Asymptote LaTeX geometry incenter parameterization function circumcircle \/closed

Hi all,

We've installed [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help]Asymptote[/url] on the site, so you can use it with $ \text{\LaTeX}$ to generate images. Of course some of you may have some questions, and we will try to answer them in here.

Click on the above name to read a very good tutorial on how to use this. I recommend you try to create the images on your local computer, and then paste the correct code inside asy tags:[code][asy]<-- insert code here -->[/asy][/code]

The official webpage for this program is: http://asymptote.sourceforge.net/ You can read the Wiki tutorial on how to download and install on your computer (however you don't need to download / install anything to actually use it on the site, just as with $ \text{\LaTeX}$).

As usual I will ask you to [b]NOT ABUSE[/b] this new feature in any way. Any attempts will be logged and punished.

You can use the [url=http://www.mathlinks.ro/Forum/index.php?f=224]Test Forum[/url] to test some pictures. Here's a demo:

[asy]size(100); pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); I=incenter(A,B,C); draw(A--B--C--cycle); dot(O); dot(I); draw(circumcircle(A,B,C)); draw(incircle(A,B,C)); label("I",I,W); label("O",O,S);[/asy]
The image above was produced using the following code:
[code][asy]size(100);
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
O=circumcenter(A,B,C);
I=incenter(A,B,C);
draw(A--B--C--cycle);
dot(O);
dot(I);
draw(circumcircle(A,B,C));
draw(incircle(A,B,C));
label("I",I,W);
label("O",O,S);[/asy][/code]

Make sure you use the proper size argument (around 100 - 200 should be fine), don't make unnecessary long pictures.
Please ask [b]ALL THE QUESTIONS REGARDING PROGRAMMING (CREATING PICTURES) WITH ASYMPTOTE IN THE [url=http://www.mathlinks.ro/Forum/index.php?f=123]LaTeX and Asymptote Forum[/url][/b].

[quote="jhredsox"]Wow, cool. I think the administraters on this forum really go above and beyond for this forum, even though they don't have to. It is great that we can have all the features right here and available.

Thank you![/quote] :blush: You're welcome. Anyway more stuff is coming in the future so stay tuned :P

Oh, no I don't have a good excuse to not put pics with geometry stuff...jk. This looks awesome! Hopefully I'll get around to learning it way faster than I did $\LaTeX$...

woah hahaaaa i didn't even know this was possible :jump: now we don't have to post the picture separately as an attachment :)
thank you!

I knew about Asymptote but never tried learning it (too lazy), but I will definately make an effort later on.

With Asymptote programmed into the forums, it will definately help with the geometry problems people post and the confused questions that might be asked. Thanks admins!

It's always easy to learn by testing simple stuff (putting a dot, creating a line, a circle) and by looking at other people's code. If you install it on your computer as well you can practice easier. In a couple of hours you will realize you can create powerful pictures ;)

Wow, great addition!

Nice job admins!

We should get someone to make some beginners guides to help most of the population here.

Are there going to be asy tags up soon?

[quote="math92"]Are there going to be asy tags up soon?[/quote]No they will not appear on the list of tags. If you cannot write the tags manually then it's doubtful you'll be able to write any good code anyway :)

[quote="DPatrick"][quote="math92"]We should get someone to make some beginners guides to help most of the population here.[/quote]A good place to start is the [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Basics]Asymptote Basics[/url] page on the AoPSWiki.

Note that the [url=http://asymptote.sourceforge.net/]official Asymptote project page[/url] likes our wiki so much that they link to it directly! :)[/quote]

I checked out the Asymptote website and looked at their introduction manual. You can really do a lot with Asymptote.

And @Valentin: I know that writing asy between two []'s is really easy, but using the tags at the top automatically closes them if you forget or something.  If someone messed up the tags and doesnt go back to look at their post, then their work becomes useless until they go back and realize it.

Still, you guys did a nice job on getting this whole thing working!  :D

Can we draw commutative diagrams in Asymptote or is there some other package installed?

You can draw commutative diagrams using $\LaTeX$.

[quote="dragonx"]this looks really complicated so can we have a test page sort of like teXer[/quote]I strongly recommend you install it on your personal computer to make tryouts and learn. It's the easiest fastest and simplest way to do it.

It seems like everyone really likes Asymptote a lot. I cant wait for everyone to become experts and start using it in almost every post.

[quote="i_like_pie"]I tried making some of the pictures in the Wiki in Asymptote, but it wouldn't compile. Will this every be possible?[/quote]yeah.. up to a while ago i don't think you could use some of the macros and packages but i think most of them seems to work at the moment. like graph.asy, math.asy, and three.asy work at least.. but you can't make [code][asy]import olympiad;
pair A,B,C,O,I;
A=origin; B=2*right; C=1.5*dir(70);
O=circumcenter(A,B,C); // olympiad - circumcenter
I=incenter(A,B,C); // olympiad - incenter
draw(A--B--C--cycle);
dot(O);
dot(I);
draw(circumcircle(A,B,C)); // olympiad - circumcircle
draw(incircle(A,B,C)); // olympiad - incircle
label("I",I,W);
label("O",O,S);
[/asy]
[/code]
which should produce
[img]http://www.artofproblemsolving.com/Wiki/images/f/fe/Figure8.gif[/img]
but i don't think the administrators have included Maria Monk's "olympiad" package yet  :(

EDIT:  :blush: o wait it does work  :huh: [asy]pair A,B,C,O,I; A=origin; B=2*right; C=1.5*dir(70); O=circumcenter(A,B,C); /* olympiad - circumcenter */ I=incenter(A,B,C); /* olympiad - incenter */ draw(A--B--C--cycle); dot(O); dot(I); draw(circumcircle(A,B,C)); /* olympiad - circumcircle */ draw(incircle(A,B,C)); /* olympiad - incircle */ label("I",I,W); label("O",O,S);[/asy]
i guess just don't include the word "import" and it will work maybe..
also you could just look up the code for each of the functions in the packages but that is tedious hehe

[quote="Valentin Vornicu"]I said above the olympiad package is included by default. All other packages must be used with import command at the beginning (the import command will not work for anything else but the default packages).[/quote]oh ok thanks :)

[quote="i_like_pie"]I tried drawing a line (nothing special, just a line) in the Wiki, but it just ignored the [asy] tags.[/quote]Wiki doesn't recognize the forum tags.

WOW! this is really cool! how come ive never heard of it before?

As of now by clicking on (most of) the asy pictures (and if you have Javascript on - 95% of the current browsers have) you will have a popup with the code for that picture.

After clicking on the asy pictures, is the current page supposed to go "up" (I mean that it has the same effect as pressing [b]Home[/b])?

BTW, very nice addition to the forum! Congrats :D

[quote="hmmm"]Great move, AoPS!  I have a feeling this will be much more convenient than uploading image files all the time, especially for little quick diagrams.

-hmmm[/quote]

To give credit where credit is due, thanks to you for introducing us to Asymptote, and for writing up the how-to pages in AoPSWiki!

Wow, the official tutorial even has a link to AoPS!  :o

And and

[quote]
After clicking on the asy pictures, is the current page supposed to go "up" (I mean that it has the same effect as pressing Home)? [/quote]

I think that was because what it did before was <a href="#" onClick="blahblahblah">, and an empty anchor links to the top of the page.


Source :  2006 AMC 12A, Problem 23

Given a finite sequence $S=(a_1,a_2,\ldots ,a_n)$ of $n$ real numbers, let $A(S)$ be the sequence
\[ \left(\frac{a_1+a_2}{2},\frac{a_2+a_3}{2},\ldots ,\frac{a_{n-1}+a_n}{2}\right)  \]

of $n-1$ real numbers.  Define $A^{1}(S)=A(S)$ and, for each integer $m$, $2\le m\le n-1$, define $A^{m}(S)=A(A^{m-1}(S))$.  Suppose $x\ge 0$, and let $S=(1,x,x^2,\ldots,x^100)$.  If $A^{100}(S)=(1/2^50)$, then what is $x$ ?

\[ \text{(A) }1-\frac{\sqrt{2}}{2}\qquad \text{(B) }\sqrt{2}-1\qquad \text{(C) }\frac{1}{2}\qquad \text{(D) }2-\sqrt{2}\qquad \text{(E) }\frac{\sqrt{2}}{2}  \]

(And by the way, where is the source bar?)

[hide="solution"]
i think this question was a bit easier than a usual #23...
just try finding $A(S),A^2(S), A^3(S)$
there is a  CLEAR pattern

$A^{100}(S)=\frac{(1+x)^{100}}{2^{100}}$ from here it is trivial, set it equal to $\frac{1}{2^{50}}$ and you get $\boxed{B}$[/hide]

This was clever, I liked it
but it was too intimidating during the test... I got it afterward


real analysis
integration
calculus
derivative
real analysis unsolved

Let f is lebesgue measurable and $ \int_{a}^{b}fdm\ge 0$ for all a, b in R , then show that f(x)>=0 a.e.

Do you know the Lebesgue theorem on the differentiation of the integral?


geometry
rectangle
circumcircle
geometry theorems

[color=darkblue]Let $ABC$ be the $A$- rectangle triangle with $m(\widehat{BCA})=15^{\circ}$. Denote : the altitude $AE$, where $E\in BC$ ; the bisector ray $[AD$ of the angle $\widehat{BAC}$, where $D\in BC$ ; the its circumcircle $w=C(O,R)$, i.e. $a=2R$ and the point $O$ is the middlepoint of the side $[BC]$ ;
the line $AD$ meets again the circle $w$ in the point $E$. Suppose $AD=1$. Then : $4\cdot AE=BC$ ; $R=\sqrt 3\ ;\ DO=1\ ;\ DE=2\ .$[/color]

Also, if $AH$ is the altitude now: $AH= \frac{BC}{4}$

See the attached image, it seems no words are necessary. $ D'$ is the simmetrical of $ D$ about the midpoint of $ AE$.

Best regards,
sunken rock


trigonometry
limit
inequalities
function
algebra
domain
real analysis

Let $x_{1}<x_{2}<...$ be all positive roots of equation $x=\tan{x}$. Find $\lim_{n\to+\infty}(x_{n}-x_{n-1})$.

First we prove that $n\pi <x_{n}<\left(n+\frac{1}{2}\right)\pi.$ :wink:

Examine the increase and decrease of the function $x-\tan x.$

If exist $\lim_{n\to+\infty}(x_{n}-x_{n-1})$ then by Stolz-Cesàro theorem we have $\lim_{n\to+\infty}(x_{n}-x_{n-1})=\lim_{n\to+\infty}\frac{x_{n}-x_{n-1}}{n-(n-1)}=\lim_{n\to+\infty}\frac{x_{n}}{n}={\pi}.$

Can anyone solve this problem without using Stolz-Cesaro? :D

[quote="N.T.TUAN"]But at this time, I don't know exist or NOT exist limmit, also Stolz-Cesàro theorem can't work.[/quote]

1.Let f(t)=t/tant, then f(t)↓

2.x(n) - x(n-1) = tan[x(n) - x(n-1)]*[1+x(n)*x(n-1)],
so [x(n) - x(n-1)]/tan[x(n) - x(n-1)] < [x(n+1) - x(n)]/tan[x(n+1) - x(n)]

3. x(n) - x(n-1) > x(n+1) - x(n) >0, so that limmit exits.

Rewrite the equation $x=\tan x$ as $x\cos x-\sin x=0.$

Let $f(x)=x\cos x-\sin x.$ Then $f((n+1/2)\pi)=(-1)^{n}$ and $f'((n+1/2)\pi)=(-1)^{n}(n+1/2)\pi.$

Then one step of Newton's method gives as an approximation of the root of $f$ the number

$a_{n}=(n+1/2)\pi-\frac{1}{(n+1/2)\pi}.$

We can now show that on the interval $[a_{n},(n+1/2)\pi],$ $f''(x)$ has the opposite sign to that of $f((n+1/2)\pi).$ Examining the picture of Newton's method (or any other means of analysis of the error in Newton's method) shows that $f$ must have a zero in the interval $(a_{n},(n+1/2)\pi).$

Using other methods (such as what kunny suggested above), we can show that there are no other zeros of $f,$ although the existence of a root in $(0,\pi/2)$ gives the root in $(a_{n},(n+1/2)\pi)$ the label $x_{n+1}.$ Thus:

$(n-1/2)\pi-\frac{1}{(n-1/2)\pi}<x_{n}<(n-1/2)\pi$

$(n+1/2)\pi-\frac{1}{(n+1/2)\pi}<x_{n+1}<(n+1/2)\pi$

Subtracting,

$x_{n+1}-x_{n}<(n+1/2)\pi-\left[(n-1/2)\pi-\frac{1}{(n-1/2)\pi}\right]$ and

$x_{n+1}-x_{n}>(n+1/2)\pi-\frac{1}{(n+1/2)\pi}-(n-1/2)\pi$

Thus, $\pi-\frac{1}{(n+1/2)\pi}<x_{n+1}-x_{n}<\pi+\frac{1}{(n-1/2)\pi}.$

The limit is clear from there. If I were asked to produce a table of approximate values of $x_{n}$ (one could certainly produce eigenvalue problems arising from PDE's - perhaps Schrödinger's equation - that call for this), I would phrase it as finding the zeros of $f(x)$ and run Newton's method forward from each $a_{n}.$

I claim that $x_{n}\approx \left(n-\frac12\right)\pi-\frac{1}{\left(n-\frac12\right)\pi}$ is a valid asymptotic formula. More detailed examination of the method should be capable of producing an asymptotic formula with more terms.

[quote="N.T.TUAN"]But $x_{n+1}-x_{n}$ and $x_{n}-x_{n-1}$ both aren't lies domain monotone of $\frac{x}{\tan{x}}$.[/quote]

Yes, they do.
f'(t) = (sin2t - 2t)/[2(sint)^2] <0 for all t in (Pi/2, 3Pi/2)
and Pi/2 < x(n) - x(n-1) < 3Pi/2.

Thanks, you are right  :blush:
[quote="liaoge"]
Yes, they do.
f'(t) = (sin2t - 2t)/[2(sint)^2] <0 for all t in (Pi/2, 3Pi/2)
and Pi/2 < x(n) - x(n-1) < 3Pi/2.[/quote]
@Kent Merryfield: I will check your post later :) thanks.


$ \triangle ABC$, O is the incenter. A line through O intersects AB, AC at M and N respectively. Prove that $ \frac {BM.CN}{AM.AN} \leq \frac {BC^{2}}{4.AB.AC}$

I really need you help in this problem.
Can anyone give me the solution?

Let $\mathcal{H}$ be the circumconic of $\triangle ABC$ tangent to lines $BO,CO.$ Define $P \equiv CM \cap BN$ and $Q \equiv AP \cap BC.$ Since $N(B,C,Q,M) \equiv N(P,A,Q,M)=-1,$ it follows that $MO$ is the polar of $Q$ WRT $\mathcal{H}$ $\Longrightarrow$ $P \in \mathcal{H}.$ Trilinear equation of the circumconic $\mathcal{H}$ tangent to $\alpha-\beta=0$ and $\alpha-\gamma=0$ is then $\alpha^{-1}=\beta^{-1}+\gamma^{-1}.$ Since $P(\alpha,\beta,\gamma)$ lies inside $\triangle ABC,$ then $\alpha,\beta,\gamma >0.$ Therefore, by GM-HM we have:

$\sqrt{\beta\gamma} \ge \frac{2\beta\gamma}{\beta+\gamma}=2\alpha  \Longrightarrow  \frac{1}{4} \ge \frac{\alpha}{\beta} \cdot \frac{\alpha}{\gamma}=\frac{ \sin \widehat{PCB}}{\sin \widehat{PCA}} \cdot \frac{ \sin \widehat{PBC}}{\sin \widehat{PBA}} $

Then substituting the expressions,

$\frac{BM}{AM}=\frac{BC}{AC} \cdot \frac{ \sin \widehat{PCB}}{\sin \widehat{PCA}} \ , \ \frac{CN}{AN}=\frac{BC}{AB} \cdot \frac{ \sin \widehat{PBC}}{\sin \widehat{PBA}} $

$\Longrightarrow  \frac{1}{4} \ge \frac{BM}{AM} \cdot \frac{CN}{AN} \cdot \frac{AB \cdot AC}{BC^2}  \Longrightarrow  \frac{BC^2}{4 \cdot AB \cdot AC} \ge \frac{BM}{AM} \cdot \frac{CN}{AN}.$


trigonometry
logarithms
integration
calculus
calculus computations

I can't get any of the answer choices for this differential equation problem:

Find the solution to the differential equation $\frac{dy}{dx}=\frac{\sin{x}}{e^{y}}$, where $y\left(\frac{\pi}{4}\right)=0$.

$(A)\ y=\ln\left(\cos{x}-\frac{\sqrt{2}}{2}\right)$

$(B)\ y=\ln\left(-\cos{x}+\frac{\sqrt{2}}{2}\right)$

$(C)\ y=-\frac{\cos{x}}{e^{y}}$

$(D)\ y=\ln(-\cos{x})$

$(E)\ y=\ln(-\cos{x})+\frac{\sqrt{2}}{2}$

[hide="heres what i did"]$e^{y}dy=\sin{x}dx$

$\int e^{y}dy=\int \sin{x}dx$

$e^{y}=-\cos{x}+C$

$y=\ln(-\cos{x})+C$

But this gives $0=\ln\left(-\frac{\sqrt{2}}{2}\right)+C \implies C=\ln(-\sqrt{2})$, which doesn't give one of the answer choices...[/hide]

Am I doing something wrong here, or is there a mistake in the problem?

[quote="DanK"]$e^{y}=-\cos{x}+C$

$y=\ln(-\cos{x})+C$
[/quote]
$y=\ln (C-\cos x)$ and $C-\cos x>0$.

[quote="DanK"]I thought about that, but that gives $0=\ln\left(C-\frac{\sqrt{2}}{2}\right) \implies C-\frac{\sqrt{2}}{2}=1 \implies C=\frac{2+\sqrt{2}}{2}$, which isn't an answer choice...[/quote]

I think all the answer choices are incorrect. For one thing, $y\left(\frac{\pi}{4}\right) \neq 0$ in any of them.

That's what I was thinking, but I wanted to make sure it wasn't just me being dumb or anything.  Thanks for the help...

Yes I get $y=\ln\left({\frac{2+\sqrt2}{2}-\cos{x}\right)}$

But that could be possible for some complex x right?


MATHCOUNTS AMC USA(J)MO USAMO ARML email

[quote="indianamath"]anybody know the scores that Indiana individuals received? i heard that Justin got 1st written with 41(wow-didn't expect that) and Richard got 14th with a 35. ANybody know some other scores?[/quote]
Lyndon got 27th, I dont know what the fourth place person got but I bet that [i]might[/i] be a reason why Indiana didnt get 1st on team..
I didnt even know the guy who got 1st nationally, never heard of him..but now all hes going to have to do is improve on his speed and ability to read and comprehend the questions because a 7th grader written is pretty impressive.

Oh yea..does anyone have a copy of the questions?  I just want to see how I would do and its good AMC practice..

hahaha a white kid won......................whos going to cry?

well thats not nice to say...
I'm gonna find out what the other member of the team got tomorrow. hehe Mr. Fisch is either gonna be really happy because 2nd is good or he'll be mad because SOMEONE brought down the team average.

[quote="Fanatic"]You might want to edit that out lol. HINT HINT NUDGE NUDGE DO IT NOW EDIT.

We weren't even close to texas, and we only beat new jersey by one sprint rond question (.25 team points).

Justin:

25 + 16
1st

Richard:

25 + 10 (ouch ouch ouch ouch ouch 10)
14th

Lyndon:

23 + 10
27th

Andrew:
16 + 10
Something between 98th and 108th.[/quote]

What was your Team Round score? Texas? New Jersey

We had 10 on the team round, same with texas... not sure about NJ, but they were really close to beating us

Do you know how many teams got 10 on team round?

According to CircleSquared in the Mc forum, only 2 teams did. Texas and Indiana. Don't know if that's right though.

Edit: NVm, now someone says the statistics were printed incorrectly, so many teams did get 10 or 9

I thought the 4th person on the team was named anderw?.. :maybe:

on a sidenote the top 5 teams all got 10 and new jersey and pennsylvania deadlocked for 3rd and did a team tiebreaker

[quote="Fanatic"]Not trying to offend you guys, but you guys seemed very. very. very. bitter towards 4th place. Whereas we also got 2nd place, albeit by a much larger margin, but none of us are blaming andrew for anything...[/quote]
The comment in question is in no way representative of any non-empty sample of the views of the 05 IN team.

Fine, maybe it isn't. That was just my perception. Heck, even Dr. Chi seems very bitter... he brings it up all the time.

[quote="mathcrazed"]David!! You have an Aops account?? Since when lol? Nice job on the USAMO btw[/quote]
[quote="area on left of screen"]Joined: 10 Apr 2004[/quote]
Yes!! I have an Aops account!! Since then lol. Nice job on Mathcounts btw

:)

Ya i noticed -_-... u're going to ARML right? Unless there's like MOSP those days

[size=75]The sole point of that was to mimic your post as closely as possible. :P[/size]

Anyway, no, I am not going to ARML... as of... some time after the first and before the second practice. I made USAICO this year (yay! :D) and they are mutually exclusive. In fact, they are so more-or-less by definition. Both are scheduled so as not to conflict with MOP[size=59]1[/size] and stuff, which leaves pretty much only one place for them to be.

(Your dad already knows. I sent him an email a while back)

[size=59]1[/size] Don't believe what the official documents tell you. The S is a by-product of a deranged imagination.

I know that NJ got 10 on team round - as did PA - but I thought NJ lost to you (Indiana) by more than a little -- when Fanatic says you only won by a single sprint question - did you/he mean, on average, if EACH of NJ team members got one more question right (or if 2 kids each got 2 more sprint right) - you'd be tied??   Or did only ONE NJ kid need to get One more sprint right?  Are you SURE this was the case? How do you know for sure -- were the actual results made known to your coach -- and then to you? If the coach is reading this, please feel free to answer as well!! (And Congrats to you and your whole team!)

And while everyone was guessing that you guys (in Indiana) and Texas would rank high - why do you think people SO underestimated states like NJ and PA which were in contention for both high spots teamwise and individually?  Just wondered....... I mean in the last few years, not one, but two NJ guys took first, and NJ came close to having a 3rd First place winner this yr! (and PA came close too!)

[quote="Fanatic"]Uh, you do realize that no one thought indiana would do well? 9/10 of Indiana didn't even think indiana would do well. People were thinking we'd be lucky to get in the top 10.[/quote]

i was in the 9/10

[quote="solafidefarms"][quote="nat mc"][quote="indianamath"]i'm pretty sure *some dude* screwed up the team.[/quote]
sounds familiar... *cough* 2005 *cough*[/quote]

This isn't my forum, but I am going to ask that you think about how that 4th person feels and what they would think were they to read that. It wouldn't help their feelings any.[/quote]

edited

[quote="Fanatic"]I'd just like to ask the moderators not to get mad at the entire Indiana forum because of the behavior of this ONE idiot  :roll:[/quote]

OK, I erased it...my god.

PEOPLE HERE CANT HANDLE A JOKE..

Not only did you fail to actually annoy anyone, you also failed to be funny.

Great joke you got there

hey lyndon would you mind posting here what you told me and youkow on sunday?

this has something to do with mathcounts nationals indiana team, and i wont elaborate.

I'm so late but...

Yeah would you mind? I've heard 10 different stories from 10 different people, and I'd like to get the original story from the original's mouth.


For what integers $ n$ can we triangulate a regular polygon with $ n$ sides into isosceles triangles using non-intersecting diagonals? ([i]Non-intersecting diagonals[/i] mean diagonals intersecting only at the vertices of the $ n$-gon.)

No one? :(  I am sure some of you saw something similar...at least this summer :P
Is this problem too esay? I think it is nice but i have second thoughts when no one replies :mad:

Still no one? What is wrong eith this problem? :(  :(


function
real analysis
real analysis solved

hi all,

here is a very nice pb:
find all functions f:R->R so that if the series sum(u_n,n=1..infinity) converges then the series sum(f(u_n),1..infinity) converges too.

bye.

we can prove that f must be linear on a open ball containing 0....

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31422[/url] for (a sketch of) a solution.


vector
combinatorics unsolved
combinatorics

Find closed formula for number of 0 - 1 matrices dimension n times m, such that no two ones are adjacent (share common edge).
For m = 1, we get Fibonaci numbers :-)
Thanks!

Nobody helped me  :(
I thought that this was an easy problem, but I couldn't solve it for hours...

I doubt there is a neat closed form for the general $n\times m$ case.  I think the best you can do is probably a reccurence relation.

For some $m$, let $v_1,v_2,\cdots,v_{F_{m+1}}$ be the $F_{m+1}$ different row vectors with the property that they contain no two adjacent ones.  Let $V(n,i)$ be the number of $n\times m$ matrices whose last row is $v_i$.  Then you get a reccurence of the form $V(n+1,i)=\displaystyle\sum_{j=1}^{F_{m+1}}a_jV(n,j)$, where $a_j=0$ if the scalar product of $v_i$ and $v_j$ is non-zero, and $a_j=1$ otherwise.

Then the total number of $n\times m$ matrices of this form will be $\displaystyle\sum_{i=1}^{F_{m+1}}V(n,i)$.

It might be possible to extract a neat closed form from these recurrences, but I doubt it.


[color=darkblue]Find:
 $ lim \displaystyle \left ( \frac{1}{2}\plus{}\frac{1}{2.4}\plus{}\frac{1}{2.4.6}\plus{}...\plus{}\frac{1}{2.4.6....(2n)} \right )$[/color]

Can you post full solution? How you find it.

$ e^x \equal{} 1 \plus{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} ... \plus{} \frac {x^n}{n!} \plus{} ...$

$ \sqrt {e} \equal{} e^{\frac {1}{2}} \equal{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{2!}\cdot\frac {1}{2^2} \plus{} \frac {1}{3!}\cdot\frac {1}{2^3} \plus{} ... \plus{} \frac {1}{n!}\cdot\frac {1}{2^n} \plus{} ...$

easy to see that our sum is $ \sqrt {e} \minus{} 1$

P.S. $ 2\cdot 4\cdot 6\cdot ...\cdot 2n\equal{}2^{n}n!$  :wink:


Does anyone know where I can find past IMO best solution?
I wanna see it.

Thanks.

my opinion:

No, probably you can't find them in the Internet. But there are some in some websites, yeah i remember i had posted here one of them. You can look for it here in this forum! Best Regards. If you will need a help then tell ... contact me by pm

davron latipov

What do you mean by "best" solutions?

Here on this forum, [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=16]all the IMO problems have been posted[/url] (click on the year and then on the problem number to go to the respective thread). Not all of them are solved, but thanks to everybody being able to reply, you will find some solutions here which you won't find in standard literature.

Kalva's site also has [url=http://www.kalva.demon.co.uk/imo.html]an IMO section[/url] with all problems up to 2003 solved. Alas, the site has never been updated since 2003...

The [url=http://www.imo.org.yu/]IMO Compendium[/url] contains all IMO problems solved, but only some sample chapters are free to download.

Finally, there are the three famous books: [url=http://www.amazon.com/International-Mathematical-Olympiads-1959-1977-Library/dp/0883856271/sr=8-3/qid=1167478062/ref=sr_1_3/104-6381389-1878340?ie=UTF8&s=books]IMO 1959-1977[/url], [url=http://www.amazon.com/International-Mathematical-Olympiads-Supplementary-1978-1985/dp/088385631X/sr=8-5/qid=1167478062/ref=sr_1_5/104-6381389-1878340?ie=UTF8&s=books]IMO 1978-1985[/url], [url=http://www.amazon.com/International-Mathematical-Olympiads-1986-1999-Problem/dp/0883858118/sr=8-1/qid=1167478062/ref=sr_1_1/104-6381389-1878340?ie=UTF8&s=books]IMO 1986-1999[/url], with solutions, some extra exercises and theory explanations.

  Darij

[quote="darij grinberg"]What do you mean by "best" solutions?[/quote]
I mean the best solution prize in IMO. Maybe, it is called a special prize.
Thank you.

[quote="OHO"]I mean the best solution prize in IMO. Maybe, it is called a special prize.[/quote]

There is a subtle difference - special prize solutions are not "best" solutions, they are in some way the most surprising or creative solutions. I don't think they even have to be shorter than the proposed solutions, let alone more conceptual.

Alas, I don't think the special prize solutions have been collected anywhere. The three books above contain some of them, probably the ones by the US team. However, here on MathLinks you are most likely to find nonstandard solutions to numerous IMO problems...

  Darij

The special-prize solution of the IMO 2005 by Iura exist in the forum . :wink:
There was a special-prize-solution in the IMO 2006 ?

ha-ha
It is a really good idea to collect them and make a pdf file etc. To bring that solutions in one - sounds cool!

p.s.: why don't we do this stuff?

davron latipov

[quote="silouan"]
There was a special-prize-solution in the IMO 2006 ?[/quote]

I think there was no Special Awards given in Slovenia... Correct me if I'm wrong...

Yes, you are right! [there was NO!]

davron latipov


[quote="sylow_theory"][quote="EquinoX"]1. An aquarium pool has volume 2*10^6 liters. The pool initially contains pure fresh water. At t =0
   minutes, water containing 10 grams/liter of salt is poured into the pool at a rate of 60 liters/minute.
   The salt water instantly mixes with fresh water, and the excess mixture is drained out of the pool at
   the same rate (60 liters/minute)
   a. Write a differential equation for S(t), the mass of salt in the pool at time t.[/quote]

$\frac{dQ}{dt}= 600-\frac{60Q}{2\cdot 10^{6}}, Q(0)=0$[/quote]

can you tell me where this answer comes from??


$Fe2O3$ your told to find the fourmla mass would it be $2*(55) + 3*(16)$ then you add...?

my teacher gave us this pattern: Particles $6.02*10^{23}$ ----> Mole ----> Mass

is that right?

right....for the particles, you start with a number of particles, the use stoichiometry with avogadros number to conver to mols then to grams

You don't have to add anything to the sum of the atomic weights time the number of atoms in the molecule.  The result is the weight (in grams) of a gram-mole of molecules (6.02X10^23).  It happens also to be the weight (in pounds) of a pound-mole of molecules (multiply the 6.02X10^23 by the number of grams in a pound).

Most folks just use the atomic weght and end up with the weight of the molecule in the desired units (grams, kg, pounds, tons, etc)  The will react with the same ratio of other reactant molecules in the weight (mass) ratio that is the same as the molecular weight calculated this way.

Hope this helps.


It would be nice if one could hide threads in certain forums in the "View posts since last visit" list.

Agreed.  The FTW threads, for example, can get very cluttering.

[quote="isabella2296"]I think that's a good idea, too. But rather than a "hide" option, maybe it would be better to have a "choose" option.[/quote]
I agree--you could make it like the main screen of AoPS...click the arrows to hide/open the sections of the forum

This time, I even agree with myan... ;)

  dg

Alright I will think about this, and see how to implement it. It was on the list of things to do, but unfortunately doesn't have that much of a priority.


geometry
Euler
vector
inequalities
geometric transformation
rotation

Can anyone give me a link, a document or any other material, teaching of using complex numbers in geometry problems? Thank you in advance.

I'm looking for books, articles, etc that would expand my knowledge of using complex numbers in geometry  :D I mean so texts beyond basics. I'll be glad to see any link

How about this book?

[b]Complex Numbers and Geometry [/b]written by Liang-shin Hahn

From the text of introducing author:

The purpose of this book is to demonstrate that complex numbers and geometry can be blended together beautifully,resulting in easy proofs and natural generalizations of many theorems in plane geometry---such as the theorems of Napoleon,Ptolemy---Euler,Simson,and Morley.

kunny

[b]kunny[/b] I have nothing against this book :) [I like the introduction] but there's no chance I get any title unless it is published on Internet freely - so that's why I'm asking for links.

Thanks for showing your interest though. :)

You can try 南盘江's post in the pre-olympiad section:

http://www.mathlinks.ro/Forum/viewtopic.php?t=41553

What you are referring to I assume is the geometry of complex numbers.
Preparatory things include learning displacement vectors, inversion of geometry in complex plane, Mobius transformations etc.
I think a nice starting point is proving Ptolemy's inequality with it.

Bomb

Well,

   if you need such problems with solutions you can read the new(and very nice book) by Titu Andreescu and Dorin Andrica << Complex Numbers from A to Z >>, Edition birhauser.

You can find this book searching in

                         http://www.amazon.com

    Put   ''  Titu andreescu ''  , ''  Dorin Andrica  '' and you will have the book.
  See also my post in High School section , Books and Journals. I have the link there. The best book solving difficult problems in geometry with comples numbers is << Modenov : Problems in geometry , edition MIR >> ,you can fie book only in big limbraries.

  Of course there are also and other books but you can not buy them( they are not available now).

[u] babis[/u]

90% of geometry problems can be solving with useing [b]complex numbers[/b] :lol: (at laste for me ;) ),but for same problems that solutions is wery long :( .
For example,the 5th problem of the lest IMO I solved with conplex numbers,and I did have 6 point for that problem :lol:.(they take 1 point ,becaus in my solution was be samething juste a little not clear :blush: )

I know that - everything can be solved using complex numbers, but more often than not the "complex" solution is literally [b]complex[/b]  ;) I meant not very hard problems that I could use to gain some skill in doing geometry using complex numbers. Any likns/problems?

We can write here some tricks :). For example in complex numbers inversion of point $z$ in $(0,0)$ circle (with unit radius) is $\frac{1}{x\bar}$

[quote="stergiu"] The best book solving difficult problems in geometry with comples numbers is << Modenov : Problems in geometry , edition MIR >> ,you can fie book only in big limbraries.
[u]Babis[/u][/quote]
I have that book :P  :D .
Also ther is too nice book"Yaglom:Geometry and complex numbers" ;)

In the section "Geometry" are many solutions using the complex numbers for proposed problems. Read them!

[quote="stergiu"]I have seen this book in the literatur , but can I find it in english or only in russian?

                 "Yaglom:Geometry and complex numbers"

  [u]babis[/u][/quote]
If you want I can say you how to find  Russian wersion of  that boof in internet ;) ,but I dont know is ther that book in english :( .

Given a point $P_0$ in the plane of the triangle $A_1A_2A_3$. Define $A_s = A_{s-3}$ for all $s \geq 4$. Construct a set of points $P_1, P_2, P_3, ...$ such that $P_{k+1}$ is the image of $P_k$ under a rotation center $A_{k+1}$ through an angle $120^{\circ}$ clockwise for $k = 0, 1, 2, ...$ . Prove that if $P_{1986} = P_0$, then the triangle $A_1A_2A_3$ is equilateral. (IMO 86)

It is given a convex pentagon $ABCDE$, such that $AB=BC=CA$ and $CD=DE=EC$. Let $S$ be the medicentre of $\triangle ABC$ and $M$ and $N$ - the midpoints of $BD$ and $AE$ respectively. Prove that triangles $SME$ and $SND$ are similar.

A non-isosceles triangle $A_1A_2A_3$ has sides $a_1, a_2, a_3$ with ai opposite $A_i$. $M_i$ is the midpoint of side ai and $T_i$ is the point where the incircle touches side $a_i$. Denote by $S_i$ the reflection of $T_i$ in the interior bisector of $\angle A_i$. Prove that the lines $M_1S_1$, $M_2S_2$ and $M_3S_3$ are concurrent. (IMO 82)

Anyways, Tiks is so right.. Most of the geometry problems [i]must[/i] have a solution with complex numbers

if u know russian go to http://ilib.mirror0.mccme.ru/ from there you can download Yaglom's book(a classic)

if u dont know, take my advice: Try to learn!

You can study the book written by Arthur Engel, from GIL, Romania.

[quote="Xixas"]There are two articles in the Mathematical Excalibur on this theme.[/quote]
I googled it and found these two articles:
[url]http://www.math.ust.hk/excalibur/v1_n3.pdf[/url]
[url]http://www.math.ust.hk/excalibur/v9_n1.pdf[/url]
I hope it will help. :)

yes, complex number in geometry is very powerful, but I don't know how can we use it in locus, maxima or minima. who can help me and give me some example. And who can give me some tips when solve the problem with circle (because when solve, it's so complex  :lol: ) thanks.

I'm getting familiar with mobius transformation right now but I haven't seen any applications of it on olimpiad problems. Does anybody know a problem that could be solved using it ?


you got %DELTA ABC m%angle A = 48 , m%angle C = 18 find m%angle DBC if D %element AC and AB = DC

Problem: In $ \triangle ABC$, $ \angle A=48^\circ,\angle C=18^\circ$.Point $ D$ is on $ [AC]$ such that $ AB=DC$.Find measure of $ \angle DBC$.

Answer: $ 12^\circ$

Let's take a point $ D'$ on $ AC$ such that $ \angle D'BC=12^\circ$.We will prove $ D=D'$.Let's assume that $ AB=DC=1$.
$ \frac{AB}{\sin30}=\frac{BD'}{\sin48}...(1)$ and $ \frac{D'C}{\sin12}=\frac{BD'}{\sin18}...(2)$ .
from $ (1)$ and $ (2)$,$ D'C=\frac{2.\sin48.\sin12}{\sin18}=\frac{\cos36-\cos60}{\sin18}=1$.Therefore $ DC=D'C=1$ and $ D=D'$.Finally, $ \angle DBC=12^\circ$
Lokman GÖKÇE

nice solution by trigonometry, does anybody got a pure geometry solution?


Simplify the expression:

$ \frac {a^7}{(a \minus{} b)(a \minus{} c)} \plus{} \frac {b^7}{(b \minus{} c)(b \minus{} a)} \plus{} \frac {c^7}{(c \minus{} a)(c \minus{} b)}$

Thanks in advance, need help with this one.

Thank you for the reference.
I saw how you can express those sums, but I don't see how it helps with simplifying it, for $ k \equal{} 7$.
I've worked it out to this point:

$ S \equal{} \frac {a\cdot\sum^{5}_ {i \equal{} 1}a^ib^{5 \minus{} i} \minus{} b\cdot\sum_{i \equal{} 1}^{5}c^ib^{5 \minus{} i}}{(b \minus{} c)(a \minus{} b)}$

If anyone could continue from there or some other way...

And latex question, why is the sum not showing the numbers above/below it?

Have you seen Darij Grinberg's post:[url]http://www.mathlinks.ro/viewtopic.php?p=143167#143167[/url]?

[code]\sum_{i=1}^n a_n[/code]

You would see that $ \sum_{i\equal{}1}^n a_n$.

I saw, but how does that help? We get

$ a^7(b \minus{} c) \plus{} b^7(a \minus{} c) \plus{} c^7(a \minus{} b)$ and it still needs simplification..

And I don't see how writing it as:

$ [a^2(b \minus{} c) \plus{} b^2(a \minus{} c) \plus{} c^2(a \minus{} b)]\cdot C_5(x,y,z)$

helps either. If you see how, could you post the solution?

I still can't get the sum to look right, is it because it's on a fraction? The same code gives me:

$ \sum^{5}_ {i \equal{} 1} a^ib^{5 \minus{} i}$

Maybe you would factorize $ a^7(b\minus{}c)\plus{}b^7(c\minus{}a)\plus{}c^7(a\minus{}b)$.

I tried that, and the problem is a factorizing problem, so that's the idea.

I have spent a lot of time trying to factor this, which is why I'd like help with it.

Here's where I think the solution is headed

$ \frac{a^7b\minus{}a^7c\plus{}b^7a\minus{}b^7c\plus{}c^7a\minus{}c^7b}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$

$ \equal{}\frac{ac(c^6\minus{}a^6)\minus{}b^7(c\minus{}a)\minus{}b(c^7\minus{}a^7)}{(a\minus{}b)(b\minus{}c)(c\minus{}a)}$

$ \equal{}\frac{\cancel{(c\minus{}a)}(ac\cdot \sum_{i\equal{}0}^{5}c^ia^{5\minus{}i}\minus{}b^7\minus{}b\cdot\sum_{i\equal{}0}^{6}c^ia^{6\minus{}i})}{(a\minus{}b)(b\minus{}c)\cancel{(c\minus{}a)}}$

And in that way cancelling the denominator... but I get lost in those sums, so I hope to see a solution continuing where I left off or in some other way..

hello, i have got this
$ a^5\plus{}a^4 b\plus{}a^4 c\plus{}a^3 b^2\plus{}a^3 b c\plus{}a^3 c^2\plus{}a^2 b^3\plus{}a^2 b^2 c\plus{}a^2 b c^2\plus{}a^2c^3\plus{}a b^4\plus{}a b^3 c\plus{}a b^2 c^2\plus{}a b c^3\plus{}a c^4\plus{}b^5\plus{}b^4 c\plus{}b^3 c^2\plus{}b^2c^3\plus{}b c^4\plus{}c^5$
Sonnhard.


I have the following recurrence that I am trying to come up with atleast a simplified version if not a closed form.

$ T(n) \equal{} T(n\minus{}1) \plus{} \sum_{i\equal{}1}^{(n\minus{}1)/2} [(n\minus{}(i\plus{}1)) * (i\minus{}1) * 2 \plus{} 2]$

in addition if n is even I must add the following to T(n)
$ ((n/2) \minus{} 1)^2$

If any of you can help that would be awesome.

Motivation?  Is this from a counting problem?

$ T(2k \plus{} 1) \equal{} T(2k) \plus{} 2\sum_{i \equal{} 1}^{k} [(2k \minus{} i) * (i \minus{} 1) * 2] \plus{} 2k$
$ T(2k) \equal{} T(2k \minus{} 1) \plus{} 2\sum_{i \equal{} 1}^{k} [(2k \minus{} i \minus{} 1) * (i \minus{} 1) * 2] \plus{} 2k \minus{} 2$

Still working

Turns out I was able to enter the first few results here and come up with what the pattern was.

http://www.research.att.com/~njas/sequences/

Thanks for trying though guys.


Find the remainder obtained for (5555)^2222 + (2222)^5555 being divided by 7.

http://www.mathlinks.ro/viewtopic.php?t=180357
http://www.mathlinks.ro/viewtopic.php?t=128920

Search function makes wonders.

Thanks , i just wanted to know the samething , if this is just another modulo problem.
Is there any way of doing it through binomial expansion?? or would that be super lengthy??


Let $ F_1$ and $ F_2$ be families of subsets of set of $ n$ elements such that $ |A\cap B|$ is odd for all $ A\in F_1$ and $ B\in F_2$. Prove that $ |F_1|\cdot |F_2|\leq 2^{n\minus{}1}$.

http://www.mathlinks.ro/viewtopic.php?t=290708

  darij


Fie $\ a,b,c,d$ numere reale strict pozitive astfel incat $\ a<c\leq\ d<b$ si $\ ab<cd$.Daca $\ a,b$ reprezinta lungimile laturilor dreptunghiului $\ D_1$, iar $\ c,d$ lungimile laturilor dreptunghiului $\ D_2$, sa se arate ca $\ D_1$ poate fi introdus in $\ D_2$ daca si numai daca avem inegalitatea $\ (b^2-a^2)^2\leq\ (bc-ad)^2+(bd-ac)^2$.

     Eu unul n-am stiut s-o fac... :(  Poate da cineva o solutie frumoasa pentru problema asta??Macar sa fie de nivel de clasa a VIII-a.

ce inseamna "poate fi introdus" ?
eu am inteles ca exista o izometrie $f$ astfel incat $f(D_{1})\subseteq D_{2}\cup \mbox{int}(D_{2})$.
totusi problema de pe shortlist..deci cred ca e destul de grea. o sa ma gandesc la ea diseara.

Ei utilizeaza coordonatele...nu prea stiu asa de bine.


Can you edit font color, table color, and etc in the ljdigital template? I edit thing and submit them and no changes are made.

[quote="4everwise"]It's one of the 3 templates that you can base your blog on.[/quote]The blog system is not (yet) very well done. Some errors might appear, but usually it should work :) If it doesn't then I'll add this to the bug list, but we can't deal with it right now.


On the mathcounts countdown round tests you often see problems that are like: "please answer as a decimal to the nearest hundreth"..

Lets say an answer is 11.34 the correct answer is eleven and thirty four hundrendths. But if I said eleven point three four, would it be right?

4.5 whereas its four point five ; four and a half.... would any of those be right?

Thanks

If it says to the nearest hundredths, then say 11 point whatever.
-It's easier than saying 11 and whatever hundredths
-I think they will not accept your answer if you say a fraction, but I wouldn't know     because I've never tried it before :D

I know that they will not accept your answer if you put it in fraction form when they as for a decimal. When it is 13.56 you can just say thirteen point five six?

How do you say 11(3/5)? I know how to write the answers.....just not sure how to say them.

You mean like eleven and three fifths?

And, yes, they do want it as a decimal in the form of 13.56.


wow my first functional equation!

if $x=y=0 \rightarrow f(0)=f^2(0) \rightarrow f(0)=0 \vee 1$

$y=0 \rightarrow x^2+2f(0)=f(x)(f(x)+f(0))$
for $f(0)=0 \rightarrow f^2(x)=x^2$

so one case is $f(x)=\pm x$
is it right?

Put x = y = 0, we will obtain $f(0)=0$ or $f(0)=1$.
Put y = 0, $x^2+2f(0)=f(x)(f(x)+f(0))$
Case 1 $f(0)=0$
Then $f(x)=x^2$. Therefore, $f(x)=x$ or $f(x)=-x$.
Case 2 $f(0)=1$
Then $x^2+2=(f(x))^2+f(x)$
$(f(x)+\frac 1 2)^2=x^2+\frac 9 4$.
$f(x)=\pm \sqrt{x^2+\frac 9 4}-\frac 1 2$.

By direct checking, $f(x)=x$, $f(x)=-x$, $f(x)=\sqrt{x^2+\frac 9 4}-\frac 1 2$ or $f(x)=-\sqrt{x^2+\frac 9 4}-\frac 1 2$ all satisfy the given condition.

Remark:
We have found four solutions only. There could be other solutions unless we can prove the uniqueness.
Probably the given condition will imply some nice property of $f$, like continuity or monotonicity?

There arent any more solutions because the method used guarantees uniqueness.

Common error:
$f(x)^2=x^2$ does NOT imply that $f(x)=x$ or $f(x)=-x$.  What it implies is that for each $r \in \bf R$, $f(r)=\pm r$.  But there are many (more that the continuum-many) other functions with this property.

That is why I claimed that we have found four solutions and it doesn't imply they are all the solutions.

[quote="violetcraze"]If $x^2 + y^2 + 2f(xy) = f(x + y). (f(x)+f(y))$
Find all functions $f: R \to R$[/quote]

$x = y = 0 \implies f(0) = 1$ or $f(0) = 0$.

[b]Case[/b] $f(0) = 0$:

$y = 0 \implies f(x)^2 = x^2$
$y = -x \implies 2x^2 + 2f(-x^2) = 0 \implies f(-x^2) = -x^2$ or in other words $f(x) = x$ for non-positive $x$.

Take $x = s, y = -t$, where $s$ and $t$ are positive, and $s \ne t$. Then

$(s-t)^2 = f(s-t)(f(s) - t)$
$(s-t)^4 = (s-t)^2[f(s)^2 - 2tf(s) + t^2]$
$(s-t)^2 = s^2 - 2tf(s) + t^2$
$f(s) = s$

Hence we can conclude $f(x) = x$.

[b]Case[/b] $f(0) = 1$:

$y = 0 \implies x^2 + 2 = f(x)[f(x) + 1] \implies \left( f(x) + \frac{1}{2} \right)^2 = x^2 + \frac{9}{4}$
$x = 1, y = -1 \implies 2 + 2f(-1) = f(1) + f(-1) \implies f(1) = 2 + f(-1)$

We test the possible values of $f(1)$ and $f(-1)$ and we get a contradiction.

[quote="luimichael"]
By direct checking, $f(x)=x$, $f(x)=-x$, $f(x)=\sqrt{x^2+\frac 9 4}-\frac 1 2$ or $f(x)=-\sqrt{x^2+\frac 9 4}-\frac 1 2$ all satisfy the given condition.

[/quote]

[quote="probability1.01"]
Hence we can conclude $f(x) = x$

[/quote]

oh well......is it just one solution, or four, or infinitely many? which one is correct....?   :?

I checked again. My solution is INCORRECT.

$f\ : \ R\rightarrow R$ so that for any $x\in R$ , $\left[f(x)\right]^2=x^2$ $\Longleftrightarrow$ there is $E\subset R$ so that for any $x\in E$ , $f(x)=x$ and for any $x\not\in E$ , $f(x)=-x\ .$ Therefore, the solution of this functional equation is :

$\Phi=\left\{f_E: R\rightarrow R\ ,\ E\subset R\ |\ f(x)=x\ ,\ x\in E\ ;\ f(x)=-x\ ,\ x\not\in E\right\}\ .$

[u]Examples :[/u] $f_R(x)=x\ ;\ f_{\emptyset}(x)=-x\ ;\ f_{[0,\emptyset)}(x)=|x|\ ;$

$f_{\{1\}}(x)=1$, if $x=1$ and $f_{\{1\}}(x)=-x$ if $x\ne 1$ a.s.o.


geometry
circumcircle
incenter
trigonometry
Gauss
geometry proposed

Let $ \triangle ABC$ be a triangle.Let $ \omega$,$ \zeta_B$,$ \zeta_A$ are the circumcircle,$ B$-exincircle,$ A$-exincirlce,respectively and $ B_1$,$ A_1$ be the tangency points of the $ \zeta_B$,$ \zeta_A$ with the sides $ AC$,$ CB$ respectively.
Let $ I$ be the incenter of the $ \triangle ABC$.Denote $ P\in CI\cap\omega$.Let $ l$ be a line through $ P$,perpendicular to $ CP$.Denote $ Q\in\l\cap AB$.
Prove that $ QI\|B_1A_1$.
[img]http://img212.imageshack.us/img212/283/1stproblemjt6.png[/img]

Let $ w$ be a incircle of $ ABC$ and it touches $ BC, CA, AB$ at $ D,E,F$ repectively..
The circumcircle of $ ABC$ and $ AEF$ intersect at $ X(\not \equal{} A)$.
Because $ I,D,P,Q$ is concyclic, $ \angle IQD \equal{} \angle IPD$.
It is easy to prove that $ XB : XC \equal{} FB: EC$
But $ FB: EC \equal{} AB_{1}: AA_{1}$, $ XBC \sim AB_{1}A_{1}$.
It ie enough to show that $ \angle AA_{1}B_{1} \plus{} \angle IQD \equal{} \angle B$. To prove this,
Since $ \angle AA_{1}B_{1} \equal{} \angle XCB \equal{} \angle XPB \equal{} \angle B \minus{} \angle XPA$, we will show that $ \angle XPA \equal{} \angle DPI$,
or $ X, D, P$ is colinear.
But $ XB: XC \equal{} BF: CE \equal{} BD: CD$, $ XD$ is a bisector of $ \angle BXC$..... Now it is obvious.

Oh sorry.... $ A\rightarrow C, B\rightarrow A, C\rightarrow B$...

can be solved by trigonometry too....

when we put $ X$ a intersection point of $ CI$ and $ A_1B_1$

and assume that $ BC > CA$
just calculate $ \tan PIQ$ and $ \tan CXB_1$

then two are same to $ \dfrac{\sin \dfrac{C}{2}}{\sin \dfrac{A \minus{} B}{2}}$

To Little Gauss:
Actually,an easier version of your proof can be obtained,if we prove that $ \triangle CB_1Q\sim\triangle BDP$,where $ D$ is th feet of the perpindicular from
$ I$ to $ AB$ and $ Q\in CI\cap A_1B_1$.


inequalities
number theory unsolved
number theory

For any nonnegative integer $n$, let $f(n)$ be the greatest integer such that $2^{f(n)}|n+1$. A pair $(n,p)$ of nonnegative integers is called [i]nice[/i]  if $2^{f(n)}>p$. Find all triples $(n,p,q)$ of nonnegative integers such that pairs $(n,p),(p,q)$ and $(n+p+q,n)$ are all nice.

A solution is $(0,0,0)$. If $n=1$, we have $f(n)=f(1)=1$. This case is simple, and if I didn't skip anything, we get solutions $(1,1,1)$ and $(1,0,0)$.
Suppose $n>1$. We have the inequalities $p<2^{f(n)}\leq n+1$, so $p\leq n$ and $q\leq p$. We have $\frac{n+p+q+1}{2^{f(n+p+q)}}<\frac{n+p+q+1}{n}\leq\frac{3n+1}{n}$. Since $t$ is a positive integer, we must have $t$ equals either $3$ or $1$.
If $t=3$, then we must necessarily have $n=p=q$, so $n=2^{r}-1$. But then $n+p+q+1=3n+1$ is not divisble by 3, however $n+p+q+1=t\cdot2^{f(n+p+q)}=3\cdot2^{f(n+p+q)}$. Impossible.
So $t=1$. Let $n+1=2^{r}\cdot s$, with $s$ odd. We then have $n+p+q+1=(n+1)+p+q=2^{t}$. So $2^{r}\cdot s+p+q=2^{t}$. Assume either $p$ or $q$ is not 0, so $t>r$, and hence $p+q$ is divisible by $2^{r}$. Because $(n,p)$ is nice, $p\leq2^{r}$. Let $p+1=2^{r}-2^{k}\cdot m$. Assume that $m>0$, then $q<2^{k}$. But then $p+q<2^{r}-2^{k}\cdot m-1+2^{k}\leq 2^{r}-1<2^{r}$, so $p+q$ cannot be divisible by $2^{r}$. Let $m=0$, then $p+1=2^{r}$. To have $p+q$ divisible by $2^{r}$, we must have $q-1$ dibisible by $2^{r}$. But $q<p+1=2^{r}$. We deduce $q=1$. Then $n+p+q+1=2^{r}\cdot s+(2^{r}-1)+1=2^{r}(s+1)=2^{t}$. Then $s+1$ is a power of 2.
Solutions are $(n,p,q)=(2^{r}(2^{b}-1)-1,2^{r}-1,1)$.
We are left with the case $p=q=0$, which gives $2^{r}\cdot s=2^{t}$, so $s=2^{l}$, so another solution is $(n,p,q)=(2^{a},0,0)$, $a>0$.


Cada casilla de un tablero de 8x8 está coloreada de blanco o de negro, cada operacion permitida consiste en elegir en el tablero cualquier rectangulo de 3x4 o de 4x3 que cubran exactamete 12 casillas y cambiar simultaneamente los colores de esas 12 casillas( las blancas pasan a negras y las negras pasan a blancas).
Diremos que una coloracion es buena si es posbile, mediante una sucesion de operaciones permitidas, obtener a partir de ella un tablero que tiene todas las casillas blancas en caso contrario, la coloracion es mala.

Demostrar que el número de coloraciones es mayor o igual que 15 veces el numero de coloraciones buenas.

en el tablero de 8*8 hay 60 maneras posibles de ubicar un tablero de 4*3 o 3*4, si cambio los colores de uno de dichos tableros 2 veces lo estaria dejando como al principio(para formalizar esto es posible hacer una invariante), luego en el tablero a lo mas se pueden hacer 60 cambios, de un conjunto de 60 cambios es posible escoger 1, 2, 3, ....; esto es exactamente la cantidad de subconjuntos=2^60, entonces habran a lo mucho esta cantidad de tableros mediante el cual se llega a una coloracion buena de un total de 2^64=posibles coloraciones. la cantidad total de coloraciones es a lo mas 16 veces el numero de coloraciones buenas. [u]Ni se te ocurra presentar exactamente la misma solucion en la tarea![/u]


function
integration
calculus
calculus computations

Is the fourier transform of $ f(x) \equal{} e^{ \minus{} |x|}$ the zero function ?
I calculated $ \int_{\mathbb{R}}e^{ \minus{} |x|}e^{ \minus{} itx}\ dx$ and my solution was $ 0$. Is this right ?

answer will be $ \frac{2}{t^{2}\plus{}1}$
since :$ \int_{\mathbb{R}}e^{\minus{}|x|}e^{\minus{}itx}dx\equal{}2Re \left( \int_{0}^{\infty}e^{\minus{}x(\minus{}it\plus{}1)}dx\right)$
it can't be zero since if you put $ t\equal{}0$ then you will have $ \int_{R}e^{\minus{}|x|}dx$ which will be positive since $ e^{\minus{}|x|}$ is positive .

After my 3 calculation I have the same, I made a stupid mistake everytime. Thanks


calculus
integration
trigonometry
function
algorithm
calculus computations

$ \int \frac {1}{x} \tan^{ \minus{} 1} x dx \equal{}$

hello, Mathematica gives the following answer
$ \frac{i (\text{Li}_2(\minus{}i x)\minus{}\text{Li}_2(i x))}{2}$
Sonnhard.

thanks..

how can we know that there is no elementary function that is the solution of the integration?

hello, i think this will help you
http://en.wikipedia.org/wiki/Risch_algorithm
Sonnhard.


the book says that gravity exits in orbit around the earth. WHy then does an astronaut feel weightlessness in orbit?

The astronaut feels [i]apparently[/i] weightlessness, the only force acting on the astronaut is the gravitational force, therefore the astronaut cannot feel her weight, and appear to be weightless.


prove that:
for p prime p>3
[tex]p^{2} \equiv 1 \mod 12[/tex]

[hide]12 | (p+1)(p-1)

p is prime, so we can write p as: 2k+1 with k integer.  2 | (p+1) AND 2 | (p-1)

So, 4 | (p+1)(p-1).

Let's prove: 3 | (p+1)(p-1)

p prime so, working modulo 3:

p=2 [3] OR p=1 [3]

So: p-2=p'+1= 0 [3]  OR p-1=0 [3]

3 | p+1 OR 3 | p-1, and  4| (p+1)(p-1)

Then, 12 | p^2 - 1   [/hide]

HUGO

Mod edit--I put this solution in spoiler. Please take the time to do this! It's appreciated at the Getting Started and Intermediate levels. -JSRosen3

This is a very good problem, but because of the notation, and because we now have FIVE levels of problem-solving forums, I can't regard this as a Getting Started problem. I will move it to Intermediate Forum, which is the natural level of problems using secondary school (or early competition program) math such as this.

Thanks for sharing the problem and the solution. We generally ask that solutions be posted in spoiler text, so that more than one participant may attempt to post a solution in each thread.

[quote="icare"]attempt[/quote]

?? :?: :?

what does that mean?

lol.  I tried to write something in "spoiler" text to see.

You can prove that $p^2 \equiv 1 \pmod{24}$.

jhaussmann5 wrote:You can prove that .

You weren't asking a question but I ll prove this anyway

[hide]Assuming p>3

We want

or

Which means that p^2-1=(p-1)(p+1) must be a factor of 24.  We know that p is prime and greater than 3.  Thus it is not a multiple of two.  Using this logic p-1 and p+1 are both multiples of 2 and one of them is a multiple of 4.  So p^2-1 is a multiple of 8.  Then since p is prime either p+1 or p-1 will be a multiple of 3.  So p^2-1 is also a multiple of 3 and thus a multiple of 24.  [/hide]

shorter:

[hide]p = 1 mod 2 so p = 1 mod 4

p = 1 or 2 mod 3 so p = 1 mod 3

combining gives p = 1 mod 12[/hide]

Mod edit--I put this solution in spoiler. Please take the time to do this! It's appreciated at the Getting Started and Intermediate levels. -JSRosen3


Let m be a positive integer and let {z(k)} be a sequence of n integers such that m =< n .
Prove the statements :
1) There is {u(i)} sub-sequence of {z(k)} such that  : Sum{u(i)} = 0 (modulo m)
2) If each z(k) is coprime with m  then there is {v(j)} sub-sequence of  {z(k)} such that : Product{v(j)} = 1 (modulo m)

1) Let v(i)=[sum of first i members of z(k)] mod m.

If v(j)=0 for some j then sum of z(1) to z(j)=0 mod m.

If not, then we have n>=m elements v(j) taking m-1 possible values, so for some j1 and j2, v(j1)=v(j2).
In which case sum of z(j1+1) to z(j2)=0 mod m.

2) Similar argument, but with v(i)=[product of first i members of z(k)] mod m. Either some v(j)=1, or v(j1)=v(j2) for some j1 and j2. If the latter, we can conclude that the product of z(j1+1) to z(j2)=1 mod m because all z(k) and hence all v(j) are co-prime to m.


Prove that there are no pairs of positive integers $a,b$ that solve the equation $4a(a+1)=b(b+3)$.

[i]This problem was solved by only one competitor...I got 2 out of 8 points.[/i]

[hide]I think I did this right...
Expand to get $4a^{2}+4a=b^{2}+3b$ which is equivalent to $4a^{2}+4a-(b^{2}+3b)=0$.  Solving for $a$ we get $a=\frac{-4\pm\sqrt{16+4(b^{2}+3b)}}{8}$ which simplifies to $-\frac{1}{2}\pm\frac{\sqrt{b^{2}+3b+4}}{4}$.  For $a$ to be an integer we must have $\sqrt{b^{2}+3b+4}$ be an integer.  But, $(b+\frac{3}{2})^{2}<b^{2}+3b+4$.  And, since $b>0$ we have $b^{2}+3b+4<(b+2)^{2}$ implying that $b+\frac{3}{2}<\sqrt{b^{2}+3b+4}<b+2$.  So we have it that $\sqrt{b^{2}+3b+4}$ is not an integer for integer $b$ implying that $a$ is not an integer implying that there are no solutions.
[/hide]


Let $ I$ be the incenter of triangle $ ABC$. Prove that
\[ 3(IA^2\plus{}IB^2\plus{}IC^2)\ge AB^2\plus{}BC^2\plus{}CA^2\]

Let $ AB\equal{}x\plus{}y,AC\equal{}x\plus{}z,BC\equal{}y\plus{}z$ and $ r$ be the inradius.

The inequality becomes
\[ 3(3r^2\plus{}\sum x^2)\ge \sum(x\plus{}y)^2
\]

We find $ r^2\equal{}\frac{xyz}{x\plus{}y\plus{}z}$. Plugging this in, this inequality is merely Schurs

That's how I solved it too. I wonder how it can be solved a bit more geometrically...

[quote="nayel"] [color=darkred]Let $ I$ be the incenter of triangle $ ABC$. Prove that $ \sum IA^2 \ge \frac 13\cdot\sum a^2\ .$[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] For $ X: \equal{} I$ in the [b]Lagrange's identity[/b] $ \sum XA^2 \equal{} 3\cdot XG^2 \plus{} \frac 13\cdot\sum a^2$ obtain :[/color]
[hide="Proof of the Lagrange identity."][color=darkblue]===========================================================
Denote the midpoint $ M$ of $ [BC]$ . Apply the [b]theorem of median[/b] in $ \triangle\ BXC$ :

 $ 4\cdot XM^2 \equal{} 2\left(XB^2 \plus{} XC^2\right) \minus{} a^2\ (*)$ . Apply the [b]Stewart's theorem[/b] $ \triangle\ AXM$ :

$ 9\cdot XG^2 \plus{} 2m_a^2 \equal{} 3\cdot XA^2 \plus{} 6\cdot XM^2$ . Using the relation $ (*)$ obtain :

$ 18\cdot XG^2 \plus{} 4m_a^2 \equal{} 6\cdot XA^2 \plus{} 6\cdot\left(XB^2 \plus{} XC^2\right) \minus{} 3a^2$ . But $ 4m_a^2 \equal{} 2(b^2 \plus{} c^2) \minus{} a^2$ .

Thus, $ 6\left(XA^2 \plus{} XB^2 \plus{} XC^2\right) \equal{} 18\cdot XG^2 \plus{} 2\left(a^2 \plus{} b^2 \plus{} c^2\right)$ , i.e. the required relation. ===========================================================[/color][/hide][color=darkblue]$ \sum IA^2 \equal{} 3\cdot IG^2 \plus{} \frac 13\cdot\sum a^2\ge \frac 13\cdot\sum a^2$ with equality iff $ I\equiv G$ , i.e. $ a \equal{} b \equal{} c$.
[b]Remark.[/b] For any $ X$ we have $ \sum XA^2 \ge \frac 13\cdot\sum a^2$ with equality if and only if $ X: \equal{} G$ .  [/color]

A bit more geometrically, the inequality can be simplified to $ IA^2\plus{}IB^2\plus{}IC^2 \ge 4[ABC]$ by using cross products. Now we just have to show that.


trigonometry
real analysis
real analysis unsolved

Is the sequence $(\sin n)^{n}$ dense in $[-1,1]$?

Hi,
Which AMM issue have you taken the sequence?
Didi

I don't have the issue, in my paper it is just marked AMM.

harazi - could you post your prof of this problem, please? :)


It was the most amazing Halloween party ever. The theme this year was super heroes and villains. The party was going great, everyone was having a great time. Everyone was dancing, flirting and doing math problems. However, the mysterious host had better things in store for you all. You see, he was very angry…very very angry, because not everyone showed up in a costume. Some people just showed up in “normal street clothes”. This severely angered him. As the night was edging midnight and everyone was preparing to leave, all of a sudden the guards of the house shut the doors and locked them from the outside. Everyone was confused, they didn’t understand what was going on. You heard Bugi shout, “We’re being held hostage! We’re all going to die!” With this revelation came much murmuring. However that was instantly silenced when the host of the party’s voice came from seemingly out of nowhere. He said, “Many of you are not wearing costumes to my wonderful party. Due to these certain people who aren’t dressed up I have decided to punish all of you!” Then from out of nowhere he swept over them on his broomstick and waved his magical wand. Instantly, everyone turned into the costume they were wearing, and if they didn’t have a costume they didn’t change forms. The host talked again, “I am now making you fight for your lives because of those selfish people who decided not to dress up for the party. I noticed numerous of you who were wearing costumes had on villain costumes. You may notice a sudden urge to kill people.” Everyone looked at the cheerleaders who were at the party. They had decided to come dressed up as The Joker and his henchmen. All of a sudden they didn’t look quite so human anymore. In fact, it seemed they were wielding knives and daggers. The voice continued, “Each day you must vote on killing one person. Then it will turn to night, during night is when these savage killers will strike. However, do not worry, you have people protecting you. I will let all the remaining alive go when they are the only group left.”  With that his voice drowned out. Pythag011 then said, “Well I guess we don’t have much of a choice, so let’s get lynching!”

[quote="Townie PM"]You are a [b]citizen[/b]. You have no special powers. You win when all scum are dead.[/quote]

Some basic rules:
•	Votes must be in bold. If you do not bold your vote, it will not be counted.
•	The game is not to be discussed outside of the thread unless your role specifically states that you may do so.
•	Once your death scene has been posted, you’re dead. Stop typing. The only thing you can type is a "bah" post.
•	Don’t edit/delete previously submitted posts.
•	Don’t quote any PMs from me. Neither true, nor false. Seriously.
•	If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.
•	Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.
•	If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough! No choice will be made.

[b]The Alive:[/b]
1. Wickedestjr
2. budi713
3. 142857
4. alexhhmun
5. AIME15
6. mz94
7. crazypianist1116
8. Peter
9. 1=2
10.erabel
11.mewto55555
12. zapi2007
13. Brut3Forc3
14. chenhsi
15. Dojo
16. pythag011
17. hedgehog12
18: setheus_soul
19: Garyzx
20. alekim
21. SuperMathBoy (also known as SMB)
22. BOGTRO
23. aleph0
24. Hamster1800
25. hongsquare
26. jhredsox93
27. Bugi
28. 0.o
29. ffao
30. eggylv999
31. pacman2812

[b]The Dead:[/b]
None...yet

[b]It is now day, 16 votes to lynch.[/b]

[b] Vote: Joker [/b]

 :P  :P  :P

BTW, this happened in the Simpsons holloween special.

confirm.

[b]vote: joker and his henchmen[/b]

[quote="BOGTRO"]Oh also define "Quoting PM's".

Does it mean I can post:

I'm <random person here>. I have <random powers here>. Every night I <random thing here>

But I can't post

[quote="randompersonhere"] I'm <random person here>. I have <random powers here>. Every night I <random thing here> [/quote][/quote]

Quoting a PM is if it is word for word the PM I gave you. Or if you put your role in quoted form. I'm fine with anyone rewording their PM as long as it doesnt say I said it or if it's exactly what I wrote. For instance this is not allowed:

[quote="Crazychinesemaniac"] Don't put it in this form[/quote]

However this is allowed as long as it isn't word for word, the role PM:

This is my role...blah blah blah.

Confirm.

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

BAH

I confirm! Awesome theme! Maybe i won't get totally owned like in previous games...

Very intesesting. It seems the scum are joker and henchmen? Great...

confirm. lalalala message to short lalala

[b]random vote: aleph0[/b]

Let's get this started

[b]mrifnoc[/b]

[b]FoS: pythag011[/b] for being in the storyline.

[b]Vote: Batman[/b]

mrifnoc???

What's that? or are you just typing random letters?

Was he the smart one who kept banging the red guy on the rock and then flew him over a mountain?

yes i suppose you could put it that way...

Well, I guess we could have enforced stricter rules (such as the 72 hour rule). But I digress... Go town!

빅뱅 > 에픽하이

I lost ALL of my respect for you. Map the Soul > Lollipop. Period. It's been proven by EVERYONE.

dude if you ask me

1) map the soul is liek the only good epik high song
2) what does that hafta do with the game

[b]Vote: SuperMathBoy[/b] for only having posted once in the past 2 months.

Should we just call this player abandoned?

I've decided that I'm not going to play this anymore. If you want to continue the game, find a replacement for me.

UNLESS 3 (different) players and the mod object within the next week, this can officially be called both mod-abandoned and player abandoned.

I'm indifferent, either ways ok, i guess

lastpost (this is used to preserve order of locked topic)


[quote="sharkman"]A 5 horsepower (hp) pump can empty a pool in 5 hours.  A smaller, 2 hp pump empties the same pool in 8 hours.  The pumps are used together to begin emptying this pool.  After 2 hours, the 2 hp pump breaks down.  How long will it take the larger pump to empty the pool?[/quote]
Wait, so you mean to calculate the time left [b]after[/b] the two pumps have been pumping 2 hrs. and one breaks down, right?

There are too many answers here.

Which is the correct reply?

Hunter34's solution is wrong (I think) because the equation should be
[hide]$\frac{2}{5}+\frac{2}{8}+\frac{1}{5}x=1$, which means $x=\frac{7}{4}$[/hide]

[quote="alkjash"][hide]
The 2hp pump empties 1/8 of the pool in one hour, while the 5hp pump empties 1/5 of the pool in one hour.
After two hours, they will have emptied $\frac{1}{4}+\frac{2}{5}= \frac{13}{20}$ of the pool.
Then, the 2hp pump breaks down, and it takes the 5 hp pump $\frac{\frac{7}{20}}{\frac{1}{5}}= \boxed{\frac{7}{4}}$ more hours to finish emptying the pool.[/hide][/quote]
Oh, I see now.


If $ x,y,z$ are nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 3$, then

$ 4(\sqrt x \plus{} \sqrt y \plus{} \sqrt z) \plus{} 15 \le 9(\sqrt \frac {x \plus{} y}{2} \plus{} \sqrt \frac {y \plus{} z}{2} \plus{} \sqrt \frac {z \plus{} x}{2})$

[hide]The inequality's true when $ x\equal{}3,y\equal{}z\equal{}0$. If no two of $ x,y,z$ are 0,  set $ x\equal{}a^2$ etc. it becomes
\[ 8(a\plus{}b\plus{}c)\plus{}10\sqrt{3a^2\plus{}3b^2\plus{}3c^2}\leq 9\left( \sqrt{2a^2\plus{}2b^2}\plus{}\sqrt{2b^2\plus{}2c^2}\plus{}\sqrt{2c^2\plus{}2a^2}\right)\]
\[ \iff 10\left( \sqrt{3a^2\plus{}3b^2\plus{}3c^2}\minus{}(a\plus{}b\plus{}c)\right) \leq 9\sum_\text{cyc} \sqrt{2a^2\plus{}2b^2}\minus{}(a\plus{}b)\]
\[ \iff 10\sum_\text{cyc} \frac{(a\minus{}b)^2}{a\plus{}b\plus{}c\plus{}\sqrt{3a^2\plus{}3b^2\plus{}3c^2}} \leq 9\sum_\text{cyc} \frac{(a\minus{}b)^2}{a\plus{}b\plus{}\sqrt{2a^2\plus{}2b^2}}\]
\[ \iff \sum_\text{cyc} \left( \frac{9}{a\plus{}b\plus{}\sqrt{2a^2\plus{}2b^2}}\minus{}\frac{10}{a\plus{}b\plus{}c\plus{}\sqrt{3a^2\plus{}3b^2\plus{}3c^2}}\right) (a\minus{}b)^2\geq 0\]
Now each term is nonnegative, for in fact,
\[ 9(a\plus{}b\plus{}\sqrt{3a^2\plus{}3b^2})\geq 10(a\plus{}b\plus{}\sqrt{2a^2\plus{}2b^2})\]
because
\[ \left( \frac{9\sqrt{3}}{\sqrt{2}}\minus{}10\right) \sqrt{2a^2\plus{}2b^2}>\sqrt{2a^2\plus{}2b^2}\geq a\plus{}b\][/hide]

Very nice solution, scorpius119! :lol:

But I think your solution doesn't work for the best (greatest) $ p$ such that

$ \sqrt x \plus{} \sqrt y \plus{} \sqrt z\minus{}3 \le p(\sqrt \frac {x \plus{} y}{2} \plus{} \sqrt \frac {y \plus{} z}{2} \plus{} \sqrt \frac {z \plus{} x}{2}\minus{}3)$.

The best is $ p\equal{}(\sqrt 3\minus{}1)(\sqrt 3\plus{}\sqrt 2)\approx2.303$.

I think that is, sioluan.

[quote="Vasc"]Very nice solution, scorpius119! :lol:

But I think your solution doesn't work for the best (greatest) $ p$ such that

$ \sqrt x \plus{} \sqrt y \plus{} \sqrt z \minus{} 3 \le p(\sqrt \frac {x \plus{} y}{2} \plus{} \sqrt \frac {y \plus{} z}{2} \plus{} \sqrt \frac {z \plus{} x}{2} \minus{} 3)$.

The best is $ p \equal{} (\sqrt 3 \minus{} 1)(\sqrt 3 \plus{} \sqrt 2)\approx2.303$.[/quote]
I haven't any nice solution to it, but there is a not hard solution by the mixing variables method, which means that
If we put $ f(x,y,z)\equal{}RHS\minus{}LHS$, then $ f(x,y,z) \ge 0$ if and only if $ f(1,1\plus{}x,1\minus{}x) \ge 0 \ \forall 0 \le x \le 1$ and $ f(x,1,1) \ge 0 \ \forall x \ge 0$.

[quote="silouan"]Vasc which is your solution to the initial one ? Is it elementary ?
For the one you posted I think you have a solution with (n-1)EV  :)[/quote]
By either RCF/LCF Theorem or EV Theorem we can solve the more general problem:

[i]Let $ k > 0$ and let $ x_1,x_2,...,x_n$ be nonnegative real numbers such that $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} n$.
Find the maximum and minimum values of[/i]

$ E_k(x_1,x_2,...,x_n) \equal{} \frac {x_1^k \plus{} x_2^k \plus{} ... \plus{} x_n^k \minus{} n}{(\frac {n \minus{} x_1}{n \minus{} 1})^k \plus{} (\frac {n \minus{} x_2}{n \minus{} 1})^k \plus{} ... \plus{} (\frac {n \minus{} x_n}{n \minus{} 1})^k \minus{} n}$.

A reverse inequality is the following:

If $ x,y,z$ are nonnegative real numbers such that $ x \plus{} y \plus{} z \equal{} 3$, then

$ \sqrt x \plus{} \sqrt y \plus{} \sqrt z \plus{} 36 \ge 13(\sqrt \frac {x \plus{} y}{2} \plus{} \sqrt \frac {y \plus{} z}{2} \plus{} \sqrt \frac {z \plus{} x}{2})$.


ceiling function
floor function
modular arithmetic
geometry
rectangle
number theory proposed
number theory

assume that $a,b$ are positive integer .prove that $a+(-1)^b\cdot \sum_{m=0}^{a}(-1)^ {[\frac{bm}{a}]}\equiv b+(-1)^a\cdot \sum_{n=0}^{b}(-1)^{[ {\frac{an}{b}]}}$(mod 4).([] means ceil) :blush:

Maybe you meant something else instead of $(-1)^{\frac{an}b}$? I saw a \ceil in your coding. Did you mean to write integer parts instead of exponents? :?

I'll sketch it, although it's a bit hard for me to explain.

First of all, since $\lceil x\rceil+\lfloor -x\rfloor=0$, we reduce it to the following, easier to state problem:

$a+\sum_{m=0}^a (-1)^{\lfloor\frac {bm}a\rfloor}\equiv b+\sum_{n=0}^b (-1)^{\lfloor\frac {an}b\rfloor}\pmod 4$.

Now notice that all we need to do in order to prove that is show that if $k_1$ is the number of even numbers of the form $\lfloor\frac{an}b\rfloor,\ n\in\overline{0,b}$, and $k_2$ is the number of even numbers of the form $\lfloor\frac{bm}a\rfloor,\ m\in\overline{0,a}$, then $k_1+k_2$ is even.

Draw a rectangle with width $a$ and height $b$. Now, on each segment having one end on one of the sides $(0,0)(a,0)$ or $(0,0)(0,b)$ and the other end on the diagonal, count the lattice points. $k_1+k_2$ is the number of such segments on which there's an odd number of points. On the other hand, the total number of points is $(a+1)(b+1)+d+1$, where $d=(a,b)$. It's easy to see that this number is even, so the number of segments on which there's an odd number of points must be even.


Find all infinite sequences $q_1$, $q_2$, $q_3$, ... of natural numbers such that for any n > 3, we have

$q_n=\frac{q_{n-3}+q_{n-1}}{q_{n-2}}$.

[b]PS.[/b] The proposed solution (in Russian) can be found at

http://kvant.mccme.ru/1992/03/resheniya_zadachnika_kvanta_ma.htm

  Darij


[quote="Geeimatree"]Can someone tell me how to find the upper and lower bounds of polynomial roots? I think you need to use descartes' rule of signs and synthetic division but I'm having trouble proving it.[/quote]
[hide]You can use descartes' rule of signs to find the parity of the number of positive and negative roots. To find an upper and a lower bound, use that fact that a polynomial is a continuous function and that if $ f(x)<0<f(y)$ and $ x<y$ there exists some z such that $ x<z<y$ and $ f(z)\equal{}0$.[/hide]


quadratics
algebra
polynomial
calculus
derivative
superior algebra
superior algebra unsolved

Let $K/F$ be a separable quadratic extension such that $F$ has characteristic 2. Then $K=F(\alpha)$ for some $\alpha\in K$ such that $\alpha^{2}+\alpha\in F$.

If $K$ is a quadratic extension of $F$, then there exists some $\gamma\in K$ such that $K=F[\gamma]$ where the minimal polynomial of $\gamma$ over $F$ is $f(x)=x^{2}+bx+c$.  Then the derivative of $f$ is $f'(x)=b$ since we are in a field of characteristic 2, so we must have that $b\ne 0$ in order for the extension to be separable.

Thus $b^{-1}$ exists in $F$, and an easy calculations shows we can take $\alpha = b^{-1}\gamma$.


Two circular arcs are inscribed in a square of side length 1, as shown.  Find the area of the shaded region.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/eedf4be5d2da7e1742ffe9c0e3808c49.png[/img]

[hide]
call $A,B$ the vertices on the lower part of the square, $CD$ the vertices on the upper part, and $P$ the point where the two circular arcs meet themselves. Clearly, $AB=BP=AP=1$, because they are all radii of the same circle. So triangle $ABP$ is equilateral, and his area is
$\frac{\sqrt{3}}{4}$. We know now that $\angle DAP= \angle PBC = \frac{\pi}{2} - \frac{\pi}{3}=\frac{\pi}{6}$. So we can find the area of the two sector of the circle $DAP$ and $PBC$ cause $\frac{\pi}{A_{DAP}}=\frac{2\pi}{\frac{\pi}{6}} \rightarrow A_{DAP}=\frac{\pi}{12}$. The area of the shaded region is $1-\frac{\sqrt{3}}{4}-2*\frac{\pi}{12}=\frac{12-3\sqrt{3}-2\pi}{12}$[/hide]

[hide]Construct the eqquilateral triangle with two vertices the corners at the bottom, and the other on the intersection of the arcs. We get a equilateral triangle $=\frac{\sqrt{3}}{4}$, and two 30 degree sectors $=\frac{\pi}{6}$.

$1-\frac{\sqrt{3}}{4}-\frac{\pi}{6}=\boxed{\frac{12-3\sqrt{3}-2\pi}{12}}$[/hide]


A rising number, such as 34689, is a positive integer each digit of which is larger than each of the digits to its left. There are $ {9\choose 5}\equal{}126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, find the 97th number in the list.

[hide="Hidden by mod"]
Total no of such numbers = $ \binom {9}{5}$
Out of these ,total numbers which start with one are $ \binom{8}{4} \equal{} 70$
Now, number which begin with 2 : $ \binom{7}{4} \equal{} 35$

But total = 105

So, number that have 2,3 in their first two place = $ \binom{6}{3} \equal{} 20$
Total numbers = 90

So,
91 : 24567
92 : 24568
93 : 24569
94 : 24578
95 : 24579
96 : 24589
97 : 24678   Ans.
[/hide]

guru_bhai, could you use the "[hide]" tag on your answers next time?

sry ,
i will remember to do it next time.....


In the set $ \{a, b, c\}$, the members $ a, b$ and $ c$ are distinct positive integers such that $ a < b < c < 50$, and the mean of the set is an integer value. What is the greatest possible mean?

We see that the numbers $ \{47, 48, 49\}$ fit, so the solution is $ 48$. I think.....  :maybe:


a pipe is 150 feet long, 4 feet in diameter

consistently full of water

water flows through at a rate of 1 foot/second

how much water enters the pipe in one day, report your answer in the nearest 10,000 cubic feet

It would matter if it starts out with full water.

so i'd assume that it's full to begin with.

as a last question, do you mean 1 foot/second or 1 cubic foot/second

The problem does not say cubic feet, it wants answer in cubic feet.

Hot or cold water and temperature, science does matter!

[hide]Expansion and contraction[/hide]

lol xD

[hide]The distance of the pipe doesn't matter because it asks how much water [b]enters[/b] the pipe.

there are 86,400 seconds in a day, which means the amount of water could fill an 86,400 ft long pipe. enter in formula for a cylinder and you get A=86,400(Pi)(R^2)=345,600(Pi)

which is 1,085,733 cubic ft or 1,090,000 rounded[/hide]


function
calculus
derivative
real analysis
real analysis unsolved

Let $ f$ be a finite real function of one variable. Let $ \overline{D}f$ and $ \underline{D}f$ be its upper and lower derivatives, respectively, that is, \[ \overline{D}f\equal{}\limsup_{{h,k\rightarrow 0}_{{h,k \geq 0}_{h\plus{}k>0}}} \frac{f(x\plus{}h)\minus{}f(x\minus{}k)}{h\plus{}k}\] ,
\[ \underline{D}f\equal{}\liminf_{{h,k\rightarrow 0}_{{h,k \geq 0}_{h\plus{}k>0}}} \frac{f(x\plus{}h)\minus{}f(x\minus{}k)}{h\plus{}k}.\] Show that $ \overline{D}f$ and $ \underline{D}f$ are Borel-measurable functions. [A. Csaszar]

It is obviously sufficient to prove the assertion for $ \overline{D}f$. It is also clear that
\[ \{x : \overline{D}f(x)>c\} \equal{} \bigcup_{i\equal{}1}^\infty \bigcap_{k\equal{}1}^\infty A_{ik}\]
for all real values $ c$, where $ A_{ik}$ denotes the union of intervals $ [a, \ b]$ shorter than $ 1/k$ satisfying $ (f(b) \minus{} f(a))/(b\minus{}a) > c \plus{} 1/i$. We show that $ A_{ik}$ is a set of type $ F_\sigma$ which implies the assertion.

More generally let $ \{ I_\gamma\} \ (\gamma \in \Gamma)$ be a system of open intervals, and let $ A \equal{} \bigcup_{\gamma \in \Gamma} \overline{I}_\gamma$ (where $ \overline{I}_\gamma$ stands for the closure of $ I_\gamma$). Obviously,
\[ A \equal{} B \cup C \cup D,\]
where $ B \equal{} \bigcup_{\gamma \in \Gamma} I_\gamma$, $ C$ is the set of all points that are left-hand endpoints of at least one $ I_\gamma$ but are not interior points of any of them, and $ D$ is the set of those points that are right-hand endpoints of at least one $ I_\gamma$ but do not lie in the interior of any $ I_\gamma$. Clearly, $ B$ is an open set. If $ x \in C$, let $ r_x$ be a rational point in the interior of the interval $ I_\gamma$ that has the left-hand endpoint $ x$. Points $ r_x$ corresponding to distinct points $ x$ are distinct since in the case $ x<y$, $ r_x \equal{} r_y$, the point $ y$ would lie in the interior of the interval $ I_\gamma$ starting at $ x$. Consequently, $ C$, and similarly $ D$, is countable. So each of the sets $ B$, $ C$, $ D$, and therefore $ A$ as well is of type $ F_\sigma$.


The point at $ (a, b)$ on a Cartesian plane is reflected over the y-axis to the point at $ (j, k)$. If $ a \plus{} j \equal{} 0$ and $ b \plus{} k \equal{} 0$, what is the value of $ b$?

When a coordinate is reflected over the y-axis, the coordinate $ (a, b)$ becomes $ (\minus{}a, b)$. Note that $ b$ stays the same! So if $ b \plus{} k \equal{} 0$, and we know $ b\equal{} k$, clearly $ b$ must be equal to $ \boxed{0}$.


A graphic art designer's annual salary is a whole number of dollars between $ \$62,\!400$ and $ \$62,\!600$. If the hundreds, tens and units digits in her salary are all different and in descending order, how many possibilities exist for her salary?

The hundreds place is either 4 or 5.
If 4: Then there are 4 choose 2 = 6 ways to pick last two digits
If 5: Then there are 5 choose 2 = 10 ways to pick last two digits

So, our answer is 16.


Hello,

$ i)$ find all function $ f$ verifying : $ f(x\plus{}1)\minus{}f^2(x)\equal{}x$

$ ii)$ solve in $ \mathbb{R}$ the equation : $ (\frac{1}{4})^{x\plus{}13} \plus{} (\frac{1}{4})^{4\minus{}x}\equal{}3$

[quote="linda2005"]Hello,

$ i)$ find all function $ f$ verifying : $ f(x \plus{} 1) \minus{} f^2(x) \equal{} x$

$ ii)$ solve in $ \mathbb{R}$ the equation : $ (\frac {1}{4})^{x \plus{} 13} \plus{} (\frac {1}{4})^{4 \minus{} x} \equal{} 3$[/quote]

[b]i)[/b] Do you have other constraints? Or there are infinitely many solutions.

[b]ii)[/b] It's easy to see that we can multiply both sides by $ (\frac{1}{4})^x$ to get a quadratic equation.

But I don't think anyone will have the patience to do it by hand. And the answer may be something like $ log k$

Question 1 only makes sense if the domain of $ f$ is the integers (or possibly the positive or nonnegative integers), and I assume there is also supposed to be some restriction on the range.  What are all the details?

stephencheng is right that the answer to question 2 is extremely unpleasant.

[quote="linda2005"]Hello,

$ f^2(x)$

Is it $ (f(x))^2$ or $ fof(x)$ ?

[quote="linda2005"]Hello,

$ i)$ find all function $ f$ verifying : $ f(x \plus{} 1) \minus{} f^2(x) \equal{} x$

 [/quote]

For $ z \in C$
 If $ f(z \plus{} 1) \minus{} fof(z) \equal{} z$

Did you try $ f(x)\equal{}ax\plus{}b$ with $ a,b \in C$


how to solve for natural numbers x and y ?

x^2-56y^2=105

[hide]Consider the equation mod 3: $ x^2\plus{}y^2\equiv 0\pmod 3$. Then $ x,y\equiv 0\pmod 3$. However that means LHS of the original equation is divisible by 9, while 105 is not. Therefore, no solutions.[/hide]


inequalities
trigonometry
vector
geometry
parallelogram
triangle inequality
complex numbers

[quote="eddie77"]For $a,b,c,d > 0$, prove that \[\sqrt{(a+b)^{2}+(c+d)^{2}}\leq \sqrt{a^{2}+c^{2}}+\sqrt{b^{2}+d^{2}}\] [/quote]That's just Minkowsky  :huh:

[hide="Other method"]Square to get the equivalent inequality $ab+cd \leq \sqrt{\left(a^{2}+c^{2}\right)\left(b^{2}+d^{2}\right)}$. And that's just plain Cauchy.[/hide]

We can also normalize and use trig in the obvious way.

[quote="FieryHydra"]This one uses the idea of Complex Numbers.

Let $z_{1}=a+ic$ and $z_{2}=b+id$ be two numbers on the complex plane.

Then, by the triangle inequality, we have
$|z_{1}+z_{2}| \leq |z_{1}|+|z_{2}|$
$\Rightarrow |(a+b)+i(c+d)| \leq |a+ic|+|b+id|$
$\Rightarrow \sqrt{{(a+b)}^{2}+{(c+d)}^{2}}\leq \sqrt{a^{2}+c^{2}}+\sqrt{b^{2}+d^{2}}$[/quote]

Can you post other inequalities that can be shown using complex numbers :?:

[quote="M4RI0"]
Can you post other inequalities that can be shown using complex numbers :?:[/quote]
Right now, I can think of an elementary number theory problem that one can solve easily using complex numbers. Try this: Let $N_{1}$ and $N_{2}$ each be the sum of squares of two integers. Show that $N_{1}N_{2}$ itself is the sum of squares of two integers.

Since complex numbers can be shown to have a one-to-one correspondence to vectors on the the Euclidean plane (2-D plane),  we can prove quite a few results in inequalities using complex numbers or vectors. More on this later though I admit I am not an expert on this.

[quote="1234567890"]Can anyone prove this geometry theorem without using analytic geometry (can anyone find a pure geometric solution?)?

In triangle $ABC$, $M$ is the midpoint of $BC$. Prove that $2AM \le AB+AC$.[/quote]

Construct point $D$ such that $\overline{AMD}$ and $AM = MD$.  Then $ABDC$ is a parallelogram, so $BD = AC$ and apply the triangle inequality to $\triangle ABD$.


[quote="Rushil"]Please recommend some online, free software that can be used for drawing Diagrams - geometry, 3D, vector etc. diagrams - that can be later be inserted in Latex/Pdf documents using texniccenter???[/quote]Use the serach function on this site. This question has been discussed many times before on this forum.

The other programs I have found are based on prgrmas similar to writing documents in LAtex. Can you plz suggest a WYSIWYG Geomwtry editor whose files are easily integrable with Latex in Texniccenter. Of course it should be free online!!!

I'm sure you can find useful information in this very forum if only you look properly... (as Valentin said), it's not hard!


How many different permutations of the numbers {1, 2, ..., n} have the following property: For every number i, except of the one standing on the very first place, at least one of the numbers i - 1 and i + 1 stands on the left of i.

[b]PS.[/b] I was asked to post some Kvant problems here. Sorry if they were already posted. For those speaking Russian, the solution can be found here:

http://kvant.mccme.ru/1989/11/resheniya_zadachnika_kvanta_ma.htm

[b]EDIT:[/b] Sorry, I see the problem has been already discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=1164 .

  Darij

I had begun by asking whether it's $2^{n-1}$ or not, but then I clicked that link and, although I can't speak or read Russian, I saw that the answer is, indeed, $2^{n-1}$.

At each step (when we add another number), the set of numbers we have so far must be a block, so we have to count the permutations with this property: for any $1\le k\le n$, the first $k$ terms form a block.

Given $a,b$, let $f(a,b)$ be the number of ways in which we can complete a block of length $a+b+1$ starting from a number which has $a$ numbers on the left and $b$ on the right, by making a step to the left or right at each time (it's obvious that $f(b,a)=f(a,b)$). Clearly, the number we're looking for is $f(0,n-1)+f(1,n-2)+\ldots+f(n-2,1)+f(n-1,0)$.

From $f(0,a)=f(a,0)=1,\ \forall a$ and the recurrence relation $f(a,b)=f(a-1,b)+f(a,b-1)$ we find $f(a,b)=\binom{a+b}a$, so the number we're looking for is $\sum_{i=0}^{n-1}\binom{n-1}i=2^{n-1}$.


so did anyone here write the final round of the bc math contest today?

aww, man I'm so annoyed by that.
Last year I did, I wrote the Junior contest and won for the district, but this year I didn't even make it from my school. It really peeves me actually. My math buddy's probably gonna tell me he won tomorrow too. Argh

how come ive never heard of such a contest?

are you from B.C.?

[url=http://cariboo.tru.ca/advtech/math-sta/mathcompetition/mathcompOverview.html]cariboo.tru.ca/advtech/math-sta/mathcompetition/mathcompOverview.html[/url]


For any equation of rectangular hyperbola $xy=a^2$ , $a\in\mathbb{R}$, what is the foci point of the locus ?

More generally , if the rectangular hyperbola has equation $(x-p)(y-q)=a^2$ , $a,p,q\in\mathbb{R}$, what is its foci point ?

Hint
[hide]The difference of the distances from a point on the graph to the two foci is constant.[/hide]

[quote="jhcreinhardt"]The difference of the distances from a point on the graph to the two foci is constant.[/quote]

Actually, the [b]sum[/b] of the distances from a point on the graph to the two foci is constant.

If we denote the foci as $F(f,f)$ .Try to prove that $f^2=2a^2$ where $xy=a^2$ .


What is the largest possible area, in square centimeters, of a right
triangle with one side of length 12 cm and another side of length
20 cm?

We want the largest possible area, so we will assume that 20 is the length of one the legs, meaning that the hypotenuse is longer and therefore the triangle is bigger. So we use the formula, $ \frac{1}{2}bh$, and get $ \frac{1}{2} \cdot 12 \cdot 20 \equal{} \boxed{120}$.

[quote="AIME15"]Let these be the two smalles sides, so 12*20=240 divide by 2=120

ps i got this one in about 3 sec.[/quote]

DUDE!  STOP SPAMMING!

[size=59][color=white]Sorry for spamming.[/color][/size]

It is, but that's not what the question asks.

[quote="AIME15"]Let these be the two smalles sides, so 12*20=240 divide by 2=120

ps i got this one in about 3 sec.[/quote]

I answered this a day ago. If you bothered to actually read posts, you would realize that I wrote basically the EXACT SAME THING. Do I need to recite the spiel again?

Only make another post if:

1. You found another way.
2. You got a different answer.
3. You noticed a mistake.

[quote="myyellowducky82"][quote="isabella2296"]We want the largest possible area, so we will assume that 20 is the length of one the legs, meaning that the hypotenuse is longer and therefore the triangle is bigger. So we use the formula, $ \frac {1}{2}bh$, and get $ \frac {1}{2} \cdot 12 \cdot 20 \equal{} \boxed{120}$.[/quote]
is it possible to find the smallest possible area[/quote]

The smallest possible answer would be really tiny.

[quote="mewto55555"]SMallest possible would be 0 but then the trianlge would be degenerate. Smallest possible for right triangle is 96. I'll leave you to figure out why[/quote]

But wouldn't that triangle not exist...

But maybe...

Actually if you want to be technical, there is no minimum value!

Wouldn't it be 0? You can't have negative area, can you?

Yes but if it's 0, it's no longer a triangle. As the largest angle approaches 180 degrees, the area approaches 0, but if the largest angle is 180, then it has area of 0, but isnt triangle.


People have complained that people only rate posts spam, and not higher ratings (which is false).

To counter people who have generally given: 1) consistent ratings (either always spam or always good) and 2) inaccurate ratings (those who post 6 cubes even though, say, 10 others say it is a 2 cuber), it might be necessary to adopt a formula in the likeness of this:

We want to normalize ratings first. Thus, for a person who normally rates posts 1-4 (and rarely the 5 or a 6, like me), it would be fair to make that commonplace 1 and 2 less weighted than a person who is more fair than a person who votes evenly from 1~6.

We first calculate the average of all post ratings given. Then we find the difference of AoPS's average rating votes to that user's average rating votes. So, for a person who votes from 1~4, has their votes (approximately) raised by 1. (again this is one solution, I'm just saying that this type of rating normalization is a way to stop those who say that everyone votes for spam)

Second is inaccuracy. Perhaps we can once again, compare a person's ratings to the other ratings for that post, and if it deviates significantly (3 cubes or more difference is significant), then it gets weighed less - the weight lessening is more drastic the more cubes it differs. In addition, this also affects that person's global voting (albeit just one inaccurate vote should not sway their global voting capability by too much, but many inaccurate votes should)

I think that's a great idea james41!  But in your proposal, there'd still be a weighting so those who have higher ratings will have more of a say in the rating, right?

Also, how would you make an algorithm for that?

Yes, those with higher ratings will have higher weightings as before; you can tack on these extra weightings as well.

In all honesty I don't know what the exact formula may be, or if they already implement these ideas. Just throwing some ideas around.


$ x,y$ both are positive integers. find all solutions of $ (x , y)$ that satisfy the given equation
$ x^3 \equal{} y^3 \plus{} 2y^2 \plus{} 1$

[quote="cosinerburc"]y^3 < y^3 + 2y^2 + 1 < (y+1)^3 if y is a positive integer.[/quote]

saol dostum guzel cozum olmus.

sıkıştırma sorusu antalya 1.aşama 2007 :D

x^3=y^3+2y^2+1   y^3+2y^2+1 >y^3  y^3+3y^2+3y+1 küçük eşit y^3+2y^2+1 olmlı bu durumda ; y^2+3y büyükeşit 0 olur
buradan y byük eşit -3  , y küçük eşit 0 y=-3 ise x=-2, y=-2 ise x=1, y=-1 se çözüm yok ,y=0 sa  x=1
(-2,-3),(1,-2),(1,0)
bu arada antalya 2006:D
CÜNEYD ÖZTÜRK