calculus
geometry
geometric transformation
reflection

How many people are taking AP Physics C? Do you feel prepared? Are you taking both exams, or just one? Also, people that took it last year, do you have any advice? I'm kind of worried.

Question: Do you guys know what the curve usually is for ap physics c mechanics? I just took the test today and I think I got all the free response, but multiple choice was pretty sketchy and I got anywhere from 2-~12 wrong.

I'd be happy if I ONLY got 12 wrong, much less 2 wrong on the Physics B exam.

Don't sweat it; if that's seriously how you performed, then there is no chance you didn't get a five.

i felt fairly confident for the mc mech. hopefully i did well there. the free response for the mech was not too bad, except that last problem got me.

i didn't pace myself well for the mc e&m. i felt ok about the e&m free response though.

hopefully i did well overall. now i have to worry about ap chem

and, we are allowed to be discussing this? Probably not.

[quote="agolsme"]er well, the physics C test only has 35 multiple choice questions :P[/quote]

Oh, well, bummer. Guess you should pray your FRQ get graded good.

You guys are smart....

I just took Physics B. I'm having a headache now...

BTW, is it true that if u get about 2/3 of MC and half of the Free

Response then u can get 5 (in Physics B)?

Physics B has a pretty loose curve, from what I've heard. In the 1998 exam, a 59% overall was a five.

[quote="filletwho"]I heard Physics C curves were the lowest out of all the AP's....somewhere around 35%? or was it 45%? I'm definently taking it next year and guessing on the multiple choice.[/quote]
Does your school offer a class for Physics C?? or do u self-study?

My school doesn't offer, so I'm thinking of self-stuydying...

My school offers it. I'm actually taking the class next year...but w/e...guessing is always fun.

im taking that next yr, dunno if im gonna survive :?

I heard Physics C + Bio + English is the worst possible line-up of AP's.

I took Physics C; both Mech and E&M. I thought Mechanics was a cakewalk; there weren't like any really challenging questions. E&M was easier than I expected, too; no hardcore calculus, probably since most people can't be cool like that. I did manage to misread one part of a question on the FR section in the last minute and I was like :mad: but whatever.

And I think it's ok if we discuss the general difficulty of the test? As long as there are no specific MC questions and no FR stuff before like tomorrow.

As for worst line of AP's, someone at my school has Physics C, American/Comp Government, Chemistry... that's a pretty intense three days.

[quote="filletwho"]Actually, American and Chemistry aren't hard at all (I'm saying this after I took them...it would be the exact opposite if this was before I took them). Comp Sci I heard was hard. Bio is right up there with Physics C.[/quote]

Oh, I see, you meant difficulty-wise. Yeah, Chemistry today wasn't hard (I thought it was a little harder than Mechanics, though); nothing compared to E&M.

I've beaten that.  Monday I had Bio in the morning, then both Physics C tests in the afternoon.  That's 3 AP's in one day, although none of them were difficult.

[quote="filletwho"]Wait, dont' they have a standard schedule for the AP's nationwide?[/quote]

Yeah...  :huh:

[quote="filletwho"]Well, the tests are given two per day here...I thought it would be like that everywhere.[/quote]

You can take AP Bio, AP Physics C: Mech, AP Physics C: E&M all in one day, though.

Hey guys,
I have a general question about grading for Free-response for Phyiscs..

Say on the first part of the problem, you got a wrong answer and u had to use that wrong answer to solve other parts...So it's like you knew how to do the problem, but messed up the answers becz of the first part. How would this be scored?? Will points be deducted from only the first part and u will get full credit for other parts?
Thanks,

A student shall not be penalized for using an incorrect answer from one part to answer another part, should that answer be required for the next part  (no double jeopardy).

I think that's pretty close to the actual wording.

That's why my physics teacher made us take practice tests and go over them.  You can get wrong answer for one part and get counted off for it, but as long as you use the wrong answer [i]correctly[/i] for the next part, you're fine.  Sometimes, if they ask you a multiple choice/yes-no question on the free response and justify your answer, you can pick the wrong answer, but if you justify it well and clearly, you can snag some points from it.

[quote="drunner2007"]i'm pretty sure we can talk about the test now...i'm trying to remember the e&m.[/quote]

The free response questions are posted on the collegeboard website two days after each exam, btw.

[quote="drunner2007"]i'm pretty sure we can talk about the test now.

what did you guys think about the free-response for mech. and e&m? for the mech. i felt ok, but I did not figure out part (a) of the rolling ball problem. that reflects a lack of mech. practice recently. what were the other two again?

i'm trying to remember the e&m. I really enjoyed doing the flux problem with the spring etc. The problem with the electric fields was the easiest I think (hopefully I didn't mess it up). I can't remember the other problem.[/quote]

I'm pretty sure I aced the Mech FRQ's; they were all standard exercises, nothing with like torsional pendulums and weird stuff like that. E&M had strange, but also easy questions - the electric field one was just a joke (I did manage to misread something through), the RC circuit was mostly standard (adding in the second resistor in parallel made me think a little through), and the wire loop in the magnetic field with the spring was like the only interesting problem. I thought the FRQ's this year were really short, too. Not sure why.

ok, ready to discuss now :)

mechanics was easy. I somehow managed to convince myself e&m was hard but now that I think about it, it was easy too

hmm, for hard AP lineups, I had french lang, comp sci, stat in a row . . . wait stat was trivial . . . scratch that then

and bio is usually pretty easy, isn't it? Unless the difficulty of the tests vary from year to year (which they might, I dunno).

statistics was fairly easy. i have forgotten almost all statistics that is not very basic. maybe i should go back and review.

are any of you guys taking a subsequent physics course in high school, or in college any time soon? if so, which ones? i'm considering a second year mech. or e&m class, or a modern physics class.

I'm taking modern physics next year.  I may self-study more advanced mechanics/E&M outside of school as well.


geometry
3D geometry
geometric transformation
reflection
inequalities
inequalities proposed

[quote="toanhocmuonmau"]This is a problem in Mathematical Reflections so we are not allowed to discuss about it now.[/quote]

No. The deadline is Nov.15  :)

[quote="Summerburn"]Let $ a,b,c\geq 0$ and no two being zero together.

Show: $ \sqrt [3]{\frac {a^{2} \plus{} bc}{b^{2} \plus{} c^{2}}} \plus{} \sqrt [3]{\frac {b^{2} \plus{} ca}{c^{2} \plus{} a^{2}}} \plus{} \sqrt [3]{\frac {c^{2} \plus{} ab}{a^{2} \plus{} b^{2}}}\geq \frac {9\sqrt [3]{abc}}{a \plus{} b \plus{} c}$[/quote]
See also here: http://can-hang2007.blogspot.com/2010/01/inequality-41-p-h-duc.html


calculus
integration
abstract algebra
trigonometry
complex numbers
real analysis
real analysis unsolved

I did not know exactly where to put this. Move it if you want to.

Consider the following integral:

$ u(a) \equal{} \frac{1}{2 \pi} \int_0^{2\pi} \left(\frac{e^{it} \plus{} e^{\minus{}it}}{2}\right)^2 \frac{e^{it} \plus{} a}{e^{it} \minus{} a} \, dt$, $ |a| < 1$

That is, if I wrote this correctly the real part of this solves the Dirichlet problem in the unit disc with boundary condition $ u(e^{i \theta}) \equal{} cos^2 \theta$. Now, this integral is a little messy but fairly easy to evaluate to $ \frac{a^2 \plus{} 1}{2}$. However, both Maple 9.5 and Mathematica 5 will answer with a value for this integral, but they are both wrong. Does anyone know why this is so?

In any case, is there an easier way to find the solution to the Dirichlet problem than to evaluate this integral?

Mathematica ver 6 gives the correct answer.  If you use Mathematica a lot, you may wish to join the Mathematica forum and ask the questions pertaining to Mathematica directly to the pros:

http://mathforum.org/kb/forum.jspa?forumID=79

[quote="mlok"]Given a boundary condition $ u(e^{i\theta}) \equal{} \sum_{n\in\mathbb Z} c_n e^{i n\theta}$, you can solve the Dirichlet problem for Laplace equation as follows: $ u(re^{i\theta}) \equal{} \sum_{n\in\mathbb Z} c_n r^{|n|} e^{i n\theta}$.

One can say that the Poisson kernel has multiplier $ r^{|n|}$.[/quote]
How come this is so? I have completely missed fourier analysis in my education so I don't know what to do. I'm back at the forums; never knowing anything  :lol:

On the formal level, this is easy to check: each term of the series solves the equation, and the boundary condition is satisfied as well. Of course, these statements require justification when the series is infinite, but in your case $ \cos^2 \theta$ has a finite Fourier series.


let a,b,c be three sides of a triangle, prove:
a^2 x b x (a-b)+b^2 x c x (b-c)+c^2 x a x (c-a) >= 0

Let $ a \equal{} x\plus{}y$, $ b \equal{} y\plus{}z$ and $ c \equal{} z\plus{}x$ for some $ x,y,z > 0$. The equation becomes equivalent to $ x^{3}z\plus{}y^{3}x\plus{}z^{3}y\geq xyz(x\plus{}y\plus{}z)$. But according to AM-GM we have \[ \begin{aligned}& 4x^{3}z\plus{}2y^{3}x\plus{}z^{3}y\geq 7x^{2}yz\\ & x^{3}z\plus{}4y^{3}x\plus{}2z^{3}y\geq 7xy^{2}z\\ & 2x^{3}z\plus{}y^{3}x\plus{}4z^{3}y\geq 7xyz^{2}\end{aligned}\]Sum them up and we're done.


let's say a function f(x) is defined for [-2, 6]. and it is not defined at 0 (removable discontinuity). If the question asks: find all points on which f(x) is not continuous, is the only point zero, or -2, 6 also discontinuities? because the definition of continuity states that 1) limit of f(a) should exist, 2)f(x) is defined at x=a, and 3)limit of f(x) at a is equal to f(a). the end point clearly doesn't satisfy the first condition, (only one side limit exist, therefore limit does not exist). is it correct to say there is a discontinuity at f(-2) and f(6)

thanks ina dvance.

Please look at the announcement at the top of the forum before posting in the future. Calculus is not considered a high school topic.

The answer depends on context; the one-sided limit is sometimes good enough and sometimes not.


What is the result when we apply PCC to a primary alcohol with R - CDHOH, where the carbon is attached to the deuterium, hydrogen, and hydroxyl. Two major products are formed, but is the primary one the aldehyde with the D or the H? I would think H is easier to eliminate then D.

the product RCDO will be formed i think coz the leaving ability of H is better than D comparitively as it is easier to abstarct H than D as D is heavier!!! :)


Find all positive integers n such that n 2 is palindrome.

[Note: A palindrome number is the same when read from left to right and from right to left. (i.e. 282 is and 239 is not.)]

I asked Michael about this, but neither of us came up with useful things...difficulty is at that n is not necessarily palindrome...

I need some help...  :(

I think it would be an extremely difficult and most of all ugly problem. It interferes with the decimal system so you'd get lots of difficult 10's involved etc.


trigonometry
calculus
integration
logarithms
calculus computations

The integral is:

$\int{\frac{dx}{\sqrt{x^{2}-9}}}dx$

We are supposed to use a trig substitution for this problem apparently because next to the problem, the author of the book im using (Thomas' 11th edition), says - hint: Trig.

So the substitution im thinking is let $x = a\sec{u}$, so $dx = a\tan{u}\sec{u}du$

In our case, $a = 3$, and x is just x. Now what to do from here?

[b]FORMULA :[/b] $a\ne 0\Longrightarrow \int\frac{1}{\sqrt{x^{2}+a}}\ dx=\ln\left|x+\sqrt{x^{2}+a}\right|+C\ .$

[b]Remark.[/b] If $a>0$, then can without "modul".


counting
derangement
calculus
integration
absolute value
number theory
prime factorization

You randomly scramble the integers from 1 to 2009, and denote them $ a_1, a_2, \cdots, a_{2009}$.

How many sequences $ a_1, a_2, \cdots, a_{2009}$ are there such that $ |a_1\minus{}1|\equal{}|a_2\minus{}2|\equal{}\cdots\equal{}|a_{2009}\minus{}2009|>0$?

perhaps none.
Note that $ a_n \minus{} n \equal{} \pm(a_m \minus{} m)$ implies $ a_n \equal{} n \pm (a_m \minus{} m)$

not sure... without the condition $ > 0$ one solution would be $ a_n \equal{} n$.  Is this the only solution?

The greater than 0 condition tells us that solutions must be derangements of the set $ {1,2,3 \ldots 2009}$.

idea: let $ c$ be the common value.  if $ c$ is $ 1$, $ a_1 \equal{} 2$, $ a_2 \equal{} 1,3$, $ a_3 \equal{} 4$ (can't be 2 because $ a_1$ is), $ a_4 \equal{} 3,5$ $ a_5 \equal{} 6$ (in general $ a_n\equal{}n\plus{}1$ for $ n$ odd)... to $ a_{2009} \equal{} 2010$, contradiction.  And I'll bet other values of $ c$ will produce similar contradictions.

btw, this belongs in Intermediate.

No, you're right. Suppose $ a_n \equal{} n\plus{}k$. Then, $ |a_n \minus{} n| \equal{} |k|$. The only index $ j$ such that $ a_j \equal{} n$ must then satisfy $ |n\minus{}j| \equal{} |k|, \; j \neq n\plus{}k \implies j \equal{} n\minus{}k$. We repeat this process until we reach an index that is either less than $ 1$ or greater than $ 2009$.

We have that $ |a_i\minus{}i|\equal{}m$ for all $ i\equal{}1,2,...,2009$ and $ 2009>m>0$. From here it follows that $ a_i\minus{}i \equal{} \plus{}m$ or $ a_i\minus{}i \equal{} \minus{}m$. However, if we sum up all terms without using the absolute value: $ (a_1\minus{}1)\plus{}(a_2\minus{}2)\plus{}...\plus{}(a_{2009}\minus{}2009)\equal{}(a_1\plus{}a_2\plus{}...\plus{}a_{2009})\minus{}(1\plus{}2\plus{}...\plus{}2009)\equal{}0$. This is true because $ a_1,...,a_{2009}$ is a permutation of $ 1,...,2009$. Therefore half of the terms should be equal to $ \plus{}m$ and the other half should be $ \minus{}m$, which is impossible because $ 2009$ number of elements cannot be split into two equal parts.

[quote="vvs"]We have that $ |a_i \minus{} i| \equal{} m$ for all $ i \equal{} 1,2,...,2009$ and $ 2009 > m > 0$. From here it follows that $ a_i \minus{} i \equal{} \plus{} m$ or $ a_i \minus{} i \equal{} \minus{} m$. However, if we sum up all terms without using the absolute value: $ (a_1 \minus{} 1) \plus{} (a_2 \minus{} 2) \plus{} ... \plus{} (a_{2009} \minus{} 2009) \equal{} (a_1 \plus{} a_2 \plus{} ... \plus{} a_{2009}) \minus{} (1 \plus{} 2 \plus{} ... \plus{} 2009) \equal{} 0$. This is true because $ a_1,...,a_{2009}$ is a permutation of $ 1,...,2009$. Therefore half of the terms should be equal to $ \plus{} m$ and the other half should be $ \minus{} m$, which is impossible because $ 2009$ number of elements cannot be split into two equal parts.[/quote]

Ohh thanks :D

So the solution is 'none' because there are odd number of elements, right?

Then, how many solutions are there if there are even number of elements, say, 2010?

starting with vvs's method:

$ a_i \minus{} i \equal{} \pm m$.  Now if we let $ a_1 \equal{} m \plus{} 1$, $ a_2 \equal{} 2 \pm m$, $ a_3 \equal{} 3\pm m$ etc.  Then if $ a_i \equal{} i \minus{} m$ and $ m \plus{} i \equal{} j \minus{} m$ then $ a_j \neq j \minus{} m$, in which case $ j \minus{} i \equal{} 2m$.  Now if $ m$ is greater than $ (2010 \minus{} 1)/2 \equal{} 1004.5$, this is always true.  But $ m$ is at most $ 1005$ because $ 2010 \ge 1005 \pm m > 0$ which for the negative gives $ m < 1005$ and for the positive gives $ 1005 \ge m$.  Thus $ 1 \le m \le 1005$.

Now lets try to find a sequence for each $ m$.  (We are looking for a pattern, not trying to do the whole problem by casework.)
$ m \equal{} 1$ gives $ a_1 \equal{} 2$, $ a_2 \minus{} 2 \equal{} \pm 1 \implies a_2 \equal{} 1,3$; $ a_3 \equal{} 3 \pm 1 \equal{} 2,4$ but $ a_3 \neq a_1 \implies a_3 \equal{} 4$, etc.  Notice the domino effect that occurs: $ a_n \equal{} n \plus{} 1$ for all $ n$ odd.  Additionally, $ a_n \equal{} n\pm1$ for all $ n$ even implies $ a_2 \equal{} 1$ since it is the only term that can be $ 1$, and this makes $ a_n \equal{} n \minus{} 1$ for all $ n$ even.  Thus our sequence goes $ 2,1,4,3,6,5 \ldots 2010,2009$ and we have $ 1$ solution.

In fact, for any $ m$ we have $ a_1 \equal{} 1 \pm m > 0$ implies $ a_1 \equal{} 1 \plus{} m$.  The domino effect then proceeds with $ a_{1 \plus{} 2m} \equal{} 1 \plus{} 2m \pm m \neq 1 \plus{} m \implies a_{1 \plus{} 2m} \equal{} 1 \plus{} 3m$, and in general $ a_{1 \plus{} 2km} \equal{} 1 \plus{} (2k \plus{} 1)m$ for all integers $ k \ge 0$.

Now in the same way we have $ a_2 \equal{} 2 \pm m \implies a_2 \equal{} 2 \plus{} m$ if $ m > 1$ so in general $ a_{2 \plus{} 2km} \equal{} 2 \plus{} (2k \plus{} 1)m$.  But at this point we really don't have ways to rule out the minus.

Now $ a_{1 \plus{} m}$ is the only term which can be $ 1$, hence $ a_{1 \plus{} m} \equal{} 1$ which implies $ a_{1 \plus{} (2k \plus{} 1)m} \equal{} 1 \plus{} 2km$
$ a_{2 \plus{} m}$ is the only term which can be $ 2$, so $ a_{2 \plus{} m} \equal{} 2$ which implies $ a_{2 \plus{} (2k \plus{} 1)m} \equal{} 2 \plus{} 2km$

we now rule out $ a_{b} \equal{} b \minus{} m$ for $ 1 \le b \le m$, which implies $ a_{b \plus{} 2km} \equal{} b \plus{} (2k \plus{} 1)m$ and note if $ m < b \le 2m$ we have a term of the form $ a_{b' \plus{} (2k \plus{} 1)m}$ for $ 1 \le b' \le m$ which falls in the other category, since these are the only candidates for $ 1,2,3,4 \ldots$.  Thus we must have:

$ a_{b \plus{} 2km} \equal{} b \plus{} (2k \plus{} 1)m$ for $ 1\le b \le m$ and
$ a_{b \plus{} (2k \plus{} 1)m} \equal{} b \plus{} 2km$ for $ 1 \le b \le m$

now exactly half the terms must be of one form and half of the other.  Since we start in the first form, we must end with the last possible term of the second form, thus for the last $ k$, $ (2k \plus{} 2)m \equal{} 2010$ gives $ 1005 \equal{} m(k \plus{} 1)$.  Since our only restrictions are that $ k$ is a nonnegative integer and $ m\le 1005$ is a positive integer, there are as many valid $ m$'s as integral divisors of $ 1005 \equal{} 3*5*67$ thus $ 1005$ has $ 2^3 \equal{} 8$ factors, and there are $ 8$ sequences.

In general, there are $ \tau (n/2)$ sequences when $ n$ is even, where $ n$ is the number of terms in the sequence. ($ \tau(x)$ denotes the number of factors of $ x$.  If $ p_1^{e_1}p_2^{e_2} p_3^{e_3} \cdots$ is the prime factorization of $ x$, then $ \tau(x) \equal{} (e_1 \plus{} 1)(e_2 \plus{} 1)(e_3 \plus{} 1) \cdots$)


The arithmetic mean (ordinary average) of the fifty-two successive positive integers beginning with 2 is:
\[ \text{(A)}\ 27\ \ \ \ \text{(B)}\ 27\frac{1}{4}\ \ \ \ \text{(C)}\ 27\frac{1}{2}\ \ \ \ \text{(D)}\ 28\ \ \ \ \text{(E)}\ 28\frac{1}{2} \]

[hide]The sum of the first 52 integers starting with 2 would be $\frac{(52)(55)}{2}$.  To find the mean, we divide by $52$ which gives $\frac{55}{2}=27.5$.  $C$.[/hide]

[hide]
if 2 is the first term, the last term will be 2+51 or 53
answer=(53+2)/2 or [b]27.5 which is C[/b][/hide]


What is a coefficient? Also how do you rearrange linear equations?

Hi. I'm not sure how to answer this one.

In the expression $ 3x \plus{} 7$, $ 3$ is the coefficient of $ x$: it is the number in front of $ x$.

In the expression $ 4x^{2} \minus{} 7x \plus{} 2$, $ 4$ is the coefficient of $ x^{2}$ and $ \minus{} 7$ is the coefficient of $ x$.

As to rearranging linear equations, I don't see how I can answer it without giving lots of examples. But I think the following might be helpful: http://www.youtube.com/watch?v=nNfweUGCgOc

The coefficient is the number by which a variable is multiplied by.

Rearangement example: see $ 3x\plus{}8\equal{}7x\plus{}4$

1) Subtract $ 4$ from both sides: $ 3x\plus{}4\equal{}7x$
2) Subtract $ 4x$ from both sides: $ 4x\equal{}4$
3) Divide both sides by $ 4$: $ x\equal{}1$

You're done

Coefficient is the number that is multiplied by the variable or the number in front of the variable

Number multiplied by variable.]

For xample in 6t

6 is the coefficient

[quote="ernie"]The coefficient is the number by which a [color=red]variable[/color] is multiplied by.[/quote]

necessarily a variable?

Well actually, no.  But in algebra, I believe so.  A coefficient can be a constant multiplied to any specific object.

Umm... I'm not sure what you are talking about when you say rearrangement, but I think you might be talking about rewriting this;

$ A \equal{} Bx \plus{} y$

where you know the values of $ x$ and $ y$. For example let's say $ x \equal{} 5$ and $ y \equal{} 10$, then;

$ A \equal{} 5B \plus{} 10$

Then you isolate B;

$ A \minus{} 10 \equal{} 5B$

$ \frac{A \minus{} 10}{5} \equal{} B$

Is that what you meant?

There are several different types of equations you can rearrange. (By "rearrange", I assume you mean isolate, or solve for, a variable.)

A general rule is to undo all the operations that have been performed on the variable.

To keep things simple, in linear equation, you would add and subtract terms as necessary to isolate the terms containing that variable on one side, and the rest of the terms on the other. Then divide by the variable's coefficient to solve.

Examples:

\[ 5x\minus{}10\equal{}0\]
Add $ 10$ to both sides.
\[ 5x\equal{}10\]
Divide both sides by $ 5$.
\[ x\equal{}2\]

Or to solve for other variables:

\[ 5x\plus{}2y\minus{}10\equal{}7y\] Solve for $ x$.
Subtract $ 2y$ from both sides.
\[ 5x\minus{}10\equal{}5y\]
Add $ 10$ to both sides.
\[ 5x\equal{}5y\plus{}10\]
Divide both sides by $ 5$.
\[ x\equal{}y\plus{}2\]

If you have any particular examples, feel free to post them.


find all integers positives $(n,m)$satisfaies $\frac{n^3+m^3-n^2m^2}{(n+m)^2} \in N$

Writing $n+m=S$, $nm=P$ we see that $\frac{n^3+m^3-n^2m^2}{(n+m)^2}=S-\frac{P(3S+P)}{S^2}$, and this is an integer if and only if $S | P$.  So an equivalent condition for the expression to be an integer is $m+n | mn$.

One way to characterise all such positive integers $m$, $n$ is:  $m=k+a$, $n=\frac{k^2}{a}$, where $k$ is any positive integer and $a$ is any divisor of $k^2$.

If you want the expression to be a natural number, there also needs to be a bound placed so that $n^3+m^3>n^2m^2$; for example, we need $S^3>P(3S+P)$, or $0>P^2+3SP-S^3=(P+\frac{3}{2}S)^2-(S^3+\frac{9}{4}S^2)$.


[quote="satsun"]None of the digits of a 3 digit number N is zero. when the leftmost and rightmost digits are swapped, the resulting 3 digit number is N + 792. Find all such numbers N.[/quote]

Notice how 792 = 800 - 8.

Notice how the pattern for the answers is 1-digit-9.

Notice how 9 - 1 = 8.


There exist infinite primes of the form $2^n+1$?
What their distribution (how many primes of this form are $<x$?
Is it possible to answer?

Bye, Uber

It's an open problem iff there are $\infty$ such primes.
Some results (not hard to show) are:
- When it is prime, then $n$ itself is a power of two.
- Every divisor of $2^{2^k}+1$ is $\equiv 1 \mod 2^{k+2}$

All known primes (Fermat primes) are those for $k=0,1,2,3,4$.

Thus it is a Fermat's prime... Just five are known!

[quote="ZetaX"]It's an open problem iff there are $\infty$ such primes.
Some results (not hard to show) are:
- When it is prime, then $n$ itself is a power of two.
- Every divisor of $2^{2^{k}}+1$ is $\equiv 1 \mod 2^{k+2}$

All known primes (Fermat primes) are those for $k=0,1,2,3,4$.[/quote]

how do you prove that if $a^{n}+1$ is prime for some $n \ge 1$ and $a \ge 2$, then $n$ must be a power of $2$.

Supose that n is not a power of 2.
n=c*b for some c and b whit c odd and greater than 1
Then
$a^{cb}+1=(a^{b})^{c}+1^{c}=((a^{b}+1)((a^{b})^{c-1}-(a^{b})^{c-2}+...+1)$
So, it is not prime.


advanced fields
advanced fields unsolved

Hello,

If I'm given the value of the following sum, how do I explicitly solve for a?

x + x*(x-a) + x*(x-a)*(x-2a) + x*(x-a)*(x-2a)*(x-3a) +x*(x-a)*(x-2a)*(x-3a)*(x-4a)......etc until ....*(x-na)

a is a constant close to zero.

Thanks for any help.

Please don't double-post.  Or triple-post, in this case.


function
trigonometry
geometry
geometric transformation
rotation

$ f(x)\equal{}\dfrac{x\sqrt{3} \minus{} 1}{x\plus{}\sqrt{3}}$
$ Determine  g(x) \equal{} fofofo...f(x) 2006 times.$

[hide]Note that $ f^6 (x)=x$ where $ f^6(x)$ represents $ f\circ f\circ \cdots f(x)$ 6 times.  Since $ 2006 = 6\times 334 + 2$, $ f^{2006}(x) = f^2(x) =\frac{x-\sqrt3}{\sqrt{x}+1}$[/hide]

fofofofofofof(x) = f(x) ?

Edit :^beat me to it

[hide]We first note that $ f(x) \equal{} \minus{} \cot(\tan^{ \minus{} 1} x \plus{} \pi/3) \equal{} \minus{} \tan(\pi/6 \minus{} \tan^{ \minus{} 1} x) \equal{} \tan(\tan^{ \minus{} 1} x \minus{} \pi/6)$ (there's definitely a faster way to arrive at that formula...)

Therefore, if we let $ x \equal{} \tan \theta$, we see that $ f(x) \equal{} \tan(\theta \minus{} \pi/6)$. It follows easily that $ f^{n}(x) \equal{} \tan(\theta \minus{} n\pi/6)$ for any positive integer $ n$. Therefore,  $ f^{2006}(x) \equal{} \tan(\tan^{ \minus{} 1} x \minus{} 2006 \pi/6) \equal{} \tan(\tan^{ \minus{} 1}(x) \minus{} \pi/3) \equal{} \frac {x \minus{} \sqrt {3}}{x \sqrt {3} \plus{} 1}$. [/hide]

What inspired you to substitute so many times and hope for a repeat? In an actual competition, I'd die trying. 6 is too much for me. >_<

Thank you Zhero! Its a amazing solution!
How do you note this?

Thanks. :)

Basically, I noted that $ f$ greatly resembled the tangent angle sum formula, and that $ \sqrt{3}$ was the tangent of a nice angle, which inspired me to do what I did. The final formula, $ \tan(x \minus{} \pi/6)$ wasn't so easily notable for me from the original equation. However, if we take the reciprocal of $ f$ and take its opposite, and if we write $ \sqrt{3} \equal{} \tan \pi/3$, we get that $ \minus{}1/f \equal{} \frac{x \plus{} \tan \pi/3}{1 \minus{} x \tan \pi/3} \equal{} \tan(\tan^{\minus{}1} x \plus{} \pi/3)$. Manipulations followed.

[quote="Zhero"]What inspired you to substitute so many times and hope for a repeat? [/quote]
Composition of fractional linear transformations behaves like composition of matrices, and $ \left[ \begin{array}{cc} \sqrt {3} & \minus{} 1 \\
1 & \sqrt {3} \end{array} \right]$ is (a multiple of) a rotation matrix.  Your tangent summation observation follows.

[b]Edit:[/b]  This comes up a lot, so I think it's worth explaining.  Define a map $ \phi : \mathbb{R}^2 \minus{} \{(0, 0) \} \to \mathbb{R} \cup \{\infty \}$ by sending $ (x, y)$ to $ \frac {x}{y}$ and sending $ (x, 0)$ to $ \infty$ for every $ x \neq 0$.  (There's a fancy name for this map called "projectivization," but for now think of it as "taking the inverse slope.")  A linear transformation $ \mathbf{M} : \mathbb{R}^2 \to \mathbb{R}^2$ acts generically by sending

$ (x, y) \mapsto (ax \plus{} by, cx \plus{} dy)$

and taking $ \phi$ of both sides lifts $ \mathbf{M}$ to a function on $ \mathbb{R} \cup \{ \infty \}$ sending

$ t \mapsto \frac {at \plus{} b}{ct \plus{} d}$

where $ t \equal{} \phi(x, y)$; by convention,

$ \frac {a \infty \plus{} b}{c \infty \plus{} d} \equal{} \frac {a}{c}$.

This is the "nice" way to look at fractional linear transformations in that it completely explains their [url=http://en.wikipedia.org/wiki/Projective_linear_group]group structure[/url] and in that on $ \mathbb{R} \cup \{ \infty \}$ (better known as the [url=http://en.wikipedia.org/wiki/Projective_line]projective line[/url]) all fractional linear transformations are honestly bijections:  the preimage of $ \infty$ is $ \minus{} \frac{d}{c}$.

You only need to substitute twice because $ f(f(f(x))\equal{}\minus{}\frac{1}{x}$, so obviously $ f^{6}{x}\equal{}x$.


calculus
derivative
topology
vector
linear algebra
matrix
advanced fields

$ g$ is Riemannian metric tensor of manifold $ M$ of dim $ n$.(i.e. $ g$ is symetric covariant tensor filed on $ M$, which defines a Riammanian G-structure on $ M$).
Write $ (h_{x},k_{x})_{=}<g(x), h_{x}\otimes k_{x}>$. Define a nondegenerate symetric bilinear form $ g_{r}(x)$ on the space of $ \bigwedge^{r}T_{x}(M)$ by:
$ <g_{r}(x), (h_{1}\wedge... \wedge h_{r}) \otimes (k_{1}\wedge...\wedge k_{r}){>}=det((h_{i}, k_{j}))$. This form is positive definite since $ g$ is positive definite (Riemannian, i.e. signature is (n,0)). (refer to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=161659]positive definite[/url])

When $ r=n$ and M is oriented, it is possoble to define a positive n-form (volume form) $ v_{g}$ on $ M$ by:
$ <v_{g}(x), h_{1}\wedge... \wedge h_{r}>{=}<g_{r}(x), (h_{1}\wedge... \wedge h_{r}) \otimes (h_{1}\wedge...\wedge h_{r})>^{1/2}$

Consider the covariant derivative (along vector field $ Z$ on $ M$) relative to Levi-Civita linear connection on $ M$.

(for covariant derivative along a vector field, refer to
[url]http://en.wikipedia.org/wiki/Connection_(vector_bundle)[/url]

First of all, we have $ \nabla_{X}\bullet g_{r}=0$ for any field $ X$.

For if $ Y_{i}$ and $ Z_{i}$ are $ 2r$ vector fields on $ M$, then since Lie derivative $ \theta_{X}$ is a derivative, we have $ \theta_{X}\bullet det((Y_{i}, Z_{j}))$ is the sum of the determinants of the $ r$ matrices obtained by replacing the entries in the $ j$th column of the matrix $ ((Y_{i}, Z_{j}))$ by $ \theta_{X}\bullet (Y_{i}, Z_{j})$. Then by the following properties of covariant derivative:

$ \theta_{X}\bullet det((Y, Z))=(\nabla_{X}\bullet Y, Z)+( Y, \nabla_{X}\bullet Z)$
$ \nabla_{X}\bullet (Y \wedge Z) =(\nabla_{X}\bullet Y) \wedge Z+\ Y \wedge (\nabla_{X}\bullet Z)$ also true if replacing $ \wedge$ by $ \otimes$
$ \theta_{X}\bullet <g, Y>{=}<\nabla_{X}\bullet g, Y>+<g, \nabla_{X}\bullet Y>$

and with a moment's reflextion we have $ <\nabla_{X}\bullet g, (Y_{1}\wedge... \wedge Y_{r}) \otimes (Z_{1}\wedge...\wedge Z_{r})>{=}0$, since $ Y_{i}$, $ Z_{i}$ are arbitrary, QED.

Now how to prove $ \nabla_{X}\bullet v_{g}=0$ for any field $ X$?
my book said proof is analogous and use $ det((Y_{i}, Y_{j}))=(<v_{g}, Y_{1}\wedge... \wedge Y_{n}>)^{2}$.

My thinking is that to prove $ <\nabla_{X}\bullet v_{g}, Y_{1}\wedge... \wedge Y_{n}>{=}0$, it is equivalent to prove:
$ \theta_{X}\bullet det((Y_{i}, Y_{j}))=2<v_{g}, Y_{1}\wedge... \wedge Y_{n}><v_{g}, \nabla_{X}\bullet (Y_{1}\wedge... \wedge Y_{n})>$   *********************** (#1)

But LHS of (#1) =
$ <g_{r}, \nabla_{X}\bullet ((Y_{1}\wedge... \wedge Y_{n}) \otimes (Y_{1}\wedge... \wedge Y_{n}))>{=}2<g_{r}, (\nabla_{X}\bullet (Y_{1}\wedge... \wedge Y_{n})) \otimes (Y_{1}\wedge... \wedge Y_{n})>$ ********************(#2)

by symetry of $ g_{r}$.

RHS of (#1) =

$ 2(<g_{r}, (Y_{1}\wedge... \wedge Y_{n}) \otimes (Y_{1}\wedge... \wedge Y_{n})> <g_{r}, (\nabla_{X}\bullet (Y_{1}\wedge... \wedge Y_{n})) \wedge (\nabla_{X}\bullet (Y_{1}\wedge... \wedge Y_{n}))>)^{1/2}$      ********************(#3)

Now need to prove (#2)=(#3), by suqare both side and let $ \nabla_{X}$ to act on each individual $ Y_{i}$.... we have on bothside sum of $ n^{2}$ number of determinants of matrix. And trying to prove
these matrix are same by adjust order of terms due to symetry of forms ..., failed.

Did I miss something obvious? I feel this proof should be simple since it is analogous to given proof.

IMHO, here's a more conceptual way to answer the question.

If $ e_{1},\ldots,e_{n}$ is an ON basis of sections over some open set $ U$, then $ e_{1}\wedge\ldots\wedge e_{n}$ is equal to the volume form over $ U$.

In order to show that $ \nabla_{X}v_{g}=0$ at some point $ p$, just choose an ON basis that is parallel at $ p$, and now the result follows from the Leibniz rule.

(It's an exercise in linear algebra to show that your definition of the volume form agrees with $ e_{1}\wedge\ldots\wedge e_{n}$.  You also have to understand why you can choose an ON basis that is parallel at $ p$.)

[quote="yenlee"]IMHO, here's a more conceptual way to answer the question.

If $ e_{1},\ldots,e_{n}$ is an ON basis of sections over some open set $ U$, then $ e_{1}\wedge\ldots\wedge e_{n}$ is equal to the volume form over $ U$.

In order to show that $ \nabla_{X}v_{g}\equal{} 0$ at some point $ p$, just choose an ON basis that is parallel at $ p$, and now the result follows from the Leibniz rule.

(It's an exercise in linear algebra to show that your definition of the volume form agrees with $ e_{1}\wedge\ldots\wedge e_{n}$.  You also have to understand why you can choose an ON basis that is parallel at $ p$.)[/quote]

I think you mean $ e_{1}^{*}\wedge\ldots\wedge e_{n}^{*}$, this is a n-form. The space of n-forms dim=1. Can you tell me what is "parallel" ON basis? parallel with what?

Also, I don't think $ \nabla_{e_{i}}(e_{1}^{*}\wedge\ldots\wedge e_{n}^{*}) \equal{}0$ (covariant dirivative)
Did you confuse with $ d(e_{1}^{*}\wedge\ldots\wedge e_{n}^{*}) \equal{}0$ (exterior dirivative)?

Yes, I should have said ON basis of sections of the *cotangent bundle*, or I should have at least used the correct index notation, i.e. $ e^{1},\ldots,e^{n}$.

Parallel means that the covariant derivative vanishes, which is a very strong condition.  But I only said "parallel at $ p$."  In other words, you want $ (\nabla_{X}e^{i})(p)\equal{}0$ for each $ i$.  This can be achieved by starting with $ e^{1},\ldots,e^{n}$ at $ p$, and then parallel-translating along $ X$.  Then it's an exercise to show that $ e^{1},\ldots,e^{n}$ remain ON at the other points.


[quote="maggot"]What is a harmonic quadrilateral? How is it useful?[/quote]
An inscribed quadrilateral $ABCD$ in which $AB*CD=AC*BD$.It has a lot of other properties,which  worth to remember, such as that the tanghanity lines to the circumscribed circle of $ABCD$ at $A$ and $D$ intersecting at $BC$.Also see the problem Q.4 of the IMO 2003 :lol: .


I keep getting problems that I've already seen before and they all happen to be only 1 point. This has been happening for a while now and I don't know if this is normal.

You have exhausted all the problems in the database. You are now seeing problems which you originally got incorrect.


inequalities
quadratics
function
algebra
inequalities open

If x, y, z are side length of a triangle, then there exist 3 non-negative numbers a, b, c such that $x=b+c, y=c+a, z=a+b$.
Like this, I have been thought about this.
it is following
for non-negative real numbers x, y, z, What condition will be need there? to satisfy
$x=a^{2}+bc$, $y=b^{2}+ca$, $z=c^{2}+ab$ for some non-negative numbers a, b, c.

If $x=a^{2}+bc$, $y=b^{2}+ca$, $z=c^{2}+ab$, I would like to know (x, y, z) will be satisfied with what condition.

This is one of my result -> [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=144942[/url]
equality occurs iff $(a, b, c)=(t, t, t), (t,t,0), (t,t,(\sqrt{2}-1)t)$ and their permutations.
( first I guess this inequality, and try to prove. but my solution is long and ugly, anyone have a nice solution?)

Not much to go on, but this problem seems hard.
for example, (x,x,kx) can be always expressed when k>min((root(65)-1)/8,(root(109)-7)/10)) which is about 0.6. (when k is smaller, it cannot be expressed)
contrary to the triangle, which has some constraints on the lengths of the sides.

Well, I did the same thing
(a=b+c or a=b) I must have made a mistake somewhere.
I was working on deternimant of quadratic equation on b.
The point is, unlike the lengths of a triangle, a^2+bc
cannot easily bounded above by functions of (b^2+ac,c^2+ab)

come to think of it,there are 5 days till the school test starts. What am I doing here?


geometry
function
vector
Pythagorean Theorem
geometry unsolved

Can someone help me out.... It has been 3 years since I have done geometry and there is a small chapter Euclidean geometry in the course I'm doing now. It is not really advanced, but I can't solve it. Can it be done synthetic? analytic?

Well, here is the problem:

"Consider a triangle $ABC$ in $\mathbb{E}^{2}$ and an inscribed circle $\mathcal{C}$. Let $A'$ be the point were $\mathcal{C}$ is tangent to $BC$, $B'$ the point were $\mathcal{C}$ is tangent to $CA$ and $C'$ the point were $\mathcal{C}$ is tangent to $AB$, Show that $AA'$, $BB'$ and $CC'$ are concurrent."

thanks in advance

BA'=BC', AB'=AC', CA'=CB', so

$\frac{AC'}{C'B}\cdot \frac{BA'}{A'C}\cdot \frac{CB'}{B'A}=1$ thus AA', BB' and CC' are concurrent by Ceva.

Edit: What does $E^{2}$ mean? Am I considering the problem too easy?

[quote]
Edit: What does  $\mathbb{E}^{2}$ mean? Am I considering the problem too easy?[/quote]

Thanks alot

No, it is fine. $\mathbb{E}^{2}$ just stands for the Euclidean plane. If you have an 'affine space $\mathbb{A}^{n}$' which is just a space where you can add, subtract n-tuples
$(x_{1},x_{2},...,x_{n})$ on such a way that they follow some axioms (commutativity,associativity,etc...) and you add an extra structure to this space: a distance function, you get an Euclidean space. The concept distance in an Euclidean space is defined by the scalar product. If you take two 'elements', lets say $v$ and $w$, of your space (also called 'vectors', since  the affine space is also a vector space http://en.wikipedia.org/wiki/Vector_space) and apply the scalar product on them you get a scalar (= a real number). Let the function's name be $\langle ., . \rangle$ then $\langle v, w \rangle = k$ with $k \in \mathbb{R}$.

For example, if you take an element of your space, lets say $v=(v_{1},v_{2})$ in $\mathbb{E}^{2}$ and you want to know the distance of this vector relative to the origin $(0,0)$, you can apply the Pythagorean theorem such that  $d(v,0) = \sqrt{(v_{1}-0)^{2}+(v_{2}-0)^{2}}$. Since the scalar product in any Euclidean space for two elements $v=(v_{1},....,v_{n})$ and $w=(w_{1},...,w_{n})$ is defined as $\langle v,w \rangle = v_{1}w_{1}+v_{2}w_{2}+...+v_{n}w_{n}$, you can use this to measure distance, i.e. $d(v,0)= \sqrt{ \langle v, v \rangle }= \parallel v\parallel = \sqrt{v_{1}^{2}+...+v_{n}^{2}}$.

excuses if you already knew this ;)


How many solutions does the equation |x + 2| = 2x have?

2 at most since when you graph them one is a v and the other is a line.
only 2 works
\------/-
-\----/--
--\--/---
---\/----
----/----
--/------

I only got one real solution. (One extraneous)

True, |x+2|=2x would bring about two solutions, one negetive, one positive (all absolute value equations do)

However, only the positive is correct. That is because no matter what ends up within the absolute value symbols, it will turn out positive. So we have a positive number on the left side. But if you multiply a negetive number by 2, it will stay negetive. Now we have a negetive number on the right side. This is impossible, so the negetive solution (-2/3) is extraneous. The only answer is 2.


Let $ x,y\in \mathbb{R^+}$ such that: $ x^m+y^n\le k$ ($ k,m,n$ are constants). Find the maximum value of:
\[ m\sqrt[n]{x}+n\sqrt[m]{y}\]

My brother, will you solve it at least one minute? :D

More general:
Let $ x_1,x_2,...,x_n\in \mathbb{R^\plus{}}$ such that $ x_1^{m_1}\plus{}x_2^{m_2}\plus{}...\plus{}x_n^{m_n}\le k$ and $ m_{n\plus{}1}\equal{}m_1$. Find the maximum value of:
\[ \huge V\equal{}m_1\sqrt[m_2]{x_1}\plus{}m_2\sqrt[m_3]{x_2}\plus{}...\plus{}m_{n}\sqrt[m_{n\plus{}1}]{x_n}\]

Let kill it  :wink:


sallam doostan,

in masalaro negah konin va hall konin.

Two circles $\omega_{1}$ and$\omega_{2}$ touch internally the circle$\omega$  in $M$ and $N$ and the center of$\omega_{2}$ is on $\omega_{1}$. The common chord of the circles$\omega_{1}$  and $\omega_{2}$ intersects $\omega$ in$A$  and$B$ .  $MA$and$MB$  intersects$\omega_{1}$  in$C$  and$D$ . Prove that $\omega_{2}$  is tangent to$CD$ .

sallam azizam.in rahehalle mane:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=19775&highlight=]Here[/url]


A quarter, 2.5cm in diameter, is dropped randomly on the table top shown so that at least half of the coin lies on the table top. What is the probability that the quarter lies on one fo the segments AB, CD, or EF? Express your answer as a common fraction.

Sorry, the picture is demented  ;) The numbers are 3cm, 6cm, 6cm, and 3cm.

That's a really good question. Can you give a hint?

My hint is that the answer to this problem is far off what I thought.  :P
I don't know the solution...

How would you know that the answer is far off what you thought if you don't know the solution?   :|

Umm.  what is the width opf the table, the probability depends on it as well.

Just create a bunch of rectangles of where the center of the coin can be.

[quote="4everwise"]Ok, I think that the answer is 30/41.

First of all, the quarter can land less than 2.5 cm away from either if the three lines in order to be touching it. That yields a 15cm width . Now, divide by the the total width that the quarter may land on, which is 3+6+6+3+2.5. The 2.5 comes from the fact that half the quarter may be outside of the table.

This gives 15/20.5 or 30/41

 ;)[/quote]

That's what I got, but for some reason, it isn't correct.

You don't need to know the width of the table.

Oh yeah.   It thought that it included the lines from the edge of the table

Do you want me to give out the answer, and work from it?  :)

Hres what I think it is:
[hide]Think of it in terms of where the center is.  In order for at least half to be on the table, the center can be from edge to edge, so the width of the table is the total probability, 3+6+6+3=18.  For it to land o one of the lines, the center must be less than or equal to 1.25 away from the line.  Since it can be on each side the width for one line is 1.25*2=2.4.  There are three lines, 2.5*3=7.5.  SO the probability is 7.5/18=15/36=5/12.[/hide]

Correct nat mc!

Now, you used the center of the coin as a refrence point, now why can't you use the entire coin, like 4everwise's solution?

Because the coin can lie off the table partially.


geometry
3D geometry
sphere
integration
calculus
calculus computations

You have a sphere, of radius 6m, filled with water. A hole is drilled in the bottom of the sphere so that the volume of the water in the sphere decreases by 3m^3/min.  Find the rate of change of the height of the water in the sphere when the height is 4m.

Let $V$ be the volume of water, and $h$ be the hight of water at the time $t$.
Set the sphere $x^2+(y-6)^2=6^2$ on $x-y$ plane,

We have $V=\int_0^h \pi\{6^2-(y-6)^2\}\ dy=\frac{4}{3}\pi\cdot 6^3-3t,$ thus

by $\frac{dV}{dt}=\frac{dV}{dh}\cdot \frac{dh}{dt},$

$\pi\{6^2-(y-6)^2\}\frac{dh}{dt}=-3\Longleftrightarrow \frac{dh}{dt}=-\frac{3}{\pi}\{6^2-(h-6)^2\}\ m^3/min$

The desired answer is $\left[\frac{dh}{dt}\right]_{h=4} =-\frac{3}{\pi}\{6^2-(4-6)^2\}=\boxed{-\frac{96}{\pi}}\ m^3/min.$

[hide="solution"]
Well I just found out Kunny beat me to the punch, but I get a slightly different answer.  Kunny did everything right until the $\pi {6^2-(y-6)^2}\frac{dh}{dt}=-3 \Longleftrightarrow \frac{dh}{dt}=-\frac{3}{\pi}(6^2-(h-6)^2)$m^3/min, where he made a slight mistake.  He should have written $\frac{dh}{dt}=-\frac{3}{\pi (6^2-(h-6)^2)}$ m^3/min.  Thus his answer is dimensionally incorrect, so you could have checked it that way if you properly traced the units.
Answer:
$\boxed{-\frac{3}{32 \pi}}$m^3/min

Of course I could have made a mistake too.
[/hide]

FMako, yes, you are right. Thank you for correcting my answer.

kunny


algorithm
logarithms
HCSSiM
pigeonhole principle

Is there an integer n such as decimal expansion of 2^n begins 2007.... ???
I have no idea, any body helps.

I wrote a little program for a brute force attack and the smallest results are.
2^4954=2.0072363783736 * 10^1491
2^7090=2.0075633454119 * 10^2134
2^9226=2.0078903657113 * 10^2777

To prevent working with extremely large numbers, i used the fact that $ 2^n \equal{} 10^{log(2^n)} \equal{} 10^{(n \cdot log2)} \equal{} 10^{fpart(n \cdot log2)} \cdot 10^{ipart(n \cdot log2)}$, where  fpart is the fractional part of the decimal expansion and ipart the integer part. Since an integer power of 10 does nothing but ad zero's at the end. The algorithm only has to look for an integer $ n$ for which $ 10^{fpart(n \cdot log2)} \equal{} 2007...$

There are many more results, see:
http://mats.stijnshome.be/powerof2.php

I believe the solution which is being sought is one which shows that 2^n can start with any sequence

Wouldn't that require proving essentially that any sequence of integers can be found in the decimal expansion of $ \log_{10} 2$?  It certainly may be true, but I'm skeptical that an elementary proof is within reach.  I'd love to see someone prove me wrong though. :)

As for the original question, Mathematica does the trick:

k = 1; While[k > 0, If[Floor[1000*2^k/10^(Floor[Log[10, 2^k]])] == 2007, Print[k]; k = -1]; k++]
4954

Actually I've seen an elementary proof that 2^n starts with every possible string of digits, it was a talk given at HCSSiM two years ago.  I don't remember much about it though, but it was a really nice because I remember going ahh! at various parts. If I can remember the details I'll post it later.

ok now I remember the general argument, its actually not terribly bad:
[hide]
Let b be the string of digits that you want 2^n so we want b*10^k<2^n<(b+1)*10^k where k is larger than the length of b. Now take the log of this and get:
k+log(b)<nlog(2)<k+log(b+1)

Now since log(2) is irrational we know that the fractional parts of log(2), 2log(2), 3log(2),... are dense in (0,1), the result follows.

Of course for full rigor you need to justify alot more and take into account a few details and special cases (like what if b=10^m-1), but the idea is there.
[/hide]

I believe with enough searching you can find that proof (meeting your requirements) in the Olympiad Number Theory fora.

I hope some of you can offer me an elementary solution.This has bothered me for months, I have a headache now because of it.

[quote="JoeBlow"]Nice proof!  Now, can you prove by elementary means the statement

"...since log(2) is irrational we know that the fractional parts of log(2), 2log(2), 3log(2),... are dense in (0,1)..."

That is, without the words 'harmonic analysis' or the names Weyl or Fourier appearing in the proof?  This is, after all, a high school forum  :wink:[/quote]

I dont think thats very hard at all let x be irrational:

It suffices to show that there is a multiple of x with fractional part between k/n and (k+1)/n (for arbitrary integers k>=0 and n>0).

First of all letting k=0 it reduces to a trivial application of the pidgeonhole principle and the fact that x is irrational. So now we have a j such that 0<{jx}<1/n where {a} denotes the fractional part of a.  Now consider {jx}, {2jx},{3jx},... since {jx} > 0 it keeps increasing, and since {jx}<1/n it cannot miss any of the intervals (1/n,2/n), (2/n,3/n),...


inequalities
number theory proposed
number theory

Find, with proof, all positive integers $n$ such that the equation $x^{3}+y^{3}+z^{3}=nx^{2}y^{2}z^{2}$ has a solution in positive integers.

nice!

in my solution, i prove that $n \leq 3$ and see many cases!

the basic idea is use inequalities!  :)

OK. $ n\leq 3 $, but you have to verify that whether solutions exist for those values of $ n $.


Let $ a_1,a_2,...,a_n$ and $ b_1,b_2,...,b_n$ be two finite sequences of numbers. Then
$ a_1b_2\plus{}a_2b_2\plus{}...\plus{}{a_n}{b_n}\equal{}(a_1\minus{}a_2)b_1\plus{}(a_3\minus{}a_2)(b_1\plus{}b_2)\plus{}...
\plus{}(a_{n\minus{}1}\minus{}a_n)(b_1\plus{}b_2\plus{}...\plus{}b_{n\minus{}1})\plus{}a_n(b_1\plus{}b_2\plus{}...\plus{}b_n)$

Abel summation can be proven using a nice general summation concept that was introduced to me by Nathan Pflueger:  [i]to help evaluate a single summation, write it as a double summation and then sum it in a different way.[/i]

This concept can be used to solve a number of Olympiad and Putnam-level summations, although I cannot find any good examples at the moment.

Edit:  See:
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1117941940&t=193921
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1117941940&t=72892
- http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1117941940&t=13664
- http://wangsblog.com/jeffrey/?p=328

Will one see this on USAMO or APMO because in the book i'm reading, the only olympiad example was a very old IMO problem.

Offhand, I'd say you're more likely to find use for Abel summation on the Putnam than on the USAMO.  However, it's a powerful technique to keep in mind; summations such as $ \sum kx^k$ can also be evaluated in this manner.

Edit:  So what I mean is that it is very unlikely that you'll find a problem that can [i]only[/i] be solved with Abel summation, but keeping it in mind can help you find a faster solution than you otherwise might.


[quote="apollo"](Vasile Cirtoaje)(a,b,c >0)
prove that $ \sum_{cyclic}(a\plus{}\frac{1}{b}\minus{}1)(b\plus{}\frac{1}{c}\minus{}1)\geq 3$[/quote]
Let $ x \equal{} a\plus{}\frac{1}{b}\minus{}1,y \equal{} b\plus{}\frac{1}{c}\minus{}1,z \equal{} c\plus{}\frac{1}{a}\minus{}1$
Suppose $ x\ge y\ge z$

We have : $ (x\plus{}1)(y\plus{}1)(z\plus{}1)\ge 5\plus{}x\plus{}y\plus{}z$  so $ xy\plus{}yz\plus{}zx\plus{}xyz\ge 4$

We consider two cases :

Case 1 :$ x > 0,yz\le 0$ we get $ xy\plus{}yz\plus{}zx\ge 4 > 3$

Case 2 . $ x,y,z > 0$ we have by AM-GM, $ \sqrt [3]{(xyz)^{2}}\le xy\plus{}yz\plus{}zx$, so $ t^{3}\plus{}3t^{2}\ge 4( t \equal{} xy\plus{}yz\plus{}zx)\Leftrightarrow (t\minus{}1)(t\plus{}2)^{2}\ge 0$ hence $ t\ge 1$  :)

Oh,I don't think so with you in case 2
Case2 $ x,y,z>0$ by AM-GM we have  $ 3\sqrt [3]{(xyz)^{2}}\leq xy\plus{}yz\plus{}zx$ ($ xy\plus{}yz\plus{}zx\equal{}t$ )
So,$ t^{3}\geq 27(4\minus{}t)^{2}\Leftrightarrow (t\minus{}3)(t\minus{}12)^{2}\geq0\Rightarrow t\geq3$


who can participate in IMC ??

what is the criterion of national team selection?

Everyone and none. You can come as you want, as often as you want.

The competition is 'intended' for people in their first to fourth year of university, but that's not a firm rule.


brilliant drawings  :D  join the club  :D


I need to do this exercise for a programming class and I don't fully understand the algorithim to convert an expression written in infix to postfix.  The textbook is too brief and searching Google, I cannot find a good site that gives a clear process to convert and enough examples.

Here's the problems:

4 a. (A + B) * (C + D) - E =
I came up with:

AB+CD+*E-

right or wrong?

b. A - (B + C) * D + E / F =

hmm dunno.. guessing

ABC+*-DEF/+

c. ((A + B) / (C - D) + E) * F - G =

dunno again.. guessing

AB+CDE-/FG*-

d. A + B * (C + D) - E / F * G + H =

guessing again:
ABCD+*+EF/-GH*/

no clue really... I need a "clear" procedure that tells me how to do this.

thanks

There's a simple recursive algorithm that you can use to convert an expression from infix to postfix. Here's a basic outline:

Base case: If the expression has no operations, remove all parentheses and you have your answer.
1. Find the operator in the expression that is performed last. (First find the operators with the most parentheses. Then if there are ties, go by order of operations (* before +, etc.).
2. Take the part of the expression before the operator and convert it to postfix. Next take the part after the operator and convert it to postfix as well. Your answer is the first part + the second part + the operator.

An example: (A + B) * (C + D) - E
The "-" symbol is the one that goes last, so we now have to solve the expressions on both sides of it. For (A + B) * (C + D), the "*" goes last (note the parentheses), so we now solve (A + B) and (C + D). There is only one operator in each, so our answers are AB+ and CD+. Now we have to solve E (from earlier), and this is clearly already solved, so our overall answer is
AB+CD+*E-.

I think that the easiest way to convert infix to postfix is to draw a syntax tree of the expression. I hope that the following example will be enough to explain what a syntax tree is:

[img]http://www.gamca.sk/~bus/syntax_tree.jpg[/img]
Syntax tree for expression $((A+B)/(C-D)+E)*(F-G)$.

You get the postfix notation from this picture by traversing the tree in the postfix order. (You'll also get the infix notation by traversing it in infix order and prefix notation by traversing it in prefix order.)

Looking at the examples you gave I see another problem to consider though - what to do with expressions where it's not clear which operation is executed first e.g. $A+B+C-D$? I believe that the correct way to handle this is to do the leftmost operation first which means putting brackets around the letters like this: $(((A+B)+C)-D)$. That should produce the following postfix notation: $AB+C+D-$.

thanks a lot for the help.... I understand it now...


algebra
polynomial
geometry proposed
geometry

Find a 3rd degree polynomial whose roots are $r_a$, $r_b$ and $r_c$ where $r_a$ is the radius of the outer inscribed circle of $ABC$ with respect to $A$.

Use $\frac{r_a}{r}=\frac{p}{p-a}$.

easily we'll get that:

$\ \ \ \ \ \ f(x)=x^3-(4R+r)x^2+p^2x-p^2r$


this is the 1000th post on the mathcounts part of the forum. mathcounts ownz you

yeah mc rocks [hide]lolt lolt lolt lolt[/hide]

grrrrrr...you have no morale (with the "e" :) ) as a comoderator, do you.
I mean, you're telling me not to spam and you spam in your own forum! Geez-Loiz! :)

this is not spam, this is showing how popular the mathcounts forum is

*sigh* OK, but does it have to be at the top? I mean, people usually don't like the top of the forum crammed. (like Mr. R, for example)

grrrrr...this is an important moment, but fine.  :evil:

Fine, keep it at the top...*sigh*

Current irritation level: 69%

I unstuck a couple posts at the top - try not to litter (or this forum in general) the top of the site with so much non-math stuff.

Thanks.


n-variable inequality
inequalities
function
algebra
IMO Shortlist

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

Standard Cauchy induction works just fine. We first prove it for two numbers: It takes the form $a,b\ge 1\Rightarrow \frac 1{a^2+1}+\frac 1{b^2+1}\ge \frac 2{ab+1}$. After expanding everything, this reduces to $2ab(ab-1)\le (a^2+b^2)(ab-1)$, which is true by AM-GM $(*)$.

The next step is proving it for $n=2^k$. This is done easily by induction: we add two inequalities with $2^{k-1}$ variables each, and then apply the inequality for two variables to get the result $(**)$.

After this, we prove that $P(n)\Rightarrow P(n-1)$. Assuming the inequality is true for $n$ variables, take $r_n=\sqrt[n-1]{r_1r_2\ldots r_{n-1}}$. We get $\sum_{i=1}^{n-1}\frac 1{r_i+1}+\frac 1{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}\ge \frac n{\sqrt[n-1]{r_1\ldots r_{n-1}}+1}$. This proves the inequality for $n-1$ variables $(***)$.

$(*),(**),(***)$ complete the induction.

[quote]Jensen inequality works even easier[/quote]

How?
would you send your solution I couldn't solve it with jensen

Yes . Jensen for the function $f(x)=\frac{1}{x+1}$  doesnot work.

I have just seen [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=113248]this theorem[/url] and I think I can prove the inequality by showing that:

$\frac{1}{r_{1}+1}+\frac{1}{r_{2}+1}\geq \frac{1}{\sqrt{r_{1}r_{2}}+1}+\frac{1}{\sqrt{r_{1}r_{2}}+1}$

$(r_{1}+r_{2}+2)(\sqrt{r_{1}r_{2}}+1)\geq 2(r_{1}+1)(r_{2}+1)$

$(r_{1}+r_{2}-2\sqrt{r_{1}r_{2}})\sqrt{r_{1}r_{2}}\geq (r_{1}+r_{2}-2\sqrt{r_{1}r_{2}})$

$\sqrt{r_{1}r_{2}}\geq 1$

(True by hypotesis)

Changing the minimum and the maximum element with their GM $\infty$ times all the elements become equal, so I have:

$\sum_{i=1}^{n}\frac{1}{r_{i}+1}\geq \sum_{i=1}^{n}\frac{1}{\sqrt[n]{x_{1}...x_{n}}+1}=\frac{n}{\sqrt[n]{x_{1}...x_{n}}+1}$

Yeah....grobber's solution is just like the A-M G-M induction proof...Great job!!.....Anyway the proof using jensen's inequality is an elegant one and easier to realise.

@sayantanchakraborty basically that's how one can prove Jensen's inequality.

Smoothing also works, note that when $x_1x_2>1$ is fixed, the function $\frac {1}{x_1+1}+\frac {1}{x_2+1}=\frac {x_1+x_2+2}{x_1x_2+x_1+x_2+1}$ decreases as $x_1+x_2$ decrease. This happens when we apply the replacement for $r_i<r'<r_j$($r'=\sqrt[n]{r_{1}r_{2}\cdots r_{n}}$) as $(r_i,r_j)\rightarrow (r',\frac {r_ir_j}{r'})$.

Regular induction also works; it's so easy it feels like cheating.  The $n=2$ case is equivalent to $(r_1r_2-1)(r_1-r_2)^2\ge 0$ which is obvious and has equality $r_1=r_2$.  For $n$ numbers, assuming the $n-1$ case is correct, order them such that $r_n$ is the least, and let $k=r_1r_2\dots r_{n-1}$.  Now by the $n-1$ case on all of them except $r_n$ we would like to show that $\tfrac{n-1}{\sqrt[n-1]{k}+1}+\tfrac{1}{r_n+1}\ge \tfrac{n}{\sqrt[n]{kr_n}+1}$.  This follows since $\tfrac{n-1}{\sqrt[n-1]{k}+1}\ge \tfrac{n-1}{\sqrt[n]{kr_n}+1}$ and $\tfrac{1}{r_n+1}\ge \tfrac{1}{\sqrt[n]{kr_n}+1}$ both reduce to $k\ge r_n^{n-1}$, which is true. :D

  Sorry to revive . I think that there is a mistake here ! Please check it.
This follows since $\tfrac{n-1}{\sqrt[n-1]{k}+1}\ge \tfrac{n-1}{\sqrt[n]{kr_n}+1}$ and $\tfrac{1}{r_n+1}\ge \tfrac{1}{\sqrt[n]{kr_n}+1}$ both reduce to $k\ge r_n^{n-1}$, which is true. :D[/quote]

[quote=Vrangr][quote=sqing]Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For  $\lambda\ge 1$, prove that

\[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \][/quote]

Jensen with $f(x) = \frac1{\lambda e^x + 1}$,
$f''(x) = \frac{\lambda e^x(\lambda e^x-1)}{(\lambda e^x+1)^3} > 0$.[/quote]
Thanks.

[quote=sqing]
Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For  $\lambda\ge 1$, prove that

\[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \][/quote]

Take in the initial inequality $r_i=\lambda x_i$ where $\lambda\ge 1\wedge x_i\ge 1$ for all $i\in\lbrace 1,2,...,n\rbrace$. Actually weaker assumptions: $\lambda>0\wedge \forall_{i\in\lbrace 1,2,...,n\rbrace}\ x_i\ge \frac{1}{\lambda}$ are sufficient. Then you get \[ \frac{1}{\lambda x_{1} + 1} + \frac{1}{\lambda x_{2} + 1} + \cdots +\frac{1}{\lambda x_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{x_{1}x_{2} \cdots x_{n}}+1}. \]

For storage (same smoothing proof as others have pointed out).

[quote=1998 A2]Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that

\[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \][/quote]

Cute :)

We prove the following interim step.

[b]Lemma.[/b] Let $a, b \geqslant 1$ be real numbers. Then $$\frac{1}{a^2+1}+\frac{1}{b^2+1} \geqslant \frac{2}{ab+1}.$$

[i]Proof.[/i] Clearing denominators, we need to show that $(2+a^2+b^2)(1+ab)-2(1+a^2)(1+b^2) \geqslant 0$. Notice that $(1+a^2)(1+b^2)=(1+ab)^2+(a-b)^2$, so the previous expression factors as $(ab-1)(a-b)^2$ and the claim follows. $\blacksquare$

Now we prove the result by induction on $n \geqslant 2$, with base case following from the earlier lemma. Assume the result holds upto all integer less than $n$. For $(r_1, \dots, r_n) \in \mathbb{R}^n$ with $r_i \ge 1$, define $f_n(r_1, \dots, r_n)=\left(\sqrt[n-1]{r_2\dots r_n}, \dots, \sqrt[n-1]{r_1 \dots r_{n-1}}\right)$ and $g(r_1, \dots, r_n)=\sum^n_{i=1} \frac{1}{r_i+1}$.

[b]Claim.[/b] For any such $n$-tuple, we have $g(r_1, \dots, r_n) \geqslant g(f_n(r_1, \dots, r_n))$.

[i]Proof.[/i] By induction hypothesis, we have the $n$ inequalities for $j=1, \dots, n$: $$\sum^n_{i=1, i \ne j} \frac{1}{r_i+1} \ge \frac{(n-1)}{\sqrt[n-1]{r_1\dots r_{j-1}r_{j+1}\dots r_n}+1}$$ and adding them all proves the claim. $\blacksquare$

Thus, for any $k \in \mathbb{N}$ we have $g(r_1, \dots, r_n) \geqslant g(f_n^k(r_1, \dots, r_n))$ (here $f^k$ is the composition $k$ times). Taking limits as $k \rightarrow \infty$, since $\frac{\max(r_1, \dots, r_n)}{\min(r_1, \dots, r_n)}$ converges to $1$, we see that the RHS converges to $g(r, \dots , r)$ where $r^n=r_1\dots r_n$.

Thus, the desired inequality follows. $\blacksquare$

It's easy to see that the inequality works for $r_1 = r_2 = ... = r_n$. Let $C$ is $\sqrt[n]{r_{1}r_{2}...r_{n}}$. Suppose that not all $r_i$ are equal $C$. We will make operations to increase the number of $r_i = C$. So, we can take $r_i < C$ and $r_j > C$. Let's exchange $r_i$ for $kr_i$ and $r_j$ for $\frac{r_j}{k}$ where $k$ is $\frac{C}{r_i}>1$. To prove this problem we only need to prove this inequality:
$$ \frac{1}{r_1 + 1} + \frac{1}{r_2 + 1} \geq \frac{1}{kr_1} + \frac{1}{\frac{r_2}{k} + 1}$$
$$ \frac{r_1+r_2+2}{(r_1+1)(r_2+1)} \geq \frac{kr_1+\frac{r_1}{k}+2}{(kr_1+1)(\frac{r_2}{k}+1)}$$
$$(r_1 + r_2 + 1)(r_1r_2 + kr_1 + \frac{r_2}{k} + 1) \geq (kr_1 + \frac{r_2}{k} + 2)(r_1r_2 + r_1 + r_2 + 1)$$
$$r_1^2r_2 + r_1r_2^2 + kr_1 + \frac{r_2}{k} \geq kr_1^2r_2 + \frac{r_1r_2^2}{k} + r_1 + r_2$$
$$(r_1r_2 - 1)(k - 1)(\frac{r_2}{k} - r_1) \geq 0$$
So, we can get $r_1 = r_2 = ... = r_n = C$, Q.E.D.

We will prove this by induction on $n$. First, we prove the base case $n=2$. Let $r_1=x_1^2, r_2=x_2^2$. Then the inequality we must prove is $$\frac{1}{x_1^2+1}+\frac{1}{x_2^2+1}\ge \frac{2}{x_1x_2+1}.$$ Multiplying everything by $(x_1^2+1)(x_2^2+1)(x_1x_2+1)$ gives us we must prove $$(x_1x_2+1)((x_1^2+1)+(x_2^2+1))\ge 2(x_1^2+1)(x_2^2+1).$$ Subtracting the RHS from the LHS and factoring gives $$(x_1x_2-1)(x_1-x_2)^2$$ which is obviously always nonnegative.

Now we use smoothing.
[b][color=#f00]Claim:[/color][/b] Let $\epsilon$ be any real number greater than or equal to $1$, and let $x,y$ be greater than or equal to $1$ such that $y\ge x \epsilon^2$. Then $$\frac{1}{x+1}+\frac{1}{y+1}\ge \frac{1}{x\epsilon+1}+\frac{1}{\frac{y}{\epsilon}+1}.$$
[i]Proof.[/i] Rearranging and combining fractions, we want to prove $$\frac{1}{x+1}-\frac{1}{x\epsilon+1}\ge \frac{1}{\frac{y}{\epsilon}+1}-\frac{1}{y+1} \Rightarrow \frac{x\epsilon -x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon}-y}{(\frac{y}{\epsilon}+1)(y+1)}.$$ Cancelling a factor of $\epsilon-1$ on the numerator gives $$\frac{x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon}}{(\frac{y}{\epsilon}+1)(y+1)}.$$ However, notice that the derivative of $\frac{x}{(x+1)(x\epsilon+1)}$ is equal to $-\frac{\epsilon x^2-1}{(x+1)^2(\epsilon x+1)^2}\le 0$ (for $x\ge 1$), so the LHS is nonincreasing. As $x\le \frac{y}{\epsilon^2}$, $$\frac{x}{(x+1)(x\epsilon+1)}\ge \frac{\frac{y}{\epsilon^2}}{(\frac{y}{\epsilon^2}+1)(\frac{y}{\epsilon}+1)}$$ so we simply need to prove that $$\frac{\frac{y}{\epsilon^2}}{(\frac{y}{\epsilon^2}+1)(\frac{y}{\epsilon}+1)} \ge \frac{\frac{y}{\epsilon}}{(\frac{y}{\epsilon}+1)(y+1)}\Rightarrow \frac{1}{\frac{y}{\epsilon}+\epsilon}\ge \frac{1}{y+1}\Rightarrow \frac{1}{\epsilon}(\epsilon-1)(y-\epsilon)\ge 0,$$ which is clearly true. $\blacksquare$

Returning to the problem, suppose $n=k$, let $G=\sqrt[k]{r_1r_2\ldots r_k}$ and WLOG $r_1\le G\le r_2$. Then apply the claim with $x=r_1, y=r_2, \epsilon=\text{min}\left(\frac{G}{r_1}, \frac{r_2}{G}\right)$. Then notice that one of $r_1,r_2$ has become equal to $G$, and the other has been turned into $\frac{r_1r_2}{G}$ (as the product of the numbers stays the same). Thus we need to prove $$\frac{1}{G+1}+\frac{1}{\frac{r_1r_2}{G}+1}+\frac{1}{r_3+1}\ldots +\frac{1}{r_k+1}\ge \frac{k}{G+1}\Rightarrow \frac{1}{\frac{r_1r_2}{G}+1}+\frac{1}{r_3+1}\ldots +\frac{1}{r_k+1}\ge \frac{k-1}{G+1}.$$ However, as this is a case of the inequality with $n=k-1$, this directly follows from the induction hypothesis, and we are done.

Here's my awful solution.

Firstly WLOG $r_n\geq  r_{n-1}\geq \ldots\geq r_2\geq r_1$.
We use induction.
We claim that for any $1\leq i\leq n$,
$$\frac{1}{r_i+1}+ \frac{i-1}{ \sqrt[i-1]{r_{1}r_{2} \cdots r_{i-1}}+1}\geq \frac{i}{ \sqrt[i]{r_{1}r_{2} \cdots r_{i}}+1}.$$
Let $r_i=x^i$ and $\sqrt[i-1]{r_{1}r_{2} \cdots r_{i-1}}=y^i$. Hence, we need
\begin{align*}
\frac{1}{x^i+1}+ \frac{i-1}{ y^i+1}&\geq \frac{i}{xy^{i-1}+1}\Longleftrightarrow \\
((i-1)x^i+y^i+i)(xy^{i-1}+1)&\geq i(x^i+1)(y^i+1)\Longleftrightarrow \\
xy^{2i-1}+ixy^{i-1}+(i-1)x^{i+1}y^{i-1}&\geq ix^iy^i+x^i+(i-1)y^i\Longleftrightarrow \\
y^{i}+i+(i-1)x^{i}&\geq ix^{i-1}y+\frac{x^i+(i-1)y^i}{xy^{i-1}}.
\end{align*}
Now notice that \begin{align*}x^{i}+i+(i-1)y^{i}&\geq ixy^{i-1}+\frac{x^i+(i-1)y^i}{xy^{i-1}} \Longleftrightarrow\\  (x^{i}+(i-1)y^{i})\left(1-\frac{1}{xy^{i-1}}\right)+i(1-xy^{y-1})&\geq 0  \Longleftrightarrow\\
(xy^{y-1}-1) (\frac{x^{i}+(i-1)y^{i}}{xy^{i-1}}-i)&\geq 0,
\end{align*}which holds by AM-GM and as $x,y\geq 1$.

Now, I show that always
\begin{align*}
y^{i}+i+(i-1)x^{i}- ix^{i-1}y-\frac{x^i+(i-1)y^i}{xy^{i-1}}&\geq x^{i}+i+(i-1)y^{i}- ixy^{i-1}-\frac{x^i+(i-1)y^i}{xy^{i-1}}\Longleftrightarrow\\
y^{i}+(i-1)x^{i}- ix^{i-1}y&\geq x^{i}+(i-1)y^{i}- ixy^{i-1}\Longleftrightarrow\\
(i-2)(x^{i}-y^{i})&\geq ixy(x^{i-2}-y^{i-2}).
\end{align*}
If $i=1$ and $i=2$, then the inequality above is obvious. Now for $i\geq 3$,
\begin{align*}(i-2)(x^{i}-y^{i})\geq ixy(x^{i-2}-y^{i-2})\Longleftrightarrow\\
(x-y)((i-2)(x^{i-1}+x^{i-2}y+\ldots+xy^{i-2}+y^{i-1})-i(x^{i-2}y+x^{i-3}y^2+\ldots+x^2y^{i-3}+xy^{i-2}))\geq 0\Longleftrightarrow\\
(x-y)((i-2)(x^{i-1}+y^{i-1})-2(x^{i-2}y+x^{i-3}y^2+\ldots+x^2y^{i-3}+xy^{i-2}))\geq 0,
\end{align*}which is true by separate Muirhead's and as $x^i=r_i\geq \sqrt[i-1]{r_1r_2\ldots r_{i-1}}=y^i$.

Hence, we build up our desired inequality. Note that the equality holds iff $r_1=r_2=\ldots=r_n$.

Let $a_i=\ln r_i\ge0$ and $f(x)=\frac1{1+e^x}$, then we want to prove that:
$$\sum_{i=1}^nf(a_i)\ge nf\left(\frac1n\sum_{i=1}^na_i\right),$$
which is trivial by Jensen since $f''(x)=\frac{e^x\left(e^x-1\right)}{\left(1+e^x\right)^3}\ge0$.

Let function $f(x)=\tfrac{1}{e^x+1}$. We have $\tfrac{1}{r_1+1}=f(\ln(r_1)).$ Since $r_1\ge 1$, $\ln(r_1)$ is nonnegative. When $x\ge 0$, we have
\begin{align*}
f'(x) &= \frac{-e^x}{(e^x+1)^2} \\
f''(x) &= \frac{a(a-1)}{(a+1)^3}
\end{align*}
where $a=e^x$. Since $x\ge 0$, we have $a\ge 1$, so $f(x)$ is convex. Therefore, $f(\ln(r_1))+f(\ln(r_2))+\dots+f(\ln(r_n))\ge nf(\ln(\sqrt[n]{r_1r_2r_3\dots r_n}))$ which is the desired inequality.

Let $r_i=e^{a_i}$. Note that the condition says that all $a_i$ are nonnegative. Then define $f(x)=\frac{1}{e^x+1}$ and note that it's enough to show that $f(x)$ is convex on the interval $[0, \infty)$ and we will be done by Jensen's Inequality. Note that $$f''(x)=e^x(2(e^x+1)^{-3}e^x-(e^x+1)^{-2})\ge 0$$
so we are done. $\blacksquare$

Define the $n$-variable polynomial $$P(r_1, r_2, \cdots, r_n) = \sum_{i=1}^n \frac 1{r_i+1} - \frac n{\sqrt[n]{r_1r_2\cdots r_n} + 1}.$$ Suppose there existed some $r_i \neq r_j$, for instance $r_1$ and $r_2$ without loss of generality. I claim that $$P(r_1, r_2, \cdots, r_n) \geq P(\sqrt{r_1r_2}, \sqrt{r_1r_2}, \cdots, r_n).$$ This is not difficult to prove because the second term and most of the first summation remains constant, and $$\frac 1{r_1+1} + \frac 1{r_2+1} = 1-\frac{r_1r_2-1}{r_1r_2+r_1+r_2+1} \geq 1-\frac{r_1r_2-1}{r_1r_2+2\sqrt{r_1r_2} + 1}$$ by AM-GM and because $r_1r_2 \geq 1$, which is precisely what we wanted. Observe that $\sqrt{r_1r_2} \geq 1$ as well.

The operation can be applied as long as the $r_i$ are not all equal. As a result, $$P(r_1, r_2, \cdots, r_n) \geq P(\sqrt[n]{r_1r_2\cdots r_n}, \sqrt[n]{r_1r_2\cdots r_n}, \cdots, \sqrt[n]{r_1r_2\cdots r_n}) = 0,$$ so we are done.

We use Cauchy induction. We first prove the inequality for $n=2$. Let $r_1=a^2$ and $r_2=b^2$ so that $a,b \ge 1$. We have the inequality is $$\frac{1}{a^2+1}+\frac{1}{b^2+1} \ge \frac{2}{ab+1} \iff (ab+1)(a^2+b^2+2) \ge 2 (a^2+1)(b^2+1) \iff a^3b+ab^3-2a^2b^2 \ge a^2+b^2-2ab \iff (ab-1)(a-b)^2 \ge 0,$$ which is true so we are done for $n=2$. We use standard induction to prove the inequality for $n=2^k$. The base case was proven above. Assume it is true for $n=2^k$. Then we have $$\frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} + \frac{1}{r_{n+1} + 1} + \frac{1}{r_{n+2} + 1} + \cdots +\frac{1}{r_{2n}+1} \ge \frac{n}{\sqrt[n]{r_1r_2 \cdots r_n} +1} + \frac{n}{\sqrt[n]{r_{n+1}r_{n+2} \cdots r_{2n}} +1} \ge \frac{2n}{\sqrt[2n]{r_1r_2 \cdots r_{2n}} +1},$$ so we have finished the inequality. To finish the Cauchy induction, we will prove that the inequality being true for $n$ means that it is true for $n-1$. Taking $r_n=\sqrt[n-1]{r_1r_2 \cdots r_{n-1}}$, we have $$ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n-1}+1}+\frac{1}{\sqrt[n-1]{r_1r_2 \cdots r_{n-1}}+1} \ge \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n-1} \sqrt[n-1]{r_1r_2 \cdots r_{n-1}}}+1} \iff \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n-1}+1} \ge \frac{n-1}{\sqrt[n-1]{r_1r_2 \cdots r_{n-1}} +1},$$ which finishes the Cauchy induction so we are done.

Let $a_i=\ln r_i$ so that all $a_i$ are nonnegative, and let $$s=\frac{a_1+a_2\cdots + a_n}{n}.$$ Then, we want to show $$\sum_{i=1}^n\frac{1}{e^{a_i}+1}\geq \frac{n}{e^s+1}.$$

Claim: $$f(x)=\frac{1}{e^x+1}$$ is convex for $x\geq0$. This is because $$f''(x)=2(e^x+1)^{-3}e^{2x}-(e^x+1)^{-2}e^x=e^x\cdot \frac{e^x -1}{(e^x+1)^3},$$ which is nonnegative when $x\geq0$.

Therefore, by Jensen's, we are done.

[quote=OGGY_666]By Titu Inequality
we have: LHS[u]>[/u]
n^2/(n+r[size=55]1[/size]+...r[size=55]n[size=100])[/size].
.[size=100]which is By applying AM-GM in the denominator[u]>[/u]
  n^2/(n(r[size=55]1...[size=100]r[/size][size=55]n[/size][/size])^(1/n)+n)
which is equivalent to:
n/((r[size=55]1...[size=100]r[/size][size=55]n[/size][/size])^(1/n)+1)=RHS.
Hence Proved.[/quote]

I almost had this solution initially, but it is a fakesolve since you flipped a sign (because you are AM-GMing the denominator so its actually <= in that step)

Fix the product. We claim the left hand side is minimized when all the variables are equal. Assume not, then there are two variables $x\neq y$. It remains to prove
\[
\frac{1}{x+1} + \frac{1}{y+1} \geq \frac{2}{\sqrt{xy}+1}.
\]
Expanding, it is equivalent to proving $(x+y+2)\sqrt{xy} \geq 2xy+x+y$, which can be factored as $(\sqrt{xy}-1)(\sqrt{x} - \sqrt{y})^2\geq 0$, which is obviously true. Then it remains to check the case when the variables are equal, which is trivial.

By Jensen's this is just asking us to prove that $f(x)=\frac{1}{e^x+1}$ is convex on $[0,\infty)$, which follows since its second derivative is $\frac{e^x(e^{2x}-1)}{(e^x+1)^4}$.

[quote=sqing][quote=orl]Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. Prove that

\[ \frac{1}{r_{1} + 1} + \frac{1}{r_{2} + 1} + \cdots +\frac{1}{r_{n}+1} \geq \frac{n}{ \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \][/quote]
Let $r_{1},r_{2},\ldots ,r_{n}$ be real numbers greater than or equal to 1. For  $\lambda\ge 1$, prove that

\[ \frac{1}{\lambda r_{1} + 1} + \frac{1}{\lambda r_{2} + 1} + \cdots +\frac{1}{\lambda r_{n}+1} \geq \frac{n}{\lambda \sqrt[n]{r_{1}r_{2} \cdots r_{n}}+1}. \][/quote]

Substituting $r_1=\dfrac{p_1}{\lambda}$, we get that this isn't really a generalization because one can applicate the original inequality to $p_i$'s.
We get
$$\dfrac{2^k}{\sqrt[n]{p_1p_2\cdots p_n}+1}=\dfrac{2^k}{\lambda \sqrt{r_1r_2\cdots r_n}+1}$$
While generalizing inequalities, these transformations should have been tried to check whether this is a real generalization or a fake one trivial by original problem.


I have studied the AoPS I on my own since Jan.  I've done up to chapter 11.  Should I take  Intro IS A or B?

If you feel you have completely understood all the problems in those 11 chapters, then you will be ok enrolling in B (though we will suggest some reading and problems to fill the small gap between those chapters and Intro B).

You can always switch after the first couple weeks if you find the work too easy or too challenging.

Is the IS B harder than A? Also, does this apply to the Olympiad IS?

Yes, the B's are harder than the A's.  There's not yet an olympiad B course.  When there is one, it will likely be closer to Olympiad A than Intermediate B is to Intermediate A.


geometry
articles
Ross Mathematics Program

I think I should have posted this in another forum but since this college forum has more mature and experienced people than any other forum in here, I need some suggestions in studying my new studying book "Geometry Revisited." I'm very surprised to this book has something I've never heard and I don't know where to start.  Any suggestions?

Geometry Revisited... it says under the resources page of this web site that it is recommended for pros.  That doesn't seem to be your level.  You probably should have tried a different book.  Btw,  your right this should be in a different forum probably 'other problem solving topics' or something.  Most "mature ppl" on this site look at multiple forums so it doesn't really matter which one you post it in because they will probably see it.  Just some friendly advice.

You might try [u]Challenging Problems in Geometry[/u], by Posamentier and Salkind, first.  Also look in the Resources section of this website, as joml mentioned.

joml88 is correct about which forum is the most appropriate for that kind of question, and, no, you shouldn't revisit geometry until you have visited it the first time. What have you, Silverfalcon, used to learn geometry so far? Can you please post the exact titles of any other geometry books you have used?

(After edit:) Because this was first posted in a forum where I have moderator powers, I will move it to the correct forum (Other Problem Solving Topics) myself.

Having studied most of Geometry Revisited, I think I'll give my 2 cents. Geometry Revisited is a wonderful exposition into many different kinds of geometries, and even deeper into classical euclidian. But be warned. It is probably not a good first source for much of the material in there. It introduces topics in a very non-standard way, which can be difficult to understand if you have not had any previous exposure to that field of study. For example, the chapter on inversion is very poorly written. It is very difficult to understand, alot of it is simply useless, and much of it barely relates to inversion at all! Furthermore, it gives very FEW problems to go along with the MANY concepts and theorems it offers.

Don't get me wrong. I'm not trying to badmouth Geometry Revisited. It IS a great book, and I learned alot from it. But I had to be very selective about what I chose to truly study and what I merely skimmed through. You've been warned. Good luck!

[quote="JSRosen3"]zabelman, what geometry books would you recommend for an intermediate (and willing to try some harder stuff) problem sovler in lieu of "Geometry Revisited"? I've heard good things about some Dover books; do you know any specific ones that are particularly good? Thanks.[/quote]

The only BOOKS that I have for geometry are my 8th grade geometry class textbook (which is horrid) and geometry revisited. I learned most of my geometry from working alot of problems, reading a bunch of stuff online, and mainly [i]taking the olympiad geometry class[/i]. I highly recommend it.

[quote="JSRosen3"]zabelman, what geometry books would you recommend for an intermediate (and willing to try some harder stuff) problem sovler in lieu of "Geometry Revisited"?[/quote]

I'm not zabelman, who has already answered you, but I will second JBL's recommendation of Challenging Problems in Geometry, which is an inexpensive Dover reprint. I think Excursions in Geometry, by Ogilvy, another Dover reprint, is also more accessible than Geometry Revisited. I own all three of the books mentioned in this paragraph. My son is taking EPGY's geometry course (VERY GOOD) right now, and I will suggest to him that he work through those books in the order Excursions, then Challenging Problems, then Revisited (possibly starting each next book before finishing the previous book). And of course there are plenty of good geometry problems and a good bit of geometry explanation in the two AoPS volumes, which problem-solvers of Silverfalcon's age should definitely work through.

Hope this helps!

I bought Geometry Revisted a while ago and I think it's a pretty good geometry book. Although I lack much formal education in geometry, I can recognize several of the concepts they teach. I will say that the problems are pretty difficult (for me at least) and the solutions require a lot of thinking to understand (they make assumptions not evident at first sight). I just move through it slowly, trying to gather everything together into a whole. So I would get this book, but familiarize yourself with the concepts before just diving in (I look online a lot for definitions/theorems that they use).

You might try [url=http://www.amazon.com/exec/obidos/tg/detail/-/0471504580/qid=1087617968/sr=8-3/ref=sr_8_xs_ap_i3_xgl14/102-6400132-3721748?v=glance&s=books&n=507846]this other Coxeter book[/url] instead. I find it a good one.

[quote="tokenadult"]I'm not zabelman, who has already answered you, but I will second JBL's recommendation of Challenging Problems in Geometry, which is an inexpensive Dover reprint. I think Excursions in Geometry, by Ogilvy, another Dover reprint, is also more accessible than Geometry Revisited. I own all three of the books mentioned in this paragraph. My son is taking EPGY's geometry course (VERY GOOD) right now, and I will suggest to him that he work through those books in the order Excursions, then Challenging Problems, then Revisited (possibly starting each next book before finishing the previous book). And of course there are plenty of good geometry problems and a good bit of geometry explanation in the two AoPS volumes, which problem-solvers of Silverfalcon's age should definitely work through.

Hope this helps![/quote]

I have a feeling your son will be the next 4-year IMO gold medalist with the sort of preparation that he is getting right now.

[quote="mathnovice"]I have a feeling your son will be the next 4-year IMO gold medalist with the sort of preparation that he is getting right now.[/quote]

The crucial difference between my son and any previous multi-year IMO gold medalist I have heard of is that my son doesn't have a parent with higher education in mathematics. I have to SCRAMBLE in my self-education just to know enough to know who a good mentor might be, or what a good summer program might be. (Thanks, by the way, to certain unnamed forum participants here who PMed me about various summer programs.) My last math class was a precalculus class (it was called "analysis") I took in high school in the 1970s, and I didn't do well in that class. I have some aptitude, if my SAT math score is to be believed, but very little background. My wife had reasonable math background (about the same as mine, but probably with better instruction and certainly more practice) in Taiwan, but her passion is musical performance so neither of us can model doing mathematics at a level higher than the level my son is already at.

The acid test will be next year: I've shown my son the [url=http://www.cebiz.org/cds/ericsson_charness.pdf]Ericsson and Charness article[/url], which he hasn't read page-by-page yet, and after his two summer programs he will have to decide for himself if he has enough fire in the belly to put in consistent daily practice to get better the way that many of the AoPS participants are. Bibliography I know cold: I can always find my children good books, but I have to manage our schedules better to give each child, and especially each older child, enough peace and quiet to practice thoroughly. When I hear about a conspicuously early precocious achiever I always want to know how many younger siblings that person has. Only each learner can do the work, and all I can try to do is provide peace and quiet, and possibly a negative example: "Don't do what I did as a teenager, but what those other guys are doing."

Thanks for all the recommendations, everyone! I'm going to get some of them off of Amazon today.

What exactly is in the other Coxeter book (Introduction to Geometry, 2nd Edition) ComplexZeta mentioned? There isn't a lot of information at amazon. Is it like Geometry Revisited but a little easier and with more topics? I don't think I'll buy it (for now, as I'm planning on working through Challenging Problems in Geometry and some other stuff first), but I'm rather curious as I've seen it before.

It has the Geometry Revisited stuff and the lower-level geometry stuff as well, and then it has some other stuff.

Since we're talking about geometry books, do any of you know what Honsberger's "Episodes in 19th & 20th Century Euclidean Geometry" is like? Is it any good?

[quote="ThAzN1"]Since we're talking about geometry books, do any of you know what Honsberger's "Episodes in 19th & 20th Century Euclidean Geometry" is like?[/quote]

I own it. It's all right. If you like the style of most of Ross Honsberger's other books I think you would like this one fairly well. There is some topic similarity with the books already mentioned in this thread.


AIME 1993 #12

The vertices of $\triangle ABC$ are $A = (0,0)$, $B = (0,420)$, and $C = (560,0)$. The six faces of a die are labeled with two $A\'s$, two $B\'s$, and two $C\'s$. Point $P_1 = (k,m)$ is chosen in the interior of $\triangle ABC$, and points $P_2,P_3,P_4,\ldots$ are generated by rolling the die repeatedly and applying the rule: If the die shows label $L$, where $L \in \{A,B,C\}$, and $P_n$ is the most recently obtained point, then $P_{n+1}$ is the midpoint of $\overline{P_nL}$. Given that $P_7 = (14,92)$, what is $k + m$ ?

[hide="Cats, Dogs, and ..... Cows??!"]Let $P_b=(x,y)$. Then $P_{b+1}=$ either:
$=(\frac{x}{2},\frac{y}{2})$, (Transform Meow)
$=(\frac{x}{2}+280, \frac{y}{2} )$, (Transform Woof)
$=(\frac{x}{2}, \frac{y}{2}+210)$. (Transform Moo)

Now we will use an algorithmic approach.
If: $P_{b+1}=(\text{x-coordinate less than }280 ,\text{y-coordinate less than } 210)$, we go from $P_b$ to $P_{b+1}$using the inverse of Trasform Meow.
If: $P_{b+1}=(\text{x-coordinate greater than }280 ,\text{y-coordinate less than } 210)$, we go from $P_b$ to $P_{b+1}$ using the inverse of Trasform Woof.
If: $P_{b+1}=(\text{x-coordinate less than }280 ,\text{y-coordinate greater than } 210)$, we go from $P_b$ to $P_{b+1}$ using the inverse of Trasform Moo.
If: $P_{b+1}=(\text{x-coordinate greater than }280 ,\text{y-coordinate greater than } 210)$, then we've clearly made a mistake, since $P_b$ will be outside the triangle formed by the three points $A,B,C$ in any transform (try each transform if you don't believe me).

Working backwards,
$P_7=(14,92)\rightarrow P_6=(28,184)$ (from Transform Meow).
$P_6=(28,184)\rightarrow P_5=(56,368)$ (from Transform Meow).
$P_5=(56,368)\rightarrow P_4=(112,316)$ (from Transform Moo).
$P_4=(112,316)\rightarrow P_3=(224,212)$ (from Transform Moo).
$P_3=(224,212)\rightarrow P_2=(448,4)$ (from Transform Woof).
$P_2=(448,4)\rightarrow P_1=(336,8)$ (from Transform Woof).

So $(m,n)=(336,8) \Rightarrow \boxed{m+n=344}$[/hide]


A 'guest' (i.e. an AoPS user who was not logged in at the time) posted in a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20787&start=0&postdays=0&postorder=asc&highlight=]thread[/url] that is under my moderatorship, then logged in and posted after it 'whoops, that was me'.  I tried to fix it by replacing the 'whoops' message with the guest's message, then deleting the guest's message, but I can't delete the guest's message.  When I go to edit this guest message and click 'delete this post', a notice says 'you must enter a message when posting'.  For the record, putting in a random message (e.g. 'a') doesn't make it work either.

So the problem is, I can't edit or delete a guest's posts even though I can do it to a normal user's posts.

Also, shouldn't guests not be allowed to post anyway?

Solved the guest editing bug :)
Guest can now only post in the Test Forum, just to test out LaTeX stuff.


analytic geometry
geometry proposed
geometry

Circles $ S_1$ and $ S_2$ intersect at points $ A$ and $ B$, with the center $ O$ of $ S_1$ lying on $ S_2$. A chord $ AC$ of $ S_1$ intersects $ S_2$ again at $ D$. Prove that the segments $ OD$ and $ BC$ are perpendicular.

$ \angle BOD \equal{} \angle BAD \equal{} \frac{1}{2} \angle BOC$ and the result follows since the triangle $ OBC$ is isosceles.

Straightforward solution with coordinates, too.


function algebra polynomial quadratics trigonometry

The expression x^5 + y^5 can be written as the product of x + y and another factor. Find that
other factor and show your work.

Using $ a^n \plus{} b^n \equal{} (a \plus{} b)(a^{n \minus{} 1} \minus{} ba^{n \minus{} 2} \plus{} ... \minus{} ab^{n \minus{} 2} \plus{} b^{n \minus{} 1})$ where $ n$ is odd!
$ x^5 \plus{} y^5 \equal{} (x \plus{} y)(x^4 \minus{} x^3y \plus{} x^2y^2 \minus{} xy^3 \plus{} y^4)$

An unrelated question.  Can this be factored further?

Agh... I was gonna put this on my mock test...

No, it can't be factored over the integers.

What is its factorization over the reals, then?  :)

[hide]
If $ x^4 \minus{} x^3y \plus{} x^2y^2 \minus{} xy^3 \plus{} y^4 \equal{} 0$, we can divide both sides by $ x^2y^2$ to obtain

$ \frac {x^2}{y^2} \minus{} \frac xy \plus{} 1 \minus{} \frac yx \plus{} \frac {y^2}{x^2} \equal{} 0$.

Letting $ z \equal{} \frac xy \plus{} \frac yx$, we find that

$ z^2 \minus{} 2 \equal{} \frac {x^2}{y^2} \plus{} \frac {y^2}{x^2}$, so our equation is equivalent to

$ z^2 \minus{} z \minus{} 1 \equal{} 0$, which has solutions

$ z \equal{} \frac {1\pm\sqrt5}2$.
[/hide]

This gives us the two equations

$ \frac xy \plus{} \frac yx \equal{} \frac {1 \plus{} \sqrt5}2\implies x^2 \minus{} xy\left(\frac {1 \plus{} \sqrt5}2\right) \plus{} y^2 \equal{} 0$

and

$ \frac xy \plus{} \frac yx \equal{} \frac {1 \minus{} \sqrt5}2\implies x^2 \minus{} xy\left(\frac {1 \minus{} \sqrt5}2\right) \plus{} y^2 \equal{} 0$.

These are not factorable over the reals, so our answer is

$ x^5 \plus{} y^5 \equal{} \boxed{(x \plus{} y)\left(x^2 \minus{} xy\left(\frac {1 \plus{} \sqrt5}2\right) \plus{} y^2\right)\left(x^2 \minus{} xy\left(\frac {1 \minus{} \sqrt5}2\right) \plus{} y^2\right)}$.

Haha...I just realized how long that took to type. :P

Correct :)  What does this tell you about the value of trigonometric functions of $ \frac{\pi}{5}$?

ernie:  if you want the answer to the yes/no question alone, one way to state the Fundamental Theorem of Algebra is that every polynomial can be factored into quadratic or linear factors over the reals, so we should expect that $ x^5 \plus{} y^5$ factors, at worst, into a linear and two quadratic factors, which is indeed the case.

All of the factors are of the form $ (x\plus{}ya)$, where $ a$ is a fifth root of unity (i.e. $ \operatorname{cis}\left(\frac{2k\pi}5\right)$), giving us

$ \cos\left(\frac{2\pi}5\right)\plus{}\cos\left(\frac{8\pi}5\right)\equal{}\phi\minus{}1$ and

$ \cos\left(\frac{4\pi}5\right)\plus{}\cos\left(\frac{6\pi}5\right)\equal{}\minus{}\phi$.

I'm not really sure what else though...

Right.  Now recall that $ \cos x \equal{} \cos (2\pi \minus{} x)$...

Um, I think you mean $ \cos x \equal{} \cos (2\pi \minus{} x)$ or $ \cos x\equal{}\minus{}\cos(\pi\minus{}x)$...

Anyhow, that means $ \cos\left(\frac {8\pi}5\right) \equal{} \cos\left(\frac {2\pi}5\right)$ and $ \cos\left(\frac {4\pi}5\right) \equal{} \cos\left(\frac {6\pi}5\right)$, so

$ \cos\left(\frac {2\pi}5\right) \equal{} \frac {\phi \minus{} 1}2 \equal{} \frac { \minus{} 1 \plus{} \sqrt5}4$ and

$ \cos\left(\frac {4\pi}5\right) \equal{} \minus{} \frac\phi2 \equal{} \frac { \minus{} 1 \minus{} \sqrt5}4$.


We can expand the right part to get:
$ 2(a \plus{} b \plus{} c)\left(\frac a{b^2 \plus{} c^2} \plus{} \frac b{c^2 \plus{} a^2} \plus{} \frac c{a^2 \plus{} b^2}\right)\equal{}2\left[\sum \frac{a^2}{b^2\plus{}c^2}\plus{}\sum \frac{a(b\plus{}c)}{b^2\plus{}c^2}\right]$
and the left part as:
$ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\equal{}\frac{a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}\plus{}2\left(\frac b{a} \plus{} \frac c{b} \plus{} \frac a{c}\right)$

Now, use the fact, $ \sum \frac{a}{c} \geq \sum \frac{2ab}{b^2\plus{}c^2}$ and $ \sum \frac{a}{b} \geq \sum \frac{2ac}{b^2\plus{}c^2}$
It remains to prove that $ \sum \frac{a^2}{b^2}\plus{}\sum \frac{b}{a} \geq \sum \frac{a}{b}\plus{}\sum \frac{a^2}{b^2\plus{}c^2}$.
 :)

[quote="Sunkern_sunflora"]We can expand the right part to get:
$ 2(a \plus{} b \plus{} c)\left(\frac a{b^2 \plus{} c^2} \plus{} \frac b{c^2 \plus{} a^2} \plus{} \frac c{a^2 \plus{} b^2}\right) \equal{} 2\left[\sum \frac {a^2}{b^2 \plus{} c^2} \plus{} \sum \frac {a(b \plus{} c)}{b^2 \plus{} c^2}\right]$
and the left part as:
$ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2 \equal{} \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \plus{} 2\left(\frac b{a} \plus{} \frac c{b} \plus{} \frac a{c}\right)$

Now, use the fact, $ \sum \frac {a}{c} \geq \sum \frac {2ab}{b^2 \plus{} c^2}$ and $ \sum \frac {a}{b} \geq \sum \frac {2ac}{b^2 \plus{} c^2}$
It remains to prove that $ \sum \frac {a^2}{b^2} \plus{} \sum \frac {b}{a} \geq \sum \frac {a}{b} \plus{} \sum \frac {a^2}{b^2 \plus{} c^2}$.
 :)[/quote]
I think need prove
 $ \sum \frac {a^2}{b^2} \plus{} \sum \frac {b}{a} \geq \sum \frac {a}{b} \plus{} 2\sum \frac {a^2}{b^2 \plus{} c^2}$.

But it 's wrong . Try $ a\equal{}9999;b\equal{}3;c\equal{}2$ :wink:

The inequality:

       \[ (\frac a{b} + \frac b{c} + \frac c{a})^2\ge 2(a + b + c)(\frac a{b^2 + c^2}
+ \frac b{c^2 + a^2} + \frac c{a^2 + b^2} )\]

 does not hold( a counter-example:a=100,b=5,c=1)!

         And  we have

\[ (\frac a{b} + \frac b{c} + \frac c{a})^2\ge 2(a + b + c)(\frac a{a^2 + b^2}
+ \frac b{b^2 + c^2} + \frac c{c^2 + a^2} ).\]

[quote="fjwxcsl"]
         And  we have
\[ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\ge 2(a \plus{} b \plus{} c)(\frac a{a^2 \plus{} b^2} \plus{} \frac b{b^2 \plus{} c^2} \plus{} \frac c{c^2 \plus{} a^2} ).
\]
[/quote]
But it is easy We have $ \sum \frac{a^2}{b^2}\ge \sum \frac{a}{b}$
$ \sum \frac{a}{c}\ge 3$
since them and AMGM we have the proof

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

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

I posted it [url=http://www.mathlinks.ro/viewtopic.php?t=199832&start=20]here[/url] but don't receive any solutions yet.[/quote]
This inequality is wrong, for example $ a\equal{}12,b\equal{}2,c\equal{}1$. but it is true if $ a,b,c$ are the side lengths of a triangle and it follows from the following inequalities of mine
\[ (ab^2\plus{}bc^2\plus{}ca^2)^2 \ge 3abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\]
:)

[quote="fjwxcsl"]The inequality:
\[ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\ge 2(a \plus{} b \plus{} c)(\frac a{b^2 \plus{} c^2} \plus{} \frac b{c^2 \plus{} a^2} \plus{} \frac c{a^2 \plus{} b^2} )
\]
does not hold( a counter-example:a=100,b=5,c=1)!

         And  we have
\[ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\ge 2(a \plus{} b \plus{} c)(\frac a{a^2 \plus{} b^2} \plus{} \frac b{b^2 \plus{} c^2} \plus{} \frac c{c^2 \plus{} a^2} ).
\]
[/quote]
This inequality is quite easy, since
\[ 2\sum \frac{a}{a^2\plus{}b^2} \le \sum \frac{1}{a}\]
Hence, it suffices to prove that
\[ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})^2 \ge (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\]
\[ \Leftrightarrow \sum \frac{a^2}{b^2} \plus{}\sum \frac{b}{a} \ge \sum \frac{a}{b} \plus{}3\]
which is true since by Cauchy Schwarz Inequality, we have
\[ \sum \frac{a^2}{b^2} \ge \frac{1}{3}(\sum \frac{a}{b})^2 \ge \sum \frac{a}{b}\]
and the AM-GM gives us
\[ \sum \frac{b}{a} \ge 3\]
And we have done.
:)

[quote="onlylove_math"][quote="Allnames"][quote="fjwxcsl"]
         And  we have
\[ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\ge 2(a \plus{} b \plus{} c)(\frac a{a^2 \plus{} b^2} \plus{} \frac b{b^2 \plus{} c^2} \plus{} \frac c{c^2 \plus{} a^2} ).
\]
[/quote]
But it is easy We have $ \sum \frac {a^2}{b^2}\ge \sum \frac {a}{b}$
$ \sum \frac {a}{c}\ge 3$
since them and AMGM we have the proof[/quote]
 Can U explain it ? :wink:[/quote]
Can_hang helped me .Thanks
And more(still easy)
       $ (\frac a{b} \plus{} \frac b{c} \plus{} \frac c{a})^2\ge 2(a \plus{} b \plus{} c)(\frac a{a^2 \plus{} b^2} \plus{} \frac b{b^2 \plus{} c^2} \plus{} \frac c{c^2 \plus{} a^2} ).\plus{}\sum \frac {(a\minus{}b)^2}{2ab}$

[quote="dduclam"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be positive real numbers. Prove that

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

I posted it [url=http://www.mathlinks.ro/viewtopic.php?t=199832&start=20]here[/url] but don't receive any solutions yet.[/quote]
This inequality is wrong, for example $ a \equal{} 12,b \equal{} 2,c \equal{} 1$. but it is true if $ a,b,c$ are the side lengths of a triangle and it follows from the following inequalities of mine
\[ (ab^2 \plus{} bc^2 \plus{} ca^2)^2 \ge 3abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)
\]
:)[/quote]

Oh, I just see that. And I also is want to saying that $ a,b,c$ are the side lengths of a triangle. Can you post your solution for it, Can ?

Beside, the following inequality's also true for all $ a,b,c$ are the side lengths of a triangle, but easy
$ \frac a{b} \plus{} \frac b{c} \plus{} \frac c{a}\ge a\sqrt {\frac2{b^2 \plus{} c^2}} \plus{} b\sqrt {\frac2{c^2 \plus{} a^2}} \plus{} c\sqrt {\frac2{a^2 \plus{} b^2}}$[/quote]
My inequality has a very nice and interesting solution posted by VasC here, my friend. My solution is longer so I dont want to posted it here, you may see it in my book. :)
for the second inequality, it still also follows from my inequality, my friend. :)

[quote="can_hang2007"][/quote][quote="dduclam"][quote="can_hang2007"][quote="dduclam"]Let $ a,b,c$ be positive real numbers. Prove that

$ (\frac a{b} + \frac b{c} + \frac c{a})^2\ge 2(a + b + c)(\frac a{b^2 + c^2} + \frac b{c^2 + a^2} + \frac c{a^2 + b^2} )$

I posted it [url=http://www.mathlinks.ro/viewtopic.php?t=199832&start=20]here[/url] but don't receive any solutions yet.[/quote]
This inequality is wrong, for example $ a = 12,b = 2,c = 1$. but it is true if $ a,b,c$ are the side lengths of a triangle and it follows from the following inequalities of mine
\[ (ab^2 + bc^2 + ca^2)^2 \ge 3abc(a + b + c)(a^2 + b^2 + c^2)
\]
:)[/quote]

Oh, I just see that. And I also is want to saying that $ a,b,c$ are the side lengths of a triangle. Can you post your solution for it, Can ?

Beside, the following inequality's also true for all $ a,b,c$ are the side lengths of a triangle, but easy
$ \frac a{b} + \frac b{c} + \frac c{a}\ge a\sqrt {\frac2{b^2 + c^2}} + b\sqrt {\frac2{c^2 + a^2}} + c\sqrt {\frac2{a^2 + b^2}}$[/quote][quote="can_hang2007"]
My inequality has a very nice and interesting solution posted by VasC here, my friend. My solution is longer so I dont want to posted it here, you may see it in my book. :)
for the second inequality, it still also follows from my inequality, my friend. :)[/quote]

The second, It's true from $ \sum \frac a{b}\ge \sum\frac {2a}{b + c}\ge \sum a\sqrt {\frac2{b^2 + c^2}}$ , but I would like to see a more simple solution  :wink:

[quote="manlio"]Is this easy to prove, dduclam ?

$ \sum \frac a{b}\ge \sum\frac {2a}{b \plus{} c}$[/quote]

Yes, this's  :)
[hide="hint"]Use SOS method  :wink: [/hide]

[quote="manlio"]Is this easy to prove, dduclam ?

$ \sum \frac a{b}\ge \sum\frac {2a}{b \plus{} c}$[/quote]
It is equivalent to
\[ \sum \frac{a}{b} \plus{}2\sum \frac{b\plus{}c\minus{}a}{b\plus{}c} \ge 6\]
By Cauchy Schwarz Inequality, we have
\[ \sum \frac{b\plus{}c\minus{}a}{b\plus{}c} \ge \frac{(a\plus{}b\plus{}c)^2}{\sum (b\plus{}c\minus{}a)(b\plus{}c)} \equal{}\frac{(a\plus{}b\plus{}c)^2}{2\sum a^2}\]
Hence, it suffices to prove that
\[ \sum \frac{a}{b} \plus{}\frac{(a\plus{}b\plus{}c)^2}{\sum a^2} \ge 6\]
Now, using the old inequality of mine
\[ \sum \frac{a}{b} \ge \frac{9(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\]
It suffices to prove that
\[ \frac{9(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2} \plus{}\frac{(a\plus{}b\plus{}c)^2}{\sum a^2} \ge 6\]
which is obviously true by AM-GM inequality. :)

[quote="can_hang2007"][quote="manlio"]Is this easy to prove, dduclam ?

$ \sum \frac a{b}\ge \sum\frac {2a}{b \plus{} c}$[/quote]
It is equivalent to
\[ \sum \frac {a}{b} \plus{} 2\sum \frac {b \plus{} c \minus{} a}{b \plus{} c} \ge 6
\]
By Cauchy Schwarz Inequality, we have
\[ \sum \frac {b \plus{} c \minus{} a}{b \plus{} c} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum (b \plus{} c \minus{} a)(b \plus{} c)} \equal{} \frac {(a \plus{} b \plus{} c)^2}{2\sum a^2}
\]
Hence, it suffices to prove that
\[ \sum \frac {a}{b} \plus{} \frac {(a \plus{} b \plus{} c)^2}{\sum a^2} \ge 6
\]
Now, using the old inequality of mine
\[ \sum \frac {a}{b} \ge \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}
\]
It suffices to prove that
\[ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {(a \plus{} b \plus{} c)^2}{\sum a^2} \ge 6
\]
which is obviously true by AM-GM inequality. :)[/quote]
Very nice  :lol: ,
Assum $ c \equal{} max${a;b;c}

$ \sum \frac a{b}\ge \sum\frac {2a}{b \plus{} c}$

$ \leftrightarrow (\frac {1}{ab} \minus{} \frac {2}{(a \plus{} c)(b \plus{} c)})(a \minus{} b)^2 \plus{} (\frac {1}{ac} \minus{} \frac {a \plus{} b \plus{} 2c}{(a \plus{} b)(b \plus{} c)(c \plus{} a)})(a \minus{} c)(b \minus{} c) \ge\ 0$
Which is obvious  :)