I solved it.But ı cant write my solution.because ı cant use mathematical signs in this forum. please help

Check the "Latex Help" button at the top of the page.

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[quote="Nexmus"]No i think the best YouTube ever is Yu-Gi-Oh the Abridged Series.

"Screw the rules, I have money!" -Seto Kiba (Yu-Gi-Oh: The abridged series episode 1)
"(flashback) Screw the rules, I have green hair. Hmmm...maybe I should cut down on the drugs." -Seto Kiba (Yu-Gi-Oh: The abridged series episode 10 (i think))

"Super-Altra-Awesome-Choclately-Fudge-Coated-Super-Card" - Yugi (the abridged series episode 1)
"I activate the Super-Altra-Awesome-Choclately-Fudge-Coated-Super-Card. Yugi, that's gettin' old. How dare you, you get back here. Nanabooboo." -Joey and Yugi (episode 19)[/quote]

Well, I would have to disagree...

This song is pretty awesome. (I'm not saying that it's the best youtube video, but it's quite good)

This, no doubt, is by far my favorite youtube. You've probably watched this before.

 [youtube]PKnloiM-0Ns[/youtube]

[quote="Nexmus"]No i think the best YouTube ever is Yu-Gi-Oh the Abridged Series.

"Screw the rules, I have money!" -Seto Kiba (Yu-Gi-Oh: The abridged series episode 1)
"(flashback) Screw the rules, I have green hair. Hmmm...maybe I should cut down on the drugs." -Seto Kiba (Yu-Gi-Oh: The abridged series episode 10 (i think))

"Super-Altra-Awesome-Choclately-Fudge-Coated-Super-Card" - Yugi (the abridged series episode 1)
"I activate the Super-Altra-Awesome-Choclately-Fudge-Coated-Super-Card. Yugi, that's gettin' old. How dare you, you get back here. Nanabooboo." -Joey and Yugi (episode 19)[/quote]

Yeah those are funny.

Funny, but not as funny as the video i posted  :D  :D

Lmao... I've seen that vid. before... The one nerd of the ages posted

Uh dude that isn't funny. I had to go though that. Except my mom threw webter's dictionary at my head, sprayed me with cleaning spray.

well its funny in a cruel kind of way  :|  its definitely not funny when you're that guy though... but its amusing to see this parody of chinese adults...

[quote="azhang"]r u being racist or something. Whats wrong with Chinese adults? their accent? i could come up with 10 things on the spot that's wrong with ur parents[/quote]

He meant that in general, parents pertaining to an Asian culture tend to radically emphasize the importance of academic success. It's still a big generalization, though. Methinks there are too many stereotypes portrayed in the media, which negatively affects the impressionable consumers.

man, take a chill pill azhang.

And learn how to take a joke

ok my bad but seriously, im just tired of online communities hatin on azns. (no offence) my apologies to u sir nerd_of_the_ages

It's not that I hate Asians, as a matter of fact, I am one. My parents aren't like this, its just that some parents are...

[quote="Some person my mom knows"]OMFG, MY SON GOT A B+! HE CAN'T GO TO BERKELY ANYMORE[/quote] The same person also stopped her car in the middle of a crossroad to shout at her son...


linear algebra
matrix
function
search
group theory

In olympiad lit. i have sometimes seen the notation which stands for the permutation of the set {1,2,3,4,5}. An example is
$ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\
2 & 5 & 4 & 3 & 1\end{pmatrix} \equal{} (2,5,4,3,1) \equal{} \begin{pmatrix}1 & 2 & 5 \end{pmatrix} \begin{pmatrix}3 & 4 \end{pmatrix} \equal{} \begin{pmatrix}3 & 4 \end{pmatrix} \begin{pmatrix}1 & 2 & 5 \end{pmatrix}$

What does it mean when you see a function as a product of two permutations like this (?):
$ v \equal{} v\begin{pmatrix} \alpha & \beta \\
\beta & \alpha \end{pmatrix} \begin{pmatrix} \beta & \gamma \\
\gamma & \beta \end{pmatrix}$, where $ v$ is a function?

my guess is that this means a permutation $ \begin{pmatrix} \alpha & \beta \\
\beta & \alpha \end{pmatrix}$ followed by another permutation $ \begin{pmatrix} \beta & \gamma \\
\gamma & \beta \end{pmatrix}$

Los

If I came across that, I would immediately think that those are matrices, which don't have to do with combinatorics.  That $ v$ thing indicates matrix multiplication to me.

there is not much explanation at http://en.wikipedia.org/wiki/Permutation , but says under [u]Notation [/u]that this definitely is also used for permutations.

From the above example I see now the answer to my question, but I dont see why this is [b]usefull[/b]!

cool...i JUST needed to look this up. search cycle notation

[quote="Los"]From the above example I see now the answer to my question, but I dont see why this is [b]usefull[/b]![/quote]

Being able to express what a permutation does to a set of objects is important because sometimes it's the permutations themselves you want to do operations on. Group Theory is mostly about this.

one can classify symmetries using this notation...ex. if you have a bracelet, with five beads, a,b,c,d,e, then you can use cycle notation to look at the effect of rotating and flipping, for an awesome application, look up polya counting or whatever it is called


it was found that when isobutylene reacts  with $Cl_{2}$ under $0^{o}C$ in the absence of light it gives more methallyl chloride(3-chloro,2-methyl propene) than the addition product.bubbling oxygen through the reaction mixture doesn't affect the product formation.
thsi reaction is carried with labelled iso-butylene with the $C-1$ labelled as $14C$.now the reaction product was made to go through ozonolysis and it was found that the majority of radioactive carbon was present in chloroacetone(about 97%) and very few in formaldehyde.

a)give the structure of methallyl chloride obtained and mention the position of the radioactive carbon.

[hide="my ans"] $3-chloro,2-methyl but-1ene$ where $C-3$ is radioactive. [/hide]

b)is the reaction ionic or free radical?

[hide="my ans"]ionic[/hide]

c)outline a mechanism

[hide="my ans"]first the chlorine attacks the double bond one atom  gets captured and a tertiary carbocation is formed whic elimnates it nearby hydrogen to form propene.[/hide]

d)suggest a reason why isobutylene is more prone than 1-2-butene?

e)under similar conditions in the presence of oxygen $3,3-dimethyl-1-butene$ yeilds more addition product but also a small yield of $4-chloro-2,3-dimethyl-1-butene$.how do you fit this fact with the mechanism u proposed and also account for the minor product


What are your favorite bands?

Mine's are (in no particular order)

Tool
Rage Against the Machine
Sublime
Iron Maiden
Led Zeppelin
Queen
System of a Down
Primus
The Eagles

Yours?

Check these strange bands:

Groundation
Steel Pulse
Cultura Profetica
Jarabe de Palo

I like anything, but these are my favorites I guess

Linkin Park
Evanescence
The Killers
Sugarcult
Green Day

Lifehouse
Nickel Creek
Gaelic Storm
Smashmouth
Ween
Train
Coldplay

(im going to add performers too)
Michael Tolcher
Josh Groban
Norah Jones

Angra (Brazilian Metal Band, very,very nice)
Iron Maiden
Led Zeppelin
Pink Floyd

feeuhhhh, let's see if I can name them all, with no particular order,

Soda Stereo
Cramberries
Caifanes
Savatage
Live
Depeche Mode
Orgy
Pet Shop Boys
The Cure
The Doors
Alice in Chains
Epic
Errasure
New Order
Pink Floyd
Placebo
REM
Smaching Pumpkings
Smachmouth
Sunder Garden
Tears for Fear
The Element
Wheatus
10000 Maniacs
Blind Melon
Cardigans
Colective Soul
Eagle Eye Cherry
General Public
Goo goo dolls
Savage Garden
Linkin Park
U2
Green Day
No Doubt
OffSpring
Queen
UB40
The Wailers
Aerosmith
Guns n' Roses
Iron Maiden
Korn
Metallica
Motley Crue
Nine Inch Nails
Red Hot Chilli Pepers
Men at work
Perl Jam
Toreros Muertos
69
El Parque
Inconciente Coletivo
Ghandi
Tango India
Aterciopelados
Autenticos Decadentes
Cafeta Cvba
Maldita Vecindad
Enanitos Verdes
Heroes del Silencio
Duncan Dhu
Angeles del Infierno
Celtas Cortos
Hombres G
La Ley
Los prisioneros
Azucar Moreno
Calle Dolores
Jarabe de Palo
Los Rodriguez
Todos Tus Muertos
Celtic Mystique
Tatu
Zebda
Rammstein

And I am not including any solist! You can say that I have a variated taste.

Best regards,

The New York Philharmonic
Eastman Wind Ensemble
Led Zepplin
Oracle
Bela Fleck and the Flecktones - Victor Wooten is awesome
Pat Metheny Group
Eurythmics
Earth, Wind, and Fire
Billy Joel

rage against the machine and audioslave
black sabbath
led zeppelin
nine inch nails
AC/DC
nirvana
red hot chili peppers
and Lil Jon  :lol:

Oh man, there's so many good bands.... Let's see:

AC/DC
Beatles
Beach Boys
Byrds
Creedence Clearwater Revival
The Clash
Eagles
Led Zeppelin
Nirvana
Pink Floyd
Queen
Ramones
Rolling Stones
Sex Pistols
Simon and Garfunkel
U2

There's probably bands that I forgot since there's so many bands I like.... If you have not heard of the bands I listed above.... SHAME ON YOU!  :P jk

Simple Plan
Green Day
Yellow Card

they are my top favs, at least for right now... changes with time....

[quote="manuel"]David which solists do you like...
I like silvio rodriguez do you like him?
I like enanitos verdes, the wailers, soda stereo also...[/quote]

OK, let's see, Solists that I like,

Fito Paez
Joan Manuel Serrat
Joaquin Sabina
Pedro Guerra
Perales
Silvio Rodriguez
Mercedes Sosa
Atahualpa Yupanki
Facundo Cabral
Esteban Monge
Maria Pretiz
Ruben Blades
Aleks Syntek
Pedro Suares
Franco de Vita
Ramazotti (the old ones)
Miguel Rios
Rosana
Soledad Bravo
Chichi Peralta
Juan Luis Guerra
Coolio
Billy Joel
Daniel Amos
James Brown
Jerry Lee Lewis
Jimmy Hendrix
Joe Coker
John Lennon
John Mellecamp
Ben E. King
Fiona Apple
Lisa Loeb
Joan Osborne
Kelly Osborne
Merideth Brooks
Kendall Payne
Nathalie Imbruglia
Chris Isaacs
Sinead O'Conor
Leonard Cohen

And of course, I really like Geri Halliwell (But not her music  :D )

Best,

greenday (fav. song- boulevard of broken dreams)
linkin park (um.... dunno like them all i guess...)
aerosmith (dream on or maybe kings and queens...)
Queen (bohemian rhapsody)
are my fav. bands

elvenchamp777

I like a lot of the same bands as you, AznChristmas.

Tool
System of a Down
Rage Against the Machine

I like a lot of the bands Magnara mentioned and yes, Bright Eyes rocks. Conor Oberst has a very sexy voice. And Something Corporate...

[quote="zerovolume"]I like:

Marilyn Manson
NIN
Rammstein
Cradle of Filth
Dashboard Confessional
Iron Maiden
Disturbed
Slipknot

but the BEST song hands down is Baby Got Back.[/quote]

Marilyn Manson= Creep/ Weirdo/Freak/........

[quote="zerovolume"]
but the BEST song hands down is Baby Got Back.[/quote]
*agrees  ;)

Taking Back Sunday
With Broken Wings
Yellowcard
Simple Plan
Mest
Blink
GC
Green Day
Jay Chou
BoA

I have to add the Mamas and the Papas, Aerosmith, the Who, the Police to my list! I don't really like modern rock bands...... Although, I kind of like Greenday, Good Charlotte, Maroon 5, and Coldplay. Still, oldies are much better! (in my opinion of course.....)

Good Charlotte DEFINITELY my fave, also green day (sort of-only song i've heard is blvd of broken dreams) linkin park, simple plan, bowling for soup, the killers, papa roach(scars is only one i've heard), 3 doors down, trapt, blink 182,newfound glory,... im forgetting somebody... OH YEAH! did i mention The Calling? only a fraction of the groups i like... :lol:

In order

Gym Class Heroes
Incubus
Pink Floyd [just got into them this past year]
Aerosmith

[quote="zerovolume"]I like:

Marilyn Manson
NIN
Rammstein
Cradle of Filth
Dashboard Confessional
Iron Maiden
Disturbed
Slipknot

but the BEST song hands down is Baby Got Back.[/quote]
I like NIN, Iron Maiden, and Cradle of Filth, but Marilyn Manson is just too... weird.

Linkin Park- FAINT, Breaking the Habit, Paperkut, etc

Green Day- Boulevard of Broken Dreams, American Idiot.

Rock is the best type of music

[quote="Magnara"]Taking Back Sunday, Thursday, Postal Service, Straylight Run, Bright Eyes.  You guys should all go listen to Bright Eyes.  It's amazing.  Nice acoustic songs.[/quote]

oh, the postal service, i love them

perhaps the beatles. i love their yesterday, heyjude and so on...

[quote="Jhyon8149"]I have to add the Mamas and the Papas[/quote]

Omg, I was so scared of them in 6/7th grade.

[color=red][b]Dream Theater[/b][/color]
Metallica
Marilyn Manson
Muse
Red Hot Chili Peppers
System Of A Down
The Beatles


Functional Analysis
advanced fields
advanced fields unsolved

Let $ A$ - isometry in a Hilbert space. Show that if $ A(0)\equal{}0$ then $ A$ is linear operator.

Every help will be appreciated:)

If the space is over complex scalars, you must also allow for conjugate-linear $ A$. Let's consider the real case.

1) The only isometric embedding of $ \mathbb R$ into a Hilbert space is the obvious (linear) one. This follows from strict convexity: the images of points $ a$ and $ b$ determine where the points between them go. As a consequence, $ A(\lambda x)\equal{}\lambda Ax$ for all $ \lambda\in \mathbb R$.

2) To prove $ A(x\plus{}y)\equal{}Ax\plus{}Ay$, consider the distance from $ x\plus{}y$ to various points on the lines $ \langle x\rangle$ and $ \langle y\rangle$.


is this question too hard? cuz no one answer it yet... I can make it multible choice

[quote="KBabe"][hide]is this it? the words weren't very clear in the question... Chris and Brittany??[/hide][/quote]

nope

[quote="shinwoo"]The question needs more info

like, you need at least one settled info..[/quote]

what does settled info mean, one that's given like anyone can figure out, well, there is one: JC is on the farthest left.

[quote=RC-7th wrote]Does three spots away mean there are two spots in between them? Or three? [/quote]

this is where most people gets confuseed

[quote="LuCy4EvA"][quote="RC-7th"]Does three spots away mean there are two spots in between them? Or three?

[hide="Assuming the former is true..."]There is only one order: [color=white]JC,Chris,Lance,Brittney,Justin,Joey[/color][/hide][/quote]

well, I think it's it's two between them, but it was from one of the old tournaments that we went to, so not even our teacher's sure... but the answer is Christ and Brittany.[/quote]

So am I right?

[quote="KBabe"][quote="LuCy4EvA"]

well, I think it's it's two between them, but it was from one of the old tournaments that we went to, so not even our teacher's sure... but the answer is Christ and Brittany.[/quote]

wut??? but i said Chris and Brittany and you said it wasn't right!!!!!![/quote]

o, sorry, I redid it and got a different answer than the key I got from Mrs.Mills, and I couldn't exactly go back and delete the post. SORRY!!!!!

[quote="RC-7th"]So three spots away means that there are three spots in between them?[/quote]

2. Wait, did I do it wrong?

[quote="somedude"]Ok well I got it, but to make sure people aren't discuraged (sp?) here is a little hint :P

[hide]start by saying "who is on the far left" than by saying who has to be within two spots of that person?  Put all the spots where each person could possibly be located and then it's pretty easy  :P [/hide]
^--hint (if you need it)[/quote]

wait, since you got it, can you tell what the answer is? I'm really not sure right now.

[quote="KBabe"]to RCV: it means 2 spaces in between.. the words weren't very clear...
to Lucy: no worries.. you did it right....[/quote]

Assuming RCV is RC, what's wrong with my solution then? Solution here : [color=white]JC,Chris,Lance,Brittney,Justin,Joey[/color]

Wait... if chris and brittney can't be next to each other... how come they're third and fourth in line?

3 spots away means 2 in between them I think.

[quote="shinwoo"]To : LuCy4EvA

Did you make up this question?

did you understand what RC-7th meant?

could you rewrite the question for me?[/quote]

nope, it was from a past tournament I got

[quote="KBabe"][quote="LuCy4EvA"]Britney, JC, Justin, Lance, Chris, and Joey are standing in line together at the unemployment office. No one else is in line with them. Chris is not next to Britney. He is, however, within two spots of JC. JC is on the farthest left. Justin and Joey are next to each other. Britney, who is directly to the right of Lance, is not on either end of the line. Lance is exactly three spots away from Joey. From the left, which two people are standing 3rd and 4th in line?

sry if there's any typos, but I'm typing in a hurry.  :D[/quote]

yea lucy, how could the answer be Chris and Brittany if they aren't standing next to each other???

YAY!! NO SCHOOL MONDAY!![/quote]

okay, I'm have no idea what the answer is, think I did something wrong AGAIN, I'm gonna ask someone else to do this at school and I'll check with the teacher. It just so happens that we don't have the answer key to this one so we had to make one out of the top student's answers. I'll ask...

[quote="shinwoo"]Yeah,, check your wordings..

again, you need a settled info...[/quote]

what's settled info?

a settled info....

People gets confused because everything you said was being mixed up.. At least one of them has to be settled..

doesn't "Chris is not next to Britney. He is, however, within two spots of JC. JC is on the farthest left." count as a settled fact? it tells you that Chris is on the farthest left.


group theory
superior algebra
superior algebra unsolved

Show that $ S_n\equal{}\left\langle(1\,2\,\cdots\, n), (1\,2) \right\rangle$, where $ S_n$ is the symmetric group of degree $ n$.

[quote="ifai"]Show that $ S_n \equal{} \left\langle(1\,2\,\cdots\, n), (1\,2) \right\rangle$, where $ S_n$ is the symmetric group of degree $ n$.[/quote]

Let $ S\equal{}(1\,2\,\cdots\, n)$ and $ T\equal{}(1\,2)$. We have that $ S^{k\minus{}1} T S^{\minus{}k\plus{}1} \equal{}  (k\,k\plus{}1)$. Now for $ 3 \le i \le n$, we have that $ (1\,2\,\cdots\,i) \equal{} (1\,2)(2\,3)\cdots(i\minus{}1\,i)$. Using conjugation by $ S$, we can get all $ i$-cycles of consecutive numbers. Now suppose $ 1 \le i < j \le n$. Let $ R \equal{} (i\,i\plus{}1\,\cdots\, j)$. Then $ (i\,j) \equal{}R^{\minus{}1} (i\,i\plus{}1) R$. Thus we can get all $ 2$-cycles and hence all of $ S_n$.

[b]Convention Note[/b]: For $ A,B \in S_n$, $ AB$ is the permutation obtained by first applying $ B$ and then $ A$.


algorithm
search
number theory
prime numbers
number theory unsolved

What d you mean?
Do you need numerical algorithm? Try to search net :)

I think he meant that he would like to make some money by breaking codes  :D

Pierre.

[quote="ramox"]how to find the largest prime factor p of an integer n .[/quote]

Is he asking how to find the largest prime factor $p$ of an integer $n$?

Usually, wheneve I tried to find some factor, I just take the square root and divide by each prime numbers until I reach and say "Nope" then it's prime and if I reach something, then I say, "yes!."

Divisibility in mod would be useful also.

ex. 16^{1/2} = 4 try 3,2 and 2 works so 2 is the smallest one. Oh, question is asking for largest. blah. Wait, 2 IS the smallest AND largest :P

I think there is a more efficient way...  :D


Let $A\subset \{1,2,\ldots,4014\}$, $|A|=2007$, such that $a$ does not divide $b$ for all distinct elements $a,b\in A$. For a set $X$ as above let us denote with $m_{X}$ the smallest element in $X$. Find $\min m_{A}$ (for all $A$ with the above properties).

Let $a_{1}<a_{2}<...<a_{n}<2n$ be positive integers such that $a_{i}\not | a_{j}\;\forall i<j.$ Assume that $k\in\mathbb{N}\; : 3^{k}<2n<3^{k+1}$. Prove that $a_{1}\geq 2^{k}$.

[hide]
Partition the set $\{ 1,2,3,\cdots ,4014\}$ into 2007 parts $P_{1}\cup P_{2}\cup P_{3}\cup \cdots \cup P_{2007}$ such that $P_{a}$ contains all numbers of the form $2^{b}(2a-1)$ where $b$ is a nonnegative integer.

$A$ cannot have two elements from the same part (otherwise one would divide the other by a power of 2). So $A$ must have exactly one element from each part.

Let $t_{a}$ be the element of $A$ contained in $P_{a}$. Then consider $t_{1},t_{2},t_{5},\cdots t_{1093}$, each being a product of a power of 3 and a power of 2. The highest power of 2 dividing $t_{1}$ must be strictly greater than the highest power of 2 dividing $t_{2}$ (otherwise $t_{1}$ divides $t_{2}$). Similarly, the highest powers of 2 dividing $t_{1},t_{2},t_{5},\cdots t_{1093}$ must be a strictly decreasing sequence. In particular, there are 8 numbers in the sequence, so $t_{1}$ must be at least $2^{7}=128$. In a similar manner, if $1338<3^{b}(2a-1)<4014$, then $t_{a}\geq 2^{b}(2a-1)$. (Here we look at the numbers where the odd factors are $(2a-1)$ times a power of 3.)

Suppose one of the $t_{a}$ is less than 128. Then we have
\[1338<3^{b}(2a-1)\leq (3/2)^{b}t_{a}<128(3/2)^{b}\]
so that $b\geq 6$. However, this means either $a=1$ (we know $t_{1}\geq 128$) or $t_{a}\geq 3\cdot 2^{6}$, which is a contradiction. Therefore, every element must be at least 128.

We could attain a minimum by setting $t_{a}=2^{f(a)}(2a-1)$ where $f(a)$ satisfies $1338<3^{f(a)}(2a-1)<4014$ if we can show it works. To do this, we must show that $t_{x}$ cannot divide $t_{y}$ for any $x\neq y$. Suppose there were such $x$ and $y$. Then
\[2^{u}(2x-1)|2^{v}(2y-1)\]
where $u=f(x)$ and $v=f(y)$. In particular, $u\leq v$ and $(2x-1)|(2y-1)$.

In order to have $(2x-1)|(2y-1)$, we must have $2y-1\geq 3(2x-1)$, but this implies
\[3^{u+1}(2x-1)>4014>3^{v}(2y-1)\geq 3^{v+1}(2x-1)\]
and $u>v$, a contradiction! Therefore $\boxed{128}$ is the minimum.
[/hide]

Compare with this: http://www.mathlinks.ro/Forum/viewtopic.php?p=333498#333498


inequalities
search
inequalities unsolved

[quote="turcas_c"]Let $a,b,c > 0$ with $a^{2}+b^{2}+c^{2}=3$ .

Prove that :$a+b+c \ge abc+2$ .[/quote]
It's wrong. Try $b=c\rightarrow0^{+}$ and $a\rightarrow\sqrt3^{-}.$

Maybe the corect condition is $a^{2}+b^{2}+c^{2}=2$. Equality occurs when one of the numbers is really close to $0$, and the other two are really close to $1$. If I remember correctly, the problem was solved by Darij, using a very tricky Cauchy-Schwarz. I solved by it using mixing variables, and again, If I remember correctly, scorpius solved it in a nice way. Also, the inequality with $a^{2}+b^{2}+c^{2}=2$ holds for real a,b,c.

Edit: the inequality has a reverse sign, and I found the link eith darij`s Cauchy solution : http://www.mathlinks.ro/Forum/viewtopic.php?search_id=72355616&t=32399

This was not solved by me and my solution is not [i]my[/i] solution... ;)

[b]EDIT:[/b] Exact sources are
http://www.mathlinks.ro/Forum/viewtopic.php?t=149554
http://www.mathlinks.ro/Forum/viewtopic.php?t=54136
One time with equality case, the other time without. Actually both topics are results of mass merging...

  Darij


Anyhow, it is the year 2409

You are a pilot.

Your mission: To survive

Your weapons: A laser

Your Shields: A shield

Your ship: Average fighter

You must fight your way to the top as you move from lowly pilot to Spaceforce Commander!

Basically, its another game of activeness.

I will give you an event, you are to respond to it within 3 days or the event's effect will take place.

For every event yoy succeed in, you earn 100 dollars, unless I say "this is worth 200" or something.

It's very simple.

Space for 15 people.

Join while you can! ([b]JOIN BY PRIVATE MESSAGE[/b])

SHOP:

refer to this anytime needed, copy and paste or quote if you will.

SHIPS:

[hide="class 1"]

Average fighter - 1000 dollars

Good fighter - 2000 dollars

Great fighter! - 3000 dollars[/hide]

[hide="class 2"]

Average cruiser - 10,000 dollars

Good cruiser - 20,000 dollars

Great Cruiser! - 35,000 dollars[/hide]

[hide="class 3"]

Average Battlecruiser - 100,000 dollars

Good battlecruiser - 210,000 dollars

Great Battlecruiser - 400,000 dollars[/hide]

SHIELD SHOP:

shield - 1000 dollars

good shield - 10,000 dollars

great shield! - 100,000 dollars

WEAPON SHOP: note: all weapons never run out of ammo

laser - 500 dollars

Mass Driver - 10,000 dollars

Magnetic accelerator - 15,000 dollars

Energy Pulse - 30,000 dollars

Missile - 10,000 dollars

Protius Missiles - 50,000 dollars

Antimatter pulse - 500,000 dollars (OMG!)

[b]UPDATE: all events are now 1000 dollars each[/b]

So when are we going to start?

Pilot:
I have a laser, shield, average fighter, and 0 dollars
How many people have joined?

As a general suggestion, its better to have people sign up by posting rather than pm's.  This way everyone knows how many have signed up, and the thread stays near the top of the forum so it continues to grab peoples interest (seeing as this and knights have started to die).

I'll join btw :wink:

I guess i should update my signup list more often :P

5 signups

spaces left:

10

We have 8 more spaces left!

 :ewpu:  :jump:  :noo:

From now on, I guess you sign up by posting...
Shouldn't this be in the Fun Factory?

[quote="louiethechewy123"]
Shouldn't this be in the Fun Factory?[/quote]

Not quite... the events can possibly eliminate people so there is a possible winner and therefore an end to the game.

OOPS  :blush:

I forgot to mention that the winner is the last one standing. Sorry.

How many spots left now? I'm IMPATIENT!!!

[b] GAME STARTS[/b]

EDIT II:

[b] EggyLv.999, it's your turn[/b]

After travelling, EggyLv.999 finds a floating mine in space. He sends out a probe to it and realized that it is deactivated. He puts the mine in his fighter's cargo hold. Soon after, he discovers an asteroid made of completely iron ore. It's too big to put in his cargo hold, so he must reactivate his mine and blow up the asteroid. This quest is worth 3000 credits since you sell the ore after completing the quest.

You have 3 days to respond, EggyLv.999.

You have earned 3000 credits :D

EggyLv.999's stats:
1 shield
1 laser
average fighter
3000 credits

Now its ytrewq's turn.

At a space station, you meet up with police. They said they'll pay you if you hunt down a fugitive.
Soon after, a mysterious looking man passes by. But before that, you find a taser on the ground.
Will you taze the man? And will it be the right person? WORTH: 1000 credits.

you have 3 days to respond.

About 5000 creds, and Once someone else is eliminated, people can buy properties which give u money per day :D

Update:

Pilot: I have a laser, shield, average fighter, and 7000 dollars

[quote="tcjy"]Anyhow, it is the year 2409

You are a pilot.

Your mission: To survive

Your weapons: A laser

Your Shields: A shield

Your ship: Average fighter

SHOP:

refer to this anytime needed, copy and paste or quote if you will.

SHIPS:

[hide="class 1"]

Average fighter - 1000 dollars

Good fighter - 2000 dollars

Great fighter! - 3000 dollars[/hide]

[hide="class 2"]

Average cruiser - 10,000 dollars

Good cruiser - 20,000 dollars

Great Cruiser! - 35,000 dollars[/hide]

[hide="class 3"]

Average Battlecruiser - 100,000 dollars

Good battlecruiser - 210,000 dollars

Great Battlecruiser - 400,000 dollars[/hide]

SHIELD SHOP:

shield - 1000 dollars

good shield - 10,000 dollars

great shield! - 100,000 dollars

WEAPON SHOP: note: all weapons never run out of ammo

laser - 500 dollars

Mass Driver - 10,000 dollars

Magnetic accelerator - 15,000 dollars

Energy Pulse - 30,000 dollars

Missile - 10,000 dollars

Protius Missiles - 50,000 dollars

Antimatter pulse - 500,000 dollars (OMG!)

[b]UPDATE: all events are now 1000 dollars each[/b][/quote]

[b]I buy a great fighter and sell my average fighter.[/b]

Can you sell things? If so, what percentage of the original price do you get?

You sell for the original price,

BUT, WEAPONS AND SHIELDS ARE SOLD FOR HALF PRICE!

Oh, and Mewto55555 has died, shot down by space police.

So I have 3000 credits, a laser, a great fighter, and a regular shield. Is that right?

Whose turn is it?

@tcjy: send me a PM when it is my turn.

Let me check:

It's EggyLv.999's turn.

Far away from planet mars, there is a planet called umgar.

It will be invaded in 3 days. The people on mars know when it's going to happen, and what angle the aliens will attack from.

You must get to umgar in 3 days to deliver the news since satelite messaging does not work for this distance:

Reward: Unknown.

THIS IS A SPECIAL QUEST:

you have 3 days instead of 2 days.

Here is your reward:

[hide="click here for your reward"][url=http://techsmartly.net/lol/3e8fbd41a7]Click here :D[/url][/hide]

It is louiethechewy123's turn.

THIS IS A VERY SPECIAL QUEST:

you get 10K of credits for doing it within 1 day of this post. You get 5K if you do it within 2 days. If not, you will be eliminated. REPLY QUICK!

The following message was found just now.
\Message:-

From Somewhere in the ort cloud:

HELP! Alien are attacking! I'll give you 15K if you help fend off the aliens!!! -\End-Message

You got something.

You gotz PWN3D.

BY YOUR OWN ROLL.

And since u actually clicked it, you get 2K of credits. :D

you know you can just use aops.tk to get there...

[quote="tcjy"]Uh... get where?

Anyhow, for those joining other games, YOU HAVE BEEN ELIMINATED.

Eggylv.999 is dead.

Spaceguy524 is dead.

louiethechewy123 is dead.

abcak is dead.

The winner is!:

[hide="here"]nobody![/hide][/quote]
what the heck!!!
to get to my customroll, use [url]aops.tk[/url]
:(


Ascertain $ \{m,n\}\subset N^{*}$ so that the equations $ \{\begin{array}{c}x^{m+1}-x^{n}+1=0\\\ x^{n+1}-x^{m}+1=0\end{array}$ have at least a common root.

I have a question, what does the asterisk mean for $ N^{*}$?

And I'm not sure if this solution works, but...
[hide]
Let $ r_{i}$ be the common root, $ P(x)=x^{m+1}-x^{n}+1$ and $ Q(x)=x^{n+1}-x^{m}+1$

Because $ P(r_{i})=0$ and $ Q(r_{i})=0$,
$ P(r_{i})-Q(r_{i})=0$.
$ x^{m+1}+x^{m}-x^{n+1}-x^{n}=0$
$ x^{m+1}+x^{m}=x^{n+1}+x^{n}$
$ x^{m}(x+1)=x^{n}(x+1)$
$ x^{m}=x^{n}$
$ m=n$

Is this correct...?[/hide]


Prove that for all $ \mathbb{N} \ni n > 1$ the inequality

\[ \prod_{k\equal{}2}^{n} \left( 1\minus{} \frac{1}{k^3} \right) > \frac{1}{2}\]

holds.


In the game \textit{Five Flying Fish}, a goal is worth seven points and a basket is worth three points. There are no other ways to score points in this game. What is the largest score that cannot be obtained?

Chicken McNugget theorem gives $ (7)(3)\minus{}(7\plus{}3)\equal{}21\minus{}10\equal{}\boxed{11}$ points, which is the largest score that can't be obtained.

The formula is $AB-(A+B)$ this is Chicken McNugget.

[quote="AKAL3"]How can you prove the $ Chicken McNugget $ theorem?[/quote]
http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem#Proof

[quote="tmathman"]LOL this was revived twice.[/quote]
Well this is the FTW problems forum and someone might have a question.


* I am a four-digit positive integer such that the sum of my digits is 18 and my digits reversed are exactly four times greater than myself. What number am I?

Because the reversal of the digits is four times [b]greater[/b] than the original number, this should mean that the number is five times the original number.  Thus, the original number must begin with 5 or 0 (so the reversal will be divisible by 5).  But the original number cannot begin with 0 (because then the number will not have four digits, right?).  So, the number must begin with 5, but this is impossible, because for five times the number to be expressed as four digits, the number must be < 2000.

Can someone help me find the error in my logic? and a solution....

Your logic is faulty. By 4 times greater, it means 4 times as much. For example, 48 is 4 times greater than 12.

For a number to be 4 times another, it must end in 0,2,4,6, or 8.

In this case our original number must begin with 0,2,4,6,8. As we want 4 times the first digit to be less than 10, it must begin with 0 or 2. As it ais a 4 digits number, it must begin with 2.

2xxx
xxx2

When we multiply our first number by 4, it ends in a 2. Thus its last digit must be 3 or 8.

If it ends in 3 our numbers are

2xx3
3xx2 which is clearly not 4 times the other.

Thus our numbers are 2xx8 and 8xx2.

Now for 4 times a number to be less than 9000, it must be less than 2250. Thus the 2nd digit of the first number is 0,1, or 2.

For our 2nd number to be divisble by 4, it must end in 2 digits who are divisble by 4 (like 28, or 72). Thus, the 3rd digit of our 2nd number must be odd. Combining these facts, the 2nd digits of the first number is 1.

21x8
8x12

Now we can solve an equation.

4(2108+10x)=8012+100x

8432=8012+60x

x=7

Thus our numbers are 2178 and 8712.

Hello
[hide="Answer"]
2178
[/hide]
[hide="Explanation"]
As u said the reversed number is 4 times of original number than we must conclude that the unit digit of reversed number is even or original number's thousand's place number is even also it should be two only.bcoz if it some other number then reversed number would become five digit number.
till now our original number is
[b]2_ _ _[/b]
multiplying 2 by four we get eight therefore the units digit should be 8
[b]2 _ _ 8[/b]
Listen we don't want that if we multiply 4 by hundredth place number the resultant should be more than nine otherwise original will more than eight in thousandth place.
so third place should be either 1 or 2.
As total sum is 18.The respective tenth place number are 7 and 6.
as 2178 satisfy all the condition.[/hide]
Thank u.

Thanks for the replies.  I understand your interpretation.....but still, I think that the problem could have been worded more clearly.

Anyway, with your interpretation, the easiest way i see to solve the question is using the divisibility rule for 4.  (the last two digits of the number must be divisible by 4.)

The original number must be <2500 (because four times the number can be expressed in four digits)

Multiples of four, that when reversed yield are less that 25:

12, 32 (reversed 21, 23)

All viable numbers with digits that add up to eighteen paired with 21 are: 2187, 2169, 2196, 2178
All viable numbers with digits that add up to eighteen paired with 23 are: 2394, 2385, 2376, 2349, 2358, 2367

And the only one that works it 2178 (and its pair is 8712)

I don't understand, why must the sum be 18?

[quote="mismatchtea"]* I am a four-digit positive integer [b]such that the sum of my digits is 18[/b] and my digits reversed are exactly four times greater than myself. What number am I?

Because the reversal of the digits is four times [b]greater[/b] than the original number, this should mean that the number is five times the original number.  Thus, the original number must begin with 5 or 0 (so the reversal will be divisible by 5).  But the original number cannot begin with 0 (because then the number will not have four digits, right?).  So, the number must begin with 5, but this is impossible, because for five times the number to be expressed as four digits, the number must be < 2000.

Can someone help me find the error in my logic? and a solution....[/quote]


inequalities
MATHCOUNTS
email
number theory
\/closed

The Spring class schedule has been updated.  We will be offering the Olympiad Inequalities class on Sundays in the Spring.  The Intermediate Counting/Probability class will be on Mondays.  The Introductory Number Theory Course has been changed back to a one day course on Mondays only.

For more details about the AoPS Spring Class schedule, visit our [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Class Enrollment page[/url].

I would like to apply for the Mathcounts Problem Series class, and it says on the enroll page that these classes "designed for strong math students in grades 7 through 12". I am not a strong math student, should I enroll?

This depends on what you mean by 'not a strong math student'.  I knew IMO team members who did not consider themselves strong math students.  If you like, you can email me your background in mathematics at rusczyk@artofproblemsolving.com and I can let you know if the class is appropriate.  If you are going to be participating in MATHCOUNTS, the class is probably appropriate for you.

Also, please note that you can drop the course before the third class and get a full refund.  So you can try it out and see if it suits you.

[quote="answerseeker95"]When will you offer Olympiad Algebra (I think that Mr.Crawford mentioned that it was scheduled as one of the Spring classes)?[/quote]

We have have not written an Olympiad Algebra class and we do not plan to offer one in the Spring.


Let $ z\in\mathbb{C}$ such as $ z^2\equal{}\overline{z}$ $ 2^{1\minus{}|z|}$ then find $ z$

[hide="Solution"] $ z^3 \equal{} |z|^2 2^{1 \minus{} |z|}$, hence $ z^3$ is purely real.  We can write $ z \equal{} re^{i \frac {2 \pi k}{3}}, k \equal{} 0, 1, 2$ and we therefore only need to solve

$ r \equal{} 2^{1 \minus{} r}$

The LHS increases monotonically while the RHS decreases monotonically, so this only occurs when $ r \equal{} 1$; hence $ z$ is a third root of unity. [/hide]

Στέλνω τη λύση στο συνημμένο έγγραφο (word) γιατί δεν ξέρω latex.

If $ |z|\equal{}0$, then $ z\equal{}0$, which is a solution. Otherwise, assume
$ |z|>0$, then multiply both sides by $ z$ and proceed with Torajirou's solution.


After homogenization, I mean multiplication of left side by $ab+bc+ca=1$ this becomes equivalent to Iran 1996 so please see:
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=3547&start=20[/url]


Using two facts:
$\sin^{4}x=\frac{3}{8}-\frac{1}{2}\cos2x+\frac{1}{8}\cos4x$
And $\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2}$
Combine the ones we get $A=\frac{21}{16}$


I plan to launch a competition hear. Write as much as you can about a topic with the best formatting spelling and creativness as you can. A picture can accompany you. I suggest you put it in attachments on the forum. Use any program. There will be ten rounds each taking a week. After each round I will go around checking the quality of what has been done and I will announce the winner. That winner gets to have chose the next topic. PM me if you have got a topic. I cannot acces outside web pages so please do not post it outside of the forums.Anyone can join at any time. First topic will be posted at ten people ready.

[quote="alkjash"]Wow, can I just submit stuff I did already for school and stuff, because I have some ok writing and a few good pictures I did on Mic. Paint (took me hours).[/quote]

You could but I still want to see some creativness there.

In reply to a privite message that alkjash sent me, I think that animals is a good idea. On further note I would like to ask someone to post if they are interested on the forum. They can send me a PM though.

we seem to have two competitors eigt more to go before we want to start. If you wish to start nowyou can post. If you do not we will have wait.


number theory
least common multiple
arithmetic sequence

An arithmetic sequence with first term $ 1$ has a common difference of $ 6$. A second sequence begins with $ 4$ and has a common difference of $ 7$. In the range of $ 1$ to $ 100$, what is the largest number common to both sequences?

The first sequence is 1,7,13,19,25....

2nd is 4,11,18,25,32,39,46

We see 29 is common to both of them. However, the LCM of 6 and 7 is 42 so we can keep adding 42.

29,71,113....

The largest term in the range [1,100] is 71.

I just asked this question and this thread came up - the correct answer is 67.

It would appear the term common to both of them is 25 not 29, so it would be 25+42=67.


Two questions: How do I clear the screen in Python, and how do I call reference to a different file?
This is for an international currency converter. I'll post it here later when I've got it done.

Can you be more specific?  Is this a text based program or a GUI?  By 'call reference to a different file' do you mean that you have a text file and you want to input it or is it another python file whose methods you want to use (or something else...)?

1) Are you talking about IDLE or are you talking about the graphics window? If it's the former, I don't think you can (unlike C++ which has clrscr()). As for the graphics screen, you can use undraw().

2)I'll assume you are referring to reading text files that contain data. You can use

<filevar> = open("<filename>", "r")

then any of the three

<filevar>.read() (Reads through the whole file as one string)
<filevar>.readline() (Reads a particular line)
<filevar>.readlines() (Reads all the remaining lines one by one)

for that purpose.


Let $ a,b,c$ are positive real numbers such that :$ a^2\plus{}b^2\plus{}c^2\equal{}3$. Find the least value of
$ A\equal{}\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}$

Notice that (by $ AM\geq GM$) we have:

$ \sum (\frac {a^{2}}{b} \plus{} \frac {a^{2}}{b} \plus{} a^{2}b^{2})\geq 3(a^{2} \plus{} b^{2} \plus{} c^{2}) \equal{} 9$   so

$ 2A\geq 9 \minus{} (a^{2}b^{2} \plus{} b^{2}c^{2} \plus{} c^{2}a^{2})\geq 9 \minus{} \frac {(a^2 \plus{} b^2 \plus{} c^{2})^{2}}{3} \equal{} 6$ so $ A\geq 3$ and equality is achieved when $ a \equal{} b \equal{} c \equal{} 1$


[b]1.[/b] $ (x,y)$ is a [i]lattice point[/i] if both $ x$ and $ y$ are integers.  Let $ T(a,b)$ be the set of lattice points strictly inside (not on the boundary) of a triangle whose vertices are $ (0,0), (a,0),$ and $ (0,b)$ where $ a$ and $ b$ are positive integers.  There are several $ T(a,b)$ for which $ (3,6) \in T(a,b)$ but $ (4,6), (1,7) \notin T(a,b)$.  Compute the difference between the largest and smallest numbers of lattice points in such $ T(a,b)$.

[hide="Answer"]$ 15$[/hide]

[b]2.[/b] Two two-digit positive integers, not necessarily different, are chose at random.  What is the probability that the product of these integers is a four digit number whose leftmost digit is $ 9$?

[hide="Answer"]$ \frac{1}{180}$[/hide]

Thanks in advance!

2. Note that $ \max(ab) \equal{} 99^2 \equal{} 9801$. Therefore, the problem is equivalent to finding the probability that $ ab \geq 9000$. Fortunately, there are relatively few pairs that work.

(99, 91)-(99, 99) (17)
(98, 92)-(98, 98) (13)
(97, 93)-(97, 97) (9)
(96, 94)-(96, 96) (5)
(95, 95) (1)

So the number of desired possibilities is $ 17\plus{}13\plus{}9\plus{}5\plus{}1 \equal{} 45$, while the total number of possibilities is $ 90^2 \equal{} 8100$, so the probability is just $ \frac {45}{8100} \equal{} \frac 1{180}$.

[color=darkred]Edit:Something tells me I fail at counting... :maybe:[/color]

2. There are $ 99 \minus{} 10 \plus{} 1 \equal{} 90$ two digit numbers.  We are looking for all $ ab$ such that $ a$ is less than or equal to $ b$ and $ ab \in [9000,10000)$.  Since $ \frac {9000}{99}$ is equal to $ 90.9090...$, the smallest such $ a$ would be $ 91$.  We have that $ 91 \bullet 98 \equal{} 8918$, so we have only one possibility for this case: $ 91 \bullet 99$.  Similarly, we have that $ 92 \bullet 97 \equal{} 8924$, so for this case, we have two possibilities: $ 92*99$ and $ 92*98$.  We have $ 3$ possibilities for $ 93$, $ 4$ possibilities for $ 94$, $ 5$ possibilities for $ 95$, and then, since $ a$ has to be less than or equal to $ b$, we only have $ 4$ possibilities for $ 96$, $ 3$ for $ 97$, $ 2$ for $ 98$, and $ 1$ for $ 99$.  In total, we have $ 25$ possibilities.  Of these, $ 5$ are squares $ 95 \minus{} 99$.  The other $ 20$ possibilities we have to multiply by $ 2$ (why), to get $ 45$ final possibilites. Our answer is thus $ \frac {45}{90^{2}} \equal{} \boxed{\frac {1}{180}}$.

[hide="Hint to #1"]Consider the line going through $ (a,0)$ and $ (0,b)$. Now we can use inequalities to bound $ a$ and $ b$.[/hide]

[quote="Yongyi781"]Therefore, the problem is equivalent to finding the probability that $ ab \geq 9000$.[/quote]

I don't get why that is.  Could you please explain?  Thanks!


algebra
polynomial
function
calculus
derivative
symmetry
real analysis

1)
Let $ P(x)$ be a non-constant polynomial such that $ P(n) \equal{} P(\minus{}n)$  for all $ n \in N$
Then $ P'(0)$.

1) $ \equal{}1$
2) $ \equal{}0$
3) $ \equal{}\minus{}1$
4) can not be determined from given data

[hide="click"]On drawing the graph of the polynomial,the answer seems to be $ (4)$.But how to formally prove it[/hide]

2) Let $ f : R \to R$ be a differentiable function such that $ f'$ is continuous. Define the function
     $ G(x,y) \equal{} f( \sqrt{x^2\plus{}y^2} )$ for all $ (x,y) \in R^2$
Then,

1) $ \frac{\partial G}{\partial x}$ and $ \frac{\partial G}{\partial y}$ are always continous at each $ (x,y) \in R^2$

2) $ \frac{\partial G}{\partial x}$ and $ \frac{\partial G}{\partial x}$ always exist but are not continuous at some point.

3) $ G$ is always continuous on $ R^2$.

4) The continuity of $ G$ depends on the choice of $ f$.

[quote="guru_bhai"]1)
Let $ P(x)$ be a non-constant polynomial such that $ P(n) \equal{} P( \minus{} n)$  for all $ n \in N$
Then $ P'(0)$.

On drawing the graph of the polynomial,the answer seems to be $ (4)$.But how to formally prove it

[/quote]

my  god!  how r u getting  4  ?

 take  p(x)=x^2      :roll:

1) the ans is 0.as f(n) += f(-n) therefore it is an even function. so consider any even function then we see that the ans is 0

Let $ P(x)\equal{}a_1x^n\plus{}a_2x^{n\minus{}1}\plus{}a_3x^{n\minus{}2}...\plus{}a_{n\minus{}1}x\plus{}a_n$

We note that if any of the non-zero terms contain an $ x$ to an odd power, $ P(n) \neq P(\minus{}n)$

As such, $ a_{n\minus{}1}\equal{}0$.

Thus, the derivative of $ P(x)$ contains terms which each have at least one $ x$ in them. As such, $ P'(0)\equal{}0$.

The answer to #2 is 3, I'm pretty sure, because $ G'$ exists and is continuous, which I'm pretty sure means that G exists and is continuous.

This is actually true for any differentiable function. Just do this:

$ f(x)=f(-x) / (\frac{d}{dx})|_{x=0}$

$ f'(x)|_{x=0}=-f'(-x)|_{x=0}$

$ f'(0)=-f'(0)$

$ f'(0)=0$

As for 2, the answer is (3), $ G$ is continuous. That's true because $ G$ is the composition of two globally continuous functions, $ f$ and $ (x,y)\mapsto\sqrt{x^2\plus{}y^2}.$

If $ f(r)\equal{}r^2,$ then (1) would be true: $ G_x$ and $ G_y$ continuous, but

if $ f(r)\equal{}r,$ then (1) and (2) would both be false, as $ G_x$ and $ G_y$ would fail to exist at the origin. So neither (1) nor (2) are always true.

First of all thank you all for the response.

Now the question says that $ P(n) = P( - n)$ for all natural numbers.    not for all real numbers ,So how could, $ P(x)$ be an even function.

[quote="milin"]This is actually true for any differentiable function. Just do this:

$ f(x) = f( - x) / (\frac {d}{dx})|_{x = 0}$

$ f'(x)|_{x = 0} = - f'( - x)|_{x = 0}$

$ f'(0) = - f'(0)$

$ f'(0) = 0$[/quote]

That, seems to be an intresting formula, but it was all greek and latin to me.....plz elaborate a bit more abt it.

[quote="Kent Merryfield"]As for 2, the answer is (3), $ G$ is continuous. That's true because $ G$ is the composition of two globally continuous functions, $ f$ and $ (x,y)\mapsto\sqrt {x^2 + y^2}.$

If $ f(r) = r^2,$ then (1) would be true: $ G_x$ and $ G_y$ continuous, but

if $ f(r) = r,$ then (1) and (2) would both be false, as $ G_x$ and $ G_y$ would fail to exist at the origin. So neither (1) nor (2) are always true.[/quote]

Sry but could not get that $ f(r)$ part. what is $ r$ doing here?

If you know something about a polynomial (in one variable) at infinitely many points, you know it everywhere. $ P(x)\minus{}P(\minus{}x)$ is a polynomial with infinitely many zeros, so it must be the zero polynomial, and $ P$ is an even function.

Just try to differentiate the equation

$ f(x)\equal{}f(\minus{}x)$

and plug in $ x\equal{}0$ after you do.

[quote="jmerry"]If you know something about a polynomial (in one variable) at infinitely many points, you know it everywhere. $ P(x) \minus{} P( \minus{} x)$ is a polynomial with infinitely many zeros, so it must be the zero polynomial, and $ P$ is an even function.[/quote]

Thanks.....

The original problem gives us a function $ f$ such that $ f(n) \equal{} f(\minus{}n)$ for all [i]integers[/i] $ n$.  So, the solution requires first that we show (following jmerry) that because our function is a polynomial, it is even, and then use your argument (which shows that any even function has zero derivative at zero).  The given symmetry isn't strong enough in the absence of polynomiality.  (Also, using words in place of the big slash would make what you write a lot clearer ;).)

guru_bhai, Kent is just using $ r$ as the variable the function $ f$ acts on.  It's standard functional notation.  He gives two functions, one of which is a counter-example to the first answer and the other of which is a counter-example to the second answer.


[img]http://bbs.pep.com.cn/attachments/Radical%20Tower_TzcOIBCFL4Er.jpg[/img]

What's the domain of $x$ ,the tower converge

[hide="Might be the wrong approach"]
Let $y=x^{z}$, where $z= \frac{1}{x^{\frac{1}{x^{\frac{1}{x^{...}}}}}}$, then $x^{\frac{1}{y}}=y$
Then $x= y^{y}$.
[/hide]

Your method correctly determines the lower bound, but I don't think that works fully, since the sequence (it's really a sequence of numbers) oscillates for large numbers.

Here's what I did:
[hide]Rewriting with exponents, this becomes:

$x^{\left( \frac{1}{x^{\left( \frac{1}{x^{\left( \frac{1}{...}\right)}}\right)}}\right)}$

By properties of exponents, this becomes:
$x^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{...}}}}$

Hence, the value converges iff.
$\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{...}}}$
converges.

Thus, $1/x$ must fall into the domain of a standard power tower; $[1/e^{e}, e^{1/e}]$, and hence $x \in [e^{-1/e},e^{e}]$[/hide]

The Zuton Force you are right , both Radical Tower and Power Tower converge to $[1/e,e]$

I need two graphs of these functions , my computer can't figure them out , Could you give me . Thanks

$f(x)=x^{x^{x^{...}}}\ ; \ x\in[1/e^{e},e^{1/e}]$

$f(x)=x^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{...}}}}\ ; \ x\in[e^{-1/e},e^{e}]$

[quote="The Zuton Force"]$x^{\left( \frac{1}{x^{\left( \frac{1}{x^{\left( \frac{1}{...}\right)}}\right)}}\right)}$

By properties of exponents, this becomes:
$x^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{\left(\frac{1}{x}\right)^{...}}}}$
[/quote]

What property of exponents did you use?

[b]@236factorial:[/b]I used the property that $\frac{1}{x^{n}}=\left( \frac{1}{x}\right)^{n}$ for all $n$.

[b]@kg2100:[/b]
[img]http://img.photobucket.com/albums/v194/mukmasterxps/pt.gif[/img]
[img]http://img.photobucket.com/albums/v194/mukmasterxps/rt.gif[/img]

The following definitions were used:
$PowerTower(x)=\frac{W\left( \ln{\frac{1}{x}}\right)}{\ln{\frac{1}{x}}}$
$RadicalTower(x)=\frac{\ln{x}}{W(\ln{x})}$

Where $W$ is the Lambert/ProductLog function, the "inverse" of $xe^{x}$.

These definitions hold true [i]only[/i] inside the domain over which the towers converge, and nowhere else. Also, these definitions produce a removable discontinuity at $x=1$ because of the use of $log$, which is not present in the actual towers.


[quote="parrot"]find three natural numbers consecutives n, n+1, n+2 so that 27 divides n, 25 divides n+1 and 121 divides n+2[/quote]

this is just a mod problem...
$n\equiv 0 \mod 27$
$n\equiv -1 \mod 25$
$n\equiv -2 \mod 121$

these problems are pretty trivial as there is universal way to solve them...these kinds of problems have been discussed before...i don't know the link though


limit
logarithms
function
real analysis
real analysis unsolved

Let $u_n $ a sequence such that :
$ \lim_{n \rightarrow + \infty} \frac{x_n}{n}=a (0<a<+\infty) $
Find :
$ \lim_{n \rightarrow +\infty} [\prod_{k=1}^n(u_n+\frac{k}{u_n})]^\frac{1}{u_n \ln u_n} $

Hmm... $e^{1/a}$?
I have a bad feeling... :?

I guess theres some problem we are to find the limit of some complicated function of u_n.But given is some limit of (x_n)/n.Then what is then x_n.

I am totally confused.. :(

i've also got the same thing as Myth did....

Yes , Myth is right .
$ \lim_{n\rightarrow +\infty}[\prod_{k=1}^n ((u_n)^r+\frac{k^r}{u_n})]^\frac{1}{u_n \ln u_n} =e^{r/a} $
$ (r \geq 1) $
It have a simple solution . Try !

Indeed, it was easy. Too easy... That is why I had some doubts about it.


Let x,y,z be three real pozitive numbers so that xyz=1. Prove that:

sum(x^2)+3>=sum(xy)+sum(x).

thank you.

If  two of x, y, z are greater than or equal to 1, one is less than or equal to 1.

Then (x - 1)2 + (y - 1)2 + (z - 1)2 :ge: 0 :ge: (x - 1)(y - 1)(z - 1) = xy + yz + zx - x - y - z.

If 2 :ge: x :ge: 1 :ge: y, z, then (x - 1)2 + (y - 1)2 + (z - 1)2 :ge: (y - 1)2 + (z - 1)2 :ge: (y - 1)(z - 1) :ge: (x - 1)(y - 1)(z - 1).

If x :ge: 2, 1 :ge: y, z, then (x - 1)2 + (y - 1)2 + (z - 1)2 :ge: (x - 1)2 :ge: (x - 1) :ge: (x - 1)(y - 1)(z - 1)

Expand out both sides and bring over (2x + 2y + 2z) to the right to get the desired result.


geometry
inequalities
circumcircle
trigonometry
geometry solved

Let H be the orthocenter of an acute-angled triangle ABC.
The lines (AH), (BH) and (CH) intersect with the circumcircle of triangle ABC at A', B' and C', respectively.
Show that $\frac{AA^{\prime}\cdot BB^{\prime}\cdot CC^{\prime}}{AH\cdot BH\cdot CH}\geq 8$.

[color=red][Moderator edit: I've corrected the problem - the condition that triangle ABC must be acute-angled was forgotten.][/color]

Let AA', BB' and CC' cut BC,CA and AB at A",B" and C" respectively.

Then $\angle HBC=90^o-\angle ACB=\angle A'AC=\angle A'BC$. Similarly, $\angle HCB=\angle A'CB$. Thus $\triangle A'BC\cong \triangle HBC$ and therefore HA"=A"A'. Similarly HB"=B"B' and HC"=C"C.

Let $x=\frac{AA'}{AH}$, $y=\frac{BB'}{BH}$, and $x=\frac{CC'}{CH}$. Thus we are to prove $xyz \geq 8$

Note that $\frac{HA\"}{AH}=\frac{HA'}{2AH}=\frac{1}{2} (\frac{AA'}{AH}-1)=\frac{x-1}{2}$ and $\frac{AA\"}{AH}=1+\frac{HA\"}{AH}=\frac{x+1}{2}$. Thus  $\frac{HA\"}{AA\"}=\frac{HA\"/AH}{AA\"/AH}=\frac{x-1}{x+1}=1-\frac{2}{x+1}$. Similarly, $\frac{HB\"}{BB\"}=1-\frac{2}{y+1}$ and $\frac{HC\"}{CC\"}=1-\frac{2}{z+1}$.

So,
$(1-\frac{2}{x+1})+(1-\frac{2}{y+1})+(1-\frac{2}{z+1})=\frac{HA\"}{AA\"}+\frac{HB\"}{BB\"}+\frac{HC\"}{CC\"}=\frac{[HBC]}{[ABC]}+\frac{[HCA]}{[ABC]}+\frac{[HAB]}{[ABC]}=\frac{[ABC]}{[ABC]}=1$
, implying $\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}=1$. Cross multiplying both sides and simplifying gives $xyz=x+y+z+2$. Thus it remains to prove $x+y+z \geq 6$

By AM-HM or Cauchy Schwartz, $1=\frac{1}{x+1}+\frac{1}{y+1} +\frac{1}{z+1}\geq \frac{9}{(x+1)+(y+1)+(z+1)}$, i.e., $x+y+z \geq 6$ and we are done

I have another solution of the original problem.

[color=blue][b]Problem.[/b] If H is the orthocenter of an acute-angled triangle ABC, and if the altitudes AH, BH, CH of triangle ABC intersect the circumcircle of triangle ABC at the points A', B', C', respectively (apart from the points A, B, C), then prove the inequality

$\frac{AA^{\prime}\cdot BB^{\prime}\cdot CC^{\prime}}{AH\cdot BH\cdot CH}\geq 8$.[/color]

[i]Solution.[/i] If X is the foot of the altitude AH of triangle ABC, then the triangle AXC is right-angled, and thus < XAC = 90° - < XCA. In other words, < A'AC = 90° - C. Since the point A' lies on the circumcircle of triangle ABC, the chordal angle theorem yields < AA'C = < ABC = B. Thus, by the sum of angles in triangle AA'C, we have < ACA' = 180° - < A'AC - < AA'C = 180° - (90° - C) - B = 90° - (B - C). Thus, sin < ACA' = cos (B - C).

Since the length of a chord in a circle equals twice the radius of the circle times the sine of the chordal angle of the chord, we have AA' = 2R sin < ACA', where R is the radius of the circumcircle of triangle ABC. Thus, AA' = 2R cos (B - C). After http://www.mathlinks.ro/Forum/viewtopic.php?t=24870 post #3 Lemma 1, since A < 90° (this is because the triangle ABC is acute-angled), we have AH = 2R cos A. Thus, $\frac{AA^{\prime}}{AH}=\frac{\cos\left(B-C\right)}{\cos A}$.

Now, we have

sin (2B) + sin (2C) = sin ( (B + C) + (B - C) ) + sin ( (B + C) - (B - C) )
= ( sin (B + C) cos (B - C) + cos (B + C) sin (B - C) ) + ( sin (B + C) cos (B - C) - cos (B + C) sin (B - C) )
= 2 sin (B + C) cos (B - C) = 2 sin A cos (B - C),

where I have used the fact that sin (B + C) = sin A (since B + C = 180° - A, because A + B + C = 180° by the sum of angles in triangle ABC). On the other hand, sin (2A) = 2 sin A cos A. Thus,

$\frac{\sin\left(2B\right)+\sin\left(2C\right)}{\sin\left(2A\right)}=\frac{2\sin A\cos\left(B-C\right)}{2\sin A\cos A}=\frac{\cos\left(B-C\right)}{\cos A}$.

Comparing this with $\frac{AA^{\prime}}{AH}=\frac{\cos\left(B-C\right)}{\cos A}$, we get $\frac{AA^{\prime}}{AH}=\frac{\sin\left(2B\right)+\sin\left(2C\right)}{\sin\left(2A\right)}$. Similarly, $\frac{BB^{\prime}}{BH}=\frac{\sin\left(2C\right)+\sin\left(2A\right)}{\sin\left(2B\right)}$ and $\frac{CC^{\prime}}{CH}=\frac{\sin\left(2A\right)+\sin\left(2B\right)}{\sin\left(2C\right)}$. Thus,

$\frac{AA^{\prime}\cdot BB^{\prime}\cdot CC^{\prime}}{AH\cdot BH\cdot CH}=\frac{AA^{\prime}}{AH}\cdot\frac{BB^{\prime}}{BH}\cdot\frac{CC^{\prime}}{CH}$
$=\frac{\sin\left(2B\right)+\sin\left(2C\right)}{\sin\left(2A\right)}\cdot\frac{\sin\left(2C\right)+\sin\left(2A\right)}{\sin\left(2B\right)}\cdot\frac{\sin\left(2A\right)+\sin\left(2B\right)}{\sin\left(2C\right)}$.

Now, since the triangle ABC is acute-angled, its angles A, B, C lie in the interval ]0°; 90°[, and thus, the angles 2A, 2B, 2C lie in the interval ]0°; 180°[. Thus, the sines sin (2A), sin (2B), sin (2C) of these angles are positive. Hence, we can apply the AM-GM inequality to get $\sin\left(2B\right)+\sin\left(2C\right)\geq 2\sqrt{\sin\left(2B\right)\sin\left(2C\right)}$, $\sin\left(2C\right)+\sin\left(2A\right)\geq 2\sqrt{\sin\left(2C\right)\sin\left(2A\right)}$, $\sin\left(2A\right)+\sin\left(2B\right)\geq 2\sqrt{\sin\left(2A\right)\sin\left(2B\right)}$. Thus,

$\frac{AA^{\prime}\cdot BB^{\prime}\cdot CC^{\prime}}{AH\cdot BH\cdot CH}=\frac{\sin\left(2B\right)+\sin\left(2C\right)}{\sin\left(2A\right)}\cdot\frac{\sin\left(2C\right)+\sin\left(2A\right)}{\sin\left(2B\right)}\cdot\frac{\sin\left(2A\right)+\sin\left(2B\right)}{\sin\left(2C\right)}$
$\geq \frac{2\sqrt{\sin\left(2B\right)\sin\left(2C\right)}}{\sin\left(2A\right)}\cdot\frac{2\sqrt{\sin\left(2C\right)\sin\left(2A\right)}}{\sin\left(2B\right)}\cdot\frac{2\sqrt{\sin\left(2A\right)\sin\left(2B\right)}}{\sin\left(2C\right)}=8$.

And we are done.

  Darij


I have two questions in geometry that I have been unable to figure out.

1. How do you construct a pentagon?

2.Suppose equilateral triangle ABC is inscribed in a circle. Points D and E are the midpoints of AB and AC, respectively. Suppose ray DE is extended, and the point at which this ray hits the circle is F. Find the ratio of DE to EF. The answer was surprisingly phi ((1+sqrt5)/2), and I can see this clearly algebraically (using point of a point about point E). Does anyone see an intuitive reason as to why this ratio must be phi (i.e. can someone see the golden rectangle or the spiral of Archimedes somewhere in the diagram described?)

I would appreciate any help. Thanks

I have seen many ways to construct a pentagon on the internet. One website actually had about 9 different ways to do it. However, I have not yet seen one that proves the construction after describing the construction. I need help on how to prove the construction works (I would appreciate a proof of any construction, I don't care which one. Also, if the proof/construction could be one that does not involve trig, since I do not know trig, that would also be helpful).

Once again, I would appreciate any help.


Hi!
I´m looking for elementary solutions for the next problem, i mean, by a construction or by a good idea! Anyway, all the solutios are welcome.
The problem says:
Let ABC a triangle with B=60°. Prove that $ 2b>\equal{}a\plus{}c$

Well if $ \angle B \equal{} 60 \text{degrees}$, then the sum of the other two angles is $ 180 \minus{} 60 \equal{} 120$.
$ 2(\angle B) \equal{} 120 \therefore 2b \equal{} a\plus{}c$

by cosine rule $ a^{2} \plus{} c^{2} \minus{} ac \equal{} b^{2}$
$ \Rightarrow b^{2}\ge ac$ (AM-GM)
$ \Rightarrow 3b^{2} \ge 3ac$
$ \Rightarrow 4b^{2}\ge (a \plus{} c)^{2}$
$ \Rightarrow 2b\ge (a \plus{} c)$

An obvious fact.
Let D be the point on the ray AB such that BD=BC and AD>AB. We simply have that angle ADC equals to 30 degrees.
Then, let H be the projection of point A to the line CD.
2AC>=2AH=AD=AB+BD=AB+BC. Q.E.D.


function
inequalities
vector
geometry
3D geometry
tetrahedron
geometry proposed

Let $ P_1,P_2,P_3,P_4$ be points on the unit sphere. Prove that $ \sum_{i\neq j}\frac1{|P_i\minus{}P_j|}$ takes its minimum value if and only if these four points are vertices of a regular pyramid.

Official Solution:
From the QM-HM inequality for the six numbers $ |P_i\minus{}P_j|$ we have \[ \sum_{i\neq j}\frac{1}{|P_i\minus{}P_j|}\geq C\left(\sum_{i\neq j}|P_i\minus{}P_j|^2\right)^{\minus{}\frac 12}\]where $ C$ is a constant. The equality holds iff $ |P_i\minus{}P_j|$'s are all equal, which is satisfied when $ P_i$'s are the vertices of a regular tetrahedron.

We now try to find an upper bound for $ \sum_{i\neq j}|P_i\minus{}P_j|^2$. Considering $ P_i$'s as vectors in 3D space, then this sum becomes $ 3\sum_i(P_i.P_i)\minus{}2\sum_{i\neq j}(P_i.P_j)$. Adding and subtracting $ \sum_i(P_i.P_i)$ we get $ 4\sum_i(P_i.P_i)\minus{}(\sum_iP_i).(\sum_iP_i)$. But $ P_i.P_i$ is a constant. Therefor the expression is maximized when $ \sum_iP_i\equal{}0$ for which it is equal to $ 4$. This condition is again satisfied when $ P_i$'s are the vertices of a regular tetrahedron.

Therefore, the condition of minimality of $ \sum_{i\neq j}\frac 1{|P_i\minus{}P_j|}$ becomes that, $ |P_i\minus{}P_j|$'s should be equal for different $ i,j$'s and that $ \sum_iP_i\equal{}0$. But if $ |P_i\minus{}P_j|$ has a constant value, then every face of the tetrahedron is an equilateral triangle, which itself shows that the tetrahedron is regular.

[quote="Omid Hatami"]Official Solution:
[i]...
We now try to find an upper bound for $ \sum_{i\neq j}|P_i \minus{} P_j|^2$. Considering $ P_i$'s as vectors in 3D space, then this sum becomes $ 3\sum_i(P_i.P_i) \minus{} 2\sum_{i\neq j}(P_i.P_j)$...[/i][/quote]
I don't see this; isn't it $ 2\left(\sum_i|P_i|^2\minus{}\sum P_i\cdot P_j\right)$?

I guess it doesn't really matter in your proof.
Mine was similar, except I used Cauchy engel form:
\[ \sum (P_i\minus{}P_j)^{\minus{}1} \leq \frac{C}{\sum |P_i\minus{}P_j|}\]
so we need upper bound for $ \left(\sum |P_i\minus{}P_j|\right)^2\leq C' \sum |P_i\minus{}P_j|^2$ then end same as above

Suppose $P_i=(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})$ for all $i\in\{1,2,3,4\}$. Now call $f(i,j)=|P_i-P_j|$.

Certainly we need to minimize $\sum \frac {1}{f(i,j)}$.

$\sum f^2(i,j)=\sum(\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}-\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2+\sum (\frac {c_i}{\sqrt{a^2_i+b^2_i+c^2_i}},\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$.

And now so $\sum f^2(i,j)=4-\sum (\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}})^2$.

Now, $(\sum f^2(i,j))(6)\geq (\sum f(i,j))^2\geq (\frac {36}{\sum \frac {1}{f(i,j)}})^2$-------(**)

Now so $\sum \frac {1}{f(i,j)}$ is minimized when $\sum \frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {b_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=\frac {a_i}{\sqrt{a^2_i+b^2_i+c^2_i}}=0$.

Now so clearly we've two cases for when $\sum \frac {1}{f(i,j)}$ can be minimized, one if all of $P_i$ lies on a circle and vertices of a square, and if not then vertices of a regular tetrahedron. If first case would br true then, equality wouldn't hold for the inequality of (**). Thus we must have those are vertices of  regular tetrahedron since $f(i,j)$ is constant for any $i,j$.

[hide]Sol is more or less same as above two, so,I'm really interested to see a non algebraic  method[/hide]


The state income tax where Kristin lives is levied at the rate of $ p \%$ of the first $ \$28000$ of annual income plus $ (p \plus{} 2) \%$ of any amount above $ \$28000$. Kristin noticed that the state income tax she paid amounted to $ (p \plus{} 0.25) \%$ of her annual income. What was her annual income?

$ \textbf{(A)} \ \$28000 \qquad \textbf{(B)} \ \$32000 \qquad \textbf{(C)} \ \$35000 \qquad \textbf{(D)} \ \$42000 \qquad \textbf{(E)} \ \$56000$

[hide="Solution"]Simply set up an equation:
$28000p + (p + 0.02)(x - 28000) = (p + 0.0025)x \Rightarrow x = 32000$
[/hide]


Isn't moems like math olympiads level(US), by george lenchner?

Are you talking about a book? I think Math olympiad and MOEMS are two different things.

He's probably talking about one these books : http://www.fetchbook.info/search_George_Lenchner/searchBy_Author.html

Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=47710]this[/url] for the difficulty level.

I meean in this site, http://moems.org/, these books: http://www.moems.org/Books.htm

covers:

[img]http://www.moems.org/Books_files/CPSsm.jpg[/img][img]http://www.moems.org/Books_files/MOPsm.jpg[/img]

MOEMS is the acronym for Mathematical Olympiads for Elementary and Middle Schools, or just Math Olympiads. Founded by George Lenchner in 1979 and directed by Richard Kalman since 1995, it offers 5 contests ("Olympiads") per year. The book is a compilation of the MOEMS contest problems and solutions for the first 16 years. Another book, Creative Problem Solving in School Mathematics (2nd Edition) by Dr. Lenchner, is being printed now and a sequel to the book you mention (1995-2005 problems) should be ready in about 2 years.

YAY!!! Our moderator is back!

Do they have more samples than the one on the website?

So i can post problems from the above books?

Good. I'lll post some here, after all, this forum is lately inactive.

You can post problems from either Lenchner book or from later MOEMS contests to this Forum. Better — why not try to make up your own problems, using past MOEMS problems as a guide. The level is fundamentally pre-algebra.

[color=red][b]BUT: please do not post anything about MOEMS contest problems within 2 weeks of a contest.[/b][/color]

New sample problems will be posted to [url]http://www.moems.org[/url] soon.

of course, or others might copy and get it right,and i don't want that.


Understanding the above paper (PDF document, Gold Bach Conjecture proof) requires a good/great deal of mathematical knowledge, and  a cursory reading isn't going to help in the process of understanding of the proof; I feel the Gold Bach  conjecture is true, but I may be mistaken.

You should have posted this in the thread you are referring to, since:
1. This is directly related to that topic
2. This is a statement and not a point of discussion

Also, if you refer to another topic, link it instead of referring to it as the "above paper", since the order of threads in a forum is constantly changing.


search
number theory unsolved
number theory

When 4444^4444 is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B.

http://www.mathlinks.ro/Forum/viewtopic.php?search_id=449346122&t=158349
Posted many times before

Daniel


calculus
integration
inequalities
triangle inequality

$ \triangle ABC$ and $ \triangle DBC$ share $ BC$. $ AB \equal{} 5\ \text{cm}$, $ AC \equal{} 12\ \text{cm}$, $ DC \equal{} 8\ \text{cm}$, and $ BD \equal{} 20\ \text{cm}$. What is the least possible integral number of centimeters in $ BC$?

[asy]size(100); import graph; currentpen = fontsize(10pt);
pair B = (0,0), C = (13,0), A = (-5,7), D = (16,10);

draw(B--A--C--cycle); draw(B--D--C);
label("$A$",A,W); label("$B$",B,W); label("$C$",C,E); label("$D$",D,E);[/asy]

By the Triangle Inequality, we have

$ 5\plus{}x>12\implies x>7$,
$ 5\plus{}12>x\implies17>x$,
$ 8\plus{}x>20\implies x>12$, and
$ 8\plus{}20>x\implies28>x$.

Combining these, we have $ 12<x<17$, so our answer is $ \boxed{13}$.


Let $ x;y$ be real numbers satisfying $ x^{2}\plus{}xy\plus{}y^{2}\equal{}1$
Find min and max of $ M\equal{}x^{2}\minus{}xy\plus{}2y^{2}$

[hide="wrong answer"]min = 1 at points $ (\pm 1, 0)$ and max=4 at points $ (\pm 1,\mp 1)$[/hide]

You are right. I've mistaken in determinant of the matrix
$ \left({\begin{array}{*{20}c}2\minus{}2\lambda &\minus{}1\minus{}\lambda\\ \minus{}1\minus{}\lambda & 4\minus{}2\lambda\\ \end{array}}\right)$  :D

The correct calculation leads to the same roots.


For non-negative real numbers x,y,z satisfying  x + y + z = 5, find max. value of $ x^{4}y \plus{} y^{4}z \plus{} z^{4}x$

[hide]
Since expression is cyclic, WLOG let $ x\equal{}\max \{ x,y,z\}$. If $ x\geq z\geq y$, then $ x^4y\plus{}y^4z\plus{}z^4x\leq x^4z\plus{}y^4x\plus{}z^4y$, since the right hand side minus the left hand side is
\[ x^4(z\minus{}y)\plus{}y^4(x\minus{}z)\plus{}z^4(y\minus{}x)\equal{}(y^4\minus{}z^4)(x\minus{}z)\plus{}(x^4\minus{}z^4)(z\minus{}y)\]
\[ \equal{}(x\minus{}z)(z\minus{}y)\left((x\plus{}z)(x^2\plus{}z^2)\minus{}(y\plus{}z)(y^2\plus{}z^2)\right) \geq 0\]
so we can switch two variables and reduce the inequality to one of the form $ x\geq y\geq z$. So now assume $ x\geq y\geq z$.

We show $ x^4y\plus{}y^4z\plus{}z^4x\leq (x\plus{}z)^4y$. To do this, $ (x\plus{}z)^4y\minus{}x^4y\equal{}y((x\plus{}z)^2\minus{}x^2)((x\plus{}z)^2\plus{}x^2)\equal{}yz(2x\plus{}z)(2x^2\plus{}2xz\plus{}z^2)$ so it's equivalent to
\[ y(2x\plus{}z)(2x^2\plus{}2xz\plus{}z^2)\geq y^4\plus{}z^3x\]
But the left hand side is certainly at least $ y(x^3\plus{}z^3)\equal{}y^4\plus{}z^3x\plus{}y(x^3\minus{}y^3)\plus{}z^3(y\minus{}x)$, and the difference is $ (x\minus{}y)(y(x^2\plus{}xy\plus{}y^2)\minus{}z^3)\geq 0$.

So we get
\[ x^4y\plus{}y^4z\plus{}z^4x\leq (x\plus{}z)^4y\equal{}y(5\minus{}y)^4\equal{}\frac{1}{4}(4y)(5\minus{}y)^4\]
\[ \leq \frac{1}{4}\left( \frac{4y\plus{}4(5\minus{}y)}{5}\right)^5\equal{}\boxed{256}\]
which we can get by $ x\equal{}4,y\equal{}1,z\equal{}0$.
[/hide]

Thats an interesting solution. A few questions/comments:

I don't quite get what you did when you say now we can switch two of the variables. I don't think the logic is so clear here...

So you do if $ x\ge z\ge y$, then $ f(x,z,y)\ge f(x,y,z)$? So it isn't really switching, but if we want to find a max, then we always look at the $ x\ge y\ge z$ case because when the variables are sorted in that order, it is greater than the other order.

I liked the identity:$ x^4z\plus{}y^4x\plus{}z^4y\minus{}x^4y\minus{}y^4z\minus{}z^4x$
$ \equal{}(x\minus{}y)(z\minus{}y)(x\minus{}z)(x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx)$

A cyclic sum is the product of a cyclic identity and a symmetric sum.

Also, where did the $ (x\plus{}z)^4y\ge x^4y\plus{}y^4z\plus{}z^4x$ come from? It is pretty tricky that equality occurs when $ z\equal{}0$...

But again, nice solution.


[quote="math92"]A rectangular chalk board is 3 times as long as it is wide. If it were 3 metres shorter and 3 metres wider, it would be square. What are the dimensions of the chalk board?[/quote]

[hide="answer"]

$3x=y$

$x+3=y-3$
$x+6=y$
So the answer is:
$3x=x+6$
$x=3$
$\framebox{3, 9}$[/hide]


algebra
polynomial
inequalities proposed
inequalities

let $ a^2\plus{}^2\equal{}1$,$ a,b\in R$
find the max and min of $ (2a\plus{}1)(2a\plus{}b)$
[hide]i use lagrange multipliers to solve this problem but i can't solve the equation so i can't find the maximum.[/hide]

[quote="Squall-Leonhart"]let $ a^2 \plus{} ^2 \equal{} 1$,$ a,b\in R$
find the max and min of $ (2a \plus{} 1)(2a \plus{} b)$
[hide]i use lagrange multipliers to solve this problem but i can't solve the equation so i can't find the maximum.[/hide][/quote]

         $ f_{min} \equal{} \minus{} 1{\le}{f \equal{} (2a \plus{} 1)(2a \plus{} b)}\le{6.4241845266...} \equal{} f_{max},$

where  $ \\
\ f_{max}$is a root of the polynomial  $ {64x^3 \minus{} 544x^2 \plus{} 853x \plus{} 3}.$

hello, i have found $ \minus{}0.350914229205462726E\minus{}2$ as the minimum by a numerical way.
Sonnhard.

f min=-1 is true :D
but i can't prove that f max=6,4214... how????


Hi everyone,

I am writing a book and I have a need to
 set a new counter for equation every once in a while.
There is a running counter for all the equations book.
However, when I write a solution for the examples I
would like to have a new number for the equations that appear in
that the solution. This counter should not interfere with with the regular
counter for the regular equations.  I also would like this "solution" counter to be reset to 1
every time I start new "solution''.

If possible the example have their counter and to  have the solution counter to be slave counter
of the example master counter.

that is
regular counter
...........
\begin{align}}
bla bla
\end{align}}

\begin{align}}
bla bla
\end{align}}

\solution {

solution counter (start from 1 and possible example.1)
...........
\begin{align}}
example bla
\end{align}}

example.2
\begin{align}}
more example bla
\end{align}}

}
I wonder how to do it.
Is their a package that handle it?
any not any suggestion how to tackle this problem?

Genick Bar-Meir
http://www.potto.org
where free textbooks roam!

When you define a new counter, there's an optional parameter to link it to an already defined counter. Thus
\newcounter{counterA}[counterB]
means that a new counter, counterA, is linked to an already existing counterB. Whenever counterB is incremented by \stepcounter, counterA is reset to 0.
This is how the subsection counter works, which is reset to 0 every time the section counter is incremented.

Thanks stevem,

That is great .
I think that now it will work for most part but there might
be another problem.

In my document I am using the book style.
Thus the counter appears \thechapter.\theexample e.g. 1.2.
how to get ride of the chapter first part or how to access the counter of the example.

Second how to define the new equation so that it will not used the old regular
counter of \theequaiton.

that is it will be using \theExampleEquation counter or something like this.

Thanks,

Genick

You can redefine the output of any counter to show whatever you like.
For example,
\renewcommand{\theequation}{\arabic{equation}}
or
\renewcommand{\theequation}{(\arabic{equation})}
which gives you (1) instead of 1.
You can put anything you like in the last set of curly brackets so
\renewcommand{\theequation}{Equation \arabic{equation}}
gives Equation 1
etc etc

Thanks Steven for your help so far,

Here is what is understand from your suggestion

\newtheorem{examl}{Example}[chapter]
\newcounter{exEq}[examl]
\interfootnotelinepenalty=10000
\newcommand{\latexsolution}[1]{\par \parindent 0pt
   \normalfont
   \setcounter{exEq}{1}
   \renewcommand{\theexEq}{\arabic{exampl}.\arabic{exEq}}
   \underline{\scshape Solution}\\[3mm]
    \parindent 25pt
    {#1}\par
   \centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}  }
}
Which provide me a new counter exampl.eq
inside the latexsolution

[b]however, it complain about [/b]

! Missing number, treated as zero.
<to be read again>
                   \c@exampl

[b]which I take as the counter exampl is not defined inside the latexsolution.[/b]

The other issue remained:  how to center equation between \$ bla bla \$  and put the counter on the right hand side and
to center the height of my counter on the right hand side relative the equation.

Can used the definition of alian and reduplicate  it with the my counter. if so where is the definition
of alian.

Thanks,

Genick

[quote="genick"][b]however, it complain about [/b]

! Missing number, treated as zero.
<to be read again>
                   \c@exampl

[b]which I take as the counter exampl is not defined inside the latexsolution.[/b][/quote]
I can't reproduce this error so please give a complete (including preamble) minimal document that shows this.

[quote="genick"]The other issue remained:  how to center equation between \$ bla bla \$  and put the counter on the right hand side and to center the height of my counter on the right hand side relative the equation.[/quote]
Use \begin{equation}...\end{equation} to centre the equation and \begin{split}...\end{split} to centre the counter.
For example
[code]\begin{equation}
\begin{split}
x=1 \\
y=2
\end{split}
\end{equation}[/code]
[quote="genick"]Can used the definition of alian and reduplicate  it with the my counter. if so where is the definition
of alian.[/quote]I don't understand this question.

[quote="genick"]! Missing number, treated as zero.
<to be read again>
                   \c@exampl

[b]which I take as the counter exampl is not defined inside the latexsolution.[/b][/quote]
I see the error now. The first two lines say
\newtheorem{[i]examl[/i]}{Example}[chapter]
\newcounter{exEq}[ [i]examl[/i]]
and so uses [i]examl[/i] whereas
\renewcommand{\theexEq}{\arabic{[i]exampl[/i]}.\arabic{exEq}}
uses[i] exampl[/i]

\documentclass[10pt]{book}

\newtheorem{examl}{Example}[chapter]
\newcounter{exEq}[examl]
\newcommand{\latexsolution}[1]{\par \parindent 0pt
   \normalfont
   \setcounter{exEq}{1}
   \renewcommand{\theexEq}{\arabic{exampl}.\arabic{exEq}}
   \underline{\scshape Solution}\\[3mm]
    \parindent 25pt
    {#1}\par
   \centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}  }
}

\begin{document}

\latexsolution{
\theexEq{} is not counter
}
\end{document}

thank you for the last comment

For example
Code:
\begin{equation}
\begin{split}
x=1 \\
y=2
\end{split}
\end{equation}

Yet it again using the equation counter.
but when I use the split between \$ and \$ it will give
! Package amsmath Error: \begin{split} won't work here.

thanks,

[quote="genick"]
but when I use the split between \$ and \$ it will give
! Package amsmath Error: \begin{split} won't work here.[/quote]
split must be used inside an equation or align environment. \$...\$ is only meant for inline equations (equations in a line of text) where you wouldn't have an equation number. You need to use a displayed equation which is on a new line and centred.

Thanks,

that is the core of the problem.
I cannot use \begin{equation} or \begin{align} because the counter
will  be used and affected  the equation counter outside the latexsolution environment.

that i would like to have
\begin{equation}
bla bla
\end{equaiton}

yields

bla bla                                              (chapterNumber.1)

\begin{equation}
bla bla
\end{equaiton}

yields

bla bla                                              (chapterNumber.2)

but when
\latexsolution{
\begin{equation}
something do with the ex
\end{equation}
should yield now

something do with the ex                                         (examleNumber.1)

and will not change the counter of chapterNumber.
so if I will return to ouside and  do

\begin{equation}
back to the chapter eq
\end{equation}

[b]back to the chapter eq                                                                                (chapterNumber.3)
and not chapterNumber.4! [/b]

I have seen a package that supposed to it but it is way beyond me to understand
it.

I do not understand the remady
[code]
\renewcommand{\theexEq}{\arabic{exampl}.\arabic{exEq}}
uses exampl
[\code]

thanks
genick
http://www.potto.org

I really don't understand what you are doing. The equation and align environments use the standard equation counter, but \latexsolution uses a different counter so they don't affect one another.

Every time you use \latexsolution you redefine \theexEq to be exactly the same as before so it should be defined separately. You reset \exEq to 1 but then you never use \theexEq anywhere.

Please show me a complete example tex file where you are getting the wrong numbers that you mention in your post. Also include where \newtheorem{exampl}{Example}[chapter] is used as a theorem environment \begin{exampl}...\end{example}

Does this do what you want? It uses two different counters, one for \latexsolution and one for myequation environment so they don't interfere with each other. The normal counter is used for \latexsolution so you can use split, but \tag is used to give a different counter for equations.

[code]\documentclass[10pt]{book}
\usepackage{amsmath}
\newcounter{myequation}
\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
\renewcommand{\themyequation}{\arabic{chapter}.\arabic{myequation}}

\newcommand{\latexsolution}[1]{
\par \parindent 0pt
\normalfont
\underline{\scshape Solution}\\[3mm]
\parindent 25pt
{#1}\par
\centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}}
}

\newenvironment{myequation}{\stepcounter{myequation}\begin{equation}}{\end{equation}}

\begin{document}
\chapter{First Chapter}
\begin{myequation}
\text{equation } 1 \tag{\themyequation}
\end{myequation}

\begin{myequation}
\text{equation } 2 \tag{\themyequation}
\end{myequation}

\latexsolution{
\begin{equation}
\begin{split}
x=1\\
y=2
\end{split}
\end{equation}
}

\latexsolution{
\begin{equation}
\begin{split}
x=3\\
y=4
\end{split}
\end{equation}
}

\latexsolution{
\begin{equation}
\begin{split}
x=5\\
y=6
\end{split}
\end{equation}
}

\begin{myequation}
\text{equation } 3 \tag{\themyequation}
\end{myequation}

\begin{myequation}
\text{equation } 4 \tag{\themyequation}
\end{myequation}
\end{document} [/code]

I think that you almost got it.
the counter in the latexsolution should be reset.

if it will be something like this

\documentclass[10pt]{book}
\usepackage{amsmath}
\newcounter{myequation}
\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
\renewcommand{\themyequation}{\arabic{chapter}.\arabic{myequation}}

\newcommand{\latexsolution}[1]{
\par \parindent 0pt
\normalfont
\underline{\scshape Solution}\\[3mm]
\parindent 25pt
{#1}\par
\centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}}
}

\newenvironment{myequation}{\stepcounter{myequation}\begin{equation}}{\end{equation}}

\begin{document}
\chapter{First Chapter}
\begin{myequation}
\text{equation } 1 \tag{\themyequation}
\end{myequation}

\begin{myequation}
\text{equation } 2 \tag{\themyequation}
\end{myequation}

\latexsolution{
fkdfjkdfj
\begin{equation}
\begin{split}
x=1\\
y=2
\end{split}
\end{equation}
fkdkfjdkfjdf k
\begin{equation}
\begin{split}
x=3\\
y=4
\end{split}
\end{equation}
}

now the numbers should reset for the latexsolution.
and it should 2.1 2.2. etc.

\latexsolution{
\begin{equation}
\begin{split}
x=5\\
y=6
\end{split}
\end{equation}
{\begin{align}
        3+ 7 = 10
\end{align}}
}

\begin{myequation}
\text{equation } 3 \tag{\themyequation}
\end{myequation}

\begin{myequation}
\text{equation } 4 \tag{\themyequation}
\end{myequation}

\end{document}

So you don't want the latexsolution counter to show the chapter? In that case:
[code]\newcounter{solution}
\renewcommand{\theequation}{\arabic{solution}.\arabic{equation}}
\renewcommand{\themyequation}{\arabic{chapter}.\arabic{myequation}}

\newcommand{\latexsolution}[1]{
\stepcounter{solution}
\setcounter{equation}{0}
...[/code]
will do it

Steven,

First,  thank for all the advises and work that you put into this.

I think that you close in the last post.

My book organization is
example
which is followed by
the latexsolution.
Thus the counter is of exampl(e)
in the latexsolution there are several equations
that needed to refered to.
Now I relized that counter should be chaper.exampl.Ex_equation.

I was playing around with your ideas
but I still do not get it right.
The missing part that it does not reset to 1 everytime
the \latexEx is issue.

this the test file
*****************************
\documentclass[10pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
%\usepackage{potto}
\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\newtheorem{examl}{Example}[chapter]
\newcounter{myequation}[examl]
\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
\renewcommand{\themyequation}{\arabic{chapter}.\arabic{examl}.\arabic{myequation}}

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}

\newcommand{\latexsolution}[1]{
\par \parindent 0pt
\normalfont
\underline{\scshape Solution}\\[3mm]
\parindent 25pt
{#1}\par
\centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}}
}

\newenvironment{myequation}{\stepcounter{myequation}\begin{equation}}{\end{equation}}

\begin{document}
\chapter{First Chapter}
\begin{equation}
\text{equation } 1 %\tag{\themyequation}
\end{equation}

\begin{equation}
\text{equation } 2 %\tag{\themyequation}
\end{equation}

\theexaml
\latexEx{a}{
}
rest the counter

\latexsolution{
\begin{myequation}
\begin{split}
x=1\\
y=2
\end{split}
\tag{\themyequation}
\end{myequation}
\begin{myequation}
\begin{split}
x=6\\
y=7
\end{split}
\tag{\themyequation}
\end{myequation}

}

\latexEx{b}{
}
reset the  counter but it does not work!

\latexsolution{
\begin{myequation}
\begin{split}
x=3\\
y=4
\end{split}
\end{myequation}
\begin{myequation}
\begin{split}
x=3\\
y=4
\end{split}
\tag{\themyequation}
\end{myequation}
}

\latexsolution{
\begin{myequation}
\begin{split}
x=5\\
y=6
\end{split}
\tag{\themyequation}
\end{myequation}
}

\begin{equation}
\text{equation } 3
\end{equation}

\begin{equation}
\text{equation } 4
\end{equation}
\end{document}

I'm not clear as to which counter you want to reset. In your document could you add by hand what you want the counter to say where it's wrong. For example
[code]\latexEx{b}{
}
reset the counter but it does not work! Counter should be (a.b) not (1.3) [/code]

Steven,

First thanks, it is working now. :lol:
Now i can remove the ugly things  from my book.

I wonder if there is a way to put the \tag{\themyequation} in a definition
so that I do not have to retype it.
Again thank you very much for your help.

genick
http://www.potto.org
where free textbooks roam!

See below.

\documentclass[10pt]{book}
\usepackage{amsmath, amsthm}
\usepackage{latexsym}
%\usepackage{potto}
\newtheoremstyle{myExampl}{\baselineskip}{\baselineskip}{\itshape} {}{\sffamily}{:}{\newline}{}
\newtheorem{examl}{Example}[chapter]
\newcounter{myequation}[examl]
\renewcommand{\theequation}{\arabic{chapter}.\arabic{equation}}
\renewcommand{\themyequation}{\arabic{chapter}.\arabic{examl}.\arabic{myequation}}

\newcommand{\latexEx}[2]{
\begin{examl}
\label{#1}
#2
\end{examl}
}

\newcommand{\latexsolution}[1]{
\par \parindent 0pt
\normalfont
\underline{\scshape Solution}\\[3mm]
\parindent 25pt
{#1}\par
\centerline{\rule{2in}{2pt} \tiny End solution \rule{2in}{2pt}}
}

\newenvironment{myequation}{\stepcounter{myequation}\begin{equation}}{\end{equation}}

\begin{document}
\chapter{First Chapter}
\begin{equation}
\text{equation } 1 %\tag{\themyequation}
\end{equation}

\begin{equation}
\text{equation } 2 %\tag{\themyequation}
\end{equation}

\latexEx{a}{
}

\latexsolution{
\begin{myequation}
\begin{split}
x=1\\
y=2
\end{split} \tag{\themyequation}
\end{myequation}
\begin{myequation}
\begin{split}
x=6\\
y=7
\end{split} \tag{\themyequation}
\end{myequation}

}

\latexEx{b}{
this is the second example to followed by the solution with
number 1.2.1 and not 1.2.3.
}
\latexsolution{
\begin{myequation}
\begin{split}
x=3\\
y=4
\end{split} \tag{\themyequation}
\end{myequation}
\begin{myequation}
\begin{split}
x=3\\
y=4
\end{split}
\tag{\themyequation}
\end{myequation}
}
\latexEx{c}{
this is the second example to followed by the solution with
number 1.3.1 and not 1.3.4.
}

\latexsolution{
\begin{myequation}
\begin{split}
x=5\\
y=6
\end{split}
\tag{\themyequation}
\end{myequation}
}

\begin{equation}
\text{equation } 3
\end{equation}

\begin{equation}
\text{equation } 4
\end{equation}
\end{document}

What you can do is replace two environments (myequation and split) by one. It's a bit tough as amsmath environments like split behave oddly. But this should work.
Add
[code]\newenvironment{spliteq}
  {\stepcounter{myequation}\equation\aligned}
  {\endaligned\tag{\themyequation}\endequation} [/code]
then replace
[code]begin{myequation}
\begin{split}
x=1\\
y=2
\end{split} \tag{\themyequation}
\end{myequation}[/code]
by
[code]\begin{spliteq}
x=1\\
y=2
\end{spliteq}[/code]

You won't then need [code]\newenvironment{myequation}{\stepcounter{myequation}\begin{equation}}{\end{equation}}[/code]


A value of $ x$ satisfying the equation $ x^2 \plus{} b^2 \equal{} (a \minus{} x)^2$ is:

$ \textbf{(A)}\ \frac{b^2 \plus{} a^2}{2a}\qquad
\textbf{(B)}\ \frac{b^2 \minus{} a^2}{2a}\qquad
\textbf{(C)}\ \frac{a^2 \minus{} b^2}{2a}\qquad
\textbf{(D)}\ \frac{a \minus{} b}{2}\qquad
\textbf{(E)}\ \frac{a^2 \minus{} b^2}{2}$

hello, solving the equation we have $ x^2\plus{}b^2\equal{}a^2\plus{}x^2\minus{}2ax$ from here we get $ x\equal{}\frac{a^2\minus{}b^2}{2a}$ if $ a\ne0$.
Sonnhard.


prove that the diameter of a square are equal.

[quote="ashegh"]if i can use pytagoras theorem,here is my solution:

AD=DC=BC=AB

AD.AD+DC.DC=AC.AC

AB.AB+AD.AD=BD.BD

then we have :

                     AC=BD[/quote]
yes i think you can use it but this problem is not for you because you are so cool and this problem is easy for you.
but if you can solve it without using pytagoras theorem please send your solution.

u know little guys,i think i should talk about symmetry. :blush:

but u can think about it alittle...

i think because of the symmetry of square,the diameters should be also equal. ;)

yes you are right but here we should prove it nice and easy. :D

The triangle ABC is equal to the triangle BCD then AC=BD

we can prove some problems ,and after that the special form of them will also be prove.


$ 1 \le n \le 455$ and $ n^3 \equiv 1 \pmod {455}$. The number of solutions is ?

$\textbf{(A)}\ 9  \qquad\textbf{(B)}\ 6  \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 1  \qquad\textbf{(E)}\ \text{None}$

[hide]Taking mods $ 5,7,13$, we find that $ n\equiv 1\pmod{5},\equiv 1,2,4\pmod{7},\equiv 1,3,9\pmod{13}$, so there are $ 1\cdot 3\cdot 3\equal{}9$ solutions and the answer is $ A$.[/hide]


I noticed some of the pages have a link to a proof while others have the proof included on the page. Is there a preferred way for this?

I'm inclined towards putting the proof on the page.


If the positive integer solutions to $ \sqrt [n]{x} \equal{} \log_{n}{k}$ are $ (x_1 , n_1), (x_2 , n_2),.... (x_k , n_k)$ for $ x_i,n_i < 1000$ where $ 1\le i \le k$  Find the value of $ \displaystyle\sum_{i \equal{} 1}^{k}(x_i \plus{} n_i)$,

i came up with this problem, and i wrote a program and found the answer, but i don't know how to do it.

I think there have to be restrictions on (x_i,n_i), like x_i>x_i-1 and same with n or something...otherwise I think there's a lot of possible solutions.

well all terms are less than 1000 and $ (x_i, n_i)$ represents a solution for x and n.

Sorry, I'm not sure that I understand the question, but what is $ k$?

Edit: I mean, is $ k$ a constant or an arbitrary positive integer?

$ k$ is nothing special: $ (x_1 , n_1), (x_2 , n_2), ....., (x_k , n_k)$ are the solutions of the equation and $ (x_i , n_i)$ is an arbitrary solution.

Um, you have (x_k, n_k) which appears to mean that there are at most k solutions, then you have k=n^(x^(1/n)), which means k would be limited by x and n, but you have a bunch of possible values for both variables, and therefore many different values of k.

OMG i am really sorry, i mistyped the question, it is actually

If the positive integer solutions to $ \sqrt [n]{x} \equal{} \log_{n}{x}$ are $ (x_1 , n_1), (x_2 , n_2),.... (x_k , n_k)$ for $ x_i,n_i < 1000$ where $ 1\le i \le k$  Find the value of $ \displaystyle\sum_{i \equal{} 1}^{k}(x_i \plus{} n_i)$

Could you post the answer - I'm still not sure I understand completely? (why is the answer not just $ 2^2 \plus{} 2 \plus{} 3^3 \plus{} 3 \plus{} 4^4 \plus{} 4 \plus{} 2^4 \plus{} 2 \plus{} 2^4 \plus{} 4$?)

that is actually the correct answer except you forgot $ (1,1)$ so i got 336 and you got 334, how'd you do it?

I don't believe that $ \log_1 1$ is defined.

Let $ y \equal{} \sqrt [n]{x} \equal{} \log_n x \Longrightarrow x \equal{} y^n \equal{} n^y$. So either $ y \equal{} n$ (yielding $ \boxed{(x_{i \minus{} 1},\, n_{i \minus{} 1}) \equal{} (i^i, i)}$), or without loss of generality let $ y > n > 1$. It follows by examining their prime factorizations that $ n|y$, so we let $ y \equal{} zn,\ z > 1$, and we get $ n^{zn} \equal{} (zn)^n \Longrightarrow (n^{z \minus{} 1})^n \equal{} z^n \Longrightarrow n^{z \minus{} 1} \equal{} z$. For $ z \equal{} 2$ we find $ n \equal{} 2 \Longrightarrow \boxed{(x_{k \minus{} 1,k},n_{k \minus{} 1,k}) \equal{} (16,2)(16,4)}$, and for $ z \ge 3$ we have $ n^{z \minus{} 1} \ge 2^{z \minus{} 1} > z$ (which follows by induction etc).

Edit: I believe there's a problem on the last page of the intermediate problem marathon that generalizes to rationals.


We have a circle $C: x^2+y^2=r^2$, and a point $P: (x_0,y_0)$. (P is out of the circle.)
Find the equations of the two tangent lines from P to circle C.

in terms of $x_0, y_0$ and $R$?  if thats the case let the tangent line be $y-y_0=m(x-x_0)$, plug this into the circle equation, then you'll get a quadratic in $x$.  since it is a tangent, it intersects the circle in 1 point $\rightarrow$ the discriminant is 0.  So doing that you'll have a quadratic in $m$, and you can solve that in terms of $x_0, y_0$ and $R$.

calculus method is to differentiate implicitly, that'll get you the slope... i believe at least, i dont know calculus that well.

Use that $ S S_1 = T^2 $ where $S = x^2 + y^2 - r^2 $  $S_1 = x_0 ^2 + y_0 ^2 - r^2 $ and $T = x x_0 + y y_0 - r^2 $

This is a basic result and gives the combined eqn. for both the tangents....

[quote="shobber"]Yep, I used your method to get this equation:
\[m^2(r^2-x_0^2) +2x_0y_0m+(r^2-y_0^2)=0 \]
But it turns out to be ugly from here so now I begin to wonder if there is a more easier way to find the equations?[/quote]

write $m = \frac{{y - y_0 }}{{x - x_0 }}$ and simplify.


Circles centered at $ A$ and $ B$ each have radius $ 2$, as shown. Point $ O$ is the midpoint of $ \overline{AB}$, and $ OA \equal{} 2\sqrt {2}$. Segments $ OC$ and $ OD$ are tangent to the circles centered at $ A$ and $ B$, respectively, and $ EF$ is a common tangent. What is the area of the shaded region $ ECODF$?
[asy]unitsize(6mm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;

pair O=(0,0);
pair A=(-2*sqrt(2),0);
pair B=(2*sqrt(2),0);
pair G=shift(0,2)*A;
pair F=shift(0,2)*B;
pair C=shift(A)*scale(2)*dir(45);
pair D=shift(B)*scale(2)*dir(135);

pair X=A+2*dir(-60);
pair Y=B+2*dir(240);

path P=C--O--D--Arc(B,2,135,90)--G--Arc(A,2,90,45)--cycle;

fill(P,gray);
draw(Circle(A,2));
draw(Circle(B,2));
dot(A);
label("$A$",A,W);
dot(B);
label("$B$",B,E);
dot(C);
label("$C$",C,W);
dot(D);
label("$D$",D,E);
dot(G);
label("$E$",G,N);
dot(F);
label("$F$",F,N);
dot(O);
label("$O$",O,S);
draw(G--F);
draw(C--O--D);
draw(A--B);
draw(A--X);
draw(B--Y);
label("$2$",midpoint(A--X),SW);
label("$2$",midpoint(B--Y),SE);[/asy]$ \textbf{(A)}\ \frac {8\sqrt {2}}{3}\qquad \textbf{(B)}\ 8\sqrt {2} \minus{} 4 \minus{} \pi \qquad \textbf{(C)}\ 4\sqrt {2}$
$ \textbf{(D)}\ 4\sqrt {2} \plus{} \frac {\pi}{8}\qquad \textbf{(E)}\ 8\sqrt {2} \minus{} 2 \minus{} \frac {\pi}{2}$

[hide]
$\angle CAE = 90-\angle CAO = 45^\circ$

$[\text{sector}(CAE)] = (4\pi)\left(\frac{45}{360}\right)=\frac{\pi}{2}$

$[CAO] = \frac{1}{2}\left(2^{2}\right) = 2$

$\therefore (2)\left(4\sqrt{2}\right)-2\left(2+\frac{\pi}{2}\right)$

$=8\sqrt{2}-4-\pi$

$\boxed{\text{B}}$
[/hide]

Just divide the figure in to simple triangles and rectangle :wink:

[hide=Solution] It's easy to see the area of triangle OEF is $4\sqrt2$, but the shaded area is a little less than triangle OEF, that will eliminate $A$, $C$ and $D$.  $4\sqrt2-(8\sqrt2-2-\pi/2) = 2+\pi/2 - 4\sqrt2 < 0$, that leaves only $\boxed{B}$ as the correct answer.

[/hide]


integration
limit
function
geometry
rectangle
calculus
calculus computations

If $ f$ is Riemann integrable on [0,1], then

$ \int_{0}^{1} f \equal{} \lim\displaystyle{\frac{1}{n}\sum_{k\equal{}1}^{n} f(\frac{k}{n})}$

I honestly have no idea where to even start. could i get some hints?

I'm not sure if you're asking to prove it or just what it means.  I can explain what it means.

$ \int_a^b f(x) ~dx \equal{} \lim_{n\to \infty} \sum_{k\equal{}1}^n f \left( a\plus{} k \cdot \frac{b\minus{}a}{n} \right) \cdot \frac{b\minus{}a}{n}$

Thus:

$ \int_0^1 f(x) ~dx \equal{} \lim_{n\to \infty} \sum_{k\equal{}1}^n f(x_k) ~\Delta x$

and $ \Delta x \equal{} \frac{b\minus{}a}{n} \equal{} \frac{1\minus{}0}{n} \equal{} \frac{1}{n}$ and $ x_k \equal{} a\plus{}k \cdot \Delta x \equal{} 0\plus{}k \cdot \frac{1}{n} \equal{} \frac{k}{n}$, we get:

$ \int_0^1 f(x) ~dx \equal{} \lim_{n\to \infty} \frac{1}{n} \sum_{k\equal{}1}^n f \left( \frac{k}{n} \right)$

And if you want to prove it, try the following: Draw a function $ f(x)$ (any function, it's used for theoretical purposes only) on the interval $ [0,1]$. Now, divide the interval $ [0,1]$ into $ n$ new intervals of the same length. At the beginning of each interval $ [a_i,b_i]$, draw a rectangle that's set to have vertexes in $ (a_i,0)$, $ (a_i,f(a_i))$, $ (b_i,0)$, $ (b_i,f(a_i))$. This way you can say that the approximation to the surface below $ f(x)$ could be represented as the sum of all the surfaces of all the rectangles. Also, when the number of rectangles $ \rightarrow\infty$, the sum equals the surface! I hope this helped!

yeah I m proving it.

i appreciate the help.

This is the [i]definition[/i] of a Riemann integral, not a problem.  What are you trying to "prove"?

More precisely, it's a consequence of the definition of the Riemann integral. That definition calls for a limit to be taken over all shapes of partition, provided the mesh tends to zero, and all choices of a point of evaluation in each subinterval. The above limit samples only a restricted few of those, namely the equal-width partitions with the right hand endpoint of each subinterval chosen.

If the function is Riemann integrable, than this restricted choice of Riemann sums, being in effect a subsequence of the set of all Riemann sums, must converge to the integral.

On the other hand, the existence of this limit does not imply Riemann integrability. (That old standby, the characteristic function of the rationals, should serve as a counterexample.)


How do you make tree diagrams in $ \text{\LaTeX}$?

In Asymptote, you can use the tree, binarytree, or drawtree packages.  See the Asymptote manual or read the associated asy files.

Using LaTeX there are packages providing features aimed at this:

[list][*] [url=http://www.ctan.org/pkg/pgf]pgf/TikZ[/url], see some examples [url=http://www.fauskes.net/pgftikzexamples/tag/trees/]here[/url], the manual shows simpler examples,
[*] [url=http://tug.org/PSTricks/]PSTricks[/url], see [url=http://tug.org/PSTricks/main.cgi?file=pst-tree/pst-tree]trees examples[/url].[/list]
Usually I prefer pgf/TikZ because it directly produces pdf output and because of my personal taste.

Stefan


Jackson and Lisa divide some money in a ratio of $ 3: 2$, respectively.  If Lisa gives Jackson $ \$6$, then Jackson will have twice as much money as Lisa will have.  How much money did Lisa have after the original division of money?

Let x = the amount Lisa has.

Since Jackson and Lisa divided their money in a ratio of 3 : 2, Jackson would have 3x/2.

If you add 6 to Jackson's amount, you have to subtract 6 from Lisa's.

So they Jackson and Lisa would have (3x+12)/2 and 2x-12, respectively.

Since Jackson would have twice what Lisa has, you would multiply Lisa's amount by two and set that and Jackson's amount equal.

So (3x+12)/2 = 4x-24.

Multiply by two, add 24, and subtract 3x and you would get: x=36.

So the answer is 36.