geometry
3D geometry
tetrahedron
symmetry

Hi.I'm a new guy here.

Find the electric field at the center of a charged hollow tetrahedron with side lenght a and surface charge density σ .

If I've got the question right, we want the field at the center of a regular tetrahedron.
By symmetry, this is obviosly zero.

I'm sorry.The problem is:

Find the electric field at the center of the hollow tetrahedron whose [b]3 sides are charged with surface density σ, the other side side charged with 2σ?[/b]
[/b]

By superposition principle the problem boils down to calculating the field due to a single triangular sheet of charge density $ \sigma$ at the centre of the tetrahedron.

I do not know how we can calculate that without brute-force integration. A [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186306]similar problem[/url] was posted earlier.


Does anyone know how to do this? I want to rename the appendix sections to Appendix A, Appendix B,... instead of just A,B.. I also want this change to be visible in the table of contents.

I have found code that does this (below, courtesy of http://www.cse.iitd.ernet.in/~anup/homepage/UNIX/latex.html#newappendix ), but using this, the appendix' section titles gets mixed up with the section header..

\appendix
\renewcommand\thesection{Appendix \Alph{section}}
\makeatletter
\def\@seccntformat#1{Appendix\ \csname the#1\endcsname\quad}
\makeatother

Please help!!

Sorry. That was actually two ways..

\appendix
\renewcommand\thesection{Appendix \Alph{section}}

is the one that mixes the section name with Appendix....

I don't know exactly what is the problem. I just compiled a document I did, and if I write

[code]
\appendix
\section{Results of Simulation}
     .
     .
     .
Bla bla bla
     .
     .
     .
\section{Proof of Estimation Theorems}
     .
     .
     .
Bla bla bla
     .
     .
     .
[/code]

Those last two sections are numerated

Appendix A: Results of Simulation
Appendix B: Proof of Estimation Theorems!

I don't need to rename any command!

On my computer, it doesn't display "Appendix" neither.  I built this litle solution :

\newcommand{\annexe}[1]{%
\addtocounter{section}{1}
\addcontentsline{toc}{section}{Appendix \thesection \quad #1}
\section*{Appendix \thesection \quad #1}}

but the thing is that you have to use the command [b]\annexe[/b] instead of [b]\section[/b] (annexe is french for appendix).  I tried to use [b]\renewcommand{\section}...[/b] with the same definition then above but I had an error "TeX capacity exceeded, sorry..." and I don't know what it means and what to do about it.  If anyone can fill me up about that...

So if you didn't wrote everything, it ain't that bad but if you have 12 appendix, it's long to chnage it all but that's all I can do for the moment.

ledurt

p.s.: I didn't test the behavior of the [b]\ref[/b] command with my command [b]\annexe[/b]

Thank you Ledurt!!! It works beatifully!!

[quote="smilet"]Maybe it depends on the type of report...

I have:
\documentclass[a4paper, 12pt, titlepage]{article}

If I do like you did, with

\appendix
\section{any title}

I just get (in the toc and in the report) A any title..[/quote]

Also, you if it is an article, you may try to use also amsart (AMS Article), it looks nice, and it name things with the Appendix A as you want!, and I like how it looks.

Best,

Ok, thanks, I will try that in the next one (starting today...)

What packages do you need for that? (Apart from the usual ones..)
Is it enough with

\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}

or am I overdoing it?

Thanks both of you for your help!!

amsart automatically imports many of the ams packages, including amsmath and amsthm, not sure about amssymb

This is what I use as the start of my header.

[code]\documentclass[11pt,reqno]{amsart}
\usepackage[english]{babel}
\usepackage{amssymb, bezier, latexsym, amsfonts}
[/code]


In a plane there are an even number of points no three of which are on a straight line. Half of the points is red, the other half is blue. Prove that there is a line which connects one red point and one blue point so that in each of the half-planes bounded by the line the number of red points is equal to the number of blue points.


69
 in line with italian previous results...

 69/6=11+ (as fph said)

 marco g. 11 -> honorable mention (1st exercise solved, with inversion)
 luca barbieri 20 -> bronze
 rosati 19 -> bronze
 the other 11, 6 , 6

 bye, ciao brend

Hi, fra_sca...
methinks there should be an error: 20+19+11+11+6+6=73,
though we obtained only 69 pts.. maybe Rosati's done 15 points? itamiiru ashikarazu ai

sorry, the problem is that i've done only 2 points.... :blush:

sorry, when marco told me 20 19 11 2 6, i was thinking not 2 AND 6 but 6*2..
never mind simone, next year will be better!
-f-


How many integer solutions for $0<x<2^{13}$ satisfy the equation below?
\[ x^2\equiv 5 \text{ (mod } 2^{13}-1)  \]
Assume that $2^{13}-1$ is prime.

Masoud Zargar

Maybe I'm misremembering how "mod" works, but isn't 5 (mod 213-1) always equal to 5? Did you perhaps mistype the equation?

Using the mod operator, yes that's what it is.  But here, the mod means mod for both sides of the equation.  So it's basically saying $x^2 \text{ (mod } 2^{13}-1) = 5 \text{ (mod } 2^{13}-1)$.  That's determined by using the $\equiv$ sign instead of the $=$.


[quote="joml88"]As some of you may know, my blog has been the SHS Math Team Blog for a while now.  However, it's growing a little inactive as there's not much notable happening on our team currently.  So I'm thinking maybe changing it to the SC All State Math Team blog and letting all the members of the team who are on the forum be contributors.  Like the idea?[/quote]

Why change the SHS blog, that's for the Summerville Dynasty. How bout we get one of the coaches...(Mr. R or Mr. Neal) to make a blog (since they don't have one) and then do that idea.

[quote="mysmartmouth"]I'll be a contributor.  Why not get chinagrl to make a blog?[/quote]

Because she's from NC and we're talking about SC... I think the coaches making one is the best idea

OK, someone contact Mr. R then.  This idea is dying.  Dying!

Yes!  The SC blog is alive!  Click Mr. R's username for details...


[quote="mdk"]20 pairwise distinct positive integers are all < 70. Prove that among their pairwise differences, there are four equal numbers.[/quote]
This is how I understand the problem.
Let, $ a_{1},a_{2},...,a_{20}$ be distinct positive integers. With $ 1\leq a_{i}\leq 69$ so, their differences are $ 1\leq |a_{i}-a_{j}|\leq 68$. Now there are $ {{20}\choose 2}= 190$ possible pairs. By the strong form of Pigeonhole we have $ [ 190/68]+1=3$ are the same. How do you get 4?

[quote="sylow_theory"]By the strong form of Pigeonhole we have $ [ 190/68]+1=3$ are the same. How do you get 4?[/quote]

You don't have $ 68$ different differences! Think about it. You have $ 20$ dinstinct numbers $ a_{1},\cdots,a_{20}$ satisfying the property $ 1\leq |a_{i}-a_{j}| \leq 68, \ \ i\neq j$ and $ 1\leq i,j\leq 20$ . Is it possible to have $ 68$ different differences? This is the main point of this problem! Count the upper bound of those different differences!

Alexandros


[quote="bryantacademy"]:oops:
How do you calculate all that base 6, 12, 10 stuff?[/quote] You can expand any number in the following form: Consider 347 for example. By 347 we mean $347_{10} = a_2\cdot 10^2 + a_1\cdot 10^1 + a_0\cdot 10^0$

In this case, $ a_2 = 3\; a_1 = 4 \; a_0 = 7$

Now let's evaluate the same number base-6,

$347_{10} = b_3\cdot 6^3 + b_2 \cdot 6^2 + b_1 \cdot 6^1 + b_0\cdot 6^0$

$\frac{347}{6^3} = 1\; R\; 131 \Rightarrow b_3 = 1$

$\frac{131}{6^2} = 3\; R\; 23 \Rightarrow b_2 = 3$

$\frac{23}{6^1} = 3\; R\; 5 \Rightarrow b_1 = 3$

$\frac{5}{6^0} = 5\; R\; 0 \Rightarrow b_0 = 5$

$\therefore \boxed {347_{10} = 1335_6}$


Consider a set $ P$ of parallel lines, and another set $ Q$ consisting of parallel lines which are not parallel to those in $ P$.  Define the set of points $ X \equal{} P \cap Q$.
Let $ n \in \mathbb{N}$.  Color each point in $ X$ one of $ n$ colors.  (Each color may be used once, more than once, or not at all.)
It is known that for any such coloring of the points in $ X$, there exist two lines in $ P$ and two lines in $ Q$ such that their four points of intersection are all of the same color.
Given that $ |P| \equal{} n \plus{} 1$, show that the minimum possible value of $ |Q|$ is $ \frac {n^3 \plus{} n^2}{2} \plus{} 1$.

[hide]Let $ |Q|\equal{}m$.  Label the lines of $ P$, $ 1,2,...,n\plus{}1$ and the lines of $ Q$, $ 1,2,...,m$.  We thus label each point as an ordered pair $ (p,q)$ which is the intersection of $ p$ which is a line from $ P$ and $ q$ which is a line from $ Q$.  Thus, we are looking for $ p_1,p_2,q_1,q_2$ such that $ (p_1,q_1),(p_1,q_2),(p_2,q_1),(p_2,q_2)$ are all colored the same color.

First, we prove that $ m\equal{}\frac{n^3\plus{}n^2}{2}\plus{}1$ is sufficient.  In each line of $ Q$, there are $ n\plus{}1$ intersection points.  Thus, there are at least 2 points, $ (p_i,q_k),(p_j,q_k)$ of the same color.  There are $ \binom{n\plus{}1}{2}$ possible choices for $ i,j$.  Thus, there will be $ \lceil\frac{q}{\binom{n\plus{}1}{2}}\rceil\equal{}n\plus{}1$ lines in $ q$ for which the choice of $ i,j$ is the same.  Now consider the common color of the two points.  Since there are $ n$ colors, there exist 2 of these lines in $ Q$ for which the common color is the same.  Thus, we have our 4 points.

It is easy to show that $ \frac{n^3\plus{}n^2}{2}\plus{}1$ is necessary by creating a configuration where lines $ q_1$ through $ q_{\binom{n\plus{}1}{2}}$ each have exactly two points of the same color $ c_1$, but all in different combinations.  Similarly create lines $ q_{\binom{n\plus{}1}{2}\plus{}1}$ and $ q_{2\binom{n\plus{}1}{2}}$ with color $ c_2$.  Continue like this and you have a configuration without the desired 4 points.[/hide]


trigonometry
calculus
integration
logarithms
function
derivative
calculus computations

Find the following  integrals  :)

[1] $ \int \frac{sinx}{cosx\plus{}1}dx$

[2] $ \int \frac{cosx}{cos2x}dx$

[hide="Solution"]1. Taking the sub $ \cos x \equal{} t$ we quickly get the solution: $ \minus{} \ln(\cos x \plus{} 1) \plus{} C$, or, simplified, $ \minus{} 2\ln(\cos \frac {x}{2}) \plus{} C$.

2. Taking into account that $ \cos 2x \equal{} 1 \minus{} 2\sin^2 x$, and taking the sub $ \sin x \equal{} t$, we obtain the subintegral function $ \frac {1}{1 \minus{} 2t^2}$ which can be split into two fractions:  $ \frac {1}{2 (1 \plus{} \sqrt 2 t)} \minus{} \frac {1}{2 (\sqrt 2 t \minus{} 1)}$, and then the solution is a simple logarithm,

$ \frac {\ln (\sqrt 2 \plus{} 2 \sin x) \minus{} \ln(\sqrt 2 \minus{} 2 \sin x)}{2 \sqrt 2} \plus{} C$
[/hide]

1) Let u = cosx +1 and it should be straightforward after

2) Use $ cos2x \equal{} 1\minus{} 2sin^2(x)$ and let $ u \equal{} \sqrt{2}sinx;$ to get:
$ \int\frac{cosx}{cos2x}dx \equal{} \int \frac{cosx}{1\minus{}2sin^2(x)}dx$ $ \equal{}\frac{1}{\sqrt{2}}\int\frac{1}{1\minus{}u^2}du$
$ \equal{}\frac{\sqrt{2}}{4} [\int \frac{1}{1\minus{}u}du \plus{} \int \frac{1}{1\plus{}u}]du$

For one I just not that the derivative of cos(x)+1 is -sin(x). so, the integral is $ I\equal{}\minus{}\ln{(cos(x)\plus{}1)}\plus{}C$. Your answer is extremely different. Why? I'm confused.

$ cos(x) \equal{} cos(2.\frac{x}{2}) \equal{} cos(\frac{x}{2})^2 \minus{} sin(\frac{x}{2})^2 \equal{} 2cos(\frac{x}{2})^2 \minus{} 1$

So that $ cos(x) \plus{} 1 \equal{} 2cos(\frac{x}{2})^2$
And you are right the integral is $ \minus{}ln(cosx \plus{}1) \plus{} C \equal{} \minus{}ln(2cos(\frac{x}{2})^2) \plus{}C \equal{} \minus{}(ln2 \plus{} 2ln(cos(\frac{x}{2}))) \plus{} C$

Yes [b]igiul[/b], the $ \sin$ was a typo in the [1] (edited). And as far as simplification is concerned, you can take the $ \ln$ 2 from [b]akesh's[/b] simplification into the constant $ C$, and you get my solution.

This might be a good time to know that in the integration of $ \sqrt{a^2\plus{}x^2}$ and other like functions you actually let $ \minus{}\ln{a}$ or something become part of $ C$ or you can't derive the familiar indefinite integral of a hyperbola. Don't forget your C's.  :wink:

Nice !

Try the following

[3] $ \int \frac{1\minus{}cosx}{1\plus{}cosx}dx$

hello,
$ \int\frac{1\minus{}\cos(x)}{1\plus{}\cos(x)}dx\equal{}\int\tan^2\left(\frac{x}{2}\right)dx\equal{}
\int\tan^2\left(\frac{x}{2}\right)\plus{}1dx\minus{}\int 1dx\equal{}$
$ 2\tan\left(\frac{x}{2}\right)\minus{}x\plus{}C$.
Sonnhard.

Multiply by the denominators conjugate.

\begin{eqnarray*}I
&=&\int (\csc{x}-\cot{x})^2dx\\
&=&\int 2\csc^2{x}dx-\int \cot{x}\csc{x}dx-\int dx\\
&=&\csc{x}-2\cot{x}-x+C\\
\end{eqnarray*}


geometry
perimeter
number theory
relatively prime

Find primitive Pythagorean triples whose perimeter is a square.

[hide="hint"]Let $ a,b,c$ be the sides of the right $ \triangle ABC$ right angled at $ C$.
Now let \[ a\equal{}p^2\minus{}q^2\] \[ b\equal{}2pq\] \[ c\equal{}p^2\plus{}q^2\] for some positive integers $ p,q$.
Therefore, $ a\plus{}b\plus{}c\equal{}2p^2\plus{}2pq\equal{}2p(p\plus{}q)$
Now, $ p/2$ and $ p\plus{}q$ are relatively prime and their product is a perfect square $ \Longrightarrow$ Both are perfect squares.(Here we have WLOG assumed that p is the one which is even out of the two)[/hide]


algebra
polynomial
limit
integration
function
calculus
real analysis

If f is continuous and $\forall x \in R$,we have:
$\lim_{h \to 0}\frac{\int_{0}^{h}f(x+u)+f(x-u)-2f(x) du}{h^{3}}=0$
Prove: f is a polynomial of degree 1.

I suppose we can assume that $f$ is integrable? If so, we have that

${\int_{0}^{h}f(x+u)+f(x-u)-2f(x) du}= \int_{-h}^{h}f(x+u)-f(x) du$

I feel like we should be able to show that $f'$ exists and is constant from here, but I can't think of how.

Change of variables $u =-h+2ht$ gives

$\int_{-h}^{h}f(x+u)-f(x) du = 2h\int_{0}^{1}f(x-h+2ht)-f(x)dt$

Seems so very close to something useful.

Sorry,f is continuous.But I can't see anything with what you wrote.
Can you explain more,thank you very much,

I don't know how to proceed either, but in one way or another it must be possible to prove that $f$ is differentiable in every point and that it has a constant derivative. $f(x+u)-f(x)$ is the numerator of the differential quotient. I was just throwing out some ideas.

By a corollary to the mean value theorem there is a function $\xi(h)$ with $|\xi(h)| \le h$ and
$\int_{-h}^{h}f(x+u)-f(x) du = 2h\left( f(x+\xi(h))-f(x)\right)$

I am thinking a proof similar to the fundamental theorem of calculus, but I don't know how to make it work.

Yes,May it is a problem seems easy...
But indeed it is hard to prove...

Where is this problem from? Is there a solution? I'd definitely like to see it. I have worked a lot with this problem but I can't get anywhere.

It is from a Chinese forum asked by someone,But there is also no proof...
I also don't know how to consider this problem.


It will be held in Greece, Rhodes. From 26th April to 2nd May: almost a month from now. Enough time to start talking about it!

Italy is going to be there (for the first time), and as you can guess, $\mbox{me}\in (\mbox{italian team})$. And of course, I'm very happy; it's the first international competition that I'll join!  :)

Who is coming :?:

[quote="edriv"]$\mbox{me}\in (\mbox{italian team})$. [/quote]
You? :P Congrulation  :) You will go to IMO 2007?

Well, no... the italian national olympiad is from 11 to 13 May, and the team selection test for the IMO I don't know, but probably near the end of May. In conclusion, I can't be sure I won't be there, but there are many smart and nice people who will try to go to your country in july... hope dies last  :)

[quote="edriv"]
Italy is going to be there (for the first time), and as you can guess, $\mbox{me}\in (\mbox{italian team})$. And of course, I'm very happy; it's the first international competition that I'll join!  :)

Who is coming :?:[/quote]

Well...Azerbaijan is going to be there too and first time as Italy :lol: I think, I will be there .Good luck to everybody

Azerbaijan,Italy,Great Britain...
BMO becomes very popular ! :lol:

I will be in Rodos  but at the second team . :wink:

Is there any official BMO site? I couldn't find it... :ninja:

[quote="edriv"]I don't think so... we didn't know if would go to Balkan until our team leaders told us it. Before that, it was a comic situation; at the top of the papers with the problems of the Balkan TST, the title was litteraly "Balkan (?!?) TST"  :rotfl:

To silouan: will Greece partecipate with two teams? I don't understand!

To hana: well, it's true that BMO becomes very popular, but it's also true that the Balkans now occupy a wider area.
See this recent map:
[img]8307[/img]

To cefer: (I'm reading backwards) Wow, I like Azerbaijan, although I have heard of it very few times  :D[/quote]

According to BMO 2005 you must add Kazakhstan and Yakutia to this map :rotfl:
P.S.And all part of Turkey

To [b]edriv[/b] : You coloured only part of Serbia !   :mad:

Well, he original map is from wikipedia... I have just coloured Italy, Great Britain and Azerbaijan, don't be angry with me!

Ok, i am coming to Rhodes :D  and  we applied for visa today.See you there... :wink:

How many student get medal in BMO? Do anyone know cutoffs of medals in previous years?

What about Romanians? You have done the TST, haven't you?

And italians? Andrea, Denis, Maria, Mattia, Roberto... write in this forum!

I'll arrive to Rhodes' airport at 17:30, see you there  :D

You can check out the webpage of the Balkan Mathematical Olympiad 2007at

http://www.hms.gr/bmo2007/main.html

You can also download the programme of the contest and useful information about it.

Alexandros

Now, who can post all problems from Balkan Mathematical Olympiad 2007? Thanks  :wink:

I have posted them already at noon !
See the olympiad section , usolved problems. :)
 Babis

Good morning !

 Does any one have the results for every contestant  and for each problem ? I saw the results for the italian team in their oli site.

 Babis

You can find the results and the official solutions of the BMO 2007 in the official webpage of the BMO

http://www.hms.gr/bmo2007/main.html

Alexandros


geometry
trapezoid
ratio
trigonometry
geometry unsolved

From the given relation, if $AP=2m$, then $AT=5m$, $AB=2BT=2m \sqrt{10}$, which gives $m(\widehat{ABC})=60^\circ$, hence $AB=2CD$.

Best regards,
sunken rock

Actually , i cannot find any pure geometric method to this problem , i would very appreicate any method related !

Here is mine , trigonometric , obviously .

Let AP = 2k , AT = 5k then BT = sqrt((2k)(5k)) = sqrt(10) k and then AB = 2sqrt(10)k

so we have a triangle ABT with side-ratio , AB : BT : TA = 2sqrt(10) k: sqrt(10) k: 5k = 2sqrt(2) : sqrt(2) : sqrt(5) , by law of cosine , we have cos(ABT)  = ( 8+2-5)/8 = 5/8 , then tan^2( ABT/2) = ( 1 - 5/8)/(1+5/8) = 3/13 .

AB : CD = 1/(tan^2(ABT/2)) = 13/3 .

In general , if AP : AT = a:b , here 2:5 , then AB : CD = (9a-b):(b-a)

[quote="uglysolutions"]Isosceles trapezoid $ ABCD$, with $ AB \parallel CD$, is such that there exists a circle $ \Gamma$ tangent to its four sides. Let $ T \equal{} \Gamma \cap BC$, and $ P \equal{} \Gamma \cap AT$ ($ P \neq T$).
If $ \frac{AP}{AT} \equal{} \frac{2}{5}$, compute $ \frac{AB}{CD}$.[/quote]

Let $M,N, R$ the points where $\Gamma$ touch $DA,AB, CD$, respectively, and let $I$ the center of $\Gamma$. Because $ABCD$ is an isosceles trapezoid with bases $AB,CD$, it follows that  $AM=AN=BN=BT$ and $DM=DR=CR=CT$ .

Assume that $AP=2a$, $AT=5a$ for a fix $a\in \mathbb {R}^{+}$. We have that $AN^2=AP\cdot AT\Rightarrow AN=a\cdot \sqrt{10}$. Then $AB=2a\cdot \sqrt{10}$ and $BP=a\cdot \sqrt{10}$. Let $Q=AP\cap DC$. Note that $\triangle APB\sim \triangle QPC$ (*), and in consequence $CQ=2\cdot CP=CD$. Assume that que $CD=b$ (where $b\in \mathbb {R}^+$). By the similitude in (*) $PQ=b\cdot \sqrt{5/8}$. Then $QR^2=QT\cdot QP$, that is:

$\dfrac{9}{4}b^2=b\cdot \sqrt{\dfrac{5}{8}}\cdot \left(b\cdot \sqrt{\dfrac{5}{8}}+3a\right)$

It follows that $b=\dfrac{3a}{13}\cdot \sqrt{40}$. Because $AB=2\cdot AN=a\cdot \sqrt{40}$, we can conclude that: $\dfrac{AB}{CD}=\dfrac{13}{3}$. $\blacksquare$

I have posted this solution in [url]http://www.fmat.cl[/url] forum :D


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

$ \sum (\frac{a^2\plus{}b^2}{b}) \geq 2 \sqrt{(a^2\plus{}b^2\plus{}c^2)(\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a})}$


geometry
trigonometry
circumcircle
perpendicular bisector

Let be $ ABC$ a triangle with $ M$ the midpoint of $ BC$ and $ \angle ABC\plus{}\angle MAC\equal{}90^\circ$. Prove that $ \angle BAC\equal{}90^\circ$.

Let $ \angle MAC \equal{} \alpha$, $ \angle MAB \equal{} \beta$. Then $ \angle ABC \equal{} 90 \minus{} \alpha$ and, as the angles of a triangle add to 180, $ \angle ACB \equal{} 90 \minus{} \beta$.

Because $ M$ is the midpoint of $ BC$, $ [AMB] \equal{} [AMC]$. Hence, using area=1/2*a*b*sin(C),

$ a\sin(\alpha) \equal{} b\sin(\beta)$ (using sides AM,AC and AM,AB) and $ a\sin(90 \minus{} \beta) \equal{} b\sin(90 \minus{} \alpha)$ (using sides CM,CA and BM,BA).

Hence $ \sin(\alpha)\cos(\alpha) \equal{} \sin(\beta)\cos(\beta) \Leftrightarrow \sin(2\alpha) \equal{} \sin(2\beta)$. So, either $ \alpha \equal{} \beta$ or $ \alpha \plus{} \beta \equal{} 90$.

(Note that BAC doesn't have to be right if $ AB \equal{} AC$  :wink: )

[quote="Mîndrilă Claudiu"]Let be $ ABC$ a triangle with $ M$ the midpoint of $ BC$ and $ \angle ABC \plus{} \angle MAC \equal{} 90^\circ$. Prove that $ \angle BAC \equal{} 90^\circ$.[/quote]
[hide="Solution"]
Extend $ AM$ to meet the circumcircle of $ \triangle ABC$ at $ P$.  Then, we have that $ \angle ABP \equal{} \angle ABC \plus{} \angle PBC \equal{} \angle MAC \plus{} \angle ABC \equal{} 90$.  Thus, the circumcenter of $ \triangle ABP$ lies on $ AP$, so the circumcenter of $ ABPC$ lies on $ AP$.  Therefore, the intersection between $ AP$ and the perpendicular bisector of $ BC$ is the circumcenter of $ ABPC$.  Notice that if $ AP$ is the perpendicular bisector of $ BC$, then $ AP$ and the perpendicular bisector of $ BC$ coincide, $ M$ is not the intersection between the perpendicular bisector of $ BC$ and $ AP$.  Yet, this would imply that $ A$ lies on the perpendicular bisector of $ BC$, so $ AB \equal{} AC$ which contradicts the problem statement.  Now, $ M$ is the midpoint of $ BC$, so $ M$ lies on both the perpendicular bisector of $ BC$ and on $ AP$.  Thus, $ M$ is the circumcenter of $ \triangle ABC$, so $ BC$ is a diameter of the circle, so $ \angle BAC \equal{} 90$. [/hide]

And now, my proof.

1. Let be $ D \in (BC)$ such that $ AD \perp BC$.
2. Because $ \angle ABC\plus{}\angle MAC\equal{}\angle ABC \plus{} \angle BAD\equal{}90^ \circ$ it`s obvious that $ AD$ and $ AM$ are the trisectors of $ \angle BAC$.
3. Using Steiner`s theorem we get: $ \frac{AB^2}{AC^2}\equal{}\frac{BD \cdot BM}{CD \cdot CM}\equal{}\frac{BD}{DC}$.
4. Because $ AB^2\equal{}AD^2 \plus{}DB^2$ and $ AC^2\equal{}AD^2\plus{}DC^2$ we get the following chain of equvalents:
\[ \frac{AB^2}{AC^2}\equal{}\frac{BD}{DC} \Leftrightarrow AB^2 \cdot CD\equal{} AC^2 \cdot BD \Leftrightarrow (AD^2\plus{}DB^2) \cdot CD\equal{}(AD^2\plus{}CD^2) \cdot BD \Leftrightarrow AD^2 \cdot DC\minus{}BD \cdot DC^2\minus{} BD \cdot AD^2\plus{} BD^2 \cdot DC\equal{}0 \Leftrightarrow (DC\minus{}BD)(AD^2\minus{}BD \cdot DC)\equal{}0.\]

Now there are two cases.

$ a)$ If $ BD\equal{}DC$ we get that $ AB\equal{}AC$, which isn`t true because in the problem is mentionated that $ AB \neq AC$.
$ b)$ If $ AD^2\equal{}BD \cdot DC$ then we obtain that $ \angle ACB\equal{} \angle BAD$ and we denote $ \angle BAC\equal{} 90^\circ$

It is clear the problem is true if the triangle is right. But if the triangle wasn't right, we could find point D on AM so BDC is right. Assume angle A is acute (it is easy to use a similar argument if it's obtuse). In that case,
DBM+MDC=90
DBM+MAC+ACD=90
ABC-ABD+MAC+ACD=90
ABC+MAC+(ACD-ABD)=90
So a non right triangle would only work if ACD=ABD. This is true if the triangle is isoceles, but can someone else help finish this method up and show this is never true in any other triangle?


limit
logarithms
real analysis
real analysis unsolved

$\phi = \frac{1+\sqrt 5}{2}$ (the golden ratio).
We now define the sequence $(x_{n})$
$x_{1}= 1$
$x_{n+1}= \sqrt{1+x_{n}}$ for all positive integers n.
Show that $\lim_{n \rightarrow \infty}(2\phi)^{n}(\phi-x_{n})$ exists and compute its exact value (its numerical value is 0,678998075...).

for the convergence, notice that $\frac{\phi-x_{n+1}}{\phi-x_{n}}= \frac{1}{\phi+x_{n+1}}$, hence $\frac{\phi-x_{n+1}}{\phi-x_{n}}\to \frac{1}{2 \phi}$. This shows that $\ln(\phi-x_{n}) =-n \ln(2 \phi)+o(n)$. Now, if one call $L_{n}=\ln((2 \phi)^{n}(\phi-x_{n}))$, the same calculation shows that
$L_{n+1}-L_{n}= \ln(\frac{2 \phi}{\phi+x_{n+1}}) \approx \frac{(2 \phi)^{-n-1}e^{o(n)}}{2 \phi}= o(\frac{1}{n^{2}})$. This shows the convergence, if I haven't make mistakes.


Use rules of logarithms to find the exact value of each expression WITHOUT a calculator. Note: The underscore symbol indicates the base of the log function. For example,
log_2 means "log base 2."

(1) 5^[log_5 (6) + log_5 (7)]

Note: For question (1), the exponent is the expression
[log_5 (6) + log_5 (7)]. Is this clear?

(2) e^[log_e^2 (9)]

Note: For question 2, the exponent is the expression
[log_e^2 (9)], which is read "log base e squared nine."

$ log_a(b) \plus{} log_a(c) \equal{} log_a(bc)$
and $ a^{log_a{bc}} \equal{} bc$
Also $ log_(a^n)(b^m) \equal{} m/n log_a(b)$
:)

Number 1:  $ 6\times 7\equal{}42$, so the fives cancel out to get 42.

Number 2:  $ e^{log_{e^2} 9}\equal{}e^{log_e 3}\equal{}3$

Why didn't you just use Latex?  All you had to do was surround your expressions with $ \$$ and replace [ with {.

Some online math courses don't really use LaTeX, maybe?  :wink:

It is a lot clearer, and it's not hard to learn at all


A sonologist assists the oncologist by reading angles and velocities of ultrasound graphs. Recently, Pat, a sonologist, discovered an angle of a triangle which measured at least $ \frac{2}{3}$ of its supplement and at most 5 times its complement. Find the difference between the largest and smallest possible degree measures of the angle.

[quote="GameBot"]A sonologist assists the oncologist by reading angles and velocities of ultrasound graphs. Recently, Pat, a sonologist, discovered an angle of a triangle which measured at least $ \frac {2}{3}$ of its supplement and at most 5 times its complement. Find the difference between the largest and smallest possible degree measures of the angle.[/quote]

Let the angle be $ x$. We know $ 5(90 \minus{} x) \ge x \ge \frac {2}{3}(180 \minus{} x)$. Simplifying yields $ 450 \minus{} 5x \ge x \ge 120 \minus{} \frac {2}{3}x$ and we now have two cases to consider - $ 450 \minus{} 5x \ge x$ and $ x \ge 120 \minus{} \frac {2}{3}x$. The first case yields $ x \le 75$ and the second $ x \ge 72$. Since $ x$ must satisfy both conditions, the smallest possible measure it could have is 72 and the largest 75. Thus, our answer is $ \boxed{3}$.


function
logarithms
calculus
derivative
real analysis
real analysis theorems

I have having issues deducing the Maclaurin Series of a function $f(x) = x^n \log (1+x^n)$.

If I instead compute the Maclaurin Series of the function $g(x) = x \log (1+x)$ and replace $x$ with $x^n$, I get the right series: $x^{2n} - x^{3n}/2 + x^{4n}/3 - ...$. However, simply computing the Maclaurin Series of $f(x)$ (using the definition of Maclaurin Series) yields a different series expansion. Why is this?

Any help would be appreciated, thanks.

What with all of the chain rules, quotient rules, and product rules, actually trying to calculate the derivatives of this function leads to an intractably ugly mess. Substitution, as you describe, would appear to be the only practical method.

If you think you're getting wrong answers (rather then simply thowing up your hands and gettng no answers) by taking the derivatives, then you've probably taken the derivatives incorrectly.


Of 100 citizens surveyed in a certain town, only $ 45\%$ were in favor of the city council's proposition. $ 40\%$ were opposed and $ 15\%$ were undecided. If the survey holds and 40,000 citizens vote, what percent of the undecided voters must vote in favor of the proposition in order for it to pass with $ 51\%$ of the votes?

For 45 percent to become 51 percent, then need 6 percent more of the 15 percent undecided. 6/15 = [b]40[/b] percent.


geometry
circumcircle
trigonometry
geometry proposed

[color=darkblue]Let $ABC$ be an acute triangle with the centroid $G$.
The its incircle $C(I,r)$ touches the sides $AB$, $AC$ in the points $E$, $F$.
Denote the second intersection $M$ between the line $AI$ and the circumcircle $C(O,R)$.
The line $d$ for which $d\parallel BC$ and $G\in d$ meets the circumcircle in points $P$, $Q$. Prove that :
$1.\blacktriangleright$ There is the ([b]interesting ![/b]) relation $MA^{2}=ME^{2}+p(p-a)$, where $2p=a+b+c\ .$
$2.\blacktriangleright$ The points $E,F,P,Q$ are conciclycally if and only if $\cos A=\frac{1}{3}\ .$[/color]

[color=darkblue][b]Generalization.[/b] Let $ABC$ be a triangle. The its incircle touches the sides $AB$, $AC$ in the points $E$, $F$.
Let $d$ a line for which $d\parallel BC$ and separates the points $A$ and $B$ The line $d$ meets the circumcircle of the given triangle in points $P$, $Q$.
Prove that the points $E,F,P,Q$ are conciclycally if and only if the distance between the lines $d$ and $BC$ is equal to $\delta=h_{a}\sin^{2}\frac{A}{2}$,
where $h_{a}$ is the distance of the point $A$ to the line $BC$.What is the geometrical significance of the value $\delta$ ?

[hide="Here is the significance."]If $D$ is the projection of the point $A$ to the line $BC$, $X\in (DC)$ is the point so that $m(\widehat{DAX})=\frac{A}{2}$
and $L$ is the projection of the point $D$ to the line $AX$, then $\delta$ is the distance of the point $L$ to the line $BC$, i.e. $L\in d\ .$[/hide]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=106637 [/color]


geometry
perimeter
inequalities
triangle inequality

The lengths of two sides of a triangle are 7cm and 3cm. If the number of centimeters in the perimeter is a whole number, what is the number of centimeters in the positive difference between the greatest and least possible perimeters?

By the Triangle Inequality, the minimum is $ 7\minus{}3\plus{}1\equal{}5$ and the maximum is $ 7\plus{}3\minus{}1\equal{}9$, giving us $ 9\minus{}5\equal{}\boxed4$.

Its also 2n-2 where n is the small number.


inequalities
limit
algebra unsolved
algebra

From Cauchy - Schwarz inequality we have :

$\\ n \cdot (1+2+...+n)^2 >n \cdot a_{n}^n \geq (1+2+..+n)^2 \Rightarrow \sqrt[n] {\frac{n^{2} (n+1)^2}{4} } > a_{n} \geq \sqrt[n] {\frac{n (n+1)^2}{4} }$

and

$\lim_{n \to \infty}\sqrt[n] {\frac{n (n+1)^2}{4} } = \lim_{n \to \infty}\sqrt[n] {\frac{n^{2} (n+1)^2}{4} } =1$

therefore $\lim_{n \to \infty} a_{n} =1$ .

 :)

[quote="dieuhuynh"]you need edit your solution.
here my solution:
 I have :if $a_n\Rightarrow 0$ for all $n$ and $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=a$
then ${\lim_{n\to\infty}a_{n}^{1/n}}=a$[/quote]

I don't understand your solution  :D   :)

first, I am sorry, your solution is right.
    about my solution :I have       $lim_{n\to\infty}\frac{1^2+...+(n+1)^2}{1^2+...+n^2}=1$
 :arrow: $lim_{n\to\infty}(1^2+...+n^2)^{\frac{1}{n}}=1$


MIT
college
inequalities proposed
inequalities

Let $ a,b,c\geq 0$prove that
$ \sum \frac {2a^2 \minus{} bc}{\sqrt {5(a^2 \plus{} 2b^2 \plus{} 2c^2)}} \ge \frac {a \plus{} b \plus{} c}{5}$

[quote="Allnames"]Let $ a,b,c\geq 0$prove that
$ \sum \frac {2a^2 \minus{} bc}{\sqrt {5(a^2 \plus{} b^2 \plus{} c^2)}} \ge \frac {a \plus{} b \plus{} c}{5}$[/quote]
Do U mistake ? I think U should check carefully. If not :

$ LHS\equal{} \frac {2(a^2\plus{}b^2\plus{}c^2) \minus{}(ab\plus{}bc\plus{}ca)}{\sqrt {5(a^2 \plus{} b^2 \plus{} c^2)}}$

$ \rightarrow LHS\ge \frac {a^2\plus{}b^2\plus{}c^2}{\sqrt {5(a^2 \plus{} b^2 \plus{} c^2)}} \ge\ \frac {a \plus{} b \plus{} c}{5}$  :wink:

Yes,i have some mistake,i have edited it, sorry

Your solution have a bit  wrong ;) .  ( If $ 2a^2 \ge bc$)
;)
By chebysev and am-gm ,we have :
$ LHS \ge \sum \frac{a^2}{\sqrt{5(a^2\plus{}2b^2\plus{}2c^2)} } \ge RHS$
;)

[quote="mitdac123"]Your solution have a bit  wrong ;) .  ( If $ 2a^2 \ge bc$)
;)
[/quote]
Oh  yes. You are right. Thank you! The stupid mistake! :(

Can you deyail it mitdac,my proof is quite long

[quote="mitdac123"]$ LHS \ge \sum \frac {a^2}{\sqrt {5(a^2 \plus{} 2b^2 \plus{} 2c^2)} } \ge RHS$
;)[/quote]
How about it,mit dac
My proof is very long,which is used CS,if you are true,thank for your nice solution


Murphy's sells notebooks at a cost of five for $\$2, Right-Ade sells six for $\$3 and Callahan's sells twelve for $\$5.  What is the number of cents in the cost per notebook for the least expensive store?

[hide]i would set up some ratios
5/2
6/3
12/5
then find the least common denominator, 30
75/30
60/30
72/30
Right-Ade(weird name) sells them the cheapest.
6/3=.5
the answer is $\boxed{50  \text{cents}}$[/hide]

I disagree with Math92's answer. Here's how I came to my solution.

[hide]I would also set up some ratios. The variables will be the price of each salesman's notebook.
Murphy: $5a=2$
Right-Ade: $6b=3$
Callahan: $12c=5$

We can immediately rule out Right-Ade, whose price is obviously higher, and determine if Murphy or Callahan's price is better.
Murphy)
$5a=2$
$a=\frac{2}{5}$
$a=.40$
Callahan)
$12c=5$
$c=\frac{5}{12}$
$c=.417$

Therefore, Murphy has the best price at 40 cents a notebook.

[/hide]

math92, you may want to read the question more carefully, specially the part
[quote="MCrawford"]...cost per notebook...[/quote]

[hide]the three rates are
1/2, 2/5, and 5/12.
Immdiately eliminate 1/2 as the lowest because it is the only one equal to or greater than half. now we nave 2/5 and 5/12.

If you don't know that 5/12 is about .42, you can always cross multiply to find that 5/12 is greater than 2/5!

Murphy gives the best deal (lowest cost per notebook)[/hide]


We have $\binom {n}{p} = \frac{n-p+1}{p} \times \binom {n}{p-1}$

So inequality is ${\frac{k}{2\,n+2-k}}+{\frac{2\,n+1-k}{k+1}}-2\,{\frac{n+1}{n+2}} \geq 0$
which nicely factors to :

$2\,{\frac{\left(2\,n+3\right ) \cdot \left (n+1-k\right ) \cdot \left (n-k\right ) } {\left (2\,n+2-k\right ) \cdot \left (k+1\right ) \cdot \left (n+2\right )}}$

Now $\left (n+1-k\right ) \cdot \left(n-k \right ) \geq 0$

So ineq is true and there is equality iff $n= k-1$ od $n = k$


[quote="Kalimdor"]At first, I tried this way $\ \frac{a}{b^{2} + c^{2}} + \frac{b}{a^{2} +c^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{9}{2 * (a + b + c)} \rightarrow \sum \frac{2a}{(b+c)^2} \geq \frac{9}{(a+b)+(b+c)+(c+a)}$
[/quote]
  Something is wrong here! :lol: (Or u want to try to prove a weaker inequality?)And I don't understand your sols either . Can u write clearly?  Sorry for my stupid math :D

[quote="leepakhin"][quote="Kalimdor"]In the case of : $\ a=b=1, c=0$ the statement is not true.
But I tried it anyways
At first, I tried this way $\ \frac{a}{b^{2} + c^{2}} + \frac{b}{a^{2} +c^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{9}{2 * (a + b + c)} \rightarrow \sum \frac{2a}{(b+c)^2} \geq \frac{9}{(a+b)+(b+c)+(c+a)}$
Use Cauchy Schwartz, $\ \sum \frac{2a}{(b+c)^2} * \sum (a+b)^2 \geq \sum \frac{2a(b+c)^2}{(b+c)^2} = 2(a+b+c)$
Which finally the origin question turns out to be: [color=red]Prove the following:$\ 2(a+b+c) \geq \frac{9}{2(a+b+c)}*\sum (a+b)^2$
Then we get: $\ 4(a+b+c)^2 \geq 9\sum (a+b)^2$[/color]
$\ \rightarrow (\frac{2}{3}(a+b+c))^2 \geq \sum (a+b)^2$  When any of the ${a,b,c}$ is $0$, the statement is clearly not true.
We may change the number $9$ on the right side of the origin ineq into $1$
So we can get $\ (2(a+b+c))^2 \geq \sum (a+b)^2 \Rightarrow$
$\ LHS = ((a+b)+(b+c)+(c+a))^2 = (\sum (a+b))^2 \geq \sum(a+b)^2 = Right$[/quote]
How can you prove a wrong inequality?[/quote]
No,I think Kalimdor just show us that the inequality weaker than the organal inequality is wrong...
ps:there may have a little mistake

Adding that $a,b,c>0$ doesn't help at all. We can take $a,b\rightarrow 1$, $c\rightarrow 0$.

Oh, sorry for my stupidity. :oops:
I just want to say that the origin one is not true in some cases, meanwhile, if you convert the origin one into $\frac{a}{b^{2} + c^{2}} + \frac{b}{a^{2} +c^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{1}{2 (a + b + c)}$
Which I've already proved.
And if you can find some mistakes in it, that could occur if I messed up with $\geq$ and $\leq$ :|

What's more, [quote] I'm sorry, it's $a,b,c > 0$[/quote]
I don't think this will change the result greatly, take a number which appoarches $0$, like $0.00001$

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167445

[hide=**][quote=chidat]If $a,b,c \geq 0$ prove or disprove that
$\frac{a}{b^{2} + c^{2}} + \frac{b}{a^{2} +c^{2}}+\frac{c}{a^{2}+b^{2}} \geq \frac{9}{2 * (a + b + c)}$[/quote]
[quote=hungkhtn]It should be:

$\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2} \ge \frac{4}{a+b+c}$

Which this one I had post in the forum![/quote][/hide]
Let  $a,b,c \geq 0 .$ Prove that
$$\frac{3}{2}\left(\frac{a}{b^2+bc+c^2}+ \frac{b}{c^2+ca+a^2}+\frac{c}{a^2+ab+b^2}\right)\geq\frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2} \ge \frac{4}{a+b+c}$$
$$\frac{a}{b^2-bc+c^2}+ \frac{b}{c^2-ca+a^2}+\frac{c}{a^2-ab+b^2} \ge \frac{4}{a+b+c}$$
Let $a,b,c$ be positive real numbers . [url=https://artofproblemsolving.com/community/c4h1061866p5345313]Prove that[/url]
\[\dfrac{a}{b^2+bc+c^2}+\dfrac{b}{c^2+ca+a^2}+\dfrac{c}{a^2+ab+b^2}\geq \frac{a+b+c}{ab+bc+ca}\geq \frac{3}{a+b+c}.\]
[url=https://artofproblemsolving.com/community/c6h325259p1744191]h[/url]
[url=https://artofproblemsolving.com/community/c6h325259p1744138]h[/url]

Let $a, b, c$ be nonnegative real numbers such that $a + b + c = 2. $ Prove that$$\frac{a}{b^2+bc+c^2}+\frac{b}{c^2+ca+a^2}+ \frac{c}{a^2+ab+c^2} +8\geq \frac{10}{ab+bc+ca}$$
(Marius Stănean, Zalău, România)


When you login, there is always an option to log in automatically. However, I clicked that quite some time ago, and it always worked, except recently, I tried to go on and it didn't log me on. Is there a reason behind this?

If you haven't been on AoPS for long time or if they shut down the site and restarted it between your visits, the automatic login doesn't work and you have to enter your user name and password once again. You may again click on the automatic login button and everything will be fine until the next period of long absence or the next site shutdown.

i recommend NOT clicking it to avoid problems.


At the end of December, 2005, about 6,000 pilots of financially troubled Delta Airlines accepted a 14% pay cut. Delta announced that this would save them approximately 143 million annually. Based on this information, the average annual pay of a Delta pilot before the cut is ________________.  After the cut it is ____________________

here it is
[hide]let the payment before the cut be $ x$. using the information given, we get the equation $ 6000(0.86x)\equal{}143,000,000$. solving for $ x$ we get that the payment of each pilot before the cut is approximately $ \$$27713.20 from there it is easy to get the paycheck after the cut. right now im too tired, just got back from school. [/hide]

First you gotta ask yourself if that's a realistic answer. What kind of pilot would accept a paycheck cut if he's only making 28k a year?

You made a mistake - $ \frac{143\cdot10^{6}}{6\cdot10^{3}}\equal{} 23833\frac{1}{3}$ dollars is the amount of the average pilot's paycut.

Therefore each pilot annually made 170238.10 dollars prior to and 146404.76 dollars after the 14% cut.

oh..yeah..im so stupid..i don't know what i was thinking. thanks for the fix lazarus.
i see my mistake, but i don't know, maybe there are some pilots that make 28k a year and are getting a pay cut. i wouldn't know that. :oops:


MATHCOUNTS
function
geometric series
algebra
binomial theorem

The trick is to use generating functions.

I'm sure someone here will post a better response.

u could also make a very organized table...there's a similar post about this in the mathcounts forum...about the different ways to make a particular amount of money...there's this whole long formula that i don't think is worth memorizing, but understanding..you should check it out if it interests u

thats what i always do, but its very time consuming

It may be easier to think of the question as: How many ways can you make [i]85[/i] cents and then add two pennies to every solution as $ 87 mod 5\equal{}2$.

still, it doesnt help much, but it does save time when making a chart.

I'll post more in depth now.

Basically, we use a generating function to encode data in the coefficients of a series.

For example, the series $ P(x) \equal{} 1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}x^{4}...$ encodes the number of ways to make change using ONLY pennies for a given amount of cents (the exponent).

It can be shown with a geometric series arguement that this is equivalent to $ \frac{1}{1\minus{}x}$

We can write an equivalent series for using only nickels:

$ N(x) \equal{} 1\plus{}x^{5}\plus{}x^{10}\plus{}x^{15}\plus{}x^{5k}\equal{}\frac{1}{1\minus{}x^{5}}, k\in\mathbb{Z}$ This is clearly true because for each integer multiple of 5 cents, there is only ONE way to make that much money using only nickels.

By similar logic, we have the functions $ D(x) \equal{} 1\plus{}x^{10}\plus{}x^{20}\plus{}x^{30}... \equal{}\frac{1}{1\minus{}x^{10}}$
and
$ Q(x) \equal{} 1\plus{}x^{2}5\plus{}x^{5}0\plus{}x^{7}5 ... \equal{}\frac{1}{1\minus{}x^{25}}$

(I have named the functions after the first letter of the value of currency they represent)

(I'll also assume that half dollars are not allowed, but if they are, you'll be able to work that out as an extension.)

By the multiplication principle of counting, it turns out the number of ways to make $ R$ cents is given by the coefficient $ a_{n}\cdot x^{R}$ when we take the product of these functions:

Therefore, we have the function $ F(x) \equal{} P(x)N(x)D(x)Q(x) \equal{}\frac{1}{(1\minus{}x)(1\minus{}x^{5})(1\minus{}x^{10})(1\minus{}x^{25})}$

So now, all you have to do is expand this function, and find the coefficent of the term $ a_{n}\cdot x^{87}$

Ok, Bleh.

[quote="DiscreetFourierTransform"]$ F(x) \equal{} P(x)N(x)D(x)Q(x) \equal{}\frac{1}{(1\minus{}x)(1\minus{}x^{5})(1\minus{}x^{10})(1\minus{}x^{25})}$

So now, all you have to do is expand this function, and find the coefficent of the term $ a_{n}\cdot x^{87}$
[/quote]

I'm not sure I understand you... Do you mean find the coefficient of $ x^{87}$ on the [i]bottom[/i] of the function (if so, isn't it pointless for it to be a fraction with a one on the top)? Or do you mean to find the coefficient of $ x^{87}$ [i]after[/i] expanding the function (if so, there isn't one)?

Note how he was using sum of geometric sequences, so the coefficient of $ x^{87}$ in $ \frac{1}{(1\minus{}x)(1\minus{}x^{5})(1\minus{}x^{10})(1\minus{}x^{25})}$ is the same as the coefficient of $ x^{87}$ in $ (1\plus{}x\plus{}x^{2}\plus{}\cdots)(1\plus{}x^{5}\plus{}x^{10}\plus{}\cdots)(1\plus{}x^{10}\plus{}x^{20}\plus{}\cdots)(1\plus{}x^{25}\plus{}x^{50}\plus{}\cdots)$.

When this is expanded, you will get every possible combination of 1s, 5s, 10s, and 25s as exponents of $ x$, and the coefficients show how many ways yield that particular exponent.

I'm sorry to reawaken this topic...but how would you go about finding the term with 87 as the exponent? That seems pretty crucial to me, but I'm at a loss as to finding that.

The extended binomial theorem is handy to invoke here, I think.
It's a nice shortcut to getting the expansion on all of those polynomials.

If someone wants to do a demonstration, that would probably be useful.

[quote=centslordm]/bump sad[/quote]

this is 13 years old no need to bump -.-

Oh I finally understand how to do this type of question
generating functions....

when you realize how old AoPS is
and how old you are

Bruh the thread is older than me :rotfl:

[quote=sharkman]I have a math question dealing with money. The question is "How Many Ways Are There to Make 87 Cents?" I know there is a formula to get the answer, but I don't what the formula is. Can you help?[/quote]

I doubt there is a formula

This thread was made when I was months old :P
You don't need a formula
Just use casework to solve

This was made before I was born lol. Use casework.

Also this user isn't even active anymore

[quote=Possible]
I doubt there is a formula[/quote]

As bjc noted, you can use the recursive definition to solve for a formula, although it seems to be extremely messy. (also note that order does not matter, so the formula above might not work)

Maybe in another 13 years someone is going to bump it...
Let this thread rest in peace.

[quote=bjc]this is from 13 years ago
it's a recursive formula
probably like $f(n)=f(n-1)+f(n-5)+f(n-10)+f(n-25)+f(n-50)$
(assuming half-dollar coins)[/quote]

This is wrong; order does not matter.


inequalities
geometry
inradius
set theory
geometry proposed

Prove that with any triangle contained an unit square then inradius of it less than $\frac{\sqrt{5}-1}{4}$.

Really interesting problem... I'm assuming when you say "contains" you mean "has area".

I'm just confused: does the triangle have unit area of does it physically contain a unit square?

oh thanks for your interest, I think I mean all point of triangle belong to that square, by set theory $\triangle\subset\square$

Oh sorry... I think you had a typo: "contained an square" I think you meant "contained in square"... now it's completely different.
Thanks for clarification!

ok I must write clearly
Prove that with any triangle is contained in an unit square then inradius of it less than $\frac{\sqrt{5}-1}{4}$
 :)

Here's what I have so far:
1) we can assume one vertex lies on a corner of the unit square
2) we can assume other vertices are equal distance from the two corners adjacent to the first vertex, and lie on the sides of the cube.

The equation reduces to: find the maximum value of:
\[\sqrt{\frac{\frac{(1-x)^{2}}{2}\cdot \left(\sqrt{1+x^{2}}-\frac{1-x}{\sqrt{2}}\right)}{\sqrt{1+x^{2}}+\frac{1-x}{\sqrt{2}}}}\]

because the triangle has sides $\sqrt{1+x^{2}}, \sqrt{1+x^{2}}, \sqrt{2}(1-x)$

There is a solution to this problem at http://www.mia-journal.com/files/4-2/full/04-30.pdf

Also a shorter solution appears on page 220 of the book Geometric Problems on Maxima and Minima, by Titu Andreescu, Oleg Mushkarov and Luchezar Stoyanov (published by Birkhauser).


Hi folks,

If you made the ARML team, congratulations. If you didn't make the team, but would like to go, then, you CAN! We are looking for 15 students to form a Georgia C team. (Or even a D team if we have enough folks.) Teams are comprised of 15 people, so that's a minimum to field a team. You will be required to attend the practices on the next 3 Sundays and Memorial Day and be at UGa on Thursday, May 29 through Saturday, May 31.

If you enjoy math team and problem solving (why are you here otherwise?), ARML is a great competition to be with some of the best folks in Georgia and compete against the best in the nation.

Email me for more information. The first practice is this coming Sunday, May 4. I believe it's at Lassiter and starts at 1:30 but need to confirm that.

Tom

[quote="gtf"]The first practice is this coming Sunday, May 4. I believe it's at Lassiter and starts at 1:30 but need to confirm that.
[/quote]

That's right.

Um, Mr. Fulton, I don't think Arvind and I will be able to attend the first meeting at Lassiter cuz I have a MC state team practice that day at 2:00 in the Ocee Regional Library.  Seeing as Lassiter and the library are considerably far apart from one another, I guess I'll only be able to go to the other practices.  Sorry that I forgot to put that in the schedule conflicts part of the form that we had to fill out. :oops:

We will miss you, but it's probably just fine. Enjoy the MathCounts practice and we will see you soon. And good luck in Denver!

Actually Sitan

We mite be able to rearrange MC meeting.

So yea.

What time are your flights coming back from Denver?

Um, no we can't rearrange it.  The library [i]opens [/i]at 2:00, according to Ms. Harris, so we can't reschedule it.

I have a mandatory mock AP History Exam on the first date (This Sunday). Is it ok if I miss out on that practice?

- Oliver Huang

Ah, my stupidity has blinded me... :)

Sitan is right. We can't make the first meeting...

um....so i forgot that i had a piano recital this sunday....so i don't think i can make practice...

ZOMGz!  ARML kinda, sorta, almost is tomorrow!  Good luck to everyone!

PS: OMGzOMGzOMGzOMGzOMGzOMGzOMGzOMGzOMGzOMGz, need to study m...o...r...e!


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry proposed

Points $M$ and $M'$ are isogonal conjugates in triangle $ABC$. We draw perpendiculars $MP, MQ, MR$ and $M'P', M'Q', M'R'$ to the sides of $ABC$ where $P,P'$ are in $BC$, $Q,Q'$ are in $AC$ and $R,R'$ are in $AB$. Let $QR$ and $Q'R'$ intersect at $E$, $PR$ and $P'R'$ intersect at $F$, $PQ$ and $P'Q'$ intersect at $G$. Show that $AE, BF$ and $CG$ are all parallel.

We will prove that all those three lines are perpendicular to $MM'$

It is well-know that $Q'R' \bot AM$ (simple angle-chasing)
Analogically $QR \bot AM'$

Let $Q'R' \cap AM = X$ and $QR \cap AM' = X'$
Thus quadrilateral $AX'EX$ is cyclic.

Points $P,P',Q,Q',R,R'$ all lie on one circle (well-known property or simple angle-chasing).

Furthermore $\triangle ARX' \sim \triangle AM'R'$ and $\triangle AQ'X \sim \triangle AMQ$

Now from the power of the point and above similarity we obtain that $AX'*AM'=AR*AR'=AQ'*AQ=AX*AM$
Thus quadrilateral $X'M'MX$ is cyclic.

Let $MM' \cap AE = E'$

Therefore $\angle E'AX' = \angle EAX' = \angle X'XE$
and $\angle MM'X' = \pi-\angle X'XM = \angle AXX'$

Thus $\pi-\angle AE'M' = \angle E'AM'+\angle AM'E' = \angle X'XE+\angle AXX' = 90^{\circ}$

Then $AE \bot MM'$

Analogically we can prove that $BF \bot MM'$ and $CG \bot MM'$

Thus $AE \| BF \| CG$

WOW! That was a simple, really nice solution, congrats on finding it  :)  :) it was great. i'll outline my solution, it's actually a bit complicated.

Lemma $1$: In a cyclic quadrilateral $ABCD$, $DB$ intersects $AC$ at $E$, $AB$ intersects $CD$ at $G$, $AD$ intersects $BC$ at $F$. Then the orthocentre of $FEG$ is the centre of the circumcircle.
Proof: This result is very well-known, and the shortest proof i know is still complicated. Basically it uses basic polar results ($A$ lies on the polar of $B$ iff $B$ lies on the polar of $A$) and Brianchon's theorem with a degenerate hexagon - an inscriptible quadrilateral - to glue the proof together.
Lemma $2$: In a cyclic quadrilateral $ABCD$ with $AB$ intersects $CD$ at $P$, Suppose perpendicular to $AB$ thru $A$ intersects perpendicular to $CD$ through $D$ at $U$, and similarly for pair $(B,C)$ is $V$. Then $UV, AC, BD$ are concurrent.
Proof: Let perpendicular from $P$ to $AD$ and $BC$ be $K$ and $L$. Its easy to see the quadrilaterals $UKAB$,$UKCD$,$LUCD$,$LUAB$ are concyclic so it follows under an inversion centre $P$ and radius $\sqrt (PA)(PB)$, the point pairs $(A,B),(C,D),(U,K),(L,V)$ go to each other. Indeed $AC$ goes to the circumcircle of $PAC$, and similarly for $BD$. So if $AD$ intersect $BC$ at $Z$, it is easy to see $K$ and $L$ lie on the circle diameter $PZ$. So it reduces to prove the three following circles pass thru a point: $PZ$ diameter, $PAC$,$PBD$. Now i suggest you redraw the diagram to suit the problem right now. Suppose the circumcircles of $PAC$ and $PBD$ intersect at $F$. If $O$ is the circumcentre of circle through $A,B,C,D$, from IMO $1985/5$ we have $OF$ is perpendicular to $FP$. By Lemma 1, we have $FZ$ is perpendicular to $FP$ since $F,P,Y$ are collinear by Radical Axis Thm, where $Y$ is $AC$ intersect $BD$. So thus $ZF$ is perpendicular to $FP$ as required, proving that those three circles are indeed passing through one point, finishing of this proof.
Now to the problem:
it suffices to prove $CG$ is perpendicular to $MM'$. indeed, by Lemma 1, $CG$ is perpendicular to $ON$ where $N$ is the intersection of $QP'$ and $PQ'$. It's easy to see $N$ is the midpoint of $MM'$. Indeed, by Lemma 2, $M,M', N$ are collinear. So thus $CG$ is perpendicular to $MM'$ as required.

[quote="vinoth_90_2004"]
 It's easy to see $N$ is the midpoint of $MM'$. [/quote]
Actually sorry that was a typo, i meant to say $O$ is the midpoint of $MM'$.
Just a comment: about the hexagon $PP'QQ'RR'$ it suffices to prove that hexagon's Pascal Line passes through the centre using Lemma $1$ repeatedly, and we actually prove the Pascal line of that hexagon is indeed $MM'$. So, with a little more work, it's not hard to see that a Pascal line of a hexagon passes through the centre iff the lines perpendicular to every alternate sides through every second vertice, those three lines are concurrent. It's an interesting result I didn't know prior to this problem.
Anyway this is a cool problem.
Another comment: about Lemma $2$, is it true that the quadrilateral $MM'PZ$ is an orthocentric quadrilateral? i think it's quite likely, given that $MM'$ is perpendicular to $PZ$, from applying Lemma $1$ to Lemma $2$. I can't see why it's false and can't prove it either.

[quote]Lemma : In a cyclic quadrilateral ,  intersects  at ,  intersects  at ,  intersects  at . Then the orthocentre of  is the centre of the circumcircle.
Proof: This result is very well-known, and the shortest proof i know is still complicated. Basically it uses basic polar results ( lies on the polar of  iff  lies on the polar of ) and Brianchon's theorem with a degenerate hexagon - an inscriptible quadrilateral - to glue the proof together.
[/quote]
As far as I know this is Brocard's theorem and it has a simple (elementary) proof


In LaTeX, how do I put a custom equation number to the left of an equation or eqnarray? Say I want it to show (*) or (3').

I'm not sure if you're question is about the left side or changing the labels.  To put labels on the left side, use the option leqno, i.e.

[code]\documentclass[12pt,leqno]{article}[/code]

To vary the labels, try this example:
[code]\begin{align}
x_1 &= y_1 \\
x_2 &= y_2 \tag{$*$} \\
x_3 &= y_3 \tag{$2'$}
\end{align}[/code]


I've been writing REALLY slowly so it's neat, and why spend time learning latex when I could be doing a problem?  I hardly can fit in the time to solve the USAMTS problems, but I really want to get a commended solution or two.

Edit:  Do I write out the problem like this?

Problem:
fhsdjkfs
sdfdsgh

Solution:
dfklghil
dfiojklld

Should I write "USAMTS Problem x/x/19" at the top of each new page of my solution?

You do not need to restate the question.
Under the "Forms" tab in "My USAMTS" on http://www.usamts.org , there is a Solution Form. There are spaces to put the round number and problem number.

[quote="LawOfSigns"]I've been writing REALLY slowly so it's neat, and why spend time learning latex when I could be doing a problem?[/quote]
Handwriting is just fine, so long as we graders can read it. If you are going to fax your handwritten solution, use dark ink so that the fax image is good and clear.

[quote]Edit:  Do I write out the problem like this?[/quote]
No particular format is required. What is important is making your reasoning clear.

[quote]Should I write "USAMTS Problem x/x/19" at the top of each new page of my solution?[/quote]
Putting the problem number and the page of the solution, such as "Problem 5/1/19, Page 1 of 2," on every page is a good idea in case the pages somehow get out of order, but it is not required.

Erin Schram
USAMTS grader


Two persons $A,\ B$ play a game many times repeatedly. The probability that $A$ wins for $B$ is $p$ regardless of the result so far. Let $q$ be the probability that $A$ wins at least one time before $B$ wins two times.Note that you may not consider a draw game. If $q=\frac{1}{2},$ then find the value of $p.$

[quote="Treething"]This basically means that B doesn't win two games in a row, so the probably of A winning is $\frac{\sqrt2-1}{\sqrt2}$.[/quote]

Okay, how the heck did you arrive to that number.

[hide="mathgeniuse^ln(x)"]

Well the original problem says that $q$ is the probability that A wins at least once before B wins twice.

Note, Treething translated that into the fact that $q$ is the probability that B doesn't win twice in a row.

So, $1-q$ is the probability that B wins twice in a row.

Probability that B wins one game = $\sqrt{1-q}$

Probability that A wins one game=  $1-\sqrt{1-q}$

I think this is what he means.[/hide]


geometry
vector
function
calculus
derivative
advanced fields
advanced fields unsolved

so this was a question on my differential geometry homework that i just turned in, and it is bothering me so much because i know that i didn't do it correctly. it's a very very easy question, i'm just having a huge mental block. it's not even really about differential geometry; this is more like review for differential geometry. anyhow, here's that stupid question:

Let [i]alpha(t)[/i] be a parametrized curve which does not pass through the origin.  If [i]alpha(t_0)[/i] is the point of the trace of [i]alpha[/i] closest to the origin and [i]alpha'(t_0)[/i] does not equal 0, show that the position vector [i]alpha(t_0)[/i] is orthogonal to [i]alpha'(t_0)[/i].

i messed with dot products for a good while but i am just being extremely dense right now. please help me. i neeeeeeeeeeeeeeeed to know how it's done or else it's going to bug me for like three days straight.

The function $\parallel \alpha(t)\parallel =\alpha(t)\cdot\alpha(t)$ attains its minimum at $t=t_{0}$. Since the dot product is a bilinear operation, the product rule for derivatives applies: $\frac{d}{dt}(\alpha(t)\cdot\alpha(t))=\alpha'(t)\cdot \alpha(t)+\alpha(t)\cdot \alpha'(t)=2\alpha(t)\cdot \alpha'(t)$. This vanishes at $t_{0}$.

it's always nice when you forget calculus I.

derivatives vanish at minima and maxima...................... boy, i'm pretty slow  :blush:

thanks!!!!!


How do I get the vert line to be full size, and how do you control it.

\textbf{Ax} = \left[\left{}
\begin{matrix}
   a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\
   a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\
   \vdots  & \: & \ddots & \vdots \\
   a_{n,1} & a_{n,2} & \cdots & a_{n,n}
\end{matrix}
\right{\Biggr\vert}
\begin{matrix}
   \, & x_1 & \, \\
   \, & x_2 & \, \\
   \, & \vdots & \,\\
   \, & x_n
\end{matrix}
\right]

Thanks.

So how do you partition a matrix so that it has both horizontal and vertical partitions???

Have a matrix parted into quadrants???

Just add \hline wherever you want the horizontal line $ \textbf{Ax}= \left [ \begin{array}{cccc|c}a_{1,1}& a_{1,2}& \cdots & a_{1,n}& x_{1}\\ a_{2,1}& a_{2,2}& \cdots & a_{2,n}& x_{2}\\ \hline \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{n,1}& a_{n,2}& \cdots & a_{n,n}& x_{n}\end{array}\right]$
[code]\textbf{Ax} = \left [
\begin{array}{cccc|c}
a_{1,1} & a_{1,2} & \cdots & a_{1,n} & x_1 \\
a_{2,1} & a_{2,2} & \cdots & a_{2,n} & x_2 \\ \hline
\vdots & \vdots & \ddots & \vdots  & \vdots \\
a_{n,1} & a_{n,2} & \cdots & a_{n,n} & x_n
\end{array}
\right] [/code]

Thanks

Even one better

\textbf{Ax} = \left[\:\,
\begin{matrix}
   a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\
   a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \hline
   \vdots  & \: & \ddots & \vdots \\
   a_{n,1} & a_{n,2} & \cdots & a_{n,n}
\end{matrix}
\:\,\left |
\begin{matrix}
   \, & x_1 & \, \\
   \, & x_2 & \, \\
   \, & \vdots & \,\\
   \, & x_n & \,
\end{matrix}
\right{}
\right]

You don't need such complication. Replace \hline by \cline{1-4} $ \textbf{Ax}= \left [ \begin{array}{cccc|c}a_{1,1}& a_{1,2}& \cdots & a_{1,n}& x_{1}\\ a_{2,1}& a_{2,2}& \cdots & a_{2,n}& x_{2}\\ \cline{1-4}\vdots & \vdots & \ddots & \vdots & \vdots \\ a_{n,1}& a_{n,2}& \cdots & a_{n,n}& x_{n}\end{array}\right]$
[code]\textbf{Ax} = \left [
\begin{array}{cccc|c}
a_{1,1} & a_{1,2} & \cdots & a_{1,n} & x_1 \\
a_{2,1} & a_{2,2} & \cdots & a_{2,n} & x_2 \\ \cline{1-4}
\vdots & \vdots & \ddots & \vdots  & \vdots \\
a_{n,1} & a_{n,2} & \cdots & a_{n,n} & x_n
\end{array}
\right][/code]


Hey guys, here is an AIME level problem I find interesting. Answer is bw 000 and 999.

In a round robin tournament, a cyclic 3-set occurs occasionally; that is, a set {a,b,c} of three teams where a beats, b beats c, and c beats a. If 23 teams play a round robin tournament (w/o ties), what is the largest number of cyclic 3-sets that can occur?

Try this problem and happy holidays.  ;)

By Round-Robin you mean every team plays every other team exactly once, right?

This is a very nice problem!

[hide]We analyze the 8 possible outcomes of games played within a group of three teams. This group is a cyclic 3-set if and only if there is no team that won both of its games or lost both of its games in such as set. Moreover, all of the non-cyclic 3-set outcomes involve one team that won both games, one team that lost both games, and a team that won a game and lost a game.  Call an instance where within such a three set, a team either won both of lost both of its games a [i]fluke[/i].  All non-cyclic 3-sets have exactly two flukes.

Now, we consider the number of flukes that involve some given team T. Suppose that T lost some $x$ of its games and won the remaining $22-x$ games. Flukes for team T include $\frac{x(x-1)}{2}$ instances where T lost two games and $\frac{(22-x)(21-x)}{2}$ instances where T won both games. Therefore, T is involved in

$\frac{x(x-1)}{2} + \frac{(22-x)(21-x)}{2} = x^2-22x+231 = \left(x-11\right)^2 + 110$

flukes. It is clear that the number of flukes involving T is minimized at 110 if and only if T wins half of its games and loses the other half of its games. It is clear that there are at least $23\cdot 110$ flukes in any distribution of games, meaning that there are at least $23\cdot 55$ non-cyclic 3-sets. Since there are ${23 \choose 3} = \frac{1}{6} \cdot 23 \cdot 22 \cdot 21 = 23 \cdot 77$ 3-sets, we conclude that there are at most $77 \cdot 23 - 55 \cdot 23 = 22 \cdot 23 = 506$ cyclic 3-sets. Furthermore, this occurs if and only if each teams wins 11 games and loses 11 games. Obviously, this is possible.

What's interesting is that any distribution of 11 wins and 11 loses for each team obtains this answer, not just symmetric cases.[/hide]

Excellent solution, Mildorf. 506 is the correct answer. I'm glad someone has actually solved it.  :)

I did it a different way, but I was wondering if my solution looked valid.

[hide] Given a triple of three teams $a, b, c$, the triple fails to be a cyclic 3-set iff one of the three teams beats the other two. Let the 23 teams be $a_1, a_2, \ldots, a_n$.

Consider team $a_1$. Let the number of teams $a_1$ beat be $k_1$. Then if $a_i, a_j$ are both teams beaten by $a_1$, the triple $(a_1, a_i, a_j)$ will not be a cyclic 3-set. Therefore, there $\displaystyle k_1\choose 2$ of these cases.

If we continue with this method of defining the number of teams beaten by team $a_i$ being equal to $k_i$, we see that there are a total of \[\displaystyle{k_1 \choose 2} + {k_2 \choose 2} + \ldots + {k_{23} \choose 23} = \frac{k_1^2 + k_2^2 + \ldots + k_{23}^2}{2} - \frac{k_1 + k_2 + \ldots + k_{23}}{2}\] triples that are not cyclic 3-sets. To maximize the number of cyclic 3-sets, we must minimize the value of the above expression.

Note $k_1 + k_2 + \ldots +  k_{23} = 253$, since there are a total of $\displaystyle {23\choose 2} = 253$ games played. Therefore, minimizing $\displaystyle {k_1 \choose 2} + {k_2 \choose 2} + \ldots + {k_{23} \choose 23} = \frac{k_1^2 + k_2^2 + \ldots + k_{23}^2}{2} - \frac{k_1 + k_2 + \ldots + k_{23}}{2}$ is equivalent to minimizing $k_1^2 + k_2^2 + \ldots + k_{23}^2$. Since the sum of the $k_i$'s is 253, minimizing the sum of the squares of the $k_i$'s is done by setting them all equal. This fact is not hard to prove, and can be done in a variety of ways. From here, we see that each $k_i = 11$. Then, $\displaystyle {k_1 \choose 2} + {k_2 \choose 2} + \ldots + {k_{23} \choose 2}$ equals $\displaystyle 23 \cdot {11\choose 2} = 1265$. Notice there are a total of $\displaystyle {23 \choose 3} = 1771$ possible triples, so our answer is $1771 - 1265 = {\bf 506}$.
[/hide]

I believe this is sufficient, since each triple that is not a cyclic 3-set must fall into the category of having one team beat the other two, so all remaining triples must be cyclic 3-sets. Can anyone read through my work and confirm this?

If my work is correct, it generalizes for any situation naturally.

Edit: Looking over Mildorf's solution again, I notice that our approaches really aren't significantly different. However, his looks a lot less scary. Kudos to the poser of this problem, as well as Mildorf for his excellent solution.

Yeah, this is a pretty nice problem.  I remember seeing it somewhere; I think it was [i]A Path to Combinatorics for Undergraduates[/i] (Andreescu/Feng) and the solution they gave there is basically the same.

It seems like there should be a solution involving Hamiltonian cycles and digraph...I tried and couldn't find such a solution but it would be nice if somebody could.

[quote="darktreb"]\[\displaystyle{k_1 \choose 2} + {k_2 \choose 2} + \ldots + {k_{23} \choose 23} = \frac{k_1^2 + k_2^2 + \ldots + k_{23}^2}{2} + \frac{k_1 + k_2 + \ldots + k_{23}}{2}\] [/quote]

You mean

\[\displaystyle{k_1 \choose 2} + {k_2 \choose 2} + \ldots + {k_{23} \choose \mathbf{2}} = \frac{k_1^2 + k_2^2 + \ldots + k_{23}^2}{2} - \frac{k_1 + k_2 + \ldots + k_{23}}{2}\]

but your solution looks good to me!

Yes that's what I mean :). Thanks for pointing that out - the woes of fast latex :(. Let me edit it.


modular arithmetic
number theory unsolved
number theory

Need help with an exercise.

Prove tha 9^1980 - 7^1980 is congruent with 0mod130.

Can you help?

hello, prove that $ 9^6\equiv 1 \mod 130$.
Sonnhard.

hello, and prove that $ 7^{12}\equiv 1 \mod 130$.
Sonnhard.

ok... If we prove that 9^1980=1mod130 and 7^1980=1mod130 then 9^1980 - 7^1980 = 0mod130.

I took 9^1980 = (9^6)^330 = 1mod130.
We now that (a^k)^d=1modn, so it is enough to so tha 9^6=1mod130. The last one is easy to prove with a calculator since 9^6mod130 = 1mod130 = 1 (here "=" is equivalence)

We do the same thing with 7^1980.

And after that we see that 9^6-7^12 = 0mod130.

 ("=":congruence)

Am i right? Or i miss something?

$ \lambda(130)\equal{}12$, and, very conveniently, $ 12|1980$, so your problem reduces to $ 9^0\minus{}7^0\equiv 0\pmod{130}$, which is obviously true.

$ \lambda(n)$ is Carmichael's lambda function.


probability
geometry
combinatorics unsolved
combinatorics

A horizontal flat square board 6 inches width is ruled with a grid of fine lines, 2 inches apart and has a vertical rim like a carrom board, all around the edge. If a coin of diameter 2/3 inches is tossed on to the board. What is the probability that it rests without crossing a line?

The options given are:
A) 0.5625
B) 0.6525
C) 0.6255
D) 0.5562
E) None of these

Consider where the centre of the coin can land.

For each square (2x2) it's centre must be at least $ \frac{1}{3}$ from each side, this gives an area of

$ (2\minus{}\frac{2}{3})^2 \equal{} \frac{16}9$ per square.

There are 9 squares which gives a 'safe' area of 16

Total area is

$ (6\minus{}\frac{2}{3})^2 \equal{} 256/9$

The probability is therefore

$ 16/ \frac{256}{9} \equal{} 9/16 \equal{} 0.5625$

[quote="SimonM"]Consider where the centre of the coin can land.

For each square (2x2) it's centre must be at least $ \frac {1}{3}$ from each side, this gives an area of

$ (2 \minus{} \frac {2}{3})^2 \equal{} \frac {16}9$ per square.

There are 9 squares which gives a 'safe' area of 16

Total area is

$ (6 \minus{} \frac {2}{3})^2 \equal{} 256/9$

The probability is therefore

$ 16/ \frac {256}{9} \equal{} 9/16 \equal{} 0.5625$[/quote]

Thanks! :) The $ (6 \minus{} \frac {2}{3})^2$ part was difficult for me to understand at first. Now I realise that it is because of the rim that surrounds the edge, isn't that right?


Find all integers a; b; c; d satisfying the following relations :
(i) 1 <= a <= b <= c <= d;
(ii) ab + cd = a + b + c + d + 3.
where <= stands for smaller than or = to....i was just able to get the ordered pairs (2,2,2,5) and (2,2,3,3) which satisfy this...by trial and error..can any1 plz post a robust and mathematical solution to this???

write it as $ (a\minus{}1)(b\minus{}1) \plus{} (c\minus{}1)(d\minus{}1) \equal{} 5$
then it is easier to go through the possible cases.

[quote="sahilsharma94"]thanks for giving a generalised Eq[/quote]
Thats not a generalised equation......
That IS the only way u can answer this question easily..

ya whatever!!!!!how does the terminology matter????


mewto55555 started something called CTC to control the amount of tournies being created.
I'll post a link to site later

edit: here's the site
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214516[/url]

yeah! CTC is an awesome idea! [color=darkblue][size=150]VOTE MEWTO 4 PREZ!![/size][/color]

Oh come on, I even made the font extra large...

[quote="fedja"]And most importantly,
[size=150]
[b]
3) Abstain from questions about opinions on things that are clearly of little or no interest to other people.
[/b]
[/size]
[/quote]

Locked.


calculus
integration
trigonometry
calculus computations

Im having trouble with this one, I know the answer but I cant get to it.

The problem is the Indefinite integral of

  [(e^ :sqrt: r)/ :sqrt: r]dr

 I know that the integral of e^u is (e^u)du
 but Im getting lost somewhere, what do I do with the denominator?

Hmm... It is a standart substitution.
\[\int \frac{e^{\sqrt{x}}}{\sqrt{x}}dx=\int 2e^{\sqrt{x}}d\sqrt{x}=2e^{\sqrt{x}}.\]

Im a little confused, what exactly are you substituting?
 e^ :sqrt: x
?

I don't know how it is eactly sounds in english. I think I should say "method" instead of "substitution".

Here's what Myth did: let $u=\sqrt{x}.$ Then $du = \frac{dx}{2\sqrt{x}}.$

That's exactly what he meant when he replaced $\frac{dx}{\sqrt{x}}$ with $2d\sqrt{x}.$

[quote="adidasty"]yeah thats right. you really dont have to put $du$ in if you dont want to. its a differential so as long as it works, it works. like if i say (just for example):

$\int \sin{(\sqrt{x})}dx$

If I want to make the u-substitution as $u=\sqrt{x}$ just to do it I can say $\int\sin{u}du$ OR i can say $\int\sin{(\sqrt{x})}d\sqrt{x}$[/quote] :?

$\int \frac{e^{\sqrt{r}}}{\sqrt{r}}dr=2\int e^{\sqrt{r}}(\sqrt{r})'dr=2e^{\sqrt{r}}+Const.$


Do you know where I could find a program or website that lets you enter in data points and functions and graphs them?

I'm thinking of something similar to a ti graphing calculator option that lets you type in data into two separate lists and can determine a parabola or cubic that roughly fits the data.

Here is an online grapher which takes in equations (single variable):

[url]http://www.tutor-homework.com/grapher.html[/url]

Mathematica has a fairly powerful graphing mechanism that's a little more versatile than some online ones. If you don't have Mathematica, you can request a trial version: http://www.wolfram.com/products/mathematica/experience/request.cgi.  :)

If you have a TI-83+ series calculator, press [STAT].  Then choose 1.EDIT.  Then, enter the data into whatever lists you want (the defaults are L1 to L6).

Now, return to the homescreen.  Press [STAT], then [LEFT], then select 5 for quadratic reg, or whatever you want.  Then, type L1, L2 (if those are your list names) and hit enter.

Just an addendum to that.
If you want to be able to graph it, you need to do everything B-flat said, then after typing "L1, L2"

add the Y1 (or Y2, or whatever) variable to the list. This automatically stores the values of the constants generated by the regression into the y1 graphing variable.

So it might look something like
QuadReg(L1, L2, Y1)


Hello, this is just a tip I wanted to post. See the following diagram.[asy]pair A,B,C,D,E,F;
D = (0,0); E = (-1,1); F = (0,2);
C = (sqrt(2) ,0); B = sqrt(2)*dir(35) + C;
A = sqrt(2)*dir(65)+B;
draw(F--E--D--C--B--A--F--D);
draw(F--C--B--F);
label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SE);
label("$D$",D,S); label("$E$",E,W); label("$F$",F,N);[/asy]

MATHCOUNTS has a few questions like this I believe. Instead of wasting time trying go through the repeated application of Pythagorean Theorem, just make a generalized formula for it beforehand. This way, when you take a test (whether it be practice or real), use your formula!

I would appreciate it if y'all posted some advice for those new to MATHCOUNTS (btw, I'm not experienced that much).

This will save time, reduce careless mistakes, improve your understanding, and maybe even make your teacher proud. Generalize problems known to usually be on MATHCOUNTS tests (I mean similar problems; no problem is used twice); this will save a lot of time so you can check your work (when I was in MATHCOUNTS, I never had time to check my work which is why I made so many careless mistakes).

Well, you're not going to have a formula for everything, are you? :wink:

That's where the [b]problem solving aspect[/b] comes in handy. :P  :)

He means like really frequent stuff and it takes time to prove things.

While I admit that formulas are useful for speed in MC, I might also add that just studying a lot of formulas and not working on problem solving enough can lead to forgetting which formula is which, using the formulas as a crutch to problem solving (i.e. nothing but formulas in one's arsenal), or both. So even though such problems as the one in the OP do occur occasionally, it's not worth it to memorize a formula for it since the problem's so specific. If it's something that has a very wide range of uses or reduces work [b]substantially[/b], then it's a good idea to know it.


[quote="pvthuan"]Zhaobin (from China) is one active member of this box. I have not seen Zhaobin for ages in this web since 22 Feb.

Do you know where he is now? Is there anything wrong with him?[/quote]

Actually you are right. His last post was on Thu Feb 23, 2006 3:40 am. It is very sad, that he didn't enter this forum since this time, because he helped us here with many problems...

Let's hope he is OK and he will return back..

conceive

[quote="pvthuan"]Zhaobin (from China) is one active member of this box. I have not seen Zhaobin for ages in this web since 22 Feb.

Do you know where he is now? Is there anything wrong with him?[/quote]

  Well, I suggest that moderate should put this one to a stickey so that when Zhaobin comes back, he sees our messages. Since there always many posts every day, this post probably will be lost in the tons of math problems. Isn't it? :roll:

Yes, [b]Gibbenergy[/b], that's a good idea. :)

And, I was also wondering about that... I really hope everything is fine with him, and it is just that he has too much homework to do, or that his internet connection is temporarily down.
So, yeah, I hope to see him back soon!

:thumbup:

Proof of the vasc's problem by using zhaobin's method. (Dedicated to Zhaobin.)
Let $a,$ $b,$ $c$ are non - negative real numbers.
Prove that $(a+b+c)^5\geq12(a^4b+a^4c+b^4a+b^4c+c^4a+c^4b).$
Let $a\leq b\leq c$ and $a+b+c=1.$ Then our inequality equivalent to $a^4-a^5+b^4-b^5+c^4-c^5\leq\frac{1}{12}.$
Let $f(x)=x^4-x^5.$
Hence, $f''(x)=12x^2-20x^3=-20x^2(x-\frac{3}{5}).$
Id est, $f$ is convex on $[0,\frac{3}{5}]$ and $f$ is cocave on $[\frac{3}{5},1].$
Case 1. $a\leq b\leq c\leq\frac{3}{5}.$
Since $f$ is convex on $[0,\frac{3}{5}]$ then $f(a)+f(b)+f(c)\leq f(0)+f(\frac{2}{5})+f(\frac{3}{5})=\frac{42}{625}<\frac{1}{12}.$
Case 2. $a\leq b\leq\frac{3}{5}\leq c.$
Hence, $a+b<\frac{3}{5}.$
Hence, $f(a)+f(b)\leq f(0)+f(a+b)=f(1-c).$
Remain to prove that $f(1-c)+f(c)\leq\frac{1}{12}.$
But $f(1-c)+f(c)\leq\frac{1}{12}\Leftrightarrow(1-c)^4-(1-c)^5+c^4-c^5\leq\frac{1}{12}\Leftrightarrow(6c^2-6c+1)^2\geq0.$ :)

Well guys, he has returned!! We can see his posts here:

http://www.mathlinks.ro/Forum/viewtopic.php?p=509775#p509775

 :)

hi,everyone:
     Now I am very exciting,moving,happy :lol:  and surprised.And also I am so sorry :oops: .Sorry to all mathlinkers for my going away without saying goodbye.
     In the beginning of this term I planned to learn my own course in the university with my whole heart(I am a student in Mechanical institute) and other annoying things,so I planned not to go to mathlinks.Just learn  My specialty first.But I find this plan is too wrong,is too absolute.Because I like maths. :) And I can come to mathlink sometimes,and also learn my specialty well.
    So I am back now,and thank you to everyone.And sorry again.

Can you solve the general problem :
Let $a,b,c$ be positive numbers have sum 3. Find the maximum value of $P=a^{k}(b+c)+b^{k}(c+a)+c^{k}(a+b)$, where $k$ is a real number ?

@ ehku :we already have this problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=30732&highlight=maximum]here[/url] :wink:
together with a solution.

[quote="arqady"]Proof of the vasc's problem by using zhaobin's method. (Dedicated to Zhaobin.)
Let $a,$ $b,$ $c$ are non - negative real numbers.
Prove that $(a+b+c)^5\geq12(a^4b+a^4c+b^4a+b^4c+c^4a+c^4b).$
Let $a\leq b\leq c$ and $a+b+c=1.$ Then our inequality equivalent to $a^4-a^5+b^4-b^5+c^4-c^5\leq\frac{1}{12}.$
Let $f(x)=x^4-x^5.$
Hence, $f''(x)=12x^2-20x^3=-20x^2(x-\frac{3}{5}).$
Id est, $f$ is convex on $[0,\frac{3}{5}]$ and $f$ is cocave on $[\frac{3}{5},1].$
Case 1. $a\leq b\leq c\leq\frac{3}{5}.$
Since $f$ is convex on $[0,\frac{3}{5}]$ then $f(a)+f(b)+f(c)\leq f(0)+f(\frac{2}{5})+f(\frac{3}{5})=\frac{42}{625}<\frac{1}{12}.$
Case 2.[b] $a\leq b\leq\frac{3}{5}\leq c.$
Hence, $a+b<\frac{3}{5}.$[/b]
Hence, $f(a)+f(b)\leq f(0)+f(a+b)=f(1-c).$
Remain to prove that $f(1-c)+f(c)\leq\frac{1}{12}.$
But $f(1-c)+f(c)\leq\frac{1}{12}\Leftrightarrow(1-c)^4-(1-c)^5+c^4-c^5\leq\frac{1}{12}\Leftrightarrow(6c^2-6c+1)^2\geq0.$ :)[/quote]
Why? $a\leq b\leq\frac{3}{5}\leq c.$
Hence, $a+b<\frac{3}{5}.$
If $a=b=\frac{3}{5}.$ => $a+b=\frac{6}{5}>\frac{3}{5}.$

[quote="NYY"]
Why? $a\leq b\leq\frac{3}{5}\leq c.$
Hence, $a+b<\frac{3}{5}.$
[/quote]
Because $a+b=1-c\leq1-\frac{3}{5}=\frac{2}{5}<\frac{3}{5}$.

[quote="arkanm"]Can anyone post 'zhaobin's method'?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=64933


Show that $e=\sum^{\infty}_{n=0} \frac{1}{n!}$ is irrational.

Here's an outline of a proof :

We establish that for any natural integer $n, e = \sum_{k=0}^n \frac 1{k!} + \int_0^1 \frac {(1-t)^n}{n!}e^t dt,$ then that [i]for any[/i] positive integer $n, 0 < e - \sum_{k=0}^n \frac 1{k!} < \frac 3{(n+1)!}.$ We then proceed by contradiction.

[b]Historical remark :[/b] Hermite has proved in 1873 that $e$ is transcendental. That's historically the first "real" number for which we managed to prove the transcendental character.

Sure. :)

We will show that first relation by induction.
The base case is easy and left to the reader.
Let $n \geq 0.$ Let us assume that $e = \sum_{k=0}^n \frac 1{k!} + \int_0^1 \frac {(1-t)^n}{n!} e^t dt.$
An integration by parts yields :
$\int_0^1 \frac {(1-t)^n}{n!} e^t dt = \frac 1{(n+1)!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt,$
and so $e = \sum_{k=0}^n \frac 1{(n+1)!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt = \sum_{k=0}^{n+1} \frac 1{k!} + \int_0^1 \frac {(1-t)^{n+1}}{(n+1)!} e^t dt.$
Thus our result is proved by induction.
Let $n$ be a positive integer. From the above,
$0 < e - \sum_{k=0}^n \frac 1{k!} = \int_0^1 \frac {(1-t)^n}{n!}e^t dt < e \int_0^1 \frac {(1-t)^n}{n!}dt = \frac e{(n+1)!} < \frac 3{(n+1)!}.$
Let us then assume for sake of contradiction that $e$ is rational. Then, there exist $(a, b) \in \mathbb{N}^2 / e = \frac ab.$
Let $n$ be an arbitrary positive integer. From the above, we have $0 < \frac ab - \sum_{k=0}^n \frac 1{k!} < \frac 3{(n+1)!},$ which rewrites $0 < an! - b\sum_{k=0}^n \frac {n!}{k!} < \frac {3b}{n+1}.$ Especially, for $n = 3b,$ we have $0 < a(3b)! - b\sum_{k=0}^{3b} \frac {(3b)!}{k!} < \frac {3b}{3b+1} < 1.$
But this is impossible since $an! - b \sum_{k=0}^{3b} \frac {(3b)!}{k!}$ is an integer.
Hence, $e$ is irrational.

However, if we need just the [b]irrationality[/b] there is a completely elementary proof.
Assume $ e \equal{} \frac{p}{q}$ for some $ p, q \in \mathbb{N}$ with $ (p, q) \equal{} 1$ and $ q \ge 2$ - it is trivial to check, by an easy estimation, that $ e \in (2, 3)$.
This leads to
$ p \cdot (q \minus{} 1)! \equal{} \sum_{k\equal{}1}^q \prod_{j \equal{} k \plus{} 1}^q j \plus{} \sum_{k \ge q \plus{} 1} \frac{q!}{k!}$,
thus $ \sum_{k \ge q \plus{} 1} \frac{q!}{k!} > 0$ is an integer. Hence let's keep in mind the relation $ \sum_{k \ge q \plus{} 1} \frac{q!}{k!} \ge 1$.
On the other hand,
$ \sum_{k \ge q \plus{} 1} \frac{q!}{k!} \equal{} \sum_{t \ge 1} \frac{1}{(q \plus{} 1) \cdots (q \plus{} t)} \le \frac{1}{q \plus{} 1} \plus{} \sum_{t \ge 2} \frac{1}{(q \plus{} t \minus{} 1)(q \plus{} t)}$,
and by reducing terms,
$ \frac{1}{q \plus{} 1} \plus{} \sum_{t \ge 2} \left( \frac{1}{q \plus{} t \minus{} 1} \minus{} \frac{1}{q \plus{} t} \right) \equal{} \frac{2}{q \plus{} 1} \le \frac{2}{3} < 1$,
a contradiction.


calculus
derivative
integration
function
calculus computations

I believe that something not go so well. If we set $ G(x)$=$ f(x) * h(x)$, where $ f(x) \equal{} x^{3}$ and $ h(x) \equal{} \int\limits_{{x^{2}}}^{1}{\left({u \minus{} 1}\right)du}$ from the laws of derivation for a product of functions I found a different solution.

The answer mathxl has can be achieved in two ways:
1. By computing the given integral and then differentiate the result using the product rule of differentiation [which mathxl did].

2. By differentiating and then integrate the part with integral. Differentiation in this second approach requires the product rule and also some version of the Fundamental Theorem of Calculus.

Given $ G(x) \equal{} x^{3}*\int_{x^{2}}^{1}(u \minus{} 1)\,du$, one has
$ G'(x) \equal{} 3x^2 \int_{x^{2}}^{1}(u \minus{} 1)\,du \plus{} x^3\frac {d}{dx}\left(\int_{x^{2}}^{1}(u \minus{} 1)\,du\right)$
        $ \equal{} 3x^2( \minus{} \frac {1}{2} \plus{} x^2 \minus{} \frac {x^4}{2}) \plus{} x^3( \minus{} (x^2 \minus{} 1)2x)$
       $ \equal{} \minus{} \frac {3}{2}x^2 \plus{} 3x^4 \minus{} \frac {3}{2}x^6 \minus{} 2x^6 \plus{} 2x^4$
      $ \equal{} 5x^4 \minus{} \frac {7}{2}x^6 \minus{} \frac {3}{2}x^2$

yes you are right. I forgot a derivative. I am sorry. :(


What is the minimum number of triangles into which a convex decagon can be divided by drawing some of its diagonals?

[hide]Draw only diagonals from one point.
10 sides in convex polygon= 8 triangles created[/hide]


inequalities
trigonometry
trig identities
Law of Cosines

Show that for any triangle $ABC$ , the following is true...

$ a^2 + b^2 + c^2 > \sqrt{3} max ( |a^2 -b^2 |, |b^2 -c^2 | , |c^2 -a^2 |) $
where $ a, b, c $ are the sides...

We suppose that $a, b, c > 0$ and $c < a < b$ so the $max$ is $|b^2 - c^2|$

For the law of cosines we have $a^2 = b^2 + c^2 - 2bc\cos \alpha$ since
$2((b + c)^2 - 2bc) - 2bc\cos \alpha > \sqrt{3}(b^2 - c^2)$

$-2bc(2 + \cos \alpha) > \sqrt{3}(b^2 - c^2) -2(b + c)^2$

$\cos \alpha < \frac{b^2(2 - \sqrt{3}) + c^2(2 + \sqrt{3})}{2bc}$

$\cos \alpha < \frac{b}{c}(\frac{2 - \sqrt{3}}{2}) + \frac{c}{b}(\frac{2 + \sqrt{3}}{2})$

We have that $\alpha \in ]0, \frac{\pi}{2}[$ so its $\cos$ is $> 0$
and we have that $\frac{b}{c} > \cos \alpha$

If we put $\frac{b}{c} = \cos \alpha$, as we have in a rectangular triangle, we obtain
$\cos^2\alpha < \frac{2 + \sqrt{3}}{\sqrt{3}}$ that is always true.

In the other cases the fact that $\frac{b}{c} > \cos \alpha$ makes the inequality stronger.


Prove or disprove that :
$ x^{10}\plus{}x^5\plus{}1\equiv 0 \pmod{x^7\plus{}x^5\plus{}1}$

[quote="Obel1x"]Prove or disprove that :
$ x^{10} \plus{} x^5 \plus{} 1\equiv 0 \pmod{x^7 \plus{} x^5 \plus{} 1}$[/quote]

Just put $ x\equal{}2$ we know it's wrong. But it holds for $ x\equal{}\minus{}1,0,1$ :rotfl:

[quote="stephencheng"][quote="Obel1x"]Prove or disprove that :
$ x^{10} \plus{} x^5 \plus{} 1\equiv 0 \pmod{x^7 \plus{} x^5 \plus{} 1}$[/quote]

Just put $ x \equal{} 2$ we know it's wrong. But it holds for $ x \equal{} \minus{} 1,0,1$ :rotfl:[/quote]

I think you didn't understand.The problem is when $ x^{10}\plus{}x^{5}\plus{}1$ is divided by $ x^{7}\plus{}x^{5}\plus{}1$ the roof is equal to 0 or isn't equal to 0,(you have to prove this) !

If the statement is true over $ \mathbb{Q}[x]$, it's true over $ \mathbb{Z}[x]$.  In other words, the quotient, if it exists, is also an integer polynomial, so has an integer value at $ x \equal{} 2$.


Hi,

I realise this is a common question on these forums but I really just can't get my head around it. I'm talking about this:
[img]http://xs100.xs.to/xs100/06193/definition.jpg[/img]
Do you randomly choose a value for epsilon bigger than 0 and work from there? What are you actually trying to prove? The example I have in my textbook doesn't help me a great deal (prove that lim (x->1) 3x-1 = 2).

Thanks alot,
Nick

As far as I know about epsilon-delta proofs is that you arbitrarily pick $\epsilon$ and just show that for any $\epsilon$ you pick, no matter how small, you can find a $\delta$ satisfying the property described. For example, to show that

$\lim_{x \to 1} (3x-1) = 2$,

we have $|3x-1-2| < \epsilon \Rightarrow |3x-3| < \epsilon \Rightarrow |x-1| < \frac{\epsilon}{3}$. Thus if we choose $\delta = \frac{\epsilon}{3}$ (and we can obviously choose such a $\delta > 0$ for any $\epsilon > 0$), we will always have

$0 < |x-1| < \delta$,

which means $\lim_{x \to 1} (3x-1) = 2$.

I think you just have to show that $\delta$ always exists for any $\epsilon$.

[quote="JBL"]Your presentation is backwards.  That's how we figure out what $\delta$ should be, but when we present our proof, it has to be as follows:

If $|x - 1| < \delta = \frac{\varepsilon}3$ then $|(3x - 1) - 2| < \varepsilon$, so for each $\varepsilon$ we have the required $\delta$ (namely $\frac\varepsilon3$.[/quote]

Oh, whoops. Good point - I should learn how to do epsilon-delta proofs again...

I mean, most of the time it won't matter (since everything will be reversible, as in this case), but for explaining it to someone, it's much more helpful if he can see what the final product should look like.


The positive integers $ x$ and $ y$ satisfy the equations $ x^{4}\minus{}y^{4} \equal{} 2009$ and $ x^{2}\plus{}y^{2} \equal{} 49$. Find the value of $ y$.

$ x^4\minus{}y^4\equal{}2009$

$ \therefore\;(x^2\plus{}y^2)(x^2\minus{}y^2)\equal{}2009$

$ \therefore\;49(x^2\minus{}y^2)\equal{}2009$

$ \therefore\;x^2\minus{}y^2\equal{}41$

$ \therefore\;\begin{cases}x^2\plus{}y^2\equal{}49\\ x^2\minus{}y^2\equal{}41\end{cases}\;\Rightarrow\;2y^2\equal{}8$

$ \fbox{y\equal{}2}$

But x is proven to be an irrational number?

[hide="Why"]$ (x\plus{}y)^2\minus{}2xy \equal{} 49 \Leftrightarrow (x\plus{}y)(x\plus{}y\plus{}x\minus{}y)\minus{}2xy \equal{} 90 \Leftrightarrow x^2 \equal{} 45$ [/hide]

all it is is just that the wording wasn't quite correct; they don't have to be integers. y=2 works.


linear algebra
matrix
linear algebra unsolved

Let $A,B \in \mathbb{R}^{n\times n}$ such that $AB+B+A=0$. Prove that $AB=BA$.

we have $(A+I)(B+I)=A+B+AB+I =0+I=I$
then $A+I$and $B+I$ are inverses of each other
then $(A+I)(B+I)=(B+I)(A+I)$
then $AB+A+B+I=BA+B+A+I$
$AB=BA$

[u][b]My solution:[/b][/u]
We have: $AB+B+A=0\Leftrightarrow AB+B+A+I=I;$
Where $I$ is the [b]identity matrix.[/b]
$\implies (A+I)B+A+I=I \Leftrightarrow (A+I)(B+I)=I$;  $(1)$
and $BA+B+A+I=B(A+I)+(A+I)=(B+I)(A+I)$;    $(2)$
[b][i]Note that[/i][/b]: the matrices $(A+I) $ and $(B+I)$ are [i][b]converses[/b][/i]. $(3)$
From $(1);(2)$ and $(3)$: $\implies (A+I)(B+I)=(B+I)(A+I) \implies \boxed{AB=BA}.$

Is there a reason to post an exactly identical solution again?

When I post my solution, I do not pay attention to his answers above. Perhaps, it occurred almost at the same time, my answer and his answers.
However, you should read this carefully, it is the difference. The overlap is normal, because the same solution to a problem, so instead of comment on other people's answers.
 Better, if [b]you can make a soluion for the problem above.[/b]. That will help everyone [b]have a look at various angles about the problem[/b], and how to use a polite thing you lack.
Perhaps, before you post a comment, [b]you should think carefully [/b]and to avoid the above.
Because, it shows you are [b]thinking simply[/b], and restrictions on behavior.
One more thing. The problem of completely simple, no matter what not difficult for everyone. Anyone can do it, maybe so. So again, I'd say that you should think carefully when you have to post comments, because this is a [b]problem of the Day[/b]. You do not seem to realize that.
That is the reason why I do not want to argue with you.Thanks.

[quote="tuan119"]When I post my solution, I do not pay attention to his answers above. Perhaps, it occurred almost at the same time, my answer and his answers.[/quote]

Well, I just think that your "same time" has a difference of "Jun 20, 2011, 10:47pm" minus "Oct 14, 2005, 3:44 pm" which is about five years and eight months according to my calendar system.  It's not about whether you could solve it.  I simply view that if you have solved the problem, you could spend a little time checking whether someone has done it the same way you have.  Then, posting the same solution won't be necessary.  That's all.  I didn't mean to be impolite, I was just extremely curious.

Best regards...

PS: I can't realize anything you have said except for a few sentences.  Your words are too lyrical for me to comprehend.

Oh, really good if you could have a solution here, instead of an act of a person, who compare as drunk.
And I have a big surprise, I found [b]a guy like that children[/b], maybe so. And so I do not want to talk to a boy. :D
[b]Give me an answer, if you can. [/b]
I'm waiting for here and I will waiting for your solution [b]all today[/b]. I stopped arguing with a boy here!
The End.
[b]Sincerely![/b]P/S: Sorry dear moderators!
But again, I repeat that I do not pay attention to his answers, and if you do not understand what I wrote, because it lyrical? I do not think so,  because you do not understand what you are writing. :( .

Go ahead and wait.  I won't be posting solutions.  What would I gain if I win the challenge?  Your acceptance?  Why do I have to worry about that then?

PS: Sorry dear moderators.  I was having a little too much fun.  This is the last reply concerning the mess I have started. :P

The same proof (towards $AB=BA$) works if $AB+B+A=N$, where $N$ is a nilpotent matrix that is known to commute with $A$ or $B$.

We also have that $AXB+A+B=0$ implies $AXB=BXA$ (being posted before).

I just can prove that $BXA=AXB$ if $X$ is invertible.
For this, multiply the given equality by $X$:
$AXBX+AX+BX=0$
Then, $(AX+I)(BX+I)=I$ and so, we get that $AXBX=BXAX$. If $X$ is invertible, we get that $AXB=BXA$.

Anyone can prove this in the case that $X$ isn't invertible?
Maybe using that for some small $t$, $X-tI$ is invertible.

Noting that $(A + I)(B+I) = A + AB + I + B = I$, we conclude that $(A+I)$ and $(B+I)$ are inverses of each other.  Thus $(B+I)(A+I) = BA + B + A +I = I$, which implies that $AB = BA.$


Let positive real number $ a,b,c$ satisfy $ abc\equal{}1$ .Prove that:
       $ \sqrt[3]{(1\plus{}a)(1\plus{}b)(1\plus{}c)}\geq \sqrt[4]{4(1\plus{}a\plus{}b\plus{}c)}$

[quote="math10"]Let positive real number $ a,b,c$ satisfy $ abc \equal{} 1$ .Prove that:
       $ \sqrt [3]{(1 \plus{} a)(1 \plus{} b)(1 \plus{} c)}\geq \sqrt [4]{4(1 \plus{} a \plus{} b \plus{} c)}$[/quote]
$ \Leftrightarrow (2\plus{}p\plus{}q)^4 \ge 64(1\plus{}p)^3$

$ q^2 \ge 3pr \equal{}3p$ , ($ p\ge 3$)

$ f(p)\equal{}(2\plus{}p\plus{}\sqrt {3p})^4 \minus{} 64(1\plus{}p)^3 \ge 0$


Let be $ d_1,d_2,...,d_n$ all the natural divisors of the number $ n$ such that $ 1\equal{}d_1<d_2<...<d_k\equal{}n$ . Find all the numbers $ n$ with $ k\ge 4$ such that :

$ n\equal{}d_1^2\plus{}d_2^2\plus{}d_3^2\plus{}d_4^2$

Suppose n is divisible by 4 (n=4m), then we have
 $ 4m\equal{}1^2\plus{}2^2\plus{}3^2\plus{}4^2$ or $ 4m\equal{}1^2\plus{}2^2\plus{}4^2\plus{}d_4^2$
Both of these possibilities are impossible:
$ 1^2\plus{}2^2\plus{}3^2\plus{}4^2\equal{}30$ - not divisible by 4
$ 1^2\plus{}2^2\plus{}4^2\plus{}d_4^2\equiv2(mod4)$ if $ d_4$ - odd number, $ 1^2\plus{}2^2\plus{}4^2\plus{}d_4^2\equiv1(mod4)$ if $ d_4$ - even number - contradiction.

Suppose n is not divisible by 2, then all divisors are not divisible by 2:
$ n\equal{}d_1^2\plus{}d_2^2\plus{}d_3^2\plus{}d_4^2\equiv1\plus{}1\plus{}1\plus{}1\equiv0(mod2)$ - contradiction.

So n is divisible by 2 and not divisible by 4 (n=4m+2)
$ 4m\plus{}2\equal{}1^2\plus{}2^2\plus{}d_3^2\plus{}d_4^2$
$ d_3^2\plus{}d_4^2\equiv1(mod4)$
One divisor is odd and the other is even.
We can easily prove that $ d_4\equal{}2d_3$.
$ n\equal{}1^2\plus{}2^2\plus{}d_3^2\plus{}4d_3^2\equal{}5(1\plus{}d_3^2)$ n is divisible by 5.
If n is divisible by 3, then $ d_3\equal{}3$,$ d_4\equal{}6>5$ - this case is wrong.
So $ d_3\equal{}5,d_4\equal{}10$
$ n\equal{}1^2\plus{}2^2\plus{}5^2\plus{}10^2\equal{}\boxed{130}$

Check: $ 130\equal{}2\cdot5\cdot13$
Smallest divisors - $ 1,2,5,10$ - correct!

Nice problem  :coolspeak:

Good work . Your solution is nice as well .

Cute.  Is there a group of order $ 130$ with four irreducible representations?


geometry
angle bisector
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle with altitude $ AH$. Its incircle touches the sides $ BC$, $ CA$, $ AB$ in $ D$, $ E$, $ F$. The angle bisector $ w_{\alpha}$ of $ \angle BAC$ meets $ DE$ and $ DF$ in $ X$ and $ Y$, respectively. Prove that $ BC$ bisects $ \angle XHY$

Let $ (I)$ be the incircle of $ \triangle ABC$. Let $ C_0\equiv ED\cap AB$. THen $ (C_0FBA)\equal{}\minus{}1\Longrightarrow (YC_0,YF,YB,YA)\equal{}\minus{}1$. But $ Y\in$ the perpendicular bisector of $ EF$, hence $ FA$ is the bisector of $ \angle FYE\Longrightarrow YB$ is the bisector of $ \angle FYC_0$, which implies that $ YB\bot YA$. With the same arguments, we also have $ CX\bot AX$. Now, since $ \angle AXC\equal{}\angle AHC\equal{}90^{\circ}\Longrightarrow AHXC$ is concyclic $ \Longrightarrow \angle CHX\equal{}\angle CAX\equiv \angle CAI\equal{}\angle BAI\equiv \angle BAY$. But $ \angle AHB\equal{}\angle AYB\equal{}90^{\circ}\Longrightarrow AYHB$ is concylic, which implies $ \angle BAY\equal{}\angle YHC$. Therefore, $ \angle CHX\equal{}\angle CHY$, which implies $ BC$ bisects $ \angle XHY$.
Our proof is completed then.

Yes, exactly the same to mine
just notice A,H,X,B  and A,Y,H,C are concyclic the problem is trivil

Dear Mathlinkers,
some facts appear regularly... like the perpendicularity of BX and CY with AI. This result from Lascases is well knowed...
see for example
http://perso.orange .fr /jl.ayme    vol. 4   An unlikely concurrence, revisited and generalized p. 4.
Sincerely
Jean-Louis


Let $ABCDEF$ a cyclic hexagon and circumcircble. The bisectors of the angles $B, D, F$ interrsect sides $AC, CE, EA$ in $M, N, P.$ Prove that the lines $AN, CP, EM$ are concurrent. ;)

A simply problem with applications of angle bisector theorem, Ceva, Brianchon and Ceva again ;)

by Ceva and Angle Bissector Theorem, we only have to prove that $\frac{AB.CD.EF}{BC.DE.FA} = 1$.

By Briachon, we know that $AD$,$BE$ and $CF$ conccur. let $M$ this point.

$ABCDEF$ is cyclic, so, the pairs of triangles $(ABM,MED)$,$(BCM,MEF)$,$(CDM,EAF)$ are similars.

the rest is easy  :)


Suppose $\tan{\alpha}=\frac{p}{q}$ , where $p,q\in\mathbb{Z}$. Prove that number $\tan{\beta}$ for which $\tan{2\beta}=\tan{3\alpha}$ is rational only $p^{2}+q^{2}$ is the square of an integer.

Let $t = \tan\beta$. We know that $\frac{2t}{1-t^{2}}= \tan 2\beta = \tan 3\alpha = \frac{3pq^{2}-p^{3}}{q^{3}-3p^{2}q}$, which reduces to $(3pq^{2}-p^{3})t^{2}+2(q^{3}-3p^{2}q)t-(3pq^{2}-p^{3}) = 0$, which is a quadratic in $t$ with discriminant $\Delta = 4(p^{2}+q^{2})^{3}$. It's easy to conclude from this point.


Nice.
Schur's inequality of third degree works for $a+b+c<\frac{45}{13}$.

[quote="arqady"]Let $a,$ $b$ and $c$ are non-negative numbers, such that $ab+ac+bc=3.$
Prove that $13(a+b+c)+6abc\geq45.$
There is a nice proof. :wink:[/quote]
I think it's not hard but nice
We have $3abc(a+b+c)\ge (ab+bc+ca)^{2}=9\Rightarrow abc\ge\frac{3}{a+b+c}$
Let $x=a+b+c\Rightarrow x\ge 3$ . We have to prove $13x+\frac{18}{x}\ge 45\Leftrightarrow (x-3)(13x-6)\ge 0$

[quote="Nameless"]
We have $3abc(a+b+c)\ge (ab+bc+ca)^{2}=9\Rightarrow abc\ge\frac{3}{a+b+c}$
[/quote]
I think, $3abc(a+b+c)\le (ab+bc+ca)^{2}.$ :wink:

Mixing variables  :lol:
Assume $a=min (a,b,c)$, then $a \le 1$.
Put $f(a,b,c)=13(a+b+c)+6abc-45$.
We call $t$ is a positive real numbet such that $2at+t^{2}=3$.
We have $a(b+c-2t)+bc-t^{2}=0$, so $\frac{b+c}{2}\ge t \ge \sqrt{bc}$.
Hence $f(a,b,c)-f(a,t,t)=13(b+c-2t)+6a(bc-t^{2})=(b+c-2t)(13-6a^{2}) \ge 0$.
We need only prove $f(a,t,t)=13a+26t+6at^{2}\ge 0$. This is equivalent to $13.\frac{3-t^{2}}{2t}+26t+3t(3-t^{2})-45 \ge 0$ (because $a=\frac{3-t^{2}}{2t}$), which is equivalent to $(t-1)^{2}(2t^{2}+4t-13 \le 0$, which is true because $t \le \sqrt{3}$.

All values of $k$ satisfy the inequality $k(a+b+c)+abc \ge 3k+1$ holds are $k \ge \frac{3+2\sqrt{3}}{3}$.


calculus
integration
function
real analysis
real analysis unsolved

1. Show that if sum of a[size=75]n[/size] diverges with sum of a[size=75]n[/size] not= -infinity then there is a rearrangement {b[size=75]n[/size]} of {a[size=75]n[/size]} such that sum of {b[size=75]n[/size]} = +infinity.

2. If f(x) is positive and continuous, and integral from 0 to infinity of f(x) converges then sum of f(n) from 0 to infinity converges

[quote="newbiemathie"]
2. If f(x) is positive and continuous, and integral from 0 to infinity of f(x) converges then sum of f(n) from 0 to infinity converges[/quote]
It's false. Let $ f(n)\equal{}1$ for integer $ n$ and $ \int_{k\minus{}1}^{k} {{f(x)}dx}\equal{}\frac{1}{2^k}$ for $ k\equal{}1,2,3...$. It's easy to show, that this continuos function exists, integral is converges, but $ f(1)\plus{}f(2)\plus{}f(3)\plus{}...$ diverges.

The key hypothesis for the integral test? It's a decreasing function.

Ask Stephen New... He has office hours on Tuesday.
I am sure he has a solution to this question. :D


I got this question in class...

The bisectors of the exterior angles at the vertices of the acute angles of a right triangle intersect at an angle of what measure?

I know that exterior angles add to 360, and with disregarding the rt angle...the ext. angles add to 270. A bisector cuts an angle in half so then 135.

But then from there, I'm lost...the intersection part.

[hide="Solution"]As you stated, the sum of the exterior angles is $ 270$ degrees, so after bisecting each exterior angle, the sum of one of each angle is $ 135$ degrees. Therefore, the third angle in a triangle consisting of that triangle would be $ \boxed{45}$ degrees, since the sum of the angles in a triangle must sum to $ 180$ degrees.[/hide]