axis of the two circles is equidistant to the parallel lines $S_bT_b,S_cT_c$, it means that it must pass through the midpoint of $P_bP_c$, which is fixed. It means that all that remains to be done is proving $(*)$.
We'll show that $P_b$ is the projection of $A$ on the bisector of $\angle B$. Call this point $M$ for now, and let $K=AM\cap BC$. $M$ is the midpoint of $AK$, so if the incircle of $ABX$ touches $AB$ in $R_b$, then $(T_bR_b,T_bM;TA,TX)=-1\ (\#)$. On the other hand, if $U=T_bR_b\cap AX$, then $BS_b$ is both the polar of $U$ wrt the incircle of $ABX$ and the polar of $U$ wrt the angle $\angle ABX$, so $(T_bU,T_bS_b;T_bA,T_bX)=-1$. This, together with $(\#)$, shows that $T_bM$ and $T_bS_b$ are one and the same line, i.e. $T_bS_b$ always passes through $M$, as desired.

To show that $M\in T_bS_b$ we can apply Menelos in triangle $AKX$ and secant $S_bT_b$ and using the fact that $AR_b=T_bK$ and $XS_b=XT_b$. We will find that $S_bT_b$ cuts $AK$ in two equal segments. No projective geometry.

BTW. What means $(AB,AC,AD,AE)=-1$ i'm not aquainted with this notation.

[quote="grobber"]Let's assume first that we've shown the following:

If the incircle of $ABX$ touches $AX,BX$ at $S_b,T_b$ respectively, then $S_bT_b$ passes through a fixed point $P_b\ (*)$.
We'll show that $P_b$ is the projection of $A$ on the bisector of $\angle B$. Call this point $M$ for now, and let $K=AM\cap BC$. $M$ is the midpoint of $AK$, so if the incircle of $ABX$ touches $AB$ in $R_b$, then $(T_bR_b,T_bM;TA,TX)=-1\ (\#)$. On the other hand, if $U=T_bR_b\cap AX$, then $BS_b$ is both the polar of $U$ wrt the incircle of $ABX$ and the polar of $U$ wrt the angle $\angle ABX$, so $(T_bU,T_bS_b;T_bA,T_bX)=-1$. This, together with $(\#)$, shows that $T_bM$ and $T_bS_b$ are one and the same line, i.e. $T_bS_b$ always passes through $M$, as desired.[/quote]
Clearly $\widehat{AIB}=180^0-\frac{1}{2}(\widehat{ABC}+\widehat{BAC})=90^0-\frac{1}{2}\widehat{AXB}=\widehat{AS_bP_B}$, so $A,I,P_b,S_b$ are concyclic. Q.E.D


function
integration
real analysis
real analysis theorems

Let $ f$ be a nonnegative integrable function. Show that the function $ F$ defined by $ F/(x )$ $ = \displaystyle{\int_{-x}^{x} f}$ is continuous.

For a given $ x,$ choose an arbitrary sequence $ x_n$ such that $ x_n\to x.$

Define $ f_n(t)\equal{}f(t)\chi_{[\minus{}x_n,x_n]}(t)\equal{}
\begin{cases}f(t),&|t|\le x_n\\0,&|t|>x_n\end{cases}$

Define $ g(t)\equal{}f(t)\chi_{[\minus{}x,x]}(t)\equal{}
\begin{cases}f(t),&|t|\le x\\0,&|t|>x\end{cases}$

We note that $ f_n(t)$ converges pointwise to $ g(t)$ everwhere except possibly at $ |t|\equal{}x,$ which is almost everywhere. Note also that $ f_n(t)\le f(t)$ and $ f$ was given as integrable.

Apply the Dominated Convergence Theorem to get that

$ \int f_n\to\int g,$ which gives us that  $ F(x_n)\to F(x).$

This is the required continuity.

Variations on this:

We can weaken $ f$ integrable to $ f$ locally integrable (integrable on bounded sets).

We don't need $ f$ to be nonnegative; drop that hypothesis and use the same proof, with $ |f|$ as the dominating function.

Or you could keep the global integrability of $ f$ as a hypothesis and prove that $ F$ is uniformly continuous.


function
algebra
polynomial
floor function

Consider an equilateral triangle of side x made up of other equilateral triangles of side 1. A triangle with side length 1, obviously, will have only 1 triangle. Duh. A triangle with side 2 will have 5 triangles in it: the four triangles of side 1 and the big triangle of side 2. Count carefully. The upside-down ones are included.

OK. Enough babbling about triangles. Find a formula in x for the total number of triangles in the big equilateral triangle.

I don't think there is a formula; it depends on $x$.

[hide]yoou could try: first count the right side up ones:
2: 1+3
3: 1+3+6
4:1+3+6+10
...

Now upside down:

2:1
3:3
4:6+1
5:10+3
6:15+6+1

For right side up we see it is sum of the 1st,2nd,3rd,... xth triangular number or ${{x+2}\choose{3}}$

For upside down it is the sum of the x-1 th, x-3 th, x-5th... until it is less than or equal to zero.

So if x is odd the formula is:

${{x+2}\choose3}+\sum_{i=1}^{\frac{x-1}2}{i{(2i+1)}}$

(gotten from triangular numbers of $2i$, this is simplified from ${{x+2}\choose3}+\sum_{i=1}^{\frac{x-1}2}{\frac{(2i){(2i+1)}}{2}}$ )

which simplifies to be
 $\frac{2x^3+5x^2+2x-1}{8}$

If x is even:

${{x+2}\choose3}+\sum_{i=1}^{\frac{x}2}{i{(2i-1)}}$

(gotten from triiangular numbers of $2i-1$, this is simplified from ${{x+2}\choose3}+\sum_{i=1}^{\frac{x}2}{\frac{(2i-1)(2i)}{2}}$

which simplifies to be
 $\frac{2x^3+5x^2+2x}{8}$    how familiar.

You could just say the formula is

$\boxed{\left[\frac{2x^3+5x^2+2x}{8}\right]}$  where $[n]$ is the greates integer function[/hide]

EDIT :  Natmc is perfectly correct.  I had the wrong answer.
Ahh, correct me if I'm wrong, but

Sum of first $n$ perfect squares=$\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$
$=\frac{2n^3+3n^2+n}{6}$
$=\frac{(n)(2n+1)(n+1)}{6}$.

You can validate this using factoring of polynomials and induction.
You can show that it is the sum of the first $n$ perfect squares using the same thing you would with an array of squares.

[quote="Boy Soprano"]Ahh, correct me if I'm wrong, but

Sum of first $n$ perfect squares=$\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n$
$=\frac{2n^3+3n^2+n}{6}$
$=\frac{(n)(2n+1)(n+1)}{6}$.

You can validate this using factoring of polynomials and induction.
You can show that it is the sum of the first $n$ perfect squares using the same thing you would with an array of squares.[/quote]
This is not squares, it is triangles.  My formula works

EDIT :  This is also wrong.  I neglected to see that upside-down triangles of length 2 wouldn't start appearing until the third row, and that makes this entire thing worthless.
[hide="Please do not click here to reveal very embarrassing example of what happens when I do not check my work"] Excuse me for pointing this out, but when you put in 1, your formula generates $\frac{9}{8}$.  As a matter of fact, your formula only generates an integer result when $x$ is even, because if $x$ is odd, then you have $\frac{2(1 \mod{2})+5(1 \mod{2})+2(1 \mod {2})}{8}=\frac{1 \mod{2}}{8}$, which cannot be an integer.  Also, it [b]is[/b] a sum of perfect squares, because :
To find the number of equilateral triangles of length $1$ in an equilateral triangle of length $x$, the solution is :
right-side up triangles = $x$ triangle = $\frac{(x)(x+1)}{2}$
+
upside-down triangles = $(x-1)$ triangle = $\frac{(x-1)(x)}{2}$
=
$\frac{x^2+x}{2}+\frac{x^2-x}{2}=x^2$
Then we find the number of equilateral triangles of length $2$, which is (x-1)^2.  Then equilateral triangles of length $3$, triangles of length $4$, etc., so that it soon becomes apparent that the solution is the sum of the first $x$ perfect squares.  The formula for that is $\frac{1}{3}x^3+\frac{1}{2}x^2+\frac{1}{6}x=\frac{(x)(x+1)(2x+1)}{6}$, which is what I posted above.
If you want a proof of that, [hide="click here."]
Let us call the function that generates the sum of the first $x$ perfect squares $f(x)$.  We know that $f(x+1)-f(x)=x^{2}+2x+1$.  Now, if we take the function $g_0(x)=x^3$, we know that $g_0(x+1)-g_0(x)=3x^2+3x+1$.  Thus if we have $f(x)=\frac{1}{3}x^3$, we have $f(x+1)-f(x)=x^2+x+\frac{1}{3}$.  However, this still leaves $x+\frac{2}{3}$ for which we must account.  Now, we can take $g_1(x)=x^2$, and $g_1(x+1)-g_1(x)=2x+1$.  Therfore, we add $\frac{1}{2}x^2$ ti=o the function, and we only have $\frac{1}{6}$ left of the difference, which we can remedy by adding $\frac{1}{6}x$ to the function, which is thus complete :  $f(x)=\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+\frac{1}{6}x$, which we can validate by subtracting $f(x)$ from $f(x+1)$.  (We know that there is no zero-degree co-efficient by noting that $f(1)=1$.  Factored and put over a common denominator, the final result is $\frac{(x)(x+1)(2x+1)}{6}$.  We can verify that this always gives an integer result when $x$ is an integer by noting that if $x$ is not divisible by 2, then $(x+1)$ must be, and that if $x$ is not divisible by three, either $(x+1)$ or $(2x+1)$ must be.[/hide][/hide]

You obviously didn't notice the greatest integer function in his answer.

Also, your equation is not correct. In a triangle with side length $5$, there are $13$ triangles of side length $2$, and $13$ is obviously not a perfect square; therefore, the sum is not the sum of the first $x$ perfect squares.

[quote="Boy Soprano"]Excuse me for pointing this out, but when you put in 1, your formula generates $\frac{9}{8}$.  As a matter of fact, your formula only generates an integer result when $x$ is even, because if $x$ is odd, then you have $\frac{2(1 \mod{2})+5(1 \mod{2})+2(1 \mod {2})}{8}=\frac{1 \mod{2}}{8}$, which cannot be an integer.  Also, it [b]is[/b] a sum of perfect squares, because :
To find the number of equilateral triangles of length $1$ in an equilateral triangle of length $x$, the solution is :
right-side up triangles = $x$ triangle = $\frac{(x)(x+1)}{2}$
+
upside-down triangles = $(x-1)$ triangle = $\frac{(x-1)(x)}{2}$
=
$\frac{x^2+x}{2}+\frac{x^2-x}{2}=x^2$
Then we find the number of equilateral triangles of length $2$, which is (x-1)^2.  Then equilateral triangles of length $3$, triangles of length $4$, etc., so that it soon becomes apparent that the solution is the sum of the first $x$ perfect squares.  The formula for that is $\frac{1}{3}x^3+\frac{1}{2}x^2+\frac{1}{6}x=\frac{(x)(x+1)(2x+1)}{6}$, which is what I posted above.
If you want a proof of that, [hide="click here."]
Let us call the function that generates the sum of the first $x$ perfect squares $f(x)$.  We know that $f(x+1)-f(x)=x^{2}+2x+1$.  Now, if we take the function $g_0(x)=x^3$, we know that $g_0(x+1)-g_0(x)=3x^2+3x+1$.  Thus if we have $f(x)=\frac{1}{3}x^3$, we have $f(x+1)-f(x)=x^2+x+\frac{1}{3}$.  However, this still leaves $x+\frac{2}{3}$ for which we must account.  Now, we can take $g_1(x)=x^2$, and $g_1(x+1)-g_1(x)=2x+1$.  Therfore, we add $\frac{1}{2}x^2$ ti=o the function, and we only have $\frac{1}{6}$ left of the difference, which we can remedy by adding $\frac{1}{6}x$ to the function, which is thus complete :  $f(x)=\frac{1}{3}x^{3}+\frac{1}{2}x^{2}+\frac{1}{6}x$, which we can validate by subtracting $f(x)$ from $f(x+1)$.  (We know that there is no zero-degree co-efficient by noting that $f(1)=1$.  Factored and put over a common denominator, the final result is $\frac{(x)(x+1)(2x+1)}{6}$.  We can verify that this always gives an integer result when $x$ is an integer by noting that if $x$ is not divisible by 2, then $(x+1)$ must be, and that if $x$ is not divisible by three, either $(x+1)$ or $(2x+1)$ must be.[/hide]
[size=167]Q.E.D.[/size][/quote]
my formula was:$\boxed{\left[\frac{2x^3+5x^2+2x}{8}\right]}$ where $[n]$ is the greatest integer function as stated above.

It is that because the formula for an odd number $n$ as stated above is $\frac{2x^3+5x^2+2x-1}{8}$

while for an even $n$ as stated above is
$\frac{2x^3+5x^2+2x}{8}$

therefore we can just put the greatest integer function on the even formula to make a general formula for all numbers

My general formula is not  $\frac{2x^3+5x^2+2x}{8}$.

Count even a 3x3 triangle
1x1 s:9
2X2 s:[b]3[/b]
3x3 s:1
that obviously does not work.

Your formula works if you want to count squares in an $n$ by $n$ grid, but mine works for an $n$ by $n$ triangle

[quote="mathnerd314"]I'm getting progress.

Let's categorize the triangles into two types: the upside-down ones and the rightside-up ones. The formula for the upside-down triangles is:

n + 2(n-1) + 3(n-2) + ...... + 3(n-2) + 2(n-1) + n

That's the only formula I could get. Now, the PATTERN (not the formula) for the upside-down triangles is:

For the n'th triangle, it's the sum of the n'th triangular number, plus the (n-2)'th triangular number, plus... and so on.

You can prove this.[/quote]
If you look at my post, you can see how I got a formula out of those patterns

Make it the ceiling, then; instead of rounding down, round up.

[quote="mathnerd314"]Just noticed, nat mc... the greatest int of

$\displaystyle \frac{2x^3+5x^2+2x-1}{8}$

is not $\displaystyle \frac{2x^3+5x^2+2x}{8}$ because you subtract 1/8 from the odd formula and get one less.[/quote]
no...
when n is odd, that formula gives youa number that is 9 mod 8. [9 mod 8]= 8 mod 8.

$\displaystyle \frac{2x^3+5x^2+2x}{8}\text{ when odd}>\displaystyle \frac{2x^3+5x^2+2x}{8}\text{ when even}$
because you have to subtract 1/8 to make it a whole number meaning it was greater than that whole number, so you use the greatest integer function to round down.

Or, to get from the odd formula to the general formula, you add 1/8, so we want to take that back off by rouunding down


You might remember my bug problem before. This problem is based on that one.

The old bug problem(solved in another topic):
There are $4$ bugs on a unit square. Suddenly, each bug begins to chase its counterclockwise neighbor. Assuming each bug travels at $1$ unit per minute, how long will it take them to crash?

Generalize to an $n$-gon.

The new bug problem(solve it here):
As the bugs travel, they "turn". For example, if one bug starts out facing due north, but then gradually comes to face due west, it will have rotated $90$ degrees. It may even be that the bugs rotate $360$ degrees. How much does each bug turn (in degrees) before they crash?

Generalize to an $n$-gon.

[quote="krustyteklown"][hide]do they ever stop turning?[/hide][/quote]

when they crash

Wait, how do they crash? Aren't their distances always going to remain the same if they all travel at the same speed in the same direction?

they keep chasing its counterclockwise neighbor, so eventually they will all crash/meet at the center

[hide="hint"]
Consider the symmetry of the n-gon, and find an invariant over time that follows from the symmetry.
[/hide]


[quote="Beat"]Prove that if $a,b,c$ are non-negative real numbers such that $a+b+c=3$, then $(a^{2}-ab+b^{2})(b^{2}-bc+c^{2})(c^{2}-ca+a^{2})\le 12$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62544

In this step, \[=((a+b)^{2}-3ab)a^{2}b^{2}\leq(9-3ab)a^{2}b^{2}\leq12\]
how is that determined? Is there any way besides homogenizing?

A famous problem of Hungkhtn appeared just as the first problem of chapter II in his book.


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that mayb no spam...but i hav no patience to read such a long post :rotfl:

Do they also give credit for referrals...it seems so , otherwise you wouldn't have provided the link:(http://cashcrate.com/1318960)
This link would have been sufficient:
[url]http://cashcrate.com/[/url]

 :D

so true........hes obv gettin money fr referrals.....but all this is a damn waste of tym... a friend of mine wasted so much tym of his on such sites but dint get a single cent,paisa,penny...nthn
 :rotfl:
@ d pic above :  :rotfl:  :rotfl:

good pic, darij

I dunno what it implies though :roll:

I'll pay you 10K bucks if you can manage to earn even 10 paise from that site!  :rotfl:  :rotfl:  :rotfl:  :rotfl:
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par used to hav a real difficult time in deciding wat is not spam..

so he left(virtually) aops :rotfl:


function
calculus
derivative
geometry
complex analysis
topology
complex analysis unsolved

assume that non constant entire function $f$ is said $g-periodic$(g is analytic) if $f(g(z))=f(z)$ .does it implies $g$ should be linear.

Yes, it does. One possible approach is to show that if $g$ is not linear and $f$ is not constant, then, for all sufficiently big $R$, we have $\max_{|z|\le R}|f(g(z))|>\max_{|z|\le R}|f(z)|$ though the proof of this statement is not very simple. I'm not posting a complete proof in hope that, perhaps, somebody will be able to offer a simpler approach...

I have some questions about this:

i) Is g entire? There seems to be a natural way to extend g to the entire plane if it isn't, but will it still be entire if we do so?

ii) If f has period g, does it follow that f' (the derivative) has period g too? It seems to do so for "normal" periods (periods of the form w, 2w, 3w and so on, like in sin, cos and exp), but does it still hold in the general case? If it does, a straightforward proof follows from the chain rule.

iii) Are there any functions with periods of the form az for any a in C? (By this I mean that f(z) = f(az) = f(a^n z) for some a in C)? It's clear that no such non-constant function exists if |a|<1 (the set {f(a^n z) : n in N} - for a constant z - has a point of accumulation at f(0)), but what about |a|>1?

[quote="Gunni"]

i) Is g entire? There seems to be a natural way to extend g to the entire plane if it isn't, but will it still be entire if we do so?
[/quote]

I assumed that g is also entire in my solution. Otherwise just take any $f$ that is not identically constant and put $g(z)=f^{-1}(f(z))$. This function is well-defined at least in some open set and, of course, it may be not linear if you choose a branch of $f^{-1}$ such that $f^{-1}(f(z_0))\ne z_0$ for some $z_0\in \mathbb C$. The simplest example of this kind is $f(z)=z-z^3$ and $g(z)$ defined as the root of $w-w^3=z-z^3$ close to $1$ when $z$ is close to $0$, i.e., $g(z)=-\frac z2+\sqrt{1-\frac 3 4 z^2 }$

[quote] ii) If f has period g, does it follow that f' (the derivative) has period g too? It seems to do so for "normal" periods (periods of the form w, 2w, 3w and so on, like in sin, cos and exp), but does it still hold in the general case? If it does, a straightforward proof follows from the chain rule.[/quote]

Of course not! The same chain rule shows that you get an extra factor $g'$ when differentiating the composition.

[quote]
iii) Are there any functions with periods of the form az for any a in C? (By this I mean that f(z) = f(az) = f(a^n z) for some a in C)? It's clear that no such non-constant function exists if |a|<1 (the set {f(a^n z) : n in N} - for a constant z - has a point of accumulation at f(0)), but what about |a|>1?[/quote]

Yes: just take $f(z)=z^n$ and any $n$-th root of unity $a$. Note, by the way, that, if $az$ with $|a|>1$ is a period, then so is $\frac 1a z$, so the case $|a|>1$ is impossible too.

would mind fedja releaseing your solution? thanks

[quote="sam-n"]would mind fedja releaseing your solution?[/quote]

Here goes. Let $D(R)$ be the closed disk of radius $R$ centered at $0$.  Consider the image $g(D(R))$. It is a compact set. Assume that the union of $g(D(R))$ and all bounded components of the open set $\mathbb C\setminus g(D(R))$ does not contain the disk $D(2R)$ (otherwise, we, clearly, have $\max_{D(R)}|f\circ g|\ge \max_{D(2R)}|f|>\max_{D(R)}|f|$ (if $f$ is not constant) and we are done. Then there exists a curve $\Gamma$ joining the circumference $\{|z|=2R\}$ and $\infty$ such that $g(D(R))\cap \Gamma=\varnothing$. Now let $\psi$ be the conformal mapping from the disk $D(R)$ to $\mathbb C\setminus \Gamma$ such that $\psi(0)=0$ and $\psi'(0)>0$. The non-trivial theorem I'm going to use is that the image $\psi(D(R/2))$ is then contained in $D(AR)$ for some absolute constant $A>0$. If we believe this, the rest is easy. Applying the Schwartz lemma to $\psi^{-1}\circ g$, we see that   $\psi^{-1}\circ g(D(R/2))\subset D(R/2)$ and, thereby, $g(D(R/2))\subset D(AR)$. If this hapens for arbitrarily large $R>0$, we can conclude that $g$ is linear.

Basically, that's it. The problem is that the full proof of the theorem I used is rather involved. Of course, it follows easily from the Koebe theorem and the Koebe theorem follows from the estimate for the second coefficient of a univalent function, which is a consequence of the area principle but I'm too lazy to rewrite 3 pages from a complex analysis textbook here. You can sort of convince yourself that the theorem holds by considering the case when $\Gamma=[2R,+\infty)$ (which is actually the worst possible case). In this case you can easily write an explicit formula for $\psi$ and check all the properties I referred to.

Hope this helps a bit :).


number theory
greatest common divisor
algorithm
function
Euclidean algorithm
algebra
rational function

[quote="mgao"]Prove that the fraction $\frac{a^{3}+2a}{a^{4}+3a^{2}+1}$ cannot be reduced, for any integer.  Not just algebraically, in real life.

 :P[/quote]

[hide]You just have to prove that the GCD of $a^{3}+2a$ and $a^{4}+3a^{2}+1$ is 1.

You just keep using the Euclidean algorithm and you will eventually arrive at $1$ and $a$ whose GCD is 1.[/hide]

If a rational function is reducible, then so is its reciprocal.  Hence, you could find the reciprocal and show that it can't be reduced and there you have it.

[quote="DanK"][hide]It's pretty easy to see that the denominator can't be factored, and that proves it right away, since you can't reduce a fracion with a prime numerator or denominator, but I'm not sure how you would write a proof showing that the denominator can't be factored.  I'm not very good at proof-writing...[/hide][/quote]
Well, the denominator of $\frac{2x}{x^{2}+1}$ can't be factored, but it can be reduced for $x=1, 3, etc.$.

Sorry, stupid mistake on my part.  Forget what I said then...


I need to convert this matrix
$ [ \left( \begin{array}{ccc}0 & 1 & 1 \\1 & 1 & 0 \\1 & 0 & 1 \end{array} \right)\]$
so that the first column is zero. I think the first entry in the first column is okay to be 1. I've tried p2:=p2-p1 but don't know where to go from here.

Okay so my code for the matrix didn't work. The matrix is first row: 011, second row: 110, third row: 101

Convert?

What exactly are you trying to do? Solve a system? Find an inverse? Find a determinant? Matrices cannot be freely converted out of context.

[quote="JasonSullivan"]$ \left( \begin{array}{ccc}0 & 1 & 1 \\
1 & 1 & 0 \\
1 & 0 & 1 \end{array} \right)$[/quote]

I should learn Linear Algebra...

So that the first column is zero. The first step to echelon form.

The first step to echelon form would be that the portion of the first column below the pivot element is zero. You can't zero the whole column.

One possible procedure (not unique): For your first step, interchange the first and second rows. For your second step, subtract what has become your new first row from the third row and place that result in the third row. Do that, and what do you have?


Assume time t runs from zero to 2pi and that the unit circle has been labled as a clock.
Match  with the best description of the curve from the following list.

A. Starts at 12 o'clock and moves clockwise one time around.
B. Starts at 6 o'clock and moves clockwise one time around.
C. Starts at 3 o'clock and moves clockwise one time around.
D. Starts at 9 o'clock and moves counterclockwise one time around.
E. Starts at 3 o'clock and moves counterclockwise two times around.
F. Starts at 3 o'clock and moves counterclockwise to 9 o'clock.

____ x=sin(t); Y=cos(t)
____ x=cos(t/2); y=sin (t/2)
____ x=-sin(t); y=-cos(t)
____ x=cos (2t); y=sin(2t)
____ x=-cos(t); y=-sin(t)

whoever solves this question first get a prize from the god.

Um I don't really understand what all the sin/cos stuff is for at the bottom...could you explain some more please?

You are suppose to match those equation with the best statement and enter the letter (a, b, c, d, e,f) in the blank.

Also, please explain how you get the answers or how to tackle this kind of problems.

First, orient yourself.  0 degrees correspond to 3 o'clock, 90 deg to 6 o'clock, 180 deg to 9 o'clock, 270 deg to 12 o'clock.

Then, work out what happens to x and y right at the beginning, at t = 0.

(1)     x=sin(t); Y=cos(t) -> x=0, y=1
(2)     x=cos(t/2); y=sin (t/2) -> x=1, y=0
(3)     x=-sin(t); y=-cos(t) -> x=0, y=-1
(4)     x=cos (2t); y=sin(2t) -> x=1, y=0
(5)     x=-cos(t); y=-sin(t) -> x=-1, y=0

What do we have?  For case (1), x is at zero, but y is at 1.  Imagine the x-y axis - the horizontal direction is x, the vertical direction is y.  So we have 0 in the horizontal direction, but 1 in the vertical direction.  Therefore the result is something that points straight up, at 12 o'clock.   We look for the option that starts at 12 o'clock - it is A.

Now look down to (3).  Same thing, but now y points straight down, to 6 o'clock.  B starts at 6 o'clock.

Now look at (2).  We have the expression (t/2) inside the sine and cosine.  That means at t = 2 pi, for example, we have sin(Pi) and cos(Pi).  In other words, the clock is going twice as slow as t.  Therefore it is the option where the clock only moves half a circle, from 3 to 9 o'clock - it is too slow to move a full circle.

Same thing for (4).  Now the clock moves twice as fast as t.  It moves two times round - option E.

Only one to go!  In (5), the clock starts at 9 o'clock.  It then moves counterclockwise - why?  Because at t=Pi, for example, we get -sin(Pi) and -cos(Pi).  Now, cos(Pi)=-cos(Pi), so that's okay, but the -sin(Pi) moves the clock in the opposite direction from t.

I hope that helps.  If you're still unclear, try working out a few values of x and y with different angles - 0, Pi/2, Pi, etc. and drawing a graph.

The key is that the clock has two parts.  x is the part of the clock in the horizontal direction (if y=0, then x=1 is 3 o'clock, x=-1 is 9 o'clock), and y is the vertical part (if x=0, y=1 is 12 o'clock, y=-1 is 6 o'clock)- you have to look at both two parts to get the final motion.


[quote="ifai"]Find the square roots of $-7+24i$.[/quote]

[hide]
$(x+yi)^2=x^2-y^2+2xyi$
So $2xy=24 \rightarrow xy=12$
I try $(x,y)=(3,4)$:
$9-16+24i=-7+24i$
We could also have $-3-4i$

So the answer is $\pm(3+4i)$.
[/hide]


function
algebra
functional equation
algebra solved

Find all the functions f:N*->N* such that
f(mn)=f(m)+f(n)+7f(m)f(n). for every m,n naturals.

Put m = n = 1, then f(1)(7f(1) + 1) = 0 which is impossible since you assume that f(1) > 0.

If you replace the consition f(n) > 0 by f(n) \geq 0, then there are an infinite number of solutions :
The functional equation may be rewritten as
7f(mn) + 1 = 7f(m) + 7f(n) + 49f(m)f(n) + 1
that is 7f(mn) + 1 = (7f(m) + 1)(7f(n) + 1).
Let g(n) = 7f(m) + 1.
Then the problem is equivalent to find the function g : |N <sup>*</sup>  - > 7|N + 1 such that g(m)g(n) = g(mn) (that is g is completely multiplicative. Note that the product of any two number of the form 7n+1 is also a number of that form).
Thus any such a multiplicative function leads to a solution, and is defined by arbitrary choices on the set of the prime numbers....

Pierre.


It is given triangle $ABC$. Circle $k$ through $B$ and $C$ meet $AB$ and $AC$ at $C_1$ and $B_1$ respectively. If $H_1$ and $H$ are orthocenters of triangles $AB_1C_1$ and $ABC$ respectively prove that $BB_1$, $CC_1$ and $HH_1$ are concurent.

http://www.mathlinks.ro/Forum/viewtopic.php?t=29893

  darij


inequalities
quadratics
inequalities proposed

A joke ?
By Am-Gm .we have:
$ LHS \ge \frac {64}{(a \plus{} b \plus{} c \plus{} d \plus{} 4)^2} \ge 1$

[quote="mitdac123"]A joke ?
$ LHS \ge \frac {64}{(a \plus{} b \plus{} c \plus{} d \plus{} 4)^2} \ge 1$[/quote]
What is the name of the inequality,that you've used...
Could you prove it?

Did you use Cauchy's?
It looks like Cauchy's than the AM-GM.  :huh:

Actually problem can be easily solved applying following "chain" of inequalities to $ LHS$

Quadratic mean $ \geq$ Arithmetic mean $ \geq$ harmonic mean and after that we use given condition.


Find all positive integers $n$ such that \[n=d_{6}^{2}+d_{7}^{2}-1\] where $1=d_{1}<d_{2}<...<d_{k}=n$ are all positive divisors of the number $n$.

[quote="PenguinIntegral"]This is an interesting problem. Could you explain that last condition more, I don't *quite* understand it. I have a few interpretations though...[/quote]

Eg. for $n=12$:
$d_{1}= 1, d_{2}= 2, d_{3}= 3, d_{4}= 4, d_{5}= 6, d_{6}= 12$.


A carton is to be designed in the shape of a cuboid that holds $ 8000$ cm$ ^3$ of material. How many square centimeters are in the surface area of a carton that uses a minimum amount of cardboard?

We want this in the shape of a cube, so each side has length $ \sqrt[3]{8000} \equal{} 20$. Thus, we find that the surface area is $ 20 \times 20 \equal{} 400 \times 6 \equal{} \boxed{2400}$ square centimeters.

Wait but wouldn't a carton have one open side?


For $ A \equal{} \varnothing$ (for instance) this obviously fails...

He's probably asking to show that metric spaces are compactly generated, which they are. It should say "where $ A$ ranges over all compact subsets of $ X$; otherwise the converse makes no sense.

[quote="HFL"]Let $ (X,d_{1}),(Y,d_{1})$ are metric space. Prove: $ f: X\rightarrow Y$ continous $ \Leftrightarrow$ $ f|_{A}$ continous.
Where $ A$ be a subset compact of $ X$[/quote]

Yes,where $ A$ ranges over all compact subsets of $ X$

[quote="HFL"]Let $ (X,d_{1}),(Y,d_{1})$ are metric space. Prove: $ f: X\rightarrow Y$ continous $ \Leftrightarrow$ $ f|_{A}$ continous.
Where $ A$ be a subset compact of $ X$[/quote]

($ \Rightarrow$), evident!
($ \Leftarrow$), $ \{x\}$ compact in $ X$ ($ \{x\}$ is the set with an element $ x$), $ f|_{\{x\}} \equal{} f(x)$, $ f|_{\{x\}}$ continous at $ x \Rightarrow f(x)$ continous at $ x, \forall x\in X \Rightarrow f(x)$ continous on $ X$

[quote="limsup"][quote="mitdacvn"]$ f|_{\{x\}}$ continous at $ x \Rightarrow f(x)$ continous at $ x$[/quote]
Can you justify this implication?[/quote]

 :lol: I've corrected.

That's not true; any function out of a single point set is continuous. You haven't used the metric anywhere. That's important.

Being compactly generated is equivalent to (or defined as) having the property that a set $ A$ is closed if and only if $ A \cap K$ is closed
for every compact $ K \subset X$.   Going right, this is clear, as every compact set in a metric space is necessarily closed since metric spaces are Hausdorff. Going left, it suffices to show we have "enough" compact sets.

Specifically, a set $ A$ is closed iff it contains all its limit points. Let $ x_1,x_2,\dots$ be a sequence converging to a point $ x \in X$; suppose furthermore that each $ x_i \in A$. We want to show that $ x \in A$. Let $ K \equal{} \{x_i\}_{i\equal{}1}^\infty \cup \{x\}$. Then $ K$ is compact, for any open set containing $ x$ contains as well all but finitely many points $ x_i$. Then if $ K \cap A$ is a closed subset of $ K$, implying that $ x \in A$. Applying this for any limit point $ x$ of $ A$ gives that $ A$ is itself closed.


geometry
vector
geometric transformation
rhombus
3D geometry
prism
geometry solved

Hi...
the problem is following:

Let's imagine n points in the plane such that there are exactly 3 neighbouring points with distance d = 1 from each point (or each point has exactly these 3 neighbours). What values can n have?

My observation:
I'm able to show the existention for n = 6k, where k \in {3, 4, 5, ...}. See this:

Now we define a "special polygon" P:
Imagine the square ABCD with side of length d = 1. Then we construct (out of the square ABCD) points E and F as vertexes of equilateral triangles ABE and CDF. This hexagon is our P. Each point of A, B, C, D has exactly 3 demanded neighbours and only points E, F have 2 demanded neighbours, because of this we'll call the points E, F as "tips".

It's easy to see that if we link two different P by abscissa of length d = 1 between their tips we obtain the structure "2P" that has 2 tips again. We can eliminate the tips if we make a cycle of i*P structures. It's impossible to make a cycle from two P, but for more P it's possible always.

But what about remaining values of n?

Thanks

A  nice question, kind of combinatorial geometry I like.

a) We prove that n has to be even :
Let E be a such with the desired property, and |E| = n. Now, form the graph whose vertices are the elements of E and the edges are the segments of length 1. Then, each of vertices has degree 3. But, it is well-known that in any graph, the number of vertices with odd degree is even. Thus, n is even.

b) We have n > 4.
Clearly, we must have n > 3. Moreover, if n = 4, then each of the points A,B,C,D must be a neighbour of each of the others. It forces that ABC and BCD has to be equilateral triangles, and then AD =/= 1.

c) Each n = 6k, for some positive integer k is admissible :
Start from an equilateral triangle ABC with side 1. Let u be a unit vector not colinear to AB,BC or CA, and A'=t(A), B'=t(B), C'=t(C), where t is the translation with vector u. Then, it is easy to verify that E = {A,B,C,A',B,'C'} is an admissible set (it looks like the projection of a right prism). Then, for each k, consider k copies of such a set, sufficiently far from each other to ensure that they do not interfer.

d) Each n = 8k, for some positive integer k is admissible :
Now, start from a segment AB of length 1. Then use a translation t with unit vector u not colinear to AB. As above, let C = t(A) and D = t(B). It is possible to choose u such that BC =/= 1 and AD =/= 1 (note that ABDC is a rhombus with sides 1, which is not form by two equilateral triangles). Then, as above, consider an unit vector v and a translation T with vector v, such that :
- For each of the point M in {A,B,C,D}, let M' = t(M),
- And, v has been chosen such that if M =/= n then MN' =/= 1 (this condition eliminates a finite number of unit vectors, so it is always possible to do such a choice).
Then, E = {A,B,C,D,A',B',C',D'} is an admissible set (it looks like the projection of a 3-d cube).
As above, you get an admissible set with n = 8k elements by k copies of E sufficientely far from each others.

e) As above, using copies, you may obtain an admissible set E with n = 8a + 6b elements for any nonnegative integers a,b such that (a,b) =/= (0,0).
Another well-known result, is that each integer p  \geq 7 may be written as p = 4a + 3b for some nonegative integers a,b. Thus, each even integer n  \geq 14 is admissible.
Since, n = 6, 8 and 12 are admissible, it follows that :
The admissible n are the even integers not less than 6, except perhaps n = 10.

I do not have an admissible construction for n = 10. Maybe later...

Pierre.

I'm so blind sometimes...
In fact, it is easier than above.

Let n = 2p, for some integer p  \geq 3.
Consider a regular convex p-gon P = A_1...A_p.
Now choose an unit vector u such that, for each i u = A_iB_i where B_i satifies B_iA_j =/= 1 for each j =/=i (it is possible because these conditions eliminate only a finite number of choices for u).
Now consider the translation t with vector u. Let A'_i = t(A_i) for each i.
Now, it is easy to see that E = {A_1,...,A_p,A'_1,...,A'_p} is an admissible set for n = 2p (it looks like the projection of a prism with basis P).
Thus, each even integer n  \geq 6 is admissible and, from above, there is no other admissible values, so we are done.

Pierre.


Compute the number of sets of three distinct elements that can be chosen from the set $\{2^1,2^2,...,2^{2006}\}$ such that the three elements form an increasing geometric progression.

So the number of tripels $(x,y,z) \in \{1,2,3,...,2006\}^3$ with $x<y<z$ and $x+z=2y$ is wanted. But thats just some simple calculation.
Chose any $x$, then there are $1002$ possibilities for $z$ with the same parity, thus there are $1002 \cdot 2006$ such pairs. But half of them must be killed since they fulfill $z<x$, thus there are $1002 \cdot 1003$ such tripels in total.

[quote="ZetaX"]So the number of tripels $(x,y,z) \in \{1,2,3,...,2006\}^3$ with $x<y<z$ and $x+z=2y$ is wanted.[/quote]

This is my solution(I find it,today  :) )

we must have $y\in \{2,3,...,2005\}$.If $k\in\{2,3,...,1003\}$ then we have $k-1$ tripels such that $y=k$.If $k\in\{1004,1005,...,2005\}$ then we have $2006-k$ tripel such that $y=k$.Final answes :$1+2+...+1002+1002+1001+...+1=1002.1003$  ;)


using Eulers theorem
$ R\left(\dfrac{3^{55}}{140}\right) \equal{} 87$

because $ (3,140) \equal{} 1$ so eulers theorem applies $ a^{\phi(n)}\equiv 1\mod{n}$
$ 3^{\phi(140)}\equiv 1\mod (140) \equal{} 3^{48}\equiv 1\mod (140)$
$ \implies 3^{48 \plus{} 7}\equiv 3^{48}\times 3^7\equiv 1\times 3^7\equiv 2187 \mod(140)$
$ \implies 2187 \equiv 87 \mod(140)$
so the remainder when $ 3^{55}$ is divided by $ 140$ is $ \boxed{\boxed{\boxed{\boxed{\color{red}87}}}}$

[quote="iin77"]No calculator.

What is the remainder when $ 3^{55}$ is divided by $ 140$?[/quote]

[hide="Answer"]Using the Totient theorem, we have $ a^{\phi(n)}\equiv 1 \mod n$. We plug in what we have into this: $ 3^{\phi(140)}\equiv 1\mod 140\implies 3^48\equiv 1\mod 140$. Since we had $ 3^55$ in the original, we have to add 7 to the LHS $ \implies 1(3^7)\cong 1(2187)\mod 140$. Note how you can multiply both sides by $ 3^7$. Using Euler's theorem we have $ R\left(\dfrac{3^55}{140}\right)\equal{}87$. Thus, we have $ 2187\equiv 87\mod 140$. Since 2100 is divisilbe by 140, we have $ 1\cong 87\mod 140$, so the remainder will be $ \boxed{87}$.[/hide]


The goal is to change the previous word. You can add a letter, remove a letter, or change a letter to form a new word. Keep track of the words that we have gone through. You are not allowed to repeat words. If a word is not changed for an entire week, whoever came up with it will win. The number of points the winner will gain is equal to the total number of words for that round. That user will start the next round. You are not allowed to win at the beginning of a round, and you are not allowed to make consecutive word changes. I will start.

[size=200]ROUND 1[/size]

sun

stun

 The message is too small. Please make the message longer before submitting.

spun

The message is too small. Please make the message longer before submitting.

sun
stun
spun
shun
shin
shiny
shine
[b]shrine[/b]

sun
stun
spun
shun
shin
shiny
shine
shrine
[b]shrines[/b]

sun
stun
spun
shun
shin
shiny
shine
shrine
shrines
shrinks
shrink
[b]shrunk[/b]

sun
stun
spun
shun
shin
shiny
shine
shrine
shrines
shrinks
shrink
shrunk
[b]shrank[/b]

[hide="start hiding"]sun
stun
spun
shun
shin
shiny
shine
shrine
shrines
shrinks
shrink
shrunk
shrank
shank[/hide]

sun
stun
spun
shun
shin
shiny
shine
shrine
shrines
shrinks
shrink
shrunk
shrank
shank
spank

NO ERNIE :P

[hide="5"]
Dog
hog
bog
blog
blogs[/hide]

[hide="6"]

Dog
hog
bog
blog
blogs

logs[/hide]

[hide="8"]

Dog
hog
bog
blog
blogs
logs
hogs
dogs[/hide]

[hide="9"]Dog
hog
bog
blog
blogs
logs
hogs
dogs
digs[/hide]

Hey I just realized I haven't been on this thread since the last chain was on 110 or something.

[hide="10"]Dog
hog
bog
blog
blogs
logs
hogs
dogs
digs
pigs[/hide]

[hide="11"]
Dog
hog
bog
blog
blogs
logs
hogs
dogs
digs
pigs
pegs
[/hide]

[hide="12"]
Dog
hog
bog
blog
blogs
logs
hogs
dogs
digs
pigs
pegs
pets
[/hide]

[hide="13"]
Dog
hog
bog
blog
blogs
logs
hogs
dogs
digs
pigs
pegs
pets
gets
[/hide]

mods please lock this its too old, and i am not doing it anymore. it is a bad thing when the creator does not post in here for a while


√(〖1-((.5)/φ)〗^2 )  ×  1/φ=

What does this simplify to?  If it can be simplified.

φ = phi = irrtiational # approx 1.618

[quote="jli"][quote="chickendude"]LaTeX is useful
Just toss latex code in dollar signs

$\sqrt{ [1-\frac{.5}{\phi}]^{2}}\cdot \frac{1}{\phi}$

[hide]
If you look under the radicand, inside the parentheses, you'll notice that

$1-\frac{.5}{\phi}$ has a positive value, so the squaring and the square root will cancel without sign issues

$\left(1-\frac{.5}{\phi}\right) \cdot \frac{1}{\phi}$

is that what you're looking for?
[/hide]

EDIT:
oops, I didn't realize it was a double post
I'm sorry[/quote]

[hide]Remember that $\phi^{2}=\phi+1$

The radicand is positive so it becomes $\left(1-\frac{.5}{\phi}\right)\cdot \frac{1}{\phi}$

Becomes $\frac{1}{\phi}-\frac{.5}{\phi^{2}}=\frac{1}{\phi}-\frac{1}{2\phi+2}$

$=\frac{\phi+2}{2(\phi^{2}+\phi)}=\frac{\phi+2}{2(2\phi+1)}$

I think thats its most simplified form[/hide][/quote]

Rationalize the denominator...   :P


in adad gong ast chon hich vaght dor gardesh nadard!


[quote="mathubu"]$ABCD$ is a  quadrilateral.$X,Y,Z,T$ are the arbitrary points on $[AB],[BC],[CD],[AD]$ respectively.$[XZ]\cap[YT]$=$M$ and $R$ and $S$ are the points on $[MT]$ and $[MY]$ respectively. $R$ and $S$ satisfied that  $RA,BS,XZ$ are meet on same point  also  $RD,CS,XZ$ are meet on same point.

Prove that $R$ and $S$ can be found for all  $X,Y,Z,T$

sorry to my bad english :blush:[/quote]

Look,if I understood the problem well, I can say that it is a false statment.

Let's suppose that $Z$ and $X$ are such a points that $ZX$,$AD$ and $BC$ are concurent(you obviously can choose them that way).

So, if the statment of the problem is correct then there are two points $R$ and $S$ on the segments $TM$ and $MY$ so that $AR$ and $BS$ intersect at $XZ$ ,and $DR$ and $DS$ intersect on $XZ$.But as we supposed $AD$ and $BC$ intersect on $XZ$ too,therefore, for the triangles $ADR$ and $BCS$ from Desarg's theorem we can say that $AB$,$RS$ and $DC$ are concurent.So,the lines $AB$,$TY$ and $DC$ are concurent too,which contradic to the statment that $T$ and $Y$ are  randomly selected on the segments $AD$ and $BC$ :wink: .

You are right.I am sorry to wrong statement...It seems was true but I realized that it is false after your answer
Thanks :oops:


Find all positive integrers $x,y,z,m,n$, $(m,n)\ge 3$ such that

$x^m+y^m=z^m$
$x^n+y^n=z^n$

I have alredy posted this problem, but I have no answer......
You can't use Fermat Last Theorem, because this problem was proposed before Wiles' demonstration.......

We can suppose that $m\geq n.\;$ Then we have \[ z^m=(x^n+y^n)^{m/n}=x^m+y^m+\alpha=z^m+\alpha \] so $\alpha=0,\;$ that is, we have either $x=y=0\;$ or $m=n\;$ which is a pretty hard case  ;)

[quote="bodan"] $\[ z^m=(x^n+y^n)^{m/n}$[/quote]

I don't get this part.....

[quote="ZetaX"]You have $z^n=x^n+y^n$, now take it to the $m$-th power and take the $n$-th root. Done.[/quote]

Got it


trigonometry
calculus
integration
function
Galois Theory
calculus computations

Prove the following equations:

1.    $y^{'}+y=\frac{1}{1+x^{2}}\ \ \ \ ,\ \ \ \ y(0)=0$

2.    $\frac{dy}{dt}=\left(\tau\cos t+T\right)y-y^{3}$
($\tau$ and $T$ are constants)

in#1 just using the technique of integration factor may be enough. The tough part is to determine $\int_{0}^{x}{\frac{e^{x+c}}{1+x^{2}}dx}$

FOR (1) MULTIPLE
:e.^x .in two phase of the equations then you want to evalute :U= int{/frac{{e.^{x}}/1+x.^2}. and your answer is:e.^(-x)*U

For (2) is the case of the bernoulies equation then you assume :z=y.^(-2)
at the end of the case you should evalute:int{e.^{sin(t)} dt}.
If you want to find an analytical solution for equations this is method and if you want to find solutions you can find a series  for its solutions. :idea: [/code]

what about the result of the integral in #1?Seems very hard to get an answer. can anyone post any result? i couldn´t get anywhere until now.

[quote="fedja"]Not "hard" but "impossible": the antiderivative is not an elementary function. So, don't waste your time trying to find an explicit formula.[/quote]
Do you actually know this somehow or is it an educated guess?

Well... there is an [url=http://en.wikipedia.org/wiki/Differential_Galois_theory]entire theory[/url] dedicated to that.

If one introduces the integral into Mathematica, one gets that the antiderivative is a combination of [b]ExpIntegralEi[/b]s. If this integral would be elementary, then [u]probably[/u] ExpIntegralEi would be an elementary function, which is known not to be true (I'm guessing here).

[quote="Kalle"]
Do you actually know this somehow or is it an educated guess?[/quote]
In this case I "know it somehow".  Read [url=http://math.hunter.cuny.edu/ksda/papers/churchill.pdf]this[/url] and you'll know it too ;).


Harvard
college
MIT
pigeonhole principle
binomial coefficients
complementary counting

Here is the problem:

Four students from Harvard, one of them named Jack, and five students from MIT, one of them named Jill, are going to see a Boston Celtics game. However, they found out that only 5 tickets remain, so 4 of them must go back. Suppose that at least one student from each school must go see the game, and at least one of Jack and Jill must go see the game, how many ways are there of choosing which 5 people can see the game?

There are a number of posted answers. I've got two different solutions and cannot figure out why they are not equal.

[hide]
1. Use inclusion/exclusion:

Jack, someone from MIT, 3 more $ \equal{} 1 \cdot \binom{5}{1} \binom{7}{3}$ = 175

+

Jill, someone from Harvard, 3 more $ \equal{} 1 \cdot \binom{4}{1} \binom{7}{3}$ = 140

-

Jack, Jill, 3 more $ \equal{} 1 \cdot 1 \cdot \binom{7}{3}$ = 35

So 175 + 140 - 35 = 280.

2. Use three mutually exclusive groups

Jack but not Jill $ \equal{} 1 \cdot \binom{4}{1} \binom{6}{3}$ = 80

+

Jill but not Jack $ \equal{} 1 \cdot \binom{3}{1} \binom{6}{3}$ = 60

+

both Jack and Jill $ \equal{} 1 \cdot 1 \cdot \binom{7}{3}$= 35

So 80 + 60 + 35 = 175.
[/hide]

[hide="Explanation"]Well, both answers are obviously wrong in that there can be only $ \binom{9}{5} \equal{} 126$ ways to distribute the tickets independent of any restrictions, and both your solutions present a number greater than $ 126$.

You cannot multiply the binomial coefficients as you have done because the two events are linked; you overcount on these situations because you assume that the "someone from MIT/Harvard" is special, but this is never the case. You would be safer using complementary counting; there are $ \binom{8}{4}$ ways to distribute the tickets once you give one to Jack. In fact, all of those ways are valid, since by pigeonhole, at least one student will have to be from MIT.

Then you do the same with Jill, except one of those ways is banned because all $ 5$ students would be from MIT.

Then you have $ \binom{7}{3}$ ways where both Jack and Jill can attend.

$ 2\binom{8}{4} \minus{} 1 \minus{} \binom{7}{3} \equal{} \boxed{104}$[/hide]

You, sir, darkdieuguerre, are a genius.
Your brief explaination cleared the complexity of the problem with facility.
Thank you for making it clear to me.

[hide="Alternatively"]
You can use complementary counting since there are only two restrictions in this problem.
To make 5 people with NEITHER Jack nor Jill is simple $ \binom{7}{5}$ and there is only one way for all of the five people to be from the same school (MIT). Note that these two conditions do not overlap because obviously Jill must be included in the same-school selection because there are only 5 people.

So its just:

$ \binom{9}{5} \minus{} \binom{7}{5} \minus{}1 \equal{} \boxed{104}$[/hide]