Find all real numbers $a$ for which the equation $x^2a- 2x + 1 = 3 |x|$ has exactly three distinct real solutions in $x$.

Cute Problem :D
[hide]We will divide the problem into 3 cases: $x=0,x>0,x<0$
-Case 1: $x=0$
We have $1 = 0$, clearly not true; no value of $a$ will satisfy
-Case 2: $x>0$
We have $ax^2-5x+1$
$x=\frac{5\pm\sqrt{25-4a}}{2a}$           (1)
-Case 3: $x<0$
We have $ax^2+x+1$
$x=\frac{-1\pm\sqrt{1-4a}}{2a}$            (2)
In order for the equation to have only 3 real values: we should have two scenarios:
1. The discriminant of (1) must be 0 and the value of x from that corresponding a-value must be positive in (1). Then plugging in the a-value into (2), we must have two real, negative roots.
In (1)
$a=\frac{25}{4}$, but in (2), we have two imaginary roots
2. The discriminant of (2) must be 0 and the value of x from that corresponding a-value must be negative in (2). Then plugging in the a-value into (1), we must have two real, positive roots.
$a=\frac{1}{4}$; in (1), x is positive and real.

So $a =\frac{1}{4}$[/hide]

[u][i][b]FIRST OF ALL - THE POSTED PROBLEM IS WRONG!!!![/b][/i][/u]

the problem goes as such
$ {x^2} \plus{} \left( {a \minus{} 2} \right)x \plus{} 1 \equal{} \left| x \right|$
without the a in beginning

[b][i][u]<NOTE THE SOLUTION MAY LOOK COMPLEX BUT IS REALLY VERY SIMPLE, IT LOOKS LENGTHY BECAUSE I HAVE EXPLAINED EVERY STEP TO GIVE A COMPREHENSIVE SOLUTION>
<SOLUTION CONTINUES TO NEXT ARTICLE>[/u][/i][/b]

The solution to this problem would be -

$ {\left( {{x^2} + \left( {a - 2} \right)x + 1} \right)^2} = {x^2} \\
\implies{\left( {{x^2} + \left( {a - 2} \right)x + 1} \right)^2} - {x^2} = 0 \\
\implies \left( {{x^2} + \left( {a - 2} \right)x + 1 - x} \right)\left( {{x^2} + \left( {a - 2} \right)x + 1 + x} \right) = 0 \\
\implies \left( {{x^2} + \left( {a - 3} \right)x + 1} \right)\left( {{x^2} + \left( {a - 1} \right)x + 1} \right) = 0$

each solution from each quadratic equation must be real for there to be three real solutions as complex roots always occur in pairs

Condition 1 ------
$ D \ge 0{\rm{ in both expressions}} \\
\Rightarrow {\left( {a - 3} \right)^2} - 4 \ge 0{\rm{ AND}} \\
\Rightarrow {\left( {a - 1} \right)^2} - 4 \ge 0 \\
{\left( {a - 3} \right)^2} - 4 \ge 0 \\
\Rightarrow \left( {a - 5} \right)\left( {a - 1} \right) \ge 0 \\
\Rightarrow a \in \left( { - \infty } \right.,\left. 1 \right] \cup \left[ 5 \right.,\left. \infty \right)$

$ {\left( {a - 1} \right)^2} - 4 \ge 0 \\
\Rightarrow \left( {a - 3} \right)\left( {a + 1} \right) \ge 0 \\
\Rightarrow a \in \left( { - \infty } \right.,\left. { - 1} \right] \cup \left[ 3 \right.,\left. \infty \right) \\
\Rightarrow a \in \left( {\left( { - \infty } \right.,\left. { - 1} \right] \cup \left[ 3 \right.,\left. \infty \right)} \right) \cap \left( {\left( { - \infty } \right.,\left. 1 \right] \cup \left[ 5 \right.,\left. \infty \right)} \right) \\
 \\
\Rightarrow a \in \left( { - \infty } \right.,\left. { - 1} \right] \cup \left[ 5 \right.,\left. \infty \right)$

Condition 2------------------

Now, If there are only three solutions then it may be that only one of the polynomials may have equal roots
\[ \& {\rm{For : - }}{x^2} + \left( {a - 3} \right)x + 1 = {\left( {x - \alpha } \right)^2} \\
\& {\left( {a - 3} \right)^2} - 4 = 0 \\
\& \Rightarrow \left( {a - 5} \right)\left( {a - 1} \right) = 0 \\
\& \Rightarrow a \in \left\{ {5,1} \right\}\]

The problem is now also wrong.  The problem is

X^2+(a-2)x+1=3|x|

[quote=Maharjun]Similarly,
$ {\rm{For : - }}{x^2} + ( {a - 1} )x + 1 = {( {x - \alpha } )^2} \\
{( {a - 1} )^2} - 4 = 0 \\
\Rightarrow ( {a - 3} )( {a + 1} ) = 0 \\
\Rightarrow a \in \{ {3, - 1} \}$

Thus we have as of now,
$ \& \Rightarrow a \in ( {\{ {5,1} \} \cup \{ {3, - 1} \}} ) - ( {\{ {5,1} \} \cap \{ {3, - 1} \}} ) \\
\& \Rightarrow a \in ( {\{ {5,1} \} \cup \{ {3, - 1} \}} ) - \phi \\
\& \Rightarrow a \in \{ {5,1} \} \cup \{ {3, - 1} \}$

3rd Condition-------------------

both the polynomials have only one root common -

f $ \alpha$  is the common root of :-

$ \& {x^2} + ( {a - 3} )x + 1{\rm{ AND}} \\
\& {x^2} + ( {a - 1} )x + 1$

$ \& \Rightarrow {\alpha ^2} + ( {a - 3} )\alpha + 1 = 0{\rm{ AND}} \\
\& \Rightarrow {\alpha ^2} + ( {a - 1} )\alpha + 1 = 0$

Subtracting, we get -
${ \& \Rightarrow \alpha = 0}$

cross multiplying & subtracting, we get :-
$ {\alpha ^2} = {{a - 3 - a + 1} \over {a - 1 - a + 3}} = - 1 \ne {0^2} \ne {( \alpha )^2}$

Hence, for no value of a, the two polynomials have common root -

Therefore $ a\in\phi$

NOW, Conditions 1,2 and 2,3 were simultaneous conditions but 2,3 were alternate conditions(either one of them was sufficient to have only 3 solutions).

thus $ -$ $ a \in (\{5,1,3, - 1} \}) \cap ( {( - \infty, - 1] \cup [5,\infty))$

so $ a \in \{5, - 1\}$

so a can be [u][i][b]only 5 or -1[/b][/i][/u][/quote]

The two equations can't have a common root as
$x \ge 0$ and $x < 0$ can't occur simultaneously.


beddoes left? huh why?
not sure if jls has any worthy teacher left
hm i heard peterson was good, but probably not in the problem solving way

Mr. Peterson. Plus the meetings will be on Thursday, unlike before.

Ms. Beddoes moved to Utah because she got married.

[quote="bookaholic"]Mr. Peterson. Plus the meetings will be on Thursday, unlike before.

I believe Ms. Beddoes moved to Utah because she got married.[/quote]
how did i guess
i am so awesome

alright you guys need some beastly problem sets now
gunn's lacking in (young) beasts, we need to train some

edit: whats a utah

[quote="CheeseIsGood"][quote="junggi"][quote="bookaholic"]Mr. Peterson. Plus the meetings will be on Thursday, unlike before.

I believe Ms. Beddoes moved to Utah because she got married.[/quote]

edit: whats a utah[/quote]

I seriously hope that was a joke/sarcasm.

I'm not even sure that I'm going to Gunn. I want to, but I live in the Paly area. My parents want to move to a Gunn area soon though.
And I suck at math. Either that, or I just don't study enough. 15-60 minutes a day in the summer, lol.[/quote]
the fact that yo did math everyday is really impressive actually
unless it was a parental thing

[quote="CheeseIsGood"]My parents want me to study, but they don't actually enforce it. I really didn't learn anything new over the summer. For like a whole week all I did was simple problems involving triangle congruency. And I got a 36/46 on a chapter MathCounts thing, because I missed 2 target and 2 sprint problems with stupid mistakes. I always lose at least 1 point because of stupid mistakes.  :wallbash_red:[/quote]
then do more sprints
inever actually did those  because they are very uninteresting, but i heard they improve the mistakes thing
as you may have noticed i still suffer from stupid mistakes thing

edit
[url]http://www.artofproblemsolving.com/Resources/AoPS_R_A_Mistakes.php[/url]

I think part of what I failed to provide last year was non-beastly problem sets--I kept on putting in lots of cool problems without realizing that what the majority of people really need are problems that are more representative of typical School/Chapter rounds. This a) scared a lot of people off and b) didn't give them a chance to practice at the right level. This year I'm going to tone down the first 2/3 or so of the problem sets and see if that helps.

CheeseIsGood: hey, 36 is pretty good. Any daily work on math is extremely good--I for one haven't pulled it off yet. Everybody makes stupid mistakes, so don't get too frustrated about it. You'll improve, especially if you can maintain the amount of studying you did during the summer.

[quote="CheeseIsGood"]My parents want me to study, but they don't actually enforce it. I really didn't learn anything new over the summer. For like a whole week all I did was simple problems involving triangle congruency. And I got a 36/46 on a chapter MathCounts thing, because I missed 2 target and 2 sprint problems with stupid mistakes. I always lose at least 1 point because of stupid mistakes.  :wallbash_red:[/quote]
As bookaholic said, 36 is a pretty good score. Don't think of yourself as a person who can't achieve anything because that's how it'll turn out. Think positive, and positive things will happen. Studying is necessary and practice is the key to success once you know what you're doing.

[quote="CheeseIsGood"]So... I don't think that anyone that still goes to JLS uses this forum, but I'll ask anyway. Since Ms. Beddoes left, who's the new coach?[/quote]

WHAT???? SHE LEFT??????
not cool. she was the pro math teacher at JLS


let $f: \mathbb{R}^2 \longrightarrow \mathbb{R}^2$  be a continous fuction s.t: $fof=id$
show that $f$ have at least one fix point.
I hope find a elementery solution without using brouwer theorem

Good luck

But can you post the solution using Brower's theorem? :)


geometry
3D geometry
octahedron
pyramid
tetrahedron

An octahedron is 2 square pyramids glued together. So what is a figure with 2 tetra hedrons glued together called?

[quote="Ubemaya"]it's a cube[/quote]
NO ITS A SQUARE!!!!
Well if it's regular tetrahedrons, then you get a hexahedron ;) Cubes have square faces.

It is a hexahedron. But not a regular hexahedron. A regular hexahedron is a cube.

I'm having a hard time imagining a tetrahedron with square faces.  What you want is the [url=http://en.wikipedia.org/wiki/Triangular_dipyramid]triangular dipyramid[/url].  (There is more than one [url=http://en.wikipedia.org/wiki/Hexahedron]hexahedron[/url].)

Thanks. :D

(Cubes have square faces, while tetrahedrons have triangular faces, so it can't be a cube)

[quote="Ubemaya"]
it's a cube
[/quote]

[quote="serialk11r"]
NO ITS A SQUARE!!!
Well, if it's regular tetrahedrons, then you get a hexahedron
[/quote]

[quote="Ubemaya"]
A HEXAHEDRON IS A CUBE
[/quote]

[quote="hunter34"]
A regular hexahedron is a cube
[/quote]

epic fail

No. I no epic fail.

http://dictionary.die.net/regular%20hexahedron

6 sides = hexahedron = cube
I no epic fail either.

this is random

an octahedron is an [url=http://en.wikipedia.org/wiki/Antiprism]antiprism.[/url]


inequalities
algebra
polynomial
linear algebra
matrix
geometry
linear algebra unsolved

a) Let $A,B\in\mathcal{M}_3(\mathbb{R})$ such that $\mathop\textrm{rang} A> \mathop\textrm{rang} B$. Prove that $\mathop\textrm{rang}{(A^2)} \geq \mathop\textrm{rang}{(B^2)}$;

b) Determine the non-constant polynomials $f$ with real coefficients, with the property that for any matrices $A,B\in\mathcal{M}_4(\mathbb{R})$ with $\mathop\textrm{rang}A > \mathop\textrm{rang}B$ we have
\[\mathop\textrm{rang}{f(A)} \geq \mathop\textrm{rang}{f(B)}. \]

a)Suppose $Rang(B^2)\geq 1$

$Rang(A^2)\geq 2Rang(A)-3\geq 2Rang(B)-1\geq Rang(B^2)$

Can I ask a stupid question, does there exists some simple inequalities
characterizing $ \geq Rang(X+Y)\geq $ with $Rang(X), Rang(Y)$

For a/

Case $rank(A)=3$

$A$ is invertible $A^2$ also, $rank(A^2)=3\geq rank(B^2)$

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

Case $rank(A)=2$

$rank(A)=2>rank(B)$ then $rank(B)=0$ or $1$
With Sylvester inequality $rank(XY)\geq rank(X)+rank(Y)-n$ we have

$rank(A^2)\geq 2rank(A)-3=1$

If $rank(B)=0$ then $B=0$ and $rank(A^2)\geq rank(B^2)=0$

If $rank(B)=1$ then $B^2=(trB)B$ and $rank(A^2)\geq 1=rank(B^2)=rank(B)$

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

Case $rank(A)=1$

then $rank(B)=0=rank(B^2)$
$rank(A^2)\geq rank(B^2)$

[quote="prowler"]
 does there exists some simple inequalities
characterizing $ \geq Rang(X+Y)\geq $ with $Rang(X), Rang(Y)$[/quote]

$|rang(X)-rank(Y)|\leq rank(X+Y) \leq rank(X)+rank(Y)$

$x<y$  imply that $x \leq y$

For b/

$f(X)=X$

Any other suggestion for b/ ?

thx myth,

$f(x)=kx$ where $k\in R-(0)$  in question b/ I don't see

the reason of restriction to matrix size 4x4,  $M_4(R)$

it works for matrices size nxn, $M_n(R)$

Yes, [b]M[/b]oubi... Rrrr...

I also don't see why we need the restriction (because I have not the solution :P). These area is quite dark for me. But I am trying to think about it.

Suppose $\deg f\geq 2$ and $f(x)$ has a nonzero root (real or complex). Then $f(x)$ has a factor of the form $x^2+bx+c$, there $b\neq 0$. We can now easily find $A\M_2(\mathbb R)$ s.t. $f(A)=0$ (use the fact that $P(A)=0$, where $P$ is the charecteristic polynomial). It shows that such polynomials $f$ cannot satisfy the condition.
Therefore we are to check $f(x)=x^k$. I guess there is a similar trick for $k\geq 2$. :?

Sorry, if it is nonsense...  :blush:


[quote="Nplusnequals"]Well, I recently encountered triangular numbers, eg. the number of points you can arrange into an equilateral triangle. The first triangular numbers are 1, 3, 6, 10, 15, ....
They can be found with the formula n(n+1)/2.
Then threre are square numbers, the number of points you can arrange into a square, their formula is obviosly n^2.
Then there are pentagonal numbers, the number of points you can arrange into a regular pentagon. The first such numbers are 1, 5, 12, 22, .... Their formula, significantly less obvious, is n(3n-1)/2.
I set out to find a formula that could find the nth x-gonal number....and I found one! This quite interesting formula is

[n(s-2)(n-1) + 2n]/2

where s is the number of sides of the x-gon.
For example, using the formula, we can fin the 1000th 100-gonal number, which is

48,952,000.[/quote]

o.0 Nice
I'm not sure if this belongs here but is it possible to prove this formula


I could use some help with this question:

Let A and B be finite sets. Prove that the (cardinality of A x B) = (cardinality of a) x (cardinality of b)

Could someone give me a hint or point me in the right direction? Thanks!

Which are the elements of $ A \times B$ ? In how many ways can one choose the first element in the ordered pair ? and the second ? Now put all together !


floor function
number theory unsolved
number theory

From the list of positive integers in increasing order, delete all multiples of 4 and all numbers 1 more than a multiple of 4. Let $ S_n$   be the sum of the first $ n$ terms in the sequence which remains. Compute  $ \lfloor\sqrt {S_1}\rfloor\plus{}\lfloor\sqrt {S_2}\rfloor\plus{}\cdots\plus{}\lfloor\sqrt {S_n}\rfloor$

let $ a_n$ such that $ a_{2k}\equal{}4k\plus{}2,a_{2k\plus{}1}\equal{}4k\plus{}3$
we have $ S_n\equal{}\sum_{i\equal{}0}^{n\minus{}1}a_i$
then $ S_{2k}\equal{}\sum_{i\equal{}0}^{k\minus{}1}a_{2i}\plus{}\sum_{i\equal{}0}^{k\minus{}1}a_{2i\plus{}1}\equal{}(2k)^2\plus{}k$ gives $ [\sqrt{S_{2k}}]\equal{}2k$
and $ S_{2k\plus{}1}\equal{}S_{2k}\plus{}a_{2k\plus{}1}\equal{}(2k)^2\plus{}5k\plus{}3$ gives $ [\sqrt{S_{2k}}]\equal{}2k\plus{}1$
then $ [\sqrt{S_n}]\equal{}n$ and the sum is $ n(n\plus{}1)/2$


For $\alpha \in (0,1)$ we consider the equation $\{x\{x\}\}= \alpha$.

a) Prove that the equation has rational solutions if and only if there exist $m,p,q\in\mathbb{Z}$, $0<p<q$, $\gcd(p,q)=1$, such that $\alpha = \left( \frac pq\right)^2 + \frac mq$.
b) Find a solution for $\alpha = \frac {2004}{2005^2}$.

For (b): $\alpha = \frac{\sqrt{2004}}{2005}$. I heard that the jury gave points for that since it said nowhere that the solution must be rational.

During the exam, I also thought of this, but I decided not to write anything at (b). Didn't miss much, though.

(a): $ \{x\{x\}\} = x(x + a) + b$ for some integers $ a,b$
(b): $ \alpha = \frac {822}{2005} - 498$

Extension: in part $ (b)$, for which of the following values of $ \alpha$ is there a solution:
1. $ \frac{6}{49}$
2. $ \frac{136}{137^2}$
3. $ \frac{1800}{1801^2}$
[hide="Answer"]
Only in 2. and 3.
[/hide]


trigonometry
geometry
circumcircle
geometry proposed

Let $ABC$ be an acute triangle with altitudes $AD,BE,CF$ and let $O$ be its circumcenter. Let $M,N,P$ be the midpoints respectively of the segments $BC,CA,AB$. Let $D_{1},E_{1},F_{1}$ be the reflexions respectively of $D$ in $M$, of $E$ in $N$, of $F$ in $P$. Prove that $O$ lies inside the triangle $D_{1}E_{1}F_{1}$.

We have $AE_{1}= CE = a \cos C,$ $AF_{1}= BF = a \cos B.$ By the sine theorem for the $\triangle AE_{1}F_{1},$

$\frac{\sin \widehat{AF_{1}E_{1}}}{\sin \widehat{AE_{1}F_{1}}}= \frac{AE_{1}}{AF_{1}}= \frac{\cos C}{\cos B}$

Let AO meet $E_{1}F_{1}$ at K. Then $\angle KAE_{1}= \angle OAC = 90^\circ-B$ and $\angle KAF_{1}= \angle OAB = 90^\circ-C.$ By the sine theorem for the $\triangle AKE_{1},\ \triangle AKF_{1},$

$\frac{KE_{1}}{KF_{1}}= \frac{\sin (90^\circ-B)}{\sin \widehat{AE_{1}K}}\cdot \frac{\sin \widehat{AF_{1}K}}{\sin (90^\circ-C)}= \frac{\cos B}{\cos C}\cdot \frac{\cos C}{\cos B}= 1$

so that K is the midpoint of $E_{1}F_{1},$ AK the A-median of the $\triangle AE_{1}F_{1},$

$AK^{2}= \frac{2AE_{1}^{2}+2AF_{1}^{2}-E_{1}F_{1}^{2}}{4}=$

$= \frac{a^{2}}{4}\ (\cos^{2}B+\cos^{2}C+2 \cos A \cos B \cos C) = \frac{a^{2}\sin^{2}A}{4}$

$AK = \frac{a \sin A}{2}= R \sin^{2}A,\ \ OK = AO-AK = R \cos^{2}A > 0.$

($\angle A$ is acute). Thus K lies on a circle $(O,\ \rho)$ with center O and radius $\rho = R \cos^{2}A$ concentric with the triangle circumcircle (O, R). As long at the $\triangle ABC$ is acute, $E_{1},\ F_{1}$ lie inside the segmets AC, AB and K lies inside the segment AO. The line $E_{1}KF_{1}$ then intersects the $\triangle AOC,\ \triangle AOB$ inside the segments AC, AO and AB, AO, respectively, hence it separates their common vertices A, O. In exactly the same way, $F_{1}D_{1}$ separates B, O and $D_{1}E_{1}$ separates C, O. As a result, O lies inside the $\triangle D_{1}E_{1}F_{1}.$


topology
Functional Analysis
advanced fields
advanced fields unsolved

Is the set of invertible operators on a complex Hilbert space connected in the topology induced by the norm of operators?

(edited to add the word [i]complex[/i])

The experience of the finite dimensional case suggests that it may matter whether you take the scalars of your Hilbert space to be real or complex.

The set of invertible complex matrices is connected, but the set of invertible real matrices splits into two connected components according to whether the determinant is positive or negative.

Take the scalars to be complex. Sorry for not mentioning it.

I think one can do as follows:

suppose that $A$ is an invertible continuous operator. Then $A=UH$ where
$U$ is a unitary operator and $H$ is a positive hermitian ( It suffices to take
$H=\sqrt{A^{*}A}$ ) Hence to show that the set of invertible operator is connected
it suffices to show that the set of invertible positive hermitian operators is connected
and that the set of unitary operators is also connected:
1/ the set of invertible positive hermitian operators is convex
2/ suppose that $U$ is a unitary operator. Then if $\phi_{t}: \mathbb{U}\to \mathbb{U}$ is defined by
$\phi_{t}(e^{i \theta})=e^{i \theta t}$, $\theta \in ]-\pi; \pi]$ then
$[0;1] \to \phi_{t}(U)$ is a path between $Id$ and $U$.


calculus
derivative
function
trigonometry
calculus computations

Here are some of the problems that I got stuck:

Find the derivatives of the functions:

$ f(x) \equal{} (3x^2 \plus{} \pi)(e^x \minus{} 4)$

$ f(t) \equal{} 2 \cdot e^{\minus{}2e^{2t}}$

$ f(x) \equal{} \sin(\sin x \plus{} \cos x)$

$ f(x) \equal{} \sin(2x) \cdot \sin(3x)$

[quote="jeez123"]
$ f(x) = (3x^2 + \pi)(e^x - 4)$  [/quote]
Applying product rule, i get $ f'(x)=6x(e^x-4)+e^x(3x^2+\pi)$
[quote="jeez123"]
$ f(x) = \sin(\sin x + \cos x)$
 [/quote]
$ f'(x)=\cos(sinx+cosx).(cosx-sinx)$

[quote="jeez123"]
$ f(x) = \sin(2x) \cdot \sin(3x)$[/quote]
$ f(x)=\frac{1}{2} ( \cos x-\cos5x)$
Now this can be differentiated easily

[hide="For the second"]For $ f(t) \equal{} 2 \cdot e^{\minus{}2e^{2t}}$ we have $ f'(t) \equal{} 2 \cdot \left(\minus{}2e^{2t}\right)' \cdot e^{\minus{}2e^{2t}} \equal{} 2 \cdot (\minus{}4e^{2t}) \cdot e^{\minus{}2e^{2t}} \equal{} \minus{}8e^{2t \minus{} 2e^{2t}}$.[/hide]


algebra
polynomial
calculus
integration
logarithms
function
rational function

Prove that there are no polynomials $ P,Q\in\mathbb{R}[X]$ such that
\[ \int_{0}^{\log n}\frac {P(x)}{Q(x)}dx \equal{} \frac {n}{\pi(n)},\forall n\geq 1,
\]
where $ \pi(n)$ is the prime counting function.

This is quite clear, right? Let $ R$ be that rational function and put $ f(x)\equal{}\frac{1}{x}\cdot\int_{0}^{\ln(x)}{R(t)dt}$. Then you have $ f(n)\equal{}f(n\plus{}1)$ whenever $ n\plus{}1$ is composite, thus $ f'$ vanishes infinitely many times by Rolle, so there is a sequence $ c_n$ between $ n$ and $ n\plus{}1$ whenever  $ n\plus{}1$ is composite such that $ f'(c_n)\equal{}0$. Since $ f(x)\plus{}xf'(x)\equal{}\frac{1}{x}R(\ln(x))$, this gives $ R(\ln(c_n))\equal{}\int_{0}^{\ln(c_n)}{R(t)dt}$, which is certainly a bit difficult using asymptotic study!

[quote="harazi"]$ R(\ln(c_n)) \equal{} \int_{0}^{\ln(c_n)}{R(t)dt}$, which is certainly a bit difficult using asymptotic study![/quote]
why?
if $ R(t) \approx 1$ for big $ t$ then there is no contradiction in asymptotic study  :P

Try to think a bit more and you'll see that your case is absolutely not contradictory with what I said.

If I undersand right your last statement about
$ R(\ln(c_n))\equal{}\int_{0}^{\ln(c_n)}{R(t)dt}$ that this can't be true for all $ c_{n}$ where $ R(t)$ is rational function.
This statement is false because choose $ R(t): \equal{}1$ then contradiction with your statement. I know that such $ R(t)$ don't satisfies condition of problem , but satisfies your statement.  :maybe:

I think that when you integrate 1 from $ 0$ to $ a$ you get $ a$. So surely by choosing $ R(t)\equal{}1$ you cannot have the equality you suggest. On the other hand, it is really a trivial matter to realize that the two parts cannot have the same asymptotic behaviour unless $ R\equal{}0$, which is definitely not acceptable.

sorry, I'm so stupid  :blush: . I was thinking that : if $ R(t) \equal{} 1$ then $ R(\ln c_{n}) \equal{} \ln c_{n}$ :lol:


1. Three straws are chosen from a set of nine straws whose lengths are 1 inch, 2 inches, 3 inches, . . . , 9 inches. What is the probability that the three straws, placed end to end, will form a triangle?

2. Find three consecutive odd numbers, the sum of whose squares consists of four identical digits.

[hide="Solution to 2"]
Let $ a$ be the middle number. Then the three consecutive odd numbers are $ a\minus{}2,a,a\plus{}2$.
Thus, $ (a\minus{}2)^{2}\plus{}a^{2}\plus{}(a\plus{}2)^{2}\equal{}3a^{2}\plus{}8\equal{}1111,2222,...,9999$. By trial and error, we find that only $ 3a^{2}\plus{}8\equal{}5555 \Rightarrow a\equal{}43$ produces an integer $ a$. Thus, the three consecutive odd integers are $ \boxed{41,43,45}$.

[/hide]


articles
AMC
AIME
function
AoPSwiki
geometry
power of a point

Isn't it possible for articles to have sub-articles?  For example, rather than having "Power of a Point Theorem Introductory Problem 4" on the page "Power_of_a_Point_Introductory_Problem_4", we could have it on the page "Power_of_a_Point/Introductory_Problem_4".  It's not a huge issue, obviously, but I think it would both be neater and easier to follow if example problems weren't their own articles.

I agree.  In fact, I think it would be nice in general if this were true of the problem pages like AMC, AIME, etc.  Otherwise, the "Random Page" function becomes very useless, because there are so many pages devoted to competition questions that you never get to anything interesting.  I was going to complain about each of the different states having its own page, too, but those 50 pages will eventually be dwarfed by the mathematical content.  People don't create new states as frequently as they create new contests.

Yes, that's obviously better.  Don't be afraid to speak up....you shouldn't assume that I (or anyone else for that matter) know all the in's and out's of wikis.

Edit:

Is the AoPSWiki capable of fully handling subarticles yet? When you use the syntax, it doesn't seem to do anything different.   [url=http://meta.wikimedia.org/wiki/Help:Link#Subpage_feature]Here[/url] it says that activation depends on the project/namespace.  I'm not sure if AoPSWiki has it activated.


We have a cylindrical test tube with the inner radius $r$ and the outer radius $R$. The tube is filled with fluorescent liquid. Knowing the indices of refraction of glass and the liquid, for what ratio $r/R$ we won't see the glass of the test tube at all? This is a very well-known problem, but I'm puzzled with its solution... :maybe:

is it r/R =< n_2/n_1 where n_1 is the refractive index of the glass and n_2 is that of the air?


[quote="manlio"]For $ a,b,c$ positive reals prove that

$ \sum ab \geq \sum \frac {4a^3(b \plus{} c \minus{} a)}{(b \plus{} c)^2}$

See also http://canhang2007.files.wordpress.com/2008/12/topic-of-luat.pdf

at pag.11[/quote]

I can't open your link.

and for this ineq, I try to use tangent line method to solve it but failed, because when I assume that $ a\plus{}b\plus{}c\equal{}1$, ineq becomes

\[ \sum {\frac{{a^2 \left( {1 \plus{} 6a \minus{} 15a^2 } \right)}}{{\left( {1 \minus{} a} \right)^2 }}}  \le 1.\]

for $ a\equal{}\frac{1}{3}$ the tangent line is $ 2a\minus{}\frac{1}{3}$, but ineq $ \frac{{a^2 \left( {1 \plus{} 6a \minus{} 15a^2 } \right)}}{{\left( {1 \minus{} a} \right)^2 }} \le 2a\minus{}\frac{1}{3}$ dose not hold,  :( ...

[quote="geniusbliss"]i am sorry to revive this but what exactly is the tangent line method? can someone provide me with a link or something..[/quote]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=149168]See here.[/url]

I also failed by tangent line method. :blush:


logarithms
quadratics
modular arithmetic
algebra
polynomial
Vieta
search

Find the last three digits of the product of the positive roots of \[ \sqrt{1995}x^{\log_{1995}x}=x^2. \]

[hide="Solution"]Take the log base 1995 of both sides
\begin{eqnarray*} \log_{1995}\left(\sqrt{1995}x^{\log_{1995}x}\right) &=& \log_{1995}(x^2)\\ \log_{1995}1995^{1/2}+(\log_{1995}x)^2 &=& 2\log_{1995}x\\ 2(\log_{1995}x)^2-4\log_{1995}x+1 &=& 0\\ \Rightarrow \log_{1995}x=\frac{2\pm \sqrt{2}}2 \end{eqnarray*}

Both $\frac{2+\sqrt{2}}2$ and $\frac{2-\sqrt{2}}2$ are positive, so they both produce a solution.  So then $x_1=1995^{(2+\sqrt{2})/2}$ and $x_2=1995^{(2-\sqrt{2})/2}$ where $x_1, x_2$ are solutions to the original equation.  The product $x_1x_2$ is therefore equal to $1995^2.$  We only want the last three digits, which is the same as taking it mod 1000.  So our answer is $1995^2\equiv (-5)^2\equiv 025\mod 1000.$[/hide]

[hide]Because $1995$ is popping up all over the place, it seems like a good idea to take $\log_{1995}$ of both sides of the equation.

$\log_{1995}\sqrt{1995}x^{\log_{1995}x}=\log_{1995}x^2$
$\log_{1995}\sqrt{1995}+\log_{1995}x^{\log_{1995}x}=\frac12 +(\log_{1995}x)^2=2\log_{1995}x$

Now, we can apply the quadratic formula to get $\log_{1995}x=\frac{2\pm \sqrt2}{2}\implies x=1995^{\frac{2\pm \sqrt2}{2}}$.

When we multiply these two values of x together, we get $1995^2$. Our answer is therefore $1995^2\equiv (-5)^2\equiv \boxed{025} \pmod{1000}$.[/hide]

how does $\log_{1995}x^{\log_{1995}x}$ = $(\log_{1995}x)^2$ ?
sorry for the crappy question =p

logs and exponents: $\log_a b^c=c\log_a b$

So $\log_{1995} x^{\log_{1995} x}=\log_{1995} x\times\log_{1995} x=\displaystyle \left(\log_{1995} x \right)^2$

it is true that you can find the roots, but can you find the faster way with vieta (search before asking what vieta is) (be careful with the logarithms too!)

My solution is a little more elegant.  I didn't actually find the roots of the equation.  :D

[hide]Taking the $\log_{1995}$ of both sides, we have

$\log_{1995}\sqrt{1995}+\log_{1995}x^{\log_{1995}x}=\log_{1995}x^{2}$
$\frac{1}{2}+(\log_{1995}x)^{2}=2\log_{1995}x$

The above equation is a quadratic, so there are two roots.  Let $r_{1}$ and $r_{2}$ be the two roots of the equation.  Let $y=\log_{1995}x$, so that $x=1995^{y}$.  The above equation becomes

$\frac{1}{2}+y^{2}=2y$
$2y^{2}-4y+1=0$

Let $s_{1}$ and $s_{2}$ be the two roots of this equation, so that $s_{1}+s_{2}=\frac{4}{2}=2$.  We want

$r_{1}r_{2}=1995^{s_{1}}1995^{s_{2}}=1995^{s_{1}+s_{2}}=1995^{2}$.

The answer is $1995^{2}\equiv(-5)^{2}\equiv \boxed{25}\pmod{1000}$.[/hide]

sorry for a question after a decade lol, but how did u guys find the last 3 digits using mod of 1995^2

If you're wondering how the $1995$ appeared, the solutions explains it pretty well. But the idea is that we can find the last three digits of something by taking $\pmod{1000}.$ This is because we can write any number as $1000x+\overline{abc}$ where $x$ is some integer and $\overline{abc}$ is a $3$-digit number. Taking $\pmod{1000}$ gets rid of that $1000x$ term and just leaves us with $\overline{abc}$, which are the last three digits.

Is this what you're confused about?


number theory
relatively prime
number theory proposed

Let $ a,b$ are distinct positive numbers.Prove that there exist infinitely many positive integer $ n$ such that $ a\plus{}n$ and $ b\plus{}n$ are relatively prime

[quote="Niichavo"]Let $ a,b$ are distinct positive numbers.Prove that there exist infinitely many positive integer $ n$ such that $ a \plus{} n$ and $ b \plus{} n$ are relatively prime[/quote]
Just set $ n \equal{} ka\minus{}(k\plus{}1)b\plus{}1$ then we have $ (a\plus{}n,b\plus{}n) \equal{} (a\minus{}b,k(a\minus{}b)\plus{}1) \equal{} (a\minus{}b,1) \equal{} 1$


Two dogs, each traveling 10 ft/sec, run toward each other from 500 feet apart. As they run, a flea flies form the nose of one dog to the nose of the other at 25 ft/sec. The flea flies between the dogs in this manner until it is crushed when the dogs collide. How far did the flea fly?

source: MAO 1992

[hide]It takes 25 seconds for the dogs to collide, and the flea travels 25ft/sec, therefore the flea travels 25*25= 625 feet.[/hide]

thats right!!!  :winner_first:

anyone else agree?

[hide]Since 500/20=25 seconds, it means the fly travels 25*25=625 feet[/hide]

@Buzzer

Where do you get these old Mu Alpha Theta problems from?

thanks.

try the MAO website:

http://www.mualphatheta.org/Problem_Corner/MAO_Problem_Corner_links.htm

[hide="easy way"]
The dogs take off 20 feet of the distance each second, so they collide after 500/20=25 seconds. Distance = rate * time so the flea travels 25 ft/s * 25 s = 625 feet.
[/hide]
[hide="hard way"]
Say the flea starts from the nose of dog #1. The first distance is 500. After the flea reaches dog #2's nose, it turns around and there is some distance to the other dog. Say the previous distance was x feet. The flea reaches to other nose in x/35 seconds (the flea goes 25 feet per second and the dogs reduce the distance at 20 feet per second , but only one dog decreases the distance the flea goes) The dogs have then decreased the distance by 20x/35 = 4x/7 feet. Therefore, each distance is 3/7 that of the last one. So we have the infinite series:

\[ 500 + 500(\frac{3}{7}) + 500(\frac 37)^2 + 500(\frac 37)^3 + \cdots = \frac{500}{1-\frac{3}{7}} = \frac{500}{\frac{4}{7}} = 875 \]

However, the fly does not travel the whole distance each time. It only travels 5/7 of the distance so the final answer is $\frac{5}{7} \cdot 875 = \boxed{625\ feet}$
[/hide]


What is the value of the mean of the set $ \{4, 5, 7, 2, 6, x\}$ if the range is 10, the median is 5.5 and $ x$ is an integer?

If the range is $ 10$, then $ x$ must equal $ 12$. $ \frac{4\plus{}5\plus{}7\plus{}2\plus{}6\plus{}12}{6}\equal{}\boxed{6}$


Circles $ O_1$ and $ O_2$ intersect at points $ A$ and $ B$, and the circle $ O_3$ tangents them from outside at the points $ C$ and $ D$ respectively. The line $ AB$ intersects the circle $ O_3$ in the points $ P$ and $ Q$. Prove that $ \angle PKC \equal{} \angle QKC$.


If there is an even number of distinct elements in a set, then how do you find the median?
If there is two elements that appears the same amount of times and appears the most frequent, then how do you find the mode?
What is the mean of an empty set?
If all elements in a set are equal, what is the range?

OK, those were definition questions, so here's another practice problem involving all four of these concepts:
Out of a set of eleven natural-number scores on a 10-point quiz, the mean is 8, the median is 8, and the mode is 7. Find the maximum number of perfect scores possible on this test.

Assuming the set is ordered, the answer is 3.

1)  What do you mean, "assuming the set is ordered?"  I don't see how that would affect the answer at all.
2) Do you not give solutions?  Why doesn't anyone give solutions?  Don't give an answer -- give a solution!

I give solutions.

Quote:Out of a set of eleven natural-number scores on a 10-point quiz, the mean is 8, the median is 8, and the mode is 7. Find the maximum number of perfect scores possible on this test.

[hide]Let's assume there are 11 perfect scores.  But we know that can't be right, because the mode is 7 (among other problems).  So there can be at most 5 perfect scores, since that would allow there to be 6 7's (thereby satisfying the mode requirement).  But the 6th-largest (or 6th smallest, for that matter) member must be an 8, so we need to take off a 7 and add an 8, like so:

7,7,7,7,7,8,10,10,10,10,10

Now there's another problem: there are 5 7's and 5 10's, so the mode is the average of the two, i.e. 8.5  That is bad.  So we need to take off a 10 so that there are still a majority of 7s.  Add an 8 to replace the 10.

7,7,7,7,7,8,8,10,10,10,10

Now for the last part: the mean.  Since the mean is 8, the numbers' sum must be 8*11 = 88.  But this sequence's sum is 91.  If we lower one of the 8s to a 5 to get 88, 8 is no longer the median.  Instead, we must lower one of the 7s to a 4.

4,7,7,7,7,8,8,10,10,10,10

This looks great!  Yeah!

Uh-oh.  Problem: the mode is now 8.5 again, since there are 4 10s and 4 7s.  You know what to do...take away a 10.  Make it a 9.  Then make the 4 a 5 (to rebalance; since we dropped the sum by one, we have to increase it by one again).

5,7,7,7,7,8,8,9,10,10,10

Thus there are three 10s maximum.[/hide]

[quote="mathfanatic"]Now there's another problem: there are 5 7's and 5 10's, so the mode is the average of the two, i.e. 8.5[/quote]

[quote="mathfanatic"]
Uh-oh. Problem: the mode is now 8.5 again, since there are 4 10s and 4 7s.[/quote]

No, your modes would be: in quoteA, 7 & 10. and in quoteB 7 & 10. You can have more than one mode...you don't average the two, but I don't think that affects the outcome of the problem...I was simply pointint it out.

Mode:The number(s) in a set that occur(s) the most.
So, since 8.5 isn't a number in the set, it can't be the mode.

Exactly why I made this problem in the first place, people (well, most) often screws up on the naming of these four (well mostly mean and mode) so they gets points subtracted, and it's annoying (I've got it wrong numerous times :( ), so just wanted people to be careful about these.
Whoops, he's actually correct since he took the average of the two most frequently appearing number, so (7+10)/2=8.5 Course I think teachers that point out people misused mode/mean [b]every single day[/b] (even after you graduate and is just visiting :lol: ), would be "mode"...er, "mean" :lol:
Just a little mode/mean joke, I'm in the mode today :lol:

As for the answer, it's correct.

No -- for mode you do not average the most frequently appearing numbers.  For median, you average if necessary.  For mode, you do not -- you say that there are several modes, for example 7 and 10.


Can someone post the site to the ARML books? I went onto the website and clicked on the links, but couldn't find them  :(

http://www.arml.com/books.htm

You might also want to look on Amazon or E-bay.

Does anyone know the prices of the books that are sold at the ARML site?
(I tried Amazon but couldn't find any...)

The word is that they are intentionally hard for students to obtain, because this ensures that coaches will have material that students have not seen before, which is obviously important in such a team setting.

If I remember correctly, the 1995-2003 book costs 20 bucks and the 1989-1994 book costs 10 bucks (at the Iowa site).


function
floor function
trigonometry
algebra

Ok I guess this is elementary over here, but could some one explain why are the periods of the functions:

$f(x) = \frac{|sinx+cosx|}{|sinx|+|cosx|}$ given as $\pi$

and

$f(x) = x-\lfloor x-b \rfloor$ given as $1$ ?

[hide="1"] We have the identities $\sin (x+\pi) =-\sin x$ and $\cos (x+\pi) =-\cos x$, so

$f(x+\pi) = \frac{|-\sin x-\cos x| }{ |-\sin x|+|-\cos x | }= f(x)$ [/hide]
[hide="2"] Note that the floor function $\lfloor a \rfloor$ satisfies

$\lfloor a+n \rfloor = \lfloor a \rfloor+n$

For all integers $n$.  Thus

$f(x+1) = (x+1)-\lfloor x+1-b \rfloor = (x+1)-\left( \lfloor x-b \rfloor+1 \right) = f(x)$ [/hide]

Surely if f(x + pi) = f(x), then we cannot deduce that the period is pi. A period of pi/2 would satisfy this for instance.

The period for (1) can't be smaller because $f(x)=0$ if and only if $x=\frac{3\pi}{4}+n\pi$ for some integer $n$.

The period for (2) can't be smaller because $f(x)=b$ if and only if $x=b+n$ for some integer $n$.


how do you get to proceed to the chapter competition??? :?:  :huh:

If you're homeschooled or your school doesn't have MATHCOUNTS program, you can compete as an individual.

[quote="mathluver2.718"]You can go to chapter MathCounts by being in the top 8 people at your school.[/quote]

not necessarily. some schools (my school, for example) only send one team of four, without any individs.

[quote="echen"]Alternates are also allowed in most years so your school can send 10-12.[/quote]

Um, is it not $ 8$ people at most?

i believe 8 is the max-4 person team and 4 individs.

Nah, the team I was on in sixth grade also brought two alternates. I don't think that their scores counted at all though, so bleh. :P

[quote="athunder"]if for some odd reason some guy didnt make the school team can he still register for mathcounts as an individual...[/quote]
Yes. Each school can send 4 individuals. If you don't make those, go through a different school.


modular arithmetic
number theory unsolved
number theory

$\\a^{x}+x \equiv b \pmod c \\(a^{x}+x-b)/c = k \\a^{x}= kc+b-x \\kc+b \equiv x \pmod a \\(kc+b-x)/a = p \\x = kc+b-ap$

Where k, c, b, a, and p are all integers.
Therefore, x is also integer.
When $kc+b > ap$, then $x > 0$.

I know that something has to be wrong with this proof. What is it?

[quote="car202"]$kc+b \equiv x \pmod a$[/quote]

Not necessarily, if $x = 0$.

Thanks. One more question though.
Is it complete if I prove the case of $x=0$?
And also, since we only have to prove in case of $x>0$, why does it matter?
Sorry for my ignorance.

What exactly is the problem being posed?

I think the problem is http://www.mathlinks.ro/Forum/viewtopic.php?p=354533#p354533

The logic is still iffy; it runs in the wrong direction.  You start by assuming that $x$ exists, and then shows that it has some form in terms of $k, p$.  But if $x$ does not exist then neither do $k$ or $p$!  Circular.


plz explain or link to me which about m.v

hello, see here
http://www.reflections.awesomemath.org/2006_6/2006_6_mixing.pdf
Sonnhard.


[quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum \frac {a^3}{b^4 \plus{} c^4} \ge \frac {a \plus{} b \plus{} c}{2}$
------------------------------------
[/quote]
Maybe  $ \sum \frac {a^4}{b^3 \plus{} c^3} \ge \frac {a \plus{} b \plus{} c}{2}$?
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216986

[quote="Pain rinnegan"][quote="frankvista"][quote="bigbang195"]$ a,b,c > 0$. Prove that !
$ \sum \frac {a^3}{b^4 \plus{} c^4} \ge \frac {a \plus{} b \plus{} c}{2}$
[/quote]
\[ \Leftrightarrow (\sum {{a^{10}}} \minus{} \sum {{a^6}{b^3}c} ) \plus{} (\sum {{a^{10}}} \minus{} \sum {{a^6}b{c^3}} ) \plus{} (\sum {{b^3}{c^3}({b^4} \plus{} {c^4})} \minus{} \sum {{b^4}{c^4}({b^2} \plus{} {c^2})} ) \ge 0\]
$ 6a^{10} \plus{} 3b^{10} \plus{} c^{10} \ge 10a^6b^3c$ (AM-GM)

So

$ \sum {{a^{10}}} \minus{} \sum {{a^6}{b^3}c} \ge 0$

Similarly,

$ \sum {{a^{10}}} \minus{} \sum {{a^6}b{c^3}} \ge 0$

Hence

$ \sum {{b^3}{c^3}({b^4} \plus{} {c^4})} \minus{} \sum {{b^4}{c^4}({b^2} \plus{} {c^2})} \equal{} \sum{b^3c^3(b \minus{} c)^2(b^2 \plus{} bc \plus{} c^2)} \ge 0$,

We've done. #[/quote]

But that inequality is clearly wrong.[/quote]

Couldn't you think the problem in the topic is only a slip of the pen?

Couldn't you understand the meaning? Only a swap of 3 and 4.

Hence arqady's post!!!

[quote="Pain rinnegan"][quote="arqady"]

Maybe  $ \sum \frac {a^4}{b^3 \plus{} c^3} \ge \frac {a \plus{} b \plus{} c}{2}$?
[/quote]

Is the following one true?

Consider $ a$, $ b$, $ c$ three positive arbitrary real numbers. Prove that
\[ \frac {a^{4}}{a^{3} \plus{} b^{3}} \plus{} \frac {b^{4}}{b^{3} \plus{} c^{3}} \plus{} \frac {c^{4}}{c^{3} \plus{} a^{3}}\geq\frac {2(a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 3abc}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}\]
[/quote]

As well,does it work well?

Consider $ a$, $ b$, $ c$ three positive arbitrary real numbers. Prove that
\[ \frac {a^{4}}{b^{3} \plus{} c^{3}} \plus{} \frac {b^{4}}{c^{3} \plus{} a^{3}} \plus{} \frac {c^{4}}{a^{3} \plus{} b^{3}}\geq\frac {2(a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 3abc}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}\]

[quote="frankvista"]

Consider $ a$, $ b$, $ c$ three positive arbitrary real numbers. Prove that
\[ \frac {a^{4}}{b^{3} \plus{} c^{3}} \plus{} \frac {b^{4}}{c^{3} \plus{} a^{3}} \plus{} \frac {c^{4}}{a^{3} \plus{} b^{3}}\geq\frac {2(a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 3abc}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}\]
[/quote]

It follows from the one i proposed because by rearrangement \[ \sum_{cyc}\frac{a^4}{b^3\plus{}c^3}\ge \sum_{cyc}\frac{a^4}{a^3\plus{}b^3}\]

[quote="arqady"][quote="Pain rinnegan"]
Is the following one true?

Consider $ a$, $ b$, $ c$ three positive arbitrary real numbers. Prove that
\[ \frac {a^{4}}{a^{3} \plus{} b^{3}} \plus{} \frac {b^{4}}{b^{3} \plus{} c^{3}} \plus{} \frac {c^{4}}{c^{3} \plus{} a^{3}}\geq\frac {2(a^{3} \plus{} b^{3} \plus{} c^{3}) \plus{} 3abc}{2(a^{2} \plus{} b^{2} \plus{} c^{2})}\]
[/quote]
It seems true. But I still can't  prove it.  :([/quote]

I also tried can_hang's ideas for the first one but i failed. I couldn't prove it even using the Buffalo way.


inequalities
induction
function
inequalities proposed

[hide="Solution"]
It's clear that if $A$ is a finite set of non-negative reals, then $1+\sum_{x\in A} x\le \prod_{x\in A}(1+x)$.

$\sum_{i=1}^n a_i = n+ \sum_{i=1}^{n}(a_i -1) = n+ \sum_{a_i\le1}(a_i -1) +\sum_{a_i>1}(a_i -1)$

$\le n + \sum_{a_i>1}(a_i -1) \le (n-1)+ \prod_{a_i>1}a_i = (n-1) + \prod_{i=1}^{n}{\max{\{1,a_i\}}}$

[/hide]

[hide]
Let $b_{k}=\max \{ 1,a_{k}\}$, so that
\[\sum_{k=1}^{95}a_{k}\leq \sum_{k=1}^{95}b_{k},\ 94+\prod_{k=1}^{95}\max \{ 1,a_{k}\}=94+\prod_{k=1}^{95}b_{k}\]
So we will be done if we can prove that
\[\sum_{k=1}^{95}b_{k}\leq 94+\prod_{k=1}^{95}b_{k}\]
knowing that $b_{k}\geq 1$ for each $k$. Of course, this suggests
\[\sum_{k=1}^{n}b_{k}\leq n-1+\prod_{k=1}^{n}b_{k}\]
is true for any positive integer $n$. We show that this is in fact true, using induction. The base case is clearly true. For the inductive step,
\[n+b_{1}b_{2}b_{3}\cdots b_{n+1}=n+b_{1}b_{2}b_{3}\cdots b_{n}+b_{1}b_{2}b_{3}\cdots b_{n}(b_{n+1}-1)\]
\[\geq 1+(b_{1}+b_{2}+b_{3}+\cdots+b_{n})+b_{1}b_{2}b_{3}\cdots b_{n}(b_{n+1}-1)\geq b_{1}+b_{2}+b_{3}+\cdots+b_{n+1}\]
where we have simply used $b_{n+1}\geq 1,\ b_{1}b_{2}\cdots b_{n}\geq 1$ for the last step. This proves the inequality for any positive integer $n$; in particular, it works for $n=95$.
[/hide]

if $a_i\le 1$
then $LHS=\sum_{k=i}a_k+a_i\le\sum_{k=i}a_k+1$
RHS remains unchanged.
so let us assume $a_i\ge 1$
notice that $L-R$ is a linear function for all $a_i$
so when $a_i=1$,it attains minimum $94$
QED


determine the values of $ x$ where the series below converges conditionally or absolutely:

$ \sum_{n\equal{}1}^{\infty}\frac{1}{\sqrt{n}}(\frac{x\minus{}1}{x\plus{}1})^n$

Just for fun, I'm going to interpret this as a complex series. (The result for real series will be included in what I find.)

For what $ z\in\mathbb{C}$ does $ \sum_{n\equal{}1}^{\infty}\frac1{\sqrt{n}}\left(\frac{z\minus{}1}{z\plus{}1}\right)^n$ converge?

The key is that $ \left|\frac{z\minus{}1}{z\plus{}1}\right|<1$ precisely when $ \text{Re }z>0$ - that is, the right hand half plane. In that region, the series converges absolutely. (The root test should do nicely.)

Now look at points on the boundary. Suppose $ \text{Re }z\equal{}0.$ Then $ \frac{z\minus{}1}{z\plus{}1}\equal{}w,$ where $ |w|\equal{}1$ and $ w\ne 1.$ For such points, in absolute terms, the series does not converge. However the terms are not all positive.

We have the following theorem, which is a generalization of the alternating series theorem:

If $ b_n$ is positive, decreasing, and tends to zero, then $ \sum_nb_nw^n$ converges for all $ |w|\equal{}1,w\ne1.$

So our answer is this: the series converges absolutely in the open right hand half plane, converges conditionally on the imaginary axis, and diverges on the open left hand half plane.


Assume every $ 7$-digit whole number is a possible telephone number except those which begin with $ 0$ or $ 1$.  What fraction of telephone numbers begin with $ 9$ and end with $ 0$?

\[ \textbf{(A)}\ \frac{1}{63} \qquad
\textbf{(B)}\ \frac{1}{80} \qquad
\textbf{(C)}\ \frac{1}{81} \qquad
\textbf{(D)}\ \frac{1}{90} \qquad
\textbf{(E)}\ \frac{1}{100}
\]

Out of 2-9, there is a $ \frac18$ chance you get $ 9$.  Out of 0-9, there is a $ \frac1{10}$ chance you get a $ 0$.  So the answer is simply $ \frac18 \times \frac1{10}$, or $ \boxed{\textbf{(B)}\ \frac{1}{80}}$.

8*10^6 total possible numbers, 10^5 numbers that start with 9 and end with 0, 10^5/8*10^6=1/80--> (B)

We already have a solution. If you have nothing new to post, don't revive a 5 year old thread.

Yeah. Also, how did they even find this post?

The forum rules say that it is perfectly fine to revive a thread, especially from the contest section (1985 AMC 8 #22) [b] iff you have something more to contribute [/b]. His solution, while not LaTeXed does show a different mindset in solving the problem. The first one being done by multiplying fractions and his being done with counting. So in all honesty, he isn't doing anything wrong, but reviving and speculating on how the thread was revived is simply cluttering up the forums a lot more.

[quote="tmathman"][quote="FAQ"][size=200][b]How long of a non-commented thread is considered reviving?[/b][/size]
If any post is still on-topic and isn't spammy or anything, it isn't considered reviving. The definition of reviving in the Games forum is 1 month. However, everyone has a different period of time that they consider reviving. In general, [b]apply common sense[/b].[/quote]

Although I do agree with mathman523 that he [i]did[/i] post a new solution, five years is a bit long to be bringing back.  I get one, two, maybe even three years, but 5 years?  One almost has to be purposely looking through old topics to find a suitable topic to revive.  Also, it seems a bit suspicious that this user only has 3 posts, as if it were a multi who didn't want to get their main account butchered.

[size=85](this isn't backseat modding, is it?)[/size][/quote]
Well, I don't think someone would make a multi solely to revive a thread, that too one year in advance, eh? :)

I don't think it's a multi, and I think mathman523 is completely right about his post.

1. Chances are he/she isn't a multi and recently discovered the forum because the account was created an year ago.
2. Chances are that ChessBen has discovered the forum recently from 1985 AMC 8 and just wants to give his input in a few questions.
3. What he did in this thread (haven't looked at others) is definitely not reviving due to the change of the solution
4. We are now cluttering up the thread by arguing about whether he did anything wrong or not --> it is quite obvious reading the forum rules he didn't.
5. Where did being a multi even come from?
6. We should use the report button rather than posting on the forum. If the report is invalid then it will be dismissed.
7. We should seriously stop posting on this thread because now everyone who will look at all the solutions to the 1985 AMC 8 will see this.


Given two vectors with equal magnitutide. Add together. What is the probability that the magnitude of the resultant result is bigger than the previous magnitude?

isn't it $1$? ;)
or am i missing something?

No, you can have vector $\vec{a}$ and then add $-\vec{a}$ which would create zero and yet it would satisfy the conditions of the problem.

[edit: I think the answer is [hide]$\frac 23$[/hide]]

[hide="solution"]
Consider $\vec{AB}$ in the following figure:

[img]http://img.photobucket.com/albums/v213/joml88/vectorprobability.png[/img]

The circle centered at $B$ shows all the possible vectors resulting after you add a vector of the same magnitude.  The circle centered at $A$ shows is the region which if the resultant vector falls on equals the original magnitude.  So we the degree measure of the larger arc $DC$ is what we want.  $\triangle ABC$ and $\triangle ABD$ are equilateral making the larger arc $DC$ equal to $240^\circ$.  Thus the probability is $\frac{240^\circ}{360^\circ}=\frac 23.$[/hide]

yes the answer is $\frac{2}{3}$
here is another way to show it:
assuming the angle between the two vecors is $\theta$ $(0\le\theta\le\pi)$
then the magnitude of the resulting vector would be:
$\sqrt{2a^2+2a^2\cos{\theta}}=\sqrt{2a^2(1+\cos{\theta})}$
we want this to be bigger than the original vector so:
$\sqrt{2a^2(1+\cos{\theta})}>\sqrt{a^2}$ therefore
$2(1+\cos{\theta})>1$
$\cos{\theta}>\frac{-1}{2}$
$\theta<\frac{2\pi}{3}$ so
probibilit would be
$\frac{\frac{2\pi}{3}}{\pi}=\frac{2}{3}$ :)


The points $E$ and $D$ lie in the interior of sides $AC$ and $BC$, respectively, of a triangle $ABC$. Let $F$ be the intersection of the lines $AD$  and $BE$.Show that the area of the traingles $ABC$ and $ABF$ satisfies:

		$ \frac{S_{ABC}}{S_{ABF}} = \frac{\mid{AC}\mid}{\mid{AE} \mid} + \frac{\mid{BC}\mid}{\mid{BD}\mid} - 1$.

$\frac{S_{ABC}}{S_{ABF}}=\frac{AC}{AE}+\frac{BC}{BD}-1?$

And this is equivalent to:

$\frac{S_{ABC}}{S_{ABF}}=\frac{S_{ABC}}{S_{ABE}}+ \frac{S_{ABC}}{ S_{ABD}}-\frac{S_{ABC}}{S_{ABC}}$

$\frac{1}{S_{ABF}}=\frac{1}{S_{ABE}}+\frac{1}{S_{ABD}}-\frac{1}{S_{ABC}}$

$\frac{1}{S_{ABF}}-\frac{1}{S_{ABD}}=\frac{1}{S_{ABE}}-\frac{1}{S_{ABC}}$

$\frac{S_{BDF}}{ S_{ABF} \cdot S_{ABD}}= \frac{S_{BCE}}{S_{ABE} \cdot S_{ABC}}$

$\frac{S_{BDF} \cdot S_{ABE} \cdot S_{ABC}}{S_{ABF} \cdot S_{BCE} \cdot S_{ABD}}=1$

$\frac{FD}{AF} \cdot \frac{AE}{CE} \cdot \frac{BC}{BD}=1$

And this is correct due to Menelaus theorem for triangle $ADC$ and line $EFB$
MT


Hello

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Thank u
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For having every information about it u can follow
[url=http://twitter.com/Astro_Mike]Mike Massimino[/url] and [url=http://twitter.com/HubblePAO]Hubble[/url] in twitter .
Thank u
Have fun with space .  :alien:


WITH FULL SOLUTIONS PLEASE

Fractions a/b and c/d are called neighbor fractions if their difference (ad - bc)/(bd) has a numerator of positive or negative 1, that is, ad - bc = positive or negative one. Prove that

1. In this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

2. If a/b and c/d are neighbor fractions, then (a + b)/(c + d) is between them and is a neighbor fraction for both a/b and c/d; moreover,

3. no fraction e/f with positive integer "e" and "f" such that f is less than b + d is between a/b and c/d.

Double post: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=204464

[color=blue]Mods - Please delete.[/color]


integration
calculus
calculus computations

This is a problem regarding differential equations. The text has been translated from another language, sorry if its not clear.

A fish stock in a lake has stopped reproducing due to emission from a nearby factory.
This happened at time t = 0. Calculations show that the alteration in this fish stock pr. time units is proportionally inverse to the square root of the fish stock.
There were 900 fish in the lake when the emission happened and 657 days after there were 441 fish alive.
How many days will it take until all of the fish are dead?
(Answer: 1000 days)

thanks!

$ N\equal{}N(t) \equal{}$ number of fish at time $ t$.

$ \frac{dN}{dt} \equal{} \frac{\minus{}k}{\sqrt{N}}$, where $ k$ is the constant of proportionality.

$ \int N^{\frac{1}{2}} dN \equal{} \int \minus{}k dt$

$ \frac{2}{3} N^{\frac{3}{2}} \equal{} \minus{}kt \plus{} c$

$ \frac{2}{3} (900)^{\frac{3}{2}} \equal{} c$

$ c\equal{}18000$

$ \frac{2}{3}N^{\frac{3}{2}} \equal{} \minus{}kt \plus{} 18000$

$ \frac{2}{3}(441)^{\frac{3}{2}} \equal{} \minus{}657k \plus{}18000$

$ k\equal{}18$

$ N \equal{} 0 \implies 18t \equal{} 18000$.

Therefore $ t\equal{}1000$ days.


induction
number theory unsolved
number theory

For a nonempty set $\, S \,$ of integers, let $\, \sigma(S) \,$ be the sum of the elements of $\, S$.  Suppose that $\, A = \{a_1, a_2, \ldots, a_{11} \} \,$ is a set of positive integers with $\, a_1 < a_2 < \cdots < a_{11} \,$ and that, for each positive integer $\, n\leq 1500, \,$ there is a subset  $\, S \,$ of $\, A \,$ for which $\, \sigma(S) = n$.  What is the smallest possible value of $\, a_{10}$?

[hide]If we take the numbers 1,2,4,8,16...,1024, it is similar to the binary number system.  Any other number system cannot reach all the numbers up to 1500.

But the set 1,2,4...1024 can reach all the numbers under 2048, which is more than we need.  If we decrease the larger numbers, a10 will be smaller, but we can't sum to as many numbers.  If we are able to only sum to 1500 and less with a10 as small as possible, we have reached the minimum a10 possible.

The set above can reach all the numbers from 1 to 2047 which is 547 greater that we need.  If we decrease the 10th number by 1 we cannot make 1+2+4...+512=1023 anymore.  But if we decrease both the 10th number and the 11th number by one, that number can still be reached.  Seeing as a10>a9, if we decrease a10 all the way to 257, a11= 769.

now the set is 1,2,4,8,16,32,64,128,255,256,769 and can make all numbers from 1-1537

We cannot decrease a10 anymore without decreasing a9.  If we decrease a9 by 1 and a10 by 1, we cannot make 767 or 768 anymore.  So we would need to decrease a11 by 2.  That would make a set that could reach all numbers from 1-1535.  Repeating this set 8 more times we have the set:

1,2,4,8,16,32,64,128,247,248, 751 which can make all the numbers from 1-1501.  We cannot repeat the set above further as that would lower the range to 1-1499.

Assuming it is possible for a10 to be 247, then we would need at minimum 8 numbers to reach all numbers below it (1,2,4,8...,128).  To get the maximum possible range of sums we would have to maximize a9.  So a9=246.  Now to maximize the possible range of sums we would have a11=a1+a2+a3...+a10+1.  So a11=750.  Then a1+a2...+a11=1499 and 1500 cannot be reached.

The minimum value of a10 is 248.[/hide]

I am new to proofs so feel free to point out any errors or post a better one

For your first proof, not bad!  However, while $248$ is correct, you only showed that it suffices; your proof for why it is minimial needs to be attacked from a different angle.  The part where you say "Any other number system cannot reach all the numbers up to 1500" needs to be proved, and in fact this is the toughest part of the problem.

If you would like to see a nice solution for this, here is the link to do so:
[url]http://www.kalva.demon.co.uk/usa/usoln/usol923.html[/url]

Keep it up!


Functional Analysis
topology
real analysis
real analysis solved

Here there are some exersices on functional analysis:
1). Let $E$- be separable metric space. Point $x\in E$ is said to be condensation point of the set $A$ in $E$ if in every it's neighborhood there are uncountable number of point from $A$.
Prove that if $A$ hasn't condensation points then $A$ is countable

2). Prove that the set of isolated points of separable metric space is at most countabe

(1) Each point which is not a condensation point belongs to some basis element which contains countably many points of $A$ (we choose a countable basis for the space; since it's a separable metric space, it has a countable basis). Eliminate all the points of $A$ which belong to such basis elements. We have eliminated all the non-condensation points, but, at the same time, we have eliminated countably many points, so except for countably many points, $A$ consists only of condensation points, which is even more than what we wanted to prove.

(2) Pretty much the same as above.

Here is another one:
1).Let $A$ be non-empty subset of real line and $B$- the set of $x\in \bar{A}$ for which there exists interval $(x,y)$ which has void intersection with $A$. Prove that $B$ is at most countable

For each such $x$ we can, in fact, find an interval $(x,y_x)$ which has void intersection with $\bar A$ (this is pretty clear, I think). These intervals cannot intersect each other, so we have an injection from our set of points to a collection of disjoint open intervals, and such a collection is countable.


[quote="GameBot"]In this figure, each pair of adjacent sides is perpendicular.
The figure is to be divided into 12 equal pieces having the
same size and shape. The area of one of these pieces can
be expressed in the form $ ax^n$. What is the product of $ a$ and
$ n$, expressed as a common fraction?

[asy]draw((0,0)--(-10,0)--(-10,10)--(-20,10)--(-20,20)--(0,20)--cycle);
label("$x$",(-5,0),S);
label("$x$",(-10,5),W);
label("$x$",(-20,15),W);
label("$2x$",(-10,20),N);[/asy][/quote]

This has area $ 3x^2$. twelve peices each have area $ \frac{1}{4}x^2$, so 1/2.


i have a strange feeling that nobody will read the last post in a thread to get the rules, so I've decided to create another topic

-no more than 4 tourneys are allowed at any one time, this will (hopefully) help in making sure tourneys are completed
-a person may not join more than 2 tourneys at any given time, even though I'm sure many of you can keep up with your games, i feel this rule might be able to help
-make sure you have enough time to devote to playing your games
-if any person is inactive, the creator of the tourney should write down his/her name and pm to a mod, if a person is inactive in too many tourneys, he/she will be not allowed to join any more tourneys (maybe for just an amount of time, if people think ever is too harsh)
-new users must play more than 50 games before they can join a tourney, this is to prevent new users to sign up and never come on again (special thanks to jjx)
-if anybody does not fulfill his/her tourney modding requirements, he/she cannot mod another tourney for a month (special thanks to jjx)
-tourneys can have a maximum of 16 people (special thanks to AIME15)
-players who are not currently in a tourney have priority when sign ups start for a different tourney (special thanks to AIME15)

if i think of any more, or if anyone else has more, please post here
if anyone disagrees with a rule for any reason, please post here
these rules can still be changed

do you have a suggestion for how many games?

[quote="jjx1"]I think that we should also do a rule like a new user can't join a tourney until they have played a certain number of games. Now, I know lots of people will post that thats not fair and such and such. But most new users that join are inactive. I think they jsut join things they see and don't visit it.[/quote]

What makes you think that?

well I see lots of new users that are inactive in games

I would suggest like 50 games or so.

[quote="jjx1"]I think that we should also do a rule like a new user can't join a tourney until they have played a certain number of games. Now, I know lots of people will post that thats not fair and such and such. But most new users that join are inactive. I think they jsut join things they see and don't visit it.[/quote]

Gotta agree with this.

I think that's a good idea as well. I remember several new people who join and see a tournament and think, "Hey, this sounds cool! I'd like to be in it." Some of these people do play all their games, but I've seen too many newbies join a tournament and then never play.

I know that a lot of chess.com tournaments implement this rule as well. But of course, for FTW, anyone can play a large amount of 2/5 games in a short period of time, so maybe it should be more like be a fairly active member for two weeks or so. This too has flaws(how do you judge fairly active?) but that was just an idea I threw out.

[quote="isabella2296"]But of course, for FTW, anyone can play a large amount of 2/5 games in a short period of time[/quote]

:)

How could we check how many games someone played if they're not on FTW?  Can moderators do that?  I doubt levans wants to spend time check the amount of games played.

no, we can't check how many games people play
i'm not sure if levans can, but i highly doubt he will

it doesn't really matter though, if a person has enough time to devote to 50 2/5 games, i'm pretty sure he/she will have enough time to play a tourney

also, if he/she is inactive the 4th rule will come into affect =)

My point was that there will be people who say, "Oh! I need 50 games? Okay, I'll make a 2/5 game, walk away and do something else, come back, make another, etc."

lol . I doubt someone will spend that effort jsut to get in a tourney. We could always raise the limit too.

So, yeah. When this rule is in effect and someone makes a tourney. They should post Please post how many games you have finished when you sign up. If someone doesn't post that we will just ignore the post until they change it. idk

I think some set of rules for in-game play are necessary too.

For example, what if someone encounters flashies or a blank problem?

Or "flashies" and "blank problems" :wink:

I guess. But if that rule does happen you can only say you have flashies when you're winning a cd. Then the other person has to leave.

How about in order to participate in a tourney, you need to be a "qualified user," and have like, for example, three other players sign to insure you are often active as a requirement for you to join the "qualified users club". The most probable problem with this is that that might prompt someone to make multiple accounts to recommend themselves...but I'm quite sure that this will, if everyone goes honestly (which, alas, probably won't happen), filter out the inactive people, as users would easily remember users that frequently are in FTW as opposed to others.

As has been stated in another post (but probably should have been posted here as well, my fault), the rules for establishing tournaments are currently suspended. Since there are so few tournaments going on and seemingly very little interest in creating tournaments, there is no reason to have a strict set of rules regarding how people are allowed to run their tournament. So basically, do whatever you want with your tournament as long as it complies with the Terms of Service and the laws of your country, state, city, etc.

[quote="jeff10"]I think the maximum is 16 players only if it is a quick one-day tourney. [/quote]

You'd be surprised how long a 16 player tournament takes to finish...

It doesn't take as long even as a one-day tourney.

I am thinking of starting a tourney...
Do i have to get permission from you guys?

Thanks

[quote="Mrdavid445"]
Do i have to get permission from you guys?
[/quote]

Nope. In general, whatever restrictions that are on these stickies can now be disregarded, as we don't have the overwhelming number of tournaments anymore. Perhaps we should unsticky these....

As vallon22 said, we realize that people never create tournies, and so the stickies are no use. Just do whatever you want, but tournaments are all FTW though.

erm...

type:

/join highest rating: xxx
current rating: yyy

[quote="vallon22"]
-new users must play more than 50 games before they can join a tourney, this is to prevent new users to sign up and never come on again (special thanks to jjx)
[/quote]
Do we still have to follow that rule?

Why not? The rule is still fine, since we still have many users forfeiting, these days.

I disagree with the first rule. Tourneys can be very long, and it could take a while for somebody to be able to participate. I suggest that the limit for the number of tourneys be raised to 8-10

That's way too many. It would spam up the forums too.

[quote="ReciterOfPi"]If there's an inactive person who signs up, what would we do to find a replacement?[/quote]

You could eliminate that match; and pair the other party up with someone else; if they're willing.

You should get a very high prize rating.


I've never done MC before, and I probably will next year.

Can someone tell me about qualifying and the rules on what places go to chapter, state and nats?

Also, I'd like to know the format of the tests, such as the rules of rounds like sprint and target. I'm not exactly sure what they are... the time periods, number of questions and perfect scores.

Thanks.

You can find everything [url=http://www.mathcounts.org]on their website[/url]. Go to "Rules and Procedures", located [url=https://mathcounts.org/Page.aspx?pid=286]here.[/url]

Thanks (even though i'm not Dojo) cuz i might take MC in 5th/6th grade. :|

The rounds:

[b]Sprint: [/b]No calculator, individual, written, 30 problems, 40 minutes
[b]Target: [/b]Calculator allowed, individual, written, you get 2 problems at a time, and 6 minutes for each pair, and there are 4 pairs (8 problems)
[b]Team: [/b]Calculator allowed, team test, written, 10 problems, 20 minutes
[b]Countdown: [/b]No calculator, individual, oral, a head to head competition (first to answer right gets a point), like FTW, 45 seconds per problem
[b]Masters: [/b]Calculator, individual, essay, 30 minutes to write an essay about a given topic

Qualifying:

Your school chooses 8 people, 4 individual and 4 on a team, to go to the chapter competition. The top two teams and the top 8 individuals (can be on a school team NOT in the top 2, or an individual to start with) go to the State competition. There, the top 5% or top 12, whichever one is less, from written individual tests compete in the countdown round. In some states, the top 4 in the countdown go to Nats, and in some states the top 4 written go to Nats. If your team gets first place, you coach goes to Nats. At Nats, the top 12 written compete in an official countdown. There is also an extra Masters Round, where you have to write an essay in 30 minutes.

Uh, some corrections to point out:

The Masters Round involves making an oral presentation to a panel of judges, not writing an essay.

For chapter-->state progression, it varies from chapter to chapter.

Not all schools have to send 8 people, it's just that a maximum of 4 people can be representing the team (yes, some teams can have less than 4 people), and a maximum of 4 individuals whose scores don't contribute to the team score can compete just for fun.

Also, to add to the countdown rules: If the countdown is official, a ladder format is used. If unofficial, anything can be used, such as a bracket single elimination.

[quote="PowerOfPi"]The rounds:

[b]Sprint: [/b]No calculator, individual, written, 30 problems, 40 minutes
[b]Target: [/b]Calculator allowed, individual, written, you get 2 problems at a time, and 6 minutes for each pair, and there are 4 pairs (8 problems)
[b]Team: [/b]Calculator allowed, team test, written, 10 problems, 20 minutes
[b]Countdown: [/b]No calculator, individual, oral, a head to head competition (first to answer right gets a point), like FTW, 45 seconds per problem
[b]Masters: [/b]Calculator, individual, essay, 30 minutes to write an essay about a given topic
[/quote]

Hmmm a few things: The team round doesnt count toward your individual score, just your team score, so really theres no pressure during the team round except for in Nats (unless you really really really want your teacher to got to nats). Um masters is only done in some states (very few), but its always done at nats. Also, qualifying scores for the next round depends on your state. For example, if you live in california or Texas, you have a really low chance at making nats since Cali. and texas have extremely strong students. If you lived in an esier state, qualifying scores for to advance would be much lower. Hmmm, if im not mistaken, Washington is a pretty strong state too, right?

EDIT: Yep, I just checked the site, and washington is a [b]REALLY[/b] strong state. (placed 2nd and 3rd the past 2 years)

[quote="gauss1181"]
For chapter-->state progression, it varies from chapter to chapter.
 [/quote]
If I'm correct, I believe it ranges from the top 5-10. (but it depends on # of team, etc.)

Chapter => state depends on the size of your chapter. We had a big chapter, so the top four teams were the first to go to states. Then after that was chosen, the top four individuals who werent on a team advanced. So in our pretty big chapter, 20 people went to states. Ive heard people say that as few as the top 2 individuals and teams advanced to states.

States=> nats is always top 4, but the process of choosing the top 4 could vary. Some states have an unofficial countdown, where the top 4 written advance, while others have official CD, where top 4 after CD advance. Since the CD in official CD states is ladder style (10th written vs 9th, winner vs. 8th....winner vs. 1st), the top 3 automatically make it to nats, even if they lose the CD. (because if 3rd loses, he will be 4th and still make it). So basically in official CD states, it's only 4th place who has to fight for his spot in nats.


ABCD is a quadrilateral, E,X,F,Y are midpoints of AD,AB,BC,CD, AF∩BE=M,DF∩CE=N,
Prove：MN∥XY


geometry
circumcircle
ratio
Euler
geometry unsolved

[b][size=100][color=DarkBlue]The line connecting the isotomic point of the Lemoine point, of a given triangle, with its orthocenter, is parallel to the line connecting the Lemoine point, with its circumcenter[/color][/size][/b].

Kostas Vittas.

I know this problem very good. :D
It was on Ukrainian national math fights tournament.
I solved it using 'weight' method.

Construct a triangle $ DEF$ such that $ A,B,C$ are the midpoints of its sides $ EF,FD,DE$ respectively.
Let $ G$ be the centroid of triangle $ ABC$, two triangles $ DEF$ and $ ABC$ are homothetic, $ G$ is its homothetic center, its similitude ratio is equal to $ 2: 1$.
 Let $ M$ be the Lemoine point of triangle $ DEF$, thus $ MG \equal{} 2GL$, by Euler theorem, $ HG \equal{} 2GO$, hence, we have $ HM\parallel LO$.
Let $ PQR$ be  the tangential triangle of triangle $ DEF$, thus $ M \equal{} QE\cap RF$.
Let $ X \equal{} QE\cap CA, Y \equal{} RF\cap AB,$ since $ T$ is the isotomic point of the Lemoine point of triangle $ ABC$,
hence $ T \equal{} BX\cap CY$.
Let $ Z \equal{} BC\cap XY$, By Menelaus theorem for the triangle $ ABC$ cut by the transversal $ XYZ$,
$ \frac {ZC}{ZB} \equal{} \frac {CX}{AX}\cdot\frac {AY}{BY} \equal{} \frac {AB^2}{AC^2}$ (1).
Let $ Z' \equal{} BC\cap QR$, By Menelaus theorem for the triangle $ BCH$ cut by the transversal $ QRZ'$,
$ \frac {Z'C}{Z'B} \equal{} \frac {RC}{RH}\cdot\frac {HQ}{QB} \equal{} \frac {CD^2}{DH^2}\cdot\frac {DH^2}{BD^2} \equal{} \frac {AB^2}{AC^2}$ (2),
by (1) and (2), we have $ Z \equal{} Z'$, thus two triangles $ XYM$ and $ BCH$ are perpective, hence by Desargue's theorem, $ BX,CY,HM$ are concurrent, then $ H,T,M$ are collinear, as a result, $ HT\parallel LO$, as desired.

[b][size=100]EXCELLENT !!![/size][/b]

Thank you a lot dear [b][size=100]Fang-jh[/size][/b], for an important stuff.

This kind of solution is lovely to me and let me please to add some more details, about the relations $ (1),$ $ (2),$ you mentioned.

$ \bullet$ So, because of $ BX',$ where $ X'\equiv AC\cap BL,$ is the $ B$-symmedian of $ \bigtriangleup ABC,$ we have $ \frac {AX'}{CX'} \equal{} \frac {(AB)^{2}}{(BC)^{2}}$ $ ,(1a)$

Because of the point $ X,$ is the isotomic point of $ X',$ with respect to the side-segment $ AC$ $ ($ = the symmetric point of $ X,$

with respect to the midpoint of the side-segment $ AC$ $ ),$ we have $ \frac {AX'}{CX'} \equal{} \frac {CX}{AX}$ $ ,(1b)$

From $ (1a),$ $ (1b)$ $ \Longrightarrow$ $ \frac {CX}{AX} \equal{} \frac {(AB)^{2}}{(BC)^{2}}$ $ ,(1c)$

By the same way, we have $ \frac {AY'}{BY'} \equal{} \frac {BY}{AY} \equal{} \frac {(AC)^{2}}{(BC)^{2}}$ $ \Longrightarrow$ $ \frac {AY}{BY} \equal{} \frac {(BC)^{2}}{(AC)^{2}}$ $ ,(1d)$ where$ CY',$ is the $ C$-symmedian of $ \bigtriangleup ABC$ and $ Y'\equiv AB\cap CL.$

From $ (1c),$ $ (1d)$ $ \Longrightarrow$ $ \frac {CX}{AX}\cdot \frac {AY}{BY} \equal{} \frac {(AB)^{2}}{(AC)^{2}}$ $ ,(1e)$

Hence, from $ \frac {ZC}{ZB}\cdot \frac {BY}{AY}\cdot \frac {AX}{CX} \equal{} 1$ $ \Longrightarrow$ $ \frac {ZC}{ZB} \equal{} \frac {CX}{AX}\cdot \frac {AY}{BY} \equal{} \frac {(AB)^{2}}{(AC)^{2}}$ $ ,(1)$

$ \bullet$ From the similar right triangles $ \bigtriangleup CDR,$ $ \bigtriangleup DHR,$ $ \Longrightarrow$ $ \frac {RC}{DR} \equal{} \frac {CD}{DH}$ $ ,(2a)$

From the similar right triangles $ \bigtriangleup CDH,$ $ \bigtriangleup DHR,$ $ \Longrightarrow$ $ \frac {RH}{DH} \equal{} \frac {DR}{CD}$ $ \Longrightarrow$ $ \frac {RH}{DR} \equal{} \frac {DH}{CD}$  $ ,(2b)$

From $ (2a),$ $ (2b)$ $ \Longrightarrow$ $ \frac {RC}{RH} \equal{} \frac {(CD)^{2}}{(DH)^{2}}$ $ ,(2c)$

By the same way, from the similar right triangles $ \bigtriangleup BDQ,$ $ \bigtriangleup DHQ$ and $ \bigtriangleup BDH,$ $ \bigtriangleup DHQ$ we have $ \frac {QH}{QB} \equal{} \frac {(DH)^{2}}{(BD)^{2}}$ $ ,(2d)$

From $ (2c),$ $ (2d)$ $ \Longrightarrow$ $ \frac {RC}{RH}\cdot \frac {QH}{QB} \equal{} \frac {(CD)^{2}}{(BD)^{2}}$ $ ,(2)$

Hence, from $ \frac {Z'C}{Z'B}\cdot \frac {QB}{QH}\cdot \frac {RH}{RC} \equal{} 1$ $ \Longrightarrow$ $ \frac {Z'C}{Z'B} \equal{} \frac {RC}{RH}\cdot \frac {QH}{QB} \equal{} \frac {(CD)^{2}}{(BD)^{2}} \equal{} \frac {(AB)^{2}}{(AC)^{2}}$ $ ,(2)$

So, from $ (1),$ $ (2)$ $ \Longrightarrow$ $ \frac {ZC}{ZB} \equal{} \frac {Z'C}{Z'B}$ $ \Longrightarrow$ $ Z'\equiv Z$ and after that, the wonderful perspective triangles $ \bigtriangleup XYM,$ $ \bigtriangleup BCH.$  :)

$ \bullet$ I will preserve this nice proof with your name in the future, as a beautiful present to all of us, from a Chinese friend.

Thank you again and best regards, Kostas Vittas.

Hi,Dear Kostas Vittas
Thank you very much for your interests and adding some more details that are easy to follow.
Best regards
Fang-jh

This problem can be generalized as following:

Let $ O, H $ be the circumcenter, orthocenter of $ \triangle ABC $ .
Let $ D=AP \cap BC, E=BP \cap CA, F=CP \cap AB $ .
Let $ X, X', Y, Y', Z, Z' $ be the midpoint of $ AD, BC, BE, CA, CF, AB $ .
Let $ Q=XX' \cap YY' \cap ZZ' $ and $ P^* $ be the isotomic conjugate of $ P $ .

Then $ OQ \parallel HP^* $ ( Just notice that $ Q $ is the isotomcomplement of $ P $ :) )

Dear Mathlinkers,
just a point of history : this problem comes from Brocard.
Sincerely
Jean-Louis


[hide]Let $\frac{a}{b}= m^{3}$ and $\frac{b}{a}= n^{3}$. Cube both sides of the inequality to get the equivalent \[2(m^{3}+n^{3}+2) = 2(a+b)\left(\frac{1}{a}+\frac{1}{b}\right) \geq \left(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\right)^{3}= m^{3}+n^{3}+3(m+n).\] So we would like to prove that $\boxed{m^{3}+n^{3}+4 \geq 3(m+n)}$. But $mn = \sqrt[3]{\frac{a}{b}\cdot\frac{b}{a}}=1$, so \[\begin{aligned}m^{3}+n^{3}+4-3(m+n) & = (m+n)(m^{2}-mn+n^{2})-3(m+n)+4 \\ & = (m+n)(m^{2}+n^{2}-4)+4\end{aligned}\] Now remark that according to AM-GM we have that \[m+n = \sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}\geq 2\sqrt{\sqrt[3]{\frac{a}{b}}\sqrt[3]{\frac{b}{a}}}= 2.\] Use AM-GM to prove that $m^{2}+n^{2}\geq 2$. So now we have that $(m+n)(m^{2}+n^{2}-4)+4 \geq 2\cdot(2-4)+4 = 0$, so we're done![/hide]

Does anyone have a shorter, nicer proof?

[quote="D.P.L"]Correct me if i m wrong.

Can you use the fact that if a+b+c=0
then a^3+b^3+c^3=3abc.[/quote]
That is indeed true!

You may use the fact that $a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca).$

[quote="sqing"][quote="parinaz.sa"]let$a,b>0$prove this:
$\sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}$[/quote]
By AM--GM , we have
$4+\frac{a}{b}+\frac{b}{a}=(1+1+\frac{a}{b})+(1+1+\frac{b}{a})\ge 3\sqrt[3]{\frac{a}{b}}+3\sqrt[3]{\frac{b}{a}}$
$\Rightarrow 4+2(\frac{a}{b}+\frac{b}{a})\ge (\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})^3\Leftrightarrow \sqrt[3]{2(a+b)(\frac{1}{a}+\frac{1}{b})}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}$.[/quote]

Can you generalise your solved ineq. up to that for n=3 or above?

[quote="arqady"]
2)Let $a,$ $b$ and $c$ are positive numbers. Prove that
$\sqrt[3]{3(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\geq\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}.$[/quote]
$\Leftrightarrow 3+2\sum \frac{a}{b}+3\sum \frac{b}{a}\ge 3\sum \sqrt[3]{\frac{a^2}{bc}}+3\sum \sqrt[3]{\frac{bc}{a^2}}$.

By AM--GM , we have

$3+2\sum \frac{a}{b}+3\sum \frac{b}{a}\ge 6+2\sum \frac{a}{b}+2\sum \frac{b}{a}$

$=\sum (1+\frac{a}{b}+\frac{a}{c})+\sum (1+\frac{b}{a}+\frac{c}{a})\ge 3\sum \sqrt[3]{\frac{a^2}{bc}}+3\sum \sqrt[3]{\frac{bc}{a^2}}$.

[quote="arqady"]1)Let $a$ and $b$  are positive numbers. Prove that
$\sqrt[3]{\frac{4(a+b)(\frac{1}{a}+\frac{1}{b})+8}{3}}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}.$[/quote]
By AM--GM , we have
$ (\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})^3+16= (\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})^3+8+8\ge 12 (\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})$

$\Leftrightarrow \frac{a}{b}+\frac{b}{a}+16\ge 9 (\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}})\Leftrightarrow\sqrt[3]{\frac{4(a+b)(\frac{1}{a}+\frac{1}{b})+8}{3}}\ge\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{a}}.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=365861&p=2750213&hilit=SRMC#p2750213]International competition SRMC 2006 P-2:[/url]
For positive $a,b,c$, such that $abc=1$ prove the inequality$4(\sqrt[3]{\frac{a}{b}}+\sqrt[3]{\frac{b}{c}}+\sqrt[3]{\frac{c}{a}}) \leq 3(2+a+b+c+\frac{1}{a}+\frac{1}{b}+ \frac{1}{c})^{\frac{2}{3}}$


Consideram un trapez isoscel $ABCD\ (AB\parallel CD)$ in care notam $AB=a$, $CD=b\ (b\le a)$, $AD=BC=c$, $AC=BD=d$. Sa se arate ca pentru orice numar natural $n$ exista relatia
$(2n+1)d^2\le (n+1)a^2+nb^2+(2n+1)c^2\ .$ [hide="Here is a indication."] The Euler's relation in a isosceles triangle : $d^2=ab+c^2\ .$[/hide]

[color=blue][b]Considerăm un trapez isoscel $ABCD\left( {\left. {AB} \right\|BC} \right)$ in care notăm
$AB = a$, $CD = b\left( {b \le a} \right),AD = BC = c,AC = BD = d$.
Să se arate că pentru orice număr natural   există relaţia:
(*)    $\left( {2n + 1} \right)d^2 \le \left( {n + 1} \right)a^2 + nb^2 + \left( {2n + 1} \right)c^2$.
[/b][/color]

[u][b]Soluţie:[/b][/u]
Avem succesiv:
$\left( {2n + 1} \right)d^2 \le \left( {n + 1} \right)a^2 + nb^2 + \left( {2n + 1} \right)c^2 \Leftrightarrow$

$\Leftrightarrow 2nd^2 + d^2 \le na^2 + a^2 + nb^2 + 2nc^2 + c^2 \Leftrightarrow$

$\mathop \Leftrightarrow \limits_{\mathop {d^2 = ab + c^2 }\limits^ \uparrow } 2nab + 2nc^2 + ab + c^2 \le na^2 + a^2 + nb^2 + 2nc^2 + c^2 \Leftrightarrow$

$\Leftrightarrow ab - a^2 \le n\left( {a - b} \right)^2$ evident adevărată $\forall n \in \left( {0, + \infty } \right)$

[u][b]Observaţie:[/b][/u]
Identitatea $d^2 = ab + c^2$ rezultă şi din [color=red][b]teoremele lui Ptolemeu[/b][/color]
in patrulaterul inscriptibil.


inequalities geometry Euler algebra polynomial Vieta circumcircle

We write $r$ the rayon of the incircle of the triangle $ABC$. Let $r_a$, $r_b$, $r_c$ the rayons of the exincircles of the triangle $ABC$.

Show that $r_a+r_b+r_c>=9r$

Since $\sum \frac{1}{r_a} = \frac{1}{r}$ The question is equivalent to $(\sum r_a)(\sum \frac{1}{r_a}) \ge 9$ which is amhm.

Very  easy  one....use the  fact  that  [tex]r_a+r_b+r_c=4R+r[/tex]  and  [tex]R\geq2r[/tex]  (Euler)

It is very easy.
It is well known that the $r_a,r_b,r_c$ are roots of the equation
$x^3-(4R+r)x^2+s^2x-s^2r$ where s the semiperimeter.
Now from Vieta $r_a+r_b+r_c=4R+r$

It's well known, maybe, but how would you [u]prove[/u] that they are indeed roots of that equation? The proof of $\sum\frac{1}{r_a} = \frac{1}{r}$ is easy though. Use $S = sr$ and the similar relations, where $s$ is the semiperimeter and $S$ is the area.

To prove that $4R = r_1 + r_2 + r_3-r$ we can do the following
We have to check that $p(p-b)(p-c)+p(p-c)(p-a)+p(p-a)(p-b)-(p-a)(p-b)(p-c)=abc$ where $p$ is the semiperimeter.
Note that $abc=4RS$, where $S$ is the area and $R$ the circumradius of the triangle.
We have $p(p-b)(p-c)+p(p-c)(p-a)+p(p-a)(p-b)-(p-a)(p-b)(p-c)=4RS$
Recall Heron's formula and also $S=r_1(p-a)=r_2(p-b)=r_3(p-c)$ to get
$4SR=\frac{S^2}{p-a}+\frac{S^2}{p-b}+\frac{S^2}{p-c}- \frac{S^2}{p}=S(r_1+r_2+r_3-r)$
so that indeed $4R= r_1 + r_2 + r_3-r$ :10:


Hi! I qualified for the state algebra I competition at UNCA on May 3rd. I need of these state algebra I tests so I can practice. Can someone show me the websites with the tests and answers or PM me? Thanks! I need major practice! :)

[quote="b-flat"][url=http://courses.ncssm.edu/goebel/statecon/tests/tests.htm]website[/url]

The website above contains all (I think) of the old practice tests and solutions.  It also contains the results.[/quote]

Thank you so much b-flat!!!!!


Solve each system:
$ 2x + 3y - z = - 2$
$ x + 2y + 2z = 8$
$ 5x + 9y + 5z = 22$

I found that this has infinitely many solutions, but when I tried to put it in parametric form with three variables like I'm supposed to, I couldn't figure out what exactly I am supposed to do. As in, do I have to first let $ x = t$? Does it matter what variable I cancel out?  Here is what I did:

I ended up with $ y + 5z = 18$ to start with from subtracting the 2nd from the 1st equation, and 3rd from 2nd.

$ 2x + 3y - z = - 2$
$ 2x + 4y + 4z = 16$
-------------
$ - y - 5z = - 18$ -> $ y + 5z = 18$

$ 5x + 10y + 10z = 40$
$ 5x + 9y + 5z = 22$
-------------
$ y + 5z = 18$

So I knew that $ 18 = 18$, which means infinitely many solutions.

Then, I decided to let $ x = t$, and then rearranged the second equation $ x + 2y + 2z = 8$, as well as using the $ y + 5x = 18$ to make it $ y = - 5x + 18$:

$ t = - 2y - 2z + 8$
$ t = - 2( - 5z + 18) - 2z + 8$
$ t = 10z - 36 - 2z + 8 = 8z - 28$

Solving for $ z$:

$ t = 8z - 28$ -> $ t + 28 = 8z$ -> $ \frac {t + 28}{8} = z$

Then, solving for $ y$ in terms of $ t$:

$ y + 5(\frac {t + 28}{8}) = 18$
$ y + \frac {5t + 140}{8} = 18$
$ y = \frac {144 - 5t + 140}{8}$ - > $ \frac {4 - 5t}{8}$

$ {(x,y,z) = (t,\frac {4 - 5t}{8}}$, $ \frac {t + 28}{8}, t = reals)$

I messed up working on this problem the first and second times around, but that was because I knew my answer was incorrect. =// To clarify, I did not get the right answer on the 1st two tries.

It doesn't matter which variable you choose to parametrize on, as long as you keep your calculations correct. Hence it would've been perfectly OK to eliminate let's say $ y$ and then parametrize on $ z$. (Final expressions would be different, of course, but that's inconsequential.)

(And you have a typo in the last row: $ z$ shoud be $ {t \plus{} 28\over 8}$)

[quote="i_like_pie"]It would be a lot easier to use [url=http://en.wikipedia.org/wiki/Cramers_rule]Cramer's Rule[/url].[/quote]

Um.... no? I think you would get division by $ 0$, since the matrix has a determinant of $ 0$. Note that equation 1 + 3*equation 2 = equation 3.

thanks everyone! I guess I will work at a decent pace and try to make sure I do all my calculations correctly!


Hi Friends,

In Knuth they somewhere count the number of distinct number of binary trees that are possible when we r given n distinct nodes. They use generating functions. Can someone give a different proof for the same???

Thank You and have a nice day :)

why dn't u write down the recurrence and prove that by induction?


Are there any rather simple cases that maths logical and reason in maths can lead to a stupid awnser like:
4 - 1 = 5
with a piece of paper that is true...cut out a corner and u then have 5. are there any more senarios like this?
thanks

Well if you cut out a corner then yuo get two pieces and you do not include the fact that you have a second piece. If you get something like this you have obviously made a mistake somewhere.

do you mean if you cut off one corner, then you end up with 5 corners?  you lose one corner, but you add 2. so you should really say 4-1+2=5, which makes sense.  it's just a matter of looking at it right.

like this problem: three guys went out for dinner, and the waiter told them it would cost 30 dollars.  so they each gave him 10 dollars.  then the waiter found out it only cost 25, so he decided to give one dollar back to each guy, then keep the other 2.  so, adding this all up, each guy paid 9 dollars, for a total of 27, plus the other 2 dollars that the waiter kept, so it adds up to 29.  what happened to the other dollar?

@perfectnumber, that's not math. That's just a brainteaser. There is no reason that the amount the guys paid + the tip the waiter has should add up to 30. The 3 guys paid 27, subtract 2 for the waiter, and the cost is 25.

There are algebraic anomalies that come up in math, i.e.

$1+2+3 ... = \frac{-3}{12}$

$1^{2}+2^{2}+3^{2}... = 0$.

Also, there are tricks to complex number arithmetics: $i^{2}= \sqrt{-1}* \sqrt{-1}= \sqrt{-1 *-1 }= \sqrt{1}= 1$. That's why such operations are forbidden.

[quote="lotrgreengrapes7926"][quote="chess64"][quote="Phelpedo"]Also, there are tricks to complex number arithmetics: $i^{2}=\sqrt{-1}*\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}=1$. That's why such operations are forbidden.[/quote]

WRONG. $\sqrt{1}=\pm 1$.[/quote]
#1: The square root symbol only takes the principal square root, the positive one.
#2: It would still be incorrect, $i^{2}\ne\pm1$. :D[/quote]
:huh:
$i^{2}$=-1

If $i^{2}=\sqrt1$ and $\sqrt1=-1$ and $\sqrt1=-1$, then by transitivity, we have $i^{2}=1$ and $i^{2}=-1$, the first is obviously not true.

[quote="chess64"][quote="Phelpedo"]Also, there are tricks to complex number arithmetics: $i^{2}=\sqrt{-1}*\sqrt{-1}=\sqrt{-1*-1}=\sqrt{1}=1$. That's why such operations are forbidden.[/quote]

WRONG. $\sqrt{1}=\pm 1$.[/quote]

WRONG. The $i \neq \pm 1$. The point is, math is perfectly and universally laid out, so humans have to improvise, define exceptions, etc.

[quote="Phelpedo"]Also, there are tricks to complex number arithmetics: $i^{2}= \sqrt{-1}* \sqrt{-1}= \sqrt{-1 *-1 }= \sqrt{1}= 1$. That's why such operations are forbidden.[/quote]

That's because the identity $\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ only works when $a$ and $b$ are positive real numbers.

How do you do $1^{2}+2^{2}+3^{2}+...=0$? :?

And I thought it was $1+2+3+...=\frac{-1}{12}$ not $\frac{-3}{12}$...

You may be right. Either way, it's clearly wrong :wink:

$1^{2}+2^{2}+3^{2}+\cdots = \zeta(-2) = 0$

and

$1+2+3+\cdots = \zeta(-1) =-\frac 1{12}$

and you could keep going with the zeta's.  (In general $1^{2k}+2^{2k}+3^{2k}+\cdots = 0$, for example).

[quote="Chooka"]are there any more senarios like this?
thanks[/quote]

If you define symbols to mean things other than what they "really mean," it is of course possible to construct statements whose "new meaning" is that of a true statement but whose "real meaning" is that of a false statement.  In your particular example, the symbol "a - b" doesn't mean "the result of subtracting b from a in the conventional way," it means "the largest number of vertices we can create if we use b cuts on an a-gon."  Expressed in this way, the type of things you're looking for seem much less exciting, I think.  (Many, many brainteasers are based on this substitution of one, less likely meaning for another.  As are many good crossword clues.)

i have a really great proof but this thread is too short for it^^

no, i have really one. i will show that 1=3 and then i onkly have to add 2 on eaxh side.

1.) one cat has one tail more than no cat
2.) no cat has 2 tails

so one cat has three tails and one tail so 1=3 holds.

the rest ist trivial...

Naphthalin

[quote="OHO"][quote="ShoeFactory"][quote="Phelpedo"]That's not a logic error, that's just taking advantage of a wording thingy.[/quote]

duh.

i think he is able to realize it is so.[/quote]I think Phelpedo realizes that he is able to realize it is so.[/quote]

I think everyone realizes that everyone else is able to realize that is so, and furthermore, this is :spam: and probably should stop.

How do you figure the zeta function? (Probably the wrong forum)

Like here: http://mathworld.wolfram.com/RiemannZetaFunction.html ;)


I'm very pleased to anybody translates the following link into english
[url=http://www.pandd.demon.nl/]http://www.pandd.demon.nl/[/url]

 :blush: I dont know in which language it was writen

thank a lot

Well, according to the ".nl", it should be in Dutch, but, since I don't know it, I won't bother looking at it... :)


logarithms
inequalities
number theory solved
number theory

Let $S_1$ and $S_2$ denote the sum of the numbers of odd resp. even divisors of the numbers 1, 2, ..., n. Prove that :$|S_1-S_2-n.\ln 2|<\sqrt{n}+1$

Lemma: $k|n\Leftrightarrow \left[\frac{n}k\right]-\left[\frac{n-1}k\right]=1$.
Proof: Exercise (I always wanted to do this to someone).

Let $D_n=S_1(n)-S_2(n)$.
It is easy to see, using the lemma that
$\displaystyle D_n=\sum_{k=1}^{n}((-1)^{k+1}(\sum_{m=k}^{n}(-1)^{n-m}\cdot\left[\frac{m}k\right]))$.

We have to prove that $|D_n-n\cdot \ln 2|<\sqrt{n}+1\Leftrightarrow n\cdot \ln 2-\sqrt{n}-1<D_n<n\cdot \ln 2+\sqrt{n}+1$.[b][1][/b]
We attempt to prove this by induction.

$\displaystyle D_{n+1}=\sum_{k=1}^{n+1}(-1)^{k+1}\left[\frac{n+1}k\right]-D_n$
Let $f(n)=\sum_{k=1}^n(-1)^{k+1}\left[\frac{n}k\right]$.
So we would like to prove:
$(n+1)\cdot \ln 2-\sqrt{n+1}-1+D_n<f(n+1)<(n+1)\cdot \ln 2+\sqrt{n+1}+1+D_n$.
It would be sufficient to prove
$(2n+1)\ln 2-\sqrt{n+1}+\sqrt{n}<f(n+1)<(2n+1)\ln 2+\sqrt{n+1}-\sqrt{n}$.[b][2][/b]

For the second inequality in (2):
$f(n+1)=\sum_{k=1}^{n+1}(-1)^{k+1}\left[\frac{n+1}k\right]<(n+1)(1-\frac{1}2+\ldots+(-1)^n\frac{1}{n+1})+\left[\frac{n}2\right]+1<(n+1)\ln\left(\frac{n+1}{[\frac{n}2]+1}\right)+\left[\frac{n}2\right]+1<(2n+1)\ln 2$ (the last is true for n large enough and we can check (1) for small n).

Hmmm the first inequality in (2) doesn't seem to hold.

Well, I found this a very hard problem, with a very nice idea. Too bad I know the solution given in Komal. Anyway, this estimation helps us to prove another inequality with fractional part that hxtung insisted to solve.


What is the highest common factor of $ n^{5}\minus{}n$ for positive integers $ n$?

Obviosly $ 2|n^5\minus{}n,3|n^5\minus{}n,5|n^5\minus{}n$ or $ 30|n^5\minus{}n$. But $ 3\equal{}2^5\minus{}2$. Therefore $ gcd(n^5\minus{}n,n\in N)\equal{}30$