AMC AIME USA(J)MO USAMO Mafia geometry inequalities

Yes, I'm wondering as well, what is the best way to make the transition from doing AIME type questions to olympiad questions?
The difficulty difference is much greater than that between AIME and AMC.

AoPS Vol. II -> Zeitz and Engel, possibly.

[quote="JSteinhardt"]Well, there's always http://activities.tjhsst.edu/vmt/pages/tjusamo[/quote]

Wow, that's amazing! No wonder Thomas Jefferson is such an amazing school. :)

do problems

olympiad problems

read books

do problems

you don't need to read that much if you can pick things up from the forum, etc, and you are able to generalize things yourself

Hmm well I've been preparing somewhat over this year, though not that much. I'm probably going the next month going through some practice USAMOs, which, although it does not constitute serious 7-12+ month practice, should significantly improve my score by giving my mind some experience going through actual olympiad problems/USAMOs.

Later, for future years, I plan to get Engel's Problem Solving Strategies, do many more practice olympiads (as I only have time for a few over the next month), and do some more of the WOOT message board problems, which I haven't had time to do due to being busy this year.

summer camps are good

i need to get engel...does anybody perhaps have a rapid share link :P

[quote="xxazurewrathxx"]See if you can steal Arnav's DNA and infuse it with your own.[/quote]

Errr what that is kind of disturbing to think about...  :maybe:

[quote="roadnottaken"]I'm almost afraid to ask this, but who exactly is Arnav? :blush:[/quote]
 :o

You should be very ashamed of yourself...

And after 231 posts?? SHAME ON YOU.  :P
Nah, he's MysticTerminator and a genius.

hero worship is unhealthy; with that being said, arnav worship is even worse...j/k lol

Mafia worship is way worse.

And even worse... UNDERCOVER MAFIA.

[quote="xxazurewrathxx"]See if you can steal Arnav's DNA and infuse it with your own.[/quote]
I do not want to grow hair on the undersides of my feet thank you

In response to the OP: It's hard. Have you tried MOT and MOC? They might help you bridge the gap by providing problems clustered in subjects, so you don't need to worry about technical stuff as much as how to approach the problem...

The best way might be to hire Titu, Zuming, and the rest of the MOP faculty as private tutors. If you can't do that, then Larson, Zeitz, and Engel are good ways to go, depending on where you are (Larson < Zeitz < Engel, in terms of difficulty)

Just for fun, you might want to read Count Down by Steve Olson. Even though it's probably not going to help you much as far as problem solving goes, it has some interesting information about USAMO and IMO as well as how the Olympians work their way through to perfect solutions.

On an obviously tangent question, what is typical day of MOP like and where do they draw the problems from?

[quote="101101101"]...practice writing proofs.[/quote]
There is no substitute for having your work graded- and that's not easy to find. It's obviously too late to start now if you're looking at this year, but mail-in contests such as USAMTS are a very good resource for this.
This is also one of the things MOP offers- many olympiad-style tests which are graded and returned within days.

Personally, I think that you should start by doing CMO (Canada) problems.

When you have familiarized yourself with those, move on to USAMOs, Russian MOs, IMOs, and IMO Shortlists.

By the way, I think that many MOP homework problems are from ISLs.

[quote="101101101"]I would take a day of the USAMO--spending at least a couple hours, even if you can't make any progress--and then spend a long time going over the solutions. That is, try reading only the approach and see if that was the approach you took. If it wasn't, then consider why you didn't find that approach, and then spend some more time with the problem using the new approach. Then continue reading bit by bit, each time considering why you failed to make some observation, what might have motivated you to make said observation, and then trying to continue after being handed the observation. It is also important to try and translate things from the approach given in the solution into your approach. That is, if they did something complete different than you, try and see if you could have made the same critical step in your different approach, just in a way that looked different.

One problem you will probably encounter is that a solution will hinge on a theorem you do not know. This will happen mostly in geometry, but sometimes in NT or algebra. There are lists elsewhere of what theorems you need to know; I would advise finding one of the shorter lists as most of the stuff people say you "need" to know is useful only rarely and can normally be done without. It is fine to write down the theorems somewhere and, if you get really stuck, trying to think of how relevant ones might apply to the situation. If you see a theorem or fact you don't know used in a proof make note of it and then return to the problem using that fact.

That takes a really long time and it is fairly tiring, but I think it is quite productive. Once you finish all the USAMOs, IMOs are equally useful. You should not be discouraged if you answer exactly 0 of the questions your first 4 or 5 tries; each failure should provide you with an idea of how to change your mindset to make finding a solution more likely.

As a side note, it is important to write up solutions exactly as you would on a formal test. This tends to illuminate holes in proofs, gives you practice writing proofs, and highlights any horribly inelegant solutions. If your solution is unduly horrible or sketchy, it is advisable to continue searching later for a nicer one (especially if you have dumbassed an inequality or found a horrendous analytic solution).

That said, I am not the utmost authority on the topic. I am inclined to think this is substantially more useful than reading books, but there are many who are more successful at olympiad math than I who might differ. It also varies a great deal from person to person, and unlike me you might have a life which interferes with that kind of time commitment.

Edit: It also isn't too late to get in substantive practice. When I passed the AIME last year by a fluke, my average USAMO score rounded to 0. By the time the test came around, my average was a little below 21.[/quote]

So would you say that just going over old past USAMOs a couple of hours a day would help immensely in addition to the stuff you said?

What about other olympiads?

[quote="101101101"]There are lists elsewhere of what theorems you need to know; I would advise finding one of the shorter lists as most of the stuff people say you "need" to know is useful only rarely and can normally be done without. [/quote]

Any idea where one could obtain such a list?

I would personally advise against searching for said list. Any important enough theorems will be given as an introduction to whatever subject material you are dealing with. If you just memorize theorems, you will have no idea how to apply the theorems (for example the first time I learned Fermat I forgot it really quickly because I had no clue how to use it). That said, I'll contradict myself by giving a list:

Fermat's Little Theorem / Euler's Totient Theorem
Wilson's Theorem

AM-GM Inequality / Power Mean Inequality / Muirhead's Inequality
Cauchy-Schwartz Inequality
Jensen's Inequality

Power of a Point
Inscribed Angle Theorem

Pigeonhole Principle
Well-ordering Principle / Induction

Um...I can't really think of anything else super-important. I probably started Olympiad Math with less than this and did fine.


inequalities solved
algebra
Inequality
inequalities proposed

Assume a + b + c = 1 to get

product((1+3a)/(1-a)) > 25. Then multiply by (1-a)(1-b)(1-c).

Then we get 13abc + 1 > 4 \sum ab (*).

But by Schur's, 1 + 9abc \geq 4 \sum ab.

Since a, b, c > 0, equality in * can never hold. Hence result.

How  do  you  get  1+9abc>=4Σab  ?
Thanks

Schur's yields $a^3 + b^3 + c^3 + 3abc \ge a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2$. To use the $a + b + c = 1$ condition, we add thrice the RHS and $6abc$ to both sides to get $(a+b+c)^3 + 3abc \ge 6abc + 4(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)$. To factor the RHS, we again add $6abc$ to both sides, giving $1 + 9abc \ge 12abc + 4(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)$ $= 4(a^2b + ab^2 + abc + a^2c + ac^2 + abc + b^2c + bc^2 + abc) = 4(ab(a+b+c) + ac(a+b+c) + bc(a+b+c)) = 4(ab+ac+bc)$, as desired.

[quote=sqing]Show that $ \forall a,b,c>0$ we have:
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$[/quote]
It's Schur inequality.

[quote=Lagangia]Show that $ \forall a,b,c>0$ we have:
$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)>25 $

cheers! :D :D :cool:[/quote]

A generalization:
$ ( \frac{na}{b+c}+1)( \frac{4nb}{c+a}+1)( \frac{nc}{a+b}+1)>(n+1)^2 $
for n $\ge$ 1
which can be prove by Jensen on
f(x)=ln ( $\frac{4nx}{1-x}+1 )$ as we put $\sum a = 1$

[quote=sqing][quote=Lagrangia]Show that $ \forall a,b,c>0$ we have:
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)>25 $$

cheers! :D :D :cool:[/quote]
Show that $ \forall a,b,c>0$ we have:
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$[/quote]

UVW will kill this one!!!

[quote=sqing]
Show that $ \forall a,b,c>0$ we have:
$$ ( \frac{4a}{b+c}+1)( \frac{4b}{c+a}+1)( \frac{4c}{a+b}+1)\geq 25+\frac{16abc}{(a+b)(b+c)(c+a)}. $$[/quote]
[url]https://artofproblemsolving.com/community/c6h1660525p10538912[/url]
[url]https://artofproblemsolving.com/community/c6h517577p13276698[/url]
$$\left(\frac{2a}{b+c}+1\right)\left(\frac{2b}{a+c}+1\right) \left(\frac{2c}{a+b}+1\right) \leq  4+\frac{(a+b)(b+c)(c+a)}{2abc}\leq 2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+1\right)    $$


Hello
Can anyone solve this problem?
For how many ways can be divided convex decagon for triangles inside it with diagonals, which do not intersect themselves.
Sorry for my english.

http://en.wikipedia.org/wiki/Catalan_numbers

Look at the fourth proof.


Asymptote AoPSwiki articles integration trigonometry LaTeX

It seems that when making multiple Asymptote diagrams, they always get put on separate lines (never lined up horizontally). How do you make it so that you can half one diagram on the left and one on the right?

This is the same for any graphics where you want them side by side, you use minipages. In the case of asymptote you won't need the figure environment. Revamping one of the examples on AoPSWiki, here is a sample with two smiley faces side by side, with optional text between them.
[code]\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Here are two Asymptote images, side by side \\
\begin{minipage}{3cm}
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{minipage}
$\displaystyle\int_0^{\pi}{\sin{x}}\,dx=2$
\begin{minipage}{3cm}
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{minipage}
\end{document}[/code]


algebra
polynomial
number theory unsolved
number theory

Is that true that the coefficient of the polynomial $ 1 \plus{} (x \plus{} 1) \plus{} (x \plus{} 1)^2 \plus{} ... \plus{} (x \plus{} 1)^{p \minus{} 1}$ is all multiples of p (except the p-1 exponent one) for p prime :maybe:


Prove that if 2*(sqr2)*[4^(k-1)]-sqr(3)*[9^(l-1)] >0 (k,l are positive integers), then it's greater than 1 .


inequalities
integration
trigonometry
real analysis
real analysis unsolved

$ b \geq a \geq 0$, then

$ |\int_a^b \frac {\sin x}{x}|\leq 3$

It's sufficient to prove $ \int_0^\pi \frac {\sin x}{x}dx < 3$ (because $ \left \lbrace \int_{k\pi }^{(k \plus{} 1)\pi}\frac {\sin x}{x}\right \rbrace _k$ is a sequence alternating in sign with terms strictly decreasing in absolute value).  You can do that by using the Taylor series as an estimate:
\[ \frac {\sin x}{x} < 1 \minus{} \frac {x^2}{6} \plus{} \frac {x^4}{120}\Rightarrow \int_0^\pi \frac {\sin x}{x}dx < \pi \minus{} \frac {\pi ^3}{18} \plus{} \frac {\pi ^5}{600}\]
This actually proves $ \left \vert \int_a^b \frac {\sin x}{x}dx\right \vert < 2$ for all $ b\geq a\geq 0$, which one can get by using a crude estimate on $ \pi$, say $ 3.14 < \pi < 3.15$.

Small mistake: should read $ \frac{\pi^6}{600}$.


[quote="chim3y"]SURE TO MAKE U LAUGH

My friend and I made this random pictures web page... ENJOY

[url]http://www.sitesled.com/members/randompictures/index.html[/url][/quote]

lol

And yes, those are quite random.

And nothing obscene either, that's a plus...

Some of the pictures are huge though........they take quite a bit of time to load with the connection that I'm using..... :(

Wow, those are huge pictures, my computer can barely generate them


Find the least value and the greatest value of the funcition
$y=\sin^{3}x+2\sqrt{2-\sin^{2}x}-2$

[hide="Answer"]The greatest value is $1$, at $x=\frac{\pi}{2}$ (there are more places, since this is periodic).

The smallest value is $-1$, at $x=-\frac{\pi}{2}$ (there are more places, since this is periodic).[/hide]

when you sub in

[hide]
$\frac{-\pi}{2}$
you get $\boxed{-1}$ not $-2$...
[/hide]


I think this problem be easy with you.And we need prove:
$\frac{1}{sinC}+\frac{2\sqrt2}{sinA}\geq\frac{1+2\sqrt2}{cos\frac{B}{2}}$

The solution 2 of this problem:(I have found)
We have:
$(3\sqrt3)^2=(\sqrt3+2\sqrt3)^2=(\frac{\sqrt{3a}}{\sqrt{a}}+\frac{\sqrt[4]{8}}{\sqrt{c}}.\frac{2\sqrt{3c}}{\sqrt[4]{8}})^2\leq(\frac{1}{a}+\frac{2\sqrt2}{c}).(3a+3\sqrt2.c)$
Apply Bunhiacowski Inequality again,we have:
$(\frac{1}{a}+\frac{2\sqrt2}{c}).(3a+3\sqrt2.c)\leq(\frac{1}{a}+\frac{2\sqrt2}{c}).(\sqrt{(9+18).(a^2+c^2)})\leq(\frac{1}{a}+\frac{2\sqrt2}{c}).((3\sqrt3).b)$
$\rightarrow \frac{1}{a}+\frac{2\sqrt2}{c}\geq\frac{3\sqrt3}{b}$
The sign"=" occur when:$B=\frac{\pi}{2},c=\sqrt2.a,b=\sqrt3.a$
Problem be proved completely

[quote="Frey"]$\frac{1}{sinC}+\frac{2\sqrt2}{sinA}\geq\frac{1+2\sqrt2}{cos\frac{B}{2}}$[/quote]

Are you sure that this is true? (I get problems with A = 15°, B = 153°, C = 12°.)

Anyway, just for the sake of completeness, here is a slightly shorter solution of the problem using Hölder's inequality:

[color=blue][b]Problem.[/b] Let ABC be a triangle with $B\geq 90^{\circ}$. Show that $\frac{1}{c}+\frac{2\sqrt2}{a}\geq\frac{3\sqrt3}{b}$.[/color]

[i]Solution.[/i] Since $B\geq 90^{\circ}$, we have $c^2+a^2\leq b^2$. Now, by the Hölder inequality,

$\left(\frac{1}{c}+\frac{2\sqrt2}{a}\right)^2\left(c^2+a^2\right)\geq\left(\sqrt[3]{\left(\frac{1}{c}\right)^2\cdot c^2}+\sqrt[3]{\left(\frac{2\sqrt2}{a}\right)^2\cdot a^2}\right)^3$
$=\left(1+2\right)^3=3^3$,

so that

$\left(\frac{1}{c}+\frac{2\sqrt2}{a}\right)^2\geq\frac{3^3}{c^2+a^2}\geq\frac{3^3}{b^2}$

(since $c^2+a^2\leq b^2$), and thus $\frac{1}{c}+\frac{2\sqrt2}{a}\geq\sqrt{\frac{3^3}{b^2}}=\frac{3\sqrt3}{b}$, and the problem is solved.

  Darij

[quote=hien]Oke, i post my solution as following:
Put $CA=b, AB=c, CB=a$
Thus $B\geq90^0$, deduced that:$b^2-a^2-c^2\geq0\Rightarrow b\geq\sqrt{a^2+c^2}$.
Next, we have: $\frac{3\sqrt{3}}{CA} = \frac{3\sqrt{3}}{b}=\frac{9}{\sqrt{3}b}\leq\frac{9}{\sqrt{3.(a^2+b^2)}}=\frac{9}{\sqrt{(2+1)(a^2+c^2)}}\leq\frac{9}{a+c\sqrt{2}}$.
From Cauchy inequality, arrive to: $\frac{1}{a}+\frac{4}{c\sqrt{2}}\geq\frac{(1+2)^2}{a+c\sqrt{2}}=\frac{9}{a+c\sqrt{2}}$.
From this we have conclusion fro that problem.
The sign "=" occur when: $B=90^0, c=\sqrt{2}.a,b=\sqrt{3}.a$[/quote]
Nice.
 $B\geq90^0 \Rightarrow b\geq\sqrt{a^2+c^2}$.
 $ \frac{3\sqrt{3}}{b}\leq\frac{9}{\sqrt{3(a^2+c^2)}}=\frac{9}{\sqrt{(2+1)(a^2+c^2)}}\leq\frac{9}{a+c\sqrt{2}}=\frac{(1+2)^2}{a+c\sqrt{2}}\leq \frac{1}{a}+\frac{4}{c\sqrt{2}}$ .
The equality occur when $B=90^0, c=\sqrt{2}a,b=\sqrt{3}a.$

[quote=darij grinberg]
[color=blue][b]Problem.[/b] Let ABC be a triangle with $B\geq 90^{\circ}$. Show that $\frac{1}{c}+\frac{2\sqrt2}{a}\geq\frac{3\sqrt3}{b}$.[/color][/quote]
By AM-GM ,
$\frac{1}{c}+\frac{2\sqrt2}{a}\geq3\sqrt[3]{\frac{1}{c}\cdot\frac{\sqrt2}{a}\cdot\frac{\sqrt2}{a}}=\frac{3\sqrt2}{\sqrt[6]{2c^2a^4}}\geq\frac{3\sqrt2}{\sqrt{\frac{2c^2+a^2+a^2}{3}}}\geq\frac{3\sqrt2}{\sqrt{\frac{2b^2}{3}}}=\frac{3\sqrt3}{b}.$

[quote=Frey]Let $ABC$ be a triangle satisfying $\hat{B}\geq \frac{\pi}{2}$. Prove that:
$\frac{1}{AB} +\frac{2\sqrt2}{BC}\geq\frac{3\sqrt3}{AC}$[/quote]

Let $ABC$ be a triangle satisfying $\hat{B}\geq \frac{\pi}{2}$. Prove that:

\[b^2(\frac{1}{c}+\frac{2}{a})^2 \geq 6\sqrt[3]{2}+3\sqrt[3]{4}+5\]

\[{\frac {b}{b+c}}+{\frac {b}{a+b}}\geq 4-2\,\sqrt {2}\]


2. Positive integer will be called white, if it is equal to $1$ or is a product of even number of primes (not necessarily distinct). Rest of the positive integers will be called black. Determine whether there exists a positive integer which sum of white divisors is equal to sum of black divisors

[quote="Rust"]It is obviosly. Answer - never.[/quote]

Please show your proof, Rust  :wink:

Let \[n=\prod_{i}p_{i}^{k_{i}}.\]
Let d is divisor, then $d=\prod_{i}p_{i}^{m_{i}}, m_{i}\le k_{i}.$
Consider all divisors d, suth that $d=p_{i}^{l_{i}}d_{i}, p_{i}\not |d_{i}$. Let $s(d)=0$ if d - white divisor and $s(d)=1$ if d black divisor.
We have $(-1)^{s(d)}=(-1)^{l_{i}}(-1)^{s(d_{i})}$. Let $S(n)=S_{W}(n)-S_{B}(n)$, then
$S(n)=\sum_{j}(-p_{i})^{j}S(n_{i})=S(p_{i}^{k_{i}})S(n_{i}),n_{i}=\frac{n}{p_{i}^{k_{i}}}, p_{i}\not |n_{i}.$
It mean $S(n)$ is multiplicative. Therefore sufficiently calculate $S(n)$ for $n=p^{k}$.
In these case all white divisors $p^{2j},0\le j\le [k/2]$ and all black divisors $p^{2j+1},0\le j\le [(k-1)/2].$
Therefore $S(p^{k})=1-p+p^{2}-...+(-p)^{k}=\frac{1+(-1)^{k}p^{k+1}}{1+p}.$ (I am sorry for my mistake.)
It give \[S(n)=\prod_{i}\frac{1+(-1)^{k_{i}}p^{k_{i}+1}}{1+p_{i}}\not =0.\]

Suppose that there is a positive integer $n$ such that the sum of its white divisors is equal with the sum of its black divisors. It is obviously that $n$ is not equal with $1$. So there is a prime number $p$ such that $p\mid n$. Let $n=kp$. Let $W$ and $B$ be the sum of the white and black divisors of the number $k$ respectively. The sum of the white divisors of the number $n$ that $p$ does not divide them is $W$. The sum of the white divisors of the number $n$ that $p$ divides them is $p.B$. This is because that when we factorize $p$, the color of this numbers will turn in black and they will be the black divisors of $k$. So when we factorize $p$, the sum of the rest part will be $B$. So the sum of the white divisors of $n$ is $W+p.B$. In the same way we get that the sum of the Black divisors of $n$ is $B+p.W$. So we have $B+p.W=W+p.B$ or $p(B-W)=B-W$. $p$ is a prime number and is not equal with $1$. So $B=W$. So we can replace $n$ with $k$ and do the same things for it. We do this until we get that the sum of the white divisors of $1$ is equal with the sum of its black divisors. which is not true. So there is not a positive integer which the sum of its white divisors is equal with the sum of its black divisors.

ATimo
I have an idea, where n is the smallest, that is, the sum of the white divisors and the sum of the black divisors are equal. Then, since K is a divisor, W and Bs are not equal and p is greater than 1 so this contradicts and it follows that such a number does not exist.I liked this solution.
[url][/url]


What is the median of the composite integers that are greater than 20 and less than 35?

[hide="Solution"]

Just list them out in order:

21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34

Find the median, which is $ \boxed{27}$.[/hide]

If you're like me and were unsure what composite means, mentally replace it with nonprime :(

That definition "works" for every natural number except $ 1$, since $ 1$ is neither prime nor composite.   :wink:


group theory
abstract algebra
modular arithmetic
superior algebra
superior algebra unsolved

After this one
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21942
----------------------------------------------------------------------------

Prove that there are no sipmle group of order $300$

Just consider the conjugation of 5-Sylow-subgroups by the elements of G.
It is only an easy application of Sylow theorems and Lagrange.

Misha

The number of Syllow $5$ subgroups is a divisor of $\frac {300}{25}=12$, and must also be $1\pmod 5$, so it's $6$. As Misha said, $G$ acts on the set of Syllow $5$ subgroups by conjugation, adn we get a homomorphism between $G$ and $S_6$, taking $g\in G$ to the permutation of the Syllow $5$ subgroups induced by conjugation with $g$.

The kernel of this homomorphism must either be trivial or the whole group $G$ (otherwise it's a normal proper subgroup of $G$). Clearly, it can't be the whole group, so it must be trivial, meaning that the homorphism is injective, so $G$ can be imbedded in $S_6$. We thus get $300|6!=720$, which is absurd.


hey guys i need help for a question:

If              log 5 = m
and          log 7 = n

Determine:

Log (35/10) in terms of m and n

Thanks

$\log\frac{35}{10}=\log\frac{7}{2}=n-\log 2$

$log 2=\log 10 -\log 5 = 1-m$

Our answer is thus $m+n-1$.

thanks man, one more question though:

If i encounter more of these questions in the future how should i go about solving them thanks

Let me use an example to help explain. :)

If $\log 36=a$ and $\log 125=b$, express $\log \frac{1}{12}$ in terms of $a$ and $b$.

First, try to see how you can form the argument of the logarithm with the things you given. In this case, the argument is $\frac{1}{12}$.

To form $\frac{1}{12}$, we'll probably need a 2 and a 6, or a 3 and a 4. Let's see if we can do that with the stuff given.

Well, 6 is easy, because $\log \sqrt{36}=\log 6 = \frac{a}{2}$. To form 2, let's use the fact that $\frac{10}{5}=2$. $\log 2=\log 10-\log 5=1-\frac{b}{3}$.

Just examine the things you're given, what you need to find in order to get the answer, and try to find those things. Same goes for all other problems as well.


geometry
3D geometry
calculus
integration

Find the dimensions of all right circular cones with positive integral height and radius, whose total surface area and volume are numerically equal.

we know that the volume of the right cone is $\frac{\pi r^2h}{3}$ and the surface area is $\pi rl + \pi r^2$, where $l$ is the slanted height.  Since it is a right cone, $l = \sqrt{r^2 + h^2}$. equating them and simplify:

\begin{eqnarray*} \frac{\pi r^2h}{3} &=& \pi r\sqrt{r^2 + h^2} + \pi r^2\\ \frac{rh}{3} &=& \sqrt{r^2 + h^2} + r\\ rh - 3r &=& 3\sqrt{r^2 + h^2}\\ r^2h^2 - 6r^2h + 9r^2 &=& 9r^2 + 9h^2\\ (r^2 - 9)h^2 - 6r^2h&=& 0\\ hr^2 - 6r^2 -9h &=& 0\\ r^2 &=& \frac{9h}{h-6}\\ \end{eqnarray*}

So $\frac{9h}{h-6}$ must be a perfect square. Since $9$ is a perfect square already we can only consider $\frac{h}{h-6}$.  So:

\begin{eqnarray*} \frac{h}{h-6} &=& p^2\\ h &=& hp^2 - 6p^2\\ \frac{6p^2}{p^2 - 1} &=& h\\ \end{eqnarray*}
we can disregard $p^2$ because it is integer already, and consider only $\frac{6}{p^2-1}$. the only value for $p$  that makes this an integer is $2$. plug back in to find $h = 8$ and $r = 6$

[quote="pkerichang"]
\begin{eqnarray*} \frac{\pi r^2h}{3} &=& \pi r\sqrt{r^2 + h^2} + \pi r^2\\ \frac{rh}{3} &=& \sqrt{r^2 + h^2} + r\\ rh - 3r &=& 3\sqrt{r^2 + h^2}\\ r^2h^2 - 6r^2h + 9r^2 &=& 9r^2 + 9h^2\\ (r^2 - 9)h^2 - 6r^2h&=& 0\\ hr^2 - 6r^2 -9h &=& 0\\ \end{eqnarray*}[/quote]

Here's a quicker way to go from here:

$(h-6)(r^{2}-9)=54$

Looking at the factors of 54, we see that 2,27 is the only set of factors that works (because it makes $r$ an integer), so $h=8$, and $r=6$


geometry
circumcircle
cyclic quadrilateral
radical axis
geometry proposed

Given triangle $ ABC$. Let $ M,N$ be points on $ AB, AC$ such that $ BMNC$ is a cyclic quadrilateral; $ L,K$ be points on $ MN$ such that $ BLKC$ is a cyclic quadrilateral. $ (ALM)\cap (AKN)\equal{}\{A,X\}$, $ (ALM)\cap (BMNC)\equal{}\{M,Y\}, (AKN)\cap (BMNC)\equal{}\{N,Z\}$.  Prove that the circumcenters of triangle $ AMN, XYZ$ and the radical center of three circles $ (XYZ)$ are collinear.

We don't need the quadrilateral $ BLKC$ to be cyclic. Let $ T \equiv AX \cap MY \cap NZ$ be the radical center of $ \omega_1 \equiv \odot(AML),$ $ \omega_2 \equiv \odot(ANK)$ and $ \omega_3 \equiv \odot(BMNC).$ Inversion centered at $ T$ with power equal to the power of $ T$ with respect to $ \omega_1,\omega_2,\omega_3$  transforms them into themselves and maps $ X,Y,Z$ into $ A,M,N$ respectively $ \Longrightarrow$ $ \odot(AMN)$  is the inverse of $ \odot(XYZ)$ under the referred inversion, thus their centers $ O_1$ and $ O_2$ are collinear with the center $ T$ of the inversion.


geometry
3D geometry
rotation
combinatorics unsolved
combinatorics

4x6x9 rectangular parallelepiped has been made on 216 unit cubes. Two people splits the cube by a plane into two smaller parallelepiped. After player has splitted the parallelepiped, it is next player's turn. The player who makes tha last split is a winner. Which one has the winning strategy.

WLOG always rotate the parallelepiped for axbxc with $ a \le b \le c$

The first one has a winning strategy,
First move leave 4x6x7

after any move of the second player the first one can split to some of the following winning positions:
Leaving 1xaxa (winning by always keeping 1xkxk)
Leaving 2x3x4 (after any move split to the previous case)
Leaving 2x5x6 (after any move split to any of the previous cases)
Leaving 3x5x7 (after any move split to any of the previous cases)


projective geometry
geometry theorems
geometry

i saw two versions of harmonic division on the line (ABCD) recently.

one says that it refers to CA/CB=-DA/DB

another one in wikipedia refers to BA/BC=DA/DC

are they equivalent? i am confused... :(

and is it equivalent to say (ABCD),(ACBD),(ABDC) for the same harmonic division?

[quote="morheli"]one says that it refers to CA/CB=-DA/DB

another one in wikipedia refers to BA/BC=DA/DC[/quote]

cause I think, the first considered the point in this order: A--C--B--D and the second in this: A--B--C--D
is not correct to use BA/BC=DA/DC on the first order


calculus
integration
algebra
function
domain
polynomial
Ring Theory

Let $ k$ be a field of characteristic $ \neq 2$. Let $ f \in k\left[x_1,...,x_n \right]$ be a square-free nonconstant polynomial. Let $ A \equal{} k\left[x_1,...,x_n,z \right]/(z^2 \minus{} f)$. Show that $ A$ is an integrally closed ring.

More generally, let $ A$ be an integral domain with field of fractions $ K$, and $ L$ be a finite extension of $ K$. We need two results:

(1) Let $ B$ be the set of elements in $ L$ which are integral over $ A$. Then $ B$ is integrally closed.

[i]Indeed, if $ x\in L$ were integral over $ B$, then $ x$ is also integral over $ A$, and so must lie in $ B$).[/i]

(2) Let $ s\in L$ be an element which is integral over $ A$ and let $ f(x)$ be its minimal polynomial over $ K$. If $ L/K$ were separable, then the coefficients of $ f(x)$ must lie in $ A$.

[i]Indeed, let $ L'$ be the Galois closure of $ L/K$; then the conjugates $ s_i$ of $ s$ in $ L'$ are likewise integral over $ A$; since the coefficients of $ f(x)$ are symmetric polynomials in the $ s_i$'s they are also integral over $ A$; since these coefficients are in $ K$ and $ A$ is integrally closed in $ K$, the result follows.[/i]

So let's make use of these results: let $ A \equal{} k[x_1, \dots, x_n]$, $ K \equal{} k(x_1, \dots, x_n)$, $ L \equal{} k(x_1, \dots, x_n)[z] / (z^2 \minus{} f)$. By (1), we wish to show that $ B \equal{} k[x_1, \dots, x_n, z]/(z^2 \minus{} f)$ is the integral closure of $ A$ in $ L$. To this end, let $ s\in L$ be an element which is integral over $ A$. Write $ s \equal{} a \plus{} bz$, where $ a,b \in K$. Its minimal polynomial over $ K$ is then given by:

\[ T^2 \minus{} 2a\cdot T \plus{} (a^2 \minus{} b^2 f) \equal{} 0.\]

By (2), we know that the coefficients are in $ A$, so $ 2a \in A$, $ a^2 \minus{} b^2 f \in A$. Since the char is not 2, this means $ a\in A$ and $ b^2 f\in A$. Since $ f$ is square-free, we must have $ b \in A$. This implies that $ s\in B$ as we claimed.

An alternative solution is here: http://www.mathlinks.ro/Forum/viewtopic.php?t=331949

There is missing assumption in (2), that $ A$ is integrally closed, but this is the case for $ A \equal{} k[x_{1},\ldots, x_{n}],$ so the proof still works fine.


we have :
 $ \sum_{i=2}^{n}(1-x_{i})^{2}\geq \frac{ n-1-\sum x_{i}}{n-1}$
note that :$ x_{1}= 2(n-1)-\sum_{i=2}^{n}x_{i}$
 ....  i think the way is true but i busy. i'll solve it later


[img]http://randomroms.com/upload/misc/triangle.gif[/img]

If you can't get the image, click [url=http://randomroms.com/upload/misc/triangle.gif]here[/url]

I still receive just a Forbiden message when I try to see any of the links.

Best,

http://www.mathlinks.ro/Forum/viewtopic.php?t=64298

Wrong forum in my opinion.


:D  :ninja:  :starwars:

What should be our strategy, subtract PCC or add FL?

it doesn't matter which one we do, we'll end up with gaining 3 points on them either way. If we don't think we'll win soon enough but want to drag it longer until we get more active, then let's just focus on gaining. Otherwise, focus on making them lose more points.

Are you sure 'bout that strategy ProtestanT? Let's say we have a turn equally often as PCC. Assume they keep hurting us. If we keep hurting them, we will reach -75 sooner. If we keep healing ourselves, it will be a standstill... just trying to spark some debate here...  :lol:

Maybe we should keep our score _just_ below theirs, so someone will join our team instead of theirs. And then we GOGOGOGOGO WIN!!!!!!


Find all continuous functions $ f : R \to R$ which is satisfy the requirements :
$ 2f(x)\equal{}f(\frac{x}{x^{2}\plus{}1})\plus{}f(\frac{1\minus{}x^{2}}{2})$   $ \forall x \in R$

Any idea ?  :)
I found that : $ f(x)\equal{}const$  $ \forall x \in [\minus{}\sqrt{2}\minus{}1;\sqrt{2}\plus{}1]$

looks suspiciousally tangently like.......


In the grid below, place one digit in each of the 24 non-shaded squares such that each of the 14 numbers below occurs either vertically or horizontally in adjacent squares. What is the number $A,BCD,EFG$ where each letter denotes the digit in the so letter square?

[list]
22
23
25
27
78
232
298
965
2237
2829
2837
2936
3286
66236[/list]

[img]7354[/img]

[hide]There's only one space for the 5-digit one, and then everything just fits. The answer is $8222657$.[/hide]

Ah, I wasted so much time on this one (though I got it right), which caused me to scramble during the last 10 problems and get only about 4 of em right.

That is about what I did (scramble to finish the last 10), I just glanced at the last page and remember something similar to 30 and finished 30 as time expired.  I remember spending so much time on that ice cream cone problem.

But you weren't like me and missed #1 and #3 on the sprint round.

yeah, i was like "this is 4?" it took like 2 minutes.


function calculus integration parameterization trigonometry derivative logarithms

Hi everyone,

First time here to suggest a problem I'm stuck with for 1 week. As you will see it might be obvious for a mathematician, but I'm only a teorician physicist :-)

Well : the deal here is to evaluate analytically the integrals of the form :

(hope you can read it properly)

$\int_{r=0}^{+\infty}e^{i\,t(\omega-i\,\eta)}\,j_0^2(\mu\,t)\,dt$

involving the square of the spherical bessel function j_0, \eta being a small number to move the associated pole away from the real axis (typical "diagrammatic" trick in physics)

Actually, the final goal would be to evaluate the more genral form :

$\int_{r=0}^{+\infty}e^{i\,t(\omega-i\,\eta)}\,j_l(\mu\,t)\,j_m(\mu\,t)\,dt$

but I think once you figure out the trick for the first one it could be easy then....
My guess would be a simple use of Residue theorem but from this point I'm stuck...

Any help would be greatly appreciated

Cheers

I would try to use the formula $e^{t(z-1/z)/2}=\sum_{n=-\infty}^\infty z^n J_n(t)$. Then, disregarding $\mu$ (the linear change of variable is trivial), we find that, at least for large $\nu>0$,
\[ \sum_{n,m=-\infty}^{\infty}z^n w^m \int_0^\infty J_n(t)J_m(t)e^{(-\nu+i\omega)t}\,dt= \frac 1{\nu-i\omega-\frac1 2(z-1/z)-\frac 1 2(w-1/w)}\,. \]
Now the task is reduced to finding the Laurent coefficients of a rational function of two variables. To do it, just assume that $w$ is constant first and find the Laurent coefficients in $z$ as functions of $w$. Then decompose those functions of $w$. It seems doable in theory for each fixed pair $n,m$, but I'm afraid that the analytic formulae will get more and more complicated as $m,n\to\infty$. I'll try to work out a few examples when I have more time. Hope I didn't make some stupid mistake so far :).

Seems like I was overly optimistic when saying that finding the Laurent coefficients is "possible in principle" :blush:. Even when $n=0$, we arrive at the task of decomposing $(1+F(w)^2 )^{-1/2}$ where $F(w)=\nu-i\omega-\frac 1 2 (w-1/w)$ and my analytical skills are insufficient to do it. Still, if you would agree to settle for less than an exact formula, namely, for an asymptotic expansion of your integrals in inverse powers of $\nu-i\omega$ valid for large $\nu$ or $\omega$, this approach gives you the result immediately :). If it is insufficient for you, tell what exactly is needed from these integrals and we'll try to go from there.

Hi and thank you very much for your time.

As I said the parameter $\eta$ (I'm assuming it's your $\nu$ is that it ?) is just something we introduced in reference to our classical diagrammatic point of view to solve the pole problem, but this guy should be taken to zero, or even removed if you don't need it absolutely.
Now if I look at what I want exactly, we're not quite sure right now if the asymptotic expansion in powers of inverse of $\eta-i\omega$ would be enough, I think so but i'd like to see wether we can find something exact, in a way something like  a function $f(\mu/omega)$ would be great. But on the other hand your expression looks good in a sens where we can relate that to some diagrams I think...

Now to be clear, using physical arguments we know for sure we want to find a least the expression for $j_0^2(\mu\,t)$, $j_1^2(\mu\,t)$ and $j_0(\mu\,t)j_1(\mu\,t)$ as they correspond to the first orders in a bigger picture expansion in short time, so we could settle with this I think

I just want to point out that in the $j_0^2(\mu\,t)$ case we find an anlytical result using mathematica involving $log$ functions, but as we are not that confident we'd like to make the exact derivation, but this may give some leads.

So if you think it's possible to carry out the expansion simply this could be a good start if you can show me how you do it...

Anyway thank you very much for your time

Best regards

[quote="Rotiti"]
I just want to point out that in the $j_0^2(\mu\,t)$ case we find an anlytical result using mathematica involving $log$ functions, but as we are not that confident we'd like to make the exact derivation, but this may give some leads.[/quote]
I would be most grateful if you could show me the analytic formula that you got for $J_0$. Maybe I'll be able to see where my problem is then because when I tried to get it, I faced the integral $\int_{-\frac\pi 2}^{\frac\pi 2}\frac{d\theta}{\sqrt{1+\eta^2-\sin^2\theta+2i\eta\sin\theta}}$, which is not exactly the complete elliptic integral, but something quite close, so I assumed that it is not an elementary function of $\eta$.
On the other hand, while I still have doubts that it is possible to write a simple closed formula, we can, no doubt, figure out any relevant information about the behaviour of this function for the three products you are interested in. Just tell what exactly you want to know.

Hi

Thnaks again for your time

If I ask mathematica to compute :

(And now I feel bad I realize I missed a minus sign in the exponential in the formula I sent first, my deepest apologies for that)

$\int_{t=0}^{+\infty}e^{-\imath\omega t+\nu t}j_0^2(\mu t)$

I get, if I don't make any mistakes

$\frac{1}{\mu} \, Arctan\left(\frac{2\mu}{-\nu+\imath\omega}\right)+\frac{1}{4\mu^2}(\nu-\imath\omega)Log\left(1+\frac{4\mu^2}{(\imath\omega-\nu)^2}\right)$

if $\mu\ge 0$ and $Im[\omega]+Re[\nu]\ge 0$, which is Ok I guess (the sign of $\nu$ is arbitrary I would guess, and $\omega$ is real, being an energy, and the arctan can be recast into logarithms.

I still have to make sure of the sign for $\eta$ but I think it does not matter, it's just the way you write you problem in terms of diagrams and should not be relevant here

Thanks anyway

Let's run a simple test on this formula: put $\omega=0$ and let $\nu\to 0$. Since the Bessel functions squared decay like $1/t$ at $+\infty$, we should expect a logarithmic singularity in $\nu$. Meanwhile, the formula you gave tends to a finite limit under these assumptions. This makes me very suspicious about it. By the way, can you put into Mathematica the integral I mentioned in my previous post? I wonder what it will say...

No I tried to run your integral and nothing comes out.

But when I look at your argument, I wonder if I made a point clear : I'm considering spherical bessel functions, eg for instance $j_0(t)=\frac{sin(t)}{t}$. So I would guess that the square is continuous for t going to zero and goes like $\frac{1}{t^2}$ for large t, giving a definite integral, am I right ?

That's why I didn't rule out my result when I did the same kind of limit you did, but I may be wrong....

[quote="Rotiti"] I'm considering spherical bessel functions, eg for instance $j_0(t)=\frac{sin(t)}{t}$. So I would guess that the square is continuous for t going to zero and goes like $\frac{1}{t^2}$ for large t, giving a definite integral, am I right ?
[/quote]
Oh, my God! I didn't realize that (all I said concerned usual Bessel functions :( ). Well, back to the board then...

No it's my fault, I'm way too used to work with them in my field (we never use the standard BEssel functions, for instance those spherical Bessel come from a plane wave expansion), I should have emphasized this point.

I hope I didn't make you lose too much time

Well, those are much easier and, indeed, can be expressed in an elementary way, though I still have to think of whether we can get a simple formula for arbitrary $m$ and $n$. The trick is to consider $z=\nu-i\omega$ and to write the product $j_n(t)j_m(t)$ as the sum of expressions of the kind $t^ke^{i\ell t}$ divided by $t^{m+n+2}$. Now, for $Re\,z>0$, we can find the $m+n+2$-nd derivative of the desired integral with respect to $z$ by just integrating and adding the expressions of the kind $t^k e^{-(z-i\ell)t}$, which is a piece of cake. Then we'll have to integrate the sum of fractions $\frac1{(z-i\ell)^{k+1}}$ several ($m+n+2$) times taking into account that the value of all derivatives at $+\infty$ is $0$. That's all. Since I wasted so much of your time because of my misunderstanding your original post, let me (almost) compute the one corresponding to $j_0j_1$. We have
\[ j_0(t)j_1(t)=\frac{\sin^2 t-t\sin t\cos t}{t^3}=\frac 14\frac{2-e^{2it}-e^{-2it}+it(e^{2it}-e^{-2it})}{t^3}\,. \]
Forgetting $t^3$ for a second and integrating the rest against $e^{-zt}$, we get
\[ \frac 14\left[\frac 2 z-\frac1{z-2i}-\frac 1{z+2i}+\frac i{(z-2i)^2}-\frac i{(z+2i)^2}\right] \]
Integrating once, we get
\[ \frac 14\left[2 \log z-\log(z-2i)-\log (z+2i)-\frac i{z-2i}+\frac i{z+2i}\right] \]
Integrating once more, we get
\[ \frac 14\left[2 z\log z-(z-2i)\log(z-2i)-(z+2i)\log (z+2i)- i\log(z-2i)+i\log(z+2i)\right] \]
(a few terms happily cancelled out here :)).
Now, integrating the last time, we get
\[ \frac 14\left[z^2\log z-\frac{(z-2i)^2}2\log(z-2i)-\frac{(z+2i)^2}2\log (z+2i)- i(z-2i)\log(z-2i)+i(z+2i)\log(z+2i)+C\right] \]
where $C$ should be chosen to make the expression tend to $0$ at $+\infty$ (a lot of terms cancelled out and I ignored the constants). I leave finding $C$ to you. Once you do it, you'll get the answer up to the minus sign (it appears because each differentiation of the integral with respect to $z$ multiplies the integrand by $-t$, not just $t$).

Hope I haven't made a mistake in my computations. It is clear what to do in the general case too, but the answers we get are extremely ugly for large $m$ and $n$ and there seems to be no nice generating function to round them up. For $j_0$ it is an easy exercise in integrating $1/z$ twice and the formulae for $j_1^2$ won't fit the screen...

Well that should be it !

Just to be sure : I think we should consider $z=\imath\omega-\nu$ instead of the opposite to perform te integration with respect to $e^{-zt}$ as you suggest ?

I just don't get exactly your point when you say you can find the $m+n+2^{nd}$ derivative the way you suggest.... you seem to discard in your example your $1/t^3$ factor and just integrate your numerator...

Other than that it looks like it's working, I'm just wondering if any place you use really $\nu$ to avoid your pole on the real plane ? Because at the end we can without any problem take $\nu\rightarrow 0$, so maybe we can remove it from the beginning....

So thank you very much for your help, I guess I just have to perform the variable change for $\mu$ and look closely at the methosd you suggest but looks like it's working. If I can find the general expression that's cool, otherwise the first 3 orders I mentioned will be enough.

One thing I like in your expression is that at some point (second expression) you have exactly what we might expect, I mean fractions with increasinfg order in the denominator, and we will surely want to cut the second order  (squares in the denominator) as they correspond to higher order diagrams which I love !

So thanks again (just if you can find just 5 minutes to answer my 2 / 3 questions above... otherwise don;t worry you already did a LOT)

[quote="Rotiti"]
Just to be sure : I think we should consider $z=\imath\omega-\nu$ instead of the opposite to perform te integration with respect to $e^{-zt}$ as you suggest ?[/quote]
That depends on how you write the integral (with $+$ or $-$ on $\nu$ and $\omega$)
[quote]
I just don't get exactly your point when you say you can find the $m+n+2^{nd}$ derivative the way you suggest.... you seem to discard in your example your $1/t^3$ factor and just integrate your numerator...[/quote]
And that gives you the third derivative of the answer. If you differentiate with respect to $z$ three times, you get the integrand multiplied by $-t^3$, that's why it is enough to integrate the numerator with respect to $t$.
[quote]
Other than that it looks like it's working, I'm just wondering if any place you use really $\nu$ to avoid your pole on the real plane ? Because at the end we can without any problem take $\nu\rightarrow 0$, so maybe we can remove it from the beginning....[/quote]
You'll get a few divergent integrals if you remove it from the beginning. It is true that the singularities will all cancel out, but still it makes sense to keep $\nu$ when deriving the formulae. Of course, once you have the answer, you may use it with $\nu=0$.

Ok, I guess all is left is write all of these very thoroughfully, I'll di that tomorrow.
In the example you got I guess your constant $C$ is automatically equal to zero, if I take the asymtotics of each term they cancel out.... si that means I have my solutions.

So thank you very much for the valuable contribution, your help was very needed and appreciated

Now I can go back to more physics :-)


trigonometry
geometry
rectangle
angle bisector
geometry unsolved

Let $ ABC$ be an acute-angled triangle; $ AD$ be the bisector of $ \angle BAC$ with $ D$ on $ BC$; and $ BE$ be the altitude from $ B$ on $ AC$.
Show that $ \angle CED > 45^\circ .$
[b][weightage 17/100][/b]

we must show:$ \sin \angle CED> \sin \angle DEB$.
$ \frac{CD}{DB}\equal{}\frac{\sin \angle CED}{\sin \angle DEB}.\frac{CE}{EB} \implies \frac{\sin \angle CED}{\sin \angle DEB}\equal{}
\frac{CD}{DB}.\frac{EB}{EC}$.But,since $ ABC$ is an acute-triangle we have

$ \angle ABC>90^\circ\minus{}\angle ACB \implies \sin \angle ABC>\sin (90^\circ\minus{}\angle ACB)$.so we get

$ \frac{CD}{DB}\equal{}\frac{CA}{AB}\equal{}\frac{\sin \angle ABC}{\sin \angle ACB}>\frac{\sin (90^\circ\minus{}\angle ACB)}{\sin \angle ACB}\equal{}\frac{CE}{EB}$.which leads to $ \frac{CD}{DB}.\frac{EB}{EC}\equal{}\frac{\sin \angle CED}{\sin \angle DEB}>1$

 so we get $ \sin \angle CED>\sin \angle DEB \implies \angle CED>45^\circ$.

Project the point D on the lines AB,BE and AC as L,M and K respectively. This is basically done to get the AAS congruence as we already have angle equality with the given angle bisector condition.
Triangles ALD and AKD are congruent which gives us that LD=KD.
From here the result is obvious as LD >DM which gives EM > DM which gives  angle CED > 45 degrees.

Draw DL perpendicular to AB; DK perpendicular to AC; DM perpendicular to BE.

Then EM = DK. angle LAD = angle KAD. And angle AKD = angle ALD = 90.
Hence triangles AKD and ALD are congruent. This gives DL = DK.

Now DL less than DM,  since BE lies inside the triangle.
Thus EM less than DM.

This implies angle EDM greater than angle DEM = 90 - angle EDM.
Hence angle EDM greater than 45.
Also angle CED = angle EDM.

Hence the result. :)
 :evilgrin:

[hide="@zzzzz"]You've simply copied the official answers..isn't it?[/hide]

[hide="Own Solution"]
Let $ X$ be a point on $ BC$ (extended) such that $ XA \perp AC$.Also, let the angle bisector of $ \angle BEC$ meet $ BC$ in $ Y$.
Now, the only pre-requisites we need to know is the angle bisector theorem and that the sin of an angle is constant.

Therefore,
$ \frac {BD}{DC} \equal{} \frac {AB}{AC} < \frac {AX}{AC} \equal{} \frac {EB}{EC} \equal{} \frac {BY}{YC}$

Therefore,
$ \frac {BD}{DC} < \frac {BY}{YC}$

Adding $ 1$ to both sides proves that
$ DC > YC$
$ \Rightarrow \angle CED > \angle CEY \equal{} 45$
We are done... :lol:
$ Q.E.D.$
[/hide]

[Moderator edit: also see http://www.mathlinks.ro/viewtopic.php?t=239360 ]

Suppose $DK$ perpendicular to $AC$,$DM$ to $BE$ and $DN$ to $AB$.we can easily see that $DN=DK$,also $DN>DM$ because $BNMD$ is cyclic with diameter $BD$,but $DMEK$ is a rectangle,hence $DM=EK$ but $DK>EK$ because $DK>DM$.so ${\angle}KED>45$.


trigonometry
integration
calculus
calculus computations

Solve $ y''\plus{}y\equal{}\cos x\plus{}3\sin 2x$

Characteristic equation is
\[ \lambda^2+1=0 \ \text{i.r.} \ \lambda=\pm i.\]
So, let
\[ y = C_1(x) \cos x + C_2(x) \sin x.\]
and assume that
\[ C_1'(x) \cos x + C_2'(x) \sin x = 0.\]

So,
\[ y' = -C_1(x) \sin x + C_2(x) \cos x,\]
\[ y'' = -C_1'(x) \sin x + C_2'(x) \cos x - C_1(x) \cos x - C_2(x) \sin x.\]

Therefore,
\begin{eqnarray*}
(-C_1'(x) \sin x + C_2'(x) \cos x - C_1(x) \cos x - C_2(x) \sin x) + (C_1(x) \cos x + C_2(x) \sin x)
& = & - C_1'(x) \sin x + C_2'(x) \cos x \\
& = & \cos x + 3 \sin 2x.
\end{eqnarray*}

So,
\[ \left\{\begin{array}{l}
C_1'(x) \cos x + C_2'(x) \sin x = 0 \\
- C_1'(x) \sin x + C_2'(x) \cos x = \cos x + 3 \sin 2x.
\end{array}\right.\]

Therefore, $ C_1'(x) = - \sin x \cos x (1+6\sin x)$, $ C_2'(x)= \cos x (\cos x +3 \sin 2x).$
Thus,
\[ C_1(x) = \int (-\sin x \cos x (1+6\sin x)) dx = \frac{1}{2}(\cos^2 x - 4 \sin^3 x) + C_1,\]
\[ C_2(x) = \int \cos x (\cos x +3 \sin 2x) dx = \frac{1}{4}(2x - 8 \cos^3 x + \sin 2x) + C_2.\]

Thus, general solution is
$ \left(\frac{1}{2}(\cos^2 x - 4 \sin^3 x) + C_1\right) \cos x + \left(\frac{1}{4}(2x - 8 \cos^3 x + \sin 2x) + C_2\right) \sin x$
$ = \frac{1}{2}\cos x + \frac{1}{2} x \sin x - \sin 2x + C_1 \cos x + C_2 \sin x.$

That's correct. :)

You can summarize as $ y\equal{}A_1\cos x\plus{}A_2\sin x\plus{}\frac{1}{2}x\sin x\minus{}\sin 2x$.


Solve in $ \mathbb{Z} \times \mathbb{Z}$ the equations:

$ a)$ $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{3}$

$ b)$ $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{5}$

$ c)$ $ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{1}{7}$

[hide]a.) multiply it by 3xy

$ 3y\plus{}3x\equal{}xy\\
xy\minus{}3x\minus{}3y\equal{}0$

using sfft

$ xy\minus{}3x\minus{}3y\plus{}9\equal{}9$
$ (x\minus{}3)(y\minus{}3)\equal{}9\equal{}1\cdot9\equal{}3\cdot3\equal{}9\cdot1$

It follows that the ordered pairs $ (x,y)$ are $ \boxed{(4,12),(6,6),(12,4)}$.

b. and c. are solved the same way.[/hide]


There are 6 dice, 5 normal and one with its 6 replaced by a 1. If you pick one at random and roll it twice and its a 1 both times, what is the chance if you roll it again youll get a 1 again?

@chessplayingballer: That's incorrect. The chance that you get one of the normal dice and have double ones is $ \frac56\left(\frac16\right)^2 \equal{} \frac5{216}$, and the probability that you get the other die and have double ones is $ \frac16\left(\frac26\right)^2 \equal{} \frac4{216}$.

Since we are given that one of these must happen, the chance that we have a normal die before our final roll is $ \frac {\frac5{216}}{\frac5{216} \plus{} \frac4{216}} \equal{} \frac59$, and the chance that we have the die with two ones is

$ 1 \minus{} \frac59 \equal{} \frac49$.

Thus, our probability is $ \frac59\cdot\frac16 \plus{} \frac49\cdot\frac26 \equal{} \frac5{54} \plus{} \frac8{54} \equal{} \boxed{\frac {13}{54}}$.

Edit: Hm...our solutions are essentially the same. :P  :wink:

Yes, they are independent events, but the fact that two one's were rolled first does matter, since it increases the chances that the die chosen was the one with 2 one's.

[hide]The total probability of picking a random die and rolling two one's in a row is $ 5\cdot \left(\frac {1}{6}\right)^2 \plus{} \left(\frac {1}{3}\right)^2 \equal{} \frac {5}{36} \plus{} \frac {1}{9} \equal{} \frac {1}{4}$. The probability of having picked one of the 5 normal dice given that two one's were rolled in a row is thus $ \frac {\frac {5}{36}}{\frac {1}{4}} \equal{} \frac {5}{9}$, and the the probability of having picked the die with two one's is thus $ \frac {4}{9}$. The probability of rolling a one next in the former case is $ \frac {1}{6}$, and for the latter $ \frac {1}{3}$. Thus the total probability of rolling a one next is $ \frac {1}{6}\cdot \frac {5}{9} \plus{} \frac {1}{3} \cdot \frac {4}{9} \equal{} \boxed{\frac {13}{54}}$.[/hide]

 :maybe:


ratio
percent
geometry
geometric transformation
reflection

I want to have some idea as to the error margin of AoPS polls, so I offer this ridiculous question to get a sense of how many people are lying.  :lol:

Popular in what sense?  As in well known and liked?  Or respected?

[quote="frankthetank"]I'd say I'm one of the 800 most popular people at my school.[/quote]
How many people are in your school?

By the way, the choice "I don't go to school" was intended for the adults in this forum, so it is still a respectable answer.

It looks like a lot of people are homeschooled here  :lol:

or homeschoolers!!!i would like to say first of all....i am the nicest...best looking...nicest.....most popular......greatest boy in my school!! Unfortunatly....im also the dumbest...most ugly...weakest...most hated kid

Well since there are 5 "yes" answers, I guess so far 5 people have lied :D.

It's 6 now... maybe those polls of the AMC/AIME scores aren't so accurate either  :rotfl:

[quote="h_s_potter2002"]Um, why can't someone be one of the most popular people at their school?[/quote]
Because they go on this forum.

And what does voting on this forum have to do with popularity?

Or what if that person went to TJHSST or some other magnet / boarding school?

well even if they were, the ratio from mostpopularpeople/allpeople should definitely be less than 1/3

for a good example of a large AoPS error margin, just take a look at the 2004 US democratic primary election poll (joe lieberman, who really got about 4 percent of the real vote, outscored the next leading candidate by 300 percent on the AoPS poll)

This poll does not at all reflect the overall accuracy of AoPS polls in general.  For one, it's in Games and Fun.  Secondly, people are more prone to lie in this poll where they think it would be funny or whatever.  Third, you can't really say that people who put that they are the most popular at their school are lying.  What if they go to like TJ or something...I dunno...maybe the mathletes are idolized there or something :P Anway, there might also be people in self denial :D

...or maybe they ARE cool in their school...like me, for instance  :yup: (jk...i put no...).


geometry
parallelogram
ratio
trapezoid
geometric transformation
dilation
geometry proposed

Let $P$ be a point in the interior of $\triangle{ABC}$ such that $\angle{BPC}=90^{\circ}$ and $\angle{BAP}=\angle{BCP}$. Let $M$, $N$ be the midpoints of sides $AC$, $BC$ respectively. Suppose $BP=2PM$. Prove that $A$, $P$, $N$ are collinear.

[quote="prime04"]Let P be a point in the interior of triangle ABC such that [b]angle(BPC)=90'.[/b]
Also,[b]angle(BAP)=angle(BCP)[/b].Let (M,N) be the mid-pts of (AC,BC) respectively.Suppose [b]BP=2PM[/b].Prove that [i]A,P,N[/i] are [b]Collinear[/b].[/quote]
Let I be the midpoint of PC. We have $ \angle NPC\equal{}\angle PCN\equal{}\angle BAN$
It's easy to show that $ \angle BAN\equal{}\angle NMI hence \angle NPC\equal{}\angle NMI$
So PMIN is a cyclic quadrilateral.
Since $ NI\equal{}\frac{1}{2}BP\equal{}PM$ we get PMIN is a scale trapezium.
Therefore $ MI//PN$
On the other hand $ MI//AP$ so A,P,N are collinear. (QED)

First, let’s properly construct the figure:2 equal circles, of centers N and O, secant at B and P, C opposite to B in circle N, then NP cuts again O at A, suppose P is on (AN). Construct the parallelogram ABCD, M is the common point of its diagonals, Q the middle of CP, R opposite to B in circle O. If Q is midpoint of PC, dilating by ratio 2 the triangles MQN and APB will map each other, hence PMQN is isosceles trapezoid and MQ\\PN, but MQ\\AP, so we were right in our construction and A, P and N are collinear. Obvious, APBR is isosceles trapezoid, the same previous dilation will map MQNP and APBR, hence AB = PR, but
PR = PC, so PC = AB. Again obvious, PD and PG are the angle bisectors of $ \angle BPM$, hence DP is tangent to circle N.

Best regards,
sunken rock


Prove that a decimal is rational if and only if it repeats.

Technically, that should be "repeats after a finite sequence of digits or terminates."

[hide="Solution"] First off, this is true in an arbitrary base $b$.

Showing that every repeating decimal is rational is easy, and I won't bother to do it.

Showing that every rational is a repeating decimal is only somewhat harder.  Let our rational be $\frac{p}{q}, gcd(p, q) = 1$.

Let $q = mn$ where $m$ is the product of all primes in the prime factorization of $q$ that are in $b$, so that $\frac{p}{m}$ has a terminating base-$b$ representation (because $m$ must divide some power of $b$).

Then $gcd(b, n) = 1$.  This implies that $n | b^{ \phi(n) }-1$, so that

$\frac{1}{n}= \frac{ \text{Something}}{ b^{\phi(n)}-1}= \text{Something}\left( \frac{1}{b^{\phi(n)}}+\frac{1}{b^{2\phi(n)}}+... \right)$

Which is a repeating decimal (unless $n = 1$, in which case $\frac{1}{n}= 1$).  The product of $\frac{1}{n}$ and $\frac{p}{m}$ is either terminating or repeating.  QED. [/hide]


The length, width, and height of a cubical shipping container are all decreased by 1, thereby decreasing the volume of the cube by 37.  What was the volume of the original container?

Masoud Zargar

[hide]For cube I, we have:

$a_I=a$ and $V_I=a_I^3=a^3$

For cube II, we have:

$a_{II}=a_I-1=a-1$ and $V_{II}=a_{II}^3=(a-1)^3$ and $V_{II}=V_I-37$

Therefore

$(a-1)^3=a^3-37 \Rightarrow a^3 - 3a^2 + 3a - 1 = a^3 - 37 \Rightarrow 3a^2-3a-36=0 \Rightarrow$

$a^2-a-12=0 \Rightarrow \{ \begin{array}{ll} a=4 \\ \textrm{or} \\ a=-3 \end{array}$

Since $a>0$, $a=4$.

So $V_I=4^3=64$.[/hide]

I am currectly like  :o, because this is so easy.  My mistake was that I forgot that it was a cube.  I approched it as if it was a rectangular prims having different distinct dimensions.  I feel so stupid.  :wallbash_red:

dont fell too bad i made the same mistake. but then the work got really ugly so i reread the question and laughed at myself when i saw that it was a cube.  :oops:  :blush:


Who can solve Cuong's problem by Cauchy ?

Not good at all! Please post it in Anh Cuong problem's topic. :arrow:


geometry
circumcircle
geometric transformation
reflection
ratio
homothety
parallelogram

let $ A$, $ B$, $ C$, $ D$ be points on circle centered at $ O$.(in that order)
let $ AB$ and $ CD$ meet at $ E$.
Let the perpendicular from $ E$ to $ BC$ meet $ AD$ at $ X$,
and the perpendicular from $ E$ to $ AD$ meet $ BC$ at $ Y$
Prove that the midpoint of $ XY$, $ E$, and $ O$ are collinear.

Let $ F$ be, the intersection point of $ AD,\ BC$ and we denote the points $ X'\equiv BC\cap EX$ and $ Y'\equiv AD\cap EY.$

From the complete quadrilateral $ ABCDEF,$ we have that the point $ E'\equiv CD\cap FP,$ where $ P\equiv AC\cap BD,$ is the harmonic conjugate of $ E,$ with respect to the points $ C,\ D.$

Similarly, from the complete quadrilateral $ XX'YY'EF,$ we have that the point $ E''\equiv YY'\cap FK,$ where $ K\equiv XY\cap X'Y',$ is the harmonic conjugate of $ E,$ with respect to the points $ Y,\ Y'.$

So, we conclude that the line segment $ E'E'',$ passes through the point $ F,$ as the intersection point of $ YC,\ Y'D$ $ ($ from $ (E,C,E',D) \equal{} (E,Y,E'',Y')$ we conclude that the points $ E'',\ E',\ F$ collinear $ ).$

$ \bullet$ Because of the cyclic quadrilateral $ ABCD,$ we have that the line segment $ FP\equiv FE',$ is the polar of $ E,$ with respect to the circumcircle $ (O)$ of $ ABCD$ and so, we conclude that $ EO\perp FP$ $ ,(1)$ where $ O,$ is the circumcenter of $ ABCD.$

Because of the cyclic quadrilateral $ XX'YY',$ we have that the line segment $ FK\equiv FE'',$ is the polar of $ E,$ with respect to the circumcircle $ (K)$ of $ XX'YY'$ and so, we conclude that $ EM\perp FK$ $ ,(2)$ where $ M,$ is the midpoint of the segment $ XY,$ as the circumcenter of $ XX'YY'.$

From $ (1),$ $ (2)$ and because of $ FP\equiv FK,$ we conclude that $ EO\equiv EM$ and the proof is completed.

Kostas Vittas.

Consider the following points:
$ F \equal{} BC\cap{AD}$, $ G \equal{} BD\cap{AC}$, $ X_1 \equal{} EX\cap{BC}$ and $ Y_1 \equal{} EY\cap{AD}$
$ E$ is the pole of $ FG$, so $ EO\perp{FG}$. The quadrilateral $ XX_1YY_1$ is cyclic, and $ \angle{YX_1X} \equal{} \angle{XY_1Y} \equal{} \frac {\pi}{2}$ so the midpoint of $ YX$ is its circumcenter, say $ O_1$. Call $ H \equal{} {XY}\cap{X_1Y_1}$ and using the same fact, $ E$ is the pole respect to circle $ XYX_1Y_1$ of $ FH$ so, $ EO_1\perp{FH}$. So, what we need to prove is that $ F$, $ G$ and $ H$ are collinear. But that is true since: $ G{'} \equal{} FG\cap{CD}$ follows: $ H[DC,G{'}E]$ and $ H{'} \equal{} FH\cap{XX_1}$ follows $ H[XX_1,H{'}E]$ and since the reflections from $ F$ of $ X$, $ X_1$ are $ D$ and $ C$ respectively, and the cross ratio is conserved, then, the reflection of $ H{'}$ (in $ CD$) must be harmonic with $ E$, therefore it must be $ G{'}$ this means $ F$, $ G$ and $ H$are collinear. Then $ EO_1\perp{FG}$ and as $ FG\perp{OE}$, $ EO$ and $ EO_1$ coincide. Therefore $ E$, $ O$ and $ O_1$ are collinear

I wanted to explain perfectly the last part :)

I was writing my solution when vittasko posted his :(

[quote="Nemion"] I was writing my solution when vittasko posted his :([/quote]
Don’t worry dear [b][size=100]Nemion[/size][/b].

We had the same approach to solve this sweet problem and we posted the solutions simultaneously (only ten minutes between our posts). :)

Kostas Vittas.

After the collinearity of the points $ E'',\ E',\ F,$ $ ($ [b][size=100]schema t=195986[/size][/b] $ ),$ we can also complete the solution, based on the well known theorem which has already been presented in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=110887]Triangle, perpendicular lines[/url].

So, in the cyclic non-convex quadrilateral $ ADBC,$ we have that the line segment $ OE,$ where $ O$ is the circumcenter and $ E$ is the intersection point of its diagonals, is perpendicular to the line $ FP,$ where $ F,\ P,$ are the intersection points of its opposite sidelines.

That is, we have $ OE\perp FP$ and similarly $ ME\perp FK,$ from the cyclic non-convex quadrilateral $ XY'X'Y$ and the proof is completed.

Kostas Vittas.

Let the circumcenters of $ \triangle{OAD}$ and $ \triangle{OBC}$ be $ O_1$ and $ O_2$, respectively.  Notice that $ BC$ and $ AD$ are antiparallel: hence, $ EX$ goes through $ O_1$ and $ EY$ goes through $ O_2$.  Moreover, because the triangles are similar, we have a homothety about $ E$ mapping $ XY$ to $ O_1O_2$.  So it suffices to show $ E$, $ O$, and the midpoint of $ O_1O_2$ are collinear.

But notice that $ EO_1 \perp BC \perp OO_2$ and $ EO_2 \perp AD \perp OO_1$, and so $ EO_1OO_2$ is a parallelogram, which yields the desired conclusion.

[quote="ywl"][color=darkred]Let $ ABCD$ be a cyclical convex quadrilateral with the circumcircle $ w = C(O)$ for which $ AB\cap CD\ne\emptyset$ .

Denote $ \{\begin{array}{c} E\in AB\cap CD \\
\ X\in AD\ ,\ EX\perp BC \\
\ Y\in BC\ ,\ EY\perp AD\end{array}$ . Prove that the midpoint $ M$ of the segment $ [XY]$ belongs to the line $ EO$ .[/color][/quote]
[color=darkblue][b][u]Proof[/u][/b] (similarly to [b]K8107[/b]'s). Denote the circumcenters $ O_1$ , $ O_2$ of the triangles $ EAD$ , $ BCE$ respectively. Observe that

$ \{\begin{array}{c} EAD\sim ECB \\
 \\
\widehat {AEX}\equiv\widehat {DEY}\end{array}$ and $ \{\begin{array}{ccccc} O_1\in EX & ; & OO_1\perp AD & \implies & OO_1\parallel EY \\
\ O_2\in EY & ; & OO_2\perp BC & \implies & OO_2\parallel EX\end{array}$ . Thus, $ EO_1OO_2$ is a parallelogram,

i.e. the midpoint $ N$ of the segment $ [O_1O_2]$ belongs to the line $ EO$ .  Since $ EAD\sim ECB$ $ \implies$ $ \frac {EO_1}{EX} = \frac {EO_2}{EY}$ $ \implies$

$ O_1O_2\parallel XY$ obtain that the midpoint $ M$ of the segment $ [XY]$ belongs to the line $ EN$ , i.e.  $ M\in EO$ .

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=196404]See here an its classical particular case ![/url] [/color]

[quote="staretak"][quote="K81o7"]Let the circumcenters of $ \triangle{OAD}$ and $ \triangle{OBC}$ be $ O_1$ and $ O_2$, respectively.  [/quote]

Do you mean the circumcenters of $ \triangle EAD$ and $ \triangle EBC$?[/quote]

Oh right...yeah I do...

Can't edit it anymore  :(


Ring Theory
topology
superior algebra
superior algebra theorems

This is a basic question from "A Singular Introduction to Commutative Algebra" by Greuel and Pfister, Section 1.4. I'm somewhat confused as to the details of the following and was hoping for more explanation.

First the basic definitions:

Let $S \subset A$ be multiplicatively closed. We define the localization or ring of fractions $S^{-1}A : = \left\{ \frac{a}{b}\mid a \in A, b \in S\right\}$ where $a/b$ denotes the equivalence class $(a,b) \in A \times S$ with respect to the following equivalence relation:

$(a,b) \sim (a', b') : \Leftrightarrow \exists s \in S$ such that $s(ab'-a'b) = 0$

Now the point of concern:

The map $j : A \rightarrow S^{-1}A, a \rightarrow a/1$ is a ring homomorphism satisfying

a) $j(s)$ is a unit in $S^{-1}A$ if $s \in S$,
b) $j(a) = 0$ if and only if $as = 0$ for some $s \in S$,
c) $j$ is injective if and only if $S$ consists of non-zerodivisors,
d) $j$ is bijective if and only if $S$ consists of units.

So my question is, what exactly does the $a \rightarrow a/1$ part mean? Because $j(a) = a/1$ is not necessarily always true? So what exactly is the map? Part (a) seems to make sense since $\frac{1}{s}$ is an element of $S^{-1}A$, and part (b) because $j(a)j(s) = j(as) = j(0) = 0$ and $j(s)$ is a unit so it can't be a zerodivisor so $j(a)$ must equal 0 (but then what about $a \rightarrow a/1$?) From (b) we can conclude (c) since we lose injectivity if $S$ contains zerodivisors as we'd have $j(0) = j(a) = 0$ assuming $as = 0$ for some $s \in S$ and we can check distinctness of the image of distinct non-zerodivisors. Finally (d) also seems to make sense since if $S$ consists only of units, then all the "denominators" in $S^{-1}A$ are actually units which means $\frac{a}{b}$ is essentially equivalent to $ab^{-1}$ in $A$ so we do have surjectivity and (c) gave injectivity.

Well it seems I have kind of half BSed my way through this but I don't really understand the map $j$ itself still, specifically the $a \rightarrow a/1$ part and what it's referring to. Any help would be greatly appreciated.

I will try, please ask when it is not clear.

The map $j: a \to \frac{a}{1}$ just gives you to opportunity not to "loose" the old ring when looking at the ring of fractions.
It's just that one wants to preserve the old $A$ by seeing it as the fractions $\frac{a}{1}$  with denominator $1$; these fractions = the image of $j$ is not always isomorphic to $A$ by $j$, only when $j$ is injective, but at least it's the canonical and often also the only way to do this. There is simple no other/better way since all elements must be some fraction.
Look at the integers and the rationals: the integers are then just considered as those rationals with denominator $1$ (and one often neglects the denominator and does not write it then; but purely fomraly, it should be there).

Your approaches for a) till d) look fine for me (only the "only if" part of d) is missing).

Hmm I guess what I am fuzzy on is, am I allowed to just assume for instance that $j(g) = g/1$ or $0$? It seems that if $g$ is not a zerodivisor with respect to some element of $S$, then $j(g) = g$.

Maybe looking at it in the context of this problem will make my question clearer:

Let $A$ be a ring, $I \subset A$ an ideal and $f \in A$. Prove that $IA_{f}\cap A$ is the saturation of $I$ with respect to $f$, that is, equal to $I: \lan f \ran^{\infty}= \{ g \in A \mid \exists n$ such that $gf^{n}\in I \}.$ (There is also a remark included: Note that $IA_{f}\cap A = j^{-1}(IA_{f})$ where $j(a) \ra f{a}{1}$)

Now here is my attempt at a solution: First consider some $x \in IA_{f}\cap A$. Then $x = i\frac{a}{f^{n}}$ for some $i \in I, a \in A, n \ge 0$. But of course $x \cdot f^{n}= ia \in I$. Conversely, consider some $g$ such that there exists $n \ge 0$ such that $gf^{n}\in I$. Note that $IA_{f}\cap A = j^{-1}(IA_{f})$ where $j(a) \ra \frac{a}{1}$. Now $j(g) = \frac{g}{1}$ and $\frac{g}{1}\cdot f^{n}= i$ for some $i \in I$. Equivalently, this implies $\frac{g}{1}= \frac{i}{f^{n}}$. Hence, if we take $j^{-1}(\frac{g}{1}) = g = j^{-1}(\frac{i}{f^{n}})$, we see that $g$ is indeed an element of $j^{-1}(IA_{f})$ as desired, and so we are done.

Is this ok, and if not, how exactly have I missed the point behind the $j$ map. I do understand (at least I think I do) that $j$ is just saying that you don't want to lose the old ring when you are essentially expanding it with respect to some multiplicatively closed set (which makes up the denominators). Just that $j(a) = 0$ at times....

I don't really understand your problem. $j$ is just the canonical map $A \to S^{-1}A$, which is defined by $j(a) = a/1 = \frac{a}{1}$.

Your proof is almost correct. You should write $\frac{g}{1}* f^{n}= \frac{i}{1}$ instead of $\frac{g}{1}* f^{n}= i$, and $j^{-1}(\frac{g}{1})=g=j^{-1}(\frac{i}{f^{n}})$ is nonsense (think a little bit of it ... you'll see it).

Er, I should have wrote $g \in j^{-1}\left(\frac{i}{f^{n}}\right)$ since we don't necessarily have injectivity.


Can anybody explain to me how to add marginal notes in the math environment? (specifically in eqnarray)
For example, if I write the following equation
\[ xy = 2x \]
\[ y = 2 \]
I wish to write in the margin (right side) of the second row $[x \neq 0]$.

Is the only way to do that is to use \mbox or is there another way?
This is because I prefer not to use \mbox here as it does not automatically align everything.

Please help.

do you want to have the notes just aside or all the way in the right margin?  If you want the notes just aside, try this:

\begin{eqnarray}
xy&=&2x \\
y&=&2 \qquad [x\neq0]
\end{eqnarray}
\begin{eqnarray} xy&=&2x \\ y&=&2 \qquad [x\neq0] \end{eqnarray}

you can have a better result like this, but i'm sure there's a better way to take off those huge spaces..

\begin{eqnarray}
xy&\!\!\!\!=\!\!\!\!&2x \\
y&\!\!\!\!=\!\!\!\!&2 \qquad [x\neq0]
\end{eqnarray}
\begin{eqnarray} xy&\!\!\!\!=\!\!\!\!&2x \\ y&\!\!\!\!=\!\!\!\!&2 \qquad [x\neq0] \end{eqnarray}

The command works but my only problem is the alignment. I wanted all these "notes" or explanations to be aligned under each other regardless of the length of the row.

Thanks a lot.

even in different equation?  I mean if there is text between two equations?  I'll try something...


complex numbers
inequalities proposed
inequalities

Let $z_{0},z_{1},\cdots,z_{n}$ be $n+1$ complex numbers lying in the closed left half plane ${\rm Re}(z)\le 0.$ Prove that
\[\sum_{k=0}^{n}\binom{n}{k}\left(\frac{|1-z_{k}|}{1+| z_{k}|}\right)^{2}\ge 2^{n-1}.\] When does equality occur?


[quote="SUPERMAN2"]Prove that the equation $ x^3 \plus{} y^3 \equal{} z^4 \minus{} t^2$ has an infinte number of solutions in the set of positive integer numbers so that $ gcd(x,y,z,t) \equal{} 1$.[/quote]

$ (u\minus{}1)^3\plus{}(2u\minus{}1)^3\equal{}(\frac{3u^2\minus{}1}{2})^2\minus{}\frac 94(u\minus{}1)^4$

So, let $ u,v$ positive integers such that $ 3u^2\minus{}2v^2\equal{}1$, then :
1) : $ u$ is odd and so $ u\minus{}1$ is even
2) : $ \gcd(u\minus{}1,2u\minus{}1)\equal{}1$
3) : $ (u\minus{}1)^3\plus{}(2u\minus{}1)^3\equal{}v^4\minus{}(\frac 32(u\minus{}1)^2)^2$

And the equation $ 3u^2\minus{}2v^2\equal{}1$ has infinitely many solutions since :

$ (1,1)$ is solution
$ (u,v)$ solution $ \implies$ $ (5u\plus{}4v,6u\plus{}5v)$ solution

Here are the first solutions found with this method :

$ (1,0,1,0)$ (not to be kept since zeroes values)
$ (17,8,11,96)$
$ (177,88,109,11616)$
$ (1761,880,1079,1161600)$
$ (17441,8720,10681,114057600)$
$ (172657,86328,105731,11178785376)$
$ (1709137,854568,1046629,1095429699936)$


function
set theory
number theory unsolved
number theory

In Set Theory, a Universal Set is defined as a set that contains all the sets.

A UNIVERSAL SET CANNOT EXIST BECAUSE IT DOES NOT CONTAIN THE UNIVERSAL SET.

Suppose U is a Universal Set.

Then it contains all the sets in the universe.

But a set cannot contain itself -> U does not contain U.

We found a set that is not contained in U, a contradiction.

Thus a universal set cannot exist.

What are your views on this?

This is not the correct formulation of Russell's paradox.  See http://en.wikipedia.org/wiki/Russell's_paradox .

Russell's paradox only occurs in what is called [url=http://en.wikipedia.org/wiki/Naive_set_theory]naive set theory[/url].  The actual standard model of set theory is [url=http://en.wikipedia.org/wiki/ZFC]Zermelo-Fraenkel set theory with the Axiom of Choice[/url], in which universal sets cannot be defined at all.

I don't know if there's anything preventing a set from containing itself.

However, $ U$ couldn't exist anyway because $ U$ would have to contain all of its subsets. Conversely, every set in $ U$ is a subset of $ U$. So we must have $ U\equal{}\mathcal{P}(U)$ (the set of subsets of $ U$). However, there's a well-known theorem due to Cantor that states that there can be [b]no[/b] onto function from [b]any[/b] set to its powerset, yet there's an obvious onto function from $ U$ to $ \mathcal{P}(U)$.

As I understand it, universal sets can't exist in ZFC because none of the axioms guarantee that "the set of all elements with Property X" exists.  See [url=http://en.wikipedia.org/wiki/Axiom_schema_of_specification]the axiom schema of specification[/url].

[quote="scorpius119"]I don't know if there's anything preventing a set from containing itself.[/quote]
Axiom of foundation.

I think Russel's paradox has something to do with the following famous problem:

A barber in a certain town shaved all those who did not shave themselves and only those. Does the barber gets shaved or not?

Now,

If he shaves himself then he would shave a person that shaved himself, a contradiction.

If he does not shave himself then he is not shaving a person who does not shave himself, a contradiction again.

So, what does the barber do?


A particle moves in a circular orbit under the action of a force
f(r)=-(k/r^2).If k is suddenly reduced to half its value, what would be the nature of the orbit?

I used e=sqrt[1+(2*L^2*E)/(mk^2)]

My attempt:
Clearly,the particle moves under attractive central force.Now,for the circular orbit,eccentricity e=0 and as the motion is bound,the energy is negative.
If k is reduced to k/2, eccentricity changes to
e=1+(8*L^2*E)/(mk^2)=6*L^2*E/(mk^2)

Since e becomes negative as E is negative,no motion is possible.

Am I correct?

[quote="kolahalb"]no motion is possible.[/quote]

Intuitively, the particle has a tangential velocity that doesn't just go away if the force is suddenly cut in half.  It seems to me that the particle should move in a hyperbolic orbit, but I am not very experienced with this sort of stuff yet.

When I say no motion is possible,I meant that under the condition no motion is possible.However,When you SWITCH from original condition to given one,the motion is perturbed and the circular motion should tend towards no-motion.

I just want to know whether my deduction of "no motion" is correct.

if k is reduced to half you must reach that the total energy is 0.no motion would mean that KE is 0 but here PE is equal to KE so e=1 then the body will move on a parabola

My calculation was possibly wrong...however,I am not shifting from my earlier result.

e=sqrt[1+{(2*L^2*E)/(mk^2)}]
For circle, e=0 and E=(-mk^2)/(2L^2)

For the 2nd case, e=sqrt[1+{(2*L^2*E)/(m(k/2)^2)}]

Put the value of E (which is constant)
e=sqrt[-3]--- no orbital motuion possible.

why should E->0 as k is reduced to k/2?

E is not constant because the force is changing.E=F*r and if F is half-reduced the PE is also half-reduced.the KE is not changing when the modification is made

2a=k/E where a is the semi major axis...It is a standard result.For all planets,E depends on the semi major axis.For earth it is
E=-(GmM/2a) where k=-GmM

So,reducing k should not change E.
In central force motion the total energy is given by:

E=(1/2)m(dr/dt)^2 +[{(L^2)/(2mr^2)}+V(r)]

Redcing k to k/2 merely changes V(r) to V(r)/2.

So what?Does it mean E---->0?

I do not think so.

2a=k/E where a is the semi major axis...It is a standard result.For all planets,E depends on the semi major axis.For earth it is
E=-(GmM/2a) where k=-GmM

this is for eliptical movements.you know circular but there is also parabolic and hyperbolic movement
   You know E=KE+PE but when the shift is happening KE remains constant and PE is reduced to half but PE is also negative while KE remains positive so if E=0 then PE=KE in module,of course.there will still be movement because KE won't be 0 and neither PE.you should deduce the expression of E in circular movement(begin with centripetal force=F(gravity force)) then try the same when the shift is made.

Though the formula is for elliptical orbits,circle is a mere special case of ellipse.

[quote] E=KE+PE but when the shift is happening KE remains constant and PE is reduced to half but PE is also negative while KE remains positive so if E=0 then PE=KE in module,of course.[/quote]

if E=0,then KE becomes equal to PE.But,what you are saying is not enough to prove that E=0.As I mentioned earlier,the total energy in a cental force motion is given by(I need not find energy of circular path) to :

E=(1/2)m(dr/dt)^2 +[{(L^2)/(2mr^2)}+V(r)]

here V(r)= -int{F.dr}

Please contradict working with this formula.Also show why I went wrong in the calculation involving e=sqrt[1+{(2*L^2*E)/(mk^2)}]

The formulae I am using are all standard ones and working with those should help...

OK,radac_mail,I got the answer...what you were saying was correct.
I did not understood it from your posts...However,this is how one may be clear about it.

A particle moving in a field like that would have a circular velocity corresponding to v = sqrt(k/r).The escape velocity at a certain distance from the central point is v_e = sqrt(2)*sqrt(k/r)

(verify putting k=GM)

Considering the changes in the values of k,
v = sqrt(k_old/r) = sqrt(2*k_new/r) = sqrt(2) * sqrt(k_new/r)

We have shown the particle will attain the escape speed...Now given a particle has attained escape speed,its energy must be zero.Then its path is parabolic.

I didn't know the condition for escape but from what you said it's logical that it only happens when E=0.
But you could have prooved E=0 without knowing and involving escape speed:
-in the circular movement:E=mv^2/2-GMm/r but from Fcentripetal=Fgravitational v^2=GM/r then
E=-GMm/2r
-when the shift is happening then in exactly that point KE is the same but PE is changing because:PE=F*r (in module) and if F is half-reduced PE will also be half reduced so the new PE in exactly that point will be PE'=-GMm/2r
-the total energy in that point will now be: E'=GMm/2r-GMm/2r=0 and since the energy is conserved before the drift and after then after E keeps being 0

It is perfectly clear now...earlier I could not follow your approach...sorry for that.Perhaps I needed the concrete calculation as you posted this time to appreciate this properly.


At a speed of 70 miles per hour, how many minutes will it take you
to travel 30 miles? Express your answer rounded to the nearest
whole number.

We have $ 60 \times \frac{30}{70} \approx \boxed{26}$.


Call a positive number [i]special[/i] if it has a decimal representation that consists entirely of digits 0 and 7. For example, $\frac{700}{99}=7.070707...$ and 77.007 are special numbers. What is the smallest $n$ such that 1 can be written as a sum of $n$ special numbers?

A) 7
B) 8
C) 9
D) 10
E) 1 cannot be represented as a sum of finitely many special numbers

[quote="mathcnt2005"][hide] common sense tells me it's E. ( I think ). [/hide][/quote]

AHSME number 29 would never let you get away with solving the problem  :D

It looks like someone moved my problem....


I got this question on my homework, and my book doesn't adequately explain how to solve it.  I need to know how to solve it in case there is a question on the test like it.  The question goes like this:

Braving Blizzard Conditions on the planet Hoth, Luke Skywalker sets out at top speed in his snow speeder for a rebel base 4800 mi away.  He travels into a steady head wind and makes the trip in 3 hr.  Returning, he finds that the trip back, still at top speed but now with a tailwind, takes only 2 hr.  Find the top speed of Lukes snow speeder and the speed of the wind.

Can anyone help me with this?

[hide]
distance=rate*time
$ 48000\equal{}(x\minus{}w)(3)$
$ 48000\equal{}(x\plus{}w)(2)$
(x=top speed, w=wind speed)[/hide]

Thank you, With that I'm up to

4800 = (x-w)3
4800 = (x+w)2

Which turns to

1600 = (x-w)
2400 = (x+w)

But how do I get the answer from there?

$ x \minus{} w \equal{} 1600$
$ x \plus{} w \equal{} 2400$

Adding the equations together, we obtain $ 2x \equal{} 4000$, or $ x \equal{} 2000$. Plugging it back into either equation, we find that $ w \equal{} 400$ and we are done.

Thank you very much, you've been a great help.


Could anyone tell me what a ring means in algebra ?? And where rings are actually used ??

[quote="boxedexe"]It's a system where it works under two binary operators wish are addition and multiplacation.  It is called a ring because under a certain number of operations, it returns to it's original value, like a ring.

EDIT: Algebraic rings are used to prove Fermat's Last Theorem and many other number theoretical problems.  It was also developed to actually to prove this theorem.

Masoud Zargar[/quote]

The integers form a ring under addition and multiplication, but I don't think that adding, say, $1$ to itself repeatedly brings you back to $1$ eventually.  I believe that the phrase 'ring' is derived from the Germen 'Zahlring' (number ring) which was coined by Hilbert.  If you like the metaphor of going around in circles perhaps what we call 'finite fields' should really be called rings.

I read some thing on Mathworld, but just skimmed through it.  It seems interesting. :)

They are :)  The place to start, if you want to learn, is really with groups, which are better behaved (identities exist, inverses exist, etc).  Rings inherit their additive structure from an abelian group, and add the multiplication structure.

They seem pretty interesting.....but then could anyone give an example of a sum using rings and also it's solution ??


As everyone knows, there is a nice section called Resources (or Contests in AoPS Skin). There are over 3904 problems from various countries. Now, I'm main contributor of AMC problems and I want to ask AoPS visiters for the solutions of these problems. From several users over time, I got messages that it is hard to study using the problems in resources pages because everytime you solve problem(s), you always need to open that window to view the solution. So, for old AHSME, you have to do that 30 times, and they find it that sometimes, the solution is too vague or not even answered perfectly.

So, I'm going to do a poll on this matter. [b]Please note that this only matters to AMC/AHSME now (so AIME and USAMO are not covered in this issue)[/b]. If there are many people who feel that they want to have solutions as well, I'll start writing up solutions and putting them up on PDF. However, this is [i]not[/i] an easy task since it takes quite amount of time on simply writing them (not to mention putting them in computer with diagrams, etc). So, please take time on doing this.

[size=150]This poll will be valid only from June 8th to June 28th[/size]

Are you planning to write your own solutions to all of the problems, or would you copy the posted solution into a PDF?  Writing up your own solutions would take a lot of time.

I think that a nice improvement would be to have a list of correct answers for each year.  Some AMC problems that are posted either do not get answered, or there is a disagreement.  It would be convenient to see all of the correct answers so you can figure out your score.

Maybe it would be better to set up the AMC solutions so that you can click on the "S" to the right of the problem and see a solution, like with some AIME and USAMO solutions.  That wouldn't require a ton of work, and it would be convenient.

For what it's worth, I've begun writing up solutions to the 1997 AHSME. The thread is here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=530624#p530624

I really haven't thought it through completely, but the AoPSWiki may make it easier to compile nice solutions.

I think having a solutions section to go along with the oodles of problems on the resources page is a really good idea (kalva's page has the solutions).  I don't think it'd be necessary for you silverfalcon to compile all the answers yourself.  As for collecting/typing up the solutions, I'm sure there are plenty of people on this site that would help you with that.  Perhaps you could ask for solution submissions and you could choose the most clearly written ones to be posted on the solutions page.

[quote="matt276eagles"][quote="SnowStorm"]Yea.  It ought to make it a lot easier for them to be added user-by-user so we can end up with more docs like 4everwise's AIME write ups.  Those are awesome.[/quote]

What are those AIME write ups you're talking about?[/quote]

Matt, I believe SnowStorm's talking about [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=AIME+Solution&t=66843]here[/url].

I'm going to end the poll now.

The result is: YES

So, during this summer, I'll start on working on solutions. This will take time but it will happen.

I think the popular opinion is that yes, of course it would be nice to have solutions for AMC problems.  However, no one is forcing you to write up solutions on your own.  If you want to do it, that's great.  Writing up solutions to AMC problems will help you understand the problems and solutions, and it will help people when studying for the AMC.  Just don't feel like anyone's pressuring you to do this.

Of course.

Mr.Crawford and I are working (well he's working majorly on developing) on using wiki for this. As soon as I get some help, I should be able to start writing solutions and etc. It's just fun thing.

Here's your starting point:

http://www.artofproblemsolving.com/Wiki/index.php/AMC_Problems_and_Solutions

It's both good organization and the ability to edit/improve at the same time (which is a really nice bonus for AoPSers).


[quote="shobber"][hide]
Is that $3$?[/hide][/quote]
I made a big mistake... :blush:
Yes, it should be:
[hide]$99$[/hide]

[quote="G-UNIT"]LOL, yea how did u get 3?[/quote]
I don't know... I was thinking that when x and y approach 50, the fraction ---> 0. So i just chose the smallest integers that i thought that would be the correct ones....

[quote="math92"]$x$ and $y$ are two different numbers selected from the first fifty counmting numbers from 1 to 50 inclusive.

What is the largest value that $\frac{x+y}{x-y}$ can have?[/quote]

[hide]To maximize the value of a fraction, you must maximize the numerator and make the denomitor as small as possible.

To do that, $x=50$ and $y=49$

So the biggest value is $99$[/hide]

[hide]

You have to have the biggest values possible, so:

$  \frac{50 + 49}{50 - 49} = \frac{99}{1}  $

The answer is 99[/hide]

[quote="pascal_1623"][hide]

You have to have the biggest values possible, so:

$  \frac{50 + 49}{50 + 49} = \frac{99}{1}  $

The answer is 99[/hide][/quote]

I hope you mean $\frac{50+49}{50-49}$

[quote="math92"]$x$ and $y$ are two different numbers selected from the first fifty counmting numbers from 1 to 50 inclusive.

What is the largest value that $\frac{x+y}{x-y}$ can have?[/quote]

The answer is 99, since x=50 and y=49.

and 236 factorial, you are right.

[hide]numerator=big
denominator=small

50+49/50-49
99/1=99[/hide]

[quote="@nn@"]well x = 50 and y = 49 because
50 + 49 = 99 (the biggest number that we can get witn two numbers from 1 to 50)
and 50 - 49 = 1 (the smallest number we can get)

 :maybe:[/quote]

Woah, how did you dig this one up? it's three years old.  :o


[quote="GameBot"]Alice and Bob play a game with a baseball. On each turn, if Alice has the ball, there is a 1/2 chance that she will toss it to Bob and a 1/2 chance that she will keep the ball. If Bob has the ball, there is a 2/5 chance that he will toss it to Alice, and if he doesn't toss it to Alice, he keeps it. Alice starts with the ball. What is the probability that Alice has the ball again after two turns?[/quote]

Case 1: Tosses to bob and recives it back: $ \frac{1}{2} \cdot \frac{2}{5}\equal{}\frac{1}{5}$

Case 2: Keeps for helself twice: $ \frac{1}{2} \cdot \frac{1}{2}\equal{}\frac{1}{4}$.

The sum is $ \frac{9}{20}$.


[quote="happyme"]The sum of the squares of the lengths of the three sides of a right triangle is $300 cm^{2}$. How many cm are in the length of the hypotenuse? Express your answer in simplest radical form.[/quote]

[hide]From Pythagorean Theorem,

The square of the length of the hypotenuse is $\frac{300}{2}$ cm$^{2}$, which is $150$ cm$^{2}$.

Take positive square root: $\sqrt{150}=\boxed{5\sqrt{6}}$.[/hide]

[quote="happyme"]The sum of the squares of the lengths of the three sides of a right triangle is $300 cm^{2}$. How many cm are in the length of the hypotenuse? Express your answer in simplest radical form.[/quote]
[hide]a^2+b^2+c^2=300, so c^2=300-a^2-b^2. Add c^2=a^2+b^2 to that and get 2c^s=300, c^2=150. so the hapotnuse is 5 radical 6.[/hide]
I seriousely need to learn LaTex.

[quote="happyme"]If you put a dollar sign before and after your equations, most of them will come out right (including exponents), I'm not sure about the radical though.[/quote]

radical: use \sqrt{xy} between dollar signs for $\sqrt{xy}$ to be displayed


Ryan has 3 red lava lamps and 3 blue lava lamps. He arranges them in a row on a shelf randomly, and then randomly turns 3 of them on. What is the probability that the leftmost lamp is blue and off, and the rightmost lamp is red and on?

From the problem, there is a $ \frac {1}{2}$ chance that the leftmost lamp is blue and a $ \frac {1}{2}$ chance that it is off. Now, there are 5 lamps left, three of them red, and three of them must be turned on, so the odds that the last lamp is red is $ \frac {3}{5}$, and the odds it is turned on is also $ \frac {3}{5}$ (color and turned on or off are independent). So the odds of the leftmost lamp being blue and off and the rightmost lamp being red and on is $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{5} \cdot \frac{3}{5} \equal{} \boxed{\frac {9}{100}}$.

[hide=solution]
We have $3$ red lamps and $3$ blue ones. Three are on; three are off. Since the successful outcome is when there is a blue lava lamp to the left that is off and a red lava lamp to the right that is on, the middle can be arranged in $$\binom{4}{2}^2$$ ways. This is because you choose $2$ lamps to be on and choose $2$ to be red.

Doing the same for the total outcomes results in the desired answer as $$\frac{\binom{4}{2}^2}{\binom{6}{3}^2} = \frac{9}{100}$$ ways.[/hide]

How come for the first solution (Textangle's), you don't have to consider the middle four lamps? Does the probability change if you "place" the middle lamps first as opposed to the ends first?

The probability to obtain the correct color combination is $\frac12\cdot \frac35$ because the leftmost lamp has a probability of 3/6 that it is blue and the rightmost lamp then has probability of 3/5 to be red (since one blue lamp is already used/taken).

Similarly, the probability to obtain the correct lighting combination is $\frac12 \cdot \frac 35$ because the probability the leftmost lamp is off is 3/6=1/2 and then the probability that the rightmost lamp is on is 3/5.

Thus, the answer is $\left(\frac12 \cdot \frac 35\right)\cdot\left(\frac12 \cdot \frac 35\right)=\frac{3}{10}\cdot\frac{3}{10}=\boxed{\frac{9}{100}}$.


For every integer $ n \geq 3$ find all sequences of real numbers $ (x_1, x_2, \ldots, x_n)$ such that
$ \sum_{i \equal{} 1}^{n}x_i \equal{} n$ and $ \sum_{i \equal{} 1}^{n} (x_{i \minus{} 1} \minus{} x_i \plus{} x_{i \plus{} 1})^2 \equal{} n$,
where $ x_0 \equal{} x_n$ and $ x_{n \plus{} 1} \equal{} x_n$

By Cauchy,
\begin{align*}
\left( \sum_{i=1}^n {1} \right) \left( \sum_{i = 1}^{n} (x_{i - 1} - x_i + x_{i + 1})^2 \right)
&\geq \left( \sum_{i = 1}^{n} (x_{i - 1} - x_i + x_{i + 1}) \right)^2 \\
&= \left( \sum_{i = 1}^{n} x_i \right)^2 \\
&= n^2.
\end{align*}
Thus $ \left( \sum_{i = 1}^{n} (x_{i - 1} - x_i + x_{i + 1})^2 \right) \geq n$ and we need the equality condition of Cauchy to hold. This only happens when $ x_1 = x_2 = \ldots = x_n$, and it's obvious they all need to be $ 1$ from the first equation.

[quote="MellowMelon"]Thus $ \left( \sum_{i \equal{} 1}^{n} (x_{i \minus{} 1} \minus{} x_i \plus{} x_{i \plus{} 1})^2 \right) \geq n$ and we need the equality condition of Cauchy to hold. This only happens when $ x_1 \equal{} x_2 \equal{} \ldots \equal{} x_n$, and it's obvious they all need to be $ 1$ from the first equation.[/quote]This is not the only solution. I did it (correctly ;) ) on that MO (tommorow results of that second round...), but about 70% of people who noticed usage Cauchy or QM-AM next wrote that "obviously" the only sollution is $ (1,1,\ldots,1)$. There should be probably 0 or 2 points for such solution (we have marks: 0,2,5,6), so better keep attention  :P

Hint:
We have set of equations: $ x_{i \minus{} 1} \minus{} x_i \plus{} x_{i \plus{} 1} \equal{} 1$, try to get: $ x_i \equal{} x_{i \plus{} 6}$ and consider cases: $ (n,6) \equal{} 6, \ (n,6) \equal{} 3, \ (n,6) \equal{} 2, \ (n,6) \equal{} 1$ (i passed over (n,6)=2, so I probably won't have 6 points for that problem, lucky that condition is very easy to check ;) ).

Oops, somehow I thought I was doing Cauchy on $ (1,1,\ldots ,1)$ and $ (x_1, x_2, \ldots x_n)$; wasn't trying to do any "obviously" stuff, just being stupid. :D

Yeah, I'm in the final of LX Polish MO :D :D :D (5/6 points for that problem, as I thought ;) )

So I'm gonna post the solution  :lol:

We have: $ 2 \equal{} 1 \plus{} 1 \equal{} (x_i \minus{} x_{i \plus{} 1} \plus{} x_{i \plus{} 2}) \plus{} (x_{i \plus{} 1} \minus{} x_{i \plus{} 2} \plus{} x_{i \plus{} 3}) \equal{} x_i \plus{} x_{i \plus{} 3}$, so: $ x_i \equal{} 2 \minus{} x_{i \plus{} 3} \equal{} (x_{i \plus{} 6} \plus{} x_{i \plus{} 3}) \minus{} x_{i \plus{} 3} \equal{} x_{i \plus{} 6}$ and:

1) $ (n,6) \equal{} 1$, we immediately have that the period of sequence $ x_k$ equals 1, so $ x_i \equal{} 1$

2) $ (n,6) \equal{} 2$ - the period of sequence $ x_k$ equals 2 so we have set of equations:
$ \begin{cases}a \minus{} b \plus{} a \equal{} 1 \\
b \minus{} a \plus{} b \equal{} 1 \end{cases} \Rightarrow a \equal{} b \equal{} 1$, so: $ x_i \equal{} 1$

3) $ (n,6) \equal{} 3$ - the period of sequence $ x_k$ equals 3 so we have:
$ \begin{cases}a \minus{} b \plus{} c \equal{} 1 \\
b \minus{} a \plus{} c \equal{} 1 \\
c \minus{} b \plus{} a \equal{} 1 \end{cases} \Rightarrow a \equal{} b \equal{} c \equal{} 1$, so $ x_i \equal{} 1$

4) $ (n,6) \equal{} 6$ - the period of sequence $ x_k$ equals 6, so we have:
$ \begin{cases}x_1 \minus{} x_2 \plus{} x_3 \equal{} 1 \\
x_2 \minus{} x_3 \plus{} x_4 \equal{} 1 \\
x_3 \minus{} x_4 \plus{} x_5 \equal{} 1 \\
x_4 \minus{} x_5 \plus{} x_6 \equal{} 1 \\
x_5 \minus{} x_6 \plus{} x_1 \equal{} 1 \\
x_6 \minus{} x_1 \plus{} x_2 \equal{} 1 \end{cases} \Rightarrow  \\ \Rightarrow (x_1,x_2,x_3,x_4,x_5,x_6) \equal{} (a,b,1 \minus{} a \plus{} b,2 \minus{} a,2 \minus{} b,1 \plus{} a \minus{} b) \wedge x_i \equal{} x_{i \plus{} 6} \wedge a,b \in \mathbb{R}$


quadratics
modular arithmetic
number theory unsolved
number theory

Is it possible that $ \left(\frac{n}{m}\right)\equal{}1$ while the congruence $ x^{2}\equiv n\pmod{m}$ has no solutions?  Prove your answer.

There are no specifications on $ m$ and $ n$.

That depends on whether you're using the Legendre symbol or the Jacobi symbol.

Unfortunately, it doesn't say.  If it's Legendre, I would assume the answer is no.  If it's Jacobi...

By definition, for a prime $ m$, $ \left(\frac {n}{m}\right) \equal{} 1$ implies the existence of a solution to $ x^{2}\equiv n\pmod{m}$. So it is safe to assume that $ m$ need not be a prime.

It -is- possible (which sort of reduces the power of Jacobi Symbol), and it is sufficient to find distinct prime $ p, q$ and some $ n$ such that $ m \equal{} pq$, $ \left(\frac {n}{p}\right) \equal{} \left(\frac {n}{q}\right) \equal{} \minus{} 1$ (the multiplicity of the symbol will imply $ \left(\frac {n}{m}\right) \equal{} 1$).

My first try - $ p \equal{} 5, q \equal{} 7, m \equal{} 35$. The non-quadratic residues mod $ 5$ are $ 2,3 \pmod 5$ and the non-quadratic residues mod $ 7$ are $ 3,5,6 \pmod 7$. So taking $ n \equal{} 3$ is OK - $ \left(\frac {3}{35}\right) \equal{} 1$, but if there's a solution to $ x^{2}\equiv 3 \pmod{35}$ then there's also a solution to $ x^{2}\equiv 3 \pmod{5}$ which isn't true.