Is there a way to prevent functions in LaTeX to work. For example you just want to show the function
\frac{}{}.

For a small amunt of text use \verb using a delimiter such as =[code] \verb=\frac{}{}= [/code] which gives $ \verb= \frac{ }{ }=$.
For longer text use the verbatim environment instead.
[code]\begin{verbatim}
\textbf{bold text} \\
\frac{}{}
\int_0^1 x\;dx
\end{verbatim}[/code] which gives $ \par\begin{verbatim}
\textbf{bold text} \\
\frac{1}{2}
\int_0^1 x\;dx
\end{verbatim}$
(ignore the \par which is only to make it work on the forum. In a document the above will show as 3 lines of text).
The starred versions show a $ \verb*= =$ for spaces.


King Arthur's knights fire a cannon from the top of the castle wall. The cannonball is fired at a speed of 53  and an angle of 38. A cannonball that was accidentally dropped from the top of the castle wall hit the moat below in 1.6 .

What's the question? (and I think you forgot the units: 53 m/s, 30 degrees, 1.6 s?)

hey man where's the question??? :rotfl:  :rotfl:  :P

Same doubt here!!!!
Also the units........


function
Support
real analysis
real analysis unsolved

This is, probably, a classical one but I do not remember it posted. Let $x_{k}$ be a sequence of pairwise distinct real numbers and let $d_{k}$ be any sequence of reals tending to $0$. Prove or disprove that then there always exists a function $f : \mathbb R\to\mathbb R$ such that it is continuous at every point except the points $x_{k}$ and at each point $x_{k}$ the left and the right limits exist and the difference between them (the "jump") is exactly $d_{k}$.

Yes, there is such a function. We can construct a sequence of functions $f_{n}$ converging uniformly to $f$ with the property that $f_{n}$ is continuous except at $x_{1},x_{2},\dots,x_{n}$ with jumps $d_{1},d_{2},\dots,d_{n}$ at these points. Verifying that the limit has the right jumps and is continuous elsewhere given these properties is a routine exercise.

Let $g(x)=\begin{cases}0& x\le 0\\ 1-x&0<x\le 1\\ 0&x>1\end{cases}$. Let $g_{k}(x)=d_{k}g(\frac{x-x_{k}}{c_{k}})$, where $c_{k}$ are positive constants to be chosen later. Then we will have $f_{n}(x)=g_{1}(x)+g_{2}(x)+\dots+g_{n}(x)$.

Let $N_{0}=0$, and choose an increasing sequence $N_{i}$ such that $d_{k}<\frac1{2^{i}}$ for $k>N_{i}$. If $N_{i-1}<k\le N_{i}$, choose $c_{k}$ so that either $x_{j}\le x_{k}$ or $x_{k}+c_{k}<x_{j}$ for each $j$ between $N_{i-1}$ and $N_{i}$. Now the supports of $g_{k}$ for $N_{i-1}<k\le N_{i}$ are disjoint, and $\sum_{k=N_{i-1}+1}^{N_{i}}|g_{k}(x)|<2^{i-1}$ for each $i$ (Well, that's $\max_{k}|d_{k}|$ rather than 1 for $i=1$). This implies uniform convergence, and the limit is our desired $f$.


Hello all. Let me introduce myself here. I am a 23 yr old graduate student pursuing Masters in some applied math course(Actuarial Science). I came into problemsolving quite late in my life.  I became seriously interested in math when i was 18(which i am sure is quite late as compared to most of the students on this forum) as a result of which i couldn't possible be involved with any kind of competetions or olympiads( IMO obviously ruled out).  But  there was always something in me telling me to persue math. I realised i had some mathematical talent which due to lack of proper surroundings and right friends got unnoticed and sidetracked.
   But i became very determined not to let anyone intrude on my life or my mathematics and started fresh from the basics of olympiad mathematics when i was 20.  Now i am considerably better at math than i used to be. I occasionally solve some problems of USAMO level of difficulty.  I generally get so deeply involved in a problem that i re-discover some formulae/theorems that are well known while  trying to understand the problem in the first place. As a consequence of which i digress a lot and waste a lot of time to re-learn some basic theorems. Had i started early i would not have wasted so much time in trying to learn the theorems by actually proving them and then knowing them.  I realise its such a waste to time to keep learning things in such a harsh way.

   But the main reason why i am unable to really accomplish my long yearned dream to successfully solve a USAMO paper or an IMO problem is that i do not have time. I just don't have enough time to concentrate for hrs/days on problems, which i feel is the very essence of learning to solve difficult problems. This i attribute to  graduate school demands  like Teaching responsibilites(T.A.), maintaining good grades , job search, etc.. etc..
      Right now i am really worried about my destiny, which is math. I    am beginning to wonder if i have become too old to pursue problemsolving???  Though problem solving is not what mathematics  is all about.  But i have a strong belief that only a good problem solver can become a good mathematician. Hence i realised it is my duty to equip myself first with all the necessary tools before i make any worthwile contribution to Mathematics and human kind.
  All i am earnestly seeking is a suggestion or advice as to what i must aim for ? Am i being irrational about persuing math at olympiad level at this age? Or should i continue to believe in myself and try to acheive some personal satisfaction atleast as an adult??? Is age necessarily a factor to influence development of  extra-ordinary mathematical abilities?

You're never too old to pursue the career you feel you're best suited for.  Mankind is short on good mathematicians as it is, so from the sound of it, you've got a hoppin' good chance of reaching fame and fortune.

Still, given the minimal window of time you describe, my recommendation would be to simply wait a bit until your position is a little more stable.  Math may be one of the most important things you'll ever have in your life, but food on the table comes first.  Wait until you find a job and get settled into it, then pursue your chosen field of mathematics.  Good luck to you, and I hope I speak for more than a few of us on these forums when I say that we hope to see your name in lights (or at least in the annals of mathematical history) someday.

You aren't ever wasting your time if you are improving your understanding of the field you will major in. I wish I had the concentration to rediscover theorems while solving problems.

to the original questioner,

IMO-level problem solving is much like a 3-4 year
bachelor's degree in mathematics, but with a different
curriculum.  So if you earned one degree you'll have
a general idea of the effort needed to reach
USAMO/IMO level as a hobby.  Are you talking
about solving a paper completely in the allotted time,
or being able to solve all the problems at your leisure?

There's also a problem-solver culture of people who
submit solutions (and problem proposals) to the
problem columns of mathematical magazines.  Many of
those people are or were involved in the national and
international olympiads as competitors or coaches.
This is another way to pursue a serious interest in
problem solving after the teenage years.


Since $ \frac1{b\plus{}\frac1{c\plus{}\frac1d}}>0$, and $ 1\le a\in\mathbb N<2$, $ a\equal{}1\implies b\plus{}\frac1{c\plus{}\frac1d}\equal{}\frac{25}{22}$
$ \implies b\equal{}1\implies c\plus{}\frac1d\equal{}\frac{22}3\implies c\equal{}7\implies d\equal{}3$
$ \implies3a\minus{}2b\plus{}c^2\equal{}3(1)\minus{}2(1)\plus{}(7)^2\equal{}\boxed{50}$.


I got a 42, but I know somebody who got a 45. I beat him in the Countdown Round though :) .

Please just refer to the other topic that is up for Chapters, since it already has a lot of posts. Thanks.

I got a 36, sad I know. I didn't even get to Countdown. Yet out of the ashes of this tradegy our team made it to state and so I can still compete in the team and individual again. Luck is with me.

I missed out (barely) on countdown with a 34. I'm not too ashamed of that as a sixth grader :).

I really messed up on 4 points worth of problems. I got swamped on the sprint, and the target was like, the easiest target i've ever taken chapter level (I only got 5 in shcool lol, and I got 7 here with 1 stupid mistake)

The target was too easy, but the sprint was really hard. I got 18/7(32), and got 9 on the team round.

I do agree that the target was easy, but the sprint round I thought was harder than the target round!!!
My score: 15/30 (sprint round), 12/16 (target) = 27/46, getting seventh inmy chapter. I was expecting higher...

About the target round... I think I solved all but one by hand... b/c they reminded me of the first pages of a State Sprint... ya but I think the fact that the CT competitions are so early in the morning makes it a bit tough to be fresh for the Sprints... in hindsight even the spring (except for 24) was pretty easy.. but it had too much combinatorics for my own good. ;)


I want to activate HTML in my posts, and I did so in my profile. However, when I am posting, under the Options box it still says HTML is off. This is what I get when I post HTML:

My School

You have to do it every time you post. Did you press Save in your profile? (sometimes I don't and my selections don't save)

yes, I pressed save, even before my last post...

HTML is not allowed on th site. You can use BBCode to make the same thing:
[code]
[url=http://ligon.wcpss.net/]My School[/url]
[/code]
gives you

[url=http://ligon.wcpss.net/]My School[/url]


Solve the following equation over rationals
$ a^b \equal{} b^a$

Well, it has surely infinitely many solutions since $ f(x) \equal{} x^{\frac {1}{x}}$ is continous and is strictly monotonic (rising) on $ (0,e)$ and monotonic (deacreasing) on $ e, \plus{} \infty)$, thus using the Darboux property  $ \forall_{a\in (0,e)} \ \exists_{b\in (e, \plus{} \infty)} \ f(a) \equal{} f(b)$  :wink:
Also look here:
http://www.mathlinks.ro/viewtopic.php?t=150682

Similarly, one finds that the set of real solutions is given by $ (x, y) \equal{} \left( k^{ \frac{1}{k\minus{}1} }, k^{ \frac{k}{k\minus{}1} } \right)$.


A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper-left corner, so that the first row is numbered $1, 2, \ldots , 17$, the second row $18, 19, \ldots , 34$, and so on down the board. If the board is renumbered so that the left column, top to bottom is $1,2, \ldots , 13$, the second column $14, 15, \ldots , 26$ and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).

\[ \text{(A) } 222 \qquad \text{(B) } 333 \qquad \text{(C) } 444 \qquad \text{(D) } 555 \qquad \text{(E) } 666  \]

[hide]
If we label the rows x and the columns y, we can create expressions for each number which end up being $13(y-1)+x$ and $17(x-1)+y$.
Equating these:
$13y - 13 + x = 17x - 17 + y$
$12y - 16x = -4$
$4x - 3y = 1$

Next, we look for x and y values which satisfy this equation which can be easily done by graphing the function (and going to the table to look for integer values).  These end up being (1,1); (4,5); (7,9); (10,13); (13,17).
We get a sum:
$1 + 56 + 111 + 166 + 221 = 555$
$\mbox{D}$
[/hide]

[hide="Answer"]Any numbers of this form are such that $(n-1)x+m=(m-1)y+n$, so $13(n-1)+m=17(m-1)+n\Rightarrow 13n+m-13=17m+n-17\Rightarrow$
$12n+4=16m\Rightarrow 3n+1=4m$. Therefore, our possible answers are $(1,1)$, $(5,4)$, $(9,7)$, $(13,10)$, and $(17,13)$. Substituting those into the original equation, we have $1$, $56$, $111$, $166$, and $221$, which sum to $555\Rightarrow \boxed{D}$.[/hide]


Let $p_{i}$ be the $i^{th}$ prime.

$p_{1}p_{2}p_{3}\cdots p_{n}=p_{x}$

where $x>n$

Does $x=2n$?

[quote="matt276eagles"]Huh?  By definition a prime can't be expressed as a product of primes.[/quote]

I think he meant to add $1$ to the product.

Rofl. I didn't look at the second part of the question.

I guess not...

[quote="1=2"]sorry. I meant p_1p_2...p_n +1 = p_x.[/quote]
Uhh, there's no guarantee that the LHS is prime
$2\cdot 3 \cdot 5\cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19+1=1487\cdot 593$

[quote="Derek"]diophantient, check your work[/quote]
oops, thanks for pointing that out.
2*3*5*11*13+1=7*613


Number of posts: 11182
Posts per day: 29.81

Number of topics: 2793
Topics per day: 7.45

Number of users: 432
Users per day: 1.15

Board started: 03 Feb 2003 11:16 pm
Database size: 11.92 MB

if latex is available, i think, the size of database will grow faster.
You need to store many gif files.
and all latex-to-html software converts duplicated formulae into different pictures - that's bad.

[quote]all latex-to-html software converts duplicated formulae into different pictures - that's bad[/quote]That's not quite true. Latexrender is very efficient and caches images so duplicated formulae re-use the same images.

Whoops, forgot to log in before posting that message :?

one month has passed, so I will update the stats

Number of posts: 14080
Posts per day: 34.81

Number of topics: 3576
Topics per day: 8.84

Number of users: 583
Users per day: 1.44

Database size: 15.24 MB

another month has passed, so I will again update the stats

Number of posts: 16038
Posts per day: 36.72

Number of topics: 3959
Topics per day: 9.06

Number of users: 748
Users per day: 1.71

Database size: 17.73 MB

more stats this month :)

Number of posts: 18090
Posts per day: 38.83

Number of topics: 4453
Topics per day: 9.56

Number of users: 878
Users per day: 1.88

Database size: 20.65 MB

Number of posts: 21016
Posts per day: 41.84

Number of topics: 5050
Topics per day: 10.05

Number of users: 1065
Users per day: 2.12

Database size: 24.77 MB

now some nice graphs and we'll all clap our hands :D

well seems like the site is nicely growing! Now too bad there's no statistic to see how much of the post are really mathematical and how much are just gossip :P


inequalities
induction
inequalities proposed

I'm interested in finding out some new proofs for the next inequality.
$ \frac{a^{n\plus{}1}}{b^n}(a\minus{}b)\plus{}\frac{b^{n\plus{}1}}{c^n}(b\minus{}c)\plus{}\frac{c^{n\plus{}1}}{a^n}(c\minus{}a)\geq0$, where: $ a,b,c>0$ and $ n\in N$.

I'd like to give you a huge hint:
$ LHS\equal{}\frac{(a^2)^{n\plus{}1}}{(ab)^n}\plus{}\frac{(b^2)^{n\plus{}1}}{(bc)^n}\plus{}\frac{(c^2)^{n\plus{}1}}{(ca)^n}\stackrel{(*)}{\geq}\underbrace{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}_{\geq1}\cdot\left(\frac{(a^2)^n}{(ab)^{n\minus{}1}}\plus{}\frac{(b^2)^n}{(bc)^{n\minus{}1}}\plus{}\frac{(c^2)^n}{(ca)^{n\minus{}1}}\right)\geq RHS$
Any remarks? I'm interested to find out your oppinions related to the inequality $ (*)$.

I am trying not to be offensive but a little bit more constructive here. How can you get the inequality (*) ? Please restate the used inequality and where you applied it.

Regards,

$ \frac{a^{n\plus{}1}}{b^{n}}(a\minus{}b)\plus{}\frac{b^{n\plus{}1}}{c^{n}}(b\minus{}c)\plus{}\frac{c^{n\plus{}1}}{a^{n}}(c\minus{}a)\geq0 \Longleftrightarrow$
$ \Longleftrightarrow\ \frac{a^{n\plus{}2}}{b^n}\plus{}\frac{b^{n\plus{}2}}{c^n}\plus{}\frac{c^{n\plus{}2}}{a^n}\geq\frac{a^{n\plus{}1}}{b^{n\minus{}1}}\plus{}\frac{b^{n\plus{}1}}{c^{n\minus{}1}}\plus{}\frac{c^{n\plus{}1}}{a^{n\minus{}1}}$
We obtain:
$ LHS\equal{}\frac{(a^{2})^{n\plus{}1}}{(ab)^{n}}\plus{}\frac{(b^{2})^{n\plus{}1}}{(bc)^{n}}\plus{}\frac{(c^{2})^{n\plus{}1}}{(ca)^{n}}\stackrel{(*)}{\geq}\underbrace{\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{ab\plus{}bc\plus{}ca}}_{\geq1}\cdot\left(\frac{(a^{2})^{n}}{(ab)^{n\minus{}1}}\plus{}\frac{(b^{2})^{n}}{(bc)^{n\minus{}1}}\plus{}\frac{(c^{2})^{n}}{(ca)^{n\minus{}1}}\right)\geq\frac{(a^{2})^{n}}{(ab)^{n\minus{}1}}\plus{}\frac{(b^{2})^{n}}{(bc)^{n\minus{}1}}\plus{}\frac{(c^{2})^{n}}{(ca)^{n\minus{}1}}\equal{} RHS$
Try to prove the following result, and use it in $ (*)$ for $ n\equal{}3$. I haven't seen this result before anywhere. I think and hope it's something new. I have already a new article concerning this inequality, and its derived results.
[b]MY LEMMA:[/b]

If: $ a_i\geq0,b_i>0,i\equal{}1,2,\dots,k$ and $ k,n\in N,k\geq1$, then:
$ \frac{{a_1}^{n\plus{}1}}{b_1^n}\plus{}\frac{a_2^{n\plus{}1}}{b_2^n}\plus{}\cdots\plus{}\frac{{a_k}^{n\plus{}1}}{b_k^n}\geq\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_k}{b_1\plus{}b_2\plus{}\cdots\plus{}b_k}\cdot\left(\frac{{a_1}^n}{b_1^{n\minus{}1}}\plus{}\frac{{a_2}^n}{b_2^{n\minus{}1}}\plus{}\cdots\plus{}\frac{{a_k}^n}{b_k^{n\minus{}1}}\right)$
This is my solution. I tought that someone could give me a different solution. I'm sure that everything is OK now, [b]great math[/b]!

Thanks for the lemma. I haven't proven it yet but I have another proof for the given inequality

Using mathematical induction, it suffices to prove that

\[ \frac{a^{n\plus{}2}}{b^n} \plus{}\frac{a^n}{b^{n\minus{}2}} \ge 2\sum_{cyc} \frac{a^{n\plus{}1}}{b^{n\minus{}1}}\]

and

\[ \sum_{cyc}\frac{a^{n\plus{}1}}{b^{n\minus{}1}}   \ge \sum_{cyc}\frac{a^n}{b^{n\minus{}2}}\] (due to mathematical induction)

Hence, we complete our proof here.

[quote="marin.bancos"]I'm interested in finding out some new proofs for the next inequality.
$ \frac {a^{n \plus{} 1}}{b^n}(a \minus{} b) \plus{} \frac {b^{n \plus{} 1}}{c^n}(b \minus{} c) \plus{} \frac {c^{n \plus{} 1}}{a^n}(c \minus{} a)\geq0$, where: $ a,b,c > 0$ and $ n\in N$.[/quote]
This inequality is true according to the obvious inequality
$ \frac{2a^{n\plus{}1}}{b^n} (a\minus{}b) \ge a^2\minus{}b^2.$

Thank you, [b]great math[/b]!  Thank you, [b]can_hang2007[/b]! Very nice ideas! Using your solution, this inequality could be generalized:
$ \frac{a_1^{n\plus{}1}}{a_2^{n}}(a_1\minus{}a_2)\plus{}\frac{a_2^{n\plus{}1}}{a_3^{n}}(a_2\minus{}a_3)\plus{}\cdots\plus{}\frac{a_k^{n\plus{}1}}{a_1^{n}}(a_k\minus{}a_1)\geq0$, where: $ a_1,a_2,\dots,a_k\geq0$ and $ n,k\in N, k\geq2.$
A very beautiful expression.
I'm sure that my lemma is also a valuable inequality because of its well known and interesting derived results.

Some friends of MathLinks asked me to post my proof for MY LEMMA. You can see it here. I'm very interested in your oppinions. What do you think about it, [b]Manlio[/b]? Owing to [b]can_hang2007[/b]'s idea, we can give a new generalization for my initial inequality. Try to prove it.
Prove that: $ \frac{a_{1}^{n}(a_{1}\minus{}a_{2})}{a_{2}^{n\minus{}k\plus{}1}}\plus{}\frac{a_{2}^{n}(a_{2}\minus{}a_{3})}{a_{3}^{n\minus{}k\plus{}1}}\plus{}\cdots\plus{}\frac{a_{p}^{n}(a_{p}\minus{}a_{1})}{a_{1}^{n\minus{}k\plus{}1}}\geq0,$ where: $ a_{1},a_{2},\dots,a_{p}\geq0$ and  $ n,k\in N, n\geq k\geq1.$

First, let me recall your lemma:
$ \frac {{a_{1}}^{n \plus{} 1}}{b_{1}^{n}} \plus{} \frac {a_{2}^{n \plus{} 1}}{b_{2}^{n}} \plus{} \cdots \plus{} \frac {{a_{k}}^{n \plus{} 1}}{b_{k}^{n}}\geq\frac {a_{1} \plus{} a_{2} \plus{} \cdots \plus{} a_{k}}{b_{1} \plus{} b_{2} \plus{} \cdots \plus{} b_{k}}\cdot\left(\frac {{a_{1}}^{n}}{b_{1}^{n \minus{} 1}} \plus{} \frac {{a_{2}}^{n}}{b_{2}^{n \minus{} 1}} \plus{} \cdots \plus{} \frac {{a_{k}}^{n}}{b_{k}^{n \minus{} 1}}\right)$

I see your proof is nice, but I think we can get a simpler proof as follows:

Because the inequality is homogeneous in $ a_1,a_2,\ldots, a_k$ and $ b_1,b_2,\ldots, b_k,$ we may assume that $ a_1 \plus{} a_2 \plus{} \cdots\plus{} a_k \equal{} b_1 \plus{} b_2 \plus{} \cdots \plus{} b_k \equal{} k.$ Then, we can write our inequality as
$ \sum^k_{i \equal{} 1}\frac {a^{n \plus{} 1}_i}{b^n_i} \ge \sum^k_{i \equal{} 1}\frac {a^n_i}{b^{n \minus{} 1}_i},$
or
$ \sum^k_{i \equal{} 1}\frac {a^{n \plus{} 1}_i}{b^n_i} \plus{} \sum^k_{i \equal{} 1}(b_i \minus{} a_i)\ge \sum^k_{i \equal{} 1}\frac {a^n_i}{b^{n \minus{} 1}_i}.$
This inequality is true because
$ \frac {a^{n \plus{} 1}_i}{b^n_i} \plus{} b_i \minus{} a_i \minus{} \frac {a^n_i}{b^{n \minus{} 1}_i} \equal{} \frac {a^n_i(a_i \minus{} b_i)}{b^n_i} \plus{} b_i \minus{} a_i \equal{} \frac {(a^n_i \minus{} b^n_i)(a_i \minus{} b_i)}{b^n_i} \ge 0.$

Very nice proof, [b]can_hang2007.[/b] Thanks for it. What do you think about the final generalization of my initial inequality?
I'm waiting for new proofs or new results, my friends!


Find all x such that:

$ \sqrt{cos2x\minus{}sin4x}\equal{}sinx\minus{}cosx$

[quote="MR.129"]Find all x such that:

$ \sqrt {\cos2x \minus{} \sin4x} \equal{} \sin x \minus{} \cos x$
 [/quote]
[hide="solution"]After squaring we attain:
$ \cos 2x \minus{} 2 \sin 2x \cdot \cos 2x \equal{} (\sin x \minus{} \cos x)^2 \implies  \sin 2x\plus{}\cos2x\equal{}2\sin 2x \cdot \cos 2x$

Substituting $ \cos^2 2x\equal{}1\minus{}\sin^2 2x$ and calculating we get:
\[ 4 \sin^4 2x \cdot (\sin^2 2x\minus{}1)\equal{}0\]

and therefore the solutions are:
$ x_1\equal{}k\cdot \pi$

$ x_2\equal{}\frac{\pi}{4}\plus{}k\cdot \pi$

$ x_3\equal{}\frac{3\pi}{4}\plus{}k\cdot \pi$

where $ k \in Z$[/hide]

hello, $ x\equal{}2\cdot\pi$ is not a solution of your equation.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, $ x \equal{} 2\cdot\pi$ is not a solution of your equation.
Sonnhard.[/quote]

I didn't actually check solutions properly but let's see

$ \sqrt{\cos2 \cdot 2\pi \minus{} \sin 4\cdot 2\pi}\equal{}\sin 4\pi \minus{} \cos 4\pi$

$ \sqrt{1}\equal{}\minus{}1$  .. what;s wrong ?

hello, $ \sqrt{1}\equal{}\minus{}1$ is definitively wrong.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, $ \sqrt {1} \equal{} \minus{} 1$ is definitively wrong.
Sonnhard.[/quote]

why ?

By definition, $ \sqrt{x}$ denotes the nonnegative real number $ y$ such that $ y^2 \equal{} x$ (if it exists). Therefore, $ \sqrt{x} \ge 0$ for all $ x$ and hence $ \sqrt{1} \not\equal{} \minus{}1$.

hello, see here
http://en.wikipedia.org/wiki/Square_root
Sonnhard.

Got it.

So $ x = k\cdot \pi$ it's correct for the equation
\[{ \cos2x - \sin 4x} = (\sin x - \cos x)^2\]
but not for
\[ \sqrt {\cos2x - \sin4x} = \sin x - \cos x\]
that's cool  :ninja:

hello, ok, one more problem is solved.
Sonnhard.

I think the most important thing to solve this trigonometric equation is all real numbers x that cause expression under the root is less than 0 must be eliminated from the set of possible solutions.

I have another solution that is quite clearly and completed.

[b]Answer.[/b]

$ \sqrt {cos2x \minus{} sin4x} \equal{} sinx \minus{} cosx$

$ \Leftrightarrow \left\{\begin{array}{l} \sin x \minus{} \cos x \ge 0 \\
\cos 2x \minus{} \sin 4x \equal{} 1 \minus{} \sin 2x \\
\end{array} \right.$

$ \Leftrightarrow \left\{\begin{array}{l}\sin (x \minus{} \frac {\pi }{4}) \ge 0 \\
(\cos 2x \plus{} \sin 2x)(\cos 2x \plus{} \sin 2x \minus{} 1) \equal{} 0 \\
\end{array} \right.$

$ \Leftrightarrow \left\{\begin{array}{l}k2\pi \le x \minus{} \frac {\pi }{4} \le \pi \plus{} k2\pi \,(k \in Z)\,\,\,\,\,\,\,(*) \\
\left[ \begin{array}{l} \sin (2x \plus{} \frac {\pi }{4}) \equal{} 0\,\,\,\,\,\,\,\,\,\,\,\,\,(1) \\
\sin (2x \plus{} \frac {\pi }{4}) \equal{} \frac {{\sqrt 2 }}{2} \equal{} \sin \frac {\pi }{4}\,\,\,\,\,\,\,\,\,\,\,\,\,(2) \\
\end{array} \right. \\
\end{array} \right.$

We have

$ (1): \sin (2x \plus{} \frac {\pi }{4}) \equal{} 0$

$ \Leftrightarrow x \equal{} \minus{} \frac {\pi }{8} \plus{} \frac {{l\pi }}{2}\,\,\,(l \in Z)$

Because the condition (*) must be satisfied by x, therefore :

$ k2\pi \leq \minus{} \frac {\pi }{8} \plus{} \frac {{l\pi }}{2} \minus{} \frac {\pi}{4} \leq \pi \plus{} k2\pi$    $ (l,k \in Z)$

$ \Rightarrow \frac {3\pi}{8} \leq \frac {l \pi}{2} \minus{} k2\pi \leq \frac {11\pi}{8}$

$ \Rightarrow \frac {3}{8} \leq \frac {l}{2} \minus{} 2k \leq \frac {11}{8}$

$ \Rightarrow l \equal{} 2(2k \plus{} 1) \equal{} 2a (a \in Z)$

$ \Rightarrow x \equal{} \minus{} \frac {\pi}{8} \plus{} a\pi (a \in Z)$

We have

$ (2): sin(2x \plus{} \frac {\pi}{4}) \equal{} sin\frac {\pi}{4}$

$ \Leftrightarrow \left[ \begin{array}{l} x \equal{} m\pi \\
x \equal{} \frac {\pi }{4} \plus{} m\pi \\
\end{array} \right.(m \in Z)$

Similarly, we obtain $ x \equal{} \frac {\pi}{4} \plus{} m\pi$ and $ x \equal{} (2k \plus{} 1)\pi$ where $ k, m\in Z$

Conclusion, the solutions for the given equation are: $ x \equal{} \minus{} \frac {\pi}{8} \plus{} a\pi, x \equal{} \frac {\pi}{4} \plus{} m\pi, x \equal{} (2k \plus{} 1)\pi$ where $ a, m, k \in Z$.


MATHCOUNTS
analytic geometry
geometry
email
complex numbers
NSML

Any Illinois AoPSers attending the NSML Meet #1 on September 30 at Naperville North?

If so, any plans/meetups/thoughts/whatever?

I'm going, and I'm a (fre)shmen from IMSA. I will be competing in the freshman division (which I have to), but my school's tryouts haven't been complete yet, so I'm not sure fully yet whether or not I'm also doing sophomore or senior division. (Obviously, if I make the team for senior, I'll do senior) I'm pretty excited in meeting everyone there! :D

Sadly my school didn't get invited so i can't go.

maybe Aryth is going.

Yeah, I'm pretty sure he is. But for the first four meets, your school doesn't have to get "invited" to go there, it just has to be initially registered. Did your school not register or something?

It's just for the [b]NSML All-Conference Meet[/b](that's at Evanston Township High School) after all four meets are done (probably around March, I think) that you have to be invited to in order to go there (and that's based on your scores from the first four regular meets).

Our school got a new principal and our main math person is very old (I'd say about 65) so he can't do much.
All we do is the amc's and mathcounts and also wyse.

Wait, since you do Mathcounts, I'm guessing you're in middle school, right?

FYI, only high school students can compete on a NSML team.

dude we should play like fantasy NSML math team  :rotfl:

hm I'm gonna be NSMLing this year as an alternate for Payton since my school (Northside) doesn't do it. I have no idea what the topics are or what the questions look like, but this could be fun I guess...

These are the topics for the first meet:

Freshmen: Sets and Venn Diagrams
Sophomores: Sets, Logic, and Venn Diagrams
Juniors: Coordinate Geometry
Seniors: Algebra of Complex Numbers

What grade levels are you competing for? I'm doing (fre)shmen (sophomores who skipped ninth grade) and senior teams for IMSA.

also, we IMSA students are gonna see you, aryth. :D (yes, we're competing at Naperville North, and the schools traveling there are IMSA, Naperville Central, Glenbrook North, and Stevenson)

NCHS calls dibs on the chipotle in downtown naperville.

Lol... Do IMSA people know if we still have to eat Sodhexo on meet nights?

Uh....idk yet. wait until our coaches send an email for more info.

@Aryth: Yet fortunately I do not believe IMSA is gonna eat at chipotle. If we did that'd be funny. :rotfl:  :rotfl:  :rotfl:

rofl.

I can see it now.. an epic battle between the Naperville Central and IMSA math teams for who gets the Chipotle...or maybe it could be resolved with an extreme duck, duck, goose tournament.

lol...But at least IMSA can hope for dinner at the cafe before leaving for the bus. :rotfl:  :rotfl:  :rotfl:

Is anyone going to the one at Hinsdale Central or is that a different contest?

One set of schools is competing at Hinsdale Central today. Wait, echen, aren't you still in middle school? Because NSML is only for high school students (or actually, only high school students can compete as actual school contestants)

just FYI guys, AceofDiamonds is saszs  :D

LOL RLY?

wow.

a spammer coming back to the forum= :o

By the way, the results are up on the NSML homepage:

[url]http://www.nsml.org/statistics[/url]

No i'm in 9th grade but I didn't go to the first meet.

i went... hinsdale south meet... soph team... epic failed...

dude I am going there.  well not.  but good luck to people that are.

oh no seriously aceofdiamonds is saszs?

[quote="echen"]Is all conference the one at Evanston cause my coach said that everyone goes to that one.[/quote]

@echen: Yes. Every year, the final fifth meet, the All-Conference meet, is at Evanston Township High School.

However, participation is by invite-only, you have to be in the top scoring students from the first four regular meets to get invited to the Conference meet. And yes, all 50something schools in the meets participate in that final meet, same place, same time.

@Truffles: yes. I know. it's scary. :o

[quote="gauss1181"]you should set your goal to make all conference this year so that you can get a chance to meet me.

also, what was the hardest sophomore question that you couldn't do?[/quote]

haha they were surprisingly really easy. but for the last question it was multi choice and u had to put all of the ones that applied... but i only put one of them >.< and i don't remember what the other one i missed was about...


Find the positive real root of $ x^4 \minus{} 2x^2 \minus{} x \minus{} 1 \equal{} 0$.


algebra
function
domain
advanced fields
advanced fields theorems

The following theorem is from Harvey Cohn's "A Classical Invitation to Algebraic Numbers and Class Fields" (p.48).

Theorem 6.18: Let $A$ be a non-zero ideal in a Dedekind domain $R$, and $C|A$. Then there exists $b$ in $R$ such that $C=(A,(b))=A+(b)$.

Proof: Factor $C=P_{1}^{s_{1}}...P_{t}^{s_{t}}$ and $A=P_{1}^{r_{1}}...P_{t}^{r_{t}}$ so $s_{i} \leq r_{i}$. Let $C_{0}=P_{1}^{s_{1}+1}...P_{t}^{s_{t}+1}$ and $C_{j}=C_{0}/P_{j}$. Now $C_{j}|C_{0}$ (strictly) so there exists $b_{j} \in C_{j}-C_{0}$. We assert that $b=b_{1}+...+b_{t}$ will satisfy the requirement, because ... Hence $(b)=C.B$ where no $P_{i}|B$. By (6.6c) $A+(b)=C$.

I believe that the assertion "Hence $(b)=C.B$ where no $P_{i}|B$" deserves a non-trivial proof. Can anybody prove that? I have a proof but would like to see yours first.

Best regards,
Murat Aygen

[quote="Murat Aygen"] Hence $(b)=C.B$ where no $P_{i}|B$" deserves a non-trivial proof.[/quote]

Could you please give a definition of $B$ ?

--
Fabrice

[quote="fmathurin"]
Could you please give a definition of $B$ ?[/quote]

$A,B,C$ are ideals in the Dedekind domain $R$. They need not be principal ideals. Principal ideals are denoted with $(b)$ where $b$ is a generator. In such a domain each ideal is equal to the product of some prime (thus maximal) ideals. Therefore there exists an ideal $B$ such that $(b)=C.B$ whenever $C|(b)$ (This is the contention of Remark 6.8 on page 44).

Best regards,
Murat Aygen


induction
calculus
geometry
3D geometry
arithmetic sequence

The set of consecutive odd numbers 1,3,5,7,9...,N has a sum of 400. How many numbers are in the set?

I'll be posting the answer in about a week

you posted it twice :dry:

I know the solution, but surely I am very old to answer it.

[hide="REALLY strong hint, do not open if you want to do it yourself"] 20^2=400  [/hide]

[hide] since the sum of an odd number sequence is the number of terms squared, 20^2=400.[/hide]

do it like this with the help of arithmetic progression
S=N/2[2A+(N-1)*D]
400=N/2[2+(N-1)*2]
800=N[2+2N-2]
800=2N^2
N^2=400
N=SQRT(400)
N=20
AND I HOPE 20 IS UR ANSWER!

[hide]the sum of the first n odd numbers is n^2, so sqrt of 400=20[/hide]

hey gogators my method is correct or not!

Um no, N is the 20th odd number, not 20.

Um, he meant that the number of terms, which was what the problem was asking for, was 20.

he was using the formula for the sum of the first N terms of an arithmatic series, with N as n in the formula S=n[2a+(n-1)d]/2, as you should hae seen.

rt_08: your method IS correct, although somewhat unneccesary as you could simply take the square root of 400.

as gogators did, and as you did at the end.

Yes; I recently thought about this. When there's a hard topic, few people reply because most people don't want to look stupid or anything. Then, when there's an easy topic, people always overcomplicate things to show off and stuff because it's easy and they look smart (to the extent where people use calculus to find the cube root of 2...in the MC forum...xD) :rotfl:  :rotfl: .

i totally agree with yongyi look friends the place where i study we r taught this method
i am getting used to the new method.hope so ill do this pretty quickly
anyways thanx for correcting me !
hope ill learn a lot from u all!

$ \forall$ odd $ L$
$ S\equal{}1\plus{}3\plus{}5\plus{}7\cdots\equal{}\sum_{L\equal{}1}^{n}L \equal{}n^2$
$ \implies S\equal{}n^2\equal{}400$
$ \implies n\equal{}20$

therefore the answer is $ \boxed{\boxed{\boxed{\color{red}20}}}$

hey thaNKS again
also
plz answer my question that is
can u tell me how should i study for olympiads and wich is the best prescribed book for it
also i submitted an online form for KVPY nd now thay are asking to submit the pay in slip but wen i opened the link to 'pay in slip'
it read the message that "the file is damaged"
could any of u plz help
plz........
thanx!

Using induction isn't hard, and using it to prove something such as the formula for the sum of the first $ n$ odd numbers isn't really necessary.  I'll prove it in a more leisurely way:

First, I'll prove what the formula for the sum of the first $ n$ natural numbers is:
$ S \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} (n \minus{} 1) \plus{} n$
$ S \equal{} n \plus{} (n \minus{} 1) \plus{} \cdots \plus{} 2 \plus{} 1$
When we sum these two expressions for the sum of the first $ n$ natural numbers, we can view it as $ n$ pairs of two numbers that sum to $ n \plus{} 1$:
$ 2S \equal{} (n \plus{} 1) \plus{} (n \plus{} 1) \plus{} \cdots \plus{} (n \plus{} 1) \plus{} (n \plus{} 1) \equal{} n(n \plus{} 1)$
We seek this sum divided by $ 2$ (which is equal to $ S$):
$ S \equal{} \frac {n(n \plus{} 1)}{2}$

Now I'll show what the sum of the first $ n$ even numbers is:
$ S \equal{} 2 \plus{} 4 \plus{} \cdots \plus{} 2(n \minus{} 1) \plus{} 2n$
I divide by $ 2$ to get a familiar expression and then multiply by $ 2$ again:
$ \frac {S}{2} \equal{} 1 \plus{} 2 \plus{} \cdots \plus{} (n \minus{} 1) \plus{} n \equal{} \frac {n(n \plus{} 1)}{2}$
$ S \equal{} n(n \plus{} 1)$

At long last, the sum of the first $ n$ odd numbers can be expressed in a compact form (a formula, in this case):
$ S \equal{} 1 \plus{} 3 \plus{} \cdots \plus{} (2n \minus{} 3) \plus{} (2n \minus{} 1)$
I add $ 1$ to each of the $ n$ terms of this sum (which is basically adding $ n$ to the sum) to get a familiar expression and then subtract $ n$ to get back to the original sum:
$ S \plus{} n \equal{} 2 \plus{} 4 \plus{} \cdots \plus{} 2(n \minus{} 1) \plus{} 2n \equal{} n(n \plus{} 1) \equal{} n^2 \plus{} n$
$ S \equal{} n^2$

So, to solve the problem, we know that $ n^2 \equal{} 400$, and obviously the answer to the problem (the number of members of the set of the first $ n$ odd numbers) is $ n$.  Certainly, $ n$ can't be negative, so we can take the positive square root of the equation we made to get $ n$: $ n \equal{} 20$.  The answer is thus $ \boxed{20}$.

Notice how, in proving the formulas for the various sums of the first whatever numbers, we built each following formula by the one that preceded it.  All it took was a little creativity; you don't need to hustle through induction (and it'll probably help your problem solving skills more in the long run to [i]not[/i] use induction on finding formulas for sums such as these).

[hide="Another way to come up with this formula"]
Let $ S \equal{} 1 \plus{} 3 \plus{} 5 \plus{} \dots \plus{} (2n \minus{} 1)$. $ S$ can also be rewritten as follows:
$ S \equal{} (2n \minus{} 1) \plus{} (2n \minus{} 3) \plus{} (2n \minus{} 5) \plus{} \dots \plus{} 1$

Now we pair up (similar to what MathAndKnowledge said). Notice that each pair adds up to $ 2n$, and there are $ \frac n2$ pairs, so multiplying, we get $ 2n \cdot \frac n2 \equal{} \boxed {n^2}$.[/hide]

[quote="Yongyi781"][quote="Math Geek"]It's kinda funny that like 50 ppl are responding to this cuz it's an easy question. :rotfl:

@YongYi, yes, we all know you can use induction.  Stop showing off. :)[/quote]

What I find funny is that you didn't contribute anything to this topic. :|

The reason I took the steps to [i]prove[/i] it was because it was necessary to show that it is true, rather than just "spoon-feeding" them the "formula" [size=59](Kinda like what they do in middle schools nowadays)[/size]. :)

And if you think this is showing off, then you have no idea of the effort I took to make this as simple as possible. :([/quote]

Well, I wasn't making fun of you for showing off.  I was just implying that you're too smart for your own good. :P   (Though proving it with induction isn't difficult.)

well i dont think yongyi was showing off!
|

$ S \equal{} 1 \plus{} 3 \plus{} 5 \plus{} 7\cdots \equal{} \sum_{L \equal{} 1}^{n}2L\minus{}1$

$ \implies S\equal{}\sum_{L \equal{} 1}^{n}2L\minus{}1\equal{}\sum_{L \equal{} 1}^{n}2L\minus{}\sum_{L \equal{} 1}^{n}1$

$ \implies S\equal{}2\sum_{L \equal{} 1}^{n}L\minus{}\sum_{L \equal{} 1}^{n}1\equal{}2\times\frac{1}{2}n(n\plus{}1)\minus{}n$

$ \implies S\equal{}n^2\plus{}n\minus{}n$
$ \implies S\equal{} n^2$
$ \implies S \equal{} n^2 \equal{} 400$
$ \implies n \equal{} 20$

therefore the answer is $ \boxed{\boxed{\boxed{\color{red}20}}}$

all of u got it right...but i dont even know what induction is...so...

Induction is basically:
[list=1]
[*]Showing it works for a starting value, typically $ 1$.
[*]Showing that [b]if[/b] it works for general $ n$, it works for $ n \plus{} 1$, by manipulation, etc.[/list]

The second step, combined with the first, essentially says that if it works for 1, it works for 2. Since it works for 2, it works for 3. Etc, etc.

well thats basically wat mathematic
al induction is?

[quote="Yongyi781"]

The reason I took the steps to [i]prove[/i] it was because it was necessary to show that it is true, rather than just "spoon-feeding" them the "formula" [size=59]([b]Kinda like what they do in middle schools nowadays[/b])[/size]. :)

[/quote]

You've been reading the Middle School Teacher Forums!  :P

hey i still didnt whether i got my ans in the right way or noT!

[quote="gogators"]all of u got it right[/quote]

u got it right


你好

你知不知道我可以在哪裡找到趣味高等代數 ,三角問題?


Find integers $ m$, $ n$ such that $ m^3\plus{}m^2\plus{}m\plus{}1 \equal{} n^3$.

For $ m>0$, $ m^3<m^3\plus{}m^2\plus{}m\plus{}1<m^3\plus{}3m^2\plus{}3m\plus{}1\equal{}(m\plus{}1)^3$, so there are no solutions in $ m,n$.
For $ m\equal{}0$, $ m^3\plus{}m^2\plus{}m\plus{}1\equal{}1\equal{}1^3$, so $ n\equal{}1$.
For $ m\equal{}\minus{}1$, $ m^3\plus{}m^2\plus{}m\plus{}1\equal{}0\equal{}0^3$, so $ n\equal{}0$.
For $ m<\minus{}1$, $ m^3<m^3\plus{}m^2\plus{}m\plus{}1<m^3\plus{}3m^2\plus{}3m\plus{}1\equal{}(m\plus{}1)^3$, so there are no solutions in $ m,n$.

Thus, the ordered pair solutions are $ \boxed{(m,n)\equal{}(0,1),(\minus{}1,0)}$.


[quote="Silverfalcon"]If the points (2,-3), (4,3) and (5, k/2) lie on a line, k equals

$ \textbf{(A)}\ 6 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$[/quote]

[hide]
The slope is 3, obviously
so when x increases by 1 from 4 to 5
y increases by 3 from 3 to 6
ans: (D)12

[/hide]

[hide]The points must be collinear, so their slopes must be the same:
\[ \frac{\minus{}3\minus{}3}{2\minus{}4}\equal{}\frac{3\minus{}\frac{k}{2}}{4\minus{}5} \\
\Rightarrow 3\equal{}\frac{k}{2}\minus{}3 \\
\Rightarrow 6\equal{}\frac{k}{2} \\
\Rightarrow 12\equal{}k \\
\Rightarrow \boxed{D}\][/hide]


calculus
integration
logarithms
trigonometry
calculus computations

hello, do you mean
$ \int_{0}^{\frac{\pi}{3}}x\ln\left(2\sin\left(\frac{x}{2}\right)\right)\,dx$?
Sonnhard.


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let E/F be a normal extension whose Galois group is A5. Show there is no
intermediate field F included in K included in E with [B:F]=2

The corresponding group of automorphisms is a subgroup of index $ 2$, thus normal, contradicting the fact that $ A_5$ is simple.


Let $ x,y,z>0$.Prove that :
$ 3(\sqrt{x(x\plus{}y)(x\plus{}z)}\plus{}\sqrt{y(y\plus{}z)(y\plus{}x)}\plus{}\sqrt{z(z\plus{}x)(z\plus{}y)})^2\leq 4(x\plus{}y\plus{}z)^3$

By Cauchy-Schwarz ineq , we have :
$ LHS \leq 3(x \plus{} y \plus{} z)( \sum x^2 \plus{} yz \plus{} zx \plus{} xy)$
Then we prove that :
$ 4(x \plus{} y \plus{} z)^2 \geq 3[ (x \plus{} y \plus{} z)^2 \plus{} xy \plus{} yz \plus{} zx]$
$ \leftrightarrow \sum (x \minus{} y)^2 \geq 0$
Done :D


probability
calculus
integration
search
Pascal\u0027s Triangle

First of all, what is [b]Intermediate[/b] Counting and Probability doing here?

What are proofs? :o :? Well, it's when you prove something. For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?

[quote="chess64"]What are proofs? :o :? Well, it's when you prove something. [/quote]

 :rotfl: duh

[quote="chess64"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?[/quote]

How would you prove that?

[quote="chess64"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :? [/quote]

how would you do that? what are integral squares

I think integral squares are squares of integers

[quote="236factorial"][quote="chess64"]For example, prove that 2002 can be expressed as the sum of two integral squares. Haven't you ever seen that kind of problem before? :?[/quote]

How would you prove that?[/quote]

:mad: this will help on my competition :mad:

[hide]
All perfect squares are $\equiv 0$ or $1 \mod {4}$

$2002 \equiv 2 \equiv 1+1 \mod {4}$
Therefore it is possible. Or you could just find the integer squares.
[/hide]

It will help on your competition  :D  Thanks  :rotfl:

Just kidding, I'm still a little confused  :?  NOT  :D

[quote="chess64"]Aha, but [hide]can you prove that all perfect squares are $\equiv 0,1 \mod {4}$?[/hide] :idea: :D[/quote]

yeah  :roll:
maybe.

Maybe it comes from the following:

For all integers $n$, we have $n\equiv 0,1 \bmod{2}$. Therefore, $n^2\equiv 0^2,1^2 \bmod{2^2}$.

Does that work?

Questions about classes should be asked in the Classes Information section. And please do a Google define: search before posting questions such as "what is pascal's triangle" and 'what is a proof'. Type this into google: define:pascals triangle or define:math proof.

Topic locked


Here is a problem an interesting problem on Fibonacci Sequences I have tried but have been unable to solve it, any help given would be appreciated since I think its a very interesting problem.

The Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...) is defined where $ F_1 \equal{} 0, F_2 \equal{} 1$ and $ F_n \equal{} F_n \minus{} 1 \plus{} F_n \minus{} 2$ for n≥2.

Show that $ F_n(F_n \minus{} 1 \plus{} F_n \minus{} 2)$ is a Fibonacci number for all values of n.

Show that $ (F_n \minus{} 1)^2 \plus{} (F_n \minus{} 2)^2$ is a Fibonacci number for all values of n.

[quote="The Anomaly"]Here is a problem I could remember for my Cambridge Interview Exam this past winter, I have tried but have been unable to solve it, any help given would be appreciated since I think its a very interesting problem.

The Fibonacci sequence (0, 1, 1, 2, 3, 5, 8...) is defined where $ F_1 \equal{} 0, F_2 \equal{} 1$ and $ F_n \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2}$ for $ n\ge 2$.

Show that $ F_n(F_{n \minus{} 1} \plus{} F_{n \minus{} 2})$ is a Fibonacci number for all values of n.

Show that $ (F_{n \minus{} 1})^2 \plus{} (F_{n \minus{} 2})^2$ is a Fibonacci number for all values of n.[/quote]

I'll assume you meant it this way  :)

First one doesn't make sense, since $ F_n \equal{} F_{n \minus{} 1} \plus{} F_{n \minus{} 2}$ so it's basically $ F_n^2$, and that's never a Fibonacci number (except for $ F_n \equal{} 1$).
The second one is correct, and moreover it's $ (F_{n \minus{} 1})^2 \plus{} (F_{n \minus{} 2})^2 \equal{} F_{2n \minus{} 3}$ in this setting of indices. Now you know the exact formula, you can (probably) try proving it using strong induction, which would eventually get you to products of Fibonacci numbers. But we can do better than that:

[hide="Solution"]
[quote="matt276eagles"]
Lemma: $ f_{i} \equal{} f_{j}f_{i \minus{} j \minus{} 1} \plus{} f_{j \plus{} 1}f_{i \minus{} j}$.

Proof: We induct on j. For j=1, the lemma becomes $ f_{i} \equal{} f_{i \minus{} 2} \plus{} f_{i \minus{} 1}$, which is just the definition of Fibonacci numbers. Now assume the lemma is true for j=k, so that $ f_{i} \equal{} f_{k}f_{i \minus{} k \minus{} 1} \plus{} f_{k \plus{} 1}f_{i \minus{} k}$. Then

$ f_{i} \equal{} f_{k \plus{} 1}(f_{i \minus{} k \minus{} 2} \plus{} f_{i \minus{} k \minus{} 1}) \plus{} f_{k}f_{i \minus{} k \minus{} 1}$
$ f_{i} \equal{} f_{k \plus{} 1}f_{i \minus{} k \minus{} 2} \plus{} (f_{k} \plus{} f_{k \plus{} 1})f_{i \minus{} k \minus{} 1}$
$ f_{i} \equal{} f_{k \plus{} 1}f_{i \minus{} k \minus{} 2} \plus{} f_{k \plus{} 2}f_{i \minus{} k \minus{} 1}$

This is our lemma for $ j \equal{} k \plus{} 1$, so the induction is complete, and the lemma is proved for all i and j.

Now, let $ i \equal{} 2n \minus{} 3$ and $ j \equal{} n\minus{}2$ in our lemma. We get $ f_{2n \minus{}3} \equal{} f_{n\minus{}2}^{2} \plus{} f_{n\minus{}1}^{2}$[/quote]

[/hide]

Sorry my bad, I made an error in writing the first one, I meant to say this:

Show that $ F_n(F_n \minus{} 1 \plus{} F_n \minus{} 1)$ is a Fibonacci number for all values of n.


vector
topology
real analysis
real analysis unsolved

Let $ (X,d)$ be a metric space. We said $ (X,d)$ has the homogeniety property if there exists $ N>0$ such that for any open ball $ B(x,r)\equal{}\{y: d(y,x)<r\}$ then there are at most $ N$ points in $ B(x,r)$ for which $ d(x_i,x_j)\geq \frac{r}{2}$ for all $ i\neq j$ from $ 1$ to $ N$.
Here're some questions to discuss:
1) Is $ {\mathbb R}^n$ with metric Euclidean has homogeniety or not? Explain your answer.
2) Give some examples in which $ (X,d)$ has not homogeniety.
3) Give some more examples in which $ (X,d)$ has homogeniety.
Happy working :)

1. Yes. It basically means we can find a finite subcover of balls of radius $ \frac {r}{2}$ to cover ball of radius $ r$ "uniformly". We find N for $ r \equal{} 1$, and since any other balls are just translates or rescaling, the same N works.

2. discrete metric space with infinitely many points.

3. Any normed vector space should do.

[quote="pluricomplex"][quote="Soarer"]
3. Any normed vector space should do.[/quote]
Could you explain this more detains?

Now is the fouth question:

4) Let $ \mu$ is doubling nontrivial Borel measure on $ X$, ie... for any ball $ B(x,r)$ we must have $ \mu(B(x,r))\leq c\mu(B(x,r/2))$, where $ c > 0$ is some constant. Show that $ X$ has the homogeniety property defined above.[/quote]

I'm sorry, I meant to say proper normed vector space (those with $ B(x,r)$ compact)

May I know what is doubling measure?

The definition of doubling measure I gave it above  :roll:

Let $ x \in X$ and $ r > 0$. If $ y \in B(x, r)$ we note that
$ \mu(B(x,r)) \le \mu(B(y, r \plus{} d(x,y)) \le \mu(B(y, 2r)) \le c^3\mu(B(y,r/4))$.

Now let $ x_1, \ldots, x_N \in B(x,r)$ so that $ d(x_i, x_j) \ge r/2$, $ i \ne j$.  We note that the balls
$ B(x_i, r/4)$ are disjoint and $ B(x_i, r/4) \subset B(x, 5r/4) \subset B(x, 2r)$. Now we get
$ N\mu(B(x,r)) \le c^3\sum_{i\equal{}1}^N \mu(B(x_i, r/4)) \le c^3\mu(B(x,2r)) \le c^4\mu(B(x,r))$ whence
$ N \le c^4$.

Okey it's time for the fivth question

5) Give some examples for which  $ \mu$ is not a doubling measure.


[quote="Phelpedo"]Eleven teachers run a conference. Every hour, one or more teachers give one-hour presentations, while all of the other teachers observe the presentations. Find the least amount of time during which it is possible for each teacher to observe all other presentations at least once.[/quote]
We can solve recursively. Divide the eleven teachers into one group of six and one of five. In the first hour the group of five give presentations to the group of six. In the second hour that is reversed. Now each teacher needs to see the presentations of his own group only. Divide the group of five into one group of three and one of two. Divide the group of six into one group of three and another of three. Now we spend two more hours with one group presenting to the other group, and then reversing. Now we have each teacher needing to listen to two other presentations. This can be done in three hours: just have each teacher stand up and present in turn. This takes a total of $7$ hours. I haven't come up with a better method.

Hmm, I think this question needs a little explanation.  WatsonLadd's solution assumes that the observing teachers can watch more than one presentation at a time, but I interpreted it meaning each hour, each teacher can either present or observe 1 presentation, which would make the answer much larger...

That was also how I interpreted it, although I'm not sure "much larger" is really accurate.  After all, the simplest-looking solution is just to have one person present each hour and have all 10 others watch, for a total of 11 hours.  WatsonLad's interpretation seems to be for a more natural question, though.

The problem is obviously trivial with that interpretation, so we are probably meant to assume that the non-presenters observe all presentations simultaneously.


calculus
counting
derangement
search
calculus computations

Find the general fomular of $ x_n$:
$ x_n \equal{} (n \minus{} 1)(x_{n \minus{} 1} \plus{} x_{n \minus{} 2}), n \geq 4 \\
x_2 \equal{} 1, x_3 \equal{} 2$

Has got what to do with calculus, exactly?  It's counting derangements; the numbers themselves are subfactorial or rencontres numbers; the sequence is [url=http://www.research.att.com/~njas/sequences/A000166]A000166[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  A search for any of the key words in this post will find you plenty of information.


This one is famous, but neat:

Find the unique 9 digit number containing each of the digits 1 through 9 such that:

- the whole number is divisible by 9
- when the last  digit (on the right) is removed, the remaining number is divisible by 8
- when the last 2 digits are removed, the remaining number is divisible by 7
- when the last 3 digits are removed, the remaining number is divisible by 6, and so on....

number is [size=75][color=darkred](hidden by moderator)[/color][/size]
[hide]
                 381654729

i have find it with lots of efforts.[/hide]

:o Hide your answer so others can try it!!!

Before you start typing whatever it is you want to hide, click the button that says "Hide" just above the text window at the right end of the list of buttons.  When you are done with whatever you want to hide click the hide button again.


I'm very bad at inequalities. Is there some elegant way to prove that $5x^{7}+2>3x$ as $0\leq x\leq 1$?

I think Ravi B means

[hide]Apply the AM-GM inequality to
\[5 x^{7}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\, . \][/hide]

Oops, I can't add.  SplashD is correct about what I meant.


can we find the solution of the following equation in positive integers
a^2+b^2-ab=2005.
best regards
(hope this is not too easy.returning to mathematics after quite some time)

I suppose I ignored the case where 5|(a+b) and 5|(a-b), but that should be okay because then $25|a^{2}+b^{2}-ab=2005$ a contradiction.

Generalisation:
Every integer $n$ can be written as $n=x^2-xy+y^2$ iff every prime divisor $p,p\equiv 2 \mod 3$ occurs an even number of times in the prime factorisation of $n$.
More generally, there are $ 6 \cdot \left( \sum_{\substack{ d \in \mathbb{N} \\  d \mid n \\ d \equiv 1 \mod 3}} 1 - \sum_{\substack{ d \in \mathbb{N} \\  d \mid n \\ d \equiv 2 \mod 3}} 1 \right)$ different ways to write $n$ as $n=x^2-xy+y^2$ (the pairs $(x,y),(y,x),(-x,-y)$ etc. are considered as different)


algebra
system of equations
algebra unsolved

solve th system of equations :
                                               $\sqrt{3x}(1+\frac{1}{x+y})=2$
                                                $\sqrt{7y}(1-\frac{1}{x+y})=4\sqrt2$

This is a problem in Vietnam National Olympiad. It is a problem 1 in year 1996

ýes ,all right .can you póst solution ?  :P


Vietnamese Undergraduate Mathematical Competition 2004 was held in Quy Nhon last week. My students won 3 gold, 3 silver and 5 bronze. I will post all problems (two sections - Analysis and Algebra) time by time. Enjoy the promlems!

Namdung

Congratulations :) but which of your students are on ML? :) so that we can congrat them too :P


limit
trigonometry
calculus
calculus computations

hi!!
evaluate this limit without using Taylor series or hopital's rule. :

$ \lim_{x\to0}\frac{x\minus{}\sin{x}}{x\minus{}\tan{x}}$

thanx.

[quote="kunny"]$ \tan x \equal{} \frac {\sin x}{\cos x}$[/quote]

hi! , can you explain more ?  :blush:

thanx again :)

divide numerator and denominator by x^3
then let x = 3y
first solve lim (x-sinx)/x^3 . let x=3y
lim [3y-sin(3y)]/3y ,sin3x= 3sinx-4sin^3(x)
also:lim(x-tanx)/x^3=lim[3y-tan3y]/3y
=lim(3y- [3tany-tan^3y](1-3tan^2y))/27y^3

Why not use Taloy series or H'ospital rules ? :?:


ratio
geometry
probability
3D geometry
number theory
prime numbers

1. How many positive three-digit integers less than 400 satisfy each of the following characteristics simultaneously?(#7 97 team)
A. Each of the three digits is prime
B. The sum of the three digits is prime
C. The number is prime.
[hide="Answer"]5 numbers[/hide]
2. Points $ A(0,0), B(9,6)$ and $ C(6,12)$ are vertices of triangle ABC. Point D is on segment $ AB$ such that $ 2(AD)\equal{}DB$, point $ E$ is on segment $ BC$ such that $ 2(BE)\equal{}EC$ and point $ F$ is on segment $ CA$ such that $ 2(CF)\equal{}FA$. What is the ratio of the area of triangle $ DEF$ to the area of triangle $ ABC?$ Express your answer as a common fraction.(#7 03 team)
[hide="Answer"]$ \frac13$[/hide]
3. An integer between 0 and 5000 inclusive is selected at random and is found to be a perfect cube. What is the probability that it is also a perfect fourth power? Express your answer as a common fraction. (#5 97 target)
[hide="Answer"]$ \frac16$[/hide]

#1:

[hide]The only prime numbers are 2,3,5, and 7 that are one-digit numbers.
The number can't end with 5 either because numbers that end with 5 are multiples of 5.
The numbers are :
223, 227, 337,353,373
Edit: changed one number to 373
[/hide]

[quote="ac-king"]#1:

[hide]The only prime numbers are 2,3,5, and 7 that are one-digit numbers.
The number can't end with 5 either because numbers that end with 5 are multiples of 5.
The numbers are :
223, 227, 337,377, and 353[/hide][/quote]

How do you know did you that those are the only ones that are prime, besides the fact that the answer is 5? I can't recognize primes quickly.

All the numbers below 400 that follow all the restrictions (except for the one with the number being a prime number) are:
223, 227, 335, 337,353, 373, 377.
335 ends with 5 so it can't be one.

For the rest of them, you have to check them out with the divisibility rules.

Then you have to check the 7,13,17, and 19.
It isn't necessary to check the ones any higher because for any pair of numbers that multiply up to these numbers, the lower number can't be higher than 19 because all of these numbers are lower than 400.
You can find that 377 is 29*13 and the rest are prime.

[quote="JRav"]3) Well there are only 18 perfect cubes including 0 and 5000, and of those, the only ones are $ 0$, $ 1$, and $ 4096$.[/quote]
here's a potentially more thorough solution-you can't just check every cube up to 5000 for 4th powered-ness  :rotfl: i think this is a  bit more practical...i did this problem timed yesterday
if a number is a cube and a 4th power simultaneously, then it is a 12th power(make sure you understand this)

well what is probability?
probability is defined as the number of successes divided by the total number of outcomes.
how many outcomes are there?
well, how many cubes couuld we have picked less than 5000? 18(possible outcomes)(please verify)
how many 12th powers are there(the successes) only 3.
so the answer is successes divided by outcomes=3/18=1/6

#2 (03 nat Team 7)
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=816546&search_id=1702538419#816546[/url] for the solution.


MATHCOUNTS
blogs
parameterization
algebra
system of equations

The sides of an equiangular hexagon measure $ 10, 6, 12, 14, x,$ and $ y$
units, in the order given. Find $ x$ and $ y$.

This was actually very similar to a National MATHCOUNTS Sprint round problem. (Actually, it was #28 in the 2008 competition). :)

[hide="The General Result"]Suppose the side lengths are a, b, c, d, e, and f.

Orient the hexagon so that sides of length a and d are both horizontal. Then, the "height" of the hexagon in terms of b and c is [color=blue][b]sin60*b+sin60*c=sin60*(b+c)[/b][/color]. The height of the hexagon in terms of e and f is, similarly, [b][color=blue]sin60*(e+f)[/color][/b].

Since under the same orientation, the height is the same, we can equate to get [b][color=blue]sin60*(b+c)=sin60*(e+f)[/color][/b], or [b][color=blue]b+c=e+f[/color][/b].

Similarly, considering the other two orientations, we get the system of equations:
[b]
[color=blue]a+b=d+e
b+c=e+f
c+d=f+a[/color]
[/b][/hide]

[hide="Particular solution"]We have the system of equations:
10+6=14+x
6+12=x+y
12+14=y+10

This system is overdetermined -- we can solve it using the first and third equations only, and use the second to check our answer. We find that [color=green][b](x,y)=(2,16)[/b][/color]

[/hide]

Just extend the sides to form an equilateral triangle.

[quote="The Zuton Force"][hide="The General Result"]Suppose the side lengths are a, b, c, d, e, and f.

Orient the hexagon so that sides of length a and d are both horizontal. Then, the "height" of the hexagon in terms of b and c is [color=blue][b]sin60*b+sin60*c=sin60*(b+c)[/b][/color]. The height of the hexagon in terms of e and f is, similarly, [b][color=blue]sin60*(e+f)[/color][/b].

Since under the same orientation, the height is the same, we can equate to get [b][color=blue]sin60*(b+c)=sin60*(e+f)[/color][/b], or [b][color=blue]b+c=e+f[/color][/b].

Similarly, considering the other two orientations, we get the system of equations:
[b]
[color=blue]a+b=d+e
b+c=e+f
c+d=f+a[/color]
[/b][/hide]
[/quote]
[hide="More concisely..."]
$ |a\minus{}d|\equal{}|b\minus{}e|\equal{}|c\minus{}f|$.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=152452]my old blog post[/url].  Verify that the parameterization I gave is equivalent to The Zuton Force's conditions.


I just downloaded some movies via torrent, in the dvdrip.xvid format. But the only thing I see, is a bunch of .rar files.
How can I get the movies to play?

Thanks in advance :)


search
trigonometry
geometry unsolved
geometry

Let $ (O)$ be a circle and a point $ S$ in a plane.Draw $ SA;SB$ is tangent of $ (O)(A;B \in (O))$ .$ AB$ cuts $ SO$ at $ H$
Let $ d$ be midperpendicular of $ SH$.$ M$ is a point on $ d$.Draw $ MN;MP$ is tagent of $ (O)$$ (N;P \in (O))$
$ NP$  cuts $ d$ at $ E$
Prove that $ \angle MSE \equal{} 90^o$

Midpoint of $ [SH]$ is $ K$. $ PMKO$ , $ PKNO$ , $ MKNO$ are cyclic quadrilaterals.

$ |OH|.|OS|\equal{}|ON|^2 \Longrightarrow$  $ m(ONH)\equal{}m(OSN)\equal{} \gamma$ , $ m(KPN)\equal{}m(KON)\equal{}\beta$ , $ m(MPK)\equal{}m(MNK)\equal{}m(MOK)\equal{}m(NEK)\equal{}\alpha$.

We will prove that $ |MN|\equal{}|MH|$. Let $ |NO|\equal{}x$ so $ |MN|\equal{}xtan(\alpha\plus{}\beta)$ , $ |NH|\equal{}\frac{xsin\beta}{sin(\beta\plus{}\gamma)}$. If we prove that

$ \frac{sin\beta}{sin(\beta\plus{}\gamma)tan(\alpha\plus{}\beta)}\equal{}2sin\gamma$ solution is done.

We will prove that $ tan(\alpha\plus{}\beta)\equal{}\frac{sin\beta}{2sin\gamma.sin(\beta\plus{}\gamma)}$

Lemma: Triangle $ ABC$ is given. $ m(ABC)\equal{}\alpha$ , $ m(ACB)\equal{}\beta$ , $ D$ is the midpoint of $ [BC]$ and $ m(ADC)\equal{}\theta$

$ cot\theta\equal{}\frac{sin(\beta\minus{}\alpha)}{2sin\beta.sin\alpha}$ If we prove this lemma solution is done because in triangle $ SNO$, $ |SK|\equal{}|KH|$ and

$ m(NKO)\equal{}90\minus{}\alpha\minus{}\beta$ so $ cot(90\minus{}\alpha\minus{}\beta)\equal{}\frac{sin\beta}{2sin\gamma.\sin(\beta\plus{}\gamma)}$.

Let prove this lemma:

$ |BD|\equal{}|DC|$ hence $ \frac{sin\alpha}{sin(\theta\minus{}\alpha)}\equal{}\frac{sin\beta}{sin(\beta\plus{}\theta)}$

$ \frac{sin(\theta \minus{}\alpha)}{sin(\beta \plus{}\theta)}\equal{}\frac{sin\alpha}{sin\beta}$

$ \frac{sin\theta.cos\alpha\minus{}cos\theta.sin\alpha}{sin\beta.cos\theta\plus{}cos\beta.sin\theta}\equal{}\frac{sin\alpha}{sin\beta}$

$ cot\theta\equal{}\frac{sin\beta.cos\alpha\minus{}cos\beta.sin\alpha}{2sin\alpha.sin\beta}\equal{}\frac{sin(\beta\minus{}\alpha)}{2sin\alpha.sin\beta}$ and solution is done.  :wink:


Let $ n_1,n_2,n_3...$ be the odd numbers none of which has a prime divisors greater than 5, prove that

$ \frac{1}{n_1}\plus{}\frac{1}{n_2}\plus{}\frac{1}{n_3}\plus{}... <\frac{15}{8}$

[hide]Any odd number that has no prime divisor greater than $ 5$ is in the form $ 3^a \cdot 5^b$ where $ a,b \in \mathbb{N}_0$

Thus:
$ \displaystyle \sum_{i\equal{}1}^{\infty}\dfrac{1}{n_i} \equal{} \sum_{a\equal{}0}^{\infty}\sum_{b\equal{}0}^{\infty}\dfrac{1}{3^a \cdot 5^b} \equal{} \sum_{a\equal{}0}^{\infty}\dfrac{1}{3^a}\sum_{b\equal{}0}^{\infty}\dfrac{1}{5^b} \equal{} \left(\sum_{a\equal{}0}^{\infty}\dfrac{1}{3^a}\right)\left(\sum_{b\equal{}0}^{\infty}\dfrac{1}{5^b}\right)$ $ \equal{} \dfrac{1}{1\minus{}\frac{1}{3}} \cdot \dfrac{1}{1\minus{}\frac{1}{5}} \equal{} \dfrac{3}{2} \cdot \dfrac{5}{4} \equal{} \dfrac{15}{8}$[/hide]


Let $A=12345...20062007$   (Write 2007  numeral from 1 -->2007 )
Prove : $A$   not  $\vdots$ 9

(We can calculate the sum of numerals from A to prove ?)
 ( :maybe: )
Louis Latin and Vicky

[quote="Vicky and Louis Latin"]
(We can calculate the sum of numerals from A to prove ?)
 [/quote]

Yes, you can do that easily... :wink:

Pierre.

Oh , no !
It very dificult because numerals from A not to keep to law !
It to be different from find the sum of sequence !
( :maybe: )
[i]Louis Latin and Vicky[/i]

Let s(A) is sum of digits A. Because $n=s(n)(mod 9)$ we have
$s(A)=\frac{2007*2008}{2}=0(mod 9)$.

Rust you are amazing,but could you proove what you just said ?
maybe it is obvious to you but not to me


Try this... after some experimenting with factorials I came up with this. perhaps it;'s beginner level, I've never done any American contests.

Simplify: [((c^d)-1)!((c^d)+1)]/[((c^2d)-1)((c^d)-2)!]

btw plz spoiler answers. thankyou.


In how many ways can $n$ people be seated in a row of $r$ chairs given that no two people may sit next to one another?

[hide]
Assuming that each person is indistinct:
Sit each person down in a chair. You have used up $n$ chairs. Put a chair in between each person to make sure they aren't sitting next to each other. You have used up $n-1$ more chairs.
Now you're left with $r-n-(n-1)=2n-1$ chairs. You can add these chairs in between any two people or on either edge, so there's $n+1$ spots where you can add the rest of the chairs.
Using the balls & urns method, you get:
$\binom{r-2n+1+n}{n}= \boxed{\binom{r-n+1}{n}}$
If each person is distinct, you can just permute each of the possibilities to get $\binom{r-n+1}{n}n!$
[/hide]

[quote="computer_nuke"][quote="scorpius119"][hide]
No two people can sit next to each other if nobody sits in the seat immediately after each person. So if we label empty chairs by $E$'s and chairs with a person sitting in them by $S$, we can split a string of $E$'s and $S$'s into $SE$'s and $E$'s...

[b]unless[/b] the last seat is taken, in which case there is no next seat. But we can deal with this anyway by just adding another seat (so now there are $r+1$) and requiring that no person sits in the last seat. Now the string of length $r+1$ can be split into $n$ $SE$'s and $r+1-2n$ $E$'s, for a total of $r+1-n$ sections.

The number of seatings is just the number of ways of choosing the positions for $n$ $SE$'s out of $r+1-n$ possibilities, which is $\boxed{\binom{r+1-n}{n}}$.
[/hide][/quote]

That's what I got, but note that r>2n-1 for it to be possible to arrange n people into a row of r chairs such that no two people may sit next to each other.[/quote]

it should be $r \geq 2n-1$. if the people are distinct then of course this becomes [hide]$\binom{r-n+1}{n}n!$.[/hide]

Whats the balls and urns method? I don't really understand how you got (r-2n+1+n)C(n).


Consider a real poylnomial $p(x)=a_nx^n+...+a_1x+a_0$.
(a) If $\deg(p(x))>2$ prove that $\deg(p(x)) = 2 + deg(p(x+1)+p(x-1)-2p(x))$.
(b) Let $p(x)$ a polynomial for which there are real constants $r,s$ so that for all real $x$ we have \[ p(x+1)+p(x-1)-rp(x)-s=0 \]Prove $\deg(p(x))\le 2$.
(c) Show, in (b) that $s=0$ implies $a_2=0$.

for a) let $n=deg(p(x))$. then let us look at $x^n,x^{n-1},x^{n-2}$ and their coefficents.
the result of $p(x+1)+p(x-1)-2\cdot p(x)$ has $a_n+a_n-2a_n=0$ many $x^n$, $({n\choose 1} \cdot a_n+a_{n-1})+(-{n\choose 1}a_n+a_{n-1})-2a_{n-1})=0$ many $x^{n-1}$, and $({n\choose2}a_n+{n-1\choose1}a_{n-1}+a_{n-2})+({n\choose2}a_n-{n-1\choose1}a_{n-1}+a_{n-2})-2a_{n-2}=2\cdot {n\choose2}a_n$ many $x^{n-2}$ in it. since $n(n-1)a_n\neq 0$ because $n>0$ and $a_n\neq 0$ it follows that $deg(p(x+1)+p(x-1)-2\cdot p(x))=n-2$ and hence $deg(p(x))=2+deg(p(x+1)+p(x-1)-2\cdot p(x))$.

for b) let us assume $n=deg(p(x))>2$.
then we have $a_nx^n+a_nx^n-ra_nx^n=$ so $r=2$.
from a) we saw that $deg(p(x+1)+p(x-1)-2p(x))=p(x)-2>0$, so just adding a constant cant result that the sum of plynoms is zero for all real x. contradiction -> the opposite is true

for c)
[quote="Peter VDD"]
(c) Show, in (b) that $s=0$ implies $c=0$.
[/quote]
what is $c$ ?

c) has been edited. (also for the post below)

Are you sure that is the correct version Peter?

I think that if $s = 0$, then the polynomial must be constant or linear, you see... :)

i guess that makes sense ... (that $a_2=0$, which implies that $p$ is either linear or constant)


LaTeX
geometry
3D geometry
tetrahedron
AMC 10
AMC

I know joml88 is doing a mock AMC H tomorrow (which I already did) and other ppl (including myself) did mock AMC, which were mostly on level of AMC-12. Of course, Mildorf did a wonderful job on the mock AIME.

But I wonder if anyone can host a mock AMC-10. I did all the past tests and some AHSMEs and I wonder if I can have some extra tests, if possible.

Thank you.

:(

No reply yet.... :(

I wonder whether this thread would ever get replies...

I guess it would be gone soon...  :(

The AMC's are on Febuary 1st and 16th, so having a mock one on January 30th should be okay with me.

[quote="Lucky707"]I collaborated on a Mock AMC with Singthesorrow (who doesn't seem to be around) back in August. I still have the questions but would have to create "fake" answers and then put it into LaTEX (which I don't know how to do)... It's about the level of a Mock AMC 10, maybe slightly higher like an 11...

I have exams coming up though so I'm not too sure I would be able to administer it within January (my last exam is on the 27th, I might be able to do the 29th or 30th)

When are the AMCs?[/quote]

I'd be willing to put it into LaTeX and proofread it, since I'm not sure it would help me too much if I took it.

I'm willing to help as well. I'm a bit old to do the AMC-10 this year.

Yeah, just send the test to tetrahedr0n and he will tex it for you.

Also, when he tex it, I'm pretty sure he'll check for any errors.

 :)

Believe it or not I just finished it...

It's on its way for correcting as you read.

I sent it to tetrahedron yesterday...

Would this be the 10th mock AMC? because we just had H and H=8... so would this be Mock AMC 9 or will someone make one before the 30th? Shall I call it Mock AMC 9 because Mock AMC I looks like Mock AMC 1

Actually it would be the eleventh (1, 2, A-H)


number theory
least common multiple
function

pls help to find all the possible solutions of x & y where

x - y^3 = LCM(x, y)

Is there a quick way to solve the quest instead of trial and error?

I don't know if it might help, but:

$ xy \equal{} LCM (x,y) * GCM (x,y)$.

I got this info. Thanks.

I wrote a vb script in microsoft excel to find the answer since excel provides a lcm function to ease my life. But I am interested in mathematical method to solve it.

cheers

The important observation is that $ y | \text{lcm}(x, y) \Rightarrow y | x$.  From here it's straightforward.

I'm confused, isn't $ lcm(x,y) \ge x$? Then y would have to be 0.....
And what do you mean, t0rajir0u? Doesn't y always divide lcm(x,y)?

We are given that $ x \equal{} y^3 \plus{} \text{lcm}(x, y)$.  Since $ y | y^3$ and $ y | \text{lcm}(x, y)$ it follows that $ y | x$.  The problem is straightforward from here.

[quote="mathwizarddude"]Are x and y allowed to be negative (I guess not...)?[/quote]
I don't think so.

Hmm.  Well, if $ x$ and $ y$ are both non-negative then the question is trivial since $ \text{lcm}(x, y) \ge x$, but if either $ x$ or $ y$ is allowed to be negative then the definition of $ \text{lcm}$ is ambiguous up to sign.

Is $ \text{lcm}(x,0)=0$?
How about $ \gcd(x,0)$?
Is lcm allowed to be negative (I guess not...)?

$ \gcd(x, 0)$ is $ x$ because every integer divides $ 0$.  $ \text{lcm}(x, 0)$ is $ 0$ because every multiple of $ 0$ is $ 0$.  (By the way, the standard compact notation is $ \text{lcm}(a, b) \equal{} [a, b]$; parentheses denote the $ \gcd$ without further qualification.)


algebra
binomial theorem
number theory unsolved
number theory

prove :

if $2^n+1$ is prime then $n=2^k$

That is easy. You should know the identity that x+y divides x^k+y^k if k is odd. The proof to this identity is a consequence of the binomial theorem: consider  x^k=[(x+y)+(-y)]^k and expand.

The proof to your question follows from this identity. Suppose n is not a power of 2, then it has an odd factor greater than 1; this is your k. Hence 2^n+1 is composite.

Therefore if 2^n+1 is prime then n=2^m for some positive integer m. These are called Fermat numbers and there is a conjecture that there are only finite many Fermat numbers which are prime.

By the way the nos. of the form 2^n+1 where n=2^k are called Fermat's Numbers

Yes, [e^pi], your conjecture is true, but it's an open problem if $2^{2^n}+1>2^{32}$ is always composite.


Let triangle ABC and H is its orthocenter.BH cuts AC at Q, CH cuts AB at P. M is the symmetric point of H wrt AB. K is the symmetric point of B wrt CH. Let point E such that $ \angle EMA\equal{}\angle EHB\equal{}90^o$. Prove that AE,PQ,CK are concurrent.

I prove your problem in few steps:
1.If {$ R$}$ \equal{} PQ\cap CK$ and {$ V$}$ \equal{} AR\cap KQ$ then $ HV\parallel{}AB$.
2.If $ BE'\parallel{}KQ$ $ (E'\in AR)$, {$ S$}$ \equal{} HE'\cap AB$, {$ T$}$ \equal{} HV\cap BE'$
  then $ \angle E'HB \equal{} 90^{\circ}$.
3.$ \angle E'MA \equal{} 90^{\circ}$
     So $ E \equal{} E'\Rightarrow AE,PQ,CK$ are concurrent.

1.It's easy to prove that the quadrilateral $ AQRK$ is cyclic

$ \frac {VQ}{VK} \equal{} \frac {AQ\cdot sin\angle QAR}{KR\cdot sin\angle ARK}$     (a)

$ \frac {QA}{QC}\cdot \frac {PK}{PA}\cdot \frac {RC}{RK} \equal{} 1\Rightarrow \frac {AQ}{KR} \equal{} \frac {PA}{PK}\cdot \frac {QC}{RC}$     (b)

It's easy to prove that $ \frac {sin\angle QAR}{sin\angle ARK} \equal{} \frac {QR}{AK}$     (c)

(a),(b),(c)$ \Rightarrow \frac {VQ}{VK} \equal{} \frac {PA}{PK}\cdot \frac {QC}{RC}\cdot \frac {QR}{AK}$     (d)

But $ \frac {PA}{PK} \equal{} \frac {PA}{PB} \equal{} \frac {HQ}{HB}\cdot \frac {CA}{CQ}$     (e)

(d),(e)$ \Rightarrow \frac {VQ}{VK} \equal{} \frac {HQ}{HB}\cdot \frac {AC}{AK}\cdot \frac {RQ}{RC} \equal{} \frac {HQ}{HB}\Rightarrow HV\parallel{}AB$

2.$ VT\parallel{}AB\Rightarrow \frac {SA}{HV} \equal{} \frac {E'S}{E'H} \equal{} \frac {SB}{HT}\Rightarrow \frac {SA}{SB} \equal{} \frac {HV}{HT}$     (f)

But $ BT\parallel{}VQ\Rightarrow \frac {HV}{HT} \equal{} \frac {HQ}{HB}$     (g)

(f),(g)$ \Rightarrow \frac {SA}{SB} \equal{} \frac {HQ}{HB}\Rightarrow SH\parallel{}AQ\Rightarrow HE'\parallel{}AC$

$ HE'\parallel{}AC$ and $ BQ\bot AC\Rightarrow BQ\bot HE'\Rightarrow \angle E'HB \equal{} 90^{\circ}$

3.It's easy to prove that $ MA\bot CK$. {$ D$}$ \equal{} MA\cap CK$

$ \frac {AE'}{AR} \equal{} \frac {AH\cdot sin\angle AHE'}{sin\angle AE'H}\cdot \frac {sin\angle ARK}{AD}$

$ AH \equal{} AM$, $ \angle AHE' \equal{} \angle HAC \equal{} 90^{\circ} \minus{} \angle ACB$ and $ \angle AE'H \equal{} \angle RAQ$

$ \Rightarrow \frac {AE'}{AR} \equal{} \frac {AM}{AD}\cdot \frac {cos\angle ACB\cdot sin\angle ARK}{sin\angle RAQ}$

Let's prove that $ \frac {cos\angle ACB\cdot sin\angle ARK}{sin\angle RAQ} \equal{} 1$.

It's easy to prove that $ \frac {sin\angle ARK}{sin\angle RAQ} \equal{} \frac {AK}{QR}$     (h)

$ \frac {AK}{AP}\cdot \frac {QP}{QR}\cdot \frac {CR}{CK} \equal{} 1\Rightarrow \frac {AK}{QR} \equal{} \frac {AP}{QP}\cdot \frac {CK}{CR} \equal{} \frac {AC}{BC}\cdot \frac {BC}{CR} \equal{}$
$ \equal{} \frac {AC}{CR} \equal{} \frac {CK}{CQ} \equal{} \frac {CB}{CQ} \equal{} \frac {1}{cos\angle ACB}$     (i)

(h),(i)$ \Rightarrow \frac {cos\angle ACB\cdot sin\angle ARK}{sin\angle RAQ} \equal{} 1$

So $ \frac {AE'}{AR} \equal{} \frac {AM}{AD}\Rightarrow ME'\parallel{}CK$. But $ MA\bot CK$, so $ MA\bot ME'\Rightarrow \angle E'MA \equal{} 90^{\circ}$.   :)

P.S.    $ M$ is the symmetric point of $ H$ wrt $ AB$ $ \Rightarrow$
         $ \Rightarrow$ the point $ M$ is on the circumcircle of the triangle $ ABC$.

Dear Petry,
You have a wonderful proof, however the step 3 in your post can be solved by a faster way (download my file).
This problem is a lemma of another my own problem, it was posted at [url]http://www.mathlinks.ro/viewtopic.php?t=216885[/url]
Sincerely, mr.danh


Find the radius and interval of convergence for the following power series.

$ \sum_{n \equal{} 2}^{\infty}\frac {(1 \plus{} 2cos\frac {\pi n}{4})^n}{lnn}x^n$

Attempt:

$ R \equal{} \frac {1}{\lim_{n\rightarrow\infty}\sqrt [n]{\frac {(1 \plus{} 2cos\frac {\pi n}{4})^n}{lnn}}} \equal{} \lim_{n\rightarrow\infty}\frac {e^{\frac {ln(lnn)}{n}}}{(1 \plus{} 2cos\frac {\pi n}{4})}$ In answers $ R \equal{} \frac {1}{3}$

$ \lim_{n\rightarrow\infty}e^{\frac {ln(lnn)}{n}} \equal{} 1$.Then is

$ \lim_{n\rightarrow\infty}(1 \plus{} 2cos\frac {\pi n}{4}) \equal{} 3$? :|

[quote="azatkgz"]Find the radius and interval of convergence for the following power series.

$ \sum_{n = 2}^{\infty}\;\frac {(1 + 2\,\cos\frac {\pi n}{4})^n}{\ln\,n}\,x^n$

answer:

$ R = \dfrac {1}{\limsup_{n\rightarrow\infty}\; {\bigg|\; \frac {(1 + 2\,\cos\frac {\pi n}{4})^n}{\ln\,n} \;\bigg| }^{\dfrac{1}{n}} } \; = \; \liminf_{n\rightarrow\infty}\; {\bigg|\; \frac {\ln\,n}{(1 + 2\,\cos\frac {\pi n}{4})^n}\;\bigg| }^{\dfrac{1}{n}}$

$ = \; \liminf_{n\rightarrow\infty}\;{\bigg|\; \frac {e^{\frac {\ln\,(\ln\,n)}{n}}}{1 + 2\,\cos\frac {\pi n}{4}} {\bigg|\;}}$

now,
$ \liminf_{n\rightarrow\infty}\;\;e^{\frac {\ln\,(\ln\,n)}{n}} \;=\; \lim_{n\rightarrow\infty}\;\;e^{\frac {\ln\,(\ln\,n)}{n}} \;=\; 1$

and
$ \limsup_{n\rightarrow\infty}\;\;1 + 2\,\cos\frac {\pi n}{4} = 3$

so, yes, answer is  $ R \;=\;\bigg|\; \frac {1}{3} \;\bigg| \;= \frac {1}{3}$ [/quote]

you are sort of on the right track, what you really need is
$ R \; = \; \dfrac{1}{\limsup_{n\to\infty} \;\left|a_n\right|^{\frac {1}{n}}}$

what i mean is your answer is correct, you just needed to justify the steps properly (see what i wrote above)


geometry
trigonometry
inequalities
power of a point
radical axis
geometry solved

Let $H$ be the orthocenter of the acute triangle $ABC$. Let $BB'$ and $CC'$ be altitudes of the triangle ($B^{\prime} \in AC$, $C^{\prime} \in AB$). A variable line $\ell$ passing through $H$ intersects the segments $[BC']$ and $[CB']$ in $M$ and $N$. The perpendicular lines of $\ell$ from $M$ and $N$ intersect $BB'$ and $CC'$ in $P$ and $Q$. Determine the locus of the midpoint of the segment $[ PQ]$.

[i]Gheorghe Szolosy[/i]

Note that $\{ M,P,N,B'\}$,$\{ A,C'N,Q\}$ are concyclic, so $C'H\cdot CH=MH\cdot NH=B'H\cdot BH$, Thus B',C',P,Q are also concyclic. $\angle HBC=\angle HC'B'=\angle HPQ$, which implies PQ//BC. Let D be the midpoint of BC. Then the midpoint of PQ must lie on HD.
However, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\frac{2\cos A}{\cos A+1}\le\frac{HP}{HB}\le1$. The answer is trivial then.

[quote="mecrazywong"]Note that $\{ M,P,N,B'\}$,$\{ A,C'N,Q\}$ are concyclic, so $C'H\cdot CH=MH\cdot NH=B'H\cdot BH$, Thus B',C',P,Q are also concyclic.[/quote]

I think you wanted to say:

Note that the points M, P, N and B' are concyclic, and that the points N, Q, M and C' are concyclic, so that $PH\cdot B^{\prime}H=MH\cdot NH=QH\cdot C^{\prime}H$. Thus, the points B', C', P, Q are also concyclic.

All the rest of your solution is correct and very nice.

  darij

Darij, thanks for reminding me :)  and sorry for the typos :blush:

[quote="mecrazywong"]However, not all the points on HD satisfy the conditions. By simple trigo computation, we obtain $\frac{2\cos A}{\cos A+1}\le\frac{HP}{HB}\le1$. The answer is trivial then.[/quote]

I can't understand this part of the solution ....   :(

This problem is a "painting" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\parallel BC$. The its prove is identically with the Darij's solution.   It is a pity for the author of the initial problem!

Virgil Nicula, do you know what mecrazywong meant by that trig at the end of his solution?

We are finding the locus of the midpoint of PQ. The first part of my solution only proves that the midpoint of PQ lies on HD. Since not every points on HD does belong to the family of the loci, we must eliminate them and the trigo inequality at the end of my solution serves for this purpose.

[quote="perfect_radio"]mecrazywong, i also deduced that. but how did you reach that trigo inequality? that was what i was asking[/quote]

?

[quote="Virgil Nicula"]This problem is a "painting" of a wellknown problem which has the same hypothesis but the conclusion is $PQ\parallel BC$. The its prove is identically with the Darij's solution.   It is a pity for the author $\boxed {S.\ J.\ Weyl}$ of the initial problem![/quote][color=red]This problem  (R.M.O. $\boxed {2005}\ ,\ 9^{th}$-grade) appears in my book (only with $PQ\parallel BC$, without geometrical locus) " [b]Geometrie plana. Culegere de probleme, Ed. GIL,[/b] $\boxed {2002}$ (the page 51 with the number 5.17). Am I right ?[/color]\[ \boxed {\ 2005\ -\ 2002\ =\ 3\ >\ 0\ } \][b]Question:[/b] [u]Who is the author of this problem from R.M.O., 2005 ?[/u]

I don't know who is the author of this problem. My teacher gave me a year ago a "Gazeta Matematica", which contained the problems from RMO 2004 and their authors. But he took it back shortly after.

Do you know how mecrazywong reached that trig inequality? I can't seem to obtain anything similar. (The other part with $PQ \| BC$ I understood)

The limiting position of midpoint of PQ happens when

  it is colliner with the radical axis of the Droz-Farny cirles.

   T.Y.

   M.T.


[quote="e.lopes"][quote="NguyenDungTN"]Let a,b,c be the side of a triangle.Show that:
$\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\geq \frac{17}{8}abc+\frac{7}{8}\prod (b+c-a)$[/quote]

I think that this is wrong![/quote]

Example?

[quote="NguyenDungTN"]
Example?[/quote]
Try $a=11,$ $b=10$ and $c=2.$ :wink:

[quote="NguyenDungTN"]Let a,b,c be the side of a triangle.Show that:
$\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\geq \frac{17}{8}abc+\frac{7}{8}\prod (b+c-a)$[/quote]
$\frac{abc(a+b+c)^{2}}{a^{2}+b^{2}+c^{2}}\geq 2abc+\prod (b+c-a)$ is true already.  :wink:

[quote="arqady"]
$ \frac {abc(a \plus{} b \plus{} c)^{2}}{a^{2} \plus{} b^{2} \plus{} c^{2}}\geq 2abc \plus{} \prod (b \plus{} c \minus{} a)$ is true already.  :wink:[/quote]
We can suppose that $ abc\equal{}1$ and $ c\equal{}min\{a;b;c\}$
We can rewrite this ineq as:
$ (a\minus{}b)^2M\plus{}N(a\minus{}c)(b\minus{}c) \geq 0$
with $ M\equal{}a\plus{}b\minus{}c\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}$
$ N\equal{}c\minus{}\frac{2}{a^2\plus{}b^2\plus{}c^2}$
$ M \geq 0$ <=>$ a^3\plus{}b^3\plus{}ab^\plus{}a^2b\plus{}ac^2\plus{}bc^2 \geq a^2c\plus{}b^2c\plus{}c^3\plus{}2$
Note that $ ab^2\plus{}ac^2 \geq 2$
$ a^3\plus{}b^3 \geq a^2c\plus{}b^2c$
$ a^2b \geq c^3$
So $ M \geq 0$
$ N \geq 0$<=>$ a^2c\plus{}b^2c\plus{}c^3 \geq 2$
This ineq true because $ a^2c\plus{}b^2c \geq 2$
Done!

[quote="arqady"]
$ \frac {abc(a + b + c)^{2}}{a^{2} + b^{2} + c^{2}}\geq 2abc + \prod (b + c - a)$ is true already.  :wink:[/quote]
We can rewrite this ineq as:
$ \frac {abc(2ab + 2bc + 2ca - a^2 - b^2 - c^2)}{a^{2} + b^{2} + c^{2}}\geq \prod (b + c - a)$
We can suppose that $ a = x + y, b = y + z, c = z + x$ where $ x,y,z > 0$
Then we have:
${ \frac {(x + y)(y + z)(z + x)(4xy + 4yz + 4zx)}{2x^{2} + 2y^{2} + 2z^{2} + 2xy + 2yz + 2zx}}\geq 8xyz$
I.e.
${ \frac {(x + y)(y + z)(z + x)(xy + yz + zx)}{x^{2} + y^{2} + z^{2} + xy + yz + zx}}\geq 4xyz$
We can rewrite this ineq as:
$ \sum_{sym}x^{3}y^{2} + \sum_{sym}x^{3}yz + 2\sum_{sym}x^{2}y^{2}z\ge2\sum_{sym}x^{3}yz + 2\sum_{sym}x^{2}y^{2}z$
But last ineq is true by Muirhead's ineq
Done!  :)


Given a pentagon $ABCDE$ has all sides equal $a$. Construct another pentagon $A'C'E'B'D'$ such that $DC=A'B',AB=C'B',ED=C'D',CB=D'E',AE=E'A'$. If sum of square of diagonals of the first pentagon is $s^{2}$. What's sum of square of sides of the second pentagon. Or find counterexample to show that we have not enough informations.

[img]http://www.vcharkarn.com/vcafe/dekvit/uploaded_pics/CS22320x13.gif[/img]

We have not enough informations. I will just show how to construct a dynamic sketch of the situation: Draw a point E' in the plane, and two points A' and D' on the circle with center E' and radius $a$. Draw the circle with center A' and radius $a$, and choose a point B' on this circle. Let the circles with centers B' and D' and with radius $a$ intersect each other at a point C'. Then, the pentagon A'C'E'B'D' will have all of its diagonals equal to $a$. There are some troubles - sometimes, the circles with centers B' and D' and with radius $a$ don't intersect, and often the pentagon A'C'E'B'D' won't be convex, though it is clear that this doesn't really matter - but you can evade them all by moving around the points A', E', D' and, most importantly, B'. Now move B' along the circle with center A' and radius $a$, and you will see that the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' is variable. This means that we cannot compute it from $a$ and $s$.

A more interesting question is whether we can give some bounds. E. g., is the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' always $\geq 5\left(\frac{1-\sqrt5}{2}\right)^{2}a^{2}$ ?

[b]EDIT:[/b] Yes, it is! It follows from the right part of the inequality http://www.mathlinks.ro/Forum/viewtopic.php?t=112482 .

  darij

[quote="darij grinberg"]Draw a point E' in the plane, and two points A' and D' on the circle with center E' and radius s. Draw the circle with center A' and radius s, and choose a point B' on this circle. Let the circles with centers B' and D' and with radius s intersect each other at a point C'. Then, the pentagon A'C'E'B'D' will have all of its diagonals equal to s. [/quote]

But the problem told that the pentagon A'C'E'B'D' have all of its diagonals equal to $a$,is'n it?

[quote="mathpk"]But the problem told that the pentagon A'C'E'B'D' have all of its diagonals equal to $a$,is'n it?[/quote]

I had confused a with s. Now I have corrected my post. Anyway, the knowledge of s doesn't change anything, as we can still vary the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' without changing the pentagon ABCDE (so that both values a and s remain constant).

  Darij

[quote="mathpk"]How you see that the sum of the squares of the sidelengths of the pentagon A'C'E'B'D' is variable?[/quote]

Dynamic computer geometry, as it would be quite a pain to do this by hand.

  darij

I see... btw this problem is in my national competition. How to do this by hand?

ps. Thanks for helping me :)  I had learned from you how to do geometry problems by using dynamic computer geometry. :D


[quote="dheera"]Hi all,
Just wanted to note that we created an online graphing calculator:
  [url]http://fooplot.com/[/url]
which should work in most modern browsers (except Safari). Hope you find it useful - share it, let us know if you find it useful, see any bugs, or have any suggestions about what would be useful. It's primarily targeted to help the educational community.

Also, if you ever need a quick plot of a function, just type in the function right after the URL:
  [url]http://fooplot.com/exp(-x^2)[/url][/quote]

grr I hate users who make an account just to advertise things.....burn him!

[quote="7h3.D3m0n.117"][quote="mysmartmouth"]He is a new user, but he's an MIT friend of Joe Laurendi, (joml88), and this really is a good website if you were looking for a quick way to graph things (which I was a few days ago).[/quote]

I know but at least some people would have the decency to put them in the right place or something....[/quote]
Mods, please send this topic to 'High School>>>>Other Problem Solving Topics Forum'. Thank you in advance (if this ever gets sent there :P)


geometry
rectangle
combinatorics unsolved
combinatorics

In a multiple choice test there were 4 questions and 3 possible answers for each question. A group of students was tested and it turned out that for any three of them there was a question which the three students answered differently. What is the maximum number of students tested?

I put in a 3x4 rectangle the answer of 3 students
I get:
  OOO
  OOO
  OOO
  OOO
then one line have to contain 3 different answer thus 3! possibilities
in the other space can be any answer, thus 3^9 possibilities
--> 3! * 3^9

this is the maximum possibilities of the question of 3 generic students that are binomial(x,3)
--> binomial(x,3) < 3! * 3^9 --> the maximum x is x=27

the answer is 27 students

Sorry for my bad math and my worst english

z.


The solution they give makes very little sense to me.  Could someone explain it, especially how they determined that the points must be one of the 5 given ones.

Thank you all
BTW I'm new to this site, so if I'm doing anything incorrectly, please let me know.

JB

Hey John,

One of the great things you'll learn to love about AOPS is the math jams.  Look at the transcript from this night's math jam.  You can click the tab next to Forum.


topology
Ring Theory
advanced fields
advanced fields unsolved

Let $ M$ compact Hausdorff topological space. For $ t\in M$ we define $ \chi_t: C(M)\rightarrow \mathbb{C}$ as evaluation at point $ t$: $ \chi_t(f)\equal{}f(t),f\in C(M)$. Prove that $ t\rightarrow \chi_t$ is homeomorphism for $ M$ to $ \sigma(C(M))$. If by that homeomorphism we identify $ t$ with $ \chi_t$, prove that Gelfand transformation is an identity.

[hide="what I have done so far"] I proved that $ t\rightarrow \chi_t$ is continuous, and that $ Ker \chi_t$ is maximal ideal in $ C(M)$ ( since it codimension equals 1 ). I don't know how to continue from that, it's told that we have to notice that every maximal ideal is of that form, to prove that, we have to assume otherwise and use the fact that $ M$ is compact.[/hide]

Again any hints are welcome!

if $ \mathfrak{m} \in \sigma(C(M))$ is not of the form, for every $ t \in M$ there is some $ f \in \mathfrak{m}$ such that $ f(t) \neq 0$, i.e. $ f$ does not vanish on an open neighborhood of $ t$. now use compactness to construct a unit in $ \mathfrak{m}$, which is impossible.

the openness of $ t \to \chi_t$ is directly verified or by using that this is a continuous bijection from a compact space into a hausdorff space. have a look at the gelfand transform, then it's clear that it becomes the identity here.


induction
geometry
parallelogram
analytic geometry
combinatorial geometry
combinatorics unsolved
combinatorics

Suppose there are $997$ points given in a plane.  If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least $1991$ red points in the plane.  Can you find a special case with exactly $1991$ red points?

I'm curious as to whether or not this proof works...

We will prove by induction that given $n \ge 2$ points in a plane, there are at least $2n-3$ distinct midpoints.

Base Case: $n = 2$, we clearly have one midpoint.

Induction Step: Call one of the points on the convex hull of the $n+1$ points $P$. Then take the convex hull of the other $n$ points not including $P$. We know $P$ is outside of the convex hull. By our inductive hypothesis, these $n$ points have at least $2n-3$ distinct midpoints. But there must exist two (consecutive) vertices $A$ and $B$ of the convex hull such that the midpoints of $AP$ and $BP$ lie outside of the convex hull and cannot be the same as any of the other midpoints (which clearly have to lie inside the convex hull). Hence there are at least $2n-3+2 = 2(n+1)-3$ distinct midpoints, completing the induction. (For the case that all the points are collinear, just take $A$ and $B$ the closest points to $P$ and we create two new midpoints.)

So for $n = 997$, we have at least $2(997)-3 = 1991$ midpoints.

And if we take the points $(0,0); (2,0); \ldots; (1992,0)$, we have exactly $1991$ distinct midpoints, namely $(1,0); (2,0); \ldots; (1991,0)$.

I have not yet got the complete solution. But I believe this is the first step.

Consider a point A1, when it is joined with the rest of the 996 points, no two lines shall intersect. Therefore there shall be 996 red points. And I noticed that 1991=996 +995.
So, I am trying on these lines.

Let points $A_i$ and $A_j$ be such that segment $A_iA_j$ has the greatest length of all other segments that are connecting some two points. Then construct segments $A_iA_1$, $A_iA_2$ ... $A_iA_n$ and $A_jA_1$, $A_jA_2$ ... $A_jA_n$. Those segments have distinct midpoints, because if any segment $A_iA_x$ have the same midpoint with some $A_jA_y$ then quadriliteral $A_iA_jA_xA_y$ is parallelogram with diagonals $A_iA_x$ and $A_jA_y$ so $A_iA_x>A_iA_j$ or $A_jA_y>A_iA_j$. So we have at least $995\cdot 2 +1=1997$ distinct midpoints ("1" is the midpoint of $A_iA_j$).

[color=blue][b]Other solution:[/b][/color]
Sort points in the coordinate plane by their $y$ coordinates (points with same $y$ coordinate are sorted by $x$ coordinate). Then take this segments $A_1A_2$, $A_2A_3$... $A_{n-1}A_n$ and $A_1A_3$, $A_3A_5$  and $A_2A_4$, $A_4A_6$...
It isn't hard to show that all this segments have distinct midpoints and so on
(I am not so sure about this second solution, because I solve this problem in that way a few months ago and I wrote this right from my head, so maybe something is missing, but I don't think so ... :) )

Special case is when all points are on a line and $A_1A_2=A_2A_3=...=A_{n-1}A_n$

bye

My attempt at a solution:

As before, plot the points on a coordinate plane. Now, choose the uppermost point, and if not unique the leftmost of these to be $P_1$. Similarly, choose the lowermost, rightmost point to be $P_2$. Consider all the segments with $P_1$ as a vertex. There are $996$. Similarly, there are $996$ with $P_2$, and only one is double counted, i.e. $P_1P_2$. Thus, there are atleast $1991$ unique red points, since all of these midpoints must be unique. (Proof is left as a simple exercise in inequalities..) Hence, we are done.

Note: This also proves it for the more general $2n-3$, as stated above.

Fun fact: This problem was unknowingly solved while doing the famous "windmill problem".. :P

Impose a Cartesian coordinate system on the set of points such that they have all distinct $y$-coordinates; let these be $y_1, y_2, \cdots, y_{997}$ in increasing order. Furthermore, let $A_i$ be the point with $y$-coordinate $y_i$.

Now, consider the midpoints of $\overline{A_iA_{i+1}}$ for $1 \leq i \leq 996$; there are $996$ such midpoints, and by the order none of them have the same $y$-coordinate, and thus they do not coincide. Furthermore, consider the midpoints of $\overline{A_iA_{i+2}}$ for $1 \leq i \leq 995$. These cannot coincide with any of the former midpoints because the $y_i$ are all distinct. This yields $996+995=1991$ distinct midpoints, as desired.

For a construction, take all the points to be equally spaced along a line.

Let, the points, be $(a_1, b_1), (a_2, b_2),....., (a_{997},b_{997}).$Now, take the midpoints of the points, $(a_i, b_i), (a_{i+1}, b_{i+1}).$ There, are $996,$ of these, and these do not coincide. Now, take the points, $(a_i, b_i), (a_{i+2}, b_{i+2}),$ which holds, $995,$ midpoints. These, again can not coincide, hence, this is $996+995=1991,$ hence, proved.


Solve this ecuation :

$ (3x\minus{}m\minus{}n\minus{}p)(\frac{1}{x\minus{}m}\plus{}\frac{1}{x\minus{}n}\plus{}\frac{1}{x\minus{}p})\equal{}1$

$ m,n,p\in \mathbb{R}\ ,\ m\neq n\neq p\neq m$ are constants .

hello, your equation is equivalent with $ \frac{(m\plus{}n\minus{}2x)(m\plus{}p\minus{}2x)(n\plus{}p\minus{}2x)}{(m\minus{}x)(n\minus{}x)(p\minus{}x)}\equal{}0$, so we have the following solutions $ x_1\equal{}\frac{m\plus{}n}{2}$, $ x_2\equal{}\frac{m\plus{}p}{2}$, $ x_3\equal{}\frac{n\plus{}p}{2}$.
Sonnhard.


function
topology
calculus
calculus computations

We know that continuous functions map compact sets to compact sets.

1) Whats an example of a closed set $ S\subset \mathbb{R}$ and a continuous function $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(S)$ is not closed?

2)  Whats an example of an open set $ U\subset \mathbb{R}$ and a continuous function $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(U)$ is not open?

[quote="wizard"]1) Whats an example of a closed set $ S\subset \mathbb{R}$ and a continuous function $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(S)$ is not closed?[/quote]

\[ \left[ {1, \plus{} \infty } \right)\xrightarrow{{f\left( x \right): \equal{} \frac{1}
{{\left| x \right| \plus{} 1}}}}\left( {0,\frac{1}
{2}} \right].
\]

[quote="wizard"]2)  Whats an example of an open set $ U\subset \mathbb{R}$ and a continuous function $ f: \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(U)$ is not open?[/quote]
\[ \left( { \minus{} 1,1} \right)\xrightarrow{{f\left( x \right): \equal{} \left| x \right|}}\left[ {0,1} \right).
\]


superior algebra
superior algebra theorems

Hi, I was wondering how Frey's Equation, the equation that describes FLT as an elliptic curve,                y^2 = X(X-x^n)(X-y^n), is derived from the general form of elliptic curves, y^2 =x ^3 +ax + b.

See G. Frey, [i]Links between stable elliptic curves and certain diohantine equations[/i] Ann. Univ. Sarav. Ser. Math. (1986), 719-722.


[quote="jarod"]This problem was proposed on KBAHT (maybe it's a problem in Russian MO) :
 From $a+b+c+d=S$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=S$ , we infer : $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d} =S$ . Find $S$ ?
  Can you please help me to find a solution of this problem ? Thanks .[/quote]

It'a a problem from the 4th step(fo 5 steps) of RMO 2005.
Solution:

From $a+b+c+d=S$ and $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=S$ , we infer : $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}+\frac{1}{1-d} =S$
also $\frac{1}{1-\frac{1}{a}}+\frac{1}{1-\frac{1}{b}}+\frac{1}{1-\frac{1}{c}}+\frac{1}{1-\frac{1}{d}} =S$
But $\frac{1}{1-t}+\frac{1}{1-\frac{1}{t}}=1$
So $2S=4 \Rightarrow S=2$

Examples of such numbers:
$a=2+\sqrt3$
$b=2-\sqrt3$
$c=d=-1$


A three-digit  bumber  $\overline{abc}$ (in  decimal  representation)  is such  that

(i) its  hundreds  digit  is  equal  to the sum  of the  other two  digits, and

(ii)$b(c+1)=52-4a$.

Find all  such numbers.


Let $ \{A_1,A_2,...\}$ be a sequence of finite sets. Is the product $ P \equal{} \Pi_{i \equal{} 1}^{\infty}A_i$ countable?

Is the product $ \{0,1\}^{\mathbb N}$ countable?


$ x,y,z>0$,then:

$ 1\minus{}\frac{8xyz}{(y\plus{}z)(z\plus{}x)(x\plus{}y)}\geq \frac{z\minus{}y}{y\plus{}z}\plus{}\frac{x\minus{}z}{z\plus{}x}\plus{}\frac{y\minus{}x}{x\plus{}y} \geq  \frac{8xyz}{(y\plus{}z)(z\plus{}x)(x\plus{}y)}\minus{}1$;

Let $ x_1,x_2,...,x_n > 0,x_{n \plus{} 1} \equal{} x_1,n \geq 3$,prove that:

$ (n \minus{} 2)(1 \minus{} \frac {2^n\prod_{i \equal{} 1}^n{x_i}}{\prod_{i \equal{} 1}^n{(x_i \plus{} x_{i \plus{} 1})}})\geq \sum_{i \equal{} 1}^n{\frac {x_i \minus{} x_{i \plus{} 1}}{x_i \plus{} x_{i \plus{} 1}}} \geq \minus{} (n \minus{} 2)(1 \minus{} \frac {2^n\prod_{i \equal{} 1}^n{x_i}}{\prod_{i \equal{} 1}^n{(x_i \plus{} x_{i \plus{} 1})}})$

BQ

[size=150]I said,once:
Our vision should be to wide and wide,
From a small place,
fond out fine and nice![/size]

[size=200]yes ,
Our vision should be wide and wide,[/size]
as Follows:

Let $ x_1,x_2,...,x_n > 0,x_{n \plus{} 1} \equal{} x_1,n \geq 3,t$is odd number ,prove that:

$ (n \minus{} 2)(1 \minus{} \frac {2^n\prod_{i \equal{} 1}^n{x_i}}{\prod_{i \equal{} 1}^n{(x_i \plus{} x_{i \plus{} 1})}})\geq \sum_{i \equal{} 1}^n{\frac {(x_i \minus{} x_{i \plus{} 1})^t}{(x_i \plus{} x_{i \plus{} 1})^t}} \geq \minus{} (n \minus{} 2)(1 \minus{} \frac {2^n\prod_{i \equal{} 1}^n{x_i}}{\prod_{i \equal{} 1}^n{(x_i \plus{} x_{i \plus{} 1})}})$

BQ


this somehow relates to my other question, anyways,

Consider a uniform distribution of mass on the solid ball of radius $ r$ about the origin.  Show that the gravitational field at point $ \mathbf{x}$ is the same as if the mass closer to the origin than $ \mathbf{x}$ were all located at the origin and the mass farther from the origin than $ \mathbf{x}$ (if any) were absent.

This is the Shell Theorem from physics.  Look in any calculus-based physics text on mechanics and there should be a proof.


Show that for all integer k, there are infinitely many natural numbers n such that k*2^2^n+1 is composite.

I know that it is an old problem from PEN but I can't catch the solution gave in the original.Can somebody post a more detail and complete solution?
Thank you very much!

If a solution is not clear, ask in the corresponding thread. So post your questions in the PEN subforum.

You are right ZetaX :)
But do you have any idea towards this problem?I think that it worth discussing here :)

I am curious about this result.Can somebody give me a proof?Or just a "given k" such as $ k\equal{}1$?

we can always find a prime p such that 2^2^n+1=p-1.

[quote="hxy09"]I think that it worth discussing here[/quote]
No. Just revive the PEN topic; spreading around discussion of one problem in multiple topics is one of the biggest pains in the ass about this forum.


A point P is given inside a triangle ABC. A line l through P not con-
taining A intersects AB and AC at Q and R, respectively. Find the line
l such that the perimeter of △AQR is least possible.

[url]http://www.imocompendium.com/othercomp/Nd/NdMC1990.pdf[/url]

To construct a line $ l$ through $ P,$ such that the $ \triangle AQR$ has a given perimeter $ p,$ draw a circle centered at $ A$ with radius $ s \equal{} \frac{p}{2}$ intersecting the rays $ (AB, (AC$ at $ S, T,$ then a circle $ (E)$ tangent to $ (AB, (AC$ at $ S, T$ and finally, a tangent from $ P$ to $ (E),$ intersecting $ (AB, (AC$ at $ Q, R.$ $ (E)$ is the A-excircle of the $ \triangle AQR$ with the given perimeter $ p.$

For minimum perimeter $ p$, $ P$ has to be the tangency point of $ QR$ with $ (E).$ Draw a circle $ (E)$ tangent to $ (AB, (AC$ and passing through $ P$ and then draw a tangent of $ (E)$ at $ P$ intersecting $ (AB, (AC$ at $ Q, R.$ 2 circles exist, tangent to $ (AB, (AC$ and passing through $ P.$ Pick the one, which lies outside of the resulting $ \triangle AQR.$

[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14460[/img]

(point line line)  Apolonius problem

 :thumbup:    :clap:

[url]http://garciacapitan.auna.com/bella/htm/apo-ppr.htm[/url]