probability integration limit quadratics function topology vector

IMC 2010 comes back to Blagoevgrad! :wink:

http://www.imc-math.org.uk/

[b]17th IMC 2010
Blagoevgrad, Bulgaria
24th - 30th July 2010[/b]

[quote]This year's competition is co-organized by [url=http://www.ucl.ac.uk/]University College London[/url] and [url=http://www.aubg.bg/]American University in Bulgaria, Blagoevgrad[/url]. It will take place in Blagoevgrad, Bulgaria, from the 24th to the 30th of July, 2010.[/quote]

[b]Schedule[/b]
First Day, Saturday, 24th July - Arriving and Accommodation, Registration
Second Day, Sunday, 25th July - Opening Ceremony, Additional Registration, Meeting of the Jury
Third Day, Monday, 26th July - First Exam Day
Fourth Day, Tuesday, 27th July - Second Exam Day
Fifth Day, Wednesday, 28th July - Meeting of the Jury
Sixth Day, Thursday, 29th July - Closing Ceremony, Final Dinner
Seventh Day, Friday, 30th July - Departure

So it also comes back to seven-day IMC (as in 2008, in contrary to the six-day one in 2009).

In English:
IMC 2010 group in Facebook: http://www.facebook.com/group.php?gid=122393464444895
Event: http://www.facebook.com/event.php?eid=118098911546611

In Russian:
IMC 2010 event in vKontakte: http://vkontakte.ru/event18749503

Students from Kyiv Taras Shevchenko National University:
Ivan FESHCHENKO, Faculty of Mechanics and Mathematics, 4th year student (1st prize at IMC 2009, 1st prize at IMC 2008, gold medal at SEEMOUS 2008, 1st prize at IMC 2007)
Danylo RADCHENKO, Faculty of Mechanics and Mathematics, 3rd year student (1st prize at IMC 2009, 1st prize at IMC 2008, gold medal at SEEMOUS 2008, gold medal at IMO 2007, bronze medal at IMO 2006)
Ivan IURCHENKO, Faculty of Mechanics and Mathematics, 4th year student (1 st prize at IMC 2009, 1 st prize at IMC 2008, gold medal at SEEMOUS 2008, 1st prize at IMC 2007)
Iuliia SEMIKINA, Faculty of Mechanics and Mathematics, 2nd year student (bronze medal at IMO 2008)
Vitalii SENIN, Faculty of Mechanics and Mathematics, 1st year student (gold medal at IMO 2009, bronze medal at IMO 2008)
Iurii SHYSHATSKYI, Faculty of Mechanics and Mathematics, 2nd year student (1 st prize at IMC 2009, gold medal at IMO 2008, bronze medal at IMO 2007)
Oleksandr SHAMOV, Faculty of Mechanics and Mathematics, 3rd year student (1st prize at Kolmogorov Probability Theory Olympiads in Moscow State University in 2010, 2009 and 2008)
Mykyta MAKAROV, Faculty of Mechanics and Mathematics, 1st year student
Andrii KOROTKOV, Faculty of Cybernetics, 3rd year student
(Leaders: Andrii BONDARENKO, Dmytro MITIN)

Other Ukrainian participants:
Vira SEMENOVA, 2nd year student, Kyiv-Mohyla Academy and Kyiv Taras Shevchenko National University
Pylyp Fil, 3rd year student, Kharkov Polytechnic Institute

[quote="Moubinool"]Do not forget  to post the problems of IMC 2010

Thank you

m.o.[/quote]

I will try tomorrow. Today it was meeting of Jury with selecting the problems. Tomorrow it will be the 1st day of competition.

I have the problems now (1st day). I will post them as soon as the exam finishes.

Problems of the 1st day:

http://www.mediafire.com/?73dz9o29lgz10j8

Please could somebody TeXify them and post as separate threads for each problem?
Thanks.

I downloaded the file and TeXified it =)

[b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=358763]Problem 1.[/url][/b] Let $0<a<b.$ Prove that
\[\int_a^b(x^2+1)e^{-x^2}dx\geq e^{-a^2}-e^{-b^2}.\]

[b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=358764]Problem 2.[/url][/b]Compute the sum of the series
\[\sum_{k=0}^{\infty} \frac1{(4k+1)(4k+2)(4k+3)(4k+4)}=\frac{1}{1\cdot 2 \cdot 3 \cdot 4} +\frac{1}{5\cdot 6 \cdot 7 \cdot 8} +\dots .\]

[b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=358765]Problem 3.[/url][/b] Define the sequence $x_1,x_2,\dots$ inductively by $x_1=\sqrt{5}$ and $x_{n+1}=x^2_n-2$ for each $n\geq 1.$ Compute
\[\lim_{n\to \infty} \frac{x_1\cdot x_2\cdots x_3\dots x_n}{x_{n+1}}.\]

[b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=358768]Problem 4.[/url] [/b]Let $a,b$ be two integers and suppose that $n$ is a positive integer for which the set
\[\mathbb{Z}\setminus \{ ax^n+by^n\, | \, x,y \in \mathbb{Z} \}\] is finite. Prove that $n=1.$

[b][url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=358770]Problem 5.[/url] [/b]Suppose that $a,b,c$ are real numbers in the interval $[-1,1]$ such that
\[1+2abc\geq a^2+b^2+c^2.\]
Prove that \[1+2(abc)^n\geq a^{2n}+b^{2n}+c^{2n}\] for all positive integers $n.$

Duration is probably 5 hours and each problem is given 10 points.

Not any linear algebra?

This is not level university  math problems

problem 1: It is 12th grade level high school

problem 2: Look like first year university level.

problem 3: This one was already discussed and solved on the forum
It is 11th grade level high school.

problem 4 and 5: Looks like more olympiad high school problem.

 I don't understand the choice of these problems for day 1.

For a university  International Math Competition ( first until four years )
it would be natural to choose problems among the following :

group , ring , field theory

linear algebra, matrices (diagonalisation , trigonalisable, others stuff with Jordan , Dunford decomposition ...)

Bilinear algebra ( quadratic form , isotropy ...)

Linear , non linear differential equation , properties of Cauchy problem

Fourier series

Power series

function with several variables

Sequence and series functions ( simply uniform convergence ...)

Topology , vector normed space (continuity of linear map in normed vector space,  $|||f|||$ ...)

Metric space , completion

Some problems using Baire theorem

Integral with parameter ( derivative , differential equation ...)

Double , triple , multiple  integral

Hilbert space

Euclidean space

Hermitian space (hermitian matrices ...)

Analytique functions

complex functions

functional analysis

Lebesgue integral , measure,

etc ....

[quote="Moubinool"]I could not open your file dmitin.

Could post the problems  in PDF ?

thank you[/quote]

It's just a JPEG.

[b]rustam[/b]
Thank you very much.

[quote="rustam"]
Duration is probably 5 hours and each problem is given 10 points.
[/quote]

Yes, surely.

[quote="rustam"]By the way, on which site the partial results would be available tomorrow? ))
Last year it was http://imc2009.elte.hu/[/quote]

I would be very surprised ;)

I will try to post them.

[quote="Moubinool"]For a university  International Math Competition ( first until four years )
it would be natural to choose problems among the following :
[/quote]

Not everything you mentioned would be a good choice for first year students ;)

Thanks for the problems. Actually I didn't expect this kind of problems for the IMC  :trampoline: . May be there are different problems for tomorrow.

I agree with most of you. It does not look like an University Level Olympiad.

Problems 1 and 2 are very easy, problem 3 was discussed a lot in the forum. In my opinion 5 is easy as well(at least for the Nr5 of IMC). This is just what I have done   :wink: :

Solution:[hide]Obviously at least one of the numbers $ab$, $bc$ and $ca$ is nonnegative, wlog say $ab$. Since $c\in [-1,1]$ by Cauchy-Schwarz

$L=\left(\sum_{k=0}^{n-1}(ab)^{n-1-k}c^{k}\right)^2\leq\left(\sum_{k=0}^{n-1}(ab)^k\right)^2\leq\left(\sum_{k=0}^{n-1}a^{2k}\right)\left(\sum_{k=0}^{n-1}b^{2k}\right)=R$. Note that $(ab-c)^2\leq (1-a^2)(1-b^2)$ .

 Therefore  $((ab)^n-c^n)^2=(ab-c)^2 L\leq(1-a^2)(1-b^2)R=(1-a^{2n})(1-b^{2n})$
i.e. $1+2(abc)^n\geq a^{2n}+b^{2n}+c^{2n}$.  Equality holds if only if $a=b, c=1$. (or for their permuations)[/hide]

I agree with the above opinions. I've finished high school this year and the problems were, aside from the fact that they didn't require much university-knowledge, quite easy... at least the first three problems. But let's see which problems are proposed for the second day :)

Preliminary results (before coordination) of the day 1:

http://www.mediafire.com/?cs76g486137m4oh

(JPEG.RAR)

I will post the problems of the day 2 as soon as the exam finishes.

Problems of the 2nd day:

http://www.mediafire.com/?ncd38b806d7vndz

(JPEG)

Could somebody kindly TeXify them and post as separate threads for each problem?
Thanks.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=358973][b]Problem 1.[/b][/url] $(a)$ A sequence $x_1,x_2,\dots$ of real numbers satisfies
\[x_{n+1}=x_n \cos x_n \textrm{ for all } n\geq 1.\]
Does it follows that this sequence converges for all initial values $x_1?$  (5 points)

$(b)$ A sequence $y_1,y_2,\dots$ of real numbers satisfies
\[y_{n+1}=y_n \sin y_n \textrm{ for all } n\geq 1.\]
Does it follows that this sequence converges for all initial values $y_1?$ (5 points)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=358980][b]Problem 2.[/b][/url] Let $a_0,a_1,\dots,a_n$ be positive real numbers such that $a_{k+1}-a_k \geq 1$ for all $k=0,1,\dots,n-1.$ Prove that
\[1+\frac{1}{a_0} \left( 1+\frac1{a_1-a_0}\right)\cdots\left(1+\frac1{a_n-a_0}\right)\leq \left(1+\frac1{a_0}\right) \left(1+\frac1{a_1}\right)\cdots \left(1+\frac1{a_n}\right).\] (10 points)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=61&t=358975][b]Problem 3.[/b][/url] Denote by $S_n$ the group of permutations of the sequence $(1,2,\dots,n).$ Suppose that $G$ is a subgroup of $S_n,$ such that for every $\pi\in G\setminus\{e\}$ there exists a unique $k\in \{1,2,\dots,n\}$ for which $\pi(k)=k.$ (Here $e$ is the unit element of the group $S_n.$) Show that this $k$ is the same for all $\pi \in G\setminus \{e\}.$ (10 points)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=349&t=358977][b]Problem 4.[/b][/url] Let $A$ be a symmetric $m\times m$ matrix over the two-element field all of whose diagonal entries are zero. Prove that for every positive integer $n$ each column of the matrix $A^n$ has a zero entry.(10 points)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=358979][b]Problem 5. [/b][/url]Suppose that for a function $f: \mathbb{R}\to \mathbb{R}$ and real numbers $a<b$ one has $f(x)=0$ for all $x\in (a,b).$ Prove that $f(x)=0$ for all $x\in \mathbb{R}$ if
\[\sum^{p-1}_{k=0}f\left(y+\frac{k}{p}\right)=0\]
for every prime number $p$ and every real number $y.$ (10 points)

Duration: 5 hours

Question 1
[hide](a) False, Take $x_1=\pi $[/hide]

Preliminary results (before coordination) of the days 1 and 2:

http://www.mediafire.com/?uiaat94ahdx45pv

(JPEG.RAR)

Preliminary results of Israel:
  6. Dan Carmon                      69 points
20. Alexey Gladkich                 56 points
29. Inna Entova-Aizenbud      53 points
71. Itamar Hason                    44 points
79. Gal Dor                              42 points
110. Ohad Livne Bar On          37 points
256. Shahar  Papini                 21 points
Sum   (3+...)                             214 points

See also [url]http://www.mathproblems123.wordpress.com[/url] for problems and solutions of the IMC 2010 Blagoevgrad.

Final results will be after closing ceremony.

Results

http://www.mediafire.com/?iuc863x5yvxp2un

This is unofficial results (but they are supposed to be almost correct). The official results will be published at http://imc-math.org.uk in few days.

Can anybody give another link for results, please. This one doesn't work on my computer for some reason :(

Congratulations to Ukraine!
Israel - the fourth place!!! :lol:

Does anybody know what books were given to the first prizers? I know that Proofs From the Book was one of them. I'm interested in the other two I think. Thank you.

Hi, not all of the frist prizes get a book.
Anyway the other book is "Dude, Can You Count? Stories, Challenges and Adventures in Mathematics".

i heard all contestants were given a MATHEMATICA software license this year, does anybody know for how long the license given?

i didn't get it when i participated three years ago :(

I think it started from the last year IMC. I got the licence for Mathematics for Students 7. I can't remember where I read it, but I think it is valid as long as you are a student. Don't know how they will figure it out though ;)

Problem 2 may be very famous problem.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1980228#p1980228[/url]

[quote]Dear IMC students -- just wanted to let you know that Wolfram Research hopes to release all IMC Mathematica orders soon. We are awaiting the official list of participants but as soon as we receive it, all Mathematica for Student licenses will be sent to those of you who have watched the screencast and completed the webform. Please feel free to contact me with any questions.

[url=http://www.facebook.com/carol.cronin]Carol Mabus Cronin[/url]
Wolfram Research, Inc.
Public Relations
Champaign, Illinois
[/quote]

http://www.facebook.com/group.php?gid=122393464444895

Sorry to revive this thread, but I'm surprised no one mentioned that Problem 5 of Day 1 is trivial by the Schur Product Theorem.


Expand the function $ f(\alpha ,z)\equal{}e^{\frac{1}{4}\left( 2\alpha z\minus{}z^2\right) }$ in a power series in $ z$.

$ f(\alpha ,z) \equal{} e^{\frac {1}{4}\left( 2\alpha z \minus{} z^2\right) }$
               $ \equal{} e^{\left( \alpha \frac {z}{2} \minus{} (\frac {z}{2})^{2}\right) }$
               $ \equal{} e^{\left( \alpha u \minus{} u^{2}\right) }$
               $ \equal{} e^{\alpha u} *e^{\minus{} u^{2}}$, where $ u \equal{} \frac {z}{2}$
Note: $ e^{u} \equal{} \sum_{n \equal{} 0}^{\infty}\frac {u^{n}}{n!}$

\[ f(\alpha ,z) \equal{} e^{\frac {1}{4}\left( 2\alpha z \minus{} z^2\right) } \equal{} \sum_{n\equal{}0}^{\infty}\frac{z^n\left( 2\alpha \minus{} z\right)^n }{4^{n}n!}\]

\[ \equal{} \sum_{n\equal{}0}^{\infty}\frac{z^n}{4^{n}n!} \sum_{k\equal{}0}^{n}(\minus{}1)^{k} \binom{n}{k}(2\alpha)^{n\minus{}k}z^{k} \equal{} \sum_{n\equal{}0}^{\infty}\frac{(2\alpha z)^n}{4^{n}n!} \sum_{k\equal{}0}^{n}(\minus{}1)^{k} \binom{n}{k}\left(\frac{z}{2\alpha}\right)^{k} \]

I got this far, but I need to be able to read off coefficients of $ z^n$ as the function is a generating function in my work.

Follow akech's suggestion: making the substitution $ u \equal{} z/2$, we want to calculate the coefficient of $ u^n$ in $ e^{\alpha u \minus{} u^2} \equal{} e^{\alpha u}\cdot e^{\minus{}u^2}$.  $ e^{\alpha u} \equal{} \sum_{n \geq 0} \frac{\alpha^n u^n}{n!}$.  $ e^{\minus{}u^2} \equal{} \sum_{n \geq 0} \frac{(\minus{}1)^n u^{2n}}{n!}$.  Then the product is $ \sum_{n \geq 0} u^n \sum_{\substack{2i \plus{} j \equal{} n \\ i, j \geq 0}} \frac{(\minus{}1)^i\alpha^j}{i!j!}$.  This is reminiscent of the exponential generating function for involutions, and it may be that this can be expressed in some nicer form (e.g. as a recursion).


function
trigonometry
Euler
absolute value
real analysis
real analysis unsolved

how to prove that :
for even $ n\geq 2$ we have :
$ \zeta(n) \equal{} \frac {|B_{n}|2^{n \minus{} 1}\pi^{n}}{n!}$
and also how to find  taylor expansion in  point $ 0$ of function
$ \frac {1}{\sin x}$?

For the second problem.

try to find the Taylor Series of $ \frac{x}{\sin(x)}$

So the limit of $ f \to 0$ will exist and be non-zero. After that you can so divide by x.

To determine the sequence of exact coefficient is rather difficult.

[quote="EulerIntegral"]For the second problem.

try to find the Taylor Series of $ \frac {x}{\sin(x)}$

So the limit of $ f \to 0$ will exist and be non-zero. After that you can so divide by x.

To determine the sequence of exact coefficient is rather difficult.[/quote]
I know all of them but I need a proof Why coeficient in Taylor exprassion can be expressd by Bernuli number? or How can we do it?
What is most simple way to find all coeficient?

[quote="Carcul"][quote="Extremal"]What is most simple way to find all coeficient?[/quote]

Divide 1 by the expansion of sin(x).[/quote]
Yes there is something like this method i can suppose how to do this. What else? What's about complex analisys?

[quote="Carcul"][quote="Extremal"]What is most simple way to find all coeficient?[/quote]

Divide 1 by the expansion of sin(x).[/quote]
Can you show?

[quote="Extremal"]how to prove that :
for even $ n\geq 2$ we have :
$ \zeta(n) = \frac {|B_{n}|2^{n - 1}\pi^{n}}{n!}$
[/quote]

Hey Extremal,  remember that problem in the CA forum:

$ \frac {1}{e^z - 1} = - 1/2 + 1/z + \sum_{k = 1}^{\infty}\frac {( - 1)^{k + 1}\zeta(2k)}{2^{2k - 1}\pi^{2k}}z^{2k - 1}$

Now, the Bernoulli numbers is by definition (one of them) is:

$ B_n = \frac {n!}{2\pi i}\oint \frac {dz}{z^n(e^z - 1)}$

Alright then, how about multiplying both sides of the first expression by $ \frac {n!}{2\pi i z^n}$ and then integrating both sides over the unit disc?   I get:

$ B_n = n!\mathop\text{Res}\limits_{z = 0}\left\{ - 1/(2z^n) + 1/(z^{n + 1}) + \sum_{k = 1}^{\infty}\frac {( - 1)^{k + 1}}{2^{2k - 1}} \frac {\zeta(2k)}{\pi^{2k}}\frac {z^{2k - 1}}{z^n}\right\}$

which for even $ n\geq 2$ reduces to your expression (with the $ (-1)^{\frac{n}{2}+1}$ taking the place of the absolute value).  This is so not regular calculus.  Can anyone explain how Euler did it?

[edit] corrected formula to connect absolute value and power of -1

[quote="shawtoos"]
Now, the Bernoulli numbers is by definition (one of them) is:

$ B_n \equal{} \frac {n!}{2\pi i}\oint \frac {dz}{z^n(e^z \minus{} 1)}$

[/quote]
Thank you, for your help.
but main moment is Why definition of bernoullii number by this way $ B_n \equal{} \frac {n!}{2\pi i}\oint \frac {dz}{z^n(e^z \minus{} 1)}$
 and by this way :$ C_{n \plus{} 1}^{0}B_{0} \plus{} C_{n \plus{} 1}^{1}B_{1} \plus{} ... \plus{}B_{n} C_{n \plus{} 1}^{n} \equal{} 0$ and $ B_{0} \equal{} 1$
is same ?

Actually by definition, the Bernoulli numbers are defined as the numerator appearing in the power series:

$ \frac{z}{e^z\minus{}1}\equal{}\sum_{n\equal{}0}^{\infty}\frac{B_n}{n!}z^n$

Alright, so using Cauchy's Theorem:

$ B_n\equal{}\frac{n!}{2\pi i}\oint \frac{s}{(e^s\minus{}1)s^{n\plus{}1}}ds$

but why definition by this way $ C_{n \plus{} 1}^{0}B_{0} \plus{} C_{n \plus{} 1}^{1}B_{1} \plus{} ... \plus{}B_{n} C_{n \plus{} 1}^{n} \equal{} 0$ and by this way :
$ \frac{z}{e^z\minus{}1}\equal{}\sum_{n\equal{}0}^{\infty}\frac{B_n}{n!}z^n$ is same ?

[quote="Extremal"][quote="shawtoos"][quote="Extremal"]but why definition by this way $ C_{n \plus{} 1}^{0}B_{0} \plus{} C_{n \plus{} 1}^{1}B_{1} \plus{} ... \plus{} B_{n} C_{n \plus{} 1}^{n} \equal{} 0$?[/quote]

Since this is a real analysis forum, how about I re-word it in terms of that:

Given the real-valued function:

$ f(x) \equal{} \frac {x}{e^x \minus{} 1} \equal{} \minus{} 1/2 x \plus{} 1 \plus{} \sum_{u \equal{} 1}^{\infty}\frac {( \minus{} 1)^{u \plus{} 1}\zeta(2u)}{2^{2u \minus{} 1}\pi^{2u}}x^{2u}$

Show:

$ \sum_{k \equal{} 0}^{n \minus{} 1}\left\{\frac {n!}{(n \minus{} k)!k!}\,\frac {d^k}{dx^k}\left[ \minus{} 1/2 x \plus{} 1 \plus{} \sum_{u \equal{} 1}^{\infty}\frac {( \minus{} 1)^{u \plus{} 1} \zeta(2u)}{2^{2u \minus{} 1}\pi^{2u}}x^{2u}\right]_{x \equal{} 0}\right\} \equal{} 0$

for odd $ n\ge 3$.[/quote]
We can't show it i we don't use the fact that:
$ \zeta(n) \equal{} \frac {|B_{n}|2^{n \minus{} 1}\pi^{n}}{n!}$
 but we can't use this fact because  we need to prove this fact. but when you prove this fact you use the fact that :
$ \sum_{k \equal{} 0}^{n \minus{} 1}\left\{\frac {n!}{(n \minus{} k)!k!}\,\frac {d^k}{dx^k}\left[ \minus{} 1/2 x \plus{} 1 \plus{} \sum_{u \equal{} 1}^{\infty}\frac {( \minus{} 1)^{u \plus{} 1} \zeta(2u)}{2^{2u \minus{} 1}\pi^{2u}}x^{2u}\right]_{x \equal{} 0}\right\} \equal{} 0$
but you can't use this fact because  to prove this fact you need to use :
$ \zeta(n) \equal{} \frac {|B_{n}|2^{n \minus{} 1}\pi^{n}}{n!}$ ... ... ...[/quote]
 :D

You know guys . . . here's a show in Mathematica code:

[code]
In[249]:= n = 11;
Sum[(n!/((n - k)!*k!))*D[(-2^(-1))*x + 1 +
           Sum[(((-1)^(u + 1)*Zeta[2*u])/(2^(2*u - 1)*Pi^(2*u)))*
               x^(2*u), {u, 1, 20}], {x, k}] /. x -> 0, {k, 0,
  n - 1}]

Out[250]= 0
[/code]

That's certainly "show"ing it's zero for $ n \equal{} 11$.  Note I didn't say "prove" . . . just a "show" for starters to see if at least the expression is right.  If we can do more, or if someone gives a proof of the expression:

$ \sum_{k \equal{} 0}^{n \minus{} 1}\left(\begin{array}{c} n \\
k \end{array}\right) B_k \equal{} 0$

for all odd $ n\geq 3$,  that would be nice. :)  I can't, not yet anyway.

Let $ f(z) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {B_n}{n!}z^n$, and write out the generating function for $ f(z)\cdot (e^z \minus{} 1)$. Equate coefficients, and there's your identity.

Also, $ \frac {z}{e^z \minus{} 1} \plus{} \frac12z \equal{} \frac {z(1 \plus{} \frac12e^z \minus{} \frac12)}{e^z \minus{} 1} \equal{} \frac {z}{2}\frac {e^z \plus{} 1}{e^z \minus{} 1} \equal{} \frac {z}{2}\frac {\cosh \frac {z}{2}}{\sinh\frac {z}{2}}$. That's an even function, so all of the other odd terms are zero.

[quote="jmerry"]Let $ f(z) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {B_n}{n!}z^n$, and write out the generating function for $ f(z)\cdot (e^z \minus{} 1)$. Equate coefficients, and there's your identity.

Also, $ \frac {z}{e^z \minus{} 1} \plus{} \frac12z \equal{} \frac {z(1 \plus{} \frac12e^z \minus{} \frac12)}{e^z \minus{} 1} \equal{} \frac {z}{2}\frac {e^z \plus{} 1}{e^z \minus{} 1} \equal{} \frac {z}{2}\frac {\cosh \frac {z}{2}}{\sinh\frac {z}{2}}$. That's an even function, so all of the other odd terms are zero.[/quote]
yes it works. but you need instead $ f(z) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {B_n}{n!}z^n$ this function :
$ f(z) \equal{} \sum_{n \equal{} 0}^{\infty}\frac {B_n}{n!}z^n$
and after direct calculation  we get that :
$ f(z)(e^{z} \minus{} 1) \equal{} z$ if we use that :$ B_{0}C_{n \plus{} 1}^{0} \plus{} ...B_{n}C_{n \plus{} 1}^{n} \equal{} 0$ and $ B_{0} \equal{} 1$ so now we have expression for zeta function at point $ 2n$
or
$ \zeta(2n) \equal{} \frac {|B_{2n}|\pi^{2n}2^{2n \minus{} 1}}{(2n)!}$

Thanks guys.  Well I don't see it . . . direct calculations leads me through:

\begin{align}
\frac{x}{e^{x}-1} (e^x-1)&= x\\
&=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n
\sum_{k=1}^{\infty}\frac{x^k}{k!} \\
&=\sum_{n=0}^{\infty}
\sum_{k=0}^{n}\frac{B_{n-k}}{(n-k)!}\frac{x^{n+1}}{(k+1)!}
\end{align}

Equating coefficients:

$ \sum_{n=0}^{\infty}\sum_{k=0}^{n}
\frac{B_{n-k}}{(n-k)!}\frac{x^{n+1}}{(k+1)!}=x$

or:

$ B_0=1$

and:

$ \sum_{k=0}^{n}\frac{B_{n-k}}{(n-k)!(k+1)!}=0;\quad n>0$

which I suppose is equivalent to the above expressions (I'm pretty good with plots though).


2 similar cylindrical vases have diameters of 6in. and 8in.
1a)What is the ratio of their surface areas?
1b What is the ratio of their volumes :)  :lol:

I don't think Nick42 is exactly right :wink:

To find the ratio of the volumes of the two vaces, you first need to know the formula for the volume of a cylinder. This is..
$ \pi r^2 height$
Assuming the heights of the two cylinders/vaces are the same, we could brute force it.

$ diameter 6:$ $ \pi 3^2 height \equal{} 9pi$
$ diameter 8:$ $ \pi 4^2 height \equal{} 16pi$

The ratio of the two is $ \boxed{\dfrac{9}{16}}$ which was perfect.

Now for the surface areas. The equation for the surface are of a cylinder is as follows:
$ 2\pi r^2\plus{} 2\pi r h$

We can brute force it. But wait, there are some common things we can get rid of. The $ 2\pi$ and the other $ 2\pi$ r common in both equations. In this particular problem, the $ h$ is also common so we can get rid of that too. This makes the math much simpler.

$ diameter 6:$ $ 3^2\plus{}3\equal{}9\plus{}3$
$ diameter8:$ $ 4^2\plus{}4\equal{}16\plus{}4$

From this we get $ \dfrac{9\plus{}3}{16\plus{}4}\equal{}\dfrac{12}{20}\equal{}\boxed{\dfrac{3}{5}}$

What do you mean "get rid of?" You can't "get rid of" $ 2\pi r$ because $ r$ is different.

Also, the problem is faulty. Different heights yield different ratios.

sigh, i phailed. And I took all that time writing it up while trying to learn $ LaTeX$

[quote="RoFlLoLcOpT"]

Also, the problem is faulty. Different heights yield different ratios.[/quote]

You actually know the ratio of the heights, since he said that the cylinders are similar.

Ah didn't read that part :) Then it should be \[ \left(\frac{6}{8}\right)^2\equal{}\left(\frac{3}{4}\right)^2\equal{}\boxed{\frac{9}{6}},\] as nick42 already mentioned.

typo, $ \dfrac{9}{16}$  :wink:

But I said that as well, the part where we have different areas is the surface area.

Just cube the ratio in the dimensions. That would be \[ \left(\frac{6}{8}\right)^3\equal{}\left(\frac{3}{4}\right)^3\equal{}\boxed{\frac{27}{64}}.\]

Hmm...It really isn't that hard.

#1. The ratio of the surface areas is basically just the ratio of their diameters squared....

So 6*6 over 8*8 would give us 9/16...so that is the ratio of the surface areas.

#2. Same thing, only it's the ratio of the diameters CUBED...

6^3 over 8^3, simplified, is 27/64.


$ X_1,X_2, ...$ i.i.d. with $ EX_1^\plus{} < \infty$, $ Y_n \equal{} \max_{1 \le m \le n} X_m \minus{}cn$
a) if $ T\equal{} \inf \{ n: X_n > a \}, p\equal{}P(X_n >a)$, compute $ EY_T$
b)if $ \alpha$ is the solution to $ E(X_1\minus{}\alpha )^\plus{}\equal{}c$, show $ EY_T\equal{}\alpha$
what i did for a:

$ P(T\equal{}n)\equal{}(1\minus{}p)^{n\minus{}1}p$
$ Y_T\equal{}\sum_{n\equal{}1}^\infty 1_{T\equal{}n}(X_n\minus{}cn)$
$ EY_T\equal{}\int \sum_{n\equal{}1}^\infty  (X_n\minus{}cn )1_{T\equal{}n}dP\equal{}\sum_{n\equal{}1}^\infty \int [X_n\minus{}cn] (1\minus{}p)^{n\minus{}1}pdP$
$ \equal{}p\sum_{n\equal{}1}^\infty [\int X_n(1\minus{}p)^{n\minus{}1}dP\minus{}cn(1\minus{}p^{n\minus{}1}]\equal{}p\sum_{n\equal{}1}^\infty [E(X_n)\minus{}cn](1\minus{}p)^{n\minus{}1}$
$ \equal{}p\sum_{n\equal{}1}^\infty [E(X_1)\minus{}cn](1\minus{}p)^{n\minus{}1}\equal{}E(X_1)\plus{}p\sum_{n\equal{}1}^\infty cn(1\minus{}p)^{n\minus{}1}$
$ \equal{}E(X_1)\minus{}\frac{c}{p}$
the fact that this doesnt seem to help me for part b leads me to believe im already off?

tried something different..
$ Y_T\equal{}\sum_{n\equal{}1}^\infty 1_{T\equal{}n}([(X_n\minus{}a)^\plus{}\plus{}a]\minus{}cn)$\\
$ EY_T\equal{}\int \sum_{n\equal{}1}^\infty  (([X_n\minus{}a)^\plus{}\plus{}a]\minus{}cn)1_{T\equal{}n}dP\equal{}$\\
$ \sum_{n\equal{}1}^\infty \int [[(X_n\minus{}a)^\plus{}\plus{}a]\minus{}cn] (1\minus{}p)^{n\minus{}1}pdP$\\
$ \equal{}p\sum_{n\equal{}1}^\infty [\int ((X_n\minus{}a)^\plus{}\plus{}a)(1\minus{}p)^{n\minus{}1}dP\minus{}cn(1\minus{}p)^{n\minus{}1}]$\\
$ \equal{}p\sum_{n\equal{}1}^\infty [(E(X_n\minus{}a)^\plus{}\plus{}a)\minus{}cn)](1\minus{}p)^{n\minus{}1}$\\
$ \equal{}p\sum_{n\equal{}1}^\infty [(E(X_1\minus{}a)^\plus{}\plus{}a)\minus{}cn)](1\minus{}p)^{n\minus{}1}$\\
$ \equal{}E(X_1\minus{}a)^\plus{}\plus{}a)\minus{}p\sum_{n\equal{}1}^\infty cn(1\minus{}p)^{n\minus{}1}$\\
$ \equal{}E(X_1\minus{}a)^\plus{}\plus{}a\minus{}\frac{c}{p}$\\

still dont know how to do part b

I suppose what goes wrong in your calculations is the step where you integrate out the indicator
$ 1_{T\equal{}n} \equal{} 1_{X_1 \le a} \ldots 1_{X_{n\minus{}1} \le a} 1_{X_n > a}$ over $ P$. Only
the first $ n\minus{}1$ factors are independent to the rest of the integrand
(so you can integrate those out), but the last factor is not! (It seems to me that
you wanted to use independence of $ X_n$ and $ T$, which is not true.)

However, this means that the result of your second computation is only off by a factor
of $ 1/p$, so this still doesn't  help for (b). So either I messed up the calculation aswell,
or there is a mistake in the statement of the problem.

ahhhh you made it all work out thank you thank you

$ EY_T\equal{}\int \sum_{n\equal{}1}^\infty  (([X_n\minus{}a)^\plus{}\plus{}a]\minus{}cn)1_{T\equal{}n}dP$
$ \equal{}\sum_{n\equal{}1}^\infty \int [[(X_n\minus{}a)^\plus{}\plus{}a]\minus{}cn] (1\minus{}p)^{n\minus{}1}1_{X_n >a}dP$
$ \equal{}\sum_{n\equal{}1}^\infty \int ((X_n\minus{}a)^\plus{}\plus{}a)(1\minus{}p)^{n\minus{}1}1_{X_n >a}dP\minus{}c\sum_{n\equal{}0}^\infty n(1\minus{}p)^{n\minus{}1}\int 1_{X_n >a}dP$
$ \equal{}\sum_{n\equal{}1}^\infty (1\minus{}p)^{n\minus{}1}\int ((X_n\minus{}a)^\plus{}\plus{}a)1_{X_n >a}dP\plus{}cp\frac{d}{dp}(\sum_{n\equal{}0}^\infty (1\minus{}p)^n)$
$ \equal{}\sum_{n\equal{}0}^\infty (1\minus{}p)^n \int ((X_1\minus{}a)^\plus{}\plus{}a)dP\plus{}cp\frac{d}{dp}(\frac{1}{p})$
$ \equal{}\frac{1}{p}\int ((X_1\minus{}a)^\plus{})dP\plus{} \frac{1}{p}\int a 1_{X_n > a}dP\minus{}\frac{c}{p}$
$ \equal{}\frac{E(X_1\minus{}a)^\plus{}}{p}\plus{}a\minus{}\frac{c}{p}$

[quote="kenn4000"]
$ EY_T \equal{} \int \sum_{n \equal{} 1}^\infty (([X_n \minus{} a)^ \plus{} \plus{} a] \minus{} cn)1_{T \equal{} n}dP$
$ \equal{} \sum_{n \equal{} 1}^\infty \int [[(X_n \minus{} a)^ \plus{} \plus{} a] \minus{} cn] (1 \minus{} p)^{n \minus{} 1}1_{X_n > a}dP$
[/quote]

Can you justify the second statement from the first?


geometry
combinatorics unsolved
combinatorics

Given a regular 2007- gon , find the smallest positive integer k such that in any set of k vertices of the polygon, there are four vertices which form a quadrilateral with three edges of the 2007-gon.

This same problem is currently among the 10 most recent threads on this forum.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165471

What is the source?

It seems to be part of some ongoing Indian contest. There are Indians screaming about it - 'rik_sengupta', 'pardesi' and 'SOUMYASHANT NAYAK' I think this is a good time to reinforce the dictum - ML is NOT meant for cheating.

In India,once you clear the INMO, the next year you dont need to sit for the regional level.

They send you postal tests and if you pass them satisfactorily you get invited to next year's camp without needing to appear for the INMO.

So obviously its mean and unethical for an IMOTC member to take help in public foums like this.


integration
calculus
calculus computations

Find the mass of the lamina cut from the circle $x^{2}+y^{2}=36$ by the line $x=3$ if its density is given by $\partial(x,y)=\frac{x^{2}}{x^{2}+y^{2}}$.

Okay so I get that the equations above corelate to the circle $r^{2}=36\Rightarrow r=6$ and the line $rcos\theta=3$. I come up with this:
$\int_{-tan^{-1}\sqrt{5}}^{tan^{-1}\sqrt{5}}\int_{\frac{3}{cos\theta}}^{6}\frac{r^{2}cos^{2}\theta}{r^{2}}\cdot rdrd\theta$

But $tan^{-1}\sqrt{5}$ seems a little clunky. I based in on the point of intersection of the line and the cirle. Is there a better term to work with for this?

I think I identified my error:

$\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int_{\frac{3}{cos\theta}}^{6}\frac{r^{2}cos^{2}\theta}{r^{2}}\cdot rdrd\theta= \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int_{\frac{3}{cos\theta}}^{6}rcos^{2}\theta drd\theta$
$\Rightarrow \text{lots of steps later}\Rightarrow \frac{3}{2}(3\sqrt{3}+2\pi)$


An edge of a given cube is 1 inch. A space diagonal of this cube is used as an edge to form a second cube. A space diagonal of this second cube is used as an edge to form a third cube. The process is repeated. How many inches are in the length of an edge of the fifth cube.

The space diagonal of a cube is equal to $ \sqrt{3}$ of its side length. Thus, the fifth cube has a sidelength of $ (\sqrt{3})^4\equal{}\boxed{9}$


Consideremos los números $A=(10^2)!$ y $B=(10^2)^{100}.$ ¿Cuántos ceros tiene al final, el número que resulta de la diferencia entre $A$ y $B$?

Upss!, I missed for one, jeje, $\frac{100}{25}\neq5$  :blush:

[hide="La Verdadera Respuesta"]24[/hide]


geometry
trapezoid
trigonometry
LaTeX
similar triangles
trig identities
Law of Cosines

Let $ABCD$ be a trapezoid, where $AB||CD$, $AC\perp BD$, $AC\cap BD=P$, and $AD\cap BC=Q$. If $AB=4$, $CD=10$, and $\angle AQD=45^{\circ}$, the area of the trapezoid can be expressed as $\frac{a}{b}$, where $\gcd (a,b)=1$.  Find $a+b$.

Masoud Zargar

[quote="M4RI0"]What's $Q$ for :?:[/quote]

$P$ and $Q$ are intersection points.  It's... odd notation, but it makes sense.

Yeah, I understand that but I don't understand why the stament gives point $Q$ if it doesn't use it.

Err... well, the only thing I can think of is that the problem impiles $AD$ isn't parallel to $BC$.

All I can find is that the maximum value of the area is 49. :D

Since h═(AB+CD)⁄2   A(ABCD)═h∙h   h═(AB+CD)⁄2 ═(10+4)⁄2═7  Then A(ABCD)═h∙h═49cm squares. Since      A(ABCD)═a⁄b═49⁄1 ,  so a+b═50[/url][/u][/b]

The point $Q$ can both be considered as a hint or something to confuse you.  Make good use of it in order to acquire a nice solution.  For brevity, I could tell you that the intersection of $AD$ and $BC$ at $Q$ makes a $45^o$ angle.  I guess I forgot to give you this information.  :oops:

Masoud Zargar

you still have to fix the problem statement, A,Q,D are collinear

anyway,

[hide="solution"]$BQ=4x$, then $CB=6x$ and $AQ=4y$, then $AD=6y$ by similar triangles

since $ABCD$ is orthogonal, $10^2+4^2=36(x^2+y^2)$, or $x^2+y^2=\frac{29}{9}$

by the law of cosines, $10^2=(10x)^2+(10y)^2-2(10x)(10y)\cos 45^{\circ}$, so $1=(x^2+y^2)-\sqrt{2}(xy)$ so $xy=\frac{10\sqrt{2} }{9}$

using the sine area formula, $[ABCD]=\frac{1}{2}\sin 45^{\circ}xy(10^2-4^2)=\frac{140}{3}$

therefore, $a+b=143$[/hide]

[quote="nebula42"]Yeah, the equation should read "$\angle AQB = 45 ^ \circ$."
By the way, is there way to write the degree symbol in $LaTeX$ other than "^ \circ"?[/quote]
You write "^o" ;)


Given parallelogram $ ABCD$ such that $ AB\equal{}a$, $ AD\equal{}b$. The first circle has its center at vertex $ A$ and pases through $ D$, and the second circle has its center at $ C$ and passes through $ D$. A circle with center $ B$ meets the first circle at $ M_{1}$, $ N_{1}$, and the second circle at poinst $ M_{2}$, $ N_{2}$. Evaluate the ratio $ M_{1}N_{1}/M_{2}N_{2}$.


Fie patrulaterul convex ABCD cu O intersectia diagonalelor.Fie $\ r_i$ razele cercurilor inscrise triunghiurilor OBC,OAD,OCD si OAB, cu $\ i=1,2,3,4$.Demonstrati ca suma lungimilor inaltimilor coborate din O pe laturile triunghiurilor este mai mica decat $\frac{1}{4}$($\ OA*OB+OB*OC+OC*OD+OA*OD$), stiind ca min $\ r_i$>1.

N-o incearca nimeni?Macar sa vad si eu daca e prea grea sau prea usoara pentru clasa a VI-a...


[hide]
\frac{1}{6}^{17}

yea, that is a scary number.

[/hide]

[hide] The choices are independant thus the answer is $( \frac{1}{6} )^{17}$ [/hide]

[quote="Drunken_Math"]What are the odds that you roll the same number 17 times on a standard 6 sided die?  :)[/quote]

[hide]

$Odds = \frac{1-p}{p} \approx \frac{1}{p} = 6^{17}:1$

[/hide]

[quote="Drunken_Math"]What are the odds that you roll the same number 17 times on a standard 6 sided die?  :)[/quote]
If this means "what are the odds that you roll the same number 17 out of 17 times on a fair, standard 6-sided die", then I disagree with all previous answers and solutions.

Oh my god! G-UNIT pawned and owned all of you guys! Good job G-UNIT!  :)

[quote="G-UNIT"]haha, the answer is $ 6^{16}:1$         Why? because the first roll can be anything, then the probability of each, next roll being the same is $\frac{1}{6}$[/quote]
nice job G-UNIT, your answer makes sense. :lol:

Yep, he's right. It would be 1/(6^16). (Just in case the few previous repliies weren't enough...)


Find the smallest integer x greater than 1 such that x^2 and x end in the same 3 digits.

[hide]

$ x^{2}\equiv x \mod{1000}$

$ x^{2}-x \equiv 0 \mod{1000}$

$ x(x-1) \equiv 0 \mod{2^{3}\cdot5^{3}}$

x(x-1) must have a factor of 125 and a factor of 8 within it

we know that either $ x$ or $ x-1$ must be a multiple of 125 from this

125: neither 124 or 126 are multiples of 8
250: not a multiple of 8
375: 376 is a multiple of 8

thus 375 and 376 are x-1 and x
$ x = 376$
[/hide]


Εστω δυο ανισοι κυκλοι Γ1 (Κ,ρ1) και Γ2 (Ο,ρ2) ωστε ο ενας να βρισκεται στο εξωτερικο του αλλου. Εστω μια μεταβλητη ευθεια που τεμνει τους κυκλους κατα σειρα στα σημεια Α, Β, Ε, Ζ ωστε Α, Β σημεια του Γ1 και Ε, Ζ σημεια του Γ2 και ΑΒ=ΕΖ. Εστω οτι οι εφαπτομενες του Γ1 στο Α και του Γ2 στο Ζ τεμνονται στο Μ. Να βρεθει ο γεωμετρικος τοπος του Μ.

[quote]Έστω δύο άνισοι κύκλοι $ (O_{1}),\ (O_{2}),$ ο ένας στο εξωτερικό μέρος του άλλου. Έστω μία μεταβλητή ευθεία που τέμνει τους κύκλους κατά σειρά στα σημεία $ A,\ B,\ E,\ Z,$ ώστε τα $ A,\ B,$ σημεία του $ (O_{1}),$ τα $ E,\ Z,$ σημεία του $ (O_{2})$ και $ AB = EZ.$ Έστω ότι οι εφαπτόμενες του $ (O_{1})$ στο $ A$ και του $ (O_{2})$ στο $ Z$ τέμνονται στο $ M.$ Να βρεθεί ο γεωμετρικός τόπος του $ M.$[/quote]
Έστω $ X,\ Y,$ τα μέσα των $ AB,\ EZ$ αντιστοίχως και $ N$ το μέσον του $ AZ.$

Από $ AB = EZ$ και $ NA = NZ,$ συμπεραίνουμε ότι είναι και $ NX = NY$ και άρα η δια του σημείου $ N$ κάθετη ευθεία επί την $ AZ,$ ως μεσοκάθετη του $ XY,$ περνάει από το μέσον $ K$ της διακέντρου  $ O_{1}O_{2},$ των δοσμένων κύκλων $ (O_{1}),\ (O_{2})$ ( αποδεικνύεται εύκολα με το [b]θεώρημα Θαλή[/b] ).

Θεωρούμε τώρα το τρίγωνο $ \bigtriangleup MAZ,$ όπου επί των πλευρών του $ MA,\ MZ,$ έχουμε αντιστοίχως τα ορθογώνια τρίγωνα $ \bigtriangleup AMO_{1},\ \bigtriangleup ZMO_{2},$ προς το εσωτερικό μέρος του $ \bigtriangleup MAZ$ και για τα οποία έχει αποδειχθεί ότι η μεσοκάθετη της πλευράς του $ AZ,$ διχοτομεί το τμήμα $ O_{1}O_{2},$ που συνδέει τις ελέυθερες κορυφές τους ( που δεν ανήκουν στο $ \bigtriangleup MAZ$ ).

$ \bullet$ Σύμφωνα με το αντίστροφο του παρακάτω Λήμματος, συμπεραίνουμε ότι $ \angle AMO_{1} = \angle ZMO_{2}$ $ ,(1)$ και άρα τα ορθογώνια τρίγωνα $ \bigtriangleup AMO_{1},\ \bigtriangleup ZMO_{2}$ είναι όμοια.

Από την ομοιότητα αυτών των τριγώνων προκύπτει επομένως ότι  $ \frac {MO_{1}}{MO_{2}} = \frac {O_{1}A}{O_{2}Z} = \frac {r_{1}}{r_{2}}$ $ ,(2)$

 Από $ (2)$ συμπεραίνουμε ότι το σημείο $ M$ ανήκει στο [b]Απολλώνιο κύκλο[/b] του τριγώνου $ \bigtriangleup MO_{1}O_{2},$ δηλαδή στον κύκλο $ (L)$ διαμέτρου $ SS',$ όπου $ S,\ S'$ είναι σημεία επί της ευθείας $ O_{1}O_{2},$ αρμονικά συζυγή των $ O_{1},\ O_{2}$ και τέτοια ώστε $ \frac {SO_{1}}{SO_{2}} = \frac {S'O_{1}}{S'O_{2}} = \frac {MO_{1}}{MO_{2}} = \frac {r_{1}}{r_{2}}$ $ ,(3)$

Αλλά από την $ (3)$ συμπεραίνουμε ότι ο $ (L)$ είναι γνωστός κύκλος, αφού τα σημεία $ S,\ S'$ ταυτίζονται αντιστοίχως, με το εσωτερικό και εξωτερικό κέντρο ομοιότητας των δοσμένων κύκλων $ (O_{1}),\ (O_{2}).$

Άρα ο γεωμετρικός τόπος του σημείου $ M,$ είναι ο κύκλος $ (L)$ διαμέτρου $ SS'$ και το πρόβλημα έχει λυθεί.

$ \bullet$ Το σημείο $ M',$ τομής των εφαπτομένων των κύκλων $ (O_{1}),\ (O_{2}),$ στα σημεία $ B,\ E$ αντιστοίχως, ανήκει επίσης στον ίδιο γεωμετρικό τόπο.

[b][size=100][color=DarkBlue]ΛΗΜΜΑ. – Δίνεται τρίγωνο $ \bigtriangleup ABC$ και έστω τα ορθογώνια τρίγωνα  $ %Error. "bigtriangle" is a bad command.
ABD,\ \bigtriangleup ACE,$ επί των πλευρών του $ AB,\ AC$ αντιστοίχως, προς το εξωτερικό ( ή το εσωτερικό ) μέρος του $ \bigtriangleup ABC,$ με $ \angle ABD = \angle ACE = 90^{o}$ και $ \angle ADB = \angle AEC = \angle \omega.$ Αποδείξτε ότι η μεσοκάθετη της πλευράς $ BC,$ διχοτομεί το τμήμα $ DE.$[/color][/size][/b]

Μία πολύπλοκη απόδειξη του Λήμματος έχει δημοσιευτεί στο δικό μας φόρουμ http://www.mathlinks.ro/Forum/viewtopic.php?t=188591, καθώς επίσης και μία απλούστερη απόδειξή του, στο αγγλικό φόρουμ http://www.mathlinks.ro/Forum/viewtopic.php?t=250255 .

Το αντίστροφο του παραπάνω Λήμματος, αποδεικνύεται εύκολα ( αν η μεσοκάθετη της πλευράς $ BC$ του $ \bigtriangleup ABC,$ διχοτομεί το τμήμα $ DE,$ τότε τα ορθογώνια τρίγωνα $ \bigtriangleup BAD,\ \bigtriangleup CAE,$ με $ \angle ABD = \angle ACE = 90^{o},$ είναι όμοια ).

Κώστας Βήττας.

ΥΓ. Σάκη ωραίο πρόβλημα, το απόλαυσα. Μου λες σε παρακαλώ την πηγή του και αν υπάρχει επίσημη λύση ; Σε ευχαριστώ εκ των προτέρων.

K. Βηττα, η λυση σας ηταν εξαιρετικη οπως παντα και ευχαριστουμε για το πολυ ενδιαφερον λημμα. Η λυση μου ειναι καπως διαφορετικη, μαλλον λιγο απλουστερη.

Θεωρω την καθετη απο το Μ στη ΑΖ που την τεμνει στο P. Τοτε MPA τριγωνο ομοιο με το O1XA τριγωνο. Αρα $ \frac{MP}{AX}\equal{}\frac{MA}{R1}$. Ομοιως τα τριγωνα ΜΡΖ και ΥΖΟ2 ειναι ομοια. Αρα $ \frac{MP}{YZ}\equal{}\frac{MZ}{R2}$. Επειδη $ AX\equal{}YZ$ ισχυει $ \frac{MA}{MZ}\equal{}\frac{R1}{R2}$.
Τοτε ομως τα ορθογωνια τριγωνα ΜΑΟ1 και ΜΖΟ2 ειναι ομοια διοτι εχουν ισους λογους καθετων πλευρων. Τοτε $ \frac{MO1}{MO2}\equal{}\frac{R1}{R2}$.
Αρα οπως εχει εξηγησει αναλυτικα ο κ. Βηττας ο γεωμετρικος τοπος ειναι ο Απολλωνιος κυκλος.

Σάκη, πιο πολύ ευχαριστήθηκα τη λύση σου, γιατί ''[b]το σοφό είναι και απλό[/b]'' όπως έγραψε κάπου στον ΑΠΟΛΛΩΝΙΟ ο [b]Θεόφιλος Χρυσοστομίδης[/b], που έφυγε πρόωρα.

Αυτό είναι που με γοητεύει πιο πολύ στη Γεωμετρία, το ότι θα υπάρχει πάντα μία απλούστερη λύση με στοιχειώδη μέσα, κατανοητή στο μέσο νου.

Συνήθως όμως, τέτοιες όμορφες λύσεις ( και δεν αναφέρομαι μόνο στη Γεωμετρία ), επιτυγχάνονται από ευέλικτα νεανικά μυαλά και  αυτό δίνει χαρά στους μεγαλύτερους.  :)

Σε ευχαριστώ, Kώστας Βήττας.


Ring Theory
superior algebra
superior algebra unsolved

Prove that the ring of all the rational numbers whose denominators is odd is a local ring whose unique maximal ideal is the principal ideal generated by 2.

obviously, the complement of the unit group is the ideal generated by $ 2$, that's it.


DO NOT DISCUSS THE PROBLEMS BEFORE SATURDAY, OCTOBER 31.

The problems are due to be postmarked by Friday, October 30 so waiting for Saturday would be the safest bet. This is subject to further notice from Charles Koppelman. Hope you guys have fun with the test.  :)

DO NOT DISCUSS THE PROBLEMS BEFORE SATURDAY, OCTOBER 31.

Yay so it's Saturday.

I answered all of them but got one wrong, for a theoretical total of 144. Coincidentally, two other unnamed individuals at Walton also got a 144 (we all got different questions wrong). Not so coincidentally, all of the mistakes were really stupid :P

Walton has an answer key compiled from our answers (which is how I know my theoretical score). If anyone wants and it's not published already somewhere else, I could post it here.

I missed 2, and I'm not thrilled about it.  I missed number 24 and that one about the 3-element subset and its probability of being "proper" or something.  The former, I didn't have the patience or cleverness to do, but the latter, I made the mistake of thinking the first 11 positive integers have five odds, not six...  Oh well...

So a theoretical total of 138.

Did anyone think this year's was kinda, sorta, maybe a bit more difficult that last year's?  Cuz I thought last year's was pretty easy and this year's was okay.  Yes, there is a difference.

@Sitan: coincidentally, Boru missed the exact set of questions as you (though he might put blank for 24).

For #24, the "easy" way is to first make the plus-19 substitution both in the diagram and in the expression to be found, then try to obtain that expression (plus some constant) from a linear combination of the "corner" sums.

Also, anyone know of a not-so-bruteforce solution for the coin problem?

[quote="timwu"]For #24, the "easy" way is to first make the plus-19 substitution both in the diagram and in the expression to be found, then try to obtain that expression (plus some constant) from a linear combination of the "corner" sums.[/quote]

That's what I tried to do, but I couldn't find the correct linear combination.  :(

Darn Sitan that's not what I got... but then again my method was to let d = 0...

[hide="As for the coin problem:"]

Assume Debs has $ n_1,d_1,q_1$ nickels, dimes, and quarters respectively, and assume Don has $ n_2,d_2,q_2$ nickels, dimes, and quarters

Then, we have:

$ n_1 \plus{} 2d_1 \plus{} 5q_1 \equal{} 100$

$ n_1 \plus{} d_1 \plus{} q_1 \equal{} 40$

Take these mod 4 to get $ d_1 \equiv 0 \; \; (mod \; 4)$. Similarly, $ d_2 \equiv 0 \; \; (mod \; 4)$.

Moreover, eliminating n with these equations, we get

$ d_1 \plus{} 4q_1 \equal{} 60$, and similarly:
$ d_2 \plus{} 4q_2 \equal{} 60$

Using that $ q_1 \equal{} 2q_2$, we arrive at

$ 2d_2 \minus{} d_1 \equal{} 60$

Now, we can only have three ordered pairs $ (d_1,d_2)$ that satisfy this: $ (40,20),(36,12),(32,4)$.

However, in the first 2 pairs, $ n_1 \plus{} d_1 \plus{} q_1 > 40$, thus we have $ d_1 \equal{} 32, d_2 \equal{} 4$, and we solve from there.

[/hide]

Edit: cleaned up some code

Yay that's what I got.

Just saying, but is there any reason why you changed your sn to alt, Oliver?

alt, is your head O.K.?

[hide="No casework for the coin problem :"]Following alt's notation, $ d_{1}\plus{}4q_{1}\equal{} 60.$  Since $ d_{1}$ is a non-negative integer, $ q_{1} \le 15.$ Also, since $ d_{1}\plus{}q_{1} \le 40,$ then $ 3q_{1} \ge 20.$   The possible range of $ q_{1}$ is $ 7 \le q_{1} \le 15.$     $ q_{2}$ has the exactly same condition, and $ q_{2}\equal{}2q_{1}$.  Therefore, the only pair of integer solutions is $ q_{1}\equal{}7$ and $ q_{2}\equal{}14.$   Remaining portion is straight forward such that $ n_{1}\equal{}1.$[/hide]

[hide="#24"]Substitute for e, g, b, and f their corresponding variable (either a,h,i, or d)+19.
It should be
 a           |        i+19        |         c
 d           |       a+19        |       d+19
h+19       |          h          |         i
Then, create equations with the four squares.
$ a\plus{}i\plus{}19\plus{}d\plus{}a\plus{}19\equal{}77 \rightarrow 2a\plus{}i\plus{}d\equal{}39$
$ i\plus{}19\plus{}c\plus{}a\plus{}19\plus{}d\plus{}19\equal{}94 \rightarrow a\plus{}i\plus{}d\plus{}c\equal{}37$
$ d\plus{}a\plus{}19\plus{}h\plus{}19\plus{}h\equal{}86 \rightarrow a\plus{}d\plus{}2h\equal{}48$
$ a\plus{}19\plus{}d\plus{}19\plus{}h\plus{}i\equal{}83 \rightarrow a\plus{}d\plus{}h\plus{}i\equal{}45$
(We didn't actually need the second equation, but oh well :D)

Note that the thing you want to find, $ a\plus{}e\plus{}f\plus{}g$, can be written as $ 2a\plus{}d\plus{}h\plus{}57$
Now, subtract the fourth equation from the third. You should get $ h\minus{}i\equal{}3$.
Add this new equation to the first, to get $ 2a\plus{}h\plus{}d\equal{}42$
Now add 57 both sides. $ 2a\plus{}h\plus{}d\plus{}57\equal{}99$[/hide]

Yay for not posting in a long time!(a year!) :)

For the coin problem, oh, nvm, yeah, what Edward/Allen said is what I got.  But what Edward/Allen said was kinda cold. :P

For #24, oh yeah, it was 99.  Sorry, 109 was what I guessed.

That was completely Edward.

[quote="Math Geek"]That's what I tried to do, but I couldn't find the correct linear combination. :([/quote]

If you don't want to guess and check, then you have to create some more variables:
The expression $ a\plus{}e\plus{}f\plus{}g\minus{}57\equal{}2a\plus{}d\plus{}h$ can (theoretically) be written as a linear combination of the equations. If you're observant, you can notice that you didn't actually need the second equation since it has a c variable, the expression you want doesn't have a c variable, and no other equation have a c variable. Let the first equation be multiplied by x, the third by y, and the fourth by z. Then \[ 2a\plus{}d\plus{}h\equal{}x(2a\plus{}i\plus{}d)\plus{}y(a\plus{}d\plus{}2h)\plus{}z(a\plus{}d\plus{}h\plus{}i)\equal{}39x\plus{}48y\plus{}45z\]
Simply let the coefficients of a be equal, the coefficient of d be equal, etc. and solve for x, y and z. Then you can get $ 2a\plus{}d\plus{}h\equal{}39x\plus{}48y\plus{}45z$ and add 57 to get $ 2a\plus{}d\plus{}h\plus{}57\equal{}a\plus{}e\plus{}f\plus{}g\equal{}$ what you want.

@Sitan: not really.

@1077400: it's how I solve USAMO problems, but we should keep this a secret.


[quote="Interval"]NOTE: A = the hypotenuse.

The areas of two smaller semicircles are 18pi and 32pi square units.  Find A.

NOTE: There are 3 semicircles around a right triangle.[/quote]

[hide]This means the third semicircle is $50\pi$, so the radius of the circle is 10, and hence the hypotenuse is 20.

$20$[/hide]

doesnt $r^{2}=50$ so $r=5\sqrt{2}$...or am i misinterpreting something...

sorry i didnt know it was asking for area of the circle..

[quote="Interval"]Mathgeniuse,

I appreciate your help but please provide me with steps that teach how to find the answer on my own.[/quote]

My Bad.

In a right triangle, when three similar figures are placed on the sides of the triangle, the sum of the areas of the two smaller figures will equal the area of the larger figure.

I agree I messed up, because I found the hypotenuse.

The asnwer is just $18\pi+32\pi=50\pi$

$50\pi$

And both of the previous posters.

They are semicircles so the area of the semicircle will be $\frac{\pi r^{2}}{2}$

O right .. so ur solution before for the hypotenuse was also correct (since its a semi-circle not a whole circle)...darn i should read the question more carefully..

[quote="ckck"][hide]So we have two semi-circles around two legs of a right triangle. So the diameter of the circles is the length of the legs. Solving for the diameter you get $6\sqrt{2}$ and $8\sqrt{2}$. Then solving for the hypotenuse by using the 6-8-10 properties the length of the hypotenuse is $10\sqrt{2}$ and the radius of the last semi-circle is $5\sqrt{2}$. So the area of that semi-circle would be 50$\pi$.[/hide][/quote]

A remark; it's not necessary to find the side lengths.

Let the side lengths be $2a, 2b, 2c$ so that the semicircles have radii $a, b, c$ and areas $\frac{\pi a^{2}}{2}, \frac{\pi b^{2}}{2}, \frac{\pi c^{2}}{2}$.  Because

$4a^{2}+4b^{2}= 4c^{2}$

We can multiply by $\frac{\pi}{8}$ to give

$\frac{\pi a^{2}}{2}+\frac{\pi b^{2}}{2}= \frac{\pi c^{2}}{2}$

We are given the two smaller areas, which are $18 \pi, 32 \pi$, so the larger one is

$\frac{\pi c^{2}}{2}= 18 \pi+32 \pi = 50 \pi$.

[quote="nutz_for2.718281828"][quote="Interval"]NOTE: A = The area of one of the semicircles.

The areas of two smaller semicircles are 18pi and 32pi square units.  Find A.

NOTE: There are 3 semicircles around a right triangle.[/quote]

Please tell us: Is A the smallest area, or the biggest?[/quote]

He said smaller, but if he said the two smaller it would be less ambiguous.

So A is the biggest.

[quote="SM4RT"]he did say [b]the two smaller...[/b]
[quote="Interval"]
The areas of two smaller semicircles are 18pi and 32pi square units. Find A.
[/quote][/quote]

keyword: "[b]the[/b]"


These things always suggest geometrical interpretation, and I'm going to guess enneagons.?

meh general solution for these problems:

demoive's for a cos(nx) formula, replace all the sin(x)^2 with 1-cos(x)^2, not you have a polynomial in cos(x), set it equal to 1 to get your roots of cos(2kpi/n), then square root and vietas sums.  Its kindof long for n > 7 or so so I dont want to go through it.

It's very easy buy multiplying two hand by $\sin\frac{2\pi}{9}$ and continie expanding we get the rezult $-\frac{1}{2}$

Yummy Identities!
[quote="Wikipedia"]
\[1+2\cos(x)+2\cos(2x)+2\cos(3x)+\cdots+2\cos(nx) = \frac{ \sin\left(\left(n+\frac{1}{2}\right)x\right) }{ \sin(x/2) }\][/quote]

Plugging in our numbers gives \[A= \frac{1}{2}\left( \frac{ \sin \left( \frac{9 \pi}{9}\right) }{ \sin \left( \frac{ \pi}{9}\right) }-1 \right) = \boxed{-\frac{ 1}{2}}\].


For all positive numbers x, simplify

$\dfrac{1}{\log_3x}+\dfrac{1}{\log_4x}+\dfrac{1}{\log_5x}$

$x\neq1$
$\dfrac{1}{\log_3x}+\dfrac{1}{\log_4x}+\dfrac{1}{\log_5x}=\dfrac{\log3}{\log x}+\dfrac{\log4}{\log x}+\dfrac{\log5}{\log x}= \dfrac{\log3+\log4+\log5}{\log x}=\dfrac{\log60}{\log x}=\boxed{\log_{x}60}$

[quote="RobertN"]$x\neq1$
$\frac{1}{\log_3x}+\frac{1}{\log_4x}+\frac{1}{\log_5x}=\frac{\log3}{\log x}+\frac{\log4}{\log x}+\frac{\log5}{\log x}= \frac{\log3+\log4+\log5}{\log x}=\frac{\log60}{\log x}=\boxed{\log_{60}x}$[/quote]

Is it not supposed to be $\log_x 60$ instead of $\log_{60} x$.  For example, $\log_2 4=\log 4/\log 2$, not $\log 2/\log 4$. If we say that $\frac{\log60}{\log x}$, we take x as the base.  Therefore, $\frac{\log60}{\log x}=\log_x 60$.

EDIT: It seems that I can't see or you just editted it.
EDIT 2: Now I realize you editted it... :P

Masoud Zargar

[quote="4everwise"]For all positive numbers x, simplify

$\frac{1}{\log_3x}+\frac{1}{\log_4x}+\frac{1}{\log_5x}$[/quote]

[hide]$\frac{1}{\log_3x}+\frac{1}{\log_4x}+\frac{1}{\log_5x}$

${\log_x{3}}+{\log_x{4}}+{\log_x{5}}$

$\log_x{3*4*5}$

$\log_x(60)$[/hide]


[size=150]Let $ x,y,z > 0$,
$ f(t) \equal{} \frac {y \plus{} z}{(y \plus{} z)^2 \plus{} tx(x \plus{} y \plus{} z)} \plus{} \frac {z \plus{} x}{(z \plus{} x)^2 \plus{} ty(x \plus{} y \plus{} z)} \plus{} \frac {x \plus{} y}{(x \plus{} y)^2 \plus{} tz(x \plus{} y \plus{} z)} \minus{} \frac {18}{(4 \plus{} 3t)(x \plus{} y \plus{} z)}.$

if $ t \geq 2$,then:
$ f(t) \geq 0$

else
if   $ \frac {5}{3}\geq t\geq \frac {52}{57} \minus{} \frac {16}{57}\sqrt {7}$
then:

$ f(t) \le 0.$[/size]

BQ

[size=150]new problem,please give a proof.
BQ[/size]

[quote="xzlbq"][size=150]Let $ x,y,z > 0$,
$ f(t) \equal{} \frac {y \plus{} z}{(y \plus{} z)^2 \plus{} tx(x \plus{} y \plus{} z)} \plus{}$ $ \frac {z \plus{} x}{(z \plus{} x)^2 \plus{} ty(x \plus{} y \plus{} z)}$ $ \plus{} \frac {x \plus{} y}{(x \plus{} y)^2 \plus{} tz(x \plus{} y \plus{} z)} \minus{} \frac {18}{(4 \plus{} 3t)(x \plus{} y \plus{} z)}.$

if $ t \geq 2$,then:
$ f(t) \geq 0$

else
if   $ \frac {5}{3}\geq t\geq \frac {52}{57} \minus{} \frac {16}{57}\sqrt {7}$
then:

$ f(t) \le 0.$[/size]

BQ[/quote]

$ f(t) \equal{} \frac {y \plus{} z}{(y \plus{} z)^2 \plus{} tx(x \plus{} y \plus{} z)} \plus{} \frac {z \plus{} x}{(z \plus{} x)^2 \plus{} ty(x \plus{} y \plus{} z)}$ $ \plus{} \frac {x \plus{} y}{(x \plus{} y)^2 \plus{} tz(x \plus{} y \plus{} z)} \minus{} \frac {18}{(4 \plus{} 3t)(x \plus{} y \plus{} z)}.$

if $ t \geq 2$,then:
$ f(t) \geq 0$

$ f(t) \geq 0\Longleftrightarrow (6t^2 \minus{} 10t)\sum(x^4(x \minus{} y)(x \minus{} z)) \plus{}$ $ ( \minus{} 33t \plus{} 13t^2 \plus{} 3t^3 \plus{} 4)\sum(x^3(y \plus{} z)(x \minus{} y)(x \minus{} z))$

$ \plus{} (6 \minus{} 32t \minus{} 13t^2 \plus{} 15t^3)(x \minus{} y)^2(y \minus{} z)^2(z \minus{} x)^2 \plus{}$ $ (8 \minus{} 36t \minus{} 80t^2 \plus{} 48t^3)\sum(y^2z^2(x \minus{} y)(x \minus{} z))$

$ \plus{} ( \minus{} 146t \plus{} 12t^2 \plus{} 36t^3 \plus{} 24)xyz\sum(x(x \minus{} y)(x \minus{} z))$ $ \plus{} (12 \minus{} 63t \minus{} 36t^2 \plus{} 36t^3)xyz\sum((y \plus{} z)(x \minus{} y)(x \minus{} z))\geq 0$

easy prove   $ t\geq 2$

$ 6t^2 \minus{} 10t\ge 0$

$ \minus{} 33t \plus{} 13t^2 \plus{} 3t^3 \plus{} 4\geq 0$

$ 6 \minus{} 32t \minus{} 13t^2 \plus{} 15t^3\ge 0$

$ 8 \minus{} 36t \minus{} 80t^2 \plus{} 48t^3\geq 0$

$ \minus{} 146t \plus{} 12t^2 \plus{} 36t^3 \plus{} 24\ge 0$

$ 12 \minus{} 63t \minus{} 36t^2 \plus{} 36t^3\geq 0$

so  $ t \geq 2$
$ f(t) \geq 0$.

we can prove $ 0.1773381693...\le t \le \frac {5}{3}$

$ 6t^2 \minus{} 10t\le 0$

$ \minus{} 33t \plus{} 13t^2 \plus{} 3t^3 \plus{} 4\leq 0$

$ 6 \minus{} 32t \minus{} 13t^2 \plus{} 15t^3\le 0$

$ 8 \minus{} 36t \minus{} 80t^2 \plus{} 48t^3\leq 0$

$ \minus{} 146t \plus{} 12t^2 \plus{} 36t^3 \plus{} 24\le 0$

$ 12 \minus{} 63t \minus{} 36t^2 \plus{} 36t^3\leq 0$

so   $ 0.1773381693...\le t \le \frac {5}{3}$

$ f(t) \leq 0$

but

$ \frac {52}{57} \minus{} \frac {16}{57}\sqrt {7} \equal{} 0.1759049835...$.


Suppose $ a,b,c$ are distinct nonzero real numbers that satisfy $ a\plus{}\frac{1}{b}\equal{}b\plus{}\frac{1}{c}\equal{}c\plus{}\frac{1}{a}$. Find the value of $ |abc|$.

Let $ m\equal{}abc$
Then $ ma\plus{}ac\equal{}mb\plus{}ab\equal{}mc\plus{}cb$
$ \Leftrightarrow m\equal{}\frac{a(b\minus{}c)}{a\minus{}b}\equal{}\frac{b(c\minus{}a)}{b\minus{}c}\equal{}\frac{c(a\minus{}b)}{c\minus{}a}$
Then $ m^3\equal{}\frac{a(b\minus{}c)}{a\minus{}b}\frac{b(c\minus{}a)}{b\minus{}c}\frac{c(a\minus{}b)}{c\minus{}a}\equal{}abc\equal{}m$
Hence $ m^2\equal{}1\hspace{1cm}$ and $ |abc|\equal{}1$.


1. What are the elusive characters? Missing two letters or numbers?
WATE_ M_LON

2. What are the two missing numbers and why?
4 3 3 3 7 7 _ _

3. Can you figure out the missing letters in this sequence?
_ _ F O T M L I T S

4. This sequence is based on another very common one, as common as 1, 2, 3... What are the next two numbers?
3, 3, 5, 4, 4, 3, 5, _, _

[quote="mad_skillz_aops"]1. R and E[/quote]

Wrong. Think more laterally.

[hide="hint"]The question...[/hide]

[quote="mad_skillz_aops"]Oh, letters then.[/quote]

:rotfl: :rotfl: You take things too literally...

You do have to identify the particular letters, but the word doesn't spell WATER MELON.

Yeah, I'm just too literal.  Is the first letter U?

[hide="4"]
one two three...

number of letters in each word of 1,2,3.. written out.[/hide]

[quote="rem"][hide="4"]5 and 4: use the number of letters in words:
one, two, three, four, five, six, seven...
 :wink: [/hide]
[hide="3"]C and Y: use the beginning letters of words in the sentence :wink: [/hide]
[hide="2"]3 and 3: use the number of letters of words in the sentence :wink: [/hide]
[hide="1"]C and T: use the beginning letters of words in the sentences :wink: [/hide]
 :D Once you get one of them you will easily get the rest :wink:[/quote]
Oh wow, I am TOO literal.


Consider sequences of positive real numbers of the form $ x,2000,y,...,$ in which every term after the first is 1 less than the product of its two immediate neighbors. For how many different values of $ x$ does the term 2001 appear somewhere in the sequence?

$ \textbf{(A)} \ 1 \qquad \textbf{(B)} \ 2 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 4 \qquad \textbf{(E)} \ \text{more than 4}$

Totally misread it the first time.

[hide]

Ok, we start with a generic sequence with the given rules.

So we have . By the definition we know that , or . We wish to eliminate teh  variable in order to work with only two variables in hopes that there is some sort of recursion here (otherwise, the answer is E).

So we have . We can compute the fourth term by simply plugging it into the rule . Computing the fifth term, we find it is , and then we find the sixth term to be .

We now have the sequence . Since we are given  already, we only consider four possible distinct sequences, since anything after the fifth term is repeating, and a different sequence is determined by where the second term is in the sequence. We see that we only have four distinct terms to choose  to be, so there are only four distinct sequences containing . D. [/hide]

[quote="matt276eagles"]I tried to use a similar method, except I used 2000 instead of a variable and the terms only got more and more complicated.  Did I make a mistake?  I think it should have worked the same way... :?[/quote]

It should work the same way...

[hide]
The sequence goes $x$, $2000$, $y$, $\frac{y+1}{2000}$, $\frac{y+2001}{2000y}$, $\frac{2001}{y} = x$, $2000$, $y$, $\ldots$[/hide]

Ohh, I misunderstood what the problem was saying. :roll: That's why that method wouldn't work, at least for me... (I tried the same thing.)


Find $ limu_n:$
    $ u_n\equal{}\sum^n_{i\equal{}1}\frac{1}{i^2}$

You can find some solutions here: http://en.wikipedia.org/wiki/Basel_problem


Prove that the equation $ 3^n\plus{}1\equal{}2^m$ , where $ n\geq 2$ has no solution in $ \mathbb{N}$.

I know that this is true by [url=http://en.wikipedia.org/wiki/Mih%C4%83ilescu%27s_theorem]Mihailescu's Theorem[/url], but I was wondering if there is an elementary way to show this.

We note that 3´n+1=(2,4)mod 8. but 2`m=0mod 8 for m>2. If n>=2, 3`n>=9----->m>2. Hence this equation has no solutions in N for n>=2


inequalities
three variable inequality
calculus

Prove that if x, y, z > 0, then:
\sum (y+z)/x \geq 4\sum x/(y+z).

Let's write AM-HM inequality: 2/(1/y+1/z) \leq (y+z)/2 =>
4/(y+z) \leq 1/y+1/z, so

\sum 4x/(y+z) \leq \sum x(1/y+1/z) = x/y+y/x+z/x+x/z+y/z+z/y = \sum(y+z)/x

Well,I don't understand what you wrote,but i think you are talking about Levinson inequality:

        $\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\geq4(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y})$. ;)

[quote="Gil"]Prove that if x, y, z > 0, then:
$ \sum (y+z)/x \geq 4\sum x/(y+z).[/quote] $

[b]Generalization:[/b]
For $x,y,z,x',y',z'  \in \mathbb{R}^+$. We have:$\frac{y'+z'}{x}+\frac{z'+x'}{y}+\frac{x'+y'}{z} \ge 4 (\frac{x'}{y+z}+\frac{y'}{z+x}+\frac{z'}{x+y})$
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=56207&hilit=Inequality]here[/url]

Let $x,y$ and $z$ be positive real numbers. Prove that

        $$\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\geq4\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)+1-\frac{8xyz}{(x+y)(y+z)(z+x)}$$(Cezar Lupu and Cosmin Pohoata)

If $ x, y$ and $z$ are positive numbers, prove that$$
\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}\ge 2(x+y+z)(\frac{1}{ x}+\frac{1}{y }+\frac{1}{z }).$$
([size=50]WISCONSIN MATHEMATICS, SCIENCE & ENGINEERING TALENT SEARCH PROBLEM SET III (2011-2012) DECEMBER 2011[/size])
[size=75][/size]
 https://www.math.wisc.edu/talent/?q=problem-sets
For $ x, y, z > 0 $ real numbers, [url=https://artofproblemsolving.com/community/c6h1317330p7082918]prove that:[/url]
\\
\\
$ \sum_{cyc} \sqrt{ \frac{y+z}{x} } \le \sqrt{ 2(x+y+z) \left( \frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) } \ \ ; $


Mr. C will be running the Math Seminar again this year, at Manzano HS, most Saturdays 1130-1300.  We'll cover math contests, mostly the AMC->12 and AIME, during math contest season.  The general topic for this year will be applications of Group Theory.

Students should have had (or be taking) honors Alg. II, but an interest in mathematics is more important.

I know this is late, but is it still going on?

Yes; we will resume the 2nd Saturday in January.


integration
trigonometry
limit
calculus
calculus computations

Can we compute $ \int_{0}^{\infty} \frac{\sin x}{x} dx$

Do you know complex analysis?
[hide="Hint"]$ \oint \frac{e^{iz}}{z}dz\equal{}0$[/hide]

no i don't know it . is it needed for this problem sir ??

For non-negative $ \alpha ,\beta$

$ \int_{0}^{\infty }e^{\minus{}\alpha x}\cos \beta xdx\equal{}$ $ \frac{\alpha }{\alpha ^{2}\plus{}\beta ^{2}}$

$ \int_{0}^{1}\left( \int_{0}^{\infty }e^{\minus{}\alpha x}\cos \beta xdx\right) d\beta \equal{}$ $ \int_{0}^{1}\frac{\alpha }{\alpha ^{2}\plus{}\beta ^{2}}d\beta$

$ \int_{0}^{\infty }e^{\minus{}\alpha x}\left( \int_{0}^{1}\cos \beta xd\beta \right) dx\equal{}$ $ \tan ^{\minus{}1}\frac{1}{\alpha }$

$ \int_{0}^{\infty }e^{\minus{}\alpha x}\frac{\sin x}{x}dx\equal{}$ $ \tan ^{\minus{}1}\frac{1}{\alpha }$

$ \int_{0}^{\infty }\frac{\sin x}{x}dx\equal{}\lim_{\alpha \rightarrow 0\plus{}}\int_{0}^{\infty }e^{\minus{}\alpha x}\frac{\sin x}{x}dx\equal{}\frac{\pi }{2}$

A possible elementary approach is presented [url=http://www.math.binghamton.edu/paul/papers/LoyaMathMag.pdf]here[/url].

cf.
Complex Variables
http://www.scribd.com/doc/14666189/Complex-Variables
Example 7.9

Put $ J\left( \lambda \right) \equal{} \int_0^{ \plus{} \infty } {e^{ \minus{} \lambda x} \frac {{senx}} {x}dx}$

Note that $ J\left( 0 \right) \equal{} \int_0^{ \plus{} \infty } {\frac {{senx}} {x}dx}$

Then $ \underbrace \Rightarrow _{Leibniz}J'\left( \lambda \right) \equal{} \minus{} \int_0^{ \plus{} \infty } {e^{ \minus{} \lambda x} senxdx}$ $ \equal{} \minus{} \Im \int_0^{ \plus{} \infty } {e^{ \minus{} \lambda x} e^{ix} dx} \equal{} \minus{} \Im \int_0^{ \plus{} \infty } {e^{ \minus{} x\left( {\lambda \minus{} i} \right)} dx} \equal{} \frac {{ \minus{} 1}} {{\lambda ^2 \plus{} 1}}$

$ \Rightarrow J\left( {\mathop {\lim }\limits_{\lambda \to \plus{} \infty } \lambda } \right) \minus{} J\left( 0 \right) \equal{} \minus{} \int_0^{ \plus{} \infty } {\frac {{d\lambda }} {{\lambda ^2 \plus{} 1}}}$

Finally, $ \int_0^{ \plus{} \infty } {\frac {{senx}} {x}dx} \equal{} \frac {\pi } {2}$

Bye! :)


An old car has to travel a 2-mile route, uphill and down. Because it is so old, the car can climb the first mile - the ascent - no faster than an average speed of 15 mph. How fast does the car have to travel on the second mile inorder to achieve an average speed of 30 mph for the trip?

I don't know if this is right, but...
[hide]the car ascends the one-mile at 15 mph. So it takes it 4 minutes. In order to average 30 mph, his average mile has to take 2 minutes. So the overall trip must take 4 minutes. But the car has already used up the 4 minutes in his ascent. So it is impossible. (unless it uses teleport :D )[/hide]

[hide]Not possible... assuming that the answer isnt something like he could screw up the spacetime and thus actually he was travelling 100 mph to such and such reference point... but assuming standard conditions not possible.[/hide]

[hide="urhg"]
Has to travel infinitely fast =p .
[/hide]

Not infinitely fast, only the speed of light!

[hide]we need to pay attention that the way it goes down, its speed go up...[/hide]
I don't want to spam, but, what is Newton's Sums?
I know it's was on this forum a lot of times, but the search gives nothing  :(  tnx

[hide]
Use the harmonic mean, and find that theres no solution. It's impossible!
[/hide]

[hide]It's impossible to do so.  In order to be able to do that, you'd have to travel infinitely fast so the time of descent is 0.  Since $D=rt$ (distance=rate x time), $r=\frac{D}{t}$.  Since the time must be 0, We have $r=\frac{D}0$.  And according to the Ten Commandments of Mathematics, we can't do that, because "thou shalt not divide by zero".
*insert Chuck Norris joke here*[/hide]

[hide]I used harmonic mean and I found out that it is impossible.[/hide]


Prove that if x,y,z>0 and x+y+z=1, then:
   sum xy/(1-x^2y^2)<7/20.

I would like to see your solution, Harazi.

Thank you very much.

My solution ia based on one of my favourites inequalities:
  If a,b,c>=0 have sum 1, then for all n>=2 we have (ab)^n+(bc)^n+(ca)^n<=1/4^n. Now, sum xy/(1-x^2*y^2)=sum xy(1+(xy)^2+(xy)^4+...)=
sum xy+ sum (xy)^3+sum (xy)^5+...<=1/3+1/4^3+1/4^5+...=7/20.

Very nice solution for a very hard inequality.

Thank you very much Harazi.

I don't know a proof for "one of your favourite inequalities" you mentioned
above.
Is it a classical inequality?
Have you a simple proof for it?

Thank you very much.

Well, it is my favourite inequality since I " discovered" it. It may appear somewhere, but I didn't see it. It has a very simple proof:
   Let a>=b>=c. Then 1/4^n>=a^n*(b+c)^n>=a^n*b^n+a^n*c^n+a^n*b^n-1*c>=a^n*b^n+b^n*c^n+c^n*a^n. Don't you remember I have also proved it in a post of mine called: Simetry is sometimes tricky. The problem was max (ab)^n/(1-ab)+(bc)^n/(1-bc)+(ca)^n/(1-ac) when a+b+c=1.

You are right, now I remember it.

Thank you, harazi.

sorry , i don't understand this part.

a^n*(b+c)^n>=a^n*b^n+a^n*c^n+a^n*b^n-1*c>=a^n*b^n+b^n*c^n+c^n*a^n.

what does a^n*b^n-1*c mean?

a^n*b^n-1*c^n means a^n*b^(n-1)*c^n, where a^k means a*a*...*a. And it comes from the binomial expansion.


modular arithmetic
number theory
greatest common divisor
number theory unsolved

If $ m,n$ are positive integers such that the number $ 6m^2 n^2 \minus{}m^4 \minus{}n^4$ is a perfect square, prove that $ m\equal{}n$

$ 6m^{2}n^{2}\minus{}m^{4}\minus{}n^{2}\equal{}(2mn)^{2}\minus{}(m^{2}\minus{}n^{2})^{2}\equal{}(m^2\plus{}2mn\minus{}n^2)(n^2\plus{}2mn\minus{}m^2)\equal{}p^{2}$
$ (m,n)\equal{}d$ ,$ m\equal{}dx,n\equal{}dy$  $ \implies$  $ (x,y)\equal{}1$ and $ (x^2\plus{}2xy\minus{}y^2)(y^2\plus{}2xy\minus{}x^2)\equal{}q^{2}$
$ t|x^2\plus{}2xy\minus{}y^2,t|y^2\plus{}2xy\minus{}x^2$  $ \implies$  $ t|4$
$ i)t\equal{}1,4$
$ \implies$  $ x^2\plus{}2xy\minus{}y^2\equal{}u^2,y^2\plus{}2xy\minus{}x^2\equal{}v^2$
$ \implies$  $ 4xy\equal{}u^2\plus{}v^2$  $ \implies$  $ 2|u,2|v$
$ \implies$  $ 4|x^2\plus{}2xy\minus{}y^2$ ,$ 4|y^2\plus{}2xy\minus{}x^2$
$ \implies$  $ 2|x,2|y$  $ \implies$  $ x\equal{}2l,y\equal{}2k$
$ l^2\plus{}2kl\minus{}k^2$ and $ k^2\plus{}2kl\minus{}l^2$  must be perfect square(infinite descent)
$ ii)t\equal{}2$
$ \implies$  $ x^2\plus{}2xy\minus{}y^2\equal{}2u^2,y^2\plus{}2xy\minus{}x^2\equal{}2v^2$
$ 2\parallel{}x^2\plus{}2xy\minus{}y^2$  or $ 2\parallel{}y^2\plus{}2xy\minus{}x^2$
if $ 2\parallel{}x^2\plus{}2xy\minus{}y^2$  $ \implies$ $ x^2\plus{}2xy\minus{}y^2\equiv 2\pmod{4}$
$ (x\plus{}y)^2\equiv 2y^2\plus{}2\pmod{4}$  $ \implies$  $ y^{2}\equiv 1\pmod{4}$  and $ x^{2}\equiv 1\pmod{4}$
$ \Delta\equal{}4y^2\plus{}4(y^2\minus{}2u^2)\equal{}8(y^2\minus{}u^2)$ $ \implies$  $ (y\minus{}u)(y\plus{}u)\equal{}2n^{2}$
similarly $ (x\minus{}v)(x\plus{}v)\equal{}2m^2$  $ \implies$  $ 8|y^2\minus{}u^2$ and $ 8|x^2\minus{}v^2$
$ x^{2}\equiv y^{2}\equiv u^{2}\equiv v^{2}\equiv 1\pmod{8}$
I didn't solved this part,Help me please :blush:

sorry but i dont understand, if $ m\equal{}n$, then $ 6m^{2}n^{2}\minus{}m^{4}\minus{}n^{4} \equal{}6m^4\equal{}x^2$ for some $ x$ according to to question, which is absurd.

[quote="phymax"]sorry but i dont understand, if $ m \equal{} n$, then $ 6m^{2}n^{2} \minus{} m^{4} \minus{} n^{4} \equal{} 6m^4 \equal{} x^2$ for some $ x$ according to to question, which is absurd.[/quote]
err...the question asks us to prove that if $ 6m^{2}n^{2} \minus{}m^{4}\minus{}n^{4}$ is a perfect square then $ m\equal{}n$ ,it doesnt say that if $ m\equal{}n$ $ 6m^{2}n^{2} \minus{} m^{4} \minus{}n^{4}$ is a perfect square . Sorry if I am wrong .

[quote="phymax"]sorry but i dont understand, if $ m \equal{} n$, then $ 6m^{2}n^{2} \minus{} m^{4} \minus{} n^{4} \equal{} 6m^4 \equal{} x^2$ for some $ x$ according to to question, which is absurd.[/quote]

if $ m\equal{}n$ then $ 6m^2n^2\minus{}m^4\minus{}n^4\equal{}4m^4$ which is a perfect square  it's not $ 6m^4$  :wink:

we can assume that $ \gcd(m,n)\equal{}1$.
now we want to prove that $ m\equal{}n\equal{}1$ is the only solution.
we have
$ 6m^2n^2\minus{}m^4\minus{}n^4\equal{}z^2 \implies (2mn)^2\minus{}(m^2\minus{}n^2)^2\equal{}z^2 \implies (2mn\minus{}m^2\plus{}n^2)(2mn\plus{}m^2\minus{}n^2)\equal{}z^2$.
now the gcd of the two expressions in the left side of the equality is obviously $ 1$ or $ 2$.
if it is one (it means that $ m,n$ don't have the same parity) then we must have :
$ 2mn\minus{}m^2\plus{}n^2\equal{}t^2,2mn\plus{}m^2\minus{}n^2\equal{}s^2,ts\equal{}1$.
but we see that $ t^4\plus{}s^4\equal{}2(m^2\plus{}n^2)^2\equal{}2l^2$.
but u can easily prove by infinite descent that this equation(i mean $ t^4\plus{}s^4\equal{}2l^2$) has no solutions other than $ t\equal{}s\equal{}1$(note that $ (s,t)\equal{}1$).
now let's check the second case,we have
$ \gcd(2mn\minus{}m^2\plus{}n^2,2mn\plus{}m^2\minus{}n^2)\equal{}2$
which implies
$ 2mn\minus{}m^2\plus{}n^2\equal{}2t^2,2mn\plus{}m^2\minus{}n^2\equal{}2s^2,2st\equal{}z$.
now we have $ 4t^4\plus{}4s^4\equal{}2(m^2\plus{}n^2)^2\equal{}2l^2$.
which implies that $ 2t^4\plus{}2s^4\equal{}l^2 \implies 2 | l,2l'\equal{}l \implies t^4\plus{}s^4\equal{}2l'^2$.
so we have again the same equation and we know that the only answer is $ t\equal{}s\equal{}1$.
which implies $ 2mn\minus{}m^2\plus{}n^2\equal{}2mn\plus{}m^2\minus{}n^2 \implies m\equal{}n$.
we r done!

i think u have to prove that gcd(m,n)=1.... can you just assume it . imean it took me a while to prove it.

[quote="Abdek"][quote="phymax"]sorry but i dont understand, if $ m \equal{} n$, then $ 6m^{2}n^{2} \minus{} m^{4} \minus{} n^{4} \equal{} 6m^4 \equal{} x^2$ for some $ x$ according to to question, which is absurd.[/quote]

if $ m \equal{} n$ then $ 6m^2n^2 \minus{} m^4 \minus{} n^4 \equal{} 4m^4$ which is a perfect square  it's not $ 6m^4$  :wink:[/quote]

eh.. :oops:
was out of my mind


induction
limit
algebra unsolved
algebra

sequence $ (u_n)$ defined by$ u_1 \equal{} 2009, u_{n \plus{} 1} \equal{} \frac {2n \plus{} 1}{3n \plus{} 1}(u_n \plus{} 1)$ , $ n \ge 1$

compute Lim $ U_n$

Did you change the problem? I saw something here a while ago including pi.

Take $ u_n \equal{} x_n \plus{} 2$; then $ x_1 \equal{} 2007$ and $ x_{n\plus{}1} \equal{} \frac {2n\plus{}1} {3n\plus{}1} x_n \plus{} \frac {1} {3n\plus{}1}$.

Let us prove $ x_{n} > \frac {1} {n}$. This is clear for $ n\equal{}1$, and by induction $ x_{n\plus{}1} \minus{} \frac {1} {n\plus{}1} \equal{} \frac {2n\plus{}1} {3n\plus{}1} x_n \plus{} \frac {1} {3n\plus{}1} \minus{} \frac {1} {n\plus{}1} > \frac {2n\plus{}1} {3n\plus{}1} \cdot \frac {1} {n} \plus{} \frac {1} {3n\plus{}1} \minus{} \frac {1} {n\plus{}1} \equal{} \frac {1} {n(n\plus{}1)} > 0$.

Now, $ x_{n\plus{}1} \minus{}x_n \equal{} \minus{}\frac {nx_n \minus{} 1} {3n\plus{}1} < 0$, so $ (x_n)_{n\geq 1}$ is decreasing and lower bounded by $ 0$, hence convergent to some limit $ \ell \geq 0$.

Taking the limit in the recurrence relation yields $ \ell \equal{} \lim_{n \to \infty} x_{n\plus{}1} \equal{} \lim_{n \to \infty} \left (\frac {2n\plus{}1} {3n\plus{}1} x_n \plus{} \frac {1} {3n\plus{}1} \right ) \equal{} \frac {2} {3} \ell$, whence $ \ell \equal{} 0$. Therefore $ \lim_{n \to \infty} u_{n} \equal{} 2$.


search
real analysis
real analysis unsolved

Hello everyone,
Given a set $ S$, a subset $ A$ of $ P(S)$ (where $ P(S)$ denotes the power set of $ S$) is called an [i]antichain[/i] if no element of $ A$ is a subset of any other element of $ A$. Does $ P(N)$, the power set of the natural numbers, contain an uncountable antichain?

I can't understand what do you want to say.
[quote]an antichain if no element of A is a subset of any other element of A[/quote]

Of course.
First notice that if we restrict ourselves to only infinite subsets of $ \mathbb{N}$ we still have uncountably many of them. (Why?)
next consider the correspondence $ \{a_i\}_{i \equal{} 1}^{\infty} \rightarrow \{\prod_{j \equal{} 1}^i p_{a_j}\}_{i \equal{} 1}^{\infty}$ where $ p_{k}$ denotes the $ k$'th prime.
after this transformation this uncountable set is easily seen to be an antichain in $ \mathcal{P}(\mathbb{N})$

Or do a search for the words "uncountable antichain" to find the simple example posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18439]here[/url].  (Uncountable chains are more popular, apparently: see for example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276788]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159144]this[/url].)

thanks all! i had been stuck on this for a long time haha :blush:


symmetry
number theory proposed
number theory

Suppose that $k, l$ are positive integers such that $(11m-1, k) = (11m-1, l)$
for every $m \in \mathbb{N}$. Prove that there exists an integer $n$ such that $k = 11^nl$.

Let be $p^n$ any prime power ($p \neq 11$) that divides $k$. Then there is an $m$ with $11 m \equiv 1 \mod p^n$, since there is an inverse of $11$. But then $p^n | (11m-1,k) = (11m-1,l) \implies p^n |l$ and by symmetry it follows that $k,l$ have the same prime powers dividing them, except powers of $11$. But this is the desired result.


A tub can be filled in 15 minutes by a hot water faucet and in 12 minutes by the cold water faucet. The drain can empty that same tub in 20 minutes. How many minutes will it take to fill the tub if both faucets are turned on and the drain is open?

[quote="7h3.D3m0n.117"][quote="max1546"][hide="answer"]hot water faucet=$x$, cold water faucet=$y$, drain=$z$
tub's total volume=$1$. time taken=$t$
therefore $x=1/15$, $y=1/12$, $z=1/20$
${1/15+1/12-1/20}*t=1$
${{4+5-3}*t=60}$
$6t=60$
Therefore $t=10minutes$[/hide][/quote]

hmm making it easier to read use the

[code]
$\frac{x}{y}$
[/code]

LaTeX[/quote]
Also, you need to use the actual sign for parentheses () in LaTeX, not just brackets {}.

sorry just learnt latex about a week ago :P

Another thing that will help you with LaTeX is this site: [url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]


Has there been an occasion where all the prizes ran out before everyone got to choose at least one?

I don't believe so. They usually make it so that everyone (yes, including honorable mentions) receives at least one prize, I think...

Wait, really? Awesome. I think i'll get bronze, but I might get 58 or 59.

I guess it's that the gold and silver winners get to pick two prizes while bronze and honorable mention pick 1. The gold winners have their prize selection opportunity open first (yes, they get more choices), then the silver winners get to pick from what's left, followed by bronze, and then honorable mention. But all winners from all 4 categories get a t-shirt.

DP sorry

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

What happened on AIMEII dude?
yay i get mathematica, 2 olympiad books (!), and a T-shirt

Last year, if I remember correctly, it was gold 3, silver/bronze 2, HM 1.

[quote="123456789"][quote="mathemonster"]What happened on AIMEII dude? [/quote] I really really screwed up...big time. Totally ruined my year.[/quote]
just from a math standpoint or in general?


Solve the equation:
 $ \sqrt{13x^2\plus{}8x\plus{}5}\plus{}\sqrt{29x^2\minus{}24x\plus{}5}\equal{}2\sqrt{12x^2\plus{}4x\minus{}1}$

Xantos C. Guin's solution is my solution :)


I'm sure those of you who are interested in the contest are starting to wonder how to register. (In fact, this post was prompted by one of you who had that very question.) In fact, we will have an online registration form available at some point in the reasonably near future. This form will have detailed instructions regarding registration and payment.

Feel free to ask any questions; we'll do our best to clear things up. :)

It will be within a few days (like 1 or 2). Everything's ready; we just have to get the form online and stuff.

Sorry for the delays. The registration form is now available online:
http://www.math.princeton.edu/~plazonic/TST/

Please PM or email me any questions.

Please see [url=https://cgi.math.princeton.edu/mathclub/index.php/Registration_Guidelines]our website[/url] for registration guidelines before you submit the form.

2007 registration will be handled with an all-new, sophisticated database system. Please go to our website, https://cgi.math.princeton.edu/mathclub/index.php/Princeton_Math_Competition for instructions.


Find all positive integer such that $ n^3\minus{}8n\plus{}13$ is a perfect square.

[quote="basketball9"]The above link doesn't really provide a solution but only an answer. Can someone please provide a solution.[/quote]

When a link to an older post on the same topic is given, the rule is : please follow on in the previous thread, so that there will be a unique thread later in forum and archive.

If you have comments, just post in the older thread. It will come on the the top of active threads.

We would like to know the source of your problem. so that we can guage the extent of mathematical background required to attack the problem. like wether it can be solved using IMO level (High school olympiad level) mathematics. or wether it requires the use of elliptic curves and their properties.

and if you have solved it, we (as in we mathlinkers) will be grateful if you could take some time in posting the solution.

[quote="pco"]
please follow on in the previous thread, so that there will be a unique thread later in forum and archive.
[/quote]

@pco

im sure you meant this only as a little advice, but ironically this does happen to be te earlier thread.
anyway the thread in the link is more active. im posting this post here as it is adressed to kplg

by the way pco, could you help us solve this problem?

I saw this problem times and times here,in M.L.

but I didn't find any solutions for it.

By [url=http://www.wolframalpha.com]Wolfram|Alpha[/url] we get [url=http://www.wolframalpha.com/input/?i=n%5E3-8n%2B13%3Dm%5E2]This[/url].which means the only solution is $ n\equal{}3$.

Could you please help me for a complete solution?? thanks.

I am a real beginner in problems of this kind. I just pose the problem as I think it should be a problem.

@kplg
You mean that this is your own problem!!! if so then please ascertain that you have a solution before posting. or else post this in the algebra open questions forum

you may have also meant that you have picked this from someone who may have told you this problem or from another post. if so, please do find out the source(if someone has told you ask him/her for it, if it is another math-links post, please post the link), and post it. it will definitely be of great help.


function
geometry
geometric transformation
algebra
polynomial
calculus
calculus computations

Let $ a$ be a real constant.Show that the equation $ x^3\minus{}3x\plus{}a\equal{}0$ has just one real solution in $ ]\minus{}1,1[$

Since the extremes are at $ x\equal{}\pm 1$ and the function is continuous, we know $ f$ can have a certain value between $ f(\minus{}1)$ and $ f(1)$ only for one $ x \in ]\minus{}1,1[$, so there is only one such that $ f(x)\equal{}0$.

Rolle's Theorem says that between any two distinct zeros of the function must lie at least one zero of the derivative.

If $ f(x)\equal{}x^3\minus{}3x\plus{}a,$ then how many zeros does $ f'(x)$ have in the open interval $ (\minus{}1,1)?$

if we do $ f'(x)\equal{}0, f'(x)\equal{}3(x^2\minus{}1)\equal{}0 \Rightarrow x\equal{}\pm 1$ but $ f(x)$ is not defined in $ \pm 1$ so it doesn't even has one solution in $ ]\minus{}1,1[$ , I'm $ \emph{confused}$

$ f(x)$ is defined everywhere - it's a polynomial, after all.

If $ f(x)\equal{}0$ did have two (or more) roots in the closed interval $ [\minus{}1,1],$ then by Rolle, $ f'(x)\equal{}0$ would have a root in the open interval $ (\minus{}1,1).$ But since $ f'(x)\equal{}0$ has no roots in that open interval, we conclude that $ f(x)\equal{}0$ cannot have more than one root in the closed interval.

And that's really all we can conclude: that $ f(x)\equal{}0$ has [i]at most[/i] one root in $ [0,1].$ Without information on $ a,$ we don't know whether or not it has that one.

General principles for counting solutions of $ f(x)\equal{}0$ for "nice" functions $ f:$

To show that this has [i]at most[/i] a certain number of roots, use principles related to the Mean Value Theorem: Rolle, MVT, statements that the function is monotone. (milin was using the last one - he was using that the function is decreasing on $ [\minus{}1,1].$ That's not all that different theoretically from the argument I was making - it rests on the same principles.)

To show that this has [i]at least[/i] a certain number of roots, use the Intermediate Value Theorem.

Because of the behavior as $ x\to\infty$ and $ x\to\minus{}\infty,$ we can say that $ f(x)\equal{}0$ has at least one real root. (We can say that about any real cubic polynomial.) But we don't know whether or not it has a root in $ [\minus{}1,1].$


limit
function
calculus
calculus computations

[quote="mhar_teens"]can anyone help me to find this:
$ \lim_{n\to \plus{} \infty} \frac {1}{2} \plus{} \frac {2}{2^2} \plus{} \frac {3}{2^3} \plus{} \frac {4}{2^4} \plus{} ... \plus{} \frac {n}{2^n}$

thanks before :)[/quote]

[u][b][color=red][size=150]Hint[/size][/color][/b][/u]

[color=darkblue]Let $ f(x) \equal{} x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n$

We get $ f'(x) \equal{} 1 \plus{} 2x \plus{} 3x^2 \plus{} ... \plus{} nx^{n \minus{} 1}$

So $ f'\left(\frac {1}{2}\right) \equal{} 1 \plus{} 2.\frac {1}{2} \plus{} 3.\frac {1}{2^2} \plus{} ... \plus{} n.\frac {1}{2^{n \minus{} 1}}$

$ \Longrightarrow \frac {1}{2}f'\left(\frac {1}{2}\right) \equal{} \frac {1}{2} \plus{} 2.\frac {1}{2^2} \plus{} 3.\frac {1}{2^3} \plus{} ... \plus{} n.\frac {1}{2^{n}}$

But $ f(x) \equal{} x \plus{} x^2 \plus{} x^3 \plus{} ... \plus{} x^n \equal{} \frac {x(x^n \minus{} 1)}{x \minus{} 1}$

Then easy we get $ \frac {1}{2}f'\left(\frac {1}{2}\right)$ from $ f(x) \equal{} \frac {x(x^n \minus{} 1)}{x \minus{} 1}$.

Hence problem is easy to done.[/color]

[quote="mhar_teens"]can anyone help me to find this:
$ \lim_{n\to \plus{} \infty} \frac {1}{2} \plus{} \frac {2}{2^2} \plus{} \frac {3}{2^3} \plus{} \frac {4}{2^4} \plus{} ... \plus{} \frac {n}{2^n}$

thanks before :)[/quote]

It's easy to see that the limit exists.

Then let the limit be $ x$, so $ x\minus{}\frac{x}{2}\equal{}\frac{1}{2}\plus{}\frac{1}{2^2}\plus{}\frac{1}{2^3}\plus{}....\equal{}1$

So the limit $ x\equal{}2$

Hi :) !!!

other way:

Let $ f$ analatyc function definite by $ \forall |x| < 2$ $ f(x) \equal{} \sum_{n \geq 1} \frac{x^n}{2^n}$

than $ f$ is derivable $ \forall |x| <2$ and $ f(x) \equal{} \frac{x}{2\minus{}x}$

than $ f'(x)\equal{} \sum_{n \geq 1}n \frac{x^{n\minus{}1}}{2^n}$ also $ f'(x) \equal{} \frac{2}{(2\minus{}x)^2}$

finaly : $ \sum_{n \geq 1} \frac{n}{2^n} \equal{} f'(1) \equal{} 2$

and thanks :)


trigonometry
inequalities
inequalities solved

[quote="quantum leap"]Show that, for all positive numbers X,Y and $\alpha$,

$X^{\sin^2\alpha}Y^{\cos^2\alpha}<X+Y$[/quote]

hope following is true:

$X^{\sin^2\alpha}Y^{\cos^2\alpha}\le ({\sin^2\alpha}) X+({\cos^2\alpha}) Y<X+Y$

 :)

It's easy.You only use that sin^2a+cos^2b=1.


After showing $x$ exists, use $x^{2}-6=x$.

if you square 3 and subtract 6, you always get 3, so repeating that will keep the answer as 3

set \[ \sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...}}}}= x \]
so it becomes \[ \sqrt{6+x}= 3 \]
square both sides got 6+x=9 thus x=3


conics
parabola
algebra unsolved
algebra

A parabola with the equation y= ax² + bx + c, a>0 touches the two parabolas p1 and p2 with the equations y= -x²+(b1)x + (c1) and y= -x² + (b2)x + (c2) at the points A and B. One has to show that the common tangent of the parabolas p1 and p2 is parallel to the straight line AB.

Could you please please help me on this one? I am totally stuck. Thank you!!! :)

[b]This is a problem from the current [u]running[/u] 1. round of the German Mathematical Olympiad![/b]

[b]Please don't answer here.
Moderators: Please lock it![/b]


Let $f(x)=x^4+2x^2-4x+8.$

(1) Find a pair of integers $t,p,q$ such that $(x^2+t)^2-f(x)=(px+q)^2$ is an identity with respect to $x.$

(2) By solving the equation $(x^2+t)^2=(px+q)^2$ in terms of the values of $t,p,q$ obtained in $(1),$ find all solutions of the equation $f(x)=0.$

[hide="1"] By expanding and setting coefficients equal, $2t-2=p^2$, $2=pq$, and $t^2-8=q^2$.

$p$ and $q$ are either both positive or both negative, and it won't matter which, so WLOG assume both positive. Since $p$ is then positive, is even from the first equation, and divides 2 from the second equation, $p=2$, which gives $t=3$ and $q=1$.[/hide]

[hide="2"] $f(x)=0\Rightarrow (x^2+3)^2=(2x+1)^2\Rightarrow (x^2+2x+4)(x^2-2x+2)=0$. Using quadratic formula on both these factors, the roots are $\boxed{-1\pm i\sqrt{3},\ 1\pm i}$.
[/hide]

Yeah I got the same as scorpius, except that I solved the first problem algebraically.
Once you have the three equations defining p,q, and t, then you can multipy the expression for q^2 and p^2 together, and compare this to the square of 2pq, giving an easily solved equation in t. Then you plug in and go for the other values.

[quote="demji"]doesn't pq=-4?[/quote]

Setting the $x$ coefficients equal gives $-(-4)=2pq$. This reduces to $pq=2$ after simplifying.


While browsing through the AOPS forums, I've become aware some unexplained recurrences in the viewing habits the the AOPS community. Here is my question:

Why do many similar topics on the same boards have radically different numbers of viewings?

For example, on the advanced "proposed/own problems" geometry board there are a pair of topics by Omid Hatami created on the same exact day (Sept. 18, 2004), whose titles vary by a single character ("Easy; Iran 2004" as opposed to "Easy; Iran 2002"), and one of them has nearly twice as many views as the other! Not to mention that the "hotter" topic was posted later in the day.
Or, look up at the top of the board at a pair of adjacent topics by grobber. Why does "projections which lie on a circle; Dan Barbilian (famous Romanian mathematician)" have 102 views whereas "ruler construction; G.M. - no. 5 - 1980" has just 10?

These seemingly arbitrary patterns appear in every part of the AOPS forums I have seen, which, granted, is not the forums' entirety, but it is still a good chunk. Does anyone have some insight into this?

p.s. Sorry if the numbers are a bit off- they were accurate at the time of this posting.

All the stuff is weird. I've always wondered, why do topics often die right after I post? I swear, I've killed at least 40 topics. Maybe because I'm from Area 51...  :D

Anyways, who knows? Maybe its the supernatural. Or just some people like to look at a post with more responses than one with less.

[quote] Anyways, who knows? Maybe its the supernatural. Or just some people like to look at a post with more responses than one with less. [/quote]

I doubt that this is a major factor, because how would you explain a recent topic that has more posts than one that's been around awhile?

Because the ones that have been around for a while tend to be lower on the forum listing without recent posts, resulting in a decreased view count.

But they were at the top at some point, and the later topics didn't even exist then.

[quote="RC-7th"]Maybe the were posted right before a time when lots of people went online, and therefore made new threads. Or perhaps if one person thinks a topic is interesting, so will many other people. Then maybe if you have a choice between two topics, a person who has not previously looked at any of them will choose the one which seems more popular.[/quote]

ahh yes, "marketplace theory" at work.

Shouldn't the later topic have more attention? If I go onto a forum and just look at the new topics and don't see anything that stands out, I'd click the first one. That way, the later one got more attention.

Anyways, if you're noticing things like this, you've got waaaaaaay too much free time. Go play a sport or get a nice hobby. I personally suggest running or baseball.

[quote="ZennyK"][quote]Anyways, if you're noticing things like this, you've got waaaaaaay too much free time. Go play a sport or get a nice hobby. I personally suggest running or baseball.[/quote]

Mathematics is, of course, my favorite pass-time, but I also do track and field (javelin) in the spring and I'm taking foil fencing classes right now. I play 7 instruments and am learning another one, and I hold a world record.

I think I already have plenty of nice hobbies... ;) I just notice these things as I'm looking for interesting topics on here.[/quote]

Hehe, I know what you mean. I notice all kinds of crazy stuff, just doing normal, erm semi-normal stuff.

Wow, 7 instruments! Is that counting stuff like singing, the kazoo, the sounds you can make with your armpits, etc.?

Question is, how well do you know them?  Are you a Renaissance man (woman?) or a Jack (Jill?)-of-all-trades?


Numerele reale pozitive a, b, c verifica relatia 2abc+ab+bc+ca=1.
Sa se arate ca:
rad(ab)+rad(bc)+rad(ca)<=3/2.

Vrei sa spui ca avem $ 2abc+ab+bc+ca=1$ si vrei sa demonstram ca
\[ \sqrt{ab} + \sqrt {bc} + \sqrt {ca} \leq \frac 32 \ ? \] :)

[hide="In cazul asta problema este usoara"]
Sa notam cu $\sqrt {ab} = x$, $\sqrt {bc} = y$, $\sqrt{ca} = z$. Avem $2xyz + x^2 +y^2 +z^2 =1$, asadar $x=\cos A$, $y=\cos B$, $z=\cos C$, unde $A,B,C$ sunt masurile unghiurilor unui triunghi. Inegalitatea de demonstrat devine astfel
\[ \cos A + \cos B + \cos C \leq \frac 32 \] ce rezulta evident din inegalitatea Euler $2r\leq R$ si relatia binecunoscuta $\cos A + \cos B + \cos C = 1 + \dfrac rR$. [/hide]

Asa e, cum ai rescris.
Eu am gasit si o solutie algebrica.

Marius,presupun ca solutia aceea algebrica se bazeaza pe substitutiile $a=\frac{x}{y+z}$ ,$b=\frac{y}{z+x}$ si nu in ultimul rand $c=\frac{z}{x+y}$. Oare ma insel? :?

Cezar, nu pe ideea asta am mers.
Te las sa te mai gandesti, sau vrei sa scriu solutia?
Oricum, voi incerca si pe ideea ta; suna interesant!

Ma rog,nu ma mai chinui in a ghici solutia ta,daca vrei poti sa o postezi. ;)

[quote="Valentin Vornicu"]  A,B,C sunt masurile unghiurilor unui triunghi. Inegalitatea de demonstrat devine astfel
\[ \cos A + \cos B + \cos C \leq \frac 32 \] ce rezulta evident din inegalitatea Euler $2r\leq R$ si relatia binecunoscuta $\cos A + \cos B + \cos C = 1 + \dfrac rR$. [/quote]

More direct : $cos(x)$ is concave ..:D
 :cool:

Nu sunt ghicitor asa ca acum chiar m-ai facut curios.Marius,poti posta solutia ta? ;)

Folosim notatiile:

$ab=x^2/4$, $bc=y^2/4$, $ca=z^2/4$ cu $x, y, z>0$

si conditia data devine

(*) $xyz+x^2+y^2+z^2=4$, de unde avem $x, y, z$ $\leq2$.

Trebuie sa demonstram ca:

$x+y+z<3$.

Fie $m, n>0$ astfel incat

$x=m+n, y=m-n$.

Atunci conditia (*) se scrie:

$(m^2-n^2)z+2(m^2+n^2)+z^2=4$, de unde

$(z+2)(m^2+z-2)=n^2(z-2)$ $\leq0$ si rezulta

$m^2+z$ $\leq2$.

In final:

$x+y+z=2m+z$ $\leq$ $m^2+1+z$ $\leq3$

si solutia se incheie.


geometry
3D geometry
sphere
pigeonhole principle
geometry solved

Choose 400 different points inside an unit cube. Show that four of these points lie inside some sphere of radius 4/23.

[i]Kmal, March 1983[/i]

We deal with a cube with side length 4*\sqrt(2) / 23, since it can be inscribed within a sphere of radius 4/23 and many of them can be conveniently placed inside the 1x1x1 cube. If we prove that four points belong within one of these cubes, the desired result follows immediately.

Now, we see that 4 of these small cubes can be lined on the side of the big cube, but a little space is left. So in total, there are 5x5x5 = 125 regions defined by the placement of these small cubes. Each of these regions fit within a sphere of radius 4/23.

By the Pigeonhole Principle, since 400 = 125 * 3 + 25 > 125 * 3, there exist 3 + 1 = 4 points that belong to one of the regions. Hence result.


What volume of ammonia gas is needed to produce 40g of ammonium nitrate in the following reaction???

NH[size=75]3[/size]+HNO[size=75]3[/size]->NH[size=75]4[/size]NO[size=75]3[/size]

1 mole of NH3 for every mole  of NH4NO3 produced. What is molar mass of NH4NO3? What is molar volume of gas at STP?


hi all
i wana some proofs with continuity ,someone can help me?

please i want a problem in geometry that can be solved by continuity. :oops:

Here is a nice one.Prove that for each convex figure in the plane there exist two concurent lines that divide the area in four equal parts. :lol:


Do any of you know any way to improve your problem solving skills?

Thanks.

Wow confuted, that was a really friendly, helpful answer.  A great way to help him out, too.

If you don't want to be helpful, fine, but there's no reason to be a jerk.

My personal reccomendation is just to work through lots and lots of problems -- history class, for example, provides an excellent opportunity to improve your problem solving skills, an opportunity which is passed up by far too many of America's youth.  :)  But, seriously now, my vote is for doing tons and tons of problems.

About history class, could you explain that?  I used to think that way... in middle school one of my teachers tried to teach us background information, then on questions that were detail-specific we could try to figure out what people would have done in that situation.  But in high school all the history classes i've had were nothing like that.  Is that what you meant, or am I talking about something different?

Well, I meant you could sit in the back of the class and work on math problems ... ;)

Actually, though, there are definite modes of thought that one uses in math that can be used in other non-mathy fields, such as the idea of proof when writing an essay arguing a certain point (if it wouldn't make a good math proof, it probably won't be a good essay).  What you were talking about sounds interesting, but it wasn't quite what I meant :)

I agree with what you say about how math helps you organize arguments, but do you think that history class is good for problem solving outside of essays?

Haha, I answered the question that was asked, JBL ;)

Art of Problem Solving is quite good. Read through the books, and do the problems. Also, do as many other problems as you can find, and read solutions after you've tried problems for a while - even if you got the answer. Learn, and practice.

Thanks for all your replies.

As for the AoPS book, have you found it helpful because I'm really interested in buying one?

i'd guess that most people here started with these books... i think it gives you a really solid foundation.  Definitely get them if you can, i think they'll be real helpful.

[quote]i think it gives you a really solid foundation. [/quote]

I agree. After you have master the contents in there, you can read Engel while branching off to specific subjects (ie polynomials, functional equations, etc..). You definitely want to get know the basics very very well first though.

AoPS was the only reason I'm any good now...otherwise I would still be learning just curricular math. Definitely buy them. Just work through the books in the "Resources" section, and look around for any other recommendations scattered about (such as on Other Problem Solving Topics)

I'm in high school right now.  What books do you recommend?

Both of them or Book 1 or Book 2?

Thanks.

[quote="kool_dudy"]I'm in high school right now.  What books do you recommend?

Both of them or Book 1 or Book 2?[/quote]

If you're already in high school, get 'em both. I did that for a current sixth grader who is on the math team I coach (mostly because I'm such an inexperienced coach, and thus my team needs all the help it can get). Volume 2 has a FEW topics that are approachable without working through Volume 1 first, and all the topics in the whole set should be mastered before one leaves high school--so if you're in high school, dig in now.

There was [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=8674]an earlier thread[/url] with suggestions of other books, which are books most high schoolers ought to know if they are into math.

Hope this helps! There's no substitute for actually working problems, which is why the best young guys on the team I coach should always be ahead of me in problem-solving ability.