geometry inequalities algebra polynomial function probability MIT

Hi,

IMO2005 in Mexico is coming soon. This topic follows the tradition of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6480]IMO 2004 here.[/url] I am asking for your opinion regarding certain issues:

I.) What will the questions, topic distribution, be at IMO this year. My suggestion of the topics for IMO 2005:

1. Geometry
2. Algebra, Inequality
3. Combinatorics, Graphs
4. Number Theory
5. Algebra, Polynomial
6. Geometry

II.) What will be the cut-off for gold, silver and bronze ?

last year' cutoffs:

gold medal: 32, silver medal: 24 and bronze medal: 16

[quote="erdos"]Well, as I have heard. Mexican math olympiads is essentially composed from Geometry and Combinatorics problems. That's not a good thing for me because I am not very good at combinatorics  :? . However I like geometry very much  :lol: .

So, I think there will be two combinatorics problems, two geometry probems, one Algebra and one Number theory.

what do the others think about that ??[/quote]

And what kind of algebra, combinatorics etc. ? But also note that team leaders and their deputies decide which problems get selected for the contest and not some Mexican committee. Though they can influence the Shortlist a lot as often there are lots of local mathematician in the Problem Selection Committee who work out the ShortList from the proposed problems, e.g. the IMO LongList.  :)

OK here is my guess :
1- combinatorics (maybe graph theory) [i]I don't like that [/i] :(
2- Geometry (probably a circle :) ) [i]I like that [/i]  :P
3- Number theory [i]I find it interesting too[/i]  :P
4- Geometry (a triangle or a quadrilateral)
5- A nice inequality  :)  [i]I like inequalities[/i]
6- Combinatorics (a partition of a set or something like that )  :(

My guess

1 - Number theory (good :))
2 - Combinatorics (bad :( - I don't like combinatorics much)
3 - Geometry  (bad :( - I like geometry, but not in 3)
4 - Geometry ( good  :) )
5 - Functional Equation (That's OK  :) )
6 - Combinatorics or number theory

I think that the cutoffs will be pretty low, since 10 more countries are joining the IMO !

My guess :
1-Combinatorial geometry
2-Function
3-Number theory
4-Discrete algebra
5-Game(NT and C at once)
6-Euclidean geometry

I guess
1.- Number theory :)
2.- Geometry  :)
3.- Combinatorics  :|
4.- Games, Inequalities or combinatoric Geometry ( :lol: ,  :| ,   :| )
5.- Inequalities or Combinatorics ( :| , :| )
6.- Very hard Geometry  :(

Thanks for your replies though I am waiting for further ones.  :)

BTW when making a guess don't forget to consider history and for which type of problem it might be time to pop up again.  ;)

The best should be to ask Arne who almost right last year.  :D

I hope
1. Geometry
2. Algebra
3. Number Theory
4. Inequality
5. Combinatorics
6. Geometry

[quote="Igor"]Hmm ... Since last year we had number theory only in 6, I think that number theory will appear in 1,2 or 4...[/quote]
That's good to have nt in 6 :P
since I [b]MUST[/b](yes, really must...) not be able to solve it if it appears to be 2,3,5,6...

number theory was in  problem 6 in (2001,2003,2004) and in problem 3 in 2002 so...

[quote="lightrhee"]I think the cut-off would be lower than last year. Last year seemed to be ridiculously high.[/quote]The tendency I noticed is for the cut-off to increase each year. People are getting smarter  :blush:  :P

[quote="Arne"]Hehe, okay Orl, I will think about it, just give me some time to give you my prognosis :D[/quote]

Anybody with a last-minute guess before the first day's problems are released?  :)

[quote="Fedor Petrov"][quote="orl"]
Anybody with a last-minute guess before the first day's problems are released?  :)[/quote]

Where are they released? Some time, at least a minute, after your question already passed![/quote]

I think it is shortly past 11 am in Mexico. And I guess the contest is over at 12 noon. So I guess the questions will arrive during afternoon Mexico time. So it is up to to make a last guess, Fedor!  :)

1.Geometry
2.Game
3.Number Theory
4.Algebra
5.Geometry
6.Combinatorics (but very hard)

1.combinatorial game
2.ineq
3. geo

4. functional eq.
5. number th.
6. combinatorics

[quote="Fedor Petrov"]1. geometry
2. inequality
3. combi

4. number theory
5. algebra (functional equation, may be)
6. geometry[/quote]

Fedor, I see you agree with me more or less.  ;)

My guess:

1. Combinatorics
2. Geometry
3. Algebra (probably a sequence or functioal equation)
4. Geometry
5. Inequality
6. Number theory (a very hard one)

[quote="harazi"]1. number theory
 2. geometry
 3. combi
 4. geometry
 5. inequality or functional equation
 6. number theory   :)[/quote]Wow, harazi was very close. A permutation of his version is the IMO 2005 subject :)

[quote="Arne"]Right, here I go, I don't think the chance of me guessing anything correctly is big, but I'll make my prognosis anyway :)

DAY 1
1) Combinatorics
2) Geometry
3) Algebra
DAY 2
4) Geometry
5) Number theory
6) Combinatorics[/quote]

good, you have 3 correect guesses ($2,3,6,$) and you have guessed the $4,5$ good too (just to replace them :lol: )

I hope that this is not off topicl
As an American Albanian mathematician who was Norbert Wiener's student in 1960 at MIT
I was intrigued the first time I saw a commutative diagram on the black board in 1956.
Retired at 69 I am pusuing the application of Topos to Cognitive systems.

My parents were from Korce, my grandparents from Dishnitz and Orgotska [forgive spelling
as I forgot my Albanian over the years].

What mathematics are being taught in Albania?

Best wishes to my mathematical brethran.


How did other Carolinians do on the AMC and AIME?
I got 126.5 on AMC and 8 on AIME for an index of 206.5.

i did very poorly :help:
amc - 129
aime - 3
index - 159
and why isn't there a crying smiley ?
and since when is there a smiley liimit?

it's okay furious...my amc 12 score was worse than yours although my AIME was better. All in all, i would've gotten the same index as you

oh wow...you're only a freshman right?  :omighty:

Yep. I seriously doubt I'll get more than 1 problem on the USAMO, though, and there's a pretty small chance of that much.

AMC 10=120.5
AIME=3

Index = 114.4 (due to AMC 10)

[quote="Iversonfan2005"]Yay...a bragging thread!! :clap:

I think the AMC and AIME tests are a joke.

I mean, I took the AMC 12 was so easy...I got all 25 questions done with like 1 hour to spare so I got a 150.

I got all 15 AIME problems in like the first 30 minutes, so I got bored since i had two and a half hours so I decided to find atleast three solutions to each problem

Overall, I easily made the USAMO with an index of 300.

The USAMO will also be a joke. maybe i'll prove Reimann for fun now :P[/quote]

Are you serious?

[quote="origamiidiot"]AMC 10 = 143
AIME = 9[/quote]

Awesome. And welcome to the SC forum Harry!!!

[quote="origamiidiot"]AMC 10 = 143
AIME = 9[/quote]

Good luck at USAMO...make SC look good  :thumbup:


geometry
3D geometry
sphere
geometry solved

[u][b]The author of this posting is : theking[/b][/u]
____________________________________________________________________

a sphere is cut by parallel planes into n pieces. It turns out that the total surface area of these pieces is exactly 2004 times the area of the sphere. What is the smallest possible value of N

[u][b]The author of this posting is : Myth[/b][/u]
____________________________________________________________________

Total surface area of these pieces is Sphere area + 2*Sum of cuts. So Sum of cuts = $2\cdot 2003\cdot \pi R^2$. Since area of cut is at most $\pi R^2$ we conclude that there are at least $2\cdot 2003$ cuts, but area of all cuts can't be the same, so there are at least $2\cdot 2003 + 1$ cuts.
You can easily construct example of such cuts.


AMC
AMC 8
MATHCOUNTS
AIME
USA(J)MO
USAMO
AMC 10

As the new moderator for the Ohio local communities board, I'd just like to introduce myself. Feel free to add introductions of yourself too.

I'm currently in 8th grade, attending Reynoldsburg Junior High School. I haven't really done any competitions yet, but I've done the AMC 8, and got a 21. I'm going to take the AMC 10, and hopefully I'll qualify for the AIME. I don't really know about any local contests in Ohio, but I'd like to learn about them.

And please remember that this board is in the problem solving section, meaning that you can't post about how the Browns really suck this year, even though it's about Ohio. The purpose of this board is to meet other math enthusiasts in Ohio, learn about local contests, and basically anything else that has to do with math and Ohio, not to discuss football teams from Ohio.

I live in CA now but I used to live in OH!!!!
honestly OH was MUCH better....too much competition here :wink:
I pwned everyone at contests in my school at OH(I went to Jones middle school in upper arlington)
19(1st in 7th) on AMC8 in 7th when I barely understood any english...
37/40(1st in 7th or maybe it was 1st in whole school) on this random contest I don't even remember... :D
oh I forgot
GO OHIO!!!!!!!!!

i still live in ohio, and teh winters are cold here

I live in Ohio,and always have. Last year I got the top Ohio score on the ACT in...Math!(go figure)

Hi, I live in Cincinnati.

I am in 8th Grade, and am attempting Math Counts for the first time. (Just found out about it this year - long story - don't ask :P )

Hey, I'm a junior, and I live in Ohio too, currently in Mount Vernon, and formerly in Worthington.  I've lived in Ohio since I was about 6 months old...

Anyway, I'm the school winner for the AMC 3 years running (AMC 10 freshman year and 12 the last two years), and this year I've been coaching our middle school's first ever Mathcounts team.  My little bro go 4th at Chapter and is going to State, so that's cool.

And I don't know what I got on the ACT yet, but I took it a couple weeks ago, and I aced the math on the practice, so I can hope...

Man, this is a long post, isn't it?  I'll let you get back to doing something worthwhile now...

I guess I should introduce myself. Not really too much to say... I made the AIME and stuff, I did really well in some contests. My OML scores were only 3-4-6-3 though. In rocket city I got 23 for the first round... hmm that's a brief introduction of me

Hi, I'll list the results of major tests last two year, good or bad.
2005
3rd in state mathcounts
55th in nats
134.5 AMC 10A
4 AIME
2006
150 on AMC 10A
121.5 on AMC 12B
Really wants to make USAMO but knows he will mess up

There is somebody that has taken first in Mathcounts State twice, qualified for MOP in his freshman year or last year, and taken first in the OCTM last year in my school.  He has won all competitions he has participated so far (as meaning the results have come back) this year.

[quote="h_s_potter2002"]And please remember that this board is in the problem solving section, meaning that you can't post about how the Browns really suck this year, even though it's about Ohio. The purpose of this board is to meet other math enthusiasts in Ohio, learn about local contests, and basically anything else that has to do with math and Ohio, not to discuss football teams from Ohio.[/quote]

Yeah, sports are kind of boring.


geometry
quadratics
function
logarithms
complex numbers
imaginary numbers

Ok i was a little disappointed on my first day of ASHME madness. I took the 1950 exam and i only missed 2 (Geometry, of course) i guess i can suggest it to anyone that would like a source of easy problems to start off with. Its book#5 in ANML library (MAA). A few of the more interesting problems including doing the following:

Factoring (X^2 + 4) into real factors.

Solving X^X^X^X^X^X^X^X....=2

Figuring out the least value of a quadratic function (written in standard form or Ax 2 + Bx + C (of course a>0)

 I guess trying to do all of the ASHME problem sets is a waste of time, i ordered the AOSP I,II and hope it isnt too hard to go through. Meanwhile i will do part III problems from the first ASHME book.

"Factoring (X^2 + 4) into real factors."
Huh?  Wouldn't it factor to $(x-2i)(x+2i)$?

Solving X^X^X^X^X^X^X^X....=2
I don't think this is solvable.  If 0<x<1 then the left side goes to 0.  If x is negative you have a number that is going to osciliate between negative and positive.  If x=1 then it's 1.  And if x>1 it keeps getting larger and larger... I think.  If x>1 then x^x>x, and then (x^x)^x>x^x and so on.

"Figuring out the least value of a quadratic function (written in standard form or Ax 2 + Bx + C (of course a>0)"
This one you can do by completing the square.  The minimum is the point $(-B/2A, (4AC-B^2)/(4A^2))$.

I believe that the answer to x^x^x^x...=2 is $\sqrt{2}$.  I did it by taking log_x of both sides.  Then you have x^x^x^x^...=log_x(2).  We know that the left side is 2 so we have log_x(2)=2.  So then x^2=2 and x= rt(2) (I m pretty sure the negative root doesn't work).

I would assume that was $x^4+4$, not $x^2+4$. The factorization
$x^4+4 = (x^2-2x+2)(x^2+2x+2)$ is one of those things that's somewhat useful to have memorized.

X^4 it is. And  :sqrt: 2 is correct although another approach is possible. The third is fairly easy but its fun to derive anyway.

For the x^4+4 one you can just look at it as (x^2)^2+2^2 and then it comes easily.

For the other one, x^x^x^x^x^x^x^...=2, I don't think that the square root of 2 is right.  I still maintain that there is no solution.  Maybe someone can clairify?
Following you're procedure of taking the log_x of both sides to give x^x^x^x^x^x^x^...=log_x(2), then you say log_x(2)=2 so x=sqrt(2).
However, if you took the log_x of both sides again you'd get x^x^x^x^x^x^x^...=log_x(log_x(2)), and THAT by you're assumption would also =sqrt(2).  log_x(log_x(2))?log_x(2), there must be a flaw in the logic, arising from the fact that you're dealing with an infinite thing.

There is also my argument that if x>1, then x^x>x, and therefore x^x^x^x>x^x and so the left hand part x^x^x^x^x^x^x^... would not converge to a single number if x>1.  But x must be >1 if it is to equal 2, but that cannot be.

I didn't say that x^x^x^x...=sqrt(2)  because it doesn't!!!  I said that [b]if[/b]  x^x^x^x^...=2 then x=sqrt(2). Look:

$x^{x^{x^{x\cdots}}}=2$

Take the $\log_x$ of both sides:

$x^{x^{x^{x\cdots}}}=\log_x{2}$

We know that the left side equals 2.  So we sub that in and solve to get $\sqrt{2}$.

If you take the log_x again it still would work but you are complicating things too much.

Maybe the way BadApproximation did it is more understandable.

First off, for clarification, when you evaluate something like this:
2^3^4 you really mean 2^(3^4), correct?  And then 2^3^4^5 would mean 2^(3^(4^5)).  That's how you're doing it, right?

[quote="Wumbate"]For the x^4+4 one you can just look at it as (x^2)^2+2^2 and then it comes easily.
[/quote]

Really, what do you do after you get $(x^2)^2+2^2$?
What I did was
$x^4+4=x^4+4+4x^2-4x^2$

$=(x^2+2)^2-4x^2$

$=(x^2+2)^2-(2x)^2$

$=(x^2+2+2x)(x^2+2-2x)$

After $(x^2)^2+2^2$ you can have it become $(x^2+2i)(x^2-2i)$.

For the x^x^x^x^... question, I still think that the question would need parentheses for your solution to be correct.
It would have to look like this:
x^(x^(x^(x^...=2
Otherwise, you'd take it like this:
Whatever x^x is
and then that ^x and then that ^x etc
Each time it will get larger.
You see what I mean?  The problem would have to get parentheses.

Edit:
This is a related link: http://www.cut-the-knot.org/arithmetic/anEquation.shtml
Does the order of operations work the way in that you'd find 2^3^4^5=2^(3^(4^5))?  I thought it went left to right so you'd do ((2^3)^4)^5.  Is it different with exponents?

The usual convention in the US is that exponentiation is right associative.  In other words, a^b^c without parentheses means a^(b^c).

[quote="BadApproximation"]The intended solution is:

since X^X^X^X^X^X^...=2 lets look at it as X^Y where Y= X^X^X^X^X^X... now its clear that Y=2 from the question, so sub back in and you get X 2 = 2, X =  :sqrt: 2[/quote]

I don't get how you got from $Y=2$ to $x^2=2$
I also got the answer from Joml's method, and it looks like that's the method used to solve the other problem.

x^x^x^x...= x^y = x^2 = 2

I agree with ravi b, the parentheses that you talk of are implied by the convention (exponents are right associative).

I ll try to explain Bad Approximations method even more in case ppl still don t understand it.

We are trying to find x where:

$x^{x^{x^{x...}}}=2$

For clarification let's call it y:

$y=x^{x^{x^{x...}}}=2$

Notice the right hand side.  If you take away the bottom x you still have the same thing.  Likewise, if you kept on putting x's below it or kept taking them away you would still have the same expression.  This is because the expression has an infinite number of x's and since infinity - 1 =infinity and infinity plus/minus/times/divided by any real number(exluding 0) still equals infinity.  So let's exploit this by making the substitution y=x^x^x... :
$x^y=2$

We could have made this substition in other ways like:  $x^{x^{y}}$ but we are trying to simplify the equation as much as possible.

So we have $x^y=2$.  But we already know that y=2 so subbing that in we get $x^2=2$ or $x=\sqrt{2}$

(Btw, for those continued exponents does anyone know how to make ascending dots?  Like  $\ddots$ but going upwards)

$x^4 + 4 = (x^2 + 2)^2 - 4x^2 = (x^2 + 2 + 2x)(x^2 + 2 - 2x) = (x^2 + 2x + 2)(x^2 - 2x + 2)$

Oops someone else already mentioned that but isn't that called the Sophie-Germain Identity when $(x^4 + 4y^4) = (x^2 + 2xy + 2y^2)(x^2 - 2xy + 2y^2)$ ?

[quote="Wumbate"]
Solving X^X^X^X^X^X^X^X....=2
I don't think this is solvable.  If 0<x<1 then the left side goes to 0.  If x is negative you have a number that is going to osciliate between negative and positive.  If x=1 then it's 1.  And if x>1 it keeps getting larger and larger... I think.  If x>1 then x^x>x, and then (x^x)^x>x^x and so on.
[/quote]

You are correct.

Now, can you (or anyone) give a rigorous proof of why

x^x^x^x^ ..... x converges to 0 for 0 < x < 1 and diverges for x < 1?

[quote="beta"][quote="Wumbate"]After $(x^2)^2+2^2$ you can have it become $(x^2+2i)(x^2-2i)$.
[/quote]
Original Question:

Factoring (X^4 + 4) into [u][b][i][size=200]REAL[/size][/i][/b][/u] factors.[/quote]

This can be done through that if we break up the two resulting quadratics into linear complex roots and combine them into different quadratics.

[quote="BadApproximation"]You are correct.

Now, can you (or anyone) give a rigorous proof of why

x^x^x^x^ ..... x converges to 0 for 0 < x < 1 and diverges for x < 1?[/quote]

dont you mean x<0 ?[/quote]

No, but I did mean x > 1.

[quote="blahblahblah"][quote="BadApproximation"]You are correct.

Now, can you (or anyone) give a rigorous proof of why

x^x^x^x^ ..... x converges to 0 for 0 < x < 1 and diverges for x < 1?[/quote]

dont you mean x<0 ?

No, but I did mean x > 1.[/quote]

How would you calculate it if x<0? Wouldn't that give you an imaginary number?

[quote="BadApproximation"]Ok i was a little disappointed on my first day of ASHME madness. I took the 1950 exam and i only missed 2 (Geometry, of course) i guess i can suggest it to anyone that would like a source of easy problems to start off with. Its book#5 in ANML library (MAA).[/quote]

I looked into some ASHME problems which are used in Salkind's book. And I do not see any sense in these tasks. Often problems in school were harder. These problems just test your skills to solve routine problems very fast and without mistakes but it does test problem-solving skills. What do you think ?

[quote="orl"]I looked into some ASHME problems which are used in Salkind's book. And I do not see any sense in these tasks. Often problems in school were harder. These problems just test your skills to solve routine problems very fast and without mistakes but it does test problem-solving skills. What do you think ?[/quote]

I haven't seen the particular book mentioned in this thread (I have Contest Problem Book V and Contest Problem Book VI at home), but I think what you say is correct. I was quite surprised to see any mention of AHSME problems in a thread in the Getting Started forum, but I haven't "promoted" the thread because some of the problems really do look pretty easy, by today's standards (although perhaps their best level for discussion as challenge problems rather than informational questions would be Pre-Olympiad forum). When high school math contests began in America, and that is the period when those early books were published, there wasn't a lot of "competition culture" here, and problem-solving contests looked more like school mathematics. Today it is a lot tougher to be a top performer on the AMC 12/AIME combination than it was to be a top performer on the AHSME in the early 1950s. For that matter, I have read here on AoPS-MathLinks that IMO problems used to be a lot easier than they are today.

problems then were much easier, infact too easy for most people but there are a few problems that you can learn something from. I think the level of these problems is about SAT I , II . Good for a bridge to real problems. Anyway just my take.

 If x=-1 wouldnt the expression be -1? (cant you also evaluate it for other odd negative integers, or negative rationals with odd denominators).

[quote="BadApproximation"]Try this one:

Evaluate  :sqrt: (6+ :sqrt: (6+:sqrt: (6+ :sqrt: (6+:sqrt: (6+ :sqrt: (6+...[/quote]

let x = :sqrt: (6+ :sqrt: (6+:sqrt: (6+ :sqrt: (6+:sqrt: (6+ :sqrt: (6+
then x = :sqrt: (6+ x)
squaring both sides gives you:
x^2 = 6+x
solve the quadratic to give you x = 3 or x = -2
but it can't be negative so x = 3

Imagine the alternative.  x^y^z=(x^y)^z=x^(yz), and that wouldn't be any fun at all.


Let x,y,z be positive real numbers satisfying the condition.

\[x^2 + y^2 + z^2 + 2xyz =1\]

Prove that \[xy+xz +yz \leq \frac12 + 2xyz\]

this is the solution I have so far:
[hide]
  Prove \[xy+xz+yz \leq \frac12 + 2xyz\]

  Since \[x^2+y^2+z^2 +2xyz =1 \]

 Therefore, x,y,z can not all be greater than $\frac 12$ because $x^2+y^2+z^2 +2xyz$ will be greater than 1. x,y,z can not all be smaller than $\frac 12$ because $x^2+y^2+z^2 +2xyz$ will be smaller than 1.

 Therefore, we have at 2 of x,y,z which are both $\geq \frac 12$ or are both $\leq \frac 12$

 Let the two variables be x,y.

 Thus, \[(2x-1)(2y-1) \geq 0\] because (2x-1) and (2y-1) will have the same sign

 Expand and simplify:
 \[4xy - 2x-2y +1 \geq 0 \]

 \[2x+2y-4xy \leq 1 \]

 \[x+y-2xy \leq \frac 12 \]

 Since we need to prove that $xy+xz+yz \leq \frac 12 + 2xyz$

 We do some arrangements.

 \[z(x+y-2xy) + xy \leq \frac 12\]

 \[z(x+y-2xy) \leq \frac 12 - xy\]

 Since $x+y-2xy \leq \frac 12$

 Then, it is sufficient to prove $z \leq 1-2xy \Rightarrow 1-2xy-z \geq 0$

 Use AM-GM

 \[x^2+y^2 \geq 2xy\]

 \[x^2+y^2+z^2+2xyz \geq 2xy +z^2 +2xyz\]

 Since the LHS=1, therefore

 \[1 \geq 2xy +z^2 +2xyz\]

 Then, I got stuck

[/hide]

[quote="liangchene"]$1\geq 2xy +z^2 +2xyz$
Then, I got stuck
[/quote]
You got stuck [b]just[/b] before you finish the proof. :)
$1\geq 2xy +z^2 +2xyz$
$\Rightarrow 1-z^2\geq 2xy(1+z)$
$\Rightarrow 2xy\leq 1-z$

I am so going to cry, how come I didn't see that.

Sigh. :P


geometry
circumcircle
incenter
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, (0) its circumcircle, I its incenter,
A' the midpoint of the arc BAC,
A* the second intersection of A'I with 0,
P a point on the arc BC which doesn't contain A,
X, Y the incenters wrt triangles PAB, PAC
and (1) the circle passing through A*, I, C.
Prove that BI is tangent to (1) at I.
Sincerely
Jean-Louis

[quote="jayme"]let ABC be a triangle, (0) its circumcircle, I its incenter,
A' the midpoint of the arc BAC,
A* the second intersection of A'I with 0,
P a point on the arc BC which doesn't contain A,
X, Y the incenters wrt triangles PAB, PAC
and (1) the circle passing through A*, I, C.
Prove that BI is tangent to (1) at I.[/quote]

$ \angle BIA^* \equal{} \angle A'BI \plus{} \angle BA'I \equal{} \angle A'BC \minus{} \angle IBC \plus{} \angle A^*CB$
$ \equal{} 90 \minus{} \frac {\angle A}{2} \minus{} \frac {\angle B}{2} \plus{} \angle A^*CB \equal{} \frac {\angle C}{2} \plus{} \angle A^*CB \equal{} \angle A^*CI$
Hence result.
P.s. Dear Jean-Louis, what is the rule of the incenters X,Y in your post ?

Dear Mr. Danh,
I wanted just to make an indirect link with the A-mixtilinear incircle of ABC and the problem "a fixed point"  http://www.mathlinks.ro/Forum/viewtopic.php?t=228640
Sincerely
Jean-Louis


A square piece of paper, 12 cm on each side, is folded in half vertically. Both layers are then cut in half parallel to the fold, resulting in three new rectangles: a large one and two identical small ones. What is the ratio of the perimeter of one of the new small rectangles to the perimeter of the new large rectangle? Express your answer as a common fraction.

The smaller rectangles have a width of $ (12/2)/2\equal{}3)$ Their length is $ 12$. Therefore, the perimeter is $ 2(12\plus{}3)\equal{}30$

The larger rectangle has a width of $ 12/2\equal{}6$ the length is again, $ 12$. Therefore, its perimeter is $ 2(12\plus{}6)\equal{}36$

The ratio is then $ \frac{30}{36}\equal{}\boxed{\frac{5}{6}}$


MATHCOUNTS
AMC
AIME
geometry
inequalities
rotation
number theory
\/closed

Could the instructors possibly give us a list of possible classes that are currently being brainstormed or planned?

This way the members might be able to point out which ones we think are more important and would have higher enrollments. I realize that planning and running these classes is a lot of work. Perhaps it might be better if we could identify which of the classes more people would be able to benefit from and be interested in signing up for and run those first. This would also guide the instructors in preparing cirriculums and such.

Just a thought...

We have a very tentative plan to offer the following next year:

Problem Series
-------------------
MATHCOUNTS - probably in summer and fall
AMC - fall
AIME - fall/winter

Subject courses
--------------------
MATHCOUNTS/AMC level: Counting (Jan, already listed & probably in fall), Num. Theory (spring), Geometry (summer)

AMC/AIME (and a little beyond) level: Counting (summer), Num. Theory (summer)

Olympiad level: Geometry (Feb or Mar), Inequalities (Feb or Mar), General Olympiad techniques (fall)

Again, this is a very tentative schedule, and depends on instructor availability, student interest, and a number other other items.

For the Olympiad level classes, what do you recommend knowing?

Concerning the dates you listed...you said "next year"...does that mean year of 2004 or school year 04-05?

2004 calendar year.

For the olympiad courses, we'll post here a list of things you should know before taking the course once we schedule the class.

Are there gonig to be "course re-runs"? ...I'll be out of the country for most of next summer so I'll be missing a lot of potentially excellent Number Theory...

Yes - most of our courses will be offered every 9-18 months.

[quote="ml2"]is numbertheory and counting pretty much the same for the amthcounts thing?[/quote]

The Mathcounts/AMC level number theory and couting courses cover all the general curriculum and problems solving techniques we have noticed relating to Mathcounts if that's what you're asking.  They also go more in-depth into the ideas behind that set of problem solving techniques.

No - the two classes intersect very little (nearly not at all).

[quote="ml2"]What i meant is do they intersect in material covered greatly?[/quote]

While they may intersect some, the subject courses go into far greater detail.  Although our problem series courses seem to be popular, I would personally recommend the subject courses to students who are very curious about mathematics.  The difference is between learning how to accurately work problems that typically take 15 seconds to 3 minutes versus a more thorough exploration of the subject.

I am a bit confused, i think i didn't phrase this right, i meant the subject courses MC/AMc counting and MC/AMC number theory, are they overlapping. I wasn't refering to problem series, but thanks for clearing that up anyways.

My answer referred to that question: the number theory and the counting classes have very little overlap (almost none).

Do you perhaps have another tentative schedule so I can decided if I should take the inequalities course or not?

i have a motto. i take all the classes AOPS offers.

We do not yet know when we will be scheduling the inequalities course again, but you can be sure that we will rotate our classes back around once in a while.  I think we'll host that course at least once a year, depending on demand.

Honestly, I think that you're ready for most of the material in the course, but there are a few concepts that will likely be over your head.  I do think that in the end you'd be surprised as to how much of the subject of inequalities is accessible to a good problem solver with no more than an algebra I background.

That said, don't expect to master it all in 12 weeks.  There will be problems that you can solve and there will be those that challenge Nukular.

why do my classes for a particular class have to be taken down after a class ends. because i spent like an hour saving every file and picture so i can read it during the summer.


The masses are a**es
[url=http://www.putfile.com/media.php?n=axel_300k]This is proof.[/url]

[quote="nonie"]THat is SOO wierd....[/quote]
I especially like the fact that he has no pants on :cursing:

combination of  :huh:  :what?:  :rotfl: lol


algorithm
combinatorics proposed
combinatorics

I am pretty sure that the following problem is very known,

Let $ a_1,a_2,...,a_n$ be $ n$ arbitrary weights. What is the (best) algorithm to partition this sequence into two smaller ones $ A$ and $ B$ such that

\[ \left|\sum_{x\in A} x\minus{}\sum_{x\in B} x\right|\]

is the smallest possible?

PS. Please don't use recursion! This takes a year when $ n$ is maybe 20!

This is almost equivalent to the [url=http://en.wikipedia.org/wiki/Knapsack_problem]knapsack problem[/url]. If you ever find a good solution, let us know. As of today, I doubt anybody on Earth can help you :lol:


Find an equation that expresses the area of a square as a function of the dimension below.

1. the diagonal is (a)
so how could I write it as an equation if I knew that the diagonal is (a)

From number sense the area of a square given the diagonal is 2 times the diagonal squared.

do you mean that the square area is 2 (diagonal)^2. = (2a^2) no, I think it's not correct

$ a=l\sqrt{2}$, where $ l$ is the side length.
So $  l=\frac{a}{\sqrt{2}}\Rightarrow A=l^{2}=\frac{a^{2}}{2}$

wooooooooow, yes it's correct than you so much. My appreciation


AMC USA(J)MO USAMO trigonometry AIME MATHCOUNTS geometry

who here are in the USAMO, and MOSP! Not your real name, we can find them on the internet -___- we know you better by your forum name too =)
I know one person that comes here that lives near me, and that made both =)

This is for those people who are too lazy to spend 3 hours looking through their friend's profiles and then matching them to the website and then coming back to look at their name..

[b][u]USAMO PARTICIPANTS[/u][/b]

not_trig

[b][u]MOSP PARTICIPANTS[/u][/b]

not_trig

dude? WHAT IS UP WITH THIS? WHY ARE YOU DOING THIS?

and PLEASE DON'T FAIL AIME NEXT YEAR :wink:

dude i should give a list of those i know from mop

hamster1800
klebian
hexahedron
dnas
nealth
nebula42
vfiriou
mysticterminator (hmmmmmmm)
probability1.01
treething

hmm i lost... this idea is pretty SILLY

[quote="13375P34K43V312"]You won't be able to edit your post after 24 hours...[/quote]

the power of reposting =o

[quote="Karth"][quote="MysticTerminator"]also there is no S

why is my name being involved in a topic with a suspicious amount of "S"-ness[/quote]

Because S is involved in RSTLNE, which is *VERY OFTEN* used in wheel of fortune. Also, was Daesun the kid who won MathCounts (TM) a couple of years back?[/quote]

Yes

[quote="Karth"][quote="MysticTerminator"]also there is no S

why is my name being involved in a topic with a suspicious amount of "S"-ness[/quote]

Because S is involved in RSTLNE, which is *VERY OFTEN* used in wheel of fortune. Also, was Daesun the kid who won MathCounts (TM) a couple of years back?[/quote]

*last year

[quote="mihail911"]wait, treething, you're in MOP right? didn't you say something about how you don't like it or something?[/quote]

it's bollocks

COLLEGE POINTS ++ though

If this thread is to continue, maybe it should instead be to correlate real names with account names...

you know, this thread is pretty silly, but i have more usernames if you want :ninja:

[quote="PI-Dimension"]
This is for those people who are too lazy to spend 3 hours looking through their friend's profiles and then matching them to the website and then coming back to look at their name..

[/quote]
Why not spend 3 hours working on a few Olympiad problems?

sure

but he is in mop

and according to you that doesn't count huh

and please remove my name from aops

david bay rush

[quote="mdk"][quote="shawn1"][quote="bookaholic"]Evan O'Dorney is very very cool[/quote]

rofl, no offense... i admit hes amazing at math (probalby multiple IMO gold medals, he can beat me in any academic related activity) but I saw him on the Jimmy Kimmel show...wow..cant answer the most basic questions..

He needs to spend his time not studying for USAMO or any academic competition next year (because he can easily pass) but go outside of his house and be more socially active.

then with his social abilities plus his amazing mental abilities he will defiently succeed, otherwise he will probalby invent something worth 10 million dollars and get it scammed by someone for 10 thousand...seriously..

and plus why are we talking about friends?

oh yea and its not like im going to make MOSP...so yea[/quote]

he can't beat me in geography, science, track, soccer or basketball. Voila.[/quote]

Actually, he's probably beast at science. And it's better to be beast at math, music, and spelling (and by beast, I mean nationwide #1) than to be mediocre at a bunch of stuff.

i think its peculiar that this is the most popular topic (right now) in the AMC competitions forum..
considering its just a bunch of off-topic posts.. :P

[quote="Phelpedo"][/quote][quote="mdk"][quote="shawn1"][quote="bookaholic"]Evan O'Dorney is very very cool[/quote]

rofl, no offense... i admit hes amazing at math (probalby multiple IMO gold medals, he can beat me in any academic related activity) but I saw him on the Jimmy Kimmel show...wow..cant answer the most basic questions..

He needs to spend his time not studying for USAMO or any academic competition next year (because he can easily pass) but go outside of his house and be more socially active.

then with his social abilities plus his amazing mental abilities he will defiently succeed, otherwise he will probalby invent something worth 10 million dollars and get it scammed by someone for 10 thousand...seriously..

and plus why are we talking about friends?

oh yea and its not like im going to make MOSP...so yea[/quote]

he can't beat me in geography, science, track, soccer or basketball. Voila.[/quote][quote="Phelpedo"]

Actually, he's probably beast at science. And it's better to be beast at math, music, and spelling (and by beast, I mean nationwide #1) than to be mediocre at a bunch of stuff.[/quote]

so according to your definition, he is neither beast at math nor at science, so ... ok?

[quote="MysticTerminator"][/quote][quote="Phelpedo"][/quote][quote="mdk"][quote="shawn1"][quote="bookaholic"]Evan O'Dorney is very very cool[/quote]

rofl, no offense... i admit hes amazing at math (probalby multiple IMO gold medals, he can beat me in any academic related activity) but I saw him on the Jimmy Kimmel show...wow..cant answer the most basic questions..

He needs to spend his time not studying for USAMO or any academic competition next year (because he can easily pass) but go outside of his house and be more socially active.

then with his social abilities plus his amazing mental abilities he will defiently succeed, otherwise he will probalby invent something worth 10 million dollars and get it scammed by someone for 10 thousand...seriously..

and plus why are we talking about friends?

oh yea and its not like im going to make MOSP...so yea[/quote]

he can't beat me in geography, science, track, soccer or basketball. Voila.[/quote][quote="Phelpedo"]

Actually, he's probably beast at science. And it's better to be beast at math, music, and spelling (and by beast, I mean nationwide #1) than to be mediocre at a bunch of stuff.[/quote][quote="MysticTerminator"]

so according to your definition, he is neither beast at math nor at science, so ... ok?[/quote]

er...in his grade????? And for that matter, the grade above him too.

[quote="MysticTerminator"][quote="ordinary boy"][quote="SuperThread"]what is this "beast"[/quote]

DOnt RUSH to judgement[/quote]

I know, it sucks to BAY to conclusions[/quote]
dude wtf at least his made sense to the DAVID person

[quote="Klebian"]Wait, ok, how many posters in this thread actually went to MOP?

Well, actually, I know most of them, but what about:
-ordinary boy
-mdk
-spell it out 11[/quote]

I didnt go to MOP

[quote="da ban man"]Klebian, your a mod.  I think you should lock this..[/quote]
dude, wtf
klebian is not a mod of THIS forum, so he can't do anything


integration
complex analysis
function
calculus
complex analysis unsolved

Can you say that:
\[ \int_{S}\frac{g'(z)}{(z\minus{}a)^{n}\bar{g}^{n}(z)}dz\equal{}0~?\]
where $ g(z) \equal{} (z\minus{}a)\bar{g}(z)$ and $ n > 1$. $ S$ is a closed path

why?

is there a wiki where I can study this subject?

Thank you!

I guess you assume for instance $ g$ to be holomorphic and $ S$ a "small" path around $ a$ ? ... then let $ g$ have $ a$ as a zero of order $ m$, where $ m\equal{}0$, if $ a$ is not a zero. Then the function $ f(z)\equal{}\frac{g'(z)}{(z\minus{}a)^{n}g^{(n)}(z)}$ has $ a$ with multiplicity/order of $ \minus{}n\plus{}\max\{m\minus{}1,0\}\minus{}\max\{m\minus{}n,0\}\equal{}\minus{}n\plus{}\min\{n\minus{}1,\max\{m\minus{}1,0\}\}\leq\minus{}1$. The problem is that for $ n\leq m$ we get a pole of 1st order, i.e. the integral does not vanish, e.g., $ n\equal{}2$ and $ g(z)\equal{}(z\minus{}a)^{2}$ imply $ f(z)\equal{}\frac{1}{z\minus{}a}$, for which the integral on a "small closed path" is equal to $ 2\pi i$

P.S. I have seen similar formular in my school education, I guess you got one of the exponents wrong in the statement of your formula

Sorry, I just noticed the $ \bar{g}$-notation in your original statement. You can apply similar reasoning, get similar results and you can construct a similar counterexample, e.g., let $ n\equal{}2,b\not\equal{} a$ and set $ \bar{g}(z)\equal{}(z\minus{}a)(z\minus{}b)$, which implies $ f(z)\equal{}\frac{g'(z)}{(z\minus{}a)^{n}\bar{g}^{(n)}(z)}\equal{}2\,\frac{z\minus{}b}{z\minus{}a}\plus{}1$, i.e. an integration around a closed path around $ a$ gives the result $ 4\pi i (a\minus{}b)\not\equal{} 0$


AMC
USA(J)MO
USAMO
AMC 12
AIME
AMC 10
AIME II

So at my school, two students who qualified the USAMO in the past screwed up AIME2 badly.  One of them had 144 AMC12 but only scored a 5 on the AIME2, another with 150 AMC 10 and 4 on the AIME2.   The problems themselves were not super hard but still we all thought it was way tricker/unconventional than the past "normal" AIMEs.  So I was wondering what were the real reasons that made these kids not scoring so well on the test?

I know we have discussed this before, but people do tend to do worse on the AIME II?
Why?
First case: The problems ARE harder! The AMC committee say they are the same level but they could be wrong. Remember, they are a group of advanced mathematicians.

Second case: Due to lower scores on AIME II's there is psychological effect on the test-taker.

Statistically, despite the tests being chosen randomly, one test is bound to be harder than the other, and there is a small chance that that harder test is consistently chosen to be the AIME II.  Also, psychological factors could be at play as smartguy says.

I have seen no statistically significant correlation between taking the AIME I/II and scores. In a given year, there may be a difference, but averaged over all years it's negligible.

The paradox is that most people think the test they took was the harder one. You can be getting 15s regularly in practice and walk in on the testing day and get an 8. I've seen it happen. Sometimes, you just don't have a good day.

Actually, I think I saw somewhere in the forum that the average AIME II score is actually lower than the average AIME I score considering a span of many years.

Last year's AIME II average was higher than AIME I, though.

[quote="worthawholebean"]Then it's a coincidence. We know how they generate the tests.[/quote]

Agreed 100%. It depends on what types of problems people prefer to.

I have heard some say that AIME II numbers are distorted toward the high end due to many prep schools (Exeter, etc.) and the entire state of Texas taking the II since 2006.

First of all, the OP is referring to this *particular* AIME II. It happens. I died on it too, despite being a 3-time USAMO qualifier and scoring 139.5 on the AMC12. Personally I found certain questions unnerving, like #7 and #9. When I got $ 2^{4010}*1$ on #7, I second-guessed myself because I thought they wouldn't ask for $ (ab)/10$ if $ b$ was $ 1$. I also thought the answer wouldn't be $ 000$ on $ 9$. (It's one of these hazy points on the AIME...for example, if the problem asks for something mod 1000, and your something isn't >1000, you "know" you're wrong, even though it's not explicitly stated). That, plus a few silly mistakes like missing #1, led to a very low score.

[quote="worthawholebean"]

The paradox is that most people think the test they took was the harder one. You can be getting 15s regularly in practice and walk in on the testing day and get an 8. I've seen it happen. Sometimes, you just don't have a good day.[/quote]

I don't think it's so much of a paradox as a psychological defense mechanism. Of course we want to believe that our AIME was the harder one.

Hmm a few reasons I think why AIME II was more difficult...7-9 were not as straightforward as their counterparts on AIME I. Also, while 11 and 14 were unusually straightforward on the AIME I, it was less so for II.

If you take one of the exams and find it tougher than expected, I imagine you'll probably find reasons to conclude that the other exam was easier. As the data suggests, it's a crap shoot every year in terms of which exam is "harder". We all do this... it's hard to shake the stress of taking the exam while it's "official".

Things are always easier in practice, and people tend to forgive any (if not all) arithmetic errors they make in practice. On the AIME, we of course know that rote calculations are just as important as the problem-breaking insights, and I think we all too often discount how important it is to be able to do "simple" calculations mistake-free. Locating a simple mistake embedded within pages of routine calculations is often just as difficult as cracking a USAMO problem.

I'm on the AIME committee and at no point are we instructed to make one exam harder than the other. In fact, I do not know which exam is the I and which is the II until after they are administered.

[quote="Quickster94"]AIME Data:
Avgs I/II

2004: 2.246/1.959
2005: 2.75/2.43
2006: 2.711/2.867
2007: I can't find data
2008: 4.774/5.273

So, 2 years the II has been harder, 2 years the I[/quote]
You can't just take the averages. To find the difficulty levels between them, you'd have to take the upper quartile or so.

You could also look at the index cutoff and subtract the average AMC score among those who qualified.

I don't think we want to get into all of that...
Regardless, the point is, AIME I and II have basically the same difficulty level


Hello!

I got a problem with my TI84 Plus calculator. There is a problem with the connection (USB). I can't put any program down on my calculator.

Can anybody Help me?

Trinity.

Is it when you try to connect the computer to the calculator? The computer-calculator connection is famously unreliable. :( Some people have it working wonderfully; some can' ever get it to work(I'm in the last group)

Billy

Now the connection is OK, but there is something wrong with the memory. Even if I delete a lot of things, there are not enough memory on my TI for even a little game or program. ERR: MEMORY is the message.

Hold kæft det er irriterende!

I simply had to press that **** cable into the calcutator. But your'e right, Billy. It is very unreliable! The memory is still a problem...


limit
calculus
real analysis
real analysis unsolved

Answer: 1/e.
Hint:
[hide]Apply Stirling's Formula for n!.
http://www.sosmath.com/calculus/sequence/stirling/stirling.html
[/hide]

It is easy to show that if  lim{x(n+1)/x(n)}=a then lim{{x(n)^{1/n}}=a.
If we to setx (n)=n!/(n^n) then we get result. Spix you are right

How to show that $\lim_{n\to\infty}\frac{x_{n+1}}{x_n} = a \implies \lim_{n\to\infty}x_n^{\frac{1}{n}} = a$?

Let $z_n=\frac{x_n}{x_{n-1}}$ and use the fact that $\lim_{n\to\infty} \sqrt[n]{z_1z_2...z_n}=\lim_{n\to\infty} \frac{z_1+z_2+...+z_n}{n}=\lim_{n\to\infty}z_n$.

Thanks!
Just reviewed my Calculus book and found the proof.


People please send some 4-grade problems.

Evaluate the following expression if $ x\equal{}2$
$ x^2\plus{}3x\minus{}5x\plus{}16x^{x\plus{}2}$

[quote="alanchou"]hmm, 4th grade... no

here's a lot of sheets though: [url]http://www.homeschoolmath.net/worksheets/grade_4.php[/url][/quote]

erm ok but i was thinking something a bit harder. :wink:

Oh, wait no, the answer's $ \boxed{256}$.

hey i get it but was in a hurry and thats y i spoiled it
so sorry for it!


If u_n  is a Cauchy sequence in a metric space ,is the series ∑_(n=0)^∞▒〖d(u_n,u_(n+1))〗  covergent?

This is false. Example: in $ \mathbb{R},$ let $ u_n \equal{} \sum_{k \equal{} 1}^n\frac {( \minus{} 1)^k}{k}.$

$ u_n$ is a convergent sequence, hence a Cauchy sequence. But the sum of the distances that you ask for is $ \sum_{n \equal{} 1}^{\infty}\frac1n \equal{} \infty.$


Let $n\geq3$ be an integer and suppose the sequence $a_1,a_2,...,a_n$ satisfyies  $a_{i-1}+a_{i+1}=k_ia_i$ for positive integers $k_i$. Prove that:

$2n\leq k_i \leq 3n$

I don't get something. Take $a_x$ the maximum of the $a_i$, then $k_x$ is at most $2$. Maybe we shall prove that there is i such that ...

I think you typed the problem incorrectly.  The correct problem and the solution can be found on http://camel.math.ca/Competitions/MOCP/2003/sol_aug.html , problem 227.

thx mak. Probably maverick forgot to write \sum....


[quote="AYMANE"]Find all real fonctions$g$ such that:
$a$ real.
$g$:$]0,1[$->$R$
$g(x)+g(1-x)=a$[/quote]

Clearly $g(1/2) = a/2$.  For $1/2 < x < 1$, define $g(x)$ to be anything you want.  Then the $g$ on the interval $(0,1/2)$ is determined by $g(x) = a - g(1 - x)$.


Let $C$ be a circle. Choose points $A,B,D$ and $E$ on the circumference of $C$ such that $AB||DE$. Let $F$ and $G$ be the middle points of $AB$ and $DE$ respectively. Prove that line through $F$ and $G$ passes also through the center of the $C$.

Please, here is the Olympiad Section ! Must create a new section with the absolute geometry (without the parallel axiom). In this geometry $AB\parallel CD\Longleftrightarrow AB\cap CD=\emptyset\ .$

[quote="Virgil Nicula"]$AB\parallel CD$[/quote]
$C$ is the circle

[quote="Virgil Nicula"][hide=" For Pavel Kozlov."]O.K. You are right ! But I made a allegory (maybe unsuccessful) and you are too seriously. Sorry for my words. You offer here a solution for this [i]difficult[/i] problem, if you have time. See that appears some cases a.s.o.  I don't.[/hide][/quote]
I certainly understand you, Virgil. Maybe my mistake is serious attitude to it.

No solutioms yet. Maybe I should give a hint.
[hide]Let $ O$ be the middle point of the circle and $ H$ be the middle point of $ FO$. It is sufficient to prove that $ H$ is the middle point of $ DE$.[/hide]


[quote="babyloverain"]Find $ x > 0$ and
$ \frac {2x \plus{} 1}{(2x \minus{} 1)^3} \plus{} \frac {1}{x^2} \equal{} \frac {1}{4}$[/quote]

Where is this problem coming from ?

Basic computation gives $ 2x^5\minus{}3x^4\minus{}\frac{17}2x^3\plus{}\frac{43}4x^2\minus{}6x\plus{}1\equal{}0$ whose roots are $ 2.515979352...$, $ \minus{}2.09787885...$, $ 0.262606739...$ and two complex roots.

Looking at continued fractions of these roots seems to show they cant be given with square roots.
Maybe explicit solutions may be found using clever changes in variable (using trig funcions or maybe hyperbolic trig functions).

[quote="b555"]may  i  know  how  did  u  so precisely  find  the  roots?[/quote]

Standard computer spreadsheet and Newton method


I want to display a bunch of equations, nicely centered. the equationarray environment either aligns them with the `=' as the point of reference, or I have to manually specify the alignment. I don't like this kind of fine-tuning, so I was wondering if there way anyway I could just have the equations automatically aligned centre without fussing about with the specific formatting.


Hi, I am a Grade 12 student from Toronto.

I was going to rewrite the Chemistry test in November and then do the Reasoning test in December. Unfortunately the test center in my area will be closed in November  :( , so I just registered for the December Reasoning Test. Right now I don't what to do - if I write the Chem test in January will the score be sent before the February deadline? If not, any suggestions as to what I should do?

Thanks!

It should be ok. Assuming it's for college apps, usually the college will say the last SAT testing date they accept. Good luck!


The following topic, in the Colorado Forum, is displaying some garbage.  It looked fine a few days ago, and I don't believe it has been modified:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21617[/url]

The problem is that the : followed by a ( is being interpreted as a :(

Not sure how to fix it...you'll just have to live with it for now.

Forum mods can fix this problem by putting an extra space around the colons in LaTeX.  I've taken care of the CO one.


[quote="Beat"]Prove that
$\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}\le \sqrt{c (ab+1)}$

for $a, b, c \ge1$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=58255
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2609


[i]Let $ a;b;c$ be positive real numbers such that :   $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{}1$

   Prove that :  $ 8(a\plus{}b\plus{}c)^3 \ \leq \ 10(a^3 \plus{}b^3 \plus{}c^3 ) \plus{} 11 (a\plus{}b\plus{}c)(1\plus{} 4abc) \minus{} 12abc$[/i]

[hide=" Big hint "] [i]Using the lemma :
If $ x ; y ; z$ are real numbers such that : $ x^2 \plus{} y^2 \plus{} z^2 \plus{} 2xyz \equal{}1$ then there exist a triangle $ ABC$ such that :

       $ cosA \equal{} x ; cosB \equal{} y ; cos C \equal{} z$[/i][/hide]


Έστω ότι βρισκόμαστε σε ένα τηλεπαιχνίδι  :thumbup:  και ο παρουσιαστής μας καλεί να παίξουμε το εξής παιχνίδι:
Πάνω σ'ένα τραπέζι  :ddr:  υπάρχουν δύο φάκελοι,που ο καθένας έχει μέσα μια επιταγή με ένα ποσό. Επιλέγουμε έναν από τους δύο φακέλους, τον ανοίγουμε και βλέπουμε τί έχει μέσα. Στη συνέχεια ο παρουσιαστής  :diablo:  μας δίνει την δυνατότητα να επιλέξουμε αν θα κρατήσουμε τον φάκελο που διαλέξαμε ή αν θέλουμε να τον αφήσουμε και να πάρουμε τον άλλο φάκελο. Εμείς  :gleam:  κάνουμε κάτι από τα δύο και ότι ποσό αναγράφει η επιταγή του φακέλου που διαλέξαμε τελικά, γίνεται δικό μας.

Θεωρητικά, αν διαλέξουμε στην τύχη ποιον φάκελο θα πάρουμε, οι πιθανότητες να έχουμε το μεγαλύτερο από τα δύο ποσά που υπάρχουν στους φακέλους είναι 50%.  :yup:
Να βρεθεί μια στρατηγική που μπορούμε να ακολουθήσουμε έτσι ώστε να αυξήσουμε τις πιθανότητές μας να κερδίσουμε το μεγαλύτερο ποσό, δηλαδή να τις κάνουμε περισσότερες από 50%.  :read:  :ewpu:  :bruce:

 :bye:

Stedes μήπως εννοείς το εξής παιχνίδι (και ταυτόχρονα παράδοξο) με την παρακάτω extra υπόθεση:

Δηλαδή ότι ένας από τους δύο φακέλους περιέχει τη διπλάσια ποσότητα χρημάτων από αυτή που περιέχει ο άλλος...

[u]Link[/u]

http://www.maa.org/devlin/devlin_0708_04.html

 :play_ball:

Όχι mostel. Το πρόβλημα είναι έτσι όπως το γράφω και δεν έχει να κάνει με παράδοξα. Αυτό που αναφέρεις το είχα δει και στο παρελθόν. Είναι αρκετά ενδιαφέρον, αλλά δεν έχει σχέση με το συγκεκριμένο πρόβλημα. :wink:

Καλώς, πάντως σε λίγο μας βλέπω να "τρώμε" μήνυση από κάποιον τηλεοπτικό σταθμό για παραβίαση του απορρήτου  :rotfl:

Btw, ωραία emoticons χρησιμοποίησες  :coolspeak:

Lol!  :lol:
Δικό τους πρόβλημα. Ας πρόσεχαν να μην βάζαν παιχνίδια στα οποία μπορείς να αυξήσεις τις πιθανότητές σου.  :P

Όσο για τα emoticons, και που να δεις πως ήταν το αρχικό μου μήνυμα.  :P

πολυ παραξενο να υπαρχει στρατηγικη το προβλημα,(νομιζω οτι εσεις οι δυο Στελιοι πρεπει να πατε σε κανενα τηλεπαιχνιδι χαραμιζεστε) :D

Στέλιο, μόνο δύο είναι οι φάκελοι;
Γιατί αν είναι έτσι, δεν μπορώ να καταλάβω πώς είναι δυνατόν να αυξήσουμε τις πιθανότητες

Έχει καμία σχέση με το Monty Hall;

Ναι Λουκά, οι φάκελοι είναι δύο. Ουσιαστικά αυτό που εκμεταλευόμαστε είναι ότι γνωρίζουμε το ποσό του ενός φακέλου και με βάση αυτό οργανώνουμε τη στρατηγική μας. Αν δε το ξέραμε δε θα μπορούσαμε να κάνουμε τίποτα.
Εκ πρώτης όψεος το πρόβλημα φαίνεται παράλογο, αλλά υπάρχει λύση για να αυξήσουμε τις πιθανότητές μας, ελάχιστα βέβαια, αλλά μπορούμε.  :wink:

Και ένα hint:[hide] Αν θέλετε μπορείτε να θεωρήσετε το ποσό της κάθε επιταγής θετικό ακέραιο αριθμό (π.χ. εκφρασμένο σε ευρώ) και στη συνέχεια να το γενικέψετε για κάθε πραγματικό...(ακόμη και αρνητικό... :o ).[/hide]
Α, και δεν έχει σχέση με το Monty Hall, τουλάχιστον άμεση.  :)

Υποθέτουμε ότι τα ποσά που προσφέρει το τηλεπαιχνίδι ανήκουν στο διάστημα $[0,a]$

Έτσι, αν $x$ είναι το ποσό του φακέλου μας, τότε:

Αν $x\leq \frac{a}{2}$ τότε μας συμφέρει να αλλάξουμε
Διαφορετικά δε μας συμφέρει

Αλλά πώς μπορούμε να ξέρουμε το μέγιστο δυνατό ποσό $a$?

Όλα τα παραπάνω ισχύουν αν ακολουθείται ομοιόμορφη κατανομή στην εκλογή των φακέλων.

Τι συμβαίνει όμως αν τα ποσά ακολουθούν πχ εκθετική κατανομή (στην οποία και πάλι δε γνωρίζουμε τις παραμέτρους) ??

Λουκάς

[quote="darksaga"]εγω δεν καταλαβα καλα το προβλημα...αφου παιρνουμε το φακελο,κοιταμε τι εχει μεσα και μετα διαλεγουμε τον αλλο αν θλουμε...τοτε πρεπει να ειμαστε πολυ ηλιθιοι για να μην παρουμε τα λ7...αν δουμε και εχει μσα την επιταγη τον κραταμε , αν οχι τον δινουμε και περνουμε τον αλλο...πληρης ελεγχος της καταστασης...μαλλον δεν εχω καταλαβει το προβλημα μηπως θα μπορουσες να μου διευκρινησεις αυτο το σημειο?[/quote]

poios sou eipe oti koitame ti exei mesa o fakelos?

[quote="stedes"]... τον ανοίγουμε και βλέπουμε τί έχει μέσα....

 :bye:[/quote]

Μα και οι δύο φάκελλοι έχουν μέσα λεφτά! Ο Νίκος προφανώς αναφερόταν στο γεγονός ότι δεν μπορείς να δεις το ποσό που κρύβει ο δεύτερος φάκελλος!

το ποσό μπορεί να είναι οποιοδήποτε...
ή εχει καποιο όριο :?:
Όπως λέει κι ο πόντιος αν δεν υπάρχει "μέγιστο δυνατό ποσό" μου φαίνεται πολύ απίστευτο να υπάρχει στρατηγική :maybe:
Απ'την άλλη για να το λες είμαι σίγουρος ότι υπάρχει αλλά θα μπορούσες να βάλεις τη λύση το συντομότερο γιατί με τρώει η περιέργεια :stink:

Sorry που άργησα ν'απαντήσω.

Να μια πιθανή λύση του γρίφου:
[color=brown]Στρατηγική[/color]: Έστω ότι ο φάκελος που διαλέγουμε στην αρχή έχει μέσα μια επιταγή των $n$ €. Για ευκολία θα υποθέσουμε ότι ο $n$ είναι φυσικός,διαφορετικά μπορούμε να πάρουμε το ακέραιο μέρος του $n$. Ρίχνουμε ένα νόμισμα $n$ φορές. Αν τύχει και τις $n$ φορές κορώνα τότε αλλάζουμε τον φάκελο, διαφορετικά κρατάμε αυτόν που έχουμε.

Ας υπολογίσουμε τώρα τις πιθανότητες στο τέλος να έχουμε μείνει με τον φάκελο που έχει το μεγαλύτερο ποσό. Έστω ότι οι δύο φάκελοι έχουν δύο ποσά των $m$ και $k$ € με $m>k$. Αν έχουμε διαλέξει τον φάκελο με το ποσό $m$ ,οι πιθανότητες να τον κρατήσουμε ακολουθώντας την παραπάνω στρατηγική είναι $1-\frac{1}{2^{m}}$. Αν έχουμε διαλέξει τον φάκελο με το ποσό $k$ ,οι πιθανότητες να τον αλλάξουμε είναι $\frac{1}{2^{k}}$. Συνολικά δηλαδή οι πιθανότητες να έχουμε στο τέλος τον φάκελο με το μεγαλύτερο ποσό είναι $\frac{1}{2}\cdot\left(1-\frac{1}{2^{m}}\right)+\frac{1}{2}\cdot\frac{1}{2^{k}}=\frac{1}{2}-\frac{1}{2^{m+1}}+\frac{1}{2^{k+1}}>\frac{1}{2}=50$%. Δηλαδή πετύχαμε το ζητούμενο!

Προφανώς μπορεί να υπάρχουν πολύ περισσότερες στρατηγικές που να πετυχαίνουν το ζητούμενο αποτέλεσμα, απλά ανέφερα μία από αυτές. Πρακτικά, αυτό που γίνεται με αυτή τη στρατηγική είναι το εξής, όσο μεγαλύτερο ποσό δούμε στον φάκελο, τόσες περισσότερες πιθανότητες υπάρχουν να τον κρατήσουμε. Είναι δηλαδή αυτό που θα κάναμε και διαισθητικά...

mou fainetai fovero  :wink:
poly oraio provlima me kali lysi :) (kai praktiko... :D )
na rotiso omos...gti pol/siazoyme tis pi8anotites me 1/2(sto telos)?
einai typos gia tis pi8anotites :?:
episis an tin proti fora tyxei grammata tote ton kratame apey8eias i
rixnoyme antistoixa na doyme an tyxei kai tis n fores grammata???

$\frac{1}{2}$ είναι η πιθανότητα να διαλέξουμε αρχικά τον πρώτο φάκελο,αφού στην αρχή διαλέγουμε τυχαία.
$1-\frac{1}{2^{m}}$ είναι η πιθανότητα αν έχουμε διαλέξει τον πρώτο φάκελο, να τον κρατήσουμε.
[color=darkblue][u]Πολλαπλασιάζοντας[/u][/color] αυτές τις δύο πιθανότητες βρίσκουμε την πιθανότητα να διαλέξουμε τον πρώτο φάκελο [color=darkblue][u]και[/u][/color] να τον κρατήσουμε, δηλαδή $\frac{1}{2}\cdot\left(1-\frac{1}{2^{m}}\right)$.
Όμοια βρίσκουμε την πιθανότητα να διαλέξουμε τον δεύτερο φάκελο [color=darkblue][u]και[/u][/color] να τον αλλάξουμε που είναι $\frac{1}{2}\cdot\frac{1}{2^{k}}$.
Τέλος [color=brown][u]προσθέτουμε[/u][/color] τις δύο πιθανότητες, $\frac{1}{2}\cdot\left(1-\frac{1}{2^{m}}\right)+\frac{1}{2}\cdot\frac{1}{2^{k}}$,  για να βρούμε την πιθανότητα να συμβεί [color=brown][u]ή[/u][/color] το ένα [color=brown][u]ή[/u][/color] το άλλο.

Για την περίπτωση τώρα που τύχει από την πρώτη φορά γράμματα ,δεν νομίζω να υπάρχει λόγος να συνεχίσουμε να ρίχνουμε το κέρμα. Απλά κρατάμε τον φάκελο που έχουμε. :wink:


Can someone please explain what
$\prod$
and
$\sum$
mean, and how they work.

I've done math for years and really loved it, but have only recently discovered how much I suck and how little I know. The B.C. school curriculum teaches very little math.

$\sum$ indicates a summation, and $\prod$ indicates a product. This is the only difference between them, so it will suffice to explain the $\sum$ symbol. There was a similar question about this a while ago so I'll just copy my old (and slightly edited) response.

[quote="AntonioMainenti"][quote="Tokyo"]i dun even know wat that E means, can some1 explain it to me?[/quote]$\Sigma$ is the capital Greek letter sigma, and it is used to shorten (on paper) summations that are either long and tedious to write out, contain a variable number of terms, or are infinitely long (i.e. impossible to write). Here are some examples to explain the notation..

$\sum_{i=m}^n{a_i}=a_m+a_{m+1}+a_{m+2}+\ldots+a_{n-1}+a_n$.

The $i$ is called the $\text{index}$, and it has absolutely no significance to the sum; it's just a dummy variable. Its starting value is defined by $i=m$ ($m$ is an integer) and it increases by $1$ as each term is added until it reaches $n$, which of course is greater than (or equal to I guess..) $m$.

$f(x)=\sum_{n=0}^\infty\frac{(x-2)^n}{n!}=1+(x-2)+\frac{(x-2)^2}{2!}+\frac{(x-2)^3}{3!}+\frac{(x-2)^4}{4!}+\ldots$

Here $n$ ranges from $0$ to $\infty$; there is an infinite number of terms being added.

$\sum_{\text{prime }p}\frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\ldots$

(This summation has no value; it is infinite.) Sometimes there will only be stuff written on the bottom. It will give specific conditions that must be met for a term to be included in the summation. In this case, $\frac{1}{p}$ is added if and only if $p$ is a prime number. Another example of this:

$\sum_{\substack{2\le{j}\le4\\0\le{k}\le1}}x^jy^k=x^2y^0+x^3y^0+x^4y^0+x^2y^1+x^3y^1+x^4y^1=(1+y)(x^2+x^3+x^4)$

This summation adds all possible terms $x^jy^k$ such that both $2\le{j}\le4$ and $0\le{k}\le1$. And finally:

$\sum_{\text{people}}1=\text{the total number of people in the world,}$

because the summation has the term $1$ for every person in the world, so this represents the world population.[/quote]

I get it now,

thanks, I've been seeing that a lot recently


Find all function $ f$ :$ R^\plus{} \to R^\plus{}$ such that
$ f(ab)f(bc)(f(ca)f(a\plus{}b)f(b\plus{}c)f(c\plus{}a)\equal{}2008$

Put $ a \equal{} b \equal{} c$, we get $ f(k^2)f(2k) \equal{} ^3\sqrt {2008}$ for all $ k$
Put $ a \equal{} b \equal{} k$, we get $ [f(k^2)f(2k)] f(c \plus{} k)^2 f(ck)^2 \equal{} 2008$
So $ f(ck)f(c \plus{} k) \equal{} ^3\sqrt {2008}$ for all $ c,k$. Now given $ ck \equal{} x$, we have $ c \plus{} k$ can take any value $ \ge 2\sqrt {x}$.
As $ x$ can be made arbitrarily small, we have $ f(x) \equal{} m$, where $ m$ is a constant , for all $ x$.

So $ f(x) \equal{} ^6\sqrt {2008}$ is the only function satisfying the condition.


What is the minimum number of the six shapes
shown here that must be rotated so that all
arrows are pointing in the same direction?

[asy]import graph;
draw(Circle((0,0),1.5));
draw((1.5,0)--(7,0),Arrow);
draw(Circle((8,8),1.5));
draw((6.5,8)--(1,8),Arrow);
draw(Circle((20,8),1.5));
draw((18.5,8)--(13,8),Arrow);
draw(Circle((7,-4),1.5));
draw((5.5,-4)--(0,-4),Arrow);
draw(Circle((12,-4),1.5));
draw(Circle((19,4),1.5));
draw((12,-2.5)--(12,3),Arrow);
draw((19,2.5)--(19,-3),Arrow);[/asy]

It's $ \boxed{3}$, to make all the arrows point left. Note that we choose to make them point left because there's already three pointing left, but only one pointing in any other direction, and we're trying to rotate the least possible number of arrows to make them point in the same direction.


In linear measure, 7 palms equal 1 cubit, and 28 digits equal 1 cubit. What is the number of cubits in 8 palms, 6 digits? Express your answer as a fraction.

$ 1 \text{ palm} \equal{} \frac{1}{7} \text{ cubit}$

$ 1 \text{ digit} \equal{} \frac{1}{28} \text{ cubit}$

You have $ 8$ palms, or $ \frac{8}{7}$ ths of a cubit.

You have $ 6$ digits, or $ \frac{3}{14}$ ths of a cubit.

Add these up, to recieve the final answer of $ \frac{19}{14}$.


This is problem 24 from last year's AMC12A:

24. A plane contains points A and B with AB = 1. Let S be the union of all disks of radius 1 in the plane that cover AB. What is the area of S?

(A) 2(pi) + sqrt(3) (B) 8(pi)/3 (C) 3(pi) - sqrt(3)/2 (D) 10(pi)/3 - sqrt(3) (E) 4(pi) - 2sqrt(3)

I have seen some good solutions to this but I had this solution which I can't see a mistake.

S is the the set of all points that are distance 2 or less from both A and B. This means the union of circles with radius 2 centered at A and B. Let the intesections of these two circles be C and D.
Draw CD and call its intesection with AB, E. Looking at triangle ACE we see that cos A=AE/AC=.5/2=.25. Thus A=Cos^(-1).25 :approx: 75.52248781. Therefore angle CAD :approx: 151.0449756

The area of S is the area of sector ACD- Area of trangle CAD, all times 2. Sector ACD has area 2 :^2: * :pi: * 151.0449756/360 :approx: 5.272464287. Area of triangle CAD is 2*2*Sin 151.0449756 /2 :approx: .9682458366.
Thus the area of S is 2* (4.30421845) :approx: 8.6084369.

The correct answer is C which is :approx: 8.558752557.

I used a graphing calculator for all the calculations (without rounding between steps) and thus the only rounding should have been the 10th digit of my final answer. Yet my answer is about .04 off.

What is wrong?

Thanks!

The problem lies in the very first statement,

[quote]S is the the set of all points that are distance 2 or less from both A and B.[/quote]

Consider the point C such that AC = BC = 2.  This point is not in set S (since AC and BC cannot be in circles of radius 1, since they are both non-diameter chords with length 2) , but it fits your definition.

Instead, try considering the possible locations of the center, and then try calculating the area from there.


geometry
3D geometry
algebra
polynomial
advanced fields
advanced fields solved

Let $p \in V \subset k^{n}$ - an affine variety.Prove that the tangent cone $C_{p}(V) \subset T_{p}(V)$.

* Here is the definition of a tangent cone i'm using

Let $V$ be an affince variety and $I(V) = (f_{1}, f_{2},..., f_{r})$ where $r > 1$,
any polynomial $f$from $I(V)$ can be rewritten as a polynomial of $(x_{i}-p_{i})$ for a given point $p = (p_{1},p_{2},...,p_{n})$ on $V$. Then for $f$ we define $f_{p,min}$ as a homogeneous component of $f$ of a minimal total degree.
Then the tangent cone at $p$ is $C_{p}(V) = V( (f_{i})_{p,min}, f \in I(V) )$.

I will denote $\bar f=f_{p,min}$. It is an easier notation. Also, I will assume $p=0$.

$T_{0}V$ is given by $\sum_{k=1}^{n}\frac{\partial f_{i}}{\partial x_{k}}(0)\cdot x_{k}$ for all $i=\overline{1,r}$.
For $f\in I(V)$, we have $f(0)=0$ because $0\in V$ by the assumption $p=0$, so $deg\bar f\geq 1$.

We must prove that if $x=(x_{1},\ldots,x_{n})\in V(\bar f| f\in I(V))$, then $x\in T_{0}V$.

It is clear from the definition of $C_{0}(V)$ (it is given by homogeneous polynomials) that it is a cone with vertex in $0$.

Let $x\in C_{0}(V)$.
Consider for $f\in I(V)$, $F(f)$ the polynomial in $k[t]$ given by $F(f)(t)=f(tx)$. It is easy to see that $x\in T_{0}V\Leftrightarrow \frac{\partial F(f_{i})}{\partial t}(0)=0\ \forall i=\overline{1,r}$ i.e. the line $0x$ cuts $V$ at $0$ with multiplicity at least $2$.
But $F(f_{i})(t)=t^{deg \bar f_{i}}\bar f_{i}(x)+\mbox{ terms of higher degree in }t$. Since the degree of $\bar f_{i}$ is at least $1$ and since $\bar f_{i}(x)=0$ for all $i$, $F(f_{i})$ is a polynomial whose all appearing degrees of $t$ have degree at least $2$. So $t=0$ is a multiple root and $\frac{\partial F(f_{i})}{\partial t}(0)=0$ for all $i$.


function
real analysis
real analysis unsolved

How to show that if $X$ is a metric then the conditions 1 and 2 are equivalent for a function $f: X\to (-\infty,\infty]$:

1. $\forall t\,\, <f(x) \exists \delta>0$ s.t. $f(y)>t$ for all $y\in B(x,\delta)$
2. $\forall t\in \mathbb{R}\, f^{-1}((t,\infty])$ is open.

3. $f$ is lower semicontinous in $x\in X$ iff $\liminf_{y\to x}f(y)\geq f(x)$.

We have the following result: $f$ is upper semi-continuous iff $\left\{ \left. x \in X \right| f(x) < a \right\}$ is open for all $a \in \mathbb R$ (it's the same thing).

Take a look at my post [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=574519#574519]here[/url] for the $\Rightarrow$ implication.

"$\Leftarrow$"
Let's prove the contrapositive. So, $\exists x_{0}, \, \exists \epsilon > 0, \, \forall U \in \mathcal V \left( x_{0}\right), \, \exists x \in U$ s.t. $f(x) \geq f(x_{0})+\epsilon$. It follows that $\left\{ x \in X | f(x) \geq f(x_{0})+\epsilon \right\}$ is not closed and that's the contradiction we were looking for.

These are easy things. Only definitions are required plus some little theorems about open sets.

PS: I hope I'm not proving myself wrong by making a mistake :oops:

[quote="perfect_radio"]These are easy things.[/quote]
I agree  :lol: I just saw the definition first time and got confused when I didn't maged to prove that. I'm going to practice my mathematical thinking to ask less stupid questions.


geometry
parallelogram
circumcircle
analytic geometry
geometry unsolved

Let AC be a line segment in the plane and B a point between A and C . Construct isoceles triangles  PAB and QBC on one side of the segment AC such that angle APB =angle BQC = 120 degree and an isoceles triangle RAC on the other side of AC such that angle ARC=120 degree.Prove that triangle PQR is equilateral

Dear Mathlinkers,
this is an interesting degenerate case of Napoleon's theorem.
Sincerely
Jean-Louis

It can also be solved with help of analytic geometry.

[quote]just use COS law
[/quote]

and

[quote]It can also be solved with help of analytic geometry[/quote]

Can you  both please elaborate ? I fail to understand .  :?:

Let us produce $ AP$ and $ CQ$ to a point $ T$ . Now, also , Let $ Z$ be chosen as shown in [i]Plane geometry's[/i] diagram above such that $ R$ is the circumcentre of $ \triangle AZC$.
Now, brief angle chasing, will tell us that $ \angle PAZ \equal{} \angle TAZ \equal{} 90^{o}$ and $ QCZ \equal{} \angle TCZ \equal{} 90^{o}$.
This shows that $ TAZC$ is a cyclic quadrilateral. Since $ \angle AZC \equal{} \frac{1}{2} \angle ARC  \equal{} 60^{o}$ one has $ \angle ATC \equal{} 120 ^{o}$.
Now since $ ABC$ is a straight line; $ \angle PBC \equal{} 120^{o}$.
Notice also by linear pair result, $ \angle TPB \equal{} \angle TQB \equal{} 60^{o}$. Hence $ PTQZ$ is a parallelogram, forcing $ QPB \equal{} 30^{o}$.
Similarly $ \angle PQB \equal{} 30 ^{o}$ Hence $ PB \equal{} QB$ (Sides opposite to equal angles are equal).
A similar argument shows that $ BR \equal{} BP$
and hence that
$ B$ is the circumcenter of $ \triangle PQR$ and hence that $ \angle PRQ \equal{} 0.5 * 120^{o}$
Similar arguments lead to  the proof.
[hide]Albeit convoluted, this proof uses only elementary geometry  :P
[/hide]
 :)

Let B be the point (0,0). We can assume BC=2a, AB=2b and $ b\geq a$. Then our points P,Q,R have coordinates
$ Q\equal{}(a,\frac{a\sqrt{3}}{3})$
$ P\equal{}(\minus{}b,\frac{b\sqrt{3}}{3})$
$ R\equal{}(a\minus{}b,\minus{}\frac{(a\plus{}b)\sqrt{3}}{3})$ and now we can easily compute the lenghts of PQ,QR,PR ;)


Prove that, among $39$ consecutive natural numbers, necessarily, there is such a number whose  sum of digits is divisible by $11$


trigonometry
calculus
integration
function
derivative
calculus computations

Evaluate: $ \int_{0}^{\frac{\pi}{2}}\sqrt{\cos x} dx$

[quote="mlok"]$ 1.198140235\ldots$[/quote]

About the way u do it, i think everyone can do as u. How did u do it? Could u tell me about your solution?
Did u use calculator? :rotfl:

The integral can be expressed in terms of beta function as follows:

$ \int\limits_{0}^{\pi/2}\sqrt{\cos x}dx\equal{}\frac{1}{2}\beta\left(\frac{3}{4},\frac{1}{2}\right)$.

[quote="didilica"]The integral can be expressed in terms of beta function as follows:

$ \int\limits_{0}^{\pi/2}\sqrt {\cos x}dx \equal{} \frac {1}{2}\beta\left(\frac {3}{4},\frac {1}{2}\right)$.[/quote]

But i don't know beta function !
Can you show me the other answer ?

As a side note, the function to be integrated cannot be expressed as a finite sum of elementary functions. Therefore, to compute it, whether you use 1) some trick, 2) a calculator or computer, or 3) special functions.

One note about numerical integration (which is what mlok did): The original function is continuous but has an unbounded derivative as $ x\to\frac {\pi}2.$ The error estimates for most of the methods of numerical integration involve upper bounds for various higher derivatives, so an unbounded first derivative is bad.

Here's one way (among many possibilities) to prepare for numerical integration:

First, I'm going flip the quadrant around to write the integral as $ \int_0^{\frac {\pi}2}\sqrt {\sin x}\,dx.$

I didn't have to do that, but given the chance I'd prefer to move the exceptional point to zero where I can see it better.

Make the substitution $ x \equal{} u^2.$ The integral becomes:

$ \int_0^{\sqrt {\frac {\pi}2}}2u\sqrt {\sin u^2}\,du.$

Now we have an integrand that should be $ C^{\infty},$ so that we can trust our error estimates.

Using a Romberg integration with 65 function evaluations, I get $ 1.19814023473521$. That's probably about as good as I'm going to get using a spreadsheet.

[quote="Kent Merryfield"]One note about numerical integration (which is what mlok did): The original function is continuous but has an unbounded derivative as $ x\to\frac {\pi}2.$ The error estimates for most of the methods of numerical integration involve upper bounds for various higher derivatives, so an unbounded first derivative is bad.

Here's one way (among many possibilities) to prepare for numerical integration:

First, I'm going flip the quadrant around to write the integral as $ \int_0^{\frac {\pi}2}\sqrt {\sin x},dx.$

I didn't have to do that, but given the chance I'd prefer to move the exceptional point to zero where I can see it better.

Make the substitution $ x \equal{} u^2.$ The integral becomes:

$ \int_0^{\sqrt {\frac {\pi}2}}2u\sqrt {\sin u^2}\,du.$

Now we have an integrand that should be $ C^{\infty},$ so that we can trust our error estimates.

Using a Romberg integration with 65 function evaluations, I get $ 1.19814023473521$. That's probably about as good as I'm going to get using a spreadsheet.[/quote]
Thank you very much!

hello, Maple 11 gives
$ \minus{}\sqrt{2}EllipticK\left(\frac{1}{2}\sqrt{2}\right)\plus{}2\sqrt{2}EllipticE\left(\frac{1}{2}\sqrt{2}\right)$ as an answer.
Sonnhard.

[quote="Prime factorization"]There's no nice solution to this? On the surface, it seems like an easy problem, even though I'm inexperienced with integrals :S[/quote]
Believing that nice-looking formulas "ought" to have other nice-looking formulas as antiderivatives is indeed a sign of inexperience.


algebra
polynomial
Gauss
email
complex analysis

What does it mean?

Mine: dragon because... they're awesome? and 96 for birth year. 96 is also half-symmetrical, making it look nice in mathfrak. :P

"isabella" is my name, though everyone calls me "Izzy", 22 is a cool number, and 96 is my birthyear.

I forgot.

Seriously.

And that scares me.

Anyone want to help me find what it might mean?

Mine because my dad is a professor and my (coming up in $ 6$ years) professor will probably be a dad.

Um... it's my first initial and last name.

This one girl at my school called me Dojo, because it sounds like DG, which is my name [size=59](*cough* not Dee Kay or Decay, izzy and dragon *cough*)[/size].

Although, I did find out my name means a place where martial arts is practiced.

Fly Agaric is a kind of mushroom that is red with white spots (see signature)...

Meewhee009...
welll i have this thing of extending the word "me"
then at the time when I made my AoPS I said "whee"
then "009" are the last three digits in 2009.
AND somehow all of that put together is my username,  :P

My nickname bleh... and then 123 is an arithmetic sequence.

I like space. :D

As for the rest, those were random things I decided to add.

huh, another "dragon".

sdkudrgn88 = sdku + drgn + 88.

sdku = sudoku (because I like sudoku)
drgn = dragon (because I also like dragons)
88 = a weird meme from our gifted class. We had a teacher who told us that her "4+" (the equivalent of an A+ in America) would be a maximum of 88. We found this to be weird, and so ever since that, 88 has been a meme.

@r15: It's probably some weird polynomial in a complex analysis problem. (*doi* just pointing things out)

uhh... friend made it up? Although it happens to be some sort of cat brand or something... who knows.

Gauss is my favorite great mathematician :D
and 1181 is just some random number...idk why I picked that, lol.

Eggy=nickname

guess what Lv. means

999 will make sense if you know what Lv. means

Needed a short username, and 1=2 is that classic false proof theorem.

Little problem:  When I save the image on my Windows XP computer, it prints not just the image, but the image on a full sized white sheet(as seen on Microsoft Paint), making the image too big.  Is there any way I can get around this?

NO reason for my name.

I saw a falcon flying above so i made that my name.
1 um... first natrual number?

[b]D[/b] Dreams
[b]A[/b] Ate
[b]J[/b] Journal
[b]O[/b] Of
[b]E[/b] Elements

Making perfect sense.... :D

Uhm... my email is dreambold@gmail.com, so I decided that I'd use that for consistency, but the first account I made didn't work for some reason, so I remade it and added 22 at the end 'cause it's my favorite number. Dunno why.

As for for "dreambold" came to be, I'm not really sure. I like the word "dream", and eventually I added "bold" for "dream bold". I also have other accounts that are dream + another word, such as Dreammaker, Dreamlight, Dreamthin, Dreamer, PerpetualDreams, KuroDreams, etc. Dreambold just happened to be the firswt thing that popped into my mind.

I have no idea why I picked out my username...I wish I knew...I want to change mine.
Besides, 82 is a cool number.

MyYellowDucky82, you CAN change your username.  Go to profile and you'll find it.

@PowerOfPi: exactly. that's why i'm taking so long to change it. blah. i need the perfect username. :D

I choose my username because I act like a monkey (I am very Crazy) I am a girl and I am 12 and 23/24

$ remy1140$

Remy is my first name and I know it is very rare in the US, and you can easily track me down  :o
(In case you were wondering how I got this name, it comes from my Baptismal name as I'm a Catholic-Chrstian.)

And 1140...lol

It's not random numbers, because it all started when the last 4 digits of my home phone # is 1140. And then my dad's workplace # ends with 1140, as well as both of my parents' cell phone numbers. And I also have a few relatives with phone # ending with 1140. (Don't even think about prank-calling -- they're in Korea anyways, so you'll have to pay quite a lot to check every possible 1,000,000-ish numbers. :wink: )

And then lol, try sending an email to an address starting with remy1140 in Yahoo, Google, Hotmail, anywhere. Chances are...it's me  :o  :o  :o That's how uncreative I am.

well, this all started when my sister was watching sesame street.  ernie is a pretty cool guy. (and he has brown skin and black hair, see the connection?)

I like elmo better since he's a pretty cool guy. Plus he has a pretty cool voice and pretty cool hair color and pretty cool mouth.

My username probably makes people think that I am some sort of psychopath, but it's just a nice gaming handle, and I think changing it would be pretty wierd.


quadratics
search
modular arithmetic
algebra
Divisibility Theory
pen

The integers $a$ and $b$ have the property that for every nonnegative integer $n$ the number of $2^n{a}+b$ is the square of an integer. Show that $a=0$.

$2^n a+b=x_n^2$
then $x_{n+2}^2-b=4(x_n^2-b)$
$x_{n+2}^2-(2x_n)^2=-3b$
but equation$x^2-y^2=c$ has finite solution.thus $x_k=x_l$ for some $k\not= l$ hence $a=0$

A proof from first principles.

Let $(a,b)$ be a counterexample with smallest $|a|+|b|$.

[b]Theorem[/b]: b is odd.

[b]Proof[/b]

1. $b\ne 0$.

For otherwise $a$ is a non-zero square with $2a$ square. This contradicts the irrationality of the square root of 2.

2. $b$ is not divisible by 4

If it were, then for all n, $2^na+\frac{b}4=\frac{2^{n+2}a+b}4$ would be square, so $(a,\frac{b}4)$ would be a smaller counterexample.

3. $b$ is odd

Otherwise $4a+b$ is a square which is divisible by 2 but not 4.

[b]QED[/b]

Then suppose that for all $n$, $2^na+b=t_n^2$ where $t_n\ge 0$.

[b]Theorem[/b]: For all $n$ with $2^{n-2}>|a|$ and $n>0$ that there exists some odd factor $x$ of $a$ such that $t_n=2^{n-2}y-x$ where $y=\frac{a}{x}$.

[b]Proof[/b]

Note that $(t_{n+1}+t_n)(t_{n+1}-t_n)=t_{n+1}^2-t_n^2=2^na$. Furhter, since $b$ is odd, $t_n$ and $t_{n+1}$ are both odd. Thus both of $t_{n+1}+t_n$ and $t_{n+1}-t_n$ are even, but only one is a multiple of 4. Thus there exist $x,y$ with xy=a and with x odd such that $\{t_{n+1}+t_n,t_{n+1}-t_n\}=\{2x,2^{n-1}y\}$. However, $2^{n-1}y\ge 2^{n-1}>2|a|\ge 2x$. Thus $t_{n+1}+t_n=2^{n-1}y$ and $t_{n+1}-t_n=2x$. Subtracting gives the required expression for $t_n$.

[b]QED[/b]

[b]Theorem[/b]: For any odd factor x of a, there can be at most 2 n such that $(2^{n-2}y-x)^2=2^na+b$ where $y=\frac{a}{x}$.

[b]Proof[/b]

To see that there are at most 2 such n, expand and multiply by 16 to get $2^{2n}y^2-2^{n+3}a+x^2=2^{n+4}a+b$. Letting $z=2^n$, this gives $y^2z^2-(24a)z+x^2-b=0$, which is a nondegenerate quadratic equation with at most 2 solutions.

[b]QED[/b]

We have shown that for all sufficiently large n there is such an x. However, there are only finitely many such x, and each corresponds to at most 2 n. This contradiction shows that there are no counterexamples.

Can you write your solution a bit better, Hawk Tiger? If it is correct, it's very short. :)

Probably this is what Hawk Tiger meant, written out.

[quote="edriv"]For each n:
$ 2^{n}a \plus{} b$ is a perfect square $ \Rightarrow 4(2^{n}a \plus{} b) \equal{} 2^{n \plus{} 2}a \plus{} 4b$ is a perfect square.
Also $ 2^{n \plus{} 2}a \plus{} b$ is a perfect square.
Then we have two perfect squares at distance $ 3b$. We con construct arbitrary large perfect squares with distance $ 3b$. But this is impossible, then:
- the distance is 0, $ b \equal{} 0$, but then $ a,2a$ are both perfect squares $ \Rightarrow a \equal{} 0$
- then our squares are not arbitrary large: $ a \equal{} 0$.[/quote]

Trying to solve this problem, i found some curious facts about sequences that satisfy the recursive equation
$X_{n+2}=3X_{n+1}-2X_n$.

It's easy to see that the sequence of $2^n a+b$ where $n$ spans the nonnegative integers satisfy the equation.
It's also easy to prove that $X_n$ can be written as $\alpha_nX_1+\beta_nX_0$ where $\{\alpha_n\}$ and $\{\beta_n\}$ are also such sequences. Namely, $\alpha_n=2^n-1$ and $\beta_n=2-2^n$.
Another curious fact is that $X_n \equiv X_{n+j} \pmod{\alpha_j}, j\geq 2$

Nevertheless, this facts didn't help me to solve the problem... ;(

[hide=basically just post #2]Plugging in $n=0$ into the original expression gives $2^0a+b=m^2 \implies 2^2a+4b=(2m)^2$, where $m$ is an integer, Also, plugging in $n=2$ gives $2^2a+b=m_1^2$, where $m_1$ is an integer. Subtracting the two equations, we get that $3b=(2m)^2-m_1^2$. This is equivalent to finding two perfect squares with difference $3b$.

However, we can scale $n$ up to an arbitrarily large number. With this process, we can show that the closest perfect squares to the desired ones will be arbitrarily far apart, and thus larger than $3b$. With this construction, the conditions will not hold unless if $a=0$.  [/hide]

There is an extremely nice generalization of this result:
Let $W(x)$ be a polynomial with integer coefficients such that $W(2^{n})$ is a perfect square for all positive integers $n$. Then there exist another polynomial $Q(x)$ with integer coefficients such that $W(x)=Q(x)^{2}$

if $a$ is nonzero then $2^n{a}+b$ can be any big which implies :

$x^2=2^n{a}+b , y^2=2^{n+2}{a}+b \implies 4x^2-y^2=3b$

which means we got infinite pair of positive integers like $(2x,y)$ such as that the difference of their squares is a constant which a contradiction...


let a sequence be given in a euclidean space,say R^n,such that it is finitely decomposable into disjoint convergent subsequences.we then call it FINITELY DECOMPOSABLE.my question is-Is the sum and product of finitely many finitely decomposable sequences also finitely decomposable?if no plz provide counterexamples. :blush: (i already posted this in the unsolved problems forum but have not elicited a reply from anyone.)

Suppose $ (a_{n})_{n \equal{} 1}^{\infty}$ is finitely decomposed into convergent subsequences $ (a_{n_{1,k}})_{k \equal{} 1}^{\infty},\cdots, (a_{n_{ m,k}})_{k \equal{} 1}^{\infty}$.

Let $ \alpha_{i}\equal{}\lim_{k\to\infty}a_{n_{i, k}}$ and define an auxillary sequence $ b_{n}$ by

$ b_{n}\equal{}\alpha_{i}\quad\quad\text{if }n \equal{} n_{i, k}\text{ for some }k$

Then $ b_{n}$ is a well-defined sequence such that $ a_{n}\minus{}b_{n}$ is convergent to 0. Note that $ b_{n}$ has finite range.

Conversely, if $ b_{n}$ is a sequence with finite range and $ a_{n}$ is a sequence convergent to 0, then $ a_{n}\plus{}b_{n}$ is obviously finitely decomposible.

This means that we only have to study the behavior of sequences with finite range in order to examine the property of finitely decomposible sequences.

Let $ A$ be the set of sequences convergent to 0 and $ B$ be the set of sequences with finite range. It is easy to verify that

(1) If $ a_{n}, b_{n}\in A$, then $ a_{n}\plus{}b_{n}$ and $ a_{n}b_{n}$ are also in $ A$.
(2) If $ a_{n}, b_{n}\in B$, then $ a_{n}\plus{}b_{n}$ and $ a_{n}b_{n}$ are also in $ B$.
(3) If $ a_{n}\in A$ and $ b_{n}\in B$, then $ a_{n}b_{n}\in A$.

Therefore, the sum and product of finitely decomposible sequences is also finitely decomposible.

Another approach: 1) a sequence is finitely decomposable if and only if it has finitely many cluster points; 2) any such sequence is bounded; 3) for bounded sequences, each cluster point of $ (a_{n}\plus{}b_{n})$ is the sum of a cluster point of $ (a_{n})$ and a cluster point of $ (b_{n})$; 3) same as 2 but for products.

I wonder why the title mentions metric spaces :maybe: you can't add or multiply elements there.

I wonder why the title mentions metric spaces :maybe: you can't add or multiply elements there.[/quote]
you are right!i meant euclidean spaces.SOrry!


algebra
polynomial
calculus
integration
algebra theorems

Is this correct?
     [b]Every polynomial with integral roots is monic.[/b]
    Ashwath

P(X)=2X has only integral roots but is not monic.

Sorry! then I'll specify it for polynomials with degree greater than 1. now I think it ought to hold good.
   Ashwath

Consider $f(x) = 2x^3-2x^2+x-1$. Then f is neither monic nor an integral multiple of a monic polynomial but its only real root is 1.

Man, is this a joke? If this is a joke, it's not very good. A monic polynomial has leading coefficient 1. There is absolutely no condition on roots that should guarantee the polynomial is monic. So you can give 100000 counter-examples.


On any given USAMTS problem, I have a question about grading.

Will a "brute force" type of solution get full points if it follows the rules of USAMTS solution-writing, has no mathematical errors, and yields the correct solution?  For example, lets say I split a problem up into many cases, but perhaps there is an easier way to solve it.  Will the casework approach still be able to get a perfect score on that problem?

A complete solution with an inelegant method will still receive full credit.

[quote="cognos599"]again this has nothing to do with any one problem, but if on a problem, a perfect solution is written using advanced topics such as calculus or even higher level algebra, would that also receive full points. (you can tell this has nothing to do with any problem because i don't even know alot of calculus.  :) )[/quote]

As far as I know any complete and correct solution is worth full credit.

[quote="cognos599"]kind of off topic, but this contest is different from USAMO (besides the difficulty of the problems) in that in USAMO a complete correct solution may not get you a 7/7, but a complete correct solution will give you a 5/5 in USAMTS right?[/quote]

umm a complete correct solution on USAMO gets you a 7/7, no question. It is the partial credit that's a LOT stricter.

The USAMTS is much easier in terms of grading for partial credit.

[quote="cognos599"]so if you have a brute force solution that gives you the right answer and there are no errors in the brute forcing, then you will get a 7/7 on USAMO?????

the main point i am trying to get at is, which is graded more harshly, USAMO or USAMTS??[/quote]

USAMTS in more generous with partial  credit, but a correct brute force will get you a 7/7 on the USAMO, yes.


AMC AIME conics ellipse quadratics Putnam MATHCOUNTS

Does anyone have a problem that is their absolute favorite? I don't really.

My favorite problem that I have ever seen on a math contest is probably this one. But before I state the problem, note that I like it so much because my solution was so incredibly sneaky and not because the problem is really all that clever.

Find the largest real root of x^4-4x-1=0, and write your answer in the form 1/sqrt(a)+sqrt(sqrt(b)-1/c).

Try it! It's not that hard the ``normal'' way, but I don't think anyone solved it the way it was supposed to be solved.

Well I guess you could solve it by letting x^4-4x-1=(x^2+ax+b)(x^2+cx+d), expanding and equating coefficients, but that doesn't look particularly appealing.

Hmm, is your approach still...algebraic?

My approach isn't algebraic if by that you mean that I did a bunch of steps and found out what the roots were.

I have gone through over half of the AIME problems, and my favourite AIME problem was '85 #11:  given the foci of the ellipse (9,20) , (49,55) and it touches the x-axis.  find the length of the major axis

i must say, I have never seen any roots like: 1/sqrt(a)+sqrt(sqrt(b)-1/c)

i am almost inclined to work backwards on this one

I think it's written that way to allow a, b, and c to be positive integers.

I solved it by essentially working backwards, but ComplexZeta's comments make me think that maybe there's a shorter way.

Yeah, a, b, and c are supposed to be integers.

My way wasn't short at all. The normal solution is very short and easy, but mine was much messier. I imagine I did it the way gauss202 did.

Here's my rather messy solution:

[hide]

I assumed that x = 1/sqrt(a)+sqrt(sqrt(b)-1/c) was a root and knew that x' = 1/sqrt(a)-sqrt(sqrt(b)-1/c) must also be a root by conjugacy. So x2 - 2x/sqrt(a) + (1/a - sqrt(b) + 1/c) is a factor of x4 - 4x - 1.

Then using Viete's Theorem (or conjugacy again if you like), I know that x2 + 2x/sqrt(a) - 1/(1/a - sqrt(b) + 1/c) is the other quadratic factor of x4 - 4x - 1.

Let D = 1/a - sqrt(b) + 1/c.

Then equating coefficients of x and x2 when I multiply the two quadratic factors, I get:

(i)  D + 1/D = -2sqrt(a), and

(ii) D - 1/D = 4/a.

Solving these gives D = 2/a - sqrt(a) and 1/D = -2/a + sqrt(a).

Multiplying them together gives: -4/a2 + a = 1.

And it's easy to find that the only real solution is a = 2.

This makes D = 1/2 - sqrt(b) + 1/c = 1 - sqrt(2).

So sqrt(b) - 1/c = sqrt(2) - 1/2.

Therefore, (a, b, c) = (2, 2, 2).

(And this is the largest root because the other ones are just it's conjugates, which have minus signs in them).

[/hide]

ComplexZeta, I'm not sure I understand what you mean by the normal solution.  I ended up solving it like gauss202.

That was what I did on the contest. I was really pleased with myself for that. I told the guy who wrote the contest how I solved it shortly after and asked him for the normal solution, but he didn't give it to me. (I think he liked my solution though.)

Here's the normal solution:

x4-4x-1=0
x4+2x2+1=2x2+4x+2.

(I leave the rest as an exercise to the reader.)

I think this was a Putnam question:

Prove that for all composite numbers n > 3, we can find positive integers a, b, c such that n = ab + bc + ca + 1.

A simple little problem, but definitely one of my favorites.

At the getting started level, my favorite problem is any problem about a journey taken in two segments (e.g., going out and coming back) at two different speeds. Whether the question is "Find the average speed," or "How long did Jim-Bob travel," this problem seems guaranteed to flummox people until they REALLY understand problems of rate, time, and distance. In most of today's curricula in the United States, and in all curricula I have examined from Asia, such a problem is strictly a PREalgebra problem, and therefore a problem that ought to be easy for anyone participating in Mathcounts or AMC 8, but a problem like this occurs in almost every AMC 8 and in almost every Mathcounts school meet precisely because it is such a guaranteed stumper for the unwary.

At the advanced level--difficulty level unbounded above--my favorite problem is problem 4.73 (page 152) in the book Concrete Mathematics by Graham, Knuth, and Patashnik. It takes a real math guru to solve that one.

[quote]At the advanced level--difficulty level unbounded above--my favorite problem is problem 4.73 (page 152) in the book Concrete Mathematics by Graham, Knuth, and Patashnik. It takes a real math guru to solve that one.[/quote]

Would you mind posting the problem?

That looks really cool! I'm going to work on it!

For those who don't have the book, here's the problem:

73. If the Phi(n)+1 terms of the Farey series

Fn=<Fn(0), Fn(1), Fn(2), ..., Fn(Phi(n))>

were fairly evenly distributed, we would expect Fn(k):approx:k/Phi(n). Therefore the sum D(n)=:sum:(|Fn(k)-k/Phi(n)|,k=0 to Phi(n)) measures the ``deviation of Fn from uniformity.'' Is it true that D(n)=O(n1/2+epsilon) for all epsilon>0?

Phi(n)= :Sigma: ( :phi: (k), 1<=k<=n).

[hide]Unfortunately, as the answer in the back says, "This is known to be equivalent to the Riemann hypothesis".  So don't spend too much time on it.[/hide]

[Put in spoiler at tokenadult's request.]

Ah, I love that Putnam problem!  One of my favorites is also an easy Putnam question, number 2 from 2 years ago I think:

Given a sphere with 5 points on the surface.  Prove that 4 of these points lie in a closed hemisphere.

Hm if it's equivalent to the Riemann Hypothesis, I'll probably spend even more time on it. But at least I'll try not to allow it to keep me from doing my homework. I should have realized that. Almost anything with O(n1/2+epsilon) is equivalent to the Riemann Hypothesis.

[quote]Unfortunately, as the answer in the back says, "This is known to be equivalent to the Riemann hypothesis". So don't spend too much time on it.[/quote]

Hahaha, the reader must feel shafted or something at this point after having spent 3 hours on the problem. :mrgreen:

[quote]Given a sphere with 5 points on the surface. Prove that 4 of these points lie in a closed hemisphere.[/quote]

Yes, the solution to this one is VERY short. Pigeonhole.

Try wyra's, if you haven't seen it before -- it's a wonderful problem.

For wyra's problem:

[hide]If x is a divisor of n (other than 1 or itself) let a = (n-x)/x; b = x - 1; and c = 1.

The equation given in the problem can be expressed as:

a(b + c) + bc + 1 = n

by plugging in the values for a, b, and c we get:

a(b + c) + bc + 1

= (n - x)/x * (x - 1 + 1) + x - 1 + 1

= (n - x)x/x + x

= n - x + x

= n[/hide]

is that right?

Yes, although it's easier just to say [hide]let c = 1, and then we have n = ab + a + b + 1 = (a + 1)(b + 1), which is a general form for any positive composite number.[/hide]

Nice. I think I can actually see how to derive the solution to this additive number theory problem myself...unlike with some other problems. With some of the problems, the kalva site gives solutions that go somewhat like "well we obviously let a=(10^(3^n+1) - 1)/2 and b=2^n*10^(3^n-1), and we are done"  :mrgreen:

[quote="Ravi B"]Unfortunately, as the answer in the back says, ". . . ".  So don't spend too much time on it.[/quote]

Ah, Ravi, I was hoping you'd at least put that in spoiler. Maybe you still can, with an edit, so I omitted the spoiling line from the quotation above. That's why that is my favorite problem: the book poses the problem, and then the hints in the back give the punchline.

Why not rather spend a lifetime on it? It is a big problem, after all.

Quote:Yes, although it's easier just to say [hide]let c = 1, and then we have n = ab + a + b + 1 = (a + 1)(b + 1), which is a general form for any positive composite number[/hide].

Doh! -- That was the first thing i tried but I somehow found that it didnt work .... maybe this has something to do with the fact that I can barely see any white left on my scratch paper through all my random scribbling.  Oh well, nice problem

One of my favorites is from an old Putnam.  1998 B-2.

Given a point (a,b) with 0 < b < a, determine the minimum perimeter of a triangle with one vertex at (a,b), one on the x-axis, and one on the line y=x.  You may assume that a triangle of minimum perimeter exists.

Another one that I enjoy is: give a bijection between (0,1) and [0,1].

--Dan

In the same spirit as petra's second one,

Find a function from f from [0,1] to the real numbers so that f is continuous at all irrational points and discontinuous at all rational points.

It's a bit too well-known though.

ComplexZeta -- Yeah, if I hadn't known the answer to that one already, I'm not sure if I ever would have been able to guess it.  It is a cute function, though.

Ian -- your answer wasn't really all that different than mine -- all you did was essentially keep going a little bit.  It just wasn't necessary, that's all.

By the way, I think this is a great thread.

Aw, no one else is going to post here :(

Here is a nice, but rather open-ended sort of problem:

Given an integer n, consider the set [n] (the integers 1 to n).   Write down all subsets of [n] containing exactly k elements in a row, in lexicographic order - (that is, it comes earlier if it contains smaller elements: {1, 2} is before {1, 5} is before {2, 3}, for instance).  Write down, in a second row, the subsets of [n] of size k+1 in lexicographic order.  Then define the (partial) map f as follows:  consider every set c of size k in order (in the order they were written).  Let f(c) be the first set (first, again, in the order in which the sets were written) of size k+1 that contains c and is not crossed out, if such a set exists; then cross out this set.  Repeat with the next set c, etc.  For example:

n = 4, k = 2.

{1, 2} {1, 3} {1, 4} {2, 3} {2, 4} {3, 4}

{1, 2, 3} {1, 2, 4} {1, 3, 4} {2, 3, 4}

f({1,2}) = {1, 2, 3}
f({1, 3}) = {1, 3, 4}
f({1, 4}) = {1, 2, 4}
f({2, 3}) = {2, 3, 4}
f({2, 4}) and f({3, 4}) are not defined.

Find a "nice" description of the action of f(c).  More specifically, perhaps, find an O(n) algorithm for determining f(c), given n and the set c.

(Really Hard - as in, I don't know how to solve it - and even more open-ended): for what kinds of orderings does this algorithm yield the same function?

Bump.  Don't let this thread die!

The problem I gave is loads of fun, if you are willing to spend time on it.  :)

Here's mine:

A number of books are divided into m different piles (not necessarily all of the same size).  The books are then reorganized and mixed up in an arbitrary way so that they end up in n piles, where n < m.  Prove that there are at least m - n + 1 books that end up in a larger pile than the one they started in.

I just put one of my favorites atop the forum (AoPS challenge 17).  Sam Vandervelde came up with this one.

Nice problem, Richard.

Alison, in your problem, is n - m + 1 supposed to be m - n + 1?

Yes.  Edited.

[quote="Ravi B"]
Alison, in your problem, is n - m + 1 supposed to be m - n + 1?[/quote]


Let $ a, b, c$, be distinct positive real numbers. Prove that

$ \frac {a(b \plus{} c \plus{} 2a)}{(b \minus{} c)^2} \plus{} \frac {b(c \plus{} a \plus{} 2b)}{(c \minus{} a)^2} \plus{} \frac {c(a \plus{} b \plus{} 2c)}{(a \minus{} b)^2}\geq6$

[quote="Ligouras"]Let $ a, b, c$, be distinct real numbers. Prove that

$ \frac {a(b \plus{} c \plus{} 2a)}{(b \minus{} c)^2} \plus{} \frac {b(c \plus{} a \plus{} 2b)}{(c \minus{} a)^2} \plus{} \frac {c(a \plus{} b \plus{} 2c)}{(a \minus{} b)^2}\geq6$[/quote]

distinct real numbers?

don't need positive?

Dear Ligouras, your ineq is only trues with $ a, b, c \ge 0$  :wink: . It follows from 2 well-known result:
$ \sum \left(\frac{a}{b\minus{}c}\right)^2 \ge 2$ and $ \sum \frac{a(b\plus{}c)}{b^2\plus{}c^2} \ge 2$

[quote="nguoivn"]Dear Ligouras, your ineq is only trues with $ a, b, c \ge 0$  :wink: . It follows from 2 well-known result:
$ \sum \left(\frac {a}{b \minus{} c}\right)^2 \ge 2$ and $ \sum \frac {a(b \plus{} c)}{b^2 \plus{} c^2} \ge 2$[/quote]

Thanks my Friend