A, B, C, D, and E are collinear in that order such that AB = BC = 1, CD = 2, and DE = 9.
If P can be any point in space, what is the minimum possible value of AP^2 + BP^2 + CP^2 +
DP^2 + EP^2?

[hide="My solution"]Let the altitude from $ P$ onto $ AE$ at $ Q$ have lengths $ PQ=h$ and $ AQ=r$.  It is clear that, for a given $ r$ value, $ AP$, $ BP$, $ CP$, $ DP$, and $ EP$ are all minimized when $ h=0$.  So $ P$ is on $ AE$, and therefore, $ P=Q$.  Thus, $ AP$=r, $ BP=\abs{r-1}$, $ CP=\abs{r-2}$, $ DP=\abs{r-4}$, and $ EP=\abs{r-13}$.  Squaring each of these gives:

$ AP^2+BP^2+CP^2+DP^2+EP^2=r^2+(r-1)^2+(r-2)^2+(r-4)^2+(r-13)^2=5r^2-40r+190$

This reaches its maximum at $ r=\frac{40}{2\cdot 5}=4$, at which point the sum of the squares of the distances is $ \boxed{110}$.[/hide]


The integers 1,2,3,...,100 are written on the board. What is the smallest number of these integers that can be wiped off so that the product of the remaining integers ends in 2?

[hide]pretty clear that if $0 \mod 2$, and $2 \mod 5$, then $2 \mod 10$, giving us our answer

looking at the digits mod 5, we have

$0^{20}*1^{20}*2^{20}*3^{20}*4^{20}$

so for it to be $2 \mod 5$, we must get rid of the 20 multiples of 5

$2^{60}*3^{20} \mod5$, use euler's generalization to have $a^4 \equiv 1$

$1*3^4$, and we have to erase 1 of the 3's to get it 2 mod 5, so we erased $20+1=\boxed{21}$[/hide]


modular arithmetic
quadratics
number theory unsolved
number theory

Prove that if
$ 6x\plus{}1$
$ 7x\plus{}4$
$ 8x\plus{}9$
are all perfect squares, then $ 20\mid x$.

First note that $ 6x \plus{} 1$ is a perfect square, and all perfect squares are congruent to $ 1\pmod{4}$ so $ 4|6x$ so $ 2|x$. Then, $ 2|7x$ and as $ 7x \plus{} 1$ is a perfect square, $ 2|7x$ so $ 7x \plus{} 1\equiv{1}\pmod{4}$ so $ 4|(7x)$ so $ 4|x$.
On the other hand, the set of quadratic residues of $ 5$ is $ 0,1,4$. Have $ 6x \plus{} 1\equiv x \plus{} 1\pmod 5$ so $ x$ can be $ 0,3$ or $ 4\pmod {5}$. If $ x\equiv {3}\pmod{5}$ get $ 7x \plus{} 1\equiv {2}\pmod{5}$ which can never be a perfect square. If $ x\equiv {4}\pmod{5}$ get $ 8x \plus{} 1\equiv {3}\pmod{5}$ which can never be a perfect square. Hence $ x\equiv {0}\pmod {5}$ so $ 5|x$.
Therefore, $ 5|x$ and $ 4|x$; $ \gcd(4,5) \equal{} 1$ so $ 20|x$ QED.


A fair, six-sided die is rolled, and used as the thousands digit for a 4 digit #, another for the hundreds digit, and so on. What is the probability that this # is divisible by 11?

Now, same thing, with 6 dice, and a 6 digit #.

I bet ill be way off but since no one else has tried ill give it a shot

chance of integer being divisible by 11 is 1/11(+1 if it starts and ends with a number dividible by 11)

6667>x>1110  6 chances for each number 6^4=1296

1296/11= 117 9/11/1296= 1/11

i bet 6-digit is 1/11 too

I did this all in my head so there might be computation mistakes.

[quote="archimedes1"]Is it [hide]73/648[/hide]?[/quote]
[quote="stupidityismygam"]i got [hide]$\frac{73}{648}$[/hide] as well[/quote]
[b]Start posting solutions, not just answers![/b]

[quote="CheeseIsGood"]How did you get that?
 :huh:[/quote]
[hide="sorry for not posting solution earlier"]For a number to be divisible by $11$, the difference between the sum of the odd-positioned digits and the sum of the even-positioned digits has to be $11$.  For example, $1210$ is divisible by $11$ because $(1+1)-(2+0)=0$, which is divisible by $11$.  So, we have four digits.  _ _ _ _  Now since with our number constraints it is impossible to get a difference of more than $(6+6)-(1+1)=10$, we only need to focus on getting a difference of $0$.  So, let's start with the possible sum of $2$ for the first and third digits.  There is one way out of $36$ to get a sum of $2$, so the probability that we have the first and third digit add up to $2$ is $\frac{1}{36}$.  Now, We need to have the second and fourth digits also add up to $2$ in order for the difference to be $0$, so once again the probability is $\frac{1}{36}$.  So the probability that our number has the odd/even sum of $2$ and is divisible by $11$ is $(\frac{1}{36})^{2}$.  Now, lets see what happens when we make the sum $3$.  There are two ways to have 2 dice add up to $3$, so we get $\frac{2}{36}$.  And there is the same probability for the other pair of digits to add to three, so we multiply them and get $(\frac{2}{36})^{2}$.  We proceed in the same manner towards the other possible sums, all the way up to $12$.  It might help to draw a table of possible dice sums here.  At the end we get $(\frac{1}{36})^{2}+(\frac{2}{36})^{2}+(\frac{3}{36})^{2}+(\frac{4}{36})^{2}+(\frac{5}{36})^{2}+(\frac{6}{36})^{2}+(\frac{5}{36})^{2}+(\frac{4}{36})^{2}+(\frac{3}{36})^{2}+(\frac{2}{36})^{2}+(\frac{1}{36})^{2}=\frac{73}{648}$.[/hide]
You can use that method for the second problem too.

That's correct. You can use a similar method for the second one.


analytic geometry
superior algebra
superior algebra unsolved

Find the point of intersection of the line L through (1,0) and (0, -1) and the circle C with center (0, 0) and radius 3. Identify a square root extension field Q[c] of the field Q of rational numbers that contains the coordinates of this point of intersection.

[quote="LydianRain"]Where are you stuck?[/quote]
I found the intersections with the circle at
A ((1+sqrt(17))/2, (-1+sqrt(17))/2), and B((1-sqrt(17))/2, (-1-sqrt(17))/2),
but I was not sure if the extension  Q(sqrt17) is a correct one.

It is.  The field $ \mathbb{Q}(\sqrt{17})$ consists of all elements of the form $ a \plus{} b\sqrt{17}$, where $ a$ and $ b$ are rational numbers, and the coordinates of those points can be written in that form.


The Round 4 scores, and thus the total scores for the year, are now available on the [url=http://www.usamts.org]USAMTS website[/url].

We will be making announcements about prizes sometime in the next couple of weeks.  [b]Please do not ask when the prizes will be available, or what they will be[/b]: we are still working out the details.

Southern California ARML team:

Two scores $=100.$
Three scores $\ge98.$
Four scores $\ge94.$
And twenty commended problems among those four.


ratio
geometric sequence
combinatorics proposed
combinatorics

Given four weights in geometric progression and an equal arm balance, show how to find the heaviest weight using the balance only twice.

[quote="Albanian Eagle"]let the weights be a,b,c,d
weigh a,b vs c,d
the heavier side will contain the heavier weight. then compare the two ones that are left.[/quote]

I totally agree with Albanian Eagle.. Let weights be $ x, xr, xr^2$ and $ xr^3$ where $ r$ is common ratio of our geometric progression.
Doesn't matter whatever you choose, the side contain $ xr^3$ will be greater than the other side..
Proof: Let us the consider this case:
if we put the $ xr^3$ and $ x$ to one side and $ xr$ and $ xr^2$ to another side, Then first side will weigh more ,
$ xr^3 \plus{} x \geq xr \plus{} xr^2$
that is because
$ xr^3 \plus{} x \minus{} xr \minus{}xr^2 \equal{} xr^2(r\minus{}1) \minus{}x(r\minus{}1) \equal{} x(r\minus{}1)(r^2\minus{}1)\equal{}x(r\minus{}1)(r\minus{}1)(r\plus{}1)\equal{}x(r\minus{}1)^2(r\plus{}1) \geq 0$  :lol:

[quote="JBL"][quote="shaaam"]Doesn't matter whatever you choose, the side contain $ xr^3$ will be greater than the other side.[/quote]  What if we split the weights as $ (x, xr)$ against $ (xr^2, xr^3)$, with $ r \equal{} \frac {1}{2}$?  :wink:[/quote]

But in that case the heaviest weight would be $ x$ since $ r$ is less than one.  :)


ratio
geometry
3D geometry
Pythagorean Theorem

Using the pythagorean theorem, $a^2+b^2=c^2$, find the length of the diagnol of a square with each of the sides are 3.  What about a square with the sides 5?  Or a square with sides n?

Using the pythagorean theorem, a^2+b^2=c^2, find the length of the diagnol of a square with each of the sides are 3.

[hide]2(3^2) = 18, the square root of 18 is 3 root 2[/hide]

 What about a square with the sides 5?

[hide]2(5^2) = 50, the square root of 50 is 5 root 2[/hide]

o, cool, there's a pattern, a square with sides n would have a diagonal of n root 2

thanx!!
 :D

[hide]1.) 3*(square root of 2)[/hide]
         [hide]2.) 5*(square root of 2)[/hide]
Since a square has four equal sides, drawing one of its diagonals creates two
45-45-90 triangles.  In a 45-45-90 triange, the ratio of the sides are
a leg to a leg to the hypotenuse to 1:1:the square root of 2 respectively.

Good job both of you, I turn the general formula into LaTex.

For a square with a side length $s$, it's diagonal length is $s\sqrt{2}$

I thought of an extension.

What is the length of the diagonal of a cube if the side length is n?

[quote="jimli"]the length of a space diagonal of a cube with sides of n is $\sqrt{3}n$ because it has a diagonal of $\sqrt{2}n$ which squared is $2n^2$ and when added to $n^2$ is $3n^2$ take the square root and you get $\sqrt{3}n$[/quote]

you mean n root 2, and n root 3


real analysis
inequalities
vector
algebra
polynomial
complex analysis
triangle inequality

In Rudin's 3rd edition of [i]Real and Complex Analysis[/i], it is written that $|P(z)|>|P(0)|$ whenever $|z|>1+2|a_{0}|+|a_{1}|+\dots+|a_{n-1}|$, where $P(z)=z^{n}+a_{n-1}z^{n-1}+\dots+a_{0}$.

Such a precision is obviously useless (I'd even say ridiculous) in order to prove the Fundamental theorem of algebra, but Rudin decided to include it, and I can't see where it comes from (probably some triangle inequality applied somewhere, but I can't see it at first sight).

[quote="julien_santini"][prove] that $|P(z)|>|P(0)|$ whenever $|z|>1+2|a_{0}|+|a_{1}|+\dots+|a_{n-1}|$, where $P(z)=z^{n}+a_{n-1}z^{n-1}+\dots+a_{0}$.  [/quote]

For a large enough value of $|z|$ (we will find it later), the $z^{n}$ term dominates the sum of the others.  In that case, the smallest absolute value of $P(z)$ would occur if all of the $a_{i}z^{i}$ are vectors in the opposite direction to $z^{n}$, i.e.  $|z^{n}|-\Sigma |a_{i}z^{i}|$.  (It is not guaranteed that this configuration is possible for a given $P$, but the inequality will be true in any case).
So we require that
$|z^{n}| > |P(0)|+\Sigma |a_{i}z^{i}|$.
For $|z| > 1$  we can divide by $|z^{n-1}|$ to see that this is the case if
$|z| > |a_{0}|+|a_{n-1}|+\dots+|a_{1}|+|a_{0}|$.
Thus, if $|z|$ exceeds this number and also exceeds $1$, it suffices.
Rudin's bound is just the sum of these two numbers.

I would add that I've seen perhaps a dozen different proofs of the FTA, using very different techniques. Some revel in analysis; some seek to minimize analysis to utilize more algebra. But every single proof I've seen uses the fact that for large values of the variable, a polynomial's behavior resembles that of its highest degree term. Sometimes it's just to show that the values are large in absolute value, sometimes a more detailed version of the resemblence is needed.

The bound or resemblence need not be particularly sharp and can be proved crudely - as fleeting_guest showed. The details of that proof aren't particularly important.

For instance: can you prove that an odd degree real polynomial must have at least one real root? This will follow from the Intermediate Value Theorem, but to set up the use of the IVT we must find one value where the polynomial is positive and one where it is negative - and that requires us to look at large positive and negative values of the argument.

There are proofs of the FTA that need no more analysis than that.

There is a proof that does not involve evaluation of the polynomial at any point at all. You can find a sketch on [url=http://groups.google.com/group/sci.math.research/browse_thread/thread/b86a628ce3009224/af068751c86c0dfb?lnk=gst&q=fundamental+theorem+of+algebra&rnum=2#af068751c86c0dfb]sci.math.research[/url]
(The proof would be simpler over $\mathbb{C}$, still without any evaluation.)

[quote="mlok"]There is a proof that does not involve evaluation of the polynomial at any point at all. [/quote]

Which proof did you mean?  There are several at that link and it is not clear whether they depend on pointwise evaluation.

[quote="fleeting_guest"]Which proof did you mean? [/quote]
The one that I sketched in that thread: it interprets the multiplication of polynomials as a map from $\mathbb{R}^{k}\times\mathbb{R}^{m}\to \mathbb{R}^{k+m}$. (For real polynomials; the same argument works in the complex case.) Another poster remarked that the idea goes back at least to Hilbert. It seems that this proof is not widely known.

Thank you. That thing got me a little irritated. I prefer the more usual proof using only Liouville's theorem. I do not understand why Rudin insists on seeing Liouville's theorem, the maximum principle and Cauchy's estimates as applications of Parseval's theorem ...

[quote="Kent Merryfield"]But every single proof I've seen uses the fact that for large values of the variable, a polynomial's behavior resembles that of its highest degree term. [/quote]

Actually, what I quite didn't get is why Rudin strives to find an explicit bound, while it would be enough to say that such a bound exists (this is obvious enough by factorization).

For some people it is not enough to know that the object of their interest exists -- they also want to know where it is. For example, hunters tend to act this way. Rudin's estimate gives an idea of where the roots are located.


Let $ \{a_i\}_{i\equal{}1}^n$ be a sequence of real numbers and $ \{b_i\}_{i\equal{}1}^n$ a permutation of $ \{a_i\}_{i\equal{}1}^n$. Prove that
\[ \sum_{i<j}(a_i\minus{}b_i)(a_j\minus{}b_j)\le 0.\]

[quote="nayel"]Let $ \{a_i\}_{i \equal{} 1}^n$ be a sequence of real numbers and $ \{b_i\}_{i \equal{} 1}^n$ a permutation of $ \{a_i\}_{i \equal{} 1}^n$. Prove that
\[ \sum_{i < j}(a_i \minus{} b_i)(a_j \minus{} b_j)\le 0.\]
[/quote]

Let $ x_i\equal{}a_i\minus{}b_i$, then we have $ x_1\plus{}x_2\plus{}...\plus{}x_n\equal{}0$

So $ 0\equal{}(x_1\plus{}x_2\plus{}...\plus{}x_n)^2\equal{}(x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2)\plus{}2 \sum_{i<j}(x_ix_j) \ge 2 \sum_{i<j}(x_ix_j)$

$ \Rightarrow \sum_{i<j}(x_ix_j) \le 0$, done.

Expanding we see that it is equivalent to:
$ \sum_{i\not\equal{} j}a_ia_j \le \sum_{i\not\equal{} j} a_ib_j$ denoting $ \sum a_i \equal{}S$
this is in turn equivalent to
$ \frac12\sum a_i(S\minus{}a_i)\le \sum a_i(S\minus{}b_i)$
or $ \sum a_i^2\geq \sum a_ib_i$
or $ \frac12\sum (a_i^2\plus{}b_i^2) \geq \sum a_ib_i$ which is clear by Am-Gm

I solved a problem equivalence this problem someday ago.
http://www.mathlinks.ro/viewtopic.php?p=1545378#1545378


What is the largest $x$ for wich
$4^{27}+4^{100}+4^{x}$
is a perfct square?

 :wink:

[quote="Rust"]It is posted before, x=172.[/quote]

No, $x=1972$
 :wink:

[quote="Rust"]If $x>100$ then
\[
4^{27}+4^{100}+4^{x}=(2^{2}7k)^{2}
[/quote]
Why?   :blush:

I will prove a more general results: The maximum value of $x$ such that
$4^{a}+4^{b}+4^{x}$
is e perfect square  is $x=2b-a-1$, where $b>a$.

Proof: If $x=2b-a-1$, then $4^{a}+4^{b}+4^{2b-a-1}=(2^{a}+2^{2b-a-1})^{2}$. Thus i can assume that $x>2b-a-1/geq b$. Now, $4^{a}+4^{b}+4^{x}$ is a perfect square $/Leftrightarrow/$ $1+4^{b-a}+4^{x-a}$ is e perfect square.
Let $1+4^{b-a}+4^{x-a}=s^{2}$, where $s$ is odd. Then $4^{b-a}=2^{2b-2a}$ divides $s^{2}-1$, and no higher power of 2 divides $(s-1)(s+1)$ since $x>b$. Then , since $s-1$ and $s+1$ are two consecutive even numbers, one of $s-1$ or $s+1$ is divisible by $2^{2b-2a-1}$
Let $s=2^{2b-2a-1}r+-1$, where $r$ is odd.Then
$s^{2}=4^{2b-2a-1}r^{2}+-2^{2b-2a}r+1=1+4^{b-a}+4^{x-a}$ $/Leftrightarrow/$ $4^{b-a}(+-r-1)=4^{2b-2a-1}(4^{x-(2b-a-1)}-r^{2})$ $/Leftrightarrow/$ $2^{x-(2b-a-1)+r\leq |+-r+1|$ $/Leftrightarrow/$
$2^{x-(2b-a-1)\leq 1$ $/Leftrightarrow/$  $x\leq 2b-a-1$
contradicting $x\geq 2b-a-1$. Thus $x=2b-a-1$ is maximal.
$a=27$, $b=1000$ and so maximal value of $x$ is 2000-27-1=1972.
 :wink:

you must have had a typo because on your first post you have $b=100$


search
number theory unsolved
number theory

[quote="lambruscokid"]http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1736432107&t=81843[/quote]

thank you... I haven't seen it...


function
algebra
polynomial
complex numbers

Does nth root of (a^m) equal (nth root of a)^m where a is all real number, n is greater than or equal to 2, and m is an integer?
If not, I cannot express x^(m/n) as (nth root of a)^m, right?
If so, how can I prove that they are equal?

Yes, $ \sqrt[n]{a^m} \equal{} (\sqrt[n]{a})^{m}$.

$ \sqrt[n]{a^m} \implies a^{\frac{m}{n}} \implies (\sqrt[n]{a})^m$

Didn't you say for all $ \mathbb{R}$?  $ i$ isn't a real number...

a=-1 is a real number, and square root of -1 is i.

Uhhh, you aren't disproving either the positive or the contrapositive...

The short answer is that this just isn't true generally.

The long answer is that this rule is valid in the following situations:

- $ a$ is a non-negative real and all roots taken are non-negative.
- $ n$ is an odd integer and all roots taken are real.

It all boils down to how you define $ \sqrt[n]{a}$ as a [i]function[/i] (as distinct from the set of $ n^{th}$ roots of $ a$).  The above scenarios are two good ways to make sense of this function in certain contexts, but for all reals and all integers you run into problems, as you can see.

So, I can't really say that (n th root of x)^m = n th root of (x^m), right?

It would be more correct to say that the $ m^{th}$ powers of the roots of the polynomial $ t^n \equal{} x$ are precisely the roots of the polynomial $ t^n \equal{} x^m$.  If you try to say anything more specific, you end up with problems like your second post (where you're confusing different values of a "multi-valued" function).

So if nth root of (x^m) is in the equation to begin with, I can't really change it to (nth root of x)^m, correct?
Thanks for the reply by the way.

[quote="t0rajir0u"]this rule is valid in the following situations:

- $ a$ is a non-negative real and all roots taken are non-negative.
- $ n$ is an odd integer and all roots taken are real.[/quote]
If you're trying to solve an equation that has a term like that in it, it really shouldn't.  I don't like questions written like that when it's not clear what branch to take.  Could you be more specific about what you want to do with this information?

I was just wondering if those two would equal; I am not trying to do anything with those, just curious. :)

Your argument (I'm talking to you, rOadToAIME) with $ i$ does not work; $ i$ as well as the other nonreal complex numbers have their own separate rules from those of the real numbers.  It is just "defined" that you can't perform such operations.
Look at this: $ \minus{}1\equal{}i^2\equal{}(\sqrt{\minus{}1})^2\equal{}\sqrt{(\minus{}1)^2}\equal{}\sqrt{1}\equal{}1$.  There is nothing really wrong with this argument, but clearly it is not true that $ \minus{}1\equal{}1$.  So we have to make it a rule that manipulations such as $ (\sqrt{\minus{}1})^2\equal{}\sqrt{(\minus{}1)^2}$ don't hold.

If you don't carefully observe the rules of nonreal complex numbers, then you could easily get contradictions because you overlooked that you did something "illegal."

[quote="rOadToAIME"]I'm sorry, but what argument are you exactly talking about?
By the way, there is something wrong in your argument:     [code]    -1=i^2=(\sqrt{-1})^2=\sqrt{(-1)^2}=\sqrt{1}=1[/code]. [code](\sqrt{-1})^2[/code] cannot be manipulated into [code]\sqrt{(-1)^2}[/code] since the -1 is less than zero.[/quote]
Yes, that is exactly what I'm saying (the fact that the manipulation is illegal since $ \minus{} 1 < 0$)!

And the argument I'm talking about is [quote="rOadToAIME"]Hmmm.......
You are saying that (square root of -1)^2 equals square root of (-1)^2, which is impossible: i^2= -1, square root of 1 = 1[/quote].

But perhaps I misunderstood what you were saying, because I can see that you understand more than I thought.


linear algebra
matrix
algebra
polynomial
combinatorics unsolved
combinatorics

Let m and n be natural numbers. Two mxn matrices having  0 and 1 as elements are called equivalent iff one can be obtained from the other by a cyclic permutation of its rows and/or columns. Find the maximal number of non-equivalent mxn matrices (the number of equivalence classes, thus).

Supplemental question : he didn't ask me this, but how about replacing "cyclic" with "arbitrary"? What is the answer then?

What I say may look funny, but for such problems, there is a special theorem, by Polya.
If you have n elements and c colors, and any two colorings of the elements are equivalent if they can be obtained by some permutations, that of course form a group, then we make a special Polinomial P(x) for this group, and the answer will be P(c), c - # of colors.
Some patience will bring you the anwer.....


inequalities
geometry
circumcircle
trigonometry
inequalities solved

Let ABC be a triangle. Prove that
    \sum(m_a/l_a)<=1/2*( \sum a/b+a/c).
 m_a,m_b,m_c are medians of ABC
 l_a,l_b,l_c are bisectors of ABC.

Treegoner, could you please post your solution to this problem? I'm astonished, I tried a lot and couldn't find a solution.

I'm really SORRY when I forgot the condition that ABC is not obtuse. :  (  :(
  I will find out if this one is right for obtuse triangle.
  I don't know if with this condition you can solve it. Harazi, please try and I'll post my solution later.

Dear Treegoner, please believe me that I ask to see the solution of a problem only after an intensive effort, which is the case for this one. All the possible tricks fail for this one.

In this case, it's very convinient to use geometry.
 Let D',M' be the intersections of bisector AD , median AM and the circumcircle of ABC.
   It's easy to see that AD'<=AM' since ABC is an acute triangle.
    But AD'= bc/l_a,AM' = (b^2 +c^2)/2m_a. Hence, m_a/l_a <= 1/2(b/c+c/b).
    In the same manner we have our result. :D
    Though it looks simple,the direction is quite difficult to find. :(

BEAUTIFUL! And again BEAUTIFUL! Congratulations, treegoner, you create extraordinary problems. I must say that I do not like those trigo inequalities you posted (I don't say they aren't interesting, but I dislike trigonometry), but this one is a real diamond. Congratulations!


geometry
geometric transformation
reflection
circumcircle
rotation
vector
analytic geometry

here is the most difficult and the most beautiful problem occurs in 21th iranian (2003) olympiad
assume that P is n-gon ,lying on the plane ,we name its edge 1,2,..,n.
if S=s1,s2,s3,.... be a finite or infinite sequence such that  for each i, si is in {1,2,...,n},
we move P on the plane according to the S in this form: at first we reflect P through the s1
( s1 means the edge which iys number is s1)then through s2 and so on like the figure below.
a)show that there exist the infinite sequence S sucth that if we move P according to S we cover all the plane
b)prove that the sequence in a) isn't periodic.
c)assume that P is regular pentagon ,which the radius of its circumcircle is 1,and D is circle ,with radius 1.00001 ,arbitrarily in the plane .does exist a sequence S such that we move P according to S then P reside in D completely?

I want to give a hint from the proposer of problem :D  u can use {na} is dense in [0,1] :?

I give up :)
Anyone else ? :)
I wud like 2 see the solution.
I thought a little. btw, {na} is dense only for a - irrational :)

I would like to find out the solution.
I thinks {na} is dense only for a - irrational

I think I have a solution for a,b. I didn't understand c.
But I see not many people visit this forum last days.
If anyone except DusT wants to see my solution, please send me a message or reply.
DusT I will show you at school

iura you may rest asure that people will see your post. there are aproximatively 150 registred users checking in each day, 100 checking in once a week, plus many other guests that do not log on :)

Well, I think the idea for (c) is this one (I'm just explaining it; I don't have a soln or anything like that :)):

You are given P=a regular pentagon somewhere in the plane, s.t. its circumradius is 1. You are also given a circle of radius r>1 (I don't think it's important that it's 1.0001 or 1.0000000001 or anything else, as long as it's >1). Is there a sequence of moves S (defined at the beginning) s.t. the image of the regular pentagon,  when it's moved in the way described in the text, is completely inside the given circle?

I will think about c).
The proof of a) is long and tricky, some things can't be explained without a good picture. So I present it briefly.
Lemma 1) Let ABC.. be a position of the polygon, P a point in the plane and K on [AB], PK perp AB .Let PK meet the polygon at L too. Then we can find c such that KL>=c any P,AB..
Proof. Just suppose l is the shortest side u is the closest to 90 angle of the polygon( but not equal to 90). Then we can take c=l cos u.
If PK contans a side of the polygon, we put L the other extremity.

Lemma 2) Let ABC.. be a position of the polygon, d(P) the smallest distance from P to a point on the perimeter. Then for a fixed P, we can perform a move such that the polygon either contains P, or d(P) decreases with at least C, for some C>0.
Proof. Let PK =d(P). if K is not a vertex, then it satisfies conditions in lemma 1 and we just reflect by syde containing K.
Now suppose K=A is a vertex. (this is the part where you need a picture).
If the angle at A is 120 deg, we can reflect by one of sides containing A.
We see that there is  a constant c in this case.
If A is not 120 deg, then we reflect by a fixed side containing A a lot of times. If B is the other extremity of the side, B will describe either a regular polygon with at least 4 sides, or will rotate densely on a circle.
In each case, polygon will either contain P or the distance from P to B will decrease by at least a constant c.( Sorry I can't give a good picture to explain this)
Lemma is proved.
Lemma 3) For any P, we can make some moves the polygon to contain P in its interior.
Proof. By Lemma 2, we can perform moves such that either d(P) decreases by at least C, or P gets into the polygon. Since d(P) is finite, sometime we will reach the second case.
Now we can choose u very small, divide the plane into u-sided squares.
We can move the polygon to contain the center of any square, the if u is sufficiently small, we can reflect in such a way the polygon to pass thrpugh all points of square.
Now as N*N is countabel(in bijection with N), we can label the squares u1,u2..un.
Now we visit u1,u2 etc. and here is our desired sequence of moves.
b)Suppose Sn is periodic. Suppose the period is even.
By any even sequence of moves,  the plygon will change by a translation and a rotation. If we fix K a vertex of polygon and pass to complex numbers, K to have afix z, then z will change to z+d, then z+d+dw,
the z+d+dw+dw^2 etc., where d is the length of translation, w is a number of modulus 1 corresponding to the angle of rotation.
Obviously w=1 is not the case So if  w<>1 then after kT moves (T-period)
K will have afix z[k]=z+d(w^k-1)/(w-1).
So |z[k]-z|<d*2/(w-1) so k will dever exit some disc and the polygon can not visit all the plane

As I read your solution for b, you say that if there is a cycle, then the polygon doesn't exit a disk, which is wrong.
Take a square, and reflect it by 2 opposite sides.

Dust I said w=1 is not the case and your example matches w=1.
I also found a sol for c). Not that all even number of mover are just translations of polygon.
Lemma: Suppose we can translate a point with vectors (+-a,+-b),(+-c,+-d),
where a,b,c,d are non-zero real numbers with a/c,b/d irrational. Then we can move the point as close to a fixed point as we wish.
Proof of lemma. By Kronecker's Theorem, if the fixed point has x,y coordinates,
we can find ma+nc close to x/2, pb+qd close to y/2, where m,n,p,q are integers.
As + and - cancel each other, we can easily construct some moves to reach (2(ma+nc),2(pb+qd)). The lemma is proven.
To prove the problem, we can move the polygon by vector u, taking sides 1,3.
By radial symmetry, we can also move by 72,144,216,288 rotations of u.
Moveing backwards( i.e. 3 1) we can move to 180, 36,108 etc. rotations of u(backwards means rotation by 180. please draw a picture with trigonometric circle to understand better).The taking OX||u, we can easily find two quadruples of vectors satisfying the lemma. So we can move the center of polygon as close to the center of given disk as we wish, and so we are done

Definitely, this problem is far from beautiful for me. It is just opinion.


Michael spins the spinner twice. All three of the larger sectors
are equal in area and have central angles of $ 90^{\circ}$. The two
smaller sectors have equal area. What is the probability that he
will ''WIN'' on both spins? Express your answer as a common
fraction.
[asy]import graph;
draw(Circle((0,0),15));
draw((0,0)--(-10.6,10.6));
draw((-15,0)--(15,0));
draw((0,-15)--(0,15));
draw((2,0)--(2,-2)--(0,-2));
draw((0,0)--(4,4),Arrow);
label("LOSE",(4,4),NE);
label("LOSE",(4,-4),SE);
label("LOSE",(-4,-4),SW);
label("LOSE",(-3,3),SW);
label("WIN",(-1,6.7),NW);[/asy]

The central angle of $ \text{WIN}$ is $ \frac{90^{\circ}}{2}\equal{}45^{\circ}$.  Since a circle is $ 360^{\circ}$, the probability that Michael will land on $ \text{WIN}$ once is $ \frac{45}{360}\equal{}\frac{1}8$.  Thus, the probability that he will land on $ \text{WIN}$ on both spins is $ \frac{1}8\cdot\frac{1}8\equal{}\boxed{\frac{1}{64}}$

a circle is 360 degress

each 1/4 of a circle is 90 degrees   therefore the 2 small area  angles =  45 degrees

so the fraction is   45/360     =1/8

multiply 2 times of that   1/8 *1/8   =1/64

[size=200][color=#4080FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!!!!!!   [/u][/i][/b][/color][/size]

 :starwars:  :starwars:  :starwars:  :starwars:

[size=150][color=#FF40FF][b][i][u]MAY THE FORCE OF AoPS REST WITHIN YOU!!!![/u][/i][/b][/color][/size]


Does anyone know where can I find the solutions of Romanian Team Selection Tests ? The "A.Jorza" web site has only solutions of them from 1998 to 2003 , but I want all of them .


The problem is let a,b,c be positive numbers such that a+b+c=1 prove that for every real number x>=1 we have x^a+x^b+x^c=<3x^(1/3)
but this seems to not work for arbitrarily large x and a arbitrarily close to one. Also it says that f(x)=a^x is concave for a<1 but this seems wrong.

If you think about it, the problem is easily fixed.

$ f(x)\equal{}a^x$, then $ f'(x)\equal{}\ln a a^x$ and $ f''(x)\equal{}(\ln a)^2 a^x>0$ so $ f$ is convex.

Let $ g(a)\equal{}x^a$. Now by jensen of whatever they call it, $ x^a\plus{}x^b\plus{}x^c\equal{}g(a)\plus{}g(b)\plus{}g(c)\ge 3g\left(\frac{a\plus{}b\plus{}c}{3}\right)\equal{}3\cdot x^{\frac{1}{3}}$

So the sign is flipped...If you want to check an example, take $ a,b,c\equal{}\frac{1}{2},\frac{1}{3},\frac{1}{6}$ and $ x\equal{}2^6$.

Then $ 8\plus{}4\plus{}2\ge 3\cdot 4$ which is true. Whenever you look at an inequality, you should check some values.

Now also note that if $ x\equal{}2^{\minus{}6}$, then

$ \frac{1}{8}\plus{}\frac{1}{4}\plus{}\frac{1}{2}\ge 3\cdot\frac{1}{4}\iff \frac{1}{8}\ge 0$.

It is a bit strange that the problem is true for all $ x>0$.

Example 7.2 goes as follows
Let a,b,c be positive real numbers Prove that
$ a^3 \plus{}b^3 \plus{}c^3 \plus{}3abc \ge a^2b\plus{}b^2a\plus{}b^2c\plus{}c^2b\plus{}c^2a$
The solution goes let $ a\equal{}e^x b\equal{}e^y c\equal{}e^z$ with $ x \ge y \ge z$
and then says WLG let $ x\plus{}z \ge 2y$ can you actually say that or is Titu just writing the solution for one of the several cases.


Solve the equation

$ \left\{{1\over x}\right\}\equal{}{3x\minus{}1\over 4}$


I thought it was a bit harder, but I did 12 points better than the 10A anyways. :D

I thought 10B was a little harder. The hardest 10B problem was easier than the hardest 10A problem but everything else was easier on 10A.

I felt the 10B was about the same. I scored a 144 on the A, due to reading the last problem incorrectly. I scored a 138 on the B, due to deciding a line with ABCD has length 4, and also forgetting to divide by 2 when calculating the area of a triangle. Overall I missed points just due to stupid things so I think the tests were pretty even.

[quote="#H34N1"]I thought it was much easier although I messed up on 2 problems[/quote]

Exactly the same for me.  I definitely should have got a perfect score.

The overall difficulty of10B was about the same as 10A and the years before, but the 10B lacked one or two hard questions that are usually present on the 10 contest to make reaching a perfect score a harder task to do (usually, questions 24 and 25 in the previous years stood out in that it require a signifcantly harder/trickier problem solving technique)
I thought the difference in difficulty between the level 10 and 12 contests became greater starting this year.
Some of the problems on this year's 12B were really tough. True for 12A, too: for questions 24 and 25, I had no idea what to do.

i think the amc 10 B this year was one of the easiest ones ever

[quote="263kou"]This AMC10B was so easy that I even finished up half an hour before the due-time...

And I got 145.5 due to an omitted question (#23).[/quote]

why would you omit a question if you finished half an hour early?


[quote="¬[ƒ(Gabriel)³²¹º]¼"][quote="rachid"]$ ABCD$ is a square.$ ADI$ and $ ABJ$ are two equilateral triangles.Prove that (AC) and (IB) and (JD) are concurents[/quote]
the 2 equilateral triangle must be all in the square or all out and AC, IB, JD concur cause the figure is symmetric w.r.t. AC.[/quote]
They must be all out and AC, IB, JD concur.


geometry
geometric transformation
homothety
geometry proposed

Let $ d$ be a line and points $ M,N$ on the $ d$. Circles $ \alpha,\beta,\gamma,\delta$ with centers $ A,B,C,D$ are tangent to $ d$, circles $ \alpha,\beta$ are externally tangent at $ M$, and circles $ \gamma,\delta$ are externally tangent at $ N$. Points $ A,C$ are situated in the same half-plane, determined by $ d$. Prove that if exists an circle, which is tangent to the circles $ \alpha,\beta,\gamma,\delta$ and contains them in its interior, then lines $ AC,BD,MN$ are concurrent or parallel.

nice and simple problem:

it's equivalent to prove that $ \frac{\overline{AM}}{\overline{BM}}\equal{}\frac{\overline{CN}}{\overline{DN}}$ and it's equivalent to this:

Let a cicle $ \Gamma$ with center O and radius R, a chord EF of $ \Gamma$ with distance $ d$ from O and a point P on EF. Let the 2 circle $ \Gamma_a$ and $ \Gamma_b$ tangent to EF on P and tangent respectively to the minor and bigger arch EF and with radius respectively $ R_a$ and $ R_b$. Then $ \frac{R_b}{R_a}$ not dipend from choice of P.

Call A and B the center of $ \Gamma_a$ and $ \Gamma_b$. Call K the projection of O on AB.

Then we have:

$ \overline{OA}^2 \minus{} \overline{AK}^2 \equal{} \overline{OB}^2 \minus{} \overline{BK}^2$

$ (R\minus{}R_a)^2 \minus{} (R_a \plus{} d)^2 \equal{} (R\minus{}R_b)^2 \minus{} (R_b \minus{} d)^2$

$ \frac{R_b}{R_a} \equal{} \frac{R\plus{}d}{R\minus{}d}$

[hide="Solution"]
Let $ \omega$ be the circle tangent to $ \alpha$, $ \beta$, $ \gamma$, and $ \delta$ and let its center be $ O$.  Also, let the tangency points between $ \alpha$, $ \beta$, $ \gamma$, and $ \delta$ with $ \omega$ be $ W$, $ X$, $ Z$, and $ Y$ respectively.  Let $ XM$ and $ YN$ meet at $ T$ and $ WM$ and $ ZN$ meet at $ U$ and let $ MN$ meet $ \omega$ at $ P$ and $ Q$ so that $ MP < NP$ and $ NQ > MQ$.  Consider the homothety around $ X$ that maps $ \beta$ to $ \omega$.  Let it map $ d$ to $ d'$.  Notice that since $ d$ is tangent to $ \beta$, we have that $ d'$ is tangent to $ \omega$.  Let the tangency point between $ d'$ and $ \omega$ be $ T'$.  Also, consider two points $ K$ and $ L$ on $ d'$ so that $ L$ is on the same half-plane as $ A$ and $ B$ and $ K$ is on the same half-plane as $ D$ and $ C$.  We have that $ \angle T'QP \equal{} \angle LT'P$ because of the tangency between $ \omega$ and $ d'$.  We also have that $ \angle T'PQ \equal{} \angle LT'P$ since $ d'\parallel PQ$.  Thus, $ \angle T'PQ \equal{} \angle T'QP$, so $ T'$ is the midpoint of arc $ PWQ$.  Therefore, $ XT'$ passes through the midpoint of arc $ PWQ$, so $ XM$ passes through the midpoint of this arc because of the homothety.  Similarly does $ YN$, so $ T \equal{} T'$, so $ T$ is the midpoint of arc $ PWQ$.  Similarly, $ U$ is the midpoint of arc $ PXQ$.  Thus, we have that $ \angle TQP \equal{} \angle TPQ \equal{} \angle TYQ$ and $ \angle NTQ \equal{} \angle YTQ$, so $ \triangle NTQ\sim \triangle QTY$.  Thus, $ \frac {NT}{QT} \equal{} \frac {QT}{TY}\implies QT^2 \equal{} TY\cdot NT$.  Similarly, we can find that $ TM\cdot TX \equal{} TP^2 \equal{} QT^2 \equal{} TY\cdot NT$ and $ UM\cdot UW \equal{} UP^2 \equal{} UQ^2 \equal{} UN\cdot UZ$.  Now, the homotheties imply that $ \frac {MX}{XT} \equal{} \frac {MB}{OT}$ and $ \frac {WM}{WU} \equal{} \frac {AM}{OU} \equal{} \frac {AM}{OT}$.  Thus, $ \frac {AM}{MB} \equal{} \frac {WM}{MX}\cdot \frac {XT}{WU}$.  Similarly, we can find that $ \frac {CN}{DN} \equal{} \frac {ZN}{NY}\cdot \frac {TY}{ZU}$.  Notice that since $ WTUZ$ and $ TZYU$ are cyclic, we have that $ \frac {WM}{MX} \equal{} \frac {MT}{MU}$ and $ \frac {ZN}{NY} \equal{} \frac {NT}{NU}$.  This means that $ \frac {AM}{MB} \equal{} \frac {MT}{MU}\cdot \frac {XT}{WU} \equal{} \frac {TP^2}{UP^2}$ and
\[ \frac {CN}{DN} \equal{} \frac {NT}{NU}\cdot \frac {TY}{ZU} \equal{} \frac {QT^2}{UQ^2} \equal{} \frac {TP^2}{UP^2} \equal{} \frac {AM}{MB}
\]
Now, $ AM\perp MN$ since and $ CD\perp MN$ since $ \alpha$ and $ \beta$ are tangent to $ MN$ at $ M$ and $ \gamma$ and $ \delta$ are tangent to $ MN$ at $ N$.  Thus, $ AB\parallel CD$.  Also, we have just shown that $ \frac {AM}{MB} \equal{} \frac {CN}{DN}$.  From this, it is well known that $ AC$, $ MN$, and $ BD$ must concur. [/hide]


induction
number theory proposed
number theory

Given $2n-1$ irrational numbers, prove that it is possible to select $n$ numbers $x_{1},...,x_{n}$, such that if $a_{1},...,a_{n}$ are arbitrary non-negative rational numbers that are not all $0$ then $a_{1}x_{1}+...+a_{n}x_{n}$ is irrational.
It is cool,isn't it? :)  :)

Yeah, it's cool. It's also on the forum, since I distinctly remember solving it, but I enjoyed thinking about it again :).

We prove the ssertion by induction on $n$. The base step ($n=1$) is obvious, so we assume it's proven for all $n$'s smaller than the current one. We say that a set of numbers in our list with the proeprty required in the conclusion has property $(P)$.

By the induction hypothesis, we ca find $n-1$ numebrs in our list, say $x_1,\ldots,x_{n-1}$, with $(P)$. Suppose now that no $n$-element subset of our list has $(P)$. This means that for $j=\overline{n,2n-1}$ we can find $\alpha_j^i,\ i=\overline{1,n-1},\ \beta_j\in\mathbb Q_+$, not all zero, such that $\sum_{i=1}^{n-1}\alpha_j^ix_i+\beta_jx_j\in\mathbb Q\ (*)$. Clearly, all the $\beta_j$'s are actually positive. By our assumption, we can find non-negative rationals (not all zero) $\gamma_j,\ j=\overline{n,2n-1}$ such that $\sum_{j=n}^{2n-1}\gamma_jx_j\in\mathbb Q$. This, together with the relations $(*)$, gives a relation of the type $\sum_{i=1}^{n-1}\delta_ix_i\in\mathbb Q$ for non-negative rational $\delta_i$ (not all zero), which contradicts our choice of the $x_i$'s.


calculus
integration
function
logarithms
complex analysis
complex analysis unsolved

I want to solve a definite integral in my research work.

integration from -infinite to +infinite
exp(-ax^2) / (b + exp(-x))

there are infinite # of residues for the function, so i don't know how to deal with it. Do you think if it has a analytical solution?

Thank you!

It goes without saying that $b>0,$ right?

A first thought: what if you use a contour that consists of most of the real axis, most of the line $z=x+2\pi i$, with connecting crosspieces off in the distance.

This would trap exactly one pole, the one at $-\ln b+\pi i.$

On the real axis, the integral would be

$\int_{\infty}^{\infty}\frac{e^{-ax^{2}}}{b+e^{-x}}\,dx$

On the upper line, the integral would be

$\int_{\infty}^{\infty}\frac{e^{-ax^{2}}e^{-2\pi iax}e^{2\pi a^{2}}}{b+e^{-x}}\,dx.$

That's not quite the same and it is not at all clear how to proceed from here - but does that give you any ideas?


inequalities
Cauchy Inequality
inequalities unsolved

[quote="Twins"][quote="arqady"]If $x>0$ and $0<y<1$ then $x^{y}>\frac{x}{x+y}.$ :wink:[/quote]
Can you post how to prove it?[/quote]
$x^{y}>\frac{x}{x+y}\Leftrightarrow x^{y}+yx^{y-1}>1.$
Let $f(x)=x^{y}+yx^{y-1}.$ Then $f'(x)=yx^{y-2}(x+y-1).$
Hence, $x_{min}=1-y.$
Thus, $f(x)\geq f(1-y)=(1-y)^{y-1}>1.$

[quote="Twins"][quote="arqady"][quote="Twins"]$a,b,c>0$. Prove that $a^{b+c}+b^{c+a}+c^{a+b}>1$[/quote]
If $x>0$ and $0<y<1$ then $x^{y}>\frac{x}{x+y}.$ :wink:[/quote]
Can you post how to prove it?[/quote]

we can suppose that $x<1$ and $y<1$ cause the other case is obvious
then study $f_{y}(x) =yln(x)-ln(x)+ln(x+y)$
$f_{y}(x)$ is decreasing $f_{y}(x)>f_{y}(1)=ln(1+y)>0$
then conclude.

[quote="kenshin_x"][/quote][quote="Twins"][quote="arqady"][quote="Twins"]$a,b,c>0$. Prove that $a^{b+c}+b^{c+a}+c^{a+b}>1$[/quote]
If $x>0$ and $0<y<1$ then $x^{y}>\frac{x}{x+y}.$ :wink:[/quote]
Can you post how to prove it?[/quote][quote="kenshin_x"]

we can suppose that $x<1$ and $y<1$ cause the other case is obvious
then study $f_{y}(x) =yln(x)-ln(x)+ln(x+y)$
$f_{y}(x)$ is decreasing $f_{y}(x)>f_{y}(1)=ln(1+y)>0$
then conclude.[/quote]
how do u know it's decreasing?

See my proof.  :D  :D
WLOG, we only need to consider the case $a,b,c \le 1$, we have two cases
+, If $a+b+c \le 1$, then by Bernoulli inequality, we have
\[\frac{1}{a^{b+c}}\le 1+\frac{(b+c)(1-a)}{a}\le \frac{a+b+c}{a}\]
Hence
\[a^{b+c}\ge \frac{a}{a+b+c}\]
Similar for $b,c$, we have the result.
+, If $a+b+c \ge 1$, also by Bernoulli inequality, we have
\[\frac{1}{a^{b}}\le 1+\frac{b(1-a)}{a}=\frac{a+b(1-a)}{a}\]

\[\frac{1}{a^{c}}\le 1+\frac{c(1-a)}{a}=\frac{a+c(1-a)}{a}\]
Therefore,
\[a^{b+c}\ge \frac{a^{2}}{(a+b(1-a))(a+c(1-a))}\]
We need to prove
\[\sum \frac{a^{2}}{(a+b(1-a))(a+c(1-a))}\ge 1 \]
By Cauchy inequality, it suffices to prove
\[(a+b+c)^{2}\ge \sum(a+b(1-a))(a+c(1-a)) \]
Or
\[(ab+bc+ca)(a+b+c-1)+abc(3-a-b-c) \ge 0 \]
Which is true. The inequality is proved. :)


inequalities
search
inequalities unsolved

$x,y,z > 0$
can we prove
$(x+y+z)^2(\frac 1{x^2+y^2}+\frac 1{y^2+z^2}+\frac 1{z^2+x^2}) \geq 10$
If it is right,it is nice.
if it is wrong,find the minimum of $LHS$

set x+y+z=2
first we prove x^2*y^2 + y^2*z^2 + z^2*x^2<=1
set x>=y>=z
L.H.S=x^2*(y+z)^2 + y^2*z^2-2yz*x^2<=x^2*(y+z)^2<=1

we know another famous inequality:
if a,b,c are non-negative reals. ab+bc+ca=1
then 1/(a+b) + 1/(b+c) + 1/(c+a) >=5/2
obviously it also holds when ab+bc+ca<=1.

This is the proof.

[quote="zhaobin"]$x,y,z > 0$
can we prove \[ (x+y+z)^2(\frac 1{x^2+y^2}+\frac 1{y^2+z^2}+\frac 1{z^2+x^2}) \geq 10 \]
If it is right,it is nice.
if it is wrong,find the minimum of LHS[/quote]

write it as $\sum \frac{(x+y+z)^2}{x^2+y^2} \geq 10$.

it's equivalent to $\sum \frac{x^2+2yz+2zx+2xy}{y^2+z^2} \geq 7$.

use a famous ineq (from vasc i think), $\sum \frac{x(y+z)}{y^2+z^2} > 2$.

we are left with $\sum \frac{x^2+2yz}{y^2+z^2} > 3$. this seems true, but i can't prove it right now. i remembered that vasc had an ineq similar to this, but it was with $\sum \frac{2x^2+yz}{y^2+z^2}$ :(

try prove the last ineq, zhaobin !

thanks.everybody who know wei_wenzhe9170's meaning please explain,
thanks again

[quote="perfect_radio"]$\sum \frac{x^2+2yz}{y^2+z^2} > 3$[/quote]

did anyone manage to prove it?

It is too hard for me  :?  :(  :oops:  :blush:

Using the idea of Perfect_Radio, the desired inequality can be obtained by summing the inequalities:
$2\sum\frac{a(b+c)-bc}{b^2+c^2} \ge 3$;
$\sum 1 \ge 3$;  ;)
${\sum \frac{a^2+4bc}{b^2+c^2} \ge 4}$.
The first and the third ineq. have been posted on this forum long ago. Actually, the last ineq. is a particular case
of the inequality
${\sum \frac{a^2+qbc}{b^2+c^2} \ge 2+q/2}$,
where $1/2 \le q \le 16$.
Nice inequality, Zhaobin! :lol:

[quote="Vasc"]The first and the third ineq. have been posted on this forum long ago. Actually, the last ineq. is a particular case
of the inequality
${\sum \frac{a^2+qbc}{b^2+c^2} \ge 2+q/2}$,
where $1/2 \le q \le 16$.[/quote]

I was sure that will come with an excellent generalization (btw, is there an inequality to which you couldn't find a generalization?). For $q=4$, we get the inequality you used, and for $q=2$ we get the inequality I can't prove  :P

How long ago was this ineq discussed? I didn't find it  ;)

I think the generalization Vasc mentioned is posted somewhere in proposed problem section. In fact at that time the minimum of $\sum \frac{a^2+pbc}{b^2+c^2}$ was discussed, but I forgot where the post is...

BTW, does anyone have some topics trackers or whatever? It is difficult for me to search for the correct topic....

[quote="perfect_radio"]If you want, you can add the important links to favorites.

And i also think that it is very hard to find topics concerning some particular inequality. I just don't know what keywords to use[/quote]

In fact I have added around 50 already. But you know, very often I have to find some topics that I am not that interested.. and if I add all of them.... maybe I have to add 100-200 links.. :?  :(

thank you all.
Vasc,can you give me the links,
i can't find it.it is tooooooo hard

This is my first reply in this forum
I suppose this inequality is not that difficult,especially for Chinese student?
Let's suppose x>=y>=z and x+y+z=1
F(x,y,z)=Left hand side
F(x,y,z)>=F(x+z/2,y+z/2,0)
For x+y=1,1/(x^2+y^2)+1/x^2+1/y^2>=10 is trivial(You can even use Brutal Force)

[quote="Soarer"]http://www.mathlinks.ro/Forum/viewtopic.php?t=35502[/quote]
thanks but what about the first one?


We have now listed the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]Special AIME Problem Seminar[/url] for enrollment.  The seminar will take place on Saturday and Sunday, February 26 and 27 from 3:30 PM to 6:30 PM each day with a half hour break during each session.  This year the seminar will also include a full practice AIME exam and brief solution sketches for each problem.

Will they be the same problems as last year?

It's quite a bit less than the normal classes. I'm not sure about this year's rates though.

Sorry, I botched the cost field due to a new system we have for assigning database values.  Oops.

The cost for the seminar is 40 dollars.

[quote="rrusczyk"]They will not be the same questions as last year.[/quote]

Also, none of the questions will be from the current AIME Problem Series.

Can I enroll after AMC has ended, say like Feb. 20th or something?

I don't want to enroll unless I had qualified for AIME (I WISH@!!!).

Thank you very much.

We list enrollment deadlines for all of our classes on the enrollment page.  For this one it's Feb 26, so you can register right up until class time.

That's not during class is it? (the mock AIME)

No.  We'll post it after the second class.  Students can take whichever 3 hours to work on it that they choose.  Obviously nobody has any real incentive to cheat.  We'll post solution sketches a couple of days after that in a document in the My Classes area.

[quote="MithsApprentice"]Right...very good idea that. *nods*

Now is there going to be a board for this class?[/quote]

It says on the Enrollment page that there will not be a board.

We'll change that - there will be a board.

We have decided that there will be a Message Board.  There won't be Message Board problems, but we will post the Mock AIME there for discussion.

i'm sure anyone who hits the wall at the AIME would benefit from the class. the class would also be good for people who can do at least 1 or 2, as well as 3 to 6, in my opinion

[quote="h_s_potter2002"]What level is the seminar inteneded for?  For example, would someone who can't answer problems benifit from the seminar, or someone who gets about a 3-6 score range?[/quote]

A person who cannot answer any at all, but can make it to the AIME might benefit a little, but a student like that should consider taking the AMC 10 and then AMC 12 problem series and/or our Introductory classes.  If a student soaks up our Intro classes, they should be ready to get at least 2 or 3 problems correct on the AIME.

The student who will benefit the most from this seminar are those who might make 2 to 12 on the AIME.  The problems vary greatly in range, many techniques are discussed, and the solutions presented are often helpful even to a student who might already get the problem correct because they learn a new method or quicken their problem solving pace.

[quote="Silverfalcon"][quote="MCrawford"]All originals.[/quote]

Does that mean real problems from past AIME?

 ;) I thought you were making them. Like getting ideas from the past ones.[/quote]

The mock AIME problems are not from old AIMEs.  We wrote most of them.

[quote="ml2"]Will class questions be old AIME or something else?[/quote]

Most will be from old AIMEs, some will be from other sources.

You said that the Mock AIME would be posted after the  second class, but it is already in the handouts section... is that supposed to happen?  Should we take the Mock AIME before or after we had the course (or both)?

We changed our mind on that - you can take the Mock AIME now or wait until after the classes; your choice!

[quote="Silverfalcon"][quote="MCrawford"]All originals.[/quote]

Does that mean real problems from past AIME?

 ;) I thought you were making them. Like getting ideas from the past ones.[/quote]Actually some of the problems are really original :) (ie created by us for this seminar :) ).


calculus
derivative
function
integration
real analysis
real analysis theorems

Can anyone help me find the derivative of the function

$ g(x) \equal{} \int^{1}_{0} max(t,x)f(t)dt$

first split up g(x)=integral over [0,x] of xf(t)dt + integral over [x,1] of tf(t)dt
then differentiate wrt x and get
g'(x)=(x(f(x))'+xf(x)=(x+1)f(x)+xf'(x)

eh?

boy... definitely needs casework

if $ 0<x<1$ then $ g(x) \equal{} x\int_0^x f(t) dt \plus{} \int_x^1 tf(t) dt$ so $ g'(x) \equal{} \frac{d}{dx}\bigg[x\int_0^x f(t) dt \plus{} \int_x^1 tf(t) dt\bigg]$.  Note that since the $ x$ is in the limits of the integrals, we cannot simplify it further without knowing something about the function.

if $ x>1$ then $ max(x,t)\equal{}x$ so $ g(x) \equal{} x \int_0^1 f(t)dt$ and hence $ g'(x) \equal{} \int_0^1 f(t)dt$.

if $ x<0$ then $ max(x,t)\equal{}t$ so $ g(x) \equal{} \int_0^1 tf(t) dt$ and thus $ g'(x)\equal{}0$


induction
inequalities
algebra unsolved
algebra

I did spend some minutes searching for this but I couldn't find anything :( Though a lot of questions with the Harmonic series has come up :P

Anyways,

Prove $ \forall n \in \mathbb{N}\geq 2, \sum \limits_{i\equal{}1}^{n} \frac{1}{n} \neq k \in \mathbb{N}$

So although the series diverges, it is never an integer except its first sum :)

The standard proof is showing that its [url=http://mathworld.wolfram.com/p-adicNorm.html]$ 2$-adic norm[/url] is greater than $ 1$.  A simple argument:

Write as $ \frac {\sum \frac {n!}{i} }{n!}$.  Among the terms $ \frac {n!}{i}$, there is a unique one with largest $ 2$-adic norm (which will occur where $ i$ is the greatest power of $ 2$ less than $ n$).  When we divide out by the appropriate power of $ 2$ to make that term odd, the other terms will be even but the denominator will also be even.  Hence the numerator is odd.

Similar ideas can be used to show that the partial sums of $ \frac{1}{1} \plus{} \frac{1}{3} \plus{} \frac{1}{5} \plus{} ...$ are never integer.  (We use the $ 3$-adic norm.)

hi this is my solution and i will be so glade if we discussed ,

supposing the exesting of $ n'$ such $ S(n')  \in \mathbb{N}$

so $ S(n')\equal{}1\plus{}(1/2)\plus{}..........\plus{}(1/n') \in \mathbb{N}$

we know that $ n!\equal{}p1^{a1}p2^{a2}.......pk^{ak}$  , $ pi$ is prime number

let take $ pi\equal{}max(pk)$ and $ m\equal{}\frac{n!} {pi^{ai}}$   don't forget that  $ (ai\equal{}min(ak))$

$ m*S(n') \in \mathbb{N}$ so $ m\plus{}\frac{m} {2} ........\plus{}\frac{m} {pi} ......\plus{}\frac{m} {n'}  \in \mathbb{N}$

that mean $ \frac{m} {pi} \in \mathbb{N}$ and this is impossible so $ n'$ DOES NOT exist

[size=150]aplication[/size]

$ S(6)\equal{}1\plus{}(1/2)\plus{}(1/3)\plus{}(1/4)\plus{}(1/5)\plus{}(1/6)$

$ 6!\equal{}1 *2^{4}*3^{2}*5$ and $ m\equal{}(6!/5)$ so $ m\equal{}144$

$ 144S(6)\equal{}144\plus{}72\plus{}48\plus{}36\plus{}(144/5)\plus{}24  \in \mathbb{N}$ ==> $ (144/5) \in \mathbb{N}$ impossible

I read a while ago there was a proof for this using combinatorics. I didn't understand it so I don't remember anything :(

You can prove it by mathematical induction, actually there are other ways that I am going to try this evening but here we do it by applying the principle of mathematical induction:

suppose $ S(n)$ is the statement which we are going to prove and that is:

$ S(n): \forall n\in\mathbb{N}\geq2,\qquad\sum_{i \equal{} 1}^n \frac {1}{n}\not \equal{} k\in\mathbb{N}$

$ S(2): \sum_{i \equal{} 1}^2\frac {1}{n} \equal{} 1 \plus{} \frac {1}{2} \equal{} \frac {3}{2}$ which is clearly true because we know that $ \frac {3}{2}$ is not a natural number.

Let's suppose that $ S(n)$ is true for some $ m\in\mathbb{N}$

$ S(n): \sum_{i \equal{} 1}^m\frac {1}{m}\not \equal{} P,\qquad\text{where }P\text{ is some natural number.}$

let's add $ \frac {1}{m \plus{} 1}$ to both sides of the expression above so we have

$ \sum_{i \equal{} 1}^m\frac {1}{m} \plus{} \frac {1}{m \plus{} 1}\not \equal{} p \plus{} \frac {1}{m \plus{} 1}$

Now I know that if there exist another natural number say $ Q$, then $ p \plus{} \frac {1}{m \plus{} 1}$ is clearly not a natural number and so we have $ Q\not \equal{} P \plus{} \frac {1}{m \plus{} 1}$.

Now we know that $ \sum_{i \equal{} 1}^{m \plus{} 1}\frac {1}{m}\not \equal{} P \plus{} \frac {1}{m \plus{} 1}$ and also have that $ P \plus{} \frac {1}{m \plus{} 1}\not \equal{} Q$ So we can obviously conclude that $ \sum_{i \equal{} 1}^{m \plus{} 1}\frac {1}{m}\not \equal{} Q$

Hence by applying the principle of mathematical induction, the statement $ S(n)$ is proved.

I am currently at uni, but when I get home, I will post some other approaches as well.

[quote="t0rajir0u"][quote="toosi"]Now we know that $ \sum_{i \equal{} 1}^{m \plus{} 1}\frac {1}{m}\not \equal{} P \plus{} \frac {1}{m \plus{} 1}$ and also have that $ P \plus{} \frac {1}{m \plus{} 1}\not \equal{} Q$ So we can obviously conclude that $ \sum_{i \equal{} 1}^{m \plus{} 1}\frac {1}{m}\not \equal{} Q$[/quote]
Er... except inequality isn't transitive.  The correct implication is $ a \neq b, b \equal{} c \implies a \neq c$.  Proving that something [i]isn't[/i] true is rarely something you can prove inductively without strengthening the inductive hypothesis.[/quote]

What It really meant was that we know that $ \sum_{i\equal{}1}^{m\plus{}1}\frac{1}{m}\not\equal{}p\plus{}\frac{1}{m\plus{}1}$, we know that $ p\plus{}\frac{1}{m\plus{}1}$ can not be a natural number therefore $ \sum_{i\equal{}1}^{m\plus{}1}\frac{1}{m}$ also can not be a natural number.

Er... if something isn't something that can't be a natural number, that says nothing about whether the original quantity can be a natural number or not.  For example, $ 3 \neq 5.5$ and $ 5.5$ is not a natural number; but $ 3$ is obviously a natural number.

there is an easy prof. Find all the numbers, less than N, which is the form $ p^n$, where p is prime. There must be some one since N is not 1. Among them there is a biggest one, say $ q^n$. No Other numbers less than N contains nth or more than nth powers of q. So the numerator of $ \sum_{1}^{N}\frac{1}^{n}$ is kq+1, but the denominator of it is $ q^n$


Ummm.. i didn't go to the math jam.. but when i actually took the AIME, i pretty much brute forced number six until i found a pattern (which didn't take very long at all) and then.. by the pattern, i was able to solve it...

beta had a really nice explanation for #4: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11457&postdays=0&postorder=asc&start=15#76991[/url]

For #4, just get out a ruler...draw a 2X2 (inches) square...and put the 0" mark on one side and the 2" mark on another side, now draw a point at the 1" mark...shift the ruler a bit (but make sure the 0" and 2" marks are still on those sides) and make another mark at the 1" point...do this a few times at different points, connect the dots, and u'll know what we're talking about...if u dont, IM me and I'll get u a diagram...

#6
for a,b,c,d that 1 :le: a <b<c<d :le: 9, there are 5 ways to arrange them like a snakelike number(acbd, adbc, bcad, bdac, cdab). There are total 9C4 number of (a,b,c,d). The total number of snakelike number is 5(9C4). If one of a,b,c,d is 0, it has to be in a_3 position.

For number p,q,r that 1 :le: p<q<r :le: 9,there are 3 ways to arrange them to be a snakelike with a_3 is 0(pq0r, pr0q, qr0p).
Total number of snakelike numbers with one digit 0 is 3(9C3)
5(9C4)+3(9C3)=882


Given any three-digit number $ x$, define $ A(x)$ to be $ x$ minus the sum of the squares of the digits of $ x$. What is the maximum possible value of $ A(x)$?

[hide="Hint"]Let $ x \equal{} 100p \plus{} 10q \plus{} r$. Then, $ A(x) \equal{} (100p \plus{} 10q \plus{} r) \minus{} (p^2 \plus{} q^2 \plus{} r^2) \equal{} (100p \minus{} p^2) \plus{} (10q \minus{} q^2) \plus{} (1 \minus{} r^2)$[/hide]

Thanks TZF: [hide=" answer"]

$ A(x) \equal{} 844$, for $ x\equal{}950$ or $ 951$:

Looking at the graphs of $ 100p\minus{}p^{2} \equal{} p(100\minus{}p), 10q\minus{}q^{2} \equal{} q(10\minus{}q)$ and $ r\minus{}r^{2} \equal{} r(1\minus{}r)$,

we see that, since $ 0 \le x\le 9$, this is a maximum for $ p\equal{}9$, $ q\equal{}5$, and $ r\equal{} 0$ or $ 1$.  [/hide]


calculus
integration
function
logarithms
real analysis
real analysis unsolved

An integral $ \int_0^\infty \frac{dx}{\Gamma (x)}$ seems to converge (but I'm not sure)... so is it possible to express it in nicer form (using math constants, etc.)? :huh:

It doesn't converge, because it's comparable to $ \frac1x$ near zero.

Well... near 0 $ \frac{1}{\Gamma (x)}$ is like $ x$.

We have: $ \frac{1}{\Gamma (x)}\equal{}xe^{\gamma x}\prod_{n\equal{}1}^{\infty} \left(1\plus{}\frac{x}{n}e^{\minus{}x/n}\right)$
From this formula $ \frac {1}{\Gamma (x)} \equal{} x \plus{} \gamma x^2 \plus{} O(x^3)$ near $ x \equal{} 0$.
Value of integral is $ \approx 2.807770242$.

Oops- I got $ \Gamma$ and its reciprocal confused there. Near $ \infty$, of course, it converges easily.

There is a formula due to Ramanujan, whose particular case is
\[ \int\limits_{0}^{\infty}\frac{dx}{\Gamma(x)}=e-\int\limits_{0}^{\infty}\frac{e^{-t}dt}{t(\log^2(t)+\pi^2)}.
\]
It can be derived by applying
\[ \frac{1}{\Gamma(s)}=\frac{1}{2\pi{i}}\int\limits_{\gamma}e^{u}u^{-s}du,
\]
where the contour $ \gamma$ goes along the ray $ \arg(u)=-\alpha$, $ |u|\geq{\epsilon}$, along the arc $ -\alpha<\arg(u)<\alpha$, $ |u|=%Error. "eplison" is a bad command.
$ and finally along the ray $ \arg(u)=-\alpha$. the angle $ \alpha$ can be chosen from $ \pi/2$ to $ \pi$ and $ \epsilon>0$ is arbitrary.

According to [url=http://en.wikipedia.org/wiki/Frans%C3%A9n-Robinson_constant]Wikipedia[/url] $ F \equal{} \int_{0}^\infty \frac {1}{\Gamma(x)}\, dx$ equals $ e \plus{} \int_0^\infty \frac {e^{ \minus{} x}}{\pi^2 \plus{} \ln^2 x}\, dx$ .

Maybe you meant this formula.

$ I: \equal{} \int_{0}^{\infty}\frac {e^{ \minus{} t}dt}{t(\log^2(t) \plus{} \pi^2)}$ is positive, hence $ F \equal{} 2.80777 > e > e \minus{} I$ .

What is the general formula due to Ramanujan?

[quote="http://en.wikipedia.org/wiki/Fransén-Robinson_constant"]It is however unknown whether F can be expressed in closed form in terms of other known constants.[/quote]
So F is an answer.
Thanks [b]Doe John[/b] for your link.

Sorry for this long delay, but anyway the correct Ramanujan's formula is
\[ \int\limits_{0}^{\infty}\frac{z^tdt}{\Gamma(1\plus{}t)}\equal{}e^{z}\minus{}\int\limits_{0}^{\infty}\frac{e^{\minus{}z\tau}d\tau}{\tau(\log^2(\tau)\plus{}\pi^2)},\,\,\, Re(z)\ge{0},
\]
so it is clear where the mistake was in my first incorrect version.


[hide]a square is a parallelogram because it has 2 pairs of parallel sides.[/hide]

[quote="236factorial"]Is a square a parallelogram? explain.[/quote]

a square is a parallelogram. A parallelogram is a figure where 2 sides are parallel and there are 2 sets of them. and the same with a square, except tha tin a square, all 4 sides are parallel.

[quote="236factorial"]Is a square a parallelogram? explain.[/quote]

[hide]Yes, a square is in fact a parallelogram.
Not all parallelograms are squares.

A square is a quadrilateral that has 4 right angles and it's sides are all equal in length.
A parallelogram is any given figure with two pairs of parallel sides.

A square has two pairs of parallel sides; since all the angles are 90 degrees.

Meanwhile, not all parallelograms are squares.

Why?

Because not all parallelograms have 4 90 degree angles. Not all parallelograms have four sides equal in length. That's just like a square and a rectangle. :D[/hide]

[hide][quote="236factorial"]Is a square a parallelogram? explain.[/quote]

yes because a paralellogram is a four sided polygon in which the opposite sides are parallel.  a square is the same with the extra requirements that the four sides have to be of equal length and all four angles must be 90 degrees[/hide]

[hide]
Yes, because a square has two parallel sides [/hide]

i think you really dont understand the meaning of square and parallelogram ,it is good that you ask this basic question but the meaning of something like that should be in your heart not only in your mind! :idea:  (or may be we think this is asimple and easy question but in fact it is not.)

[hide]a square is a parallelogram because it has 2 pairs of parrellel sides :) [/hide]

a square is a parallelogram with 4 right angles and equal sides

a square is a parallelogram cuz opposite sides are equal and parallel.
it also has 4 sides.just in case you didn't know. :rotfl:

it is a parallogram. its four sides are equal


geometry
geometric transformation
reflection
circumcircle
parallelogram
perpendicular bisector
geometry proposed

Let $ AX$ be a diameter of $ k \equiv (O),$ let $ AC, XD$ meet at $ Y,$ let $ CD$ meet $ AO$ at $ P.$  $ C, D$ are A-, X-altitude feet of the $ \triangle AXY,$ $ B$ is the Y-altitude foot, because $ \angle ABD \equal{} \angle XBC.$ Therefore, $ \frac {\overline{PX}}{\overline{PA}} \equal{} \minus{} \frac {\overline{BX}}{\overline{BA}} \equal{} 3,$ $ \overline{PA} \equal{} \frac {\overline{PX}}{3} \equal{} \frac {\overline{PA} \plus{} \overline{AX}}{3},$ $ \overline{PA} \equal{} \frac {\overline{AX}}{2} \equal{} \overline{AO}.$ The angle $ \angle OEP$ is right and $ \overline {PO} \equal{} \overline {AX}$ $ \Longrightarrow$ $ E$ is on the reflection $ (O')$ of $ (O)$ in the perpendicular bisector of $ AO$ through $ B.$ This perpendicular bisector, the Y-altitude $BY$ of the $ \triangle AXY,$ bisects the $ \angle CBD.$ Therefore, reflection $ E'$ of $ E$ in $ BY$ is on $ (O)$ and $ BE, BE'$ are median and symmedian of the $ \triangle BCD,$ the segments $ BE' \equal{} BE$ are equal. If $ C', D' \in (O)$ are reflection of $ C, D$ in $ AO,$ then $ BF$ is symmedian of the oppositely congruent $ \triangle BC'D'.$ Therefore, $ BF$ is a reflection of $ BE'$ in $ AO$ and $ BF \equal{} BE' \equal{} BE.$

[quote="delegat"]Given is a circle $ k$ with center $ O$. Denote by $ B$ midpoint of radius $ OA$. On circle $ k$ take two points $ C$ and $ D$, on same side with the respect to $ OA$, such that $ \angle ABD \equal{} \angle OBC$.

Denote by $ E$ midpoint of $ CD$ and let $ F$ be an intersection point of line $ EB$ wih $ k$. Prove that $ B$ is midpoint of segment $ EF$.[/quote]
[hide="Solution"]
Reflect $ O$ into $ A$ to form $ X$, let the circumcircle of $ \triangle BCD$ meet $ OA$ at $ O'$, and let the midpoint of $ OE$ be $ P$.  Notice that $ \angle O'DC \equal{} \angle OBC \equal{} \angle O'BD \equal{} \angle O'CD$, so $ O'C \equal{} O'D$, so $ O'$ is where the perpendicular bisector of $ CD$ meets $ AO$.  Yet, this is $ O$ since $ O$ is the center of $ k$.  Thus, $ OBCD$ is cyclic.  Notice that the power of $ X$ with respect to the circle is $ XO^2 \minus{} R^2 \equal{} 3R^2$, where $ R$ is the radius of $ k$.  Yet, $ XB\cdot XO \equal{} \frac {3R}{2}\cdot 2R \equal{} 3R^2$, so if $ D'$ is the intersection between $ XC$ and $ k$, we would have that $ CD'OB$ is cyclic since $ XC\cdot XD' \equal{} 3R^2 \equal{} XB\cdot XO$, meaning that $ D \equal{} D'$, so $ X$ is the intersection between $ AO$ and $ CD$.  This means that $ A$ is the midpoint of $ XD$, so $ AP\parallel XE$, so $ PO \equal{} PE$.  Furthermore, the parallelism gives that $ AP\perp XE$ since $ OE\perp XE$.  This means that $ AE \equal{} OA \equal{} OF$.  Now, let $ F'$ be the point so that $ EAF'O$ is a parallelogram.  This means that $ F'O \equal{} EA \equal{} R$, so $ F'$ lies on $ k$ (on the opposite side of $ AO$ as $ E$.)  This implies that $ F \equal{} F'$ (since $ EB$ only meets $ k$ at one point that is not on the same side of $ OA$ as $ E$), so $ EOFA$ is a parallelogram, so we conclude that $ EB \equal{} FB$. [/hide]

Working little bit on [b]yetty[/b]'s: after having seen that $A$ is the midpoint of $OP$ and $OE\bot CP$ we get $AE=OA$ and, $AEOF$ being a parallelogram, $OF=AE=OA$ and F belongs to the circle $O$.

Best regards,
sunken rock


function
combinatorics unsolved
combinatorics

I needed to find the mobius function for the inversion poset shown in the picture below.  Does my graph below correctly list all the mobius functions?
[img]http://i220.photobucket.com/albums/dd160/jrb599/sc0028016d-1.jpg[/img]

Yes, except that many of the 0s should be "undefined"s -- $ \mu(x, y)$ is only defined if $ x \leq y$.


Let $ a,b,$ and $ c$ be positive real numbers

(i) Prove that $ 4(a^3 \plus{}b^3) \geq (a\plus{}b)^3$
(ii) Prove that $ 9(a^3 \plus{} b^3 \plus{} c^3) \geq (a \plus{} b\plus{}c)^3$

It's too easy if you know HOLDER'S inequality! Applying it and you get the result directly!

[hide="But just for fun..."] I'll demonstrate a straightforward way to do (i).  (ii) can be done similarly.  The given is equivalent to

$ 3a^3 \plus{} 3b^3 \ge 3a^2 b \plus{} 3ab^2$

which follows from the use of AM-GM twice:

$ a^3 \plus{} a^3 \plus{} b^3 \ge 3a^2 b$
$ a^3 \plus{} b^3 \plus{} b^3 \ge 3ab^2$.

Alternately, both problems follow pretty straightforwardly from Rearrangement. [/hide]

So we can (obviously) use Holder to generalize this to

\[ n^2(a_1^3\plus{}\ldots\plus{}a_n^3) \ge (a_1\plus{}\ldots\plus{}a_n)^3\]

[hide="Or..."]
Working backwards, this is equivalent to:
$ a^3\plus{}b^3\ge ab(a\plus{}b)$
$ a^2\minus{}ab\plus{}b^2\ge ab$
$ (a\minus{}b)^2\ge 0$.

Though that's probably the same thing as t0rajir0u's.
[/hide]

Jensen's works well too.

$ \frac {a^3 \plus{} b^3}{2}\ge\left(\frac {a \plus{} b}{2}\right)^3$

$ 4(a^3 \plus{} b^3)\ge (a \plus{} b)^3$

[quote="hpoincare"]I think this chain of inequalities is more precise in some way, hope you understand what I mean...

$ 9(a^3 \plus{} b^3 \plus{} c^3) \geq 3(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \geq (a \plus{} b \plus{} c)^3 \geq 27\ abc$

(1) $ 9(a^3 \plus{} b^3 \plus{} c^3) \geq 3(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)$ is true by Chebyshev

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

$ LHS \minus{} RHS \equal{} (a \plus{} b \plus{} c) \bigl ( (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2 \bigr ) \geq 0$

(3) $ (a \plus{} b \plus{} c)^3 \geq 27\ abc$ by AM-GM.[/quote]
Note that (2) is also a Chebyschev application.  :wink:

[quote]Note that (2) is also a Chebyschev application. Wink[/quote]

You are right Dijkschneier.

My intrerest was not in the proof in the chain of inequalities but to put somethin "between" $ 9(a^3\plus{}b^3\plus{}c^3)$ and $ (a\plus{}b\plus{}c)^3$.

 :)


The problem is find p'(0), if p(x) = f(g(x)).  How would I go about doing this?


Σε καθε τριγωνο ισχύει

$(h_a-2r)(h_b-2r)(h_c-2r)\leq r^3$ (συνήθεις συμβολισμοί) !

 :)

[quote="silouan"]Καλή άσκηση. Έχουμε ότι $h_a=\frac{2E}{a}$ όπου Ε το εμβαδό. Όμοια και για τα άλλα.
Επίσης $2r=\frac{2E}{t}$ όπου τ η ημιπερίμετρος. Όμοια και για τα άλλα. Αντικαθιστώντας κάνοντας απαλοιφή παρονομαστών και χρησιμοποιώντας το γεγονός ότι
$E^3=r^3t^3$  αρκεί νδο $(a+b-c)(c+b-a)(c+a-b)\leq abc$  η οποία είναι γνωστή.
Πώς σας φαίνεται?[/quote]

Very nice!!


Find all functions $f$ from $R$ to $R$ satisfying
              $f(x^{2}+xy+f(y)) = (f(x))^{2}+xf(y)+y$
for all $\large x,y \in R$.

I’m sorry about may poor English!!

the only solution is $f(x)=x$...

[b]proof:[/b]first of all assume that $f(0)=c$

we claim that $c=0$

[hide]put $x=0$ we get that:

$f(f(y))=c^{2}+y (I)$

so if we put $y=0$ we get that $f(c)=c^{2}$

now if we put $y=0$ in the original equation we get that $f(x^{2}+c)=f(x)^{2}+cx$

now in the last equation if we put $-x$ instead of $x$ the LHS doesnt change and the RHS becomes $f(-x)^{2}-cx$ so we get
that:$f(x)^{2}+2cx=f(-x)^{2}$ so if we put $x=c$ we get that $f(c)^{2}+2c^{2}=f(-c)^{2}(II)$

now in the original equation put $y=-x$ we get that $f(f(-x))=f(x)^{2}+xf(-x)-x$

according to $(I)$ we get that:

$c^{2}-x=f(x)^{2}+xf(-x)-x$

so $c^{2}=f(x)^{2}+xf(-x)$ so if we put $x=-c$ we get that $c^{2}=f(-c)^{2}-cf(c)$

now according to $(II)$ we get that:

$c^{2}=2c^{2}+f(c)^{2}-cf(c)$

so we get that $f(c)^{2}-cf(c)+c^{2}=0 (*)$ so assume that $p(x)=x^{2}-cx+c^{2}$ so we get that $\Delta_{p}=c^{2}-4c^{2}=-3c^{2}$

so we get that $\Delta_{p}\leq 0$ but if $\Delta{p}<0$ by $(*)$ we get that $f(c) \notin \mathbb{R}$ which is a contradiction
so we get that $\Delta_{p}=0$ so $-3c^{2}=0 \Rightarrow c=0$ so we get that $f(0)=0$[/hide]

------------------------------------------------------------------------------------

now assume that $f(0)=0$ by putting $x=0$ we get that $f(f(y))=y$ now if for a nonzero number $k$ we had $f(k)=0$ then we get that $f(f(k))=f(0)$ so we get that $k=f(0)=0$ contradiction so
for every nonezero number $k$ we get that $f(k) \not=0$

now by putting $y=0$ in the original equation we get that $f(x^{2})=f(x)^{2}$

now let $x \not= 0$ now if we put $-x$ instead of $x$ in the last one we get that $f(x)^{2}=f(-x)^{2}$ so we get that $f(x)=\pm f(-x)$ now if $f(x)=f(-x)$
then we get that $f(f(x))=f(f(-x))$ so $x=-x$ contradiction...

so $f(-x)=-f(x)$

now put $y=-x$ we get that $f(f(-x))=f(x)^{2}+xf(-x)-x$

so we get that $f(x)^{2}=-xf(-x)\Rightarrow f(x)^{2}=xf(x)$

so we get that $f(x)=0$ or $f(x)=x$ but the first one gives us a contradiction (because $x \not= 0$) so we get that $\forall x\in \mathbb{R}f(x)=x$...


Does anyone know of a really good calculus book that is not watered down with exercises and is not a super-long textbook?

Calculus by Michael Spivak is a pretty rigorous introduction to Calculus and the exercises are not "watered-down". They were considerably harder(the later ones especially) than ones we used in Calc BC class. The book is about 670 pages or so. It doesn't include multivariable stuff though.

Yes, Michael Spivak's [url=http://www.amazon.com/exec/obidos/tg/detail/-/0914098896/learninfreed]Calculus[/url] is the royal road to calculus, the one book that every calculus student should own. Another really good book is Apostol's Calculus, volumes [url=http://www.amazon.com/exec/obidos/tg/detail/-/0471000051/learninfreed]one[/url] and [url=http://www.amazon.com/exec/obidos/tg/detail/-/0471000078/learninfreed]two[/url].

A calculus book that isn't a monstrous textbook? Nope!  :D

the one i like is Calculus and Analytic Geomety by George Thomas Vol I, II.

[quote="amirhtlusa"]the one i like is [u]Calculus and Analytic Geomety[/u] by George Thomas Vol I, II.[/quote]

That is a very good textbook for an introduction to Calculus. I also like [u]Single Variable Calculus:  Concepts and Contexts[/u] by James Stewart.

Books like Spivak and Apostol are good, but not as stand-alone introductions to Calculus. They're good as supplements to textbooks like Thomas or Stewart.

Most Calculus textbooks are long because they merge single variable calculus and multivariable calculus into one book. But you can usually get just the single variable or multivariable parts as separate books.


Let (x,y) be defined as the GCF of x and y , where x and y are integers.

For positive integers a and b,

is (a+b, ab) always equal to (a,b)?

if so please give proof (i think i proved it, but i am very doubtful...)

tnx!  :)

$ 4$ and $ 6$ work.
$ GCD(4\plus{}6,(4)(6))\equal{}GCD(10,24)\equal{}2\equal{}GCD(4,6)$

As for the proof try to use division algorithm.

[quote="DonkeyKong"]NO.

Try 4 and 6. They don't work out.

EDIT: Never mind.

YES they all work, because if primes work, they all work.
Because primes are the components that makes the integers.

So YES[/quote]
That is not a proof,just to tell you.For example, if I said all primes can be expressed as $ 6n\pm{1}$, where the primes are >3,, that doesn't mean all integers can. As for the proof,
[hide="hint"]Note that: gcf|ab tells us that some factors $ a_1$ and $ a_2$ where $ a_1a_2 \equal{} gcf$, that $ a_1|a$ and $ a_2|b$.[/hide].
EDIT:The proof shouldn't be in Mathcounts.

[quote="t0rajir0u"]The statement is false.  For example, $ \gcd(10, 15) \equal{} 5$ but $ \gcd(10 \plus{} 15, 10 \cdot 15) \equal{} 25$.[/quote]Oh...You are right. Nice catch.

@xpmath: I know it wasn't a proof.. I couldn't come up with one, so I only posted explanations.,which happend to be a false.

[quote="DonkeyKong"]I couldn't come up with one, so I only posted explanations.,which happend to be a false.[/quote]

The statement is true if $ \gcd(a, b) \equal{} 1$ (a subcase: $ a, b$ are distinct primes), which seems to be the case you were investigating.  This is because when $ \gcd(a, b) \equal{} 1$, any divisor of $ ab$ must be a divisor of either $ a$ or $ b$, not both, and therefore cannot be a divisor of $ a \plus{} b$.

Unfortunately, $ \gcd(a, b) \equal{} 1$ is a rather small special case.  You can't, in general, extrapolate from special cases to the rest.  There are certain number-theoretic statements that can be proven true for all integers given that they are true for prime powers, but this is not one of them.

[b]Salvage:[/b]  Prove that $ \gcd(a, b) | \gcd(a \plus{} b, ab) | \gcd(a, b)^2$.


Show that any three pairwise distinct reals a, b, c satisfy

$ \left(\frac {c \minus{} a}{a \minus{} b}\right)^2 \plus{} \left(\frac {a \minus{} b}{b \minus{} c}\right)^2 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^2\geq 5$.

I know that the first thing you are going to do is to factor this on a CAS. This is not wrong, but after you have the factorization, the interesting part is to give a nice proof. In fact, there is one ;) .

  Darij

Okay, you got the factorization. Yet this is not the nice proof ;) .

(Hint: Don't work with x and y; substitute back to a, b, c for symmetry.)

  darij

[quote=darij grinberg]Show that any three pairwise distinct reals a, b, c satisfy
$ \left(\frac {c \minus{} a}{a \minus{} b}\right)^2 \plus{} \left(\frac {a \minus{} b}{b \minus{} c}\right)^2 \plus{} \left(\frac {b \minus{} c}{c \minus{} a}\right)^2\geq 5$.[/quote]
From Cauchy-Schwarz inequality, we have
\[\sum \frac{(a-c)^2}{(a-b)^2} \geqslant \frac{\displaystyle \left[\sum (a-c)(2a+b-3c)\right]^2}{\displaystyle \sum (a-b)^2(2a+b-3c)^2} = \frac{25(a^2+b^2+c^2-ab-bc-ca)^2}{5(a^2+b^2+c^2-ab-bc-ca)^2} = 5.\]


geometry
parallelogram
geometry unsolved

a convex hexagon ABCDEF is given on a plane such that BCEF is a parallelogram and ABF is an equilateral triangle. Known that BC=1, AD=3, CD+DE=2, find the area of ABCDEF.

i'm very2 sorry... but, did you have any idea on that?

This problem is similar to 2006 Korean MO First Round Problem 17. The difference is just $BC=2, AD=2, CD+DE=4$ in the problem.


I ended up doing it like this, got to a product of logarithms, wrote the arguments as powers of primes, wrote a log3 5 as x, replaced and got -3... could you tell me if that's right?

[quote="Gyan"][quote="xelngath"]Compute:

$\frac{log_{3}{135}}{log_{75}{3}}-\frac{log_{3}{45}}{log_{675}{3}}$

I converted everything to base 3 and ended up with an ugly quadratic. Just wondering if it stays that way or maybe there's a way to get a nicer result.[/quote]
BTW if there is a typo  above and the expression is:
$\frac{log_{3}{135}}{log_{75}{3}}-\frac{log_{3}{45}}{log_{75}{3}}$
then the answer would be something simple like 75...[/quote]

Such as it is, the result is $-3$. And in your case the result would be $\log_{3}75=1+2\log_{3}5$

Disregard, my  earlier post , sorry for the confusion.
The above  post is correct.


1. Find the greatest number that will divide 55, 127 and 175 so as to leave the same remainder in each case.

2. A number N when devided by 4, 7 gives 1, 4 as remainders respectively. Find the minimum possible value of N.

3. A number N when divided by 3,4,7 gives 2,1,4 as remainders respectively. Find remainder when N is divided by 84?

Edit: Never mind lol. Made a mistake reading question.

1.) It's $GCD(72,48,120)$ whatever that is :lol:

2.) If $x$ is the number then $x-1$ is divisible by 4 and is $3(mod7)$, the smallest number is therefore 25.

3.) Note $84=3*4*7$ therefore $N$ has to be of the form $84x+r$ for some $x$ and the desired remainder $r$. Since $r$ is $2(mod3)$, $1(mod4)$, and $3(mod7)$ (as well as $r<84)$ it's easy to see that $r=53$ is the remainder.

[hide="Number 2"]

The answer is 25.

[/hide]

[quote="ddziabenko"]1.) It's $GCD(72,48,120)$ whatever that is :lol:[/quote][hide="to those who may be wondering why this is right"]If $m$ and $n$ give the same remainder upon division by $d$, then the difference $m-n$ is divisible by $d$ (that shouldn't be too hard to see... $m=q_1d+r, n=q_2d+r\Rightarrow{m-n}=(q_1-q_2)d$). Since all the three of the given numbers $55,127,175$ give the same remainder upon division by $d$, the difference between any two of them (the differences are $72,48,120$) must be divisible by $d$. The greatest such $d$ will of course be the $\text{greatest common divisor}$ of the numbers, or $\gcd(72,48,120)=\gcd(2^3\cdot3^2,2^4\cdot3,2^3\cdot3\cdot5)=2^3\cdot3=24$[/hide]

[quote="Done"]1. Find the greatest number that will divide 55, 127 and 175 so as to leave the same remainder in each case.
[/quote]

Unless I misread this, there is no highest number. Just take n(55)(127)(175), as n approaches infinity. I probably did misread it, as somebody said "why this is right" to somebody's different answer, so...

excellent explanations (and correct answers)!


geometry
incenter
calculus
geometry unsolved

Hello to all you would leave me help on the following: Let ABC be a triangle with sides a, b, c opposite the vertices A, B, C respectively. If I incenter of ABC and G centroid of ABC determine the area of the triangle IBG. thanks

Greetings  :D

The answer is:(c-a)A/(6s) where A=area ABC and s=semiperimeter ABC

Please please tell me how you did
I appreciate that.

Greetings  :lol:

The coordenates of I=[s-b,r] and G=[(a+h)/3,h/3].The area triangle BIG is determined through the calculus of the determinant of the coordentes of the vertices B,I,G.With some arrangements you find that result


For a number n we denote bz s(n) its sum of digits.Find all the numbers n formed with three digits such that s(91n)=5

Hello.
I have a solution,but it is not elegant.
   Obviously "91n" is a five digit.Let's call a number "good" if it consists of 5 digits with sum of digits=5,"excellent" if it is divisibleby7. We can work with 4 cases.
1case) The first digit of it is= 4 gives 4 "good" numbers 40001,40010,40100,41000.None of them is "excellent".
2case)The first digit is=3 gives 8 "good" numbers.30002 is "excellent"
3case)The first digit is =2 gives 19 "good" numbers.20111,20300,22001 are "excellent".
4case)The first digit of it is=1 gives 33 "good" numbers .10031,10211,11102,12110,14000 are "excellent".

Now from "excellent" numbers only 20111 and 11102 are divisible by 13 hence satisfy the conditions of the problem. :maybe: ;)

[quote="BekJum"]Obviously "91n" is a five digit.[/quote]
This is not true for all 3-digits $n$: if $100\leq n\leq109$, $91n$ is a four digit number.
You should say: Since $100\cdot91=9100$ and all 4-digit numbers greater than 9100 have $s(n)>5$, we should consider only 5-digit numbers so $n\geq110$.
There are some not well-known tests of divisibility for 7 and 13 ($91=7\cdot13$).
I'll see if that works. If it does, I'll post a solution later today.
This is a public library computer, I cannot login.

Hi, I just came back from dinner, and as I promised, I have a solution using tests of divisibilities for 7 and 13:

We now consider only those numbers with 5 digits (from the above short discussion). Let those 5-digit numbers be $d_5d_4d_3d_2d_1$.

Condition: $d_5+d_4+d_3+d_2+d_1=5$

Test for 7: $-3d_5-d_4+2d_3+3d_2+d_1\equiv0~(mod~7)$
Test for 13: $3d_5-d_4-4d_3-3d_2+d_1\equiv0~(mod~13)$

Combinations: Find the partitions of 5, 5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1.

Result: Testing all those combinations (yes I know it's ugly) we get from each test:

7: 30002, 31010, 22001, 20111, 20300, 14000, 10031, 10220, 12110, 11102
13: 40001, 22100, 20111, 10400, 11102

Numbers appearing on both lists are those divisible by 91. Hence,
$91n=20111,~11102$ and so $n=221,~122$

[i]Remark[/i]: A short way would be to find those divisible by 13 first (we got 5 of them), and then test quickly which of these are also divisible by 7.

[quote="BekJum"]4case)The first digit of it is=1 gives 33 "good" numbers .10031,[b]10211[/b],11102,12110,14000 are "excellent". [/quote]
[b]10211[/b] is [b]not[/b] divisible by 7. And you missed two "excellent" numbers, [b]31010[/b] and [b]10220[/b].  :D  :D  :D  :cool:


limit
integration
function
real analysis
real analysis unsolved

If $ f\in C^1([ \minus{} 1,1])$, then
\[ \lim_{\varepsilon\downarrow 0} \int_{|x| > \varepsilon} \frac {f(x)}{x}dx
\]
exists and is finite, since we can write the above limit as
\[ \lim_{\varepsilon\downarrow 0} \int_{|x| > \varepsilon} \frac {f(x) \minus{} f( \minus{} x)}{x}dx
\]
In fact if $ f\in C^{0,\alpha}([ \minus{} 1,1])$ for some $ 0 < \alpha\leq 1$ then we still should be able to control the integrand above.

What if $ f$ is merely continuous?

(maybe related:  Does there exist $ f$ such that $ f\in C([ \minus{} 1,1])$ but $ f\not\in C^{0,\alpha}([ \minus{} 1,1])$ for any $ 0 < \alpha\leq 1$?)

Let $ f(x)\equal{}\begin{cases}\frac{\text{sign}(x)}{1\minus{}\ln|x|}&x\ne0\\0&x\equal{}0\end{cases}.$

This function is continuous on $ [\minus{}1,1],$ does not belong to any of your Hölder classes, and

$ \text{p.v.}\int_{\minus{}1}^{1}\frac{f(x)}{x}\,dx\equal{}\infty.$


arithmetic sequence
geometry proposed
geometry

[quote="communist"]problem: ABC is a triangle with sides in arithmetic progression with difference d. the medians intersect at point M inside ABC forming the median triangle from the foot of medians respectively. Let m(a), m(b), m(c) be the distance between M and the three foot of medians respectively. Thus prove that m(a)^2+m(b)^2+m(c)^2 >_ (mean side* d)/ sq.root 6.
thanks[/quote]

Let $ a \ge b \ge c$.
Then, to prove that

$ \sum{m(a)^2} \equal{}
(1/4) (b^2 \minus{} 2 b d \plus{} d^2/3) \plus{} (1/4)(b^2 \plus{} 2 b d \plus{} d^2/3) \plus{} (1/4)(b^2 \plus{} (4 d^2)/3) \ge \frac{bd}{\sqrt {6} }$

$ \leftrightarrow \frac{3b^2 \plus{} 2d^2}{4} \ge \frac{bd}{\sqrt {6}}$


On each of $ 12$ points around a circle we place a disk with one white side and one black side. We may perform the following move: select a black disk, and reverse its two neighbors. Find all initial configurations from which some sequence of such moves leads to the position where all disks but one are white.

Let define black as $ 1$ and white as $ -1$
and divide the vertices in to sets $ E$ for the even ones and $ O$ for the odd ones.
Each move change the sign of two elements of the same set from
$ (1,1)\rightarrow(-1,-1)$
$ (1,-1)\rightarrow(-1,1)$
$ (-1,1)\rightarrow(1,-1)$
$ (-1,-1)\rightarrow(1,1)$
So the product of all the vertices of $ E$ and the product of all the vertices of $ O$ is constant.
Then it is necessary that the initial product will be either $ (1,-1)$ or $ (-1,1)$

we denote white as $1$,black as $0$
then we have $(1,0,0)\rightarrow (0,0,1)$...(a)
so if there are odd 0s between two 1s as $(1,0,0,...,0,0,1)$
we use (a) several times,we could have $(0,0,0,...,1,0,1)\rightarrow (0,0,0,...,0,0,0)$...(b)
Obviously we have the number of 0 has the same parity after moves
so it is necessarily that the number of 0 in the initial is odd
Now we prove that it is sufficient
Consider the initial configuration $P$ has odd 0s
then consider all configurations which we could get by making several moves from $P$
it is obvious that it is finite
so we could choose a configuration $Q$ that has the least 1s (if there are more than one,choose any)
if it has at least two 1s
then because it has odd 0s
there exists two 1s that there are odd 0s between them
by (b) we could reduce the number of 1s,which is a contradiction
so $Q$ has only one 1
and we could prove by induction that
we could make some moves from $(0,0,...,0,0)$(odd 0s) to get a configuration that has only one 0

We will prove the statement for $2n$ instead of $12$.
Label the points with $A_1, A_2 , \cdots, A_{2n}$. Assume WLOG they determine a regular $2n$-gon. Let $O$ be the midpoint of $A_1A_{2n}$.Define a block, a sequence of consecutive black points, and define the length of a block, the number of black points it consists of.
The number of black points is invariant $\pmod{2}$ under the move described in the problem, so it is necessary to have an odd number of black points to achieve a configuration with exactly one black point. Now we prove that this condition is sufficient. We will start an algorithm.
Step 1: If there are more than $1$ block proceed to Step 2, otherwise stop the algorithm.
Step 2: Find the black with the smallest length, and that is the closest to $O$ if we would move clockwise. Now choose the endpoint of this block that is the furthest from $O$. Perform the move. Reset.
It is easy to see that the algorithm will "unify" all black points in a single block, call it $\omega$. Now we start one more algorithm:
Step 0: If there isn't exactly one black point, proceed to Step 1.
Step 1: Choose the 2 black points closest possible to the end-points of $\omega$ and perform the move with them.
Step 2: Perform the move with the end points of $\omega$ until we don't have a configuration that contains exactly one block, call this new block $\omega$.

We prove a more general statement, namely that for given $2n$ disks around a circle, it is necessary and sufficient for the configuration to have an odd number of black disks. First, observe that in terms of moves, we have the following options
$$WBW\rightarrow BBB \quad BBW\rightarrow WBB \quad BBB\rightarrow WBW$$
Since the parity of the number of black disks is invariant, we see that a position which one can win from, which we will denote a [i]winning position[/i], must contain an odd number of black disks. We now show that an odd number of disks is sufficient for a position to be a winning position. Observe that the desired position is
$$\cdots WWWBWWW\cdots$$
Now observe that for any odd integer $k$, we can move backwards from our winning position to a position with $k$ black disks, all of which are in a row. To see why this must be true, consider the following set of moves.
$$\cdots WWBWW\cdots \rightarrow \cdots WBBBW\cdots \rightarrow \cdots WBBWBWW\cdots \rightarrow \cdots WBBWBWW\cdots \cdots WBBBBBW\cdots$$
In general, number the $B$'s $B_1$, $B_2$...$B_k$, where $k$ is odd, by performing moves on $B_1$,$B_3$...$B_k$, we can obtain a row of $B$'s of length $k+2$. Because moves are the same forward and backward, i.e. performing a move twice undoes it, it suffices to show that we can transform any position with an odd number of black disks into a row of an odd number of black disks. Fortunately, this is easy. Since there are an odd number of disks that are black, there must exist some clump of odd size, hence we can just take this clump, and literally keep expanding it by the above algorithm, and by absorbing even blobs, until it encapsulates all of the black disks, or we obtain the following situation. Consider the following position
$$WBBB\cdots BBBBWBBBB\cdots W$$
Where WLOG, the blob on the left is of odd size, and the blob on the right is of even size. However note that performing the same algorithm that enlarges a $k$ blob to a $k+2$ blob, if applied on an odd blob, simply shifts it to the left one disk, or to the right one disk (depending on which side one starts from), when applied to an even blob. Hence we may just shift the even blob one to the left, and obtain a blob of odd size, which we already know can be reduced down to the desired position, so we win. $\square$

Solution:
Ans - $s_b \equiv 1(mod 2)$
Proof:
Let's prove the necessity,
We will use binary code i.e. $0$ = white and $1$ = black and denote the no. of white and black disks by $s_w$ and $s_b$ respectively.
Observe that we can perform  the following set of moves - $A$ : $111 \mapsto 010$ , $B$ : $010 \mapsto 111$ and $C$ : $110 \mapsto 011$.
Easy to  see that the parity of $s_b$ remains invariant under these set of moves.As the final config. has $s_b \equiv 1(mod 2)$ so must the initial.

Now  for the sufficiency,
Define a [b]$k$-chain[/b] to be a continuous sequence of $1$ of length $k$  and denote such a sequence by $1^k$. Eg: $01110$ is a $3$-chain or $1^3$.We may call $1$-chain as isolated points.

Algorithm 1 (the destroyer).
[b]every $k$-chain  can be reduced to $1,2$ - chain.[/b].In some sense we can [b]break down all chains to a bunch of isolated points and  2-chain.[/b]
Proof:
By induction on $k$.
Base step being trivial as usual , Inductive step: $01^{k}0 \mapsto^{A} 00101^{k-3}0$.

Algorithm 2 (cleaner).
Well after applying the destroyer we are left with only a bunch of isolated points and 2-chains but that's not the req'd config.
So, consider the following set of moves
[b]1.[/b] $010^k 11.. \mapsto^{C} 010^{k-1}110.. \cdots \mapsto^{C} 01110 \mapsto^{destroyer} 010$.
[b]2. [/b] same as[b] 1. [/b]but with $0110$ instead of $010$ the resultant is $0110$

This allows us to get rid of all unwanted 2-chains and so we end up with a configuration of only isolated points.If there is only 1 isolated point we are done or else consider the following

Algorithm 3 (transporter)
[b]3.[/b]$10^k 10 \mapsto^{B} 10^{k-1}111 \mapsto^{cleaner} 010^k1$.

By applying ,transporter, we are able to move the block $10^k10$ along the circle clockwise.
Now,
Select one such block (if none exist then we have reached the req'd config) and move it along the circle clockwise until the right end hits an isolated point. Such a hit is guaranteed by the fact that $s_b \equiv 1(mod 2)$ and so there has to be atleast 3 black disks in the circle.But then the block forms a config of form $010^k 110$ which can be reduced to $010$ by cleaner.
Repeat this algorithm , but at each stage we are reducing $s_b$ so this algorithm must terminate and we are done!!

My solution, as well as the algorithms presented in all of the solutions above, seem to only require an odd number of black starting disks, and the condition that we have $2n$ disks total is not used. Shouldn't this be true for any $n \ge 3$ disks around the circle?

I think it is enough that number of black discs is odd,the problem writter has covered with even number of people,and there couldnt be question about all configs if he told that number is odd.

Generalisation for any $n$:

We shall assign a weight to each white and black disk. Say that a black disk has a weight of 1 unit and a white disk has no weight.
We have the following cases:
[b]Case 1. [/b] One black disk is surrounded by $2$ white ones:
Performing a move on the black disk turns the other $2$ white disks to black and hence increases the weight by $2$.

[b] Case 2. [/b] One black disk is surrounded by one black and one white:
Performing a move on the center does not affect the weight since the black one turns white and vice versa.

[b] Case 3. [/b]3 black disks in a row:
Performing a move on the center decreases the weight by 2 since the white disks change to black.

So we notice that the only weight changes are $+2$ and $-2$. The desired configuration of only one black disks requires a weight of $1$, so we notice that we must have an odd number of black disks in the initial configuration.
This condition is necessary, but is it sufficient? We will now prove that it is sufficient.
Now we propose the following algorithm:
First note that we will only consider cases where the initial configurations have an odd number of black disks since we already proved that the others don't work so we may dismiss them.

[b]Claim: [/b]There is never a case where only Case 2 is present.

[b]Proof: [/b]We assumed that the number of black disks is odd initially. Now suppose there is a case where only Case 2 is present. If we perform the move on the center one, it preserves the parity of the black disks and hence, is an invariant. If only Case 2 is present, there must be a total even number of black disks, but the final configuration requires an odd number. Since the parity is invariant, we have a contradiction.
Therefore, there is never a case where only Case 2 is present.

Now if we have only case 1:
There are 2 configuration:
[b]Configuration 1:  [/b]$W B W B W$:
We may perform a move on the first black disk and then on the disk on the right of the first black disk to reduce the total number of black disks by $2$.

[b]Configuration 2: [/b] $W B W W ... W W B W$:
For convenience, we label them $D_1, D_2 \cdots D_r$. We may perform the following moves:
$$D_2, D_3, \cdots D_{r-2}$$Notice that this again reduces the number of black disks by $2$.

Finally, we can finish the problem by proposing the algorithm:
[b]ALGORITHM: [/b] In the case where Case 3 is present, we may perform the move on the center one to reduce the number of black disks. If Case 3 is not present, then Case 2 is present with the above $2$ configurations, so we may perform the same set of moves as proposed above. In each case, the number of black disks is reduced by $2$. So, the number of black disks is a strictly decreasing monovariant and hence will, at some point, reach $1$ as had initially assumed an odd number of black disks, it must reach the configuration where only $1$ black disk is present.
Therefore, all initial configurations where an odd number of black disks is present can result in the final configuration.
$\blacksquare$

Represent disks with black faced up as [b]B[/b], and with white [b]W[/b]. Also represent by $N(B)$ number of Bs, similarly with W.
Clearly $N(B)$’s parity remains constant. Because we want $N(B)=1$ at the end, initially it should have been odd too. So $N(B)_{\text{initial}} \equiv 1 \pmod{2}$ is necessary.
Now, we prove that it is sufficient. As long as any $B$ is surrounded by 2 $W$s, keep applying the given algorithm, until (i)$N(B)$ becomes $1$, or (ii) Any $B$ is surrounded by at most one other $B$.
Now, if (i) happens, we are done. If (ii) happens, we claim that we can bring all the $B$s together. Indeed, the B’s must be found as $B \dots B W \dots W \dots B \dots B$ now, and applying the algorithm on the left sequence of B’s, starting from the rightmost B, decreased the distance of the sequences of the two by 1. Keep applying it that way, and we can join them together.
Now, we claim that if all B’s are together, we can apply the algorithm in some way so that there’s only one B left. To do that, choose the middlemost B. If $N(B)=2(2k+1)+1$, apply the algorithm, starting from the middlemost B, and keep applying it on the next B, in each direction. This reduces the number of $B$s by 2 (1 in each direction). If $N(B)=2(2k)+1$, apply the algorithm on the two B’a adjacent to the middlemost one ; continue doing that with the next B in each direction. In this case too, the number of B’s decrease by 1 on each side, while preserving the property that all the B’s are together. Therefore, applying this consecutively, we’ll reach a stage where only one B is left, as desired

[quote=SerdarBozdag]@above I did not understand the algorithm. If we have $BBBBBBB \implies BBWBWBB$. What to do next?[/quote]

Then you apply it on the second and second last B :P
Basically keep applying it on the first B on both sides

We just prove it generally.
Let $1$ denote black and $0$ denote a white.

We have by a trivial parity of sum invariant that #no. of blacks is odd.
Now we just prove it sufficient .
WE now define some algorithms.

[color=#c00]ALGORITHM 1:[/color] Due to the given operation if we have,then it is possible that.
$(1 ,1,0,0...0,1) \mapsto (0,0,0,0...0,1,0)$

[color=#c00]ALGORITHM 2:[/color]Again due to the given operation we have,
$(0,1,0,0,...0,1) \mapsto (1,0,0,0...1,0)$.
Now, time to finish,

[color=#c00]ALGORITHM 3:[/color] Apply algorithm 1 when possible.
If not choose two arbitrary $1's$ and then consider the block with the two $1's$ and shift this block repeatedly with algorithm 2 until we get the setup required for algorithm 1, then apply it.

Now,we are just done by repeated applications of algorithm 3.
Also note that all algorithms are  guaranteed to exist.

Just to add some remarks,I solved a nice  problem after being extremely ill for a while(not recovered yet),so happy.
Firstly I thought that the invariant was quite weak and I should think of a better one but then I gave up this idea and moved ahead with small cases and they worked.
Then I played around with $1's$ till I found the algorithms.


Let a,b,c be positive real numbers. Prove that
$ \sum_{cyc}\frac{a^3}{a^2\plus{}ab\plus{}b^2}\geq\sum_{cyc}\frac{a^3}{2a^2\plus{}b^2}$

[quote="Sudoku"]Let a,b,c be positive real numbers. Prove that
$ \sum_{cyc}\frac {a^3}{a^2 \plus{} ab \plus{} b^2}\geq\sum_{cyc}\frac {a^3}{2a^2 \plus{} b^2}$[/quote]
$ \sum_{cyc}\frac {a^3}{a^2 \plus{} ab \plus{} b^2}\geq\sum_{cyc}\frac {a^3}{2a^2 \plus{} b^2}\Leftrightarrow\sum_{cyc}\frac{a^4(a\minus{}b)}{(a^2\plus{}ab\plus{}b^2)(2a^2\plus{}b^2)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac{a^4(a\minus{}b)}{(a^2\plus{}ab\plus{}b^2)(2a^2\plus{}b^2)}\minus{}\frac{a\minus{}b}{9}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a\minus{}b)^2(7a^3\plus{}5a^3b\plus{}ab^2\plus{}b^3)}{(a^2\plus{}ab\plus{}b^2)(2a^2\plus{}b^2)}\geq0.$

[quote="arqady"][quote="Sudoku"]Let a,b,c be positive real numbers. Prove that
$ \sum_{cyc}\frac {a^3}{a^2 \plus{} ab \plus{} b^2}\geq\sum_{cyc}\frac {a^3}{2a^2 \plus{} b^2}$[/quote]
$ \sum_{cyc}\frac {a^3}{a^2 \plus{} ab \plus{} b^2}\geq\sum_{cyc}\frac {a^3}{2a^2 \plus{} b^2}\Leftrightarrow\sum_{cyc}\frac {a^4(a \minus{} b)}{(a^2 \plus{} ab \plus{} b^2)(2a^2 \plus{} b^2)}\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(\frac {a^4(a \minus{} b)}{(a^2 \plus{} ab \plus{} b^2)(2a^2 \plus{} b^2)} \minus{} \frac {a \minus{} b}{9}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac {(a \minus{} b)^2(7a^3 \plus{} 5a^3b \plus{} ab^2 \plus{} b^3)}{(a^2 \plus{} ab \plus{} b^2)(2a^2 \plus{} b^2)}\geq0.$[/quote]
I think it is equivalent to
$ \sum_{cyc}(a\minus{}b)^2.\frac{7a^3\plus{}5a^2b\plus{}2ab^2\plus{}b^3}{(a^2\plus{}ab\plus{}b^2)(2a^2\plus{}b^2)}\geq 0$