geometry
parallelogram
geometry unsolved

[quote="orl"]The circles $ c_1$ and $ c_2$ are tangent at the point $ A.$ A straight line $ l$ through $ A$ intersects $ c_1$ and $ c_2$  at points $ C_1$ and $ C_2$ respectively. A circle $ c,$ which contains $ C_1$ and $ C_2,$ meets $ c_1$ and $ c_2$ at points $ B_1$ and $ B_2$ respectively. Let $ \omega$ be the circle circumscribed around triangle $ AB_1B_2.$ The circle $ k$ tangent to $ \omega$ at the point $ A$ meets $ c_1$ and $ c_2$ at the points $ D_1$ and $ D_2$ respectively. Prove that

[b](a)[/b] the points $ C_1,C_2,D_1,D_2$ are concyclic or collinear,
[b](b)[/b] the points $ B_1,B_2,D_1,D_2$ are concyclic if and only if $ AC_1$ and $ AC_2$ are diameters of $ c_1$ and $ c_2.$[/quote]
There are so many circles that it is almost evident that one has to invert it, actually the diagram pleads for an inversion. :)

So let us do that. WLOG let $\text{rad.}(c_1)\leq \text{rad.}(c_2).$ Invert everything about $A$ to map points to their respective primes. Now, note that $c_1,c_2$ map to $\overleftrightarrow{B_1'C_1'D_1'}$ and $\overleftrightarrow{B_2'C_2'D_2'}.$ The circles $\odot(AB_1B_2)$ and $\odot(AD_1D_2)$ will map to $\overleftrightarrow{B_1'B_2'}$ and $\overleftrightarrow{D_1'D_2'}.$ Obviously $B_1'B_2'D_2'D_1'$ is a parallelogram.
We also have that
$\begin{aligned}\angle B_1'B_2'C_2'&=\angle AB_2'C_2'-\angle AB_2'B_1'=\angle AC_2B_2-\angle AB_1B_2\\&=\angle C_1B_1B_2-\angle AB_1B_2=\angle AB_1C_1\\&=\angle AC_1'B_1'.\end{aligned}$
Therefore $B_1',B_2',C_1',C_2'$ are concyclic. Note that $D_1', D_2'$ are points on $B_1'C_1'$ and $B_2'C_2',$ produced, such that $B_1'B_2'\parallel D_1'D_2',$ and so using [i]Reim's theorem[/i](or simple angle-chase) on the points $B_1, B_2, C_1, C_2$ we obtain $D_1'D_2'C_2'C_1'$ to be cyclic. Since inversion preserves a circle not passing through the centre of inversion, we see that $C_1, C_2, D_1, D_2$ are concyclic.
If, however, $D_1',D_2',C_2',C_1'$ happen to be collinear, then the story would be the degenerate case, ie $D_1, D_2, C_1, C_2$ will be collinear. Hence $(a)$ is proved.$\Box$

If the points $B_1, B_2, D_1, D_2$ happen to be concyclic, then since their images are the vertices of a parallelogram, it implies that the angles must be $\frac{\pi}{2},$ leading to $\angle AB_1C_1=\frac{\pi}{2}=\angle AB_2C_2$ (This can be verified easily). Therefore $(b)$ is also proved, and we are also done!$\Box$
 :lol:
[asy]import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ffqqtt = rgb(1,0,0.2); pen ffcccc = rgb(1,0.8,0.8); pen qqzzzz = rgb(0,0.6,0.6); pen qqwwww = rgb(0,0.4,0.4);
draw((0,0)--(-2.11,-2.11)--(5.02,2.22)--cycle); draw((0,0)--(-0.96,1.77)--(0.28,1.28)--cycle); draw(circle((-2.11,0),2.11),ffqqtt); draw(circle((3,0),3),ffqqtt); draw((-5.05,(-0-2.04*-5.05)/2.65)--(6.33,(-0-2.04*6.33)/2.65),blue); draw(circle((1.22,0.43),4.19),ffqqtt); draw((0,0)--(-2.11,-2.11),blue); draw((-2.11,-2.11)--(5.02,2.22),blue); draw((5.02,2.22)--(0,0),blue); draw(circle((7.05,-9.16),11.56),ffqqtt); draw(circle((-2.89,-4.92),6.96),ffqqtt); draw((0,0)--(-2.11,-2.11),blue); draw((-2.11,-2.11)--(5.02,2.22),blue); draw((5.02,2.22)--(0,0),blue); draw((0,0)--(-0.96,1.77),blue); draw((-0.96,1.77)--(0.28,1.28),blue); draw((0.28,1.28)--(0,0),blue); draw(circle((-0.62,0.81),1.02),ffqqtt); draw(circle((0,0),2.4),qqzzzz); draw((-5.05,(-1.11+1.16*-5.05)/1.1102230246251565/10^16)--(6.33,(-1.11+1.16*6.33)/1.1102230246251565/10^16),qqwwww); draw((-5.05,(+3.28+2.41*-5.05)/2.220446049250313/10^16)--(6.33,(+3.28+2.41*6.33)/2.220446049250313/10^16),qqwwww); draw((-5.05,(+8.23+1.78*-5.05)/2.32)--(6.33,(+8.23+1.78*6.33)/2.32),qqwwww); draw((-5.05,(-0.73+1.78*-5.05)/2.32)--(6.33,(-0.73+1.78*6.33)/2.32),qqwwww);
dot((-2.11,0),ds); label("$O_1$", (-2.04,0.12),NE*lsf); dot((0,0),ds); label("$A$", (-0.29,-0.01),W*lsf); dot((3,0),ds); label("$O_2$", (3.07,0.12),NE*lsf); dot((-2.65,2.04),ds); label("$C_1$", (-2.9,2.3),N*lsf); dot((3.77,-2.9),ds); label("$C_2$", (3.55,-3.25),S*lsf); dot((-2.11,-2.11),ds); label("$B_1$", (-2.17,-2.44),S*lsf); dot((5.02,2.22),ds); label("$B_2$", (5.09,2.33),NE*lsf); dot((-0.96,1.77),ds); label("$D_1$", (-0.9,1.89),NE*lsf); dot((0.28,1.28),ds); label("$D_2$", (0.43,1.26),NE*lsf); dot((0.96,-0.74),ds); label("$C_{2'}$", (1.03,-0.64),NE*lsf); dot((-1.36,-1.36),ds); label("$B_{1'}$", (-1.7,-1.23),NE*lsf); dot((0.96,0.42),ds); label("$B_{2'}$", (0.96,0.15),NE*lsf); dot((-1.36,1.05),ds); label("$C_{1'}$", (-1.29,1.15),S*lsf); dot((0.96,4.29),ds); label("$D_{2'}$", (1.09,4.1),NE*lsf); dot((-1.36,2.5),ds); label("$D_{1'}$", (-1.76,2.64),NE*lsf); clip((-5.05,-4.18)--(-5.05,4.94)--(6.33,4.94)--(6.33,-4.18)--cycle);[/asy]


MATHCOUNTS geometry AMC USA(J)MO USAMO ARML

While some of the local communities are quite large with as many as 100 or so users, a few are quite small at this time.

Moderators who are interested in building up their local community might consider advertising its existence with their local math teams and friends who might also be interested.  I think the reason some of the local communities have thrived with dozens of posts is that two or three groups of math teamers have joined and were more willing to post there than other places in the AoPS forums.  This seems to be true to a great degree in the Alabama community where some people have posted a great deal, though scarcely at all in any other forums.  Indeed, some of the forums can be quite intimidating for even the above average math teamer.

[quote="EFuzzy"]Should we just go up to the local coaches and tell them about AoPs?[/quote]That would be a great idea :)

[quote="EFuzzy"]Should we just go up to the local coaches and tell them about AoPs? Can you give some advice on how we should do it? I'm not a good talker, and I don't believe anyone has visited the Maryland forum's yet.[/quote]

If you are interested in helping build your community by making coaches aware of it, a good problem solving exercise is finding the form of communication that you are most comfortable with.

Hi all,

I'm the moderator of the new, and as yet empty (other than myself) Vermont forum.

My plan is to start by inviting the math club kids from my own children's school to join, and send home an information notice for their parents.

I also planned to make up a little information sheet to take to MathCounts chapter competition to hand out to the other coaches there.  Perhaps I'll make extras to send along with any willing coaches headed to the state competition, since I doubt we'll be sending anyone this year.  ;-)

One thing I noticed here was the suggestion to have the kids post the name of their school -- one of the things I recommended in my introductory guidelines was that they *not* post their full name or school name, etc.  Am I just too paranoid? ;-)  AoPS forums seem really safe to me, but you just never know who might be lurking.

I can understand your concerns about such introductions.  Perhaps it's not the wisest idea.  Most of the students are high school age and I don't think it's quite such an issue at that point.

[quote="Valentin Vornicu"][quote="mathmom"]One thing I noticed here was the suggestion to have the kids post the name of their school -- one of the things I recommended in my introductory guidelines was that they *not* post their full name or school name, etc.  Am I just too paranoid? ;-)  AoPS forums seem really safe to me, but you just never know who might be lurking.[/quote]There is *no-one* lurking, other than people yourself let in :)[/quote]

To be fair, you never know who you let in.

I also seeded many of the communities with users who listed their city/state in their profile.  For instance, I added over 100 members to the Bay Area community alone.  While the great majority we know are students, we can't vouch for everyone.

However, in a small community, introductions would be safer.

[quote="Valentin Vornicu"]Yes what I doubt is that anyone is *gonna come for you* because you posted your school in a hidden forum on a math board :D ;)[/quote]

It's best to err on the side of safety in the internet age.

[quote="Valentin Vornicu"]There is *no-one* lurking, other than people yourself let in :)[/quote]

Well, sure, but the 13yo "kid" who tells me he is from the middle school halfway across the state may really be some 40yo pervert who gets off on geeky 13yo boys. :(    Assuming he's a good actor, I'd have no reason to be suspicious and would let him in with no concern.

I really think that AoPS is one of the last places something like that would happen, but it does happen (in general, not necessarily here) and I'd feel horrible about putting anyone in danger by encouraging them to post identifying information about themselves.

(That's what you get for inviting a paranoid mom to be a local moderator :P)

[quote="Valentin Vornicu"][quote="orl"]I feel it is quite overwhelming how many competitions you have. I just would like to know many competitions the students participate in, e.g. in Alabama which has dozens of competitions. I think you must be really proud of your competition structure in the U.S.A !!!  :)[/quote]It's a big country. We have (in Romania which is almost as large as Alabama) around 40 competitions per year, why shouldn't they have $ 52 \cdot 40 = 280 $ different competitions per year? :)[/quote]

Well, I thought about Germany. And I just know 2-3 competions: DEMO, BWM and Jugend Forscht, Sektion Mathematik. That's all.  :?

Methinks you have to live there or once lived there or something to request it.

[quote="Mgccl"]I think there should be a Long Island community.
Long Island is in New York State, but not part of NYC or Upstate New York
I can hardly find people in Long Island that does math...[/quote]
aren't u contradicting urself...first u say u need a community for long island then u say people there don't do math then what do u need a community for :rotfl:  :rotfl:  :rotfl:

Maybe he wants to encourage math among people of long island!

Or maybe he wants a community for other subjects? :maybe:

I was wondering if I could become a Mod for the canadian forum. I post there quite a lot, and I am online a lot as well.

[quote="pardesi"]
aren't u contradicting urself...first u say u need a community for long island then u say people there don't do math then what do u need a community for :rotfl:  :rotfl:  :rotfl:[/quote]
Uhh...  there are always exceptions like me. long islanders don't know where to go when they go into the forum...
Maybe change NYC into New York Metropolis area or NYC and Long Island.
Long Islander in Suffolk and Nassau don't consider themselves part of New York City area.

How much activity is required for a new community? I am thinking of starting one for my state.

Is ANYBODY responding to my question in this topis?

I am trying to make a Puerto Rico forum.We already have a list of Puerto Rican users. Here is the list. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?f=242&t=381924[/url]

Hi

Thanks for this topic , i want to start an Arabic speaking community on $\text{Aops}$ , and i have a question :

How can one type text (arabic) from right to left , align it from right to left as in MSWORD for example.

and is it possible to increase the police size for arabic script , i made a test and noticed that letters display smaller than

they should appear compared with latin letters for example.

Here is an example (if your computer is able to view these characters correctly) : بين أنه مهما يكن .

Now , what is the procedure the start the community  :roll:

new communities are no longer being made, sadly. if there was an older, archived version of the community you want created, try to access it.

Why can't I make a new community forum for my state?


algebra
polynomial
abstract algebra
real analysis
real analysis unsolved

let's $ f(x) \equal{} \sum_{\mathbb{Z}} c_{k}e^{ikx}$

Let's $ g(x) \equal{} \sum_{\mathbb{Z}}c_{3k}e^{3ikx}$
can $ g(x)$ expressed as the finite linear combination of $ f$ ?more exactly like this $ g(x) \equal{} \sum_{j\equal{}finite} a_{j}(x)f(c_{j}x \plus{} b_{j})$
where $ a_{j}(x)$ is finite trigonometric polynomial and $ c_{j},b_{j} \in \mathbb {R}$.
I think NO, but I can't easy   explain why..

The answer is actually "Yes". Consider the shifts by $ 0,\frac{2\pi}3,\frac{4\pi}3$ ;)

:oops:  I was thinking about Dirichlet Kernel $ D_{n}(x)$, So problem became to express $ D_{N}(3x\minus{}t)$ as linear combination of kernels such as  $ D_{3N}(x\minus{}t\plus{}b_{j})$ So it was not easy for me,   how do this ..
Thank you fedja


geometry
perpendicular bisector
national olympiad

I can't exactly remember Questions 3 and 4 (I never really got too far into them...), but the first two questions were of the senses given below. If somebody wants to edit this to include all the problems later please do.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=420940#p420940]Question 1[/url]
Given $x$ and $y$ are reals, find the minimum possible value for $x^2 + y^2$ satisfying the equation $xy(x^2 - y^2) = x^2 + y^2$ where $x \neq 0$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=420943#p420943]Question 2[/url]
Q2. Find all the possible pairs of positive integers $(x,y)$, both with no prime factor greater than 5, such that $x^2 - y^2 = 2^k$ where $k$ is a non-negative integer.

My solutions were really bad, but I'm only in Y11 and still fairly inexperienced at maths at this level, so I'm not too surprised.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=421055#p421055]Problem 3[/url]

Let ABC be a triangle with AC > AB. The point X lies on the
side BA extended through A, and the point Y lies on the side CA in
such a way that BX = CA and CY = BA. The line XY meets the
perpendicular bisector of side BC at P. Show that

BPC + BAC = 180.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=421056#421056]Problem 4[/url]

An exam consisting of six questions is sat by 2006 children. Each
question is marked either right or wrong. Any three children have
right answers to at least five of the six questions between them. Let N
be the total number of right answers achieved by all the children (i.e.
the total number of questions solved by child 1 + the total solved by
child 2 + · · · + the total solved by child 2006). Find the least possible
value of N.

I solved everything, I'm quite pleased with myself  :P

By the way, who are you Ilthigore? Anyone I know?

[quote]By the way, who are you Ilthigore? Anyone I know?[/quote]

A Y11 from BGS in Bristol, and no - I don't think I know you. I'm not really IMO standard, certainly not this year (if I'm really lucky I got about 10 points). I was mainly aiming to beat 2/40 in the BMO2 which was the school record last year :). Well done in getting all four btw!

Added in the resources pages at http://www.mathlinks.ro/Forum/resources-179-66-2006.html


Hi, I'm a beginner in Matrix and Linear Algebra. Could you help me to solve this problem?  :P
Given $ A \equal{} \left(\begin{array}{cc}a & b \\
    b&a \end{array}\right)$. Calculate $ A^{k}$

Thanks in advance.

Hint: Put $ X \equal{} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. Then $ X^2 \equal{} I$. What can we say about $ A^k \equal{} (aI \plus{} bX)^k$?

hello, i have
$ A^2 \equal{} \left(\begin{array}{cc}a^2\plus{}b^2 & 2ab \\
2ab&a^2\plus{}b^2 \end{array}\right)$
$ A^3 \equal{} \left(\begin{array}{cc}a^3\plus{}3ab^2 & 3a^2b\plus{}b^3 \\
3a^2b\plus{}b^3&a^3\plus{}3ab^2 \end{array}\right)$
and so on.
Sonnhard.

Let $ A \equal{} \left[\begin{array}{cc}a & b \\
b & a \end{array}\right]$. As the sum of each line (or column) is a+b, then a+b is an eigenvalue of A, and as the trace is 2a, the other eigenvalue is a-b.  The eigenvector associated with a+b is $ [1\,\, 1]^T$, and the eigenvector associated with a-b is $ [ \minus{} 1\,\, 1]^T$. Now consider the matrix P whose columns are these eigenvectors, $ P \equal{} \left[\begin{array}{cc}1 & \minus{} 1 \\
1 & 1 \end{array}\right]$. It's easy to see that $ P^{ \minus{} 1} \equal{} \left[\begin{array}{cc}1/2 & 1/2 \\
\minus{} 1/2 & 1/2 \end{array}\right]$. Therefore we have $ A \equal{} PDP^{ \minus{} 1}$, where D is a $ diag(a \plus{} b,a \minus{} b)$, or

$ \left[\begin{array}{cc}a & b \\
b & a \end{array}\right] \equal{} \left[\begin{array}{cc}1 & \minus{} 1 \\
1 & 1 \end{array}\right] \left[\begin{array}{cc}a \plus{} b & 0 \\
0 & a \minus{} b \end{array}\right] \left[\begin{array}{cc}1/2 & 1/2 \\
\minus{} 1/2 & 1/2 \end{array}\right]$

and so

$ A^k \equal{} \left[\begin{array}{cc}1 & \minus{} 1 \\
1 & 1 \end{array}\right] \left[\begin{array}{cc}(a \plus{} b)^k & 0 \\
0 & (a \minus{} b)^k \end{array}\right] \left[\begin{array}{cc}1/2 & 1/2 \\
\minus{} 1/2 & 1/2 \end{array}\right]$

$ \equal{} > A^k \equal{} \left[\begin{array}{cc}\frac {(a \plus{} b)^k \plus{} (a \minus{} b)^k}{2} & \frac {(a \plus{} b)^k \minus{} (a \minus{} b)^k}{2} \\
\frac {(a \plus{} b)^k \minus{} (a \minus{} b)^k}{2} & \frac {(a \plus{} b)^k \plus{} (a \minus{} b)^k}{2} \end{array}\right]$.


trigonometry
function
algebra
polynomial
complex numbers

Let a real number $x$ such that $-1 \leq x \leq 1$ and a positive integer $n$. Show that the function $f_n(x) = cos(n.\arccos{x})$ can be written as a polynomial $P(x)$ such that $\deg{P(x)} = n$ and the coefficient of highest degree monomial of $P(x)$ is equal to $2^{n-1}$.

[b]Is this partial solution correct  :?:  :?: [/b]

[hide="Finding expression for cos nx"]

Consider $cis(x) = \cos x + i \sin x$

We have:
$cis^n(x) = cis(nx)$
$\cos(nx) = \mathfrak{R}\left[cis^n(x)\right]$
$\cos(nx) = \mathfrak{R}\left[(\cos x + i\sin x)^n\right]$

Expanding Binomial:
$cis^n(x) = \sum_{k=0}^n \binom{n}{k} \cos^{n-k}(x).\sin^{k}(x).i^k$

Separating odd and even expoents of $i.\sin(x)$:

$cis^n(x) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).\sin^{2h}(x).i^{2h} + \sum_{h=0}^{\frac n2} \binom{n}{2h+1} \cos^{n-(2h+1)}(x).\sin^{2h+1}(x).i^{2h+1}$

Look that $i^{2h} = (i^2)^h = (-1)^h$, therefore:

$cis^n(x) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).\sin^{2h}(x).(-1)^{2h} + \sum_{h=0}^{\frac n2} \binom{n}{2h+1} \cos^{n-(2h+1)}(x).\sin^{2h+1}(x).(-1)^h.i$

The first sum is always a real number and second is non-real, then:

$\mathfrak{R}[cis^n(x)] = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).\sin^{2h}(x).(-1)^{h}$
$\cos(nx) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).\sin^{2h}(x).(-1)^{h}$
$\cos(nx) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).[\sin^{2}(x)]^h.(-1)^{h}$
$\cos(nx) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(x).[1 - \cos^{2}(x)]^h.(-1)^{h}$

[/hide]

Then...

$f_n(x) = cos(n.\arccos{x})$
$f_n(x) = \sum_{h=0}^{\frac n2} \binom{n}{2h} \cos^{n-2h}(\arccos x).[1 - \cos^{2}(\arccos x)]^h.(-1)^{h}$
$f_n(x) = \sum_{h=0}^{\frac n2} \binom{n}{2h} x^{n-2h}.(1 - x^2)^h.(-1)^{h}$
$f_n(x) = \sum_{h=0}^{\frac n2} \binom{n}{2h} x^{n-2h}.[1 - x^2 + x^4 - x^6 + \dotsb + (-x^2)^h].(-1)^{h}$

Now it's easy to see that is a polynomial, and it degree is n. But I can't prove scrupulously...

I'm confused... :?

im not sure i really understand the problem.  "and the coefficient of greater degree monomial of $P(x)$..." what does that mean?  to find the polynomial you can take the real part of $(\cos{\theta}+i\sin{\theta})^n$, where $\theta=\arccos{x}$.

It means "The coeficient of the highest degree term". Example: $p(x) = 2x^2 + 3x - 1$ the term of highest degree is $2x^2$ and it's coefficient is $2$.

oh ok.  I don't think the problem's right, it should be $2^{n-1}$.  anyways  i think you're on the right track, your problem now is showing $\binom{n}{0}+\binom{n}{2}+\ldots+\binom{n}{n}$ for even $n$, and a similar expression for odd.  to see why it's $2^{n-1}$ play around with the identity $(1+1)^n=2^n=\binom{n}{0}+\binom{n}{1}+\ldots+\binom{n}{n}$ to try to get what you want.

Really. I've got a writing mistake... It shall be $2^{n-1}$, just get the example for $n=1$. But the question is for every $n \in \mathbb{N}$

i know it's all N, the only difference between the expression i gave for odd integers is the bottom index on the rightmost term will be $n-1$ instead of $n$. thats why i said 'a similar expression for odd', instead of writing it out again.  the last identity's also true for all $n$.  my hint works, but theres a few  ways to prove the identity you want.


inequalities
logarithms
function
inequalities unsolved

If $ x,y,z > 0$ Show that
$ \left( {\frac{{x^2  \plus{} y^2  \plus{} z^2 }}
{{x \plus{} y \plus{} z}}} \right)^{x \plus{} y \plus{} z}  \geqslant x^x y^y z^z  \geqslant \left( {\frac{{x \plus{} y \plus{} z}}
{3}} \right)^{x \plus{} y \plus{} z}$

Jensen works well here :)
I.

$ \left( {\frac {{x^2 \plus{} y^2 \plus{} z^2 }} {{x \plus{} y \plus{} z}}} \right)^{x \plus{} y \plus{} z} \geqslant x^x y^y z^z$

After taking $ \ln$ of both sides and dividing by $ x\plus{}y\plus{}z$ we get

$ \ln ( \frac{x}{x\plus{}y\plus{}z} \cdot x \plus{}\frac{y}{x\plus{}y\plus{}z} \cdot y \plus{}\frac{z}{x\plus{}y\plus{}z} \cdot z ) \ge \frac{x}{x\plus{}y\plus{}z} \cdot \ln x \plus{}\frac{y}{x\plus{}y\plus{}z} \cdot \ln y \plus{}\frac{z}{x\plus{}y\plus{}z} \cdot \ln z$

and this is true because $ f(x)\equal{}\ln x$ is concave on $ (0, \plus{} \infty)$

II.

$ x^x y^y z^z \geqslant \left( {\frac {{x \plus{} y \plus{} z}} {3}} \right)^{x \plus{} y \plus{} z}$

just do the same things, but take a function $ f(x)\equal{}x \cdot \ln x$ which is convex on $ (0, \plus{} \infty)$

:)


[quote="mdk"]For each integer n>1, prove that n^n−n^2+n−1 is divisible by (n−1)^2[/quote]
$ n^{n}-n^{2}+n-1=(n-1)(n^{n-1}+n^{n-2}+..+n+1)-n(n-1)=(n-1)(n^{n-1}+n^{n-2}+...+n^{2}+1)$
now we only have to prove that $ n^{n-1}+n^{n-2}+...+n^{2}+1$ is divisibled by $ n-1$
notice that $ n^{k}\equiv 1 (mod n-1)$ and the sum $ n^{n-1}+n^{n-2}+...+n^{2}+1$ has $ n-1$numbers ^^

[hide="Also"] Simply expand, and it is easy to see that
$ (1+m)^{m+1}-(1+m)^{2}+m$ has no constant or $ m$ term and  thus is divisible by $ m^{2}$
[hide="This is"]$ =1+(m+1)m+(\dots)m^{2}-(1+2m+(m^{2}))+m$[/hide]
(m=n-1  is what we used above)
 [/hide]


calculus
integration
trigonometry
calculus computations

[quote="AndrewTom"]$ \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac {\cos 2x}{\cos^{2}x}\ dx.$[/quote]

the given integral equals

 $ \int_{\frac {\pi}{4}}^{\frac {\pi}{3}} \frac{\cos^2 x\minus{}\sin^2 x}{\cos^{2}x}dx\equal{}\int_{\frac {\pi}{4}}^{\frac {\pi}{3}} (1\minus{}\tan ^2 x) dx\equal{}
\int_{\frac {\pi}{4}}^{\frac {\pi}{3}} (2\minus{}\sec^2 x)dx\equal{}2(\frac {\pi}{3}\minus{}\frac {\pi}{4})\minus{}(\tan \frac {\pi}{3}\minus{}\tan\frac {\pi}{4})\equal{}\frac{\pi}{6}\plus{}1\minus{}\sqrt{3}$


Rocks on an n*n*n chessboard.Obviouly n is the smallest number of rooks which can dominate an n*n chessboard. But what is the number R(n) of rooks, which can dominate an n*n*n chessboard?

No idea, but since each rook can cover at most $ 3n \minus{} 2$ tiles, we at least certainly have the bounds
\[ \frac {1}{3}n^2 < R(n) < n^2
\]
Some (probably crude) upper bounds based only on numerical experiments:
$ R(5)\leq 14$; $ R(6)\leq 18$; $ R(7)\leq 27$; $ R(8)\leq38$; $ R(9)\leq 47$; $ R(10)\leq 59$


I'll try to solve it, hungkhtn. Thank you!

I don't think that you can find the maximum value without using computer , hungkhtn ! So , do you have an exactly result , hungkhtn ?

Of course, the exact value, with classical method, even no computing  :lol:

I am really curious about the solution by Cauchy, as toanhocmuonmau say. It is really special.

Yet I ask VASC as the last time before I must forget this very nice result  :blush: Actually, the solution is easy.

[quote=hungkhtn]This problem I created one year ago and intended to make it as a part of VIF Contest, but now it is not. This is very the same as Jack's one.

Let $x,y,z$ be non-negative real numbers with sum 1. Find the maximum value of
\[\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}}. \]
The result is a magic arrangement between $17$ and $71$.[/quote]
The answer is $$\sqrt{\frac{71-17\sqrt{17}}{32}}.$$
There is a nice solution.

We may assume WLOG that $x \ge y \ge z$ or $x \le y \le z$. For the case $x \le y \le z$, we have $$\sum_{cyc}\frac{x-y}{\sqrt{x+y}} \le 0.$$Hence, the left is to prove for the case $x \ge y \ge z$.

First, we claim that $$\sum_{cyc}\frac{x-y}{\sqrt{x+y}} \le \frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x}$$
[hide=Proof.]From $x\ge y \ge z \ge 0$, we have $$\frac{x-y}{\sqrt{x+y}}+\frac{y-z}{\sqrt{y+z}}+\frac{z-x}{\sqrt{z+x}} \le \frac{x-y}{\sqrt{x+y}}+\frac{y}{\sqrt{y+z}}-\frac{x}{\sqrt{z+x}}$$
The left is to prove that $$\sqrt{x}-\frac{x}{\sqrt{x+z}} \le \sqrt{y}-\frac{y}{\sqrt{y+z}}$$or $$\frac{\sqrt{x}}{(x+z)+\sqrt{(x+z)x}} \le \frac{\sqrt{y}}{(y+z)+\sqrt{(y+z)y}}$$or $$(y+z)\sqrt{x} + \sqrt{(y+z)xy} \le (x+z)\sqrt{y}+\sqrt{(x+z)xy}$$or $$\left(\sqrt{x}-\sqrt{y}\right)(\sqrt{xy}-z)+\sqrt{xy}\left(\sqrt{x+z}-\sqrt{y+z}\right) \ge 0$$, which is obviously true.[/hide]
Let $t=\frac{x}{x+y} \Rightarrow \boxed{\frac{1}{2} \le t \le 1}$. Notice that \begin{align*}\frac{x-y}{\sqrt{x+y}}+\sqrt{y}-\sqrt{x} &= \sqrt{x+y}\left(\frac{x-y}{x+y}+\sqrt{\frac{y}{x+y}} -\sqrt{\frac{x}{x+y}}\right) \\&= \sqrt{x+y}\left[(2t-1)+\sqrt{1-t}-\sqrt{t}\right] \\&\le (2t-1)+\sqrt{1-t}-\sqrt{t}\end{align*}
By method of calculus, we know that the maximum value of $f(t) = (2t-1)+\sqrt{1-t}-\sqrt{t}$ is $\sqrt{\frac{71-17\sqrt{17}}{32}}$ for $t \in \left[\frac{1}{2},1\right]$.


Prove that in any triangle we have:

sum(sa*ma)>=1/4(sum (a^2))+sum (b^2c^2/(b^2+c^2))

where sa is the simedian from A.

cheers ! :D :D

I guess that if one proves that:

[b] sa*ma>=1/4(b^2+c^2-a^2)+b^2c^2/(b^2+c^2) [/b]

then we are done.. by writing the ineq for sb*mb and sc*mc.

cheers! :D

may i know wt's simedian?
sorry i don't know... :(
my math is poor... :(

So it's the isogonal conjugate of the median if I'm not mistaken ???

[quote="Lagrangia"]or perhaps my english is not good to translate that simedian word!

Okay!

simedian means : the simetric of a median from the internal bisector that goes through the same vertex of the triangle

I hope you understand!

cheers![/quote]

wt's simetric... :?
sorry, i haven't met this word be4

So probably the main difficulty is calculating the length of the symedians.
I'll think about it ... :D


[hide="Solution"]

$ \mid 5\minus{}x\mid \le 2$

$ \mid x\minus{}5 \mid \le 2$

$ \minus{}2 \le x \minus{} 5 \le 2$

$ 3 \le x \le 7$

[/hide]

[hide="Another way"]Case 1: $ x\le5$ Then $ 5-x\le2$ and $ x\ge3$
Case 2: $ x\ge5$ Then $ |5-x|=x-5,\le2$ so $ x\le7$

Therefore $ 3\lex\le7$[/hide]

[hide="Two further solutions"]

We could also draw the graphs of $ y \equal{} \mid 5 \minus{} x \mid$ and $ y \equal{} 2$ and see where they meet.

This occurs where $ y \equal{} 2$ meets $ y \equal{} \minus{} x \plus{} 5$ and where $ y \equal{} 2$ meets $ y \equal{} x \minus{} 5$, that is, at $ (3, 2)$ and at $ (7, 2)$ so that the graph of $ y \equal{} \mid 5 \minus{} x \mid$ is below the line $ y \equal{} 2$ for $ 2 < x < 7$.

Alternatively, we could square both sides of the inequality to get

$ 25 \minus{} 10x \plus{} x^{2} \le 4$

$ x^{2} \minus{} 10x \plus{} 21 \le 0$

$ (x \minus{} 3)(x \minus{} 7) \le 0$

A rough sketch of the graph of $ y \equal{} x^{2} \minus{} 10x \plus{} 21$, that is, $ y \equal{} (x \minus{} 3)(x \minus{} 7)$ shows that it is beneath the $ x$-axis for $ 3 < x < 7$, that is it is negative for $ 3 < x < 7$.

[/hide]

thank you very much to
   1)  AndrewTom
   2)  AwesomeToad
your answers are true


[quote="hetten"]Let $ x,y,z$ be positive real numbers such that $ x^2 \plus{} y^2 \plus{} z^2 \ge 12$. Find the minimum of:
$ S \equal{} \frac {{{x}^{6}}}{xy \plus{} 2\sqrt {1 \plus{} {{z}^{3}}}} \plus{} \frac {{{y}^{6}}}{yz \plus{} 2\sqrt {1 \plus{} {{x}^{3}}}} \plus{} \frac {{{z}^{6}}}{zx \plus{} 2\sqrt {1 \plus{} {{y}^{3}}}}$[/quote]
It's very easy!
We have $ 2 \sqrt {1 \plus{} x^3} \leq x^2 \plus{} 2$
then, use CS=> min=96/5

[quote="2424"]
then, use CS=> min=96/5[/quote]
Maybe are you mean Holder? :wink:


[quote="tiredepartment"]Ron has 28 markers, som red (worth 1 dollar each), some green (worth 5 dollars each), some yellow (worth 20 dollars each) and some blue (worth 50 dollars each); the total worth of all the markers is 430 dollars.  If he had twice as many red markers and twice as many green markers, and half as many yellow markers, and half as many blue markers, the total worth would be 290 dollars.  Given that Ron has twice as many green markers as yellow markers, find the number of blue markers Ron has.[/quote]

i spent literally an hour trying to solve this problem algebraically, but i lost myself somewhere...so i guessed and checked and in my first try, i got it (i kinda used common sense too...)

[hide]r=10
b=6
y=4
g=8[/hide]

[hide="Sorta solution."]

Let r, b, y, g equal the number of red, blue, yellow, and green markers respectively.

$r+g+y+b=28\\r+5g+20y+50b=430$

We're also given that Ron has twice as many green markers as yellow markers, or $g=2y$. Substituting,

$r+3y+b=28\\r+5(2y)+20y+50b=430\\r+30y+50b=430$

Now, going back to the original equations, we apply the conditions:

$2r+10g+10y+25b=290\\2r+30y+25b=290$

Now we should have three equations with three unknowns:

$r+3y+b=28\\r+30y+50b=430\\2r+30y+25b=290$

Solving this gives us kuhrazy fractions, so I don't know where I went wrong. Eek.
[/hide]

-BLt

[hide="I also did it with equations and it worked out fine..."] Putting $g=2y$ gives us the following 3 equations:

$r+3y+b=28$   ...(1)
$r+30y+50b=430$   ...(2)
$2r+30y+25b=290$   ...(3)

From (1), $r=28-3y-b$.
Substituting this into (2) and (3),
$27y+49b=402$   ...(4)
$24y+23b=234$   ...(5)

From (5), $y=\frac{234-23b}{24}$.
Substituting this into (4), $27\left({\frac{234-23b}{24}} \right) +49b=402$ and $b=6$. So he has $6$ blue markers :) [/hide]

my answers is the next

 [hide]r=10
b=6
y=4
g=8
the this is that all numbers must be pairs the second is that b must be less than 8
and y must be less tha 6 son then the unique number that satisfy the ecuations are thats

 [/hide]

Good job all ... If you used systems of equations, it shouldn't take longer than 10 min.  But otherwise ...

[hide]b=6 was correct[/hide]

uh...i tried the system of equations but i lost myself somewhere...for an HOUR.

then i tried guess and check and got the answer in a minute (literally)

What's your "favorite" way of solving systems of eqns?

[quote="thebrokenlight"]Ack. I solved the system using Matrices, and it is waaaay to easy to make stupid careless errors >_<.

When I FINALLY got it right, I got r=10, y=4, [b]b=6[/b].[/quote]

Matrices? Doesn't that just make it even longer to solve...? all that expansions and additions and subtractions...


2. A student is asked to design a circuit to supply an electric motor with 1.0 mA of current at 3.0 V potential difference.
(c) The motor can lift a 0.012kg mass a distance of 1.0m in 60s at constant velocity. Determine the efficiency of the motor.

Thanks for any help,

The velocity of the mass the motor lifts is $v = s/t$, where $s = 1\text m$ and $t = 60 \text s$. The power of the motor is then $P_1 = Fv$, where $F = mg$ and $m = 0.012\text{kg}$. The electric power spent by the motor is $P_2 = IU$, where $I = 1\text{mA}$ and $U = 3\text V$. The efficiency is $\eta = \frac{P_1}{P_2} = \frac{mgs}{IUt} \approx 0.67$.


Prove this sequence $ x{}^{x{}^x^{...}}mod(n)$ become constant.


[quote="4everwise"]1) Simplify:   (1+3+5+...+199)/(2+4+6+...+200)

2) Find the value of x if the sequence  4x-1, 2x+2, and 2x-3 is an arithmetic progression[/quote]

[hide]
1:
The sum of the first n odd numbers = $n^2$...
..so $100^2 = 10000$.
The sum of the first n even numbers = $n^2 + n$...
..so $100^2 + 100 = 10100.
10000/10100 = 100/101

2:
(2x + 2) - (2x - 3) = 5
..so it is progressing by 5.

Then, we simply solve for:
4x - 1 = 2x + 2 + 5
x = 4
[/hide]

[hide]  1. The sum of N odd numbers is n^2.  There are 100 odd numbers thus their sum is 100^2=10000.  Consequently each even term is 1 bigger than each odd term, so its sum is 10100.  Our fraction is now: 10000/10100=100/101

2. [2x+ 2] -  [2x-3]  =5,  meaning our progression is increasing by 5.  Thus,

2x + 2  + 5 = 4x-1

2x + 7= 4x -1

2x + 8 = 4x

8=2x

x=4[/hide]

New question!

What is the sum of the natural numbers that are not perfect squares from 1 to 1000. Give your full solutions.

Sum of all numbers form 1 - 1000 is
Sn=n(n+1)/2
sum of numbers form 1 -1000 is 500500
Sum of first 31 perfect squares is given by
1/6n(n+1)(2n+1)=10416

Therefore ths sum of t5he given question become 500500-10416=490084

omg... i spent 15 minutes trying to figure out how to latex this and make it all nice and pretty using like sigma signs and all that.. then i messed up and clicked back and it said no post mode something...  :mad:


Nickolas,Your inequality is equal to:
(a-b) 2 [(a+b) 2 +(a+b)/2+ (\sqrt a 2 +b 2 )/2] \geq 0.
And this one is not a joke,ok....? :)

Ahh, I must be missing something, Anh Cuong how did you derive your inequality from nickolas'?. I don't see it though :? . Can someone expand this a little bit? I know it's a silly request... :blush:  :blush:  :blush:


At Forlorne Middle School, a student needs a 94\% average to earn an A. Ty Wan has 4 test scores in mathematics: 94\%, 93\%, 96\%, and 91\%. What is the lowest percent score he could earn on his fifth test and still earn an A?

Let $ x$ be his fifth score.

$ 94\plus{}93\plus{}96\plus{}91\plus{}x\equal{}94(5)\equal{}470$
$ 374\plus{}x\equal{}470$
$ x\equal{}96$

Ty Warn needs a $ \boxed{96}$ on his fifth test to get an A.


The following shapes $ A,B,C$ have the same area. Which letter corresponds to the shape which has the smallest perimeter?
[asy]import graph;
size(150);
pair A=(0,0), B=(2,0.6),C=(4,0.6),D=(4,-0.6),E=(2,-0.6),F=(5,1),G=(7,1),H=(7,-1),I=(5,-1);
draw(Circle(A,1));
label("$A$",A);
draw(B--C--D--E--B);
label("$B$",(3,0));
draw(F--G--H--I--F);
label("$C$",(6,0));[/asy]

A circle, which is choice $ \fbox{A}$, always has the smallest possible perimeter for given area. To test this out, you can plug in some values and check.

This question is flawed because the rectangle could have any perimeter. For example, the longer side could be 1000000000000 but the shorter side could be 0.00001 or something.

But im that case the perimeter would be huge and the problem is asking for the smallest perimeter thus not affecting the problem

Oh, whoops! I was thinking the wrong way


calculus
derivative
calculus computations

ok.
yes, this is a homework problem, no, it is not graded - I'm practicing for a test.

given the equation y = sin (kt), I need to find the nonzero values of k that satisfy the differential equation y'' + 9y = 0.

I differentiate y = sin (kt) to get y' = cos(kt) * k via Chain Rule.

I differentiate y' to get y'' = (k^2) * -sin(kt) + 1 * cos(kt) via Product Rule.

at this point I get stuck, and can't reason out how to isolate the value of k - I can't figure out how to k out of the "sin (kt)" so I can solve for k without giving a solution that includes k.

I am sure I am overlooking something brutally obvious, but I can't see it. Can you pull out the sin (kt) into k sin (t) or something, or is there a way to use the double angle formula? you can't know if it's a double, though.

I know the answer is +/- 3, but I want to know how to get there.

since y'' / y = -9, I can get { (k^2) * sin(kt) + cos(kt) } / sin (kt), and get -9 = (k^2) + tan (kt).. but I still can't get rid of the k in tan(kt)...

[quote="adamcannon"]I differentiate y = sin (kt) to get y' = cos(kt) * k via Chain Rule.

I differentiate y' to get y'' = (k^2) * -sin(kt) + 1 * cos(kt) via Product Rule.[/quote]
That second derivative is incorrect. $k$ is a constant, and the derivative of a constant is zero.

The correct second derivative is $y''=-k^2\sin(kt).$


geometry
circumcircle
projective geometry

Quadrilateral $ ABCD$ is cyclic. The tangents to the circumcircle of $ ABCD$ at $ A, B, C, D$ are, respectively, $ t_A, t_B, t_C, t_D$.

$ t_A \cap t_C \equal{} P_1$

$ t_B \cap t_D \equal{} P_2$

$ AB \cap CD \equal{} P_3$ (the lines)

$ BC \cap DA \equal{} P_4$ (the lines)

Prove $ P_1, P_2, P_3, P_4$ are collinear.

It's not too tough, but I thought it was a pretty nice property.

[hide="Solution"]

Lemma: Consider points $ LMNPQR$ on a circle. By Pascal's theorem, $ LM \cap PQ$, $ MN \cap QR$, $ NP \cap RL$ are collinear. If we force $ L$ and $ M$ to coincide and call the point $ S$, and we force $ P$ and $ Q$ to coincide and call that point $ V$, then we get $ t_S \cap t_V$, $ SN \cap VR$, $ NV \cap RS$ are collinear.

If we apply the lemma, we find that $ t_A \cap t_C, AB \cap CD, BC \cap DA$, or $ P_1, P_3, P_4$, are collinear.

Applying the lemma again, we find that $ t_B \cap t_D, AB \cap CD, BC \cap DA$, or $ P_2, P_3, P_4$, are collinear.

Since two points determine a line, $ P_1, P_2, P_3, P_4$ are all collinear, and we are done.

[/hide]

dude kevin hu this is extremely well-known

HOWEVER

it is very useful

your proof is the standard proof I think

related things to look at -

applying inversion to this lemma

applying pole-polar to this lemma

it gives some pretty cool results

like here: [url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=214263[/url]

oh man also the exact lemma is here

[url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=177222[/url]

try inverting it

there was a problem I had that was just the inverted version of this  :wink:


a,b,c are integers nonnegatives . such that :
$ a|b^2$    and  $ b|c^2$     and    $ c|a^2$
prove that :  $ abc|a^7\plus{}b^7\plus{}c^7$

Even easier than http://www.mathlinks.ro/viewtopic.php?t=83640 .


:D  ya que mi novia no entra al msn colgare problemas, que tal?

[color=blue][b]Problema 1.[/b] Dado $\triangle ABC$ recto en C, sea D un punto en AB tal que los triangulo ACD y DCB tienen el mismo inradio, demostrar que $[ABC]=CD^2$ ($[ABC]$ significa area del triangulo ABC)

[b]Problema 2. [/b]Dado un triangulo, demostrar que sus puntos de Fermat, el centro de su circunferencia de Euler y el circuncentro estan en una circunferencia. [/color]

(vean en [url]http://http://centros5.pntic.mec.es/~marque12/matem/fermat.htm[/url])

[color=blue][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es 4 y que el área es 2. Según lo anterior se pide calcular la altura.

[b]Problema 4.[/b] Sea ABCDE un pentágono convexo de área 1 (las diagonales quedan dentro del pentágono). Sean P, Q, R y S los baricentros de los triángulos ABE, BCE, CDE y DAE, respectivamente. Indicar la naturaleza del cuadrilátero ABCD y dar su área.

[b]Problema 5. [/b]La circunferencia inscrita en el triángulo ABC tiene centro O y es tangente a los lados BC, AC y AB en los puntos X, Y y Z, respectivamente. Las rectas BO y CO intersecan a la recta YZ en los puntos P y Q, respectivamente. Si  XP y XQ tienen la misma longitud, Indicar la naturaleza del triangulo ABC.

[b]Problema 6. [/b]Los vértices de un triangulo ABC pertenecen a la hipérbola $xy=1.$ Demostrar que el ortocentro también pertenece a la hipérbola.[/color]

(Sug. Usar tangentes)

[color=blue]Problema 7. En ABC, sean $AD,$ $BE,$ $CF$ cevianas concurrentes en M. Si los triángulos BDM,
CME y AMF tienen la misma área y el mismo perímetro, entonces $\triangle ABC$ es equilátero.[/color]

[color=red][b]Problema 8. [/b](Dedicado a $Tipe$) Los números enteros del 1 al 2002, ambos inclusive, se escriben en una pizarra en orden creciente 1, 2, . . . , 2001, 2002. Luego, se borran los que ocupan el primer lugar, cuarto lugar, séptimo lugar, etc., es decir, los que ocupan los lugares de la forma 3k + 1. En la nueva lista se borran los números que están en los lugares de la forma 3k +1. Se repite este proceso hasta que se borran todos los números de la lista. ¿Cuál fue el último número que se borró?  :dry: (detesto la aritmetica)[/color]

Espero manden sus soluciones prontito  ;)

Cuidense

José Carlos

:D jaja... el problema dedicado a mí es el problema numero 1 de la iberoamericana del 2002... dicho sea de paso fue propuesto por un peruano  :D

$Tipe$

:!:  ya sabemos que esta dedicado a ti pero manda la solucion pe  :P
La esperamos

Porque tan apurado  :lol:  ... hay que esperar un tiempo prudente como para que los demás intenten el problema... no crees? :?:

$Tipe$

En el problema 4 debe ser: Indique la naturaleza del $PQRS$ ....  :?

$Tipe$

[hide="prob5"]
$m<ABP=\frac{B}{2}$, tambien como ZAY:isosceles $m<AZY=90-A$, luego en el triangulo BPZ, $m<BPZ=\frac{C}{2}=<YPO$, tambien $m<OCY=\frac{C}{2}$, luego $90=m<OYC=<OPC=<OXC$, luego $OPCY: inscriptible$ entonces $\frac{C}{2}=<OCX=<OPX$, luego $<QPX=C$, analogamente: $<QXP=B$, luego los triangulos QXP y BAC son semejantes, como QX=XP, entonces BA=AC.[/hide]
[hide="prob8"]
formare las listas de los numeros que va quedando y sea x_n el primero de la n-esima lista:
$x_{1}=1, x_{2}=2, x_{3}=3, x_{4}=5, x_{6}=8, x_{7}=12$, es posible elaborar una recurrencia pero por falta de time dire lo sgte. x_n=(3/2)x_(n-1), si x_n-1 es par, x_n=(3/2){x_(n-1)-1}+2, si x_{n-1} es impar(demostrar por induccion), luego la sucesion sera 1,2,3,....,1065,1598, es la rpta. luego lo termino...[/hide]


Most students found this year's exam harder than other recent exams.

Well this year's AMC 10 was harder than both of last year's. I barely improved my score (124)

Josh

[quote="Rep123max"]A 106. I can only compare it to one of the tests i took this year (only took one full length practice) and on that I got like a 127 or something so I did terrible.[/quote]

I gave a practice exam to a student the other day who scored 40 points higher on a past AMC than they did yesterday.  If you judge that 21 point differential as "terrible" then I don't think you're being fair to yourself.  I think this year's exams will see far fewer high scorers than in recent years.

I hope so (got a 125.5). Do you think the top 1% will be less than 120?

It will be very likely if all the best people get less than 130 on this test (by the way, I got 110). I'm hoping that the qualifying score is low enough for me to get into AIME.... oh well, I'll just do AMC 10B then...

[quote="Dandelion"]could someone please post the unofficial answers?

Thanks[/quote]

I don't have the answer key in front of me, but we are currently working through the problem in a Math Jam (free class).  Feel free to join us.

Oh yeah, and I got a 138.  :(  Missed two that I should have gotten #2 and #24.)

I got 134.5 :D but i missed #3 which i should've obviously got and #19 cuz i added wrong :( but i skipped #23 which turns out to be not too hard... btw, someone else in my school got a 140.5 :shock:

hehe...this kid near me got 150.  GRRR!  btw, I think I got 138 or worse if I made stupid errors...don't have the answer sheet though.

-interesting_move

I made the two stupid mistakes of the two lines intersecting the 3 concentric circles and the game one...counted permutations, not number of steps, which got me 38 instead of 37.

-interesting_move

there were too many geometry questions near the end... ended up leaving 3 blank - all circle/sphere ones

ehh.... jus checked and i got 139.5  :o

My score was 104.  I'm kinda embarrassed to admit that was the highest score for my high school!

I had a 130 or a 124, I only did about 20 of them and then did the others that I thought would take a while at the end.  I got all the geometry problems right though.  Oh and Southpaw, you definitely made AIME if you had higher than a 120.

yea... i got 140.5.... it was easier than i thought... and i took some advice from Mr. crawford on how to be prepared... i ate candy before too!

i should prepare for AIME now... i won the school winner even though im only a freshman in Washington and other ppl are btter than me in our shcool... now i goota work harder..

xxreddevilzxx, GOOD JOB! What have you been doing to improve your math skills since last year's AMC? Many freshmen experienced lower score than their 8th grade. Are you taking Geometry class this year at school?

It seems that every person who scored well on this test commented on the easiness on the geometry problems. Maybe it's personal opinion, but I think every problem OTHER than geometry problems can be rather easily solved. There are a couple geometry problems that required a bit of calculation work (such as #21 and #23) and #25 is pretty complicated, in my opinion. Maybe I missed some shortcuts to these problems.....

xxreddevilzxx, you live in washington? who are you and what school? and congrats on the score; tied for the best i've seen in washington and my school (not me though :(...)

paladin8, u live in washington, too?
i go to thomas jefferson hs.. in auburn, if u heard of it...
where do u go ?

ahh... thomas jefferson? then perhaps you saw me at NWMC last saturday... im from bellevue, we got 1st place 9/10 team :D

yea... i did bad there... i lost my contacts and other things happened and my condition was horrible... i got 7th though, which isnt bad considering my condition...


function
algebra
polynomial
real analysis
real analysis unsolved

Let $ f \in C[0,1]$, prove that sequence of functions
$ f'_n(x) \equal{} \bigg(\sum\limits_{k \equal{} 0}^n C_n^kf( \frac {k}{n} )x^k (1 \minus{} x)^{n \minus{} k}\bigg)'$
uniform convergence to $ f'(x)$ on $ [0,1]$

[quote="JRav"]What you are referring to is the [url=http://en.wikipedia.org/wiki/Bernstein_polynomials]Bernstein polynomials[/url].[/quote]
Could you write to me the solution?


in $\triangle ABC$ $\angle A=90$, $BE$ and $CD$ are bisector of $\angle B$ and $\angle C$ respectively.If $BE=n_{_{B}}, CD=n_{_{C}}$, $AB=c, BC=a$ and $AC=b$ prove that $\frac{a+b}{a+c}=(\frac{b.n_{_{B}}}{c.n_{_{C}}})^{2}$

[quote="shobber"]Since $n_{B}= \frac{c}{\cos{\frac{B}{2}}}$ and $n_{C}=\frac{b}{\cos{\frac{C}{2}}}$, so: \[\left( \frac{b \cdot n_{B}}{c \cdot n_{C}}\right)^{2}=\frac{\cos^{2}{\frac{C}{2}}}{\cos^{2}{\frac{B}{2}}}=\frac{1+\cos{C}}{1+\cos{B}}=\frac{a+b}{a+c}\] [/quote]
what does the [i]n[/i] represent?

but it is clear from the first post, anyway n_i is the bisector line coming from the angle i.

davron


trigonometry
function
calculus
calculus computations

Show that there is exactly one solution on the equation cos(x)=0.5(x^2 + 1)   on [0, pi/2]

Hints:

1. Subtract to define a function. Let $g(x)=\cos x-\frac12\sqrt{x^2+1}.$ You're looking for zeros of $g.$

2. Showing that a function has at least one zero calls for the use of the Intermediate Value Theorem.

3. Showing that a function has a most a certain number of zeros calls for the use of the Mean Value Thoerem or Rolle's Theorem, or, more simplistically but still from the same family of theorems, an examination of intervals of increase and decrease.

Note that $g$ is both continuous and differentiable everywhere.

What's $g(0)?$

What's $g(\pi/2)?$

Find $g'(x).$ Does it keep the same sign on $(0,\pi/2)?$

What do the answers to these questions tell you?


Let $ a_i, b_i, c_i\ge0\ \ \ (i\equal{}1,...,n)$ such that $ \sum a_i\equal{}\sum b_i\equal{}\sum c_i\equal{}1$. Prove (or disprove)

\[ \sum_{i\equal{}1}^n\frac{(n\minus{}1)^2}{(1\minus{} a_i)(1\minus{} b_i )(1\minus{} c_i)}\ge\frac{n^2(n\plus{}1)}{1\minus{}\sum a_i b_i c_i}\]
 :)  :maybe:

BUMP this topic... Wow, this one was unsolved in two years... :?

[quote="myapsid"]This inequality doesn't hold.

counter-example:
 let a_i=b_i=0 (i=1,...,n-1), a_n=b_n=1 and c_i=1/n(i=1,...,n),
then  left =(n-1)^3/(1-1/n), right =n*n*(n+1)/(1-1/n), obviously left<right.[/quote]
If $a_n\to 1$, LHS tends to $+\infty$, you have omitted the last term. OK, to make the statement waterproof, let's add the rather obvious condition $a_i,b_i,c_i<1$.

say $x_{100}=50.5$. Yes, you are right. I see that putting $x_{1}=x_{2}=...=x_{n-1}\approx.685$ (worst case) makes it false for all $n>30$. What a pity!

Well, it seems to hold for small $n$ though... [b]fjwxcsl[/b], any idea how far it holds :?:

i see that the left hand side's degree should be bery large :maybe: ,or maybe it actually won't be true ...


does anyone know what we use this formula for ?

$=(sum of digits)(n-1)!(\frac{10^{n}-1}{10-1})$

and

$n![1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+...+(-1)^{n}\frac{1}{n!}]$

its in olympiad form they havnt said how it works,  i didnt get it how it works.ill be happy if u explain fully please.

I'm not quite sure what you're saying.
$=(sum of digits)(n-1)!(\frac{10^{n}-1}{10-1})$ Is not a complete formula.

[url=http://mathworld.wolfram.com/Derangement.html]the second equation[/url] look for equation (1)
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=623983]a derangement topic[/url]

the formula is n this page, check it out http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87557

i dont get it, how do we use it ?


trigonometry
ratio
geometry
angle bisector

Suppose triangle $ABC$ has the property that $\angle{C}=2\angle{A}$. If side $a=49$ and side $b=72$, find side $c$.

[hide="Non-trig solution"]
Draw point $D$ on $AB$ such that $CD$ bisects $\angle{C}$. Then $\angle{BCD}=\angle{DCA}=\angle{A}.$

A triangle's external angle is equal to the sum of the two opposite internal angles. So, applying this to $\triangle{DCA}$, we get $\angle{CDB}=2\angle{A}=\angle{BCA}$. By angle-angle, We have $\triangle{ABC}\sim\triangle{CBD}$.

Using the similarity we get the ratio $\frac{BD}{49}=\frac{49}{AB}$. By the angle bisector theorem $BD=\frac{49}{49+72}AB$. Substituting this into the ratio and solving for $AB$ gives

$\frac{\frac{49}{121}AB}{49}=\frac{49}{AB}$
$\frac{49}{121}AB^{2}=49^{2}$
$AB^{2}=49\cdot 121$

$AB=77$[/hide]


What does a bot and an administrator do?  I read some posts from Yahoo!Slurp and he or she said there is a lot of restrictions.  What are they?

The bots are used for search engine indexing. They're here to read and archive things, not to post.
Administrators run the forum- you'll note that all of them work for AOPS.

work for AoPS? Are they on a salary for administrating the forums?

No, they're paid for other things, or are in charge of the company as a whole.


If $f$ is analytic on $(a,b)$ prove that for every $x_{0}$ in $(a,b)$ there exists a neighborhood of $x_{0}$ and constants $M$ and $r$ s.t.

$|f^{(k)}(x)|\leq Mk!r^{k}$ for all $k$ and $x$ in the neighborhood.

Hint: extend $f(x)$ to a complex neighborhood of $x_{0}$. Let $M$ be the supremum of $f(x)$ in the neighborhood.


logarithms
function
inequalities
inequalities proposed

Let $ a,b,c$ be given positive numbers and let
$ E_{n}=a^{n}+b^{n}+c^{n}-a^{n}b^{n}c^{n}-2$.
If $ E_{1}=0$, then $ E_{n}$ has the same sign for any positive integer $ n$.

Nice problem again!  :lol:

I'd better hide my solution (hope correct)
[hide]$ abc+2-a-b-c=(ab-1)(c-1)+(a-1)(b-1)$. Write $ X=(ab-1)(c-1)$, $ Y=(a-1)(b-1)$, $ Z=a^{n}b^{n}$.
Then $ a^{n}b^{n}c^{n}+2-a^{n}-b^{n}-c^{n}=(a^{n}b^{n}-1)(c^{n}-1)+(a^{n}-1)(b^{n}-1)$
$ =Xf(ab)f(c)+Yf(a)f(b)$
$ =Z(Xf(\frac{1}{ab})f(c)+Yf(\frac{1}{a})f(\frac{1}{b}))$

where $ f(x)=1+x+...+x^{n-1}$. We need some property of $ f$ here:
1. $ f$ is increasing. 2. $ \log f$ is concave down. 3. $ \log f(e^{t})$ is convex as a function of $ t$.

Suppose $ a,b\ge 1\ge c$. Then $ Y\ge 0$, $ X\le 0$. We want to prove $ a^{n}b^{n}c^{n}+2\ge a^{n}+b^{n}+c^{n}$ for $ n>1$. It is equivalent to prove that $ f(\frac{1}{a})f(\frac{1}{b})\ge f(\frac{1}{bc})f(c)$. Since $ c\le 1\Rightarrow 1+abc\le a+b\Rightarrow \frac{1}{ab}+c\le\frac{1}{a}+\frac{1}{b}$. Moreover $ \frac{1}{ab}\le\frac{1}{a},\frac{1}{b}$. Hence by property 1. and 2. above we are done.

Suppose $ a,b\le 1\le c$. Then $ Y\ge 0$, $ X\le 0$. We want to prove $ a^{n}b^{n}c^{n}+2\le a^{n}+b^{n}+c^{n}$ for $ n>1$, i.e. $ f(a)f(b)\le f(ab)f(c)$. But $ abc\ge ab$ and $ c\ge a,b\ge ab$. Hence it follows from property 1. and 3.[/hide]

I wonder if Vasc is going to show us something after these problems :D

Have not tried to prove this one, but I am just very curious of the motivation of this kind of problem, VASC? How could you find this?  :blush: Where is its origin?

[quote="hungkhtn"]I am just very curious of the motivation of this kind of problem, VASC? How could you find this?  :blush: Where is its origin?[/quote]

I found these inequalities in my inequality archive. The origin of these inequalities is an article published in Gazeta Matematica A, 3, in 1989.
My main result in this paper is the following rough statement:
  [i]Let $ f$ be a second order convex function (with $ f'''\ge 0$). If $ x+y+z=a+b+c$ and $ x^{2}+y^{2}+z^{2}=a^{2}+b^{2}+c^{2}$, then the expressions
$ F=f(x)+f(y)+f(z)-f(a)-f(b)-f(c)$  and $ D=\min\{x,y,z\}-\min\{a,b,c\}$ have the same sign.[/i]

I think this article is also the origin for EV-Theorem. And conversely, EV-Theorem permits to refine this inequality type. An example is the following.
[i]Let $ a,b,c,d$ be given positive numbers and let $ E_{n}=a^{n}+b^{n}+c^{n}+d^{n}-a^{n}b^{n}c^{n}d^{n}-3$.
If $ E_{1}=0$, then $ E_{n}$ has the same sign for any positive integer $ n$.[/i]

[quote="Darkseer"] where $ f(x)=1+x+...+x^{n-1}$. We need some property of $ f$ here:
1. $ f$ is increasing. 2. $ \log f$ is concave down. 3. $ \log f(e^{t})$ is convex as a function of $ t$.[/quote]
I think $ \log f$ is not concave. See $ f(x)=log(1+x+x.^{2})$. :D

edited: I've been absolutely stupid  :blush: You're right. Thank you Vasc.

by the way, that is a very interesting result, Vasc :)

[quote="Darkseer"]  It is $ (1+x+x^{2})(1+y+y^{2})\le(1+\frac{x+y}{2}+\frac{(x+y)^{2}}{4})^{2}$, isn't it?[/quote]
No, it isn't.
Try $ x=0$ and $ y=0.5$. :P

The first case is very hard.
The second case is easy, by Cebisev.


Dinetai trigono ABC kai o perigegramenos kiklos tou C.Theoroume tis dixotomous tis A, AD,AE me D,E anikoun BC.Theoroume ton kiklo C1 diametrou DE o opoios temni ton C sto R kai i AR temnei tin BC sto K.H efaptomeni tou kiklou C sto A temni thn BC sto L.Theoroume metablito kiklo C2 pou dierxetai apo ta K,L.    [b]N.d.o kath' oli thn kinisi tou C2,o rizikos aksonas ton C1 kai C2 dierxetai apo to meso M tis BC.[/b]

Λοιπόν προηγουμένως δεν είχα διαβάσει καλά την εκφώνηση  :blush:
Τώρα η άσκηση βγαίνει δείχνοντας ότι $MD\cdot ME=MK\cdot ML$  το οποίο μπορεί να βγει κατά διαφόρους τρόπους  :wink:

Επίσης μια άλλη λύση είναι αυτή που χρησιμοποιεί αναλυτική γεωμετρία και σε αυτή πρέπει να χρησιμοποιήσουμε τον τύπο του ριζικού άξονα οποίος προκύπτει από την αφαίρεση των εξισώσεων των δύο κύκλων ..... :wink:


trigonometry
modular arithmetic
algebra unsolved
algebra

Let $ \alpha\equal{}\frac{\pi}{7}$. Prove that
\[ \frac{1}{\sin\alpha}\equal{}\frac{1}{\sin(2\alpha)}\plus{}\frac{1}{\sin(3\alpha)}\]

equivalent to $ sin2xsin3x\equal{}sin2xsinx\plus{}sin3xsinx$,

right hand side = $ sinxsin2x\plus{}sinxsin4x\equal{}2sinxcosxsin3x\equal{}sin2xsin3x$

we use the fact that $ 3x\plus{}4x\equal{}180$

A completely brute force method: let $ x \equal{} e^{i\alpha}$, $ x$ is a primitive root of unity of order 14. Using the representation $ \sin(x) \equal{} \frac {e^{ix} \minus{} e^{ \minus{} ix}}{2i}$ yields:
$ \frac {1}{\sin\alpha} \equal{} \frac {1}{\sin(2\alpha)} \plus{} \frac {1}{\sin(3\alpha)}\leftrightarrow \frac {1}{x \minus{} x^{ \minus{} 1}} \equal{} \frac {1}{x^2 \minus{} x^{ \minus{} 2}} \plus{} \frac {1}{x^{3} \minus{} x^{ \minus{} 3}}$. Expanding it gives $ x^6 \minus{} x^5 \plus{} x^4 \minus{} x^3 \plus{} x^2 \minus{} x \plus{} 1 \equal{} 0$, or: $ x^7 \equal{} \minus{} 1, x \neq \minus{} 1$, which is true. Also follows that the only $ \alpha$s that satisfies the identity are of the form $ \frac {\pi(2t \plus{} 1)}{7}, t \neq 3 \pmod 7$.
Trigonometrical identities and the representation I used are very close so it might be equivalent to greentreeroad's way.

Geometrically:
[hide]Let $ A_1 A_2 \dotsb A_7$ be a regular heptagon, and apply Ptolemy's theorem to quadrilateral $ A_1 A_2 A_3 A_5$.[/hide]


trigonometry
modular arithmetic
geometry
geometric transformation
calculus
calculus computations

An analysis exam is fast approaching and whilst preparing for it I have found something that left me completely clueless:

Determine whether the following series converges pointwise/uniformly on $ \mathbb R$: \[ \sum_{n \ge 1} (\minus{}1)^{n\minus{}1} \cdot \frac{n \cdot \sin (nx)}{n^2 \plus{} a^2} ,\] where $ a$ is some given natural number.

[hide="First part."]
It took me quite a while to notice that [b]pointwise convergence[/b] follows from Abel's (or Dirichlet's or however you call it) Test.
[/hide]

Now comes the second part, [b]uniform convergence[/b]. No idea. The answer should be negative, though.
(If the papers that I have are to be trusted.)

at first we must note that we only need to consider serie $ \sum( \minus{} 1)\frac {\sin nx}{n}$
we know that $ \sum \frac {\sin nx}{n}$ don''t convergent uniformaly on $ R$ because problem is in point which is closely $ 0$ or $ x$ like thi $ \frac {1}{n}$
we can find sum of your serie $ \sum_{1} ( \minus{} 1)\frac {\sin nx}{n} \equal{} \minus{} \sum _{1}\frac {\sin (2k \plus{} 1)x}{2k \plus{} 1} \plus{} \sum_{1} \frac{\sin 2kx}{2k}$
and
$ \sum _{1}\frac {\sin (2k \plus{} 1)x}{2k \plus{} 1} \equal{} \frac {\pi}{4} sgn x$ $ (|x| < \pi)$
and
$ \sum_{1} \frac{\sin 2kx}{2k} \equal{} \frac {\pi \minus{} x}{2} \minus{} \frac {\pi}{4} sgn x$ $ (|x| < \pi)$
in finally
$ \sum_{1} ( \minus{} 1)\frac {\sin nx}{n} \equal{} \minus{} \frac {\pi}{2} \plus{} \frac {\pi \minus{} x}{2} sgn x$ $ (|x| < \pi)$
and if we put $ x \equal{} 0$ we see that there is problem therefore we have not uniform convergent.

Let $ f(x) \equal{} \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n \minus{} 1}\frac {n\sin nx}{n^2 \plus{} a^2}.$

Since I'm annoyed with the $ ( \minus{} 1)^{n \minus{} 1},$ I'll get rid of it:

Let $ g(x) \equal{} \minus{} f(x \plus{} \pi) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {n\sin nx}{n^2 \plus{} a^2}.$

Now I note that $ \frac {n}{n^2 \plus{} a^2} \equal{} \frac1n \minus{} \frac {a^2}{n(n^2 \plus{} a^2)}.$

Thus, $ g(x) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {\sin nx}{n} \minus{} \sum_{n \equal{} 1}^{\infty}\frac {a^2\sin nx}{n(n^2 \plus{} a^2)} \equal{} h(x) \minus{} s(x).$

$ h(x)$ is a fairly well-known sum: $ h(x) \equal{} \frac {\pi \minus{} x}2$ for $ 0 < x < 2\pi,$ extended to be periodic of period $ 2\pi,$ with $ h(0) \equal{} 0.$ In particular, $ h$ has a jump discontinuity at $ x\equiv 0\pmod{2\pi}.$

$ s(x)$ is given by a series that is absolutely and uniformly convergent by the Weierstrass M-test. Hence $ s(x)$ is continuous.

So $ g(x)$ has a jump discontinuity at $ x \equal{} 0$ and is continuous elsewhere. Undo the translation I did at the beginning, and we conclude that $ f(x)$ has a jump discontinuity at $ x \equal{} \pi$ and is continuous elsewhere. But since $ f$ is discontinuous, the series that converges to it cannot converge uniformly.

The way I've written it also suffices to prove the first part, the pointwise convergence. I this case the summation-by-parts/Dirichlet-test argument is hiding inside my claim that $ h(x)$ is a well-known sum.

[It should be possible to find a closed-form formula for $ f(x),$ although I haven't bothered to.]

[color=green][Somehow, I hadn't noticed didilica's post before I wrote that. Of course, this is exactly the argument he proposed.][/color]


trigonometry
function
inequalities
calculus
inequalities unsolved

[i][u]Prove that : [/u]

  with $ A , B , C$ are three degrees of the triangle $ ABC$

We have :

  $ \ cos{\frac{A}{12}}\plus{}cos{\frac{B}{12}}\plus{}cos{\frac{C}{12}} > \frac{8\plus{}\sqrt{6}\plus{}\sqrt{2}}{4}$[/i]

[quote="Mij"][hide]Lemma: For any angles $ \alpha$, $ \beta$, $ \theta$ satisfying $ 0 \le \alpha < \beta < 90^\circ$, $ 0 < \theta \le \dfrac{\beta \minus{} \alpha}{2}$, the following holds:
$ \cos(\alpha \plus{} \theta) \plus{} \cos(\beta \minus{} \theta) > \cos \alpha \plus{} \cos \beta$
Proof: Consider the function $ f(x) \equal{} \cos x \minus{} \cos(x \minus{} \theta)$, where $ 0 < \theta < x < 90^\circ$.
$ f'(x) \equal{} \minus{} \sin x \plus{} \sin(x \minus{} \theta)$, and since $ \sin x$ increases over the interval $ (0, 90^\circ)$, $ \sin x > \sin(x \minus{} \theta)$, and $ f'(x) < 0$. $ f(x)$ is decreasing.
Since $ \alpha \plus{} \theta < \alpha \plus{} \dfrac{\beta \minus{} \alpha}{2} \equal{} \dfrac{\beta \plus{} \alpha}{2} < \beta$,
$ \cos (\alpha \plus{} \theta) \minus{} \cos \alpha > \cos \beta \minus{} \cos(\beta \minus{} \theta)$
$ \cos(\alpha \plus{} \theta) \plus{} \cos(\beta \minus{} \theta) > \cos \alpha \plus{} \cos \beta$
thus proving the lemma.
By the lemma, $ \cos \dfrac{A}{12} \plus{} \cos \dfrac{B}{12} > \cos 0 \plus{} \cos \dfrac{A \plus{} B}{12}$
and
$ \cos \dfrac{A \plus{} B}{12} \plus{} \cos \dfrac{C}{12} > \cos 0 \plus{} \cos \dfrac{A \plus{} B \plus{} C}{12}$
Adding together these two inequalities, we get
$ \cos \dfrac{A}{12} \plus{} \cos \dfrac{B}{12} \plus{} \cos \dfrac{A \plus{} B}{12} \plus{} \cos \dfrac{C}{12} > \cos 0 \plus{} \cos \dfrac{A \plus{} B}{12} \plus{} \cos 0 \plus{} \cos \dfrac{A \plus{} B \plus{} C}{12}$
$ \cos \dfrac{A}{12} \plus{} \cos \dfrac{B}{12} \plus{} \cos \dfrac{C}{12} > 2\cos 0 \plus{} \cos \dfrac{A \plus{} B \plus{} C}{12} \equal{} 2\cos 0 \plus{} \cos 15^\circ \equal{} \dfrac{8 \plus{} \sqrt {6} \plus{} \sqrt {2}}{4}$
as desired.[/hide]
How would I prove my lemma without calculus?[/quote]
It isn't a proof without calculus, but since $ \cos x$ is concave in the given interval and $ (\beta \minus{} \theta,\alpha \plus{} \theta) \prec (\beta,\alpha)$ the lemma follows by Karamata.

To prove it without calculus you should try to use the addition formulas. I ended up with something like $ \frac {\sin \theta}{1 \minus{} \cos \theta} \ge \frac {\cos \alpha \plus{} \cos \beta}{\sin \beta \minus{} \sin \alpha}$.

Edit:
My proof:
Since $ (\frac {A}{12}, \frac {B}{12}, \frac {C}{12}) \prec (15^\circ,0,0)$ and $ \cos x$ is concave on the given interval it follows by Karamata that $ \ cos{\frac {A}{12}} \plus{} cos{\frac {B}{12}} \plus{} cos{\frac {C}{12}} > cos(15^\circ) \plus{} 2 \cdot cos(0) \equal{} \frac {8 \plus{} \sqrt {6} \plus{} \sqrt {2}}{4}$ And this should do it :) (I haven't checked the last equality)


From condition 1 deduce $x,y,z$ are the same sign.
Consider $x,y,z>0$, let $x=\tan\frac{A}{2},y=\tan\frac{B}{2},z=\tan\frac{C}{2}$
From the condition 2, we have $A,B,C$ are three angles of triangle $ABC$
From the condition 2, too, by tranformating we get:
$3\left(\frac{x^{2}+1}{x}\right)=4\left(\frac{y^{2}+1}{y}\right)=5\left(\frac{z^{2}+1}{x}\right)$
$\Rightarrow \frac{3}{\sin A}=\frac{4}{\sin B}=\frac{5}{\sin C}$
Hence $x=\tan\frac{A}{2}=\frac{1-\cos A}{\ sinA}=\frac{1}{3}$
Similarly $y=\frac{1}{2},z=1$
Therefore all roots of system are $(x,y,z)=\left(\frac{1}{3},\frac{1}{2},1\right)=\left(-\frac{1}{3},-\frac{1}{2},-1\right)$

[quote]
$\therefore \ (x,y,z)=\left(\pm\frac{1}{3},\pm\frac{1}{2},\pm1\right), (\pm i, \mp i, \pm i), (\pm i, \mp i, \mp i), (\pm i, \pm i, \mp i),$
(Eight solutions)[/quote]

But you don't check some of solution, for example $(x,y,z)=(i,i,i)$, i see that isn't satified the system[/quote]

[hide]
We have $x+\frac{1}{x}=\frac{x^{2}+1}{x}=\frac{x^{2}+xy+yz+zx}{x}=\frac{(x+y)(x+z)}{x}$, so that
\[\frac{3(x+y)(x+z)}{x}=\frac{4(x+y)(y+z)}{y}=\frac{5(x+z)(y+z)}{z}\]
If any of $x+y,x+z,y+z$ equal 0, then two of them equal 0, so two variables are equal and one is the additive inverse of the other two. For the case $x=y,z=-x$, substituting into condition gives $x^{2}=-1$ so $x=\pm i$ (if nonreal solutions allowed); the other two cases are similar. This gives 6 solutions: $(i,i,-i)$, $(-i,-i,i)$ and permutations.

Otherwise, divide by $(x+y)(y+z)(x+z)$ to get
\[\frac{3}{x(y+z)}=\frac{4}{y(x+z)}=\frac{5}{z(x+y)}\Rightarrow \frac{3}{1-yz}=\frac{4}{1-xz}=\frac{5}{1-xy}\]
so the quantities $1-yz,1-xz,1-xy$ are in a 3-4-5 ratio and have a sum of 2. It follows that these quantities are 1/2, 2/3, 5/6. This gives
\[yz=\frac{1}{2},\ xz=\frac{1}{3},\ xy=\frac{1}{6}\]
Multiplying the equations gives $(xyz)^{2}=\frac{1}{36}$ so $xyz=\pm \frac{1}{6}$. When $xyz=\frac{1}{6}$, solution is (1/3,1/2,1) and when $xyz=-\frac{1}{6}$, solution is (-1/3,-1/2,-1). The conditions are met by both these solutions.

Therefore all solutions are $\left( \pm \frac{1}{3},\pm \frac{1}{2},\pm 1\right)$ and (if nonreal solutions allowed) permutations of $(\pm i,\pm i,\mp i)$.
[/hide]

[quote="hien"]But you don't check some of solution, for example $(x,y,z)=(i,i,i)$, i see that isn't satified the system[/quote]
Notice that $(i,i,i)$ is not part of his solutions. Pay attention to the $\pm$ and $\mp$.


linear algebra
matrix
linear algebra unsolved

A hint:

It´s necessary prove that if $J$ is a Jordam matrix then exist a matrix $K$ such that $JKJ=J$. Then the general case is a consequence of a decomposition by a matrix $A$ into it´s Jordan form.

maybe there is no Jordan form in this field ..

[quote="alexilic"]For every matrix $A$ dimension $n \times n$, there exist matrix $B$ with same dimensions, such that
\[ ABA = A.  \]

Prove it...[/quote]

It is known fact that if $A$ is a matrix with rank $r$ then there exist non-singular matrices $P$ and $Q$ such that $PAQ=I_r$, where $I_r$ is $n \times n$ matrix, which has identity $r \times r$ matrix in the left-upper corner and zeros elsewhere. Now from this repseresantation $A=P^{-1}I_rQ^{-1}$, so for such a $B$ we must have $P^{-1}(I_rQ^{-1}BP^-1I_r)Q^{-1}=P^{-1}I_rQ^{-1}$ and it is enough to put $B=PI_rQ$


Find all continuous function $ f: R^\plus{}\to R^\plus{}$ satisfy
$ f(a)\plus{}f(b)\plus{}f(c)\equal{}f(a)f(b)f(c)$
with $ ab\plus{}bc\plus{}ca\equal{}1$

Let $ f(tg x) \equal{} ctg \ g(x), x\in (0,\frac {\pi}{2})$.
Then $ c \equal{} \frac {1 \minus{} ab}{a \plus{} b} \equal{} ctg(x \plus{} y),a \equal{} tgx,b \equal{} tgy$ and $ f(c) \equal{} \frac {f(a) \plus{} f(b)}{1 \minus{} f(a)f(b)} \equal{} \minus{} ctg(\frac {\pi}{2} \minus{} g(x) \minus{} g(y))\to g(z) \plus{} g(x) \plus{} g(y) \equal{} \frac {\pi}{2}$ if $ x \plus{} y \plus{} z \equal{} \frac {\pi}{2}$, were $ c \equal{} tg z$. Let $ g(x) \equal{} \frac {\pi}{6} \plus{} t(x \minus{} \frac {\pi}{6})$, then $ t(x) \plus{} t(y) \plus{} t(z) \equal{} 0$ if $ x \plus{} y \plus{} z \equal{} 0, \ x,y,z\in ( \minus{} \frac {\pi}{6},\frac {\pi}{3})$. Because $ t(x)$ continiosly it give $ t(x) \equal{} Cx$. Because $ f(x)$ positive $ \minus{} \frac {1}{2} \le C\le 1$.
Therefore $ f(x) \equal{} ctg(\frac {\pi}{6} \plus{} C(arctg(x \minus{} \frac {\pi}{6})), \ \frac { \minus{} 1}{2} \le C\le 1$.
If C=1 we get $ f(x) \equal{} \frac {1}{x}$.


search
combinatorics unsolved
combinatorics

Let $n\ge 3$ be an integer and let $A_{1}, A_{2},\dots, A_{n}$ be $n$ distinct subsets of $S=\{1, 2,\dots, n\}$. Show that there exists $x\in S$ such that the n subsets $A_{i}-\{x\}, i=1,2,\dots n$ are also disjoint.

what i have is [hide="this"]we may assume that the union of the $A_{i}$s is $S$. [/hide]


Find the smallest value of $ n$ such that there exist $ n$ positive real numbers $ x_1,x_2,\ldots,x_n$ satisfying the following inequality
$ x^2_1\plus{}x^2_2\plus{}\cdots \plus{}x^2_n \le \frac{x_1\plus{}x_2\plus{}\cdots \plus{}x_n}{2}\le \frac{x^3_1\plus{}x^3_2\plus{}\cdots \plus{}x^3_n}{4}.$

The answer is 516.
 26-th Tournament of Towns, 2006/2007.


Could someone explain why there are two amc 10s and two amc 12s in a year?

For example, on AoPSWiki, I saw

#  2003 AMC 10A
# 2003 AMC 10B

Are we allowed to take both and if so, how? Are we allowed to take the amc 12 and the amc 10 (12a and then 10b or something?)

Also, if I have to take one or the other, which should I chose? In my 06-07 school year, I got a 22 on the AMC 8 and a 96 on the AMC 10.

What are some advantages of taking both the amc 10 and amc 12, or taking the amc10a and b or the twelve a and b?

I think you are allowed to take both because I did this year. If you take AMC 12 and AMC 10, one must be A and the other must be B, they can't both be A. If you are just looking for a high score, the AMC 10 would be the best since it is easier, however, the AMC 12 almost always has a lower score needed to qualify for AIME.
So, basically, if you are looking to get into AIME:AMC 12, to get a high score:AMC 10. I don't think A or B make too much of a difference, but for some reason I always do worse on B's...

As xpmath explained, the AMC permits taking one contest on the A date and one on the B date.  You can take two 10's or two 12's or one of each as long as you have a place to take them.  However, many schools offer just one AMC date and some don't offer any.  Some schools have additional restrictions on who can take the contests and which contests one can take.  If these restrictions negatively impact you, there are some colleges offer the AMC's for students attending any school.

The main reason for two dates is to minimize conflicts with other activities.  The AMC site and several other threads on this forum discuss related issues.  Here's one with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150092]suggestions[/url] for encouraging schools to provide the contests as flexibly as possible.

You are definitely allowed to take both.

I heard once that B's were purposely made to be harder in order to make up for the extra prep time that you earn. Hm. Seems a little fishy, but hey, why not?

I would take the AMC 12 twice, but that's just me.

No, you can take either 2 Amc 10s, two AMC 12s, or 1 Amc 10 and 1 Amc 12.

The A tests (both 10A and 12A) are administered on the same day and at the same time, so it's not exactly physically possible to take both.  Similarly with the B tests.

There are usually some common problems between the two A tests; same for B

[quote="calc rulz"]The AMC has stated over and over that the two tests are of equal difficulty. In fact, the AMC writes the two tests before deciding which is which, and then flips a coin to decide which is the AMC A and which is the AMC B.[/quote]

Okay, explain why AMC 10B #25, was the same as AMC 12B #24?

It had to have been one of the hardest contests in the past years.

And notice that the bulk of the perfect scorers always come from b than a.

But yeah, the contests seem to be of equal difficulty only if you get perfect on both.

And, you can't take two of the contests on the same day.

[quote="mathgeniuse^ln(x)"]Okay, explain why AMC 10B #25, was the same as AMC 12B #24?[/quote]

Answered in the post above yours.

[quote="mathgeniuse^ln(x)"]And notice that the bulk of the perfect scorers always come from b than a.[/quote]

The pool of people who take the B tests is both smaller than the A pool and more driven.  And, although this might not be a significant factor, this year the B test occurred during our winter break, so I, personally, got much more sleep before the B test than the A test.  You also seem to be contradicting your previous line, which seemed to be implying that the B test tends to be harder than the A test.  What, exactly, are you trying to say?

[quote="mathgeniuse^ln(x)"]But yeah, the contests seem to be of equal difficulty only if you get perfect on both.[/quote]

The issue isn't whether a given year's B test is easier or harder than a given year's A test.  If the test is selected by a coin toss, then the difficulties should fluctuate evenly.  Are you disputing the evenness of a coin toss?

Ok, I guess you are right.

And I meant, the bulk got a than b.


vector
trigonometry
complex numbers
complex analysis
complex analysis unsolved

is there any physical meaning of e^ix ?????

Well it's unit vector in the complex plane, forming an angle $ x$ with the real axis.  But yes, complex numbers are very important, especially in physics.  For example, if we're considering an AC circuit, so voltage is given by $ V=\cos{\omega t}$, it is often times easier to express quantities as the real parts of some complex number.  So for example, voltage can be seen to be $ \text{Re}{e^{i\omega t}}$, and the current can be expressed as $ \text{Re}\frac{\alpha e^{i\omega t}}{R_{\text{eff}}}$, where $ R_{\text{eff}}$ is the effective resistance.  It's actually easier to solve such circuit analysis problems when treating everything as a complex exponential than it is to go through the drudgery of solving massive differential equations.


Given set A={1,2,..,n}.Show that the number of subset of A suchthat its average(arithmetic of it elements) is interger is parity with n/

Let $S$ be a subset of $A$ with an integer average. Either $S$ contains the average (as one of its elements, I mean) or it doesn't.  If it does, we can remove the average and if it doesn't, we can add in the average (bijection!). Thus, there is an even number of subsets $A$ as long as $|A| \geq 2$. All that remains are the one-element subsets, and this clearly has the same parity as $n$.  :wink:


modular arithmetic
combinatorics proposed
combinatorics

Given a 3 edge colored MPG (maximal planar graph, all regions are triangles) with [b]r[/b]ed, [b]g[/b]reen and [b]b[/b]lue.
The number of triangles with rgb order of the edges in clockwise direction, has the same parity as the number of vertices (odd-odd, even-even). Prove it?
kweenie

A triangle is a maximal planar graph.  We can three-color it so that the edges are R-G-B reading counter-clockwise.  $ 0 \not \equiv 3 \pmod 2$.   :| Perhaps you meant something else?   :maybe:

Taking your example with 3 vertices, and a 3-coloring of the edges of the inner triangle in  (counter)clockwise rgb order (then in the "infinite" region the rgb order is opposite). We have then 1 triangle with clockwise rgb order and 3 vertices and both are odd.
Coloring a MPG on 4 vertices gives 4 (or 0) triangles with clockwise rgb order thus both are even. They have the same parity and this is so for any 3-edge-colored MPG with v vertices.
kweenie

Original question:
"Given a 3 edge colored MPG (maximal planar graph, all regions are triangles) with red, green and blue.
The number of triangles with rgb order of the edges in clockwise direction, has the same parity as the number of vertices (odd-odd, even-even). Prove it? "

As there are no answers, some hints:
Put in all triangles with clockwise rgb direction a vertex and join it with the three corner-vertices. Colour (proper) the new edges with rgb. All triangles have now counterclockwise rgb direction.
Prove now that in this new MPG:
1) all vertices have a degree divisable by 3.
2) such MPG has an even number of vertices (that's the hard part)
Then it's clear that the number of added vertices has the same parity as the original number of vertices

Here is the proof:
1) Put in all v' triangles with clockwise rgb direction a vertex and join it with its three corner-vertices. Give the new edges a proper rgb coloring. All triangles have now counterclockwise rgb direction. The total number of vertices is now v+v' and the number of bleu edges is equal to v+v'.
2) The edges around each vertex are now colored red, green and bleu in clockwise rgb direction, and then the degree of each vertex must be a multiple of 3.
3) Delete all red edges. Each vertex has now an even degree and the regions can be two colored.
4) All faces are now quadrangles (two triangles sharing a red edge), and each quadrangle has two bleu and two green edges.
5) Color this quadrangles with black and white. The number of bleu edges is now equal to 2 times the number of black quadrangles. Thus v+v' is even.
6) As v+v' is even, v' must have the same parity as v.
Done.

kweenie

PS: It's also a proof that all maximal planar graphs with degrees divisable by 3, have an even number of vertices.

I am still interested in another proof for this problem instead of my "tricky" one.
Especially for the more general case:
" A maximal planar graph with degrees divisable by 3, has an even number of vertices "

kweenie


Which are the products of this chemical reaction?

(NH4)2Cr2O7(s) + O2 (g) -->

In general - no reaction.

Perhaps in very acidic solutions ammonia can be oxidized to nitric acid - but even if so, there is no need for oxygen to do it - dichromate alone may be sufficient.


Hey, it's been a while since I've posted, hope I can still get some help on these questions....Any info is appreciated

1.
A sample of $S_{8(g)}$ is placed in an empty rigid container at 1325 K at an initial pressure of 1 atm, where it decomposes to $S_{2(g)}$ by the reaction:

$S_{8(g)} === 4S_{2(g)}$

At equilibrium, the partial pressure of $S_8$ is 0.25 atm. Calculate $K_p$ for this reaction at 1325 K.

I tried to do it with an ICE box, but I have two unknown variables in one equation.

2.
Consider the reaction:

$Fe^{3+}_{(aq)} + SCN^-_{(aq)} === FeSCN^{2+}_{(aq)}$

How will the equilibrium position shift if

a. Water is added, doubling the volume
b. $AgNO_{3(aq)}$ is added (AgSCN is insoluble)
c. $NaOH_{(aq)}$ is added ($Fe(OH)_3$ is insoluble)
d. $Fe(NO_3)_{3(aq)}$ is added

3.
The compound $SbCl_{5(g)}$ decomposes at high temperatures to gaseous antimony trichloride and chlorine. When 89.7 g of $SbCl_{5(g)}$ is placed in a 15.0 L container at 180 degrees Celsius, the $SbCl_{5(g)}$ is 29.2% decomposed (by moles) after the system has reached equilibrium.

a. Calculate the value K for this reaction at 180 C
b. Determine the number of moles of chloride gas that must be injected into the flask to make the new equilibrium pressure of antimony trichloride half that of the original equilibrium pressure of antimony trichloride in the original experiment.

Again, any help is greatly appreciated. I spent a long time on these and it's sorta frustrating me because I am lost. Thanks in advance

ah, damn

I looked at these again in the morning, and they were easier

I shouldn;t post problems late at night when my brain doesnt function


IN THE NAME OF GOD

[size=150][color=red]Solve 4^(2x^2+2x )= 8. [/color][/size]

THANKS
SALAH

OK THANK YOU VERY MUCH

Obel1x

bbgun34

THIS IS TRUE SOLUTION

SALAH


given the triangle ABC with its median is CM ( M is the midpoint of AB ) , ME and MF are two bisectors of angle AMC and BMC . N is a point on segment CM , let the intersection of AN and ME be P and the intersection of BN and MF be G
Prove that BG=BF if and only if AP=AE

Very easy, but nice problem; suprinsingly, not given a try yet!
Sketch of a solution:
Let$\{R\}\in CM$ so that $BR\bot MF$ and, as $MF$ is the bisector of $\angle BMC\implies MR=MB\ (\ 1\ )$ and $BFRQ$ is a rhombus, hence $RQ\parallel BC\ (\ 2\ )$. Obviously, $MR=MB=MA$ and $AR\bot BR\ (\ 3\ )$.
Easy to see, $EF\parallel AB\parallel PQ\ (\ 4\ )$ and $\frac{BF}{BC}=\frac{AE}{AC}\ (\ 5\ )$.
With $(3), (4)$ we get $\frac{RQ}{BC}=\frac{NQ}{NB}=\frac{PQ}{AB}\ (\ 6\ )$, so $\Delta RPQ\sim\Delta CAB$ and $RP\parallel AC$, plus $\frac{RP}{AC}=\frac{RQ}{BC}=\frac{BF}{BC}\implies PR=AE$ and $APRE$ is a rhombus (do not forget, $AE=ER$ as well).

Best regards,
sunken rock


If I divide the year of the birth of one of my ancestors by 11, 12, and 13, and I add the remainders, I get 33.  In what year was my ancestor born.

  This is a really good problem, it requires a good deal of investigation and insight.

[hide="Answer"]The only way the remainders can add to $33$ is with the maximum possible remainder after each division is performed. $LCM(11,12,13)=1716 \Rightarrow 1716-1=1715$, which, upon checking, yields remainders of $10$, $11$, and $12$ when divided by $11$, $12$, and $13$, respectively, so $1715$ is the correct answer.[/hide]

Well, that's basically right. However, the question didn't specify that you want the youngest age the ancestor could be. So, the answer is of the form 1716x - 1 for any positive integer x.

[quote="nat mc"]Umm... I don't think G-UNIT has ancestors from the future... :lol:[/quote]

He doesn't have any. It asked for the [b]year[/b]  :lol:

[quote="nat mc"][quote="G-UNIT"]WHO SAID!!!???

  My ancestors are from a parralell universe where time travels faster!!! :rotfl:  :rotfl:  :rotfl:  :rotfl:[/quote]:rotfl:  :rotfl:  :rotfl:  :rotfl:[/quote]

 :rotfl:  Where'd you think of that?

Someone said the year would be 1716x-1, not just 1715.  Every other one comes to be greater than 2005.  Ancestors from the future, AHH :rotfl: :rotfl:

Sorry. Yeah, I thought the question said [i]age[/i] instead of [i]year[/i]. That's why I said 1716x - 1.

He means that's how old they would be if they were still alive. For example, someone born in 934 B.C. would be 2939 years old, if still alive today.


[quote="Jongao"]this is a AMC 10 base problem:
What is the probability that a 3-digit base 10 number is also 3 digits in base 9 and 11?[/quote]
[hide]We want these three inequalities to be true:
$ 100_{9}\le x\le 999_{9}$
$ 100_{10}\le x\le 999_{10}$
$ 100_{11}\le x\le 999_{11}$

Expressing in base ten,
$ 81_{10}\le x\le 729_{10}$
$ 100_{10}\le x\le 999_{10}$
$ 121_{10}\le x\le 1330_{10}$

Hence,
$ 121\le x\le 729$

There exist $ 609$ numbers that satisfy this.
Hence, our probability is:

$ \frac{609}{900}= \boxed{\frac{203}{300}}$[/hide]


Find all pairs positive integers $(a, b)$ such that \[ a^b|b^a - 1 \]

1.$a=1,b-all$,
2.$a-all,b=1$,
3.$a=2,b=3$.
4. Difficult case $2\le b<a$. I think have not another solutions.


logarithms
quadratics
algebra unsolved
algebra

[size=150]Hey guys, I appreciate the help on these logs and I ran into another snag

I tried to use logarit for these but I keep getting the wrong answer

1. 3log(base 2) x + 1/2 log(base 2) x - 3log (base 2) x - 5

2. If log (base b) 3 = 0.56 and log (base b) 2 = 0.36 evaluate...

log (base b) square root of 2/3

3. Solve the equation for x. Give an exact answer and a four place decimal approximation

log (2x+1) = -0.9

4. Solve

log (base 2) x + log (base 2) (3x-2) = 3

PLEASE PLEASE PLEASE HELP!!! Answers are greatly appreciated!

[/size]

[/code]

Here's 3 and 4.

3) $ 10^{\minus{}0.9}\equal{}2x\plus{}1\Rightarrow x\equal{}\frac{10^{0.9}\minus{}1}{2}$ which you need a calculator to do.

4) Since they are in the same base, we can use a property of logs and combine them: $ \log_{2}(x(3x\minus{}2))\equal{}\log_{2}(3x^{2}\minus{}2x)\equal{}3\Rightarrow 2^{3}\equal{}8\equal{}3x^{2}\minus{}2x\Rightarrow 3x^{2}\minus{}2x\minus{}8\equal{}0$.  So now just solve this quadratic, which is easy.


For what it's worth, it should be abelian, and therefore isomorphic with $ H_1$ provided it is connected. So you'd just
have to show that $ H_1$ is finitely generated for a connected lie group -- I would suspect that if it's true for lie groups,
it's probably true for all topological groups, so that might be a good place to start.

EDIT: Scratch that, you definitely need some notion of finite-dimensionality, otherwise $ S^1 \times S^1 \times \dots$ is a counterexample.

[quote="jmerry"]Yeah, I've been staring at this and not seeing anything for a while. One more cheap step: the fundamental group of a $ n$-dimensional connected abelian Lie group is free abelian on at most $ n$ generators.[/quote]

Just curious, where does that come from? My study of topology has been sadly manifolds-deficient ;).

abelian lie groups are tori times euclidean spaces  :wink:

if i'm not mistaken, this should follow from the fact that the exponential map is a group homorphism from the tangent space $ \mathbb{R}^n$ to the group $ G$, so $ G$ is the quotient of $ \mathbb{R}^n$ by a discrete subgroup, which has to be a discrete lattice, so it has $ k\le n$ generators. quotienting out gives you $ T^k\times\mathbb{R}^{n\minus{}k}$

[quote="MysticTerminator"]Is it true that the fundamental group of a Lie group is finitely generated ?[/quote]Yes - if the Lie group is compact. And finite dimensional, of course. ;)

Note that the fundamental group of a Lie group is always abelian.

It's true for any finite-dimensional Lie group, for the inclusion of a maximal torus induces a surjection on fundamental groups.


superior algebra
superior algebra unsolved

$ a_1 , a_2, ... , a_n \in Q^*_\plus{} , n \in N^*$ Prove that $ \sqrt{a_1} \plus{} \sqrt{a_2} \plus{}...\plus{} \sqrt{a_n} \in Q \Leftrightarrow \sqrt{a_1}, \sqrt{a_2} , ... , \sqrt{a_n} \in Q$


calculus
derivative
function
logarithms
trigonometry
limit
real analysis

How do you find the derivative of x^(root(2))?

The result is unsurprising: $ \frac{d}{dx}x^{\sqrt{2}}\equal{}\sqrt{2}x^{\sqrt{2}\minus{}1}.$

What is more interesting is the question of how you even define that function in the first place. What do we mean by $ a^b$ when $ b$ is irrational? The answer is that we first define the function $ e^x$ and its inverse, the natural logarithm function, and then (as long as $ a>0,$ we say that $ a^b\equal{}e^{b\ln a}.$

Then $ \frac{d}{dx}x^r\equal{}\frac{d}{dx}e^{r\ln x}\equal{} e^{r\ln x}\cdot\frac{d}{dx}(r\ln x)\equal{}\frac{r}{x}e^{r\ln x}\equal{}rx^{r\minus{}1}.$

[quote="Ram"]Can u comment anything for negative x? ie like (-1)^(root2)[/quote]

The power function is defined for positive numbers only in general, so..

It's best not even to try to define that.

There is a possible approach in complex variable theory, which relies on choosing a "branch" of the logarithm function. If we choose a branch of the logarithm such that $ \log(\minus{}1)\equal{}i\pi,$ then we'd have $ (\minus{}1)^{\sqrt{2}}\equal{}\cos\left(\pi\sqrt{2}\right)\plus{}i\sin\left(\pi\sqrt{2}\right).$

If you're still at the beginning calculus level, I'd suggest staying away from that for now. (Which means that I agree with Soarer's post.)

And also can u tell me the proof for the parametric differentiation rule becos for eg,like if you have y=f(t) and x=g(t), and you need to find dy/dx,then we use the fact that dy/dx=(dy/dt)/(dx/dt) but how do we even know that the function y=f(x) is differentiable??

We don't.

For instance, let $ x \equal{} t^3,y \equal{} t^2.$ Both of these functions are differentiable for all $ t,$ but the resulting curve, which can be written as $ y \equal{} x^{2/3},$ does not have a tangent line at the origin.

Now, we can prove that if $ x(t)$ and $ y(t)$ are both differentiable functions of $ t,$ then $ \frac {dy}{dx}$ does exist at any point at which $ x'(t)\ne 0.$

[quote]Now, we can prove that if x(t) and y(t) are both differentiable functions of t, then \frac{dy}{dx} does exist at any point at which y'(t)\ne 0.
[/quote]
And what exactly is the proof ??(does it involve undergrad math??)[/list]

(That should be $ x'(t)\neq 0$)

The proof belongs to multivariable calculus; it's another member if the family that includes the Inverse Function Theorem and the Implicit Function Theorem. That is certainly undergrad level- just a year or so ahead of where you are now.

We can squeeze it down to a single-variable theorem if necessary. The key is to show that the function $ x\equal{}X(t)$ has a local inverse $ T(x)$, which is differentiable. Then we get $ y\equal{}Y(T(x))$, and the chain rule gives $ \frac{dy}{dx}$.

[hide="Single-variable Proof"]The hypotheses: $ X(a)\equal{}b$, $ X'(a)\equal{}c\neq 0$ and $ X$ is continuously differentiable in some neighborhood of $ a$.
Choose $ \epsilon$ so that $ |X'(t)\minus{}c|\le\frac{|c|}{2}$ for all $ x\in (a\minus{}\epsilon,a\plus{}\epsilon)$. By the Mean Value Theorem, $ \frac12|c|\cdot|t_1\minus{}t_2|\le |X(t_1)\minus{}X(t_2)|\le \frac32|c|\cdot|t_1\minus{}t_2|$ for all pairs $ t_1,t_2\in (a\minus{}\epsilon,a\plus{}\epsilon)$. This means that $ X$ is injective on that interval; by continuity, we get a unique inverse function $ T(x)$ defined on at least the interval $ (b\minus{}\frac12|c|\epsilon,b\plus{}\frac12|c|\epsilon)$, and $ \frac2{3|c|}|x_1\minus{}x_2|\le |T(x_1)\minus{}T(x_2)|\le \frac{2}{c}|x_1\minus{}x_2|$. We want to show that $ T$ is differentiable at $ b$ with derivative $ \frac1c$.
For each $ n\ge 3$, let $ \epsilon_n$ be such that $ \left|\frac{X(t)\minus{}b}{t\minus{}a}\minus{}c\right|\le\frac{c}{n}$ for all $ |t\minus{}a|\le\epsilon_n$; this exists by the derivative at $ a$. We have $ |T(x)\minus{}a|\le\frac{2}{c}|x\minus{}b|$, so $ |T(x)\minus{}a|\le \epsilon_n$ whenever $ |x\minus{}b|\le\frac{c}{2}\epsilon_n$. If $ x$ is this close, we then get $ \frac{T(x)\minus{}a}{x\minus{}b}\equal{}\frac{T(x)\minus{}a}{X(T(x))\minus{}b}$, and this difference quotient is between $ \frac{n}{n\plus{}1}\frac1c$ and $ \frac{n}{n\minus{}1}\frac1c$.

Taking $ n\to\infty$, $ \lim_{x\to b}\frac{T(x)\minus{}a}{x\minus{}b}\equal{}\frac1c$, and $ T$ is differentiable at that point. The rest follows routinely.[/hide]


A mutual fund report reads “Fund A is perfectly correlated with Fund B,
but Fund B is more risky, since, whenever Fund A moves 5%, Fund B moves 15%.”
Write a brief explanation, for someone unfamiliar with statistics, of how this can happen.
Include a sketch to illustrate your explanation.

Hi ProblemSets

If the performance of fund A follows say a centered gaussian  variable $ X$with Variance $ V^2$
and fund B has a performance following a centered gaussian variable  $ \sqrt {3}.X$ (so its variance is $ 3.V^2$ )

Then you have two funds perfectly correlated but with different variances (or risk as may rather say)

you can relax the gaussian hypothesis as long as you have existence of a second moment

Hope it helps

Fund A has 80% in cash and 20% in stocks and fund B has 40% in cash and 60% in exactly the same stocks.


Find all the functions f and g from R to R such that f(0)+g(0)=0 and f(xg(y+1))+y=xf(y)+f(x+g(y)) for all x,y real numbers.

[quote=flavian]it is so difficult this problem?[/quote]
See [url=https://artofproblemsolving.com/community/c6h380918]here[/url], [url=https://artofproblemsolving.com/community/c6h1162840]here [/url]or [url=https://artofproblemsolving.com/community/c6h1328378]here [/url]for a complete solution.


Let an = 1/(nC0) + 1/(nC1) + 1/(nC2) + ... + 1/(nCn).

(C is the choose function)

Is it bounded? Does it converge for n-> :inf: ? What to?

It is bounded because 0<an<=1/n+1/n+(n-2)/nC2=2/n+2*(n-2)/(n*(n-1)), which is bounded. This is true because we isolate 1/nC0 and 1/nCn, which are both 1/n, and the rest of the terms of the sum are all <=1/nC2, which yields a sum=(n-2)/nC2.

Haven't thought about the rest.

I think you are close.  But aren't nC0 and nCn both 1, not n?

Sorry! It was about 3 AM in Romania when I posted that..:) So I guess we have to isolate 1/nC0, 1/nCn AND 1/nC1 and 1/nC(n-1), so we will eventually get 0<an <=2+2/n+(n-4)/nC2.

Yes, that's a lot better.  I think the only thing you need to fix now is to change (n-4) to (n-3).

And we have an > 2 so lim an = 2  :)

(binomial coeffs are increasing until C(n,[n/2]) then decreasing)

We also have :
a(n+1) = 1 + (n+2)/2(n+1) a(n)

(from (n + 2)/C(n,k) - (2n+2)/C(n+1,k) = (n-k)/C(n,k+1) - (n- (k+1))/C(n,k) -  :wink:)

which also proves that a(n) is decreasing (for n > 2) and lim=2


[color=blue][size=150]Assume $ A \in M_n(\mathbb{K})$

show that: if $ AB\minus{}BA\equal{}A$, then $ Tr(A^k)\equal{}0$ for $ \forall k \in \mathbb{Z}^\plus{}$[/size][/color]

[quote="Xingyuan"][color=blue][size=150]Assume $ A \in M_n(\mathbb{K})$

show that: if $ AB \minus{} BA \equal{} A$, then $ Tr(A^k) \equal{} 0$ for $ \forall k \in \mathbb{Z}^ \plus{}$[/size][/color][/quote]
   Proof :
 We use If $ X,Y \in M_n(\mathbb{K})$ then $ Tr(XY) \equal{} Tr(YX)$
 We have : $ A^{k \minus{} 1}(AB \minus{} BA) \equal{} A^{k}$ or $ A^kB \minus{} A^{k \minus{} 1}BA \equal{} A^k$ then $ Tr(A^k) \equal{} Tr(A^kB) \minus{} Tr(A^{k \minus{} 1}BA) \equal{} Tr(A^kB) \minus{} Tr(A^kB) \equal{} 0$

[quote="Carcul"]Which, in other words, means that $ A$ is a nilpotent matrix.[/quote]

Not over every field  :P

(And I have no idea whether any matrix $ A$ satisfying $ AB\minus{}BA\equal{}A$ for another matrix $ B$ must necessarily be nilpotent. If the characteristic is $ \neq 2$, this probably has to do with representations of $ \mathfrak{sl}\left(2\right)$, but I don't understand them well enough.)

  darij

1) If $ AB\minus{}BA$ and $ A$ commute then for any $ k$, $ Tr((AB\minus{}BA)^k)\equal{}0$.
2) If $ K\equal{}\mathbb{C}$ and $ AB\minus{}BA\equal{}\alpha{A}\plus{}\beta{B}$ then $ A,B$ are simultaneously triangularizable, which implies that $ AB\minus{}BA$ is a nilpotent matrix.

About the Darij's question (post #4). Let $ K$ be the considered field.
If $ K\equal{}F_2$ then the result is false. Take $ A\equal{}\begin{pmatrix}0&1\\1&0\end{pmatrix}, B\equal{}\begin{pmatrix}1&0\\0&0\end{pmatrix}$. $ AB\minus{}BA\equal{}A$ and $ A$ is invertible.
But we have the following result:
Assume $ K$ is an algebraically closed field of characteristic $ 0$, $ \alpha,\beta\in{K}$, $ A,B\in\mathcal{M}_n(K)$ satisfy $ [A,B]\equal{}AB\minus{}BA\equal{}\alpha{A}\plus{}\beta{B}$. Then $ A,B$ are simult. triangularizable.
Proof: Let $ G$ be the Lie subalgebra (sub-vectorspace of matrices that is  stable under $ [.]$ ) that is spanned by $ A,B$. $ G\equal{}\{\lambda{A}\plus{}\mu{B}|\lambda,\mu\}$.
The $ G$'derived series is as follows:
$ G_1\equal{}[G,G]$ the sub-vectorspace spanned by the  $ [g_1,g_2]$ where $ g_1,g_2\in{G}$. $ G_1$ is the line $ \{\lambda(AB\minus{}BA),\lambda\}$.
$ G_2\equal{}[G_1,G_1]\equal{}\{0\}$ and $ G$ is solvable.
 By the Lie's theorem (valid only if $ K$ is algebraically closed and of characteristic $ 0$), $ G$ is triangularizable that is $ A,B$ are sim. triang..
Remark: there exists a much more elementary proof of this result (but it's a long way).


function
real analysis
real analysis unsolved

could you prove that the sum of two porous sets is porous?

1) What is your definition of a porous set?
2) By "sum" do you mean the union or the Minkowski sum?

our real analysis course is based on Thomson's book <<Real Functions>>

for $ E\subset R$, $ a,b\in R$ Thomson defines $ \lambda(E;a,b) \equal{}$ the length of the largest open subinterval of $ (a,b)$ that contains  no points of $ E$.

next he puts $ p(E;x) \equal{} \limsup_{h\to 0}\frac {\lambda(E;x,x \plus{} h)}{|h|}$

and now:
$ E$ is porous at $ x$ iff $ p(E;x) > 0$,
$ E$ is porous iff it is porous at each of its points.

the sum of sets means "union", sorry, Thomson uses $ \Sigma$ instead of $ \bigcup$

could you answer my question now?
these are completely new things for me (this part of real analysis) and I don't "feel" it :wink:

thanks

I can answer your question now but you'll not like the answer :(. With your definitions, there is no way to prove it because it is false.

could you give an example of porous $ E,F$ s.t. $ E\cup F$ is not porous ?

Here goes: Consider the set $ E$ consisting of all inverse positive integers. It is porous according to your definitions because it is porous at each its point (with $ P(E,x)\equal{}1$ for all $ x\in E$). Now let $ F\equal{}\{0\}$ (what can be more porous than a finite set?). $ E\cup F$ is not porous at $ 0\in E\cup F$.


calculus
integration
analytic geometry
geometry
3D geometry
sphere
calculus computations

totally lost on this question, Someone please give me some hits... Thanks alot.

Using spherical coordinates:

Find the volume of the solid that lies within the sphere X^2+Y^2+Z^2=4, above the xy plane, and below the cone Z^2=X^2+Y^2.

Okay.  This shape is easy, so we just need to find the bounds.  We're integrating around a full $2\pi$.  That takes care of $\theta$.  We need to find the range of $\phi$.  However, it's clear that the angle the cone forms with the $z$-axis is just $\frac {\pi}{4}$.  Clearly, $\rho$ goes from $0$ to $2$.  Finally, the sphere is the upper curve, the cone is the lower curve.  This should give you enough info to finish it.


If $ x\plus{}y\plus{}z\equal{}0$, Prove that:
\[ \sum_{cyc} \frac{x^2}{2x^2\plus{}yz}\equal{}1\]

$ 2x^2\plus{}yz\equal{}2x^2\plus{}y(\minus{}x\minus{}y)\equal{}2x^2\minus{}xy\minus{}y^2\equal{}(x\minus{}y)(2x\plus{}y)\equal{}(x\minus{}y)(x\minus{}z)$ and

$ (x\minus{}y)(y\minus{}z)(z\minus{}x)\equal{}x(y^2\minus{}z^2)\plus{}y(z^2\minus{}x^2)\plus{}z(x^2\minus{}y^2)$  so

$ \sum_{cyc} \frac{x^2}{2x^2 \plus{}yz}\equal{}\sum_{cyc} \frac{\minus{}x^2 (y\minus{}z)}{(x\minus{}y)(y\minus{}z)(z\minus{}x)}\equal{}\frac{\sum_{cyc} x(y^2\minus{}z^2)}{(x\minus{}y)(y\minus{}z)(z\minus{}x)}\equal{}1$

:)

[quote="gokussj3"]Provided
 x,y,z are not $ 0$ :wink:[/quote]
Sorry I didn't get your point  :(

If $ x\equal{}y\equal{}z\equal{}0$, then the expression is undefined (note that this was not set as a condition).

Sorry for the trouble, but this is very clear and I thought that there is something wrong with that problem or solution not just that small note. Anyway, it was a nice remark and thanks for your clarification.


HI,

I tried to solve problem 6 but didn't find a solution and the solution provided is certainly not sufficient.
Does anyone have a solution for that?

Problem 6. Show that it is not possible to partition the set $ {1, 3, 5, . . . , 2007}$ into
two sets $ {x_{1}, x_{2}, . . . , x_{502}}$ and $ {y_{1}, y_{2}, . . . , y_{502}}$ such that
$ \sum_{i=1}^{502} x_i^2 = \sum_{j=1}^{502} y_i^2$

yours Klemens

I moved this to Pre-Olympiad because it wasn't getting any responses in Other US Contests.

-Sean

[quote="MathEcon"]HI,

I tried to solve problem 6 but didn't find a solution and the solution provided is certainly not sufficient.
Does anyone have a solution for that?

Problem 6. Show that it is not possible to partition the set $ {1, 3, 5, . . . , 2007}$ into
two sets $ {x_{1}, x_{2}, . . . , x_{502}}$ and $ {y_{1}, y_{2}, . . . , y_{502}}$ such that
$ \sum_{i = 1}^{502} x_i^2 = \sum_{j = 1}^{502} y_i^2$

yours Klemens[/quote]

[hide]Observe that the sum $ 1^{2}+3^{2}+...+2007^{2} \equiv 0(mod 8)$.  So, each sum must be congruent to either $ 0(mod 8)$ or $ 4(mod 8)$.

But since, all odd numbers are congruent to 1(mod 8), then this means the total sum of both sides is congruent to 502(mod 8) or 6(mod 8).

Contradiction.[/hide]


bacheha shoma kheili herfee tashrif darin va mesl inke kheili baretoone. lotfan be man komak konid va manabe [u]koob[/u] ro moarefi konid man ham mikham riazi bekhoonam vali tooy in hameh ketab gij shodam. lotfan ketabha farsi basheh.
pishapish mer30

be nazare man ye sar boro entesharate fatemi va khoshkhan + sharigin(entesharate mobtakeran) + serpinski  :wink:

khob man daghighan az kodoomeshoon shoroo konam? akhe ina khodeshoon koli ketabe :huh:

bebin sharigin va serpinski faghat masale daran...
vase khoondane darsam mitooni az ketabe mirzakhani (fatemi),ketabe alipoor(fatemi),ketabe siamake ahmadpoor (khoshkhan),jabre safa (entesharate khoshkhan) shoroo konio dar kenaresham az nazarie adade safa(khoshkhan) va sharigino sepinski masale hal koni...
hendese court (fatemi) ham bad nis

man ye seri az manabe khoob ro migam:
Nazarie Adad : nazarie adade mirzakhani(fatemi) , masaeli dar nazarie adad (khoshkhan)
Tarkibiat : Tarkibiate Alipour (fatemi)
Hendesse : Hendeseye Mosatahe (khoshkhan) , Sharigin (mobtakeran)
Jabr : Jabr dar olampiade riazi (khoshkhan)
albatte age ketabe khoob vase masale va makhsoosan vase tarkibiat mikhay ina hastan :
olampiyade riazie leningrad (kheiliam kamyabe alan) , va tornomente shahrha (fatemi)

bacheha vaghean mamnoon ke komak mikonid.


I gave a very long solution to the following problem with. My question is: is there a short solution? The problem is difficult, but not impossible:
   Find all functions defined on the positice reals with values in positive reals such that for all x,y>0 we have f(f(x)+y)=xf(1+xy).

In fact, I am the guest that presented the problem. I don't knoe where I have found it, but it is quite challenging.