inequalities
function
inequalities proposed

If $a,b,c$ are positive numbers such that $a+b+c=3$, then
$8(\frac 1{a}+\frac 1{b}+\frac 1{c})+9 \geq 10(a^2+b^2+c^2)$.

It is nice with numbers $8,9,10$, but when does equality sign occur? Vasc

It is the solver task to find the equality case.

equality holds when $a=2, b=c=\frac{1}{2}$
We can use mixing variable method.

Assume $a \ge b \ge c$, $f(a,b,c) = 8(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})+9 - 10(a^2+b^2+c^2)$
Case 1: $abc \leq \frac{4}{5}$
It can be checked that $f(a,b,c) \ge f(a,\frac{b+c}{2},\frac{b+c}{2})$
${f(a,b,c) - f(a,\frac{b+c}{2},\frac{b+c}{2}) \\ = 8\left(\frac{1}{b}+\frac{1}{c}-\frac{4}{b+c}\right)-5(b-c)^2 \\ = \frac{8(b-c)^2}{bc(b+c)} -5(b-c)^2 \\ \ge \frac{8(b-c)^2}{2abc} - 5(b-c)^2} \ge 0$
since $abc \leq \frac{4}{5}$
Case 2: $abc \ge \frac{4}{5}$
It can be checked that $f(a,b,c) \ge f(a+b-c,c,c)$
$f(a,b,c)-f(a+b-c,c,c) \\ = 8\left(\frac{1}{a} + \frac{1}{b} - \frac{1}{a+b-c} - \frac{1}{c}\right) - 10(a^2+b^2-(a+b-c)^2-c^2) \\ = \frac{-8(a+b)(a-c)(b-c)}{abc(a+b-c)} + 20(a-c)(b-c) \\ = \frac{2(a-c)(b-c)}{abc(a+b-c)} (-4(a+b)+10abc(a+b-c)) \\ \ge \frac{4(a-c)(b-c)}{abc(a+b-c)} (-4(a+b)+5abc(a+b)) \\ \ge 0$
since $abc \ge \frac{4}{5}$

This means we can assume the smaller two variables to be equal
After expansion, we see that this is equivalent to $(2b-1)^2(10b^2-25b+16) \ge 0$ which is true.

[quote="Soarer"]Vasc, may I know your solution?[/quote]I created this inequality as an application to the following statemets in Algebraic Inequalities book:

[b]Left Concave - Right Convex Function Theorem [/b](LCRCF - Theorem). Let $a<c<b$  be real numbers,
and let $f$ be a continuous function on $I=[a,b)$, concave on [a,c] and convex on $[c,b)$.
If  $x_1,x_2,...,x_n \in I$ such that  $x_1+x_2+...+x_n=S=constant$, where  $S<(n-1)c+b$, then the expression
 $E=f(x_1)+f(x_2)+...+f(x_n)$
is maximal for $x_1=x_2=...=x_{n-1}\leq x_n$.

[b]Single Inflexion Point Theorem[/b] (SIP-Theorem). Let $f$  be a twice differentiable function on $R$  with a single inflexion point, let $S$  be a fixed real number and let $g(x)=f(x)+(n-1)f(\frac{S-x}{n-1})$.
If $x_1,x_2,...,x_n$ are real numbers such that $x_1+x_2+...+x_n=S$, then
       $\inf_{x\in R}{g(x)} \leq f(x_1)+f(x_2)+...+f(x_n) \leq \sup_{x\in R}{g(x)}$.

What is the proof of the first theorem??? :?


[quote="micromath"]Somehow i never mastered this and in competitions like MC it can be a big problem.
Well can someone explain the basics of sci. notation and also help me with this problem?
Problem: What is 0.0005407 in Scientific Notation?

Thanks in Advance  :D[/quote]
$ 5.407\cdot10^{ \minus{} 4}$

[quote="7h3.D3m0n.117"]The idea with scientific notation is to write it in a form of $ a \plus{} 10^b$, where $ 1 \le a \le 9$ and $ b$ represents the power of ten.[/quote]

Not exactly...

It's $ a \times 10^n$ (I like $ n$ better), where $ 1 \le a < 10$, and, yeah, your exponent is the power of 10.

[quote="distracted523"]It's $ a \times 10^n$ (I like $ n$ better), where $ 1 \le a < 10$, and, yeah, your exponent is the power of 10.[/quote]

$ 1 \le a \le 9$ and $ 1 \le a < 10$ is basically the same thing...

9.5 says hello. And you don't add, you multiply. Two mistakes, D3m0n...

Wow didn't know why I put the addition sign.  And yes, I do see the nice counterexample. Probably my 10th stupid mistake this week.  :blush:

Just make sure your scientific notation is in the correct format; it'll help you save time for calculations and if they're incorrect, your answer will be too.

Scientific notation is pretty easy to under stand.  Say we have $ 5.4\bullet10^2$.  This is the same as $ 540$.  Here is why:

We take 5.4 and basically multiply it by $ 10^2$, or 100.  So we get 540.  Or we could do this: The exponent is 2, so we would move the decimal point right two times.  You will see that the second time you can't move it, so we add a zero.

Try this:

What is 567000 in scientific notation?

[hide="Answer & Solution"]  In scientific notaition, this is how the thing should look: $ a.b\bullet10^c$.  There has to be one digit left of the decimal when we have a.b, and as many numbers right of the decimal point.  So in this problem, we would have 5.67 as a.b.  Then to figure out c.  We take 5.67 and make it the real number, only with the decimal point: 5.67000, we have to move  the decimal point right 5 places to make that number an integer.  So 5 would be c, making the answer :$ 5.67\bullet10^5$[/hide]

To make an scientific notation a rational number, we must do this:

Take $ 5.67\bullet10^5$, our previous example.  We put 5.67, then take the decimal poitns 5 points to the right.  We get: 567000!!

If c is a negative number ($ 10^{\minus{}3}$), then instead of moving the decimal right, we move it left!  This, thus, makes the answer a decimal.  Please reply if not understood, or if there are any mistakes in this.

[hide]
567000 in scientific notation.
The number multiplied by
a power of ten has to be
less than ten, and at least
one.

5.67*10^5
[/hide]

Guys, that's not the problem.
THIS IS THE PROBLEM!
[quote]Problem: What is 0.0005407 in Scientific Notation?
[/quote]

[quote="micromath"]
Well can [b]someone explain the basics of sci. notation[/b] and also help me with this problem?
[/quote]

yeah, dude, [b]read[/b] before saying something is unrelated to the topic

Yeah, and that's the problem, not "express 54600000 in scientific notation."
Which is what everyone else seems to be solving.
So I think I'm missing something.

[quote="istos"]Sorry for getting off topic.
ajain asked a question,
and so I answered it.
I think we can lock this
topic now?[/quote]

That question wasn't supposed to be answered.  It was a sample question to explain micromath how scientific notation works.  Thats why I provided the answer and a full explanation!!!

Mods: Please lock

The idea of scientific notation is to be able to write very large

(or small) numbers easily. In Scientific notation, we can write

$ 35000000000$ as $ 3.5 \cdot 10^10$. This is the same as just

moving the decimal point $ 10$ places to the right. We can also

do $ .00000000035$ as $ 3.5 \cdot 10^{\minus{}10}$. All numbers in

scientific notation are in the form $ a \cdot 10^b$, where

$ 0 < a < 10$, and b is an integer.

So, $ .0005407$ would be $ 5.407 \cdot 10^{\minus{}4}$. If you don't see

where this is coming from, ask or PM me, and I'll try to explain

more. $ 10^{\minus{}n}$ means that we move the decimal point $ n$

to the left. $ 10^{n}$ means that we move the decimal point $ n$

places to the right.


algebra
polynomial
probability
expected value
algebra open

A monic polynomial $P(x)$ of degree $n$ takes on the integer values $1 \le P(x) \le n$ for $1 \le x \le n$ - in random order.  What is the expected value of $P(n+1)$?

I am not sure if this problem has a simple solution, or a solution at all  :|

Note that $n + 1 - P$ has the same distribution as $P$.  Thus the expected value of $P(n + 1)$ is $\frac{n+1}{2}$.

Are you sure?  This doesn't hold for $n=1$...

The problem seems to have changed since I last saw it.  I will think about the revised problem.

Okay, here is the answer to the revised problem.  Define the polynomial
\[ Q(x) = n + 1 - P(x) + 2(x-1)(x-2) \ldots (x-n) \, .  \]
Note that $Q$ is a monic polynomial of degree $n$ with the same distribution as $P$.  Thus the expected value of $Q(n+1)$ is the same as the expected value of $P(n+1)$.  If we plug in $x = n+1$ to the equation above, we get
\[ Q(n+1) = n + 1 - P(n+1) + 2 n! \, .  \]
Taking expected value of both sides shows that the expected value of $P(n+1)$ is $\frac{n+1}{2} + n!$.


[quote="wonderful"]Given a > 0, b>0, c>0 and $ a^4 \plus{} b^4 \plus{} c^4 \equal{} 3$

Prove:$ \frac{a^2}{b\plus{}c} \plus{}\frac{ b^2}{c\plus{}a} \plus{} \frac{c^2}{a\plus{}b} \geq \frac{3}{2}$

Have A Great Day![/quote]

Is a problem by Alexei Gladkich.

  We show that  ${ (\sum {a^2})^3\geq9\sum a^2b^2}$  (***)

  (***) is equivalent with:  $ (\sum {a^2})^3\geq 3(\sum {a^2b^2)\cdot\sqrt {3\sum {a^4}}}$,

 or   $ x^3\geq3y\sqrt {3(x^2 - 2y)}$,  where $ x = \sum {a^2}$ and $ y = \sum {a^2b^2}$.

 By squaring,  $ x^6 - 27x^2y^2 + 54y^3\geq 0$ or, $ (x^2 - 3y)^2(x^2 + 6y)\geq 0$ wich is true.

  So , by  Holder's Inequality, we have

 $ (\sum {\frac {a^2}{b + c})(\sum {\frac {a^2}{b + c}})(\sum {a^2(b + c)^2)\geq(\sum {a^2})^3}}$

 Therefore, is suffices to show that

 $ (\sum {a^2})^3\geq \frac {9}{4}\sum {a^2(b + c)^2}$.

  Using the above inequality(***), we still have to prove that

 $ \sum {a^2b^2}\geq \frac {1}{4}\sum {a^2(b + c)^2}$ which is equivalent to $ \sum {a^2b^2}\geq \sum {ab^2c}$ which is true.

Excellent Marius! Your solution is very elegant. Well done!

Have A Great Day!


for all real $x$   find all $P(x)$ such that
$P(x).P(x+1) = P(x^{?})$

This does not seem to be very tough, but quite nice. First, it is clear that any root either has modulus 1 or is zero. Next, take a root "k". Then $ (k-1)^2 $ is also a root, thus it is either 0 or has modulus 1. So, k is either 0, either 1 or k has modulus 1 and k-1 has modulus 1. It is easy to check that in this case k is the opposite of a third root of unity. Thus, these are the only possible roots. From now on we do not have problems, just factorise and identify the orders of multiplicites of the roots.

$P(x)=0$
$P(x)=1$
$P(x)=(x^2-x)^q$  q  positive integer

a similar one with a similar solution:
find all polynomial Psuch that
P(x-1)*P(x-1)=P(x^2)

bye


[color=blue][b]
Prove that $\forall x,y,z > 0$ and $xyz = 1$ we have:
$x + y + z \ge \frac{{1 + yz}}{{1 + z}} + \frac{{1 + zx}}{{1 + x}} + \frac{{1 + xy}}{{1 + y}}$
[/b][/color]

Well, taking $x=\frac{a}{b},y=\frac{b}{c},z=\frac{c}{a}$ the inequality reduces to this one:

     $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq\frac{a+b}{a+c}+\frac{b+c}{a+b}+\frac{c+a}{b+c}$ ;) , which is not hard to prove. In fact it is quite easy. ;)

Yeah. I wasn't getting anywhere with the

$x=\frac{a^2}{bc}, y=\frac{b^2}{ac}, z=\frac{c^2}{ab}$

umm can someone show me a nice proof for the thing cezarlupu was talking about?


Colin has a 1/2 probability of waking up each time his alarm rings in the morning. His alarm continues going off at regular intervals. How many times must his alarm ring for his mom to figure there is at least an 85% probability that he is awake?

I don't think this belongs here...

85% chance that he is awake means 15% chance he isn't awake.

There is a $ \frac{1}{2}$ chance of him not waking each time the alarm rings, so we are solving for the smallest (whole number) n such that $ \frac{1}{2^n}<0.15$. So, we get n=3. The alarm must ring three times, for an 87.5% chance.


A person weighing 120 pounds burns about 7.3 cal/min jogging and 11.3 cal/min running. Kirsten, who weighs 120 pounds, wishes to exercise by jogging and running. She has 40 minutes to exercise and wishes to burn exactly 352 calories. How many minutes should she spend jogging?

Let $ x$ = the time she spends jogging and $ y$ = the time she spends running. Then we have this system of equations:

                                                $ 7.3x \plus{} 11.3y \equal{} 352$
                                                $ x \plus{} y \equal{} 40$

Since we are interested in $ x$, we multiply the bottom by $ \minus{} 11.3$ and add accordingly.

                                                $ \minus{} 11.3x \minus{} 11.3y \equal{} \minus{} 452$
So,
                                                $ \minus{} 4x \equal{} \minus{} 100 \implies x \equal{} \boxed{25}$.


a,b,c$ > 0$ Show that :

$ \sqrt {a^4 \plus{} a^2b^2 \plus{} b^4} \plus{} \sqrt {b^4 \plus{} b^2c^2 \plus{} c^4} \plus{} \sqrt {c^4 \plus{} c^2a^2 \plus{} a^4}\geq a\sqrt {2a^2 \plus{} bc} \plus{} b\sqrt {2b^2 \plus{} ca} \plus{} c\sqrt {2c^2 \plus{} ab}$

Aπο την BCS εχουμε

$ (a^2\plus{}b^2\plus{}c^2)(2a^2\plus{}2b^2\plus{}2c^2\plus{}ab\plus{}bc\plus{}ca)\geq$ $ \left(a\sqrt {2a^2 \plus{} bc} \plus{} b\sqrt {2b^2 \plus{} ca} \plus{} c\sqrt {2c^2 \plus{} ab}\right)^2$

$ 3(a^2\plus{}b^2\plus{}c^2)^2\geq$ $ \left(a\sqrt {2a^2 \plus{} bc} \plus{} b\sqrt {2b^2 \plus{} ca} \plus{} c\sqrt {2c^2 \plus{} ab}\right)^2$

Οποτε αρκει να δειξουμε οτι

$ \sqrt {a^4 \plus{} a^2b^2 \plus{} b^4} \plus{} \sqrt {b^4 \plus{} b^2c^2 \plus{} c^4} \plus{} \sqrt {c^4 \plus{} c^2a^2 \plus{} a^4}\geq$ $ \sqrt{3}(a^2\plus{}b^2\plus{}c^2)$

το οποιο ισχυει αφου $ \sqrt{a^4\plus{}a^2 b^2\plus{}b^4}\geq \frac{\sqrt{3}}{2}(a^2\plus{}b^2)$  qed.


integration
trigonometry
calculus
calculus computations

is it possible to evaluate $ \int_{0}^{4} x^{2} d([\sin(x)])$? if so, how would someone do it?

By drawing the graph of $ [\sin x]$ on that interval (begin with $ \sin x$ itself) and forming a sum with terms $ x^2\cdot (\pm 1)$, where $ \pm 1$ is the amount by which $ [\sin x]$ jumps at $ x$. Note that some of the jumps are immediately reversed and therefore do not contribute to the integral.

The integer multiples of $ \pi/2$ are relevant here, not the integers.

(Sorry, I edited your last post by mistake and then deleted it. Enough posting for me tonight.)


Hi all,

We have a temporary issue with displaying [b]new[/b] $\LaTeX$ formulaes on the site. This requires some maintainance that will be done in a few hours. Until then you can post your messages on the board, but you will see this: $\dfrac test$

Again after we fix the problem your messages should look fine.

OK, problem is fixed now :P This message will self distruct in 5 minutes :)

Ok, ok I admit ... it didn't self distruct  :(  :blush:


algebra
polynomial
floor function
absolute value
algebra solved

Prove that if a polynomial $P$ in $\mathbb{Z}[X]$ has $n$ distinct integer roots, then the polynomial
\[ P^2+1 \] has a irreducible factor of degree at least $\left\lfloor \frac {n+1}2\right\rfloor$ in $\mathbb{Z}[X]$.

[quote="Myth"]I hate to say these words :blush:, but it is an obvious problem.

Let's organize test: who can solve it in two minutes. :P[/quote]It's not hard I know ... but many of the Romanian students who took the test yesterday did not solve it :(

Well, to put it out of its misery :)

Can't we show even more? That $P^2+1$ has no factors of degree $<\left\lfloor\frac{n+1}2\right\rfloor$? Take any factor of $P^2+1$. That factor has absolute value $1$ in $n$ distinct points, meaning that it takes the same value (one of $\pm 1$) in at least $\left\lfloor\frac{n+1}2\right\rfloor$ points.

Grobber! You killed my idea about test  :(  ;)

Yet, I am sure someone told me that in an AMM there has been an extremely difficult problem very similar to this one. Of course, it cant be that, which is trivial, but I would be interested to see the difficult variant.


MIT college calculus integration email inequalities topology

This might be another emotional and sad question but one which has been my mind lately.

I am a high school student in India. I simple love Mathematics and Physics for umpteen reason ( I could give 500 posts on that alone!!) With my passion for Science, My Dream has been MIT and still is. That's a different matter whether I get in, I want to give it my best try. ANd related to that this question has cropped up in my mind!(just establishing context! :lol: )

I have already asked about the experiments (unique ones) which I can perform on my own in India (with not over - expensive equipment). Also, I have been seeing on all sites that High school research in US (like in Intel STS and RSI (by MIT)) is an integral part and an important factor in admission. I also want to do research. But plz realise that my OLNY resource is the net..(you know the sysytem in India!). [b]What can I do!!?? What can I do research on?? What can I think about and develop upon??? Will you help me??!!![/b]( research in Mathematics and Physics (in maths, it may be easier for my grandpa is a professor of MAthematics(topology I think))

Hey man, you know I'm going to MIT this year to intership Prof. Marzari  :lol: , I'm going to enter there next year and this will help me to succeed. I also like physics and mathematics very much wish me good luck in this job...

internship??? You' re going there for PhD???

hey rushil, sorry I never got your PM, just an Email saying I got one, but when I checked my inbox it was empty

so you're a rising senior?  I am too, but I'm going to school in the states.  You seem like you're really knowledgeable about math+physics, It would be a real shame if you didn't go in.  I'm also applying to MIT.  heh.

Also, you dont HAVE to do research.  there's no given formula for getting in.  just do what you love.  I guess.  otherwise, well, pandering to admissions really doesnt seem worth it.

and maybe we'll both go there someday,
good luck man.
p.s.
You haven't posted your work for that problem, with the two colliding masses+the inequality?  I dont see why it's not immediately obvious that it'll go upwards, heh.  silly me.

Rushil... Why don't you talk to your grandpa? I mean it's a little late now, you have to apply in what, 2 months? But what you COULD do is very quickly learn some UBER topology and then ask if there is anything you can work on him with. Get your name on a paper, that'll definitely impress them. Topology is hard though... ESPECIALLY at the research level, so I don't know how that would work. That's what I'd do, anyway.

The problem is he's retired now so getting my name on a rsearch paper would be next to impossible (he's been out of touch for sometime)

Probably all I can do is make something unique( a paper etc.) on my own and send it along with the application.I want to do research not only for the appl. but also for my interest!!!

All though this is a bit off topic(research prospects at high school level in india),
I got selected by IISc , Bangalore and IIt Bombay under KVPY [url=http://www.iisc.ernet.in/kvpy/]KVPY[/url]. Just 40 Young scientists were selected from all over India for encouragement to pursue Science. We even had a summer camp in Bangalore!!

You think this is a good achievemnt? How does it compare with other science scholarships and summer programs in the US(like Intel STS , RSI etc)

So daveed, have you done any research???

Any other recommendations for MAth and Physics  self-research???

I havent done any physics-related research, but I'm interning in the biology field.... studying these little amoebas and their signaling pathways, heh.
for self-research, though.  It's still cool  :) everything is wonderful.

As for your self-research, though, I don't know if you could do much aside from rediscovering old results, at least without big, expensive equipment.  It would be fun to do some of those things, though.
IISc does sound quite impressive, though.  How was it(if it has already happened) and what did you do there?

I really doubt that you "have" to do some kind of research to get into MIT, though.  They know that some people dont have the opportunity(like living right next to gov. facilities)

the IISc is basically THE premier Science research institute of our country. ALL (and I mean ALL) of our greatest scientists - Raman, S.N.Bose, G.N. Ramachandran...... were form there.Anyways, this is besides the point.

I was selected under this scholarship througha lenght and tough written exam followed by an interview. I suggest you visit the link I gave you to get more details of the scholarship. Just 40 ( from God knows how many applied - India is huge) were selected this year!!

At our summer camp, we had lectures on Physic, Chem, MAth and Bio. We had lab visits , we saw radio telescopes, laser experiments, Mose-Einstein COndensate, Nano-tech equipment and loads more.... Is that like all other camps and competitons in the US?? Can anybody make a good comparison of the difficulty(selection , camp) etc...

what??? I don't understand your short forms. Please elaborate!!!

Can somebosy plz suggest other possible resarch topics in Physics/math. I was thinking (interested in) applications of Topology to Physics. Any suggestions on web resource???

Please respond to my orig. question. Thanks


inequalities
calculus
derivative
inequalities solved

if $a,b,c,d>0$ then prove this inequality:
$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}} \leq  \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$

[quote="mffmartinelli"]Let´s observe that this is an homogeneous inequality. So we´ll consider $a+b+c+d=1$.
And now let $f: R_+\to R$ such that $f(x) = (x-a)(x-b)(x-c)(x-d)$. Observe that $f$ has 4 real roots. So it´s easy to check that it´s first derivative $g$ has 3 positive real roots. Let $m$, $n$ and $p$ this roots.
It´s true that $\frac{mn+mp+np}{3} \geq (m^2n^2p^2)^{\frac{1}{3}}$.
But we can view that $f(x)=x^4-x^3+S_{2}x^2-S_{3}x+S_{4}$, where we have $S_{2}=ab+ac+ad+bc+bd+cd$ and $S_{3}=abc+abd+acd+bcd$. So if we notice that $g(x)=4x^3-3x^2+2S_{2}x-S_{3}$, we will have that $mn+mp+np=\frac{S_{2}}{2}$ and $mnp=\frac{S_{3}}{4}$. So we must have $({\frac{S_{2}}{2}^})^{3} \geq 3^3({\frac{S_{3}}{4}})^{2}$ and so
$4^2{(ab+ac+ad+bc+bd+cd)}^{3} \geq 6^3{(abc+abd+acd+bcd)}^{2}$ and it´s proved.[/quote]
wow.nice solution.in your solution there is a very good idea. :D

I believe that you are referring to a generalization of this problem, which is possible making use of this idea.

[quote="jensen"]
wow.nice solution.in your solution there is a very good idea. :D[/quote]
This is other problem. Can you?  :wink:

Let $x_{1},x_{2},...,x_{n}(n>2)$ be nonnegative reals such that

$(n-2)\sum_{1\leq i<j\leq n}x_{i}x_{j}+\sum_{1\leq i<j<k\leq n}x_{i}x_{j}x_{k}=4C_{n}^{3}$.

Prove that $\frac{\sum_{i=1}^{n}x_{i}}{C_{n}^{1}}\geq \frac{\sum_{1\leq i<j\leq n}x_{i}x_{j}}{C_{n}^{2}}$.

Proposed by Le Quang Nam (maybe?)


function
number theory
totient function
relatively prime
prime numbers
number theory proposed

[quote="mathematica"]Find all $n\in \mathbb{N}$ so that $\phi (n)+\tau (n)=n+1$, where $\phi (n)$ is Euler's totient function and $\tau (n)$ is the the number of positive divisors of $n$.[/quote]
Obviosly $n=p$ (p - prime) is solution.
If $n\ge 4$, then $\phi (n)$ is even. Therefore if n is even $\tau (n)$ odd or $n=k^{2}$ (k -even).
If $n=p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}}$, then
$n+1-\phi(n)=1+\prod_{i}p_{i}^{k_{i}}-\prod_{i}(p_{i}^{k_{i}}-p_{i}^{k_{i}-1})\ge 1+\prod_{i}p_{i}^{k_{i}-1}$.
Obviosly, in these case $p_{i}^{k_{i}-1}\ge k_{i}+1$ ($k_{i}\ge 2$), because $f(x)=p^{x-1}-x-1, f'(x)=p^{x-1}ln(p)-1>0, x\ge 2,p\ge 2$. Therefore $n+1>\phi(n)+\tau(n)$ for these case if n>4. $n=4$ is solution.
If n is odd, then $p_{i}\ge 3$. Consider $n=n_{1}n_{2},rad(n_{1})=n_{1},(n_{1},n_{2})=1$. Then $\tau(n_{1})=2^{m_{1}}, \phi(n_{1})=\prod_{i}(p_{i}-1)$.
Obviosly, if $n_{1}+1>\phi(n_{1})+\tau(n_{1})$, when $m_{1}>1$ ($n_{1}$ is not prime).
Therefore $n+1-\phi(n)-\tau(n)=n_{1}n_{2}+1-\phi(n_{1})\phi(n_{2})-\tau(n_{1})\tau(n_{2})>0$, when $n_{1},n_{2}>1$.

Here's another solution:

First of all, note that since $\phi (n)+\tau (n)=n+1$, every natural number smaller or equal to $n$ is either a divisor of $n$, or it's relatively prime to it. It's obvious that $1,4$ and the prime numbers are a solution. I shall now prove that no composite number larger than $4$ is a solution to the above equation.

Suppose that $n$ is a solution and $n$ is composite. By definition, there exist two natural numbers like $a,b>1$, such that $n=ab$. Because $n$ is bigger than $4$, we could also assume that $b>2$.  Now consider the number $k=a(b-1)$. It's natural and smaller than $n$. According to what I said above it should either be a divisor of $n$ or it should be relatively prime to it. But neither of this is true since $\gcd(b,b-1)=1$ and $b-1>1$ therefore $k$ is not a divisor of $n$. On the other hand, $\gcd(n,k)=a>1$, so $k$ is not relatively prime to $n$ either, which is a contradiction.


[quote="flip2004"]Assume that $a$ and $b$ are real numbers (calculated with  $15$ decimals). Denote their " midd-point" by $Mp$, i.e. $Mp: =\frac{a+b}{2}\; .$
It's clear that $Mp=a+\frac{b-a}{2}\; .$
[b][color=green]Question:[/color] From "precision" point of view,  how is better to write $Mp$ :
\[Mp: =(a+b)*0.5\; \; \; \; \mbox{or}\; \; \; \; Mp: =a+(b-a)*0.5\; \; \; ? \]
[/b][/quote]

I expect (a+b)/2, because you're dividing a bigger number there and thus I don't think there's as much error.

I would think that dividing a bigger number would lead to more error, since you have to keep track of more digits.

However, I am not completely sure =/. Have you checked Numberical Recipes in C to see if they have anything to say?

Well, since you divide by TWO and the floating point numbers are usually stored in base TWO, you shouldn't lose any precision by doing it either way. Assuming you divide by two and not multiply by 0.5. (Division by two shouldn't change the mantissa, only decrease the exponent by one.)

But by adding the two numbers before dividing, if the exponent increases you lose the extra bit at the end of the mantissa, as you need to add one more digit at the beginning of your number (most significant bit).

Would left-shifting by 2 be even better then?

(110.101 + 1.10111) = 1000.01011

1000.01011 / 2 = 100.001011

In the division by two, mantissa never changes! It is 100001011 both before and after the division, the only thing that changes is the exponent.

So if you lost a bit during the addition, it's only because it wouldn't fit in the result of division by two anyway.

[quote="nebula42"]Actually, it is sometimes better to use a + (b - a) / 2, because this helps to avoid overflows (which could be caused by adding a and b in (a + b) / 2).[/quote]
I agree with this.


basketball..well it's easy to play and you don't need much, just a ball and a hoop

basketball definitely, followed by football.

[quote="samath"]Who's my tennis buddy?[/quote]

Here is the god of tennis,well i was really serious about tennis 2 years ago but now it is all studies and there wasn't any good coach here...

Now i like soccer more....so i guess it's soccer for me...i am the heir of zinedine zidane...

[quote="samath"]Who's my tennis buddy?[/quote]

'Tis me.

I "started" playing at a really long age but then I took like a 5 year break (yeah, I know) and I started back up about 6 months ago. I've got my back/forehand down very well and my serve is coming around (just need a bunch of repetition), and hopefully I'll make my school team (they placed like 3rd in the state so I hear its pretty competitive, but I'm working fairly hard on my tennis skills so I'm confident I can make it).

What about ya? :P

I play on the varsity tennis team for a junior high in Washington. we were one match away from a cochampionship this year  :( (which we've never come close to in the past six years :lol: )

I've played tennis kind of for fun since I was six, but I've only been playing competitively for three years.

I agree with MathFiend, soccer is fun.  I like tennis just as well, and my tennis team is a lot better than my soccer team, so I chose tennis (everybody likes to win :lol:)

Mainly soccer, but basketball and tennis is still cool.

basketball!

it's the most accessible sport for an urbanite like me.

[quote="riddler"]I cant believe CRICKET the GOD of all sports, only joking  :lol: , isnt up there :o[/quote]

It was, but I had to take it off because they only let me put ten options.  I originally had like 25.  I dont know why I had to take it off though  :huh:

[quote=gogiants][quote=abk2015][quote=TopNotchMath]Other is Cricket[/quote]

Cricket is also really awesome.[/quote]

I have created a cricket forum which can be accessed here: [url]http://www.artofproblemsolving.com/community/c374448_cricket_forum[/url][/quote]

I forgot if I can, but I think since I'm a mod if you make me mod of that forum I can put it in the sports collections

Ok thanks. I have added you as a moderator. Do I need to add you as an admin?

[quote=SFGiantsFan]Ok, so I added it to the Sports collection that I made, the original one, but I guess I have no control over it now that it got moved over to the Games forums because of the way it was added.....the original collection wasn't added but individual forums were. So I can't add it to the collection that's under games.[/quote]

That's ok. Thanks anyway.

Skiing and swimming at a very close second...

My favorite is wrestling. I also love football, basketball, swimming, and cross country.

Soccer! Soccer is the best in my opinion. It is the most popular sport in the world

[quote=samath]Who's my tennis buddy?[/quote]

That's definitely me.

i voted football, but i think i'm starting track tomorrow


Who else is going this Saturday? Also, are calculators allowed during any part of the tournament?

Yeah, after posting this, I saw your thing in the iTest forum about why calculators should be banned and got the impression that you don't like calculators in competitions. :lol:

Really? ;) Anyway, I'll be there for obvious reasons, though you may not see me till awards. Lassiter will assuredly be there too, as they're bringing the grading machine ;).

Excellent!  (Note the contrast with the situation at state last year.)

Sooo.... what did you guys think?

Attached are the tiebreakers; it took 7 tiebreakers to find 4th and 5th place ciphering. I thought #7 and #5 were the best ones, but nobody got #5 and only one person got #7. I'll leave the answers to your imagination. On the first 5 tiebreakers, you got 2 minutes, on next two, 3 minutes, the last was to have been time-based.

I'll post the other tests soon.

Great tournament. I especially liked the unique format of the tournament (differently weighted questions, ARML style ciphering, and team questions dependent on one another). I also liked number 11 and 14 on the written test. I wish I had had time to look at number 15 because it was a lot easier than some of the ones before it.
I do have one problem: Miles and Santhosh protested question number two on the ciphering. I think they retracted their protest, but I think that they may have been right. The last line of the solution says "The next integers are $ 1011'1', 1011'0, 1011'1,$ and $ 10100$." However, the third number is $ 81\plus{}9\minus{}3\plus{}1 \equal{} 88$ and the fourth number is $ 81\plus{}9 \equal{} 90$. I think that the actual last number should be $ 10101' \equal{} 81\plus{}9\minus{}1 \equal{} 89$. This would make the final answer $ 1'$.
I know that if this is correct, it is far too late to correct the problem, and I missed it either way because my answer was $ 1$, but I just wanted to point this out.
Also, number 4 on the ciphering was very guessable. :lol: Everyone sitting near me guessed it right.

The other tests:

Indiv was 1 hour.
Ciphering was 10 minutes a pair
Team was 40 minutes, 4-person teams.

No answers yet ;), I'll post solutions Tuesday.

The individual has one problem; #7 was thrown out and the points given to everyone. Also, I can't add; everyone got 3 points on the individual additional.

The team had one problem; infinity was also accepted as a good answer to 3a.

After about 30 minutes, I finally solved number six ciphering. Very tough. Also, what was the problem on number seven individual?

[quote="solafidefarms"]Also, I can't add; everyone got 3 points on the individual additional.
[/quote]

If it's any consolation, I did exactly the same thing on the power question at NFVIMT last year.

Overall, though, great tournament today! I'm like, really disappointed that I'm not gonna be able to do it again  :( . Not even kidding. Ah well.

[quote="Twiz"]what was the problem on number seven individual?[/quote]

The correct answer choice wasn't in the answers, I forgot to add 3.

I noticed that, and then guessed that you forgot to add 3, and put down B.

Now that I look at the 2nd tiebreaker more closely, It does look like the radical was on just the top. I guess it is hard to read that when you have 2 mins.

Great competition.

Great tournament Billy, The fact that there was only one problem with the three tests was commendable. And personally, I think that a hard test on a Saturday tournament is fine. (Of course, the Fall test writer for the Cobb County tournament could stand to learn that lesson!  :) )

The tournament was also well organized. Something that isn't always true!

Tom

Oh, wow.  It's an XKCD problem.  The funny thing is, I've considered writing just such a problem, but I could never come up with a good one.

Also,
[hide]#4 was possibly the hardest problem with an easy answer I've ever seen.[/hide]

[i]Soli deo gloria[/i] is a rather ambiguous name.
[i]Glory to god alone[/i]?
[i]Alone in the glory of god[/i]?
[i]Glory to the sun-god[/i]?
I am confused.

For the Glory of God Alone is the desired meaning.

Eh? Shouldn't that be "soli deo gloriae" rather than "soli deo gloria"? But if you insist on using "gloria", I am pretty sure you would need a preposition. I think you should change the translation to "glory to the sun god" or one of the other aforementioned possibilities.

Don't get me wrong. This is a pretty good test (whereupon I am presently working). However, I just wanted to point out that it is probably bad to have a contest with such an ambiguous title, especially when said ambiguity is so very contrary to the desired meaning.


linear algebra
matrix
Putnam
linear algebra unsolved

we have a $m*n$ matrix with $0,1$
such that each $2$ rows have more than $[\frac{n}{2}]$ diffrences
prove that $m\leq 2n$

move this one to college playground/ matrix problem
thank you

This solution is inspired by one of the 2005 Putnam problems.

Replace all zero entries with $-1$. Now, the condition is that the inner product of any two rows is negative.

Let $v_1,v_2,\dots,v_m$ be the rows. We have $\|v_1+v_2\dots+v_m\|^2\ge0$. We also have
$\|v_1+v_2\dots+v_m\|^2=\sum_{j=1}^m \|v_j\|^2+\sum_{i<j}2v_i\cdot v_j$
$= mn+\sum_{i<j}2v_i\cdot v_j\le mn-m(m-1)$ since $v_i\cdot v_j\le -1$. This gives $n\ge (m-1)$, which is even more that we needed.

right
but I have another solution for that
I solved it this way some hours ago:
if $n$ was odd then:
$\binom{m}{2}\frac{n+1}{2}\leq$the numbers of$\begin{array}{c}{1cr}1\0\end{array}$or$\begin{array}{c}{1cr}0\1\end{array}\leq \frac{m^2}{4}n$
$\Rightarrow (m-1)(n+1)\leq mn$
$\Rightarrow m\leq n+1$
now if $n$ was even then prove it by contradiction
assume that $m\geq 2n+1$
we know that the first column have at least $n+1$ same elements
now assume that these elements are the first $n+1$ elements
now remove the other $m-(n+1)$ rows
the new matrix without the first column is a $(n+1)(n-1)$ matrix
now by the assumption we know that $(n-1)$ is odd
there for according to the last part $(n+1)\leq n$
and this is a contradiction


search
topology
advanced fields
advanced fields unsolved

Let $X$ be a separable complete metric space, $\mu$ a measure on the Borel sets of $X$, and $\mu(X)=1$. Prove that $X$ is $\sigma$-compact, i.e. a union of compacts. Can we drop the appaerence of a measure such that the statement is still true?

The statement is false with or without $\mu$. For example, let $X$ be a separable infinite-dimensional Banach space. Then $X$ is not $\sigma$-compact because any compact subset of $X$ is of the first category. Also, we can take  $\mu$ to be a unit point mass: $\mu(A)=1$ if $0\in A$ and $\mu(A)=0$ otherwise.

More interesting question is whether there always exists a $\sigma$-compact subset $E\subset X$  such that $\mu(X\setminus E)=0$ ;)

fedja's question has been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=351883469&t=55675]discussed before[/url]. Short statement: finite Borel measures on Polish spaces are regular.


geometry
geometric transformation
Asymptote
LaTeX

Hello,
I need help with hyperref within a spanish document.
I've seen lots of manuals which explain how to use hyperref package, but this doesn't work in spanish. I know how to use the spanish package so there's no problem with that.
Now, I have two questions:
The first one is how can I put a link into my document for jumping to another place. A section or a figure, for example. (in spanish, hyperref package works perfectly with url's)
The another question is how can I make reference regarding a figure ... I've been tried with labels and \ref's but the output file always shows me "Figure ???" ..
If anyone can help me, I'll appreciate it.

It would be helpful if you could post a complete minimal document with what you've tried but which doesn't work. This will make it much easier to help you.

Well ... I think that I solved the second problem, I just had to compile twice ... I didn't look over the warnings.
I applied the same method for referring to the section, but it's not like I wanted.
You'll see.

The text is a translation from AoPs information about Asymptote ...
See page number 2 : "Seguir las instrucciones en la página de Descarga e Instalación (Ver 2.1). "
I want to change that for a hyperref ...

[quote="jonysatie"]See page number 2 : "Seguir las instrucciones en la página de Descarga e Instalación (Ver 2.1). "I want to change that for a hyperref ...[/quote]But it is already a link. If you click on 2.1 in that line it does go to subsection 2.1. Isn't that what you wanted?
I am attaching a copy of the compiled document in case it differs from yours in some way. You'll notice that the 2.1 is in red showing it is a hyperlink.

Yeahhh I already saw that pdf (I have it too) ..
but I wanted to click over the phrase "Descarga e instalación" (for example) and jump to the section. How do you do that?

This is a little bit different than what I did (a number is automatically set)...

For internal links use \hyperlink and \hypertarget. So at the section heading put[code]\hypertarget{Descargas}{\subsection{Descargas e Instalación}}[/code]and the link to it is[code]\hyperlink{Descargas}{\emph{Descarga e Instalación}}[/code]so you have[code]\item Seguir las instrucciones en la página de \hyperlink{Descargas}{\emph{Descarga e Instalación}}(Ver \ref{primero}).[/code]

Ohhh .. it finally works ... THANK YOUUUU


I am playing mafia for the first time in game 18, and I am really liking it. But there is space for improvement. Here we shall discuss hoew the game of mafia can be improved. I expect views from both experienced mafia players and novices. I won't oput any of these into action until I moderate a game, which may not be in a year or so. But other moderators may us then if they like to.

We can change the name to a nicer thing so my parents will let me play.

Good idea! We can change it to [i]How to be a Watchful Citizen[/i]. That way your parents would not mind. (But really, do they always watch over your shoulders to see what you are doing??)

But I meant something on a more serious note.

Heh, look at the thing I posted on the first page, he's 14.

But I think the game of Mafia is fine as it is now. It has been around for a LONG TIME and we play different variations and it seems to work fine. The only thing I would like is for the game to go faster and that means more people active.


Hello,

I don't know what the difference between a concrete population and a hypothetical population. Google was of no help. Can someone tell me the difference and give me an example or two of each?

It would, probably, help if you could tell us the context in thich these terms appeared. As it is now, I can, probably, attach some meaning to the word combination "hypothetical population" but the words "concrete" and "population" put together do not seem to make much sense in standard English and deinitely refer to some particular definition of terms that is used not very widely (possibly, nowhere outside your textbook).


A man has part of $ \$4500$ invested at $ 4\%$ and the rest at $ 6\%$. If his annual return on each investment is the same, the average rate of interest which he realizes of the $ \$4500$ is:
$ \textbf{(A)}\ 5\% \qquad\textbf{(B)}\ 4.8\% \qquad\textbf{(C)}\ 5.2\% \qquad\textbf{(D)}\ 4.6\% \qquad\textbf{(E)}\ \text{none of these}$

[hide="Click for solution"]
$ \frac{1}{x}\equal{}\frac{1}{2} \left (\frac{1}{6}\plus{}\frac{1}{4} \right ) \implies x\equal{}4.8$, or $ \boxed{\textbf{(B)}}$.
[/hide]


Find the branch points of $f(z) = \sqrt{(1-z^2)(1-k^2z^2)} (0<k<1)$. Prove that it is analytic outside the segments of
\[-\frac{1}{k}\leq x\leq -1\quad 1\leq x\leq \frac{1}{k}\]
and find the sheet of it on which $f(0)>0$.

$f(z) = |(1-z^2)(1-k^2z^2)|^\frac{1}{2}e^{i(Arg(1-z)+Arg(1+z)+Arg(\frac{1}{k}-z)+Arg(\frac{1}{k}+z))}$
(where $Arg(z)$ is multi-valued)
I find out that $-1,1,-\frac{1}{k},\frac{1}{k}$ are branch points.


A vendor has six jugs of punch with capacities of 13, 15, 16, 21, 21 and 22 gallons.  One jug, which he keeps for himself, is full of blue punch.  The other five jugs are full of red punch.  Without opening any of the jugs, he sells 1/3 of the red punch to Randy and the remaining 2/3 of the red punch to Lisa.  How many gallons of the blue punch did he keep for himself?

[hide="Solution"]Let x be the number of gallons in the jug with blue punch.

13+15+16+21+21+22-x is divisible by 3.
13+16+22-x is divisible by 3.
51-x is divisible by 3.
x is divisible by 3.
x is 15 or 21.

It looks like the answer is either 15 or 21. However, because the vendor didn't open the jugs, the answer can't be 21 because if it was, the vendor wouldn't be able to distinguish between the 21-gallon jug with red punch and the 21-gallon jug with blue punch.[/hide]

[hide="My Answer."]  Since the problem says that he sold $\frac{1}{3}$ without opening any jugs, the number of gallons must be divisible by 3.  This means that the only jugs he could keep are either 15 or 21.  If he keeps the 21, then the sum of the gallons of the red punch is $13+15+16+21+22=87$.  To separate the parts into $\frac{1}{3}$ and $\frac{2}{3}$, we have 29 and 58 gallons.  Therefore, this works, because 13+16=29, so you can separate it into the two sections.  21 is the answer.[/hide]

[hide]He has 108 gallons of punch. $(108-$his bottle$)\equiv  0$mod$3$ so that he can split it 3 ways. $108\equiv 0$mod$3$, so his bottle must also be $\equiv 0$mod$3$. The only bottles that work are 15 and 21.

However, we must be able to split it into (1/3)/(2/3) evenly.

:arrow: If blue one has capacity 15 gallons: There are 93 gallons of red. We must now test and see if  we can make 31 and 62 gallons evenly, which we cannot.

:arrow: If blue has capacity 21 gallons: There are 87 galloons of red. We test and see that we can make (13,16) and (15,21,22)

Therefore, he kept 21 gallons of blue. Now how long will it take him to drink that much!?
[/hide]
Billy

[quote="rcv"][quote="RC-7th"][hide="Solution"]. . .[/hide][/quote]
[hide="Please continue with your solution"]How much punch did he sell to Lisa?[/hide][/quote]

I see my mistake. :blush: Forgot to reality check...


In the game of Yahtzee, a person throws five dice.  The person can then rethrow any of the dice a second time, and rethrow any of the dice a third time.  The goal is to get the same number on all five dice (called a "Yahtzee").  (The actual game is more complicated, but for the purposes of this problem we only care about scoring a Yahtzee)

Example:

On the first throw, I get 4 5 2 5 5

I keep the three fives, and rethrow the 4 2.  I get 5 1.

Now I keep the 4th 5, and rethrow the 1.  I get a 5.

Yahtzee!

Assuming an optimal strategy, what is the probability of throwing a Yahtzee?

Some easier warm-up problems:

- What's the probability of a Yahtzee on the first throw?

- What's the probability of a Yahtzee after 2 throws?

I'm horribly unsure about these, but...

Yahtzee in one throw: [hide]there are 6^5=7776 combinations of dice results during the first throw, of which 6 combinations (11111, 22222, 33333, 44444, 55555, and 66666) work. So the probability is 6/(6^5)=1/(6^4)=1/1296.[/hide]

Yahtzee in two throws: [hide]You have 1*1/6*1/6*1/6*5/6=5/(6^4) chance of getting exactly four dice the same, multiplied by 1/6 chance of getting the fifth right on the second throw, so that's 5/(6^5). You have 1*1/6*1/6*5/6*5/6=25/(6^4) chance of getting three dice the same and two different from those three (whether or not the two are the same is irrelevant), with a 1/36 chance of getting both dice right on the second throw, so that's 25/(6^6). You have 1*1/6*5/6*4/6*5/6=100/(6^4) chance that you'll have two the same and no more than two others equal to each other, and a 1/(6^3) chance of getting the remaining three right on the second throw, so that's 100/(6^7). Finally, you have a 1*5/6*4/6*3/6*2/6=120/(6^4) chance that none of the dice are the same, and a 1/(6^4) chance that four dice will be corrected the second time, so that's 120/(6^8). So, the total is

,

whatever that is.[/hide]

Actually I'm pretty sure the second one is wrong, and the first one has a good chance of being wrong as well. What's the correct solution?

Lets break down the problem.

Throw 1
# of same #'s:

0 - Keep 1:roll 4 dice on role 2, Yahtzee 1/1296, 1 match - 1/6, 2 match - 1/36, 3 match - 1/216, no match - 1037/1296: roll 3, Yahtzee -  5(1/1296)
no yahtzee - 1291/1296: total chance 5/1296

2 - Keep 2:roll 3 dice on roll 2, Yahtzee - 1/216, 1 match - 1/6, 2 match - 1/36, 0 match - 173/216: roll 3, Yahtzee - 4(1/216), no yahtzee 53/54: total chance - 1/54

3 - Keep 3:roll 2 dice on roll 2, Yahtzee - 1/36, 1 matches - 1/6, 0 match - 29/36: roll 3, Yahtzee - 3(1/36), no yahtzee 11/12: total chance - 1/12

4 - Keep 4: roll 1 die on roll 2, 1/6: role 1 die on roll 3, 1/6 = 2(1/6) = 1/3

5 - 1

total possible chance with best play:(1/3)+(1/12)+(1/54)+(5/1296) = 569/1296

[hide]

#1.

The first die can be any number.

The second to fifth dies must be the first number.

1/(6^4) = 1/1296

Not that great odds...

[/hide]


Let f(k) be the number of integers n that satisfy the following conditions:
- 0<= n <= 10^k, so n has exactly k digits with leading zeros allowed
- the digits of n can be permuted in such a way so that they yield an integer divisible by m.
Prove f(2m) = 10*f(2m-1).

Hard indeed, the only combinatorics problem from the 2003 shortlist I could not solve.


1)Nα προσδιορισετε ολους τους διψηφιους αριθμους που ισουνται με το αθροισμα των τετραγωνων των ψηφιων τους
2)Nα προσδιορισετε ολους τους διψηφιους αριθμους που ισουνται με το αθροισμα των κυβων των ψηφιων τους
3)Nα προσδιορισετε ολους τους τριψηφιους αριθμους που ισουνται με το αθροισμα των τετραγωνων των ψηφιων τους

και τελος παρατηρηστε $153=1^{3}+5^{3}+3^{3}$και
4)προσδιοριστε ολους τους τριψηφιους αριθμους που ισουνται με το αθροισμα των κυβων των ψηφιων τους

Τα 3 πρωτα εχουν πολυ ευκολες αποδειξεις...Το 4 θελει λιγο δουλεια...
(Made in RAKAR....)

 :D  :D  :D

program askisirakar4;
 var a,b,c,d: integer;
begin
  for a:=1 to 9 do
   begin
    for b:=1 to 9 do
     begin
      for c:=1 to 9 do
       begin
        d= a*a*a+b*b*b+c*c*c
        if d=100a+10b+c then writeln d;
       end
     end
   end
end.

epeidi o compiler mou de dexetai copy paste as to kanei kapoios...
(omoios kai ta 1,2,3 :rotfl:)

E-L-E-O-S!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!......... :o
AMA itan na ftiaksis program....as mu estelnes pm na su elega tin lisi....... :wink:
pantos gia ta 3 prota min ftiaksis program......8a xriastis poli ligotero xrono gia na ta lisis me to xeri........ :P
oxi oti afto pu eftiakses exi xristiki aksia.....(mpori na lisi monoafto to provlima +kanena alo.. ine......programa mias xriseos... )
pu na to ekteleso??g;osoma8ia,fortran,i c++???? :rotfl:
den ftei o compiler su.....mia xara ine.....meta to if ...vale kai kanena else......kai kana end if........+8a to kani+copy+paste..... :rotfl:

 :D  :D  :D

me mia apli allagi omws mporei na lisei ola ta provlimata autou tou tipou

:o  :o ...apisteyto...!!!
o ipologistis vrike +ali lisi

[color=blue]$371=3^{3}+7^{3}+1^{3}$[/color]

 :D

αυτο το ερωτημα 4 ειναι πολυ ενδιαφερον.προσπαθω να βρω ενα τροπο χωρις τη χρηση υπολογιστη πχ χρησιμοποιοντας ενα λημμα που λεει οτι ενας τετοιος αριθμος δεν ειναι της μορφης 9k+4 ή 9k+5. αλλη ιδεα ?και κατι αλλο ,αν δεν ισχυει η ισοτητα τοτε ποιος αριθμος ειναι μεγαλυτερος?

[quote="kostas"]χρησιμοποιοντας ενα λημμα που λεει οτι ενας τετοιος αριθμος δεν ειναι της μορφης 9k+4 ή 9k+5.αλλη ιδεα ?και κατι αλλο ,αν δεν ισχυει η ισοτητα τοτε ποιος αριθμος ειναι μεγαλυτερος?[/quote]
ti lei akrivos kai pianu ine to lima mipos gnorizis??...
i ari8mi poy exun ayti tin idiotita legonte ari8mi amstrong..(etsi tus exo vri tulaxiston..)
oso gia to pio apo ta dio meli ine megalytero o ari8mos i i kivi... den fenete na yparxi kapiois kanonas...o ipologistis edikse oti se 463 ari8mi i kivi ine megaliteri apo ton ari8mo eno 433 ari8moi ine megaliteri apo tus kivus ton yifion tus...+2 ari8mi ine isi me aytus....

 :D

το λημμα λεει οτι η παρασταση a^3+b^3+c^3 den einai pote =4,5 mod9
αποδ:θεωρουμε τα δυνατα υπολοιπα ενος κυβου mod 9

aa..ayto enouses?...ayto prokipti apo to oti o kivos ka8e akereu ine isoipolipos -1,0,1 mod 9...sosta?den nomizo omos oti voi8ai kai poli.... :wink:
ego paratirisa oti kai stis dio lisis 153,371 ta yifia ine ola periti ari8moi...isos ayto na simeni kati va8itero....

 :D

edit:
i eksisosi
 $100a+10b+c=(2k+1)+a^{3}+b^{3}+c^{3}$ poles lisis..opos p.x
$437=3+4^{3}+3^{3}+7^{3}$ kai poles ales...exo vri tulaxiston 12..

ala i eksisosi $100a+10b+c=2k+a^{3}+b^{3}+c^{3}$mexri stigmis den exo vri na exi lisis...mpori na simeni ayto kati gia tin artiotita ton yifion tu ari8mu abc??...

(edit)^2:krima...pai to 8evrimataki... :P
exi lisi ...+malista exo vri dio..
$226=-6+2^{3}+2^{3}+6^{3}$
kai $372=-6+3^{3}+7^{3}+2^{3}$......

to provlima vevea genikeyete os eksis...
 na vre8un i k-yifii ari8mi i opii isunte me to a8risma ton k-ston dinameonton yifion tus....

p.x
      $1634=1^{4}+6^{4}+3^{4}+4^{4}$

 :D

[quote="RAKAR"]1) Nα προσδιορισετε ολους τους διψηφιους αριθμους που ισουνται με το αθροισμα των τετραγωνων των ψηφιων τους

3) Nα προσδιορισετε ολους τους τριψηφιους αριθμους που ισουνται με το αθροισμα των τετραγωνων των ψηφιων τους[/quote][url=http://www.mathematica.gr/forum/viewtopic.php?f=111&t=6477]εδώ[/url]


Alexio has 100 cards numbered 1-100, inclusive, and places
them in a box. Alexio then chooses a card from the box at
random. What is the probability that the number on the card he
chooses is a multiple of 2, 3 or 5? Express your answer as a
common fraction.

Using PIE, the number of cards that are a mutliple of 2, 3, or 5 is $ 50 \plus{} 33 \plus{} 20 \minus{} 16 \minus{} 10 \minus{} 6 \plus{} 3 \equal{} 74$. Thus the probability is $ \boxed{\frac{37}{50}}$

How come you add three and not subtract it?

They are the multiples of 30 (2*3*5) that you left out before.


geometry
geometric transformation
reflection
incenter
trigonometry
circumcircle
perpendicular bisector

Let ABC be an isosceles triangle $(AB=AC)$ with $\angle BAC = 20^\circ$. Point D is on side AC such that $\angle DBC = 60^\circ$. Point E is on side AB such that $\angle ECB = 50^\circ$. Find with proof, the measure of $\angle EDB$.
[hide="hint"]Looks can be deceiving :evilgrin: [/hide]

Indeed interesting
[hide="solution"]
Reflect point $D$ to $D'$ on $\overline{AB}$ such that $AD' = AD$. Now draw $D'C$. Call the intersection of $\overline{D'C}$ and $\overline{DB}$ point $P$. Since everything is symmetrical, $\angle D'CB = \angle DBC = 60^\circ$. Now it is obvious that $\triangle PBC$ is equilateral. Similarly, $\triangle PD'D$ is also equilateral. From the information given, we can calculate $\angle BEC = 50^\circ = \angle BCE$. Therefore, $BE = BC = BP$. We now know $\triangle BEP$ is isosceles. Therefore, $\angle BEP = \angle BPE = 80^\circ$. This implies $\angle DPE = \angle DD'E = 100^\circ$. Since $\triangle PD'D$ is equilateral, $DP = DD'$. This means $\angle D'PE = \angle PD'E = 40^\circ$. Therefore, $EP = ED'$. Hence, $\triangle DPE \cong \triangle DD'E$ by the $SAS$ conjecture. Finally, $\angle EDB = \angle D'DP / 2 = 60^\circ / 2 = 30^\circ$.
[/hide]

[quote="stupidkid"]Indeed interesting
[hide="solution"]
$\angle D'CB = \angle DBC = 60^\circ$
[/hide][/quote]
This is confusing. We know $AB=AC$, and $\angle BAC = 20^\circ$. This tells us, $\angle ABC = 80^\circ=\angle ACB=\angle DBC$. Please explain. :huh:
I think you made a mistake somewhere.
[quote="shobber"]This is surely not an intermediate problem.

http://www.mathlinks.ro/Forum/viewtopic.php?t=101987
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=angle+solutions&t=39387
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=incentre+reflect&t=64298
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=angle+reflect&t=71052[/quote]
Thanks for the links shobber. :)
Could a mod please move this, thanks.

I don't know how to explain it. I reflected $D$ to $D'$, so everything is symmetrical. Meaning $D'C = DB$ and such. Maybe this isn't rigorous enough...

[hide="hint"]Looks can be deceptive. This isn't as elementary as it looks, as shobber said.[/hide]
I will post a very big hint soon. (besides what I just said). :P

[color=darkblue][b]A syntetical proof.[/b] Denote the point $F\in (AC)$ so that $m(\widehat{CBF})=20^{\circ}$. Prove easily that $BE=BC=BF=EF=DF$,
the triangle $DFE$ is isosceles ($FD=FE$) and $m(\widehat{DFE})=40^{\circ}$. Therefore, $m(\widehat{BDE})=30^{\circ}\ .$[/color]

Even thought the problem has been solved (thanks to Virgil Nicula and stupidkid :) ), I will post the promised hints.
[hide="1"]Draw segment DF parallel to BC with F on AB. Draw CF intersecting BD at G. Now find the equilateral triangles and isosceles triangles.[/hide]
[hide="2"]Use the Law of Sines in triangle BED and triangle BCD. Use BE = BC to connect the results. Simplify and solve for $\angle EDB$.[/hide]
[hide="3"]Draw lines through D and B parallel to BC and DC, respectively, intersecting at H. Draw CG with G on BD and $\angle GCB = 60^\circ$. Show E is the incenter of triangle BDH.[/hide]
[hide="4"]Mark K on AC such that $\angle KBC = 20^\circ$. Draw KB and KE. Show $BE= BC= BK= KE= KD$.[/hide]
[hide="5"](Maria Gelband) Reflect E through AC to point H. Show D is on the circumcircle of triangle BEH.[/hide]
[hide="6"](Sergei Saprikin) Let the bisector of $\angle ABC$ intersect AC at point T. Show D is an excenter of triangle BET.[/hide]
[hide="7"](Alexey Borodin) Let O be the circumcenter of triangle DEC. Show BD is the perpendicular bisector of EO.[/hide]
[hide="8"](Alexander Kornienko) Reflect triangle ABC through AB to triangle ABC' and also relect it through AC to triangle ACB'. Show that C', E, and D are collinear.[/hide]
[hide="1-4"]These were provided by Tom Rike.[/hide]
Hope these help. :)

[hide="solution"]From the given information we know that $\angle ABC = \angle ACB = 80^\circ$. Since angles in a circle add to $180^\circ$ we know that $\angle BEC = 50^\circ$. So $\angle BEC = \angle BCE = 50^\circ$, and $\triangle BCE$ is isosceles with $BC = BE$.
Construct $F$ on $AC$ so that $\angle BFC = 80^\circ$. So $\angle BFC = \angle BCF = 80^\circ$, and $\triangle BCF$ is isosceles with $BC = BF$.
We can find $\angle BDF = 40^\circ$, and $\angle FBD = 40^\circ$. So $\triangle BDF$ is isosceles with $BF = DF$.
Since $\angle EBF = 60^\circ$ and $BE = BF$, then $\triangle BEF$ is equilateral, so $EF = BE = BF = DF$. Thus $\triangle FDE$ is isosceles, and we find $\angle DFE = 100^\circ-60^\circ = 40^\circ$ then $\angle EDC = 70^\circ$. Since $\angle BDF = 40^\circ$, the remainder $\angle EDB = 30^\circ$.
[/hide]

[color=darkblue][b]Lemma.[/b] If given is a convex quadrilateral $ABCD$, then there is the following trigonometrical relation:
$\boxed{\sin\widehat{ABD}\cdot\sin\widehat{BCA}\cdot\sin \widehat{CDB}\cdot\sin\widehat{DAC}=\sin\widehat{BAC}\cdot\sin \widehat{CBD}\cdot\sin \widehat{DCA}\cdot \sin\widehat{ADB}}\ \ (1)\ .$
[b][u]Proof.[/u][/b] Denote the intersection$P\in AC\cap BD$. Apply the Sinus's theorem in the triangles$PAB$,$PBC$,$PCD$,$PDA$:
$\frac{PA}{PB}=\frac{\sin \widehat{ABD}}{\sin \widehat{BAC}}\ ;\ \frac{PB}{PC}=\frac{\sin \widehat{BCA}}{\sin\widehat{CBD}}\ ;\ \frac{PC}{PD}=\frac{\sin\widehat{CDB}}{\sin \widehat{DCA}}\ ;\ \frac{PD}{PA}=\frac{\sin\widehat{DAC}}{\sin\widehat{ADB}}\ .$From the product of these relations obtain the relation$(1)$.

[b]The (ugly) trigonometrical proof of the proposed problem.[/b] Denote $x=m(\widehat{BDE})$ and apply the above lemma to the convex quadrilateral $BCDE$ : $\sin x\sin 50^{\circ}\sin 60^{\circ}\sin 30^{\circ}=\sin 40^{\circ}\sin (110^{\circ}-x)\sin 20^{\circ}\sin 50^{\circ}$ $\Longleftrightarrow$ $\sin x\cos30^{\circ}\sin 30^{\circ}=\sin 40^{\circ}\sin20^{\circ}\sin (70^{\circ}+x)$ $\Longleftrightarrow$ $\sin x\sin 60^{\circ}=\cos (20^{\circ}-x)(2\cos 20^{\circ}-1)$ $\Longleftrightarrow$ $\sin (x+30^{\circ})+\sin (x-30^{\circ})=\cos (40^{\circ}-x)+\cos x-\cos (20^{\circ}-x)$ $\Longleftrightarrow$ $\sin (x-30^{\circ})=[\sin(90^{\circ}-x)-\sin (x+30^{\circ})]+[\sin (50^{\circ}+x)-\sin (70^{\circ}+x)]$ $\Longleftrightarrow$ $\sin (x-30^{\circ})=2\sin (30^{\circ}-x)\cos 60^{\circ}-2\sin 10^{\circ}\cos (60^{\circ}+x)$ $\Longleftrightarrow$ $\sin (30^{\circ}-x)=0\ \ \vee\ \ 0=1+2\cos 60^{\circ}-2\sin 10^{\circ}$ $\Longleftrightarrow$ $x=30^{\circ}$, i.e. $m(\widehat{BDE})=30^{\circ}\ .$[/color]


geometry
rhombus
trigonometry
trig identities
Law of Sines

Find the area of rhombus $ABCD$ given that the radii of the circles circumscribed around triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively.

I liked this problem, so I'll post a solution.  :lol:

[hide="Solution"]The circumradii give us an inclination that trig is involved. We start with the Law of Sines. Let $\angle BAC=x\implies \angle ABD=90-x$ and let the intersecton of the diagonals be $E.$

Using the Law of Sines on triangle ABD,
\[25=\frac{AD}{\sin (90-x)}=\frac{AD}{\cos x}. \]
Similarly, Law of Sines on triangle ACD yields
\[50=\frac{AD}{\sin x}. \]
Note that dividing the first equation by the second tells us that $\tan x= 1/2 \implies BE/AE=1/2.$

Now, let $BE=n.$ It follows that $AE=2n$ (we just found tan x) and $AB=n\sqrt{5}.$ (pythagoras) Note that we can now find the area of triangle $ABD$ (and the rhombus) using the area formula $[ABC]=2R^{2}\sin A \sin B \sin C.$

The area of the rhombus is
\begin{eqnarray*}[ABCD]=2[ABD]&=&2[2(12.5)^{2}\sin \angle BAD \sin \angle ABD \sin \angle BDA] \\ &=& 4 (12.5)^{2}(2\sin x \cos x) ( \cos x)( \cos x) \\ &=& 2\cdot 25^{2}\sin x \cos^{3}x \\ &=& 2\cdot 25^{2}(1/\sqrt{5})(2/\sqrt{5})^{3}\\ &=& 400. \end{eqnarray*}
[/hide]

Here is my solution on the Aops wiki:
The diaganols of the rhombus perpendicularly bisect each other. Call half of diagonal BD a and half of diagonal AC b. The length of the four sides of the rhombus is $ \sqrt{a^2\plus{}b^2}$. The area of any triangle can be expressed as the abc/4R, where a,b, and c are the sides and R is the circumradius. Thus, the area of triangle ABD is $ ab\equal{}2a(a^2\plus{}b^2)/(4\cdot12.5)$. Also, the area of triangle ABC is $ ab\equal{}2b(a^2\plus{}b^2)/(4\cdot25)$. Setting these two expressions equal to each other and simplifying gives b=2a. Substitution yields a=10 and b=20, so the area of the rhombus is $ 20\cdot40/2\equal{}400$.

Don't know I am correct or not...Here is the solution ...plz verify this....
[hide]Let $M$ be the midpoint of the side $AB$. Now the perpendicular bisector of $AD$ is $MN$ which intersects $AC$ and $BD$ at points $P$ and $N$  respectively . Again assume that $AC \cap BD =Q $ and
So, from our construction above , it is clear that $AP$ and $AN$ are the circumradii of $ABD$ and $ADC$ respectively.
For $\triangle ABD$,we can write, $$[ABD]= \frac{AD^2\cdot BD}{4AP}$$
And for, $\triangle ADC$,
$$[ADC]= \frac{AD^2\cdot2AQ}{4AN}$$
Since $[ABD]$ and $[ADC]$ are equal, we get, $$BD=AQ$$. Now notice that, $\triangle AMN \sim \triangle APN \sim  \triangle ADQ $. And $\frac{AP}{AN}=\frac{12.5}{25}= \frac{1}{2}$. Therefore,$$AM=\frac{MN}{2}  \Rightarrow MN=AD$$
Now, using Pythagorean theorem in $ \triangle AMN$,we get,$$ AM^2+MN^2=AM^2 \Rightarrow \frac{5AD^2}{4}=625\Rightarrow AD=10\sqrt{5}$$
Applying the same to $\triangle ADQ$ , we get, $AQ=20$. Now, $[ABCD]$ should be $AQ^2$ since $AQ=BD=\frac{AC}{2}$. Therefore the answer is $400$.
[/hide]
Edit: I didn't notice the previous solution before my post and now I see I made it a bit complicated :)


There was nothing in the problem that suggested that the expression must be a sum of a finite number of terms. Consequently, my solution was simply an infinite sequence. I did give an expression for the nth term in the sequence. Any chance that I'm going to get marked off for this? I recognize that my solution was simpler to arrive at than the actual solution (which most of you posted), however, I don't see anything that makes my solution worse, as the problem did not specific indicate one solution over another.

This question was addressed here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=375710#p375710


Carson flips over the cards of a standard 52-card deck one at a time. What is the probability that he flips over the ace of spades before any face card (jack, queen or king)?

Let us consider a deck with just the Ace of Spades and all the other face cards.  Then there is a $ \frac{1}{13}$ chance that the Ace of Spades is on the top.  Now we can insert the rest of the deck in any way, without affecting the $ \frac{1}{13}$ chance, so that is indeed the answer.

Awesomely simple solution. I trudged my way along a scenic route using the hockey stick identity!

[quote=ilikecake]So I want to make sure I know this correctly: so the stack of cards (the whole deck) doesn't matter. Only the order or part of the 13 relevant cards matter. Like if there were 1 card between it, it would be the same as 10 cards between them. So basically whenever I see a problem like this, I only do or care about what it is refering to. Right??[/quote]

Yes, the rest of the stack of cards doesn't matter. Only the order of the 13 relevant cards do.
[size=50]Please don't tell me not to revive because reviving to answer a question is okay.[/size]

My first solution was that there's 1/4 chance for an ace to be picked (from the group ace, jack, queen, king), and  1/4 chance for that ace to be spades. So it would be 1/16, but thats wrong so where did I go wrong?

[quote=ehz2701][hide = the bash sol (which isn’t actually that bad.)]Label the cards from $1$ to $52$, and let the ace of spades be in position $1\leq j\leq 52$, where $j$ is of course, an integer. Then, for some fixed position $j$, the $12$ face cards have $\binom{52-j}{12}$ choices. Thus, the total amount of arrangements is just $\binom{52-1}{12} + \binom{52-2}{12}+ \cdots + \binom{12}{12} = \binom{51}{12}+\binom{50}{12} + \cdots + \binom{12}{12} = \binom{52}{13}$, by Hockey Stick Identity. Then, the total number of arrangements is $52 \binom{51}{12}$, since there are $52$ places the ace of spades can go to, and then the remaking $51$ cars the $12$ face cards. Thus, our probability is $\frac{\binom{52}{13}}{52\binom{51}{12}} = \frac{1}{13}$[/hide][/quote]

[hide]simple way 1/13[/hide]


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

* The index is calculated as follows: AMC score + 10 x(AIME score).

USAMO qualifers must attend a US or Canadian school or be a US Citizen.  Students who wrote the AMC 12 and achieved an index of at least 197.5  qualified for the USAMO.  This was a total of 325 students.

Students in 10th grade or below who scored 6 or higher on AIME
AND
have an index of  at least 183.5 also qualified for the USAMO.
This was a total of an additional 171 students.

A student with the highest index in a state not already represented was also invited to write the USAMO.   This was 9 students (inlcuding one pair of tied students) from 8 states.

Total number of invited students is 505.

The list of invited stdents will be on our webpage, http://www.unl.edu/amc
by sometime on Friday, April 6, 2007.

Emails to invited students are now being sent, emails to teachers of invited students will be going out later today, Thursday, April 5, 2007.

Steve Dunbar
Director, American Mathematics Competitions

Thanks for the info, and the generosity of CAMC! Both the floor and the index were much lower than expected, and the expansion of 500 was barely exceeded...wow

Apparently I was the first response last year too. Why am I always the first one to see this message?

Wow, very low cutoffs. Interesting year.

So the scores for the AIME I and II were statistically the same?

e-mails to invited students? I don't remember giving my e-mail...was that on the form?

So does this mean all the whining/complaining/begging/wishing ppl have done on this forum are over

Since I believe many of them had overestimated the index
and
have just now received emails asking for them to write the USAMO

 :rotfl:

These are amazingly low cutoffs. I'm just wondering. Are these indices posted on the AMC website yet or are we getting a look first? I do agree that we all overestimated the cutoffs. I was seriously expecting the index for AIME II to be somewhere in the 220s, and for it to be under 200 is insane.

I'm still anxiously awaiting my e-mail...I hope I get it soon :w00t:

Second Thought:

Don't many people own the AMCDirector aka Steve Dunbar
an apology for these reasons:
1) Insulting the AMC method of selection for USAMO
2) Complaining/whining/begging/wishing to retake the AIME ( :rotfl:  a topic that turned into an 18 page BEAST)
3) Over estimating the index before getting results

Edit:
Yeah my reason number 3 is bad
i apologize
i just wanted three reasons :ninja: (don't know why)

The AIME II was not really easier than the AIME I at all if any. Most people who took it unofficially need to realize that it's because they took it unofficially.

[quote="Aznwithabrain"]Second Thought:

Don't many people own the AMCDirector aka Steve Dunbar
an apology for these reasons:
1) Insulting the AMC method of selection for USAMO
2) Complaining/whining/begging/wishing to retake the AIME ( :rotfl:  a topic that turned into an 18 page BEAST)
[b]3) Over estimating the index before getting results[/b][/quote]

That makes absolutely no sense. Why would you apologize for being bad at estimating the index?

[quote="Hamster1800"][quote="kstan013"]Has anyone else that qualified received their e-mail yet? I'm starting to get a little nervous...[/quote]

Not me.

Zenith Xan just got his and his number was pretty low, so I think they're just starting.[/quote]

Mm yeah, it's good to exercise patience. They're being sent individually (although automatically?), so it'll still take some time.

Hmm..so the AIME I and AIME II were both statistically equal??  Judging from what I heard, I expected the AIME II cutoff to be higher (maybe even way higher) than the AIME I cutoff..but I guess not.

[quote="MysticTerminator"]I think freshmen are automatically in red.

not sure why this is[/quote]

There are actually many exceptions (okay, not many, but definitlely some).

Last year there were students who made the blue MOP cutoff but were placed in red MOP.  It may be possible to make black MOP, but not blue.

Eric Larson and Delong Meng both had good enough scores to make Blue (Larson made it in 8th grade, in fact). Perhaps they were just saving room for more 10th graders.

[quote="nealth"]Phelpedo, how did you know that Eric and Delong were good enough for Blue? Were you in MOP last year?
Btw, both Eric and Delong are slackers this year, so neither of them would make MOP. Yichen and Dnas are going to make Black MOP this year.[/quote]
Dude stop slacking then... anyhow considering the competition I think black MOP may be reserved for juniors and seniors this year.

Zhai wrote:
[quote]Dude stop slacking then... anyhow considering the competition I think black MOP may be reserved for juniors and seniors this year.
[/quote]
Unfortunately, if you do the same operations for this year's USAMO!(add 1 and flip the digit), then many sophomores will be in Black MOP, and my score will not even qualify for Blue.
Btw, you should tell Lars and Delong to stop slacking instead of me.

dude who?

I know the IMO team

but not the rest of black MOP

[quote="Teki-Teki"]where did this 'add 1 and flip the digits' come from?  :huh:

i have to agree with jb05, the slacking is not funny.

wow - last 3 posts were NCians.[/quote]

I think he's (she?) probably referring to probability1.01, who, I believe, got 13 when he was a freshman and then 41 when he was a sophomore.

[quote="jb05"][quote="nealth"]Unfortunately, if you do the same operations for this year's USAMO!(add 1 and flip the digit), then many sophomores will be in Black MOP, and my score will not even qualify for Blue.
Btw, you should tell Lars and Delong to stop slacking instead of me.[/quote]

Err... what are you talking about?????

And the slacking thing isn't really funny.[/quote]

Quit being a party pooper, it's freakin hilarious!

Where can I find the AIME score distributions for 2007?

They're not out yet. It will be on the AMC website when it comes out.

OMG

I know that kid from Waterloo (I lived there up until this year).

OMG he/she was probably the kid I lost to on the Gauss math contest and on Pythagorous. OMG 2nd both times to a 3rd grader. AHHH!!

ok [/end spaz]

yup, the kid is a complete genius (these are all rumors from my chinese friends from Waterloo...apparently he/she knows basic Calculus and was taught trig as a first-grader OMG)

[quote="nealth"]
Btw, you should tell Lars and Delong to stop slacking instead of me.[/quote]
Speaking in 3rd person is a sign of a disturbed mind........ :lol:


Propose a mechanism for the Koenigs-Knorr Reaction:

http://en.wikipedia.org/wiki/Koenigs-Knorr_reaction

Don't cheat! :D

An important thing you should know is that $ \alpha$ and $ \beta$ anomers give the same product. So an extra question is: 'Is the resulting product an $ \alpha$ or $ \beta$ glycoside?

Well, at least the mechanism is not there.

Yes the relevant information is there
[quote="wikipedia"]The presence of the neighboring group at C2 lends to anchimeric assistance, resulting in the formation of a 1,2-trans stereochemical arrangement.[/quote]
After that 3-membered cyclic intermediate the ROH attacks equatorially because the axial is hindered by the ring.

Yes, that explanation seems to be right.


LaTeX logarithms limit induction calculus function derivative

Show that $\sqrt2^\sqrt2^\sqrt2^\mbox{.}^\mbox{.}^\mbox{.} $$=2$

Let your expression be x.

Then ${\sqrt{2}}^x=x$

Therefore, $x=2$

Q.E.D.

WOOT!!  I'm getting good a $\LaTeX$

[edit] I'm an idiot--see next post.

[re-edit] I'm not an idiot--I'm right.  YEAH!

Of course, it's a completely absurd statement.  That expression diverges to infinity.

oh really?  :blush:

[edit] oh nevermind--of course it does.  1.414^1.414^1.414 could never be 2.

[re-edit] exponentiation is a right-associative operator--I'm right.   WOOT!

It's like saying $\infty^{\infty}=\infty$, so on taking the base $\infty$ logarithm, we get $\infty=1$.  Don't worry, lots of people make this mistake.  A more constructive exercise is to show that

$a^{a^{a^{\ldots}}}=\infty$ for $a>1$

Actually the original post is correct -- the infinite expression does converge to 2.  Remember that exponentiation is parsed as a right-associative operator.  In other words, $a^{b^c}$ means $a^{(b^c)}$.

oh yeah--ha ha--I was right!!!  GO ME!!!

Your answer is right, although your proof needs work.  For one thing, you haven't actually proved convergence.  At best, you have shown that if the expression converges, then it converges to 2.  Even assuming convergence, though, you haven't really shown that the limit is 2.  The equation ${\sqrt{2}}^x = x$ has two positive solutions, $x = 2$ and $x = 4$.  You need to rule out 4 as the answer.

hmm.  I'm struggling as to how to write this expresssion as a limit because exponentiation is a right-association operator....

maybe $\lim_{n\to\infty}\sqrt{2}^\sqrt{2}... n times$ ????  And where do I go from there...

Yes, that's a more formal way of expressing what is being asked.  Maybe you can use induction on $n$ to show that the expression is at most 2?

maybe.  lol.

ok, here goes.  The base case is trivial.  Now, let's say that it works when $n=k$.

Now, if we have $n=k+1$, we just take $\sqrt{2}$ to the power of the thingy--aren't I articulate--for $n=k$.

By the induction hypothesis, we know that thingy is less than 2.

Now, ${\sqrt{2}}^m$, where m is less than 2, is obviously less than 2.

Q.E.D.

Okay, you have now shown the sequence is bounded above by 2.  You still need to prove the sequence converges (to 2).  One way is to show that the sequence is increasing.

[quote]
A more constructive exercise is to show that

$a^{a^{a^{\ldots}}}=\infty$ for $a>1$[/quote]

This is not correct. The general question of determining when $x^{x^{x^\ldots}}$ converges is an interesting one. You could try doing some calculator/programming exercises to make a good guess, but proving it is probably beyond the scope of this board.

It would make an interesting project to work on for students of advanced calculus though.

[hide="The answer"]
$\displaystyle x^{x^{x^\ldots}}$ converges for $\frac{1}{e^e} \leq x \leq e^{\frac{1}{e}}$[/hide]

By the way bubala, you could write your sequence recursively as $\displaystyle a_1 = x$, and $\displaystyle a_n = x^{a_{n-1}}$ for $n \geq 2$. Then you're looking for $\lim_{n \to \infty} a_n$

this might be waaay wrong, but maybe we can treat $a_n$ as a function fo $a_{n-1}$?

In other words, we have $a_n={\sqrt{2}}^{a_{n-1}}$.

Now, if we take the derivative, we get $ln\sqrt{2}*{\sqrt{2}}^{a_{n-1}}$, which of course, is positive for all $a_{n-1}$.

Therefore, $a_n$ is always more than $a_{n-1}$--so the sequence is increasing.  Q.E.D.

Sorry, that argument didn't make sense.  Let $f(x) = {\sqrt{2}}^x$.  You want to show that $f(x) \ge x$ for all $x \le 2$.  Then you can conclude that the sequence is increasing.

[quote="gauss202"]
This is not correct.[/quote]

mm, old assumptions die hard, I guess.

My guilty conscience prompted me to post a solution for writing nonsense earlier.

[hide]Okay, monotone bounded sequences converge to their sup.  Set $a_0=\sqrt{2}$, $a_{i+1}=\sqrt{2}^{a_i}$.

Let's start by proving monotonicity.

A lemma: if $a>1$ and $b>c$, then $a^b>a^c$.  Proof: $a^x$ is increasing for $a>1$.

We claim that $\sqrt{2}^{a_i}\geq a_i$.  Proof: induction.  Clearly $\sqrt{2}^1\geq 1$, so that $\sqrt {2}^{\sqrt{2}}\geq \sqrt{2}$, and the result follows by induction.  By this, since $\sqrt{2}^{a_i}=a_{i+1}$, we get monotonicity.

Now for boundedness:  We claim the sequence is bounded above by $2$

For boundedness, we have $a_0<2$, and if $a_i<2$, then $a_{i+1}=\sqrt{2}^{a_i}<\sqrt{2}^2=2$.

Next.  Monotone bounded sequences have limits.  Hence we must have $\lim a_i=\sqrt{2}^{\lim a_i}$.  This has two solutions - $\lim a_i=2$ or $\lim a_i=4$.  We reject the latter because the sequence cannot converge to a value greater than $2$ (since $2\geq \sup_ia_i\geq \lim a_i$ for this sequence)[/hide]

ok, well although blahblahblah has already solved it, I am going to finish mine.

Mr. B (not finishing your name due to restrictions of AoPS privacy policy), are you sure that my last post made no sense?  I basically did what you did, only maybe my notations was wrong.

[hide="last step in "my" solution"]I basically did the same thing as you did, except I had $a_n={\sqrt{2}}^{a_{n-1}}$ rather than $f(x)={\sqrt{2}}^x$ and used $a_n$ as a function of $a_{n-1}$.

Anyhow, using your way, we have $\frac{d}{dx}f(x)=ln\sqrt{2}*{\sqrt{2}}^x$, which is positive for all $x$, so $f(x)$ is increasing.  From this and other posts, q.e.d.  [/hide]

btw, blahblahblah, I used to think that composite numbers didn't include sqares like 9 and 16, and I even wrote my proof to USAMO #1 using my old definition.  OOPS! :blush:


integration geometry circumcircle calculus inequalities topology

Let $V$ be a compact convex body in $R^{n}$. Prove the existence of a unique real number $r$ such that for any finite set of points in $V$ there is a point in $V$ such that the arithmetic mean of distances from this point to the points of the finite set equals $r$.

The case $n=1$ is just at the limit of my abilities.  :|  Let $V=[a,b]$; I claim that $r=(b-a)/2$. Indeed, if $c_{1},\dots,c_{n}\in V$, then $\int_{a}^{b}\left\{n^{-1}\sum_{i}|x-c_{i}|\right\}\,dx=(b-a)^{2}/2$, which implies the existence of $x\in [a,b]$ with $n^{-1}\sum_{i}|x-c_{i}|=(b-a)/2$.

For uniqueness set $c_{1}=a$ and $c_{2}=b$.

I read in Guy's "Unsolved Problems in Geometry" that the same statement, for any compact connected subset $V$ of $\mathbb R^{d}$, is due to Gross. Apparently, it's even true for all compact connected metric spaces. The number is usually called the [i]magic number[/i] of $V$. Szekeres gave a simple proof of the fact that if $V$ is a convex body, then this number must be equal to the circumradius of $V$ (so it is indeed $\frac{b-a}2$ when $V$ is an interval $[a,b]$ in $\mathbb R^{1}$).The results were just stated, the proofs weren't even discussed, so I don't really know how difficult they are.

I can only prove the existence part, but the proof works for any compact connected metric space $(V,d)$:

For any finite set $\mathcal F\subset V$, denote by $A_\mathcal F$ the map sending every point in $V$ to its average distance to the elements of $\mathcal F$. The existence part of the problem is simply the fact that $\bigcap_\mathcal FA_\mathcal F(V)\ne\emptyset\ (*)$. Because $V$ is a compact connected metric space, $A_\mathcal F(V)$ is a compact interval for all finite sets $\mathcal F\subset V$, so $(*)$ follows if we manage to prove that for two finite sets $\mathcal F_{1},\mathcal F_{2}\subset V,\ A_{\mathcal F_{1}}(V)\cap A_{\mathcal F_{2}}(V)\ne\emptyset\ (**)$.

Let $\mathcal F_{1}=\{p_{i}\}_{i=\overline{1,n}}$ and $\mathcal F_{2}=\{q_{j}\}_{j=\overline{1,m}}$. Now consider the number $\frac 1{mn}\sum_{i}\sum_{j}d(p_{i},q_{j})$. On the one hand, it's equal to $\frac{1}{m}\sum_{j}A_{\mathcal F_{1}}(q_{j})$, i.e. a convex combination of points in the interval $A_{\mathcal F_{1}}(V)$, so it belongs to this interval, and, on the other hand, it's $\frac{1}{n}\sum_{i}A_{\mathcal F_{2}}(p_{i})$, so it belongs to $A_{\mathcal F_{2}}(V)$. This proves $(**)$.

I'll try to prove the uniqueness part as well.

P.S.

[quote="mlok"]$\int_{a}^{b}\left\{n^{-1}\sum_{i}|x-c_{i}|\right\}\,dx=(b-a)^{2}/2$[/quote]

Why is that? :? The integral of $|x-c|$ over $[a,b]$ is not always $\frac{(b-a)^{2}}2$; it depends on $c$.

I will sketch a proof of uniqueness. Let $R$ be the circumradius of $V$. And let $B(M, R)$ be the ball of radius $R$ that contains $V$. Clearly, $M\in V$.

First of all $r\leq R$, since the distance to $M$ is at most $R$ (i.e. the average distance to points of $\{M\}$ is at most $R$, and there should be a point $X\in V$ s.t. $d(X,M) = r$).

Here is a sketch of the proof that $r\geq R$.
1) There exists a finite set of points $X_{1}, \dots, X_{k}\in V$ s.t. $B(M,R)$ is the ball of minimum radius that contains all of them.

2) Consider the map $D: V\to{\mathbb R}^{k}$ defined as
\[D: X\mapsto (d(X, X_{1}), \dots, d(X, X_{k})). \]
In particular, $D(M) = (R, \dots, R)$. Let $H$ be the convex hull of $D(V)$. Note, that the set $H$ and the set $\{a_{1}, \dots, a_{k}: \forall i\ a_{i}< R\}\subset{\mathbb R}$ are disjoint. Indeed, otherwise, for some points $P_{1},\dots, P_{m}$ and coefficients $\lambda_{1}, \dots, \lambda_{m}\geq 0$ we have
\[\forall j\in\{1,\dots, k\}\ \ \ \sum_{i=1}^{m}\lambda_{i}d(P_{i}, X_{j}) < R, \]

\[\sum_{i=1}^{m}\lambda_{i}= 1. \]
Then, by the triangle inequality, we get that for the baricenter $P$ of points $P_{i}$ with weights $\lambda_{i}$,
\[\forall j\in\{1,\dots, k\}\ \ \ d(P, X_{j}) \leq \sum_{i=1}^{m}\lambda_{i}d(P_{i}, X_{j}) < R. \]
This contradicts to (1).

3) Since the sets $H$ and $\{a_{1}, \dots a_{k}: a_{i}< R\}$ are disjoint, there exists a separating hyperplane between them. I.e. there exist coefficients $\mu_{1}, \dots, \mu_{k}$ s.t.
\[\forall (y_{1}, \dots, y_{k})\in H\ \ \ \sum_{i=1}^{k}\mu_{i}y_{i}\geq R \]

\[\sum_{i=1}^{k}\mu_{i}= 1, \]

\[\mu_{i}\geq 0. \]
4) By the condition, for every finite set $S$ there exists a point $X$ s.t. the average distance from $X$ to elements of $S$ is $r$. We can generalize this condition. First, the condition holds if $S$ is a multiset (for every point $P$ in $S$ pick its "duplicate" $P'$ in a sufficiently small neighborhood of $P$, then consider the limit). Therefore, we may assume that points in $S$ have rational weights. Considering the limit, we may assume that points in $S$ have positive real weights.

5) In particular, there exists a point $Y\in V$ s.t.
\[\sum_{i=1}^{k}\mu_{i}d(Y, X_{i}) = r. \]
I.e. for $y_{i}= d(Y, X_{i})$,
\[\sum_{i=1}^{k}\mu_{i}y_{i}= r, \]
By (3), we conclude that $r\geq R$. This finishes the proof.

Some time ago, I proved the following generalization (I read the statement in an online encyclopedia article on metric spaces):

Let $X$ be a Hausdorff, compact and connected topological space, and let $f: X\times X\to\mathbb R$ be a continuous symmetric function (that is, $f(x,y)=f(y,x),\ \forall x,y\in X$). There is, then, a unique real number $r$ such that for all regular Borel probability measures $\mu$ on $X$ we can find $x\in X$ with $\int_{X}f(x,y)\ \mbox d\mu(y)=r$.

I'm posting a proof for the existence of $r$ (which mimics the one in my message above), and later, when/if I manage to reconstruct my proof of the uniqueness, I'll either modify this message or, if it's too late and I can't do that anymore, I'll post another one. All I can remember right now is that I used a device similar to the one in Yury's message (separating hyperplanes, an application of Minkowski-Farkas, something in that area)

Denote $f(x,\cdot): X\to\mathbb R$ by $f_{x}$, and, given the regular Borel probability measure $\mu$ on $X$, define $\varphi_\mu: X\to\mathbb R$ by $\varphi_\mu(x)=\int_{X}f_{x}\ \mbox d\mu$. Finally, let $I_\mu=\varphi_\mu(X)$. Notice that $\varphi_\mu$ is continuous. In fact, the map $x\mapsto f_{x}$ is a continuous map from $X$ into the Banach space $\mathcal C(X,\mathbb R)$ of continuous real-valued functions on $X$ normed with the supremum norm, and $\varphi_\mu$ is just the composition between this map and $\mu$, the latter being regarded as a continuous linear functional on $\mathcal C(X,\mathbb R)$. Since $X$ is compact and connected, $I_\mu$ must be a compact interval of the real line. Hence, in order to prove that all $I_\mu$ intersect, it suffices to show that each two do.

Let $\mu,\nu$ be measures (of the kind we're interested in; regular, Borel, etc.). Then, on the one hand, the number $\int_{X}\varphi_\mu\ \mbox d\nu$ belongs to $I_\mu$, being an average of a function with values in $I_\mu$. On the other hand, that integral is equal to $\int_{X\times X}f(x,y)\ \mbox d\mu(y)\ \mbox d\nu(x)$. After replacing $f(x,y)$ with $f(y,x)$ and exchanging the integration order (Fubini), we get $\int_{X}\varphi_\nu\ \mbox d\mu$, which belongs to $I_\nu$ (same argument as a few lines above, with $\mu$ and $\nu$ interchanged). Hence every two $I_\mu,\ I_\nu$ intersect, and the existence part of the proof is over.

ArXiv has [url=http://arxiv.org/find/grp_math/1/ti:+AND+Rendezvous+numbers/0/1/0/all/0/1]three recent articles[/url] on "rendezvouz numbers" (by the same authors).

Thanks for the link.

Now, on with the proof of uniqueness for $r$.

Assume that, contrary to what we want to prove, there is an interval $[a,b],\ a<b$ contained in all intervals $I_{\mu}$ (same notation as above). Because, as mentioned in my previous post, $x\mapsto f_{x}$ is continuous, the set $\{f_{x}\ |\ x\in X\}$ is compact in the metric space $\mathcal C(X,\mathbb R)$. It follows that for any $\varepsilon>0$, we can find finitely many points $x_{1},\ldots,x_{n}$ such that every $f_{x}$ is within $\varepsilon$ from some $f_{x_{i}}$: $\forall x,\ \sup_{t\in X}|f_{x}(t)-f_{x_{i}}(t)|<\varepsilon$. It follows from this inequality that for every probability measure $\mu$, the value of $\int f_{x}\ \mbox d\mu$ differs from $\int f_{x_{i}}\ \mbox d\mu$ by at most $\varepsilon$. If $\varepsilon$ is small enough, we'll be able to find $c<d,\ [c,d]\subset[a,b]$ such that for every $\mu$, there are $i,j\in\overline{1,n}$ with $\varphi_{\mu}(x_{i})<c<d<\varphi_{\mu}(x_{j})\ (*)$. In particular, this applies to all $\mu$ supported by $\{x_{i}\}_{i=1}^{n}$.

Now we look for $\mu$ supported by the $x_{i}$ (i.e. convex combinations of the finitely many points $x_{i}$) which contradict $(*)$. Assume WLOG that $d=0$. We forget all about $X$ and work only with the points $x_{i}$. Instead of $X\times X$, we now have the matrix $(x_{i},x_{j}),\ i,j=\overline{1,n}$. We know from $(*)$ (with $d=0$) that for all $a=(a_{j})_{1}^{n},\ a_{j}\ge 0,\ \sum a_{j}=1$, we can find $x_{i}$ such that $\sum a_{j}f(x_{i},x_{j})>0$. In other words, if $A$ is the matrix with entries $f(x_{i},x_{j})$ and $a$ is any non-zero column vector with non-negative entries (by rescaling the vector we can disregard the fact that initially its entries summed up to $1$), the row vector $a^{t}A$ has at least one positive entry. If $A$ were any matrix, it would follow from the Farkas-Minkowski Lemma that we can find a column vector $b$ with non-negative entries such that $Ab$ has positive entries. Because $f$ is symmetric, however, $A$ is a symmetric matrix, and hence $Ab$ having positive entries is equivalent to $b^{t}A$ having positive entries. This, however, contradicts the existence of $x_{i}$ satisfying the first inequality in $(*)$.