The largest one of numbers $ p_1^{\alpha_1}, p_2^{\alpha_2}, \cdots, p_t^{\alpha_t}$ is called a  $ \textbf{Good Number}$ of positive integer $ n$, if $ \displaystyle n\equal{} p_1^{\alpha_1} \cdot p_2^{\alpha_2}  \cdots p_t^{\alpha_t}$, where $ p_1$, $ p_2$, $ \cdots$, $ p_t$ are pairwisely different primes and $ \alpha_1, \alpha_2, \cdots, \alpha_t$ are positive integers. Let $ n_1, n_2, \cdots, n_{10000}$ be $ 10000$ distinct positive integers such that the $ \textbf{Good Numbers}$ of $ n_1, n_2, \cdots, n_{10000}$ are all equal.

Prove that there exist integers $ a_1, a_2, \cdots, a_{10000}$ such that any two of the following $ 10000$ arithmetical progressions $ \{ a_i, a_i \plus{} n_i, a_i \plus{} 2n_i, a_i \plus{} 3n_i, \cdots \}$($ i\equal{}1,2, \cdots 10000$) have no common terms.

Much easier than #2?

We'll show that for any set $S = \{s_1, s_2, \cdots, s_n\}$ of distinct positive integers sharing the same Good Number $t$, there exist disjoint arithmetic progressions of common differences $s_1, s_2, \cdots, s_n$ which are pairwise disjoint. This would clearly suffice. By convention, let's set the Good Number of $1$ to be $1.$

Before proving this, observe that if there exist pairwise disjoint arithmetic progressions of common differences $d_1, d_2, \cdots, d_k$, then there exist pairwise disjoint arithmetic progressions of common differences $xd_1, xd_2, \cdots, xd_k$ which are all contained in $x \mathbb{Z} + y$ for any $x, y \in \mathbb{N}.$

Let's now proceed by strong induction on the Good Number $t$ of the positive integers in $S$.

When $t = 1$, $\mathbb{Z}$ is an arithmetic progression that works.

Suppose the result holds when $t < z.$ When $t = z$, consider the set $T = \{\frac{s}{z} | s \in S\}.$ Notice that every number in $T$ has a Good Number which is strictly less than $z$. Therefore, we can let $T_1, T_2, \cdots, T_{z-1}$ be pairwise disjoint sets such that $T = T_1 \cup T_2 \cup \cdots \cup T_{z-1}$ and each number in $T_i$ has Good Number $i$ for each $1 \le i < z.$ By our observation in the second paragraph, there exists pairwise disjoint arithmetic progressions with common differences $z \cdot T_i$ which are all contained in $z \cdot \mathbb{Z} + i,$ for each $1 \le i < z.$ Taking the union of all these progressions over $1 \le i < z$ finishes. The induction is complete, and hence so is the problem.

$\square$


[b]If $ q$ is an integer that can be expressed as the sum of two integer squares, show that both $ 2q$ and $ 5q$ can also be expressed as the sum of two integer squares.[/b]

[b]Generalization:[/b]  If $ p, q$ are two integers that can be expressed as the sum of two integer squares, $ pq$ can also be expressed as a sum of two integer squares.

[hide="Proof"] Let $ p \equal{} a^{2}\plus{}b^{2}, q \equal{} c^{2}\plus{}d^{2}$.  Then

$ pq \equal{} (ab\pm cd)^{2}\plus{}(ad\mp bc)^{2}$ [/hide]
This result is well-known.

This problem was already posted before in the Bay Area AIME/AMC math marathon. And the answer is also a very well-known result. For these kinds of problems its usually good  to try out many things.

And its in the USSR Olympiad Problem Solving Book

those identities are established nicely using complex numbers and the fact that the norm is multiplicative


Four numbers, $a,b,c,d$ are picked, each from the set ${1,2,4,8}$, not necessarily distinct.  What is the probability $\frac a b = \frac c d$

Three distinct points are chosen on a regular nonagon.  What is the probability the triangle made with those vertices is scalene?

[hide="hint to 1"]take logs, then the problem is equivalent to choosing from 0,1,2,3 and a-b=c-d[/hide]

[hide="hint to 2"]how many isosceles are there?[/hide]


function
calculus
derivative
real analysis
real analysis unsolved

Hey guys...I got this pretty interesting problem for my Multivariable Calc class, but I can seen to solve it. Been trying to figure it out for a while now, any help would really be appreciated!

[img]http://www.eden.rutgers.edu/~afiedler/problem13.png[/img]

I think I need to show <y(s),y'(s)> = 0 (inner product) for some 0<s<1, but I can't figure out how to do that. I also thought maybe this has something to do with the fundamental theorem of calc, because if I integrate y'(t)dt from 0 to 1, I know that equals y(0) - y(1), which equals 0 because y(0)=y(1), but I can't think of how to relate that to the function and the derivative being tangent.

Well, thanks for any help!

Okay...
Suppose $f(t) = (f_1(t), f_2(t), \dots, f_n(t))$.
then $f'(t) = (f_1'(t), f_2'(t), \dots, f_n'(t))$
So
$\langle f(t),f'(t)\rangle = \sum f_i(t)f_i'(t)$
which reminds us that it is the derivative of $\frac{1}{2}\sum f_i(t)^2$.

Now we get the proof.
Let $g(x) = \|f(x)\|^2$, $g(0)=g(1)$ and by Rolle's Mean-value Theorem, there exists $\xi$ such that $g'(\xi) = 2\langle f(x),f'(x)\rangle = 0$


Let $ABC$ be an equilateral triangle. Let $P$ be a point on the side $AC$ and $Q$ be a point on the side $AB$ so that both triangles $ABP$ and $ACQ$ are acute. Let $R$ be the orthocentre of triangle $ABP$ and $S$ be the orthocentre of triangle $ACQ$. Let $T$ be the point common to the segments $BP$ and $CQ$. Find all possible values of $\angle CBP$ and $\angle BCQ$ such that the triangle $TRS$ is equilateral.

http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.APMO2002Problem3

Vo Duc Dien

We use turning homegenity   AEDB is cyclic so <ABE=<ADE.    and ADCF is cyclic so <ADE= <ACF. Absolutely the triangle ABE is as same as the triangle ACF. And The vertex A is the homegenity center E goes to F. and B goes to C  and we know if X and Y be on BC and EF such that       BX÷ XC  =  EY ÷YF We have AY is altitude XY.  So M and N have the situation of X and Y


Let $ a,b,c,d\in \mathbb{R}$:

Given that $ \left\{ \begin{array}{c}a^{2}+b^{2}= 1\\ c^{2}+d^{2}= 1\\ ac+bd = 0 \end{array}\right.$

Show that $ \left\{ \begin{array}{c}a^{2}+c^{2}=1\\b^{2}+d^{2}= 1\\ab+cd = 0 \end{array}\right.$

[hide]$ ac+bd=0 \implies |\frac{a}{b}|=|\frac{d}{c}|$
This, combined with the sums of squares equations yields $ a^{2}=d^{2}, b^{2}=c^{2}$.  From here it's easy.[/hide]

Or you can just try the obvious way:

[hide]From the first and second equations $ b=\sqrt{1-a^{2}}$ and $ d=\sqrt{1-c^{2}}$.  If we plug this into the third equation, we get:

$ ac+\sqrt{1-a^{2}}\sqrt{1-c^{2}}=0$
$ ac=-\sqrt{1-a^{2}}\sqrt{1-c^{2}}$
$ a^{2}c^{2}=(1-a^{2})(1-c^{2})$
$ a^{2}c^{2}=1-a^{2}-c^{2}+a^{2}c^{2}$
$ 0=1-a^{2}-c^{2}$
$ a^{2}+c^{2}=1$

Similarly, we can substitue $ a=\sqrt{1-b^{2}}$ and $ c=\sqrt{1-d^{2}}$ into the third equation and get $ b^{2}+d^{2}=1$.

Now, because $ a^{2}+b^{2}=1$ and $ a^{2}+c^{2}=1$, $ b=\pm c$.  If $ b=c$, then we can subtract $ ac+bc$ from $ ab+cd$:

$ ab+cd-ac-bd=ab+cd$
$ (a-d)(b-c)=ab+cd=0$

If $ b=-c$, we can add $ ac+bc$ to $ ab+cd$:

$ ab+cd+ac+bc=ab+cd$
$ (a+d)(b+c)=ab+cd=0$[/hide]

[quote="JBL"]There's a nice geometric interpretation, as well.[/quote]

Good point: now that you mention it, there IS a nice complex numbers way. ([hide]ie $ a+bi, d+ci$ each have norm $ 1$, and have real product...this occurs iff their imaginary parts are the same in absolute value, giving the desired result.[/hide]


calculus
integration
function
limit
real analysis
real analysis unsolved

Hallo to everybody, is it true the following statement?
Let be $ f: [a,b] \to[c,d]$ an invertible and Riemann integrable function on $ [a,b]$, so $ f^{\minus{}1}: [c,d] \to [a,b]$ is an integrable function on $ [c,d]$
I now that it is true in the case that f is a continuous function, but in general is still true?
Thanks to everybody bye

$ f$ is continuous at $ x_0$ $ \Leftrightarrow$ $ f^{ \minus{} 1}$ is continuous at $ y_0 \equal{} f(x_0)$.

Indeed, because $ f$ is surjective we have that

$ \lim_{y\to y_0} f^{ \minus{} 1}(y) \equal{} \lim_{x\to x_0} f^{ \minus{} 1}(f(x)) \equal{} x_0$.

and the other implication in the same way.

$ f$ is integrable $ \Leftrightarrow$ $ f$ is continuous almost everywhere.But since $ f$ and $ f^{ \minus{} 1}$ have the same number of discontinuities it follows that $ f^{ \minus{} 1}$ is continuous almost everywhere, therefore $ f^{ \minus{} 1}$ is also integrable.


Let $ ABC$ be an isosceles triangle with $ AB\equal{}AC$ and $ \angle A\equal{}20^\circ$. On the side $ AC$ consider point $ D$ such that $ AD\equal{}BC$. Find $ \angle BDC$.

This is a very well-known problem.

[hide="A pure geometry approach"]
Take point $ E$ on the opposite of $ D$ with respect to $ AB$ such that $ \triangle EDA\cong\triangle ABC$.

Note that $ \angle BAE\equal{}60^{\circ}$ and hence $ \triangle ABE$ is equilateral. Also note that $ \angle DEB\equal{}40^{\circ}$. Since $ ED\equal{}EB$, we have $ \angle EDB\equal{}70^{\circ}$.

It is easy to see that $ \angle BDE\equal{}30^{\circ}$.[/hide]

Take F inside the triangle so that BCF is equilateral, F belongs to the bisector of the angle <BAC, hence <BAF = 10 degs  ( 1 ) and
<ABF = <ABC - <FBC = 20 degs ( 2 ), then see that triangles ABF and BAD are congruent (s.a.s.), thence <ABD = <BAF = 10 degs and <BDC = 30 degs, q.e.d.

Best regards,
sunken rock


calculus
integration
analytic geometry
symmetry
geometry
calculus computations

consider the solid S, where the base of S is a circular disk of radius r. Parallel cross-sections perpendicular to the base are isosceles triangles with height h and unequal side in the base. Find the volume of S by setting up and solving the integral.[/quote][/hide][/code]

Set up the coordinate system so that the center of the base is at the origin, and the line along which $ S$ is highest is on the $ x$-axis.  Consider only the volume of $ S$ that's in the first quadrant, as by symmetry this is exactly one fourth the total volume.  The height is $ h$ on the $ x$-axis, $ 0$ along the edge of the circle, and changes linearly in between, so at $ (x,y)$ the height is $ h\left(1 \minus{} \frac {y}{\sqrt {r^2 \minus{} x^2}}\right)$.  Thus, the integral we're looking for is
\[ V \equal{} 4h\int_0^r\int_0^{\sqrt {r^2 \minus{} x^2}} \left( 1 \minus{} \frac {y}{\sqrt {r^2 \minus{} x^2}} \right) dydx \equal{} \frac {1}{2}h\pi r^2
\]
Could you have gotten this answer without integrating at all, using only geometry?


number theory
prime numbers
prime factorization
Dirichlet s Theorem
IMO
IMO 1977

Let $n$ be a given number greater than 2. We consider the set $V_n$ of all the integers of the form $1 + kn$ with $k = 1, 2, \ldots$ A number $m$ from $V_n$ is called indecomposable in $V_n$ if there are not two numbers $p$ and $q$ from $V_n$ so that $m = pq.$ Prove that there exist a number $r \in V_n$ that can be expressed as the product of elements indecomposable in $V_n$ in more than one way. (Expressions which differ only in order of the elements of $V_n$ will be considered the same.)

Nice :)

Lemma: there are $ \infty$ many prime numbers $ p \not\equiv 1 \mod n$.
Proof:
Assume that there are only finetely many of them and let their product be $ k$.
Then all prime factors of $ nk \minus{} 1$ must be $ \equiv 1 \mod n$ since this number is coprime with $ k$. But that would mean that $ \minus{} 1 \equiv nk \minus{} 1 \equiv 1 \mod n$ which is impossible for $ n > 2$.

Now we can tackle the problem:
There are only finetely many residue classes $ \mod n$, thus we can find two primes $ p,q$ with $ p \equiv q \not\equiv 1 \mod n$ and coprime with $ n$.
Let $ s$ be the smallest positive integer with $ p^s \equiv 1 \mod n$, then the same property holds for $ q$ too.
Now we have that $ p^s, q^s, pq^{s \minus{} 1} , p^{s \minus{} 1}q \equiv 1 \mod n$ are all indecomposable since all their nontrivial divisors are $ \not\equiv 1 \mod n$. But the product $ p^sq^s \equal{} (pq^{s \minus{} 1})(p^{s \minus{} 1}q)$ gives a number that is represented in two ways.

Similar to above but there are a little difference.By Dirichlet's theorem there are infinity many primes congurent to $-1$ modulo $n$ take two of them like $p,q$ then $(pq)(pq)=(p^2)(q^2)=p^2q^2 \in V_n$ obviously $pq,p^2,q^2$ are indecomposable in $V_n$ so we are done.

[quote=aakhri2205]hello!
my observation   [hide]if a number is said to be decomposable then
say 1+pn is decomposable  into (1+nk_1) and (1+nk_2) then [b]k_1 +k_2 divides k_1*k_2[/b][/hide]
this observation helps in the second part(which I proved similarly like @above :-D  :P[/quote]

is there some technical error in the obs pls reply


[quote="maryam rashin"]given triangle $ABC$,let $D$,$E$ be any points on $BC$.a circle through $A$ cuts the lines $AB$,$AC$,$AD$,$AE$ at the points $P$,$Q$,$R$,$S$ ,respectively.prove that
$\frac{AP.AB-AR.AD}{AS.AE-AQ.AC}=\frac{BD}{CE}$

rasti age masale bedim lozoomi dare ke manbaesho begim? :?:[/quote]
here is my solution:
albateh kheili kootah va kholasash kardam,va faghat ghadam haye akharesho tozih dadam.kafist sabet konim:$\frac{AP}{AQ}(\frac{AB-AX}{AT-AC})=\frac{BD}{CE}$
dar natije kafist sabet konim:$\frac{BX.sin(AQP)}{TC.sin(APQ)}=\frac{BD}{CE}$ $\rightarrow$ kafist neshan dahim $sin APQ=sinMBX=sinTEC$ hal neshan midam ke in do zavie barabarand.pass az kami mohasebe be in rabete miresim $ADE+AED+DAE=180$ (?) ke be vozooh dorost ast.
(tozihat:T mahal barkhorde dayereye mohitie QSE ba khate AC ast.va N mahale barkhorde haman dayere ba khate BC.X mahale barkhorde dayereye mohitie PRD ba khate BC va X mahale barkhorde an ba khate AB ast.)
age jaeesho motevajjeh nashodid begin tozih midam.


In quadrangle $ABCD$ we have equalities:
$\angle BAC = \angle DAC = 90\circ-\angle BCD$.
Prove that:
$\angle ABD = 2\angle ACD$.

$AC$ is bisector of $\angle BAD$ and $\angle BCD=90^{0}-\angle BAC=90^{0}-\frac{\angle BAD}{2}$. Hence $C$ is the excenter of $ABD$. So: $\angle ABD=2\angle ACD$.


function
real analysis
real analysis unsolved

Let $ f: R\to R$ be a function with this condition

If $ \sum a_n$ is convergent then $ \sum f(a_n)$ is convergent too.

show that there is a $ k\in R$ and a neighborhood at zero such that  in this  neighborhood $ f(x) \equal{} kx$

Taking $ a_n\equal{}0$, we see $ f(0)\equal{}0$.

1) There exists $ A>0$ such that $ f(\minus{}x)\equal{}\minus{}f(x),\forall x\in(0,A)$
Suppose not. We can find $ a_n\downarrow0$ with $ c_n\equal{}f(a_n)\plus{}f(\minus{}a_n)\ne0$.
By taking subsequence, we can assume WLOG $ c_n>0$.
Let $ b_n\in\mathbb N$ such that $ b_nc_n>1$.
Now we construct $ x_n$ as follows:
repeat $ a_1,\minus{}a_1$ by $ b_1$ times; repeat $ a_i,\minus{}a_i$ by $ b_i$ times, and so on.
It is easy to see $ \sum x_n\rightarrow0$, but $ \sum f(x_n)>\sum_{n\equal{}1}^k b_nc_n>k,\forall k\in\mathbb N$ diverges. Contradiction!

2) There exists $ A>0$ such that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y),\forall x,y>0,x\plus{}y<A$.
Suppose not. We can find $ a_n\downarrow 0$ and $ b_n\downarrow0$ with
$ c_n\equal{}f(a_n\plus{}b_n)\minus{}f(a_n)\minus{}f(b_n)\ne0$.
Let $ d_n\in\mathbb N$ such that $ d_n|c_n|>1$.
Now we construct $ x_n$ as follows:
if $ c_i>0$ repeat $ a_i\plus{}b_i,\minus{}a_i,\minus{}b_i$ by $ d_i$ times;if $ c_i<0$ repeat $ \minus{}a_i\minus{}b_i,a_i,b_i$ by $ d_i$ times, and so on.
It is easy to see $ \sum x_n\rightarrow0$, but by using 1), we have
$ \sum f(x_n)>\sum_{n\equal{}1}^k d_nc_n>k,\forall k\in\mathbb N$ diverges. Contradiction!

3) $ f(x)$ is continuous at $ 0$.
Suppose not. There exists $ A>0$ and series $ a_n\downarrow0$ such that $ |f(a_n)|>A,\forall n$.
We can replace $ a_n$ with $ \minus{}a_n$  and taking subsequence when necessary.
Then we can assume $ f(a_n)>A$ and $ |a_n|\le\frac1{n^2}$.
Now we see $ \sum a_n$ converges but $ \sum f(a_n)>kA,\forall k\in\mathbb N$ diverges. Contradiction!

With Cauchy equation and continuity, we can derive $ f(x)\equal{}kx$ at neighborhood of zero.
Q.E.D.


function
algebra
polynomial
number theory
greatest common divisor
calculus
relatively prime

Let $\frac{f}{PQ}$ be rational functions such that deg f < deg PQ and $P,Q,F$ are [u][b]relatively prime[/b][/u]

[Unique polynomials A and B should be found such that $\frac{f}{PQ}$=$\frac{A}{P}+\frac{B}{Q}$ where deg A < deg P and deg B < deg Q ]

Since P and Q are relatively prime , it is possible to find polynomials such that  [i][u][b]$CP+DQ = 1$[/b][/u][/i] [C,D are not unique] $\leq ====???$

What does relatively prime mean and also how did the obtain the step in arrow

.
.

If two numbers are relatively prime, then they do not share any common factors.  The same applies to polynomials.

For example:  24 and 49 are relatively prime, but 49 and 84 are not.

$x^{2}+2x+1$ is relatively prime to $x+4$; $x^{3}+3x^{2}+3x+1$ is not relatively prime to $x+1$.

If I'm not mistaken, $CP+DQ = 1$ represents a linear diophantine equation. Since P and Q are relatively prime and their gcd equals the right hand side of the equation, it is possible to find integers C and D such that the equation holds (and there are infinitely many solutions)

[quote="bos1234"]ok thanks.

So how would I be able to split this into partial fractions?

$\int \frac{du}{u^{2}-1}\,dx$ where u = $e^{-}x$[/quote]

This is calculus, as far as I know, so I have split it into the calculus section.

An example of a partial fraction:

$\frac{4x+5}{x^{2}+5x+6}=\frac{A}{x+2}+\frac{B}{x+3}$

$4x+5=A(x+3)+B(x+2)$

$(A+B)x+(3A+2B)=4x+5$

Here, we have a system of equations to solve.

$A+B=4$ and $3A+2B=5$

$2A+2B=8$ and $3A+2B=5$

$A=-3$ and $B=7$

Therefore, $\frac{4x+5}{x^{2}+5x+6}=\frac{7}{x+3}-\frac{3}{x+2}$.


induction
absolute value
algebra proposed
algebra

Let $A\subset \{x|0\le x<1\}$ with the following properties:
1. $A$ has at least 4 members.
2. For all pairwise different $a,b,c,d\in A$, $ab+cd\in A$ holds.
Prove: $A$ has infinetly many members.

I'm trying induction but it isn't going well...

For distinct $a,b,c,d$ we have

$\begin{cases}ab+cd\in A\\ ac+bd\in A\\ ad+bc\in A\end{cases}$

We'll prove that at least one of those three numbers must always be different from $a,b,c,d$, which will automatically imply that $A$ is infinite (since we'll always be able to construct a number not "previously" contained in it).

Statement 1: Numbers $ab+cd, ac+bd, ad+bc$ are pairwise distinct.
Proof: Assume $ab+cd=ac+bd$. That gives $ab-ac=bd-cd\iff a(b-c)=d(b-c)\iff (a-d)(b-c)=0$. That's, however, impossible, since $a\neq d$ and $b\neq c$. Similarly for other combinations.

WLOG assume $ab+cd=a$.

Statement 2: $ab+cd=a\implies ac+bd\neq d$.
Proof: Assume $ac+bd=d$. Then $a-d=(ab+cd)-(ac+bd)=a(b-c)-d(b-c)=(a-d)(b-c)$. That would imply either $a-d=0$, which is impossible as $a\neq d$, or $b-c=1$, which is impossible as $0\leqslant b<1,0\leqslant c<1$.

Therefore, in the "worst case scenario", we can have the following possibilites: (1) $ac+bd=b\land ad+bc=c$, (2) $ac+bd=b\land ad+bc=d$, (3) $ac+bd=c\land ad+bc=b$, (4) $ac+bd=c\land ad+bc=d$. Case (1) is impossible as $ab+cd=a\land ad+bc=c\implies (a-c)(b-d)=a-c$ - confer the proof of Statement 2. Case (4) is impossible for the similar reason: $ac+bd=c\land ad+bc=d\implies (a-b)(c-d)=c-d$.

As for Case (2), we have

\begin{eqnarray}ab+cd &=& a\\ ac+bd &=& b\\ ad+bc &=& d\end{eqnarray}

Subtracting $(2)$ from $(1)$, $(3)$ from $(1)$ and $(3)$ from $(2)$ we get

$\begin{cases}(a-d)(b-c)=a-b\\ (a-c)(b-d)=a-d\\ (a-b)(c-d)=b-d\end{cases}$

Multiplying all the equations and cancelling $(a-b)(a-d)(b-d)\neq 0$, we get

$(b-c)(a-c)(c-d)=1$

which is impossible since the absolute value of all the factors is strictly less than $1$.

Similar reasoning for the remaining case.

Thus is proven that at least one of $ab+cd, ac+bd, ad+bc$ must always be different from any of $a,b,c,d$, hence $A$ is infinite.

[quote="Farenhajt"]
Thus is proven that at least one of $ab+cd,ac+bd,ad+bc$ must always be different from any of $a,b,c,d$, hence $A$ is infinite.[/quote]

This implies that $|A| > 4$ but why does it have to be infinite? Why couldn't it be equal to $e$ and the same thing with $e$ only gives values among $a,b,c,d,e$?

There is maybe some maximality/minimality argument missing.

Pierre.


[quote="BanishedTraitor"][hide=" Or"] Rearrangement Inequality[/hide][/quote] how exactly?

Rearrangement:

$\frac{1}{x_{1}}\le \frac{1}{x_{2}}$ etc.

I don't think rearrangement works here.
It's true that you have $\frac{1}{x_{1}}\leq \frac{1}{x_{2}}...\leq \frac{1}{x_{n}}$
but $x_{n}\leq x_{n-1}...\leq x_{1}$ and neither sum on each side of inequality is directed or reversed.

[quote="srulikbd"]
$\Large\text{Prove that, for real numbers }x_{1}\ge x_{2}\ge \cdots \ge x_{n}> 0,$\[\Large\frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+\cdots+\frac{x_{n-1}}{x_{n}}+\frac{x_{n}}{x_{1}}\le \frac{x_{2}}{x_{1}}+\frac{x_{3}}{x_{2}}+\cdots+\frac{x_{n}}{x_{n-1}}+\frac{x_{1}}{x_{n}}.\]there is an easy solution with [hide]induction[/hide] any other solution?[/quote]
The result follows from the identity: \[\sum_{cyc}\frac{x_{1}}{{x_{2}}}-\sum_{cyc}\frac{x_{2}}{x_{1}}=\Large\left|\frac{\prod_{cyc}(x_{1}-x_{2})}{\prod_{cyc}x_{1}}\right|\]


[hide="Hint 1"]For $ a\plus{}b\plus{}c\equal{}\pi$, we have $ \tan a \tan b \tan c \equal{} \tan a \plus{} \tan b \plus{} \tan c$.[/hide]
[hide="Hint 2"]The given equation is equivalent to
$ \tan 2x \cdot \tan (\pi \minus{} 3x) \cdot \tan (\pi \minus{} 5x) \equal{} \tan 2x \plus{} \tan (\pi \minus{} 3x) \plus{} \tan (\pi \minus{} 5x)$.[/hide]

[quote="archimedes1"][hide="Hint 1"]For $ a \plus{} b \plus{} c \equal{} \pi$, we have $ \tan a \tan b \tan c \equal{} \tan a \plus{} \tan b \plus{} \tan c$.[/hide]
[hide="Hint 2"]The given equation is equivalent to
$ \tan 2x \cdot \tan (\pi \minus{} 3x) \cdot \tan (\pi \minus{} 5x) \equal{} \tan 2x \plus{} \tan (\pi \minus{} 3x) \plus{} \tan (\pi \minus{} 5x)$.[/hide][/quote]

Thank archimedes1 very much.

[quote="archimedes1"][hide="Hint 1"]For $ a \plus{} b \plus{} c \equal{} \pi$, we have $ \tan a \tan b \tan c \equal{} \tan a \plus{} \tan b \plus{} \tan c$.[/hide][/quote]
A stronger statement is that $ a \plus{} b \plus{} c \equal{} n \pi \Leftrightarrow \tan a \tan b \tan c \equal{} \tan a \plus{} \tan b \plus{} \tan c$.  Your direction doesn't give the left arrow, which is necessary to find all solutions.

[quote="thanhnam2902"]$ \tan2x.\tan3x.\tan5x \equal{} \tan2x \minus{} \tan3x \minus{} \tan5x$[/quote]

$ \tan2x.\tan3x.\tan5x \equal{} \tan2x \minus{} \tan3x \minus{} \tan5x\equal{}\minus{}\tan3x\minus{}\frac{\sin3x}{\cos2x\cos5x}$ ----->

$ \sin2x\sin3x\sin5x\equal{}\minus{}\sin3x(\cos3x\plus{}\cos5x\cos2x)$ -----> $ \sin3x(\sin2x\sin5x\plus{}\cos2x\cos5x\plus{}cos3x)\equal{}\sin3x(\cos(2x\plus{}5x)\plus{}\cos3x)\equal{}2\sin3x\cos5x\cos2x\equal{}0$  ----->
$ \cos2x\not \equal{}0, \cos3x \not \equal{}0,cos5x\not \equal{} 0$ ----->
 $ \sin3x\equal{}0$ ----> $ x\equal{}\frac{\pi k}{3}$


1) Find the minimum value of

$(u-v)^{2}+(\sqrt{2-u^{2}}-\frac{9}{v})^{2}$

for 0 < u < sqrt 2 and v>0.

Comment: This was in the zeitz book, so is there a "clean" way to do this that is clever and doesn't just brute force it with partial derrivatives?

2) Let $a_{0}$ be any real number greater than 0 and less than 1. Then define the sequence $a_{1}, a_{2}, a_{3}..$,  by $a_{n+1}=\sqrt{1-a_{n}}$. Show that the limit as n goes to infinity is $\frac{\sqrt 5-1}{2}$, no matter what value was chosen as $a_{0}$.

3) Find the minimum value of $\sum_{k=1}^{10}\sqrt{(2k-1)^{2}+{a_{k}}^{2}}$ where $a_{1}+a_{2}...+a_{10}=17.$
Comment: what does the expression in the sum remind you of geometrically?

[quote="me@home"][hide="The idea that solves problem 1 is very similar to the idea that solves a problem in the Green Book"]
Let $D=\sqrt{(u-v)^{2}+(\sqrt{2-u^{2}}-\frac{9}{v})^{2}}$, i.e. $D$ is the distance from a point with x coord u on the circle $x^{2}+y^{2}=2$ and a point with x coord v the hyperbola $xy=9$.  Clearly, the minimum is distance is $3-\sqrt{2}$, so the min of $D^{2}=(3-\sqrt{2})^{2}$[/hide][/quote]
Wow, I need to go relearn the distance formula...

I tried the exact same thing, but really messed up the conics. :roll:

Can the same idea be used to solve 3 though?

[quote="Phelpedo"]3 is a very well known AIME problem from 1991.

#2- the problem becomes fairly trivial if you can prove that the limit exists because $\lim_{n \rightarrow \infty}a_{n}= \lim_{n \rightarrow \infty}a_{n+1}$.[/quote]
Thanks for having that exchange with towersfreak, it cleared up some of my questions about that problem.

for #3 think of the sequence $b_{n}$ such that $b_{n}=a_{n}+(2n-1)i$ then think about the triangle equality.


algorithm
modular arithmetic
induction
number theory
relatively prime

You wish to color each of the integers $ 0,1, 2, \dots,n$ red or blue, insuch a way that
[list]a) both colors are used, and
b) integers that differ by $ 7$ or $ 11$ have the same color.
[/list]
Can you do this for all $ n$? If not, what is the largest value of $ n$ for which it is possible? [i]Prove your answer.[/i]

Here's the general case.

[hide]First, say $ m$ and $ k$ are not relatively prime, so they share some common divisor $ g>1$. Then, we can color all multiples of $ g$ red and all non-multiples of $ g$ blue, which uses both red and blue since $ g>1$. Then, any two integers that differ by a $ m$ or $ k$ differ by a multiple of $ g$, and in the coloring, any two integers that differ by $ g$ are the same color, so it follows that any two integers that differ by a multiple of $ g$ are the same color. Thus, this coloring is valid, and we can color for all positive integers $ n$.

Now, assume that $ m$ and $ k$ are relatively prime. We claim that we have a coloring if and only if $ n\le m\plus{}k\minus{}3$. Without loss of generality, let $ m<k$ ($ m\equal{}k$ is possible only when $ m\equal{}k\equal{}1$ since they are relatively prime, in which case the claim is trivial). We will show that if $ n\ge m\plus{}k\minus{}2$, we have that $ 0,1,\ldots,m\minus{}1$ are all the same color, which implies that only one color can be used.

By the Division Algorithm, we can write $ k\equal{}am\plus{}b$, and since $ m<k$ we have $ a\ge1$. Note that $ \gcd(b,m)\equal{}1$, which follows by the Euclidean Algorithm. Therefore, $ \{0,b,2b,\ldots,(m\minus{}1)b\}$ is a complete set of residues modulo $ m$. Consider $ i$ such that $ ib\equiv\minus{}1\pmod{m}$ (in other words, the inverse of $ b$ modulo $ m$), which must exist since there are $ m$ terms in $ \{0,b,2b,\ldots,(m\minus{}1)b\}$, which are pairwise distinct modulo $ m$. Then, apply a frame shift on this set so that the terms are ordered as follows: $ \{(i\plus{}1)b, (i\plus{}2)b,\ldots,(m\minus{}1)b,0,b,2b,\ldots,ib\}$. Start by coloring an arbitrary term that is congruent to $ (i\plus{}1)b$ modulo $ m$, without loss of generality let it be red. This will then make all terms congruent to $ (i\plus{}1)b$ modulo $ m$ red. Consider the remainder $ r$ upon division by $ m$, between $ 0$ and $ m\minus{}1$ inclusive. Note that this is not $ m\minus{}1$, since $ ib\equiv m\minus{}1\pmod{m}$. Thus, $ r\le m\minus{}2$. We now have that $ r\plus{}k\equal{}am\plus{}b\plus{}r\le am\plus{}b\plus{}m\minus{}2 \equal{} m\plus{}k\minus{}2\le n$. Additionally, $ r\plus{}k\equiv (i\plus{}1)b\plus{}b\equiv (i\plus{}2)b\pmod{m}$. Thus, we get that all of the terms congruent to $ (i\plus{}2)b$ modulo $ m$ are red. We can show in a similar way that if all of the terms congruent to $ zb$ modulo $ m$ are red, where $ z\neq i$, so are the terms congruent to $ (z\plus{}1)b$ modulo $ m$. Now, a simple induction shows that all terms congruent to $ (i\plus{}3)b,(i\plus{}4)b,\ldots,(m\minus{}1)b,mb\equal{}0,b,\ldots ib$ modulo $ m$ are red as well, from which we get that every term is red (since these multiples of $ b$ are pairwise distinct modulo $ m$). Thus, we cannot have a valid coloring when $ n\ge m\plus{}k\minus{}2$.

Now, consider $ n\le m\plus{}k\minus{}3$; we claim that we can find a valid coloring. As before, let $ k\equal{}am\plus{}b$ with $ a\ge1$ and $ \gcd(b,m)\equal{}1$, and consider $ \{0,b,2b,\ldots,(m\minus{}1)b\}$, which are pairwise distinct modulo $ m$. Apply a similar frame shift so that we have $ \{(i\plus{}1)b, (i\plus{}2)b,\ldots,(m\minus{}1)b,0,b,2b,\ldots,ib\}$ defining $ i$ in the same way as above. Now, define $ j$ such that $ 0\le j\le m\minus{}1$ and $ jb\equiv\minus{}2\pmod{m}$. Consider the coloring such that every integer whose modulo $ m$ residue is $ (i\plus{}1)b,(i\plus{}2)b,\ldots,jb$ is red, and every integer whose residue is $ (j\plus{}1)b,\ldots,ib$ is blue. We claim that this is valid. Clearly, we're using both colors, because $ i\neq j$ (again, the set of multiples of $ b$ is a complete set of residues modulo $ m$). Also, any two integers that differ by $ m$ are the same color, so it is left consider any two integers that differ by $ k\equal{}am\plus{}b$. Note that if two integers differ by $ am\plus{}b$, one is congruent to $ zb$ and the other $ (z\plus{}1)b$ modulo $ m$, which are the same color for all $ z$ except $ z\equal{}i$ and $ z\equal{}j$. However, note that the smallest integer congruent to $ ib$ modulo $ m$ is $ m\minus{}1$, and $ m\plus{}k\minus{}1$ is not in the set of numbers we wish to color. Thus, there do not exist two integers, the first congruent to $ ib$ and the second $ (i\plus{}1)b$ modulo $ m$ that differ by $ am\plus{}b$. Similarly, since the smallest integer congruent to $ jb$ modulo $ m$ is $ m\minus{}2$, and $ m\plus{}k\minus{}2$ is not in the set of numbers we wish to color, and the same is true for two integers congruent to $ jb$ and $ (j\plus{}1)b$ modulo $ m$. Therefore, this is a valid coloring, and we're done.[/hide]


Find the positive integers(b,n) with n is as small as possible to satisfy:
log(3)(2008)(base 3)=log(b)n(base b)
PS: I found out b must be=3, n=2008 is the smallest out of possible values of n! But I'm not sure since it seems to be obvious!

Since this question seems obvious, I'll give an overly verbose way to solve it:

We want $ b^{2008} \equal{} 3^n$ with $ n$ as low as possible. Since $ b$ and $ n$ are positive integers, the key is to invoke the following corollary of the FTA: Two positive integers are equal if and only if their prime factorizations are the same. If $ b$ is the product of at least two distinct prime numbers, then the LHS is a product of two prime powers and the RHS is a prime power, contradiction. So $ b$ has to be a prime power, namely a power of $ 3$. Let $ b \equal{} 3^k$, $ k$ a positive integer.

We get $ 3^{2008k} \equal{} 3^n$. Equivalently, $ 2008k \equal{} n$. $ k$ is a positive integer, and its minimum value is $ 1$. This gives the minimum value of $ n$: $ 2008$. Consequently $ b \equal{} 3^1 \equal{} 3$.

Thanks for your solution, Mellow Mellon! My solution is slightly different from you but the same result. And how do you prove the property:b^2008=3^n? I don't think it's obvious or I'm too stupid to realize that! :) . I substituted like this: log(3)(2008)*log(2008)b=log(2008)(n). Now I have: log(3)b=log(2008)n and the smallest value of n is: 2008(since n is a positive integer)


Calculate $ \tan 84^{\circ}\ \cdot\ \tan 24^{\circ}\ \minus{}\ \tan 48^{\circ}\ \cdot\ \tan 12^{\circ}$.

I think it's 4. According to this,

[url]http://www27.wolframalpha.com/input/?i=tan+%2884%29+tan+%2824%29+-+tan+%2848%29+tan+%2812%29[/url]

lol

It is very nice that programme, [b]modularmarc101[/b].
In my proof I use only the known formulas.

$ \left\|\begin{array}{cc} \tan 84^{\circ}\cdot \tan 24^{\circ} \equal{} \frac {\sin 84^{\circ}}{\cos 84^{\circ}}\cdot \frac {\sin 24^{\circ}}{\cos 24^{\circ}} \equal{} \frac {\cos 60^{\circ} \minus{} \cos 108^{\circ}}{\cos 60^{\circ} \plus{} \cos 108^{\circ}} \equal{} \frac {\frac 12 \minus{} \frac 14(1 \minus{} \sqrt {5})}{\frac 12 \plus{} \frac 14(1 \minus{} \sqrt {5})} \equal{} 2 \plus{} \sqrt {5} \\
 \\
\tan 48^{\circ}\cdot \tan 12^{\circ} \equal{} \frac {\sin 48^{\circ}}{\cos 48^{\circ}}\cdot \frac {\sin 12^{\circ}}{\cos 12^{\circ}} \equal{} \frac {\cos 36^{\circ} \minus{} \cos 60^{\circ}}{\cos 36^{\circ} \plus{} \cos 60^{\circ}} \equal{} \frac {\frac 14(1 \plus{} \sqrt {5}) \minus{} \frac 12}{\frac 14(1 \plus{} \sqrt {5}) \plus{} \frac 12} \equal{} \sqrt {5} \minus{} 2\end{array}\right\|\Longrightarrow$ $ \tan 84^{\circ}\cdot \tan 24^{\circ} \minus{} \tan 48^{\circ}\cdot\tan 12^{\circ} \equal{} 4$


[quote="csbonny"]Let a,b,c be positive real numbers
prove that
\[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \le \frac {9}{4}
\]
[/quote]
Try $ a\equal{}b\equal{}c\equal{}\frac{1}{4}.$ :wink:

[quote="csbonny"]Let a,b,c be positive real numbers a+b+c=1
prove that
\[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \le \frac {9}{4}
\]
[/quote]
Try $ a\equal{}b\rightarrow\frac{1}{2}$ and $ c\rightarrow0^\plus{}.$ :wink:

[quote="arqady"][quote="csbonny"]Let a,b,c be positive real numbers a+b+c=1
prove that
\[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \le \frac {9}{4}
\]
[/quote]
Try $ a \equal{} b\rightarrow\frac {1}{2}$ and $ c\rightarrow0^ \plus{} .$ :wink:[/quote]

sorry again it must be \[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \ge \frac {9}{4}
\]

[quote="csbonny"]Let a,b,c be positive real numbers a+b+c=1
prove that
\[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \ge \frac {9}{4}
\]
[/quote]
\[ L.H.S. \equal{} \sum_{cyc}\frac {a^2}{ab^2 \plus{} ac} \geq \frac {(a \plus{} b \plus{} c)^2}{ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} ab \plus{} bc \plus{} ca}
\]
It's remain to prove: $ 4(a \plus{} b \plus{} c)^3 \geq 9(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} 9(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)$
$ \Leftrightarrow 4\sum_{cyc}a^3 \plus{} 3\sum_{cyc}a^2b \geq 6\sum_{cyc}a^2c \plus{} 3abc$
but it's obviously true by am-gm; $ 3a^3 \plus{} 3ac^2 \geq 6a^2c$ and $ a^3 \plus{} b^3 \plus{} c^3 \geq 3abc$

apollo, I think, for the following inequality your method doesn't work.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ a\plus{}b\plus{}c\equal{}6.$ Prove that
\[ \frac{a}{b^2\plus{}c}\plus{}\frac{b}{c^2\plus{}a}\plus{}\frac{c}{a^2\plus{}b}\geq1\]

[quote="csbonny"]Let a,b,c be positive real numbers a+b+c=1
prove that
\[ \frac {a}{{\mathop b\nolimits^2 \plus{} c}} \plus{} \frac {b}{{\mathop c\nolimits^2 \plus{} a}} \plus{} \frac {c}{{\mathop a\nolimits^2 \plus{} b}} \ge \frac {9}{4}
\]
[/quote]

Where you getting all these nice problems from?


If $ z$ is a complex number satisfying the equation $ |z \minus{} (1\plus{}i)|^2 \equal{} 2$, and $ w \equal{} \frac{2}{z}$, what is the locus produced by $ w$ in the complex $ w$-plane?

[hide="Solution"]Let $ z\equal{}a\plus{}bi$.

$ |(a\minus{}1)\plus{}(b\minus{}1)i|\equal{}\sqrt{2}$

If we think of $ (a\minus{}1)\plus{}(b\minus{}1)i$ as $ a\plus{}bi\equal{}z$ shifted down one and to the left one, we can see that the locus of $ z$ is a circle centered at $ \minus{}1\minus{}i$ in the complex plane, with radius $ \sqrt{2}$.

Now notice that the locus of $ w$ is simply the conjugate of the locus of $ z$ inverted around the unit circle and dilated by a factor of $ 2$. If we pick a random complex number $ z$, like $ z\equal{}2\plus{}2i$ (which is valid, because $ |1\plus{}i|\equal{}\sqrt{2}$), and invert it, its magnitude becomes $ \frac{\sqrt{2}}{2}$. If we consider the center of the circle, $ \minus{}1\minus{}i$, we see upon inversion and conjugation that it maps to $ \minus{}\frac{1}{2}\plus{}\frac{i}{2}$.

Therefore, the locus of $ w$ is a circle centered at the complex point $ \minus{}1\plus{}i$ in the complex plane, with radius $ \sqrt{2}$ (where I have dilated by 2).[/hide]

I'm not entirely sure that's correct (need to actually try drawing a diagram :D) so I'll get back to this once I've verified my answer.

[hide]Putting $ z\equal{}{2\over w}$ we get

$ \left|{2\over w}\minus{}(1\plus{}i)\right|^2\equal{}2\iff \left|{2\over (1\plus{}i)w}\minus{}1\right|^2\equal{}{2\over|1\plus{}i|^2}\iff\left|{1\minus{}i\over w}\minus{}1\right|^2\equal{}1$

Hence $ |(1\minus{}i)\minus{}w|\equal{}|w|$, which means $ w$ is equidistant from $ 0$ and $ 1\minus{}i$, therefore the desired locus is the perpendicular bisector of the segment delimited by those two points, or more specifically, the line passing through $ (1,0)$ and $ (0,\minus{}1)$, whose equation is $ y\equal{}x\minus{}1$.[/hide]

Essentially, this is saying that the inversive image of a circle passing through the center of inversion is a line.

Nice manipulation, Fahrenhajt :P I have to take some lessons on inversion now!


This inequality is from a friend of mine:

Let a, b, c be positive reals such that a^2 + b^2 + c^2 = 3.

Show that a^3 + b^3 + c^3 + 7(a + b + c) + 2(1/a + 1/b + 1/c) >= 30.

Here is my sln

We have $a^4 - 4a^3 + 7a^2 - 6a + 2 = (a-1)^2(a^2-2a+2) \ge 0$
Dividing both sides by a, we get
    $ a^3 + 7a + \frac {2}{a} \ge 4a^2 + 6$
Similarly for b, c.
Suming up these ineqs and using $a^2 + b^2 + c^2 = 3$ we have done.

I submit this sln as a gift for my friends Manlio and Harazi. I hope it's correct.

Namdung

Thanks a lot, Namdung, a very beautiful solution for a quite hard problem.

There's a little story behind this inequality.  It was originally going to appear on the 2004 ELMO at MOP but the night before the contest I found an unintended Jensen solution almost trivializing the problem:

Let [tex]x = a^2, y = b^2, z = c^2[/tex] and [tex]f(x) = x^{\frac{3}{2}} + 7x^{\frac{1}{2}} + 2x^{\frac{-1}{2}}[/tex].  Then [tex]x + y + z = 3[/tex], but [tex]f[/tex] is convex, since
[tex]f'(x) = \frac{3}{2}x^{\frac{1}{2}} + \frac{7}{2}x^{\frac{-1}{2}} - x^{\frac{-3}{2}}[/tex]
[tex]f''(x) = \frac{3}{4}x^{\frac{-1}{2}} - \frac{7}{4}x^{\frac{-3}{2}}[/tex] [tex]+ \frac{3}{2}x^{\frac{-5}{2}} = \frac{1}{4}x^{\frac{-5}{2}} \cdot \left( 3x^2 - 7x + 6\right) > 0[/tex]
For all positive [tex]x[/tex]. Hence,

[tex]\displaystyle{
\sum_{Cyc} a^{3} + 7a + \frac{2}{a} = \sum_{Cyc} f(x) \ge 3f(1) = 30
}[/tex]

The problem was then edited to make this approach less appealing and was actually written as: [tex]a, b, c \in \mathbb{R^{+}}[/tex] such that [tex]a^{3} + b^{3} + c^{3} = 3[/tex].  Prove that

[tex]\displaystyle{
a^4 + b^4 + c^4 + 7(a^2 + b^2 + c^2) \ge 6 + 6(a + b + c)
}[/tex]

Does the above still hold if [tex]a, b, c \in \mathbb{R}[/tex]?

The intended solution was very similar, although the Jensen solution became substantially more difficult.

[quote="Mildorf"]
The problem was then edited to make this approach less appealing and was actually written as: [tex]a, b, c \in \mathbb{R^{+}}[/tex] such that [tex]a^{3} + b^{3} + c^{3} = 3[/tex].  Prove that

[tex]\displaystyle{
a^4 + b^4 + c^4 + 7(a^2 + b^2 + c^2) \ge 6 + 6(a + b + c)
}[/tex]

Does the above still hold if [tex]a, b, c \in \mathbb{R}[/tex]?

The intended solution was very similar, although the Jensen solution became substantially more difficult.[/quote]

The solution is still the same as what [b]Namdung[/b] observed, that is

$(a-1)^4 + (a-1)^2 = a^4-4a^3+7a^2-6a+2\geq 0$.

By the way, just curious, how did you know the problem before the contest? :) Maybe I am missing something here.

Heh, I suppose my phraseology elicits some interesting thoughts in the minds of those who have not been to MOP. ;)  The ELMO is a traditional exam compiled from various submissions by the veterans at MOP. Only the rookies actually take the ELMO.  As 2004 was my second non-huge MOP, I was involved in reviewing these submissions; hence I knew about the question before the exam was administered.  Hope that clears things up. :)


geometry
3D geometry
complex analysis
Additive Number Theory

Show that any positive rational number can be represented as the sum of three positive rational cubes.

This is very hard. Does anyone have a solution?

An easier version is when we cancel the word "positive". Then we can solve with the identity $ \left(\frac{\minus{}x^3\plus{}3x\plus{}1}{x^2\plus{}x\plus{}1}\right)^3 \plus{} \left(\frac{3x^2\plus{}3x}{x^2\plus{}x\plus{}1}\right)^3 \plus{} (x\minus{}1)^3 \equal{} 9x$.

For the given version, I have no idea, can anyone type a proof here?

It's easy using your identity:
To represent some $ y>0$, it suffices to represent some $ yr^3$ for at least one $ r>0$. Now choose $ r$ such that $ x\equal{}\frac{yr^3}9$ has the property that all three terms on your RHS are positive (for example if $ x \in ]1, \frac 32]$). $ r$ can be choosen as the cubes are dense.
Thus $ 9x \equal{} yr^3$ is a sum of three positive cubes, thus so is $ y$.

Hmm... Good point. Didn't see that.  :blush:
Weird, I vaguely remember seeing a hard proof with complex analysis, probably I'm confusing things then.


What is the M Theory???

My friend told me about it...something about membranes colliding to form another universe...

Is this true?

The online explanations are way too advanced for me :(

Please help.

Thanks,
hwenterprise

Well here's a bit of background.
string theory is a new (in a way) theory being developed by theoretical physicists and mathematicians across the world. in short, string theory is an attempt to create a mathematical framework that unifies the physics of general relativity (einstein's stuff, black holes, etc.) and quantum mechanics. supposively, the mathematics involved with string theory is extremely difficult.
there are 5 distinct versions of string theory. all of them are fundamentally different from other modern views of the universe in that fundamental particles or building blocks of the universe are viewed as vibrating strings as opposed to particles (quarks).

M-theory is a newer theory that connects the 5 versions of string theory. physicists first became interested in this new M-theory when they realized that many of the distinct versions of string theory had striking similarities; people were looking at the same physics and same problems from different points of view.

according to theoretical physicist brian green, noone really know what the $M$ in M-theory stands for. mysterious, magnificent, ...

[quote="drunner2007"]according to theoretical physicist brian green, noone really know what the $M$ in M-theory stands for. mysterious, magnificent, ...[/quote]

My friend says it stands for "membrane".

I saw a documentary on this and someone sugguested that it could stand for Matrix theory

it definatly not membrane in fact its matrix. :lol:

[quote]Something I totally don't understand...[/quote]

Well, join the club, this is something that nobody REALLY understands.

http://en.wikipedia.org/wiki/M-theory

that gives a lot of info  :wink:

is anyone intrested in quantum mechanics if so then plz PM me about some book . if some one is then i will start posting problems.

There is a very cool NOVA documentary about string theory called The Elegant Universe. It is hosted by Brian Greene, author of a book on string theory, by the same name. You can watch all 3 hours of it on the website. http://www.pbs.org/wgbh/nova/elegant/

Haha yeah that was the documentary i saw. I watched it in my AP Physics class  :P . I have the book as well. I started reading it earlier this summer but i haven't gotten around to reading it much, I should do that.


The expression below equals what?

\[\frac {\sqrt 3+1}{\sqrt 2+1}-\frac {\sqrt2(2-\sqrt 2)}{\sqrt 3-1}\]

A) 0
B) -1
C)  :sqrt: 3 - :sqrt: 2

D) \[\frac {1}{\sqrt 6+\sqrt 3-\sqrt 2-1}\]

E) None of these

[hide]cross multiply to combine the fractions
numerator = ( :sqrt: 3 + 1)( :sqrt: 3 - 1) - ( :sqrt: 2)( :sqrt: 2 + 1) (2 - :sqrt: 2) = (3-1) - (4-2) = 0 (A)[/hide]

Yup.

This question doesn't look like it's going to be that answer but truly, it is.

Silverfalcon, wouldn't it be better to put all the AMC Challenges in the same topic like joml88's AMC A Day questions?  Then you wouldn't have to keep making a new topic for every question, which is a lot easier.

Good suggestion.

Yep. I'll do that way.

P.S. I'm kind of busy now looking and solving some problems so the new problems challenging would be sometimes later.


Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?

$\text{(A)}\ \$3.63 \qquad\text{(B)}\ \$5.13\qquad\text{(C)}\ \$6.30 \qquad\text{(D)}\ \$7.45 \qquad\text{(E)}\ \$9.07$

[hide]Only when he changes 1 penny into 5 quarters does his money amount change, and the change is +1.24.

Thus we want to look at an answer in the form .01+1.24n, which 7.45 is the only one. [/hide]

What kind of answer is that, .01 + 1.24, you mean that any number times (.01+1.25) will be in the answer is correct?
That is what it is right.

Um, I think 236factorial is right, we're looking for numbers in the form $.01+1.24n$, not as you said, $1.26n$.

Does anyone have a better solution -- this is more like guess and check. What I did was notice that the 1 penny goes to 125, and since the answers were relatively small, I could take away pennies and replug them into the above equation, and add to the remaining pennies. Taking away $5$ gives us $7.45$ but this is not rigorous

Please help

Putting a penny in the slot increases the amount of money we have by $1.24$. Putting in any other coin keeps the amount of money we have the same. Thus the amount of money we will have is always of the form $.01 + 1.24n$. Simply check.

[hide=Solution]Note that his money always stays the same unless he puts in a penny, in which case he gains $24$ cents. Taking $363, 513, 630, 745, 907 \pmod {24}$ gives $3,9,6,1,19$ respectively, so only the fourth option or $\boxed{\text{(D)}}$ works.[/hide]


Any conlangers here? (Look it up yourself. There's quite a community.) What do you think of [url=http://home.inreach.com/sl2120/Ithkuil/]Ithkuil[/url]?

*sigh* I guess everyone is too lazy.

[url=http://www.geocities.com/Athens/Acropolis/9219/conlangfaq.html]Start here.[/url]

[quote]For me, the greater goal is to attempt the creation of what human beings, left to their own devices, would never create naturally, but rather only by conscious intellectual effort...[/quote]

It is much more interesting to create a completely natural seeming language without relying too heavily on real languages. To my knowledge, only Tolkien has even come close to such an achievement. It seems like the question is the same as the divide between natural universes and man-made universes, fantasy and sci-fi, ultimate humility and ultimate arrogance. Of course, that is a philosophical pondering, not an accusation :). This is still interesting.

I keep finding things I want to complain about, but then I realize they all fall into the natural vs. artificial argument.

Also, I would really like to see a normal dialogue in this language (and by normal I do not mean "As it turned out, the snake-handler apparently began trapping each mouse in a container, one at a time like clockwork," which, by the way, confuses me as to why the author chose to use the idiomatic "like clockwork").

Building conlangs is an interesting intellectual exercise, about which there are some very good websites.

http://www.zompist.com/kit.html

Personally, I prefer to spend my time learning languages that are native languages of people in various places, especially languages that don't belong to the language family of my native language.

I tried to learn Quenya when I was in high school (and did a fairly good job too!), but then became more interested in non-constructed languages. This may have been a spark for me for going into math, for what is math but a conlang.

How big does a language have to be to be a conlang? My sister created a language. It has logical grammar (more than English) and it has over 1,500 words. However, only about three people know any of it.

If someone deliberately created the language, it's a conlang.

Is anyone here trying to learn the Na'vi language?

Nënom, nënom tokëseo "Na'vi." Tëroś, arëerto lësox.
No, I don't speak Na'vi. Yes, I make languages.
(Am I supposed to give translations out?)


Let $ M_{2x2}$ denote the vector space of complex $ 2x2$ matrices .Let
                    $ A\equal{}\left[\begin {array}{cc} 0 & 1 \\ 0 & 0 \end {array} \right]$
and let the linear transformation $ T: M_{2x2} \to M_{2x2}$ be defined by $ T(X)\equal{}AX\minus{}XA$.
Find the jordan canonical form for T
 :starwars:

[quote="JRav"]You've now posted around 20 problems that are not related to each other.  Are you just posting to post or do you actually need to solve all of these problems?  I'm guessing the former, since you haven't made an attempt to solve any of them.[/quote]
I remember This ! Thanks for you ! don't  :mad: I will Stop here !!! Oke !
You can delete all anything as I wrote ! Thanks!  :gathering:


[quote="grobber"]Let $ ABC$ be a non-equilateral triangle. For a point $ P$ in the plane of $ ABC$, we denote by $ A^{\prime}$, $ B^{\prime}$, $ C^{\prime}$ the intersections of $ PA$, $ PB$, $ PC$ with the circumcircle of triangle $ ABC$. Show that there are exactly two positions of $ P$ for which triangle $ A^{\prime}B^{\prime}C^{\prime}$ is equilateral, and show that the line determined by them contains the circumcenter of $ ABC$.[/quote]

Well, these two positions of P are the two isodynamic points of triangle ABC.

In fact, if P is any point in the plane of triangle ABC, and the lines AP, BP, CP meet the circumcircle of triangle ABC at the points A', B', C' (apart from A, B, C), and X, Y, Z are the orthogonal projections of the point P on the sides BC, CA, AB of triangle ABC, then the triangles XYZ and A'B'C' are directly similar.

[i]Proof.[/i] We will use directed angles modulo 180°. Now, since < PYA = 90° and < PZA = 90°, the points Y and Z lie on the circle with diameter PA, so that we have < ZYP = < ZAP, and thus < ZYP = < BAP = < BAA' = < BB'A' (cyclic). Similarly, < XYP = < BCC' = < BB'C' (cyclic again). Hence,

  < XYZ = < XYP - < ZYP = < BB'C' - < BB'A' = < A'B'C',

and similarly < YZX = < B'C'A' and < ZXY = < C'A'B'. Thus, triangles XYZ and A'B'C' are directly similar.

Now there exist exactly two points P in the plane of triangle ABC such that the triangle XYZ is equilateral, namely the isodynamic points of triangle ABC. Hence, these isodynamic points are also the two (only) points P such that triangle A'B'C' is equilateral.

Now, the line joining the two isodynamic points of a triangle is known to pass through the circumcenter of the triangle. Hence we are done.

  Darij

I had the feeling you'd solve it right away :). Anyway, the two points are more than just collinearwith $O$; they're the inverse of one another wrt the circumcircle.

[quote="grobber"]I had the feeling you'd solve it right away :).[/quote]

Oh yes, I did... actually I knew the solution, it is a kind of folklore, at least among triangle geometers.

  Darij


Let a1; a2; a3; a4 be the lengths of the sides of a quadrilateral and let s be its semiperimeter.
Prove that
1/(s+a1)+...+1/(s+a4)<=(2/9)sum of(1/(s-ai)(s-aj))with(1=<i<j=<4).

When does equality happem.

Bomb

It can be found on the forum already. As far as I remember, the problem was given in the Romanian TSTs in 2004 or 2003.

EDIT: I found the link now to the problem, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=5448]www.mathlinks.ro/Forum/viewtopic.php?t=5448[/url]


abstract algebra
trigonometry
real analysis
limit
integration
function
real analysis unsolved

remember the Fejer kernel:
$F_{N}(x)=\frac{1}{2 \pi (N+1)}(\frac{\sin(\frac{(N+1)x}{2})}{\sin(\frac{x}{2})})^{2}$
Suppose that $\mu$ is a measure on $\mathbb{T}$ with Radon-Nikodym decomposition $d \mu = f dm+\mu_{s}$ where $m$ is the Lebesgue measure. Show that for (Lebesgue)almost every $x \in \mathbb{T}$,
$\lim_{N \to \infty }\int_{\mathbb{T}}F_{N}(x-y) d \mu(y) = f(y)$

The first step is to observe that $\int F_{N}(x)dx=1$ and $\int_{-\delta}^{\delta}F_{N}(x)dx\underset{N\to\infty}\to 1$ for any $\delta>0$.

Now, I have only a proof in the case where $\mu$ is absolutely continuous with respect to the Lebesgue measure. Note that even this partial result is quite impressive because it says that if you know the Fourier coefficients $a_{k}$ of a function $f \in L^{1}(\mathbb T)$, then you can reconstruct $f$ by looking at the Cesaro means of its Fourier coefficients..


Hi,

just a curious question to all of you. I am interested who of you is an IMO (deputy leader), observer, medal winner (which medal ?), participated in IMO or maybe still wants to qualify. I hope the information is not too private ?!  :D

I could have submitted more but I choosed observer. Also trainer of the national team, bronze medalist and participant (that is Honorable Mention :) ).


Solve the following quadratic equations by completing the square

      a) 4x^2 + x = 3

      b) 3x^2 = 3 − 4x

of course this is helpful.

thank you AndrewTom

You are intelligent

good evening

Here's a good free website for this:
[url]http://www.purplemath.com/modules/sqrquad.htm[/url]

Just for reference, since I'm assuming you have to solve it by completing the square, if you just do: -b/2a, this will give you the X value of the vertex...


vector
advanced fields
advanced fields unsolved

Let $ n\in \mathbb{N}$ construct a vector field differentiable in $ S^{2}$ with exactly n singularities. :?:

I suppose that by a singularity you mean a point where the vector field vanishes. First do the case $ n\equal{}1$. To create additional singular points, multiply the field by $ |x\minus{}a|$ for some $ a\in S^2$.


if x varies as the cube of y, and y varies as the fifth root of z, then x varies as the nth power of z, what is n? (AHSME 1954).

Varies? Proportionaly inverse or directly? huh?


superior algebra
superior algebra solved

hi all.
here is a nice one:
let E a linear space of finite dimension and u an endomorphism of E.
Prove that if trace(u^k)=0 for every k then there exists p>=1 such that
u^p=0

Well known, can be proven with matricial approach
$A \in M_n(K)$ where $K$ is field of characteristic 0
suppose $\forall k, tr(A^k)=0$ with Cayley-Hamilton...


articles
Support
inequalities
trigonometry
MATHCOUNTS
number theory
prime factorization

[b]Summary of what has been decided in this thread by Temperal: Proper nouns shall be capitalized. Theorems, formulas, postulates, lemmata (the correct plural for "lemmas"), and axioms are considered proper nouns. Everything that is not a proper noun is not capitalized. All section headers are capitalized regardless of content. See [[A:NAME]] for more.[/b]

The MediaWiki software is case sensitive in a particular way.  The very first letter of an article is NOT case sensitive, but the rest of the letters ARE case sensitive.

What this means is that articles Prime factorization and Prime Factorization are different articles.  The first can be linked to with [[prime factorization]], but the second cannot.

Since prime factorization is NOT a proper noun, the article name should be Prime factorization and NOT Prime Factorization.

Furthermore, we need to discuss a standard for naming articles such as Euler's Totient Theorem and Euler's polyhedral formula.  Should we view these as proper names?  Wikipedia does NOT view them as proper names.  Either way, we need a standard because articles are being created with and without capitalization of all the words.

What are your opinions?  Mention if you are a Wikipedia contributor and have opinions as to how the system works there.  Thanks.

Well, I guess I was the one that started the issue with my two articles of Prime Factorization and Prime factorization :D
Anyway, I personally think it looks nicer if we capitalize all when in doubt...
in general, it might work just to capitalize formula names (so it would be Euler's Polyhedral Formula), but capitalize only the first word for all other articles

I changed about a dozen others before the prime factorization example.  Don't feel bad, I made similar mistakes while learning to use Wikipedia.

But we can't just keep everything capitalized because of the link issue.  We need [[prime factorization]] to take us to the prime factorization page, and the only way to work around that would be to use capitals with every internal link -- that certainly won't look pretty and could cause confusion at times (over whether or not the topic is a proper noun for instance).

[quote="chess64"]I think only theorems/formulas should be capitalized.[/quote]

Wikipedia doesn't do that, and I'm trying to research the reason why.  My only moderator friend hasn't been available for me to ask, though he might not know anyway.

Also, I'm certainly inclined to keep capitalization for names like 'Paul Zeitz' and I'm mildly inclinded to keep capitalization for names of tournaments, scholarships, and universities.  I believe this is also in-line with Wikipedia convention.

To me, the only fuzzy line is names of theorems/formulas.  It seems to me they would be proper nouns, but Wikipedia convention defies that notion for some reason, and they have a lot of experience doing what they do.

Another equally disturbing issue is singular versus plural. I would go for singular whenever possible though other people may have different opinions. So far, most of the articles that have titles in plural seem to arise from following red links in which the authors just bracketed whatever grammatical form was needed in the sentence (thank God, English doesn't have different word forms for different cases!). When setting the links, it is always a good idea to think of the appropriate names for the articles you want to refer to and to use the piped link format [[actual name | words you want to highlight]] whenever needed (I should confess here that I myself learned this trick only this morning :blush::)). There is also a convenient shortcut like [[real number]]s that will create a link to Real number while highlighting "real numbers".

As to capitalization, I would prefer the theorem names without it but it is just my personal taste and I have no compelling reason to support it.

http://dictionary.reference.com/browse/proper%20noun

I think theorems are unique enough to be classified as proper nouns.

Here's a discussion at Wikipedia about the subject at hand:

http://en.wikipedia.org/wiki/Wikipedia_talk:Naming_conventions_(theorems)

Right now, Wikipedia chooses not to capitalise theorem names as if they were proper nouns.  Right now, I feel slightly inclined to buck that trend and capitalise them, but I could still be argued down from that perspective.

But that brings up another question, should we capitalise names of mathematical fields (Number Theory or number theory)?

This is a software design issue, and trying to fix them socially is really hard on everyone, admins and users. The quickest way to fix it would be to have a handful of standard naming conventions, then a redirection script that would would handle the captilization on internal links.

I think mediawiki is open source, so get Valentin to script up a quick fix.

I think theorems and formulas are a difficult question -- on the other hand, I would never write Number Theory instead of number theory in the middle of a sentence.

Right now, Val has bigger fish to fry.

I agree with JBL for now.  That will be at least policy unless the issue is raised again.

I wouldn't write Number Theory, but I would write Euler's Totient Theorem.

What I generally do (at least, for theorems/formulas) is sort of what I did when writing the Rearrangement Inequality article.

I capitalized both "Rearrangement" and "Inequality", and then made a redirect from "Rearrangement inequality" to "Rearrangement Inequality".

As for building links, I often do something like [[Link Name | link name]], so that it shows up as lowercase in the text.

But that's just what I do.  Maybe there's a way you could turn the case-sensivity off?

(by the way, I'd write both Combinatorics and Fermat's Little Theorem)

[quote="PenguinIntegral"][quote="dts"]
As for building links, I often do something like [[Link Name | link name]], so that it shows up as lowercase in the text.[/quote]
Do this. It seems to be pretty cool. Just Capitalize everying in the title and make it appear as it needs to.[/quote]

No, not every first letter in every first word in every title needs to be capitalized.  That would frustrate too many first-time users as they fumble over creating internal links with common nouns in the middle a sentence.

I can't really imagine what reasoning was used to make the MediaWiki software this way, but it's definitely turning out to be a hassle for us.

Here are some of my personal preferences that we should perhaps adopt (unless there is objection):

* For laws, capitalize all important words, e.g. Law of Cosines and Law of Definite Proporitions.
* Capitalize all important words in a term involving a proper noun, e.g. Riemann Hypothesis and Euler's Polyhedron Theorem.
* Theorems/formulas without proper nouns should not be capitalized, e.g. change of base theorem, half angle identity.

Anything else?

[quote="joml88"]* For laws, capitalize all important words, e.g. Law of Cosines and Law of Definite Proporitions.
* Capitalize all important words in a term involving a proper noun, e.g. Riemann Hypothesis and Euler's Polyhedron Theorem.
* Theorems/formulas without proper nouns should not be capitalized, e.g. change of base theorem, half angle identity.[/quote]

I think those are perfect!

I disagree with the third.  I don't see any reason why to capitalize the word "theorem" in some theorems and not others.  Either they're proper nouns or they're not.  I lean toward thinking they are proper nouns, as they are given distinctive names.  This would cover all laws and identities as well.

Yes, as dts said [[Article you want the link to lead to|the text you want displayed]] will work for links and as an example, for an article like "MathCounts" you'd also want to create the page Mathcounts, MATHCOUNTS, Math Counts with content "#REDIRECT [[MathCounts]]" so if you link to Mathcounts, MATHCOUNTS, or Math Counts, it'll bring you to MathCounts. However, if a link has to go through two redirects, it won't and you'll go to the page the 1st redirect was pointing to and the content would be an arrow and a link to the page that redirect was pointing to.

Click "What links here" on each page to see what pages link, or redirect to that page.

As for my opinion on the Manual of Style, I believe Theorem's should be capitalized.

I think that theorems should be capitalized, especially as the "laws" joml88 mentioned are really theorems.  (Actually, I never understood why we call them laws, other than the fact that "cosine theorem" and "theorem of cosines" don't sound quite as good.)

"Vertical Line Test" or "Vertical line test?"  (Right now we have both.)  I like the latter.

[quote="JBL"]Does the AoPS wiki have "inequalities" or "Inequalities"?  Right now there seem to be a lot of both -- consider the list [url=http://www.artofproblemsolving.com/Wiki/index.php/Algebra#Olympiad_Topics]here[/url], for instance.[/quote]

I just changed a few.  A specific inequality such as the Power Mean Inequality is considered a proper noun.  However, trigonometric inequalities is descriptive of a set of nouns, though the nouns themselves may be proper in some cases.

This is obsolete.  Locked.  Please start a new topic and refer to this topic if you have a question or comment.


[hide="solution"] Multiply by four so $ 16a^2 \plus{} 16a \equal{} 4b^2 \plus{} 4b$. Completing the square yields $ (4a \plus{} 2)^2 \equal{} (2b \plus{} 1)^2\plus{}3$. Rearranging and factoring gives $ (4a \minus{} 2b \plus{} 1)(4a \plus{} 2b \plus{} 3) \equal{} 3$.

Since three is prime and clearly $ 4a \plus{} 2b \plus{} 3$ is positive and greater than $ 4a \minus{} 2b \plus{} 1$, we must have $ 4a \plus{} 2b \plus{} 3 \equal{} 3$, which can't be true.[/hide]

Another way
Complete the square on the LHS: $ 4a^2\plus{}4a\plus{}1\equal{}b^2\plus{}b\plus{}1$. Then $ (2a\plus{}1)^2\equal{}b^2\plus{}b\plus{}1$. However, $ b^2<b^2\plus{}b\plus{}1<b^2\plus{}2b\plus{}1\equal{}(b\plus{}1)^2$.
$ b^2<(2a\plus{}1)^2<(b\plus{}1)^2$ is a contradiction.

Yet another way. Very bashy, bet eh:

Let $ 2a \equal{} y$. Then $ (y)^2 \plus{} 2y \equal{} b^2 \plus{} b$
so: $ y(y \plus{} 2) \equal{} b(b \plus{} 1)$
Intuitively, there are no solutions for this, but to prove it:

Case 1: $ y > b$. Then $ y \equal{} b \plus{} x$ with $ b$, $ y$ and $ x$ being positive integers.
then: $ (b \plus{} x)(b \plus{} x \plus{} 2) \equal{} b^2 \plus{} b$
$ b^2 \plus{} x^2 \plus{} 2bx \plus{} 2b \plus{} 2x \equal{} b^2 \plus{} b$ and since $ y, b,$ and $ x$ are positive integers,
$ b^2 \plus{} x^2 \plus{} 2bx \plus{} 2b \plus{} 2x > b^2 \plus{} b$ because $ x^2 \plus{} 2bx \plus{} b \plus{} 2x > 0$

Case 2: $ b > y$. Then $ b \equal{} y \plus{} x$ again with $ b, y$ and $ x$ as positive integers. So: $ y^2 \plus{} 2y \equal{} y^2 \plus{} x^2 \plus{} 2xy \plus{} y \plus{} x$ and again, since $ y, b,$ and $ x$ are positive integers, $ y^2 \plus{} 2y < y^2 \plus{} x^2 \plus{} 2xy \plus{} y \plus{} x$ because $ x^2\plus{}2xy\minus{}y\plus{}x>0$ and $ 2xy>\minus{}y$

Case 3: $ y \equal{} b$ then $ b^2 \plus{} 2b \equal{} b^2 \plus{} b$ but since b is a positive integer $ b^2 \plus{} 2b > b^2 \plus{} b.$

And we are done?

Yeah, but instead of going through all those cases, just rearrange to $ \frac{y}{b}\equal{}\frac{b\plus{}1}{y\plus{}2}$. Now you can easily observe that this can't be true.


superior algebra
superior algebra unsolved

question:
what will happen if we apply the unity-adjunction procedure to $ C^*$ algebra having its own unity ?

erm... we will have new algebra with new (adjoined) unity.........
the old unity becomes an usual element ?

what can be written as an answer to this question ?

$ \tilde{A} \equal{} A$ if $ A$ is unital. this follows from the universal property of $ \tilde{A}$.

if you use the explicit construction as $ A \plus{} \mathbb{C}$, define $ A \plus{} \mathbb{C} \to A , a \plus{} \lambda \mapsto a \plus{} \lambda 1_A$ and show that it's an isomorphism.

[quote="-oo-"]
if you use the explicit construction as $ A \plus{} \mathbb{C}$, define $ A \plus{} \mathbb{C} \to A , a \plus{} \lambda \mapsto a \plus{} \lambda 1_A$ and show that it's an isomorphism.[/quote]

thanks :)

ok, I have already proved it.

and the proof of this fact will be a sufficient answer or I should say something else ?

I know that this question isn't precisely stated...

I had on mind the following thing:

is there anything else (property, theorem) what can be said in the most general case about the $ C^*$-algebra with its own unity, after an application of this construction ?
(not in special cases like e.g. in $ C^*$ of matrices or continuous linear operators)

so you ask: what can we say about A, besides it is A, ... hmmm!


quadratics
induction
modular arithmetic
symmetry
Congruences

During a break, $n$ children at school sit in a circle around their teacher to play a game.  The teacher walks clockwise close to the children and hands out candies to some of them according to the following rule. He selects one child and gives him a candy, then he skips the next child and gives a candy to the next one, then he skips 2 and gives a candy to the next one, then he skips 3, and so on.  Determine the values of $n$ for which eventually, perhaps after many rounds, all children will have at least one candy each.

We identify the children with the residues $\mod n$.
It starts with turn $0$ when child $0 \mod n$ gets a candy.
In turn $k$ the teacher gives a candy to the child $1+2+...+k =\frac{k(k+1)}{2} \mod n$.
We are thus asked for what $n$ the map $\sigma :k \mapsto \frac{k(k+1)}{2} \mod n$ is surjective.

If $p$ is an odd prime dividing $n$, we would need that $\sigma$ is surjective $\mod p$. But $\sigma \mod p$ only depends on the residue classes $\mod p$. By $\sigma(-1) \equiv \sigma(0) \mod p$ we get that it's not injective, thus not surjective.

Thus $n$ is a power of $2$, $n=2^k$. We show that it works for all those $n$:
We want $k$ with $\frac{k(k+1)}{2} \equiv a \mod 2^k \iff (2k+1)^2 \equiv 8a+1 \mod 2^{k+3}$ for all $a$.
But it's well known that $\mod$ any power of two, exactly the residues $\equiv 1 \mod 8$ are those that are quadratic residues, thus we are done.

[quote="ZetaX"]We identify the children with the residues $\mod n$.
But it's well known that $\mod$ any power of two, exactly the residues $\equiv 1 \mod 8$ are those that are quadratic residues, thus we are done.[/quote]

is there a better way to see this fact? (better than using an induction-like argument like the following sketch:

Let $k = 2^n.$

For the roots of the quadratic residues, it suffices to check the squares of the odd integers up to $k/4-1$ (denote this set $L$), since $i^2 \equiv (k/2-i)^2$ and obviously $i^2 \equiv (-i)^2 \equiv (k/2+i)^2$, both of which follow from binomial expansion and using $k=2^n$. In fact, these must be exactly the set of residues $\equiv 1 \pmod 8)$ since there are $k/8$ of them. We just need to show that $L$ maps injectively and surjectively to the residues $\equiv 1 \pmod 8$.

Now we induct by noting the symmetry of (2k)/2-i and i

It's easy to check the base case.
Suppose this is true for $k=2^n$. Then we need to check it's true for the odd integers up to $2k/4-1  = k/2-1$. But $(k/2-i)^2 \equiv i^2 +(2k)/2 \pmod {2k}$ since $k=2^n$ by definition. Thus we can extend this bijective map and the inductive step is done.

Induction is one way, another goes as follows:
- Check that $x^2 \equiv 1 \mod 2^n$ has at most $4$ solutions.
- Conclude that $x^2 \equiv a \mod 2^n$ has at most $4$ solutions.
- Check that only the residues $\equiv 1 \mod 8$ can be quadratic residues.
- Compare the last two using that there are $2^{n-1}$ possible residues to square.

Ahh that is much better. THanks!. I will write this for future reference:

$x^2\equiv 1 \pmod{2^n}$ has at most 4 solutions: $x^2-1=(x+1)(x-1) \equiv 0 \pmod{2^n}$. so either one of the factors evenly divides 2^n or one exactly divides 2^(n-1) and the other is 2 mod 4. (I think this is easiest way to think about this?). So $x$ is $\pm 1, 2^{n-1} \pm 1$

$x^2 \equiv a$. If $a$ is a q.r., then $a=k^2$ and the solutions should be $\pm k , 2^{n-1} \pm k$, which is multiplying above by k (mod 2^(n-1)). Basically same reasoning.

The resides mod 8 are 0,1, 4

There are $2^{n-1}$ odd numbers,  and the first two points check that for every four odd numbers, there is at least one quadratic residue, while the third says exactly one/eight of the residues can be quadratic residue, so the odd quadratic residues are exactly those 1 mod 8


Ontws basiki askisi...[hide]
Gia ola ta $ i$ einai $ x_i$ diaforo tou 0.Episis an $ y$ einai riza tote kai $ 1/y$ einai riza. Opote estw $ x_1 \equal{} 1/x_2$,   $ x_3 \equal{} 1/x_4$,  $ x_5 \equal{} 1/x_6$. Epidi $ (y^2 \plus{} 1)(\frac {1}{y^2} \plus{} 1) \equal{} (y \plus{} \frac {1}{y})^2$ exoume oti

$ P \equal{} [(x_1 \plus{} x_2)(x_3 \plus{} x_4)(x_5 \plus{} x_6)]^2$, alla

$ (x_1 \plus{} x_2)(x_3 \plus{} x_4)(x_5 \plus{} x_6) \equal{} \Sigma_{i < j < k}x_ix_jx_k \minus{} 2 \Sigma x_i \equal{} \minus{} c \plus{} 2a$,

afou  $ x_1x_2x_k \equal{} x_5x_6x_k, k \equal{} 3,4$ etc.[/hide]

[quote="stergiu"][b][i]Από την ισπανική κοινότητα επέλεξα αυτή τη βασική άσκηση : [/i][/b]

Αν $ x_{1},\ x_{2},\ldots,\ x_{6}$ είναι οι ρίζες του πολυωνύμου$ P(x) \equal{} x^{6} \plus{} ax^{5} \plus{} bx^{4} \plus{} cx^{3} \plus{} bx^{2} \plus{} ax \plus{} 1$ , να αποδειχθεί ότι :
$ \[ \prod_{k \equal{} 1}^{6}(x_{k}^{2} \plus{} 1) \equal{} (2a \minus{} c)^{2}$

 Μπάμπης[/quote]

[b] Bάζω τη μία λύση της  ισπανικής κοινότητας και αξιολογείστε την.Νομίζω πως είναι επίσης καλή :[/b]

$ \prod_{k \equal{} 1}^{6}(x_{k}^{2} \plus{} 1) \equal{} \prod_{k \equal{} 1}^{6}(x_{k} \plus{} i)(x_{k} \minus{} i) \equal{} \prod_{k \equal{} 1}^{6}(x_{k} \plus{} i)\prod_{k \equal{} 1}^{6}(x_{k} \minus{} i) \equal{} P(i)P( \minus{} i)$
Επομένως :
$ P(i)P( \minus{} i) \equal{} ( \minus{} 1 \plus{} ai \plus{} b \minus{} ci \minus{} b \plus{} ai \plus{} 1)( \minus{} 1 \minus{} ai \plus{} b \plus{} ci \minus{} b \minus{} ai \plus{} 1) \equal{} (ai \minus{} ci \plus{} ai)( \minus{} ai \plus{} ci \minus{} ai) \equal{} (2a \minus{} c)^{2}$

Βγαίνει και με Vieta αν σπάσουμε το 2α σε α+α
(το ένα του συντελεστή χ^5, και το άλλο του συντελεστή χ^2).

[quote="stergiu"][b][i]Από την ισπανική κοινότητα επέλεξα αυτή τη βασική άσκηση : [/i][/b]

Αν $ x_{1},\ x_{2},\ldots,\ x_{6}$ είναι οι ρίζες του πολυωνύμου$ P(x) \equal{} x^{6} \plus{} ax^{5} \plus{} bx^{4} \plus{} cx^{3} \plus{} bx^{2} \plus{} ax \plus{} 1$ , να αποδειχθεί ότι :
$ \[ \prod_{k \equal{} 1}^{6}(x_{k}^{2} \plus{} 1) \equal{} (2a \minus{} c)^{2}$

 Μπάμπης[/quote]
Μηπως υπάρχει κπ γενικευση :?:


calculus
integration
function
geometry
algebra
binomial theorem
calculus computations

Let $ f(x)$ be the continuous function at $ 0\leq x\leq 1$ such that $ \int_0^1 x^kf(x)\ dx\equal{}0$ for integers $ k\equal{}0,\ 1,\ \cdots ,\ n\minus{}1\ (n\geq 1).$

(1) For all real numbers $ t,$ find the minimum value of $ g(t)\equal{}\int_0^1 |x\minus{}t|^n\ dx.$

(2) Show the following equation for all real real numbers $ t.$

$ \int_0^1 (x\minus{}t)^nf(x)\ dx\equal{}\int_0^1 x^nf(x)\ dx$

(3) Let $ M$ be the maximum value of the function $ |f(x)|$ for $ 0\leq x\leq 1.$

Show that $ \left|\int_0^1 x^nf(x)\ dx\right|\leq \frac{M}{2^n(n\plus{}1)}$

[hide]Not quite sure about that, but... just a try.

$ 0 < x < 1$. Assume that $ 0 < t < 1$.
It's easy to see why that.
If $ t < 0$ the function is crescent. $ (x \minus{} t)^n$. The less $ t$ is, greater is the area.
If $ t > 1$ the function is decrescent $ (t \minus{} x)^n$.The greater $ t$ is, greater is the area.

The local maximum of the function is on $ x \equal{} 0,1$, so if we draw a line between those 2 maximuns, the area of this will be less than the area of the integral.

$ \int_0^1 |t \minus{} x|^n \, dx\leq \left|\frac {1}{2} (f(0) \plus{} f(1))\right| \equal{} \frac {1}{2} \left((1 \minus{} t)^n \plus{} t^n\right)$

This equation is symmetrical on $ x \equal{} \frac {1}{2}$. Easy to proove.

If a continuous function is symmetrical to a point, this point is a local minimum/maximum.

So $ t \equal{} \frac {1}{2}$

$ \int_0^1 \left|\frac {1}{2} \minus{} x\right|^n \, dx \equal{} \int_0^{\frac {1}{2}} \left(\frac {1}{2} \minus{} x\right)^n \, dx \plus{} \int_{\frac {1}{2}}^1 \left(x \minus{} \frac {1}{2}\right)^n \, dx \equal{} 2 \int_0^{\frac {1}{2}} \left(\frac {1}{2} \minus{} x\right)^n \, dx \equal{} \frac {2^{ \minus{} n}}{n \plus{} 1}$

[/hide]
[hide]Using binomial Theorem and interchanging the integral with the sum.

$ \int_0^1 (x \minus{} t)^n f(x) \, dx \equal{} \int_0^1 x^n f(x) \, dx \plus{} \sum _{i \equal{} 1}^n \left( \begin{array}{c} n \\
i \end{array} \right) ( \minus{} t)^i \int_0^1 x^{n \minus{} i} f(x) \, dx$

The second integral is zero, according to $ \int_0^1 x^kf(x)\ dx \equal{} 0$ , $ n \minus{} 1\geq k\geq 0$

This integral is the Cauchy integral.
http://en.wikipedia.org/wiki/Cauchy_formula_for_repeated_integration
[/hide]
[hide]
By Hölder's inequality.

$ \left|\int_0^1 x^n f(x) \, dx\right|\equal{}\left|\int_0^1 (x\minus{}t)^n f(x) \, dx\right|\leq \left(\int_0^1
   \left|(x\minus{}t)\right|^n \, dx\right) \left(\int_0^1 |f(x)| \, dx\right)$

The rest I'm not quite sure, since $ \frac{2^{\minus{}n}}{n\plus{}1}$ is a minimum, not a maximum, or I made a mistake
[/hide]

Hi Kunny!This is a nice problem!Can someone post the official solution?Please!!!Thanx!!! :lol:


trigonometry
inequalities unsolved
inequalities

Prove that in any triangle $ABC$ we have

$\sin A +\sin B+ \sin C  \geq  (\sin A)^3 + (\sin B)^3 +(\sin C)^3 + \sin A \sin B \sin C$

It's "Mathematical and Young",a math magazine in VietNam.


In triunghiul $\ ABC$ consideram $\ A',B',C'$ mijloacele segementelor $\ BC,CA,AB$.Dreapta $\delta$ intersecteaza $\ (AB)$ si $\ (AC)$ in $\ P$ si $\ R$.Sa se arate ca dreapta $\delta$ contine centrul cercului inscris triunghiului $\ A'B'C'$ daca si numai daca are loc relatia $\ (BC+AB)\frac{PA}{PB}+(BC+AC)\frac{RC}{RA}=AC+AB$.

[color=darkblue][b]Lemma.[/b] Fie un triunghi $ABC$, un punct $P$ interior triunghiului si doua puncte $M\in (AB)$, $N\in (AC)$ astfel incat punctul $P$ sa apartina dreptei $MN$.
Notam intersectia $D\in BC\cap AP\ .$ Atunci exista relatia $\boxed{\ DC\cdot\frac{MB}{MA}+DB\cdot\frac{NC}{NA}=BC\cdot\frac{PD}{PA}\ }\ .$

[b]I. Demonstratia problemei propuse (pentru preolimpici).[/b] Presupunem fara a restrange generalitatea ca $b>c\ .$ Pentru centrul $I'$ al cercului inscris triunghiului $A'B'C'$ notam intersectiile $T\in AC\cap A'I'$ si $S\in BC\cap AI'$. Se observa usor ca $B'A'=B'T=\frac{c}{2}$ si $AT=\frac{b-c}{2}\ .$
$\blacktriangleright$ $\frac{I'A'}{I'T}=\frac{B'A'}{B'T}\cdot \frac{\sin(\widehat{I'B'A'})}{\sin (\widehat{I'B'T})}=$ $\frac{\sin\frac{B}{2}}{\sin\left(C+\frac{B}{2}\right)}=$ $\frac{2\sin\frac{B}{2}\cos\frac{B}{2}}{2\sin\left(C+\frac{B}{2}\right)\cos\frac{B}{2}}=$ $\frac{\sin B}{\sin (B+C)+\sin C}=$ $\frac{\sin B}{\sin A+\sin C}=$ $\frac{b}{a+c}$ $\Longrightarrow$ $\boxed{\ \frac{I'A'}{I'T}=\frac{b}{a+c}\ }\ .$
$\blacktriangleright$ Aplicam teorema lui Menelaus transversalei $AI'S$ in trunghiul $A'CT$ : $\frac{AT}{AC}\cdot\frac{SC}{SA'}\cdot \frac{I'A'}{I'T}=1$ $\Longrightarrow$ $\frac{b-c}{2b}\cdot \frac{SC}{SA'}\cdot \frac{b}{a+c}=1$ $\Longrightarrow$ $\frac{SC}{2(a+c)}=\frac{SA'}{b-c}=\frac{\frac{a}{2}}{2a+b+c}=\frac{A'B}{2a+b+c}=\frac{SB}{2(a+b)}$ $\Longrightarrow$ $\boxed{\ \frac{SC}{a+c}=\frac{SB}{a+b}=\frac{a}{2a+b+c}\ }\ .$
$\blacktriangleright$ Aplicam teorema lui Menelaus transversalei $A'I'T$ in triunghiul $ASC$ : $\frac{A'S}{A'C}\cdot\frac{TC}{TA}\cdot\frac{I'A}{I'S}=1$ $\Longrightarrow$ $\frac{b-c}{2a+b+c}\cdot \frac{b+c}{b-c}\cdot\frac{I'A}{I'S}=1$ $\Longrightarrow$ $\boxed{\ \frac{I'S}{I'A}=\frac{b+c}{2a+b+c}\ }\ .$
$\blacktriangleright$ Din lema mentionata mai sus aplicata punctului $I'$ interior triunghiului $ABC$ rezulta relatia $SC\cdot \frac{PB}{PA}+SB\cdot\frac{RC}{RA}=BC\cdot\frac{I'S}{I'A}$ $\Longrightarrow$ $\frac{a(a+c)}{2a+b+c}\cdot \frac{PB}{PA}+\frac{a(a+b)}{2a+b+c}\cdot \frac{RC}{RA}=a\cdot \frac{b+c}{2a+b+c}$ $\Longrightarrow$ $\boxed{\ (a+c)\cdot \frac{PB}{PA}+(a+b)\cdot\frac{RC}{RA}=b+c\ }\ .$[/color]

O alta solutie este sa aflam vectorul de pozitie al centrului cercului inscris in triunghiul A'B'C' pt triunghiul ABC si dupa care aplicam din nou teorema transversalei.

Chiar e frumoasa solutia domnului Virgil Nicula.Sincer, si eu m-am gandit sa fac cu Teorema Transversalei, dar m-am impotmolit dupa aceea. :oops:

[color=darkblue][b]II. Demonstratia problemei prouse (pentru olimpici).[/b] Notam simetricul $R$ al punctului $I$ fata de punctul $A'$ si intersectia $S_{a}\in BC\cap AR$. Din omotetia triunghiurilor $ABC$ si $A'B'C'$ rezulta ca punctul $I'$ este mijlocul segmentului $AR$, $S_{a}\in BC\cap AI'$ si centrul de grutate $G$ (comun celor doua triunghiuri) apartine segmentului $II'$ astfel incat $GI=2\cdot GI'$. Vom utiliza calculul vectorial. Astfel, $2pI=aA+bB+cC$, $2A'=B+C=I+R$. Asadar, $I'=\frac{1}{2}(A+R)=$ $\frac{1}{2}(A+B+C-I)=$ $\frac{1}{2}\left[A+B+C-\frac{1}{2p}(aA+bB+cC)\right]=$ $\frac{1}{4p}\left[(b+c)A+(c+a)B+(a+b)C\right]$ $\Longrightarrow$ $\frac{S_{a}B}{S_{a}C}=\frac{a+b}{a+c}$. Analog, pentru punctele $S_{b}\in CA\cap BI'$ si $S_{c}\in AB\cap CI'$ avem relatiile $\frac{S_{b}C}{S_{b}A}=\frac{b+c}{b+a}$ si $\frac{S_{c}A}{S_{c}B}=\frac{c+a}{c+b}$. Urmeaza sa aplicam lema mentionata in postul precedent si vom obtine relatia din concluzia problemei propuse.
[b]Observatie.[/b] Aceasta demonstratie imbina in mod fericit geometria sintetica (omotetie indeosebi) cu geometria vectoriala. Exista si o rapida demonstratie cu ajutorul coordonatelor baricentrice, insa nu o voi prezenta deoarece coordonatele baricentrice sunt mai putin cunoscute, ceea ce este pacat, mai ales pentru geometria triunghiului, ca in problema noastra.
[/color]


algebra
polynomial
geometry
algebra unsolved

Let $P,Q$ be polynomails with coefficients in the set $\{1, 2002 \}$.show:
if $P$ devides $Q$ then [b]degP[/b]$+1$ divides [b]degQ[/b]$+1$

I have bad memories about this problem, since when I took the TST in which it was given, I had no idea about polynomials reduced modulo primes and finite fields. I look now at this problem and realize it is trivial. Reduce the polynomials modulo 3, that is look at the polynomials $f_1, g_1$ obtained by taking the coefficients modulo 3. Since by taking the coefficients mod 3 we have a morphism, $f_1$ will divide $g_1$ in $ Z/3Z[X]$ and of course in this field we have $ f_1=1+x+...+x^{degf}, g_1=1+x+...+x^{degg}$. Since $ f_1|g_1$, we have $ x^{degf+1}-1|x^{degg+1}-1$ from where it is trivial to conclude.

:o what a strong tool is this  :o
(some how its power is like invertion  in geometry)
Harazi , if you were in that exam , may i ask you to give us the other problems of it ?
thanks for your strong tool :o

It seems it is classical tool, it's surely not me who found it. Unfortunately, I forgot the other problems given in that exam, but I believe they were already posted on the forum.

harazi , are these, the other two problems of that exam ?
1.[url]=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43423[/url]
2.Find all pairs of positive integers $(m,n)$ so that $m$ devides $a^n -1$ for $a=1,2,3,...,n$

No, I remember something about some international conference and some languages and also about the sum of digits of $[an+b]$.

you mean this:
At an international conference there are $4$ officail languages , for any two participants there is a language among the $4$ officail ones that both of them know . prove that there is a language known by at least 60% of the participants.

Yup, and the other one was to prove that for any positive real numbers $a,b$ there is a subsequence of $ [an+b]$ in which each term has the same sum of digits.

thanks a lot, i think , i found a great exam to start working on .
Romania problems are great for getting a little advance :lol:


Treating an alcohol with acetic anhidride in excess leads to the growth of it`s molecular mass with 126g , the growth representing 136,95. What is the alcohol?

can you write the type of acetic anhidride ?


if n is not power of a prime.
prove:
$ \prod_{1\le k\le n\minus{}1\ ,\gcd(k,n)\equal{}1}\sin \frac{k\pi}{n}\equal{}\frac{1}{2^{\varphi(n)}}$

It followed from $ \prod_k sin \frac{k\pi}{n} \equal{}2^{1\minus{}n}$.

the product is ;
$ \prod_{1<\equal{}k<\equal{}n\minus{}1} sin \frac{k\pi}{n} \equal{}\frac{n}{2^{n\minus{}1}}$

[quote="drapt@"]the product is ;
$ \prod_{1 < \equal{} k < \equal{} n \minus{} 1} sin \frac {k\pi}{n} \equal{} \frac {n}{2^{n \minus{} 1}}$[/quote]
I am sorry, you are right.
Therefore \[ \prod_{(k,n)\equal{}1,1\le k<n}sin\frac{k\pi }{n}\equal{}\prod_{d|n}(\frac{n/d}{2^{n/d\minus{}1}})^{\mu(d)}.\]

[quote="Amir.S"]if n is not power of a prime.
prove:
$ \prod_{1\le k\le n \minus{} 1\ ,\gcd(k,n) \equal{} 1}\sin \frac {k\pi}{n} \equal{} \frac {1}{2^{\varphi(n)}}$[/quote]
we take
$ f(n)\equal{}\prod_{1\le k\le n \minus{} 1\ ,\gcd(k,n) \equal{} 1}\sin \frac {k\pi}{n}$ and $ g(n)\equal{}\prod_{1\le k\le n\minus{}1 }\sin \frac {k\pi}{n}$

$ g(n)\equal{}\prod_{1\le k\le n\minus{}1 }\sin \frac {k\pi}{n}$
$ \equal{}\prod_{d|n}\prod_{1\le k\le n,gcd(k,n)\equal{}d}sin\frac{k\pi}{n}$
$ \equal{}\prod_{d|n}\left(\prod_{1\le k'\le n',gcd(k',n')\equal{}1}sin\frac{k'\pi}{n'}\right)$ (where $ k'\equal{}\frac{k}{d},n'\equal{}\frac{n}{d}$)
$ \equal{}\prod_{d|n}f(n')$
$ \equal{}\prod_{d|n}f(\frac{n}{d})$
$ \equal{}\prod_{d|n}f(d)$

then
$ \prod_{1\le k\le n \minus{} 1\ ,\gcd(k,n) \equal{} 1}\sin \frac {k\pi}{n}\equal{}f(n)$
$ \equal{}\prod_{d|n}g(\frac{n}{d})^{\mu(d)}$
$ \equal{}\prod_{d|n}\frac{(n/d)^{\mu(d)}}{2^{(\frac{n}{d}\minus{}1)\mu(d)}}$
$ \equal{}\frac{1}{2^{\varphi(n)}}$
because $ n/n\equal{}\prod_{d|n}1$ gives $ 1\equal{}\prod_{d|n}(n/d)^{\mu(d)}$
and $ \sum_{d|n}(\frac{n}{d}\minus{}1)\mu(d)\equal{}\sum_{d|n}\frac{n\mu(d)}{d}\equal{}\varphi(n)$


I want to put the label at the end of the line instead of at the midpoint.  How would I do this?

[code]import graph;

unitsize(.5inch);

real x(real t){return t;}
real y(real t){return t;}

guide g=graph(x,y,-2,2);

xaxis("$x$",-3,3,LeftTicks(-2,2,1),EndArrow);
yaxis("$y$",-3,3,RightTicks(-2,2,1),EndArrow);
draw(g,Arrows);
label("$y=x$",g);
[/code]

Sorry, I don't know how to put the image into the post, but I think the code should be sufficient.  Thanks!


function
combinatorics proposed
combinatorics

Let $ B\equal{}${$ 2,3,4....$}
Find all  the subsets $ A$ of $ B$ verifing:

For every $ n \in A$:
-$ n \in A \implies n^2 \plus{}4 \in A$
-$ n \in A \implies [\sqrt{n}]\plus{}1 \in A$

Nobody untill here  :o  :o

where are you?

Notice that  $ [\sqrt {n^2 \plus{} 4}] \plus{} 1 \equal{} n \plus{} 1$ $ \forall n \ge 2$
and from $ f(n)\equal{}[\sqrt {n}]\plus{}1$ is a strictly decreasing function $ \forall n \ge 3$
then the only subset is $ B$

That's it.

$ n \in A \implies n^2\plus{} 4 \in A \implies [\sqrt{n^2\plus{}4}]\plus{} 1 \in A \implies n\plus{} 1 \in A$

and 2 is the smallest number in A


hey!

Is there any solution of that?
1.Ιf y≠0 and n=2k (even).
Show that: x^n + y^n = (x + y)^n
only when x=0

2. Ιf y≠0 and n=2k+1 (odd)
Show that: x^n + y^n = (x + y)^n
only when x = 0 or x = −y

$ n\equal{}0$, no solution

$ n\equal{}1$ all $ x,y$ are solutions

$ n\equal{}2$, $ xy\equal{}0$

$ n\equal{}3$, no positive integer solutions by fermat's last theorem, unless
i) $ x$ or $ y\equal{}0$
ii)$ x\equal{}\minus{}y$, and $ n$ is odd greater than $ 1$.

thank you for your help!
is there any proof of that,
I mean" when the n=2k (even ) the only solution is x=0 (of course when ψ≠0)

and when n=2k+1 (odd) the only two solutions are the x=0 and ψ=-x (of course when ψ≠0 again)

Thank you for your time!

I don't think $ x$ and $ y$ have to be integers. The Fermat's Last Theorem argument shouldn't work.

These proofs have handwaves that I don't care to smooth out.

1) If $ x$ and $ y$ are both positive, then doing a binomial expansion of $ (x \plus{} y)^n$ and cancelling gets $ 0 \equal{} \text{blah}$ where blah is a bunch of binomial coefficients and $ x$s and $ y$s, all positive, so contradiction. If $ x$ and $ y$ are both negative, it reduces to the previous case. If $ x$ is negative and $ y$ is positive, then $ y^n > (x \plus{} y)^n$, but since $ x^n$ is positive and $ x^n \plus{} y^n \equal{} (x \plus{} y)^n$, $ y^n < (x \plus{} y)^n$ by definition. Contradiction. This only leaves one of $ x$ and $ y$ being $ 0$.

2) We get contradictions if $ x$ and $ y$ are both nonzero with the same sign by the logic above. We get solutions if either or both is $ 0$. So it only remains to check them having different signs. WLOG $ x$ is positive and $ y$ is negative, and $ |x| \geq |y|$. (if not, switch and negate the variables) Let $ z \equal{} \minus{}y$ and rewrite the equation as $ x^n \minus{} z^n \equal{} (x \minus{} z)^n$ with $ x$ and $ z$ both positive, so $ x\minus{}z > 0$. Then we have $ x^n \equal{} (x\minus{}z)^n \plus{} z^n$. Replacing $ a \equal{} x\minus{}z$ and $ b \equal{} z$, we get $ (a\plus{}b)^n \equal{} a^n \plus{} b^n$ with $ a, b > 0$. We already said that was impossible unless $ a$ or $ b$ is $ 0$. If $ b \equal{} 0$, $ z \equal{} 0$, which we don't allow, so $ a \equal{} 0$. Thus $ x \equal{} z$, or $ x \equal{} \minus{}y$.

[hide="Without handwaves"] When $ n$ is even we can assume WLOG that $ x, y$ are positive.  Then the binomial argument suffices.

When $ n$ is odd and $ x, y$ have the same sign, the previous argument suffices.  When $ x, y$ have opposite signs, assume without loss of generality that $ x \plus{} y$ is positive.  Then either

$ x^n \equal{} (x \plus{} y)^n \plus{} (\minus{}y)^n$

or

$ y^n \equal{} (x \plus{} y)^n \plus{} (\minus{}x)^n$

and the previous argument gives the only equality case as $ x \plus{} y \equal{} 0$.  [/hide]


How does one solve the differential equation
\[ \frac{dy}{dx}\equal{}\frac{2xy}{x^2\minus{}y^2}?\]

Verify that $ \frac {d}{dx}\left(\frac {x^2 \plus{} y^2}{y}\right) \equal{} 0$ so $ \frac {x^2 \plus{} y^2}{y} \equal{} 2c$ and therefore $ \boxed{
y \equal{} c\pm \sqrt {c^2 \minus{} x^2}}$ for $ x\in [ \minus{} c,c]$.

Sorry could you write this out in detail?

[quote="123456789"]How does one solve the differential equation
\[ \frac {dy}{dx} \equal{} \frac {2xy}{x^2 \minus{} y^2}?
\]
[/quote]

$ y(x)\equal{}xz(x)$.

Then $ xz'\plus{}z\equal{}\frac{2z}{1\minus{}z^2}$.

Hello
[hide="solution "]
As it is a homogeneous equation u can substitute $ y \equal{} vx$ after that ur eq become
$ \frac {(1 \minus{} v^2)dv}{v^2(v^2 \plus{} 1)} \equal{} dx$
which u can do easily.
[/hide]

[hide="One nice thing"]
If u can just rearrange  ur answer ,u will see the equation is of d circles which pass through d origin and whose centers lie on y-axis.[/hide]
Thank u


geometry
3D geometry
trigonometry
complex numbers

Hiya I have this problem, and just dont know where to start!

Find all the roots of the equation z^4 - 27z = 0 in both polar and cartesian form.

Can anyone help me?? thanks!

I haven't studied complex numbers so sorry if I do something wrong,

$z^{4}-27z=0 \implies z(z^{3}-27)=0 \implies z(z-3)(z^{2}+3z+9)=0$

We have two real roots $0, 3$ and equation $z^{2}+3z+9=0$.

$z^{2}+3z+9=0$, $D=9-36=-27=(3\sqrt{3}i)^{2}$
$z=\frac{-3\pm 3\sqrt{3}i}{2}$

ArticMonkey is correct, and his answer gives the cartesian form of the answers. Now, how about the polar form? (Don't bother with that for $z=0$; just do the polar form for the three cube roots of 27.)

[hide="polar form"]
$z =  \frac{-3 \pm 3\sqrt{3}}{2}$

We can rewrite this as $3(\cos \frac{2\pi}{3}+i \sin \frac{2\pi}{3})$ and $3(\cos \frac{4\pi}{3}+i\sin \frac{4\pi}{3})$. So the polar form representations would be $(3, \frac{2\pi}{3})$ and $(3, \frac{4\pi}{3})$.
[/hide]

[hide="Adunakhor:"]You forgot about $(3,0)$  :P [/hide]

Isn't it trivial? $(3,0)$ is just a real number $3$, so it has polar representation the same as cartesian: $(3,0)$, because $|3|=3$ and $\text{Arg}(3)=0$.


LaTeX
superior algebra
superior algebra unsolved

I have no idea which forum this really belongs to, but since I got the idea from a book on algebra, I figured it'd go here first, and the mods can sort things out if I've missed.

First, I came into discussion with a couple of classmates where I made a statement (after drinking) that there are more irrational numbers than there are rational ones. And the day after I started thinking it over. Seeing as $\text{Q} \cup \text{I} =\text{R}$, where $\text{I}$ is the set of all irrational numbers (how do I write the standard notation for those sets in Latex anyways?), and $|\text{Q}|=\aleph_0$, then $|\text{I}|\neq\aleph_0$, since if $|\text{I}|$ was $\aleph_0$, then $|\text{R}|$ would be $2\aleph_0=\aleph_0$, which we know is not true. But we do know that $\aleph_0+c=c$ (where $c$ is the continuum, $|\text{R}|=c$), which leads me to believe that $|\text{I}|=c$, but I can't actually prove it. Any ideas?

The second thing is a rather curious result, that $|\text{R}^n|=c$ also, for positive integers $n$. Mathworld mentions that $\text{R}$ can be put in a one-to-one correspondence with $\text{R}^n$. Just how is this done? For example, the way I see it, it boils down to finding a (not necessarily continuous) parametrized curve that covers every point in $\text{R}^n$. I can't see how this can possibly be done, though. Ideas here as well?

Thanks

For the first question (showing that there are $c$ irrationals) we could do this: extract a sequence of distinct irationals $(a_n)_{n\ge 0}$. Now, this sequence can be put in a bijective correspondence with $(a_n)\cup (r_n)$, where $(r_n)$ is an enumeration of the rationals (because $(a_n)$ and $(a_n)\cup (r_n)$ are both countably infinite). Now $I=(a_n)\cup (I\setminus(a_n))$ can be put in bijective correspondence with $(a_n)\cup(r_n)\cup(I\setminus (a_n))=\mathbb R$.

As for the second part, you can find a bijection from $I=(0,1)$ to $I^2$, for instance, by writing $x=0.a_1a_2\ldots$, where $a_i$ are binary digits, and making $f(x)=(0.a_1a_3\ldots,0.a_2a_4\ldots)$. You can use this to construct a surjection from $\mathbb R$ to $\mathbb R^2$ (and apply it repeatedly to construct one from $\mathbb R$ to $\mathbb R^n$).


[quote="GoBraves"][quote="juicybooty911"]i found this on ebaumsworld.com :D[/quote]

lmao[/quote]

what's Imao? :?

[quote="Scrambled"]mind i ask what was there?[/quote]

Yes. :-)

It was a wanted ad. Go figure the rest.

Mith can you pm that to me pls :D

elvenchamp777

[quote="elvenchamp777"]Mith can you pm that to me pls :D [/quote]

Unthinkable. How could you ask such a thing. Tsk...

[quote="furious"]too bad it had to be deleted  :(
i thought it was really funny.[/quote]

Me Too

[quote="MathFiend"]LMAO stands for [b]L[/b]aughing [b]M[/b]y "Butt" [b]O[/b]ff

I think you can get it from there :)[/quote]

 :D  :D  :D  :lol:  :lol:  :P  ;)

[quote="MithsApprentice"]Come on now guys...watch your posts. Juicy, no more of those ads...(the one I deleted I mean).[/quote]

sorry, no offense, it was just too funny :)


advanced fields
advanced fields unsolved

[quote="amfulger"]Consider the sequence $(t_n)_{n\in\mathbb N}$ defined by
\[ t_1=t_2=t_3=t_4=t_5=1  \]
\[ t_{n+5}=\frac{t_{n+1}t_{n+4}+t_{n+2}t_{n+3}}{t_n}.  \]

Prove that $t_n\in\mathbb N$ for all $n\geq 1$.[/quote]
This is the [url=http://mathworld.wolfram.com/SomosSequence.html]Somos-5 sequence.[/url]

Having Don Zagier's name associated with this problem is honorable, but the person to come up with this stuff is [url=http://cis.csuohio.edu/~somos/home.html]Michael Somos[/url]. A site dedicated to Somos sequences is [url=http://www.math.wisc.edu/~propp/somos.html]here.[/url] There's also a link on the latter site that gets you to Don Zagier's (sketch of a) solution.

Looks more like a research program to me :).


geometry
perpendicular bisector
geometry unsolved

Let $ABC$ be a triangle, with $A = 2C, 2B = A+C$ ($A, B$ and $C$ are the angles). The bisector of $C$ is $CP$. The middle point of $AP$ is $R$.  The height relative to $BC$ is $AQ$. The perpendicular bisector of $RQ$ intersects $AC$ at point $M$. Prove that $AM = CM$.

only $A=B+\frac{C}{2}$ is enough...
in $\triangle ACP$ we get that $\angle CPA=B+\frac{C}{2}=A$ so $\triangle MAP$ is isosceles,so we get that $CR \bot BA$ and also in another hand we had $AQ \bot BC$ so we get that points $C,Q,R,A$ lie on a circle with center $M$ where $M$ is the midpoint of $AC$,the rest is trivial,because we know that the perpendicular bisector of chord $RQ$ passes thruogh the center of the circle which is the midpoint of $AC$...

[quote="ffao"][quote="BaBaK Ghalebi"] so we get that points $C,Q,R,A$ lie on a circle with center $M$[/quote]

Why does this follow? Probably a stupid question, though...[/quote]
well we have proved that $\angle CRA=\angle CQA=90$ so $CQRA$ is cyclic,name this circle $w$,so from $\angle CQA=90$ we get that $AC$ is the diameter of $w$ wo $M$ is the center


If a fair coin will be flipped three times, what is the probability of flipping at least two heads in a row? Express your answer as a common fraction.

The three possibilities that work are HHT, THH, and HHH.  There are $ 8$ total, so $ \boxed{\frac{3}{8}}$

1.head,head,tail
2.tail,head,head
3.head,head,head
so, is 3/8
...
ezpz :dry:

I think the answer is 1/2
The probabilities of getting heads:
No head: 1/8 (T T T)
1 head:    3/8 (H T T, T H T, T T H)
2 heads:  3/8 (H H T, H T H, T H H)
3 heads:  1/8 (H H H)
And we have 3/8 + 1/8 =1/2 is the answer.
On the other hand, we can also do
1/2*1/2 *1/2*(2C6) +1/2*1/2*1/2=1/2

[quote="GameBot"]If a fair coin will be flipped three times, what is the probability of flipping [size=150]at least two heads in a row?[/size] Express your answer as a common fraction.[/quote]
Therefore, the possibilities are, like the posts above, HHH, HHT, and THH. $\frac{3}{2^3}=\boxed{\frac{3}{8}}$