The answer sheet says 93 points is the correct solution, but I do not get it.  The largest value for a unique mode of frequency 3 I can get is 84.

70 70 72 84 84     84 82

Can anyone help me with this?
Thanks.

The problem statement is wrong. This is the correct problem:

[quote]Reid took seven tests. On the first five tests that he took, he averaged 86 points. On the last three tests, he averaged 95 points. If he averaged 88 points on all seven tests, how many points did he average on the last two tests?[/quote]

And the answer to that is indeed 93.

Thanks,

I just found the search button too.  I am more used to the standard search entry box.

It is funny to think that all my answers are right (only some of the questions are wrong).  :)

Okay so can someone explain the actual corrected question that was posted anyway?

Thanks


topology
inequalities
real analysis
real analysis unsolved

Suppose $0 < r < 1 , \; y_{0}= 1/2$ and \[y_{n+1}= y_{n}^{2}+r^{ 1/2^{n}}-1\] Prove that there exists a negative term $y_{n}<0$ for some $n > 0$.

For example if $r = 1-10^{-60}$ then $y_{8}$ is negative.

source: Peter Walters, An introduction to ergodic theory ( For those who has read the book, this problem is related to the disk homeomorphism example where $\Omega_{2}\neq \Omega$ is claimed)

EDIT: the sequence is clearly decreasing for positive terms. But it may not be decreasing over all terms.

These days my questions are being all self-answered. I am interested in other's solution though.
Suppose every term is non-negative. \[y_{n+1}\leq y_{n}^{2}\]  \[y_{n}\leq y_{0}^{2^{n}}\] Let $a_{n}= 1/2^{n}, \; \delta = 1-r$ then \[r^{a_{n}}= (1-\delta)^{a_{n}}\leq 1-\delta a_{n}\]  \[y_{n+1}\leq y_{0}^{2^{n}}-\delta a_{n}\] Look at the right hand of the above inequality, the first term is double-exponentially approaching 0+ while the second term is exponentially approaching 0+, that means the right hand side is negative for some n.


What is the pattern for $ x^n\minus{}y^n$?  Thanks in advance!

Do you mean $ x^n\minus{}y^n\equal{}(x\minus{}y)(x^{n\minus{}1}\plus{}x^{n\minus{}2}y\plus{}...\plus{}xy^{n\minus{}2}\plus{}y^{n\minus{}1})?$

Pattern ?!
didn´t you mean the factored form ?  :huh:


let $ f$:  $ \mathbb{N}.\mathbb{N} \to \mathbb{N}$
            $ (x,y) \to y\plus{} \frac{(x\plus{}y)(x\plus{}y\plus{}1)}{2}$

prove that f is a bijection and find $ f^{\minus{}1}$

$ f$ is only a bijection if $ 0\in\mathbb{N}$.

[hide="Then"]Obviously $ \binom{x\plus{}y\plus{}1}{2} \le f(x,y) < \binom{x\plus{}y\plus{}2}{2}$. This implies $ f(x_1,y_1)\neq f(x_2,y_2)$ for $ x_1\plus{}y_1\neq x_2\plus{}y_2$. Now consider $ x_1\plus{}y_1\equal{}x_2\plus{}y_2$ with $ y_1\neq y_2$; then $ f(x_1,y_1)\minus{}f(x_2,y_2) \equal{} y_1\minus{}y_2\neq 0$, so $ f$ is injective.

Let $ n\in\mathbb{N}$ be given. Then there is a number $ k$ with $ \binom{k}{2}\le n < \binom{k\plus{}1}{2}$. Set $ y\equal{}n\minus{}\binom{k}{2}\implies 0\le y < k$. Then $ x\equal{}k\minus{}1\minus{}y\ge 0$ and $ f(x,y) \equal{} y\plus{}\frac{(k\minus{}1)k}{2} \equal{} n\minus{}\binom{k}{2}\plus{}\binom{k}{2} \equal{} n$, so $ f$ is surjective.

The second paragraph describes how to construct the inverse - one could of course compute $ k$ in terms of $ n$ and then get an explixit form.[/hide]

http://www.mathlinks.ro/viewtopic.php?t=223472
http://www.mathlinks.ro/viewtopic.php?t=222124

Coincidence: It's the example I presented to my Real Analysis students a week ago  :lol:

The problem seems to be in books around the world  :lol:


I need a little help with this proof (I think I am missing something really simple):

Let $x,y\in \lbrack 0,1]:$

How would I prove that the elements of

$%
\begin{bmatrix}
x & 1-x \\
1-y & y%
\end{bmatrix}%
^{2}$

are all an element of $[0,1]$

I've thought about doing a case by case approach, but that seems too long.

[quote="towersfreak2006"]One of $\{x, 1-x\}$ is zero, and one of $\{y,1-y\}$ is zero. Thus, the operation $(x\qquad 1-x)\begin{bmatrix}
x & 1-x \\
1-y & y%
\end{bmatrix}$ returns either the first row or the second row, as does the operation $(1-y\qquad y)$. Thus, $%
\begin{bmatrix}
x & 1-x \\
1-y & y%
\end{bmatrix}%
^{2}$ only has entries of $0$ or $1$.[/quote]

I'm sorry if I confused you. The notation [0,1] means what blahblahblah has suggested.
Thank you for your help :)

Luckily, the casework is painless.  Each entry is linear in either $x$ or $y$, so it suffices to check the results over all combinations of $x\in\{0,1\}$ and $y\in\{0,1\}$.  This implies the result quite easily, as we have $e_i\cdot e_j\leq 1$, $i,j\in\{1,2\}$, where $e_i,e_j$ are the standard basis vectors for $\mathbb{R}^2$, which is easily seen to be equivalent to the desired result.

Thanks for all your help!
I've got it done using the cases :P

If you think you have a more eloquent solution, please let me know.


A positive integer $ K$ is given. Define the sequence $ (a_n)$ by $ a_1\equal{}1$ and $ a_n$ is the $ n$-th natural number greater than $ a_{n\minus{}1}$ which is congruent to $ n$ modulo $ K$.
$ (a)$ Find an explicit formula for $ a_n$.
$ (b)$ What is the result if $ K\equal{}2?$


[quote="specialme"]A publicação de anúncios no redeparede.com é inteiramente grátis e cada anúncio permanece no site de classificados por até 90 dias. Neste período, o anúncio pode ser encontrado através de buscas realizadas nos classificados ou por meio das consultas às diversas categorias de cada cidade disponível no RedeParede. Este último, inclusive, é um grande diferencial do redeparede.com[/quote]

Can someone translate to English plz?

Source: [url=http://www.google.com/language_tools?hl=en]Google Language Tools[/url]

Portuguese to English:

The publication of announcements in redeparede.com is entirely gratis and each announcement remains in the site of classified by up to 90 days. In this period, the announcement can be found through searches carried through in the classified ones or by means of the consultations to the diverse categories of each available city in the RedeParede. This last one, also, is a great differential of redeparede.com

Agreed. Some mod should lock/delete this topic.


function
inequalities proposed
inequalities

[quote="kunny"]Let $a,\ b,\ c,\ d$ be real numbers. Find the minimum value of $k$ satisying $\sqrt{2-a^{2}}+\sqrt{2-b^{2}}+\sqrt{2-c^{2}}+\sqrt{2-d^{2}}+|a|+|b|+|c|+|d|\leq k$.[/quote]
kunny, $\sqrt{2-a^{2}}+|a|\leq2.$ :wink:

Oh, I see, if you write $|a|=\sqrt{a^{2}}$ it follows nicely :P


There are 49 persons in a competition(P1,P2,...,P48 and P49).P1 is better than P2,P2 is better than P3,...and P48 is better than P49,they compete on seven by seven,what is the chance of P7
 be one of the seven finalists ?Sorry, my english is awful. :(

the  question  is unclear to me.  what are  P1,P2,P3,P4,....... ?


function
Asymptote
complex numbers
complex analysis
complex analysis unsolved

Find the image of the following region:

[img]http://i452.photobucket.com/albums/qq244/mingcai6172/2-1.png[/img]

under the action of
$ f(z)\equal{}z^5$
and
$ f(z)\equal{}e^{\bar{z}}$
(e raised to z conjugate)

I am lost on this one too.  Thanks for any advice with this one.

$ e^{\bar z}$ is nice and easy. Just note that the real part of $ z$ is responsible for $ |e^{\bar z}|$ and the imaginary part of $ z$ is responsible for the argument, so you get a sector of an annulus.

As to $ z^5$, the person who assigns such exercises was just born too late: he would be perfectly fit for torturing the heretics in the Spanish inquisition chambers back in the 13th century :P. Just for the fun of it, here is the asymptote picture for the image of the boundary with $ a \equal{} 0, b \equal{} 1, c \equal{} d \equal{} 2$:
[asy]size(200);
path P(pair A,pair B)
{
path Q=A^5;
for(int k=1;k<101;++k) Q=Q--(A+0.01*k*(B-A))^5;
return Q;
}
draw(P((1,-2),(1,2)));
draw(P((1,2),(0,2)));
draw(P((0,2),(0,-2)));
draw(P((1,-2),(0,-2)));[/asy]


[quote="Jan"]Given a triplet of numbers, $(a,b,c)$.

Now, every move, you can choose two of them, say $a$ and $b$, and replace them by $\frac{3a+4b}{5}$ and $\frac{3b-4a}{5}$.

Can you turn the triplet $(3,4,12)$ into $(5,7,10)$ after a finite number of moves?[/quote][hide]Note that the transformation preserves the distance from the origin: \[a^{2}+b^{2}+c^{2}=\left(\frac{3a+4b}{5}\right)^{2}+\left(\frac{3b-4a}{5}\right)^{2}+c^{2}.\] However, $3^{2}+4^{2}+12^{2}\neq 5^{2}+7^{2}+10^{2}$, so it is not possible.[/hide]


calculus
integration
limit
inequalities
geometry
rectangle
real analysis

Just need to be told if this is right or not:

Proof of the fundamental theorem of calculus (pt.1)

Let $ g(x) \equal{} \int_{a}^{x} f(t) \; dt$. Thus:

$ g'(x) \equal{} \lim_{\triangle x \rightarrow 0} \frac{\int_{a}^{x\plus{}\triangle x} f(t) \; dt \minus{} \int_{a}^{x} f(t) \; dt}{\triangle x}$

Using properties of integrals we derived from Riemann sum manipulations, we then have:

$ g'(x) \equal{} \lim_{\triangle x \rightarrow 0}  \frac{\int_{x}^{x\plus{} \triangle x} f(t) \; dt}{\triangle x}$

Now by extreme value theorem, we have $ [u,v]$ contained in $ [x,x\plus{}\triangle x]$ (both intervals of which are contained in the interval $ [a,b]$ ) such that

$ \triangle x f(u) \leq \int_{x}^{x\plus{} \triangle x} f(t) \; dt \leq \triangle x f(v)$

(the inequality is attained from rectangle widths in a Riemann sum). Dividing by non-zero $ \triangle x$ gives us:

$ f(u) \leq \frac{\int_{x}^{x \plus{} \triangle x} f(t) \; dt}{\triangle x} \leq f(v)$

Taking the limit as $ \triangle x \rightarrow 0$ gives:

$ \lim_{\triangle x \rightarrow 0} f(u) \leq \lim_{\triangle x \rightarrow 0} \frac{\int_{x}^{x\plus{} \triangle x} f(t) \; dt}{\triangle x} \leq \lim_{\triangle x \rightarrow 0} f(v)$

Since the interval $ [x, x\plus{} \triangle x]$ simply becomes $ x$ when computing the limit, we have:

$ f(x) \leq \lim_{\triangle x \rightarrow 0} \frac{\int_{x}^{x\plus{} \triangle x} f(t) \; dt}{\triangle x} \leq f(x)$

By squish theorem, this means $ \lim_{\triangle x \rightarrow 0} \frac{\int_{x}^{x\plus{} \triangle x} f(t) \; dt}{\triangle x} \equal{} f(x)$ but the L.H.S. is equivalent to $ g'(x)$ thus;

$ g'(x) \equal{} f(x)$ which can be re-written as:

$ \left(\int_{a}^{x} f(t) \; dt \right)' \equal{} f(x)$

$ \mathbb{Q.E.D}$

I'm a little concerned as to where mean value theorem comes in here, as well as some of the subtleties about the intervals and such.

[quote="Prime factorization"]
Taking the limit as $ \triangle x \rightarrow 0$ gives:

$ \lim_{\triangle x \rightarrow 0} f(u) \leq \lim_{\triangle x \rightarrow 0} \frac {\int_{x}^{x \plus{} \triangle x} f(t) \; dt}{\triangle x} \leq \lim_{\triangle x \rightarrow 0} f(v)$

Since the interval $ [x, x \plus{} \triangle x]$ simply becomes $ x$ when computing the limit, we have:

$ f(x) \leq \lim_{\triangle x \rightarrow 0} \frac {\int_{x}^{x \plus{} \triangle x} f(t) \; dt}{\triangle x} \leq f(x)$[/quote]

This part is a little bit non-rigorous (although it's the same proof that I give in my class). To be fully rigorous, more should be said about why $ \lim_{\triangle x \rightarrow 0} f(u) \equal{} f(x)$. The assertion is not completely trivial as u depends in an unspecified fashion on $ \triangle x$.


For the sake of completeness, here are the results of Toronto students:

Students of groups I, II, III and IV  are awarded with Certificates  of Tournament of Towns Winners.

In addition, the students of  group I are awarded with $50. checks and the
students of  group II and  group  III  are awarded with books.

[b][u]Winners of Tournament of the Towns, season 2003/2004[/u][/b]

[b]GROUP I[/b]

Jacob Tsimerman 10 University of Toronto Schools
Janos Kramar 11 University of Toronto Schools
Dziabenko Dimitry 8 Don Mills Middle Scool
Yufei Zhao 10 Don Mills Collegiate Institute
Oleg Ivrii 11 Don Mills Collegiate Institute
Wenxin Xu 11 Don Mills Collegiate Institute
Richard Peng 10 Vaughan Road Academy
Yi Song 9 St. Andrews
Amirali Modir Shanechi 12 Don Mills Collegiate Institute
Chen Shen 12 A.Y. Jackson Secondary School

[b]GROUP II[/b]

Yueyang Guan 8 Jack Miner Public School
Leo Anghelescu 9 Victoria Park SS
Chao Qu 11 Dr. Norman Bethune C.I.
Viktoriya Krakovna      10 Vaughan Road Academy
Yichun Li 12 Victoria Park Secondary School
Peipei Liu 12 George. S Henry AC
Yehua Wei 12 York Mills Collegiate Institute
Jimmy Ye 11 Sir John A. MacDonald C.I.
Ryan Zhou 10 Adam Scott CVI
Howard Park 11 Holy Trinity School
Taylor Cao Tian Shi 8 Henry Kelsey Senior Public School
Jaeseung Kim 10 Westmount Collegiate Institute
Yue Jiang 10 A.Y. Jackson Secondary School
Alex Wice 10 Leaside H.S.
David Ruggiero Michael 12 Michael Power-St. Joseph High School
Tom Yue 12 A.Y. Jackson Secondary School
Ti Yin 12 William Lyon Mackenzie Collegiate Instit
Jeffrey Baer David 9 Academy for Gifted Children

[b]GROUP III[/b]

NanQian Lin 12 Albert Campbell Collegiate Institute
John Lee Chun Ming 12 The Woodlands School
Zaishi Shen Kerry 11 A. Y. Jackson S. S.
Arash Joushaghani 12 Don Mills  C.I.
Yi Sun 12 Earl Haig Secondary School
Yun Hui Lin 12 Harbord Collegiate Institute
Shafqat Ahmed 7 Churchill Heights P.S.
Arthur Ching 10 Don Mills Collegiate Institue
Jenny Wang 11 Don Mills Collegiate Institute
Kathy Au 11 Earl Haig Secondary School
Akanksha Madan 9 University of Toronto Schools
Michael Leung 12 Earl Haig SS
Vinh Tran Tuan 12 Cambridge International College of Canad
Jeffrey Chen 10 Marc Garneau Collegiate Institute
Beckett Zhou Yu Xiang 11 Richmond Hill High School
Jay Jin 11 Rick Hansen Secondary School
Alexander Ip Chi Fai 11 Agincourt C.I.
Gregory Chambers 11 Western Technical-Commercial School
Michael Dai Hui Yu 10 Earl Haig Secondary School
Vishvajit Singh 12 Markville Secondary School
Cat Le Chinh 12 Western Technical Commercial
Ling Cen 12 Earl Haig S.S.
Patrick Kaifosh William 10 University of Toronto Schools
Ashkan Eshaghbeigi 12 Earl Haig Secondary School
Ke Lu 10 Northern Secondary School
Quentin Shao Hang 10 Marc Garneau
Theodore Lui 10 Upper Canada College
Alana Zamrij Krystyna   9 Humberside Collegiate Institute
Zachary Rong 9 Zion Heights JHS
Martin Ho 9 University of Toronto Schools

[b]GROUP IV[/b]

Natalie Chan Hoi Man 11 Dr. Norman Bethune C.I.
Steven Wu 11 A.Y.Jackson
Sandy Liu 11 MartinGrove College Institute
Andy Wu Chien-Wei 11 Thornlea Secondary School
Johnny Zhang 11 Albert Campbell
Jimmy Deng 11 Ernie
Lucy Lu 11 Marc Garneau Collegiate Institute
Patrick Ng 11 Woburn CI
Zhuo Wang 11 Earl Haig Secondary
Jonathan Chan James 10 Upper Canada College
Eunse Chang 10 Don Mills Collegiate Insititute
Ivy Zou 12 Earl Haig Secondary School
Stefan Kissiov Miroslavov 12 Marc Garneau Collegiate Institute
Jaehun Kim 12 Bayview Secondary School
Mark Zhou Wanghao 10 Erindale Secondary School
Yi Anthony An 12 Georges Henry AC
Michael Huang 11 A.Y. Jackson S.S.
Kai Yang 11 Dr. Norman Bethune C.L.
Rui Chen 11 Albert Campbell C.I.
Ying Ting Angela Cheng 11 York Mills Collegiate Institute
Dejan Graovac 11 Lawrence Park Collegiate Institute
Joshua Wong 11 St. Francis Xavier Secondary School
Thao Phan Phuong 11 Columbia International College
Bo Deng Hong 10 Javis Collegiate Institue
Zhi Bin Zhou 12 Earl Haig S.S.
Adam Chen Yi 12 Jarvis CI
Benson Lung Chek-Long 8 P.A.C.E. The Academy for Gifted Children
Geng Dai 12 Parkdale Collegiate Institute
Daniel Lam Hon-Yiu 12 Albert Campbell C. I.
Heymian Wong 10 Woburn C.I.
Helen He Yuqiong 10 North Toronto C.I
Nisanthini Ravichandiran10 Woburn C.I.

[quote="Osiris"]So who's going to be at the ceremony? Yufei, I hope you're going.[/quote]

Probably.

Weird thing, the Junior Math Battle runs at the same time as the ceremony. How do they expect us to be at two places at once?

[quote="Osiris"]Argh, I only got 11 for Senior A... I wonder how they marked it... I want to have a look at my paper.[/quote]

I certainly would want to have a look at [i]my[/i] paper.  I must have written two solutions for nothing...

Come on, my school name is constantly spelled wrong.  I requested change several times already.

Did you write it wrong because Adam Chen got his school name written correctly? Heh

Adam Chen?  He didn't even know how to spell "Collegiate Institute". 8-)  At least I wasn't lazy, but I accidentally typed everything wrong.  I don't even know how I can be as careless as that.  I guess my English has improved a lot over the last year. :mrgreen:

[quote="kool_dudy"]So does anyone know when next year's tournament of towns will take place?

I only knew about this contest this year and it was too late for me to apply.[/quote]

The contest is held four times a year, two in Oct. and two in Feb.

Check the [url=http://www.math.toronto.edu/oz/turgor/index.php]website[/url] for details.


vector
real analysis
real analysis theorems

Hello all,

For the past several month I've been looking for sources on using the Navier-Stokes equations... unfourtunatly I've found next to none.

The Navier-Stokes are a set of differential equations that describe fluid flow and movement... they are very often used in aerodynaimics engeneering.

I would like to know, has anyone here had experience using them? Could I speak to you?

Or, if not, could someone please explain to me how to use the "upside down triangle vector"?

Thanks a lot all,

Hello. [url=http://en.wikipedia.org/wiki/Navier-Stokes_equations]This Wikipedia article[/url] ha detailed information about the equations. The symbol you are talking about is the [url=http://en.wikipedia.org/wiki/Del]del[/url] differential operator. I hope this helps.

The Mathworld articles [url=http://mathworld.wolfram.com/Navier-StokesEquations.html]Navier Stokes equations[/url] and [url=http://mathworld.wolfram.com/Del.html]del[/url] might also be useful to you; mathworld is sometimes considered more reliable, but these particular articles contain less information. [url=http://scienceworld.wolfram.com/physics/Navier-StokesEquations.html]Scienceworld[/url] also has a detailed article.


modular arithmetic
induction
number theory proposed
number theory

Given positive integers $a,m$, let $a_1=a,\ a_{n+1}=a^{a_n}$. Show that the sequence $(a_n)$ eventually becomes constant $\pmod m$. This solves [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15089]another problem[/url].

Suppose that there is a counterexample.  Then there is a counterexample for which $ m$ is minimal.  Trivially, $ m\neq 1$.

Note that the sequence $ \{ a^j\}_{j\equal{}0}^{\infty}$ eventually becomes periodic mod $ m$; let the length of this period be $ k$.  Evidently, $ 1 \le k \le m\minus{}1$.  Since
\[ \{a_j\}_{j\equal{}2}^{\infty} \equal{} \{ a^{a_j}\}_{j\equal{}1}^{\infty} \]
does not become constant mod $ m$, it follows that $ \{a_j \}_{j\equal{}1}^{\infty}$ does not become constant mod $ k$.  Thus $ k$ is a smaller counterexample than $ m$.  Contradiction.  $ \blacksquare$

A less clever solution:

Use induction on m. If $gcd(a,m)=1$, then by the induction it holds for $\phi(n)$, and we find that for sufficiently large $i$ we have $a_{i+1}=a^{a_i} \equiv a^c \pmod{m}$.

If $gcd(a,m)>1$, then write $m=k_1 \cdot k_2$, where no prime divisors of $m$ are contained in $k_2$, and $k_1$ is composed of only prime divisors of $m$. Clearly, for sufficiently large $i$, $a_i$ eventually becomes constant mod $k_1$. Similarly, it eventually becomes constant mod $k_2$ by the induction. Therefore, since $gcd(k_1, k_2)=1$, by CRT $a_i$ eventually becomes constant mod $m$, and we are done.


(Me) If rhombus is inscribed in a circle, which of the following is NOT true?

A) It is a square
B) Its diagonals are same
C) Its area is less than the circumference of circle is the radius of the circle is less than $ \pi$
D) The product of diagonals of rhombus is always less than the area of the circle
E) The ratio of the side of rhombus to the length of radius of circle is $ \sqrt {2}$.

Could you clarify point c?  I'm not quite sure what it says.

[hide]If a quadrilateral is inscribed, it is cyclic and its opposite angles must add to 180 degrees.  Since a rhombus has congruent opposite angles, it follows that they are 90 degrees, and since rhombuses have congruent sides, the quadrilateral is a square.  Therefore the diagonals are the same.  For conditions D and E, let us consider an inscribed unit square.  The radius of the circle is $ \sqrt {2}/2$ and the area of the circle is $ \pi/2$.  Since the product of the diagonals is 2, D is false.  The ratio of the side length of the rhombus to the radius of the circle is $ 1: \sqrt{2}/2\equal{}\sqrt{2}: 1$.  This suggests that E is true and the answer is D.[/hide]

Thanks. I forgot to type the word "circumference."

An interesting point about why choice D is wrong.

[hide="Click"]
Note that the area of inscribed diagonal is ALWAYS less than the area of circle. So, if you just scanned through the choices and thought that "product of diagonals" (without dividing by 2!) was the area of the rhombus, you could've thought the choice D was right.

An important lesson on how important it is to read choices CAREFULLY. [/hide]


quadratics
complex numbers
algebra proposed
algebra

[quote="nrs.sowrabh"]Solve $(x^{2}+3x+4)^{2}+3(x^{2}+3x+4)+4=x$[/quote]

A lot easier perhaps if that was a -4.  Still...  sub y for the quadratic in the brackets, and factor in terms of y, equating to the right side.  Messy in complex numbers, but possible.  You're sure that wasn't a -4 ??

Ya sure it wasn't $-4$.My $sol_{n}$ is similar to that of N T TUAN's.


Let $ P$ be a polyhedron whose all edges are congruent and tangent to a sphere. Suppose that one of the facesof $ P$ has an odd number of sides. Prove that all vertices of $ P$ lie on a single sphere.

[hide="Hint"]
Let $ O$ be the center of the sphere to which the edges are tangent, and let $ R$ be its radius. Let $ D$ be the side length of an isosceles triangle whose base is the length of an edge of $ P$ and whose height is $ R$.

for each vertex $ A$ of $ P$, color it black, white or red according to wheter $ AO$ is less than, greater than or equal to $ D$. Prove that a neighbor of a red vertex must be red, a neighbor of a black vertex is white, and a vertex of a white vertex is black. This proves that if there's one red vertex, they must all be red, so the sphere with center $ O$ and radius $ D$ passes through all the vertices. If there is no red vertex, they are all black/white, which is impossible since there's an odd cycle.
[/hide]


I'd say like Mad Dash for XBOX or somethin' :D

[quote="G-UNIT"]undisputed, [b] Counterstrike: Source[/b] for PC[/quote]

^^ Yes it's a great game. :P I also like Warcraft III: The Frozen Throne, Starcraft, Diablo II: Lord of Destruction. All Blizzard games are good. :)

counterstrike is awesome.. i love halo (xbox) probably Half Life 2 for PC... and uh GTA San Andreas for PS2.. no real favorite for Gcube..

Here's my top 10:

1)Grand Theft Auto San Andreas
2) Final Fantasy X
3) Kingdom Hearts
4) NBA Live 2005
5) Madden 2005
6) Project Gotham Racing 2
7) Tekken Tag Tournament
8) Dark Cloud 2
9) Final Fantasy X-2 (I woulda put this higher but I already got FFX up there)
10) Grand Theft Auto Vice City (I woulda put this higher put already got San Andreas)

forza motorspot (currently level 40)
fable
project gothem racing 2
halo
halo 2 (hmm, maybe i should finish the storyline sometime soon)

Gonna post my top 10. It's a little bit different then other peoples...

1. Ikaruga
2. Rez
3. Gradius V
4. Katamari Damacy
5. Guilty Gear X2
6. DDR!!!!
7. Half Life 2/Counterstrike
8. Alien hominid
9. Metal Slug 4
10. Burnout 3

I liked FF7-FFX2 but didn't like FF9 as much.

[quote="Iversonfan2005"][quote="G-UNIT"]undisputed, [b] Counterstrike: Source[/b] for PC[/quote]

Hey G-Unit (50 Cent, LLoyd Banks, and Young Buck), I like your avatar, but which South Park character is that?[/quote]

 The rock  :)

[quote="juicybooty911"]FFx-2 is awesome, i'm playing it right now[/quote]

Yeah, but FFX was better. It's definitely the greatest FF so far (I can't wait for XII, though). Counter-Strike is the best multiplayer FPS out there. Kingdom Hearts is probably one of the coolest games every; it's battle system and storyline are both exceptional.

my favorite games are runescape and battlefront :)

[quote="NoSoupForYou"]My top 3:

1. Neverwinter Nights
2. Timesplitters 2
3. Warcraft 3

My interests are different from most people's  :D[/quote]

Mine too. Especially Warcraft 3.

[quote="MithsApprentice"]FFX and Kingdom Hearts hands down. FFXII needs to hurry up and get released in the US.[/quote]

*agrees* I've been waiting for FFXII to come out for a while now.

Yeah, but I don't think that they will be as good as the originals. Ahhh...such classic games....I must've played FFX and Kingdom Hearts for over 100 and some hours...Me and one of my friends got Yuna's sphere grid complete and got Sora to lv. 100.

[quote="sirkuku"]FF sucks (xbox fanboy spirit comes) Knights of the old republic is SO MUCH better than it KOTOR>>>>>>>>>>>>>>FF oh yes xbox>ps2 (starts evil game war) mwuahhaha[/quote]

English please

[quote="dangvh"]lol i loved FFX a lot lol i got almost everything in that game... dodging the lightning was a real pain... and so was catching all those monsters.. overall i have around 175 hours in that game because i got everybody around the spheregrid and i filled up every node on the grid with something =) but in my opinion i like FFVII better... it's a classic[/quote]

You got EVERYONE'S sphere grid full!!! :o  :o  :o  :o  That's amazing...I thought I really accomplished something when I got Yuna's. You must've been able to beat that Giant Slug thing in the Monster's Arena, right? I hate that thing, it's pretty much impossible to beat it.


inequalities
function
real analysis
complex analysis
complex analysis unsolved

If $ f$ is analytic on $ D$ and $ |f(z)|\leq 1$ on $ D$ such that $ f$ fixes two points, then $ f(z)\equal{}z$.

I've noticed the obvious that if $ f(0)\equal{}0$ is one of the fixed points, then I can apply Schwarz's Lemma and am basically done. Is there a reason I can make that assumption? It seems that the standard trick of subtracting the constant term could lead to messing up the modulus... The details that complex analysis texts leave out never seem obvious to me  :(

Compose a suitable disc automorphism with $ f$ and apply Schwarz lemma to it. This is an important and often used technique (see for example my post about an inequality for univalent functions )

Compose your $ f$ with a Mobius transformation of the form $ \psi_a: z \mapsto \frac {\alpha \minus{} z}{1 \minus{} \bar{\alpha}z}$ with $ |\alpha| < 1$. Prove that this is an automorphism of $ D$. If one of your fixed points is $ \alpha$ then define $ \kappa \equal{} \psi_{ \minus{} \alpha} \circ f \circ \psi_\alpha$ and check that: i) $ \kappa(0) \equal{} 0$, ii) $ \kappa$ is analytic on $ D$, and iii) $ |\kappa(z)| \leq 1$ on $ D$.

I forgot what Schwarz's Lemma was exactly about. But if your suggestion is correct, and we can now conclude that $ \kappa$ is constant, then it follows easily that $ f$ is constant too.

Ah. That makes a lot more sense. To finish it out from there, the Schwarz Lemma says that $ |\kappa(z)|\leq |z|$ for $ z$ on $ D$ and if equality occurs somewhere in $ D$, then $ \kappa(z)\equal{}\lambda z$ for some $ | \lambda|\equal{}1$, but the other fixed point will give that $ \lambda\equal{}1$. So $ \kappa(z)\equal{}z$ and now just transform back to get $ f(z)\equal{}z$.

I wish tricks like this were more explicit in texts. Does anyone know of a good complex text that is thorough on the details (but is a grad text that covers approximately the same as the second half of Rudin)?

Ruding actually has an explicit discussion of exactly this under the Schwarz lemma (pp 254-255 in my book)


The unknown real numbers $x_1,x_2,\dots,x_n$ satisfy $x_1 < x_2 < \cdots < x_n,$ where $n \geq 3$. The numbers $s$, $t$ and $d_1,d_2,\dots,d_{n - 2}$ are given, such that \[ \begin{aligned} s & = \sum\limits_{i = 1}^nx_i, \\ t & = \sum\limits_{i = 1}^nx_i^2,\\ d_i & = x_{i + 2} - x_i,\ \ i = 1,2,\dots,n - 2. \end{aligned}  \] For which $n$ is this information always sufficient to determine $x_1,x_2,\dots,x_n$ uniquely?


function
trigonometry
real analysis
real analysis solved

Find the limit of this sequence:$u_n=  \prod_{i=1}^n(1+\frac1{2^i})$

Proposed by darkmaster long time ago.

First I wiil show that
\[
(1+2^{-1})(1+2^{-2})(1+2^{-3})\cdots = 1+1+\frac {1} {1\cdot 3}+\frac {1} {1\cdot 3\cdot 7}+\frac {1} {1\cdot 3\cdot 7\cdot 15}+\cdots
\]
Put
\[
\begin {array} {ll}
u=(1+tu)(1+t^2u)(1+t^3u)\cdots (1+t^nu) \\
=1+a_1u+a_2u^2+a_3u^3+\cdots +a_nu^n+\cdots
\end {array}
\]
where $ a_1, \quad a_2, \quad \cdots, \quad a_n, \cdots $ are functions of t only and  must be deteminated.  Let $ u \to tu $ , we get
\[
\frac {u} {1+tu} = 1+a_1tu+a_2t^2u^2+a_t^3u^3+\cdots +a_nt^nu^n+\cdots
\]
Hence
\[
1+a_1u+a_2u^2+\cdots=(1+tu)(1+a_1tu+a_2t^2u^2+\cdots )
\]
and so, by comparison of coefficients, we get
\[
\begin {array} {ll}
a_1 =\displaystyle \frac {t} {1-t}, \quad a_2 = \displaystyle \frac {t^3} {(1-t)(1-t^2)}, \\
a_3 =\displaystyle \frac {t^6} {(1-t)(1-t^2)(1-t^3)}, \cdots \\
 \cdots a_n = \displaystyle \frac {t^{\frac {n(n+1)} {2}}} {(1-t)(1-t^2) \cdots (1-t^n)}, \cdots
\end {array}
\]
Also
\[
1+\frac {t} {1-t} z+\frac {t^3} {(1-t)(1-t^2}z^2+ \cdots  \displaystyle \frac {t^{\frac {n(n+1)} {2}}} {(1-t)(1-t^2) \cdots (1-t^n)} z^n+ \cdots
 \[
=(1+tu)(1+t^2u)(1+t^3u)\cdots (1+t^nu)
\]
Putting t = 1/2 and u = 1 we get what we wanted.
Then note
\[
1+1+\frac {1} {1\cdot 3}+\frac {1} {1\cdot 3\cdot 7}+\frac {1} {1\cdot 3\cdot 7\cdot 15}+\cdots < 2+\frac {1} {3} (1+\frac {1} {6} + \frac {1} {6^2} + \cdots) = \frac {12} {5}
\]
Hence the sequence converges.

Nene you have proved that it converge,how can you found his limit??

[quote="harazi"]Silviu, I am afraid you cannot find a closed expression for the limit.[/quote]
Yes harazi
The product
\[
(1+2^{-1})(1+2^{-2})\cdots =\frac {3\cdot 5\cdot 9\cdot 17\cdots} {2\cdot 4\cdot8\cdot 16\cdots}
\]
lies  between 2 and 12/5. We must remember also that the quozient of the product of all even numbers divided  by the product of all odd numbers is $ \sqrt {\frac {\pi} {2}} $.
( I think this fact is named Wallis formula, but I am not sure)

[quote="Nené"][quote="harazi"]Silviu, I am afraid you cannot find a closed expression for the limit.[/quote]
lies  between 2 and 12/5. We must remember also that the quozient of the product of all even numbers divided  by the product of all odd numbers is $ \sqrt {\frac {\pi} {2}} $.
( I think this fact is named Wallis formula, but I am not sure)[/quote]
This comes from $\sin\pi x=\pi x%
%TCIMACRO{\dprod \limits_{i=1}^{\infty}}%
%BeginExpansion
{\displaystyle\prod\limits_{i=1}^{\infty}}
%EndExpansion
\left(  1-\frac{x^{2}}{i^{2}}\right)  $

Please read [url]http://mathworld.wolfram.com/InfiniteProduct.html[/url] eqn38
the result is eqn41/eqn39

Yes, if there is some approach for this problem it should be related to complex analysis.


umm this is first time writing in this forum so i hope i find an answer from this :D even though i dont see how that has to do anythign with me being first timer here

anyways what is 0/0 anyways?
is it 1? 0? or undefined??

many ppl says its 0
soem says its undefined
and this weird guy says its 1 cus x/x = 1 o.O;;

$ \frac {0}{0}$ is undefined because division by zero is undefined.  This is probably one of the most common topics on the forum, and if you did a search you would find dozens of other threads (at least) where people who don't know what they're talking about have debated this issue.

My immediate reaction? Not again...
I do like the official "Not a Number" here. A search for NaN turns up [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=123083]two[/url] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=86053]discussions[/url] which at least have some content.[/youtube]

I would say it is technically 0. This is because, suppose you have 0 people sharing 0 slices of cake in a room. Since there is nothing in the room it would be 0. This is just a theory...

No, it's not a theory, but some useless way of viewing it  :wink:
I recommend locking this, we don't need another thread about this.

I second that motion, but x/0 is still fun to explain; there should be a folder devoted entirely to conspiracy theories about dividing by zero. :D ...

The reason we don't define zero divided by zero is that it could "be" anything: We would want zero divided by zero to be that number y such that zero multiplied by y equals zero, but that means y could be [i]any[/i] number, since any number, when multiplied by zero, equals zero.

The reason we don't define x divided by zero for x not equal to zero is that such an x could not be a number: if there were a number y such that zero times y equaled x, then, because zero times anything is zero, so zero times y equals zero, so x equals zero; but x, we said, couldn't be zero, and so there is no number that can be x/0.

Anyways, everybody knows that:
0/0= an owl with a big nose
x/0= an owl with a big nose and one empty eye socket
1/0= an owl with a big nose and a scar on his eye

[quote="crazychinesemaniac"]suppose you have 0 people sharing 0 slices of cake in a room. Since there is nothing in the room it would be 0. [/quote]

Actually, since there is nothing in the room you can't talk about "the number of slices of cake each person got" because there are no people and no slices of cake to talk about!  From a programmer's point of view, no value has been set for the corresponding variables.

Another example that will hopefully make this point clear:  suppose I tell you that there are $ 0$ horses in a stable, and then I ask you, "what color are the horses in the stable?"

Please just read the threads to which jmerry linked instead.


analytic geometry
graphing lines
slope
geometry
rectangle

Can anyone figure this out?

[img]http://farm1.static.flickr.com/147/371723025_d5769c9b24_o.gif[/img]

The slopes of the green and red triangles don't match up.  The fact that they're very close is a consequence of the fact that all the lengths involved are Fibonacci numbers.

Yep, this is a tricky one.
The slope of the green triangle is $ 2/5$ or $ .4$ while the slope of the red triangle is $ 3/8$ or $ .375$. The slopes being very close creates the illusion that they are the same.

It's true that the slopes are different. Also if you take out the triangles, the areas of the rectangles differ by 1. Making a space of 1 square unit.


This is a natural result from the famous three-variable inequality. Sorry if it was posted.

Let $a,b,c,d$ be positive numbers. Prove that: \[{ab \over c}+{bc \over d}+{cd \over a}+{da \over b}\geq a+b+c+d.\]

[quote="Lovasz"]

Let $a,b,c,d$ be positive numbers. Prove that: \[{ab \over c}+{bc \over d}+{cd \over a}+{da \over b}\geq a+b+c+d.\] [/quote]
It is wrong: $a=23,$ $b=2,$ $c=5$ and $d=1.$ :wink:


Give a triangle ABC with $AB=AC$. $M$ is the midpoint of side $BC$. $N$ is a point on $AC$ such that $MN$ perpendicular to $AC$. $P$ is the midpoint of $MN$. $Q$ is a point on $AC$ such that $BQ \parallel MN$. Let $AP$ intersect $BN$ at $R$. Prove that $\angle MAP= \angle MQR$

[quote="j555"]Nice problem.
Let $X$ be a midpoint of $MC$. We have
$CX\cdot CB=CM^{2}=CN\cdot CA$, so points$B$, $X$, $N$, $A$ lie on a circle. Thus
$\angle MBR=\angle XBN=\angle XAN=\angle MAR$, so points $B$, $M$, $R$, $A$, $Q$ lie on a circle. Thus
$\angle MAP=\angle MAR=\angle MQR$. :)  :lol:  :D[/quote]
Nice solution!!!

The problem is equivalent to prove that $MRQA$ is cyclic, but $BMQA$ is cyclic so we need to prove that $BMRA$ is cyclic, in other words $BR\perp AP$. So let $T$ be the midpoint of $NC$. Triangles $MNC$ and $ANM$ are similar and their corresponding sides are perpendicular so their corresponding medians $MT$ and $AP$ must be perpendicular, finally $MT\parallel BN$ then the result follows.


analytic geometry
algebra
function
domain
topology

Prove that if F: M -> N is an immersion, then it is a local embedding.

So, we need to show that for each point p in M there is a neighborhood U such that F_U: U -> N
is an embedding.

...this should be a two-liner. By the Rank Theorem locally we can have coordinate representation of
F as F: (x_1,..., x_m) -> (x_1,...,x_m, 0,...,0). So U we can take a chart where we have the above coor. representation.

...am I going in the right direction or it's simpler than that?

The key issue is this: local in which space? In the codomain $ N$, it's false, even if we additionally assume $ F$ is injective. In the domain $ M$, it's immediate, and your argument covers it.
We do need to verify that we can't curl back to make it a non-homeomorphism in an arbitrarily small neighborhood, and straightening it out with the rank theorem is a reasonably simple way to do that.

thank you.

When you say "local in codomain $ N$" do you mean that there is a neighborhood $ W$ of $ F(p)$ in $ N$
such that $ W\cap N$ is embedded?

and the counterexample to this would be the figure "8" $ \gamma : (a,b) \minus{}> {\mathbb R}^2$ which is
an injective immersion but not a local embedding in $ {\mathbb R}^2$?

...I think where I am confused is the approach... to prove that $ {F|}_U$ is an embedding
I am trying to show that $ {F|}_U$ is an injective immersion which is also a topological embedding.
Using the coordinate representaion I see that $ F$ is 1-1 on $ U$..but I am not sure why it is a homeomorphism onto it's image.

...or should I, instead, show that there is a slice chart for $ U$ in $ N$?

An injective topological embedding is a homeomorphism onto its image and vice versa. In this case- what's the induced topology on a hyperplane slice of $ \mathbb{R}^n$?

Yes, the figure 8 is one counterexample to an injective immersion being an embedding locally in the codomain. The map $ f$ from $ \mathbb{R}$ to $ S^1\times S^1$ given by $ f(x)\equal{}(x,\sqrt{2}\cdot x)$ is another.

oops..clear regarding the homeomorphism.

..by $ f(x) \equal{} (x, \sqrt{2}\cdot x)$ I guess you mean $ f(x) \equal{} (e^{2\pi i x}, e^{2\pi i \sqrt{2} x})$,
 and it is not an embedding (local in $ T^2$) because the image is dense in $ T^2$.

thanks!


A number is randomly selected from the set of positive integers less than 100. What is the probability that it is an even number? Express your answer as a common fraction.

From the numbers $ 1\rightarrow99$ there are $ 99$ positive integers and $ 49$ even positive integers so the probability is $ \boxed{\frac {49}{99}}$.


Harvard college MIT Putnam Princeton Duke Stanford

Since MIT is a science and engineering-oriented school, it is clear that MIT has, and probably always will have, more "smart math kids" (e.g Top 500 Putnam) than Harvard.  And since Harvard is Harvard, it probably is not even interested in having so many mathy students in its student body.

However, there is a real competition between the two schools when it comes to *top* math students (e.g. Harvard's top 10 scorers versus MIT's top 10 scorers).  What is apparent from 2006's results is that MIT dominated Harvard (as well as everyone else).  Important disclaimer:  I am going to ignore the official rankings, since they are essentially meaningless to this discussion because of the team-choosing issue.  I'm going to stick to "moral rankings" of schools, which is the general impression one gets from looking at the published list of winners.

So was 2006 an aberration, or is it the beginning of a new age of dominance for MIT?

A bit of history:  Harvard won the Putnam "officially" from 1985 to 1992, and Harvard was very likely the "moral winner" in each of those years (not sure because those results aren't easily Google-able).  Harvard continued to be dominant until 1999, when MIT started becoming as good as Harvard.  Finally, in 2004, MIT performed better than Harvard, and now in 2006, it's not even close.

So what the heck happened?  The obvious explanation is that hot-shot math competitors now prefer MIT, whereas they used to prefer Harvard.  But why?  A less likely explanation is that it has to do with whom the schools are admitting.

(In case you're wondering why I've ignored all other schools, it's because no other school has ever had enough sustained Putnam success to claim to be the dominant magnet of math talent in North America.  The irony of my question about whether 2006 marks the rise of MIT is that 2006 is the first time Princeton has ever won the Putnam.)

I don't buy it: from 1994 through 2006, Harvard has been in the top 5 teams 12 years, MIT 10 years, Princeton and Duke 9 years each, and Waterloo and Caltech 5 years each.  If we extend to Honorable Mention teams, Harvard and MIT each have 13 appearances in the top 10 in 13 years, Princeton and Duke have 12 each, Toronto and Waterloo have 11 each and Stanford and CIT have 9 each.  (If we're willing to go a little further down, Chicago and Washington University in St. Louis have 6, and Harvey Mudd has 5, 4 coming since 2001, including a top-5 finish.)

I'm aware that my metric and your metric don't match up.  I chose mine because it was easy for me to read the list of the top 10 teams from each of the past 13 years.  It strikes me as a relatively good proxy for your metric, because it's substantially less sensitive to "picking the wrong team" then the actual final team ranks are (after all, MIT still places in the top 5 despite missing most of its Putnam fellows from the last 2 years).

Harvard and MIT have more people among the top 25 because Harvard and MIT have dozens of individuals take the contest.  Mudd or Caltech will never perform as well "morally" because Mudd and Caltech simply don't have enough students to do so.

Here's a count I made from the 2005 results.

MIT: 3 Fellows, 4 N1, 2 N2, and 14 HM.
Harvard: 2 Fellows, 2 N1, 0 N2, and 4 HM.
Princeton: 1 Fellow, 0 N1, 1 N2, and 5 HM.

Now do that for this year, the 2006 results:

MIT: 3 Fellows, 5 N1, 4 N2, and 13 HM.
Harvard: 1 Fellow, 1 N1, 0 N2, and 13 HM.
Princeton: 2 N1, 1 N2, and 6 HM.

My impression is that yenlee's claim is MIT and Harvard do so well (by his standard) because they are better at attracting large numbers of people who excel at contest math, and that in particular MIT has been attracting more of these people in recent years.

I think I'd agree with him. Part of the reason is that Harvard and MIT seem to have the strongest reputations in Europe (although Princeton isn't too far behind), so the good contest math people from there who want to come to the USA usually apply to those universities. My suspicion, although this is not well grounded, would be that the farther east you go in the rest of the world, the more highly esteemed are Western USA universities like Caltech and Stanford.

It's interesting to note the influence of the international math contest community on this. Math contest type students (particularly international ones) looking to study math in the USA will ask the USA contest math people they know or the older international students who are already studying in the USA. If the people giving advice are overwhelmingly at MIT, where do you think the younger students will apply? This is why so many of the Romanians come to Princeton: a ton of the Romanian IMO veterans are here, so their younger colleagues follow in their footsteps.

So in conclusion, I think that future trends in Putnam performance would in fact be positively correlated with past trends (although the correlation might not be extremely strong).

If we look at the Putnam stats per capita, then Cal Tech looks pretty strong.  My guess is that the probability that a random Cal Tech student made Honorable Mention or above is higher than for Princeton or Harvard, second only to MIT.  (By random student, I mean counting all undergraduates, whether they took the test or not.)

I personally prefer to define the "league" for comparing my school to as "state-supported universities located in California." In that grouping, there's UC Berkeley and there's everyone else. And it wasn't that great a year for Cal: one student in group II, four others in the top 500.

The only other such school with two students in the top 500 was San Jose State, with one Group II and one other top 500. The only other Group II was from UC Santa Barbara (and you'd recognize his name.) With one each in the top 500: UCLA, UC Davis, Cal Poly Pomona, and us: Cal State Long Beach.

I guess I might as well toss in one more thing: certainly, MIT has dominated the Putnam the past couple of years.  However, predicting future Putnam success is very, very risky: the 2010 Putnam results rest on the independent choices of hundreds of students, the oldest of whom are just at this moment deciding where to go to college.

The reason that MIT students do better these days is perhaps due to the college-picking culture that exists among the top mathletes (in the US at least). I believe that the "go where the other mathletes go" is a strong force, mainly because it's fun to go to school and take classes with all those guys who lived far away from you in high school and didn't get much contact other than competitions and mop. Therefore, come college decision time, MIT is a good pick because some years ago, all the mathletes decided on MIT, whereas it was Harvard in the past. So it could just be a matter of whim: in the future, who knows, maybe they'll all decide to love UChicago one year.

I partly agree with JYTang, the Cambridge factor of having the majority of "math Oly people" go to Harvard/MIT definitely helps attract more of the top tier each year.

Overall all I have to say is using team rank is a total junk way to measure the schools.

[quote="yenlee"]I definitely agree that math people tend do go where other math people go.  However, since this causes a positive feedback loop, there should be some explanation for why math people *seemingly* started preferring MIT over Harvard.

A much less likely possible explanation would be that math people still prefer Harvard, but Harvard is rejecting so many of them that MIT now has better top Putnam people.[/quote]

I'm not gonna name names but that's actually very very true with some pretty prominent people that have done well on Putnam for MIT.

I have some problems with this argument:

1. There is much more to assessing the ability of a mathematician than the Putnam.

2. Its not clear to me that looking at only the top 3 (or even 10) students from each school should be a good measure of the quality of mathematics students at that school as a whole. Nor a good way to compare schools.

3. I don't think its safe to assume that Harvard was what you call the "moral winner" in all of those years that they were the official winner -- which makes the premise that there's been some kind of shift in the overall quality of math students from one school to the other unclear.

(btw, Princeton probably would have won the Putnam officially in some previous years if they were more consistent in choosing their top three scorers as their team representatives.)

MIT gets lots of international kids too.. don't forget that. (2 of their fellows this year).


The title says it all. Heres the template

Applied to:
Accepted to:

SAT:
SAT II:
ACT:
GPA:
Rank:
APs taken/scores:
Highest AMC/AIME/USAMO:
Supplementary Material:
Location:
High School Type:
Ethnicity:
Gender:
Extracurriculars:
Awards/hooks:
Comments?:

First, you'll probably be able to find this kind of information at [url=http://talk.collegeconfidential.com/]College Confidential[/url].

Also, many colleges (HYP included) have not yet released their (regular) decisions for the class of 2013.


no offence, but why is this in olympiad section? :?

Yes, you are right....Sorry....I will place it in the right category right now.

[quote="symmetry"]Find c.

|-6 + c| = -8[/quote]

Just wondering... but I thought absolute values couldn't be negative... Were you meaning positive 8? :?

Yes...absolute value is the distance from any number on the real line and zero.  Distance cannot be negative.  So, absolute value is always positive.

[hide]no solution, the positive value of something cannot be equal to -8, because -8 is negative, and the [b]positive[/b] value of something must equal a positive number ;) sorry, this is an impossible problem. If you meant 8 instead of -8, the answer is $\fbox{-2,14}$ :idea: [/hide]


How would I find the fconjugate acid-base pairs based on their equations?

For example:

a. $ HC_{2}H_{3}O_{2} \plus{} H_{2}O  <\equal{}> H_{2}O^{\plus{}} \plus{}C_{2}H_{3}O_{2}^{\minus{}}$
b. $ NH_{4}^{\plus{}} \plus{} OH^{\minus{}} <\equal{}> NH_{3} \plus{} H_{2}O$

Sorry about the poor chemistry formatting.  There should be a [img]http://upload.wikimedia.org/math/2/4/0/240800af7a8fdc7875e40cec17bf7d34.png[/img] instead of a $ <\equal{}>$.

Thanks in advance!

[quote="Carcul"]A conjugate acid-base pair differ by one proton.

a. acetic acid/acetate, hydronium ion/water

b. ammonium ion/ammonia, water/hydroxide[/quote]

How do you know if an acid and base differs by one proton?  Thanks!

Refer back to one of the acid /base definitions

Acids produce H+
Bases Take in H+ or give off OH- (whichever floats ur boat)

so take a look at example A.
We have some acetic ACID disolving in water.... sounds like a give away to me,,, but lets say we didnt know how to name acetic acid.... then u could see that if u assume that the reaction goes to completion (let there be some time that this occurs, even if only for a small second before equilibrium is established) u are left with a solution that has excess H+ (btw is H3O+ not H2O+, and H3O+ = H+). Therefore, something clearly had to give off that proton.... so what produces protons? thats right... acids!!.Therefore, that HC2H3O2 substance MUST be the acid and if it loses the proton, then the resulting compound is C2H3O2-. We call this compound a CONJUGATE base, since, once again, consider that point in time when all acid ---> acetate + H+.... At this time, the only things in solution are the protons and the acetate, so the acetate will tend to follow teh reverse reaction and TAKE IN the H+ to go back to an acidic compound. If it TAKES IN H+... gues what it is.... oh yea... BASE!!

Example II:
ammonium and hydroxide ion react to from ammonia and water, there USED to be OH in solution, but the hydroxide TOOK IN a proton and became h2O... once again... if the OH TAKES IN  and proton... by our definitions stated above... its basic! On the other hand... the ammonium gave OFF a proton above, and this resulted in ammonia. Key word, ammonium GAVE OFF a proton to the hydroxide, and therefore it is the acid. Obviously the NH3 is therefore its conj base, and the H2O is teh conjugate acid of OH-.

Lemme know if u didnt understand somethign I wrote

If I have notation that looks like $ 2H_{2}O$, is it releasing or acquiring hydrogen ion?  How do I know?

Thanks!

[quote="feynman_wannabe"]Im sorry I dont understand your question. What is 2H2O?[/quote]

$ 2H_{2}O$ is some random molecule which has a hydrogen in it. (We don't need to know the exact name for the molecule- or do we?)  How would I know if it would release or acquire a hydrogen ion?

That is not a molecule, but [b]two[/b] molecules of water.

[quote="Elcric"]How would I know if it would release or acquire a hydrogen ion?[/quote]

You can only know that by inspection of the relevant reaction equation: then, it will be obvious wether your molecule is acting as a base (gaining a proton) or as an acid (delivering one proton).


search
real analysis
real analysis unsolved
Brazilian Undergrad MO

Prove that there are real numbers $ a_1, a_2, ..$ such that:

i) For all real numbers x, the serie $ f(x) \equal{} \sum_{n \equal{} 1}^\infty a_nx^n$ converge;
ii) f is a bijection of R to R;
iii) f'(x) >0;
iv) f(Q) = A, where Q is the set of rational numbers and A is the set of algebraic numbers.

I think I am missing something, what about $ f(x) \equal{} x^3$?

Identity.

Maybe to make it difficult you can change into to onto.

I'm trying to solve this problem since saturday, when the test happened, and, until now, i didn't get good results.

Did anybody get something?

bump. I thought about it a bit but had no result.

Sometimes I wonder if anybody reads the posts at least on the first page (I do not even want to mention something as advanced as the search button :roll:). The last time the question was asked [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244318]here[/url].


induction
calculus
calculus computations

given $ x_0\equal{}0$ define: $ x_{n\plus{}1}\equal{}\frac{1}{p\plus{}x_n}$ for some $ p>0$.
Show that the sequence converges and find the limit.

[hide="The limit"]As a first remark, this sequence is not monotonic, since $ x_2 < x_1$ and $ x_3 > x_2$. Anyway, if the sequence converges then its limit L can be found from $ L \equal{} \frac{1}{p \plus{} L}$, from which we get $ L \equal{} \frac{\sqrt{p^2\plus{}4}\minus{}p}{2}$.[/hide]

[quote="Carcul"][hide="The limit"]As a first remark, this sequence is not monotonic, since $ x_2 < x_1$ and $ x_3 > x_2$. Anyway, if the sequence converges then its limit L can be found from $ L \equal{} \frac {1}{p \plus{} L}$, from which we get $ L \equal{} \frac {\sqrt {p^2 \plus{} 4} \minus{} p}{2}$.[/hide][/quote]
It is not a good solution.
You must prove that $ \{x_{2n}\},\{x_{2n\plus{}1}\}$ converge at a same point.

I know its not a complete solution. That's why I wrote "The limit" and not "Solution".

I prove that $ {x_{2n\plus{}1}}$ and $ {x_{2n}}$ are converges.
By using that induction we prove that $ x_1>x_3>...>\frac{1}{2p}$ and $ x_2<x_4<...<\frac 1p$. Suppose that $ x_2<x_4$($ x_1>x_3$) and $ x_{n\minus{}2}<x_n$when $ n$ is even($ x_{n\minus{}2}>x_n$ when $ n$is odd) .We have $ x_{n\plus{}2}\equal{}\frac{1}{p\plus{}\frac{1}{p\plus{}x_n}}$ and $ x_{n\minus{}2}<x_n$($ x_{n\minus{}2}>x_n$) then $ x_n<x_{n\plus{}2}<\frac{1}{p\plus{}\frac{1}{p\plus{}\frac 1p}}\equal{}\frac {p^2\plus{}1}{p^3\plus{}2p}<\frac 1p$($ x_n>x_{n\plus{}2}>\frac {1}{2p}$).
Then suppose that $ x_{2n\plus{}1}\rightarrow A$ and $ x_{2n}\rightarrow B$ and $ B<A$ ,I don't know how we can prove that $ A\equal{}B$.  :?:

It is not very hard. :wink:
$ \{x_{2n}\}\to A$ and $ \{x_{2n \plus{} 1}\}\to B$ then
$ A \equal{} \frac {1}{q \plus{} B}\Rightarrow A(p \plus{} B) \equal{} 1$
$ B \equal{} \frac {1}{q \plus{} A}\Rightarrow B(A \plus{} p) \equal{} 1$
Imply that $ A \equal{} B$
So $ A \equal{} \frac {1}{p \plus{} A}$
$ A \equal{} \frac { \minus{} p \plus{} \sqrt {p^2 \plus{} 4}}{2}$

TTsphn, your q's should be p.

I think you should be able to use the thm that:  $ \{a_n\}$ is divergent if and only if $ \{\frac{1}{a_n}\}$ converges to zero. (here, $ \{\frac{1}{a_n}\}\equal{}p\plus{}a_{n\minus{}1}$ and $ a_n \geq 0$ for all $ n$.)

on the other hand, I was talking to Jmerry regarding this problem and he gave a solution by constructing a contraction map, and showing that if $ f(x)\equal{}\frac{1}{x\plus{}p}$ then $ |f(x)\minus{}f(y)|<c|x\minus{}y|$ by the mean value theorem and using the fixed point of the function.

$ f'(x)\equal{}\frac{\minus{}1}{(x\plus{}p)^2}$
So how do you do to prove that exist $ c<1$ satisfy this $ |f(x)\minus{}f(y)|<c|x\minus{}y|$
Your is not true because
Consider the sequence $ x_{n}\equal{}(\minus{}1)^n$
Although it divergent  but the sequence $ \frac{1}{x_n}$ not tend to 0.


inequalities
vector
inequalities unsolved

Let a,b,c be positive real numbers.Prove that
$\sqrt{a^2+b^2+\frac{(a-b)^2}{12}}+\sqrt{b^2+c^2+\frac{(b-c)^2}{12}}+\sqrt{c^2+a^2+\frac{(c-a)^2}{12}}$
$\ge (a+b+c)\sqrt{2}$

Or I am idiot, or this is direct from the fact that $ 2(a^2+b^2)$ is greater than $(a+b)^2$. In fact, you find an even better inequality than the one you posted (you don't need that thing with $ (a-b)^2/12$-which reminds me of a terrible problem posted not very long time ago).

You are right Harazi
Maybe it is not the origin problem which my friend gave me last year and at that time i can not find the solution.
I'll check again to find it

THIS IS HARAZI THINK! MAYBE :D

USING COSI WE HAVE
$a^2+b^2\geq\(a+b)^2/2$;
after that:

$lhs\geq\sqrt{

using cosi we have:
$2(a^2+b^2)\geq\((a+b)^2$
 and

$lhs\geq\sum\sqrt{(a+b)^2/2+\frac{(a-b)^2}{12}}$

after that using MINSKOWSKY INEQ OR VECTOR IEQ:
$\sum\sqrt{(a+b)^2+\frac{(a-b)^2}{6}}\geq\sqrt{4(a+b+c)^2}$
we got this

maybe HARAZI THINK SO :D
BUT ABOUT THE IEQ YOU MANSION HAVE THIS STYLE $(A-B)^2/12$
 ACTUALLY THIS COMES FROM VMEO ON http://www.DIENDANTOANHOC.NET:

THEY GIVE THEIR RESULT AND THIS MUST PROVE A STRONG INEQ

AND THIS WAY DOESN'T LIKE DARKSEER WAY
 THAT SITE IN VIETNAMESE ,TOO UNCOMFORTABLE :P ,MAYBE SOME VN MATHLINKERS
STRANSLATE IT IN ENGLISH
MAYBE  VUHUNG :D

I think Harazi said that
$LHS\ge \sum \sqrt{a^2+b^2}\ge \sum \sqrt{(a+b)^2/2}=RHS$
[quote="cuong"]
BUT ABOUT THE IEQ YOU MANSION HAVE THIS STYLE $(A-B)^2/12$
 ACTUALLY THIS COMES FROM VMEO ON http://www.DIENDANTOANHOC.NET
[/quote]
I can't find it ;)

This may be the origin problem, i found it in my memmory :D
Let a,b,c be positive real numbers.Prove that
$\sqrt{a^2+b^2+\frac{(a-b)^2}{12}}+\sqrt{b^2+c^2+\frac{(b-c)^2}{12}}+\sqrt{c^2+a^2+\frac{(c-a)^2}{12}}$
$\le \sqrt{6(a^2+b^2+c^2)}$

this is what harazi mansioned:

with x,y,z>0 and x+y+z=1
prove that:

$\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(x-z)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\leq\sqrt{3}$

when you go to this site, see in your lhs and you can see vmeo

about your last ineq:

using their suppose ineq like this:

$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\leq\sqrt{2(a^2+c^2)+(b+d)^2}$

apply this:  $lhs\leq\sqrt{6(a^2+b^2+c^2)}$

apply this ineq
 ineq, harazi mansion can be prove easily

[quote="cuong"]this is what harazi mansioned:

with x,y,z>0 and x+y+z=1
prove that:

$\sqrt{x+\frac{(y-z)^2}{12}}+\sqrt{y+\frac{(x-z)^2}{12}}+\sqrt{z+\frac{(x-y)^2}{12}}\leq\sqrt{3}$

[/quote]
I don't understand what you mean :(

sory
i mean the ineq which harazi mansion is
$\sum\sqrt{x+\frac{(y-z)^2}{12}}\leq\sqrt{3}$
and this is the ineq of VMEO:

to prove it: must prove the ineq which i give

afterthat :

i use this ineq to prove your problem  :D  ;)


What is the area of the region bounded by \[ y\equal{}\sqrt{25\minus{}x^2}\] and the x-axis?

Well that is the equation for the top half portion of a circle centered at the origin. So the area of the region is $ \frac {1}{2}\cdot \pi(\sqrt {25})^2 \equal{} \boxed{12.5\pi}$.

Though if we're to assume that you mean $ \pm\sqrt{25\minus{}x^2}$, then it's the whole circle, so the answer would be $ 25\pi$.
Unless the area is signed; then its $ 0$.

[quote="grn_trtle"]Unless the area is signed; then its $ 0$.[/quote]
Do you mean that the negative $ y$-side of the circle would have negative area and the positive $ y$-side of the circle would have positive area and hence would cancel out? In which scenarios do we consider signed areas?

[quote="sunehra"]Do you mean that the negative $ y$-side of the circle would have negative area and the positive $ y$-side of the circle would have positive area and hence would cancel out? In which scenarios do we consider signed areas?[/quote]

Yeah, that's what I mean.
Typically one only considers signed area when working with calculus.
That is, $ \int_{\minus{}5}^{5}\pm\sqrt{25\minus{}x^2}\equal{}0$.


vector
topology
real analysis
real analysis unsolved

Open balls are not in general convex right? If you consider p-adic spaces? A set $ E \subset \mathbb{R}^{k}$ is convex if $ \lambda \bold{x}\plus{}(1\minus{} \lambda) \bold{y} \in E$. But this does not hold in p-adic spaces?

The definition of convexity only makes sense in real vector spaces. (Complex vector spaces count, because you can consider them to be vector spaces over the reals.) Note that you didn't quote the definition of convexity correctly. You left one a crucial part of the definition. Here's the correct definition:

$ E$ is convex iff for all $ x,y\in E$ and for all $ \lambda\in[0,1],$ then $ \lambda x\plus{}(1\minus{}\lambda) y\in E.$

You left out the restriction $ 0\le \lambda\le 1$ - and that only makes sense if $ \lambda$ is real.

Since $ p$-adic spaces are not real vector spaces, the definition is moot.

Now, for "open balls are in general convex" to make any sense at all, we must have a set which is both a metric space and a vector space and the metric space structure and the vector space structure must cooperate appropriately.

Any open ball in a real [b]normed[/b] vector space is convex.

Consider the space $ \mathbb{R}^2$ with the metric $ d(x,y)\equal{}\sqrt{|x_1\minus{}x_2|}\plus{}\sqrt{|y_1\minus{}y_2|}.$ This is a metric space and it is a real vector space, but every nontrivial ball is not a convex set. Needless to say, this is not a normed space.


Hello I am Julie Chen, and I am inviting you to apply for AoPS: Big Brother. The rules of this game are simple and the goal is to be the last person remaining in the house. Each week there will be one houseguest crowned the [b]Head of Household[/b]. He or she will have to make two [color=red]Nominations[/color]. However, those nominations can be changed through the [color=blue]Power of Veto[/color]. The winner of the Power of Veto may save one nominated person or leave the nominations the same. In this case, the Head of Household will have to nominate another person and he or she cannot nominate the Power of Veto holder. When only two people remain, the last seven houseguests to be evicted will vote on who they believe should win Big Brother.

Either 12 or 14 houseguests will enter the house depending on interest and we will need at least one alternate. Please only apply if you will have regular access to a computer. It is also recommended you have AIM as you will need to strategize a lot as well as for the ease of some competitions.

There will not be many twists as this is the first time Big Brother has been run on AoPS but if this works out well, look out for future seasons with twists!

To apply, [b]please send me a PM [/b]with the following information:
What is your AIM name? (if applicable):
Age:
Is the best player one who remained loyal or one who deceived?:
Favorite number?:
Favorite color?:
Favorite food?:
Are you more likely to fly under the radar or start up drama?:
Interesting fact about yourself?:
How often can you check a computer?:

I hope this goes better than my failed attempt at this last(?) year...

I don't this person really even gets on...
Forget this


inequalities
AMC
USA(J)MO
USAMO
inequalities unsolved

$ a,b,c \succ0$,prove that:
$ (a^2\plus{}b^2\plus{}c^2)(a^4\plus{}b^4\plus{}c^4)\ge((a\minus{}b)(b\minus{}c)(c\minus{}a))^2$

Cauchy-Schwarz?
$ LHS \equal{} (\sum_{cyc}c^{2})(\sum_{cyc}(a^{2})^{2})\geq (a^{2}c \plus{} ab^{2} \plus{} bc^{2})^{2}\geq (a^{2}c \plus{} ab^{2} \plus{} bc^{2} \minus{} (b^{2}c \plus{} a^{2}b \plus{} ac^{2}))^{2} \equal{} RHS$
Is it that easy or am I missing something?

[quote="polskimisiek"]
$ (a^{2}c \plus{} ab^{2} \plus{} bc^{2} \minus{} (b^{2}c \plus{} a^{2}b \plus{} ac^{2}))^{2} \equal{} RHS$
[/quote]
yes,with your way,it's easy :blush:  .

[quote="rachid"]$ a,b,c \succ0$,prove that:
$ (a^2 \plus{} b^2 \plus{} c^2)(a^4 \plus{} b^4 \plus{} c^4)\ge((a \minus{} b)(b \minus{} c)(c \minus{} a))^2$[/quote]
$ (a^2 \plus{} b^2 \plus{} c^2)(a^4 \plus{} b^4 \plus{} c^4)\ge((a \minus{} b)(b \minus{} c)(c \minus{} a))^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6\plus{}2a^4bc\plus{}2a^3b^3\minus{}2a^3b^2c\minus{}2a^3c^2b\plus{}2a^2b^2c^2)\geq0,$ which obviously true.

The polskimisiek's solution is right :) .
$ a^2c\plus{}b^2a\plus{}c^2b\minus{}a^2b\minus{}b^2c\minus{}c^2a\equal{}(b\minus{}a)(c\minus{}a)(c\minus{}b)$
;)

Another approach:
Let f(a,b,c)=(a^2+b^2+c^2)(a^4+b^4+c^4)-[(a-b)(b-c)(c-a)]^2
We will prove: f(a,b,c)>=f(a-c,b-c,0)(c=min(a,b,c))
It's obviously true since (a^2+b^2+c^2)>=(a-c)^2+(b-c)^2
                                 (a^4+b^4+c^4)>=(a-c)^4+(b-c)^4
Therefore, f(a,b,c)>=f(a-c,b-c,0).
Now it's sufficient to prove the inequality above in the case when c=0.
It's equivalent to prove:(a^2+b^2)(a^4+b^4)>=[(a-b)(ab)]^2
Using AM-GM we have:a^4+b^4>=2a^2b^2
Soo, the LHS>=2a^2b^2(a^2+b^2)>=(a^2b^2)(a-b)^2[true since (a-b)^2=<2(a^2+b^2)]
The equality doesn't occur.

[quote="mitdac123"]The polskimisiek's solution is right :) .
$ a^2c \plus{} b^2a \plus{} c^2b \minus{} a^2b \minus{} b^2c \minus{} c^2a \equal{} (b \minus{} a)(c \minus{} a)(c \minus{} b)$
;)[/quote]

|x|\ge |x-y| is not generally true...

Say $ c\equal{}\min \{a,b,c\}$. Then decrease all variables by $ c$, and the LHS is the same, RHS decreases...

So we have to prove $ (a^2\plus{}b^2)(a^4\plus{}b^4)\ge (a\minus{}b)^2(ab)^2\iff (a^3\plus{}b^3)^2\ge 0$

[quote="Altheman"][quote="mitdac123"]The polskimisiek's solution is right :) .
$ a^2c \plus{} b^2a \plus{} c^2b \minus{} a^2b \minus{} b^2c \minus{} c^2a \equal{} (b \minus{} a)(c \minus{} a)(c \minus{} b)$
;)[/quote]

|x|\ge |x-y| is not generally true...

Say $ c \equal{} \min \{a,b,c\}$. Then decrease all variables by $ c$, and the LHS is the same, RHS decreases...

So we have to prove $ (a^2 \plus{} b^2)(a^4 \plus{} b^4)\ge (a \minus{} b)^2(ab)^2\iff (a^3 \plus{} b^3)^2\ge 0$[/quote]
I think you have a mistake! The LHS decreases, but the RHS is the same. And it's also the solution I mentioned above :wink: .
PS: How is USAMO going


probability
geometry
3D geometry
MATHCOUNTS
quadratics
inequalities
number theory

#6 sprint: A point E is chosen at random from within square ABCD. Express as a decimal to the nearest hundredth the probability that triangle ABE is obtuse.

#8 sprint: Charlene's calculator displays the digits 0, 1, 6, 8 and 9 so that, when the calculator is held upside-down, these digits appear to be 0, 1, 9, 8 and 6 respectively. How many three-digit numbers look the same upside-down and right-side-up when using these five possible digits? (A three-digit string that begins with either 0 or 00 is not considered a three-digit number.)

#11 sprint: The diagram is made of unit cubes and is hiding no holes. The cubes are rearranged to create a rectangular solid with the smallest possible surface area. How many square units are in this surface area? THERE IS A PICTURE OF 64 CUBES, GO FIGURE..

#13 sprint: What is the probability that a randomly chosen factor of $ 15{}^9{}^5$ is a multiple of $ 15{}^6{}^5$ ? Express your answer as a common fraction.

#20 sprint: What's the largest number $ c$, such that $ 2x^2 + 5x + c = 0$ has at least one real number solution. Express your answer as a common fraction.

#24 sprint: How many distinct triangles can be constructed by connecting 3 vertices of a cube?

#30 sprint: How many four-digit integers greater than 5000 are there for which the thousands digit equals the sum of the other three digits?

Out of these questions, i can answer about 80% of them, but i know there is a better/faster way, and i want you to share. Thanks! :censored: [/hide]

Hint for number 6: Try a few points.

Hint for number 8: Find digits that correspond to other digits when turned upside-down.

Hint for number 11: Extreme principle.

Hint for number 13: Factor.

Hint for number 20: Discriminant.  Remember what you were taught in algebra class.

for 24:u just basically try a couple triangles..and well idont have paper right now but its 3 i think

and 30 :series of casework...

uhh, u wanna tell me, that the "extreme principle" and actually tell me how to solve them fast, lol. thx.

You can use balls and urns for 30.
[hide] Let's say the hundreds digit is x, tens is y, and ones is z.

Cases:
If the thousands digit is 5:
x+y+z=5. You have two separators and 5 two give out, so you have 7C2.
If the thousands digit is 6:
x+y+z=6. You get 8C2
If the thousands digit is 7:
x+y+z=7. You get 9C2

It's easy to see now that your answer is 7C2+8C2+9C2...11C2
Getting 21+28+36+45+55=185.

I bet I got it wrong.
[/hide]

That's correct; look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1426236#1426236]here[/url]. As for the surface area question, the minimum will always be achieved when the side lengths are as close to each other as possible by AM-GM.

[hide="Problem 6"]
A useful theorem to know is that an angle inscribed in a semicircle is always right.  Thus, we can conclude that point $ E$ must be located inside the semicircle, which has an area of $ \frac {s^2 \times \pi}{8}$, where $ s$ is the side length of the square.  The total area of the square is $ s^2$.  Thus, the answer is $ \frac {\frac {s^2 \times \pi}{8}}{s^2} \equal{} \frac {\pi}{8} \approx \boxed{0.39}$.
[/hide]

I hope this helped!

[hide="Problem 13"]We know that $ 15^{95}\equal{}3^{95}\cdot5^{95}$, so it has $ (95\plus{}1)(95\plus{}1)\equal{}96\cdot96\equal{}96^2$ factors. Now we just have to figure out how many factors of this are multiples of $ 15^{65}$. $ 15^{65}\equal{}3^{65}\cdot5^{65}$ Just "divide" the prime factorization of $ 15^{95}$ by $ 15^{65}$, and we get $ 3^{30}\cdot5^{30}$, which is basically the maximum number you can multiply $ 15^{65}$ by to get $ 15^{95}$. The factors of this number are also all numbers by which you can multiply $ 15^{65}$ to get a factor of $ 15^{95}$. So, $ 3^{30}\cdot5^{30}\equal{}(30\plus{}1)(30\plus{}1)\equal{}31^2$ factors. The probability is $ \frac{31\cdot31}{96\cdot96}\equal{}\frac{961}{9216}$[/hide]
probably got that wrong...

and sorry, im not that good at explaining stuff.

Well, the solution is correct, but the problem actually reads "what is the probability that a randomly chosen factor of $ 15^{90}$ is a multiple of $ 15^{65}$," which has a much nicer answer. By the way, these are from the 1996 National Sprint, not the 1995 National Sprint.

To clear this up, these problems are from the [b]1996 National Sprint[/b], not the 1995 National Sprint.  I have verified this using the All-Time Greatest MATHCOUNTS Problems book, as well as my copies of the tests.

Usually people refer to the year by the latter year, so I'm pretty sure this is 1995-1996.

This is a 1995-1996, i thought it was held in '95, so i just put 95.

[hide="Problem 20"]
For a quadratic equation $ ax^2+bx+c=0$ to have at least one real solution, we must have $ b^2-4ac \ge 0$.  In this case, we have $ a=2, b=5$, so:

\begin{align*}
b^2-4ac &\ge 0
\\ 25 - 8c & \ge 0
\\ -8c & \ge -25
\\ c & \le \frac{25}{8}
\end{align*}

The largest possible value of $ c$ that satisfies this inequality is obviously $ c=\boxed{\frac{25}{8}}$.
[/hide]

[quote="ReseesPieces"]remember this problem...its 95 btw.....uhh i just finished 96-97 nats

neways

15^90=5^90*3^90 so 91*91 factors

then 90-65=25....so it has 26*26 factors...the 15 thing again

from which u get 26^2/91^2 which is 2^2/7^2 =4/49.....the 13 comes out(i remember i forgot to simplify the 13 when i did this sprint....)[/quote]

Erm, it was 15^95.

oh whoa ok...well then i failed....i think i memmed the solution to this.....which is why i solved it like that...


geometry
AMC
AIME
trigonometry
vector
AoPS Books
analytic geometry

I am not that good at geometry, but I would like to improve, so I'm considering buying the AoPS Introduction To Geometry book. I already have AoPS volumes 1 & 2 though. Is there any material in the geometry book that's not in the other AoPS books? Would buying the geometry book help me more than just the geometry in the AoPS books? Thanks.

[quote="rzsolt"][quote="surge"]I'm sure there's extra material in the 'Intro to Geometry' book, but if you go through the Geometry chapters in AoPS Vol. 1 and 2 your "not good at geometry" problems will be solved![/quote]

all of my geo experience is from aops 1. I've done up the quadliteral chapter. I still can't do any aime geo at all, its costing me 4 points per exam =(, will the rest of aops 1 and 2 geo finish me off, or is the intro geo book really that awesome?[/quote]

I don't have Intro to Geometry, so I can't really compare them. But AoPS 2 has a lot of geometry. Ch. 2 is all about trig and Ch. 3 is completely about triangles. That material will probably be much more useful than the stuff from AoPS 1. Plus, there's cyclic quadrilaterals, vectors, analytic geometry, collinearity, concurrency, etc.

http://www.artofproblemsolving.com/Books/AoPS_B_TOC.php?page_id=2

When I worked though AoPS 1, I could not really get the geometry in it.  However, intro to geometry has taught me a lot.  I think it is easier than AoPS 2, but it goes deeper into geometry than either AoPS 1 or 2.  If you are good enough to do AoPS 2, I think that the intro to geometry book would not be very hard for you.  However, I recommend it if you are a beginner in geometry  :D


quadratics
function
algebra
polynomial
Rational Root Theorem

There is an explicit formula for the zeroes of a quartic, but it is EXTREMELY nasty.  If you want to look for zeroes over $ \mathbb{Q}$, use the rational root theorem.

[quote="Alpine"]Thanks for the reply. I had forgotten that the rational root
theorem applies to more than just quadratic equations. So the candidates
would be -1, 1, -2, 2, -4 and 4 right? I tried all of them and none worked so
does that mean that all of the real zeroes are irrational?[/quote]

Alpine, those are the correct candidates.
Yes, it means that all of the real zeroes,
if there are any real zeroes, are irrational.

If you make a list of consecutive integers, you can
see that there is a sign change in the value of the
function between x = -2 and x = -1.  Also there is
a sign change in the value of the function between
x = 0 and x = 1.

These are the two locations where there are
irrational roots.

You can approximate the roots of this fairly easily.

Re-arrange the equation as $ x \equal{} \frac{4\plus{}4x^3\minus{}x^4}{16}$

From here, create a sequence of guesses $ x_{n\plus{}1} \equal{} \frac{4\plus{}4x_n ^3\minus{}x_n ^4}{16}$

Guess $ x_0 \equal{} 0$, giving $ x_1 \equal{} \frac{1}{4}$ and $ x_2 \equal{} \frac{1039}{4096}$, which is pretty close.

Unfortunately, this recurrence doesn't give the other root, because it tends to diverge away from that root. Fortunately, there's a quick-fix for this:

$ x \equal{} 2x\minus{}x \equal{} 2x \minus{} \frac{4\plus{}4x^3\minus{}x^4}{16} \equal{} \frac{\minus{}4\plus{}32x\minus{}4x^3\plus{}x^4}{16}$

This gives us another recurrence equation $ x_{n\plus{}1} \equal{} \frac{\minus{}4\plus{}32x_n \minus{}4x_n ^3\plus{}x_n ^4}{16}$. Any appropriate guess (-2, -1, -3/2, -7/4, whatever) will cause it to converge on the second root.


geometry
3D geometry
sphere
trapezoid
dodecahedron
pyramid

is there a good way to calculate the shapes and lines on the surface of a football?
I know there is a very famous formula to solve this kind of problems but, what is it?

What do you mean?  American football, or soccer?

Aren't there different number of lines in each football, depends which one: NFL, College, etc.

[quote="h_s_potter2002"]I blieve he was talking about a soccer ball, and not an American football.  Like the one which is basically a tesselation of pentagons and  hexagons.[/quote]
that is... a soccer ball with pentagons and hexagons in white and black!

I'm not sure that this is the right formula, but the pentagons' surface areas could be calculated by dividing it into a trapezoid and a triangle.  Likewise, the hexagons would be 2 trapezoids of equal area.

There are 12 pentagons and 20 hexagons on a (at least my) soccer ball, and all regular with 4.3 cm sides.

[hide]Total Surface Area = 12p * 20h

p = Tra1 + Tri
p = [h/2 * (b1 + b2)] + [(b*h)/2]
p = [4.3/2 * (4.3 + 7)] + [(7*2.5)/2]
p = [24.295] + [8.75]
p = 33.045 cm2

h = 2Tra2
h = 2[h/2 * (b1 + b2)]
h = 2[4.3/2 * (8.8 + 4.3)]
h = 2[28.165]
h = 56.33 cm2

TSA = 12(33.045) * 20(56.33)
TSA = 446,741.964 cm2
[/hide]

i am wondering if there is a good way to calculate the number of the pentagons and hexagons and their sides using some formulas. If we are given a football, and we know it has pentagons and hexagons on it, is there a possible way to know the number 12 and 20 without looking at the soccer ball??

[quote="Zyloch"]If you know the total amount of hexagons and pentagons, then there is a way. Suppose there are 12 pentagons and 20 hexagons (like 2kid*bari2 said), then the total amount of sides for the pentagons are $12\cdot5=60$ and the total amount of sides for the hexagons are $20\cdot6=120$ making the total amount of lines $180$, but each line is counted twice since every shape shares a side with another shape so there are $\frac{180}{2}=90$ different lines.[/quote]
that is a good point...got something to do with $C_{60}$

But if i do not know how many pentagons and haxegons there are, can i find the number by mathmatical theorems or formulas?

[quote="dk2"]I'd think technically the answer is no, because you could have a dodecahedron with only pentagons.
But if you need to use both hexagons and pentagons, I'm not quite sure.[/quote]
yes, we have both hexagons and pentagons to form the surface of a soccer ball.

This as far as I got:

Let a be the number of pentagonal faces and b the number of hexagonal faces,

F=a+b
E=(5a+6b)/2
Since F-E+V=2,
V= (3a+4b+4)/2

Three faces must meet at each vertex (because the 4 internal angles of a pentagon/hexagon must be less than 360 when we unfold the polyhedron. Two faces cannot meet at one vertex because then the polyhedron isn't 3D).
Then V=(5a+6b)/3
Equating (3a+4b+4)/2=(5a+6b)/3 we get
9a+12b+12=10a+12b
This nicely cancels out to give a=12.
Plugging this in, we have b=20
This is the standard soccer ball
Therefore the soccer ball is the only polyhedron that can be made by using both and only hexagons and pentagons.  :lol:

Here's a (unfortunately pretty easy if you know the formula) question involving a foot ball (soccer ball):

Find the surface area of 2kid*bari2's soccer ball (in case you don't know, although it was posted above, his soccerball is covered with regular pentagons and hexagons with 4.3 cm sides)

Also, does any body know any way to calculate the volume of the soccer ball? This is probably pretty hard so I apologize if its above the level of this forum.

[quote="tarquin"]Bucky ball![/quote]
Is that hamster/dinner called bucky

can you tell me what you used to get the answer?

If you're talking about my question,

[hide]
Use the formula for area of a regular polygon. It's been posted on this forum a couple times now.
[/hide]

Even though my answer doesn't make complete sense, I already posted it as [hide]446,741.964 cm2[/hide]

i'm not sure if this is right, but hint for volume:
[hide]try finding the areas of a bunch of pyramids with pentagonal and hexagonal bases[/hide]
do you know the width of the ball? (one for pentagonal, one for hexagonal)

[quote="dk2"]Plugging this in, we have b=20
[/quote]
How do you pluge this in?  :?


At 1:00 p.m. Sue left her house and began walking at 4km/h toward Sandy's house. Fifteen minutes later, Sandy left her home and walked at 8km/h toward Sue's house. If Sue lives 10 km from Sandy, at what time did they meet? Who walked farther?

4km/h means that in 15 mintues she walked 1 km.

now the distance between them is 9 km.

now the distance would get closer by 4+8km/h or 12km/h

now find how much time u need to walk 9km in 12km/h which is 45 minutes. It should be 1:45 pm.

Who walked farther? I didnt use any formulas but it's Sandy because

8x 30minutes and 4x 45minutes.. find the larger one..

i think i made a mistake somewhere some1 correct me plz lol

Sue has already been walking for 15 minutes when the 45 minutes starts, so it's 2:00pm when they meet. And Sandy walked for 45 minutes and Sue for an hour  :wink:


function
inequalities proposed
inequalities

Find all real-valued functions f(x) on the reals such that
$\frac{1}{2}f\left(x^2+y^2\right) + \frac{1}{2}f\left(x^2+z^2\right) - f(x) f(yz) \geq \frac{1}{2}$ for all $x, y, z \in \mathbb{R}$.

[quote="Arne"]Take $x = y = z = 0$, then $f(0) - f(0)^2 \geq \frac{1}{2}$, or \[\left(f(0) - \frac{1}{2}\right)^2 + \frac{1}{4} \leq 0,\] which is impossible. So, the problem is false.[/quote]
Well, to be a little bit more precise, the problem is not false but only such function does not exist :P

I saw a similar (Vietnamese) problem once which asked to find all functions satisfying

\[f(xy) + f(xz) - 2f(x)f(yz) \geq \frac{1}{2},\;\forall x,y,z \in \mathbb{R}.\]

That one is easy though.


Are there $1000$ consecutive positive integer numbers with only $5$ primes among them?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=window&t=100038]It's a classic[/url] :D

PS: 2000th post


If a book says
"Unless otherwise specified, all rings in this book will be supposed commutative and with unit."
and there is a sentence "Let $ A$ be a commutative ring", then does it mean that $ A$ don't necessary have the unit as there is the word commutative again?

I think it means that the second operation is commutatuive but the ring need not have a unit(y). For example the commutatuve ring of even integers under addition and multiplication.


I often see problems like "The product of two numbers is 4 and their sum is 5. What is the sum of their reciprocals?"

Do I have to use two-variable linear equations to solve this?

Use this:

$ \frac{1}{x}\plus{}\frac{1}{y}\equal{}\frac{y}{xy}\plus{}\frac{x}{yx}\equal{}\frac{x\plus{}y}{xy}$

Just insert sum into numerator and product into denominator, it's that simple.


Prove that for all real $ a,b$ with $ 0<a,b<1$,
\[ a(1\minus{}b^2)\plus{}b\sqrt{1\minus{}a^2}\ \leqslant\ \sqrt{1\minus{}a^2b^2}
\]

I think the inequality should be:
for all real $ a,b$ with $ 0 < a,b < 1$,
\[ a\sqrt{1 \minus{} b^2} \plus{} b\sqrt {1 \minus{} a^2}\ \leqslant\ \sqrt {1 \minus{} a^2b^2}
\]

[quote="Sunkern_sunflora"]
\[ a\sqrt {1 \minus{} b^2} \plus{} b\sqrt {1 \minus{} a^2}\ \leqslant\ \sqrt {1 \minus{} a^2b^2}
\]
[/quote]
This one isn't true ($ a\equal{}b\equal{}\frac{2}{3}$)

There is no mistake with the inequality I originally posted. :)

[quote="Mary Barton"]Prove that for all real $ a,b$ with $ 0 < a,b < 1$,
\[ a(1 \minus{} b^2) \plus{} b\sqrt {1 \minus{} a^2}\ \leqslant\ \sqrt {1 \minus{} a^2b^2}
\]
[/quote]

[quote="Mary Barton"]Prove that for all real $ a,b$ with $ 0 < a,b < 1$,
\[ a(1 \minus{} b^2) \plus{} b\sqrt {1 \minus{} a^2}\ \leqslant\ \sqrt {1 \minus{} a^2b^2}
\]
[/quote]
The inequality is equivalent to
\[ a^2(1\minus{}b^2)^2 \plus{}2ab(1\minus{}b^2)\sqrt{1\minus{}a^2} \plus{}b^2(1\minus{}a^2) \le 1\minus{}a^2b^2\]
\[ \Leftrightarrow a^2(1\minus{}b^2)^2 \plus{}2ab(1\minus{}b^2)\sqrt{1\minus{}a^2} \le 1\minus{}b^2\]
\[ \Leftrightarrow a^2(1\minus{}b^2) \plus{}2ab\sqrt{1\minus{}a^2} \le 1\]
Now, by AM-GM Inequality, we have
\[ 2ab\sqrt{1\minus{}a^2} \le a^2b^2\plus{}1\minus{}a^2\]
Hence
\[ a^2(1\minus{}b^2)\plus{}2ab\sqrt{1\minus{}a^2} \le a^2(1\minus{}b^2) \plus{}a^2b^2\plus{}1\minus{}a^2 \equal{}1\]
We have done. :)

Very nice! :clap2:

[hide="My solution"]By Cauchy–Schwarz,
\[ a\sqrt{1\minus{}a^2b^2}\plus{}\sqrt{1\minus{}a^2}\cdot ab\ \leqslant\ \sqrt{a^2\plus{}(1\minus{}a^2)}\sqrt{(1\minus{}a^2b^2)\plus{}(ab)^2}\ \equal{}\ 1
\]
Multiplying by $ \sqrt{1\minus{}a^2b^2}\minus{}b\sqrt{1\minus{}a^2}$ (which is positive) we have
\[ \sqrt{1\minus{}a^2b^2}\minus{}b\sqrt{1\minus{}a^2}\ \geqslant\ a\left(\sqrt{1\minus{}a^2b^2}\plus{}b\sqrt{1\minus{}a^2}\right)\left(\sqrt{1\minus{}a^2b^2}\minus{}b\sqrt{1\minus{}a^2}\right)
\]
\[ \equal{}\ a\left((1\minus{}a^2b^2)\minus{}b^2(1\minus{}a^2)\right)
\]
\[ \equal{}\ a(1\minus{}b^2)
\][/hide]
:wink:


induction
arithmetic sequence
algebra proposed
algebra

Why does this remind me of Cauchy-Davenport ??  :lol:

I don't understand what you said.  :maybe:

You can simply prove it by induction
you get that $ u_{1},u_{2},...,u_{n-1}$ satisfies the condition thus is in arithmetic progression also $ u_{2},u_{3},...,u_{n}$ are in arithmetic progression. Thus we only need to prove it for $ n=3$ but this is trivial.
I said that the fact that $ |X-X|\geq |X|-1$ somehow reminds me of the other statements about sum/difference sets such as Cauchy Davenport (Of course in this case the fact and its case of equality are much easier)

[quote="N.T.TUAN"]Let $ u_{1},u_{2},...,u_{n}$ be an increasing sequence and let $ A_{n}$ be the set of all positive reals of the form $ u_{i}-u_{j}(1\leq j<i\leq n)$. Prove that if $ A_{n}$ has fewer that $ n$ elements. Then $ u_{1},u_{2},...,u_{n}$ forms an arithmetic sequence.[/quote]

here's solution :
[hide]Consider $ n-1$ positive integers :$ u_{2}-u_{1}<u_{3}-u_{1}<...<u_{n}-u_{1}$ .Because $ A_{n}$ has fewer than $ n$ elements so $ A_{n}=\{u_{2}-u_{1},u_{3}-u_{1},...,u_{n}-u_{1}\}$ .
Then consider $ n-2$ positive integers $ u_{3}-u_{2}<u_{4}-u_{2}<...<u_{n}-u_{2}$ are $ n-2$ elements of $ A_{n}$, and $ u_{3}-u_{2}<u_{4}-u_{2}<...<u_{n}-u_{2}<u_{n}-u_{1}$ so we must have that :$ u_{2}-u_{1}=u_{3}-u-2,u_{4}-u_{2}=u_{3}-u_{1},...,u_{n-1}-u_{1}=u_{n}-u_{2}$.
Hence we have done .[/hide]


function
induction
algebra unsolved
algebra

Find all functions $f: Q^{+}\rightarrow R$ such that

$f(x)+f(y)+2xyf(xy)=\frac{f(xy)}{f(x+y)}$ for all $x,y\in Q^{+}$

[quote="AYMANE"]If $f(1)=0$[/quote]

If $f(1) = 0$, so set $x = y = 1/2$. Which is implies $2f(\frac12) + \frac12. f(\frac14) = \frac{f(\frac{1}{4})}{0}$. But, it is impossible since the kodomain is real number.


Solve in $ \mathbb{R}^*$ this equation :

$ (ab\plus{}bc\plus{}ca)^2\equal{}3(a^4\plus{}b^4\plus{}c^4)$

[quote="alex2008"]Solve in $ \mathbb{R}^*$ this equation :

$ (ab \plus{} bc \plus{} ca)^2 \equal{} 3(a^4 \plus{} b^4 \plus{} c^4)$[/quote]

just expand both sides.you´ll see that equality only occurs when a=b=c(direct consequence of rearrangement inequality) :wink:

[quote="ocha"][hide="solution"]
By AM-GM

$ \sum a^4 \equal{} \sum \frac {a^4 \plus{} b^4}{2} \ge \sum a^2b^2 \qquad(1)$

$ \sum a^4 \equal{} \sum \frac {a^4 \plus{} a^4 \plus{} b^4 \plus{} c^4}{4} \ge \sum a^2bc \qquad(2)$

From (1) and (2)

$ 3\sum a^4 \ge \sum a^2b^2 \plus{} 2 \sum a^2bc \equal{} (ab \plus{} bc \plus{} ca)^2$

There is only equality in AM-GM when $ a \equal{} b \equal{} c$, and that is the solution

Note: even if $ a,b,c\not > 0$ the inequality still holds because $ a^4,b^4,c^4 > 0$
[/hide][/quote]
Nice but there is no need using all this...
$ 3[(a^2)^2\plus{}(b^2)^2\plus{}(c^2)^2] \geq (a^2\plus{}b^2\plus{}c^2)^2 \geq (ab\plus{}bc\plus{}ca)^2$,and we finished


For a constant number $ \theta \left(0 < \theta < \frac {\pi}{2}\right)$, consider a sequence $ a_n \equal{} \sin n\theta\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$. Prove that there exist constant numbers $ p,\ q$ which don't depend on $ n$ such that $ a_{n \plus{} 2} \equal{} pa_{n \plus{} 1} \plus{} qa_n\ (n \equal{} 0,\ 1,\ 2,\ \cdots)$.

[hide="Hint"]$ \sin n \theta \equal{} \frac{1}{2i} (e^{i \theta} )^n \minus{} \frac{1}{2i} (e^{\minus{}i \theta})^n$[/hide]

Alternately, one can use the determinant identity given [url=http://qchu.wordpress.com/2009/06/07/the-catalan-numbers-regular-languages-and-orthogonal-polynomials/]here[/url].

[hide="I think this is correct"]
$ e^{i\theta}\sin (n\plus{}1)\theta\equal{}\frac{1}{2i}(e^{i\theta})^{n\plus{}2}\minus{}\frac{1}{2i}(e^{\minus{}i\theta})^{n}$

$ e^{\minus{}i\theta}\sin (n\plus{}1)\theta\equal{}\frac{1}{2i}(e^{i\theta})^{n}\minus{}\frac{1}{2i}(e^{\minus{}i\theta})^{n\plus{}2}$

Summing gives us:

$ (e^{i\theta}\plus{}e^{\minus{}i\theta})\sin (n\plus{}1)\theta\equal{}\sin (n\plus{}2)\theta \plus{} \sin n\theta$

$ a_{n\plus{}2}\equal{}a_{n\plus{}1}(e^{i\theta}\plus{}e^{\minus{}i\theta})\minus{}a_n$[/hide]

[quote="Flame"]
$ e^{i\theta}\sin (n \plus{} 1)\theta \equal{} \frac {1}{2i}(e^{i\theta})^{n \plus{} 2} \minus{} \frac {1}{2i}(e^{ \minus{} i\theta})^{n}$

$ e^{ \minus{} i\theta}\sin (n \plus{} 1)\theta \equal{} \frac {1}{2i}(e^{i\theta})^{n} \minus{} \frac {1}{2i}(e^{ \minus{} i\theta})^{n \plus{} 2}$

Summing gives us:

$ (e^{i\theta} \plus{} e^{ \minus{} i\theta})\sin (n \plus{} 1)\theta \equal{} \sin (n \plus{} 2)\theta \plus{} \sin n\theta$

$ a_{n \plus{} 2} \equal{} a_{n \plus{} 1}(e^{i\theta} \plus{} e^{ \minus{} i\theta}) \minus{} a_n$[/quote]

Yes, $ (p,\ q) \equal{} (2\cos \theta ,\ \minus{} 1).$

My Solution:

By Sum to product formula,

$ \sin (n \plus{} 2)\theta \plus{} \sin n\theta \equal{} 2\sin (n \plus{} 1)\theta \cos \theta \Longrightarrow (p,\ q) \equal{} (2\cos \theta ,\ \minus{} 1).$


With k a positive integer, prove that (1−k^−2)^k ≥ 1−1/k.

Um kinda easy?

Solution:
[hide]

Mere simplification gives k>=1, hence a positive integer.
[/hide]


inequalities parameterization quadratics function

solve with parameter $m$:
$x^{2}-(2+\sqrt{1-m})x+2\sqrt{1-m}>0$

I know it seems to be quite easy providing $t=\sqrt{1-a}\ \geqslant 0$
but I can't come up with the final answer...

It's just like a quadratic

[hide]
The function can be factored as \[(x-2)(x-\sqrt{1-m})>0\]The zeroes are at $x=2, \sqrt{1-m}$.
So when $m<-3$ we have the solution set to be \[x\in (-\infty,2)\cup(\sqrt{1-m},\infty)\]
When $m=-3$, the solution set is \[x\in (-\infty,2)\cup(2,\infty)\]
And when $m>-3$, we have the solution set to be \[x\in (\sqrt{1-m},2)\]
I think this is how you're supposed to do these problems.

 [/hide]

[quote="NeverOddOrEven"]It's just like a quadratic

[hide]
The function can be factored as
\[(x-2)(x-\sqrt{1-m})>0 \]
The zeroes are at $x=2, \sqrt{1-m}$.
So when $m<-3$ we have the solution set to be
\[x\in (-\infty,2)\cup(\sqrt{1-m},\infty) \]
When $m=-3$, the solution set is
\[x\in (-\infty,2)\cup(2,\infty) \]
And when $m>-3$, we have the solution set to be
\[x\in (\sqrt{1-m},2) \]
I think this is how you're supposed to do these problems.

 [/hide][/quote]
think the last part shud be corrected to
[hide]And when $-3<m \le 1$, we have the solution set to be

$x\in (-\infty,\sqrt{1-m})\cup(2,\infty)$[/hide]


trigonometry
quadratics
number theory
relatively prime
algebra
quadratic formula

Given that $(1+\sin t)(1+\cos t)=5/4$ and \[ (1-\sin t)(1-\cos t)=\frac mn-\sqrt{k}, \] where $k, m,$ and $n$ are positive integers with $m$ and $n$ relatively prime, find $k+m+n.$

I expanded the left-hand side of both and added the equations together (1).  Then I multiplied the original two together (2).  I was able to write (1) and (2) in terms of $\sin{2t}$, and was able to eliminate it, coming up with:

$\left(\frac{3}{4}+\sqrt{k}-\frac{m}{n}\right)^{2}=5\left(\frac{m}{n}-\sqrt{k}\right)$

I don't have the time now to do anything else (I don't even know if what I did [i]is[/i] correct thus far), but I will try to post later...

Multiplying the first equation by the second gives us $(1 - \sin^2 t)(1 - \cos^2 t) = \frac{5}{4} ( \frac{m}{n} - \sqrt{k})$.  Simplifying gives $\sin^2 2t = 5 (\frac{m}{n} - \sqrt{k}) = 5(1 - \cos t)(1 - \sin t)$.

I think this equation is solvable algebraically, but I am unsure how.

yeah i used t0rajir0u's method also..

Multiply both equations, get:
$(1 - \sin^2 t)(1 - \cos^2 t) = \frac{5}{4} ( \frac{m}{n} - \sqrt{k})$
$\cos^2 t \sin^2 t = \frac{5}{4} ( \frac{m}{n} - \sqrt{k})$
let $x=( \frac{m}{n} - \sqrt{k})$
$\cos t \sin t = \pm \sqrt{\frac{5}{4} x}$
Substituting back into the original equations, we get
$\sin t + \cos t = \frac{1}{4} \mp \sqrt{\frac{5}{4}x}$
$\sin t + \cos t = 1 \pm \sqrt{\frac{5}{4}x}-x$
$\frac{1}{4} \mp \sqrt{\frac{5}{4}x}=1 \pm \sqrt{\frac{5}{4}x}-x$
EDIT: (for this line the $\pm$ on both sides cannot both be positive or negative, because that would get a negative x)
$x \pm \sqrt{5x}-\frac{3}{4}=0$
$5x=x^2-\frac{3}{2}x+\frac{9}{16}$
$x=\frac{13}{4} \pm \sqrt{10} = \frac{m}{n} - \sqrt{k}$
EDIT: (Here, $\pm$ sign is turned into a negative because x cannot exceed 1)

$m+n+k=13+4+10=\boxed{27}$

EDIT: thanks paladin8 for pointing out correct answer  :oops:

[hide]
Hmm, I'm sure there's a better solution, but here's a mess of algebra:

Let $x = \sin{t}+\cos{t}$ and $y = \sin{t}\cos{t}$.

We have $1+x+y = \frac{5}{4}$ and $x^2-2y = 1$ (pythagorean identity).

So $y = \frac{x^2-1}{2}$, substitute back into the first equation to get a quadratic in $x$:

$2x^2+4x-3 = 0 \Rightarrow x = -1+\frac{\sqrt{10}}{2}$.

So then $y = \frac{\left(-1+\frac{\sqrt{10}}{2}\right)^2-1}{2} = \frac{5}{4}-\frac{\sqrt{10}}{2}$

We want to find

$1-x+y = 1-\left(-1+\frac{\sqrt{10}}{2}\right)+\left(\frac{5}{4}-\frac{\sqrt{10}}{2}\right) = \frac{13}{4}-\sqrt{10}$

so $k+m+n = 10+13+4 = 27$. [/hide]

[hide]$(1+\sin{t})(1+\cos{t})=\frac{5}{4}$
$(1-\sin{t})(1-\cos{t})=c$

Multiply it out.

$1+\sin{t}\cos{t}+\cos{t}+\sin{t}=\frac{5}{4}$
$1+\sin{t}\cos{t}-\cos{t}-\sin{t}=c$

From here we can subtract or add.

$2+2\sin{t}\cos{t}=\frac{5}{4}+c$
$2\cos{t}+2\sin{t}=\frac{5}{4}-c$

Factor out the two and square the second equation because that makes us happy.

$2+2\sin{t}\cos{t}=\frac{5}{4}+c$
$4+8\sin{t}\cos{t}=\frac{25}{16}-\frac{5}{2}c+c^2$

Multiply the first equation by 4 and subtract the second one from it.

$4=5+4c-\frac{25}{16}+\frac{5}{2}c-c^2$
$= c^2-\frac{13}{12}c+\frac{9}{16}=0$
$c=\frac{13}{4}\pm\sqrt{10}$

Because $(1-\sin{t})(1-\cos{t})$ has a maximum of $4$

$c=\frac{13}{4}-\sqrt{10}$

$13+4+10=\boxed{27}$[/hide]

I think my solution requires less algebra/work. :)

[hide="Solution"]We're given that \begin{eqnarray*}(1+\sin t)(1+\cos t) &=& 1+(\sin t+\cos t)+\sin t\cos t = \frac{5}{4}\\ (1-\sin t)(1-\cos t) &=& 1-(\sin t+\cos t)+\sin t\cos t = \frac{m}{n}-\sqrt{k}.\end{eqnarray*} The difference between the two equations is $2(\sin t+\cos t),$ so we want to find that value. Note that we can eliminate the $\sin t \cos t$ term with the Pythagorean Identity; $(\sin t+\cos t)^{2}=1+2\sin t \cos t.$

Multiplying both sides by $2$ and substituting, \[(\sin t+\cos t)^{2}+2(\sin t+\cos t)+1=\frac{5}{2}.\] Applying the quadratic formula, \[\sin t+\cos t=\frac{-2\pm \sqrt{10}}{2}.\] The negative root is extraneous, so $\sin t+\cos t=\frac{-2+\sqrt{10}}{2}$ and \begin{eqnarray*}(1-\sin t)(1-\cos t)&=&(1+\sin t)(1+\cos t)-2(\sin t+\cos t)\\ &=& \frac{5}{4}-(2+\sqrt{10})\\ &=&\frac{13}{4}-\sqrt{10}\implies k+m+n=27. \end{eqnarray*} [/hide]

Multiplying out, we have
$ 1\plus{}sint \cdot cost\plus{} sin t \plus{} cos t\equal{}1\plus{} \sqrt{1\plus{}sin(2t)}\plus{}\frac{1}{2} \cdot sin (2t)\equal{}\frac{5}{4}$
Let $ sin (2t)\equal{}x$. This simplifies to
$ 16(x\plus{}1)\equal{}(2x\minus{}1)^2$, $ x\equal{}\frac{5}{2}\minus{}\sqrt{10}$ (since  $ \minus{}1 \le sin(t) \le 1$)
The answer is $ \frac{5}{4}\minus{}2\sqrt{1\plus{}x}\equal{}\frac{5}{4}\minus{}2\sqrt{\frac{7}{2}\minus{}\sqrt{10}}\equal{}\frac{5}{4}\minus{}\sqrt{14\minus{}4\sqrt{10}}\equal{}\frac{5}{4}\minus{}(\sqrt{10}\minus{}2)\equal{}\frac{13}{4}\minus{}10$

I thought I had a pretty easy solution:
[hide="Solution"]
So we have that:
$1+sin(t)+cos(t)+sin(t)cos(t)=5/4$
Say that:
$1-sin(t)-cos(t)+sin(t)cos(t)=x$
Now, let us say that:
$sin(t)+cos(t)=v$
Then we have that:
$5/4+x=1+v^2$
We also have that:
$5/4-x=2v$
Adding the two equations, we get:
$5/2=1+2v+v^2$
$5/2=(1+v)^2$
$sqrt(5/2)=1+v$
So,
$v=sqrt(5/2)-1$
We also have that:
$2v=5/4-x$
Solving, we get that:
$x=13/4-sqrt(10)$
so the answer is:
$13+4+10=27$.
[/hide]

p12: Expand to $\sin t + \cos t + \sin t \cos t = \frac{1}{4}$, or $\sqrt{1+\sin(2t)} + \frac{\sin(2t)}{2} = \frac{1}{4}$. Multiplying by $4$ gives $4\sqrt{1+\sin(2t)} + 2\sin(2t) = 1$. Denote $\sin(2t) = x$, so $4\sqrt{1+x} + 2x = 1 \implies 16(1+x) = (1-2x)^2$, so $x = \frac{5}{2} \pm \sqrt{10}$. The positive solution is extraneous, so $x = \frac{5}{2} - \sqrt{10}$. Note that our answer is just $\frac{5}{4} - 2(\sin t + \cos t) = \frac{5}{4} - 2 \sqrt{1+\sin(2t)}$. We have $1+\sin(2t) = \frac{7}{2} - \sqrt{10}$. Square rooting gives $\frac{\sqrt{10}-2}{2}$, and multiplying by $-2$ gives $2-\sqrt{10}$. Adding $\frac{5}{4}$, we get $\frac{13}{4} - \sqrt{10} \implies \boxed{27}$.

We have $$(1+\sin t)(1+\cos t)-(1-\sin t)(1-\cos t)=2(\sin t+\cos t)\ \quad \text{(I)}$$

Now we are to find the value of $\sin t+\cos t$. Let $\sin t+\cos t=x$. We have $x^2=1+2\sin t\cos t$, so we obtain $\sin t\cos t=\frac{x^2-1}{2}$.

Therefore $$(1+\sin t)(1+\cos t)=1+\sin t+\cos t+\sin t\cos t \iff \frac{5}{4}=1+x+\frac{x^2-1}{2}\iff 2x^2+4x-3=0$$

Since $(\sin t+\cos t)^2+(\sin t-\cos t)^2=2$, we have $(\sin t+\cos t)^2 \leqq 2 \iff -\sqrt{2}\leqq x \leqq \sqrt{2}$, yielding $x=\frac{\sqrt{10}-2}{2}$. Conseqently from $\text{(I)}$:

$$\frac{5}{4}-(1-\sin t)(1-\cos t)=2\cdot \frac{\sqrt{10}-2}{2}\Longleftrightarrow (1-\sin t)(1-\cos t)=\frac{13}{4}-\sqrt{10}$$

Therefore $k+l+m=10+13+4=27$.

[hide]Expanding the first equation, we get $$\sin t\cos t+\sin t+\cos t=\tfrac14.$$ Let $P=\sin t\cos t$ and $S=\sin t+\cos t$. Then, the equation turns into $P+S=\tfrac14$ Expanding $(1-\sin t)(1-\cos t)$, we see that we are asked to find $$\sin t\cos t-(\sin t+\cos t)+1.$$ However, this is simply equal to $P-S+1$. By the Pythagorean identity, we can deduce that $S^2=1+2P \implies P=\tfrac{S^2-1}2$. Plugging this into out first equation, we get that $(S,P)=\left(-1\pm\sqrt{\tfrac52},\tfrac54\pm\sqrt{\tfrac52}\right)$. From here, it is easy to find that $P-S+1=\tfrac{13}4-\sqrt{10}$ to fit with the desired answer form, so $k+m+n=\boxed{027}$.[/hide]


x,y and z positive reals verifying: $xyz=1$ . Given a real t as follow: tx2y2 +tx2z2 +ty2z2 =1.
Show that we have:
xy+xz+yz >= t :sqrt: t(x 2 +y 2 +z 2 )

Put $xy=\sqrt{a} ; yz=\sqrt{b} ; zx=\sqrt{c}$
then $x=\sqrt{ac} ; y=\sqrt{ab} ; z=\sqrt{bc}$
The inequality becomes:
$\sqrt{a}+\sqrt{b}+\sqrt{c}>=\frac{ab+ac+bc}{(a+b+c)^\frac{3}{2}}$
As this inequatity is homogeneous we can suppose that:
$a+b+c=1$
wich is already discussed.


[quote="LideR_M"]See here:
http://en.wikipedia.org/wiki/Billion[/quote]

i ve already seen that.

i want know in mathematics ,how is defined 1 billion.
if i ask one mathematician,what will be his answer?

In my opinion it depends from which country this mathematican is :P

In my oppinion, this question hasn't even to do with math (expecially not Number Theory), but with linguistics ;)

[quote="Prime factorization"]The answer is 1 000 000 000.
A thousand thousands is a million
A thousand millions is a billion
A thousand billions is a trillion[/quote]

Like LideR_M and ZetaX say, this is more of a linguistic issue than a mathematical one.  See http://en.wikipedia.org/wiki/Long_and_short_scales .

I think this is one of the most confusing metrics difference we have.

For the first 20 years of my life a billion was 1,000,000,000,000

Ever since I moved to the US it "Shrunk" to 1,000,000,000

I think the only good thing is that it's such a high number, it's not used very often.

Can you imaging if it was the case with 100 or 1,000  :maybe:


Could anyone with 'scientific technological idea,'
along with descriptive sketches,and 'scientific explanatory terms
relevant to said 'SCIENTIFIC TECHNOLOGICAL IDEA' be submitted for 'PEER REVIEW?'How about if such presented 'tech-
nological ideas' don't have accompanying 'quantitative mathema-
tical description? Can the author submitted it for "peer analytical
commentary?

                                               Respectfully,
                                              A,Eddington

Hello,
     Of course it all depends on what your 'scientific technological idea' is all about.If it is based on already well established physical principles and mathematical backings,then your idea,if original,is worth publishing.However,if it is a new 'physical idea' i.e., a new phenomenon or so to speak,then physicists would expect a mathematical treatment and not merely descriptions.
Sathej