Prove that the number permutations $ \alpha$ of $ \{1,2,\dots,n\}$ s.t. there does not exist $ i<j<n$ s.t. $ \alpha(i)<\alpha(j\plus{}1)<\alpha(j)$ is equal to the number of partitions of that set.

These are the permutations that avoid the [i]generalized permutation pattern[/i] $ 1\minus{}32$.  There's a lot of literature on generalized permutation patterns -- a nice survey paper is at [url=http://arxiv.org/abs/0801.2412]arXiv:0801.2412v2[/url].  This is a nice easy problem, and I'll leave the proof for others to find.

Cool problem.

Call permutations satisfying the condition of the problem [b] tasty. [/b]

We'll establish a bijection between partitions of the set $[n]$ and the tasty permutation of $[n]$.

For the forward direction, consider some partition $[n] = S_1 \cup S_2 \cup \cdots \cup S_k$, where the sets $S_i$'s are in increasing order based on their minimal element. Then, map this partition to the permutation which consists of the elements of $S_1$ in descending order, then the elements of $S_2$ in descending order, then the elements of $S_3$ in descending order, etc. For example, if our partition is $\{1, 3, 4\} \cup \{2, 5\} \cup \{6, 9\} \cup \{7, 8\}$, then it is mapped to the permutation $431529687.$

It's not hard to verify that the permutations obtained in this manner are all tasty, and that no two partitions map to the same tasty permutation. This implies that the number of tasty permutations is at least the number of partitions of $[n]$.

For the reverse direction, consider a permutation $\alpha$ of $n$ and draw bars in the spaces where the permutation increases. For instance, if our permutation is $431529687$, then we add in spaces like so: $431 | 52 | 96 | 87.$ Now treat the bars as dividers, and consider the partition of $[n]$ which simply comprises of the sets of numbers delineated by the bars. In the case of our example, $431529687$ is mapped to $\{4, 3, 1\} \cup \{5, 2\} \cup \{9, 6\} \cup \{8, 7\}.$

It's not hard to verify that this operation is the exact reverse of the first operation, and this shows that the number of tasty permutation is exactly equal to the number of partitions of $[n]$.

$\square$

Consider an arbitrary partition of $[n]$, with each part sorted in increasing order, and the parts sorted from largest to smallest minimum element. Then, concatenating the parts in their current order gives a unique, valid $\alpha$. To reverse this process, consider the set of all elements in $\alpha$ such that they are smaller than any number before them. Note that if at any point $\alpha$ decreases over consecutive terms, it must be at one of these points, or else the condition is violated. So, taking these points to be the start of our parts in the partition gives all parts to be in increasing order, and hence our initial partition. As there exists a bijection, these two quantities are equal, as desired.


If n is a positive integer prove that

[color=blue][b] [n/(n+1)]^n  * [(2n+1)/3]^n      >      (n ! ) ^ 2[/b][/color]


I M Inviting 9th Graders to post their Problems {Only Of 9th Standard} Here and I'll Try my Best to answer them!
Thanks
Piyush

Can we call it another spam???
Its so out of the way and strange..... :huh:

Anurag I M Not understanding these questions which the people R posting here

They R Just Out Of My View!

 U Can call my post , a spam
 but can u post me some maths' and Physics' Questions in this spam!

welcome Piyush!!

well it wont do u much good as ppl here dont think of 9th level at all (excluding me) :P
even the so called 9th level probs will trbl u out of ur wits (u may certainly b an exception :oops: )

Ritu don't try to be too modest.
U have cleared the CBSE GMO-top 30 in the ciubtry.

And also to the thread starter
U can discuss 9th std related stuff in the Middle school section ok.

I think there are some forums(esp for math) for middle school and high school check them once before opening this.
Ofcourse i am not banning u from posting this thread but unnecessary waste of space.

[quote="Aravind Srinivas L"]U have cleared the CBSE GMO-top 30 in the [b]ciubtry[/b].
[/quote]
wats this????? :rotfl:  :rotfl:

neway its not an achievement in front of u guys :)

Sorry. country--typing mistake
It is a really big achievement to come in top 30 in the [b]country[/b]

i-o and b-n are adjacent oon the keyboard :wink:

[quote="Aravind Srinivas L"]
It is a really big achievement to come in top 30 in the [b]country[/b][/quote]

i've been AIR 1 plenty of times
[hide]will u call it an achievement--if i m AIR 1 in 100 students :wink: [/hide]

Don't try to hide your achievements.You rock man
As for piyush bhatia well chemrock is correct,although if there are some doubts you can check some of the topics in maths section or make your topic. :)


trigonometry
vector
function
inequalities proposed
inequalities

The maximum is $ \sqrt{an}$ and it is realised, e.g., for $ \sin 2x_i\equal{}\sqrt{\frac{a}{n}}\leq 1$ with $ x_i\in \left[0,\frac{\pi}{4}\right]$. Clearly, we can assume $ n>1$.

Denote vectors by $ \bar{x}\equal{}(x_1,\ldots,x_n)$ and $ f(\bar{x})\equal{}\sum_{i\equal{}1}^n\sin^22x_i$ and $ g(\bar{x})\equal{}\left|\sum_{i\equal{}1}^n\sin 2x_i\right|$. Since $ \sin^2$ is a symmetric function, we can always assume $ \sin 2x_i\geq 0$ in order to maximise $ g(\bar{x})$. For any such vector let $ M\equal{}\max\{\sin 2x_i\,|\,i\equal{}1,\ldots,n\},m\equal{}\min\{\sin 2x_i\,|\,i\equal{}1,\ldots,n\}$ and assume $ M\equal{}\sin 2x_r,m\equal{}\sin 2x_s$ with $ r\not\equal{}s$. Set $ z\in \left[0,\frac{\pi}{4}\right]$ implicitly via $ \sin^22z\equal{}\frac{m^2\plus{}M^2}{2}\leq 1$ and define $ x_i'\equal{}\left\{\begin{array}{cc}z & \mbox{if}\;i\equal{}r,s\\x_i & \mbox{if}\; i\not\equal{}r,s\end{array}\right.$

Direct computation gives $ f(\bar{x'})\equal{}a$ and $ g(\bar{x'})\equal{}g(\bar{x})\minus{}(m\plus{}M)\plus{}2\sqrt{\frac{m^2\plus{}M^2}{2}}\geq g(\bar{x})$ with equality iff $ m\equal{}M$. In particular, if such $ \bar{x}$ maximises $ g$ we must have $ m\equal{}M$, hence, $ g(\bar{x})\equal{}\sqrt{an}$.

I founded: $ 2\sqrt{an\minus{}a^{2}}$  :wink:
by cauchy-Schwarz we get:
$ \left(\sum\sin{x_{i}}\cos{x_{i}}\right)^{2}\le\left(\sum\sin^{2}{x_{i}}\right)\left(\sum\cos^{2}{x_{i}}\right)$

$ \Longrightarrow2\left|\sum\sin{x_{i}}\cos{x_{i}}\right|\le\sqrt{4\sum\sin^{2}{x_{i}}\sum\cos^{2}{x_{i}}}$

$ \Longrightarrow$$ \left|\sum\sin{2x_{i}}\right|\le\sqrt{4an\minus{}4a^{2}}$

so: $ max{\left|\sum\sin{2x_{i}}\right|}\equal{}2\sqrt{a\left(n\minus{}a\right)}$  :wink:

Sorry  :blush: , I have added $ 2$ in the first formul...
it was edited.

thanx for your solution :)


function
real analysis
real analysis unsolved

Prove that any monotone function $f: \mathbb{R}\to\mathbb{R}$ is Borel measurable. ;)

The set of discontinuities of such a function is at most countable (and a countable set, along with each of its subsets, is Borel-measurable with measure $0$).

or, easierly, consider any over-level (i mean the set $\{ y|f(y)>t \}$).
$f$ is measurable just if $\{ f>t \}$ is measurable for any $t$.
but it's easy to show that each of those sets is connected, then borelian.

IIRC, f: R->R is Borel-measurable iff
(1) f-1(B) is a Borel-subset whenever B is an intervel of the form (a,\infty)
iff
(2) f-1(B) is a Borel-subset whenever B is a Borel-subset.

So I think it follows easily from (1) that any monotone function f is Borel-measurable.

Set-union, Set-intersection and Set-Complement operations commute with the set-inverse operation: B -> f-1(B). This establishes the equivalence of (1) and (2), since the intervals of the form (a,\infty) generates the whole collection of Borel subsets of R.

_________________
One minute prayer for peace in korea


logarithms
integration
calculus
calculus computations

$ y^{2}(x\minus{}1)dx\equal{}x(xy\plus{}x\minus{}2y)dy$
I'm sure there is some trick but I can't find it  :maybe: ..

[quote="Extremal"]$ y^{2}(x \minus{} 1)dx \equal{} x(xy \plus{} x \minus{} 2y)dy$[/quote]

$ \log\frac{x\minus{}y}{y}\plus{}\frac{1}{y}\plus{}\frac{y\minus{}1}{y\minus{}x}\equal{}C$.

Is that the solution and if so, is your solution too long to include in this thread?

I find trick  :roll: :
it's obviusly find that solution is also $ x\equal{}0,$ $ y\equal{}o$
after we suppose $ x,y$ is not equal to $ 0$.
we have:
$ y^{2}xdx\minus{}y^{2}dx\equal{}x^{2}ydy\plus{}x^{2}dy\minus{}2xydy$
therefore
$ \minus{}yx^{3}d\frac{y}{x}\equal{}(x^{2}\minus{}xy)dy\minus{}yx^{2}d\frac{y}{x}$
or
...
$ (yx\minus{}x^{2}y)d\frac{y}{x}\equal{}(x\minus{}y)dy$ I note that easy seen here we have solution $ y\equal{}x$
let's do substitution:
$ \frac{y}{x}\equal{}\alpha$
and
$ y\equal{}\beta$
therefore we have :
$ (\beta\minus{}\frac{\beta^{2}}{\alpha})d\alpha\equal{}(1\minus{}\alpha)d\beta$
or
$ \beta ' \equal{}\frac{\beta}{1\minus{}\alpha}\minus{}\frac{\beta^{2}}{\alpha(1\minus{}\alpha)}$
it's particular case of Rikati equation when $ c(x)\equal{}0$
or we have bernulii equation if we put $ \frac{1}{\beta}\equal{}z$ after is simpl.
and in finally we get:
answer:
1)$ x\equal{}0$, $ y\equal{}0$;
2)$ y\equal{}x$
3) (after solving bernulii equation we have)
$ xy\minus{}x\equal{}y(y\minus{}x)\ln |\frac{Cy}{y\minus{}x}|$
done

Here's how I would solve it (which is essentially the method used by Extremal). [hide="Click here"]
We have $ \frac{dy}{dx} \equal{} \frac{y^2(x\minus{}1)}{x(xy\plus{}x\minus{}2y)}$. Let's now make $ y \equal{} u(x) \cdot x$ (and so y' = u'x + u):

$ u'x \plus{} u \equal{} \frac{u^2x^2(x\minus{}1)}{x(ux^2\plus{}x\minus{}2ux)} \equal{} \frac{u^2(x\minus{}1)}{ux\minus{}2u\plus{}1}$

$ \equal{}> u'x \equal{} \frac{u^2\minus{}u}{ux\minus{}2u\plus{}1} \equal{}> \frac{du}{dx} \equal{} \frac{u(u\minus{}1)}{x(ux\minus{}2u\plus{}1)}$

$ \equal{}> \frac{dx}{du} \equal{} \frac{x(ux\minus{}2u\plus{}1)}{u(u\minus{}1)} \equal{}> x' \equal{} \frac{x^2}{u\minus{}1} \minus{} \frac{x(2u\minus{}1)}{u(u\minus{}1)}$

$ \equal{}> x' \plus{} \frac{2u\minus{}1}{u^2\minus{}u}\,x \equal{} \frac{x^2}{u\minus{}1}$

which is a Bernoulli differential equation. Now making z = 1/x we get $ z' \minus{} \frac{2u\minus{}1}{u^2\minus{}u}\,z \equal{} \frac{1}{1\minus{}u}$ which is a first order linear equation. Since $ \minus{} \int\frac{2u\minus{}1}{u^2\minus{}u}\,dx \equal{} \minus{} \ln|u^2\minus{}u|$, it's solution is:

$ z \equal{} (u^2\minus{}u) \left(C \minus{} \int \frac{du}{u(u\minus{}1)^2}\right) \equal{} (u^2\minus{}u) \left(C \minus{} \ln\left|\frac{u}{u\minus{}1}\right|  \plus{} \frac{1}{u\minus{}1}\right) \equal{}$

$ \equal{} u\left[C(u\minus{}1) \minus{} (u\minus{}1) \ln\left|\frac{u}{u\minus{}1}\right| \plus{} 1\right]$.

Now substituting z = 1/x and u = y/x we get:

$ \frac{1}{x} \equal{} \frac{y}{x}\left[\left(\frac{y\minus{}x}{x}\right)\left(C \minus{} \ln\left|\frac{y}{y\minus{}x}\right|\right) \plus{} 1\right]$

$ \equal{}> y\left[\left(\frac{y\minus{}x}{x}\right)\left(C \minus{} \ln\left|\frac{y}{y\minus{}x}\right|\right) \plus{} 1\right] \equal{} 1$.
[/hide]


N people are planning on having a meeting. The person in charge ordered n chairs, all of different sizes, ranked 1 to n. Each person has an equally likely chance of having each size. So someone has a 1 in n chance of fitting the smallest chair, a 2 in n chance of being able to fit the second smallest, and so on. However, multiple people can be the same size.
When meeting time comes, the members arrive one at a time and sit in the smallest unoccupied chair that is big enough for them and remain sitting for the entire meeting. If there is no available seating, they go home. For example, the worst case scenario is that everyone is size n and all but the first person go home. This has a $ \frac{1}{n^n}$ chance of occuring. Under the obscure, impractical rules of the committee, quroum is only reached when every member is present. The question is;

  What is the largest number of people there can be with at least a 1% chance of obtaining quroum?

So noone's going to even try my problem? oh well


group theory
abstract algebra
superior algebra
superior algebra solved

$G$ is a group and $H$ is a subgroup so that the index $(G:H)$ is finite.
Show that there exists a subgroup $I$ of $H$ with $(G:I)$ finite and $I$ normal in $G$.

It's not too hard ;)

Misha

I think I can remember you posting something very similar, and even essentially the same :?. I remember showing (almost showing, in fact leaving out some easy to check details) that for the largest subgroup of $H$ which is normal in $G$, which is $H_G=\bigcap_{g\in G}gHg^{-1}$, there is in injective homomorphism mapping $G/H_G$ to $S([G:H])$, where $S(A)$ denotes the set of bijections from $A$ to itself. In particular, it means that $|G/H_G|\le [G:H]!$ (here [$G:H$] is the index of $H$).

[quote="grobber"]I think I can remember you posting something very similar, and even essentially the same :?. [/quote]

Sorry if it is true but I just can't remember it  :blush:

Your solution is of course correct.

Misha


vector
function
advanced fields
advanced fields unsolved

We have the following hypothesis.

Let $ V \equal{} C([ \minus{} 1,1])$ be the set of continuous complex functions defined on $ [ \minus{} 1,1].$
Let $ x_1,x_2 \in [ \minus{} 1,1].$
Let $ M_1 \equal{} \{f | f \in V, f(x_1) \equal{} 0\},$ $ M_2 \equal{} \{f| f\in V, f(x_2) \equal{} 0\}.$

Then we have the conclusion.

Then $ V \equal{} M_1 \plus{} M_2.$ That is, each $ f\in V$ can be written in the form $ f \equal{} g \plus{} h$ for some $ g \in M_1, h\in M_2.$

Can anyone see why this is true?

Let $ P(x)$ be any continuous function that satisfies $ P(x_1) \equal{} \frac {f(x_1)}{2}$ and $ P(x_2) \equal{} \minus{} \frac {f(x_2)}{2}$.  (For example, there is a linear function with that property.)  Then let $ g(x) \equal{} \frac {f(x)}{2} \minus{} P(x)$ and $ h(x) \equal{} \frac {f(x)}{2} \plus{} P(x)$.


If sinx +siny +sinz=0  prove that

|sin3x+sin3y+sin3z| \leq 12|xyz|

Indeed, it is very simple:
  Use the fact that sin3x=3sinx-4sin^3x and the fact that if a+b+c=0 then a^3+b^3+c^3=3abc. The inequality becomes |sinx*siny*sinz|<=|xyz|, which is true since |sinx|<=|x| and so on.


If $ a\plus{}b\plus{}c\equal{}0$ ,prove the identity:

$ (\dfrac{a}{b\plus{}c}\plus{}\dfrac{b}{a\plus{}c}\plus{}\dfrac{c}{a\plus{}b})(\dfrac{b\plus{}c}{a}\plus{}\dfrac{a\plus{}c}{b}\plus{}\dfrac{a\plus{}b}{c}\plus{}2)\equal{}3$

[hide="Solution"]
$ a\plus{}b\equal{}\minus{}c$, $ b\plus{}c\equal{}\minus{}a$, and $ c\plus{}a\equal{}\minus{}b$, so this is $ (\minus{}1\plus{}\minus{}1\plus{}\minus{}1)(\minus{}1\plus{}\minus{}1\plus{}\minus{}1\plus{}2)\equal{}\minus{}3\cdot\minus{}1\equal{}3$, as desired.
[/hide]


Something related to this [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=845756#845756]problem[/url]:
Prove that for a set $ S\subset\mathbb N$, there exists a sequence $ \{a_{i}\}_{i \equal{} 0}^{\infty}$ in $ S$ such that for each $ n$, $ \sum_{i \equal{} 0}^{n}a_{i}x^{i}$ is irreducible in $ \mathbb Z[x]$ if and only if $ |S|\geq2$.
[i]By Omid Hatami[/i]

if $ |S|<2$
we must have $ |S|\equal{}1$ ($ S\equal{}\{a\}$)
we have for $ n\equal{}3$: $ \forall\{a_{i}\}_{i\in\mathbb{N}}\in S\equal{}\{a\}:\sum_{k \equal{} 0}^{3}a_{k}x^{k}\equal{}a(x^{3}\plus{}x^{2}\plus{}x\plus{}1)$ isn't irreducible in $ \mathbb{Z}[X]$.

if $ |S|\ge 2$
we take $ a,b\in S$ such that $ a\equal{}a'd,b\equal{}b'd,a'>b',gcd(a',b')\equal{}1$
we take $ \{a_{i}\}_{i\in\mathbb{N}}\in S:\ a_{0}\equal{}a,\forall i>0,a_{i}\equal{}b$
then we have $ \forall x\in\mathbb{Z}:\forall n\in\mathbb{N}\frac{\sum_{k\equal{}0}^{n}a_{k}x^{k}}{d}\equiv a'\ (mod\ b')\not\equiv 0\ (mod\ b')$
then $ \forall n\in\mathbb{N},\forall x\in\mathbb{Z}:\sum_{k\equal{}0}^{n}a_{k}x^{k}\neq 0$
so $ \forall n\in\mathbb{N}:\ \sum_{k \equal{} 0}^{n}a_{k}x^{k}$ is irreducible in $ \mathbb{Z}[X]$.

Obviously above solution for $\mid S\mid \geq 2$ is wrong
It’s the right time for “BUMP”.

[quote=Iranian National Olympiad (3rd Round) 2007]Prove that for a set $ S\subset\mathbb N$, there exists a sequence $ \{a_{i}\}_{i = 0}^{\infty}$ in $ S$ such that for each $ n$, $$ \sum_{i = 0}^{n}a_{i}x^{i}$$ is irreducible in $ \mathbb Z[x]$ if and only if $ |S|\geq2$.[/quote]

Here's a proof for the $|S| \ge 2$(the 'if' direction')
Choose primes \begin{align*} p_1=a_0 10^2+a_1 \\ p_2=a_0 10^3+a_1 10^2+a_2 \\ . \\.\\.\\.\\p_n=\sum_{i = 0}^{n}a_{i}10^{i} \end{align*}
which by Dirchlet's theorem exist, hence by [url=https://en.m.wikipedia.org/wiki/Cohn%27s_irreducibility_criterion]Cohn's irreduciblity Criterion  [/url] $$ \sum_{i = 0}^{n}a_{i}x^{i}$$ is irreducible.$\blacksquare$


function
symmetry
calculus
derivative
geometry
analytic geometry
vector

Let a, b, and c be real numbers, with a > b > c > 0, and, in $R^{3}$ (reals), write S for the compact set $S = \left\{{(x,y,z)|\frac{{x^{4}}}{{a^{2}}}+\frac{{y^{4}}}{{b^{2}}}+\frac{{z^{4}}}{{c^{2}}}= 1}\right\}$.

Let $f: R^{3}\to R$ (reals) be the function $f(x,y,z) = x^{2}+y^{2}+z^{2}$. The restriction $f|_{S}$ of f to (the smooth constraint set) S has 26 critical points.

Locate these points (multitude of them is caused by symmetry) and determine which of them are local maximum or local minimum points for $f|_{S}$.

Denoting $x^{2}=X$, etc., we see that the problem is [b]almost[/b] reduced to finding the critical points of a linear function on the ellipsoid, which is a routine exercise included into almost any textbook (Lagrange multipliers + a linear system, which in this case is even with a diagonal matrix). and which can be easily visualized in geometric terms. We'll have just one point where the function takes its maximum (the minimum corresponds to negative values of $X,\dots$), which will correspond to $8$ choices of signs for square roots. I said "almost", because the mapping $x\to x^{2}$ has a point with $0$ derivative, so one will have to consider separately the cases when one or two variables are $0$. So, let's count: $8$ points where all three variables are non-zero, $3\times 4=12$ points where one variable is $0$ and $3\times 2$ points where two variables are $0$. Seems like we have found all $26$ :lol: It remains to determine which of them are local maxima or minima. Definitely, the $8$ ones corresponding to the maximum of the linear form with all non-zero variables will give the global maximum. Those with one variable equal to $0$ will give the maximum in two other variables but not a maximum in any two variables including the one that is $0$, so they are all saddles. At last, it is not hard to check that the points where two variables are $0$ are local minima: just change $X$ and $Y$ from $0$ to some small positive values and note that $Z$ drops by not much more than $X^{2}+Y^{2}$, so the sum goes up.

Generally, the problem seems to be a standard exercise in multivariate calculus/convex geometry involving no tricks or unexpected twists. :sleeping:

Is there anyway you can expand on how the equations work out in this problem, I understand the description of where the points are coming from, but the restriction idea is confusing me and finding that they are local max and min is hard to see also. Thanks.

My whole point was that the problem could be solved geometrically without writing a single formula but, if you prefer the analytic approach, here it goes. First, a few generalities. Suppose we have to find and classify the critical points of $F(x)$ considered as a function on a compact smooth surface in $\mathbb R^{3}$ given by $\Phi(x)=0$ such that $\nabla\Phi(x)\ne 0$ on the surface. The key word is [b]Lagrange multipliers[/b]. In this case, it means that we have to solve a system of equations $\Phi(x)=0$, $\nabla F(x)=\lambda \nabla\Phi(x)$ with $4$ equations and $4$ unknowns (3 coordinates of $x$ and $\lambda$). In our particular case, the system is
\[\left\{ \begin{aligned}&\frac{{x^{4}}}{{a^{2}}}+\frac{{y^{4}}}{{b^{2}}}+\frac{{z^{4}}}{{c^{2}}}= 1}\\ &2x=\lambda\cdot 4\frac{x^{3}}{a^{2}}\\ &2y=\lambda\cdot 4\frac{y^{3}}{b^{2}}\\ &2z=\lambda\cdot 4\frac{z^{3}}{c^{2}}\end{aligned}\right.\]
I'll leave the algebra to you. After you solve the system, you'll get your 26 points. One of them is   $x=0$, $y=\frac b{\sqrt{2\lambda}}$, $c=\frac c{\sqrt{2\lambda}}$ with $\lambda=\frac{\sqrt{b^{2}+c^{2}}}2$. We'll try to determine the charachter of this particular point. The other 25 (actually, 2 :)) cases are similar. Here the Lagrange multipliers are useful again. In order to determine the nature of the critical point $x_{0}$, we have to consider the Hessian of $F(x)-\lambda\Phi(x)$ at $x_{0}$ and determine whether it is positive definite, indefinite, or negative definite on the tangent plane to the surface at $x_{0}$ (more precisely, on the plane given by $\langle x,\nabla\Phi(x_{0})\rangle=0)$. (We'd better first pray that it won't be degenerate there, of course, but I'll leave this part to you as well :P). In our particular case, the Hessian is easy to find: it is just $2(dx)^{2}-(dy)^{2}-(dz)^{2}$. The basis of the tangent plane is given by the vectors $(1,0,0)$ and some other vector with first component $0$ and 2 other components nonzero (you may find it exactly, if you wish, but it doesn't matter in the slightest). Considering the Hessian on the linear combination of these vectors with coefficients $t$ and $s$, we get $2t^{2}-As^{2}$ where $A$ is the square of the norm of the second basis vector. This is an indefinite form. Hence, the point is a saddle.


function
combinatorics unsolved
combinatorics

Ummm...

How many ways can 100 represented with power of 3s?

for example, 100=81+9+9+1=81+9+3+3+1+1+1+1=...

Canonical (probably unhelpful) generating functions response:
[hide]
\[ [x^{100}] \prod_{i\equal{}0}^\infty \frac{1}{1\minus{}x^{3^i}}.\]
[/hide]

I'm really sorry, but can you tell me where I can read something about this thing you mentioned ... ?  :oops:

A good (and easily available, and free) reference is Wilf's [url=http://www.math.upenn.edu/~wilf/DownldGF.html]generatingfunctionology[/url].

The associated sequence is [url=http://www.research.att.com/~njas/sequences/A062051]A062051[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  Since for numbers congruent to 1 or 2 modulo 3 the answer is always the same as the next-smallest multiple of 3, entries of that sequence are all repeated three times.  Without triplication, the sequence is [url=http://www.research.att.com/~njas/sequences/A005704]A005704[/url].


suppose that X1,X2,...,X2n-1 are 2n-1 different real number which they are not in Q.prove that there are Xi1,Xi2,...,Xin (they should be different)
so that for each a1,a2,...,an: a1*X1+...+an*Xn is not in Q.
(ai should be in Q(ai>=o) and they are not 0 at the same time) :wink:


If $ \frac{a}{b}\plus{}\frac{b}{a}\equal{}2$ prove that $ a\equal{}b$

[hide="solution 1"]Rearrange to get $ a^2 \minus{} 2ab \plus{} b^2 \equal{} 0$, so $ (a \minus{} b)^2 \equal{} 0$, so a=b.[/hide]
[hide="solution 2"]Equality condition of 2 variable Am-GM, but thhe proof is the same, so...[/hide]

Fixed.

Yeas but to soution 1 $ a^{2}\minus{}2ab\minus{}b^{2}\not\equal{}(a\minus{}b)^{2}$  :P

Although this proof is almost trivial, I still don't think it belongs here.

Anyways, I think the first solution is better.


Show that for positive reals $a, b, c, d$ we have
\[ \left(\dfrac{ab + ac + ad + bc + bd + cd}{6} \right)^3 \geq \left(\dfrac{abc + abd + acd + bcd}{4}\right)^2 \]

This problem is nice !
 Put $A = ab+ac+ad+bc+bd+cd$ ; $B = abc+abd+bcd+cda$
 Consider $P(x)= (x-a)(x-b)(x-c)(x-d)$
 $P(x)$ has 4 real roots $a,b,c$ and $d$ .
 So $P'(x)= 4x^3-3(a+b+c+d)x^2+2Ax-B =$
 $=4x^3-4(m+n+p)x^2+4(mn+np+mp)x-4mnp$
 where $m,n,p$ are three roots of $P'(x)$
  The original ineq is equivalent to $\dfrac{(mn+np+mp)}{3}^{\dfrac{1}{2}} \geq (mnp)^{\dfrac{1}{3}}$ which is obviously true by AM-GM

Please show your Cauchy (edit your previous post)

Maybe ehsan2004 has made a mistake,
this is part of Maclaurin Inequality.

which is the generalisation of the AM-GM inequality, right?

[quote="ehsan2004"]Write the inequality in this form:

$\sqrt [3]{\dfrac{abc+abd+acd+bcd}{4}}\leq \sqrt {\dfrac{ab+ac+ad+bc+bd+cd}{6}}$

then use the Newton inequality . :D[/quote]

yes with using macLaurins ineq it is true but how did you rewrite the ineq in this form?

[quote="nttu"]This problem is nice !
 Put $A = ab+ac+ad+bc+bd+cd$ ; $B = abc+abd+bcd+cda$
 Consider $P(x)= (x-a)(x-b)(x-c)(x-d)$
 $P(x)$ has 4 real roots $a,b,c$ and $d$ .
 So $P'(x)= 4x^3-3(a+b+c+d)x^2+2Ax-B =$
 $=4x^3-4(m+n+p)x^2+4(mn+np+mp)x-4mnp$
 where $m,n,p$ are three roots of $P'(x)$
  From here , what we want to prove is followed by Cauchy  ;)[/quote]

  In my country , Cauchy means AM-GM  ;)
 The original ineq is equivalent to $\dfrac{(mn+np+mp)}{3}^{\dfrac{1}{2}} \geq (mnp)^{\dfrac{1}{3}}$ , which is obviously true by AM-GM

  I think Maclaurin's ineq = Newton's ineq  and its proof is longer than this ineq's   :)


search
geometry
geometric transformation

Having spent a hell lotta time on learning nature's language and computer's language
Now I want to learn some human language...

Do you have any idea abt free ebooks on learning French or Spanish through English with pronunciation.....

These few months are going to be quite boring for me as I barely have any entrance exam to write
( just one more i.e. NEST for NISER,CBS- DAE on 20 Apr)

So thought that I would learn some of these languages along with doing some maths and physics...

[quote="anitsahu"]wanna learn german
i can teach u some :rotfl:[/quote]

But I guess its said that German is lot tougher than these...

As we can see from the very origin of these languages from Celtic, Germanic tribe and corrupted forms of Latin and Greek....

So any corrupted language is bound to be easier than original one...
(just like Sanskrit and Devnagri) :P

I know just one thing in German,
Achtung Minen == Beware of mines  :rotfl:   :rotfl:

the problem with european languages is that, even if you learn them, to be able to have a regular conversation with a local is really quite tough.
the speed with which they speak is astounding  :roll:

[quote="Chronoz"]Hmm... I'll PM you if I find something interesting. But do want to learn [u][b]spain[/b][/u] because of IMO?[/quote]

Chronoz, i think you are rather tensed, go get some sleep :D I believe he wants to learn spanish and not spain :rotfl:

madness I think you are right. Going to sleep now.

hey do any of you here know portugese very well???? :roll:  :D

[quote="Chronoz"]Hmm... I'll PM you if I find something interesting. But do want to learn spanish because of IMO?[/quote]

From where did IMO come into picture.. :|
 :|

I am actually more keen on learning French on which I have a 'book' which is too hard to comprehend and I want to confirm the pronunciation (lalfabe- franse- Pron-onsyasyon-)

I downloaded a free software yesterday and its quite good...

[quote="SOUMYASHANT NAYAK"][quote="Chronoz"]Hmm... I'll PM you if I find something interesting. But do want to learn spanish because of IMO?[/quote]

From where did IMO come into picture.. :|

[/quote]

Chumma Jujubi :rotfl:  :rotfl:  IMO 2008 is in Spain I believe?

Hey,

EBooks for Spanish? I know of one, but I am not sure if access is granted generally. I will give you my user id for the book, use it if you want to..

But I have to search for that though!

i received an unknown mail--wat's the language???????? :rotfl:  :rotfl:
Vindem baza de date cu 1.000.000 adrese de e-mail, cuprinzand atat adrese ale persoanelor juridice cat si ale persoanelor fizice cu domiciliul in Romania, la cel mai mic pret de pe piata!
Va puteti promova acum afacerea sau ideile dumneavoastra, prin mail, in cel mai ieftin mod cu putinta, cu impact maxim.
La achizitionarea pachetului veti primi in mod GRATUIT si un program de trimitere a e-mailurilor care poate trimite pana la 50.000 de mailuri pe ora, program care in mod normal ar costa 100 Euro.
Aveti asadar toate premizele unei promovari IEFTINE si de succes, tinand cont cati bani ar trebui sa investiti in publicitatea clasica, asa incat reclama dumneavoastra sa fie vazuta de 1 milion de oameni.

Pretul bazei de date care include cele 1 milion de mailuri si al programului de trimitere oferit gratuit este 150 RON.

Pentru comenzi si detalii ne puteti scrie la adresa de mail: echipa.promovare@yahoo.com

Telefon : 0751.709848

i think Romanian -but i dont kno the translation

do ne 1 know ne1 knowing Romanian

I know for sure that its not Spanish, Italian or German... Dont think its French either. But Romanian? No clue! Will ask a couple of my friends, and let you know...

hey ppl dont worry there is srini to rescue use this site !!!!
http://www.freetranslation.com/

its cool no???? i use it on orkut!!!! :D

[quote="Albert Einstien"]its cool no???? i use it on orkut!!!! :D[/quote]

that explains  :rotfl:   :D   :P

[url=http://www.ielanguages.com/languagetutorials.html]This is very useful.[/url]

[quote="madness"][quote="Albert Einstien"]its cool no???? i use it on orkut!!!! :D[/quote]

that explains  [/quote]
what da??? :D

[quote="rituraj007"]fizice cu domiciliul in [u]Romania[/u], la cel mai mic pret de pe piata!
[/quote]
this explains

hey guys i only know french hope it might help ne1
so i can teach him just a few basics

[quote=M7pj]If $S_r \equiv \sum_{i=1}^{r} a_i $ mod($m+1$) for $r=1,2,...,m$;  and  $S_r \neq 0$  (mod $m+1$)
And $ 0<a_i<m+1$ for all $0<i<m+1$
And all $S_r$ are distinct  (mod $m+1$) .
Then prove that all $a_i$ are equal.[/quote]

nice problem @m7pj
now for even m we can say that
 $ S_1 \euiv  a_{i} mod (m+1)$
now  $ a_i \equiv (m+1)-k(mod m+1) \equiv(- a_j) (mod m+1)
so in genral
for some $ i ,j$
we have
$a_i + a_j \equiv 0 mod(m+1)$
as given we have for some  p and q
$S_p + S_q \equiv 0 (mod m+1)$...... (1)
 so let us define a set of n integers following given condition
 so we have
$ T_1 , T_2............. T_m $  such m terms
and from (1) adding all such p`s and q`s we get
$ S_1 + S_2 +............................ S_m \equiv 0 mod(m+1) $
which can be written as
$m(T_1) + (m-1)(T_2)+.......... (m-k)(T_{k+1}) .... + T_m \equiv o (mod m+1)$.. (2)
as given that  each $S_i$ is unique module (m+1)
we can write (2) as
 $ (S_p)(T_1) +...................... (S_o)(T_m) \equiv  o (mod m+1)$
 now assume that  $T_j$ is 2nd maximum term in sequence and $T_y$ be maximum term
now subtract (1) from 2


There have been a lot of problems like:
Prove that the equation
$ x^{2}\equal{}y^{5}\plus{}7$ or
$ x^{2}\plus{}1\equal{}y^{3}$ or
$ x^{2}\equal{}y^{5}\minus{}4$
hasn't got any solution in natural numbers. If we write generalized form of these equations - $ x^{2}\equal{}y^{m}\plus{}r$ we will see that the solution of the above problems almost always uses that the equation $ x^{2}\plus{}r\equal{}y^{m}$ has solution (for example for the equation  $ x^{2}\equal{}y^{5}\plus{}7$ we use $ 5^{2}\plus{}7\equal{}2^{5}$).
And so I suppose that if r is such integer number that the equation $ x^{2}\plus{}r\equal{}y^{m}$ has natural solution, then the equation $ x^{2}\equal{}y^{m}\plus{}r$ has no natural solution.
Could we prove or disprove this. Could we say that this is true also for arbitrary taken powers of x (not only second power).

[quote="SAPOSTO"]And so I suppose that if r is such integer number that the equation $ x^{2} \plus{} r \equal{} y^{m}$ has natural solution, then the equation $ x^{2} \equal{} y^{m} \plus{} r$ has no natural solution.
Could we prove or disprove this.[/quote]
Are there any restrictions on $ m$, $ r$? If not, then your assertion is not correct, since, i. e.:
• $ x^2\plus{}3\equal{}y^2$ has solution $ (x,y)\equal{}(1,2)$, and $ x^2\equal{}y^2\plus{}3$ has solution $ (x,y)\equal{}(2,1)$ (it is similar whenever $ m\equal{}2$, or even $ 2\mid m$);
• $ x^2\plus{}15\equal{}y^3$ has solution $ (x,y)\equal{}(7,4)$, and $ x^2\equal{}y^3\plus{}15$ has solution $ (x,y)\equal{}(4,1)$;
• $ x^2\plus{}18\equal{}y^3$ has solution $ (x,y)\equal{}(3,3)$, and $ x^2\equal{}y^3\plus{}18$ has solution $ (x,y)\equal{}(19,7)$;
• $ x^2\plus{}32\equal{}y^5$ has solution $ (x,y)\equal{}(88,6)$, and $ x^2\equal{}y^5\plus{}32$ has solution $ (x,y)\equal{}(8,2)$.

Is that it?


trigonometry
geometry
perpendicular bisector
geometry proposed

In a quadrilateral $ABCD$ sides $AB$ and $CD$ are equal, $\angle A=150^\circ,$ $\angle B=44^\circ,$ $\angle C=72^\circ.$
Perpendicular bisector of the segment $AD$ meets the side $BC$ at point $P.$
Find $\angle APD.$

[i]Proposed by F. Bakharev[/i]

We have that $ \angle{D} \equal{} 94$, and denote $ \angle{PDA} \equal{} \angle{PAD} \equal{} x$, therefore $ \angle{APD} \equal{} 180 \minus{} 2x$.

Therefore we have that:

1) $ \angle{CDP} \equal{} 94 \minus{} x$, so $ \angle{CPD} \equal{} x \plus{} 14$,

2) $ \angle{BAP} \equal{} 150 \minus{} x$, so $ \angle{APB} \equal{} x \minus{} 14$.

Now from the sin law we have:

1) $ \frac {AP}{\sin{44}} \equal{} \frac {AB}{\sin{(x \minus{} 14)}}$.

2) $ \frac {DP}{\sin{72}} \equal{} \frac {CD}{\sin{(x \plus{} 14)}}$.

By dividing them we obtain: $ \sin{72}\cdot \sin{(x \minus{} 14)} \equal{} \sin{44}\cdot \sin{(x \plus{} 14)}$

$ \iff \cos{(86 \minus{} x)} \equal{} \cos{(x \minus{} 30)}\iff 2x \equal{} 116 \iff x \equal{} 58$.

Thus, $ \angle{APD} \equal{} 180 \minus{} 116 \equal{} 64$., and the problem is solved.

Let $A'$ be a point on $BC$ such that $AA'=AB$. Then $\angle BAA' = 180-2\cdot 44 = 92^\circ$
Similarly
Let $D'$ be a point on $BC$ such that $DD'=CD$. Then $\angle CDD' = 180-2\cdot 72 = 36^\circ$

Let $E$ be the intersection point of  $AA',DD'$

We have $\angle EAD = 150-92 = 58^{o}$ and $\angle EDA = 94-36 = 58^{o}$

So the triangle $EAD$ is isosceles $\Rightarrow EA=ED \ (1)$
Also $AB = CD \Rightarrow AA' = DD' \ (2)$

We'll prove that $E \in BC$.

Assume that $E\not\in BC$

The line $BC$ bisects the plane into two half-planes
If $E$ is on the same half-plane with $A,D$ then $EA' = AA'-EA$ and $ED' = DD'-ED$
If $E$ is on the other half-plane with $A,D$ then $EA' = EA-AA'$ and $ED' = ED-DD'$

At any case, from $(1)\wedge(2)$ we get $EA'=ED'$ which yields that $\triangle EA'D'$ is isosceles.
But this is not possible because the lines $AA',DD'$ make different angles with line $BC \ \ (44^\circ \mbox{ and }72^\circ$)

We got in a contradiction because we assumed that $E\not\in BC$. So the only possible case is that $E\in BC$ (this means that it is the point $P$)
The segments $EA',ED'$ are equal in this case too, but this is because they have zero length  :D

So $\angle APD = 64^\circ$

[u]Note[/u]
We can generalize the problem:

[color=brown][size=150]Theorem[/size]
Let $ABCD$ be a convex quadrilateral with $A-D = 2(C-B)>0$. The perpendicular bisector of $AD$ intersects $BC$ at $P$. If $AB=CD=x$  then $x=AP=DP$ and consequently $\angle APD = 180-B-C$[/color]

Solution: Just follow the same steps (a little attention when $C>90^\circ$)


geometry proposed
geometry
Mass Points
center of mass
similar triangles

Let $ABC$ be a triangle. $D$ is the midpoint of $AB$, $E$ is a point on the side $BC$ such that $BE = 2 EC$ and $\angle ADC = \angle BAE$. Find $\angle BAC$.

Let CD, AE meet at P. Putting masses $m_A = 1, m_B = 1, m_C = 2$ at the vertices A, B, C, the center of mass is at P, so that $m_C \cdot CP = (m_A + m_B) \cdot DP,\ CP = DP$. The triangle $\triangle ADP$ is isosceles with equal base angles by the problem condition, hence DP = AP. Therefore AP = CP = DP. AP is the A-median of the triangle $\triangle ADC$ equal to half the side CD, so that the angle $\angle CAB \equiv \angle CAD = 90^\circ$ is right.

If we apply Menelaus's theorem in the triangle $BCD$ with $A-F-E$, where $F$ is the intersection of $AE$ and $CD$, then
$\frac{AD}{AB} \cdot \frac{EB}{EC} \cdot \frac{FC}{FD}=1 \Leftrightarrow FC = FD.$
Therefore, $\angle BAC = 90^{\circ}$.

[quote=baleanu]Let a poin $ M$ in the plane i.e. $ DCMB$ is parallelogram. Let $ T\equal{}AM\cap BC$ .  $ \triangle{ATB}$ and $ \triangle{MTC}$ are similar and we have $ AB/CM\equal{}BT/TC\equal{}2$ an we have $ E\equal{}T$ $ \equal{}>$ $ A,E,M$ are collinear.$ \angle ADC\equiv\angle ABM\equiv\angle BAM$ $ \equal{}>$ $ \triangle{ABM}$ is isosceles $ \equal{}>$ $ MD\perp AB$ . WE have $ AC\parallel DM$ $ \equal{}>$ $ \angle BDM\equiv\angle BAC\equal{} 90^\circ$ .[/quote]

What was the motivation to take a parallelogram?


pigeonhole principle
geometry
3D geometry
rectangle
sphere

Here's a problem to ponder...

In a 1 by 1 box, we trap 26 point bugs moving about constantly.

Prove that there always exists two bugs that are no more than 2/7 apart.

[quote="Syntax Error"]hm, just use pigeonhole principle.  Partition the box into 26 square things each with sidelength of rt26/26.  The diagonal of each square is rt13/13, which is less than 2/7.  The diagonal would be the farthest apart the bugs could be, so I guess that's it.

I've never really used pigeonhole, so don't mind the unsurity of it all.[/quote]

First of all, its a box. Secondly, your reasoning doesn't work because what if two bugs are in different sectors, but are both really close to the same point (between the two sectors)?

not exactly sure but here goes.

[hide]On one inner face of the cube, the bugs need to be 1/7 away from the edge so they arent 2/7 from another bug on a different face.  There are four bugs on one face.(thats the maximum) because thats how many bugs >2/7 apart that can fit.  4*6=24.  Thats how many bugs can be over 2/7 away from eachother.[/hide]

Is this a 1x1 rectangular surface or a 1x1x1 cube?

It's a square.  The isn't true for a cube -- partition the cube into a 27 1/3 x 1/3 x 1/3 cubes, and put each bug at the center of one of the cubes (or lower-left hand corner, or whatever you want).  Then any two bugs are at least 1/3 = 2/6 > 2/7 away from each other.

As for your solution, Syntax, I'd like to see you disect a square into 26 squares of area 1/26.  Furthermore, in your setup, there is no need to have two bugs in the same sector.  But the fact that it's impossible to divide up a square as you suggested is a bigger problem.

Scrambled, your solution, even if you weren't trying to prove something that isn't true, has serious issues with it.  In general, it is very difficult to prove that something is a "best-case scenario."  You're better off staying away from terms or ideas like that.

okay. assuming it is a 1*1 rectangle:

[hide]

i think that we should partition the box into 25 squares, each of side 1/5. i dont see how making 26 squares is possible, not to mention that it wont help, since we dont have 2 falling into the same box.

if we make it 25 squares, then some two will fall into the same square, so the greatest poss. dist. between them will be sqrt(2/25) = sqrt(2)/5 which is less than 2/7, so we are done.

this also takes care of the moving about constantly thing.

  [/hide]

of course, im assuming that it is a square. a box complicates things. and what is a 1 * 1 box?

Spoiler?  (And see above message for comments on the box.)

And another problem: given 9 bugs in a 1 x 1 x 1 box, show that there exist 2 of them of a within a distance of :rt3:/2 of each other.

If there are 109 bugs in the 1 x 1 x 2 box, show that there are 3 of them contained in a sphere of diameter 1/ :rt3:.

Would it be logical to argue with area, or does this have to be done pigeonhole style?

[hide]26 circles of radius 1/7 has combined area 1.666 approximately. This is way more than an 9/7 by 9/7 square has (81/49 or 1.65 approximately). The separate circles layout would be the layout of maximum distance apart of the circles minus the distance between the circles, but since 1.666 is already greater than 1.65, it works.[/hide]

Nice thought, rlee, but the circles could overlap.  Usually bounds like this are much too tight for an area argument to work.  Pigeonhole is pretty much the only way, most of the time.

By the way, both of my above posts have been edited several times, so if you saw them once, you might want to go check them again.

Good job to everyone who solved it.

Apologies for suggesting that it was a "box". I need to work on my problem phrasing...


Hi.

I'm having trouble understanding a few steps in the proof for vieta's formula.

"We thus have that
$ a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}a_{1}x\plus{}a_{0}\equal{} a_{n}(x\minus{}r_{1})(x\minus{}r_{2})\cdots(x\minus{}r_{n}).$

Expanding out the right hand side gives us

$ a_{n}x^{n}\minus{}a_{n}(r_{1}\plus{}r_{2}\plus{}\!\cdots\!\plus{}r_{n})x^{n\minus{}1}\plus{}a_{n}(r_{1}r_{2}\plus{}r_{1}r_{3}\plus{}\!\cdots\!\plus{}r_{n\minus{}1}r_{n})x^{n\minus{}2}\plus{}\!\cdots\!\plus{}(\minus{}1)^{n}a_{n}r_{1}r_{2}\cdots r_{n}."$

How do you expand to get that and how can you prove that for polynomials in general?

? I don't quite understand what you mean.

you just multiply it out?

How can you prove it works for any degree polynomial?

It's just a consequence of multiplying the expression out. You don't need to prove it or anything.

Yes, I am thinking of induction.

Actually, what will help me learn the formula the best? - Seeing the proof or just multiplying a bunch of polynomials and then saying, "Aha! It works for this one! And this one! And... this one two!"... till I'm convinced the formula works.

Ok, I think I see what you mean, are you not sure how to expand it?  Because after expanding, it should explain itself very well.

Try to expand $ (x \minus{} r_1)(x \minus{} r_2)$, then expand $ (x \minus{} r_1)(x \minus{} r_2)(x \minus{} r_3)$, and finally expand $ (x \minus{} r_1)(x \minus{} r_2)(x \minus{} r_3)(x \minus{} r_4)$.  (Each expansion can use your previous expansion.)  That should imprint the formula in your mind.


vector
geometry
geometric transformation
rotation
linear algebra
matrix

Time is scarlar or vector.

Canyou think in your mind that the particle can be the wave and the wave can be particle(de Broglie :duality)?
i know that question 2 can't not be express clearly

in my opinion , the duality can be known by the properties of wave and particle

HE want me false. :D

Vector is something with 3 components that transform as $v_i' = C_{ij} v_j$ under space rotations, where $C_{ij}$ is the orthogonal matrix $C_{ij}C_{kj} = \delta_{ik}$ or $CC^T = I$. Time does not transform at all, but remains the same, so it is a scalar. In relativity, time transforms as a component of 4-vector under rotations of spacetime (i.e., under Lorentz transformations), so it is a component of 4-vector - neither a scalar, nor a vector.

Electrons, photons, ... behave like waves when they propagate through space and as particles when they strike matter, but never simultaneously. A classical object behaving this way is not known, so it is difficult to imagine. But surely we can imagine an object always looking one way when viewed from one angle and always differently when viewed form another angle, but never looking both ways simultaneously, i.e., from from the same angle. We could consider a paricle something with definite position x and momentum p. But $\triangle x\ \triangle p \ge \frac \hbar 2$. If we measure p of an object with high accuracy (by counting many wavelengths as it propagates), $\triangle x$ is large, so we consider the object a wave, but we may not know where it is after the measurement. If we measure x with high accuracy (by a small detector), $\triangle p$ is large, so we consider the object a particle, but we may not know where it goes after the detection.

i also answer like you, time is scalar in GALILLE frame but not always in EINSTEIN frame( MINKOWSKY space) but he don 't believe

the second question , i answer that it is the symetry  of light and matter , when we see in different angle :it can be wave or particle.
he doesn't like this answer but he can't give his result
note that he is not in physics department of NTU :D
(HUMANITY department)

thank you for your answer :D


geometry
3D geometry
sphere
tetrahedron
search
circumcircle
geometry proposed

Let $ABCD$ be a tetrahedron and let $E,F,G,H,K,L$ be points lying on the edges $AB,BC,CD$ $,DA,DB,DC$ respectively, in such a way that
\[AE \cdot BE = BF \cdot CF = CG \cdot AG= DH \cdot AH=DK \cdot BK=DL \cdot CL.\]
Prove that the points $E,F,G,H,K,L$ all lie on a sphere.

Thank you pohoatza for a nice problem.

I think that the points $E,$ $F,$ $G,$ $H,$ $K,$ $L$ and also the points $E',$ $F',$ $G',$ $H',$ $K',$ $L',$ as their isotomic points with respect to side segments $AB,$ $BC,$ $AC,$ $AD,$ $BD,$ $BC$ respectively, lie on a sphere taken the same center as the circumsphere of the given tetrahedron.

I will search my notes and if I am not mistaken, I will post here later the proof I have in mind.

Kostas vittas.

I think solution is simple as following:
Let $(O)$ be circumsphere of $ABCD$ we have
$AE \cdot BE =-P_{E/(O)}=R^{2}-OE^{2}$ similarly for $F,G,H,K,L$ ($P_{X/(O)}$ is power of $X$ w.r.t $(O)$ )thus
$R^{2}-OE^{2}=R^{2}-OF^{2}=....=R^{2}-OL^{2}\Rightarrow OE=OF=...=OL$ thus they lie on a sphere center $O$
@Vittasko:what's isotomic points with respect to side segments ?

Let the triangle $\bigtriangleup ABC$ be, as the face of the given tetrahedron $ABCD.$

Let also $E,$ $G$ be, the points on side segments $AB,$ $AC$ respectively, as the problem states, such that $(EA)\cdot (EB) = (GA)\cdot (GC) = \kappa^{2}$ $,(1)$

We denote as $P,$ $P',$ the intersection points of the circumcircle $(O_{1})$ of $\bigtriangleup ABC,$ from the segment line $EG.$

So, we have $(EP)\cdot (EP') = (EA)\cdot (EB) = (GA)\cdot (GC) = (GP')\cdot (GP)$

$\Longrightarrow$ $(EP)\cdot (EP') = (GP')\cdot (GP)$ $\Longrightarrow$ $\frac{EP}{GP}= \frac{GP'}{EP'}$

$\Longrightarrow$ $\frac{EP}{GP-EP}= \frac{GP'}{EP'-GP'}$ $\Longrightarrow$ $\frac{EP}{EG}= \frac{GP'}{EG}$

$\Longrightarrow$ $EP = GP'$ $\Longrightarrow$ $O_{1}E = O_{1}G$ $,(2)$

Because of $(EP)\cdot (EP') = \kappa^{2}= R_{1}^{2}-(O_{1}E)^{2}$ where $R_{1},$ is the circumradius of $\bigtriangleup ABC,$ we conclude that $O_{1}E = O_{1}G = \sqrt{R_{1}^{2}-\kappa^{2}}$ $,(3)$

That is the points $E,$ $G,$ lie on a circle $(O_{1}),$ centered at $O_{1},$ with radius given by $(3).$

By the same way we can also prove that the point $F$ lies on $BC$ and such that $(FB)\cdot (FC) = \kappa^{2},$ belongs to $(O_{1}).$

Also the points $E',$ $F',$ $G',$ as the isotomic points of $E,$ $F,$ $G,$ with respect to $AB,$ $BC,$ $AC$ respectively, lie on the circle $(O_{1}).$

$\bullet$ We consider now, the triangle $\bigtriangleup BCD,$ as the other face of the given tetrahedron $ABCD,$ whose the side segment $BC,$ is the intersection line of the planes $[ABC]$ and $[BCD].$

We denote as $O_{2}$ the circumcenter of the triangle $\bigtriangleup BCD$ and it is easy to show that the center $O,$ of the circumsphere of $ABCD,$ is the intersection point of the lines through the points $O_{1},$ $O_{2}$ and perpendicular to the planes $[ABC],$ $[BCD],$ respectively.

Let $K,$ $L$ be, the points on $BD,$ $CD$ respectively, such that $(KB)\cdot (KD) = (LC)\cdot (LD) = \kappa^{2}$ and also let $K',$ $L'$ be, their isotomic points with respect to $BD,$ $CD,$ respectively.

Because of $O_{1}E = O_{1}E' = O_{1}F = O_{1}F' = O_{1}G = O_{1}G'$ and $OO_{1}\perp [ABC],$ we conclude that $OE = OE' = OF = OF' = OG = OG'$ $,(4)$

By the same way we have that  $OK = OK' = OF = OF' = OL = OL'$ $,(5)$

From $(4),$ $(5)$ we conclude that the points $E,$ $F,$ $G,$ $K,$ $L$ and also the points $E',$ $F',$ $G',$ $K',$ $L',$ as their isotomic point with respect to $AB,$ $BC,$ $AC,$ $BD,$ $BC$ respectively, lie on a sphere $(S'),$ taken the same center $(O),$ as the circumsphere $(S)$ of the given tetrahedron $ABCD.$

By the same way ( consider the triangles $\bigtriangleup ABD,$ $\bigtriangleup ACD,$ as the other two faces of $ABCD$ ), we can prove that the points $H,$ $H',$ isotomic conjugates with respect to $AD$ and such that $(HA)\cdot (HD) = (H'A)\cdot (H'D) = \kappa^{2},$ lie on the sphere $(S')$ and the proof is completed.

Kostas Vittas.

PS. We say two points as isotomic conjugates with respect to the corrspondent sidesegment of a given triangle, if they are symmetrical each other wrt its midpoint.

I have inspired a similar problem as this problem
Given regular tetrahedron $ABCD$ inscribed $(O)$ and $A',B',C',D'$ are arbitrary points on (inside) the face $BCD,CDA,DAB,ABC$ inface $(BCD)$ let  $[A'BC]=S_{ad},[A'CD]=S_{ab},[A'DB]=S_{ac}$ similarly in others face, prove that $A',B',C',D'$ lie on a sphere center $O$ iff $S_{ad}^{2}+S_{ab}^{2}+S_{ac}^{2}=S_{bc}^{2}+S_{bd}^{2}+S_{ba}^{2}=S_{cd}^{2}+S_{cb}^{2}+S_{ca}^{2}=S_{da}^{2}+S_{db}^{2}+S_{dc}^{2}$


Pencils come in a package holding 10 pencils. Rina's teacher bought 23 packages. How many pencils did she buy?

10 packages of 23 pencils=10*23= a total of 230 pencils


This was shown in my chemistry textbook and there wasn't any example question that followed. Could someone please explain this and give an example. Thanks.

H2CO3 +H2O <-> H3O+ + HCO3-

Ka1 = [H3O+][HCO3-]/  [H2CO3]

HCO3- +H2O <-> H3O+ + CO3 2-

Ka2= [H3O+][CO3 2-]/ [HCO3-]

We can express the composition of the solution in terms of the fraction, a(X), of carbonate present as each species X, where X could be H2CO2, HCO3-, or CO3 2- and

a(X) =                [X]/[H2CO3]+[HCO3-][CO3 2-]

It will turn out to be helpful to express a(x) in terms of the ratio of each species to the intermediate species, HCO3-. So we divide the numerator and denominator by [HCO3-] and get

a(x) =        [X]/[HCO3-]/  ([H2CO3]/[HCO3-]+1+([CO3 2-]/[HCO3-])

All three concentrations ratios can be written in terms of the hydronium ion concentration. We simpy rearrange the expressions for the acidity constants:

[H2CO3]/[HCO3-]   = [H3O+]/  Ka1

[CO3 2-]/ [HCO3-]  = Ka2/ [H3O+]

Therefore, the three fractions are

a(H2CO3) = [H3O+]2 /f

a(HCO3-) = [H3O+]Ka1/f

  a(CO3 2-) = Ka1Ka2/f

where
f = [H3O+]2 +[H3O+]Ka1 + Ka1Ka2

Thanks in advance.

Look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=153278]this post[/url] where a general diprotic acid H2A is considered and the expression for a(HA-) is derived. The expressions for a(H2A) and a(A2-) are derived in a similar way.

Thanks for your help but I am wondering if there is a formula if we also take into account a changing pH. How about this question that I found as an example please. Thanks in advance

When H2S is dissolved in water it is about saturated at a concentration near 0.1M. Three species containing sulfur exist in this solution, their proportions depending on the acidity of the solution: H2S, HS- and S2-.Given the following equilibrium constants;

H2S -> H+ +HS-    Ka1 = 1.0X 10^-7
HS- -> H+ +S2-    Ka2 = 1.3 X 10^-13

a) What is the concentration of the sulfide ion, [S2-], in a saturated 0.100M solution of H2S when for the solution the pH =2.0?

[hide="Answer"]First, the concentration of sulphide ion (S2-) is expected to be very small due to the low pH. We have a diprotic acid (H2S) of concentration C = 0.100 M, and so the fraction of this acid that exists as S2- is given by

$ \alpha(S^{2\minus{}}) \equal{}\frac{[S^{2\minus{}}]}{C}\equal{}\frac{K_{1}K_{2}}{[H_{3}O^{\plus{}}]^{2}\plus{}K_{1}[H_{3}O^{\plus{}}]\plus{}K_{1}K_{2}}$.

Substituting by the data given we get $ \alpha \equal{} 1.3\times 10^{\minus{}16}$ and so $ [S^{2\minus{}}] \equal{}\alpha C \equal{} 1.3\times 10^{\minus{}17}\,\, M$, a neglegible concentration, as expected.
[/hide]

Thank you so much for your help. I also got that answer when I used the equilibrium table method but when I actually tried the formula method, I was off by 0.10 and now I know I actually have to divide by C.


function
calculus
derivative
limit
integration
real analysis
real analysis unsolved

anyone ever seen this fubini theorem?

let f_n be series of nondecreasing functions on [0,1] convering everywhere. set $f(x) = \sum f_n(x)$. then we have

$\displaystyle \sum_{n=1}^\infty f^\prime_n(x) = f^\prime(x)$

does anyone have proof?

Do you want the identity for the derivatives to hold pointwise, almost everywhere or in some other sense? My point is that it is false for some meanings (for instance, it is false if you want pointwise convergence: the sum may fail to be differentiable at $x$ even if all functions $f_k$ are differentiable at $x$ and even if it is, the derivative of the sum may be strictly greater than the sum of the derivatives).

Of course, he means it holds almost everywhere. I will use increasing for 'non-decreasing'.

First, $\sum_{n=1}^\infty f_n$ is increasing, and it is differentiable almost everywhere.
$\sum_{n=1}^\infty f_n = \sum_{n=1}^N f_n + R_N$, $R_N = \sum_{n=N+1}^\infty f_n$
$R_N$ is differentiable almost everywhere, too.
Thus we have
$\frac{d}{dx}\sum_{n=1}^\infty f_n = \sum_{n=1}^N \frac{d\ f_n}{dx} + \frac{d}{dx}R_n$, a.e.$x\in [a,b]$
Now it suffices to show that $\lim R_n'(x) = 0$ a.e.

Since $f_n'(x)\geq 0$ a.e., we have
$R_N'(x)=f_{N+1}'(x)+R'_{N+1}(x)\geq R'_{N+1}(x)$ a.e.
So there exists $R(x)=\lim R_N'(x)\geq 0$

$\int_0^1 R(x)dx = \int_0^1 \lim R_N'(x)dx \leq \liminf \int_0^1 R_N'(x)dx$(Fatou's lemma)
$\leq \liminf (R_N(1)-R_N(0)) \to 0$ (Lebesgue's Theorem)
hence $R(x)=0$ a.e. $x\in [0,1]$.


[hide="hint"]The sum can be written as $\sum_{k=1}^{1319}\frac{1}{k}-2\left(\sum_{k=1}^{659}\frac{1}{2k}\right)$,
which then can be written as $\sum_{k=660}^{1319}\frac{1}{k}$[/hide]

[hide="Bigger Hint"]Some intuition shows that the above sum is equal to $\sum_{660}^{989}\left(\frac{1}{k}+\frac{1}{1979-k}\right)$[/hide]

wow, very nice solution, thanks a lot!  :)

How can you solve this...
Two circles intersect at points P and Q. A line l that intersects PQ  intersect the two circles at the points A, B,C, and D( in that order along the line). It is known that <APB= 27 . Find the degree measure of <CQD.


Since the AP season is coming up, I was wondering if people would like to post difficult AP level problems they encountered during the school year, so we can get some practice.

Heres a reasonably hard one:

How many millimoles of NaOH must be added to 50.0 milliliters of .20 molar HOCl to obtain a buffer solution that has a pH of 7.49? Assume that the addition of solid NaOH results in a negligible change in volume.

Given that for HOCl: $K_a = 3.2 \times 10 ^{-8}$

This is (I think) the hardest part of a 5 part free-response problem. (Sorry, I do not have the year)

Solution
[hide]
pH = pKa + log([base]/[acid])   this is the Henderson-Hasselbalch Equation (sp?)

7.49 = -log(3.2x10^-8) = log([base]/[acid])

[base]/[acid] = 0.99

so, enough base must be added so that [OCl-]/[HOCl] is 0.99 (I actually assume that the AP exam meant for it to be 1, but not rounding probably gave me the 0.99)

to find the molarity of NaOH needed:
x/(0.20-x) = 0.99
x=0.099 M

(y mmol)/(50 mL) = 0.099
y = 4.95 mmol... and using sig figs... 5.0 mmol.
[/hide]

[hide]What's the reaction that occurs?
[hide]OH- + HOCl $\Rightarrow$ H2O + OCl-
[hide]How much acid is left, and how much conjugate base is created?
There were .05 L * .2 M = .01 mol = 10 mmol HOCl to begin.
There are $10 - x$ mmol HOCl and $x$ mmol OCl- (the conjugate base), where $x$ is mmol of NaOH added.
What equation do we use with buffers?
Henderson-Hasselbach: $pH=pKa + \log\left(\frac{[base]}{[acid]}\right)$
$7.49=-\log(3.2\cdot10^{-8}) + \log\left(\frac{x}{10-x}\right)$
$x=5$
5 mmol NaOH[/hide][/hide][/hide]

Alternative method:
[hide]We observe that 7.49 is the pKa of HOCl: $-\log\left(3.2\cdot10^{-8}\right)=7.49$,
[hide]so we are searching for the half-equivalence point of HOCl.
[hide]We calculate that there are .05 L * .2 M = .01 mol = 10 mmol HOCl.
At the half-equivalence point, we want half of that amount of base, or
$\frac{10}{2}=5$
5 mmol NaOH[/hide][/hide][/hide]


A charge $Q$ is placed on a corner of a cube with side length $a$. What is the net flux through the cube?

consider a cube of length $2a$ and place this charge at the centre of this big cube now it's flux through the entire cube is sum of the fluxes through the 8 similar smaller because at the surfaces of the cubes(internal) field is perpindicular to area vector and opposite area vectors cancel out each othe so effectively by symmetry we have flux through each smaller cube is same and is equal to $\frac{Q}{8\epsilon_{0}}$

Ok that's what i thought. Needed confirmation though  :)


see here:

http://www.mathlinks.ro/viewtopic.php?t=206627

and, please learn to use TeX to write down your problem here.


1.A question paper consists of 2 parts  A and B,Part A has 5 questions with 1 alternate each and part B has 4 questions without any alternate.Find the no. of ways by which one can attempt the paper when atleast one question must be attempted from each part???

3.Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as thier vertices and having exactly 2 sides common  with the polygon is????????

Also,Number of quadrilaterals that can be made using the vertices of a polygon of 10 sides as thier vertices and having exactly 1 side common  with the polygon is????????

4.P is a set containing n elements.A subset A of P is selected at random.The set P is reconstructed by replacing its elements.A subset B of P is again selected at random.

a.The no. of ways of selecting the subsets A and B such that 	A$ \cup$B is equal to P is???

b.The no. of ways of selecting the subsets A and B such that 	A$ \cap$B is equal to $ \phi$ is???

[hide="Solution to 1"]If I understand you correctly, then for each question in Part A, we can  do one of three things: attempt the question, attempt its alternate, or not attempt it at all. Thus, there are $ 3^5\minus{}1\equal{}242$ ways to attempt Part A, since we must attempt at least one question.

For Part B, there are two choices for each question, resulting in $ 2^4\minus{}1\equal{}15$ ways to attempt Part B. In total, we have $ 242*15\equal{}3630$ ways to attempt the paper[/hide]

Thats the correct answer!!
It took 2 hours for me to solve the problem.
Its a great solution
Thankyou very very much.
Please try others also.


function
algebra
domain
conics
parabola
special factorizations

Any help on these questions will be very much appreciated:

1. If g(t) = t/t-1, find g(x+1)

2. Given f(x)= ax^2 + b and f(2)= 5, f(3)= 20 find [i]a[/i] and [i]b[/i]

3. What are the zeros of function y = -(square-root 9- x^2)?

4. Describe in words the locus of a point which moves so as to trace out a path represent by each of these equations:

a) x^2 + y^2 - 2x + 7y +2 = 0

b)x +4y - 1 = 0

c)y= (x -6)^2

5. Here is a continuous piecewise function:

f(x)= | x+6 for x < -2
        | x^2 for -2 <_ x <_ 3
        | 9 for x > 3

a) Find f(-6), f (1), f(5) - (already done)

b) Sketch the graph of y = f(x)

c) State it's domain and range

Thanks in advance!~

1.

$ g(x\plus{}1) \equal{} \frac{x\plus{}1}{x\equal{}1\minus{}1} \equal{} \frac{x\plus{}1}{x}$.

2.

$ f(2) \equal{} 5$ gives $ 4a \plus{} b \equal{} 5$

$ f(3) \equal{}20$ gives $ 9a \plus{}b \equal{} 20$

Adding, $ 5a \equal{} 15$

$ a\equal{}3$

Substituting, $ b\equal{}\minus{}7$.

3.

$ x\equal{}\pm3$.

4.

(a) Completing the square, we get $ (x\minus{}1)^{2} \plus{} (y\plus{}\frac{7}{2})^{2} \equal{} \frac{45}{4}$, a circle with centre at $ (1, \frac{\minus{}7}{2})$ and radius $ \frac{3\sqrt{5}}{2}$.

(b)  A straight line passing through $ (1, 0)$ and $ (0, \frac{1}{4})$.

(c) A parabola touching the x-axis at $ (6, 0)$

5.

(a) $ f(\minus{}6) \equal{} 0, f(1) \equal{} 1, f(5) \equal{} 9$.

(b) line through $ (\minus{}6, 0)$ and $ (\minus{}1, 4)$ parabola and line $ y\equal{}9$.

(c) domain is all the real numbers, range is $ y\le 9$.

[quote="AndrewTom"](b) line through $ ( \minus{} 6, 0)$ and $ ( \minus{} 1, 4)$ parabola and line $ y \equal{} 9$.

(c) domain is all the real numbers, range is $ y\le 9$.[/quote]

Cheers for response, I really appreciate the help.

On your answer for this, what are exact co-ordinates and how many lines will there be? I know there will be a line at y=9, but you say there will be parabola with only 2 values x and 2 values of y?

Sorry, just a little confused~ ;\

Thanks for help  :blush:

Hi Follz20.

The graph of the function is piecewise continuous and consists of three sections:

For $ x < \minus{} 2$, we have the part of the line $ y \equal{} x \plus{} 6$ to the left of the point $ ( \minus{} 2, 4)$. It passes through the points $ ( \minus{} 3, 3), ( \minus{} 4, 2), ( \minus{} 5, 1), ( \minus{} 6, 0), ( \minus{} 7, \minus{} 1), ( \minus{} 8, \minus{} 2)$ and so on.

For $ \minus{} 2 \le x \le 3$, we have the part of the parabola $ y \equal{} x^{2}$ (a U-shaped curve) between $ ( \minus{} 2, 4)$ and $ (3, 9)$. It passes through $ ( \minus{} 1, 1), (0, 0), (1, 1), (2, 4)$ and $ (3, 9)$.

Then for $ x > 3$ we have the part of the horizontal line $ y \equal{} 9$ to the right of the point $ (3, 9)$. It passes through $ (4, 9), (5, 9), (6, 9), (7, 9)$ and all points $ (x, 9)$ where $ x > 3$.

It never goes above $ y\equal{}9$ but goes to $ \minus{}\infty$ as $ x$ goes to $ \minus{}\infty$ and so the range is $ y\le 9$.

I hope this is more helpful (as I don't know how to post a graph).

If you're still a bit confused, just ask again.


[b][color=cyan]Here are problems from the 6-th Kyiv mathematical festival (6.05.07)
Any comments are welcome![/color][/b]

[b][size=200][color=orange]The sixth Kyiv mathematical festival[/color][/size][/b]

[b][size=150][color=indigo]8-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837042#837042]1.[/url] Is it possible to cut the table of size $2007\times2007$ into figures shown here, if one has to use at least one figure of each sort? $\begin{picture}(45,25) \put(5,5){\put(0,0){\line(1,0){16}}\put(0,8){\line(1,0){24}}\put(0,16){\line(1,0){24}}\put(8,24){\line(1,0){16}}\put(0,0){\line(0,1){16}}\put(8,0){\line(0,1){24}}\put(16,0){\line(0,1){24}}\put(24,8){\line(0,1){16}}}\put(35,5){\put(0,0){\line(1,0){8}}\put(0,8){\line(1,0){8}}\put(0,16){\line(1,0){8}}\put(0,24){\line(1,0){8}}\put(0,0){\line(0,1){24}}\put(8,0){\line(0,1){24}}}\end{picture}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837043#837043]2.[/url] Find all pairs of positive integers $(a,b)$ such that $\sqrt{a-1}+\sqrt{b-1}=\sqrt{ab-1}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837046#837046]3.[/url] The vertices of 100-gon (polygon with 100 sides) are colored alternately white or black. One of the white vertices contains a checker. Two players in turn do two things: move the checker into other vertice along the side of 100-gon and then erase some side. The game ends when it is impossible to move the checker. At the end of the game if the checker is in the white vertice then the first player wins. Otherwise the second player wins. Does any of the players have winning strategy? If yes, then who?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837053#837053]4.[/url] The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837051#837051]5.[/url] One has a set of stones with weights $1, 2, \ldots, 20$ grams. Find all $k$ for which it is possible to place $k$ and the rest $20-k$ stones from the set respectively on the two pans of a balance so that equilibrium is achieved.

[b][size=150][color=indigo]9-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837043#837043]1.[/url] Find all pairs of positive integers $(a,b)$ such that $\sqrt{a-1}+\sqrt{b-1}=\sqrt{ab-1}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837053#837053]2.[/url] The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837051#837051]3.[/url] One has a set of stones with weights $1, 2, \ldots, 51$ grams. Find all $k$ for which it is possible to place $k$ and the rest $51-k$ stones from the set respectively on the two pans of a balance so that equilibrium is achieved.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837046#837046]4.[/url] The vertices of 100-gon (polygon with 100 sides) are colored alternately white or black. One of the black vertices contains a checker. Two players in turn do two things: move the checker into other vertice along the side of 100-gon and then erase some side. The game ends when it is impossible to move the checker. At the end of the game if the checker is in the white vertice then the first player wins. Otherwise the second player wins. Does any of the players have winning strategy? If yes, then who?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837061#837061]5.[/url] Let $a,b,c>0$ and $abc\ge1.$ Prove that $\left(a+\frac{1}{a+1}\right)\left(b+\frac{1}{b+1}\right) \left(c+\frac{1}{c+1}\right)\ge\frac{27}{8}.$

[b][size=150][color=indigo]10-th form[/color][/size][/b]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837043#837043]1.[/url] Find all pairs of positive integers $(a,b)$ such that $\sqrt{a-1}+\sqrt{b-1}=\sqrt{ab-1}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837053#837053]2.[/url] The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837051#837051]3.[/url] One has a set of stones with weights $1, 2, \ldots, 51$ grams. Find all $k$ for which it is possible to place $k$ and the rest $51-k$ stones from the set respectively on the two pans of a balance so that equilibrium is achieved.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837061#837061]4.[/url] Let $a,b,c>0$ and $abc\ge1.$ Prove that
$27(a^{3}+a^{2}+a+1)(b^{3}+b^{2}+b+1)(c^{3}+c^{2}+c+1)\ge$ $64(a^{2}+a+1)(b^{2}+b+1)(c^{2}+c+1).$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=837046#837046]5.[/url] The vertices of 100-gon (polygon with 100 sides) are colored alternately white or black. One of the vertices contains a checker. Two players in turn do two things: move the checker into other vertice along the side of 100-gon and then erase some side. The game ends when it is impossible to move the checker. At the end of the game if the checker is in the white vertice then the first player wins. Otherwise the second player wins. Does any of the players have winning strategy? If yes, then who?

[hide="[color=cyan"]authors[/color]]
The problems are [supposed to be] by:
V.Brayman (8.1, 8.2=9.1=10.1, 8.3$\sim$9.4$\sim$10.5, 8.4=9.2=10.2, 9.5$\sim$10.4),
O.Rudenko (8.5$\sim$9.3=10.3)[/hide]

Here is another problems from sixth KMF. They was proposed for 10-th form on speaking olympiad.
1. In table 2007*2007 consider centers of all squares 1*1.What is minimal number of lines (neizer gorizontal nor vertical)we have to draw so that all centers of little squares are on the lines.

2. On the list of paper there is a convex poligon which vertises are in integer points. There is no vertical or gorizontal sides.Consider the gorizontal and vertical segments which lie in the poligon.Prove that sum of length of gorizontal segments is equal to sum of length of vertical segments.

3. ${a_{n}}$ is a sequanse of positive integers. $a_{n+1}=a_{n}^{3}+2007$. Prove that there is no more then two  squares in it.

4. x,y,z are positive.Prove:
$(1+\frac{x}{y})(1+\frac{y}{z})(1+\frac{z}{x}) \ge 2+\frac{2(x+y+z)}{(xyz)^{\frac{1}{3}}}$.

5. Circles $w_{1}$ and $w_{2}$ with centers in A and B respectively are tangent in a point X.Circle  $w_{3}$ with center C is tangent to  $w_{1}$ and $w_{2}$ in points Y and Z respectively.X is inside  $w_{3}$.Tangent line to $w_{1}$ and $w_{2}$ in point X intersect $w_{3}$ in points P and Q.M is midle of PQ.Prove
that $\measuredangle YMZ=\measuredangle ACB$.

6.Find all functions $f: N \to N$ which satisfy.
   (1) if x<y then f(x)<f(y);
   (2) if f(yf(x))=x^{2}f(xy).

7. ABCD is table n*n.We call diagonal of square 1*1 positive if it is paralel to diagonal AC.What is a list number of coins we have to put in some squares 1*1 so that for all squares 1*1 there is a coin in it or there is a coin in same line or colon or positive diagonal.

My english is bad so please ask quastions.

To moderators:
The problems of 6-th Kyiv Math Fest [url=http://www.mathlinks.ro/Forum/resources.php?c=180&cid=108&year=2007]here[/url] don't correspond somewhere with problems in the first message of this topic. Please fix it


Excuse me, is the trigonometry text book finished yet? If so, where can I find it?

Nope, it's coming out hopefully before year's end.

[quote="rrusczyk"]That is hopefully correct.

I think I'll go work on it now...[/quote]

Thank you that's good. I am terrible at trig :blush:

Wait!  I thought the "Trig" book was really Pre-Calc?  Is It?

[quote="Ihatepie"]Yeah, it's called Pre-Calculus and covers all of Pre-Calculus not covered in the Algebra 3 class, which to my knowledge is trigonometry and Complex Numbers.[/quote]

A large part of Vectors and Matrices is also included (which I'm horrible at  :D ).


Let $ ABC$ be a triangle with circumcircle $ (O)$ and incircle $ (I)$ . Denote :

 $ \boxed P$ for which $ PA\perp PI\parallel BC$ ; $ \boxed R\in (O)$ for which $ RA\perp RI$ ;

second intersection $ A'$ of $ AI$ with $ (O)$ ; second intersection $ \boxed S$ of $ A'C$

with circumcircle of $ \triangle AIC$ . Prove that quadrilateral $ CPRS$ is cyclic.

[quote="Virgil Nicula"]... $ \boxed R\in (O)$ for which $ AR\perp AO$ .... [/quote]
Did you mean $ AR \perp RI$????

Dear Mathlinkers,
1. Let 0 be the circumcircle of ABC, 1 the circle with diameter AI (it goes through R and P) and 2 the circumcircle of AIC.
2. According to a converse of Reim's theorem applied to 0 and 1,            P, R and A' are collinear
3. According to a converse of the pivot theorem applied to 0, 1 and 2,     P, I and S are collinear
4. According to a converse of the pivot theorem,                                   C, P, R and S are cocyclics.
Sincerely
Jean-Louis


This game will be similar to the card game "In a Pickle", if anyone has played it. The rules are this: the first person posts a noun, and the next person posts a noun that can contain that noun.

For example:

Player 1: Book
Player 2: Bookshelf
Player 3: Room
.
.
.
.
And so on like that. If someone posts, and within 2 days, another person doesn't post a noun that works, the person who stumped everyone wins and gets a point. A noun can only be used once per round.

Another rule is that you may be creative and think outside of the box. For example:

Player 1: Peanut
Player 2: Elephant
Player 3: Dictionary

That's pretty much it, but remember, above all that, no nouns may be an abstract noun, like feelings, and you can not post a noun that was already posted this round. If you have any questions, PM me.

I will give the first word:

[b]Pickle[/b]

Note: You don't have to say: My house, or your city, just say "house" or "city".

[b]State[/b]

Heh...I dont get it....so if no one posts next then who gets a point...
Also....are you allowed to go from North America to Universe? If so...then I can just say "Universe" after someone says Pickle Jar...

[b]Heaven[/b]?

[b]Building[/b]
[hide="used words"]Pickle
Jar
House
City
State
Country
North America
Universe
Heaven
Bible
Bookstore
Mall
Dictionary
Library
University
Thesaurus
Box
Truck
Parking Lot
Picture
Frame
Room
Hotel
Town
County
Mountain
Earth
Solar System
Galaxy
Space
Painting
People
Building[/hide]

Wait..polkadotman did it wrong :(

He made 2 mistakes:

1.A painting cant be in people
2.A building cant be in a window

Let's let it slide...... for now. >:)
Block. (as in like "2 blocks away")
[hide="used words"]Pickle
Jar
House
City
State
Country
North America
Universe
Heaven
Bible
Bookstore
Mall
Dictionary
Library
University
Thesaurus
Box
Truck
Parking Lot
Picture
Frame
Room
Hotel
Town
County
Mountain
Earth
Solar System
Galaxy
Space
Painting
People
Building
Window
Office Building
Printer
Street
Block[/hide]

Oh... yea. I was thinking the other way around.

@meewhee: Is it a city block or just a normal block?

Oh well, city block

[b]Street[/b]

Rockies

I assume you were talking about [url=http://sports.espn.go.com/mlb/players/profile?playerId=6175]Huston Street[/url]. :D

Hrmm...
[b]Stadium[/b]
[hide="used worddds"]Pickle
Jar
House
City
State
Country
North America
Universe
Heaven
Bible
Bookstore
Mall
Dictionary
Library
University
Thesaurus
Box
Truck
Parking Lot
Picture
Frame
Room
Hotel
Town
County
Mountain
Earth
Solar System
Galaxy
Space
Painting
People
Building
Window
Office Building
Printer
[b]Street[/b]
Block
[b]Street[/b]
Rockies
Stadium[/hide]
@polkadotman1 Eh hem... ?

laboratory
[hide="used words"]
Pickle
Jar
House
City
State
Country
North America
Universe
Heaven
Bible
Bookstore
Mall
Dictionary
Library
[b]Laboratory[/b]
University
Thesaurus
Box
Truck
Parking Lot
Picture
Frame
Room
Hotel
Town
County
Mountain
Earth
Solar System
Galaxy
Space
Painting
People
Building
Window
Office Building
Printer
Street
Block
Street
Rockies
Stadium
[/hide]

which is used.

Complex (scientific complex, although it has other meanings :D)


find a sequence of positive integer $f(n)$ ($n \in N$) such that:
           i)$f(n) \leq n^{8}$ for all $n \geq 2$
           ii)For any distinct $a_{1},a_{2},...,a_{k},n$ then $f(n)$ different to $f(a_{1})+f(a_{2})+...+f(a_{n})$

Don't think i'm getting this...
Can't we say that :
$a_i=1$
$f(n)=n$

I think this works.


[quote="CircleSquared"][quote="worthawholebean"]([b]6[/b]) Determine the number of ways to select a sequence of $ 8$ sets $ A_1,A_2,\ldots,A_8$, such that each is a subset (possibly empty) of $ \{1,2\}$ and $ A_m$ contains $ A_n$ if $ m$ divides $ n$.[/quote]

stupid question:
if m divides n, does that mean m must be divisible by n or n must be divisible by m

i think its the 2nd one, but that would make the problem easier than the first, and obviously nobody wants that[/quote]

n must be divisible by m


function
number theory
algebra proposed
algebra

Find all functions f from the integers with values in the same set and with the properties:
 a) f(xy)=f(x)f(y);
b) 2f(x^2+y^2)-f(x)-f(y) is 0 or 1 for all x and y.

Just the beginning:

f(0)=f(0)f(x) for all x, so there are 2 cases: if f(0)=/=0 then f(x)=1 for all x, and this is a soln. Now let's assume f(0)=0. We have f(x)=f(x)f(1) for all x, so there are, again, 2 cases: if f(x)=0 for all x then this is another soln. We assume there exists an x s.t. f(x)=/=0. We get f(1)=1.

We make x=y in (a) to find f(x 2 )=f 2 (x). We make y=0 in (b) to get 2f(x 2 )-f(x) is 0 or 1 for all x, but we know that f(x 2 )=f 2 (x), so we have
2f 2 (x)-f(x) is 0 or 1 for all integer x, which, in turn, means f(x)=0 or 1 for all x.

I didn't continue, but I think this is where the interesting part begins  :D

Very interesting problem. In short list IMO 2002 there is a similar question, but for function on R. Harazi question, after algebraic consideration of Grobber becomes number theory problem:
Let A = {x from Z| f(x) = 0} B = {x from Z| f(x) = 1}
then by Grobber, A  \cup B = Z. We wont consider the trivial constant-cases. Then we have f(0) = 0, f(1) = 1 and
1) x in A, y in Z => yx in A
2) x in B, y in B => xy in B
3) x in A, y in A => x^2+y^2 in A
4) x in B, y in B => x^2+y^2 in B
5) x in A, y in B => x^2+y^2 in B

In other hand, each partition of Z into 2 subset A, B with abovementioned will define one solution of functional equation.

I didnt get a final result, but here is some useful fact:
1) - 1 in B => A, B symmetric under 0 subset.
2) If n mininal positive integer beloging to A then n is prime number of the for 4k+3.
3) Partition A = {multiples of 3} , B = Z \ A satisfying all 5 conditons and we get one example of nonconstant solution for initial equation.

I think, the other members of forum can help me to finish this nice question.

Namdung

You are very close to the solution Namdung. You've got the idea!


[quote="Rajiv"][quote="MNrule"]A circular table with smooth horizontal surface is rotating at an angular speed ω about its axis. A groove is made on the surface along a radius and a small particle is gently placed inside the groove at a distance l from the centre. Find the speed of the particle with respect to the table as its distance from the centre becomes L.[/quote]
Does the particle have any mass? For this question, you need mass of table as well as that of particle. If not mass of table, atleast the information that the table keeps rotating with $ \omega$ even when particle is placed, else the Moment of Inertia will vary according to distance of particle from center.[/quote]

This is a kinematic question, the implicit assumption is that the mass of the particle is negligible compared to the mass of the table -- the moment of inertia of the table/particle system is unaffected by the addition of the particle.

As a kinematic question its solution is simple:

\[ v_{r}^{2}(L) \equal{} v_{r}^{2}(\ell) \plus{} \omega^2(L^2 \minus{}{\ell}^2) \]

One might well ask what happened to the factor of "2" one typically sees in the linear kinematic equation: $ v_{2}^2 \equal{} v_{1}^2 \plus{} 2a\cdot x$ ? The answer is that it vanishes for the rotational case as a consequence of the calculus, and as such discussion of why is inappropriate for [b]Introductory Physics[/b].


Find all polynomials $ P(x)\in \mathbb R\left[x\right]$ satisfying $ P((x \plus{} 1)^3) \equal{} (P(x) \plus{} 1)^3$, $ \forall x\in\mathbb R$

let $ Q(x)\equal{}P(x\minus{}1)\plus{}1$.
$ Q(x^3\plus{}1)\equal{}Q(x)^3\plus{}1$  (*)

Let $ w$ be the root to $ w^3\equal{}1$.
From (*), we have
$ Q(wx)^3\equal{}Q(x)^3$.
Since $ Q$ is polynomial, one of the following three equation always holds:
$ Q(wx)\equal{}Q(x)$; $ Q(wx)\equal{}wQ(x)$; $ Q(wx)\equal{}w^2Q(x)$.

1) $ Q(wx)\equal{}wQ(x)$ or $ Q(wx)\equal{}w^2Q(x)$.
Obviously $ Q(0)\equal{}0$.
If we define $ x_0\equal{}0$ and $ x_n\equal{}x_{n\minus{}1}^3\plus{}1$, we see $ x_n\rightarrow\infty$ and $ Q(x_n)\equal{}x_n,\forall n$ .
Hence $ Q(x)\equal{}x$. This is consistent with $ Q(wx)\equal{}wQ(x)$, but inconsistent with $ Q(wx)\equal{}w^2Q(x)$.

2) $ Q(wx)\equal{}Q(x)$.
We can write $ Q(x)\equal{}h(x^3)$ for some polynomial $ h$.
$ Q(x\plus{}1)\equal{}Q(\sqrt[3]{x})^3\plus{}1\Rightarrow h((x\plus{}1)^3)\equal{}h(x)^3\plus{}1\Rightarrow Q(x)\equal{}h(x\minus{}1)^3\plus{}1$
Let $ H(x)\equal{}h(x\minus{}1)$, we have
$ Q(x)\equal{}H(x)^3\plus{}1$ and
$ H(x^3\plus{}1)\equal{}H(x)^3\plus{}1$. In other words, $ H$ satisfies (*) again.

Hence, we have $ Q_1(x)\equal{}x$ and $ Q_n(x)\equal{}Q_{n\minus{}1}(x)^3\plus{}1$.
It is easy to get
$ P_1(x)\equal{}x$ and $ P_n(x)\equal{}(P_{n\minus{}1}(x)\plus{}1)^3$.

Using the method, we should be able to generalized to $ G(x)\equal{}(x\plus{}1)^k$ for natural $ k>1$.

[quote="mr.danh"]$ \forall x\in\mathbb R$[/quote]
This implies that your solution made some mistakes. You used the complex number $ wx$ in $ Q(wx)$

No.
Note $ P$ is polynomial. If that equation holds for all real, it will hold for all complex.

Thanks mr.danh for nice problem and thanks so much for xxp2000 for nice proof.
[quote="xxp2000"]Note $ P$ is polynomial. If that equation holds for all real, it will hold for all complex.[/quote]
Can you give some explainations or links which proves or states what you said?

Assume $ P(x)\equal{}\sum_ia_ix^i$. That equation implies
$ \sum_ia_i(x\plus{}1)^{3i}\equal{}(\sum_ia_ix^i\plus{}1)^3,\forall x\in\mathbb{R}$ .
If you treat the difference of both sides as a new polynomial, it has infinitely many roots. That tells you both sides are exactly the same polynomials. Then you can plug in $ x\in\mathbb{C}$ as well.

Am I right?

[i]Thank , xxp2000 , very nice solution :lol:

    I also like problem which involve in complex numbers[/i]

[i]But , i think , there is a mistake in your  solution , Although i don't find out this mistake :

  For example : We can easily see that the equation have a root :

  $ Q(x) \equal{} c$ with c is the real root of the eqation $ x^3 \minus{} x \plus{} 1 \equal{} 0$ :o  [/i]

  [i]Solve that equation is not hard but i think the problem should be added the condition $ degP \geq 1$[/i]

[quote="xxp2000"]let $ Q(x) \equal{} P(x \minus{} 1) \plus{} 1$.
$ Q(x^3 \plus{} 1) \equal{} Q(x)^3 \plus{} 1$  (*)

Let $ w$ be the root to $ w^3 \equal{} 1$.
From (*), we have
$ Q(wx)^3 \equal{} Q(x)^3$.
Since $ Q$ is polynomial, one of the following three equation always holds:
$ Q(wx) \equal{} Q(x)$; $ Q(wx) \equal{} wQ(x)$; $ Q(wx) \equal{} w^2Q(x)$.

1) $ \boxed{Q(wx) \equal{} wQ(x)}$ or $ Q(wx) \equal{} w^2Q(x)$.
Obviously $ Q(0) \equal{} 0$.
If we define $ x_0 \equal{} 0$ and $ x_n \equal{} x_{n \minus{} 1}^3 \plus{} 1$, we see $ x_n\rightarrow\infty$ and $ Q(x_n) \equal{} x_n,\forall n$ .
Hence $ Q(x) \equal{} x$. This is consistent with $ Q(wx) \equal{} wQ(x)$, but inconsistent with $ Q(wx) \equal{} w^2Q(x)$.

2) $ Q(wx) \equal{} Q(x)$.
We can write $ Q(x) \equal{} h(x^3)$ for some polynomial $ h$.
$ Q(x \plus{} 1) \equal{} Q(\sqrt [3]{x})^3 \plus{} 1\Rightarrow h((x \plus{} 1)^3) \equal{} h(x)^3 \plus{} 1\Rightarrow Q(x) \equal{} h(x \minus{} 1)^3 \plus{} 1$
Let $ H(x) \equal{} h(x \minus{} 1)$, we have
$ Q(x) \equal{} H(x)^3 \plus{} 1$ and
$ H(x^3 \plus{} 1) \equal{} H(x)^3 \plus{} 1$. In other words, $ H$ satisfies (*) again.

Hence, we have $ Q_1(x) \equal{} x$ and $ Q_n(x) \equal{} Q_{n \minus{} 1}(x)^3 \plus{} 1$.
It is easy to get
$ P_1(x) \equal{} x$ and $ P_n(x) \equal{} (P_{n \minus{} 1}(x) \plus{} 1)^3$.

Using the method, we should be able to generalized to $ G(x) \equal{} (x \plus{} 1)^k$ for natural $ k > 1$.[/quote]

[i]   The content in box is not true if $ Q(x)$ is a constant :lol:[/i]


factorial
floor function
logarithms
inequalities
Putnam

Well, $ n\equal{}561$ is definitely an answer...

We need $ \left \lfloor \frac {n!}{10^{\lfloor \log _{10} n!\rfloor \minus{} 3}} \right \rfloor \equal{} 2007$

This was an Olympiad problem, how do you prove it without a calculator?

[quote="aidan"]Prove that for some natural number $ n$, $ n!$ begins with the digit sequence $ 2007$.

How do we even begin here? Lol..[/quote]

This is true for any beginning sequence : Let $ S$ the beginning sequence and $ k$ its length (number of digits). Here $ S\equal{}2007$ and $ k\equal{}4$.

Let $ p>k\plus{}3$, $ N\equal{}10^{2p}\minus{}1$, $ m\equal{}$the number of decimal digits of $ N!$ and $ n\equal{}3\times 10^p$.
Let then the sequence $ a_i\equal{}(N\plus{}i)!10^{\minus{}(m\minus{}k\plus{}2ip)}$ for $ i\in\{0,...,n\}$
$ \lfloor a_0\rfloor$ has exactly $ k$ digits.

$ \frac{a_n}{a_0}\equal{}\frac{(N\plus{}n)!}{N!}10^{\minus{}2np}$ $ \equal{}\prod_{k\equal{}0}^{n\minus{}1}(1\plus{}k10^{\minus{}2p})$
Using inequality $ x\minus{}\frac{x^2}{2}<\ln(1\plus{}x)<x$, it's rather easy to show that $ 10<\frac{a_n}{a_0}<100$

Then, $ \forall i\in\{1,...,n\}$, $ a_{i}\minus{}a_{i\minus{}1}\equal{}a_{i\minus{}1}(i\minus{}1)10^{\minus{}2p}<a_n n10^{\minus{}2p}$ $ < 100a_0 3\times 10^p 10^{\minus{}2p}$ $ <300 \times 10^{k\minus{}p}<1$  and so the $ \lfloor a_i\rfloor\equal{}\lfloor a_{i\minus{}1}\rfloor$ or $ \lfloor a_i\rfloor\equal{}\lfloor a_{i\minus{}1}\rfloor\plus{}1$.

So the sequence $ \lfloor a_i\rfloor$ is a sequence of integers, each equal to the previous or to the previous$ \plus{}1$, beginning with a number of $ k$ digits and ending with a number greater than $ 10$ times the first one.
Then one of these numbers, say $ \lfloor a_q\rfloor$ must be $ S$ or $ 10S$.

Then the $ k$ first digits of $ (N\plus{}q)!$ are the required sequence $ S$.

Im not sure if anyone has heard of the "elegant solution" for this problem.

consider 1001000000!,1001000001!,...,1001009999!

i think the solution is self explanatory.

Ok.. good one.  :P  However, that's not easy to see.. So by giving an instance of a factorial beginning with 2007, the solution is considered complete?

[quote="aidan"]So by giving an instance of a factorial beginning with 2007, the solution is considered complete?[/quote]

Of course: if you are asked to show that something exists, giving an explicit example is sufficient (indeed, even ideal in some sense).

[quote="MellowMelon"]powerchess, I believe pco generalized that solution in the post before yours.[/quote]

Yes, I did not really read his solution but I think my quotes should explain how displeased I am at the existence of that solution. Its because I spent so much time at the actual test doing this one question before giving up. Its like Putnam B5 this year.


This was problem G1 on the IMO Shortlist 2005.

In a triangle ABC satisfying $AB+BC = 3AC$, the incircle has centre $I$ and touches the sides $AB$ and $BC$ at $D,E$ respectively. Let $K$ and $L$ be the symmetric points of $D$ and $E$ with respect to $I$. Prove that the quadrilateral $ACKL$ is cyclic.

I think this problem is very nice  :)
Sorry if this was posted before, but I was unable to find it with the search function.

See these posts:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76458[/url]
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=82813[/url]

Thank you nsato.  :)

 Now I see why I couldn't find it... Searching on TeX-code is nearly impossible and both posts didn't mention the fact it came from the Shortlist.


Let $ a$ , $ b$ be two positive constants and let $ x$ , $ y$ be two real variables for which $ \{x,y\}\subset (0,1)$

and $ a\cdot\frac {x}{1\minus{}x}\plus{}b\cdot\frac {y}{1\minus{}y}\equal{}1$ . Ascertain the maximum value of the product $ xy$ .

[hide="Answer."]

$ \frac {x_m}{b\plus{}1}\equal{}\frac {y_m}{a\plus{}1}\equal{}\frac {1}{(a\plus{}1)(b\plus{}1)\plus{}\sqrt {ab(a\plus{}1)(b\plus{}1)}}$ .

$ \sqrt {xy}\ \le\ \sqrt {x_my_m}\ \equal{}\ \frac {1}{\sqrt {ab}\plus{}\sqrt {(a\plus{}1)(b\plus{}1)}}$ .[/hide]


hi ;
determine according to the values of the reality a the limit of the real continuation(suite) (u_{n}) defined by: u_0= a , and u_{n+1}= u_n+E(u_n)
thank you

The whole part(party)
 E(x) £ Z  ,  and  E(x)=<x < E(x)+1


calculus
derivative
function
limit
real analysis
real analysis unsolved

$n \geq 1$ an integer and $f: \mathbb{R} \to  \mathbb{R}$ a $C^{\infty}$ function satisfying
$\lim_{-\infty} \frac{f(x)}{x^n}=\lim_{+\infty} \frac{f(x)}{x^n}=0$. Prove that for all $r \geq n+1$, $f^{(r)}$ has at least a zero in $\mathbb{R}$.

Suppose WLOG that $f^{(n+1)}>0$ on the real line. Then $f^{(n)}$ is increasing, so either it becomes positive for large enough $x$, or negative for small enough $x$. In the latter case, we can replace $f(x)$ by $-f(-x)$, and nothing changes, so we may assume that for some $b\in\mathbb R,\ a_n>0$, we have $x\ge b\Rightarrow f^{(n)}(x)>a_nn!$, which means that $x\ge b\Rightarrow (f(x)-a_nx^n)^{(n)}>0$. This says that $(f(x)-a_nx^n)^{(n-1)}$ is increasing on $[b,\infty)$, so we can find $a_{n-1}$ such that $x\ge b_{n-1}\Rightarrow (f(x)-a_nx^n)^{(n-1)}>a_{n-1}(n-1)!$, i.e. $(f(x)-a_nx^n-a_{n-1}x^{n-1})^{(n-1)}>0$ on $[b,\infty)$. We repeat this procedure until we find, in the end, $f(x)>a_nx^n+\ldots+a_1x+a_0,\ \forall x\in[b,\infty)$. If we divide this by $x^n$ and make $x\to\infty$, we get $\liminf_{x\to\infty}\frac{f(x)}{x^n}>a_n>0$, contradiction.


Call a set A of positive integers "good" if it has the following properties:
i) 1 and 2 belong to A
ii) if x belongs to A and y belongs to A then (3x+y) belongs to A.
Let B be the set of all positive integers not divisible by 3. Show that B is "good" and that any good set has to contain B.

1st part: Easy see that $1,2\in B$ . If $3|3x+y$ then $3|y$ therefore if $x,y\in B$ then $3x+y\in B$.
2nd part: Induction. Assume that $X$ is good and  $1,2,...,3k+1,3k+2\in X$ we'll prove that $3k+4,3k+5\in X$. If $3\not | k$ then we're done because $k,4,5\in X$, if $3|k$ then we're so done because $k+1,1,2\in X$.


Prove that there exist an integer $ n$ such that $ 2^n\plus{}3^n$ have exactly 23 prime factors.

Where does this problem come from, please ?


vector
superior algebra
superior algebra solved

Suppose $\mathbb {F}$ is an infinite field and $V$ is a vector space on $\mathbb {F}$.We know $V_1,...,V_n$ are sub-spaces of $V$ that:
\[V= V_1 \cup ... \cup V_n\]
Prove that there exist $i$ that $V_i=V$

suppose n minimal and consider a point $a$ in $ V_1 \cup ... \cup V_{n-1}$ but not in $V_n$ and $b$ in $V_n$. Then the line through $a$ and $b$ is not included in $V_i$ so it cuts $V_i$ in at most one point. Thus the line has at most $n$ points. Contradiction because the field is infinite.


function
calculus
derivative
integration
vector
real analysis
real analysis unsolved

Say that $ \varphi\in C^{\infty}_c(\Omega)$ for some open and bounded $ \Omega \subset \mathbb{R}^n$, and that
\[ \int_{\Omega} \varphi dx \equal{} 0
\]
Fix $ 1\leq i\leq n$.  When can we assert that $ \frac {d}{dx_i}\psi \equal{} \varphi$ for some $ \psi \in C^{\infty}_c(\Omega)$?  Certainly we can for $ n \equal{} 1$, but I am not sure about higher dimensions.

e: fixed

I think you meant $ \frac{\partial}{\partial x_i}\psi\equal{}\varphi$. The way you wrote it looks backwards.

Look at it through the Fourier transform. The condition that the integral over $ \mathbb{R}^n$ is zero gives us a single point: $ \widehat{\varphi}(0)\equal{}0.$

But if $ \frac{\partial}{\partial x_i}\psi\equal{}\varphi$, then $ \widehat{\psi}(\xi)\equal{}\frac{\widehat{\varphi}(\xi)}{2\pi i\xi_i}.$

And that's the problem: we need more than a single point of zero to keep that quotient from having singularities along the line $ \xi_i\equal{}0.$

So (assuming I understand the question you meant to ask), the answer is no in general.

Taking that same fix, there's an obvious candidate for $ \psi$: $ \psi(x)\equal{}\int_{\minus{}\infty}^0 \varphi(x\plus{}t\hat{x_i})\,dt$, where $ \hat{x_i}$ is the unit vector in the $ x_i$ direction.

This is compactly supported iff $ \int_{\minus{}\infty}^{\infty} \varphi(x\plus{}t\hat{x_i})\,dt$ is zero for all $ x$. Those are integrals along lines, not all of space.


Guys please help me with this question.  I have seen the solution given by the author, but it doesn't satisfy me :)
[b][i]
The space between two coaxial cylinders with radia R1 and R2 (R2>R1) is filled with a dielectrical(polar) substance with the absolute permitivity k.
The cylinders are placed in an external field with strength E.  What is the strength of the electrical field on the cylinders' axe? [/i][/b]

Thanx

what is a cylinder's axe? Is that uniform field? If so, is it perpendicular to the cylinder?

I meant axis, I think. I am a bit confused with this english terminology.

ok, that's what I figured, I just wanted to make sure. What about the uniform field? And the direction of the field?

oh, right. I forgot about the field.
Well, you were right,  it's uniform and perpedicular to the axis.

So you're looking for the E field inside the inner cylinder? And what book is this from?

I doubt that you have ever heard of it before.
It's a romanian book containing questions from the contests held from 1986 through 1993 in Romania.

blarg, you didn't answer the more relavent question of the two. You never specified an orientation so I am not exactly sure what you mean by E field on the axis... Do you refer to inside the inner cylinder?

In any event, this is a fairly difficult problem that I would solve with a solution to Laplace's Equation in cylindrical coordinates and then apply boundary conditions.

The field is uniform and perpedicular to the cylinders' axis. I think there is only one variant which satisfies this condition.

And we must find the strength of the electric field on the cylinders' axis. Given their radia R1,R2 and length L, relative permitivity of dielectric k, and external eletric field strength E.

Actually I have already found the distribution of charges in the dielectric. But my result is a bit different from the result given in the "official" solution, and I don't understand why.


12/1457, (it seems really bad compared to the guy above me :D )

15/526  I've only been to one tournament. I'm going to one probably in January or February though.

[quote="BOGTRO"][quote="dbkarp"]I want to find an age to USCF rating ratio for AoPS, please post the info!
I'm 13:1100[/quote]

11/1810[/quote]

Erm, what country do you live in? If it was either the US of Canada, then you should be in those lists at the back of chess magazines.

[quote="Temperal"]12/1680

I think it got messed up when I moved; the USCF still thinks I live in Michigan...[/quote]

You're not registered on the USCF. Sure there's a guy in your name in Michigan but he's only rated 960  :roll: .

[quote="7h3.D3m0n.117"]You're not registered on the USCF. Sure there's a guy in your name in Michigan but he's only rated 960  :roll: .[/quote]

Um, then that's not me, obviously.  :roll: Mayhaps you didn't check carefully enough?

There were only 3 people listed under your name when I searched. Unless you're no longer active?

Hey, I don't know much about the USCF but what's up with those ratings? I was rated FIDE for some time and back then it was 1800+ only, now it's 1600+ I think. Does USCF not go by elo points? How much does a GM like Seirawan, Gata, or Nakamura have?

FIDE and USCF both use basically the same rating system. USCF is separate because they are weird. Those GMs are about 2600-2700 in USCF and FIDE.

11/834

I've only been in two tournaments and both of them were when I was 9 years old.

My friend beats u : 13/2112
i beat him usually though  :wink:  mind games  :P

Who's your friend, what state, I might know him. (BTW I'm actually master strength 2200, I just haven't played the tournaments)

if i may make predictions, supermathboy is from MA, while mathemonsters friend is from GA? or FL?

[quote="mathemonster"]14/1930
#35 U14 nationally, Bogtro thats pretty gud. however math is more useful for college thats why im not real
ly trying for expert/master....how much do the gud ppl study daily? when i try about 1 hr for me...[/quote]

What's good?

Anyway, update:

12/1911

I'm underrated thought. I should be more like 2000.

[sarcasm] wow, how'd you guess [/sarcasm]

[quote="AIME15"]If I actually started playing a bunch, I'd probably be ~ 1000-1100.[/quote]

I would be 900-1000...

how can I go to tournaments? How would I find out where they are? I don't have a chess club at my school and they're not willing to start one so I need help :).

my rating is 1280, age 11.................i prbly can get to 1400 soon

supermathboy, what is your rating now?

Who's going to super nationals? if i registered i would be the highest rated in my section!

14/1950. I haven't played in about two years, but I'm going to supernationals.

12/1650.  Probably underrated, hopefully 1800 in a few tourns.

Woot, top 50 U14 Worldwide :P (by fide rating)

Too bad i failed at supernationals, where they were giving a scholarship :huh:


Let $ ABC$ be a (non-degenerate) triangle and $ a, b, c$ the lengths of the sides opposite $ A, B, C$ respectively. Show that there is a triangle $ A', B', C'$ with corresponding sides $ \sqrt a , \sqrt b, \sqrt c$. Show further that $ \angle B'A'C' > 1/2 \angle BAC.$

Let $ a\ge b\ge c$

To show: $ \sqrt{c}\plus{}\sqrt{b} \ge \sqrt{a}$

Which follows from squaring.


Prove that for every integer $ n>1$,

$ n((n\plus{}1)^{\frac{2}{n}}\minus{}1)<\displaystyle\sum_{i\equal{}1}^{n}\frac{2i\plus{}1}{i^2}<n(1\minus{}n^{\minus{}\frac{2}{n\minus{}1}})\plus{}4$.

The greater than part (first inequality) is simple.

Note that $ \sum_{i\equal{}1}^{n}{\frac{2i\plus{}1}{i^2}} \equal{} \sum_{i\equal{}1}^{n}{\left(\frac{i\plus{}1}{i}\right)^2\minus{}1}$.

Hence it suffices to prove that

$ n((n\plus{}1)^{\frac{2}{n}}) < \sum_{i\equal{}1}^{n}{\left(\frac{i\plus{}1}{i}\right)^2}$

which is true by AM-GM.

That's ok. But how do you prove the right side of inequality?


Let 20 numbers are distinct and no number equal 0 . we can divede them into 10 pairs such that with each pair sum and product of these numbers are rational numbers .prove that square of each number in 20 numbers is also rational number.

This is (1) not combinatorics and (2) false: let the 20 numbers be $ 1 \plus{} \sqrt 2$, $ 1 \minus{} \sqrt{2}$, and the integers from 1 to 18.


This question, I remember, is from AHSME. If you recognizes this, please leave to beginners to try it.

Solve this:

The three sides of a right triangle have integer lengths which form an arithmetic progression. Which of the following could be one of the sides?

A) 81
B) 100
C) 121
D) 151
E) 169

Also, tell why the answer is ALL if it wasn't asking for integer value.

Show your work.

Silverfalcon

Hint: Think of what arithmetic means and translate them in terms of pythagorean theorems :)

You don't necessarily have to answer the second question.

Just try to do the first one.

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a^2 + (a+x)^2 = (a+2x)^2

a^2 + a^2 + 2ax + x^2 = a^2 + 4ax + 4x^2

2a^2 + 2ax + x^2 = a^2 + 4ax + 4x^2

a^2 - 2ax - 3x^2 = 0

(a - 3x)(a + x) = 0

a = 3x

a = -x

a can't equal -x (then one side would be zero)

a = 3x

a, 4a/3, 5a/3

where a, 4a/3, and 5a/3 are all integers

81 is the only value divisible by 3, so 81 is the answer.

Check:

81^2 + (81 + 27)^2 = (81 + 54)^2

6561 + 11664 = 18225

18225 = 18225

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Not a bad question  .

Simple solution below:

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3, 4, and 5 make a pythagorean triple that's an arithmetic sequence. Multiply it by 27 and voila, you get 81, 108, and 135.

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Very good..

Your answer is very good, and I rather have myself working in the way you can always be sure rather than just triplets.. Of course, triplet is an easier way though..


I don't actually know how interesting this question will be.

For a positive integer $ n = \prod_{k=1}p_{k}^{e_{k}}$, define

$ \text{deg}(n) =\sum_{k=1}p_{k}e_{k}$
$ \text{rad}(n) = \prod_{k=1}p_{k}$

[b]Conjecture:[/b]  $ a, b, c \in \mathbb{N}, (a, b) = 1, a+b = c \implies \text{deg}(c) \le \text{deg}( \text{rad}(abc) )+1$

I believe the equality case occurs when $ a = 1, b = 8, c = 9$, the prime powers that differ by $ 1$.  Mihailescu's Theorem strongly suggests to me that no other equality cases exist, hence the possible truth of the conjecture.

It fails, for example, at a=4, b=8, c=12. A more general example occurs when a is divisible by 2 but not 3 and is not squarefree, b=2a, and c=3a. Or... something like that.

[quote="jb05"]It fails, for example, at a=4, b=8, c=12[/quote]

Oops, sorry.  $ (a, b) = 1$.


superior algebra
superior algebra solved

Do you have an elementary proof for it? :D

someone gave me that problem - i couldn't solve it :D


Same problems as already mentioned ( http://www.mathlinks.ro/Forum/viewtopic.php?t=49135 ) !!!