Suppose not, for a contradiction.

Let A and B "share" sleeps, if for the pair of people (A,B), there exists a moment where they were both asleep.
Let B "drop" A, if there is a time where A starts sleeping, then B starts and then wakes up, then A wakes up.

Case 1:  There exists A, B so that B drops A.
Then B shares with one person during that sleep,  and on his other sleep must share with the other three people.
It can only work if C drops B, for some $C \ne A$.  Then C must share with the other 3 people on his other sleep,
etc.  so that we cannot fuffill it.  Contradiction.

Case 2:  There doesn't exist A, B so that B drops A.
Then during one sleep, A needs to share with 2 other people (only 2 sleeps, maximum 2 shares per sleep).
Consider the person A that woke up last.  He can only share with one person.  Contradiction.

[quote="liangchene"]During a certain lecture, the caterers didn't bring enough coffee, so each of five mathematicians fell asleep exactly twice. For each pair of these mathematicians, there was some moment when both were sleeping simultaneously. Prove that at some moment, some three were sleeping simultaneously.[/quote]

This is at the olympiad level, though the solution is elementary:

http://www.kalva.demon.co.uk/usa/usoln/usol862.html


calculus
integration
logarithms
search
calculus computations

Arn't Integrals part of calculus?

This should be moved to a higher level.

btw I noticed that your new and you probably didn't know, greetings to AoPS.

Define:

$ \int ^1_ 0 \frac{x\minus{}1}{\ln x} \, dx \equal{} \minus{} \int ^1_ 0 \frac{x\minus{}1}{\ln x} x^a \Big|^\infty_{a \equal{} 0} \, dx \equal{}  \minus{}\int ^1_ 0 \int ^\infty _0 (x\minus{}1) x^a \,da \,dx\equal{}$

$ \equal{}\int ^\infty_0 \int^1_0 (1\minus{}x)x^a\,da\,dx\equal{}\int ^\infty_0 \left(\frac{x^{a\plus{}1}}{a\plus{}1} \minus{} \frac{x^{a\plus{}2}}{a\plus{}2}\right)\Big|^1_{x\equal{}0}\,da\equal{}$

$ \equal{}\int ^\infty_0 \left(\frac{1}{a\plus{}1} \minus{} \frac{1}{a\plus{}2}\right)\,da\equal{} \ln\left(\frac{a\plus{}1}{a\plus{}2}\right)\Big|^\infty_{a\equal{}0} \equal{} \ln 2$

Or:

Make substitution:

$ \int^1_0\frac{x\minus{}1}{\ln x} \equal{} \Big| x \to e^{\minus{}t} \Big|  \equal{} \minus{}\int ^0 _ \infty \frac{e^{\minus{}t} \minus{} 1}{\minus{}t}e^{\minus{}t}\,dt \equal{} \int ^\infty _ 0 \frac{1\minus{}e^{\minus{}t}}{t}e^{\minus{}t}\,dt \equal{}$

Now, we have:

$ \frac{1\minus{}e^{\minus{}t}}{t} \equal{} \frac{1}{t}\left(1\minus{}\sum^\infty_{n\equal{}0} \frac{(\minus{}t)^n}{n!}\right)\equal{} \frac{1}{t}\sum^\infty_{n\equal{}1}\frac{(\minus{}t)^n}{n!} \equal{} \sum^\infty_{n\equal{}1}\frac{(\minus{}t)^{n\minus{}1}}{n!}$

So:

$ \int ^\infty _ 0 \frac{1\minus{}e^{\minus{}t}}{t}e^{\minus{}t}\,dt \equal{}\sum^\infty_{n\equal{}1}\frac{(\minus{}1)^{n\minus{}1}}{n!} \int ^\infty _ 0\frac{t^{n\minus{}1}}{e^t}\,dt \equal{} \sum^\infty_{n\equal{}1}\frac{(\minus{}1)^{n\minus{}1}}{n!} (n\minus{}1)! \equal{}$

$ \equal{} \sum^\infty_{n\equal{}1}\frac{(\minus{}1)^{n\minus{}1}}{n} \equal{} \ln 2$

since

$ \sum^\infty_{n\equal{}1}\frac{x^n}{n} \equal{} \minus{}\ln(1\minus{}x)$

Or (the simplest):

Define: $ F_x(a) \equal{} \int^1_0 \frac{x^a\minus{}1}{\ln x} \,dx$

Then:

$ \frac{d F_x}{d a} \equal{} \int ^1_0 x ^a \, dx \equal{} \frac{1}{a\plus{}1}$

So:

$ F_x(a) \equal{} \int \frac{da}{a\plus{}1} \equal{} \ln (a\plus{}1)$

and we have:

$ F_x(1) \equal{} \int^1_0 \frac{x\minus{}1}{\ln x} \,dx \equal{} \ln 2$

Hey [b]millin[/b],

You forgot to enter [url=http://www.mathlinks.ro/viewtopic.php?search_id=1380427866&t=236045]this[/url] approach, man...

Haha.. I must have missed it.. But a reference is as good, I suppose ! :)


An endless chord consists of two portions of lengths $ 2l$ and $ 2l'$ respectively, knotted together. The mass per unit length of each string is $ m$ and $ m'$. It is placed in stable equilibrium over a smooth peg and slightly displaced. Find time period of oscillation.

Answer:
[hide]$ T \equal{} 2\pi\sqrt{\frac{ml \plus{} m'l'}{(m' \minus{} m)g}}$[/hide]

How to do it?

Solution

[hide]

Assumption: negligible rotational inertia of pulley, and small sized pulley.  Would the period change if the size of the pulley differed?  Assumption, m'>m

Call $ M_1$ the mass on one side, $ M_2$ the mass on the other side. $ x$ is small displacement
$ M_1 \equal{} m(l \minus{} x) \plus{} m'(l' \plus{} x) ; M_2 \equal{} m (l \plus{} x) \plus{} m' (l' \minus{} x)$
$ M_1 \minus{} M_2 \equal{} 2x (m' \minus{} m)$

Equation of motion:

$ \vec F \equal{} 2 (ml \plus{} m'l')\ddot x \equal{} \minus{} 2x (m' \minus{} m)g$
Rearrange.
$ \ddot x \plus{} \frac {(m' \minus{} m)g}{(ml \plus{} m'l')} x \equal{} 0$
$ \omega \equal{} \sqrt \frac {(m' \minus{} m)g}{(ml \plus{} m'l')}$
\[ T \equal{} 2 \pi \sqrt \frac {(ml \plus{} m'l')}{g(m' \minus{} m)}
\]
[/hide]


search
number theory
relatively prime
prime numbers
number theory proposed

Find all positive integers $ n$ with the following property: whenever $ a$ and $ b$ are integers such that $ ab\plus{}1$ is multiple of $ n$, then $ a\plus{}b$ is also multiple of $ n$.

Obviosly $ n\equal{}1$ solution. Let $ n>1$.
Obviosly $ n|ab\plus{}1\to (b,n)\equal{}1$ and for any $ (b,n)\equal{}1$ exist a, suth that $ n|ab\plus{}1$.
But $ n|ab\plus{}1\to n|a\plus{}b$ equavalent to $ n|ab\plus{}b^2$ or $ n|b^2\minus{}1$ for any $ (b,n)\equal{}1$. Therefore for any $ p|n$ we get or $ p\equal{}2$ or $ p|b\minus{}1$ or $ p|b\plus{}1$ for $ p\not |b$. It give $ n\equal{}2^k3^m$ and $ n|5^2\minus{}1\equal{}24\to k\le 3,m\le 1$.

Assume both are true to get information on n:
$ ab \equiv \minus{}1 \mod n$ implies that (a,n)=(b,n)=1, and for all a such that (a,n)=1 there exists b such that this is true (and vice versa).
We also have that $ a \equiv \minus{}b \mod n$ meaning that $ a^{2} \equiv 1 \mod n$.
So, since a can be anything relatively prime to n, the only n for which this works is n such that all relatively prime numbers squared are equal to $ 1 \mod n$.
$ x^{2} \equiv 1 \mod n$
$ (x\plus{}1)(x\minus{}1) \equiv 0 \mod n$
Bashing small cases, I get n=1, 2, 3, 4, 6, 8, and 12, their prime factors are limited to 2 and 3, but I can't find a way to prove that they are the only ones.

Solutions = divisors of 24.

But search PEN please.

A solution can be found here: http://www.mathlinks.ro/viewtopic.php?p=121227#121227


When I was typing up a LaTeX document for an AoPS class, I really liked the formatting that \begin{equation}----\end{equation} gave me but the numbers on the side were really unnecessary for my purposes.  Is there a way you can get rid of them?

Use [code]\[...\][/code] or \begin{displaymath}...\end{displaymath}.
Most people use the first for obvious reasons :)

Ooo...that's convenient.  Thanks so much  :thumbup:


logarithms
algebra
function
domain
calculus
calculus computations

Solve the DE, $ \frac{dy}{dx}\equal{}e^{y\minus{}x}$

$ e^{\minus{}y}dy \equal{} e^{\minus{}x}dx \to e^{\minus{}y}\equal{}C\plus{}e^{\minus{}x}$

$ y\equal{}x\minus{}\ln(Ce^x\minus{}1)$

Shouldn't the $ \minus{}1$ in the $ ln()$ be $ \plus{}1$?

$ e^{\minus{}y}dy \equal{} e^{\minus{}x}dx$

$ <\equal{}> \minus{}e^{\minus{}y} \equal{} \minus{}e^{\minus{}x} \plus{} C$

$ <\equal{}> e^{\minus{}y} \equal{} e^{\minus{}x} \minus{} C \equal{} e^{\minus{}x} \plus{} C'$

$ <\equal{}> \minus{}y \equal{} \ln(e^{\minus{}x} \plus{} C') \equal{} \ln\left(\frac{1 \plus{} C'e^x}{e^x}\right) \equal{} \ln(C'e^x \plus{} 1) \minus{} x$

$ <\equal{}> y \equal{} x \minus{} \ln(C'e^x \plus{} 1)$.

oops typo, ty for fix

ps i love the iff arrows you ghetto rigged.

hello, i have
$ y(x)\equal{}x\minus{}\ln(1\minus{}Ce^x)$ under the condition $ 1>Ce^x$
Sonnhard.

Just clarifying that the "condition" $ 1 > C e^x \iff x < \minus{} \log C$ is a statement of the domain of $ y(x)$, and not a condition on $ C$.


Refer to http://www.archive.org/details/solutionsofexamp00hallrich for a full solution book.

[b]Solution[/b]: Utilize the following trigonometric identity:

sec^4 A  = 1 + 2tan^2 A + tan^4 A

which leads us to consider the equation whose roots are tan (pi/7)r, where r  = 1, 2 and 3.

In order to find this equation, first we will look for the equation whose roots are

tan (pi/7)r, where r  = 1, 2, 3, 4, 5 and 6

We will reason this way: If tan 7A = 0. Then, 7A =(pi)r, where A = (pi/7)r

Using De Moivre, we have: 7tanA - 35tan^3 A + 21tan^5 A – tan^7 A = 0

So then, factoring tan A, we have: 7 – 35 tan^2 A + 21 tan^4 A – tan^6 A = 0, let x = tan A, then rearrange.

We have, x^6 – 21x^4 + 35x^2 - 7 = 0, where tan (pi/7)r, where r = 1, 2, 3, 4, 5, & 6.

Let x^2 = y, thus y^3 – 21y^2 + 35y – 7 = 0.

Then the 3 roots are tan^2 (pi/7)r, where r = 1, 2 & 3.

We are NOW ready to calculate the value of sec^4 pi/7 + sec^4 2pi/7 + sec^4 3pi/7 by using the above identity

                             sec^4 A  = 1 + 2tan^2 A + tan^4 A

Therefore, sec^4  pi/7 + sec^4 2pi/7 + sec^43pi/7

= 3 + 2[tan^2 pi/7 + tan^2 2pi/7 + tan^2 3pi/7] + [tan^4 pi/7 + tan^4 2pi/7 + tan^4 3pi/7]

= 3 + 2(21) + [21^2 – 2(35)]

= 3 + 42 + 371

= 416


How do you find out what you got on the AIME?
I forgot all of my answers.

A score report was sent to your school.  Ask your teacher who was in charge of the contest.

If you live in California, I received a listing of all AIME scores (well, all scores of at least 2) in the state.


Determine the vector $ \vec {v}$, parallel to the vector $ \vec {u} \equal{}
(1, \minus{}1,2)$ such that $ \vec {u} \,. \, \Vec {v} \equal{}\minus{} 18$

[quote="brunA"]Determine the vector $ \vec {v}$, parallel to the vector $ \vec {u} \equal{} (1, \minus{} 1,2)$ such that $ \vec {u} \,. \, \Vec {v} \equal{} \minus{} 18$[/quote]
[hide]Denote the vector that is parallel $ <a,\minus{}a,2a>$. Since the dot product is -18 we have the following equation: $ a\plus{}a\plus{}4a\equal{}\minus{}18$. Solving we get that $ a\equal{}\minus{}3$, so the vector $ \vec {v}$ that is parallel so $ \vec {u}$ is $ (\minus{}3,3,\minus{}6)$[/hide]


trigonometry
inequalities unsolved
inequalities

[quote="can_hang2007"]Dear friend, take a look here: http://www.mathlinks.ro/viewtopic.php?t=16351[/quote]
can you post full solution?

I don't see any solution in the links.....
Can you post your solution... :maybe:


logarithms
inequalities
trigonometry
function
calculus
calculus computations

Discuss the convergence of the series with nth term
a 2x4x.....(2n)/5X7x.....(2n+3)
b n^n/(n+1)^(n+1)

given summation an is convergent,either prove summation bn converges or give a
counter example, when we define bn by
c, an*sin(n)
d, sqrt(an)/n ,(an>0)
e, an* n^(1/n)
f   an/(1+abs(an))

$\sum_{n=1}^{\infty}\frac{2\cdot4\cdots(2n)}{5\cdot7\cdots(2n+3)}$ converges by comparison to $\sum\frac1{n^p}$ with $p$ near $\frac32.$

$\sum_{n=1}^{\infty}\frac{n^n}{(n+1)^{n+1}}$ diverges by limit comparison to the harmonic series.

In the first case, let $a_n=\frac{2\cdot4\cdots(2n)}{5\cdot7\cdots(2n+3)}.$

Then $\ln(a_n)=\sum_{k=1}^n\ln\left(1-\frac3{2n+3}\right).$ This can be reasonably approximated by

$-\sum_{k=1}^n\frac3{2n+3},$ which can in turn be reasonbably approximated by $-\frac32\ln n.$

We really should pay more attention to how good those approximations are, but the essence of the result is there.

In the second case, $\frac{n^n}{(n+1)^{n+1}}=\left[\left(\frac n{n+1}\right)^n\right] \left(\frac1{n+1}\right)$ and as $n\to\infty,\ \left(\frac n{n+1}\right)^n\to\frac1e.$

As for parts (c) through (f): in all of those cases, if $\sum a_n$ converges absolutely, then $\sum b_n$ would converge absolutely. In some cases, conditional convergence could make things more delicate.

To address one case in particular, (d): there it is specified that $a_n>0$ so we do have absolute convergence. By Cauchy-Schwarz (in this case, call it simply Cauchy's Inequality),

$\sum_n\frac{\sqrt{a_n}}n\le\left(\sum_n a_n\right)^{\frac12} \left(\sum_n\frac1{n^2}\right)^{\frac12},$ so it converges.

Here's a counterexample for (f):

Let $a_{3k-2}=a_{3k-1}=\frac{-1}{2\sqrt{k}}$ and $a_{3k}=\frac1{\sqrt{k}}$

Then every third partial sum of the series $\sum_{n=1}^{\infty}a_n$ is zero, and since the terms go to zero, that means the series converges (to zero).

Now let's look at $b_n=\frac{a_n}{1+|a_n|}.$

$b_{3k-2}=b_{3k-1}=\frac{-\frac1{2\sqrt{k}}}{1+\frac1{2\sqrt{k}}}= \frac{-1}{2\sqrt{k}+1}$

$b_{3k}=\frac{\frac1{\sqrt{k}}}{1+\frac1{\sqrt{k}}}=\frac1{\sqrt{k}+1}$

$b_{3k-2}+b_{3k-1}+b_{3k}=\frac1{\sqrt{k}+1}-\frac2{2\sqrt{k}+1}$

$=\frac{(2\sqrt{k}+1)-(2\sqrt{k}+2)}{(2\sqrt{k}+1)(\sqrt{k}+1)} = \frac{-1}{(2\sqrt{k}+1)(\sqrt{k}+1)}$

That is comparable to $\frac{-1}{2k}$ and the series $\sum_nb_n$ diverges to $-\infty$ by comparison to the harmonic series.

It would be worth pursuing examples based on the same idea for (e).

For (c): let $a_n=\frac{\sin n}n$

It can be shown that $\sum_{n=1}^{\infty}\frac{\sin n}n$ converges, but

$\sum_{n=1}^{\infty}\frac{\sin n}n\cdot\sin n=\sum_{n=1}^{\infty}\frac{\sin^2n}n$

is a positive term series that diverges by comparison to the harmonic series.

The general results here, applicable to parts e and f:

Let $f$ be a function from $\mathbb{R}$ to $\mathbb{R}$.

The series $\sum f(a_n)$ converges for all convergent series $\sum a_n$ if and only if there exist $c,\epsilon$ such that $f(x)=cx$ for all $|x|<\epsilon$.

The series $\sum f(a_n)$ converges absolutely for all absolutely convergent series $\sum a_n$ if and only if there exist $c,\epsilon$ such that $|f(x)|\le c|x|$ for all $|x|<\epsilon$.


geometry
trapezoid
trigonometry
ratio
trig identities
Law of Cosines

In trapezoid ABCD, AB || CD, AB = 4, and CD = 10.  Suppose that lines AC and BD intersect at right triangles, and that lines BC and DA, when extended to point Q form an angle of 45 degrees.  Compute the area of ABCD.

My solution is pretty long.

[hide]Let $ AC$ and $ BD$ intersect at $ P$. Let $ M$ be the midpoint of $ CD$, $ N$ the midpoint of $ AB$. Since $ \triangle QAB \sim \triangle QDC$, $ \frac {QA}{AD} \equal{} \frac {QB}{BC}$. Thus, $ \frac {QA}{AD} \cdot \frac {DM}{MC} \cdot \frac {CB}{BQ} \equal{} 1$, so $ AC$, $ BD$, and $ QM$ concur by Ceva. Since $ \triangle QAB\sim \triangle QDC$, $ QM$ also passes through $ N$, and $ \frac {QA}{QD} \equal{} \frac {2}{5}$, or $ \frac {AD}{AQ} \equal{} \frac {3}{2}$. Since $ PAB$ and $ PDC$ are right triangles, $ PN \equal{} 2$ and $ PM \equal{} 5$. By Menelaus, $ \frac {CM}{CD} \cdot \frac {DA}{AQ} \cdot \frac {QP}{PM} \equal{} 1$. From here, we find $ QP \equal{} \frac {20}{3}$, or $ QM \equal{} \frac {35}{3}$. Let $ QD \equal{} x$, $ QC \equal{} y$. By Stewart's theorem on $ \triangle QDC, x^2 \plus{} y^2 \equal{} \frac {2900}{9}$. By the law of cosines on $ \triangle QDC$, $ 100 \equal{} x^2 \plus{} y^2 \minus{} \sqrt {2}xy$. Combining them, $ \frac {\sqrt {2}}{4}xy \equal{} \frac {500}{9}$, and this is the area of $ \triangle QDC$ by the formula $ \frac {1}{2}xysinDQC$. The area of $ \triangle QAB$ will be $ \frac {4}{25}$ of this, so subtracting it away, the final answer is $ \frac {140}{3}$.[/hide]

[hide=" click to reveal hidden content"]

I get $ \frac{140}{3}$:

Let the point of intersection of the diagonals be $ E$. Triangles $ ABE$ and $ CDE$ are similar and their sides are in the ratio $ 4: 10\equal{}2: 5$, so let $ AE\equal{}2a, EB\equal{}2b, DE\equal{}5b, EC\equal{}5a$. Then the area of the trapezium is $ \frac{1}{2}AC\times BD\equal{}\frac{49ab}{2}$. Also Pythagoras' theorem in triangle $ ABE$ gives $ 4(a^{2}\plus{}b^{2})\equal{}16 \implies a^{2}\plus{}b^{2}\equal{}4$.

From the angle sums of triangles $ DQC$ and $ DEC$, we see that angles $ QDB$ and $ QCA$ sum to $ 45^{\circ}$. Call these angles $ \alpha$ and $ \beta$ respectively. Since they sum to $ 45^{\circ}$, we can use the formula $ \tan(\alpha \plus{} \beta) \equal{} \frac{\tan \alpha \plus{} \tan \beta}{1 \minus{} \tan \alpha \cdot \tan \beta}$ to get $ 1 \equal{} \frac{\frac{2a}{5b}\plus{}\frac{2b}{5a}}{1\minus{}\frac{2a}{5b}\cdot \frac{2b}{5a}} \equal{} \frac{10(a^{2}\plus{}b^{2})}{21ab}\equal{} \frac{40}{21ab}$.

Therefore $ ab\equal{} \frac{40}{21}$ and the area of the trapezium $ ABCD$ is $ \frac{49ab}{2} \equal{} \frac{49\times 40}{2 \times 21} \equal{} \frac{140}{3}$

[/hide]

Thanks dbgreen801. It took me quite some time. (As we had different answers, I presumed mine was probably wrong somewhere.) Now I see they're the same. I didn't think of using Ceva'a theorem although I like it. Happy New Year.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

[color=darkblue]Show that there don't exist invertible matrices $ B \in M_{2n}(\mathbb{R})$ such that: $ B\plus{}B^{\minus{}1}\minus{}I_{2n}\equal{}O_{2n}$[/color]

Isn't this a counterexample: $ M \equal{} \frac {1}{4}\left( \begin{array}{cc}2 & \minus{} 3 \\
4 & 2\end{array}\right)$?

EDIT: Typo of coefficients.

[color=darkred]It seems to be correct, maybe there is something incorrect in the problem itself, but this is the form it appears in my book. [/color]

Even simpler: $ M\equal{}\begin{bmatrix}0&1\\\minus{}1&1\end{bmatrix}.$

But all solutions are similar to each other over the reals, aren't they? ;) Even for arbitrary $ n>0$.

Perhaps the intended problem was about matrices of odd dimension? Then, non-real eigenvalues come in pairs.

An example of solutions which aren't similar over $ \mathbb{R}$:
Let $ A\equal{}\frac12\begin{bmatrix}1&\sqrt{3}\\\minus{}\sqrt{3}&1\end{bmatrix}$ and $ C\equal{}\begin{bmatrix}0&1\\1&0\end{bmatrix}$. Both $ B\equal{}\begin{bmatrix}A&0\\0&A\end{bmatrix}$ and $ B'\equal{}\begin{bmatrix}A&C\\0&A\end{bmatrix}$ are $ 4\times 4$ solutions.

[quote="jmerry"]Perhaps the intended problem was about matrices of odd dimension? Then, non-real eigenvalues come in pairs.[/quote]This is not possible since the minimal polynomial of $ M$ is $ X^2 \minus{} X \plus{} 1$ which is irreducible over the reals, hence the dimension must be even.
[quote]An example of solutions which aren't similar over $ \mathbb{R}$:
Let $ A \equal{} \frac12\begin{bmatrix}1 & \sqrt {3} \\
\minus{} \sqrt {3} & 1\end{bmatrix}$ and $ C \equal{} \begin{bmatrix}0 & 1 \\
1 & 0\end{bmatrix}$. Both $ B \equal{} \begin{bmatrix}A & 0 \\
0 & A\end{bmatrix}$ and $ B' \equal{} \begin{bmatrix}A & C \\
0 & A\end{bmatrix}$ are $ 4\times 4$ solutions.[/quote]I must admit that I have not gone through the examples yet in detail, but here just my thoughts on similarity. If the reasoning is correct, the examples given should be similar:
Since $ M$ has the minimal polynomial as given above, the matrix $ M$ is similar (over the reals) to a matrix with real 2x2-matrices on the diagonal satisfying the minimal polynomial and zeros elsewhere. [b]If[/b] these 2x2-matrices are similar to each other, then so should be every pair of these matrices of dimension $ 2n$, shouldn't they?

EDIT: And the proof for the [b]If[/b]: In dimension 2 every $ M$ is similar to its companion matrix, hence all those 2x2-matrices are similar.

So, what is wrong here? ;)

You're right that they should be similar- I was thinking of characteristic polynomials.
If that were $ I$ rather than $ C$ up there, they wouldn't be similar, but $ 2\times 2$ block form has a little more flexibility than Jordan form above the diagonal. They're $ 4\times 4$ matrices, by the way- we can't even think of something like this in two dimensions.

So, there is just one single similarity class in every dimension $ 2n>0$. Please see my EDIT above.

More generally, let $ p$ be an irreducible polynomial of degree $ r$ over a field $ K$. Then $ p$ has only simple roots over $ \bar {K}$, the algebraic closure of $ K$, even if $ charac(K)$ is finite.
Let $ E\equal{}\{A\in\mathcal{M}_n(K); p(A)\equal{}0\}$. $ A$ is necessarily diagonalizable over $ \bar {K}$  and necessarily the characteristic polynomial of $ A$ is a power of $ p$. Thus necessarily $ r$ is a divisor of $ n$ or $ n\equal{}r.s$. Finally $ E$ is a single similarity class, that is all solutions are similar to $ diag(C_1,\cdots,C_r)$ where $ C_i$ is the companion matrix of $ p$.


sper ca imi amintesc bine cum se scrie in latex:D

fie sirurile a_n, b_n n>=1.
$ a_{n}$ apartine $ (0;\frac{1}{2})$ oricare ar fi n mai mare sau egal cu 1.
$ b_{3k\plus{}1}\equal{}a_{3k\plus{}2}$
$ b_{3k\plus{}2}\equal{}a_{3k\plus{}3}$
$ b_{3k\plus{}3}\equal{}a_{3k\plus{}1}$, oricare ar fi k natural

definim sirul x_n astfel:

$ x_{n}\equal{}a_{n}(\frac{1}{2}\minus{}b_{n})$ oricare ar fi $ n$ natural. sa se calculeze $ lim_{n\minus{}>\infty}$
produs de la k=1,3n din $ x_{k}$

astept solutiile voastre:P

de unde ai scos ultima inegalitate? aia cu $ <\frac{1}{n^{6n}}$?

Am editat. Era 2 in loc de n. Inegalitatea reiese din $ a_{n}\in (0,1/2)$ si $ (1/2\minus{}a_{n})\in (0,1/2)$,$ \forall n\in N$

Solutiile de la etapa locala din Galati 2008 sunt pe site-ul http://www.mategl.com la lick-ul "Download"


Let $a,b,c$ be positive real numbers such that $ab+bc+ca+abc=4$. Prove that
$3\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)^{2}\ge (a+2)(b+2)(c+2)$

P/s: Why I can't post in Solved Problems box?

some idea: $ab+bc+ca+abc=4 \ let \ a=2x,b=2y ,c=2z\ then\ the\ condition\ becomes:$
\[\(\sqrt{xy})^{2}+(\sqrt{yz})^{2}+(\sqrt{zx})^{2}+2\sqrt{xy}\sqrt{yz}\sqrt{zx}=1 \]

\[\iff \ there\ exists\ a \ triangle\ such\ that\ cosA=\sqrt{yz}\ cosB=\sqrt{zx}\ cosC=\sqrt{xy}\implies a=2(cosB.cosC)/cosA . \ similarly \ for\ b\ and\ c \]
substituting, the given inequality turns to proving:
\[3(cosA+cosB+cosC)^{2}\ge 16\prod(cosB.cosA+cosC) \]
now forget the above varibles so that we can use them again.
\[if\ p\ q\ r \ be\ the\ sides\ of\ this\ triangle. \]
then using the following identity:
\[\prod(cosA.cosB+cosC)=\frac{(p+q-r)^{3}(-p+q+r)^{3}(p-q+r)^{3}(p+q+r)^{3}}{4^{3}.p^{4}.q^{4}.r^{4}}. \]
and
\[cosA+cosB+cosC=\frac{(p+q-r)(p+r-q)(r+q-p)+2pqr)}{4pqr}: \]
now substitute:
\[p=y+z;\ q=z+x;\ r=x+y;\ \]
so that it becomes:
\[3((x+y)(y+z)(z+x))^{2}(8xyz+2(x+y)(y+z)(z+x))^{2}\ge8^{4}x^{3}y^{3}z^{3}(x+y+z)^{3}\]
$\iff$
\[3*64x^{2}y^{2}z^{2}((x+y)(y+z)(z+x))^{2}+12((x+z)(x+y)(y+z))^{4}+32*3xyz((x+z)(x+y)(y+z))^{3}\]

\[\ge8^{4}x^{3}.y^{3}.z^{3}(x+y+z)^{3}\]
after  this   the  thing  that  wilL give   solution   is expanding. but i  belive  that much labour can be reduced by some sharp AM  GM s like
\[2 x^{6}y^{3}+x^{3}y^{3}z^{3}\ge 3x^{5}y^{3}c. \]
anuj


Is this just me who's confusing or is this whole forum a new one?

There was a suggestion to Mr.Ruscyzk that we should have a forum for Canadian contests discussion as they are being overcrowded in Other US Contests....


MATHCOUNTS
number theory
relatively prime
prime factorization

Heh, I tried a little investigation but my results were not helpful.  I just substituted numbers that have that many divisors for each one but I did not see a pattern. Oh well, waiting for the solution.

[color=cyan]Ripper, try more examples.  Make sure that they follow the givens of the question, though, particularly the very first bit.  By the time you've tried about ten or so cases, I think a pattern should emerge.  [/color]

not too hard

Solution:

[hide]

a) 8

b) 112

c) 5,000,000

I determined the pattern from example a.  Look at the following numbers:

3*8 = 24  factors: 1,2,3,4,6,8,12,24  number of factors: 8

5*8 = 40  factors: 1,2,4,5,8,10,20,40 number of factors: 8

7*8=56 factors: 1,2,4,7,8,14,28,56 number of factors: 8

5*6=30 factors: 1,2,3,5,6,10,15,30 number of factors: 8

The number of factors of m*n is obviously the product of the number of factors of m and n.

[/hide]

I hope I did well for my first post here.

Answers look fine to me.. but.....

(highlight the below text)

[hide]

To make it a proper 'proof', you should really explain why the number of factors *obviously* is the product of the number of factors of m and n, not just by looking at 4 examples.

[/hide]

hmmmmmmmm...i used a different way. same answers though.

Solution:

[hide]1)8

2)112

3)5,000,000

take for example the prime factorization of a number. take 2^a. the number of factors it has is a+1. or if the prime factorization is 2^a x 3^b

then the number of factors is (a+1)(b+1).

a)a+1=2, so a=1, thus it is prime as stated. b+1=4, so the exponent is 3. multiplying both numbers, m^1 x n^3. so (1+1)(3+1)=8

b)same thing as above. except with a=7 and b=13

c)once again, same concept, but with a=999 and b=499

Quote:For each of the following problems let m and n be two relatively prime natural numbers.

(a) If m has 2 positive divisors (m is prime) and n has 4 positive divisors, how many positive divisors does the product mn have?

(b) If m has 8 positive divisors and n has 14 positive divisors, how many positive divisors does the product mn have?

(c) If m has 1000 positive divisors and n has 5000 positive divisors, how many positive divisors does the product mn have?[/hide]

Quote:3*8 = 24 factors: 1,2,3,4,6,8,12,24 number of factors: 8
5*8 = 24 factors: 1,2,4,5,8,10,20,40 number of factors: 8
7*8=56 factors: 1,2,4,7,8,14,28,56 number of factors: 8
5*6=30 factors: 1,2,3,5,6,10,15,30 number of factors: 8

i think you meant 5*8=40. you may want to edit that.

hope i helped. happy solving

it would have 4, but if m and n are the same number, then it's 3.

m^1 x n^1 has 4 factors cause 2:^2:=4

but if the numbers were m^1 x m^1=m^2, then the number of factors would be 2+1=3

unless you were meaning to imply that they were relatively prime again

General solution:

Say m and n are relatively prime.

Then in their prime factorizations,

m = p1^e1 * p2^e2 * ... * pk^ek
n = q1^f1 * q2^f2 * ... * qp^fp

Where pi doesn't equal qj for all applicable i,j.  This ensures that none of the exponents add in the product, so that the number of factors of the product is:

(e1+1)(e2+1)...(ek+1)*(f1+1)(f2+1)..(fp+1)

Which is the product of the number of factors of the original numbers.

"Nice work you guys. What if m and n had exactly 2 common divisors (including 1)?"

^ how to go about solving the extension  :?:  :?:


One cube has a volume of 343 cubic centimeters. The edge length of a second cube is three times the edge length of this first cube. What is the positive difference between the volume of the second cube and the volume of the first cube, in cubic centimeters?

so length of first cube is $ \sqrt[3]{343}\equal{}7$
so the length of teh 2nd cube is $ 21$
so we do $ 21^3\minus{}7^3\equal{}8918$


calculus
integration
function
algebra
polynomial
limit
real analysis

If $ f$ is continuous on [0,1] and if $ \int_0^1{f(x)x^n}dx\equal{}0, (n\equal{}0,1,2\dots)$,
prove that $ f(x)\equal{}0$.

I have a solution, but there is one step that is confusing me.

By Weierstrass's theorem, there is a sequence of polynomials $ \{P_n\}$ such that $ \{P_n\}\to f$ uniformly.

So we can easily show using the hypothesis that $ \int_0^1{f(x)P_n(x)}dx\equal{}0.$

Now we take the limit $ \lim_{n\to\infty}{\int_0^1{f(x)P_n(x)}dx}\equal{}\int_0^1{f(x)\lim_{n\to\infty}P_n(x)}dx\equal{}\int_0^1{f(x)^2}dx\equal{}0$ and it follows.

I don't understand why we can do:
$ \lim_{n\to\infty}{\int_0^1{f(x)P_n(x)}dx}\equal{}\int_0^1{f(x)\lim_{n\to\infty}P_n(x)}dx$

I know the limit of a sequence of functions and integration can be reversed if the sequence is uniformly convergent, but how do we know $ f(x)P_n(x)$ is uniformly convergent? Or does it not matter?

Any insight would be great. Thanks

$ [0,1]$ is compact, hence $ f$ is bounded.  This implies that $ f(x)P_n(x)$ also converges uniformly, and you can pull the limit inside the integral.


According to the following table, at what pH would the indicators bromthymol blue, thymol blue, and methyl orange all appear as yellow?

$ \textbf{(1)} \ 1.9 \qquad \textbf{(2)} \ 2.9 \qquad \textbf{(3)} \ 4.7 \ \qquad \textbf{(4)} \ 9.8$

I believe it is 4.7.  At this pH, Bromthymol blue will retain its yellow color, as will thymol blue.  Methyl orange will have already passed its range so it will be yellow.  I haven't taken chem in 2 years though so take my answer with a grain of salt.

Yes, the correct answer is 4.7: methyl orange is yellow above 4.4 and bromothymol blue is yellow bellow 6, so the required pH must lie in the range [4.4, 6].

This appears to be a question from a NYS Regents.  I seem to recall taking a final exam in AP Chem, and this question showed up.


For integers $ a$ and $ b$, what is the smallest possible positive value of $ 8a\plus{}12b$?

$ \gcd (8, 12) \equal{} \boxed{4}$.
Because $ \minus{}8\plus{}12\equal{}4$.


trigonometry
function
algebra unsolved
algebra

solve,
$sin(sin(sin(\theta)))=cos(cos(cos(\theta)))$ for $0\le\theta\le2\pi$

My work so far:

$\frac{\sin (\sin (\sin (\theta)))}{\cos (\cos (\cos (\theta)))} = 1 = \frac{\sin \frac{\pi}{4}-x}{\cos \frac{\pi}{4}+x}$

where x is a number to be determined
$\sin (\sin (\theta)) = \frac{\pi}{4}-x$

$\cos (\cos (\theta)) = \frac{\pi}{4}+x$

We must determine x such that the above equations have a solution.

Equivalently,

$sin(sin(\theta))=\frac{\pi}{2}-cos(cos(\theta))$

This can be obtained by using the identity $cos(x)=sin(\frac{\pi}{2}-x)$.

(By the way, this leaves out no real solutions.)

This could be simplified (?) further by using the identity $cos(x)=\pm\sqrt{1-sin^2(x)}$

Let $u=sin(x)$.

Then the equation becomes $sin(u)=\frac{\pi}{2}-cos\left(\pm\sqrt{1-u^2}\right)$

This smacks of being transcendental.  An approximate solution is 0.9602464211106627, but that doesn't look like a particularly nice number; again, it's probably transcendental.

Ya, its almost certainly transcendal (these are trig functions), but that of itself isn't so bad; the solution to $cos(x)=0$ is trandscendal.

However, it might also be a number without a closed form using the typical constants (such as $\pi$).

I confirm [u]kokosai[/u]'s solution to the listed places and add the only other other solution on $[0,2\pi]$

x=2.181346232479131


Prove or disprove: $(a^2+2ax+x^2)^n>4a+x$, where  $a>1, a\in \mathbb{R}$,  $n\ge 2, n\in \mathbb{N}$ and $x>0, x\in \mathbb{R}$.

This seems to be wrong: Choose $x : = \epsilon >0$ and $a : = 1 + \epsilon$,
then for every natural $n$ the left hand side tends to $1$ for $\epsilon \to 0$
and the right hand side tends to $4$ for $\epsilon \to 0$.
So for $\epsilon$ small enough this gives a counterexample.


1.If  $ a,b,c$ are positive integers such that $ a$ divides$ b^2$,$ b$ divides $ c^2$ and $ c$ divides $ a^2$.

   Prove that $ abc$ divides $ (a\plus{}b\plus{}c)^7$.

2.If $ n$ is any natural number greater than $ 1$,then prove that

   $ \frac{1}{2}<\frac{1}{1\plus{}n^2}\plus{}\frac{2}{n^2\plus{}2}\plus{}\frac{3}{n^2\plus{}3}\plus{}.....\plus{}\frac{n}{n^2\plus{}1}<\frac{1}{2}\plus{}\frac{1}{2n}$

[quote="pankajsinha"]1.If  $ a,b,c$ are positive integers such that $ a$ divides$ b^2$,$ b$ divides $ c^2$ and $ c$ divides $ a^2$.

   Prove that $ abc$ divides $ (a \plus{} b \plus{} c)^7$.

for problem1 , $ (a \plus{} b \plus{} c)^7$ can be expanded by maple,first term is $ 7bc^{6}$.let $ b^{2}\equal{}k_{1}a,c^{2}\equal{}k_{2}b,a^{2}\equal{}k_{3}c$.by $ b^{2}\equal{}k_{1}a$,we have $ k_{2}^{2}b^{2}\equal{}k_{2}^{2}k_{1}a$.then,by $ c^{2}\equal{}k_{2}b$,we have $ k_{2}^{2}k_{1}a\equal{}c^{4}$,so $ abc$ divides the first term,other terms can be done in the same way.


Euler
combinatorics unsolved
combinatorics

Let G be a non-trival and eulerian, and let v be any vertex of G. Show that (i) and (ii) are equivalent:

(i) Every trail with initial vertex v can be extended to an euler circuit of G;
(ii) G - v is a forest


geometry
3D geometry
sphere
real analysis
real analysis unsolved

Some comments:

(i) I think that's called a regular simplex.

(ii) The volume follows by definition.

(iii) The hyper-volume of the sphere has been computed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=26191]here[/url] by Kent Merryfield.


quadratics
limit
abstract algebra
algebra
polynomial
analytic geometry
algebra proposed

Post your solutions to this nice problem.
Let $a,b,c,A,B,C$ real numbers
If   $\left| ax^2+bx+c\right| \leq \left| Ax^2+Bx+C\right|$ for all $x\in R$  then prove

that $\left|b^2-4ac \right| \leq \left|B^2-4AC \right|$

Let $P(x) = Ax^2+Bx+C$ and $Q(x)=ax^2 + bx +c$

If $P$ have roots, then they are also roots of $Q$ because:
$\mid P(x_1)\mid =0 \ge \mid Q(x_1) \mid$ then $Q(x_1) =0$
$\mid P(x_2)\mid =0 \ge \mid Q(x_2) \mid$ then $Q(x_2) =0$

Then Proove $A\ge a \ge 0$ :
1° $\displaystyle\lim_{x\to\infty}{\mid P(x)\mid} \ge \displaystyle\lim_{x\to\infty}{\mid Q(x)\mid}$ so $\mid A\mid\ge \mid a\mid$
2° because of the module we can multiple the polynomials by $-1$ if necessary to have the first coefficient positiv so $A\ge a \ge 0$

Because $A$ and $a$ are $\ge 0$ and $P$,$Q$ have the same roots then both have a minimum with the same coordonates $(x,y) = (\frac{-b}{a},\frac{- \Delta_a}{4a})$ but the minimum of $P$ (in the module so we can remove the $-$) is bigger than $Q$

so $\frac{\Delta_A}{4A} \ge \frac{\Delta_a}{4a}$ but $4A \ge 4a$ then $\Delta_A \ge \Delta_a$ and that's what we have to prove.

[quote="Agrippina"]I don't believe this is a complete solution.  Indeed, if $P$ has a real root $\alpha$ then as $|Q(\alpha)|\leq|P(\alpha)|=0$ $\alpha$ must be a root of $Q$ also; however, this doesn't necessarily hold for any complex roots of $P$, so in this case it need not be true that $-\frac{b}{a}=-\frac{B}{A}$.[/quote]

If $P$ has a real roots this root must be its minimum and also the minimum of $Q$. So in this value $P=Q$ but $P$ grows faster than $Q$ because his first coefficient $A \ge a$...

Of course.  I mean that the problem isn't yet solved for the case when $P$ has only complex roots.

Okey... Let's take the case when $A \ge a \ge 0$ and consider the minimum of the 2 paraboles (because of the complex roots, $P$ and $Q$ are always positiv). So $\forall x \to P(x) \ge min_{Q}$ so also $min_{P(x)} \ge min_{Q(x)}$ so we fall again in the first case with $\frac{-\Delta_{A}}{4A} \ge \frac{-\Delta_{a}}{4a}$...


summer program
Mathcamp
AMC
USA(J)MO
USAMO
AMC 10

How many questions do most people who get in answer on the qualifying quiz?    Does Mathcamp take in students who really enjoy math, but arn't exceptional?  I think I got around a 100 on the AMC 10.   I think I can do the first four problems on the quiz.

 thanks :)

Define exceptional. Well, let's just say that most kids at MathCamp are not USAMO qualifiers, but qualifiers aren't an insignificant minority either. Generally, the kids at MathCamp are just fun people who also happen to be interested in math and - more or less - have some talent in it. The average, as far as I know, is probably around 7 problems but 5 or 6 may do it.

You still have several months. I'd say work until the very last day and squeeze in as many solutions or partial solutions as you can.

Well, I was in this same position last year. I would not suggest to make mathcamp the only camp you apply to, because if you don't get in, you won't go anywhere for a camp. However, I would suggest applying even if you only solve just half of them.  People do get in with fewer solutions. I got in with 5 complete and a few partials, even if I was deferred. :lol:

So bottom line: apply, do the best that you can, but have a backup camp in case you don't get in.

(deleted for obseleteness), but what you get is the poster with qualifying quiz problems, which you can get from elsewhere. The information on it is also on http://www.mathcamp.org.

[quote="Hellstar"]What do you have to get on the AMC to get an invitation from them? My friend got a 120.5 on AMC 10 last year and he got invited. I dunno if its the same for AMC 12. Do you have to get above 120 on AMC 12 to get invited, or is it above 100 for AMC 12?[/quote]

Yes, your score has to be higher to get a brochure via AMC 10 than via AMC 12.  We send to specific numbers of people rather than to specific score cutoffs, so the scores vary from year to year.  Of course anyone can apply, this is just who we send our advertising to.


The numbers a, b, c and d are positive. Prove that among the inequalities
a + b < c + d;
(a + b)(c + d) < ab + cd;
(a + b)cd < ab(c + d)
at least one is false

Let $S=a+b,T=c+d,P=ab,R=cd$.
Let $P(x) = (x-a)(x-b)(x+c)(x+d)$.
Then :
$P(x) = x^4 + (T-S)x^3 + (P+R-ST)x^2 + (TP-RS)x + PR$
Since $PR = abcd > 0$ and that $P$ is not strictly positive on $[0;+ \infty[$, at least one of the coefficients of $x^3,x^2,x$ is non-positive, and we are done.

Pierre.


function
real analysis
real analysis unsolved

To prove the Cauchy Mean Value Theorem, I took the function h(x)=g(x)(f(b)-f(a))-f(x)(g(b)-g(a)) on the interval [a,b], where f and g are continuous functions.  Then h(a)=h(b), so if h(x) is constant, then the theorem is proved.  However, if h(a)=h(b) and h(x) is not constant, do we use Rolle's Theorem to prove that h'(x)=0 for some x in [a,b]?

Because $h(a)=h(b)$, use Rolle's Theorem : Exist $c\in (a,b)$ such that $h'(c)=0$. But $h'(x)=(f(b)-f(a))g'(x)-(g(b)-g(a))f'(x)\forall x\in (a,b)$ therefore we're done!


algebra
polynomial
quadratics
AMC
AIME
HMMT
complex numbers

Let $ P(x)\equal{}x^{2007}\plus{}17x^{2006}\plus{}1$.
$ Q(x)$ is a minimal-degree polynomial for which $ z\plus{}\frac{1}{z}$ is a root of $ Q$ for every root $ z$ of $ P$.

Compute $ \frac{P(1)}{P(\minus{}1)}$

What do we need Q for? P(1)=19, P(-1)=17, the result follows...

Perhaps we want $ \frac{Q(1)}{Q(\minus{}1)}$.

[quote="cancer"]Let $ P(x) \equal{} x^{2007}\plus{}17x^{2006}\plus{}1$.
$ Q(x)$ is a minimal-degree polynomial for which $ z\plus{}\frac{1}{z}$ is a root of $ Q$ for every root $ z$ of $ P$.

Compute $ \frac{Q(1)}{Q(\minus{}1)}$[/quote]

transform the polynomial so that when you input Z+1/Z into P(?), you get P(Z)=0.
then new P will be Q(x).

Not necessarily; in particular, if $ Q(x)$ turns out to have repeated roots it will not be of minimal degree.  This will occur whenever two roots of $ P(x)$ are multiplicative inverses of each other, which is the content of the problem.

This actually wasn't that trivial. I don't know how to show that there are no double roots, but we can check that there are no reciprocal roots: $ P(1/x) x^{2007}\equal{} P(x)\plus{}17(x\minus{}x^{2006})\implies z \equal{} z^{2006}\implies z\equal{}0,\; z^{2005}\equal{} 1$. Clearly, no such $ z$ solves the equation. That being said, here it goes...

[hide="Possible Solution"]For $ j\in\mathbb{N}^{\leq 2007}$, we have $ Q(x) \equal{} n\prod (x\minus{}z_{j}\minus{}1/z_{j})$

We somehow want to transform this into being in terms of $ P$, since we know what that is, explicitly (and we know it has only three terms!)

$ \frac{Q(1)}{Q(\minus{}1)}\equal{}\frac{n\prod (1\minus{}z_{j}\minus{}1/z_{j})}{n\prod (\minus{}1\minus{}z_{j}\minus{}1/z_{j})}\equal{}\prod\frac{z_{j}\plus{}1/z_{j}\minus{}1}{z_{j}\plus{}1/z_{j}\plus{}1}\equal{}\prod\frac{z_{j}^{2}\minus{}z^{j}\plus{}1}{z_{j}^{2}\plus{}z^{j}\plus{}1}$

Now, you have quadratics in the numerator and denominator. Each factor is a linear in $ z_{j}$, which means that using the associative property, you can write this in terms of $ P$! Except it's not [b]that[/b] cool, because we have to use complex numbers? Let $ (\minus{}1)^{2/3}\equal{}\omega$ be our third root of unity. We can get all the sixth roots of unity by reexpressing that.

We can factor this as follows; $ \prod\frac{(z_{j}\plus{}\omega) (z_{j}\plus{}\omega^{2})}{(z_{j}\minus{}\omega)(z_{j}\minus{}\omega^{2})}\equal{}\frac{\prod (z_{j}\plus{}\omega)\prod (z_{j}\plus{}\omega^{2})}{\prod (z_{j}\minus{}\omega)\prod (z_{j}\minus{}\omega^{2})}$

And then this happens:
$ \equal{}\frac{\left(n\prod (z_{j}\plus{}\omega)\right)\left(n\prod (z_{j}\plus{}\omega^{2})\right)}{\left(n\prod (z_{j}\minus{}\omega)\right)\left(n\prod (z_{j}\minus{}\omega^{2})\right)}\equal{}\frac{\left(n\prod (\minus{}\omega\minus{}z_{j})\right)\left(n\prod (\minus{}\omega^{2}\minus{}z_{j})\right)}{\left(n\prod (\omega\minus{}z_{j})\right)\left(n\prod (\omega^{2}\minus{}z_{j})\right)}\equal{}\frac{P(\minus{}\omega) P(\minus{}\omega^{2})}{P(\omega) P(\omega^{2})}$

Now we can evaluate it directly (not shown) using the fact that $ \omega^{3}\equal{} 1$ and $ \omega^{2}\plus{}\omega\plus{}1 \equal{} 0$ and it happens to simplify nicely to a real number, $ \boxed{\frac{289}{259}}$[/hide]

[i]This looks AIME-esque... where did you get it?[/i]

[quote="The Zuton Force"][i]This looks AIME-esque... where did you get it?[/i][/quote]

It's from HMMT 2007.  I very distinctly recall mixing up a plus sign and minus sign, and thus getting the wrong answer...


Suppose that (m, n) be odd integers satisfying: m>n>1 and m^2 is divided by m^2 - n^2 + 1. Show that m^2/(m^2 - n^2 + 1) is square of an integer.

<i>Source: Selection Test of Hanoi University of Natural Sciences 2003 </i>

Namdung

how about letting m-n=2a, m+n=2b?
Then it resembles the imo q

Let m=x+y , n=x-y
Then m^2/(m^2-n^2+1) = (x+y)^2/(4xy+1)=k_(x,y)
We have the equation :
  x^2 + 2xy(1-2k_(x,y)) +y^2 - k_(x,y)=0 (1)
+let x0,y0 be the root of (1) with x0+y0 is min . Fix k_(x,y)=k_(x0,y0)
Note that (x',y) be the root of (1)
where x'=2y0(1-2k_(x0,y0))-x
But 0<=x'< x so x'=0 => k_(x0,y0)=y^2 is square of an integer
+Let x1,y1 be the root of (1) with x1+y1 is min and x1+y1 > x0 +y0
Similarly we have K_(x1,y1)=y1^2
So For all xn,yn be the roots of (1) then
     K_(xn,yn)=yn^2


limit
search
Functional Analysis
real analysis
real analysis unsolved

Let $ X$ is a Banach space, $ X^*$ is its dual and $ f: X\to \mathbb R$ differentiable Frechet, bouded from below on all bouded subset of $ X$ and such that $ \lim_{|x|\to\infty}\frac{f(x)}{\|x\|}\equal{}\plus{}\infty$. Prove that the set $ \{f'(x): \;x\in X\}$ is dense in $ X^*$.

I'm very glad to see any proof of this problem :)

Let $ y \in X^*$ be given. Set $ f_1(x) \equal{} f(x) \minus{} <y,x>$. Then $ f_1$ satisfies the same assumptions as $ f$, and $ f_1'(x) \equal{} f'(x) \minus{} y$ for all $ x \in X$.

Now if $ X$ is finite-dimensional, we can find the absolute minimum $ x_0$ of  $ f_1$ and see that $ f_1'(x_0) \equal{} 0 \equal{} f'(x_0) \minus{} y$. Hence $ f'$ is actually onto if $ dim \, X < \infty$.

If $ dim \, X \equal{} \infty$, we use [url=http://planetmath.org/encyclopedia/EkelandsVariationalPribciple.html]Ekeland's variational principle[/url] to find $ x_\epsilon \in X$ such that $ f_1(x_\epsilon\plus{}z)  \ge f_1(x_\epsilon) \minus{} \epsilon \|z\|$ for all $ z \in X$, for any given $ \epsilon > 0$. Then for $ \|z\| \equal{} 1$ and $ t > 0$
\[ f_1(x_\epsilon) \plus{} t <f'(x_\epsilon),z> \plus{} o(t) \equal{} f_1(x_\epsilon\plus{}tz)  \ge f_1(x_\epsilon) \minus{} \epsilon t\]
implying $ | <f_1'(x_\epsilon),z>| \le \epsilon$ for all $ z$ with $ \|z\| \equal{} 1$. Hence $ \|f'(x_\epsilon )\minus{} y\|_* \equal{} \|f_1(x_\epsilon)\| \le \epsilon$. Since $ \epsilon$ was arbitrary, we are done. :)

I think this problem belongs in the advanced section.

for a discussion about the Ekeland principle:
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=129187535&t=127075


integration
function
complex analysis
complex analysis theorems

Let $f : \Omega \longrightarrow \mathbb{C}$ be a continuous function such that
\[ \oint_{\gamma} f ( z ) \mbox{dz} = 0  \] for any closed path $\gamma : [ a, b ] \longrightarrow \Omega .$
Prove that $f$ has a primitive on $\Omega$.

Doesn't this follow from Morera's theorem?

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


It seems very wrong, since it is not homogenous.


Well, quite extraordinary proof. A few minutes ago I found between my papers a very similar (nearly the same) proof by Lev Kourliantchik. Anyway, thanks ;)


Find all positive integers $ n$ such that, if $ 1\equal{}d_1<d_2<...<d_k\equal{}n$ are the positive divisors of $ n$, then $ d_2\minus{}d_1,d_3\minus{}d_2,...,d_k\minus{}d_{k\minus{}1}$ form an increasing sequence.

If you mean all divisors and k>2, then $ d_{k\minus{}1}\le d_k/2$, therefore $ d_k\minus{}d_{k\minus{}1}\ge d_{k\minus{}1}\ge d_2\minus{}d_1$.
Therefore in these case $ k\equal{}2$ or $ n\equal{}p$ prime.

[quote="Rust"]If you mean all divisors and k>2, then $ d_{k \minus{} 1}\le d_k/2$, therefore $ d_k \minus{} d_{k \minus{} 1}\ge d_{k \minus{} 1}\ge d_2 \minus{} d_1$.
Therefore in these case $ k \equal{} 2$ or $ n \equal{} p$ prime.[/quote]

There might be some misunderstanding.

For instance n=10 is not a prime, but its divisors 1,2,5,10 yield the
sequence 2-1, 5-2, 10-5, which is increasing and satsifies all
conditions of the problem statement.

I am sorry, I solve $ d_2\minus{}d_1\ge d_3\minus{}d_2\ge ...ge d_k\minus{}d_{k\minus{}1}$ (or decrease sequence).
For increase sequence very many n satisfyed your condition.
For example $ n\equal{}p^k$ or $ n\equal{}p^kq$ if $ p>2q\minus{}1$ or $ p^l<q<p^l\frac{p\minus{}1}{2}$.


vector
algebra
polynomial
group theory
complex numbers
set theory
number theory

How we could define the multiplication on $ \mathbb{R}^3$ , to have :

$ (\mathbb{R}^3 ,\plus{},\cdot)$ is a Field

 :?:

This is a little cheap, but identify $ \mathbb{Q}^3$ with any algebraic number field of degree $ 3$ (for example, $ \mathbb{Q}[ \sqrt [3]{2} ]$) and extend the multiplication there to the reals.

In other words, define $ (a, b, c) \cdot (x, y, z) \equal{} (ax \plus{} 2cy \plus{} 2bz, ay \plus{} bx \plus{} 2cz, ax \plus{} cx \plus{} by)$.  The result is a field with multiplicative identity $ (1, 0, 0)$.

I'm curious, though; for an appropriate choice of algebraic field, can we define a multiplication that is symmetric in $ a, b, c$ and $ x, y, z$?

O.k. ..thank you .

Could you generalize it to $ \mathbb{R}^n$.

$ \mathbb{Q}[\sqrt [n]{2}]$ is an algebraic field of degree $ n$ for any natural $ n$.  The multiplication is not hard to generalize; the coefficient of any term is either $ 1$ or $ 2$.

Could you write please...

I need it so much , ad I haven't a chance o make it wrong.

Thank you

$ (0,0,0)(a,b,c)\equal{}(0,0,0)$ must hold

we gotta have an identity, so
$ (x_o,y_o,z_o)(a,b,c)\equal{}(a,b,c)$

for each $ (a,b,c)$ there must exist $ (a',b',c')$
$ (a,b,c)(a',b',c')\equal{}(x_o,y_o,z_o)$

so lets study a basis $ e_1 \equal{}(1,0,0), e_2\equal{} (0,1,0), e_3 \equal{} (0,0,1)$

then by distributivity and all that we get

$ (a,b,c)\equal{}ae_1\plus{}be_2\plus{}ce_3, (a',b',c')\equal{}a'e_1\plus{}b'e_2\plus{}c'e_3$
all we must study then is how the basis vectors interact (and they are commutative)

$ e_je_k \equal{} \sum_{l\equal{}1}^3\alpha_{j,k,l}e_l$ with $ \alpha_{j,k,l}\equal{}\alpha_{k,j,l}$

so our identity give us

$ (x_oe_1\plus{}y_oe_2\plus{}z_oe_3)e_j\equal{}x_o\sum_{l\equal{}1}^3\alpha_{j,1,l}e_l\plus{}y_o\sum_{l\equal{}1}^3\alpha_{j,2,l}e_l
\plus{}\sum_{l\equal{}1}^3\alpha_{j,3,l}e_l\equal{}$
$ \sum_{l\equal{}1}^3 \left(x_o\alpha_{j,1,l}\plus{}y_o\alpha_{j,2,l}\plus{}z_o\alpha_{j,3,l}\right)e_l \equal{} e_j$

implies $ v_i\equal{}(x_o,y_o,z_o)$ dot $ (\alpha_{j,1,k},\alpha_{j,2,k},\alpha_{j,3,k})$ = $ \delta_{j,k}$

letting $ A_{j,k}\equal{}(\alpha_{j,1,k},\alpha_{j,2,k},\alpha_{j,3,k})$ and $ B_j\equal{}(A_{j,1},A_{j,2},A_{j,3})$

then $ B_j v_i \equal{}e_j \to v_i \equal{} B^{\minus{}1}_je_j \forall j$

so you will have a field when you select your 27 different $ \alpha_{j,k,l}$ s.t.

$ B^{\minus{}1}_1e_1\equal{}B^{\minus{}1}_2e_2\equal{}B^{\minus{}1}_3e_3$

maybe?

t0rajir0u's way does not work:
There are not always inverses. In fact the triple $ (a,b,c)$ is invertible, iff $ a^3 \plus{} 2b^3 \plus{} 4c^3 \equal{} 6abc$. In $ \mathbb Q$ this has only the trivial solution, but in $ \mathbb R$ there are much more.

Another more complicated reason why this cannot be: the only finite dimensional commutative real division algebras (meaning: a field structure on $ \mathbb R^n$ such that the operations are compatible with scalar multiplication as a real vector space) are $ \mathbb R$ and $ \mathbb C$. If we drop commutativity, only $ \mathbb H$ joins.

For similar reasons, kenn4000's way cannot work.

To get a working version:
- prove that there is a group isomorphism $ i: \mathbb R^3 \to \mathbb R$ (regarding the additive groups).
- for $ a,b \in \mathbb R^3$ define $ a \cdot b : \equal{} i^{ \minus{} 1}( i(a) \cdot i(b) )$.
- by this, we get $ \mathbb R^3$ equipped with a multiplication, and in fact this is a field.

[Edit: small correction, see below]

[quote="ZetaX"]

To get a working version:
- prove that there is a group isomorphism $ i: \mathbb R^3 \to \mathbb R$ (regarding the additive groups).
- for $ a,b \in \mathbb R$ define $ a \cdot b : \equal{} i^{ \minus{} 1}( i(a) \cdot i(b) )$.
- by this, we get $ \mathbb R^3$ equipped with a multiplication, and in fact this is a field.[/quote]

you meant $ a,b \in \mathbb R^3$.

Could you prove 1st statement please. or give me such an idea how to start.

Thank you.

dimension over $ \mathbb{Q}$ and cardinal arithmetic.

[quote="ZetaX"]

To get a working version:
- prove that there is a group isomorphism $ i: \mathbb R^3 \to \mathbb R$ (regarding the additive groups).
- for $ a,b \in \mathbb R^3$ define $ a \cdot b : \equal{} i^{ \minus{} 1}( i(a) \cdot i(b) )$.
- by this, we get $ \mathbb R^3$ equipped with a multiplication, and in fact this is a field.

[/quote]

some of Algebraist answer for this argument as follows :

[quote]
Your argument is correct as far as it goes ($ \mathbb R^3$ and$ \mathbb R$ are both $ Q\minus{}vector$ spaces of
dimension $ 2^{\aleph_0}$ so have isomorphic additive groups). What you've shown is
that a multiplication $ *$ can be defined on the additive group $ (\mathbb R^3,\plus{})$ such that
$ (\mathbb R^3,\plus{},*)$ is a field. It is, of course, isomorphic as a field to the real numbers.

What cannot be done is to make $ \mathbb R^3$ into a field compatibly with the $ R\minus{}vector$ space
structure: i.e., there is no field (or even division algebra) which is an $ R\minus{}algebra$
of dimension $ 3$.
[/quote]

What is your opinion ZetaX .

I believe we can't define the multiplication on $ \mathbb R^3$.since we know the complex field is closed and we can't make an extension of it (using the fundamental theory of algebra).Nevertheless,  there is a hypercomplex numbers which is an extending for complex numbers, but to make a field on $ \mathbb R^3$ is an impossible mission .

Opinion guys  :huh:  :huh:

There is a multiplicative field structure on $ \mathbb R$ compatible with the canonical addition (as described above), but yes, there is no multiplication compatible with the vector space structure.
If you would have one (completely compatible with scalar multiplication), you would get an algebraic (note that every finite dimensional field extension is algebraic) field extension of $ \mathbb R$ (identify $ \mathbb R$ with the real multiples of $ 1$ in this field) of degree $ 3$, and such one doesn't exist. To see this, either use that $ \mathbb C$ is closed (and has already smaller dimension), or juse the intermediate value theorem:
Assume that you have such a field $ K$ (of dimension $ 3$ over the reals) and thus an $ a\in K \backslash \mathbb R$. Then $ 1,a,a^2$ is a basis of $ K$, so there is some irreducible real polynomial $ f(x)$ of degree exactly $ 3$ and $ f(a)\equal{}0$. But such a polynomial doesn't exist, e.g. by the intermediate value theorem.

Sorry for annoying ; :blush:

Again to discussion  :

Look at this :

[quote]
The group $ (\mathbb{R}, \plus{})$ is isomorphic to the group $ (\mathbb{C}, \plus{})$ of all complex numbers with addition.

The wikipedia said we need to depend on axiom of choice ..

>>>>How we could do that???

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

Q) The answer :

Both are vector spaces over the field Q, and they have the same dimension.
So they are isomorphic (use the base of R over Q, and C over Q, that exists by the axiom of choice;
you cannot write it out explicitly....)

Need the procedure of that.

Q) The answer :

Let V be a vector space (over some field K)
Use Zorn's lemma [there is no procedure for finding bases in general vector spaces]:
the set of all linearly independent sets is such that if we have a chain of them, the union
is an upperbound. So Zorn applies; there is a maximal linearly independent subset and
this can be shown to be a base for the vector space.

[/quote]

=========================================================

My question ; I know the dimension of $ \mathbb C$ and $ \mathbb R$ are not the same .

Both of them have the same dimension over $ \mathbb Q$ ?????

=========================================================
[quote]

Is $ \mathbb R^3$ and $ \mathbb R$ Isomorphic ??

Yes ;

take a basis e_i (i in I) for R as a Q vector space.
take a basis f_j (j in J) for R^3 as a Q vector space.
Then, as Q is countable, and |R| = |R^3| we have that the dimensions are the same
so |I| = |J|.

Take any bijection g between I and J.
define i as F(e_i) = f_g(i), and extend by linearity (as every x in R is by definition
a finite Q linear combination of e_i's, so x = sum_{i in F(x)} q_i*e_i [F(x) subset I finite, q_i in Q)
we define F(x) to be sum_{i in I]} q_i*f_g(i)...)

Q) Are R^n and R isomorphic also?????

Yes, in the same manner[/quote]

Now what I know the number of basis of $ \mathbb R^3$ is not same as the number of basis $ \mathbb R$ , Or we have the same number of basis , since they are vector space over $ \mathbb Q$.??

If you have some theorem for the vector spaces over $ \mathbb Q$ for dimensionality ..Please write here.

Waiting your opinion ZetaX and guys.

 :huh:

This requires some set theory, especially cardinality.

Dimension doesn't only depend on the vector space $ V$, but also from the base field $ K$. For example, as a complex vector space $ \mathbb C$ has dimension $ 1$, as a real vector space it has dimension $ 2$ and as a rational vector space it has dimension $ \infty$.
The effect that $ \mathbb R \cong \mathbb R^3$ as $ \mathbb Q$-vector spaces (not as $ \mathbb R$-vector spaces!) is similar to the "fact" there are equally many prime numbers than integers (there are simply countable infinite many).

Let $ V,W$ be vector spaces over a field $ K$ and having bases $ \mathcal V \equal{} \{v_i\}$ and $ \mathcal W \equal{} \{w_j\}$ (both of them possibly infinite). If there is a bijection $ b: \mathcal V \to \mathcal W$, then this gives an isomorphism $ \bar b$ of $ K$-vector spaces, defined by $ \bar b \left( \sum k_i v_i \right) \equal{}  \sum k_i b(v_i)$ for $ k_i \in K$ (where the sum is finite/all but finitely many summands are $ 0$; note that a basis is defined to reach any elements as a finite sum).
So vector spaces are isomorphic if a set of bases is isomorphic (which for sets means: there is a bijection). Now let $ K$ be an infinite field and $ V,W$ be $ K$-vector spaces such that $ |K| < |V| \equal{} |W|$ (here $ |\cdot |$ means cardinality, so this means: there is a bijection $ V \to W$ as sets [not necessarily as vector spaces!] and there is an injection $ K \to V$ but not an injection $ V \to K$ [again as sets, not spaces!]). Then by the axiom of choice and some set theory, any $ K$-basis $ \mathcal V$ of $ V$ has the same cardinality than $ V$ itself, so $ |\mathcal V | \equal{} |V|$. Similar, $ |\mathcal W | \equal{} |W|$, thus $ |\mathcal V | \equal{} |V| \equal{} |W| \equal{} |\mathcal W |$, meaning that there is a bijection between the bases. By the previous said, this means that $ V$ and $ W$ are isomorphic as $ K$-vector spaces.
This applies for $ K\equal{}\mathbb Q$, $ V\equal{}\mathbb R^m$, $ W\equal{}\mathbb R^n$ as well.

[quote="ZetaX"]This requires some set theory, especially cardinality.

Dimension doesn't only depend on the vector space $ V$, but also from the base field $ K$. For example, as a complex vector space $ \mathbb C$ has dimension $ 1$, as a real vector space it has dimension $ 2$ and as a rational vector space it has dimension $ \infty$.
The effect that $ \mathbb R \cong \mathbb R^3$ as $ \mathbb Q$-vector spaces (not as $ \mathbb R$-vector spaces!) is similar to the "fact" there are equally many prime numbers than integers (there are simply countable infinite many).

Let $ V,W$ be vector spaces over a field $ K$ and having bases $ \mathcal V \equal{} \{v_i\}$ and $ \mathcal W \equal{} \{w_j\}$ (both of them possibly infinite). If there is a bijection $ b: \mathcal V \to \mathcal W$, then this gives an isomorphism $ \bar b$ of $ K$-vector spaces, defined by $ \bar b \left( \sum k_i v_i \right) \equal{} \sum k_i b(v_i)$ for $ k_i \in K$ (where the sum is finite/all but finitely many summands are $ 0$; note that a basis is defined to reach any elements as a finite sum).
So vector spaces are isomorphic if a set of bases is isomorphic (which for sets means: there is a bijection). Now let $ K$ be an infinite field and $ V,W$ be $ K$-vector spaces such that $ |K| < |V| \equal{} |W|$ (here $ |\cdot |$ means cardinality, so this means: there is a bijection $ V \to W$ as sets [not necessarily as vector spaces!] and there is an injection $ K \to V$ but not an injection $ V \to K$ [again as sets, not spaces!]). Then by the axiom of choice and some set theory, any $ K$-basis $ \mathcal V$ of $ V$ has the same cardinality than $ V$ itself, so $ |\mathcal V | \equal{} |V|$. Similar, $ |\mathcal W | \equal{} |W|$, thus $ |\mathcal V | \equal{} |V| \equal{} |W| \equal{} |\mathcal W |$, meaning that there is a bijection between the bases. By the previous said, this means that $ V$ and $ W$ are isomorphic as $ K$-vector spaces.
This applies for $ K \equal{} \mathbb Q$, $ V \equal{} \mathbb R^m$, $ W \equal{} \mathbb R^n$ as well.[/quote]

This post carries the helpful for me .

Thank you :-)

This is a nice discussion

I have the following comment :
If we find the isomorphism , then we could define the multiplication on $ \mathbb R ^3$ by $ \varphi ^ { \minus{} 1} ( \varphi (a) . \varphi (b) )$ , as ZetaX suggested.
At the end of the day, we end up with a field defined on $ \mathbb R ^3$

BUT ..

Isn't this a field extension of $ \mathbb C$ which contradicts with the Fundamental Theorem of Algebra ??

Regards ,

No it isn't. The reason is simple: Our isomorphism defines a completly new multiplication which has nothing to do with the multiplication on $ \mathbb{C}$, so $ \mathbb{R}^3$ is no field extension of $ \mathbb{C}$.

You are right !

I had  in the back of my mind the idea of the impossibility of defining a field structure on $ \mathbb R ^ n$ , n > 2 .

Actually I have read it in "Elements of Real Analysis' , by R. Bartle !


integration
calculus
calculus computations

calculate $ \int\sin(x^2) dx$ .(does it exist?)

Yes, it exists.  No, it is not an elementary function.

See the [url=http://en.wikipedia.org/wiki/Fresnel_integral]Fresnel integral[/url].


Show that the binomial coefficient $ \binom{2n}{n}$ is divisible by $ n+1$ for any positive integer $ n$

Well we know that every binomial is an integer. This is because of the recursive definition:
\[ \binom{n}{r}+\binom{n}{r+1}= \binom{n+1}{r+1}\]
(ok, i know this isn't rigorous, can someone help me make this rigorous?)

Then we have that \[ \binom{2n}{n+1}= \binom{2n}{n}\left( \frac{n}{n+1}\right)\]

so because the LHS of the above equation has to be an integer \[ n+1 | \binom{2n}{n}\] because $ gcd(n, n+1) = 1$

[quote="agolsme"](ok, i know this isn't rigorous, can someone help me make this rigorous?)[/quote]

That was totally rigorous (at least, when you toss in the fact that we have the initial conditions $ \binom{n}{0}=\binom{n}{n}= 1$ for all $ n$; it's just an induction over two variables), and a very nice solution.

Another way to state the fact that all binomials are integers is to say, "because of the well-known combinatorial interpretation of those numbers."  In that case, of course, one can say the same thing about the original question, since the sequence $ \frac{1}{n+1}\binom{2n}{n}$ is the Catalan numbers.


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

Hello,

I have another algebra problem.

Let $R$ be an integral domain, and let $F$ its fraction field.

Consider an element $f=\frac{a}{b}\in F$, and assume that this holds :

for every maximal ideal $M$ in $R$, $f$ can be written as $\frac{r}{m}$ where $m\notin M$.

Prove that $f$ can be written as $\frac{x}{1}$ (or thus : prove that it is an "integer")

Now I did manage to prove "something" : if $b$ is not a unit, every maximal ideal containing $b$ will also have to contain $a$.  But I need more than that : I need that $b$ divides $a$.
Can anyone help me? :huh:

Thanks!

[quote="-oo-"]of course, we may assume $a=1$. then we have to show that $b$ is a unit. if it is not, find a maximal ideal $M$ containing $b$. obviously, $1/b$ has not the form $r/m$ where $m \notin M$.[/quote]
This might be dumb, but why can $a=1$ be assumed? :blush:

sorry, we may not assume $a=1$. the proof is not easy and makes use of the local-global-principle of commutative algebra, which I will explain in detail. we have to show, that $R$ is the intersection of the localizations $R_{m}$, where $m$ run's through the maximals ideals of $R$, where the one inclusion is obvious. for the converse, it is enough to prove the following:

[b]1[/b]. let $N$ be a submodule of $M$ and $a \in M$. then $a \in N$ is equivalent to $\iota(a) \in N_{m}$ for all maximal ideals $m$, where $\iota : M \to M_{m}$.

if we consider $\phi : R \to M/N, 1 \to \overline{a}$, this means that $\phi = 0$ if and only if $\phi_{m}=0$ for all $m$. so it suffices to prove:

[b]2[/b]. a module-homomorphism $\phi : M \to N$ equals $0$ if and only if so does $\phi_{m}$ for all maximal ideals $m$.

to prove this, consider the exact sequence $0 \to ker(\phi) \to M \to N$. then also $0 \to ker(\phi)_{m}\to M_{n}\to N_{n}$ is exact, hence $\ker(\phi)_{m}=0$ by assumption for all $m$. thus, finally we have to prove:

[b]3[/b]. for every module $M$ we have $M=0$ if and only if $M_{m}= 0$ for all maximal ideals $m$.

to prove this pick an element $0 \neq x \in M$. then $Rx \cong R / I$ for the ideal $I=Ann(x)$. find a maximal ideal $m$ over $I$. consider the homomorphism $Rx \to R/I \to R/m =: F$, where $F$ is a field. then also $(Rx)_{m}\to F_{m}\to 0$ is exact. we may embed $F$ into $F_{m}$ (because $F$ is a field, there are no zero divisors outside of $m$). in particular, $F_{m}\neq 0$. thus, also $(Rx)_{m}\neq 0$. but since $Rx \to M \to 0$ is exact, also $(Rx)_{m}\to M_{m}\to 0$ is exact, which yields $M_{m}\neq 0$, as desired.

:maybe: What you say about the intersections is in fact what my book says (I just used different terms to make the question a bit more clear)

It seems my book has a different definition of "a simple algebraic fact" :!:
Thanks, I will look at this proof carefully, and I might have to come back if I  don't get it.

I will later (try to) go through -oo-'s proof (am I right to assume it uses the axiom of choice?), but what about the following?

For the sake of laziness, consider $R$ as a subring of $F$, instead of working with the canonical embedding $a\mapsto \frac{a}{1}$. We want to show that $f\in R$.

Let $S$ be the set of all elements $s$ of $R$ such that $sf\in R$. Then, $S$ is an ideal of $R$ (obvious: if $u\in S$ and $v\in R$, then $vu\in S$ because

$u\in S\Longrightarrow uf\in R\Longrightarrow vuf\in R\Longrightarrow vu\in S)$

; besides, if $u\in S$ and $v\in S$, then $u+v\in S$ because

$u\in S\wedge v\in S\Longrightarrow uf\in R\wedge vf\in R\Longrightarrow uf+vf\in R$
$\Longrightarrow (u+v)f\in R\Longrightarrow u+v\in S$

). If $1\in S$, then $1f\in R$, so that $f\in R$ and the proof is complete.

Hence, it is enough to show that $1\in S$. In order to prove this, we assume the contrary. Thus, assume that $1\notin S$. Then, $S$ is a proper subset of $R$. Since $S$ is also an ideal of $R$, we thus conclude, using the axiom of choice, that there exists a maximal ideal $T$ of $R$ such that $S\subseteq T$. According to the condition of the problem, $f$ can then be written as $f=\frac{r}{m}$ for some $r\in R$ and $m\notin T$. Thus, $mf=r\in R$, so that $m\in S$. Since $S\subseteq T$, this yields $m\in T$, what contradicts with $m\notin T$. Hence, our assumption that $1\notin S$ was wrong. Therefore, $1\in S$ is proven, qed.

  Darij

Okay I see it now, Darij's proof is actually quite logic, but you need to know which maximal ideal to use. :|

-oo's approach is more general.

So thanks to all of you  :) , but I still don't think this is "a simple algebraic fact" :o

$ {\rm denominators}(f) \equal{} \{d \in R : d\, f \in R\}$  is an $ \bf{ideal}$ not contained in any maximal ideal so it is 1.
E.g. in a Dedekind domain, where "contain" = "divides", it says: the denominator ideal has no prime divisor so is 1.


geometry
cyclic quadrilateral
projective geometry
geometry unsolved

[color=darkred]The sides $AD$ and $CD$ of a tangent quadrilateral $ABCD$ touch the incircle $\varphi$ at $P$ and $Q,$ respectively. If $M$ is the midpoint of the chord $XY$ determined by $\varphi$ on the diagonal $BD,$ prove that $\angle AMP = \angle CMQ.$[/color]

This problem has already been discussed here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=113117[/url] :)

After showing $ \angle PMD \equal{}\angle QMD$ the problem is equivalent, by duality, to the following problem:

[i]Let $ PQRS$ be a cyclic quadrilateral and let:
$ M'\in RS\cap QP$
$ T\in SP\cap RQ$
$ l\bot OM'$ Y $ l$ and $ M'\in l$

$ X\in SP\cap l$
$ Y\in RQ\cap l$

Show $ XM' \equal{} YM'$. [/i]

(Showing $ XM' \equal{} YM'$ is equivalent to $ \angle XOM' \equal{}\angle YOM$)

if $ M$ is the pole of $ l$, $ TM$,$ TX$,$ TM'$ and $ TY$ cut lines in harmonic conjugate (easely proven by Ceva and Menelao on the $ TSQ$) and $ TM\parallel l$ ($ TM\bot MM'$) so $ XM' \equal{} YM'$.

by Googo's theorem,$\angle AMD=\angle CMD$
since $DPOMQ$cyclic,PD=DQ
so $\angle PMD=\angle QMD$
hence $\angle AMP=\angle CMQ$.

about Googo's theorem,see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=425471

Nice one!

Let $\varphi$ touch sides $CB$ and $AB$ at $R$ and $S$, respectively. Let $I$ be the center of $\varphi$. Obviously, $IM\perp BD$. Then $D\widehat{M}I=D\widehat{P}I=D\widehat{Q}I=\frac{\pi}{2}$, therefore $D$, $P$, $M$, $I$, $Q$ are concyclic. This yields $P\widehat{M}D=D\widehat{M}Q$, since $DP=DQ$, and now we must show that $A\widehat{M}D=C\widehat{M}D$.

Invert around $\varphi$; for each point $X$ denote its inverse by $X^\ast$. Then $A^\ast$ and $C^\ast$ are the midpoints of $PS$ and $QR$, respectively. Observe that $BD$ is the polar of $K=PQ\cap RS$, therefore $M$ is the inverse of $K$. Since $IM\perp BD$, what we want to prove is equivalent to $\angle(AM, MI)=\angle(MI, MC)$, or, equivalently, $\angle(A^\ast I, A^\ast K)=\angle(C^\ast K, CI)$. Notice that $\angle(A^\ast I, PS)=\angle(QR, C^\ast I)=\frac{\pi}{2}$, therefore subtracting $\frac{\pi}{2}$ from both sides reduces the problem to showing that $\angle(PS, A^\ast K)=\angle(C^\ast K, QR)$. But this is easy: observe that triangles $KPS$ and $KRQ$ are similar and $A^\ast$ and $C^\ast$ are the midpoints of $PS$ and $QR$, respectively, and this yields the desired conclusion.

What a nice problem! :)

Let $S,R$ be the tangency points of the incircle with $AB,BC$ respectively. Let $AC \cap BD = L$ and $I$ the center of the incircle. Furthermore let $M' = SR \cap PQ$ and $SP \cap QR = T$.

Note that $BD$ is the polar of $M'$ because of the tangents and since $AC$ is the polar of $T$ by Brokard's theorem we must have $M'$ is on line $AC$. Since $IM \perp BD$ we must also have $I,M,M'$ are collinear.

Brianchon's theorem on degenerate hexagon $APDQCB$ gives $AQ,PC,DB$ concurrent. Therefore $(A,C;L,M') = -1$. Thus the pencil $MA,MC,ML,MI$ is harmonic and $\angle IML = 90^{\circ}$ so $MD$ bisects $\angle AMC$.

To conclude note that $\angle IMD = \angle IPD = \angle IQD = 90^{\circ}$ so $PMIQD$ is cyclic and $DP=DQ$ meaning $\angle PMD = \angle DMQ.$

Subtracting this from $\angle AMD = \angle DMC$ we get the desired $\angle AMP = \angle CMQ.$


Hi everyone,

I recognize the fact that this paradox/question has been discussed in other posts. This post is distinct from the others because I am trying to explain my proof instead of answering questions posted by others. Hope this point is well understood.

 :)

There is a decently well known paradox which goes like this:

[i]A man is guarunteed by a truthful woman that a certain event will happen within a certain time period. He is also guarunteed by this woman that he will not know that this event will occur until the time it begins happening. From these two facts he comes to the conclusion that this event will never happen within the given time period because he knows that the event cannot happen at the last possible (this varies depending on the problem) time, because then he would expect it. He also knows that it will not happen the first unit of time before the last possible time, and so on and so forth until he reasons that this event will not happen.[/i]. Sometimes this paradox is simplified by furthur guaruntees that for example, the event can only happen at 8:00AM on the morning of a Monday, or something else of the sort, which makes it easier to reason (i.e. It cannot happen on the last Monday of the month at 8:00, because then I'll know, etc.etc.etc.)

What makes this paradox so shocking is simply the different between the "unordinary-ness" of the circumstances and the result. To clarify, we do not find the circumstances that unusual, of course not taken literally, but nevertheless; but we find the result highly unexpected. Had the circumstances been less "ordinary" we as readers would have simply ignored this paradox, claiming that it was highly unlikely to ever happen.

But what makes this paradox indeed interesting is that it is not highly unlikely, but [i]appears[/i] to be quite a normal situation.

I believe that I have found 3 proofs which shows that there is no paradox, or any "closed circle".

Definition: All statements such as "this statement is false" are [i]closed circle[/i] statements, because they will simply circle around forever. Forgive the redundancy of the [i]closed[/i] circle - it is not that I do not ([i]2N[/i]) recognize that all circles are indeed closed.

1. If the man is familiar with this "paradox" then he will not expect the event at all, and the event will always be unexpected.

2. Such a guaruntee can never be made truthfully. (Debatable...)

3. Although at any given point in time, there corresponds a year, day, hour, minute, second, etc. (this continues indefinitely), this [i]does not mean[/i] that two given points in time that are seemingly identical from these measurements (however far/detailed they go) are indeed the same. Because of this, there are infinite slices of time within each nano-second, pico-second, and so forth. This means that the man can never expect when the event will happen.

[i]Although the event may last longer than these infinite slices of time do, that is irrelevant because all we care about is the beginning of the start of the event[/i].

Questions? Comments?

Open for Cross-X.

In the mathematical sense (assuming time is continuous) your statement falls apart when you say 'the first unit of time before the last possible time'.  This is analogous to saying 'the smallest number greater than zero' or 'the biggest integer'.

Otherwise, I don't think anyone really cares about pointless philosophical quasi-paradoxes.


ARML probability geometry trigonometry MATHCOUNTS

1.
When the sum of the first $k$ terms of the series

\[ 1^2 + 2^2 + 3^2 + \cdots + n^2 + \cdots \]
is subtracted from the sum of the first $k$ terms of the series

\[ 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) + \cdots, \]
the result is 210.  Find the numerical value of $k$.

2.
Three coplanar circles are concentric, and the lengths of their respective radii are 10, 20, and 30.  A square is circumscribed about the smallest circle, and squares are inscribed in the other two circles.  Find the length of the shortest line segment that could contain a point on each square.

3.
Find the numerical value of $k$ for which

\[ \frac{7}{x+y} = \frac{k}{x+z} = \frac{11}{z-y}. \]

4.
Spender Sam always carried only dollars and pennies, but he never carried more than 99 pennies.  When leaving home one day, Sam noted that the number of dollars he had was 150% of the number of pennies he had.  After spending 10 dollars and 78 cents, Sam discovered that the number of dollars he had remaining was 1 less than the number of pennies he then had.  With how much money did Sam leave home that day?

5.
Two cards are randomly selected from a standard 52 card deck of playing cards, without replacement.  Find the probability that at least on king and at least one heart are selected (though not necessarily on different cards).

6.
Find the ordered pair of positive integers $(a, b)$, with $a<b$, for which
\[ \sqrt{1+\sqrt{21+12\sqrt{3}}} = \sqrt{a} + \sqrt{b}. \]

7.
The length of the base of an isosceles triangle is 20.  In the plane of the triangle, a point 4 units from this base is 10 units from each leg.  Find the area of the triangle.

8.
If Arc denotes the principle value, find all values of $x$ which satisfy

\[ \sin{(4\text{Arc}\tan{x})} = \frac{24}{25}. \]

I haven't checked these, but if you want them:

[hide]1. 20
2. $ 5 \sqrt{2} $
3. 18
4. 96.64
5. $ \frac7{26} $
6. (1,3)
7. $ \frac{20 \sqrt{391} -160}3 $ Seems too messy to be correct.
8. -3, $ \frac13 $, -2, $ \frac12 $[/hide]

hmm it sounds like there is a competition this saturday???

[quote="nirvanatear"]buddy are you in the washington team?
im competing u on saturday
im from BC[/quote]

Oh, is this a reference to the Mathcounts intramural?


If a function f(x)=(x^2+kx+1)/(x^2+x+1),x belongs to R . Given that there are 3 real numbers a,b,c such that                       f(a)+f(b)<f(c),calculate the range of k.

[hide]$ f(x)=1+(k-1)\frac{x}{x^{2}+x+1}$.  But $ -1 \le \frac{x}{x^{2}+x+1}\le \frac{1}{3}$, by AM-GM.  This is because, by AM-GM,
\[ x^{2}+x+1 \ge 2|x|+x\]
$ = 3x, x>0$ or $ = |x|, x<0$.  (the $ x=0$ case can be disregarded because it yields $ \frac{x}{x^{2}+x+1}=0$.)
The given inequality turns into, by the identity,
\[ \frac{1}{k-1}<\frac{c}{c^{2}+c+1}-\frac{a}{a^{2}+a+1}-\frac{b}{b^{2}+b+1}\le \frac{1}{3}+1+1=\frac{7}{3}\]

Therefore, $ k-1>\frac{3}{7}\implies k>\frac{10}{7}$.

Indeed, for $ k>\frac{10}{7}$, letting $ a$ and $ b$ get arbitrarily close to $ -1$ and letting $ c$ get arbitrarily close to $ 1$ yields the desired inequality.[/hide]


[quote="Ligouras"]Prove that

$ \frac {[(2a + b - c)(2b + c - a)(2c + a - b)(2a + c - b)(2b + a - c)(2c + b - a)]^2}{2{}^1{}^1a{}^4b{}^4c{}^4} + \frac {(a + b)(b + c)(c + a)}{(a + b - c)(b + c - a)(c + a - b)}\geq10$[/quote]

Algebraization  Can be shown using the Muirhead theorem.

 :D


limit
trigonometry
calculus
calculus computations

Evaluate:
\[ z\cdot\prod_{k \equal{} 1}^{\infty}{\left(1 \minus{} \frac {z}{(2k \minus{} 1)\pi}\right)\left(1 \minus{} \frac {z}{2k\pi}\right)\left(1 \plus{} \frac {z}{k\pi}\right)}\]

$ \prod_{k \equal{} 1}^{\infty}{\left(1 \minus{} \frac {z}{(2k \minus{} 1)}\right)\left(1 \minus{} \frac {z}{2k}\right)\left(1 \plus{} \frac {z}{k}\right)}$
$ \equal{} \lim_{n \to \infty} \left(\prod_{k \equal{} 1}^n \left(1 \minus{} \frac {z}{k}\right)\left(1 \plus{} \frac {z}{k}\right) \times\prod_{i \equal{} n}^{2n}\left(1 \minus{} \frac {z}{i}\right)\right)$
$ \equal{} \frac {\sin \pi z}{\pi z} \times 2^{ \minus{} z}$
$ \equal{} \frac {\sin \pi z}{2^z \pi z}$

Thus,
\[ z\cdot\prod_{k \equal{} 1}^{\infty}{\left(1 \minus{} \frac {z}{(2k \minus{} 1)\pi}\right)\left(1 \minus{} \frac {z}{2k\pi}\right)\left(1 \plus{} \frac {z}{k\pi}\right)} \equal{} \frac {\sin z}{2^\frac {z}{\pi}} \ \ \left( \equal{} 2^{ \minus{} \frac {z}{\pi}} \sin z\right)\]


In a certain colony of bacteria, the number of bacteria doubles every day. The colony starts with 3 bacteria, and has 6 at the end of day 1, 12 at the end of day 2, and so on. What is the number of the first day which ends with the colony having more than 100 bacteria?

In this problem, I find it helps to know the powers of 2. We know that on the end of day [color=blue]2[/color], there are 12 bacteria. Our goal would be to multiply 12 by some power of 2 that can get the bacteria count as close to 100 and possible. 2^[color=red]3[/color] (or 8) fits the job perfectly.

Thus, on day 5 ([color=blue]2[/color]+[color=red]3[/color]) there would be 96 bacteria. The next day would be the first greater than 100, for [b]day 6[/b].

As quick verification:
[hide]
Day 0- 3
Day 1- 6
Day 2- 12
Day 3- 24
Day 4- 48
Day 5- 96
Day 6- 192
[/hide]


The Selective Examination  (1) for the National Team of China for IMO 2005
2005年中国数学奥林匹克国家集训队选拨考试（一）

Thanks. Could you post more as they come?

[quote="Valentin Vornicu"]Could someone translate them in English? :) Thanks![/quote]

Sure, I'll do it.

Here're the problems in English.

Problem 1 http://www.mathlinks.ro/Forum/viewtopic.php?t=30718
Problem 2 http://www.mathlinks.ro/Forum/viewtopic.php?t=30719
Problem 3 http://www.mathlinks.ro/Forum/viewtopic.php?t=30720
Problem 4 http://www.mathlinks.ro/Forum/viewtopic.php?t=30722
Problem 5 http://www.mathlinks.ro/Forum/viewtopic.php?t=30725
Problem 6 http://www.mathlinks.ro/Forum/viewtopic.php?t=30724

[quote="shobber"]开始考啦,效率可不低哪.......
呃呃......这些题我一道都不会作(也许第一题有点希望) :blush: .....
不过这些都是我没学过的东西,我还年轻,来日方长嘛! :lol:[/quote]
Young? How old are you?

其实对于竞赛来说,我已经很老了,都15了......有好多东西我都没学过,我不知道什么是数论,不知道排列组合......不过我只是很喜欢数学,所以我不在乎能否参加竞赛. 我每天抽出一些时间学我感兴趣的东西,所以我相信有一天我也能够解决那些问题!(尽管那时我可能已经大学毕业了......) :D

请问feizhenpeng，您是不是东北育才学校的费振鹏老师？


$\left(\sqrt{3+\sqrt{2}}\ \right)^x +\left(\sqrt{3-\sqrt{2}}\ \right)^x = 4$.
Find x.

[i]Edited by mod: $\LaTeX$[/i]

Can you post a full solution rather than just the answer?

Are you sure it is NOT   $ ( \sqrt 3 + \sqrt 2)^x + ( \sqrt 3 - \sqrt 2)^x = 4 $ ?
(In that case it will be not so hard )

Edited: please use $\LaTeX$, it makes the equations look much nicer :)

Added Later: Thank you. I am trying to learn $\LaTeX$ .. (See the picture on the right) But it is so hard to  use the key board with paws as it is... :)

That might be my fault. I'm 99% sure that I copied it correctly, I did check it carefully before editing it..

I will go by his original problem.

let a, b be $3+\sqrt{2}$ and $3-\sqrt{2}$  respectively.
Note ab = 7, a+b = 6.

(im not sure if this proof is right, but)

Suppose we have soln x.

by amgm, we can get at $16 \ge 7^x$.

Using $6^{x/2} = (a+b)^{x/2} \ge a^{x/2} + b^{x/2} = 4$ we can get at $6^x \ge 16$.

Now we have $6^x \ge 7^x$ contradiction.

Not a contradiction if $x \leq 0$.  There must be something wrong with your proof, though, because a graph shows a solution for x > 0.

[quote="Singular"]
$6^{x/2} = (a+b)^{x/2} \ge a^{x/2} + b^{x/2}$
 [/quote]

Thats not necessarily true (consider when $0<x<2$).


hi all
please anyone can help me to solve this limit
[img]http://www.mathsland.com/cgi-bin/mimetexcache/66813a5cced94114d3d3f9e130734b13.gif[/img]
thanks


calculus
integration
function
linear algebra
matrix

Can someone compare Mathematica, Maple, and Matlab so I know which to get?
Which is better at algebraic manipulation and indefinate and exact integration?
Thanks.

I have used MathCad for last 8 years... I tried Mathematica this winter but I failed to understand its basic conceptions.

At first, Mathematica is designed for solving differential equations. So it has powerful functions on symbolic computations. Maple can also perform symbolic computations well and it has much more pacakages than Mathematica.
Matlab is designed for matrix computations (MATtrix LABorlotary) and it cannot do symbolic computations. But it is very great at numerical computations, it can compute the inverse, determinant of a very very big matrix... Mathematica and Maple cannot do this.

Thanks, for what I need I think I'll get Mathematica.

Make sure you get the student version! ;)

For what you want, just about any CAS can do the job.  Even my Ti-89 can do algebra and differentiate/integrate.  I'm partial to Maple, but that's probably because I live on the West Coast.


geometry
circumcircle
parallelogram
geometry unsolved

ABC is a triangle.
AD, BE and CF are the median of the triangle.
AD, BE and CF are extended to point P, Q, and R, respectively.
If point C is contained on line PQ, point B is contained on line PR, and point C is contained on line PQ, then prove or disprove, that PA, QB and RC are the median of triangle PQR.

Using a parallel projection, project the general $ \triangle ABC$ into an equilateral $ \triangle ABC'.$ the midpoint $ F$ of $ AB$ stays in place, the midpoints $ D, E$ of $ BC, CA$ go into the midpoints of $ D', E'$ of $ BC', C'A,$ and the centroid $ G$ of $ \triangle ABC$ goes into the centroid $ G'$ of the equilateral $ \triangle ABC'.$ Let $ R'$ on the extension of $ C'F$ be arbitary, let $ R'A$ meet $ BE'$ at $ Q'$ and let $ Q'C'$ meet $ R'B$ at $ P'.$ Then

$ 180^\circ \equal{} \angle Q'AC' \plus{} \angle C'AB \plus{} \angle BAR' \equal{} 90^\circ \minus{} \frac {\angle P'Q'R'}{2} \plus{} 60^\circ \plus{} 90^\circ \minus{} \frac {\angle Q'R'P'}{2}\ \Longrightarrow$

$ \angle P'Q'R' \plus{} \angle Q'R'P' \equal{} 120^\circ,\ \ \angle R'P'Q' \equal{} 60^\circ$

$ P'$ is on the circumcircle of the $ \triangle BC'G',$ which cuts the extension of $ AD'$ at exacly one point, when the $ \triangle P'Q'R'$ is also equilateral, anticomplementary to the $ \triangle ABC'.$ Consequently, the original $ \triangle PQR$ (before the parallel projection) is anticomplementary to the $ \triangle ABC$ and $ PA, QB, RC$ are its medians.

[hide="Solution"]
Applying Menalaus's Theorem on $ \triangle AFG$ with transversal $ PBR$ yields that
\[ \frac {AB}{FB}\cdot \frac {FR}{RG}\cdot \frac {GP}{AP} \equal{} 1
\]
and Menalaus's Theorem on $ \triangle FBG$ with transversal $ RAQ$ yields that
\[ \frac {AB}{AF}\cdot \frac {RF}{GR}\cdot \frac {GQ}{BQ} \equal{} 1
\]
Yet, since $ AF \equal{} FB$, we have that $ \frac {AB}{FB} \equal{} \frac {AB}{AF}$, so
\[ \frac {AB}{FB}\cdot \frac {FR}{RG}\cdot \frac {GP}{AP} \equal{} 1 \equal{} \frac {AB}{AF}\cdot \frac {FR}{GR}\cdot \frac {GQ}{BQ}\implies \frac {GQ}{BQ} \equal{} \frac {GP}{AP}
\]
and thus $ AB\parallel QP$.  Simiarly, $ BC\parallel RQ$, so $ AQCB$ is a parallelogram.  Thus, $ BC \equal{} AQ$.  Similarly, $ RACB$ is a parallelogram, so $ RA \equal{} BC \equal{} AQ$, giving us that $ PA$ is a median of $ \triangle PQR$.  Similarly, $ RC$ and $ BQ$ are medians. [/hide]


With $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$. Find all values of $ k$ satifying this inequality is true
           $ (k\plus{}ab)(k\plus{}bc)(k\plus{}ca) \geq (k\plus{}1)^3$

[quote="2424"]With $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Find all values of $ k$ satifying this inequality is true
           $ (k \plus{} ab)(k \plus{} bc)(k \plus{} ca) \geq (k \plus{} 1)^3$[/quote]
Just a couple of notes.

1) We can set $ a \equal{} \frac{2x}{y\plus{}z}, b \equal{} \frac{2y}{x\plus{}z}, c \equal{} \frac{2z}{x\plus{}y}$.

2) Using $ 3u\equal{}a\plus{}b\plus{}c,3v^2\equal{}ab\plus{}bc\plus{}ca,w^3\equal{}abc$, the constraint becomes $ 3v^2\plus{}w^3\equal{}4$ and the ineq: $ k^3\plus{}3k^2v^2\plus{}w^3\plus{}9kv^4\minus{}6kw^3u \ge (k\plus{}1)^3$. Fixing $ v^2,w^3$ we only have to prove it when $ u$ attains max or min(because it is linear), so wlog $ a\equal{}b$. (This means $ x\equal{}y$ in 1)

3) Setting $ a\equal{}0,b\equal{}c\equal{}2$ it becomes $ k^2(k\plus{}4) \ge (k\plus{}1)^3 \iff k^2\minus{}3k\minus{}1 \ge 0 \iff k \ge \frac{3\plus{}\sqrt{13}}{2}$ or $ k \le \frac{3\minus{}\sqrt{13}}{2}$

4) We easily see $ \frac{4}{3} \ge \frac{ab\plus{}bc\plus{}ca}{3} \ge 1 \ge abc \ge 0$. That is $ \frac{4}{3} \ge v^2 \ge 1 \ge w^3 \ge 0$ in 2.

5) If $ a\equal{}b\equal{}x, c\equal{}y$ the constraint becomes $ yx \equal{} 2\minus{}x$, and the ineq: $ (k\plus{}x^2)(k\plus{}2\minus{}x)^2 \ge (k\plus{}1)^3$ when $ 0 \le k \le 2$.

My solution is:
$ (k\plus{}x^2)(k\plus{}2\minus{}x)^2 \ge (k\plus{}1)^3 \iff$
$ (x\minus{}1)^2(x^2 \minus{} 2(k \plus{} 1)x \plus{} k^2 \plus{} k \minus{} 1) \ge 0 \iff$
$ (x\minus{}(k\plus{}1))^2 \minus{} k\minus{}2 \ge 0$
If $ k\plus{}1 \in [0;2]$ we will require $ \minus{}k\minus{}2 \ge 0 \iff k \le \minus{}2$. But then $ k \notin [0;2]$. So the min of $ (x\minus{}(k\plus{}1))^2\minus{}k\minus{}2$ is attained either when $ x\equal{}0$ or when $ x\equal{}2$. $ x\equal{}0$ requires $ k^2\plus{}k\minus{}1 \ge 0$ and $ x\equal{}2$ requires $ k^2\minus{}3k\minus{}1 \ge 0$

So all $ k$ such that $ k^2\minus{}3k\minus{}1, k^2\plus{}k\minus{}1 \ge 0$. That is:
$ k \ge \frac{3\plus{}\sqrt{13}}{2}$ or $ k \le \frac{\minus{}1\minus{}\sqrt{5}}{2}$. QED :)

Let $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$. Prove that
  $ (ab\plus{}c)(bc\plus{}a)(ca\plus{}b) \geq 8$

[quote="2424"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Prove that
  $ (ab \plus{} c)(bc \plus{} a)(ca \plus{} b) \geq 8$[/quote]
Remark: find all the k such that
$ (ab\plus{}kc)(bc\plus{}ka)(ca\plus{}kb) \geq (k\plus{}1)^3$  with $ a,b,c>0$  such that $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$ :P

[quote="2424"]Let $ a,b,c > 0$ such that $ ab \plus{} bc \plus{} ca \plus{} abc \equal{} 4$. Prove that
  $ (ab \plus{} c)(bc \plus{} a)(ca \plus{} b) \geq 8$[/quote]
we have $ abc(a \plus{} b \plus{} c) \leq ab \plus{} bc \plus{} ca$ :P


Given $a,b,c$ positive real numbers, Prove that
$\max (|{{a-b} \over 2}|^3,|{{b-c} \over 2}|^3,|{{c-a} \over 2}|^3) \geq \sum {{a^4} \over {a+b}} - \sum {{a^6} \over {a^3+b^3}}$.


Find all integers $a,b,c,x,y,z$ such that \[a+b+c=xyz, \; x+y+z=abc, \; a \ge b \ge c \ge 1, \; x \ge y \ge z \ge 1.\]

Without loss of generality assume $ a\plus{}b\plus{}c\ge x\plus{}y\plus{}z$ (otherwise just switch $ (a,b,c)$ with $ (x,y,z)$), then $ a\plus{}b\plus{}c\ge abc$. We then have two cases:
[b]1. [/b]$ c\equal{}1$. Then, $ a\plus{}b\plus{}c\ge abc\Leftrightarrow 1\plus{}a\plus{}b\ge bc\Leftrightarrow 2\ge ab\minus{}a\minus{}b\plus{}1$
$ \Leftrightarrow 2\ge (a\minus{}1)(b\minus{}1)$. Then, because $ a\ge b$, we have either $ b\equal{}1$, or $ b\equal{}2$ then $ a\equal{}2$ or $ a\equal{}3$. Therefore, the only triplets $ (a,b,c)$ satisfying $ c\equal{}1$ and $ a\plus{}b\plus{}c\ge abc$ are $ (a,1,1), (2,2,1), (3,2,1)$.
[b]2. [/b]$ c\ge 2$. Then, $ a\plus{}b\plus{}c\ge abc\rightarrow a\plus{}b\plus{}c\ge 2ab\Leftrightarrow 1\ge ab\minus{}a\minus{}b\plus{}1\plus{}ab\minus{}c\leftrightarrow 1\ge (a\minus{}1)(b\minus{}1)\plus{}(ab\minus{}c)$  (*)
However, as $ a\ge b\ge c\ge 2$ then $ (a\minus{}1)(b\minus{}1)\ge 1$ and $ ab\minus{}c>1$ so (*) is false, so the case when $ c\ge 2$ fails.

Now, if $ (a,b,c)\equal{}(2,2,1)$ have $ x\plus{}y\plus{}z\equal{}4, xyz\equal{}5$ but then one of $ x,y,z$ is equal to $ 5$ and as $ x,y,z\ge 1$ we have $ x\plus{}y\plus{}z>4$ so we cannot have $ (a,b,c)\equal{}(2,2,1)$.
If $ (a,b,c)\equal{}(3,2,1)$ have $ x\plus{}y\plus{}z\equal{}6,xyz\equal{}6$ then $ x,y,z\in \{1,2,3,6\}$; none of $ x,y,z$ is $ 6$ as otherwise $ x\plus{}y\plus{}z>6$. Then one of $ x,y,z$ is $ 3$; as $ x\ge y\ge z$, $ x\equal{}3$ then $ y\plus{}z\equal{}3,yz\equal{}2$ so $ y\equal{}2,z\equal{}1$. Hence get $ (x,y,z)\equal{}(3,2,1)$.
If $ (a,b,c)\equal{}(a,1,1)$ have $ x\plus{}y\plus{}z\equal{}a$ and $ xyz\equal{}a\plus{}2$.  Because $ x\ge y\ge z$ and $ x\plus{}y\plus{}z\equal{}a$ we must have $ a\ge \frac{a}{3}$ then, if $ yz\ge 5$ then $ x\le \frac{a\plus{}2}{5}$ so $ \frac{a}{3}\le\frac{a\plus{}2}{5}$ so $ a\le 3$ which is impossible as $ a\plus{}2\equal{}xyz\ge 5x>5\rightarrow a>3$. Therefore, $ yz\le 4$
If $ yz\equal{}4$, we can have $ y\equal{}z\equal{}2$ giving $ x\plus{}4\equal{}a, 4x\equal{}a\plus{}2$ so $ x\equal{}2,a\equal{}6$ and get $ (a,b,c)\equal{}(6,1,1), (x,y,z)\equal{}(2,2,2)$, or we can have $ z\equal{}1,y\equal{}4$ then $ x\plus{}5\equal{}a,4x\equal{}a\plus{}2\rightarrow 3x\equal{}7$ which does not give an integer value for $ x$, so $ (y,z)\equal{}(4,1)$ is impossible.
If $ yz\equal{}3$, $ y\equal{}3,z\equal{}1$ then $ x\plus{}4\equal{}a, 3x\equal{}a\plus{}2$ so $ x\equal{}3,a\equal{}7$ and get $ (a,b,c)\equal{}(7,1,1), (x,y,z)\equal{}(3,3,1)$.
If $ yz\equal{}2$, $ y\equal{}2,z\equal{}1$ then $ x\plus{}3\equal{}a,2x\equal{}a\plus{}2$ so $ x\equal{}5,a\equal{}8$ so get $ (a,b,c)\equal{}(8,1,1), (x,y,z)\equal{}(5,2,1)$.
If $ yz\equal{}1$, $ y\equal{}z\equal{}1$ then $ x\plus{}2\equal{}a, x\equal{}a\plus{}2$ which is impossible.

Therefore, if $ a\plus{}b\plus{}c\ge x\plus{}y\plus{}z$ we get the following six-tuplets $ (a,b,c,x,y,z)$: $ (3,2,1,3,2,1);(6,1,1,2,2,2);(7,1,1,3,3,1);(8,1,1,5,2,1)$.If $ x\equal{}y\plus{}z\ge a\plus{}b\plus{}c$ by symmetry of the system and teh conditions on $ a,b,c,x,y,z$ we just switch $ (a,b,c)$ with $ (x,y,z)$.
Therefore, all the six-tuplets satisfying the conditions of the problem are:
$ (3,2,1,3,2,1);(6,1,1,2,2,2);(7,1,1,3,3,1);(8,1,1,5,2,1), (2,2,2,6,1,1);(3,3,1,7,1,1);(5,2,1,8,1,1)$

Consider $ (ab\minus{}1) (c\minus{}1)$ , $ (a\minus{}1)(b\minus{}1)$, $ (xy\minus{}1)(z\minus{}1)$, $ (x\minus{}1)(y\minus{}1)$.

When you add up these 4 numbers, we get that $ (ab\minus{}1) (c\minus{}1) \plus{} (a\minus{}1)(b\minus{}1) \plus{} (xy\minus{}1)(z\minus{}1) \plus{} (x\minus{}1)(y\minus{}1) \equal{} 4$.
Their non negativity is assured since we are given that \[ a\plus{}b\plus{}c\equal{}xyz,\; x\plus{}y\plus{}z\equal{}abc,\; a\ge b\ge c\ge 1,\; x\ge y\ge z\ge 1.\].
So we just have to check for I think  $ \binom {4\plus{}4\minus{}1}{4\minus{}1} \equal{} \binom {7}{3} \equal{} 35$. We can eliminate a lot of cases though using parity and primes condition etc.
Then we get the 7 6tuplet solutions.


ARML AMC AIME Asymptote trigonometry email USA(J)MO

yeah, here is a discussion topic for the 2008 ARML above i added a poll just to jazz it up.  :)

yes i am going

[size=25][color=yellow]not[/color][/size]

i doubt dr. ji, dr. chi, and dr. rickert will allow me to go because of my amazing 1 last year. but i think i've gotten a little better and maybe this year, if its a really ez test, i might pull a 2 or mmmmaaaayyyybbbbbeeee a 3 but unlikely.

[quote="cognos599"]um...... because shawn told me too.[/quote]

really?  too bad you have to proof  :jump:  :heli:  :lol:

yea ill go (like 100%)...maybe ill do better than last year.

This is weird, i only have three smiles but it keeps on saying i have four...

hmm odd...maybe ill just say some more words....

lalala...

I MAYYY go cause I MAYYYY actually come back to the U.S. within this year. Probably after the ARML though :(

hehe who else noticed the tiny "not" in unimpossible's post?

SOO HYUN COME BACK!!

and cognos, they'll let you go for sure. dont even say that  :roll:

I wishhhhh

RICHARD WE HAVE UNFINISHED KEMPS BUSINESS TO ATTEND TO

Whats the equavalent to ARML?  Like AIME questions?

[quote="cognos599"]and jeffrey bleeds too much ( as in shows his cards).[/quote]
:lol:

ANIRUDH, IT'S TIME FOR BED

[quote="the future"][quote="cognos599"]i guess they're easier... mainly cuz you only have like 5min/prob while in AIME you have 12 min/prob. and alot of ppl score 5+ on ARML.

btw fanatic: if we play kemps this year, you are not allowed to be with mwpl11, cuz there is no competition if you are. no offense mathcrazed, your signals stunk

"GODZILLA GODZILLA."

and jeffrey bleeds too much ( as in shows his cards).[/quote]

wont you guys have an odd number of people this year?
this indiana forum is making my posts go up...and thats making other people get the impression that im good at math, which im not compared to them.[/quote]
seconded :lol:

do you guys know asymptote? bahh random image!

[asy]unitsize(.5cm);
draw((5,0)--(0,0)--(5/2,5sqrt(3)/2)--(5,0)--(15/2,5sqrt(3)/2)--(5/2,5sqrt(3)/2));
draw((0,0)..(2,4)..(15/2,5sqrt(3)/2),orange+dotted+linewidth(2));
draw(scale(2.5)*unitcircle,blue+dashed+linewidth(1));
label("Hi!",(5,0),SE);[/asy]

Of course

[asy]unitsize(.5cm);
draw((5,0)--(0,0)--(2/2,5sqrt(3)/10)--(5,0)--(15/2,5sqrt(2)/2)--(5/2,5sqrt(2)/2));
draw((0,10)..(2,4)..(15/2,5sqrt(10)/2),blue+dotted+linewidth(10));
draw(scale(1.2)*unitcircle,blue+dashed+linewidth(15));
label("Whats the value of $\cos{\frac{\pi}{7}}\cos{\frac{2\pi}{7}}...\cos{\frac{6\pi}{7}}$",(10,0),SE);[/asy]

The quote button works nicely.
Hmm now its screwed up picture.  Oh well.

Whats alkashi bridge?  Famous bridge in Japan?

Hmm i dont care if you do those problems.  Its no big deal, but if you attempt it[hide="warning"]If you find yourself spending more than 10 mins doing this probelm^, please do not waste your time (like i spent more than 20 mins on it and i couldnt even get started)[/hide]

Wait and neat, isnt it like 12:10 PM in Japan?
Hmm that means its 12:10 AM here...dang I think I have to go to sleep.

My dad isn't going cause he has a "conference" in China... cough cough vacation at some hidden resort.

[quote="cognos599"]logic, don't lie, joseph had a 3 not a 1 which shows how much you lie, i doubt that you got a 2, since you lie so much. and plus i just looked at the team results. you told me that your team got an 18 on power and that you got 15 points and yushi got 3. which was of course a lie, because your actual score was a 13 and i wouldn't be surprised if yushi got 11 or 12 of the 13 points, because he is smarter than me let alone you.  :)[/quote]

I'm sorry for butting in here since I didn't go, but I would have to say OWNED.

Ah, but telling me that joseph had a 1 was a lie on your part.
Plus, I estimated that our team had a 18, I told you this before the results even came out.
Apparently I was pretty off...
All yushi did on power was check everyone's work and make a solution for #5, which
turned out to be a bogus proof. (I say bogus by incomplete)
You call that 11 or 12 points?

Anyways back on subject here...

I just retook that test and -----------------

I overestimated the difficulty of number 6.  Its actually only like 20 secs long question but you have to realize the trick right away.

Looking back at it, my score of 5 couldve easily been a 7 (number  2 and 6).
But then again its easier to say it looking back on it where theres no pressure and youve already seen the problems once...

but yea i personally was disappointed at my score (same score as last year and a 6 wouldve gotten me a medal lol).

Yeah, I totally p00ned ARML this year (p00ned is not a word, and therefore I never went to ARML).

But anyways, I'm pretty sure I'm going to ARML next year, as well as MustIt.

You guys will then have two more people to help you encourage *cough*make fun of*cough* logic812 (jk, jangsta).  And cognos and I can pull off our awesome Kemps signal ("I love Kristin!").  Saying that makes you feel awkward.

And I want to see what the future (the AoPS user, not the literal future) is like; he's one of the few on the Indiana forum that I don't know.

Come to think of it, it seems like most people on this forum are from Indiana!  (Note the sarcasm in that statement.)

hmm still waiting for ARML to upload copies of this year's tests..

and Taiwan kinda sorta but not really pwned on the relays in division A?

lmao "jangsta" XD <_<

on the oregon forum, i believe that they have the team and power. i can check if i have the individual, round and i guess i can like, um, .... , well idk asymptote but i can send you the problems and i can make the pictures on word or something and send them to you. but that is if i still have all of the problems, which i don't know if i do.  :(

I have all the problems (indiv.) so if anybody wants them, I'll post them.
(I love Kristin as a signal? pretty lame, if you ask me.)
MustIt is probably gonna get like a 1, maybeee a 2 (but I doubt it)
jangsta   :lol:
 :arrow: P.S.: MathAnd, please check your PMs

the questions are now available on the [url=http://arml.com/2008contest.php]ARML website[/url]

Cool, now I don't have to type/scan every problem. :lol:

Over in the grading room at Las Vegas, we were joking that the relays must have been easier in Chinese. They were hard for everyone, but the top Taiwan team had the single highest score anywhere on each relay.
In the overall results, Indiana was 22nd (behind 20 Div. A teams and Hong Kong). Before relays, that was a 3-way tie for 16th-18th.

What happened to the other teams? You've gone from three Indiana teams to one in just two years.

two Ind teams, except the former Indiana Silver lacked so many people (there were about 7 people I believe :lol: ) and the Silver team was put together with leftover Wisconsiners and Indiana Silver became Idis-consin.   :rotfl:   Thier actual name was Alternate Iowa I (Roman numeral).

[quote="unimpossible"]huh there was only one team?[/quote]
there was unofficially 2, but officially 1


Proton, kalosirthes!
defteron, einai kataplhktiko alla kai spanio se mathites pou den asxolountai me diagonismous na aresei h geometria. den tha arxiso na leo giati h geometria voithaei sth mathimatikh skepsi kai sthn katanohsh dyskolon enoion giati den tha teleiosoume pote. an thes th gnomh mou, doulepse sklhra pano sthn geometria-einai panemorfh!
triton, lypamai pou sto leo...alla prepei na allakseis eikona!oi palioteroi isos katalavainoun giati to leo....O nikos o romanidihs exei apo palia th sygekrimenh foto, ki epeidh einai kai argyros olybionikhs prepei na ton sevastoume...Plaka kano, o nick den tha exei provlhma!
tetarton, kapou yparxei ena topic opou o kyrios babhs (stergiu) exei proteinei arketa vivlia. borei na sou proteinei polla vivlia geometrias ki oxi mono. o kyrios babhs einai o katallhloteros na sou proteinei vivlia,epishs yparxei kai to http://www.stfu.gr (an leitourgei) checkare to.

auta

nick

[quote="george2"]Αγαπητοί φίλοι των μαθηματικών γεια σας!!!
Σήμερα βρήκα τη σελίδα και πραγματικά χάρηκα.Βέβαια δεν ασχολούμαι με τα μαθηματικά όσο εσείς (η φυσική είναι το χόμπυ μου) αλλά πιστεύω ότι μπορώ να μάθω πολλά στην παρέα σας.Επίσης αν και δεν έχω ασχοληθεί πολύ(μόνο την ύλη της Α' Λυκείου έχω κάνει από το σχολικό βιβλίο) μου αρέσει πάρα πολύ η γεωμετρία.Είδα ότι στη σελίδα υπάρχουν πάρα πολλά θέματα τα οποία και θα μελετήσω τις επόμενες ημέρες αλλά θα ήθελα να μου προτείνετε και κάποια καλά βιβλία γεωμετρίας (ελληνικά ή αγγλικά) ώστε σιγά σιγά να βελτιωθώ.

Φιλικά,
Γιώργος[/quote]

 Αγαπητέ Γιώργο , καλώς όρισες !

 Χαίρομαι πολύ που αγαπάς τη γεωμετρία ! Λοιπόν , επειδή τώρα δεν είμαι σπίτι αλλά σε διακοπές , τέλος Αυγούστου να μου γράψεις στο προσωπικό  email

  stergiu@otenet.gr

για να σου πω τι θα κάνεις. Φυσικά μέχρι τότε μπορεί να σου απαντήσουν και άλλα μέλη . Ό,τι σου προτείνουν , να ξέρεις ότι θα είναι καλό.
 Μπορείς επίσης να γυρίσεις λίγο πίσω σε αυτή τη σελίδα και να βρεις ένα topic με θέμα '' Βιβλία προετοιμασίας- Ολυμπιάδες'' , ΄οπου θα δεις και άλλα πολλά πράγματα.

  (Αυτά που σου είπε ο Νίκος Ράπανος παραπάνω να τα σκεφτείς σοβαρά  :) )

Ξεκίνα να διαβάζεις και τις ασκήσεις γεωμετρίας που βάζουμε για το βιβλίο. θα περιμέμουμμε με χαρά και νέες λύσεις ή βελτιώσεις.
 Σε χαιρετώ - Μπάμπης

Ευχαριστώ για το καλωσόρισμα
Δυστυχώς το link http://www.stfu.gr δεν λειτουργεί.Βέβαια στο topic "Βιβλία προετοιμασίας- Ολυμπιάδες" βρήκα αρκετά βιβλία αλλά το πρόβλημα με τα βιβλία γεωμετρίας είναι ότι τα περισσότερα είναι εξαντλημένα.Θα περιμένω λοιπόν μέχρι τον Αύγουστο...
Κύριε stergiou επειδή θα λείπω κι εγώ την τελευταία εβδομάδα του Αυγούστου μάλλον θα στείλω e-mail αρχές Σεπτεμβρίου.

Φιλικά,
Γιώργος