[b]Hey everyone ..
i need your help in finding a good idea for a graduation project ..
my school makes my life complicated so i need to make a project that would shut them up :P
i would be thankful if you can relate it with mathematics .
thanks in advance guys :) [/b]

Would giving a presentation on the history of mathematics?
Where it came from... the founders and such?

It would definitely help if you gave some guidelines about this project. Like, is it a research project? A demonstration? A proof that you learned something throughout the years?

no .. that wouldn't be appreciated by my school !!
usually students do projects in how develop a part of the city .. with full designing..
i meant by this example .. that our school needs something unique .. or unusual ,
so the project needs to be beneficial for the society or let's say the society of the students ..
i thought to make a pack of visual proofs .. designed by adobe flash ..
but really need better ideas .

[quote="BadrGhamdi"]
usually students do projects in how develop a part of the city .. with full designing..
[/quote]
Really? I doubt any of them is able to design a single traffic light that would work properly :P (I have to confess that I'll not be able to do it either but at least I don't pretend that I can do things I know next to nothing about).

Moral: stick to what you can do well or can learn in some reasonable time. Don't try to impress other people with "great designs" that will pop up like soap bubbles at the first attempt to implement them. And, above all, change your attitude. An attempt to do something with the main purpose to "show them" and to "make them shut up" has never resulted in anything good for either the person who was making such attempt, or anybody else. If you want to do something useful for the school and have proper skills, go to your teachers and ask them how your skills can be put at work. Don't look down at "small" things. There is infinitely more value in, say, competent decorating a classroom for Christmas than in making an incompetent project for the lunar base (not because Christmas decorations have so big value, but merely because the ratio in question has strictly positive numerator and zero denominator).

If you pick up something math. related and feel like you need to ask for help with something specific, this forum may be the right place to ask your questions. As to the question you posed, I doubt anybody can really answer it without knowing what you are really good at and what the society/school/whatever community you want to make your project for really needs.

At last, try to use normal grammar and fonts in your posts. :).

[quote="fedja"][quote="BadrGhamdi"]
usually students do projects in how develop a part of the city .. with full designing..
[/quote]
Really? I doubt any of them is able to design a single traffic light that would work properly :P (I have to confess that I'll not be able to do it either but at least I don't pretend that I can do things I know next to nothing about).

.[/quote]

if you can't .. that doesn't mean other people can't too !
and i'm serious , they design it ..
for example , some students worked to make a community service center ..
they designed the structure of the building and of the employees ..
maybe they failed at something , but they learned ..

 [quote]
Moral: stick to what you can do well or can learn in some reasonable time. Don't try to impress other people with "great designs" that will pop up like soap bubbles at the first attempt to implement them. And, above all, change your attitude. An attempt to do something with the main purpose to "show them" and to "make them shut up" has never resulted in anything good for either the person who was making such attempt, or anybody else. If you want to do something useful for the school and have proper skills, go to your teachers and ask them how your skills can be put at work. Don't look down at "small" things. There is infinitely more value in, say, competent decorating a classroom for Christmas than in making an incompetent project for the lunar base (not because Christmas decorations have so big value, but merely because the ratio in question has strictly positive numerator and zero denominator).
[/quote]

i don't agree with you , it's a graduation project .. i "have"to do it .. i dont "love" to do it .. so if you can help me by giving a nice idea , please do .. or you can't help me any other way .

[quote]If you pick up something math. related and feel like you need to ask for help with something specific, this forum may be the right place to ask your questions. As to the question you posed, I doubt anybody can really answer it without knowing what you are really good at and what the society/school/whatever community you want to make your project for really needs.[/quote]
sure,

[quote]
At last, try to use normal grammar and fonts in your posts. :)[/quote]
ok :)

and thanks for the reply

[quote="BadrGhamdi"]
 [quote]
Moral: stick to what you can do well or can learn in some reasonable time. Don't try to impress other people with "great designs" that will pop up like soap bubbles at the first attempt to implement them. And, above all, change your attitude. An attempt to do something with the main purpose to "show them" and to "make them shut up" has never resulted in anything good for either the person who was making such attempt, or anybody else. If you want to do something useful for the school and have proper skills, go to your teachers and ask them how your skills can be put at work. Don't look down at "small" things. There is infinitely more value in, say, competent decorating a classroom for Christmas than in making an incompetent project for the lunar base (not because Christmas decorations have so big value, but merely because the ratio in question has strictly positive numerator and zero denominator).
[/quote]

i don't agree with you , it's a graduation project .. i "have"to do it .. i dont "love" to do it .. so if you can help me by giving a nice idea , please do .. or you can't help me any other way .

[/quote]

If you don't love to do it, then there is no way that you are going to make it so spectacular that it'll "shut them up." You're apathy for the project will certainly show.
And I am still confused on the guidelines. You mentioned a project that benefits society, but what exactly are the parameters and such?

So you just have to design something? (Don't get me wrong, designing is not easy, but it's definitely more doable than also building whatever you design)

The actual building process doesn't seem involved with your example.

[quote="fedja"][quote]
if you can't .. that doesn't mean other people can't too ! and i'm serious , they design it .
[/quote]

Well, OK. Then try to design a traffic light on a simple crosswalk. You know, the one that goes red-green-yellow-red for the cars and green-flashing red-red-green for the pedestrians. The task is to minimize the mean time wasted on the intersection or some other reasonable functional of similar nature. You have also to guarantee that there are no accidents unless someone violates the rules. Your task is to tell the overall period and the timings of the signals within the period. To figure out what data are relevant to the question and how to get them is a part of the problem (all you will need can be observed in almost any city with a piece of paper, a pencil, a clock and a measuring tape; you may look quite funny to bypassers standing on the intersections and making notes, of course, but it is a part of the job :P).

To be sure, this has most likely been done before (but so were community center projects and other things you mentioned). It is an interesting mixture of pure mathematics with real process modelling, certainly doable within a semester time if you really put your effort into it. Once you get your theory, go around the city and
see if your answers agree with how the actual lights work.

If you do not like this particular project, it can still give you an idea: just look at anything around you and ask why it works the way it does and what will happen if you change some parameters. That's how many of the serious "applied" math. projects started. Another way is just to talk to engineers and see what they are doing and what gives them a headache. Occasionally they even come to AoPS fora to ask their questions though it didn't happen often lately.[/quote]

thanks ,, that really looks interesting ..
but how about making a timing for the traffic lights just to decrease the traffic ..
you impressed me .. :)

[quote]
If you don't love to do it, then there is no way that you are going to make it so spectacular that it'll "shut them up." You're apathy for the project will certainly show.
And I am still confused on the guidelines. You mentioned a project that benefits society, but what exactly are the parameters and such?[/quote]
i don't like doing projects in general .. unless they are interesting
and that's why i came here to ask
i want an idea that i can do it with my full desire ..
and because i love math .. i wanted to do something related to math ..
but i didn't find that fascinating idea .. :)

[quote="BadrGhamdi"]
thanks ,, that really looks interesting ..
but how about making a timing for the traffic lights just to decrease the traffic ..
[/quote]
The intersection of two roads light is a harder problem with more parameters involved, but you can try that too. The standard pattern then is red-left green-left yellow-straight green-straight yellow-red (but there are other patterns that are used as well). In addition to finding the signal timing, you may ask here how long the left turn pockets should be, etc.

It is interesting that many today's trafic lights have cameras and can react to the actual traffic on the road. So far they change from their standard patterns to custom ones only in very simple situations when the right decision is a "no-brainer" but the trend is there.

Needless to say, finding the optimal regime for a real road network with many lights is a total nightmare. There are thick books on traffic control theory but it is still an art as much as a science.

[color=green]Moderator says: there is never a good reason to block-quote the entire post immediately preceding your own.  (You are responding to it by default.)[/color]

nice information .. thanks :)


Let $\mathcal{D}$ be the set of positive divisors of a given natural number $n$.

Show that $n = \sum_{d \in \mathcal{D}}\varphi(d)$.  (We define $\varphi(1) = 1$.)

Nice proof is given [url=http://mcraeclan.com/mathhelp/BasicNumberCoprimesTotientFunction.htm]here[/url]

It's a nice proof, but it lacks generality.  Here is the one-liner I used at PROMYS, entirely algebraic ;)

[hide="Beautiful"] $n = p_{1}^{e_{1}}p_{2}^{e_{2}}... = \left( \varphi(p_{1}^{e_{1}})+\varphi(p_{1}^{e_{1}-1})+...+\varphi(1) \right) \left( \varphi(p_{2}^{e_{2}})+\varphi(p_{2}^{e_{2}-1})+...+\varphi(1) \right) ...$ [/hide]
Edit:  I say that this approach is more general because it suggests the following very interesting property:

[hide="1"] Let $f(k)$ be an arithmetic function.  We say that $f(k)$ is "multiplicative" if $gcd(a, b) = 1 \implies f(ab) = f(a) f(b)$.

Say that $f(k)$ is multiplicative.  Then $F(k) = \sum_{d | k, d > 0}f(k)$ is also multiplicative. [/hide]
Which is in turn a special case of this:

[hide="2"] Let $f(k), g(k)$ be two multiplicative functions.

Then $H(k) = \sum_{d | k, d > 0}f(k) g( \frac{d}{k})$ is also multiplicative.  This is known as the Dirichlet convolution of $f, g$. [/hide]


LaTeX
geometry
3D geometry
sphere
analytic geometry
graphing lines
slope

okay

2 to the 2006th power divided by 7 equals how much? I don't know about this one. :o

Um........

The number, or the remainder?

Remainder:

Take mod 7. Powers of 2 mod 7 = 2, 4, 1, 2, 4, 1...

2006 (mod 3) = 2. So, remainder is 4...

(apologizes for lack of LaTeX)

yeah i meant the remainder . its really 4? yay! thats what i got, yet I seriously doubted myself

okay how about this problem.

A chord 16 cm long is 15 cm from the center of a circle. What is twice the difference of the area of the circle and its curcumference to the nearest tenth?

ooh nice one,,, i like it.

anyone know this one? A sphere with radius 3 inches in made into 72 identical disks that are circular, each with radius 2 in. How many inches thick is each disk?

USE ANALYTIC GEO!!!

WOOT!

(Too lazy to post answer)

[quote="miyomiyo"]This is a thread for math team people to post, explain, talk about, think about, question about, expand, and do math league problems.

Starting with - I don't know if anybody else got this problem - the mandelbrot problems with taking out powers of 2 and stuff.[/quote]

yeah i know what your talking about. it was kind of hard.

[quote="now a ranger"]anyone know this one? A sphere with radius 3 inches in made into 72 identical disks that are circular, each with radius 2 in. How many inches thick is each disk?[/quote]
[hide] The volume of the sphere is $36\pi$. If the height of the disks is $h$, the volume is $72\cdot4h\pi$. $h=\frac{1}{8}\text{ inch}$ [/hide]

[quote="not_trig"]Diameter splits chords in 2 when perpendicular.

$16^{2}=15x \Rightarrow x=256/15$

where $x$ is the distance from the intersection point to the place where the diameter, when extended, hits the circle. (Whew.)

So, $r=\frac{256+225}{15}= \frac{481}{15}$...

And the rest is obvious[/quote]
I think you made a mistake.
[hide]The shortest segment from the center to the chord is a perpendicular bisector. The distance is 15 and half the chord is 8, giving an 8-15-17 triangle. So the radius is 17...[/hide]

Given ABC triangle with points A (-4,0) B (8,0) C (0,10)

If the equation of the altitude from side BC is written in standard form Ax+By=C, what is A+B+C?
show methods and steps

[hide="Solution"]The equation of the line $\overline{BC}$ is $y=-\frac{5}{4}x+10$. The altitude connecting $A$ to $BC$ has to be perpendicular to $BC$. Thus the slope of the altitude is $\frac{4}{5}$. Substituting, we get $b=\frac{16}{5}$. So the altitude is $y=\frac{4}{5}x+\frac{16}{5}$. Or in standard form, $4x-5y=-16$. $A+B+C=-17$, if I didn't make mistakes...[/hide]


calculus
integration
function
LaTeX
calculus computations

For what value of K>0, does S(4kx-5k)dx=x^2 ?
S(4kx-5k)dx is an integral

There is no such number, but there is a function $ k(x)$ with that property:

$ \int(4k(x)x\minus{}5k(x))dx\equal{}x^2$

$ \left(\int(4k(x)x\minus{}5k(x))dx\right)^{'}\equal{}2x$

$ 4k(x)x\minus{}5k(x)\equal{}2x$

$ k(x)\equal{}\frac{2x}{4x\minus{}5}$

I'm sorry I typed in the question incorrectly.
It should be....

For what value of K>0, does S(4kx-5k)dx= [b]k^2[/b]?

By the way how do you type the mathematics out like the post above this one?

As written, that's nonsense. The indefinite integral of a nonzero function can't be a constant.

You probably mean to ask about a specific definite integral- so what are the limits?

The [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]pretty math[/url] on this forum is $ \text{\LaTeX}$.


Sally has a digital alarm clock that automatically resets to 12:00 midnight when there is a power outage. When she returned home from school, she discovered the power had been out, but was on again. Her clock read 4:13 a.m. when the actual time was 3:10 p.m. At what time did the power return? (Assume she left for school at 7:50 a.m.)

so 15 hours and 10 minutes had elapsed.

and 4 hours and 13 minutes had elapsed since the outage.

15:10-4:13=14:70-4:13=10:57

answer : 10:57


A circle with radius $ 5$ and center $ (a,b)$ is tangent to the lines $ y\equal{}6$ and $ y\equal{}0.75x$. Compute the largest possible value of $ a\plus{}b$.

[hide="solution"]$ a \plus{} b$ will be maximized if the line $ y \equal{} 6$ is tangent to the bottom of the circle and if the lines $ y \equal{} 6$ and $ y \equal{} \frac {3}{4}x$ intersect to its left. This intersection occurs at $ (8,6)$. We will refer to this point as $ C$.

We draw the given circle and its two tangents in the manner described above. Call the pts. of tangency of $ y \equal{} \frac {3}{4}x$ and $ y \equal{} 6$ $ D$ and $ E$ respectively. Draw a perpendicular to $ y \equal{} 6$ at $ E$ through center $ O$ to a point $ F$ on $ y \equal{} \frac {3}{4}x$. Draw a perpendicular to $ y \equal{} \frac {3}{4}x$ at $ D$ through center $ O$ to a point $ G$ on $ y \equal{} 6$. Note that $ CDO$ and $ CEO$ are congruent and $ CFO$ and $ CGO$ are congruent as well.

Using the slope of line $ DO$ and the radius, we can calculate $ EG$ and $ DF$ to be $ \frac {15}{4}$.

$ CF^2 \equal{} CE^2 \plus{} EF^2$
$ EF \equal{} \frac {3}{4}CE$ so $ CF^2 \equal{} CE^2 \plus{} \frac {9}{16}CE^2 \equal{} \frac {25}{16}CE^2$
$ CF \equal{} \frac {5}{4}CE$

We can calculate $ OG$ and $ OF$ to be $ 25/4$ with Pythagorean.

We can now also use the Pythaogrean theorem on triangle $ CEF$ to calculate the center as $ (8 \plus{} 15,6 \plus{} 5)$ or $ (23,11)$

The largest possible value for $ a \plus{} b$ is $ 34$.[/hide]


[quote="Jutaro"]For every positive integer $n$ we define $a_{n}$ as the last digit of the sum $1+2+\cdots+n$. Compute $a_{1}+a_{2}+\cdots+a_{1992}$.[/quote]

Let $S_{n}=1+2+\cdots+n=\frac{n(n+1)}{2}$. It's immediate to see that $S_{n+20}=S_{n}$ (mod $10$).
So $a_{1}+a_{2}+\cdots+a_{1992}=99(a_{1}+a_{2}+\cdots+a_{20})+a_{1}+a_{2}+\cdots+a_{12}=6984$.

--
Patrick

We try to find a pattern by listing the last digit of the first sums, and notice that $a_{20}=0$, so beyond that point the cycle will repeat every 20 numbers and the sum of each cycle is equal to 70, thus $a_1 +a_2+...a_{1980}=99 \cdot 70$ adding the last 19 numbers we get that the sum is equal to  $99 \cdot 70 +54=6984$

Also 2017 AIME I #3 except slightly different. :wink:

I participated at this event in Venezuela competing on behalf of Puerto Rico.  A lot of people got a score of 8 out of 10 on this problem including myself because although my answer was correct, they took 2 points of because like all of the solutions posted here, merely showing that the pattern repeats after 20 was not enough.  "see" that the pattern repeats, or explaining that it repeats after 20 is not sufficient proof according to the graders.  All of the answers provided above would have also gotten 8 out of 10.  I've included the proof here:
https://artofproblemsolving.com/wiki/index.php/1992_OIM_Problems/Problem_1


calculus
integration
trigonometry
logarithms
calculus computations

i)
$ \int_0^{\frac {\pi}{3}} \tan x \ln \cos x\, dx$

$ \int_0^{\frac {\pi}{3}} \frac {\sin x}{\cos x} \ln \cos x\, dx$

Use the substitution $ u \equal{} \cos x$

-----

ii)
$ \int \frac {\cos^3 x}{\cot^2 x}\, dx$

$ \int \cos x \sin^2 x\, dx$

Obviously, use the substitution $ u \equal{} \sin x$

-----

iii)
1- $ \tan^2 \frac {x}{2} \equal{} \frac {1 \minus{} \cos x}{1 \plus{} \cos x}$
2- if you can't see that identity, multiply both numerator and denominator by $ 1 \minus{} \cos x$ to get the form $ \int \frac{(1\minus{}\cos x)^2}{\sin^2 x}\,dx$

It's easy to finish using one of the ways above.

-----

iv)
$ \int \frac {x^3}{(1 \plus{} x)^6}\, dx$

Substitute $ u \equal{} 1 \plus{} x$, then expand and integrate term by term.

[hide="Number 1"]$ \int_0^{\pi/3} \tan x \ln(\cos x)\,dx \equal{} [\minus{} \ln(\cos x) \cdot \ln(\cos x)]_0^{\pi/3} \minus{} \int_0^{\pi/3} [\minus{} \ln(\cos x)] \frac{\minus{} \sin x}{\cos x}\,dx$

$ \equal{}> I \equal{} [\minus{} \ln^2(\cos x)]_0^{\pi/3} \minus{} I \equal{}> 2I \equal{} \minus{} \ln^2 \left(\cos \frac{\pi}{3}\right) \equal{} \minus{} (\ln 2)^2$

$ \equal{}> I \equal{} \minus{} \frac{(\ln 2)^2}{2}$.[/hide]

hello, to number one:
By substituting $ u \equal{} \ln(\cos(x))$ we get: $ du \equal{} \minus{}\tan(x)dx$ and we have:
$ \int_0^{ \minus{} \ln(2)}udu \equal{}  \frac {1}{2}\ln(2)^2$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, to number one:
By substituting $ u \equal{} \ln(\cos(x))$ we get: $ \color{red}du \equal{} \tan(x)dx$ and we have:
$ \int_0^{ \minus{} \ln(2)}udu \equal{} \minus{} \frac {1}{2}\ln(2)^2$.
Sonnhard.[/quote]
Two typos.
First is if $ u\equal{}\ln \cos x$, then $ du \equal{} \color{red}\minus{}\color{black} \tan x$.
Second is, $ \int_0^{ \minus{} \ln(2)}udu \equal{} \color{red}\plus{}\color{black} \frac {1}{2}\ln(2)^2$ which is not the correct answer because of the 1st typo.

hello, thank you for the hint, i have it corrected.
Sonnhard.

hello, $ \int\frac{\cos^3(x)}{\cot^2(x)}dx\equal{}\int \cos(x)\sin^2(x)dx$.
By substituting $ t\equal{}\sin(x)$ we have $ dt\equal{}\cos(x)dx$ and so we get
$ \int t^2dt\equal{}\frac{t^3}{3}\plus{}C\equal{}\frac{\sin^3(x)}{3}\plus{}C$.
Sonnhard.

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

@wingless: in proving the identity $ \tan^2(x/2) \equal{} \frac{1\minus{}\cos x}{1 \plus{} \cos x}$, I don't see how multiplying top and bottom by $ 1\minus{}\cos x$ simplifies the problem; one can alternately use the identity $ \cos x \equal{} 1 \minus{} 2\sin^2 x \equal{} 2\cos^2 x \minus{}1$ to solve the problem.


[quote="Laputa"]Prove that:

1) |(a-b)/(a+b) + (b-c)/(b+c) + (c-a)/(c+a)| < 1/8 (| means the absolute value)
     with a,b,c are three sides of a triangle

2) With a,b,c > 0 and a+b+c <= 1
     (1+1/a)(1+1/b)(1+1/c) >= 64[/quote]

1)
$|(a-b)/(a+b) + (b-c)/(b+c) + (c-a)/(c+a)|=\br |\frac{(a-b)(b-c)(c-a)}{(a+b)(b+c)(c+a)}|<\frac{abc}{(a+b)(b+c)(c+a)}<\frac{1}{8} \br$

because $|a-b|<c$ (triangle ineq) and $(a+b)(b+c)(c+a)\geq 8abc$, which is easy to prove.

2)
Using Holder's ineq or whatever you get $(1+\frac{1}{a})(1+\frac{1}{b})(1+\frac{1}{c})\geq(1+\frac{1}{\sqrt[3]{abc}})^3\geq (1+\frac{3}{a+b+c})^3\geq (1+3)^3=64$
taking into account AM-GM...

I hope you liked my approach!!  :)  :D

BTW learn to type $\text{\LaTeX}$...

How do you type | ??

Can you show me how to prove that

(a+b)(b+c)(a+c) >= 8abc  ?

And is there an english name for holder inquality? (I think it's Cauchy).

Thanks

If you want to type |, hold Shift and hit the key on top right of Enter.

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

   Let $a, b, c>0$. Prove the following inequality: \[ (a+b)(b+c)(c+a)\geq 8abc  \].
[/color]

[i] Solution [/i]

  Let us apply AM-Gm inequality: $a+b\geq 2\sqrt{ab}$. now by writing the analogues ones and by multiplying these three inequalities, we get the inequality given. :)   ;)

I didn't find much when searching the web for Huygens.  Any one know its English name?

Just use | to type $|$ in $\LaTeX$.  If you have something that is tall (like a fraction) you'll want to put \left before the first | and \right before the second | to make it look nicer.  Example:  \[ \left|\frac{x-1}{x+1}\right| \]

I'm pretty sure that the english name for Holder is, er...well Holder.  It's not Cauchy since Holder is a generalization of Cauchy.

[quote="cezar lupu"]And by the way, Socrates it's not Holder inequality it is Huygens. ;)[/quote]

Cezar thanks for the info.

I believe that Huygen's ineq is a particular case of Holder's one.

However, you know much more than me.... ;)


Let $ a,$ $ b$ and $ c$ are non-negative numbers for which any two of them are not equal to zero
and $ a^2\plus{}b^2\plus{}c^2\equal{}2(ab\plus{}ac\plus{}bc).$ Prove that
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{a\plus{}c}\plus{}\frac{c}{a\plus{}b}\geq2\]

[quote="arqady"]Let $ a,$ $ b$ and $ c$ are non-negative numbers for which any two of them are not equal to zero
and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2(ab \plus{} ac \plus{} bc).$ Prove that
\[ \frac {a}{b \plus{} c} \plus{} \frac {b}{a \plus{} c} \plus{} \frac {c}{a \plus{} b}\geq2
\]
[/quote]

[hide="approach"]
try adding 3 to both sides
$ \frac {a \plus{} b \plus{} c}{b \plus{} c} \plus{} \frac {a \plus{} b \plus{} c}{a \plus{} c} \plus{} \frac {a \plus{} b \plus{} c}{a \plus{} b}\geq5$

i'll edit if i think of something else

[/hide]

edit: i'm being stupid again, aren't i.  see below.
[hide]
$ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ac}\equal{}2$
$ \frac{a^2}{ab\plus{}bc\plus{}ac}\le\frac{a^2}{ab\plus{}ac}$
$ \frac{a^2}{ab\plus{}bc\plus{}ac}\le\frac{a}{b\plus{}c}$
repeat for other 3 terms, k

[/hide]

Use Cauchy to get $ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq \frac{(a\plus{}b\plus{}c)^2}{a(b\plus{}c)\plus{}b(c\plus{}a)\plus{}c(a\plus{}b)}$. Given $ a^2\plus{}b^2\plus{}c^2\equal{}2(ab\plus{}bc\plus{}ca)$, the right hand side equals 2.

Yes, scorpius119! It's also my proof. The Rust's proof:
$ \sum_{cyc}\frac{a}{b\plus{}c}\equal{}\sum_{cyc}\frac{a^2}{ab\plus{}ac}\geq\sum_{cyc}\frac{a^2}{ab\plus{}ac\plus{}bc}\equal{}2.$ :lol:


Let $ a,b,c$ be three positive real numbers such that : $ \frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ca}\plus{}\frac{c}{1\plus{}ab}\ge 3$ . Find the minimum value of :

$ P\equal{}\frac{a}{a\plus{}bc\plus{}1}\plus{}\frac{b}{b\plus{}ca\plus{}1}\plus{}\frac{c}{c\plus{}ab\plus{}1}$

We'l prove that the minimum of $ P$ is $ 3/4$.
If one of the terms, say $ \frac {a}{a \plus{} bc \plus{} 1}$ is greater than or equal to $ 3/4$, then the inequality is proven. So assume $ \frac {a}{a \plus{} bc \plus{} 1} < \frac {3}{4} \Leftrightarrow 3(bc \plus{} 1) > a$ and likewise $ 3(ca \plus{} 1) > b$ and $ 3(ab \plus{} 1) > c$. But then we have $ \frac {a}{a \plus{} bc \plus{} 1} \ge \frac {a}{4(1 \plus{} bc)} \Leftrightarrow a(3(bc \plus{} 1)\minus{}a) \ge 0$, which is true. Hence the inequality follows.

But the problem suggests that $ a,b,c$ be positive reals .
 :surf:

Not again. If $ a$, $ b$, $ c$ are positive reals, then there is no minimum, but you can get infinetely close to $ 3/4$.

I meant $ a,b,c$ are non-negative numbers .

Felix, I agree with you, it gets closer to $ \frac{3}{4}$ doesn't mean that the minimum exists and I just confuse that the problem should state clearler or otherwise there will be no equality case.


1. For any set S of five points in the plane, no three of which are colinear. Let M(S) and m(S) denote the greastest and smallest areas, respectively, of triangles determined by three points from S.
Find the minimum possible value of M(S)/m(S).

Interesting problem!
I think that it woudl be a regular pentagon.  Then it would be (tan72)/2(sin72)
which is about 1.618034..
I'm guessing that's right...regular pentagon seems logical, after the consluion that it should be a convex polygon.  But it's still a guess...correct me if i am horribly wrong


A world-class sprinter accelerates to his maximum speed in 4.0s. He then maintains this speed for the remainder of a 100-m race, finishing with a total time of 9.1s.

a) What is the runner's average acceleration during the first 4.0s?

b) What is his average acceleration during the last 5.1s?

c) What is his average acceleration for the entire race?

For some reason, every time I do this problem I get an answer different from the one in the answer key. It's not even a hard problem... :huh:

Well, is the runner accelerating uniformly?

Else, we need some function to find the velocity against time during the first 4 seconds.

If he accelerates uniformly then:
s = (0.5)(4s)(v) + (5.1s)(v) = 100m
.'. v = 100m / (7.1 s) = 14.08m/s
hence :
a(4s) = (v-u)/4s = 3.52 m.s^(-2) in the direction of motion

a (5.1s) = (v-u)/5.1s = 0

a (total) : a = (v^2 - u^2)/ 2s = 0.991 m.s^(-2) in the direction of motion


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

If I am not wrong,
$x^n - a^n = (x-n)(x^(n-1)a+.......+xa^(n-1)$

Using this the problem can be solved.

this is the same thing as $1+5^{25}+5^{50}+5^{75}+5^{100}$ so at least it is an integer...

Write $x=5^{25}$ and simplify a bit. Now show its composite (You will need the fact that $5/x$)

Here's another dumb method:
Let$\frac{5^{125}-1}{5^{25}-1}=p$(p is a prime number except 2 since it's obviously not)
${5^{125}-1}=5^{25}p-p$
${5^{125}-1}=5^{25}p-(p-1)-1$
but $5^{25}p-(p-1)$ is not equal to$5^{125}$ since it is not a multiple of 5
Therefore${5^{125}-1}\neq5^{25}p-(p-1)-1$ and it contradicts to the last equation.
Therefore$\frac{5^{125}-1}{5^{25}-1}=$composite number.

[quote="Eru"]
but $5^{25}p-(p-1)$ is not equal to$5^{125}$ since it is [b]not a multiple of 5[/b][/quote]

Why not?   say $p=31$ the expression above is in fact a multiple of 5. ..
(BTW - If I am not missing something obvious NONE of the posts above [b]prove [/b] that the fraction is indeed a composite number, please check and make sure that there are no "holes" in the proof.

Also, IMO  the  problem should probabbly go in the advanced topics because, again I think, it is not that easy.
[hide=" Here   are  the two factors"]$(5^{50}+5^{38}+3\cdot 5^{25} +5^{13}+1)(5^{50}-5^{38}+3\cdot 5^{25} - 5^{13} +1)$[/hide]

$x=5^{25}$

$\frac{x^{5}-1}{x-1}=x^4+x^3+x^2+x+1=(x^2+3x+1)^2-5x(x+1)^2$

At $x=5^{25}$ this becomes the difference of two squares, both parts greater than 1.

Yes  that was the idea behind the above tow factors ... Also see a similar problem and discussion ... [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=x%5E2%2B3x%2B1&t=21174]Here[/url]


I don't see how all the USAMTS latex downloads interact with ech other.  I've saved all of them in the same file.. woo.  So, what do I do now?  I tried reading the instructions in the forms, but it doesn't tell you how to put them with the other two.  So, could someone explain how I actually fill out the Latex forms?

Mr.Ocax

I'm not exactly sure what your question is, but I'll give it a shot...

Assuming you're trying to write up the solutions, you need to download 3 files, and they need to remain separate files: usamts.tex, solutions.tex, and fancyhdr.sty.  (There are the three links at the bottom of [url]http://www.usamts.org/MyUSAMTS/U_MyForms.php[/url].)  They should all be in the same directory.  The only one you should edit is solutions.tex.

Open up solutions.tex using your favorite editor (such as TeXnic Center), and follow the instructions in the file.  Note that until you actually type in a solution, you'll get blank output.

So we don't need to do anything with the other 2 files?

That is correct.

Must we submit a file named solutions.tex? Currently, I am using that as a template so I will not have to download again in the future. My actual work is in another file, although it would be no inconvenience at all just to move it all back...

[quote="towersfreak2006"]Must we submit a file named solutions.tex? Currently, I am using that as a template so I will not have to download again in the future. My actual work is in another file, although it would be no inconvenience at all just to move it all back...[/quote]It doesn't matter -- if you want to rename the file, that's fine.


For a positive number $ x$, we know $ (x^2)^3 \equal{} \frac{1}{64}$. What is the value of $ x$? Express your answer as a common fraction.

We know that x^2^3 is the same as x^6 and we know that 2^6=64 so its just 1/2


Let $ a,b,c$ be the sides of a triangle .Prove that:
$ \frac{a}{\sqrt{5a^2\plus{}13bc}}\plus{}\frac{b}{\sqrt{5b^2\plus{}13ca}}\plus{}\frac{c}{\sqrt{5c^2\plus{}13ab}} \ge \frac{1}{\sqrt{2}}$
 :)

Rewrite the inequality as
$ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} \lambda bc}} \ge \frac {3}{\sqrt {1 \plus{} \lambda }},$
where $ \lambda \equal{} \frac {13}{5}$.
That's a little generalisation of IMO01/2. For $ 0 < \lambda < 8$ the inequality is valid only for the sides of a triangle. For $ \lambda \ge 8$, the inequality is true for all positive reals $ a,b,c$.  :)

edit: it's false to say only for the sides of a triangle, but you'll get some trouble when for example $ a\equal{}1, b\equal{}c\equal{}\frac{1}{n}, n \to \infty$.

[quote="Mathias_DK"][quote="quykhtn-qa1"][quote="FelixD"]Rewrite the inequality as
$ \sum_{cyc} \frac {a}{\sqrt {a^2 \plus{} \lambda bc}} \ge \frac {3}{\sqrt {1 \plus{} \lambda }},$
where $ \lambda \equal{} \frac {13}{5}$.
That's a little generalisation of IMO01/2. For $ 0 < \lambda < 8$ the inequality is valid only for the sides of a triangle. For $ \lambda \ge 8$, the inequality is true for all positive reals $ a,b,c$.  :)

edit: it's false to say only for the sides of a triangle, but you'll get some trouble when for example $ a \equal{} 1, b \equal{} c \equal{} \frac {1}{n}, n \to \infty$.[/quote]
this inequality is true :wink: with $ a,b,c$ be the side of a triangle[/quote]
You really need to learn proper english! He didn't say otherwise. He said that the inequality may hold for $ a,b,c$ where $ a,b,c$ aren't sides of a triangle :roll:

Btw. It can be solved the same way as IMO01/2.. By Jensen/Cauchy-Schwartz we know that $ LHS \ge \frac {(a \plus{} b \plus{} c)^{3/2}}{\sqrt {5a^3 \plus{} 5b^3 \plus{} 5c^3 \plus{} 39abc}} \ge \frac {1}{\sqrt {2}} \iff$
$ 2(a \plus{} b \plus{} c)^3 \ge \sum_{cyc} 5a^3 \plus{} 13abc \iff$
$ 3(b \plus{} c \minus{} a)(a \minus{} b)(a \minus{} c) \plus{} 3(3a \minus{} b \minus{} c)(b \minus{} c)^2 \ge 0$. Which is obvious when you assume $ a \equal{} max \{ a,b,c \}$[/quote]
Nice solution, :lol:
After use Cauchy- Schwarz,we must prove that: $ 2(ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)) \ge a^3\plus{}b^3\plus{}c^3\plus{}9abc$.
It very easy and nice :)

[quote="quykhtn-qa1"]
 $ 2(ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a)) \ge a^3 \plus{} b^3 \plus{} c^3 \plus{} 9abc$.
It very easy and nice :)[/quote]
It's my old problem for 2 years ago. See it here: (problem 12)
http://www.maths.vn/forums/showthread.php?t=9244


quadratics
Gauss
polynomial
function
modular arithmetic
symmetry
Euler

[quote="Omid Hatami"]I know that Gauss gave 8 proofs for Quadratic Reciprocity law, and there has been over 200 proofs for this theorem. Does anybody know all of the Gauss's proofs, or a place where I can find other proofs.
I know a few proofs by Gauss Sums, Gauss Lemma, a proof by Eisenstein, a Fourier Analysis one, etc.[/quote]
I know two, I know one by Eisenstein - the geometric proof. And I know one using finite fields.

[url=http://www.rzuser.uni-heidelberg.de/~hb3/fchrono.html]here[/url] there is a source to find most of the proofs.

One of my favorite proofs, using symmetric functions:

Let $ \omega\equal{}e^{\frac{2\pi i}{p}}$ and $ \delta_p\equal{}(p\minus{}1,p\minus{}2,\dots,1,0)$. First note that for any partition $ \lambda$ of length $ \le p$ we have $ s_{\lambda}(1,\omega,\dots,\omega^{p\minus{}1})\in \{\minus{}1,0,1\}$.
Now since $ s_{\lambda}(1,\omega,\dots,\omega^{p\minus{}1})\equiv s_{\lambda}(1,1,...,1)\pmod{p}$ it follows that for $ \lambda \equal{}(q\minus{}1)\delta_p$ we have $ s_{\lambda}(1,1,...,1)\equal{}\left(\frac{q}{p}\right)\pmod{p}$ and therefore
\[ s_{\lambda}(1,\omega,\dots,\omega^{p\minus{}1})\equal{}\left(\frac{q}{p}\right)\]
Let $ G_p$ (resp $ G_q$) denote the set of complex $ p$th roots of unity (respectively $ q$th roots of unity) and choose $ S\subset G_p,T\subset G_q$ such that
\[ \frac{x^p\minus{}1}{x\minus{}1}\equal{}\prod_{\sigma\in S}(x\minus{}\sigma)(x\minus{}\sigma^{\minus{}1})\]
\[ \frac{x^q\minus{}1}{x\minus{}1}\equal{}\prod_{\tau\in T}(x\minus{}\tau)(x\minus{}\tau^{\minus{}1})\]
We deduce using $ s_{\lambda}(x_1,x_2,\dots,x_p)\equal{}\prod_{i<j}\frac{x_i^q\minus{}x_j^q}{x_i\minus{}x_j}$
\[ \left(\frac{p}{q}\right)\equal{}\prod_{\sigma,\tau}(1\minus{}\sigma \tau)(\tau^{\minus{}1}\minus{}\sigma^{\minus{}1})\]
and by symmetry
\[ \left(\frac{q}{p}\right)\equal{}\prod_{\sigma,\tau}(1\minus{}\sigma \tau)(\sigma^{\minus{}1}\minus{}\tau^{\minus{}1})\]
from which
\[ \left(\frac{p}{q}\right)\equal{}(\minus{}1)^{\frac{(p\minus{}1)(q\minus{}1)}{4}}\left(\frac{q}{p}\right)\]
follows.

Sorry but could you explain why for $ \lambda \equal{} (q \minus{} 1)\delta_p$ we have $ s_{\lambda}(1,1,...,1) \equiv \left(\frac {q}{p}\right)\pmod{p}$ ? Otherwise it looks like a nice proof.

  darij

From
\[ s_{\lambda}(x_{1},x_{2},\dots,x_{p}) \equal{} \prod_{i < j}\frac {x_i^q \minus{} x_j^q}{x_i \minus{} x_j}\]
we get
\[ s_{\lambda}(1,1,\dots,1)\equiv q^{\frac {p(p \minus{} 1)}{2}}\equiv\left(\frac{q}{p}\right)\pmod{p}\]

There is a book dedicated to that subject: Franz Lemmermeyer, Reciprocity laws: From Euler to Eisenstein


real analysis
geometry
3D geometry
calculus
derivative
limit
integration

Suppose $ S\subset\mathbb{R}^2$ measurable such that for every $ x\in S$ we have a sequence $ Q_{\varepsilon_k}(x)$ of cubes of length $ \varepsilon_k>0$ such that $ \varepsilon_k\to 0$ and $ \big|S\cap Q_{\varepsilon_k}(x)\big|\leq \frac{1}{2}\big|Q_{\varepsilon_k}(x)\big|$.

Prove: $ |S|\equal{}0$.

This seems pretty intuitive, but I don't see how to get a handle on it. It seems like you could argue to consider only rationals by epsilon-cube neighborhoods to get only a countable consideration or something. This seems to be completely stumping me.

P.S. It is an old U Texas prelim exam question.

I assume that $ Q_{\epsilon_k}(x)$ contains $ x$ - otherwise this makes no sense. (Perhaps the cube is even centered at $ x$ - a convenience, but we don't have to have that.)

I don't think that any naive and intuitive approaches are actually going to arrive at a valid proof. You need some serious tools.

This is a consequence of the Lebesgue Theorem on the Differentiation of the Integral. That's a big theorem, and its proof requires a covering lemma (Vitali, or something weaker than Vitali, depending on your proof - but you need something.)

The theorem states that for locally integrable $ f,$

$ \lim_{\epsilon\to0}\frac1{m(Q_{\epsilon}(x))}\int_{Q_{\epsilon}(x)}f\equal{} f(x)$ for almost every $ x.$

Now let $ f$ be the characteristic function of $ S$ and apply this theorem.

Oh  :(  Here is what was driving me crazy. I could have sworn that that theorem only held for $ f\in L^1$, and I couldn't figure out how to show that without somehow assuming something about $ S$ that made the problem trivial.

Lebesgue's Differentiation Theorem holds for $ f\in L^1_{loc}$, which the characteristic function of a measurable set certainly is.

Not to belabor the point, since I got it once I realized that fix, but it seems that the $ 1/2$ in the problem is completely arbitrary, since we could fix $ \alpha\in (0,1)$ and say that $ \big|S\cap Q_{\varepsilon_k}(x)\big|\leq \alpha\big|Q_{\varepsilon_k}(x)\big|$ in the same way.

Let $ \chi_S \equal{} f$, then a.e. $ f(x) \equal{} \lim_{\epsilon\to0}\frac1{m(Q_{\epsilon}(x))}\int_{Q_{\epsilon}(x)}f \equal{} \lim_{\epsilon\to 0}\frac {|S\cap Q_\epsilon|}{|Q_\epsilon|}\leq \alpha$. Since $ f$ is a characteristic function, as long as $ \alpha$ is less than 1, we have that it is 0, and since this is almost everywhere $ |S| \equal{} 0$.

Of course, being a little more careful and taking the limit along $ \varepsilon_k$.


induction
inequalities
limit
algebra
binomial theorem

Prove that $ 2\leq (1 \plus{} \frac {1}{n})^n < 3\ (n\in N)$

I've never proved anything before but, if we plug in 1
we get 2 = (1+1)^1 and because its well known that
Lim (1+1/n)^n as n approaches positive infinity is equal to "e" then
Throught induction it can be proved that the inequality above is true

Proof of the left part:

Let $ a_n \equal{} \left(1 \plus{} \frac {1}{n}\right)^n$, is $ a_n$ increasing sequence or decreasing sequence?

Proof of the right part:

Here is a problem from Japan.

1981 Kyusyu Univesrity entrance exam/Humanities

Answer the following questions 1, 2, 3.

1. Use induction to prove that $ k!\geq 2^{k \minus{} 1}$ for positive integers.

2.  For integers $ k,\ n$ such that $ 1\leq k\leq n$, prove that $ \frac {_nC_k}{n^k}\leq \frac {1}{2^{k \minus{} 1}}$

3. For positive integers $ n$, prove that $ \left(1 \plus{} \frac {1}{n}\right)^n < 3$.

[quote="kunny"]
Let $ a_n \equal{} \left(1 \plus{} \frac {1}{n}\right)^n$, is $ a_n$ increasing sequence or decreasing sequence?
[/quote]

Edit: I was just expanding on this, because this is clearly not trivial.

Fail, or you could note that $ \left(1 \plus{} \frac {1}{n}\right)^n \ge 1 \plus{} {n \choose 1} \frac {1}{n} \equal{} 2$. The right-hand side can also be proven by some bounding on the binomial expansion (for example, proving that $ \left(1 \plus{} \frac {1}{n}\right)^k < 1 \plus{} \frac {k}{n} \plus{} \frac {k^2}{n^2}$ by induction).

Edit: [quote="Yongyi781"](2) $ \lim_{n \to \infty} a_n \equal{} e$.[/quote]Hmm wouldn't this be circular logic; $ e$ being defined as $ \lim_{n \to \infty} \left(1 \plus{} \frac 1n\right)^n$ while the question here is asking to show that $ a_n$ is bounded, which would then point to the existence of a limit.

[quote="azjps"]Clearly $ a_n \equal{} \left(1 \plus{} \frac {1}{n}\right)a_{n \minus{} 1} > a_{n \minus{} 1}$ is strictly increasing, or you could note that $ \left(1 \plus{} \frac {1}{n}\right)^n \ge 1 \plus{} {n \choose 1} \frac {1}{n} \equal{} 2$.[/quote]

I'm sorry, I did not follow that. I see that you are using the Binomial Theorem to some extent, but you must note that $ a_{n \plus{} 1} \equal{} \left(1 \plus{} \frac 1{n \plus{} 1}\right)^{n \plus{} 1}$, not $ \left(1 \plus{} \frac 1n\right)^{n \plus{} 1}$.

On another note, we can prove that $ a_n < 3$ in an interval $ I$ if we prove the following three statements.

(1) $ a_n$ is strictly increasing in $ I$.
(2) $ \lim_{n \to \infty} a_n \equal{} e$.
(3) (Almost forgot) $ a_n$ is continuous in $ I$ :!:

Because if for some $ k$, $ a_k > e$, then by the Intermediate Value theorem, there must be some $ m$ in the interval $ [k, \infty)$ such that $ e < a_m < a_k$, and therefore $ a_m < a_k$ when $ m > k$, and it would be decreasing. But that is a contradiction!

We can also further the restriction to $ 2 \le a_n < e$ when $ 1 \le n$.

So what do you think?
It's very easy! :D
[hide="Solution"]If n=1, the inequality is right.
If $ n\geq 2$, we have $ (1 \plus{} n)^n \equal{} \sum_{i \equal{} 0}^n{n \choose i}n^{n \minus{} i} \equal{} 2n^n \plus{} \sum_{i \equal{} 2}^n{n \choose i}n^{n \minus{} i}$
$ \therefore (n \plus{} 1)^n > 2n^n (obvious) \Rightarrow (1 \plus{} \frac {1}{n})^n > 2$
On the other hand, we have
$ {n \choose i}\cdot n^{n \minus{} i} \equal{} \frac {n(n \minus{} 1)(n \minus{} 2)...(n \minus{} i \plus{} 1)}{i!}\cdot n^{n \minus{} i} < \frac {n^n}{i!}$
$ \Rightarrow (1 \plus{} n)^n < n^n\cdot\sum_{i \equal{} 0}^n\frac {1}{i!} < 3n^n$
$ \Rightarrow (1 \plus{} \frac {1}{n})^n < 3\ (shown)$[/hide]


calculus
derivative
LaTeX
calculus computations

Sorry,but is this too easy? :?  :!:  :?:  :huh:

Cezar, first of all I think the right place is computations. Secondly, what is so hard? We need Leibnitz+ derivatives of $(1+x)^k$. I bet you can find the last one and you know the first one. :P

[quote="harazi"]Cezar, first of all I think the right place is computations. [/quote]
I agree, so be it moved there  :)

I has a big patience !
$f'(x)=\frac 32 (x+1)^{\frac 12}-\frac 12 (x+1)^{-\frac 12}$ a.s.o. In finale, for any n>2, \[f^{(n)} (x)=(-1)^n \frac{1}{2^n}(2n-5)!!\cdot \frac{\sqrt {x+1}}{(x+1)^{n-1}}(3x+2n).\] You can verify for n=3, n=4 a.s.o. You see that I know Latex better. Ave Caesar ! :spider:


probability
calculus
integration
pifinity is ze best
69
eeeeeeeee

The integers $ r$ and $ k$ are randomly selected, where $ \minus{}3 < r < 6$
and $ 1 < k < 8$. What is the probability that the quotient $ r \div k$ is
an integer value? Express your answer as a common fraction.

there are 5-(-2)+1=8 posibilities for r and 7-2+1=6 possibilities for k

so there are 8*6=48 combos

[hide="EPIC BRUTE FORCE!!!"]
r/k is integral if k is a multiple r.

when k=2 there are 4 possibilities for r (evens)
when k=3 there are 2 possibilities for r (0&3)
when k=4 there are 2 possibilities for r (0&4)
when k=5 there are 2 possibilities for r (0&5)
when k is 6 or 7 theres only 1 possibiility (0)[/hide]

(4+2+2+2+2)/48=1/6

How can there be six choices for [i]k[/i] when [i]r[/i]=0 if zero isn't an integer? 0/[i]k[/i]=0

pretty much list everything that works
skull


how can you tell whether a graph is even or odd by just looking at it?

An even function is a function $ f(x)$ such that $ f(x) \equal{} f(\minus{}x)$. This means that if $ f(x) \equal{} y$, $ f(\minus{}x) \equal{} y$. Therefore, for every coordinate pair $ (x,y)$ where $ x > 0$ in the graph of $ f(x)$, there exists a corresponding ordered pair $ (\minus{}x, y)$. This means that if you only consider the positive part of the graph (the part that is to the right of the $ y$-axis), then you can flip it over the $ y$-axis to get the remaining part of the graph. Then you just have to plot $ f(0) \equal{} y$, because we have ignored it throughout so far.

Using this process, can you figure out yourself how an odd function will look like given that a function is odd if $ f(x) \equal{} \minus{}f(\minus{}x)$?


You have three points to use.
You lose when your team has -50.
Use (^_^)'s thread for all other rules.

Cinnamon Toast Crunch: 2
ernie

Froot Loops: 0

Cheerios: 0

Raisin Bran: -1

mehh.

Cinnamon Toast Crunch: 4
ernie

Froot Loops: 0

Cheerios: 0

Raisin Bran: -2

mehh.

Cinnamon Toast Crunch: 4
ernie

Froot Loops: 2
pacman2812

Cheerios: 0

Raisin Bran: -4

You have to join Raisin Bran b/c they have lowest number of points
Cinnamon Toast Crunch: 4
ernie

Froot Loops: 0
pacman2812

Cheerios: 0

Raisin Bran: -2
pacman2812

me, not joining

Then I don't join.

mehh.

Cinnamon Toast Crunch: 5
ernie

Froot Loops: 0

Cheerios: 0

Raisin Bran: -4

Nope, you have to be there. :P

Cinnamon Toast Crunch: 6
ernie

Froot Loops: 0

Cheerios: 0

Raisin Bran: -6
pacman2812

Wow.  I'm the only one playing. xD

WHO DOESN'T?!

Yeah, you should kill this and make it a fruit ToW.

Banana - 0

Mango - 3
1=2

Peach - 0

Apple - 0

[b]Banana - 0[/b]

[b]Mango - 3[/b]
1=2

[b]Peach - 3[/b]
ernie

[b]Apple - 0[/b]

[quote="ernie"][quote="1=2"]I'm not joining anybody but cinnamon toast crunch.[/quote]

Whoa.  Should I change the themes and kill this tug-of-war?  Everyone loves cinnamon toast crunch. xD[/quote]

Actually, I really hate Cinnamon Toast Crunch.

Banana: 3
kimmystar94

Mango: 3
1=2

Peach: 3
ernie

Apple: 0

please do not use dashes. dashes make people think "ooh. negative"

Banana: 3
kimmystar94

Mango: 3
1=2

Peach: 6
ernie

Apple: 0

Banana: 6
kimmystar94

Mango: 3
1=2

Peach: 6
ernie

Apple: 0

Banana: -46
kimmystar94
dreambold22
james4l

Mango: -18
1=2
crazychinesemaniac
chenhsi[/quote][/quote][/quote]

Banana: -49
kimmystar94
dreambold22
james4l

Mango: -18
1=2
crazychinesemaniac
chenhsi[/quote][/quote][/quote][/quote]

Mango: Wins
1=2
crazychinesemaniac
chenhsi

New Game:

Team 1: 3
chenhsi

Team 2: 0

Team 3: 0

Feel free to rename the team names.

Team 1: 3
chenhsi

Team 2: 0

Riemann Hypothesis: 3
james4l

chenhsi/AoPSWiki WoW: 3
chenhsi

Complex conjugate: 0

Riemann Hypothesis: 0
james4l

chenhsi/AoPSWiki WoW: 6
chenhsi

Complex conjugate: 0

Riemann Hypothesis: 0
james4l[/quote]

chenhsi/AoPSWiki WoW: 9
chenhsi

Complex conjugate: 0

Riemann Hypothesis: 0
james4l[/quote][/quote]

chenhsi/AoPSWiki WoW: 9
chenhsi

Complex conjugate: 0

Riemann Hypothesis: 3
james4l

chenhsi/AoPSWiki WoW: 9
chenhsi

BUTTERCUP!!! :D   : 3
1=2

Riemann Hypothesis: 0
james4l

chenhsi/AoPSWiki WoW: 9
chenhsi

BUTTERCUP!!! :D   : 0
1=2

Riemann Hypothesis: 0
james4l[/quote]

chenhsi/AoPSWiki WoW: 9
chenhsi

BUTTERCUP!!!  : 3
1=2
Crazychinesemaniac

Riemann Hypothesis: 0
james4l[/quote]

chenhsi/AoPSWiki WoW: 9
chenhsi

BUTTERCUP!!!  : 0
1=2
Crazychinesemaniac

Riemann Hypothesis: 0
james4l[/quote][/quote]

chenhsi/AoPSWiki WoW: 12
chenhsi

BUTTERCUP!!!  : 0
1=2
Crazychinesemaniac

Riemann Hypothesis: 0
james4l[/quote][/quote][/quote]

chenhsi/AoPSWiki WoW: 15
chenhsi

BUTTERCUP!!!  : 0
1=2
Crazychinesemaniac

Riemann Hypothesis: 0
james4l[/quote][/quote][/quote][/quote]


[quote="#H34N1"]This question should have a tag that says "Answer in latex"[/quote]

I just answered that one a+b\sqrt{c} without dollar signs and it worked out fine.

[quote="#H34N1"]This question should have a tag that says "Answer in latex"[/quote]

Please read [url=http://www.artofproblemsolving.com/Edutainment/g1/Pages/Documentation.php]How to Play[/url] or see the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=177887]Forms of Answers[/url] topic for directions on how to answer problems.


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

Find a polynomial in $Q[x]$ with galois-group $Z_3$.

Note: I'm really doubtful about even the existence of such a creature, but being wrong wouldn't be bad...

i think x*x*x-3*x+1 would be a good polynomial

Any galois extension of degree $3$ will do, since in the case of a galois extension, the degree of the exnension equals the order of the galois group, and there's only one group of order $3$.

We want an irreducible cubic with square discriminant. fredbel's example has discriminant $-4\cdot(-3)^3-27\cdot1^2=81$, so it works.


Well with all this talk about Ameem coming back to life and beating people up I'm scared to vote for you    :P

Hmm... Who are you running against? Bush? Then yes, I would.

yup, I agree jb05.

If you didn't run against Bush, then you'd have to convince me that you are a good candidate.  What are you like?  What are your views on things?  I would guess that most AoPSers, unlike most Americans, actually want to know things about a candidate before voting for him/her.

[quote="yif man12"][quote="Valentin Vornicu"][quote="bubala"]I would guess that most AoPSers, unlike most Americans, actually want to know things about a candidate before voting for him/her.[/quote]Take a closer look at the current status of the poll  :icecream:[/quote]

Awesome new emoticons!!!!![/quote]Took you guys some time to realize it  :yankchain:

maybe
:pilot:  :starwars:  :stretcher:  :cleaning:  :  :spider:

Cool i mean  :clap2:

Now, play with them, but don't abuse them too much. I'm watching you :police:

oh. my.  :censored: gosh.

there's a:  :starwars: !!!!

ok, i just totally wanted to use those. heehee.  :evilgrin:

[quote="mkkool64"]I would vote for you if you brought ddr to every school in America

 :ddr:

(Lame excuse to post coolest new emoticon)[/quote]

I would vote for you if you brought awesome star wars lightsaber battles to every school in America

 :starwars:

But I already voted for you because I felt like it.

[quote="furious"]who would elect a braves fan into the white house?  :P[/quote]I would! :winner_first: Braves are the best team!

You can't be any worse thean politicians right now, so sure, I would vote for you! :diablo:

I voted. no.  Sorry, you're not old enough.  :chief:  :starwars:  :surf:  :ddr:  :gathering:  :chief:  Wow this is fun.  :roll:

I just remembered a random quote from some random book: "My dad was a conservative. If an ape ran for president, he would vote for it as long as it ran as a republican."

Again, you can't just ask "would you vote for me". When Nixon was president, I would have said I would have voted for Bush. I may not have said I would have voted for Clinton. Who knows? You can't just ask if we would vote for you. I said yes anyway, we need someone who isn't completely stuck on  :starwars: and  :football: for a change. Academics are good.

[quote="juicybooty911"]pay meeeeeeeeeeeeeeeeeeee :rotfl:[/quote]

ditto. or i will  :ninja:  you.  :P

:10: I voted for you!  I saw you post a lot, and I think you are cool :D.   :coolspeak:

[quote="yif man12"]Hopefully you guys are kidding, because you've got to be a semi-big  :censored: to honestly believe an 8th grader can do a better job as president than our current one. Just shows how nutty some people are...[/quote]

Yeah, 8th grade might be a little young. Perhaps a senior. Though it isn't hard to do a better job.

[quote="mkkool64"]only if i was guaranteed a position of power in your new American empire ;)[/quote]

spoils system?
cool!

although the age limit is like 30 something

i think it's 35.  i just took American Gov but i don't listence too much in class :?

maybe well vote for u like 20 years later although i highly doubt i will

Probably not.  Of course, I've gotten kind of fed up with this country's politics, so I probably wouldn't vote unlesss you were running against someone I [i]really[/i] disliked (no names, please); then I suppose I'd vote for you.

By the way, guys, enough with the new emoticons.  Unless I'm greatly mistaken, there's already a topic for them.

just becuz im texan doesnt mean im for bush okkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkk :D

[quote="bleumoose"][quote="nr1337"]Considering this community is largly anti-Bush, I'd say no :)[/quote]

that's rather funny, because I have considered AoPS to be mostly conservative. There are a LOT of Texans in this forum.[/quote]I'm a Texan, yet I'm not exactly a fan of Bush...probably a good 60-70% of my school is though.  :?

Haha totolee~ Though everyone here's pretty smart so I guess it would be a safer decision than to vote for Bush

 :juggle: OMG these emos are so :censored:  :coolspeak: I lurve this :spider:  :starwars:

[quote="EUCLA"]what are braves? :ninja:[/quote]
an american baseball team

[quote="noneoftheabove"]hmmm go braves it doesnt seem like your getting a lot of popularity[/quote]

Are you kidding me??? I got $\infty$ times more votes than i thought i would

that is because 15 out of the 16 who voted for u were probably joking


[b]20[/b] objects are arranged in a row.[b]A subset of 4 elements [/b]is to be selected such that [b]no two [/b]of the selected elements are consecutive in the given arrangement.Then determine the number of such subsets.

Its' a fairly simple question ....

picking 4 objects from a 20 consecutive articles is same as saying putting 4 objects in between 1g objects such that only one can be put between any two.
therefore we have 17 spaces:
_0_0_0_0_0_0_0_0_0_0_0_0_0_0_0_0_

where 0 denotes object and _ space.

therefore the answer is 17C4. :wink:


:D

$ v_x\equal{}v_0\cos(\theta)\equal{}\text{const}$
$ v_y\equal{}v_0\sin(\theta)\minus{}gt$

So:

$ x(t_0)\equal{}v_0\cos(\theta) t_0\equal{}L$
$ y(t_0)\equal{}v_0\sin(\theta) t_0\minus{}\frac{gt_{0}^2}{2}\equal{}0$

This gives: $ \sin(2\theta)\equal{}\frac{Lg}{v_{0}^2}\equal{}\frac{(2500\plus{}300)\cdot 9.81}{250^2}\equal{}0.439488$. This yelds $ \theta\equal{}13.035^{\circ}$, but $ \theta\geq\arctan\frac{1800}{2500}\equal{}35.7538^{\circ}$ and we will be never able to get in the second ship.


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{2}} \frac {x^2}{(\cos x \plus{} x\sin x)^2}\ dx$[/quote]

let us note $ (\cos x \plus{} x\sin x)\,'\equal{}x\cos x$
and $ (\sin x \minus{} x\cos x)\,'\equal{}x\sin x$

now $ x^2\equal{}x^2\sin^2\,x\plus{}x^2\cos^2\,x$
$ \equal{}x\sin x(\cos x \plus{} x\sin x)\minus{}x\cos x(\sin x \minus{} x\cos x)$
$ \equal{}(\sin x \minus{} x\cos x)\,'(\cos x \plus{} x\sin x)\minus{}(\cos x \plus{} x\sin x)\,'(\sin x \minus{} x\cos x)$

since $ d\left(\dfrac{u}{v}\right)\equal{}\dfrac{v\,du\minus{}u\,dv}{v^2}$,

$ I\equal{}\int_0^{\frac {\pi}{2}}\;d\left(\frac{\sin x \minus{} x\cos x}{\cos x \plus{} x\sin x}\right)\equal{}\boxed{\frac{2}{\pi}}$

[url=http://www.mathlinks.ro/viewtopic.php?t=106333]More Wonderful Collections of Virgil Nicula[/url]

[quote="kunny"]Evaluate $ \int_0^{\frac {\pi}{2}} \frac {x^2}{(\cos x \plus{} x\sin x)^2}\ dx$

[color=darkblue]Virgil Nicula have already posted the integral[/color] :oops:[/quote]

two solutions for this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=296&t=368881


Show that the number of permutations $\pi$ of $\{ 1,2,3,\cdots,n\}$ with no triple $i<j<k$ such that $\pi(i)>\pi(k)>\pi(j)$ is $C_n = {1\over n+1}{2n \choose n}$.

Suppose $\pi(n)=m$. We will show that $\pi(m),\pi(m+1),...,\pi(n-1)\in\{m+1,...,n\}$ $(*)$. Indeed, otherwise there is an $j\geq m$ s.t. $\pi(j)<m$, but then there is an $i<m$ s.t. $\pi(i)>m$. So $i<j<n$ and $\pi(i)>\pi(n)>\pi(j)$ -- contradiction.
Using $(*)$ we can easily see that $\pi(m),\pi(m+1),...,\pi(n-1)$ can be an arbitrary proper map to $\{m+1,...,n\}$ and $\pi(1),\pi(2),...,\pi(m-1)$ also can be an arbitrary proper map to $\{1,2,...,m-1\}$. Therefore $T_n=T_0T_{n-1}+T_1T_{n-2}+...+T_{n-1}T_0$, where $T_n$ is the required number of permutations, $T_0=1$ and $T_1=1$. It means, $T_n$ are Catalan numbers.

I would like to see a combinatorical solution for this problem.


Let $n, k$ be positive integers and let
$F(n,k)$
$=1$ when $n|k$, $n<k$
$=2$ when $k|n$, $k<n$
$=3$, otherwise
Let $m$ be a positive integer, find $\sum_{1\leq n,k \leq m} F(n,k)$

[hide]All the values where the function equals 1 are

n = 1, k = 2 to m ($\displaystyle\lfloor \frac{m}{1} \rfloor -1$ times)
n = 2, k = 4 to m or m - 1 ($\displaystyle\lfloor \frac{m}{2} \rfloor - 1$ times)
$\vdots$

All the values where the function equals 2 can be obtained by switching the values of n and k when the function equals 1.

The total number of possibilities shown so far is
$2(\displaystyle\lfloor \frac{m}{1} \rfloor + \lfloor \frac{m}{2} \rfloor + \cdots + \lfloor \frac{m}{m} \rfloor - m)$

To get the number of values where the function equals 3, we subtract this from the total, which is $m^2$.

Adding all these values, we get

$3(\displaystyle\lfloor \frac{m}{1} \rfloor + \lfloor \frac{m}{2} \rfloor + \cdots + \lfloor \frac{m}{m} \rfloor - m) + 3(m^2 - 2(\lfloor \frac{m}{1} \rfloor + \lfloor \frac{m}{2} \rfloor + \cdots + \lfloor \frac{m}{m} \rfloor - m)$
$= 3(m^2 - (\displaystyle\lfloor \frac{m}{1} \rfloor + \lfloor \frac{m}{2} \rfloor + \cdots + \lfloor \frac{m}{m} \rfloor) + m)$[/hide]


calculus
integration
calculus computations

prove :D

$ \int_{0}^{\infty}\;\;\left(\frac{e^{\minus{}x}}{x}\;\minus{}\;\frac{e^{\minus{}5x}}{1\minus{}e^{\minus{}x}}\right)\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{25}{12}\minus{}\gamma}$

More generally, let $ k\geq 2$ be a natural number. Then

$ I_{k}\equal{}\int_{0}^{\infty}\left(\frac{e^{\minus{}x}}{x}\minus{}\frac{e^{\minus{}kx}}{1\minus{}e^{\minus{}x}}\right)dx\equal{}H_{k\minus{}1}\minus{}\gamma$,

where $ H_{k}$ is the $ k^{th}$ harmonic number. To prove this formula we will use the following well-known definition of

$ \gamma$:

$ \int_{0}^{\infty}\left(\frac{e^{\minus{}x}}{1\minus{}e^{\minus{}x}}\minus{}\frac{e^{\minus{}x}}{x}\right)dx\equal{}\gamma$.

Thus, we get that

$ I_{k}\equal{}\minus{}\int_{0}^{\infty}\frac{e^{\minus{}kx}\minus{}e^{\minus{}x}}{1\minus{}e^{\minus{}x}}dx\minus{}\gamma$.

Use the substitution $ e^{\minus{}x}\equal{}t$ to get that

$ I_{k}\equal{}\int_{0}^{1}\frac{1\minus{}t^{k\minus{}1}}{1\minus{}t}dt\minus{}\gamma\equal{}\int_{0}^{1}\left(1\plus{}t\plus{}\cdots\plus{}t^{k\minus{}2}\right)dt\minus{}\gamma\equal{}H_{k\minus{}1}\minus{}\gamma$.

For $ k\equal{}5$ we get that

$ I_{5}\equal{}H_{4}\minus{}\gamma\equal{}\frac{25}{12}\minus{}\gamma$.


search
geometry
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral, $ P\equal{}AD\cap BC, Q\equal{}AB\cap CD$. The point E is satisfying $ EA\parallel BC, EC\parallel AB$. PQ meets CE at F. Show that $ D,E,F,Q$ lie on a circle.

[quote="mr.danh"]Let $ ABCD$ be a cyclic quadrilateral, $ P \equal{} AD\cap BC, Q \equal{} AB\cap CD$. The point E is satisfying $ EA\parallel BC, EC\parallel AB$. PQ meets CE at F. Show that $ D,E,F,Q$ lie on a circle.[/quote]

 Well , it is posted before at

http://www.mathlinks.ro/viewtopic.php?search_id=1271717223&t=191681

 Babis


integration
function
calculus
analytic geometry
real analysis
real analysis solved

This one might not be too tough, but here it goes:

Prove that for any positive t  \int {0->t} sin(x^2)dx >0

[quote="grobber"]
Prove that for any positive t  \int {0->t} sin(x^2)dx >0[/quote]

f(t)= \int {0->t} sin(x^2)dx, f(0)=0, f'(t)=sin(t) the variations of f is
f is increasing on [ \sqrt (2n \pi ); \sqrt (2(n+1) \pi )]
f is decreasing on [ \sqrt (2n-1) \pi ; \sqrt (2n \pi )] with n \in N
Let have a look of the sign of
f( \sqrt (2n \pi ))
= 1/2. \int _{0 to 2n \pi }siny/y dy (by change variable)
=1/2\sum_{k=0 to 2n-1} \int _{k \pi ;(k+1) \pi }siny/y dy
=1/2\sum_{p=0 to n-1} \int _{0 to  \pi }siny( 1/ \sqrt (2p \pi +y) - 1/ \sqrt ((2p+1) \pi +y))dy>0
So for t>0,f(t)>0 and f(0)=0
 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \emptyset  \lambda

Is it from GM grobber ?
Myth my proof is 10 lines, do you have a shorter one ?

I no need 10 lines to solve this problem but just one my insight. Simplicity can't be expressed by number of lines.

[quote="Moubinool"][quote="grobber"]
Prove that for any positive t  \int {0->t} sin(x^2)dx >0[/quote][/quote]

Sorry i forgot  \sqrt the correct exprressions are
.....
= 1/2. \int _{0 to 2n \pi }siny/ \sqrt y dy (by change variable)
=1/2\sum_{k=0 to 2n-1} \int _{k \pi ;(k+1) \pi }siny/ \sqrt y dy
......

Great Moobi, very good .. And you can generalize easily. Let F,G be two functions such that :
- F: R ---> R ; periodic, odd, positive on ]0,T/2[ where T is the period
- G : [0,+oo[ ---> [0,+oo[ ; bijective, derivable, convex
Then S(t) = Integral(0 =< x =< t)){F(G(x)).dx} is positive for all t > 0

Sorry for making it so easy for you, guys! :D I still think it's cute, though...

I followed some ideas in entrance problem to
ENTP(Ecole Nationale Travaux Public) 1996 to answer the grobber question

Here the part I of the problem:

Let C Cornu spiral points M(t) with coordinates (X(t);Y(t))
X(t)=\int_{0 to t }cos(u)du, Y(t)=_int_{0 to t }sin(u)du

1 Show that X,Y are C^oo on R(for any n\in N, X^(n)(t),Y^(n)(t) exist
and are continous)
What is the parity of X,Y?
Does the curve C has some symetry?

2 Show that lim_{t->+oo}X(t),lim_{t-->+oo}Y(t) exist ?
What are the consequence of the limit X,Y for the curve C

Let K( \sqrt (2 \pi )/4; \sqrt (2 \pi )/4)
3 What are the variations of the functions X, Y ?

4 Let F(t)=(\int_{t to +oo}cos(u)du)+(\int_{t to +oo}sin(u)du)
What is the geometric interpretation of F(t).
Show that F is C^1 (derivable with F' continous) on R+
and F'(t) = -\int_{0 to +oo}cos(w)/ \sqrt (w+t) dw

5 Let a>0 fixed I=\int_{0 to +oo}cos(w)/ \sqrt (a+w) dw
5a) if f is continous function strictely decreasing on [0;pi] show that
\int_{0 to pi}f(u)cos(u)du>0
5b) Study the convergence of serie
 \sum_{n \geq 0}\int_{npi to  (n+1)pi}cos(w)/ \sqrt (a+w)dw , and deduce that I>0
5c) What can you say about the function F ?

6) Does C has  double points ?

7) Show that C has a unique "point d'inflexion"(i don't know how to translate )

8 By a similar method as in I.5 study the sign of Y(t) for t>0
What is the sign of X(t) for t>0

9 Draw the curve C

 \cap  \cap  \emptyset  \cup \beta  \int  \cap  \emptyset  \emptyset  \lambda


quadratics
inequalities
algebra
algebra proposed

Find the  number of positive solution of the system    $ \left\{\begin{array}{l}

x\plus{}y\plus{}z\equal{}1 \\x^3\plus{}y^3\plus{}z^3\plus{}xyz\equal{}x^4\plus{}y^4\plus{}z^4\plus{}1

\end{array}\right.$
________________________________
Azerbaijan Land of FIre :D

i have a solution for your nice problem
from the first equation we get $ x\plus{}y\equal{}1\minus{}z$
so we have
$ x^3\plus{}y^3\equal{}(1\minus{}z)^3\minus{}3xy(1\minus{}z)$ and
$ x^4\plus{}y^4\equal{}(1\minus{}z)^4\minus{}4xy(1\minus{}z)^2\plus{}2x^2y^2$
substitute to the second equation we obtain
$ 2x^2y^2\minus{}(2z\minus{}1)^2xy\plus{}2z^4\minus{}4z^3\plus{}3z^2\minus{}z\plus{}1\equal{}0$
consider this is a quadratic equation with xy this equation has$ \Delta\equal{}\minus{}7$ so our system has no solution

$ 3.x^4\plus{}1\geq4.x^3$  by A-G-M inequality and equality is obtained when x^4=1 if we sum up
so if we add first inequality for x,y,z we reach $ 3.(x^4\plus{}y^4\plus{}z^4\plus{}1)\geq4.(x^3\plus{}y^3\plus{}z^3)$
and $ x^4\plus{}y^4\plus{}z^4\plus{}1\geq 4xyz$ comes from am-gm again so if we sum up these two

$ x^4\plus{}y^4\plus{}z^4\plus{}1\geq x^3\plus{}y^3\plus{}z^3\plus{}xyz$ and inequality case is seen if only x=y=z=1 so there is no solution.

dear mehmetcantu i think your solution is nice but with my solution it is no need to require positve numbers

actually in $ 3x^4\plus{}1\geq4x^3$ and $ x^4\plus{}y^4\plus{}z^4\plus{}1\geq4xyz$ there is no need for x to be positive because $ 3x^4\plus{}1\geq4{|x|}^3\geq4x^3$


analytic geometry
graphing lines
slope
trigonometry
geometry
angle bisector

Two distinct lines are of slopes $m_{1}$ and $m_{2}$, where $m_{1}\neq m_{2}$. What are the slopes of their two angle bisectors?

:maybe: [hide="Maybe"]
$\text{tan}\left( \frac{| \text{tan}^{-1}\ (m_{1})-\text{tan}^{-1}\ (m_{2}) | }{2}+\text{tan}^{-1}\ (m_{1}) \right) \ = \ x$
Where $m_{1}< m_{2}$
And $y$ should be perpendicular, $y = \frac{-1}{x}$
Where $x$ and $y$ are the slopes of each bisector.
[/hide]

[hide="I think you have a few mistakes in there, actually"]If $b_{1}$ and $b_{2}$ are the slopes of the two bisectors:
$b_{1}= \tan\left(\frac{\tan^{-1}m_{1}+\tan^{-1}m_{2}}{2}\right)$
$b_{2}= \tan\left(\frac{\tan^{-1}m_{1}+\tan^{-1}m_{2}+\pi}{2}\right)$
[/hide]

[hide]Draw a triangle with points $O(0,0)$, $A(1,m_{1})$, and $B(1,m_{2})$, and WLOG, let $m_{1}>m_{2}$.  So $OA=\sqrt{m_{1}^{2}+1}$ and $OB=\sqrt{m_{2}^{2}+1}$.  Draw the angle bisector of $\angle O$ to $M$, on $\overline{AB}$.

Let $x=BM$, hence $AM=m_{1}-m_{2}-x$.

By the angle bisector theorem, $\frac{\sqrt{m_{2}^{2}+1}}{\sqrt{m_{1}^{2}+1}}=\frac{x}{m_{1}-m_{2}-x}$.
Solving for $x$ gets $x=\frac{(m_{1}-m_{2})\sqrt{m_{2}^{2}+1}}{\sqrt{m_{1}^{2}+1}+\sqrt{m_{2}^{2}+1}}$.

Since $M$ is $x$ units above $B$, $M$ is at $\left(1,m_{2}+\frac{(m_{1}-m_{2})\sqrt{m_{2}^{2}+1}}{\sqrt{m_{1}^{2}+1}+\sqrt{m_{2}^{2}+1}}\right)$.  So its slope is $\boxed{m_{2}+\frac{(m_{1}-m_{2})\sqrt{m_{2}^{2}+1}}{\sqrt{m_{1}^{2}+1}+\sqrt{m_{2}^{2}+1}}}$.

Of course, the other slope can be found by taking the negative reciprocal of that.[/hide]

[quote="AstroPhys"]
And $y$ should be perpendicular, $y = \frac{-1}{x}$
[/quote]

Thought so, the angle bisectors being perpendicular.  :)
Then my solution should still be valid  :maybe: .

Can you please show your work, AstroPhys?  :)

And lingomaniac, your $m$ should be changed to $m_{2}$  :)

For my solution, I just found the angle of each slope then found the angle bisector, by finding the difference and halving it.
Then to get its slope, added the smaller slope or $m_{1}$ and took the tangent to get from angles to slope.
The slope of the other bisector follows from geometric rules.

angle bisector is the locus of points equidistant from those two lines...thus we apply the point line distance formula (our line is ax+by+c=0:
$m_{1}x-y=0$ and $m_{2}x-y=0$
$\frac{|m_{1}x-y|}{\sqrt{m_{1}^{2}+1}}=\frac{|m_{2}x-y|}{\sqrt{m_{2}^{2}+1}}$...put this in y=kx form...


Evaluate the following
$ (1\plus{}\cos\frac{\pi}{8})(1\plus{}\cos\frac{3\pi}{8})(1\plus{}\cos\frac{5\pi}{8})(1\plus{}\cos\frac{7\pi}{8})$

[hide="Solution"]
$ \cos ( \pi \minus{} \theta) \equal{} \minus{}\cos \theta$, so the expression is equivalent to:
$ (1 \plus{} \cos \frac{\pi}{8} )(1 \minus{} \cos \frac{\pi}{8} )(1 \plus{} \cos \frac{3\pi}{8} )(1 \minus{} \cos \frac{3\pi}{8} )$.

Simplifying, we obtain:
$ (1 \minus{} \cos ^ 2 \frac{\pi}{8} )(1 \minus{} \cos ^ 2 \frac{3\pi}{8} )$
$ \sin ^ 2 \frac{\pi}{8} \sin ^ 2 \frac{3\pi}{8}$

$ \sin ^ 2 \frac{\pi}{8} \cos ^ 2 \frac{\pi}{8}$

By double angle formulas, we have:
$ \frac{1 \minus{} \cos \frac{\pi}{4}}{2} \cdot \frac{1 \plus{} \cos \frac{\pi}{4}}{2}$
$ \frac{1 \minus{} \cos ^ 2 \frac{\pi}{4}}{4}$
$ \frac{\sin ^ 2 \frac{\pi}{4}}{4}$
$ \frac{\frac{1}{2}}{4}$

So, the answer is $ \boxed{\frac{1}{8}}$.
[/hide]

My solution after which i got stuck,the given expression equals
$ [(1 \plus{} \cos\frac {\pi}{8})(1 \plus{} \cos\frac {3\pi}{8})]^{2}$

let z=$ \cos\frac {\pi}{8} \plus{} i\sin\frac {\pi}{8}$

so $ \cos\frac {\pi}{8}$=$ \frac {1}{2}(z \plus{} \frac {1}{z})$
likewise $ \cos\frac {3\pi}{8}$=$ \frac {1}{2}(z^{3} \plus{} \frac {1}{z^{3}})$

now  (1+$ \cos\frac {\pi}{8})$$ (1 \plus{} \cos\frac {3\pi}{8}$)=$ \frac {(3 \plus{} z^{2})(3 \plus{} z^{6})}{4z^{4}}$

also note that $ z^{8} \equal{} 1$ so simplifying we get
$ \frac {10 \plus{} 3z^{2}(z^{4})}{4z^{4}}$
multiplying and dividing by $ (z^{4} \minus{} 1)$ we get
$ \frac {10 \plus{} 3z^{2}(z^{8} \minus{} 1)}{4z^{4}(z^{4} \minus{} 1)}$ , not able to proceed further plz help!! :)

plz tell me where i am wrong :(

[quote="RAGHAV GROVER"]My solution after which i got stuck,the given expression equals
$ [(1 \plus{} \cos\frac {\pi}{8})(1 \plus{} \cos\frac {3\pi}{8})]^{2}$
[/quote]
I think the problem is that the expression is not equal to $ [(1 \plus{} \cos\frac {\pi}{8})(1 \plus{} \cos\frac {3\pi}{8})]^{2}$.
It is actually equal to $ (1 \plus{} \cos \frac{\pi}{8})(1 \plus{} \cos \frac{3\pi}{8})(1 \minus{} \cos \frac{3\pi}{8})(1 \minus{} \cos \frac{\pi}{8})$.
This is because $ \cos (\theta \minus{} \pi ) \equal{} \minus{}\cos\theta$, so $ \cos\frac{5\pi}{8} \equal{} \minus{}\cos\frac{3\pi}{8}$ and $ \cos\frac{7\pi}{8} \equal{} \minus{}\cos\frac{\pi}{8}$.

Idea: construct a polynomial with the given roots and use that to calculate the product.

Let $ r_k\equal{}\cos \frac{(2k\minus{}1)\pi}{8}$. We want $ (1\plus{}r_0)(1\plus{}r_1)(1\plus{}r_2)(1\plus{}r_3)$. If $ t\equal{}\frac{(2k\plus{}1)\pi}{8}$, then $ \cos 4t\equal{}\cos k\pi \plus{}\frac{\pi}{2}\equal{}0$.

Now $ \cos 4t\equal{}8x^4\minus{}8x^2\plus{}1\equal{}f(x)$ where $ x\equal{}\cos t$.

Now clearly, $ f(x)$ has the roots $ r_k$ $ k\equal{}0,1,2,3$ and since $ f$ has degree four, it has precisely these roots. We can factor $ f$ as follows:
\[ f(x)\equal{}8(x\minus{}r_0)(x\minus{}r_1)(x\minus{}r_2)(x\minus{}r_3)\] It is easy to verify that\[ (1\plus{}r_0)(1\plus{}r_1)(1\plus{}r_2)(1\plus{}r_3)\equal{}\frac{f(\minus{}1)}{8}\equal{}\frac{1}{8}\]

My solution
the given experssion equals $ (1 \minus{} \cos^{2}\frac {\pi}{8})(1 \minus{} \cos^{2}\frac {3\pi}{8})$

set z=$ \cos\frac {\pi}{8} \plus{} i\sin\frac {\pi}{8}$

Note the following points,
$ z^{8} \equal{} 1$

$ \frac {1}{2}(z^{6} \plus{} \frac {1}{z^{6}}) \equal{} \frac { \minus{} 1}{\sqrt {2}}$

$ \frac {1}{2}(z^{2} \plus{} \frac {1}{z^{2}}) \equal{} \frac {1}{\sqrt {2}}$

substitute this ones in the expression to get $ \frac {1}{8}$
The easiest one , no need to remember any formulas.. :P


The maximum speed of a 2.7 kg mass attached to a spring is 0.73 m/s, and the maximum force exerted on the mass is 13 N.
(a) What is the amplitude of motion for this mass? (m)
(b) What is the force constant of the spring? (N/m)
(c) What is the frequency of this system? (Hz)

why don't u write down the relevant formulaes.These are direct questions

because i do not know the formula for amplitude that uses the given information.

[hide="My Solution for (a)"]
(a)
Conservation of energy: [Subcript m - max]
$ \frac {1}{2} mv_m^2 = \frac {1}{2} kA^2 $
$ mv_m^2 = F_mA $
$ A = \frac {mv_m^2}{F_m} $
$ A = 0.11 m $
[/hide]
[hide="My Solution for (b)"]
$ F_m = kA $
$ k = \frac {F_m}{A} $
$ k = 120 N/m $
[/hide]
[hide="My Solution for (c)"]
$ f = \frac {1}{2\pi} \sqrt {\frac {k}{m}} $
$ f = 1.0 Hz $
[/hide]

[quote="pardesi"]$ v_{max}\equal{}A\omega$
$ \omega\equal{}\sqrt{\frac{k}{m}}$
$ F_{max}\equal{}kA$[/quote]

Is $ \omega $ equal to $ 2*\pi*f $?

[quote="gaussintraining"]Yes, the angular speed $\omega=\frac{2\pi}{T}=2\pi f$.[/quote]
Thanks

[quote="gaussintraining"]Yes, the angular speed $\omega=\frac{2\pi}{T}=2\pi f$.[/quote]

You might be interested in: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=636&t=351053

[quote="gaussintraining"][quote="gaussintraining"]Yes, the angular [b]frequency[/b] $\omega=\frac{2\pi}{T}=2\pi f$.[/quote]

Corrected a mistake  :blush:  (I can no longer edit).[/quote]

You made a mistake?  :)


The Möbius bundle is [0,1]xR/~ where ~ identifies (0,y) with (1,-y) and the projection map p send [(x,y)] down to exp{(2pi)ix}.

How to show it is nontrivial? That is there doesn't exist a homeomorphism between  [0,1]xR/~ and $ S^1$xR

I would rather view both as quotients of $ \mathbb{R}^2$: the Möbius bundle is the quotient when we identify $ (x,y)$ with $ (x\plus{}1,\minus{}y)$ and the trivial bundle on $ S^1$ is the quotient when we identify $ (x,y)$ with $ (x\plus{}2,y)$. With this labeling of the circle having length 2, we get a natural 2-1 covering map from the trivial bundle to the Möbius bundle.

The tool I'd use here is orientability. There are sections of the trivial bundle which are nowhere zero, but not of the Möbius bundle.

There not existing a homeomorphism between these spaces is stronger, I believe, than saying they non-isomorphic vector bundles. Which do you need to show? The existence of
a nonzero section is certainly a bundle property, but I don't think it's a topological property.

jmerry, does your argument with coverings answer the question?

anyway, there's a boundary-like argument which should conclude easily: $ S^1\times\mathbb{R}$ has disconnected "boundary", while the moebius band has connected "boundary". here by "boundary" i just mean taking the space of vector of lenght 1 (with respect to any metric), or the boundary of any (sufficiently nice) compact neighbourhood of the zero-section.

In the Möbius strip there is an injective loop representing twice a generator of the fundamental group. In the cylinder, no such loop exists.

or, again, by jordan's theorem, every simple closed curve in the cylinder disconnects, while the core of the moebius band doesn't.


[quote="sharkman"]Assume body weight and height are related by the formula W = 2H + 13, where W represents weight in pounds and H represents height in inches.  What is the height, in pounds, of a person whose height is (2x + 1) inches, when x = 12?[/quote]
ummm.....wtf???????????
Am I being stupid or did I read the question wrong?
Just plug in 12.......
Why is this in hsb?????????
This should be in classroom math.

EDIT: I just read the question, and realized.... how can height be in pounds??? You mean weight right? Even though, this should be in classroom math.

Sorry.  I will put it in classroom math.


the sum from 0 to infinity of

((x+3)^k)/(k^8)(6^k)

find the integral of convergence
estimate S(2) within .001 of its exact value

That's "interval" of convergence, not integral.

And this better not be a sum from $k=0$ to $\infty$ because the $k=0$ term is undefined.

Anyway, if you fix that (say, by starting the sum at $k=1$), the radius of convergence is 6, the behavior on the ends of the interval of convergence are controlled by $\frac1{k^8}$ which gives an absolutely convergent series, and thus the interval of convergence is $[-9,3].$

As for the approximate sum: you have the Alternating Series Test on your side, and the terms should decrease quite rapidly. Just add up terms until the next one is smaller than your error tolerance - that should happen quickly.

[Problems like this are best placed in the "Computations and tutorials" part of the Calculus forum.]


induction
function
LaTeX
algebra unsolved
algebra

Show that for all non negative x holds: $ \sqrt {1 \plus{} x\sqrt {1\plus{} (x\plus{}1)\sqrt {1 \plus{} (x\plus{}2)\sqrt {1\plus{}(x\plus{}3) \sqrt {1\plus{}\ldots}}}}} \equal{} x \plus{} 1$  :D

Thanks for the correction

I don't know if I'm wrong, but apparently $ LHS>1$ whereas $ RHS\equal{}1$ for $ x\equal{}0$.

While this is not a solution, it is some progress.
[hide]
Let $ f(x) \equal{} \sqrt {1 \plus{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} (x \plus{} 2)\sqrt {1 \plus{} (x \plus{} 3)\sqrt {1 \plus{} \ldots}}}}}$.

Then we have:
$ f(x)^2 \equal{} 1 \plus{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} (x \plus{} 2)\sqrt {1 \plus{} (x \plus{} 3)\sqrt {1 \plus{} \ldots}}}}$

$ f(x)^2 \minus{} 1 \equal{} x\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} (x \plus{} 2)\sqrt {1 \plus{} (x \plus{} 3)\sqrt {1 \plus{} \ldots}}}}$

$ f(x)^2 \minus{} 1 \equal{} xf(x \plus{} 1)$

$ f(x \plus{} 1) \equal{} \dfrac{f(x)^2 \minus{} 1}{x}$

If $ f(n) \equal{} n \plus{} 1$ for some $ n \in \mathbb{N}$, then $ f(n \plus{} 1) \equal{} \dfrac{(n \plus{} 1)^2 \minus{} 1}{n} \equal{} n \plus{} 2$

While it is easy to show that $ f(0) \equal{} 1$, it is impossible to use this as a base case for induction.

Also, the induction would only show that $ f(x) \equal{} x \plus{} 1$ for $ x\in \mathbb{N}$ not all reals.
[/hide]

I'm not to sure on how to proceed from here.

Nice!

The functional relation is f(x)=√(1+xf(x+1)) with f(x) non decreasing.
We want to show that f(x)=x+1. Suppose that we have found reals 0<a<1<b s.t. ax+1 ≤ f(x) ≤ bx+1 for all 0 ≤ x.

We see that the recursive relation defined by t_0=b and t_{n+1}=1/2*(t_n+1) for all n in N converges to 1.
At the same way we see that the n-root of a positive real tends to 1 for n big enough.

If a exists then f(x)=√(1+xf(x+1))≥ √(1+x(a(x+1)+1)) ≥ 1+x√a.
If b exists then f(x)=√(1+xf(x+1))≤ √(1+x(b(x+1)+1)) ≤ 1+x((b+1)/2).

So we have only to show that the couple (a,b) exist so that x+1 ≤ f(x) ≤ x+1.
a exists, infact f in non decreasing and we have f(x)=√(1+xf(x+1)) ≥ √(1+xf(x))  that implies a=1/2.
also b exists in fact f(x)=√(1+xf(x+1)) ≤ √((1+x)f(x+1)) = (√(1+x))*√√(2+x)f(x+2)... and for PMI it holds f(x) ≤ Π(x+i)^{2^(-i)} where i goes from 1 to infinity. We have only to show that Π(x+i)^{2^(-i)} is bounded by some linear function: Π(x+i)^{2^(-i)} ≤ Π(2ix)^{2^(-i)} forall x>1 (if 0<x<1 it sufficient to take the bound with x+1). Now 2x is constant, then (2x)Π(i)^{2^(-i)} ≤ 2xΠ(i)^{(i-1)*2^(-i)} that is not greater than 4x: yeah, b=4, then f(x)=x+1.

I hope that i have not made stupid mistake.

[quote="jin thynj"]Thanks for the correction[/quote]

You're welcome.

[quote="EastyMoryan"]LHS=$ \sqrt {1 \plus{} 0\sqrt {1 \plus{} (0 \plus{} 1)\sqrt {1 \plus{} ...}}} \equal{} \sqrt {1} \equal{} 1$[/quote]

Now it is :lol: .

Actually whether this is a correct way and is this what u expect i dont know.this is very famous and was proposed by ramanujan if my knowledge is correct.

now,

      x = $ \sqrt {x^{2}}$
          = $ \sqrt {x^{2} \plus{} 1 \minus{} 1}$
         = $ \sqrt {1 \plus{} (x \minus{} 1)*(x \plus{} 1)}$    [b]  1[/b]

now x+1 = $ \sqrt {(x \plus{} 1)^{2} \plus{} 1 \minus{} 1}$

substitute this for x+1 in [b] 1[/b]

doing this we get finally x =  $ \sqrt {1 \plus{} x \minus{} 1\sqrt {1 \plus{} (x)\sqrt {1 \plus{} (x \plus{} 1)\sqrt {1 \plus{} (x \plus{} 2)\sqrt {1 \plus{} \ldots}}}}}$

now for our required equation put x -1 as y so it becomes $ \sqrt {1 \plus{} y\sqrt {1 \plus{} (y \plus{} 1)\sqrt {1 \plus{} (y \plus{} 2)\sqrt {1 \plus{} (y \plus{} 3)\sqrt {1 \plus{} \ldots}}}}} \equal{} y \plus{} 1$

and this is what is required.
i hope this is clear,sorry for not writing the entire solution as i am poor at LATEX


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ f\in K[X]$ be irreducible and let $ L/K$ be a normal field extension. Show that if $ g,h$ are normalized prime factor of $ f$ in $ L[X]$, there exists $ \sigma$ in $ G(L/K)$ (set of all homomorphisms form $ L/K$ to $ L/K$) such that $ \sigma g\equal{}h$

Let $ E$ be the Galois closure of $ L$ (the smallest Galois extension of $ K$ containing $ L$). Then since $ G(E/K)$ acts transitively on the roots of $ f\equal{}\prod (x\minus{}a_i)\prod (x\minus{}b_j)\prod (x\minus{}c_k)$ we can choose $ \rho\in G(E/K)$ that sends the roots of $ g\equal{}\prod (x\minus{}a_i)$ to the roots of $ h\equal{}\prod (x\minus{}b_i)$. Thus $ \rho(g)\equal{}h$. Now we just restrict $ \sigma\equal{}\rho|_{L}\in G(L/K)$. This is well-defined since the extension $ L$ is normal so $ L$ is stable (i.e. for any $ rho\in G(L/K)$ we have $ \rho(L)\subset L$).

A problem occurs if $ deg(g)<deg(h)$. There is a reason they have to be of same degree, but I can't think of one right now.

Just rushing through your solution ... and the following popped my eye:

[quote="HilbertThm90"]Let $ E$ be the Galois closure of $ L$ (the smallest Galois extension of $ K$ containing $ L$). [/quote] What happens if $ E$ collapses to $ K$? Does your reasoning still work in this case - which is not evident to me in this very moment?
[quote]Then since $ G(E/K)$ acts transitively on the roots of $ f \equal{} \prod (x \minus{} a_i)\prod (x \minus{} b_j)\prod (x \minus{} c_k)$ [/quote] Why does $ f$ have necessarily a root in $ E$? I am racking my brain how the a,b,c's are allocated to f,g,h ... :(
[quote]we can choose $ \rho\in G(E/K)$ that sends the roots of $ g \equal{} \prod (x \minus{} a_i)$ to the roots of $ h \equal{} \prod (x \minus{} b_i)$. Thus $ \rho(g) \equal{} h$. Now we just restrict $ \sigma \equal{} \rho|_{L}\in G(L/K)$. This is well-defined since the extension $ L$ is normal so $ L$ is stable (i.e. for any $ rho\in G(L/K)$ we have $ \rho(L)\subset L$).[/quote]Frankly speaking I cannot grasp that but it is already late in my time zone ... ;)
[quote]A problem occurs if $ deg(g) < deg(h)$. There is a reason they have to be of same degree, but I can't think of one right now.[/quote]Right, this is invoked in your argument. The other case is open.

If $ L|K$ is separable:
One may want to take $ E$ to be a Galois extension of $ K$ such that $ L \subset E$ and $ f$ splits completely (exists, but we don't care which one we take), now everything works out fine:
If $ g,h$ are two irreducible factors of $ f$ over $ L$, then let $ a,b \in E$ be roots of $ g,h$, respectively. $ G(E|K)$ acts transitively, thus there is a $ \sigma \in G(E|K)$ with $ \sigma a\equal{}b$. Now $ \sigma$ permutes irreducible factors of $ f$ over $ L$ (by it's action on $ L[x]$ as $ L$ is normal), so $ \sigma g$ has the same degree as $ g$ and is again irreducible. But as $ (\sigma g)(b) \equal{} (\sigma g)(\sigma a) \equal{} \sigma(g(a)) \equal{}0$, we get that $ \sigma g | h$ (using that $ \sigma g$ is irreducible), thus $ \sigma g \equal{}h$ as $ h$ is irreducible, too.
For general extensions, we have $ f(x) \equal{} \bar f(x)^{p^s}$ for some separable polynomial $ \bar f$, thus we can reduce it to the separable case.

Note that one needs $ L|K$ to be normal: Take $ K\equal{}\mathbb Q, L\equal{}\mathbb Q[\sqrt[3]2], f(x)\equal{}x^3\minus{}2$. It factors in $ L$ as $ x^3\minus{}2 \equal{} (x\minus{}\sqrt[3]2)(x^2\plus{}\sqrt[3]2y\plus{}\sqrt[3]2^2)$, and surely factors of different degrees cannot be mapped onto each other.

Oops. True. I was thinking to take the splitting field, but I wrote Galois closure thinking it would be the same thing. Sorry...


quadratics
analytic geometry
graphing lines
slope
geometry
geometric transformation
reflection

What's the equation of the line that passes through $(10,-1)$ and is perpendicular to $y = \frac{1}{4}x^2-2$?

[quote="Hokkage"]$y=-2x+19$
Wait, not sure if this is right, checking.[/quote]

I don't think so  :)

[quote="JavaMan"][hide]
y = -x/5 + 1
[/hide][/quote]

Try again...

What do you mean by "perpendicular"? This is a quadratic function. Almost every line that passes through this point is perpendicular to a tangent of  the function. . .

[hide]
is there a way to do it without calc?

$\frac{dy}{dx}=\frac{1}{2}x$(tangent slope)
$\frac{-2}{x}$ = Normal slope (negative reciprocal)
Point-slope formula:
$y+1=m(x-10)\\ (\frac{1}{4}x^2-2)+1=(\frac{-2}{x})(x-10)\\ x^3+4x-80=0\\ (x-4)(x^2+4x+20)=0\\ x=4\\ y+1=(\frac{-1}{2})(x-10)\\ y=-\frac{1}{2}x+4$

[/hide]

Ok, this branches off of this problem but it's definitely related. Reflect point $(x_1,y_1)$ over some equation. Give the method. This isn't a specific case. I've done this before, but I'm extremely tired and can't seem to solve your problem.

There are two answers y=9.898979486x-99.98979486.
y=.1010205144x-2.010205144

To solve use this:
y=mx+b
y=ax^2+Bx+c

(B and b are different varaibles)
Set them equal to each other, set it equal to zero, use the quadratic formula to solve x.  We want the roots of x to be equal so set them equal. Set it to zero.  We get.

((B-m)^2-4a(c-b))/a^2=0  (These corespond to the coeffiecints in the upper part.)

Now plug your x and y points into y=mx+b.

So we get  -1=10m+b

And this is true  ((0-m)^2-4(1/4)(-2-b))/(1/4)^2=0
Now solve for m and b using these to equations.


What is the additive inverse of:
A)0
B) -7/3
C) n-2

What is the mulpicative inverse of:
D) 2/3
E) -1
F) n-2

[hide]

 the additive inverse of an element a is the element a^' such that a+a^' = a^'+a = 0

Arrrrrrggghhh. Sorry guys but i gotta go. Someone please finish this problem for me or I'll do this tomorrow.[/hide]

Hahahaha now that I look at it, it's so easy! :D

Haha, indeed, Sinistra, no need to take out the 'big formula' :P

http://en.wikipedia.org/wiki/Additive_inverse
http://en.wikipedia.org/wiki/Multiplicative_inverse

:)

[quote="mathmanman"]Haha, indeed, Sinistra, no need to take out the 'big formula' :P
:)[/quote]
i agree


ceiling function
function
combinatorics proposed
combinatorics
Probabilistic Method

At a tourist camp, each person has at least $50$ and at most $100$ friends among the other persons at the camp. Show that one can hand out a t-shirt to every person such that the t-shirts have (at most) $1331$ different colors, and any person has $20$ friends whose t-shirts all have pairwisely different colors.

[quote="ychjae"]I think using Hall's theorem and the following lemma will be OK:
Lemma: if each vertex of a graph have degree between $n$ and $m$, then we can delete some edges of the graph such that each vertex in the new graph have degree between $n-k$ and $m-k$, while $k \leq n,m$ .[/quote]

Can you show in detail?

Let $G$ is a graph with vertices $v_1,v_2\ldots ,v_n$ which satisfies $50\leq deg(v_i)\leq 100$ for each
$i,1\leq i\leq n.$
$Lemma:$ There exist an edge set $E=\{e_1,e_2,\ldots ,e_x\}$ on $G$ such that if we delete $e_1,e_2,$
$\ldots ,e_x$ from $G$; for each vertice $v_i$ from the remaining graph, $20\leq deg(v_i)\leq 70$.
$Proof$ $of$ $Lemma:$ For each vertice $v_i$ from $G$, $50\leq deg(v_i)\leq 100$. If $deg(v_i)\leq 70$ for
each $i,1\leq i\leq n$, then we're done, so suppose the contrary. Let $T=\{v_1,v_2\ldots ,v_k\}$ is the set
of vertices $v_i$ such that $deg(v_i)\geq 71$. Call an edge ${v_i}{v_j}$ $good$ if and only if $deg(v_i)\geq 71$
and $deg(v_j)\geq 71$. If there is a good edge on $G$, then delete it. After this operation if there is a good
edge on the remaining graph, delete it. By repeating this procedure, after a while we won't be able to find a
good edge. For this moment $deg(v_i)\geq 50$ for each $v_i$ from $G$. If at this moment $deg(v_i)\leq 70$
for each $i,1\leq i\leq n$, then we're done, so suppose the contrary. Let $M=\{m_1,m_2,\ldots ,m_l\}$ is
the set of vertices  $v_i$ such that $deg(v_i)\geq 71$ at this moment. For a subset $S$ of $M$ let $f(S)$ is
the set of friends of vertices from $S$. Consider a subset $S$ of $M$. Since for each $v_i\in S$, $deg(v_i)\geq 71$
and for each $v_i\in f(S)$, $deg(v_i)\leq 70$; $|f(S)|>|S|$. Then by Hall's Theorem, there exist vertices
$p_1,p_2,\ldots ,p_l\in f(M)$ such that $p_i\neq p_j$ for $i\neq j$ and $p_i$ is a friend of $m_i$ for each $i,1\leq i\leq l$.
Now delete edges $p_1m_1,p_2m_2,\ldots ,p_lm_l$. If there is still a vertice $v_i$ on $G$ such that $deg(v_i)\geq 71$,
repeat this procedure again for the new set $M$. Since for each vertice $v_i$ from $G$, $deg(v_i)\leq 100$; after at
most $30$ operations, for each vertice $v_i$ on $G$, $deg(v_i)$ will be $\leq 70$. Since for each $v_i$ on $G$, $deg(v_i)$
can be decrease by at most $1$ on an operation, after all operations $deg(v_i)$ will be $\geq 50-30=20$ for each $v_i$
from $G$. $\Box$
Now color vertices $v_1,v_2\ldots v_n$ with colors $c_1,c_2,\ldots ,c_{1331}$ by an arbitrary way. Let $A_i$ is the set of
friends of $v_i$. Let $R_i$ is the set of colors of $A_i$. If for each $i,1\leq i\leq n$, $|R_i|\geq 20$, then we're done, so
suppose the contrary. Let $|R_s|\leq |R_i|$ for each $i,1\leq i\leq n$. Let $Q=\{i|1\leq i\leq n, |R_i|\leq 20, i\neq s\}$.
Let ${R_s}'=\{c_1,c_2,\ldots ,c_{1331}\}-{R_s}$. Since $|R_s|\leq 19$ and $|A_s|\geq 20$, there exist two different vertices
$v_i,v_j$ on $A_s$ such that $v_i$ and $v_j$ are in same color. If we recolor $v_i$ with a color from ${R_s}'$, then $|R_s|$
will be increase. Suppose that for each color $c_k$ from ${R_s}'$, if we recolor $v_i$ with $c_k$, this operation decreases
a $|R_j|$ which satisfies $j\in Q$. Then $v_i$ has at least $\lceil \frac{1331-19}{20-1} \rceil +1=71$ friends, which is
impossible. So, by repeating this procedure, after a while $|R_i|$ will be $\geq 20$ for each $i,1\leq i\leq n$, so we're done.

[quote="fagot"]More interesting result:
At a tourist camp, each person has at least 50 and at most 100 friends among the other persons at the camp. Show that one can hand out a t-shirt to every person such that the t-shirts have (at most) 761 different colors, and any person has 20 friends whose t-shirts all have pairwisely different colors.
(Generalization of ARO-2006 problem).[/quote]

Sorry to revive, but can anybody achieve this bound of 761? Or at least improve from 1331?

[quote="abconjecture"]Sorry to revive, but can anybody achieve this bound of 761? Or at least improve from 1331?[/quote]
A [url=http://www.wolframalpha.com/input/?i=%28e*100%5E2%29%28sum+k%21*%2836+choose+k%29*StirlingS2%5B50%2Ck%5D%2C+k%3D1+to+19%29%2F36%5E50]direct application[/url] of the [url=http://en.wikipedia.org/wiki/Lov%C3%A1sz_local_lemma]Lovász local lemma[/url] gives a bound of 36. :wink: (If there are $C$ colors, we have $d\le 100^2 - 1$ and $p\le C^{-50} \sum_{k=1}^{19} \binom{C}{k} k! S(50,k)$, where $S(n,k)$ is a Stirling number of the second kind. Of course, the weaker bound $p\le C^{-50} \binom{C}{19} 19^{50}$ is more than enough, but it only gives 38. For contest purposes, the bound $\binom{C}{19} \le \left(\frac{eC}{19}\right)^{19}$ from Stirling's approximation gets 49.)

[hide="proof of lemma for completeness"]In the notation of the Wikipedia article, we want to prove \[ p(\overline{A_1}\cap\cdots\cap\overline{A_k}) = p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}}) - p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}}\cap A_k) \] is positive, so we upper bound the second term by
\begin{align*}
p(\overline{A_1}\cap\cdots\cap\overline{A_{\max(0,k-d-1)}}\cap A_k)
&= p(\overline{A_1}\cap\cdots\cap\overline{A_{\max(0,k-d-1)}}) p(A_k) \\
&\le p(\overline{A_1}\cap\cdots\cap\overline{A_{\max(0,k-d-1)}})\cdot p,
\end{align*}where WLOG $A_k$ is independent of (all unions, or equivalently by PIE, all intersections, of) $A_1$ through $A_{k-d-1}$. (Here we interpret "empty probabilities" as $1$.)

This "approximately" gives us $f(k) \ge f(k-1) - p f(k-d-1)$ where $f$ is a lower bound for $k$ sets. Since we want to prove $f$ is always nonzero (i.e. positive), it's natural to look for $r>0$ such that $f(k) \ge r f(k-1)$ for all $k$ (heuristically, since $1,d+1$ are independent of $k$, $f$ "satisfies" $f' \ge -pf$, which has super-exponential solutions, and more concretely, $f$ should be lower-bounded by an exponential function of the largest positive real root of $x^k - x^{k-1} + p$, provided such a root exists).

We cannot directly work with the $f(k)$, but it will clearly suffice to prove $p(\overline{A_{i_1}}\cap\cdots\cap \overline{A_{i_k}}) \ge r\cdot  p(\overline{A_{i_1}}\cap\cdots\cap \overline{A_{i_{k-1}}})$ instead (where for $k=1$, this simply means $p(A_i) \ge r$ for all $i$). But any $r\in(0,1)$ with $1 - \frac{p}{r^d} \ge r$ will work, since from the first paragraph we have
\begin{align*}
p(\overline{A_1}\cap\cdots\cap\overline{A_k})
&\ge p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}}) - \frac{p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}})}{r^{\min(k-1,d)}}\cdot p \\
&\ge p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}}) - \frac{p(\overline{A_1}\cap\cdots\cap\overline{A_{k-1}})}{r^d}\cdot p
\end{align*}for $k\ge1$. (Note that $\min(k-1,d) + \max(0,k-d-1) = k-1$.)

The rest is easy: by AM-GM, $r = (pd)^{1/(d+1)}$ minimizes $\frac{p}{r^d} + r \ge (d+1)\sqrt[d+1]{\frac{p}{d^d}}$, so we're done if $1 \ge \frac{(d+1)^{d+1}}{d^d} p$. But $e > (1+1/d)^d$, so we're done.[/hide]

However, for the version with 20 and 70 (or anything like $20+x,70+x$ with $x$ small), I don't think the probabilistic method will work nearly as well.

If a vertex has degree $50$, then its $d$ is actually bounded by $5000$. Using this observation and the general version of the local lemma (with $x_i = \frac{1}{5000}$), we can improve the bound to 35 colors.

The lower bound is perhaps also interesting. It is not even easy to construct such a graph that forces the use of $>20$ colors.


Find a general solution to the equation:
$ a^2\plus{}b^2\equal{}2c^2$
and $ a$, $ b$, $ c$ are all non-negative integers.

I know a bit for Pythagorian numbers $ a^{2} \plus{}b^{2}\equal{}c^{2}$ but it seems they are not usually in this case. :wink:

[hide="Solution"]
So easy.
Has anyone thought of a=b=c?[/hide]

@Bugi:
 I think it is just a case of general form (if it exist)

[hide="Hint"]If $ a^2 \plus{} b^2 \equal{} 2c^2$, then
\[ \left( \frac{a \plus{} b}{2} \right)^2 \plus{} \left( \frac{a \minus{} b}{2} \right)^2 \equal{} c^2.\]
[/hide]

[hide]
This really gave it away.
Now we have a Pythagorean triple, and use $ m^2\minus{}n^2,2mn,m^2\plus{}n^2$ formula  [/hide]

You need to show that $ a, b$ have the same parity.

In this problem $ 2$ can be replaced by $ 5$ or $ 13$ or any other sum of two squares and the solution method is essentially the same, although to discuss the general case it is easiest to work on the [url=http://en.wikipedia.org/wiki/Gaussian_integers]Gaussian integers[/url].


geometry
parallelogram
geometry unsolved

Let the intersections of $\odot O_1$ and $\odot O_2$ be $A$ and $B$. Point $R$ is on arc $AB$ of $\odot O_1$ and $T$ is on arc $AB$ on $\odot O_2$. $AR$ and $BR$ meet $\odot O_2$ at $C$ and $D$; $AT$ and $BT$ meet $\odot O_1$ at $Q$ and $P$. If $PR$ and $TD$ meet at $E$ and $QR$ and $TC$ meet at $F$, then prove: $AE \cdot BT \cdot BR = BF \cdot AT \cdot AR$.

Let's look for case when $R$ is on the arc $AB$ not containing by $O_2$ and $T$ is on the arc $AB$ not containing by $O_1$
Firstly $\angle CRE=\angle ARP=\angle ABT=180-\angle RCT$ $\Longrightarrow RE\Vert FT$
Similarly $RF\Vert ET$ $\Longrightarrow FRET$ is a paralelogram.
$\angle QFT+\angle QBT=180-\angle FRE+\angle QBT=\angle PBQ+\angle QBT=180$$\Longrightarrow QFTB$ is cyclic.
Similarly $RBCF,RADE$ and $PATE$ are cyclic.
Using all these circles we get
$\angle RCD=\angle ACD=\angle ATD=\angle ATE$
$\angle CRD=\angle ARD=\angle AED=\angle AET \Longrightarrow \bigtriangleup RCD \sim \bigtriangleup ETA$ $(1)$
Similarly
$\angle BFR=\angle RCB=\angle ACB=\angle ATB=\angle QTP$
$\angle BRF=\angle BCT=\angle BAT=\angle QPT \Longrightarrow \bigtriangleup BFR \sim \bigtriangleup QTP$ $(2)$
From $(1)$ and $(2)$
$\frac{RD}{AE}=\frac{RC}{ET}$ and  $\frac{BF}{QT}=\frac{FR}{TP}$
Multiplying side by side and usinf fact $RF=TE$ we get
$BF.RD.TP=AE.QT.RC$ $(3)$
And from the power of point we have
$QT.TA=TB.TP$ $(4)$
$RC.RA=RD.RB$ $(5)$
Multiplying $(3),(4),(5)$ we get
$AE.TB.RB=BF.TA.RA$
I think other cases are similar to this one. ;)

My solution is very similar to cefer's one, in fact the first few steps are just the same; but I finished it off in some other way. So, let me post a solution I got some while ago.
[hide="Solution/very indifferent to cefer's one"]
Note that from some angle chasing we have
$\angle APE=\angle ABR=\angle ACD=\angle ATD,$ so that $AETP$ is found to be a cyclic quadrilateral. By similar means, $AERD, BFRC, BFTQ$ are all cyclic quadrilaterals.
So $\angle QRB=\angle FCB=\angle TDB,$ so that $QR\parallel DT;$ and similarly $CT\parallel PR.$
Also note that $\angle TEA=\pi-\angle APT=\pi-\angle TQB=\angle BFT,$ and $\angle ATE=\angle TQF=\angle TBF,$ leading to $\triangle AET\sim\triangle BFT.$
This implies $\frac{AT}{BT}=\frac{AE}{TF}.$ By similar means we have $\frac{AR}{BR}=\frac{AE}{FR}.$
Therefore we obtain that
\[\frac{AT}{BT}\cdot\frac{AR}{BR}=\frac{EF}{BF}\cdot \frac{AE}{FR}=\frac{AE}{BF},\]
Since $ETFR$ is a parallelogram $\implies ET=FR.$
We are done. $\Box$[/hide]
Sorry for posting a similar solution after $5$ years. :oops:


Anyone going to Destination Imagination Global Finals next week?


I did not know where to post this:

Can someone explain to me what the Mellinium problems are?  Are they the rankings (p vs. np, birch & swinnerton dyer)?  Thanks for any explanations!

the millenium problems are 7 of the most famous unsolved problems in mathematics and yes, they are the rankings here. These are:
Birch and Swinnerton-Dyer Conjecture
Hodge Conjecture
Navier-Stokes Equations
P vs NP
Poincaré Conjecture
Riemann Hypothesis
Yang-Mills Theory

It's kinda like a challenge and whoever solves or proves one of these, will get 1 million dollars USD

Go to http://en.wikipedia.org/wiki/Clay_Math_Institute, it has a lot of in-depth info.

[quote="Inspired By Nature"]Go to http://en.wikipedia.org/wiki/Clay_Math_Institute it has a lot of in-depth info.[/quote]

fixed link.  ;)

Now that it seems like the Poincaré Conjeture it's proved, if it's true it has to be replazed, maybe Goldbach could be a good candidate.

The Poincaré conjecture was proved in 2003 by Grigori Perelman of the Steklov Institute of Mathematics in Saint Petersburg, Russia.  As a matter of fact, it was only earlier this very month that a team of Asian mathematicians put the finishing touches on Perelman's proof.  The Clay Mathematics Institute has yet to award the prize to Perelman, or to announce whether the problem will be replaced on the list, and if so what will replace it.

The other 6 Millenium Problems are

Birch and Swinnerton-Dyer conjecture
Complexity classes P and NP
Hodge conjecture
Navier-Stokes equations (solutions and smoothness)
Riemann hypothesis
Yang-Mills existence and mass gap

Information from Wikipedia.

I don't think the rank of Poincaré will be changed. It's a millenium problem nontheless.

I also believe Goldbach's conjecture is a consequence of Riemann's hypothesis.

As of June 6th (that was yesterday, right?), the revised Poincaré conjecture proof still needs to be completely reviewed. Goldbach and Riemann are not currently known to be at all related.


A machine generates 0's and 1's one after another with the same probability 1/2 to produce a sequence of 1's and 0's. Which string is more likely to appear first: 011 or 110? A rigorous solution is preferred.

[hide]If the first number is $ 0$, $ 011$ is bound to appear before $ 110$

If the first number is $ 1$, and the second number is $ 0$, $ 011$ is bound to appear first too.

If the first two numbers are $ 1$, $ 110$ is bound to appear first.

Hence the probability of $ 011$ appearing first is $ 0.75$  [/hide]

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=315143[/url]