inequalities
geometry
incenter
inequalities unsolved

Let $ a,b,c$ be sides of a triangle. Show that:

$ 2(ab^2 \plus{} bc^2 \plus{} ca^2) \ge a^2b \plus{} b^2c \plus{} c^2a \plus{} 3abc$

[hide="My Progress"]
Okay guyz I got it down to this:

$ a(b \minus{} a)(b \minus{} c) \plus{} c(a \minus{} b)(a \minus{} c) \plus{} b(c \minus{} a)(c \minus{} b)\ge 0$.

There seems to be something obvious I am missing to make it into Schur.. :(

Another idea:

Let $ x \equal{} b/c$, $ y \equal{} c/a$, $ z \equal{} a/b$. This turns it into

$ 2(x \plus{} y \plus{} z)\ge (xy \plus{} yz \plus{} zx) \plus{} 3$

for $ xyz \equal{} 1$

But I don't think this is useful either... :(

It seems much more likely that I should sub in $ a \equal{} x \plus{} y$, etc. but I have not much luck here either.
[/hide]

Could someone give me a hint to follow up on either of these approaches? OR atleast a sign if one of them is a bad step and I should redo it?

Thanks :)

It's late at night, so I really may be missing something obvious :)

Your second approach yields an inequality that can be factored and rearranged into

$ (2\minus{}x)(2\minus{}y)(2\minus{}z)\le 1$

now can you prove this?

Hmm I cant figure that out, but I got something that might prove it...
[hide]
From triangle ineqs:
$ x\plus{}\frac{1}{y}\ge 1$
$ y\plus{}\frac{1}{z}\ge 1$
$ z\plus{}\frac{1}{x}\ge 1$

Adding these we get $ (x\plus{}y\plus{}z)\ge 3\minus{}(\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})\equal{}3\minus{}(xy\plus{}yz\plus{}zx)$ so we add this to $ 2(x \plus{} y \plus{} z)\ge (xy \plus{} yz \plus{} zx) \plus{} 3$ and we have $ 3(x\plus{}y\plus{}z)\ge 6$, but $ x\plus{}y\plus{}z\ge 3$ by AM-GM, so we are done?
[/hide]

Can someone check please?

what do you mean add the inequalities...thats clearly wrong. and also equality doesn't hold...

does $ \minus{} 3\ge 0$ hold... How about lets add $ 5\ge 0$ to both sides to get $ 2\ge 0$ so clearly $ \minus{} 3\ge 0$?!

Also why would the substitution: $ x \equal{} \frac {b}{c}$ help? This just makes the problem harder... You don't do random substitutions and hope that they work. You use a substitution because it makes the problem simpler.

You should do the $ a \equal{} y \plus{} z$ substitution because $ x,y,z$ are positive reals and it removes the tough condition that we have sides of a triangle.

Then it is equivalent to prove that

$ x^3 \plus{} x^2 y \minus{} 2 x y^2 \plus{} y^3 \minus{} 2 x^2 z \plus{} y^2 z \plus{} x z^2 \minus{} 2 y z^2 \plus{} z^3\ge 0$

which you should try and prove. (AM-GM!)

Oh I see my flaw  :blush:

Well the substitution wasn't random, I had played with it and it atleast make the problem statement [i]look[/i] simpler. That's why I suggested it...  :(

The fact that I did other things to make it look simpler was simply because I thought there would be a less of a messy solution than the large inequality you just posted. It isn't that hard with AM-GM, but I thought their would be a more elegant way  :)

Thanks anyway :)

[quote="cosinator"]Your second approach yields an inequality that can be factored and rearranged into

$ (2 \minus{} x)(2 \minus{} y)(2 \minus{} z)\le 1$

now can you prove this?[/quote]

Here's how I got here:
$ 2(x \plus{} y \plus{} z)\ge(xy \plus{} yz \plus{} zx) \plus{} 3$

$ \Leftrightarrow 4(x \plus{} y \plus{} z)\ge2(xy \plus{} yz \plus{} zx) \plus{} 6$

$ \Leftrightarrow \minus{} 8 \plus{} 4(x \plus{} y \plus{} z) \minus{} 2(xy \plus{} yz \plus{} zx) \plus{} xyz\ge \minus{} 1$

$ \Leftrightarrow (x \minus{} 2)(y \minus{} 2)(z \minus{} 2)\ge \minus{} 1$

$ \Leftrightarrow (2 \minus{} x)(2 \minus{} y)(2 \minus{} z)\le1$

Now here's what you do:
If the LHS is negative we're done.  If not, then either all terms are positive or only one is.  If all terms are positive, then by AM-GM

$ (2 \minus{} x)(2 \minus{} y)(2 \minus{} z)\le\left(\frac {6 \minus{} x \minus{} y \minus{} z}{3}\right)^3\le\left(\frac {6 \minus{} 3\sqrt [3]{xyz}}{3}\right)^3 \equal{} 1$

Now, suppose only one term is positive.  Without loss of generality, let $ 2\le x$, $ 2\le y$.  Then we have $ 2c\le b$ and $ 2a\le c$.  Thus, we have $ a\le c\le b$ with $ a \plus{} c\le2c\le b$ which contradicts the triangle inequality.

That's the clever solution I was looking for :). I almost had it :(. I really hate dumbassing stuff with the incenter substitution+...

Thanks a lot man :)

Once you had that it was $ \sum_{cyc}a(b\minus{}a)(b\minus{}c)$ it is very easy to apply the substitution...there is practically no expanding.

Since the difference $ a\minus{}b\equal{}(y\plus{}z)\minus{}(z\plus{}x)\equal{}y\minus{}x$, etc

That's very very true actually, but in the end you still have something of a mess to prove. I do not have much experience in inequalities, so I always shy away from large intimidating expressions. I would have done it eventually, but I wanted to see if what I had yielded anything :)

So, if I'm not mistaken:
Since $ a,b,c$ are the sides of a triangle, put $ a\equal{}y\plus{}z$, $ b\equal{}x\plus{}z$, $ c\equal{}x\plus{}y$, where $ x,y,z$ are not negative.
Expand everything, you get the inequality:
$ x^3\plus{}y^3\plus{}z^3\plus{}xy^2\plus{}yz^2\plus{}zx^2\minus{}2x^2y\minus{}2y^2z\minus{}2z^2x\ge 0$
Which is true, since from using $ AM\minus{}GM$ 3 times, you get:
$ x^3\plus{}xy^2\minus{}2x^2y\ge 0$ and $ y^3\plus{}yz^2\minus{}2y^2z\ge 0$ and $ z^3\plus{}zx^2\minus{}2z^2x\ge 0$


Here the problems of the Schweitzer math competition 2002
http://www.cs.elte.hu/~schw02/

oubi

It is a Hungarian open book exam for university students and you have ten days to complete it. There are typically 1-3 first prizes, 0-5 second prizes and 0-7 third prizes. The famous mathematician Bela Bollobas got an honourable mention in 1962, and a first prize in each of 1964, 1965 and 1966. It's used to discover research talents. There are two volumes.
The first book is out of print  :(  but may be obtained in some good university libraries. The second can be found in any English bookstore, e.g. http://www.amazon.com .  :)

(1)

Title Details: Contests in higher mathematics : Hungary, 1949-1961 :
in memoriam Mikls Schweitzer / editorial board : G. Szsz and others.
Publisher: Budapest : Akadmiai Kiad, 1968.
Subject: Schweitzer, Mikls.
Other Names: Szsz, Gabor.
pages: 260

(2)

Title Details: Contests in higher mathematics : Mikls Schweitzer competitions, 1962-1991 / Gbor J. Szkely, editor
Series: Problem books in mathematics
Publisher: New York : Springer, c1996
Physical desc.: vii, 569 p : ill ; 25 cm, New York : acid-free paper
ISBN/ISSN: 0387945881 (New York : acid-free paper)
0387945881 (hbk)
Note: Includes index
Subject: Mathematics - Problems, exercises, etc
Mathematics - Competitions - Hungary
Other Names: Schweitzer, Mikls, 1923-1945
Szkely, Gbor J., 1947-


Can someone help me in proving the following?  In all of these problems, all variables are considered positive real numbers.

1.) $a+b+c=1$.  Prove $(1-a)(1-b)(1-c) \geq 8abc.$

2.) $(ab+cd)(ac+bd) \geq 4abcd.$

3.) $x^5+y^5 \geq x^4y+xy^4 \geq x^3y^2 + x^2y^3.$

4.) $[(ab+cd)(ac+bd)(ad+bc)]^2 \geq (4abcd)^3$.

Any assistance is greatly appreciated.  :D

AM-GM!

1)[hide]
$(1-a)(1-b)(1-c) = (b+c)(a+c)(a+b)$

Apply AM-GM to each term and multiply them all together to get the result.

[/hide]

2) [hide]
Just multiply the terms out and use AM-GM on the resulting four summands. This gives the result. [/hide]

3)[hide]

Let's do the first part first:

$x^5+y^5 \geq x^4y+xy^4$

Multiply by 5:

$5x^5 + 5y^5 \geq 5x^4y + 5xy^4$

Conveniently split up the terms on the LHS

$(x^5+x^5+x^5+x^5+y^5)+(y^5+y^5+y^5+y^5+x^5)$

Apply AM-GM to each parentheses to get

$\geq 5x^4y+5y^4x$

Now the second part: Start similarly by multiplying by 3:

$3x^4y+3xy^4 \geq 3x^3y^2 + 3x^2y^3$

Split up terms again

$(x^4y+x^4y + y^4x) + (xy^4+xy^4+yx^4)$

Apply AM-GM to both to get

$\geq 3x^3y^2 + 3x^2y^3$

[/hide]

4 doesn't work, a=b=c=d=1/3 is a counterexample.
Maybe that first exponent is supposed to be a 2? If so then...

[hide]Applying AM-GM each term (ab+cd, ac+bd, and ad+bc) is greater than or equal to 2sqrt(abcd). Multiplying the three terms and squaring gives the result[/hide]

[quote="scorpius119"]Maybe that first exponent is supposed to be a 2? If so then...[/quote]

Yeah it is, sorry about that.  Thanks for the solutions!

For $3$, try using the

[hide]Rearrangement Inequality[/hide]


[quote="darkquantum"]The complex sequence z0, z1, z2, ... is defined by z_0 = i + 1/137 and z_(n+1) = (z_n + i)/(z_n - i). Find z_2002.[/quote]

this repeats with period 3.
let z_0 be a. Then z_1 by definition will be (a+i)/(a-i). z_2 will then be ((a+i)/(a-i)+i)/((a+i)/(a-i)-i)=(a+i+ai+1)/(a+i-ai-1). z_3 will be ((a+i+ai+1)/(a+i-ai-1)+i)/((a+i+ai+1)/(a+i-ai-1)-i) = (a+i+ai+1+ai-1+a-i)/(a+i+ai+1-ai+1-a+i) = (2a+2ai)/(2+2i)=a=z_0.
Therefore, it repeats with period 3, and z_2002=z_1=(2i+1/137)/(1/137)=1+274i.

I wonder why this was #12. I suppose it was meant to just intimidate people and that's it.

yea
its really not that hard once u actually do it
anyway
just keep going and see when it repeats


inequalities
search
inequalities unsolved

Let three nonnegative real numbers $a,b,c$ satisfy $a^{3}+b^{3}+c^{3}=3$. Prove that: \[a^{2}b^{4}+b^{2}c^{4}+c^{2}a^{4}\leq 3.\]

[quote="hungkhtn"]It's follow from VASC's most famous inequality. Now, we have \[P=b^{\frac{9}{2}}a^{\frac{3}{2}}+c^{\frac{9}{2}}b^{\frac{3}{2}}+a^{\frac{9}{2}}c^{\frac{3}{2}}\le \frac{1}{3}(a^{3}+b^{3}+c^{3})^{3}\] furthermore, by Holder inequality \[P.P.(b^{3}+a^{3}+c^{3}) \ge (b^{4}a^{2}+c^{4}b^{2}+a^{4}c^{2})^{3},\] and we are done.[/quote]I don't understand how you use Hölder here. There is a problem with the degrees. But it can be repaired.

If I'm right, Hölder gives us $P.P.(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}) \ge (b^{4}a^{2}+c^{4}b^{2}+a^{4}c^{2})^{3},$ doesn't it?
But we have $a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3}\le \frac13(a^{3}+b^{3}+c^{3})^{2}=3$ so it works. Brillant idea!

[hide]Typo in your first line, it should rather say$P\le\frac{1}{3}(a^{3}+b^{3}+c^{3})^{\boxed2}=3.$[/hide]
Note that interestingly enough,
$b^{\frac{9}{2}}a^{\frac{3}{2}}+c^{\frac{9}{2}}b^{\frac{3}{2}}+a^{\frac{9}{2}}c^{\frac{3}{2}}\ge b^{4}a^{2}+c^{4}b^{2}+a^{4}c^{2}$ does not hold (see http://www.mathlinks.ro/Forum/viewtopic.php?t=103761), so Hölder is really needed  :)  :)

Sorry, you are right. I'm so hurry, so I was edit it.

[quote="hungkhtn"]_________________
See here for SMV-theorem

http://www.mathlinks.ro/Forum/viewtopic.php?t=113248

Up to 1000 posts, I will leave :Sad:

[/quote] :yankchain: ?? rather give us something nice for your 1000th post  :lol: :harhar:

can you tell me more about those inequalities
What is VASC
thanks a little

[quote="nttjuvwamsncc"]can you tell me more about those inequalities[/quote]
You should search for them in this forum.

[quote="nttjuvwamnscc"]What is VASC [/quote]
Vasc is a member of this forum and he has many excellent inequalities. (And a book!)


Prove that among any 7 integers there exist three numbers $ a,b,c$ such that $ a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ac$ is divisible by 7.

Equivalently, $ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2$ is divisible by $ 7$.  Note that $ a^2 \equiv 0, 1, 4, 2 \bmod 7$, so a sum of three squares is divisible by $ 7$ if and only if either each term is divisible by $ 7$ or if the squares are congruent to $ 1, 2, 4 \bmod 7$.  The rest is casework.


[quote="BanishedTraitor"]Prove that for any positive numbers $a,b,c,d$

$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\geq \frac{2}{3}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131608

[quote="navid"]Use jensen inequality for :f(x)=1/x
By supposing: a+b+c+d=1.[/quote]
Can you please post your solution?

$x=b+2c+3d$ , $y=c+2d+3a$ , $z=d+2a+3b$ , $w=a+2b+3c$ then above equation can be written as:

$-\frac{5}{6}+\frac{1}{24}\sum_{cyclic}\frac{7y}{x}+\frac{z}{x}+\frac{w}{x}$ now use AM-GM.

[quote="navid"]HI :Dear Alexandr:
Im sorry because my:La Tex .was damaged post my solution by
word document here:[/quote]
Thank you very much, navid, it's a beautiful application of Jensen.
I solved this problem in the same way as Gabriel(fr).


Let triangle $ ABC$ . $ X,Y,Z\in BC,CA,AB$ such that $ \Delta ABC$ and $ \Delta XYZ$ are similar. $ S_{ABC}\equal{}S$. Find the minimum of $ S_{XYZ}$

[quote="mathangel"]$ \frac {1}{4}S$[/quote]
Why? :o  :o

Denote by $ H$, $ K$ the orthogonal projections of $ Y$, $ Z$ on $ BC$, respectively. And let $ L$ be the second intersection of the circumcircle of triangle $ XYZ$ with $ BC$.
We have $ \angle ABC\equal{}\angle XYZ\equal{}\angle ZLB$, it implies that $ ZB\equal{}ZL$, i.e. $ AK\equal{}KL$. Similarly, $ CH\equal{}HL$. Hence we conclude that $ HK\equal{}\frac{1}{2}\cdot BC$. And so, we get $ \frac{S_{XYZ}}{S}\equal{}\left(\frac{YZ}{BC}\right)^2\ge \left(\frac{HK}{BC}\right)^2\equal{}\frac{1}{4}$


Ken wants to fence in a square region that shares a side with his house, so only three sides of the square will be made by the fence. The area of the fenced-in region measures 900 square feet. How many linear feet of fencing does he need?

rt900=30 so the sides are 30, so 3 sides is 90.

As the region must be a square, but one of the sides is the house wall (no fencing), we need to fence three equal sides with one of them squared equaling $900$ square feet. Since we know that a square's area is $s^2$, with $s=$ the side length, we know that $900=r^2$, so $r=30$. We need to fence three sides with $30$ feet of fencing, so $3 \times 30 = \boxed{90}$.


Scientist have found a new type of active galaxy.

[url=http://www.astronomy.com/asy/default.aspx?c=a&id=5881]The website[/url]

[quote="Astronomy"]
According to popular models, AGNs are surrounded by a donut-shaped ring of material, which partially obscures our view of the black hole. Our viewing angle with respect to the donut determines what type of object we see. But team member Richard Mushotzky, also at NASA Goddard, thinks these newly discovered AGN are completely surrounded by a shell of obscuring material. "We can see visible light from other types of AGN because there is scattered light," says Mushotzky. "But in these two galaxies, all the light coming from the nucleus is totally blocked."
[/quote]

Very interesting.  :)  [i]Astronomy[/i] magazine is quite good.

More information:

http://www.scienceblog.com/cms/new-type-active-galaxy-spotted-13818.html

http://www.sciencedaily.com/releases/2007/07/070730123803.htm


You are given 2 bottles of equal volume having 2 clear liquids. You CANNOT open the bottles and CANNOT do any chemical experiment. Heating etc. is not advised for the chemical might react vigorosly to something. How will you find which bottle contains $ H_2 0 $ and which has $ H_2 S O_4 $ ??

How tough did you consider the question???

Darn thats a though one.
And im on my second year of chemistry in high school.

I'd actually love to find the answer for this.

Could you PM me the answer.

Unless its labeled on the bottle what is what, then i find it next to impossible to find out what is what, you're not even allowed to do experiments with the 2 liquids?

Don't worry.I'm sure somebody will come up. [hide]Actually it;s quite simple[/hide] I'll PM to you if I don't get any other response.!!!

[hide]
$H_2SO_4$ is denser than $H_2O$.  We have equal volumes.  Thus measure the mass.[/hide]

Exactly. It's that simple!!!

Does anybody know some excellent , tough questions like the bove or other related ones???

Doesn't anybody have any other good, similar questions as above???

Come on people. A bit of frustration is setting in!!! (Please don't mind)

I need such questions for a paper I am mahing. Can you please provide interesting , out of the box questions in Chemistry(may involve some good math - integration etc)!!!!

interesting heh? :huuh:
take a  look [url=http://www.funtrivia.com/quizzes/sci__tech/chemistry/chemistry_mix.html]here[/url]

The quiz is OK(probably because we do much greater detail in India) but I do find them good!!!

My question was a bit different. SOme tough (OUT OF THE BOX) questions???

you can always look at ICHO problems.. ;)

I may be repeating(and spamming - moderators) but I'm not asking for questions you'll find in a final exam of MIT or an Olympiad. I'm asking for question which might be asked in a quiz or a science interview. I want such questions for a quiz.

I also would appreciate if you can give me the link for ICho and other olympiad problems!!Thanks


I did USAMTS past two years. So, I just set my grade to 12 (I'm senior now) and that's it? Or is there any paper form I need to mail?

You need to resubmit [url=http://usamts.org/Forms/USAMTSEntryForm.pdf]this[/url] form

Thanks. I thought I needed to turn in Entry Form so thanks for confirmation.


Recently, I have become curious about the underlaying psychology of problem solving. My current theory is that practice is the main element to being a good problem solver once the concepts are mastered. I am also interested in how a problem is solved, and how other people do it, and what main methods there are. My method for solving problems is largely based on my experience. I know theorems and factorizations and techniques for cleverly munipulating things, and the first thing I do when seeing a problem is to apply this gathered intuition and look for solutions or something that would lead to a solution, following by an extened query of my mental database. If this doesn't work, I play around with the problem for 10 or 20 minutes, massaging the terms, looking at the why the specific problem is different from the general case, what makes this problem hard, and some other common strategies and techniques, like induction and adding lines to geometry problems etc.

Now, my method of problem solving gets interesting when we hit olympiad problems. These are unique, and specially made so they don't fall to quick, clever solutions. How does one, over an extended period of time, come up with clever solutions?

[b]Big Question 1: How are clever and unique solutions for olympiad problems generated? Are they just harder versions of problems on the AIME and AMC tests, where having a degree of logic and cleverness, though nothing earth shattering(like recognizing some terms are squares, or a the equations are almost like something you have seen before, etc, logical comon sense type stuff, nothing requiring any large IQ) and remembering practice problems, techniques, formulae, and above all, easily aquired intuition will get you through? Do they require a different method of thinking?[/b]

[b]Big Question 2: How do you solve problems?[/b]

[quote="probability1.01"]Huh? I think it is a mistake to think of solving a problem as something that can be done using this "toolbox" thing. While over time you can build up experience/intuition, you should have good ideas pop up no matter what kind of problem you face. It's kind of like art -- you can get better at technique, but that alone doesn't get you a "Starry Night."[/quote]
So how do you get these good ideas to pop up? Why and how do they pop up? That is sort of what I am concerned with.

It's because of fairies that exist in the 7th dimension that carry ideas and they randomly shift dimensions (a la easter bunny) into the 3rd dimension to collide with objects, thus transfering the ideas they hold.

yif man, I do believe you are referring to the White Mice

Personally? I'm not very good with problem solving, but
try everything you know,
and go with what seems to work

[quote="PenguinIntegral"][quote="yif man12"]No, it's different. Perhaps I should explain it more in depth.

Ok, there exists information carrying fairies in the 7th dimension that for the most part float around. They can shift dimensions like the easter bunny does on easter each year, and they transfer information by colliding with other objects. If a fairy that has a information particle that's useful to you hits you, you usually get part of the idea. That's why you need more than one to hit you to get the complete solution. But if too many fairies get offtrack and hit random object, it results in the object gaining heat from the information. That's one of the causes of global warming, random information fairies having too many collisions with objects.

One question that'd often asked is why do people get ideas more when they're taking tests? That's because high levels of concentration and thought attracts fairies to the vicinity, sorta like high reiatsu and hollows. That's why you really do get smarter if someone smart sits next to you. But if you're in a room with a lot of people, the information fairies you attract might go to other people. It works both ways, so learn to mooch of smarter people!

Wow, I just referenced the easter bunny, bleach, and global warming in one post. That sarcasm prize is SO mine again.[/quote]

This isn't sarcasm. This trying to create funny where there is no funny, and failing horribly.[/quote]

pwned.


Local communities serve several purposes.  Students can discuss problems and problem solving ideas, exchange information about local tournaments, work with their local ARML teams, and get to know other enthusiastic math students in their own area.

The discussion in these communities might also be less intimidating to some students who might be shy about public posting.

[quote="Silverfalcon"]I saw this new forum but I still don't understand what this really is. If you become a member of that state's forum, do you like able to see a lot of math problems or basically, [b]WHAT"S THE BENEFIT?[/b].

Can anyone give me a good explanation? Thanks.

Regards, :)[/quote]It's not really about only more math problems. You can talk about other things too, even organize local meetings between people lying in a near (geographical) location to you. Dunno, it's a new feature, you (users) must put it to a good use :P :-)

The national forums can be used in the local languages, so there's a clear purpose there.

[quote="MithsApprentice"][quote="Valentin Vornicu"]The national forums can be used in the local languages, so there's a clear purpose there.[/quote]

The foum can support different languages?[/quote]It will. As of now, it also (partially) supports German and (very soon) Romanian.

We plan to offer support for all the languages for which there exist forums (the ones in my signature), but we will need help from users in certain languages (Chinese, Korean, Vietnamese, Spanish, Hungarian, etc.) to make it ok.

South Carolinians use the forum to conduct online ARML practices using AIM as well.  They aren't really official yet, but they help a lot.  The All State coaches also occasionally post information and problems in the forum.  And we can also say what we want to people we know personally, and get a little bit more off topic than we would in, say, Getting Started or The Round Table.


3 non overlapping regular plane polygons, at least two of which are congruent, all have sides of length 1. The polygons meet at A in such a way that the sum of the 3 interior angles at a is 360. Thus the 3 polygons form a new polygon with A as an interior point.What is the largest possible perimeter that this polygon can have

[hide="Solution"] Let the original polygons have $a, b, b$ sides.  Then the sum of the interior angles is

$540-\frac{360}{a}-\frac{360}{b}-\frac{360}{b}= 360 \implies$
$1 = \frac{2}{a}+\frac{4}{b}\implies$
$ab = 2b+4a \implies$
$(a-2)(b-4) = 8$

We wish to maximize the perimeter, which is given by $a+2b-6$.  So let's maximize $b$.  At $b = 12, a = 3$ we have $p = \boxed{ 21 }$. [/hide]
[b]Follow-up:[/b]  What is the largest possible perimeter if we remove the constraint that two of the polygons must be congruent?

awww... I found the smallest, :oops:  I think:

[hide="solution"]Let's say that all of the polygons are congruent. That means that the polygons are hexagons. There are 12 sides showing, so 12.[/hide]


Consider a circle of any radius(immaterial) with centre $O$. Consider a point $X$ inside the circle. Also, let $P$ be a point on the circle. Join $OP$, $OX$ and $PX$. Let $PQ$ be a line perpendicular to $OP$ ( in the direction of velocity when particle is moving anticlockwise) . Let the acute angle (simple case ) between $OP$ and $OX$ be $\theta$. Also, let the angle between $PQ$ and $XP$(extended) (again acute) be $\alpha$. Prove, without using ANY calculus or variational principles or maxima/minima etc, that $\alpha$ is minimum when $\alpha = \theta$. i.e. when $P$ lies on perpendicular to $OX$ through $X$.

Instead of bumping up a thread after one day, please better clarify your problem. For in its current form, the angles $\alpha$ and $\theta$ can be more or less arbitrary, provided that $90^{\circ}\geq\alpha>\frac{\theta}{2}$, but this is probably not what you want. I think you have forgotten to give some conditions, particularly to say what is fixed and what is variable. Maybe you should just give the original physical problem where this geometry problem comes from.

  darij

Sorry, I was abit impatient.

What is fixed is basically the circle and the 'observer' point $X$. The physical connection was related to Doppler's effect. the source (Pt $P$) is moving around a circle and point $X$ is the listener. Basically, after doing all the physics, the problem reduces to the fact that $\alpha = \theta$ as defined above. Of course this is the simple case. Actually, in the absolutely general case, we have that, in order to meet the 'physics' of the problem, $P$ must lie on the perpendicular to $OX$ through $X$ ! ( I got this from calculus.)

I hope it is clear now. As far as I know, nothing seems to be missing


[quote="silouan"]Consider the positive real numbers $x,y,z$ so that $x^{2}+y^{2}+z^{2}=xyz$
Prove that $xy+yz+zx\geq 4(x+y+z)-9$[/quote]
Let $a=\sqrt{\frac{x}{yz}}$$b=\dots,c=\dots$
then the condition change to $\sum a^{2}=1$
the inequality changes to $\sum ab+9abc\ge 4abc(a+b+c)\leftrightarrow [510]-\frac32[411]-2[321]+[330]+\frac32 [222]\geq $ \[\leftrightarrow\sum (a-b)^{2}(ca^{3}+cb^{3}+ac^{3}+ab^{3}-2abc^{2}+\frac12 cab(a+b))\ge0\] Apply AM_GM $ca^{3}+cb^{3}+ac^{3}+ab^{3}-2abc^{2}\geq 2a^{2}b^{2}+2a^{2}c^{2}-2abc^{2}\geq 2abc^{2}$
I hope it is correct


Let $ ABC$ is a triangle, let $ AB\equal{}c, BC\equal{}a, CA\equal{}b, a\plus{}b\plus{}c\equal{}2p$. Prove that: $ cosA\plus{}cosB\plus{}cosC \le \frac{2p\minus{}a}{2p\plus{}a}\plus{}\frac{2p\minus{}b}{2p\plus{}b}\plus{}\frac{2p\minus{}c}{2p\plus{}c}$

$ \sum(\frac{2p\minus{}a}{2p\plus{}a}\plus{}1)\equal{}\sum\frac{4p}{2p\plus{}a}\geq \frac{36p}{8p}\equal{}\frac{9}{2}$
$ \equal{}> \sum\frac{2p\minus{}a}{2p\plus{}a}\geq\frac{9}{2}\minus{}3\equal{}\frac{3}{2}\geq cosA\plus{}cosB\plus{}cosC$


linear algebra
matrix
linear algebra unsolved

If a matrix $ A_1,A_2,\cdots,A_n$ statisfies that $ A_j^2 \equal{} A_j$ for all $ j$ and $ A_1 \plus{} A_2 \plus{} \cdots \plus{} A_n \equal{} I$, simplify the expression :
$ det(a_1.A_1 \plus{} a_2.A_2 \plus{} \cdots \plus{} a_n.A_n)$, where $ a_1,a_2,\cdots,a_n$ is a complex number. Thanks in advance.

It depends on the size of the matrices, which you haven't specified.

No he missed $ A_iA_j\equal{}A_jA_i$ and characteristic $ \equal{}0$.
Then the determinant is simply $ a_1\cdots a_n$.

Just a word about the "missing conditions":

If we are working over a field of characteristic zero, then by [url=http://www.mathlinks.ro/viewtopic.php?t=167879]this problem[/url], we have that $ A_j^2 \equal{} A_j$ and $ \sum A_j \equal{} I$ imply $ A_i A_h \equal{} 0$ for all $ i \neq h$. (That is, the commutativity condition isn't really missing.)

Using this, you can easily achieve simultaneous diagonalization, in which case the answer is something like $ \prod a_j^{\text{rank} \left( A_j \right)}$.

Thank you for your corrections.
I was looking in vain for a counterexample in characteristic $ 0$. :oops:


topology
function
group theory
abstract algebra
geometry
geometric transformation
real analysis

Suppose  $ (M,d)$ is a compact metric homogeneous space of the group $ G$, and define a semi metric on $ G$ by $ d'(g,h) \equal{} max_{t \in M}d(gt,ht)$ Then $ G$ is compact . Is there some simple explanation of this statement ? :maybe:

Just identify any two elements of $ G$ with the same action and make it a metric. Simplicity is good.

This is the topology of uniform convergence, which is a very sensible one for functions on a compact metric space. We would use uniform convergence on compact sets if the base space were merely locally compact.

The [url=http://en.wikipedia.org/wiki/Homogeneous_space]homogeneous space[/url] structure matters here, because we don't get that compactness with mere continuous functions, or even functions that form a group under composition. The functions $ x^{2^n}$ on $ [0,1]$ are an infinite discrete group.

OK, let's try an example. The group of linear fractional transformations $ f_{a,b}(z)\equal{}a\frac{z\minus{}b}{1\minus{}\overline{b}z}$ with $ |a|\equal{}1,|b|<1$ acts on the unit circle. The metric for the group is a mess, but the topology is that of a product of the open disk and a circle, as should be easy to see. It's definitely not compact. The proposition is false.

Thank  you jmerry  :) .
if I understand everything right , in your examples the groups is not homogeneous space (homogeneous space = i.e  group  act as isometry and also  the action of group on compact is transitive  ) and therefore they are not compact  in your examples ($ x^{2^{n}}$ and $ f_{a,b}(z)$) , Hence as you said
[quote="jmerry"]...The [url=http://en.wikipedia.org/wiki/Homogeneous_space]homogeneous space[/url] structure matters here... [/quote]
but anyway I don't see  proof why in homogeneous space  with semi metric,  group will be compact ..  :(

Isometry? There's nothing in the definition I was using about isometries. My first example was not transitive, of course. My second was a homogeneous space in the topological sense I was using.

All right, new thoughts. As before, we assume a faithful action.
Pick a point $ x_1$. We have a continuous map $ F$ from $ G$ to $ M$ given by $ F(h)\equal{}h(x_1)$. The preimages of each point $ p$ are cosets of the subgroup $ G_1\equal{}F^{\minus{}1}(x_1)$, and it seems reasonable to look for compactness in $ G_1$. Now, $ G_1$ acts faithfully on the compact set of points at distance $ d$ from $ x_1$; we can repeat the process on this set by choosing a point $ x_2$ and the group $ G_2\subset G_1$ that fixes it. This process may terminate wit enough points to determine everything by triangulation, or it may continue indefinitely with points $ x_i$ such that $ d(x_i,x_j)$ depends only on the minimum of $ x_i$ and $ x_j$.

Hmm, that sounds like a $ p$-adic structure. What can we do with $ M$ as the space of $ p$-adic integers? The most obvious isometries are translation by anything and multiplication by anything not divisible by $ p$. The former gives transitive coverage, but is compact; we need more.
Let $ G$ be the group of maps of the form $ f_{a,n}(x)\equal{}q^n x\plus{}a$, where $ q$ is some prime other than $ p$, $ n$ is an integer, and $ a$ is a $ p$-adic integer. We have $ f_{a,n}(x)\minus{}f_{a',n'}(x)\equal{}(q^n\minus{}q^{n'})x\plus{}(a\minus{}a')\le \max{|q^n\minus{}q^{n'}|,|a\minus{}a'|}$. If $ p\ge 3$, then we always have equality at either $ x\equal{}1$ or $ x\equal{}2$. The $ |q^n\minus{}q^{n'}|$ is simpler than it looks, and generally comes down to a multiple of $ |n\minus{}n'|$; for $ p\equal{}3,q\equal{}2$, we get that $ |q^n\minus{}q^{n'}|\equal{}\begin{cases}1&n\minus{}n'\text{ is odd}\\\frac13|n\minus{}n'|&n\text{ is even}\end{cases}$.
The resulting topology is the product of $ \mathbb{Z}_p$ and $ \mathbb{Z}$ with the $ p$-adic norm. The latter is not compact, since it is not complete.

Still false.

closed of $ G$ will be compact by Hausdorf Theorem. (if for all $ \epsilon$ there exist finite $ \epsilon$-net...)
indeed:
There is no improtant that the group act as  transitive , or that it's group . important is isometry .
Let's choose in $ M$ $ \epsilon$- net say $ \{x_{i} \}_{i\equal{}1}^{n}$ and in $ M^{n}$ say $ \{A_{i}\}_{i\equal{}1}^{m}$
we can do this since $ M$ is compact .
Cosnider : $ x_{0} \equal{} (x_{1},x_{2},...,x_{n}) \in M^{n}$
define $ gx_{0} \equal{} (gx_{1},gx_{2},...,gx_{n}) \in M^{n}$ for all $ g \in G$  (note that $ G$ act as isometry on $ M^{n}$)
consider all $ gx_{0}$ when $ g \in G$ we say $ g_{1} \equal{} g_{2}\equal{}g^{(j)}$ if there exist $ j$ such as $ g_{1} x_{0} \in B(A_{j},\epsilon)$ and $ g_{2}x_{0} \in B(A_{j},\epsilon)$
thus we hvae finite $ g^{(j_{1})}, g^{(j_{2})},...,g^{(j_{N})}$ where $ N \leq m$ and $ \{j_{1},...,j_{N}\}$belong to the set $ \{1,2,...,m \}$
then lets' prove that $ g^{(j_{1})}, g^{(j_{2})},...,g^{(j_{N})}$ will be $ const \epsilon$-net for $ G$
really consider $ g \in G$ then  there exist $ A_{j}$such as  $ gx_{0}\in B(A_{j},\epsilon)$.
take any $ x \in M$ then there exist $ i$ such as $ d(x,x_{i})<\epsilon$
thus :
$ d(gx,g^{(j)}x) < d(gx,gx_{i}) \plus{} d(gx_{i},g^{(j)}x_{i})\plus{}d(g^{(j)}x_{i},g^{(j)}x) <3\epsilon$
since $ G$ is isometry. done

Why do you insist on "proving" something for which you already have a counterexample? A metric space is compact if and only if it is complete and totally bounded. Your argument appears to show that $ G$ must be totally bounded. This is not enough. My example shows that $ G$ can fail to be compact by not being complete.

I absolutely agree with you   :)  . $ G$ is not compact.
in my "proving" I only note that  $ G$ always will be rather compactly (I don't know how it will be in English exactly ) . I mean $ Cl (G)$ will be compact I take $ Cl$ rather metric $ d$ ($ Cl$ is closure )

You mean the completion, not the closure. "Totally bounded" is the property you're looking for.


Ally has 50 dimes and 20 quarters. What percent of the value of her money is quarters?

[quote="professordad"]First, let's find the total value of Ally's coins.  We have $ 50 \cdot 10 \equal{} 500$ cents in dimes, and $ 20 \cdot 25 \equal{} 500$ cents in quarters.  Thus, Ally has 1000 cents in total.  Our final answer is $ \frac{500}{1000} \equal{} \boxed{\frac{1}{2}}$.[/quote]

[hide="Yes, but...."]
If you reread the question, the problem asks for percent, so our answer is actually $\boxed{50\%}$[/hide]


hello
I believe base is $ 10$.
u know two points of line that are $ (0,1)$ and $ (log3,3)$
so ur equation::
$ \frac{y\minus{}1}{x}\equal{}\frac{2}{log3}$
thank u


AMC
USA(J)MO
USAMO
geometry
3D geometry
tetrahedron
vector

I know this problem is probably easy to most, but it is the second olympiad problem I have started work on. Does anyone have a solution to 1973 USAMO Problem 1?
Problem:
Two points P and Q lie in the interior of a regular tetrahedron ABCD. Prove that angle PAQ<60^o.

Thanks everyone, I tried getting it into vector form and also tried toying with law of sines, cosines and slopes. Nothing seemed to work for me, maybe I'm approaching it incorrectly. Nevertheless, I promised myself that I'd work for at least 2 hours before asking for help.

If someone gets a solution, I'd prefer to them to post a hint or two and then the solution. It would help me and others much more.

And if anyone has any tips about preparing for the USAMO (I'm preping for #1 and #4 as thats what matters for Red), I would love to recieve it. (e.g. learn how to use these formulas or think about this in this situation.

Thanks,
Shri

[hide="OK, hint"]As for many 3-D geometry problems, it often helps to turn it into a 2-D problem. So... look for ways to do so.[/hide]

And a solution can be found [url=http://web.archive.org/web/20040405065644/http://www.kalva.demon.co.uk/index.html]here[/url].

lol, wow. I had a very similar solution at first, but I thought it wasn't rigorous enough... xD
And thanks, much appreciated.


AMC
AIME
USA(J)MO
USAMO
geometry
analytic geometry
AMC 10

So how did you all do on the 2009 AIME  I?

I failed.

Interestingly, I could solve 15 but not 5.

yea same for me
i got a 2 =.=

and so far i made FOUR careless mistakes :wallbash:

im starting to notice a pattern:
last year 7th grade i bombed amc 10 (i went from 99 (6th grade) [color=cyan]down to a 75 (7th)[/color] and way up to 138(8th grade))
this year i bombed the aime (2 in 8th grade)
so hopefully the pattern continues  :D and i fail usamo next year which means i still qualify and thats my goal which i failed to reach this year

*sigh* o well  :( theres always next year......and the rest of high school.....

I did a lot better on the practices though O.o which is why im mad
and im also mad that i made four careless mistakes which i didnt bother to check because they were on easy problems.  But even if i did get a 6 i dont think a 198 amc 10 index is high enough to qualify for USAMO.

EDIT: my bad, I swear you told me you got 14 of them...

[quote="DataBox"]How did you do Ubey?  15?

Yeah, I failed too...Two arithmetic errors  =([/quote]

no

Ahaha Taylor is funny. But you really, really should have taken AIME I, you know...

Grrr.

Computation error on #14.

Down to 10. And will be 10.

ughhh 204.5 amc 10 index might not qualify.. dang my crappy amc 10 score and the 4 dumb mistakes on aime.

[quote="junggi"]yo i slept at 4am that night
no way[/quote]
Oh hmm I guess that changes things.

waste your time trying to find out (and dont quote, it ruins the fun[hide]2 or 3, i forget[/hide]

I didn't quote, I viewed the source of the page.

Also, did anyone else feel that #15 was easier than #5?

Whoa, no? Actually I didn't work on #15 at all--drew a quick diagram, didn't see the solution immediately, and decided it wasn't worth my time...so I can't really judge. But #5 was really easy. I was already going to calculate an answer assuming LP was parallel to AM before I noticed the parallelogram.

Huh, thats sort of the opposite of what happened to me.

I got an epic fail of...

[hide=":("]
6
[/hide]

Meh, and I should have had at least 9.  I got 6 last year, and I hadn't even done any math outside of Algebra 1 at school....

The answer of #12 minus the question whose answer was 697. Or like, the same thing I got last year.

My answers were 1 off on 2 separate questions. One of them was a guess though, and I had about a 1/60 chance to get it right.

DataBox had me convinced for a few hours that I had guessed #5 right, but it turns out he had the wrong answer too. Oh well.

[quote] DataBox had me convinced for a few hours that I had guessed #5 right, but it turns out he had the wrong answer too. Oh well. [/quote]

And that is why when doing geometry problems, one must learn to make neat diagrams and not assume that triangles are similar... :rotfl:

darn so I felt that #4 was the hardest problem out of 1-12 (ignoring calculation time)

hmm maybe tied with 10 for longest time it took to find a solution

also 13 was pretty good but too easily guessable

14-15 were lame

lol i tried to coordinate bash number 4

that was amusing

attempting to take the square root of 10 digit numbers...:rotfl:

[quote="AIME15"]Heh.  On problem 5, I drew an un-neat diagram and assumed similar triangles, which gave me the correct answer in about 2 minutes :D[/quote]

On number 5, I drew an incorrect diagram, assumed stuff, and got the correct answer. However, I thought 72 was wrong.

hey guys
did you know that Harker has some pretty tricky indian kids

Failed - 6
I did 41+49=89 on 1 problem. That was lame.
lol and that was by luck too
I wrote out the factoring for 2009 and I plugged numbers in.

ugh. total AIME bomb this year.
[hide="please dont look"][hide="i said dont look"][hide="why are you looking?"]FAIL--8[/hide][/hide][/hide]

anyways. missed #1 and #5...  :mad:


[b](1) Let a and b be  two natural numbers. Can both a^2+4b and b^2+4a be  perfect squares?[/b]

Let $ a^2\plus{}4b\equal{}(a\plus{}2k)^2, \ b^2\plus{}4a\equal{}(b\plus{}2m)^2$, then $ b\equal{}k^2\plus{}ak>a, a\equal{}m^2\plus{}bm>b.$.


[quote="vipCD_A1"][b]Problem:[/b] (vipCD) For positive numbers  $ x,y,z$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that :

$ {\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2} \geq} \frac {3\sqrt [4]{3}}{2\sqrt {x^2 \plus{} y^2 \plus{} z^2} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}$

This problem is very interested and difficult. I hope to will have beautiful demob profit from you[/quote]

I don't think so,because your problem's wrong,try $ a\equal{}b\equal{}c\equal{}\frac1{3}$  :wink:

[quote="vipCD_A1"][b]Problem:[/b] (vipCD) For positive numbers  $ x,y,z$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that :

$ {\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2} \geq} \frac {3\sqrt [4]{3}}{2\sqrt {x^2 \plus{} y^2 \plus{} z^2} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}$

This problem is very interested and difficult. I hope to will have beautiful demob profit from you[/quote]
The problem must be
\[ {\frac {1}{(x \plus{} y)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(z \plus{} x)^2} \geq} \frac {2}{\sqrt {3(x^2 \plus{} y^2 \plus{} z^2)} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)}\]
Using the known
\[ (a\plus{}b\plus{}c)^2 \ge 3(ab\plus{}bc\plus{}ca)\]
we get
\[ \left( \frac{ab}{c}\plus{}\frac{bc}{a}\plus{}\frac{ca}{b}\right)^2 \ge 3\left( \frac{ab}{c}\cdot \frac{bc}{a}\plus{}\frac{bc}{a}\cdot \frac{ca}{b}\plus{}\frac{ca}{b}\cdot \frac{ab}{c}\right) \equal{}3(a^2\plus{}b^2\plus{}c^2)\]
Applying this inequality with $ a\equal{}y\plus{}z,b\equal{}z\plus{}x,c\equal{}x\plus{}y$, we get
\[ \frac{(x\plus{}y)(x\plus{}z)}{y\plus{}z}\plus{}\frac{(y\plus{}z)(y\plus{}x)}{z\plus{}x}\plus{}\frac{(z\plus{}x)(z\plus{}y)}{x\plus{}y} \ge \sqrt{3[ (x\plus{}y)^2\plus{}(y\plus{}z)^2 \plus{}(z\plus{}x)^2]} \ge [(x\plus{}y)\plus{}(y\plus{}z)\plus{}(z\plus{}x)] \equal{}2\]
\[ \Rightarrow \frac{1}{(x\plus{}y)^2} \plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(z\plus{}x)^2} \ge \frac{2}{(x\plus{}y)(y\plus{}z)(z\plus{}x)} \ge \frac{2}{\sqrt{3(x^2\plus{}y^2\plus{}z^2)} (x\plus{}y)(y\plus{}z)(z\plus{}x)}\]
With equality if and only if $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$.


If $ x \plus{} y \equal{} 7$ and $ x \minus{} y \equal{} 9$, find the product of x and y.

hello, adding the two given equations we get $ 2x\equal{}16$ or $ x\equal{}8$, from here we get $ y\equal{}\minus{}1$ so we find their product $ xy\equal{}\minus{}8$.
Sonnhard.


[color=darkred]Fie triunghiul $ABC$ pentru care notam aria $S=[ABC]$, perimetrul $2p$, circumraza $R$ si inraza $r$. Sa se arate ca :

$\boxed{\ a^{2}+b^{2}+c^{2}\ge bc+ca+ab+S\cdot\left(1-\frac{2r}{R}\right)\ }\ge bc+ca+ab$.

[u][b]Observatie.[/b][/u] Aici este compararea a "doi infiniti mici" in vecinatatea unui triunghi echilateral : $2S\cdot\left(1-\frac{2r}{R}\right)\le \sum (b-c)^{2}$.[/color]

Inegalitatea este echivalenta cu $p^{2}\geq 3r^{2}+12Rr+pr-\frac{2pr^{2}}{R}$.

Dar $p^{2}\geq 16Rr-5r^{2}$, deci ne ramane: $4R\geq 8r+p \cdot (\frac{R-2r}{R})$

$\iff 4R\geq p$, care e evident adevarata, caci $p\leq \frac{3R\sqrt{3}}{2}< 4R$. (s-a presupus ca $R>2r$.)

In inegalitatea propusa egalitatea are loc daca si numai daca $a=b=c$, i.e $R=2r$.


abstract algebra
algebra
polynomial
modular arithmetic
number theory solved
number theory

I've just seen a quite useful result:
let p>=3 a prime number and n>=2. Then (Z/p^nZ)*, the group of invertible elements in Z/p^nZ), is cyclique. Any proof  ? :)

This is simply the statement that $p^n$ has a primitive root. I'm pretty sure it's been proven before on the forum.

for those who don't want to search(the result is very classical):
by the structure theorem of finitely generated abelian groups, it is isomorphic to $\mathbb{Z}/d_1\mathbb{Z} \times \mathbb{Z}/d_2\mathbb{Z} \times ... \times \mathbb{Z}/d_n\mathbb{Z}$ with $d_1 | d_2 | ... | d_n$. it follows that all units $x\in \mathbb{Z}/p^n\mathbb{Z}$ fulfill $x^{d_n}=1$ in $\mathbb{Z}/p^n\mathbb{Z}$. but this polynomial has at most $d_n$ roots, so $d_n=p^n-p^{n-1}$ and we're done.

There's something I don't quite get: you say that $x^{d_n}-1$ has at most $d_n$ roots, but this only holds in fields, while $\mathbb Z/p^n\mathbb Z$ is not a field. I think I'm misunderstanding something here.

Here's another idea: first show that there is a primitive root $\pmod{p^2}$, and then show that if $a$ is a primitive root of $p^2$, then it's also a primitive root of $p^n,\ n\ge 3$.


I am trying to do long division and compile it, I still don't know how to arrange it in the correct places.   :wallbash_red: Is there a package that enables you to do it, or is it just plain obvious that I should wait for $\text{\LaTeX} 3\epsilon$?

This question has been asked before:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19859

Please do a search before asking a specific question like this.  All you had to do was search for "division" in this forum to get the above topic and two others.


Suppose you roll a die many times. Each time you roll you add the throw. You stop rolling the die whenever the accumulated throw exceeds 15. What is the probability that the accumulated throw is exactly 17? Is there any quick way of doing this problem? Thanks!

When do you stop throwing the dice? whenever you have just passed 17?

if so, defining $ P(n) \equal{}$ the probability of the accumulated throws is $ n$ and
$ p(n) \equal{}$ probability of having a $ n$ after one throw.
Clearly $ p(n) \equal{} \frac {1}{6}$ for $ n \in \{1,2,3,4,5,6\}$

then $ P(1) \equal{} p(1) \equal{} \frac {1}{6}$
$ 2 \equal{} 1 \plus{} 1 \equal{} 0 \plus{} 3 \Rightarrow P(2) \equal{} P(1) p(1) \plus{} p(2) \equal{} \frac {7}{36}$
$ 3 \equal{} 2 \plus{} 1 \equal{} 1 \plus{} 2 \equal{} 0 \plus{} 3 \Rightarrow P(3) \equal{} P(2) p(1) \plus{} P(1) p(2) \plus{} p(3) \equal{} \frac {7 \plus{} 6 \plus{} 36}{6^3} \equal{} \frac {49}{6^3}$
similarly for $ P(4), P(5),P(6)$ and after that the following recursion holds:
$ P(6 \plus{} n) \equal{} P(n) p(6) \plus{} P(n \plus{} 1) p(5) \plus{} \cdots \plus{} P(n \plus{} 5) p(1) \equal{}$
$ \equal{} \frac {1}{6} (P(n) \plus{} P(n \plus{} 1) \plus{} \cdots \plus{} P(n \plus{} 5))$

This was an ARML power contest question, I think, right? Either that or something close, I can't seem to remember.

Hmm.. For your solution, mszew,

we need $ P(17)\equal{}P(11)\plus{}P(12)\plus{}P(13)\plus{}P(14)\plus{}P(15)$ (note the lack of the 16)


calculus
integration
logarithms
calculus computations

i am looking for the integral of x^x as shown in the file....

I am not sure what you are asking.
Maybe this.
http://en.wikipedia.org/wiki/Sophomore%27s_dream

this is what i was asking for...thanks...double posting won't happen again...


calculus
integration
limit
real analysis
real analysis unsolved

just the answer not the proof
[hide] $e^{\frac{1}{3}}$ please write a proof ? [/hide]

are you trying to make people hurry, moubinool? :wink:

i have a pretty ugly solution, by now...
anyway, i'll post a sketch of it blanked, since i think there's a more elegant one...

[quote][color=white]so... consider f:[0,1]->R continuous. there exist f_k a sequence such that f_{k+1}<f_k for every n, that converges (uniformly, if needed) to f. now, for every fixed k, the limit you're asking for (with f_k instead of x^2) is the (weighted) geometric mean of e^f_k, so the exp(int_0^1 f_k). after working a bit on it, you can show all limits can commute, and actually show that the limit you look for is exp(int_0^1 f). hence your answer.[/color][/quote]

just write
$e^{\frac{x^{2}}{n}}= 1+\frac{x^{2}}{n}+B(x)(\frac{x^{2}}{n})^{2}$
where $B(x)$ is bounded on $[0,1]$. Integrate and use the fact that $(1+\frac{1}{3n})^{n}\to e^{\frac{1}{3}}$ ??

lol!
did anyone notice that i don't like taylor? :lol: :blush:


Prove that the equation $ x^3\minus{}3x\plus{}1\equal{}0$ has 3 solutions $ x_1<x_2<x_3$ satisfying $ x_3^2\equal{}x_2\plus{}2$

Let $ x\equal{}2\cos \theta$, we have...


OK, this game is:

You post a word in English. The next poster traduce the word to another language and post a new word in english. The next post traduce both words in another language and post another, and so on. If one day passes without posts, we start another "pyramid".

Let's begin!

1: Ice cream

Is Piglatin allowed? because I know n0o other languages.

Ice cream: Helado, мороженое
Yes: Дa
No:

We can use translators, do you know that? :wink: I didn't; I knew it off the top of my head.

[quote="1=2"]Ice cream: Helado
Yes: Дa
No:

We can use translators, do you know that? :wink: I didn't; I knew it off the top of my head.[/quote]

You must traduce Ice cream, too.

1. Ice cream: Helado, Dondurma
2. Yes: Дa, Evet
3. No: Hayir
4. Face:

1. Ice cream: Helado, Dondurma, Bin ja lin (how do you type chinese in the forum?)
2. Yes: Дa, Evet, oui
3. No: Hayir, non
4. Face:चेहरा
5. Sleep:

Go Google translator! (for the hindi, I knew the others)

[quote]Go Google translator! (for the hindi, I knew the others)[/quote]

You know turkish? O_O

1. Ice cream: Helado, Dondurma, Bin ja lin, Roomijs
2. Yes: Дa, Evet, oui, sí
3. No: Hayir, non, 不
4. Face:चेहरा, lian (san sheng)
5. Sleep: 睡觉
6. Assassinate:

Remember - All traductions in the same language *coughmathlete9999cough*

well 3 days have passed without a single post so in accordance with the rules

Eat:

Eat: ecT, ჭამა, manger
Hello:გამარჯობა, bonjour
mother: mère

Some of the languages don't show up properly on my computer.

Eat: ecT, ჭამა, manger ,: кушать
Hello:გამარჯობა, bonjour: приветствую
mother: mère : мама
earth

Eat: ecT, ჭამა, manger ,: кушать jesti
Hello:გამარჯობა, bonjour: приветствую zdravo
mother: mère : мама majka
earth zemlja
love

Eat: ecT, ჭამა, manger ,: кушать jesti comer
Hello:გამარჯობა, bonjour: приветствую zdravo hola
mother: mère : мама majka mamá
earth: zemlja tierra
love: amor
spin:

Eat: ecT, ჭამა, manger ,: кушать jesti comer , 吃
Hello:გამარჯობა, bonjour: приветствую zdravo hola, 你好
mother: mère : мама majka mamá , 妈妈
earth: zemlja tierra 地
love: amor 爱
spin:转
zygote:

Suggestion, post the language it is in.

In Danish - [b]På Dansk[/b]

Eat - ecT, ჭამა, manger , кушать jesti , comer , 吃 , [b]spiser[/b]
Hello - გამარჯობა, bonjour , приветствую zdravo hola, 你好, [b]Hej[/b]
mother - mère , мама majka mamá , 妈妈 , [b]mor[/b]
earth - zemlja tierra 地 , [b]jord[/b]
love - amor 爱 , [b]elske[/b]
spin - 转 , [b]spinde[/b]
zygote - [b]zygote (?)[/b]
book - [b]bog[/b]
word -


There are $ n$ piles of book, each pile with at least one book. Lauren comes along and rearranges the books into $ n\plus{}1$ piles, each with at least one book. Call a book "lucky" if it ends up in a pile with fewer books than it was before. Prove that there are at least two lucky books.

Assume the piles are always arranged in decreasing order of number of books.  Suppose that the old piles held $ \lambda_1 \geq \lambda_2 \geq \ldots \geq \lambda_n > 0$ books, while the new piles hold $ \mu_1 \geq \mu_2 \geq \ldots \geq \mu_n \geq \mu_{n \plus{} 1} > 0$ books.  Since $ \sum_i \lambda _i \equal{} \sum_j \mu_j$, we have $ \sum_{i\equal{}1}^n \lambda_i > \sum_{j \equal{} 1}^n \mu_j$.  Let $ k$ be minimal so that $ \lambda_1 \plus{} \ldots \plus{} \lambda_k > \mu_1 \plus{} \ldots \plus{} \mu_k$.  Minimality implies that $ \lambda_k > \mu_k$.  Every book that was in one of the first $ k$ piles originally began in a pile of size $ \lambda_k$ or larger.  If such a book is not now in the first $ k \minus{} 1$ piles, it is in a pile of size $ \mu_k$ or smaller, and thus it is lucky.  Thus, the number of lucky books is at least $ \lambda_1 \plus{} \ldots \plus{} \lambda_k \minus{} (\mu_1 \plus{} \ldots \plus{} \mu_{k \minus{} 1}) > \mu_k \geq 1$, and we're done.


trigonometry
limit
real analysis
real analysis unsolved


induction
quadratics
algebra
algebra proposed

Let $\{a_n\}_{n\geq 1}$ be a sequence with $a_1=1$, $a_2=4$ and for all $n>1$, \[ a_{n} = \sqrt{ a_{n-1}a_{n+1} + 1 } . \]
a) Prove that all the terms of the sequence are positive integers.

b) Prove that $2a_na_{n+1}+1$ is a perfect square for all positive integers $n$.

[i]Valentin Vornicu[/i]

Okay first thing to admit is that this problem (point a) ) follows from one implication of BMO 1986 / problem 3. I found this out after creating the problem. Also point b) is rather similar with BMO 2002 / problem 2.

However, given all these facts, the problem was still though enough for most of the students not to solve it or make mistakes at it.

The main idea to a) and b) is to find out that $a_{n+1}=4a_n-a_{n-1}$

Then we can prove $a_{n+1}^2-4a_{n+1}a_n+a_n^2=1$ by induction.
So $2a_na_{n+1}+1=(a_{n+1}-a_n)^2$.is a perfect square.

[quote="Flatter"]The main idea to a) and b) is to find out that $a_{n+1}=4a_{n}-a_{n-1}$[/quote]

How did you introduce this ? Please tell me ,Flatter or someone can help me this  :)

My solution also uses several inductions:

1. By induction show that $a_{n-1}^{2}\equiv 1 \mod a_{n}$ and $a_{n}^{2}\equiv 1 \mod a_{n-1}$ simultaneously. Deduce a).

2. By induction show that $a_{n+1}=4a_{n}-a_{n-1}$ and $a_{n}^{2}=4a_{n}a_{n-1}-a_{n-1}^{2}+1$ simultaneously. Deduce b).

I wonder how one comes up with such problems (a weird recurrence + initial terms = something is always integer/square)? Does it follow from some deeper theory which suggests more ingenious ways leading to soution than dumb induction (e.g. Pell equations or something)?

Hi, Xixas! I didn't solve the problem in the contest, but after that I found something interesting: different types of recurrence relations that are in fact equivalent.

(1) $x_{n}^{2}=x_{n-1}x_{n+1}+k$
(2) $x_{n+1}=2x_{n}+\sqrt{3x_{n}^{2}+k}\implies{x_{n+1}^{2}-4x_{n}x_{n+1}+x_{n}^{2}=k}$
(3) $x_{n+1}=4x_{n}-x_{n-1}$

In a contest you usually are given one of these recurrences, and must use the other  :lol: . eg. prove that all of the terms of a type 1 or 2 recurrent sequence are integer. Here are the methods:

(1)$\implies$(3): write (1) for $n,n+1$ substract, get ${\frac{x_{n}}{x_{n+1}+x_{n-1}}=constant}$. To determine the constant, use the initial conditions.

(1)$\implies$(3): write (3) in the form ${\frac{x_{n}}{x_{n+1}+x_{n-1}}=4}$ so ${\frac{x_{n}}{x_{n+1}+x_{n-1}}=\frac{x_{n-1}}{x_{n}+x_{n-2}}}$ equivalent to $x_{n}^{2}-x_{n-1}x_{n+1}=constant$.

(1)(3)$\implies$(2): substitute $x_{n+1}$ from (3) into (1).

(2)$\implies$(1)(3):  write (2) for $n,n+1$, consider it as a quadratic equation which roots are $x_{n+2},x_{n}$. Use Viete's relations.

[hide=My solution]
a) By calculating with small values, I claim that $a_{n+1}=4a_n-a_{n-1}$, then the statement trivially follows. I'll prove this by induction.

[b]Base case:[/b] For $n=2$, $a_2^2=a_1\cdot a_3+1\implies a_3=15=4a_2-1.$

[b]Inductive step:[/b] Assume that this holds for all positive integers less than or equal to $n$. Now, we first prove that for all postitive integers $n$, we have $a_{n+1}^2=4a_{n+1}a_n-a_n^2+1$ by... induction.
[list]
[*] [i]Base case:[/i] For $n=1$, we have $a_2^2=4=4\cdot 4\cdot 1-1+1$.

[*] [i]Inductive step:[/i] Assume that this holds for all positive integers less than or equal to $n$. Then
\begin{align*}
a_n^2&=4a_na_{n-1}-a_{n-1}^2+1\\
\iff 9a_n^2-6a_na_{n-1}+a_{n-1}^2&=8a_n^2-2a_na_{n-1}+1\\
\iff (3a_n-a_{n-1})^2&=2a_n(4a_n-a_{n-1})+1\\
\iff (a_{n+1}-a_n)^2&=2a_{n+1}a_n+1\\
\iff a_{n+1}^2&=4a_{n+1}a_n-a_n^2+1.
\end{align*}
[/list]
Ok now let's prove $a_{n+2}=4a_{n+1}-a_{n}$. Indeed, we have
\begin{align*}
a_{n+1}^2-1&=a_na_{n+2}\\
\iff 4a_{n+1}a_n-a_n^2&=a_na_{n+2}\\
\iff a_{n+2}&=4a_{n+1}-a_{n}
\end{align*}Done.

b) Indeed, since $a_n^2=4a_na_{n-1}-a_{n-1}^2+1$, hence $2a_na_{n+1}+1=(a_{n+1}-a_n)^2$.[/hide]


[b]Problem. [/b] Let $\{F_k\}$ be the Fibonacci sequence. Prove that the number $\sum_{k \geq 0} \frac{1}{k!F_{2^k}}$ is [i](a)[/i] irrational (trivial  :) ); [i](b)[/i]* transcendental (but not a Liouville number).

[b](Semi)-Joke.[/b] How about $\sum_{k \geq 0} \frac{1}{F_{2^k}}$?  ;)

Let's do the first step :)
[quote="vess"]Prove that the number $A=\sum_{k \geq 0} \frac{1}{k!F_{2^k}}$ is irrational[/quote]
Since $F_{2^k}\mid F_{2^{k+1}}$,  we easily get $k!F_{2^k}A$ is not an integer for all $k$.

[quote="Myth"][quote="vess"]How about $B=\sum_{k \geq 0} \frac{1}{F_{2^k}}$?[/quote]
And again $0<B-\sum_{k=0}^{n} \frac{1}{F_{2^k}}=B-\frac{b_k}{F_{2^k}}<\frac{c}{F_{2^k}^2}$ for all $k$, $b_k\in\mathbb N$. It is impossible if $B$ is a rational number.[/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

$B=1+\frac{\sqrt{5}}{x^2-1}=1+\frac{\sqrt{5}}{x}=4-x$, where $x=\frac{1+\sqrt{5}}{2}$.

:D  That's right! :P  :cool:  :rotfl:

The non-trivial thing is to prove the transcendence of $\sum \frac{1}{k!F_{2^k}}$. Along these lines, it has been shown that the sum $\sum_{n \geq 1} \frac{1}{F_n}$ of the inverses of all Fibonacci numbers is irrational (though it's not known whether that number is transcendental), but this is much more difficult than to prove that $\sum \frac{1}{k!F_{2^k}}$ is transcendental!

Here is another similar (but not so straightforward) problem (yet incomparably easier than that of the irrationality of the sum of the inverses of [b]all[/b] Fibonacci numbers):

[b]Problem. [/b] Prove the irrationality of the number $\sum_{k \geq 0} \frac{1}{F_{2^n + 1}}$. (Note: don't look for a simple closed form evaluation! It has been shown that this number is in fact transcendental!)

[quote="vess"][b]Problem. [/b] Prove the irrationality of the number $ \sum_{k \geq 0} \frac {1}{F_{2^n \plus{} 1}}$. (Note: don't look for a simple closed form evaluation! It has been shown that this number is in fact transcendental!)[/quote]
the index should be $ n$ :wink:

could you please write a proof :idea: I'm not familiar with suth things ...

[quote="Myth"][quote="Myth"][quote="vess"]How about $ B \equal{} \sum_{k \geq 0} \frac {1}{F_{2^k}}$?[/quote]
And again $ 0 < B \minus{} \sum_{k \equal{} 0}^{n} \frac {1}{F_{2^k}} \equal{} B \minus{} \frac {b_k}{F_{2^k}} < \frac {c}{F_{2^k}^2}$ for all $ k$, $ b_k\in\mathbb N$. It is impossible if $ B$ is a rational number.[/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

$ B \equal{} 1 \plus{} \frac {\sqrt {5}}{x^2 \minus{} 1} \equal{} 1 \plus{} \frac {\sqrt {5}}{x} \equal{} 4 \minus{} x$, where $ x \equal{} \frac {1 \plus{} \sqrt {5}}{2}$.[/quote]
Sorry, would you please explain more? I don't understand the first step   :blush:

But I think there is an identity $ \sum_{k\equal{}0}^n \frac {1}{F_{2^k}} \equal{} \frac{3F_{2^n}F_{2^n\plus{}1}\minus{}1\minus{}F^2_{2^n}}{F_{2^n}F_{2^n\plus{}1}}$   :roll:

I really don't know how I obtained the answer 3 years ago. I took some time to get a closed form for $ B$ again.

Note that $ F_{2^n}\equal{}\frac{x^{2^n}\minus{}(\minus{}1/x)^{2^n}}{\sqrt5}$, where $ x\equal{}\frac{1\plus{}\sqrt5}{2}$. So for $ n\geq 1$:
\[ \frac{1}{F_{2^n}}\equal{}\frac{\sqrt5}{x^{2^n}\minus{}(\minus{}1/x)^{2^n}}\equal{}\frac{\sqrt5}{x^{2^n}(1\minus{}(1/x)^{2^{n\plus{}1}})}\equal{}\sqrt5\sum_{k\equal{}0}^{\infty}\frac{1}{x^{2^n(2k\plus{}1)}}.\]
It is easy to see now that $ \sum_{n\equal{}1}^{\infty}\frac{1}{F_{2^n}}\equal{}\sum_{m\equal{}1}^{\infty}\frac{\sqrt5}{x^{2m}}\equal{}\frac{\sqrt5}{x^2\minus{}1}.$ It follows $ B \equal{} 1 \plus{} \frac {\sqrt {5}}{x^2 \minus{} 1}.$

Prove the irrationality of the number $ \sum_{n \geq 0} \frac {1}{F_{2^n \plus{} 1}}$.

could anyone give a hint how to deal with it ?


linear algebra
matrix
algebra
polynomial
induction
vector
quadratics

This is the well-known Vandermonde matrix,

\[ \left(\begin{array}{ccc}1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\end{array}\right).\]

Show that the matrix has the determinant $ (b\minus{}a)(c\minus{}a)(c\minus{}b)$.

The challenge is to do so conceptually without computing the determinant.

The first step is to show that the determinant is a polynomial in $ c$(or $ a$, $ b$) of degree $ 2$ and i'm stuck here.  The steps following which are left for the reader to figure out  :) .

I had to do this problem earlier this semester, only I had to prove it for the general case, which requires induction and some annoying algebra.  Do you have anything so far?

There is no algebra... Seen as a polynomial in $ c$, it has zeros in $ a$ and $ b$, and similar statements viewing it as a polynomial in $ a$ and $ b$ gives the leading coefficient $ b \minus{} a$.

Ok, you want to know why it is a second degree polynomial as well. Let $ d(x) \equal{} d_0 \plus{} d_1x \plus{} d_2x^2$. Suppose $ a,b,c$ are distinct. If multiplying the vandermonde matrix with the vector $ d_0, d_1, d_2$ gives $ 0$ then this means $ d$ has zeros $ a,b,c$ and hence $ d_0 \equal{} d_1 \equal{} d_2 \equal{} 0$. So the matrix is invertible when the $ a,b,c$ are distinct. Hence the determinant is not zero except when $ c\equal{}a$ or $ c\equal{}b$. So going over to the complex numbers, this shows the polynomial has degree 2

[quote="mao_zai"]This is the well-known Vandermonde matrix,
\[ \left(\begin{array}{ccc}1 & a & a^2 \\
1 & b & b^2 \\
1 & c & c^2\end{array}\right).
\]
Show that the matrix has the determinant $ (b \minus{} a)(c \minus{} a)(c \minus{} b)$.

The challenge is to do so conceptually without computing the determinant.

The first step is to show that the determinant is a polynomial in $ c$(or $ a$, $ b$) of degree $ 2$ and i'm stuck here.  The steps following which are left for the reader to figure out  :) .[/quote]

$ \left(\begin{array}{ccc}1 & x & x^2 \\
1 & b & b^2 \\
1 & c & c^2\end{array}\right)$

Informally, look at the first row. The $ x$ and $ x^2$ never multiply one another and hence the highest degree of x in the determinant expansion is 2.
This the polynomial is a quadratic. Clearly x=b and x=c are roots since they make two rows identical. And these are the only roots since a quadratic has at most two roots.Thus

$ \left(\begin{array}{ccc}1 & x & x^2 \\
1 & b & b^2 \\
1 & c & c^2\end{array}\right) \equal{} \lambda(x\minus{}b)(x\minus{}c)$

Do this with (a,c) and (a,b) and finally figure out the determinant

Alternately:  it's easy to prove that Lagrange interpolation works without inverting the Vandermonde matrix.  The form of Lagrange Interpolation immediately tells you what the inverse of the Vandermonde matrix is, so computing any entry of the adjugate matrix gives you the answer by induction (compute the top leftmost entry).

Kalle, isomorphism, how do u know that the determinant is a polynomial in $ a$, $ b$ and $ c$? I understand that the determinant as a function over $ a$, $ b$ and $ c$ has zeroes when $ a\equal{}b$, $ b\equal{}a$, $ c\equal{}a$ and nowhere else.

It's possible to notice the determinant is a polynomial through computation (since, say, only the first row contains $ a$).  But without computation?

Well, the determinant is always a multivariable polynomial in its entries. This is certainly not hard to prove. Now several entries are polynomial functions of the same variable, but when you substitute a polynomial into a polynomial you still get a polynomial.

Actually performing the computation is not the same thing as noting that the computation always produces a polynomial.  The second is an observation of the structure of the computation.


[quote="mcalderbank"]Soccor?? lmao. I follow soccer a lot, so I guess it is my favorite sport. I like playing ultimate frisbee though.[/quote]

hehe....i always spell that wrong for some reason

[quote="djimenez"]GoBraves,

(1) It is soccer and not soccor.
(2) I believe it has been already discussed in the forum.
(3) You forgot at least to mention Martial Arts. I mean, there are thousands of martial arts, but at least mention them as a group.[/quote]

1. I know its soccer, i just wasnt thinking when i wrote that.
2. I checked...nvr been discussed before
3. Thats in Other

soccEr all the way! woohoo! btw, this has been discussed in the "polls" forum. anyways, yes i love soccer. it is amazing. :)

Gah, how can you not mention HOCKEY?!  Shame on you!

speaking of non-mentioned sports, how about tennis???

i play a ton of sports, mostly tennis and track. i play basketball and soccer with friends, but tennis with my local team tennis and track in hs.

this is somewhat related (not really) but is curling a sport???

(the "sport" where the people roll that rock on the ice and other people have brooms and make a path for the rock to make it to some scoring section...)

[quote="mcalderbank"]usa vs. mexico today!!!!!!!!!!! world cup qualifier[/quote]

I KNOW! Really excited right now.

Anyways, I *love* soccer, especially DC United in the MLS (who won the MLS Cup 2004!).  :lol:

SHAME ON YOU!!! you should slap urself for not putting martial arts as a option!!!  that's why i voted other and it's almost the leader :D

elvenchamp777

[quote="mathgeek2006"]If you added Ultimate Frisbee, I believe that that would get a few more votes. . .[/quote]

yay for frisbeeness. Although ameem hates it, so w/e

I'm surprised Ameem didn't include Cricket

you know, cause we're indian

Either way, I choose basketball

[quote]------------------------------------------------------------------------dam ppl quit telling me to add stuff. If its not on there then choose "OTHER"[/quote]

no wonder other is winning  :D

elvenchamp777

[quote="djimenez"][quote="sky9073"][quote="mcalderbank"]usa vs. mexico today!!!!!!!!!!! world cup qualifier

they lost 2-1  :([/quote]
As a 3+year resident of the winning country... Yay! (runs for cover)

btw, that "other" option really needs to be expanded... I like badminton.[/quote]

QUE VIVA MEXICO!! (Or may be is that I like when a latinamerican country wins :D )

 :first:[/quote]

technically, mexico is SUPPOSED to win since they have a higher world rank, isn't that correct???

basketball all the way baby. anyone have a favorite for ncaa?? i was rooting for duke, but..... *sniff... they couldn't pull it off...

[quote="plokoon51"][/quote][quote="djimenez"][quote="sky9073"][quote="mcalderbank"]usa vs. mexico today!!!!!!!!!!! world cup qualifier

they lost 2-1  :([/quote]
As a 3+year resident of the winning country... Yay! (runs for cover)

btw, that "other" option really needs to be expanded... I like badminton.[/quote]

QUE VIVA MEXICO!! (Or may be is that I like when a latinamerican country wins :D )

 :first:[/quote][quote="plokoon51"]

technically, mexico is SUPPOSED to win since they have a higher world rank, isn't that correct???[/quote]

but USA has beat them in six of their last eight games. The game was played in Stadio Azteca, in Mexico, yet USA has never won there in a really long time

now and always soccer or [b]football[/b]

[quote="plokoon51"][quote="Rodolphe2005"]table tennis for me[/quote]

gosh, i almost forgot about table tennis...

Table Tennis is the best.[/quote]

Eitherway, I still think that there is nothing better than martial arts, specially not competitive martial arts  :D , the instructors there worry less on teaching you very fancy and stetical techniques, and teach you dirty tricks :ninja:

[quote="ankur87"]I'm surprised Ameem didn't include Cricket

you know, cause we're indian

[/quote]

I love cricket. But sometimes, since its so long, it gets tiring to me.  :( I dunno, that's just me.


limit
trigonometry
real analysis
real analysis unsolved

find this limit

$ \lim_{x\rightarrow \frac {\pi}{4} }\frac {tanx \minus{} 1}{2sinx \minus{} \sqrt {2}}$ x $ \frac {2cos^2x \minus{} 1}{\sqrt {tanx} \plus{} tanx \minus{} 2 }$

$ \lim_{x\rightarrow \frac {\pi}{4}} \frac {tanx \plus{} cosx \minus{} sinx \minus{} 1}{cos2x \plus{} sin2x \minus{} 1}$

$ \lim_{x\rightarrow 0^ \plus{} } \frac {tanx \sqrt {tanx} \minus{} sinx \sqrt {sinx}}{x^3\sqrt {x}}$

$ \lim_{x\rightarrow 0} \frac{sin^3x\plus{}sin^32x\plus{}sin^33x}{sin^4x\plus{}sin^42x\plus{}sin^43x}$

[hide="Number 3"]For small x, $ \tan x \equal{} x \plus{} \frac{x^3}{3}$, and so

$ \tan x \sqrt{\tan x} \equal{} (\tan x)^{3/2} \equal{} \left(x \plus{} \frac{x^3}{3}\right)^{3/2} \equal{} x^{3/2}\left(1 \plus{} \frac{x^2}{3}\right)^{3/2}$

$ \equal{} x^{3/2}\left(1 \plus{} \frac{x^2}{2}\right)$

In a similar fashion, for small x, $ \sin x \equal{} x \minus{} \frac{x^3}{6}$, and

$ \sin x \sqrt{\sin x} \equal{} (\sin x)^{3/2} \equal{} \left(x \minus{} \frac{x^3}{6}\right)^{3/2} \equal{} x^{3/2}\left(1 \minus{} \frac{x^2}{6}\right)^{3/2}$

$ \equal{} x^{3/2}\left(1 \minus{} \frac{x^2}{4}\right)$

Therefore,

$ \lim_{x \to 0^\plus{}} \frac{\tan x \sqrt{\tan x} \minus{} \sin x \sqrt{\sin x}}{x^3 \sqrt{x}} \equal{} \lim_{x \to 0^\plus{}} \frac{x^{3/2}\left(1 \plus{} \frac{x^2}{2} \minus{} 1 \plus{} \frac{x^2}{4}\right)}{x^{7/2}} \equal{} \frac{3}{4}$[/hide]

[hide="Number 4"]$ L \equal{} \lim_{x \to 0} \frac{x^3(1 \plus{} 8 \plus{} 27)}{x^4(1 \plus{} 16 \plus{} 81)} \equal{} \infty$[/hide]

[hide="Number 2"]$ \frac{\tan x \plus{} \cos x \minus{} \sin x \minus{}1}{\cos(2x) \plus{} \sin(2x) \minus{} 1} \equal{} \frac{\tan x \minus{} 1 \minus{} (\sin x \minus{} \cos x)}{\minus{}2 \sin x (\sin x \minus{} \cos x)} \equal{} \frac{2 \minus{} \tan x}{2 \sin x}$

Therefore, your limit becomes $ L \equal{} \lim_{x \to \frac{\pi}{4}} \frac{2 \minus{} \tan x}{2 \sin x} \equal{} \frac{1}{\sqrt{2}}$[/hide]

[quote="Carcul"][hide="Number 4"]$ L \equal{} \lim_{x \to 0} \frac {x^3(1 \plus{} 8 \plus{} 27)}{x^4(1 \plus{} 16 \plus{} 81)} \equal{} \infty$[/hide][/quote]

What? I haven't the faintest clue how you got x terms out of a sinusoidal function.


Hello Everybody ,,, !

I am a noobie in C programming ,,, !  :blush:
I only know baby basic knowledge of this High Level Language ,,, !

And my exams are on june ,,, next month ,,, ! i'm kinda tense about it ,,, !

It seems impossible for me to mug all about C from text books before its exams as I also have other subjects to prepare ,,, !

Guys ,,, you know reading from text book and then thinking about 'how' those function works ,,, takes kinda a long time ,,, ? Cuz my text books are like that ,,, it always hints me ,,, and tells 'why' and 'how' ,,, that baaad ,,, !!

Hence ,,, [b]I'm looking for a good C knowledge teacher to tutor me[/b] whenever I have such a doubts C ,,, !

Kindly private message me your ID if you are interested to teach me all about C before 20th June ,,, !

Thanks ,,, !

Oops ,,, !
It is supposed to be C ,,, !

[b]Mod[/b] ,,, please remove the '#' from this topic title ,,, !

[b] Point 1 [/b]

I think it would be highly beneficial for you if you adhere to the 'rules underlying English grammar'; It would be much better if you could make your posts more legible by using proper diction.

[b] Point 2 [/b]

Since AoPS isn't a "chat forum", it would be much better if you desisted from using "chat speak lingo". (say using - "cuz", "kinda" and other "wannabe" stuff)

The only reason I believe you hadn't received any substantive replies to your question was that you had grossly violated the aforementioned points (i.e. point 1 and point 2).

[quote]  Hence ,,, I'm looking for a good C knowledge teacher to tutor me whenever I have such a doubts C ,,, !  [/quote]

I would like to paraphrase this horrible sentence (it may slightly distort the original meaning) and it is as follows-- " Since I'm not proficient in 'C' language, I would like to hone my skills by seeking the  help of a tutor who is very proficient at 'C'. I also would like to resolve my doubts in 'C'.

[quote]Kindly private message me your ID if you are interested to teach me all about C before 20th June ,,, !  [/quote]

You should realize that AoPS is more of a 'math-oriented forum', so, maybe this isn't an appropriate place for finding tutors in 'C'. Of course, you can always post your programming problems in here, and we would be more than happy to post solutions, but I personally think the probability of finding someone here who is going to tutor you is pretty less.


function
floor function
calculus
integration
algebra unsolved
algebra

Find every function $ f: \mathbb{Q}\to\mathbb{Q}$ satisfy:
$ f(f(x)\plus{}x\plus{}y)\equal{}x\plus{}f(x)\plus{}f(y),\forall x,y\in\mathbb{Q}$.

[hide="This problem is yuck"]Let $ g(x) \equal{} f(x)\plus{}x$. The equation reduces to $ g(g(x)\plus{}y) \equal{} 2g(x)\plus{}g(y)$. In particular, for any $ a$ in the range of $ g$, $ g(a\plus{}y) \equal{} 2a\plus{}g(y)$. If there is an infinite sequence of positive rationals in the range of $ g$ that converges to $ 0$, then that equation implies that $ g$ is monotonic. Additionally, $ g(a) \equal{} 2a \plus{} g(0)$ for every $ a$ in the range of $ g$. By monotonicity and the fact that $ a$ can be arbitrarily small, $ g(x) \equal{} 2x \plus{} g(0)$ for all $ x$.

Otherwise, either $ g(x) \equal{} 0$ or there is a minimal positive element $ r$ in the range of $ g$. Let $ h(x) \equal{} g(rx)/r$. Then $ h(h(x)\plus{}y) \equal{} 2h(x) \plus{} h(y)$ just like $ g$, and the minimum positive element of its range is $ 1$. Then $ h(1\plus{}y) \equal{} 2\plus{}h(y)$ (*). Suppose $ s$ is some noninteger in the range of $ h$. If $ \lfloor s \rfloor$ is even, then applying (*) a few times gets an element of the range between $ 0$ and $ 1$, contradicting $ 1$ minimal. If it's odd, then we get an element $ t$ between $ \minus{}1$ and $ 0$. But now $ h(t\plus{}y) \equal{} 2t\plus{}h(y)$. Pick $ y$ such that $ h(t\plus{}y) \equal{} t$, and we get $ h(y) \equal{} \minus{}t$, between $ 0$ and $ 1$. Contradiction.

We have that $ h$ can only be an integer, and that $ h(k\plus{}y) \equal{} 2k\plus{}h(y)$ for all integral $ k$, rational $ y$. This is the best we can do: pick an arbitrary function $ h^\prime : ( \mathbb{Q} \cap [0,1) ) \rightarrow \mathbb{Z}$, and we can extend it to an $ h$ that works.

Reversing these transformation, the solutions of $ f$ are: $ \minus{}x$, $ x \plus{} c$, and the big mean set just described plus $ x$ and times some constant.[/hide]


abstract algebra
geometry
geometric transformation
reflection
complex analysis
complex analysis unsolved

Find all the conformal mappings between {$r_1<module(z)<r_2$} and {p_1<module(z)<p_2}.
I think it's a classic problem . I try to use the carathodory,but not succee :( d
Can anyone help me?

What do you mean by $\mbox {module}(z)$

If you mean by $|z|$, then We could prove that if $f: R\longrightarrow S$ is a conformal isomorphism form $R=\{z\in\mathbb C|r_1<|z|<r_2\}$ to $S=\{z\in\mathbb C|s_1<|z|<s_2\}$ then $\frac{r_1}{r_2}=\frac{s_1}{s_2}$ or $\frac{r_1}{r_2}=\frac{s_2}{s_1}$ also all of these mappings are Mobius mappings.

[quote="sam-n"]by reflection principle you can extend the conformal isomorphism to whole complex plane origin deleted but in neighbor of origin is bounded so it should be $az+b$[/quote]

[url=http://en.wikipedia.org/wiki/Schwarz_reflection_principle]Schwarz reflection principle[/url] can only be used when we know that our map is real-valued on the boundary, so it is not applicable in all cases.


s`a dat anu trecut mi se pare, la admitere la poli :
fie $x$ real cu $|x|>1$.
calculati
$\lim_{n\rightarrow\infty}\bigg( \sum_{k=1}^{n}\frac{k(k+1)}{2x^{k-1}}\bigg)$.

cred ca raspunsul este
$\frac{x^{3}}{(x-1)^{3}}$

yes, the answer is
$\frac{x^{3}}{(x-1)^{3}}$.
when i saw the problem, i used same notation, $y=1/x$, and i denoted the sum with $S_{n}$.
i computed
$\iint S_{n}\,dy\,dy$ because it gets very nice :) then i simply took two derivatives of the obtained result, and it was a looong thing there but most of the terms were $y^{n}$, and they go to $0$. all that remained was the desired result.


algorithm
combinatorics proposed
combinatorics

in a lake lives 2 species of unidimensional fishes, green and black fishes. the lenght  sum of  each species is 1.A fish is slim until hadnt eat another fish.Each fish gest can eat slim fishesof another species and the lenght of the eaten fishes musnt exced his leght. Prove it is olways possible that they can eat each other, such that the lenght  of the survivors is less or equal to 3/2.

thank yu iura,
i had proved it by induction.
i have introduced new fishes and said that they can eat no more than their lenght

[quote="iura"]Consider the following algorithm: Pick up the largest fish in the lake at the moment, and let it eat other fish of the other species, in descending order of length, while it can. Call it a "predator" fish, and the fish eaten "prey" fish.Repeat this step while we can. We stop when one of the species becomes extinct. Now call the fish which ate the last representatives of this species the "killer" fish. Let's prove the total length of prey at this moment is at least $\frac{1}{2}$, which solves the problem.

First remark: if the predator is not the killer, then the prey eaten by it has total length at least half of the length of the predator. Indeed, if one of the fish eaten has length more than half of the length of the predator, we are done. Otherwise, all fish that could be eaten by the predator at that moment had length less than half. Since eating one more fish is not possible because the limit would be exceeded, we conclude that the length of the fish already eaten must be more than half.

Next, look at the extinct species. Assume that the prey from this species has total length $m$. Then the predators from the species have length $1-m$, and they are not killers, so they ate fish of total length at least $\frac{1-m}2$. So the total length of the prey is at least $m+\frac{1-m}2=\frac{1+m}2\geq \frac{1}{2}$.[/quote]


1) If (-2,7) is the maximum point on the graph of y=-2x^2-4ax+k, what is k?

2) Jones runs at a speed of 12 miles per hour. Smith runs at a speed of 10 miles per hour. They start at the same place on a circular track 1/4 mile long and and run in opposite directions. How far has Jones run when they meet for the 20th time?

3) What is the maximal number of points of intersection of a fourth degree polynomial and a fifth degree polynomial?

4) Given a square, another square is formed by joining the midpoints of the sides of the original square. Starting with a square 2 inches on 1 side, one repeats the above process to get a sequence of 16 squares. What is the length of a side of the last square so formed?

5) If x and y are positive integers and x^2-y^2 is prime, what does x-y equal?

theone gets # 2

Numbers 3 & 4 are still unanswered.

That's correct, theone. Give me a few minutes to post some more or PM me if you don't want me to.

Um, how did you do 3? I got 10 this time (I thought the graph could be sideways the first time)

If you think about it, when it is a fifth degree, x^5, then it can be factored into at most 5 other polynomials. So at the fourth degree, x^4, it can only be factored into at most 4 other polynomials, so the most that can be in common is 4!

Overall, they travel a combined 5 miles. Jones runs 12 mph while Smith runs 10 mph. That means they run at a combined 22 mph. This means Jones runs 12/22 of the total distance, or 6/11. 6/11*5=30/11, or 2+8/11


AMC USA(J)MO USAMO AIME inequalities geometry

Hey, I was just wondering what score was needed this past year to qualify for MOP?

I think it was 21 on USAMO for 10th grader and up, and 7 for the freshmen.

Well you have to be top 12 as a senior, otherwise you're not a candidate for the IMO team. If you're not a candidate then why would they send you to MOP?

only a 7 for freshmen?? my mom told me I had to get 11 and I've been studying for the USAMO like crazy.  :mad:

normally its not even 7
sometimes less than 30 freshmen qualify, so all go

Talking about qualification scores on the USAMO is somewhat ridiculous. Unlike the AIME, where you really want to be safe and sometimes it's more important to ensure you didn't make silly mistakes than to go for those extra problems, on the USAMO there should be time for both.

If qualification is even an issue for you, odds are #3 and #6 will not be too approachable (NOTE: This is NOT always the case, see USAMO 2005 #3, IMO 2004 #6 (ridiculously easy), or USAMO 2003 #3, which got me into MOP that year). The point is, you can only check your solution to #1 or #4 so many times before you're pretty much done. Go on to #2, or #5. It's not easy to judge USAMO scoring without a decent amount of experience, experience that usually comes with going to MOP. Therefore, it's probably more important to actually keep going for the whole 4 1/2 hours (far easier said than done) than to worry about qualification scores. I mean, you want the highest score you can get right?

Finally, saying "only a 7 for freshmen?", is a little ridiculous as well. See USAMO 2004, for a good example of a test that would have been quite difficult for freshmen, in my opinion. The "easiest" problem that year might have been #4, and not #1, since #4 was a problem you needed no theorems to solve, and even then it was not easy to see the "mirror match" style solution that was most elegant.

[quote="MysticTerminator"][quote="darktreb"]NOTE: This is NOT always the case, see USAMO 2005 #3, IMO 2004 #6 (ridiculously easy)[/quote]

IMO05/3, IMO05/6 ... ;)[/quote]

Well... they were easier than most questions 3 and 6 but I wouldn't call them RIDICULOUSLY easy. Look at how many solved those question: not thát many...

What do you think the cutoff is going to be this year for blue group? Is it still going to be around 3 problems(19-23) or will it go up(various reasons for this possibility, WOOT is going to be one of them for sure)?

[quote="beta"](for blue group)[/quote]

what's that? :?

bluegroup = nonfreshmen, nonwinners (there are freshmen who qualify normally, but it's not the "freshmen" group)

I think cutoff will be ~ the same

[quote="sr6622"]so, darktreb, what if you were an honorable mention as a senior? Would you then get to go to MOP even though you were not in the top 12?

Like for this year if a senior had a 26, he would not be invited to MOP and would not take the TST?[/quote]

I believe they go to MOP, but then they get to leave early or something. There are people who went to MOP as HM.

Of course, some senior HMers didnt get to take TST. Mildorf was an exception ----- he got the top scoring HM, and took TST becuase there were Canadians in the top 12, so he took their testing spot.

No. You don't actually get to go to MOP as a senior Honorable Mention (think about it, it DOESN'T MAKE SENSE).

You get to go to MOP as a senior if you're TOP TWELVE. But since Canadians were Winners this year, the "winners" and the "top 12" are two different things. So technically there were a couple of "top 12ers" who were also technically "honorable mentions". But they were top 12 out of US students.

There was absolutely no leaving early. This isn't nearly as complicated as we've managed to make it sound.

ohhh.....I see.......

Just wanted to know....

hhhmmm....thanx darktreb, I'm glad it seems like ur not in a bad mood  :roll:

[quote="sr6622"]thanx darktreb, I'm glad it seems like ur not in a bad mood  :roll:[/quote]

lol :rotfl: I think he just wanted to drive home a point that it seemed like you guys weren't getting :)

[quote="darktreb"][quote="MysticTerminator"][quote="darktreb"]NOTE: This is NOT always the case, see USAMO 2005 #3, IMO 2004 #6 (ridiculously easy)[/quote]

IMO05/3, IMO05/6 ... ;)[/quote]

I didn't include those because in general, freshmen can't solve Inequalities (which are a relatively simple subject but one that REQUIRES training/experience, as opposed to certain other subjects), and because three out of the six people on this year's IMO team that still placed second in the world were unable to produce full solutions to this year's #6. (Though I do admit I found it pretty straightforward, but once again having a decent background of standard techniques helps). The aforementioned problems (2003 #3, IMO 04 #6, 2005 #3) didn't require such a background necessarily. (in other words, simple experimentation can go a loooooong way)[/quote]

good point. USA05/3 does require some standard geometry knowledge though, I guess. Eh, probably not that much.

Worrying too much about a qualifying USAMO score at any level is silly, because the difficulty of the test (and the balance between easy/hard questions) can vary dramatically from year to year (probably more so than AMC/AIME), as you can see from recent statistics.

I really didn't know Alison, I'm rather pathetic

Caps does not necessarily equate to "bad mood". Caps could be HEY LOOK AT THIS, PAY SPECIAL ATTENTION TO THIS BECAUSE IT'S IMPORTANT, if you know what I mean.

And Alison has it exactly right. What's the point of worrying about scores? You want to do as well as you can, it's not like in AMC where if you're a sophomore or below you don't want to throw away a qualification for AIME.

First of all, darktreb, it was a joke....As in it was meant to make you laugh and ease up. Obviously, it didn't work.

Second, I was curious and there is nothing wrong with that. I just wanted to know what the mark was.

Seems like you still haven't read what I said in either post.

Hahaha, this is rather amusing but let's stay on topic guys.... :lol: As has been mentioned before I wouldn't worry too much aobut USAMO scores since the questions and question difficulty vary from year to year and from student to student. With that said, I think solving problems in general helps you develop THINKING skills (haha)  as well as familiarizes you with certain approaches that may come in handy when faced with a certain type of problem. Anyway, I think I am going to do some preparation for AIME so I can be certian of making it to USAMO rather than prepping for USAMO and then screwing up on the AIME because of carelessness, etc.

ok I'm done darktreb.... i read your post.....I'm glad you can caps lock...... :D


continued fraction
number theory proposed
number theory

Show that there exists a positive constant $ c$ with the following property: To every positive irrational $ \alpha$, there can be found infinitely many fractions $ \frac{p}{q}$ with $ (p,q)\equal{}1$ satisfying
$ \left|\alpha\minus{}\frac{p}{q}\right|\le \frac{c}{q^2}$

I wonder why they put such a well known theorem as a contest problem...
To solve it with $ c\equal{}1$, take a continued fraction of $ \alpha$  and $ \frac pq$ as any of the corresponding rational approximations.
The best possible $ c$ is $ \frac{1}{\sqrt 5}$, which one gets if one chooses the right out of three continued fraction approximations.

That best $\dfrac {1} {\sqrt{5}}$ is known as Hurwitz' theorem; see [url]http://www.artofproblemsolving.com/Wiki/index.php/Rational_approximation#Hurwitz.27s_theorem[/url].

[quote=ZetaX]I wonder why they put such a well known theorem as a contest problem...
[/quote]
Maybe in 1950 it wasn't so well known among undergraduates as it's now.
https://en.wikipedia.org/wiki/Dirichlet%27s_approximation_theorem


Let $ x$ and $ y$ be positive real numbers such that $ x \plus{} y\le 3$ and $ xy\ge 2$.

a] Determine the maximun value of $ xy(x\plus{}y)$

:)

hello, by AM_GM we have $ x\plus{}y\geq2\sqrt{xy}$ and since $ xy\geq2$ we get $ xy(x\plus{}y)\geq 2\sqrt{(xy)^3}\equal{}4\sqrt{2}$.
Sonnhard.

Thanks Dr Sonnhard Graubner, I suddenly noted that for the question that was on the post before.

I have changed the question.

:)

hello, we have $ xy\le \left(\frac{x\plus{}y}{2}\right)^2$ and from here we get $ xy(x\plus{}y)\le\frac{(x\plus{}y)^3}{4}\equal{}\frac{27}{4}$.
Sonnhard.