In a circle with radius $1$, draw a chord that subtends a minor arc $\widehat{AB}$ with measure $\theta$. If a triangle is drawn, such that two vertices are on the minor arc $\widehat{AB}$ and the third vertex is on the segment $\overline{AB}$, what is its maximum possible area?

[EDIT: Oh, oops; I posted this in the wrong forum. If no one comes up with a solution, can the mods move this to the Intermediate Forum?]

[quote="Ponnamperuma"]Excuse me, what do you mean for "a minor arc $\widehat{AB}$ with measure $\theta$"? Perhaps you identify an arc $\widehat{AB}$ and $\theta$ is one of the angles on the circle which insist on the arc $\widehat{AB}$?
Thank you in advance...  :)[/quote]

I mean that the central angle which subtends $widehat{AB}$ is $\theta$. Usually I think that saying these two things are equivalent... but I hope this clears that up.


vector
induction
linear algebra
linear algebra unsolved

I have been working on this question and I was wondering if you someone can help me out. Thanks.

Prove that if {w1, w2, .... , wN} is an orthogonal set of nonzero vectors, then the vectors, v1, v2, .... vN derived from the Gram-Schmidt process satisfy vi=wi for i=1,2,...., n. Hint: Use mathematical induction.
[/code]

Basis: Since the first step of Gram-Schmidt process is $ v_1 \equal{} w_1$, we have $ v_1 \equal{} w_1$ is trivial.

Induction: Suppose that $ v_k\equal{}w_k$ for all $ 1 \leq k \leq i\minus{}1$. We have
$ v_i \equal{} w_i \minus{} \sum_{j\equal{}1}^{i\minus{}1}{\frac{w_i \cdot v_j}{\| w_i \| ^ 2} v_i}$

Note that since $ \lbrace w_1, ..., w_n \rbrace$ is an orthogonal basis and $ v_j\equal{}w_j \forall 1 \leq j \leq i\minus{}1$ as the inductive hypothesis, we derive that $ w_i \cdot v_j \equal{} 0 \forall j\equal{}\overline{1,i\minus{}1}$. Hence, it follows that $ v_i \equal{} w_i$.

By principle of mathematical induction, we conclude that $ v_i \equal{} w_i \forall i$.

Note that it is not necessary to start with $ w_1$. We could start with any $ w_j$ but the result as stated does not hold. In fact, we still have the set $ \lbrace v_1, v_2, ..., v_n \rbrace$ is equal to the set $ \lbrace w_1,w_2,...,w_n \rbrace$.

[quote="mathangel"]
$ v_i \equal{} w_i \minus{} \sum_{j \equal{} 1}^{i \minus{} 1}{\frac {w_i \cdot v_j}{\| w_i \| ^ 2} v_i}$[/quote]
What's this? :huh:
A correct formula would be
$ v_i\equal{}{w_i \minus{} \sum_{j \equal{} 1}^{i \minus{} 1}\langle w_i,v_j\rangle v_j\over \|w_i \minus{} \sum_{j \equal{} 1}^{i \minus{} 1}\langle w_i,v_j\rangle v_j\|}$.

There are two interpretations of the Gram-Schmidt process. In the first, the goal is an orthogonal but not necessarily orthonormal set. In the second, the goal is an orthonormal set, and normalization is pursued at every step.  mathangel was using the first interpretation, olorin the second.

I'm teaching an elementary linear algebra course now, and I 'm using the first interpretation. To be honest, the arithmetic is a lot nicer that way in concrete examples, and if you want orthonormal, you can always normalize as the very last step.

I can't make sense of this.
The a correct formula for the orthogonal basis would be
$ v_i \equal{} w_i \minus{} \sum_{j\equal{}1}^{i\minus{}1}{\frac{w_i \cdot v_j}{\| v_j \| ^ 2} v_j}$,
which is nowhere near $ v_i \equal{} w_i \minus{} \sum_{j\equal{}1}^{i\minus{}1}{\frac{w_i \cdot v_j}{\| w_i \| ^ 2} v_i}$.
Even $ v_i \equal{} w_i \minus{} \sum_{j\equal{}1}^{i\minus{}1}{\frac{w_i \cdot v_j}{\| w_j \| ^ 2} v_j}$ would be false, as $ \|v_j\|\not\equal{}\|w_j\|$. (look at $ j\equal{}2$)

That was perhaps a little sloppy of mathangel, but in this very particular case, it truly does not matter. Since the $ w_i$ started out orthogonal, it turns out that $ w_i\equal{}v_i$ all along.


inequalities
trigonometry
inequalities proposed

[color=purple]Hi all Mathlinker ! Here are three problem of me in March and i hope you 'll kill it before i come back ML. :D .Prove or disprove that :
   1-[Murray Klamkin]
    $\frac{1}{x}sin A+\frac{1}{y}sin B+\frac{1}{z}sin C\leq\frac{1}{2}(xy+yz+zx)\sqrt{\frac{x+y+z}{xyz}}$
   2- [A Mathlinker] (I don't remember )
    $\frac{1}{x}sinA+\frac{1}{y}sinB+\frac{1}{z}sinC\leq\frac{3\sqrt{3}(x+y+z)}{2(xy+yz+zx)}$
   3-[evarist]
    $\frac{1}{x}sin A+\frac{1}{y}sin B+\frac{1}{z}sin C\leq\frac{9}{2(xy+yz+zx)}$
  Thank very much if you can show me the author of 2nd problem.And if you get the stronger please post them :D .Thanks and I hope you'll have fun ! :D [/color]

[quote="skywalker"][quote="evarist"]Hi all Mathlinker ! Here are three problem of me in March and i hope you 'll kill it before i come back ML. :D .Prove or disprove that :
   1-[Murray Klamkin]
    $\frac{1}{x}sin A+\frac{1}{y}sin B+\frac{1}{z}sin C\leq\frac{1}{2}(xy+yz+zx)\sqrt{\frac{x+y+z}{xyz}}$
   2- [A Mathlinker] (I don't remember )
    $\frac{1}{x}sinA+\frac{1}{y}sinB+\frac{1}{z}sinC\leq\frac{3\sqrt{3}(x+y+z)}{2(xy+yz+zx)}$
   3-[evarist]
    $\frac{1}{x}sin A+\frac{1}{y}sin B+\frac{1}{z}sin C\leq\frac{9}{2(xy+yz+zx)}$
  Thank very much if you can show me the author of 2nd problem.And if you get the stronger please post them :D .Thanks and I hope you'll have fun ! :D [/quote]
It is a pity for March First and Third is wrong   :D because it isn't homogeneous.
First must be
$\sum x\sin A\le\frac{xy+yz+zx}{2}\sqrt{\frac{x+y+z}{xyz}}$  :D
Prove it quite easily
$(\sum yz\sin A)^{2}\le (\sum yz)(\sum yz\sin^{2}A)\le \frac{(\sum yz)(x+y+z)^{2}}{4}$
$\Rightarrow \sum yz\sin A \le\sqrt{\sum yz}\frac{x+y+z}{2}$
Now replace $yz\rightarrow x,zx\rightarrow y,xy\rightarrow z$ we will get mine not yours  :D
And for the second It should weaker than mine if it isn't it will be false!
That's all  :D[/quote]
  [color=purple] Thank you skywalker ! Good proof but is the it the solution of Klamkin? Sorry because i always want to prove a problem by many way.[/color]
 [quote="skywalker"] It is a pity for March First and Third is wrong   :D because it isn't homogeneous.[/quote]
[color=purple]  Why please give me an example for third .What about the 2nd ? [/color]

[quote="skywalker"]
It is a pity for March First and Third is wrong   :D because it isn't homogeneous.
[/quote]

You don't undertand anything I said, do you understand the word homogeneous, ok if you want an example it is easy try when $x=y=z=2$ thus yours become
$\sum \sin A\le \frac{3}{4}$ it is wrong  now try with $A=B=C=\frac{\pi}{3}$  :D  :D
[quote="skywalker"]
And for the second It should weaker than mine if it isn't it will be false!
[/quote]
Indeed from 1st $\sum\frac{\sin A}{x}\le \frac{(x+y+z)\sqrt{\sum yz}}{2xyz}\le\frac{3\sqrt{3}(x+y+z)}{2(xy+yz+zx)}$
and the solution for 1st is of me!
Before reply you should read carefull!

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=78742 for no. 1, and please correct no. 3 (it is inhomogeneous in x, y, z).

  darij

[color=purple]Any one can post the solution ? Dariji and skywalker ?[/color]

[color=purple] Here is 27/04 and no one can't help me. [/color]

[quote="evarist"][color=purple]Sorry it has a mistake skywalker.The 2nd must be
  $\frac{1}{x}sinA+\frac{1}{y}sinB+\frac{1}{z}sinC \leq \frac{3\sqrt{3}(x+y+z)}{2(xy+yz+zx)}$
      Thank for your replying dariji the 3rd is not correct and you don't need think about it. [/color][/quote]
What happen if you let $z\rightarrow 0$?  :P

[color=purple] Ok thanks Vo Quoc Ba Can i am wrong . [/color]


Asymptote AoPSwiki articles AMC AIME geometry trapezoid

Why does this not work?  When I try to compile it, the image just doesn't show up in the PDF.
I set everything up correctly and successfully compiled the sample document from the AoPSWiki using Asymptote.  But for some reason, this particular diagram won't work.

[code]
\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage{array}
\usepackage{xy}
\usepackage{fancyhdr}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}

\pdfpagewidth 8.5in
\pdfpageheight 11in
\oddsidemargin 0.0in
\textwidth 6.1in
\pagestyle{fancy}
\headheight 35pt

\rhead{AIME 2008II \#5}
\lhead{\Huge Problem 17}

\begin{document}
\noindent\textbf{Problem:} In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $m$ and $n$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$.
\vskip .5cm
\noindent\textbf{Solution:}

We extend segments $AB$ and $CD$ to meet at $E$, which is a $90^\circ$ angle.

\begin{figure}[h]
 \begin{asy}
  include graph;
  size(250);
  defaultpen(0.7+fontsize(10));
  real f=100, r=1004/f;
  pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180);
  pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251;
  pair[] M = intersectionpoints(N--E,B--C);
  draw(A--B--C--D--cycle);
  draw(B--E--C,dashed);
  draw(M[0]--N); draw(N--E,dashed);
  draw(rightanglemark(D,E,A,2));
  picture p = new picture;
  clip(p,A--D--D+(0,20)--A+(0,20)--cycle);
  add(p);
  label("\(A\)",A,SW);
  label("\(B\)",B,NW);
  label("\(C\)",C,NE);
  label("\(D\)",D,SE);
  label("\(E\)",E,NE);
  label("\(M\)",M[0],SW);
  label("\(N\)",N,S);
  label("\(1004\)",(N+D)/2,S);
  label("\(500\)",(M[0]+C)/2,S);
\end{asy}
\end{figure}

Since $M$ and $N$ are midpoints, $M$ is the circumcenter of $\triangle EBC$ and $N$ of $\triangle EAD$. Thus, $EN = ND = 1004$ and $EM = MC = 500$, so $MN = EN - EM = \boxed{504}$.
\cfoot{Solution was written by Patrick Hulin and compiled from Art of Problem Solving Forums.}
\end{document}
[/code][/code]

I didn't realize we had a Asymptote forum now...
Moderator please move this.  Thanks!

1. This is an asymptote problem as compiling with Asymptote gives an error code of:
[code]20.6: no matching variable 'rightanglemark'[/code]so no picture is generated.
So I have transferred this post to the Asymptote forum for the experts there to tell you the solution.

2. When posting documents with dollar signs in, use Code rather than Quote and you will find that the dollar signs remain, which makes it easier for others to copy and paste into a doment.

3. Use the geometry package to set margins; it's simpler and takes care of all the other variables that need setting.

[quote="stevem"]1. This is an asymptote problem as compiling with Asymptote gives an error code of:
[code]20.6: no matching variable 'rightanglemark'[/code]so no picture is generated.
[/quote]
No wonder: [b]rightanglemark[/b] is a command from the Olympiad package, not the standard Asymptote one. Add the line
import olympiad;
to the beginning of the code and you should be fine (provided that you have the Olympiad package, of course).


summer program
MathZoom
AMC
AIME
AwesomeMath
USA(J)MO
USAMO

Who's going to go the MathZoom summer 2009?

I just want to know to see who I'll meet.

[quote="AIME15"]I might.  I looked for the admissions test, but couldn't find it...[/quote]
If you qualified for AIME, or had AMC 8 score 20+, you don't need to take the admission test.  There are some other criteria for automatic qualification.  Check the MathZoom.org site for details.

Anyway, the test will be online soon.

[quote="AIME15"]Phew.

Can't decide whether to go to MathZoom or AwesomeMath...at least I have automatic qualification xD[/quote]
It depends on your summer schedule.  Unlike last year, we are holding the program at different dates than AwesomeMath this year.  Also some people want to go with their old friends from last year, and some people enjoy meeting new ones.  It's up to you :)

If you come to Math Zoom, I believe you qualify for the Black Group, which is the most advanced group.  For the Black Group we focus on USAMO and IMO level training.  I also keep the Black Group small, so the instructors can keep the contents at around the Red MOP level.

Uh...I barely made AIME last year :blush:

Anyway, I might have a schedule conflict with AMSP...so... :)

And then I can meet my AIME Problem Series B teacher!  Yay!

Im probably going as well
First time for me, sounds like its going to be fun  :lol:

[quote="turak"]if anyone cares, i (sarah) am coming back[/quote]
I am happy to see you back Sarah!
We will have more fun this year. :D

[quote="helagha"]I'm going, although I'm sure no one knows me :P[/quote]

hahaha, same heree

I might go next year, but not this year.

i'm going!

any idea of how many ppl are attending this year?

[quote="helagha"]What exactly do we do in the study session :?:[/quote]

The "supervised self study" is the time right before bed (8:30-9:30) used for activities of your choice. This time is used by most people (or was last year anyway) to talk to friends, play minature versions of sports/games, play SET (I plan to bring a set of cards 1st and 3rd week if I remember to), play videogames, read, study, etc.

Basicly do what you want, as long as you don't wander too far off.

[quote="suma_milli"]I'll probably go...[/quote]
Wait, who are you?


Starting from toluene and any appropriate aliphatic compound, synthesise novocaine

[hide]1) Nitrate, and separate the para product.

2) Oxidize with hot, alkaline potassium permanganate.

3) React with thionyl chloride (reflux, half an hour).

4) Treat the acyl chloride with $ \ce{HO\minus{}CH2CH2N(CH2CH3)2}$

5) Perform selective reduction of nitro group to amino group.[/hide]


any schools here taking AMC B?  i'm looking for a place to take it.

I think Lexington, Chelmsford, and Waltham offers B, I'm not sure though.

Worcester Academy participates in both sittings, and we allow visitors at the B sitting.  The standard AMC visitor fee (last year, anyway) is $\$32$ per person.  We charge $\$15$.

(We register visitors as a separate team.  Last year, it wasn't clear whether the per-person cost was going to be over $\$10$ or under $\$10$, which is why I picked $\$15$.  It actually cost us just under $\$9$ per visitor.  The other money went to defray our own costs for the second administration -- of course we don't charge our own students.)


This is very interesting

For $a,b,c$ positive reals prove that

$\sum \frac{2a^2+bc}{b^2+c^2} \geq \frac 92$

Manlio, you could remember that one: http://www.mathlinks.ro/Forum/viewtopic.php?t=26104 ...

  Darij

Sorry, I think it is Alzheimer but I don't remember recent  events :(

Sorry again.

What Is VGI?
What is the general problem?


function
inequalities
trigonometry
algebra unsolved
algebra

For given a,b,c,d in R, consider function f(x) = acos(bx)+ ccos(dx)
(1). Prove that for any a,b,c,d in R, the maximum of f(x) exist.
(2). Find the maximum value of f(x).

Using Cauchy-Schwalz's inequality, $|a\cos x+b\sin x|\leqq \sqrt{a^2+b^2}$.

The equality holds when $(a,b)\parallel (\cos x, \sin x)$.


[quote="Ligouras"]Let $ a,b,c,d$ non-negative reals such that $ a \plus{} b \plus{} c \plus{} d \equal{} 1$. Prove that

$ 5(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 12(abc \plus{} bcd \plus{} cda \plus{} dab)\geq1$[/quote]

the inequality is equivalent to
$ 5(a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3) \plus{} 12(abc \plus{} bcd \plus{} cda \plus{} dab)\geq (a \plus{} b \plus{} c \plus{} d)^3$

assume  $ a\geq b \geq c \geq d$
substitution  $ a \equal{} x \plus{} y \plus{} z \plus{} d,b \equal{} y \plus{} z \plus{} d,c \equal{} z \plus{} d$
$ x,y,z\geq 0$

expand
then equivalent to
$ 4d^3 \plus{} (9c \plus{} 3a \plus{} 6b)d^2 \plus{} (12bc \plus{} 6ac \plus{} 6ab \plus{} 9c^2 \plus{} 6b^2 \plus{} 3a^2)d \plus{} 4a^3 \plus{} 3ab^2 \plus{} 2b^3 \plus{} 9a^2b \plus{} 6a^2c \plus{} 6abc \plus{} 6b^2c\geq 0$


calculus integration geometry perimeter percent

It took me a few minutes (more like 10) to solve this one, I'm not too good with these kinds of maximization problems.

1. When a survey of $ 800$ people was done, $ 55\%$ of those surveyed said they ate candy, $ 65\%$ said they ate chips and $ 90\%$ said they ate hot dogs. What is the smallest possible number of people who said they ate all three?

2. what is the sum of the $ 13$ smallest positive palindromes that have a tens digit of $ 3$ and a ones digit of $ 7$?

3. (really easy) The side lengths of a right triangle are each an integral number of units. If one of the legs is $ 13$ units, what is the perimeter of the triangle?

[hide="#3"]Well, when you say leg, I assume you mean hypotenuse, so a 5-12-13 triangle would satisfy it, with perimeter 5+12+13=17+13=[u]30[/u][/hide]

If it means leg, you can have a 13-84-85 right triangle, and thus it has a perimeter of [b]182[/b].

Is #1 some sort of a trick question?
Because as I see it, the answer would be $ .55\times800$
But it can't be that easy....
 :|

Heheh...whoops.

The question should be: "what is the smallest possible that ate all three?"

Sorry, I'll change that.

And yes, by leg it means leg, and so you could use: $ p, \frac{p^2\plus{}1}{2}, \frac{p^2\minus{}1}{2}$

[hide="New"]If leg means leg, then we have:
x^2+13^2=y^2==>13^2=y^2-x^2=(y-x)(y+x)
hence, either y-x and y+x are both 13, which doesn't work because it doesn't make integral numbers, or y+x=169 and y-x=1 which yields y=85 and x=84.  85+84+13=$ \boxed{182}$.[/hide]

[hide="#2"]First, we focus on three-digit palindromes, of the form -37.  Obviously, the blank will have to be occupied by 7.  Now, four-digit palindromes which are in the form --37.  Once again, the blanks will have to be occupied by 73 respectively.  Five digit palindromes are in the form ---37, or 73-37==>the blank will have to be occupied by either 0-9 for a total of 9-0+1=10 possibilities.  As for six-digit palindromes, in the form 73--37, the smallest will have the middle blanks occupied by 0's.  Now, we have a total of 1+1+10+1=13 palindromes, namely:
730037, 7337, 737, 73037, 73137, 73237, 73337, 73437, 73537, 73637, 73737, 73837, 73937 which add to $ \boxed{1,472,981}$ whether by straight addition or manipulation.[/hide]

[quote="happyme"][quote="abacadaea"][quote="Z-coli"]No, the answer would not be .55 times 800 because the question is asking for people who only eat candy, not who eat candy and possibly something else.[/quote]

i think it is...its asking for all 3.[/quote]

That was before I edited the question.

 :ninja:[/quote]
Weird enough I get $ 0$...

Yeah, It's wrong..

[quote="George S Bashour"]Number 1 is relatively easy. Since 90 percent ate cotton candy, or something like that, and 65 ate hot dogs (sorry, items are probably wrong) a minimum of 55 percent ate both. Express this as a seperate value. Now, easy Venn problem. 55 percent ate on thing, and 55 percent ate cotton candy and hot dogs. 10 Percent ate everything, so 80 people had massive indigestion problems the next morning.[/quote]

I followed you till "55 percent ate one thing, 55 ate cotton candy and hot dogs. 10 percent ate everything."

Your answer is right.

[quote="unimpossible"][hide="#1"]its just 65%-55%=10%[/hide][/quote]

Reasoning?

[quote="George S Bashour"]Number 1 is relatively easy. Since 90 percent ate cotton candy, or something like that, and 65 ate hot dogs (sorry, items are probably wrong) a minimum of 55 percent ate both. Express this as a seperate value. Now, easy Venn problem. 55 percent ate on thing, and 55 percent ate cotton candy and hot dogs. 10 Percent ate everything, so 80 people had massive indigestion problems the next morning.[/quote]

Sorry if I'm spamming, but why did you post this "relatively easy solution" twice?  Just curious. :lol:

[hide="#1"] For the one with 55% and the one of 65%, imagine a number line of 100. The 55 starts at zero and moves towards 100, while the 65 starts at 100 and moves towards 0. Thus the overlap (minimum who ate both the hotdog and whatever) is 55-35=20. Now we do this with the cotton candy thingy (90%). The overlap is 90-80 = 10%.[/hide]

[quote="minicon"]His Internet connection probably was flaky when he tried to post resulting in two submissions.  :)[/quote]
That happens to me sometimes, but you can prevent it by clicking preview before you submit. Sometimes, it just submits by itself, I think..

#1: I get number one. I just basically analyzed what George S Bashour had said.
[hide="Faciliated Version"]What he did was he took the minimum possible between 2 items, which were chips(65%) and hot dogs(90%). It came out to be [u]55%[/u].

Then he compared the percentages of people who ate both chips and hot dogs(55%) and candy(55%).

So it's basically [b](55%+55%)-100%=10%[/b]

10% of 800 is $ \boxed{80}$[/hide]

edited

Summarizing DK's explaination (of #1) gives you this formula (since you now understand why):

|A∩B| = |A| + |B| - |AUB|

to apply on this problem

which lead to this formula for three sets:

|AUBUC| = |A| + |B| + |C| - (|A∩B| + |B∩C| + |C∩A|) + |A∩B∩C|

In practice however, most of these intersections are either empty or have the same size and it is more efficient to simply analyze the over-counting and under-counting within the context of the particular problem. The principle of inclusion/exclusion is the key idea to remember, not the formulas. (ocw.mit.edu)

That's PIE for 2 sets.  #1 is like PIE with three sets.

[quote="abacadaea"][hide="#2 seems quite tedious...."]
theres
737
7337
73*37 where * can be any digit
730037
so we have $ 737 \plus{} 7337 \plus{} (73037 \plus{} 100\sum_{n \equal{} 0}^{9} {n}) \plus{} 730037 \equal{} 811148 \plus{} 4500 \equal{} 815648$
[/hide][/quote]

Using your numbers, I think we can just add 9 more 73037s to your total sum to get $ \boxed{1472981}$.  Or, at least I [i]think [/i]that's right. :maybe:


algebra polynomial trigonometry function AMC

Prove that the polynomial $p$ with degree not exceeding $n$ that assumes the values $y_k$ at the values $x_k=\cos u_k$ where $u_k=\frac{(2k+1)\pi}{2k+2}$ with $k\in \mathbb{N}$ is given for $x\not =x_k$ for any $k$ by $p =\frac{N }{D }$ where $N =\sum_{k=0}^n\frac{(-1)^ky_k\sin u_k}{x-x_k}$ and $D =\sum_{k=0}^n\frac{(-1)^k\sin u_k}{x-x_k}$

Interesting. The $x - x_k$ makes it look "interpolation-ish" to me (along with the whole, "takes on values at blah" part). I'll look at it when I have time.

[quote="eXistenZ"]Prove that the polynomial $p$ with degree not exceeding $n$ that assumes the values $y_k$ at the values $x_k=\cos u_k$ where $u_k=\frac{(2k+1)\pi}{2n+2}$ with $k\in \mathbb{N}$ is given for $x\not =x_k$ for any $k$ by $p =\frac{N }{D }$ where $N =\sum_{k=0}^n\frac{(-1)^ky_k\sin u_k}{x-x_k}$ and $D =\sum_{k=0}^n\frac{(-1)^k\sin u_k}{x-x_k}$
[b] Remark: the original proposed question is corrected  ![/b] [/quote]
Let us consider following table of experimental data : $ \;  \begin{array}{||c||c|c|c|c|c|} \hline\hline
 x & x_0 &x_1& x_2&\cdots & x_n\\  \hline
 f(x)& y_0 &y_1&y_2&\cdots &y_n\\
 \hline\hline
\end{array} \; $  where $  x_0,x_1,...,x_n $ are supposed to be distinct and $  y_k:=f(x_k). $ It's well-known that there exists an unique polynomial function $ L_n(f;x)\equiv L_n(x_0,x_1,...,x_n;f|x) $ , of degree $ \le n \; ,\; $ which "interpolates " $ f $ at the "knots" $ \{ x_0,x_1,...,x_n\} $, that is $ L_n(f;x_j)=f(x_j)\; \; ,\; \; \forall j\in \{0,1,...,n\}. $ This polynomial ,  named  [i]Lagrange interpolation polynomial [/i] , may be written in different ways. More precisely, let us introduce  following notation and terminology:
\[\omega(x):=(x-x_0)(x-x_1)\cdots (x-x_n)\; \; \; \; \;
\mbox{\"knot-polynomial\"}\]
\[(*)\; \; \; \; \; \;
l_k(x):=\frac{\omega(x)}{(x-x_k)\omega^{\prime}(x_k)}=\prod\limits_{\nu=0\;
,\; \nu\ne k}^{n}\frac{x-x_\nu}{x_k-x_\nu} \; ,\; 0\le k\le n,\;
\; \; \mbox{\"fundamental interpolation polynomials\"}.
\]
Some important properties of $l_0(x),\ldots,l_n(x) $ are :
i) for each $ k\in \{0,1,...,n\}, $ $ l_k(x) $ is a polynomial of exact degree $n ,$
ii) $  l_k(x_j)=\left\{\begin{array}{lcl} 0 & , & 0\le k\ne j \le
n\\
1 &,& 0\le k=j\le n \; .
\end{array}\right.
$
It may be verified that   $\; \; \; (1)\; \; \; \; \; \; \; \; \begin{array}{|c|} \hline \displaystyle
 L_n(f;x)=\displaystyle  \sum\limits_{k=0}^{n}l_k(x)f(x_k)=\displaystyle
\sum\limits_{k=0}^{n}l_k(x)y_k \\ \hline
\end{array} \; .  $  This is the so-called [b]Lagrange form of the interpolation polynomial[/b]. Now suppose that $ p $ is a polynomial function of degree $ \le n . $ Then $ L_n(p;x)=p(x) . $ The reason is that  both are polynomials of degree $ \le n $ which coincide at  $ n+1 $ points $ x_0,x_1,...,x_n . $ Therefore,  according to (1) we may write
$\; \; \; \; \; (2)\; \; \; \;  p(x)=\displaystyle  \sum\limits_{j=0}^{n}l_j(x)p(x_j)\; . $  In (2) let us select $ p(x)=1 . $ One finds  $\; \; \; \; \; \; \; \; \;  (3)\; \; \; \; \; \; 1=\displaystyle  \sum\limits_{j=0}^{n}l_j(x)\; .  $  Using (1) and (3) we obtain   $\; \; \; \; \; \; (4)\; \; \; \; \; L_n(f;x)=\frac{L_n(f;x)}{1}=\frac{
 \sum\limits_{k=0}^{n}l_k(x)f(x_k) }{  \sum\limits_{j=0}^{n}l_j(x)}\; \; . \; $  Taking into account (*), after some simplification, one obtains
\[(5)\; \; \; \; \;\begin{array}{|c|}
\hline
 L_n(f;x)=\frac{\displaystyle
 \sum\limits_{k=0}^{n}\frac{y_k}{(x-x_k)\omega^{\prime}(x_k)}}
 {\displaystyle \sum\limits_{j=0}^{n}\frac{1}{(x-x_j)\omega^{\prime}(x_j)}}\\
\hline
\end{array}\; ,\; \; x\not \in \{x_0,x_1,...,x_n\},\; y_k:=f(x_k).
\]
Some numerical analysts named  (4) or (5)  as [b]barycentric-interpolation formula[/b]. This is the key of your posted question. From (2) and (5) , for $ p $ a polynomial function of degree $ \le n $, one finds
\[
(6)\; \; \; \; \; \;  \; \;\begin{array}{||c||} \hline \hline
 p(x)=\frac{\displaystyle
 \sum\limits_{k=0}^{n}\frac{y_k}{(x-x_k)\omega^{\prime}(x_k)}}
 {\displaystyle \sum\limits_{j=0}^{n}\frac{1}{(x-x_j)\omega^{\prime}(x_j)}}\\
\hline\hline
\end{array}\; ,\; \; x\not \in \{x_0,x_1,...,x_n\}\; ,\; \;
y_k:=p(x_k).
\]
Further select in (6) the knots $ x_k=\cos{u_k}\; ,\;
u_k=\displaystyle \frac{(2k+1)\pi}{2n+2} , \; \; k\in
\{0,1,...,n\}.$ [b] [Remark: Perhaps here, in your original posted question, there is a minor misprint].[/b] The  numbers $ x_0,x_1,...,x_n $ there are the roots of Chebychev-polynomial (of the first kind) $ T_{n+1}(x) $ defined on  $ [-1,1] $ as $ T_{n+1}(x)=\cos\left((n+1)\arccos{x}\right)=2^nx^{n+1}+\ldots . $ Therefore, in this particular case  $ \omega(x)=\displaystyle \frac{1}{2^n}T_{n+1}(x) $ and because $ T_{n+1}^{\prime}(x_k)=\displaystyle
\frac{(-1)^{k}(n+1)}{\sin{u_k}} \; ,\; $ we give $\omega^{\prime}(x_k)= \displaystyle
\frac{(-1)^{k}(n+1)}{2^n\sin{u_k}} $. Using (6) we conclude with your proposed identity, that is
\[
\begin{array}{|c|} \hline
 p(x)=\frac{\displaystyle
 \sum\limits_{k=0}^{n}\frac{(-1)^{k}y_k \sin{u_k}}{x-\cos{u_k}}}
 {\displaystyle \sum\limits_{j=0}^{n}\frac{(-1)^j\sin{u_j}}{x-\cos{u_j}}}\\
\hline
\end{array}\; ,\; \]
where $  \; x\not \in \{\cos{u_0},...,\cos{u_n}\}\; ,\; \;  u_k=\frac{2k+1}{2n+2},\; \; y_k= p\left(\cos{u_k}\right),
$  $ p $ being any polynomial of degree $\le n .$


a triangle ABC,D ∈ AC, m< A = 4x°, m< C = 3x°, m< CBD = 5x°,

 and AB = CD. Find x°.

Use sine rule for the two triangles i.e. ABD and BDC.

For triangle ABD we have AB/sin8x = BD/sin4x. Similarly for triangle BDC we have CD/sin5x = BD/sin3x

Hence, sin8x/sin4x = sin5x/sin3x...

Solving we get, sin7x - sin5x = sinx

Therefore, 2.cos6x.sinx = sinx. Hence, cos6x = 1/2

cos6x = cos 60 degrees

x = 10 degrees

Is this problem from some contest?Because one guy asked me to solve this problem by pm...Maybe moderators knows something about it...


[b]In  a triangle ABC if AD is the altitude to BC then prove that : $ \frac {AH}{HD}$ = $ \frac {Tan(B)  \plus{} Tan(C)} {Tan(A)}$ where $ H$ is the orthocentre.[/b]

hello,let $ \alpha_2=\angle DCH$ then we have
$ \overline{AH}=\frac{AE}{\cos(\alpha_2)}$
${ \overline{HD}=\overline{CD}\tan(\alpha_2}$
from here we get
$ \frac{\overline{HD}}{\overline{HD}}=\frac{\overline{AE}}{\overline{CD}\sin(\alpha_2)}$
further
$ \overline{AE}=b\cos(\alpha)$
$ \overline{CD}=b\cos(\gamma)$
$ \sin(\alpha_2)=\frac{\overline{BE}}{a}$
$ \overline{BE}=a\cos(\beta)$
inserting this in the formula above we get
$ \frac{\overline{AH}}{\overline{HD}}=\frac{\cos(\alpha)}{\cos(\beta)\cos(\gamma)}$
it is easy th show that $ \frac{\cos(\alpha)}{\cos(\beta)\cos(\gamma)}=\frac{\tan(\beta)+\tan(\gamma)}{\tan(\alpha)}$
Sonnhard.

WHy do you need all that Sonnhard?$ AH\equal{} 2R cosA$ and $ HD\equal{} 2R cosB cosC$.


geometry
geometric transformation
homothety

This is the 9-point circle problem. It has been solved a number of times on this forum.

[quote="liangchene"]Prove that the feet of the altitudes, the midpoints of the sides, and the midpoints of the segments connecting the orthocenter to the vertices are concyclic. (meaning one circle goes through them all).[/quote]

This circle is called Nine Point Circle.

Hey Nikhil Im just asking How old are you?

This is to say:
In $\triangle{ABC}$, $H$ is the orthocentre. $AD \perp CB$, $BE \perp AC$, $CF \perp AB$. Construct a circle through D,E and F.
Prove the circle passes through the mid points of the sides and the midpoints of AH, BH and CH.

Theres a few more not so well know points it goes through aswell

the solution of the problem uses homothety, if i remember correctly

what are the not-so-well-known points the circle passes through?

nine-point circle.
Can you please show what other points it goes through?


Hello,

rewrite it in form $ ab\plus{}bc\plus{}ac\plus{}a\plus{}b\plus{}c\ge 3\sqrt[3]{a^2b^2c^2} \plus{} 3\sqrt[3]{abc}$

Now $ a\plus{}b\plus{}c\ge 3\sqrt[3]{abc}$ my $ AM\minus{}GM$

and $ ab\plus{}bc\plus{}ac\ge 3\sqrt[3]{a^2b^2c^2}$also by $ AM\minus{}GM$ and we are done.


calculus
integration
analytic geometry
trigonometry
rotation
logarithms
calculus computations

I know that $\int_{0}^{1}\int_{0}^{1}\frac{dxdy}{1-xy}=\sum_{n=1}^{\infty}\frac{1}{x^{2}}=\frac{\pi^{2}}{6}$.

our teacher asked us to prove this by evaluating the integral, with out introducing any sigmas.
[hide="what i have so far"]
let $u=x+y$ and $v=x-y$ so $x=\frac{u+v}{2}$ and $y=\frac{u-v}{2}$

applying this and a change of variable to the initial integral gives us $2\iint_{D}\frac{dudv}{4-(u^{2}+v^{2})}$.
At this point i would like to convert to polar coordinates but do not know how to find the range of $r$.
[/hide]

The region of integration is the square bounded by the lines $u+v=0$, $u-v=0$, $u+v=2$, and $u-v=2$. Since it is symmetric about the $y$-axis, and the integrand is even with respect to $x$, we can integrate over the right half of the square and multiply the result by $2$. The line $u+v=2$ has polar equation $r(\cos\theta+\sin\theta)=2$, hence the upper limit for $r$ is $2(\cos\theta+\sin\theta)^{-1}$.  :huh:  you can either use trigonometric identities to simplify this, or rotate the half-square by $45^{o}$ clockwise, which leads to the integral
$\int_{0}^{\pi/4}\int_{0}^{\sec\theta}\frac{r}{4-r^{2}}drd\theta =-\frac12\int_{0}^{\pi/4}(\log(4-\sec^{2}\theta)-\log 4)d\theta$ :what?:


a cowboy ride into a hotel on Friday stays for 3 days and leaves on Friday. how does he do that

[hide] Friday is the name of his horse. It really doesnt matter when he left or came to the hotel.[/hide]

Obviously, the hotel is actually onboard a space ship traveling at relativistic speeds, so that when three days pass in the space ship, a week has already passed in the outside world...

[hide]Who names their horse Friday? And why did the hotel staff let him ride into the hotel, instead of making him get off his horse before coming in? :D[/hide]

would it be better if its name was tuesday? :D

no....its none of those answers..

the cowboy is old and takes forever to ride to the hotel. then, thinking its only 3 days, it was really a week.

or maybe timezones?


Επιλέγουμε τυχαία τρεις(διαφορετικές) κορυφές ενός κανονικού εξαγώνου. α) Ποια είναι η πιθανότητα να είναι τα σημεία που επιλέξαμε, κορυφές ισόπλευρου τριγώνου? β)Ποια είναι η πιθανότητα ώστε το τρίγωνο να είναι ισοσκελές και ποια να είναι ορθογώνιο?

Για το (α) ερώτημα  ( μπορεί να κάνω και λάθος γιατί τα προβλήματα των πιθανοτήτων καμιά φορά οδηγούν σε λανθασμένες λογικές ... :)  )

Νομίζω ότι είναι  $ \frac{1}{10}$

Ονομάζω Ν τον πληθικό αριθμό των συνόλων, Ρ την πιθανότητα μιας ιδιότητας.

α) Ν(τριγώνων) = $ \binom{60}{3} \equal{} 20$

Είναι δυνατή η κατασκευή 2 ισοπλεύρων τριγώνων,
άρα Ν(ισόπλευρων τριγώνων) = $ 2$

Είναι Ρ(ισοπλεύρων τριγώνων) = Ν(ισοπλεύρων τριγώνων)/Ν(τριγώνων) = $ \frac {2}{20} \equal{} \frac {1}{10}$

β) Διαδικασία για κατασκευή ορθογωνίων τριγώνων:
επιλέγω μια διαγώνιο (3 δυνατότητες)
διαλέγω μια από τις κορυφές που ορίζονται από τα δύο τετράπλευρα που δημιουργούνται (4 δυνατότητες)

Εφαρμόζοντας τώρα τη θεμελιώδη αρχή της απαρίθμησης είναι:
Ν(ορθογωνίων τριγώνων) = $ 3 \cdot 4 \cdot 1 \equal{} 12$

Είναι Ρ(ορθογωνίων τριγώνων) = Ν(ορθογωνίων τριγώνων)/Ν(τριγώνων) = $ \frac {12}{20} \equal{} \frac {3}{5}$

Εύκολα βγαίνει ότι κάθε τρίγωνο που σχηματίζεται είναι ή ορθογώνιο ή ισοσκελές άρα
Ρ(ορθογωνίων τριγώνων) + Ρ(ισοσκελών τριγώνων) = 1 $ \Leftrightarrow$
Ρ(ορθογωνίων τριγώνων) = $ \frac {2}{5}$

Πείτε μου για πιθανά λάθη.


LaTeX
inequalities unsolved
inequalities

Converted to readable LATEX:
[quote="Thao"]Let $x,y,z \ge 0$ and $x+y+z = 2$. Prove that
  $x^{t}(y+z)+y^{t}(z+x)+z^{t}(x+y) \leq 2$ $\forall t \in[2;3]$[/quote]


calculus
integration
vector
absolute value
calculus computations

I've been trying to solve this problem my professor posted. From what I understand of the problem, I think it has to do with applying the divergence theorem to a large ball D and letting its radius go to infinity.

If f(x) is continuous and |f(x)|≤ 1/(|x|^3+1) for all x, show that the triple integral over the whole space of divf dx is equal to 0.

Thanks for any help.

[color=green][Moved into the calculus section, where it should have been posted in the first place.][/color]

It will have to be understood here that $f(x)$ is a vector field, right?

[quote]I think it has to do with applying the divergence theorem to a large ball D and letting its radius go to infinity. [/quote]

Indeed, that's what you'll use. Why don't you try writing out the steps of this  - assume a ball of radius $R$ centered at the origin, estimate how big (in absolute value) the surface integral of $\vec{f}\cdot\vec{n}$ can be over the surface of that ball. In the end, we're going to take the limit as $R\to\infty.$


geometry perimeter circumcircle trigonometry

if the in circle of a quadrangle $ABCD$ has radius $r$, then prove that:

$AB+CD$>_ $4r$

As proven some place else in this forum by silouan I think, (sorry if I'm wrong) the minimum perimeter of a quadrilateral with an incircle is a square.
Therefor, the minimum perimeter of a quadrilateral with an incircle is $8r$.
Now, in this kind of quadrilaterals, it is true that \[ AB + CD = AD + BC  \]
So this gives us that \[ AB + CD + BC + AD \geq 8r  \]
\[ 2AB + 2CD \geq 8r  \]
\[ AB + CD \geq 4r  \]

That theorem is also true, but the theorem I used is not about circumcircles, but about incircles.
It says that among all quadrilaterals with an incircle radius $r$, the square is the one with the smallest perimeter $(8r)$. Sorry, but I'm afraid you misread my post.

You are right .I my misunderstanding was about circumcircle and incircle

My problem is, how to prove: $AB+BC+CD+DA \geq 8r$.

What is $r$?
radius of the circle circumscribing the quadrilateral $ABCD$?

Read the problem more carefully. It said the radius of the incircle.

Hmm, well drawing the incircle and all the radii to the tangent points (call $P$ the tangent point on $AB$, $Q$ on $BC$, $R$ on $CD$ and $S$ on $DA$). $O$ is the incenter.

Call $\angle AOP = \angle AOS = \alpha_1$, $\angle BOQ = \angle BOP = \alpha_2$, $\angle COR = \angle COQ = \alpha_3$, and $\angle DOS = \angle DOR = \alpha_4$. Note that $\alpha_1+\alpha_2+\alpha_3+\alpha_4 = 180$ and $0 < \alpha_i < 90$.

We find that $AB = r(\tan{\alpha_1}+\tan{\alpha_2})$, $BC = r(\tan{\alpha_2}+\tan{\alpha_3})$, $CD = r(\tan{\alpha_3}+\tan{\alpha_4})$, $DA = r(\tan{\alpha_4}+\tan{\alpha_1})$. So it remains to show that

$AB+BC+CD+DA = 2r\sum \tan{\alpha_i} \ge 8r$, or $\sum \tan{\alpha_i} \ge 4$.

But since $\tan{x}$ is convex on $(0,90)$, by Jensen's with equal weights $w_i = \frac{1}{4}$ we get

$\tan{\alpha_1}+\tan{\alpha_2}+\tan{\alpha_2}+\tan{\alpha_2} \ge 4\tan{\left(\frac{\alpha_1+\alpha_2+\alpha_3+\alpha_4}{4}\right) = 4\tan{45} = 4}$ as desired.

Does that work?


A $ 12' \times 15'$ rectangular room has $ 8'$ high walls. Ignoring any doors or windows, how many one-quart cans of paint must you buy to pain the walls, provided that each quart will cover $ 100$ square feet of surface?

$ \frac{2(12\plus{}15)\times{8}}{100}\equal{}4.32$. Thus, $ \boxed{5}$ cans are needed.


Let $ a,b,c$ be positive real numbers such that $ a\plus{}b\plus{}c\equal{}1$.
Find the minimum value of the following expression:

      $ \frac{a\plus{}b}{a^3\plus{}b^2\plus{}c}\plus{}\frac{b\plus{}c}{a\plus{}b^3\plus{}c^2}\plus{}\frac{c\plus{}a}{a^2\plus{}b\plus{}c^3}$

The minimum value is 2, which is attained for example for $ (a,b,c) \equal{} (1,0,0)$.

To prove the inequality
$ \sum_{cyc} \frac{a\plus{}b}{a^3\plus{}b^2\plus{}c} \geq 2$,
observe simply that if $ a\plus{}b\plus{}c\equal{}1$, that means $ a,b,c \leq 1$, so $ a^3 \plus{} b^2 \plus{} c \leq a \plus{} b \plus{} c$.
Therefore, the sum is
$ \sum_{cyc} \frac{a\plus{}b}{a^3\plus{}b^2\plus{}c} \geq \sum_{cyc} \frac{a\plus{}b}{a\plus{}b\plus{}c} \equal{} 2$.


inequalities
function
calculus
rearrangement inequality

If a, b, c, d, e are arbitrary positive reals prove that:

(a + b + c + d + e) :^2:   :ge:  4(ab + bc + cd + de + ea)

I tried to prove this inequality for half an hour and kept thinking about it for 30 minutes everyday for a few days but I still can't prove it. I'll appreaciate it if someone ends my misery and finds a way to prove it.

Calculus?  That's no fun.  What if we homogenized somehow?  Maybe that's the way to do this one ... grrrr.

It is easy to prove if a :ge: b :ge: c :ge: d :ge: e :ge: 0
but I haven't been able to do it for arbitrary numbers
(a + b + c + d + e) :^2:  :ge: 4(ab + bc + cd + de + ea)
This is what I have done so far:

since

(a+b)/2 :ge:  :sqrt: ab
(b+c)/2 :ge:  :sqrt: bc
(c+d)/2 :ge:  :sqrt: cd
(d+e)/2 :ge:  :sqrt: de
(e+a)/2 :ge:  :sqrt: ea

Then (a+b+c+d+e)  :ge: :sqrt: ab + :sqrt: bc + :sqrt: cd + :sqrt: de + :sqrt: ea (This can also be obtained from the re-arrangement inequality)
and (a+b+c+d+e) :^2:  :ge: (:sqrt: ab + :sqrt: bc + :sqrt: cd + :sqrt: de + :sqrt: ea) :^2:

So if ab = A bc = B cd = C de = D ea = E

I have to prove (A+B+C+D+E) :^2:   :ge: 4(A :^2: + B :^2: + C :^2: + D :^2:  + E :^2: )

Maybe there's an inequality for these but I don't know it so I can't get farther than this. If you know any inequalities leading to the proof of this, then please tell me.

By The Way: This problem is from one of the ALL SOVIET UNION MATH OLYMPIADS but I can't remember which one. But my point is, the kids that participated in these competitions were about 14-15 so I don't think calculus is required to solve these.

evarol, I thought about that, but the inequality you want to change it to isn't true -- make one of the variables very very large and keep the others fixed, for example.

Simon -- I don't think it's true if you change 4 to 5.  For example, what happens if you take a = b, c = d = e = 0?  It only works for 4, not 5.

[quote="ComplexZeta"]There's something fishy about this problem (not that it's false or anything), but if we change 4 to 5, isn't it still true by the rearrangement inequality or one of its close relatives?[/quote]
I thought it would be 5 as well, but let a=b=1, and the others 0. It has to be 4..

Maybe you need to group 5 terms on the left to get bigger than one of those on the right?  But which 5?

Also, evarol, you could just try looking at all 24 orderings of the numbers -- that might do it for you.

Hmm, I shouldn't have put the "just" in that last sentence.

But anyway, using simple calculus (knowing that a^2 differentiates to 2a and that ac differentiates to c) gives a very very simple answer.. takes about a minute to solve.

I found a simple way to prove it for 4 terms

If A, B are positive then

(A+B) :^2:  :ge: 4(AB) by AM-GM Inequality

So,

(a + b + c + d) :^2: = [(a + c) + (b + d)] :^2:

So (a + b + c + d) :^2: :ge: 4(a + c)(b + d)
which is
(a + b + c + d) :^2: :ge: 4(ab + bc + cd + da)

I've seen this one before, but since the OP needed a solution, here goes:

[hide]

(a + b + c + d + e)^2 >= 4(ab + bc + cd + de + ea)

(a + b + c + d + e)^2-4(ab + bc + cd + de + ea) >= 0

(a - b + c - d + e)^2 + 4ad + 4be - 4ae >= 0

(a - b + c - d + e)^2 + 4ad + 4e(b-a) >= 0

which is clearly true if b>=a. If a>b, then instead write

(b - a + c - d + e)^2 + 4bd + 4(a-b) >= 0

[/hide]

TripleM wrote:Yeah, it looks like Cauchy -Schwartz. However, I haven't been able to solve it that way yet, I do have a horrible solution though if you want it. in spoiler (highlight to read)
[hide]

Put everything on the left hand side, call that function f(a,b,c,d,e), we want to prove f(a,b,c,d,e) >= 0;

If you fix four of the variables and differentiate w.r.t to other one, you get the five equations:

2a = 3e + 3b - c - d
2b = 3a + 3c - d - e
2c = 3b + 3d - e - a
2d = 3c + 3e - a - b
2e = 3d + 3a - b - c

Theres only one solution to these, which is 0,0,0,0,0. This makes the minimum value of f(a,b,c,d,e) = 0.
[/hide]

You've shown that (0,0,0,0,0) is a critical point, but you haven't shown that it produces a minimum for f on the set that's asked for. That's going to require a lot more work.

As I showed in my previous post

(a+b+c+d) :^2: :ge: 4(ab + bc + cd + da)

Is it possible to use this and prove the case for 5 terms?

If a + b + c + d = n, ab+bc+cd+de = Q

n :^2: :ge: 4Q
If we add the 5th term,

We have to prove
n :^2: + 2ne :^2: + e :^2: :ge: 4Q + 4ea + 4de - 4da

[quote="TripleM"]Why not? If a,b,c,d,e isn't 0,0,0,0,0, then we can change one of them to get a smaller value for f(a,b,c,d,e)...[/quote]

You have to be careful with multivariable functions, they can be tricky. For instance, look at the function f(x,y) = xy - x - y. It has a critical point at (1,1), but that point doesn't produce a minimum value for f (even on x, y :ge: 0).

First  case :
Let  b>=a . Then  we  have
(a+b+c+d+e)^2 - (a-b+c-d+e)^2=...=4(ab+bc+cd+de+ea)+4e(b-a)+4ad>=4(ab+bc+cd+de+ea)
Second case:
Let  b<a . Then  we  have
(a+b+c+d+e)^2 - (a-b-c+d-e)^2=...=4(ab+bc+cd+de+ea)+4c(a-b)+4bd.


I wonder whether (a+b) can be further factorized other than this one.

(a+b) =  (a^1/2+ib^1/2)(a^1/2-ib^1/2)

You can further factorize a^1/2+ib^1/2 using i=(omega_8)^2 with omega_8 standing for the eighth root of unity.

a^1/2+ib^1/2=(a^1/4+(omega_8)^3b^1/4)(a^1/4-(omega_8)^3b^1/4)

and so on.

But I see no use of this factorization.

Misha


Ring Theory
superior algebra
superior algebra solved

Let $R$ be a ring with unity 1 and let $G=\{n\cdot1 | n \in Z\}$ , show that $G$ is a subring of $R$

the only thing you have really to prove is that $(m \cdot 1)(n \cdot 1)=(mn)\cdot 1$ and this is simple use of distributivity


[quote="frogfucius"]verify that the equation is true:

16(sin ^ 5 x + cos ^ 5 x)= 11 (sin x + cos x)[/quote]

Pick any x such that sin x = 1 and cos x = 0.  Then LHS = 16 and RHS = 11, contradiction.

[quote="frogfucius"]eh??
sin ^5 x
what i meant was sin x to the fifth
as in sin x^5
not sin ^ 5X[/quote]

If sin x = 1, then sin^5 x = 1 as well.. and cos^5 x = 0.. so we get 16=11.

I think he means sin (x^5), though he really raped the notation.

Edit: Not that I checked it to see if that was true....

So maybe I didn't think so hard.  Or at all.  I had a long day.  Math and swimming don't go along together very well.

[tex]16\left( {\sin ^5 x + \cos ^5 x} \right) = 11\left( {\sin x + \cos x} \right) \Leftrightarrow [/tex]
[tex] \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {16\left( {\sin ^4 x - \sin ^3 x\cos x + \sin ^2 x\cos ^2 x - \sin x\cos ^3 x + \cos ^4 x} \right) - 11} \right] = 0 \Leftrightarrow [/tex]
[tex] \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {16\left( {\sin ^4 x - \sin x\cos x + \sin ^2 x\cos ^2 x + \cos ^4 x} \right) - 11} \right] = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {16\left( {1 - 2\sin ^2 x\cos ^2 x - \sin x\cos x + \sin ^2 x\cos ^2 x} \right) - 11} \right] = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {5 - 16\sin x\cos x - 16\sin ^2 x\cos ^2 x} \right] = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {\sin x + \cos x} \right)\left[ {4\sin ^2 2x + 8\sin 2x - 5} \right] = 0 \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {\sin x + \cos x = 0} \right) \vee \left( {4\sin ^2 2x + 8\sin 2x - 5 = 0} \right) \Leftrightarrow [/tex]

[tex] \Leftrightarrow \left( {\sin x + \cos x = 0} \right) \vee \left( {\sin 2x = \frac{1}{2}} \right) \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = 0 \vee \sin 2x = \frac{1}{2} \Leftrightarrow [/tex]

[tex] \Leftrightarrow x \in \left\{ { - \frac{\pi }{4} + k\pi } \right\}_{k \in \mathbb{Z}}  \cup \left\{ {\left( { - 1} \right)^k \frac{\pi }{{12}} + k\frac{\pi}{2} } \right\}_{k \in \mathbb{Z}} [/tex]

yea..i meant (sin x)^5
and is there no easier way to verify this?
metru's way is too complicated for my scope of trig so far =(

The thing is: "verify" means that you're required to show that it works for ALL values of x, which it doesn't. Metru solved for the values of x for which the equation holds.

Which meaning do you want?

It must have been intended as "Solve for x where..." because the original equation clearly does not hold for all values of x...try x=0.


Bob and Alice plans a meeting. They prepare to meet each other at a time between 12am-1am :ninja: . If the first person arrives first, he/she waits for 20 minutes before leaving. What is the probabillity that Bob and Alice meet (hmmmmmmmmmmm a bit similar to 9 in mock aime)

I risk to say that it equals $\frac{5}{9}$ although I've not caught what does mean "a bit similar to 9 in mock aime"


algebra polynomial Gauss calculus derivative inequalities logarithms

Please write in this topic about properties of critical point of complex polynomial that you known.
Thank you.

Saman Gharib

A nice and simple result is the Gauss-Lucas Theorem, stating that the complex roots of the derivative $p'$ of a complex polynomial $p$ lie in the convex hull of the roots of $p$:

Let $p(z)=(z-z_0)\cdot\ldots\cdot(z-z_n)$. We then have $\frac{p'(z)}{p(z)}=\frac 1{z-z_1}+\ldots+\frac 1{z-z_n}$. The zeros of $p'$ which are also zeros of $p$ are of no interest to us (they clearly lie in the convex hull of $\{z_i\}$), so let $\omega$ be a root of $p'$, different from all the $z_i$. We have $\sum_{i=1}^n\frac 1{\omega-z_n}=0\Rightarrow \sum_{i=1}^n\frac{\bar\omega-\bar z_i}{|\omega-z_i|^2}=0$, which, after conjugation, becomes $\omega=\frac{\sum_{i=1}^n\frac 1{|\omega-z_i|^2}\cdot z_i}{\sum_{i=1}^n\frac 1{|\omega-z_i|^2}}$, and this is what we wanted.

Assume that $f(z)$ is a real polynomial and thus that its non-real zeros occur in conjugate imaginary pairs. Let us construct the circles  whose diameters are the line-segments joining the pairs of conjugate imaginary zero of $f(z)$. These circles we shall call Jensen circles of $f(z)$.
  We shall state a theorem which was announced without proof by Jensen in 1913. It was proved by Wlsh in 1920 and later by Echols and Nagy.

  Jensen Theorem. Every non-real zero of the derivative of a real polynomial $f(z)$ lies in or on at least one of the Jensen circles of $f(z)$.

Saman Gharib

[quote="grobber"]We then have $\frac{p'(z)}{p(z)}=\frac 1{z-z_1}+\ldots+\frac 1{z-z_n}$.[/quote]

The $LHS$ is $(\ln p(z))'$. Is there going to be a problem with the formula when $p(z) \in \mathbb{R}^-$? (I ask this because in that case $\ln$ is not continuous)

No. This function doesn't have a global antiderivative, but we don't need anything like that. The principal branch isn't the only definition of the logarithm anyway.

Sorry, but I don't understand what you are saying. ( :( ) So the formula works for all $p(z) \neq 0$? (using that argument)

[color=blue]EDIT: I understand it now...  :)[/color]

On the other hand, that's just a polynomial identity which can easily be checked, so I don't really need to introduce $\ln$.


If we deactivate our account, can we register with a different username using the same email address? Also, how should we ask to deactivate our account?

I can change your username if you want. And no, you can't do that.

[quote="Dojo"]I thought the username policy was only if you used your real name in your account.[/quote]

No, if you provide a legit reason as to why you want to change your username, then you can.

I might change mine; depending on what the current rules are but I'm not too pushy about it.

I will send a pm to you, Valentin, about changing my username.


I mad some careful figures and I think your conclusion is not true !
Or I'm wrong ? :?

$H$ is the middle of $AC$, it's right ?
I found also $\widehat{DHA}>\frac{\pi}{2}>\widehat{BHA}$... :D

$(I)$ is the inscribed circle means $(I)$ inside the quadrilateral and tangent to $AB,BC,CD,DA$

[quote="James Potter"]Let be given a quadrilateral $ABCD$ and a circle $(I)$ inscribed this quadrilateral.Let $H$ be a point on $AC$ such that $IH$ and $AC$ are perpendicular.Prove that $\widehat{BHA}=\widehat{DHA}$[/quote]
I think we have discussed this problem here[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=69101]www.mathlinks.ro/Forum/viewtopic.php?t=69101[/url]
You don't think so? :?  ;) .


inequalities
function
LaTeX
inequalities unsolved

Let be $ a,b,c \geq 0$ with $ a\plus{}b\plus{}c\equal{}1$. Prove that \[ \frac{1}{3}\leq \frac{a}{a^2\plus{}a\plus{}1}\plus{}\frac{b}{b^2\plus{}b\plus{}1}\plus{}\frac{c}{c^2\plus{}c\plus{}1}\leq \frac{9}{13}\]

[i]Ion Nedelcu, Romanian Shortlist, 2008[/i]

[quote="Mîndrilă Claudiu"]Let be $ a,b,c \geq 0$ with $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac {a}{{a^2 \plus{} a \plus{} 1}} \plus{} \frac {b}{{b^2 \plus{} b \plus{} 1}} \plus{} \frac {c}{{c^2 \plus{} c \plus{} 1}} \le \frac {9}{{13}}
\]
[i]Ion Nedelcu, Romanian Shortlist, 2008[/i][/quote]
By the AM-GM and Cauchy Schwarz inequality we have
\[ \sum {\frac {a}{{a^2 \plus{} a \plus{} 1}}} \le \sum {\frac {{9a}}{{15a \plus{} 8}}} \le \frac {9}{{13}}
\]
[quote="Mîndrilă Claudiu"]Let be $ a,b,c \geq 0$ with $ a \plus{} b \plus{} c \equal{} 1$. Prove that
\[ \frac{a}{{a^2  \plus{} a \plus{} 1}} \plus{} \frac{b}{{b^2  \plus{} b \plus{} 1}} \plus{} \frac{c}{{c^2  \plus{} c \plus{} 1}} \ge \frac{1}{3}
\]
[i]Ion Nedelcu, Romanian Shortlist, 2008[/i][/quote]
By the Cauchy Shcwarz inequality we have
\[ \sum {\frac{a}{{a^2  \plus{} a \plus{} 1}}}  \ge \frac{{\left( {\sum a } \right)^2 }}{{\sum {a^3 }  \plus{} \sum {a^2 }  \plus{} 1}}
\]
It suffices us to show that
\[ 3\left( {\sum a } \right)^2  \ge \sum {a^3 }  \plus{} \sum {a^2 }  \plus{} 1
\]
\[ \Leftrightarrow 2\left( {\sum {a^2 } } \right)\left( {\sum a } \right) \plus{} 6\left( {\sum {ab} } \right)\left( {\sum a } \right) \ge \sum {a^3 }  \plus{} \left( {\sum a } \right)^3
\]
\[ \Leftrightarrow 5\sum {ab\left( {a \plus{} b} \right)}  \plus{} 12abc \ge 0
\]
which is obviously true. Equality holds if and only if $ \left( {a^2  \plus{} b^2 } \right)\left( {b^2  \plus{} c^2 } \right)\left( {c^2  \plus{} a^2 } \right) \equal{} 0$, :).

function $ f(x)\equal{}\frac{x}{x^2\plus{}x\plus{}1}$ is concave on interval $ (0,1)$ therrefore we have by Karamata-majorization inequality
$ \frac{1}{3}\equal{}f(1)\plus{}f(0)\plus{}f(0)\leq f(a)\plus{}f(b)\plus{}f(c)\leq f(\frac{1}{3})\plus{}f(\frac{1}{3})\plus{}f(\frac{1}{3})\equal{}\frac{9}{13}$

Give function:  f(a) = a./(a.^2+a+1)   and g(a) = 15./169+72./169.*a  then:f(a)-g(a)=-1/169*(8*a+15)*(-1+3*a)^2  is negative
use this inequality for a,b,c then add and proof is complited.

nice navid. but for the next time please use $ latex$.


For $ k \ge 4$, $ k! \ge [(1)(k)][(2)(k\minus{}1)]>k^2$, impossible.

Check $ k\equal{}1,2,3$, we get $ k\equal{}1,3$ are the solutions.

[quote="RisingFlame"]For $ k\ge 4$, $ \sum_{i \equal{} 1}^k i! \equiv 1 \plus{} 2 \plus{} 6 \plus{} 24 \equiv 3 \mod 10$.
Since $ 3$ is not a quadratic residue $ \mod 10$, we only need to check for $ k \equal{} 1,2,3$. $ 1! \equal{} 1^2$ and $ 1! \plus{} 2! \plus{} 3! \equal{} 9 \equal{} 3^2$ so $ \boxed{k \equal{} 1}$ and $ \boxed{k \equal{} 3}$ are the only solutions.[/quote]
Just wondering, in the technique of quadratic residues, how do we choose which modulo to work with? (so in this case how do we know that mod 10 is going to work beforehand?) Do we just guess a modulo and see if it works and just hope to get lucky? ... or is it that we have to memorize that residues?

Well you could have chosen other other mod as well, for example $ \pmod 5$ would work just as good as $ \pmod {10}$. It's mostly intuition you are relying on in choosing mods, but you don't have to remember quadratic residues. Just type down all possible residues of $ \pmod m$ $ (0, 1, 2, ... m\minus{}1)$ and square them.
Finding $ \pmod {10}$ is equivalent to finding last digit of a number so it's practical here since most of high school students are familiar with units digit of squares.


:?:
Hi, can anyone tell me how the Arithmetic Mean-Geometrig Mean Inequality is derived?

I would really appreciate your help :lol:

Hello, prototypex24. :)

$(a+b)^2-(a-b)^2=4ab\Longleftrightarrow \left(\frac{a+b}{2}\right)^2 -\left(\frac{a-b}{2}\right)^2=ab$

$\Longleftrightarrow \left(\frac{a-b}{2}\right)^2=\left(\frac{a+b}{2}\right)^2 -ab$

For real numbers $a,b$, since $\left(\frac{a-b}{2}\right)^2 \geqq 0$, we have $\left(\frac{a+b}{2}\right)^2 \geqq ab$.

Especially for [color=red]positive numbers a,b[/color], $\frac{a+b}{2}\geqq \sqrt{ab}$, which is called

[color=blue]A[/color]rithmetic [color=blue]M[/color]ean $\geqq$ [color=blue]G[/color]eometic [color=blue]M[/color]ean. The equality holds when $a=b$.

Somebody will explain this theme geometrically.

kunny

Thanks kunny,
 But what I meant was the more general inequality:
$\displaystyle \frac tion {1}/{n} \cdot (a_{1}+a_{2}+a_{3}+...+a_{n})\geq(a_{1}\cdot a_{2}\cdot a_{3}\cdot ...\cdot a_{n})^{\frac tion{1}/{n}}$

There's a proof for AM-GM, found by Cauchy. It's a very difficult proof, with two steps.

1) Proof that AM-GM holds when $n$ is a power of $2$.
2) Proof that AM-GM holds for any $n$ such that $2^k<n<2^{k-1}$, for any $k$.

Yeah, something like that. There's a section about it in Zeitz's [i]Art and Craft of Problem Solving[/i] if you want more information.

i dont know if this is appropriate for Getting Started....


Does anybody know where I can find a copy of the AIME rescore request form that I can print out?  I've scoured the AMC website (including the AIME teachers' manual) but have been unable to find it.

you mean this?
http://www.unl.edu/amc/d-publication/pubform.html

No - that only has publications of contest and practice problems.  Can somebody please help me out with this?

[quote="joml88"]I don't see anything online about that.  Are you sure they rescore AIMEs?  You should call up their office.[/quote]
Hmm... I thought so, but I did more searching, and I did find an AMC rescore form, so possibly I was thinking about that instead.  I've been sending e-mails to the AMCinfo e-mail address, but they haven't been very helpful (they did direct me to their website and told me to fill out a rescore form, but they may have been thinking about the AMC rather than the AIME).  Maybe I'll try calling their office, thanks.


Greetings all!
i wanted to alert you to an on-site competition happening in India in December.  i attended the last such contest in May of 2002 and plan to go back this year.  You can find out some information (but unfortunately not much about the structure of the contest itself) at [url]http://www.cmseducation.org/iymc/[/url]
i am in contact with the organizers to get more information about the structure (a proposal to use tests similar to [url=http://mathleague.org]GPML[/url]/[url=http://www.arml.com]ARML[/url] failed) and hope to have that soon.  In the meantime, anyone interested in participating in the contest is invited to contact me and we can see about organizing a team from the US or a subset thereof.

Tim Sanders
tim@mathleague.org

I admit to not having looked carefully at the site, but about what level are the problems?

trust me, it's not on the site.  :)  i would say the problems will range in difficulty from early AMC to early AIME, about what you might expect at a typical high school contest.  since i've been to the contest before, i can answer specific questions people might have, and when i find out more about the format i'll post that here as well


Find all prime numbers p , q such that : $ 2^p + 2^q $ is divisible by pq

Do you know solution, or you left this part for Pierre?  :D

[quote="nttu"]I have a soluiton :
 $ p=q=2 : OK $
 $ p>2:
  2^p + 2^q = 2^q(2^{p-q} + 1) $ is divisible by $ p $
  So $ 2^{p-q}+1 $ is divisible by $ p $
  Let d be a minimal possitive integer such that : $ 2^d $ is divisible by $ p $
$ ==> 2(p-q) $ is divisible by $ p $ .
$ ..... $ the rest is for you , Myth :lol:  :lol:[/quote]
Do you mean $2(p-q)$ is divisible by $d$.  :D
And, of course, it is an obvious step in solution...

yeah , I mean $ 2(p-q) $ is divisible by $ d $
[quote] And, of course, it is an obvious step in solution... [/quote]
  Are you sure ???

first, let $p\geq q$. if $q=2$, then $p|2^p+4$, but $2^p+4\equiv 2+4=6\mod p$, so $p=2,3$ giving the solution $(2,2),(3,2),(2,3)$.

now, assume $q>2$; thus $pq|2^{p-q}+1$. let $a$ be minimal s.t. $2^a+1\equiv 0\mod p$, analogous $b$ for $q$. then we have $\frac{p-1}{a}\in 2\mathbb{Z}$, analogous for $b$. but we also have $\frac{p-q}{a}\in 2\mathbb{Z}+1$, $\frac{p-q}{b}\in 2\mathbb{Z}+1$. let $v,w$ be the exact powers of 2 in $a,b$, respectively. we have $\frac{q-1}{a}=-\frac{p-q}{a}+\frac{p-1}{a}\in 2\mathbb{Z}+1$ and $\frac{q-1}{b}\in 2\mathbb{Z}$; therefore $v>w$. analogously we get $w>v$, contradiction.

Peter


USA IMO team is announced.  See the official IMO site.  Evan O'Dorney made it.

Wait, where is it posted?  I can't seem to find it.

http://www.imo-official.org/year_reg_team.aspx?code=USA

i suppose that's what he's referring to.

aw, pythag didn't make it. well, he still has 5 more years.
congrats and good luck to the team!

Oh, cool. The 12 USAMO winners had to take another test, TST. Hmm I had thought they didn't do anything before MOP. lol Urc.

The TST is the first few days of MOP, which just happened.

Evan O'Dorney wasn't even HM. I'm quite surprised he was taking the TST at all.

[quote="orl"][quote="dnkywin"][url=http://www.imo-official.org/year_reg_team.aspx?code=USA]USA team[/url]

Yay a fellow North Carolinian!

John Berman
Wenyu Cao
Eric Larson
Delong Meng
Evan O'Dorney
Qinxuan Pan

Congrats!!![/quote]

Congrats to everybody on the team!  :)

Can somebody please post the members with ML/AoPS user name to the [url=http://www.mathlinks.ro/viewtopic.php?t=277976&start=20]IMO 2009 thread?[/url][/quote]

Ummm.... we're not supposed to reveal the true identity of any Aopser without their permission...

[quote="orl"]Don't worry about the IMOers!  :D Just put the name as a link to the user profile!  :)[/quote]No, the first poster was right.  Don't connect anyone's real name to their username [b]without their permission[/b].  Some people prefer to remain "anonymous" on AoPS.

[quote="mathemonster"]congrats everyone. yeah why was o'dorney taking the TST[/quote]There is precedent for past IMO participants to take the TST.  See also the last paragraph of Section VIII of the [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/TM-USAMO-2009.pdf]USAMO Teachers Manual[/url], as I believe that's what happened here.

Apparently all of blue took it this year.

Blue was supposed to be only taking it for practice, but then some people did extremely well, and their USAMO + TST score overcame several winners despite not having a USAMO score at the winner cutoff. (One of the alternates is from blue as well.)

Hi

Does anyone have the problems from the 2009 USA TST?

Thanks in advance.

[quote="In another topic, I"]The TST will very likely not be posted until after this year's IMO; this is the norm at least.[/quote]
I plan to post it fairly soon afterwards.

[quote="gauss1181"]No, it's vice versa. You make IMO from making black MOP if you take the USA TST (Team Selection Test) and get in the top six of scorers. However, in addition, blue MOP people (in which Evan was in) also get to take the TST and have the opportunity to make IMO if they score high enough, but I think only one blue MOPper can make IMO. Also, students who have made IMO in the previous year are invited to take the TST and get a chance to be selected, even if they miss the black MOP cutoff from USAMO.[/quote]

Just so you know, you're right if we're just talking about this year, but the TST has not been offered to any Blue MOPpers in previous years.  Everything after the word "However" in your post is only true of this year, and none of this has anything to do with standard rules and procedures.  It's all within the rules, however, because of that clause that Sly Si and orl pointed out.

[quote="sophia"][quote="matt276eagles"]Just so you know, you're right if we're just talking about this year, but the TST has not been offered to any Blue MOPpers in previous years.[/quote]

This is not correct.   Blue MOPpers have taken the TST in several previous years, and sometimes 12th grade honorable mention students have been invited to MOP specifically to take the TST, especially in cases when several of the top 12 students were Canadians or otherwise unable to accept a place on the USA IMO team (e.g., due to RSI.)  The most notable example would be Thomas Mildorf, a member of the USA 2005 IMO team who was an honorable mention on the USAMO that year--and he went on to be a very valuable member of the US team, bringing home a gold medal with score of 39 points.[/quote]

aha, but there's a catch. This only happens if that honorable mention actually scored on the winner cutoff, but was pushed to HM through tiebreaks. (This has happened to Brian Hamrick this year, Palmer last year, and Thomas in 2005. all Virginians, as noted by Brian ;) ) So this is rare, and as far as I know only happens in case of ties.

[quote="not_trig"]

aha, but there's a catch. This only happens if that honorable mention actually scored on the winner cutoff, but was pushed to HM through tiebreaks. (This has happened to Brian Hamrick this year, Palmer last year, and Thomas in 2005. all Virginians, as noted by Brian ;) ) So this is rare, and as far as I know only happens in case of ties.[/quote]

Many non-winners who were far from borderline tiebreak cases took the TST in 2004 as well--an AoPS [url=https://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1651125370&t=13444]thread from moppers that year[/url] reported that Brian Lawrence, a 9th grade mopper who had not even made honorable mention on the USAMO that year managed to beat all the USAMO winners on the TST that year, and wound up as an alternate to the 2004 IMO team.

[quote="JSteinhardt"]Actually, in 2006 many non-winners took the TST due to multiple Canadians/RSI students (and were also invited to Black MOP). Peter Diao is one such example (yet another Virginian).[/quote]
And Zeb Brady, who wound up on the IMO team.

[quote="sophia"][quote="matt276eagles"]Just so you know, you're right if we're just talking about this year, but the TST has not been offered to any Blue MOPpers in previous years.[/quote]

This is not correct.   Blue MOPpers have taken the TST in several previous years, and sometimes 12th grade honorable mention students have been invited to MOP specifically to take the TST, especially in cases when several of the top 12 students were Canadians or otherwise unable to accept a place on the USA IMO team (e.g., due to RSI.)  The most notable example would be Thomas Mildorf, a member of the USA 2005 IMO team who was an honorable mention on the USAMO that year--and he went on to be a very valuable member of the US team, bringing home a gold medal with score of 39 points.[/quote]

What I was trying to say was that in some previous years, the TST has not been offered to any Blue MOPpers.  Although, I didn't know about the situations you're talking about, and in that light, my post does look blatantly wrong.  :P   Thanks.

How much (if any) harder is the IMO than the TST?

It would vary since neither is the same difficulty from year to year, but at least for the last two years, the TST was harder mostly because there were (usually) no easy problems. The hardest problems on each contest are probably comparable, but all 3 problems in a day being of substantial difficulty really cuts into your time.

[quote="gauss1181"]No, it's vice versa. You make IMO from making black MOP if you take the USA TST (Team Selection Test) and get in the top six of scorers. However, in addition, blue MOP people (in which Evan was in) also get to take the TST and have the opportunity to make IMO if they score high enough, but I think only one blue MOPper can make IMO. Also, students who have made IMO in the previous year are invited to take the TST and get a chance to be selected, even if they miss the black MOP cutoff from USAMO.[/quote]
Sorry for the bump, but do past IMO students really get to take the TST? I was always under the impression that Zeb Brady was not able to take the TST in his 12th grade year.


inequalities
geometry
angle bisector
inequalities unsolved

[quote="mathson"][quote="tuandokim"]prove that:
$ \sum \frac {h_{a}}{l_{a}}\leq \sqrt {6 \plus{} \frac {5r}{R} \plus{} \frac {2r^2}{R^2}}$[/quote]

What is $ l_a$?[/quote]
I think $ l_a$ is the length of the angle bisector from $ A$ to $ BC$, I mean that If the angle bisector of $ \angle A$ intersects $ BC$ at $ D$, the $ AD\equal{}l_a$. :maybe:

Am I correct? :blush:


function
integration
calculus
derivative
real analysis
trigonometry
calculus computations

Let f be a periodic function with period $ 2 \pi$.

(a) Knowing that the coefficients of the Fourier series of f are

$ a_0 \equal{} 2 \pi$,           $ a_n \equal{} 0$,          and  $ b_n \equal{} \frac{\minus{}2(\minus{}1)^n}{n}$

justify which of the following expressions may define f in $ [\minus{} \pi, \pi[$:

$ (i)\,\,\,f(t) \equal{} |t|$

$ (ii)\,\,\, f(t) \equal{} \minus{}t$

$ (iii)\,\,\, f(t) \equal{} t \plus{} \pi$

(b) Let g(t) be the sum of the Fourier series of f. Find $ g(0)$ and $ g(\pi)$.

$ a_n \equal{} 0$ for $ n > 0$ means that we can subtract off a suitable constant and get an odd function.

$ |t|$ is an even function. Hence it is clearly wrong. The the other two answers are still possible.

$ \minus{} t$ is an odd function, and has an average value of zero. $ t \plus{} \pi$ has an average value of $ \pi.$

There are two possible conventions for what we could mean by the notation $ a_0.$ In one convention, $ \frac {a_0}2$ is the average value of the function, and in the other convention, $ a_0$ is the average value of the function.

We conclude that you must intend the first convention. That means that answer (iii), $ t \plus{} \pi$ is the only possible correct answer.

That still doesn't mean that it is the correct answer. Let me do a couple of plausibility checks:

The function which is $ t \plus{} \pi$ on $ ( \minus{} \pi,pi)$ has a jump discontinuity at $ \pi.$ Functions with a jump discontinuity should have coefficients whose absolute values are $ O(n^{ \minus{} 1})$ (and not better than that); this passes that test.

At this point, I would recommend computing the Fourier series of the function in (iii) to confirm that it is the correct answer. (Hint: it is.)

As for question (b): for "nice enough" functions, which this is, the sum of the Fourier series is the average of the right and left hand limits at any point. Once we know that the function $ g$ is $ \pi\plus{}t$ on $ (\minus{}\pi,\pi),$ then we know that $ g(0)\equal{}\pi$ (as a point of continuity) and that $ g(\pi)\equal{}\pi$ (the average of the right and left hand limits of the periodic function at that point of jump discontinuity.)

Professor Merryfield, what is the justification for "Functions with a jump discontinuity should have coefficients whose absolute values are $ O(n^{ \minus{} 1})$(and not better than that)"?

I was also assuming that $ f$ was well-behaved between jump discontinuities - say, piecewise $ C^1$ or something like that.

Let me express the Fourier coefficients in complex form:
\[ \widehat{f}(n)=\frac1{2\pi}\int_{\mathbb{T}} f(x)e^{-inx}\,dx\]
where by $ \mathbb{T}$ I mean the circle group ${ \mathbb{R}/(2\pi\mathbb{Z}}.$ In practice the integral can be from $ -\pi$ to $ \pi,$ or $ 0$ to $ 2\pi,$ or any other interval of length $ 2\pi.$ There's a straightforward linear mapping from $ \widehat{f}(n)$ and $ \widehat{f}(-n)$ to $ a_n$ and $ b_n,$ and vice versa.

Assume $ f$ is smooth and integrate by parts to find that

$ \widehat{f'}(n)=\frac1{2\pi}\int_{\mathbb{T}} f'(x)e^{-inx}\,dx =-(-in)\cdot\frac1{2\pi}\int_{\mathbb{T}}f(x)e^{-inx}\,dx =in\widehat{f}(n).$

Now consider $ f$ to be a (periodic) distribution (that is, generalized function). The identity above still holds, with the derivative being the weak derivative.

Now suppose $ f$ is piecewise $ C^1$ but has one or several jump discontinuities. Then $ f'$ can be represented as the sum of a piecewise continuous function and a finite linear combination of translates of the Dirac delta function. The Fourier transform of $ \delta(x-a)$ is simply $ e^{-ina}.$ The Fourier transform a a finite linear combination of such functions is a finite linear combination of functions of the form $ e^{-ina}$ for various $ a.$ Such a function of $ n$ remains bounded but does not go to zero as $ |n|\to\infty.$ The piecewise continuous portion of $ f'$ has a Fourier transform that does tend to zero as $ |n|\to\infty,$ by the Riemann-Lebesgue Lemma. Hence, $ \widehat{f'}(n)$ is bounded but does not tend to zero as $ |n|\to\infty.$ Then we divide that (for $ n\ne 0$) by $ in$ to get $ \widehat{f}(n),$ and conclude that that behaves as $ O(n^{-1})$ but not better.

Now let's turn this distributional language into an actual computation. I'll go back to the real (cosine-sine) form for this.

Let $ f(x)\equal{}\pi\plus{}x$ on $ (\minus{}\pi,\pi),$ extended to be periodic.

Then $ f'(x)\equal{}1\minus{}2\pi\delta(x\minus{}\pi).$ The delta is there because of the jump discontinuity at $ \pi.$ (I don't mention the discontinuity at $ \minus{}\pi$ because $ \pi$ and $ \minus{}\pi$ are the same point; we only count what goes on there once.)

$ f'$ is even, so we compute only its $ a_n$. For $ n>0,$

$ a_n\equal{}\frac{1}{\pi}\int_{\minus{}\pi}^{\pi} (1\minus{}2\pi\delta(x\minus{}\pi))\cos nx\,dx$

$ \equal{}\frac1{\pi}(0\minus{}2\pi\cos n\pi)\equal{}\minus{}2(\minus{}1)^n.$

Hence $ f'(x)\equal{} \sum_{n\equal{}1}^{\infty}\minus{}2(\minus{}1)^n\cos nx.$

(A word about that equal sign: of course the series does not converge in any classical sense. I mean to imply that the series converges weakly as a distribution to the distribution $ f'.$)

Integrate that term by term:

$ f(x)\equal{}\frac{a_0}2\plus{}\sum_{n\equal{}1}^{\infty}\frac{\minus{}2(\minus{}1)^n}{n}\, \sin nx.$

We can never compute $ a_0$ from any knowledge of the series for $ f'.$ That requires a separate computation. But $ \frac{a_0}2$ is the average value of the function, and by inspection of the graph, the average value of $ f$ is $ \pi.$


There's a carpet with unknown thickness.Its mass initial radius is $R$.Total length is $L$.We give the initial velocity $v_{0}$ to the center of the carpet as in the figure.Find the opening time of the carpet.

I'm not completely sure about it, but:
[hide="my answer"]$T=\frac{L}{4v_{0}}$[/hide]

I don't know the answer.Can you post your solution?

I found a mistake. I'll post my new solution:
Let $d$ be the thikness of the carpet. The total surface of its boundary (the surface seen in your picture) can be given in two ways: $\pi R^{2}$ or $dL$, from which we get:
\[d=\frac{\pi R^{2}}{L}\]
Consider now a moment when some part (of lenght $L-l$) of the carpet is lying on the floor and the rest (of lenght $l$) is still rolling. Again we can calculate the surface of the rolling part in two ways:
\[l=\frac{\pi r^{2}}{d}=\frac{Lr^{2}}{R^{2}}\]
The angular momentum of the carpet counted from some point on the floor level is constant (torque of the friction force is $0$). As the mass of the rolling part is proportional to the surface of its boundary, so from conservation of angular momentum we get:
\[Mv_{0}R=M\frac{\pi r^{2}}{\pi R^{2}}vr\]
We also have got:
\[v=-\frac{dl}{dt}\]
and by integrating the second equation:
\[dl=\frac{2Lrdr}{R^{2}}\]
Putting that into the angular momentum conservation we get:
\[\frac{v_{0}R^{5}}{2L}dt=-r^{4}dr\]
And by integrating:
\[T=\frac{v_{0}R^{5}}{2L}\int_{0}^{T}dt=-\int_{R}^{0}r^{4}dr=\frac{R^{5}}{5}\]
Finally:
\[T=\frac{2L}{5v_{0}}\]

I got a different answer. here is my solution:

let the thickness of the carpet be $s$, its mass density $\rho$ its width $l$.
hence the mass of the whole carpet: $M=\rho L l s$

the velocity of the middle of the rolling part after some time $v$, the radius of the rolling part $r$, the length of the carpet wich already rolled off (is horizontaly on the floor) $x$.
the mass of the rolling part :
$m=M-\rho xls=\rho ls(L-x)$
the link between $r$ and $x$:
$r=\left ( 1-\frac{x}{L}\right ) R$

so the conservation of energy:
$\frac12 Mv_{0}^{2}+\frac12 \frac12 M R^{2}\frac{v_{0}^{2}}{R^{2}}+MgR= \frac12 mv^{2}+\frac12 \frac12 mr^{2}\frac{v^{2}}{r^{2}}+mgr$

and also :
$v=\frac{dx}{dt}$
so the total time needed for rolling out:
$T=\int_{0}^{L}\sqrt{\frac{3L(L-x)}{3v_{0}^{2}L^{2}+4Rgx(2L-x)}}dx$
now this integral looks quite nasty, but I think we can probably neglect the effect of gravity.
if we do that then in the equation for the conservation of energy the gravitational potential energy parts fall out so we get for $v$:
$v=\cfrac{v_{0}}{\sqrt{1-\frac{x}{L}}}$
so the time needed:
$T=\frac{1}{v_{0}}\int_{0}^{L}\sqrt{1-\frac{x}{L}}dx= \frac{2L}{3v_{0}}$

[quote="Thaakisfox"]the link between $r$ and $x$:
$r=\left ( 1-\frac{x}{L}\right ) R$
[/quote]
That's not true. It would mean that if you double the lenght of the carpet the radius will double too. But as the radius is getting bigger also the circumference gets bigger - so the relation cannot be linear.
[quote="Thaakisfox"]but I think we can probably neglect the effect of gravity[/quote]
Well, that's some kind of an approximation, I'm not sure if we can do that...  :huh: we should have $v_{0}^{2}\gg gR$ to do that and I think that this relation is not very well satisfied...

I dont understand, if the length grows then of course the radius grows. whats the problem with that? :huh:

But the radius grows slower then the lenght because to increase the radius by $s$ (the thickness of the carpet) we need to increase the lenght by $2\pi r$ ($r$ is the actual radius). The relation should be $r^{2}=\left ( 1-\frac{x}{L}\right ) R^{2}$

Thaakisfox's answer $/frac{2L}{3v_{0}}$ is correct.


[quote="tranthanhnam"]find  value maximal of K such that forall $x,y,z >0$and $xyz=1$ then ineq true :
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2} +3K \geq (K+1)(x+y+z)$[/quote]
maxK=1

To prove that the inequality is true for $K=1$, we can let $a^3=\frac{yz}{x}$ and etc. Then we may apply \[ a^3+b^3+c^3+3abc \ge ab(a+b)+bc(b+c)+ca(c+a)  \]
followed by AM-GM. The inequality will then be proved.

To prove that 1 is the largest value of $K$, we can let $x=y$. If $K=1+e$, $e>0$, the inequality will be false when x tends to 0.


I was playing black.

_ R _ _ _ _ _ _
P _ _ Q _ _ _ P
_ _ P K _ _ _ _
_ _ _ P _ [b]Q[/b] _ _
_ _ _ _ _ _ _ _
_ [b]P[/b] [b]N[/b] _ _ _ _ [b]P[/b]
[b]P[/b] _ [b]P[/b] _ _ [b]P[/b] _ _
_ [b]K[/b] _ _ [b]R[/b] _ _ _

(black's move)

I had to go somewhere, so I was going to stop the game.
but white said, "I'll surrender. I can checkmate me in two moves or so"
well, I see its almost my game, but I dont see a brilliant checkmate move.

Do you?

I tried to transfer your text diagram into a picture. I had to invert the sides. Did I get it right? Is that the position you had? (The Chess Captor Label is partially covering a black king at the top)

thank you.
I dont know how to post a picture yet. :blush:

yes, the diagram is right, and it was my turn to move.
What I would have done is trading queens, but I dont think that was what my opponent was saying.

i don't see the mate in two moves, but you have better options than trading queens.
for example put your queen at d4. then the only possible moves for whit was Kf4. then you could take his rook and so on,.

thanks.
Of course taking free rook is better than queen trade.

how about Qc3? anything after Kf2 or Kf4?
how about Qc4? then the King is forced to Kf4?

I agree with amirhtlusa, 1...Qd4 is definitely your best move - though Qc3 isn't bad, it's not as strong as Qd4. After White flees with 2. Kf4, you could also consider 2...Rd5 which cuts off that rank for the black king. I'm not sure if this line will end in checkmate so the safer move probably would be to just take the free white rook.

Qd4+ and Qxe2+ should both win very quickly. You might have mate after Qd4+ (although I don't see it), but my chess philosophy is that there's no point in looking for mate when it doesn't take any effort to win normally.

i haven't completely calculated this yet, but

1. Nd5+
If: King moves, but doesn't go to f2, you win the king with discovered check
If: 1....exd5, Re8+ skewers king and queen, followed by you winning the rook
1.............Kf2
2. Qc5+
If: 2....Kg2, Nf4 wins queen
2.............Ke1 or Kf1
3. Qc1+   Kf2
4. Haven't calculated furthur yet, will probably edit post.....

I think I Qd4+ works...
1Qd4+ Kf4  forced
2Qxg1  I dont think black can defend against Nh5 + Ke5 Qg5++

What about Qxe4+ Kf2 Qh4+?  Then white is screwed.  It could then continue Ke3 Qd4++.  Or it could go Rg3 Ng4 Kg2 Nxh2 or something like that.
Eh...Qd4 with rook capture is probably best.

[quote="K81o7"]What about Qxe4+ Kf2 Qh4+?[/quote]

White would probably play fxe4 rather than Kf2.  ;)

[quote="jhollenbeck"]i think that Qd4+ winning the rook is the best possible move.[/quote]

I agree.  At this stage, loss of a rook would basically be checkmate against a good opponent.

I solved the whole thing.
Q+c5 not d4 :!:
Kf4(forced)
N+h5
Kg4(forced)
Q+g1
Now 2 possibilities:
1. Kh3
N+f4
Kh4(forced)
pXg5 it's over

2.Kh4
Nf4
King can't do anything, so:
if not Qg2 or Qb5, pXg5 we are done;
if Qb5, next pb5, king is still stranded, so next black's move pXg5
if Qg2, next Qg2, king is still stranded, so next black does pXg5. :D


Probably AwesomeMath is the best preparation for the AIME.

Indeed, it was pretty helpful... like doubled my AIME score? Although I suspect that that was not [i]all[/i] AMP...


Given a set of $n(n\geq 10)$persons
It is known that
1)Each person is acquainted with at least $[\frac{n+2}{3}]$others
2)For any 2 persons A,B,either thay are acquainted or there is a chain of persons $A_1,A_2,...,A_k$such that A_i acquainted with A_{i+1}(A_0=a,A_{k+1}=B)
3)IT is impossible to arrange n persons in a line so that any 2 adjacent persons are acquainted
Prove that:The et can partitionned into 2 groups so that:
i)the members of group 1 can sit round a roung table any 2 adjacent pesons being acquaited
ii)the members of group 2 are pairwise unacquainted


$a,b,c$ real positives, then prove
$\frac{1}{\sqrt{4a^{2}+bc}}+\frac{1}{\sqrt{4b^{2}+ac}}+\frac{1}{\sqrt{4c^{2}+ab}}\geq \frac{4}{a+b+c}$

[quote="hien"]I think it relate with the following problem at:
http://www.mathlinks.ro/Forum/viewtopic.php?p=119168#p119168[/quote]
Show  please as. Thanks.

Really thx for that. But I still can't find the equality......  :(

The inequality holds when a=b, c=0. If c is a positive real, therefore this ineq doesn't have the equality.

Then, can u give me the proof ??
Will that be the same with hien's suggestion ??  :maybe:

Don't woory, maybe this inequality haven't been solved or noone was interested in it!
http://www.mathlinks.ro/Forum/viewtopic.php?t=63645


geometry
geometric transformation
rotation
conics
ellipse
quadratics
dilation

Find , to the nearest integer, the area of the region enclosed by the graph of $ 13x^2 \minus{} 20xy \plus{} 52y^2 \minus{} 10x \plus{} 52y \equal{} 563$

Since we're dealing with area, we can ignore the $ xy$ term.  The $ xy$ term is basically a rotation of the figure, and since rotating preserves area, we can just ignore the $ xy$ term.

Complete the square, and figure out what shape it is.

[hide]
It's an ellipse, I think. $ \rightarrow ab\pi$
[/hide]

The cross term does matter; the scaling from the quadratic terms looks like a discriminant. Include it when completing the square.

My bad, I didn't really look into it, I just looked online about the $ xy$ term and it just told me that it relates to the rotation of the figure...nothing about changing the size though.

A rotation changes all of the terms, not just the $ xy$ term.  Conversely, modifying the $ xy$ term does more than just rotate.  Consider, for example, the difference between the graphs of $ x^2 \plus{} y^2 \equal{} 1$ and $ x^2 \plus{} 2xy \plus{} y^2 \equal{} 1$.

I did this fast, so I might've made an arithmetic mistake, but the general procedure is what you would follow. The first step is key.

[hide="Procedure"]

[hide="1 Skew"]Note that if we replace $ y$ with $ y \minus{} ax$, the area does not change.

So let's go do this:
$ 13x^2 \minus{} 20x(y \minus{} ax) \plus{} 52(y \minus{} ax)^2 \minus{} 10x \plus{} 52(y \minus{} ax) \equal{} 563$

We want to make the $ xy$ term equal to $ 0$. To do this, let's extract the $ xy$ terms from each of the summands above, and set the sum equal to 0:
$ 0 \minus{} 20xy \minus{} 104axy \plus{} 0 \plus{} 0 \equal{} ( \minus{} 20 \minus{} 104a)xy \equal{} 0$

$ \implies \minus{} 20 \minus{} 104a \equal{} 0 \implies a \equal{} \minus{} \frac {5}{26}$

Under a skew transformation with this $ a$ value, the original equation becomes:
$ \frac {144}{13} x^2 \plus{} 52 y ^2 \plus{} 52 y \equal{} 563$[/hide]
[hide="2 Translate"]$ \frac {144}{13} x^2 \plus{} 52 \left( y \plus{} \frac {1}{2} \right)^2 \equal{} 576$

Now we can translate by 1/2 in the y direction to get:
$ \frac {144}{13} x^2 \plus{} 52 y ^2 \equal{} 576$[/hide]
[hide="3 Dilate"]We want to get rid of these ugly 13's. If we compress by a factor of $ \sqrt {13}$ in the $ x$ direction and expand by the same factor in the $ y$ direction, we get:
$ 144 x^2 \plus{} 4 y^2 \equal{} 576 \iff (12x)^2 \plus{} (2y)^2 \equal{} (24)^2$[/hide]
[hide="4 Area of an ellipse"]The area of this ellipse is $ \frac {1}{12} \cdot \frac {1}{2} \cdot \left( 24 ^2 \pi \right) \equal{} \boxed{24 \pi }$[/hide][/hide]

I can't edit my previous post right now, I forgot you wanted the answer to the nearest integer. So in that case:
[hide="the answer is"]75[/hide]


IN THE NAME OF GOD
HI

What is the number you started with?

regards
salahcool

Hi salahcool.

Working backwards, I get 45.

Any more?

to find the answer just do the following.
Start from end and moving backwards do the division of number when it is written multiply and do multiplication when it is written division. because if a=x*b then x=a/b.

Let the original number be $ x$ Then $ \frac{308\cdot.1x}{3}\equal{}462$ so you just solve for $ x$.

so you have to work backwards.
462x3/10x308
which is 45

[hide="The answer"]

You just work backwards so you multiply by 3...

$ 462\cdot3\equal{}1,386$

Then you Divide by .10...

$ 1,386\div0.10\equal{}13,860$

Then you divide by 308...

$ 13,860\div308\equal{}...$

$ \boxed{45}$[/hide]

PEOPLE
[quote="cgyao(the really annoyed one)"]
Seriously. STOP it. don't post the SAME solution OVER and OVER again.[/quote]
READ


calculus
derivative
analytic geometry
graphing lines
slope
quadratics
geometry

Every time Allen Yuan plays piano, Carl Lian assigns a "show-off index" to his playing, based on how blatantly Allen is showing off his technique (or lack thereof). The show-off index ranges from 1, from Bach Minuets, to 10, somewhere near whatever Allen did during the talent show. The technical effectiveness of his performance is defined as the product of the show-off index and fifty minus `the number of wrong notes per minute, which is known to be $ 4c\minus{}3$, where $ c$ is the show-off index. For what $ c$ is the technical effectiveness of his performance maximized?

So maximizing 53c-4c^2?
What happens if he's playing Feux Follets?  :roll: Does the show off index become 12?

[quote="FantasyLover"][quote="serialk11r"]So maximizing 53c-4c^2?
[/quote]

if the question is asking for maximum value of $ 53c \minus{} 4c^2$, $ 53 \minus{} 8c \equal{} 0$ so $ c \equal{} \frac {53}{8}$?[/quote]

Why is 53-8c=0?  Why does that (if it even does) maximize the value of $ 53 \minus{} 4c^2$?

Extension:  Generalize - If when Carl Lian is playing, the wrong notes per minute is $ ac\minus{}b$ and technical effectiveness is defined as before, find the optimal $ c$ in terms of $ a$ and $ b$.

uhhh thats the derivative of $ 53c\minus{}4c^2$...

when the slope of a point is 0, that point is maximum of minimum value of a graph.

and the slope of $ 53c\minus{}4c^2$ is $ 53\minus{}8c$, and that should be 0 bc we're supposed to find the maximum value.

...i hate my explaining skills

[quote="FantasyLover"][quote="serialk11r"]So maximizing 53c-4c^2?
[/quote]

if the question is asking for maximum value of $ 53c \minus{} 4c^2$, $ 53 \minus{} 8c \equal{} 0$ so $ c \equal{} \frac {53}{8}$?

...didnt read any of the question[/quote]

wouldnt it be $ \frac{53}{4}$?

$ 53c\minus{}4c^2\equal{}0;$ $ c(53\minus{}4c)\equal{}0$ $ c\equal{}{0,\frac{53}{4}}$

i doubt you need calculus for a mc level problem

stop showing off jp  :P

[quote="FantasyLover"]no thats just the expression of some graph.

$ 53c \minus{} 4c^2$ doesnt equal 0.

[quote="mz94"]
i doubt you need calculus for a mc level problem

stop showing off jp  :P[/quote]

 :rotfl:[/quote]

oh oops  :blush:

then yea, its 53/8

BUT you dont need frikin derivatives

53c-4c^2 is a quadratic, maximized when c=-b/2a, where in the form ac^2+bc+d

so c=-53/2(-4)=53/8

For quadratics, $ \frac{\minus{}b}{2a}$ is good enough to find the local min and max.

For higher powers, you use derivatives to find the local max and min when the derivative of the curve is set to 0. Then you use the second derivatives as well. With all this and the basic x-intercepts, you can now draw the graph accurately. That's how much I got out of graphing using calculus anyways.

I've seen calculus used in MC but it is obviously never needed. I remember seeing a problem with area bounded by 3 lines. I was shown us a calculus solution (but didn't understand it).

Extension question: Is the show-off index still 1-10?

[quote="FantasyLover"]I DONT CARE

I LIKE MY WAY  :rotfl:  :rotfl:  :rotfl:[/quote]

JP NOBODY CARES THAT YOU USE CALCULUS TO SOLVE MC LEVEL PROBLEMS!!! :D

its gonna take you forever if you try that on a real MC test

:dry:

how bout on thursday? it took u 5min to use calculus to find that length in the square when i jsut reasoned and got it a LOOOONNNNGGGGG time before you ddi :P

and it took me <1 sec to find -b/2a of 53c-4c^2 :P

uh that was an excuse...

i made a mistake...
ps. my rating is now 2.6 lol

[quote="FantasyLover"]uh that was an excuse...

i made a mistake...[/quote]

idc jp

it still took me less than 1 sec to do

-53/2(-4) in my head

so since {>1sec}<{abt 1.8 sec}

my way>your way

[quote="FantasyLover"]ANYWAY YOU COULDNT SOLVE MY PROBLEM!!!!!

 :rotfl:

hm only me and you are posting in this thread[/quote]

and you couldnt solve mine...

and since {basic probability}<{calculus}

you<me

"You must first learn to crawl before you can run"
JP :P

learn probability before jumping into calc :P

ugh counting and probability is just not my thing... :wink:

EDIT : ok lets stop posting on this thread

lets pm each other

[quote="FantasyLover"]ugh counting and probability is just not my thing... :wink:[/quote]

thats your problem

now that we've settled that {my way}>{your way}

lets let this thread die  :P

DONT USE CALCULUS! GAH!

First, as you said, just taking a derivative ruins the problem, and that's definitely not the point of this problem (Mathcounts is written without assumed knowledge of even trig).  Second, it doesn't really help you to just calculus bash everything you see - on olympiad math, calculus bash makes graders not like you, and the majority of problems can't be calculus bashed anyways.

I like moogra's method...

There's also quite a bit of spam in this thread...

Here's another way. We want to maximize c(53-4c). Equivalently, we can maximize 4c(53-4c) and divide stuff away later. But note that we can just directly apply AM-GM here, and we find that the equality case is when 4c=53-4c. Of course, we have to check that 4c and 53-4c are positive, but this is clear as 1<=c<=10. If we weren't so fortunate, however, we could just note that if one's positive and one's negative, we can throw it away as we get something negative and negatives are smaller than positives. If they're both negative, we can just flip the signs and use AM-GM on the positive stuff without changing the value of the whole expression (actually it's not even possible for them both to be negative, but just in general...)

Well, you generally don't need AM-GM in MathCounts, and AM-GM is basically the same completing the square...

ehh sorry about the spam, CatalystOfNostalgia.

is $ \frac{53}{8}$ right answer btw?

[quote="FantasyLover"]it took me like 1.8 sec to figure out that the derivative of $ 53c \minus{} 4c^2$ is $ 53 \minus{} 8c$

 :rotfl:

ps. ...just noticed that my rating is 2.8...

hm i finally have rating!![/quote]
hehehe, no offense, but it is quite fast to find the derivative of $ 53c\minus{}4c^2$, but why use such hard math?

[quote="FantasyLover"]ehh sorry about the spam, CatalystOfNostalgia.

is $ \frac {53}{8}$ right answer btw?[/quote]

No one will know..........

[quote="Carl"]Away from 7/17-8/7[/quote]

[quote="Math Geek"]Hint: There trying to act smart and impress people by showing that they know how to find the derivative of a simple polynomial. :wink: (And spamming while doing so.)[/quote]

you are adding to the spam
as am i :P


geometry
rectangle
calculus
integration
combinatorics unsolved
combinatorics

Suppose that a rectangular board (composed of unit squares) can be covered by L-Tetrominoes. Show that the number of unit squares is a multiple of 8.

The problem is solved now :)

If anybody wants the solution then do let me know.

Hi rahulakaneo;

Can you post your solution, please, if you have time.

Thanks.

Yeah sure!

Here's the solution: Note that a tetromino occupies 4 unit squares (You may see the picture of an L-Tetromino here http://en.wikipedia.org/wiki/Tetromino), so at least one side of the tiled rectangle is even. Now we color odd rows of the rectangle with white, and the even rows with black. Since at least one side of the rectangle is even, the total number of white squares equal the total number of black squares. Now any L-Tetromino occupies either 3 white or 1 black, or 3 black and 1 white squares, and hence an even number of L-Tetrominoes are required to maintain the property that there are equal number of black and white squares. So the total number of unit squares is an integral multiple of 8.

Hi;

Thanks for showing me your solution and for the link.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Show that if a group $ G$ has proper subgroups $ H_1$ and $ H_2$, such that every proper subgroup of $ G$ is contained either in $ H_1$ or in $ H_2$, then $ G$ must be cyclic of finite order and for which orders is this possible?

1. if there isn't any x generate G, then $ \forall x\in H_1$ or $ x\in H_2$.
2. set $ a\in H_1\minus{}H_2$,  $ b\in H_2\minus{}H_1$.  consider c=ab.  if $ c\in H_1$ then $ a,b\in H_1$, contradiction.

such cyclic group G, $ |G|\equal{}p_1^{e_1}q_2^{e_2}$. since all possible maximal group for a cyclic group is pG, where p|G is a prime. And for any p|G, such group is maximal, so |G| consists only 2 primes.


Find functions f and g such that f(g) = H if H(x) = (x^2 + 1)^50

[hide]
$H(x)=(x^{2}+1)^{50}$ is made up of the two functions "square x, add 1" and "raise to the power of 50".

We seek $f$ and $g$ such that $f(g(x))=H(x)$.

It is clear that $f(x)=x^{50}$ and $g(x)=x^{2}+1$.[/hide]

[quote="Interval"]So, do multiply f(x) by g(x).

Now, f(x) = x^50 and g(x) = x^2 + 1

So, do I do this:

x^50 (x^2 + 1) =[/quote]
$f(g(x))$ (function composition) does not mean $f(x)\cdot g(x)$.

$f(x)=x^{50}$ and $g(x)=x^{2}+1$.

So, $f(g(x))=f(x^{2}+1)=(x^{2}+1)^{50}=H(x)$.

I see...it all ends with H(x).

This is similar to a prove question, right?

Thanks!

note that there is not only one answer...

f(x)=x^25, and g(x)=(x^2+1)^2, gives f(g(x))=(x^2+1)^50 as well


A wooden $ n$x$ n$ board is divided into unit squares by lines parallel to the sides. Two players alternately make cuts of length $ 1$ along these lines, starting from the border or from a point at a previously made cut. The player after whose move the board breaks loses the game. Which player has a winning strategy, the one who goes first or the other one?

$ \star$ [u]Note[/u]
In the following, I assume that there are $ n\times n$ dots. And $ (n-1)\times(n-1)$ small squares

[u]Definitions[/u]
A state is said to be a [b]W[/b]-state if it can lead to an [b]L[/b]-state
A state is said to be an [b]L[/b]-state if it can't lead to an [b]L[/b]-state

When some player makes a cut from the point $ (a,b)$ to the point $ (c,d)$, we will say that the point $ (c,d)$ is now [u]visited[/u]

At the beginning, the points that are "visited" are the points:
$ (1,k)$ and $ (n,k)$
$ (k,1)$ and $ (k,n)$, where $ k=1,2,...,n$
(namely, those are the points at the edges of the square)
$ \hline$

Let's see the case $ n=3$
[img]9483[/img]

There is only one point which is unvisited.
Te first player makes a cut from the bounds of the square to it.
After that, the second player has not a choice. Every move he makes he will break the board

See the case $ n=4$. Here are some possible results, after 4 moves:
[img]9484[/img]

After each of those situations, the first to make a cut loses. We'll see why

We assume that, both the players play in the best way. So each player will choose the move that doesn't break the board.
Always there is such move, if there are unvisited points (just join some visited with some unvisited)
But when all the points are visited, then every move you do will break the board. Indeed:

We'll say that two points $ A_{1},A_{k}$ are [u]connected[/u] if there is a path $ A_{1}A_{2}A_{3}...A_{k}$ such that $ A_{i},A_{i+1}$ are adjacent $ \forall i$ and there is a cut along the segment $ A_{i}A_{i+1}$

If some point is visited, this means that there is a path from this point to the bounds of the square
So, every two visited points are connected
(accurately, the set of the visited points is connected)

So, when we join two adjacent points we create a closed path, hence we broke the board.
[b]So, a state with 0 unvisited points is definitely an L-state[/b]

As we can see at the last diagram, it is not important which exactly are the cuts, but [u]how many[/u] are the unvisited points

Let's see the case $ n=5$
The state below has $ 0$ unvisited points, so it's an [b]L[/b]-state
[img]9486[/img]

The states with $ 1$ unvisited point are the states below.
We can visit the unvisited point and go to the state of the previous diagram, so they are all [b]W[/b]-states
[img]9487[/img][img]9488[/img][img]9489[/img]

If we continue this procedure, we'll see that:
- every state with $ 2m$ unvisited points is an [b]L[/b]-state
- every state with $ 2m+1$ unvisited points is a [b]W[/b]-state
$ \hline$

[size=150]Conclusion[/size]

At the beginning there are $ n\times n$ dots, so there are $ (n-2)^{2}$ total unvisited dots

- If $ n$ is odd then the 1st player wins
- If $ n$ is even then the 2nd player wins

For $n$ even, player 1 wins.
For $n$ odd, player 2 wins.

We view this as a graph with cells as vertices. Two vertices are connected with an edge if their cells are adjacent (that is, they share an edge). Then each move consists in deleting an edge from this graph, and the player who disconnects the graph loses.

Note that there are $2n(n-1)$ edges at first, and the minimum number of edges for the graph to be connected is $n^2-1$, which is when the graph is a tree and has no cycles.

So the maximum number of moves that preserves connectedness is $2n(n-1)-n^2+1=(n-1)^2$. Now we show this is always attainable. The condition that cuts are made from the edge of the board or from a previously made cut translates to the following:

Place the graph in the plane. Suppose there is an ant with a knife wandering on the outside. It may not pass through edges, but may cut any edge that it can access.

Now suppose that for every edge the ant can see, removing it disconnects the graph, but there is an edge the ant cannot see such that removing it does not disconnect the graph. If the ant cannot reach this edge, it must be enclosed in some region. But the boundary of this region is a cycle! So the player can remove one of the edges from this cycle.

It follows that the player who makes the $(n-1)^2+1$th move loses, so if $n$ is odd, player 2 wins, while if $n$ is even, player 1 wins.


[quote="arkhammedos"]if $x+y+z=1$ and $x,y,z\gtr0$ prove that
$\frac{xy}{\sqrt{xy+yz}}+\frac{yz}{\sqrt{yz+zx}}+\frac{zx}{\sqrt{zx+xy}}\leq\frac{\sqrt{2}}{2}$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88439


linear algebra
matrix
calculus
integration
vector
derivative
quadratics

How do I compute $ \int_{-\infty}^\infty \frac{|A|^{-1/2}}{(2\pi)^{n/2}}e^{-\vec{x}^\top A^{-1}\vec{x}/2}\left(-\vec{x}^\top B^{-1}\vec{x}/2\right) \mathrm{d}^{n}\vec{x}$, where $ \vec{x}$ is an $ n\times 1$ vector and $ A$ and $ B$ are $ n\times n$ matrices?

I don't understand how to take the derivative of the dot product $ \vec{x}^\top A^{-1}\vec{x}$.

I'll assume that the integral is meant to be taken over $ \mathbb{R}^{n}$ and not just $ (-\infty,\infty).$ I'll also assume that $ A$ is symmetric and positive definite - without those assumptions, it gets quite awkward. In particular, if $ A$ is not positive definite, then the integral diverges.

Make the substitution $ \vec{x}=A^{1/2}\vec{u}.$ The Jacobian gives us that $ d\vec{x}=|A|^{1/2}d\vec{u}.$

Make that change of variables and the integral becomes

${ \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}e^{-|\vec{u}|^{2}/2}(-\vec{u}^{T}A^{1/2}B^{-1}A^{1/2}\vec{u}/2})\,d\vec{u}.$

At this point, we separate out all of the terms of the quadratic form. We note that all of the off-diagonal terms result in an integral of zero. As for the terms on the main diagonal, we note that

$ \frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^{n}}u_{j}^{2}e^{-|\vec{u}|^{2}/2}\,d\vec{u}=1.$

This gives us that the integral equals

$ -\frac12\,\text{trace}(A^{1/2}B^{-1}A^{1/2})=-\frac12\,\text{trace}(AB^{-1}).$

Clearly this integral arose in a probability or statistics context. Can you tell us any more about that context?

This problem is very similar to problem 19.4 in Prasolov's problem book in Linear Algebra.  It's wording is as follows:

[quote]Let A be a positive definite matrix.  Prove that:

$ \int_{-\infty}^{\infty}e^{-x^{T}Ax}dx = (\sqrt \pi)^{n}|A|^{-\frac{1}{2}}$[/quote]