Please post your opinion:

(Translated by AltaVista)

The kidnapped young person identifies itself with his detector Natascha Kampusch, that now is 18 years old, remained captive from the 10 in the cellar of a house in the outskirts of Vienna. His kidnapper, an electrician of 44 years killed itself throwing itself underneath a train. According to the experts, the young sample symptoms of the "syndrome of Stockholm". Austria does not leave the astonishment and the commotion by the case of Natascha Kampusch, a young person who had been kidnapped with 10 years of age and reappeared yesterday, eight years later, with 18. According to one inquired, after the appearance of young the his detector was killed throwing itself under the routes of the train, to the north of Vienna. On the other hand, the psychologists who attend the Natascha detailed that he shows symptoms of the "syndrome of Stockholm", a phenomenon that usually is observed in victims of kidnappings and that imply the development of affection and attachment with the detector. Natascha had disappeared in March of 1998. A companion hers related then that boarded and was put in a light truck to the force by a stranger when it went to the school. She never knew herself more of her. Until now. According to one inquired, the young person she escaped yesterday and she managed to hide in the garden of a house in Strasshof, to the north of Vienna, near the house that became its jail during eight years. She was found by a woman who alerted to the police, after which the pale Natascha told him who was and that its adolescence catched in a cellar had lived all. "I am Natascha Kampusch", said to him to the authorities; and in addition, it revealed that his kidnapper, identified like Wolfgang Priklopil, a technician of electricity of 44 years, had left towards Vienna in his BMW 850. Shortly after one knew that the man had killed itself throwing itself under the routes of a train. Last night the police registered the house and found the hiding place. It had three meters in length, 1.6 meters wide and 2 of depth, constructed from the grave of an accessible garage and by a hollow of 50 by 50 centimeters that were closed with an electronic system. The hiding place had been transformed into a species of dormitory equipped for the young person. There was a bed and a small bookcase with infantile books, one radio and a television set. In that reduced space, Natascha lived at least eight years, although lately it could have left some times. The first declarations of the young person threw that it was not always locked up, but that in the last years his kidnapper allowed him to accompany it to the supermarket, and even, in some trip of vacations or strolls by the garden of the house. Nevertheless, Natascha seems to have undergone strong pressures or threats so that it is not contacted with strangers. Everything aims at that his kidnapper was practically the only person with whom he maintained relation in these eight years. The parents of the young person recognized it in a meeting that made cry to the victim. In a letter directed to the press and spread today by average Austrians, the parents requested "patience", and three days of tranquillity. For that reason, one hopes that the details of the case begin to know themselves slowly. In as much, the victim was under the supervision of psychologists of the police, who have adopted a strategy of extreme caution in their interrogations. The drama of Natascha would have arrived at its aim, although parececía to be to the doors of another one.

interesting... bad grammar by AltaVista though

just to let you know, AltaVista isn't a user on some random forum, it's an online translating device :p

[url=http://www.cnn.com/2006/WORLD/europe/08/25/austria.kidnap/index.html]CNN[/url] and BBC (which I apparently can't get access to right now) offer articles in English.

The story is amazing. What I would like to know is "WHY"... why did he, the kidnapper, do what he did. Maybe he wanted to have someone that can love him (that what he thought)

[url=http://news.bbc.co.uk/2/hi/europe/5285290.stm]BBC article in English[/url]

Well, I think it's the same as Belgium's Dutroux : paedophilia.
[url=http://news.bbc.co.uk/2/hi/europe/3829075.stm]BBC article on Dutroux[/url]

This girl in Austria caused quite a stir because Europe (and lots of countries in the world) were still shocked about the Dutroux case.

[quote="fredbel6"][quote="kostas"]well jose I don't kwow what kind of love is this.pure Natasha[/quote]
 :huh:  pure Natasha??[/quote]

Maybe "poor Natascha"

[quote="Kurt Gödel"][quote="fredbel6"]Well, I think it's the same as Belgium's Dutroux : paedophilia.[/quote]
Is this a good comparision? Of course, this was a kidnapping, but Dutroux kidnapped 6 kids and killed 4 of them, no? And this one's also different because of the Stockholm-syndrom-stuff. Really, I think the Dutroux-story was on a completely different level.[/quote]
Of course!

But we can say that now, but at that time in Austria in 1998, those people didn't know that.

[quote="Xevarion"]Is it just me, or is crime in the USA much more horrific than in other countries?[/quote]
You mean perception or the crime itself?

[color=white]No, fred, he means that the kidnapping of an American person is much more horrifying than the kidnapping of some Polish lady. The Holocaust should've wiped them all out, he's cleaning up the stragglers. That's what he's trying to say, fred, that's what he's trying to say.[/color]

Foresight tells me I should either delete the above or post a big [color=red]I'M BEING SARCASTIC, THIS IS WHITE HUMOR, HE MEANS THAT CRIME IN AMERICA IS THOUGHT OF MUCH MORE HORRIFIC-LY THAN IT IS IN OTHER COUNTRIES[/color] notice below my post. That or white it out.

Actually, what I meant was that there are crimes that were so much more violent, cruel, disturbing, or whatever in America that I am mildly surprised by the idea of there being a large amount of national interest in something as "tame" as kidnapping a girl for 8 years. (I recall at least two crimes like that in America, one of which was resolved relatively recently. The other was a girl called Elizabeth Smart, I think.)

edit: I should specify that I may be much more jaded than the average US citizen, and I have very little capacity for empathy. So it might just be me. On the other hand, the USA has way higher rate of murders per capita than most other countries in the world....

[quote="PenguinIntegral"]I can't tell if this is crude spam or some kind of subtle jab.

You're frick'in awesome when you do that.[/quote]

crude spam

man you're subtle too

i'm not sure if high-profile crimes tend to be worse in the u.s.
there is some pretty cruel and weird stuff that goes on in europe and other parts of the world.
consider this example. the movie hostel was based off of some underground hotel in thailand (i think) that advertised torturing people for an hourly rate.
of course thats just one example, but i do not think that high-profile crimes in the u.s. are necessarilly worse than those in other parts of the world.

[quote="Treething"]yeah... they're just covered more

like take that terry schiavo thing, in most any other country they would've let her go and made no big deal about it... here? people get upset about letting a vegetable die...[/quote]
Well, at least the USA is not forcing abortions on people with too many kids.  But yes, Terry Schiavo would have received euthanasia.  I never got the moral ethics.  It's not okay to let someone fall asleep forever peacefully, but it's okay to let them die of thirst?  Do it and do the job well or don't do it.

The worst things happen in Africa, there you have children who are abducted when they're 10, and force to serve as sex slaves for rebel leaders and fight (even kill their own parents).  Very notorious for this is the Lord's Resistance Army in Uganda.  It is often said that nobody cares about this due to Missing White Woman Syndrome.

[quote="Xevarion"]I am thinking of places where the government is stable enough that these kinds of crimes are due to a few individuals, not to the army or to large organizations like drug cartels.

I recall some incident in Germany where two guys met in an online chat room or something and met so taht one of them could eat the other... the Rammstein song Mein Teil is based on that event.

But does Europe have many people like Ed Gein, Jeffrey Dahmer, the Boston Strangler, the BTK dude (Suicide Commando has an album with that title; some of the songs are rather disturbing), Ted Bundy, Gacy, ...
Maybe it is just b/c in the USA, only USA crimes make big news most of the time, but the only one outside of US that I can think of is Jack the Ripper.[/quote]
You've been brought up in the US, of course you're going to remember mainly famous US crimes.

There's definately a lot of criminals in every country.

Actually I usually read BBC news when I read news at all, but that's not going to help me learn about crimes in the past.

And what about those boys that went into their school and killed lots of people (in Germany)??? that was terrible too

USA does it better, USA does it first.
http://en.wikipedia.org/wiki/School_massacre#Infamous_school_massacres

Actually, the most deadly one on that list is the Russia incident from 2 yrs ago, but that wasn't done by students.


calculus
real analysis
real analysis theorems

Has anybody used [i]Multivariable Calculus [/i]by Theodore Shifrin. If so, what did you think about the book?

Not sure if this is in the right forum (the calculus forum is centered more on analysis, which this book does not cover)

Well, I think it does belong in calculus/analysis, as an "other question." Personally, I'm not familiar with the book.


floor function
combinatorial geometry
combinatorics unsolved
combinatorics

how many convex 4-gons can be formed if we are given n non-colinear points in 2D?

The minimum number is $ f(n)\cdot\binom{n}{4}$, where $ f$ has the following properties: it's nondecreasing, positive for $ n\ge 5$, and bounded above by some $ c<1$. There is a limit, and I have no idea how to calculate it.

[quote="JBL"]Well, if the points are in convex position (e.g., along a circle) then the answer is just $ \binom{n}{4}$, but of course the answer can be smaller.  With five points in general position, there is always at least one convex quadrilateral (this is the [url=http://en.wikipedia.org/wiki/Happy_Ending_problem]Happy Ending problem[/url]), and I don't have any idea what the lower bound should be in general.[/quote]

if you put one point inside the convex polygonal region formed by n points, then the number of convex 4-gons formed is $ C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3)$+$ \binom{n}{3}*(n\minus{}3)/4$

if you put two points inside the convex hull formed by the aforementioned n points, that is, total n+2 points, then the total number of convex 4-gons formed by all those points is at least $ \lfloor_{n/2}\rfloor*n$ $ \minus{}$ $ \displaystyle\sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t}$ $ \plus{}$ $ 2*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3)$ + $ \binom{n}{3}*(n \minus{} 3)/4$ and at most $ \lfloor_{n/2}\rfloor*n$ $ \minus{}$ $ \displaystyle\sum_{t \equal{} 1}^{\lfloor_{n/2}\rfloor}{2t}$ $ \plus{}$ $ 2*C_{n \minus{} 2}/C_{n \minus{} 3}*(n \minus{} 3)$+$ \binom{n}{3}*(n \minus{} 3)/4$ + $ \displaystyle\sum_{k \equal{} 2}^{\lfloor_{n/2}\rfloor}{t}$.

dear petronius, how on earth did you derive those eequations?  Can you expalin to me, please?  And what does $ C_{n\minus{}2}$ mean?  What does C stand for?  Thank you so much if you could explain to me.  *Big kiss*

Petronius has pseudo-responded here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=317887


The digits 1, 2, 3, 4, 5, 6, 7, and 9 are used to form four two-digit prime numbers, with each digit used exactly once. What is the sum of these four primes? \[ \text {(A) } 150 \qquad \text {(B) } 160 \qquad \text {(C) } 170 \qquad \text {(D) } 180 \qquad \text {(E) } 190  \]

[hide]2, 4, 5, and 6 must be the tens digits, since the numbers would not be prime if they were the units digits. Thus, 1, 3, 7, and 9 are the units digits. $20+40+50+60+1+3+7+9=190\implies\boxed E$[/hide]

[hide="Answer"]Since they are all two-digit, the units digits must be the number 1, 3, 7, 9. Therefore, we have $1+3+7+9+20+40+50+60=190\Rightarrow \boxed{E}$.[/hide]

Ahh... the above methods were excellent, but I did it the dumb way (which was a lot more fun).

[hide]We know that the tens digits have to be 2, 4, 5, and 6, since if they were units digits, the resulting number would be composite. Playing with resrictions on units digits and placing digits, we get four numbers: 23, 41, 67, and 59. This sum is 190. [/hide]

And catcurio, you had the right answer, but the wrong answer choice...

[quote="Farenhajt"][quote="236factorial"]...we get four numbers: 23, 41, 67, and 59...[/quote]

Actually, there are four possibilites: (23, 41, 59, 67), (23, 47, 59, 61), (29, 41, 53, 67), (29, 47, 53, 61). So the above methods cover for all of them.[/quote]

Yeah I was suspecting that but I didn't really want to go through all that... I'm really tired.  :sleeping:

[quote="236factorial"]And catcurio, you had the right answer, but the wrong answer choice...[/quote]
Oops... edited. :oops:


I recently attempted to typeset a document in LaTeX involving arcs of circles.  However, I was unable to find any means of creating the arc symbol.  I know that it must be possible since it is done in the [i]Art of Problem Solving[/i] books, but I have no idea how.  Can anybody help?

You can do it crudely with $\overset{\frown}{ABC}$, but you'd want to make the frown wider. I'll look into that. If you find it, please do tell me.


[hide="A bit inelegant, but works"]Clearly, only the last non-zero digit of each factor in the factorial is significant.  Thus, if $n$ is the last non-zero digit of $9!$, then we want the residue of $n^{11}$ mod 10 (10 $n$s for the non-multiples of 10; 1 $n$ for the multiples of 10, as 100 obviously has no effect whatsoever).  We can easily see that $n=8$, so by Fermat's Little Theorem, the answer must be congruent to 2 mod 5 and 0 mod 2, i.e., it must be $\boxed{2}$, Q.E.D.[/hide]


let ABC be a triangle and (C)be a circle with the diameter AB.(C)intersects CB and CA in points Pand Q,respectivly.

and let the tangents to (C),from Pand Q ,intersect each other in point X.

prove:CX is prependicualr to AB.

$\angle{C}=\frac{1}{2}(180^2-\angle{POQ})=\frac12 \angle{PXQ}$ and $PX=QX$, So $X$ is the circumcentre of $\triangle{CPQ}$.

$CX \perp AB \Longleftrightarrow \angle{PCX}+\angle{A}=90^o$
Since $\angle{PCX}=\angle{CPX}$, $\angle{A}=\angle{APO}$, so:
\[\angle{PCX}+\angle{A}=\angle{CPX}+\angle{A}=90^o\]


How many prime numbers less than 100 have a units digit of 3?

You can quickly see that the only values that aren't prime that have 3 as a unit's digit are the ones with multiples of 3 as the ten's digit. Thus there are $ 10\minus{}3\equal{}\boxed{7}$ such primes.

Hello!

[hide=My answer]There are 10 integers that have a units digit of 3 that are less than 100.
The only integers that have a units digit 3 that are NOT prime are $\boxed{33}$$\boxed{63}$$\boxed{93}$
Notice that the tens digit is a multiple of 3.
$10-3=7$
Answer: $\boxed{7}$[/hide]

THANKS!!! {:)+<:


function
trigonometry
algebra
polynomial
logarithms
domain
complex analysis

Is there some sort of general way to show that a function is analytic? For example, how would you show that $ \cos^2(x)$ is analytic?

I'm assuming you mean "analytic" as "locally given by a power series."

The simplest way to do that is to explicitly find a power series, although that's not always available. It is here:

$ \cos^2x\equal{}\frac12(1\plus{}\cos 2x)$

$ \equal{}\frac12\left(1\plus{}1\minus{}\frac{2^2x^2}2\plus{}\frac{2^4x^4}{4!}\plus{}\cdots\right)$

which you can then simplify. This is, in fact, an entire function (given by a power series with an infinite radius of convergence.)

In general, we need some theorems:

1. The sum of analytic functions is analytic (where both defined - this is understood below).

2. The product of analytic functions is analytic.

3. The quotient of analytic functions is analytic as long as we stay away from the zeros of the denominator.

4. The composition of analytic functions is analytic.

5. Many familiar functions are analytic: polynomials; the exponential (and sine and cosine), the logarithm (in the interior of its domain), the inverse trig functions (in the interiors of their domains).

In this case, we can see that the square of a cosine is the composition of two analytic functions, hence analytic.


AMC AIME USA(J)MO USAMO quadratics algebra

What do you guys think about the CNMLs this year, compared to last?
I think they're quite a bit easier, last year I was getting 2's and 3's, but this year I've got two 5's and 6.

Also, what did you think of today's CNML? I messed up on the one with the irregular numbers, put 60 instead of 36.  :(

I think you should wait about a week next time before discussing answers.

But yeah, I <almost> made the same mistake you did.

I don't really see a change in difficulty though, since last year I got nothing but 5's and 6's and this year I have 6/5/6.

Maybe you're just getting better.

This round was really really really easy, relative to the previous rounds. I got a 6. :)

Does the school co-ordinator give out free (or at a cost) solutions after the math contest session?

I'm doing the cnml as well. my score so far is 5/4/5

The CNML is the worst contest that has ever existed in the history of human-kind :furious:
I do each contest in under 10 min.
Basically all of the problems IN HARDER FORM :rotfl:  have appeared somewhere before.
It is sooo easy to cheat (though i don't do that), but at least in my school, some people write the same thing on different days. :!:
I know some incidents when the wrong score was recorded for some people. :mad:
About 80 % of the problems can be done win nearly completely trial and error. :mad:
The wording of the problems is  horrible.:stretcher:
There many other bad things about this contest. :mad:

yeah CNML does suck, I made a mistake in the first one, with negative signs, no more perfect. I don't even think you win anything anyway, it's not challenging enough.

it's just a series of fun contests... nothing more...

Well, it's just kinda a funny tool to get little kids involved in the mathematics. :D

It is, but i think it makes high school students who take it seriously a bit more stupid. :(

Well...I usually use them as a practice of avoiding "2+5=8" type mistakes.
But yeah...it is sort of stupid to take it too seriously.

[quote="Jeff219"][quote="liudaveliu"]Did anyone get a certificate for being the top student in their school?[/quote]
My school hands out a certificate and book prize to the top student in each grade at the end of the school year. :)[/quote]

My school gives book prizes to the top two of all grades...
BTW, I got perfect this year! lol... :rotfl:

I got 35. btw cnml 6 was such a joke. i want my school to give out prizes too but it doesn't :(

[quote="mnica"]i agree with alex, cnml is such a trashy contest, the only difficult in it is a) understanding what they want you to do (the questions are poorly worded) and b) making sure that your TI89 has batteries (and c)  making sure you dont make adding mistakes, but if you take care of b) you should be ok with that). If you go to their website, you will see that people frequently "challenge" the questions (they try to agrue that their incorrect answers actually deserve the marks) in a comicaly hillarius fasion.[/quote]
Yay. You joined :)

is it for high school students?
which level is it?

It's intended for grades 11 and 12, but it's a very easy contest compared to AIME, USAMO, IMO, etc...very easy.  Grades 9 and 10 may also write it if their teachers allow them.

oh really?
sounds lik euclid.
i wanna have a try hehe

[quote="Effervescence"]CNML is to Hypatia as Hypatia is to Part B of the Canadian Open as Part B of the Canadian Open is to the .. IMO?[/quote]

Too true.
But still, CNML is as fun and as stupid as you want it to be... :roll:

Just looking over this message board here and decided to sign up (after years of viewing the AoPS forums).  CNML really is a joke contest and frankly it disgusts me how some schools (cough Pushkin) use it as a means of making a reputation for themselves.  I was one of four students to score perfect in 2003-04, but 2 of the students were from Pushkin which is now not allowed to participate in the CNML.  I am a member of the Euclid marking committee and remember hearing numerous teachers and other members of the math community discuss how Pushkin ruined the contest by cheating.  Anyone could cheat on the contest because the moderator has a copy of the solutions and could tell any of the students them, further it is marked by the moderator who could submit any grades he wishes.  It is amusing however how pushkin was able to have like 10 perfect students last year, but yet still unable to have one student score over 70 on Euclid!

CNML questions can be worded weirdly, and they leave a lot of room for ambiguity with solutions because they say "all equivalent values are acceptable".  I remember hearing of a question that asked you to find the roots of this one quadratic, and one student wrote the quadratic equation and did not simplify but received credit because it was equivalent to the solution (i think the solution was a and b, but his enormous values received credit).

As for prizes, if you finish first overall, you receive a shiny plaque with your name on it! lol.  That's about that though.

[quote="jonnyg"]
As for prizes, if you finish first overall, you receive a shiny plaque with your name on it! lol.  That's about that though.[/quote]

lol...wonder where those book prizes came from then...


function
integration
algebra
polynomial
number theory
relatively prime
real analysis

Let $f$ be a continuous function defined on $[0,1]$ and let for all $n$ the sequence $r_1, r_2,..., r_{\phi(n)}$ be those numbers smaller than $n$ and relatively prime to it. Find the limit of $\frac{ f(\frac{r_1}{n})+...+f(\frac{ r_{\phi(n)}}{n})}{\phi(n)}$

I don't get it :?. What if $f$ is constant with value $1$, for example? For every $n$, that number is $\frac{\varphi(n)}n$, and this sequence certainly doesn't have a limit as $n\to\infty$. In fact, the closure of its set of values is the whole interval $[0,1]$. What am I missing?

Terribly sorry, I forgot that at denominator we have $\phi(n)$. Sorry again.

It's clear that it cannot be anything else (just consider prime $n$). But how about the proof?  ;)

I think that a solution would be more indicated than 19 posts, don't you think? I really don't think the problem is obvious.

It isn't obvious (as Myth told me once, nothing is :) ) but it is still quite manageable. The solution I found runs like this. Let $S(n,f)$ be our sum. First of all, since $a$ is relatively prime to $n$ if and only if $n-a$ is, we see that the  $S(n,f)=0$ for any function $f$ satisfying $f(1-x)=-f(x)$. This allows to reduce the problem to the case when $f(0)=f(1)$. Now, using the Weierstrass theorem on density of trigonometric polynomials, we see that it is enough to consider the case $f(x)=e_k(x)=e^{2\pi i kx}$. It is clear that for $k=0$, we have $S(n,e_0)=1=\int_0^1 e_0$ for all $n$. Now, it is a straightforward exercise in elementary number theory to show that $S(n,e_k)$ is a multiplicative function of $n$ for fixed $k$. Now observe that, for any prime number $p$,  $S(p^m,e_k)=0$ whenever $p^{m-1}>k$. Also, $S(p,e_k)=-\frac1{p-1}$ for $p>k$. From here it follows that, for any fixed $k\ne 0$, we have $S(n,e_k)\to 0=\int_0^1e_k$ as $n\to\infty$  and we are done. What is your solution, harazi?

Very nice. I also used Weierstrass theorem, but for polynomials. In order to prove the result for $x^k$ I used the inclusion-exclusion principle. The computations are however quite messy.


calculus
integration
trigonometry
logarithms
derivative
calculus computations

$ \int \frac{1}{\sin x \sqrt{\sin^2 x \plus{} a^2}} dx$.

I am looking for the most elegant and compact answer, so different methods will be appreciated.

i don't know about   elegant, but i know the following method and another one using DuIs.

$ \int \dfrac { \sin \, x \; dx } { \sin^2 \, x \cdot \sqrt { a^2 \plus{} \sin^2 \, x } } \;\equal{}\;  \int \dfrac { \sin \, x \; dx } { ( 1 \minus{} \cos^2 \, x ) \cdot \sqrt { (a^2 \plus{} 1) \minus{} \cos^2 \, x } }$

$ \equal{}\; \int \dfrac { \sec^2 \, x \;\tan\, x \; dx } { ( \sec^2 \, x \minus{} 1  ) \cdot \sqrt { (a^2 \plus{} 1 ) \sec^2 \, x  \minus{} 1} }$

now put $ u^2 \;\equal{}\;(a^2 \plus{} 1 ) \sec^2 \, x  \minus{} 1$

$ I\; \equal{}\;\dfrac { 1 } { a^2 \plus{} 1} \; \int  \dfrac { u\; du } { ( \dfrac { u^2 \plus{} 1} { a^2 \plus{} 1}  \minus{} 1 ) \cdot u } \;\equal{}\; \int \dfrac { du } { u^2\minus{}a^2}$

$ \;\equal{}\; \dfrac {1} {2a} \ln  \bigg| \dfrac {u\minus{}a} {u\plus{}a } \bigg | \qquad \text{or} \qquad \minus{}  \frac {1}{a}   \text{coth}^ {\minus{}1} \; \dfrac { u} { a}$

$ \;\equal{}\;  \dfrac {1} {2a} \ln  \bigg| \dfrac { \sqrt {  a^2 \plus{} \sin^2 \, x  } \minus{}a\cos\, x } {\sqrt {  a^2 \plus{} \sin^2 \, x  } \plus{}a\cos\, x } \bigg | \qquad \text{or}$

$ \minus{} \frac {1}{a}  \text{coth}^ {\minus{}1} \left(  \dfrac { \sqrt {   a^2 \plus{} \sin^2 \, x } } { a \cos\, x } \right)$

Thanks, that method is indeed nice. Could I see the Differentiation under the integral sign method?

[quote="Gib Z"]$ I \equal{} \int \frac {1}{\sin x \sqrt {\sin^2 x \plus{} a^2}}\ \mathrm {dx}\ ,\ a > 0$ .[/quote]
[color=darkblue][b][u]Proof.[/u][/b]  I"ll use the substitution $ \left\|\begin{array}{c} \boxed {t \equal{} \phi (x) \equal{} \cos x} \\
 \\
\phi'(x) \equal{} \minus{} \sin x \\
 \\
\left(\ b \equal{} \sqrt {a^2 \plus{} 1}\ \right)\end{array}\right\|$ $ \Longrightarrow$

$ I \equal{} \int \frac {(\cos x)'}{\left(\cos^2 x \minus{} 1\right) \sqrt {b^2 \minus{} \cos^2x}}\ \mathrm {dx} \equal{}$ $ F(\cos x) \plus{} \mathcal C$ , where

$ F \equal{} \int\frac {1}{\left(t^2 \minus{} 1\right)\sqrt {b^2 \minus{} t^2}}\ \mathrm {dt}$ . Now I"ll use the [b]Abel[/b]'s substitution

$ \left\|\begin{array}{c} \boxed {y \equal{} \psi (t) \equal{} \frac {t}{\sqrt {b^2 \minus{} t^2}}}\ ,\ t^2 \equal{} \frac {b^2y^2}{y^2 \plus{} 1} \\
 \\
\psi'(t) \equal{} \frac {b^2}{\left(b^2 \minus{} t^2\right)\sqrt {b^2 \minus{} t^2}} \\
 \\
\left(\ t^2 \minus{} 1 \equal{} \frac {\left(b^2 \minus{} 1\right)y^2 \minus{} 1}{y^2 \plus{} 1}\ ,\ b^2 \minus{} t^2 \equal{} \frac {b^2}{y^2 \plus{} 1}\ \right)\end{array}\right\|$ $ \Longrightarrow$

$ F \equal{} \int\frac {1}{\left(t^2 \minus{} 1\right)\sqrt {b^2 \minus{} t^2}}\cdot\frac {\left(b^2 \minus{} t^2\right)\sqrt {b^2 \minus{} t^2}}{b^2}\cdot \psi '(t)\ \mathrm {dt} \equal{}$

$ \int\frac {b^2 \minus{} t^2}{b^2\left(t^2 \minus{} 1\right)}\cdot \psi '(t)\ \mathrm {dt} \equal{}$ $ G(\frac {t}{\sqrt {b^2 \minus{} t^2}}) \plus{} \mathcal C$ ,

where $ G \equal{} \int\frac {1}{\left(b^2 \minus{} 1\right)y^2 \minus{} 1}\ \mathrm {dy} \equal{}$ $ \frac {1}{2a}\cdot\ln\left|\frac {ay \minus{} 1}{ay \plus{} 1}\right| \plus{} \mathcal C$ a.s.o.

[b]Remark.[/b] You can use directly the substitution $ t\equal{}\frac {\cos x}{\sqrt {b^2\minus{}\cos x}}\ .$[/color]


analytic geometry
graphing lines
slope
LaTeX

[quote="grizzland"]A 10-foot plank is used to brace a basement wall during construction of a home. The plank is nailed to the wall 6 feet above the floor. The slope of the plank is ________[/quote]

[hide]Its a 6-8-10 triangle so $\frac{3}{4}$[/hide]

[quote="Temperal"][hide][latex]\frac{3}{4}[/latex][/hide]

I don't know why everyone hides their answers, but I guess I'll do it too.[/quote]
the [latex] tags are for the wiki I think, use dollar signs for latex on the forums.

you need to hide it so other people dont accidentally look at your solution.  technically, only the first couple posts should be hidden, but still hide all your solutions anyways.  also, if they werent hidden, the page would get so long and it would take longer to load the page

[hide]Depending on how the coordinate system is placed, the slope could be either $\frac34$ or $-\frac34$.  The sides of the triangle are 6, 8, and 10, so the slope is rise over run = $\frac68$.[/hide]

[quote="Teki-Teki"]no, for the wiki you have to use <math> and </math>

by the way, when i was putting up an old IMO problem, i got errors when i did <math>x^2 + y^2 + z^2 = b^2</math>
but not when i did <math>x^2</math><math>+ y^2 + z^2 = b^2</math>.  any idea why?[/quote]
Probably because of missing brackets {}, but still, that's very odd  :huh:

[quote="b-flat"][quote="Temperal"]Oh, yeah. Another math forum I go to uses [latex][/latex], so...[/quote]

Really?  Which forum (I'm interested just to see)?[/quote]
http://www.xpmath.com/forums
It's a small forum run off VBulletin.
But this is off topic.

I don't use Latex :P

[hide]3/4 because the sides are 6,8, and 10, so 6/8 which is 3/4.[/hide]
Yeah http://www.xpmath.com is awesome!

You should. Otherwise you can't write stuff like summation symbols.


What is the value of \\
$ \frac{1}{2}\times4\times\frac{1}{8}\times16\times\frac{1}{32}\times64\times\frac{1}{128}\times256\times\frac{1}{512}\times1024$?

You basically group the numbers by 2s, and you should notice that the product of every two is always 2. Since there are 5 pairs of numbers/fractions, the answer is $ 2^5\equal{}32$.


A grocer stacks oranges in a pyramid-like stack whose rectangular base is $ 5$ oranges by $ 8$ oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?

$ \textbf{(A)}\ 96 \qquad
\textbf{(B)}\ 98 \qquad
\textbf{(C)}\ 100 \qquad
\textbf{(D)}\ 101 \qquad
\textbf{(E)}\ 134$

[hide]
you notice that each new stack decreases in both dimensions by 1.
that means:
5*8 + 4*7 + 3*6 + 2*5 + 1*4 =100, C
 [/hide]

[quote="#H34N1"]A grocer stacks oranges in a pyramid-like stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges in the level below. The stack is completed by a single row of oranges. How many oranges are in the stack?

(A) 96
(B) 98
(C) 100
(D) 101
(E) 134[/quote]

[hide="solution"]$5\cdot 8+4\cdot 7+3\cdot 6+2\cdot 5+1\cdot 4=\boxed{100}$[/hide]


calculus
derivative
real analysis
real analysis unsolved

This is probably easy, but I can't find the subject in my calculus book.  Here's an example from an assignment of mine...

C(X) = 15 + 7 log8 X, where X>1

x is number of years on market
rate of sales for years 4,8,10

alright, I figured the derivitive as

$\ c'(x)=\frac8{xln(5)}$

now I need for find the derivitive when X is 4,8, and 10

yet I don't know how to find derivitives for the fraction
I've seen fraction examples but the answers make no sense to me, help!

you just have to put in your values for x, if f(x) = x^2, at x=3 it equals 9.

$c'(4)= \frac{7}{4ln(8)}$
$c'(8) = \frac{7}{8ln(8)}$
$c'(10) = \frac{7}{10ln(8)}$


R : advance to the right
D : advance to the down
The route supports that I arrange five R and fore D to one line.
Therefore,
$ \dfrac{(5 \plus{} 4)!}{5! \ 4!} \equal{} \binom{9}{5} \equal{} \boxed{126}$

[hide="My Solution"]
The number of steps it takes to go from $E$ to $F$ is 4 and the number of steps to $G$ from $F$ is 5.

There are $\dbinom{4}{1}=\dfrac{4\cdot3\cdot2\cdot1}{1}=4$ ways to go from $E$ to $F$.

There are $\dbinom{5}{2}=\dfrac{5\cdot4\cdot3\cdot2\cdot1}{2\cdot1}=10$ ways to get from $F$ to $G$.

Multiplying these gives: $4\cdot10=40$.

[hide="answer"]There are $\boxed{40}$ 9-step paths.[/hide][/hide]

Why is it in the official solution $\binom{9}{4}$?  Why is it selecting four steps from a group of 9 steps?  Don't we need all 9 steps?

I think kangaroos is right because doesn't the combo rule only apply when you can only go downwards and right for this problem? But this problem let's you go up down right left, right?

you can't go left or up because then you take too many steps to get to G.  Also, $\binom{4}{1}=\frac{4!}{3!1!}=\frac{4}{1}=4$
and
$\binom{5}{2}=\frac{5!}{3!2!}=\frac{5\cdot4}{2}=10$.

Nvm about my questions got them answered

I believe Kouichi Nakagawa is correct. It would be $\binom{9}{5}=\binom{9}{4}$ because we have to move 9 steps in all, 5 rights and 4 downs. If we choose where to put the rights, the placement of the downs is decided. Similarly, if we choose the downs first, then the rights are decided.

I used adding (like Pascal's triangle). There is one way to stay at E, one way to go east, and one way to go south. There are two ways to go to the southeast square from the start, which comes from 1+1. (This is a proven, yet not stated method. Aops didn't explain this solution because it didn't want you cheating like this.)

[hide] Since there are 9 steps and 4 of them have to be downwards, the answer would be $C(9,4)=\boxed{126}$ ways.[/hide]


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let $ A \equal{} <x_1> \times \cdots \times <x_r>$ be a finite abelian group with $ |x_i| \equal{} n_i$ for all $ 1 \leq i \leq r$. Prove that if $ G$ is any group containing commuting elements $ g_1, \cdots, g_r$ such that $ g_i^{n_i} \equal{} 1$ for all $ 1 \leq i \leq r$, then there is a unique homomorphism from $ A$ to $ G$ which sends $ x_i$ to $ g_i$ for all $ i$.

I believe the assumption that the order of $ g_i$ divide $ n_i$ is unnecessary. Can someone come up with a counterexample? If so, I'll post my proof, since I can't find an error in it.

Without that order condition, you don't get a homomorphism. We have $ e\equal{}f(e)\equal{}f(x_1^{n_1})\equal{}f(x_1)^{n_1}\equal{}g_1^{n_1}$ for any homomorphism $ f$.


[quote="FantasyLover"]Hi, I have a question about inequalities.

If an inequality is symmetric for variables $ a_1, a_2, \cdots a_n$, we may assume that $ a_1\ge a_2\ge \cdots \ge a_n$.

Is there such assumption we may make for cyclic inqualities?

Thanks. :)[/quote]

If an inequality is cyclic for variables $ a_1, a_2, \cdots a_n$, we may assume that $ a_1\equal{}\max \{a_1, a_2, \cdots ,a_n \}$ or $ a_1\equal{}\min \{a_1, a_2, \cdots ,a_n \}$.


#2. [color=white]The answer is 0 because it's like matching.  You can't get only one wrong.[/color]

May I ask the purpose of this poll?

The reason is we have enough new getting started stuffs and we don't want anything that's not "math."

If not replied in next two days, it will be removed.

Regards,

Silverfalcon

You're answering number 2 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18791]this thread[/url], right? When you have the answer to a question, you post it directly in the question thread. There's no need to start a new topic. Or if it's really necessary, which I don't think it is, at the very least post the original question so people know what you're talking about.

Sorry,  you can delete the poll please, or tell me how.

I didn't know the rule about having polls and I was just interested in the poll feature (and also wondered if most people thought the problem was easy).  I also accidentally pressed the wrong button to create a new topic.

I will repost my answer to the question in the proper place.


inequalities
function
inequalities unsolved

[tex] a;b;c [/tex] are positive real numbers.Prove that
[tex]\frac{a^3}{(a+b)^3}+\frac{b^3}{(b+c)^3}+\frac{c^3}{(c+a)^3}\ge \frac{3}{8}[/tex]

Put
    $\displaystyle x=\frac{b}{a}$
    $\displaystyle y=\frac{c}{b}$
    $\displaystyle z=\frac{a}{c}$
We have $xyz=1$
Now must prove
$\displaystyle \frac{1}{(x+1)^3}+\frac{1}{(y+1)^3}+\frac{1}{(z+1)^3} \ge \frac{3}{8}$

It quite easy because This one is more strong:

$\displaystyle \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge \frac{3}{2}$

 :D  :D  :D

[quote="hardsoul"][quote="Viet Math"]
It quite easy because This one is more strong:

$\displaystyle \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge \frac{3}{2}$

 :D  :D  :D[/quote]

Well your inequality is false for $x\longrightarrow0,y=z\longrightarrow\infty$.
But the inequality is true for the power 2( http://www.mathlinks.ro/Forum/viewtopic.php?t=32031  and http://www.mathlinks.ro/Forum/viewtopic.php?t=30724)
The cubic power can be reached by using Power Mean Inequality.[/quote]

Oh,i'm sorry  :blush:  :blush:

It's easy to prove :
 $ \frac{1}{(1+x)^2} + \frac{1}{(1+y)^2} \geq \frac{1}{1+xy} = \frac{a}{c+a} $
 $ \frac{a}{c+a} + \frac{c^2}{(c+a)^2} \geq \frac{3}{4} $
  After that , applying Power mean inequality ==> Q.E.D

Does anyone have the full contest?
Please post it on th forum
Thank you

technically, no proof has been shown, so i will quickly show one

as before, we reduce to xyz = 1, need to show:

$\sum \frac{1}{(x+1)^3} \geq \frac{3}{8}$

By jensen, $LHS \geq 3f(\frac{x+y+z}{3}) \geq 3f(\sqrt[3]{xyz}) = RHS$, where $f(x) = \frac{1}{(x+1)^3}$ is concave.

---

[quote="Singular"]$3f(\frac{x+y+z}{3}) \geq 3f(\sqrt[3]{xyz}) = RHS$, ---[/quote]
I don't think it true ;)

Yes I agree, $f$ is a decreasing function, hence $f(\frac{x+y+z}{3}) \leq f(\sqrt[3]{xyz})$ Which means that your proof is wrong  :lol:
Does anyone has a right Proof ??  :?

I think you maybe see the problem following:
$a,b,c,d$ are four positive numbers such that $abcd=1$
then we have $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2}+\frac{1}{(1+d)^2} \geq 1$
then  let $d=1$,we have $\frac{1}{(1+a)^2}+\frac{1}{(1+b)^2}+\frac{1}{(1+c)^2} \geq \frac{3}{4}$
use power mean inequality. :)

[quote="minhkhoa"]I think the stronger ineq. holds
$\frac{a^3}{(a+b)(a^2+b^2)}+\frac{b^3}{(b+c)(b^2+c^2)}+\frac{c^3}{(c+a)(c^2+a^2)}\ge \frac{3}{4}$[/quote]
can you post your solution for this?
thanks

I don't want to say this but I don't have the solution yet :blush:  :P

This inequality has been posted before:
http://www.mathlinks.ro/Forum/topic-17246-20.html


Let $ a,b,c>0$. Prove that:
$ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}(a\plus{}b)^3}}\geq 1$

[quote="Stephen"][quote="dvtvd"]Hint
\[ \sqrt {\frac {{{a^3}}}{{{a^3} \plus{} {{\left( {b \plus{} c} \right)}^3}}}} \ge \frac {{{a^2}}}{{{a^2} \plus{} {b^2} \plus{} {c^2}}}
\]
[/quote]

Why is it? I had started expanding...it won't work.[/quote]
Because it 's equivalent to:$ 2a^2(b^2\plus{}c^2)\plus{}(b^2\plus{}c^2)^2 \ge a(b\plus{}c)^3$
Which is obvious true by AM-GM :wink:

[quote=tdl]Let $ a,b,c>0$. Prove that:
$$ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}(a\plus{}b)^3}}\geq 1$$[/quote]
Let $a,b,c\in R^{+}$. [url=https://artofproblemsolving.com/community/c6h1998437p13957733]Prove the following inequality:[/url]
$$\sqrt{\frac{a^{3}}{b^{3}+(c+a)^{3}}}+\sqrt{\frac{b^{3}}{c^{3}+(a+b)^{3}}}+\sqrt{\frac{c^{3}}{a^{3}+(b+c)^{3}}}\geq1$$

[quote=tdl]Let $ a,b,c>0$. Prove that:
$$ \sqrt{\frac{a^3}{a^3\plus{}(b\plus{}c)^3}}\plus{}\sqrt{\frac{b^3}{b^3\plus{}(c\plus{}a)^3}}\plus{}\sqrt{\frac{c^3}{c^3\plus{}(a\plus{}b)^3}}\geq 1$$[/quote]
for $a,b,c>0.$ [url=https://artofproblemsolving.com/community/c6h62256p372570]Prove that[/url]
$$1 \leq \sqrt{\frac{a^3}{a^3+(b+c)^3}}+ \sqrt{\frac{b^3}{b^3+(c+a)^3}}+\sqrt{\frac{c^3}{c^3+(a+b)^3}} < \sqrt 2$$

Does anybody have solution for this using Holder's Inequality? :icecream:


There is a now a [url=http://www.artofproblemsolving.com/Forum/index.php?f=320]Canadian sub-forum[/url]! :D

You also access the forum from the index page under
National & Local Communities > North America > Canada

I've moved all the past Canadian topics over there. And please continue our Canada-related discussion in the new location.

I was the one who asked Valentin for it  :D  :D  :D

Enjoy it, but keep the spam at low levels, ok? :)


How can I find how fast this sum diverges in terms of x?

$ \sum_{n \equal{} 0}^\infty \sum_{k \equal{} 0}^{n} [{2n\choose2k} {(2k \plus{} 1)}^{ \minus{} x}] / 2^{2n \plus{} 1}$

0<x<1

You will get a more meaningful response if you post this in [url=http://www.artofproblemsolving.com/Forum/index.php?f=296]Calculus Computations and Tutorials[/url].


I wrote an article in my college paper about the millenium problems. I will give the link here, you can post any positive or negative comments you have about it.

http://www.thetriangle.org/media/paper689/news/2005/05/27/SciTech/Difficult.Solutions.Yield.Monetary.Rewards-954396.shtml

I was wondering why nobody was reading and responding to my article?

Well, you should probably give people a little more time to respond. I read it...to be honest it was OK, not great though.

The info and analogical explanations are all good, in my opinion (maybe take a look at the paragraph on Birch and Swinnerton Dyer).  The style, however, is fitting for the setting of the article.  I would try to take a more formal, academic tone (rid of first person pronouns, sequential references, etc.), especially considering this is a news article.  I would highly advise you to take your work to a competent editor familiar with journalistic writing style (which shouldn't be too hard to find, considering the fact that you are at a university).  You have the right idea about having outside people read it.  This often seems to be the greatest help to my own writing, because I often miss simply improved or corrected blights that I would never have seen on my own.

Good luck in the future with that sort of thing

Thats a lot for your praise and constructive criticism, the editor of the Sci-Tech section was supposed to edit this one, I agree its my fault that there was grammar mistakes as I am not good in grammar, not sure why it wasn't corrected. I tried to make it in a non-formal style on purpose, thought it would be more interesting. Not sure if that is the case or not, it was intended for a non-mathematical audience.


$2^{48}-1$ has $6$ divisors between $50$ and $100$. What is their sum?

I did that but I found $5$ divisors and not $6$.

A notice : $2^6+1=65$ and so it's not divisible by $11$

woops, i did that wrong, i thought of 32+1, but that is obviously wrong

[quote="Labiliau"]Perhaps, but could you explain please ... (the answer don't interest me, but the demarch yes!)[/quote]

perhaps this help!
The divisors are 51,63,65,85,91,97

51=3*17
63=3*21
65=5*13
85=5*17
91=7*13
97 is prime.
http://mathworld.wolfram.com/Divisor.html


the cube root of 11015 is 22.2499....., so the only possible values for a or b are ((1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22))
by working each of these we find that none of them work, therefore the answer is  (NONE)

so is there no way to do it without spending a lot of time? (Let's pretend it's a sprint question)

there probably is a quicker way, BUT i cant think of it               where did this problem come from?

[quote="erdogankerem123"]Find all ordered pairs (a,b) that a^3+b^3=11015. (where a and b are positive integers). I have no idea how to solve this, so can someone help?[/quote]
[hide="Hint"]Factor out $a^{3}+b^{3}$ to get $(a+b)(a^{2}-ab+b^{2})$. Then factor out 11015 and start guessing and checking. It shouldn't be that hard once you prime factorized 11015.[/hide]

nice nice,,,,, didn't realize that at first

I think it would be a better idea to convert all the cubes to a different mod, like mod 9. Then, it will be easy to determine which cubes even have a possibility to add up to 11015.

[hide]Notice that $11015 \equiv 4(\mod 7)$

Next, notice that perfect cubes are either $-1,0,1 (\mod 7)$

So, this means that the sum of two perfect cubes is $-2,-1,0,1,2 (\mod 7)$, none of which is 4, so no solutions. [/hide]


geometry
trigonometry
angle bisector
power of a point
radical axis
geometry solved

Let AB be the diameter of a circle [tex]\omega[/tex]. Let [tex]{\ell}_a[/tex] and [tex]{\ell}_b[/tex] be the lines tangent to [tex]\omega[/tex] at A and B, respectively. Let C be a point on [tex]\omega[/tex] and let line BC intersect [tex]{\ell}_a[/tex] at point K. The angle bisector of angle CAK meets line CK at H. Also, let M be the midpoint of arc CAB and let S be the second intersection of line HM and [tex]\omega[/tex]. If T is the intersection of [tex]\ell_b[/tex] and the line tangent to [tex]\omega[/tex] at M, then prove that S, T, K are collinear.

The key observation that helped me to solve the problem is
that $MT\parallel BC$ (which in fact is obvious, both
perpendicular to the perp. bisector of $BC$). The rest is
kitchen-work so don't even boder to read... :maybe:  :D

So we will prove that $ \frac {HK}{HK+MT}=\frac{SH}{MH}$(fact
that will lavishly prove the problem) , but
$\frac{SH}{MH}=\frac{SH\cdot MH}{MH^2}=\frac{HC\cdot
HB}{HM^2}$ (the power of point $H$ with regard to circle $\omega$)
(1)

Let $BK=1$ (by scaling) and $a=\angle ABK$ so $AB=\cos(a)$ and
$AK=\sin(a)$. Then $BC=\cos^2(a)$ and  by a small(microscopic)
calculation $CH=\cos(a)-\cos^2(a)$(2) and $HK=1-\cos(a)$(3) so
$BH=cos(a)$(4).

On the other hand we easily find that in isosceles triangle $MBC$
we have that $\angle MBC=\frac{90+a}2$ so $
MB=\frac{BC}{2\cos(\frac{90+a}2)}=\frac{\cos^2(a)}{2J}$(5) ($J$ is
a notation for $\cos(\frac{90+a}2)$). The same in isosceles
triangle $ TMB$: $
TM=TB=\frac{MB}{2J}=\frac{\cos^2(a)}{4J^2}$(6)

 What else to do now than unite results (2), (3),
(4), (5) and (6) in a funky way! Because (1) becomes:

\[\frac {HK}{HK+MT}=\frac{HC\cdot
HB}{HM^2}\Leftrightarrow
\frac{1-\cos(a)}{1-\cos(a)+\frac{\cos^2(a)}{4J^2}}=\frac{\cos(a)\cdot(1-\cos(a))\cdot
\cos(a)}{HM^2}\Leftrightarrow \] :lol:

\[\Leftrightarrow HM^2=\cos^2(a)\cdot(1-\cos(a)+\frac{\cos^2(a)}{4J^2})\]

But now try to express $HM^2$ by the cosine law in triangle $MBC$
with $\angle MBC=\frac{90+a}2$... Surprise! Identity! Finnito! :(

Here's a different approach:

It's easy to show by angle chasing in triangle $ABH$ that it's isosceles, namely $BA=BH$, so $H$ is on the circle centered at $B$ of radius $BA$. Let $B'$ be the intersection of this circle with the line $\ell_b$. Let $A'=BS\cap \ell_a$. I want to show that $B'H\cap BS=A'$. In order to do that we observe that $\angle MSB=\angle MCB=\angle MBC=\angle MBT\Rightarrow \angle B'BH=\pi-2\angle MBC$, but, at the same time, $\angle B'BH=\pi-2\angle BB'H\Rightarrow \angle MBC=\angle BB'H=\angle B'HB\ (*)$. From here we find that $\angle MSB=\angle BB'H\Rightarrow \angle BB'H+\angle BSH=\pi\Rightarrow$ the quadrilateral $BSHB'$ is cyclic. If we put $A\"=B'H\cap BS$ we have $A\"H\cdot A\"B'=A\"S\cdot A\"B\Rightarrow A\"$ is on the radical axis of the two circles, which is $\ell_a$, so $A\"=A'\Rightarrow B'H\cap BS=A'$.

At the same time, from $(*)$ we can derive the fact that $B'H||BM\Rightarrow A'H=B'A'||BM$.

We now have: $KH||TM$ (as Kappa said), $KA'||TB,\ A'H||BM\Rightarrow$ the triangles $HA'K,\ MBT$ are homothetic (in this order of the vertices), so the lines $HM,\ A'B,\ KT$ concur in one of the similarity centers, so $KT$ passes through $S=HM\cap A'B$, Q.E.D.


Let $ p$ be the perimeter of a regular polygon of $ n$ sides inscribed in a circle of radius $ r$. Prove that $ p<7r$.

$ p<7r\Longleftrightarrow n\sin \frac{\pi}{n}<\frac{7}{2}.$

[quote="LoboSolitario"]Let $ p$ be the perimeter of a regular polygon of $ n$ sides inscribed in a circle of radius $ r$. Prove that $ p < 7r$.[/quote]

Quite clearly $ p < 2\pi r$. Thus we need to check if $ 2\pi < 7\iff \pi < 3.5$, which I think needs not be proven. :wink:

PS. My lucky 777th post! :P

:rotfl:  i was thinking that it was hard. I need vacation ...


[quote="José"]Find all triplets of reals numers $x, y, z$ suxh that: $x(x+y+z)=26, y(x+y+z)=27, z(x+y+z)=28$.[/quote]

When adding all three equations, we have $x(x+y+z)+y(x+y+z)+z(x+y+z)=(x+y+z)^2=81$, since this is a quadratic, we have two distinct sets of solutions. But since we are looking for triples, there are $2 \cdot 3!=\boxed{12}$ triples.

Iversonfan...

$x,y,z$ are set because $x(x+y+z)=26, y(x+y+z)=27..$, so $x,y,z$ are not interchangeable variables.

oh oops...sorry about that :blush:

my mistake


If $ a*b\equal{}b\sqrt{a^b}$, what is the value of $ 2 * (3 * 2)$?

First, $ 2 \sqrt{3^2} \equal{} 6$

Then, $ 6 \sqrt{2^6} \equal{} 6 \cdot 8 \equal{} \boxed{48}$

Yes, we just simply plug in the numbers and solve.


A rectangular sheet of paper has dimensions $ 11''$ by $ 12''$. A triangular piece is cut from the corner to leave a pentagon with sides of lengths $ 11,12,7,5$ and $ 9$ inches. How many square inches are in the area of the pentagon.

11 and 12 must be from the original sheet of paper. 7 is from 11, and 9 is from 12 (3-4-5), leaving 5.

Thus a 3-4-5 triangle (with area 6) is cut out of a rectangle with area 132, so the area of the pentagon is 126.


dwx, both 4-digit numbers have to be even, if I am understanding the problem correctly.

SOURCE: from workout 6, 2004-2005     (jli was hiding it with small size fonts)

DWX: you almost solved it. Consider mathmom's comments

dwx314, you are very close.

hint: [hide]rearrange the digits[/hide]

my answer: [hide]2232, 1116[/hide]

[quote="jli"][size=59]from workout 6, 2004-2005[/size]

A positive, even four-digit integer contains only two different digits, and the sum of its digits is 9.  If this integer is exactly twce another number also satisfying these conditions, what is the integer(s)?[/quote]

[hide]2232-1116=1116[/hide]

yeah, there was only one answer in the answer book.  it should be 2232

[quote="dwx314"]*laughs nervously*
*Hits himself for his dub mistake*
[hide="-"]*Adds that to his endless list of dumb mistake made*
*Sighs, and resolves to start reading the problem*
*(In vain)*[/hide][/quote]

Dub?

Better hit yourself again

[hide]I got $2232$ and therefore $1116$ as the second number (logic; the first number has to be a multiple of four)[/hide]


A square is circumscribed about a circle of radius $ r$. What is the number of square units in the area of the square? Express your answer in terms of $ r$.

[quote="dwei1019"]That's what I got, 2r^2, but the system said it was incorrect....
it said that it was $ 4R^2$[/quote]

Report an error.

No, it is not an error.  Because the square is circumscribed about the circle, its sidelength is twice the radius of the circle.  Thus, the side of the square is $ 2r$.  Therefore, the area of the square is $ 2r\cdot 2r \equal{} \boxed{4r^2}$.  The picture looks like this:
[asy]pair A=(0,0), B=(2,0), C=(-2,-2), D=(-2,2), E=(2,2), F=(2,-2);
draw(C--D--E--F--cycle);
draw(Circle(A,2));
dot(A);
draw(A--B);
label("$r$", (1,0), S);
label("$2r$", (0,-2), S);[/asy]

FAIL.
It's circumscribed, not inscribed.
[quote="FantasyLover"]
draw a picture.

a length of the diagonal is $ 2r^2$.

so the area of the square is $ 2r*2r/2\equal{}2r^2$

answer : $ 2r^2$
[/quote]
and that just barely makes any sense :)

[b]this is how i did it

we know that the radius is $r$

the radius is tangent to 1 side of the square so the length of the square of 1 side will be $2r$

and the equation to find the area for a square is

$2r$*$2r$=4r^2

so the answer is $\boxed{4r^2}$ [/b]


geometry perimeter rhombus rectangle circumcircle

Take a polygon with the perimeter of 1. Prove that the more sides the polygon has, the more area it has.

Hint: [hide]Use the fact a rhombus is less efficient than a rectangle.[/hide]

eum... perhaps you could tell us what you're talking about? :?

"efficience"? :?

Are you talking about regular polygons, becuase a triangle with sides all 0.3333... has a bigger area than a rectangle with two sides 0.49 and two sides 0.01.

And what does this problem have to do with circles?

yes.
a circle has infinitly many sides so thats why the topic is so titled

Considering the case of a regular polygon,

one can arrrive at  the following formula for area

A = (nR^2/2)* sin(2*pi/n)

where R = radius of the circumcircle of the polygon

and n = no. of sides of the polygon.

Again , the expression for perimeter of a polygon is

P = 2nR* sin(pi/n)

Substituting this in the expression for Area, we get

A = (PR/2)*cos(pi/n)

Here P =1
So
A(n) = (R/2)*cos(pi/n)

Now cos(x) is decreasing in [0,pi/2]

and pi/n belongs to [0,pi/2] for n>=3

So cos(pi/(n+1))  > cos(pi/n)

=> A(n+1) > A(n)

That is the Area sequence is an increasing one


Xinran walks 3 miles per hour (mph) uphill, 4 mph on a flat surface and 5 mph downhill. If he hikes one mile uphill, then one mile on the level and then returns by the same route to his starting point, how many minutes does he walk?

He walks 1 mile uphill at a rate of 3 mph so it took him 1/3 of an hour which is 20 minutes.

He then walks 1 mile on the level at a rate of 4 mph so it would be 1/4 of an hour which is 15 minutes.

He returns the same way so he walks 1 mile on a flat surface for another 15 minutes and walks downhill at a rate of 5 mph so he walked for 1/5 of an hour which is 12 minutes.

Add the minutes up and you get 62 minutes.

Alternatively, his average speed is the harmonic mean of 3,4,4,5.

$ \frac{4}{\frac13\plus{}\frac14\plus{}\frac14\plus{}\frac15}\equal{}\frac{4}{\frac{20}{60}\plus{}\frac{15}{60}\plus{}\frac{15}{60}\plus{}\frac{12}{60}}\equal{}\frac{4}{\frac{62}{60}}\equal{}\frac{4\cdot60}{62}$

Thus he moves on average 240 miles in 62 hours. Dividing top and bottom by 60, he moves 4 miles in $ \boxed{62}$ minutes.

Refer to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1281660#1281660]this[/url] for the reason that the harmonic mean works here.


Elizabeth visits her friend Andrew and then returns home by the same route. She always walks 2km/h when going uphill, 6km/h when going downhill and 3km/h when on level ground. If her total walking time is 6 hours, then what is the total distance she walks in km?

[quote="ritchjp"]hi math92:

Are you sure you gave us all the information?

What we don't know is for how long she walked going uphill, going downhill or on level ground.

You have three unknowns. How are these unknowns related to each other?[/quote]
i think ritchjp is right. u hav at least 3 unknowns that r necessary 4 this prob.

[hide]We're not given how much of the different elevation changes there are, so let's assume it doesn't matter. If it doesn't matter, let's say that he's walking on all flat ground. Then, he would have to go for 6*3 = $\boxed{18\text{ km}}$

[/hide]

[hide="Hint"]The amount of uphill, downhill or level ground doesn't matter[/hide]

oops, overlooked some info, yeah it is solvable.

the level ground amount only matters, so it is 3km*6hours to get [b]18[/b]

[hide="Answer"]

$a$=distance uphill   ,  $b$=distance downhill   ,  $c$=distance level ground

  In an eqation her total walking time looks like:

$6=(\frac{a}{2} + \frac{b}{6} + \frac{c}{3}) + (\frac{a}{6} + \frac{b}{2} + \frac{c}{3})$

  However on her trip back, uphill turns to downhill and downhill to uphill.  So we get:

$6(\frac{a}{2} + \frac{b}{6} + \frac{c}{3} + \frac{b}{2} + \frac{a}{6} + \frac{c}{3})=36$

  Simplifying we get:

$4(a+b+c)=36$

  And see see $x+y+z$ equals 9.

  Plugging this in we see that she walks a total of $\boxed{18}$ kilometres.  [/hide]

I guess the question is how much of that distance is level and how much of that distance is not level. On the round trip, for the up and down distance, you would be averaging 4 km/hr and on the level ground you would be walking 3 km/hr. So how much of that time would you be averaging 4 km/hr and how much of that time would you be walking 3 km/hr?

[hide]
Since the problem doesn't give the time that Elizebeth was walking uphill or downhill, or on flat ground, assume that it doesn't matter. So the answer is 3*6=18[/hide]


[quote="mathdog15"]Mr. Jones has eight different children of different ages.  On a family trip his oldest child, who is 9, spots a license plate with a 4-digit number in which each of two digits appears two times.  "Look, daddy!"  she exclaims.  "That number is evenly divisible by the age of each of us kids!"  "That's right," replied Mr. Jones, "and the last two digits just happen to be my age."  Which of the following is not the age of one of Mr. Jones's children?
A) 4
B) 5
C) 6
D) 7
E) 8[/quote]

[hide]It is B) 5.  I used divisibility rules and guess and check basically, but the license plate is 5544.  This is divisible by 9, 8, 7, 6, 4, 3, 2, and 1.  [/hide]

All of the AMC problems have already been solved:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433616#433616
http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2006


mathleague.org
ARML
ratio
geometry
3D geometry
octahedron
calculus

Who all knows about this contest? The web site (mathleague.org) is fairly limited. Looks like it's just Kansas, Missouri, and Iowa. Tim Sanders (ARML president) ran this contest for a while, I'm not sure if he started it. I just won the Missouri state contest on Saturday, with scores of Target 80/80 (I protested number 8, it was thrown out) and Sprint 110/120 (used 3x/8 instead of 5x/8 for the ratio in the infinite series on 14, and they rejected my protest on number 16, the instructions don't say anything about what to do if there's not enough information), so total 190/200, second place was 170. Does anyone know the criterion for being invited to Regional Championship? It's at KU on May 14, I'm just curious who all else (and about how many people) will be there, and winning scores in Kansas and Iowa.

hehe, u did pretty well...

95 on sprint (missed 16 as well; stupid mistakes on #2 and #4, the border problem and the octahedron problem)

60 on target (missed 45 degrees, and the geometric series)...yea, number 8 on that test was sketchy...almost half my team, including me, put down "no positive integral solution"...i think they should have just counted that as the correct answer, instead of tossing out the question entirely...

anyways, cya at regionals...7 from my school qualified, ~5 or 6 ppl will be there

oh yea, criteria:
1.) top 3 in grade level in each indiv. event
2.) top 1 in each "team" event (i.e. team, relay, power)
3.) i think, but not sure, top sweepstakes?

i got 2nd in target, 3rd in sprint, 1st team, 2nd relay, 3rd power, team got 1st overall sweeps...

btw, the test (every round, except maybe the team round) this year was WAAAY too easy...far easier than it had been in previous years...i didnt like it very much...


search
function
geometry
3D geometry
octahedron
\/closed

The search function brings up "No topics or posts met your search criteria" no matter what I search for (I tried generic things that are very common).

I just tried now a few different searches and all worked. What exactly did you search for?

Edit: You are right, the search doesn't look into 2 or 3 letter words. I will try to fix that this weekend.

And it doesn't look into some longer words like "value", too...

Here's an interesting one- wildcards at the beginning of words don't work. For example, a search for *hedron generates only six results, all of which have "hedron" somewhere in them. Words like "octahedron" and "polyhedron" are highlighted, but don't count as a match to generate results in the first place.

Yes the index has to be rebuilt. I will try to do that this weekend sometime. It seems that it got corrupted during the move.


I have been doing many problems in AoPS recently. The USA uses imperial units as their primary measurement system. However, I live in Singapore where we only use SI units. Thus I know almost nothing about the imperial units.

[quote]The rails of a railroad are 30 feet long. As a train passess over the point where the rails are joined, there is an audible click, The speed of the train in mph is approximately the number of clicks heard in how many seconds?[/quote]

That was from the 1955 AHSME. The maths is very easy, but the terminology, no.

According to Wikipedia, 1 international mile is 5280 feet. Do I use nautical or the international mile?[/quote]

"Mile" without qualification pretty much always refers to the international mile in a general context.

Even in the USA, the nautical mile is never used, unless specified.

You can always interpret "mile" as the international mile, if nothing is there to tell you otherwise.

I have been having similar problems as i generally use metric measurements.
Also, I think I have it sorted now but I was having trouble with some of the money related problems.

Watch out for tons (metric ton = 1000 Kg; ton = 2000 lb in the US) and billions (1 billion = 1000 millions in the US).

I don"t know of any other country where 1 billion = 1000 millions, do you?

In my country, 1 billion is equal to a thousand millions. ($ 10^9$)

[quote="herefishyfishy1"]I'm pretty sure a billion is defined as $ 10^9$ for everyone.[/quote]
Until I came to the US, 1 billion was the 12th power of 10, one trillion the 18th power, and so on. One n-illion would be the 6n power of ten (simpler than the 3n+3 power).
I remember reading that to the French, 1000 millions is a "milliard" and one million millions is a "billion".
I also read (recently) an article titled "when a billion is not a billion" or some close rewording of that. I'll try to google my way back to it.

Here are the references I could google:

http://faculty.kutztown.edu/schaeffe/Trivia/trivia.html
http://en.wikipedia.org/wiki/Long_and_short_scales#Short_scale_countries


For what units digit $ d$ is the five-digit number $ 23,45d$ a multiple of 9?

We must have \[ 2\plus{}3\plus{}4\plus{}5\plus{}d\equiv 0\pmod{9} \Rightarrow d\equal{}\boxed{4}\]


calculus
derivative
number theory proposed
number theory

If $a$ is a real root of $x^5 - x^3 + x - 2 = 0$, show that $[a^6] = 3$.

Indian 2004!

$P(x)=x^5-x^3+x-2$ is increasing because its derivative is positive $P'(x)=5x^4-3x^2+1 >0 ,\forall x\in R$
(One)   $P(1.201) < 0 \Longrightarrow a > 1.201 \Longrightarrow a^6 > 3 $ (make the calculation :D )
(Deux) $P(1.25) > 0  \Longrightarrow a < 1.25 \Longrightarrow a^6 < 4$ (make the calculation :D )
(Trei) Trivial conclusion : $[a^6]=3$

 :cool:  Do you dislike this kind of proof   :cool:

It isn't too ugly!  :)  :)

I was trying to prove it by proving that $4<a^6+1<5$
obviously $a is \ not 0$ so $a^4-a^2+1=\frac{2}{a}$

$a^6+1=(a^2+1)(a^4-a^2+1)=2\frac{a^2+1}{a}\geq2*2=4$

for the other using the same way we have to prove that $\frac{1}{2}<a<2$
something I haven't  done yet ........
Could anyone help?


trigonometry algebra polynomial function calculus integration

Is there a formula for soving at^3+bt^2+ct+d=0?

I think it's in AOPS1 ... one of the Big Ideas. either after Chapter 6 or 7

There was something like this mentioned in AoPS 1, in the "Big Picture" section after the chapter on quadratics. It's called a cubic or something.

there is, but it's incredibly horrible. btw, there is no general method for higher degree equations.

Wow, yes there is.  It's really nasty, though, in radicals.  Look up "cubic formula" on mathworld.wolfram.com/search/ if you're interested in seeing it.  The method of solving is roughly thus:

first, make it look like
x^3 + j*x^2 + k*x +l = 0

Then, make a substitution x = y + m for some number m (that you can solve for) to make it look like this:

y^3 + p*y + q = 0.

Now, it turns out that you can actually solve things like that in radicals.

On a side note, you can also solve it not in radicals if you let
y = n*sin:theta:, choosing n judiciously so that you get something involving the triple angle formula of for sine, which lets you write your answer as an arcsin function.

its okay... I stated it wrong anyways.... it was big picture... not big idea.

don't feel bad about posting what other people just said.... think of it as confirming something... and thanks for correcting the big picture thing again.

[quote="Fierytycoon"]Yeah, there's a formula, but don't bother memorizing it for competition purposes.[/quote]

Haha, that's the understatement of the week!

[quote]Haha, that's the understatement of the week[/quote]

8-)

On the other hand, memorizing the formula for one root of the depressed cubic (that is, x^3+px+q) would not be a bad idea. It's hard to derive and not that hard to memorize. (Now I'm awaiting the mass of criticism which is sure to follow this comment.)

[quote]can it be proven that no formula for larger degree polynomials can exist?[/quote]

Yes, for polynomials with degrees above 4.

Well, sort-of right.  No formula in radicals exists for a general polynomial of degree 5 or greater.  Equations do exist for the cubic and quartic equation.  They are both very very messy, far too difficult to write up here.  If you want a full proof of the Cubic Equation, go here:
http://mathworld.wolfram.com/CubicEquation.html

If you just want the answer, go here:
http://www.math.vanderbilt.edu/~schectex/courses/cubic/

Abel proved in 1826 that the quintic has no solution in radicals.  However, it does have solutions, which means that it must in some sense be "solvable."  For example, (this I learned from that second site I posted), if you let f(x) be a function with the property that
f(z^5 + z) = z for all z, you can use f in combination with radicals to express all the solutions to a general quintic equation.

I hope that answers most questions.

[quote]? i thought a general formula does not exist for polynomials of degree greater than or equal to four[/quote]

That's what I said.

[quote]? i thought a general formula does not exist for polynomials of degree greater than or equal to four[/quote]

Change "greater than or equal" to "greater than".  Polynomials of degree 4 (quartics) have a general formula.

I believe that there is a general formula for all polynomials, but it isn't in terms of radicals.

sorry to bring this up again, but Abel's Impossibility Thorem states otherwise.

[url]http://mathworld.wolfram.com/AbelsImpossibilityTheorem.html[/url]

gives some information.

No, Abel's impossibility theorem says there isn't a general formula in terms of a finite number of additions, subtractions, multiplications, divisions, and roots.  That doesn't mean that there isn't a general formula.  In fact, I'm reasonably certain there is.  You just can't express it with those 5 operations.  Just as an example, if you introduce a number as a root of x:^5: - x = 0, which can't be expressed in terms of roots etc., you can form a general quintic equation.  (I read this somewhere online, I don't remember where.)

Also, if you are allowed to evaluate elliptic integrals as an operation, then I believe there is a formula for 5th-degree polynomials with that operation and the original operations.

I think that's right -- at least, I definitely remember hearing it before.  What is an elliptical integral?

Here's one example: the integral of :sqrt:(x3 + 1).  You can find a definition at  [url]http://mathworld.wolfram.com/EllipticIntegral.html[/url]

There is some information about quintic polynomials at
[url]http://mathworld.wolfram.com/QuinticEquation.html[/url]

I think I can remember reading that it can be proven that no formula exists for n>=5, but what abt n=4?

There's a solution for the quartic in terms of radicals.

That won't work in general, of course, since not all cubics or quartics have rational roots.


In solving a problem that reduces to a quadratic equation one student makes a mistake only in the constant term of the equation and obtains $ 8$ and $ 2$ for the roots. Another student makes a mistake only in the coefficient of the first degree term and find $ \minus{}9$ and $ \minus{}1$ for the roots. The correct equation was:
$ \textbf{(A)}\ x^2\minus{}10x\plus{}9\equal{}0 \qquad\textbf{(B)}\ x^2\plus{}10x\plus{}9\equal{}0 \qquad\textbf{(C)}\ x^2\minus{}10x\plus{}16\equal{}0\\
\textbf{(D)}\ x^2\minus{}8x\minus{}9\equal{}0 \qquad\textbf{(E)}\ \text{none of these}$

[hide]

The first statement says that the sum of the roots is 10, since the coefficient of x is correct.
The second statements says that the product of the roots is 9, since the constant term is correct.

Therefore, the polynomial is $ x^2 \minus{} 10x \plus{} 9 \equal{} 0\implies\boxed{\text{A}}$[/hide]


Prove that $ 2^n \minus{} 1$ cannot divide $ 3^n \minus{} 1$ for any $ n > 1$

If $ n$ is even then $ 2^n\minus{}1$ is a multiple of $ 3$ and $ 3^n\minus{}1$ is not; so $ n$ is odd
Then you can see a more general problem here:
http://www.mathlinks.ro/Forum/viewtopic.php?p=1148144#1148144

Daniel


trigonometry
calculus
integration
function
real analysis
real analysis unsolved

Show that the series $ \sum_{n\equal{}1}^{\infty} \frac{\sin n}{n}$ is conditionally convergent.

This follows from the Abel-Dirichlet criterion ( http://en.wikipedia.org/wiki/Dirichlet%27s_test ) since
I. $ 1/n$ is a decreasing sequence that tends to $ 0$.
II.
\[ \left|\sum_{n \equal{} 1}^k \sin n \right| \equal{} \left|\frac {\cos\frac {1}{2} \minus{} \cos\frac {k \plus{} 1}{2}}{\sin\frac {1}{2}}\right| \text{ is bounded.}
\]

OK, that's a proof of convergence (summation by parts; it has appeared plenty of times before).

Now, can you find the exact sum?

The proof of Abel-Dirichlet is about as simple as it gets; it's not a deep proof at all.

[quote="Timestopper_STG"]I wonder whether we've a simpler proof for it's convergence or not :?:[/quote]

I'd guess that my proof is the simplest  :D

If you are willing to use complex or Fourier analysis you can get a [i]shorter[/i] proof, and even compute the sum (which is $ (\pi \minus{} 1)/2$). Whether this proof is [i]simpler[/i] depends on your taste.  :)

Since I said [b]conditional[/b] convergence you also have to prove that $ \sum_{n\equal{}1}^{\infty} \frac{|\sin n|}{n}$ is divergent. Or am I missing something?

[quote="Timestopper_STG"]I wonder whether we've a simpler proof for it's convergence or not :?:[/quote]

Well, there is another option, though maybe not simpler.
First, we use integral test. So we want to prove the integral converges. But just split the integral at each 0 of the function, so we have an alternating sequence whose terms go to 0 which converges.

EDIT: As said below, this argument is wrong. Sorry.

The integral test does not apply. One of its hypotheses is that the function is nonnegative and decreasing.

The integral test only works for decreasing, nonnegative sequences.  The given sequence is neither ;)

Edit: too slow.

[quote="jmerry"]

Now, can you find the exact sum?[/quote]

Use Fourier serie

[hide]$ \sum_{n\equal{}1}^{\plus{} \infty} \frac{\sin(n \theta)}{n} \equal{} \frac{\pi \minus{} \theta}{2}$

See, for example, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110277]here[/url].[/hide]

[quote="Yury"]That is trivial.[/quote]
Can you please explain?

One way is to show that for every real number $ n$, either $ |\sin n| \ge 0.1$ or $ |\sin(n \plus{} 1)| \ge 0.1$.

[b]Hint 1[/b]: Prove that for every $ x$, either $ |\sin x| \geq \sin (1/2)|$ or $ |\sin (x \plus{} 1)| \geq \sin(1/2)$ (this is what Ravi B suggested);

or

[b]Hint 2[/b]: Prove that $ |\sin x| \plus{} |\sin (x \plus{} 1)| \geq \sin 1$;

or

[b]Hint 3[/b]: Note that $ |\sin x| \plus{} |\sin (x \plus{} 1)|$ is a continuous function, and therefore attains its minimum on $ [0, 2\pi]$ and thus (why?) on $ \mathbb R$. On the other hand, $ |\sin x| \plus{} |\sin (x \plus{} 1)| \neq 0$. Therefore, $ |\sin x| \plus{} |\sin (x \plus{} 1)| \geq \min_x (|\sin x| \plus{} |\sin (x \plus{} 1)|) \equiv c > 0$.

Hint 4: use the equidistribution theorem.


If 1 ounce is equivalent to approximately 28.35 grams, how many ounces are in 541 grams? (Answer to the nearest tenth of an ounce.)

If $ \frac{a}{b} \equal{} \frac{c}{d}$, then $ ad \equal{} bc$.

And also, the answer would be $ \boxed{19.1}$, because it says to the nearest [b]tenth[/b].


Three pies and four cakes sell for $ \$35$ while four pies and five cakes sell for $ \$44.50$.  What is the cost to purchase one pie and one cake?

hello
$ 4x\plus{}3y\equal{}35$
$ 5x\plus{}4y\equal{}44.50$
$ x\plus{}y\equal{}9.5$
thank u


function
calculus
integration
algebra proposed
algebra

Let $f$ be a function defined on the positive integers, taking positive integral values, such that

$f(a)f(b) = f(ab)$ for all positive integers $a$ and $b$,
$f(a) < f(b)$ if $a < b$,
$f(3) \geq 7$.

Find the smallest possible value of $f(3)$.

ok ill post my solution:

Now clearly $f(x)=x^2$ is a function that satisfies all the conditions, yielding $f(3)=9$

But with this conjecture we need to prove that $f(3)\neq7, 8$

Separate this into two cases.

CASE 1 $f(3)=7$

$\implies 2\leq f(2) \leq 6$

We need to check that for each of these possible values of $f(2)$, we arrive at a contradiction.

If $f(2) = 2$

$\implies f(4)=4, f(3)=7$. contradiction because $f(a)<f(b)$ if $a<b$

Now, if $f(2)=3$

$\implies f(9)=49, f(12)=63$

$\implies 50\leq f(10) \leq 62$ but we know $f(10)=3\cdot f(5)$

$\implies 17\leq f(5) \leq 20$

$\implies f(15) \geq 7\cdot 17 = 119$ but $f(16) = f(2)\cdot f(2)\cdot f(2)\cdot f(2)=81$ contradiction.

If $f(2)>3$

then $f(8)>64$ but $f(9) = 49$ contradiction.

CASE 2 is similar to CASE 1. you know how to do it :P im too lazyyy

and after proving all this we realise that the minimum value of $f(3)=9$ :D

I remember that it have been  in a Chinese competition.
we claim that all the function satisfied the given condition is $f(x)=x^a$
where a is a real number.

But this is  a function from the positive integers to the positive integers.

This gives that $a$ is a positive integer, which solves the problem!

Does it? I mean, we know that multiplicative functions defined on the reals are of the form $f(x) = x^a$, but does this also hold for multiplicative functions defined on the positive integers?

No, the original problem (the one from China) was also from $\mathbb{N}$ to some set ($\mathbb{R}$ or so), and had also the condition that the function is increasing.

Oh, I see, didn't know that :) the fact that $f$ is increasing changes a few things indeed.

Btw, here it is: http://www.mathlinks.ro/Forum/viewtopic.php?t=55870

So observe that $f(x)=x^2$ is a solution.We will prove that $f(3)\neq7,8$
For that we see that f(1)=1 and $f(2)\ge 3$.Also $f(16)<f(18)$ so $f(2)<4$ so $f(2)=3$.But also $f(27)<f(32)$ which is equivalent with $343,512<243$ which is a cotradiction.

[quote=ZETA_in_olympiad]Since strictly increasing $\ln f(e^x)=cx$. Hence $f(3)=3^k$, so minimum is $\boxed 9$.[/quote]

I slightly doubt that that solution works, because although $g(x) = \ln(f(e^x))$ does indeed satisfy Cauchy's Functional Equation, it can only take in inputs satisfy the property that $e^x$ is an integer, that is to say that the domain of $g(x)$ is only a specific subset of the real numbers. In which case it's unclear whether the linearization to Cauchy's FE still works. Please correct me if your reasoning still works.

Anyway, by using a limit argument you can prove that $f(2) = f(3)^{\log_3 2}$ which is an integer iff $f(3)$ is a power of $3$, in which case the minimum is $9$. This is a little overkill but it generalizes nicely.

Actually, I guess you need to prove that $9$ is attainable. But that’s clear from the function $f(x) = x^2$.


[quote="orl"][quote="Arne"]And Peter and Iurie will be in high school for one more year...[/quote]

I don' think so. Iurie will go to $DRAVRAH^{-1}$ I assume.[/quote]

Trying to be discreet, orl? :wink:

[quote="shobber"][quote="orl"][quote="Arne"]And Peter and Iurie will be in high school for one more year...[/quote]
I don' think so. Iurie will go to $DRAVRAH^{-1}$ I assume.[/quote]
What's this?[/quote]

I would suppose Harvard ... is that some sort of curse word here?

last time i talked to him he told me he'll stay for one more year in moldova. i think..

[quote="Quique"]Another example is:
Gabriel Carroll. He got a gold medal in 98, then a silver in 99, then didn't attend 2000, then got a perfect exam in 2001.[/quote]
Well,it doesn't count then.
 :)

[quote="orl"][quote="darktreb"]I think [3 of the best USA IMO champions, none of them Chinese] are all pretty fluent in Mandarin Chinese (I know [number 3] is for a fact as I was able to converse with him!)[/quote]

Very open-minded people. [/quote]

Tangible motivation helps.

[quote="MysticTerminator"][/quote][quote="shobber"][quote="orl"][quote="Arne"]And Peter and Iurie will be in high school for one more year...[/quote]
I don' think so. Iurie will go to $DRAVRAH^{-1}$ I assume.[/quote]
What's this?[/quote][quote="MysticTerminator"]

I would suppose Harvard ... is that some sort of curse word here?[/quote]

Why is that your opinion? IMO Harvard is a great university. First of all doing this it should be a surprise. And secondly it should not be easily found by search engines but now this state can only be reached by erasing your posts.  :D Of course, for the IMO records it would be very nice having Iurie participating one more time as he is so incredibly strong but I am not sure he will...  :)

[quote="orl"][/quote][quote="MysticTerminator"][/quote][quote="shobber"][quote="orl"][quote="Arne"]And Peter and Iurie will be in high school for one more year...[/quote]
I don' think so. Iurie will go to $DRAVRAH^{-1}$ I assume.[/quote]
What's this?[/quote][quote="MysticTerminator"]

I would suppose Harvard ... is that some sort of curse word here?[/quote][quote="orl"]

Why is that your opinion? IMO Harvard is a great university. First of all doing this it should be a surprise. And secondly it should not be easily found by search engines but now this state can only reached by erasing your posts.  :D Of course, for the IMO records it would be very nice having Iurie participating one more time as he is so incredibly strong but I am not sure he will...  :)[/quote]

I think what MysticTerminator meant was that he was wondering whether or not Harvard is "frowned" upon here. By frowned upon, I don't mean academically, but there exist plenty of people (in the USA) who don't have the highest opinion of Harvard, not because of their academics (which of course are top tier) but because of other factors such as perceived arrogance/elitism, the "wealth" factor (where some students receive entrance to the school due to financial contributions of their family), as well as its ... let's call it "less than stringent" grading system. Of course, Harvard has by far the highest endowment of any university and if human history teaches us anything, it's that to be the richest, you will be hated by some. Mix in some questionable decisions from time to time and it gets even worse (see United States Foreign Policy, 2000-present :))

that's true Orlando, you have to learn something:
better than Harvard, MIT!

haha just kidding

[quote="fleeting_guest"][quote="darktreb"]he was wondering whether or not Harvard is "frowned" upon here. By frowned upon, I don't mean academically, but there exist plenty of people (in the USA) who don't have the highest opinion of Harvard, not because of their academics (which of course are top tier) but because of other factors [/quote]

Harvard academics vary widely by department and by school (compare School Of Education admissions with theoretical physics).  They have over-marketed their "brand", handing out certain degrees and honors like candy, and it has diluted the perceived value.  Larry Summers stood for the opposite trend and some 300-400 million dollars of donations have been rescinded after the coup d'etat that forced him out.   Many opinions of Harvard dropped after his ousting, particularly among the people who would be more concerned about the academics rather than the politics.[/quote]

Yes I agree about what you said about Summers. I felt that certain things he did were in the right direction from an academic standpoint. However, he also made a few mistakes (such as his comment about potentially studying women's brains with respect to math and science) which were overblown in the public eye. Personally if you examine his statement about the brains closely he simply said that perhaps with all the statistical backing it was worth looking into; he made no judgements or premature conclusions. However, it's the wrong kind of topic to make a comment about, from a political standpoint (which is sadly what counts more than anything), in this modern media starved world.


For all positive integer $ n$, prove that there exists integers $ x, y$ such that $ x^2 \plus{} y^2 \equal{} 5^n$ and $ 5\nmid x, y$.

Without using any knowledge about representations of integers congruent to $ 1$ modulo $ 4$ as sum of two squares, use induction. $ 1^2 \plus{} 2^2 \equal{} 5^1$ works for $ n\equal{}1$. Assume $ x_n^2 \plus{} y_n^2 \equal{} 5^n$; then $ 5^{n\plus{}1} \equal{} (1^2 \plus{} 2^2)(x_n^2 \plus{} y_n^2) \equal{} (x_n \plus{} 2y_n)^2 \plus{} (2x_n \minus{} y_n)^2 \equal{} (x_n \minus{} 2y_n)^2 \plus{} (2x_n \plus{} y_n)^2$. If $ 5$ divides one of the terms in one of the sums, it must divide the other too. Assume both sums have terms divisible by $ 5$; then  $ 5 \mid (x_n \plus{} 2y_n) \plus{} (x_n \minus{} 2y_n) \equal{} 2x_n$, so $ 5 \mid x_n$, wrong. Take then that sum with terms not divisible by $ 5$ to build $ x_{n\plus{}1}$ and $ y_{n\plus{}1}$.


geometry
geometric transformation
reflection

Suppose distinct lines l and m are parallel, and p is a point. Let q be the reflection of p with respect to l, and then let r be the reflection of q with respect to m. Suppose PR=10. Find the distance between the two lines.

[hide=":ddr:"]
let the distance from P to l be x and the distance from q to m be y.  it is clear that we want x+y.  From PR=10, we have 2x+2y=10.  so, x+y=5[/hide]


How would you use De Moivre's theorem to find the real part of an expression?

I used it to solve $\sum_{k=0}^{90} \cos k$.  I added the $i \sin \theta$ term to each $\cos \theta$.  Then we can express the sum as $1 + w + w^2 + w^3 + \cdots + w^{90}$, where $w = \cos 1 + i \sin 1$.  Do you see the use of de Moivre there?  From here, we can do the sum of geometric series and conjugates to find the desired value.  Then you just take Re of that, and that's the answer.  deMoivre made that problem easy, although there are other ways without it.

$\left(\begin{array}{c} {100} \\ {0} \end{array}\right)-\left(\begin{array}{c} {100} \\ {2} \end{array}\right)+\left(\begin{array}{c} {100} \\ {4} \end{array}\right)-\left(\begin{array}{c} {100} \\ {6} \end{array}\right)+...+\left(\begin{array}{c} {100} \\ {98} \end{array}\right)-\left(\begin{array}{c} {100} \\ {100} \end{array}\right)$

Thats the problem that I was shown...

If you look it really equals $Re(1+i)^{100}$  I was told you could use De Moivre's to find that

De Moivre can be used because $1+i=\sqrt{2}e^\frac{\pi i}{4}$, where $e^{\theta i}=\cos \theta +i\sin \theta$. It's much easier to raise the exponential form to a large power than trying to expand $(1+i)^{100}$.

[quote="Farazy"]
If you look it really equals $Re(1+i)^{100}$  I was told you could use De Moivre's to find that[/quote]
We want to find a $w$ such that $rw = 1 + i$.  Since the Re = Im here, we know the angle must be $\frac \pi 4 + k\pi$, and $r = \sqrt 2$.  So $(rw)^{100} = r^{100} w^{100} = 2^{50}(\cos \frac{100 \pi}{4} + i \sin \frac{100\pi}{4}) = 2^{50}(\cos \pi + i \sin \pi) = \boxed{-2^{50}}$

However, in the problem you posted, if you note that $\binom{a}{b} = \binom{a}{a - b}$, you'll see that everything cancels and the answer is 0.

[quote="Farazy"]$\left(\begin{array}{c} {100} \\ {0} \end{array}\right)-\left(\begin{array}{c} {100} \\ {2} \end{array}\right)+\left(\begin{array}{c} {100} \\ {4} \end{array}\right)-\left(\begin{array}{c} {100} \\ {6} \end{array}\right)+...+\left(\begin{array}{c} {100} \\ {98} \end{array}\right)-\left(\begin{array}{c} {100} \\ {100} \end{array}\right)$

Thats the problem that I was shown...

If you look it really equals $Re(1+i)^100$  I was told you could use De Moivre's to find that[/quote]

$\LaTeX$ tip:  you can use \binom{n}{k} for $\binom{n}{k}$ instead of using an array.  Also, don't forget to put exponents which are longer than 1 character in brackets.


Άν κ λίγο καθυστερημένα αυτή είναι η ανισότητα που σου ελεγα Σιλουανέ:

Αν $ abc \equal{} 1,  a,b,c >0$ νδο :

$ \frac {a}{a^2 \plus{} 2} \plus{} \frac {b}{b^2 \plus{} 2} \plus{} \frac {c}{c^2 \plus{} 2} \leq 1$

Μιχάλη a,b,c θετικοί????

Αυτή είναι από Baltic Way. Εύκολη είναι...

Ένα μικρό hint:[hide]$ a/(a^2\plus{}2)<\equal{}a/(2a\plus{}1)$[/hide]

Ναι ευτυχώς που το  είπες Μήτσο... Το διόρθωσα :wink:

Ευχαριστώ για την αφιέρωση ... Όπως είπε ο Κώστας παραπάνω

[hide]$ \frac{a}{a^2\plus{}2}\leq\frac{a}{2a\plus{}1}$ άρα μένει νδο $ \sum\frac{a}{2a\plus{}1}\leq 1$ που ισοδυναμεί με

$ a\plus{}b\plus{}c\geq 3$ που ισχύει qed [/hide]

ήταν λίγο ευκολάκι.... δες την άλλη που έβαλα τώρα :)


geometry
rectangle
analytic geometry
graphing lines
slope
geometric transformation
reflection

Let $ n$ be a positive integer. Consider a rectangle $ (90n\plus{}1)\times(90n\plus{}5)$ consisting of unit squares. Let $ S$ be the set of the vertices of these squares. Prove that the number of distinct lines passing through at least two points of $ S$ is divisible by $ 4$.

Hope I didn't make some mistake again...

First of all, put origin of cartesian coordinate system in lower left corner.
There are four types of these lines: horizontal, vertical, with positive slope, and with negative slope.
There are $ 180n + 8$ horizontal and vertical lines, so its divisible with 4. Number of lines with positive slope is the same as the number of lines with negative slope, by simetry, so the number of all lines is divisible with 4 iff number of lines with positive slope is even. We are going to prove that:

The slopes of these lines are rational numbers of the form: $ \frac ba, a\in {1,...,90n + 1\}, b\in \{1,...,90n + 5\}, gcd(a,b)=1}$, and let $ A$ be the set of all the slopes of these lines (there is no line with slope smaller then $ \frac 1{90n + 1}$ or bigger then $ 90n + 5$). It's not hard to se that number of lines with slope $ \frac ba$ is $ b\cdot (90n + 1 + 1 -a) + a\cdot (90n + 5 + 1 - b) - ab = b(90n + 2 ) + a(90n + 6) - 3ab\equiv_2 ab$. If $ \frac ba \in A$ and $ (a,b)\not = (1,1)$ and $ b\leq 90n + 1$ then $ \frac ab$ is also in $ A$, so the number of lines with slope $ \frac ba$ + the number of lines with slope $ \frac ab$ is $ 2ab \mod 2$ so it's even.

Now we have to count lines with slopes $ 1, \frac {90n + 2}a, \frac {90n + 3}a, \frac {90n + 4}a, \frac {90n + 5}a$. There are $ 180n + 5$ lines with slope 1, and there are even number of lines with slopes of the form $ \frac {90n + 2}a$ and $ \frac {90n + 4}a$, so we have to deal with slopes $ \frac {90n + 3}a$ and $ \frac {90n + 5}a$.

We saw that if $ a$ is even, then number of lines with slope $ \frac {90n + 3}a$ is even, but when $ a$ is odd, then there is odd number of those lines. But there are even number of $ a$'s which are relatively prime with $ 90n + 3$ and odd ($ 4|\varphi(90n + 3)$), so the number of all lines with slope $ \frac {90n + 3}a$ is even.
On the other hand, there is odd number of odd $ a$ which are relatively prime with $ 90n + 5$ and not bigger then $ 90n + 1$ (again $ 4|\varphi (90n + 5)$ but $ a$ can't be $ 90n+3$) so there is odd number of lines with slopes of the form $ \frac {90n + 5}a$, but the number of lines with slope $ 1$ is also odd. So, the number of all lines with positive slope is even. QED

hmmm.

can't you just take the two perpendicular bisectors of the sides of the rectangle as your x and y axis? then for each such line except the lines passing through the origin and the lines parallel to the axis there will be 4 images (reflection on x axis, y axis 180 degrees rotation by the origin and the line itslef) that are distinct and this gives a partition into four-tuples for these lines. together with the 180n+8 horizontal or vertical lines we only need to show that the number of lines passing through the origin is divisible by 4.
now this number corresponds to twice the number of rational numbers $ \frac{a}{b}$ with $ a\in \{1,3,5,\dots, 90n\plus{}1\},b\in \{1,3,\dots,90n\plus{}5\}$ which we have to show is even.
as we can pair $ \frac{a}{b}$ with $ \frac{b}{a}$ then we only need to consider $ b\in \{90n\plus{}3,90n\plus{}5\}$ which gives us
$ \varphi (90n\plus{}5)\plus{}\varphi (90n\plus{}3)$
the first term is divisible by 4 as $ 5|90n\plus{}5$ and the second term is divisible by four as $ 90n\plus{}3$ contains at least two odd prime divisors

Well, yeah, yours is a bit smarter :) When I solved it, i didn't try to improve it.
I am just not sure, how do you exclude cases $ b\in \{90n \plus{} 2, 90n \plus{} 4\}$?

note that in my coordinate system the points can not have integer coordinates, instead they are of the form $ (\frac{2p\plus{}1}{2},\frac{2q\plus{}1}{2})$
do you see it now?

dude no guys

this method fails.

here's a hint: are you sure (Albanian Eagles) that what you want to consider is phi(90n+5)+phi(90n+3) ?

no dude listen

you want to consider gamma(93)+gamma(95), where the gamma function is the number of ODD numbers relatively prime to whatever.

btw if you use your method you end up with 2 mod 4, since the line y=x is paired with y=x when you pair reciprocal slopes.

that's why you actually need to show gamma(93)+gamma(95) is odd
rather than phi(93)+phi(95) is even

another way:(Hope i don't mistake .please check it)
you can take perpendicular bisector of the sides of rectangle the $X,Y$ axes.the number of vertical and horizontal lines are $180n+8$ that is divisible by $4$.the number of lines that pass the origin is $2(45n+3)+2(45n+1)$ that is divisible by $4$.rest of the lines can be partitioned into groups such that the number of each group is $4$.because we can consider of reflections of each line through the axes$X,Y$. and that is exactly $4$.


This is the thread for the posting of the American Mathematics Competitions 10. Please read over the directions and submit answers by Private Message to [b]anirudh[/b]. Please include all answers in a readable answer with solutions.

Thank you. :)

Sorry for the Mis-typed file name. Some of the problems are probably too hard though.

Sign up in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125029.

Download the problems from http://www.artofproblemsolving.com/Forum/viewtopic.php?mode=attach&id=6273 if you cannot access the attachment.

Alternate Download: [url=http://www5.upload2.net/download/2fed2efdeec894596fb24c9f6c0848e0/4588b97e/X09B1MPZEX7P25e/AMC10.pdf]here[/url]

Please do not post in this thread. Do all posting in the other post.

Update: ERRATA HAS BEEN FIXED!!! DOWNLOAD THE NEW VERSION!!!

[size=200]As posted in the other thread, as not to cause confusion, no more solutions are required.[/size]

My original plan was to take people's best solutions and put them together in a solution booklet, but I'm just going to create the solutions myself.

Unfortunately, I can't edit the above posts, so please just ignore the part about solutions.


trigonometry
trig identities
tangent half angle

could someone please give me some trig problems for practice? i just started to really learn about it yesterday. i know a bit about graphs and some identities. thanks

Show that $ \tan{\frac{x}{2}}\equal{}\frac{\sin{x}}{1\plus{}\cos{x}}$.
There are many ways to do this problem so see if you can find them.  Oh, and you can't use any formula that you don't derive (except sum and difference formulas).

[hide="One Proof"]First, we use the double angle formulas for cosine, which are easily derived from the general cosine sum formula and the identity $ \sin^2 \theta + \cos^2 \theta = 1$:
[b]Sine:[/b]
\begin{align*} \cos 2\theta = 1 - 2\sin^2 \theta & \Rightarrow 2\sin^2 \theta = 1 - \cos 2\theta \\
& \Rightarrow \sin^2 \theta = \frac {1 - \cos 2\theta}{2} \\
& \Rightarrow \sin \theta = \pm \sqrt {\frac {1 - \cos 2\theta}{2}} \end{align*}
[b]Cosine:[/b]
\begin{align*} \cos 2\theta = 2\cos^2 \theta - 1 \\
& \Rightarrow 2\cos^2\theta = 1 + \cos 2\theta \\
& \Rightarrow \cos^2 \theta = \frac {1 + \cos 2\theta}{2} \\
& \Rightarrow \cos \theta = \pm \sqrt {\frac {1 + \cos 2\theta}{2}} \end{align*}
[b]Tangent Half-Angle:[/b]
If we let $ x = \frac {\theta}{2}$, and use $ \tan \theta = \frac {\sin \theta}{\cos \theta}$, we have (the plus-or-minus sign drops out because the the numerator and denominator will be positive for any angle $ x$):
\[ \tan \frac {x}{2} \& = \sqrt {\frac {1 - \cos x}{1 + \cos x}} \\
\& = \frac {\sqrt {1 - \cos x}}{\sqrt {1 + \cos x}} \\
\& = \frac {\sqrt {1 - \cos^2 x}}{1 + \cos x} \\
\& = \frac {\sqrt {\sin^2 x}}{1 + \cos x} \\
\& = \boxed{\frac {\sin x}{1 + \cos x}}
\]
[/hide]

[hide="Another simpler one"]
Write $ \sinh = 2 \cdot \sin {\frac {x}{2}} \cdot \cos {\frac {x}{2}}$
And $ 1 + \cos x = 2 \cdot cos^2 {\frac {x}{2}}$  :)
[/hide]

1. Given $ \sin ^2 \plus{} \cos^2 \equal{} 1$, derive the other Pythagorean Identities (yes this is a simple one).


inequalities
algebra
polynomial
inequalities proposed

For $a,b,c$  positive real numbers, prove that

$\frac{a^2+2bc}{(b+c)^3}+\frac{b^2+2ca}{(c+a)^3}+\frac{c^2+2ab}{(a+b)^3}\ge \frac{27}{8(a+b+c)}$

I dedicate the idea of this proof to darij grinberg.

By generalized schur inequality (Vornicu-schur :D),
\[
\sum {\frac{{(a - b)(a - c)}}{{(b + c)^3 }}}  \ge 0
\]
So we have to prove,
\[
\left( {a + b + c} \right)\left( {ab + bc + ca} \right)\sum {\frac{1}{{(a + b)^3 }}}  \ge \frac{{27}}{8}
\]
Now,
\[
\begin{array}{l}
 \left( {a + b + c} \right)\left( {ab + bc + ca} \right)\sum {\frac{1}{{(a + b)^3 }}}  \\
  \ge \frac{1}{3}\left( {a + b + c} \right)\left( {ab + bc + ca} \right)\sum {\frac{1}{{(a + b)^2 }}} \sum {\frac{1}{{(a + b)}}}  \\
  = \frac{1}{3}\left( {a + b + c} \right)\sum {\frac{1}{{(a + b)}}} \left( {ab + bc + ca} \right)\sum {\frac{1}{{(a + b)^2 }}}  \\
  \ge \frac{1}{3}\left( {\frac{9}{2}} \right)\left( {\frac{9}{4}} \right) \\
  = \frac{{27}}{8} \\
 \end{array}
\]
by chebyshev, am-hm and iran 1996.

[quote="manlio"]I don't remember this one

$\sum {\frac{{(a - b)(a - c)}}{{(b + c)^3 }}} \ge 0$

Where did you find it?[/quote]

it is a special case of vornicu-schur inequality.

[quote="manlio"]Where did you find it?[/quote]
http://www.mathlinks.ro/Forum/viewtopic.php?p=156212#p156212  :D

Sorry, I didn't remember it  :blush:  :blush:

With the same conditions, the generalized inequality

$ \frac {a^2 \plus{} kbc}{(b \plus{} c)^{3}} \plus{} \frac {b^2 \plus{} kca}{(c \plus{} a)^{3}} \plus{} \frac {c^2 \plus{} kab}{(a \plus{} b)^{3}}\geq\frac {9(1 \plus{} k)}{8(a \plus{} b \plus{} c)}$

holds if and only if $ \minus{} 219.06\cdots \equal{} k_3\leq k\leq \frac {23}{7} \equal{} 3.2857\cdots,$

where $ k_3$ is the least real root of the following irreducible polynomial over $ \mathbb{Q}:$

$ 4k^5 \plus{} 875k^4 \minus{} 280k^3 \minus{} 274k^2 \plus{} 28k \minus{} 49;$

with equality if $ k \equal{} k_3,b \equal{} c \equal{} 1,a \equal{} 6.7506\cdots$ is the greatest real root of the following irreducible polynomial

$ 2a^5 \plus{} a^4 \minus{} 56a^3 \minus{} 226a^2 \minus{} 354a \minus{} 199.$

I'm expecting nice solutions for $ k \equal{} \frac {23}{7}:$

$ \frac {7a^2 \plus{} 23bc}{(b \plus{} c)^{3}} \plus{} \frac {7b^2 \plus{} 23ca}{(c \plus{} a)^{3}} \plus{} \frac {7c^2 \plus{} 23ab}{(a \plus{} b)^{3}}\geq\frac {135}{4(a \plus{} b \plus{} c)}$

and $ k \equal{} \minus{} 219:$

$ \frac {a^2 \minus{} 219bc}{(b \plus{} c)^{3}} \plus{} \frac {b^2 \minus{} 219ca}{(c \plus{} a)^{3}} \plus{} \frac {c^2 \minus{} 219ab}{(a \plus{} b)^{3}}\geq\frac { \minus{} 981}{4(a \plus{} b \plus{} c)}.$


modular arithmetic
induction
function
number theory
totient function

Prove that $ 2^{3^n} \plus{}1$ is divisible by $ 3^{n\plus{}1}$ for all $ n$.

so you need to show that

$ 2^{3^n} \equiv \minus{} 1 \pmod{3^{n \plus{} 1}}$

First, by euler's totient theorem

$ a^{3^n(3 \minus{} 1)} \equiv 1 \pmod{3^{n \plus{} 1}}$

for all $ a$ not divisible by $ 3$, and therefore also for $ a \equal{} 2$. Now let $ 2^{3^n} \equal{} x$.  Our relation is then

$ x^2 \equiv 1 \pmod{3^{n \plus{} 1}}$

and hence $ 3^{n \plus{} 1} | x^2 \minus{} 1 \equal{} (x \minus{} 1)(x \plus{} 1)$.  Therefore at least one of these factors must be divisible by $ 3$.

Case 1: $ 3|x \minus{} 1$ which implies $ x \equiv 1 \pmod{3}$, $ x \plus{} 1 \equiv 2 \pmod{3}$ and therefore $ 3$ does not divide $ x \plus{} 1$.  Thus, $ 3^{n \plus{} 1} | x \minus{} 1$ implies $ x \equiv 1 \pmod{3^{n \plus{} 1}}$.

Case 2: $ 3|x \plus{} 1$ which implies $ x \equiv \minus{} 1 \pmod{3}$, $ x \minus{} 1 \equiv 1 \pmod{3}$ and therefore $ 3$ does not divide $ x \minus{} 1$.  Thus, $ 3^{n \plus{} 1} | x \plus{} 1$ implies $ x \equiv \minus{} 1 \pmod{3^{n \plus{} 1}}$, as desired.

Our proof will be complete if we can show the impossibility of case 1.  That is, we only need to show that $ x \equal{} 2^{3^n} \equiv 2 \pmod{3}$ for any $ n$.  This is simple: Since $ 3^n$ is always odd, there exists an integer $ k$ such that $ 2k \plus{} 1 \equal{} 3^n$.  Therefore, $ 2^{3^n} \equiv 2^{2k \plus{} 1} \equiv 4^k \cdot 2 \equiv 2 \pmod{3}$, exactly as desired.

How would we prove the Euler's Totient Theorem and the congruence $ 2\cdot 4^k\equiv 2\mod 3$?

EDITED TYPO

Induction: $ 2^{3^n} \plus{} 1 \equal{} \left(2^{3^{n \minus{} 1}} \plus{} 1\right) \left[\underbrace{\left(2^{3^{n \minus{} 1}}\right)^2}_{\equiv 1 \pmod{3}} \minus{} \underbrace{\left(2^{3^{n \minus{} 1}}\right)}_{\equiv \minus{} 1 \pmod {3}} \plus{} 1\right]$.

[quote="mathwizarddude"]Could would we prove the Euler's Totient Theorem and the congruence $ 2\cdot 4^k\equiv 2\mod 3$?[/quote]

'Prove [[Euler's Totient Theorem]]' as in a proof of the theorem?

$ 2 \cdot 4^k \equiv 2 \cdot 1^k \equiv 2 \pmod{3}$.

On [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_totient_function#Formulas]this page[/url] where it says
"$ p_1^{e_1\minus{}1}p_2^{e_2}\cdots p_m^{e_m} \plus{} p_1^{e_1}p_2^{e_2\minus{}1}\cdots p_m^{e_m} \plus{} \cdots \plus{} p_1^{e_1}p_2^{e_2}\cdots p_m^{e_m \minus{} 1}.$

We can factor out, though:

$ p_1^{e_1\minus{}1}p_2^{e_2\minus{}1}\cdots p_m^{e_m\minus{}1}(p_1\plus{}p_2\plus{}\cdots \plus{} p_m).$"

I don't think the factorization is correct.

[quote="mathwizarddude"]azips: Is there an elementary proof of Euler's Totient Theorem?[/quote]

Yes there is;
Euler's Totient theorem is a generalisation of Fermat's little theorem, so i suggest you proove that one fist.
[url=http://www.cut-the-knot.org/blue/Fermat.shtml//url]Fermat's Little Theorem[/url]
[url=http://www.cut-the-knot.org/blue/Euler.shtml//url]Euler's Totient Function[/url]

EDIT: sorry the links don't work very well, they will only take you to the website index, but the proofs are easy enough to find from there.

[quote="mathwizarddude"]azips: Is there an elementary proof of Euler's Totient Theorem?[/quote]

Also added a quick proof to wiki - [[Fermat's Little Theorem]] which can be generalized to Euler's.


when i first registered for usamts last year, i was under 13. i did not participate, however. this year i am over 13 but it says that i need to submit the under 13 form on the usamts website. i have changed by grade under my profile. do i have to send the under 13 permission form?

Send an email to usamts@usamts.org about your age; include your USAMTS Id #.  Don't worry about the form.