calculus derivative function videos MIT college Princeton

hello everyone! i am really happy there is a physics forum now!
I am preparing a contest of Machanics now and the range of the problems is from Newtons Law of Motion to Momentum, including Uniform Circular Motion, Motion in space and Energy.
So any formulas and nice tools about those stuff?
I shall post some most basic formulas:

Newton's Second Law: $F=ma$
Uniform Circular Motions: $F=m \cdot a_n = m\cdot \frac{v^2}{r} = m \cdot {\omega}^2 r$
Kinetic Energy: $\frac12 m {v_t}^2 - \frac12 m {v_0}^2 =E$

[quote="towersfreak2006"]The formulas that College Board provides on the AP tests are usually pretty helpful in some contests.

http://www.collegeboard.com/prod_downloads/ap/students/physics/info_equation_tables_2002.pdf[/quote]

yea.. especially all of USAPhO pretty much... they even have part of that on the actual test.

it is pretty good!
i noticed that it also contains some calculus stuff, but it doesn't tell us how to use them. I shall post some:
If we have $s=f(t)$
Then $v=f'(t)$
Then $a=f''(t)$.

that just means that velocity is the first derivative of position with respect to time and acceleration is the second derivative of position with respect to time (or the first derivative of velocity with respect to time).  s denotes position, v denotes velocity, and a denotes acceleration.

In many mechanics books the notation for derivative is dot above the function [not exactly above the $f(t)$ like shobber wrote, but above $x$]. And it is rather denoted by $x$ not $s$. I'm not the expert, but just wanted to share what I know.

often in math and in physics, there is more than one way to express the same thing.  in my calculus bc class, we have always used the prime symbol, but if you have seen the video lectures of walter lewin on the mit website, you will see that he uses the dot above the letter.  either way is perfectly fine and shows the same thing.

actually, as i'm studying for the physics c ap exam which is only 10 days away, i read a sidenote in the princeton review book about the dot and prime issue with derivatives.  it says that a time derivative (a derivative with respect to time) is often written with a dot.  any derivative can be written with a prime if it's in a function format such as f'(x).  but, a shorthand only for time derivatives can be expressed with a dot over the variable, such as x with a dot on top equaling v, v with a dot on top equaling a, p with a dot on top equaling F, etc.


A 86 kg astronaut and a 1700 kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.24 m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut?

Since the system is initially at rest, $ p_{0}\equal{}0$ (units omitted).  Since $ p_{0}\equal{}p_{f}$, we conclude that $ m_{\mbox{a}}v_{\mbox{a}}\plus{}m_{\mbox{s}}v_{\mbox{s}}\equal{}0$.  So $ 1700\times .24\equal{}\minus{}86\times v_{\mbox{a}}\rightarrow v_{\mbox{a}}\equal{}\minus{}4.744$ (negative sign just indicates that direction is opposite).  Since we can assume constant velocity (nothing tells us otherwise), we use $ v\equal{}\frac{d}{t}\rightarrow d\equal{}35.6\mbox{m}$.


geometry
rectangle
geometric transformation
reflection
perpendicular bisector
geometry unsolved

Let $ ABCD$ be a rectangle and $ M,N$ are 2 points out side $ ABCD$ such that $ \widehat{AMB}\equal{}\widehat{CND}$. $ MN$ intersects $ AB,CD$ at $ P,Q$.$ DP$ intersects $ MB$ at $ E$ and $ BQ$ intersects $ ND$ at $ F$. Prove that $ AC,MN,EF$ concur.

Take a square $ ABCD$ and erect equilateral triangles $ \triangle ABO_1, \triangle CDO_2$ outward of the square. $ (O_1) \cong (O_2)$ are congruent circles with centers $ O_1, O_2$ and radii $ O_1A \equal{} O_1B \equal{} O_2C \equal{} O_2D.$ $ BD$ cuts $ (O_1)$ at $ B$ and again at $ M.$ Perpendicular bisector of $ CD$ cuts $ (O_2)$ at $ N$ outside of the square $ ABCD.$ Then $ \angle AMB \equal{} \angle CND \equal{} 30^\circ.$ $ \triangle ACM$ is equilateral $ \Longrightarrow$ $ M, C, N$ are collinear $ \Longrightarrow$ $ Q \equiv C \equiv MN \cap CD$ are identical. Whatever $ P \equiv MN \cap AB$ is, $ DP$ cuts $ MB \equiv BD$ at $ D,$ so that $ E \equiv D$ are identical. Since $ C \equiv Q$ are identical, $ BQ \equiv BC$ cuts $ ND$ at the reflection $ F$ of $ D$ in $ N.$ Consequently, $ MN$ cuts $ EF \equiv DN$ at $ N,$ which is definitely not on $ AC.$ The problem is wrong.


1.  Please send all emails about the AMC to
amcinfo@maa.org
Ms. Jensen in our office answers that email daily and re-directs the email to the proper person for an answer if she does not have the immediate answer.

For almost all questions, it speeds up the answer immensely if you include your school CEEB number, your school name (or the name of the location where you took a contest if not at your school), school address, city and state.

It also helps to include your full name (as used on the AMC answer form) and your address, including city and state.  Many names are similar, and additional identifying information helps us find the proper record in order to correctly answer your question.

2.  Since sometime on Friday, April 4, emails addressed to maa.org have been bouncing  This is almost certainly the result of a mistake in a re-setting of the configuration files done at the MAA headquarters mail server on Friday.  The proper person is aware of it, and has informed me he is working on the fix.  I have an indication that it got fixed while I was writing this message.

3.  We answer all legitimate email requests, but it sometimes may take up to a day or more, especially from late January through early May when we are busiest.   If your question is complicated or involves more than one staff person it will take longer.  Be assured that we take the email we receive seriously, and answer it.

4.  Many questions can be answered by checking with your teacher, math coach, or contest manager to determine if they have received any communication form our office.  They will get the notifications of scores first.  Many simple questions can also be answered by first reading the Teachers' Manual and our Frequently Asked Questions.

5.  Email addressed to my personal email address will be answered when I can.  It may not be soon.  Please use
amcinfo@maa.org.

Steve Dunbar
MAA Director, American Mathematics Competitions

I just sent an email - a fairly urgent one - to amcinfo@maa.org. It bounced - "no such user."

amcinfo@maa.org just got fixed.

Thanks for calling it to our attention.

Steve Dunbar
MAA Director, American Mathematics Competitions

The email I recieved informing me of USAMO qualification stated I could mail it to the AMC office. What's the address of the AMC office?


[quote="tridecalogism_lover"]hey, can someone help me solve this problem? we did this during our MC lesson, but I want to get a thorough, clear understanding b4 the competition......here goes:

********************************************
Alex owns 3 times as many brown shoes as red shoes, twice as many black shoes as brown shoes, and four times as many white shoes as red shoes. What is the ratio of the number of white shoes to the number of black shoes he owns? Express your answer as a common fraction.

I would be eternally grateful if you helped :D ! thanx[/quote][/quote]

[hide="Hint"]Okay. Let red=number of red shoes. Now write the other statements in the problem as equations. Substitute some of the variables in. What can you deduce?[/hide]

are you saying to set up linear equations?

yup a good idea for MC problems is to resort to equations  :D

[quote="tridecalogism_lover"]hey, can someone help me solve this problem? we did this during our MC lesson, but I want to get a thorough, clear understanding b4 the competition......here goes:

********************************************
Alex owns 3 times as many brown shoes as red shoes, twice as many black shoes as brown shoes, and four times as many white shoes as red shoes. What is the ratio of the number of white shoes to the number of black shoes he owns? Express your answer as a common fraction.

I would be eternally grateful if you helped :D ! thanx[/quote]
[hide]r=red shoes
b=brown
w=white

b=3r
w=4r
$\frac{w}{b}=\frac{4r}{3r}=\frac{4}{3}$[/hide]

Quevvy, you may need to try that again.
[hide]
Say that $l$ is the number of black shoes, $b$ is the number of brown shoes, $w$ is the number of white shoes, and $r$ is the number of red shoes.  We have $3r=b$ and $2b=l$.  Therefore, we have that $6r=2b=l$, and thus, $l=6r$.  We are also given that $4r=w$, so $\frac{w}{l}=\frac{4r}{6r}=\frac{2}{3}.$[/hide]

I'm sorry, I didn't see the black shoes...
[hide]br=3r
bl=2br=6r
w=4r
$\frac{w}{bl}=\frac{4r}{6r}=\frac{2}{3}$[/hide]

ok, so you can do this problem a lot of different ways, right? :maybe:

well not really. You just have to use the equations. Quevvy and The QuattoMaster 6000 just used different variables but essentially the same concept to solve the problem. However, The QuattoMaster 6000's answer is better because it only uses one variable per thing. Of course you can use Quevvy's way but you might mistaken $bl$ as two variables instead of one.

just use ratios

[hide]If there is one red shoe, there there must be three brown shoes. If there are three brown shoes, then there must be six black shoes. If there is one white shoe, then there must be four red shoes.

$4/6=2/3$[/hide]

[quote="slamchess"][quote]If there is one white shoe, then there must be four red shoes. [/quote]

kheg, u mite wanna switch that...[/quote]
Actually, he's right...

"Alex owns 3 times as many brown shoes as red shoes, twice as many black shoes as brown shoes, and four times as many white shoes as red shoes. What is the ratio of the number of white shoes to the number of black shoes he owns? Express your answer as a common fraction."
I did this problem on s school or chapter sprint round. This one is easy, as long as you know how to use algebra equations.
Put X as the number of red shoes.(the smallest amount in relation to the other shoes)
4X= the number of white shoes.
3X= the number of brown shoes
(3*2)X=the number of black shoes(6X)
and, we have all the equations we need(plus some that we can throw out)
Amount of white shoes/Amount of black shoes=4X/6X=4/6=2/3
2/3 is the final answer. If you don't get this, then I probably made a mistake =P or, just read the problem again.


pigeonhole principle
vector
inequalities
algebra
polynomial
absolute value
number theory

This is the last pigeonhole I'll post. I guess I posted enough of them now, although I still have many other pigeonhole problems.

Let x1, x2, ... , xn be real numbers satisfying x12 + x22 + ... + xn2 = 1.

Prove that for every integer k :ge: 2 there are integers a1, a2, ... , an, not all zero, such that
(01) |ai| :le: k - 1 for all i, and
(02) |a1x1 + a2x2 + ... + anxn| :le: (k - 1):sqrt:n/(kn - 1).

Nice huh ?

It's not easy. A hint :

[hide]Use the pigeonholeprinciple   Joking, sorry[/hide]

Here is my solution.

[hide]

Temporarily, I will change the condition on the a's to .  For k odd, I will require a_i to be an integer; for k even, I will require a_i - 1/2 to be an integer.  Either way, there are k solutions to .  (I will also temporarily ignore the condition that not all the a's are zero.)  There are k^n vectors a that satisfy these conditions.  The absolute value of the sum of a_i x_i will be at most (k-1)/2 times the sum of |x_i|, which by Cauchy-Schwarz is at most .  In other words, we have k^n sums in the interval .  Split that interval into k^n - 1 subintervals of equal size.  By the pigeonhole principle, two of those sums belong to the same subinterval, which means that two sums are within  of each other.  Suppose that a' and a'' are the two vectors corresponding to those two sums.  Let a = a' - a''.  Then a obeys all the original conditions, so we are done.

   [/hide]

Edit: Replaced old AoPS commands such as :le: with LaTeX.

Nice ... this is IMO 1987 problem 3.

http://www.kalva.demon.co.uk/imo/isoln/isoln873.html

Well ... I posted inequalities and pigeonholes now.
Any suggestions for a new theme ?

cant wait till i learn what pigeonhole is :), so i can look back at those problems and try to figure them out... this is all so new to me, i have to learn (a) completely new concept(s) each time you post diff tyeps of problems :), but keep going.

[quote="Arne"]Any suggestions for a new theme ?[/quote]

Anything you want but inequalities  :wink:

I like very much combinatorics or polynomial problems  :)

I was thinking of number theory ... is that OK ?

There are not many interesting polynomial problems.
And combinatorics is my weakest point ...
But I'll post also some combinatorics problems, OK ?


Let $p=a^{2}+b^{2}+c^{2}+d^{2}$. All numbers $a,b,c,d$ are positive, so $p\geq 4$. Assume by contradiction, that $p$ is a prime number.
Notice, that:
\[p=(a+b+c+d)(a+d-b-c)\] , so the second factor must be equal to $1$, thus $a^{2}+b^{2}+c^{2}+d^{2}=a+b+c+d$, but this equality holds only for $a=b=d=c=1$. We got contradiction with the condition $ad=b^{2}+bc+c^{2}$

barasawala see:  [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136345[/url]

btw. prolly easiest problem on polish mo, ever.

but i didn't consider the case a=b=c=d=1 (i thought it is obvious) haha

No, the first was definitely easier. Here you had to think for while whereas in the first one it sufficed to write down assumptions and the problem was solving itself.

Can you post all problems from Polish MO.
Thanks!

In this topic you'll find all problems

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=135374[/url]


Ορίστε ωραία άσκηση στον προσδιορισμό ρίζας συνάρτησης:

Έστω $ g(y) \equal{} f(y) \minus{} f(x)$. Να βρεθούν οι ρίζες της αν:
- $ x, y$ είναι πραγματικοί και ο $ x$ είναι μη μηδενική σταθερά,
- η $ f$ είναι συνάρτηση,
- είναι γνωστό το $ f(1)$
- η $ f(y)$ να είναι διάφορη του$ 0$ αν το $ y$ δεν είναι $ 0$,
- $ f( \minus{} y) \equal{} \minus{} f(y)$ για κάθε $ y$ πραγματικό,
- $ f(x \minus{} 1) \equal{} f(x)$
- $ 1 \plus{} yf( a \plus{} y ) \equal{} af(a)( f(1) \plus{} f(a) )$ για κάθε $ a, y$ πραγματικούς και το $ a$ μη μηδενικός.

Θα δώσω την απάντηση κρυφή για όσους βαριούνται να κάνουν την άσκηση:
[hide]
για $ y \equal{} 0$ είναι $ f( \minus{} 0) \equal{} \minus{} f(0) < \equal{} > f(0) \equal{} 0$
για $ y \equal{} 0$ είναι επίσης $ 1 \plus{} 0 \equal{} af(a)( f(1) \plus{} 0)$
 $ < \equal{} > f(a) \equal{} \minus{} \frac {1}{f(1)a}(I)$
για $ y \equal{} \minus{} 1$ είναι $ 1 \minus{} f( x \minus{} 1 ) \equal{} xf(x)( f(1) \plus{} f( \minus{} 1) ) < \equal{} > 1 \minus{} f( x \minus{} 1 ) \equal{} xf(x)0$
 $ < \equal{} > 1 \minus{} f( x \minus{} 1 ) \equal{} 0 < \equal{} > f( x \minus{} 1 ) \equal{} 1 (II)$
$ (I), (II) \equal{} > 1 \equal{} \minus{} \frac {1}{f(1)x} < \equal{} >$ [u]$ x \equal{} \minus{} \frac {1}{f(1)}$[/u]
Μία ρίζα της $ g$ είναι προφανώς η $ y \equal{} x \equal{} \minus{} \frac {1}{f(1)}$

Θα δείξουμε ότι η $ g$ δεν έχει άλλη ρίζα. Για να δειχτεί αυτό αρκεί να δείξουμε ότι η $ g$ είναι $ 1 \minus{} 1$. Επειδή η $ g$ είναι το άθροισμα της $ f$ με ένα σταθερό όρο $ \minus{} f(x)$, αρκεί να δείξουμε ότι η $ f$ είναι $ 1 \minus{} 1$.
Έστω $ a1, a2$ τέτοια ώστε $ f(a1) \equal{} f(a2)$ τότε
$ (I) \equal{} > f(a1)a1f(1) \equal{} 1$ και $ (I) \equal{} > f(a2)a2f(1) \equal{} 1$ άρα
$ f(a1)a1f(1) \equal{} f(a2)a2f(1)$ και επειδή είναι $ f(a1) \equal{} f(a2)$ βγαίνει $ a1 \equal{} a2$.
Άρα πράγματι η $ f$ είναι $ 1 \minus{} 1$ άρα και η $ g$ είναι $ 1 \minus{} 1$. Άρα η $ g$ έχει μοναδική ρίζα.

Άρα η $ g$ έχει τη μοναδική ρίζα $ y \equal{} \minus{} \frac {1}{f(1)}$.
[/hide]

Δεν έχω διερευνήσει ιδιαίτερα την άσκηση, αλλά πιστεύω πως δεν έχει περιττά στοιχεία και δεν είναι αυτοαναιρούμενη. Μη σπαταλήσετε πολύ ώρα σε αυτήν, γιατί χρησιμοποιεί κυρίως κλασικές τεχνικές. Πάντως έχει πλάκα!

Για να βοηθηθείτε στη λύση, σε κάθε σχέση βάλτε διαφορετική μεταβλητή.


$f\in Z[X]$, $f=(x^2+x)^{2^n}+1, n\ge 1$. Prove that $f$ is irreducible over $Z[X]$.


function
integration
trigonometry
real analysis
real analysis unsolved

Suppose that $1>\alpha>\frac{1}{2}$ and $f$ is a real function with complex values, $2\cdot\pi$ periodic and such that $ |f(x)-f(y)|\leq |x-y|^{\alpha}$ for all $x,y$. Prove that the series of Fourier coefficients of $f$ is absolutely convergent. Does the result remain true if $\alpha=\frac{1}{2}$?

for any $h > 0$, the fourier coefficient of $D_{h}(t)=f(t-h)-f(t)$ are $(e^{-inx}-1)\hat{f}(n)$. For example, if one chooses $h=\frac{2 \pi}{ 2^{n+2}}$, for every $2^{n}\leq |m| \leq 2^{n+1}$ one has $|e^{-imx}-1| \geq \sqrt{2}\geq 1$. So:
$\sum_{2^{n}\leq |m| \leq 2^{n+1}}|\hat{f}(m)|^{2}\leq \sum_{m \in \mathbb{Z}}|e^{-imx}-1|^{2}|\hat{f}(m)|^{2}$
Hence
$\sum_{2^{n}\leq |m| \leq 2^{n+1}}|\hat{f}(m)|^{2}\leq \frac{1}{2 \pi}\|D_{h}\|_{2}^{2}\leq \|D_{h}\|_{\infty}^{2}\leq (\frac{2 \pi}{ 2^{n+2}})^{2 \alpha}=C^{2}2^{-2n \alpha}$
Cauchy-schwartz allows to conclude that:
$\sum_{2^{n}\leq |m| \leq 2^{n+1}}|\hat{f}(m)| \leq C 2^{n(\frac{1}{2}-\alpha)}=\frac{C}{\lambda^{n}}$ with $\lambda > 1$
which imply the desired conclusion.

Extend $\sqrt{|x|}$ on $[-\pi,\pi]$ to an even $2\pi$-periodic function. I think $\int_{0}^{\pi}\sqrt{x}\cos nx\,dx$ is of size $\frac{1}{n}$, but don't have a clean proof.

Edit: $a_{n}: =\int_{0}^{\pi}\sqrt{x}\cos nx\,dx=\frac{1}{n^{3/2}}\int_{0}^{\pi n}\sqrt{t}\cos t$ by substitution $t=nx$. Since $\left|\int_{\pi n}^{\pi (n+1)}\sqrt{t}\cos t dt\right|\ge C\sqrt{n}$, it follows that $|a_{n}-a_{n+1}|\ge \frac{C}{n}$ for all $n$. Thus $\sum|a_{n}|=\infty$.

$C$ is/are some positive constant(s) that I don't care about.


Got the card now. Centre is PS :rotfl: :D

Those poor souls who got in Perambur and Thiruvattiyur must be cursing me now :D

actually i got two admit cards, and i got them long ago.

Btw who will be the invigilators? Will they be our own school teachers? :maybe: :D

Two admit cards? Well we all are getting it about now only though some of my friends got it sometime ago

my centre is some army public school. u can guess who our invigilators will be.

Still haven't got it.

Does anybody have any idea about the invigilators for JEE??? Will it be the school teachers or iit students or alumini??

I hate Billa Srinivasan.

I havent gotten it yet either. Bet my centre is going to be somewhere in Trichy.

[quote="Chronoz"]I hate [u][b]Billa Srinivasan[/b][/u].[/quote]

Just cut a word out of that for me :D :mad: I hate 'him' too :|

who's that 1

invigilators shud b good 1 to pity and tell some answers :lol:
iitians r left with nothing else to do :rotfl:

hey rohit who the hell is this billa srinivasan da??? never heard of that fellow in our gang???

and scene da we both have got our own schools da!!! how lucky!!! :rotfl:

and gokul has got a centre in perambur and Nandy has got somewhere in a village in outskirts in north madras!!!!! :D

i hope shreyas you too get PSBB and then it would be scene no!!!!!kalakku po!!!!

shre see you soon at MVM!!! :diablo:  :evilgrin:  waiting for you there!!!! :lol:

I haven't got it yet..... :rotfl:
I am so tensed....
 :P

[hide][quote]And of course, i am not supposed to get it...[/quote][/hide]

certainly u won't need an admit card--haven't u got the wildcard :rotfl:  :D

@Ritu: You just sounded like one of the judges on a popular dance show around these parts...

I got my admit card today. Some KV in Meenambakkam. Might as well board the flight to Timbuktu directly from there. Saves expenses you see

But I gotta admit, the place is just 15 mins from my house. So thats ok, considering the likes of people getting centres at Avadi and stuff!

kalaku po do you know about others from our region what centre have they got yeah hmmmm...... Ranga (valerium maximum) got AVADI as his centre i mean that place and i am waiting for shreyas to get his admit card lets see mostly you get PSBB  da shreyas :evilgrin:

me going on board to antartica!!!!  :surf:

bye!!!!   :cool:

Shreyas also got some other KV near Central if im not wrong :D

[i][b][u][color=darkblue]OH MY GOD HOW CAN SHREYAS ESCAPE FROM THE CLUTCHES OF DESTINY!!!! WHAT A BRUTAL JOKE!!!![/color][/u][/b][/i] :stretcher:

[quote="chemrock"]I got mine yesterday. My centre is JV,Ashoknagar.[/quote]

sirs school.. kalakku.

I got my card today morning on the road!!!!!(can u believe it my  card was on lying on the road from yesterday evening . i can see tyre marks on the outer cover)
Anyway i got some vani vidyalaya in virugambakkam

[quote="Nitishexpress"]  card was on lieing [/quote]

how can a card lie? :rotfl:  :rotfl:

its a conspiracy of all AIEEE admit cards. they won't tell the truth and will "lie" on the road, and in midnight, when noone's awake, they all get up and conquer the universe!!

My centre is pretty far but atleast not as far as some others.. on the outskirts of the cities.  :rotfl:  :rotfl:

What ever the conspiracy might be..... luckily a stranger took it and gave it... He saved me from a dozen of procedures for a duplicate

Prasad, got his Medium of writing as "Hindi". When he went to the regional office to complain. They asked him to go back because his was apparently a trivial problem, and many others had even got their names wrong.  :rotfl:  :rotfl:

Haha :D Medium was Hindi and they said that people had gotten their names wrong??? ROFL

[quote="Chronoz"]Prasad, got his Medium of writing as "Hindi". [/quote]

He's atleast better than me :D Imagine if i had got hindi :rotfl:  :roll:

hey so finally prasad got what center???and did they agree to change the medium??? :D

I dunno. Kaushik and Ranga are at KV Avadi. And SPD is worried that he will have to spend all his 3 hours reading the questions because of the some of the beautiful Hindi words for the commonly used mathematical and scientific terms in English. ;)

[quote="Euclidean Geometer"]oh my center is some village, dunno whether its near bangalore. its called ullal. :wink:[/quote]

Weren't you telling me that day, you are anyway not gonna write EEE, cause IIT is obvious. ;)

[quote="phoenix76"]yet not recieved my admit card
 :( really tensed[/quote]

dont worry many people havent yet got it i think because they have used normal post as means of trasnport its getting late!!! nways have a constant check out in front of your house or gate they might just drop it out there and go like it happened to nitish here in chennai!!! :o

@Nitish
[quote]Anyway i got some vani vidyalaya in virugambakkam[/quote]

Hey i too got Vani Vidyalaya in virugambakkam. :P
I have company atleast. :D

I got the other not-so-popular CV - Chinmaya Vidyalaya :D
received it today only  :)

[quote="hash_include"]I got the other not-so-popular CV - Chinmaya Vidyalaya :D
received it today only  [/quote]
hey coll hmmmm....... :lol:

I got it too........

 finally on Monday! :lol:

Ironically CBSE headquarters are just about 3 km from my place .......
I could walk to their office! Still, I'm virtually last one to get it....... :rotfl:


geometry algebra polynomial LaTeX trigonometry

Hello guys,
I'm going to post some math questions, please help me with any of these problems, with an answer and explanation:

1)  Simplify
[(1/X-1)+1] / [(X/X-1)-X]

2) Six distinct points lie in a plane such that 3 of the points are on line [i]m[/i] and 4 of the points are on line [i]n[/i]. What is the total number of lines that can be drawn so that each line passes through exactly 2 of these 6 points.

3) 2 similar polygons have a scale factor of 2 to 7. If the area of the smaller polygon is 28 sq inches, how many sq inches are in the area of the larger polygon?

4) Jason is looking down from a cliff at a friend below. The angle of depression from Jason to his friend is 53 degrees. They are each holding the end of 100 ft rope that is stretched taut. How high is the cliff to the nearest foot?

5)If P(1) = 0 for a polynomial equation P(x) = 0, then which of the following must be true?
A) -1 is a zero of P(x) = 0
B) x-1 is a factor of P(x)
C) x is a factor of P(x)
D) x+1 is a factor of P(x)

Thanks a bunch to everyone who helps out,
Peace,
Jimmy V

[hide="Number 1"]
$ \frac{\frac{1}{x-1}+1}{\frac{x}{x-1}-x}=\frac{\frac{x}{x-1}}{\frac{x-x^2+x}{x-1}}=\frac{x}{2x-x^2}=\frac{1}{2-x}$
[/hide]

Hope I interpreted your $ \text{\LaTeX}$-less post correctly.

[hide="#3"]Since it's area, multiply by the square of the scale factor. $ 28*\frac{49}{4}\equal{}\boxed{343}$[/hide]

[hide="Number 5"]
Extremely trivial... B.[/hide]

[quote="returnofkiladi"]
2) Six distinct points lie in a plane such that 3 of the points are on line $ m$ and 4 of the points are on line $ n$. What is the total number of lines that can be drawn so that each line passes through exactly 2 of these 6 points.
[/quote]
Those are 7 points, not 6; or else the answer is one line.

$ m$ and $ n$ could have one of those points in common :wink:

In that case you have $ 2$ choices for the point on $ m$ and $ 3$ choices for the point on $ n$, so $ 2\cdot 3\equal{}\boxed{6}$

[quote="007math"]Is #4 $ \cos 53 \times 100 \approx 60$?  I know trig doesn't belong in this forum...[/quote] Actually the answer is approximately 80, can anyone explain to me how this works?

because its the angle of depression, its actually $ \sin 53 \cdot100$, or $ \boxed{79.86...}$


The teacher whispers positive integer A to Annie, B to Brad, and C to Chris. The sum of A, B, and C is 14. Annie says, "I know that Brad and Chris have different numbers." Then Brad says, "I already knew that all of our numbers are different." Finally Chris says, "Now I know all three of our numbers." What is the product of ABC?

HARD???

the solution is to this is not just one
ie
since no are different
so
A     B      C        Product
1     2      11        22
1     3      10        30
1     4       9         36
1     5       8         40
1     6       7         42
2      5      7         70
and so on...........

 thats it :wink:

I will assume that all three know the sum $ A \plus{} B \plus{} C$.

Sentence: "I know that Brad and Chris have different numbers." means that Annie got an odd number, and therefore $ B \plus{} C$ is odd, so $ B\neq C$. Brad claims that he knew all three numbers are different: that would mean he's got a number which is odd and bigger than $ 7$, so $ A \plus{} C$ being odd guarantees $ A\neq C$, and $ B > A \plus{} C$ guarantees $ B\neq A, B\neq C$. So, possible combinations are:

$ (A,B,C) \equal{} (1,9,4),(3,9,2),(1,11,2)$

Since Chris knows the numbers for certain, Chris has got a $ 4$ (if he got $ 2$, he wouldn't know if it's the second or third triplet above). So, $ ABC \equal{} 36$.

@mr hsiljack
does the word different means odd :oops:

Nope, it does not, but look:

If you were Annie, and you knew that $ A\plus{}B\plus{}C\equal{}14$, and the teacher whispers $ 6$ to you. Then you know $ B\plus{}C$ is $ 8$, and as far as you know, it might be $ B\equal{}C\equal{}4$. Situation is same for any even $ A$. On the other hand, if the teacher whispered $ 7$, you'd know $ B\plus{}C\equal{}9$ and it can't be $ B\equal{}C\equal{}4.5$, with same situation for any other odd $ A$.

I thought this was easy to figure out :)

thanks , i am a fool to figure simple things


Let $ x,y,z $ be real numbers. Prove that the following conditions are equivalent:

(i) $ x,y,z $ are positive numbers and $ \dfrac 1x + \dfrac 1y + \dfrac 1z \leq 1 $;
(ii) $ a^2x+b^2y+c^2z>d^2 $ holds for every quadrilateral with sides $ a,b,c,d $.

very nice problem, i remember to have seen it in a "problems and solutions from around the world"...
the part "i) implies ii)" consists on a very simple use of the cauchy schwarz inequality... and the "quadrilateral" inequality, if it can be called like that  (a+b+c>d).
by cauchy $(a^2x+b^2y+c^2z)\geq(a^2x+b^2y+c^2z)(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})\geq(a+b+c)^2>d^2$

My solution:
$(i)\rightarrow(ii)$:
Since $ \dfrac 1x+\dfrac 1y+\dfrac 1z\leq1$ and $a+b+c>d$:
$a^2x+b^2y+c^2z\ge \left(a^2x+b^2y+c^2z\right)\left(\dfrac 1x + \dfrac 1y + \dfrac 1z\right) \ge(a+b+c)^2>d^2$.
We will prove $(ii)\rightarrow(i)$ by finding an example of $a,b,c,d$ when $(ii)$ doesn't hold if $(i)$ doesn't hold.
Motivation: since we used CS we need $a^2x^2=b^2y^2=c^2z^2$ in order to get equality. It is equivivalent to choosing $a=\frac{1}{x}$ etc.
Also, as our last step we used $a+b+c>d$ so it is logical to assume $d$ to be almost equal $a+b+c$.
Finally, $a^2x+b^2y+c^2z=\dfrac 1x+\dfrac 1y+\dfrac 1z=t$.
But $d=a+b+c-\epsilon= \dfrac 1x+\dfrac 1y+\dfrac 1z-\epsilon=t-\epsilon$.
So we only need to find $\epsilon_t>0: t^2-2t\epsilon_t+\epsilon_t^2-t>0$, which is always possible since $t>1$.
(take $0<\epsilon_t<t-\sqrt{t}$. It is possible because of $t>1\rightarrow t>\sqrt{t}$)


Let $ S$ be denote the set of all $ n$-tuples $ (n>1)$ of real numbers $ (a_1,a_2,...,a_n)$ such that
$ 3\sum^n_{i\equal{}1}a_i^2\equal{}502 (1)$
$ a)$ Prove that $ min |a_j\minus{}a_i|\leq\sqrt{\frac{2008}{n(n^2\minus{}1)}} (2)$ $ (1\leq i\leq j\leq n)$
$ b)$ Give an exampe of an-tuple $ (a_1,a_2,...,a_n)$ such that condition $ (1)$ holds and for which there is an equality an $ (2)$

If $ n \equal{} 2k \plus{} 1$ and $ |a_i \minus{} a_j|\ge h \ \forall i\not \equal{} j$, then $ 502 \equal{} 3\sum_i a_i^2\ge 6h^2\frac {k(k \plus{} 1)(2k \plus{} 1}{6}$.
Therefore $ h\ge \sqrt {\frac {502}{k(k \plus{} 1)(2k \plus{} 1)}} \equal{} \sqrt {\frac {2008}{n(n^2 \minus{} 1)}}.$
For $ n \equal{} 2k$ $ 502 \equal{} 3\sum_ia_i^2\ge (h^2/4) 6\sum_{i \equal{} 1}^k(2i \minus{} 1)^2 \equal{} k(k \plus{} 1)(2k \plus{} 1) \minus{} 3k(k \plus{} 1) \plus{} 3k/2 \equal{} 2k^3 \minus{} k/2 \equal{} \frac {n^3 \minus{} n}{4}$.
Therefore $ h\ge \sqrt{\frac{2008}{n(n^2 \minus{} 1)}}.$

[quote="Algadin"][quote="Rust"]If $ n \equal{} 2k \plus{} 1$ and $ |a_i \minus{} a_j|\ge h \ \forall i\not \equal{} j$, then $ 502 \equal{} 3\sum_i a_i^2\ge 6h^2\frac {k(k \plus{} 1)(2k \plus{} 1}{6}$.[/quote]
Why???? :wink:  :wink:[/quote]
Let $ a_1<a_2<...a_{2k\plus{}1}$, then $ a_{k\plus{}1\plus{}i}\ge a_{k\plus{}1}\plus{}ih,a_{k\plus{}1\minus{}i}\le a_{k\plus{}1}\minus{}h$. $ Min\sum_ia_i^2$, whtn $ a_{k\plus{}1}\equal{}0,a_{k\plus{}1\plus{}i}\equal{}ih,i\equal{}\minus{}k,...,k$.


[hide] $\frac{1000\ snaps}{\frac{150\ snaps}{60\ seconds}}=\frac{60000}{150}\ seconds= \boxed{400\ seconds}$ [/hide]

Another way is:
[hide] $150 times \Longrightarrow 60 seconds$
$1000 times \Longrightarrow x=\frac{1000*60}{150}=\boxed{\boxed{400 seconds}}$[/hide]

EDIT: Sorry bpms, I didn't see your post

[hide="solution"]He snaps 50 times every 20 seconds. He snaps 100 every 40 seconds. He snaps for 400 sec.[/hide]

[quote="Astronum"]The frequency of Bendan's movement is:

$f = \frac{150}{60}= 2.5 s^{-1}$

$f = \frac{times}{\Delta t}$

$2.5 = \frac{1000}{\Delta t}$

$\Delta t = 400s$[/quote]
I think thats a little too hard for classroom math.

[quote="Ignite168"]Bendan can snap his fingers 150 times in a minute. At this rate, how many seconds will it take for him to snap his fingers 1000 times?[/quote]

[hide]150/6=each second (really fast)
1000/(150/6)
6000/150
400[/hide]


inequalities
calculus
calculus computations

show that

$ e^\pi\minus{}\pi^e > \gamma$  :maybe:

where $ \gamma$ is Euler's constant

[quote="Yuriy Solovyov"]$ e^{\pi} \equal{} 23.14069264$

$ {\pi}^e \equal{} 22.45915771$

$ e^{\pi} \minus{} {\pi}^e \equal{} 0.68153493$

$ \gamma \equal{} 0.5772156649$ :D I don't see any relations between $ e,\;\pi,\; \gamma$  :([/quote]

me [i]neither.[/i]  :blush:  :(


The sum of the first n even positiive integers is h. The sum of the first n odd positive integers is k. What is the value of k-h.


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

[quote][color=darkred][b][u]An equivalent enunciation.[/u][/b] Let $ ABC$ be a triangle with the circumcircle $ (O)$ . For a mobile point $ M\in BC$ denote : the second intersection $ E$ of the line $ AM$ with the circle $ (O)$ : the point $ P\in BC$ for which $ PE\perp AM$ ; the second intersection $ N$ of the line $ AP$ with the circle $ (O)$ . Prove that the line $ MN$ passes through a fixed point.[/color] [/quote]

[color=darkblue]I think it is more nicely ! I'll come back soon. Nice problem ![/color]

[quote="chien than"]To Mods: Please lock this topic, this is a problem from M&Y, issue 1,,2008, and we can't discuss it now[/quote]
[color=darkblue][b]For nothing ![/b] They can discuss through private messages.
Particularly, appeared already a proof of it. It is ridiculously ![/color]

Let $ (O)$ be the circumcircle of triangle $ ABC$.
Denote $ MN\cap (O) \equal{} Q,AH$ be altitude of $ ABC$ we have:
$ MQ.MN \equal{} ME.MA \equal{} MH.MP$ then $ H,P,N,Q$ are concyclic.
Then $ \widehat{BHQ} \equal{} \widehat{PNQ} \equal{} \widehat{C}$
It mean $ Q$ is a fix point!

[quote="tdl"][img]http://i227.photobucket.com/albums/dd72/TPFC/namtran.png[/img]
Then $ \widehat{BHQ} \equal{} \widehat{PNQ} \equal{} \widehat{C}$
[/quote]
This is wrong ($ \widehat{PNQ} \equal{} \widehat{ACQ})$. But $ \widehat{HQC}\equal{}\widehat{C}$ since $ \widehat{BHQ}\equal{}\widehat{HQC}\plus{}\widehat{HCQ},\widehat{ACQ}\equal{}\widehat{C}\plus{}\widehat{HCQ}$

Sorry because of my mistake! But it is very lucky for us because today morning I have already found that it is a contest problem of Mathematics and Youth Magazine of Viet Nam, the dealine is 15-03-2008! Moderator, please lock this topic!

PS: I have already correct my solution! See you again!

[color=blue][b](1)[/b][/color] Let $ (Q)$ be the circle with diameter $ AP$ and $ F$ its intersection with $ BC$ other than $ P.$ Then $ AF \perp (FP \equiv BC)$ and $ F$ is fixed. Let the line $ MN$ meet the circumcircle $ (O)$ of the $ \triangle ABC$ again at $ K.$ $ ME \cdot MA  \equal{} MP \cdot MF \equal{} MN \cdot MK$ $ \Longrightarrow$ the quadrilateral $ FPNK$ is cyclic with a circumcircle $ (S).$ Inversion with center $ A$ and power $ AF^2$ takes the lines $ AP, AF$ passing through the inversion center $ A$ into themselves. The line $ BC$ tangent to the inversion circle $ (A)$ at $ F$ goes into a circle $ \Gamma_a$ with diameter $ AF$ and the circumcircle $ (O)$ of the $ \triangle ABC$ goes into some line $ o$ intersecting the circle $ \Gamma_a.$ The point $ P$ goes into the intersection $ P'$ of the line $ AP$ with the circle $ \Gamma_a$ other than $ A.$ The point $ N$ goes to the intersection $ N'$ of the line $ AP$ with the line $ o.$ Since $ AF$ is a diameter of $ \Gamma_a,$ the angle $ \angle N'P'F \equiv \angle AP'F \equal{} 90^\circ$ is right. The circumcircle $ (S)$ of the quadrilateral $ FPNK$ goes into the circumcircle $ (S')$ of the right angle $ \triangle FN'P'$ $ \Longrightarrow$ $ FN'$ is a diameter of $ (S').$ The image $ (S')$ of the circle $ (S)$ intersects the image $ o$ of the circle $ (O)$ at the image $ N'$ of $ N$ and again at the image $ K'$ of $ K.$ Since $ FN'$ is a diameter of $ (S'),$ the angle $ \angle FK'N'\equal{} 90^\circ$ is right and the point $ K'$ is fixed, regardles of the direction of the line $ AP.$ Therefore, the the original point $ K,$ the intersection of $ MN, (O)$ other than $ N,$ is also fixed.

[quote="thesunflower"]Given a triangle $ ABC$. Let $ P$ be a variable point on the line $ BC$, different from $ B$ and $ C$.
The circumcircle of triangle $ ABC$ meets the line $ AP$ at a point $ N$ and the circle whose diameter is $ AP$ at $ E$ (such that $ N$ and $ E$ are both different from $ A$).
The lines $ BC$ and $ AE$ intersect at $ M$. Prove that the line $ MN$ always passes through a fixed point![/quote]

Tdl: If I am not mistaken, you can upload attachments (like graphics) to your posts directly on the MathLinks forum. You don't have to use Imageshack or Photobucket for this!

  darij

We consider the particular case when $ P\equiv P',$ where $ P',$ is the orthogonal projection of $ A,$ on $ BC.$

Then, the line segment $ M'N',$ as the problem states, intersects the circumcircle $ (O)$ of $ \bigtriangleup ABC,$ at one point so be it $ Q$ and we will prove that the line segment $ MN,$ as the problem states $ ($ when $ P,$ is an arbitrary point on $ BC$ $ ),$ passes through the point $ Q.$

It is easy to show that $ (M'Q)\cdot (M'N') \equal{} (M'E')\cdot (M'A) \equal{} (M'P')^{2}$ $ ,(1)$

From $ (1)$ and because of the right triangle $ \bigtriangleup P'M'N',$ we conclude that $ P'Q\perp M'N'$ $ \Longrightarrow$ $ \angle P'QN' \equal{} 90^{o}$ $ ,(2)$

From the cyclic quadrilateral $ AQN'N$ we have that $ \angle N'QN \equal{} \angle N'AN$ $ ,(3)$

So, it is easy now to show that the quadrilateral $ PP'QN$ is cyclic, because of

$ \angle P'PA \equal{} 90^{o} \minus{} \angle N'AN \equal{} 90^{o} \minus{} \angle N'QN \equal{} \angle P'QN$ $ ,(4)$

Because of the circle $ (K)$ with diameter $ AP,$ passes through the point $ P'$ $ ($ because of $ AP'\perp P'P$ $ ),$ we have now three circles $ (O),\ (K)$ and $ (K')$ $ ($ as the circumcircle of $ PP'QN$ $ ),$ which taken per two of them, intersect each other and so, we conclude that the line segments $ AE,\ PP',\ QN,$ as their radical axis, are concurrent at one point.

Hence, the line segment $ QN,$ passes through the point $ M,$ as the intersection point of $ AE,\ P'P.$

That is, the line segment $ MN,$ passes through the fixed point $ Q$ and the proof is completed.

Kostas Vittas.

circle with diameter$ AP\equal{}\Omega_1$,and circum circle of triangle$ ABC\equal{}\Omega_2$,and $ Q\equal{}MN\cap\Omega_2$.and$ H\equal{}MP\cap\Omega_1,AH \perp BC$.and$ MN \cap AH\equal{}R$
$ M\in$radical axis of $ \Omega_1,\Omega_2 \Longleftrightarrow HPNR$ is cyclic $ \Longleftrightarrow AQ\equal{}$diameter of $ \Omega_2$,and $ Q$ is the fixed point!

[quote="vittasko"][quote="Ashegh"]circle with diameter$ AP \equal{} \Omega_1$,and circum circle of triangle$ ABC \equal{} \Omega_2$,and $ Q \equal{} MN\cap\Omega_2$.and$ H \equal{} MP\cap\Omega_1,AH \perp BC$.and$ MN \cap AH \equal{} R$
$ M\in$radical axis of $ \Omega_1,\Omega_2 \Longleftrightarrow HPNR$ is cyclic ....[/quote]

This is an incorrect inference.

Because of $ MQ\cdot MN \equal{} ME\cdot MA \equal{} MH\cdot MP,$ we can only say that $ HPNQ$ is a cyclic quadrilateral, but it is a not enough argument to conclude that $ Q$ is a fixed point.

Kostas Vittas.[/quote]

I will be back with a new solution :wink:

let MN intersect the circumcircle of ABC at L cause M  lies on radical axes of 2 circles ,  hence $ DPNL$ is cyclic.where D is foot of A-altitude.
Let DL intersect the circumcircle of ABC at A',and L' be a  point on this circle that $ LL'\parallel{}BC$.
$ \angle PNL\equal{}\angle A'DP\equal{}\angle L'LD$ it means that arcs $ A'L'\equal{}AL$ and cause $ LL'\parallel{}BC$ therefore $ AA'\parallel{}BC$ hence A' is a constant point and hence L is constant.

Thank you dear [b][size=100]Amir.S[/size][/b], for a nice solution.

[b][size=100][color=DarkBlue]The simple, is the most important ![/color][/size][/b]

Kostas Vittas.

Ok,an inversion,by center $ A$,and an arbitrary radius $ k$,would help me to solve this nice problem!
Consider $ H$ as foot of the altitude from$ A$ to $ BC$,and$ Q$ is the intersection of $ MN$ and the circumcircle of$ \triangle ABC$.
The changes will be found according to following relations:
$ B,C,M,H,P,E,Q,N \longrightarrow B',C',M',H',P',E',Q',N'$ (repectively)
$ BC \longrightarrow$arc $ B'C'$ of circumcircle of $ \triangle AB'C'$
$ MN \longrightarrow$arc$ M'N'$ of circumcircle of $ \triangle AM'N'$
$ Q \longrightarrow B'C' \cap$ circumcircle of $ \triangle AM'N'$
$ H \longrightarrow H'$,which $ AH'$ is the diameter of circumcircle of $ AB'C'$
the other changes would  obviously concluded, lets look at new problem,it will turn into this one:
problem:$ \triangle AB'C'$ is given with its circumcircle,$ P'$ is an arbitrary point on the circum circle,and let
$ N'$,be the intersection of $ AP',B'C'$.$ AH'$ is the diameter of the circumcircle.
$ H'P' \cap B'C' \equal{} E'$, and $ E'A$,meets the circumcircle at $ M'$,prove that the circumcircle of $ \triangle AM'N'$ goes through a fixed point.
In fact ,in the new problem we should prove that $ Q'$,which is the intersection of the circumcircle of $ \triangle AM'N'$,with segment $ B'C'$,is the fixed point.
[color=red]Solution: $ E' \in$ radical axis of the circumcircles of $ \triangle AB'C',\triangle AM'N' \Longleftrightarrow Q'N'P'H'$ is cyclic $ \Longleftrightarrow H'Q' \perp B'C' \Longrightarrow Q'$is the fixed point![/color]
I appreciate Amir.s solution,he is really a great geometer,
Dears, how good it would be, if we could change our way of seeing,
To see without comparison.
How good it would be, if the beauty could be in our eyes,not in the things which we are looking at!
A great geometer, sees every solution,beautifully.the beauty is in his eyes.the one with the power of seeing beautifuly,is the real geometer!,in fact who is the real geometer? :|


Compute $\sum_{k=0}^{359}k\cos k^\circ$.

EDIT: Fixed a typo. :oops:

Sorry, there was a typo.  I fixed it.  Otherwise it's really easy.

[hide]
Pair terms up as
\[\sum_{k=1}^{179}\left( (k+360-k) \cdot \cos k^\circ \right)+0 \cdot \cos 0^\circ+180 \cdot \cos 180^\circ = 360 \sum_{k=1}^{179}\cos k^\circ-180 \]
then note that when added up, every cosine has its opposite, reflected over the y-axis, summed as well; so that sum is equal to $0$, and the expression is equal to $\boxed{-180}$.[/hide]

[hide="solution"]Since $\cos{\left(180\theta\right)}=-\cos{\theta}$, we can manipulate the sum to get

$\left(\sum_{n=1}^{179}-180\cos{n}\right)+180\cos{180}=$

$-180\left(\sum_{n=0}^{179}\cos{n}\right)$

Then everything cancels out beatutifully and we get

$\sum_{k=0}^{359}k\cos{k^{\circ}}=-180\cos{0}=-180$[/hide]


integration
trigonometry
calculus
calculus computations

Evaluate
\[ \int_{0}^{2}\sqrt {\frac{x}{4 \minus{} x}}dx
\]

Let $ x \equal{} 4\sin ^ 2 \theta$, $ \int_0^2\sqrt {\frac {x}{4 \minus{} x}}\ dx \equal{} \int_0^{\frac {\pi}{4}} \sqrt {\frac {4\sin ^ 2 \theta}{4(1 \minus{} \sin ^ 2 \theta)}}\cdot 8\sin \theta \cos \theta d\theta \equal{} 8\int_0^{\frac {\pi}{4}} \sin ^ 2 \theta d\theta \equal{} \pi \minus{}2$.

Sorry, I did not see the square root. :). Assigning x = 4 sin[size=59]2[/size] theta is correct.

Where did the sign of square root disappear?

In the above approach x should be 4sin^2(theta) and not 2sin^2(theta). If we use 2sin^2(theta) then the denominator cannot be reduced as 4(1-sin^2(theta)). Then the whole answer will be changed. For me the answer was calculated as Pi - 2. Where sin^2(theta) = (1 - cos(2(theta))). So, the integration of 8 sin^2(theta)d(theta) to the limit 0 to pi/4 and it would be integration of 4 * (1 - cos(2(theta))) to the limit 0 to pi/4, which would be 4(theta - sin(2(theta))/2) to the limit of 0 to pi/4. Now when we apply the limits it would be 4[pi/4 - 1/2 - 0] = pi - 2.

Please correct me if I'm wrong.

Sorry, I don't know how to write using mathematical expression using this editor. If anyone can help me it would be better.


Let $ a,b,c>0$, $ abc\equal{}1$.

Prove that $ \frac{1}{(a\plus{}b)b}\plus{}\frac{1}{(b\plus{}c)c}\plus{}\frac{1}{(c\plus{}a)a}\ge\frac{3}{2}$.


A circle with a radius of 2 units has its center at $ (0, 0)$. A circle with
a radius of 7 units has its center at $ (15, 0)$. A line tangent to both
circles intersects the x-axis at $ (x, 0)$ to the right of the origin. What
is the value of $ x$? Express your answer as a common fraction.

My method at state was to find the two points on x-axes,so (2,0) and (8,0) and reasoned it was 2/(7+2) of the way from 2 to 8, which is 10/3, which is correct, but why does that work?

We use similar triangles. From the diagram, we have a simple relationship:

$ \frac {x \plus{} 2}2 \equal{} \frac {13 \minus{} x}7\implies7x \plus{} 14 \equal{} 26 \minus{} 2x\implies9x \equal{} 12\implies x \equal{} \frac43$.

Since the center of the circle with radius $ 2$ is the origin, our final answer is $ \left(2 \plus{} \frac43,0\right)\implies\boxed{\frac {10}3}$.

Edit: I only separated the $ 2$ and the $ x$, the $ 7$ and the $ 6 \minus{} x$ to make it easier to show on a diagram. One could easily let $ x$ be the final answer, and then solve

$ \frac x2 \equal{} \frac {15 \minus{} x}7\implies 7x \equal{} 30 \minus{} 2x\implies9x \equal{} 30\implies x \equal{} \boxed{\frac {10}3}$.


search
inequalities
algebra unsolved
algebra

In which set we search for solutions? If we are interested in real solutions, it is not that difficult to find that $(0,0)$ and $(-1,-1)$ are the only ones. Write the first equation as
\[ \frac{x(x^7-1)}{x-1}=y^7 \]
(clearly $x=1$ leads to a contradiction) to see that $x$ and $y$ have the same sign. If $x,y\ge 0$ simple inequalities lead to $x=y=0$. If $x,y<0$ similar (not that simple, though) arguments lead to the other stated solution.
If we look for complex solutions, the problem is far more complicated. For instance, we might have $x=y=\pm \frac{1+i\sqrt{3}}{2}$, but other solutions, not as nicely expressed, do exist.

The problem is from the USSR MATHEMATICAL OLYMPIAD 1992. The official solution is a bit complicated.

hello, $(x,y)=(0,0)$ or (x,y)=(-1,-1)$ are the only real solutions.
Sonnhard.


analytic geometry
geometry open
geometry

After a math class, a graph of $ y \equal{} 2^x$ and the $ x$-axis was remained in blackboard, but the $ y$-axis and the coordinate unit were erased. Find the $ y$-axis and the coordinate unit by only using a ruler and a compass.

Sorry for wasting your time to read this stupid post.


Let $(E, ||)$ be a normed real vectorial space and $A$ be a finite subset of $E$. Prove the existence of a number $a$ with the property: for all $n\geq 1$ and all $x_1,...,x_n\in A$  there exist $c,b \in A$ such that we have
    $\frac{1}{n}\cdot\sum_{i=1}^{n}{||c-x_i||}\leq a\leq\frac{1}{n}\cdot\sum_{i=1}^{n}{||b-x_i||}$.

I think it works also for a compact subset $ A \subset E $.
For every collection $ v = (x_{1},x_{2}, \ldots , x_{n}) $, where
$ x_{1},x_{2}, \ldots , x_{n} \in A $ define $ f_{v}(y) :=
\frac{1}{n} \sum_{j=1}^{n} ||y-x_{j}|| $. Then because $ A $ is
compact, there exist $ m_{v} := max_{ y \in A } f(y) $, $ M_{v} :=
min_{ y \in A } f(y) $. Let us show that for every two collections
$ v = (x_{1},x_{2}, \ldots , x_{n}) $, $ w = (y_{1},y_{2}, \ldots
, y_{k}) $ we have that $ [m_{v},M_{v}] \cap [m_{w},M_{w}] \neq
\emptyset $. Indeed , define $ t := \frac{1}{nk} \sum_{ 1
\leqslant i \leqslant n, 1 \leqslant j \leqslant k } ||x_{i} -
y_{j}|| $. Then $ t = \frac{1}{k} \sum_{1 \leqslant j \leqslant k}
f_{v}(y_{j}) = \frac{1}{n} \sum_{1 \leqslant i \leqslant n}
f_{w}(x_{i}) $, therefore $ m_{v},m_{w} \leqslant t \leqslant
M_{v},M_{w} $, so $ [m_{v},M_{v}] \cap [m_{w},M_{w}] \neq
\emptyset $. Therefore we have a collection of closed intervals $
[m_{v},M_{v}] $, where $ v $ runs over all collections as above,
and we know that any two intervals intersect. Therefore all the
intervals in this collection have non-empty intersection. If we
take some number $ a $ in this intersection, then it clearly
satisfies the condition of the problem.

I hope it is correct.

That's a very very nice proof, thanks a lot, Leva.  :)


geometry rhombus LaTeX rectangle probability

Although I am not writing KVPY this year but i felt the need to have this topic.....
First of all
BEST OF LUCK!!!!!!
to all the people writing the exam.....

So hows your prep... and wats your strategy for november 2..... (it sounds so kiddish  :rotfl: )

well u messed it :P :P
it shud hav been only Pre-exam discussion :rotfl: u can still edit though

my preparation is quite good to save me from dishonour of scoring 0. :(

I am sure i'll answer at-least 1 question in 3 hrs. :)

Preparation is nothing and i am just giving it as timepass

[quote]rituraj007 well u messed it
it shud hav been only Pre-exam discussion [/quote]

Why to waste the space..... ???

Hey learn as many questions you can i the exam and post them here  :P
it would be really helpful  :lol:

well i remem the simplest ques.............
1st ques of chem part 1...........

what volume of 0.5M naoh is required to completly nuetralize 10 ml of 2Mhcl------------
a)20    b)40   c)15   d)50ml

A great paper. Great questions.  :)
The problem was that I couldn't get even 1 questions out of it. :(
I remember only this one-

which one is most polar
a)butane   b)butene c)1 Butanol   d)1 butanal

Didn't knew the answer

yup.............evn i didnt remem d answer pulkit...........
i just marked butanol????????????

pl someone tell is it crrect/////////????

Thank you,........ if only I hadn't circled Butanal. :oops_sign:

ritu is correct according to me

 i think i lost in the physics an bio 2 marks section :(

How was maths....???
and cummmon u certainly remeber some other questions too.....

ya maths went very cool............ :lol:  :D
4 questions............i am tryin to remember................
Q1.
a card has 3 statements writen on it-
1)on dis card exactly 1 stat is false
2)xactly 2 are wrong
3)xactly 3 r wrong

Q:how many r rong???????

Q2)somethin lyk rhombus...

ya

in a rhombus ov area...A
1 diagonal is 2 times other..........
find side????????

Q3)wats the smallest area formed by lattice points a,b,c-a(2,3).....b(8,10)

The Rhombus question was as follows (as i remember it)

In a rhombus the area is 'A'. One diagnol is double the other one.
Find the side of rhombus in terms of 'A'?

WAS the answer- (square root of (5A))/2 ?

hello mittal that was the easiest according to me
the only one which struck me hard was this
$ (sqrt x1\minus{}1) \plus{}2(sqrt x2\minus{}4) \plus{}3(sqrt x3\minus{}9) \plus{}4(sqrt x4\minus{}16) \plus{}5(sqrt x5\minus{}25)  \equal{} x1 \plus{} x2 \plus{} x3 \plus{} x4 \plus{}x5
find : x1 \plus{} x2 \plus{} 3x \plus{} x4 \plus{} x5  /2$

sorry,i'll post it in latex soon :blush:

[quote="mittal"] if only I hadn't circled Butanal[/quote]
it [b]can[/b] be the answer :oops:

i think that on high probablity the people are bored of this joke :rotfl:  :rotfl:

or it is also probable that people are tired of your coffee drinking avatar :rotfl:  :rotfl:

my coffee drinking avatar??
no it isn't ..its just me crying after an im[b]probable[/b] result :P

Different people will probably interpret it differently. I thought it was probably Ritu being struck by lightning while reading his fav book on 'probability'!

probability of such a event tends to zero :rotfl:

But it is still probable that some probability of the event still exists.

but still the probalblity of the probablity of the event being zero is zero so the event must exist $ QED$ :rotfl:  :rotfl:

looks like the probablity of this thread continuing is decreasing
 :rotfl:

Yea.
But I want to increase that probability.
You should ask probability 1.01   about it....
The probability must be increased so.

However, I thought that probably rituraj was shocked at the taste of the milk.. :rotfl:  :rotfl:

Really I have a question for rituraj:
Why have you used the words of Swami Vivekananda and not even written his name??
Your avator is very funny... :rotfl:
................

i thought milk is sth 4 drinking and not reading :P

well, the place where i read the quote ,Swami's name wasnt written :rotfl: so i cant do nething

and moreover,what is there in the name ,his work can b compared to infinity :roll:

[quote="asish2000"]guys KVPY results are out??
who all are selected??
list starts with me
1. Asish Mahapatra, BBSR :blush:[/quote]
http://www.mathlinks.ro/viewtopic.php?t=259271


hello, we also have $ \binom{n}{n}\equal{}1$.
Sonnhard.

$\binom{n}{n}$ will always be $\boxed{1}$.

[hide=answer]it will end up into 505/505 which is 1[/hide]

think about it...
you choose 505 apples from 505 different apples only one way right? Just pick all of them!

[hide=Solution]Any number choose itself is $1.$[/hide]


At wikipedia you can pick and choose which proof you like best. http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra However, I don't think you'll find an easy proof.


Hi every one !  :)
 I think this one is really difficult and very nice:

 [b]Let $a,b,c$ are non-negative integers. Prove that:[/b]

 $\Large \frac{a^3}{2a^2-ab+2b^2}+\frac{b^3}{2b^2-bc+2c^2}+\frac{c^3}{2c^2-ca+2a^2}\geq \frac{a+b+c}{3}$

 Good Luck !  :D

this ineq is similar one ineq of Japan.
Used$SCHUR$.

[quote="altusi"]quite simple if i am not making a mistake...
by substructing (a+b+c)/2 from the two sides ,we obtain:

Inequality <=> sum[(ab(2b-a)/(4b²+4a²-2ab)]=< (a+b+c)/6

witch fellows obviously from the famous one: a²+b²>=2ab
  :D[/quote]

Sorry your solution doesn't work for the case 2b-a<0 ...
We must prove that

$\sum \frac{(2b-a)(a-b)^2}{2a^2-ab+2b^2} \geq 0$

But how to solve it in the case $b>a>c$ and $b>2c$

ok.I agree with you.
and it is easy to prove:
$\Large \frac{c^3}{2a^2-ab+2b^2}+\frac{a^3}{2b^2-bc+2c^2}+\frac{b^3}{2c^2-ca+2a^2}\geq \frac{a+b+c}{3}$
but it is not :)

oops :blush:  sorry,
in fact i was making a mistake as i thought...

[quote="Lam Trường"]
 [b]Let $a,b,c$ are non-negative integers. Prove that:[/b]
 $\Large \frac{a^3}{2a^2-ab+2b^2}+\frac{b^3}{2b^2-bc+2c^2}+\frac{c^3}{2c^2-ca+2a^2}\geq \frac{a+b+c}{3}$
[/quote]
Proof. (by Peter Shcolze method)
$\sum\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a+b+c}{3}\Leftrightarrow\sum(\frac{a^3}{2a^2-ab+2b^2}-\frac{a}{3}-\frac{a-b}{3})\geq0\Leftrightarrow$
$\Leftrightarrow\sum\frac{(a-b)^2(2b-a)}{2a^2-ab+2b^2}\geq0.$
Hence, if $\{\frac{a}{b},\frac{b}{c},\frac{c}{a}\}\subset(0,2]$ then our inequality is proved.
Let $a=max\{a,b,c\}.$
1)$a\geq b\geq c.$ Then $\frac{c}{a}\in(0,1].$
1.1 $a\geq2b.$
Then $\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a}{2}$ and $\frac{b^3}{2b^2-bc+2c^2}\geq\frac{b}{3}.$
Hence, $\sum\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a}{2}+\frac{b}{3}\geq\frac{a+b+c}{3}.$
1.2 $b\geq2c.$
Hence, $\sum\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a}{3}+\frac{b}{2}\geq\frac{a+b+c}{3}.$
2) $a\geq c\geq b.$ Thence, $\frac{c}{a}\in(0,2], \frac{b}{c}\in(0,2].$
Let $a\geq2b.$
Hence, $\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a}{2}$ and $\frac{b^3}{2b^2-bc+2c^2}\geq\frac{b}{3}-\frac{c}{9}.$ $(!)$
$(!)\Leftrightarrow2t^3-7t^2+5t+3\geq0,$ where $t=\frac{c}{b}\geq1.$ This is true.
Hence, $\sum\frac{a^3}{2a^2-ab+2b^2}\geq\frac{a}{2}+\frac{b}{3}-\frac{c}{9}+\frac{c^3}{2c^2-ac+2a^2}.$
But $\frac{a}{2}+\frac{b}{3}-\frac{c}{9}+\frac{c^3}{2c^2-ca+2a^2}\geq\frac{a+b+c}{3}\Leftrightarrow$
$\Leftrightarrow6x^3-19x^2+14x+2\geq0,$ where $x=\frac{a}{c}\geq1.$ This is true also. :)

[quote="wxydx00"]Proof. (by Peter Shcolze method)

But what is P-S method?
Is it a kind of method of undetermined coefficients?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23113
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22937

[quote="Lam Trường"]Let $ a,b,c$ are non-negative numbers. Prove that:

$ \frac {a^3}{2a^2 \minus{} ab \plus{} 2b^2} \plus{} \frac {b^3}{2b^2 \minus{} bc \plus{} 2c^2} \plus{} \frac {c^3}{2c^2 \minus{} ca \plus{} 2a^2}\geq \frac {a \plus{} b \plus{} c}{3}.$[/quote]$ \sum_{cyc}{\frac {a^3}{2a^2 \minus{} ab \plus{} 2b^2}} \minus{} \sum_{sym}\frac {a}{3}$

$ \equal{} \sum_{cyc}{\frac {7b(b \minus{} c)^2\left(8a^2b^2 \plus{} 32a^2c^2 \plus{} 103a^2bc \plus{} 108bc^3\right) \plus{} a(2b \minus{} c)^2(b \plus{} 2c \minus{} 3a)^2\left(40a^2 \plus{} 82bc \plus{} 32ca \plus{} 99ab\right)}{1176\left(2a^2 \minus{} ab \plus{} 2b^2\right)\left(2b^2 \minus{} bc \plus{} 2c^2\right)\left(2c^2 \minus{} ca \plus{} 2a^2\right)}}$

$ \geq0.$

By the way, Lam Trường's inequality holds for [b]real[/b] $ a,b,c$  with $ a \plus{} b \plus{} c\geq 0\leq bc\plus{}ca\plus{}ab.$

I have a nice proof(and short!).The inequality is equivalent with $\sum{\frac{2ab^2-a^2b}{2a^2-ab+2b^2}} \le \frac{a+b+c}{3}$,which is true by AM>GM($2a^2-ab+2b^2\ge 3ab....$)

$y\ge z \implies \frac{x}{y} \le \frac{x}{z}$ provided $x$ is non negative. Can you clarify that all the numerators are non-negative?

Yes,you're right...;)Thank you for observation!

Check out this method of solving inequalities could b helpful in many cases  (Also known as the TITU ANDRESCUUS LEMMA) ..

I think could be used in this one also and the sum could be brought down into an simpler form...

$\frac{x^2}{a}+ \frac {y^2}{b} + \frac {z^2}{c} \ge\ \frac {(x + y + z)^2}{a+b+c} $

The inequality can expressed further by a simple inductive step....

[quote=zhaobin]ok.I agree with you.
and it is easy to prove:
$$\Large \frac{c^3}{2a^2-ab+2b^2}+\frac{a^3}{2b^2-bc+2c^2}+\frac{b^3}{2c^2-ca+2a^2}\geq \frac{a+b+c}{3}$$
but it is not :)[/quote]
Let $a,b,c$ be positive real numbers. [url=https://artofproblemsolving.com/community/c6h368611p2029186]Prove that[/url]
$$\frac{a^3}{2b^2-bc+2c^2}+\frac{b^3}{2c^2-ca+2a^2}+\frac{c^3}{2a^2-ab+2b^2}\ge\frac{a^4+b^4+c^4}{a^3+b^3+c^3}$$
Let $a, b,c $ be positive reals .  Prove that
$$\frac{a^2}{2b^2-bc+2c^2}+\frac{b^2}{2c^2-ca+2a^2}+\frac{c^2}{2a^2-ab+2b^2}\geq 1$$
$$\frac{a^3}{2b^2-bc+2c^2}+\frac{b^3}{2c^2-ca+2a^2}+\frac{c^3}{2a^2-ab+2b^2}\ge\frac{\sqrt{3(a^2+b^2+c^2)}}{3}$$
Old.
Where?

[quote=sqing]
Let $a, b,c $ be positive reals .  Prove that
$$\frac{a^2}{2b^2-bc+2c^2}+\frac{b^2}{2c^2-ca+2a^2}+\frac{c^2}{2a^2-ab+2b^2}\geq 1$$
[/quote]
Let $a,b,c \ge 0$ and $p > -2$ be real number. [url=https://artofproblemsolving.com/community/c6h2425504p19992334]Prove that[/url]
$$\frac{2a^2+\left(2p+1 \right)bc}{b^2+pbc+c^2} + \frac{2b^2+\left(2p+1 \right)ca}{c^2+pca+a^2} + \frac{2c^2+\left(2p+1 \right)ab}{a^2+pab+b^2} \ge \frac{3 \left(2p+3 \right)}{p+2}.$$


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

Let $w_1=C(O_1),\ w_2=C(O_2),\ O_1\ne O_2$ be two circles,

a fixed point $A\in w_1$ and a mobile point $M\in w_2$.

I note $\{A,N\}\subset AM\cap w_1;\ P\in w_1,\ NP\parallel MM;$

$R\in PM\cap w_1;\ L\in AR\cap MM$.

Ascertain the geometrical locus of the point $L$.

[u]Remark.[/u] I note $TT$ - the tangent in the point $T\in w$ to the circle $w$.

[color=blue][b]Lemma:[/b][/color] Let ANPR be a cyclic quadrilateral (convex or reflex) with the circumcircle $(O_1)$ and M the intersection of its opposite side lines AN, PR. Let a line through M parallel to the side line NP intersect the opposite side line at a point L. The point L lies on the radical axis of the circle $(O_1)$ and the point M (circle of zero radius).

The triangles $\triangle AML \sim \triangle MRL$ are similar, because the angle $\angle MLA \equiv \angle MLR$ is common, the angles $\angle MLR \equiv \angle MLP = \angle MPN$ are equal because the lines $ML \parallel NP$ are parallel and the angles $\angle MPN = \angle NAR \equiv \angle MAL$ are equal, because the quadrilateral ANPR is cyclic. Consequently, $\frac{LR}{LM} = \frac{LM}{LA}$ or $LM^2 = LR \cdot LA$. Thus the powers of the point L to the circle $(O_1)$ and to the point M are equal, which means that the point L lies on their radical axis.

Let $(O_2)$ be an arbitrary circle tangent to the line ML at the point M. The line ML is the radical axis of the circle $(O_2)$ and the point M. According to the above lemma, the point L lies on the radical axis of the circle $(O_1)$ and the point M, hence, it is the radical center of the circles $(O_1), (O_2)$ and the point M. Therefore, it lies on the radical axis of the circles $(O_2), (O_1)$. This remains in effect when the circles $(O_1), (O_2)$ and their radical axis are fixed and the point M moves in the defined way on the circle $(O_2)$.


$ a,b,c$ are const.Let $ x,y,z$ are real numbers and $ x^2\plus{}y^2\plus{}z^2\equal{}1$.  Find Min, Max of

$ S\equal{}\sqrt{a^2x^2\plus{}b^2y^2\plus{}c^2z^2}\plus{}\sqrt{a^2y^2\plus{}b^2z^2\plus{}c^2x^2}\plus{}\sqrt{a^2z^2\plus{}b^2x^2\plus{}c^2y^2}$

[quote="Algadin"]$ a,b,c$ are const.Let $ x,y,z$ are real numbers and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$.  Find Min, Max of

$ S \equal{} \sqrt {a^2x^2 \plus{} b^2y^2 \plus{} c^2z^2} \plus{} \sqrt {a^2y^2 \plus{} b^2z^2 \plus{} c^2x^2} \plus{} \sqrt {a^2z^2 \plus{} b^2x^2 \plus{} c^2y^2}$[/quote]
Using the Cauchy Schwarz inequality we have
$ S \le \sqrt {3\left( {\sum {a^2 x^2  \plus{} b^2 y^2  \plus{} c^2 z^2 } } \right)}  \equal{} \sqrt {3\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)}$
Moreover,
$ S^2  \equal{} \sum {a^2 }  \plus{} 2\sum {\sqrt {\left( {a^2 x^2  \plus{} b^2 y^2  \plus{} c^2 z^2 } \right)\left( {c^2 x^2  \plus{} a^2 y^2  \plus{} b^2 z^2 } \right)} }$
$ \Rightarrow S^2  \ge \sum {a^2 }  \plus{} 2\sum {ab}  \equal{} \left( {a \plus{} b \plus{} c} \right)^2$
$ \Rightarrow S \ge a \plus{} b \plus{} c$
Equality hold if and only if $ x\equal{}y\equal{}z$, :).

It is old,a problem proposed by Murray Klamkin in Crux Mathematicorum,my proof is same to yours,Honey_s


superior algebra
superior algebra unsolved

How can I describe the subfield of $\bar{\mathbb{Q}}_{p}$, the algebraic closure of the p-adic numbers, generated by a finite number of its elements?

What exactly do you want¿
The algebraic closure has infinity degree, if this what it points to.

I know that the algebraic closure is the union of the finite field extensions of $\mathbb{Q}_{p}$. What is, set-theoretically, the relation between the subfield I am interested in and these finite extensions?

... the subfield of $\bar{\mathbb{Q}}_{p}$ generated by two of its elemens, say

All finite extensions of perfect fields (more generally: all finite seperable extensions) are given by one single element (called a primitive element).
Thus you need not to distinguish between the numbers of "generators".

But what exactly do you want¿ It's not clear to me what you look for (the set theoretical properties are simply those of inclusion or not, depending on what you have).


I don't think it is allowed in our forum.

bundler, sorry to disappoint you but here in mathlink-AoPS, sharing ebooks or even encouraging downloading ebooks illegally is discouraged here :wink: Just so you know, some of the most respected authors of math olympiad books are administrators and moderators here so we just have to respect that these books are their intellectual properties... But I can give you the title of the geometry books that could you in your study then do a little research in a nearby library afterwards..

[quote="bundler"]hello guys, could anyone of you recommend me a free pdf ebook for geometry, thanks a lot.

 :)  :)[/quote]
See here http://trungtuan.wordpress.com/2010/04/16/topic-42/.

[quote="math154"][quote="Jason27603"]For a really good geometry e-book that is not illegally copied, see Geometry Unbound: [url]http://math.mit.edu/~kedlaya/geometryunbound/[/url] (No diagrams though :( )[/quote]
http://artofproblemsolving.com/Forum/viewtopic.php?f=50&t=70654
 :)[/quote]
Well yeah, I downloaded those, but they're not really part of the book, which sucks since you have to find the files and open them and whatnot.


Let a,b,c,d,e be positive reals numbers, Proved:
$\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b}\ge 0$

We have by caushy-schwartz:

$(2*(a+b+c+d+e))(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b})\ge (a-b+b-c+c-d+d-e+e-a)^2$

So, $\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+e}+\frac{d-e}{e+a}+\frac{e-a}{a+b} \geq \frac{0}{2*(a+b+c+d+e)}=0$

I' m not sure , but I think that all fractions must be positive(not negative!) for to apply Cauchy - Schwarz. Or not?I ' ll check it later.

Babis

This inequality was a conjecture of Vasc. :?

[hide]Rewrite the ineq as  $\sum\frac{a+b}{b+c}-5\geq 0$[/hide]

And then what will we do, Silouan? :huh:

In this case, AM-GM would finish the job... but silouan is wrong.  It should rewrite as

[hide]$\sum_{\text{cyc}} \frac{a+c}{b+c}-5 \geq 0$[/hide] which doesn't help much...  :huh:


group theory
superior algebra
superior algebra unsolved

$G$ is a group with order $p^{4}$ such that its center has $p^2$ elements, where $p$ is prime. How many conjugacy classes does $G$ have ?

P.S: a solution to Group Theory $9$ has been posted to sci.math for those interested (it was harder than this one).

Because $|Z(G)|=p^2$ thus if $x \not \in Z(G)$ then $|C_{G}(x)|=|<x,Z(G)>|=p^3$ so $|C_x|=|G: C_{G}(x)|=p$ what means that $G$ consists of conjugacy classes with $p$ elements (there are $k$ such classes) or $1$ element ($Z(G)$).
But because $p^2+kp=p^4$ then $k=p^3-p$ so there are $p^3+p^2-p$ conjugacy classes.


If $ a$, $ b$, and $ c$ are positive real numbers such that $ a(b \plus{} c) \equal{} 152$, $ b(c \plus{} a) \equal{} 162$, and $ c(a \plus{} b) \equal{} 170$, then abc is

$ \textbf{(A)}\ 672 \qquad
\textbf{(B)}\ 688 \qquad
\textbf{(C)}\ 704 \qquad
\textbf{(D)}\ 720 \qquad
\textbf{(E)}\ 750$

[hide="cheating"]Multiplying, we have $ abc(a\plus{}b)(a\plus{}c)(b\plus{}c)\equal{}152\cdot162\cdot170$. Simply by plugging in answer choices, we find that the only value of abc that divides the right hand side is 720, so our answer is $ \text{(D) 720}$.[/hide]

^basically the same solution...but you can add up the 3, divide by 2, to get the master equation with ab+ac+bc=X then subtract the 3 equations from the master equation...anyways, then you get ab, ac, and bc.  Multiply the 3 numbers, then square root it to get the answer (and keep it in prime factorization to make it easier to square root it)

[quote="xpmath"][hide="cheating"]Multiplying, we have $ abc(a \plus{} b)(a \plus{} c)(b \plus{} c) \equal{} 152\cdot162\cdot170$. Simply by plugging in answer choices, we find that the only value of abc that divides the right hand side is 720, so our answer is $ \text{(D) 720}$.[/hide][/quote]
I think a,b,c are real number? therefore, the word "divide" shoudn't be used here...

[hide="outline of what to do"]Add all three equations up, divide by two, and subtract all 3 given equations and write out the results separately. You have ab=something, bc=something, ac=something. Multiplying and taking the square root gives abc.[/hide]

hello, solving the equation system i have found $ a\equal{}8,b\equal{}9,c\equal{}10$ or $ a\equal{}\minus{}8,b\equal{}\minus{}9,c\equal{}\minus{}10$, so the product
$ abc\equal{}720$.
Sonnhard.

Oh darn I didn't see the word real. Sorry.

@Dr Sonnhard Graubner: Would you please share with us how you solved the system of equations?

[hide]

Expanding, we have:

\[ ab\plus{}ac\equal{}152\]
\[ ab\plus{}bc\equal{}162\]
\[ ac\plus{}bc\equal{}170\]

If we add the equations, we get $ ab\plus{}bc\plus{}ac\equal{}242$. So, $ ab\equal{}72$, $ bc\equal{}90$, and $ ac\equal{}80$. Multiplying and taking the squareroot, $ abc\equal{}\sqrt{80\cdot 90\cdot 72}\equal{}\sqrt{7200\cdot 72}\equal{}\sqrt{72^2\cdot 10^2}\equal{}720\Longrightarrow\mathrm{ (D) \ }$

[/hide]

hello, from the first equation we get $ c \equal{} \frac {152}{a} \minus{} b$, setting this in the other two equations we have
$ 152\frac {b}{a} \plus{} ab \minus{} b^2 \equal{} 162$ (1)
$ 152\frac {b}{a} \minus{} ab \minus{} b^2 \equal{} 18$   (2)
From here we get $ ab \equal{} 72$, substituting this two times in (1) we get $ \frac {10944}{a^2} \minus{} \frac {5184}{a^2} \equal{} 90$ and from here we get $ a^2 \equal{} 64$ and finally $ a_1 \equal{} 8$ or $ a_2 \equal{} \minus{} 8$.The other solutions i have posted above.
Sonnhard.

[quote="AruKasera"][quote="lifeisacircle"]

If we add the equations, we get $ ab\plus{}bc\plus{}ac\equal{}242$. So, $ ab\equal{}72$, $ bc\equal{}90$, and $ ac\equal{}80$.

[/quote]

Sorry for the long revive, but, how did you get that? Does anyone have a legit solution?[/quote]

[hide]All lifeisacircle did was expand the original equations and add them together.

Upon expanding we get $ab+ac=152$  , $ab+bc=162$, and $ac+bc=170$.
Then we can take these and add them together.
$ab+ac+ab+bc+ac+bc=152+162+170$
This then simplifies to
$2(ab+bc+ac)=484$
Hence $ab+bc+ac=242$
With this information we can now find $ab$, $ac$, and $bc$.
Because $ab+bc+ac=242$, we know that
$ab+(170)=242$
$bc+(152)=242$
$ac+(162)=242$
So, $ab=72$, $bc=90$, and $ac=80$.
Now we just multiply these three equations together and take the square root.
Note that we could go further and actually find $a$, $b$, and $c$, but we don't have to.
$abc=\sqrt{ab*bc*ac}=\sqrt{72*90*80}=\sqrt{72^2*10^2}=\boxed{720}$[/hide]

[color=darkred][b][u]PP[/u].[/b] Let  $ a$, $ b$, and $ c$ be three [u]positive[/u] real numbers such that $ a(b \plus{} c) \equal{} 152$ , $ b(c \plus{} a) \equal{} 162$ and $ c(a \plus{} b) \equal{} 170$ . Find $abc$ .[/color]

[b][u]Proof[/u].[/b] Denote $\left|\begin{array}{c}
bc=x\\\
ca=y\\\
ab=z\end{array}\right|$ . Our system becomes $\left\{\begin{array}{c}
z+y=152\\\
x+z=162\\\
y+x=170\end{array}\right|\ \bigoplus\ \implies\ x+y+z=242\ \implies$

$\left|\begin{array}{c}
bc=x=242-152=90\\\\
ca=y=242-162=80\\\\
ab=z=242-170=72\end{array}\right|\ \bigodot\ \implies$ $abc=\sqrt {xyz}=$ $\sqrt{90\cdot 80\cdot 72}=$ $\sqrt {72^2\cdot 10^2}=720\implies$ $\left|\begin{array}{c}
a=\frac {abc}{bc}=8\\\\
b=\frac {abc}{ca}=9\\\\
c=\frac {abc}{ab}=10\end{array}\right|$ .

[b]Solution:[/b] Adding all three equations and dividing by $2$, we have that $ab + bc + ac = 242$. Now, subtracting each equation from this, we get that $bc = 90, ac = 80, ab = 72$. Hence, $abc = \sqrt{90 \cdot 80 \cdot 72} = \boxed{720}$.


function
algebra
functional equation
calculus
calculus computations

A function f:R->R satisfies the intermediate value theorem for all intervals $ [a,b]$ and $ c$ with $f(a) < c < f(b)$
prove or give a counterexample
that f is continuous

For example:

$ f(x) \equal{} sin x \text{ } \forall x \in \mathbb{R}^*$
$ f(0) \in [ \minus{} 1, 1]$ is arbitrary.

This function is not continuous in $ 0$, although it has the intermediate value property.

[quote="feliz"]it has the intermediate value property.[/quote]

How that?

Total overkill, but a counterexample: Take a nonlinear solution of the Cauchy functional equation $ f\left(x\plus{}y\right)\equal{}f\left(x\right)\plus{}f\left(y\right)$...

  darij

Ops... I am sorry. I meant $ f(x) \equal{} sin\frac{1}{x}$.


In the 2000 Summer Olympics, the women's 400-meter freestyle winner took 4 minutes 4 seconds to swim her race. The 200-meter freestyle winner took 1 minute 57 seconds. If the 400-meter winner had swum her whole race at the same average speed as the 200-meter winner, by how many seconds would she have reduced her time?

The $ 200$ meter freestyle swimmer would have swum for $ 117(2)\equal{}234 \text{seconds}\equal{}3\text{minutes}54\text{seconds}$. Thus, she would have reduced her time by $ 10$ seconds.


greatest common divisor
number theory
least common multiple
linear algebra
matrix

Can anyone help me with the following problem:

PROBLEM 2: Cow Imposters [Hal Burch, 2002]

FJ no longer uses the barbaric custom of branding to mark the cows
that he owns.  Instead, he creates a binary code of B (1 <= B <= 16)
bits for each cow and embosses it onto a metal strip that is fastened
to each cow's ear.

The cows have taken in a stray and wish to create an ID strip for it.
Unknown to FJ, they have created a machine that can make a new ID
strip by combining two existing ID strips using the XOR operation (ID
strips are not consumed by this machine, and the same ID strip can be
used for both inputs).

The cows wish to create a specific ID strip for the stray or at least
get as close to a desired ID as possible -- with the smallest possible
number of bits differing between the goal ID strip and the best
possible new ID strip.

Given a set of E (1 <= E <= 100) existing ID strips, the goal ID
strip, and a large number of blank ID strips to hold intermediate
results, calculate the closest possible ID strip that can be created
from the existing ID strips.

If more than one ID is closest, choose the one that can be created in
the fewest steps.  If there is still a tie, choose the `smallest' ID
(i.e., if you sorted all the IDs, the one that is first).

PROBLEM NAME: impster

INPUT FORMAT:

* Line 1: Two space-separated integers: B and E.

* Line 2: The goal ID string, represented as a string of B 0's and 1's
        (with no spaces).

* Lines 3..E+2: Each line contains an existing ID string, represented
        as a string of B 0's and 1's (with no spaces).

SAMPLE INPUT (file impster.in):

5 3
11100
10000
01000
00100

OUTPUT FORMAT:

* Line 1: A single integer that is the minimum number of steps
        required to create the best possible ID strip.

* Line 2: A single line with the best possible ID strip that can be
        created from the E existing ID strips

SAMPLE OUTPUT (file impster.out):

2
11100

[quote="bugzpodder"]what is Rob Kolstad is planning to do with those contest problems?  hide them forever?  clearly he's not putting them into trainign pages
cauchyguy, where did you get this from?  is there some sort of website that has these?[/quote]

Why won't he put them into the training pages?

I don't know why these problems are not in the training gateway per se. However, I think page 6 of the training gateway will most likely have some problems from the recent contests since it is titled 'contest practice.'

Two more problems till I can say this for certain  :)

This problem can be solved using dynamic programming. Refer to USACO Training Gateway "2.2: Dynamic Programming" & "2.5: Knapsack Problems" if you're lost by what I say below.

Basically, you use dynamic programming to find the minimum # of ID's you have to XOR to get to a given ID. The resulting values (which indicates the # of IDs used) can be kept in a table of size 2^16 since there are only a max of 2^16 IDs possible.

You then add to each entry in the table above the "closeness" of the ID that corresponds with that value. The answer is the entry in the table with the minimum value.

For example, entry 11100 will hold 0 (since it is a given ID) + 2 (its closeness)

Entry 01100 will hold will hold 1 (01000 XOR 00100) + 1 (its closeness).

Thank you, Aidin. I understand how to do the problem now.

[quote="bugzpodder"]what is Rob Kolstad is planning to do with those contest problems?  hide them forever?  clearly he's not putting them into trainign pages
cauchyguy, where did you get this from?  is there some sort of website that has these?[/quote]

I just searched for the problems on Google. Here are the rest of the February 03 problems:

PROBLEM 1: Cow Math [Vlad Novakovski, 2002]

Taking their cue from the builders of the USA's Interstate Highway
system, the cows have introduced the Interpasture Path numbering
system.  They have already numbered the N (2 <= N <= 25) pastures with
the integers 1..N and now are numbering each path between two pastures
with its own distinct Interpasture Path number in the range 1..2000
(e.g., I-9 and I-16).

In an example Interpasture Path map, four pastures numbered 1, 2, 3,
and 4 are connected by Interpasture Paths I-3, I-6, I-9, and I-16:

                  4--<I-6>--2
                 /         /
             <I-16>     <I-9>
               /         /
              1--<I-3>--3

Bessie likes to walk from pasture 1 to pasture 2 on the nifty new
Interpasture system. During each walk, she never visits the same
pasture twice, so possible walks on the sample map above are 1-4-2
and 1-3-2.

Over the years, Bessie has developed an amazing mathematical skill
that she likes to exercise.  During each walk, she enjoys finding the
greatest common factor (GCF) of the Interpasture Paths that she
traverses.  For instance, the walk designated 1-4-2 touches I-16 and
I-6 which have the greatest common factor of 2 (since 2 properly
divides into 16 and 6 but no larger integer does).

As she walks the pastures day after day, she takes all the possible
routes from pasture 1 to pasture 2 and remembers each of the GCFs.
After she has taken every possible walk once, she computes the least
common multiple (LCM) of all the GCFs.  For this example, the two GCF
values are 2 and 3 (GCF(6,16)=2 and GCF(3,9)=3), so the LCM is 6.

For large networks of paths, Bessie might get tired of all the
walking, but she really wants to know the LCM for every map.
Calculate that number for her.

PROBLEM NAME: cowmath

INPUT FORMAT:

* Line 1: N

* Lines 2..N+1: These N lines represent the symmetric Interpasture
        Path connectivity matrix of the pastures.  Line L shows the
        connectivity between pasture L-1 and the other pastures with
        its N space-separated integers.  The first integer on each
        line is the Interpasture Path number that connects pasture L-1
        and and pasture 1; the second integer is the IP number
        connecting pasture L-1 and pasture 2; etc.  If pasture A
        connects to pasture B, then pasture B connects to Pasture A.
        When no Interpasture Path is available, the integer is 0.

SAMPLE INPUT (file cowmath.in):

4
0 0 3 16
0 0 9 6
3 9 0 0
16 6 0 0

OUTPUT FORMAT:

A single line with a single integer that is the LCM of the GCFs of all
the possible walks from pasture 1 to pasture 2.  It is guaranteed that
the answer will contain 100 or fewer digits.

SAMPLE OUTPUT (file cowmath.out):

6

**********************************************************************

PROBLEM 2: Cow Imposters [Hal Burch, 2002]

FJ no longer uses the barbaric custom of branding to mark the cows
that he owns.  Instead, he creates a binary code of B (1 <= B <= 16)
bits for each cow and embosses it onto a metal strip that is fastened
to each cow's ear.

The cows have taken in a stray and wish to create an ID strip for it.
Unknown to FJ, they have created a machine that can make a new ID
strip by combining two existing ID strips using the XOR operation (ID
strips are not consumed by this machine, and the same ID strip can be
used for both inputs).

The cows wish to create a specific ID strip for the stray or at least
get as close to a desired ID as possible -- with the smallest possible
number of bits differing between the goal ID strip and the best
possible new ID strip.

Given a set of E (1 <= E <= 100) existing ID strips, the goal ID
strip, and a large number of blank ID strips to hold intermediate
results, calculate the closest possible ID strip that can be created
from the existing ID strips.

If more than one ID is closest, choose the one that can be created in
the fewest steps.  If there is still a tie, choose the `smallest' ID
(i.e., if you sorted all the IDs, the one that is first).

PROBLEM NAME: impster

INPUT FORMAT:

* Line 1: Two space-separated integers: B and E.

* Line 2: The goal ID string, represented as a string of B 0's and 1's
        (with no spaces).

* Lines 3..E+2: Each line contains an existing ID string, represented
        as a string of B 0's and 1's (with no spaces).

SAMPLE INPUT (file impster.in):

5 3
11100
10000
01000
00100

OUTPUT FORMAT:

* Line 1: A single integer that is the minimum number of steps
        required to create the best possible ID strip.

* Line 2: A single line with the best possible ID strip that can be
        created from the E existing ID strips

SAMPLE OUTPUT (file impster.out):

2
11100

**********************************************************************

PROBLEM 3: Traffic Lights [J. Kuipers, 2002]

The cows are going downtown!  Just like everyone else, they want to
optimize their driving time.

They have noted that when driving on a straight road with traffic
lights, the best strategy to get to their destination as quickly as
possible is not necessarily to drive as fast as possible to the next
traffic light, brake if it's red, wait for a green light, accelerate,
and then drive on.  It is often better to approach a traffic light
more slowly in order to have some speed when the light turns green.

The cows have observed the traffic lights for a very long time.  They
know that each traffic light behaves in the following way:
  * it is green for a certain amount of time Tg,
  * then it is red for an amount of time Tr,
  * then green again,
  * and so on.

Given
  * the integer length of the road L (1 <= L <= 100)
  * the number of traffic lights N (0 <= N <= L+1) along with
    information about each light:
     * the unique  position (0 <= position <= L)
     * Tg (1 <= Tg <= 10)
     * Tr (1 <= Tr <= 10)
     * color at t=0 (R or G)
     * Tc (the integer amount of time since the light last changed)

write a program to determine the minimal amount of time needed to get
to the end of the road.  Note that at each discrete time (starting at
t=0), a car may either change its speed (expressed in positional units
per time unit) by one or keep it constant.  The speed is always 0 or
positive, of course.  No driving backwards!

The car starts at position zero has has speed zero. The car must
complete its trip at position L, also with speed zero.  The car must
stop at all lights that are red when encountered -- be sure its speed
is 0 at the red light's position if it encounters a red light.  The
car may move when the light changes from red to green, but not when
it changes from green to red.

PROBLEM NAME: traffic

INPUT FORMAT:

* Line 1: Two space-separated integers: L and N

* Lines 2..N+1: Line Q gives the information for light Q-1.  The
        information is five space separated entities, all but the
        fourth of which are integers. The fourth entity is a single
        character, R or G.  In order, the fields are: P, Tg, Tr,
        initial color (R or G), and Tc.

SAMPLE INPUT (file traffic.in):

4 1
1 10 10 R 0

OUTPUT FORMAT:

A single line with a single integer that is the minimal time needed.

SAMPLE OUTPUT (file traffic.out):

12

**********************************************************************

PROBLEM 4: Farm Tour [Jan Kuipers, 2002]

When FJ's friends visit him on the farm, he likes to show them around.
His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first
of which contains his house and the Nth of which contains the big
barn.  A total M (1 <= M <= 10000) paths that connect the fields in
various ways. Each path connects two different fields and has a
nonzero length smaller than 35,000.

To show off his farm in the best way, he walks a tour that starts at
his house, potentially travels through some fields, and ends at the
barn.  Later, he returns (potentially through some fields) back to
his house again.

He wants his tour to be as short as possible, however he doesn't want
to walk on any given path more than once. Calculate the shortest tour
possible.  FJ is sure that some tour exists for any given farm.

PROBLEM NAME: tour

INPUT FORMAT:

* Line 1: Two space-separated integers: N and M.

* Lines 2..M+1: Three space-separated integers that define a path: The
        starting field, the end field, and the path's length.

SAMPLE INPUT (file tour.in):

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

OUTPUT FORMAT:

A single line containing the length of the shortest tour.

SAMPLE OUTPUT (file tour.out):

6

Thank you for the problems, cauchyguy. Problem 4 looks difficult (to me, at least). I am kind of new with these kind of problems. Could anyone offer some tips on #4 (Farm Tour)?


AMC
USA(J)MO
USAMO
floor function
calculus
integration
algebra

Recently the USAMO 2006 took place.  The official site is http://www.unl.edu/amc/

[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoQ.pdf]PDF Version[/url] [offsite link]
[url=http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/2006usamoS.pdf]Offical Solutions[/url] [offsite link]

Please discuss the problems in each problems thread. A .pdf-file of the problems can be retrieved [url=http://www.mathlinks.ro/Forum/resources/files/usa/27/USA-USAMO-2006-27-182.pdf]here.[/url]

\[ \textrm{Day I}  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490569#p490569]Problem 1[/url]
Let $p$ be a prime number and let $s$ be an integer with $0 < s < p.$ Prove that there exist integers $m$ and $n$ with $0 < m < n < p$ and

\[ \left \{\frac{sm}{p} \right\} < \left \{\frac{sn}{p} \right \} < \frac{s}{p}  \]

if and only if $s$ is not a divisor of $p-1$.

Note: For $x$ a real number, let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$, and let ${x} = x - \lfloor x \rfloor$ denote the fractional part of x.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490581#p490581]Problem 2[/url]
For a given positive integer $k$ find, in terms of $k$, the minimum value of $N$ for which there is a set of $2k + 1$ distinct positive integers that has sum greater than $N$ but every subset of size $k$ has sum at most $\frac{N}{2}.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490625#p490625]Problem 3[/url]
For integral $m$, let $p(m)$ be the greatest prime divisor of $m.$ By convention, we set $p(\pm 1) = 1$ and $p(0) = 1.$ Find all polynomials $f$ with integer coefficients such that the sequence
\[ \{p \left( f \left( n^2 \right) \right) \}_{n \geq 0}  \] is bounded above. (In particular, this requires $f \left (n^2 \right ) \neq 0$ for $n \geq 0.$)

\[ \textrm{Day 2}  \]

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490647#p490647]Problem 4[/url]
Find all positive integers $n$ such that there are $k \geq 2$ positive rational numbers $a_1, a_2, \ldots, a_k$ satisfying $a_1 + a_2 + \ldots + a_k = a_1 \cdot a_2 \cdots a_k = n.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490682#p490682]Problem 5[/url]
A mathematical frog jumps along the number line. The frog starts at $1$, and jumps according to the following rule: if the frog is at integer $n$, then it can jump either to $n+1$ or to $n + 2^{m_n+1}$ where $2^{m_n+1}$ is the largest power of $2$ that is a factor of $n.$ Show that if $k \geq 2$ is a positive integer and $i$ is a nonnegative integer, then the minimum number of jumps needed to reach $2^ik$ is greater than the minimum number of jumps needed to reach $2^i.$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=490691#p490691]Problem 6[/url]
Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC,$ respectively, such that $\frac{AE}{ED} = \frac{BF}{FC}.$ Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T,$ respectively. Prove that the circumcircles of triangles $SAE,$  $SBF,$ $TCF,$ and $TDE$ pass through a common point.

3. Actually, the sequence you're looking at is $\{p(f(n^2)) - 2n\}_{n \in \mathbb{Z}^+}$.  Also, $p(0) = \infty$.

5. The problem should state "where $2^{m_n}$ is the largest power of $2$ dividing $n$".


abstract algebra
superior algebra
superior algebra unsolved

I can't idenfity your formula in Word. Perhaps you mean
\[ Ann_A(M + N) = Ann_A(M) \cap Ann_A(N)  \]
where $M,N$ are submodules of a $A$-module $P$? However, this is quite trivial ... $r(M+N)=0$ implies $rM=0$ and $rN = 0$  since $M , N \subseteq M + N$. Conversely, $rM = 0$ and $rM = 0$ imply $r(M+N) \subseteq rM + rN = 0 + 0 = 0$.


Like - if your solution is appearing to be too long in math, you clearly realize that you're doing something wrong - as the solutions that come in the solutions manual are all short and elegant (especially so for the hard problems). We're also taught that "short and elegant" is always best. And indeed, most of the solutions I've seen at all levels are all very short, including the solutions I've seen for the International Math Olympiad and solution manuals I've seen for grad level textbooks.

Which then begs the question - which fields have problems with solutions that MUST be pages long? Engineering?

I am not quite sure what you mean. I mean take Wile's proof of Fermat's Last Theorem for example, it is a hundred or so pages long. Wiles worked 7 years solid on that and that doesn't mean he did something wrong.

Textbook (and math competition) problems. Most of the problems in textbook are designed to be short and elegant (you'll rarely find real world problems with long solutions in textbooks) - which then demands a conceptual leap between textbook problem solving and real world problem solving (especially when people are so overreliant on strategies that they've used specifically for textbook problems for years)


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Let $ p>0$ be a positive real number. The Liouville integral states that

$ \int_{0}^{\pi/2}\frac{\sin px\cdot\cos^{p\minus{}1}x}{\sin x}dx\equal{}\frac{\pi}{2}$.

Let $ \alpha\geq 0$. Find:

$ \int_{0}^{\pi/2}\frac{\sin px\cdot\cos^{p\minus{}1}x}{\sin x}x^{\alpha}dx\equal{}?$.

To compute Liouville's integral we need the following three formulas:

$ (1) \quad \sin px\equal{}p\,\sin x\minus{}p\, (p^2\minus{}1)\,\frac{\sin^3 x}{3!}\plus{}p\, (p^2\minus{}1)(p^2\minus{}3^2)\frac{\sin^5 x}{5!}\minus{}...$ for $ \text{Re}(p)>0$ and $ 0\le x\le \frac{\pi}{2}$

$ (2) \quad (2n)!\equal{}\frac{2^{2n}}{\sqrt{\pi}} \, n!\, \left(n\minus{}\frac12\right)!$ Legendre's duplication formula

$ (3) \quad \sum_{n\equal{}0}^\infty (\minus{}1)^n\, {x\minus{}1\choose n}\, \frac{1}{n\plus{}y}\equal{}B(x,y)$ for $ \text{Re}(x)>0$

After $ (1) \quad \sin 2px\equal{}p\sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{\left(p\minus{}\frac12\plus{}n\right)!}{ \left(p\minus{}\frac12\minus{}n\right)!}\, \frac{(2\sin x)^{2n\plus{}1}}{(2n\plus{}1)!}$

$ I: \equal{}\int_0^{\frac{\pi}{2}} \frac{\sin 2px}{\sin x}\, \cos^{2p\minus{}1} x\, dx\equal{}p\sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{\left(p\minus{}\frac12\plus{}n\right)!\,\,\, 2^{2n}}{\left(p\minus{}\frac12\minus{}n\right)!\, (2n)!}\, \frac{2}{2n\plus{}1}\, \int_0^{\frac{\pi}{2}} \sin^{2n} x\, \cos^{2p\minus{}1} x\, dx$ ,

where $ 2\int_0^{\frac{\pi}{2}} \sin^{2n} x\, \cos^{2p\minus{}1} x\, dx\equal{}B\left(n\plus{}\frac12,p\right)\equal{}\frac{\left(n\minus{}\frac12\right)!\, (p\minus{}1)!}{\left(n\minus{}\frac12\plus{}p\right)!}$ .

So $ I\equal{}p!\, \sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{2^{2n}\, \left(n\minus{}\frac12\right)!}{\left(p\minus{}\frac12\minus{}n\right)! \, (2n)!} \, \frac{1}{2n\plus{}1}$$ \stackrel{(2)}{\equal{}} \sqrt{\pi} \, p!\, \sum_{n\equal{}0}^\infty (\minus{}1)^n \, \frac{1}{\left(p\minus{}\frac12\minus{}n\right)! \, n!}\, \frac{1}{2n\plus{}1}$

$ \equal{}\frac{\sqrt{\pi}}{2}\, \frac{p!}{\left(p\minus{}\frac12\right)!} \sum_{n\equal{}0}^\infty (\minus{}1)^n\, {p\minus{}\frac12 \choose n}\, \frac{1}{n\plus{}\frac12}\stackrel{(3)}{\equal{}}\frac{\sqrt{\pi}}{2}\, \frac{p!}{\left(p\minus{}\frac12\right)!}\, B\left(p\plus{}\frac12,\frac12\right)$ $ \equal{}\frac{\sqrt{\pi}}{2}\, \Gamma\left(\frac12\right)\equal{}\frac{\pi}{2}$ .

Should there be a closed formula with elementary functions for the second integral?

\[ \int_{0}^{\pi/2}\frac {\sin px\cdot\cos^{p - 1}x}{\sin x}dx = \frac {\pi}{2}
\]
Let's do substitution :
$ e^{ix} = z$
then we get :
\[ \int_{\gamma} \frac {(z^{2p} - 1)(z^{2} + 1)^{p - 1}}{z^{2p - 1}(z^{2} - 1)2^{p - 1}}\frac {dz}{i} = \int_{\gamma} \frac {(z^{2(p - 1)} + ... + 1)(z^{2} + 1)^{p - 1}}{z^{2p - 1}2^{p - 1}}\frac {dz}{i}
\]
where $ \gamma$ is the contour on the picture. (where also radii of this "circle" tends to zero )
note that $ \int_{\gamma}\frac {dz}{zi} = \frac {\pi}{2}$
thus we get :
\[ \int_{\gamma} \frac {(z^{2(p - 1)} + ... + 1)(z^{2} + 1)^{p - 1}}{z^{2p - 1}2^{p - 1}}\frac {dz}{i} = \frac {\pi}{2}
\]
since the the coefficient of $ z^{2p - 2}$ in $ {(z^{2(p - 1)} + ... + 1)(z^{2} + 1)^{p - 1}}$ is equal to $ \sum_{k = 0}^{p - 1}\binom{p - 1}{k} = (1 + 1)^{p - 1} = 2^{p - 1}$
 :cool:

I don't  want to spend my  time   about calculation of this integral $ \int_{0}^{\pi/2}\frac {\sin px\cdot\cos^{p - 1}x}{\sin x}x^{\alpha}dx = ?$ since I don't see any sense in closed form of this integrals


do you guys have stalkers

THAT IS JUST SO COOL

wow

yes it is the return of arnav.

quagga quagga quagga


Functional Analysis
topology
advanced fields
advanced fields unsolved

Hello. I have a problem with one proof:

Let $ X,Y$ be a Banach space and $ f: X\to Y$ linear continous and surjective operator. Prove that for every sequence $ \{y_n\}_{n\in\mathbb{N}}\subset Y$ such that $ y_n\to 0$ there exists sequence $ \{x_n\}_{n\in\mathbb{N}}\subset X$ such that $ x_n\to 0$ and $ f(x_n) \equal{} y_n$.

Yes, this is just the open mapping theorem. What is your difficulty?

I don't know how describe method I choose element of $ \{x_n\}$, in formal way.

I define sequence of balls in X with radii $ \to 0$, image of each ball is open set in Y, so it's neighbourhood of $ 0$, hence almost all elements of $ \{y_n\}$ is for example in first ball, etc... ;)

You do not even need a sequence of balls here since the operator is linear. If the image of the unit ball contains the ball of radius $ \delta>0$, then the image of the ball of radius $ r$ contains the ball of radius $ \delta r$, whence you can always choose $ x_k$ with $ \|x_k\|\le \delta^{\minus{}1}\|y_k\|$.

But, of course, your argument also works. If $ B_n$ is a sequence of balls of radii tending to $ 0$ and $ C_n$ is a sequence of open balls contained in $ TB_n$, you can define $ n(k)$ as, say, the largest $ n\le k$ with the property $ y_k\in C_n$ and take any $ x_k\in B_{n(k)}$ whose image is $ y_k$.


inequalities
limit
function
inequalities solved

Show that for all positive real numbers x1,x2,... xn which satisfy x1x2...xn = 1 the following inequality holds:
1/(n-1+x1) + 1/(n-1+x2) + ... + 1/(n-1+xn) <= 1. Who solved this problems? I had a solution, and you?

I think I have a solution:

Lemma: a1,a2,....,an,x1,x2,....,xn>0 n,m>=2 n natural number, m real number

 \sum xn^m/an>=( \sum xn)^m/(n^(m-2)* \sum an)

We can observe that 1/(n-1+xn)=1-(n-2+xn)/(n-1+xn)

So we have  \sum 1/(n-1+xn)=n- \sum (n-2+xn)/(n-1+xn)

We have to prove that n- \sum (n-2+xn)/(n-1+xn)<=1

n<=1+ \sum (n-2+xn)/(n-1+xn)
Using the lemma we have:

n<=1+n(n^2-2n+ \sum xn)/(n^2-n+ \sum xn)
n^3-n^2+ n*\sum xn<=n^3-n^2+n* \sum xn+ \sum xn-n
n<= \sum xn
n*(x1*x2*...*xn)^(1/n)<= \sum xn  which is true

Hope this is the corect solution!

how do you prove the lemma? For m a natural number bigger than 2, I can apply Holder. Not sure how to do in other cases.

This problem again! Take $y_i = \frac{1}{n-1+x_i}$, and for the sake of contradiction assume $y_1 + \cdots + y_n > 1$. Since the $y_i$ are too large, the $x_i$ are too small and we shall prove $\frac{1}{x_1} + \cdots + \frac{1}{x_n} > x_1 + \cdots + x_n$. Since $x_iy_i = 1 - (n-1)y_i$, we have

$(n-i)y_i > (n-1)\left(y_i+1-\sum_{j=1}^{n} y_j\right)$
$ = (n-1)y_i - 1 + \sum_{j=1}^{n} \left(1 - (n-1)y_j\right)$
$ = -x_iy_i + \sum_{j=1}^{n} x_jy_j $
$\Longrightarrow \frac{n-1}{x_i} > -1 + \sum_{j=1}^{n} \frac{x_jy_j}{x_iy_i}$

Summing lemma this over $i$,

$(n-1)\left(\frac{1}{x_1} + \cdots + \frac{1}{x_n}\right) > \sum_{i=1}^{n} x_iy_i \left(\left(\sum_{j=1}^{n} \frac{1}{x_jy_j}\right) - \frac{1}{x_iy_i}\right)$
But by Cauchy and the lemma, we have

$\left(\sum_{j=1}^{n}\frac{1}{x_jy_j}\right) - \frac{1}{x_iy_i} \ge $ $\frac{(n-1)^2}{\left(\sum_{j=1}^{n}x_jy_j\right) - x_iy_i} > \frac{(n-1)^2}{(n-1)y_i} = \frac{n-1}{y_i}$

Whence,

$(n-1)\left(\frac{1}{x_1} + \cdots + \frac{1}{x_n}\right) > \sum_{i=1}^{n} x_iy_i\left(\frac{n-1}{y_i}\right) = (n-1)(x_1 + \cdots + x_n)$

as desired.

[quote="Maverick"]n<=1+ \sum (n-2+xn)/(n-1+xn)
Using the lemma we have:

n<=1+n(n^2-2n+ \sum xn)/(n^2-n+ \sum xn)[/quote]

here you took $m=1$, but, as you said, your lemma works only for $m \geq 2$ (consequence of holder).

in this particular case we have $\sum (n-2+x_i)/(n-1+x_i) \leq n(n^2-2n+ \sum x_i)/(n^2-n+ \sum x_i)$ by cebashev (hope i'm correct ---- anyway, by taking some simple examples i saw that lhs < rhs). so it seems that you proved only a weaker inequality :(

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mildorf, i don't understand your solution: how does the inequality you proved show the ineq??? to me it seems rather unobvious

What I did was prove the contrapositive - I proved that if the desired is false then the given is false.

the given is : [quote]positive real numbers $x_1$,$x_2$,$\ldots$,$x_n$ which satisfy $x_1 x_2 \ldots x_n = 1$[/quote].

i was asking how do you prove that (conclusion == false) implies (the given==false). this is what i didn't understand

My apologies! I was thinking of the similar problem in which it was given that $x_1 + \cdots + x_n = \frac{1}{x_1} + \cdots + \frac{1}{x_n}$. :oops_sign:

As for the one posted here, a solution is given by the fact that $f(x) = \frac{1}{k+e^x}$ has exactly one inflection point for positive $k$ (where $e^x = k$.) We consider $k = n-1$. This fact allows us to, by Jensen/Smoothing, equate $n-1$ of the $x_i$, obtaining the one variable inequality $\frac{n-1}{n-1+x} + \frac{1}{n-1+\frac{1}{x^{n-1}}} \le 1$, which after clearing denominators is equivalent to $g(x) = x(k-1) + x^{1-k} - k \ge 0$ for all positive $x$. As $g'(x) = k-1 + x^{-k} \cdot (1-k)$ is zero only at $x = 1$, it suffices to check that $g(1) = 0$, while  $\lim_{x \rightarrow 0^{+}} g(x) = \infty$, and $\lim_{x\rightarrow \infty} g(x) = \infty$.

The one-variable inequality is simply Bernoulli, eh? (I think sometimes the theorem says the exponent has to be > 1, but the proof works the same if the exponent is < 0)

[quote="Mildorf"]As for the one posted here, a solution is given by the fact that $f(x) = \frac{1}{k+e^x}$ has exactly one inflection point for positive $k$ (where $e^x = k$.) We consider $k = n-1$. This fact allows us to, by Jensen/Smoothing, equate $n-1$ of the $x_i$, obtaining the one variable ineq[/quote]

doesn't this allows only to make equal the variables smaller than $n-1$ (on the interval where it's concave)?

anyway, i have found a solution by prowler which seems to take care of this problem, but it is late in the night, so i'll check it tomorow

[quote="Mildorf"]Ah, but where it's convex one moves all of the variables apart, including setting all but one to the inflexion point, where they are then considered on the concave interval.[/quote]

let me see if i got this right:  let $f: [a,+\infty) \to \mathbb{R}$ be a convex function; also let $x \geq y \geq a$; by karamata: $f(x+y-a)+f(a) \geq f(x) + f(y)$, so we can do what you did.

very nice trick i must say !!!   :thumbup: