[quote="Ignite168"]What is the maximum number of bags that can be used to hold 190 marbles if the number of marbles in each bag must be distinct?[/quote]
Oh yes, I remember this question. In fact, there was a trick to it that allowed two possible answers depending on how you interpret the condition "the number of marbles in each bag must be distinct."
I'll let you see if you can find the trick.  :)

[hide]190 if you put bags in bags and 19 if you can't[/hide]

[quote="Ignite168"]What is the maximum number of bags that can be used to hold 190 marbles if the number of marbles in each bag must be distinct?[/quote]
[hide]190 is the sum of the first 19 numbers, so the marbles can be placed into 19 bags. Or, as SplashD said, 190 if you can put bags in bags.[/hide]

[quote="Ignite168"]What is the maximum number of bags that can be used to hold 190 marbles if the number of marbles in each bag must be distinct?[/quote]

[hide]
$190=1+2+\dots+18+19$, so 19 bags[/hide]

You should try to take note that that would be the correct answer if bags couldn't be placed inside of bags...

But that may not be the case.

[quote="Ignite168"]What is the maximum number of bags that can be used to hold 190 marbles if the number of marbles in each bag must be distinct?[/quote]

thi s wa a cheap nationals problem...
i forgot it, but it was a number that was unexpected...
190?
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=611997
also here

[quote="Ignite168"]What is the maximum number of bags that can be used to hold 190 marbles if the number of marbles in each bag must be distinct?[/quote]
[hide="solution"]
(if bags cant go inside bags,)
Since they must all be distinct, the bags will start with 1 and add one for each of the next bag. If there are leftovers, they will be added to the last bag.
However, $1+2+3+4+...+19$ equals exactly $190$. It takes 19 bags.
If a bag of 0 marbles is allowed, the maximum is 20 bags.
[/hide]

[quote="SplashD"][quote="bpms"][quote="Phelpedo"]Wrong  :wink:  You can use a lot more than 20 (and by the way, 0 marbles in a bag is not allowed).

I wonder how many people got this right on the actual test?[/quote]
Maybe frost13 or RaviB or some other experienced coach has the spreadsheet thing. My guess is <10.[/quote]

By right, do you mean the answer of 190?[/quote]
Did mathcounts have the error of 190 in mind or did someone protest it? What year is this from? What happens if someone needs to protest a countdown problem?

[quote="bpms"][/quote][quote="SplashD"][quote="bpms"][quote="Phelpedo"]Wrong  :wink:  You can use a lot more than 20 (and by the way, 0 marbles in a bag is not allowed).

I wonder how many people got this right on the actual test?[/quote]
Maybe frost13 or RaviB or some other experienced coach has the spreadsheet thing. My guess is <10.[/quote]

By right, do you mean the answer of 190?[/quote][quote="bpms"]
Did mathcounts have the error of 190 in mind or did someone protest it? What year is this from? What happens if someone needs to protest a countdown problem?[/quote]

The correct answer that Mathcounts originally intended for was 190, I think. I seriously doubt that you can protest a countdown problem as most of them are rather trivial, but you certainly can protest the problems of the other rounds.

Most countdowns are trivial, but was it a coincedence that the most interesting match-up (Nebula vs. Treething) at nationals had the hardest problems? Also, how were you supposed find the smallest possible integer spatial diagonal of a rectangular prism with all integer sides in 45 seconds??? That could easily have been one of the last ten sprints.


An archer shoots an arrow into the air such that its height at any
time, t, is given by the function h(t) = -16t^2 + kt + 3. If the maximum height of the arrow occurs at time t = 4, what is the value of k?

(1) 128     (3) 8
(2) 64       (4) 4

This seems like homework.  The key here is the minimum or maximum of a quadratic equation $ax^2+bx+c$ occurs at $x=\frac{-b}{2a}$.

you ought to post things that involve thinking, these last 2 you posted are just plug and chug problems


function
logarithms
symmetry
combinatorial geometry
real analysis
real analysis solved

$f$ is continuous on $[0;1]$ so sup of $f$ is defined, we note it $M$.
We define $g_n$ : $g_n(x)=(M+1)\sqrt[n]{x}$
Now I know that my solution is not well justified but the idea is that $g_n$ tends to the function which equals $0$ in $0$ and $M+1$ on $]0;1]$ so it exists $n_0$ such that for all $n\geq n_0$ we have $g_n(x)\geq f(x)$ for all $x$ in $[0;1]$ and $g_n$ is concave.

(edit) I have good solution, I post it in few minutes

We can consider $f$ positive so there is two cases

First case: $f(x)/x$ doesn't tend to $+\infty$ in $0$ so $\sup_{[0;1]} f(x)/x$ exists and equals $a$.
So $\forall x, ax\geq f(x)$ then we consider the point $B(1,a)$, if we construct a function which is above $[OB]$ it's done. $g(x)=a\sqrt{x}$ works

Second case: $f(x)/x$ tends to $+\infty$ in $0$. We can works on $[0;\epsilon]$, I thought I had the solution but it's not correct I just have to prove that it exists, $n$ and $k_n$ such that for all $x\in[0;\epsilon]$, $f(x)\leq  k_n\sqrt[n]{x}$

I don't think any one form will work near zero. A counterexample to yours:

$f(x)=\frac1{1-\ln x}$. For $x$ near zero, $f(x)>x^k$ for all $k>0$.

Here is a construction that actually works: Let $g(x)=\sup_{a<x<b}\frac{(x-a)f(b)+(b-x)f(a)}{b-a}$, $g(0)=f(0)$, $g(1)=f(1)$. This supremum is uniformly bounded by $\sup f$, so $g$ is finite everywhere. Concavity is natural, since for each $x$ and each $\epsilon>0$, there is a line passing through $(x,g(x)+\epsilon)$ and lying strictly above $f$, and also for each $x$ every line passing through $(x,g(x))$ intersects the graph of $x$.
Suppose for some $a<x<b$, $(x,g(x))$ lies below the segment between $(a,g(a))$ and $(b,g(b))$. Then for some $t$, $(x,t)$ lies above $(x,g(x))$ and below the line; we can find a line through $(x,t)$ that lies strictly above the graph of $f$. Since this line must pass below one of $(a,g(a))$ and $(b,g(b))$, we have a contradiction and $g$ is concave.

Continuity at the endpoints is tricky; I'll prove it at 0, and symmetry of the definition makes continuity at 1 work as well. WLOG, $f(1)=0$; if we add a linear function to $f$, the same linear function is added to $g$. In this case, clearly $g(x)\ge0$.
We want to show that for any $\epsilon>0$, there exists $\delta$ such that for $x<\delta$, $f(x)<\epsilon$. Let $a=\inf{x: f(x)\ge\frac{\delta}2}$ (If the set is empty, just take $\delta=1$). Let $b=\sup_{a\le x\le1}\frac{f(x)}x$. Then, for all $x$, $f(x)<\frac{\epsilon}2+bx$, and therefore also $g(x)\le\frac{\epsilon}2+bx$. This gives $g(x)<\epsilon$ if $x<\frac{\epsilon}{2b}$. This works for all $\epsilon$, and $g$ is continuous at 0 and 1. Continuity between those two points is automatic by concavity, and we are done.

This is the smallest possible concave function majorizing $f$.

[quote="grobber"]Here's a cute problem :):

Given a continuous function $f:[0,1]\to\mathbb R$ s.t. $f(0)=0$, show that there is a continuous concave function $g:[0,1]\to\mathbb R$ with $g(0)=0$ and such that $g(x)\ge f(x),\ \forall x\in[0,1]$.[/quote]

Take the convex hull of the graph of $f(x)$ over that interval.  The "top" of this is a continuous curve.


Prove that when one function $ f(x)$ meet her inverse $ f^{\minus{}1}(x)$ in only one point
this point there is on line y=x

assume $ f(x)\equal{}y\equal{}f^{\minus{}1}(x)$

then,

$ f(y)\equal{}x\equal{}f^{\minus{}1}(y)$

since this happens once, y=x


induction
real analysis
real analysis solved

Prove that for each rational point on the real line we can draw a circle which is tangent to the line in that point in such a way that all circles have disjoint interiors. Prove that it can't be done for all the points on the real line.

I suppose it's pretty easy, but.. :)

This one is about countability. You can do it for each of the points in any countable set (such as the rationals) and you can't do it for any uncountable set.

For the countable set (such as the rationals): The construction proceeds by induction. Order that set. Let an arbitrary circle be tangent to the line at the first point. Now, suppose we have affixed non-overlapping circles to each of the first n points. Since the closures of those circles include only one point each of the line, the (n+1)st point lies in the exterior of all previously chosen circles and at a postive distance from the nearest one. Given that, we can find a circle tangent at that point small enough not to intersect any of the previously chosen circles. Since we can do this for any n (by induction), we can accomplish the construction.

For the converse: Suppose we have a collection of non-overlapping circles tangent to the line. How many of those circles could have diameter greater than 1/n ? No more than n of them per unit length; else they would overlap. No more than finitely many circles of diameter > 1/n can be tangent to any bounded interval; since the line is a countable union of bounded intervals, the total number of circles of diameter > 1/n is countable. Since each circle must have positive diameter, the union over all n of the circles of diameter > 1/n must be all of the circles. But that is a countable union of countable sets, hence a countable set. There can be at most countably many non-overlapping tangent circles, so the set of tangent points can be at most countable.

Also see ech circle has a rational point in itself so the number of circles is countable.Contradiction.


inequalities
factorial
inequalities proposed

Prove the following inequality for positive integer $n$.

\[(n+1)\sqrt[n+1]{(n+1)!}-n\sqrt[n]{n!}\leqq{n+1}\]

It smells like AM-GM, and it is AM-GM.

\[
\frac{(n+1) +n\sqrt[n]{n!}}{n+1} \geq \sqrt[n+1]{(n+1)!}
\]

That's right! :)  moreover you can use Jacobstahl's or Rado's inequality. ;)


function
parameterization
real analysis
real analysis theorems

It is well-known that every continuous involution of the real line (that is a function $ f(x)$ defined and continuous on the whole real line $ \mathbb{R}$ and satisfying the identity $ f(f(x))\equal{}x$ for all real $ x$) has a fixed point. To prove this statement one can take the function $ g(x)\equal{}f(x)\minus{}x$ and use the mean value theorem: if for some real $ a$ we have $ f(a)\neq a$ then $ g(a)$ and $ g(f(a))$ have opposite signs and therefore there is a point $ c$ between $ a$ and $ f(a)$ such that $ g(c)\equal{}0$, so $ f(c)\equal{}c$.

But if one omits the requirement of $ f(x)$ to be continuous then one can construct an example of discontinuous involution of the real line that has no fixed points. For example, $ f(x)\equal{}\minus{}x$ for $ x$ not integer, $ f(2k)\equal{}2k\plus{}1$, $ f(2k\plus{}1)\equal{}2k$  $ (k\in\mathbb{Z})$.

It is interesting to consider a continuous one-parameter family of involutions of the real line, that is a function $ F(t,x)$ of two real variables such that $ F(t,x)$ is not constant with respect to $ t$ (that is, $ F(t_1,x_1)<F(t_2,x_1)$ for some $ t_1$, $ t_2$, $ x_1$), $ F(t,x)$ is continuous only with respect to $ t$ for all real $ t$ and satisfies the identity $ F(t,F(t,x))\equal{}x$ for all $ t$ and $ x$, and to ask such a question:
Does there exist such a function $ F(t,x)$ with the property: for all $ t$ and $ x$  $ F(t,x)\neq x$
or for every such a function $ F(t,x)$ there always exist $ t_0$ and $ x_0$ such that $ F(t_0,x_0)\equal{}x_0$ ?

As I see, this is the very interesting question...

If you ask for continuity in $ t$ only and if the "non-constant" condition is as weak as you wrote, such involutions do exist. Let again all integers be mapped in a fixed manner as you described. Now the intervals $ (2k\minus{}1,2k)$ and $ (2k,2k\plus{}1)$ can be mapped to each other in any way you wish continuously depending on $ t$. If, say, $ t\in(0,1)$ (which, for all topological purposes, is the same as $ t\in\mathbb R$), then just take the mappings $ x\mapsto x^t$ and $ x\mapsto x^{1/t}$ that map $ (0,1)$ to itself, and transfer them to the intervals mentioned above in the obvious way.

It may be harder to construct such family of mappings that is non-constant in $ t$ for [b]every[/b] $ x\in\mathbb R$, not just for some.


Let $ a,b,c,d\ge 0$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}4$. Find the maximum value of:
\[ V\equal{}abcd\plus{}a^2bc\plus{}b^2cd\plus{}c^2da\plus{}d^2ab\]

the maximum value is $ + \ infty$    (i think)

$ V = abcd + a^2bc + b^2cd + c^2da + d^2ab$

by AM-GM    $ abcd\leq1$

so : $ a^2bc\leq\frac {a}{d}$....ect .

now we have to find the maximum value of :

$ H = 1 + \frac {a}{d} + \frac {b}{a} + \frac {c}{b} + \frac {d}{c}$

if $ a = min(a,b,c,d)$  then :

$ H\leq\1 + \frac{a}{a} + \frac {b}{a} + \frac {c}{a} + \frac {d}{a} = 1 + \frac {4}{a}$

and since the maximum value of $ f(x) = 1 + \frac {4}{x}$ in $ [0, + \infty[$ is $ + \infty$ then the maximum value of $ V$  is  $ + \infty$
 :)

[quote="mehdi cherif"]the maximum value is $ \plus{} \ infty$    (i think)[/quote]

what?  the maximum value is clearly at most $ 5\cdot 4^4$, and obviously nothing near that is attainable either.


Simplify the following:
$ \frac{x^3\minus{}x\plus{}60}{x^2\minus{}4x\plus{}15}$

Let's factor the numerator first. By the Factor Theorem, we have that $ x\plus{}4$ is a factor of $ x^3\minus{}x\plus{}60$. Use synthetic division to yield $ x^3\minus{}x\plus{}60\equal{}(x\plus{}4)(x^2\minus{}4x\plus{}15)$. And hey, we don't need to factor the denominator! Thus the fraction is simplified to $ x\plus{}4$.


geometry
incenter
exterior angle
geometry proposed

Let $ABCD$ be a convex quadrilateral, $I_1$ and $I_2$ be the incenters of triangles $ABC$ and $DBC$ respectively. The line $I_1I_2$ intersects the lines $AB$ and $DC$ at points $E$ and $F$ respectively. Given that $AB$ and $CD$ intersect in $P$, and $PE=PF$, prove that the points $A$, $B$, $C$, $D$ lie on a circle.

For this is the first time to post a reply in English, I'm so ashamed of my poor English.   :blush:

To prove that the points A, B, C, D lie on a circle, we can prove that <ABD=<ACD, the same as (<ABC - <DBC) / 2 = (<BCD - <BCA) / 2 ,
so we should prove that <I1BC -<I2BC = <I2CB -<I1CB ,
the same as <I1BI2 = <I1CI2 .
To prove this, now we begin to prove that B, C, I1, I2 lie on a circle:
actually,
<PEF = <PFE , <FCI2 = <BCI2 , <EBI1 = <CBI1 lead to that
<CBI1 + <I1I2C = <CBI1 + <FCI2 + <PFE
                         = <EBI1 + <BCI2 + <PEF
                         = <BCI2 + <I2I1B
so <CBI1 + <I1I2C = 180 :^o: ,
so B, C, I1, I2 lie on a circle,
so A, B, C, D lie on a circle.

Q.E.D

(I slightly change the notations.)

[i][b]Problem:[/b] Let ABCD be a convex quadrilateral, and let I and J be the incenters of triangles ABC and DBC, respectively. The line IJ intersects the lines AB and DC at the points E and F, respectively. Given that the lines AB and DC intersect at P, and PE = PF, prove that the points A, B, C, D are concyclic.[/i]

Since PE = PF, the triangle EPF is isosceles, and thus < PEF = (180° - < EPF) / 2. But by the exterior angle theorem applied to triangle BEI, we have < BIJ = < BEI + < EBI, hence

  < BIJ + < BCJ = (< BEI + < EBI) + < BCJ
           = (< PEF + < ABC / 2) + < BCD / 2
           = ((180° - < EPF) / 2 + < ABC / 2) + < BCD / 2
           = (180° - < EPF + < ABC + < BCD) / 2.

But 180° - < EPF = < 180° - < BPC = < PBC + < PCB (sum of angles in triangle BPC), and hence

  < BIJ + < BCJ = (< PBC + < PCB + < ABC + < BCD) / 2
           = ((< PBC + < ABC) + (< PCB + < BCD)) / 2
           = (180° + 180°) / 2 = 180°,

what entails that the quadrilateral BIJC is cyclic. Hence, < BIC = < BJC.

But since I is the incenter of triangle ABC, we have

  < BIC = 180° - < IBC - < ICB = 180° - < ABC / 2 - < ACB / 2
            = 180° - (< ABC + < ACB) / 2 = 180° - (180° - < BAC) / 2
            = 180° - (90° - < BAC / 2).

In other words, < BIC = 90° + < BAC / 2. Similarly, < BJC = 90° + < BDC / 2. Hence, < BIC = < BJC yields < BAC = < BDC. And thus, the points A, B, C, D lie on one circle.

  Darij

Triangle $PEF$ is isosceles, so $\angle PEF=\angle I_2I_1B-\angle EBI_1=\angle I_2I_1B-\angle I_1BC$ and $\angle PFE=\angle I_1I_2C-\angle FCI_2=\angle I_1I_2C-\angle I_2CB$. Since they are equal, $\angle I_2I_1B-\angle I_1BC=\angle I_1I_2C-\angle I_2CB$ or $\angle I_2I_1B+\angle I_2CB=\angle I_1BC+\angle I_1I_2C$. Since the sum of angles in quadrilateral is $360^\circ$ we have $\angle I_2I_1B+\angle I_2CB=180^\circ$ or quadrilateral $I_1I_2CB$ is cyclic. $\angle BI_1C=\angle BI_2C$ (same arc), so also $\angle CAB=\angle CDB$ and quadrilateral $ABCD$ is cyclic. :)


Define a sequence:

$a_0=0$ and $a_1=1$, and $4a_{n+2}-16a_{n+1}+15a_{n}=0$ for integers $n>1$

Prove that there are a finite number of integers in the sequence.

Solving characteristic equation and .....
$a_n = (\frac 5 2 )^n -(\frac 3 2 )^ n$ for all n = 0 ,1, 2, ....
Then I don't know how to proceed. :blush:

hmm, actually, i created this problem to show that explict function had a limited number on integer solutions, this is like a tst problemt that i have been working on, also i thought that the recrusion might serve better to use in induction to perhaps prove it, but i am not sure that will work so well

So you only have to show that the function becomes periodic at a certain point?

i believe that there are only 4 terms that are integers, and it does not seem to be periodic, the only thing that i see is a general increase in the power of 2 in the denominator, but i don't know how to prove that rigirously


What is the sum of the six positive integer factors of 18?

There is a formula. As I am short on time, I will just show you the solution and not explain it. \[ 18\equal{}2\cdot3^2\]Now\[ (2^0\plus{}2^1)(3^0\plus{}3^1\plus{}3^2)\equal{}3\cdot13\equal{}\boxed{39}\]

It's probably faster to just list out the factors... I know the formula.


Let $x;y;z$ be real numbers satisfying $x^{2}+y^{2}+z^{2}= 9$. Find max $A = xy^{3}+yz^{3}+zx^{3}$

[quote="chien than"][quote="svasao_qb111"]we have
 $(x^{2}+y^{2}+z^{2})^{2}\ge 3(x^{3}z+y^{3}x+z^{3}y)$  :D[/quote]
Can you prove this?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6026


In the counting game Bing-Bong, Arlene starts counting at 1,
but skips all multiples of 3 and all numbers that contain the
digit $ ``3''$. For example, Arlene counts: $ 1, 2, 4, 5, 7, 8, 10, 11,
14, 16, \dots$. What is the $ 40^\textrm{th}$ number in this sequence?

Let's go from 1 to 40 and see how many are skipped.
We skip 3, 6, 9, 12, 15, 18, 21, 24, 27, and 13, 23, and 30-39. This is 21 numbers skipped total.
Now let's go from 41 to 61 (21 numbers total) and see how many are skipped.
We skip 42, 45, 48, 51, 54, 57, and 60, and 43 and 53. This is 9 numbers total.
Now we go from 62 to 70 (9 numbers total).
We skip 63, 66, and 69. This is 3 numbers total.
Now we can count up until we get three more... 71, (72, 73), 74, (75), 76. The answer is 76.

IS there other method beside counting? It would take a good amount of time in game.


calculus
integration
limit
trigonometry
logarithms
inequalities
real analysis

Evaluate $\lim_{x\rightarrow 0^+}\int^{2x}_x\frac{\sin^m(t)}{t^n}dt$. ($m,n\in\mathbb{N}$)

First observation: if $m$ is odd, then $\int_0^{2\pi}\sin^mx\,dx=0,$ whereas if $m$ is even, then $\int_0^{2\pi}\sin^mx\,dx=2\pi b_m$, where $b_m$ is a postive number that we could compute if we needed to.

(In fact, $b_m= \frac{1\cdot3\cdots(m-1)}{2\cdot4\cdots m}.$)

Case 1: suppose $m$ is odd and $n\in\mathbb{N}.$

Let $F(x)=\int_0^{2\pi}\sin^mt\,dt.$ Note that $F$ is periodic and bounded. Let $M$ be a number such that $|F(x)|\le M$ for all $x.$

Integrate by parts.

$\int_x^{2x}\frac{\sin^mt}{t^n}\,dt=\frac{F(2x)}{(2x)^n}-\frac{F(x)}{x^n}+ \int_x^{2x}\frac{nF(t)}{t^{n+1}}\,dt$

Then $|\int_x^{2x}\frac{nF(t)}{t^{n+1}}\,dt|\le \int_x^{2x}\frac{nM}{t^{n+1}}\,dt \le \frac{M}{x^n}.$

As $x\to\infty,$ all parts of this tend to zero and the original limit is zero.

Case 2: suppose $m$ is even and $n\ge2.$

Let $G(x)=\int_0^x(\sin^mt-b_m)dt=\int_0^x\sin^mt\,dt-b_mx.$

We claim that $G(x)$ is bounded. In fact, $G(x)$ is continous and periodic.

Integrate by parts again:

$\int_x^{2x}\frac{\sin^mt}{t^n}\,dt=\frac{G(2x)}{(2x)^n}-\frac{G(x)}{x^n}+ \frac{2b_mx}{(2x)^n}-\frac{b_mx}{x^n}+n\int_x^{2x}\frac{G(t)+b_mt}{t^{n+1}}\,dt$

The boundary terms all tend to zero (note that we need $n\ge2$ to say that). Estimate the remaining integrals:

$|\int_x^{2x}\frac{n(G(t)}{t^{n+1}}\,dt|\le\frac{M}{x^n}$

$\int_x^{2x}\frac{nb_mt}{t^n}\,dt=-\frac{nb_m}{n-1}\frac1{t^{n-1}}|_x^{2x} \le\frac{nb_m}{n-1}\frac1_{x^{n-1}}$

Again, the original limit is zero. That leaves one more case.

Case 3: suppose $m$ is even and $n=1.$

The mechanics of the integration by parts are the same as in case 2.

$\int_x^{2x}\frac{\sin^mt}{t}\,dt=\frac{G(2x)}{2x}-\frac{G(x)}{x}+ \frac{2b_mx}{2x}-\frac{b_mx}{x}+\int_x^{2x}\frac{G(t)+b_mt}{t^2}\,dt$

We can argue, as before, that $\frac{G(2x)}{2x}-\frac{G(x)}{x}$ and $\int_x^{2x}\frac{G(t)}{t^2}\,dt$ tend to zero as $x\to\infty.$

That leaves $\frac{2b_mx}{2x}-\frac{b_mx}{x},$ which equals zero, and

$\int_x^{2x}\frac{b_mt}{t^2}\,dt=b_m(\ln(2x)-\ln x)=b_m\ln 2.$

So the original limit is zero except in the case $m$ even, $n=1$, in which case it is

$\frac{1\cdot3\cdots(m-1)}{2\cdot4\cdots m}\cdot\ln2.$

Kent, in the problem we are supposed to take the limit as $ x\to 0^\plus{}$.

First of all, we have,
\[ \forall m\neq n\plus{}1,\, t\in(0,1), \,\, \frac2\pi t\le \sin(t) \le t \\ \Longrightarrow \forall x\in (0, \frac12) \,\, \,\,\, 0 \le \frac2\pi \int_{x}^{2x}t^{m\minus{}n} dt \le \int_{x}^{2x} \frac{\sin(t)^m}{t^n}dt \le \int_{x}^{2x} t^{m\minus{}n}dt\]

So by the squeeze theorem,
\[ \lim_{x\to 0^\plus{}}\int_{x}^{2x} \frac{\sin(t)^m}{t^n}dt \equal{} \begin{cases}
0 & \text{if } m\ge n \\
\infty & \text{if } n >m\plus{}1
\end{cases}\]
It remains to deal with the case $ n\equal{}m\plus{}1$: For this we use the inequalities $ x\minus{}\frac{x^3}{6}\le \sin(x) \le x$:

\[ \frac{(t\minus{}\frac{t^3}{6})^m}{t^{m\plus{}1}}\equal{}\frac1t \minus{} o(t) \equal{} \le \frac{\sin(t)^m}{t^{m\plus{}1}}\le \frac{t^m}{t^{m\plus{}1}}\equal{}\frac1t\]Integrating and again squeezing, we get $ \ln 2$. Hence

\[ \lim_{x\to 0^\plus{}}\int_{x}^{2x} \frac{\sin(t)^m}{t^n}dt \equal{} \begin{cases}
0 & \text{if } m\ge n \\
\ln 2 & \text{if } n\equal{}m\plus{}1 \\
\infty & \text{if } n >m\plus{}1
\end{cases}\]


geometry
rectangle
ratio
analytic geometry

Points A(0, 0), B(6, 0), C(6, 10) and D(0, 10) are vertices of
rectangle ABCD, and E is on segment CD at (2, 10). What is
the ratio of the area of triangle ADE to the area of quadrilateral
ABCE? Express your answer as a common fraction.

the area of rectangle ABCD is 6*10=60.

and area of the triangle ADE is 2*10/2=10.

thus the area of quadrilateral ABCE is 50.

so the ratio of the area of triangle ADE to the area of quadrilateral
ABCE is $ \frac{10}{50}$, which is $ \frac{1}{5}$.

The area of the rectangle is 60, and triangle ADE's area is 2*10/2=10, so quadrilateral ABCE's area is 50. The ratio of [ADE] to [ABCE] is 1/5

In case you don't know, to find the length of a side when you have the coordinates, like in this problem, you can use the Distance Formula:

$ \sqrt{(y_2 \minus{} y_1)^2 \plus{} (x_2 \minus{} x_1)^2}$


function
real analysis
real analysis unsolved

Suppose that $\mu ({\rm{X)}}< \infty$ and ${\rm{f}}:{\rm{X}}\times [{\rm{0}},{\rm{1}}] \to{\rm{C}}$ is a function such that ${\rm{f}}( \cdot ,{\rm{y}})$ is measurable for each ${\rm{y}}\in [{\rm{0}},{\rm{1}}]$ and ${\rm{f}}({\rm{x}}, \cdot )$ is continuous for each ${\rm{x}}\in{\rm{X}}$.
Then for any $\varepsilon >{\rm{0}}$ there is a set ${\rm{E}}\subset{\rm{X}}$ such that $\mu ({\rm{E}}) < \varepsilon$ and ${\rm{f}}( \cdot ,{\rm{y}}) \to{\rm{f}}( \cdot ,{\rm{0}})$ uniformly on ${\rm{E}}^{\rm{c}}$ as ${\rm{y}}\to{\rm{0}}$.

Consider, for every positive integer $N$,  a sequence $(a_{n,N})_{n}$ of elements $a_{n,N}\in [0,1]$ converging to $0$ and such that $\forall x \in [0,1], \exists n \in \mathbb{N}, |a_{n,N}-x| \leq \frac{1}{N}$.
Fix $\epsilon>0$. For all $N>0$ you can find, by Egoroff's theorem, a subset $E_{N}$ of $X$ such that $m(E_{N}) \leq \epsilon$ and $(f_{n,N})$ converges uniformly to $f$ on $X-E_{N}$ (*) (with respect to $n$; we defined $f_{n,N}(x)=f(x,a_{n,N})$). Consider $E = \cap E_{N}$. Clearly $m(E) \leq \epsilon$, and, using  (*), the uniform continuity of $f(x,.)$ on $[0,1]$ and the density of $\{a_{n,N}: n,N >0\}$, the result follows by using a triangle inequality.


If $ x\plus{}y\equal{}z\plus{}t$ prove that $ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}t^{2}$  is expressed as a sum of 3 perfect sqare's.  :)

$ (x\plus{}y)^{2}\plus{}(y\minus{}t)^{2}\plus{}(y\minus{}z)^{2}$ should do it.


trigonometry
algebra
polynomial
number theory proposed
number theory

Let $m,n$ be the distinct integers. Find the common roots of equations:
$x^{n+1}-x^m+1=0$ and $x^{m+1}-x^n+1=0$

[hide=" my solution"]
$x^{n+1}-x^{m}+1=0$ and $x^{m+1}-x^{n}+1=0$ so $(x^{n+1}-x^{m}+1)-(x^{m+1}-x^{n}+1)=0$ so
$x^n(x+1)-x^m(x+1)=0 <=>(x+1)x^n(x^{m-n}-1)=0$ but clearly $x=-1$ and $x=0$ don't verify the first two equalities so we have $x^{m-n}=1$
Now we must also have
$x^{n+1}-x^{m}+1=0<=>x^m(x\cdot x^{n-m}-1)=-1$ but $x^{n-m}=1$ so we have $x^m(x-1)=-1=>|x^m|\cdot |x-1|=1$ but $x^{m-n}=1$ so $|x|=1=> |x-1|=1$ so let $x=a+bi => a^2 +b^2=1$ and $ (a-1)^2 +b^2=1$ so $a=0$ so $x=i$ or $x=-i$ which don't verify so there are no solutions.
i hope this is right :?   :D [/hide]

Your solution is not correct , the two equations [u]have common roots[/u] ,let's to find it  ;)

yes sorry i made a really stupid mistake  :blush:  :oops:
now if $(a-1)^2+b^2=1$ and $a^2+b^2=1$ then $1-2a=0$ so $a =\frac 1 2$ not $a=0$ :?  and then
we have $x= \cos \frac{\pi}{3}+i\sin \frac{\pi}{3}$ or  $x= \cos \frac{5\pi}{3}+i\sin \frac{5\pi}{3}$
so $x$ is a root of the polinomial $x^2-x +1=0$ so $x^6=1$ but $x^{m-n}=1$ so $6|m-n$ and then analising the cases we get that if $m=n=1(mod 6)$ then the common roots are given by the polinomial $x^2-x+1=0$ and in the other cases there is no solution
i hope this i right now ;)


function
limit
real analysis
real analysis unsolved

Let f: R -> R be a continuous function, and suppose that n > or = 1 is an integer such that lim_x->00_(f(x)/x^n) = lim_x->-00_(f(x)/x^n) = 0. By setting g(x) = x^n + f(x), we define a function g: R -> R.

How do you show that if n is even, then there exists a real number c such that the image of g is [c,00).

For $x\ne0,\ g(x)=x^{n}+f(x)=x^{n}\left(1+\frac{f(x)}{x^{n}}\right).$

For sufficiently large $|x|,\ 1+\frac{f(x)}{x^{n}}>\frac12$ and hence, for sufficiently large $|x|,\ g(x)>\frac{x^{n}}2.$

It follows that $\lim_{x\to\infty}g(x)=\lim_{x\to-\infty}g(x)=\infty.$

Now let $b=g(0).$ (We just need one value of the function, somewhere; this is arbitrary). Find $M$ such that for $|x|>M,\ g(x)>b.$ We can do this by the limit in the previous paragraph.

Consider $g$ on the set $[-M,M].$ A continuous function on a closed bounded interval is bounded above and below and assumes its maximum and its minimum. Let $c$ be the minimum value that $g$ takes on on this interval, and let $a$ be the point such that $g(a)=c.$ Since $g(x)>b\ge c$ outside of this interval, we have that $c$ is the global minimum value.

Now suppose $y\in(c,\infty).$ Since $\lim_{x\to\infty}g(x)=\infty,$ we can find $d>a$ such that $g(d)>y.$ Now we have that $g(a)<y, g(d)>d$ and $g$ is continuous on $[a,d],$ then there exists $x\in(a,b)$ such that $g(x)=y.$

We have shown that the image of $g$ is $[c,\infty).$

Note that we have specifically quoted both the Extreme Value Theorem and the Intermediate Value Theorem.


Problem 24 on the 1986 AHSME:

Let $ p(x)\equal{}x^2\plus{}bx\plus{}c$, where $ b$ and $ c$ are integers. If $ p(x)$ is a factor of both $ x^4\plus{}6x^2\plus{}25$ and $ 3x^4\plus{}4x^2\plus{}28x\plus{}5$, what is $ p(1)$?

(A) 0
(B) 1
(C) 2
(D) 4
(E) 8

So we have $ p(x)$ is a factor of $ x^4 \plus{} 6x^2 \plus{} 25$, and $ 3x^4 \plus{} 4x^2 \plus{} 28x \plus{} 5$.
  $ p(x) \equal{} (2x^4 \minus{} 2x^2 \plus{} 28x \minus{} 20, x^4 \plus{} 6x^2 \plus{} 25) \implies (x^4 \plus{} 8x^2 \plus{} 28x \plus{} 55, x^4 \plus{} 6x^2 \plus{} 25) \implies p(x) \equal{} 2x^2 \plus{} 28x \minus{} 30 \implies p(1) \equal{} 0 \boxed{A}$

Now, we have the polynomial has to be a factor of (1 of the second polynomial)-3*(the first polynomial)=

$ \minus{} 14x^2 \plus{} 28x \minus{} 70$ so $ x^2 \minus{} 2x \plus{} 5$ has to be the factor. So the answer is D.

You could also try to factor directly --

[hide="Solution"]$ x^4 \plus{} 6 x^2 \plus{} 25 \equal{} (x^4 \plus{} 10 x^2 \plus{} 25) \minus{} (4 x^2)$ $ \equal{} (x^2 \plus{} 5)^2 \minus{} (2 x)^2 \equal{} (x^2 \plus{} 2x \plus{} 5) (x^2 \minus{} 2x \plus{} 5)$

So now, we can try out these two possibilities of $ p(x)$, and find that $ 3 x^4 \plus{} 4 x^2 \plus{} 28 x \plus{} 5 \equal{} (3x^2 \plus{} 6x \plus{} 1) (x^2 \minus{} 2x \plus{} 5)$

So, $ p(x) \equal{} x^2 \minus{} 2 x \plus{} 5 \implies \boxed{\textbf{(D)} \; 4}$[/hide]


calculus
integration
algebra
function
domain
abstract algebra
Ring Theory

R an integral domain with quotient field Q
N any R-module
QtensorN over R the module obtained from N by extension of scalars from R to Q
i: N-->QtensorN an R-module homomorphism

To show: Ker(i) = Tor(N)={n in N: rn=0 for some r in R, r not 0} (the torsion submodule of N)

it's easy to show that Tor(N) is a subset of Ker(i)
but i can't seem to show that Ker(i) is a subset of Tor(N).

if n in Ker(i), then i(n)=0.
so a sum of simple tensors, (q1 tensor n1) + ... + (qk tensor nk) = i(n) = 0
but how does i(n) relate to n? I need n somewhere so I can show there exists r in R (r not 0) such that rn=0.

obviously the integral domain thing is important, but I don't know to use it. i would appreciate any help anyone could give me.

this follows from a more general (nevertheless, easy) statement:
$ S^{ \minus{} 1} M \cong M \otimes_R S^{ \minus{} 1} R$
here $ R$ is a ring, $ S$ a multiplicative closed subset of $ R$ and $ M$ a $ R$-module. from this we optain
$ ker(M \to M \otimes_R S^{ \minus{} 1} R) \equal{} \{m \in M : \exists s \in S : sm \equal{} 0\}$

thanks.  it will take me a while to understand, but i don't doubt that you're right.


geometry
circumcircle
geometric transformation
dilation
analytic geometry
Euler
area of a triangle

Given a triangle $ ABC$, with $ H$ is the orthocenter and $ O$ is the incenter. Prove that:
$ \frac {1}{3}OH \equal{} \sqrt {(R \plus{} \frac {4}{9}r)^2 \plus{} \frac {2}{81}(r \minus{} 3p)(r \plus{} 3p)}$
where $ R$ is the radius of the circumcircle and $ r$ is the radius of the incircle of the triangle $ ABC$; $ 2p \equal{} a \plus{} b \plus{} c$ ($ a,b,c$ are three sides of the triangle $ ABC$)

That's an ugly expression.  Your best bet might just be to bash it with complex geometry, which would make it not too horrible.  Take $ O$ as the origin of the plane; then you have $ |a| \equal{} |b| \equal{} |c| \equal{} R$.  From there you're on your own, although you might be able to find a nice expression for $ H$ by considering that $ H$ is the circumcenter of the triangle $ XYZ$ which has $ ABC$ as its medial triangle.  This could be accomplished via a dilation about $ G$, which has complex coordinate $ \frac{a\plus{}b\plus{}c}{3}$.

Teki-Teki: Using the properties of the Euler line, you can determine $ H$ easily: $ OH \equal{} 3GH \equal{} 3 \frac{a\plus{}b\plus{}c}{3} \equal{} a\plus{}b\plus{}c$.

My question is how to deal with the inradius.

[quote="MellowMelon"]Teki-Teki: Using the properties of the Euler line, you can determine $ H$ easily: $ OH \equal{} 3GH \equal{} 3 \frac {a \plus{} b \plus{} c}{3} \equal{} a \plus{} b \plus{} c$.

My question is how to deal with the inradius.[/quote]

A brute-forced way of dealing with that is using $ A \equal{} rs,$ where $ s \equal{} p$ in this case, and substituting Heron's formula for $ A.$ But this gets very ugly very fast.

[quote="Teki-Teki"]That's an ugly expression.[/quote]
I just want to make it more difficult by using complex coefficents :D
Can anyone give a detailed solution to it? When you have dealt with the coefficents, everything is clear and you will find the problem a simple one!


inequalities
geometry
3D geometry
tetrahedron
trigonometry
geometry unsolved

Given a tetrahedron such that product of the opposite edges is $ 1$. Let the angle between the opposite edges be $ \alpha$, $ \beta$, $ \gamma$, and circumradii of four faces be $ R_1$, $ R_2$, $ R_3$, $ R_4$. Prove that
\[ \sin^2\alpha \plus{} \sin^2\beta \plus{} \sin^2\gamma\ge\frac {1}{\sqrt {R_1R_2R_3R_4}}
\]

$ \mbox{Let tetrahedron } A_1A_2A_3A_4. \mbox{If denote } a_{ij} \equal{} |A_iA_j|; i\ne j \\
\mbox{and } \alpha , \beta, \gamma \mbox{ angle between the opposite edges } A_1A_2 \mbox{ and } A_3A_4, A_1A_3 \mbox{ and } A_2A_4, A_1A_4 \mbox{and } A_2A_3, \mbox{ respectively, that: } \\
4.(S_1^2 \plus{} S_2^2 \plus{} S_3^2 \plus{} S_4^2) \equal{} (a_{12}.a_{34}.\sin\alpha)^2 \plus{} (a_{13}.a_{24}.\sin \beta)^2 \plus{} (a_{14}.a_{23}.\sin\gamma)^2; \ \ \ (1) \\
\mbox{and if: } a_{12}.a_{34} \equal{} a_{13}.a_{24} \equal{} a_{14}.a_{23} \equal{} 1 \mbox{, then we: } \\
\sin^2\alpha \plus{} \sin^2 \beta \plus{} \sin^2\gamma \equal{} 4.(S_1^2 \plus{} S_2^2 \plus{} S_3^2 \plus{} S_4^2) \equal{} \\
\equal{} \frac {1}{4}.\left[\left( \frac {a_{23}a_{24}a_{34}}{R_1}\right)^2 \plus{} \left( \frac {a_{13}a_{14}a_{34}}{R_2}\right)^2 \plus{} \left( \frac {a_{12}a_{14}a_{24}}{R_3}\right)^2 \plus{} \left( \frac {a_{12}a_{13}a_{23}}{R_4}\right)^2 \right]\ge \\
\ge \sqrt [4]{\left( \frac {a_{23}a_{24}a_{34}}{R_1}\right)^2 .\left( \frac {a_{13}a_{14}a_{34}}{R_2}\right)^2. \left( \frac {a_{12}a_{14}a_{24}}{R_3}\right)^2 .\left( \frac {a_{12}a_{13}a_{23}}{R_4}\right)^2} \equal{} \\
\equal{} \sqrt {\frac {\left(a_{12}a_{13}a_{14}a_{23}a_{24}a_{34}\right)^2}{R_1R_2R_3R_4}} \equal{} \frac {1}{\sqrt {R_1R_2R_3R_4}}.$
[b]HINT(for prroof (1)): [/b]
$ \mbox{If denote: } \overrightarrow{a_{ij}} \equal{} \overrightarrow{A_iA_j}, \mbox{ then: }2.S_3\equal{}|\overrightarrow{a_{12}}\times\overrightarrow{a_{14}}|,\dots \mbox{ and: } \\
a_{12}.a_{34}.\sin\alpha \equal{} |\overrightarrow{a_{12}}\times \overrightarrow{a_{34}}| \equal{} |\overrightarrow{a_{12}}\times (\overrightarrow{a_{14}} \minus{} \overrightarrow{a_{13}})| \equal{} |\overrightarrow{a_{12}}\times \overrightarrow{a_{14}} \minus{} \overrightarrow{a_{12}}\times \overrightarrow{a_{13}})| \equal{} \dots$


function
calculus
integration
induction
Functional Equations

The function $f: \mathbb{N}\to\mathbb{N}_{0}$ satisfies for all $m,n\in\mathbb{N}$: \[f(m+n)-f(m)-f(n)=0\text{ or }1, \; f(2)=0, \; f(3)>0, \; \text{ and }f(9999)=3333.\] Determine $f(1982)$.

[quote="Peter"]The function $ f: \mathbb{N}\to\mathbb{N}_{0}$ satisfies for all $ m,n\in\mathbb{N}$:
\[ f(m + n) - f(m) - f(n) = 0\text{ or }1, \; f(2) = 0, \; f(3) > 0, \; \text{ and }f(9999) = 3333.
\]
Determine $ f(1982)$.[/quote]
We show that $ f(n) = [\frac{n}{3}]$ for $ n \leq 9999$, where$ [ ]$ denotes the integral part.

We show first that $ f(3) = 1. f(1)$ must be $ 0$, otherwise $ f(2) - f(1) - f(1)$ would be negative. Hence $ f(3) = f(2) + f(1) + 0$ or $ 1 = 0$ or $ 1$. But we are told $ f(3) > 0$, so $ f(3) = 1$. It follows by induction that $ f(3n) \geq n$. For $ f(3n+3) = f(3) + f(3n) + 0$ or $ 1 = f(3n) + 1$ or $ 2$. Moreover if we ever get $ f(3n) > n$, then the same argument shows that $ f(3m) > m$ for all $ m > n$. But $ f(3.3333) = 3333$, so $ f(3n) = n$ for all $ n \leq 3333$.

Now $ f(3n+1) = f(3n) + f(1) + 0$ or $ 1 = n$ or $ n + 1$. But $ 3n+1 = f(9n+3) ≥ f(6n+2) + f(3n+1) ≥ 3f(3n+1)$, so $ f(3n+1) < n+1$. Hence $ f(3n+1) = n$. Similarly, $ f(3n+2) = n$. In particular $ f(1982) = 660$

[hide=Solution] It is trivial that $f(1)=0$ and $f(3)=1.$
If $f(3x)\geq x,$ then $$f(3x+3)\geq f(3x)+f(3)\geq x+1.$$
It can be trivially shown (by induction) that $f(3x)\geq x ~~\forall x\in \mathbb{N}.$
We can note that the inequality is strict for all integers greater than $x$ if it is strict for $x.$ It follows that $$f(9999)=3333\implies f(3x)=x ~~\forall x\leq 3333.$$
\begin{align*} \text{Now }3f(x)\leq f(3x) \leq 3f(x)+2 \\ \implies f(x)=\lfloor f(3x)/3 \rfloor =\lfloor x/3\rfloor ~~\forall x\leq 3333 \\ \implies f(1982)=660.\end{align*}[/hide]


inequalities
geometry
perimeter
inequalities unsolved

Prove that $ p\le \sqrt \frac{3(m_a\plus{}m_b\plus{}m_c)} {2R}$.

Is $ p$ the perimeter or semiperimeter? (I have seen it used as the semiperimeter...)

Anyway, I think we can use $ m_a = \dfrac{\sqrt{2b^2+2c^2-a^2}}{2}$, etc., and $ R=\dfrac{abc}{\sqrt{s(s-a)(s-b)(s-c)}$ and do some brute force :D

[quote="SonicX"]Prove that $ p\le \sqrt \frac {3(m_a \plus{} m_b \plus{} m_c)} {2R}$.[/quote]
Try $ a\equal{}b\equal{}c\equal{}2.$ :wink:

[quote]Is  the perimeter or semiperimeter?[/quote]
p is the semiperimeter

I would like to see a complete proof.  :D

SonicX your problem is incorrect,as arqady said for $ a\equal{}b\equal{}c\equal{}2$ your triangle is equilateral with $ p\equal{}3,R\equal{}\frac{2}{\sqrt{3}},m_a\equal{}m_b\equal{}m_c\equal{}\sqrt{3}$ hence according to your inequality we must have:

$ 3\leq\sqrt{\frac{3(3\sqrt{3})}{2\times \frac{2}{\sqrt{3}}}}$

$ \iff 2\leq \sqrt{3}$

which is clearly wrong...

i've always thought this problem is wrong, because the lhs has degree 1 and the rhs has degree 0, so the inequality doesn't make sense... does anybody have the correct statement?  :maybe:


Determine all positive integers k such that the sequence $r_{n}$ defined as $\binom{2n}{n}\equiv r_{n}(mod k)$ is periodic,$n \geq 1.$

By Lucas theorem, ${2n\choose n}$ is not divisible by prime $p$ if and only if all base-p digits of $n$ are less than $\frac{p}{2}.$

Let prime $p$ divide $k$ for which ${2n\choose n}\bmod k$ is periodic. Then ${2n\choose n}\bmod p$ is periodic as well. We will show that ${2n\choose n}\bmod p = 0$ for all $n\geq 1.$

Assume that there is $n_{0}$ such that ${2n_{0}\choose n_{0}}\bmod p\ne 0.$ If $t$ is a period of ${2n\choose n}\bmod p$ then ${2(n_{0}+m\cdot t)\choose n_{0}+m\cdot t}\bmod p ={2n_{0}\choose n_{0}}\bmod p\ne 0$ for all $m.$ Consider the base-p representation of $t$ and let $j$ be the smallest position at which it has a non-zero digit, which we denote $d$. Then the $j$-th base-p digit of $n_{0}+m\cdot t$ equals the $j$-th base-p digit of $n_{0}$ plus $m\cdot d$ modulo $p,$ and for some $m=m_{0}$ it will be at least $\frac{p}{2}$ (actually, it will take all possible values from $0$ to $p-1$ as $m$ grows). But then ${2(n_{0}+m_{0}\cdot t)\choose n_{0}+m_{0}\cdot t}\bmod p = 0,$ a contradiction.

Therefore, all prime divisors of $k$ must divide ${2n\choose n}$ for all $n\geq 1.$ In particular, they must divide ${2\choose 1}=2.$ Therefore, the only possible values of $k$ are powers of $2.$ But it is easy to see that ${2n\choose n}\bmod 4=0$ for $n>8$ and hence ${2n\choose n}\bmod 4$ is not periodic. Therefore, $k=1$ and $k=2$ are the only solutions and the sequence ${2n\choose n}\bmod k$ is simply zero sequence with the period $1$.


trigonometry
complex numbers
algebra open
algebra

I was wondering about the algebraic (that is, non-transcendental) outputs of $\sin \theta$ and $\cos \theta$.  For example, the unusual value $\cos (\frac{\pi}{5}) = \frac{\phi}{2}$.  Are there any other such values (aside from the well-known and trivial ones)?

Or, phrased differently, for what $q \in \mathbb{Q}$ is $\sin q\pi$ (and/or $\cos q\pi$) algebraic?

Or, phrased yet another way, given that $\sin \theta$ is algebraic, can it be shown that $\frac{\theta}{\pi} \in \mathbb{Q}$?

Indeed, $\sin(r \pi)$ is algebraical for all rationals $r$ (it's the imaginary part of a root of unity).
But the converse is not true, since e.g. $\sin(r \pi) = \frac{1}{3}$ implies $r$ to be not rational (see http://www.mathlinks.ro/Forum/viewtopic.php?t=61379 )


I'd be willing to host one sometime this week, maybe Wednesday/Thursday/Friday.Is anyone interested? If so, please post your sn and what time.
Rules would be like Klebian's.

My AIM is Kaoticity. I'd prefer it be on Wednesday or Thursday past 6:30EST.

My AIM screen name is my AoPS screen name. Wednesday afternoon 6:30 EST here too.

I don't have Yahoo Instant Messenger. :dry:

Alright guys, I'll wait for a few more posts before deciding.

Sweet! I'd like to attend.
Thursday's really good for me, but I'll check if I can make it wednesday evening.
And I'm pretty sure you know my AIM screen name, and I don't have YIM.  :)

I'll come. My AIM is my AoPS screenname with a Z at the end.

u kno my aim sn.
anyways try to make it on a weekend.
If not there's a chance i won't b there.

I'll do it. I dont have YIM and my AIM is Ddavidw111.

i would do AIM
you shoudl know my screenname :dry:

Looks like it's between wednesday and thusday...anyone else prefer any dates?
On wednesday, it'll have to start maybe around 7-8 EST though, I have a MathCounts practice. Sorry guys. :(

i would say wednesday from 8-9 EST would be my best time...

Wednesday any time between 7-10 est is good with me.  :)

Looks like I'll hold two math jams:
tomorrow at 8 EST (only time i have)
and thursday (any time in the evening).
Come online 10 minutes before to confirm, to read rules, and blahblahblah.

Scratch what I just said about mathcounts, I'll be opening a chatroom around 7 EST and I'll keep it open till about 9 EST.
Sorry about the short notice.

I'll do it today.  My sn is
insanemorgoth

[quote="biffanddoc"]i hope its 800 EST ... :dry: :nhl:[/quote]
it's 7 est. Starting about now.

Well, I didn't make it -- didn't think it'd be today.

For tomorrow, the only time I have is 6-7 EST. I'll try to host another this weekend after my state comp.


I may have missed it, but it seems there are no resources for Mandelbrot problems/solutions on the AoPS Wiki.  Is it possible for us to add Mandelbrot material? (if its already there, link please?)

Well, the wiki is still very new.  If you poke around it long enough you will find a lot of stuff is missing.

On the other hand, we'd have to get approval to post mandelbrot's in full.  Generally, we have the OK to post AMCs, AIME, national olympiads, and IMO.  But I believe  ARML isn't too keen on having their problems posted in full so we leave those alone.

By the way, you can add to the wiki whenever you like (that's the whole idea of a wiki!) just as long as you permission to post what you are posting.

I think Mandelbrots would not be good to post on the Wiki.  I remember a while back I was looking for old Mandelbrots, and I couldn't find more than one.  I asked the Mandelbrot people, and they pointed me to books I could buy (now available on AoPS!!).  So they probably don't want people posting the problem sets in full...

We can't speak for the Sam Vandervelde, so you'd have to go about getting his permission.

for the earlier mandelbrots...is that necessary?

I think all you need to do is sign up and you can post and edit


algebra
polynomial
AMC
USA(J)MO
USAMO
algebra unsolved

$P(x)$ a polynomial with integer co-efficints.

if there exists an integer $x$ such that:

$P(P(P(P(x))))=x$

Then show

$P(P(x))=x$

Hm, from $P(P(P(P(x))))=x$ it follows $P(P(P(x)))=P^{-1}(x)$, which means that $P^{-1}(x)$ is a polynomial, and that's possible only if $\deg P(x)$ is $0$ or $1$. Now it should be easy.

Apologies. I made a boo boo in the statement of the problem. (Thats how i did it when i first read it!) Ill edit it.

Here's a hint: suppose $P(x)=y\ne x$. We  must show that $P(y)=x$. Suppose $P(y)=z$ and $P(z)=t$; then $P(t)=x$. Now use the fact that for all integers $a,b$, $a-b$ divides $P(a)-P(b)$ to obtain a contradiction.


AMC USA(J)MO USAMO inequalities FTW AIME counting

Since Wanna Make Nats? failed (and it didn't have a team round), I'm making a [b]full[/b] competition, which includes Chapter, State, and National tests, each complete with Sprint, Target, and Team rounds.  This means that you will need to group yourselves in teams of four.  I will be making a sign-up thread now, sign-up and choose teams there.

Chapter will come out on Saturday or Sunday, in this thread, and it will be due January 25 at 11:59 pm EST.  I will decide who will move on based on the rank (obviously) and the number of people.  Good luck!

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=250805]Sign up here![/url]

I join.

The message is too small. Please make the message longer before submitting.

[quote="AIME15"]Join, dafjbomb hopefully joins, pythag011 hopefully joins, Bugi hopefully joins, I want them all on my team :)[/quote]

Mieh, sure.

I hope that probability1.01 or MysticTerminator joins. Probably not.

(Please join in the Sign-Ups thread now.  I don't want this to get too cluttered.)

cough COUGH HACK CHOKE DIEEEEEEEEEEEEEEEEEEEEE cough

ernie seemed rly excited about this, so I am [u][i][b]NOT[/b][/i][/u] joining...

Meh.  I'm not on past 10 pm, FYI. :P

Test coming out later today.

Without further adieu... it's the Sprint Round!

I don't have much time to type up the Target and Team rounds though, but busy yourself with the Sprint.

Any and all questions must be PM-ed.

OK, you are on Team :D.

This test will probably be due on February 7, giving you exactly 14 days.

Several issues:

#6- I typoed, I meant less than 170.

#26- It is assumed that a > b > c > d.

#27- I want the point in the first quadrant.

Thanks, 5846209328x!

Who are we supposed to send our answers to?

You?

[quote="theprodigy"]hold on i'm going to join RIGHT NOW :D[/quote]
What? theprodigy, get an  :D avatar if you wanna be on my team! PLEASE!

Yah...

Also, for #1, we have two 100's, not 3.

Guys on my team: Better submit your answers soon or we will lose!!!

wait how am i not on that list of scores :mad:

EDIT:do we put the answer to #5 in terms of pi or what

[u]Sprint Round: [/u]

gauss1181 (Team :D)- 25
dafjbomb (Teh Coordinate Bashers)- 25
12markkram34 (JUST PWN IT!)- 24
Meta_Knight (Teh Coordinate Bashers)- 24
xpmath (Individual)- 24
AIME15 (Teh Coordinate Bashers)- 22
FantasyLover (The |\/|/-\-|-|-|13375)- 22
BOGTRO (The |\/|/-\-|-|-|-|13375)- 22
r15s11z55y89w21 (Team Winners)- 21
theprodigy (Individual)- 21
mz94 (The |\/|/-\-|-|-|13375)- score to be determined later using random.org
ghjk (Team F)- pending two problems
007math (Team Winners)- score asked to be hidden
isabella2296 (The |\/|/-\-|-|-|13375)- score asked to be hidden
suma_mili (Team :D)- score asked to be hidden

[u]Target Round:[/u]

None

[u]Losers that Need to Submit the Sprint [b]and[/b] Target Rounds: [/u]

AIME_is_hard (Team :D)
athunder (JUST PWN IT!)
Discrete_Math (Team :D)
infinitypi (Team Winners)
jjx1 (Team Winners)
limabean (JUST PWN IT!)
nikeballa96 (JUST PWN IT!)
pythag011 (Teh Coordinate Bashers)
SuperMathBoy (Team F)
Twin Prime Conjecture (Team F)
weird888 (Team F)

[u]Team Round:[/u]

None

[u]Team Ranks:[/u]

1 The |\/|/-\-|-|-|13375
2 Teh Coordinate Bashers
3 Team :D
4 Team Winners
5 JUST PWN IT!
6 Team F

[quote="5849206328x"]I seem to have noticed that there isn't a deadline for the target round?[/quote]

It's the same.

OK, for #4, those four points are the vertices...

Well, people like [u]you[/u] who [b]don't submit[/b] have made this tourney simply [i]inactive[/i].

How long I have to wait more to get the official solution? Did you fix my score,ernie??

ERNIE I SUBMITTED MY TARGET ANSWERS A LONG TIME AGO!!!!

Did you receive them????? i've been waiting long for my score!

[quote="ernie"]You know what?

Screw the Chapter test, I'm making a Nats test. :)

I need help with the Master's Round though.  Pythag011? :P[/quote]

For some reason, everyone asks me to help with the master's round...

Fine.

Darn inactives! We need to ban those who are constantly inactive.


geometry
3D geometry
sphere
trigonometry
analytic geometry
Euler

A thin rod with length l, has a sphere in the end of it with mass m. In a "a" distance from the point where it's hanged it's connected with a spring with koefficient k. Write the Lagrange equation for the small waverings of the system.

How is the system positioned in the gravitational field? Is the rod initially vertical, horizontal, or at some other angle with the horizontal?

here is an illustration for the problem.
[url=http://img104.imageshack.us/my.php?image=fizikefn2.jpg][img]http://img104.imageshack.us/img104/9852/fizikefn2.th.jpg[/img][/url]

We'll suppose that when the rod is vertical the string is unstretched. Let $\phi$ be the angle the rod makes with the vertical. This will be our generalized coordinate.

The elongation of the spring is $x_{0}= a\sin \phi$, and the coordinates of the weight at the end of the rod are $x = l\sin\phi$ and $y = l(1-\cos\phi)$.
The potential energy is $V = kx_{0}^{2}/2+mgy = ka^{2}\sin^{2}\phi / 2+mgl(1-\cos\phi)$.

The velocities of the weight are $\dot x = l\dot\phi \cos \phi$ and $\dot y = l \dot \phi \sin\phi$, so the kinetic energy is $K = ml^{2}\dot\phi^{2}/2$.

The langrangian is $L = ml^{2}\dot\phi^{2}/2-ka^{2}\sin^{2}\phi / 2-mgl(1-\cos\phi)$. The Euler-Lagrange equation is
\[\frac{\text d}{\text dt}\left(\frac{\partial L}{\partial \dot\phi}\right) = \frac{\partial L}{\partial \phi},\]
so
\[ml^{2}\ddot \phi =-ka^{2}\sin\phi\cos\phi-mgl\sin\phi.\]
This is the E-L equation. By using the small-angle approximations it becomes
\[ml^{2}\ddot \phi =-(ka^{2}+mgl)\phi,\]
so the period of small oscillations is
\[T = 2\pi \sqrt{\frac{ml^{2}}{ka^{2}+mgl}}.\]


function
induction
algebra unsolved
algebra

Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that \[ f(x\plus{}f(xy))\equal{}f(x)\plus{}xf(y)\quad \forall x,y\in\mathbb{R}.\]

[quote="N.T.TUAN"]Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that
\[ f(x \plus{} f(xy)) \equal{} f(x) \plus{} xf(y)\,\forall x,y\in\mathbb{R}.
\]
[/quote]

Let $ P(x,y)$ be the property $ f(x \plus{} f(xy)) \equal{} f(x) \plus{} xf(y)$

$ 1)$ Assume it exists $ u\neq 0$ such that $ f(u)\equal{}0$.
Let then $ v\neq 0$. $ P(\frac{u}{v},v)$ gives $ f(\frac{u}{v} \plus{} f(u)) \equal{} f(\frac{u}{v}) \plus{} \frac{u}{v}f(v)$ and so $ 0\equal{}\frac{u}{v}f(v)$ and so $ f(v)\equal{}0$ and $ f(x)\equal{}0$ $ \forall x \neq 0$.
Then, if $ f(0)\equal{}a\neq 0$, $ P(a,0)$ gives $ f(2a)\equal{}f(a)\plus{}af(0)$ and so $ f(0)\equal{}0$ and contradiction.
So, if it exists $ u\neq 0$ such that $ f(u)\equal{}0$, then $ f(x)\equal{}0$ $ \forall x$ and this is a first solution to the problem.

Now, we'll consider $ f(x)$ not always 0. And so $ \boxed{f(u)\equal{}0\implies u\equal{}0}$

$ 2)$ $ P(\minus{}1,\minus{}1)$ gives $ f(\minus{}1\plus{}f(1))\equal{}f(\minus{}1)\minus{}f(\minus{}1)\equal{}0$ and so $ \boxed{f(1)\equal{}1}$ and $ \boxed{f(0)\equal{}0}$

$ 3)$ We have $ f(0)\equal{}0$ and $ f(1)\equal{}1$. Assume $ f(n)\equal{}n$. Then $ P(1,n)$ gives $ f(n\plus{}1)\equal{}n\plus{}1$ and so, with induction, $ \boxed{f(n)\equal{}n\: \: \forall n\in\mathbb{N}\cup\{0\}}$

$ 4)$ Let $ n\geq 1$ $ P(\minus{}1,\minus{}n)$ gives $ f(n\minus{}1)\equal{}f(\minus{}1)\minus{}f(\minus{}n)$ and so $ f(\minus{}n)\equal{}1\plus{}f(\minus{}1)\minus{}n$
Then $ P(\minus{}n,\minus{}1)$ gives $ f(0)\equal{}0\equal{}f(\minus{}n)\minus{}nf(\minus{}1)\equal{}1\plus{}f(\minus{}1)\minus{}n\minus{}nf(\minus{}1)$$ \equal{}(1\plus{}f(\minus{}1))(1\minus{}n)$ and so $ f(\minus{}1)\equal{}\minus{}1$ and so $ f(\minus{}n)\equal{}\minus{}n$
So $ \boxed{f(n)\equal{}n\: \: \forall n\in\mathbb{Z}}$

$ 5)$ $ P(\minus{}1,x)$ gives $ f(\minus{}1\plus{}f(\minus{}x))\equal{}\minus{}1\minus{}f(x)$ and so $ 1\plus{}f(\minus{}1\plus{}f(\minus{}x))\equal{}\minus{}f(x)$; Then $ P(1,\minus{}1\plus{}f(\minus{}x))$ gives $ \boxed{f(\minus{}f(x))\equal{}\minus{}f(x)}$
Then $ \minus{}f(\minus{}f(x))\equal{}f(x)$ and so $ f(\minus{}f(\minus{}f(x)))\equal{}\minus{}f(\minus{}f(x))$ and so $ \boxed{f(f(x))\equal{}f(x)}$

$ 6)$ $ P(1,x)$ gives $ f(f(x)\plus{}1)\equal{}f(x)\plus{}1$ and, replacing $ x$ with $ f(x)\plus{}1$, we obtain $ f(f(x)\plus{}2)\equal{}f(x)\plus{}2$ and, with immediate induction, $ f(f(x)\plus{}n)\equal{}f(x)\plus{}n$ $ \forall n\in\mathbb{N}\cup\{0\}$
$ P(\minus{}1,x)$ gives $ f(\minus{}1\plus{}f(\minus{}x))\equal{}\minus{}1\minus{}f(x)$ and so $ 2\plus{}f(\minus{}1\plus{}f(\minus{}x))\equal{}1\minus{}f(x)$ and so (since $ f(f(x)\plus{}n)\equal{}f(x)\plus{}n$) $ f(1\minus{}f(x))\equal{}1\minus{}f(x)$
So $ \minus{}f(1\minus{}f(x))\equal{}f(x)\minus{}1$ and, since $ f(\minus{}f(x))\equal{}\minus{}f(x)$ : $ f(f(x)\minus{}1)\equal{}f(x)\minus{}1$, and with induction $ f(f(x)\minus{}n)\equal{}f(x)\minus{}n$
So $ \boxed{f(f(x)\plus{}n)\equal{}f(x)\plus{}n\: \: \forall n\in\mathbb{Z}}$

$ 7)$ Now, replacing $ x$ by $ nx$ in point 6 above : $ f(f(nx)\plus{}n)\equal{}f(nx)\plus{}n$.
But $ P(n,x)$ gives $ f(n\plus{}f(nx))\equal{}f(n)\plus{}nf(x)\equal{}n\plus{}nf(x)$ and so, combining these two lines : $ f(nx)\plus{}n\equal{}nf(x)\plus{}n$ and so $ \boxed{f(nx)\equal{}nf(x)\: \: \forall n\in\mathbb{Z}}$

$ 8)$ Assume, for a given $ u$, $ f(u)\equal{}\minus{}u$, then $ f(\minus{}u)\equal{}u$ (since, according to point 7 above, $ f(nu)\equal{}nf(u)$; take $ n\equal{}\minus{}1$).
Then $ P(\minus{}1,\minus{}u)$ gives $ f(\minus{}1\plus{}f(u))\equal{}\minus{}1\minus{}f(\minus{}u)$ and so $ f(\minus{}1\minus{}u)\equal{}\minus{}1\minus{}u$
Then $ P(1,\minus{}1\minus{}u)$ gives $ f(1\plus{}f(\minus{}1\minus{}u))\equal{}1\plus{}f(\minus{}1\minus{}u)$ and so $ f(\minus{}u)\equal{}\minus{}u$ and, since $ f(\minus{}u)\equal{}u$, $ u\equal{}\minus{}u\equal{}0$
So, $ \boxed{f(u)\equal{}\minus{}u\implies u\equal{}0}$

$ 9)$ $ P(x,\minus{}1)$ gives $ f(x\plus{}f(\minus{}x))\equal{}f(x)\minus{}x$ and so $ f(x\minus{}f(x))\equal{}f(x)\minus{}x$, and so (according to point 8 above) $ x\minus{}f(x)\equal{}0$
And so $ \boxed{f(x)\equal{}x\: \: \forall x}$

So the two only solutions are :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$


Aki esta otro problema, k me parecio muy interesante resolver.

Considera una circunferencia F con diametro AB. Toma un punto H sobre F y traza HE, donde E es el pie de la perpendicular a AB. Traza una circunferencia de radio HE y centro en E, la cual corta a F en dos puntos M y N.
Demuestra que MN biseca a HE.

Espero soluciones, ya veran k bonito es...

¿la circunferencia de radio HE corta a F en M y N?, pero si corta a F en H?, explicame

Disculpa, error de redaccion, debe decir: "se traza una circunferencia de radio HE CON CENTRO EN H"

Lo demas si es correcto.
Gracias por la observacion

Llamare C a la circunferencia de radio HE, en C sea R el radio, la interseccion de MN y HE es X, a = XH, b = XE; prolongo XE hasta que toque a C en M y a F en T. Por potencia de X en C, (XM)(XN) = (XE)(XM)= (b)(a+R), en F: (XM)(XN) = (a)(b+R), de las ecuaciones: ab+aR = ab + bR de donde a=b, entonces XE=XH

me parece que tu solucion esta bien, lo unico que mencionas dos puntos con M.jeje :lol:

Aki les pongo una solucion por inversion.
Invirtamos en centro H radio HE. M, N, E se invierten en si mismos. La circunferencia F se invierte en la cuerda MN, A se invierte en el punto de interseccion de la recta HA y MN (llamemoslo A" y tenemos k B" se obtiene de un modo analogo. La recta AB se invierte en una circunferencia que pasa por H, A", B" y es tangente a AB en el punto E. Vemos entonces que HE es diametro de esta nueva circunferencia, de manera que HAE=HBE=90. Por ultimo vemos que AHB=90, de donde vemos que HA"EB" es un rectangulo, y por lo cual cumple que sus diagonales se bisecan.
K les parece??

OE QUE PADRE se me ocurrio usar inversion pero no lo hice por que pense que no iva a salir.jeje :lol:

ES MUY BONITO SABER QUE LO BISECA ME GUSTARIA QUE VEAN SU GRAFICA AQUI

http://geomtry.mundoforo.com/hola-vt3.html?sid=749f897c1d6b423e0638366ca633275b


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
AMC 12
AMC 10

[quote="09husbri"]I got a six (just problems 1-6)...I was very proud of myself, having expected myself to get only a 2 or 3  :)  I was particularly proud when I figured out problem six, but it's one heck of a bubble-burster to come on here and see people complaining about their nines that should have been elevens! Anyway, I'd like to think I didn't do too badly for an AIME first-timer fresh out of the MathCounts years!

Anyway, it doesn't seem like it from people's thoughts on other threads, but is there even the slightest remote possibility of a floor of six? Not that I would get better than a 2 on the USAMO anyway, but I think that being able to take it as a freshman would be an incredible experience for me and likely very beneficial to my results in future years (when I hope to qualify with better AIME scores). It's obviously unlikely...but possible, perhaps??

Also, would AoPS vol. 2 help next year's AIME score? What other sources might? I have vol. 1 which I love and I am leaning very strongly towards purchasing vol. 2 and working through it over the summer.

One more question: As a sophomore next year without a lot of advanced mathematical training (I just have an education up through Alg. II and the use of my great intuition), do I have a greater chance of qualifying for the USAMO if I take the AMC12 as opposed to the AMC10?

Thanks, and congrats to everyone who qualifies for the USAMO![/quote]

I'd say there is a 1% chance that someone may get a 150 on amc and 6 on aime and with the expansion, qualify when no one from the AIME II get above 3s. :)  And aops 2 is cruc to doing well on the aimes and maybe usamo.  Doing old competitions also help a great deal.  And always take the 12, it's easier to qualify(though that may change next year) and count more for usamo.

First, congratulations on making it to the AIME.  Regardless of whatever else you may read here, a 6 on the AIME is a good score, especially for a freshman.

[quote="09husbri"]Anyway, it doesn't seem like it from people's thoughts on other threads, but is there even the slightest remote possibility of a floor of six? Not that I would get better than a 2 on the USAMO anyway, but I think that being able to take it as a freshman would be an incredible experience for me and likely very beneficial to my results in future years (when I hope to qualify with better AIME scores). It's obviously unlikely...but possible, perhaps??[/quote]

I'd say there is a [i]slight[/i] possibility of that happening.  Usually, the floor is around 7 or so, and this AIME was a rather average difficulty, I'd say.  Normally, I wouldn't say there was a chance of the floor being that low, but this year, they're inviting 50% more students to take the USAMO.

[quote="09husbri"]Also, would AoPS vol. 2 help next year's AIME score? What other sources might? I have vol. 1 which I love and I am leaning very strongly towards purchasing vol. 2 and working through it over the summer.[/quote]

In fact, AoPS vol. 2 is the first place I'd recommend you to go to improve your AIME score for next year.  I highly recommend the book to you, as it not only covers most of what you could need on the AIME, but even goes into elementary Olympiad-level territory.

[quote="09husbri"]One more question: As a sophomore next year without a lot of advanced mathematical training (I just have an education up through Alg. II and the use of my great intuition), do I have a greater chance of qualifying for the USAMO if I take the AMC12 as opposed to the AMC10?[/quote]

Yes, you should take the AMC 12 rather than the 10.  You have a better chance of qualifying for the AIME via the 12, and if you take the 10, the AMC part of your USAMO index is scaled to 100, whereas if you take the 12, your AMC 12 score becomes the AMC part of your USAMO index.

Well, that's what advice I have to offer.  Once again, congratulations on your score.

[quote="09husbri"]I got a six (just problems 1-6)...I was very proud of myself, having expected myself to get only a 2 or 3  :)  I was particularly proud when I figured out problem six, but it's one heck of a bubble-burster to come on here and see people complaining about their nines that should have been elevens! Anyway, I'd like to think I didn't do too badly for an AIME first-timer fresh out of the MathCounts years!

Anyway, it doesn't seem like it from people's thoughts on other threads, but is there even the slightest remote possibility of a floor of six? Not that I would get better than a 2 on the USAMO anyway, but I think that being able to take it as a freshman would be an incredible experience for me and likely very beneficial to my results in future years (when I hope to qualify with better AIME scores). It's obviously unlikely...but possible, perhaps??

Also, would AoPS vol. 2 help next year's AIME score? What other sources might? I have vol. 1 which I love and I am leaning very strongly towards purchasing vol. 2 and working through it over the summer.

One more question: As a sophomore next year without a lot of advanced mathematical training (I just have an education up through Alg. II and the use of my great intuition), do I have a greater chance of qualifying for the USAMO if I take the AMC12 as opposed to the AMC10?

Thanks, and congrats to everyone who qualifies for the USAMO![/quote]

I literally am your twin in terms of math experience (alg II and a freshman) and stuff... I'm curretnly training for the AIME IIs and think this website has been one of the greatest factors in my improvement (I can all but guarantee that I'll do better than my 4 last year, unless the whole test happens to be trig or something) in math, especially since I'm on spring break and have had these three weeks to get ready.

I just started looking through the AoPS II books recently and have found they are amazingly helpful esp. on learning things beyond Alg II like trig and stuff, and building on what you know, with well-selected problems and clean explanations. Although intuition can carry you so far, which is pretty much what I use for all my competitions, having the theorems on paper can't hurt either... so yeah the old AIMEs are the best help, then this board, then the AoPS books (both 1 and 2... actually 1 has a lot more in it that you would think)

Hey again nice work on the AIMEs. Contrary to what this board will lead you to think, a six is a good score!

Thanks for the responses, guys. Well, no harm in keeping my fingers crossed.  :)


salam ba hama
can someone find the nth term for this sequence

1, 2, 10, 67, 467, 3268, 22876

salam.
ta hala iani bad az ie rob fekr kardan be inja residam:
a   =7(a      -1)+n             (n,n-1 andisand(excuse me))
  n        n-1
 vali rabeteie mostaghel ... :?:
Good luck.
 :?  :lol:  :blush:  :!:  :)  :roll:

motavajeh nashodam,meshe lotfan ye bar dega toze bede

salam
man tahala toonestam ie rabeteie bazgashti baraie donbale peida konam.
a[n]=7(a[n-1]-1)+n       ----(dakhele keroosheha andisan   baraie a)---
vali rabeteie mostaghel na hanooz...
fekr konam beshe tabe movaledesho nevesht.

[quote="nodet"]slam dooste aziz.
bebakhshid age narahatetoon kardam vali man nafahmidam ke che kare badi kardam,
be har hal omidvaram mano bebakhshi.man ie aze varedam ke hich chiz az riazi nemidoone .mamnoon misham age komakam konin.[/quote]

raftaret vaghan tahsin bar angizeh. bravo bravo...

azizam manzooram ba shoma nabood.

man hamchin jesarati be shoma nemikonam. va hamishe be shoma ehteram migozaram. ;)

manzooram ba shakhse digari bood.

salam.
mamnoon mano vaghean khoshhal kardin.
 :oops:  :blush:  :lol:  :)


A box contains both chewy-center and soft-center pieces of candy.  There are 8 more chewy-center pieces than soft-center pieces.  When a piece of candy is chosen at random, the probability that it has a chewy center is $\frac{3}{5}$.  How many pieces of candy are in the box?

[hide][color=indigo]there is 3/5 chewey candy and 2/5 soft candy. since there is 8 more chewey than soft, 8 = 1/5 of the candy. the total number of candies is 40.[/color][/hide]

[hide]
$\frac{3}{5}$ chewy
$\frac{2}{5}$ soft
$8$ more chewy than soft
$8=\frac{1}{5}$
$8\cdot5=40$
$40$ pieces of candy.[/hide]

[hide]The difference between the pieces of candy in terms of the whole is 1/5, and that is 8, so the total is 8*5 or 40[/hide]
I like that type of candy.  :P

[hide]MMM, candy!
Anyway, the answer is 8(5) so 40[/hide]


I'll hazard a prediction here: IF the movie Fahrenheit 9/11 can stay at number-one in box office receipts for as many weeks as the movie The Passion of Jesus Christ did earlier in the year, then George W. Bush loses the election this November. This will be difficult, because Fahrenheit 9/11 is opening on fewer screens at first, but the [url=http://movies.yahoo.com/shop?d=hv&id=1808569540&cf=critic&intl=us]reviews are mostly awesome[/url], and I think this will become the must-see movie of the summer, agree or disagree. I don't expect this movie to be a one-week wonder.

I think the modest number of swing voters left in the United States will mostly swing away from Bush (note that I did NOT say swing toward Kerry) after seeing the movie. Moore has done a phenomenal job of finding film clips that will erode Bush's support, which was his intention.

I will go see it if my parents will let me...  They think I should not have an opinion of politics yet and hate it when I bug them about political things.  ah well they are pro-bush and don't want me pro-kerry so they are afriad to let me see the movie I think.

Well, it is an R-rated movie (because it shows film of battle injuries) and I can understand why a parent wouldn't want a young person to see an R-rated movie. I think I'll see it with my wife first, and then decide if any of my children should see it at their current age.

thats not what bothers my parents.  They have let me see plenty of R rated movies before.  They think I should be aware of things and considering they are military battle injuries and stuff like that in a movie wouldn't bother them I don't think.  Oh well.  I'll probally have to wait to see if it comes out on VHS.

Really? I've done all three things with no trouble. I plan on seeing Moore's movie soon as well. Has anyone here seen it/plans to see it in the coming weekend?

[quote] Its out today! Go see it, even if you are anti Moore, pro Bush. I saw on the news one scene where he goes to Congressmen and asks if he can enlist their kids in the army, and they all decline. Only one congressmen out of 535 have a kid in Iraq. [/quote]

Of course they declined!  The people who enlist in the Army are paid and they are able/want to fight.  Just because they support the war doesn't mean they have to send their children into Iraq.

Nukular -- that site is pretty low-quality.  Anyone who acts like soft-money donations to a presidential party during an election year couldn't have [i]anything[/i] to do with the campaign is just being absurd.  Not that I'm a big fan of Moore, particularly -- I get the sense that he does have some trouble with the truth -- but that website doesn't do much better.  And it certainly doesn't represent a serious defense against any of major points that Moore makes.  At best, it points out flaws in a few pieces of evidence.

Also, he seems unable to take a joke.  He appears to treat things that were designed tongue-in-cheek as if they were serious.

[quote="Rep123max"]But Raging, do you think it is right for the people who make the decisions about us going to war, which is killing many people, to not have any of their children enlisted? I personally think that is disgusting.[/quote]

Many members of congress served in the military (and in armed conflicts) themselves. Many of them also have relatives (thought maybe not children) who are currently serving in the military. I have no problem with the fact that none of them have children over there. It's a volunteer army after all. No one had to join that didn't want to.

That's one area where I really don't like Michael Moore. Journalists and filmmakers who use "Dateline" style ambushes of people is totally uncalled for and unprofessional in my opinion. His ambush on Charleton Heston (who I don't particularly like) in Bowling for Columbine was uncalled for, and was a blight on an otherwise very good documentary.

[quote="joml88"]How will VHS let you see it?  If your parents won't let you see it on the big screen they certainly won't let you buy it.  If it is R rated you need to show ID (at most places) to get it.  In my personal experience it is easier to get into an R rated movie at a movie theatre than it is to buy a R rated movie or M rated game at a store.[/quote]
hmm, i'd have to disagree about the M rated game (my area of expertise). i've gone and bought many M rated games at local stores alone and the cashiers do not hesitate to give it to you. sure, they look at me rather suspiciously (i'm kinda short for my age), but they'll sell it to you.
about the movie, i may go see it if i have time. there's just some other movies, like dodgeball, i'd like to see before i see farenheit 9/11

Well at Wal mart and target they don't let me buy stuff like that.  I have never tried it at a video game store.  They will probably be a little more leniant at those kinds of stores.

the trick is to buy from a store run by people who also like video games a ton. it should also be a small store, because they're much more lenient

I agree. "Amazing" is probably the best word to describe it. Everyone, go see it!

I'll probably see it at some point, but I'm not rushing out to the theater.  It's not as if it will change my opinion of Bush.

Incidentally, Michael Moore says that the under 17 crowd should sneak into the theater to see the movie: [url]http://www.mtv.com/movies/news/articles/1488698/06252004/story.jhtml[/url]. I'm 18 now and never really cared to see anything R-rated before this, but I have heard that those who check ID at the movies sometimes do not look at the date of birth--they only check to see that the photo matches.

I saw Fahrenheit 9/11 tonight. Wow. Amazing. Everyone should go see it.

I saw it with someone who says, "Michael Moore picked all the bad stuff to show. It was very biased." But this viewer goes on to say that Bush was shown being Bush. That's the point. The "wow" comes from seeing the President of the United States dealing with the serious issues that the movie is about. There was some great use of news footage in the movie.

I have to agree with Michael Moore's policy point: the mostly poor young people who volunteer to serve in the all-volunteer armed forces of the United States are making a great contribution to their country, and they shouldn't be put in harm's way unless that is truly necessary for the safety of their country. This movie is definitely worth thinking about in the presidential campaign season.

The film was interesting to say the least. It will definately change many viewers political views or reinforce those currently present.

You definately have to see it. Was there bias? Sure, but if you look at where he is coming from, it's hard to disagree that his bias is unjustified. There were a few scenes that really affected me.

I don't know if it's just a San Diego thing, but it was interesting to see the reaction of people towards the movie, often expressed during the movie itself.

[quote="yif man12"]i wonder if this means a second straight oscar for michael moore[/quote]

I can't think of any other documentary that could touch Fahrenheit 9/11 for the best documentary Oscar.

I had seen all the available online film clips before seeing the whole movie, and had read lots of reviews, but there was still a lot more to see than what I had anticipated. I've been thinking about the movie overnight, and I definitely expect people to talk about it. The early reviews on Spider Man 2 are such that it will be difficult for Fahrenheit 9/11 to be the number-one movie next week, but it's a cinch to be the number-one movie this weekend. I bought my tickets for a 7:00pm showing at 3:00pm, and got to the theater at 6:30pm to see the auditorium looking mostly full already. At ten minutes to show time, the management announced that the show was sold out and asked that people near empty seats please shift in rows so that groups of friends could find seats together. Hollywood will notice that Moore's movie draws like no documentary has before.

This is kind of silly, but did anyone else find the incessant grooming kind of strange?

[quote="tokenadult"][quote="yif man12"]i wonder if this means a second straight oscar for michael moore[/quote]
Hollywood will notice that Moore's movie draws like no documentary has before.[/quote]
i know. first Bowling for Columbine, now Farenheit 9/11. he's setting the bar pretty high for next year  :D

i saw it yesterday and was totally blown away. what a great movie. my favorite part was when larry king was interviewing ambassador bendar and asked him his impression of bin laden. then it clipped to another scene to reinforce the analogy (i'll leave it out for those who haven't seen it). i've been so impressed with both of michael moore's documentaries, both 9/11 and columbine. i am really glad that there is a liberal voice that finally has some persuasion.

you should all go see it, along with bowling for columbine if you haven't already

Are you sure that line was meant to be hilarious? He seemed pretty serious when he said it.

[color=red]To state the truth, Michael Moore is a rather horizontally and mentally challenged person.  And I will state details that prove this also if you ask me to.[/color]

Well, that was a rather random ad hominem attack irrelevant to the discussion at hand ...

I had just felt like saying that even though it was irrelevant.  I just don't think his tactics are very great and I'm not a fan of him.  I think it's good to be a little liberal but when you're as liberal as he is, it's not that good.  He's too extreme.

"Kill all the extremists," eh?

Just for the record, now, "liberal" is not an extreme position.  Liberal is a relatively moderate position that has been left as the furthest-left extent of the American political spectrum with the death of serious Communist or Socialist parties.  If a fascist government had survived World War 2 as the other superpower instead of the Soviet Union, we likely live in a society today in which moderate rightists were labelled "extremists" and where the Democratic party had a small but powerful socialist wing.

Society needs extremists -- as long as they aren't blowing anything up, they help keep us balanced.

To define "liberal" and "conservative" as defined by my 10th grade history teacher:
Liberal means you want change to happen, that you want things to advance. The abolishment of slavery was a LIBERAL position. In fact, it was very far left-wing liberal, for the time period. Even the abolishment of slavery but keeping blacks as inferior was left-wing, in the 1860s. However, in the 1990s, blacks as inferior is RIGHT-WING conservative. Conservative means that you don't want things to change, that you like them just the way they are. In the 1860s, the DEMOCRATS were CONSERVATIVE. They wanted to keep slavery. In the 2000s, the REPUBLICANS are known for being  more conservative, because in many scientific arguments (like stem-cell research) they are against advancement. They do not want CHANGE to happen.
I just thought I would clarify all of that before an argument erupted.

And for the record, I really dislike "extremists" of all sorts. Extremists are the ones who kill other people. There are extreme Christians, extreme Muslims, extreme Republicans, extreme Democrats, extreme Jews. Anyone who believes so strongly about something that they are willing to kill another person is, in my opinion, extreme. So does Michael Moore actually count as an extremist?

I would quibble slightly with your terms.  A liberal does not necessarily want change.  For example, a liberal living in Canada would probably not be advocating change to the system they have there.  A progressive, on the other hand, might.  Progressivism, like conservativism, is defined in more relative terms: conservatives can be far-right (as in America today) or far-left (as in the Soviet Union in the late 1980's), and similarly for progressives.  Our usages have tended to associate conservitivism and fascism with the right of the political spectrum while associating liberalism and progressivism and communism with the left.  None of these is necessarily accurate.  (Pop quiz: when did the use of the terms "left" and "right" appear to describe political viewpoints?)

Finally, I would take issue with this statement: [quote]Extremists are the ones who kill other people.[/quote]  I think it makes more sense to say "People who kill other people for political ends tend to be either extremists or crazy."  There are lots of non-violent extremists.


There are N different types of coupons.  Each time a coupon collector obtains a coupon it is equally likely to be any one of those.  What is the probability that the collector needs more than k (k > N) coupons to complete his collection (have at least one type of each coupon)?

Here's a start:

[hide]

Inclusion-exclusion provides us with a binomialish result:

N*N^(k-1) - (N choose 2)N^(k-2) + (N choose 3)N^(k-3) - ...

= N^k - (N-1)^k?

Then divide by N^k

This probably needs some cleaning up.

[/hide]

Also, here is a related question.  What is the expected number of coupons that the collector needs to complete his collection?

Another start:

[hide]Assuming my first start is remotely correct, the pdf is geometric.  That should help some.[/hide]

Our answer's do not agree, although i like the idea of using probabilistic inclusion-exclusion.


so if bill gets one donut of each kind, then he has 2 more donuts to pick.

1) he picks 2 donuts of same kind

he can do this in 4 ways.

2) he picks 2 kinds of donuts

he can do this in $ \binom{4}{2}\equal{}6$ ways.

4+6=10

answer : 10

Bill can get at least one and at most 3 donuts of the same kind, because otherwise he wouldn't have enough space for the other three kinds of donuts. The generating function for this problem is $ (x \plus{} x^2 \plus{} x^3)^4$, and the coefficinet of $ x^6$ is 10.

just as a fun and instructive and interesting practice problem for generating functions: the numbers on the faces of 2 regular 6 sided dice are removed and replaced with 12 positive integers, one on each face, such that the probability of getting a sum k on rollling the dice before the change is the same, for all integers k. For example, the probability of getting a 2 on the new dices is still 1/36, and the probability of getting a 5 is still 4/36=1/9. Given that the new dice are different from the old dice, what numbers are on the new dice?

[quote="GameBot"]Bill is sent to a donut shop to purchase exactly six donuts. If the shop has four kinds of donuts and Bill is to get at least one of each kind, how many combinations will satisfy Bill's order requirements?[/quote]

Let's lay out the donuts like \[ o\ o\ o\ o\ o\ o\] To divide the donuts into the four categories such that there's at least one of each, we must put $ 3$ dividers in the spaces between the donuts.

There are $ 5$ spaces, so $ \binom{5}{3}\equal{}\boxed{10}$

[hide=Solution]
There are 5 dividers for six donuts.
There are 3 dividers for four donuts.
Thus, the answer is $\binom{5}{3}.$[/hide]


number theory
prime numbers
algebra unsolved
algebra

[quote="tranthanhnam"]show that if p is the product of first prime numbers , the none of the two numbers $p-1$ and $p+1$ is a perfect number.[/quote]
What do you mean by first prime numbers?

$2,3,5,7,11,13,17,19,23,...$ in that order (he means: of the first $n$ prime numbers).

EDIT: Deleted, this was total rubbish. I'm very sorry

For example, $P_n-1$ isn't prime in general. The condition you used isn't sufficient.

$x$ is a perfect number iff $x=s(x)$, with $s(x)$ is the sum of all divisors of $x$, except $x$ himself. I think it is called the aliquot sum, but I could be mistaking...

For example: $6$ is a perfect number, because $6=1+2+3$
$28$ is a perfect number, because $28=1+2+4+7+14$

Or take a look [url=http://mathworld.wolfram.com/PerfectNumber.html]here[/url].

Well if we accept the hyphotsis that there are no odd perfect numbers... :)

show that if p is the product of first prime numbers , the none of the two numbers $p-1$ and $p+1$ is a perfect square number.


I have heard that some people have gotten salmonella and died from eating bagged spinach. What are your thoughts on this?

Mmmmmhhh.......... spinach.........mmmmmmhh........with garlic.....
[hide][there was a serious but stupid thing written here due to name confusion] mmmmmmhhhh...[/hide]
mmmmhh...... [saliva pouring out].

[more saliva pouring out]

there are information in yahoo... it says that this happen because of some e.coli that here was in the spinach... anyway thats a good reason for not eating spinach... I do not like it (only raw)!

I just had some spinach and cheese pasta, and then I found out about this. It freaked me out!

As already pointed out, it was E. Coli, not salmonella.

http://www.cnn.com/2006/HEALTH/09/15/tainted.spinach.ap/index.html

So that's why spinach has been missing from my dining hall. You can't make a good salad with lettuce...

[quote="Xevarion"]On a major news network (ABC?), the reporter referred to [i]E. coli[/i] as a virus.

And people still wonder why the population of the USA does dumb stuff....[/quote]

 :rotfl:

e.coli=virus<=>0=1

In my U.S history class, my homework is to think of 5 words that people not from the U.S would use to describe the U.S. Any suggestions? :rotfl:

I'd go with something like 'arrogant',or 'dumb(probably because of bush)'

What I came up with is arrogant and wasteful, but when I said that in class, I got the nickname "Negative Boy" :mad:

and that's what you get for being different.

Really?  "Negative Boy"?
Well, that IS how it is.  Many countries probably agree that the US is the biggest threat in the world today. (which it probably is with Bush and administration in power)

Should I just make a seperate topic now?

Yeah, I heard about the bagged spinach stuff. Something with a California company


integration
trigonometry
calculus
calculus computations

hi, i just want to know how am i suppose to obtain $ \int \sqrt{4\minus{}x^2}$? please be more specific than "trig substitution" or anything like that...

thanks.

Ok: $ \int \sqrt {4 \minus{} x^2} dx \equal{} 2\int \sqrt {1 \minus{} \frac {x^2}{4}}dx$.

Now, taking the trig substitution  :)  $ \frac {x}{2} \equal{} \sin t$, you get $ 2\cos t dt \equal{} dx$, hence

$ 2\int \sqrt {1 \minus{} \sin^2 t}\cdot2\cos t dt \equal{} 4\int \cos^2 t dt$

This last integral can be solved using $ cos^2 t \equal{} \frac {\cos 2t \plus{} 1}{2}$. It gives you this: $ 2\int(\cos 2t \plus{} 1)dt \equal{} \sin 2t \plus{} 2t \plus{}C\equal{} 2\sin t\cos t \plus{} 2t\plus{}C$.

So, finally we get, using $ \sin t \equal{} x$:

$ \int \sqrt {4 \minus{} x^2} dx \equal{} \frac {x}{2}\sqrt {4 \minus{} x^2} \plus{} 2arc \sin\frac {x}{2}\plus{}C$.


Βρείτε τη μεγαλύτερη σταθερά $ K$ τέτοια ώστε $ \frac{b^{2}c^{2}}{a^{2}(a\minus{}b)(a\minus{}c)}\plus{}\frac{c^{2}a^{2}}{b^{2}(b\minus{}c)(b\minus{}a)}\plus{}\frac{a^{2}b^{2}}{c^{2}(c\minus{}a)(c\minus{}b)}\geq K$

Με ενωση των τριων κλασματων σε ενα και παραγοντοποιηση του αριθμιτη και μετα απλοποιηση, βρησκουμε οτι οταν οι τρεις μεταβλητες ειναι διαφορετικες μεταξυ τους ανα δυο τοτε το αριστερο μελος ισουτε με:

$ F(a,b,c) \equal{} \frac {a^3b^3 \plus{} a^3c^3 \plus{} a^3b^2c \plus{} a^3c^2b \plus{} a^2b^3c \plus{} a^2c^3b \plus{} a^2b^2c^2 \plus{} c^3b^3 \plus{} b^3c^2a \plus{} c^3b^2a}{a^2b^2c^2} \geq \frac {10a^2b^2c^2}{a^2b^2c^2} \equal{} 10$
απο Muirhead αφου οι τριαδες $ (3,2,1)$, $ (3,3,0)$ και $ (2, 3, 1)$ μεγιστοποιουν την τριαδα $ (2,2,2)$, με ισοτητα για $ a \equal{} b \equal{} c$.

Η συναρτηση $ F$ ομως ειναι συνεχης ως προς τις τρεις μεταβλητες και ισουτε με $ 10$ για $ a \equal{} b \equal{} c$, οποτε αν παρουμε $ b,c \rightarrow a$ τοτε το αριστερο μελος της αρχικης ανισοτητας συγκλινει στον αριθμο $ 10$, αρα το αριστερο μελος της αρχικης ανισοτητας μπορει να παρει οποιαδηποτε πραγματικη τιμη ενος ανοιχτου διαστηματος $ (10, m), m > 10$ και επηδη εχουμε δειξει οτι ειναι μεγαλυτερο του $ 10$ για καθε $ (a,b,c) \in R^3, a\neq b, b \neq c, c \neq a$ εχουμε οτι το infimum του ειναι ο αριθμος $ 10$. Αρα ειναι $ K \equal{} 10$.

Δεν είδα τις πράξεις αλλά ρε Νίκο η ισότητα δεν μπορεί να ισχύει για $ a\equal{}b\equal{}c$

[quote="bilot4"]Δεν είδα τις πράξεις αλλά ρε Νίκο η ισότητα δεν μπορεί να ισχύει για $ a \equal{} b \equal{} c$[/quote]

Οπα, εδιωξα τους παρονομαστες και το ξεχασα τελειως οτι μηδενιζονται... αλλα και παλι η κατασταση σωζεται αν παρουμε ορια για τα $ b,c$ να τεινουν στο $ a$ διοτι η απλοποιημενη παρασταση ειναι συνεχης συναρτηση ως προς τις τρεις μεταβλητες, οποτε το infimum της παραστασης ειναι ο αριθμος 10 και ειναι $ K \equal{} 10$. :wink:

ΥΓ: Solution edited.


find the greatest value of a + 1/(b+1/c) + d + 1/(e+1/f) + g + 1/(h + 1/i), where each letter represents a different non-zero digit.

[hide="Hint #1"]To find the maximum we want the following:

- Maximize $ a, d,$ and $ g$.

- Minimize $ (b \plus{} 1/c), (e \plus{} 1/f),$ and $ (h \plus{} 1/i) \implies$ Maximize $ c, f$, and $ i$, and Minimize $ b,e,$ and $ h$.[/hide]

[hide="Hint #2"]Using the hint above, we see that $ b, e$ and $ h$ must be $ 1,2,$ and $ 3$.[/hide]

[hide="Solution"]
\[ a \plus{} \frac {1}{\frac {b \plus{} 1}{c}} \plus{} d \plus{} \frac {1}{\frac {e \plus{} 1}{f}} \plus{} g \plus{} \frac {1}{\frac {h \plus{} 1}{i}}  \equal{}  a \plus{} d \plus{} g \plus{} \frac {c}{b \plus{} 1} \plus{} \frac {f}{e \plus{} 1} \plus{} \frac {i}{h \plus{} 1}\]
Because $ b$, $ e$ and $ h$ are being added to one in the denominator of a fraction, the smaller they are, the bigger the value of the fraction. all other numbers are either not in a fraction or in the numerator of a fraction, so we want to maximize those.

Since only $ b$, $ e$, and $ h$ need to be minimized, we assign $ 1$, $ 2$, and $ 3$ to $ b$, $ e$, and $ h$, in any order.

Next, we want to see if we should assign the bigger numbers to the numbers alone (call these "lonely") or to the numerators. If we assign the biggest numbers to the lonely variables, it stays the same. If we assign them to the numerators, only a fraction of them makes it. If we take a fraction of a bigger number, we lose more value than if we take the fraction of smaller number. Thus, we want to take a fraction of the smaller numbers, assigning the bigger numbers to the lonely variables.

Using the above reasoning, we assign $ 7$, $ 8$, and $ 9$ to $ a$, $ d$, and $ g$, in any order.

The only numbers left, which are $ 4$, $ 5$, and $ 6$, have to be assigned to $ c$, $ f$, and $ i$, in any order.

One more thing left: Which numerator should be paired with each denominator? Our numerators are $ 2$, $ 3$, and $ 4$ and our denominators are $ 4$, $ 5$, and $ 6$. Taking away a larger fraction of a larger number and a smaller fraction of a smaller number loses more value than taking away a smaller fraction of a larger number and a larger fraction of a smaller number. Thus, we assign the greatest numerators to the smallest denominators and vice versa, so we pair $ 6$ and $ 2$, $ 5$ and $ 3$, and $ 4$ and $ 4$.

Finally, putting everything from above into the expression, we get:
\[ 7 \plus{} 8 \plus{} 9 \plus{} \frac {6}{2} \plus{} \frac {5}{3} \plus{} \frac {4}{4} \equal{} 7 \plus{} 8 \plus{} 9 \plus{} 3 \plus{} 1 \plus{} \frac {2}{3} \plus{} 1 \equal{} \boxed{29 \frac {2}{3}}\]
[/hide][color=orange]
[size=75]
[LaTeX fixed by solafidefarms][/size][/color]


Hi AoPS/Mathlinks people;
I've been hearing about many contests and some are not even Canadian. I was just wondering how do you participate in a variety contests if your school doesn't offer much? I'm pretty sure my high school next year only does Pascal, Fryer, and Math League (in my grade, of course) and probably similar in other grades. Could someone explain to me how to get your school more interested in willing to put forth effort into math contests?

You simply have to talk to a math teacher or principal of your school.  They have to be willing to enter these contests.  If worse comes to worse, offer to pay for all the entrance fees.  I had to for a couple years since I'm the only person in my school who writes them

[quote="martha"]has anyone ever tried writing them at other schools that have them? does it work out?[/quote]

Sometimes it does.  You have to talk to your teacher, and then the prganizer at that other school and make sure it is alright with them.  Most schools don't mind, if there is an extra fee attached though offer to pay it, it'll make things easier.


Let $ x,y,z>0,x\plus{}y\plus{}z\ge 3$
Prove that $ \frac{1}{x^3\plus{}y\plus{}z}\plus{}\frac{1}{y^3\plus{}x\plus{}z}\plus{}\frac{1}{z^3\plus{}y\plus{}x}\le 1$
I have a proof but it is quite ugly,so I hope some one help me find a nicer proof
P/s:I think this problem is quite similar to problem 3 in IMO 2005

[quote="Allnames"]Let $ x,y,z > 0,x \plus{} y \plus{} z\ge 3$
Prove that $ \frac {1}{x^3 \plus{} y \plus{} z} \plus{} \frac {1}{y^3 \plus{} x \plus{} z} \plus{} \frac {1}{z^3 \plus{} y \plus{} x}\le 1$
I have a proof but it is quite ugly,so I hope some one help me find a nicer proof
P/s:I think this problem is quite similar to problem 3 in IMO 2005[/quote]

It's Vasc's problem. I think that the $ a\plus{}b\plus{}c\equal{}3$ assume that our proof quite a neat !  :roll:

[quote="Allnames"]Let $ x,y,z > 0,x \plus{} y \plus{} z\ge 3$
Prove that $ \frac {1}{x^3 \plus{} y \plus{} z} \plus{} \frac {1}{y^3 \plus{} x \plus{} z} \plus{} \frac {1}{z^3 \plus{} y \plus{} x}\le 1$
I have a proof but it is quite ugly,so I hope some one help me find a nicer proof
P/s:I think this problem is quite similar to problem 3 in IMO 2005[/quote]
I think the following inequality is true too.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x\plus{}y\plus{}z\geq3.$ Prove that:
\[ \frac {1}{x^4 \plus{} y \plus{} z} \plus{} \frac {1}{y^4 \plus{} x \plus{} z} \plus{} \frac {1}{z^4 \plus{} y \plus{} x}\le 1\]

[quote="arqady"][quote="Allnames"]Let $ x,y,z > 0,x \plus{} y \plus{} z\ge 3$
Prove that $ \frac {1}{x^3 \plus{} y \plus{} z} \plus{} \frac {1}{y^3 \plus{} x \plus{} z} \plus{} \frac {1}{z^3 \plus{} y \plus{} x}\le 1$
I have a proof but it is quite ugly,so I hope some one help me find a nicer proof
P/s:I think this problem is quite similar to problem 3 in IMO 2005[/quote]
I think the following inequality is true too.
Let $ x,$ $ y$ and $ z$ are non-negative numbers such that $ x \plus{} y \plus{} z\geq3.$ Prove that:
\[ \frac {1}{x^4 \plus{} y \plus{} z} \plus{} \frac {1}{y^4 \plus{} x \plus{} z} \plus{} \frac {1}{z^4 \plus{} y \plus{} x}\le 1
\]
[/quote]
May be,arqady,but can you show me your proof? :(

[quote="Allnames"]Let $ x,y,z > 0,x \plus{} y \plus{} z\ge 3$
Prove that $ \frac {1}{x^3 \plus{} y \plus{} z} \plus{} \frac {1}{y^3 \plus{} x \plus{} z} \plus{} \frac {1}{z^3 \plus{} y \plus{} x}\le 1$
I have a proof but it is quite ugly,so I hope some one help me find a nicer proof
P/s:I think this problem is quite similar to problem 3 in IMO 2005[/quote]
It was proposed and solved in VasC's book ;)
By the way, there exists a very nice and short solution with Cauchy Schwarz. ;) :)

Any other full proof or pure hint,I really need your proof


Hi, do you know where I can find an online source (or can you direct me to a book) of quality Electricity Magnetism problems?

I am going to take a pretty intense E&M course next fall, and my impression is I will get the full blown grad/div/curl and vector calculus tools thrown at me (the courses uses Purcell E&M).  I wonder where I can find problems of similar quality as the Purcell problems.  Irodov have challenging E&M problems, but it seems that most of the emphasis is not on vector calc, but more on complicated integrals and mathematical computations of single-variable calc.

Thanks so much!

I used the Purcell book for my E&M course and I absolutely hated it.  While the problems are good, the book doesn't prepare you for them and does a horrible job explaining the material.

I think the book by Halliday, Resnick, and Krane is a good supplement, although it doesn't use vector calculus.  For a book with nothing but problems, get [url=http://www.amazon.com/Solutions-Electromagnetism-Universities-Qualifying-Questions/dp/9810206267/ref=sr_1_1?ie=UTF8&s=books&qid=1242216387&sr=8-1]this[/url] one.  It has about 2000 problems on various topics in E&M.  They're all difficult but have complete solutions.


geometry
geometric transformation
rotation
combinatorics open
combinatorics

This is a question, which I asked at a different [url=http://www.mathlinks.ro/viewtopic.php?t=300135]topic[/url]. But it is very interesting (at least, for me), so I decided to re-post it separately.

For odd $ n$, construct a bijection between subsets $ A\subset \{1,\dots,n\}$ with sum divisible by $ n$ and two-coloured necklaces (non-free, i.e. symmetric necklaces are different) of length $ n$.

Their numbers being equal, it is, of course, possible to enumerate both and then claim that it is a bijection. But I really want a digestible construction, as one I proposed in the topic I mention.

This is very interesting.  What I want to say is that there is a bijection $ \phi$ between subsets $ A$ and all two-colored necklaces that has the following property: given a subset $ A$ and the corresponding necklace $ \phi(A)$, there is exactly one rotation of $ \phi(A)$ whose preimage $ A'$ has sum divisible by $ n$.  But the obvious bijection fails.


inequalities
algebra
polynomial
trigonometry
inequalities solved

ABC triangle. sinA*sinB*sinC=1/3
Prove that

p^3+Sr+abc>4R^2p

The inequality is equivalent with: p^2+r^2+4Rr>4R^2 or sum(sinA*sinB)>1. But this is obvious since sum(sinA*sinB)>=3[sinA*sinB*sinC]^2/3=3^1/3>1.

[quote="Maverick"]Let $ ABC$ be a triangle such that $ \sin A\ \sin B\ \sin C \equal{} \frac {1}{3},$ prove that

$ s^3 \plus{} \Delta r \plus{} abc > 4R^2 s$,

where $ a,b,c$ are the sides, $ s$ is the semi-perimeter, $ R$ is the radius of circumcircle, $ r$ is the radius of incircle, $ \Delta$ is the area.[/quote]$ \frac {s}{R}\geq 2.1360\cdots$,

which is a root of the irreducible polynomial

$ 81m^8 \minus{} 324m^6 \minus{} 1080m^5 \plus{} 72m^4 \plus{} 3456m^3 \plus{} 1728m^2 \plus{} 288m \plus{} 16$.[quote="pcalin"]The inequality is equivalent with $ \sum{\sin B\ \sin C} > 1$.[/quote]$ \sin B\ \sin C \plus{} \sin C\ \sin A \plus{} \sin A\ \sin B\geq 1.4771\cdots$,

which is a root of the irreducible polynomial

$ 11664k^8 \minus{} 23328k^7 \plus{} 11664k^6 \minus{} 41472k^5 \plus{} 68040k^4 \minus{} 31752k^3 \minus{} 9216k^2 \plus{} 51840k$

$ \minus{} 13991$.


Using three straight lines, what is the maximum number of
regions into which the enclosed
region can be divided?
[asy]size(150);
pair A = (10,4), B = (10,2), C = (10,-2), D = (10,-4);
pair O = (0,0), M = (4,0);
draw(A--B..(5,1.2)..M..(5,-1.2)..C--D..(5,-4)..O..(5,4)..A);[/asy]

$ \boxed{9}$.

I don't know how to precisely draw it with Asymptote so I don't have a solution.

It is $ \fbox{9}$. The problem must have an error. Did you report it?

The answer is 10.

Draw a vertical line.  Draw two more -- almost vertical -- lines in a way that forms a very small triangle.  You have a small triangle as well as six regions of infinite size.

Now, draw one branch of fat horseshoe to include parts of all seven regions -- your curve intersecting each line twice.  Let the other branch of the horseshoe bend around to intersect all three lines again, adding three more regions to the seven regions already enclosed.


function
integration
real analysis
inequalities
limit
real analysis unsolved

Show that if $ f(x)$ is a real-valued, continuous function on the half-line $ 0\leq x < \infty$, and \[ \int_0^{\infty} f^2(x)dx
<\infty\] then the function \[ g(x)\equal{}f(x)\minus{}2e^{\minus{}x}\int_0^x e^tf(t)dt\] satisfies \[ \int _0^{\infty}g^2(x)dx\equal{}\int_0^{\infty}f^2(x)dx.\] [B. Szokefalvi-Nagy]

We assume that the function $ f(x)$ is square integrable in the Lebesgue sense on the half-line $ (0, \infty)$.The relation
\[ f(x) - g(x) = 2e^{-x} \int_0^x e^t f(t) dt \hspace{175pt}(1)\]
implies that
\[ (f(x) - g(x))' = f(x) + g(x) \ \ \text{almost everywhere.} \hspace{100pt}(2)\]
Using the Schwarz inequality, we obtain
\begin{eqnarray*}
e^{-\omega}\left| \int_0^\omega e^t f(t) dt \right|
& \le & e^{-\omega}\left| \int_0^{\omega/2} e^t f(t) dt \right| + e^{-\omega}\left| \int_{\omega/2}^\omega e^t f(t) dt \right| \\
& \le & e^{-\omega} \left( \int_0^{\omega/2} e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_0^{\omega/2} f^2(t) dt  \right)^{\frac{1}{2}} + e^{-\omega} \left( \int_{\omega/2}^\omega e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_{\omega/2}^\omega f^2(t) dt  \right)^{\frac{1}{2}} \\
& \le & e^{-\omega/2} \int_0^\infty f^2(x) dx + \int_{\omega/2}^\infty f^2(t) dt,
\end{eqnarray*}
whence it follows that
\[ \lim_{\omega \to \infty} \int_0^\omega e^t f(t) dt = 0.  \hspace{225pt}(3)\]
Relation (1) assures that $ f(x)-g(x)$, and therefore $ (1/2)(f(x) - g(x))^2$, is absolutely continuous on each bounded subinterval of $ (0, \infty)$. By (1) and (2),
\begin{eqnarray*}
\int_0^\omega (f^2(x) - g^2(x)) dx
& = & \int_0^\omega\left( \frac{(f(x) - g(x))^2}{2} \right)' dx \\
& = & \left[ \frac{(f(x)-g(x))^2}{2} \right]_0^\omega \\
& = & 2e^{-2\omega} \left( \int_0^\omega e^t f(t) dt \right)^2 .
\end{eqnarray*}
Making use of (3), the statement follows.

My proof was similar. I set $q=\exp(-x) \int_0^{x}\exp(t)f dt$ and the problem reduces to showing that $\int_0^{\infty}q \ d q = 0$, where all that remains is establishing $\lim_{x\to\infty}q=0$.
I found this is a little bit tricky, but proceeded in a similar manner to your proof. That being said, it looks like you may have some inaccuracies in your proof.

[quote="Kouichi Nakagawa"]
Using the Schwarz inequality, we obtain
\begin{eqnarray*}
e^{-\omega}\left| \int_0^\omega e^t f(t) dt \right|
& \le & e^{-\omega}\left| \int_0^{\omega/2} e^t f(t) dt \right| + e^{-\omega}\left| \int_{\omega/2}^\omega e^t f(t) dt \right| \\
& \le & e^{-\omega} \left( \int_0^{\omega/2} e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_0^{\omega/2} f^2(t) dt  \right)^{\frac{1}{2}} + e^{-\omega} \left( \int_{\omega/2}^\omega e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_{\omega/2}^\omega f^2(t) dt  \right)^{\frac{1}{2}} \\
& \le & e^{-\omega/2} \int_0^\infty f^2(x) dx + \int_{\omega/2}^\infty f^2(t) dt,
\end{eqnarray*}
whence it follows that
\[ \lim_{\omega \to \infty} \int_0^\omega e^t f(t) dt = 0.  \hspace{225pt}(3)\]
[/quote]

I think you meant to say the following:
[quote="Is this what you meant?"]
Using the Schwarz inequality, we obtain
\begin{eqnarray*}
e^{-\omega}\left| \int_0^\omega e^t f(t) dt \right|
& \le & e^{-\omega}\left| \int_0^{\omega/2} e^t f(t) dt \right| + e^{-\omega}\left| \int_{\omega/2}^\omega e^t f(t) dt \right| \\
& \le & e^{-\omega} \left( \int_0^{\omega/2} e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_0^{\omega/2} f^2(t) dt  \right)^{\frac{1}{2}} + e^{-\omega} \left( \int_{\omega/2}^\omega e^{2t} dt  \right)^{\frac{1}{2}} \left( \int_{\omega/2}^\omega f^2(t) dt  \right)^{\frac{1}{2}} \\
& \le & e^{-\omega/2} \left|\int_0^\infty f^2(x) dx\right|^\frac{1}{2} + \left|\int_{\omega/2}^\infty f^2(t) dt\right|^\frac{1}{2},
\end{eqnarray*}
whence it follows that
\[ \lim_{\omega \to \infty} \exp(-\omega)\int_0^\omega e^t f(t) dt = 0.  \hspace{225pt}(3)\]
[/quote]