can anyone tell mw the latest research on tachyons

Or you could look it up yourself.  Try google scholar as a starting point.

hey i mean tachyonic research and non physical components having boundless energy and able to move at speed of light!
 :)

You can still look it up yourself. Google scholar has quite a bit of stuff.

tachyons r particals that r hypotesized to move at speeds greater than lights

they r trying to find to using STM and other similar particle acc processes......... :P


Derive a formula for the area of a regular dodecagon inscribed in a circle with radius r.

Isn't there a "proof without words" thing already on AOPS?

[hide]
Btw, the answer's 3r^2

Proof:

Consider the isosceles triangle formed with two of the radii and one of the sides. Use half the that triangle, which basically is a right triangle with one angle as 15 degrees.

The leg opposite the 15 degree angle has length $ r\sin15$ while the other leg has length $ r\cos15$, and the hypotenuse has length $ r$.

Now, consider the area of the original isosceles triangle is $ r^2\sin15\cos15$, so the area of the dodecagon is,
$ 12r^2\sin15\cos15$
$ \equal{} 6r^2(2\sin15\cos15)$
$ \equal{} 6r^2(\sin30)$
$ \equal{} 3r^2$

QED[/hide]

Generalization: the area of a regular n-gon with circumradius r has area: $ \frac{nr^2}{2}\sin\left(\frac {360}{n}\right)$

EDIT: Forgot to divide by 2...Thanks, igiul

[hide](r^2sin36)x 10[/hide]

We had a similar question in a recent maths test.  :cool:

The answer given by Carbon57 is correct, but $ 10\sin36^\circ\ne3$, so your answer isn't correct. :(

Can you post a solution? Your error can't be found if you don't. :wink:

can you make a figure for what you are talking about Carbon 57? :oops:

@blue: I think you were trying to find  the area of a decagon (10 sides), while the questions asks for a dodecagon (12 sides), since there's a 10 and a 36, which is 360/10, in your expression

Deriving the area of a regular n-gon is much more useful.

Partition the n-gon into n isosceles triangles as above. Bisect one of these triangles by drawing a lines through the origin and the midboint of it's base. Noting that $ r$ is the hypotenuse of the new triangle formed (there are two), we see that it's area is given by \[ \frac{r^2\sin{\theta}\cos{\theta}}{2}\] where $ \theta\equal{}\frac{\pi}{n}$. And because there are $ 2n$ of these triangles, the area of a regular n-gon taking the radius of its circumscribing circle as the argument is \[ \displaystyle A_n\equal{}\frac{nr^2}{2}\sin{\left(\frac{2\pi}{n}\right)}\]
So, $ A_{12}\equal{}6r^2\sin{\frac{\pi}{6}}\equal{}3r^2$

@Carbon57: You forgot to divide by two in your derivation of the general formula.

I am very sorry.

In my last post, I meant to write 1/2(r^2sin30) x 12

As someone here has already said, refresh the page, and look at those "animated proofs" on the left column of the page. One of them gives a visual solution to this very problem.


Can the equation be solved
             $ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} w^{2} \equal{} 19$
              $ x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} w^{3} \equal{}224$
by using the Newton series on it?

[quote="mszew"]do $ (x,y,z,w) \in \mathbb R$, or $ \mathbb Z$?[/quote]
Obviosly in R it had not solution, because $ 224>19\sqrt{19}.$


Hello, dear Mathlinkers!

Has anybody heard of [b]McLaughlin's inequality[/b] for symmetric polynimials?
There's something like that:

$ \sqrt [1]{\frac {\sigma_1}{\binom{n}{1}}}\ge \sqrt [2]{\frac {\sigma_2}{\binom{n}{2}}}\ge...\ge \sqrt [n]{\frac {\sigma_n}{\binom{n}{n}}}$ $ ?$

Does it have any practical applications? Thanks in advance.

P.S. Sorry for wrong spelling :(

You mean [url=http://en.wikipedia.org/wiki/MacLaurin%27s_inequality]MacLaurin's Inequality[/url].


salam,chejoori mishe  ie adade mokhtaleto be tavane ashari resand?(ia radikal gereft)

Age dorost mottevajje shode basham, mikhahi ke bedooni che tori mishe radicale adadhaye dar $\mathbb{C}$ peyda kard. Age $z^{n}=a+bi\implies z^{n}=\sqrt{a^{2}+b^{2}}(cos\theta+i\sin \theta)$.  $\theta$ darajeye beyne $(a,b)$ va $x-axis$e.  Mitooni rahat sabet koni ke $(\cos\theta+i\sin\theta)^{m}=\cos m\theta+i \sin m\theta$.  Pas, $z=\sqrt[2n]{a^{2}+b^{2}}\left(\cos \frac{2k\pi+\theta}{n}+i\sin \frac{2k\pi+\theta}{n}\right),\ k=\overline{0,n-1}$.:)

fagaht ie soal ,inke gofti k= 0,n-1 iani inke k, 2 meghdar dare?!!(va in ham iani inke in tabe be ezaye ie meghdar 2 meghdaro barmigardoone!!!! albate age oon 2 meghdar barabar nabashand)


conics
parabola
calculus
analytic geometry
graphing lines
slope
limit

A circular disk of gradually decreasing radius slides down a pit defined by the equation $y=x^2$, maintaining two points of contact with the pit. What is its radius when the two points of contact coalesce into one?

[hide] When at height x, the distance across the parabola is $2\sqrt{x}$, so this is the diameter of the circle.  We want $\sqrt{x}=x\Rightarrow x=0$ or $x=1$.  0 doesn't make sense, so the radius is 1.[/hide]

Why must the two points of tangency form the diameter of the circle?

[hide="UH OH I USED SOME CALCULUS  :o"]

Although there might be an easy, non-calculus way to show that the vertical distance from center of circle to point of tangency is always 1/2?

$y=x^2$ so $y'= 2x$ and since the radius of the circle is perpendicular to the tangent to the parabola, its slope is $-\frac{1}{2x}$.  So at some horizontal distance $x_0$, the slope of the radius is $-\frac{1}{2x_0}$, so the vertical distance is $\frac{1}{2}$.  Then the radius is $\sqrt{(\frac{1}{2})^2 + x_0^2}$.

We want $\lim_{x_0 \rightarrow 0} \sqrt{(\frac{1}{2})^2 + x_0^2} = \frac{1}{2}$

I hope.

[/hide]

Hey, that's what I got too, djshowdown.

[hide]
The equation of the circle (center $(0,k)$, radius $r$) is $x^2+(y-k)^2=r^2$. Let $T(t,t^2)$ be the point of tangency so that $t^2+(t^2-k)^2=r^2$, or \[ t^4-t^2(2k-1)+(k^2-r^2)=0, \] a quadratic in $t^2$. Since we are looking for the point where there is only 1 real solution to this equation, we set the discriminant equal to zero: \[ 4k^2-4k+1-4k^2+4r^2=4r^2-4k+1=0. \] At the point where there is only 1 point of tangency, that point must be the vertex (right??) so put $k=r$ to get $4r^2-4r+1=0$, so $\boxed {\frac{1}{2}}$.
[/hide]

yes, answer was 1/2. not 0 almost every candidate put :)
-kartik


16 and 76. They're just prime numbers reversed.


See [url=http://www.artofproblemsolving.com/Forum/memberlist.php]the memberlist.[/url]

Is this the user's doing or just a glitch in the server?

It's a bug which I haven't yet been able to reproduce. I can easily fix it once it appears though ... thanks for letting me know.


From point $P$ on circumcircle of triangle $ABC$ draw perpendiculars $PD\ ,\ PE$ to $AC\ ,\ BC$.

$M,N$ are midpoints of $BD\ ,\ AE$ prove that:$MN$ is perpendicular to $DE$.

This a very very nice problem  ;)

[i]Solution:[/i]

Consider the midpoints ,$ K,S,F$ in $BC,CA,AB$ respectively and it's obviously that $N\in FS$ and $M\in FK$ .We can prove that $\triangle FMN \sim \triangle PED$ then $PD \bot AC$ and $PS\bot FM$ so that $MN \bot DE$. :D

We had the same problem (under the same title ;) ) at http://www.mathlinks.ro/Forum/viewtopic.php?t=6435 .

  darij


calculus
geometry
analytic geometry
graphing lines
slope
quadratics
MATHCOUNTS

What type do you use? If you want, post what model, too.

I use a TI-30 IISX (or at least I think I do). Yep, the cheap ones that everyone has.  :oops:

Depends.

Normal stuff:TI 30xB

Calculus and beyond:Ti-89

I have a TI-89, but I haven't used it in a few weeks, mainly because

1. Pencil and paper is always a lot faster than hunting for a missing TI-89, and
2. I haven't met any math problems recently.

Oh wait I need to vote! *votes*

TI-89
pwnage right there!

i'm allowed to use it in geometry class, and my friends are all jealous because i have distance, slope, midpoint, and quadratic formula on my calculator....
they are working on one question for five minutes while i finish it in 5 seconds...
but i programmed it into my friend's TI-84... :D

TI-83+. It's over 3 years old...

I might get a better one next year but I already have a bunch of programs on there. I have distance, quadratic, last digit, and area to name the least all in one big prog.

I have a TI, but it's not very good. I don't have a calculator at home though.

TI-89, although it's Mr. Frost's

I have another TI, but it kinda sucks

TI-89 Titanium.

I also have a TI-84+, which I use in times when I can't use the 89 (for certain tests).

I have a TI-83+ and had a TI-84+ silver edition but lost it  :(

Um... you spelled calculator wrong jack...

If I have to use a calc, then TI. But I prefer to use Pencil and Paper... :P

[quote="jackdillon"]i have a ti-89 titanium, and it is 1337 compared to other calculaters[/quote]

Could you show me how it works?  :oops:

I use my brother's. (TI-83 +) In school, all we get supplied with is TI-16 Calculators.

[quote="D3m0n Shad0w"]I use my brother's. (TI-83 +)[/quote]

I think since you're going into middle school next year, Dad will get you a TI-84 (or maybe the Silver Edition). I hope he does, so I can back up some files onto the computer instead of wasting memory and creating duplicate ones  :dry:.

Personally, the TI-16 suits every need in Fifth Grade perfectly.
And if in Sixth Grade we don't do anything different, I might just buy a TI-16  :|

Don't do it! You'll disgrace the family.

You're going into MC next year and you'll need a good calculator to keep up with your progress. You're going into MC right? *prods*.

MC= MathCounts, right?  :oops:
I might not. From the stuff I see at AoPS FTW, it looks very hard for a fifth grader just starting sixth grade math and already having a few issues with sixth grade standard math..

I have a TI83+ and an old TI 83+ Silver Edition, which is currently out of batteries. I need to buy more batteries some day in the distant future... :blush:

[quote="hunter34"]Those are national mc problems D3m0n. Or a majority of them are.[/quote]

Actually, I've seen a lot of the FTW problems on old state sprints and targets.  Well, I've never done a nats test(and I never will), so I guess I wouldn't know about that. :)

I use TI-84+ silver for most things. I program on it for funs as well and the 89 programs are harder to make since I'm not familiar with the code

I use a Ti-84+ because it is what is required. And it looks better than the silver edition. Hey demon brothers why are yall talking to each other online when you probably live in the same house?

[quote="lifesapain"]Hey demon brothers why are yall talking to each other online when you probably live in the same house?[/quote]

I don't know; that's a good question.

Normally we might rarely have some input that can be suited for public purposes. It'd be different if it was completely off-topic (which it kinda was, sorry  :oops: ).

No its perfectly fine I was just wondering because it was kind of weird

well.. i use my sister's TI-84+ for calculus tests because i can't use my TI-89 because it does all the limits, derivatives, integrals, and all the other fun stuff for you.

I have a 6-year-old TI scientific calculator, and a laptop with Mathematica and Matlab  :)

[quote="The Zuton Force"]Mathematica[/quote]

I just got mathematica!!
it's awesome!!! :D

I only use my calculator for the final calculations when it asks for decimal stuff, or obscure angles that I don't want to go through and calculate by hand (sin 42 anyone?), since I'm absolutely horrible with calculators, and besides I can do all the computations without need of digits much faster than I can punch them into the calculator, which makes my teacher mad because I never have my calculator out during class.

TI-83 plus silver edition, TI 89 titanium.
Fun fun.

Mathematica.

NO COMPETITION.

In terms of a TI calculator I use the TI-84 Silver Plus. It's satisfactory, allowed in most math competitions and I don't need a higher one.

TI-89 Titanium FTW!
And a TI-30XIIS for those pesky tests like the ACT which don't allow the 89...

[quote="skyhog"]pesky tests like the ACT which don't allow the 89...[/quote]

exactly why I take SATs.. :D

Talking to demon bros.
Well I myself think that TI-16s are fine for middle schools, TI-30s/scientific are fine for middle school, and graphing (TI-89/84....) are fine for high school and college. It might help a little for MC or other competitions, but I prefer and think that scientific calculators work fine.

Anyways,  I use a TI-30Xa (although I'm getting a TI-84+ (probably) in May for MC.


function
calculus
derivative
real analysis
real analysis unsolved

Is there a real continuous function $f$ with the property that for any strictly increasing $g$ the function $ g(f(x))$ is not everywhere derivable?

I think Cantor's function should work just fine for an example.

Since $f$ is constant almost everywhere, no matter what $g$ you choose, if $g\circ f$ is to be everywhere differentiable, then its derivative is zero almost everywhere. It's well-known that such a differentiable function has to be constant, so $g$ is constant on $f([0,1])$.


Hmm...any possible way we could up the pixels and size of these? It's realli not a pressing matter or anything of huge importance...just an "if" thingy in case resources allow...

Is not the server side as much as we are worring about, but the client side. There are many dial-uppers, and increasing the sizes of the avatars would only hurt them.

Point taken.

Out of curiosity though, couldn't that be fixed by adding an option that would disable avatars?...meaning the user could choose not to load any avatars when viewing the forums if they so choose.

But I understand that can be a coding hassle...

If it would only be that simple ... :)
In plus, even dial-uppers have rights, and they should see your avatar :) ... I am sure that you can get very creative even with these restraints ... other people have, so go to work! :D


It is probably easier to think of it like this:
How many (or much) pool can Jim fill in an hour? 2.
Sue? 4/3.
Tony? 2/3.
Therefore, suppose it takes them 'x' hours.
Therefore, write an algebraic expression for how much pool Jim fills in 'x' hours (it is 2x), then do the same for Sue and Tony. Then, add them together, and put it equal to 1, and you have an equation which you can solve in a painless way!

[hide]It's incorrect because x cannot be negative, which it would be in your equation.

Jim can fill $2$ "pools" in 60 minutes.  Sue can fill $4/3$ of the pool in 60 minutes.  Tony can fill $2/3$ of the pool in 60 minutes.  Working together, they can fill

$2+4/3+2/3=6/3+4/3+2/3=12/3=4$ pools in one hour, so it takes $1/4$ of an hour to fill the pool.[/hide]

[quote="sharkman"]Jim can fill a pool carrying bucks of water in 30 minutes. Sue can do the same job in 45 minutes. Tony can do the same job in 1 ½ hours. How quickly can all three fill the pool together?

1/2 + 3/4 + 3/2 + 1/x = 1

Is this the correct set up?

Do I now solve for x?

Help!

Thank you.[/quote]
[hide]So Jim can fill the pool in 1/2 of an hour. Sue does it in 3/4 of an hour, and Tony in 3/2. So now we have the following equation $2x+\frac{4}{3}x+\frac{2}{3}x=1$. So finding x we get that $x=\frac{1}{4}$ so they can fill the pool in 15 minutes[/hide].

[hide]First, figure out how many pools each person can fill in an hour, then set up an equation: $(2+\frac{4}{3}+\frac{2}{3})$$x=1$
$4x=1$
$x=\boxed{\frac{1}{4}}$

I'm in algebra 2 and I seriously just figured out how to do this a few days ago  :blush: [/hide]


inequalities
floor function
ceiling function
inequalities unsolved

Prove that there is a constant $K$ such that the following inequality holds for any sequence of positive numbers $a_1,a_2,a_3,\dots$ :

$\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots +a_n}\le K\sum_{n=1}^{\infty}\frac1{a_n}$

your inequalities are strict, maybe they are not strong enough

well, if it finite it will be strickly less than but we sum to infinite.

now I concern about my proof, what is it mean by "number" if it's real my proof mean nothing but if it's integer than my proof is working

Another argument. It's related in some ways to mtphaovibul's argument, and comes to the same conclusion (a constant of 4), but it seems a little more straightforward. Some steps are motivated by thought that the sequence $(a_n)$ ought to be increasing, but I never actually use that as a hypothesis.

I do use the arithmetic mean - harmonic mean inequality:

$\frac{x_1+x_2+\cdots+x_m}m\ge \left[\frac1m\left(\frac1{x_1}+\frac1{x_2}+\cdots +\frac1{x_m}\right)\right]^{-1},$ or

$\frac{m} {x_1+x_2+\cdots+x_m}\le \frac1m\left(\frac1{x_1}+\frac1{x_2}+\cdots +\frac1{x_m}\right)$

We make each term of the left hand sum larger by leaving out half of the terms of the denominator:

$\frac{n}{a_1+a_1+\cdots+a_n}<\frac{n}{a_{\lfloor n/2\rfloor+1}+\cdots+a_n} \le 2\cdot\frac{n-\lfloor n/2\rfloor}{a_{\lfloor n/2\rfloor+1}+\cdots+a_n}$

$\le\frac2{n-\lfloor n/2\rfloor}\sum_{k=\lfloor n/2\rfloor+1}^n\frac1{a_k}$ by the AM-HM inequality.

Note that we can rewrite $n-\lfloor n/2\rfloor$ as $\lceil n/2\rceil.$ We have

$\sum_{n=1}^{\infty}\frac{n}{a_1+\cdots+a_n}< 2\sum_{n=1}^{\infty}\frac2{\lceil n/2\rceil}\sum_{k=\lfloor n/2\rfloor+1}^n\frac1{a_k}$

$=2\sum_{k=1}^{\infty}\frac1{a_k}\sum_{n=k}^{2k-1}\frac1{\lceil n/2\rceil} \le2\sum_{k=1}^{\infty}\frac1{a_k}\sum_{n=k}^{2k-1}\frac2k= 4\sum_{k=1}^{\infty}\frac1{a_k}$

We just estimated $\sum_{n=k}^{2k-1}\frac1{\lceil n/2\rceil}$ as less than or equal to $2.$ For large $k$, it would be closer to $2\ln2.$ None of the constants in this proof are tight.


calculus
inequalities proposed
inequalities

Find the maximum value of $k$ such that the below ineq holds for all $a,b,c >0$
 $\frac{kabc}{a+b+c} \leq (a+b)^2+(a+b+4c)^2$

$k=100$ when , for example $a=b=2,c=1$  . Mixing variables and calculus kill it !

Yes, I remember it. Can you give solution, TU?

Nice solution , Nttu ! Pvthuan , can you post the solution from Crux ?

Tu, that is a nice soluition. Send me your problems and solutions as you promised.

Anyone who has a different solution?


Post some Mao rules (that need not be valid in a real game)

It can (technically be) illegal, as with my first example:

[b]Telephone[/b]
Once at least seven cards are played, you must dial the telephone number that uses, in order, the previous seven numbers.

Be creative!

[b]Scream[/b]
Once a joker is played, everybody screams until the next three cards have been played.

Wow, you play with jokers?

[b]Checker Mao[/b]
Depending of suit color, move black or red. Alternative win condition: be the person who delivers the winning move in the checker game.

Works best with 3-4 people:
If the next card is higher than the current card, skip the next person. If the next card is lower than the current card, reverse turn order.
(optional: if the next card is the same, do both)

[quote="james4l"][b]Telephone[/b]
Once at least seven cards are played, you must dial the telephone number that uses, in order, the previous seven numbers.[/quote]

KQJ-1047

[b]Bloodshed[/b]
If an Ace of Spades is played right before a Queen of Hearts, everybody brawls until three people are left standing. They continue the game while everybody goes home. Does not apply when less than four people are in the game.

[quote="james4l"]I hadn't thought of the J/Q/K part - how about J,Q,K map to their letter code on the telephone?

What would be funny:

911-...

[b]Hari-Kari[/b]
You may commit suicide and place your body as a wild card if you have no playable cards and the draw pile is empty.[/quote]

Isn't it "Hara-Kiri"? And here are a few racy ones:

[b]Age Doesn't Matter, but Size Does[/b]
The fattest player, instead of the youngest, goes first.

[b]The Laughingstock Card[/b]
After a person draws a penalty card, they must show it around to the other players. If it turns out to be the Queen of Spades, everybody must start laughing at that player until he/she plays the Queen of Spades down legally.

[b]Cheat[/b]
Instead of cards being played face up, they are instead played face down, with the player dictating which card he or she is playing. If somebody suspects another player of cheating, he/she may flip that card up - but if it turns out to not be a cheat, that player is laughed at until he/she plays a card. If it is a cheat, the cheater draws a penalty card and is laughed at until he/she plays another card.

[b]The Game[/b]
When someone loses The Game he throws his cards at the other people playing Mao.
Note: [b]No smiling![/b]

Actual Mao rule:

If you change the suit kind you must draw a card.

[b]House of Cards[/b]
When enough cards have been played to make a good-looking card house, the last player must build one. If it falls down in the process of construction, the builder must take penalty cards. If it doesn't, then everyone else must take penalty cards.

[b]Chess[/b]
When you play a King, you must move your king. (If not possible, playing a King is not allowed.) Same for Queen. If you play a Jack, you switch colors in the chess game. If you play A-2-3-4-5-6-7-8-9-10-J-Q-K in one turn, you get to make 13 moves on the chess board.

[b]Quick Multiplication[/b]
You must multiply all previously played cards up. (J,Q,K=11,12,13, A=1). Failure to do so in 5 seconds = penalty!

[b]Speed[/b]
You may put down pairs, triples, and four-of-a-kinds as one card.

[b]Communism[/b]
Every turn, everybody's cards are shuffled together and equal numbers of cards are dealt back to each person.

[b]Slavery[/b]
One person is randomly chosen to be the slave owner and another, the slave each round. Slave owner can always see slave's cards. Also, the slave owner can trade cards or give cards to the slave in addition to his turn.

You are allowed to talk. Then people will give you penalty cards and you can give them back that penalty card.

[quote="game.guy"][b]Thank you... your welcome[/b]

If you are penalized then you have to say thank you, and the person that gave out the penalty has to say your welcome, or they can be penalized[size=9][hide="."] 5-1-20[/hide][/size][/quote]

Pretty classic rule, I have played that rule in "THE AWESOME SFBA ARML MAO GAME" which lasted about 50 rules.

[b]Cheapo[/b]
Winner doesn't have to follow any constructed rules. (Doesn't apply consecutively)

[b]Insult[/b]
You must insult the person that played right before you. If you are playing first then you must insult the person playing next.

Note: works best when the people you are playing with have never heard of it before. :)

[b]Blind[/b]
Penalty for looking at the Mao.  (This took me forever to figure out when it was usedthe one time :( )

[b]Rigged[/b]
The Mao gets to rig the deck before the next round.

[b]Whirlwind[/b]
No turn order. Everyone pounces on the pile and tries to put all of their cards on it first. Penalties for messy piles.

[b]Skip[/b]
Placing down card N means you skip N people.

[b]Memory[/b]
Each player has to put a card that can go on the card played 2 times ago. First two cards can be anything.

[b]Super Memory[/b]
Each player must recite all of the cards, in order, played as of then.

[b]Avoid Repetition[/b]
If you play the card with value N, you must take all previously played cards of value N.

[b]Blind Mao (not related to Blind)[/b]
Play cards face-down, and penalty if the card doesn't work.

[b]A Little Present[/b]

If a player forgets to say mao when they have 1 or no cards left, then they get the whole pile of cards that they had been playing on.

[b]Card Rotation[/b]

When the value of the card played is the same as the previous, rotate it 180 degrees so it is facing diametrically opposite. Similarly, rotate the card by 90 degrees counterclockwise if the value increases and 90 degrees clockwise if the value decreases. The values go from 2 to A, and the penalty would be given as "Penalty for incorrect placement of card"

Hidden for it's infamous cheesiness:
[hide]
[b]Removal of Rules[/b]

The winner of the previous game creates a rule that removes all rules every created including the original rules of Mao except for this rule and saying "Mao" or "King Mao" upon playing the last card. This rule does not apply to subsequent rules and can be removed or partially removed, restoring some or all previous rules. This rule can also be edited upon, to keep certain rules but discard all others, or discard all old rules and create a new separate one.[/hide]

[b]Egyptian War[/b]

Following the rules of Egyptian War, someone can hit the stack of cards in play if there is a pair spaced by either one or no cards at all, and the first person to hit the stack can distribute, in any proportion among the remainder of the players. Other rules used in the game apply as well, when someone plays a face card such as J, Q, K, A, the player can play 1, 2, 3, or 4 [b]LINKING[/b] cards respectively, in order to play a face card. For example, if the King of Spades is played, the subsequent player can play 3 of Spades, 3 of Diamonds, and Jack of Diamonds. If the subsequent player cannot accomplish this, then the player who played the face card can distribute the stack among remaining players. No additional cards may be drawn when playing against a placed face card. Other rules that might be used as "Ladder" meaning 3 consecutive numbers, or top and bottom can also allow the deck to be hit.

This rule best be explained in front of the players, unlike all other rules due to initial confusion and reliance on skill and understanding.

[b]Logician's Reveal[/b]

Upon having a Jack, a player at any point in the game can reveal the Jack card, but take an extra card from the pile, in return for knowing a random card of each player. For example, the player reveals the Jack card to everybody, and takes an extra card to his hand, and randomly selects a card from each person and reveals it to theirself and theirself only. Each Jack can only be used once, and multiple Jacks can be used separately. If a Jack is returned to hand, for example played and is received back when the stack is redistributed, the Jack can be replayed in the Logician's Reveal.


articles
number theory
divisibility tests
puzzles

Using the digits 1-9 exactly once to form a 9-digit number such that its first two digits must form a number divisible by 2, first three digits form a number divisible by 3, first four digits form a number divisible by 4, and so on, how many such 9-digit numbers can be formed?

[hide="comment"]

Let us denote the number by N.  Since the fifth digit from the left is divisible by 5, it follows that the fifth digit is 5. Therefore N has the form: pqrs5tuvw.

Obviously, the digits in the even positions are even, and since we have only four nonzero even digit, that is 2,4,6 and 8, it follows that the digits in odd positions are odd.

Now, rs is divisible by 4, where r is odd. This is possible, only when s = 2, or 6. Since pqrs5t is divisible by 6 and pqr is divisible by 3, it follows that s5t is divisible by 3. This is possible only when s+t = 1 mod 3. Since all the digits are distinct and t is even, we observe that : (s,t) = (2,8), (6,4)

Since pqrs5tuv is divisible by 8, and pqrs is divisible by 4, it follows that: 5tuv is divisible by 4, so that uv must be divisible by 4. Having regard to the digits being all distinct, the possibilities are: N = p4r258u6w or, p8r654u2w.

From this point onwards it really gets interesting, since divisibility by 7 is also involved.

Considering the said cases, I have found an unique solution, but would like the other members to post the same.[/hide]

[hide="answer"]381654729[/hide]

Article on the problem: [url]http://en.wikipedia.org/wiki/Polydivisible_number[/url]

The answer is not 381654729, it's 1, because the question asked how many such numbers there were!

Fine, one. But the number is 381654729.

Interestingly, there are a total of 20,456 polydivisible numbers.


Two men starting at a point on a circular $ 1$-mile race track walk in opposite directions with uniform speeds and meet in $ 6$ minutes, but if they walk in the same direction, it requires one hour for the faster walker to gain a lap.

Thanks!

Also, I sometimes have trouble with rate questions involving circular/oval tracks.  Is there a general strategy for these types of questions?  What is the rate of the slower walker?

[EDIT] I added the last part of the question, as I had originally left that out.

Say faster guy's rate is F, slower guy's rate is S

[hide="hint"]
Think about relativity

[hide="solution"]
When they're walking in opposite directions, we can write

$ (F\plus{}S \text{mph}) \times (1/10 \text{hrs. REMEMBER TO CHANGE UNITS}) \equal{} 1 \text{mile}$

and when they're walking in the same direction, we can write

$ (F\minus{}S \text{mph}) \times (1 \text{hrs}) \equal{} 1 \text{mile}$

Solving, F = 5.5 mph and S = 4.5 mph.
I'm guessing you're looking for their speeds, since you didn't specify  :wink:

[/hide]

[/hide]

When looking at circular/oval tracks, treat them the exact same way as you'd treat linear tracks. If need be, draw multiple linear tracks to represent repeatedly going around the circular track.

The key to solving racing around circular track word problems is using the following conditions to set up equations.

(1) If traveling in oppositions, they will meet when the sum of the distances covered is equal to the perimeter of the track

(2) If traveling in the same direction, the faster one will catch up with the slower one when the distance covered by the faster one minus the distance covered by the slower one is equal to the perimeter of the track.

In this case, suppose u and v are the speed of the faster walker and the slower walker respectively. We can set up two equations.

(6/60)*u + (6/60)*v = 1
1*u - 1*v = 1

Simplifying we have
u + v = 10
u - v = 1

Solving for u and v, we get
u = 5.5 (miles/hour)
v = 4.5 (miles/hour)

For more practices visit:
[url]http://www.idealmath.com/algebra/Problems/equation1variable/rate12.htm[/url]
[url]http://www.idealmath.com/algebra/Problems/equation1variable/rate12.htm[/url]


calculus
integration
algebra unsolved
algebra

Prove or disprove that if $ c$ is the hypotenuse of an integer sided right triangle iff $ c^2$ is the hypotenuse of an integer sided right triangle.

[hide="Hint"]
One direction is trivial. For the other, use the characterization of Pythagorean triplets as $ (a^2\minus{}b^2, 2ab, a^2\plus{}b^2)$.
[/hide]


[quote="xxxxx"]If $a^{2}bcd+ab^{2}cd+abc^{2}d+abcd^{2}= 27$  and $abcd=9$,
find $a+b+c+d$[/quote]

shouldn't this be in a high school board?

[quote="xxxxx"]I dunno... can someone move it then?[/quote]

i think it would be best if we let a mod move it....chance your description of the subject to say "Mod...please move to a high schoo lboard!" they'll get the picture

Nobody move this! It's not that hard![hide]This is easy! Just divide the whole thing by abcd and you get your answer of 3! [/hide]

well can u tell me how u got the answer? cuz i dunno how to do it.

[hide="Better Explanation"]divide a^2bcd by abcd. all we are left with is a! Then do this to ab^2cd and all you have is b etc. until you do that to all of them and get a+b+c+d. Since we divided everything by this, we must divide the a^2bcd+ab^2cd...which equals 27, by abcd, which is 9, to equal 3.[/hide]

[hide="Solution in LaTeX"]$a^{2}bcd+ab^{2}cd+abc^{2}d+abcd^{2}=abcd(a+b+c+d)=27$
$9(a+b+c+d)=27$
$a+b+c+d=\boxed{3}$[/hide]

lol...

[hide]

$a^{2}bcd+ab^{2}cd+abc^{2}d+abcd^{2}=abcd(a+b+c+d)$
Since $abcd=9$ we have $a+b+c+d=3$.

[/hide]

[hide="solution"]divide each side by abcd:

a+b+c+d=3[/hide]

is this really a high school level question?

technically the technique of solving this is easy but this "level" of solving doesn't appear in middle school (or at least, my middle school)

Maybe it could have been moved to MOEMS...

[quote="aahmadi"][quote="7h3.D3m0n.117"]MOEMs is a competition thou. but none the less it can be moved. remember, high school basics include algebra II (or at least...the hard parts of algebra 2)[/quote]
Eh...all you need to be familiar with in this question is the distributive property and multiplication (both of which is seen in Algebra I). Looks like a middle school question to me.[/quote]
yeah, and many of your posts that I've seen were pretty bad spam(talking to 7h3.D3m0n.117 of course).

stop talking about where to move this...i think its fine here

btw...nice signature aahmadi  :D


[color=darkred]Pentru triunghiul ABC notam : cercul inscris $C(I,r)$ ; cercul [b]A[/b]-exinscris $C(I_{a},r_{a})$, cercul [b]B[/b]-exinscris $C(I_{b},r_{b})$, cercul [b]C[/b]-exinscris $C(I_{c},r_{c})$; mijlocul $M$ al segmentului $I_{b}I_{c}$; picioarele $D$, $D'$ ale bisectoarelor (interioara si exterioara respectiv) din varful $A$. Sa se arate ca $D'I\perp I_{a}M$.[/color]

observam ca perpendicularitatea de demonstrat e echivalenta cu a demonstra ca : $\triangle{AMI_{a}}$ e asemenea cu $\triangle{ADD'}$, caci mai au 2 laturi omoloage perp.

iar acum asemanarea e echiv cu: $\angle{AI_{a}M=\angle{AD'D}\Longleftrightarrow \tan{AI_{a}M}=\tan{AD'D}}$
$\Longleftrightarrow \frac{AM}{AI_{a}}=\frac{AI}{AD'}\Longleftrightarrow AM \cdot AD' = AI_{a}\cdot AI$

dar $AI_{a}\cdot AI=AI_{c}\cdot AI_{b}$ , deorece $I$ este ortocentru in $\triangle{I_{a}I_{b}I_{c}}$, relatia rezultand din puterea punctului

deci ne-a ramas : $AM \cdot AD' =AI_{b}\cdot AI_{c}\Longleftrightarrow MA \cdot MD'=MI_{c}^{2}$, ceea ce este adevarat caci fascicolul $C(I_{b},A,I_{c},D')$ este armonic.

[color=darkblue]O.K. Faptul ca[/color][quote="Pohoatza Cosmin"]...  $I$ este ortocentru in $\triangle{I_{a}I_{b}I_{c}}$ ...[/quote] [color=darkblue]a fost esential. Astfel, problema propusa se reduce la o problema clasica (cunoscuta).
Daca o cunosti, te rog sa o prezinti in reply - ul urmator sau sa dai o adresa unde se poate accesa.[/color]

un site cu niste detalii ar fi : [url]http://mathworld.wolfram.com/ExcentralTriangle.html[/url], dar dupa cum ati spus si dumneavoastra problema este destul de cunoscuta, e destul sa stim ca $I_{a}$ este pe $l_{a}$ si $BI_{a}\perp BI_{b}$ si analoagele.

Problema cunoscuta eate urmatoarea :

[color=darkred]" Fie proiectiile $D$, $E$, $F$ ale  ortocentrului $H$ pe dreptele suport $BC$, $CA$, $AB$ ale laturilor unui triunghi $ABC$, mijlocul $M$ al laturii $[BC]$ si intersectia $L\in BC\cap EF$. Atunci $LH\perp AM$. "[/color]

In cazul nostru, $I$ este ortocentrul triunghiului $I_{a}I_{b}I_{c}$ etc.

aa, am crezut ca va referiti la altceva  :D :

problema asta se face in mod asemanator:
$LH\perp AM \Longleftrightarrow \tan{DAM}=\tan{DLH}\Longleftrightarrow \frac{DM}{AD}=\frac{DH}{DL}\Longleftrightarrow DM \cdot DL = DB \cdot DC \Longleftrightarrow MB^{2}=MD\cdot ML$


geometry
combinatorics
counting
circle
IMO Shortlist
IMO Longlist

On the circle with center 0 and radius 1 the point $A_0$ is fixed and points $A_1, A_2, \ldots, A_{999}, A_{1000}$ are distributed in such a way that the angle $\angle A_00A_k = k$ (in radians). Cut the circle at points $A_0, A_1, \ldots, A_{1000}.$ How many arcs with different lengths are obtained. ?

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)


Let $ a$ be a real number. How many real roorts can the equation : $ (x^2\plus{}ax\plus{}1)(3x^2\plus{}ax\minus{}3)\equal{}0$ have?

[hide]
If you expand it, you get
$ 3x^4 \plus{} 4ax^3 \plus{} ax^2 \minus{}2ax \minus{} 3 \equal{} 0$
Using Descarte's rule of signs, you know that there is 1 positive real root and 3 or 1 negative real roots. So there can be either 2 real roots or 4 real roots. [/hide]

Meh. probably wrong.

[hide]Obviously $ 4$... the discriminant of each of the factors should be greater than 0.  It's easy to find $ a$ to satisfy that.[/hide]

To mathcountsfreak: That counts with multiplicity. What if two or more of those roots are equal?

Also, you implicitly assumed $ a$ was positive. It might not be.

Hmm.. Maybe it sounds like cheating but if we set $ a$ to be just easy number, can't we see how many roots are there?

No.  It's not always the same number of roots.

[hide]Let $ f(x)\equal{}x^2\plus{}ax\plus{}1$ and $ g(x)\equal{}3x^2\plus{}ax\minus{}3$.

The discriminant of $ f(x)$ is $ a^2\minus{}4$, and has 0 real roots when $ |a|>2$, 1 real root when $ a\equal{}\pm2$, and 2 real roots when $ |a|<2$.

The discriminant of $ g(x)$ is $ a^2\plus{}36$, so $ g(x)$ always has two real roots.

So the desired equation can have 2, 3, or 4 real roots.

I think you'd have to actually test some values of $ a$ to confirm that each of these is possible, but I'm fairly certain that there's a case that exists for all the situations.[/hide]


Hi, I'm having a tough time solving this problem:

[b]Show that the columns of the n*p matrix A, are linearly independent exactly when there is a p*n matrix B, with BA = Ip (where p is the subscript)[/b]

I'm new to Linear Algebra and it's taking me a while to get used to it... Hopefully one you can provide your help!

If $BA=I_p$, $BA$ is injective as a linear map, and therefore $A$ is injective. For linear transformations given by matrices, this is equivalent to having linearly independent columns.

Now, the converse.
Treating $A$ as a linear transformation from $\mathbb{R}^p$ to $\mathbb{R}^n$, $A$ maps the $k$-th standard basis vector to the $k$-th column of $A$.
You want to construct a linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^p$ which maps the $k$-th column of $A$ to the $k$-th standard basis vector. To do this, extend the columns of $A$ to a basis for $\mathbb{R}^n$, and map the other basis vectors to whatever you want.

Let $P=\begin{bmatrix}A&X\end{bmatrix}$ have columns which are a basis for $\mathbb{R}^n$. The desired $B$ has the property that $BP=\begin{bmatrix}I_p&Y\end{bmatrix}$, where $Y$ may be any $p\times (n-p)$ matrix. We can get this by setting $B=\begin{bmatrix}I_p&Y\end{bmatrix}\cdot P^{-1}$


linear algebra
matrix
IMO Shortlist
system of equations
algebra
IMO Longlist

Solve the system of equations:

$
\begin{matrix}
|x+y| + |1-x| = 6 \\ |x+y+1| + |1-y| = 4.
\end{matrix}
$

Here is my solution for this problem, it's idea is to consider all cases without much effort and find the solutions
Firstly we should say that
 /x+y/<4 (1) beacuse /x+y+1/+/1-y/=4

Now consideration,
1st case, let's say x+y>=0 then
 if y<=1 then 4-y=/1-x/
  if x<=1 then x=y-3, x+y=2*y-3<0 that contradicts our assumption.
  if x>1 then x=5-y, x+y=5 contradicts (1);
 if y>1 then y+2=/1-x/
  if x<=1 then y+2=1-x, x+y=-1<0 taht contradicts our assumption
  if x>1 then x=y+3, checking we get y=1/3, x=10/3, but y<1 and again contradiction
2nd case, let's say -1<x+y<0, and opening absolute values we obtain /1-x/+/1-y/=9
 if y<=1 then 8+y=/1-x/
  if x<=1 then x+y=-7 contradicts out assumption
  if x>1 then x=9+y, checking we get y=-7, x=2 but /y+x/=5>4 contradicts (1)
 as we see the expression is symetric that means none of (x,y) is less then or equal to -1 so
  (x>1 y>1) but then x+y=11 and contradicts the assumption
3rd case, x+y<=-1 then 1+/1-x/=/1-y/
 if y<=1 then y+/1-x/=0
  if x<=1 then x=y+1 and checking we get one solution for the given system y=-7/3; x=-4/3;
  if x>1 then x+y=1 and contradiction
 if y>1 then y-2=/1-x/
  if x>=1 x+y>2 and contradiction again as well as if x<1 x+y=3

So we considered all possible cases and found the only solution for the given system (x=-4/3, y=-7/3);
  ie// :D

I graphed both relations and got two parallelograms which intersected at two points (-14/3,13/3) and (-2,-1).


[color=darkred]Let $ABC$ be a triangle. For a point $P$ of the interior of this triangle denote the intersections $D\in BC\cap AP$, $E\in CA\cap BP$, $F\in AB\cap CP$, $R\in EF\cap AD$ and the projection $H$ of the point $P$ on the side $AC$. Prove that if the quadrilateral $AEPF$ is cyclically and $\widehat {CAD}\equiv \widehat {BCF}$ then the quadrilateral $AEPF$ is a trapezoid and the ray $[HP$ is the bisector of the angle $\widehat {DHR}$.[/color]

HINT:          it si very easy  by similarity. We find AB=AC and FE parallel to BC , also  PA bisector of HPF
                        if you  go on by similaryty   you will see that ABC is an equaleteral triangle
                            Also use "SEVA   theorem"

am i missing something ? How $AEPF$ is a trapezoid ?  the pair of line $AE,FP$ cannot be parallel and $AF,EP$ also cannot be parallel . So how we can prove it is trapezoid ??


Find the maximum value of a_1*(a_1+4a_2)*(a_1+a_2+9a_3)*...(a_1+...+a_n-1+n^2*a_n), when a_1,...,a_n are positive numbers which add up to 1.

Quite nice gift. Thank you Harazi. I solved it (it reduced to some system of linear equations), but I will post the sln later. Let our other members try to solve it.

Namdung

Hey, you really made me curious about your solution. It will be a pleasure to see what splendid idea you have found. I have a simple solution to this one, but , as ussually, not very easy to find.

I just used weighted AM-GM. The answer is
        n!(n+1)!/(n+1)^n
The weights i found from a system of linear equations.

Namdung

Very nice. I started from the inequality (x_k-1)^k-1(kx_k-(k-1)x_k-1)<=(x_k)^k (which follows from Bernoully or AM-GM) and I put x_k=(a_1+...+a_k)/(k+1). Then I multiplied these inequalities and after some boring computations I obtained the problem. The equality takes place when a_1=2/(n+1), a_2=...=a_n=1/(n+1).


Let $ X$ be a normed linear space.  Prove that $ X$ is complete if only if $ \sum_{n\equal{}1}^\infty x_n$ converges in $ X$ for all sequences $ (x_n)$ that satisfy $ \sum_{n\equal{}1}^\infty\| x_n\| < \infty$

Convergent sequences are fundamental and R is complete. Given your Cauchy sequence, choose your sequence x(n) specially and use triangle inequality.

i'm sorry, i don't completely understand...can you elaborate please?


What is the value of $ x$ for which $ |3x\plus{}5|$ is not positive? Express your answer as a common fraction.

The only time when the absolute value isn't positive is when it equals 0.
Solve, 3x+5=0, 3x=-5, x=-5/3.


An unstretchable string of uniform density and constant diameter is suspended at one of its ends and is hanging. At a length of $L$ it breaks under its own weight. Is it conceivable that there is such shape of string which does not get broken under its own weight however long it is?

I am not sure of what you mean. Do you ask to find the shape of a string, say for example the diameter $d(x)$ as a function of the position on the string while it is suspended, such that
the mass and thus the wight of the string is less than what the top parts can endure?

In that case it is not possible since whatever its shape is its density per length $\rho(x)>\epsilon$
where $\epsilon$ is some physical limit (chain of quarks  :D ) so the mass which is
given by integrating this over $[0,\infty)$ will still diverge.

Well otherwise there is the [url=http://en.wikipedia.org/wiki/Gabriel_horn]Gabriel horn[/url]
just to mention a example, but its not physical.

no color you are wrong, there exists one in the way you have written it down. :D

How?
No matter the shape if it is infinite long it will get infinite mass
thus an infinite force will apply to the top due to gravity so
it breaks, unless it it can be made infinite thin..

Well to be physical none of the infinite properties above is possible.
I mean if it is infinite long, where can you find a suitable planet for the experiment?
[i]Solution[/i]: consider a homogeneous ideal field.

Also infinite mass will itself cause singularity some how.
[i]Solution:[/i] ignore this  :)

But after all we assume the rope isn't infinitely strong, so it will break.

[color=green]If physical limitations are allowed then any extremely thin shape would due,
but I assume you want some idealized solution, and probably you will
in this idealization also allow the string to have infinitesimal parts, then I have already
proposed a solution above, the [/color] [color=red]Gabriel horn[/color]
see link above (post 2)

If the diameter is constant then $L$ doesnt depend from the $A$ cross section of the string, because on the unit area of  the string with weight $AL\rho g$ the $L\rho g$ tension is the same everywhere.($\rho$ is the mass density of the string). If the diameter of the string changes it wont break, if the weight below any cross section is less than $AL\rho g$. This can be achieved if you choose the change of the diameter appropriately.(and of course the mass of the string is finite).

How about my link? It works doesn't  it?

You said one has to change the diameter appropriately, but
either you accept mathematical solutions or not.
If you do not then your string has a maximum length.
Otherwise just use the link or something similar such
that the diameter decreases so fast that the integral converges.

In my opinion this is the final answer? If you do not agree, please
answer more specific on my statements above.

Well, no your link doent work. Count over it man :D . suspend the string on a cieling(theoreticaly of course) and check if it wont break. It will. its shape is not in the $f(x)=x^{-n}$ form.

But the volume is finite! then so is the mass.
Also the curve IS on the form $x^{-n}$ with $n=1$ One could also choose $n=2$.

Or how do you mean? Please show.

Well ok,
let the area at distance $x$ beneath the suspension be $A(x)$. the tension in it is $\rho g l$ where $l \nleq L$, so the this pulling tension is smaller than the break $\rho g L$ snapping tension, so the string wont break anywhere.
The force at $x$ distance from the suspension poing is $\rho gl A(x)$ this is equal to the wieght $W$ of the string that is beneath this. A small distance $\Delta x$ lower the area : $A(x+\Delta x)$ the force $\rho gl A(x+\Delta x)$ and this is also the weight of the string under that certain area which is $W-\rho g \Delta x A(x)$
so:
$\rho gl A(x+\Delta x)= \rho gl A(x)-\rho g\Delta x A(x)$
if $\Delta x \to 0$ then :
$\frac{dA}{dx}=-\frac{A(x)}{l}$
solving the DE:
$A(x)= A_{0}e^{-\frac{x}{l}}$
Where $A_{0}$ is the area at the suspension point.
SO the area of the string changes like that. :D

Shouldnt it be $A(x)\leq A_{0}e^{-\frac{x}{l}}$ ?

No, why should it be? Look over my solution and you will see that it shouldnt :D


algebra
polynomial
Vieta
Rational Root Theorem
AMC

The first two terms of a polynomial are x^20-20x^19, and the constant term is +1. Given that the roots when the polynomial is set to 0 are postive real numbers, factor the polynomial completely.

[hide]By the Rational Root Theorem, all real roots must be +/- 1, since the coefficients of both the first and last terms are 1. Since the coefficient of x^19 is -20, the sum of all the roots is positive 20 (by viete's). Since the polynomial only has 20 roots, the polynomial is thus (x-1)^20.[/hide]

Looks tricky, but is easy.

[hide]

(x-1)^20

[/hide]

But is it given that all the polynomial's roots are rational ?
Am I missing something here ?

Fyty, are you sure you can apply the RRTh here ?

I think this is rather AM - GM then RRth.

Let the roots be x1, x2, ..., x20 (all positive). We have
x1x2...x20 = 1
x1 + x2 + ... + x20 = 20 (Viete).
So equality must occur in AM - GM which means all x's are equal.

Yeah, that would work too. I can't apply the RR theorem because the problems says "positive real roots", but once I work things out with the RR theorem, I do indeed find that the roots are all x=1, so things work out.  :D

Um, I used Fundemental Theorem.

You know the sum is 20. You know the product is 1. So they're all inverses, and there are 20 numbers. The trivial case is when they are all 1's, which is sufficient for the problem.

So Fyty, your solution was false.

And Southpaw, I'm afraid I can't follow what you are saying ...

Because the second coefficient in a polynomial is either the negative or positive sum of the roots, depending on the degree of the polynomial, you know that in this case, the sum of the roots is 20. You also know the free term in a polynomial is always the negative or positive product of the roots. So

r_1+r_2+r_3... = 20
r_1*r_2*r_3.... = 1

Now, look at the second equation. In order for the product of two ore more numbers to be 1, then all the terms must be inverses of each other. Note that 1 is always an inverse of itself.

We also know there are 20 real roots, as stated in the problem. Note that 1+1+1+1+...= 20.

Therefore, (x-1)^20 is sufficient to be considered an answer to the problem, be it that the solution may not be unique.

Yeah, of course I know Vieta's formulas.

But I think your solutions is false too.

It is obviously the only possible solution !

[quote]So Fyty, your solution was false. [/quote]

Not quite. A "solution" would only be to prove that (x-1)^20 satisfies the given conditions, which is easy. Its just that the manner in which I discovered the answer was not, err, "standard".  :D

Fyty, your solution is false since you didn't prove that it is the only possible factorization.

But the problem only asks for a factorization, doesn't it? Not all factorizations.

Hm ... this is an ambigous situation.

I think your answer is insufficient, but if you think this suffices, well ...

Yeah, I guess we are sorta arguing about something pointless. Your solution seems to be a bit more complete I guess. By the way, darkquantum, what was Linda's solution?

So here's my take on this little argument:

From a problem-solving perspective, this test is short answer.  The wording of the question clearly implies a unique polynomial.  Thus, once we have found (x-1)^20, we are done.

From a rigorous mathematical/proof-oriented perspective, you better prove everything you say.  This is generally true of any [i]solution[/i] to a short-answer question -- it should be mathematically verifiable as correct and unique.

Thus, if you're just doing the problem, it would be sufficient to get (x-1)^20 by whatever means and write it down.  If, however, you intended to explain your answer to others, you would need to verify it mathematically, in this vase via the AM-GM inequality.

My guess would be that AM-GM was her solution.


Several irrational numbers are written on a blackboard. It is known that for every two numbers $ a$ and $ b$ on the blackboard, at least one of the numbers $ a\over b\plus{}1$ and $ b\over a\plus{}1$ is rational. What maximum number of irrational numbers can be on the blackboard?

[i]Author: Alexander Golovanov[/i]

Solution...
[hide]
Let $ a, b, c, d$ are written on board. So, $ a \plus{} 1 \equal{} mb$, m - rational; $ b \plus{} 1 \equal{} nc$ (it's impossible, that $ c \plus{} 1 \equal{} nb$, because $ (a \plus{} 1)/(c \plus{} 1)$ cannot be rational); $ c \plus{} 1 \equal{} kd$ ($ d \plus{} 1 \equal{} kc$ - impossible...), n,k - rational.
It must be $ a \plus{} 1 \equal{} pc$ or $ c \plus{} 1 \equal{} pa$, p is rational. In first case we have b/c rational (impossible); in second - a/d rational. So, $ (a \plus{} 1)/d$ and $ (d \plus{} 1)/a$ are not rational. Contradiction.
So, we can see, no more 3 numbers on a board.
Answer: 3. For example: $ \sqrt2, \sqrt2 \plus{} 1, \minus{} \frac {\sqrt2 \plus{} 2}{2}$[/hide]


Let $ k \in \mathbb{N}$.
Is there infinitely many prime numbers of the form $ n^{2^k}\plus{}1$, with $ n\in \mathbb{N}$?

For $ k>0$ this is an unsolved problem I think.


Let $ a, b, c \in R$ and $ a\neq b$, $ b\neq c$, $ c\neq a$.
Prove: $ (a^2\plus{}b^2\plus{}c^2)$$ \displaystyle \left[\frac{1}{(a\minus{}b)^2} \plus{}\frac{1}{(b\minus{}c)^2}\plus{} \frac{1}{(c\minus{}a)^2} \right ] \ge \frac{9}{2}$.

$ LHS \geq 9(a^2 \plus{} b^2).\frac {1}{(a \minus{} b)^2} \geq 9/2$
 $ a \equal{} \minus{} b;c \equal{} 0$

oh! Not so.
Symbol $ " \equal{} "$ happen $ \Longleftrightarrow \frac {a \plus{} b}{a \minus{} b} \plus{} \frac {b \plus{} c}{b \minus{} c} \plus{} \frac {c \plus{} a}{c \minus{} a} \equal{} 0$.

Oh! I have got a solution for this problem, but i think it is not nice. So, I post it to this forum. You can post full solution?


Triangle ABC is inscribed in a circle of radius 3. If BC = 2, compute sin Angle BAC

Hmm... if I recall correctly (I probably don't), then

\[\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=\frac{1}{2R}\]

So $\frac{\sin A}{2}=\frac{1}{2 \cdot 3} \iff \sin A = \frac{1}{3}$.

$\arcsin {\frac{1}{3}}=19.47...$ so it seems pretty reasonable.

Editted for having said something not relating to reality.

Is $R$ not the radius of the circumscribed circle?


You have a whole pizza in the refrigerator. On your first trip to the refrigerator, you eat half the pizza. On each successive trip, you eat half of the remaining pizza. After five trips to the refrigerator, what fractional part of the pizza have you eaten?

This is $ 1\minus{}(\tfrac12\plus{}\tfrac14\plus{}\tfrac18\plus{}\tfrac1{16}\plus{}\tfrac1{32})\equal{}1\minus{}\tfrac{31}{32}\equal{}\frac{1}{32}$ of the pizza left.

[quote="GameBot"]You have a whole pizza in the refrigerator. On your first trip to the refrigerator, you eat half the pizza. On each successive trip, you eat half of the remaining pizza. After five trips to the refrigerator, what fractional part of the [b]pizza have you eaten?[/b][/quote]

This means our answer is simply $ \tfrac12 \plus{} \tfrac14 \plus{} \tfrac18 \plus{} \tfrac1{16} \plus{} \tfrac1{32} \equal{} \boxed{\tfrac {31}{32}}$
Note that the sum $ \tfrac12 \plus{} \tfrac14 \plus{} \tfrac18 \plus{} \tfrac1{16} \plus{} \tfrac1{32} \plus{} \dots \plus{} \tfrac1{2^n} \equal{} \tfrac{2^n \minus{} 1}{2^n}$


The answer is [hide]sqrt (3)[/hide].  Factor, square, expand, cancel terms, take square root, done.


Define   $m@n=\frac{m+n}{mn+4}$ where m and n are integers.

Find: $((((...........(2005@2004)@2003)@2002)............)@1)@0)$

I posted this earlier in High School Basics but nobody came up with any solution :( .

uuuh I'm confused; is there any correlation between the first and second parts of your post? It doesn't seem to be a cohesive problem...?

EDIT: aha... ok  :)

[quote="max_tm"]uuuh I'm confused; is there any correlation between the first and second parts of your post? It doesn't seem to be a cohesive problem...?[/quote]

I edited it and hopefully you understand it now. Its kind of like an operation. So if I say m$n=m+n where m and n are numbers then 1$2=1+2.

I think you mean m,n real numbers, since 2005@2004 is not an integer. Even then, there doesn't seem to be a nice solution.

[quote="jcmtnez"][hide]Is it $1/8$ ?[/hide][/quote]

I am not quite sure.....I remember the answer is a pretty clean fraction like 1/12 or 1/8 and not like $\frac{12899}{9473847}$. Can you post your solution??

I just try some examples and it let to it. For the moment is just a hypothesis, I am trying to find why  :) .

Sorry, It was $\frac{1}{12}$ I misscalculated in the final step.

[quote="Bictor717"]I spoke too soon. I should have at least tried it before speaking. :blush:
[hide]For $n=2$, $m@n=m@2=\frac{m+2}{2m+4}=\frac12$. It doesn't matter what $m$ is, so the answer is $\left(\frac12 @1\right)@0=\frac{1}{12}$.[/hide][/quote]

Oh yeah!!! Thanks everyone, this problem just looks hard but is actually quite simple.

[quote="Bictor717"]I spoke too soon. I should have at least tried it before speaking. :blush:
[hide]For $n=2$, $m@n=m@2=\frac{m+2}{2m+4}=\frac12$. It doesn't matter what $m$ is, so the answer is $\left(\frac12 @1\right)@0=\frac{1}{12}$.[/hide][/quote]

Nice solution.. you just decimated the problem  :D


trigonometry
vector
geometry
geometric transformation
rotation
homothety
ratio

Two not equal circles  $C_1, C_2$  with centers $K , L$    intersect at points $A$   and $B$  . A line through $B$   meets the first circle at point $C$   and the second at point $D$  . If  $M , N$   are the midpoints of  $KL , CD$   respectively , prove that $MA = MN$ .
[u]
  Babis [/u]

[i](Inspir[color=red]ated[/color] by a Crux Problem )[/i]

I think that it follows the fact that <CAD=<KAL.  :)

I found a metrical proof . $AM$ is median in the triangle  $AKL$  so
$AM^2=\frac{2R^2+2r^2-KL^2}{4}$  where R is the ray of the big circle $C_1$ and r from the small  $C_2$

In the triangle $KNL$  , $MN$ is the median so  $MN^2=\frac{2CM^2+2MD^2-CD^2}{4}=\frac{CK^2+CL^2-KL^2+KD^2+LD^2-CD^2}{4}$  as  CM,DM are medians in the triangles $KCL$ and $KDL$ respectively .
So $MN^2=\frac{R^2+CL^2-KL^2+KD^2+r^2-CD^2}{4}$ . (1)
From the law of cosines at the triangles CLD and  CKD we have
$CL^2=r^2+CD^2-2rCDcosD_1$ and $DK^2=R^2+CD^2-2RCDcosC_1$  where

$D_1,C_1$ are the angles $LDB$ and $KCB$ respectively .
If we add the above realtions and substuct $CD^2$ we will have
$CL^2+DK^2-CD^2=R^2+r^2+CD(CD-2RcosC_1-2rcosD_1)$ (2)
But from the law of cosines at the triangle $CKB$ we have that  $2RcosC_1=CB$
and from the law of cosines at the triangle  at the triangle $BDL$ we have that
$2rcosD_1=BD$
So the relation (2) becomes $CL^2+DK^2-CD^2=R^2+r^2$ and so
(1) becomes $MN^2=\frac{2R^2-KL^2+2r^2}{4}$
SO $MN=MA$

Silouanos

My proof involves vectors.

Since <ACD=<AKL and <ADC=<ALK, so <CAD=<KAL. Thus we have <ACK=<CAK=<CAD - <KAD=<KAL - <KAD=<LAD=<LDA. This tells us that if we let CK and DL meet at E, then A, C, D, E lie on the same circle. Therefore <CED=<CAD=<KAL.

Now we use vectors to show that MA=MN. Let $AK=r_1$, $AL=r_2$, <CED=<CAD=<KAL=$\theta$.
\[ 4 \left| \overrightarrow{AM} \right|^2 = (\overrightarrow{AK} + \overrightarrow{AL})^2=r_1^2+r_2^2+2 \left| \overrightarrow{AK} \right| \left| \overrightarrow{AL} \right|=r_1^2+r_2^2+2r_1r_2 \cos{\theta} \]
\[ 4 \left| \overrightarrow{MN} \right|^2 = (\overrightarrow{KC}+\overrightarrow{LD})^2 = r_1^2+r_2^2+2 \left| \overrightarrow{KC} \right| \left| \overrightarrow{LD} \right|=r_1^2+r_2^2+2r_1r_2 \cos{\theta} \]
Hence AM=MN.

[quote="stergiu"]Two not equal circles  $C_1, C_2$  with centers $K , L$    intersect at points $A$   and $B$  . A line through $B$   meets the first circle at point $C$   and the second at point $D$  . If  $M , N$   are the midpoints of  $KL , CD$   respectively , prove that $MA = MN$ .
[u]
  Babis [/u]

[i](Inspired by a Crux Problem )[/i][/quote]

1) $\angle CKA = 2\angle CBA = 2$arc$AD = \angle ALD$ hence triangles $CKA$ and $DLA$ are similar.
2) Hence the triangle $CAD$ is obtained from triangle $KAL$ by performing a rotation of center $A$  and angle $\measuredangle KAC$ and then an homothethy with ratio $\frac{KA}{CA}$. With this transformation, the point $M$ goes to point $N$. So the triangles $AMN$ and $CKA$ must be similar hence result.

Nice one stergiu,  ;)

Daniel

Gentlemen , nice work !!!

      [b]My aproach:[/b]

 As you allready have found , triangles  $AKL , ACD$ are similar , so the same holds for triangles $AKM , ACN$.But then
$\angle ANC = \angle AMK = \frac{1}{2} \angle AMB$
Since $MA = MB$ the above equality gives that $M$ is the circumcenter of triangle $ABN$ , so $MA = MN$ !

 [u]Babis[/u]

 [i](Silouan  must read carefully the solution by [b]dhernandez[/b] , because now  in Greece we are not taught geometrical transformation in our schools). [/i]

[quote="stergiu"] [u]Babis[/u]

 [i](Silouan  must read carefully the solution by [b]dhernandez[/b] , because now  in Greece we are not taught geometrical transformation in our schools). [/i][/quote]

I will do it .What are the comments about my solution mr Babis?

Silouan ,
your solution is [color=red]very nice[/color] and short.I have seen that in many Bulgarian Olympiads the geometry problems are officially solved with the trigonometric method. Now I recognize that it is a strong method to solve geometry and some times with an easier way than the synthetic one.I suggest you to use this method in your math competitions  , if a synthetic approach is difficult to find.
  [u]Babis[/u]

 [i](Ah ! Could you write in the high school section the problems from your school exams?-Thanks)[/i]


linear algebra
matrix
linear algebra unsolved

1. Given an n x n upper triangular matrix, K, prove that if K_{i,i} = 0 for some  1 <= i <= n , then K is singular.

2. Same matrix as above, but show that if K_{i,i} =/= 0 for some positive integer i, then K is nonsingular.

Help?

What is the determinant of an upper triangular matrix?

Alternatively: if you start with an upper triangular matrix with zero in the $ (i,i)$ position and row reduce it to row echelon form, that row echelon form won't have a pivot in the $ i$th column, right?

Also the second part is not true. It should be $ K_{i,i} \neq 0$ for [b]ALL[/b] positive integers $ i \in \{1, \dots ,n\}$.

the determinant of any triangular matrix is the product of the diagnol via Row Reduction, so if there is a zero on the diagnol of $ A$ then $ det(A) \equal{} 0$.


algebra
polynomial
function
algebra unsolved

let P be a polynomial given by $P(x)=ax^3+bx^2+cx+d$,where a b c d are integers.Suppose that xP(x)=yP(y) for infinitely many pairs of x and y with x not equal to y.Prove that the equation P(x)=0 has an integer root.

It comes from ISL 2002.
A solution is available here :
http://www.kalva.demon.co.uk/short/soln/sh02a3.html

Pierre.

I don't understand something. chaogold forgot to say that $x$ and $y$ are integer? Otherwise the statement is untrue.

So suppose $x$ and $y$ are integer in the condition. Then there are two infinite sequences $\{x_{n}\}$ and $\{y_{n}\}$ s.t. $x_{n}P(x_{n})=y_{n}P(y_{n})$, where we may assume $x_{n}$ increase and $y_{n}$ decrease.
We have \[0=x_{n}P(x_{n})-y_{n}P(y_{n})=\]  \[=(x_{n}-y_{n})(a(x_{n}^{2}+y_{n}^{2})(x_{n}+y_{n})+b(x_{n}^{2}+x_{n}y_{n}+y_{n}^{2})+c(x_{n}+y_{n})+d).\] It follows that $x_{n}+y_{n}$ is bounded. So we may consider $x_{n}+y_{n}=m$. It means the equation $xP(x)=(m-x)P(m-x)$ have infinitely many solutions. Thus $xP(x)=(m-x)P(m-x)$ for all $x$. If $m\neq 0$ then $x=m$ is an integer root of $P$. If $m=0$, then $P(x)$ is odd function, thus $P(0)=0$.


Hi, i've got an exam wednesday and i was just going through some stuff when i came across this question. Could anybody tell me how to do it?

Suppose that X is a random variable with mean 7 and variance 4, and let Y be a random variable independent of X, with mean 0 and variance 1. Let Z=X-Y. Evaluate Cov(X,Z) and hence p(X,Z).

It looks staightforward but i dont know really where to start..
Any help is appreciated!

A little bit faster: $ Cov(X,Z)\equal{}Cov(X,X\minus{}Y)\equal{}Cov(X,X)\minus{}Cov(X,Y)\equal{}Var(X)\equal{}4$

Nice going, [color=blue][b]roah[/b][/color]!  :wink:

What can you say about p(X,Z)?

Yes, it is. And it must always lie in the interval $ [\minus{}1,1].$

You were referring to $ \rho_{X,Y}$ then? O, man! :rotfl:


Hello, I am trying to make a nice table, but I have this problem:

[code]\begin{tabular} {|c|c|c|c|}
\hline
a&a&a&f^{a^2}
\end{tabular}[/code]

gives me:
\begin{tabular}{|c|c|c|c|}\hline a&a&a&f^{a^{2}}\end{tabular}
As you can see, the two and the line are to close to each other

I want that horizontal line to be a bit higher.  I tried many things including \vspace , however the latter seems to be a very "shady", "untrustworthy" command, which seems to make extra spaces in all the wrong places. :mad:  :o

I am finding this problem highly frustrating, and I come to you to ask for help, with many thanks in advance. :ninja:  :oops:

Try \renewcommand{\arraystretch}{1.25}
Also you are mixing maths and text mode here. Tabular is for text mode and array is for maths mode.
So replace tabular by array to get
[code]\renewcommand{\arraystretch}{1.25}
\[
\begin{array} {|c|c|c|c|}
\hline
a&a&a&f^{a^2}
\end{array}
\][/code]

Thank you  for that quick reply.
I am doing something which is mostly math, but there are some words in it as well, so I should switch to array now?

Another thing, isn't this quite dramatic? I mean, I noticed that this means that he will be treating arrays like this all the time after that as well? :o  :(

You can use tabular but then you need to get rid of \ [ and \ ] around it. And you would need to use dollar signs around every mathematical term which may be a chore. For text in array you use \text{}, so it depends which is easier to do.

The number in arraystretch is a multiplying factor so 1.25 means multiply the standard distance between the rows by 1.25. After this array you can return the other arrays to normal by using \renewcommand{\arraystretch}{1}.

Although \arraystretch can be confined to a single array, it will space all the rows of that array. Suppose you only want one row to be spaced? Then you take advantage of the fact that the top line must be above any text there, so tall text will give more room. But you only want f^{a^2} to be printed so you add text with \vphantom. This ensures that the vertical spacing is there but the text is not printed.
Thus you can use
a&a&a&f^{a^2}\vphantom{\int^{A}}
or
a&a&a&f^{a^2}\vphantom{\int^{x^2}}
or
a&a&a&f^{a^2}\vphantom{\int\limits^{e^{x^2}}}
or
a&a&a&f^{a^2}\vphantom{\boxed{A^1}}
or whatever you fancy. This gives
\[\begin{array}{|c|c|}\hline \text{Extra text}& \text{Effect}\\ \hline \text{none}&f^{a^{2}}\\ \hline \verb=\vphantom{\int^{A}}=&f^{a^{2}}\vphantom{\int^{A}}\\ \hline \verb=\vphantom{\int^{x^{2}}}=&f^{a^{2}}\vphantom{\int^{x^{2}}}\\ \hline \verb=\vphantom{\int^{e^{x^{2}}}}=&f^{a^{2}}\vphantom{\int^{e^{x^{2}}}}\\ \hline \verb=\vphantom{\boxed{A^{1}}}=&f^{a^{2}}\vphantom{\boxed{A^{1}}}\\ \hline \end{array}\]

Hmm, isn't that a bit "dirty"? :D
But - just a silly question - can this renewcommand trick also affect my spacing above and below my table? I have this weird unusually big (but stil reasonable) white space between my text, and the table.

arraystretch only affects the spacing of the table rows. There must be something else that's giving you the extra vertical spacing. Can you give an example?

I had the same first trouble ... thanks for the trick


Probe that for any natural $ n$, the expression $ 1^{n}+2^{n}+3^{n}+4^{n}$ is divisible by $ 5$ if and only if $ n$ is not divisible by $ 4$.

[quote="i_like_pie"]If $ n\equiv1\bmod4$, then $ 1^{n}+2^{n}+3^{n}+4^{n}\equiv0\bmod{5}$.
[/quote]
Can you explain More this, please.

[b]Fermat's Little Theorem:[/b]  For a prime $ p$, $ x^{p}\equiv x \bmod p$

[b]Corollary:[/b]  $ x^{5}\equiv x \bmod 5$

If $ x$ is not divisible by $ 5$, we have further that $ x^{4}\equiv 1 \bmod 5$.  Hence if $ n \equiv 1 \bmod 4$, let $ n = 1+4k$.  Then

$ 1^{4k+1}+2^{4k+1}+3^{4k+1}+4^{4k+1}\equiv 1^{1}+2^{1}+3^{1}+4^{1}\equiv 0 \bmod 5$

Is it true for any prime $ p$? If so, prove it.
That is, if  $ (p-1)$ does not divide $ n$ then
$ 1^{n}+2^{b}+3^{n}+\dots+(p-1)^{n}$ is a multiple of $ p$

[hide="On a related note..."] It's known that $ 1^{k}+2^{k}+...+n^{k}$ is a polynomial in $ \frac{n(n+1)}{2}$ (this is clearest for $ k = 1, 3$).  The only question is what the constant term is.  :) [/hide]


calculus
integration
trigonometry
algebra
function
domain
calculus computations

can anyone help me with this question, please
---> find the region D for which the double integral of (4-3x^2-3y^2) dA  is a maximum , find the volume of maximum
thank you ,

$ I(x,y)\equal{}\int\int(4\minus{}3x^2\minus{}3y^2)\;dx\;dy$

Easy to show that $ I(x,y)\equal{}xy(4\minus{}x^2\minus{}y^2)$

$ I_{x}'\equal{}y(4\minus{}3x^2\minus{}y^2)\equal{}0$
$ I_{y}'\equal{}x(4\minus{}x^2\minus{}3y^2)\equal{}0$

If $ x$ or $ y$ is $ 0$ then $ I\equal{}0$. Let $ x,y\ne 0$.

$ 4\minus{}3x^2\minus{}y^2\equal{}0$
$ 4\minus{}x^2\minus{}3y^2\equal{}0$

Add two this equations:

$ x^2\plus{}y^2\equal{}2$

So, we can put $ x\equal{}\sqrt{2}\cos \theta$ and $ y\equal{}\sqrt{2}\sin \theta$.

Put this into second equation and note $ \cos^{2}\theta\equal{}1\minus{}\sin^{2}\theta$:

$ 4\equal{}2(1\minus{}\sin^{2}\theta)\plus{}6\sin^{2}\theta$

or

$ \sin^2\theta\equal{}\frac{1}{2}$. This gives $ x\equal{}y\equal{}1$ or $ x\equal{}y\equal{}\minus{}1$ extremum points. We mean in this solution $ x\equal{}x_2\minus{}x_1$ and $ y\equal{}y_2\minus{}y_1$. So, maximum value of the intagral is $ I(1,1)\equal{}2$ and region $ D$ is any square with side $ 1$.

Yuriy, what on Earth are you talking about? :what?: The question was to find the domain $ D$ on the plane such that $ \iint_D (4 \minus{} 3x^2 \minus{} 3y^2) dA$ is as large as possible, not the local extrema of the function $ I(x,y)$ such that $ I_{xy} \equal{} 4 \minus{} 3x^2 \minus{} 3y^2$ (by the way, the latter cannot even be determined uniquely enough: you can always add $ F(x) \plus{} G(y)$ to it).

@manda: [color=red]value[/color] of the maximum, not "volume of the maximum"!

The first part of the question is basically equivalent to the following problem: you have a bunch of checks and bills written on your name. You can take any subset of those. Then you are paid the amounts on all checks you took and have to pay all the bills you took. Your task is to get as rich as possible. What will you take?

The second part, (once you figured out the first) is a routine integral computation in polar coordinates.


Find the smallest possible $n$ for which there exist integers $x_{1}$, $x_{2}$, $\cdots$, $x_{n}$ such that each integer between $1000$ and $2000$ (inclusive) can be written as the sum (without repetition), of one or more of the integers $x_{1}$, $x_{2}$, $\cdots$, $x_{n}$.

Of course those sets work. :) Now a proof that 10 doesn't work.

Is it correct?

Suppose that $ x_1 < x_2 < \cdots < x_9 < x_{10}$ works, there are $ 2^{10} \minus{} 1 \equal{} 1023$ possible sums.
If $ x_7 < 500$ then there are more than $ 22$ useless sums because all the sums of one or two different elements of $ \{x_1,\cdots,x_7\}$ are less than $ 1000$ $ (C_7^2 \plus{} C_7^1 \equal{} 28)$ then $ x_7 \ge 501$
But $ x_7 \plus{} x_8 \plus{} x_9 \plus{} x_{10} > 2000$ is a useless sum and all the sums with these four numbers, which  are $ 2^6 \equal{} 64 > 22$ contradiction

I didn't check the numbers, but the reasoning works for sure. :)

[quote="mszew"]
But $ x_7 \plus{} x_8 \plus{} x_9 \plus{} x_{10} > 2000$ is a useless sum and all the sums with these four numbers, which  are $ 2^6 \equal{} 64 > 22$ contradiction[/quote]

there is a mistake: if $ x_1 <0$, then $ x_1\plus{}x_7 \plus{} x_8 \plus{} x_9 \plus{} x_{10}$ may less than 2000 and become a useful sum.

my solution:
since there are at most 22 useless sums,
if there exist $ x_i$ not less than 1000 , then for any sum S among the other nine numbers, at least one of $ S$ and $ S\plus{}x_{i}$ is an useless sum. so there are at least 512 useless sums, a contradiction.

if there exist i<>j such that $ x_{i}\equal{}x_j$, then for any sum S among the other eight numbers, one of $ S\plus{}x_{i}$ and $ S\plus{}x_{j}$ is an useless sum. so there are at least 256 useless sums, a contradiction, too.

so for all i, $ x_{i}<1000$. thus the 10 sums of one number are useless.
then there are at most 12 useless sums of at least two numbers

if there are at least six numbers <500, then there are  \[ {6 \choose 2}\] =15 useless sums of two numbers, a contradiction.

if there are at least six numbers >=500, since there are at most one 500, so the sum of any four numbers among the six numbers is more than 2000. so there are  \[ {6 \choose 4}\] =15 useless sums of four numbers, a contradiction.

the rest case is that there are exactly 5 numbers <500 and 5 numbers >=500.
in this case,
for the "<500" numbers, there are \[ {5 \choose 2}\] =10 useless sums of any two numbers among them;
 for the ">=500" numbers, \[ {5 \choose 4}\] =5 useless sums of any four numbers among them.
and 10+5>12, a contradiction.


The roots of the equation $ x^2 \minus{} 2x \equal{} 0$ can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:

$ \textbf{(A)}\ y \equal{} x^2, y \equal{} 2x \qquad\textbf{(B)}\ y \equal{} x^2 \minus{} 2x, y \equal{} 0 \qquad\textbf{(C)}\ y \equal{} x, y \equal{} x \minus{} 2$
$ \textbf{(D)}\ y \equal{} x^2 \minus{} 2x \plus{} 1, y \equal{} 1 \qquad\textbf{(E)}\ y \equal{} x^2 \minus{} 1, y \equal{} 2x \minus{} 1$
[i][Note: Abscissas means x-coordinate.][/i]

[hide="Click for solution"]
$ x^3 \minus{} 2x \equal{} 0 \implies x^3 \equal{} 2x \implies x^2 \minus{} 2x \plus{} 1 \equal{} 1 \implies x^2 \minus{} 1 \equal{} 2x \minus{} 1$

Thus, the answer is $ \boxed{\textbf{(C)}}$.
[/hide]

[quote="AIME15"][hide="Click for solution"]
$ x^3 \minus{} 2x \equal{} 0 \implies x^3 \equal{} 2x \implies x^3 \minus{} 2x \plus{} 1 \equal{} 1 \implies x^2 \minus{} 1 \equal{} 2x \minus{} 1$

Thus, the answer is $ \boxed{\textbf{(C)}}$.
[/hide][/quote]
Right answer, wrong solution. (Mostly reading x^2-2x as x^3-2x...)

[hide="Shorter way"]
y=x and y=x-2 has no intersection (equal slopes; x =/= x-2), so it must be C.

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


Some potential hints:

[hide="Hint 1"]
Try equation 1 squared minus equation 2. The $ x^2 \plus{} y^2$ should then cancel out.
[/hide]

[hide="Hint 2"]Try subbing $ a \equal{} x \plus{} y$ and $ b \equal{} xy$.[/hide]

[hide="Hint 3"]Note that $ (2,4)$ and $ (4,2)$ are two of the solutions[/hide]

[quote="dgreenb801"]Square equation 1, and then let xy=a, x+y=b. We have:
$ b(b^2 \minus{} 2a^2) \equal{} 85a^2$ and
$ b^2(b \plus{} 2a) \equal{} 225a^2$.
Divide these two and simplify to get $ 45a^2 \plus{} 17ab \minus{} 14b^2 \equal{} 0$. This factors.
(5a-2b)(9a+7b)=0, so a=2b/5 or -7b/9. Then plug everything back in, the result follows ( a=-7b/9 is extraneous. (-2,-4), (-4,-2), (2,4), (4,2), (0,0) are the answers).

Whoops, this is the same as the zuton force, whatever.[/quote]

$ b \equal{} x^2 \plus{} y^2$ simple and good thanx


vector
linear algebra
matrix
Functional Analysis
advanced fields
advanced fields unsolved

let  H be an Hilbert space
and let A:H->H be a compact and a symmetric operator
let v be a vector in H which:
1. norm (v) =1

2. norm (A(v))= norm (A)

prove that v is an eigenvector

thanks....

A counterexample: $ A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $ v=\langle 1/\sqrt{2},1/\sqrt{2}\rangle$.


for wich natural Number $ a \geq 1$ does there exist $ x,y \in \mathbb{Z}$ such that $ p\equal{}x^{2}a \plus{} y^{2}$ , where $ p$ is a prime.

For at least one $ p$, or for all $ p$¿
For one $ p$, always.
For all $ p$, never.

oops sorry i didn't mention this.
actually you also have to find out for what form of $ p$.


Point E is given on the diameter AC of the circle. Draw a chord BD through E in such a way that an area of the quadrilateral ABCD is maximal.

Let the circle have center O(0,0) and radius 1. Let A(-1,0) and C(1,0). Let E(h,0) wlog h>0. Let the lines from E be $ y\equal{}k(x\minus{}h)$, then $ x \equal{} \frac{y}{k} \plus{} h$ and $ x^2\plus{}y^2 \equal{}1$ is the equation of the circle. The intersection of the circle with the line is $ y^2 \plus{} \left ( \frac{y}{k} \plus{} h \right )^2 \equal{} 1$ from wich $ y_{1,2} \equal{} \frac{\minus{}hk \pm \sqrt{k^2 (1\minus{}h^2)\plus{}1}}{k^2\plus{}1}$. Now the area is given by $ AC \cdot ( |y_1| \plus{} |y_2|)$ so we have to maximized $ |y_1| \plus{} |y_2|$. Wlog $ y_1 > y_2$, so we have $ y \equal{} |y_1| \plus{} |y_2| \equal{} y_1 \minus{} y_2 \equal{} \frac{2k \sqrt{k^2 (1\minus{}h^2)\plus{}1}}{k^2\plus{}1}$. Now do the sqare on the 2 side a sobstitude $ k^2\equal{}x$, so we have: $ y\equal{}\frac{2x(1\minus{}h^2)\plus{}1}{(x\plus{}1)^2}$ So take the derivate of y and we have $ y' \equal{} \frac{2 [1 \minus{} x(2h^2 \minus{} 1]}{(x \plus{} 1)^3}$ then $ y' \equal{} 0$ when $ x \equal{} \frac{1}{2h^2 \minus{}1}$ that is the max of y. Then $ k \equal{} \frac{1}{\sqrt{2h^2 \minus{} 1}}$.

When $ h \le 1 \minus{} \frac{1}{\sqrt{2}}$ the the max is when BD is perpendicolar to AC and when $ h > 1 \minus{} \frac{1}{\sqrt{2}}$ the costruction is this:

The circle with center E and radius EO meet the perpendicolar line from E to AC on X and Y. The circle with diameter OX meet the circle on $ B_1$ and $ B_2$. Then $ B_1$ and $ B_2$ are the 2 possibility for B cause is simple to verify that $ |\tan \angle B_1EO| \equal{} |\tan \angle B_2EO| \equal{}k$.


Two boys (Jake and Miles) and two girls (Betty and Abby) are elected to the four student council offices (president, vice-president, treasurer and secretary). If a girl is elected president and Jake is elected vice-president, in how many ways can the four students fill the four offices?

There are [b]2[/b] possibilities for president, [b]only Jake[/b] can be vice president, and there are [b]2[/b] ways to fill the rest.
So there are $ 2 * 1 * 2 \equal{} \boxed{4}$ ways for 4 students to fill the offices.


function
calculus
derivative
complex analysis
complex analysis unsolved

I've recently seen this result and I've remembered of
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=another&t=38574]this[/url],which is a straight application of the following theorem:
Let $u_n$ a sequence of real numbers.Show that there exists a function $f\in R^{(\infty)}$ such that $\forall p,u_p=f^{(p)}(0)$.

hehe .. the two problem are quite different but notations are similar ..

Let $u_n$ be an enumeration of the rational numbers.And $f$ a function which verifies the hypothesis of the theorem.Then it works for the other problem mentioned.Why are you saying they are that different?

in one exercice, $f^k(x)$ is the $k$-th derivative . In the other one, ${f^{(k)}(x)=f(f(f(..(fx))))}$

:blush: Ooops.I didn't see this.I thought that in the other problem the author meant also derivates.


[quote="Interval"]Rita is paid time-and-a-half for hours worked in excess of 40 hours and double-time for hours worked on Sunday.  If Rita had gross weekly wages of 342 dollars for working 50 hours, 4 of which were on Sunday, what is her regular hourly wage?[/quote]
If x=hourly wage
[hide]Since 4 hours were on Sunday, that would be 8x.
Since 50-40=10 there were 10 overtime hours which is 40x+1.5(10x).
$8x+40x+15x=63x$
$\frac{342}{63}=5\frac{3}{7}=5.\overline{428571}\approx\$5.43\text{dollars}$
The final answer is $\boxed{\$5.43\text{per hour}}$[/hide]

i believe this is not specific enough.  couldnt it be interpeted in either of these ways?

$1.$ The four sunday hours were also overtime hours.
$2.$ The four sunday hours were not overitme hours.

Well, think about it this way, why would an employer pay someone TRIPLE for working on Sunday?

Both could be argued correct, but I would definitely assume #2.

well,

i was thinking like after the first 40 hours. and since in the work week, sunday is the last day, those hours would be part of the 10 overtime.

Question:

Where did 63 come from?

Also, where did 8x come from?

I was wondering about the 63 also.
 I figured you had 40 straight time hours,6 overtime hours,4 doubletime hours on Sunday.
  This would translate into 40 + 9(6 ot=9 st) + 8(4dt=8st)
 That is 57 straight time hours. She made 342 dollars
342 divided by 57= 6 dollars an hour.

[quote="guess"]I was wondering about the 63 also.
 I figured you had 40 straight time hours,6 overtime hours,4 doubletime hours on Sunday.[/quote]
The 63 is actually supposed to be 63x. :blush:
I interpreted the Sunday hours as "not overtime" hours. I will edit my post.

Quevvy,

Can you please edit your post?  I am collecting all these notes to study for my state exam.

You said 63 should be 63x, right?

Yes, I said 63 x
Also, I edited my post.

[quote="Interval"]Is the answer 5.43 an hour or 6.00 an hour?[/quote] It really depends on how you interpret the question.
[hide="If Sunday isnt overtime"]
Then it's 5.43 per hour[/hide]
[hide="If Sunday is overtime"]
Then it's 6.00 per hour[/hide]
From the way it's written, I think that sunday isn't overtime...

No matter how you look at it Sunday is double time. This means that it is twice the standard rate. Therefore 4 hours double time= 8 hours straight time.  You can't have double time off of a time & a half rate. Double time is twice the standard hourly rate.


The perimeter of the rectangle to the side is 96 cm and the base measures 12 cm more than height. Calculate the area of the shaded part, knowing that P and Q are the midpoints of the sides AB and AD, respectively.

[img]http://img75.imageshack.us/img75/9803/cals.jpg[/img]

Anyone knows how to explain me how to do this question  ?
Thanks ;)

Perimeter = 2(b+h)=96
b=12+h

Solving for systems of equations, we get h=18 so b=30. Find the area of the whole rectangle and subtract the nonshaded parts, so that will be $ 30\cdot18\minus{}(\frac{1}{2}\cdot30\cdot9\plus{}\frac{1}{2}\cdot18\cdot15)\equal{}270cm^{2}$

well all you have to do is find the sides first

you use the equation 2x+12=96
that would be 2x=84
x=42

so the two sides are 42 and 54

the easier way to find the area is to find the whole thing then substract it

first you do 42x54=2268

then you djo the side that is on the right and is not shaded so you would do (1/2)(42)(54/2)

you do 1/2 times the stuff because it is the area of the triangle

and then you do 54/2 because you are using half of the longest side.

so if you compute that you get 21x27=567

then you compute the triangle that is at the bottom that would be (1/2)(42/2)(54)

that would also come out to 21x27=567

then you do 2268-567-567=1134

if my calculations ar right the answer is 1134 square units

There's a mistake, calculatorwiz. Sides of 42 and 54 would give a perimeter of 192, not 96.

Thanks pascal12.
Your answer is correct.
:)

woops my bad well if you get the perimiter and area you can do what i just

sorry for the bad calculations my bad


group theory
abstract algebra
superior algebra
superior algebra solved

Let G a group with 24 elements which has exactly 24 automorphisms. Does it follow that G is isomorphic to S_4? I don't know how to prove that it is, but I think so. Any opinions? (it is known that S_4 has these properties).

I was thinking about a counterexample. I don't know if the following works:

Take an element a of order 8 and and element x of order 3, and generate our group with a and x, following the rules:

xa4=a4x;
xax2=ax=x2a;
x2ax=xa=ax2 (actually, we only need xax2=ax, and all the things written after it derive from that).

You can see that this group has 3 subgroups of order 8 and one of order 3. It has 12 elements of order 8 and 2 of order 3. An automorphism takes a to another element of order 8 and x to another element of order 3, so it looked to me like there are 2*12=24 cases we can consider, so there are 24 automorphisms. We only need the images of a and x, since the group is generated only by those. However, this doesn't mean it works. I'm not sure, but maybe someone else could check this, because it looks Ok to me.

I didn't explain why G has 24 elmts, but that's easy to see: we take H to be the group generated by a (it has 8 elements). Then H'=xHx2 and H"=x2Hx. These 3 have 2 elements in common: e and a4, so we have 3*6+2=20 elements. Aditionally, we have 4 more elements: x and x2 and a4x and a4x2. That makes 24.


function
topology
real analysis
real analysis unsolved

Is there a enumeration of the rational numbers $r_1,...$ and a continuous real function $f$ such that $f(r_n)=r_{n+1}$/?

I believe such a function does exist, unless I'm making some stupid mistake somewhere below.

First of all, according to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=another&t=38574]this[/url], there are continuous functions $f: \mathbb R\to\mathbb R$ for which $\mathcal O_+(f,0)=\{0,f(0),f(f(0)),\ldots\}$ is dense in $\mathbb R$ (I think that set is usually called the positive semiorbit of $0$ under $f$, so that's why I denoted it by $\mathcal O_+(f,0)$). If we prove that given a dense countable real set $\mathcal S$, there is a homeomorphism $h$ of the real line carrying $\mathcal S$ onto $\mathbb Q$, we would be done, for we would then take $\mathcal S=\mathcal O_+(f,0)$ for a function $f$ as above, and we could take our continuous function to be $h\circ f\circ h^{-1}$.

Both a strictly increasing bijection from $\mathbb R$ to itself and its inverse are clearly continuous, so it suffices to prove that there is a strictly increasing bijection of $\mathbb R$ into itself carrying $\mathcal S$ onto $\mathbb Q$.

We can find a strictly increasing bijection $h$ from $\mathcal S$ to $\mathbb Q$ (we could use the construction [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=metric+space&t=48347]here[/url], post #5 for instance). If $x\in\mathbb R\setminus\mathcal S$ has a sequence $(s_n)_n\subset\mathcal S$ increasing towards it and another sequence $(s^m)_m\subset\mathcal S$ decreasing towards it, define $h(x)$ to be the unique real $y$ situated between every term of $(h(s_n))_n\subset\mathbb Q$ and every term of $(h(s^m))_m\subset\mathbb Q$. It's not hard to see that $h$ extended in this manner to the whole real line is a well-defined strictly increasing bijection from $\mathbb R$ to itself which takes $\mathcal S$ onto $\mathbb Q$, as desired.


[quote="MR.129"]Given an equilateral triangle ABC(AB=BC=AB=a).M and N be two points respectively on the side AB,AC such that:

$ \left\{\begin{array}{l} \overrightarrow {AN} \equal{} k\overrightarrow {AC} \\
\overrightarrow {AM} \equal{} \frac {{\overrightarrow {AB} }}{3} \\
\end{array} \right.$
Find all k such that:

a)$ BN \bot CM$

b) Angle created by $ \overrightarrow {BN} ,\overrightarrow {CM}$ is $ 120^\circ$[/quote]
$ a)\overrightarrow{BN}\equal{}\overrightarrow{BC}\plus{}(1\minus{}k)\overrightarrow{CA}$
$ \overrightarrow{CM}\equal{}\overrightarrow{CB}\plus{}\frac{2}{3}\overrightarrow{BA}$
$ BN\bot CM\Leftrightarrow\overrightarrow{BN}\cdot \overrightarrow{CM}\equal{}\frac{1\minus{}5k}{6}a^2\equal{}0\Leftrightarrow k\equal{}\frac{1}{5}$
b)Denote by P the intersection of BN and CM.
$ \widehat{BPC}\equal{}120^o\Leftrightarrow \widehat{NBC}\equal{}\widehat{ACM}\Leftrightarrow k\equal{}\frac{2}{3}$


algebra
polynomial
Vieta
complex numbers

Let $a,b,c,d$ be real numbers. Suppose that all the roots of $z^{4}+az^{3}+bz^{2}+cz+d=0$ are complex numbers lying on a circle in the complex plane centered at $0+0i$ and having radius 1. The sum of the reciprocals of the roots is necessarily

(A) $a$
(B) $b$
(C) $c$
(D) $-a$
(E) $-b$

[quote="scorpius119"][quote="cincodemayo5590"][hide="Solution"]
By Vieta's formulae we get that $\frac{c}{d}=\sum_{k=1}^{4}\frac{-1}{z_{k}}$. Since all of the roots have $|z_{k}|=1$, their product will have $|z_{1}z_{2}z_{3}z_{4}|=|d|=1$ and thus $\frac{c}{d}=\pm c$ since $d$ is real. Since I don't see the answer $-c$ as one of the choices (are the coefficients on the polynomial  [i]positive[/i] reals?), my answer would be $\boxed{C}$.[/hide][/quote]

Oh no, they tricked you! They never said $a,b,c,d$ had to be distinct...

[hide]
By complex conjugates theorem, if $z$ is a root of the polynomial, then so is $\bar{z}$. But $z\bar{z}=|z|^{2}=1\Rightarrow \bar{z}=\frac{1}{z}$. It follows that the reciprocals of the roots is just a rearrangement of the roots themselves, making the sum of the reciprocals of the roots equal to the sum of the roots, which is $-a\Rightarrow \boxed{D}$. (Note that $a=\pm c$ and $d=\pm 1$.)
[/hide][/quote]

Oh oops, lol. I was wondering why $-c$ wasn't up there....


[color=#6A220B]Muon-catalyzed fusion was dismissed on the grounds that it takes nearly as much (or more) energy to generate muons as is produced by the d-d or d-t fusion they catalyze. However, when the muons decay, they should release energy in the form of a gamma-wavelength photon, no? Shouldn't this energy be partially retrievable (thereby lowering the required efficiency)?[/color]

what is a muon?????

P.S are u referring to a meson?????

Muon is another elementary particle like a meson valerium maximum

For more details see http://en.wikipedia.org/wiki/Muon


Sorry, i edited
$ (1\minus{}2a)(1\minus{}2b)(1\minus{}2c)\equal{}2a^2b^2\plus{}2b^2c^2\plus{}2c^2a^2\minus{}a^4\minus{}b^4\minus{}c^4 \leq \frac{9a^2b^2c^2}{a^2\plus{}b^2\plus{}c^2}$ by Shur, so it is enough to prove that
$ (a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^2 \geq 9(a^3b\plus{}b^3c\plus{}c^3a)$

[quote="Nak"][quote="quykhtn-qa1"]let a,b,c be none-gative numbers such that :a+b+c=1.prove that:
(abc)^2≥(1-2a)(1-2b)(1-2c)(ba^3+cb^3+ac^3)[/quote]
Try $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$.[/quote]
i donot understand.it is true.

[quote="Nak"]
$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)^2 \geq 9(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]
How would you prove that....??

[quote="Nak"]Sorry, i edited
$ (1 \minus{} 2a)(1 \minus{} 2b)(1 \minus{} 2c) \equal{} 2a^2b^2 \plus{} 2b^2c^2 \plus{} 2c^2a^2 \minus{} a^4 \minus{} b^4 \minus{} c^4 \leq \frac {9a^2b^2c^2}{a^2 \plus{} b^2 \plus{} c^2}$ by Shur, so it is enough to prove that
$ (a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)^2 \geq 9(a^3b \plus{} b^3c \plus{} c^3a)$[/quote]    (1)
(1) false  (try a=0,b=c=1/2 we have 1/2<9/16)

to trannguyenquyND_VN and arqady :you can solve this inequaliti? :lol:

to nguoivn :you can solve this inequaliti by SOS?you can post it? :P

[quote="nguoivn"]Using Vasc' ineq: $ (x^2 \plus{} y^2 \plus{} z^2)^2 \geq\ 3(x^3y \plus{} y^3z \plus{} z^3x)$
And let $ x \equal{} b \plus{} c$;...  I obtain:
$ Sa \equal{} \frac {1}{8bc} \minus{} \frac {b^2 \plus{} c^2 \plus{} 2a^2 \plus{} 3bc}{3(a \plus{} b \plus{} c)(a \plus{} b)(b \plus{} c)(c \plus{} a)}$
Assume $ a\geq\ b \geq\ c \geq\ 0$, easily too see that: $ Sa \geq\ Sb \geq\ Sc$
So, we only need to prove: $ Sb \plus{} Sc \geq\ 0$
<=> $ 3(a \plus{} b \plus{} c)(a \plus{} b)(a \plus{} c)(b \plus{} c)^2 \geq\ 8abc(2a^2 \plus{} 3b^2 \plus{} 3c^2 \plus{} 3ac \plus{} 3ab)$
with $ a \geq\ b \geq\ c \geq\ 0$
But I haven't found a nice proof for this last ineq.
(if using: $ (b \plus{} c)^2 \geq\ 4bc$, the ineq 'll become not true )
@arqady: Can you finish for me this proof? Thank you.
@quykhtn-qa1: Good problem, my friend. I like this ineq.
 :)[/quote]
you can write more appearly? :)


A conference hall has a round table with $ n$ chairs. There are $ n$ delgates to the conference. The first delegate chooses his or her seat arbitrarily. Thereafter, the $ (k\plus{}1)$th delegate sits $ k$ places to the right of the $ k$th delagate, for $ 1 \le k \le n\minus{}1$. (In particluar the second delegate sits next to the first.) No chair can be occupied by more that one delegate.

Find the set of values of $ n$ for which this is possible.

$ k$-th delegate sits on $ 1\plus{}\frac{k(k\minus{}1)}{2} \ mod \ n$ chair
hence the problem comes down to finding n, such that
$ \frac{k(k\minus{}1)}{2} \not\equiv \frac{l(l\minus{}1)}{2}\ (mod \ n)$ for $ k \neq l$ $ k,l \in\{1,2,...,n\}$
it's easy to prove that the only n that satisfies this condition are $ n\equal{}2^k$

Thanks Dumel. Could you show us your proof of your last assertion? Thanks.

let $ 2n\equal{}2^{b\plus{}1}(2a\plus{}1)$
we'll show that if $ a\neq0$ then there exist different $ k,l \in \{1,2,...,n\}$ satisfying:
$ 2n|k(k\minus{}1)\minus{}l(l\minus{}1)\equal{}k^2\minus{}l^2\minus{}(k\minus{}l)\equal{}(k\minus{}l)(k\plus{}l\minus{}1) \ (*)$
1. case : $ 2^{b\plus{}1}>2a\plus{}1$
we can put:
$ k\equal{}2^b\plus{}a\minus{}1$ and $ l\equal{}2^b\minus{}a$. then (*) and $ 0<l<k \le n$
1. case : $ 2^{b\plus{}1}<2a\plus{}1$
then let's take
$ k\equal{}2^b\plus{}a\minus{}1$ and $ l\equal{}a\plus{}1\minus{}2^b$. then (*) and $ 0<k<l \le n$

if $ a\equal{}0$ then (*) would imply $ n|k\minus{}l$ or $ n|k\plus{}l\minus{}1$ which is impossible because $ k\minus{}l<k\plus{}l\minus{}1<n$


Let the set $ X = \{ x\mid x\geq0 \}$ and $f:X\to X $, $ f $ is bounded on [0,1] satisfying : $ f(x)f(y) \leq x^2f\left(\frac{y}{2}\right) + y^2f\left(\frac{x}{2}\right) $ for all $ x,y \in X$.
Prove that : $ f(x) \leq x^2 $

[i]Edited by Myth[/i]

Substituting $x=y$ we obtain $f(x)\leq x\sqrt{2f(x/2)}$ [b](*)[/b]. So assuming $x\to 0$ we see that $f(x)\to 0$, since $f$ is bounded on $[0,1]$.
Now we can extend [b](*)[/b]. Indeed, taking $x/2$ in [b](*)[/b] instead of $x$ we obtain: $f(x/2)\leq x/2\cdot\sqrt{2f(x/4)}$, so
\[f(x)\leq x\sqrt{2f(x/2)} \leq x\sqrt{2\cdot x/2\sqrt{2f(x/4)}}=x^{3/2}\sqrt[4]{2f(x/4)}.\]
And again $f(x/4)\leq x/4\cdot \sqrt{2f(x/8)}$, so $f(x)\leq x^{7/4}\sqrt[8]{f(x/8)/2^2}$ and so on.
Since $x^{(2^n-1)/2^{n-1}}\to x^2$ and $f(x/2^n)\to 0$ we conclude $f(x)\leq x^2$.


algebra
function
domain
Ring Theory
abstract algebra
calculus
integration

If a domain $ R$ is Dedekind and has Pic($ R$)=$ 0$ then it is a PID.

The only definition of a Dedekind domain I 'may' use is: a one-dimensional noetherian domain where for every prime $ p$ of $ R$, the local ring $ R_p$ is a discrete valuation ring.
I thought may be I could use the following thing:
I know from  Pic($ R$)=$ 0$ that every invertible ideal is principal. I also know that for a fractional $ R$-ideal $ I$ in a domain $ I\equal{}\bigcap_m I_m$ where the intersection is taken over the localizations at all maximal ideals $ m$ of $ R$.  How can I use these things?

Any one?

Thanks

[quote="Third Edition"]Try to show that every ideal in $ R$ is an invertible ideal. You might need that every ideal in $ R$ is a projective $ R$-module.[/quote]
Never heard of projective $ R$-modules !
Any way, I think I solved the problem:
$ I$ is invertible if and only if: $ 1$. $ I$ is finitely generated.
$ 2$. the localization $ I_m$ at each maximal ideal $ m$ of $ R$ is a principal fractional $ R_m$-ideal.
This is the case when $ R$ is Dedekind:
$ 1$. $ R$ is noetherian.
$ 2$. $ R_m$ is a discrete valuation ring.
thanks

I added the remark on projective modules to my hint since I did not know which definition you have and what theorems you have in stock.

Projective modules finitely generated over a Noetherian integral domain - this version suffices here - can be characterized by the fact that there is a natural number $ n$ such that the localisation at every prime ideal is a free module of rank $ n$. In your case $ n$ is equal to 1.

Your solution looks very ok to me.