Which is the fifth term of the sequence: $ 1,2,4,8,...?$

Each number is twice the value of the previous number, so $ 8*2\equal{}16$.

This reminds me of a counterexample to assuming that a sequence follows the obvious pattern: the maximum number of regions inside a circle obtained when $ n$ points around the circle are joined by chords is $ 1, 2, 4, 8, 16, 31, 57, 99, ...$, as far as I remember.

Also if we consider the number of factors of $ n!$ for $ n\equal{}1, 2, 3,4, ...$ we get $ 1, 2, 4, 8, 16, 30, 96, ...$.

Each number is added by itself to make the next number.
Therefore,

[b]8+8=16[/b]

The formula is 2^n-1
Therefore, the answer is 16

Solution: input 5 for n= 2^4=16

Please don't post a solution when many have been already posted  :wink:

[quote="AndrewTom"]This reminds me of a counterexample to assuming that a sequence follows the obvious pattern: the maximum number of regions inside a circle obtained when $ n$ points around the circle are joined by chords is $ 1, 2, 4, 8, 16, 31, 57, 99, ...$, as far as I remember.

Also if we consider the number of factors of $ n!$ for $ n \equal{} 1, 2, 3,4, ...$ we get $ 1, 2, 4, 8, 16, 30, 96, ...$.[/quote]

This is possible, we don't know unless the problem states it for sure


Zuleica's mother Wilma picks her up at the train station when she comes home from school, then Wilma drives Zuleica home.  They always return home at 5:00 p.m.  One day Zuleica left school early and got  to the train station an hour early.  She then started walking home.  Wilma left home  at the usual time to pick  Zuleica up, and they met along the route between the train station and their house.   Wilma picked Zuleica up and then drove home, arriving at 4:48 p.m.  For how many minutes   had Zuleica been walking before Wilma picked her up?

How would you approach this problem?  There are so many things to deal with, it makes it so confusing.  :(

[quote=sub_math]BRUh

oK AT THIS POINT I dont even care this is like a 10 year bump because this problem is STUPID

I legit bashed this with d=rt for like 3 hours
no joke !!!!!!!!

AND THEN THE ANSWER IS SOME ez logic thing
WHAT
I HATE THIS PROBLEM
oops caps
I feel everyone should know how bad this problem is
>:( :censored:  :furious:  :furious:  :wallbash_red:[/quote]

LOL yeah this was a pretty hard problem

We only need to consider when things happen to solve the problem. No need to worry about distances or speeds.

[hide]
Let's define our terms:
   Tw: when Wilma leaves home
   Th: when Wilma and Zuleica arrive home
   Tm: when Wilma and Zuleica meet each other
   Ts: when Zuleica arrive at the train station
We use these terms as is for the usual case and primed versions (T'w, T'h, T'm, T's) for this specific case.
We are asked to find x, i.e., for how many minutes Zuleica walked after arriving at the station until meeting with Wilma. So
1) x = T'm - T's
Let's state what we know:
2) Tw = T'w         (Wilma left home at the usual time)
3) Th = 5.00 pm  (They usually arrive home at 5.00 pm)
4) T'h = 4.48 pm (They arrived home at 4.48 pm this time)
5) Ts = Tm           (They usually meet as soon as Zuleica arrives at the
                               train station)
6) Ts - T's = 1 h.  (Zuleica arrived at the station 1 hour earlier this time)
Using 5 and 6 we get:
7) T's = Ts - 1h = Tm - 1h
Using 7 in 1 we get:
8) x = T'm - (Tm - 1h) = 1h - (Tm - T'm)
So now we need to find (Tm - T'm). To do so we use the information that both legs of Wilma's ride (from home to meeting place and from meeting place to home) take the same time in all cases. So for the usual case we have:
9) Tm - Tw = Th - Tm
and organizing we get:
10) 2Tm = Th + Tw
And for this specific case we have:
11) T'm - T'w = T'h - T'm
and organizing we get:
12) 2T'm = T'h + T'w
Subtracting equation 12 from 10 we get:
13) 2Tm - 2T'm = Th + Tw - T'h - T'w
Using 2, 3, and 4 in 11 we get:
14) 2(Tm - T'm) = 5.00 pm + Tw - 4.48 pm - Tw
and organizing we get:
15) Tm - T'm = 6 minutes
Using 15 in 8 we get
16) x = 1h - 6 minutes = [b]54 minutes[/b]
[/hide]

[quote=tenebrine]Sad I put 66[/quote]

66 as in [hide][hide][hide][size=50]bestzack[b]66[/b]?[/size][/hide][/hide][/hide]

and yeah im kinda too scared to even read this problem because it looks very scary and i SUCK at rate problems so yeah no comment

[url=https://artofproblemsolving.com/videos/mathcounts/mc2013/385]The MATHCOUNTS Mini with this problem.[/url]
[url=https://youtu.be/QHce_s90Ivg?t=381]6:21[/url] is when he starts on this problem. That's okay to say right? Because this is the solution page anyways.

[hide = Solution]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has save$6$ minutes each way. Therefore, the point where she picks up Zuleica early is $6$ minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station $60$ minutes earlier than she usually gets picked up. She was picked up $6$ minutes earlier than usual, so she must have reached the train station $60 - 6 = 54$ minutes before she was picked up. Therefore, se was walking for $\boxed{54}$ minutes.[/hide]

[quote=wovenfractal]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has saved 6 minutes each way. Therefore, the point where she picks up Zuleica early is 6 minutes driving time from the train station. In other words, she picks up Zuleica 6 minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station 60 minutes earlier than she usually gets picked up. She was picked up 6 minutes earlier than usual, so she must have reached the train station $60-6= 54$ minutes before she was picked up. Therefore, she was walking for $\boxed{54} minutes$ .[/quote]

[quote=wovenfractal]Rather than focusing on Zuleica, let's think about Wilma. Wilma leaves at the same time as usual, but gets home 12 minutes earlier. So, she has saved 6 minutes each way. Therefore, the point where she picks up Zuleica early is 6 minutes driving time from the train station. In other words, she picks up Zuleica 6 minutes earlier than she usually does. Because Zuleica reached the train station one hour earlier than usual, she reached the train station 60 minutes earlier than she usually gets picked up. She was picked up 6 minutes earlier than usual, so she must have reached the train station $60-6= 54$ minutes before she was picked up. Therefore, she was walking for $\boxed{54} minutes$ .[/quote]

hmmm I think thats bascially the solution in the intro to algebra solutions manual


What percent of circle B is shaded? Express your answer to the nearest tenth of a percent.
[asy]import olympiad; import geometry_dev; import graph; size(150); defaultpen(linewidth(0.8)); import patterns;
path largeArc = Arc((0,0),1,90,405); draw(unitcircle); draw(dir(90)--origin--dir(45));
add("hatchback",hatch(NW));
fill(largeArc--origin--cycle,pattern("hatchback")); dot(Label("$B$",align=S),origin);[/asy]

Checking the Asy code, the unshaded portion is $ \frac{45^\circ}{360^\circ}\equal{}\frac18$ of the circle, so the answer is $ 1\minus{}\frac18\equal{}\frac78\equal{}\boxed{87.5\%}$.

[quote="math154"][b]Checking the Asy code[/b], the unshaded portion is $ \frac {45^\circ}{360^\circ} \equal{} \frac18$ of the circle, so the answer is $ 1 \minus{} \frac18 \equal{} \frac78 \equal{} \boxed{87.5\%}$.[/quote]

LOL? Is there another solution?

[quote="GameBot"]What percent of circle B is shaded? Express your answer to the nearest tenth of a percent.[/quote]

Uh, no?

wait...... but how do you know it is 45 degrees?
like on the real problem without looking at the Asy code

There is no way. Evidently, unless I, math154, BOGTRO, you, etc., are missing something huge, some key amount of information is missing. I suppose they(as in the problem writers) want you to assume it's to scale and estimate about 7/8 are shaded.

So in real games it is impossible to find out for sure? I wonder how some people actually got it.

Well, when I first encountered the problem, i guessed it was 7/8 of the circle and I put in 87.5%...
And I got IT right!!

Thats right guessage rulez!!!!


One (well-known?) theorem in my introduction to analysis textbook states:

Let $E\subset\mathbb{R}$. If $E$ has a finite supremum (respectively, a finite infimum), then there is a sequence $x_n\in E$ such that $x_n \rightarrow \sup{E}$ (respectively, $x_n \rightarrow \inf{E}$) as $n\rightarrow\infty$.

Can I have a clarification of the definition?
I am assuming that $E$, a set, is not ordered, so a sequence of $E$ consists of elements from $E$ in arbitrary order. Of these sequences, at least one most converge to $\sup{E}$. I am also assuming that terms used in the sequence can be repeated elements from the set, or else a set like $\{1\}$ would hamper the construction of a sequence...
Is this all correct?

Yes, that is correct. We do want to say that $\sup\{-1,3,5,8\}=8$ and the only way that a sequence in that set could converge to 8 would be for it to eventually equal 8.

On the other hand, suppose you have a nonempty bounded set that does not have a greatest element. In that case, the set must be infinite, and if you wanted to, you could choose a strictly increasing sequence that tends to the supremum.

Your book singled out the case of a finite supremum. Suppose we adopt the following as a notational convention: if $E$ is not bounded above, then $\sup E=\infty$ and if $E$ is not bounded below, then $\inf E=-\infty.$ Then this theorem still works: if $\sup E=\infty$ we can find $a_n\in E$ such that $\lim a_n=\infty.$

Unlike lots of other analysis topics, which tend to cunfuse me with symbols and weird words, I actually understand this topic. I had to post because I'm glad I understand. lol


How many 3-digit numbers are there in which the digits are such that if you create all the different 3-digit numbers from them, and you add these numbers, the sum is divicible by 111?

[hide="hint"]  Think about ALL the values. [/hide]

[hide]729??
abc+cab+bac+cba+acb+bca=222(a+b+c)
a,b,c are not 0.because 0bc is not a 3-digit number.
a,b,c can be 9 different digits.
9*9*9=729[/hide]

[hide="Solution"] The sum of all the possible three digit numbers made using the digits a, b, and c is 222(a+b+c), which is a multiple of 111 no matter what. There are 9*10*10 three digit numbers, so our answer is [b]900[/b].[/hide]


In an infinite chessboard, each square is filled with a real number such that it is the arithmetic mean of the four adjacent squares. Prove that every number in the chessboard must be the same.

it's not true.fill the $ i$th infinite row with the number $ i$.
i think your numbers must be from an interval...it was a problem in the mediterranian mathematical olympiad, just with $ 3D$ coordinates and the numbers were from the interval $ [0,1]$ it was not concidered an easy problem...

It has been posted before, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5327 for example.  There are several other versions floating around, but this is the only one I could find with any solutions.  The case with positive integers has a substantially more elementary solution than what is posted.


quadratics
algebra
polynomial
factoring polynomials

I think what you said is true but I'm not a great prover so I can't prove it to you but from what the math that I've done before, you can't factor unless there is a common factor.

You can however find the roots by using the quadratic formula.

Split it up does not always work

To have [tex]a^2-b^2[/tex]
[tex]a^2-b^2+ab-ab[/tex]
=[tex]a(a-b)+b(a-b)[/tex]
[tex]=(a+b)(a-b)[/tex]
where you added ab-ab to the expression.

There will always be SOME way to split it up into two things which are factorable -- however, as Beta pointed out, that might require you to add and subtract something, split a term (for example, write 3x^2 as (4x^2 - x^2) or something), or combine various different terms with each other.  To see why this might be, just imagine a polynomial in factored form -- when you multiply it out, you can distribute one factor across each of the terms of the other.  Then you get a sum of things, each with a common factor.  However, when you combine all of them, you will probably get terms combining with each other or cancelling each other out.

Or use synthetic division.  I find it easier for me to use synthetic division when factoring polynomials with the greatest term having a degree of more than 3.  Granted, there has to be some guessing of the factor in the first steps of synthetic division, but after some practice you can often determine a factor in less than 3 tries.  And besides, it solves the problem of [tex]x^2-1[/tex].  Just use 1 0 -1 as the dividend.

This also works for [tex]a^3-1[/tex].


trigonometry
function
limit
probability
logarithms
real analysis
real analysis unsolved

Prove that $\lim_{n \to \infty}{\cos^{n}\Bigg(\frac{x}{\sqrt{n}}\Bigg)}= e^{-x^{2}/2}$ for any $x \geq 0$

Suppose $f(x)$ is positive, $C^{2},$ $f(0)=1,$ and $f(x)<1$ for $x\ne0.$ Suppose also that $f''(0)=-a\ne0.$

What is $\lim_{n\to\infty}\left[f\left(\frac{x}{\sqrt{n}}\right)\right]^{n}$ for all $x\in\mathbb{R}?$

[This is actually an intermediate step in some proofs of the Central Limit Theorem in probability theory.]

That should be raised to the $n$th power, not squared. If it's squared, it just goes to 1. [color=green][Fixed now][/color]

We have that
\begin{eqnarray*}\lim_{n \to \infty}\left[ f \left( \frac{t}{\sqrt n}\right) \right]^{n}&=& \exp \left[ \lim_{n \to \infty}\frac{\log \left( f \left( \frac{t}{\sqrt n}\right) \right)}{\frac1{n}}\right] .\end{eqnarray*}
We will compute $L = \lim_{x \to \infty}\frac{\log \left( f \left( \frac{t}{\sqrt x}\right) \right)}{\frac1{x}}$ instead.
This a $\frac00$ case, so we can apply L'Hospital (the other conditions are also satisfied):
\[L = \lim_{x \to \infty}\frac{\frac1{f \left( \frac{t}{\sqrt x}\right)}\cdot f^\prime \left( \frac{t}{\sqrt x}\right) \cdot \frac{-t}{2 x \sqrt x}}{-\frac1{x^{2}}}= \lim_{x \to \infty}\frac{f^\prime \left( \frac{t}{\sqrt x}\right) \cdot \frac{t}{2}}{\frac1{\sqrt x}}. \]
If $f$ is defined on a neighbourhood of $0$, then $f^\prime (0) = 0$, since $0$ is a point of maximum. Hence, we can apply again L'Hospital:
\[L = \lim_{x \to \infty}\frac{f^{\prime \prime}\left( \frac{t}{\sqrt x}\right) \cdot \frac{-t^{2}}4 \cdot \frac1{x \sqrt x}}{-\frac1{2 x \sqrt x}}= \lim_{x \to \infty}\left[ \frac{t^{2}}2 \cdot f^{\prime \prime}\left( \frac{t}{\sqrt x}\right) \right] = \frac{-a t^{2}}2 . \]


function
integration
real analysis
real analysis unsolved

$A$ would only be self-adjoint if $K(x,y)=K(y,x).$ That wasn't given as a hypothesis.

Oops! Had to erase all the stupidities I had written :).

[quote="alekk"]$K: [0;1] X [0;1] \to \mathbb{R}^{*+}$, a strictly positive continuous function.
For every function $u \in C^0([0;1], \mathbb{R})$, one defines:
$A(u)(x)=\int_{[0;1]}K(y,x)u(y)dy$.

Prove that there exists $c>0$ and a function $u$ such that: $A(u)=c.u$[/quote]
Perron-Frobenius. Set $u_0(x) = 1$ and $u_n(x) = \frac{v_n(x)}{v_n(1)}$ with $v_n(x) = A(u_{n-1})(x)$. Then show that $u_n$ converges to a continuous limit.


If $ a, b, c ,d$ are, in some order, $ 1, 2, 3 , 4$ . What is the greatest possible value of:

$ ab \plus{} bc \plus{} cd \plus{} da$   ?

And how to do this ? What I need to do ?

[hide="Easy solution"]Note that the expression is equivalent to $ (a \plus{} c)(b \plus{} d)$.

Then, we note that $ (a \plus{} c) \plus{} (b \plus{} d) \equal{} 10$; by AM-GM, $ \sqrt {(a \plus{} c)(b \plus{} d)} \le \frac {(a \plus{} c) \plus{} (b \plus{} d)}{2} \equal{} 5$.

Hence the maximum value is no greater than $ 25$, where equality holds if $ a \plus{} c \equal{} b \plus{} d \equal{} 5$. Solutions are $ (a,c)$ and $ (b,d)$ are permutations of $ (1,4)$ and $ (2,3)$.[/hide]

Ah ok ;)
It's a nice solution. Thanks for your help :)


In triangle ABC, AB= 85, BC= 75 and CA = 40. A semicircle is tangent to AB and AC. It diameter lies on BC. Find the radius of the semi circle.

[hide]My friend got 24, but I don't know how to do it[/hide]

This is not the right place for this kind of problems. Next time you should post at most into Intermediate.
Btw, by Heron the area is $ 1500$. Also $ 1500 \equal{}\frac{85r}{2}\plus{}\frac{40r}{2}$

Sorry I got the problem from the Australian Intermediate Olympiad Exam, so I thought Olympiad was the correct section. Thanks :)


What sort of approximations does one make when calculating concentrations of ions in water solutions where the original compound only partially ionizes?

e.g. Calculate the concentration of $ C_{8}H_{4}O_{4}^{2\minus{}}$ in $ a)$ a $ 0.010 M$ solution of $ H_{2}C_{8}H_{4}O_{4}$, $ b)$ a solution which is $ 0.010M$ with respect to $ H_{2}C_{8}H_{4}O_{4}$ and $ 0.020M$ with respect to $ HCl$.

The first and second ionization constant for phthalic acid are $ 1.3\times 10^{\minus{}3}$ and $ 3.9\times 10^{\minus{}6}$ respectively.

A very common approximation is to consider the activity coefficients of all species equal to unity, and so the thermodynamic activities equal the molar concentrations - this is equivalent to neglect the ionic strenght of the solution.
When we have a weak diprotic acid, the first step in to evaluate $ \frac{K_{1}}{K_{2}}$. If $ \frac{K_{1}}{K_{2}}\geq 10^{4}$, then we can treat the two equilibria separatly, and for $ pH <\frac{pK_{1}\plus{}pK_{2}}{2}$ only the first ionization is relevant, while for $ pH >\frac{pK_{1}\plus{}pK_{2}}{2}$ only the second equilibrium is important. This is not the case with phthalic acid, because $ \frac{K_{1}}{K_{2}}\equal{} 333$, and so it is more correct to consider both equilibria.


Just an Unsolved Problem:

Non-negative reals $a,b,c$ satisfy $a^{2}+b^{2}+c^{2}=1$, Prove:

$\frac{1}{1-ab}+\frac{1}{1-bc}+\frac{1}{1-ca}\leq \frac{9}{2}$

[quote="libra_gold"]Please give me a hint or a link, thanx.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66936
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109029


Aratati ca nu exista numere rationale $\ x$ care verifica egalitatea
$\ (2n+1)x^{m}=x^{n}-2m-1,n>m$, cu $\ m,n$ naturale nenule
Sorin Peligrad

Simpatica problema...
Fie $x=\frac{a}{b}$, $a$ si $b$ numere intregi, $b\neq\ 0$, fractie ireductibila, deci $(|a|,|b|)=1$.
Atunci:
$(2n+1)a^{m}b^{n}=a^{n}b^{m}-(2m+1)b^{m}b^{n}$
$b^{n}[(2n+1)a^{m}+(2m+1)b^{m}]=a^{n}b^{m}$
Din moment ce rezolvarea se va baza doar pe paritate si pe numarul de factori de doi, nu conteaza semnul lui $a$ si $b$, deci putem presupune $a$ si $b$ pozitive.
Daca $b$ este impar, iar $a$ este par, atunci membrul stang al identitatii este impar, iar cel drept este par, imposibil.
Daca $b$ este impar, iar $a$ este impar, atunci membrul stang al identitatii este par, iar cel drept este impar, imposibil.
Daca $b$ este par, iar $a$ este impar atunci paranteza din membrul stang este impara, deci nu intervine in stabilirea numarului de factori de $2$ prezenti. Rezulta ca $b^{n}$ si $b^{m}$ au acelasi numar de factori de doi, rezulta $m=n$, contradictie.


geometry
inequalities
email
inequalities open

Prove that :
 $ 3a^2 +2(b^2+c^2) \geq 16.S $ , where $ S $ is the area of triangle $ ABC $
  Question : Generalize this problem .

One of the strongest inequalities of this type is the following:

                $ 2(ab+bc+ca)-(a^2+b^2+c^2)\geq 4S\sqrt{3}$.
  This was given by Hadwiger-Finsler in 1937.
  In Arhimede Magazine,no.9-10/2003, I gave a new proof for this just using AM-GM.  ;)

A 'new proof'?  It is just Muirhead if we substitute $a=x+y$ and so on, or Schur if we square it and expand.

[Moderators, please leave this post here - I need to communicate with this user; as you'll see below, this is the only way.]

spdf - please change your email in your profile.  You have been subscribing to topics with a bad email in your profile, so all your topic watch emails are bouncing.  I tried PMing, but you aren't reading your PM's either.  I have deleted your current watched topics.  Please do not watch any more topics or forums until you fix your email.

to Blahblahblah: No,Schur,or Muirhead. they are too heavy for the simplity of this inequality.If you want i'll post my solution.

 By the way what's wrong with my email? :?  (I don't know if you are talking to me) ;)
 And by the way ,i'm reading my Private messages almost everyday. ;)

  I hate Muirhead. :mad:

This is what I meant:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=195906&highlight=#195906

It is 'muirhead', but really just a trivial application of AM-GM.

Excuse me, Blahblahblah, but my solution is not in all these posts and it is much more easier. ;)

Your "well-known" ineq doesn't have a connection with my problem , Cezar !
  There are many ways to solve your ineq ( as blahblahblah showed ) . We don't discuss that ineq here .

Here is the solution :
 First , prove that : $ b^2 + c^2 \geq \frac{a^2}{2} + 2h_a^2 $
 After that , prove that : $ 4(a^2 + h_a^2) \geq 16S $
  The first ineq can be proved by drawing the attitude from A and using Pythagore's theorem
  I don't know whether we can generalise this problem !


[hide]

noticing that $20+25=45$, and $\tan{45}=1$ prompts us to try this,
(let $\tan{x}=t$)

$(1+\tanh)(1+\tan{45-x})=1+\tan{x}+\tan{45-x}+\tan{20}\tan{45-x}$ and using the tangent subtraction

formulas, and $\tan{x}=1$, we get,

$1+\tan{x}+\tan{45-x}+\tan{20}\tan{45-x}=1+t+\frac{1-t}{1+t}+\frac{t(1-t)}{1+t}$ and simplifying gives

$\frac{2t+2}{t+1}$ and since we know that $\tan{20}\ne1$, we can cancel out the $t+1$
giving us $\boxed{B=2}$

[/hide]

EDIT: woops, forgot to hide...

[hide]

Since $A=20^{\circ}$ and $B=25^{\circ}$, we have:

$(1 + \tan A)(1 + \tan B) = (1 + \tan 20^{\circ})(1 + \tan 25^{\circ})$

$(1 + \tan A)(1 + \tan B) = 1 + \tan 20^{\circ} + \tan 25^{\circ} + \tan 20^{\circ} \cdot \tan 25^{\circ}$    $( \ast )$

But

$\tan (20^{\circ} + 25^{\circ}) = \frac{\tan 20^{\circ} + \tan 25^{\circ}}{1 - \tan 20^{\circ} \cdot \tan 25^{\circ}} \Leftrightarrow$

$\tan 45^{\circ} = \frac{\tan 20^{\circ} + \tan 25^{\circ}}{1 - \tan 20^{\circ} \cdot \tan 25^{\circ}} \Leftrightarrow$

$1 = \frac{\tan 20^{\circ} + \tan 25^{\circ}}{1 - \tan 20^{\circ} \cdot \tan 25^{\circ}} \Leftrightarrow$

${1 - \tan 20^{\circ} \cdot \tan 25^{\circ}} = \tan 20^{\circ} + \tan 25^{\circ}$.

Thus

${1 = \tan 20^{\circ} + \tan 25^{\circ} + \tan 20^{\circ} \cdot \tan 25^{\circ}}$.

Applying this result in $( \ast )$, we get:

$(1 + \tan A)(1 + \tan B) = 1 + \tan 20^{\circ} + \tan 25^{\circ} + \tan 20^{\circ} \cdot \tan 25^{\circ} = 1 + 1 = 2$.

[/hide]

[hide="Hopefully this is a different solution from everyone elses..."]
$(1+\tan A)(1+\tan B)=1+\tan A + \tan B + \tan A \cdot \tan B$
$\tan (A + B) = \frac{\tan A + \tan B}{1-\tan A \cdot \tan B}$
$\tan (45^\circ) = \frac{\tan A + \tan B}{1-\tan A \cdot \tan B}$
$1 = \frac{\tan A + \tan B}{1-\tan A \cdot \tan B}$
$\tan A + \tan B = 1-\tan A \cdot \tan B$
$\tan A + \tan B + \tan A \cdot \tan B =1$
$1 + \tan A + \tan B + \tan A \cdot \tan B ={\Huge \textbf{2}}$ :D
[/hide]

edit: ah, same thing as everyone else :roll:

[quote="Silverfalcon"]If $A = 20^\circ$ and $B = 25^\circ$, then the value of $(1+\tan A)(1+\tan B)$ is

A. $\sqrt{3}$
B. 2
C. $1 + \sqrt{2}$
D. $2(\tan A + \tan B)$
E. None of these[/quote]

[hide]
$1 + \tan{A} + \tan{B} + \tan{A}\tan{B} = x$
$\tan{A + B} = \frac{\tan{A} + \tan{B}}{1 - \tan{A}\tan{B}}$
$\tan{45}(1 - \tan{A}\tan{B}) = \tan{A} + \tan{B}$
$(1 - \tan{A}\tan{B} + 1 + \tan{A}\tan{B} = x$
$2$
is the answer.[/hide]

Excuse me, but posting same solutions before checking if that was posted before seems too pathetic.  :spam:


Given a 400 by 400 chessboard and a line on that cheesboard, what is the maximum amount of squares that the line could go through?

Give the spaces coordinates ranging from (1,1) to (400,400).

We move into a new square whenever either the x coordinate changes by 1 or the y coordinate changes by 1.

There are a maximum of 399 possible changes in each of these two dimensions. Since we must include the starting square, the maximum number of squares the line can pass through is:

$399+399+1=\boxed{799}$

This is indeed doable. Connect the lower left and upper right endpoints, and move the resulting line up by half a unit.

OMG I didn't think of using the coordinate plane.

So for a N by N cheesboard, that lain may at most pass through exactly 2N-1 squares? But then what happens if the board is N by 2N or 3N?

For the most part, its $length+width-1$, because there are $length-1$ vertical changes and $width-1$ horizontal changes, and we add $1$ for the starting square.


If $ \triangle{ABC}$ and $ \triangle{DEF}$ are such that
\[ \sin A\sin B\sin C \equal{} \sin D\sin E\sin F
\]

\[ \cos A\cos B\cos C \equal{} \cos D\cos E\cos F
\]
then prove the two triangles are similar.

I came up with this problem while looking at another.  The solution I found isn't very messy, but I was wondering whether there is a very clean, insightful solution...

EDIT: Fixed!  Thanks, Daan.


Prove that $4x^3-2x^2-15x+9$ and $12x^3+6x^2-7x+1$ has three distinct real roots

[hide]if u mixed up the 9 and the 1 it would be easy, but nevertheless the idea is that if $p_1(x)$ is the first and $p_2(x)$ the second polynom, then $ap_1(x_0)+bp_2(x_0)=0$ for all real $a,b$, $x_0$ being the shared root. if all solutions of $x_0$ are not root of $p_1$ or $p_2$ then we have a contadiction, and there are no shared solutions.[/hide]

I don't understand what you said. Could you elaborate?

lets say they share one root $x_0$ then $p_1(x_0)=0$ and $p_2(x_0)=0$ therefore for every real $a,b$: $a\cdot p_1(x_0)+b\cdot p_2(x_0)=0$. then choose $a$ and $b$ so that something disappears. (e.g. $a=-1$ and $b=9$) then u only add up the polynoms and find its roots and see if any of its roots would match so that it is also a root of $p_1$ and $p_2$ ...
peeta


See the second post in this thread.

I got this while posting in the Number Theory section. Some part of the post was substituted by something else (I  had some $ \text{\LaTeX}$ error first, but surely the output was not what I typed).

It also caused severe html problems, making the site look rather weird.

tjhance: this means that the mentioned field extension has degree 2, meaning that $ \mathbb Q[quote: 375f7f5700 \equal{} "t0rajir0u"]thbb Q$ vector space.

[quote="t0rajir0u"]Alternately, $ x^2 \minus{} x \minus{} 1$ is not a cyclotomic polynomial.[/quote]
...

And here is the original message (see the obvious error in line 1^^):

[code]
tjhance: this means that the mentioned field extension has degree 2, meaning that $\mathbb Q[\phi}$ is 2-dimensional $\mathbb Q$ vector space.

[quote="t0rajir0u"]Alternately, $x^2 - x - 1$ is not a cyclotomic polynomial.[/quote]
...
[/code]

While in this case the mistake is obviously user-generated, it might also be better handled by the computer. I will investigate this further.

It turned out to be also a parser error, but the user mistake was nice enough to point it out ;) Bug fixed now.


abstract algebra
vector
search
Galois Theory
superior algebra
superior algebra unsolved

$\mathbb{K}$ a field and $\mathbb{L}$ an extension of $\mathbb{K}$.
Suppose there exists $x_1,..,x_n \in \mathbb{L}$ such that : $\mathbb{L}=\mathbb{K}[x_1,x_2,..,x_n]$. Prove that $dim_\mathbb{K}(\mathbb{L}) < +\infty$

That's not exactlllllllllllllly waht you mean  :D

Is $\mathbb{L}$ supposed to be a field¿ Then it's very well know ;)

yes it is well known. But the only proof I found with a friend use Hilbert zeroes theorem. Is there a simpler way to do this ?

:?
As I see that, you just claim that adjoining finetely many algebraic numbers will give an extension of finite degree, which is rather obvious.

Now why wouldn't this work?

$\mathbb L=\mathbb K[x_1,\ldots,x_n]$ is a finitely generated module over $\mathbb K$, which is, of course, Noetherian, so $\mathbb L$ is a Noetherian $\mathbb K$-module. An infinite-dimensional $\mathbb K$-vector space cannot be a Noetherian module though, because if infinitely many elements $\{e_n\}_n\subset\mathbb L$ form a linearly independent set, then the chain $\mathbb Ke_1\subsetneq\mathbb Ke_1+\mathbb Ke_2\subsetneq\ldots$ is a stirctly increasing chain of $\mathbb K$-submodules of $\mathbb L$, contradicting $\mathbb L$'s quality of being Noetherian.

Since $\mathbb{K}$ is a field, a finitely generated module is a finite-dimensional vector space and you're assuming the conclusion.

This is a pretty subtle problem.

Yeah, in my reasoning, I foolishly mistook $\mathbb K[\{x_i\}]$ for the submodule of $\mathbb L$ generated by $x_i$. Sorry. :blush:

[quote="jmerry"]That's not it at all- $\mathbb{L}$ doesn't have to be an algebraic extension. We want to show that if some of those $x_n$ are transcendental, the ring they generate isn't the same as the field they generate.[/quote]

I actually did the same mistake of interpretation. I remember a proof for this in the particular real case (taking $\mathbb{L}=\mathbb{R}$ and $\mathbb{K}$ any subfield such that the existence of the $x_{i}$ is satisfied; in this case we can prove that $\mathbb{K}=\mathbb{R}$).

The proof uses some developments in Galois Theory (absolutely nontrivial). You can find it back by running a search on the french newsgroup fr.sci.maths (Put my name as an author, and look for "dimension finie"), and maybe you can adapt it to the more general case above.


real analysis
calculus
integration
topology
real analysis unsolved

If f is measurable on [a,b] and $ \int_{[a,c]} f\equal{}0$ for all c<b,show that f=0 almost everywhere on [a,b].

You know that $ f\equal{}0$ a.e. on any interval $ [a,b\minus{}1/n]$.  Tr applying MCT or DCT to the appropriate sequence.

$ f$ should be integrable on $ [a,b]$

Proof : suppose $ f>0$ on a set $ E$ of positive measure, we can find a closed set $ F\subset E$ with $ m(F)>0$ and $ \int_{F}f>0$ , now let  $ O\equal{}(a,b)\setminus F$

we have $ 0\equal{}\int_{a}^{b}f\equal{}\int_{O}f\plus{}\int_{F}f$ hence $ \int_{O}f\equal{}\minus{}\int_{F}f \neq 0$. Since $ O$ is open then $ O\equal{}\bigcup_{n}(a_{n},b_{n})$ of disjoint intervals

since $ f$ is integrable then  $ \int_{O}f\equal{}\sum_{n}\int_{(a_{n},b_{n})}f  \neq 0$ then there exists $ n$ st : $ \int_{(a_{n},b_{n})}f \neq 0$
but we have $ \int_{(a_{n},b_{n})}f\equal{}\int_{(a,b_{n})}f\minus{}\int_{(a,a_{n})}f \neq 0$ then either $ \int_{(a,b_{n})}f\neq 0$ or $ \int_{(a,a_{n})}f\neq 0$.  In any case we have if $ f$ is positive on a set of positive measure we can find $ x\in [a,b]$ st : $ \int_{(a,x)}f\neq 0$ .

Similarly for the case $ f$ is negative on a set of positive measure. Thus we have a contradiction


Prove that there exist infinitely many n such that the denominator of the irreducible fraction 1+1/2+...+1/n is not a prime power. I don't know how to solve this one. I've tried to take n a prime, so that the denominator should be divisible by p, but I don't know how to continue.

Take n do be prime >5 then 1+1/2+...+1/n=(n!/1+n!/2+...+n!/n)/n! .
All the terms, but n!/n, of the sum in the numerator are 0 mod p. So the numerator is not 0 mod p. If we want it to be a prime power it must be n, so we should have that (n-1)! divides n!/1+n!/2+...+n!/n.
Take this for granted: there is a prime number n/2<p<n.
So p| (n-1)! and p divides all the terms in the sum except for n!/p.
The conclusion easily follows.

First of all, there was an error in my first message. I should have said: I tried to take n of the form p-1, with p prime, such that the denominator should be divisible by p. On the other hand, I must recognise that I don't understand anything from your proof. You proved that 2*3*...*n+...+1*2*...*(n-1) is prime with n. So how can we deduce that if the denominator of 1+1/2+...+1/n is a prime power then it must be n?
 Could you please post a very detailed solution. Consider this a little help from a very stupid person and please present what values of n you want to prove are good and then prove that these are good. Thank you.

What do you understand by denominator?

I searched my dictionary and it sais
denominator (d`nmnet?) the divisor of a fraction, as 8 in 7/8. Compare numerator  (sense 1)Archaic. a person or thing that denominates or designates.

So I wanted to prove that after we reduce common factors in the sum
1+1/2+...+1/n=(n!/1+n!/2+...n!/n)/n! where [b]n is any prime[/b] larger that 5, the denominator is a multiple of n and is not a power of a prime.

[b]If it was a power of a prime, since it is divisible by n and not by n 2 , it must have been n.[/b] (it realy was not important)

But taking a prime n/2<p<n (possible thanks to Bertrand) we notice that p and n both divide the demoninator of the reduced fraction (because they don't divide the numerator of the (n!/1+n!/2+...n!/n)/n!) , so it can [b]not[/b] be a power of a prime.

P.S. I advise you to base your diet on milk and phosphorus food such as fish. If not you could always scratch the head of some matches. It's not a myth, you will have no further reason to think you are stupid.  :D
I like the problems you post.  :D

I forgot to tell harazi what to do with the result of scratching the matches. It must be eaten. It is supposed to contain a significant amount of phosphorus. It is supposed to make people smarter. Might get some ready for me now for not finishing the sentence.  :D
Hope you get it is a joke. :D


calculus
integration
LaTeX
calculus computations

[quote="7h3.D3m0n.117"][quote="Together4Ever"]Find  ∫1/(X^4+4) dx

        ∫1/(X^6-1) dx[/quote]

whoa how'd you get the integral symbol  :D

here let's put this in LaTeX:

$\int \frac{1}{x^{4}+4}dx$

$\int \frac{1}{x^{6}-1}dx$[/quote]
in first: put $x: =y(4)^{1/4}$ and your integral will take next form:
$4^{-3/4}\int\frac{dy}{1+y^{4}}$ and this integral we have solved in more good result $\int\frac{dx}{1+x^{2n}}$ it was post few days ago here . and in the seccond integral by Ostrogradski method or: $\frac{1}{x^{6}-1}=\frac{1}{(x^{3}-1)(x^{3}+1)}=1/2(\frac{1}{x^{3}-1}-\frac{1}{1+x^{3}})$ and last is easy

Thanks Extremal very much

please can you continue the answer

Sure I can. We must find $\int\frac{dx}{x^{3}-1}$ and $\int\frac{dx}{x^{3}+1}$ lets find $A,B,C$ such that $\frac{1}{x^{3}-1}=\frac{A}{x-1}+\frac{Bx+C}{x^{2}+x+1}$=$\frac{(A+B)x^{2}+(A+C-B)x+(A-C)}{x^{3}-1}$
then $(A+B)x^{2}+(A+C-B)x+(A-C)=1$ if and only if when
$(A+B)=0$
$A+C-B=0$
$A-C=1$
if and only if when $A=\frac{1}{3}$, $B=-\frac{1}{3}$ and $C=-\frac{2}{3}$
very good. in this case:$\frac{1}{x^{3}+1}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}-x+1}$=$\frac{(A+B)x^{2}+(-A+C+B)x+(A+C)}{x^{3}+1}$
$(A+B)x^{2}+(-A+C+B)x+(A+C)=1$
if and only if when
$(A+B)=0$
$-A+C+B=0$
$A+C=1$
$\Rightarrow$ $A=B=\frac{1}{3}$, $C=\frac{1}{3}$
then how to find integral
$\int\frac{Adx}{1+x}=Aln(1+x)$
$\int\frac{Adx}{x-1}=Aln(x-1)$
and how to find integral
$\int\frac{(Bx+C)dx}{x^{2}+x+1}$=$\int\frac{(Bx+C)dx}{(x-\frac{1}{2})^{2}+\frac{3}{4}}$
let $x-1/2=y$ then
$\int\frac{(Bx+C)dx}{(x-\frac{1}{2})^{2}+\frac{3}{4}}$=$B(\int\frac{ydy}{y^{2}+\frac{3}{4}}+\int\frac{1/2-C/B}{y^{2}+\frac{3}{4}}dy)$
last is easy can you do it? (in the seccon is analogysli.

Don't ever underestimate the power of this trivial fact: $a^{2}+b^{2}= (a+b)^{2}-2ab$.
Let $u = x^{2}$ in the first equation, we are able to get $x^{4}+4 = (x^{2}+2x+2)(x^{2}-2x+2)$ and then we apply the partial fraction. Factorization in the last part was obvious because it was difference of two numbers to the power of six.


Let ABC is a triangle and let point O. Three lines through O and perpendicular to the lines OA, OB, OC sequence meet BC, CA, AB at M, N, P.
Prove that these point M, N, P are in a straight line.

[size=150][color=blue]Thank you very much.[/color][/size]

Bad title, wrong forum... it's not new anyway: http://www.mathlinks.ro/Forum/viewtopic.php?t=51844 .

  dg


inequalities
ceiling function
floor function
combinatorics unsolved
combinatorics

please help me to solve this problem
For natural numbers n and N.
Find the minimum of W/ W=B1+B2+....Bn with B1,B2,....,Bn are natural numbers and B1*B2*...Bn :ge: N
thanks

hungvuong, explore what the Arithmetic Mean-Geometric Mean (AM-GM) Inequality tells you about this situation.

thanks.
Could you please post the entire solutions? Thanks

I don't think you should post this in advanced if one simple application of AM-GM is sufficient.. Besides it isn't combinatorics at all.

Anyway:

AM-GM says that $\frac{a_1+\cdots+a_n}{n}\geq\sqrt[n]{a_1\cdots a_n}$

So your requested W is $n\cdot \sqrt[n]{B}$ :maybe:

[quote="Peter VDD"]I don't think you should post this in advanced if one simple application of AM-GM is sufficient.. Besides it isn't combinatorics at all.

Anyway:

AM-GM says that $\frac{a_1+\cdots+a_n}{n}\geq\sqrt[n]{a_1\cdots a_n}$

So your requested W is $n\cdot \sqrt[n]{N}$ :maybe:[/quote]
[quote="guas"]Unfortunately $\root n \of B$ is not always an integer... which makes the problem
somewhat more advanced :-). The case $n=3$, $B=28$ may convince you that
the answer may depend of factors of numbers close to $B$. By the way, I could not solve the problem...[/quote]

This looks like a job for hill-climbing, a combinatorial technique by which a reasonably good value is improved in gradual steps to the best value.

Sippose we do have a set of natural numbers B1, B2, ..., Bn such that B1*B2*...*Bn :ge:  N. We wish to modify the set so that the sum B1+B2+...+Bn is as small as possible.

Suppose Bi :ge: Bj+2 for some i, j. Then (Bi-1)(Bj+1) = Bi*Bj + Bi - Bj - 1 :ge: Bi*Bj + (Bj+2) - Bj - 1 = Bi*Bj + 1. So by replacing Bi with Bi-1 and Bj with Bj+1 we increase the product B1*B2*...*Bn without changing the sum B1+B2+...+Bn. Our goal is to reduce the sum, but if we increase the product enough then we might be able to replace some Bk with Bk-1 and still have the product greater than or equal to N. So increasing the product is good.

It turns out we don't have to perform the actual hill climb (fortunate for us, because it is difficult to prove that a hill climb stops at the best value). The hill climb will stop when no Bi  is 2 or more greater than any other Bj. That means that either all the Bi are equal, or that the Bi have only two values among them, whose difference is 1. So our list B1,B2,....,Bn has become b, b, ... ,b, b+1, ..., b+1 for some natural number b.

Having $b < \sqrt[n]{N}-1$ is impossible, since then the product B1*B2*...*Bn :le: (b+1)(b+1)*...*(b+1) < N. If $b \geq \sqrt[n]{N}$, then b*b*...*b :ge: N, so there are no b+1 elements and $b = \lceil\sqrt[n]{N}\rceil$.  So the the only numbers we need to work with are $\lfloor\sqrt[n]{N}\rfloor$ and $\lfloor\sqrt[n]{N}\rfloor+1$.  We simply need to find the largest integer r such that

$(\lfloor\sqrt[n]{N}\rfloor)^r(\lfloor\sqrt[n]{N}\rfloor+1)^{n-r}\geq N,$

which is

$r = \left\lfloor\frac{n\log(b+1) - \log(N)}{\log(b+1) - \log(b)}\right\rfloor$

where $b=\lfloor\sqrt[n]{N}\rfloor$. This gives the sum B1+B2+...+Bn = n*b + n - r.

For example, for n = 3 and N = 28, the solution is B1 = 3, B2 = 3, and B3 = 4, so the product is 36 and the sum is 10.

Erin Schram


I was confused as to how to do this, so any help would be appreciated:

I have to see if the following series converges and justify my answer.$\sum_{j=1}^{\infty}(-1)^{j}(1-\frac{1}{j})$

[quote="Kent Merryfield"]Ignore the alternating sign for a moment. How does $\left(1-\frac1j\right)$ behave as $j\to \infty.$ Does it go to zero? If so, how fast?[/quote]

It approaches infinity as well.

The expression would become zero only if there was a 1 substituted for j in the denominator.

use the alternating series test.
first note that $\lim_{j\to\infty}\left(1-\frac{1}{j}\right)=1$
also
$\left(1-\frac{1}{n}\right)<\left(1-\frac{1}{n+1}\right)$
so both conditions are violated therefore the original series does not converge.

Let's [b]not[/b] call it the alternating series test.

First, the "decreasing in size" criterion for the the alternating series test is a sufficient condition but not a necessary condition. You're not entitled to conclude anything at all if the condition is not met - it cannot be used to prove divergence.

But back to the other condition: the terms don't tend to zero. That's the most basic test there is:

If the series converges, then the terms go to zero.

In the contrapositive, if the terms don't go to zero, then the series diverges. Period. End of story. No other tests needed.

(Just don't go trying to invoke the converse of this theorem on any test of mine - that's false.)

[quote="muckymotter"]It approaches infinity as well. [/quote]
I don't know what "it" is in that sentence, but no matter what you meant, there is no sense in which this could be true.

Let $S_{n}=\sum_{j=1}^{n}(-1)^{j}\left(1-\frac1j\right)$ be the sequence of partial sums. Then $S_{n}$ is a bounded sequence, and although $\lim_{n\to\infty}S_{n}$ does not exist, both $\lim_{k\to\infty}S_{2k}$ and $\lim_{k\to\infty}S_{2k+1}$ do exist.


What is the value of the following expression:
$ 1 \minus{} 4 \plus{} 7 \minus{} 10 \plus{} 13 \minus{} \cdots \minus{} 46 \plus{} 49 \minus{} 52 \plus{} 55$ ?

$ 1 \plus{} 7 \plus{} 13 \plus{} \cdots \plus{} 43 \plus{} 49 \plus{} 55 \equal{} \sum_{k\equal{}0}^9 (1 \plus{} 6k) \equal{} \sum_{k\equal{}0}^8 (1 \plus{} 6k) \plus{} 55$
$ 4 \plus{} 10 \plus{} 16 \plus{} \cdots \plus{} 40 \plus{} 46 \plus{} 52 \equal{} \sum_{k\equal{}0}^8 (4 \plus{} 6k) \equal{} 3 \times 9 \plus{} \sum_{k\equal{}0}^8 (1 \plus{} 6k)$
Thus $ \left( \sum_{k\equal{}0}^8 (1 \plus{} 6k) \plus{} 55 \right) \minus{} \left( 3 \times 9 \plus{} \sum_{k\equal{}0}^8 (1 \plus{} 6k) \right) \equal{} 55 \minus{} 27 \equal{} \boxed{28}$.


These questions are from 93-94 State Competition.

Any hints or solutions on #7 or #11??

#7:  It is just a matter of factoring, but you have to be careful to not make mistakes on your exponents.

#11:  What are the possible outcomes?

I figured out #11...

#7, I'll try again..


If $ \sum^{2008}_{n\equal{}1} n \cdot 3^n \equal{} \frac{a \cdot 3^b \plus{} 3}{c}$ where $ (a,c) \equal{} 1,$ then find the remainder when $ a\plus{}b\plus{}c$ is divided by 1000.

Level: Early AIME

$ \sum^{2008}_{n \equal{} 1} n \cdot 3^n \equal{} 3\frac {d}{dx}\sum^{2008}_{n \equal{} 1} \int n \cdot x^{n \minus{} 1} dx$ where $ x \equal{} 3$
$ \implies \equal{} 3\frac {d}{dx}\sum^{2008}_{n \equal{} 1} x^n$
$ \implies \equal{} 3\frac {d}{dx}\frac {x^{2009} \minus{} x}{x \minus{} 1}$
$ \implies \equal{} 3\frac {(x \minus{} 1)(2009x^{2008} \minus{} 1) \minus{} (x^{2009} \minus{} x)}{(x \minus{} 1)^2}$
$ \implies \equal{} 3\frac {2008x^{2009} \minus{} 2009x^{2008} \plus{} 1}{(x \minus{} 1)^2}$

Substituting back in $ x \equal{} 3$ gives $ \frac {(2008 \cdot 3 \minus{} 2009)\cdot3^{2009} \plus{} 3}{4} \equal{} \frac {4015 \cdot 3^{2009} \plus{} 3}{4}$ so we have $ a \equal{} 4015$, $ b \equal{} 2009$, and $ c \equal{} 4$ giving $ a \plus{} b \plus{} c \equal{} 4015 \plus{} 2009 \plus{} 4 \equal{} \boxed{6028}$

Unless there is a way to do this easily without calculus, I would not consider this "early AIME" level

Um..."remainder when divided by 1000"...

Hmm.

[hide="Why do I believe this is not too bad"]
Consider the desired sum to be $ S$. Find $ 3S$. Subtract $ S$ from $ 3S$ to get $ 2S$. Does this help with you anything? [/hide]

And your answer's correct by the way [when divided by 1000 to get remainder]

I don't know, maybe I've been used to seeing #10-15 lately and that's why I thought this was easy.  :oops: And yeah, I'm not always bright with judging level of problems I created.  :|

No this is very easy because it is so well-known.

Another method to solve this problem is write $ n\cdot 3^n$ in the form $ f(n\plus{}1)\minus{}f(n)$ so it telescopes. We can easily find constants $ a,b$ such that $ f(n)\equal{}(an\plus{}b)\cdot 3^n$.

[quote="facis"]Ok so a non-calc solution (as you are suggesting):

$ 3S \minus{} S \equal{} 2S \equal{} 3^2 \plus{} 2*3^3 \plus{} 3*3^4 \plus{} \ldots \plus{} 2007*3^{2008} \plus{} 2008 * 3^{2009} \minus{} [3 \plus{} 2*3^2 \plus{} 3*3^3 \plus{} 4* 3^4 \plus{} \ldots \plus{} 2008 * 3^{2008}] \equal{} \minus{} 3 \minus{} 3^2 \minus{} 3^3 \minus{} 3^4 \minus{} \ldots \minus{} 3^{2008} \plus{} 2008 * 3^{2009}$

$ 2S \equal{} \minus{} 3\frac {3^{2008} \minus{} 1}{3 \minus{} 1} \plus{} 2008*3^{2009}$
$ \implies \equal{} \frac {(2008*2 \minus{} 1)3^{2009} \plus{} 3}{2}$
$ S \equal{} \frac {4015 \cdot 3^{2009} \plus{} 3}{4}$ just as before.

I'd rate this problem medium level AIME (around 7-10)[/quote]

Yes, that's correct.  :)

Altheman, I'm not sure if I can follow about $ f(n\plus{}1) \minus{} f(n)$. Maybe you can expand on that little bit more? (I don't feel like doing the paper work now too but yeah  :D ).

$ n\cdot 3^n\equal{}( a(n\plus{}1)\plus{}b)\cdot 3^{n\plus{}1}\minus{}(an\plus{}b)\cdot 3^n$...solve for $ a,b$

Now find a way to evaluate this sum [i]combinatorially.[/i]  :)

[hide="Hint"] The combinatorial argument appears to be easier if you instead evaluate the sum $ \sum_{n \equal{} 0}^{2008} (2008 \minus{} n) \cdot 3^{n}$. [/hide]

Wow, nice solutions! I haven't really looked at all of them in details but they are all unique and excellent! Thanks all of you for inputs.  :)  :)  :)


What is the sum of all five-digit numbers of the form $ \underline{1wxyz}$ where $ w$, $ x$, $ y$ and $ z$ are each 0 or 1?

16 combinations, observe that every number w,x,y and z is equal to 1 in exactly 8 of these combinations.
Sum: 160000+8(1111)=168888


limit
integration
probability
probability and stats

Prove that the weak law of large numbers holds for sequence $ (X_n: n\in\mathbb{N})$ of independent random variables such that $ \lim_{a\rightarrow \infty}\max_{1\leq k\leq n}\int_{|x|\geq a}|x|d\mathbb{F}_{X_k}(x)\equal{}0$.

Any hints are welcome.

Convergence in probability, I guess, but the statement is clearly misstated :(. If $ n$ in the last formula should be $ \infty$ (i.e., you have a uniform estimate for tails), the statement is fairly simple (with the conclusion that $ \frac 1n\sum_{j\equal{}1}^n (X_j\minus{}EX_j)$ tends to $ 0$ in probability). Otherwise, I do not know what it should be.

Mr .Doe here are my two cents

Your hypothesis on the sequence $ X_n$ means that the sequence is uniformly integrable. Moreover if you suppose that every $ X_n$ is centered then you have a martingale. As your sequence is formed by independent variable the sequence formed by the partial sums is also a martingale let's call it $ S_n$

In the book of Hall "martingale limit theory and its application"
you have the following theroem :

Let $ S_n\equal{}\sum_{i\equal{}0}^{n}X_n$ be a martingale and $ b_n$ an increasing sequence of postive constants which tends  to infinity
Then S_n/n tends weakly to 0 as n tends to infinity if
1/$ Lim_{n\rightarrow \plus{}\infty}\sum_{i\equal{}1}^nP[|X_i|>b_n] \equal{}0$
2/$ 1/b_n.\sum_{i\equal{}1}^n E[X_i .1_{|X_i|<b_n}|\mathcal{F}_i]$  tends to 0 in probability.
3/$ Lim_{n\rightarrow \plus{}\infty}1/b_n^2 \sum_{i\equal{}1}^n E[X_i .1_{|X_i|<b_n}^2]\minus{}E[E[X_i .1_{|X_i|<b_n}|\mathcal{F}_i]]^2\equal{}0$

The reciprocal statement is true if the $ X_n$ are independent.

So I think that you have to check that your hypothesis implies point 1/2/ and 3/. But I think you might need some extra intregration conditon for your sequence $ X_n$

Thank you for all the help, I'm sorry if I haven't been that clear about what I'm trying to prove. I'll try to be more precise in the future.
It should be: Let $ (X_n: n\in\mathbb{N})$ be a sequence of independent r.v. such that for any given $ n\in\mathbb{N}, \lim_{a\rightarrow \infty}\max_{1\leq k\leq n}\int_{|x|\geq a}|x|d\mathbb{F}_{X_k}(x)\equal{}0$ then $ (\mathbb{P})\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k\equal{}1}^n(X_k\minus{}\mathbb{E}X_k)\equal{}0$.

@fedja: I still don't understand how do you get some bound for the variance, or at least how to show that the variance of $ \frac{1}{n}\sum_{i\equal{}1}^nX_i$ can be made arbitrary small.

suppose that all the random variables are centred to keep things simple:

can you solve the exercice if you suppose one uniform bound: there exists $ a$ such that for all $ n$, $ \| X_n \|_{\infty} \leq a$? Basically, this is what you where saying about the variance of $ \frac{1}{n} \sum_1^n X_n$

So now, if there is no uniform boundedness assumption, you can still use the hypothesis that $ \max_n E[ |X| 1_{X>a}]$ tends towards $ 0$ as $ a$ tends to $ \infty$ (umiform integrability). Then, for a fixed $ \epsilon$, you can introduced $ X^{(a)}_n \equal{} X 1_{|X|<a}$ and $ Y^{(a)}_{n}$ such that $ X_n \equal{} X^{(a)}_n \plus{}Y^{(a)}_n$. Because $ X^{(a)}_n$ is bounded by $ a$, you can do this part, and because
$ P(|  \frac{\sum_1^n X_n}{b } | > \epsilon) \leq P(|  \frac{\sum_1^n X^{(a)}_n}{n } | > \frac{\epsilon}{2})\plus{}P(|  \frac{\sum_1^n Y^{(a)}_n}{n } | > \frac{\epsilon}{2})$
it should not be hard to conclude. Do you see how ?

[quote="Mr.Doe"] such that for any given $ n\in\mathbb{N}, \lim_{a\rightarrow \infty}\max_{1\leq k\leq n}\int_{|x|\geq a}|x|d\mathbb{F}_{X_k}(x) \equal{} 0$ [/quote]
This is not a condition at all; it merely means that each variable has finite expectation, which does not really allow you to conclude anything. You need some uniformity in $ n$ for the statement to be true.


Okay. So I guess I will start.

During the 1st week of high school in early August, I remember when some upperclassmen were in the hallway, and as I was walking to class after lunch, a short, sturdy boy suddenly jumped right in front of me and let out a terribly feminine scream. As a result, I dropped my Spanish book, and it went clunk. Then, people around him burst out with laughter.

Mine's lame, but I'm sure a lot of you have much better ones.

I was rather late to class, because my locker was jammed (the school gives us half lockers and still expects us to store all of our textbooks, notebooks, backpacks, coats...  but that's another tangent) and my locker was upstairs and my class was downstairs.  So I was running, taking the stairs two at a time, and caught my heel on one or tripped over my own feet or something and suddenly I was tumbling down the stairs (of course I did this at the very top of them) and landed pretty much on my head at the bottom, with my books and papers scattered all around me.  To make matters worse, a PE class was walking outside to the fields and saw it all happen...  I won't ever forget that.

Ok,

So I was trying go to a math class, but the door was locked. So I knocked and waited for about 5 minutes. I tried knocking again and no one came to the door. Then I ran my head into the door, and at that moment the door opened, and I barely missed the teacher, ran into the podium and fell down.

Hmm...

Once in 4th grade, I was having recess (in case you forgot  :P , that's when you take a break and go on the play ground), and my pants literally ripped in front of 15 people. I had to go up to the office and wait for my mom to bring me another pair of pants.

Field trips: In second grade, it sucked. We had to go to the BFI, to learn about recycling and stuff. It was stinky.
Luckily, we also got to go to an ice cream factory, that was cool.

Science camp was cool. I injured my leg on the last day. In my cabin, there was vandalism all over, saying stuff like "Seth loves Cathy" and "Cathy is a lesbion". Coincidentally, on one of the beds was the name of one of my cabin mates. We were all joking about how someone was stalking him. One of my cabin members was really hyper, and started rolling around in his sleeping bag and was shouting "I'm a caterpillar!". For if I got bored, I brought a science magazine. There was an ad with a picture of an old grandma, which I accidentally called grandpa. We were cracking up about it the whole week.

yo

the other day in PE we were doing this thing where you have to get everybody across a rope swing and get a bucket of water across without a drop spilling which kinda requires you to hold onto the rope with one hand

and i'm somewhat athletic so i volunteered to take it across

so i get it to the other side and somebody takes the bucket of water while i'm still on the rope, and it's all cool

and i swing back to the other side while straddling the rope, and on the other side somebody gets the bright idea to pull me to safety by pulling the rope

which is between my legs

pain

One of my friends (he's a user here, but I won't name him) was in China or something at the beginning of his freshman year in high school, so he missed the first week or two of our chemistry class (I was in 10th grade and in the same chem class). So he missed the introductory lab where we learned about all the lab equipment and did some basic things, and he had to come after school to do it. He didn't know what any lab equipment was, since he also missed the class. The teacher was doing something at the front of the room and told him to light the Bunsen burner (which she had left out on the counter in the lab for him). But he didn't know what it was... the teacher turned around a minute later and saw him trying to light the $ \$$2000 digital balance on fire...


trigonometry
inequalities
geometry proposed
geometry

Prove that

$ \frac {\cos^2 A}{a^2} \plus{} \frac {\cos^2 B}{b^2} \plus{} \frac {\cos^2 C}{c^2}\geq\frac {1}{4R^2}$

We have $ \sum\frac{\cos^2 A}{a^2}\equal{}\sum\frac{1\minus{}\sin^2 A}{a^2}\equal{}\sum\frac{1\minus{}\frac{a^2}{4R^2}}{a^2}\equal{}\sum\frac{4R^2\minus{}a^2}{4R^2a^2}$

Thus, the inequality reduces to : $ \sum\frac{4R^2\minus{}a^2}{a^2}\ge 1\ \Longleftrightarrow\sum\frac{4R^2}{a^2}\ge 4\ \Longleftrightarrow\ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\ge \frac{1}{R^2}$

This results from [b]Cauchy-Schwarz[/b] inequality and $ a^2\plus{}b^2\plus{}c^2\le 9R^2\ :$ $ \frac{1}{a^2}\plus{}\frac{1}{b^2}\plus{}\frac{1}{c^2}\ge \frac{9}{a^2\plus{}b^2\plus{}c^2}\ge\frac{1}{R^2}$ .

To virgil nicula.

${ \left(\frac{a^{2}+b^{2}+c^{2}}{4S\sqrt 3}^{2}\ \right)^{2}\ }$
where does the intermediate square work?
i think you mistyped the square sign..

[quote="Virgil Nicula"]Prove that one stronger inequality $ \boxed {\ \frac {\cos^2A}{a^2} \plus{} \frac {\cos^2B}{b^2} \plus{} \frac {\cos^2C}{c^2}\ \ge\ \frac {1}{4R^2}\cdot \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{4S\sqrt 3}\ \right)^2\ }\ \ge\ \frac {1}{4R^2}$ .[/quote]
[b][u]Proof[/u].[/b] $ 4R^2\cdot\sum \frac {\cos^2A}{a^2} \equal{}$ $ \sum\cot^2A \equal{} \sum\left(\frac {b^2 \plus{} c^2 \minus{} a^2}{4S}\right)^2\ge$ $ \frac {1}{16S^2}\cdot\frac 13\cdot\left[\sum\left(b^2 \plus{} c^2 \minus{} a^2\right)\right]^2 \equal{} \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{4S\sqrt 3}\right)^2\ge 1$ .

[quote="Virgil Nicula"][quote="Virgil Nicula"]Prove that one stronger inequality $ \boxed {\ \frac {\cos^2A}{a^2} \plus{} \frac {\cos^2B}{b^2} \plus{} \frac {\cos^2C}{c^2}\ \ge\ \frac {1}{4R^2}\cdot \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{4S\sqrt 3}^2\ \right)^2\ }\ \ge\ \frac {1}{4R^2}$ .[/quote]
[b][u]Proof[/u].[/b] $ 4R^2\cdot\sum \frac {\cos^2A}{a^2} \equal{}$ $ \sum\cot^2A \equal{} \sum\left(\frac {b^2 \plus{} c^2 \minus{} a^2}{4S}\right)^2\ge$ $ \frac {1}{16S^2}\cdot\frac 13\cdot\left[\sum\left(b^2 \plus{} c^2 \minus{} a^2\right)\right]^2 \equal{} \left(\frac {a^2 \plus{} b^2 \plus{} c^2}{4S\sqrt 3}\right)^2\ge 1$ .[/quote]

Dear friend [color=red][b]Virgil [/b][/color]who is the author of the inequality? where is published? thanks

[quote="Virgil Nicula"]Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...[/quote]

Is very nice and stronger.... Congratulation my friend  :wink:

[quote="Virgil Nicula"][quote="Ligouras"][quote="Virgil Nicula"]Thank you, [b]Liguras[/b] for your words ! This stronger inequality is mine. But I think that it is easily and at most interesting ...[/quote]Is very nice and stronger.... Congratulation my friend  :wink:[/quote]
[b]Thank you ![/b] I consider that this inequality is my present for you in these days. [b]Happy New Year ![/b][/quote]

it is a honor for me dear friend....  :oops:
with great respect

[i][b]"[color=blue]T[/color][color=red]à[/color][color=blue]kis[/color]"[/b][/i]

$ \frac {\cos^2 A}{a^2} \plus{} \frac {\cos^2 B}{b^2} \plus{} \frac {\cos^2 C}{c^2}\geq\frac {1}{4R^2} \iff cotA+cotB+cotC\ge 1.$


[i]Ελπίζω αυτή η ασκησούλα να αρέσει :[/i]

Εξωτερικά ενός ορθογωνίου τριγώνου $ ABC$ με $ \angle A = 90$ θεωρούμε τρίγωνο $ BCD$ με $ AB=BD$.

Αν $ AH \perp BC$  και $ O$ είναι το περίκεντρο του $ \triangle BCD$ , να αποδείξετε ότι $ DH \perp BO$

 Μπάμπης

Afou $ BH\cdot BC \equal{} BA^2 \equal{} BD^2$,    i  $ BD$ einai efaptomeni tou perigegrammenoun kiklou tou $ \triangle HDC$ opote $ \widehat{BDH} \equal{} \widehat{BCD}$ (1).Eukola fainetai twra oti i $ \widehat{BCD}\leq\widehat{BCA}$ ara einai panta okseia ara $ \widehat{DBO} \equal{} 90^\circ \minus{} \widehat{BCD}$ (2). Apo tis (1) kai (2) exoume oti $ DH\perp BO$.

Mια άλλη λύση είναι:

Είναι γνωστό (από 2ο θέωρημα διαμέσων) ότι αρκεί $ BD^2\minus{}BH^2\equal{}OD^2\minus{}OH^2$.

Πράγματι:

$ BD^2\minus{}BH^2\equal{}AD^2\minus{} BH^2\equal{}AH^2\equal{}BH\cdot HC\equal{}R^2\minus{}OH^2\equal{}OD^2\minus{}OH^2$. (R η ακτίνα του περιγεγραμμένου του $ BCD$).


AMC
AIME
MATHCOUNTS
AMC 10
geometry
calculus
3D geometry

i hope AIME 2  is easy because i am taking that, because march 13 we are on spring break. :D

i wish we took the 2. i would do 7hrs a day of AIME and then get like 7 or 8 on the test. but no, i have to take it from 8-11(it's like 7-10 because of the time change) on March 13th. rawr

i was wondering if anyone knew,about the index. see, i here you add your 12 score with your 10 score or something and then add your AIME score. but you see i qualified for AIME through 10, i didn't even take 12, but i here something about a floor score for people under 10th grade or something so now i am really confused. could someone explain this to me.

OK, i will assume you are under 10th grade.

your index is:

AMC 10 score + 10*AIME score.

the "floor" is suppose the lowest index that qualified was 210 and that person got 150/6.

then if you scored 6+, no matter what your AMC score was, you qualify for USAMO.

(mathcountsfreak) any other scores?

[quote="mathcountsfreak"]Didn't make it!
I was hoping on doing better on 10B, but then my dad woke me up late, and I had 30 minutes less time to do the contest, so I had to think about speed a little more, and even so I had to guess on around 9.[/quote]

that's gotta hurt...but at least you don't live in a retarded school district and you got to take both. What I'd give to take the 10b also, especially since i made a careless mistake on the 10a, which kept me from getting a 141...i was soooo close!

P.S. i...[color=red][size=150]hate[/size][/color]...compliment....counting!!!! :cursing:

Mathcrazed please post the url of the indiana forum on your signature.

did i read something wrong or did someone say west lafayette doesn't allow people to take the 10B/12B?

omg, seriously! we r allowed to, technically, but our school hates anyone who goes against the flow. they hadted the whole Benjamin family cuz they went above and beyond. west lafayette is sooo lazy!

so they hate and don't do anything about it.. i really don't care..?

Can you take both 10 and 12 Undefined(do you mind if i use your name, everyone probably knows who you are by now.), do you need to ask Mr. A or something.

ok
ne ways, if i wanted to skip to high school, i could (easily), but that's only because my dad teaches me math (he's a college professor)... so... ya... could've easily gotten 9 on the AIME, makes me feel so stupid...  :mad:  :mad:  :mad:

[quote="undefined117"]heh my dad's a nuclear engineer.. and kinda leaves me alone, so yeah. i look things up in wolfram mathworld and wikipedia and stuff. i get along..[/quote]

i thought u said ur dad was lecturing u about calculus when i gave u the problem w/the sphere and cylinder. i'm not going to lie. my dad teaches me....

uhhuh.. your dad teaches you.. my dad talks/writes some stuff down on paper and leaves me to contemplate.

maybe that's why ur so good at contemplating :lol: and i'm not. i get frustrated really easily...

asian superiority (pwnage :D...) indian=asian, seriously... :D

hey, like everyone on the indiana forum is asian (except for owna i think). no wonder indiana owns (actually that isn't the main reason, but oh well :lol: )

[quote="usaha"]No, actually indians do not equal asians (to be more specific, ill tell you next saturday..)[/quote]

i take it ur indian (ur name sounds indian)?

umm,.. india is in asia, am i correct there?

yeah but usually when people say asian they mean korean/filipino/japanese/chinese

OMG if india isn't part of asia then is it antarctican or something.


limit
logarithms
geometric sequence
calculus
calculus computations

Prove that \[ \gamma\equal{}\lim_{n\to\infty}\left[\left(\frac{2^m}{1}\plus{}\frac{3^m}{2}\plus{}\cdots\plus{}\frac{n^m}{n\minus{}1}\right)\plus{}m\minus{}(S_n^0\plus{}S_n^1\plus{}\cdots\plus{}S_n^{m\minus{}1})\minus{}\ln (n\minus{}1)\right],\] where $ S_n^m\equal{}1^m\plus{}2^m\plus{}\cdots\plus{}n^m$ and $ \gamma$ is Euler's constant.

it's true for all $ m$
it's easy you only need to check that $ \sum_{0 \leq i \leq m \minus{} 1} S_{n}^{i} \equal{} m \plus{} \sum _{ 1 \leq i \leq n \minus{} 1} \frac {(i \plus{} 1)^{m} \minus{} 1}{i}$
(use geometric progression)
and in finnaly you have :
$ \lim ... \equal{}\lim _{n\minus{}>\plus{}oo} \sum _{ 1 \leq i \leq n \minus{} 1} \frac {(i \plus{} 1)^{m}}{i} \plus{} m \minus{} (m \plus{} \sum _{ 1 \leq i \leq n \minus{} 1} \frac {(i \plus{} 1)^{m} \minus{} 1}{i} ) \minus{} \ln (n \minus{} 1)$
 =$ \lim _{n\minus{}>\plus{}oo}   H_ {n \minus{} 1} \minus{} \ln (n \minus{} 1) \equal{} \gamma$ (by definition)
where $ H_{n} \equal{} \sum_{0 < i < n \plus{} 1} \frac {1}{i}$


geometry
search
AoPSwiki
trigonometry
function
Pythagorean Theorem
area of a triangle

What is the area of a equaladaral triangle :D

Please search this in AoPSWiki or Wolfram before asking it in a fourm.

[quote="123456789"]The area of an equilateral triangle with side length $ s$ is given by $ \frac { s^2\sqrt {3}}{4}$.

This can be derived by dropping an altitude, and using your familiar sine and cosine functions to find the base and height of the triangle.[/quote]
I'm pretty sure all you need is the Pythagorean Theorem.
If you have an equilateral triangle $ ABC$, with side length $ s$, the area is as follows:
Dropping an altitude, you get two right triangles.  They both have hypotenuse length $ s$ and base length $ \frac{1}{2}s$.  Using the Pythagorean Theorem, the square of the height of the altitude is $ s^2\minus{}\frac{1}{4}s^2$, so the altitude is $ \sqrt{\frac{3}{4}s^2}$  This can be simplified to $ \frac{\sqrt{3}}{2}s$.  Since the area of a triangle is $ \frac{1}{2}bh$, and the base is s, while the height is $ \frac{\sqrt{3}}{2}s$, the area is $ \frac{s^2\sqrt{3}}4$

This was on another math forum problem of the week and me and another person spent 2 hours of english class coming up with the proof for this... what a waste :P

[quote="shaun ong"]Area of equilateral triangle? I tought the formula to find equilateral triangle is the length multiply by height /2 as it is the same for all triangles?[/quote]

That is true for all triangles but what if you're only given three sides to an equliateral triangle? You won't know the height and therefore you'll have to apply different formulas for different situations.

[quote="7h3.D3m0n.117"]That is true for all triangles but what if you're only given three sides to an equliateral triangle? You won't know the height and therefore you'll have to apply different formulas for different situations.[/quote]

if it was equallateral and you knew the sidlengths couldnt you just use the pythagorean theorum to find the altitude because the altitude in an equallateral triangle bisects the base at a right angle.  so you could take the square root of half of the base squared subtracted from the side length squared and you will have the altitude.  This can be done without any knowlege of the formula $ \frac{s^2\sqrt{3}}{4}$.  Now yes the formula does save valuable time in a countdown or a sprint so it is a good thing to know!


Ring Theory
superior algebra
superior algebra unsolved

Find the cardinality of the set of all Subrings of $\mathbb{Q}$.

It seems like it should be simple, but I can't do it yet..

The cardinality is $|\mathbb{R}|=2^{|\mathbb{Q}|}$. Of course this is an upper bound, so we have to prove, that there are at least so much subrings.

For each subset $Q\subseteq\mathbb{P}$ the set
$\{\frac{r}{s}| gcd(r,s)=1 \wedge \forall p\in\mathbb{P}: p|s \Rightarrow  p\in Q\}$
is a subring. Of course these subrings are pairwise not equal.

So there are at least $2^{|\mathbb{P}|}=|\mathbb{R}|$ subrings of $\mathbb{Q}$.


The sum of the digits of a positive three-digit palindrome is equal to the square of the product of the digits. What is the palindrome?

Let the number be $ \overline{aba}$ so that $ 2a\plus{}b\equal{}a^4b^2$. It is easy to see that neither of $ a$ and $ b$ can be $ 0$, so $ a\equal{}\{1,2\}$ are the only possibilities. If $ a\equal{}2$, $ b\equal{}1$, which doesn't work, so $ a\equal{}1\implies2\plus{}b\equal{}b^2\implies b^2\minus{}b\minus{}2\equal{}(b\minus{}2)(b\plus{}1)\equal{}0\implies b\equal{}2\implies\boxed{121}$.


MATRIX (1,0,1)(2,2,6)(3,1,a)

Inverse the matrix using the gauss method.....

solve the system      x+z=1
                              2x+2y+6z=2
                              3x+y+az=3

Is this what you mean?

1 0 1 | 1 0 0
2 2 6 | 0 1 0
3 1 a | 0 0 1

1 0 1     | 1 0 0
0 2 4     |-2 1 0
0 1 (a-3)|-3 0 1

1 0 1       | 1 0 0
0 2 4       |-2 1 0
0 0 -2a+10| 4 1 -2

...proceeding to convert the left matrix into the identity matrix, at which time your inveted matrix will be on the right.

(Note that you will need to analyze two cases: $a=5$, for which the matrix won't have an inverse, and $a\neq5$, for which it will.)


What is the units digit of $ 1997^{1997}$?

It is well known that these units digits of powers cycle in 4's.

$ 1997^{1997} \equiv 7^1 \equiv 7 \pmod 10$. 7.


Let $a,b,c$ sides of a triangle.Prove the inequality:

            $a^3+b^3+c^3+9abc\leq 2[bc(b+c)+ca(c+a)+ab(a+b)] $  ;)

hm. this seems straightforward, so I most likely made a mistake.

the inequality is untrue for any positive a, b, c. Take $a=1$, and then let $b, c$ tend to $0$. We get $1 \le 0$. So presumably a, b, c are the sides of the triangle. If we substitute in a = x + y, etc, we get $\sum_{\text{sym}} x^3 + 12x^2y + 3xyz \le \sum_{\text{sym}} 2x^3 + 10x^2y + 4xyz$, which is Schur.

yes,you are right.i forgot to say that $a,b,c$ are sides of a triangle.Thanks anyway. ;)


Let n be a positive integer; if the equation 2x+2y+z=n has 28 solutions in positive integers for x,y, and z, what is n?(Give 2 possibilities)

AHSME 1989 No.27

[quote="eryaman"]Unless I'm making things too complicated, I think this should go in intermediate. [hide]Since $x$ and $y$ must be even natural numbers we can form the following generating functions for both:
$(x^2+x^4+x^6+...)$.
Since $z$ is any natural number we can form the following generating function:
$(x+x^2+x^3+x^4+...)$.

So we want the exponent of the term with coefficient $28$ in the expansion of $(x^2+x^4+x^6+...)^2(x+x^2+x^3+x^4+...)$.

Factoring, we have:
$x^5(1+x^2+x^4+x^6+...)^2(1+x^2+x^3+...)$

Expanding, we have:
$x^5(1+2x^2+3x^4+4x^6+5x^8+...)(1+x+x^2+x^3+x^4+x^5+...)$
$x^5(1+x+3x^2+3x^3+6x^4+6x^5+10x^6+10x^7+...)$

Notice that the coefficients are the triangular numbers repeated twice.
Continuing the pattern, we have $x^5(1+x+3x^2+3x^3+...28x^{12}+28x^{13}+...)$  Multiplying out (the $x^5$ term) we find the two terms $28x^{17}+28x^{18}$.

So our two answers are $17$ and $18$.  I hope I didn't make any mistakes...[/hide][/quote]

ok


geometry
circumcircle
geometric transformation
reflection
parallelogram
trigonometry
power of a point

Denote by $ O$, $ O'$ the circumcenters of triangles $ ABC$ and $ BCE$, respectively. We have $ \angle HBC\equal{}\angle HAD\equal{}\angle HEC$, it follows $ H$ lies on circles $ (BCE)$. On the other hand, we can easily (maybe well-known as well) prove that $ AOO'H$ is a parallelogram. Let $ H'$ be the reflection point of $ H$ in $ O'$, so we have quadrilateral $ AOH'O'$ is a parallelogram, too. It implies $ AH'$ passes through the middle point of $ OO'$ which is also the middle point of $ BC$. Therefore we conclude that $ AH'$ passes through $ F$, so $ HFH'\equal{}90^{\circ}$, i.e. points $ A$, $ D$, $ H$, $ F$ all lie on the circle with diameter $ AH$.

Er, so complicated. Sorry for my unclear writing.  :D

Muy fácil ese problema, nada mas que no sabia dibujarlo xD...

[quote="campos"]Let $ ABC$ be an acute-angled triangle, and let $ H$ be its orthocenter. Let $ D$ be the foot of the altitude from $ B$ to $ AC$, and let $ E$ be the reflection of $ A$ on $ D$. The circumcircle of triangle $ BCE$ intersects the median from $ A$ in an interior point $ F$. Prove that $ A, D, H$ and $ F$ are concyclic.[/quote]
We notice that $ B,H,E,C$ are concyclic as $ \angle BHC\equal{}\angle BEC\equal{}180^{\circ} \minus{}A$. It is well known that the radius of the cicumcircle of $ \triangle ABC$ and $ \triangle BHC$ are equal ($ \equal{}R$).
Let $ H',A'$ be the midpoint of $ AH,BC$ respectively and $ O'$ be the circumcenter of $ \triangle BHC$.
We have $ AH'\equal{}R \cos A$ and also $ \angle BO'A'\equal{}A$. So, $ O'A'\equal{}R\cos A$.
So, $ A'O'\equal{}AH'$ and $ AH' \parallel O'A'$. So, $ H'O' \parallel AA'$. But $ H'O'$ is perpendicular to the radical axis of the circles and so is $ AA'$. So, $ \angle AHA\equal{}90^{\circ}$. So, $ F$ lies on the circumcircle of $ \triangle ADH$ and we are done.

[quote]Let $ ABC$ be an acute-angled triangle, and let $ H$ be its orthocenter. Let $ D$ be the foot of the altitude from $ B$ to $ AC$, and let $ E$ be the reflection of $ A$ on $ D$. The circumcircle of triangle $ BCE$ intersects the median from $ A$ in an interior point $ F$. Prove that $ A, D, H$ and $ F$ are concyclic.[/quote]
[b]Another approach:[/b]
At the very first, it is very easy to see that $ \minus{} \overline {DA}.\overline {DC} \equal{} \overline {DH}.\overline {DB}$. But $ \overline {DA} \equal{} \minus{} \overline {DE}$. Hence, it follows that $ \overline {DH}.\overline {DB} \equal{} \overline {DE}.\overline {DC}\Longrightarrow B,H,E,C$ are concyclic, which implies that $ F\in (BHC)$ also. Let $ A_1,C_1$ be the projections of $ H$ onto $ BC, AB$, respectively. Consider the inversion through pole $ A$, power $ \overline {AH}.\overline {AA_1} \equal{} \overline {AC_1}.\overline {AB} \equal{} \overline {AD}.\overline {AC} \equal{} k^2$. $ \mathcal {I}(A,k^2): H\mapsto A_1$, $ B\mapsto C_1$, $ C\mapsto D$. Hence $ \mathcal {I}(A,k^2): (BHC)\mapsto (C_1DA_1)$- which is the $ 9$- point circle wrt $ \triangle ABC$. Thus $ \mathcal {I}(A,k^2): F\mapsto A'$. But $ A',B,C$ are collinear, $ \mathcal {I}(A,k^2): BC\mapsto (AHD)\Longrightarrow F\in (AHD)$. The result is lead as follow.
Our proof is completed then.

Denote the points like my figure.
Let $ \{I \} \equal{} AM \cap (BCE)$,by angle-chasing we have ABIC is parallelogram so $ \overline{AI} \equal{} 2 \overline{AM}$
We have:$ 2\overline{AF}. \overline{AM} \equal{} \overline{AF}. \overline{AI} \equal{} \overline{AE}. \overline{AC} \equal{} 2\overline{AD}. \overline{AC} \equal{} 2\overline{AH}. \overline{AK}$ so $ H,F,M,K$ are concyclic so $ \hat{HFM} \equal{} \hat{ADH} \equal{} \frac {\pi}{2}$ then $ A,H,F,D$ are concyclic.

[img]http://img37.imageshack.us/img37/9916/huhu.jpg[/img][/img]


inequalities
rotation
geometry
parallelogram
triangle inequality

For a triangle , Prove that
$ b \plus{} c \geq 2m_a$

If $ X$ is the midpoint of side $ b$, and $ Y$ is the midpoint of side $ a$, then $ XY\equal{}\frac{c}{2}$ , $ m_a\equal{}AY$ and $ AX\equal{}\frac{b}{2}$ so your inequality follows directly from a triangle inequality :)

Wow nice  .......... :lol:

Any other proofs? :wink:

yes :) if you rotate the triangle by $ 180^o$ around the midpoint of sida $ a$ you'll get a parallelogram with sides $ b$ and $ c$ and with diagonal $ 2 \cdot m_a$ so again by applying triangle inequality you get the desired result

Exactly!!!! That was the one i found (after half an hour  :maybe: )

Alternate method:
To prove:
$ b\plus{}c \ge 2m_a$
$ b\plus{}c \ge 2* 1/2 * \sqrt{2b^2 \plus{} 2c^2 \minus{} a^2}$

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

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

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

$ a \ge (b\minus{}c)$ which is true by triangle inequality

[hide=" Doubt"]
I dont think equality holds here? ( last inequality is strict ) .... can someone clarify?[/hide]


Find, with proof, the unique square which is the product of four consecutive odd numbers

[hide="Solution"]
Let the numbers be $(2n-1), (2n-3), (2n-5),$ and $(2n-7)$. Also, let $4n^{2}-16n+7 = x$ and $a^{2}$ be the square we are looking for, for some integers $x$ and positive integer $a$. So we have

\begin{eqnarray*}(2n-1)(2n-3)(2n-5)(2n-7) &=& (4n^{2}-16n+7)(4n^{2}-16n+15) \\ &=& x(x+8) = a^{2}\Rightarrow x =-4 \pm \sqrt{16+a^{2}}\end{eqnarray*}

Since $x$ must be an integer, set $16+a^{2}= k^{2}$, for some integer $k$. The only solutions in integers to this equation, in the form $(a,k)$, are $(2,-8), (4,-4),$ and $(8,-2)$. Solving gives that $a=0, \pm 3$, so the only plausible value for $a$ is $3$. Hence, $x=1,-9$.

Plugging in $x=1$ to solve for $n$ gives no solution in integers. Plugging in $x=-9$ shows that $n=2$ is the only solution for $n$. Therefore, the unique square is $(3)(1)(-1)(-3) = \boxed{9}$.
[/hide]

A different approach:

[hide]
Let $k_{n}= 2n+1$. Note that $k_{n+1}-k_{n}= 2 \not | k_{n}, k_{n+1}$, so they are relatively prime. Similarly, $k_{n}$ and $k_{n+2}$ are relatively prime. $k_{n}$ and $k_{n+3}$ differ by 6, so they are not relatively prime only if both are divisible by 3.

Now, consider $k_{n}k_{n+1}k_{n+2}k_{n+3}$

There are two cases.

[b]Case 1:[/b] $3 \not | k_{n}$, in which case all four $k$'s are relatively prime, so each term is either -1 or 1 (the product must be a square, so the product must be 1). However, this leaves 2 possibilities for 4 numbers, so clearly we can't have consecutive odd numbers.

[b]Case 2:[/b] $3 | k_{n}$. Note that $k_{n}/ 3$ and $k_{n+3}/3$ differ by 2 and are odd, so they must be relatively prime.

Then, $(9)(k_{n}/ 3) k_{n+1}k_{n+2}(k_{n+3}/3)$ is a square, implying that $(k_{n}/ 3) k_{n+1}k_{n+2}(k_{n+3}/3)$ is a square. However, since all 4 terms are relatively prime, they must all be either 1 or -1.

This leaves only one solution: $9(-1)(-1)(1)(1)=(-3)(-1)(1)(3)=9$

[/hide]

nice solution Zuton. much easier than mine  :P


conics
parabola
geometry
rectangle
ratio
calculus
calculus computations

A negative parabola is inscribed in a rectangle, toughing two of its corners. What is the ratio of the area under the parabola to the area of the rectangle?

[hide="solution"]
If the parabola is of height $ h$ and width $ 2w$, then the area underneath is $ \frac {4hw}{3}$ (use $ y\equal{}h\minus{}\frac{h}{w^2}x^2$ as the equation of the parabola by centering the origin on the midpoint of the bottom side of the rectangle).
The area of the rectangle is $ 2hw$.
So the ratio is $ \frac {2}{3}$
[/hide]


The total area of all the faces of a rectangular solid is $22 \text{cm}^2$, and the total length of all its edges is $24 \text{cm}$. Then the length in $\text{cm}$ of any one of its internal diagonal is

A. $\sqrt{11}$
B. $\sqrt{12}$
C. $\sqrt{13}$
D. $\sqrt{14}$
E. Not uniquely determined

[hide]Call the sides $a,b,c$, thus: $2bc+2ab+2ac=22$ and $4b+4a+4c=24$ so $bc+ab+ac=11$ and $a+b+c=6$. And therefore the diagonal would be $\sqrt{a^2+b^2+c^2}$. Now you can see that $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac) \Longrightarrow 36=a^2+b^2+c^2+2(11) \longrightarrow \sqrt{a^2+b^2+c^2}=\boxed{\sqrt{14}}$.[/hide]

[quote="dangvh"][quote="Iversonfan2005"][hide]Call the sides $a,b,c$, thus: $2bc+2ab+2ac=22$ and $4b+4a+4c=24$ so $bc+ab+ac=11$ and $a+b+c=6$. And therefore the diagonal would be $\sqrt{a^2+b^2+c^2}$. Now you can see that $(a+b+c)^2=a^2+b^2+c^2-2(ab+bc+ac) \Longrightarrow 36=a^2+b^2+c^2-2(11) \longrightarrow \sqrt{a^2+b^2+c^2}=\boxed{\sqrt{58}}$.[/hide][/quote]

it should be $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)=36$, not $-2(ab+bc+ac)$[/quote]

Oops..fixed it :blush:


Hey, I am just desperate to know your opinions on what would be best to place on a three-fold poster that would easily explain thermodynamics in a user-friendly manner.  And we're not talking hardcore thermo; just about the first three laws and then nicely explaining what heat is and a nice heat vs pressure explanation, also explaining how 100% efficiency cannot be achieved.  So yeah, please give your advice on what you would place on the poster that'd be fun.  And I'm not just a guy who's oblivious to thermodynamics, I just want to make a poster that is as enjoyable as possible to learn from for someone other than me.

Why don't you consult some high school books? You will find for sure some nice pictures and easy explanations.

Yeah, that's where I am currently at, I just figured someone might have something unique in their head.


I'd like to learn something about optimization, because I've heard it can be useful sometimes and I know almost absolutely nothing about it. I'll be glad if any of you could give some links to texts about it [I've read about optimization on wikipedia, but there's nothing interesting]  and could solve some problems with optimization.

I think you should read many books about this thing. Especially the book about CONVEX ANALYSIS of ROCKKAFELLER.

Good luck!


Hi Ph-An.
$(a^2+1)(b^2+1)=a^2+b^2+a^2b^2+1\geq (ab+1)^2$=>
$(a^2+1)(b^2+1)(c^2+1)\ge (ab+1)(bc+1)(ac+1)=(a+1)(b+1)(c+1)$ :)

Ph-An also looking this
[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61826]www.mathlinks.ro/Forum/viewtopic.php?t=61826[/url] :D  :D

Just take the calculation and we will use $\sum a^2b^2\geq \frac{(\sum ab)^2 }{3},\sum ab\geq 3 , \sum a^2b^2\geq \sum ab$.Simmilar we have $\sum a^2\geq \sum a$.So by adding 2 we obtain the inequality.:)

Here is [u]my solution:[/u]

$2(x^2+1)^3\ge (x+1)^3(x^3+1)\Longleftrightarrow (x-1)^4(x^2+x+1)\ge0$, what is true.

Therefore, $8\prod (a^2+1)^3\ge \prod (a+1)^3\cdot \prod (a^3+1)$.

But ([u]Huygens's inequality[/u]) $(a^3+1)(b^3+1)(c^3+1)\ge (abc+1)^3.$

Thus, $8\prod (a^2+1)^3\ge (abc+1)^3\cdot \prod (a+1)^3\Longleftrightarrow 2\prod (a^2+1)\ge (abc+1)\cdot \prod (a+1).$

For $abc=1$ results the proposed inequality.


function
real analysis
real analysis unsolved

Determine whether the following series converges:

1 + 1/2 + 1/3 - 1/4 - 1/5 - 1/6 + 1/7 + 1/8 + 1/9 - 1/10 - 1/11 - 1/12 + ... = sum_n=1 to oo { C(n)/n }, where C(n) = (-1)^j, with j = (n/3)-1, if n/3 is an integer or [n/3], if n/3 is not an integer.

This is a classical result.
A more general result holds:

If you consider the alternanting harmonic sum of the form $\frac{1}{n}$ where the first $k$ terms are considered with $+$ and the next $k$ terms with $-$ and so on, then such a series is known to converge, however if the numbers of terms with $+$ is not equal to the number of terms with $-$ then the series diverges.

Let's see what we can get from summation by parts:

Consider $\sum_{n=1}^{\infty}s(n)b(n)$ where

$s(n)=\pm1$ and $b(n)$ is a decreasing function of $n$ that tends to zero as $n\to\infty.$

Let $A(n)=\sum_{k=1}^{n}s(k).$ We have that

$\sum_{n=1}^{N}s(n)b(n)=A(N)b(N)+\sum_{n=1}^{N-1}A(n)(b(n)-b(n+1)).$

Now take the limit as $N\to\infty.$ If we want to be able to prove convergence, then a useful sufficient condition would be that $A(n)$ is bounded. That is, there exists $M$ such that $A(n)\le M$ for all $n.$ In the limit, we then have that

$\sum_{n=1}^{\infty}s(n)b(n)= \sum_{n=1}^{\infty}A(n)(b(n)-b(n+1)).$

But the sum on the right is absolutely convergent:

$\sum_{n=1}^{\infty}|A(n)(b(n)-b(n+1))|\le M\sum_{n=1}^{\infty}(b(n)-b(n+1))=Mb(1)$

(We used a telescoping series there.)

What does it take for $A(n)$ to be bounded? Long-term balance between the plus signs and the minus signs, and no arbitrarily long runs of either kind of sign. This is more or less what didilica said. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121050]this[/url] recent topic by Cezanne123 in which there were arbitrarily long runs of each kind of sign.


[url]http://en.wikipedia.org/wiki/File:Cayley_graph_of_F2.svg[/url]

could any one please explain for me why the universal covering of the figure 8 looks like above.
I read it from Hatcher 's book ( page 59) more than ten times but still don't get it . And by the way, how to find the universal covering of $ A\bigvee B$,  for example $ S^{2}\bigvee \mathbb{R}P^{2}$ ? what are the ideas behind? Thanks so much [/url]

The figure eight has a four-way fork, and each path loops back to the fork. The universal cover has to unfold all loops while preserving the local structure; this means that every path away from the fork leads to another copy of the fork. Iterate that, and you get the pictured tree.


Find the volume of a doughnut  with inner diameter $ 2$cm and boundary diameter $ 2$cm. Generalize.

[hide]$ V \equal{} \left(\pi r^2\right)\left(2\pi R\right) \equal{} 2\pi^2Rr^2$

$ R \equal{} r \equal{} 2\implies V \equal{} 2\pi^2\cdot2\cdot2^2 \equal{} 16\pi^2$[/hide]

This is a special case of [url=http://mathworld.wolfram.com/PappussCentroidTheorem.html]Pappus' Centroid Theorem[/url].

i_like_pie did it quite nicely without integrals.
[hide="With Integrals"]
Rotate a circle at $ \left( 1 , 0 \right)$ about the line $ x\equal{}\minus{}1$.
$ 4 \pi \int_{0}^{2} \left( \sqrt{1\minus{}\left( x \minus{}1 \right)^{2}} \right) \left( 1\plus{}x \right)  \ \text{d} x$.
[/hide]

That should be right.

Here's a more interesting question, then:  prove Pappus' Centroid Theorem.  :)

I did for the cylinder.  :lol:  However, I can not find a way to generalize it.


Find limit of the sum:
 $ \lim_{ x \to \infty} \sum_{n \equal{} 1}^{\infty} \frac {x^2}{1 \plus{} n^2x^2}$

PS: Sorry,  i edited it.

lim as n goes to infinity of the index? what? is that limit supposed to be there? do you mean as x goes to infinity?


Find all entires  value of $ x$ and $ y$  that sodisfy $ x^{3} \plus{} y^{3} \minus{} 3xy \equal{} 2$

(Note:entire I mean integer)

[hide]As \[ x^3\plus{}y^3\plus{}z^3 \minus{} 3xyz \equal{} (x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2 \minus{} xy\minus{}yx\minus{}zx),\] we know that \[ x^3\plus{}y^3\plus{}1^3 \minus{} 3xy \equal{} (x\plus{}y\plus{}1)(x^2\plus{}y^2\minus{}xy\minus{}x\minus{}y\plus{}1)\] for all $ x,y$. Then if $ x,y$ also satisfy $ x^3\plus{}y^3\minus{}3xy\equal{}2$, then we have \[ 3 \equal{} (x\plus{}y\plus{}1)(x^2\plus{}y^2\minus{}xy\minus{}x\minus{}y\plus{}1).\] Since $ x$ and $ y$ are integers, we must have $ x\plus{}y\plus{}1 \in \{3, 1, \minus{}1, \minus{}3\}$ or $ x\plus{}y \in \{2, 0, \minus{}2, \minus{}4\}$. In each of these cases we find that $ xy \not\in \mathbb{Z}$, so there are no integers $ x,y$ for which $ x^3\plus{}y^3\minus{}3xy\equal{}2$.[/hide]


Suppose p is an odd prime,and a,b are two positive integers.We have
[some image here before we changed servers]
Prove that a=b=1.

[mod.: don't know what it was, but below it is claimed to be something more general :D]

Have a look here for a more general proof :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=catalan&t=28571

Pierre.

i have found now , this a very much easy solution to this trivial problem:

we have that $ b > a$ (if not  then the problem is trivial)

now take $b>a>2$ and

and we have that $[(P+1)^a=\sum_{i=0}^a \binom ai (p^i)] - P^b=1$ that will lead us to

$[\sum_{i=1}^a \binom ai (P^i)] - P^b \equiv 0 $ mode P^2 and cause $b>a>2$

we have$\binom a{a-1}(P)=0$ (mode $P^2$) so $\binom a{a-1}=0$ so $a-1=0$ and $a=1$ and so $b=1$

but the only thing that remains is that what if $b=2$ and the problem in these case is trivial

really

so when we have $\binom a{a-1} P \equiv 0$ mode ($P^2$) what is the meaning of it?

[quote="ZetaX"]the relation is the Catalan conjecture (proven 2002):
the only solution of $a^b - c^d =1$ in positive integers where $b,d>1$ is $a=3, b=2 , c=2 , d=3$.[/quote]

can some one post a link for the solution of this problem, any one knows you first ask this problem?

I gave a link to a prove here:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=2002+conjecture&t=32508


What values of x satisfy the following inequality:

$\log_{x^2 - 3}27\ge 3$

Quote:What values of x satisfy the following inequality: [hide]

We have: logb(27) > 3 . . . Then: blogb(27) > b3 . . . and: 27 > b3

Then: b3 < 27 . . .  (x2 - 3)3 < 27 . . . x2 - 3 < 3 . . . x2 < 6

Hence: |x| < sqrt{6}

Since x2 - 3 is the base of a log, x2 - 3 > 0 . . . |x| > sqrt{3}

Since x2 - 3 is the base of a log, x2 - 3 is not 1 . . . |x|  is not 2

Therefore, the inequality is true for values of x on:

(-sqrt{6}, -2) U  (-2, -\sqrt{2}) U (sqrt{2}, 2) U (2, sqrt{6})[/hide]


Find $m$ and $n$ if $m$ and $n$ are both natural numbers such that
\[(m+n)^{m}= n^{m}+1413. \]
(I don't even know the answer :( )

[hide="possible hint"]Consider the equation mod m.[/hide]

Question: does $>>$ mean [i]much greater than[/i]?

[hide]In any case,  you just have to try m=1,2,3,4 only (since $5^{5}$ is already greater then $1413$. [/hide]


[quote="#H34N1"]What is the result when the sum of fifteen and ninety is divided by teh positive difference between thirty-eight and twenty-three?[/quote]
[hide] 15+90=105
38-23=15
105/15
21/3
7[/hide]

[quote="Quevvy"][quote="kstan013"][hide]7[/hide][/quote]
It would really help if you explained how you got this for the people who don't know how to do this problem.[/quote][hide]15+90/38-23=105/15=7[/hide]happy?

[hide]15+90=105
38-23=15
105/15
21/3
7
[/hide]
Not a hard question. I'm not sure why you have problems with it. And would this be considered number theory? lol...I don't think so.


geometry
geometric transformation
rotation
geometry proposed

Let $ABCD$ a square and $K$ a point in the side $AB$ and $N$ a point in the side $AD$, such that $AK*AN$=$2BK*DN$. Prove that $K, L, M, N$ lie in the same circle. Please someone now how to solve it with a rotation!? :spider:

You haven't defined points $L$ and $M$... :maybe:

:oops:
Let L be the point of intersection of BD with CK and M be the point of intersection BD with CN. :blush:


The Ackermann function is

$A(m,n)=$

$n+1$ if m=0

$A(m-1,1)$ if m>0 and n=0

$A(m-1,A(m,n-1))$ if m>0 and n>0

Compute $A(4,2)$.

Is there an easiier way than brote force, or should I put this in the olympiad forum?[/hide]

:mad: Double post detected. Topic should be locked. For a discussion, go to the post replied by t0rajir0u

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136369[/url]

Now this is very rare! I posted it yesterday, I only saw one of those topics, and still today I see only one of those topics, even though both topics are in this forum. :huh:


Well, here it goes :

Prove that :

$\sum_{k=0}^{n}k(k-1)\binom{n}{k}^2=n(n-1)\binom{2n-2}{n-2}$

Good luck ;)

EDIT :  :blush: Sorry, I completely messed up, now that's alright :)

What is $p$?  I think that matters.
And for the $\binom{k}{n}$ I think you meant $\binom{n}{k}$, considering $n \ge k$.

I've edited it now, so you can try it :)

Say we have $2n$ people, $n$ men and $n$ women. We have to choose a leader, a deputy leader and $n-2$ members for a society, and discard the rest of the people; we have the additional condition that the leader and the deputy leader must be men.

We have $n(n-1)\binom{2n-2}{n-2}$ ways of doing this: first, among the $n$ men, we choose the leader ($n$) and the deputy leader ($n-1$), and then we choose $n-2$ members among the $2n-2$ people left ($\binom{2n-2}{n-2}$).

But we can also choose them in this way: for each $k$, choose $k$ men ($\binom{n}{k}$) and $n-k$ women ($\binom{n}{n-k}=\binom{n}{k}$), and among the selected men choose a leader ($k$) and then a deputy leader ($k-1$). Thus LHS=RHS.

Combinatorics proofs are so cool  :)

Here is an other one using generating functions :

By differentiating two times $(1+x)^n$ we get :

$\sum_{k=0}^{n}k(k-1)\binom{n}{k} x^k = n(n-1)x^2 (1+x)^{n-2}$

$\sum_{k=0}^{n}\binom{n}{k} x^k = \sum_{k=0}^{n}\binom{n}{n-k} x^k = (1+x)^n$ is well known ;)

$\sum_{k=0}^{n}k(k-1)\binom{n}{k}^2$ is the "Cauchy product" of the two previous sums and we are looking for the coefficient of $x^n$ in both side.
The coefficient of $x^n$ in $n(n-1)x^2 (1+x)^{2n-2}$ is indeed $n(n-1)\binom{2n-2}{n-2}$

[quote="tµtµ"]Combinatorics proofs are so cool  :)
[/quote]
I strongly agree with it! :D

Severius, I love your proof

Tµtµ, I solved it almost the same way as you :) Moreover, I've learnt one more thing : the [url=http://mathworld.wolfram.com/CauchyProduct.html]Cauchy product[/url]

So, thanks a lot ;)


[quote="vishalarul"]Given that $y = 2x^{2}+px+16$ and that $y < 0$ when $2 < x < k$, find the range of values of $p$ and $k$.[/quote]

Can't you just find $p$ so that $2$ is a root, and extend the ranges to negative infinity for $p$ and positive infinity for $k$? I might not be understanding the question, but here is my answer (tell me if im right, ill post my solution):

[hide] $-\infty<p<-12$ and $4<k<\infty$[/hide]

[quote="vishalarul"][quote]Can't you just find $p$ so that $2$ is a root, and extend the ranges to negative infinity for $p$ and positive infinity for $k$? I might not be understanding the question, but here is my answer (tell me if im right, ill post my solution):
[hide]$-\infty<p<-12$ and $4<k<\infty$[/hide][/quote]
The problem is that $2$ doesn't necessarily have to be a root. Look at the case where $p=-18$. Then the polynomial can be factored as $2(x-1)(x-8)$, and since $2<x<k$, $k$ can take on any value between $2$ and $8$. So $k$ can be $3$, which contradicts your solution.

Counterexample: $p=-18$, $k=3$

Tell me if my counterexample is incorrect. I don't know the solution either...[/quote]

I know that $2$ did not have to be a root. (Actually if you look at my solution it's $p<-12$ so $2$ cannot be a root)  :)

But i understood $k$ to be the upper bound for $x$ such that $y<0$, not just any value that $x$ could take.

(edit: i dont know why i put infinity in my solution... wasnt thinking...   my solution is $p<-12 \ , \ k>4$)


inequalities
calculus
integration
function
linear algebra
matrix
complex numbers

I've just solved an oral problem from ENS and I saw that the conclusion is more than impressive:
   For all real numbers $a_1,...a_n$ we have $\sum_{i,j=1,n}{\frac{a_ia_j}{1+|i-j|}\geq0}$. I really could not find an elementary solution, though I tried for a long time. Does anyone have a "natural approach"? PS: it is far more difficult than the classical poland problem where $1+|a_i-a_j|$ is replaced by $i+j$.

we have to find the right scalar product isn't it ? with an integral ?

Harazi - can you show this non-elementary solution ? I'll be thankful

I have found a solution and i hope it's right

First ,the problem is obviously if a(i) =0 for all i =1,...,n .Then we can assume that not all a(i) is equal to 0

Let F(x) = :Sigma: a(i)a(j)/1+/i-j/  . x^[a(i)a(j)(1+/i-j/)] for all x >0

Then we have F'(x) > 0 for all x >0 ,so F(x) is increasing ,x>0

But now  F is continuous for x>0 and Lim F(x) = 0 when x comes to 0(+) .We get F(1) >= 0 ,and we are done .

[quote="namanhams"]I have found a solution and i hope it's right

Let F(x) = :Sigma: a(i)a(j)/1+/i-j/  . x^[a(i)a(j)(1+/i-j/)] for all x >0

Then we have F'(x) >=0 for all x >0 ,so F(x) is increasing ,x>0

But now F(0)=0 and F is continuous so F(x) is increasing for all x>=0 .We get F(1) >= 0 ,and we are done .[/quote]

how do you show F'(x) >= 0??! It is not trivial to me at all.

[quote="namanhams"]F'(x) = :Sigma: (a(i)a(j)):^2: .x^[a(i)a(j)(1+/i-j/) -1] >=0 for all x >0.

I think it's obviously[/quote]

oh, i'm sorry that i misread your solution.... :blush:

I think it's wrong, though i can't point out the fallancy of it.

Consider this,
\[
a_1  + ... + a_n
\]
where $a_i$ are reals.
Using your idea,
Construct $ f(x) = \sum {a_i x^{a_i } } $
then $ f'(x) = \sum {a_i ^2 x^{a_i  - 1} } $
For positive x, $f'(x) \ge 0$, so f is increasing if x is positive.
Since f(0) = 0, we have f(1) is positive.
However, $f(1)=a_1  + ... + a_n$
choose all $a_i$ to be negative and it leads to an absurdity....

[quote="siuhochung"]Consider this,
\[
a_1  + ... + a_n
\]
where $a_i$ are reals.
Using your idea,
Construct $ f(x) = \sum {a_i x^{a_i } } $
then $ f'(x) = \sum {a_i ^2 x^{a_i  - 1} } $
For positive x, $f'(x) \ge 0$, so f is increasing if x is positive.
Since f(0) = 0,[/quote]

The fallacy is that you can't claim f(0) = 0: In fact, if, at least, one of the reals $a_1$, $a_2$, ..., $a_n$ is negative, then $f\left(x\right)=\sum_{i=1}^n a_i0^{a_i}$ is undefined.

I don't know for sure whether Namanhams' proof of the original inequality suffers from the same lack, but it looks so.

  darij

[quote="namanhams"]Thank for your remark ,Sihochung and Darij .
I've just corrected my proof .[/quote]

I think it still has the same mistake. Grinberg not only mentioned that $0^0$ is meaningless, he also meant that $0^a$ where a is negative is also meaningless. Your solution cannot ensure a positive power.

I don't think so .In my solution ,F(x) is a function on positive real number ,not contain 0.
And my solution doesn't use F(0)  :)

consider the matrix $B(t)$ such that: $B(t)_{i,j}=t^{|i-j|}$. This matrix is positive definite for $0<t<1$ (compute principal determinants).
Then the matrix $A$ such that $A_{i,j}=\frac{1}{1+|i-j|}$ is positive definite because
$^tXAX=\int_{0}^{1} {^tXB(t)Xdt}>0$ for $X \neq 0$
conclusion follows ..

to harazi: how on earth did you find $F$ ??

I just solved one oral problem from ENS and used the result from there.  :D   Your solution is the classical one, but in fact I prefer not to compute principal determinants.  :D

[quote="harazi"]Well, this problem is far more difficult that one could think.
  To alekk: indeed, the scalar product is $(f,g)=\int_{0}^{2\pi}{conj(f)(x)g(x)F(x)}$ where $F(x)=\frac{1-r^2}{2\cdot\pi\cdot (1-2rcosx+r^2)}$, where $conj(f)$ is the complex conjugate of f.[/quote]
Hi,Can somebody tell me what does the mean of $r$,how can we use this scalar product to solve the problem?
How can $\frac{1}{1+|i-j|}$ appears in the integeral value?

to alekk:

How to compute principal determinants?

Thank you in advance.

Edit:Now I know harizi want to prove that the matrix $B(r)$ is positive defined...


When you open the Site Faq, the shortcuts are distorted and in the message area. Can you fix this?

I have Mozilla Firefox 2.0.0.5

Here too, Seamonkey 1.2 on Gentoo Linux.

Shortcuts? Message area? All I see is a page full of allowed extensions.

Same here. The problem is not there on the ML theme. Attached is a screenshot.


Let $ABC$ be a triangle and $O$ the center of its circumcircle. The tangent in $C$ at the circle meets $AB$ in $M$. The perpendicular on $OM$ in $M$ meets $BC$ and $AC$ in $P$ respectively $Q$. Prove that $MP=MQ$.

See this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=16784


Yes.  This is the same as showing that
$(x-1)*(x^1999)>(2000^1999)(2000-1)$
is true, which obviously holds if $x>2000$

EDIT: Sorry, I'm really bad with latex, for some reason it won't accept more than one character in the exponent...so i'll just write it without latex...

(x-1)(x^1999)>(2000^1999)(2000-1)


function
limit
topology
real analysis
advanced fields
advanced fields unsolved

consider the following PDE:
$\partial_{t}u(x,t)+\Delta u(x,t)=f(u(x,t))$
where $f$ is a smooth function that looks like $k.x(1-x)$, that is something positif on $(0;1)$, negatif on $\mathbb{R}\setminus[0;1]$, and $f'(0) > 0$. (Such an equation arises in biology, physic, etc ..). One would like to know if there exists a function $\phi: \mathbb{R}\to \mathbb{R}$ with $\lim_{-\infty}f=0$, $\lim_{+\infty}=1$ such that $u(x,t)=\phi(ct-x.v)$, where $v \in \mathbb{R}^{n}$, $c>0$, is a solution of the initial PDE.That is: we are looking for solutions that are travelling on the direction $v$, with speed $c$. Pluging this into the PDE, one sees that $\phi$ satisfies:
$c \phi'-\phi''=f(\phi)$
The question is: for what value of $c$ this equation has a solution satisfying the hypothesis ? The answer is for $c$ large enough, but how to prove this ?

This is a really nice problem for an intermediate (by European standards) course in ODE. Here are the steps you may try:
0) Rescale the time to reduce everything to the case $f'(0)=1$
1) Rewrite your second order ODE as a $2\times 2$ system of first order ODE's setting $x=\phi$, $y=\phi'$
2) Show that $(0,0)$ and $(1,0)$ are stationary points.
3) Show that $(1,0)$ is a hyperbolic equilibrium and conclude, using the stable manifold theorem, that there exist two solutions approaching this point at $+\infty$.
4) Use the fact that the stable manifold is tangent to the stable space of the linearized equation to conclude that one of this solutions has $x(t)<1$, $y(t)>0$ for large positive $t$.
5) Use the special kind of your system to show that, when you run time backwards, this solution preserves these properties until $x$ reaches $0$.
6) Show that $y\le \max_{[0,1]}f/c$ during this time.
7) Use the Lyapunov function $x^{2}+y^{2}-xy/c$ to show that the point $(0,0)$ has an attraction basin (for the time running backwards) of size bounded from below by some positive constant independent of $c$ when $c$ is large.
8) Happily conclude that $(0,0)$ catches your solution at the moment when it crosses the $y$ axis and finish the problem.

Hope, that I didn't make a mistake anywhere and that this will be enough of a hint for you :).

thank you Fedja, this problem appeared in my PDE course with a nasty solution. I will try your approach :)

You are welcome,alekk :). By the way, I realized overnight that steps 5-7 can be replaced by showing that, for large $c$, the trajectory with time running backwards cannot leave the triangle with vertices $(0,0)$, $(1,0)$ and $(0.5,0.5)$ (just look at the direction field on the boundary). This approach would require doing step 4 more carefully but the benefit is that you can prove now that $0\le\phi\le 1$ for all times.


find the probability that the birth day of six different persons will fall in exactly two calendar months

[hide="Basic Idea"]Consider two arbitrary months, say they have $ x$ days and $ y$ days, respectively. The probability that the condition is satisfied is equal to the probability that all six people have their birthdays within the two months minus the probability that all six people have their birthdays within just one month.

So basically, $ \left(\frac{x\plus{}y}{365}\right)^6 \minus{} \left(\frac{x}{365}\right)^6 \minus{} \left(\frac{y}{365}\right)^6$ is the desired quantity. (This assumes people can have the same birthday. If not, then you do conditional probability).

Now, sum over this with all valid combinations of months, depending on your definition of valid (two non-coinciding months, two consecutive months [the problem doesn't state this explicitly]).

[/hide]