calculus
integration
trigonometry
real analysis
limit
real analysis unsolved

I am in only Multivariable Calculus so is there a better integration technique that is taught?  Is the Lebesgue integral at all helpful?

this is actually one of ramanujan's derivations. he uses a clever formulation of an expression involving an infinite product and all it takes after that is a partial fraction decomposition and the answer follows after some simple integration.

For fixed $ \alpha,\beta>0$ define $ g(z)$ as $ \prod_{\ell\equal{}0}^n \Big((\beta\plus{}\ell)^2\plus{}z^2\Big)$ and $ f(z)$ as $ \frac{e^{i\alpha z}}{g(z)}$ .

$ g'\Big(i(\beta\plus{}k)\Big)\equal{}\prod_{\ell\equal{}0}^{k\minus{}1} \underbrace{\Big((\beta\plus{}\ell)^2\minus{}(\beta\plus{}k)^2\Big)}_{(\ell\minus{}k)(2\beta\plus{}\ell\plus{}k)}\cdot 2i(\beta\plus{}k)\cdot \prod_{\ell\equal{}k\plus{}1}^n \underbrace{\Big((\beta\plus{}\ell)^2\minus{}(\beta\plus{}k)^2\Big)}_{(\ell\minus{}k)(2\beta\plus{}\ell\plus{}k)}$

$ \equal{}(\minus{}1)^k\, k! \, \frac{(2\beta\plus{}2k\minus{}1)!}{(2\beta\minus{}1\plus{}k)!} \cdot i(2\beta\plus{}2k)\cdot (n\minus{}k)!\, \frac{(2\beta\plus{}n\plus{}k)!}{(2\beta\plus{}2k)!}$ .

$ \int_{\minus{}\infty}^\infty \frac{\cos \alpha x}{\Big(\beta^2\plus{}x^2\Big)\Big((\beta\plus{}1)^2\plus{}x^2\Big)\cdots \Big((\beta\plus{}n)^2\plus{}x^2\Big)} \, dx\equal{}\lim \oint_{\text{upper} \atop \text{half circle}} f \, dz$

$ \equal{}2\pi i\, \sum_{k\equal{}0}^n \text{res}\Big(f,i(\beta\plus{}k)\Big)\equal{}2\pi i\, \sum_{k\equal{}0}^n \frac{e^{\minus{}\alpha(\beta\plus{}k)}}{g'\Big(i(\beta\plus{}k)\Big)}$

$ \equal{}2\pi \,\sum_{k\equal{}0}^n (\minus{}1)^k \, \frac{(2\beta\minus{}1\plus{}k)!}{(2\beta\plus{}n\plus{}k)!}\, \frac{1}{k!\, (n\minus{}k)!}\, e^{\minus{}\alpha(\beta\plus{}k)}$

Now multiply both sides with $ \beta^2\, (\beta\plus{}1)^2\cdots (\beta\plus{}n)^2\equal{}\frac{(\beta\plus{}n)!^2}{\Gamma^2(\beta)}$ .

$ I_n: \equal{}\int_{\minus{}\infty}^\infty \frac{\cos \alpha x}{\left(1\plus{}\frac{x^2}{\beta^2}\right)\left(1\plus{}\frac{x^2}{(\beta\plus{}1)^2}\right)\cdots \left(1\plus{}\frac{x^2}{(\beta\plus{}n)^2}\right)}\, dx$

$ \equal{}2\pi \,\frac{\Gamma(2\beta)}{\Gamma^2(\beta)}\, \sum_{k\equal{}0}^n \underbrace{(\minus{}1)^k \, \frac{(2\beta\minus{}1\plus{}k)!}{\Gamma(2\beta)\, k!}}_{{\minus{}2\beta\choose k}}\, \underbrace{\frac{(\beta\plus{}n)!^2}{(2\beta\plus{}n\plus{}k)!\, (n\minus{}k)!}}_{A_{n,k}}\, e^{\minus{}\alpha(\beta\plus{}k)}$

$ \equal{}2\pi\, \frac{\Gamma(2\beta)}{\Gamma^2(\beta)}\, \left[\sum_{k\equal{}0}^{M\minus{}1} {\minus{}2\beta\choose k} \, A_{n,k}\, e^{\minus{}\alpha k}\plus{}\sum_{k\equal{}M}^n {\minus{}2\beta\choose k} \, A_{n,k}\, e^{\minus{}\alpha k} \right]\, e^{\minus{}\alpha \beta}$

$ \big(A_{n,0},A_{n,1},...,A_{n,n}\big)\subset [0,1]$ is a decreasing sequence with $ \lim_{n\to\infty} A_{n,k}\equal{}1$ for all $ 0\le k<M$ .

So $ I_{\infty}\equal{}2\pi \, \frac{\Gamma(2\beta)}{\Gamma^2(\beta)}\, \left(\sum_{k\equal{}0}^{M\minus{}1} {\minus{}2\beta \choose k}\, (e^{\minus{}\alpha})^k\plus{}\varepsilon_M\right) \, e^{\minus{}\alpha\beta}$ .

The term in brackets tends to the binomial series $ \left(1\plus{}e^{\minus{}\alpha}\right)^{\minus{}2\beta}$ for $ M\to\infty$ .

Now $ I_\infty$ is $ \sqrt{\pi} \, \frac{\Gamma\left(\beta\plus{}\frac12\right)}{\Gamma(\beta)} \, 2^{2\beta} \, \left(1\plus{}e^{\minus{}\alpha}\right)^{\minus{}2\beta} \, e^{\minus{}\alpha\beta}$ after Legendre's duplication formula

and $ 2^{2\beta} \,\left(\frac{1}{1\plus{}e^{\minus{}\alpha}}\right)^{2\beta}\, \frac{1}{e^{\alpha\beta}}\equal{}\left(\frac{2}{e^\frac{\alpha}{2}\plus{}e^{\minus{}\frac{\alpha}{2}}}\right)^{2\beta}\equal{}\text{sech}^{2\beta}\left(\frac{\alpha}{2}\right)$ .

So $ I_{\infty}\equal{}\sqrt{\pi}\, \frac{\Gamma\left(\beta\plus{}\frac12\right)}{\Gamma(\beta)}\, \text{sech}^{2\beta}\left(\frac{\alpha}{2}\right)$ .

$ \int_{\minus{}\infty}^\infty \frac{1}{1\plus{}\frac{x^2}{a^2}}\times \frac{1\plus{}\frac{x^2}{(b\plus{}1)^2}}{1\plus{}\frac{x^2}{(a\plus{}1)^2}}
\times \frac{1\plus{}\frac{x^2}{(b\plus{}2)^2}}{1\plus{}\frac{x^2}{(a\plus{}2)^2}}\times \cdots \,\, dx$ $ \equal{}\sqrt{\pi} \, \frac{\Gamma\left(a\plus{}\frac12\right)\cdot \Gamma(b\plus{}1)\cdot \Gamma\left(b\minus{}a\plus{}\frac12\right)}{\Gamma(a)\cdot\Gamma\left(b\plus{}\frac12\right)\cdot \Gamma\left(b\minus{}a\plus{}1\right)}$

for $ 0<a<b\plus{}\frac12$ is also an integral from Ramanujan which has a certain similarity to the upper one. Concerning this see [url=http://en.wikipedia.org/wiki/Srinivasa_Ramanujan]Wikipedia[/url].

Let's try to prove it:

For fixed $ 0<a<b\plus{}\frac12$ define $ f(z)\equal{}\prod_{\ell\equal{}1}^n \Big((b\plus{}\ell)^2\plus{}z^2\Big)$ and $ g(z)\equal{}\prod_{\ell\equal{}0}^n \Big((a\plus{}\ell)^2\plus{}z^2\Big)$ .

$ f\Big(i(a\plus{}k)\Big)\equal{}\prod_{\ell\equal{}1}^n \Big((b\plus{}\ell)^2\minus{}(a\plus{}k)^2\Big)$

$ \equal{}\prod_{\ell\equal{}1}^n \Big( (b\plus{}a\plus{}k\plus{}\ell)(b\minus{}a\minus{}k\plus{}\ell) \Big)\equal{}\frac{(b\plus{}a\plus{}k\plus{}n)!}{(b\plus{}a\plus{}k)!}\, \frac{(b\minus{}a\minus{}k\plus{}n)!}{(b\minus{}a\minus{}k)!}$

and

$ g'\Big(i(a\plus{}k)\Big)\equal{}\prod_{\ell\equal{}0}^{k\minus{}1} \underbrace{\Big((a\plus{}\ell)^2\minus{}(a\plus{}k)^2\Big)}_{(\ell\minus{}k)(2a\plus{}\ell\plus{}k)}\cdot 2i(a\plus{}k)\cdot \prod_{\ell\equal{}k\plus{}1}^n \underbrace{\Big((a\plus{}\ell)^2\minus{}(a\plus{}k)^2\Big)}_{(\ell\minus{}k)(2a\plus{}\ell\plus{}k)}$

$ \equal{}(\minus{}1)^k\, k!\, \frac{(2a\plus{}2k\minus{}1)!}{(2a\minus{}1\plus{}k)!}\cdot i(2a\plus{}2k)\cdot (n\minus{}k)!\, \frac{(2a\plus{}n\plus{}k)!}{(2a\plus{}2k)!}$

$ \equal{}i\, (\minus{}1)^k\, \frac{(2a\plus{}n\plus{}k)!}{(2a\minus{}1\plus{}k)!}\, k!\, (n\minus{}k)!$

$ \int_{\minus{}\infty}^\infty \frac{1}{a^2\plus{}x^2}\times \frac{(b\plus{}1)^2\plus{}x^2}{(a\plus{}1)^2\plus{}x^2} \times \cdots \times \frac{(b\plus{}n)^2\plus{}x^2}{(a\plus{}n)^2\plus{}x^2}\,  dx\equal{}\lim \oint_{\text{upper} \atop \text{half circle}} \frac{f}{g} \, dz$

$ \equal{}2\pi i\, \sum_{k\equal{}0}^n \text{res}\left(\frac{f}{g},i(a\plus{}k)\right)\equal{}2\pi i\, \sum_{k\equal{}0}^n \frac{f\Big(i(a\plus{}k)\Big)}{g'\Big(i(a\plus{}k)\Big)}$

$ \equal{}2\pi \sum_{k\equal{}0}^n (\minus{}1)^k\, \frac{(2a\minus{}1\plus{}k)!}{(2a\plus{}n\plus{}k)!}\, \frac{1}{k!\, (n\minus{}k)!}\, \frac{(b\plus{}a\plus{}k\plus{}n)!}{(b\plus{}a\plus{}k)!}\, \frac{(b\minus{}a\minus{}k\plus{}n)!}{(b\minus{}a\minus{}k)!}$

Now mulitply both sides with $ a^2\, (a\plus{}1)^2\cdots (a\plus{}n)^2\equal{}\frac{(a\plus{}n)!^2}{(a\minus{}1)!^2}$

and divide both sides with $ (b\plus{}1)^2\cdots (b\plus{}n)^2\equal{}\frac{(b\plus{}n)!^2}{b!^2}$ .

$ I_n\equal{}\int_{\minus{}\infty}^\infty \frac{1}{1\plus{}\frac{x^2}{a^2}} \cdot \frac{ \left(1\plus{}\frac{x^2}{(b\plus{}1)^2}\right)\,\left(1\plus{}\frac{x^2}{(b\plus{}2)^2}\right) \cdots \left(1\plus{}\frac{x^2}{(b\plus{}n)^2}\right) }{\left(1\plus{}\frac{x^2}{(a\plus{}1)^2}\right)\,\left(1\plus{}\frac{x^2}{(a\plus{}2)^2}\right) \cdots \left(1\plus{}\frac{x^2}{(a\plus{}n)^2}\right)} dx$

$ \equal{}2\pi\, \frac{b!^2}{(a\minus{}1)!^2}\, \sum_{k\equal{}0}^n \underbrace{(\minus{}1)^k\, \frac{(2a\minus{}1\plus{}k)!}{k!}}_{\frac{(\minus{}2a)!\, (2a\minus{}1)!}{k!\, (\minus{}2a\minus{}k)!}} \, \frac{1}{(b\plus{}a\plus{}k)!\, (b\minus{}a\minus{}k)!}\, \underbrace{\frac{(b\plus{}a\plus{}k\plus{}n)!\, (b\minus{}a\minus{}k\plus{}n)!}{(2a\plus{}n\plus{}k)!\, (n\minus{}k)!}\, \frac{(a\plus{}n)!^2}{(b\plus{}n)!^2}}_{A_{n,k}}$

$ \equal{}2\pi\, \frac{b!^2\, (\minus{}2a)!\, (2a\minus{}1)!}{(a\minus{}1)!^2}\, \left[\sum_{k\equal{}0}^{M\minus{}1} \frac{A_{n,k}}{k!\, (b\plus{}a\plus{}k)!\, (\minus{}2a\minus{}k)!\, (b\minus{}a\minus{}k)!}\plus{}\sum_{k\equal{}M}^n \frac{A_{n,k}}{k!\, (b\plus{}a\plus{}k)!\, (\minus{}2a\minus{}k)!\, (b\minus{}a\minus{}k)!} \right]$

$ \lim_{n\to\infty} A_{n,k}\equal{}1$ for all $ 0\le k<M$ .

So $ I_\infty\equal{}2\pi\,\frac{b!^2\, (\minus{}2a)!\, (2a\minus{}1)!}{(a\minus{}1)!^2}\, \left[\sum_{k\equal{}0}^{M\minus{}1} \frac{1}{k!\, (b\plus{}a\plus{}k)!\, (\minus{}2a\minus{}k)!\, (b\minus{}a\minus{}k)!}\plus{}\varepsilon_M \right]$ .

The term in brackets tends to $ \frac{(2b\minus{}2a)!}{(\minus{}2a)!\, (b\minus{}a)!\, (b\minus{}a)!\, (2b)!}$ for $ M\to\infty$ .

At this I used the formula $ \boxed{ \sum_{k\equal{}0}^\infty \frac{1}{k!\, (\alpha\plus{}k)!\, (\beta\minus{}k)!\, (\gamma\minus{}k)!}\equal{}\frac{(\alpha\plus{}\beta\plus{}\gamma)!}{\beta!\, \gamma!\, (\alpha\plus{}\beta)!\, (\alpha\plus{}\gamma)!} }$

Now $ I_\infty\equal{}2\pi\, \frac{b!^2 \, (2a\minus{}1)!\, (2b\minus{}2a)!}{(a\minus{}1)!^2\, (b\minus{}a)!^2\, (2b)!}$ and this is $ \sqrt{\pi} \, \frac{\Gamma\left(a\plus{}\frac12\right)\cdot \Gamma(b\plus{}1)\cdot \Gamma\left(b\minus{}a\plus{}\frac12\right)}{\Gamma(a)\cdot\Gamma\left(b\plus{}\frac12\right)\cdot \Gamma\left(b\minus{}a\plus{}1\right)}$ .

Can somebody prove the framed formula? It might have something to do with the hypergeometric series.


MATHCOUNTS probability geometry counting distinguishability

Welcome to AoPS!

I think the best thing to do for MathCounts is to do a lot of old problems and familiarize yourself with the topics. The 3 Intros are also helpful, especially Counting and Probability and Geometry I think.

Your chance of making Nats also depends a lot on which state you're from.

im guessing they had like a 38 cutoff last year...

someone confirm this...

jorian

Cutoff in Wisconsin last year was 37, so Ohio would have been a bit higher I believe

Last year, our cutoff was I believe 40 or 41

Our countdown is pretty weak though

What the score include? Sprin round + target round? Or sprind round + target round + countdown?

no countdown. Countdown only moves your place up at the end

sprint round + 2(target round)

and for team
1/4(added indi scores) + 2 (team round)

-jorian

[quote="mathmatatics"]Well, I need to know how to prepare for state MC competition. Last year, as a seventh grader, I did horribly, and I am hoping to do better this time around, and hopefully even make nationals. I do great at chapter, I get 1st every time, but last year at state I got 200th something place. Can you guys help?

Also, It's my first time posting on these boards. I've been on here doing problems before, but I never posted.[/quote]
I think you should pinpoint you're weakness, i.e. not good at geometry, etc. then work on those areas more. especially to tough problems, because they really force you to think, and even if you encounter a problem type you've never seen before, you can have lots of weapons in your arsenal to attempt the problems

What do you guys do to prepare? I just need some sources to look at, and maybe some problems to do. My goal is to get to nationals, but I don't have much time, so any help would be appreciated.

If you don't put in the time, you won't make it to Nationals. Case closed. Anyway, if you do put in the time, do old national tests, or state tests. Your MathCounts coach probably has these resources for you

i think old nationals will help more than old states, because the problems generally get harder every year.

Can someone help me with combinations? I know basic combinations but I need to work on problems like this one:

If there are 8 red cards and 7 white cards, if 3 cards are dealt, what is the probability that all 3 will be the same color?

I am going to state and I really need help as this is my first time being in MathCounts and I want to do well. I have been doing some past national sprint rounds but I need help with combinations.

adding the probability of 3 of each colour

[quote="Quevvy"][quote="jhredsox"]wouldn't the answer be

8/15*7/14*6/13+7/15*6/14*5/13

?

jorian[/quote]
How would you come to that answer? and how would you simplify it without a calculator (It's from a sprint round)?[/quote]
Simplifying would be easy if you see that mostly everything cancels out.

Okay, here's the solution to that problem:

First, we're assuming that the cards are distinguishable: In other ways, all of them are different. For example, you can say they have names, and drawing the three red cards Bob, Anna, and Carol is not the same as drawing the three red cards Joe, Billy, and Susanna.

To find the probability for either color, find the probabilities of each color and add them together.

There are $\binom{8}{3}$ ways to take 3 of the red cards.
There are $\binom{7}{3}$ ways to take 3 of the white cards.
Total, there are $\binom{15}{3}$ ways to take any three of the cards.

So we're looking for \[\frac{\binom{8}{3}+\binom{7}{3}}{\binom{15}{3}}.\]

$\binom{8}{3}= \frac{8 \times 7 \times 6}{3!}= 8 \times 7 = 56$.
$\binom{7}{3}= \frac{7 \times 6 \times 5}{3!}= 7 \times 5 = 35$.
$\binom{15}{3}= \frac{15 \times 14 \times 13}{3!}= 5 \times 7 \times 13 = 455$.

\[\frac{\binom{8}{3}+\binom{7}{3}}{\binom{15}{3}}= \frac{56+35}{455}= \frac{91}{455}= \frac{1}{5}\].

[quote="Ubemaya"]Okay, here's the solution to that problem:

First, we're assuming that the cards are distinguishable: In other ways, all of them are different. For example, you can say they have names, and drawing the three red cards Bob, Anna, and Carol is not the same as drawing the three red cards Joe, Billy, and Susanna.

To find the probability for either color, find the probabilities of each color and add them together.

There are $\binom{8}{3}$ ways to take 3 of the red cards.
There are $\binom{7}{3}$ ways to take 3 of the white cards.
Total, there are $\binom{15}{3}$ ways to take any three of the cards.

So we're looking for
\[\frac{\binom{8}{3}+\binom{7}{3}}{\binom{15}{3}}. \]
$\binom{8}{3}= \frac{8 \times 7 \times 6}{3!}= 8 \times 7 = 56$.
$\binom{7}{3}= \frac{7 \times 6 \times 5}{3!}= 7 \times 5 = 35$.
$\binom{15}{3}= \frac{15 \times 14 \times 13}{3!}= 5 \times 7 \times 13 = 455$.
\[\frac{\binom{8}{3}+\binom{7}{3}}{\binom{15}{3}}= \frac{56+35}{455}= \frac{91}{455}= \frac{1}{5}\]
.[/quote]
Thank you very much! I was doing this problem in preparation for chapter competition and I couldn't figure it out but now I know how to do these types.


symmetry
linear algebra
linear algebra theorems

I am not too sure if it is correct at all. With this:

A(transposed)    (EQUIVALENT TO)    A(transpose)(transposed again) = A

Would this actually mean that both equations' "A" is actually the same? The "equivalent to" (three little dots, right?) is not something I am so used to...

Thanks

Some people use three dots forming a triangle as a replacement for "therefore". As I wrote earlier,
[quote]English (or whatever your language is) >> any kind of symbols.[/quote]

Thx.

What I am trying to show is that two different equations are equivalent. Like...

A is transposed (= itself once transposed)
B is transposed twice = B

Both are EQUIVALENT,   THEREFORE A = B

Is it possible to do that?

Sorry, I'll try to be clearer. Let's suppose that I have A, but that it passed by two different paths through colloquial language. Let's call this Ai and Aii (instead of a problematic A and B). I wish to write that Ai and Aii are still mathematical, and actually the same by comparing a transposition and a double transposition. If they are equivalent, there should be symmetry and all. This is how I [i]think[/i] I could present it formally:

( [Aij^T = Aji]^[other transpose afterwards to do as other side ->]     EQUIVALENT TO     [(Akl^T)^T = Akl]     THEREFORE     [i = k AND j = l]   THEREFORE     (Aij^T)^T = Alk )

I am not too sure how to represent "other transpose afterwards"
I really hope that I am understandable...

PS:
Isn't the formal symbol for "therefore" "Γ" ? Or is it the three points? Or both?


On my Mathematica CD (received as an USAMTS prize), it says "MAC OS X (Universal)". Does this mean that it will not work on my Windows XP computer? If so, will I be able to return it to get a Windows version?

I have also sent an e-mail to Wolfram Research regarding this.

[quote="USAMTS"]If you use an operating system besides Windows XP and choose this prize, email your username and operating system to usamts@usamts.org[/quote]

My operating system is Windows XP, but I got a MAC OS X version...
Could this be a mixup?

It happened to me too. PM Mr. Rusczyk if he doesn't post on this thread... yet.

I'm told "Your package should have contained three different CDs-- one for Mac, one for Windows, and one for Linux."

If this is not the case, email usamts@usamts.org and I'll give you further information.

I have not received any information about Mathematica. Have all CDs been sent out?  :|

I only got my Mathematica email a couple of days ago, and it said that my copy would come next week, so I think it's likely that not all the orders have started shipping yet.

[quote="ehehheehee"]It has 3 CD's like you said.  The top CD was the mac one.  There was just a sticker that said if you open the container, you have agreed to the terms of the License Agreement.  That suggested that if we opened it and it was a mixup, we could not return it for one for a different system.[/quote]

That's what I thought, too.
So it DOES have the Windows version.
Thanks!

A mod can lock this topic since it is redundant now.

Just because I didn't want to start a new topic,
If I were to download Mathematica on my computer, then had to exchange that computer for a new computer, would I be able to download the Mathematica on the new computer?

When trying to register my copy, I got an error online that said something to the effect of "can't produce registration number, platform information not entered"

I got the same error as well...I tried 3 times and I get the same error message which is also repeated in an automatic e-mail sent (I blanked out my math ID):

Thank you for registering on the web.

Password generation errors were encountered for the following MathIDs:

****-*****-***** ERROR: Missing required field: platform.

A Wolfram Research Customer Service representative will contact
you regarding your password(s).

If you have any questions about Wolfram Research or our products,
please contact Wolfram Research Customer Service

Oh dear, I better hold off on installing this bad boy until I get my college computer.

This happened to me, I registered by calling 1-800-WOLFRAM. However, although I ultimately registered by phone, the customer support agent told me that the error may have been been caused by leaving the organization field blank (on the first screen in the program stage of registration, not yet on the web) which I had done. This may help, otherwise it takes under 10 minutes to register by calling 1-800-WOLFRAM from 7AM-5PM CST (-6 GMT).

[quote="Zellex"]I believe not; I installed Mathematica on my computer, but I reformatted it, and then I tried to install Mathematica, and it wouldn't let me. So...yeah.[/quote]

Yes even I installed it ...

sometimes I faced problem installing it.....

-----------
Online Math tutor at [url=https://www.esumz.com]Mental Math[/url] .


who will go to NTU? please tell me, and  i am really curious about the ENTRANCE test of NTU in INDIA :D  :rotfl:

i am a student of ntu and its a nice place to be.The entrance exam is not that tough if u guys r preparing 4 iit n stuff.But it is seriously good place for your future.If you do not get into iit than its the second best college according to me.

You guys should try for an American school too.

come on this is really a good college better than many american college in the world ranking of all colleges its 61st and many good american clleges are behind it and the best thing about this college is that it is affordable and not very expensive.

guys do check ou this site http://www.alchemy-education.com
its made by me and my friend would be very happy to get your valuable comments please support us and we have a bond that we are indians so atleast have a look at it.

enjoy life always

There are LOTs of good american colleges and british colleges infront of 61th in the world. And there are scholarships for people.


Can someone help me solve this non-homogeneous PDE (Poisson's equation)?

uxx + uyy = sin(x)sin(2y), 0 < x < pi, 0 < y < pi
with the conditions:
u(x,0) = sin(3x), u(x,pi) = 0, u(0,y) = 0, and u(pi,y) = sin(y)
Any help is greatly appreciated.

In russian http://u-pereslavl.botik.ru/~trushkov/pde/pde.pdf pp. 23-24


Let $ a,b,c>2$ and $ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}1$. Prove that

$ (a^2\minus{}3a\plus{}2)(b^2\minus{}3b\plus{}2)(b^2\minus{}3b\plus{}2)\leq6\sqrt{3}\minus{}10$

-----------------------
 :blush:  Panagiote Ligouras

[quote="vuthanhtu_hd"][quote="Ligouras"]Let $ a,b,c > 2$ and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 1$. Prove that

$ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(b^2 \minus{} 3b \plus{} 2)\leq6\sqrt {3} \minus{} 10$

-----------------------
 :blush:  Panagiote Ligouras[/quote]

Oh,I think

$ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(c^2 \minus{} 3c \plus{} 2)\leq6\sqrt {3} \minus{} 10$

If  $ a \equal{} b \equal{} c \equal{} 3$ then $ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(c^2 \minus{} 3c \plus{} 2) \equal{} 8 > 6\sqrt {3} \minus{} 10$ ???[/quote]

ops....... mistake. Thank you very much VUTHANHTU

[quote="Ligouras"][quote="vuthanhtu_hd"][quote="Ligouras"]Let $ a,b,c > 2$ and $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} 1$. Prove that

$ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(b^2 \minus{} 3b \plus{} 2)\leq6\sqrt {3} \minus{} 10$

-----------------------
 :blush:  Panagiote Ligouras[/quote]

Oh,I think

$ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(c^2 \minus{} 3c \plus{} 2)\leq6\sqrt {3} \minus{} 10$

If  $ a \equal{} b \equal{} c \equal{} 3$ then $ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(c^2 \minus{} 3c \plus{} 2) \equal{} 8 > 6\sqrt {3} \minus{} 10$ ???[/quote]
------------------------
P. Ligouras

ops....... mistake. Thank you very much VUTHANHTU[/quote]

CORRECT:
Let  and $ a,b,c > 2$ and $ ab\plus{}bc\plus{}ca\equal{}abc$. Prove that

$ (a^2 \minus{} 3a \plus{} 2)(b^2 \minus{} 3b \plus{} 2)(b^2 \minus{} 3b \plus{} 2)\leq6\sqrt {3} \minus{} 10$

[quote="Ligouras"][/quote][quote="Ligouras"][quote="vuthanhtu_hd"][quote="Ligouras"]Let $ a,b,c > 2$ and $ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = 1$. Prove that

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(b^2 - 3b + 2)\leq6\sqrt {3} - 10$

-----------------------
 :blush:  Panagiote Ligouras[/quote]

Oh,I think

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(c^2 - 3c + 2)\leq6\sqrt {3} - 10$

If  $ a = b = c = 3$ then $ (a^2 - 3a + 2)(b^2 - 3b + 2)(c^2 - 3c + 2) = 8 > 6\sqrt {3} - 10$ ???[/quote]

------------------------
P. Ligouras

ops....... mistake. Thank you very much VUTHANHTU[/quote][quote="Ligouras"]

CORRECT:
Let  and $ a,b,c > 2$ and $ ab + bc + ca = abc$. Prove that

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(b^2 - 3b + 2)\leq6\sqrt {3} - 10$[/quote]
The so-called "CORRECT" one is still incorrect that is just the same as the former.

[quote="Guest"][/quote][quote="Ligouras"][/quote][quote="Ligouras"][quote="vuthanhtu_hd"][quote="Ligouras"]Let $ a,b,c > 2$ and $ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = 1$. Prove that

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(b^2 - 3b + 2)\leq6\sqrt {3} - 10$

-----------------------
 :blush:  Panagiote Ligouras[/quote]

Oh,I think

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(c^2 - 3c + 2)\leq6\sqrt {3} - 10$

If  $ a = b = c = 3$ then $ (a^2 - 3a + 2)(b^2 - 3b + 2)(c^2 - 3c + 2) = 8 > 6\sqrt {3} - 10$ ???[/quote]

------------------------
P. Ligouras

ops....... mistake. Thank you very much VUTHANHTU[/quote][quote="Ligouras"]

CORRECT:
Let  and $ a,b,c > 2$ and $ ab + bc + ca = abc$. Prove that

$ (a^2 - 3a + 2)(b^2 - 3b + 2)(b^2 - 3b + 2)\leq6\sqrt {3} - 10$[/quote][quote="Guest"]
The so-called "CORRECT" one is still incorrect that is just the same as the former.[/quote]
Sorry:
$ a+b+c=abc$


The diameter of a circle is increased so that the circumference increases by $ 70\%$.  By what percent does the area increase?

Since the circumference increases by $ 70\%$, the diameter increases by $ 70\%$ and is multiplied by $ 1.7$ The new percent area increase is $ 70\%^2 \equal{} \frac {7}{10}^2 \equal{} 49/100 \equal{} 49\%$

actually, you can't square an increase and get the increase of the square (if that makes any sense)
e.g. $ (x \minus{} 100)^2\neq x^2 \minus{} 100$

in this problem the lateral length is multiplied by $ 1.7$ so it'll be multiplied by $ 2.89$ area-ly. $ 2.89 \minus{} 1$ as a percent is 189.


trigonometry
inequalities
geometry
3D geometry
algebra proposed
algebra

Given $ x,y,z\in (0,1)$ satisfying that
$ \sqrt{\frac{1 \minus{} x}{yz}} \plus{} \sqrt{\frac{1 \minus{} y}{xz}} \plus{} \sqrt{\frac{1 \minus{} z}{xy}} \equal{} 2$.
Find the maximum value of $ xyz$.

too simple...
let's make the following substitution:
$ x\equal{}\sin^2{\alpha}$ and so on...
it follows that
$ 2\sin{\alpha}\sin{\beta}\sin{\gamma}\equal{}\sum \cos{\alpha}\sin{\alpha}$
But it means that $ \alpha\plus{}\beta\plus{}\gamma\equal{}\pi$,then obviously
$ (\sin{\alpha}\sin{\beta}\sin{\gamma})^2\leq \frac{27}{64}$.

Why from $ 2\sin{\alpha}\sin{\beta}\sin{\gamma}\equal{}\sum \cos{\alpha}\sin{\alpha}$ we know that this angles are angles in some triangle(acute angled probably).?

[quote="matex (L)(L)(L)"]Why from $ 2\sin{\alpha}\sin{\beta}\sin{\gamma} \equal{} \sum \cos{\alpha}\sin{\alpha}$ we know that this angles are angles in some triangle(acute angled probably).?[/quote]

actually,this is a well-known identity,I'm pretty sure that the proof is available in any rational geometry book,for example,Prasolov.
anyway I would do that by making the following transformations(s):
$ 2\sum\cos{\alpha}\sin{\alpha}\equal{}\sum \sin{2\alpha}\equal{}2\sin\gamma(\cos{(\alpha\minus{}\beta)}\plus{}\cos\gamma)\equal{}4\prod\sin{\gamma}$

[quote="kunny"]Can't we prove the inequality without triangular substituition?[/quote]

Indeed, we can, and more importantly, the suggested solution from the organizing committee was with NO trigonometric substitution. So, is there any non-trigo answers?

[quote="KenHungKK"]Given $ x,y,z\in (0,1)$ satisfying that
$ \sqrt {\frac {1 \minus{} x}{yz}} \plus{} \sqrt {\frac {1 \minus{} y}{xz}} \plus{} \sqrt {\frac {1 \minus{} z}{xy}} \equal{} 2$.
Find the maximum value of $ xyz$.[/quote]

We have that
\[ 2 \sqrt {xyz} \equal{} \frac {1}{\sqrt {3}} \sum{\sqrt {x(3 \minus{} 3x)}} \leq \frac {1}{\sqrt {3}} \sum \frac {x \plus{} 3(1 \minus{} x)}{2} \equal{}
\]

\[ \equal{} \frac {3\sqrt {3}}{2} \minus{} \frac {1}{\sqrt {3}} \sum{x}.
\]
So $ 2 \sqrt {xyz} \leq \frac {3\sqrt {3}}{2} \minus{} \frac {1}{\sqrt {3}} \sum{x} \leq \frac {3\sqrt {3}}{2} \minus{} \sqrt {3} \cdot \sqrt [3]{xyz}.$

If we denote $ p \equal{} \sqrt [6]{xyz}$ we get that $ 2p^3 \leq \frac {3\sqrt {3}}{2} \minus{} \sqrt {3}p^2$. This is equivalent to
\[ 4p^3 \plus{} 2\sqrt {3}p^2 \minus{} 3\sqrt {3} \leq 0 \Rightarrow (2p \minus{} \sqrt {3})(2p^2 \plus{} 2\sqrt {3}p \plus{} 3) \leq 0,
\]
then $ p \leq \frac {\sqrt {3}}{2}$.

So $ xyz \leq \frac {27}{64}$. The equality holds for $ x \equal{} y \equal{} z \equal{} \frac {3}{4}$.

Ya the one turcas_c posted is the official solution and..... I think it's still natural. I in fact thought of using AM-GM in this way but then I failed to use it for a second time as the solution does.

But then I would say the trigo solution is more beautiful... which I still remember we Hong Kong team solved it in this method on the coach when we're heading to visit the Waterfall LOL

P.S. turcas_c i think you're one of the Romanian Team members participating in this competition right?

[quote="brianchung11"]

P.S. turcas_c i think you're one of the Romanian Team members participating in this competition right?[/quote]

Yap, I am the guy that you (I think) were trying to explain how to solve the rubik cube :)), and also the one that you gave your e-mail adress. I hope that we'll keep in touch.

i also solved it  using the trig substitution but i cant prove the fact that if the identity is true for some angles $ A,B,C$ then $ A\plus{}B\plus{}C\equal{}180$
can anyone provide the proof clearly?

Given $ x,y,z\in (0,1)$ satisfying that$ \sqrt{\frac{1 \minus{} x}{yz}} \plus{} \sqrt{\frac{1 \minus{} y}{xz}} \plus{} \sqrt{\frac{1 \minus{} z}{xy}} \equal{} 2$.Find the maximum value of $ x^4y^5z^6$.
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=518033


Hey guys!...

It seems to me there's quite a few Texans here...so I was wondering how many of you are going to ARML this year (just so I know who to expect from AoPS :-D )


Show that this polynominal does not have any real roots

$ x^4\minus{}2x^3\plus{}2x^2\minus{}6x\plus{}9$

Please help me.

[hide="Hint"]First write the polynomial as $ (x^4 \minus{} 2x^3 \plus{} x^2) \plus{} (x^2 \minus{} 6x \plus{} 9)$.[/hide]

[hide="Answer"]
Ravi B rewrote the polynomial as $ (x^4 \minus{} 2x^3 \plus{} x^2) \plus{} (x^2 \minus{} 6x \plus{} 9)$

This can be further rewritten as $ x^2(x\minus{}1)^2\plus{}(x\minus{}3)^2$

If this were to equal zero, then both terms would have to equal zero, since squares are always non-negative.

Hence, from the second term, we have $ x \equal{} 3$, and from the first term, we have $ x \in \{0,1 \}$, a contradiction.

Thus there are no zeros.
[/hide]


Let $ a$, $ b$, $ c$, and $ d$ be real numbers with $ |a\minus{}b|\equal{}2$, $ |b\minus{}c|\equal{}3$, and $ |c\minus{}d|\equal{}4$. What is the sum of all possible values of $ |a\minus{}d|$?

$ \textbf{(A)}\ 9 \qquad
\textbf{(B)}\ 12 \qquad
\textbf{(C)}\ 15 \qquad
\textbf{(D)}\ 18 \qquad
\textbf{(E)}\ 24$

[hide="Click for solution"]
$ |a\minus{}b|\equal{}2 \implies a\minus{}b\equal{}2,\minus{}2
\\ |b\minus{}c|\equal{}3 \implies b\minus{}c\equal{}3,\minus{}3
\\ |c\minus{}d|\equal{}4 \implies c\minus{}d\equal{}4,\minus{}4$

Thus, we can have $ |a\minus{}d|\equal{}9,5,3,1$ by testing cases, so the sum of these is $ 18$, or $ \boxed{\textbf{(D)}}$.
[/hide]

[hide]We can think of this problem using a simple line. _________________
Since $ |a \minus{} b|$ is just the distance, $ b$ can be either on the left side of $ a$ or the right side of $ a$, let's put it on the left side.
________b___a________

Simlilarly, $ |b \minus{} c|$ means that $ c$ can be on both sides of $ b$, but since we already have $ a$ on the right side of $ b$, putting it on the left side is more logical.
The same can be done for $ |c \minus{} d|$
___d___c___b___a___

We know that
$ |a \minus{} b| \equal{} 2$
$ |b \minus{} c| \equal{} 3$
$ |c \minus{} d| \equal{} 4$
Thus, $ |a \minus{} d| \equal{} 9$.
Since we know this can be done on the right side of a, we do $ 9*2 \equal{} 18$.[/hide]

Why does Chopstiks method work?  The distance and everything makes sense, but why does just doing 9 * 2 give you the sum of all possible values?

It doesn't. Imagine if all 3 values were 2. Then, the only answers would be 2,4 for a sum of 6, but Chopstik's method gives 12.


Are there any North Korean members that wish to elaborate on [url=http://www.amazon.com/Beta-Algebra-Algirdas-Javtokas/dp/1419636138]this book[/url]. I'm sure some people are aware of the fuss. I'm actually really curious what the motivation (besides the content) of banning it was. And how can a math book possibly violate a part of your culture :)

I am pretty sure North Korea is one of the most poorest countries in the world, thus somewhat difficult to have computers for people to use. With that, I am 99% certain that no North Koreans are in this forum.

It also siad that dictators hate original thinkers.

north korea isn't poor, kim jong il just takes all the money and doesn't LET them use computers

Judging by the results of some searching online, I get the impression that

(a) The book is legit and contains actual math

but

(b) the author either used a pseudonym or is not affiliated with a university that has a website,

and

(c) the author created the rumor that the book had been banned in order to increase interest in the book and in his ideas.

http://www.nigeltomm.com/betalgebra/javtokasbeta.pdf contains the axiomatic definition of his "Beta algebra," the structure in question.  This is all the project of one publicity-hungry person which suggests to me a high likelyhood of crankitude, but I haven't delved into his stuff yet to determine if it makes sense (or if it's interesting).  His website is at http://www.nigeltomm.com/betalgebra/ .  To answer your question, Fedja, it seems that he avoids your problem by making arithmetic messy between different "idenity elements," so (at least) one of the two equalities you wrote will fail to be true.  (Reading a little further ... the "identity elements" form a semigroup.)

Unrelatedly, the CIA World Factbook lists the North Korean GDP per capita as 1800 USD.  (For reference, this is about half of the India GDP per capita.)  The entry for GDP (Purchasing power parity), which lists 40 billion USD, notes: "North Korea does not publish any reliable National Income Accounts data; the datum shown here is derived from purchasing power parity (PPP) GDP estimates for North Korea that were made by Angus MADDISON in a study conducted for the OECD; his figure for 1999 was extrapolated to 2006 using estimated real growth rates for North Korea's GDP and an inflation factor based on the US GDP deflator; the result was rounded to the nearest 10 billion USD (2006 est.)"

[quote="mws21"]when I say poor, it is in terms of GDP which for North Korea is about $\$1000$[/quote]yes, well that's how much money each commoner gets. the dictator probably gets more. You also have to consider that since they're socialist it's harder to put GDP into as accurate of terms

[quote="fedja"]It is slightly off topic, but it would be nice if somebody could tell what the notion of an "identity element" in this book is (I mean, we have a proof that any two identity elements must coincide: $e_{1}=e_{1}\cdot e_{2}=e_{2}$, so, what is the book author talking about? :?[/quote]
Matrix algebra, for example, contains infinitely many identity elements, such as $\begin{pmatrix}1&0\\0&1\end{pmatrix}$ or $\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}$.  These matrices [b]cannot[/b] be multiplied together as they are of different sizes.


In a math class of 30 students, 12 out of 15 girls are freshmen and 11 out of 15 boys are freshmen. What is the probability that in a randomly selected group of five students from the class, there will be two freshmen girls and three freshmen boys? Express your answer as a decimal to the nearest thousandth.

There are 12C2 ways to pick two freshmen girls and 11C3 for freshmen boys. (12C2)*(11C3)=10890. There are 30C5 ways to choose 5 people from 30. 30C5=142506. 10890/142506 is 0.076 when rounded to the nearest thousandth.

how do you know whether a problem has distinguishable people/whatever?

[quote=ernie]This is simply $ \frac{12}{30} \times \frac{11}{29} \times \frac{11}{28} \times \binom{3}{2}$ or $ \boxed{0.179}$.[/quote]

This seems to be wrong.

We can choose 2 freshmen girls and 3 freshman boys in $\binom{12}{2}\binom{11}{3} = 10890$ ways. There are a total of $\binom{30}{5} = 142506$ possible groups of 5 students we can choose. Thus the probability of selecting a group of 5 students with 2 freshman girls and 3 freshmen boys is $\frac{10890}{142506} \approx \boxed{0.076}$.

First find the number of ways to order the two freshman girls and three freshman boys that are chosen. This is simply $\frac{5!}{2!3!} = 10$. Let us consider one of these orders, where the two girls are picked first and then the boys. To compute this probability, we compute $\frac{12}{30} \times \frac{11}{29} \times \frac{11}{28} \times \frac{10}{27} \times \frac{9}{26}$ which simplifies to approximately $.00764$. Because there are $10$ orders, so we multiply $.00764$ by $10$ to get our final answer as $\fbox{.076}$ (rounded to the nearest thousand).

Wait so are they replaced or not replaced?

[quote=Natsuki]NOOOOO! I PUT $\dfrac{66 \cdot 165}{142506}$ AND IT SAID WRONG EVEN THOUGH IT IS RIGHT!!!!!![/quote]

i think you need to simplify?

[quote=Natsuki]NOOOOO! I PUT $\dfrac{66 \cdot 165}{142506}$ AND IT SAID WRONG EVEN THOUGH IT IS RIGHT!!!!!![/quote]

Do you know what simplifying is?

[quote=centslordm]Wait, doesn't a pop up show up if you have entered an unsimplified fraction?[/quote]

Uh but (s)he did multiply...

[quote=abc2142]argh I got this problem wrong the first try because I expressed my answer as a fraction[/quote]

lol


Let $ a,b,c,d$ be positive numbers such that $ abcd\equal{}1$.prove that \[ \frac{1\plus{}ab}{1\plus{}a}\plus{}\frac{1\plus{}bc}{1\plus{}b}\plus{}\frac{1\plus{}ca}{1\plus{}c}\plus{}\frac{1\plus{}da}{1\plus{}d} \geq 4\]

[quote="apollo"]Let $ a,b,c,d$ be positive numbers such that $ abcd \equal{} 1$.prove that
\[ \frac {1 \plus{} ab}{1 \plus{} a} \plus{} \frac {1 \plus{} bc}{1 \plus{} b} \plus{} \frac {1 \plus{} ca}{1 \plus{} c} \plus{} \frac {1 \plus{} da}{1 \plus{} d} \geq 4
\]
[/quote]
Maybe you mean $ \frac {1 \plus{} ab}{1 \plus{} a} \plus{} \frac {1 \plus{} bc}{1 \plus{} b} \plus{} \frac {1 \plus{} cd}{1 \plus{} c} \plus{} \frac {1 \plus{} da}{1 \plus{} d}\geq4$ $ ?$
If so see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85389
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88280


calculus
derivative
calculus computations

The Chain Rule. Using the variables $u=x-2$ and $v=y-3$, you can write $\vec G$ in a form similar to $\vec F$. The derivatives are related as follows: $\frac{\partial P}{\partial x}=\frac{\partial P}{\partial u}\frac{\partial u}{\partial x}=\frac{\partial P}{\partial u}$, etc. It follows that $\mathrm{curl}\, \vec G$ is the same as $\mathrm{curl}\, \vec F$ with the above substitution, i.e. zero.


Solve inequality:
$ \sqrt{x\plus{}3} \plus{} \sqrt{3x\plus{}2} > 7$

Tricky problem. I would solve the problem as if it were an equality, ie transpose one term then square both sides and solve for x. Then, since the sum of two square root functions would only intersect a horizontal line once, your answer would simply be the interval $ (x, \infty)$.

I'm not sure but could you solve this by using natural log?
If this is possible any logarithm can be used it's just natural log uses less ink.

[hide](x+3)^(1/2) + (3x+2)^(1/2) greater than 7

(1/2) ln (x+3) + (1/2) ln (3x+2) greater than ln 7

ln(x+3) + ln((3x+2) greater than ln(7^2)

ln (x+3)(3x+2) greater than ln 49

ln (3x^2 +11x + 6) greater than ln 49

3x^2 + 11x -43 greater than 0

to get x greater than (-11+7(13)^1/2)/6 or (-11-7(13)^1/2)/6

If this is wrong what and were are my mistake(s)?[/hide]

$ \ln (a\plus{}b) \neq \ln a \plus{} \ln b$


Let $ x^2 \plus{} y^2 \plus{} xy \equal{} 1$
Denote $ f(x,y) \equal{} |(m \plus{} 1)x^2 \plus{} (m \plus{} 2)y^2 \plus{} (m \minus{} 1)xy|$
Find the $ Min(Max(f(x,y)))$ there Min means  the minimum value and Max means the maximum one

[quote="Allnames"]Let $ x^2 \plus{} y^2 \plus{} xy \equal{} 1$
Denote \[ f(x,y) \equal{} |(m \plus{} 1)x^2 \plus{} (m \plus{} 2)y^2 \plus{} (m \minus{} 1)xy|
\]
Find the \[ Min(Max(f(x,y)))
\]
there Min means  the minimum value and Max means the maximum one[/quote]
Sorry for this spam post,but I really need your help,It is my work at home which I must complete before March  :blush:
@ Can Mod delete my second post?


geometry
circumcircle
rotation
geometric transformation
geometry solved

A line passing through the center $O$ of an equilateral triangle $ABC$ intersects the circumcircles of the triangles $OAB,OBC,$ and $OCA$ at $K,L,$ and $M.$ Prove that
\[
OK^{2}+OL^{2}+OM^{2}=2AB^{2}.
\]

Cute little problem! :)

Rotate the circle $(OAB)$ by an angle of $\frac {2\pi}3$ so that it lands on $(OBC)$, and rotate the circle $(OCA)$ by an angle of $\frac {4\pi}3$ so that it also lands on $(OBC)$. The images $K',M'$ of $K,M$ through these rotations form, together with $L$, an equilateral triangle inscribed in $(OBC)$ (which has side equal to $AB$, of course, because their circumcircles have equal radi).

Let this triangle have center $T$. We can now use $OK^2+OL^2+OM^2=OK'^2+OL^2+OM'^2=3OT^2+TK'^2+TL^2+TM'^2$, and the result follows from $OT=TK'=TL=TM'=\frac {AB}{\sqrt 3}$.


[quote="minsoens"][quote="tunganh"][quote="silver1989"]Let a,b,c are three sides of a triangle such that:a+b+c=3.Prove that:
 $4(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 3abc+9$[/quote]
I don't think it's true :D
Try $a=2.9, b=0.5, c=0.5$[/quote]i think he means $a=2.9, b=0.05, c=0.05$[/quote]
Thanks, just what I meant :D

Yes, but: $a,b,c$ have to be the sides of a triangle.
In the above example we clearly have $a > b+c$.

[quote="silver1989"]Let a,b,c are three sides of a triangle such that:a+b+c=3.Prove that:
 $4(\sqrt{a}+\sqrt{b}+\sqrt{c})\geq 3abc+9$[/quote]

We may assume $a \geq b \geq c$
Let $F(a,b,c)= 4(\sqrt{a}+\sqrt{b}+\sqrt{c})-3abc-9$
$F(a,b,c)-F(a,\frac{b+c}{2},\frac{b+c}{2}) = 3a\frac{(b-c)^{2}}{2}+4(\sqrt{b}+\sqrt{c}-2\sqrt{\frac{b+c}{2}})$
$= (\sqrt{b}-\sqrt{c}^{2})(\sqrt{b}+\sqrt{c})(\frac{3a (\sqrt{b}+\sqrt{c})}{2}-\frac{1}{(\sqrt{b}+\sqrt{\frac{b+c}{2}})(\sqrt{c}+\sqrt{\frac{b+c}{2}})})$
But ${(\sqrt{b}+\sqrt{\frac{b+c}{2}})(\sqrt{c}+\sqrt{\frac{b+c}{2}})}) \geq (\sqrt{b}+\sqrt{c})^{2}(\leftrightarrow b^{2}+c^{2}\geq 2bc)$ and $(\sqrt{b}+\sqrt{c})^{3}\geq 1)$
It follows that $F(a,b,c)-F(a,\frac{b+c}{2},\frac{b+c}{2}) \geq 0$
We will prove inequality when $b=c$
We have $a+2b=3$ then $b=\frac{3-a}{2}$ and $a \in \ [1,\frac{3}{2})$
We need to prove: $4(\sqrt{a}+2\sqrt{b})-3ab^{2}-9 \geq 0$
$\leftrightarrow F(a)=3a^{3}-18a^{2}+27a-16\sqrt{a}-16\sqrt{6-2a}+36 \leq 0$
$F^{'}(a)=9a^{2}-36a+27-\frac{8}{\sqrt{a}}+\frac{16}{\sqrt{6-2a}}$
$F^{''}(a)=18a-36+\frac{4}{a\sqrt{a}}+\frac{16}{\sqrt{6-2a}}\geq 0$
It follows that
$F^{'}(a) \leq F^{'}(\frac{3}{2})<0$
Or $F(a) \leq F(1) =0$


Find all real solutions to
$ [x^2 \plus{} 2x] \equal{} {[x]}^2 \plus{} 2[x]$

OMG double posting within 3 mins. I don't blame you.
http://www.mathlinks.ro/viewtopic.php?p=1380211#1380211


There are 32 people sitting around a round table and each person was either one year older or one year younger than the person to his or her left. Then, is it true that there must be two people directly opposite one another having the same age?

[hide]let $ a_1, a_2, ..., a_{32}$ be the ages, in that order around the table. for each $ i$, $ i \equal{} 1, ..., 32$, let $ d_i \equal{} a_{i \plus{} 1} \minus{} a_i$ (with the convention that $ a_{33} \equal{} a_{1}$). now for each $ i$, let $ x_i \equal{} d_i \plus{} d_{i \plus{} 1} \plus{} \cdots \plus{} d_{i \plus{} 15}$. since $ d_1 \plus{} d_2 \plus{} \cdots \plus{} d_{32} \equal{} 0$, $ x_{17} \equal{} \minus{} x_1$. note that in the sequence
\[ x_1, x_2, x_3, ..., x_{17} \equal{} \minus{} x_1,\]
the difference between successive terms is $ 0$ or $ \pm 2$. moreover, note that $ x_1$ is even because it is a sum of an even number of $ \pm 1$'s. therefore, the above sequence starts and ends at even numbers of opposite sign and passes only through even numbers, so it must pass through 0: $ x_i \equal{} 0$ for some $ i$, and this means $ a_i$ and the one opposite $ a_i$ are the same.[/hide]


calculus
integration
algebra
polynomial
calculus computations

Consider the polynomial $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, with degree less than or equal to 2.

When $ f$ varies with subject to the constrain $ f(0) \equal{} 0,\ f(2) \equal{} 2$, find the minimum value of $ S\equal{}\int_0^2 |f'(x)|\ dx$.

[quote="kunny"]Consider the polynomial $ f(x) \equal{} ax^2 \plus{} bx \plus{} c$, with degree less than or equal to 2.

When $ f$ varies with subject to the constrain $ f(0) \equal{} 0,\ f(2) \equal{} 2$, find the minimum value of $ S \equal{} \int_0^2 |f'(x)|\ dx$.[/quote]

cases:
A. a>0, h=-b/2a<2     then given intregral=2
B. a>0,h=-b/2a<0      then also given intregral=2.
C.a=0, then again given=2.
D. a<0.  then given intregral=2f(h)-2.
 now, f(h)=$ \minus{}b^2/{2(1\minus{}b)}$. but, 2a+b=1 means b>1. so min(f(h))=2. so again min value= 2*2-2=2

Does this work?

$ \int_0^2 |f'(x)|\ dx \ge \int_0^2 f'(x)\ dx\equal{}f(2)\minus{}f(0)\equal{}2$ with equality iff $ f'(x)\ge 0$ in $ [0,2]$. Since $ f'(x)\equal{}2ax\plus{}b$, $ a$ and $ b$ can be arbitrarily chosen such that $ f'(x)\ge 0$ in $ [0,2]$.


may be posted before but i think yhat is a nice inequalite. here you are.
let $a$, $b$, $c$ be positive real such $a^{2}+b^{2}+c^{2}=3$.

prove that $a+b+c\ge{a^{2}b^{2}+c^{2}b^{2}+a^{2}c^{2}}$. :lol:

Cauchying on RHS $\sqrt{(ab+ac+bc)^{2}}\leq a^{2}+b^{2}+c^{2}$.

EDIT maybe the proof is wrong


integration
calculus
limit
derivative
calculus computations

$ \frac{d}{dx} \int_0 ^x {{e^{\minus{}t}}^2dt}$

How do I solve this, or any other problem with this format:

$ \frac {d}{dx} \int_a^x{f(t)}dt$

the first one equals e^(-x^2)
the second one f(x). It is called the Fundamental Theorem of Calculus
You can see the link at Wikipedia:
[url]http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus[/url]

[quote="sneezywalrus"]$ \frac {d}{dx} \int_0 ^x {{e^{ \minus{} t}}^2dt}$

How do I solve this, or any other problem with this format:

$ \frac {d}{dx} \int_a^x{f(t)}dt$[/quote]

well, isn't it $ {e^{ \minus{} x}}^2$ ?

by solve, do you mean how to prove the fundamental theorem of calculus, i.e. $ f(x)\;\equal{}\;\frac {d}{dx} \int_a^x f(t) \mathrm dt$ ?  :maybe:

suppose that $ F(x) \equal{} \int_{a}^{x} f(t) \mathrm dt$, then $ F(x_1 \plus{} \Delta x) \minus{} F(x_1) \equal{} \int_{a}^{x_1 \plus{} \Delta x} f(t) \mathrm dt \minus{} \int_{a}^{x_1} f(t) \mathrm dt \equal{} \int_{x_1}^{x_1 \plus{} \Delta x} f(t)\mathrm dt$

apply mean value theorem to get

$ \lim_{\Delta x \to 0} \frac{F(x_1 \plus{} \Delta x) \minus{} F(x_1)}{\Delta x} \equal{} \lim_{\Delta x \to 0} f(c)$

now you can either use the squeeze theorem or give a short $ \delta\minus{}\varepsilon$ proof to get the RHS limit and hence conclude  $ F'(x_1) \equal{} f(x_1)$

I found this generalized version (which can be shown using the Fundamental Theorem of Calculus and the chain rule for derivatives) useful for a couple of AP Calculus BC questions:

$ \frac{d}{dx} \int_{a(x)}^{b(x)} f(t) dt \equal{} b'(x)f(b(x)) \minus{} a'(x)f(a(x))$

where $ a(x)$ and $ b(x)$ are expressions in terms of $ x$.

Yes, if I am not mistaken this is also called Leibniz's Theorem


Hey,
how many extra curriculars are you guys or gals involved in?

In college, I have about 1: running.  I also do a bunch of other little ec's on the side: working/tutoring, UMA, hmmt, and hacking occasionally.

In high school.

Have maybe 5 or 6. President of two clubs and then mostly volunteering and 2 sports.

I'm also in a couple of extra curriculars:

Varsity b-ball
It's academic club (didn't make the team,wasn't fast enough)
recycling club
year book club
aaalead
diversity club
school newspaper
student council.

We have no math team, and I am trying to make one, but no one I've spoken to seems interested. How should I approach this?

[quote="mdk"]We have no math team, and I am trying to make one, but no one I've spoken to seems interested. How should I approach this?[/quote]

I just emailed the head of the math department, who forwarded my email to the other math teachers. One of them replied . . .

[quote="K81o7"]In order of probable time commitment...though violin and track will probably be about equal.

Violin
Track (once winter starts)
AoPS
Karate
Math club (which I'm currently running alone, trying to create interesting problem sets for hungry freshmen)[/quote]

lol, is AoPS counted as an extracurriculum and can be added to our college application resume?

[quote="kimby_102"][quote="K81o7"]No...that was more or less a joke.  But heck, I don't know if I can even put math club.  I do make problem sets and do problems with the freshmen, but we don't get funding, go on trips (ie HMMT, ARML...maybe I can pull that off next year), or even meet weekly...[/quote]

doesn't taking classes at AoPS website count as an ec?[/quote]

no, it doesn't.

[quote="kimby_102"]doesn't taking classes at AoPS website count as an ec?[/quote]

Anything you do that school doesn't force you to do could count as an extracurricular activity. (I was reading a book about college applications tonight, and the chapter I was in was the chapter on activities.) How involved you are in AoPS classes may be something that shows up in your scores on AMC tests or in the ARML competitions, so maybe it doesn't need to be mentioned separately, but no one should feel bashful about mentioning participating in AoPS classes on a college application if the mention comes up in an appropriate context.

A lot of colleges probably don't know much about AoPS.  So it might not make that much of a difference if you put it on or not. If you're applying to MIT I would definitely suggest putting AoPS down.  Matt McGann, one of the admissions officers, has done mathjams on AoPS the last couple years about MIT.  He know a lot more about AoPS than most college admissions officers.  I'd guess that the other admissions officers also know about it.


geometry
incenter
projective geometry
geometry unsolved

Let $ ABC$ be an isosceles triangle with $ AC \equal{} BC.$ Its incircle touches $ AB$ in $ D$ and $ BC$ in $ E.$ A line distinct of $ AE$ goes through $ A$ and intersects the incircle  in $ F$ and $ G.$ Line $ AB$ intersects line $ EF$ and $ EG$ in $ K$ and $ L,$ respectively. Prove that $ DK \equal{} DL.$

First, notice that $ K$ and $ L$ are both either on or both outside the segment $ AB$. Thus, it is sufficient to prove that $ AK \equal{} LB$ and to prove this, it is obviously sufficient to prove $ \angle ARK \equal{} \angle LEB$, where $ R$ is the tangency point on $ AC$ to the incircle of $ \triangle ABC$.
Let $ I$ be the incenter of $ \triangle ABC$. Then,
\[ \angle EFR \equal{} \frac {\angle EIR }2 \equal{} \angle CIR \equal{} \angle BAC,
\]
so $ A,K,F,R$ are cyclic. Thus,
\[ \angle ARK \equal{} \angle AFK \equal{} \angle GFE \equal{} \angle LEB,
\]
which solves the problem.

It is also possible to use Menelaus and power of a point, based on a solution of czech contestant Tomas Pavlik.

Yes, I know, but I can't help my self.  :D  :roll:
Although there is solution already posted
I will post my solution. It is nothing special but it is projective. :whistling:

Let incircle touch $ AC$ at $ T$ and let $ TD$ cut $ FG$ at $ X$
Since $ TD$ is polar line for $ A$ we have $ (F,G;X,A)=-1$.

Now we have:

\begin{eqnarray*}
(K,L;D,\infty) &=& (EK,EL;ED,E\infty)\\
                    &=& (EF,EG;ED,ET)\\
                    &=& (F,G;D,T)\\
                    &=& (TF,TG;TD,TT)\\
                    &=& (TF,TG;TX,TA)\\
                    &=& (F,G;X,A)=-1
\end{eqnarray*}

So $ D$ is midpoint of segment $ KL$.  Q.E.D.

Dear Number1, we have quadrilateral $ TFDG$ is harmonic, so $ E.(TFDG)$ is also a harmonic bundle. Now, notice that $ ET\parallel KL$, the result follows.

Every time there are two equal segments next to each other,
     there is some kind of asynchronous ( possibly) degenerate species of butterfly
     flying around.

    M.T.

The solution using Menelaus theorem according to Tomas Pavlik: Menelaus theorem is used twice to the triangle $AXB$ and lines $EF$ and $GE$, where $X$ is the intersection of $AF$ and $BC$. You will get to $AL*AK=BK*BL$ which we can somehow finish.

Let $T$ be the tangency point of $AC$ with the incircle. Since we have that $BC=CA$ this implies $ET \parallel BC$, thus we have that:
$$-1=(F,G;D,T) \overset{E}{=} (K,L;D,P_{\infty})$$
this gives us $KD=DL$ ... :D


for positive reals a,b,c prove ineq:
$ (a^5 \minus{} a^2 \plus{} 3)(b^5 \minus{} b^2 \plus{} 3)(c^5 \minus{} c^2 \plus{} 3) \ge (a \plus{} b \plus{} c)^3$

I tried Jensen and Minkowski but there are no results

Holder works :

$ (a^3 \plus{} 1 \plus{} 1)^{\frac {1}{3}} (1 \plus{} b^3 \plus{} 1)^{\frac {1}{3}} (1 \plus{} 1 \plus{} c^{3})^{\frac {1}{3}} \geq (a \plus{} b \plus{} c)$

For $ a > 0$ we have $ 0 \leq (a^3 \minus{} 1)(a^2 \minus{} 1) \equal{} a^5 \minus{} a^3 \minus{} a^2 \plus{} 1\Leftrightarrow a^5 \minus{} a^2 \plus{} 3 \geq a^3 \plus{} 2.$

Then it follows :

$ (a^5 \minus{} a^2 \plus{} 3)(b^5 \minus{} b^2 \plus{} 3)(c^5 \minus{} c^2 \plus{} 3) \geq (a^3 \plus{} 2)(b^3 \plus{} 2)(c^3 \plus{} 2)\ge (a \plus{} b \plus{} c)^3$

Edit : Certainly it is Holder .

[quote="alex2008"]Holder works :
$ (a^3 \plus{} 1 \plus{} 1)^{\frac {1}{3}} (1 \plus{} b^3 \plus{} 1)^{\frac {1}{3}} (1 \plus{} 1 \plus{} c^{3})^{\frac {1}{3}} \geq (a \plus{} b \plus{} c)$
[/quote]isn't it Minkowski?  :wink:  but it's not important detail. nice solution- thx!


[quote="Silverfalcon"]In a class, some pupils learn German, the other learn French. The number of girls learning French and the number of boys learning German total to 16. There are 11 pupils learning French, and there are 10 girls in the class. In addition to the girls learning French, there are 16 pupils. How many pupils are there in the class?

A. 18
B. 21
C. 23
D. 27
E. 31[/quote]

[hide]
Call the number of girls learning French X.
Then the number of boys learning German is 16-x
The number of boys learning French is 11-x
The number of girls learning German is 10-x
We know that $ 16\minus{}x\plus{}11\minus{}x\plus{}10\minus{}x\equal{}16$ by the last part.
So x=7.  That means the total number of pupils is $ 7\plus{}16\equal{}23$?

[/hide]

Nice solution! That was very fast.

[hide="Rather boring, slow method"]
When I first attempted this problem, I actually couldn't solve it right away because my four variables were $ F_g, F_b, G_g, G_b$. It didn't work out very well so here is my lesson: avoid setting up variables like that because it makes difficult to see the obvious.

Now:

Let # of girls in French = A, # of boys in French = B, # of girls in German = C, and # of boys in German = D. Then:

A+D = 16
A+B = 11
A+C = 10
B+C+D = 16

Note that RHS is same for the first and fourth equation so, we know that A = B+C. Now, add the second and third equation:

2A+B+C = 21

Using A = B+C, we see that A = 7. Following that, B = 4, C = 3, and D = 9. So, the answer is 7+4+3+9 = 23. [/hide]


all the 7-digit numbers containing each of the digits 1,2,3,4,5,6,7 exactly once, and divisible by 5, are arranged in the increasing order. Find the 2000-th number in this list.

but there are only 720 numbers in this list, so how can we find the 2000th?


Before some weeks I added to my signature the "join the Braiteaser competition http://www............ Now I want to erase it but everytime I try it says that my signature is too long. I deleted all the text but this message was still being displayed. Are we supposed to have -10 characters in our signatures? :)

There is a character limit, and it was set to 128 for some reason. Now it's back to 255 as it should be.

does my signature show in my posts? i know i have one but wheenever i post i never see it...

go to http://www.artofproblemsolving.com/Forum/usercp.php?mode=prefs&sub=posting
 and choose yes to "Always attach my signature"


$ABCD$ is a convex quadrilateral,where $AC$ and $BD$ intersects at $P$.$M$,$N$ are the midpoints of $AB$ and $CD$ respectively.Prove that the lines passing through $P$,$M$,$N$ which is perpendicular to $AD$,$BD$,$AC$ are concurrent if and only if $ABCD$ is cyclic

We denote as $A',$ $D',$ the orthogonal projections of $A,$ $D,$ on $BD,$ $AC,$ respectively.

The segment lines $AA',$ $DD',$ $PP',$ where $P'$ is the orthogonal projection of $P$ on $AD,$ as the altitudes of the triangle $\bigtriangleup PAD,$ are concurrent at one point, so be it $Q.$

We denote also as $M',$ $N',$ the orthogonal projection s of $M,$ $N,$ on $BD,$ $AC$ respectively and as $R,$ the intersection point of $MM',$ $NN'.$

Because of the quadrilateral $ABCD$ is cyclic, the triangles $\bigtriangleup PAB,$ $\bigtriangleup PCD$ are similar and so, we conclude that their pairs of homologous segment lines $AA',$ $DD'$ and $MM',$ $NN',$ intersect their homologous also side segments $PB,$ $PC,$ at points formed segments with the same ratio.

That is we have that $\frac{PA'}{A'M'}= \frac{PD'}{D'N'}$ $,(1)$

From $(1)$ we conclude that the points $P,$ $Q,$ $R$ are collinear and so, the segment lines $P'PQ,$ $PQR$ are coincided, because of they have two common points.

Hence, the segment lines $PP',$ $MM',$ $NN',$ are concurrent at one point and the proof is completed.

Kostas Vittas.

PS. It is easy to prove that the quadrilateral $ABCD$ is cyclic, when the segment lines $PP',$ $MM',$ $NN',$ are concurrent $($ We suppose that the circumcircle of $\bigtriangleup ABD,$ intersects the diagonal $AC$ at a point, so be it $C'$ and so, we conclude that the segment line $N_{1}N'_{1},$ where $N_{1}$ is the midpoint of $C'D$ and $N'_{1},$ its orthogonal projection on $AC'\equiv AC,$ passes through the point $R,$ as the intersection point of $PP',$ $MM'.$ Hence $N_{1}N'_{1}\equiv NN'$ $\Longrightarrow$ $C'\equiv C$ $).$


geometry
geometric transformation
reflection
incenter
geometry unsolved

The interior bisector of the angle $ A$ of the triangle $ ABC$ cuts the incircle at two points; the one closer to $ A$ is denoted by $ O_A$. Define the points $ O_B$ and $ O_C$ similarly on the internal bisectors of the angles $ B$ and $ C$, respectively. The circle $ c_A$ is drawn with center $ OA$ and it is touching the sides $ AB$ and $ CA$. Draw $ c_B$ and $ c_C$ similarly. Taken any two of the circles $ c_A, c_B$ and $ c_C$ consider their common external tangents different from the corresponding sides of the triangle, respectively. Prove that these three tangents are concurrent.

Denote ObOc∩BC=X , Y,Z are defined similarly

It is easy to prove OaD,OcF,ObE are concurrent at O
∠EOOa=∠OaOE ∠OOaOc=∠OcOaE  => O is the reflection of E w.r.t OaOc
Analogously, O is the reflection of F w.r.t OaOb
O is the reflection of D w.r.t ObOc
So their common external tangents are corrent

We can also prove X,Y,Z are collinear by using Menelaus

wen it comes to tangents to three circles..........
only one think comes to my mind..............
monges theorem :rotfl:  :rotfl:

but I don't think you can solve it with monge's theorem

ya i know dat but i was justt joking :rotfl:  :rotfl:  :P

http://mathworld.wolfram.com/MongesCircleTheorem.html

You can also prove that if O is the incenter of triangle DEF then points F, O are symmetric with respect to the line OaOb and D,O and E,O are symmetric with respect to ObOc and OaOc (similar problem appeared in Polish MO this year. Proof can be found on the forum).

Can you prove, that the tangent in Oa to he incircle of ABC is parallel to the comon tangent of the two circles centered at Oa and Oc, without using the fact, that these three tangents intersect in a point.


There are 5 blue chips, 4 red chips and 3 yellow chips in a bag. One chip is drawn from the bag. That chip is placed back into the bag, and a second chip is drawn. What is the probability that the two selected chips are of different colors? Express your answer as a common fraction.

Same chip probability is $ \frac{5\cdot4\plus{}4\cdot3\plus{}3\cdot2}{12\cdot11}\equal{}\frac{19}{66}$, so different chip probability is $ 1\minus{}\frac{19}{66}\equal{}\boxed{\frac{47}{66}}$.

That's incorrect. The denominator should be $12 \times 12$ or $144$.

The answer is actually [hide=solution]47/72[/hide]

Wait a second.....
I thought we can do this:

Count the probability when the two cards picked are the same:

Total # of ways for picking two of the five blue cards= $5C2$ which is $10$.

We do this with the other two and we count the total # of ways from picking two from the 7 cards which is $12C2$ that is $66$.

So the probability that the two cards are the same is:

$\frac{19}{66}$

So answer would be $1-\frac{19}{66}$ which is $\boxed{\frac{47}{66}}$

[hide=Another way to solve this.]Total is $12^2 = 144$. Now we do case work to find the desired.

Case 1: Blue then Red: There are $5$ choices for the blue chip and $4$ choices for the red chip. So, there are $5 \cdot4=20$ ways to do this.

Case 2: Blue then Yeallow: Similarly, there are $5$ choices for the blue chip and $3$ choices for the yeallow chip. So, there are $5 \cdot3=15$ ways to do this.

Case 3: Red then Yeallow: There are $4$ choices for the red chip and $3$ choices for the yeallow chip.  So, there are $4 \cdot3=12$ ways to do this.

Since order matters here (I think b/c we are not using combinations. Honestly, I really don't know why but I am pretty sure I am right)
we must multiply by $2$.

Thus, the desired is $2(20+15+12)=94$.

Our answer is $\frac{desired}{total} = \frac{94}{144}=\boxed{47/72}$[/hide]

EDIT: @above, this might help

[quote=illogical_21][hide=sol]
$$\frac{5\cdot4+4\cdot3+3\cdot2}{12\cdot12}=\frac{19}{72}\implies 1-\frac{19}{72}=\boxed{\frac{53}{72}}$$[/quote]

This is the one time I've seen a HOFer get something wrong (not saying you're bad this one was hard for me too even though I got it) lol :)
wait no it's not aditya11 made one

We may also use complementary probability, where we compute the probability that the two chips are of the same color, then subtract that from $1$.

The way to use complementary counting?

Most other solutions forgot that they put one chip back in so:
1 - (5*5 + 4*4 + 3*3)/(12*12) =
1 - (25 + 16 + 9)/144 =
1 - 50/144 =
94/144 =
47/72.

READ CAREFULLY BECAUSE

I did the same thing as what @bove did, complementary counting:

[hide=Solution]
total ways to draw two chips from that bag is $144$

ways to draw two blue chips is $5^2=25$
ways to draw two red chips is $4^2=16$
ways to draw two yellow chips is $3^2=9$
total ways to draw two chips of the same color is $25+16+9=50$

$\frac{144}{144}-\frac{50}{144}=\frac{94}{144}=\boxed{\frac{47}{72}}$
[/hide]


trigonometry
inequalities unsolved
inequalities

We shall seek to prove:
$\forall x \in \left( {0,\pi } \right) \Rightarrow \sum\limits_{k = 1}^n {\left( {n - k + 1} \right)\sin kx > 0} $


topology
geometry
3D geometry
sphere
real analysis
advanced fields
advanced fields unsolved

Hi! I know I will always be too stupid for differential topology, but I'm nevertheless interested in the following question that shouldn't be hard to solve for people in this forum:

Let $M$ be a smooth, compact, connected 2-manifold without boundary and $f: M \to S^{2}$ a smooth map of degree 2 ($S^{2}$ is the 2-sphere, degree is defined as in Milnor's "Topology from the Differentiable Viewpoint"). Can f have exactly one critical point?

I suppose the answer is no. But I don't know, how to prove it.

How would you use Sard's theorem in this context? Sard's theorem is a statement about critical [b]values[/b], not about critical [b]points[/b]. Furthermore in the situation of one critical point we know that there is only one critical value: if $x \in M$ is our only critical point, $f(x) \in S^{2}$ is the only critical value. So there is no contradiction to Sard's theorem, that asserts that the critical values have Lebesgue measure zero.

There must be a reason for the target manifold to be $S^{2}$. I suppose one has to use some property of $S^{2}$ itself or some feature of maps from a 2-manifold to $S^{2}$ that I'm not aware of to get a contradiction to the assumption of the existence of such a map with only one critical point.


Prove that:
$ r_{a}^2\plus{}r_{b}^2\plus{}r_{c}^2 \geq m_{a}^2\plus{}m_{b}^2\plus{}m_{c}^2\plus{}2(R^2\minus{}4r^2)$
(happy birthday to me)

This is my solution,it's very ugly:
Let :a=x+y;b=y+z;c=z+x;
and we get to prove this inequality:
$ 6(xyz)^2\plus{}\sum x^2y^2(x^2\plus{}y^2)\plus{}2\sum(xy)^3\geq 2xyz(\sum x^3\plus{}\sum xy(x\plus{}y))$

$ \sum_{sym}(x^4y^2 \minus{} x^4yz \plus{} x^3y^3 \minus{} 2x^3y^2z \plus{} x^2y^2z^2)\geq 0$
Which is obvious true.

Happy birthday Tuan,
Here is my gift for you :)
1/Prove that $ p^2\geq 7r^2\plus{}10Rr$
2/Prove that $ p^2\leq \frac{R(4R\plus{}r)^2}{2(2R\minus{}1)}$

[quote="Kyoshiro"]Happy birthday Tuan,
Here is my gift for you :)
1/Prove that $ p^2\geq 7r^2 \plus{} 10Rr$
2/Prove that $ p^2\leq \frac {R(4R \plus{} r)^2}{2(2R \minus{} 1)}$[/quote]
1/Remember that:
$ p^2 \geq 2R^2\plus{}10Rr\minus{}r^2\minus{}2(R\minus{}2r)\sqrt{R^2\minus{}2Rr}$
so we will prove that:
$ 2R^2\plus{}10Rr\minus{}r^2\minus{}2(R\minus{}2r)\sqrt{R^2\minus{}2Rr} \geq 7r^2\plus{}10Rr$
<=>$ R^2\minus{}4r^2 \geq (R\minus{}2r)\sqrt{R^2\minus{}2Rr}$
which is obvious true. :lol:
the second problem ,I think it's not true,(try R<1/2)
p/s:thank you for your gift.

[quote="Kyoshiro"][quote="tuandokim"]
1/Remember that:
$ p^2 \geq 2R^2 \plus{} 10Rr \minus{} r^2 \minus{} 2(R \minus{} 2r)\sqrt {R^2 \minus{} 2Rr}$ [/quote]
Where do you get it? :huh:
#2:It was my typo. :blush:
$ p^2\leq \frac {R(4R \plus{} r)^2}{2(2R \minus{} r)}$[/quote]
2.
Remember that:
$ p^2 \leq 2R^2\plus{}10Rr\minus{}r^2\plus{}2(R\minus{}2r)\sqrt{r^2\minus{}2Rr}$
We will prove that:
$ 2R^2\plus{}10Rr\minus{}r^2\plus{}2(R\minus{}2r)\sqrt{r^2\minus{}2Rr}\leq \frac{R(4R\plus{}r)^2}{2(2R\minus{}r)}$
<=>$ (R\minus{}2r)(8r^2\minus{}12Rr\plus{}r^2) \geq 4(2R\minus{}r)(R\minus{}2r)\sqrt{R^2\minus{}2Rr}$
($ R\geq 2r$)
<=>$ (8r^2\minus{}12Rr\plus{}r^2)^2 \geq 16(2R\minus{}r)^2(R^2\minus{}2Rr)$
<=>$ 320R^3r\plus{}8r^3R\plus{}16R^2r^2\plus{}r^4\geq 0$(obvious true)
done.


Int-o interventie a dl. prof. Virgil Nicula am gasit urmatoarea problema a lui Sungyoon Kim:

Fie ABC un triunghi oarecare inscris in cercul (O,R) si circumscris cercului (I,r).
Cercul (Pa,ρa) este tangent interior in A la cercul (O,R) si tangent exterior cercului (I,r),
iar cercul (Qa,ρ'a) este tangent interior in A la cercul (O,R) si tangent interior la (I,r).
Cercurile (Pb, ρb), (Pc,ρc) si (Qb, ρ'b), (Qc,ρ'c) se definesc in mod asemanator. Sa se arate
ca are loc inegalitatea:   8.|PaQa|.|PbQb|.|PcQc|<=R^3.

Eu am gasit si demonstrat insa o urmatoarea proprietate a configuratiei care apare in
aceasta problema:    r^3<=|PaQa|.|PbQb|.|PcQc|. (Mihai Miculita)

Solutia acestei noi inegalitati o gasiti in attachment!

It is http://www.mathlinks.ro/Forum/viewtopic.php?t=145852 .  :D


calculus
function
real analysis
real analysis unsolved

Say that $ \alpha,\beta\in (0,1]$ and $ x$ and $ y$ are positive.

What is a quick way (i.e. no calculus) to determine the maximum value (w.r.t $ R$) of
\[ xR^\alpha \plus{} yR^{\minus{}\beta}
\]
?

Of course it is easy using calculus but it seems that everyone I know who works in harmonic analysis or PDE can optimize in $ R$ here at a mere glance.

That's because they do not care about constant factors, so when they need to optimize the sum of an increasing and a decreasing function, they just look at the point where the terms are equal. It doesn't give you the precise minimum but the value you get is not more than twice larger than the true minimum :).


How many positive integers not exceeding 2001 are multiples of 3 or
4 but not 5?

Multiples of 3:

$ \rfloor{\frac {2001}{3}}\lfloor \equal{} 667$

Multiples of 4:

$ \lfloor{\frac {2001}{4}}\rfloor \equal{} 500$

Multiples of 12:

$ \lfloor{\frac {2001}{12}}\rfloor \equal{} 166$

Multiples of 15:

$ \lfloor{\frac {2001}{15}}\rfloor \equal{} 133$

Multiples of 20:

$ \lfloor{\frac {2001}{20}}\rfloor \equal{} 100$

Multiples of 60:

$ \lfloor{\frac {2001}{60}}\rfloor \equal{} 33$

$ 667 \plus{} 500 \minus{} (166 \plus{} 133 \plus{} 100 \plus{} 33) \equal{} \boxed{735}$


vector inequalities trigonometry algebra polynomial

Find the minimum of $\frac{1}{x}+\frac{4}{y}+\frac{9}{z}$, knowing that $x+y+z=1$.  :ninja:

Ok, I saw someone do this problem once, but...
a) What is this method called?
b) How come this works?
c) Can you prove this so that this can be used under defined conditions if a) is defined?

Cauchy-Schwarz (or just Cauchy) states that

$(x_{1}^{2}+x_{2}^{2}+\dots+x_{n}^{2})(y_{1}^{2}+y_{2}^{2}+\dots+y_{n}^{2}) \geq (x_{1}y_{1}+\dots+x_{n}y_{n})^{2}$, for all $x_{i},y_{i}\in \mathbb{R}$.

If you square root it, it is just the vector dot product with the inequality $|\cos \theta| \leq 1$.

In this case, $n=3$ and $x_{1}=\sqrt{x}, x_{2}=\sqrt{y}, x_{3}=\sqrt{z}, y_{1}=\frac{1}{\sqrt{x}}, y_{2}=\frac{2}{\sqrt{y}}, y_{3}=\frac{3}{\sqrt{z}}$.

You can also prove it by calculating the determinant of the polynomial

$P(t) = \sum_{k=1}^{n}(x_{1}t-y_{1})^{2}$

Which satisfies $\Delta \le 0$ (for obvious reasons.)


Let $ a,b,c,x,y,z \neq 0$ and $ \begin {cases} \frac{a}{x}\plus{}\frac{b}{y}\plus{}\frac{c}{z}\equal{}0 \\ \frac{x}{a}\plus{}\frac{y}{b}\plus{}\frac{z}{c}\equal{}1 \end {cases}$.

Prove that: $ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\plus{}\frac{z^2}{c^2}\equal{}1$

[hide="Proof"]
Let $ p \equal{} \frac{x}{a}$, $ q \equal{} \frac{y}{b}$, and $ r \equal{} \frac{z}{c}$ for simplicity.  Then from the first equation, we have
$ \frac{1}{p} \plus{} \frac{1}{q}  \plus{} \frac{1}{r} \equal{} 0$, so $ pq\plus{}pr\plus{}qr \equal{} 0$.
From the second equation, we have $ p\plus{}q\plus{} r\equal{}0$.
Thus, $ \frac{x^2}{a^2}\plus{}\frac{y^2}{b^2}\plus{}\frac{z^2}{c^2} \equal{} p^2 \plus{} q^2 \plus{} r^2 \equal{} (p\plus{}q\plus{}r)^2\minus{}2(pq\plus{}pr\plus{}qr) \equal{} 1$, as desired. [/hide]

Thank! it's a nice way. But I think we can done it by other way!


1. Sa se demonstreze ca :
    $\begin{enumerate} {\bullet (1+\frac{x}{n})^n\geq 1+x,\forall x\in(-n,\infty),\forall n\geq 1}\\{\bullet e^x\geq x+1}\\{\bullet \lim_{n\to\infty} \frac{1+\sqrt [n] 2+\sqrt [n] 3+\ldots +\sqrt [n] n-n}{\ln n}=1} \end{enumerate}$.

2.Se considera matricea $A\in\M_3(\Re)$,$\left ( \begin{array} {ccc} a&1&1\\1&b&1\\1&1&c \end{array}\right )$.
a) Sa se calculeze: $\det(A)$ si $A^*$;
b)Daca $abc=1$ si $a,b,c\in (0,\infty)$. Sa se arate ca: $\det(A)\leq 0$,$\det(A^*)\geq 0$ si $\det(A^*)^*\geq 0$.
c)Pentru $a=b=c$ calculati $(A^*)^n$.

3.Fie $p\in\mathcal l N^*$. Sa se determine toate matricele $A\in\mathcal l M_n(C)$ cu urmatoarele proprietati:
$\begin{enumerate} \bullet tr(A)=n\\\bullet A^p=I_n \end{enumerate}$.

4.a)Studiati convergenta sirului $(x_n)_{n\geq 0}$ cu $x_0=0$,$x_1=1$ si $(n+2)x_{n+2}-(n+1)x_{n+1} -x_n=0$
             se poate utiliza  $\lim_{n\to\infty}(1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!})=e^x$.
b)Fie $(x_n)_{n\geq 0}$ un sir de numere reale strict pozitive care satisfac urmatoarea relatie de recurenta:
     $\sqrt{x_{n+2}}= \frac{n}{2}(x_{n+1}+\frac{1}{x_n}),n\geq 1$.
                Sa se arate ca : $\lim_{n\to\infty}x_n=+\infty$.

Imi cer scuze dar pe foaia de concurs nu erau trecuti autorii.Prima problema am vazut-o propusa pe forum de d-l Virgil Nicula.

[quote="spix"]Cel mai interesant lucru la concurs a fost faptul ca la problema a 3-a conditiile erau notate cu a) si b) iar in subsolul paginii la punctaje scria : "III.a)5p;b)4p" acest lucru a fost naucitor  :roll: .[/quote]
Ce să spun? Cred că e prima dată când folosesc acest emoticon  :rotfl:

Probabil ca 1p ramas era analizarea cazului $p=1$  :lol:

erau prea mandri de proprile proprietati pe care le-au enuntat   :lol:

reinviu un pic acest subiect ... ( oricum posteaza lumea p-aici o data la saptamana :D ) spixule sigur ai definit bine sirul de la 4b) ? ca la un moment da ( scuzati-ma dar nu stiu Latex , stiu doar cs :) ) , x2=0 ... iar apoi la x4 apare 1/x2 ... deci e cam aiurea . sa imi zica si mie cineva cum se face limita de la primul punct ca m-am blocat undeva .

In exemplul tau $x_2$ nu prea e strict pozitiv.

ah, am citit eu aiurea ... zice n>=1 . sorry ! oricum, cum se face asta ?

nu cred ca ca datele de la subp a) se aplica si la subp b)...

aaaaa, nu stii s-o faci spix? :P rusine . subpunctu a) poate fi facut ... dar b) are o forma f ciudata . sa imi spui si mie spix cum se face 1c) ca ma blochez intr-un punct, nu stiu de ce ... o fi oboseala de vina

1c pare sa mearga folosind estimari cu integrale. Probabil estimarile alea pot fi gasite si la nivelul clasei a 11a.

nup . in clasa a 11-a nu se fac integrale . integralele le facem intr-a 12-a. intr-a 11-a facem derivatele . si avand in vedere data la care a avut loc concursul, oricum nu cred ca se facea nici cu derivate nici cu integrale . sunt niste metode clasice care la ora actuala imi scapa :P

de faaapt  :idea:  la un moment dupa ce aplici stolz , iti apare un raport 0/0 care poate fi facut cu l'hopital ... dar banuiesc , ca in baremul de la acest concurs era alta metoda decat cu l'hopital

Stiu si eu cand se fac integralele, ca-s doar a 11-a. Oricum, lema Stolz-Cesaro, chiar daca e "elementara", mi se pare un pic fortata pentru problemele de genul asta. Nu cred ca loveste in "esenta" problemei. Parerea mea.... hâââccc.

pai si daca esti a 11-a de unde stii integralele ? sau esti unii din aceia care invata toata materia de liceu din cls a 8-a? :) credeam ca numai la chimie exista asa ceva ... vii la nationala ?

pai ce are Stolz, altfel nu vad cum ai putea sa scapi de sirul acela de radicaluri de ordin n ... propune o alta metoda atunci . si nu este deloc fortat, ba chiar sunt indeplinite conditiile de a-l aplica, si inca toate ;)

[quote="pureblack2003"]pai si daca esti a 11-a de unde stii integralele ?[/quote]
Am deschis o carte si m-am uitat si pe ML.

[quote="pureblack2003"]sau esti unii din aceia care invata toata materia de liceu din cls a 8-a? :)[/quote]
nu

[quote="pureblack2003"]credeam ca numai la chimie exista asa ceva ... vii la nationala ?[/quote]
nu (mama lor de solutii care seamana cu cele din barem si pe care iau 1p)

Despre Stolz-Cesaro: nu inghit teorema asta... nu stiu cum sa-ti zic.

:) am luat si eu 1p pe o demonstratie imensa la subiectul 2 . a corectat un prof care cik la un moment dat intreaba pe alt prof: ma , daca lambda este valoare proprie pt A, lambda la puterea k este valoare proprie pt A^k? am folosit o chestie de genu rang(A)=rang(AB) , unde B inversabila . si mi-a cam picat tot ... probabil ca trebuia sa o demonstrez in prealabil . acuma depinde ... si la matematica se poate corecta subiectiv . obiectivitatea 100% gasesti doar la info . ce punctaj ai avut ?

Stolz este o teorema f utila ... in limitele care se fac la clasa , la bac, la admiteri etc ... utilitatea ei poate fi observata si cand calculezi limita unui sir prin intermediul unui sir cu totul altul ... am mai vazut unele exercitii . dar dak nu-ti place fata lui , ce sa-i faci ? ;)

La informatica am luat 30p.  :D  Anul viitor nu mai incerc cu alocare dinamica !!!

S-a dovedit o decizie proasta. Asa am luat 10 p la prima pb. Algoritmul pe care l-am gasit valora 50p (parcurgeam graful in latime de n ori, eliminand cate un varf) - asa zicea ca valoreaza ideea asta in solutie, dar a dat erori mari cu memoria. La 2, long live the backtracking -> 20p. Acasa am gasit un back care-ti da 60p. (si cu inverse modulo p, unde p e numarul prim 19997)

Tu cat ai luat? La amandoua adica.

Poate ar fi mai nimerit sa continuati dialogul prin mesaje personale (PM). De fiecare data cand postati, se trimit notificari prin email. Nu stiu daca toata lumea e interesata de performantele de la olimpiada de informatica. Eu nu sunt.
In plus, asta e o regula de baza aici: no post off topic

puteati sa ne dati un mesaj personal pt asta :D spuneti va rog , aveti o solutie la 1c sau 4b? as fi f recunoscator .

Cum am mai precizat prima problema este propusa de d-l Virgil Nicula si este discutata [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=74807]aici[/url].
Iar la 4b) din cate imi amintesc... am demonstrat ca de la un rang incolo $\sqrt{x_{n+1}}>x_n$ iar apoi reducere la absurd.

fi mai explicit, geniule ... referitor la 4b ... nu prea observ cum merge ce ai propus tu ...
ai fost si la concursul asta ? ca te-am vazut si pe la alexandru cojocaru, sau cum ii zice, si ai iesit primu si acolo ... sau al doilea dupa colegul tau . oricum, m-am lamurit cu limita aia.... eu am aplicat Stolz intai, si apoi l'Hopital . dar intr-adevar, solutiile acelea sunt cele mai bune ;)

Fiecare din termenii $x_1$, $x_2$ se afla in $(0,1]$ sau in $[1,\infty)$.
Analizam cazurile posibile si deducem prin inductie ca, de la un rang $n_0$ in colo, avem $x_n>1$.
Mai departe, oricare ar fi $n\geq n_0$, avem:
$\sqrt{x_{n+2}}=\frac{n}{2}(x_{n+1}+\frac{1}x_{n})>\frac{n}{2}(1+\frac{1}x_{n})>\frac{n}{2}$
si concluzia rezulta imediat.


How many ways to write a natural number $ n > 0$ as sum of some natural number $ >0$?
Example: with $ 3$ we have $ 2$ ways: $ 3 \equal{} 1 \plus{} 2,3 \equal{} 1 \plus{} 1 \plus{} 1$

What you are referring to are known as partitions of $ n$ (although typically people allow $ n$ itself to be used in the sum, so that for example there are actually 3 partitions of 3: 3=1+1+1, 3=2+1, 3=3).

The exact formula is due to Rademacher and is VERY complicated, and can be seen as equations (24) and (25) at [url]http://mathworld.wolfram.com/PartitionFunctionP.html[/url].

OK! Thanks! It is very complicated! I don't think it is very difficult like this! :o


symmetry trigonometry calculus logarithms

I am not looking for answers.  I am not in high school.  I have not been in high school in over 20 years.  I am preparing for a BIG state math exam that I need to pass to earn my certification.  Blocking my topics is really stupid.  Here's a suggestion: If you dislike my questions or think they are a repeated topic or too easy or whatever, SKIM THEM!!!

Does anyone here know how to skip questions?  I am tired of all the unnecessary criticism.  There's no need to put me down.  If you think I am lying, then what can I do?  STOP BLOCKING MY QUESTIONS! Learn to skip, okay?  Come on, it's easy to skip questions.  I am sure that someone here will provide me with the steps or hints needed for me to solve questions on my own.  For those of you who help me all the time, thank you very much.

Symmetry

If you want to prepare for an exam, try these books:

(Algebra the Easy way)
http://www.amazon.com/exec/obidos/ASIN/0764119729/artofproblems-20

(Trigonometry the Easy way)
http://www.amazon.com/exec/obidos/ASIN/0764113607/artofproblems-20

For algebra

http://www.purplemath.com/modules/index.htm

This website has a lot of algebra stuff, from easy to advanced.

Or you can go to mathworld website to check. There are tons of things there. However, mathworld is more challenging.

And if you are preparing for a particular contest, you can do some problems from the previous years. It will help a lot on what kind of questions they are going to ask.

Good luck.

well...ill be happy to unlock them after you look on mathworld

pm when you have

if you need extra help, just post something on a general topic

I feel that it has become necessary to raise a few points with regards to your posting habits.

First of all, we are sympathetic to your desire to brush up on your math skills.  Some of us are of the age where we realize how valuable it is to be proficient in math - some of us are young enough that we just take it for granted.  In any event, I (and I assume everyone else who has helped you) wish you all the best on your proficiency exam and hope that our solutions have aided you.

Unfortunately, you have posted a constant stream of problems which are decidedly not of the 'problem-solving' variety.  This is fine, but the primary purpose of this forum is to solve difficult problems, whatever 'difficult' means (depending on the ability of the person asking the question).  Your problems are unsuitable for the forums they are posted in, and they serve to clutter and detract from the expressed purpose of these forums as a mechanism for young problem solvers to learn new techniques and solve difficult problems using the techniques that they already know.  Whether your problems are homework or not is immaterial - certainly they may not be homework in the strict sense of the word, but they are of the 'homework-type-problem' genre, which is not particularly welcome in these forums.

Now, the calculus computations and tutorials section is a bit of an exception, but even there, your problems are excessively easy, and the answers could be easily found in any basic calculus textbook.  You have been using these forums as a substitute for actual textbooks and actual tutors, and that is not particularly fair to anybody.  There are many forums which are willing to offer homework help, and even if you could find better or quicker solutions here, that is not a valid reason to misuse this resource.  Now, I am not saying that every question that appears in a textbook should not be posted here - certainly, there are many challenging problems to be found in textbooks, especially at the university level.  Indeed, some of your posts have been of the appropriate level as to be posted in these forums.  But the vast majority have not, and telling us to just ignore your posts if we are unwilling to help is not an acceptable request.  The moderators here put in a significant amount of work to make sure that the right problems  are in the right forums and the right subsections, and we'd like to minimize that amount of work.

amen

also, we are not tutors, we are just people who like math, it is not our job to do this, we just enjoy solving difficult problems, and many of the problems that you have posted are quite basic, and unsatisfying to solve

[quote="blahblahblah"]Unfortunately, you have posted a constant stream of problems which are decidedly not of the 'problem-solving' variety.  This is fine, but the primary purpose of this forum is to solve difficult problems, whatever 'difficult' means (depending on the ability of the person asking the question).  Your problems are unsuitable for the forums they are posted in, and they serve to clutter and detract from the expressed purpose of these forums as a mechanism for young problem solvers to learn new techniques and solve difficult problems using the techniques that they already know.  Whether your problems are homework or not is immaterial - certainly they may not be homework in the strict sense of the word, but they are of the 'homework-type-problem' genre, which is not particularly welcome in these forums.

[/quote]

There is a classroom math section, but not a classrom high school section. Could that be a problem? If someone posted a problem that needed knowledge of synthetic division in classroom math, where would the mod move it?

[quote="236factorial"]There is a classroom math section, but not a classrom high school section. Could that be a problem? If someone posted a problem that needed knowledge of synthetic division in classroom math, where would the mod move it?[/quote]
If a problem that needed knowledge of synthetic division, then it should be moved to Getting Started because:

1) The problem posted should not be a regular homework problem recalling to perform only synthetic division, and hence it should involve some exciting interest in the problem;
2) Synthetic division is a topic that appears often in high school algebra textbooks;
3) I remember administrators said that when you are unsure of where to move high school topics to which forums (GS, IT, Pre-O, Other, etc), GS should be the starting point to move the unsure topics, and then, if it is still too difficult for GS, the mods/admins in GS would 'escalate' those topics. This is a lesson that [b][u]all[/u][/b] mods should have known, haven't you guys learned this?  :D  ;)

That's right. This is because

a) AoPS is not an interactive calculator,

b) We are not tutors,

c) Look in any good algebra II level math textbook in the world and you should find it there. :)

[quote="10000th User"][quote="236factorial"]I guess my example wasnt' as good as I thought  :D

What if the problem asked for the value of $\log_5 25$? There isn't any place for that  :([/quote]

My algebra textbook gives harder homework problems than finding $\log_5 25$... this is homework problem
The issue is, how should you handle homework problems topics? I know we can't delete it, so there...
Resolution: contact admins. Mods should not work alone, admins are there to help you.[/quote]

There shouldn't be a place for "homework problem topics." But you can have a thread concerning "general topic." For instance, if person named A wants to know about logarithrm, he/she can post the thread in Getting Started forum (because middle school competition do not involve log) with title "logarithrm" and on the thread, maybe put something like "Hi, I need help on understanding logarithrm. Can you help me understanding what it is and give some example problems?"

That is what we want.

(as opposed to posting about 10 topics with almost identical logarithm questions in them.)

[quote="Silverfalcon"]But you can have a thread concerning "general topic." For instance, if person named A wants to know about logarithrm, he/she can post the thread in Getting Started forum (because middle school competition do not involve log) with title "logarithrm" and on the thread, maybe put something like "Hi, I need help on understanding logarithrm. Can you help me understanding what it is and give some example problems?"

That is what we want.[/quote]That is the solution to this problem. symmetry, you can post all of your questions similar to "how much is $\log_5{25}$" in [b]one thread[/b], on which you may [b]not[/b] post consecutively. People will help you as long as you understand that they are not obligated to do so in any way.

I hereby authorize the moderators to instantly delete any other topic of that kind (meaningless computational like problem) on site. So as long as you post easy/silly questions in one single place (again without bumping it out yourself) and real problems in different topics as I saw you are doing now, we're all ok :)

Topic is closed.


Find the $n$ th term of the sequence $\{a_{n}\}$ such that  $\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}=n+\frac{1}{n}\ \ (n=1,\ 2,\ 3,\ \cdots).$

Yup, I see it now.

[hide="It should have been"]$a_{1}=2, a_{n}=2n-1, n\geqslant 2$[/hide]

Now your answer is perfect, Farenhajt. :)

Same solution here...this one seems significantly easier than the previous one...


I've just thought up of a new game.

In this game, someone thinks up of a word and scrambles it.  The next person unscrambles it to the original word(s), and provides a new word in scrambled form.  SCORES:  The first person gets $ 1$ point, the second person gets $ 2$ points, so on.

If no one unscrambles the word within two days, the last person who posted wins the game.

Okay, so I start:

[b]ECETNOUJRC[/b]

Answer:
[b]CONJECTURE[/b]

New Word:
[b]IYPSSHHOET[/b]

Answer:

[b]HYPOTHESIS[/b]

Problem:

[b]EPMBLOR[/b]

[hide="Scores"]

professordad 4
K Sengupta 2

[/hide]

Answer:

[b]PROBLEM[/b]

New Word:

[b]TIOSUONL[/b]

Answer:

[b]SOLUTION[/b]

Problem:

[b]OCNSUCIONL[/b]

[hide="Scores"]
professordad 9
K Sengupta 6
[/hide]

Answer:

[b]CONCLUSION[/b]

Problem:

[b]REIDVONIAT[/b]

Answer:

[b]DERIVATION[/b]

Problem:

[b]SIGSIRONDE[/b]

[hide="Scores"]

professordad 16
K Sengupta 12

[/hide]

A: [b]DIGRESSION[/b]
P: [b]TEUEIEIRRSSPQ[/b]

Answer:

[b]PREREQUISITES[/b]

Question:

[b]AOPSNOTUSEA[/b]

[hide="Scores"]
professordad 25
K Sengupta 12
james4l 8
[/hide]

Thanks, nikeballa96!  (And preventing my win  :P )

Answer:

[b]INTESTINE[/b]

Problem:

[b]CRENRONTUC[/b]

[hide="Scores"]

professordad 36
K Sengupta 12
nikeballa96 10
james4l 8

[/hide]

Answer

[b]CONCURRENT[/b]

NP

[b]EZTRIALMAEI[/b]

Answer:
[b]Materialize[/b]

New Problem:
[b]SUIISOSNDC[/b]

[hide="Scores"]

professordad 49
inoit 12
K Sengupta 12
nikeballa96 10
james4l 8

[/hide]

Answer

[b]DISCUSSION[/b]

NP

[b]DEDVCANA[/b]

Answer

[b]XANTHOSIS[/b]

NP

[b]UVRTNEEIRN
[/b]

[hide="scores"]inoit 368
professordad 310
james4l 278
yaofan 109
nikeballa96 55
kelvin 34
K Sengupta 12[/hide]

Answer

[b]TURNVEREIN[/b]

NP
[b]SPIEDPEAS[/b]

[hide="scores"]inoit 368
james4l 328
professordad 310
yaofan 109
nikeballa96 55
kelvin 34
K Sengupta 12[/hide]

Answer (I love this word)

[b]PASSEPIED[/b]

NP

[b]ZTKIIBRE[/b]

Answer

[b]KIBITZER[/b] -  :)

NP
[b]RMNTACEOREPHP[/b]

[hide="scores"]inoit 419
james4l 380
professordad 310
yaofan 109
nikeballa96 55
kelvin 34
K Sengupta 12[/hide]

Answer

[b]RAPPROCHEMENT[/b]

NP

[b]PALMETTES[/b] <-- This is a anagram

[hide="scores"]inoit 471
james4l 380
professordad 310
yaofan 109
nikeballa96 55
kelvin 34
K Sengupta 12[/hide]

Congratulations to [b]inoit[/b] for winning Round 1!

[hide="FINAL SCORES (according to SCORING RULES CHANGE):"]

inoit                 571    1st
james4l             380    2nd
professordad      310    3rd
yaofan              109    Honorable Mention
nikeballa96         55
kelvin                34
K Sengupta        12

[/hide]
[hide="SCORING RULES CHANGE"]

If no one posts a word within two days, the person who last posted gets 100 points. However, he or she [b]may not[/b] win if the first place person is 100 points or higher than the person who posted last.

[/hide]
[hide="RANKINGS:"]

There shall be a 1st, 2nd, and 3rd place and an Honorable Mention for the people who didn't get 1st, 2nd, or 3rd but still got above 100 points.  If you don't like this system, notify me.  Private Message or posting on this forum both work.

[/hide]

So still, we need a new word!

[b]DNTEEMDIRE[/b]

Answer:

[b]DETERMINED[/b]

New Problem:

[b]KAMENCGNOTELWDE[/b]

Answer
[b]ACKNOWLEDGEMENT[/b]

NP

[b]NOTALION[/b]

btw 4 the scores, do we start all over again with zero or do we continue on with the round one scores?

Answer:

[b]NOTIONAL[/b]

New Problem:

[b]EIUIPSAEDNQALS[/b]

For the scores, they start over at 1

[hide="Scores"]

professordad 3
inoit 2
yaofan 1

[/hide]

Answer

[b]SESQUIPEDALIAN[/b]

NP

[b]GCURNCOENY[/b]

[hide="scores"]professordad 3
inoit 5
yaofan 1 [/hide]

Answer:
[b]congruency[/b]
Newproblem
[b]dlruaom[/b]

Answer
[b]
MODULAR[/b]

NP

[b]NGSTUIREA[/b]

[hide="scores"]professordad 3
inoit 10
yaofan 1
SuperNerd123  4[/hide]

staphylococci

NP is....

[b]ENSARPUCE[/b]


[hide]I got the 5th year which is 1993[/hide]

[quote="cindy"]The number of y students attending Nequa Valley High School between 1988 and 1996 can be approximated by the model y= 45.28x 2 -37.6x+585, where x=0 corresponds to 1988.  Based on this model, in what year did the school have about 1550 students?[/quote]
[hide]
1550=45.8x^2-37.6x+585
45.8x^2-37.6x-965=0
Using the quadratic formula, you get
     -b(+/-)sqrt{b^2-4ac)
x=--------------------------------
      2a

= ~5, -4.33
-4.33 doesn't work, because the equation only corresponds to values between 1988 and 1996
Therefore the nearest year when the school has 1550 students is in 1993, with 1529[/hide]

The problems that cindy has been posting look suspicously like homework.  You shouldn't really respond with full answers if this is the case, only hints.


[b][A.]   Prove that the equation $\begin{array}{|c|}\hline \\ x^4+y^5=z^7\\ \\ \hline \end{array}$ has infinitely many solutions in  ${\mathbb N}^*: =\{1,2,...\}$ given by
\[ \left\{\begin{array}{lcr} x&=&x(u,v,n)=u\left(u^4+y^5\right)^{35n+5}\\ y&=&y(u,v,n)=v\left(u^4+y^5\right)^{28n+4}\\ z&=&z(u,v,n)=\left(u^4+y^5\right)^{20n+3} \end{array}\right. , \forall (u,v,n)\in {\mathbb N}^*\times {\mathbb N}^* \times {\mathbb N}.  \]
[B.]   The same question for equation   $\begin{array}{|c|}\hline \\ x^2+y^2=z^3\\ \\ \hline \end{array}$ having the solutions
\[ \left\{\begin{array}{lcr} x&=&x(u,v,n)=u\left(u^2+v^2\right)^{6n+4}\\ y&=&y(u,v,n)=v\left(u^2+y^2\right)^{6n+4 }\\ z&=&z(u,v,n)=\left(u^2+v^2\right)^{4n+3} \end{array}\right. , \forall (u,v,n)\in {\mathbb N}^*\times {\mathbb N}^* \times {\mathbb N}.  \]

[size=67][color=blue]Note: Question [B.] was raised by W.Sierpinski in a more general frame,i.e. $x^p+y^p=z^{p+1} , \; p \in {\mathbb N}^* .$ See Aufgabe  103 in Elemente der Mathematik, also "Problem of the day" , this site,see
http://www.mathlinks.ro/Forum/resources.php?potd=1&week=-1
[/color][/size]

[/b]

Well... As given here, it's just calculating...
I think it would be more interesting without already giving some infinite set.

Well, the $x^p+y^p = z^{p+1}$ remains, but that was posted before (B was that, too).

[quote="ZetaX"]Well...    I think it would be more interesting without already giving some infinite set. [/quote]
In my opinion, of interest is to justify that these parametric solutions are [color=green] [b][u]all[/u]  solutions[/b] [/color]. It's clear that  it's easy to verify that the parametric forms satisfy the equations.


I'm currently doing some number theory problems, and I have some questions on some of them.
One problem goes as follows:
[b]A positive integer leaves a remainder of 2 when divided by 9, a remainder of 5 when divided by 11, and a remainder of 8 when divided by 12. What is the remainder when the number is divided by 66?[/b]

I'm stuck on this problem. I tried setting up 3 linear congruences but those efforts were fruitless. I would appreciate any help on how to approach this problem and others of its type.

A second problem goes as follows:
[b]Given that $ 88^4\plus{}82^4\plus{}63^4\plus{}28^4\plus{}6^4\equal{}k^4$ with k an integer, find k.[/b]
I understand how to approach this problem. First I considered this equality in modulo 5. From Fermat's Little Theorem, I was able to discover that the LHS would be congruent to 0 mod 5. This came from applying Fermat to each individual term. That would imply that k is congruent to 0 mod 5 as well. However, doesn't this oppose Fermat's Little Theorem, because that would imply k^4 is 0 mod 5 also, when k^4 should be 1 mod 5. I solved the problem by these methods and than checked it on my TI-89. Both ways I got that k and k^4 are 0 mod 5. This is a somewhat strange result?

The first problem is a straightforward application of the [url=http://mathworld.wolfram.com/ChineseRemainderTheorem.html]Chinese Remainder Theorem[/url] once you find the remainder upon division by $ 2, 3$.

Fermat's Little theorem says that $ k^4 \equiv 1 \bmod 5$ when $ 5$ [b]does not divide[/b] $ k$.

The second problem just requires some trial and error.

First, look at the units digit.  Second, you should already have a general idea what n is.  (90-120 ish)

[quote="mihail911"]I'm currently doing some number theory problems, and I have some questions on some of them.
One problem goes as follows:
[b]A positive integer leaves a remainder of 2 when divided by 9, a remainder of 5 when divided by 11, and a remainder of 8 when divided by 12. What is the remainder when the number is divided by 66?[/b][/quote]

I seem to have lost the chinese remainder theorem post I made a long time ago that I used to link everyone to, so I'll have to write a new solution  :roll:

[hide]Let the number be $ N$.  We know from the problem that:

$ N\equiv2\pmod9$
$ N\equiv5\pmod{11}$
$ N\equiv8\pmod{12}$

We can also write the first two equations as:
$ N\equal{}2\plus{}9k\equal{}5\plus{}11j$
$ 9k\equiv3\pmod{11}$
$ 9k\equiv36\pmod{11}$
$ k\equiv4\pmod{11}$
Let $ k\equal{}4\plus{}11d$.  Then, we can say that:
$ N\equal{}2\plus{}9(4\plus{}11d)$
$ N\equal{}38\plus{}99d$

Equating this to the 3rd equation, we have:

$ N\equal{}38\plus{}99d\equal{}8\plus{}12e$
$ 33d\plus{}10\equal{}4e$
$ 4e\equiv10\pmod{33}$
$ 4e\equiv76\pmod{33}$
$ e\equiv19\pmod{33}$
Let $ e\equal{}19\plus{}33x$.  Thus, we know that:

$ N\equal{}8\plus{}12(19\plus{}33x)$
$ N\equiv236\pmod{396}$
Because $ 396\equal{}66\cdot6$, we also know that
$ N\equiv236\pmod{66}$
$ N\equiv38\pmod{66}$[/hide]


Can a countably infinite set have an uncountable collection of non-empty subsets such that the intersection of any two of them is finite?

This has been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156787]recently.[/url]

But then mdk doesn't read topics started by other people, does he? He's too busy posting new topics by the dozen.


Sorry for the revival, but I was wondering if you could infer that CF=CD and solve it that way? Is any chord that is intersected by a diameter cut in half? Also, how do you know for sure that the height remain constant when you transpose E to F? Can you just tell from the picture? Thnx.

[quote=JesusFreak197]This is 2005 AMC-10A? I don't remember this...

[hide="Answer"]Both triangles have a diameter as their base, so we must simply find the ratio of the heights. We draw the height $CF$ of $\triangle DCE$, then use similar triangles with $CFO$ and $CDO$, $O$ being the center of the circle. $\frac{CF}{CD}=\frac{OC}{OD}$, $OD=r$ and $BC=2\cdot AC$, so $AC=2\cdot OC\Rightarrow OC=\frac{r}{3}$, and $\frac{OC}{OD}=\frac{\frac{r}{3}}{r}=\frac{1}{3}\Rightarrow \boxed{C}$.[/hide][/quote]
Can anyone explain how to get that $CFO$ and $CDO$ are similar?
Additionally, can anyone explain the logic from $BC=2\cdot AC$, so $AC=2\cdot OC$?

can you use that angle [b]ADB[/b] is right angle to solve

Sure, but you'll need more than just that.

Assume AC=1, then BC=2. With AC/DC=DC/BC we have 1/x=x/2. So DC=sqrt(2). Then DF=2sqrt(2) and EF^2=3*3-(2sqrt(2))*(2sqrt(2))=1. So EF=1. And EF is vertical to DF. so ABD area = 3*sqrt(2)/2 and CDE=DC*EF/2=sqrt(2)/2. So the area is 1/3.

[quote=Silverfalcon]Let $ \overline{AB}$ be a diameter of a circle and $ C$ be a point on $ \overline{AB}$ with $ 2 \cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ \overline{DC} \perp \overline{AB}$ and $ \overline{DE}$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area [quote=Silverfalcon]Let $ \overline{AB}$ be a diameter of a circle and $ C$ be a point on $ \overline{AB}$ with $ 2 \cdot AC \equal{} BC$. Let $ D$ and $ E$ be points on the circle such that $ \overline{DC} \perp \overline{AB}$ and $ \overline{DE}$ is a second diameter. What is the ratio of the area of $ \triangle DCE$ to the area of $ \triangle ABD$?
[asy]unitsize(2.5cm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;

pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);

draw(A--B--D--cycle);
draw(D--E--C);
draw(unitcircle,white);
drawline(D,C);
dot(O);

clip(unitcircle);
draw(unitcircle);
label("$E$",E,SSE);
label("$B$",B,E);
label("$A$",A,W);
label("$D$",D,NNW);
label("$C$",C,SW);

draw(rightanglemark(D,C,B,2));[/asy]$ \textbf{(A)} \ \frac {1}{6} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {2}{3}$[/quote]

defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;

pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);

draw(A--B--D--cycle);
draw(D--E--C);
draw(unitcircle,white);
drawline(D,C);
dot(O);

clip(unitcircle);
draw(unitcircle);
label("$E$",E,SSE);
label("$B$",B,E);
label("$A$",A,W);
label("$D$",D,NNW);
label("$C$",C,SW);

draw(rightanglemark(D,C,B,2));[/asy]$ \textbf{(A)} \ \frac {1}{6} \qquad \textbf{(B)} \ \frac {1}{4} \qquad \textbf{(C)}\ \frac {1}{3} \qquad \textbf{(D)}\ \frac {1}{2} \qquad \textbf{(E)}\ \frac {2}{3}$[/quote]
Let the centre be denoted by O and the point where CD intersects the circle be denoted by X.
We drop a perpendicular from EY to BC.Area of COE=rCD/6 where r is the radius of the circle.Area of OCD=rCD/6. Therefore area of triangle CDE=rCD/6*2=rCD/3. Area of ADB=1/2*2r*CD=rCD
ADB/DCE=rCD/3rCD=1/3. Done.

Let $O$ be midpoint of $AB$, $D'$ be second intersection of $CD$ with circle which has diameter $AB$
We have: $\dfrac{S_{DCE}}{S_{ABD}}$ = $\dfrac{ED' . CD}{AB . CD}$ = $\dfrac{2 CO}{AB}$ = $\dfrac{2 OA - 2 AC}{AB}$ = $\dfrac{AB - BC}{AB}$ = $\dfrac{AC}{AB}$ = $\dfrac{1}{3}$

Let the center of circle be M .
Area of DCE= 1/2*b*h=1/2*2x
diameter of 2y, 2y = 3 =>y=3/2 , CM =2-y=1/2=h
=>area of DCE = 1/2x
Area of ABD =1/2*b*h=1/2*3*x=3/2x where x=CD
Area [DCE/ABD]=1/2x/3/2x=1/3.
Neat Solution :D