how could i make my TeX pdf color fonted.... i ran out of black ink....ick! ;)

use the package [b]color[/b]
There's some already defined colors like blue, red and, you know, colors!  You can define your own colors using

\definecolor{[i]name[/i]}{[i]colorsystem[/i]}{[i]values[/i]}

the color system I know is rgb (red, green, blue) and [i]values[/i] is numerical values between 0 and 1, representing the % of each basic color.

Anyway, the simple colors are probably ok for you. Then use

\textcolor{[i]colorname[/i]}{[i]text[/i]}

There's surely a way to define a new default color for the text, but I don't know it.

ledurt

I get the impression that the color package doesn't work quite perfectly. Sometimes very strange things happen when I use colors: sometimes it insists on changing the color of a whole page, and sometimes it gets rid of page numbers.

I used the color package on a very limited among of text, like URL.  I didn't have any trouble.  I quickly look on internet though and found some more infos on the color package, but with no notes on stability.

[url=http://oregonstate.edu/~peterseb/tex/samples/docs/color-package-demo.pdf]oregonstate.edu/~peterseb/tex/samples/docs/color-package-demo.pdf[/url]

ledurt

Finally found some color commands

[quote]In the following colour commands, col_spec is either the name of a defined colour, like {blue}, or [model]{spec}, like [rgb]{0,1,0}.

\pagecolor col_spec
sets the background colour for the current and following pages;
\color col_spec
is a declaration to switch to setting text in the given colour;
\textcolor col_spec{text}
sets the text of its argument in the given colour;
\colorbox col_spec{text}
sets its argument in a box with the given colour as background;
\fcolorbox col_spec1col_spec2{text}
is like \colorbox, with a frame of col_spec1 around a box of background colour col_spec2; the two specifications must either both be defined ones, or both use the same model, which is given only once; for example, \fcolorbox{red}{green}{Text} sets `Text' in the current text colour on a green background with a red frame;
\normalcolor
switches to the colour that was active at the end of the preamble. Thus placing a \color command in the preamble can change the standard colour for the whole document. This is the equivalent to \normalfont for font selection. [/quote]

this was taken from this site:

[url=http://tex.loria.fr/graph-pack/grf/grf.htm#Q1-1-21]tex.loria.fr/graph-pack/grf/grf.htm#Q1-1-21[/url]

ledurt


trigonometry
number theory proposed
number theory

Solve in rational numbers:
$y^2=x^3-3x+2$

Suppose $x\neq 1$, then we have $\left(\frac{y}{x-1}\right)^2=x+2$, so $x+2=q^2$, $q\in\mathbb{Q}$. It yields that $x=q^2-2$ and $y=\pm q(q^2-3)$.

That's all.


[quote="Stun"]Let $f,g : R \rightarrow R$ sastisfying there exist $n,k \in R+$ such that
$|f(y)-f(x)-g(x)(y-x)| \le n.|x-y|^{2+k}$   $\forall x,y\in R$
Prove that : $f(x)=ax+b$[/quote]


function
trigonometry
inequalities
inequalities unsolved

Prove that for every triangle $ ABC$ :
$ (cos A\plus{}cos B)^2 \plus{}(cos B\plus{}cos C)^2\plus{}(cos C\plus{}cos A)^2\le 3$

[quote="duongptnkvn"]Prove that for every triangle $ ABC$ :
$ (cos A \plus{} cos B)^2 \plus{} (cos B \plus{} cos C)^2 \plus{} (cos C \plus{} cos A)^2\le 3$[/quote]
This inequality is false.
Try $ A \equal{} 15^\circ$, $ B \equal{} 15^\circ$, and $ C \equal{} 150^\circ$.
$ (\cos 15^\circ \plus{} \cos 15^\circ)^2 \equal{} \left(\dfrac{\sqrt {6} \plus{} \sqrt {2}}{2}\right)^2 \equal{} \dfrac{8 \plus{} 2\sqrt {12}}{4} \equal{} 2 \plus{} \sqrt {3} > 3$


[img]http://oyak.domivka.net/photo/RockEkzystenciya-2005/images/27.jpg[/img][img]http://oyak.domivka.net/photo/RockEkzystenciya-2005/images/31.jpg[/img]

[size=150][b]Welcome to the official site of the Ukrainian ethno-art-rock performance band

"Yanka Kozyr Orchestra"![/b]

We are working on the Ukrainian stage to make new, high-quality professional modern show-product.

[b][u]"Yanka Kozyr Orchestra" is:[/u][/b]
[b]* [/b] Original stage costumes
[b]* [/b]Unique masks
[b]* [/b]Colorful treated vocal
[b]* [/b]Modern way to realize theatrical ethno-art-rock music
[b]* [/b]Frank & shocking show

The theatrical, mysterious folk action, effective & fantastic energy
of the singer,
taken together with the work of the group, impresses everybody.
It's original and new (at least the performance is like nothing in
Ukraine has ever been).
So, let's get acquainted with this band![/size]

 :arrow: [url]http://www.oyak.domivka.net/index.php?mova=eng[/url]
 :arrow: [url]http://www.oyak.domivka.net/mp3[/url]


function
logarithms
complex analysis
complex analysis unsolved

Prove that
$ cosh^{\minus{}1} z \equal{} 2\: ln \dfrac{\sqrt{z\plus{}1}\plus{}\sqrt{z\minus{}1}}{\sqrt{2}}$

I know how to derive the the standard formula which is: $ cosh^{\minus{}1} z \equal{} ln(z\plus{}\sqrt{z^2\minus{}1})$ but the above formula is used in some texts but I don't know how to derive it.

Try bringing the 2 inside the logarithm (with appropriate operations).


Someone said these weren't MC level, so I'm posting the questions here also.

They are going to be in increasing levels of difficulty.

I) I have a 10 foot flagpole. I have 3 different types of 2 foot flags and 2 different types of 1 foot flags. I have thousands of each of the types of flags. The order of the flags matter, and while flags of the same type are indistinguishable, flags of different types are distinguishable. Find the number of ways I can arrange flags to exactly fill the 10 foot flagpole.

II) Let $ a_1, a_2,...$ be a seqeunce with the following properties:

(i) $ a_1 \equal{} 1$
(ii) $ a_{2n} \equal{} n(a_n)$ for any positive integer n.

What is the value of $ a_{2^{100}}$?

III) How many 6 digit base 4 numbers have the property that they contain at least one 1 and that the first 1 is to the left of the first 0? There can be no 0's.

IV) Let $ a_1 \equal{} p$ and $ a_2 \equal{} q$, where $ p$ and $ q$ are positive integers, and let $ a_n \equal{} a_{n \minus{} 1} \cdot a_{n \minus{} 2}$ for all $ n \ge 3$. Find a formula for $ a_n$ in terms of $ n$, $ p$, and $ q$.

V) Call a set of integers [b]spacy[/b] if it contains no more than one out of any three consecutive integers. How many subsets of $ \{ 1, 2, 3, ... ,12 \}$ are spacy?

VI) Find the number of 10 digit binary sequences that have exactly one pair of consecutive 0's.

V) We have the set of numbers 1 through 12, from which we can choose no more than 4 numbers. If we consider those the balls, there must be a minimum of 2 dividers between each one. For subsets with $ n$ elements, there are $ 2(n\minus{}1)$ dividers, so we have $ \binom{(12\minus{}3n\plus{}2)\plus{}(n\plus{}1)\minus{}1}n \equal{}\binom{12\minus{}2n\plus{}2}n$ as the number of ways. Thus, our answer is $ \binom 64\plus{}\binom 83\plus{}\binom{10}2\plus{}\binom{12}1\plus{}\binom{14}0 \equal{} 129$.

[hide="II)"]we know $ a_{1} \equal{} 1$

we also know $ a_{2n} \equal{} n(a_{n})$

let's list some values

$ a_1 \equal{} 1$
$ a_2 \equal{} a_{2^1} \equal{} a_{2(1)} \equal{} 1(a_1) \equal{} 1(1) \equal{} 1$
$ a_4 \equal{} a_{2^2} \equal{} a_{2(2)} \equal{} 1(a_2) \equal{} 1(1) \equal{} 1$
$ a_8 \equal{} a_{2^3} \equal{} a_{2(4)} \equal{} 1(a_4) \equal{} 1(1) \equal{} 1$

we see that for any $ a_n$ where $ n \equal{} \textit{a multiple of 2}$, the answer is $ 1$

since $ 2^{100}$ is a multiple of $ 2$, $ a_{2^{100}} \equal{} \boxed{1}$[/hide]

[hide="IV)"]
let's list out some terms

$ a_1 \equal{} p$
$ a_2 \equal{} q$
$ a_3 \equal{} a_2\times a_1 \equal{} q\times p \equal{} qp$
$ a_4 \equal{} a_3\times a_2 \equal{} qp\times q \equal{} q^2p$
$ a_5 \equal{} a_4\times a_3 \equal{} q^2p\times qp \equal{} q^3p^2$
$ a_6 \equal{} a_5\times a_4 \equal{} q^3p^2\times q^2p \equal{} q^5p^3$

it seems to me like $ a_n \equal{} q^{(n \minus{} 1)\textit{th fibonacci number}}p^{(n \minus{} 2)\textit{th fibonacci number}}$

not exactly sure how to write that with only $ n, p, \text{and} q$, but that's what i got[/hide]

[quote="vallon22"]we know $ a_{1} \equal{} 1$

we also know $ a_{2n} \equal{} n(a_{n})$

let's list some values

$ a_1 \equal{} 1$
$ a_2 \equal{} a_{2^1} \equal{} a_{2(1)} \equal{} 1(a_1) \equal{} 1(1) \equal{} 1$
$ a_4 \equal{} a_{2^2} \equal{} a_{2(2)} \equal{} 1(a_2) \equal{} 1(1) \equal{} 1$
$ a_8 \equal{} a_{2^3} \equal{} a_{2(4)} \equal{} 1(a_4) \equal{} 1(1) \equal{} 1$

we see that for any $ a_n$ where $ n \equal{} \textit{a multiple of 2}$, the answer is $ 1$

since $ 2^{100}$ is a multiple of $ 2$, $ a_{2^{100}} \equal{} \boxed{1}$[/hide]

[/quote]

About your comment on $ a_4$, $ a_4$ = $ 2(a_2)$ which would be equal to 2. The question stated that $ a_2n \equal{} n(a_n)$.

Sorry, but it's not a trick question.

oops
that was stupid

so,
$ a_2\equal{}a_{2^1}\equal{}a_{2(1)}\equal{}1(a_1)\equal{}1(1)\equal{}1\equal{}2^0$
$ a_4\equal{}a_{2^2}\equal{}a_{2(2)}\equal{}2(a_2)\equal{}2(1)\equal{}2\equal{}2^1$
$ a_8\equal{}a_{2^3}\equal{}a_{2(4)}\equal{}4(a_4)\equal{}4(2)\equal{}8\equal{}2^3$
$ a_16\equal{}a_{2^4}\equal{}a_{2(8)}\equal{}8(a_8)\equal{}8(8)\equal{}64\equal{}2^6$
$ a_32\equal{}a_{2^5}\equal{}a_{2(16)}\equal{}16(a_16)\equal{}16(64)\equal{}1024\equal{}2^{10}$

so $ a_{2^n}\equal{}2^{\textit{the sum of the first (n\minus{}1) numbers}}$

so we want to find $ a_{2^{100}}$, which would be $ 2^{\textit{sum of the first 99 numbers}}$

the sum of the first $ n$ numbers is $ \frac{n(n\plus{}1)}{2}\equal{}\frac{99(100)}{2}\equal{}99\times50\equal{}4950$

so our answer is $ 2^{4950}$, which i don't wanna calculate

hope i didn't make a mistake there =)

[quote="TheDeity"][quote="cgyao15"]I'm pretty sure you got it this time :)
aw... dang my calculator got owerflowed...  :([/quote]

Some people can do math without calculators. Try it.[/quote]I'm pretty sure he meant calculating $ 2^{4950}$ which has over 1000 digits, exceeding most calculator's overflow threshold.


Let $ a_0\equal{}a$
$ a_n  \equal{} 2a_{n \minus{} 1} ^2  \minus{} 1$
Find which value of a that $ a_n$ have the limit :)

[hide]For |a|>1 it is clear that the absolute value of the terms of the sequence always increases, so we can assume that $ |a|\leq1$.  Let $ a\equal{}cos(\theta)$.  Then since $ 2cos^2(\theta)\minus{}1\equal{}cos(2\theta)$, we have that $ a_n\equal{}cos(2^n\theta)$.  Since cos is continuous, this having a limit means that it must tend towards a fixed point, of which there are 2: 1 and -1/2.  If $ a_i$ is close to 1, say 1-x, then $ 2(1\minus{}x)^2\minus{}1\equal{}1\minus{}4x\plus{}2x^2$.  But if x is close to 1, |4-2x|>1, so |4x-2x^2|>|x| unless x=0, so it differs from 1 more than its predecessor did.  Thus it cannot tend to 1 unless one of the terms eventually equals 1. Also, $ 2(\minus{}1/2\minus{}x)^2\minus{}1\equal{}\minus{}1/2\plus{}2x\plus{}2x^2$, which differs from -1/2 by 2x+2x^2.  Also, |2x+2x^2|>|x| if x is nonzero and small, so by the same principle, if it has a limit, the terms must equal 1 or -1/2 at some point.  Notice that the cosines corresponding to these points are $ cos(\frac{2\pi{k}}{3})$ for some positive integer k.  Thus $ \theta$ is this divided by a power of 2 i.e. some binary fraction times $ \frac{\pi}{3}$, and a is the cosine of this.[/hide]


group theory
abstract algebra
algebra
function
domain
superior algebra
superior algebra unsolved

Show the following by observing that $ G/N$ is a homomorphic image of $ G$:

a.  If $ G$ is a cyclic group and $ N$ is a subgroup of $ G$, show that $ G/N$ is a cyclic group.
b.  If $ G$ is an abelian group and $ N$ is a subgroup of $ G$, show that $ G/N$ is an abelian group.

[u]Attempted Solutions[/u]:
a.

Let $ G \equal{} <a>$, then claim $ G/N \equal{} <aN>$
		also $ aN \in G/N$ since $ G/N \equal{} \left\{gN|g \in G\right\}$ modulo the equivalent class.

		This shows that $ <aN> \subseteq G/N$

		Now let $ wN\in G/N$, then $ w \in G$ so $ w\equal{}a^{n}$ for some n.
		This means that  $ wN \equal{} (aN) ^{n}$ or $ wN \in <aN>$

		Therefore $ G/N \subseteq <aN>$.

b.

Let $ G$ be an abelian group.  Let $ N$ be a normal subgroup in $ G$, then	let    $ aN, bN \in G/N$

		then $ aNbN \equal{} abN \equal{} baN \equal{} bNaN$

		using the fact that ab=ba  (abelian); and the definition of product in $ G/N$.
		therefore $ G/N$ (factor group) is abelian.

Any advice on using the homomorphic image to solve these would be appreciated.   Thanks in advance.

Good solutions!

You've implicitly used the fact that $ g \mapsto gN$ is a homomorphism.

If $ \phi: G\to H$ is any surjective homomorphism--for example, when $ H\equal{}G/N$ and $ \phi(g)\equal{}gN$--then, assuming $ G\equal{}\langle a\rangle$, you can show that $ H\equal{}\langle \phi(a) \rangle$ exactly like you did in your solution.  Also, if $ G$ is abelian, then $ \phi(g)\phi(h)\equal{}\phi(gh)\equal{}\phi(hg)\equal{}\phi(h)\phi(g)$, just like in your solution.

Thank you for the meaningful response.  For the cyclic problem (a), can I use the exponent (j) for both $ Na$ and $ Nb$ as follows?

Let $ G \equal{} <a>$, then $ G/N \equal{} <aN>$
also $ aN \in G/N$ since $ G/N \equal{} \left\{gN|g \in G\right\}$
This shows that $ <aN> \subseteq G/N$
Now let $ wN \in G/N$, then $ w \in G$ so $ w\equal{}a^{n}$ for some $ n$.
This means that $ wN \equal{} (aN)^{n}$ or $ wN \in <aN>$
Therefore $ G/N \subseteq <aN>$

Using homomorphic image
let $ x \in G$ where $ x$ is of the form $ x\equal{}a^{j}$
let $ \varphi : G \rightarrow G/N$ such that $ \varphi(g) \equal{}N/G$ and $ \varphi(g) \rightarrow \varphi(Ng)$
using the cyclic form $ \varphi (x^{j}) \rightarrow \varphi (Nx)^{j}$ where $ <Nx>$ is the generator
let $ a,b \in G$ such that $ \varphi(Na), \varphi(Nb) \in G$
$ \varphi(Na)^{j}\varphi(Nb)^{j} \equal{} \varphi(Nab)^{j}$
$ \varphi(Na)^{j}(Nb)^{j}$ so $ \varphi(N^{j}(ab)^{j})$ as a cyclic group is abelian and using NN = N;
$ \varphi(Nab)^{j}$

$ \varphi (Na)^{j}\varphi(Nb)^{j} \equal{} \varphi(Nab)^{j}$ therefore cyclic.

Keep in mind, $ \phi(Ng)$ does not really make sense, as $ Ng$ is not in the domain of $ \phi$.

Am i understanding that it should be more like:

$ \varphi(a^{j})\varphi(b^{j}) \equal{} (Na^{j})(Nb^{j})\equal{}(Nab)^{j}\equal{}\varphi((ab)^{j})$

Where $ (ab)^{j}$ is of the cyclic form with $ <ab>$ as the generator.


analytic geometry
real analysis
real analysis unsolved

Find field lines in R3 is not obvious to me, as they were in R2..

A fluid flow in R3 has velocity field v=(2x,z,-z^2)

Find the equation of the flow line that passes through an arbitrary point (x0,y0,z0) at time t=0.

Since the field does not depend on $y$, the problem is almost two-dimensional. First, find the flow lines of the field $(2x,-z^{2})$ in the $xz$-plane. Then find the $y$-coordinate from $dy/dt=z$, where $z$ is already known.


search
calculus
integration
derivative
ceiling function
floor function
geometry

[url=http://www.mathlinks.ro/viewtopic.php?search_id=118204258&t=82801]The Original[/url].

$ p_{1} \cdot p_{\left( p_{\left( p_{1} \right) }\right)}^{p_{1}} \cdot p_{\left( p_{1}^{\left( p_{p_{1}}\right) } \right) }\plus{}\frac{p_{1}}{p_{1}}\equal{}851$.
Where $ p_{n}$ is the $ n^{\text{th}}$ prime.

[color=blue][b]8[/b][/color] 1 6
3 [color=blue][b]5[/b][/color] 7
4 9 [color=blue][b]2[/b][/color]

Tne sum of the roots of $ x^2\minus{}853x\plus{}\pi^{5238429347}\equal{}0$

The value of $ \sum^{854}_{k\equal{}1} 1$

a+b is a perfect number.
(a-b)/2 is one less than a perfect number.
Find a*b*(a-b)/2.

$ \dfrac{444\cdot 4}{\frac{4}{4} \plus{} \frac{4}{4}} \minus{} 44 \minus{} 4 \plus{} \dfrac{44 \minus{} 4}{4}\plus{}4\cdot 4\cdot 4^{\minus{}\dfrac{4}{4\plus{}4}}$
hmm apparently 4 is unlucky in china
 :o

[quote="bowei"]$ \dfrac{444\cdot 4}{\frac {4}{4} \plus{} \frac {4}{4}} \minus{} 44 \minus{} 4 \plus{} \dfrac{44 \minus{} 4}{4} \plus{} 4\cdot 4\cdot 4^{ \minus{} \dfrac{4}{4 \plus{} 4}}$
hmm apparently 4 is unlucky in china
 :o[/quote]

Actually, that's because in chinese, the word for 4 is very similar to the word for dead.

They are not similar, they're pronounced similarly. In fact they're pronounced identically, they're just stressed differently, but in Chinese stress matters a lot. There are five stresses.

In fact it's considered so unlucky some buildings even skip over it in the floor numbers.

Anyways:

Take the binary palindrome 10001. Convert to octal. Treat literally as decimal. Convert to binary.  Treat literally as decimal. Convert to hexadecimal. Treat literally as octal. Convert to decimal. Treat literally as octal. Convert to decimal.

To treat literally is to e.g. take 2555555 in octal and change to 2555555 in decimal.

[quote="math_explorer"]In fact they're pronounced identically, they're just stressed differently, [/quote]

I believe that qualifies as pronounced differently.

$ 22.7105263\cdot \frac{D(10!\minus{}2*9!\minus{}3*8!\minus{}4*7!\minus{}5*6!\minus{}6*5!\minus{}7*4!\minus{}8*3!\minus{}9*2!\minus{}10*1!)}{4}$, where $ D(x)$ is the sum of the digits of $ x$.

... I wish I knew good math... TPAM's post +1...

Thanks.  :lol:
everyone else in this thread is at least twice as good as me still, though.
$ (.5^{\minus{}1}) \equal{} 2$
so...
$ (.5^{\minus{}1})*458$
because that is 1+ TPAM's

The area of a square with side equivalent to the diagonal of a rectangular prism with dimensions 30, 5, and 1.

[hide="Why this thread is locked."]
[quote="sailingdog"]2TPAM[/quote]

[b][quote="airyork6=)"]ummm whatever he said +1[/quote][/b]

[quote="erabel"]... TPAM's post + 1...[/quote]

[quote="LevatorNasolabialis"]TPAM +1

The message is too small. Please make the message longer before submitting.
The message is too small. Please make the message longer before submitting.
The message is too small. Please make the message longer before submitting.
The message is too small. Please make the message longer before submitting.[/quote]
[/hide]


Jeremy made a Venn diagram showing the number of students in his class who own types of pets. There are 32 students in his class. In addition to the information in the Venn diagram, Jeremy
knows half of the students have a dog, $ \frac{3}{8}$ have a cat, six have some other pet and five have no pet at all. How many students have a cat, a dog and some other pet?

[asy]unitsize(50);
import graph;
pair A = (0,-1); pair B = (sqrt(3)/2,1/2); pair C = (-sqrt(3)/2,1/2);
draw(Circle(A,1.2) ^^ Circle(B,1.2) ^^ Circle(C,1.2));
label("10",A); label("2",B); label("9",C); label("$z$",(0,0)); label("$w$",(B+C)/2); label("$y$",(A+B)/2); label("$x$",(A+C)/2);
label("Cats",2.4C,C); label("Other Pets",2.4B,B); label("Dogs", 2.4A,A);[/asy]

[quote="math154"]From this information, we have $ x\plus{}y\plus{}z\equal{}6$, $ x\plus{}z\plus{}w\equal{}3$, $ y\plus{}z\plus{}w\equal{}4$, and $ x\plus{}y\plus{}z\plus{}w\plus{}9\plus{}2\plus{}10\plus{}5\equal{}32\implies x\plus{}y\plus{}z\plus{}w\equal{}6$. Subtracting the first, second, and third equations from this, we get $ w\equal{}0$, $ y\equal{}3$, and $ x\equal{}2\implies z\equal{}\boxed1$.[/quote]
Um, I don't get how you got any of the equations. $ x\plus{}y\plus{}z\equal{}6$
I don't believe that's stated in the problem, sorry for my inexperience ;)

If there are $32$ students and $\frac{1}{2}$ have dogs, then $16$ have dogs. So looking at the bottom circle, we get $x+y+z+10=16\implies x+y+z=6$.

If there are $32$ students and $\frac{3}{8}$ have cats, then $12$ have cats. So looking at the upper left circle, we get $w+x+z+9=12\implies w+x+z=3$.

$6$ students have other pets. So looking at the upper right circle, we get $w+y+z+2=6\implies w+y+z=4$.

And because there are $32$ students total, we can look at the Venn diagram as a whole (keeping in mind that there is an unwritten $5$ outside the circles, since $5$ students have no pet) and get $w+x+y+z+10+2+9+5=32\implies w+x+y+z=6$.

We subtract each of the first $3$ equations from this last one to get $w,x,y$, and then substitute into this last equation to get $z$.


geometry
3D geometry
trigonometry
calculus
calculus computations

Judging from the lack of response I assume I have posted this in the wrong place, sorry.

[quote="Dr. No"][quote="wolfuncle"]I have been working on an article on pebble bed nuclear reactors and am trying to create a visual based on the spent fuel produced by one 110MW plant in its 40 year life span. The fuel for this reactor is in the shape of a ball and is 60 mm in diameter. I have converted this to 2.36 inches and converted that to .197 feet, since I would like to have the final representation in feet.

My problem is that I would like to create a graphic representing the size of the pile (cone). The reactor will create 2,500,000 spent fuel balls in its lifetime, that are .197 feet in diameter. If I pile them into a conical pile what size will that pile be? And if I want to increase that pile to include 10, 50, 100 or more reactors would the conical pile grow in width and height simply by the multiple of the number of nuclear reactors producing waste?

Any help or direction on this basic formula would be greatly appreciated.[/quote]

wolfuncle,

It might not be the wrong place. The forum is subdivided into various areas of interest. The area you posted to is for students studying for Math Olympiads in their respective countries.  It is an interesting problem that I would help you solve myself if I had the time.  Might I suggest though that you ask one of the moderators to move your post to the physics forum where you might get a few more takers.[/quote]

Thanks for the suggestion

[quote="jmerry"]If small objects are dropped in the same place, they will form a pile in the shape of a right circular cone with sides at some fixed angle $\theta$ from the horizontal ($\theta$ depends on the shape and size of the small objects and should probably just be guessed here). If the base has radius $r$, the volume is then $Kr^3$, where $K=\frac\pi3\tan\theta$. This is a standard volume formula- it is easy to see that the height is $r\tan\theta$. To find the radius for a given volume, just solve for $r$; $r$ will be proportional to the cube root of the volume.[/quote]

The size of the balls is 60mm diameter, i would assume the pile would be stacked from the bottom up and would actually not be contained so a free standing pile of round objects I understand would created a cone with a base diameter nearly the same as the height. Thanks or the help I will work on this.


group theory
abstract algebra
superior algebra
superior algebra unsolved

hi everyone, i've got this really hard exercise set by my maths lecturer and i really have no idea how to approach it, i was wondering whether anyone could help?

(i) Let [i]"Theta" belong to S[size=75]n[/size][/i]. Show that "Theta" can be expressed as a product of transpositions in the form (1 x) where 2 <= x <= n

(ii) Show that "Theta" can be expressed as a product of transpositions of the form (x x+1) where 1 <= x <= n-1

(iii) Let Y denote the permatation (1 2 ... n). Show that Y(x x+1)Y[size=75]^(-1)[/size] = (x+1 x+2) for 1 <= x <= n-2.

(iv) By considering (Y^x)(1 2)(Y^(-x)) for 1 <= x <= n-2 or otherwise show that the smallest subgroup that contains (1 2) and (1 2 ... n) is S[size=75]n[/size]

thanks in advance :) [b]any[/b] help would be greatly appreciated

aah sorry i didn't spot that, thank you very much


AMC
USA(J)MO
USAMO
geometry
circumcircle
trigonometry
perpendicular bisector

Those lines are the symmedians of the triangle, so you bet they're concurrent! :D Try to show that $A_2$ is the intersection of the tangents in $B,C$ to the circumcircle $(O)$ of $ABC$.

[quote="paul_mathematics"]Given a nonisosceles, nonright triangle ABC, let O denote the center of its circumscribed circle, and let $A_1$, $B_1$, and $C_1$ be the midpoints of sides BC, CA, and AB, respectively. Point $A_2$ is located on the ray $OA_1$ so that $OAA_1$ is similar to $OA_2A$. Points $B_2$ and $C_2$ on rays $OB_1$ and $OC_1$, respectively, are defined similarly. Prove that lines $AA_2$, $BB_2$, and $CC_2$ are concurrent, i.e. these three lines intersect at a point.[/quote]

Since the triangle $OAA_1$ is similar to triangle $OA_2A$, we have $\frac{OA_1}{OA} = \frac{OA}{OA_2}$. On the other hand, we have OA = OB, since the point O is the circumcenter of triangle ABC, and thus this rewrites as $\frac{OA_1}{OB} = \frac{OB}{OA_2}$. Together with $\measuredangle BOA_1 = \measuredangle A_2OB$, this yields that the triangles $OBA_1$ and $OA_2B$ are similar, what, in turn, shows us that $\measuredangle OBA_2 = \measuredangle OA_1B$. But the point O is the circumcenter of triangle ABC and thus lies on the perpendicular bisector of its side BC; since the point $A_1$ is the midpoint of this side BC, we thus have $OA_1\perp BC$ and $\measuredangle OA_1B = 90^{\circ}$. Consequently, $\measuredangle OBA_2 = \measuredangle OA_1B = 90^{\circ}$, and $BA_2 \perp OB$. Since OB is a radius of the circumcircle of triangle ABC, the line $BA_2$ is therefore the tangent to the circumcircle of triangle ABC at the point B. In other words, the point $A_2$ lies on the tangent to the circumcircle of triangle ABC at the point B. Similarly, the same point $A_2$ lies on the tangent to the circumcircle of triangle ABC at the point C. Thus, our point $A_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points B and C. Similarly, the point $B_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points C and A, and the point $C_2$ is the point of intersection of the tangents to the circumcircle of triangle ABC at the points A and B.

Now, the concurrence of the lines $AA_2$, $BB_2$, $CC_2$ becomes a well-known fact (in fact, these lines concur at the symmedian point of triangle ABC); the simplest proof of this fact uses the Ceva theorem: Since the two tangents from a point to a circle are equal in length, we have $CB_2 = B_2A$, $AC_2 = C_2B$ and $BA_2 = A_2C$, so that

$\frac{B_2A}{AC_2}\cdot\frac{C_2B}{BA_2}\cdot\frac{A_2C}{CB_2} = \frac{B_2A}{C_2B}\cdot\frac{C_2B}{A_2C}\cdot\frac{A_2C}{B_2A} = 1$.

After the Ceva theorem, applied to the triangle $A_2B_2C_2$, it now follows that the lines $A_2A$, $B_2B$, $C_2C$ are concurrent. In other words, the lines $AA_2$, $BB_2$, $CC_2$ are concurrent. $\blacksquare$

As usual, Grobber was quicker than me ;) . Anyway, I have tried to give a proof as elementary and detailed as possible since I don't know how much advanced geometry you know. I also have solutions for some of the other problems you posted here, but I am quite under time pressure now and I'll look whether I will succeed to write them up.

  Darij

Let $G$ be the centroid and $H$ the orthocenter of $\triangle ABC$. Then $\angle OAA_2=\angle OA_1A=\angle A_1AH$, and $\angle BAO=90^{\circ}-C=\angle HAC$, so we have $\angle BAA_2=\angle A_1AC$. Similarly we can get $\angle AA_2C=\angle BAA_2$ and so on. By trig Ceva we have \begin{align*}\frac{\sin\angle BAA_2}{\sin\angle A_2AC}\cdot\frac{\sin\angle ACC_2}{\sin\angle C_2CB}\cdot\frac{\sin\angle CBB_2}{\sin\angle B_2BA}&=\frac{\sin\angle A_1AC}{\sin\angle BAA_1}\cdot\frac{\sin\angle B_1BA}{\sin\angle CBB_1}\cdot\frac{\sin\angle C_1CB}{\sin\angle ACC_1}\\ &=1\end{align*} since $AA_1$, $BB_1$, and $CC_1$ concur at $G$. Therefore $AA_2$, $BB_2$, and $CC_2$ are concurrent as well. $\Box$

Note that $(OA_1)(OA_2)=OA^2$, so $A_2$ is the image of $A_1$ about an inversion with respect to the circle.

Because $BA_1C$ are collinear, $OBA_2C$ are concyclic ($B$ and $C$ are their own images under the inversion centered at $O$).  Clearly there is a unique point $X$ that is not $O$ on $OA_1$ such that $OBXC$ are conyclic.  Now, the intersection of the tangents to the circumcircle at $B$ and $C$ is clearly on $OA_1$ and clearly is concyclic with $COB$ since if that points is $X$, $<XCO=<XBO=90$.  Hence, $A_2$ is the intersection of the tangents to the circumcircle at $B$ and $C$.  Thus, $AA_2$ is a symmedian by the well known symmedian lemma.  The others are similarly symmedians, so we are done and they concur by the isogonal conjugates theorem.

As a consequence, we have that the inversions of the lines $AA_2$ which are the circles through $AOA_1$, $BOB_1$, $COC_1$, concur.  In fact we have that they concur at the image of the symmedian point after an inversion about $O$.

clearly, the best way to do it is

We use a polar transformation with respect to the circumscribed circle, so we need to prove that $BB \cap AC$, $CC \cap AB$, and $AA \cap BC$ are concurrent. But this is easy through Pascal's theorem on $AABBCC$

We'll angle chase to show that $AA_2,BB_2,CC_2$ are the symmedians of $\triangle ABC$. By a well known fact we know that those intersect in the [i]Lemoine Point/Symmedian Point[/i]. (We could also easily prove it using Ceva by using the property that if $D$ lies on $BC$, such that $AD$ is the $A$-symmedian of $\triangle ABC$, then $\tfrac{|BD|}{|DC|} = \left( \tfrac{|AB|}{|CA|} \right)^2$.)

Now let $\theta := \angle C_1CO$ and $\sigma := \angle OC_1C$. By the similarity we get $\angle C_2CO= \sigma$ and $\angle OC_2C = \theta$. By simple angle sum in the triangle we get \[ \angle C_1CB = 180^{\circ} - \beta - \left(90^{\circ}+\sigma \right) = 90^{\circ}-\beta-\sigma. \] We also get \[ \angle ACC_2 = \gamma - (\sigma-\theta)-\left(90^{\circ}-\beta-\sigma \right) = \beta+\gamma+\theta-90^{\circ} \] Thus, by the definition of the symmedian it suffices to prove \[ \angle ACC_2 = \angle C_1CB \iff \beta+\gamma+\theta-90^{\circ} = 90^{\circ}-\beta-\sigma \iff 2 \beta+ \gamma = 180^{\circ}-\sigma-\theta \] after some rearranging.
But that is easy to prove. Let $C_M$ be the midpoint of the arc $\overarc{BC}$ not containing $C$. Then \[ 180^{\circ}-\sigma-\theta = \angle COC_M = 2 \angle CBC_M = 2 \left(\beta+\tfrac{\gamma}{2} \right). \] That last equality follows from simple angle chasing using that $CC_M$ is the angle bisector of $\angle ACB$.

We're done, yay!

It suffices to prove that $ \angle OBA_2 = 90^{\circ} $ to show that $AA_2$ is a symmedian. For that, our similar triangles gives us $OA_1 * OA_2 = OA^2 = OB^2$.

Specifically, we have:

$OB^2 = OA_1 \cdot OA_2$.

This can be rearranged to get:

$\frac{OB}{OA_1} = \frac{OA_2}{OB}$

This, combined with the fact that $\angle BOA_1 = \angle BOA_2$, shows that triangles $BOA_1$ and $A_2OB$ are similar. Hence, $\angle OBA_2$ = 90 degrees as desired.

Similarly, $\angle OCA_2$ = 90 degrees, and you have that $A_2B$ and $A_2C$ are tangent to (ABC). So $AA_2$ is a symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. Since it is well known that the symmedians of a triangle are concurrent, our three lines concur.

wait. does this even work?

(My previous thing before the last edit did not work. I think the present version does work.)

[quote=Delray][quote=vsathiam]It suffices to prove that $ \angle OBA_2 = 90^{\circ} $ to show that $AA_2$ is a symmedian. For that, our similar triangles gives us that $OA_1 * OA_2 = OA^2 = OB^2$. So the power of $A_1$ with respect to the circle with diameter $A_1A_2$ = $OB^2 $, so OB is tangent to this circle. so $ \angle OBA_2 = 90^{\circ} $ as desired, and we can invoke the symmedians to get concurrency.[/quote]

Wouldn't it be the power of $O$, not $A_1$?[/quote]

Yeah it's the power of O. I'll fix that.

[quote=Delray][quote=vsathiam]wait. does this even work?[/quote]

I don't think $OB^2$ being the power of O with respect to $(A_1A_2)$ implies that it is tangent to $(A_1A_2)$. This would imply that any point of (ABC) is tangent.[/quote]

I agree, but the above symmedian logic still holds by similar triangles. Specifically, we have:

$OB^2 = OA_1 \cdot OA_2$.
This can be rearranged to get:
$\frac{OB}{OA_1} = \frac{OA_2}{OB}$
This, combined with the fact that $\angle BOA_1 = \angle BOA_2$, shows that triangles $BOA_1$ and $A_2OB$ are similar. Hence, $\angle OBA_2$ = 90 degrees as desired.

Similarly, $\angle OCA_2$ = 90 degrees, and you have that $A_2B$ and $A_2C$ are tangent to (ABC). So $AA_2$ is a symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. Since it is well known that the symmedians of a triangle are concurrent, our three lines concur.

We have
 $$\angle CAA_2=\angle OAA_2+\angle OAC=\angle OA_1A_2+(90^{\circ}-\angle B)=(90^{\circ}-\angle AA_1B)+(90^{\circ}-\angle B) = \angle BAA_1$$.
Hence, $AA_2$ is symmedian. Similarly, $BB_2$ and $CC_2$ are symmedians. They concur at the symmedian point!

[hide=Solution]
Note that all of these lines are symmedians, so they are clearly concurrent at the symmedian point!.

[b]1.Inversion wrt $(ABC)$[/b]
[b]2.Symmedian Point[/b]

[hide = solution] We note that because $\triangle OAA_1 \sim \triangle OA_2A$, $$\frac{OA}{OA_2} = \frac{OA_1}{OA}.$$Noting that $OA = OB$, we see that $$\frac{OB}{OA_2} = \frac{OA_1}{OB}.$$Because $\triangle OAA_1 \sim \triangle OA_2A$, we have that $$\angle BOA_1 = \angle AOA_1 - \angle AOB = \angle A_2OA - \angle AOB = $$$$\angle AOA_2 - \angle AOB = \angle BOA_2.$$Because $\frac{OB}{OA_2} = \frac{OA_1}{OB}$ and $\angle BOA_1 = \angle BOA_2 = \angle A_2OB$, we see that $\triangle BOA_1 \sim \triangle A_2OB$. Noting that $\angle BA_1O = 90^\circ$ because the circumcenter of $ABC$ is on the perpendicular bisector of $BC$, we observe that $\angle A_2BO = 90^\circ$, and thus $A_2B$ is a tangent to the circumcircle of $\triangle ABC$. Similarly, $A_2C$ is also a tangent to the circumcircle of $\triangle ABC$, and thus $AA_2$ is the $A$-symmedian of $\triangle ABC$. Similarly, $BB_2$ is the $B$-symmedian of $\triangle ABC$ and $CC_2$ is the $C$-symmedian of $\triangle ABC$, and thus $AA_2$, $BB_2$, and $CC_2$ concur at the symmedian point. $\fbox{}$ [/hide]

[quote=ophiophagous]Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$.[/quote]

Are directed angles necessary for this solution?

[quote=Delray][quote=ophiophagous]Alternatively, you can draw the altitude $AH$. $AH \parallel OA_2$, so it's obvious that $\angle BAA_2 = \angle A_1AC$, so $AA_2$ is a symmedian, and similarly for $BB_2$, $CC_2$.[/quote]

I assume $H$ denotes the orthocenter. How does this imply that  $\angle BAA_2 = \angle A_1AC$?[/quote]

Using parallel lines and the fact that orthocentre is the isogonal conjugate of circumcentre

[hide = Definitions]
$OAA_1 \sim OA_2A \Longrightarrow \angle OAA_1 = \angle OA_2A = x$ and $ \angle OA_1A = \angle OAA_2 = y$
$\angle OAB = a$ and $\angle OAC = b$. $AA_2 \cap BC = P$
[/hide]

[b][color=#f00]Solution :[/color][/b]
$\angle APB = \angle AA_1B + \angle A_1AA_2 = 90 - y$
$2 \angle B = \angle AOC = 2(90 + x - a + y)$
$\angle AOC = 180 - 2b$
$2(90 + x - a + y) = 180 - 2b \Longrightarrow a - x = b + y $
$\therefore AA_2$ is the $\text{Symmedian}$ of $\triangle ABC$. Similarly, $BB_2$ and $CC_2$ are $\text{Symmedians}$.
Thus, $AA_2$, $BB_2$ and $CC_2$ concur at the $\text{ Lemoine Point of } \triangle ABC$

[quote=WWY][quote=GMOH]here is the generalization of above:
assume triangle $ABC$ and point $P$ inside $ABC$
and let $D,E,F$ be the feet of altitudes from $P$ to
$AB,AC,BC$ respectively . let $A',B',C'$ be points on the rays $PD,PE,PF$ respectively such that $PD.PA'=PE.PB'=PF.PC'$ . prove $AA',BB',CC'$ are concurrent[/quote]

I've been trying very hard to prove this, but then failed. Would you kindly plz gimme a hint? thanks.[/quote]

$ABC$ and $A'B'C'$ have common orthology center $P$

From similarity ratios we have $(OA_1)(OA_2)=OA^2=R^2$, so $(OA_1)(OA_1+A_1A_2)=R^2$. Thus, $(OA_1)(A_1A_2)=BA_1^2$, implying that $\triangle{OBA_2}$ is right. Similarly, we have $\triangle{OCA_2}$ is right, so $BA_2, CA_2$ are tangent to $(ABC)$, meaning that by tangent construction, $AA_2$ is the $A$-symmedian of $\triangle{ABC}$. Similarly, $BB_2$, $CC_2$ are also symmedians, and the three symmedians concur at the symmedian point, so we are done.

From the given condition of the problem we have
 $(OA_1)(OA_2)=OA^2$ and let let the tangent to $B,C$ to the circumcircle of $ABC$ meet at $A_2'$.This point lie on $OA_1$ and from $\triangle OA_1B\sim \triangle OBA_2'$ we have $(OA_1)(OA_2')=OB^2=OA^2$.Again note that $\angle A_1OA=\angle AOA_2'$
So $A_2'\equiv A_2$ and so $AA_2$ is the $A$-symmedian.

On dropping a perpendicular from $A$, such that it meets $BC$ at $D$, we observe that, since $AD \perp BC$ and $OA_1 \perp BC$ $\implies AD \parallel OA_1-(\bigstar)$. Also $\angle HAB=90^{\circ} - \angle B=\angle OAC$, so, if we prove that $\angle HAA_2 = \angle A_1AO$ then we would get: $\angle BAA_2 = \angle A_1AC$, which would ultimately imply that $AA_2$ is the symmedian.

Now note that $\angle HAA_1=\angle OA_1A$ (from $\bigstar$), also $\angle A_2AO=\angle AA_1O$ (since $\triangle OAA_1 \sim \triangle OA_2A$) $\implies \angle HAA_2=\angle OAA_1$, which is the required condition, we wanted to prove. In a similar manner, we can prove that $BB_2$ and $CC_2$ are also symmedians, and as we know that the symmedians of a triangle concur, hence we're done. $\quad \blacksquare$

Let $\angle BAA_1 = \theta$. Then

$\angle AA_1O = \angle BA_1O - \angle BA_1A = 90^\circ - (180^\circ - \theta - \angle B) = \angle B + \theta - 90^\circ$.

But by the similarity condition,

$\angle BAA_2 = \angle BAO - \angle A_2AO = \angle BAO - \angle AA_1O = (90^\circ - \angle C) - (\angle B + \theta - 90^\circ) = \angle A - \theta$

and hence $AA_2$ is the $A$-symmedian. But similarly $BB_2$ and $CC_2$ are the $B-$ and $C-$ symmedians, so the three lines concur about the symmedian point, as desired.

In fact, $\overline{AA_2},\overline{BB_2},\overline{CC_2}$ are the symmedians of $\triangle ABC$, which directly implies the concurrency, since the symmedians concur at the well-known symmedian point.
I will show that $\overline{AA_2}$ is the $A$-symmedian, which implies that the other two are as well. Observe that from the triangle similarity, we get:
$$\frac{OA}{OA_1}=\frac{OA_2}{OA} \implies OA_1\cdot OA_2=OB^2,$$
as $OA=OB$. This implies that $\overline{OB}$ is tangent to $(A_1A_2B)$ by PoP, hence $\angle OBA_2=90^\circ$ and $A_2B$ is the tangent from $A_2$ to $(ABC)$. By symmetry, $A_2C$ is also a tangent. It is well-known that the line connecting $A$ and the intersection at the tangents at $B,C$ is the symmedian, so $\overline{AA_2}$ is the symmedian as desired. $\blacksquare$

[hide=Solution]Since $\triangle OAA_1\sim\triangle OA_2 A,$ we know $\frac{A_2 O}{AO}=\frac{AO}{A_1 O},$ or $\frac{A_2 O}{OC}=\frac{OC}{A_1 O}.$ Thus, $\triangle A_1 OC\sim\triangle COA_2$ by SAS, which means that $CA_2$ is tangent to $(ABC)$ at $C.$ Similarly, $BA_2$ is tangent to $(ABC)$ at $B.$ Hence, $AA_2$ is the $A$-symmedian of $\triangle ABC.$ Similarly, $BB_2$ and $CC_2$ are also symmedian of $\triangle ABC,$ so they are concurrent. [/hide]

It got bumped ..so.....
[hide= ..]trivial by Carnots [/hide]

[hide=Solution]Since $A_2$ lies on the perpendicular bisector of $BC,$ $A_2B=A_2C.$ Also, from similar triangles, $$\frac{OA}{OA_2}=\frac{OA_2}{OA}\implies OA_2=\frac{A}{OA_1}\cdot OA=\frac{OB}{OA_1}\cdot OA=\frac{OC}{\cos{\angle{A}}}.$$ Similarly, $$OA_2=\frac{OB}{\cos{\angle{A}}}.$$ Since $\angle{BOC}=2\angle{A}=\angle{A}+\angle{A},$ quadrilateal $\angle{OCA_2}=\angle{OBA_2}=90^{\circ},$ implying that quadrilalteral $BOCA_2$ is cyclic, with $OA_2$ as its diameter. Since $\angle{BOC}=2\angle{A},$ $\angle{OCB}=90-\angle{A}.$ This means that $$\angle{BCA_2}=\angle{OCA_2}-\angle{OCB}=90-(90-\angle{A})=\angle{A}=\angle{BAC},$$ meaning that $A_2C$ is tangent to the circumcircle of triangle $ABC.$ In the same manner, we find that $A_2B$ is tangent to the circumcircle of triangle $ABC.$ Thus, $AA_2$ is a symmedian point. By symmetry, this implies that $BB_2,CC_2$ are also symmedians. In other words, $AA_2,BB_2,CC_2$ concur, namely, at the symmedian point of triangle $ABC,$ as desired.[/hide]

We claim that $AA_2$ is the $A$-symmedian, from which the desired concurrency is trivial. To show this, we will use Phantom Points. Let $A_2$ be the intersection of the tangents to the circumcircle at $B$ and $C$. We will show that $\triangle OAA_1\sim\triangle OA_2A$ in negative orientation.

Toss on the complex plane with $(ABC)$ as the unit circle. Then $A=a,O=0,A_1=\frac{b+c}2,A_2=\frac{2bc}{b+c}$. Clearly $\measuredangle AOA_1=-\measuredangle A_2OA$ so it suffices to check that $\measuredangle OAA_1=-\measuredangle OA_2A$. Hence we just need to show $$\frac{a-0}{a-\frac{b+c}2}\cdot \frac{\frac{2bc}{b+c}-0}{\frac{2bc}{b+c}-a}\in\mathbb{R}.$$ However, note \begin{align*}\frac{a-0}{a-\frac{b+c}2}\cdot \frac{\frac{2bc}{b+c}-0}{\frac{2bc}{b+c}-a}&= \frac{2a}{2a-b-c}\cdot\frac{2bc}{2bc-ab-ac}\\&= \frac{4abc}{4abc-2a^2b-2a^2c-2b^2c+ab^2+ac-2bc^2+abc+ac^2}\\&= \frac{4abc}{6abc-2abc\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}\\&= \frac{4}{6-2\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}\\&= \left(\overline{\frac{4}{6-2\left(\frac{a}{c}+\frac{a}{b}+\frac{b}{a}+\frac{c}{a}\right)+\frac{c}{b}+\frac{b}{c}}}\right)\end{align*} so we are done!

Note that $OA^2=OA_1OA_2$, so $A_2$ is the inversion of $A_1$ with respect to the circumcircle.\\

Claim: $A_2B$ and $A_2C$ are tangents to the circumcircle. We will use complex numbers with $(ABC)$ as the unit circle. Note that $A_1=\frac{b+c}{2}$. Then, $$A_2=\frac{1}{\overline{A_1}}=\frac{2}{\overline{b}+\overline{c}}=\frac{2}{1/b+1/c}=\frac{2bc}{b+c},$$ so $A_2$ is the intersection of the tangents at $B$ and $C$.\\

Thus, $AA_2$ is a symmedian, and similarly for the other vertices. Hence, they concur at the symmedian point, QED.

[b][color=pink]Claim:[/color][/b] $AA_2$ is the $A$-Symmedian and so on.
[i]Proof.[/i] Note that the condition is equivalent to $OA_1 \cdot OA_2 = R^2$. As such, $\triangle OBA_1 \sim \triangle OBA_2$ so $A_2$ lies on the tangent to the circumcircle of $B$. By symmetry it also lies on the tangent at $C$. Thus, $AA_2$ is the $A$-Symmedian. $\blacksquare$
However, the symmedians simply concur at the isogonal conjugate to the median.

[hide]
[b]Claim.[/b] $AA_2$ is a symmedian.
[b]Proof.[/b] Employ phantom points: assume $AA_2$ is a symmedian, so that we want to show $OAA_1\sim OA_2A$, i.e., $\angle OA_1A=\angle OAA_2$. If $\angle BAA_2=\angle CAA_1=\theta$, we have
\begin{align*}
\angle OA_1A&=90^\circ-\angle AA_1B\\
&=90^\circ-(180^\circ-B-(A-\theta))\\
&=90^\circ-C-\theta
\end{align*}
and
\begin{align*}
\angle OAA_2&=A-\angle BAA_2-\angle CAO\\
&=A-\theta-(90^\circ-B)\\
&=90^\circ-C-\theta,
\end{align*}
which are equal as desired. $\square$

Analogously $BB_2,CC_2$ are symmedians, and so said lines concur at the symmedian point.


limit
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real analysis unsolved

[quote="sos440"]For positive integer $ n$, define $ \sigma_{s}(n) = \sum_{d|n, \ d>0}d^{s}$. Evaluate

$ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sigma_{s}(k)}{k^{s}}\quad \quad \quad (s>0)$[/quote]

If $ n+1=p_{1}^{\alpha_{1}}\cdot p_{2}^{\alpha_{2}}\cdot...\cdot p_{r}^{\alpha_{r}}$ then
$ \sigma_{s}(n+1)=\prod_{i=1}^{r}\frac{p_{i}^{(\alpha_{i}+1)s}-1}{p_{i}^{s}-1}$

$ L=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sigma_{s}(k)}{k^{s}}=\lim_{n\to\infty}\frac{\sigma_{s}(n+1)}{(n+1)^{s}}=\lim_{\prod_{i=1}^{r}p_{i}^{\alpha_{i}}\to \infty}\frac{\prod_{i=1}^{r}\frac{p_{i}^{(\alpha_{i}+1)s}-1}{p_{i}^{s}-1}}{\prod_{i=1}^{r}p_{i}^{\alpha_{i}s}}$ where we used the Cesaro-Stolz lema and the fact that $ n+1=\prod_{i=1}^{r}p_{i}^{\alpha_{i}}$ so

$ L=\lim_{\prod_{i=1}^{r}p_{i}^{\alpha_{i}}\to \infty}{\prod_{i=1}^{r}\frac{p_{i}^{(\alpha_{i}+1)s}-1}{p_{i}^{(\alpha_{i}+1)s-p_{i}^{\alpha_{i}s}}}}$

If r is finite and $ p_{a}\to \infty\implies n+1\to \infty$ the others $ p_{i}$ are fixed or $ p_{b}\to \infty\implies n+1\to\infty$ the others $ p_{i}$ are fixed we got different limits so the limit does not exists.

The limit is $ \zeta(s+1)$. For each $ k$, $ \frac{\sigma_{s}(k)}{k^{s}}=\sum_{d|k}\frac1{d^{s}}$. Looking at this over all terms, we get a $ \frac1{d^{s}}$ term if and only if $ d|k$, and thus this term contributes $ \frac1{d^{s+1}}$ to the average of the first $ n$ terms as $ n\to\infty$. To make this rigorous, we could quote dominated convergence; the $ \frac1{d^{s}}$ terms always contribute no more than $ \frac1{d^{s+1}}$ to the average.

jmerry's answer is right.I'm not so clever as you guys.I found an answer in a book and studied it.It's as follows:
$ \sum_{k=1}^{n}\frac{\sigma_{s}(k)}{k^{s}}=\sum_{k=1}^{n}\sum_{d|k}\frac{1}{d^{s}}=\sum_{\begin{subarray}{}\,\,\,d,q\\dq\leq n \end{subarray}}\frac{1}{d^{s}}\&=\sum_{d\leq n}\frac{1}{d^{s}}\sum_{q\leq \frac{n}{d}}1\&=\sum_{d\leq n}\frac{1}{d^{s}}\left\{\frac{n}{d}+O(1)\right\}$
$ =n\sum_{d\leq n}\frac{1}{d^{s+1}}+O(\sum_{d\leq n}\frac{1}{d^{s}})$
The last term is $ O(\log n)$ if $ s=1$ and $ O(n^{t})(t=max\{0,1-s\})$ if$ s\neq1$.

You are right. The answer is $ \zeta(s+1)$. I accidently found it while solving another problem, and that problem was entirely an analytic one. So my first solution was fairly analytic, too.

Fact 1) If $ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}a_{k}$ converges to $ M$, then $ \lim_{x\to1^{-}}(1-x)\sum_{n=1}^{\infty}a_{n}x^{n}$ also converges and its value is $ M$.

Fact 2) $ \sum_{n=1}^{\infty}\frac{\sigma_{s}(n)}{n^{s}}x^{n}\; = \; \sum_{n=1}^{\infty}\frac{x^{n}}{n^{s}(1-x^{n})}$ for $ |x|<1$

So if $ \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\frac{\sigma_{s}(k)}{k^{s}}$ converges, then its value is equal to

$ \lim_{x\to1^{-}}(1-x) \sum_{n=1}^{\infty}\frac{\sigma_{s}(n)}{n^{s}}x^{n}\; = \; \lim_{x\to1^{-}}\sum_{n=1}^{\infty}\frac{x^{n}(1-x)}{n^{s}(1-x^{n})}\; = \; \sum_{n=1}^{\infty}\frac{1}{n^{s+1}}\; = \; \zeta(s+1).$

At this moment, I didn't establish the convergence of this limit. Later I found a simpler way, using the formula

$ \frac{1}{n}\sum_{k=1}^{n}\frac{\sigma_{s}(k)}{k^{s}}\; = \; \frac{1}{n}\sum_{k=1}^{n}\left\lfloor \frac{n}{k}\right\rfloor \frac{1}{k^{s}}\; = \; \sum_{k=1}^{n}\frac{1}{k^{s+1}}-\sum_{k=1}^{n}\frac{1}{n}\left( \frac{n}{k}-\left\lfloor \frac{n}{k}\right\rfloor \right) \frac{1}{k^{s}}.$

It is possible that the latter term tends to $ 0$ as $ n \to \infty$ for $ s > 0$.


This came from an SAT practice test:

[u]X[/u]         [u]f(x)[/u]
1            3
2            4
3            5
4            6
5            7

According to the table above, if f(f(x))= 7, then x=?

The answer key said 1, I said 3. Am I missing something?

$f(f(x)) = 7$
$f(x) = f^{-1} (7) = 5$
$x = f^{-1} (5) = 3$

Alternately, it's obvious $f(x) = x + 2$.  $f(f(x)) = x + 4 = 7 \implies x = 3$

Either way, the answer key is wrong.  Are you sure it's the right problem?  And are you sure there aren't three $f$s?

Yes, the key was wrong. I looked on the book's website, however, and I found a corrected answer key  :roll:


a+b+c=m
what is the maximum of a*b*c?

The max is achieved at a=b=c=m/3
But why?

And should a+b+c=m be extended to a+b+c+d+.......+n=m
what is the maximum of a*b*c*d*......*n? Why?

[quote="stellaffaith"]
And should a+b+c=m be extended to a+b+c+d+.......+n=m
what is the maximum of a*b*c*d*......*n? Why?[/quote]
The max is $ (\dfrac{m}{n\minus{}m\plus{}1})^{n\minus{}m\plus{}1}$ I think. This is for the same reason as your first example. I know this is kind of confusing but I hope it helps

For your first example, without loss of generality, let $ a \ge b$. Let the mean of $ a$ and $ b$ be $ x$, so $ a \equal{} x \plus{} y$ and $ b \equal{} x \minus{} y$. The product of these is $ (x \plus{} y)(x \minus{} y) \equal{} x^2 \minus{} y^2$. To maximize $ x^2 \minus{} y^2$, we have to minimize $ y^2$. The minimum of any real square is $ 0$, so the maximum product occurs when $ y \equal{} 0$. Thus, $ a \equal{} x \plus{} 0$ and $ b \equal{} x \minus{} 0$, so $ x \equal{} a \equal{} b$.

[quote="AIME15USAMO"][quote="stellaffaith"]
And should a+b+c=m be extended to a+b+c+d+.......+n=m
what is the maximum of a*b*c*d*......*n? Why?[/quote]
The max is $ (\dfrac{m}{n \minus{} m \plus{} 1})^{n \minus{} m \plus{} 1}$ I think. This is for the same reason as your first example. I know this is kind of confusing but I hope it helps[/quote]

Perhaps you mean $ \left(\frac{m}{n}\right)^n$.

To prove this, simply apply the AM-GM inequality $ \frac{x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n}{n} \geq \sqrt[n]{x_1 \cdot x_2 \cdots x_n}$ to all of these variables.

Hmm, it would be $ (\dfrac{m}{n})^n$ if we knew $ a\equal{}1$, but $ n\minus{}a\plus{}1$ is the number of elements in $ a, b, c, \cdots n$. I think that was my typo. *$ n\minus{}a\plus{}1$*

We can also use the method of Lagrange Multipliers here.

Let the numbers be $ a_1 , a_2 , ... , a_n$

We have the constraint $ \sum ^n_{i\equal{}1}a_i\equal{}S$ and we want to find the maximum of $ \prod ^n_{i\equal{}1}a_i \equal{} P$

Taking the partials we have the simultaneous equations:

$ \lambda\equal{}\frac{P}{a_1}$

$ \lambda\equal{}\frac{P}{a_2}$

...

$ \lambda\equal{}\frac{P}{a_n}$

Thus the extrema occurs when all the a's are equal, or when one of them is 0.

Clearly at the boundary cases, its a minimum as the product is 0.

Thus the maximum is at equality.

In addition, since the problem never specifies they are all POSITIVE, there is no true maximum, since we could have 40000000000000000, -10000000000000000, -20000000000000000.


Jeng made a purchase and received change of $ \$0.69$ in pennies, dimes, and quarters only. If the clerk gave Jeng the minimum number of coins possible in change, how many coins did Jeng receive?

Well, we do need to do *some* guessing here.  We have to maximize the number of quarters and dimes.  There can be $ 2$ quarters, $ 1$ dime, and $ 9$ pennies (for the least coins, the clerk had to use $ 9$ pennies).  Hence, the answer is $ 2 \plus{} 1 \plus{} 9 \equal{} \boxed{12}$ coins.

FTW says the answer is 9 though... Could this be a typo?

[hide="My answer"]
I found a quicker way:
If you have one quarter that would give you $\$.25$
$\$0.69-\$0.25=\$.44$ left.
We have used 1 coin so far.

We could have 4 dimes to get:
$\$.44-\$.40=\$.04$ left and we have used 4 dimes, and 1 quarter.

We have $\$0.04$ left which can be made using pennies.  Four pennies are needed.

Therefore $\boxed{\text{4 dimes, 4 pennies, 1 quarter}}$ would sum up to $\$0.69$ and the answer is $\boxed{9}$ [/hide]

Well, 1 quarter, 4 dimes, and 4 pennies is my awnswer, and (1x25)+(4x10)+(4x1)=25+40+4=69


[quote="miwalin"]Let $ A$ and $ B$ be two arbitrary compatible matrices, show that $ \rho(A^*B \minus{} B^*A)\leq \rho(A^*A \plus{} B^*B)$, where $ \rho(.)$ denotes spectral radius.[/quote]
  The fist we have $ \rho(A^*B \minus{} B^*A) \equal{} \rho[i(A^*B \minus{} B^*A)]$ put $ C \equal{} i(A^*B \minus{} B^*A)$ then $ C^{*} \equal{} \minus{} i(B^{*}A \minus{} A^{*}B) \equal{} i(A^*B \minus{} B^*A) \equal{} C$ ,$ C$ is Hermitian matrix,$ A^*A \plus{} B^*B$  is Hermitian matrix and semipositive definite.We have :
$ (A^*A \plus{} B^*B) \minus{} i(A^*B \minus{} B^*A) \equal{} A^{*}(A \minus{} iB) \plus{} iB^{*}(A \minus{} iB) \equal{} (A \minus{} iB)^*(A \minus{} iB)\geq 0$

Use [b]Hweyl Inequality[/b]  http://planetmath.org/encyclopedia/WeylsInequality.html
Implies  we have :
$ \lambda_k(i(A^*B \minus{} B^*A)) \leq \lambda_k((A^*A \plus{} B^*B) \minus{} i(A^*B \minus{} B^*A) \plus{} i(A^*B \minus{} B^*A)) \equal{} \lambda_k((A^*A \plus{} B^*B))$
 $ \rho(A^*B \minus{} B^*A) \leq \rho(A^*A \plus{} B^*B)$,
 :)

There is a gap in your proof.
It can be seen that $ (A^{*}A\plus{}B^{*}B)\plus{}i(A^{*}B\minus{}B^{*}A) \geq 0$ from your proof. Now it is sufficient.


Show that every odd prime except $5$ divides some number of the form $111...11$  ($k$ digits, all ones).

Take, in general, any integer $n$ that is not divisible by $2$ or by $5$. Now take the numbers $1, 11, ..., 11...11$ where the last one has $n$ digits. If any one of them is a multiple of $n$, we're done. If not, there are $n-1$ residues mod $n$ so two are congruent mod $n$. Then their difference $11...1100...00$ is a multiple of $n$. Since $n$ is coprime with $2$ and $5$, you can take out the 0's. Hence the result.

[hide]same as above, or particularly .. given  any $p$ coprime to  $10$ we have   $(10^{p-1}  - 1) = 9999..((p-1)times) $ is divisible by $p$ etc.. so in fact $1111 \dots (p-1) times $ would be divisible by  such $p$   (pay attention when p=3,9 etc too but it does not change anything..)[/hide]


function
topology
real analysis
real analysis unsolved

Let $ d:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ function defined with $ d(x,y)\equal{}|x\minus{}y|\plus{}|f(x)\minus{}f(y)|$, where $ f$ an arbitrary function.
Prove that metric $ d$ is equivalent (in sense of topology) to Euclid metric on $ \mathbb{R}$ if and only if $ f$ is a continuous function.

Let $ d_{e}$ be the Euclidean metric. Since $ d_{e}\le d$, it follows that any $ d_{e}$-neighborhood of a point $ x$ contains a $ d$-neighborhood of $ x$. It remains to prove that any $ d$-neighborhood of $ x$ contains a $ d_{e}$-neighborhood of $ x$. Once you write down what this means, you will see the definition of continuity.


The number 34,459,425 is the product of several consecutive positive odd numbers.  What is the greatest of these numbers?

p.s. if anyone has answers for workout 8 and 9 from 1999-2000 school handbook could you please post?

thanks for your help  :D

[hide]$ 34,459,425\equal{}3^4\times 5^2\times 7\times 11\times 13\times 17$  We obviously need numbers divisible by this in a series and close together.  The first one that comes to mind is $ 3,5,7,9,11,13,15,17$  Seeing that the product fits, the answer is $ 17$.[/hide]

thanks for help.

now does anyone have answers for workout 8 from '1999-2000 handbook?

[quote="vamathletes"][hide]$ 34,459,425 \equal{} 3^4\times 5^2\times 7\times 11\times 13\times 17$  We obviously need numbers divisible by this in a series and close together.  The first one that comes to mind is $ 3,5,7,9,11,13,15,17$  Seeing that the product fits, the answer is $ 17$.[/hide][/quote]

cough cough

He wants all the answers:
[hide="Workout 8"]
[b]1[/b]. 6144
[b]2[/b]. 3200
[b]3.[/b] 228
[b]4[/b]. 28
[b]5[/b]. 259
[b]6[/b]. 8
[b]7.[/b] 11.5
[b]8[/b]. 17
[b]9[/b]. 84
[b]10[/b]. 80525.5[/hide]
[hide="Workout 9"]
[b]1[/b]. 85
[b]2.[/b] 1021
[b]3[/b]. 567
[b]4.[/b] 76.9
[b]5[/b]. 84000
[b]6[/b]. 60.4
[b]7[/b]. 4
[b]8.[/b] 32
[b]9.[/b] 158
[b]10.[/b] $ \dfrac{3}{7}$
[/hide]

If you got a couple wrong, I have the solutions for them as well, so PM me or something :)

P.S. Be patient with responses. You only waited 2 hours to post "no one is responding". Give people some time  :wink:


group theory
abstract algebra
superior algebra
superior algebra solved

If one defines $\{ g \in G \ | \ |g| < \infty \}$ in a group $G$. When is this not a subgroup. It certainly has to be a subgroup if $G$ is abelian. But what is a counterxample for $G$ non-abelian?

Let $F=< a,b \mid a^{2}=b^{2}=1 \text{ but }ab \neq ba >$

So essentially all words made with a's and b's with $a=a^{-1}$ and $b=b^{-1}$ but they don't commute.

The element $x=ab$ has infinite order even though $a$ and $b$ are both torsion elements. Hence the set of torsion elements is not closed under composition.


I noticed that some people were able to hide their posts, so that when someone clicks on "See all posts by *username*", it comes up with "No posts match the criteria" or something like that. But when I look in the preferences, no option like that shows up. How do we change our privacy settings so this is possible?

Their posts may be from a class for which you might not be in.

It's all of their posts, though, not just some?

Quite possible some accounts are class-specific and don't ever go out of that forum. Can you provide an example of an account that doesn't follow that?

I don't think I should cite a specific account, but I'm sure it is not class-specific.

Certain forums are private and only specific members have permission to view or post in them. Class fora and private state fora are examples.

Some users have all of their posts in a class forum, some post outside of that forum as well. If someone wants to take an AoPS Class, they must have an account. Some accounts are made so that they can take the class. Otherwise, that person doesn't post outside of that class forum.

You don't have an option to hide a post or not. Some people have permission to view them and some don't.

I'm sure this is not the case for the specific account I have in mind, but I guess this is a strange glitch with the unique user...

Because he is a fairly active poster and his posts are not restricted to private forums.

Is there a post that you can see on the forums but not on the user's profile?

I haven't done that much look into the account but I have seen the posts before... and his post count is over 1000.
Now I feel sort of like a stalker :) Its just something I noticed and I was wondering about.


I heard about Lagrange Interpolations and Holder's Inequality, are they useful for MathCounts?

lagrange interpolation is useful for bashing "find the pattern" problems
i have no idea what holder's inequality is

neither of those are even remotely useful for mathcounts... as for the question what are they, there is a wonderful thing called google.

Well, these aren't useful for mc.. though I have used Lagrange Interpolation on a mc problem before (and once even Combinatorial Nullstellensatz).

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

http://en.wikipedia.org/wiki/Holder's_inequality

These are often used in olympiad problems though.

Neither are useful for any level of Mathcounts. Don't worry over them.


linear algebra
matrix
linear algebra unsolved

I am looking for a 4 by 4 symmetric matrix $A=A^{T}$ such that $A^{4}=0_{4}$. Any ideas?

Over what field? If it's the reals, then there are no nontrivial answers.

A real symmetric matric can always be diagonalized. If it's nilpotent (some power of it is zero), then all of its eigenvalues must be zero. But then, the diagonal form that it is similar to is the zero matrix, and the zero matrix is similar only to itself.


geometry
3D geometry
combinatorics solved
combinatorics

we have a cube .
we consider the points of the cube and the middles of every edge
how many distinct plans can we form with the 20 points

I do, I do.
Look at the cube with those midpoints and count all the planes (by the way this is the plural and not plans).

what do u mean with "Look at the cube with those midpoints and count all the planes "???
i don't know how to count the [b]planes[/b](thx :D )

Easy. Build a string cube. Connect all the 20 vertices and count.

There are at most 20C3 (I see this notation is becoming the standard here so "If you can't beat then, join them) planes.
This means 20*19*18/6=380*3=1140 planes.

There are 6 planes containing 8 points of the 20 (the faces of the cube).
There are 6 planes containing 6 points (for each side take the side symetrical to it relative to the center of the cube and form a plane).
There are 9 planes containing 4 points (for each diagonal of a face take the palne formed by it and its projection on the opposite side of the square - 6 palnes and the 3-planes paralel to the faces of the cube passind through the middle of some sides).

If a plane has n points then it is repeated nC3 times in the total sum of 20C3 planes, so we shouls substract (nC3)-1 from 1140 for each plane containg n points.
Thus we have 1140-6(8C3-1)-6(6C3-1)-9(4C3-1) planes.
This means 1140-8*7*6+6-6*5*4+6-9*3=1140-336-120-27+12=1140-456-15=1140-471=669 planes.
They seem to be to many, so tell me if I'm wrong.

what about the planes which are formed just from 3 points??
i think u missed them
i don't have an answer... but sir gologan should have... because he proposed this problem at the contest dan barbilian 2003 (calarasi)

I didn't miss them. For those I should have substracted q(e.g. their number)*(3C3-1).
But 3C3=1, so draw the conclusion.

there is smth i didn't understand
"If a plane has n points then it is repeated nC3 times in the total sum of 20C3 planes"

Ok we have a set of 4 points A,B,C,D that are coplanar.
Generally, 4 points determine 4C3=4 planes:
ABC,BCD,CDA,DAB. But if the points are in the same plane, all these 4 planes are the same.
This is just an example, but it should help you understand.

Didnt you forget to count the planes with 5 points each?. There are 24 planes with this property  :)

this problem is not completely solved

I will follow amfulger's idea but his solution is not complete.
Total number of planes passing through at 3 of n points in general position is nC3, where nC3 means combinations of n things taking 3 at a time. In this case we got 20C3=1140 planes in grand total, so every trio of points determining a plane is counted only once. Next step is to subtract those planes that contain more than 3 points.

There are 6 planes with 8 points: faces of the cube.
There are 6+4 planes with 6 points: 6 planes containing two opposite edges across the center of the cube and 4 planes that are perpendicular to each of the main diagonals crossing the center of the cube and 6 middle points forming a hexagon (this maybe hard to visualize, I will put up an attached figure if requested).
There are 24 planes with 5 points: planes with an edge and 2 middle points.
There are 3+12+24 planes with 4 points: 3 planes parallel to the faces of the cube and crossing the middle points, 12 planes cutting an edge out of the cube passing through all middle points closest to that cut-out edge, 24 planes passing through 2 vertices and 2 middle points in an slanted manner similar to those 12 planes that cut an edge out.

One remark should be considered cautiously: each edge contains 3 points and those trio of points are counted exactly [b]once[/b] in 20C3. When we subtract, say, 8C3 for each face of the cube, we are subtracting twice a plane formed by an edge because an edge is shared by 2 faces.

So the total is by inclusion-exclusion,
total - multiple counting + edges that were over-excluded + those planes with multiple points
1140-6(8C3)-10(6C3)-24(5C3)-39(4C3)+(12+12+24)+(6+10+24+39) = 335 planes.

PS: This amount still looks large, I believe I got it right though  :maybe:
3X.lich


In a recent poll of 1000 voters, 430 approved of the job their senator was doing, 290 disapproved, and the rest had no opinion. What percent of those polled had no opinion?

Everyone either approves of the job their senator is doing, disapproves, or has no opinion.  Thus if we let $ x$ be the number of people who have no opinion, we can write the equation $ x \plus{} 430 \plus{} 290 \equal{} 1000$, and $ x \equal{} 1000 \minus{} 430 \minus{} 290 \equal{} 570 \minus{} 290 \equal{} 280$.  $ \frac {280}{1000}$ can be simplified to $ \frac {28}{100}$, which is $ \boxed{28\%}$.


floor function
number theory proposed
number theory

[quote="pohoatza"]Find all nonzero subsets $A$ of the set $\left\{2,3,4,5,\cdots\right\}$ such that $\forall n\in A$, we have that $n^{2}+4$ and $\left\lfloor{\sqrt{n}\right\rfloor}+1$ are both in $A$.[/quote]

$n\in A$ $\Rightarrow$  $m=n^{2}+4\ \in A$ $\Rightarrow$  $\left\lfloor{\sqrt{m}\right\rfloor}+1\ \in A$. But, for $n>1$, $\left\lfloor{\sqrt{n^{2}+4}\right\rfloor}+1=n+1$.
So, $n\in A$ $\Rightarrow$ $n+1\in A$ $\Rightarrow$  $A=\{a,a+1,a+2, ...\}$ and we must have $\left\lfloor{\sqrt{a}\right\rfloor}+1\geq a\geq 2$, which means $a=2$

Then the only nonempty subset is $\left\{2,3,4,5,\cdots\right\}$ itself.

--
Patrick

$ a \ge 2 \implies \sqrt{a} \ge 1 \implies \lfloor\sqrt{a}\rfloor \ge 1 \implies \lfloor\sqrt{a}\rfloor \plus{}1 \ge 2$
So,
$ 2 \ge a \ge 2 \implies a \equal{} 2$

[quote="Math4tots"]$ a \ge 2 \implies \sqrt {a} \ge 1 \implies \lfloor\sqrt {a}\rfloor \ge 1 \implies \lfloor\sqrt {a}\rfloor + 1 \ge 2$
So,
$ 2 \ge a \ge 2 \implies a = 2$[/quote]
It's wrong. For example:
$ x>=a>=2$ and $ x=100,$ so $ x>=2 =>a=2???$

We have $ a<=\left\lfloor{\sqrt{a}\right\rfloor}+1<=\sqrt{a}+1 => a<=\sqrt{a}+1=>a>=2$

[quote="pco"][quote="earldbest"][quote]and we must have $ \left\lfloor{\sqrt {a}\right\rfloor} + 1\geq a\geq 2$, which means $ a = 2$
[/quote]

how? :([/quote]
$ A = \{a,a + 1,a + 2, ...\}$ for some integer $ a\ge 2$
$ a\in A$ $ \implies$ $ [\sqrt a] + 1\in A$ $ \implies$ $ [\sqrt a] + 1\ge a$

$ \sqrt a\ge[\sqrt a]$ so $ \sqrt a + 1\ge a$ and so $ a - \sqrt a - 1\le 0$ and so $ \sqrt a\le \frac {1 + \sqrt 5}2$ $ \implies$ $ a\le \frac {3 + \sqrt 5}2$ $ \implies$ $ 2.618\ge a\ge 2$
So $ a = 2$

Is there something you dont understand ?[/quote]

oh, I see. Thanks!  :)


On each planet of a star system,there is an astronomer observing the nearest planet. the number of planets is odd, and pairwise distances between them are different.
Prove that at least one planet is not under observation.

We prove by induction.  The case of one planet is obvious.  Suppose it is true for n planets.  Let us prove for n+2 planets.

Let A and B be the two planets with the shortest (unique) distance.  Then A and B observe each other.  Suppose there are X observations from the n other planets towards A or B.  Then there are n-X observations among the n remaining planets.  Our induction hypothesis was that n observations among n planets guarantees atleast one unobserved planet: clearly this is true for less than n observations too.  So our induction is complete.

[quote="jensen"]this problem is very nice and i saw it very very long ago and i think if you search you will find it in site. :D[/quote]

thanks for reminding me.
i found the solution on the other website.

http://www.kalva.demon.co.uk/soviet/ssoln/ssol661.html


linear algebra
matrix
vector
algebra
polynomial
quadratics
limit

All the eigenvalues of a real symmetric matrix are real. This can be seen as a corollary of the fact that all Hermitian matrices have real eigenvalues.

But isn't there a more direct proof? One which does not involve extending to complex vector spaces, but working within real numbers only.

For example, one possible approach would be to,
prove that characteristic polynomial of any symmetric matrix of 2n X 2n dimensions has atleast a real root and then proceeding by strong induction.

Thanks.

While this may be possible, it would be very hard not to see the complex numbers lurking in the background somewhere.

A key point: before you even argue that eigenvalues are real, you'll have to argue that there are eigenvalues at all. That is, we're going to need some version of the Fundamental Theorem of Algebra. That theorem can be stated in strictly real terms: every real polynomial can be completely factored into linear and quadratic factors.

So a possible plan of a attack: use that, the real version of FTA, in an inductive proof that parallels Schur's Lemma: every real square matrix is orthogonally similar to a block upper triangular matrix with $ 1\times 1$ and $ 2\times 2$ blocks on the main diagonal. Then take the transpose of that to argue that the block upper triangular matrix is really block diagonal, and that the diagonal blocks are themselves symmetrical. But since  the diagonal blocks had to be of the form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix},$ if they're symmetric, we never needed any $ 2\times 2$ blocks in the first place.

I don't see that as avoiding complex numbers since those $ 2\times 2$ fundamental diagonal blocks really are complex numbers anyway.

If $ A$ is of dimensions $ 2n\plus{}1 \times 2n\plus{}1$ we already have that the characteristic polynomial has a real root.

$ \lim_{t \rightarrow \infty} det(tI\minus{}A) \equal{} \infty$.
So we would like to find some $ t$ such that $ det(tI\minus{}A) \le 0$.

Let's check for the $ 2 \times 2$case.
$ \begin{bmatrix}t\minus{}a& \minus{}b\\\minus{}b&t\minus{}d\end{bmatrix}$.
Once we substitute $ t\equal{}\frac{a\plus{}d}{2}$, we get it to be negative.
(Of course, I just differentiated the quadratic polynomial and got that value)

I was talking of a similar approach. Maybe some weird combination of the elements of the matrix will serve as the required value of $ t$ to make the value of char. poly. -ve and the matrix being symmetric will help our cause in such an approach.


inequalities
search
inequalities unsolved

Let $ a,b,c\geq0$,s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:
$ a^3\plus{}b^3\plus{}c^3\plus{}7abc\geq10$

Use schur ,we have :$ r \ge \frac{p^3\minus{}6p}{9}$
The ineq is equivalent to :
$ (p\minus{}3)(19p^2\plus{}57p\plus{}30) \ge 0$
Done

look at problem iii here:
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1390423348&t=202298[/url]

The following inequality is also true
[img]http://s7.sinaimg.cn/middle/0018zOAxgy70JxchVLUd6&690[/img]


[quote="Yosh..."]Let $a,b,c,d \in R$ such that $a+b+c+d=0$. Prove the inequality:

$(ab+ac+ad+bc+bd+cd)^{2}+12\geq 6(abc+abd+acd+bcd)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71504

_________________

Isn't this from Kazakhstan? As far as I remember it was straightforward. In the solution I gave at the contest I just replaced d with -a-b-c...


[b]but why anyone dont type his idea guy?[/b]

[quote="Ramtin Madani"][quote="mohamadhosein"]farda hamin vagta bayad berim olamiad basij...[/quote]
movafagh baashid :rotfl:[/quote]
mamnoon be yari khoda hamamoon movafagh bashim.albate az ma ke bayide...
(ama cher enghadr bad mikhandi  :rotfl: )

2:yek gerafe namotanahi darim ke har zir gerafi az an n rang pazir ast sabet konid kole geraf n rang pazir ast.

[quote="mohamadhosein"]soalaye ghashangi hastand.chera kasi javab nemide?[/quote]
too in modati ke man ozve site boodam be ye noktei pei bordam(shayad joob zade basham)
ke kasai ke too in ja ozvan mamoolan ziad tarkibiat kar nistano tarkibiate in ja ham az vaghti man va suboy oomadim ye zare too cheshm miad vagarna hamash hendese va jabre ;)

manam hamin aghide ro daram ,bahat movafegham ,hala khodet say kon be ehtemale ziad halesh mikoni.

[quote="mohamadhosein"]salam
$a$ magmoyeyi az shebhe gozarehast.$a={a_1,..a_n}$
har $p_i$ tashkil shode az motanahi $a_i$($p_i$ ha namotanahiand)
masalan $p_i=(a_1 va a_2)ya ~a_4$
farz konim har tor $a-i$ha ra meghdar dehi konim hamzaman hameye $p_i$ ha dorost nabashand.sabet konid moshkel az motanahi$p_i$ ast.[/quote]
agha mohamad hosein vaghti migi $p_i$ha ejtemai az $a_i$hast oonvaght beine $A_i$ha $\bigvee$ gharar dare ya beimeshoon $\bigwedge$ gharar dare??
yani sharte dorost boodane $p_i$ chie??
dorost boodane tamame $a_i \in p_i$ ya inke dorost boodane hadeaghal yeki az $a_i \in p_i$??
akhe be in tip masaleha dar halate kolli fek konam $K-sat$ migan (shayad esmesho eshtebah gofte basham!!)
ke in masale daar halate kolli [b]open[/b] tashif dare vali halathaye khasi azinaro (mese in masale) ro mishe hal kard
hala javabe soale man chie??
yeni sharte dorost boodane $p_i$ ha chie??
albate man ta ye jaye kheili kami bedoone doonestane sharte dorost boodane $p_i$ ha jelo raftam:
chon tedade $p_i$ ha namotanahie pas haddeaghal yeki az $a_i$ha dar namotanahi ta az $p_i$ ha oomade
(khaste nabasham!!)
esbate in gozareham ke badihie dige!!vali man migam shayad yeki ino nadoone
tebghe asle lane kabootarie!!
yeni dar halate kolli agar namotanahi ta kabootar dar tedade motanahi ta lane beran oonvaght vojood dare loonei ke namotanahi ta kabootar dare
esbate inam ba borhane kholf mohaghagh mishe:
ba borhane kholf farz konid dar har lane tedade motanahi ta kabootar bashe
pas:
[b]tedade kabootarha=motanahi*motanahi=motanahi[/b]
pas be tanaghoz residimo hokm esbat shode
hala bebin shayad az in nokte masale be tanhai hal beshe
 :o

beine gozarehaye $P_i$ ham $\bigwedge$ ham$\bigvee$hata shart ham mitoone bashe
va dorost boodane $P_i$ha ham ba ghavanin gozareha hastesh.
omidvaram be javabet reside bashi


[quote="bryantacademy"]If a circle has a raidus of one cm, and it is roll ing ariound in a 6 by 10 rectangular box, how far will the raidus travel.

Hint: the circle always touches at least one side and does not slip.

[hide]S1- radius and S2- radius gives you the new dimensions.[/hide]

Try to solve this, it is in Math Counts. DO NOT PEEK!!!!![/quote]

If it's in mathcounts why did you post it in MOEMS??? :huh:

Umm..idont really understand the question...does the ball roll around the inside perimeter of the box?


If anyone wants to start a tournament, that would be nice. I need a little practice.

If you want to have a tournament, check out the other forums. There are several going on in MOEMS, as well as marathons. The same goes for MATHCOUNTS.


Let  $2n>k$  natural numbers and  $a_1,...,a_n$  integers such that leaving different
remainder when they are divided by $k$ . Prove that for all integers $l$  there exist
index  $i,j$  from the set  $ { 1,2,...,n }$ such that

$k|(a_i+a_j-l)$

first i want to state that the last condition is equvilent to \[ a_i+a_j -l \equiv 0\pmod k \iff a_i \equiv l-a_j \pmod k \]

then considering how many $a_i$ can exist so no $i,j$ satisfy the condition we get $n<\frac{k-1}{2}$ which is contrary to $2n>k$ ...


modular arithmetic
group theory
abstract algebra
number theory unsolved
number theory

Determine all integers $(x,y)$ that satisfy equation:
$x^{5}+y^{7}=2004^{1007}$

This is Crux problem from September 2006 issue and since all solutions to this problem should arrive until 1 March 2007 I hope so that now it is not forbidden to post this problem in this forum.

There're no integer solutions. Just consider the equation modulo 71. :D

Please can you post just a short form of your proof  :?:

One has
\[x^{5}+y^{7}\equiv8\pmod{71}. \]
But $y^{7}\in\{0,\pm1,\pm5,\pm14,\pm17,\pm25\}\pmod{71}$. For every of these residues the congruence $x^{5}\equiv8-y^{7}\pmod{71}$ is unsolvable.

OK Thank you I have to carefully read and check this solution and it will take some time.

Is there a short way to check residues of $y^{7}$  and $x^{5}$ $mod71$ or you checked for all numbers or maybe you used computer program for this?

To RIP: how did you think to look at mod 71??

[quote="RIP"]2 [b]delegat[/b]:
Of course, I used the computer. But it can be verified "by hand" in the following manner. Notice that 71 is prime and $71\equiv1\pmod7$. Hence $\{y^{7}\mid y\in(\mathbb{Z}/71\mathbb{Z})^{*}\}$ is a subgroup of $(\mathbb{Z}/71\mathbb{Z})^{*}$ of order $\frac{71-1}7=10$. Since $(-10)^{7}\equiv-5\pmod{71}$ and -5 has the order 10 modulo 71, one gets the statement about the residues of $y^{7}\pmod{71}$. Now, since $71\equiv1\pmod5$, the congruence $x^{5}\equiv a\pmod{71}$ for $a\in(\mathbb{Z}/71\mathbb{Z})^{*}$ is solvable if and only if $a^{\frac{71-1}5}=a^{14}\equiv1\pmod{71}$. That's all.

2 [b]mathisfun1[/b]
I assumed that the corresponding congruence $x^{5}+y^{7}\equiv2004^{1007}\pmod{p}$ didn't have solutions for some prime $p$ (otherwise the problem seems to be difficult). If so, then it's necessary that $p\equiv1\pmod{5\cdot7}$. The first possible choise $p=71$ was successful :lol: .[/quote]

how did u get $x^{5}+y^{7}\equiv2004^{1007}\pmod{p}\implies p\equiv1\pmod{5\cdot7}$

how did u get

[quote="Tomaths"]how did u get $x^{5}+y^{7}\equiv2004^{1007}\pmod{p}\implies p\equiv1\pmod{5\cdot7}$[/quote]
This is not what I wrote. I meant that if there're no solutions of $x^{5}+y^{7}\equiv2004^{1007}\pmod{p}$, then $p\equiv1\pmod{5\cdot7}$.

[quote]2 delegat:
Of course, I used the computer. But it can be verified "by hand" in the following manner. Notice that 71 is prime and . Hence  is a subgroup of  of order . Since  and -5 has the order 10 modulo 71, one gets the statement about the residues of . Now, since , the congruence  for  is solvable if and only if . That's all.

[/quote]

OK Thank you I didnt have such ideas so I haven't send any solution to this crux problem.


ethyne+Hgso4/H2o->Ch3cho
+aldol condensaion->Ch3Ch=ChCHO
+PCC(CH2Cl/N2 atm) allylic oxidation->CHO-Ch=Ch-CHO
protect the to aldehyde groups by forming acetals
foll by reduction to give butanedial
+ P4S10 or H2S/Hcl(Paal Knorr)->thiophene
reasonable?? :D  :wink:  :D

1) Reduce ethyne to ethene.
2) React ethyne with ethene to get cyclobutene.
3) Ozonolysis to get butanodial.
4) Heat with a sulphur reagent to get thiophene.

Your method carcul doesn't seem to be right.
you have used a photochemical reaction and more stable side products are expected.

React two moles of ethyne with sulphur(that's what my book says even though the
mechanism is unsure).

[quote="Carcul"][quote="Chemrock"]you have used a photochemical reaction and more stable side products are expected.[/quote]

So, based on your vast experience, what other stable products are expected? Also, I didn't say I have used a photochemical reaction.[/quote]

so, what type of reaction is it? if it was a photochemical reaction, cyclobutane would be another stable product(cyclobutadiene being aniaromatic)

[quote="BanishedTraitor"]I think Carcul is referring to this:

http://en.wikipedia.org/wiki/Pauson%E2%80%93Khand_reaction[/quote]

are you,carcul :?:

[quote="BanishedTraitor"]Was that sarcasm?[/quote]
no. i just wanted to know if carcul was really referring to that reaction(wats so sarcastic abt that). neways sorry if i sounded sarcastic :wallbash:  :wallbash_red:

How could  be referring to the Pauson-Khand reaction? Do I want to synthesize a five-membered ring?

Anyway, the only expected answer to the problem originally posted by Valeriummaximum is "treat ethyne with sulphur" and that's it. But what in Good's name is the usefullness of that? Treating alkynes with sulphur is not even a common method of thiophene synthesis. That's almost the same thing as "Prepare benzene from ethyne".


topology
analytic geometry
advanced fields
advanced fields solved

Is $\mathbb{R}^n-\mathbb{Q}^n$ path connected under the normal metric topology for $n \ge 2$ ?

My suspicion is yes, but I'm not sure how to prove it.

Just use paths parallel to the axes. This works in the obvious way if either end has all coordinates irrational; for the general case, take two steps, with the intermediate step a point with all irrational coordinates.

If the points $x=(x_1,\ldots x_n),y=(y_1,\ldots y_n)$ are such that $x_1\in\mathbb R\setminus\mathbb Q$ and $y_j\in\mathbb R\setminus\mathbb Q$ for some $j\ne 1$, then we can first fix $x_1$ and find any path from $(x_2,\ldots x_n)$ to $(y_2,\ldots y_n)$, and, after that, we can fix $x_2,\ldots x_n$ (of which $x_j=y_j$ is now irrational), and use any path from $x_1$ to $y_1$. If, on the other hand, $x_1,y_1\in\mathbb R\setminus\mathbb Q$ and all the other $x$'s and $y$'s are rational, then we can first use a path from $y$ to some point $y'$ with $y'_2\in\mathbb R\setminus\mathbb Q$ through the procedure above, and we can then use that again to find a path from $x$ to $y'$.

Edit: looks like I'm too late :)

Thanks guys. I actually posted the wrong problem, but thank you for the quick responses.

The problem I really wanted to ask was is $(\mathbb{R} \setminus \mathbb{Q})^n$ path connected for $n \ge 2$, where $(\mathbb{R} \setminus \mathbb{Q})^n$ means the n-fold direct product of $\mathbb{R} \setminus \mathbb{Q}$ under the normal topology of $\mathbb{R}^n$.

But this problem is even easier than that one. Its obviously not path connected by considering the projections of paths onto any component. I should think more before I post. :oops:

In this case, I don't think it's true :?. Take $n=2$ and any two points with both coordinates irrational. If their first coordinates differ, for instance, then any path from one point to another will cut a line consisting of points which have a rational first coordinate somewhere between the first coordinates of our two points. The same can be done for any $n\ge 2$. What am I missing here? :?

Moreover, $(\mathbb R\setminus\mathbb Q)^n$ is totally disconnected with the subspace topology it inherits from the standard topology on $\mathbb R^n$.

Edit: it took me a while to notice the edited post above :).

I think we can even show that $R^n-I^n$ is pah-connected under usual metric topology.
I - set of all irrationals

I think $\mathbb {R}^n-\mathbb Q^n-\mathbb I^n$ path connected.
 But i´m not sure.

That's false when $n=2$ and true for $n>2$.

Jmerry Can u giv some hints how to prove it.


who knows.......I just know the method :D I don't reallly solve sums of which i get the obvious method :roll: I think your answers are right,i remember getting 2 nice answers.What's ur method eminem?Mine wasn't too small.

And maven,how did u end up with u(u-1)(....)=120?

[quote="MathMaven53"]Let u = 12x -1 and express other factors in terms of u

 You should end up with u(u-1)(u-2)(u-3) = 120

 Expand factors (u-1)(u-2) and u(u-3)

 Then  (u^2 -3u)(u^2 -3u + 2) = 120

  This can be transformed into a quadratic equation in u^2 -3u

 (u^2-3u)^2 + 2(u^2 - 3u) -120 = 0

  Complete the square

  (u^2 -3u + 1)^2  -121 = 0

 Factor this as the difference of two squares

 (u^2 -3u -10)(u^2 -3u + 12) = 0

 (u+2)(u-5)(u^2 -3u + 12)) = 0

 Solve this for u and for each u get x from   x = (u+1)/12[/quote]

$n = u^2-3u \Longleftrightarrow n^2+2n - 120 = 0 \Longleftrightarrow (n+12)(n-10) = 0 \Longleftrightarrow n = -12,10$

$u^2-3u +12 = 0$ and $u^2-3u -10 = 0$.  Substitute $u=12x-1$ back in and have fun with those yummy quadratics and you will get $x = -\frac{1}{12}, \frac{1}{2}, \frac{5 \pm i\sqrt{39}}{24}$. Yay!

Solving it is easy but th e tricky part is arriving to u(u-1)(..)=120.....i think this is the shortest method possible.

I have 3 methods to solve this...in one method i expressed the terms in the form of (x-1/12)(x-1/6)....and then solved it...it is awesome but long...

[quote]Let u = 12x -1 and express other factors in terms of u

You should end up with u(u-1)(u-2)(u-3) = 120 [/quote]
Yes, and also goes without saying, that if you happen to notice that  $120=5! = 5 \cdot 4\cdot 3 \cdot 2$  (Think of four consecutive integers  - also, of course,  not to discriminate against negative numbers so counting $(-2,-3,-4,-5)$   too...
you get  $12x-1= (5 \ or \ -2)$ whch gives two roots, rest is easy, as now you will be left with second degree equation only.


An orthogonal mx. U is proper if |U|=1. Using the orthogonal diagonalization of symmetric mxes, show that if A is a symmetric nxn mx, then there is a proper orthogonal mx U with U^-1AU a diagonal mx.

It made perfert sence. Thanks!
[quote="fredbel6"]First find an U such that $U^{-1}AU=D$ with D diagonal
if your U was proper, then we are done
however, if it was not, call $F$ the diagonalmatrix with all diagonal elements 1, except the first, which is -1
now check that $(UF)^{-1}A(UF)=D$ with $UF$ having as determinant $(-1)(-1)=1$[/quote]


Prove that the polynomial

$f(x) = x^{n}+x^{3}+x^{2}+x+5$

Is irreducible over $\mathbb{Z}$ for $n \ge 4$.

Same way than my proof in http://www.mathlinks.ro/Forum/viewtopic.php?t=122333 (we don't need Rouche but just easy modulus observations).

We have a canonical generalisation.


Suppose A = $ \mathbb{R}$  $ \cap$ [0, 1].

For each $ \epsilon$ > 0, it is possible to create a countable series of open intervals around each rational number of A, such as the measure of the union of all these sets is less than $ \epsilon$.

Prove that for 1 > $ \epsilon$ > 0, this union of sets does not include all the irrational numbers of A.

Just wondering why you wouldn't just write A$ \equal{}[0,1]$. Is there a different meaning for how you wrote it? :S

A hint: the measure of a (countable) union is less than or equal to the sum of the measures.

Once the notion of measure makes sense- well, a set of measure 1 is bigger than a set of measure less than 1, and thus must contain some additional elements.


This is simply enamine formation, and it should be obvious which alpha proton is more acidic.


limit
geometric series
real analysis
real analysis unsolved

study the behavior of this series:

$ \displaystyle\sum{sin(x^n)},\ with \ x\in\mathbb{R} \ and\ x>0$

[hide="One case of convergence"]For 0 < x < 1 the series converges, because then $ \lim \frac {\sin(x^n)}{x^n} \equal{} 1$, and as $ \sum x^n$ is a convergent geometric series also our series converges.[/hide]

that's the trivial case; what is interesting, it's evalutate how it does behave for x>1 (x=1, the sum is +inf)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17863]This thread[/url] shows that the set of $ x>1$ for which the series converges is at most countable. Most likely, there are no such $ x$ at all but it seems to be quite a difficult question.

The Mathworld article to which Xevarion linked in that thread says
[quote]Pisot (1938) subsequently proved the fact that if $ \theta$ is chosen such that there exists a $ \lambda\ne 0$ for which the series $ \sum_{n\equal{}0}^{\infty}\sin^2(\pi\lambda\theta^n)$ converges, then $ \theta$ is an algebraic integer whose conjugates all (except for itself) have modulus $ <1$, and $ \lambda$ is an algebraic integer of the field $ K(\theta)$.[/quote]
Here we have $ \lambda\equal{}1/\pi$.

and what about $ \displaystyle\sum_{n\equal{}0}^{\plus{}\infty}{sin(n^x)}, \ with \ x\in\mathbb{R}, \ and \ x>0 \ ?$

[quote="mlok"]
Here we have $ \lambda \equal{} 1/\pi$.[/quote]
Yes, but our series is without square!


Express
  $\sum_{k=0}^{n}(-1)^k(n-k)!(n+k)!$
in closed form

Its posted by Hungkhtn,I think.Plz shearch his one if u want!

This problem is in Washington TST for 42st IMO.Can you post the solution,[b]keira_khtn[/b]

Searching for what keira_khtn, said I get http://www.mathlinks.ro/Forum/viewtopic.php?t=35403 .


inequalities
linear algebra
matrix
vector
linear algebra unsolved

Let $A_1,A_2,\cdots A_m \in M_n(R)$ such that :$A_iA_j=0, \forall i,j \in {1,2,\cdots m}$.
Prove that : $r(A_1)+r(A_2)+\cdots +r(A_m) \leq n$.

Counterexample:
Let $n=2k$. Let $N$ be the nilpotent matrix with entries 1 just above the main diagonal and zero elsewhere. Let $A_i=N^k$ for all $i$. For $i\le n$, $\mathrm{rank}(N^i)=n-i$, so $\mathrm{rank}(A_i)=k$ and the sum of the ranks is $mk$. For the other condition, $A_iA_j=N^{2k}=0$.
For $m$ large enough, $mk>n$.

Now if you had $A_i$ idempotent, this would be a different problem.

This is not true. Consider the matrices $\left(\begin{array}{cc}0&0\\1&0\end{array}\right)$ in $M_2(\mathbb R)$.

You can take any number of copies of the same non-zero matrix $A$ which satisfies $A^2=O$.

Edit: I see jmerry was here first :)

And here's a proof for the case when the $A_i$'s are idempotent:

For any $i\ne j$, we have $Im(A_i)\subseteq \ker(A_j)$, and since $Im(A_j)\cap \ker(A_j)$ only contains the null vector, the sum of $Im(A_i)$ and $Im(A_j)$ is direct. In pretty much the same way we show that the sum of all $Im(A_i)$ is direct (for example, given distinct $i,j,k$, if we apply $A_k$ to $A_i x_1+A_j x_2$ gives $0$, so every non-null combination of vectors from $Im(A_i)\oplus Im(A_j)$belongs to $\ker(A_k)$, and is thus outside $Im(A_k)$), so $\sum dim(Im(A_i))=\dim(\sum Im(A_i))\le n$.

If they are idempotent, then $A_1+...+A_n$ is also idempotent, and thus its rang is equal to its trace and the trace is equal to the sum of traces of $A_i$, which is the sum of rangs of $A_i$, from where the conclusion.


Without doing algebra one arrives at boxedexe's result by noting that any function $f(n)=ax^{n}+by^{n}$ is the solution to a second order recursion and is of the form $f(n)=m_{1}f(n-1)+m_{2}f(n-2)$
We have $217=79m_{1}+23m_{2}\qquad 691=217m_{1}+79m_{2}$ from which is it easily solved :P

edit

we wish to calculate $691m_{1}+217m_{2}$ as a function of $79m_{1}+23m_{2}, 217m_{1}+79m_{2}$

edit edit

$691m_{1}+217m_{2}=(79x_{1}+217x_{2})m_{1}+(23x_{1}+79x_{2})m_{2}$
so $691=79x_{1}+217x_{2},\qquad 217=23x_{1}+79x_{2}$
Then merely compute $691x_{1}+217x_{2}=691+612x_{1}$


modular arithmetic
combinatorics unsolved
combinatorics

Let $S$ be an infinite set of integers containing zero, and such that the distances between successive number never exceed a given fixed number. Consider the following procedure: Given a set $X$ of integers we construct a new set consisting of all numbers $x \pm s,$ where $x$ belongs to $X$ and s belongs to $S.$ Starting from $S_0 = \{0\}$ we successively construct sets $S_1, S_2, S_3, \ldots$ using this procedure. Show that after a finite number of steps we do not obtain any new sets, i.e.  $S_k = S_{k_0}$ for $k \geq k_0.$

First we can suppose WLOG that not all numbers in $S$ have a common factor greater than $1$ (if they do, divide all numbers by this factor $d$ and we get the case we will consider now). Let $r$ be the number bounding the difference of consecutive numbers.

We want to show that there are some $a_1, \ldots, a_n$ in $S$ such that some linear combination of them is equal to $1$. To find them, take any nonzero number $a_1$ in $S$; at each stage, if $d=(a_1, \ldots, a_i)>1$, take some $a_{i+1}$ not divisible by some prime factor of $d$ (it must exist as, otherwise, all numbers in $S$ would be divisible by it), until $d=1$ (this must happen because the number of prime factors of $d$ is finite and we lose at least one for each $a_i$ we add).

Now it's easy to prove that $1$ can be expressed as a linear combination of them. Suppose you have the set of all linear combinations of them, which doesn't include $1$, and take the smallest positive number $m$ in it; if $m \nmid a_i$ for some $i$, then we substract $m$s from $a_i$ till we get the residue of $a_i \pmod{m}$, which is between $1$ and $m-1$, contradicting the minimality of $m$; thus $m$ must divide all the $a_i$ and then $m=1$. Now, if
\[ \sum_{i=1}^{n} c_ia_i=1 \] for some $c_i$, then
\[ \sum_{i=1}^{n} (lc_i)a_i=l \] for each $l$ between $1$ and $r$.

Now for any number $N=rn+s$, take the largest number, not larger than $N$, that is in $S$, which will be at least $r(n-1)+s$, and call it $N_2$. Then $N=N_2+R$, where $R$ is between $0$ and $r$ and thus can be represented as a linear combination of $a_i$ with at most $rc_1+\ldots+rc_n$ terms. Then any $N$ can be written with at most $rc_1+\ldots+rc_n+1$ terms, so that the set $S_k$ with $k=rc_1+\ldots+rc_n+1$ will be the set of integers, and we're done.


let $ \alpha\equal{}e^{i2\pi/n}$ $ Z\equal{}\sum_{k\equal{}0,n\minus{}1}\alpha^{k^{2}}$
find l$ Z$l

It is standart problem. $ |Z^{2}|\equal{}n.$

[quote="drapt@"]i think it's false because z and t are not independant if you take t=k-m,z=k+m[/quote]
If n is odd $ k\equiv \frac 12 (t+z)\mod n, m\equiv \frac 12 (z-t)\mod n$ these method work.
When n is even it may be wrong. For example n=8 give $ Z=4\alpha$ and $ |Z|^2=16\not = 8$.


http://4rcanefox.deviantart.com/

Take a look everyone. ^^

[quote="buzzer11"]nice. :) what do you use to make those?[/quote]

He says Photoshop is one of his favorite things, so I would guess that.

As for the work....

Meh. The lighting and design is superb, and the texture is extraordinary, but most of the pictures are a bit too monotone. Good job; you're quite a bit better than me.  :)

I use a mix of Bryce 6.0 and Photoshop. Thanks for the comments.

Could you give a link to Bryce 6.0? I've never heard of it.


inequalities
number theory
prime numbers

Let $ p,q,r$ be 3 prime numbers such that $ 5\leq p <q<r$. Knowing that $ 2p^2\minus{}r^2 \geq 49$ and $ 2q^2\minus{}r^2\leq 193$, find $ p,q,r$.

[hide="Solution (somewhat brute force)."] We have $ 2q^2 \minus{} 193\le{r^2}\le2p^2 \minus{} 49$, so $ 2q^2 \minus{} 193\le2p^2 \minus{} 49$ and $ q^2 \minus{} p^2\le72$, so $ (q \plus{} p)(q \minus{} p)\le72$. It is clear q must be less than 37, we can easily see there are no solutions for q=31,29,23. So we try q=19 which gives p=17 and r=23 as solution (and all the inequalities are equalities). [/hide]


Consider a linear function $ f(x)\equal{}ax\plus{}b$ such that $ 3\leq f(1)\leq 6,\ 4\leq f(2)\leq 8$. Find the maximum and minimum value of $ f(5)$.

[quote="math154"]$ f(5) \equal{} 4f(2) \minus{} 3f(1)\implies(\max{f(5)},\min{f(5)}) \equal{} (4(8) \minus{} 3(3),4(4) \minus{} 3(6))$
$ \equal{} \boxed{(23, \minus{} 2)}$.[/quote]

That's correct.

First, solve for $ a,\ b$ the system of equation :$ f(1) \equal{} a \plus{} b,\ f(2) \equal{} 2a \plus{} b$.

Second, express $ f(5)$ by $ f(1),\ f(2)$.

[quote="enndb0x"][quote="math154"]$ f(5) \equal{} 4f(2) \minus{} 3f(1)\implies(\max{f(5)},\min{f(5)}) \equal{} (4(8) \minus{} 3(3),4(4) \minus{} 3(6))$
$ \equal{} \boxed{(23, \minus{} 2)}$.[/quote]

I have no idea what you did. I know you easily solve these problems ,but for me and many others here in mathlinks there is  needed some more help.[/quote]

OK.

We are given $ 3\le f(1) \equal{} a \plus{} b\le6$ and $ 4\le f(2) \equal{} 2a \plus{} b\le8$, so we can now see that $ a \equal{} f(2) \minus{} f(1)$ and $ b \equal{} 2f(1) \minus{} f(2)$ so $ f(5) \equal{} 5a \plus{} b \equal{} 5f(2) \minus{} 5f(1) \plus{} 2f(1) \minus{} f(2) \equal{} 4f(2) \minus{} 3f(1)$. To maximize this, we want the largest possible value for $ f(2)$ (since the term is positive) and the smallest possible value for $ f(1)$ (since the term is negative), and similarly to minimize this, we want the smallest possible value for $ f(2)$ (since the term is positive) and the largest possible value for $ f(2)$ (since the term is negative).

ohh thanks a lot math154 , I really learned very much by this post  :)


rotation
pigeonhole principle
geometry
geometric transformation
combinatorics unsolved
combinatorics

Kati cut two equal regular $ n\minus{}gons$ out of paper. To the vertices of both $ n\minus{}gons$, she wrote the numbers 1 to $ n$ in some order. Then she stabbed a needle through the centres of these $ n\minus{}gons$ so that they could be rotated with respect to each other. Kati noticed that there is a position where the numbers at each pair of aligned vertices are different. Prove that the $ n\minus{}gons$ can be rotated to a position where at least two pairs of aligned vertices contain equal numbers.

Fix one $ n$-gon, and rotate the other one through each of the $ n$ possible configurations. Then each pair of equal numbers will be aligned exactly once, a total of $ n$ aligned vertices through all rotations. If at one position all the vertices are different, then by Pigeonhole at least one of $ n\minus{}1$ other rotations will have two pairs of aligned vertices.


We all need some maths fun!!! :D :D

here is a very funny thing.

I think everyone knows the "proof" for 1=2.

(if a=b then we have that a^2=a^2; a^2-b^2=a^2-ab; (a-b)(a+b)=a(a-b) which gives us:
a=a+b which gives you that 1=2)

Lets "prove" now that  1=0 in another way! :D :D

lim(n-->oo) (1/n+1/n+..+1/n) (where the numbers of the terms 1/n is n)
the limit can be written like so: lim(1/n)+lim(1/n)+...+lim(1/n) =0 because lim(1/n)=0, but also lim(n-->oo) (1/n+1/n+..+1/n)=lim (n*1/n)=1 so we get that 1=0. where is the mistake?! :D :D

cheers! :D :D
this is very funny! :D :

[quote="Lagrangia"]

Lets "prove" now that  1=0 in another way! :D :D

lim(n-->oo) (1/n+1/n+..+1/n) (where the numbers of the terms 1/n is n)
the limit can be written like so: lim(1/n)+lim(1/n)+...+lim(1/n) =0 because lim(1/n)=0, but also lim(n-->oo) (1/n+1/n+..+1/n)=lim (n*1/n)=1 so we get that 1=0. where is the mistake?! :D :D

cheers! :D :D
this is very funny! :D :[/quote]

that'S cool probably the mistake is
lim ( 1/n+...+1/n) = lim 1/n + ...+ lim 1/n
 \cap \cap ou \beta i

Indeed.
The property sum of the limits is equal to the limit of the sum if and only if we have a finite number of terms! :D :D

cheers! :D


About the inequality

$  ab^{x}+bc^{x}+ca^{x}\leq 3$,

for $  a+b+c=3$ and $  x=\frac{3}{2}$.

the hard question is:

Does exist a solution using only AM-GM, Cauchy and Shur?

P.S. This question was also proposed by Vasc many years ago.  :wink:

Actually, in my book, there is a solution by AM-GM, Schur and Cauchy (not exactly, but just classical method). This is by Vo Quoc Ba Can. If $ x\le \frac{3}{2}$, we can use Holder to continue.


search
number theory unsolved
number theory

Let $ a,b\geq 2$ be integers with $ (a,b)\equal{}1$. Prove that every integer $ c\geq (a\minus{}1)(b\minus{}1)$ can be expressed as a linear combination $ c\equal{}ax\plus{}by$ with $ x,y\geq 0$. Prove also that the result is the best possible in sense that the number $ (a\minus{}1)(b\minus{}1)\minus{}1$ has no such an expression.

For some stupid reasons called "Chicken McNugget Theorem". Search it, please.

I found it. http://209.41.180.112/Wiki/index.php/Chicken_McNugget_Theorem


factorial
function
number theory unsolved
number theory

For any odd prime $p$, $p$ does not divide $0!+1!+...+(p-1)!$.

This is sometimes called left factorial (sum of factorials).
A proof was given for the 1st time this year using Bell numbers:
http://mathworld.wolfram.com/BellNumber.html
I wonder if an easier way to proof this exists?

Hi!
Can you rename this topic accordingly to http://www.mathlinks.ro/Forum/viewtopic.php?p=138806

Thank you!

I think now it is OK (I added to the topic: "left factorial")
Kurepa's left factorial function is defined by:
$K(0)=0$, $K(n)=\sum \limits_{k=0}^{n-1} {k!} $.
We have to proof that for any $p$-prime, $p$ does not divide $K(p)$.
For more details you can download:


SUPERDUPERBunny is four times as tall as SUPERBunny and nine times as tall as Bunny. SUPERDUPERUBERBunny is twenty-five times as tall as as SUPERDUPERBunny. If everything else grows in a proportion equal to that of the height growth, how many factors does the number of the magnitude of the size change from Bunny to SUPERDUPERUBERBunny have?

[quote="lotrgreengrapes7926"][quote="Ignite168"]If everything else grows in a proportion equal to that of the height growth, how many factors does the number of the magnitude of the size change from Bunny to SUPERDUPERUBERBunny have?[/quote]
Could you rephrase that?  :maybe:[/quote]

The length and width and everything grows at the same rate of the height. So if height is doubled, length is also doubled.

[hide="Umm..."] From Bunny to SUPERDUPERBunny, size factor of $9$. From SUPERDUPERUBERBunny to SUPERDUPERBunny, a size factor of $25$. So, $9\cdot25=3^{2}\cdot5^{2}$. So the number of factors is $3\cdot3=\boxed{9}$? [/hide]

Correct but please hide your answer. You too SplashD although yours is incorrect.


A sequence of positive integers is defined by $ a_0 \equal{} 1$ and $ a_{n\plus{}1} \equal{} a_n^2 \plus{} 1$ for all $ n > 1$. Find $ \gcd (a_{999}, a_{2004})$.

[quote="aidan"]A sequence of positive integers is defined by $ a_0 \equal{} 1$ and $ a_{n \plus{} 1} \equal{} a_n^2 \plus{} 1$ for all $ n > 1$. Find $ \gcd (a_{999}, a_{2004})$.[/quote]

I'll assume the problem is supposed to read "for all $ n\ge 0$."

[hide="Solution"]Define a new sequence $ b_{n} \equal{} a_{n \minus{} 1}$ for all $ n\ge 1$, so that $ b_{1} \equal{} 1$ and $ b_{n \plus{} 1} \equal{} b_{n}^{2} \plus{} 1$.  We want $ (b_{1000},b_{2005})$.

Suppose some prime power $ p^{\alpha}$ divides some term in the sequence $ \{b_{n}\}$.  Consider this sequence modulo $ p^{\alpha}$.  Some term $ b_{k}$ is 0, and then the next term is 1, so the sequence is periodic with period $ k$, and $ p^{\alpha}$ divides $ b_{n}$ iff $ k|n$.  Therefore, if $ p^{\alpha}$ divides $ b_{i}$ and $ b_{j}$, then $ k|i$ and $ k|j$, so $ k|(i,j)$, so $ p^{\alpha}|b_{(i,j)}$.  The converse also holds, so $ (b_{i},b_{j}) \equal{} b_{(i,j)}$.

Therefore $ (b_{1000},b_{2005}) \equal{} b_{(1000,2005)} \equal{} b_{5} \equal{} 677$.[/hide]

This is some (abstract?) number theory!

I think another similar problem is to prove that $ \gcd (F_m, F_n) \equal{} F_{\gcd (m, n)}$.