If $ tanx\plus{}cotx\equal{}4$, find the general solution for $ x$

The answer should be :
\[ \frac{{n\pi }}
{2} \plus{} ( \minus{} 1)^n \frac{\pi }
{{12}}
\]

[quote="Farenhajt"][hide="Continuation"]$ {1\over\cos^2x} \equal{} 1 \plus{} \tan^2x \equal{} 1 \plus{} 4\pm 4\sqrt {3} \plus{} 3 \equal{} 8\pm 4\sqrt {3}$

$ \cos^2x \equal{} {8\mp 4\sqrt {3}\over 16} \equal{} {2\mp\sqrt {3}\over 4}$

$ 2\cos^2x \minus{} 1 \equal{} \mp{\sqrt {3}\over 2}\iff \cos 2x \equal{} \pm{\sqrt {3}\over 2}$

Hence $ 2x \equal{} \pm{\pi\over 6} \plus{} 2k\pi$ or $ 2x \equal{} \pm{5\pi\over 6} \plus{} 2k\pi$, which yields

$ x \equal{} \pm{\pi\over 12} \plus{} k\pi$ or $ x \equal{} \pm{5\pi\over 12} \plus{} k\pi$

(Note 1: these are $ \pm 15^\circ, \pm 75^\circ,\pm 105^\circ, \pm 165^\circ$ angles)

(Note 2: the solutions are the same as in the previous post, only the form of the expression is different.)[/hide][/quote]
Just wondering in general how do you combine expressions like $ \pm{\pi\over 12} \plus{} k\pi$ and $ \pm{5\pi\over 12} \plus{} k\pi$ to a single expression?

[quote="Differ"][quote="addikaye03"]hardly a pre-olympiad Q one would think[/quote]

Well, the quadratic equation solves it pretty quickly.[/quote]

yeah thats what i mean, and the quadratic equation approach was VERY obvious.


From $\text{MA}\theta \text{ 1992}$
If, for all $x$, $f(x)=f(2a)^x$ and $f(x+2)=27 f(x)$, then find $a$.

The part where it says $f(2a)^x$. Is it the same as $\left( f(2a) \right)^x$ or $f \left( (2a)^x \right)$ ?

Presuming that f is a variable (not a function), it's not ambiguous.  The exponentiation by x applies to the contents of the parentheses.  The second expression you have there is correct.

Sorry, i typed it wrong, F is a function

Okay.  Now, the meaning is the other: $f(2a)^x = (f(2a))^x$


Prove that for every n>2, exists such numbers q and p for which:
2 not divide p and q
7p^2 + q^2 = 2^n

Happy New Year!  :D

The famous one,click [url=http://www.mathlinks.ro/viewtopic.php?t=161309]here[/url]


[asy]size(150);
path A=(6,0),B=(0,0),C=(3,5.196),D=(1.5,2.598),E=(4.5,2.598),F=(2,0);
draw(A--B--C--A,linewidth(2));
draw(D--E,linewidth(1));
draw(D--F--E,linewidth(1));
label("$A$",B,S);
label("$B$",A,S);
label("$C$",C,N);
label("$D$",D,W);
label("$E$",E,NE);
label("$F$",F,S);[/asy]

ABC is a equilateral triangle.
D is the midpoint of AC
F is a point on AB such that BF=2AF
Area of triangle DEF is $ \frac{\sqrt{2}}{3}$
Find the area of ABC.

Is $ E$ the midpoint of $ BC$? Otherwise answers would vary.

[hide="Assuming that $ E$ is the midpoint of $ BC$..."]
$ DE$ is $ \frac{1}{2}$ of $ AB$, and the altitude from $ D$ or $ E$ is $ \frac{1}{2}$ the altitude from $ C$ so regardless of where point $ F$ is, the area of $ \triangle DEF$ is $ \frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$. Thus, $ \frac{\sqrt{2}}{3} \cdot 4= \boxed{\frac{4\sqrt{2}}{3}}$.
[/hide]

As $ FD$ and $ BC$ are not parallel lines, and the area of the triangle $ DEF$ is $ \frac{h\cdot DF}{2}$, $ h$ changes with the movement of point $ E$. If point E is needed only to be on the line $ BC$ (not necessarily between those two points), the height changes between zero and infinity, and if it is needed that the point $ E$ is between points $ B$ and $ C$, then it changes between some minimal and maximal values (which are achieved in points $ C$ and $ B$, respectively). Those values could be calculated in function of $ a\equal{}AB$ using some basic trigonometry, but it would still give just an interval in which the wanted area is.

I dont't get it. Your problem is obscure.

That's what we are saying. Without knowing the position of $ E$, we cannot find the area.

I think you meant
[quote="Textangle"]the area of $ \triangle DEF$ is $ \frac {1}{2}\cdot \frac {1}{2} \equal{} \frac {1}{4}[ABC]$, where [ABC] is the area of triangle ABC.[/quote]


I just wanted to point out that this inequality has appeared in the British Mathematical Olympiad in 1999. It can probably be found somewhere else on the forum.

[quote="payman_pm"]a,b,c are positive and we have a+b+c=1 prove that
7(ab+bc+ac) <= 2+9abc[/quote]

I don't remember it being discussed on the forum, but an equivalent form of the inequality ($ab+bc+ca\leq\frac27+\frac{9abc}{7}$ - this is just the inequality $7\left(ab+bc+ca\right)\leq 2+9abc$ after division by 7) was solved on http://www.kalva.demon.co.uk/bmo/bsoln/bsol993.html .

  Darij

As usual, it's enough to show that

$2(a+b+c)^3+9abc\geq 7(ab+bc+ca)(a+b+c)$

This all is equivalent to

$2\sum a^3+6\sum a^2b+21abc\geq 7\sum a^2b+21abc$

So we just need to show

$2\sum a^3\geq \sum a^2b$ which is obvious by $a^3+a^3+b^2\geq 3a^2b$ and so on.

I have another ineq using Schur
Let $a,b,c\ge 0$ and $a^2+b^2+c^2=abc$ Prove that
$ab+bc+ca\ge 4(a+b+c)-9$

[quote="RaduSorici"]From the Schur inequlity we have that :
$\sum a(a-b)(a-c)\geq0$ or
$\sum a(1-2b-c)(a-c)\geq0$ or
$\sum (a^2-ac-2a^2b+2abc-a^2c+ac^2)\geq0$ or
$\sum a^2 -\sum (ab)+6abc \geq \sum a^2b+\sum a^2c = \sum a^2(1-a) = \sum a^2 - \sum a^3$ or
$a^3+b^3+c^3+6abc\geq ab+bc+ca$ or
$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+9abc \geq ab+bc+ca$ or
$9abc \geq 2(ab+bc+ca)-(a^2+b^2+c^2)=1-2(a^2+b^2+c^2)$. We will insert this result in the initial inequality and we will have to prove that :
$3-2(a^2+b^2+c^2)\geq 7(ab+bc+ca)$ or
$3(a+b+c)^2-2(a^2+b^2+c^2)\geq 7(ab+bc+ca)$ or
$a^2+b^2+c^2 \geq ab+bc+ca$  :D[/quote]
Wonderful Radu, but don't you think that this is very similar to this one: http://www.mathlinks.ro/Forum/viewtopic.php?t=28546

I didn't look in there seriously If u mean that :P


quadratics
calculus
integration
number theory
Diophantine equation
number theory unsolved

Let p be an odd prime. Show that the Diophantine equation x^2+py+a=0 where gcd(a,p)=1

has an integral solution if and only if (-a/p)=1 ( (-a/p) is a Legendre symbol.


calculus
derivative
function
real analysis
real analysis unsolved

If function $f$  is non negative on  $IR$ and $g$ positive on  $IR , f(g(x)) = e^{-2x}, g(f(x)) = e^{-x^2}$, and both have derivatives, $f(0) = f'(0) = 0 , g(0) =0, g'(0) = - 1$

is it possible to find these two function?  Or we must add some more conditions?

 [i]Of course I'm trying to prove that $f(x) = x^2 , g(x) = e^{-x}$[/i]

[u]Babis[/u]

There may be a typo because $\ (e^{-x})'(0)=-1$.

Yes!

  Thank you. I edit right now!

 Babis

If yoy find some time, please see it again.

 Babis


probability
geometry
topology
combinatorics unsolved
combinatorics

sorry I'm not sure if this question should be posted here (if it's not please don't delete and move it to the right section).  Thanks

Two points are selected randomly in the interval [0,1], what's the probability that the sum of their squares is less than 1?

any hint would be helpful.

We call one of the numbers $x$, and the other $y$, and consider all the possible selections of points $(x,y)$ in the Cartesian plane.  All possible combinations are represented by the square with vertices at $(1,1)$, $(1,0)$, $(0,1)$, and $(0,0)$.  Now in order to pick two points such that the sum of their squares is less than one, we need all points in this region such that $x^{2}+y^{2}<1$, which is...

[hide]...the interior of the quarter circle $x^{2}+y^{2}=1, x\geq 0, y\geq 0$.  Taking the area of this quarter circle, $\frac{\pi}{4}$, and dividing it by the area of our "sample space," that is, the area of the square above, which is $1$, we get a probability of $\frac{\pi}{4}$, or approximately $.7854$.[/hide]

the vertices are not necessary right? i mean they are there for the "borders" for the number to be picked between 0 and 1?

It doesn't matter, in this instance, whether you choose the number from the open set (0,1) or the closed set [0,1], because there are an infinite number of choices of $x$ and $y$.  So it doesn't matter if you include the vertices as possible choices or not, the answer remains the same as long as the least upper bound on the choice of $x$ and $y$ is 1.


search
induction
number theory unsolved
number theory

Find all primes $ p$ such that for every positive integer $ s$, $ 2^{p^s}\plus{}1$ is divisible by $ p^{s\plus{}1}$.

We know:
$ 2^p$ =2(mod p)
$ 2^p \plus{} 1$ =3(mod p)
$ 2^p \plus{} 1$ is divisible by $ p^2$ , it means p is 3.
We can show that for every s , $ 2^{p^s}\plus{}1$ is divisible by $ p^{s\plus{}1}$

You still have to show that $ 3^{s\plus{}1}|2^{3^s} \plus{} 1$. But it's a straightforward induction.

We can show it by using Gelzel Lemma ( I don't know exactly the name)

[quote="t0rajir0u"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=345802527&t=205928]lifting the exponent[/url].[/quote]

But the lifting exponent is with a minus ($ a^n\minus{}1$). This one is with a plus. Can it be used as well?

[quote="discredit"][quote="t0rajir0u"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=345802527&t=205928]lifting the exponent[/url].[/quote]

But the lifting exponent is with a minus ($ a^n \minus{} 1$). This one is with a plus. Can it be used as well?[/quote]
Yes.  The equivalent problem is to prove that $ 3^{s\plus{}1} | 4^{3^s} \minus{} 1$.  Clearly $ 2^{3^s} \minus{} 1 \equiv 1 \bmod 3$.

[quote="discredit"]Umm.. ,but does the "lifting exponent" theorem hold with a plus?[/quote]
put $ a \equal{} \minus{} 2$ in the lifting exponent lemma,we get that $ 3^{s \plus{} 1}|( \minus{} 2)^{3^s} \minus{} 1\Rightarrow 3^{s \plus{} 1}|2^{3^s} \plus{} 1$

I see. Thanks for this nice theorem, t0rajir0u  :)

[quote="t0rajir0u"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=345802527&t=205928]lifting the exponent[/url].[/quote]
I knew this is Genzel Lemma , isn't it?


Simplify
\[ \sqrt[3]{x\sqrt[3]{x\sqrt[3]{x\sqrt{x}}}}
\]$ \textbf{(A)}\ \sqrt{x} \qquad
\textbf{(B)}\ \sqrt[3]{x^2} \qquad
\textbf{(C)}\ \sqrt[27]{x^2} \qquad
\textbf{(D)}\ \sqrt[54]{x} \qquad
\textbf{(E)}\ \sqrt[81]{x^{80}}$

$ \sqrt [3]{x\sqrt [3]{x\sqrt [3]{x\sqrt {x}}}} \equal{} x^{\frac{2}{3}} \equal{} \sqrt[3]{x^2}$

so answer choice $ \boxed{B}$.

Actually:

[hide="Correct solution"]
$ \sqrt [3]{x\sqrt [3]{x\sqrt [3]{x\sqrt {x}}}}$

$ \equal{} \sqrt [3]{x\sqrt [3]{x\sqrt [3]{x^\frac {3}{2}}}}$

$ \equal{} \sqrt [3]{x\sqrt [3]{x\cdot x^{\left(\frac {3}{2}\cdot\frac {1}{3}\right)}}}$

$ \equal{} \sqrt [3]{x\sqrt [3]{x^\frac {3}{2}}}$

Apply the same logic again here to get:

$ \sqrt [3]{x^\frac {3}{2}} \equal{} x^\frac {1}{2}\implies\boxed{\text{A}}$
[/hide]

Edit:
http://www.artofproblemsolving.com/Wiki/index.php/2003_AMC_10A_Problems/Problem_9
AoPSWiki agrees.

The correct answer is $\boxed{\sqrt{x}}$

xd that 3.5 year bump is really unnecessary because it really didn't add anything to the conversation at all

We deal with each $x$ separately and then multiply them out (note: you can do this because: $\sqrt[n]{a \cdot b} = \sqrt[n]{a} \cdot \sqrt[n]{b}$). So the $x$ that is under the square root can be rewritten as $x^{\frac{1}{2}^{\frac{1}{3}^3}}$, which is $x^{\frac{1}{54}}$. Similarly, the $x$ under the three cube roots is $x^{\frac{1}{27}}$, the $x$ under the two cube roots is $x^{\frac{1}{9}}$, and the $x$ under the single cube root is clearly $x^{\frac{1}{3}}$. So we have $x^{\frac{1}{54}} \cdot x^{\frac{1}{27}} \cdot x^{\frac{1}{9}} \cdot x^{\frac{1}{3}}$, so the exponent on the $x$, it is $\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{54} \rightarrow \frac{18}{54} + \frac{6}{54} + \frac{2}{54} + \frac{1}{54}$, this is $\frac{27}{54}$, or $\frac{1}{2}$. So $x^{\frac{1}{2}} \rightarrow \boxed{\sqrt{x}}$.

The exponent of $x$ can be expressed as $$\frac {1+\frac{1+\frac{1+\frac{1}{2}}{3}}{3}}{3}=\frac{1}{2}$$
Therefore, the answer shall be $\boxed {(A) \sqrt {x}}$.

this thread has gotten bumped so many times... from 2009 to 2015, from 2015 to 2018, from 2018 to 2020 - 2021. :)

but yeah I got A


function
limit
induction
algebra
functional equation
algebra unsolved

Suppose that for a function $f: [a,b]\rightarrow \mathbb{R}$ we have:

$\forall \, x,y \in [a,b]: f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}{2}$

Is the statement

$\forall \,t\in [0,1]\, ,\,x,y\in [a,b]: f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$

necessarily true?

Yes, it is true, use that f(x) is continiosly for all $x\in (a,b)$.

We had this discussion for a dozen of times on ML I believe. See http://www.mathlinks.ro/Forum/viewtopic.php?t=46490 post #10, point [b]3.[/b].

  darij

[quote="perfect_radio"]No, it's not true, since $f$ is not continuous.[/quote]
It is easy to prove, that f(x) is continiosly for $x\in (a,b)$. But  can be $\lim_{x\to y}f(x)\not =f(y)$ for y=a and for y=b.

[quote="Rust"]It is easy to prove, that f(x) is continiosly for $x\in (a,b)$. But  can be $\lim_{x\to y}f(x)\not =f(y)$ for y=a and for y=b.[/quote]

So you claim that the wild solutions of the Cauchy functional equation are continuous on open intervals?

  Darij

[quote="Rust"]I can prove it. Can you give contrexample?[/quote]

It is a well-known exercise to show that a solution of Cauchy's functional equation $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ which is continuous on some interval must be linear. On the other hand, if you assume the axiom of choice to hold, then there are non-linear solutions of Cauchy's functional equation; these solutions must be therefore non-continous on any interval of nonzero length. And of course, they satisfy $f\left(\frac{x+y}{2}\right)\leq\frac{f\left(x\right)+f\left(y\right)}{2}$ (even with the equality sign). Even if you don't assume the axiom of choice, you can't prove that there are no such solutions, unless, of course, you work in some different system of axioms than ZF, but if this is the case you should mention it in your post.

  Darij

I am sorry, I was wrong. By axiom of choise can be constructed Hamel's basys and construct not continiysly function f, suth that $f(x+y)=f(x)+f(y)$. In this way can be consruct function f(x), suth that:
$f(\frac{x+y}{2})\le f(x)+f(y) \forall x,y\in [a,b]$, but exist $x,y\in [a,b],t\in (0,1)$, suth that $f(tx+(1-t)y)>tf(x)+(1-t)f(y)$.

From what I understood from your posts the statements

$\forall \, x,y \in [a,b]: f(\frac{x+y}{2})\leq \frac{f(x)+f(y)}{2}$

and

$\forall \,t\in [0,1]\, ,\,x,y\in [a,b]: f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$

are equivalent only if $f$ is continuous on $[a,b]$. Can someone prove this?

[quote="Rust"]Yes it is true. Fron $f(\frac{x+y}{2})\le \frac{f(x)+f(y)}{2}$ by induction get, that $f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$ for any $t=\frac{a}{2^{k}},0\le a\le 2^{k}$.[/quote]
The proof with details by Hojoo Lee is [url=http://ultrametric.googlepages.com/tin2006new.pdf]here[/url].


Have any of you ever been there? If so, where in the BWCA were you, and when did u go? How was ur trip? I just got back from a week long trip up there through my school, and was wondering if anyone on here had been there b4.

I see that there are 2 yes votes...that means that someone other htan myself has been there...who might that be?

I was there a while ago when I used to live in Minnesota. I don't remember it very well, though.


geometry
parallelogram
circumcircle
geometry solved

Let ${ABCD}$ a parallelogram and ${P}$ a point such that:

$2\angle{PDA}= \angle{ABP}$ and $2\angle{PAD}= \angle{PCD}$

Prove that $AB=BP=CP$.

This is a nice problem, though it uses a pretty standard trick (cf. http://www.mathlinks.ro/Forum/viewtopic.php?t=30661 ). But first let me rewrite your problem using directed angles modulo 180°, in order to make it less ambivalent:

[color=blue][b]Problem.[/b] Let ABCD be a parallelogram, and P a point in its plane such that 2 < PDA = < ABP and 2 < PAD = < DCP, where all angles are directed angles modulo 180°. Prove that AB = BP = CP = DC.[/color]

[i]Solution.[/i] Since the quadrilateral ABCD is a parallelogram, $\overrightarrow{AB}=\overrightarrow{DC}$. Now, let Q be a point such that $\overrightarrow{PQ}=\overrightarrow{AB}=\overrightarrow{DC}$. Then, since $\overrightarrow{PQ}=\overrightarrow{AB}$, the quadrilateral ABQP is a parallelogram, so that PQ || AB, QP = AB and BQ || AP, and since $\overrightarrow{PQ}=\overrightarrow{DC}$, the quadrilateral DCQP is a parallelogram, so that PQ || CD, QP = DC and CQ || DP. On the other hand, since ABCD is a parallelogram, BC || DA.

Since CQ || DP and BC || DA, we have < (CQ; BC) = < (DP; DA), or, in other words, < QCB = < PDA. Also, PQ || AB yields < (PQ; BP) = < (AB; BP), or, equivalently, < QPB = < ABP. Hence, the equation 2 < PDA = < ABP becomes 2 < QCB = < QPB. Now, if U is the circumcenter of the triangle BQC, that is the center of the circle through the points B, Q and C, then, by the central angle theorem, < QUB = 2 < QCB. Hence, < QUB = < QPB. Thus, the point P lies on the circumcircle of triangle QUB. Similarly, by transforming the equation 2 < PAD = < DCP, we see that the point P lies on the circumcircle of triangle QUC. But the circumcircles of triangles QUB and QUC have only two common points - namely, the points Q and U -; thus, we see that the point P coincides with either the point Q or the point U. Since the point P cannot coincide with the point Q (since $QP=AB\neq 0$), it follows that the point P coincides with the point U. Thus, the point P is the circumcenter of triangle BQC. Hence, BP = QP = CP. But since QP = AB and QP = DC, this becomes AB = BP = CP = DC, and the problem is solved.

  Darij


limit
trigonometry
calculus
calculus computations

Show that
 $\lim_{n\rightarrow\infty} cos\frac{x}{2}.\ cos\frac{x}{4}.\ cos\frac{x}{8}\ ...\ cos\frac{x}{2^n} \ = \ \frac{sin (x)}{x}$

Look at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=68475]www.mathlinks.ro/Forum/viewtopic.php?t=68475[/url] there was something similar.

[quote="NaZAf"][quote="kunny"]Multiply $\sin \frac{x}{2}$ by $\cos \frac{x}{2}\cdot \cos \frac{x}{4}\cdot \cos \frac{x}{8}\cdot\cdot\cdot\cos \frac{x}{2^n} \ ,$ then use the identy $2\sin \theta \cos \theta=\sin 2\theta.$ repeatedly.[/quote]

Can you give a formal demonstration?[/quote]

Let $A= \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^n}$
then
\begin{eqnarray*}2 \sin \frac{x}{2^n} A &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots 2 \cos \frac{x}{2^n} \sin \frac{x}{2^n}\\ &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n-1}} \sin \frac{x}{2^{n-1}}\end{eqnarray*}

Next multiply $2 \sin \frac{x}{2^n} A$ with 2, we have
\[ 2^2 \sin \frac{x}{2^n} A = \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n-2}} \sin \frac{x}{2^{n-2}}  \]
Do it $n-1$ times. we have
\[ 2^n \sin \frac{x}{2^n} A = \sin x \Longrightarrow A = \frac{\sin x}{\displaystyle\frac{\sin \frac{x}{2^n}}{\frac{1}{2^n}}} \]

So $\lim_{n \rightarrow \infty} A = \frac{\sin x}{x}$

[quote="Beginner"]
Let $A= \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^n}$
then
\begin{eqnarray*}2 \sin \frac{x}{2^n} A &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots 2 \cos \frac{x}{2^n} \sin \frac{x}{2^n}\\ &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n-1}} \sin \frac{x}{2^{n-1}}\end{eqnarray*}

Next multiply $2 \sin \frac{x}{2^n} A$ with 2, we have
\[ 2^2 \sin \frac{x}{2^n} A = \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n-2}} \sin \frac{x}{2^{n-2}}  \]
Do it $n-1$ times. we have
\[ 2^n \sin \frac{x}{2^n} A = \sin x \Longrightarrow A = \frac{\sin x}{\displaystyle\frac{\sin \frac{x}{2^n}}{\frac{1}{2^n}}}  \]

So $\lim_{n \rightarrow \infty} A = \frac{\sin x}{x}$[/quote]

Yes but Is not $2 \cdot cos\frac{x}{2^{n}} \cdot sin \frac{x}{2^{n}} = sin \frac{x}{2^{n-1}}$ ??

[quote="Beginner"]Sorry i don't understand what you mean?
$2 \sin x \cos x = \sin 2x$ ??[/quote]

Because you wrote \begin{eqnarray*}2 \sin \frac{x}{2^n} A &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots 2 \cos \frac{x}{2^n} \sin \frac{x}{2^n}\\ &=& \cos \frac{x}{2} \cos \frac{x}{4} \cdots \cos \frac{x}{2^{n-1}} \sin \frac{x}{2^{n-1}}\end{eqnarray*}

But $2 \cos \frac{x}{2^n} \sin \frac{x}{2^n} = \sin \frac{x}{2^{n-1}}$, then where does $\cos \frac{x}{2^{n-1}}$ come from?

It was there in the product all along; the ... notation merely hid it.


let $ q_i,p$ be odd primes,$ \alpha$ positive integer

solve or proof that no solutions exist for the follwing eequations simultanously

$ q_1q_2 \equal{} p^\alpha\plus{}2$

$ (q_1\minus{}1)(q_2\minus{}1)\equal{}p^\alpha\plus{}2\minus{}p^{\alpha\minus{}1}$

[quote="wauwau"]let $ q_i,p$ be odd primes,$ \alpha$ positive integer

solve or proof that no solutions exist for the follwing eequations simultanously

$ q_1q_2 \equal{} p^\alpha \plus{} 2$

$ (q_1 \minus{} 1)(q_2 \minus{} 1) \equal{} p^\alpha \plus{} 2 \minus{} p^{\alpha \minus{} 1}$[/quote]

Subtracting the second equation from the first, we get $ q_1\plus{}q_2\equal{}p^{\alpha\minus{}1}\plus{}1$ and so :

$ q_1q_2\equal{}p^{\alpha}\plus{}2$
$ q_1\plus{}q_2\equal{}p^{\alpha\minus{}1}\plus{}1$

A consequence is that $ q_1q_2\minus{}2\equal{}p(q_1\plus{}q_2\minus{}1)$ which may be written $ (q_1\minus{}p)(q_2\minus{}p)\equal{}p^2\minus{}p\plus{}2>0$

We cant have $ q_1<p$ and $ q_2<p$ since this would imply $ p^{\alpha}\plus{}2\equal{}q_1q_2<p^2$ and so $ \alpha\equal{}1$ and $ q_1\plus{}q_2\equal{}p^{\alpha\minus{}1}\plus{}1\equal{}2$ and so $ q_1\equal{}q_2\equal{}1$ not odd primes.

So $ q_1>p$ and $ q_2>p$ and then :
a) $ p^{\alpha}\plus{}2\equal{}q_1q_2\ge(p\plus{}1)^2$ and so $ \alpha>2$
b) $ (q_1\minus{}p)(q_2\minus{}p)\equal{}p^2\minus{}p\plus{}2$ implies $ q_1\minus{}p\le p^2\minus{}p\plus{}2$ and $ q_1\le p^2\plus{}2$ and so $ p^{\alpha\minus{}1}\plus{}1\equal{}q_1\plus{}q_2\le 2p^2\plus{}4$ and so either $ \alpha\equal{}4$ and $ p\equal{}2$, either $ \alpha < 4$

The case $ \alpha\equal{}4$ and $ p\equal{}2$ implies $ q_1q_2\equal{}18$ and $ q_1\plus{}q_2\equal{}9$ and so $ (q_1,q_2)\equal{}(3,6)$, not a solution.

So $ 2<\alpha<4$ and $ \alpha\equal{}3$ and the system becomes :

$ q_1q_2\equal{}p^3\plus{}2$
$ q_1\plus{}q_2\equal{}p^2\plus{}1$

And $ q_1,q_2$ solutions of $ x^2\minus{}(p^2\plus{}1)x\plus{}p^3\plus{}2$ whose discriminant is $ \Delta\equal{}(p^2\plus{}1)^2\minus{}4(p^3\plus{}2)\equal{}p^4\minus{}4p^3\plus{}2p^2\minus{}7$

But, if $ p>5$ : $ (p^2\minus{}2p\minus{}2)^2\equal{}p^4\minus{}4p^3\plus{}8p\plus{}4$ $ <p^4\minus{}4p^3\plus{}2p^2\minus{}7<(p^2\minus{}2p\minus{}1)^2\equal{}p^4\minus{}4p^3\plus{}2p^2\plus{}4p\plus{}1$ and so $ \Delta$ cant be a perfect quare.

So $ p\le 5$ and :

$ p\equal{}5$ implies $ q_1q_2\equal{}5^3\plus{}2\equal{}127$ and $ q_1\plus{}q_2\equal{}5^2\plus{}1\equal{}26$ and so no solution.
$ p\equal{}3$ implies $ q_1q_2\equal{}3^3\plus{}2\equal{}29$ and $ q_1\plus{}q_2\equal{}3^2\plus{}1\equal{}10$ and so no solution.
$ p\equal{}2$ implies $ q_1q_2\equal{}2^3\plus{}2\equal{}10$ and $ q_1\plus{}q_2\equal{}2^2\plus{}1\equal{}5$ and so no solution.

So no solution for this equation
(If I made no error)

Perfect!

any ideas on disproving the same, if instead of only $ q_1,q_2$
$ q_1,q_2,...q_k$ are in place?

Or are there any solutions for k=3,.... ?


I need a good avatar...where can I find a moving one?

Ask people who are good at making avatars?

@meewhee..how did you do that :?:
I want one like that :)

Well, I don't know how to make a moving avatar, unless your talking about smileys.  Those you just save the image and attatch for your signature.  I just did that with the wallbash because I was kinda desperate to get some sort of feature.

And sorry Powerofpi:  For about 10 seconds I stole your avatar to test to see if I could actually do something and have some sort of avatar.  It's gone now.

EDIT:  I finally got my Maybach avatar working.  Whew!

who's really good at making custom avatars?

Dojo is.

He made every one of those spinning ball avatars. I think...

@King1 Oh sure. I can make you one. Dojo's good too if you want a spinning ball.... I make avatars. Like the one I'm using... If you want I can make it flash...  :lol:. PM me if you want something specific.

for mine, i just googled some random things and i found my moving avatar...
hmm, do you need some software to create moving avatars?

Well for some I use Microsoft Digital Image Pro 9.... and.... to make them move use picture2life.com or photobucket. :)

Something like this..

[img]http://img002.picture2life.net/5624422/untitled_animation_web-large_medium.gif[/img]

That's just basic, you can get all technical and creative other than something like this.
If you want an avatar like mine
<--
PM me :)

Flashing example;
[img]http://img002.picture2life.net/5634160/untitled_animation_web-large_medium.gif[/img]
Take it if you want

Please do not revive topics that are 1 year old thanks!


Anyone know where I can find a complete list of gate commands?
the triple gates, and a proper deduction gate "$ \vdash_o$", in particular.

I don't know what you mean by 'gate commands'. \vdash is a binary relation and you'll find lots of similar ones in [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url]. But if you mean circuits then you need macros to draw them. [url=http://tug.ctan.org/cgi-bin/search.py?metadataSearch=circuit]CTAN[/url] has some packages for circuit drawing.


function
algebra
polynomial
group theory
abstract algebra
Galois Theory
advanced fields

Prove that any finite group can be obtained as a Galois group.

The Superior Algebra forum should fit better ;)

Let $G$ be any finite Group and $n=|G|$, consider the field $L=\mathbb C (x_{1},...,x_{n})$ of rational functions in $n$ variables and the field $K=\mathbb C (\sigma_{1},...,\sigma_{n})$ with $\sigma$ the symmetric polynomials in the $n$ variables $x_{i}$.
Then $L$ is Galois over $K$ with Galois group $S_{n}$ (any permutation of the $x_{i}$ gives a $K$-automorphism, and $L$is the splitting field of $x^{n}-\sigma_{1}x^{n-1}\pm ... \pm \sigma_{n}$), and as $G$ can be seen as subgroup of $S_{n}$, we find by Galois correspondence a intermediate field $M$ such that $L$ has Galois group $G$ over $M$.


Find all positive integers $(x,y,z,w)$ such that

$x>y$, $x^{2}z=y^{2}w$ and $xy+zw = 2^{99}+2^{101}$


ARML
algebra
function
domain
complex numbers

1.  Find x and y if

kx + y/k = k^2
x/k - ky = 1

2.  Let r be a root of x^2+5x+7-0.  Compute
(r-1)(r+2)(r+6)(r+3)

3.  Let f(x) = x^2 + bx + 9 and g(x) = x^2 + dx + e.  If f(x) = 0 has roots r and s, and g(x) = 0 has roots -r and -s, compute the two roots of f(x)+g(x)=0.

4.  If x = 1989(a-b), y=1989(b-c), and z=1989(c-a), compute the numerical value of (x^2+y^2+z^2)/(xy + yz +xz) given that xy + yz +xz !=0.

4. [hide]x+y+z=0. the expression is [(x+y+z):^2:-2xy-2zx-2yz]/(xy+yz+zx)=-2[/hide]

yeah.  I dont actually know how to do #1, does anyone have a solution?  It is not supposed to be hard, but i spent a good deal of time on it and could not get it.  The place i got #1 from has usually very easy questions.  I got all of the others from an ARML book.

I don't know how to do # 1 either. But I think It has something to do with the fact that the equations equal

kx+y(k^-1)=k^2

x(k^-1)-ky=k^0

just as a side note, k is a fourth root of -1, methinks. multiply the second equation by k and divide the first equation by k and set equal.

oh, alright.  thanks.  Was that one of the first things you tried?

kx + y/k = k^2
x/k - ky = 1

You're all being silly.  The solution is (k, 0), which I got by looking at it and saying, "Hey, if x = k and y = 0, then it works!"  If you're interested in a slightly more rigorous approach:

Clearly k cannot be 0, or else the givens cannot be true (since we will have an equality between a real number and an undefined expression, which is nonsense.)
kx + y/k = k:^2:
k:^3:x + ky = k:^4:
x/k - ky = 1
x(k:^3: + 1/k) = k:^4: + 1 = k*(k:^3: + 1/k).  Assuming k is a real number, which is generally to be assumed when a question does not give a specific domain to operate in, especially for linear equations, we then can divide by (k:^3: + 1/k) to get x = k and plug back in to get y = 0.

Sidenote:
If we were operating in complex numbers, we would also have to consider the case where k:^3: + 1/k = 0, which occurs when k is a fourth root of -1.  In that case, we find that
kx + y/k = k:^2:
k*x - y*k:^3: = k:^2:
x/k - yk = 1
so that the equations are in fact equivalent.  Thus, we will have in those situations an infinite set of solutions.

Second sidenote:
MysticTerminator, the reason you missed the answer is because you divided out a y without thinking about whether y was equal to 0 or not.  Shame, shame!  (j/k)


geometry
geometric transformation
reflection
geometry unsolved

[quote="nktp"]Let $ ABC$ be a triangle with the Kosnita point $ K$ (the isogonal conjugate of the nine-point center $ N$), $ P$ be the inversive image of $ K$ in the circumcircle. Denote $ P_1,P_2,P_3$ be the reflection of $ P$ by $ BC,CA,AB$.
a) Prove that $ AP_1,BP_2,CP_3$ concur at point $ Q$ in the circumcircle.
b) Prove that $ Q,K,N$ colinear.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120538

Thanks chien than! But it is not my problem. It only prove that Kosnita point is the isogonal conjugate of nine-point circle center. My problem is harder (I think)!


Call the four numbers, in order, a, b, c, and d. Therefore the final sum will be $ a \plus{} 3b \plus{} 3c \plus{}d$ (if my memory serves me right). $ a$ and $ d$ may be interchanged to give same value, so may the pair $ b$ and $ c$. So the number of distinct sums is at most $ C(4,2) \equal{} 6$. List them and find that one of the values is indeed repeated, so the answer is $ \boxed{5}$.


If the famous baseball player for the San Francisco Midgets, Larry Ponds, has a $ \frac{2}{5}$ chance of earning a walk on each plate appearance, what is the probability that he will earn a walk exactly once in his next two plate appearances? Express your answer as a common fraction.

There are two ways for Larry to earn a walk once: getting the walk the first time, or getting it the second time.

For the first way, we simply have $ \frac{2}{5} \times \frac{3}{5} \equal{} \frac{6}{25}$. Similarly, we have $ \frac{3}{5} \times \frac{2}{5} \equal{} \frac{6}{25}$ for the second way. Therefore, our answer is $ \frac{12}{25}$.


0<x,y<1. demonstrate that:
 $ {x}^{y}\plus{}{y}^{x}>1$

[quote="rts2007"]What if x=1 and y=0? Then the inequality would look like this: 1+0>1 which implies that 1>1, which is not true. Are there any restrictions for x and y?

-rts2007[/quote]
you should remember the conditions are:0<x,y<1

But my numbers work according to the conditions. Is there something I have missed?

-rts2007

Yes: the notation $ 0< x, y < 1$ is ambiguous and is intended to mean "$ x, y \in (0, 1)$," not "$ x > 0$ and $ y < 0$" (which is good, because the expression under discussion wouldn't make sense most of the time if that's what the condition meant).

This was also just reposted, with an equally pointless title and more typos: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268877


let $ABC$ be a triangle and (O) be the circumcenter of it.let $PP_'$ be a diameter of it.prove that if simson line of $P$ passes through the $p_'$ then it passes through the centroid of $ABC$ too.

Let s be the Simson line of the pole P and s' the Simson line of pole P'. Since PP' is a diameter of the circumcircle (O), the Simson lines $s \perp s'$ are perpendicular and they meet at a point K, which lies on the 9-point circle (N) of the triangle $\triangle ABC$ (a known theorem). Let H be the orthocenter of the triangle $\triangle ABC$. The Simson line s of the pole P cuts the segment PH at its midpoint M, which lies on the 9-point circle (N) and the Simson line s' also cuts the segment P'H at its midpoint M', which also lies on the 9-point circle (N) (another known theorem). Thus $MM' \parallel PP'$ are parallel, MM' being the midline of the triangle $\triangle PP'H$. Since the angle $\angle MKM' = 90^\circ$ of the 2 perpendicular Simson lines s, s' is right and $K, M, M' \in (N)$, MM' is a diameter of the 9-point circle. Let Q be the other intersection of the Simson line $s \equiv P'M$ with the circumcircle (O), different from the point P'. Since PP' is a diameter of (O), the angle $\angle PQP' = 90^\circ$ is right. Thus the right angle triangles $\triangle M'KM \sim \triangle PQP'$ are centrally similar, having the corresponding sides parallel: $PP' \parallel M'M$ ($MM'$ is the midline of the $\triangle PHP'$), $PQ \parallel M'K$  (both the Simson line $M'K \equiv s'$ and the line PQ are perpendicular to the Simson line $KM \equiv P'Q \equiv s$) and the lines $KM \equiv P'Q \equiv s$ are identical. Their coefficient of similarity is negative, because the corresponding vertices $K \Leftrightarrow Q$ and $M \Leftrightarrow P'$ follow on the Simson line s in the opposite directions. The lines connecting their corresponding vertices pass through their similarity center, in particular, the Simson line $s \equiv KQ$ passes through this similarity center. This similarity center is identical with the internal similarity center of their circumcircles (N), (O). But the internal similarity center of the 9-point circle (N) and the circumcircle (O) of the triangle $\triangle ABC$ is its centroid G, because the 9-point circle (N) is the circumcircle of its medial triangle, which is centrally similar to the triangle $\triangle ABC$ with the similarity center G and similarity coefficient $-\frac 1 2$.

nice solution but too long yetti.is there any other solution?
but i have to thank you for your solution,thanks. :D


Use numbers, written as words, to make the following sentence true: This sentence contains $ \_\_\_\_$ e's, $ \_\_\_\_$ t's and $ \_\_\_\_$ s's. What is the sum of the number of e's, t's and s's in the previous sentence?

Do each of the numbers need to be four letters long? :huh:

I'm not sure but I think it means how many E's,T's,and S's are in "This sentence contains."
If that is the case it would be 3 E's 3 T's and 3 S's which makes the sum of E's T's and S's 3+3+3=9

The sentence is    "Use numbers, written as words, to make the following sentence true" :pilot:  :pilot:  :first:
Nine E's Five T's Five S's 9+5+5=19

[quote]The sentence is "Use numbers, written as words, to make the following sentence true"
Nine E's Five T's Five S's 9+5+5=19"[/quote]

It says "the following sentence:" as in the sentence that comes after the colon.

for:
This sentence contains  e's,  t's and  s's

it is impossible, or at least the answer of 19 which is shown is incorrect because in the sentence with blanks there are 4 e's, 3 t's, and 7 s's, but if you use "seven" then there are 8 s's, but if you write "eight" then there are 7.

and there are no other numbers with "s" in them all the way up to "sixteen", but since there are 4 e's and 4 t's this is more than 19.

There might be an solution that would make this sentence correct but it might require longer numbers, but to find the answer then would require too much time for a test like this.

Got it: (assuming length does not matter)

[b]T[/b]wo e's
On[b]e[/b] t
Z[b]e[/b]ro s's

UPDATE:
answer is 19, so I guess length DOES matter


In the $ 5\times 5$ grid of dots shown, the dots are equally spaced verticaly and horizontally. Three dots are chosen as the vertices of a triangle. What is the maximum number of dots that could be located on the inside of that triangle?

[asy]dot((0,0));
dot((0,1));
dot((0,2));
dot((0,3));
dot((0,4));
dot((1,0));
dot((1,1));
dot((1,2));
dot((1,3));
dot((1,4));
dot((2,0));
dot((2,1));
dot((2,2));
dot((2,3));
dot((2,4));
dot((3,0));
dot((3,1));
dot((3,2));
dot((3,3));
dot((3,4));
dot((4,0));
dot((4,1));
dot((4,2));
dot((4,3));
dot((4,4));[/asy]

I got $ 6$, but I am not sure if it is correct.

I connected the bottom left dot, the top left dot, and the 2nd to top dot on the right. :blush:  :D

$ 6$ [i]is [/i]correct. I can't really think of a very mathematical way to do it, though.  Just test different (large) triangles, I suppose.

Let the bottom left point be (0,0) and let that be quadrant 1. The triangle obviously be one of the larger ones that have height 5 and base 5. Using the line from (0,0) to (4,0) as the base, there are only 3 triangles that would be like this. One would have the last point on (0,4), which would only have 3 points. The one with the point on (2,4) only has 5. The one with the point on (1,4) has 6 points in it.


function
number theory theorems
number theory

Hi there,
Now i'm preparing to write a technique like you see below.
[i]f[/i][i]([/i]x[i])[/i] is a function... There we have an argument x_(0) for which [i]f[/i][i]([/i]x_(0)[i])[/i] <h, here h is the smallest real number bigger than 0. So x_(0) is the value that [i]f[/i][i]([/i]x_(0)[i])[/i] =0.
In the place once I used to live in Azerbaijanian this technique is used frequently to prove a number is zero or two real numbers are equal to one another, finally, or limit of a function or a sequence is equal to a real in where the argument goes to a real. In here I wonder if the othe places, for example, in Europe, in Iran or in America, is this technique used anymore? and note that i wonder that because I saw this technique only in Azerbaijan and I didn't see this technique in other places, even in Iran and USA when I lived there.

There simply is no "smallest real number bigger than 0". Thus this reasoning is wrong.
If $ f$ is continuous, it can at least be corrected to something true, otherwise not.


calculus
integration
trigonometry
calculus computations

What's the best way to find $\int \frac{dx}{x^{3}\sqrt{x^{2}-25}}$?
Only trig substitution?

That's certainly the easiest to find. There's no reason to avoid it, since you know it will work.


calculus
integration
trigonometry
logarithms
derivative
real analysis
algebra

they are not contest problems, so I put them here  :D

the problems are for those calculus beginners

(1) Compute: [tex]\int \frac{\sin^2 x}{\cos^3 x}dx[/tex]
(2) Compute: [tex]\int e^{\sin x}\frac{x\cos^3 x - \sin x}{\cos^2 x}dx[/tex]
(3) Compute: [tex]\int x\ln\frac{1+x}{1-x}dx[/tex]
(4) $f(x)$ is continuous, show that [tex]\int_0^x f(u)(x-u)du = \int_0^x\left\{\int_0^u f(x)dx\right\}du[/tex]

[tex]\int{\frac{\sin^2{x}}{\cos^3{x}}}dx[/tex]

Since [tex]\sin^2{x}=1-\cos^2{x}[/tex], let u=cos(x), and you have:

[tex]\int{\frac{1-u^2}{u^3}}[/tex]

Now, use partial fraction decomposition:

[tex]\frac{1-u^2}{u^3}=\frac{A}{u}+\frac{B}{u^2}+\frac{C}{u^3}[/tex]

It follows that A=-1, B=0, and C=1.

Now you have:

[tex]\int{\frac{-1}{u}}+\int{\frac{1}{u^3}}=-\ln{|u|}+\frac{1}{-2*u^2}=-\ln{|\cos{x}|}+\frac{-1}{2*\cos^2{x}}+C[/tex]

(?)

A thought to go along with #4: Whenever you see an iterated (double) integral, or  double sum, or the integral of a sum or the sum of integrals, expect that you are going to change the order of integration or summation. The only question is the tactical question of when would be the best time to do the interchange. Interchanging the order of integration or summation is one of the most powerful problem solving ideas I know.

A very large fraction of what is taught at higher levels as "Real Analysis" are theorems that justify such interchanges, provided appropriate conditions are met. (Key words: Uniform convergence; Fubini's Theorem, Tonelli's Theorem, Dominated Convergence Theorem, Monotone Convergence Theorem.)

But there's a warning about #4 - its notation is really improper. With $x$ a variable that has meaning outside all of the integrals, then there should not be any integral that is $dx$. Try using a new "dummy" variable and rewriting the right hand side as
$\displaystyle \int_0^x\left\{\int_0^u f(t)dt\right\}du$.
In this particular case, note that the set in the $(t,u)$ plane over which this iterated integral is taken is a triangle. Sketching that triangle may help you determine what the limits of integration should be after you interchange the order of integration.

[quote="liyi"]Try to differential your answer, is it right?

You'd better develop a habit to remember $dx, ...$ etc.
you set $u=\cos x$. In the following steps, you assume that $du=dx$, which is wrong.[/quote]

SHOOT!

That was really stupid of me.  Now I can't figure it out though...

To Kent Merryfield:

We don't need to exchange the order of integration. The proof of the theorem is not for a calculus beginner.
Think about the primitive of $f(x)$, $F(x) = \int_0^x f(x) dx$ and the derivative of $xF(x)$  ;)

Liyi can you give me a hint on the first one?

Julie: rewrite #1 in terms of tan and sec. At that point it becomes something which, while difficult, is usually done as an example in calculus textbooks. In particular, [hide](Trig identity, integration by parts, solve linear equation for the unknown integral.)[/hide] And I would never, ever, put #2 on an exam for a course.

[tex]\int{\frac{\sin^2{x}}{\cos^3{x}}}=\int{tan^2{x}*sec{x}}[/tex]

Starting with the identity [tex]\sin^2{x}+\cos^2{x}=1[/tex], divide everything by [tex]\cos^2{x}[/tex] to get [tex]\tan^2{x}+1=\sec^2{x}[/tex], and thus [tex]\tan^2{x}=\sec^2{x}-1[/tex].  Substitute this in to the integral:

[tex]\int{(\sec^2{x}-1)(\sec{x})}=\int{\sec^3{x}-\sec{x}}=\int{\sec^3{x}-\int{\sec{x}}[/tex]

From memorization you know that [tex]\int{\sec{x}}=\ln{(\sec{x}+\tan{x})}+C[/tex].

For [tex]\int{\sec^3{x}dx}[/tex], use integration by parts:

Let [tex]u=\sec^2{x}[/tex], [tex]u'=\frac{2*\sin{x}}{\cos^3{x}}[/tex].

Let [tex]v'=\sec{x}[/tex], so that [tex]v=\int{\sec{x}}=\ln{(\sec{x}+\tan{x})}[/tex].

Since: [tex]\int{uv'}=u*v-\int{u'v}[/tex]

[tex]\int{\sec^2{x}*\sec{x}}=\sec^2{x}*\ln{(\sec{x}+\tan{x})}-\int{\frac{2*\sin{x}}{\cos^3{x}}*\ln{(\sec{x}+\tan{x})}}[/tex]

I must have made a wrong turn somewhere...
(please ignore the fact that I forgot to put dx after the integrals, sorry about that)

you didn't finish it? and unfortunately, you made it more complicated...
you get $\int \frac{\sin x}{\cos^3 x}\cdot \ln(\sec x+\tan x) dx$....

you didn't deal with $\int \sec^3 x dx$ well...
think about $\sec^2 x dx = d (\tan x)$...  ;)

On #2: Try integration by parts- you might get lucky...

 [hide]e^(sin x)(x-sec x) + C[/hide]


AMC USA(J)MO USAMO AIME MATHCOUNTS Support Harvard

I plan to organize two letters to the Prof. Steven Dunbar, AMC Director, and Prof. Elgin Johnston, CAMC Chair, regarding the non-preferencing of 10th grade and below AMC 12 takers over AMC 10 takers in terms of USAMO qualification.  The fact that the tiebreaking index among those scoring 8 on AIME did not favor AMC 12 takers is an inherent violation of rule #7 on the USAMO webpage and the USAMO teacher's manual.

Before I give more details, I'd like to take a moment to apologize to the AoPS community for the improper manner in which I conducted myself writing the last letter.  I have learned from that experience, and I assure you, it will never happen again.  Lest you be skeptical, may I remind you that I had nothing personally at stake sending the ORKIM letter, for it was a MATHCOUNTS program in far away san diego that had been cancelled whereas now it is my rightful USAMO qualification that has been denied.

Because I believe that the AMC has not adhered to rule #7, I am asking all members of AoPS to join me in peacefully, calmly, and logically protesting this unjust rule violation.  I propose that we send two letters to the aforementioned recipients: one "victims" letter signed by all AoPS affected by this apparent mistreatment of AMC 12 takers, i.e. those non-qualifiers with an 8 on the AIME and higher than 125 on the AMC 12, and one "concerned citizens" letter signed by all other members of AoPS who although not directly affected, believe that the prestige and significance of the AMC contests is such that preserving the integrity of the rules is paramount.

In each letter, we shall simply present our case that we believe the AMC has effectively disenfranchised many hard-working students deserving of USAMO qualification by refusing to strictly adhere to rule #7.  Beyond that, through rule #7 the AMC has also misled many students in 10th grade or below that taking the "more mathematically comprehensive" AMC 12 contests is beneficial when, to the contrary, it is detrimental.  For these reasons, we feel that the AMC ought to reconsider the non-qualification of the erroneously counseled individuals.  Probability1.01 and I will be writing the letters although we would appreciate the help from any AoPSer willing to assist us.

All we ask of you, however, is simply to agree to attach your name to the correct letter.  If you are willing to do so, please post in this thread and PM bubala telling me your full name and which letter you wish to sign.  Thank you very much, all of you.  I, as well as others affected by this sudden manuever, appreciate your support.

Though I see the validity of some of the many important decisions made by the AMC committee, and realize that one year's USAMO isn't likely to make or break a person's life  I am in full accordance with the rather eloquently presented thoughts above. Even if for logistical reasons the changes don't occur this year, I still feel as though this is an important issue to adress in a constructive manner, and changes should certainly be made in the selection process in future years, and hopefully this one as well.

I couldn't say it any better -- totally agree, count me in.

Of course not.  A similar situation arose in 2002 when the AMC accepted extra qualifiers because a similarly clear violation of the rules was demonstrated.

Indeed, while I appreciate what the AMC has done and realize that one year's USAMO isn't likely to make or break a person's life (contrary to how it appears on this forum), I also believe that the tiebreak system used this year was not a fair one. Thus, at the very least, I hope that we can convince the AMC to reconsider the current system for next year, if not make any adjustments to USAMO qualifications this year. I encourage everyone to support this, because I feel that it is one way to address this issue in a constructive manner (i.e. not just saying "This is so unfair! Let me make some more posts complaining that this is unfair! Let me reiterate my scores for the kth time and induct on k.").

Bubala, I agree that it is unfair that you are being punished because you took the much harder AMC12. As a sophomore, I too was encouraged to take the harder of the two, and  concur that it is a blatant violation of rule 7 on the AMC website.

[quote="Xantos C. Guin"][quote="filletwho"]Does that mean they disqualify the people who already qualified through AMC 10?[/quote]
No, just those who barely squeezed in due to the floor that wouldn't be there if they gave a tie-breaking advantage to the AMC-12 takers.[/quote]

No-- As I tried to make clear above, we are certainly not out to disqualify anyone already admitted to the USAMO.  If the AMC agreed to act on our appeal and favored AMC 12-takers, around 20 or so extra people would qualify, depending on the degree of the preferencing.  No one, under any circumstances, would lose their spot.

[quote="tcs09"]Apparently, there was no tiebreaking involved becuase we only had a floor value of  8. This floor gave 121 qualifiers, right on target for the 120 described in the USAMO teacher's manual.[/quote]

No, in fact, the floor gave far more than 120 qualifiers, thus they only accepted those 8-on-AIME scorers with USAMO indices higher than 212, weighting the AMC 10 and AMC 12 equally.  That, I believe, is unfair, and moreover, violates specifically cited AMC policy.

Please note: I need all of you in support of this effort to PM me your name and which letter you wish to sign in.  I apologize if you missed that due to the length of the introductory post.

However, I still wish for you to display your support publicly by posting a short message in this thread.

I support your letters and will send you a PM.

Although I agree that the AMC 10 takers were favored over the AMC 12 takers during the 2006 USAMO qualification, and that it would be a good idea to [i]consider[/i] doing something about it, this letter is too harsh, and in essence, unnecessary.

What is all this "blatant violation" and "victim" stuff? I mean, all that happened was that they overlooked one detail.

Furthermore, I think that this letter is completely unnecessary. Doesn't Steve Dunbar regularly view these forums? Won't he already be completely aware of the issue? Don't we trust him to make a wise and sound decision? I certainly do.

I agree to sign the letter as I support the cause, but only with the above comments attached to my name.

PS: I only got a 122 on the AMC 12 so I guess I'll be in the..."concerned members" letter :S.

[quote="yifman12"]Has anybody thought about the choice the AMC had to make? Because it's really a no-win scenario for them. What are they supposed to do, curve the AMC12 by 20 points? There'll be people complaining it's easier to get a 120 on the AMC12 than 140 on the AMC10. There will always be people griping about the choice. I bet this was an extremely tough choice for the AMC to make, and I'd bet that they went over stats and their own rule code many times before reaching the decision.

I understand the argument that students were "deceived" by the rules which stated that taking the AMC12 is advantageous to the student, but that's how things are. If you feel you were "cheated" by this selection, fine. You've got every right to feel cheated and mad, but that's what happens. Stuff happens, and sometimes, you just have to roll with the punches.[/quote]

By rolling with the punches now, not only do we prevent any action from being taken this year, but we also ensure that next year many students can be left in a similar predicament.  By kindly asking the AMC to reconsider this situation, at least for future years if not this year, the AMC may update its protocol for USAMO selection and clarify the procedure to a degree precluding all misunderstanding, or it may simply announce that it retains the right to select USAMO participants according to a small set of general guidelines and make controversial judgment calls every year.  However, confusion due to the violation of a specifically cited rule will never again occur with either scenario.

[quote="zanttrang"]

I think your language does come across as too strong - that might be why my reaction is so strong.  I agree that there can be disputes over whether the selection process was the best way possible.  However, I don't think that this makes the selection process in violation of any binding "rules."  I think the rules are actually just guidelines helped to inform students about the selection process, and can be violated.  In this case, I think you're right that the rules have misled at least a few younger students to take a harder test, resulting in no qualification, when they could have qualified using the easier test.  But again, I don't think that should be solved by trying to strong-arm the AMC director about what happened.  I think a way that would give a much better reception would be to suggest that a different method be used for future qualifications, and that the AMC look into possibly admitting more students this year if they concur with the assessment that the qualification system used was badly flawed.[/quote]

I responded to most of this in an above post, but I want to take another moment to apologize for the strong language in which the initial post was written, as it seems it has detracted from my purpose.  I am by no means attempting to strong-arm the AMC director who runs the most amazing math competition in the United States.  Instead, I simply ask that he reconsider the process by which this year's selection was done and, if not that, then at the very least, clarify the process for future years.  Please see my above post to fully understand my disagreement with your statements.

[quote="filletwho"]Eh? you guys keep talking about rule 7...but what about rule 2? I hate it when my posts get lost in all this...

Anyway, I'm just curious how this is affected by rule 2...

[quote="filletwho"]I make no claims about the fairness of the process and will not dispute about that point. I'd just like to point out that there's somewhat of a flaw in your claims about a rule violation. It states specifically in rule 2 that "Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10."

Obviously, selection to USAMO is subjective and varies year to year. It is not binding and if it were so, no one who took the 12 would be eligible for the floor based on rule 4: "The second selection of USAMO participants will be from the highest USAMO indices among students who took the AMC 10A or AMC 10B and the AIME, and got an AIME score at least as high as the floor value."

Of course, they are also violating rule 1, "The goal is to select about 250 of the top scorers"

 :P[/quote][/quote]

Sorry I have not responded to you until now.  The rules on the website are outdated--that is why they seem misleading and certainly not binding.  However, the rules in the teacher's manual, in order to preserve the AMC integrity and credibility, ought to be binding, and are quite accurate.  Feel free to refer to rule #7 or its equivalent in the teacher's manual, and you will still see the apparent violation whereas you will see that rules#1, #2, or #4 have been updated to include accurate information.

Hm, all right, wrong choice of words, I didn't really get the message I meant across. Rereading, it's not exactly what I meant to say, so here's another shot.

I think this all sort of stemmed from a misunderstanding. The first post suggesting a letter was a bit... overzealous, I think. I know it's been adressed many times already, so there's no need to go there again. According to rule 7,

[quote]The selection process is designed to favor students who take the more mathematically comprehensive AMC12A and AMC12B contests[/quote]

How much favoring is favoring? 5 points? 10 points? 20 points? All AMC12 takers get an automatic 150 in comparison with AMC10 takers? If all we're doing is asking for a clarification from the AMC on this rule, I think it's a pretty good idea. But, what seems like a good handicap one year may not work for other years. There will be years where the difficulties and scores are closer, and others where the difficulties and scores are further apart. It's perfectly fine to ask for an explanation from the AMC, but a straight-forward policy regarding the difficulties of the 2 contests may not be the right route here. I believe the AMC should examine the tests in these cases and decide on how much one test should be favored over the other. That being said, I would personally be interested if Mr. Dunbar gave a more detailed explanation of the the selection process.

Obviously, the AMC can claim as much flexibility as it wants to, yifman, but rule 7 implies that there must be some degree of favoritism.  5 points?  10 points?  15 points?  I don't know.  That's the AMC's choice.  In fact, the AMC need not even use a handicap at all, but rather alter the process in some other way as to favor the "more mathematically comprehensive" AMC 12 contests.  However, no favoritism at all, as invoked in this year's USAMO selection, does indeed violate rule 7.  That is the extent of the my empirically confirmed claim.

The AMC should have as much flexibility as it needs/wants/feels like, but it should be consistent with its own rules.  Even if Professor Dunbar refuses to clarify the rules, we will at least know that the process is, to some degree, at the whim of the AMC, rather than being under the impression that the AMC chooses people according to certain rules and unhappily surprised.

Oh and I agree with you, themonster (next post).  The difference here, is that not only do some people disagree with the selection process, it was also in violation of explicitly stated protocol.  One could not make the same claim if the AMC chose to make, say, a 5-point handicap.

[quote="filletwho"]Hmmm...rule 2 is the same in the teacher manuel.[/quote]

Upon re-checking, indeed that is true; however, that does not affect the fact that the AMC's non-preferencing violated rule 7. Treating AMC 12 and AMC 10 equally is a clear, on-face violation, and is just justified by the AMC's consistency with rule 2.

[quote="zanttrang"][quote="bubala"]
I responded to most of this in an above post, but I want to take another moment to apologize for the strong language in which the initial post was written, as it seems it has detracted from my purpose.  I am by no means attempting to strong-arm the AMC director who runs the most amazing math competition in the United States.  Instead, I simply ask that he reconsider the process by which this year's selection was done and, if not that, then at the very least, clarify the process for future years.  Please see my above post to fully understand my disagreement with your statements.[/quote]
I think we can agree to respectfully disagree.  I wouldn't have gone about this the same way you've chosen to, but I thank you for explaining your intentions to me so well and having patience with my retorts.  :)[/quote]

I also consent to respectfully disagree, and I appreciate the constructive criticism.  While I will continue to pursue this goal, unfortunately without your support, I assure you that you have helped me retool my intentions and to that extent I think you for your support for the project in that alternate sense.

[quote="themonster"][quote="pieterminate"][quote="themonster"]Although I still overwhelmingly agree with bubala's argument, I think yifman brings up an important point.  We all agree that the AMC12 is significantly more difficult than the AMC10.  However, I doubt there would be clear consensus about a quantitative measure of the difference in difficulties between the two exams.  Any given point adjustment to allow a better comparison between AMC10 and AMC12 scores would be be unattractive to at least a few people.[/quote]

So use rule 6 from usamo (i know it doesn't apply, but it would work).[/quote]

I haven't actually looked at the statistics (I don't even know if the applicable ones would be available), but I doubt this would be fair.  I would suspect that the average quality of an AMC 12 taker in 10th grade or below is significantly greater than the average quality of an AMC 10 taker.  Thus a proportion system like the one used to compare A and B exams would not really be appropriate.[/quote]

Note: there are obviously a lot more 10ers qualifying currently than 12ers.

If you're reading this post, I'm sure you realize that this topic is now locked.

While I agree that the selection process for sophomores and below was not fair to AMC 12 takers, I hardly see this as reason to start more USAMO-qualification related stress for the CAMC. As far as I'm concerned, the AMC series is basically free to almost everyone who takes it. I mean sure, we pay a few dollars for the tests and a few more for shipping, but how on earth does that compare to the thousands of hours that the CAMC puts into producing tests every year? The least we can do is be grateful and respect the sovereignty of the CAMC on USAMO selection. From threads above, I'm sure you've all noticed that there has already been dissatisfaction at UNL over the role that AoPS has played in making the selection process more difficult. Although I certainly don't claim to speak on behalf of this site's administration, the last thing AoPS needs now is to host a discussion about more ways to aggravate an already stressed out team.

If you'd like to complain, fine. But organize it somewhere else. Do I think it's unfair? Yes. Do I think it's worth complaining about? Yes. But all in all, Prof. Dunbar and the AMC office has been most generous to this site, and what I've read so far on this topic is a poor way to repay them. We've seen some significant changes in the USAMO selection this year; I for one think that the committee should be extolled for the job they've done in creating a reasonable selection process.

Again, I don't claim to speak for any of the administration. If you disagree with me and would like to unlock this thread and continue the Declaration of Dissatisfaction, then please (and by all means), ask one of the Admin to allow this thread to continue. I, of course, will yield to their final judgment on this matter. But should they agree with me and keep this discussion closed, I ask all of you to cooperate and not make this into a bigger mess than it already is.

I've had a request to unlock this forum.  I'm not going to do that, and here's why, and here's what I think you should do (which is only my advice -- you all of course are free to do what you like).

I think that writing a letter to the AMC expressing your concerns with the USAMO selection process is an excellent idea -- rather than just complaining about the perceived "unfairness" of the process on this forum and elsewhere, you're attemping to make a change for the better.  I don't consider this as an "attack" on the AMC -- any organization that deals with the public should welcome reasoned, well-thought, [i]constructive[/i] criticism of its behavior.

However, you are asking people to sign a letter sight-unseen.  I personally would never do that.  My suggestion: write the letter first, post it as an "open letter" to the AMC, then get signatures.

This thread seems to have devolved into micro-managing about what goes in this letter and/or the same general griping that has persisted in many other threads.  That's why the thread is staying locked.  If you'd like to start a new thread that channels [b]constructive[/b] criticism of the AMC and the USAMO selection process, including perhaps a letter-writing campaign, please feel free to do so.  Again, I strongly recommend that you write a letter first and show it to everybody before asking people to sign on.  If it devolves into bashing the AMC or complaining about how the process was "unfair" in a non-constructive manner, then I'll probably lock it again.  If you want to blow off steam, do it in your blog.  If you want to do something constructive, feel free to use the forums.

I think the dissatifaction that the AMC/UNL expressed over what was happening over the last couple days (prior to the USAMO announcement) was legitimate, but it is separate from this issue.

While I was writing this, the AMC Director made a post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=467557#p467557]here[/url].  I suggest that you read this post and consider it before proceeding further.


geometry
trapezoid
geometric transformation
homothety
ratio
trigonometry
analytic geometry

We have $2$ circles which intersect at $A,B$. Let $d$ be a variable line through $A$ and $M,N$ be the points where $d$ intersects again the circles. Find the locus of the midpoint of $MN$.

Nice problem. I will rewrite it with some more notations:

[color=blue][b]Problem.[/b] Let o and p be two circles, and let A be one of the points of intersection of these two circles. A variable line d through the point A meets the circles o and p at the points M and N, respectively (apart from A). Let W be the midpoint of the segment MN. Identify the locus of the point W while the line d varies.[/color]

[i]Solution.[/i] Let O and P be the centers of the circles o and p, respectively.

Since the segment AM is a chord in the circle o, its perpendicular bisector must pass through the center O of the circle o. Thus, $OR\perp AM$, where R is the midpoint of the segment AM. In other words, $OR\perp d$. Similarly, $PS\perp d$, where S is the midpoint of the segment AN. Now, denote by V the midpoint of the segment OP, and by T the midpoint of the segment RS. Since $OR\perp d$ and $PS\perp d$, the quadrilateral OPSR is a trapezoid with its bases OR and PS being perpendicular to the line d; hence, the line VT, connecting the midpoints V and T of the legs OP and RS of this trapezoid, must be parallel to its bases, and thus perpendicular to the line d as well. In other words, $VT\perp AW$.

On the other hand, the points R and S are the midpoints of the segments AM and AN, i. e. the images of the points M and N under the homothety with center A and ratio $\frac12$. Thus, the midpoint of the segment RS must be the image of the midpoint of the segment MN under this homothety. Since these midpoints are T and W, respectively, it follows that the point T is the image of the point W under the homothety with center A and ratio $\frac12$, i. e. the midpoint of the segment AW. Thus, $VT\perp AW$ yields that the point V lies on the perpendicular bisector of the segment AW. Consequently, VW = VA.

Now, this yields that the point W lies on the circle with center V and radius VA. This is a fixed circle, and when the line d performs a full revolution around the point A, the point W, lying on this line, must trace the whole circle with center V and radius VA. Thus, the circle with center V and radius VA is the locus of the point W when the line d varies.

  Darij

Very nice "synthetic" solution, Darij. I was sure that you (or someone else) will come up with something like this.

My ugly solution is analytic: (before you posted your solution, I had only partially solved it)

Let $P,Q$ be the centers of the two circles, $P(p,0)$ and $Q(q,0)$. Also, $A(0,a)$. Then, $B(0,-a)$, where $B$ is the second point of intersection of the two circles.

$m = \tan \alpha$ is the slope of $d$. (the case $d$ vertical is easily solved)

$d$: $y=mx+a$;

$\mathcal C_{P}$: $(x-p)^{2}+y^{2}= p^{2}+a^{2}$.

$\mathcal C_{Q}$: $(x-q)^{2}+y^{2}= q^{2}+a^{2}$.

Solving the system we obtain $ x = \frac{(p+q)-2am}{m^{2}+1}$ and $ y = \frac{(p+q)m+a-am^{2}}{m^{2}+1}$. (the coordinates of the midpoint)

Go back to $m = \tan \alpha$: $ x = \frac{p+q}{2}-\left( a \sin 2 \alpha+\frac{p+q}{2}\cdot \cos 2 \alpha \right)$, $ y = \frac{p+q}{2}\cdot \sin 2 \alpha-a \cos 2 \alpha$.

Let $\phi$ s.t. $ \sin \phi = \frac{\frac{p+q}{2}}{\sqrt{\left( \frac{p+q}{2}\right)^{2}+a^{2}}}$ and $ \cos \phi = \frac{a}{\sqrt{\left( \frac{p+q}{2}\right)^{2}+a^{2}}}$.

Then we have $ x = \frac{p+q}{2}-\left( \sqrt{\left( \frac{p+q}{2}\right)^{2}+a^{2}}\right) \cdot \sin \left( \phi+2 \alpha \right) \ \ y =-\left( \sqrt{\left( \frac{p+q}{2}\right)^{2}+a^{2}}\right) \cdot \cos \left( \phi+2 \alpha \right)$.

Thus, the locus is a circle of center $ \left( \frac{p+q}{2},0 \right)$ and radius $ \sqrt{\left( \frac{p+q}{2}\right)^{2}+a^{2}}$. (the vertical case also works)

Darij 's proof is very nice.
 Another way to solve this problem:
 If we denote by $\zeta(W, (O))$ and $\zeta(W, (P))$ the powers of the point $W$ wrt $(O)$ and $(P)$, then $\zeta(W, (O)) = \zeta(W, (P))$, i.e $OW^2 + PW^2 = {R_1}^2+{R_2}^2 = k$.
 Therefore, the locus of $W$ is the circle center $X$ the midpoint of $OP$ with radius $\sqrt{\frac{2k^2 - {OP}^2}{4}}$

[quote="darij grinberg"]Nice problem...

.. In other words, $OR\perp d$. Similarly, ....
  Darij[/quote]

  What happened to the 'other words'? Where are they? In what language?

  Friendly,

  M.T.

[quote="armpist"]  What happened to the 'other words'? Where are they? In what language?[/quote]

Indeed, in the mathematical language. Probably, "in other formulas" would be more precise.

  Darij

A possible generalization:

Let two secant circles in the points $A$ and $B$. Let a fixed point $C\in (AB)$. A variable line $d$ which passes through the point $C$ meets the circles in the points $M$ and $N$, both nearer by the common secant $AB$ or both farer of the same secant $AB$. Ascertain the geometrical locus of the midpoint of the segment $[MN]$.

I don't think the locus is included in a conic-line !. Can someone draw with computer this line ? Thanks.

[quote="Virgil Nicula"]Let two secant circles in the points $A$ and $B$. Let a fixed point $C\in (AB)$. A variable line $d$ which passes through the point $C$ meets the circles in the points $M$ and $N$, both nearer by the common secant $AB$ or both farer of the same secant $AB$. Ascertain the geometrical locus of the midpoint of the segment $[MN]$.

I don't think the locus is included in a conic-line !. Can someone draw with computer this line ? Thanks.[/quote]

It's an unsual locus.

Kostas vittas.


modular arithmetic
number theory proposed
number theory

Find all primes $ p$ such that $ 2^{p}+p^{2}$ is also prime

If $ p=3,$ then $ 2^{p}+p^{2}=17$ is prime. This is the only such instance.

The case of $ p=2$ give the composite number $ 8.$

Every other prime is an odd number not divisible by $ 3.$

If $ n$ is odd, then $ 2^{n}\equiv-1^{n}=-1\pmod{3}.$

If $ n$ is not divisible by $ 3,$ then $ n^{2}\equiv 1\pmod{3}.$

Hence, if $ n$ is an odd number not divisible by $ 3,$ which includes all primes $ >3,$ then $ 2^{n}+n^{2}$ is divisible by $ 3.$


how do you get the picture on the side like the one Mr. Crawford and Mr. Rusczyk?

(My best guess as to what you mean)
Go to "Profile".
Look at the "Avatar Control Panel" at the bottom of the screen.

I wish people would start reading FAQs. For example for this question: http://www.mathlinks.ro/Forum/faq.php#11


Two circles have the same center O. Point X is the midpoint of segment OP. What is the ratio of the area of the circle with radius OX to the area of the circle with radius OP? Express your answer as a common fraction.

[asy]import graph;
draw(Circle((0,0),20));
draw(Circle((0,0),12));
dot((0,0));
draw((0,0)--(20,0));
label("$O$",(0,0),SE);
label("$P$",(20,0),E);
label("$X$",(12,0),SE);[/asy]

Let $ OX\equal{}r$. We then have $ \frac{\pi(r^2)}{\pi(2r)^2}\equal{}\boxed{\frac{1}{4}}$.

Area is a squared quantity. Thus, when any linear quantity of a shape is increased by a factor of $ n$, the respective area is increased by a factor of $ n^2$. A midpoint divides a segment into two equal portions, so each of these resulting portions is $ \frac{1}{2}$ the length of the original segment. In the diagram, we are given that $ X$ is a midpoint, and thus the ratios of the circles that we are seeking is $ \left(\frac{1}{2}\right)^2\equal{}\boxed{\frac{1}{4}}$.

[hide="my carefully explained math problem"]ill answer this in a series of steps.

1.first of all we know that $ X $ is the middle of the radius $ OP $

2.so lets say the radius of $ OX $ is radius $ R $

3.having this radius $ R $ we know that the radius for $ OP $ is $ 2R $

4.now the question asked to find the ratio between the [b][i][u]small circle[/u][/i][/b] and the [b][i][u]big circle[/u][/i][/b] now we just put it into a ration.

5.$ \frac{\pi(r^{2})}{\pi(2r)^{2}} $ now that we have got this far all we need to do is just simplify.

6.[hide="Hopefully this is the correct answer"]1/4[/hide][/hide]

[hide="."]YAY!! you saw this! [size=200][color=#00BF80][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!!!![/u][/i][/b][/color][/size][/hide]


floor function
calculus
integration
induction
algebra unsolved
algebra

Let $f(0, 0) = 5^{2003}, f(0, n) = 0$ for every integer $n \neq 0$ and
\[\begin{array}{c}\ f(m, n) = f(m-1, n) - 2 \cdot \Bigg\lfloor \frac{f(m-1, n)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n-1)}{2}\Bigg\rfloor + \Bigg\lfloor\frac{f(m-1, n+1)}{2}\Bigg\rfloor \end{array}\]
for every natural number $m > 0$ and for every integer $n$.

Prove that there exists a positive integer $M$ such that $f(M, n) = 1$ for all integers $n$ such that $|n| \leq \frac{(5^{2003}-1)}{2}$ and $f(M, n) = 0$ for all integers n such that $|n| > \frac{5^{2003}-1}{2}.$

[quote="orl"]Let $f(0, 0) = 5^{2003}, f(0, n) = 0$ for every integer $n \neq 0$ and $f(m, n) = f(m-1, n) - 2 \cdot \left[ \frac{f(m-1, n)}{2}\right] + \left[\frac{f(m-1, n-1)}{2}\right] + \left[\frac{f(m-1, n+1)}{2}\right]$ for every natural number $m > 0$ and for every integer $n$.

Prove that there exists natural number $M$ such that $f(M, n) = 1$ for all integers $n$ such that $|n| \leq \frac{(5^{2003}-1)}{2}$ and $f(M, n) = 0$ for all integers n such that $|n| > \frac{5^{2003}-1}{2}.$
(Here $[x]$ denotes the integral part of real number $x$).[/quote]
The original version is rather $f: N^*\times Z\rightarrow Z$
Let $f(1, 0) = 5^{2003}, f(1, n) = 0$ for every integer $n \neq 0$ and $f(m, n) = f(m-1, n) - 2 \cdot \left[ \frac{f(m-1, n)}{2}\right] + \left[\frac{f(m-1, n-1)}{2}\right] + \left[\frac{f(m-1, n+1)}{2}\right]$ for every natural number $m > 0$ and for every integer $n$.

Prove that there exists natural number $M$ such that $f(M, n) = 1$ for all integers $n$ such that $|n| \leq \frac{(5^{2003}-1)}{2}$ and $f(M, n) = 0$ for all integers n such that $|n| > \frac{5^{2003}-1}{2}.$
(Here $[x]$ denotes the integral part of real number $x$)
----------------------------
And here is my solution (Its long, so I only show the ideas if the proof are simple)
1)First, we directly deduce the following facts:
\[ f(m,-n)=f(m,n)  \]
\[ f(m,n)=0 \forall |n|\geq m  \]
\[ f(m,n)\in\{0,1\},\textrm{ if }m,n \textrm{ are both either odd or even }  \]
\[ \sum_{n\in Z}f(m,n)=5^{2003}  \]
2)For all $m$, let $h(m)$ be the greatest number such that $f(m,h(m))\neq 0$
Trivially, $h(m+1)-h(m)\in \{0,1\}$ (why?)
And Ofcourse $h(m)< m$
3) [b]proposition[/b]: there is no $m_0\in N^*$ and $|n|\geq h(m_0$:
\[ f(m_0,n)=f(m_0,n+1)=0  \]
[b]Proof[/b] If not, we deduce$f(m,n)=f(m,n+1)=0 \ \forall m\geq m_0$
Since $h(m+1)-h(m)\in \{0,1\}$,$|n|\geq h(m_0$ and $h(1)=0$ , there is $m_1<m_0$: $h(m_1)=n$
Thus, $f(m_1,h(m_1))=f(m_1,n)=0$ contradiction to the denotation of $h(m)$
we can easily to show that $lim (h(m))<+\infty$ or there is $m_3$: $h(m)=N \forall m>m_3$
4)By induction on $i=0,1,... N$ One can infer the following Proposition
$\forall i=0,1,... N$ there is $M_i: f(m,N-i)\in \{0,1\}\forall m\geq M_i$
indeed, suppose the proposition still holds at $i-1$.
If $f(m,N-i+1)=0$ for all $m\geq M_{i-1}$
therefore, by the hypothesis formula, $[\frac{f(m,N-i)}{2}]=0\forall m>M_{i-1}$
hence, $f(m,N-i)\in \{0,1\}\forall m>M_{i-1}$
If there is $M_i$: $f(M_i,N-i+1)=1$, then $f(m,N-i+1)=1\forall m\geq M_i$ ($1\geq f(m, n) \geq f(m-1, n) - 2 \cdot \left[ \frac{f(m-1, n)}{2}\right]=1$)
thus, $[\frac{f(m,N-i)}{2}]=0\forall m>M_{i-1}$
and hence, $f(m,N-i)\in \{0,1\}\forall m>M_{i-1}$
Evidently, $f(m,n)\in\{0,1\}\forall m>M_N, n\geq0$ and also for all $n\in Z$
5) By inducton on $k=1,2,...<5^{2003}$ we can prove the following statement:
"if $0<k<5^{2003}$, then there is no $|n|\leq h(m)$, $m>M_N: f(m,n)=f(m,n+k+1)=0$ and $f(m,n+i)=1 \forall i=1,2,...,k$"

I leave 5) and many above facts for all of you. Nice proof!!

I don't know why: there is no $m_0\in N*$ and $\m_0\\geq h(m_0)$:
      $f(m_0,n)=f(m_0,n+1)=0$

Sorry for reviving this old thread.

Can someone please explain nmtruong1986's solution or post a new one? Thanks! :)

As far as I can tell, this solution is distinct from [b] nmtruong1986[/b]'s solution.

First we will eliminate $f$ by turning the problem into a combinatorics problem.

Consider an infinite row of people standing side by side, and label them with the integers in the obvious manner. Give $c = 5^{2003}$ coins to person $0$ at time $t = 0$. Every second, each person will pass as many coins to his or her neighbors so that each of them receives an equal number of coins. It's easily verified that $f(m, n)$ is the number of coins that person $n$ has at time $m.$ By symmetry, it's clear that $f(m, n) = f(m, -n), \forall m \in \mathbb{N}, n \in \mathbb{Z}.$

We want to show that, at some point in time, all of the students $-\frac{c-1}{2}, - \frac{c-3}{2}, \cdots, \frac{c-1}{2}$ will each have $1$ coin, with everybody else having $0$ coins. Let's start the proof of this.

[b] Lemma. [/b] No student with label $\ge \frac{c+1}{2}$ or $\le -\frac{c+1}{2}$ will ever receive a coin.

[hide = Proof ]

Suppose, for contradiction, that some person with label $\ell \ge \frac{c+1}{2}$ received a coin. WLOG $\ell > 0$, as the other case is analogous. Any coin which arrives to the person labelled $\ell$ must have originated from the person labeled $0$; since it travels only from one person to their neighbor, this coin must have been in the possession of the person labeled $\frac{c+1}{2}$ at some point. We will now show by induction on $i$ that for each $1 \le i \le \frac{c+1}{2}$, there was a point in time where the persons with labels at least $\frac{c+1}{2} - i + 1$ had at least $i$ coins.

We've already shown the base case $i = 1.$ Suppose it holds for $i = k$. By the inductive hypothesis, we may consider the first time $t = z$ where the persons with labels $\frac{c+1}{2}  - k + 1$ and above had at least $k$ coins. We'll show that at time $t = z  - 1$, the persons with labels $\frac{c+1}{2} -  k$ and above had at least $k+1$ coins, which would complete the induction. This is actually quite easy to show; simply note that there must have been at least one coin passed from person $\frac{c+1}{2} - k$ to person $\frac{c+1}{2} - k +1$, and so there also was at least one passed from $\frac{c+1}{2} - k$ to $\frac{c+1}{2} - k - 1.$ If we then let time flow backwards, all of the $k$ coins which were in the hands of the people of labels $\ge \frac{c+1}{2} - k$ at time $t = z$ would be in the hands of people with label at least $\frac{c+1}{2} - k - 1$ at time $t = z-1.$ Additionally, the coin(s) which was(were) passed from $\frac{c+1}{2} - k$ to $\frac{c+1}{2} - k - 1$ would return to the hands of $\frac{c+1}{2} - k$ at time $t = z-1.$ This completes the induction.

Applying the induction for $i = \frac{c+1}{2}$, we know that there was a time where there were at least $\frac{c+1}{2}$ coins in the hands of the people with positive labels. However, this would imply by symmetry that a similar statement holds for people with negative labels, for at least $2 \cdot \frac{c+1}{2} = c+1$ coins in total. There are only $c$ coins, so this is a clear contradiction.

Hence, our assumption was incorrect, and the lemma is shown.

[/hide]

From Lemma, we know that there are only a finite number of possibilities for the distribution of the coins among the people, because all but $c$ people are essentially useless. Hence, there exists a constant $C$ so that for all $t > C$, every person has the same number of coins at time $t$ as at time $t + C.$

Observe that the leftmost person (smallest label) with a coin can only move left as $t$ increases. This is because at any moment, he/she either doesn't pass a coin (so that they remain the leftmost coinholder) or makes the person left of them the leftmost coinholder.

WLOG assume that $C$ is sufficiently large so that the leftmost coinholder does not change. Let the leftmost coinholder have label $-x,$ so that the rightmost coinholder is person $x.$

Consider all transactions made in the period from time $t$ to time $t+C$, for some $t > C.$ Consider the leftmost person to have passed coins during this time. Clearly this person is not $-x$, as then $-(x+1)$ would become the leftmost coinholder, contradiction. Hence, let $y > -x$ be this leftmost person who passes coins, if exists. Since person $y-1$ never passes coins, during the time from $t$ to $t+C$, $y-1$ strictly increases his/her number of coins. However, this is a contradiction to the definition of $C$, since $y-1$ should have the same number of coins at time $t$ as at time $t+C.$ Hence, $y$ actually doesn't exist, and so nobody passes a coin from times $t$ to $t+C.$

From the definition of $C$, we know that nobody passes a coin after time $C.$ Hence, everybody has $0$ or $1$ coin, and so all coinholders have exactly one coin. This means that there are exactly $c$ coinholders at some point. From the lemma, we know that they are precisely $-\frac{c-1}{2}, -\frac{c-3}{2}, \cdots \frac{c-3}{2}, \frac{c-1}{2}.$

The problem is now solved.

$\square$


probability
rotation
geometry
geometric transformation
dilation
calculus
integration

[quote="Wickedestjr"]1/4 + 1/16 + 1/64 + 1/256 +... = x

4x = 1 + 1/4 + 1/16 + 1/64 +...

3x = 1

x = 1/3, so the answer is 1/3.[/quote]
You assumed that the particle has to take the exact same path back to get back to the origin. That's not true.

For example, if the particle moves up-right-up-left, then it can move back just by going down-down, or left-down-up-down-right-down

Then would it be;

1/4 + 1/8 + 1/16... = x
2x = 1/2 + 1/4 + 1/8 + 1/16...

x = 1/2

1/2?

Or maybe, it would just be 1/4, because there are 4 possible points that you can get to from each point.

I'm not sure where you're pulling those numbers from, but I'm pretty sure you're not right.  The probability must be [i]at least[/i] 1/4, since there's a 1/4 chance of returning to the original point after two turns.

This is similar to a problem from WOOT - one case of the problem from WOOT is the same as the one-dimensional case of this.  The probability for the one-dimensional case turns out to be 1, which means that logically the answer to this problem should be 1.  Now to prove it...   :)

Does it involve adding an infinite amount of numbers that form a geometric sequence?

I don't actually know what the probability is for sure, or how to find what it is, I was just giving an educated guess.

However, adding up a bunch of stuff doesn't seem promising since there are so many possibilities here.

So it is guaranteed to return to the origin? That is interesting.

[quote="matt276eagles"]If the particle passes through $ A$ again, then it must first return to $ A$ without passing through the origin, [...][/quote]

Can you explain why this must be? Can't the particle start at $ (1,0)$ and go U-L-D-R, which would result in the particle reaching the origin before returning to $ (1,0)$?

[quote="Nawahd Werdna"]Consider a particle at the origin, and on each turn, it must move one unit up or down, and then one unit left or right. After $ n$ moves, what is the probability that it has passed through the origin again; what about as $ n$ approaches infinity?[/quote]

Is moving up then left considered one move or two?

[quote="jvenezuela716"]

[quote="Nawahd Werdna"]Consider a particle at the origin, and on each turn, it must move one unit up or down, and then one unit left or right. After $ n$ moves, what is the probability that it has passed through the origin again; what about as $ n$ approaches infinity?[/quote]

Is moving up then left considered one move or two?[/quote]
It should be one move.

[quote="jvenezuela716"]Can you explain why this must be? Can't the particle start at $ (1,0)$ and go U-L-D-R, which would result in the particle reaching the origin before returning to $ (1,0)$?[/quote]

We are calculating $ P_{A}$, so we are assuming that the point never returns to the origin.

Also, I now realize that I misinterpreted the problem.  I assumed that on any one turn, the particle chooses U, L, D, or R, each with equal probability.  But the problem that I solved is exactly the same as the problem in this thread: starting with the problem in this thread, rotate the plane by 45 degrees and then dilate the plane by $ 1/\sqrt {2}$, and you get the version that I solved.  :)

So basically, after an infinite amount of moves, it will end up back at the origin, because it will not just avoid the origin.

Do it in $ 3$-space first, I'm sure you'll be interested in the answer :)

[quote="1=2"]I just thought of a problem: do it in $ n$-space.

Would it be the same proof, except replacing $ P_A$ with $ (1,0,0,...,0)$?[/quote]

Hmm...If i remember correctly, the probability in n space if n is at least 3 is not 1.

[quote="nebula42"][quote="matt276eagles"]Therefore, $ P_{A}\ge (1 \minus{} P_{O})P_{A} \plus{} P_{O}$.[/quote]
Shouldn't this be $ P_{A}\le (1 \minus{} P_{O})P_{A} \plus{} P_{O}$? :|[/quote]

Um, yeah, so my proof is actually completely worthless.  :P

Good call.

I think I've read somewhere before that the proof has something to do with or is somewhat similar to finding electrical resistance.

Also, there's http://en.wikipedia.org/wiki/Random_walk#Higher_dimensions

[quote="wikipedia"]
One way to prove this result is using the connection to electrical networks. Take a map of the city and place a one ohm resistor on every block. Now measure the "resistance between a point and infinity". In other words, choose some number R and take all the points in the electrical network with distance bigger than R from our point and wire them together. This is now a finite electrical network and we may measure the resistance from our point to the wired points. Take R to infinity. The limit is called the resistance between a point and infinity. It turns out that the following is true (an elementary proof can be found in the book by Doyle and Snell):

Theorem: a graph is transient if and only if the resistance between a point and infinity is finite. It is not important which point is chosen if the graph is connected.

In other words, in a transient system, one only needs to overcome a finite resistance to get to infinity from any point. In a recurrent system, the resistance from any point to infinity is infinite.
[/quote]

First note that the problem is equivalent to just moving U, D, L, or R on each move. (The particle only reaches points $ (x,y)$ such that $ x \plus{} y$ is even, which is the checkerboard thingy, blablabla you get what I mean.)

Being unable to do any real math myself, due to my own remarkable stupidity, I am forced to reading casual, recreational math books.  :oops:

In Ian Stewart's [i]How to Cut a Cake and Other Mathematical Conundrums[/i], Chapter 2 addendum, it states that two-dimensional random walks on adjacent ($ (x,y\pm1)$ or $ (x\pm1,y)$) lattice points starting at the origin are to reach the origin with probability $ 1$, eventually. However, on a three-dimensional lattice grid, the book gives

$ \frac {1}{[ 3(18 \plus{} 12\sqrt {2} \minus{} 10\sqrt {3} \minus{} 7\sqrt {6} )K(2\sqrt {3} \plus{} \sqrt {6} \minus{} 2\sqrt {2} \minus{} 3) ]^2}$

where $ K(z)$ is "$ 2/\pi$ times the complete elliptic integral of the first kind with modulus $ z^2$" (lol, I don't want to know what that is). It supposedly comes out to about $ 0.34053729551$.

And then it says that on a four-dimensional hypercubic grid the probability was "worked out" by computer to be about $ 0.193201673$ by Yuichi Tanaka.

[quote="matt276eagles"]Oh wow, I just found a nice solution.

[hide="Solution"]We say that the particle "escapes" a point $ X$ if it never passes through $ X$ on or after its first move.  Let $ P_{X}$ be the probability of the particle escaping the origin if it starts at point $ X$.  We want to evaluate $ P_{O}$, the probability escaping the origin if it starts at the origin.

Now let $ A \equal{} (1,0)$, and consider $ P_{A}$.  There are two possibilities here: the particle either passes through $ A$ again, or it doesn't.  If the particle passes through $ A$ again, then it must first return to $ A$ without passing through the origin, which occurs with a probability at most $ 1 \minus{} P_{O}$.  Then the probability of the particle escaping the origin once it passes through $ A$ for the second time is $ P_{A}$.  The probability of the particle not passing through either $ A$ or $ O$ is at most $ P_{O}$.  Therefore, $ P_{A}\ge (1 \minus{} P_{O})P_{A} \plus{} P_{O}$.  Expanding and simplifying, we get $ P_{O}P_{A} \minus{} P_{O}\ge 0$, so $ P_{O}(P_{A} \minus{} 1)\ge 0$.

Now either $ P_{O} \equal{} 0$ or $ P_{A} \equal{} 1$, otherwise the expression is negative.  But $ P_{A}$ can't be 1, since there is at least a 1/4 chance of the particle not escaping the origin if it starts at $ A$.  Therefore, $ P_{O} \equal{} 0$, so a particle starting at the origin eventually returns to the origin with probability 1.[/hide][/quote]

I didn't really read the solution but it looks like the only point you use the fact that this is two-dimensional simple random walk is at the end where you have that $ P_A<1$.  This is true for $ n\minus{}$dimensional simple random walk for $ n\geq 3$ as well, but random walk is transient in those dimensions, so your argument (if I'm reading it right), can't possibly work.  In particular, you need actual quantitative estimates.

ProtestanT mentions electrical resistance as a tool for analyzing random walks, but the main use of this technique is on nonsymmetric lattices, since it's a little more robust there.  On $ \mathbb{Z}^d$ (or some lattice isomorphic to it), usual combinatorial techniques suffice.

So here's an outline of the proof in one and two dimensions.

1)  Let $ P_n(0,0)$ be the probability that a simple random walk starting at $ 0$ returns to $ 0$ after $ n$ steps.  Then the recurrence of the random walk is determined by whether $ S \triangleq \sum_{n\geq 0} P_n(0,0)$ is infinite .

2)  $ P_{2n}(0,0)\equal{}\binom{2n}{n}2^{\minus{}2n}$ is 'about' $ 1/\sqrt{\pi n}$, so $ S\equal{}\infty$ in one dimension.

3)  Up to rotating and scaling the axes, two dimensional simple random walk is the product of two independent one dimensional random walks.

4)  $ P_{2n}(0,0)$ is about $ 1/(\pi n)$, so $ S\equal{}\infty$ in two dimensions.

5)  Simple random walk in two dimensions is recurrent.

n.b.  In $ d\geq 3$ dimensions $ P_{2n}(0,0)$ is about $ C/n^{d/2}$, so $ S<\infty$, but you need to do a little more work to get transience.

Here's a discussion of this problem that took place in the early (pre-LaTeX) days of this site: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=7554


function
advanced fields
advanced fields unsolved

I need to find an exponential generating function on the number of forests on a set with $ x$ elements.

I found a list of the first 200 numbers on MathWorld.  I also found a function there which I'm not sure about.  It says that the generating function is:

$ e^{(\sum_{k>0} B(x^k)/k )}$, where $ B(x) \equal{} x \plus{} x^2 \plus{} x^3 \plus{} 2*x^4 \plus{} 3*x^5 \plus{} 6*x^6 \plus{} 11*x^7 \plus{} ... \equal{} (C(x)\minus{}1)/x$ and $ C$ is the generating function for the number of trees.

I know that $ C(x)$ is $ x^{x\minus{}2}$.

This would make the function: $ e^{(\sum_{k>0} (x^{kx^k\minus{}2}\minus{}1)/kx^k}$.

Is that correct?  I haven't ever seen an exponential of a summation like that.  It seems that the exponential should be inside the sum.

Any insight would be helpful!

You missed something in your calculation:

$ B(x) \equal{} \frac{C(x)\minus{}1}{x}$

$ C(x)\equal{}x^{x\minus{}2}\equal{}\frac{x^x}{x^2}$

so:

$ B(x) \equal{} \frac{\frac{x^x}{x^2}\minus{}1}{x}$

and

$ B(x^k)\equal{}\frac{\frac{(x^k)^{x^k}}{(x^k)^2}\minus{}1}{x^k}\equal{}\frac{\frac{x^{k x^k}}{x^{2k}}\minus{}1}{x^k}\equal{}\frac{x^{k x^k\minus{}2k}\minus{}1}{x^k}$

Thanks for pointing that out.  Do I still have the exponential of a summation though?

Yes, your full function looks like:

$ e^{\sum_{k>0}\frac{x^{k x^k\minus{}2k}\minus{}1}{x^k}}$

The idea here is that of the [url=http://en.wikipedia.org/wiki/Exponential_formula]exponential formula[/url].


geometry
geometric transformation
rotation
geometry unsolved

Given Sector OAB (O is the center, A,B are the end points of arc AB, angle OAB is within range (0, pi) ) covers a convex polygon M.

Prove: There exist another sector O'A'B' congruent to OAB covers M such that one of M's edges is on either O'A', or O'B', or arc A'B'.

This one is pretty tough, I have been working on for one week but still did not get anything? Anyone will to try it? Thanks.

I believe it can be solved similarly to a Chinese TST problem from last year. That problem had a triangle covering the polygon instead of a sector. The basic idea was to start with ABC with points D, E, and F on its sides. Then you can find another smaller triangle A'B'C' which is similar to ABC while still containing D, E, and F. I'm not sure how to rigorously translate that to a sector, but that idea should work.

sectors don't have very nice geometric properties....that limits us on what we can do in terms of shifting, etc, if we have something like OA is parallel to a side of the polygon, i am not sure we can make the leap that we can do a parallel translation...therefore, i suggest we somehow complete the arc...we can also wlog points on the arc/sides...but yea...nothing really significant here...just thinking...

also what does it mean a segment lies on arc AB?

and convex is also a tough condition...i never really see how to "use" that piece of information effectively in these types of problems...

I have thought over using tangent line, but seems not working, because the final polygon M' after shifting or rotating may go out of the sector but still in the triangle.

[quote="probability1.01"][quote="temphan"]Yes, I knew the problem of China TST 2006. However, the situation is different here, so the rigorous proof is truely hard to derive. Especially,
how to prove that we can find a small sector by rotating the edges, arc of sector around the fixed points. I have no idea![/quote]

Perhaps you can consider the triangle such that two of its sides coincide with the sector, and the third side is tangent to the arc at the same point that the polygon meets the arc?[/quote]

ok, everyone, let us check if following method works? I directly copy an solution for triangle cover with little change.  The only problem is: how to prove that after the rotation, the new sector O'A'B' is no large then the original one. Since if it could become small, it definitely could become large if we rotate in the opposite direction:

Move the sides of the  OAB closer to the polygon, keeping them parallel to themselves, until they all touch the border of the polygon.

If one of them intersects the border along a non-degenerate segment, we're done: that side was parallel to a side of the polygon to begin with. Otherwise, the (arc AB), OA,OB meet the border of the polygon \mathcal M in the three points X,Y,Z respectively.

Now start rotating the arc AB, OA,OB at the same time, with equal angular velocities, around X,Y,Z respectively. Eventually, we'll reach a position in which one of the lines will meet the border of the polygon along a non-degenerate segment. The triangle we've obtained now is similar to the initial one, but possibly smaller. Enlarge it (move its sides away from the polygon, keeping them parallel to themselves) until it reaches the initial size. This last triangle has the desired properties.

[quote="probability1.01"]I think you only need to prove that one direction makes it smaller (which should not be hard). The idea is you rotate it until you "hit" an edge. So that gives me an idea of how to finish--suppose three consecutive vertices of the polygon are P, X, Q (using your notation). Let the perpendicular bisectors of PX and QX be $l_{1}$ and $l_{2}$, respectively. At the starting position, O lies "between" $l_{1}$ and $l_{2}$. So we simply perform the spiral similarity until either one of the sides of M coincides with one of the sides of the sector or O hits one of $l_{1}$ or $l_{2}$.[/quote]

I have been trying to prove this fact for several days, but did not make any progress. I guess maybe I do not have enough knowledge. Anyway, I will keep trying. Could anyone else also take a look at this particular problem? Any comments or suggestions would be more than welcome. Thanks.


For $a,b,c$ positive reals prove that

$\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1} \leq 1$

You will not believe it, but after computations this reduces to $ (abc-1)^2$ is at least 0.


prove that if p is prime and p|ab then p|a or p|b

Let $ p$ not divide $ a$.then $ gcd(a,p)\equal{}1$.Hence, we can write $ 1\equal{}ax\plus{}py$ where $ gcd(x,y)\equal{}1$.So,$ b\equal{}b.1\equal{}b(ax\plus{}py)\equal{}abx\plus{}bpy$. Now,$ p|bpy$ and $ p|(ab)x$. So,$ p|abx\plus{}bpy\Rightarrow p|b$


floor function
number theory proposed
number theory

Let $s(n)$ be the sum of all positive divisors of $n$, so $s(6) = 12$. We say $n$ is almost perfect if $s(n) = 2n - 1$. Let $\mod(n, k)$ denote the residue of $n$ modulo $k$ (in other words, the remainder of dividing $n$ by $k$). Put $t(n) = \mod(n, 1) + \mod(n, 2) + \cdots + \mod(n, n)$.

Show that $n$ is almost perfect if and only if $t(n) = t(n-1)$.

I use the fact that $\mbox{mod } ( n, j ) = n \mbox{ mod } j = n - j \cdot  \lfloor \frac{n}{j}  \rfloor$

The statement can be restated as
\[ \sigma ( n ) = 2 n - 1 \Longleftrightarrow \sum_{j = 1}^n n \mbox{ mod } j = \sum^{n - 1}_{j = 1} ( n - 1 ) \mbox{ mod } j  \]
We work by equivalence
\begin{eqnarray*} \sum_{j = 1}^n n \mbox{ mod } j & = & \sum^{n - 1}_{j = 1}  ( n - 1  ) \mbox{ mod } j\\  \sum_{j = 1}^n ( n - j \cdot  \lfloor \frac{n}{j}  \rfloor  ) & = & \sum^{n - 1}_{j = 1}  ( n - 1 - j \cdot  \lfloor \frac{n - 1}{j}  \rfloor )\\ n - \sum_{j = 1}^n j \cdot  \lfloor \frac{n}{j}  \rfloor & = & - ( n - 1 ) - \sum_{j = 1}^{n - 1} j \cdot  \lfloor \frac{n - 1}{j}  \rfloor\\ 2 n - 1 & = & \sum_{j = 1}^n j \cdot  \lfloor \frac{n}{j}  \rfloor - \sum_{j = 1}^{n - 1} j \cdot  \lfloor \frac{n - 1}{j}  \rfloor\\ 2 n - 1 & = & n + \sum^{n - 1}_{j = 1} j \cdot (  \lfloor \frac{n}{j}  \rfloor -  \lfloor \frac{n - 1}{j}  \rfloor ) \end{eqnarray*}
Remark that $\delta ( j | n ) =  \lfloor \frac{n}{j}  \rfloor -  \lfloor \frac{n - 1}{j}  \rfloor = \{\begin{array}{l} 1, j|n\\ 0, \mbox{otherwise} \end{array}$

Hence $n + \sum^{n - 1}_{j = 1} j \cdot (  \lfloor \frac{n}{j}  \rfloor -  \lfloor \frac{n - 1}{j}  \rfloor ) = n + \sum^{n - 1}_{j = 1} j \cdot \delta ( j|n ) = \sum_{j|n} j = \sigma ( n )$.

This finish the proof.

Because $ (n, n\minus{}1)\equal{}1$ each number $ \le n\minus{}1$ satisfies one of the following :
1) it divides $ n$, but not $ n\minus{}1$. Let's denote these numbers by $ b_k$, so $ mod(n, b_k)\equal{}0$ and $ mod(n\minus{}1, b_k)\equal{}b_k\minus{}1$.
2). it divides $ n\minus{}1$, but not $ n$. Denote these numbers by $ a_k$; $ mod(n\minus{}1, a_k)\equal{}0$, $ mod(n, a_k)\equal{}1$.
3).  it neither divides $ n\minus{}1$, nor $ n$. If these numbers are denoted $ p_k$ and the remainder when divided by $ n\minus{}1$ $ c_k$ then $ mod(n\minus{}1, p_k)\equal{}c_k$, $ mod(n, p_k)\equal{}c_k\plus{}1$.

$ t(n)\minus{}t(n\minus{}1)\equal{}(1\plus{}1\plus{}...\plus{}1) \plus{} \sum (c_k\plus{}1) \minus{} \sum (b_k\minus{}1) \minus{} \sum c_k$, where in the first paranthesis, 1 is added for every divisor of $ n\minus{}1$. Thus we have that $ t(n)\minus{}t(n\minus{}1)\equal{}2n\minus{}1\minus{}\sum b_k$ (after a little computation). Note that $ b_k$ are divisors of $ n$ so the conclusion immediately followes.

[quote=broniran][quote="{x}"]I use the fact that $ \mbox{mod } ( n, j ) \equal{} n \mbox{ mod } j \equal{} n \minus{} j \cdot \left \lfloor \frac {n}{j} \right \rfloor$
[/quote] Can you explain what do you mean by this?[/quote]

The quotient of a division of $n$ by $j$ is equal to $\left \lfloor \frac{n}{j} \right \rfloor$ (see some book or article about Euclidean Algorithm). Then the rest will be $n-j\cdot \left \lfloor \frac{n}{j} \right \rfloor$