function
integration
calculus
real analysis
real analysis theorems

Well,I have one question. I have demonstrated the second mean value theorem for Stieljes Integral.
  My question is?? It's well known or it's not. Plz,answer if someone is interrested.

Could you post the theorem?  I'm not sure what you mean by "second" mean value theorem.

First of all,I'm sorry for my bad Latex.Well,here it goes:

   Let$f,g:[a,b]\to R$ two continuous functions such that $g(x)\geq 0$ for all $x$ in $[a,b]$ and $h:[a,b]\to R$ be
   an increasing function. Prove that there is a point $c$ in $[a,b]$ such that

                          $\int^b_a f(x)g(x)dh(x)=f(c)\int^b_ag(x)dh(x)$
    I hope you do understand what i have written.

I don't know ?? What do you seem odd in my statement? I do know the theorem you wrote.Actually I think it is in
 my father's Analysis book problem.

Yes, what you have there is easy to prove.  We have $fg\in\mathcal{R}(h)$ and $g\in\mathcal{R}(h)$ by some standard theorems, and let $m=\min_{[a,b]} f$ and $M=\max_{[a,b]} f$.  By some well-known results on integrals, we have

$m\int^b_a gdh\leq \int^b_a fgdh\leq M\int^b_a gdh$, so by the continuity of $f$ there exists a $c$ such that $f(c)\int^b_a gdh=\int^b_a fgdh$

Hey,that's exactly my proof.I was wondering if it is a well known theorem?

I would think yes, if for no other reason that it's proved in exactly the same manner as the traditional mean value theorem for integrals.

The 'second mean value theorem' that I posted earlier in this thread probably has a truer claim to that name, as your theorem is referred to as the 'generalized first mean value theorem' in both a Schaum's Outline (by Murray Spiegel) and also in an analysis text (by Steven Lay) that I own (of course for Riemann integrals rather than their Stieltjes counterparts)

About that definition you gave. Is no need for that function to be continous.If the function is integrabile the
 statement is also true.

I don't understand what you're referring to.

[quote="blahblahblah"]The second mean value theorem is kind of an odd thing.  It appears in older texts, like some of the Schaum's outlines I have, but not so much anymore.  The only modern reference I have to it is in Folland, which uses it to prove a result from Fourier analysis.

Cezar's statement doesn't seem to make any sense they way it's worded though, because you are inevitably going to need to impose some continuity conditions on the functions you're using in the Stieltjes integral.  Anyways, here is a definition.

Let $\phi$ and $\psi$ be real functions on $[a,b]$.  Suppose that $\phi$ is monotone and right continuous on $[a,b]$, and that $\psi\in C([a,b])$.  Then there exists $\eta\in[a,b]$ such that

$\int^b_a \phi(x)\psi(x)dx=\phi(a)\int^\eta_a\psi(x)dx+\phi(b)\int^b_\eta \psi(x)dx$

The proof proceeds as follows.  WLOG $\phi(a)=0$, and $\phi(x)$ is increasing.  Set $\delta(x)=\int^b_x\phi(t)dt$, and perform a Stieltjes integration by parts to obtain

$\int^b_a \phi(x)\psi(x)dx=-\phi(x)\delta(x)|^b_a+\int_{(a,b]} \delta(x)d\phi(x)$

Since $\phi(a)=\delta(b)=0$, the endpoint evaluations vanish

A simple appeal to the continuity of the integral and the intermediate value theorem yields an $\eta$ with the desired properties.[/quote] About those two functions $phi$ and $psi$.You said that one of them is continuos.I said that the statement takes place if that continuous function is just integrabile. What do you think about that?

If $psi$ is only continuous a.e, I think it's fine - but really, it's not a big difference.


I don't know if this is even solvable, I was just thinking...
$\prod^{k}_{n=0}(\sin a+\sin (b+n))$
In a closed for in terms of a and b.
Cos a/b+n?
Tan a/b+n?

If this is too hard, how about if n=90?
n=60?
n=45?
n=30?
n=0?

[quote="Altheman"]random problems don't usually have nice solutions

you need to have some idea in mind when you are making the problem[/quote]
I know, but I was just wondering if there was a quick method to solve it because I was curious about this.

The sum is equal to
\[\frac{1}{2}\csc \left(\frac{1}{2}\right) \left(\cos \left(\frac{1}{2}-b\right)-\cos \left(b+k+\frac{1}{2}\right)\right)+(k+1) \sin (a)\)\]
Don't ask me how though, that's Mathematica speaking :lol:

[quote="mets501"]The sum is equal to
\[\frac{1}{2}\csc \left(\frac{1}{2}\right) \left(\cos \left(\frac{1}{2}-b\right)-\cos \left(b+k+\frac{1}{2}\right)\right)+(k+1) \sin (a)\) \]
Don't ask me how though, that's Mathematica speaking :lol:[/quote]

it is a product...

sum of sin(aritmetic) is pretty simple


calculus
integration
logarithms
derivative
calculus computations

Integrate 4/(t^2 +1) dt, upper limit is 1, lower is 0.  There are two ways i thought of to do it, but answers are different. one way i just did ln|t^2 +1|, the other way i used arctan. so which way is right? thanks a lot.

The correct answer is the one using $ \arctan$, where you essentially use the antiderivative $ 4\arctan{x}$ to get $ 4(\arctan{1} \minus{} \arctan{0}) \equal{} \pi$.

The reason $ \ln{(x^2 \plus{} 1)}$ won't work is because the derivative of $ \ln{(x^2 \plus{} 1)}$, by the chain rule, is $ (\frac {1}{x^2 \plus{} 1})(2x) \equal{} \frac {2x}{x^2 \plus{} 1}$; in this case, you don't have an $ x$ in the numerator.

I get it now, thank you so much, Elixir!


[quote="sagar_astakala"]If the positive integers are added in groups as shown:
 [b]1 + (2+3) + (4+5+6) + (7+8+9+10)......................[/b]

Find the first integers of the nth group.
Also find the sum of the integers in the nth group.[/quote]

I don´t understand what you`re asking, acutually. However, I can tell you that the sum of the first n numbers is:

1 + (2+3) + (4+5+6) + (7+8+9+10)+...+n=1+2+3+4+5+6+7+8+9+10+...+n=1/2*n*(n+1).

[quote="sagar_astakala"]If the positive integers are added in groups as shown:
 [b]1 + (2+3) + (4+5+6) + (7+8+9+10)......................[/b]

Find the first integers of the nth group.
Also find the sum of the integers in the nth group.[/quote]

[hide]If you add up the numbers in the group, you get 1, 5, 15, 34. The third difference is constant, so the equation is cubic.
$34=64a+16b+4c+d$
$15=27a+9b+3c+d$
$5=8a+4b+2c+d$
$1=a+b+c+d$
solving, $a=\frac{1}{2}$, $b=0$, $c=0$, and $d=\frac{1}{2}$
so the equation is $\boxed{y=\frac{1}{2}n^3+\frac{1}{2}n}$[/hide]

[hide]the last number in the $n$th group is $\frac{n*(n+1)}{2}$, and the number before the first term in the group would be $\frac{n*(n+1)}{2}-n$, so do the sum including the last term and lower minus the sum of the numbers from 1 below the start of the group

$\frac{ ( \frac{n*(n+1)}{2})(\frac{n*(n+1)}{2}+1)}{2}-\frac{( \frac{n*(n+1)}{2}-n)(\frac{n*(n+1)}{2}-n+1)}{2}$
$=\frac{1}{8}*[ (n^2+n)(n^2+n+2)-(n^2-n)(n^2-n+2)]$
$S_n=\frac{1}{2}*n^3+\frac{1}{2}{n}$[/hide]

[hide]

Firstly we must realise that the last term in the (n-1)th group is $\frac{n(n-1)}{2}$

Hence the first term in the nth group will be $\frac{n(n-1)}{2} + 1$

Now, the sum of the nth group is $(\frac{n(n-1)}{2} + 1) + (\frac{n(n-1)}{2} + 2)+ ... + (\frac{n(n-1)}{2} + n)$

$\implies n \cdot \frac{n(n-1)}{2} + \frac{n(n+1)}{2}$

$= \frac{n^3 + n}{2}$

[/hide]


Here is a nice solution to this nice problem.

 [color=blue] [b] Problem. [/b]

    Let $a, b, c$ three positive real numbers. Prove the inequality

 \[ a\sqrt{b^{2}+c^{2}+bc}+ b\sqrt{c^{2}+a^{2}+ca}+ c\sqrt{a^{2}+b^{2}+ab}\geq\sqrt{3}(ab+bc+ca).  \]

 [/color]

 [i] Solution. [/i]

   I'll prove that $\sqrt{a^{2}+b^{2}+ab}\geq\frac{\sqrt{3}}{2}(a+b)$. This is equivalent to
   \[ 4(a^{2}+ab+b^{2})\geq 3(a^{2}+2ab+b^{2})\Leftrightarrow  \]
   \[ a^{2}-2ab+b^{2}\geq 0\Leftrightarrow  \]
   \[ (a-b)^{2}\geq 0,  \]
    which is obvious. By writing the analogues ones and by adding them up we get the conclusion. :)  ;)


trigonometry
floor function
geometric series
real analysis
real analysis unsolved

study the nature of $\sum \sin (\pi (2+\sqrt{3})^{n})$

Note that $(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}= 2\cdot\sum_{j=0}^{\lfloor n/2\rfloor}\binom{n}{2j}2^{n-2j}3^{j}$ is an even integer.

Hence, $\sin\left(\pi(2+\sqrt{3})^{n}\right)=-\sin\left(\pi(2-\sqrt{3})^{n}\right)$

The series converges absolutely by limit comparison to the convergent geometric series $\sum (2-\sqrt{3})^{n}.$

After spending like 1 hour trying to solve it, I found the same solution as Kent, It shouldn't take so much time I think  :blush:


induction
number theory proposed
number theory

I had seen in a book a proof that, If $n$ is even, so $n+1$ is odd.

but i did not remember it. how could we proof that using postulates and others stuff.

PS: Say that $n= 2 k$ so $2 k+1$ is odd, is not a proof of nothing.

the proof I had seen was not that simple, was a beautiful and ellegant proof.

c ya

Suppose both n and n+1 are even. Then 2/(n+1)-n i.e. 2/1. Contradiction.
So if n is even n+1 is odd.

err... I dont want a random proof of it, dude.

the proof i saw is not about contradiction, induction or else.

is based in some postulates, but i cant remember...

Do you think we have any chance to guess what proof you have in mind¿ By the way, you didn't define "+",  "even" and "odd"...

i dont know the "chances", i'm bad with probabilities. sorry.

but it's a real proof. i just asking if someone have a good proof, not a random one

thanks

You didn't answer my question on "odd", "even" and "+"...
That's rather important as this depends on the level of useable stuff you want...

ZetaX. if i answer your question, there is no problem enough...

I can define "+" to you, but i cannot definie Even or Odd. because if i define it i am answering the question.

Then I define:
a) $0$ is even.
b) if $n$ is even, then $n+1$ is odd.
c) if $n$ is odd, then $n+1$ is even.

Done  :D

but you did not proof nothing.

because there is not Axiom saying "After a even number has every a odd number"
there is no such kind of axiom

and there is no axiom saying $even number is divisible by 2, odd not$
i will be a lost of time making those so much axioms

Peano Axioms say no word about even or odd

but since i dont remember de proof of the book i cannot say more words...

people here does not understand that: If i remember what the book says, there is no reason for this topic...

if you dont know a proof using some axioms, etc. like i said, dont post here. is a kind of spam, because i had said that i dond whant a random proof

May I repeat the third time: you can't decide iff a number is "even" as long as you haven't even ( ;) ) defined the meaning of this word. I gave you one, but you are not satisfied. This is mathematics and not metaphysics, thus you have to give some definitions or don't have to expect any answer.

i'll repeat the nth time.

if i define even there is no problem enough.

you dont know what i'm talking, so, dont post anymore here.

like i said, if someone read the same book i read, post here. if not, is a waste of time

i will not say it anymore the (n+1)th time...

Just to clear things up:

As it seems, you want someone to find out of what proof you talk about without saying _anything_ about that proof. Especially, you refuse to give definition of what you talk or any hints on what that book is exactly about. In my oppinion, this task is impossible.
And I don't like the way you react on people trying to help you. If I don't know what you talk about, you should rather tell me what you talk about than telling me not to post here again.
But at the moment, no helpful answer can be expected as there is no meaningful question.

okay, an other try:

you want to have proven that if $n$ is even, then $n+1$ is odd.

the problem is, that you don't say what "even" and "odd" means (i think we all know what you mean with $+$)

an example to explain the problem:
i can say that i want to have proven that if $n$ is "lkjajbk", then $n+1$ is "oirtzwo". (two invented words, btw, they have no meaning in no language, i hope). now, i say that if an number is "lkjajbk", then it is divisble by $2$ and if an number is "oirtzwo", it is not. (yes, you can guess that "lkjajbk" is what we say "even" and "oirtzwo" is odd ;-) ) now, you are able to prove what i want to have proven because you know what i mean. when i only write, that i want to have proven that if $n$ is "lkjajbk", then $n+1$ is "oirtzwo", without explanation, no one knows what to do. that's the same with your "even" and "odd", the only difference is that we already have a few definitions, so we know which numbers you mean, but we don't know which definition you mean, so we can't find the proof you want.

ZetaX

You can know all about a problem or nothing about the problem.
there is no half-know.

I cannot define the meaning of even or odd, because i didnt remember the definition of the book for that.

if i know the definition, there is no reason for this post.

this is a situation that we can no all or nothing, but not half.

I make a question about someting.
you make a question about my thing
if i answer your question I'm automatically answering mine
but this is impossible, since i dont know how to answer mine.

a -> b -> c -> a?

and the ciclic process never end

and I'm onfending anyone. If you are ofended, there is no my fault.

i'm here to make math, and only that.

2 or 3 persons had tried to solve that problem using a thing that i said they could use.

mathematicians are too fragile. perhaps, let's call this a paradox and stop talk about it. it's better


abstract algebra
superior algebra
superior algebra unsolved

Plz explain me how we solve this kind of problems. They look similar.

1. Find all finite abelian groups $ G$ such that $ p^{2} \cdot t \equal{}0$ for all $ t \in G$ where $ p$ is prime number .

2. Find all abelian groups when $ x^{p}\equal{}1$ for all $ x \in G$ where $ p$ is prime number .

Thank you.  :)

Are you familiar with the Fundamental Theorem of Finite Abelian Groups? I think that's the way to go.

Thx for the reply.

I know that for a group $ G$ if every $ t \in G$ has order a power of a prime $ p$ then $ |G| \equal{} p^{n}$.

In both cases the group has order a power of $ p$ . Now we can write $ G$ as direct sum of cyclic p - groups.

Is what i wrote correct ? Is there any way to find $ n$ ?


calculus
integration
limit
trigonometry
calculus computations

given
 $ \textbf f(x,y)\;\equal{}\;\dfrac{1}{\pi}\;\int_{\minus{}\infty}^{\infty}\;\dfrac{x\cdot\textbf g(u)}{x^2\plus{}(u\minus{}y)^2}\;\;du$

show
$ \lim_{x\to 0}\;\textbf f(x,y)\;\equal{}\;\textbf g(y)$

$ u\equal{}y\plus{}x \tan (\theta )$

$ \lim_{x\to 0} \, \frac{1}{\pi } \int_{\minus{}\infty }^{\infty } \frac{x g(u)}{x^2\plus{}(u\minus{}y)^2} \, du\equal{}\lim_{x\to 0}
   \, \int_{\minus{}\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{g(y\plus{}x \tan (\theta ))}{\pi } \, d\theta$
$ \int_{\minus{}\frac{\pi }{2}}^{\frac{\pi }{2}} \lim_{x\to 0} \, \frac{g(y\plus{}x \tan (\theta ))}{\pi } \, d\theta
   \equal{}\int_{\minus{}\frac{\pi }{2}}^{\frac{\pi }{2}} \frac{g(y)}{\pi } \, d\theta \equal{}g(y)$

The rest of the proof is just proove that the interchange of the integral and the limit is allowed, since the integral dont converge for all $ g$. I think  :P


The maximum IS 13 and the minimum is -13, unless I've gone completely crazy.

The minimum for $x,y,z\geq 0$ should also be 3.

This has to be the dumbest question ever, but...

Do x, y, and z have to be real numbers?

Do you have order properties if you're working in the complex (with non-zero imaginary part) numbers?

The answer, as you say,  is 13,    3/x = 4/y = 12/z  =  a, x^2+y^2+z^2=1 will give  the value of a etc... the plane touching the spehere ..,,,


ratio
geometry
search
function
3D geometry
probability
floor function

since this forum is pretty much dead and there already is a competition in another forum or another forum is not dead, i decided to start a competition in this forum. :lol: if anyone thinks it should be another forum, just tell me and ill put it somewhere else, otherwise ill begin.

this competition will be about proofs, the easy kinds that u should probably know. each round will consist of different problems that u must prove and pm me, but they'll be relatively easy. For example,
prove that the sum of the first n positive integers is n(n + 1)/2. i will see which proofs/solutions are the best based on certain criteria. if enough people are interested, ill start.

if this might be too hard for this forum, just tell me.

EDIT: nevermind, instead this'll be like some other competitions with problems, no proof.

I guess i'll participate. If it's going to be proofs, I'm pretty sure a mod is going to move it sooner or later.

Ummm..I'll participate. If you don't want this to be moved, you could change the subject from proofs to ordinary problems even if the proofs are easy. And I doubt they would ask for proofs in MOEMS...?

I'll do it.

Easier proofs would probably be ok for the forum, since it would help people learn to apply strategies, but it should be easier than the first formula you have up there. On another note, proofs for formulas are easily found on the www.

that's what i thought at first. ill take ur suggestion. ill make it just problems.

I'd like to participate.

Do you still want a proof for that problem?

i think we could do this with 4 people, maybe someone will join later on,

about the proof, no thanks. i'll post rules(it's basically the same as any other) and the first problem around 4:00, im in the central USA time zone.

ok, this will be really simple. i post a problem and u get two days to solve and pm me ur answers, no solutions.
The first person to get it correct will get 5 points, next two will get 4, next three will get 3 and everyone else that gets it correct gets 1.

PROBLEM NUMBER 1 (let's start easy)

Suppose a and b are positive integers where
$\frac{a}{2}+\frac{b}{7}=\frac{10}{7}$ What is $a^3+b^3?$

Darn it I never bothered to look for the problem.

The first round is now over. The correct answer was 35.
The standings are

1. samath, 5
2. nonie, 4
3. Captain Sugar, 4
4. Andrew619, 3
5. mysmartmouth, 3

The next problem (2): What is the smallest positive integer that has exactly 15 factors?

[quote="kyyuanmathcount"]Ok, not in accordance with your signature SplashD, I feel dumb.  What is it?[/quote]

 :huh: What is what? :huh:

[quote="SplashD"][quote="mathgeniuse^ln(x)"]Look at the telephone and try to relate RYAN and 7926 and NEIL 6345 and look for a pattern.  Then you will find a code.  It is really clever.[/quote]

That's cool.  :D but that can spell other things too.[/quote]

That's the point we have to figure out your name from the number itself even if there are many possibilities.  What is your number though?

[quote="mathgeniuse^ln(x)"][quote="Ignite168"]My code is 56628426[/quote]

Your name is either JONATHAN or KONATIAM.

I will pick JONATHAN.[/quote]
The other one sounds cooler. :P

[quote="mathgeniuse^ln(x)"][quote="SplashD"][quote="lotrgreengrapes7926"]Wow, this used to be a competition.  :roll:  :P[/quote]

Yeah, it's all because mathgeniuse^ln(x) never heard the phrase 'Curiosity killed the cat' :D[/quote]

I don't quite understand your phrase.  I know I am curious, but I am being curious in a good way, aren't I?[/quote]

Yeah but you're curious about wierd things. I saw that he was missing the number but the thought of posting a message asking why never occured to me.

So, any progress on the competition?  :?

[quote="SplashD"][/quote][quote="mathgeniuse^ln(x)"][quote="SplashD"][quote="lotrgreengrapes7926"]Wow, this used to be a competition.  :roll:  :P[/quote]

Yeah, it's all because mathgeniuse^ln(x) never heard the phrase 'Curiosity killed the cat' :D[/quote]

I don't quite understand your phrase.  I know I am curious, but I am being curious in a good way, aren't I?[/quote][quote="SplashD"]

Yeah but you're curious about wierd things. I saw that he was missing the number but the thought of posting a message asking why never occured to me.[/quote]

Oh, okay, but that is not being curious in a bad way, it is just realizing random facts. :roll:

[quote="mathgeniuse^ln(x)"][/quote][quote="SplashD"][/quote][quote="mathgeniuse^ln(x)"][quote="SplashD"][quote="lotrgreengrapes7926"]Wow, this used to be a competition.  :roll:  :P[/quote]

Yeah, it's all because mathgeniuse^ln(x) never heard the phrase 'Curiosity killed the cat' :D[/quote]

I don't quite understand your phrase.  I know I am curious, but I am being curious in a good way, aren't I?[/quote][quote="SplashD"]

Yeah but you're curious about wierd things. I saw that he was missing the number but the thought of posting a message asking why never occured to me.[/quote][quote="mathgeniuse^ln(x)"]

Oh, okay, but that is not being curious in a bad way, it is just realizing random facts. :roll:[/quote]

...maybe a little too random

Okay, people from my school do know that I am random just fyi.  But, I will try to be a little less too random, okay.


I did not receive my year 19 prize/T-Shirt.
I received them in year 18, without any changes in my address.
My USAMTS ID is 8277.
Is there any way for the people at Art of Problem Solving to determine if anything has gone wrong with the shipping?

Thanks,
-diophantient

I didn't receive a T-shirt. Did we have to sign up to receive this?

Send an email to usamts@usamts.org if you haven't received your prizes.


Prove that $ 39^n \plus{} 2009^n \not \vdots n$ for all odd integers $ n\neq 1$ and $ n \neq 1$

I'm extremely sorry about the wrong posted problem before :blush:

[quote="Mashimaru"]Prove that $ 39^n \plus{} 2009^n \not \vdots n, \forall n\in\mathbb{N}^*$ and $ n \neq 1$[/quote]
Have some mistaken here : $ n\equal{}1$ is the unique odd integer satisfies condition and there are infinite even numbers satisfy condition. Do you remember the following problem : Let a,b be two natural numbers such that  $ a\plus{}b$ is not a power of 2 then exist infinite natural number $ m$ such that $ m|a^m\plus{}b^m$ ?
For n odd ,only need to consider the smallest prime divisor of $ n$ and notice that $ 2009\plus{}39$ is a power of 2 .  :wink:


geometry
circumcircle
rectangle
geometric transformation
homothety
projective geometry
geometry proposed

Consider a triangle ABC,let $ O$ be its circumcenter,$ G$ be its centroid and $ AL,BM,CN$ be its median.Let the tangent line of A wrt $ (O)$ meet BC at P.$ PO$ meet $ AB$ at E and $ AC$ at F.$ EM \cap FN \equal{} {Q},BF \cap CE \equal{} {R}$.Let a perpendicular line from L to $ MN$ meet $ MN$ at S.Prove that $ Q,R,G,L$ are collinear.

How come Q, R, G, L are collinear??? :huh:
I think, it should be Q, R, G, S are collinear?? Could you check it??? :D

[b]mathVNpro[/b]: Yes, the problem has a minor typo; your version is correct.

[b]No Reason[/b]: Just out of curiosity, where did you find this problem?

[hide="Solution"]
Let $ T$ be the foot of the perpendicular from $ A$ to $ BC$, let $ AR$ meet $ BC$ and $ EF$ at $ Z$ and $ K$ respectively.  Let $ TG$ meet $ OL$ at $ Y$ and $ TL$ meet $ OL$ at $ X$.  Finally, let $ TR$ meet $ OL$ at $ Y'$ and $ PW$ meet $ AC$ at $ W$.

By considering the complete quadrilateral $ APBCFE$, we have that $ (P, B, Z, C) \equal{} \minus{} 1$.  This implies that $ \frac {BZ}{ZC} \equal{} \frac {PB}{PC} \equal{} \frac {PB}{PA}\cdot \frac {PA}{PC} \equal{} (\frac {AB}{AC})^2$ since $ \triangle PAB \sim \triangle PCA$ (by the tangent line).  This means that $ AZ$ is a symmedian and, hence, $ AZ$ and $ AL$ are isogonal conjugates.  It is known that $ AT$ at $ AO$ are as well.  Hence, $ \angle RAT \equal{} \angle LAO$.  Since $ \angle OAP \equal{} 90 \equal{} \angle OLP$, we have that $ OAPL$ is cyclic, and, therefore, $ \angle KAT \equal{} \angle OAL \equal{} \angle TPK$, implying that $ KTPA$ is cyclic, so $ \angle ATX \equal{} \angle APO \equal{} \angle ALX$.  Hence, $ ATLX$ is cyclic.  Combining this with the fact that $ \angle ATL \equal{} 90 \equal{} \angle XLT$, we have that $ ATLX$ is a rectangle, so $ AT \equal{} XL$.

Notice the from the complete quadrilateral $ APBEFC$, we have that $ P(C, W, F, A) \equal{} \minus{} 1$.  Intersecting this with $ AZ$, we have that $ P(A, L, R, Z) \equal{} \minus{} 1$ and hence $ T(A, K, R, Z) \equal{} \minus{} 1$.  Intersecting this with $ OL$, we have that $ (L, Y', X, \infty) \equal{} \minus{} 1$ and hence, $ Y'L \equal{} \frac {XL}{2} \equal{} \frac {AT}{2}$.  Now, notice that $ \triangle AGT\sim \triangle LGY$ from the fact that $ YL\parallel AT$.  Combining this with the fact that $ AG \equal{} 2GL$, we have that $ AT \equal{} 2YL$, so $ Y'L \equal{} YL$ and, thus, $ Y' \equal{} Y$, so $ T$, $ R$, and $ G$ are collinear.

Now, a homothety around $ G$ maps $ \triangle ABC$ to $ \triangle LMN$, and, hence, it maps $ T$ to $ S$, so $ T$, $ G$, and $ S$ are collinear.  Furthermore, applying Pappus's Theorem to $ NEBCWM$, we have that $ Q$, $ G$, and $ R$ are collinear.  Combining everything gives that $ Q$, $ S$, $ G$, $ R$, and $ T$ are collinear, as desired. [/hide]

About proving Q, G, R are collinear- It is a very well- known problem, we can prove it by using "8 time" Menelaus :D. I remember... I proved when I was junior, I think :D
Anyway, nice solution The QuattoMaster 6000

@Lam Minh (mathvnpro): Of course,because that is just Pappus theorem,I added it to make the problem more "beautiful". :D
@QuattoMaster6000:I found that problem when I was working on another problem.Thanks for your interest. :oops:

$BF$ cut to $(O)$ in $B$ and $Z$
$CE$ cut to $(O)$ in $C$ and $Y$
Pascal's theorem (for 5 points), it has $Z,Y$ and $P$ collinear. ie. $P$ and $R$ are conjugate points WRT $(O) \Rightarrow R$ lies on symmedian of $A$  (WRT $ \triangle ABC$)

Using barycentric coordinate in $ \triangle ABC$ and Conway notation

$R= r: b^2: c^2$
$X=0: S_C: S_B$ ($X$ is foot of the altitude from A)
$G= 1: 1: 1$
$X,G$ and $R$ are collinear $ \Longleftrightarrow\left(\begin{array}{ccc}r & b^2 & c^2\\ 0 & S_C & S_B\\  1 & 1 & 1 \end{array}\right)=0 $
$ \Longleftrightarrow r(S_B-S_C)=b^2S_B-c^2S_C$
$ \Longleftrightarrow 2r(c^2-b^2)=c^4-b^4+b^2a^2-a^2c^2=(c^2-b^2)(c^2+b^2-a^2)$
$ \Longleftrightarrow r=S_A$

$F= r: 0: c^2$
$O=a^2S_A: b^2S_B: c^2S_C$
$E= r: b^2: 0$

$F,O$ and $E$ are collinear $ \Longleftrightarrow\left(\begin{array}{ccc}r & 0 & c^2\\ a^2S_A & b^2S_B &c^2S_C\\  r & b^2 & 0 \end{array}\right)=0 $
$ \Longleftrightarrow a^2S_A=r(S_B+S_C)$
$ \Longleftrightarrow r=S_A$

For Homothecy $S,G$ and $X$ are collinear. (easy to see)
For Pappus theorem $R,G$ and $Q$ are collinear.

$ \Rightarrow Q,R,G$ and $S$ are collinear, if only if $O \in EF$

Second solution

$W$ is foot of the altitude from $A$
$ AR \cap BC =X $
$ (C,B,X,P)=-1 \Rightarrow  AX$ is A-symmedian of $ \triangle ABC$ ... $(1)$
$ FE \cap AX= Y$ and $ (X,Y,R,A) =-1 \Rightarrow AR.XY=2AY.XR$ ... $(2)$

By Menelaus theorem in $ \triangle ALX$ obtain $G,R$ and $W$ are collinear $ \Longleftrightarrow \frac{2LW}{XW}=\frac{AR}{RX}$ ... $(3)$
But $ \angle XAP= \angle ALB$ (Using $(1)$)
$K$ is the symmetric of $X$ WRT $W$, then $ \triangle AXP \cup Y \sim \triangle LKA \cup W \Rightarrow \frac{LW}{XW}=\frac{AY}{YX}$
$(2)$ and $(3)$ obtain $G,R$ and $W$ are collinear

From Homothecy $S,G$ and $W$ are collinear. (easy to see)
From Pappus theorem $R,G$ and $Q$ are collinear.

$ \Rightarrow Q,R,G$ and $S$ are collinear. Done!


Heres a old theorm that's probably in most geomerty text books, but trying proving it yourself, if you alredy haven't.

  Angle in a Semi-Circle

Angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle. So c is a right angle.

Isn't the center of the circle the midpoint of the hypotenuse?

Yeah, the circumcenter is at the midpoint of a side of a triangle iff the triangle is right

[color=green]the measure of an inscribed angle is half the central angle it intercepts, and angle c intercepts a straight line, so it is a right angle (180/2 = 90)[/color]

let's prove the moer general way. But first let us lay the foundation by proving a few things. The measure of an exterior angle of a triangle is equal to the sum of the two non-adjacent angles. To do that, use the sum of teh interior angles in a triangle. a, b, and c are the interior angles and A is the exterior angle adjacent to A

A+a = 180 (linear pair)
a + b + c = 180 (sum of interior angles)
A + a = a + b + c (substitution/transitive)
A = b + c (subtractive property of equality)

then we draw a circle, and inscribed angle, and the radii going to the places where the inscribed angle touches the circle (including the vertex)

we now have 2 isosceles triangles. say the base angles of one are a, and the base angles of the other are b. This gives teh inscribed angle a measure of a+b, and the central angle, and thus the arc, a measure of 2a+2b and thus the inscribed angle is half of the intercepted arc.

I understand that this is not exactly basics level but it is a way that can probably be understood if you think about it enough.


hi. I am ali.i live in iran .I study astronomy olympiad.i  want  new proplems to solve can you help me.? i have solved all of international astronomy olympiad.and now I dont have any problem!!pleas help me . :(

There should be some archived ones on the International Astronomy Olympiad's website.  Other than those I do not know where to get more unless there is a book of such practice problems.

You have solved all IAO problems? :o  :o

If you haven't already done so, I suggest going over your solutions with a teacher. IAO problems are often deceptive and require some ingenuinity and imagination. Several problems do not have a well-defined 'correct answer'.

As I see the IAO site doesn't have this year's papers, so you may want to look over [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=170589]this[/url]

From my experience, IAO is not uber-competitive like IMO or IPhO- not to say it isn't difficult, but you don't have to study crazy like people do for IMO. (This has the advantage of keeping the IAO much more interesting, academically)

hi.(at firs i excuse for my poor English!)dear Shreyas_patankar  and Astrophys, I have solve near 85%of IAO problem I think they are not hard  but i belive that Several problems do not have a well-defined 'correct answer'. in iran we must solve hard problem .i dont have the problem of IAO2001,2002,2004. I have 2 question,do you know a good web site that has a good problem ?because i dont have any teacher!!!! iam sad,i dont have any problem plaese help me!
also in your country  what the people study for astronomy olympiad? for example  in iran we study spherical astronomy ( w m smart),fundamental astronomy ,celastial mechanic  and etc.
(i have IAo problem 2006,2007 if you whant them i  can get them to you,)
thanks!!!!  :)

Has Iran stepped their training recently? I believe they were like #2 this year...
Well, India was #1 for four years before this, and our teams don't prepare this much, spherical trigometry is not taught at the trainig camps. My freinds who were on the team said that we beat our rivals mainly in the observation part, one of my freinds was the 'best observer' for the last three years.

Our team members study according to their interests-there is no *required* reading. Popular references include [i]Astronomy:Principles and Practice[/i] by Roy & Clark, [i]Introduction to Astrophysics[/i] by Zeilik and Smith/Gregory, [i]Astrophysics I&II[/i] by K.D. Abhyankar (a local publication), W.M. Smart etc.

I begin to follow astronomy Olympiad. where can I get IAO Problem solving?
Thanks before :)

Check [url=http://www.issp.ac.ru/iao/]this[/url] and the other link I posted

Try the astronomy related problems on physics olympiads (for example, the pink problem at IPhO 2007), and look for National Astronomy Olympiads and TSTs to IAO. IOAA site has all the last year problems with solutions. There are many nice problems, easier than IAO ones, in the MIR book, "Problems and Exercises in Astronomy" or somewhat like this, by B. A. Vorontsov-Veliamínov.

Well, I have to start solving the IAO ones to improve my skills.


probability
conditional probability
combinatorics unsolved
combinatorics

My brother gave me these problems from his mathematical statistics exam.
I find these fun. I really cant seem to get part b of the problems.

1. In a busy holiday, a fleet of ten super large ships of which three are cargo ships are approaching the South Florida coastlines. Suppose in that day,the port of Miami has room only for five more ships, the port of Fort Lauderdale has room only for 3 more ships, and the port of Key West has room only for 2 more ships.
a) How many different ways can these ships be directed among these ports? I got $ \left( \frac{10!}{5!3!2!} \right)$
b) If these ships are randomly distributed among the above ports, find the probability that all three cargo ships will be directed to the port of miami?  :wacko:

2. There are two boxes labeled box I and box II. Box I contains 4 white balls and 8 black balls, and box II contains 6 white balls and 6 black balls. Suppose you randomly select one of the boxes and then from the selected box randomly draw 3 balls.
a) Find the probability that all three selected balls are black balls,
b) If all three selected balls are black, find the probability that they are drawn from box I. (Bayes??)

3. Suppose you have n keys, wish to open a door, and try the keys, one at a time, at random till you open the door. Only one of the keys will open the door.
a) Find the probability that you open the door on your $ j^{th}$ try, $ 1\leq j \leq n$, if each unsuccessful try you eliminate the unsuccessful key from further tries.
b) Find the probability that you open the door on your $ j^{th}$ try, $ 1\leq j < \infty$,  if each unsuccessful try you do not eliminate the unsuccessful key from further tries.
 :noo:   :help: :noo:  :noo:

Wow fraenkel I cant believe you are in the fourth grade and able to comprehend these problems!

I believe you are right on 1a)

1B) seems hard to me (and im in college!)

2a) I think you can get this.
2b) is a straightforward application of conditional probability (bayes)

3a) I got $ \frac{1}{n}$
b) Confused on this one.

Ill try working them out and see if I get any.
Keep up the good work.

Thanks guys for the help. I think I have figured it out now!!!

by the way Ish part 1B is $ \frac{\binom{5}{2} \binom{10}{7}}{ \left ( \frac{10!}{5!3!2!} \right )}$

3b) $ {\left ( \frac {n\minus{}1}{n} \right )}^{j\minus{}1} \frac{1}{n}$

lol it seems so easy now  :rotfl:

For 1b, count how many ways the desired task can be done.  Then divide.  I got the numerator in the fraction to be $ \binom{7}{2}\binom{5}{2}$ (the first is the number of ways to select 2 other ships to go to miami, and then second is the number of ways to pick 2 more ships to go to the last port).

Otherwise, you seem to be fine.


integration
search
calculus
calculus computations

prove that ${ \int_{0}^{x}}\frac {sin  t }{t + 1}\, dt$ is >= 0 (greater than or equal to 0) for all x>=0.

Ah, you made me search my own posts, which can be annoying.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275552]Here,[/url] with further links contained in that thread.

Thanks Kent and sorry for the trouble

can i do it this way actually i prove its greater than 0 till $ \ 2 {pi}$ since its a period . is that enough ?


inequalities
geometry
circumcircle
geometry proposed

Let $P$ be a point in the interior of $\triangle ABC$. The lengths of the sides of $\triangle ABC$ is $a,b,c$, and the distance from $P$ to the sides of $\triangle ABC$ is $p,q,r$. Show that the circumradius $R$ of $\triangle ABC$ satisfies \[\displaystyle R\le \frac{a^2+b^2+c^2}{18\sqrt[3]{pqr}}.\] When does equality hold?

I bet you mean circumradius ;)
Denote S by the area of $\triangle ABC$. We have $2S=ap+bq+cr\ge3(abcpqr)^{\frac{1}{3}}\Rightarrow\frac{a^2+b^2+c^2}{18\sqrt[3]{pqr}}\ge \frac{(a^2+b^2+c^2)\sqrt[3]{abc}}{12S}$. Now subsitute R by $\frac{abc}{4S}$. The result follows by AM-GM.

You're right! A correction has been posted. Thanks for the solution. :D


geometry
algebra
polynomial
area of a triangle
Heron\u0027s formula

Three distinct points A, B, and C lie on a plane. Let AB = x, BC = y, CA = z.

Prove that A, B, and C are collinear if and only if $ x^4 \plus{} y^4 \plus{} z^4 \equal{} 2x^2y^2\plus{}2y^2z^2\plus{}2z^2x^2$.

[hide]expanding heron's formula for the area of the triangle gives
$ A\equal{}\frac{\sqrt{x^4\plus{}y^4\plus{}z^4\minus{}2x^2y^2\minus{}2x^2z^2\minus{}2y^2z^2}}{4}$
iff A, B, and C are non-collinear, then the area of ABC is positive
iff they are collinear, then the area of 'triangle' ABC is 0
If A, B, and C are collinear, then
$ \frac{\sqrt{x^4\plus{}y^4\plus{}z^4\minus{}2x^2y^2\minus{}2x^2z^2\minus{}2y^2z^2}}{4}\equal{}0$
$ x^4\plus{}y^4\plus{}z^4\minus{}2x^2y^2\minus{}2x^2z^2\minus{}2y^2z^2\equal{}0$
$ x^4\plus{}y^4\plus{}z^4\equal{}2x^2y^2\plus{}2x^2z^2\plus{}2y^2z^2$

if the above statement is true, then you can logically conclude , through arithmetic operations, that
$ \frac{\sqrt{x^4\plus{}y^4\plus{}z^4\minus{}2x^2y^2\minus{}2x^2z^2\minus{}2y^2z^2}}{4}\equal{}0$
which means that 'triangle' ABC has no area, meaning A, B, and C are collinear
can't think of any better way right now
[/hide]

The interior of your squareroot should be negated.

[hide]Note that $ \sum_{cyc} x^4 \minus{} 2 x^2 y^2 \equal{} \minus{} (x \plus{} y \plus{} z)(x \plus{} y \minus{} z)(x \minus{} y \plus{} z)( \minus{} x \plus{} y \plus{} z)$

Under what conditions is this product zero?[/hide]

TZF, how did you factor that expression?

He knew that stevenmeow expanded it from it's original form $ (x\plus{}y\plus{}z)(x\plus{}y\minus{}z)(y\plus{}z\minus{}x)(z\plus{}x\minus{}y)$, which is in Heron's Formula:

$ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}\equal{}\frac14\sqrt{(a\plus{}b\plus{}c)(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b)}$.

math154, are you saying that the only way to factor it is by knowing the factorization beforehand?  :maybe:

The problem basically tells you already that $ x\plus{}y\minus{}z, x\minus{}y\plus{}z, \minus{}x\plus{}y\plus{}z$ are factors of it, since the polynomial has to be $ 0$ when any of those are $ 0$ for the problem to be true. Now you can just do the division to figure out the last factor (or consider negative side lengths).


Does anyone have the most recent (2008) NEAML competition or have any NEAMLs that are not found here http://matharchive.org/neaml/ that you could share

[quote="Pineappleperson"][quote="Yongyi781"]I am going. I will not predict which of Choate or Bedford will win Medium. :maybe:[/quote]

I am going. I will not predict which of Acton Boxborough or Lexington will win Large. It's going to be a titanic struggle.[/quote]

I am going. I will not predict which of Acton Boxborough or Lexington will win Large. However, it will be fun to see. :D

How did Choate do in their state round, and how are they looking this year in general?

Choate had a 69 if I remember correctly. They haven't been looking particularly good - they lost a few strong seniors to graduation and their coach left. They won the Medium Schools division, but we had a higher score. We also beat them in the local league.

man i am SO PUMPED for the new england meet

Could someone please direct me to some NEAML practice problems asides from what I referenced in the previous post as this is the first time my school has made it to NEAML

[quote="worthawholebean"]We do have what is probably the best team in school history. Note that our school is 350 years old.[/quote]
What school do u attend? How did u do in small schools last year? yea if u have a bunch of ppl as good as u u will win.

[quote="mathemonster"]Ill be there (this is the first time we are taking a legit team even tho we've won RI 3 years in a row) I want to do well but I think im not really going to practice b/c i do best that way. Well have: Me, 2 USAMOing seniors (I usually tie or get close to them on contests cept i failed the AMC this year  :maybe: ), a decent freshman, and a decent junior. I think we will win small schools fosho'[/quote]

[img]http://img529.imageshack.us/img529/8820/wheeler.png[/img]

Last year only freshman went who weren't in the top 6 because AP BIO + Physics both had field trips so....
Our max is probably: 18,18,18,15,15 + (10)= 94 so it depends on what kind of day we have. 4 example, We beat Barrrington (#4 medium) by like 20+ this year.

[quote="perfect628"]Grr I swear if RL went to these things...[/quote]

...RL would not do so well
well actually RL would count as a small school

in which case it would do very well

then again, what if exeter went?

Look at the results from last year. Nothing wrong with Small Schools!

[quote="worthawholebean"]We do have what is probably the best team in school history. Note that our school is 350 years old. We have a USAMO qualifier who isn't even on the team.[/quote]

My school is 374 years old.

[quote="worthawholebean"]. We have a USAMO qualifier who isn't even on the team.[/quote]
Well im guessing u dont have 7 USAMO qualifiers so thats kinda dumb.

also NEAML HAS NOT been around for 40000 years so who cares?

[quote="perfect628"][quote="Smartguy"]My school is 374 years old.[/quote]
Ah, but has it been in continuous existence? :P[/quote]

............ :|  :mad:

[quote="worthawholebean"]We have a USAMO qualifier who isn't even on the team.[/quote]

darn, five of our USAMO qualifiers aren't on the team (though we don't actually have a team yet, they won't be on the team anyway).

why don't u have a team yet... im confused

I'm definitely excited. Our team usually doesn't do amazingly well, but top 5 large may be doable for us this year.

[quote="baozhale"]I'm going... and practicing because we need to get Bedford back for breaking our winning streak at MAML :wink:[/quote]

Lolololol. Don't worry, USAMO will definitely wear me down for NEAML. Good luck, but Bedford will still win! :lol:

NEAML tomorrow

...I think my team, Sharon will do decent. We had a bad MAML run but I think we'll be top 3-4 this year if we can get our heads into the rounds

ITS GOING DOWN
WHEELER FTW
PONE PONE PONE

[quote="CatalystOfNostalgia"]darn well like last year we were missing three people (two due to DOE nationals and the third due to State Science Fair), and the third we didn't actually know wouldn't be coming until about two hours before the thing started. so we had to run around looking for our alternate.[/quote]

Who won?

Large:
1. Acton-Boxborough - 134
2. Lexington

Medium:
1. Choate - 88

Small:
1. Hopkins - 95
2. Worcester Academy - 90
[b]3. Wheeler[/b]

Also I apologize for making that round 5 protest, but they were wrong, so it needed to be done.

What round 5 protest?...

Blah, Bedford failed so badly. The scores went something like:

Choate - 88
Westborough - 76
Winchester - 75
Bedford - 73

T-T This was like our worst year ever.

Oh well, I got 18, so....

Also, free 2 points on Round 4? I don't remember the second problem; I assume it was incorrect? (i.e. there are no solutions, etc)?


Compute the area enclosed by a circular pool with a circumference of 1000 feet.

a) What would be your choice of shape for the pool?

b) If rectangular, what is your preference for dimensions?

c) If your only consideration is to have a pool that encloses the most area, what shape should you use?

NOTE: The perimeter of the rectangular-shaped pool to also consider is 1000 feet.

a: Rectangular, or else its hard to do laps, but thats just me

b: You already posted this!

c: Circular. I invite you to prove this yourself :wink:

Thanks for your reply.

To prove it myself?

I could never do that!!!  LOL.

I'm pretty sure it does... polar integration, probably...  :roll:

Well, I've never taken calculus 1.

Thus, I could not prove it, right?


I'm having trouble figuring out how to submit connecting flights in the Travel Itinerary page... :oops: Can anyone help me out? Thanks!

Er.. just submit each flight separately (that's what I did).

ok thanks...but now the page is down... :(

Yeah, I had similar problems.  I had to retype info in multiple times.  I suggest that after entering each leg of your itinerary you click "next" and then come back, as this seems to save the information you entered.  If you try it a couple times eventually it should work properly.

Hm...I can't even get to http://www.mathcamp.org/registration now!  This message keeps on coming up: "Registration:1 (localhost:16500) doesn't respond"  :(


Two standard 6-sided dice are rolled.  What is the probability that the sum rolled is a perfect square?

Perfect squares are 4 and 9 as the sum is between 3 and 12 inclusive. We can get 4 in 3 ways, and 9 in 4 ways, and as we can get 36 different throws, probability is $ \frac{7}{36}$

Why is 2+2 only counted once for 4? Shouldn't it be (1,3)(3,1)(2,2)(2,2)?

(2+2) is only counted once because when you rearrange, it is the same, where (3+1) and (1+3) look different. Another way to think of this is that to get (2,2), the probability is $\frac{1}{6}\times\frac{1}{6}$. To get (3+1) or (1+3), the probability is $\frac{2}{6}\times\frac{1}{6}$, because the first one can be either 3 or 1. Hope that helps!

So how you do it, there is a formula. If you are trying to find how many possibilities there are for the sum of 2 dice rolls from 2 to 7, you subtract one. For example, if you are finding the probability of getting a sum of 3 with 2 dice rolls, 3-1=2, and 2/36=1/18 is the probability. From 7-12, you subtract the number from 13.  For example, if you are finding the probability of getting a sum of 9 with 2 dice rolls, 13-9=4, and 4/36=1/9 is the probability.


geometry
geometric transformation
rotation
trigonometry
inequalities
complex numbers
geometry proposed

[color=darkred][size=117][u]Nice property, [b]Sunken Rock[/b], I don't know it[/u].[/size][/color]

[color=darkblue]But I think that we can solve it easily with complex numbers ![/color]

and proof [quote="sunken rock"] [color=darkred]One constructs the equilateral triangles $ BCD$ , $ CAE$  and $ ABF$ outside of the triangle $ ABC$ .
If $ M$ , $ N$ and P are the midpoints of $ EF$ , $ FD$ and $ DE$ respectively, then $ AM \equal{} BN \equal{} CP$
and every two of the lines $ AM$ , $ BN$ , $ CP$ make an angle of $ 60^{\circ}$ between them.[/color] [/quote]

[color=darkblue][b][u]Proof (with complex numbers)[/u].[/b] Denote : $ X(x)$ - the point $ X$ with affix $ x$ ; rotation $ w\equal{}\cos\frac {\pi}{3}\plus{}i\cdot\sin\frac {\pi}{3}\ \ (w^3\plus{}1\equal{}0\ ,\ w^2\equal{}w\minus{}1)$ .

So $ A(a)$ , $ B(b)$ , $ C(c)$ and $ \left\|\begin{array}{c}
d\equal{}c\plus{}w(b\minus{}c)\\\
e\equal{}a\plus{}w(c\minus{}a)\\\
f\equal{}b\plus{}w(a\minus{}b)\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c}
2m\equal{}e\plus{}f\equal{}(a\plus{}b)\plus{}w(c\minus{}b)\\\
2n\equal{}f\plus{}d\equal{}(b\plus{}c)\plus{}w(a\minus{}c)\\\
2p\equal{}d\plus{}e\equal{}(c\plus{}a)\plus{}w(b\minus{}a)\end{array}\right\|$ $ \implies$ $ \left\|\begin{array}{c}
2(m\minus{}a)\equal{}(b\minus{}a)\plus{}w(c\minus{}b)\\\
2(n\minus{}b)\equal{}(c\minus{}b)\plus{}w(a\minus{}c)\\\
2(p\minus{}c)\equal{}(a\minus{}c)\plus{}w(b\minus{}a)\end{array}\right\|$ .

Show easily that $ \frac {m\minus{}a}{1}\equal{}\frac {n\minus{}b}{\minus{}w}\equal{}\frac {p\minus{}c}{w^2}$ $ \implies$ $ AM\equal{}BN\equal{}CP$ and every two of $ AM$ , $ BN$ , $ CP$ make an angle of $ 60^{\circ}$ between them.

[b][u]A metrical remark[/u].[/b] Prove easily that $ AM^2\equal{}\frac 18\cdot\left(a^2\plus{}b^2\plus{}c^2\minus{}4S\sqrt 3\right)$ (symmetrical expression in $ a$ , $ b$ , $ c$ ). Thus, $ AM\equal{}BN\equal{}CP$.

Now we can observe an [u]interesting geometrical interpretation and proof of  the remarkable inequality[/u] $ a^2\plus{}b^2\plus{}c^2\ge 4S\sqrt 3$ . [b]Nice problem !

Thank you.[/b]

V.N.[/color]

[quote="sunken rock"]One constructs the equilateral triangles BCD, CAE and ABF outside of the triangle ABC.
If M, N and P are the midpoints of EF, FD and DE respectively, then:
AM = BN = CP, they have a common point and every two of them make an angle of 60 degs between them.

Best regards,
sunken rock[/quote]
This problem is very nice and here is my solution :D
Since $ \vec{AM} = \frac {1}{2}(\vec{AF} + \vec{AE})$ we get $ AM^2 = \frac {1}{4}(AF^2 + AE^2 + 2.\vec{AF}.\vec{AE})$
${ = 1/4(b^2 + c^2 + 2bc.cos(120^o + A}) = 1/4(b^2 + c^2 - bc.cosA - bc.\sqrt {3}.sin A$
We will prove $ b^2 + c^2 - bc.cos A - bc.\sqrt {3}.sin A = a^2 + c^2 - ac.cos B - ac.\sqrt {3}.sin B (1)$
$ \Leftrightarrow a^2 - b^2 + c(bcosA - acosB) + c.\sqrt {3}(bsin A - asin B) = 0$
$ \Leftrightarrow 2ac.cosB - c^2 + bc.cosA - ac.cosB = 0$ (because $ \frac {a}{sinA} = \frac {b}{sinB}\Rightarrow bsinA - asinB = 0$)
$ \Leftrightarrow c^2 = ac.cosB + bc.cosA$
It's true because $ RHS = \frac {a^2 + c^2 - b^2}{2} + \frac {b^2 + c^2 - a^2}{2} = c^2$
Therefore (1) is true then $ AM = BN$, similarly we obtain $ AM = BN = CP$
Let K be the intersection of the arc $ 60^o$ which go through A,B and the arc $ 60^o$ which go through A,C then $ \angle BKC = 120^o$.Let M' be the intersection of KA and EF. We need to prove that $ M\equiv M'$
Let Q be the intersection of FK and (ABK)
M' is the midpoint of EF
$ \Leftrightarrow \frac {FK}{EK} = \frac {sin \angle AKE}{sin \angle AKF}$ or $ FK.sin \angle AKF = EK.sin \angle AKE$
On the other hand, $ \frac {AF}{\sin \angle FKA} = \frac {FK}{sin \angle FAK} = \frac {c}{sin \angle FKA}$
then $ FK.sin \angle FKA = c.sin \angle FAK = c.sin (\angle FAQ + \angle QAK)$
$ = c.sin (\angle QKA + \angle QAK)= c.sin \angle AQK$
We have $ \frac {AB}{sin \angle PKA} = \frac {AK}{sin \angle AQK} = \frac {c}{sin 60^o}$
So $ FK.sin \angle FKA = c.sin\angle AQK = AK.sin 60^0$
Similarly, $ EK.sin \angle EKA = AK.sin 60^o$
then $ FK.sin \angle FKA = EK.sin \angle EKA$
$ \Rightarrow M'$ is the midpoint of EF or $ M'\equiv M$
Similarly, $ BN,CP$ go through K (q.e.d)

Hi, mathlinkers,

Attached, my proof to this problem. It's little bit boring, but ... it's synthetic 100%!

Many thanks to Virgil for his fine remark!

I hope the interested ones to be able to see the attachment, as my .pdf file is very heavy, over .7 Mb

Best regards,
sunken rock


geometry
rectangle
combinatorics unsolved
combinatorics

in a 7*7 square each cell are colored white or black.
prove that there is at least 21 rectangle with the same color of it vertices ( 4 cell ) that the sides are paralel to the 7*7 square
i hope you will undrestand[/size]

Let $b(i,j)$ be the number of columns which are black in both row $i$ and row $j$.  Let $b=\sum_{i<j} b(i,j)$.  Let $w(i,j)$ and $w$ be defined similarly for white squares.

Any particular column will always have at least 9 pairs of squares in it which are the same color (Corresponding to when there are between 3-4 white squares in the column; other configurations lead to more).  It follows that $w+b \geq 63$.

Twice the number of rectangles is equal to

\[\sum_{i<j} [w(i,j) (w(i,j)-1) + b(i,j) (b(i,j) - 1)] \]

It suffices to show that this sum is at least 42 for every set of 42 nonnegative integers adding to at least 63 that we place as our $w(i,j)$ and $b(i,j)$.  To see this, we note 2 things:

1. The sum is monotone increasing in every $w(i,j)$ and $b(i,j)$, so it suffices to consider those 42-tuples adding to exactly 63.  This is a finite set, so there is a minimum somewhere.

2. If any $w(i,j)$ or $b(i,j)$ differ by at least 2, we can make the sum smaller by increasing the smaller one by 1 and decreasing the larger by 1.  Therefore at the minimum which we know exists the largest and smallest among the $w(i,j)'s$ and $b(i,j)'s$ differ by at most 1.  The only way this can happen and still have everything add up to 63 is if half of them are 2, the other half are 1.  But in this case the sum is exactly 42, so we are done.


Assume that $ a_n > 0$ and that $ \sum a_n$ diverges.
I can show that $ \displaystyle\sum\frac {a_n}{1 \plus{} n^\alpha a_n}$ converges if $ \alpha$ is a real number greater than $ 1$.
Since $ \frac{a_n}{1\plus{}n^\alpha a_n}\equal{}\frac{1}{\frac{1}{a_n}\plus{}n^\alpha}<\frac{1}{n^\alpha}$

My issue is as follows.
Assume that $ n^\alpha a_n\to 0$ and $ a_n\to 0$ as $ n\to\infty$. Then there exist $ N$ such that $ n^\alpha a_n < 1$ for $ n > N$.
Now if the sequence converges we can have that for $ n,m$ large enough

$ \epsilon > \frac {a_n}{1 \plus{} n^\alpha a_n} \plus{} \cdots \plus{} \frac {a_m}{1 \plus{} m^\alpha a_m} > \frac {a_n}{2} \plus{} \cdots \plus{} \frac {a_m}{2}$

which contradicts the fact that $ \sum a_n$ diverges. this doesn't seem to depend on what the value of $ \alpha$ is.

did I do something wrong here?

[quote="maokid7"]Assume that $ n^\alpha a_n\to 0$ ... as $ n\to\infty$.
...
which contradicts the fact that $ \sum a_n$ diverges.
...
did I do something wrong here?[/quote] If $ n^\alpha a_n \to 0$ then $ \sum a_n$ converges by limit comparison with $ \sum n^{ \minus{} \alpha}$.  What you did was had a sequence with property X, assumed the sequence had some property Y that implies not-X, and then became surprised when you could deduce not-X from Y :wink:

Thanks for the clarification. How would you handle the case where $ \alpha \le{1}$?

What about that case?  We could have divergence (e.g. if $ a_n \equal{} 1$) or convergence (e.g. if $ a_{2^n} \equal{} 1$ and $ a_m \equal{} 2^{ \minus{} m}$ otherwise), assuming I haven't done anything stupid.

It might also help if you gave more indication of what it is exactly that you're trying to solve.


Show that if a k-regular bipartite graph has bipartiton (X,Y) then |X|=|Y|

(k regular means that every vertex is of degree k)

Haven't I seen this problem before? (on the forum, I mean :)). Anyway, since it's been staying here for quite some time :), the number of edges is both $k|X|$ and $k|Y|$, so $|X|=|Y|$.


νδο ο $ 2003$ δεν ειναι αθροισμα [b]δυο [/b]τελειων τετραγωνων ακαιραιων αριθμων

[quote="Demetres"]Κάτι δεν καταλαβαίνω

$ 2003 \equal{} 1^2 \plus{} \cdots \plus{} 1^2 \equal{} 1^2 \plus{} 3^2 \plus{} 12^2 \plus{} 43^2$.[/quote]

εχεις δίκιο παλι (ως συνηθως) εκανα βλακεία στην διατυπωση  :wallbash:

Ίσως είναι για ακριβώς δύο η εκφώνηση.

Σε αυτή την περίπτωση έχουμε ότι $ 45^2\equal{}2025 > 2003$
οι αριθμοί $ a,b: a^2\plus{}b^2\equal{}2003$ που ψάχνουμε είναι $ < 45$

$ a^2 \plus{} b^2 \equiv 2003 mod 5 \Leftrightarrow a^2 \plus{} b^2 \equiv 8 mod 5$

και επειδή

$ a^2 \equiv [0,1,4] mod 5$

$ b^2 \equiv [0,1,4] mod 5$

πρέπει

$ a^2 \equiv b^2 \equiv 4 mod 5$

άρα $ a,b \equal{} [4, 9, 14, 19 ,24, 29, 34, 39, 44]$

ομοίως με τετραγωνικά κατάλοιπα του 3 είναι

$ a^2 \equiv b^2 \equiv 1 mod 3$

άρα $ a,b \equal{} [4, 19 , 34]$

εξετάζοντας τώρα περιπτώσεις (το έχω κάνει με πρόγραμμα) βγαίνει ότι κανένας συνδυασμός των $ a,b$

δεν δίνει $ a^2\plus{}b^2\equal{}2003$

[quote="maniopas"]Ίσως είναι για ακριβώς δύο η εκφώνηση.

Σε αυτή την περίπτωση έχουμε ότι $ 45^2 \equal{} 2025 > 2003$
οι αριθμοί $ a,b: a^2 \plus{} b^2 \equal{} 2003$ που ψάχνουμε είναι $ < 45$

$ a^2 \plus{} b^2 \equiv 2003 mod 5 \Leftrightarrow a^2 \plus{} b^2 \equiv 8 mod 5$

και επειδή

$ a^2 \equiv [0,1,4] mod 5$

$ b^2 \equiv [0,1,4] mod 5$

πρέπει

$ a^2 \equiv b^2 \equiv 4 mod 5$

άρα $ a,b \equal{} [4, 9, 14, 19 ,24, 29, 34, 39, 44]$

ομοίως με τετραγωνικά κατάλοιπα του 3 είναι

$ a^2 \equiv b^2 \equiv 1 mod 3$

άρα $ a,b \equal{} [4, 19 , 34]$

εξετάζοντας τώρα περιπτώσεις (το έχω κάνει με πρόγραμμα) βγαίνει ότι κανένας συνδυασμός των $ a,b$

δεν δίνει $ a^2 \plus{} b^2 \equal{} 2003$[/quote]

Δούλεψες με mod3 και με mod5. Με mod4 γιατί δεν το έψαξες  :P

Με mod 4 τα τετραγωνικά υπόλοιπα είναι 0, 1, 2. Θα έπρεπε πάλι να κάνω την ίδια διαδικασία.
Ως προς τους αριθμούς πάντως, προτιμάω να χρησιμοποιώ πρώτους γιατί είναι πιο δύσκολο να κάνει
κανείς λάθος σε πιο δύσκολα θέματα.

[quote="maniopas"]$ 6 \equiv 2 mod 4$[/quote]

Αλλά το 6 δεν είναι τέλειο τετράγωνο.  :)

Δίκιο έχεις!!! Εγώ έκανα ένα λάθος στον πίνακα που κατασκεύασα. Με mod 4 λοιπόν βγαίνει κατευθείαν.

[quote="bilot4"]Τα τετραγωνικά κατάλοιπα mod $ 4$ είναι $ 0,1$. Αρα αμέσως αποδεικνύεται ότι $ 2003$ διάφορο του $ x^2 \plus{} y^2$ αφού 2003 ισούται με $ 3$ mod $ 4$[/quote]
σωστός!!!
ή με γυμνασιακή ορολογια αρκεί να αποδείξουμε οτι το αθροισμα 2 τελειων τετραγώνων διαιρούμενο με το 4 δεν μπορεί να είναι 3 (περνονας περιπτωσεις για τους αριθμους περιττός ή άρτιος)

Ωραία. Τώρα γνωρίζουμε ότι κάθε φυσικός αριθμός της μορφής $ 3 \bmod 4$, δεν μπορεί να γραφτεί ως άθροισμα δύο τέλειων τεραγώνων. Ας την προχωρήσουμε λίγο την άσκηση.

Να δειχθεί ότι αν $ n \equal{} x^2 \plus{} y^2$ και $ p$ πρώτος της μορφής $ 3 \bmod 4$ τότε $ p|n \Rightarrow p^2|n$.

Παραμένει να λυθεί η άσκηση που έβαλε ο bilot 4, και βάζω ακόμη μία άσκηση:

Να δειχθεί ότι αν $ p$ πρώτος της μορφής $ 1 \bmod 4$ τότε ο υπάρχουν ακέραιοι $ m,n$ ώστε $ m^2 \plus{} n^2 \equiv 0 \bmod p$. (Ισχύει ότι υπάρχουν $ m,n$ ώστε $ m^2 \plus{} n^2 \equal{} p$ αλλά αυτό είναι αρκετά δύσκολο. Το ισχυρίστηκε ο Fermat το 1640 και το απέδειξε ο Euler το 1747. Ίσως βάλω κάποια από τις γνωστές αποδείξεις αργότερα.)

[quote="bilot4"][quote="Demetres"]
Να δειχθεί ότι αν $ n \equal{} x^2 \plus{} y^2$ και $ p$ πρώτος της μορφής $ 3 \bmod 4$ τότε $ p|n \Rightarrow p^2|n$.[/quote]

Αρκετά ωραίο και γνωστό λήμμα :)

Αν $ y\equiv 0 \pmod p$, τότε το ζητούμενο είναι εύκολο αφού πρέπει τότε $ x\equiv 0 \pmod p$ κι άρα $ n\equiv 0 \pmod {p^2}$.

Aν $ (p,y) \equal{} 1$ τότε υπάρχει αντίστροφο $ y'$ του $ y\pmod p$. Αρα:
$ x^2\equiv \minus{} y^2\iff (xy')^2\equiv \minus{} 1 \pmod p$. Ατοπο (αφήνω την απόδειξη αυτού για άσκηση!)[/quote]

Εστω $ xy' \equal{} k$ και $ k^2 \equiv \minus{} 1 (modp)$.
Αν $ k \equiv 1(modp)$ τοτε $ k^2 \equiv 1(modp)$, ατοπο.
Αν $ k^3 \equiv 1(modp) \Rightarrow \minus{} k \equiv 1(modp) \Rightarrow k^2 \equiv 1(modp)$, ατοπο
Αρα αφου $ k^4 \equiv 1(modp)$, αν θεωρησουμε την ομαδα $ < G, * > , G \equal{} \{1,2,..p \minus{} 1(modp)\}$ με $ p \minus{} 1$ στοιχεια, εχουμε οτι η ταξη του $ a(modp)$ (το οποιο ταυτιζεται με καποιο απο τα στοιχεια $ \{1,2,...p \minus{} 1(modp)\}$ της ομαδας) ειναι ο αριθμος 4.
Επομενος απο το θεωρημα του Lagrange εχω $ 4|p \minus{} 1$, δηλαδη $ p \minus{} 1 \equal{} 4m \Rightarrow p \equal{} 4m \plus{} 1$, ατοπο αφου $ p \equiv 3(mod4)$.
Αρα ο $ p$ δεν μπορει να διαιρει τον $ k^2 \plus{} 1$, για $ k:$ ακεραιο με $ gcd(k, p) \equal{} 1$.

ΥΓ: Γινεται και με το μικρο θεωρημα του Fermat διακρινοντας περιπτωσεις για το υπολοιπο του $ p\minus{}1$ με το 4

[quote="NickNafplio"][quote="bilot4"][quote="Demetres"]
Να δειχθεί ότι αν $ n \equal{} x^2 \plus{} y^2$ και $ p$ πρώτος της μορφής $ 3 \bmod 4$ τότε $ p|n \Rightarrow p^2|n$.[/quote]

Αρκετά ωραίο και γνωστό λήμμα :)

Αν $ y\equiv 0 \pmod p$, τότε το ζητούμενο είναι εύκολο αφού πρέπει τότε $ x\equiv 0 \pmod p$ κι άρα $ n\equiv 0 \pmod {p^2}$.

Aν $ (p,y) \equal{} 1$ τότε υπάρχει αντίστροφο $ y'$ του $ y\pmod p$. Αρα:
$ x^2\equiv \minus{} y^2\iff (xy')^2\equiv \minus{} 1 \pmod p$. Ατοπο (αφήνω την απόδειξη αυτού για άσκηση!)[/quote]

Εστω $ xy' \equal{} k$ και $ k^2 \equiv \minus{} 1 (modp)$.
Αν $ k \equiv 1(modp)$ τοτε $ k^2 \equiv 1(modp)$, ατοπο.
Αν $ k^3 \equiv 1(modp) \Rightarrow \minus{} k \equiv 1(modp) \Rightarrow k^2 \equiv 1(modp)$, ατοπο
Αρα αφου $ k^4 \equiv 1(modp)$, αν θεωρησουμε την ομαδα $ < G, * > , G \equal{} \{1,2,..p \minus{} 1(modp)\}$ με $ p \minus{} 1$ στοιχεια, εχουμε οτι η ταξη του $ a(modp)$ (το οποιο ταυτιζεται με καποιο απο τα στοιχεια $ \{1,2,...p \minus{} 1(modp)\}$ της ομαδας) ειναι ο αριθμος 4.
Επομενος απο το θεωρημα του Lagrange εχω $ 4|p \minus{} 1$, δηλαδη $ p \minus{} 1 \equal{} 4m \Rightarrow p \equal{} 4m \plus{} 1$, ατοπο αφου $ p \equiv 3(mod4)$.
Αρα ο $ p$ δεν μπορει να διαιρει τον $ k^2 \plus{} 1$, για $ k:$ ακεραιο με $ gcd(k, p) \equal{} 1$.

ΥΓ: Γινεται και με το μικρο θεωρημα του Fermat διακρινοντας περιπτωσεις για το υπολοιπο του $ p \minus{} 1$ με το 4[/quote]

Για όσους δεν έχουν ξαναδεί την απόδειξη χρησιμοποιώντας το μικρό θεώρημα του Fermat στην οποία αναφέρεται ο NickNafplio:

Αφού $ k^4 \equiv 1 \bmod p$ και $ p \equiv 3 \bmod 4$, τότε $ k^{p \minus{} 3} \equal{} (k^4)^{(p\minus{}3)/4} \equiv 1 \bmod p$. Αλλά από το μικρό θεώρημα του Fermat, $ k^{p \minus{} 1} \equiv 1 \bmod p$, άρα $ k^2 \equiv 1 \bmod p$, άτοπο.

Για το άλλο ερώτημα που έχω θέσει,
[quote="Demetres"]
Να δειχθεί ότι αν $ p$ πρώτος της μορφής $ 1 \bmod 4$ τότε ο υπάρχουν ακέραιοι $ m,n$ ώστε $ m^2 \plus{} n^2 \equiv 0 \bmod p$. (Ισχύει ότι υπάρχουν $ m,n$ ώστε $ m^2 \plus{} n^2 \equal{} p$ αλλά αυτό είναι αρκετά δύσκολο. Το ισχυρίστηκε ο Fermat το 1640 και το απέδειξε ο Euler το 1747. Ίσως βάλω κάποια από τις γνωστές αποδείξεις αργότερα.)[/quote]

[hide] Θεώρημα Wilson [/hide]


Rationalize the denominator of the fraction $ \frac{(\sqrt3\plus{}\sqrt5)(\sqrt5\plus{}\sqrt7)}{\sqrt3\plus{}2\sqrt5\plus{}\sqrt7}.$

[hide="Hint"]$ \frac {ab}{a \plus{} b} \equal{} \frac {1}{\frac {1}{a} \plus{} \frac {1}{b}}$.[/hide]

let $ a$ , $ b$ and $ c$ be two real numbers therefore we have :
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c) (a \minus{} b \minus{} c)$
we get $ a \minus{} b \minus{} c$ if we just take the conjuate (forget the spelling) of $ a \plus{} b \plus{} c$ so we have:

$ \frac {(\sqrt3 \plus{} \sqrt5)(\sqrt5 \plus{} \sqrt7)}{\sqrt3 \plus{} 2\sqrt5 \plus{} \sqrt7} * \frac {\sqrt3 \minus{} 2\sqrt5 \minus{} \sqrt7} {\sqrt3 \minus{} 2\sqrt5 \minus{} \sqrt7}$
you do the simplifying.... sorry, too lazy to get up and get a piece of paper  :D

[quote="Poincare"]let $ a$ , $ b$ and $ c$ be two real numbers therefore we have :
$ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c) (a \minus{} b \minus{} c)$[/quote]

Wrong!

yep Poincare is wrong, my solution is correct  :P

If I'm not mistaken,

$ (a\plus{}b\plus{}c)(a\minus{}b\minus{}c)\equal{}a^2\minus{}b^2\minus{}c^2\minus{}2bc$.

Perhaps you were thinking of

$ a^2\minus{}b^2\equal{}(a\plus{}b)(a\minus{}b)$?

[quote="ifai"]Rationalize the denominator of the fraction $ \frac {(\sqrt3 \plus{} \sqrt5)(\sqrt5 \plus{} \sqrt7)}{\sqrt3 \plus{} 2\sqrt5 \plus{} \sqrt7}.$[/quote]

You would multiply the whole fraction by: [b]$ \sqrt {3}\minus{}2\sqrt {5} \plus{} \sqrt {7}$[/b] [b]OR[/b] [b]$ \sqrt {3} \plus{} 2\sqrt {5}\minus{}\sqrt {7}$[/b].  You could also multiply the numerator to get: [b]$ \sqrt {15} \plus{} \sqrt {21} \plus{} \sqrt {35} \plus{} 5$[/b].  Here is the answer:

$ \frac {\sqrt {15} \plus{} \sqrt {21} \plus{} \sqrt {35} \plus{} 5}{\sqrt {3} \plus{} 2\sqrt {5} \plus{} \sqrt {7}} \bullet \frac {\sqrt {3}\minus{}2\sqrt {5} \plus{} \sqrt {7}}{\sqrt {3}\minus{}2\sqrt {5} \plus{} \sqrt {7}}$

That would get you the answer.  (I think)  I don't have paper so someone could simplify it.

[quote="ajain"][quote="azn6021023"]I'm not sure this is correct.  If we do multiply it out, we get -10+ 2 sqrt 21.

What I think is an explanation:

(a+b+c)(a+b-c) is actually a difference of two squares: (a+b)^2 -c^2.  However, since a and b are two different numbers under the radical, we will still have a radical in the denominator.[/quote]

Well, if we still have a radical, you can find teh conjuncate (something like that) and multiply it and get the final answer.[/quote]Yes it works but that would be a whole lot of computations. In general, it works but the way the problem is written, kunny's hint is a 'slick' way of solving this problem.

[quote="kunny"]$ \frac {ab}{a \plus{} b} \equal{} \frac {1}{\frac {1}{a} \plus{} \frac {1}{b}}$.[/quote]

[quote="mathemonster"]the solution was trivial after rationalizing each denominator separately. Sorry for the double post[/quote]

Thank you! When I saw this question, I thought about Kunny's formual. However it seemed useless for this question I thought. So I did nothing with it.

[quote="fishythefish"]If I'm not mistaken,

$ (a \plus{} b \plus{} c)(a \minus{} b \minus{} c) \equal{} a^2 \minus{} b^2 \minus{} c^2 \minus{} 2bc$.

Perhaps you were thinking of

$ a^2 \minus{} b^2 \equal{} (a \plus{} b)(a \minus{} b)$?[/quote]

whoops!!  :oops:

[quote="mathemonster"]Kunny's way which i used is the slick solution. Can somebody verify my answer using his method?[/quote]
$ \frac {(\sqrt3 \plus{} \sqrt5)(\sqrt5 \plus{} \sqrt7)}{\sqrt3 \plus{} 2\sqrt5 \plus{} \sqrt7}$

$ \equal{}\frac1{\frac1{\sqrt3\plus{}\sqrt5}\plus{}\frac1{\sqrt5\plus{}\sqrt7}}$

$ \equal{}\frac1{\frac{\sqrt5\minus{}\sqrt3}2\plus{}\frac{\sqrt7\minus{}\sqrt5}2}$

$ \equal{}\frac2{\sqrt7\minus{}\sqrt3}$

$ \equal{}\frac{2(\sqrt7\plus{}\sqrt3)}4$

$ \equal{}\frac{\sqrt7\plus{}\sqrt3}2$

why is it important to rationalize the denominator?

It puts radical expressions into a "standard form" where they can be compared (similar to how lines and parabolas and circles are put into a standard form).  It might be very difficult to tell that two radical expressions are in fact the same value if they are not written with their denominators rationalized.  It's also easier to add, subtract, multiply, and divide by radical expressions when their denominators are rationalized.  Try to add $ 4 \plus{} \sqrt{15} \plus{} \frac{1}{4 \plus{} \sqrt{15}}$ with and without rationalizing!


geometrically, problem 4 is 3 points on a unit circle centered at the origin. We are given that the centroid is the origin, and the circumcenter, hence we have an equilateral triangle. $ |z_{1}\minus{}\minus{}z_{2}|$ is the length of a side of the triangle, which is easily computed using 30-60-90 triangles.

5, he means that show z lies on that circle. i.e. $ |z\minus{}2|\equal{}1$, which is obvious since complex modulus is distributive over divison.

[quote="t0rajir0u"][hide="1"] Let $ a \equal{} z_{1}\minus{}z_{2}, b \equal{} z_{2}\minus{}z_{3}$.  Then

$ z_{1}^{2}\plus{}z_{2}^{2}\plus{}z_{3}^{2}\equal{} z_{1}z_{2}\plus{}z_{2}z_{3}\plus{}z_{3}z_{1}\Leftrightarrow a^{2}\plus{}b^{2}\plus{}(a\plus{}b)^{2}\equal{} 0\Leftrightarrow a^{2}\plus{}ab\plus{}b^{2}\equal{} 0\Leftrightarrow (a\minus{}\omega b)(a\minus{}\omega^{2}b) \equal{} 0$

The conclusion follows ($ a, b,\minus{}a\minus{}b$ are an equilateral triangle). [/hide]
[hide="2"] $ z_{1}^{2}\plus{}z_{2}^{2}\equal{} z_{1}z_{2}\Leftrightarrow (z_{1}\plus{}\omega z_{2})(z_{1}\plus{}\omega^{2}z_{2}) \equal{} 0$

The conclusion follows (again, $ z_{1},\minus{}z_{2}, z_{2}\minus{}z_{1}$ are an equilateral triangle). [/hide]
[hide="3"] Again let $ a \equal{} z_{1}\minus{}z_{2}, b \equal{} z_{2}\minus{}z_{3}$.  Then

$ \frac{1}{a}\plus{}\frac{1}{b}\minus{}\frac{1}{a\plus{}b}\equal{} 0\Leftrightarrow a^{2}\plus{}ab\plus{}b^{2}\equal{} 0\Leftrightarrow (a\minus{}\omega b)(a\minus{}\omega^{2}b)$

The conclusion follows (again). [/hide]
[hide="4"] $ z_{1}, z_{2}, z_{3}$ are an equilateral triangle with side length $ 1$; WLOG, let $ z_{1}\equal{} 1, z_{2}\equal{}\omega, z_{3}\equal{}\omega^{2}$.  Computation. [/hide]
What do you mean by $ M(z)$ on problem 5?[/quote]

If $ z\equal{}a\plus{}bi$, then $ M(z)\equal{}(a,b)\in\mathbb{R}^{2}$.

How can i learn advanced complex numbers so I will be able to do the qns above?


X-posted from the "other topics" thread"

Cassini's ovals are defined by locus of a point P  where product  of its distance from two  given points is contatnat. (It the sum was constant, the curve would have been an ellipse)

There is a very interesting problem associated with Cassini's ovals.

Let F and G are two "focal points" and Let P is a point on Cassini's oval.
P is defined as a locus when (P F )*(P G) is constant.

Now consider right bisectors of P F and P G and call the point Q where they intersect. What is the locus of point Q?

(Please provide the answers in  spoiler )

[hide]Q should be the midpoint of FG if i understand this problem right[/hide]

[quote="demji"]Q should be the midpoint of FG if i understand this problem right[/quote]
Q is not fixed, it depend on where P is.  Also can you put the answer is spoiler.  Thanks.


quadratics
number theory unsolved
number theory

Let $g(n)$ be defined as follows: \[ g(1) = 0, g(2) = 1 \] and  \[ g(n+2) = g(n) + g(n+1) + 1, n \geq 1. \] Prove that if $n > 5$ is a prime, then $n$ divides $g(n) \cdot (g(n) + 1).$

Note that $g(n)+1$ is the Fibonacci number $f(n-1)$ (defining $f(0) = 0, f(1) = 1$).  So we want to prove that if $p>5$ is prime then $p$ divides $f(p-1)-1$ or $f(p-1)$.

We can write $f(n) = \frac{\phi^n - \bar\phi^n}{\phi-\bar\phi}$ where $\phi = \frac{1+\sqrt5}{2}$.  Over the finite field $GF(p)$ either $\phi,\bar\phi$ are in $GF(p)$ or they are a conjugate pair in $GF(p^2)$: depending on whether 5 is a quadratic residue or non-residue of $p$.  If $\phi,\bar\phi \in GF(p)$ then $\phi^{p-1} = \bar\phi^{p-1} = 1 \in GF(p)$ and so $f(p-1) \equiv 0 \bmod p$.  If $\phi \not\in GF(p)$ then $\phi^p = \bar\phi$ and $\bar\phi^p = \phi$ since $x \mapsto x^p$ is the non-trivial automorphism $x \mapsto \bar x$ of $GF(p^2)$.  Hence $\phi^{p+1} = \bar\phi^{p+1} = \phi\bar\phi = -1$.  So $f(p) \equiv -1 \bmod p$ and $f(p+1) \equiv 0 \bmod p$ giving $f(p-1) \equiv 1 \bmod p$.  The result follows.


which function do we call as bijective?  :?

A function $f:A\rightarrow B$ that is injective ($f(a)=f(b)$ implies $a=b$) and surjective (for every $b\in B$, there exists $a\in A$ such that $f(a)=b$ is called a bijection.

Bijections are especially useful because for every bijection $f:A\rightarrow B$ there exists an inverse $f^{-1}:B\rightarrow A$ such that $f(f^{-1}(a))=a$ and $f^{-1}(f(a))=a$


Εστω παραλληλόγραμμο $ ABCD$, με εμβαδόν $ S$. Να αποδείξετε ότι για οποιοδήποτε $ P$ του επιπέδου, ισχύει:
$ AP\cdot CP \plus{} BP\cdot DP\geq S$

Θα χρησιμοποιήσω μιγαδικούς στη λύση μου . Έστω ότι τα σημεία $ A,B,C,D,P$ έχουν εικόνες $ a,b,c,p$ στο μιγαδικό επίπεδο. Τότε $ a\plus{}c\equal{}b\plus{}d$  (1).
Έτσι έχουμε :

$ K\equal{}AP\cdot CP\plus{}BP\cdot PD\equal{}\|a\minus{}p\|\|c\minus{}p\|\plus{}\|b\minus{}p\|\|d\minus{}p\|\equal{}\\
\|ac\minus{}p(a\plus{}c)\plus{}p^2\|\plus{}\|bd\minus{}p(b\plus{}d)\plus{}p^2\|$

και με τη βοήθεια της (1)

$ K\equal{}\|ac\minus{}p(a\plus{}c)\plus{}p^2\|\plus{}\|\minus{}bd\plus{}p(a\plus{}c)\minus{}p^2\|\geq \|ac\minus{}bd\|\equal{} \|ac\minus{}b(a\plus{}c\minus{}b)\|\equal{}\|(b\minus{}c)(b\minus{}a)\|\geq \sin{\angle{ABC}}\|(b\minus{}c)(b\minus{}a)\|\equal{}S$

Αυτό με το ημίτονο στο τέλος δεν το κατάλαβα... πώς προκύπτει η τελευταία ισότητα;

Η τελευταία παραάσταση ισούται με $ BA\cdot BC\cdot\sin{\angle{ABC}}\equal{}2(ABC)\equal{}(ABCD)$    :)

[quote="bilot4"]Ωραία λύση Σιλ :wink: αλλά περιμένω τώρα και καθαρά συνθετική λύση χωρίς πράξεις  :)[/quote]

Θεωρούμε σημείο $ T$ τέτοιο ώστε $ \overrightarrow{PT}\equal{}\overrightarrow{AD}$. Τότε τα τετράπλευρα $ PTDA$ και $ PTCB$ είναι παραλληλόγραμμα, επομένως $ PT\equal{}AD\equal{}BC$, $ TC\equal{}PB$, $ TD\equal{}AP$. Εφαρμόζω την ανισότητα του Πτολεμαίου στο τετράπλευρο $ PCTD$ και λαμβάνουμε:

$ PC\cdot TD\plus{}PD\cdot CT\geq PT\cdot CD$ και αντικαθιστώντας σύμφωνα με τις ισότητες που προέκυψαν από τα παραλληλόγραμμα έχουμε:
$ PC\cdot PA\plus{}PD\cdot PB\geq AD\cdot CD$ αλλά $ AD\cdot CD\geq AD\cdot CD\cdot sin(\theta)\equal{}S \Longrightarrow q.e.d.$

Μιας και εχουμε ποικιλια λυσεων, βαζω και εγω μια η οποια δουλευει μονο για $ P$ μη εσωτερικο του παραλληλογραμμου (για εσωτερικο δεν καταφερα να βρω λυση που να διαφερει απο τις προηγουμενες):

Αρχικα αποδεικνειουμε το Λημμα:
Αν $ S$ ειναι σημειο της πλευρας $ AD$ παραλληλογραμμου $ ABCD$, τοτε $ (ABCD) \equal{} 2(BSC)$
Αποδειξη:
Ειναι:
$ (ABCD) \minus{} (BCS) \equal{} (SAB) \plus{} (SDC) \equal{} \frac {1}{2}SA*AB\sin{A} \plus{} \frac {1}{2}SD*DC\sin{D} \equal{} \frac {1}{2}SA*AB\sin{A} \plus{} \frac {1}{2}SD*AB\sin{A} \equal{} \frac {1}{2}AD*AB\sin{A} \equal{} (DAB) \equal{} \frac {(ABCD)}{2} \Rightarrow (ABCD) \equal{} 2(BCS)$

Στην ασκηση τωρα, εστω χωρις βλαβη της γενικοτητας οτι το $ P$ βρησκεται σε διαφορετικο ημιεπιπεδο απο το παραλληλογραμμο ως προς την ευθεια $ AD$, εστω $ P'$ το συμμετρικο του $ P$ ως προς το σημειο τομης των διαγωνιων του $ ABCD$, τοτε
$ LHS \equal{} PC*P'C \plus{} PB*P'B \geq 2(\frac {1}{2}PC*P'C\sin{PCP'} \plus{} \frac {1}{2}PB*P'B\sin{PBP'}) \equal{} 2(PBP'C) \geq S$
με βαση το παραπανω λημμα, αφου στο εσωτερικο του τετραπλευρου $ PBP'C$ υπαρχει ενα τριγωνο το οποιο εχει για κορυφες του 2 διαδοχικες κορυφες του παραλληλογραμου και ενα σημειο πανω στην απενατι πλευρα.

ΥΓ: Η περιπτωση να εχουμε $ B,D,P$ η $ A,C,P$ συνευθειακα, δεν επηρεαζει τη λυση.

Περιτευει η δικια μου ειναι διανυσματικη (αναλυση διανυσματων με σημειο αναφορας το κεντρο Ο του παρ/μου ......) μετα απ τις κομψες σας λυσεις
(επειτα αυτη η LATEX θελει αρκετο χρονο.......)


calculus geometry trigonometry function derivative inequalities algebra

This is from my calculus textbook but I'm trying to solve it without calc.

"Find the area of the largest isosceles triangle that can be inscribed inside a circle with a radius of 4."

I've rewritten this problem as: Given $ x^2 \plus{} y^2 \equal{} 16$, find the maximum value of $ x(y\plus{}4)$.

Needless to say, the triangle is an equilateral triangle with side length $ 4\sqrt{3}$.

I got a slightly different answer
[hide="solution"]
Let the triangle have one point at $ ( \minus{} 4,0)$, clearly the maximum area will occur when the height of the triangle is greater that four, this can be seen by a simple sketch.
Let $ x \equal{} 4\cos(\theta)$ and $ y \equal{} 4\sin(\theta)$, so

$ A \equal{} (4 \plus{} 4\cos(\theta)).4\sin(\theta)$

$ \dfrac{dA}{d\theta} \equal{} 16\cos(\theta) \plus{} 16\cos(2\theta) \equal{} 32\cos^2(\theta) \plus{} 16\cos(\theta) \minus{} 16 \equal{} 0$

$ 2x^2 \plus{} x \minus{} 1 \equal{} 0 \rightarrow x \equal{} \minus{} 1$ or $ \dfrac{1}{2}$

$ y \equal{} \sqrt {16 \minus{} (2)^2} \equal{} 2\sqrt{3}$

$ A \equal{} 2\sqrt{3}. \left(4 \plus{} 4\dfrac{1}{2}\right) \equal{} 12\sqrt {3}$
[/hide]

kitsune wants the solution without using calculus. :wink:

Haha, i read that too quickly, thought it meant without a calculator  :oops:
either way we got different answers, which ones correct?

This generalized form of the original question is more likely to bring out more interesting solutions:

[i]"Three points are chosen on a circle of fixed radius, and the area of the triangle with these vertices is recorded. How should the points be arranged to maximize the area?"[/i]

[hide="Solution Sketch"]Given any two points A and B, the point C that will maximize the area [ABC] is the point on the circle with the furthest perpendicular distance to line AB. This happens to be the point on the perpendicular bisector of AB, which would make ABC isosceles wrt apex C. So let's move C there. Now, fix C and make B variable. There are two cases:

- ABC is equilateral
- ABC is not equilateral. Then, we should move B to the point on the perpendicular bisector of AC to further increase the area

Now, fix B and make A variable. There are there are two cases

- ABC is equilateral
- ABC is not equilateral. Then, we should move A to the point on the perpendicular bisector of BC to further increase the area

Note that if ABC is not equilateral, then if we execute this this three-step procedure on ABC, the area will increase. Therefore, if ABC is not equilateral, it cannot have the maximum possible area.

Since there is a maximum area, it must then be achieved when ABC is equilateral. $ \square$

[i]This maximum is attainable under your restriction forcing an isosceles triangle, and thus solves your particular problem.[/i][/hide]

In fact, we can cast the problem as a multivariable calculus problem and get a clean solution in the same spirit as the above solution:
[b](1)[/b] We can define each point by its angle wrt the x axis, write the area as a function of these three angles, and set the partial derivatives of the area with respect to the three angles equal to zero.
[b](2)[/b] We can also set three variables equal to the three angles between consecutive points and use Lagrange Multipliers under the condition that the angles add to $ 2 \pi$.

TZF, that's a really nice solution!  :)

Ocha, when I use calculus, I get $ 12\sqrt{3}$ too so I'm pretty sure that's the right answer.

But I'm still wondering if there is a way to solve the inequality problem. (pretend they never said anything about a triangle and just gave you the expression to maximize)

Oh, it occured to me that $ [ABC]\equal{}32\sin A\sin B\sin C\Longrightarrow Jensen\Longrightarrow A.M.\minus{}G.M$ Done! :lol:

[quote="TZF"]Since there is a maximum area[/quote]
This does not actually follow from what you said (it's possible that the area has a least upper bound but that it's not attainable).  What we need is that the area of the triangle is a continuous function of the locations of the points and that the set of triplets of possible locations is compact (say, in $ \mathbb{R}^6$, or even better in $ (S^1)^3$.)

Edit:  Hmm.  Actually, I guess it does, but you should elaborate.  What you're essentially showing is that the equilateral triangle is a local maximum and that there aren't any others, so it's also the global maximum.

[quote="kunny"]Oh, it occured to me that $ [ABC] \equal{} 32\sin A\sin B\sin C\Longrightarrow Jensen\Longrightarrow A.M. \minus{} G.M$ Done! :lol:[/quote]
I don't see where the 32 comes from? It should be: [ABC] = [b]2[/b]R^3.sinA.sinB.sinC, Mr.kunny : wink: .Then using the inequality: sinA + sinB + sinC = < 3.sqrt(3)/2 to get the desired result.

[b]EDIT: Yes, it should be 2R^3, Mr. Kunny[/b][/b]

Isn't $ [ABC] \equal{} 2R^2\sin A\sin B\sin C$,ghjk? :wink:

[quote="TZF"]Can't we just say the circle is a closed set and so a continuous function over the set achieves a global maximum and a global minimum at some point?[/quote]
We need both closed and bounded, which implies compact, and what we really want is the Cartesian product of three circles (since there are three points) or two circles if we fix one.

How does the idea of compact help us here?

Its not that hard to show that the domain is closed and bounded...

For example, if  $ A\equal{}(\cos a, \sin a)$, etc. Then we have $ a\plus{}b\plus{}c\equal{}\pi$... but also $ 0\le a\le \pi$...

Now the set is closed (it contains the endpoints) and it is bounded since it lies in a cube (in $ a,b,c$ space).

I suppose that we have to be a bit careful about showing the set is closed but I think that it is obvious since all the inequalities involved have $ \equal{}$ in them.

Furthermore, our objective function is continuous (its easily found with shoelace or whatever you choose that is equivalent).

So blah blah a maximum exists.

[quote="Altheman"]How does the idea of compact help us here?

Its not that hard to show that the domain is closed and bounded...[/quote]
Compactness is more fundamental than closed and bounded.  The theorem being cited is the following:

[i]If $ f : X \to \mathbb{R}$ is continuous and $ X$ is a compact metric space, then $ f$ is bounded and assumes its maximum and minimum values on $ X$. [/i]

"Closed and bounded" is equivalent to "compact" for subsets of $ \mathbb{R}^n$ by the Heine-Borel theorem, but in general the two notions are not equivalent.

There are several equivalent choices of $ X$ here - they are all homeomorphic to each other - but one in which it is easiest to see both continuity and compactness is the simplest one:  the affine variety in $ \mathbb{R}^6$ consisting of the set of points $ (x_1, y_1, x_2, y_2, x_3, y_3)$ satisfying $ x_1^2 \plus{} y_1^2 \equal{} x_2^2 \plus{} y_2^2 \equal{} x_3^2 \plus{} y_3^2$.  This is a Cartesian product of circles, which are compact, hence it is compact, and the area of the resulting triangle is polynomial, hence continuous, in the coordinates.


One circle is inscribed in a square of area 16 cm$ ^2$, and another circle is circumscribed about the square. Find the ratio of the area of the inscribed circle to the area of the circumscribed circle. Express your answer as a common fraction.

The area of the inscribed circle is simple $ 4\pi$, because the side length of the square, 4, is also the circle's diameter.

The diagonal of the square, $ 4\sqrt{2}$, is the diameter of the larger circle, so that circle's area is $ 8\pi$.

Therefore, the ratio is $ 1: 2$.


Every integer k can write with the form of x^2+y^2-z^2=k x,y,z are integers where k is the integer.

Every integer $\not \equiv 2 \mod 4$ can be written as difference of two squares (all odd numbers are allready differences of consecutive squares).
Thus it's enough that $k-x^2$ is $\not \equiv 2 \mod 4$, which can clearly be archieved.

Arne i forgot to say that x,y,z are different of zero.

Indeed, the condition doesn't change much :)


algebra polynomial calculus derivative trigonometry conics parabola

find a plynomial $ ax^2\plus{}bx\plus{}c$ that is $ \equal{}cos(x)$ for small values of $ x$

$ 1 \minus{} x^2/2$ by Taylor series, which is calculus. The best way of doing this problem is to take tangent lines, and then tangent lines of tangent lines (tangent parabolas). This is also known as taking derivatives, something calculus makes easy. Compute the leading coefficient of each resulting equation and plug that in to the appropriate coefficient.

give a proof of $ 1\minus{}x^2/2$ without using taylor series or derivates

This is rather difficult to do without calculus.

The below isn't totally rigorous, but it is a good demonstration:

[hide]Draw a picture with the right triangle in the unit circle - you know that for $ \theta$ near $ 0$, the tangent line to the circle perpendicular to the hypotenuse is approximately vertical, so the length along the arc is very close to the length opposite side of the triangle. Thus, for small angles, we can use the approximation $ \sin x \approx x$

From there, $ \cos x \equal{} \sqrt {1 \minus{} \sin ^2 x} \approx \sqrt {1 \minus{} x^2}$

Since $ x$ is really small, $ x^2 > > x^4$, so we can say that $ 1 \minus{} x^2 \approx 1 \minus{} x^2 \plus{} \frac {x^4}{4} \equal{} \left( 1 \minus{} \frac {x^2}{2} \right) ^2 \implies \cos x \approx 1 \minus{} \frac {x^2}{2}$ as desired.[/hide]

Then there's the easy way out - the Taylor series for $ \cos x$ is one of the [url=http://en.wikipedia.org/wiki/Cosine#Series_definitions]definitions[/url] of the function.

i solved in this way
from $ sinx\approx\ x$ for really small values of $ x$ we have $ cosx\approx\sqrt(1 \minus{} x^2)$
we know that $ cosx\approx\ ax^2 \plus{} bx \plus{} c$
but $ c \equal{} 1$ because the parabola meets $ x \equal{} 0$ in $ (0;1)$
$ b \equal{} 0$ because $ cosx$ is an even function
so $ ax^2 \plus{} 1\approx\sqrt(1 \minus{} x^2)$ from this we have that $ a^2x^4 \plus{} 2ax^2 \plus{} x^2 \equal{} 0$
we can cut $ a^2x^4$ and we have $ x^2(2a \plus{} 1) \equal{} 0$ that is $ a \equal{} \minus{} 1/2$

The question isn't entirely meaningful (by which I mean a rigorous answer cannot really be given) without calculus, but for the purposes of this exercise I will assume that we can assume the approximation $ \sin x \approx x$ is valid for small values of $ x$.  Then the solution is fast:

[hide="One line"] $ \cos x \equal{} 1 \minus{} 2 \sin^2 \frac {x}{2} \approx 1 \minus{} \frac {x^2}{2}$ [/hide]


[quote="mathnerd314"]A plane departs from Mathnerd314 Airport on time and flies at 540 mph to Mathcounts City, arriving 8 minutes late. Another plane departs from Mathnerd314 Airport on time and flies at 480 mph to Mathcounts City, arriving 53 minutes late. How far is Mathnerd314 Airport from Mathcounts City?

Please show your work.[/quote]
[hide]
$r=\frac{d}{t}$
$540=\frac{d}{t+8}, 480=\frac{d}{t+53}$
$540t+4320=d, 480t+25400=d$
$60t=21080$
$t=\frac{1054}{3}$
$\boxed{d=194040}$ miles [/hide]

[quote="biffanddoc"]Is that assuming that the planes are allwoed the same amount of time to fly?[/quote]
Yeah, otherwise you have a 2 variable 1 equation system with infinite sols.

[hide]This is less of an equation method, more "reasoning."

The ratio of the faster speed to the slower one is 9/8, this 1/8 makes a difference of 45 minutes.

Thus, the whole trip, 1, is 45*8=360 minutes, or 6 hours at the fast speed.

Thus, the total distance would be 6*540=3240.[/hide]


factorial
advanced fields
advanced fields unsolved

I am trying to prove $ (a \plus{} b)_n \equal{} \sum_{{j \equal{} 0}}^n {n \choose j} (a)_{n \minus{} j}(b)_{j}$, where $ (a \plus{} b)_n$ is the Pochhammer symbol denotes the falling factorial, $ (a \plus{} b)_n \equal{} (a \plus{} b)(a \plus{} b \minus{} 1)(a \plus{} b \minus{} 2)...(a \plus{} b \minus{} n \plus{} 1) \equal{} \sum_{{k \equal{} 0}}^n{s(n,k)}x^k$, where $ s(n,k)$ is the stirling number of the first kind.

I was told that:
$ (a \plus{} b)_{(n)} \equal{} \sum_{k \equal{} 0}^n s(n,k)(a \plus{} b)^k \\
\equal{} \sum_{k \equal{} 0}^n \sum_{t \equal{} 0}^k s(n,k){k \choose t} a^{k \minus{} t} b^t$
Now let s+t=k, then we have:
$ \equal{} \sum_{s \plus{} t\leq n} s(n, s \plus{} t){s \plus{} t \choose s}a^s b^t$

Now the coefficients are given by $ s(n,s \plus{} t){s \plus{} t \choose s}$, how can we further manipulate it in order to obtain the result $ \sum_{{j \equal{} 0}}^n {n \choose j} (a)_{n \minus{} j}(b)_{j}$

Can someone following this procedure to obtain the desire result, thank you very much.  :P

wrong section and double post (http://www.mathlinks.ro/Forum/viewtopic.php?t=180901, there is already an answer?!).


If three points are chosen on the circumference of a circle, what is the probability that the triangle determined by the three points contains the center?

Nope... What if you pick two points at, say, 12 and 1 o'clock of a clock, draw a diameter from 12 to 6, then pick a point at 11? It would be on the other side, but clearly, that triangle would not enclose the center.

[color=#666666]Hi..........[/color]

[quote="malcolm"]0.5 I think. Pick two points, and draw a diameter through one of them. Then if the third point is on the same side of the diameter as the second, the triangle is obtuse and does not contain the center. If it is on the other side of the diameter, the triangle is acute and does contain the center. Since the diameter encloses half the circle, the probability is 0.5.[/quote]
your idea is 50% true but you must draw two diameter through tow point that you choose and then the third point is on the one of 4 slice of circle that make with two diameter. if we choose two point and approximate and far one of them to another we can understand median of probability is 0.25.
you can obtain it. it's very easy. :roll:

[color=#666666]Bye.........[/color]

[geogebra]8e11f389f8ad8eb9a2240e2873bfc3b273af783c[/geogebra]
Clearly if the third point isn't in the sector labeled BOO, then the triangle does not contain the center.
Now imagine B is fixed, and run C through the circle counter-clockwise. The probability that the third point is in BOO increases from 0 to 1/2 linearly and then falls linearly back to 0. The 1/2 case has a negligible probability of happening, so we can say with mathematical rigor that that case is irrelevant. Now we want the 'average' of all the probabilities, so we sum all of them and divide by the number there are. Note that when we have one probability less than 1/2 (call it $ p$) we also have it's compliment (call it $ \frac{1}{2}\minus{}p$), and they both sum to 1/2. So we have a bunch of 1/2's. Let's say there are $ x$ probabilities. So the average is $ \frac{\frac{1}{2}*\frac{x}{2}}{x}$, because each probability is paired with another to get 1/2. So we have a 1/4 probability.

I expect this to not make much sense. :(

It makes perfect sense -- essentially, what you're trying to say is that if we fix one point B and let C be ccw of B (by between 0 and 180 degrees), then the third point, D, must be on the arc B'C', where B' and C' are the points on the circle opposite B and C, respectively. The average value of arc BC is 90, so the average value of arc B'C' is also 90 degrees, or 1/4 of the circle.


ANYONE PLAYED MONKEY ISLANd I,II,III or IV?

NEVER HEArD OF IT
WHY ARE WE TYPING IN ALL CAPS?

NEVER HEArD OF IT!!!!!NEVER!!!YOU'RE THE ONE WHO LIIVES IN AMERICA,THE CRAZY COUNTRY OF VIDEO GAMES AND YOU NEVER HEARD OF IT?ONE OF THE MOST AWESOME STRATEGY/COMEDY/DETECTIVE/AWESOME  GAME!!!BETTER TRY IT OUT!!!MONKEY ISLAND 4 RUUUULLEEZZZZ

It most definitely does not!
Havinh said that, I must say I have heard of its name, and once even plaed some version. Very silly. Very stupid.

I agree with bubka, monkey island is not a good game

WOAH.You guys talking abnout THE mionkey island with guybrushthreepwood?If yes,then you'd better play the 4th part,its kick nutt!!!!
Tell me if you clear even one part of it,so amazingly logical,tough+funny oh man...Don't come crying to me if you don't get a thing :D

On a serious not,play the 4th part and its awesome....

Yes, Monkey Island is excellent. The Secret and LeChuck's Revenge are true classics (the more recent ones aren't so great though). I'm amazed no one else likes it (if you asked this on any Informatics/Programming board you'd get a flood of agreement).

I've played Monkey I, II, III and IV. I must say that this is the only game that I have enjoy in last 3 years.
So now I'm waitting for part V.

I've played I, II, III, and part of IV.  The first three were good, the fourth less so.  I personally think they shouldn't have released IV on the PS2 as well as the PC, as that necessitated a wrecking the control system.

Yeah,finally some people with good taste.I liked the part in mionkey island 4,when you click on their bedroom door,he says,"This is mine and someones bedroom,ohhhh yeahhhh" :rotfl:


hey am i the only aopser psyched about spore or is there more ( i heard the second half is strategy)

spore is an awesome game the first level is like pacman kinda but u can skip it or beat it in 30 min im on the second level where u pretty much go around and kill or make friends with all the other creatures then next level your a tribe (team work) then civilization (more team work and technology) then space (go do stuff in a spaceship) the gameplay is like a (8-9)\10 but the thing that really makes the game speacial is that you design EVERYTHING your character, your homes, factories, tanks, planes, spaceship, planets you take over ...........  for crying out loud you can customize the customized parts :!:

Hrmm... I can do research!
[url=http://www.apple.com/games/articles/2008/09/spore/]Here![/url]
[url=http://en.wikipedia.org/wiki/Spore_(2008_video_game)]And Here![/url]
[url=http://www.spore.com/]And Also Here![/url]
[url=http://www.youtube.com/results?search_query=Spore&search_type=&aq=f]And Here As Well![/url]

My brother plays spore, it's kinda weird, but all his friends are obsessed w/ it, so he has to play it :D

:lol:

I'd [i]like[/i] Spore, but I'd also like to find Worms3D as well. If I buy Spore [b]and[/b] find Worms3D, then it will be a war of the universes all in one computer, and I don't think that my computer can take it.

I was semi-excited about spore, but with all the hype, I was rather expecting uber ultra leet awesome, not something really good. Right now I'm not really interested in getting spore anymore (thinking about chrono trigger when it comes out for DS and okami for wii).

Spore isn't great....

And it's so expensive...

50 bucks for my imac...no way

go to torrentz ^^ and u can get like anything free its amazing :D
one of my friends told me about it if u want a game or a movie u can get it from torrentz :)

Again, torrents are illegal.  Especially if you're getting copyrighted software for free...

And clicking on that link yields a whole bunch of xxx tags...

why would torrentz be illegal ? other people are giving u there propertiy at their free will?

It's not their right to give you their property.

well ive nvr heard of toorentz.. but i do kno that spore rocks o.o i have it...... it rocks kinda o.o i guess.... o.o i mean im only 12... in 7th grade.... of course it rocks to me P:

Spore: a brief synopsis

Spore is divided up into five parts: cell, creature, tribal, civilization, and space.

In cell stage you are a cell, you eat stuff to get DNA points. Eventually you can "spend" these DNA points to get new parts to facilitate getting more DNA points. There are cells both smaller, equal in size, larger in size, and "epic" cells - these cells are fairly hard to beat.

The creature stage is similar, except you can befriend certain groups to obtain DNA points as well.

The tribal stage involves 6 tribes. You now control multiple of the same species. In other words, you can't change the species. Permanently. Your currency is meat. You have this tribal totem, and when you ally or destroy a tribe you get another block on the totem. Complete the totem to go on.

The civ stage: there are 12 city-regions. You try to control them using ships, land vehicles, and airplanes. The vehicles are either religious, economic, or military. Religious vehicles try to convert cities: happier cities are more resistant, revolting cities will go down quickly. Economic vehicles try to buy out cities; cities cost a few thousand "Sporebucks". Oh, I forgot to mention: you no longer use meat (that's barbaric!), you use sporebucks.

The space stage is a stage where you try to conquer the galaxy. It's boring after you get about 125 cities and such. There's this thing called the Grox which inhabit the vicinity of the galactic core. There is no win situation overall.

Overall this game is only interesting in certain areas (to me it's the cell and civ) and lacks a good part of strategy.

Just noticed this on the boards so I thought I'd chime in:

I got spore the day it came out (or rather, my dad got it for me, because he thought I'd like it.) I would not really call it a strategy game, though.

Overall, Spore was okay. The main gripe I have with it is that it has ZERO replay value, in my opinion. After creature stage its really the same and designing new buildings etc. just becomes a pain. Also, the space stage becomes extremely tedious after awhile because it is too large, and it takes too long to accomplish anything without just invading everyone. (With that said, I had a great time planet-busting, antimatter bombing, and generally blowing stuff up.) The tribal stage is weak (WAY to short, I enjoyed it as long as it lasted though  :| ).

With that said though, creature stage is great (but too short). Seriously, I don't know what it is about creature stage but I enjoyed that alot. Especially hunting epic creatures... good times.

unfortunately all this fun completely wears off after first playthrough.

Its a good game, but not great. The overhyping contributed alot to the letdown factor. I would try borrowing it or torrenting it, as imo its not worth the 50 dollars (to me, a game is worth buying iff it has replay value.)


The diagonals of a rhombus measure 10 meters and 24 meters and are perpendicular bisectors of each other.  What is the perimeter of the rhombus?

divide the rhombus into 4 triangles.

its 5-12-13 right triangle.

so the hypotenuse, which is one side of the rhombus, becomes 13.

13*4=52

answer : 52


function
analytic geometry
algebra
polynomial
topology
advanced fields
advanced fields solved

Show that a function regular on $\mathbb{P}^n \setminus (1: 0: \ldots: 0)$, $n \geq 2$ is constant.

I work with homogeneous coordinates $[x_0: x_1: \ldots: x_n]$.
A function on $\mathbb P^n\setminus [1: 0: \ldots: 0]$ is regular if and only if it is regular on the affine open set $U_i=\mathbb P^n\setminus \{x_i=0\}$ for all $i=\overline{1,n}$. Since the regular functions on $U_i$ are $K[\frac{X_0}{X_i},\ldots,\frac{X_n}{X_i}]$, the set of regular functions on $\mathbb P^n\setminus [1: 0: \ldots: 0]$ is $R=\cap_{i=1}^nK[\frac{X_0}{X_i},\ldots,\frac{X_n}{X_i}]$.
Let $f\in R$. Then for all $i=\overline{1,n}$ there exist $d_i\in\mathbb N$ and a homogeneous polynomial $P_i\in K[X_0,\ldots,X_n]$ of degree $d_i$ such that $f=\frac{P_i}{X_i^{d_i}}$ on $U_i$. On $U_i\cap U_j$, $\frac{P_i}{X_i^{d_i}}=\frac{P_j}{X_j^{d_j}}\Rightarrow P_iX_j^{d_j}=P_jX_i^{d_i}\Rightarrow P_i=X_i^{d_i}Q_i$ for some homogeneous polynomial $Q_i$. Since $P_i$ is of degree $d_i$, $Q_i$ is of degree 0, hence $Q_i$ is constant. Then $f=Q_i$ on $U_i$ for all $i=\overline{1,n}$. These constants are equal on the overlaps and $U_i$ cover $\mathbb P^n\setminus [1: 0: \ldots: 0]$, so $f$ is constant.

Remark: The problem works if you remove a linear subspace of codimension greater or equal to 2.


If: $ x\plus{}y\plus{}z\equal{}9,y\minus{}f\equal{}2,f\minus{}z\plus{}d\equal{}0$ so what does $ x\plus{}2y\plus{}d\equal{}?$ equals to?

Summing all those equation you won x+2y+d=11

Edit: OH YES  :D  How I didn't see it thanks  :lol:
Solution just for general benefit:
$ x\equal{}9\minus{}y\minus{}z$
$ y\equal{}2\plus{}f$
$ d\equal{}\minus{}f\plus{}z$
So: $ x\plus{}2y\plus{}d\equal{}(9\minus{}y\minus{}z)\plus{}2(2\plus{}f)\plus{}(\minus{}f\plus{}z)\equal{}f\minus{}y\plus{}13$
Then we know that $ f\equal{}y\minus{}2$
So $ (y\minus{}2)\minus{}y\plus{}13$
$ y\minus{}2\minus{}y\plus{}13\equal{}11$

Firstly, express from the second equation, f=y-2.

Substitute the value of 'f' in the third equation to get as follows: y-2-z+d->[b]1[/b]

Add the first equation and [b] 1 [/b] to get: x+2y+d=11.

[hide="A General Approach"]
Usually in problems like this, if there doesn't exist a simple linear combination that will yield the answer, there is a non-trial-and-error method to going about it. Suppose you multiply the first equation by $ k_1$, the second by $ k_2$, and the third by $ k_3$. We obtain

$ k_1x \plus{} k_1y \plus{} k_1z \equal{} 9k_1\\
k_2y \minus{} k_2f \equal{} 2k_2\\
k_3f \minus{} k_3z \plus{} k_3d \equal{} 0$

Summing up all the equations, we obtain:

$ (k_1)x \plus{} (k_1 \plus{} k_2)y \plus{} (k_1 \minus{} k_3)z \plus{} (k_3 \minus{} k_2)f \plus{} (k_3)d \equal{} 9k_1 \plus{} 2k_2.$

Our desired expression requires the following conditions:

$ k_1 \equal{} 1\\
k_1 \plus{} k_2 \equal{} 2\\
k_1 \minus{} k_3 \equal{} 0\\
k_3 \minus{} k_2 \equal{} 0\\
k_3 \equal{} 1$

Solving this simple system, we obtain $ (k_1,k_2,k_3) \equal{} (1,1,1)$. And it follows that the answer is $ 9(1) \plus{} 2(1) \equal{} \boxed{11}$.

This is clearly not the best method for this problem, but this is just a good approach to try if you can't find the right constants to multiply to get the desired expression.
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Let $A,B,$ and $C$ are the angles of a triangle. If \[\cos A \cos B+\sin A \sin B \sin C = 1,\] show that either the triangle is right-angled or isoceles.

--Edited by 4everwise to be more aesthetically pleasing--

Are you sure this is correct?

I'm inclined to think that this only works for an isosceles right triangle.

If we had a strictly isosceles triangle, (say $A=B$), then $\cos^{2}A+\sin^{2}A \sin C=1.$ We know that $\cos^{2}A+\sin^{2}A=1,$ so our statement is true if and only if $\sin C=1,$ or $c=90.$

If we simply had a right triangle (say $C=90$), then \begin{eqnarray*}\cos A \cos B+\sin A \sin B \sin C&=& \cos A \cos B+\sin A \sin B \\ &=& \cos A \sin A+\sin A \cos A\\ &=& \sin 2A. \end{eqnarray*} Of course, this is equal to one when $2A=90$ and the triangle is isosceles.

Equation array is awesome. :D

[hide="spoiler"]
since $\sin C\leq 1$ we can write the expression as

$\cos A\cos B+\sin A\sin B\geq 1$ or equivalent to

$\cos (A-B)\geq 1$

and guess the next steps  :)
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