$ a,b,c,d$are four positive numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1$.This inequlity
$ 1\plus{}(ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd)\leq (a\plus{}b\plus{}c\plus{}d)\plus{}(abc\plus{}bcd\plus{}cda\plus{}dab)$
can be hold?

No, just take $ a\equal{}b\equal{}0$, $ c\equal{}d\equal{} \frac{1}{\sqrt{2}}$.

[quote="Jianxing113725"]$ a,b,c,d$are four [b]positive[/b] numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$.This inequlity
$ 1 \plus{} (ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd)\leq (a \plus{} b \plus{} c \plus{} d) \plus{} (abc \plus{} bcd \plus{} cda \plus{} dab)$
can be hold?[/quote]
 :wink:

But taking $ a\equal{}b\rightarrow0$ would do the trick

Yes, I know that $ a,b,c,d$ are positive numbers, but it does not matter if I say $ a\equal{}b\equal{}0$ or $ a\equal{}b \to 0$.


algebra
polynomial
inequalities
Kyiv mechmat competition
function
limit
integration

[color=orange][size=150]Open Competition for University Students[/size]
of The Faculty of Mechanics and Mathematics of Kyiv National Taras Shevchenko University[/color]

23.02.2009

[color=indigo]Problems for 1-2 years students[/color]

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417684#1417684]1.[/url][/b] Triangle $ ABC$ is inscribed into a circle. Does there exist a point $ D$ on this circle such that $ ABCD$ is a circumscribed quadrilateral?  (A. Kukush)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417685#1417685]2.[/url][/b] Let $ F_0 \equal{} 0,F_1 \equal{} 1, F_k \equal{} F_{k \minus{} 1} \plus{} F_{k \minus{} 2},\, k\ge2$ be a Fibonacci sequence. Find all positive integers $ n$ for which the polynomial $ F_n x^{n \plus{} 1} \plus{} F_{n \plus{} 1} x^n \minus{} 1$ is irreducible in the ring of polynomials with rational coefficients $ \mathbb{Q}[x]$. (R. Ushakov)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417686#1417686]3.[/url][/b] Let $ A,B,C$ be angles of acute triangle. Prove the inequalities:
a) $ \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}\ge 2;$ b) $ \dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\le\sqrt {3}.$ (A. Kukush, M. Rozhkova)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417687#1417687]4.[/url][/b] Find all positive integers $ n$ for which there exist matrices $ A,B,C\in M_{n}(\mathbb{Z})$ such that $ ABC \plus{} BCA \plus{} CAB \equal{} I.$ Here $ I$ is the identity matrix. (V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417688#1417688]5.[/url][/b] Let $ {x, y: \mathbb{R}\to\mathbb{R}}$ be a pair of functions such that $ {\bigl(x(t) \minus{} x(s)\bigr)\bigl(y(t) \minus{} y(s)\bigr)\ge0}$ for all $ {t,s\in\mathbb{R}.}$ Prove that there exist two non-decreasing functions $ {f,g: \mathbb{R}\to\mathbb{R}}$ and function $ {z: \mathbb{R}\to\mathbb{R}}$ such that $ {x(t) \equal{} f(z(t))}$ and $ {y(t) \equal{} g(z(t))}$ for all $ {t\in \mathbb{R}.}$ (J. Dhaene (Belgium), V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417691#1417691]6.[/url][/b] Let $ \{x_n,\, n\ge1\}$ be a sequence of real numbers such that there exists finite limit $ \lim\limits_{n\to\infty}\frac {1}{n}\sum\limits_{k \equal{} 1}^{n}x_k.$ Prove that for every $ p > 1$ there exists finite limit $ \lim\limits_{n\to\infty}\frac {1}{n^p}\sum\limits_{k \equal{} 1}^{n}k^{p \minus{} 1}x_k.$ (D. Mitin)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417692#1417692]7.[/url][/b] Let $ K(x) \equal{} xe^{ \minus{} x}, x\in\mathbb{R}.$ For every $ n\ge3$ determine $ \sup\limits_{x_1,\ldots,x_n\in\mathbb{R}}\,\min\limits_{1\le i < j\le n}K(|x_i \minus{} x_j|).$ (A. Bondarenko, E. Saff (USA))

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417695#1417695]8.[/url][/b] Does there exist a function $ f: \mathbb{Q}\to\mathbb{Q}$ such that $ f(x)f(y)\le|x \minus{} y|$ holds for any $ x,y\in \mathbb{Q},$ $ x\ne y,$ and for each $ x\in \mathbb{Q}$ the set $ \bigl\{y\in \mathbb{Q}\,|\, f(x)f(y) \equal{} |x \minus{} y|\bigr\}$ is infinite? (V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417697#1417697]9.[/url][/b] Find all $ n \ge 2$ for which it is possible to enumerate all permutations of set $ \{1,\ldots,n\}$ with numbers $ 1,\ldots,n!$ in such way that for any pair of permutations $ \sigma, \tau$ with adjacent numbers, as well as for pair with numbers $ 1$ and $ n!$, $ \sigma(k) \neq \tau(k)$ holds for every $ 1\le k\le n$. (O. Rudenko)

[color=indigo]Problems for 3-4 years students[/color]

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417687#1417687]1.[/url][/b] Find all positive integers $ n$ for which there exist matrices $ A,B,C\in M_{n}(\mathbb{Z})$ such that $ ABC \plus{} BCA \plus{} CAB \equal{} I.$ Here $ I$ is the identity matrix. (V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417691#1417691]2.[/url][/b] Let $ \{x_n,\, n\ge1\}$ be a sequence of real numbers such that there exists finite limit $ \lim\limits_{n\to\infty}\frac {1}{n}\sum\limits_{k \equal{} 1}^{n}x_k.$ Prove that for every $ p > 1$ there exists finite limit $ \lim\limits_{n\to\infty}\frac {1}{n^p}\sum\limits_{k \equal{} 1}^{n}k^{p \minus{} 1}x_k.$ (D. Mitin)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417688#1417688]3.[/url][/b] Let $ {x, y: \mathbb{R}\to\mathbb{R}}$ be a pair of functions such that $ {\bigl(x(t) \minus{} x(s)\bigr)\bigl(y(t) \minus{} y(s)\bigr)\ge0}$ for all $ {t,s\in\mathbb{R}.}$ Prove that there exist two non-decreasing functions $ {f,g: \mathbb{R}\to\mathbb{R}}$ and function $ {z: \mathbb{R}\to\mathbb{R}}$ such that $ {x(t) \equal{} f(z(t))}$ and $ {y(t) \equal{} g(z(t))}$ for all $ {t\in \mathbb{R}.}$ (J. Dhaene (Belgium), V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417692#1417692]4.[/url] [/b] Let $ K(x) \equal{} xe^{ \minus{} x}, x\in\mathbb{R}.$ For every $ n\ge3$ determine $ \sup\limits_{x_1,\ldots,x_n\in\mathbb{R}}\,\min\limits_{1\le i < j\le n}K(|x_i \minus{} x_j|).$ (A. Bondarenko, E. Saff (USA))

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417695#1417695]5.[/url][/b] Does there exist a function $ f: \mathbb{Q}\to\mathbb{Q}$ such that $ f(x)f(y)\le|x \minus{} y|$ holds for any $ x,y\in \mathbb{Q},$ $ x\ne y,$ and for each $ x\in \mathbb{Q}$ the set $ \bigl\{y\in \mathbb{Q}\,|\, f(x)f(y) \equal{} |x \minus{} y|\bigr\}$ is infinite? (V. Brayman)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417699#1417699]6.[/url][/b] Let $ \mu$ be a measure on Borel $ \sigma$-algebra in $ \mathbb{R}$, such that $ {\int_{\mathbb{R}}e^{ax}d\mu(x) < \infty}$ for all $ {a\in\mathbb{R},}$ and $ {\mu\bigl(( \minus{} \infty,0)\bigr) > 0,}$ $ {\mu\bigl((0, \plus{} \infty)\bigr) > 0.}$ Prove that there exists unique real $ a$ such that $ \int_{\mathbb{R}}xe^{ax}d\mu(x) \equal{} 0.$ (A. Kukush)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417703#1417703]7.[/url][/b] Let $ \{\xi_n,\, n\ge0\}$ and $ \{\nu_n,\, n\ge1\}$ be two independent sets of i.i.d. random variables (distribution may be different in each set). It is known that $ \textsf{E}\xi_0 \equal{} 0$ and $ {P\{\nu_1 \equal{} 1\} \equal{} p,}$ $ {P\{\nu_1 \equal{} 0\} \equal{} 1 \minus{} p},$ $ {p\in(0,1)}.$ Denote $ x_0 \equal{} 0$ and $ x_n \equal{} \sum\limits_{k \equal{} 1}^n \nu_k, n\ge1$. Prove that $ \dfrac{1}{n}\sum\limits_{k \equal{} 0}^n \xi_{x_k} \to 0,$ $ n\to\infty,$ with probability one.
(A. Dorogovtsev)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417704#1417704]8.[/url][/b] Let $ X_1, \ldots, X_{2n}$ be a set of i.i.d. random variables such that $ X_1\ne0$ a.s. Denote $ Y_k \equal{} \frac {\left|\sum_{i \equal{} 1}^k X_i\right|}{\sqrt {\sum_{i \equal{} 1}^k X^2_i}},\ 1\le k\le 2n.$ Prove the inequality $ \textsf{E}(Y^2_{2n})\le 1 \plus{} 4(\textsf{E}Y_n)^2.$ (S. Novak, Great Britain)

[b][url=http://www.mathlinks.ro/viewtopic.php?p=1417697#1417697]9.[/url][/b] Find all $ n \ge 2$ for which it is possible to enumerate all permutations of set $ \{1,\ldots,n\}$ with numbers $ 1,\ldots,n!$ in such way that for any pair of permutations $ \sigma, \tau$ with adjacent numbers, as well as for pair with numbers $ 1$ and $ n!$, $ \sigma(k) \neq \tau(k)$ holds for every $ 1\le k\le n$. (O. Rudenko)

[quote=Naive_boy]Problem for 3-4 years problem 1
$Trace (ABC) = Trace (BCA) =Trace (CAB) = k$ (say)
Hence we must have $3k=n$.
Thus n will be all multiples of 3.[/quote]

I think you should also show an example for each multiple of $3$ to complete the solution.

About "Problem for 3-4 years problem 1".
The difficult part is to show that, for every $n$, multiple of $3$, there are $A,B,C\in M_n(\mathbb{Z})$ s.t. $ABC+BCA+CAB=I_n$.
It suffices to show that it's true when $n=3$ (use diagonal block matrices, each block being of dimension $3$ ).
Using a computer, I found this solution
$A=\begin{pmatrix}0&1&0\\0&0&0\\-1&0&1\end{pmatrix},B=\begin{pmatrix}0&0&-1\\-1&-1&1\\0&0&0\end{pmatrix},C=\begin{pmatrix}0&-1&0\\0&1&0\\1&1&0\end{pmatrix}$.
I don't know how the competition was going; however if the students did not have access to a computer, then there must be a constructive solution.

A particular solution is obtained with $A,B,C$ all similar to the Jordan form $\begin{pmatrix}0&0&0\\1&0&0\\0&0&-1\end{pmatrix}$.
For example, if $A=\begin{pmatrix}0&0&0\\a&0&0\\0&0&-1\end{pmatrix},B=\begin{pmatrix}0&0&b\\0&-1&0\\0&0&0\end{pmatrix},C=\begin{pmatrix}-1&0&0\\0&0&0\\0&c&0\end{pmatrix}$, then $ABC+BCA+CAB=abcI_3$.

Years 3 - 4 , problem 1:
[url]https://artofproblemsolving.com/community/c7h260567[/url]


search
function
combinatorics unsolved
combinatorics

How many triples $ (x;y;z)$ do they satisfy the conditions:
$ x;y;z$ are natural numbers and $ x \plus{} 2y \plus{} 3z \equal{} n$ with n is a given positive integer?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=933734708&t=134892
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=933734708&t=147018

I found these by searching for "A001399," the number associated to this sequence by the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url].

After your hint, I tried to solved by generating function method. and I found the formula:
$ S_n\equal{}\frac{1}{3}([1\minus{}\{\frac{n}{3}\}]\plus{}\frac{1}{2}[1\minus{}\{\frac{1}{2}\}]\plus{}\frac{(n\plus{}1)(n\plus{}5)}{4}\plus{}\frac{1\plus{}(\minus{}1)^{n}}{8}$.
Is it wrong?

this problem can using method bijection  or generating function.
by method bijection,this problem is equivalent to $ a\plus{}b\plus{}c\equal{}n\plus{}3$
because my English is very bad ,so I can not write clearly.


probability
invariant
Functional Analysis
real analysis
real analysis unsolved

Let H be an infinite dimensional hilbert space and let sigma be the sigma-field generated by {Ux(U(perp)) : B a borel subset of Rn for some n, U=f(B) for some continuous open f:Rn->H}. Does there exist a countably additive probability measure mu on H which is invariant under all hilbert space automorphisms of H? (i.e. for each hilbert space automorphism phi and each A in sigma, mu(A) = mu(phi(A)).)

There are no continuous open maps $f\colon \mathbb R^{n}\to \mathcal H$. On one hand, $f(\{x\colon |x|<1\})$ is an open subset of $\mathcal H$; on another, it is contained in the compact set $f(\{x\colon |x|\le 1\})$.

Sorry, I meant what the sigma algebra is generated by all borel subsets of finite dimensional subspaces of H

If automorphisms are invertible linear maps $\mathcal H\to \mathcal H$, then they all fix $0$. The unit point mass at $0$ is invariant.

sorry i keep screwing up my problem statement,  i want "non-trivial measures", ie not the delta at 0.

For any $n\ge 1$ the group of linear automorphisms of Hilbert space $\mathcal H$ acts transitively on $n$-dimensional subspaces of $\mathcal H$. Indeed, if $\dim V_{1}=\dim V_{2}=n$, then $V_{1}^{\perp}$ and $V_{2}^{\perp}$ are isometric. To get an invertible map $F\in B(\mathcal H)$ such that $F(V_{1})=V_{2}$, define $F$ on $V_{1}$ and $V_{1}^{\perp}$ separately.

Suppose that $\mu$ is an invariant probability measure not concentrated at $0$. We may assume $\mu(\{0\})=0$, otherwise subtract a point mass and renormalize. Since $\mu$ is a nonzero measure on the $\sigma$-algebra generated by Borel subsets of finite dimensional subspaces, it follows that there exists a subspace $V$ such that $\dim V=n<\infty$ and $\mu(V)>0$. But $\mathcal H$ contains infinitely many mutually orthogonal subspaces of dimension $n$. Thus $\mu(\mathcal H)=\infty$, a contradiction.


inequalities LaTeX symmetry function geometry rectangle

This is a proof of $ a^2\plus{}b^2\plus{}c^2>\equal{}ab\plus{}ac\plus{}bc$. I can't really understand this one. It is provided in Engel's Problem Solving Strategies in the Inequalities chapter.

Let $ f(a,b,c)\equal{}a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}ac\minus{}bc$. Then we have $ f(ta,tb,tc)\equal{}t^2f(a,b,c)$. Hence, f is homogenous of degree two. For t is not equal to zero, we have $ f(a,b,c)>\equal{}0$ (two sided arrow aah sorry for no latex) $ f(ta, tb, tc)>\equal{}0$. Therefore we may make various normalizations. For example, we may set a=1, b=1+x, and c=1+y and recieve a true inequality by the trivial inequality.

Um, what? sry for little latex, but could someone plz explain this to me?

It would help if you told us which step you didn't understand.

(There is, by the way, an additional nice symmetry of the problem, which is translational:  that is, $ f(a, b, c) \equal{} f(a \plus{} t, b \plus{} t, c \plus{} t)$, which allows you to normalize even further to the case $ a \equal{} 0, b \equal{} 1$.)

true, hehe  :blush:

It's how they went from declaring that the function was homogenous with degree two, to declaring that  $ f(a,b,c)>\equal{}0$ and $ f(ta,tb,tc)>\equal{}0$.

If you understand that $ f(ta, tb, tc) \equal{} t^2 f(a, b, c)$, then if $ t \neq 0$ the statements that $ f(a, b, c) \ge 0$ and $ t^2 f(a, b, c) \equal{} f(ta, tb, tc) \ge 0$ are equivalent because $ t^2 > 0$ and inequalities are preserved when we multiply by positive reals.

There is much simpler proof (at least to me):

$ a^2\plus{}b^2\plus{}c^2 \ge ab\plus{}ac\plus{}bc$     $ /*2$

$ a^2\minus{}2ab\plus{}b^2\plus{}b^2\minus{}2bc\plus{}c^2\plus{}a^2\minus{}2ac\plus{}c^2 \ge 0$

$ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(a\minus{}c)^2 \ge 0$

There's still value in understanding how symmetry can help you simplify the proof, not least because the sum-of-squares argument doesn't immediately help you prove the generalization (although you do immediately see the translational symmetry):

[b]Follow-up:[/b]  Show that $ a, b, c \in \mathbb{C}$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} ab \plus{} bc \plus{} ca$ if and only if they form the vertices of an equilateral triangle.

Ohhhh I think I get it now! Thanks guys :D

[quote="Flame"]There is much simpler proof (at least to me):

$ a^2 \plus{} b^2 \plus{} c^2  \ge (ab \plus{} ac \plus{} bc )$*$ (2/2)$

$ a^2 \minus{} 2ab \plus{} b^2 \plus{} b^2 \minus{} 2bc \plus{} c^2 \plus{} a^2 \minus{} 2ac \plus{} c^2 \ge 0$

$ (a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (a \minus{} c)^2 \ge 0$[/quote]

Flame,

 I don't do Latex, but your top line was wrong.  You must
have grouping symbols around the trinomial "ab + ac + bc"
to show it being divided and multiplied in this horizontal
fashion.  I tried to fix it in the highlighted rectangle in
this post.

By <equation> /*2, he meant multiply (both sides of) the equation by 2.


In $ \triangle ABC$,the circle $ I$ is the incircle of $ \triangle ABC$(in addition,point $ I$ is the incenter of $ \triangle ABC$).The midpoints of $ BC,CA,AB$ are $ D,E,F$.The common point of the segment line $ AI$ and circle $ I$ is $ J$,the common point of the segment line $ BI$ and circle $ I$ is $ K$,the common point of segment line $ CI$ and circle $ I$ is $ L$.
Try to prove that $ DJ,EK,FL$ have and only have a common point.

Dear frankvista
but $ AI$ cuts the incircle of triangle $ ABC$ at two distinct points, I think you mean $ J$ is the intersection of $ AI$ with the incircle which is farther from sideline $ BC$. Namely point $ J$ lies on between $ A$ and $ I$  right?

Dear Frankvista and Mathlinkers,
you ara not alone to have a synthetic proof for this very known problem. One Mathlinker have one (it will reconized himself) and personnaly I have two, one will be edited next in Revistaoim and the second one mont later on my website.
I give you the honor to present at first your proof.
Sincerely
Jean-Louis

There were some proofs by using square method or sin,cos method.
But I don't like these proofs.I don't like prove geometry by calculating.
I prefer the pure geometry to the calculating geometry.

I think it is a probable way to prove this problem by inverse.


inequalities
algebra
polynomial
induction
LaTeX
system of equations
complex numbers

(Edited for cleaner argument.)

Note that this argument works for $x_i \in \mathbb{C}$ as an added bonus!

Let $S_k = \sum_{i=1}^n x_i^k$.  We are given $a^k = S_k$.
Let $\theta_k = \frac{1}{(n-k)!}\sum_{sym} \prod_{i=1}^k x_i$, where by convention $\theta_0 = 1$.

Consider $f(y) = \prod_{i=1}^n (y-x_i)$.  It expands to $f(y) = \sum_{j=0}^n (-1)^j \theta_j x^{n-j}$.

Since $f(y)$ is a polynomial with at most $n$ roots (namely, $\{ x_i \}_{i = 1}^n$), it follows we have

$0 = \sum_{j=0}^n (-1)^j \theta_j x_i^{n-j}$  for $i = 1,2,\dots, n$  (*)

$\Rightarrow 0 = \sum_{i=1}^n \sum_{j=0}^n (-1)^j \theta_j x_i^{n-j}$

$\Rightarrow 0 = \sum_{j=0}^n (-1)^j \theta_j S_{n-j}$

$\Rightarrow 0 = \sum_{j=0}^n (-1)^j \theta_j a^{n-j}$

$\Rightarrow f(a) = 0$

So $a$ is a root.  But $x_1 , \dots, x_n$ are the roots.  Therefore, one of the $x_i$'s equals $a$.  Wlog, $x_1 = a$.

From (*), we again have:
$0 = \sum_{i=2}^n \sum_{j=0}^n (-1)^j \theta_j x_i^{n-j}$
$\Rightarrow 0 = (-1)^n \theta_n + \sum_{j=0}^{n-1} (-1)^j \theta_j (0)$
$\Rightarrow$ one of the x_i's = 0, wlog $x_2$

The same argument shows that $x_2 = x_3 = x_4 = \dots = x_n = 0$.

Hence, the solutions are $(a,0,0,\dots,0)$ and it's permutations.

[quote="Centy"]Assume $x_1 \leq x_2 \leq \ldots \leq x_n$.

Then by Chebyshev's Inequality,

$n\sum x^k \leq \sum x \sum x^{k-1}$
$\therefore n \leq 1$

Therefore the only solution is $n = 1$, $x_1 = a$.[/quote]

Sorry, but it`s wrong. ;)

    The Chabychev`s inequatily:

$\large{{{\sum a_ib_i}\over{n}} \ge {{\sum a_i}\over{n}}{{\sum b_i}\over{n}}}$

[quote="Arne"]He IS applying Chebyshef, in fact...

However, there might be a problem. Say, for example, that $x_1 < 0 < x_2 \leq \cdots \leq x_n$ but $\left|x_1\right| > x_n$...

If $k$ is odd, we don't have $x_1^{k - 1} \leq x_2^{k - 1} \leq \cdots \leq x_n^{k - 1}$ anymore.[/quote]

  I didn´t  believe. :P

   Can you prove that $\large{n\sum\,x^k\,\leq\,\sum\,x_i.\,\,\sum\,x_i^{k-1}}$ by Chebychev?????

Molina, your latest solution doesn't work:  we could have more than one of the $x_i$ vary at once.  But, Arne, I think molina is right -- he applied Chebyshev backwards.  So, even without the problem you mentioned, it only yields $n \geq 1$, not $n \leq 1$.

I didn't even notice that because I was focusing on the other "problem" with Centy's solution :) :D

It's still wrong. Look at this: $1 \neq 2$, $4 \neq 3$ and $1 + 4 = 2 + 3$. Do you see the link?

[b]Singular, your argument is excellent! [/b]
Last night I could not sleep well because I was thinking about the same line of attack as in your proof but finally I failed:
Step I Trying to argue that any one of $x_i$ is a or zero.
Step II Then shrink the problem to a lower dimension and hence inductively solve the problem.

This problem was originally solved for the case n = 3 in my book, and the author used direct elimination to solve the problem. The method is clumsy and not applicable to other similiar  problems. With a while of thought I noticed that the problem could be generalized to a higher dimension and I think the main idea is to find a polynomial which captures all the $x_i$'s as roots.

Here is my argument:
I need to make use of a result called [b]Newton Girard Formula [/b] below:
*********************************************************************
Let $E_k$ be the elementary symmetric polynomials in $n$ variables and $S_k$ be defined by
$\displaystyle S_k(x_1, ... , x_n) = \sum_{i=1}^n{x_i^k}.$

Then the $S_k$ and $E_k$ are related as follows:
$\displaystyle S_1$ 	$\displaystyle = E_1$
$\displaystyle S_2$ 	$\displaystyle = S_1 E_1 - 2E_2$
$\displaystyle S_3$ 	$\displaystyle = S_2 E_1 - S_1 E_2 + 3E_3$
  	$\displaystyle \vdots$
$\displaystyle S_k$ 	$\displaystyle = -\left(\sum_{j=1}^{k-1} {(-1)^j S_{k-j} E_j} \right) - (-1)^k k E_k$
**********************************************************************
Clearly, we have $E_1=S_1=a$
Next, look at the second equation, plugging in $S_1,S_2$ and $E_1$, we will see that $E_2=0$.
Preceeding in this way, we will show that $E_2=E_3=...=E_n=0$.
Of course, this can be done by induction if we want.
The key point is that, by the theroy of equations, we have constructed the following polynomial which captures all the values of $x_1,x_2,...,x_n$ as roots:
$x^n-ax^{n-1}=0$
Solving this equation yields $x=a,0,.....,0$ as desried. ( in view of the case n = 3)
Hence, $(x_1,x_2,...,x_n)$ are all possible permuations for $(a,0,....,0)$.

As a final remark, as mentioned by Singular, this problem can be applied to the case for complex numbers, and probably to other fields with suitable limitations of the field.  :)

Excuse me if this solution doesn't work, can someone tell me why?

$x_1^k + x_2^k + ... + x_n^k$ defines the $k^{th}$ term of an $n^{th}$-order homogeneous linear recurrence whose characteristic equation has roots $x_i$.  The problem gives the first $n$ values of the sequence, from which all others are uniquely determined - hence, only one possible set of values of $x_i$ (that uniquely determine the coefficients of the linear recurrence), and since $a, 0, 0, ...$ and permutations works, it is the only solution.

lets construct a polynomial with roots $x_i$, using newton's sums

$x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n-1}x+a_n$ (note $a_0=1$)

$s_1=a$, $a_{1}=-a$

$s_2+(-a)(a)+2(a_{2})=0\implies a_{2}=0$

show by strong induction that the rest of the coefficients are $0$ (showing $2,3,...n-1 \rightarrow n$)

for $p<n$, we have the newton's sum statement
$\sum_{k=1}^{p} s_{p+1-k}a_{k-1}=-p*a_{p-2}$

by inductive assumption, $a_2=a_3=...=a_{p-1}=0$, and then everything cancels except

$(1)(a^p)+(-a)(a^{p-1})=-p*a_{p-2}\implies a_{p-2}=0$

so our polynomial is

$x^n-ax^{n-1}$, obviously solutions $(a,0,...)$, and permutations

[quote="gotztahbeazn"]hmm interesting... but can you prove that $(a,0,0,\ldots)$ and its permutations is the only solution altheman? or do u have to at all?[/quote]

it is done...i proved that given a nth degree polynomial with roots x_i, it is

$x^n-ax^{n-1}$, and then it is trivial to solve from there, i will edit a bit though, i messed up the latex

Chebyshev wouldn't work anyway because the order of $x^k_i$ for negative $x_i$ isn't fixed.  :(

Besides, the polynomial solution is far better.


modular arithmetic
inequalities
Congruences

Let $n_{1}, \cdots, n_{k}$ and $a$ be positive integers which satify the following conditions:[list][*] for any $i \neq j$, $(n_{i}, n_{j})=1$, [*]  for any $i$, $a^{n_{i}} \equiv 1 \pmod{n_i}$, [*] for any $i$, $n_{i}$ does not divide $a-1$. [/list] Show that there exist at least $2^{k+1}-2$ integers $x>1$ with $a^{x} \equiv 1 \pmod{x}$.

Ok, new try (after the first was spam):

Let $p_i$ be the smallest prime dividing $n_i$ ($n_i=1$ is not possible because $1|a-1$).
Claim: $p_i|a-1$, especially $p_i < n_i$ (otherwise $n_i=p_i|a-1$).
Assume the contrary:
Let $q_i$ be any prime dividing the order of $a \mod p_i$ (the order is $>1$ if $p_i \nmid a-1$).
As we have $p_i|n_i|a^{n_i}-1$ and $p_i|a^{p_i-1}-1$, we get $q_i|n_i, (p_i-1)$, contradiction since $q_i<p_i$ divides $n_i$.

For each $i$ we can choose one number from $c_i \in \{1,p_i,n_i\}$ and look at $m=\prod_{i=1}^k c_i$.

We have $a^{c_i} \equiv 1 \mod c_i$ independet what $c_i$ we took, thus also $a^{m} = \left( a^{c_i} \right)^\frac{m}{c_i} \equiv 1 \mod c_i$.
But the $c_i$ are coprime, giving $a^m \equiv 1 \mod m$ and we always get different $m$ by different choices of $c_i$.

In total we got a total number of $3^k$ such $m$ (where $m=1$ was counted). But the inequality $3^k-1 \geq 2 \cdot 2^k-2$ is trivial to check to be true.

I will post an alternative solution.

proposition：Let $c$ and $h$ be positive integers. If $a^{c^{h}}-1$ is divisible by $c^{h}$,then $a^{c^{h+1}}-1$ is divisible by $c^{h+1}$.

proof：Obviously, $a^{c^{h}} \equiv 1\pmod{c}$.
$(a^{c^{h}})^{c-1}+(a^{c^{h}})^{c-2}+\cdots+1 \equiv 1+1+\cdots+1 \equiv c \equiv 0 \pmod{c}$.
Therefore, $a^{c^{h+1}}-1 = (a^{c^{h}}-1)( (a^{c^{h}})^{c-1}+(a^{c^{h}})^{c-2}+\cdots+1) $ is divisible by $c^{h+1}$.

Now back to the problem.
This proposition implies that a positive integer $b$ such that $b>1$ if $a^{b} \equiv 1 \pmod{b}$
then $a^{b^{t}} \equiv 1 \pmod{b^{t}}$ for an arbitrary positive integer $t$.

It implies that infinity many positive integers $x$ satisfy $a^{x} \equiv 1 \pmod{x}$, of course, more than $2^{k+1} -2$.


Find m to satisfy $2^{sinx}+2^{cosx}=m$ has root x in $[0, \frac{pi}{2}]$

[quote="kunny"]Using $A.M.-G.M.$ gives answer $2\sqrt{2}\leq m\leq 2^{\frac{2+\sqrt{2}}{2}}.$[/quote]
it should be detailed,i dont think so

If you investigate the function $f(x)=2^{\sin(x)}+2^{\cos(x)}$ for $x \in [0,\frac{\pi}{2}]$, then you find that the argument of the maximum is $\pi/4$, and the minimum must be reached on the boundary of the interval. This not-so-elegant solution yields the answer $3 \leq m \leq 2^\frac{\sqrt{2}+2}{2}$.

[quote="Teki-Teki"]just in case anybody really cared, one could restrict the interval to $\left[0,\frac{\pi}{4}\right]$ without changing anything.

[hide]
here's my tip - use Jensen's to investigate the convexity/concavity of the functions.
[/hide][/quote]
$m\geq 2\sqrt{2}$ $m\leq 2^{1+\frac{1}{\sqrt{2}}}$

Well, and does $2^{\sin(x)}+2^{\cos(x)}=2\sqrt{2}$ admit a solution $x\in [0,\pi/2]$? Either I have missed something cruical, or there is no such solution since $2\sqrt{2}<3$  :o


ratio
AMC
AIME
arithmetic sequence
AIME I

When the sum of the first ten terms of an arithmetic progression is four times the sum of the first five terms, the ratio of the first term to the common difference is:

$ \textbf{(A)}\ 1: 2 \qquad\textbf{(B)}\ 2: 1 \qquad\textbf{(C)}\ 1: 4 \qquad\textbf{(D)}\ 4: 1 \qquad\textbf{(E)}\ 1: 1$

[hide="Click for solution"]
$ S\equal{}\frac{n}{2} (2a\plus{}(n\minus{}1)d))$ so $ \frac{10}{2}(2a\plus{}9d)\equal{}4 \cdot \frac{5}{2}(2a\plus{}4d)$ so $ d\equal{}2a \implies a: d\equal{}1: 2$, or $ \boxed{\textbf{(A)}}$.
[/hide]

[quote="r15s11z55y89w21"]AIME, I really feel that you take solutions way too simply. I, and probably most of other users, would not understand the bulk of your intricate formula. PLEASE PLEASE PLEASE define your variables, show the smaller steps in between, and tell WHY that is related to the problem asked.[/quote]

Sorry? The solution is VERY clear in my opinion. If you don't understand arithemetic sequences make a different topic.

If you need more of a solution,

The sequence is $ a,a\plus{}d,a\plus{}2d,...,a\plus{}9d$.

Thus, $ 10a\plus{}45d\equal{}4(5a\plus{}10d) \implies 10a\plus{}45d\equal{}20a\plus{}40d \implies 10a\equal{}5d \implies \frac{a}{d}\equal{}\frac{1}{2}\equal{}1: 2$.

Thx, BOGTRO...THAT is what I wanted...words explaining the process instead of just a bunch of numbers.


integration
limit
real analysis
real analysis unsolved

Let $ (X,\mu)$ be a measure space, $ (f_n)$ a sequence in $ L^1(\mu)$, $ f \in L^1(\mu)$. Suppose that $ f_n \to f$ a.e..
Show that
		\[ \begin{aligned}
			\liminf_{n\to\infty} \int |f_n| d\mu \minus{} \int |f| d\mu &\le \liminf_{n\to\infty} \int |f_n\minus{}f| d\mu \\
			\le \limsup_{n\to\infty} \int |f_n\minus{}f| d\mu &\le \limsup_{n\to\infty} \int |f_n| d\mu \minus{} \int |f| d\mu.
		\end{aligned}\]
------------------------------------
[b]Corollary:[/b] If $ \displaystyle\lim_{n\to\infty} \int |f_n| d\mu$ exists then $ \displaystyle\lim_{n\to\infty} \int |f_n\minus{}f| d\mu$ also exists and
		\[ \lim_{n\to\infty} \int |f_n\minus{}f| d\mu \equal{} \lim_{n\to\infty} \int |f_n| d\mu \minus{} \int |f| d\mu.\]
[b]P.S.:[/b] This corollary solves several problems on the forum.

[quote="volodja"][b]P.S.:[/b] This corollary solves several problems on the forum.[/quote]

which?

I don't remember all the topics I encountered... E.g. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146034]this one[/url] is the immediate corollary.


Could someone please provide a solution to Part D?  I got

Part A - $ 14.0$ atm

Part B - $ \frac{5}{7}$ $ \text{H}_2$

Part C - $ 5.0$ g/L

, but was stuck on D.

Thanks in advance!

[quote="Carcul"]The "until one of the reactants is entirely consumed" part should be a strong hint. How is the limiting reagent?[/quote]

The O2 is the limiting reagent, right?


calculus
integration
trigonometry
calculus computations

For integers $k\ (0\leq k\leq 5)$, positive numbers $m,\ n$ and real numbers $a,\ b$, let $f(k)=\int_{-\pi}^{\pi}(\sin kx-a\sin mx-b\sin nx)^{2}\ dx$,
$p(k)=\frac{5!}{k!(5-k)!}\left(\frac{1}{2}\right)^{5}, \ E=\sum_{k=0}^{5}p(k)f(k)$. Find the values of $m,\ n,\ a,\ b$ for which $E$ is minimized.

[quote="kunny"]For integers $ k\ (0\leq k\leq 5)$, positive numbers $ m,\ n$ and real numbers $ a,\ b$, let $ f(k)=\int_{-\pi}^{\pi}(\sin kx-a\sin mx-b\sin nx)^{2}\ dx$,
$ p(k)=\frac{5!}{k!(5-k)!}\left(\frac{1}{2}\right)^{5}, \ E=\sum_{k=0}^{5}p(k)f(k)$. Find the values of $ m,\ n,\ a,\ b$ for which $ E$ is minimized.[/quote]

any solution? :(

Let S be a logical statement.  Then let [S] = 1 if S is true, and 0 if S is false.  We have

$ f(k|a,b,m,n) \equal{} \int_{x \equal{} \minus{} \pi}^\pi (\sin kx \minus{} a \sin mx \minus{} b \sin nx)^2 \, dx$
$ \equal{} \pi(1 \plus{} a^2 \plus{} b^2 \plus{} 2ab[m \equal{} n] \minus{} 2a[m \equal{} k] \minus{} 2b[n \equal{} k])$.

It follows that

$ E \equal{} \sum_{k \equal{} 0}^5 p(k) f(k|a,b,m,n)$.

Clearly for a, b > 0, the above sum is minimized if $ m \ne n$, and $ m \equal{} k_1$ and $ n \equal{} k_2$ for $ k_1 \ne k_2$ such that $ p(k_1)$ and $ p(k_2)$ are as large as possible.  Hence we may choose without loss of generality $ m \equal{} 2$, $ n \equal{} 3$, and we obtain

$ \frac {E}{\pi} \equal{} (1 \plus{} a^2 \plus{} b^2) \left(\sum_{k \equal{} 0}^5 p(k)\right) \minus{} 2a p(2) \minus{} 2b p(3) \equal{} (1 \plus{} a^2 \plus{} b^2) \minus{} \frac {5}{8}(a \plus{} b)$.

We then collect like terms and complete the square in a and b:

$ \frac {E}{\pi} \equal{} \left(a \minus{} \frac {5}{16}\right)^2 \plus{} \left(b \minus{} \frac {5}{16}\right)^2 \plus{} \frac {103}{128}$,

from which is clear that E is minimized when $ a \equal{} b \equal{} 5/16$, and the minimum value attained is $ E \equal{} \frac {103}{128}\pi$.


[quote="Interval"]There are 12 horses in a horseshow competition.  The top three winning horses receive money.  How many possible money winning orders are there for a competition with 12 horses?[/quote]

[hide]this is a permutations problem because order matters so P(12,3)=$\frac{12!}{9!}=12\cdot 11\cdot 10=1320$[/hide]

What lead you to see that it is a permutation question?

Thanks!

Well if you don't see the permutations, just think like this:

We want to pick 3 horses out of 12 to place in the top 3.  So we do $\binom{12}{3}= 220$.  But those are in any order.  We want a specific order.  So multiply by 3! (because there are 3 spots) and we get $220 \cdot 6 = \boxed{1320}$

I like the tip given in the second reply.


trigonometry
inequalities
inequalities unsolved

Given triangle $ABC$.
What is the largest value of:
$\cos A+\cos B+\cos C$

I was thinking about doing this:
$\frac{a^{2}+b^{2}-c^{2}}{2ab}+\frac{a^{2}+c^{2}-2ab}{2ac}+\frac{b^{2}+c^{2}-a^{2}}{2bc}$
Now what?

(If you use some well-know inequality can you please post the full inequality seperate, thanks  :) )

It is $\frac{3}{2}$, and posted before :D

here http://www.mathlinks.ro/Forum/viewtopic.php?t=133464


It looks like I ended up in 16th place for regional mandelbrot. However, considering ties I ended up in like 7th place. Does that mean I will be a winner or not?

it depends on your school, i thoguht. not sure.
what region?

no they select 10 winner from each region (the contest ppl). And my region is NE


I think that he was intend to find the equation of the line passing through two inter section points of the two circles.

[quote="ckck"][quote="Treething"][quote="Flyrev"]Hello

The first circle is described by the equation:
x^2+y^2-22x+4y+1 = 0
which can also be written like this:
(x-11)^2+(y+2)^2=8^2

The second circle is described by the equation:
(x+1)^2+(y-3)^2=5^2

What point is on both circles? In Norwegian we call this [i]tangeringspunktet[/i], I have no idea what it is called in English.[/quote]

Set them equal to each other:

$(x-11)^{2}+(y+2)^{2}-64 = (x+1)^{2}+(y-3)^{2}-25$

Solve for x and y from here.[/quote]
I'm confused. Why would you set them equal to each other?[/quote]
You want to find the intersection point

[quote="Flyrev"]Hello

Yes, the intersection point is what I want to find. Will setting the equations equal and solving it do that, and if so, why?[/quote]

Yes it will.  Setting the two equations equal to each other is the same as trying to find points that satisfy both equations, i.e. points that lie on both circles.  If you end up with a line, it is the line that goes through both intersection points.  In that equation, solve for either x or y and plug it into one of the equations of the circle, then solve for the coordinates.

Hello

Thanks. But [i]why[/i] is that so? What are the underlying concepts? I don't want to just memorize it and never understand [i]why[/i] exactly [i]that[/i] gives me what I'm after.

Best regards
 Flyrev

Well, there are several ways of approaching this, but here we go.  The basic concept is that equal things are equal.  Take the following:

a+b+c=0
x+y+z=0

0=0, right?  so

a+b+c=x+y+z

Since they equal the same thing, they must themselves be equal.   This is the principal which we are using to find the intersection of the circles.  When two equations both equal 0, or 1, or anything, they equal each other.  Now, when trying to find solutions, we can do this:

x+y=4
x-y=2

Now, solve both equations SEPARATELY for y

y=4-x
y=x-2

When the lines intersect, the values of both x and y are equal.  So the y in the first equation is the y in the second equation.  Since 4-x and x-2 equal the same thing, y, they equal each other.

4-x=x-2
x=3

Does that make sense?  In the same way, you could solve both circle equations for one variable and then solve for the other variable.  Or you could make both equations equal to zero and then equate them (since 0 obviously equals 0).  You will find the values that satisfy both equations that way.

the equations with x and y taht you have are descriptions of the graphs of circles- every point on the circle has x and y which fit that description. when the x and y for one are equal to the x and y for another, the point is the same, but it lies on both circles.

does that make it a little more clear?

Yes, perfectly.

However, I get either x=-y+17 and 0y=-10 or x=(5y+38)/12 and 0y=-11

Why?

Best regards
 Flyrev

[quote="mathgeniuse^ln(x)"][quote="Flyrev"]Hello

The first circle is described by the equation:
x^2+y^2-22x+4y+1 = 0
which can also be written like this:
(x-11)^2+(y+2)^2=8^2

The second circle is described by the equation:
(x+1)^2+(y-3)^2=5^2

What point is on both circles? In Norwegian we call this [i]tangeringspunktet[/i], I have no idea what it is called in English.[/quote]

[hide]Notice that if you subtract the two equations you get

24x-120-10y-13=-39, so the line that passes through it is y=2.4x-9.4.  Now plug this into one of the equations, solve for x, then solve for y with the line and you got it.[/hide][/quote]

Hmm, how did you get that? I got [hide]y=2.4x-7.6[/hide] ...


Gauss
number theory unsolved
number theory

Find all integer  $x,y,z$ satisfying  $gcd(x,y,z) =1$ and  $x^{8}+4 y^{8}= 5z^{4}.$

[quote="Rust"] $x^{4}+2iy^{4}=i^{k}(1+2i)z^{2}$. Therefore $z=x^{2},y=x$.[/quote] Could you explain a little more?I can't understand :blush: how can you get $z=x^{2},y=x$?

[quote="Rust"]It is factorization in Gauss ring $Z[i]$.
$x^{4}+2iy^{4}=z^{2}+2iz^{2}$ mean $z=x^{2},y=x$.[/quote]

I also don't  understand  that solution.
Who have an other solution?


A 3-ounce can of tomato sauce costs $ \$$1.68. In cents, what is the price per ounce?

Yeah. I always answer 56 before the practice comes up.


modular arithmetic
calculus
integration
number theory unsolved
number theory

If $'p'$ is a prime and $p \equiv 3 \pmod8$
Then  find all integral solutions to :
$y^{2}$ $=$ $x^{3}$ $-$ $p^{2}x$

[hide="Hint"]The trivial solutions are the only solutions :D  :P [/hide]

well since this problem after being posted the third time by three different people got a solution
so [url=http://www.mathlinks.ro/viewtopic.php?t=155489]here[/url] is it :)


Let f:[a,b]->[a,b] be a function s.t. |f(x)-f(y)| < |x-y|. x1 is in [a,b].         x_(n+1) = (x_n+f(x_n))/2. Prove that lim {n->00} x_n = x_0 s.t. f(x_0)=x_0 (a fixed pt of f).

Let g(x)=(x+f(x))/2 is continuous, g([a;b]) is in [a;b]
From |f(x)-f(y)| < |x-y| g is increasing
If x_1 \leq  x_2 the sequence x_n=g(x_(n-1)) is increasing
If x_2 \leq x_1, x_n is decreasing
x_n is bounded like g, lim x_n =k exist in any case.
f is continous k=(k+f(k))/2=>f(k)=k
Moubinool Omarjee

Yeah, of course  :D This was given at some IMC (if I remember correctly).

another solution for this:
\[ |{f(x)+x \over 2}-{f(y)+y \over 2}| = {|f(x)-f(y) + x-y| \over 2} \leq {|f(x)-f(y)|+|x-y| \over 2} \leq |x-y|\]
then the result follows from theorem of contractions (since ${f(x)+x\over 2}$ is a contraction).
just, as moubinool did, let's set $g={f(x)+x \over 2}$. it's easy to prove that the sequence $x_n$ is a cauchy sequence... and being $f$ (and $g$) continuous, we have that $x_n$ tends to $x_f$, fixed point...

[quote="nishant"]the only thing worth pondering upon in this is about the function {f}.it can be proved that f{x} constant and I think that rest everything will follow.
what do u think?[/quote]

I believe nishant is right

Consider the fact that $ \frac{| F(X) - F(Y) |}{|X-Y|} \leq 0 $

Now take $ \lim_{X \rightarrow Y}\frac{| F(X) - F(Y) |}{|X-Y|} $

$ \Rightarrow | \frac{d(F(X))}{dx} | \leq 0 \Rightarrow F'(X) = 0 $

Hence F(X) is a constant function

If you had done the division right, you would have gotten

$\left|\frac {f(x)-f(y)}{x-y}\right|<1$, which doesn't help you.

OOps I read it as $ (| X - Y|)^2 $ Sorry!!!


trigonometry
inequalities proposed
inequalities

Let $x,y,z$ be real numbers with $x+y+z=0$. Prove that \[ |\cos x |+ |\cos y| +| \cos z | \geq 1 .  \] [i]Viorel Vajaitu, Bogdan Enescu[/i]

:o the problem may be wrong,let $x=y=\dfrac{\pi}{3}, z=-\dfrac{2\pi}{3}$,there is a contradiction! :)

[quote="suweijie"]:o the problem may be wrong,let $x=y=\dfrac{\pi}{3},z=-\dfrac{2\pi}{3}$,there is a contradiction! :)[/quote]You are right :) Thanks, I have corrected the statement now ;)  :blush:

let me have a try.put $a=|\cos{x}|,b=|\cos{y}|$,then $|\cos{z}|=|\cos(x+y)|=|\cos{x}\cos{y}-\sin{x}\sin{y}|\geq |ab-\sqrt{(1-a^2)(1-b^2)}|$.define $s=\arccos{a},t=\arccos{b}$,thus $|ab-\sqrt{(1-a^2)(1-b^2)}|=|\cos(s+t)|$.Since $s,t$  are in the interval $[0,\dfrac{\pi}{2}]$,so if $s+t \le \dfrac{\pi}{2}$,then $\cos{s}+\cos{t}+|\cos(s+t)|\geq\cos{s}+\cos{t}\geq\cos^2s+\cos^2t\geq 1$.
if $s+t\geq\dfrac{\pi}{2}$, then \begin{eqnarray*} \cos{s}+\cos{t}+|\cos(s+t)| &=& \cos{s}+\cos{t}-\cos(s+t) \\ \ &=& 4\sin\dfrac{s}{2}\sin\dfrac{t}{2}\cos{\dfrac{s+t}{2}}+1\\ \ &\geq& 1 . \end{eqnarray*}

We have two useful ineqs here :
  $|sin(x+y)| \leq min(|cosx|+|cosy| ,|sinx|+|siny|)$
  $|cos(x+y)| \leq min(|sinx|+|cosy| ,|siny|+|cosx|)$
 Using these ineqs , we can prove :
 $|cosx_1|+|cosx_2|+...+|cosx_{2k+1}| \geq 1$ , where $x_1+x_2+...+x_{2k+1} = 0$

i think it can be proved without any difficulty                                        take for example x=min(x,y,z)
  we have  |cos x |+ |cos y| +| cos z | >= |cos x |
  x min so 0>=3x so 0>=x so cosx>=cos(0)=1
  so it's proven

[quote="nawram88"]0>=x so cosx>=cos(0)=1
[/quote]
obviously false.


put signs but no digits(just signs): 1 1 1=6 :?: thank you

$(1+1+1)! = 6$...

(I think this should be moved to "Getting Started".)


geometry
rectangle
combinatorics open
combinatorics

There are two families of straight lines mutually perpendicular to each other, drawn on an infinite sheet of paper. They divide the sheet of paper into unit squares. Some of the unit squares are coloured red such that in any (2*3) or (3*2) rectangle, there are exactly 2 unit squares that are coloured red. Find the number of red squares in a (9*11) rectangle.

I have a solution which I shall post later.Looking forward to different solutions.

[quote="thealchemist"]Here's the solution :D , see if its right :| ,

Divide the 9*11 rectangle into 15 ,2*3 or 3*2, rectangles and one 3*3 square. In the 3*3 square by taking several cases( try it), we get that there should be exactly 3 red squares, and the other rectangles shall have 30 red squares, so the 9*11 rectangle has 33 red squares.

Try solving it any other way, please reply.[/quote]

I think you need to prove the 3x3 square case. You can do it like this:

A 3x3 square contains 4 3x2 or 2x3 rectangles, so there should be 8 red squares. But there is some double-counting going on. In particular, the grid below shoes how many times we overcount a red square for each position:

2 3 2
3 4 3
2 3 2

Now 8 = 2 + 2 + 2 + 2 = 3 + 3 + 2 = 4 + 2 + 2.

2 + 2 + 2 + 2 means placing the red squares on the corners, which does not work, because it in particular makes it impossible to tile the 5x5 grid surrounding the 3x3 square (x means red, o means empty).

? ? ? ? ?
? x o x ?
? o o o ?
? x o x ?
? ? ? ? ?

Let the upper-left corner be (0,0). Then draw the rectangle with corners at (3,1) and (4,3). 2 squares are already red so the rest must be empty:

? ? ? ? ?
? x o x o
? o o o o
? x o x o
? ? ? ? ?

But now the rectangle with corners at (2,1) and (4,2) has only one red square, so we have a contradiction.

3 + 3 + 2 means placing a red square in the lower left corner, the middle of the upper edge, and the middle of the right edge, which also works.

4 + 2 + 2 means placing a red square in the middle and on two opposite corners. This also works.

Now we have proven that the 3x3 square must have 3 red squares inside it.


quadratics
geometry
incenter
inradius
ratio
Pythagorean Theorem
similar triangles

Chords KN and ST are perpendicular to diameter CP of circle (O) at points Q and R. SQ intersects the circle in V. (K, S are on one side of CP, N and T on the other.) Let r be the radius of the circle (M) inscribed into the curvilinear triangle TQV. Prove that

1/r = 1/PQ + 1/QR.

In addition, it would be nice to have a short independent proof of the fact that r is independent of the radius of (O).

The points P, Q, R, C have to follow on the line PC in this order. Let r > q be radius of the circle (O),

$ SR^{2} \equal{} PR \cdot RC \equal{} PR (2r \minus{} PR),\ \ r \equal{} \frac {SR^{2} \plus{} PR^{2}}{2PR}$

The triangle $ \triangle STQ$ is isosceles and $ QN \parallel ST,$ hence QN bisects the angle $ \angle TQV.$ Let (M) be a circle centered on the ray (QN with radius $ q \equal{} \frac {PQ \cdot QR}{PR}$ and tangent to both QT and QV at X, Y. The right triangles $ \triangle QXM \sim \triangle SRQ$ are similar, $ QX \equal{} \frac{MX \cdot SR}{QR} \equal{} \frac{PQ \cdot SR}{PR}.$ Using Pythagorean theorem for the right triangle $ \triangle OQM,$

$ OM^{2} \equal{} QM^{2} \plus{} QO^{2} \equal{} MX^2 \plus{} QX^2 \plus{} (PQ \minus{} r)^{2} \equal{}$

$ \equal{} q^{2} \plus{} \frac {PQ^{2}\cdot SR^{2}}{PR^{2}} \plus{} PQ^{2} \minus{} 2r \cdot PQ \plus{} r^{2} \equal{}$

$ \equal{} q^{2} \plus{} PQ^{2}\cdot \frac {SR^{2} \plus{} PR^{2}}{PR^{2}} \minus{} 2r \cdot PQ \plus{} r^{2} \equal{}$

$ \equal{} q^{2} \plus{} PQ^{2}\cdot \frac {2r}{PR} \minus{} 2r \cdot PQ \plus{} r^{2} \equal{}$

$ \equal{} q^{2} \minus{} 2r \cdot \frac {PQ \cdot (PR \minus{} PQ)}{PR} \plus{} r^{2} \equal{}$

$ \equal{} q^{2} \minus{} 2rq \plus{} r^{2} \equal{} (r \minus{} q)^{2}$

As a result, OM = r - q and the circle (M) is internally tangent to the circle (O). Given collinear points P, Q, R, the point S on a normal to PR at R is arbitrary, i.e., radius r of (O) varies, while $ q \equal{} \frac {PQ \cdot QR}{PR}$ does not.

Impressive. I saw all the right and similar triangles there are, but tried to get from OM = r - q through a chain of applications of the Pythagorean theorem to "q = something" and could not.

Thank you.

Starting with $ OM \equal{} r \minus{} q,$ where $ q \equal{} MX,$

$ (r \minus{} q)^2 \equal{} OM^2 \equal{} QM^2 \plus{} QO^2 \equal{} MX^2 \plus{} QX^2 \plus{} (PQ \minus{} r)^2\ \Longleftrightarrow$

$ QX^2 \plus{} PQ^2 \minus{} 2r PQ \plus{} 2rq \equal{} 0.$

Substituting $ QX \equal{} \frac {MX \cdot SR}{QR},$

$ q^2 \frac {SR^2}{QR^2} \plus{} 2rq \minus{} PQ (2r \minus{} PQ) \equal{} 0.$

Substituting $ SR^2 \equal{} PR \cdot RC,$ $ 2r \minus{} PQ \equal{} QC,$

$ q^2 \plus{} 2rq\ \frac{QR^2}{PR \cdot RC} \minus{} \frac{PQ \cdot QR}{PR} \cdot \frac{QC \cdot QR}{RC} \equal{} 0$

Substituting $ 2r QR \equal{} (PR \plus{} RC)(PR \minus{} PQ) \equal{} PR \cdot QC \minus{} RC \cdot PQ,$

$ q^2 \minus{} q\ \left(\frac{PQ \cdot QR}{PR} \minus{} \frac{QC \cdot QR}{RC}\right) \minus{} \frac{PQ \cdot QR}{PR} \cdot \frac{QC \cdot QR}{RC} \equal{} 0$

The two roots of the quadratic are $ q \equal{} \frac {PQ \cdot QR}{PR} > 0$ and $ q' \equal{} \minus{} \frac {QC \cdot QR}{RC} < 0.$ If $ (M), (M')$ are directed incircle and directed excircle of the curvilinear $ \triangle TQV$ in the angle $ \angle TQV,$ internally and externally tangent to the directed $ (O),$ then $ (M)$ has the same direction as $ (O)$ and $ (M')$ has the opposite direction. $ q > 0, q' < 0$ are their radii.

Rescue ? If your life depended on the answer and you tried this:

[quote="armpist"]Using the two chords as bases of two triangles and completing it with the Thebault

   circles, the circle in question is the common T-circle of both triangles.

   The other T-circles have their centers on the diameter in known points.

   The triangles incenters are on a line parallel to this diameter.

   I conjecture that after applying T- formulas  that inradius is a linear combination of

   T-radii  and incenter divides the distance between T-centers in the ratio

   of  ( tan X )^2,  see

http://forumgeom.fau.edu/FG2002volume2/FG200223.pdf,

    and after  a couple of  similar triangles and a lot of complicated multiplication

    and division, the T-radii will be replaced by diameter segments of the big circle

    and the result as required will be obtained.

   I just do not have time now for it, and those  SINs and COSs  scare me a bit, sorry.

   T.Y.
   M.T.[/quote]

you would be pushing daisies.


Mr. Smith's sixth grade class has 24 students. The average weight of the girls is 96 pounds, the average weight of the boys is 108 pounds, and the average class weight is 103 pounds. How many girls and how many boys are in the class?

In a class section, the plaintiffs were collectively awarded $ \$ 36,000,000$. The money was to be equally divided among those eligible. Before the money was distributed, it was discovered that 6 additional people were eligible for compensation. Again, dividing the money equally, this reduced the amount that would be paid to each person by $ \$10,000$.
In the end, how many people receieved money and how much did each receive?

Each half hour a bus makes a 30 minute trip from A to B and starting at the same time another bus makes the 30 minute trip from B to A.
A pedestrian leaves A, following the bus route from A to B, at the same time the bus does. After walking for 25 minutes, the pedestrian meets a bus going from B to A.
How long after this will the pedestrian be overtaken by a bus going from A to B?
Assume that hte average rate of each bus over the whole trip is the same as the average rate on that portion of the trip from the time the bus leaves the terminal until it meets the pedestrian. Likewise, make a similar assumption about the average rate of the pedestrian.

A triangle is formed using 3 of hte vertices of a regular decagon. How many such triangles can be drawn having no edge in common with the decagon?

[b]Moderator Note: Fixed LaTeX error.[/b]

1)[hide] lets say number of girls is x and boys is y. Then (96x+108y)/(x+y) = 103 throught this we can find x= (5/7)y and y+(5/7)y = 24 so y = 12 and x is 10. [/hide]

2)[hide] since the pedestrian met the bus 25 minutes after the bus left, the bus travels 5 time faster than the pedestrian. So it takes the bus from B-A a minute 5 minutes to reach the destination and the pedestrian has moved 1minute worth in the bus's speed. When the bus from B-A reaches its destination another bus from A-B starts towards the pedestrian and since the bus travels 5 times faster than the pedestrian it's 6+x = 5x where x is the time it amount of time spent until the bus catches up to the person. We find for x which is 1.5 and add it to 6 and get 7.5 minutes for the total time [/hide]

3)[hide] To make a triangle we need three points and there are 10 ways to choose the first point from the decagon and 7 points for the second since we can't choose any points adjacent to it. For the third points it's 6 since we chose a point in the second. so the answer is 6*7*10= 420 [/hide]

um jhangil, I'm not sure if I follow you but I did my own algebra and found there to be ten girls and fourteen boys.

[hide="for the plaintiff problem"]Let $ x$ be the number of plaintiffs originally.  Then, each person was awarded $ \$\frac{36,000,000}x$.  The second time, they are awarded $ \$\frac{36,000,000}{x\plus{}6}$, which is $ \$10,000$ less than the original amount they would receive.  Writing this as an equation, we have:

$ \frac{36,000,000}x\equal{}\frac{36,000,000}{x\plus{}6}\plus{}10,000$
$ x(x\plus{}6)\minus{}3600(6)\equal{}0$
$ x^2\plus{}6x\minus{}21,600\equal{}0$
$ (x\minus{}144)(x\plus{}150)\equal{}0$

Clearly, $ x$ is positive, so $ x\equal{}144$.[/hide]

[hide="About problem 1"]
jhangil's is incorrect, since 12 + 10 is not 24. I think the reason jhangil got y = 12 is because after the line

$ y \plus{} \frac {5y}{7} \equal{} 24$

he said y = 12. However,

$ y \plus{} \frac {5y}{7} \equal{} \frac {12y}{7} \equal{} 24$
$ y \equal{} \frac {7*24}{12} \equal{} 7*2 \equal{} 14$

Not 12. Not sure how jhangil got from there to x = 10, though.

[hide="Solution"]
Call the number of boys and girls x and y respectively. We know the total number of students is 24, so we can easily form the equation

$ x \plus{} y \equal{} 24$

Using the knowledge from the averages, we can also for the equation

$ \frac {108x \plus{} 96y}{24} \equal{} 103$

Now, just solve the system. Begin by multiplying the second equation by 24.

$ 108x \plus{} 96y \equal{} 103*24$

Multiply the first equation by 108.

$ 108x \plus{} 108y \equal{} 108*24$

Subtract the two new equations to get

$ 12y \equal{} 5*24$

Divide both sides by 12 to get y = 5*2 = 10.

Plug back in to x + y = 24 to get

$ x \plus{} 10 \equal{} 24$

Subtract 10 from both sides to get x = 14

$ \boxed{14\text{ boys, }10\text{ girls}}$[/hide]
[/hide]

Sorry for the wrong answer(s) I think i put x=10 because I did it in my head first and try to write things down so it might be a little (or a lot) messed up. I think my algebra for number on is correct though y + 5y/7 = 24 then y=14 (I don't know why I wrote 12).

[hide]
y= #of boys
x= #of girls

Equation 1: $ \frac{96x\plus{}108y}{x\plus{}y}\equal{}103$
Equation 2: $ x\plus{}y\equal{}24$.

Equation 1 is equal to: $ \minus{}7x\plus{}5y\equal{}0$.  <-- Just for simplicity.

Solve the system and you get that x=10 and y=14. [/hide]


number theory
prime numbers
number theory unsolved

Prove  that for $ a,b$ are positive integers numbers and $ (a,b) \equal{} 1$ there exist infinite pair $ (m,n)$ are positive integers numbers such that $ n|a^{m \minus{} 1} \minus{} b^{m \minus{} 1}$ and $ m|a^{n \minus{} 1} \minus{} b^{n \minus{} 1}$.

If $ m,n$ is not different from each other then problem is trivial when we choose both of them are prime numbers .We only need to consider the nontrivial case , $ m\#n$ .
If $ a,b$ has different parity then only need to chose  $ (m,n) = (F_{t},F_{t + 1})$ where $ F(t) = a^{2^t} + b^{2^t}$ . This base on the fact that $ F_{t}\equiv 1 (\mod 2^{t + 2})$ .
If $ a,b$ has same parity we only need to consider the number $ (m,n) = (2^{2^a} + 1,2^{2^{a + 1}} + 1)$ . It it not hard to check that $ m\equiv 1 (\mod 2^{a + 2})$ and $ n\equiv 1 (\mod 2^{a + 2})$ , thus we have $ n|a^{m - 1} - b^{m - 1}$ and $ m|a^{n - 1} - b^{n - 1}$
Proof completes .


integration
calculus
real analysis
real analysis unsolved

$ \int \int \int xy^2z^3 \,dx \,dy \,dz$
 , where D is the solid in R3 bounded by the surface z = xy
and the planes x = y, x = 1, and z = 0.

I couldn't figure out the right limit.

but I tried, the inner one is from z=0 to z=xy

the middle one is from y=1 to y

the outter one is from 0 to 1

am i right?
if not , can anyone explain to me?

We can see at a glance that the middle limit can't possibly be right; $ \int_1^y \,dy$ doesn't make sense.

Draw a picture- in particular, look at it in terms of $ x$ and $ y$. The two lines $ x\equal{}y$ and $ x\equal{}1$ cut the $ xy$-plane into four unbounded regions. Which one do we want? None of them; we want a bounded region, and that means we're going to have to use the boundary conditions involving $ z$ to cut this plane again. For the surfaces $ z\equal{}xy$ and $ z\equal{}0$ to create a boundary in $ x$ and $ y$, they must intersect in some curve where $ z\equal{}xy\equal{}0$. That's the union of the $ x$ and $ y$ axes.

OK, we now have four potential boundary lines in $ x$ and $ y$: $ x\equal{}1,x\equal{}0,y\equal{}x,y\equal{}0$. What bounded region do these define, and is $ xy$ greater than zero or less than zero in this region? (If $ xy<0$, we flip the innermost integral to $ \int_{xy}^0 \,dz$.)

[quote="jmerry"]We can see at a glance that the middle limit can't possibly be right; $ \int_1^y \,dy$ doesn't make sense.

[/quote]

so y bound should be from x to 1 ?

Please don't quote the entire post immediately preceding yours- it's always redundant

What's the whole integral now? I can't interpret one isolated piece.

This is what I got now.

I think it is right.

$ \int_{x\equal{}0}^{x\equal{}1} \int_{y\equal{}0}^{y\equal{}x} \int_{z\equal{}0}^{z\equal{}xy} xy^2z^3 \,dz \,dy \,dx$


A natural number $n$ is called 3-partite if the set of its divisors can be partitioned into three subsets with the same sum of elements.

Prove that there exist infinitely many 3-partite numbers.

Claim:If natural number $n$ is 3-partite and p>n is prime then p.n is also 3-partite.
Proof: the divisors of p.n can be partitioned in two sets{divisors of n} and {p times divisors of n} since n is 3-partite each of this sets can be partitioned into three subsets with equal sums. So the divisors of p.n can be partioned into 6 subsets with equal sums. So the claim is proven.
It remains to find one  3-partite number.

120 is three partite; the sets; {120},{60,40,20} and all the other divisors
(remember 120 is three perfect). Any three perfect number-n-divisible by three is three partite, the sets being {n}, {n/2,n/3,n/6},{all the other divisors of n}. ( Of course this assumes all 3 perfect numbers are even...)


AMC
USA(J)MO
USAMO
search
combinatorics unsolved
combinatorics

There are 51 senators in a senate. The senate needs to be divided into n committees so that each senator is on one committee. Each senator hates exactly three other senators. (If senator A hates senator B, then senator B does not necessarily hate senator A.)
Find the smallest n such that it is always possible to arrange the committees so that no senator hates another senator on his or her committee.

It looks like the answer is $ n \equal{} 7$.   To show this, it would be enough to show

1) You can always put divide the Senators into $ 7$ committees (hint: try assigning the Senators to committees one by one, essentially inducting on the size of the Senate.  ).

2) You can devise a situation where $ 6$ committees do not suffice (hint: try creating a group of $ 7$ Senators such that no two Senators in the group can be on the same committee).

Alternatively, you could just check where this problem was [url=http://www.mathlinks.ro/viewtopic.php?search_id=1930038977&t=6165]posted[/url] [url=http://www.mathlinks.ro/viewtopic.php?p=371491#371491]before[/url] (I found both of these links just by doing a search for the opening sentence of the problem in this forum)

Many, thanks guys. By the way really cool and hard problem.
 :coolspeak:


The sides of a rectangle are $ 20$ cm by $ 30$ cm. The longer side is increased by $ 10$ percent and the shorter side is decreased by $ 10$ percent. Find the area of the new figure.

We know that the new side lengths are $ 30 * 1.1 \equal{} 33$ and $ 20 * .9 \equal{} 18$.

Multiplying these lengths, we have the area, $ \boxed{594}$


[color=darkblue]Given are a right triangle $ABC\ (AB\perp AC)$ and a point $D\in (BC)$. Denote :
$\bullet$ the second intersection $E$ between the line $AD$ and the circumcircle $w(O)$ of the triangle $ABC$ ;
$\bullet$ the second intersection $F$ between the line $BC$ and the circumcircle of the triangle $AOE$.
Prove that $\widehat{FAB}\equiv\widehat{DAB}\ .$[/color]

I think you have a mistake :)
I see that AB and AC are the bisectors of EAF (which is pretty obvious I think!)

Sorry, you are right ! It is evidently. I corrected above. Thanks.

I guess we should consider the angles mod 180.
Then obviously $\angle{COE}=2\cdot\angle{CAE}$ and $\angle{FOE}=\angle{FAE}$. Therefore, $AC$ is the bisector of $\angle{EAF}$ which yields the conclusion.


Today at scholl at a very boring lesson i construct this one. I hope it is correct.

Let  $x,y,z$  real numbers such that  $x^2+y^2+z^2=1$.

Prove that  $\frac{x^2+2y(y-x)}{x^4+4y^4}+\frac{y^2+2z(z-y)}{y^4+4z^4}+ \frac{z^2+2x(x-z)}{z^4+4x^4}\geq\frac{9}{5}$


trigonometry
vector
inequalities
geometry
circumcircle
function
geometry unsolved

[quote="stancioiu sorin"]How can we prove that $\sin x+\cos x\leq\sqrt 2$, where $\ x\in\left(0,\frac{\pi}{2}\right)$?[/quote]

You could always derive  :lol:. Or use $\frac{a+b}{2}\leq \sqrt{\frac{a^{2}+b^{2}}{2}}$

$\sin x+\cos x \leq 2\sqrt{\frac{\sin^{2}x+\cos^{2}x}{2}}= \sqrt{2}$

Or even geometrically, you want to find the maximum of $a+b$ in triangle rectangle.

Daniel

Once, my teacher proved this with vectors and I had a prove with Cauchy-Swarz Inequality. (but it is the same with dhernandez's one)

By Cauchy Swarz Inequality we have $(a^{2}+b^{2})(x^{2}+y^{2})\geq(ax+by)^{2}$ so we have :
$(sin^{2}x+cos^{2}x)(1^{2}+1^{2})\geq(sinx+cosx)^{2}$ thus wehave
$2\geq(sinx+cosx)^{2}\to \sqrt{2}\geq sinx+cosx$

davron

oops ...

Choose an angle $<BAC=x$ of any triangle ABC then draw its cicumcircle, we can find triangle PBC such that (point P is on the circumcircle) <CBP is right. Now it is enough to show that $sin(CPB)+cos(CPB)\leq\sqrt{2}$.
Which is equivalent to : $\frac{BC}{CP}+\frac{BP}{PC}\leq\sqrt{2}$
knowing that $CP^{2}=BC^{2}+PB^{2}$ it can be easily proven.

davron

OK, but I have another problem then.How can we prove that
$\sin x+\cos x+\tan x+\cot x\geq\ 2+\sqrt 2$?

[quote="stancioiu sorin"]OK, but I have another problem then.How can we prove that
[hide]$\sin x+\cos x+\tan x+\cot x\geq\ 2+\sqrt 2$?[/quote]
let $f(x)=\sin x+\cos x+\tan x+\cot x$ after simplifying we get $f(x)=\frac{\sin x \cos x (\sin x+\cos x)+1}{\sin x \cos x}$. Now substitute $\sin x+\cos x=u$, and note that when ${0<x<\frac{\pi}2}$ then$1<u \le \sqrt2$. Also $\sin x \cos x=\frac{u^{2}-1}2$. So we obtain $g(u)=\frac{u^{3}-u+2}{u^{2}-1}=u+\frac{2}{u^{2}-1}$. Now $g'(u)=\frac{-4u}{(u^{2}-1)^{2}}$ so g is a decreasing function or $g(u) \ge g(\sqrt 2)=\sqrt 2+2$[/hide]

$\boxed{0<x<\frac{\pi}{2}}\ ,\ t\equiv \sin x+\cos x\Longrightarrow 1<t<\sqrt 2\ .$
$\boxed{E\equiv \sin x+\cos x+\tan x+\cot x-(2+\sqrt 2)}=\frac{(t-\sqrt 2)f(t)}{t^{2}-1}$, where $f(t)=t^{2}-2t-(1+2\sqrt 2)\ .$
$\boxed{\ E\ge 0\ }$ because $t-\sqrt 2<0$, $t^{2}-1>0$ and $f(1)<0$, $f(\sqrt 2)<0\Longrightarrow \left(\forall\right)x\in \left(0,\frac{\pi}{2}\right)\ ,\ ft)<0\ .$


Let $A = \begin{bmatrix}2&2&1\\2&5&2\\1&2&2\\\end{bmatrix}$, $B = \begin{bmatrix}-x&-y&z\\0&y&2z\\x&-y&z\\\end{bmatrix}$ where $x,y,z$ are positive reals.
and Let $B^tAB = \begin{bmatrix}8&0&0\\0&27&0\\0&0&42\\\end{bmatrix}$.
Find $x,y,z$
(Note: can it be done without actually multiplying out $B^tAB$?)

Pre and post multiplication of invesre matrices, is all i can think of...

Maybe something with the multiplicative property of determinents?

probably Gauss-Jordan? or determinants, that's all I could think of right now


Let $S$ be a subset of $\{1,2,3,\ldots,1989\}$ such that no two members of $S$ differ by $4$ or $7$. What is the largest number of elements $S$ can have?

[hide="Possibly"]Making groups of 4 and then skipping the next 7 elements gives 19 groups of 4 (since [1989/11]=18 and add one for the first group of 4 before an 11) so $4*19=\boxed{76}$ elements?[/hide]
By the way, 4everwise, you should've made it 1987, (your current postcount).

[hide=" Solution"]We first show that we can choose at most 5 numbers from $\{1, 2, \ldots , 11\}$ such that no two numbers have a difference of $4$ or $7$. We take the smallest number to be $1$, which rules out $5,8$. Now we can take at most one from each of the pairs: $[2,9]$, $[3,7]$, $[4,11]$, $[6,10]$. Now, $1989 = 180\cdot 11+9$, but because this isn't an exact multiple of $5$, we need to consider the last $9$ numbers.

Now let's examine $\{1, 2, \ldots , 20\}$. If we pick $1, 3, 4, 6, 9$ from the first $11$ numbers, then we're allowed to pick $11+1$, $11+3$, $11+4$, $11+6$, $11+9$. This means we get 10 members from the 20 numbers. Our answer is thus $179\cdot 5+10=\boxed{905}$.[/hide]

Well, I had to post the problem in its original context. ;)

My solution was off by a factor of 10. (1989/11!=18) Also, it didn't include the second part of 4everwise's.
PS I can divide... I hope.

What 4everwise posted was the almost word for word kalva solution... can anyone think of another way to do this (most people won't randomly decide to look at a list of 11 numbers...)
also, can you generalize?

http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_13

Ok so what i did was in groups of 11, i chose
1,2,3,4
12,13,14,15
23,24,25,26
etc.
I found that there are 181 such groups, and each of them have 4 numbers so the answer is 181 * 4 = 724. However, the answer is 181 * 5 = 905. What am I doing wrong?

Read 4everwise's post.

It is actually possible to pick 5 numbers.

What is the motivation for looking at groups of 11?

I think its because no matter what, the number 12 can be chosen, as 1+4+7=12.

Sorry to revive an old post.

Does anybody have an idea as to why taking groups of eleven consecutive numbers works? Not that it works, or how it works, but why, intuitively, it works.

Lemme look at the beginning.  Blue is for numbers that could be +4d, purple is for both +4d and +7d, and red is for exclusively +7d.

1, 2, 3, 4, [color=#00f]5, 6, 7,[/color] [color=#9a00ff]8,[/color] [color=#f00]9, 10, 11,[/color] 12, 13, 14, 15, [color=#00f]16, 17, 18[/color][color=#9a00ff], 19,[/color] [color=#f00]20, 21, 22, [/color] 23, 24, 25, 26, [color=#00f]27, 28, 29,[/color] [color=#9a00ff]30,[/color] [color=#f00]31, 32, 33, [/color]34, 35, 36...

I think as to @above, "why" it works is because every number in the range (first number is n) n + 4 to (n + 3) + 7 -> n + 10 can be reached, and therefore, can't be in the set, so that's why taking 4 out of every 11 consecutive integers works.  Questions regarding this strategy as opposed to taking multiples of 3 (or every number of the form 3n + 1)...

Oh, oopses.  There's a different "more-optimal" strategy than this, so I'll figure it out here real quick.

[b]A (hopefully) more understandable verison of 4everwise's solution[/b]

We start by looking at groups of 11.  We look at groups of 11 because the maximum "reach" of a single number is 10, since you can inculde up to n + 3 but then that rules out n + 10.  So, the golden question...How many numbers in a group of 11 numbers can we take?

The answer is 5 (which could be shown through trial-and-error, that's how I would do it).  We can pick 1, 3, 4, 6, 9.  Then we divide 1989 by 11 to get 180 * 11 + 9, so there are 180 sets of these.  Then there are 9 left, but luckily all 9 of ours already fit into the extra 9.  Therefore, we have 181 * 5 = 905.

But still one question..... how is the maximum reach of a single number 10?


articles
geometry
perimeter
circumcircle
trigonometry
area of a triangle
Heron\u0027s formula

This is the problem in the section 8 "Following the Lemmas" of the "How to Write a Solution" article([url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteLemma.php]the article[/url]). Two solutions were provided in the article. I thought both of them are too complex.

[quote]Problem: From vertex A of triangle ABC, perpendiculars AM and AN are drawn to the bisectors of the exterior angles of the triangle at B and C. Prove that MN is equal to half the perimeter of ABC.[/quote]

Extend AM to meet BC at K and extend AN to meet BC at L (ignore AJ and KJ in the picture).
Triangles AMB and KMB are congruent since <ABM=<KBM and <AMB=<KMB=90 and MB is a common side.
So, M is the midpoint of AK and AB=KB.
Similarly, N is the midpoint of AL and AC=LC.
Therefore, MN is equal to half of KL which is equal to the perimeter of the triangle ABC.

[img]http://www.artofproblemsolving.com/Resources/Images/fangeom3.gif[/img]

Nice!  However, because this isn't related to the AMC at all, I'm moving it from AMC to Intermediate Topics.

[quote="fuzzylogic"]
 Two solutions were provided in the article. I thought both of them are too complex.
[/quote]

I'm hurt, fuzzy. :)

It wasnt that complex when I did it. I just used well known ideas.

It's well known the distance from vertex to tangent pt on the excircle is the semiperimeter. So when I saw semiperimeter, I figured some proof existed showing those two segments had equal length. Sure enough, it wasn't too hard to prove.

It's not THAT complex. But it's not easy, either. You kinda have to know the excircle property beforehand to arrive at my solution.

What is the excircle? Is it the same as the circumcircle?

No. Extend two sides. The excircle is tangent to these extensions and the third side.

Ah... so that's what it is.  Ive seen those before but didn't see the name associated.

[quote="Nukular"]It's well known the distance from vertex to tangent pt on the excircle is the semiperimeter. So when I saw semiperimeter, I figured some proof existed showing those two segments had equal length.[/quote]Great insight!

Here is a different solution to another problem in the section "Bookends" of the same [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteBookends.php]article[/url].

[quote]Problem: Let I be the incenter of triangle ABC. Prove that

(IA)(IB)(IC) = 4Rr2,

where R is the circumradius of ABC and r is the inradius of ABC.

(Problem due to Sam Vandervelde of the Mandelbrot Competition. Note that the incenter of a triangle is the center of the circle inscribed in the triangle. The inradius is the radius of this circle. The circumradius is the radius of the circle that passes through the vertices of ABC.)
[/quote]

(Revision, same idea):
By the area formula and (extended) law of Sines, we have
[tex]rs=[ABC]={1\over2}ab\sin{C}=2R^2\sin{A}\sin{B}\sin{C}[/tex]. So
[tex]\sin{A}\sin{B}\sin{C}={[ABC]\over2R^2}={rs\over2R^2}[/tex], where s is the semi-perimeter of the triangle ABC.

[tex](IA\cos{A\over2}) \cdot (IB\cos{B\over2}) \cdot (IC\cos{C\over2}) = (s-a)(s-b)(s-c)[/tex]

[tex](IA\sin{A\over2}) \cdot (IB\sin{B\over2}) \cdot (IC\sin{C\over2}) = r^3[/tex]

Multiplying the two equations above together, use [tex]\sin{x}\cos{x}={1\over2}\sin{2x}[/tex] and Heron's formula [tex](s-a)(s-b)(s-c)=[ABC]^2/s = (rs)^2/s = r^2s[/tex] we get

[tex](IA\cdot IB\cdot IC)^2 \cdot {\sin{A}\sin{B}\sin{C} \over 8} = r^3 (s-a)(s-b)(s-c) = r^5s[/tex]
Therefore [tex] (IA\cdot IB\cdot IC)^2 = (r^5s) (8) (2R^2/rs) = 16R^2r^4[/tex]
QED.

(Old version):
By the area formula and (extended) law of Sines, we have
[tex]rs=[ABC]={1\over2}ab\sin{C}=2R^2\sin{A}\sin{B}\sin{C}[/tex]. So
[tex]\sin{A}\sin{B}\sin{C}={[ABC]\over2R^2}={rs\over2R^2}[/tex], where s is the semi-perimeter of the triangle ABC.
On the other hand, we have
[tex]\tan{A\over2}\tan{B\over2}\tan{C\over2}={r\over{s-a}}{r\over{s-b}}{r\over{s-c}}= {r^3s\over {s(s-a)(s-b)(s-c)}}={r^2(rs)\over {[ABC]^2}}={r\over s}[/tex] by the well known Heron's formula.
Multiplying the two equations above together and use [tex]\sin{2x}=2\sin{x}\cos{x}[/tex], we get
[tex]8(\sin{A\over2}\sin{B\over2}\sin{C\over2})^2={r^2\over{2R^2}}[/tex]. Therefore,
[tex]\sin{A\over2}\sin{B\over2}\sin{C\over2}={r\over{4R}}[/tex] and
[tex]IA\cdot IB\cdot IC={r\over\sin{A\over2}}\cdot{r\over\sin{B\over2}}\cdot{r\over\sin{C\over2}}={r^3\over {\sin{A\over2}sin{B\over2}sin{C\over2}}}=4Rr^2[/tex]
QED.


I've been looking over the putnam stuff for a few days now..

But they're so different from regular olympiads for high schoolers, mostly because of the introduction of new topics. I've been decent in olympiads with a couple of recognitions in high school, but now i'm not sure what to do, because i have almost no calc/lin algebra background.

any suggestions for free online resources?

also it'd be helpful if someone pointed me to the right directions (topics commonly in putnam, must-know theorems etc)

thanks!

This question gets posted a lot in this forum. Look back at some of the old discussions.

If you want to know what sort of stuff is on Putnam, look at old tests. There are dozens online. (The most recent ones will be the most accurate predictors of future tests, of course.)

If I was to go through the past contests, from what year would they resemble the current contests? Like, 1st putnam is just basic calculus..


blogs
calculus
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Comment on the user above/below you

Example:
[quote]The user above me has 1337 posts.
Q.E.D.[/quote]

The user below me is bad at math because they got the first post.

The user below me finds Ubemaya extremely attractive.

yeah.. i think so

the person below me.... is eating something right now

Correct, I'm eating chocolate

The person above me is computer addicted
Q.E.D

The person below me is about to use the bathroom.

close, but false.

The person above the person above me is insulting.

The Person below me is not a straight-A student...?

that would all depend on how you look at it

the dude(tte) agrees with my sig

what dude(tte)?

The person below me is a n3wb because they dont know atleast 50 digits of pi.

the person above me is a crazii silly banana

the person above me has great spelling, i'm going to name my self funcii now.
[rant]
HEY AND WHAT HAPPENED TO THIS FORUM? WHERE ARE THE TWO PEOPLE WITH THE BLOG AND THE CALCULUS BOOK! WHAT HAPPENED HERE?
[/rant]

The person me just made a Pi symbol out of white clay that has 42 digits of pi on it :D

The person above me said nothing about the person above/below him/her.

depends.

tpbm is distressed bc facebook has just gotten blocked in China D:

No, uh-uh, nope, absolutely not, negative.

TPBM has at least 2 of the AOPS textbooks (solutions don't count!)
I have 5...

No, uh-uh, nope, absolutely not, negative. i have none:D

tpbm is manly.

LOL [b]no[/b] ... m wo-manly  :D

TPBM is an avid blogger

^irrelevant

lol@Nora9.1 - it has been decided by my social studies teacher that the girls in our class were much more manly than the guys, so even if you are female (like me) you can still be manly haha

tpbm has heard someone sing extremely badly on t.v. recently *shudder

Don't think so...

TPBM has more posts than me

um ... sometimes

TPAM is addicted to herion

Right! I lose!  :wink:

TPAM has been granted special authority to declare the loser.  :agent:


Nogte haye $M,N$ be tartib vasate zelhaye $BC$ va$AD$ az morabae $ABCD$ hastand.$K$ nogteiy delkhah roye emtedad gotre $AC$ az samte noteye $A$ ast. Pare khate $KM$ zel e $AB$ ra dar nogteye $L$ gat karde ast. Sabet konid $\angle KNA=\angle LNA$.

man in saal ro ba tajanos hal kardam mikham bebinam ke rahe hale digari darad ya na?

Raahe man ham baa tajaanose(albate be taales shabihtare!!!). vali laazeme $ABCD$ moraba baashe? baraaye mostatil ham doroste :huh:


inequalities
calculus
integration
inequalities proposed

[quote="Albanian Eagle"]doesn't this come from integrating :
$a^{3}+b^{3}+c^{3}+3abc\geq a^{2}b+b^{2}a+a^{2}c+c^{2}a+b^{2}c+c^{2}b$? :D[/quote]
Nice solution!

Nice proof! actually I was thinking about the inequality $f(x_{1})+f(x_{2})+\cdots+f(x_{n})+n(n-2)f(a)\geq \sum_{i \neq j}f(a+\frac{x_{i}-x_{j}}{n})$ where $a=\frac{x_{1}+\cdots+x_{n}}{n}$ and $f$ is convex, but there was a counterexample :(

[quote="Albanian Eagle"][url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=integration&t=22364]here[/url] is the generalisation by harazi![/quote]

What is name of this inequalities [img]http://www.mathlinks.ro/Forum/latexrender/pictures/aac0578b32304d198890118114719e35.gif[/img]?  :|

[quote="arqady"]Albanian Eagle, what about it:
$a,b,c$ are positive. Prove that
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{9}{a+b+c}\geq6\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)$ $?$
$a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+6\sqrt[3]{a^{4}b^{4}c^{4}}\geq3(a^{2}bc+b^{2}ac+c^{2}ab)$ $?$ :D  :)[/quote]

The second one has counterexample: $a=b=1, c=M$ for big enough M. The first one seems to be true, while it is an integrated form of the second one. I tried SOS on the first one and I got $\sum (b-c)^{2}(b^{2}+bc+c^{2}-a^{2})(b+c)(2a+b+c) \geq 0$.

It is better to assume, that $a+b=z,$ $a+c=y$ and $b+c=x$ and SOS. :wink:

[quote="arqady"]Albanian Eagle, what about it:
$a,b,c$ are positive. Prove that
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{9}{a+b+c}\geq6\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)$ $?$
$a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+6\sqrt[3]{a^{4}b^{4}c^{4}}\geq3(a^{2}bc+b^{2}ac+c^{2}ab)$ $?$ :D  :)[/quote]
Oh I believe that from integration we don't get [u]equivalent[/u] inequalities :wink: :)

[quote="Albanian Eagle"]
Oh I believe that from integration we don't get [u]equivalent[/u] inequalities :wink: :)[/quote]
I agree with you. You correctly believe. Even because of this example!

Yes. If the original inequality is true then the integrated form (inequality of fractions) is true. But these are not sufficient and necessary. The counterexample is the ineq arqady posted.

[quote="Sung-yoon Kim"]For positive real numbers $a,b,c$, prove that \[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{9}{a+b+c}\geq\frac{3}{2a+b}+\frac{3}{a+2b}+\frac{3}{2b+c}+\frac{3}{b+2c}+\frac{3}{2c+a}+\frac{3}{c+2a}\] [/quote]

By the way, I proved it by another way. Try this by another way!

[quote="arqady"]Albanian Eagle, what about it:
$a,b,c$ are positive. Prove that
$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{9}{a+b+c}\geq6\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)$ $?$
[/quote]
Let $a+b=x,b+c=y,c+a=z$. And SOS. A similar form comes from trthnam topic.

[quote] $a^{2}b^{2}+a^{2}c^{2}+b^{2}c^{2}+6\sqrt[3]{a^{4}b^{4}c^{4}}\geq3(a^{2}bc+b^{2}ac+c^{2}ab)$ $?$ :D  :)[/quote]
Let $x=0.1\ ,\ y=0.2\ ,\ z=3$.


Let $ x_1$, $ x_2$, $ \dots$, $ x_n$ be a sequence of integers such that
(i) $ \minus{}1 \le x_i \le 2$, for $ i \equal{} 1,2,3,\dots,n$;
(ii) $ x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n \equal{} 19$; and
(iii) $ x_1^2 \plus{} x_2^2 \plus{} \cdots \plus{} x_n^2 \equal{} 99$.
Let $ m$ and $ M$ be the minimal and maximal possible values of $ x_1^3 \plus{} x_2^3 \plus{} \cdots \plus{} x_n^3$, respectively. Then $ \frac{M}{m} \equal{}$

$ \textbf{(A)}\ 3\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 5\qquad
\textbf{(D)}\ 6\qquad
\textbf{(E)}\ 7$

Let a be the number of 2s, b the number of 1s, c the number of 0s, d be the number of -1s. a,b,c, and d are all positive. We have
2a+b-d=19
4a+b+d=99
From here we have
$ 2b\equal{}118\minus{}6a \ge 0 \implies a \le 19$
$ 2d\equal{}80\minus{}2a \ge 0 \implies a \le 40$
So $ a \le 19$. Subracting the two above equations, we also have
b-d=19-2a.
We wish to maximize
8a+b-d=6a+19
So we need only maximize a, so a=19, b=2, and d=21.  Then 6a+19=133.
We also need only minimize a, so a=0, then 6a+19=19. The answer is $ \frac{133}{19}\equal{}7$.


[quote="manlio"]For all $ a,b,c\ge0$ we have $ \sum\frac {a^2(b \plus{} c)^2}{a^2 \plus{} bc}\ge2(ab \plus{} bc \plus{} ca)$

P.S. I don't remember if it is an old or a new inequality. Sorry :blush:[/quote]
for $ a\equal{}3,b\equal{}4,c\equal{}5$ ,it's wrong

we have:

$ \text{RHS} \minus{} \text{LHS} \equal{} \sum_{cyc}\frac {(a \minus{} b)^2(a \plus{} b)(a \plus{} b \minus{} c).c^2}{(a^2 \plus{} bc)(b^2 \plus{} ac)}$

so it's obviously wrong if $ a,b,c$ are the side of triangle  :oops:

I read it at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310089


Two candles have equal lengths.  One is consumed uniformly in 6 hours and the other in 5 hours.  If they are both lit at the same time, then how many hours after lighting will one be twice as long as the other


linear algebra
matrix
algebra
polynomial

let $A$ in $M_2(K)$ with $A^2=A$, and $X$ in $M_2(K)$
Prove that AX=XA if only if exist $X_0$ in $M_2(K)$ such that:
  $X=AX_0+X_0A-X_0$

We know that for all $x \in \mathbb K^{2}$ we have either $Ax = x$ or $Ax=0$, from where we deduce that $A$ is similar to $\displaystyle \left( \begin{matrix}0 & 0 \\ 0 & 0 \end{matrix}\right), \, \left( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}\right)$ or $\displaystyle \left( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix}\right)$.

[i]Case 1.[/i] $A = S^{-1}0_{2}S = 0_{2}$
All $X \in \mathcal M_{2}\left( \mathbb K \right)$ commute with $A$, so we can take $X_{0}=-X$.

[i]Case 2.[/i] $A = S^{-1}I_{2}S = I_{2}$
All $X \in \mathcal M_{2}\left( \mathbb K \right)$ commute with $A$, so we can take $X_{0}= X$.

[i]Case 3.[/i] $A = S^{-1}\left( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}\right) S$
The minimal and the characteristic polynomials of $\left( \begin{matrix}1 & 0 \\ 0 & 0 \end{matrix}\right)$ are the same, so $m_{A}= p_{A}$. Thus, by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=73310]this topic[/url], we have that $X = \alpha A+\beta I_{2}$, for some $\alpha, \, \beta \in \mathbb K$.
Let $V$ be the space $\left\{ \alpha A+\beta I_{2}| \alpha, \beta \in \mathbb K \right\}$. Then, $AX = XA$ iff $X \in V$.
For $X_{0}= A$ we have that $X = A^{2}+A^{2}-A = A+A-A = A$ and for $X_{0}= I_{2}$ we have that $X = A+A-I = 2 A-I_{2}$.
The map $T : V \to V, \, T \left( X_{0}\right) = A X_{0}+X_{0}A-X_{0}$ is linear and $\left| \begin{matrix}1 & 0 \\ 2 &-1 \end{matrix}\right| =-1$, so $T$ is bijective, which is what we wanted to prove.

Is it OK :maybe:


function
integration
calculus
calculus computations

Let f be a continuous real-valued function defined on a closed interval [a, b]. Let F be the function defined, for all x in [a, b], by

   $ F(x) \equal{} \int_a^x f(t)\, \mathrm dt$. Then $ F$ is an antiderivative of $ f$ and  $ F(b) \equal{} \int_a^b f(t)\, \mathrm dt$.

But:  $ \int_a^b f(x)\,\mathrm dx \equal{} F(b) \minus{} F(a)$ so we get $ F(a)\equal{}0$, which is not necessary true.

I probably made some big stupid error somewhere along the way, but i just can't figure it out. I would appreciate if someone would tell me where it went wrong.

Of course it is. $ \int_a^a f(t)\,dt\equal{}0$ no matter what $ f$ is.


the answer follows from the definition of gases  and interfcaes...gases occupy the entire space they are given to and interfaces between two matters means matter of one kind is present in a certain space and of the other kind in other space...so if such a thing ahppens for gases then they won't be called gases :D

but why man,...

consider 2 diff solids..though they are in the same phasr , they do have an interface.. and i am not satisfied with ur exp[lanation..

a mixture of gases is always homogeneous, as the inter molecular forces are negligible.

i told u before....
[quote="pardesi"]interfaces between two matters means matter of one kind is present in a certain space and of the other kind in other space[/quote]
thsi happens in solids so they have interfaces  :wink:

[quote="valeriummaximum"]considering ur explanation pardesi....

the two solids are also in the same state and not of 2 diff kinds..
moreover, two diff liquids have interfaces know :huh: ????

i think this has something to do with the miscibility...

@varun-thinking abt ur explanation..[/quote]
where have i used the word 'state' in my explaination ...i have just used matter...that means they can be solid liquid or gas :huh:

two gases are always miscible in each other, as the gases act independant of each other.
now consider at the molecular level, if two liquids have an interface, it means that the molecules of one liquid do not remain surrounded by molecules of other the other liquid, as the intermolecular forces are not the same, but in case of gases that is not so.

but u havce used that they are of 2 diff kinds...
but both are made of molecules only know????

i said two matters who cares if they are different or not...i don't find anything illogical in it it's just a matter of definition :wink:

do u mean to say that 2 gases becomes diffudsed into one other that ther is no interface????

@varun- i'm sorry ,coz i don undersatnd...

no beacuse they diffuse there exists no interface :wink:

I heard chemrock saying that in the NCERT text it  is given that gases dont have an interface due to complex miscibility....

What is complex miscibility???? :huh:  :huh:  :huh:

yep, the molecules can move freely as i already mentioned, there are no forces to bind them. they are in a state of high entropy. but that is not cas with solids and liquids. thats precisely the reason why you can use gas laws for all gases irrespective of nature but not for solids and liq.

i dont know, i think it means that the gases are miscible in all proportions :huh:

all fall @ varun's feet for darshan!!!  :D  He has appeased the lord ranga the ruler of the three worlds  :P   :rotfl:

actually, most of the questions reg diff btw gases and other states can be answered by consisdering this simple point " ther can be only type of disorder whrereas many types of order"
@madness- everyone must fall @ someone else's feet not mine. not gonna mention who it is. :P

hey i beg all to stop talking like this.. :rotfl:  :rotfl:  :(
tell directly who it is...i am bored of it esp everyone :(  :(  :rotfl:  :rotfl:

this [b]is not[/b] a personal attak Varunarasu...i just wanna take this opprtunity...
what's the use of writing in such cryptic language...those who know of it know of it...those who don't know of it remain ignorant...what happens as a net result is people smells something and it's told to them that there exists omething u don't know which somone knows :rotfl:  :rotfl:  :rotfl:  :(

well , those who need to know it, will get to know it, for others its just a polite way of saying " mind your own freaking business" ( no offense meant).
@ madness see the pm ive sent you.

[quote="pardesi"]:rotfl:  :rotfl:  :rotfl:  that was even more digusting :rotfl:  :rotfl:  :([/quote]
now stop acting like a baby or you will join the ranks of madness aand ranga. :ninja:

no da it's not in response to urs comments  madness will get it... :ninja:

Ranga incase you didnt know. The answer is clearly given in first page of Surface Chemistry in NCERT text book. Gases just diffuse too fast forming a homo mixture and thereby there can;t be an interface.

that's what i asked shre...

what do u mean by complex miscibility????

[quote="valeriummaximum"]What is complex miscibility????    :huh:
 :huh:[/quote]
Excuse me I said complete miscibility!!! :mad:

And if u have any doubts as u why then it is because the entropy of the system increases when 2 different kinds of gases diffuse and so that's the driving force for them to form a homogenous solution. :wink:

hey chemrock,.. sorry t in text it s given only complete miscibility

[quote="chemrock"][quote="valeriummaximum"]What is complex miscibility????
 [/quote]
Excuse me I said complete miscibility!!! [/quote]

Ranga 'complete smell'  :P   :rotfl:  :rotfl:

ithu ellan
 CHUMMA JUJUBI  :rotfl:  :rotfl:  :rotfl:

My reply is same as that for pardesi:

[size=200][b][i]PETHAL[/i][/b][/size]


a) Show that there are no functions $f, g: \mathbb R \to \mathbb R$ such that $g(f(x)) = x^3$ and $f(g(x)) = x^2$ for all $x \in \mathbb R$.

b) Let $S$ be the set of all real numbers greater than 1. Show that there are functions $f, g : S \to S$ satsfying the condition above.

This looks interesting. Can anybody show solution?

for part [b]a)[/b] assume that such functions exist,then:

$ g(f(x)) \equal{} x^3\Rightarrow f(g(f(x))) \equal{} f(x^3)$ (I)

also from $ f(g(x)) \equal{} x^2$ we get that:

$ f(g(f(x))) \equal{} (f(x))^2$

which with (I) implies that:

$ f(x^3) \equal{} (f(x))^2$ (*)

now note that $ f$ is injective,because if $ f(x_1) \equal{} f(x_2)$ then:

$ g(f(x_1)) \equal{} g(f(x_2))\Rightarrow x_1^3 \equal{} x_2^3\Rightarrow x_1 \equal{} x_2$

now in (*),plug in $ x \equal{} 0,1, \minus{} 1$ we get that:

$ f(0),f(1),f( \minus{} 1)\in\{0,1\}$

so two of them must be equal which is a contradicion...

P.S:see [url]http://www.mathlinks.ro/viewtopic.php?p=2007#p2007[/url] for different solutions .


What is the limitation of a base?

Can it be any real number? Can it even be unreal?

I don't think bases can be negative because there would be "negative digits". Maybe decimals. Well let's try converting $20_{10}$ to base $3.4$. The largest place value would be $3.4^2 = 11.56$. $11.56$ goes into $20$ one time, so the first digit must be $1$, leaving $8.44$. $3.4$ goes into $8.44$ two times, so the second digit is $2$ leaving $1.64$. The last place value, $1$, goes into that one time, making the third digit $1$. So far we have $121$. $\frac 1{3.4}$ goes into $1.64$ five times, leaving something irrational... I think that a base can have a decimal in it... just that some of the numbers will simply go on forever if you're converting from base 10 to that base.

Even if the base you're converting to is a positive integer, it can still go on forever. There are numbers that go on forever in base 10 too, so there's nothing wrong with them.

However, from Mathworld, the base must be an integer. It does not say that it must be positive, although it does not mention how to use a non-positive number as a base.

I think the trickiest thing about doing this is figuring out what the digits should be.  It seems possible to design a system base pi, for example, if you can make a good case for what the digits are -- otherwise, it seems difficult.  rcv is right that negative numbers should work out without too much trouble, although you'll probably still want to be careful with your choices of digits to make unique expressions.

Another base-related interesting thing is factorial base -- the first place can have digits 1 or 0, for 1!.  The second place can have digits 0, 1 or 2, for 2!.  The third place can have digits 0, 1, 2 or 3 for 3!, and so on.  I don't know how you would extend it to get "factorial decimals," but it does give a unique representation to each positive integer.  For example, $124_{10} = 10020_{F}$, $18_{10} = 300_F$ and $51090942171709439999_{10} = (20)(19)(18)(17)(16)(15)(14)(13)(12)(11)(10)987654321_F$ (where the numbers in parentheses are digits, not things being multiplied together).

[quote="rcv"][quote="tarquin"]I don't think bases can be negative because there would be "negative digits" ...[/quote]
Actually, there is a reasonably well-known system (among computer software engineers) called "balanced ternary".  It is a base-3 system that uses the "digits" {-1, 0, +1}.  Every integer, positive or negative, can be expressed uniquely in this system.  The rules of "place value" apply in this base as with other base-n numbers.  For example, the number "2" is equal to $(+1)3^1+(-1)3^0$.  The number "25" is equal to $(+1)3^3+(0)3^2+(-1)3^1+(+1)3^0$.

A conventional number base is an integer greater than 1.  However, I don't see any reason why you couldn't create a "balanced ternary" system, using base $-3$.[/quote]

I've heard about that before.  There was some question on MathCounts...either it was a warmup or workout...and it asked what is the least number of numbers needed to get any number up to 100...and it is the first four powers of three.  Powers of three have the power to yeild anything and results in a very power system.

Because I was working a problem I made up and there were two solutions to the bases given.  One was positive and the other was negative so I'm thinking that it is possible...

I know this is somewhat old, but I thought I should say that there are negative bases. Run a google search on "negabinary" or "negadecimal". Even "negaternary" will get a few results. Turns out these bases have been around since 1885.


function
linear algebra
matrix
algebra
polynomial
linear algebra theorems

I don't have a solution for the following problem and it seems very interesting: what are the linear maps of $M_n(K)$ such that if $A,B$ commute so does $f(A), f(B)$?

wath is f, if you mean by f a polynom ,it's clear that f(A) and f(B) also commut !!!

I think the statement is pretty clear: f is a linear map of $M_n(K)$, that is a function such that $ f(aA+bB)=af(A)+bf(B)$ for all matrices $A,B$ and all $a,b\in K$.

Maybe of the form

$f(X)=aPXP^{-1}$

or

$f(X)=bQ{(^{t}X)}Q^{-1}$  where a,b scalar in K

i think they do...but i haven't  finished the proof yet

Well, I have something that maybe solves the problem. We know that if all  eigenvalues of matrix $A$ are different then if $A$ and $B$ commute then $B$ is a polynomial of $A$.


A cylindrical glass is half full of lemonade. The ratio of lemon juice
to water in the lemonade is 1:11. If the glass is 6 inches tall and has a diameter of 2 inches, what is the volume of lemon juice in the glass? Express your answer as a decimal to the nearest hundredth.

Volume of the lemonade:
$ V\equal{}1\cdot 1\cdot \pi\cdot 3\equal{}3\pi$
Lemon juice is 1/12 of that, so it's
$ V_L\equal{}\frac{\pi}{4}\equal{}0.79$


Rewrite the inequality in the form [tex]a^2+b^2+c^2+2\sqrt{3abc(a+b+c)}\leq (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)[/tex], or equivalently, [tex]\sqrt{3abc(a+b+c)}\leq ab+bc+ca[/tex], which is well known.

A similar problem was posted before by ThAzN1:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13767

This can be generalized to the following
[quote]If $k\geq 12$ then for any $a,b,c\geq 0$ satisfying $a+b+c=1$, the following inequality is true,
\[
a^2+b^2+c^2+\sqrt{kabc} \leq \frac{1}{3}\left(1+\sqrt{\frac{k}{3}}\right).
\]
[/quote]

When k=12, it's the inequality posted by Maverick.

When k=75, it's the inequality posted by ThAzN1.


inequalities
number theory
greatest common divisor
least common multiple
relatively prime
number theory unsolved

Let the positive integers $ a,b,c$ chosen such that the quotients $ \frac{bc}{b\plus{}c},$ $ \frac{ca}{c\plus{}a}$ and $ \frac{ab}{a\plus{}b}$ are integers. Prove that $ a,b,c$ have a common divisor greater than 1.

Let $ g_1$ be the $ gcd(a,b)$. Hence there are numbers $ a_1$ and $ b_1$ that following holds: $ a = g_1  a_1$ and $ b = g_1  b_1$. Obviously $ gcd(a_1,b_1)=1$. For $ \frac{ab}{a+b} = \frac{g_1a_1b_1}{a_1 + b_1}$ to be an integer $ a_1 + b_1$ must divide $ g_1$, since $ gcd(b_1,a_1 + b_1)$ and $ gcd(a_1,a_1 + b_1)$ are both equal to 1, because $ gcd(a_1,b_1)=1$. Therefore $ a_1 + b_1 |g_1$ and $ g_1 \geq a_1 + b_1$.
$ a_1,b_1 \geq 1$, because $ a$ and $ b$ are greater than zero. Hence $ g_1 \geq 2$. Analog $ g_2 = gcd(b,c) , g_3=gcd(c,a) \geq 2$

Now I will show by contradiction that $ g = gcd(g_1, g_3) > 1$. Let's assume $ g=1$. $ a$ can be written both as $ a=a_3 g_3$ and  $ a=a_1 g_1$. By combining both equations we obtain $ a_3 g_3=a_1 g_1$. $ gcd(g_3,g_1)= 1$ implies that $ g_1 | a_3$ and $ g_3 | a_1$. Therefore $ a_3 \geq g_1$ and $ a_1 \geq g_3$. $ c$ may also be written in the form $ c=c_3g_3$. Analog to above $ g_3 \geq a_3 + c_3$. By combining the inequalities $ a_3 \geq g_1$ , $ a_1 \geq g_3$, $ g_3 \geq a_3 + c_3$ and $ g_1 \geq a_1 + b_1$ we obtain $ a_3 \geq a_1 + b_1$ and $ a_1 \geq a_3 + c_3$. By adding the last inequalities we obtain $ 0 \geq b_1 + c_3$, which is obviously a contradiction. Therefore $ gcd(g_1,g_3) >1$ and a number $ t$ exists for which  $ t>1$ and $ t|g_1=gcd(a,b) , g_3(c,a)$. Thus $ t|a$, $ t|b$ , $ t|c$. $ \Box$

[hide="solution."]

First, we note the following straightforward facts:

[b]Fact 1[/b]. Let $ x$, $ y$ be positive integers, $ d = \gcd(x,y)$, $ x' = x/d$, $ y' = y/d$. Then $ x + y$ divides $ xy$ iff $ x' + y'$ divides $ d$.

Proof:
$ \Leftarrow$ If $ x' + y' | d$ then then $ d(x' + y') | d^2$ so $ x + y | d^2$. It follows that $ x + y | d^2x'y' = xy$. (we don't really need this direction for our problem, we note it just for the sake of completeness).
$ \Rightarrow$ Rewrite $ x + y | xy$ as $ d(x' + y') | d^2 x' y'$. It follows that $ x' + y' | d x' y'$. But $ x'$ and $ y'$ are relatively prime, so $ x' + y'$ is relatively prime to both $ x'$ and $ y'$. Therefore $ x' + y'$ must divide $ d$. QED

[b]Fact 2[/b]. Let $ x$, $ y$, $ z$ be three positive integers such that $ \gcd(x,y,z) = 1$, namely no integer greater than 1 divides all three. Write $ f = \gcd(x,y)$, $ e = \gcd(x,z)$, $ d = \gcd(y,z)$. Then $ d$, $ e$ and $ f$ are pairwise relatively prime and so there exist positive integers $ x'$, $ y'$, $ z'$ such that $ x = efx'$, $ y = dfy'$, $ z = dez'$.

Proof: Note that $ x$ is divisible by both $ e$ and $ f$. Now $ e$ and $ f$ must be relatively prime, otherwise whatever divides them would divide all of $ x$, $ y$ and $ z$. Therefore $ x$ is actually divisible by $ ef$, so $ x = efx'$ for some integer $ x'$. The other claims follow by symmetry. QED

Finally, let $ x$, $ y$, $ z$ be given that satisfy the requirements of the problem: for each pair, their sum divides their product. Assume by way of contradiction that $ \gcd(x,y,z) = 1$ and let $ d$, $ e$, $ f$, $ x'$, $ y'$, $ z'$ be as in Fact 2. Finally, assume wlog that $ d < e < f$.

We have $ y = dfy'$ and $ z = dez'$. The gcd of $ y$ and $ z$ is $ d$, so from Fact 1 we must have $ fy' + ez' | d$. But this is nonsense since both $ e$ and $ f$ are strictly larger than $ d$. This shows that the assumption $ \gcd(x,y,z) = 1$ is false. QED

It's now easy to construct such triplets: take any triplet $ a$,$ b$,$ c$, and take $ d$ to be the lcm of $ a + b$, $ a + c$ and $ b + c$. Then $ x = da$, $ y = db$ and $ z = dc$ will satisfy the requirements. A simple example is $ (4,4,12)$, but a more convincing one perhaps is $ (24, 72, 120)$.

[/hide]

i've seen this problem on mathlinks before. i think it's on the pre-olympiad forum. i remember that the problem poster said that he has managed to prove that $ \gcd(a,b),\gcd(b,c),\gcd(c,a)$ are each greater than 1, but he cannot complete the proof.

[quote="orl"]Let the positive integers $ a,b,c$ chosen such that the quotients $ \frac {bc}{b \plus{} c},$ $ \frac {ca}{c \plus{} a}$ and $ \frac {ab}{a \plus{} b}$ are integers. Prove that $ a,b,c$ have a common divisor greater than 1.[/quote]
This lemma instantly kills the problem. And it can easily be generalised to $ n$ variables.

[color=darkred][b]Lemma 1:[/b][/color] If $ x \plus{} y \mid xy$ then $ \gcd(x,y) > \sqrt {x}, \sqrt {y}$.

[color=darkblue][b]Proof: [/b][/color] Let $ d \equal{} \gcd(x,y)$ and let $ x \equal{} dx_1, y \equal{} dy_1$. Then $ x \plus{} y \mid xy \iff x_1 \plus{} y_1 \mid dx_1y_1 \iff x_1 \plus{} y_1 \mid d$ since $ \gcd(x_1y_1, x_1 \plus{} y_1) \equal{} 1$. Thus $ d \equal{} k(x_1 \plus{} y_1), k \in \mathbb{N}$. So $ x \equal{} k(x_1 \plus{} y_1)x_1, y \equal{} k(x_1 \plus{} y_1)y_1$. And hence $ \gcd(x,y)^2 \equal{} k^2(x_1 \plus{} y_1)^2 > x,y \iff k(x_1 \plus{} y_1) > x_1,y_1$ Which is true $ \Rightarrow \gcd(x,y) > \sqrt {x}, \sqrt {y}$.

Using [b]Lemma 1[/b] we obtain $ \gcd(a,b) \cdot \gcd(a,c) > a$. Since $ \gcd(a,b), \gcd(a,c) \mid a$ we get that $ \gcd(\gcd(a,b), \gcd(a,c)) > 1 \iff \gcd(a,b,c) > 1$, which concludes the solution.

Note: I did not use that $ b\plus{}c \mid bc$, so it is only nessecary that two of the quotients are integers.


Now that ARML's a few hours from over, let's start a year-long practice series. Here's the first problem. One with a solution can put another problem, but to maintain order, at most 2 problems at a time, pls.?

Here are the first two problems:

Compute the length of the tangent segment from the origin to the circle that passes through the points (3,4), (6,8), and (5,13).

If ABCDEFG is a regular heptagon and diagonals AC and BF intersect at P, compute angle APB.


Does anybody have the 1990 States Team Round page #2, the 2007 and 08 nats tests, and a bunch of nats solutions? I lost a few, but if anybody wants to trade with me, please pm so i can tell you what i have and swap with u. :)

i have 07 and 08 nats. i also have some stuff rfom the 90s (nats.) i'd like old states (90-00), do you have any? email me at nikeballa96[]yahoo.com, where[]=@


I think it doesn't converge to a finite number.

If you're so sure, then why don't you give an answer?

[quote="worthawholebean"][quote="jb05"]Consider the following sequences:
1/2
1/3+1/4
1/5+1/6+1/7+1/8
etc.

Each element of the first sequence is greater than or equal to 1/2 (clearly).

Each of the two elements of the second is greater than or equal to 1/4. Double to get 1/2.

Each of the four elements of the third is greater than or equal to 1/8. Quadruple to get 1/2.

Continuing, we determine that the infinite sum diverges.[/quote]

Nope. If the numbers you add keep gettingsmaller, it has to converge.[/quote]

No, jb05 is correct. Read his logic again, it makes complete sense. Actually, he shouldn't put or equal to, he should just put is greater than (except for the first set of elements).

Never mind, I guess you're right. My logic made complete sense to me, but it was wrong.

Yeah another classic problem that does converge is 1/2+1/4+1/8..... Notice however that 1/4+1/8+1/16...never exceeds 1/2, so it's not exactly like the other problem.

That's very strange. When graphed, it seems to converge. But it doesn't.

I believe the answer was labeled divergent or something.

No, surely this graph converges. I haven't really analyzed it yet, but if it's a geometric series, you can use the formula. It obviously isn't an arithmetic progression.

it is not geometric. It is not arithmetic. It does not converge.