inequalities
function
inequalities solved

Given a convex function $f: \left(a,b\right)\to \mathbb{R}$, where $a$ and $b$ are two real numbers. Prove that
$f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)+f\left(\frac{x_{1}+x_{2}+x_{3}}{3}\right)$
$\geq \frac{4}{3}\left(f\left(\frac{x_{1}+x_{2}}{2}\right)+f\left(\frac{x_{2}+x_{3}}{2}\right)+f\left(\frac{x_{3}+x_{1}}{2}\right)\right)$
for any three reals $x_{1},x_{2},x_{3}\in \left(a,b\right)$.

After Popoviciu's inequality (*), we have

$ f\left(x_1\right) \plus{} f\left(x_2\right) \plus{} f\left(x_3\right) \plus{} 3f\left(\frac {x_1 \plus{} x_2 \plus{} x_3}{3}\right)$
$ \geq 2f\left(\frac {x_2 \plus{} x_3}{2}\right) \plus{} 2f\left(\frac {x_3 \plus{} x_1}{2}\right) \plus{} 2f\left(\frac {x_1 \plus{} x_2}{2}\right)$.

Moreover, by Jensen's inequality,

$ f\left(x_2\right) \plus{} f\left(x_3\right)\geq 2f\left(\frac {x_2 \plus{} x_3}{2}\right)$;
$ f\left(x_3\right) \plus{} f\left(x_1\right)\geq 2f\left(\frac {x_3 \plus{} x_1}{2}\right)$;
$ f\left(x_1\right) \plus{} f\left(x_2\right)\geq 2f\left(\frac {x_1 \plus{} x_2}{2}\right)$.

Adding these four inequalities together, we get

$ 3f\left(x_1\right) \plus{} 3f\left(x_2\right) \plus{} 3f\left(x_3\right) \plus{} 3f\left(\frac {x_1 \plus{} x_2 \plus{} x_3}{3}\right)$
$ \geq 4f\left(\frac {x_2 \plus{} x_3}{2}\right) \plus{} 4f\left(\frac {x_3 \plus{} x_1}{2}\right) \plus{} 4f\left(\frac {x_1 \plus{} x_2}{2}\right)$.

Dividing this by 3, we get

$ f\left(x_1\right) \plus{} f\left(x_2\right) \plus{} f\left(x_3\right) \plus{} f\left(\frac {x_1 \plus{} x_2 \plus{} x_3}{3}\right)$
$ \geq \frac43\left(f\left(\frac {x_2 \plus{} x_3}{2}\right) \plus{} f\left(\frac {x_3 \plus{} x_1}{2}\right) \plus{} f\left(\frac {x_1 \plus{} x_2}{2}\right)\right)$.

QED.

(*) In the above, I have used the so-called Popoviciu inequality. This inequality was proven at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=430 in one way, and at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=19097 in a different way together with a strong generalization.

  darij


How do you "quote" a picture that someone else has posted? I've seen people do that, like
[b]"so and so wrote: (picture)"[/b]
Do you have to save the picture and upload it yourself? Thanks! :D

It depends on where the picture is from.

If the picture is just embedded in the post, it means that the picture was from somewhere on the internet. If you quote that type of picture, it will embed and show in the quoted part of your post.

If the picture was uploaded, there will be a box around it and a file name above it. You can't quote these attached pictures directly in your quote, but you can put it back in yourself. To do that, click on the picture (a page should come with just the picture and white space around it). Then, copy the URL (address) and paste it into your post wherever you want it to go, which can include the quote. Put [img] tags around it, like this:
[code][img]http://... (your image URL)[/img][/code]

Thank you so much! I was wondering why some worked and some didn't!


conics
parabola
analytic geometry
graphing lines
slope
parameterization
blogs

Describe the set of rational points of the equation $ y\equal{}x^2$.

Is this motivated by [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=197236]my post[/url]?  :)  Think about how to parameterize a parabola.

It is inspired by your post :) . I liked how you related the set of rational points to the set of Pythagorean triples. I figured that I could use the first method you posted as a guide to this problem. So far I have:

[hide="Solution"]
$ y\equal{}x^2$

$ x\equal{}\frac{a}{c}$ and $ y\equal{}\frac{b}{c}$ where the $ gcd(a,b,c)\equal{}1$

$ \frac{b}{c}\equal{}\frac{a^2}{c^2}$ (Should I multiply through by $ c^2$?)

I'm unsure whether this will get me anywhere. So any help will be appreciated. Also, I'm prety sure that a parabola can be parameterized by $ x\equal{}t$ and $ y\equal{}t^2$.[/hide]

[quote="JZambrano201"]I'm prety sure that a parabola can be parameterized by $ x \equal{} t$ and $ y \equal{} t^2$.[/quote]

Uh-huh.  What happens when $ t$ is rational?  Irrational?

[hide="As you can see..."] This problem is much easier than the corresponding problem for a circle because we already have a convenient parameterization.  Note that the second method in my post (fixing a point and considering lines of various slopes) still works. [/hide]

If $ t$ is irrational then the point will not be in the set. And if $ t$ is rational it will be. Correct? Do problems like these in general only require us to know the parameterization of the equation?

Also, what happens when we are given a parabola with coefficients such as $ y\equal{}3x^2\plus{}4x\plus{}1$ and are asked to describe the set of rational points? I tried it with the second method and got stuck when substituting $ tx\plus{}t$ for $ y$.

[quote="JZambrano201"]Do problems like these in general only require us to know the parameterization of the equation?[/quote]
See the discussion in the blog post.  What we are interested in is a rational parameterization, which is roughly a parameterization that gives rational points for rational $ t$.  Precisely, it is an isomorphism to the rational projective line.

[quote="JZambrano201"]Also, what happens when we are given a parabola with coefficients such as $ y \equal{} 3x^2 \plus{} 4x \plus{} 1$ and are asked to describe the set of rational points? I tried it with the second method and got stuck when substituting $ tx \plus{} t$ for $ y$.[/quote]
The set of rational points on $ y \equal{} f(x)$ is given by $ x \equal{} t, y \equal{} f(t)$ for [b]any[/b] polynomial $ f(x)$ with rational coefficients.  (Once you understand this, you will understand why I didn't consider this type of equation.)

Part of the success of the second method depends on the choice of point $ P$; pick the vertex.

Thanks, that cleared everything up. Keep up the blog, it gives me inspiration :).


Hi, I am a high school students and I am currently trying to improve my Math skills. Is there any E-books out there that relate to math? If there is, could you please tell me what it is.

Thank you very much.

http://www.kalva.demon.co.uk/aime.html

for problems with brief solutions

http://myhome.personaldb.net/ideahitme/

for some lecture notes i think

http://www.math.ust.hk/excalibur/

more math articles on like every topic

http://web.mit.edu/rwbarton/Public/mop/

for IMO level problems heh

http://web.mit.edu/~tmildorf/www/Inequalities.pdf

mildorfs lecture on inequalities

http://www.cms.math.ca/Competitions/MOCP/

stuff from Canadian MO

http://www.imo.org.yu/index.php?options=oth|other&p=rkl

more IMO problems

thats some stuff right there.... im not sure if it'll help you much depending on the level ur at

[quote="gotztahbeazn"]

more math articles on like every topic

http://web.mit.edu/rwbarton/Public/mop/

[/quote]

that one is pretty amazing...it is like mop for free, (minus the practice tests, teachers, and atmosphere...lol)


[quote="gabrielsw"]ehm, there isn't trcik for that problem? because i found that problem from high school in indonesia.[/quote]

There is a trick to find its integer part, but not the exact value..


Determine the maximum value of $p$ in this statement : If $f(x)=x^3+ax^2+bx+c$ is a third-degree polynomial which has non-negative roots, then $f(x)\geq p(x-a)^3$ for every $x\geq 0$
When is the equality occurs?

shouldn't this be:
$f(x)\geq p(x+a)^3$??


[quote="hwenterprise"]What is the sum of:

1+2+3+4+...1000?[/quote]

[hide="Solution"]
Pair up each set of numbers on either side of the expression. You get 1 and 1000. Pair up the next set. You get 2 and 999. Pair up the next set, and you get 3 and 998. Notice how all the pairs add up the 1001, and since there's 1000 numbers, there's 500 pairs. 1001 times 500 = [b]500500[/b].
[/hide]

[hide="Alternative Solution"]

$(500+501)\cdot500=\boxed{500500}$ [/hide]

Hehe, pascal_1623, you've used Gauss' method :)

1000+1=1001, then we multiply by 500 to get 500500 :idea:

[hide]you match the end with the begining, and so on, so you get 500500[/hide]


[quote]i have problems in writing constraint equations[/quote]

welcome to AoPS :)
i am not so good but i can attempt to answer your few queries about constraint equation and this help will be hopefully extended by genius minds here :D

To write constraint relatins,following methods are useful:

1. Just as given in all books,give names to lengths of different strings,etc and remember that string is inextensible,so total length is constant after adding the lengths.Then differentiate them and in cases where length decreases $ v$ becomes negative and all that.This method is discussed in detail in all books for JEE. :D

2.[b]Objective approach[/b]:Just take one block of system and check for the displacement when block is displaced for distance $ x$(if pulley is fixed from one side,it'll move $ 2x$and so on.Then superimpose all movements one by one and you'll find equation.(Its a great method if you do some practice on it,believe me,but it works well only if system is little bit less complicated meaning less forces to cause acceleration on one body)

3.[b]My favourite[/b]:Use the concept that work done by Tension in an ideal string on a system is always $ 0$.This can be proved also as mass of string is 0.So take direction of motion ($ v_A ,v_B,a_A$and $ a_B$) as per your wish.Since tension in ideal string is always same,take it as $ T$.Then take dot product $ T$ with all velocities $ (v_A,v_B,etc)$and equate them as 0(most of time dot product will be in form of $ Tv_A, or \minus{} Tv_A$ etc,but many time it may be in the form of $ Tv_Acos 30$ or $ \minus{} Tv_Acos 30$etc.Equate them as 0.Then you'll find constraint relation between velocities,differentiate them to get relations between accelerations. :D
While solving,if you get some values as negative,it means accelerations are opposite to the direction assumed.

4.[b]In case of rigid bodies[/b]:just use the fact that Work done by the normal force along perpendicular to the surfaces in contact is always 0 as bodies are rigid.Then proceed as in previous.

Hope i've helped you a little bit.I am sorry i can't explain quite properly
i am very bad in explaining
For rest wait for others to come  :D

P.S-I suppose it can be a Knowledge Base article in JOI  :rotfl:  if you all seniors can contribute something.
for even further details contact thaakisfox sir(he'ld be on AoPS as well as JOI).

thanks a lot

am waiting for seniors to come

btw,what happened to your NTSE
bad news for you

you're really genius as you at-least cleared NTSE 1st stage

where you got to know about my NTSE??????

I've somehow passed 2nd stage but flanked my interview :(
couldn't name the first Chief Minister of my state :(

and there is nothing so 'genius like' in clearing 1st round

check out your web-site :D

and there is 'genius' involved in clearing NTSE 1st round
and clearing 2nd round is even more trustworthy of applause

keep it up

i just updated my profile today and somebody visited my site :D

you are the first visitor of my website :D

then what you'll call about rituraj,da vinci,pardesi,and for that matter,everybody else??

most of students here have a lot of feathers under their belt
most of them are genius.

P.S-Check out your PM

and yes one more thing

i've taken too much pain to write that much
at-least i expect some good ratings

mine is only 2(though i think it should have been less :rotfl: )

so genuises working on constraint relations and its position effect!
very good i dunno it properly by the da i also visited ur website and its quite cool !
kk bye!
 :)

Your favourite is my fav too but its very imp to learn constraint relations.

yes,i understand it  :)

the other methods are mainly for objective examinations(they save a lot of time :D )


The sequence x_1, x_2, x_3, ... is defined by x_k = 1/(k^2 + k). A sum of consecutive terms x_m + x_m+1 + ... + x_n = 1/29. Find m and n.

I used partial fractions and got some stuff but nothing equals 1/29... anyone have any ideas?

[quote="Spoon"]The sequence x_1, x_2, x_3, ... is defined by x_k = 1/(k2 + k). A sum of consecutive terms x_m + x_m+1 + ... + x_n = 1/29. Find m and n.

I used partial fractions and got some stuff but nothing equals 1/29... anyone have any ideas?[/quote]

Wait, do you mean x_k=1/(k^2+k)? Cuz u only wrote k2+k...

1/(k^2+k)=1/k-1/(k+1). Things will telescope nicely. I don't know what the question is, but I seem to remember that the answer is [hide]840[/hide].

HOw can 840 be the asnwer?

Wait, if we expand it as 1/k-1/(k+1), then it the telescope will end up as 1/k-1/(k+n), which after we solve, get that m=1, n=28, or am I wrong?

I think i am, it's too late....

I don't think he posted the question right. It was from AIME 2002 I think.

I copy and pasted the question...

I got here:

1/m-1/(n+1) = 1/29.... no integers will make it work... the only solutions are:

m= (-1+3:sqrt:13)/2 and n=(1+:sqrt:13)/2

or hmm, that assumes m and n are equal, and i gave the n = m+1, which is wrong...

the question asks: The sum as k goes from m to n of 1/(k^2+k) = 1/29... what are m and n.

Note the construction 1/m=1/(m+1)+1/(m^2+m). Does that help?

I'm giving you a hint, and you're going to use that hint to solve it.

[color=cyan]darkquantum, you were mostly right until you said "so that means m=1 and n=29," or whatever you said that looked kind of like that.  Those values give you 28/29, which is not what you wanted.  You need to figure out how to get 1/29.

Complexzeta's answer is the correct value of m + n, by the way.[/color]


I am walking from my school to my home.It's known that on each hour of walking I meet 60 warning boards.Find the probability of no meeting any warning board in 30 minutes?

I think this is really tough with 60 warning boards...

We let the time when he meets the $ k$th warning board be $ t_k$ in the interval $ [60(k - 1),60k]$ We are required to find the probability of ${ t_{k + 1} - t_k} > 30$.

This should be straightforward with 2 warning boards, and still doable with 3 warning board (in this case use xyz-coordinates), but with 60 warning board we are extending to $ \mathbb{R}^{60}$ which I think perhaps higher level mathematics is required...  :maybe:

(btw, would someone like to try 3 warning boards?)

I think that the probability of each individual board not being in your thirty minutes is 1/2, so the chance for 60 of them not being in your thirty minutes is $ \frac {1}{2^{60}}$. So where did I mess up? :maybe:

Hmmm if we first try 2 warning board using geometry we should get a sample space of area $ 60^2 \equal{} 3600$.
now we graph $ t_2 \minus{} t_1 > 30$ with $ t_1\in(0,60),t_2\in(60,120)$.
If I didn't messed up, the probability is $ \frac {3600 \minus{} 0.5*30*30}{3600}\neq1/4$

Ah, no wonder the problem seems so complicated... I misunderstood the question (I read it as the person is meeting 60 boards with 1 board every hour, which should have been all of the 60 board in one hour)...
sorry about that...

[quote="JBL"]The random model we're using here is somewhat unclear.  Is the following question equivalent?

Choose numbers uniformly and randomly in $ [0, 1]$.  What is the probability that there is an interval of length $ \frac {1}{2}$ containing none of our chosen numbers?

Differ, you assumed that there is a [i]particular[/i] 30 minute period in which we must see no boards.

mathwizarddude, I've no idea why you're taking each board to be in a different (non-overlapping) 60-minute interval.[/quote]

I think so. Can you solve it?

Here're some threads that discuss the same problem or related problems and generalizations.  (Note that choosing $ n$ points on $ [0, 1]$ without hitting some interval of length $ \frac{1}{2}$ is the same as placing $ n \plus{} 1$ points on a circle without hitting some semicircle, or equivalently so that all points end up in some semicircle.)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5715
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247928
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=196595

I found these threads by searching for "probability semicircle" (without quotes).

For more information let's go by:
http://www.lsus.edu/sc/math/psisson/putnam/putnam-web.htm


hello.

I found this problem in this forum but i can't remember where, and now i don't know how to prove it.

Prove that for $ n>1$, the polynomial $ P(x)\equal{}x^n\plus{}5x^{n\minus{}1}\plus{}3$ is irreducible over $ \mathbb{Z}$.  Can someone help me?  thanks.   :)

thanks.   :)

but is the proof there correct?  i think i understood up to the part where all the $ b_i$ must be a multiple of $ 3$ but then i got lost at the end.   :(

can someone explain to me the solution better...?    :blush:

Let me reword this:
[quote]That is, looking for the coefficient 0 of x^{n} in p(x) and having that b_{i}=3k for all i = 0,1,...,n-1 is sufficient enough to show irreducibility if the coefficient is 0 in x^{q} [color=darkblue]and doesnt occur if q=n-1 [/color].
If that is true then consider g(x)=x+g with g not multiple of 3. so then g^n+5g^{n-1}  =g^{n-1}(g+5)=-3 for some integer g, but then g^{n-1} divides 3, so g=1 or g=+/-3,n=2 which both do not work.[/quote]

The dark blue is the only sentence which I still needs some help to clarify.
Hopefully this makes the problem easier to understand.  I don't really think induction was the right word to use, but you continue to consider that the coefficients of p(x) for the terms with degrees from 1 to n-2 are all 0.  Perhaps the 0,1,..., (_) should only be until n-2?

There's a rather complex technique which solves this rather quickly.
[hide]
We check +- 1, 3 to see that there are no integer roots.

Applying Dumas' Theorem when considering the Newton Polygon of this polynomial for 3 tells us that the only way to decompose this polynomial is into 2 polynomials, one linear and one of degree n-1, which has been shown to be impossible over $ \mathbb{Z}$.

I like irreducibility because there are so many sufficient criteria, but any that are necessary and sufficient appear to be exceedingly complex.  [/hide]

Can someone confirm this?  Would this technique require a proof during an Olympiad?

I like irreducibility problems.  Are there any reducibility criteria (besides factorization) at all?  Or necessary and sufficient conditions for irreducibility that aren't murderously complicated?
p.s.  Assuming nobody goes and looks this over by the next time I am online, I'll look at it more thoroughly, but I don't want to now because it's, well, late, and I don't want to shame myself even more than I otherwise would.

Does Dumas' theorem prove [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=150]Perron's criterion[/url]?  That's enough for this problem, and as far as analysis goes its proof isn't terribly difficult.

Come on, use Eisenstein extended to prove that P(x) has an irreducible factor of degree n-1 or greater, so it's up to prove that P(x) hasn't got linear factors, that is it hasn't got a integer root.

$ x^n\plus{}5x^{n\minus{}1}\plus{}3\equal{}0$, solve it, there are no integer roots, so the problem is done.

Eisenstein rules.


hi,
lets assume that ABC a triangle .$AB = c ,BC=a,AC=b$ and

$a+b+c=1$

prove that : $\frac{1}{2}\geq a^2+b^2+c^2+4abc$

see:
http://www.mathlinks.ro/Forum/viewtopic.php?p=126074&highlight=a%5E2%2Bb%5E2%2Bc%5E2%2B4abc#126074

anonyme, this inegality was in the last morrocan olympiads are you solved it? :roll:

this problem is very easy, this inegality was in the last morrocan [color=red]junior [/color] olympiads : $a<b+c$ so$a<\frac{1}{2}$ ...............
$(\frac{1}{2}-a)(\frac{1}{2}-b)(\frac{1}{2}-c)> 0$
say $ab+bc+ca =\frac{(a+b+c)^2 -(a^2+b^2+c^2)}{2}$

could you tchebytchev show us Morrocan Olympiad

Problems ,pleaze ?


inequalities
calculus
integration
inequalities solved

Who can solved this problems?
(n-1)(m-1)(xm+n + ym+n) + (m+n-1)(xnym + xmyn) >= mn(xm+n-1y + ym+n-1x).
I've a solution, and you?

If you denote Sk=(xk-yk)/k, the inequality comes to S1*S(m+n-1) \geq Sm*Sn.

Well done, Mihai:
   All you have to do now is to apply Cebasev integral inequality:
  integral from y to x from t^(m-1)dt* integral from y to x from t^(n-1)dt  is smaller than integral from y to x from t^(m+n-2) dt.


calculus
integration
real analysis
real analysis theorems

Do someone have any ebook, site, text, etc, about this Product Integral!?

I need to find a book about that.

on internet i just found
http://wiki.contextgarden.net/Product_integral

but this is not math at all...

and http://eom.springer.de/P/p075020.htm
and http://en.wikipedia.org/wiki/Product_integral

i'm glad if someone have someting about that...

I've found

http://math2.org/math/paper/sect1-0.htm


How many digit 1s are there in the following:

$ 9 + 99 + 999 + 9999 + 99999 + ....................... + [ (10^{2009}) - 1 ] + [ (10^{2010}) - 1 ]$

I think the answer is $ 2007$.

$ 10 + 100 + ..... + 10^{2010} - 2009$

Which is $ 1111111........109101$

There are only 3 digits that  are not 1 in the above number.  Hence, the total number of ones will be

$ 2010 - 3 \& = 2007$

Can you confirm & provide alternate ways of solving this if possible?

Thanks in advance.

$ \sum_{i \equal{} 1}^{2010} 10^i \equal{} \overline{\underbrace{11 \ldots 1}_{2010}0}$, so Your number has $ 2011\minus{}3$ ones.

Oops!  I missed out one.  Thanks very much!


A quadrilateral $ ABCD$ is inscribed in a circumference, and its diagonals intersect at $ Q$. The side $ DA$, prolonged from $ A$, and side $ CB$, prolonged from $ B$, intersect at $ P$. If $ CD \equal{} CP \equal{} DQ$, find $ \angle CAD$

Since $ CD\equal{}CP$ we have that $ \widehat{CDP}\equal{}\widehat{DCP}$, $ \widehat{CDP}$ is supplementary to $ \widehat{ABC}$, which is supplementary to $ \widehat{ABC}$, so $ \widehat{CDP}\equal{}\widehat{ABC}\equal{}\alpha$.
Since $ CD\equal{}DQ$, $ \widehat{CQD}\equal{}\widehat{QCD}$, $ \widehat{ABD}\equal{}\widehat{QCD}$ because they insist on the same arc and $ \widehat{CQD}\equal{}\widehat{AQB}\equal{}\beta$

Call $ \widehat{CAD}\equal{}\widehat{CBD}\equal{}x$, the angles $ \widehat{PAD}$ and $ \widehat{PBC}$ are equals to $ \pi$, so
\[ \widehat{PAB}\plus{}\widehat{BAQ}\plus{}x\equal{}\pi \Rightarrow (\pi \minus{} 2\alpha) \plus{} (\pi \minus{} 2\beta) \plus{} x \equal{} \pi \qquad (1)\]and
\[ \widehat{ABP}\plus{}\widehat{ABQ}\plus{}x\equal{}\pi \Rightarrow \alpha \plus{} \beta \plus{} x \equal{} \pi \qquad (2)\]

From $ (1)$ we obtain that $ \alpha \plus{} \beta \equal{} \frac{1}{2} (\pi \plus{} x)$, so substituting into $ (2)$ we have that
\[ \pi \plus{} x \equal{} 2\pi \minus{} 2x \Rightarrow x\equal{}\frac{\pi}{3} \equal{} 60^\circ\]


[quote]Problema 4
Hallar todas las parejas de enteros positivos $a$, $b$ tales que $2a-1$ y $2b+1$ son primos relativos y $a+b$ divide a $4ab+1$.[/quote]

Si no es conveniente que ponga mis intentos de solución o si debería ocultarlos para que se habrieran al hacer click, avisadme (avísenme), por favor.  :blush:

Bueno, este parece (nunca se sabe) más relajado.
Primero consideremos el caso en que $b=a+1$, lo que da $mcd (2a-1,\; 2b+1)=mcd (2a-1n\;2a+3)=mcd (2a-1;4)=1$, y $a+b=2a+1\;|\;4a(a+1)+1=(2a+1)^{2}$.
Por otro lado
$(a+b)|4ab+1 \Rightarrow [(a+b) | (4ab+a+b+1)]\wedge [(a+b)| (4ab-a-b+1)] \Leftrightarrow$
$\Leftrightarrow [(a+b) | (2a+1)(2b+1)]\wedge [ (a+b)| (2a-1)(2b-1) ]$.
Como $mcd (2a-1,\;2b+1)=mcd (2a-1;\;2a+1)=1$, de lo anterior se deduce $(a+b)|(2b-1)$. Por un lado $a+b=2b-1 \Rightarrow a+1=b$, caso que ya ha sido tratado. Por otro $\exists k>1 \;\;/\;(2b-1)=k(a+b) \Rightarrow 2(a+b)\leq 2b-1 \Rightarrow 2a\leq-1$, que es imposible. Las únicas soluciones son las del tipo $b=a+1$.

Hey! Vas a resolver toda la ibero tu solo! :) ji Muy bien!

Daniel

you can make a substution:

Let $2a-1=x$, $2b+1=y$ then:

$x+y|2[x(y-2)+x+y] \implies x+y|x(2y-4)$ so we have

$x(2y-4)=k(x+y) \implies x(2y-4-k)=ky$ for $k$ a positive integer, but since $(x,y)=1$ it implies that $x|k$ and $y|4+k$ therefore

$y+x|2y-4=y+y-4$ and $x+y|2x+4=x+x+4$ but by the existence condition we have that

$x \le y-4$ and $y \le x+4 \implies y-4 \le x \le y-4 \implies x=y-4$

aqui va otra, q creo no es la misma q las pasadas:

$a+b|4ab+1 \Leftrightarrow a+b|(a+b)^{2}-4ab-1=(a-b+1)(a-b-1)$

luego, si $(a+b,a-b-1)=d>1$ entonces $d|2a-1, 2b+1$, lo cual es una contradicción.

entonces $a+b|a-b+1$. Es facil verificar que $a+b>|a-b+1|$, lo que implica que entonces $a-b+1=0$, es decir $(a,b)=(k,k+1)$.

es facil verificar que este par ordenado cumple las condiciones del enunciado.

pd: en realidad el par ordenado es $(k,k-1)$, porque en el examen era tal que $(2a+1, 2b-1)=1$ y no $(2a-1,2b+1)=1$.


What are they? Is it some kind of code to draw images...? I'm sorry if I didn't look enough but I never knew this was a code to make images (I thought it was by Geometer's Sketchpad?)

Like

[code][asy]
size(150);
pair A=(0,0), C=(-6,0), B=(-25/6,5*11^.5/6);
draw(Circle(A,5));
draw(A--B--C--cycle);
label("\(6\)",(A+C)/2,S);
label("\(5\)",(A+B)/2,NE);
picture p = new picture;
draw(p,Circle(C,0.4));
clip(p,A--C--B--cycle);
add(p);
label("\(\theta\)",C+(.3,0),NE);
label("\(A\)",A,SE);
label("\(B\)",B,NW);
label("\(C\)",C,SW);
[/asy][/code]

by azjps on one of his recent threads produced image:

[asy]size(150);
pair A=(0,0), C=(-6,0), B=(-25/6,5*11^.5/6);
draw(Circle(A,5));
draw(A--B--C--cycle);
label("\(6\)",(A+C)/2,S);
label("\(5\)",(A+B)/2,NE);
picture p = new picture;
draw(p,Circle(C,0.4));
clip(p,A--C--B--cycle);
add(p);
label("\(\theta\)",C+(.3,0),NE);
label("\(A\)",A,SE);
label("\(B\)",B,NW);
label("\(C\)",C,SW);[/asy]

How?

More specifically, the Vector Graphics Language.
:D

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1150828185&t=148120]Thread where Asymptote implementation on AoPS forums is discussed[/url]. It is basically what you said; code which can output mathematical diagrams (avoids having to upload images/attachments etc). The wiki tutorial is very useful.

Thanks to you all.

I'll definitely have to learn this. I think this image is far better looking than KSEG.


[quote="xp"]Για $ a,b,c$ στo $ R$  ν.δ.ο.

$ \sqrt [4]{a^4 \plus{} b^4 \plus{} c^4 \plus{} 1}\geq \sqrt [5]{a^5 \plus{} b^5 \plus{} c^5 \plus{} 1}$[/quote]

Έχουμε ότι $ \sqrt [5]{\sum a^{5} \plus{} 1}\leq \sqrt[5]{\sum \left|a\right|^{5}\plus{}1}$ άρα αρκεί να αποδείξουμε την ανισότητα για $ a,b,c\geq 0$.

$ \left( \sum a^{4} \plus{} 1\right)^{\frac {5}{4}}\geq \sum a^{5} \plus{} 1\Longleftrightarrow \sum \frac {a^{5}}{\left(\sum a^{4} \plus{} 1\right)^{\frac {5}{4}}} \plus{} \frac {1}{\left(\sum a^{4} \plus{} 1\right)^{\frac {5}{4}}}\leq 1\Longleftrightarrow \sum \left(\frac {a^{4}}{\sum a^{4} \plus{} 1}\right)^{\frac {5}{4}} \plus{} \frac {1}{\left(\sum a^{4} \plus{} 1\right)^{\frac {5}{4}}}\leq 1$

Όμως $ \left(\frac {a^{4}}{\sum a^{4} \plus{} 1}\right)^{\frac {5}{4}}\leq \frac {a^{4}}{\sum a^{4} \plus{} 1}$.

Άρα $ \sum \left(\frac {a^{4}}{\sum a^{4} \plus{} 1}\right)^{\frac {5}{4}} \plus{} \frac {1}{\left(\sum a^{4} \plus{} 1\right)^{\frac {5}{4}}}\leq \frac {\sum a^{4} \plus{} 1}{\sum a^{4} \plus{} 1} \equal{} 1$

O πυρετος πυρετος και η ανισωση ανισωση....

Ρίξτε μια ματιά και εδώ για την πιο γενική ανισότητα
 http://www.mathlinks.ro/viewtopic.php?t=204688

Sil το δεν το ειχα ξεχασει οτι την ξαναεβαλα.
Την παραλλαξα με το 1 για να σπασω την ομοιογενεια (αλλιως βγαινει στο λεπτο οπως και η δικη σου λυση).


Matt is given a jar with eight marbles, six of which are red and two of which are green. Matt then randomly draws marbles without replacement. If he selects a red marble, he puts it in the first box; similarly, he continues to place all red marbles he selects into the first box until a green marble is chosen. When a green marble is chosen, he puts the first box aside. He then selects and places red marbles in the second box until the second green marble is chosen. When the second green marble is selected, he will put the second box aside and place all remaining red marbles in a third box. After all marbles are drawn, what is the probability that each box contains two red marbles? Express your answer as a common fraction.

[hide]
Matt has to draw a red,red,green,red,red,green,red,red.
The probability of that is
$\frac 68 \cdot \frac 57 \cdot \frac 26 \cdot \frac 45 \cdot \frac 34 \cdot \frac 13 \cdot \frac 22 \cdot \frac 11$.
Simplifying fractions first, we get
$\frac 34 \cdot \frac 57 \cdot \frac 13 \cdot \frac 45 \cdot \frac 34 \cdot \frac 13 \cdot 1 \cdot 1$.
$\frac{1}{28}$[/hide]


I decided to create a mock math counts. Sign up in this thread. PM me the answers. Total possibility is 46 as usual. It is attached. Sign up here.

I don't think its [i]that[/i] hard.

more info plz but ya sure ill do it. There like a deadline or anything?

Thanks I'll do it if I remember. Which I will.

I'll do it. And I hope you don't just let this competition die like you let Competition de Battle, or whatever it was called.

I'll try to compete.

Put me as a "maybe".

I'll PM you with the answers, if I do it.

EDIT: I did it.

I'll sign up but whats the difficulty? chapters? states? nationals?

I don't get # 4 target. It's like infinite. You can't get any odd number.

[quote="kheg_k"]do we use the standard 40 min., 24 min.[/quote]

I would assume as it is supposed to be like a practice real thing.

I may or may not do it, depending on time.

Does it not say there that it is 40 minutes and 6 minutes for 4 pairs :blush:

I hope it is counted as original.

I might make a team round depending on who does it

From now on, you have to create your own team.

Do I just send you the answers on this regular forum, is that the same as PM-ing?

Did you get my PM, with my answers for the mock MATHCOUNTS test?

can i sign up?
i want to join.
ill send in my answers

Yes. InsanelyBored can join.

Please stall all PM's for now.

Did you receive my PM, or did I send it to myself?

I have a question:
Since #4 is free, we get 6 min to do #3?

Okay. I really won't have itme to grade these. I'll grade the pm's I got, but I need a grader? Anyone is fine for grading.

i wanna sign up
 but what lvl is this btw (im assuming states but is that right?)

please explain how to round a scientific notation to the nearest tenth) (mathcounts rules clearly state all answers must be rounded  at the FINAL step
clearly, the final step is denotating the value in the scientific notation


number theory
least common multiple
greatest common divisor
number theory unsolved

Show that [tex]\left(\begin{array}{c}
  2 n\\
  n
\end{array}\right) | \mbox{lcm} [ 1, 2, \ldots, 2 n ][/tex] for all positive integers [tex]n[/tex].

This is problem #136 here :

http://www.animath.fr/cours/arithm.pdf

Pierre.

just count the power of primes on each side.... :)

It's all right I can read french  :P
And that's true the problem's not too hard  :oops:

Sorry for my stupidness,what does[color=red] lcm [/color]mean?

lcm = least common multiple.
another (maybe more familiar) notation is [tex][1,2,...,n] = \mbox{lcm} [1,2,...,n][/tex] just as [tex](m,n) = \mbox{gcd}(m,n)[/tex]


Compute the largest factor of $ 6^{7}\plus{}37$.

$ Source: MANDELBROT!$

EDIT: ah yes, my mistake.  Largest [i]prime[/i] factor, I meant.

I think dgreenb801 means that $ 6^7\plus{}37$ IS the largest factor, and the way the problem is stated would actually lead to this answer.  However, I'm pretty sure that the original intent of the problem was to ask for the largets proper factor of $ 6^7\plus{}37$.

[hide="Solution"]
$ 6^7 \plus{} 37 \equal{} 6^7 \plus{} 6^2 \plus{} 1$. If $ \omega^3 \equal{} 1$ and $ \omega \neq 1$, since $ \omega^7 \plus{} \omega^2 \plus{} 1 \equal{} 0$, we get that $ x^2 \plus{} x \plus{} 1$ divides $ x^7 \plus{} x^2 \plus{} 1$, so 43 divides 279973. We now check for primality of 279973 / 43 = 6511; we see that it is equal to 17 * 383. It is easy to verify that 383 is prime, so we are done (and the answer is $ \boxed{383}$).
[/hide]

I think he meant prime factor.

[quote="mathwizarddude"][quote="Zhero"][hide="Solution"]
$ 6^7 \plus{} 37 \equal{} 6^7 \plus{} 6^2 \plus{} 1$. If $ \omega^3 \equal{} 1$ and $ \omega \neq 1$, since $ \omega^7 \plus{} \omega^2 \plus{} 1 \equal{} 0$, we get that $ x^2 \plus{} x \plus{} 1$ divides $ x^7 \plus{} x^2 \plus{} 1$, so 43 divides 279973. We now check for primality of 279973 / 43 = 6511; we see that it is equal to 17 * 383. It is easy to verify that 383 is prime, so we are done (and the answer is $ \boxed{383}$).
[/hide][/quote]
hmm am not very familiar with this technique of roots of unity... could you please prove that $ \omega^7 \plus{} \omega^2 \plus{} 1 \equal{} 0\implies x^2 \plus{} x \plus{} 1|x^7 \plus{} x^2 \plus{} 1$?[/quote]
[hide]$ x^2\plus{}x\plus{}1|0$ ,which is true...[/hide]

Whoops i have been so dumb... how could i not realize that the second polynomial can be factored into the first polynomial times some other polynomial since both have the same root.... nice solution anyways...


f( x ) = x^2 + 6x - 1

I've tried to find it by factoring... but that aint factorable.
Help me as soon as possible please!!! This is really important.

[hide="hint"]Try the quadratic formula[/hide]

[hide]You plugged it in wrong. The two zeros are $-3\pm\sqrt{10}$.[/hide]

$0=x^{2}+6x-1$
$10=x^{2}+6x+1+10$
$10=x^{2}+6x+9$
$10=(x+3)^{2}$
$\pm \sqrt{10}=x+3$
$x =-3 \pm \sqrt{10}$

Now let us look at how we solved that, by looking at a general case.  (Note that this will not give us the full quadratic formula, just the formula for when the coefficient of $x^{2}$ is 1.)

$0=x^{2}+bx+c$
$-c=x^{2}+bx$
$-c+\frac{b^{2}}{4}= x^{2}+bx+\frac{b^{2}}{4}$
$-c+\frac{b^{2}}{4}= (x+\frac{b}{2})^{2}$
$\pm \sqrt{-c+\frac{b^{2}}{4}}= (x+\frac{b}{2})$
$x =-\frac{b}{2}\pm-c+\frac{b^{2}}{4}$
$x =-\frac{b}{2}\pm \sqrt{-\frac{4c}{4}+\frac{b^{2}}{4}}$
$x =-\frac{b}{2}\pm \sqrt{\frac{b^{2}-4c}{4}}$
$x =-\frac{b}{2}\pm \frac{\sqrt{b^{2}-4c}}{2}$
$x = \frac{-b \pm \sqrt{b^{2}-4c}}{2}$

Note that the real quadratic formula when we have an $a$ in $ax^{2}+bx+c$ term is:

$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$

Try to prove this yourself using a method similar to the one outlined above.

[quote="t0rajir0u"][quote="helpmeplz"]I did it by completing the square. How come it works both ways?[/quote]

They're the same method.  The quadratic formula is a general completing of the square.[/quote]

Yeah.  Basically when you complete the square, every step you do goes towards making the quadratic formula more complicated-looking, as shown in my example above.

The quadratic formula is derived from completing the square.


integration
real analysis
trigonometry
calculus
function
real analysis unsolved

I don't see where I went wrong. Someone point me out

[b]Theorem[/b]
If $0 \leq p \leq \infty$ and if $\{f_{n}\}$ is a Cauchy sequence in $L^{p}(\mu)$, with limit $f$ then ${\{f_{n}\}}$ has a subsequence which converges pointwise a.e. to $f(x)$.

So, this is what I tried:
Since $\parallel f_{n}-f\parallel_{p}\to 0$ as $n \to \infty$, there exists a subsequence $f_{\phi(n)}$ which is monotone and converges too: $\parallel f_{\phi(n)}-f\parallel_{p}\to 0$.
Assume that $|f_{\phi(n)}-f|$ does not converge pointwise a.e. Denote $A_{m}=\{x \in X \ | \ |f_{\phi(n)}(x)-f(x)|>1/m \ for \ all \ n \geq m\}$ and ${A=\bigcup_{m \in \mathbb{N}}}A_{m}$. Then $\mu(A)>0$ by assumption. Take $A' \subset A$ such that $0<\mu(A')<\infty$. Then there exists an $M>0$ such that $|f_{\phi(n)}(x)-f(x)|>1/M$ for all $x \in A'$. But then $\parallel f_{\phi(n)}-f\parallel_{p}\geq \int_{A'}|f_{\phi(n)}-f|^{p}dx > \int_{A'}1/M^{p}dx = 1/M^{p}\mu(A')$. Let $n \to \infty$. Contradiction.

I expect this to be wrong. Because Rudin solved it on a whole page. And he is pretty concise. So I really doubt this is correct too :)

[quote="Jacobson"]
Since $\parallel f_{n}-f\parallel_{p}\to 0$ as $n \to \infty$, there exists a subsequence $f_{\phi(n)}$ which is monotone and converges too:
[/quote]
What do you mean by "monotone" here? I couldn't come up with any reasonable sense of this word for which this claim would be true. It seems that what you need in what follows is that the  sequence $f_{\phi(n)}(x)$ is monotone for a.e. $x$ but you cannot really achieve this: just consider $f_{n}=\frac{1}{n}\sin nx$ on $[0,2\pi]$. Since this sequence tends to $0$ at every point, the only way for its subsequence to be monotone a.e. would be to preserve sign for a.e. point. So, you would have a sequence $n_{k}$ and a set of positive measure $A$ such that $\sin n_{k}x>0$ (or $<0$) on $A$ for every $k$. But then $\int_{A}sign(\sin n_{k}x)\,dx=|A|>0$ for all $k$. On the other hand, this integral tends to $0$ for every measurable set $A$.

Sorry, I was indeed a bit sceptical, because I noticed something could go wrong if $f_{n}(x)$ was not monotone increasing or decreasing. Because then  $|f_{n}(x)-f(x)|>1/m$ is not necessarily true for all $n \geq m$ as you pointed out. What other things can I demand to fix my proof? Or at which level does this proof not make any sence?

[quote="Jacobson"] What other things can I demand to fix my proof? Or at which level does this proof not make any sence?[/quote]
Well, it is not easy to fix your proof by any minor change. The flaw is exactly where I pointed it out. On the other hand, if you want a "one-liner" for the theorem as you stated it, here it goes (for $0<p<\infty$). Choose a subsequence $n_{k}$ such that $\int|f_{n_{k}}-f|^{p}<2^{-k}$. Then
\[\int\sum_{k}|f_{n_{k}}-f|^{p}=\sum_{k}\int |f_{n_{k}}-f|^{p}<+\infty\,.\]
Hence $\sum_{k}|f_{n_{k}}-f|^{p}<+\infty$ a.e. But, for any fixed point, this is a stronger property than $f_{n_{k}}\to f$.
On the other hand, I am afraid that you might misstate the theorem as well; if all that was given was a Cauchy sequence in $L^{p}$ and the existence of the limiting function $f$ was a part of the conclusion, the proof might well extend over a page.


geometry
pigeonhole principle
inequalities
symmetry
area of a triangle
combinatorics solved

Show that given any 9 points inside a square of side 1 we can always find 3 which form a triangle with area less than $\frac 18$.

[i]Bulgaria[/i]

Basicly, this problem is very easy . I think  it would be more suitable for Beginer's Section.
Here is the solution:
Divide the square in 4 smaller squares, of area 1/4.
By Dirichlet, we have in one of them 3 points.
Now show that the maximum area is 1/8.
Do this by drawing parallel lines to the sides of the square and move the points.
Try yourself, now that you know the ideas...

In fact I've posted it at the Advanced Section Section accidentally,although it seemed very difficult to me.
What is Dirichlet?
 :?

Dirichlet stands for the Dirichlet principle, also called Pigeonhole Principle. Roughly speaking, this principle says that if I have 15 rabbits in 14 cages, then there must be at least one cage with more than one rabbit. It's one of the fundamental principles of number theory and combinatorics.

  Darij

Well, in fact Dust you only proved that a triangle has area $ \leq \frac {1}{8}$, which is indeed easy.
The problem asked for a strict inequality, thus there is some more work to do....

Pierre.

[quote="pbornsztein"]
The problem asked for a strict inequality, thus there is some more work to do....
[/quote]
There are lots of cases!!
Well, let's call A, B, C, D the four small squares, in clockwise order, and suppose all the triangles with vertices in 3 of our 9 points have area >= 1/8. If 3 points lie inside a square, then two of them must lie on the vertices of one of its sides, while the other one must lie on the opposite side. This is easy to prove by DusT's outlined argument. If 4 points lie inside a square, then they must coincide with its vertices. More than 4 points cannot lie inside the same square, or they'll generate a too small triangle.
Now, suppose that exactly 4 points lie inside A. Then, the remaining 5 points can lie on just two sides of the big square: the one touching B and C, and the one touching C and D. By pigeonhole principle, 3 points must lie on the same side, generating a triangle with area 0. So, not more than 3 points can lie inside the same square.
Suppose that exaclty 3 points lie inside A. Then, by the above argument both B and D have at least one point in common with A (meaning that the points lie on the common sides). Then, of the remaining 6 points, not more than 2 can lie inside B, and the same for D. Since not more than 3 of them can lie inside C, then exactly two must lie inside B or D. Suppose by symmetry it's B: then they must lie on the distant vertices wrt A, and one of them must be in common with C. A similar result holds for D, and so 4 points must lie in C: contradiction.

I known there were some cases, and I was too lazy to check them by myself.
So, thanks for doing the job Mindflyer :D

Pierre.

I think it worths mentioning :

Show that given any m^2 points inside a square of side 1 we can always find 3 which form a triangle with area less than $\frac{1}{2(m-1)^2} $ .

I think it can be proven immediately because, sincethe points are IN the square, we can make a smaller square that includes the 9 points, and by Dirichlet again the area is less than 1/8
P.S mahanmath, you listen to linkin park?

[quote="mahanmath"]I think it worth mentioning :
Show that given any $m^2$ points inside a square of side $1$ we can always find $3$ which form a triangle with area less than $\dfrac{1}{2(m-1)^2} $ .[/quote]
See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1030132&sid=882f23ec68fcc2399e09ec3ff54a1394#p1030132[/url] for origin and solution.
Notice the naive attempt to again use the pigeonhole principle fails for $m>3$. On the other hand, a slight refinement of the method in the link (printed in the RMC 2008) proves that for $m=3$, i.e $9$ points, one can find a triangle of area at most $\dfrac {1} {9}$.

I don't really understand mindflyer's solution. Could someone please explain it?

Thanks.

[quote="cheeseyicecream"]Can someone please show me in more detail the argument that the maximum possible area of a triangle within one of the smaller squares is 1/8?[/quote]

Bump,  I would appreciate one too.

[asy]
import graph; size(4cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10);
defaultpen(dps); pen dotstyle = black; real xmin = -4.3, xmax = 11.92, ymin = -2.22, ymax = 6.3; pen
zzttqq = rgb(0.6,0.2,0); pen ffqqtt = rgb(1,0,0.2); draw((-2,4)--(-2,-1)--(3,-1)--(3,4)--cycle,
zzttqq); draw((-2,4)--(-2,-1), zzttqq); draw((-2,-1)--(3,-1), zzttqq); draw((3,-1)--(3,4), zzttqq);
draw((3,4)--(-2,4), zzttqq); draw((-0.24,3.82)--(-1.46,1)); draw((-1.46,1)--(1.98,-0.44));
draw((1.98,-0.44)--(-0.24,3.82)); draw((-2,1)--(3,1), ffqqtt); draw((-0.24,3.82)--(-0.24,1));
draw((1.98,1)--(1.98,-0.44)); dot((-2,4),dotstyle); label("$A$", (-1.92,4.12), NE *
labelscalefactor); dot((-2,-1),dotstyle); label("$B$", (-1.92,-0.88), NE * labelscalefactor);
dot((3,-1),dotstyle); label("$C$", (3.08,-0.88), NE * labelscalefactor); dot((3,4),dotstyle);
label("$D$", (3.08,4.12), NE * labelscalefactor); dot((-0.24,3.82),dotstyle); label("$E$", (0,3.64),
NE * labelscalefactor); dot((-1.46,1),dotstyle); label("$F$", (-1.46,1.32), NW * labelscalefactor);
dot((1.98,-0.44),dotstyle); label("$G$", (2.06,-0.32), NE * labelscalefactor); dot((-2,1),dotstyle);
label("$H$", (-2.44,0.92), W * labelscalefactor); dot((3,1),dotstyle); label("$I$", (3.26,1), NE *
labelscalefactor); dot((1.23,1),dotstyle); label("$J$", (1.3,1.12), NE * labelscalefactor);
dot((-0.24,1),dotstyle); label("$K$", (-0.16,1.12), NE * labelscalefactor); dot((1.98,1),dotstyle);
label("$L$", (2.06,1.12), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]

Suppose we have 3 arbitrary points $E, F, G$ inside (or on) the unit square. If all 3 points are on the vertices of the uniq square, then we know the area of the triangle in $\frac{1}{2}$. Otherwise, pick a point such that a horizontal or vertical line through it will cut the triangle into two triangles. For example, line $HI$ in the diagram cuts $\triangle EFG$ into $\triangle EFJ$ and $\triangle FGJ$. It is easy to see that:
\[
[EFG] = [EFJ] + [FGJ] = \frac{1}{2} FJ (EK + LG) \le \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2}.
\]

It boils down to showing that given any 3 points inside a square with area $A$, with points being allowed on the boundary [i]only[/i] in the [b][color=#FF80FF]pink[/color][/b] part (see Fig.2 attatched--bottom one), then the area of the triangle formed by those points is strictly less than $\tfrac{A}{2}$. We can then apply this reasoning to the problem by splitting the given unit square into 4 congruent ones each with area $\tfrac{1}{4}$, then by the [b]Pigeonhole principle[/b] one of these squares must contain 3 points. Now the problem states that points must lie within the unit square, so the points can other be on the pink part or inside the square (See Fig.1--top one), then by the above [b]lemma[/b] the area of any triangle formed by 3 such points is strictly less than $\tfrac{A}{2}=(\tfrac{1}{4})/2=\tfrac{1}{8}$.

divide the square into 8 parts, and the 9 points means there are 2 in a part so it's at most area 1/8

I think that we don't have to consider cases which some points lie on a side of large square.
because the problem states that points lie inside a square
if not, then what about the case which all 9 points lie on 9 vertices of small squares when we divide into 2x2

Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality.

[quote=twinbrian]@above, do you mean each must contain 3 or more points?[/quote]
Yup.

The points lie inside the square, and thus cannot be on the square. We can choose a smaller square where at least one of the points is on the square. Its area must be less than that of the square, so by the above argument, the area of the triangle is at most $\frac18$ of the area of the small square and is thus less than $\frac18$ the area of the large square.

edit: actually I think this is wrong. They can be on the square.

It says strictly inside (supposed no 3 of them are collinear)

[quote=TuZo][b]A simple solution:[/b]
Divide the square in $4$ right angled triangle (see the picture). We apply the Pigeonhole Principle. We can see that in on of the $4$ triangle there exist definitely $3$ points, let this triangle the $ABC$ triangle. But the area of the triangle $ABC$ it is exactly $1/8$[/quote]

If you divide a unit square into 4 congruent pieces, the area of each piece must be $\frac14$.

[quote=Mathlete2017][quote=Mathlete2017]Simply split the square into four smaller squares with area $\frac{1}{4}.$ By the Pigeonhole Principle, one of these squares must contain 3 or more points and the largest possible triangle in such a square has area $\frac{1}{8}.$ However, I don't get how to go from here to proving the strict inequality.[/quote]
Bump.[/quote]

in the case of largest possible triangle that is one of area 1/8 at least one of the vertices will lie on the exterior square (square of side 1) so that case will not be considered (inside means within the area enclosed by the sides and not on them) thus area will be less than 1/8

I'm kinda having problems with this? Sorry not the Pigeonhole master. I divided it into four congruent squares with area $\frac{1}{4}$ and so at least one must have $3$ points in it and the maximum area is $\frac{1}{8}$. However the problem says [i]less[/i], so what do I do?

Thanks for the help.

I solved it. Posting for storage.

[hide=Solution]Split the square into four congruent squares of side $\frac{1}{4}$. By the Pigeonhole Principle, at least one of these squares must contain $3$ points. The maximum area of the triangle formed by these $3$ points is $\frac{1}{8}$. However, to uphold this bound we'd have $3$ points in $3$ out of the $4$ squares. However this is not possible, as the maximum is formed when $2$ of the vertices are used per square and hence some vertices would be repeated, which is not allowed. This means that we can always find $3$ points forming a triangle with area less than $\frac{1}{8}$, as claimed.[/hide]

Label the square $ABCD$, and let the midpoints of $AB,BC,CD,DA$ be $E,F,G,H$ respectively, and let the center of $ABCD$ be $O$. Then by Pigeonhole we may assume WLOG that some three of the points lie either in the interior of $AEOH$ or on $EO,HO$. So the area is less than or equal to $\frac{1}{8}$, but equality cannot occur because this implies that one of the points must be $A,E,H$ by Pigeonhole, contradiction.


I wonder if it 's right or wrong. Formerely, AMC10 and AMC 12 were AHSME ?

The AMC 10 started in 2000, and that was when the AHSME became the AMC 12. According to the website, only the AMC 12 was fomerly called the AHSME.


The Favorite Subject Poll is surviving so I've decided to add this one.
I left out math. Anyone that thinks I should include math is wasting their time on this site. ;)

Biology and history.... I realise that thery're important, and I do well in those subjets, but I just dislike them, I don't know how........ the same with language (spanish for me)... it's simply not for me...

I mean, there are alot more jobs in science than there are in like, english, history (although there still are a few), or foreign languages other than teachers.

Personally, I hate Spanish, although my Spanish average was (I believe) higher than my math average this year. ...scary...

Anyway, I don't like biology much, but there are several interesting parts. Spanish is just plain boring.

English is my least favorite.  History, I had a good teacher this year, and I find history kind of interesting in general.

English though - it can be intriguing sometimes, but this year...I had the hardest teacher (gradewise and workwise).  The year could have been much better if she was more responsive to us in the amount of useless work and didn't kill everyone gradewise.

Even though I place much of the emphasis on the teacher, the subject English is in general less interesting than all the others to me.

[quote="nat mc"]i hate english. i think its pointless[/quote]Is that why you don't capitalize or punctuate properly?

[quote]I mean, there are alot more jobs in science than there are in like, english, history (although there still are a few), or foreign languages other than teachers.[/quote]That depends.  Nearly every job you're likely to undertake will require a basic grasp of English spelling, grammar and usage.  Only a very small percentage require you to apply mathematics directly.  There are many jobs and opportunities available to those who speak a foreign language -- many of them are overseas, and I think Americans in general would benefit from the broader outlook provided by going to foreign countries and actually bothering to speak enough of the language to know what's going on.  As for history, there may not be a large number of jobs which directly apply the knowledge one gains studying history, but the research and writing skills that one needs for history are significant in many careers (think law), and the basic thought-processes and knowledge are important in understanding modern-day situations (policy maker or politician).

[quote]i think that studying literature is bs.
when you look at course descriptions of english classes, they give you bs reasons for studying literature like "it gives you a connection to the rest of humanity, etc." i could watch a popular movie and then I would be able to "relate" more to humanity than by reading a "classic."
as far as english in general goes, I think that writing and speaking are the most important parts, even though i don't like either one in a class setting.[/quote]Another one, fighting back against the oppressive regime of English teachers by refusing to use the skills taught (and seriously abusing "scare quotes," too).  What your comment actually says: "I don't like reading.  They should teach writing and speaking.  But they shouldn't teach those, either."  That's a good plan you've got there.  Maybe you should run it past the local school board, see what they think.

History's my least favorite subject.  I still enjoy it on occasion, but one subject has to be my least favorite.  Behind that is english.  I find english more important than history.  You're always going to have to write letters and reports, talk to others, and read.  I also enjoy reading literature for the asthetic qualities of literary works (also on occasion).

JBL you have not taken time to differentiate between literature and an english class in general. I don't like studying (basically reading) the kind of literature taught in schools; I enjoy non-fiction and other types of writing that directly address information from the real world.

Also, I mentioned that I think that writing and speach are important, but then stated that I did not [i]like[/i] them in a class setting. Once again, you interpreted this incorrectly, infering that I think that they shouldn't be taught in schools. There's a difference between disliking something and believing it shouldn't be taught.

[quote="JBL"]That depends.  Nearly every job you're likely to undertake will require a basic grasp of English spelling, grammar and usage.[/quote]
You don't need to attend English class to be able to attain a basic grasp of the English language.

I would have to say history: fortunately, I have had good teachers this year (My english teacher is probably the most popular teacher in the school,) but regardless of how well it is taught, I have never liked history.

I'd have to say English, despite having a fun teacher. History got pretty cool this year because my teacher made it fun - plus, it's not that hard for me, so that's always a bonus. On the other hand, AP American Language test = death.

English and history.
I don't think english would be so bad if we didn't review the same grammar and literary vocab every year.
History can be interesting, but how it's taught/presented is pretty dull (depends on the teacher, i guess)

I try not to "hate" any subject as then it's easier to put forth less effort. So if I had to choose a least favorite, I guess it would be the one I have my lowest grade in, which would be English (an A-). Also, the department is really hard on grading, so it's almost impossible to get an A+ in English at my school...

[quote="nat mc"]i hate english. i think its pointless[/quote]
This is probably the funniest post ever. Nice use of not capitalizing and not puncuating.

Yeah that's kind of funny, but once again we can understand something without using the strict english rules drilled into our heads.

This is better:
[quote="Ralph Wiggum"]me fail english thats unpossible[/quote]

This year my grades were All A's and a C in English. My parents chided me even though they have difficulty in speaking English.
English was pointless. I personally don't care that the Greeks and Romans made so many myths and Some guy took 10 years to get home. My EOCT score was 92, so that's good enough for me.

I would have to say English, but that's largely because it's my second language. I was in ESL (English as a Second Language) class during my first year in the U.S. (sophomore year) and then moved to a regular English class. My teacher was really nice, but he gave us too much work...

I'm great at math yet my best other subject is history and my highest grade is English.

Math and History are my extreme fortes... what a useless combination of subjects... by that I mean math doesn't help history and history doesn't help math.

My "least favorite" has to be biology, I guess, because I suck at it. I actually like learning about biology but I just suck at it. I like everything else AND I'm good at that stuff too, so my "least favorite" subject is biology.

English Literature.

I don't mind English Language. Once I get going, creative stories start pouring out and it is actually quite enjoyable.

I loathe analyticial essay writing though. Having to present a balanced argument from different viewpoints, citing quotations and sources. I really dislike it because I can neber understand how it is marked. There's no right or wrong answer like in mathematics, and yet there is a mark scheme where you must cover each key point. It's not even like English Language, where when you re-read your story and you can already judge its merit.

 :wacko:

Why isn't math a choice?  :P   :rotfl: Anyway, I would say I like most subjects and can't really pick one out that I like least.

I wouldn't say I dislike math on the whole, but then i really find some areas of calculus to be quite frustrating...

history is nice but the way it is teached... not

I always thought the worst part of history was the exams. Learning it was fun but having to study and analyse sources was not.

I have to say History. Why do you have to know the French and Indian war? It's useless in the long run. It's just ____ stuffed into your brain that you eventually forget. It's also not going to help you in life unless you are a politician.(I think.) I am past hating history. Who here would like to join me in burning their history books?
[hide=" :furious: "]Just kidding! :P [/hide]
$\uparrow$
Click me!

There's only 2 classes on there that I take and I picked English.

I would have to say history because of my teacher.

History for me.  However, it was only my second lowest grade even though I did [i]nothing[/i] all year and still somehow pulled a 94.  I got a 92 in English, but I actually had a real teacher and learned a lot in that class, so it was hard to get an A even with effort.

I'd like to note that history depends a lot on the teacher.  It's interesting to learn about if they actually teach, but I'm not very good at memorizing facts, so tests are harder for me.  In 10th grade I had a great world history teacher though, and I also took military history as an elective at first only because it was the only one available in that slot.  But I turned out to enjoy the class, since I had an awesome teacher there as well.

But overall, I'd definitely say history is least favorite.

[quote="star99"]everything really depends on what your teacher is like, so this year its history[/quote]
I agree, our teacher wasted time telling us about useless things, e.g, the cellphone he bought the other day :!: :?:.

Our school :D

Our history teacher this year was competent although he liked to try and force his beliefs on us sometimes. Last year, the teacher got lessons off MSN Encarta and did nothing in the textbook so we basically learned nothing, instead, we had stupid assignments like make a tour of at least 10 states. Create a brochure about it including information on transportation, which states you are visiting, the capitals of those states, and one attraction in each state.

Our school has no science curriculum at all so it's up to the teachers. The one last year I think did the best she could while she wasn't really an expert in science, she did the best she could and found a lot of good resources for us so I think we still got a good year of learning. This year, we have no work at all literally. The teacher's grading method is 10% Classwork (Either you did it, or you didn't) 20% Projects, 20% Quizzes. The other 50 points is automatic, I've seen kids not hand in their projects and still get A+s easily. When I'm under a lot of stress, I just use Science as a Study Hall and I can still get an A+. We don't have tests, projects are once a quarter, and we rarely have quizzes.

I can't describe last year's English class in words, you had to be there. Let's just say we went outside for about 2 hours everyday and the teacher brought in food and movies for us every Friday and we never worked. Also, there were lots of crazy things like sticking pencils in the ceiling, busting the sink's pipes, the teacher having a period on a chair and not being told until lunch, tracking dog crap in the carpet, leaving a full milk carton in a corner of a cabinet over winter break, etc.

I would absolutely love English if it was creative writing. But if I have to read Bury My Heart at Wounded Knee again I quite possibly will go insane.

History is boring; I don't really care about it. (Although our teacher this year made it substantially interesting.) Geography, which has sort of been grouped with history, is fun for me though. I guess that's because I'm a statistics geek.

I know many students at my school would say math, but that is due in no small part to the horrendous curriculum they insist on using.

English all the way.  It's the only subject I have EVER got lower than an A in.

History's useful because you analyze it and learn from it. Definitely useful in the big picture, politics, or militarily speaking.

You could argue math is useless, but that's blatantly false because although the direct knowledge may be unnecessary, it trains your brain to think better, smarter, and faster.

The people I know that are good at math are the, in general, most overall competent and cleverest.[/code]

[quote="drunner2007"]Ignite168 wtf is up with your school? that's crazy.
anytime i have a class like that i start learning things on my own
like ap chem at my school was a class that was going at a really slow pace when it shouldn't have been. we were listening to these dumb stories about my teacher's daughter forgetting her dress at some place before her wedding. it was really frustrating when the exam was only weeks away.[/quote]

Well it is middle school, standards aren't really that defined yet. And most of it was because a new principal took over. I checked online, while the state average of teachers that are Qualified by their definition remain at a steady 96.4%, our school went from 98% to 84% then to 72% where 98% was the first year of the new principal. Our state testing scores also dropped about 20-30 percentiles too, especially in 4th grade. Anyway, our school goes for 200 days (longest in the state by at least 3 weeks, there was a newspaper article about it) and is  8-3:30, which is the same as about half the schools in the state, the other half are 8-2:30.

Well, my school day goes from 7:30 to 3:30 and 4:30 if I stay afterschool which is almost every day. I do have less school days than you do though.

My school day runs from 8:00 to 2:30. But I come early (for Jazz Band, Math Team, etc.) and stay after (with my friends) almost every day. Which means 9 hours in school. :D

Back to the topic, this year English was my favorite class (surprisingly) because of our teacher. Last year, we had a HORRIBLE English teacher and everyone disliked that class. I was only exposed to English, History (in Social Studies), and Physics and Chemistry (in Science). But, if I were to take all the classes, I guess my least favorite would be Computer Science.

It is nessecary to speak and write English fluently, but it is not that nessecary to spend alot of time on it unless you are planning to have a writing profession, or you are worried that some one on the street will come up to you, point a gun to your head and threaten to kill you unless you can find the direct object in a sentence.


geometry
geometric transformation
search
combinatorics unsolved
combinatorics

On a plane, given a finite set X that is not less than 9 points, and a triangle ABC. Any 9 points covered by two translation triangles of ABC. Prove that: X can be covered by two translation triangles of ABC.
(Translation triangle is congruent with ABC and their corresponding sides parallel.)

Looks like it's Russia 2003 to me -- a search for "triangle translation" finds it easily:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=439302164&t=293448


The product of two whole numbers is 1000. If neither of the numbers is a multiple of 10, what is their sum?

find prim factorization of 1000 and get 5^3*2^3                125*8  neither are multiples of 10 so the answer is

[hide="is this your answer or something else?"]133 it is your answer![/hide]

[size=200][color=#FF8040][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED !!!!!!!!!![/u][/i][/b][/color][/size]

:icecream: :icecream:  :icecream:  :icecream:  [size=200][color=#BFFFFF][b][i][u]IM HUNGRY FOR MATH !!!! : D [/u][/i][/b][/color][/size]


We can easily find from AM-GM that
$A=\frac{a(a+c)}{2b(b+c)}+\frac{b(a+b)}{2c(a+c)}+\frac{c(b+c)}{2a(a+b)}\geq\frac{3}{2}$

and Nesbit's inequality said that
$B=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$

Prove that  $A\geq B$

[quote="silouan"]$\frac{a(a+c)}{b(b+c)}+\frac{b(a+b)}{c(a+c)}+\frac{c(b+c)}{a(a+b)}\geq 2\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)$[/quote]
Put $x=\frac{a}{b+c},\ y=\frac{b}{c+a},\ z=\frac{c}{a+b}.$ Then
$\frac{a(a+c)}{b(b+c)}+\frac{b(b+a)}{c(c+a)}+\frac{c(c+b)}{a(a+b)}=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\frac{x+y+z}{\sqrt[3]{xyz}}$
The conclusion then follows from
$\sqrt[3]{xyz}=\sqrt[3]{\frac{abc}{(a+b)(b+c)(c+a)}}\le\sqrt[3]{\frac{abc}{8abc}}=\frac{1}{2}$

We can also use
$\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\ge\sqrt{(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)}$
and
$\sum_{cycl}\frac{a+b}{c}\ge 4\sum_{cycl}\frac{c}{a+b}$

By the way, the following stronger inequality holds:
For any positive real numbers $a,b$ and $c$, \[3+\frac{a(a+c)}{b(b+c)}+\frac{b(b+a)}{c(c+a)}+\frac{c(c+b)}{a(a+b)}\ge 4\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)\]

was already posted http://www.mathlinks.ro/Forum/viewtopic.php?t=91546
 :wink:

[quote="silouan"]Nice dunctung .
My solution uses only AM-GM in the way:
$\frac{2a(c+a)}{b(b+c)}+\frac{b(a+b)}{c(a+c)}\geq 3\sqrt[3]{\frac{a^{2}(c+a)(a+b)}{bc(b+c)^{2}}}\geq \frac{6a}{b+c}$
:wink:[/quote]

Ok, that is, in essence, proving
$\frac{X}{Y}+\frac{Y}{Z}+\frac{Z}{X}\ge\frac{X+Y+Z}{\sqrt[3]{XYZ}}$

You are invited to try the stronger version I posted above.


trigonometry
inequalities unsolved
inequalities

let $\alpha,\beta,\gamma$ be angles of a acute tirangle with $\alpha<\beta<\gamma$
prove the following:
$\sin^2{(\beta)}\cdot\sin{(\frac{\alpha}{2})}\cdot\sin{(\alpha +\frac{\beta}{2})>\sin^2{(\alpha)}\cdot\sin{(\frac{\beta}{2})}\cdot\sin{(\beta+\frac{\alpha}{2})}}$

no one can do this?
i tought this was an easy problem ;)

Easy problem,but I founde an interesting solution(that take 5 minuts) .
Her it.
Let $ABC$ is triangle white angles $\alpha,\beta$and$\gamma$respectively equals to the
angles$A,B$and$C$,and$AB=c,BC=a,AC=b$.Let $AA_1$ and $BB_1$ are angle bisectors.Then from sinus' thaeom we have that
$\frac{sin\alpha}{sin\beta}=\frac{a}{b}$
$\frac{sin\frac{\alpha}{2}}{sin(\alpha+\frac{\beta}{2})}=\frac{BA_1}{c}$
$\frac{sin\frac{\beta}{2}}{sin(\beta+\frac{\alpha}{2})}=\frac{B_1A}{c}$
=>$\sin^2{(\beta)}\cdot\sin{(\frac{\alpha}{2})}\cdot\sin{(\alpha +\frac{\beta}{2})>\sin^2{(\alpha)}\cdot\sin{(\frac{\beta}{2})}\cdot\sin{(\beta+\frac{\alpha}{2})}}$
<=>$\frac{b^2}{a^2}>\frac{B_1A}{A_1B}$ also you can easy see that $AB_1=\frac{bc}{a+c}$ and $BA_1=\frac{ac}{b+c}$=>$\frac{b^2}{a^2}>\frac{B_1A}{A_1B}$
<=>$\frac{b}{a}>\frac{b+c}{a+c}$<=>$b>a$ laste one is true because$\beta>\alpha$ ;)

[quote="10000th User"]Administrators and/or moderators, feel free to remove Tiks solution as I am made aware by [b]paladin8[/b]'s post that this is problem #4 of 2006 SIMUW application quiz[/quote]
What mean  [b]SIMUW[/b]?

[quote="Virgil Nicula"][quote="Tiks"]Easy problem,but I founde an interesting solution(that take 5 minuts) .
[/quote]
Tiks, you used the same words alike me. Therefore, and you are submissively to mistake. Even and you arn't perfectly.[/quote]
But no one is perfect :rotfl:

[quote="Virgil Nicula"]I didn't say I is perfectly. Tiks, answer me: is the your mistake (if you assume it) identicaly with my mistake (I assumed it) ?
[u]Remark.[/u] I like the your solution.[/quote]
Allrigte Virgil,about this problem I only want to say that it is easy to solve(but I haven't
say that it is not Olimpiad Problem) :lol: .
[u]Remark[/u].I am glad that you like my solution ;) .

[quote="#20002 pwns"][quote="Tiks"]$\frac{sin\frac{\alpha}{2}}{sin(\alpha+\frac{\beta}{2})}=\frac{BA_1}{c}$
$\frac{sin\frac{\beta}{2}}{sin(\beta+\frac{\alpha}{2})}=\frac{B_1A}{c}$[/quote]
I don't see how you got this, unless you used $\sin{(\pi- \theta)}=\sin{\theta}$, is this what you did? I have trouble understanding your solution. Thanks![/quote]
I am sorry ther is a little typo :oops: ,it should be
$\frac{sin\frac{\alpha}{2}}{sin(\beta+\frac{\alpha}{2})}=\frac{BA_1}{c}$
$\frac{sin\frac{\beta}{2}}{sin(\alpha+\frac{\beta}{2})}=\frac{B_1A}{c}$
Now I think you will get it. ;)


inequalities
induction
inequalities solved

I solved this one long time before. It is a generalization of Turkevici inequality, posted by manlio!

Can you give me the source of this problem , Harazi? Thank you very much

No, I can't, I just know that it was posted by Manlio long time ago.

Dear romano,

please be so kind to adopt your sources. Look a the problems posted by Darij in order to see how sources references for Kvant problems can be given. Thanks.  :)

I have the link to Manlio's original posting:

http://www.mathlinks.ro/Forum/viewtopic.php?t=5285

I don't know the number of the Kvant problem, and searching it in the archive would be a really hopeless task...

  Darij

[quote="darij grinberg"]

I don't know the number of the Kvant problem, and searching it in the archive would be a really hopeless task...

  Darij[/quote]

Well, Romano should not be forced too much. He has taken the problem from some source and I bet the authors indicated the "complete" reference for the Kvant problems !  :)

This is problems of F. Shleyfer from Kvant No 3, 1979, page 35. Published without solution after solution of Turkevici inequality.

Great! At least now we know who created first this beautiful problem.

A beautiful problem should have a beautiful solution ! :D

I have two different proofs by induction, but they are not beautiful, so I'm still waiting. By the way, was it solved after being proposed in Kvant?


geometry
geometric transformation
rotation

I have read that astronomers from three observatories (Lowell Observatory at Arizona, California Institute of Technology's Palomar Observatory in San Diego County, and one observatory at Canary Islands) have recently discovered the so far largest planet in the Universe. The planet, named TrES-4, is located at a distance of 1435 light years from the Earth in the constellation of Hercules. TrES-4 is a giant gaseous planet made of mostly hydrogen, like Jupiter, but it is much hotter, lighter, and about 70% larger - it is 20 times larger than Earth, however its mass is only about three-quarters of the mass of Jupiter, which is quite interesting. In fact, it is referred that its densitity of only 0.2 g/mL is somewhere between a wine cork and balsa wood, and so this planet could theoretically float on water. TrES-4 makes a complete revolution around its parent star every 3.55 days, so a year on this planet is shorter than a week on Earth. Also, being only 7 million kilometres from its parent star, its temperature is about 1327 ºC.

I have some questions that I would like to address:

1) How can astronomers determine the size, temperature, and period of rotation of a planet so far away?

2) I remember to have read quite a long ago that Canary Islands are a privileged site for astronomical observations because their sky is almost always without clouds. If this is true, why it is so?

1. This may not be a thorough explaination.  Astronomers measure the size and mass of exoplanets by trying to detect the smallest wobbles, shifts, or changes of brightness of a suspected star.  They do this by using either viewing thier spectra or directly thorough a telescope with computers.  From the amount and period of such changes in the suspected star they can to determine possible traits of the planet from the most probable cases that could give such data.

2. Tried to look it up, could not find anything.  Usual places for good observing are in deserts (like Lowell) or in the mountians (like the ones form CalTech and at Mauna Kea in Hawaii).

The canary islands have no clouds because of where it is where it is located. It is the same for why San Fran has lots of fog.

Thanks for the explanation AstroPhys.

[quote="Hunter34"]The canary islands have no clouds because of where it is where it is located.[/quote]

Forgive my ignorance, but that seems to me a vague explanation. Why, for example, isn't Madeira Island or Cabo Verde Islands also used, or used instead? What special geographical conditions make Canary special?

I think you may find more about this on National Geographic. This was just a summary of it.


[quote="Vasc"]If $a,b$ are positive numbers, then

$4+\sqrt{2(1+a^2)(1+b^2)(\frac 1{a^2}+\frac 1{b^2})} \geq (1+a)(1+b)(\frac 1{a}+\frac 1{b})$.[/quote]
For all $a\neq0,b\neq0$ let $\frac{1}{a}+b+\frac{a}{b}-1=u$ and $\frac{1}{b}+a+\frac{b}{a}-1=v.$
Hence,
${(1+a^2)(1+b^2)(\frac 1{a^2}+\frac 1{b^2})}=u^2+v^2$ and $(1+a)(1+b)(\frac 1{a}+\frac 1{b})-4=u+v.$
Hence, remain to prove that $\sqrt{2(u^2+v^2)}\geq u+v$, but this is obviously true. :)

This is indeed a beautiful solution! :)

But, may I ask something?
Here :

[quote="arqady"]
...
For all $a\neq0,b\neq0$ let $\frac{1}{a}+b+\frac{a}{b}-1=u$ and $\frac{1}{b}+a+\frac{b}{a}-1=v.$ ...
[/quote]

What did make you think of this substitution? :)

[quote="mathmanman"]
What did make you think of this substitution? :)[/quote]
Use $(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2$ or $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$
to  expression $(1+a^2)(1+b^2)(\frac{1}{a^2}+\frac{1}{b^2})$  the  two times so that   final expressions under squares
could be equal for $a=b=1$ ;)

[hide="Ugly solution"]
By Holder we have $(1+a)(1+b)(\frac{1}{a}+\frac{1}{b})\geq 8$, then
$\left( 2(1+a^{2})(1+b^{2})(\frac{1}{a^{2}}+\frac{1}{b^{2}})-\left( (1+a)(1+b)(\frac{1}{a}+\frac{1}{b})-4\right) ^{2}\right) a^{2}b^{2}=\left( b-1\right) ^{2}\left( a-1\right) ^{2}\left( a-b\right) ^{2}$
And the last expression is non-negative[/hide]


Let $ a,b,c$ be positive real numbers. Prove that

$ \sqrt{3(\frac a{b}\plus{}\frac b{c}\plus{}\frac c{a} )}\plus{}2\sqrt{\frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}} \ge 5$

[hide="@Loc"]Have fun ! :)[/hide]

[b]Problem.[/b] Let $ a,b,c$ are positive real numbers, prove that
\[ \sqrt {3\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right)} \plus{} 2\sqrt {\frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}} \ge 5
\]
By using the well known
\[ 2\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \plus{} 1 \ge \frac {{21\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}{{\left( {a \plus{} b \plus{} c} \right)^2 }}
\]
Setting $ x \equal{} \sqrt {\frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}} \le 1$. It suffices to show that
\[ \sqrt {\frac {{3\left( {10 \minus{} x^2 } \right)}}{{2x^2 \plus{} 1}}} \plus{} 2x \ge 5
\]

\[ \Leftrightarrow \frac {{3\left( {10 \minus{} x^2 } \right)}}{{2x^2 \plus{} 1}} \ge 4x^2 \minus{} 20x \plus{} 25
\]

\[ \Leftrightarrow \frac {{ \minus{} 8x^4 \plus{} 40x^3 \minus{} 57x^2 \plus{} 20x \plus{} 5}}{{2x^2 \plus{} 1}} \ge 0
\]

\[ \Leftrightarrow \frac {{\left( {x \minus{} 1} \right)\left( { \minus{} 8x^3 \plus{} 32x^2 \minus{} 25x \minus{} 5} \right)}}{{2x^2 \plus{} 1}} \ge 0
\]
which is clearly true.

A little suprised for me ddlam, thank you! :)

How do you think about the followed
\[ \sqrt {\frac {{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}}{3}} \left( {\frac {{\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}}}{3}} \right)\left( {\frac {{ab \plus{} bc \plus{} ca}}{{a^2 \plus{} b^2 \plus{} c^2 }}} \right) \ge 1
\]
And we also note that the followed is not true
\[ \left( {\frac{{\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a}}}{3}} \right)\left( {\frac{{ab \plus{} bc \plus{} ca}}{{a^2  \plus{} b^2  \plus{} c^2 }}} \right) \ge 1
\]
:)

Hi, akai!
I think, the brute force's way ( let $ a \equal{} \min\{a,b,c\},$ $ b \equal{} a \plus{} x$ and $ c \equal{} a \plus{} y$ ... ) helps here.
By the way, the following inequality is true too:
\[ {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \plus{} 0.7 \ge \frac {{11.1\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}{{\left( {a \plus{} b \plus{} c} \right)^2 }}\]
but the following is wrong already:
\[ {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \plus{} 0.8 \ge \frac {{11.4\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}{{\left( {a \plus{} b \plus{} c} \right)^2 }}\]
( $ a\equal{}8,$ $ b\equal{}4$ and $ c\equal{}3.$ ) :wink:

My solution: :)

By using the well known
$ (\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})(a\plus{}b\plus{}c)^2\ge 9(a^2\plus{}b^2\plus{}c^2)$
And now,we setting $ x\equal{}\sqrt \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\le 1$,we need prove that
$ 3\sqrt{ \frac{3}{2x^2\plus{}1}}\plus{}2x\ge 5$
That ineq is true with $ 1\ge x \ge 0$
The inequality hold on $ a\equal{}b\equal{}c$

[quote="Math pro"][quote="akai"][quote="Honey_S"] By using the well known

$ 2\left( {\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}} \right) \plus{} 1 \ge \frac {{21\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)}}{{\left( {a \plus{} b \plus{} c} \right)^2 }}$
 [/quote]

Dear Honey_S,
Could you post your solution for above inequality?
Thank you![/quote]
no problem.We can use half Schur-half SOS[/quote]
Are you sure? :wink:  :P


in statements and reasons prove that....

if the vertex angle of one isosceles triangle is conguent to the vertex angle of another isosceles triangle, then the triangles are similar.

umm isnt that like for triangle one let $A,B,C$ be the angles for triangle two let $D,E,F$ be the angles therefore

let $x=A=D$
so by definition $B=C$ and $E=F$
$B=C=\frac{180-x}{2}$
$E=F=\frac{180-x}{2}$

Therefore the two triangles are similar..hmm did i interpret the questions right..


prove that for any m>n we have $\sum\limits_{k=1}^{m} \frac{1}{n+k} \not\in Z$

Only one terms had maximal mean $ord_2(n+k)=r>0, (m>1)$. Therefore $ord_2(\sum )=-r<0$.


On AoPS, when it says at the top right corner of the screen;

View posts since last visit
View unanswered posts
View your posts

I know what the bottom two ones mean, but I don't know what the top one means. Please help?

Er... this would be better in the Questions forum...

But I believe that "View posts since last visit" means that you can see the links of all the new posts since your last visit to AoPS.


inequalities
quadratics
algebra
difference of squares
special factorizations
number theory proposed
number theory

prove that $\frac{5^{125}-1}{5^{25}-1}$ is a composite number. :D

DON'T DOUBLE POST.  You've been around here long enough that you should know that.

I don't think your hint is useful, Fermat -Euler.

I am not sure that U are right

$\frac{5^{125}-1}{5^{25}-1}=5^{100}+5^{75}+5^{50}+5^{25}+1$
I think this a hard Problem Khaknezad

I think $(5^5)^n = 1$ (mod $11$) so the number can be divided by 11.

thanks all of you but i want a nice solution for it becasue i know an ugly solution for it. :D

ow I'm sorry
 if $(5^5)^n=1$ (mod 11) then ${5^100}+\ldots+1=5$ (mod 11).
So the number can not be divided by 11... (it would have been too obvious, i know...)

[quote="ashegh"]no its not hard.(too ketabe mariam joone digeh ;) )[/quote]

You mean you found a factor without computer help or perhaps a proof it is composite?

[quote="Leva1980"]I think that this is true generally : if $p$ is a prime which is
$\equiv 1 (mod 4)$, the for every natural $m$, we have that $\frac{p^{mp} - 1}{p^{m} - 1}$ is a composite number.[/quote]
are you sure this is true? :?:

The solution Jensen and ashegh were refering to :

By Maryam mirzakhani (IMO hong khong and Canada gold medalist)
Let $f(x)=1+x+x^2+x^3+x^4$ ,We are to prove $f(5^{25})$ is not prime .

generally we have :

$f(x)=(x^2-3x+1)^2-5x(x+1)^2, x \longrightarrow 5^{25}=(x^2-3x+1)^2-5^{26}(x+1)^2$

I think the following is true:
 Consider a prime $p$, which is $\equiv 1 ( mod 4 )$. Consider
the $p$'s root of unity $\zeta = e^{ 2 \pi i /p }$ and a
Galois extension $\mathbb{Q}( \zeta ) / \mathbb{Q}$. Consider
also the norm of it's extension $N : \mathbb{Q}( \zeta ) \rightarrow \mathbb{Q}$. We have that $\frac{p^{mp} - 1}{p^{m} - 1} = N(p^{m} - \zeta )$. It is well-known from the proof of
quadratic reciprocity that $\pm \sqrt{p} = \sum_{x \in \mathbb{F}_{p} } ( \frac{x}{p} ) \zeta^{x}$, therefore $\sqrt{p} \in \mathbb{Z} [ \zeta ] \subset \mathbb{Q}( \zeta )$. We have
that $p^{m} - \zeta = ( \sqrt{p} )^{2m} - \zeta^{p+1} = (( \sqrt{p} )^{m})^{2} - (\zeta^{(p+1)/2})^{2} = ( ( \sqrt{p} )^{m} - \zeta^{(p+1)/2} ) ( ( \sqrt{p} )^{m} + \zeta^{(p+1)/2} )$.
Therefore if we denote $\alpha = ( \sqrt{p} )^{m} - \zeta^{(p+1)/2}$ and $\beta = ( \sqrt{p} )^{m} + \zeta^{(p+1)/2}$, then $\alpha , \beta$ are algebraic integers and $\alpha , \beta \in \mathbb{Q}( \zeta )$. Therefore $\frac{p^{mp} - 1}{p^{m} - 1} = N(p^{m} - \zeta ) = N( \alpha \beta ) = N( \alpha ) N( \beta )$, where $N( \alpha ),N( \beta )$ are positive
integers. It is remained to show that $N( \alpha ),N( \beta ) > 1$. For this if we take some automorphism $\sigma : \mathbb{Q}( \zeta ) \rightarrow \mathbb{Q}( \zeta )$, then $\sigma( \alpha )$ has the form $\pm ( \sqrt{p} )^{m} - \zeta^{j}$ for some $j$, therefore $| \sigma( \alpha ) | \geqslant ( \sqrt{p} )^{m} - 1 > 1$. This works also for $\beta$.

I hope it is correct.

[quote="lomos_lupin"]$f(x)=(x^2-3x+1)^2-5x(x+1)^2$[/quote]

Small typo? :)

nice solutions indeed but not really my idea of easy, it requires quite some trial and error to find such things. Anyhow just for your information: pc says (using calculations above as brute force would require too much time on this old thing)

\[ \frac{5^{125}-1}{5^{25}-1}=24687045214139234043375683501 \cdot 3597751 \cdot 767186663625251 \cdot 4032808198751 \cdot 28707251  \]

[quote="Arne"]By the way, this problem wasn't created by M. Mizakhani as this is from ISL 1992.[/quote]
yes i knew that but it is in her book.


parameterization
probability
function
conditional probability
probability and stats

Let $ X$ be a Poisson with parameter $ \mu$ and the conditional distribution of $ Y$ given $ X \equal{} x$ is a Binomial $ Bin(n,p)$. Show that $ Y$ is a Poisson with parameter $ \mu p$

I am confuse with the conditional distribution, can someone enlighten me how to start? Thanks

Let $ p(x,y)\equal{}P(X\equal{}x,Y\equal{}y)$ be the joint probability function.

Let $ p_1(x)\equal{}\sum_yp(x,y)\equal{}P(X\equal{}x)$ be the marginal distribution of $ X.$ We're told that it is Poisson. That is, $ p_1(x)\equal{}\frac{e^{\minus{}\mu}\mu^x}{x!}$ for $ 0\le x.$

The conditional probability function of $ Y$ given $ X$ is

$ p(y\mid x)\equal{}P(Y\equal{}y\mid X\equal{}x)\equal{}\frac{p(x,y)}{p_1(x)}.$ We can turn that around to say that

$ p(x,y)\equal{}p_1(x)p(y\mid x).$

We are told that $ p(y\mid x)$ is a binomial distribution. Here is where you must have a typo in your statement: if this distribution were $ \text{Bin}(n,p)$ for fixed $ n,$ then $ X$ and $ Y$ would be independent and the marginal distribution of $ Y$ would be $ \text{Bin}(n,p).$

So we'll have to reconstruct what you must have meant to say. I'm going to guess that you meant that the conditional distribution of $ Y$ given $ X$ is $ \text{Bin}(x,p).$ That is, $ p(y\mid x)\equal{}\binom{x}{y}p^y(1\minus{}p)^{x\minus{}y}.$ Then,

$ p(x,y)\equal{}\frac{e^{\minus{}\mu}\mu^x}{x!}\cdot \binom{x}{y}p^y(1\minus{}p)^{x\minus{}y}\equal{}\frac{e^{\minus{}\mu}\mu^x}{y!(x\minus{}y)!}\,p^y(1\minus{}p)^{x\minus{}y}.$

We now wish to find the marginal distribution of $ Y:$ $ p_2(y)\equal{}\sum_xp(x,y).$

This leads us to the sum $ \sum_{x\equal{}y}^{\infty}\frac1{(x\minus{}y)!}\,\mu^x(1\minus{}p)^{x\minus{}y}.$ Make the change of variables $ k\equal{}x\minus{}y$ and this becomes

$ \sum_{k\equal{}0}^{\infty}\frac1{k!}\,\mu^{k\plus{}y}(1\minus{}p)^k\equal{}e^{\mu(1\minus{}p)}\mu^y.$

Multiply that by the remaining factors that depend only on $ y$ to get

$ p_2(y)\equal{}\frac{e^{\minus{}\mu}}{y!}\,p^ye^{\mu(1\minus{}p)}\mu^y\equal{}\frac{e^{\minus{}\mu p}(\mu p)^y}{y!}$

which is the probability function of a Poisson distribution of mean $ \mu p$ and we are done - and it looks like I made the right correction to your statement.

Thanks for the help Prof Kent Merryfield that makes clear everything.   :)

The $ n$ is also the part Im confuse with. Now I know that it is an error . Thank you very much


trigonometry
integration
geometry proposed
geometry

If $\cot{A}\cot{B}+\cot{B}\cot{C}+\cot{C}\cot{A} = 1$, find $A+B+C$.

Isn't it a well-known formula? It's 180 if I'm not mistaken.

But that formula doesn't tell it's not true for other $A+B+C$...

[quote="levi"]\[A\ +\ B\ +\ C\ =\ k\pi,\ where\ k\in Z\][/quote]

Can you show?

$\sum ctgB.ctgC=1\ \ iff\ \sum tgA=\prod tg A\ \ iff$
$(tgB+tgC)+tgA(1-tgBtgC)=0\ \ iff\ \ \frac{tgB+tgC}{1-tgBtgC}=tg(-A)\ \ iff$
$tg(B+C)=tg(-A)\ \ iff\ \ B+C=k\pi -A\ \  iff\ \ B+C+A=k\pi,\ where\ k\in Z.$

$Ex.: \tan x\cdot \tan (2x)\cdot \tan (3x)=$$\tan (-x)\cdot \tan (-2x)\cdot \tan (3x)=-\tan x-\tan (2x)+\tan (3x).$
$Ex.: \int \tan x\cdot \tan (2x)\cdot \tan (3x)\ dx=\int [-\tan x-\tan (2x)+\tan (3x)]\ dx,\ a.s.o.$


group theory
abstract algebra
calculus
integration
number theory open
number theory

Solve in positive integers:

$x^{2}+x+1=y^{3}$

Rearrange and factor as $x(x+1)=y^{3}-1$

Obviously $y$ is odd, so set $y=2a+1$, giving, $x(x+1)=2a(2a(2a+3)+3)$

$3|x(x+1) \Leftrightarrow 3|a$, so $3\not|a \Leftrightarrow x\equiv 1\mod{3}$

Oh well, maybe that wasn't all that helpful, but I've started somebody's proof possibly.

(18,7) is a solution. Also its easy to see the equation is the same as
4y^3=a^2+3. But this doesn't help any.

[quote="clinkmath"](18,7) is a solution.[/quote]

Indeed, $(x,y)=(18,7)$ is a solution. Moreover we can say that this is the only one solution amoung all numbers $x,y\le1290$.

[quote="Parkdoosung"]Please see http://www.mathlinks.ro/Forum/viewtopic.php?p=224998[/quote]

Can it be useful in our problem?

[quote="ZetaX"]Well, it's a first partial result.
With usage of $\mathbb{Z}[\zeta_{3}]$, this can be reduced to solve $a^{3}+b^{3}-3a^{2}b=1$ (it has a solution $a=3,b=1$). But not sure if this is really easier...[/quote]

I've just seen such a problem:

Show that the only solutions of the equation $x^{3}-3xy^{2}-y^{3}= 1$ are given by $(x; y) = (1; 0); (0;-1); (-1; 1); (1;-3); (-3; 2); (2; 1)$.

I don't know a solution to it, but if we assume it's true then our problem is solved.  :)

I proved it has infinitly many rational solution.
$\frac{2x+1}{2}=p$ : $p^{2}=y^{3}-\frac{3}{4}$.
It is elliptic curve, and there is a rational solution
-> there is infinitly many solution.

Please explain why. For example there are also rational points on the curve $y^{2}=x^{3}+1$ and there are only finitely many of them.

At least its rank is 1 and the torsion subgroup trivial.

[quote="TomciO"]Please explain why. For example there are also rational points on the curve $ y^{2} \equal{} x^{3} \plus{} 1$ and there are only finitely many of them.[/quote]
Torsion points are integral, and the given point isn't.

[quote="conceive"]Solve in positive integers:

$ x^{2} \plus{} x \plus{} 1 \equal{} y^{3}$[/quote]
Letting $ a\equal{}4y, b\equal{}8x\plus{}4$, we have the Mordell curve: $ b^2\equal{}a^3\minus{}48$ whose integral points are $ (a, b) \equal{} (4, \pm 4), (28, \pm 148)$ per a table on Mordell curves. So $ (x, y) \equal{} (18, 7)$ is the only solution to the original problem.

[quote="ZetaX"]With usage of $ \mathbb{Z}[\zeta_{3}]$, this can be reduced to solve $ a^{3} \plus{} b^{3} \minus{} 3a^{2}b \equal{} 1$ (it has a solution $ a \equal{} 3,b \equal{} 1$). But not sure if this is really easier...[/quote]
To solve this Thue equation, try [url=http://www.ams.org/mathscinet-getitem?mr=86d:11023]this article[/url].


[quote="conceive"]Let $a,b>0$. Prove that

$\frac{a+b}{2}-\sqrt{ab}\ge\frac{(a-b)^2(a+3b)(b+3a)}{8(a+b)(a^2+6ab+b^2)}$[/quote]

Let $a=x^2$ ,$b=y^2$
Expand all, we will get :
$(x-y)^8 \geq 0$


There are 100 pennies on a table, 20 of which are heads up. There is an empty table next to it. You are blindfolded, and cannot feel which side of a penny is heads or tails. You can, however, transfer as many pennies from one table to the other as you want, and you can flip any amount of pennies. You must find a way to guarantee that both tables have an equal amount of pennies tails up.

[hide]
grab a coin and flip it and put it on the other table. If it was originally heads, it will add 1 tail to the other table, and if it was originally tails, it will subtract one tail for the first table. Either way, the difference between the number of tails on each table decreases by one. Do this 80 times and the difference will be 0, as desired.
[/hide]

I don't see it happening, OR I am not seeing it (the proposed solution).

Let me reduce the problem in scope in numbers and to a different ratio, but the total number of pennies is even and the total number of heads at the start is even.

Specifically, let the number of pennies be 4 and the number of heads be 2.  If I use 2 moves (or trying more) of tossing a penny, I will get different possible cases of some equal tails on each side (others not) whereby you cannot guarantee equal numbers of tails.

Expanding this to larger totals of pennies that are even and larger numbers of heads that are even, it should not work out for the original problem.

By flip, he means turn over, so if it was heads, it is now tails, and vice-versa.

I didn't have any misunderstanding with THAT part of it.

I'm challenging the solution as being wrong.

It's probably a little simpler explained like this. Choose any $80$ coins, say $m$ are heads and $n$ are tails (originally). So $m+n = 80$. Put them all in the second pile and flip them.

Then there are $80-n$ remaining tails in the first pile.

Also, there are $m$ tails in the second pile. But since $m = 80-n$, there are the same number of tails in both piles.

After the explanation was presented to me, I disagree with

tjhance's solution as being "the one" (my emphasis).

The 80 coins can just as well be moved over to the other table FIRST,

and THEN flipped over afterward to accomplish the same result.

Nobody ever said that tjhance's solution is the only one, I just said it was correct. Transferring and then flipping achieves the same effect, so it also works. Paladin8's solution is complete and should answer all your questions/doubts.


The chart here shows the prices for different seat-types at a stadium for professional soccer.  If every seat in the stadium is sold for a game, what is the average price per seat?

\begin{tabular}{|l|c|c|}\hline
 \textbf{{S}ection} & \textbf{Price} & \textbf{No.\ of Seats Avail.} \\\hline
 West, center & $ \$10$ & $ 1000$ \\\hline
 West, {e}nd & $ \$8$ & $ 1500$ \\\hline
 East, center & $ \$8$ & $ 800$ \\\hline
 East, {e}nd & $ \$6$ & $ 1700$ \\\hline

\end{tabular}

You have to find total money, and divide it by the number of seats ($ 5000$).

$ \$10 \times 1000 \text{ seats } = \$10000$

$ \$8 \times 2300 \text { seats } = \$18400$

$ \$6 \times 1700 \text { seats } = \$10200$

Total money = $ \$38600$

Average money per seat = $ \boxed{\$7.72}$


[quote="Inspired By Nature"]Could someone please explain with an example what a partition is? Any help would be greatly appreciated.[/quote]

Partitions are sets of numbers that add up to another number..so partitions of 3 are {3} {2+1} and {1+1+1} and then the function P(n) is the number of partitions in the number, so P(3)=3 since there are 3 partitions.

Could you give me a harder example? I know the basics but I just want to see how they're used.

http://www.virtuescience.com/partitions.html
http://www.math.niu.edu/~rusin/known-math/index/11PXX.html
http://mathworld.wolfram.com/PartitionFunctionP.html

Enjoy! :)


algebra
polynomial
superior algebra
superior algebra unsolved

Let f(x) = x^2 + x + 1 and let g(x) = x^3 + x + 1 be two polynomials over the finite field F_2. Let a be a root of f and b be a root of g. Then f is irreducible over F_2 and g is irreducible over F_2(b). (Checked)

Now consider the extension F_2(a,b) = F_2(a)(b) of F_2, which is of degree 6. Find a basis of F_2(a,b) over F_2 and then determine the minimal polynomial of the element a*b in F_2(a,b) over F_2.

I found a basis {1, b, b^2, a, a*b, a*(b^2)}. Also i obtained the order of a as 3 and the order of b as 7. Hence the order of a*b must be the lcm(3,7) which is 21. I don't know what to do next. Please help! Thanks!

$ g$ is irreducible over $ F_2[b]$ is obviously false.
$ f,g$ are irreducible then $ F_2\subset{F }_2[a,b]$ is of degree $ 6$ and the minimal polynomial of $ ab$ is of degree a divisor of $ 6$.
Let $ ab\equal{}c$. Then $ b\equal{}\minus{}c^3\minus{}1,a\equal{}\minus{}c^5\plus{}c^3\minus{}c^2\minus{}c$ and the minimal polynomial of $ c$ is: $ x^6\minus{}x^4\plus{}2x^3\plus{}x^2\minus{}x\plus{}1$. A basis of $ F_2[a,b]$ is $ \{1,c,c^2,c^3,c^4,c^5\}$.


geometry
3D geometry
sphere
analytic geometry
vector
geometry unsolved

Given a fixed circle S with center O and radius r, we define the inverse of a point P (P !=O) to be the point P' on the OP whose distance from O satisfies equation OP.OP' = r.r

1) Prove that two spheres which touch each other at O invert into parallel planes,

2) Let S', S'' and S''' be three spheres all touching one another. Let Q1, Q2,... be a sequence of spheres touching one another successively and all touching S', S'' and S'''. Prove that, then Q6 touching Q1, so we have a ring of 6 spheres interlocked with the original ring of 3 (hint: invert in a sphere whose center is the point of contact of S' and S'').

Proof for 1:

We have to prove that the normal to these planes are parallel....

We fix the origin at O (0,0,0)..
The centre of both spheres and O lie on a straight line..
So wlog we define our coordinate system such that,
centre of one sphere is (b,0,0) and the other is (a,0,0)...

Inversion about O takes vector Z to (Z/|Z|^2) * a^2
(a -> radius of circle of inversion)

We can write equation of position vector of each point on the sphere is
(b+x)(i) + y(j) + z(k)..(x^2 + y^2 + z^2 = b^2)
It is inverted to
[(1/2b)(i) + (y/2b(b+x))(i) + z/2b(2b+x)(k)]*a^2

Observe that the x- coordinate is constant, which mean that the normal is along X-axis...
Similar procedure proves that x-coordinate is constant for the inverted plane of the other sphere and hence the normal is along X-axis...

So, done...

P.S. We haven't proved that the inversion actually covers all points of the plane.
That can be easily done by choosing any point on the plane,
(1/2b)(i) + l(j) + m(k)

and proving that
y/2b(b+x) = l and
z/2b(b+x) = m has consistent solution for any l,m belonging to R
y/z= l/m = k,
l = + or -k/sqrt(k^2+1) *sqrt(b-x/b+x)
m= + or -1/sqrt(k^2+1) *sqrt(b-x/b+x)
And we can always find such x...and thus y and z...


Hello all, can anyone translate this PDF, i will be eternally grateful.

[spam]
Hmm... that should be [i]grateful[/i], not [i]greatfull[/i].
[/spam]

:) Thanks, I edited. I never used that word in writing.  :)

PS: This was a spam, too :)

Sorry pohoatza i forgot translate for you
If you still want it i will post at here

I'm sorry but I have to say that there really exist too many mistakes in you sentenses,are you a Chinese ?


I need an algebraic way to solve this problem other than guess and check:

The product of 4 consecutive positive integers is 1680.  What is the largest of these 4 integers?

For this one, I don't get it:

Two trains are headed directly towards each other at rates of 60 MPH and 90 MPH.  How many miles apart are they 3 minutes before impact?  Express your answer as a decimal number of miles.

[hide="Product One"]
Let the average of the four integers be $ n$.

Then we have $ 1680 \equal{} \left(n \minus{} \frac32\right)\left(n \minus{} \frac12\right)\left(n \plus{} \frac12\right)\left(n \plus{} \frac32\right)$

$ \equal{} \left(n^2 \minus{} \frac14\right)\left(n^2 \minus{} \frac94\right) < n^2\cdot n^2 \equal{} n^4\implies n^4 > 1680\implies n > \sqrt [4]{1680} > 6.4$.

It is easy to see that $ n \equal{} 6.5$ gives the working solution $ 5\cdot6\cdot7\cdot\boxed8 \equal{} 1680$.

We could use equations for this, but it's not worth it.
[/hide]
[hide="Train One"]
Instead, we imagine that they are starting at the impact point and departing (i.e. "going back in time").

Thus, they are moving apart at a rate of $ \left(\frac {60 \plus{} 90}{60}\right)\frac {\text{miles}}{\text{minute}} \equal{} 2.5\frac {\text{miles}}{\text{minute}}$, and our answer is

$ 3\,\text{minutes}\cdot2.5\frac {\text{miles}}{\text{minute}} \equal{} \boxed{7.5\,\text{miles}}$.
[/hide]

[hide="alternate solution to the product one"]
$ 1680\equal{}10\cdot 168$
$ \equal{}10\cdot 6\cdot 28$
$ \equal{}10\cdot 6\cdot 4\cdot 7$
$ \equal{}2\cdot 5\cdot 6\cdot 4\cdot 7$
$ \equal{}5\cdot 6\cdot 7\cdot \boxed{8}$
The idea is that you want to get the numbers closer to each other in the product.
[/hide]