What is the probability that the first three cards drawn from a
standard deck of playing cards will all be two's? Express your
answer as a common fraction.

There are $ \binom{4}{3}\equal{}4$ ways to choose the twos. There are $ \binom{52}{3}\equal{}22100$ ways to choose the 3 cards. Thus it is $ \frac{4}{22100}\equal{}\frac{1}{5525}$

Alternatively, the chance to get it the first times is $ \frac{1}{13}$, 2nd $ \frac{1}{17}$ and 3rd $ \frac{1}{25}$ so $ \frac{1}{13\cdot17\cdot25}\equal{}\frac{1}{5525}$


My school offers the 10A but not the 10B, so I took the 10B at another school. I got a 132 on 10B but failed 10A, so that means they will send my AIME to that other place right? My school offers AIME, so how can I contact the AMC people to send it to my school instead? (amcinfo@maa.org doesn't work)

anyone?

BTW, I asked the contest manager at my school, and he said he wouldn't have any extra copies of AIME since he only gets enough AIMEs for the people who passed the AMCs at that school. So, how do I get the AMC people to send my AIME to my school instead of that other place?

The only way is to email them. I don't know what you mean by the email address "not working."

A kinda related question:

Is it possible to have two AIME testing sites at the same school?
I mean, when I'm doing math, I talk to myself a lot (like whispering 223 plus 415 is 638 or something like that) and the people testing near me 1) find it bothersome or 2) get a free answer.

Although this wouldn't be too convenient in terms of finding proctors...

[quote]contact the AMC people to send it to my school instead? (amcinfo@maa.org doesn't work)[/quote]

The IT staff at the MAA did some work on the maa.org email accounts over the weekend of March 6-7-8.  It is possible that you tried emailing at the point when the account was temporarily down.  Try again, or use the phone number.
In any case, it should be the Contest Managers who are taking care of this, since both need to be informed of the change.

Steve Dunbar
MAA Director, American Mathematics Competitions


How do you solve this?

What is the remainder when the 100th Fibonacci number is divided by 3?

Please correct me if it said something else.

Thanks in advanced.

oh srry, I meant to say divided by 4 :blush:
the answer is 3 by the way.

[quote="shentang"]How do you solve this?

What is the remainder when the 100th Fibonacci number is divided by 4?

Please correct me if it said something else.

Thanks in advanced.[/quote]

take mod 4

1,1,2,3,1,0,1,1,...
repeats every 6, so 4th one, which is 2, I mean 3.


what is a^2 + b^2
a^3 - 2*a*b^2=44
b^3 - 2*b*a^2=8
try to do it wifout finding a and b (that involves solving cubics  :P  :P ).
the answer (wif maple) is 17.something

Are you sure about it? Is it not $3ab^2$ and $3a^b$ instead of $2$?

Pierre.

no, its not 3a*b^2 and 3*a^b but 2a*b^2 and 2b*a^2
whatever made you think of that ?

Because in my case, let $z = a+ib$.
Then it is known that $z^3 = a^3 - 3ab^2 + i(3a^2b - b^3) = 44 - 8i$.
And we are looking for $|z|^2$.

Since $|z|^2 = |z^3|^{ \frac {2} {3} }$, the problem is easy.
In another hand, I don't think about it, but the above may be adapted to produce a solution.

Pierre.


analytic geometry
vector
linear algebra
matrix
function
calculus
derivative

Suppose I have a plane(is that the word generally used?) in 6 dimensional space
u=c1*v+c2+w+c3*x+c4*y+c5*z+c6 where all the ci's are fixed, and a point of coordinates
(m,n,p,q,r,s) where the m coordinate corresponds to the u axis, n coordinate corresponds to v axis, etc.
How can I compute the distance between the point and the "plane"?
Thanks

Can it be called a hyperplane?

My conjecture is that it would be just like in 3D:

$ \left| \frac {c_1\cdot v \plus{} c_2\cdot w \plus{} c_3\cdot x \plus{} c_4\cdot y \plus{} c_5 \cdot z \plus{} c_6}{\sqrt {c_1^2 \plus{} c_2^2 \plus{} c_3^2 \plus{} c_4^2 \plus{} c_5^2}}\right|$

but I'm not quite sure. Looking at Hessian form of the 3D analogue equation (in terms of vectors), I don't see why this wouldn't be correct (as the n-dimensional form could be generalized straight from the 3D vector form).

EDIT: The solution, trivial as it is, probably isn't true at all. After I've read this

[quote="Martin Baker"]When we embed planes in four and higher dimensions the plane can't be defined by a normal vector and the plane will need more scalar values to define it than the number of dimensions.[/quote]

I think there's no Hessian form of equations in higher dimensions.  :maybe:

This can be done with linear algebra...

Suppose that $ v_1$, $ v_2$, ..., $ v_d$ are $ n$ dimensional vectors that form a basis for whatever you are trying to project on. Then let $ A$ be the matrix that has $ v_i$ as the $ i$th column. Now consider $ A x$ where $ x\in\mathbb{R}^d$. It is easy to see that $ Ax$ gives the whole plane. Suppose that we have some point that we want to find the minimal distance. Place this as a column vector, $ b$.

We need to minimize $ Ax \minus{} b$ in the sense that the sum of the squares of the components is minimized.

But we have that $ Ax \minus{} b$ is orthogonal the space that we are trying to project on (hand wave woosh its from my linear algebra book).

Anyway if a vector is orthogonal to all the columns, the dot products of the column with the vector are all zero. It is easy to see that this ends up being:
\[ A^T(Ax \minus{} b) \equal{} 0_n\implies A^TAx \equal{} A^Tb
\]
So we take the solution for $ x$ and plug it into the expression that we were trying to minimize... $ x \equal{} (A^TA)^{ \minus{} 1}A^Tb$ and the minimial distance is
\[ \parallel{} A (A^TA)^{ \minus{} 1}A^Tb \minus{} b\parallel{}
\]

Wait. Maybe I've missed something, but isn't

$ \parallel{} A (A^TA)^{ - 1}A^Tb - b\parallel{} = \parallel{} A A^{ - 1} A^T^{ - 1}A^Tb - b\parallel{} = \parallel{}b - b\parallel{} = 0$?

BTW, with this approach, I'm pretty sure we'll get to my conjectured formula

$ D=\frac{\left| \left[ \begin{array}{cccc} c_1 \\ c_2 \\ ... \\ c_n \end{array} \right]\cdot \left[ \begin{array}{cccc} x_1^0-x_1 \\ x_2^0-x_2 \\ ... \\ x_n^0-x_n \end{array} \right]\right |}{\left| \left[ \begin{array}{cccc} c_1 \\ c_2 \\ ... \\ c_n \end{array} \right]\right |}=\left| \frac {c_1\cdot x_1 + c_2\cdot x_2 + ... + c_n \cdot x_n + d}{\sqrt {c_1^2 + c_2^2 + ... + c_n^2}}\right|$

(I noticed this morning that Baker in the quoted statement didn't mean a plane in this sense (hyperplane), but plain old plane from 3D in higher dimensions).

No $ A$ is not necessarily a square matrix, so $ (A^TA)^{\minus{}1}$ cannot necessarily be simplified in that way.

If $ A$ is a square matrix, thats like saying lets project a point in 3 space into 3 space...the distance is obviously zero. But the interesting case is when the dimension of the space that you are projecting onto is less than the dimension of the space in general (i.e. point to plane, or point to line)

The term is [i]affine hyperplane.[/i] In the general setting of $ \mathbf R^n$ (or in fact any inner product space with its naturally defined norm) it is defined as the set of vectors $ \mathbf x$ for which $ \langle \mathbf a , \mathbf x \rangle \equal{} c$ or $ \langle \mathbf a, \mathbf x \minus{} \mathbf p \rangle \equal{} 0$; since $ \mathbf a$ is orthogonal to $ \mathbf x \minus{} \mathbf p$ by definition, the distance from the hyperplane to some point $ \mathbf s$ is equal to the length of the projection of $ \mathbf s \minus{} \mathbf p$ on $ \mathbf a$. The distance from a point $ \mathbf s$ to the hyperplane $ \langle \mathbf a, \mathbf x \rangle \minus{} c \equal{} \langle \mathbf a, \mathbf x \minus{} \mathbf p \rangle \equal{} 0$ is therefore
\[ d \equal{} \left|\left\langle \mathbf s \minus{} \mathbf p, \frac {\mathbf a}{\|\mathbf a\|}\right\rangle\right| \equal{} \frac {|\langle \mathbf s, \mathbf a\rangle \minus{} \langle \mathbf p, \mathbf a \rangle|}{\|\mathbf a\|} \equal{} \frac {|\langle \mathbf a, \mathbf s \rangle \minus{} c|}{\|\mathbf a\|}.
\]

Another approach, which can be helpful to find the distance from a point to an algebraic variety but works only for $ \mathbf R^n$, or in general to the vanishing (preferably convex) set of any $ C^1$ function is the following: the distance from a point $ \mathbf s$ to the hyperplane $ \langle \mathbf a, \mathbf x \rangle \equal{} c$ is the (unique) local (and therefore global) minimum $ \mathbf x_0$ of $ \|\mathbf x \minus{} \mathbf s\|$ which we find, after calculating the (vector) derivatives $ (\langle \mathbf a, \mathbf x \rangle \minus{} c)' \equal{} \mathbf a$ and $ (\|\mathbf x \minus{} \mathbf s\|)' \equal{} 2(\mathbf x \minus{} \mathbf s)$, to satisfy $ \mathbf x_0 \minus{} \mathbf s \equal{} \mu \mathbf a \equal{} \frac {c \minus{} \langle \mathbf s, \mathbf a \rangle}{\|\mathbf a\|^2}\mathbf a$. The distance from a point $ \mathbf s$ to a line $ \langle \mathbf a, \mathbf x \rangle \equal{} c$ is therefore
\[ \|\mathbf x_0 \minus{} \mathbf s\| \equal{} \frac {| \langle \mathbf a, \mathbf s \rangle \minus{} c |}{\| \mathbf a\|}.
\]
In a similar fashion we could find, say, the distance from a point to an ellipsoid (problem of Appolonius, I think).


AIME15 posted all of his tests, go get them quick! There are a few there I was missing, and then like 10 cds!

aime15mctests.tk

Maybe a link would be in due.

[url]http://www.aime15mctests.tk[/url]

OMG!!

[hide]i can't believe i frickin' fell for it[/hide]

i'm going to take down this site in 2 hrs so admins dont see it


If you want to play write your username and say "I want to play"
To see the rules go
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40669]Here[/url]
But this time we are not naming Jedi, we are naming Hogwarts Students from all 6 HArry POtter books. SO you could name Percy.(He is an example because he was in the 1st book)

The order that the rounds go in is the first person to sign up here will go first!
 :surf:  YAY!!!!!!!!!!!!!

math92
I want to play

\/\/oot!!!!! :surf:

[quote="CLee"]I'll play.  I don't know Harry Potter nearly as well as SW, though. ;)[/quote]
yeah, you are amazing at Star Wars...now i have a chance to win!!! (Evil laugh goes here)!!! :lol:

well, when I get bored, I don't reread Harry Potter books... maybe I can still beat you!  (evil laugh back at math92 - glaring contest begins)

EDIT: plus - I'm at a disadvantage because I haven't yet read Harry Potter 6.  Still, I might hear about some of the characters, so don't all jump on me if I do name a Book 6-only char.

[quote="L_Li"]i'm in... there should be a LOT more harry potter characters than jedi, right?[/quote]
probably, and plus more people know harry potter then SW.

Ah,what the heck! Let's kick off round 1!!!!!! I'll keep track of the chacters named for you. the order that we go in is:
math92
mathclass
CLee
Drunken_Math
 Let us start!  :wow:  VV00T!!!!!

i think we should name 3 people from HW like L_Li said.  We will wait for colts18 for confirmation.

[quote="L_Li"]hey dont i count?

just a suggestion: wouldn't it be easier if we did this game in a blog? players comment in new characters and the host can keep a mutable list of named characters by updating the actual blog entry at the top.
other suggestion: can we name 3 or 5 characters per turn instead? with the huge hp world it'll take forever otherwise... i can think of more muggles from the books than i can jedi... and that's nothing compared to the number of wizards, past and present...[/quote]
Well, I confirm but you know I could specify the characters like instead of just characters, just muggles,or name shops from diagon alley(Though THAT would be hard) but since its only round 1 I'm giving you all a warm up.
I agree we should play it in a blog but we are playing in MY BLOG :!:  And yes, L_Li you count i was reading the topic review and didn't see you. So my blog!!!!!! and others who saw this and want to play tell me at my BLog. math92,you start!

:rotfl:
she's not on often.
we will give her two days.
on Sept 7 she will be eliminated if she doesn't go.

do you not see the sign next to "gender" on every single one of my posts?!?!?

anyway, mwahahaha should know better, cuz i was gone for most of Labor Day weekend and so was he - we were at the same friggin places

Black, Sirius
Lupin, Remus
Pettigrew, Peter

sorry there's this girl in my school who's last name is li. :blush:
sorry.

hey...i thought we were only supposed to name ppl who were at hogwarts as students while harry was...oh well...

severus snape
bill weasley
charlie weasley

Rubeus Hagrid
Bellatrix Black
Luna Lovegood

McGonagall, Minerva
Flitwick, Filius
Dumbledore, Albus

madam rosmerta
prof. sprout
Lord Voldemort

:blush: how do we know that rosmerta, sprout, flitwick, dumbledore were students? does it say that somewhere?

hmmm......i don'[t know......can somebody prove that they were students at hogwarts?

i thought we struck out the must-be-hogwarts-student rule when you named bellatrix (who, by the way, should be a Lestrange, since she's been married the entire duration of the series)

just name human characters in the series, ok?

Zacharias Smith
Hannah Abbott
Susan Bones

is this game over, or are we gonna play?

All teachers at hogwarts came from hogwarts school but flitwik didn't because he isn't a magician (or a teacher).


Why is the AMC website missing links to 2007 state statistics,  perfect score, answers etc. for the AMC 8? I'm pretty sure it was up there earlier. Sorry if this would be considered spam. I'm just wondering if there's a technical glitch or if there was a previously given reason.

If you're wondering why I care, I'm trying to digitally send a picture of myself to mathcounts for nationals, and I don't have that picture on the computer anymore, but I remember it was up there on the AMC 8 website and it would be convenient for me to copy/paste >_<.

Or send someone else's picture... haha.

Look [url=http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2006-8a/2006-amc8Perfect.shtml]here.[/url]

The AMC frequently reorganizes its website, but it seldom throws anything away. Note that since the AMC 8 is a fall event, the 2007 AMC 8 hasn't happened yet; hence I assume you are referring to the 2006 AMC 8.

Wow. Yeah I'm dumb. I kept thinking of it as the 2007 AMC 8

I'm too crazy after state round. lock/delete this thread, please.  :blush:


vector
combinatorics unsolved
combinatorics

We call n-tuple $x=(x_1;x_2;...;x_n)$ with $x_i\in R$ a real vector. The set of all real vectors is usually denote by $R^n$. If x;y are two vectors in $R^n$ : $x=(x_1;x_2;..;x_n) ;y=(y_1;y_2;...;y_n)$ , we define the product of x;y as the real number $x.y=x_1y_1+x_2y_2+...+x_ny_n$.
Find the largest positive integer k such that there exist a set A contains k vectors in $R^n$ satisfying the following condition: if $x;y \in A$ then $x.y\le 0$.


Let $a_{n}$ and $b_{n}$ ($n\in\mathbb{N}$) be two infinite seqences of positive integers. Show the existence of two numbers $p$ and $q$, $p<q$ both such that
\[a_{p}\leq a_{q}\quad\mbox{and}\quad b_{p}\leq b_{q}\]

This is my solution:
Define the sequence $n_{i}$ as the following:
$n_{0}=1$,$a_{n_{i+1}}=\min_{k>n_{i}}\{a_{k}\}$
then $a_{n_{1}}\leq a_{n_{2}}\leq ......$,$n_{1}<n_{2}<....$
and we have that for all $k>n_{i-1}$ ,$a_{k}\geq a_{n_{i}}$
Let us consider $b_{n_{i}}$
there must exist $t$ such that $b_{n_{t+1}}\geq b_{n_{t}}$
The let $p=n_{t},q=n_{t+1}$
we have that $n_{t}<n_{t+1}$ $a_{n_{t}}\leq a_{n_{t+1}},b_{n_{t}}\leq b_{n_{t+1}}$
,
EDIT:Think you ,pco.I have edited.

Nice demo.

Just a little correction :

[quote="Hawk Tiger"]...we have that $n_{t}<n_{t+1}$ $a_{n_{t}}<a_{n_{t+1}},b_{n_{t}}<b_{n_{t+1}}$,[/quote]

$a_{n_{t}}\leq a_{n_{t+1}},b_{n_{t}}\leq b_{n_{t+1}}$

--
Patrick


[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {3}{2}. \frac {a \plus{} b \plus{} c}{\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}}$
 :)[/quote]

Okay.By Am-Gm We have:
$ \frac {3}{2}. \frac {a \plus{} b \plus{} c}{\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}} \le \frac {9}{2(a \plus{} b \plus{} c)}$
By cauchy-swarchz:
$ LHS \ge \frac {9}{2(a \plus{} b \plus{} c)}$
Done! :lol:

[quote="arqady"][quote="caubetoanhoc94"]

Okay.By Am-Gm We have:
$ \frac {3}{2}. \frac {a \plus{} b \plus{} c}{\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}} \le \frac {9}{2(a \plus{} b \plus{} c)}$
[/quote]
Are you sure?[/quote]
I'm sorry.I'm wrong.I will try  :(

I came back :)
Setting:
$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ca \equal{} 1;r \equal{} abc$
Inq
$ < \equal{} > \frac {p^2 \plus{} 1}{p \minus{} r} \ge \frac {3p}{2\sqrt [3]{p^2 \minus{} 2}}$
We have:
$ r \le \frac {p}{9}$
We need prove that:
$ 18(p^2 \plus{} 1).\sqrt [3]{p^2 \minus{} 2} \ge 24p^2$
Setting:$ \sqrt [3]{p^2 \minus{} 2} \equal{} a \ge 1$
So:We prove that:
$ 18(a^3 \plus{} 3).a \ge 24(a^3 \plus{} 2)$
We have:(by Am-Gm)
$ 9a^4 \plus{} 9a^4 \plus{} 9a \ge 27a^3$
And:$ 3a^3 \plus{} 45a \ge 48$
We have done.
Very interesting  :P

[quote="caubetoanhoc94"]$ < \equal{} > \frac {p^2 \plus{} 1}{p \minus{} r} \ge \frac {3p}{2\sqrt [3]{p^2 \minus{} 2}}$
We have:
$ r \le \frac {p}{9}$
We need prove that:
$ 18(p^2 \plus{} 1).\sqrt [3]{p^2 \minus{} 2} \ge 24p^2$
[/quote]
You was wrong again  :wink:
If use pqr for this problem, you should use  Schur ineq.
@ caubetoanhoc94: [hide]Em bi. nguoc. dau' roi` do'  :) [/hide]

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that: $ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \geq\ \frac {3}{2}. \frac {a \plus{} b \plus{} c}{\sqrt [3]{(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}}$
 :)[/quote]
Using the uvw-method we only have to prove it for $ b\equal{}c\equal{}1$ and when $ c\equal{}0$.

I'm sorry.Thanks;nguoivn.I edited
Setting:
$ p \equal{} a \plus{} b \plus{} c;q \equal{} ab \plus{} bc \plus{} ca \equal{} 1;r \equal{} abc$
Inq
$ < \equal{} > \frac {p^2 \plus{} 1}{p \minus{} r} \ge \frac {3p}{2\sqrt [3]{p^2 \minus{} 2}}$
We have:
Case 1:$ p^2 \le 4$
$ r \ge \frac {p(4 \minus{} p^2)}{9}$
So: we need prove that:
$ 6(p^2 \plus{} 1)\sqrt [3]{p^2 \minus{} 1} \ge p^4 \plus{} 5p^2$(1)
Setting:
$ 1 \le \sqrt [3]{p^2 \minus{} 1} \equal{} a \le \sqrt [3]{2}$
$ (1) < \equal{} > 6(a^3 \plus{} 3)a \ge 5(a^3 \plus{} 2) \plus{} (a^3 \plus{} 2)^2$
$ < \equal{} > (14 \minus{} a^5 \minus{} a^4 \plus{} 5a^3 \minus{} 4a^2 \minus{} 4a)(a \minus{} 1) \ge 0$
It's true :P
Case 2:
$ p^2 \ge 4$
$ \sqrt [3]{p^2 \minus{} 1} \equal{} a \ge \sqrt [3]{2}$
We need prove that:
$ 2(p^2 \plus{} 1)\sqrt [3]{p^2 \minus{} 1} \ge 3p^2$
$ < \equal{} > 2(a^3 \plus{} 3).a \ge 3(a^3 \plus{} 2)$
It's true by Am-Gm  :P
P/s:2 lần em đều ngược dấu :(.

@caubetoanhoc94: Besides, you should think about the solution with Am-Gm for this ineq because it's very weak.
And will your way work for this similar result?  :P
http://www.mathlinks.ro/Forum/viewtopic.php?t=282291


algebra
system of equations
algebra unsolved

4. Find all real solutions of the following system of equations:
$ 72x^3 \plus{} 4xy^2 \equal{} 11y^3$,
$ 27x^5 \minus{} 45x^4y \minus{} 10x^2y^3 \equal{} \minus{} \frac {143}{32}y^5$.

What exactly is THAT problem doing in the geometry section? :P

[b]HINT:[/b]
1).$ y \equal{} 0\Leftrightarrow x \equal{} 0.$
2).$ \mbox{If } y\neq 0, \mbox{ then: }$
$ 72x^3 \plus{} 4xy^2 \equal{} 11y^3\Leftrightarrow 72.\left( \frac {x}{y}\right)^3 \plus{} 4.\frac {x}{y} \minus{} 11 \equal{} 0; \mbox{ and if: } \frac {x}{y} \equal{} t$
$ 72t^3 \plus{} 4t \minus{} 11 \equal{} 0 \mbox{ and } t \equal{} \frac {1}{2} \mbox{ solution }\Rightarrow x \equal{} 2y \dots$


geometry
trigonometry
trig identities
Law of Cosines

The quadrilateral $ ABCD$ is cyclic and $ AB=BC=AD+CD$. Given that $ \angle BAD=\alpha$, and the diagonal $ AC=d$, find the area of triangle $ ABC$.

Notation: $ AB=BC=x$; $ CD=b$, $ AD=a$. $ a+b=x$.

Law of cosines:

$ BD^{2}=x^{2}+a^{2}-2ax\cos\alpha=x^{2}+b^{2}-2bx\cos\pi-\alpha$

$ (a-b)(a+b-2x\cos \alpha)=0$

i) if we have $ a=b$, then $ \alpha =\frac{\pi}{2}$.
ii) $ a+b=x\implies \cos \alpha =1\implies \alpha =\frac{\pi}{2}$.

Hence we have $ \alpha=\frac{\pi}{2}$

I bet there there is a nicer argument...

Anyway, we have that $ a=b$, $ a+b=x\implies a=b=\frac{x}{2}$. We have that $ d(A, BD)=\frac{d}{2}$. The problem is trivial now.

$ (x)(0.5 x)=(0.5d)(x0.5*\sqrt{5})\implies d=\frac{2\sqrt{5}}{5}x$

It trivially follows that $ [ABC]=\frac{2}{5}x$.

[Sorry if I made a computation mistake.]

i think you made a factorization mistake here:
[quote="Altheman"]Law of cosines: $ BD^{2}=x^{2}+a^{2}-2ax\cos\alpha=x^{2}+b^{2}-2bx\cos\pi-\alpha$
$ (a-b)(a+b-2x\cos \alpha)=0$[/quote]\begin{eqnarray*}x^{2}+a^{2}-2ax\cos\alpha&=&x^{2}+b^{2}-2bx\cos(\pi-\alpha)\\ a^{2}-2ax\cos\alpha&=&b^{2}+2bx\cos\alpha\\ a^{2}-b^{2}&=&2x(a+b)\cos\alpha\text{, }a+b=x>0\\ a-b&=&2x\cos\alpha \end{eqnarray*}

but otherwise i don't have a solution  :blush:


I've been wondering, as I get constantly confused when I shift alphabetical orders to look up words in my English, Greek, and Sanskrit dictionaries-- where does the 'alphabetical order' of a writing system come from?  (I put 'alphabetical order' in quotes because I want to include all writing systems, not just alphabets, in this discussion.)  I can think of three possibilities:

[b](1)  'Alphabetical order' comes from phonetic properties of the letters.[/b]
Devanāgarī (used for Sanskrit, Hindi, and a few other Indian languages) is a great example-- the letters are ordered by type of sound and place and means of articulation.

[b](2)  'Alphabetical order' comes from some aspect of the shapes of the signs.[/b]  I don't know of any, but some modern arrangements of undeciphered scripts have been made along these lines.

[b](3)  'Alphabetical order' is based off a poem/song/story which uses all of the letters once.[/b]  This would be the Japanese [i]iroha[/i], a poem used as the order for the [i]katakana[/i] syllabary.

[b](4)  'Alphabetical order' doesn't correspond to anything and is used for arbitrary/historical reasons.[/b]  The English alphabet, of course.    Alphabetical order in English traces its origins way way way back-- this order appears as far back as the Ugaritic alphabet ca. 1400 BCE, possibly farther than that.

Can you think of any other ideas, or any other examples of these?

Actually, the most conventional order for the kana syllabaries--the one I have seen in all Japanese dictionaries I own--groups the characters by common initial sounds. I haven't seen a whole book organized in the iroha poem order.


function
algebra
polynomial
calculus
derivative
functional equation

Find all functions $f$ from the real numbers to themselves satisfying \[ x[f(x)+f(y)]=f(x)f(x+y) \] for all real numbers $x$ and $y$.

[quote="shobber"]Put $x=y$, we get $f(x) \left[ f(2x) - 2x \right] = 0$, thus $f(x) \equiv 0$ or $f(x) = x$.[/quote]

And what about

$f(x)=\begin{cases}0 & x\in\mathbb{Q}\\ x & x\notin\mathbb{Q}\end{cases}$

or

$f(x)=\frac{x+|x|}{2}=\begin{cases}0 & x<0\\ x & x\geq0\end{cases}$

or whatever...

Hmmm......I guess the problem meant continuos functions.......

It's nowhere said, and yet if it IS implied, the second example is continuous. It doesn't satisfy the initial equation, but it IS a solution to YOUR equation.

[color=blue][b]HINT[/b][/color]

  [hide]For  $x = y = 0$ we find f(0) = 0. Let for some a we have f(a) = 0 and a is not equal to 0. Puting x = a we take f(y) = 0 for all y, that is f is constant  and equal to 0. If  f(x)  is not equal  to 0  for every x not equal to 0 , then we put  $y = x$ to take $f(x) (f(2x) - 2x)$ or finally $f(x) = x , \forall x \in IR$ (since $f(0) = 0$)[/hide]
   [u]Babis[/u]

I think that to solve a functional equation, we are to find "all" of its roots!

And indeed "all" solutions of this functional equation are $f(x)\equiv x$ and $f(x)\equiv 0$.  Stergiu's solution suffices, I think.  Alternately, here is my own proof of the difficult bit:
[hide]We know that for each $x$ either $f(x)=x$ or $f(x)=0$.  Suppose that, for some (nonzero) number $k$, $f(k)\neq k$.  We know then that $f(k)=0$.  Letting $x=k$, we see that \[ k[f(k)+f(y)]=f(k)f(k+y)  \] for all $y$, or, rearranging and substituting 0 for $f(k)$, that \[ f(y)=\frac{0}{k}=0.  \]  So the only possible solution to the equation apart from the identity function is the zero function.[/hide]

But if check Farenhajt's functions, they are also correct.

As Farenhajt says,

[quote="Farenhajt"]It doesn't satisfy the initial equation, but it IS a solution to YOUR equation.[/quote]

[quote="10000th User"][quote="Farenhajt"][quote="Agrippina"]Stergiu's solution is correct; it is not necessary to assume continuity or similar to rule out Farenhajt's "crazy" solutions.  :)[/quote]

My "crazy" solutions have another point, which happens to be the reason for many mistakes: From $(g(x)-p)(g(x)-q)=0$ it CAN'T be derived $g(x)\equiv p\lor g(x)\equiv q$[/quote]
Farenhajt is right, the implication is not generally correct [u]unless[/u] we assume $f$ is continuous. Same thing with zero divisors that appear in mod arithmetic $\mathbb{Z}_k$ when $k$ is not prime.[/quote]

And even with the continuity assumption, the implication may not hold (cf. my example of $f(x)=(x+|x|)/2$, which is continuous, though not differentiable). The thing is in the well-known moot point in formal quantifier algebra: $(\forall z)[P\lor Q]\nRightarrow [(\forall z)P]\lor[(\forall z)Q]$ Of course, wrong reasoning leads to a correct result often enough, but that doesn't make it less wrong.

Indeed, this logical distinction is the point of the problem.   [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=17&year=2001&p=119165]ISL 2001 A4[/url] is an example of a problem in which failing to make it leads to an incorrect answer.

Well , this comunication is very usefull and makes clear the fact that

         $f(x) g(x) = 0 , x \in I$ does not imply

      $f(x) = 0 , x \in I or g(x) = 0 , x \in I$

    But this is true , if $f , g$  are [b]polynomials[/b]. Even  $f , g$  have derivative ,the proposition is wrong!

       See also my problem :
                 http://www.mathlinks.ro/Forum/viewtopic.php?highlight=find&t=70005

  Farhenhajt has there good examples. The problem is now solved and I hope Mlinkers will study these posts carefully.

    [u]Babis[/u]


When forming covalent bonds, which atom can have the more than eight valence electrons?

a) H
b) N
c) F
d) Cl

[hide]
D
[/hide]

can someone provide an explanation as well?

an atom can have more than 8 electrons only when it can have a d-shell.  The first 3 answer choices all have principal quantum number n = 1 or n = 2 in their highest electron energy state.  Cl on the other hand, has electrons w/ principal quantum number n = 3.  This allows Cl to have d orbitals, which are responsible for the "expanded octet"

wah? the electron configuration is

$1s^22s^22p^63s^23p^5$ isnt it? or maybe its $1s^22s^22p^63s^13p^6$ but either way i dont think theres a d-subshell..

dragon slayer means in n=3...there's the s, p, and d, subshells

oh... so it has a d-subshell even tho its usually empty... gotcha


vector
function
trigonometry
integration
linear algebra

let's consider the real vector field of continuos functions $ f: \mathbb{R}\minus{}>\mathbb{R}$:
how can I prove that the set $ \{sin(nx): n\in\mathbb{N}, n\neq0\}\bigcup\{cos(nx): n\in\mathbb{N}\}$ is linearly independent?

There are many possible proofs. One possibility is to introduce an inner product, even though the space wasn't given as an inner product space.

A finite linear combination of these functions is a trigonometric polynomial.

Let $ T(x)\equal{}\sum_n(a_n\cos nx\plus{}b_n\sin nx)$ where this is a finite sum.

Suppose $ T(x)$ is the zero function - that is, the zero element of the vector space. Then

$ \int_0^{2\pi}[T(x)]^2\,dx\equal{}0.$

But also, by direct computation,

$ \int_0^{2\pi}[T(x)]^2\,dx\equal{}2\pi a_0^2\plus{}\pi\sum_{n>0}(a_n^2\plus{}b_n^2)$

Since this is a sum of squares, the only way for the sum to be zero if for each and every coefficient to be zero. Hence $ a_0\equal{}a_1\equal{}a_2\equal{}\cdots\equal{}b_1\equal{}b_2\equal{}\cdots\equal{}0.$

Hence the set is independent.

Language note: this is a vector space, not a vector field. The term "vector field" denotes a vector-valued function on a manifold.

thanks a lot; another question, is this a finite or infinite space?

It's infinite (as is every vector space of positive dimension over an infinite field), but the much more interesting claim is that it's infinite-dimensional.  This follows immediately from the result of the original post.


April. May, and Jun are siblings. April's present age is the sum of the present ages of May and Jun. May's age at the time Jun reaches the present age of May will be twice the present age of Jun. If the sum of the present ages of April and May is 24, what is the sum of present ages of April, May, and Jun?

this is not me speaking its my little sister! im just letting her use my account!

[color=darkblue]Solution[/color]

[hide]If A+M=24
And M+J=A
Their ages must be
A=15
M=9
J=6[/hide]


The COP The Third will take place in here at March 7th.

The students who want to participate in this must sign up till March 7th. If you happen to miss the date, however, you may still send the answers and it will be counted as official answer.

Good luck to you all!

Have any questions?

Ask here!

Practice 5 COP questions are posted in my blog.

Please visit my blog to solve them.

Thanks.


algebra
polynomial
boundary
complex analysis
Convex hull

Let $f: \mathbb{C}\rightarrow\mathbb{C}$ be the polynomial . Prove that all roots of derivate of $f$ are element of convexhull of roots of $f$ and if roots of $f$ are not colinear they can not be on the boundary of convexhull .[excause me for useing ''of'' almost after every word :oops: ]

Saman Gharib

Are your sure with the boundary part¿
E.g. take $(x^3-1)^2$, it's roots are not colinear, but some roots of the derivate are on the boundary.

excause me .we must add this condition that [for boundary part] the root of $f'$ must be simple[not multiple].
Thank you ZetaX for your attention.

Saman Gharib

In my very first [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=12915]post[/url] on this forum, I stated and proved this theorem as a tool for the problem. The key? Look at half-planes.

I didn't expect you to know that- that was before the AOPS-Mathlinks merge, and you didn't even join until seven months later. I didn't know the name of the theorem when I made that post, either- a search might have turned it up since the name was mentioned later in the thread, but the time sort would put that very low on the list of places to check.

It's a bit of nostalgia for me- back when I was just another new member.


stevem told me before to use \cfrac for large fractions, but I've recently read elsewhere that \dfrac also works. What's the difference between the two? So far I haven't noticed anything..

\dfrac stands for "displayed fraction".  Here is an example:
\[ \dfrac{1}{2 \plus{} \dfrac{1}{3}}
\]
\cfrac stands for "continued fraction".  Here is an example:
\[ \cfrac{1}{2 \plus{} \cfrac{1}{3}}
\]
See the difference?  I don't know which one I prefer in this case.  Does anyone have a preference?

hmm.. I see now, but.. then what makes \dfrac different from \frac, then? :blush:

\dfrac means "displayed fraction", and looks the same (large) in display mode or inline mode.  \tfrac means "text fraction", and looks the same (small) in both modes.  Compare the dfrac $ \dfrac{1}{2}$ and the tfrac $ \tfrac{1}{2}$.

In displayed math, \frac is the same as \dfrac.  In inline math, \frac is the same as \tfrac.  In my documents, I tend to use \frac, which lets LaTeX decide the proper size.

Note: On this forum, I believe \frac always means \dfrac.


real analysis
algebra
polynomial
integration
real analysis unsolved

Let $ m^*$ denote Lebesgue outer measure on $ \Re^2.$
If $ p(x,y)$ is a nonzero polynomial and $ Z\equal{}\{(x,y)\in \Re^2 | p(x,y)\equal{}0\},$ show that $ m^*Z\equal{}0.$

Let $ Z_{x_0} : \equal{} \{y \in \mathbb{R} \  | \  p(x_0, y) \equal{} 0\}$. There are two cases which depend on $ x_0$: $ Z_{x_0}$ is finite or $ Z_{x_0} \equal{} \mathbb{R}$. Let $ C \subset \mathbb{R}$ be the set of those $ x_0$ for which $ Z_{x_0} \equal{} \mathbb{R}$. $ C$ is a finite set. Therefore, from Fubini theorem we have

\[ m(Z) \equal{} \int_{\mathbb{R}^2} \chi_Z \equal{} \int_\mathbb{R}\int_C \chi_Z \plus{} \int_\mathbb{R}\int_{\mathbb{R}\setminus C} \chi_Z \equal{} 0 \plus{} \int_{\mathbb{R}\setminus C}\int_\mathbb{R} \chi_Z \equal{} 0.\]


Lines $l,m,n$ are parallel to each other with $m$ between $l$ and $n$.The distance between $l$ and $m$ is $p$, and that between $m$ and $n$ is $q$.Take points $A,B$ and $C$ on $l,m$ and $n$ respectively such that triangle $ABC$ is an equilateral triangle.Let $D$ be the intersection point between $m$ and $AC$.
Express the following values in terms of $p$ and $q$.

(1) Length of side $AB$
(2) Length of segment $BD$
(3) $\sin \angle ADB$

[quote="levi"]$3\cdot AB^2=4(p^2+q^2+pq).$ a.s.o.[/quote]

Your answer is correct,levi.How did you solve it?


Does anyone know who actually started Mathcounts? I can't find that anywhere... (read: on the first page of google results)

everyone knows it was Frederick W. Mathcounts. ;-)
jk. seriously though, i don't know. it may have just been started by an association rather than an individual person.

I don't believe that MATHCOUNTS was founded by one individual.
Rather, it was founded by a program and/or sponsered by some engineer program/thing.
From there, it all gets kinda fuzzy for me.
Hope that helped. A little. ;)

Well, thanks for the response, but I really need the specific program/association that founded it, along with a reliable source. It's for a Wikipedia article...

Ok...?
Well umm, I guess you could try to find out when MathCounts started and go from there.

I already know: 1982.

That doesn't help.

Google.  :lol:
[quote="[url=http://www.gm.com/explore/education/news/2007/mathcounts_061907.jsp]GM.com[/url]"]
The National Society of Professional Engineers, National Council of Teachers of Mathematics and CNA Foundation founded MATHCOUNTS.

Additional information on MATHCOUNTS is available at http://www.mathcounts.org.
[/quote]

Though a couple links say the National Society of Professional Engineers founded it and no other organizations.


pigeonhole principle
combinatorics unsolved
combinatorics

In a party there are $ n$ guests. Each of them belongs to exactly one [i]clan[/i] and one [i]group[/i]. There are $ 10$ groups: $ G_1, G_2,...,G_{10}$ and $ 10$ clans: $ C_1,C_2,...,C_{10}$. Find the minimum of $ n$ such that there exists at least $ 11$ guests such that everyone of them belongs to the group and the clan with the same identified number.

I am not sure that I understand this problem. You are either missing a condition (because, for instance, every guest might belong to $ G_1$ and $ C_2$ (where $ 1\neq 2$) or you meant to say that $ \geq 11$ guests belong to the same $ (G_a, C_b)$, in which case the answer follows by trivial Pigeonhole ($ 1001$).

Thank you Mr.Vworldv, you have the same thinking with me. Actually I do not sure I did translated this problem into English right $ 100$%. It is one of the homework problems of my friend but I do not understand in the same way with you. His teacher hinted that the answer is $ n\geq 101$ (not $ 1001$ :( ). I think he was wrong...


function
integration
logarithms
calculus
derivative
inequalities
analytic geometry

[i] Prove that
\[ \frac{\pi (x)}{x}+\int_2^x\frac{\pi (u)}{u^2}\mathrm{d}u>\ln\ln x-1 \]
[/i]

Masoud Zargar

we should proove a lemma:
$\displaystyle \sum_{p \leqslant x} \frac{1}{p} \geqslant \ln \ln x - 1.$

+we can proove easily that $\displaystyle ln(x) \leq \sum_{n \leqslant x} \frac{1}{n} \leqslant \prod_{p \leqslant x} \left ( 1 + \frac{1}{p} + \frac{1}{p^2} + ... \right ).$
so $\displaystyle ln(ln(x)) \leq ln \left(\sum_{n \leqslant x} \frac{1}{n}\right) \leqslant ln \left(\prod_{p \leqslant x} \left ( 1 + \frac{1}{p} + \frac{1}{p^2} + ... \right )\right).$
so $ln(ln(x)) \leq \sum_{p \leqslant x} ln \left(\left ( 1 + \frac{1}{p} + \frac{1}{p^2} + ... \right )\right).$
so $ln(ln(x)) \leq 1+ \sum_{p \leqslant x}\frac1p$

So lemma prooved

come back to the exercice
let $\pi(x)=n$
$\int_{0}^{x}\frac{\pi(u)}{u^2}du=\sum_{i=0}^{n}\int_{p_i}^{p_{i+1}}\frac{\pi(p_i)}{u^2}+\int_{p_n}^{x}\frac{\pi(p_n)}{u^2}du^$
$=\sum_{i=1}^{n-1}i(\frac{1}{p_i}-\frac{1}{p_{i+1}} )+n(\frac{1}{p_n}-\frac1x)$
$\int_{0}^{x}\frac{\pi(u)}{u^2}du +\frac{\pi(x)}{x}=\sum_{p \leq x}\frac{1}{p} \geq ln(ln(x))-1$
%

marocleverness, putting the word "easily" in your proof is not a benefit to the reader, because it's your interpretation.

The correct form of the words that you repeatedly spelled wrong that deal with  math are "proved" and "exercise."  Please learn
to spell these.

Note:  Just for this use, let pi*(x) = 1st derivative of pi(x), where
pi(x) is the number of prime numbers less than or equal to x.

Take derivatives of each side of the given inequality:

[xpi*(x) - pi(x)(1)]/x^2 + pi(x)/x^2 > (1/lnx)(1/x)

[xpi*(x) - pi(x) + pi(x)]/x^2 > 1/(xlnx)

xpi*(x)/x^2 > 1/(xlnx)

Cancelling an x out on the left hand side leaves:

pi*(x)/x > 1/(xlnx)

Multiply both sides by x (x is a positive number):

p*(x) > 1/lnx  ###

Establish ###.

x  1  2  3  4 .......k-1.........k.....
y  0  1  2  2 ...pi(k-1)....pi(k)....

pi*(k) = [y(sub 2) - y(sub 1)]/[x(sub 2) - x(sub 1)]    <The slope>

pi*(k) = [pi(k) - 0]/(k - 1)

pi*(k) = pi(k)/(k - 1)

Take derivatives of both sides of ###:

[(k-1)pi*(k) - pi(k)(1)]/(k - 1 )^2 > -1/[x(lnx)^2]

Replace pi*(k) with pi(k)/(k - 1):

{(k-1)[pi(k)/(k-1)] - pi(k)}/(k-1)^2 > -1/[x(lnx)^2]

[pi(k) - pi(k)]/(k-1)^2 > -1/[x(lnx)^2]

0 > -1/[x(lnx)^2]

Because x is positive, the right-hand side is always negative.

$\pi(x)$ is not differentiable!  Furthermore, where it is differentiable, it is zero (I'm not counting in delta functions for obvious reasons) which makes inequalities like $\pi'(x)>(\ln x)^{-1}$ patently false.

blahblahblah,

make up your mind!

First you state that it's not differentiable.  Then in the next sentence you ALLOW that it is in certain places.  Can't have it both ways.  I volunteer to knock down your 5 star rating.

There's an understanding in that post that a differentiable function refers to a function that is differentiable over its entire domain.  But I can hardly expect you and your two alter egos to pick up on something so subtle.

To moderator
move this to College Playground /Analysis/ Proposed problem

thank you

It is NOT "an understanding in that post" that it refers to its entire domain; it is a FACT!  To state it is an understanding, is to interject your opinion, just as using the bogus "easily."  I can play Devil's Advocate with the material.

YOU "can hardly expect" because you are weak-minded.

Describing something as "so subtle" is, AGAIN, another subjective hole you've fallen into because yo have a perpetual
sickness that you have shown no improvement of recovering from.

Two alter egos?  Don't pat yourself on the back!
You can't count them on one hand.

Judgment:  You get even MORE stars subtracted from under your
name than before!

You misspelled 'you' and misused the word 'improvement'.  If you're going to compose posts devoid of mathematical content, at least try and keep them grammatically correct.

Wait - do I lose more stars for that?  Is it possible for me to have a negative number of stars?  Are you going to create another screenname to insult me?  Will you call me weak-minded again?  Please, anything but that.  :(  :(  :(


calculus
inequalities unsolved
inequalities

Given $ a, b, c > 0$ satisfy $ abc \equal{} 1$. Prove that:
$ \frac{1} {3a \plus{} 2b \plus{} c} \plus{} \frac {1} {3b \plus{} 2c \plus{} a} \plus{} \frac {1} {3c \plus{} 2a \plus{} b} \leq\ \frac {1} {2}$

[quote="nguoivn"]Given $ a, b, c > 0$ satisfy $ abc \equal{} 1$. Prove that:
$ \frac {1} {3a \plus{} 2b \plus{} c} \plus{} \frac {1} {3b \plus{} 2c \plus{} a} \plus{} \frac {1} {3c \plus{} 2a \plus{} b} \leq\ \frac {1} {2}$[/quote]
$ <\equal{}> 11(a^2\plus{}b^2\plus{}c^2)\plus{}25(ab\plus{}bc\plus{}ca) \le \frac{1}{2}(54\plus{}6(a^3\plus{}b^3\plus{}c^3)$ $ \plus{}25(a^2b\plus{}b^2c\plus{}c^2a)\plus{}23(b^2a\plus{}a^2c\plus{}c^2b))$
$ LHS \le 12(a\plus{}b\plus{}c)^2$ , $ 3(a\plus{}b\plus{}c)^3\plus{}27 \le RHS$ ;
$ t \equal{} a\plus{}b\plus{}c \ge 3$
so we must prove : $ 12t^2 \le 3t^3\plus{}27$ for $ t \ge 3<\equal{}> (t\minus{}3)(t^2\minus{}t\minus{}3) \ge 0$ for $ t \ge 3$ :)

Thank you. I can solve it easily by AM-GM.
Indeed, I don't like to calculus :D

And the stronger:
$ \frac{1}{(3a\plus{}2b\plus{}c)^2} \plus{} \frac{1}{(3b\plus{}2c\plus{}a)^2} \plus{} \frac{1}{(3c\plus{}2a\plus{}b)^2} \leq\ \frac{1}{12}$

[quote="nguoivn"]And the stronger:
$ \frac {1}{(3a \plus{} 2b \plus{} c)^2} \plus{} \frac {1}{(3b \plus{} 2c \plus{} a)^2} \plus{} \frac {1}{(3c \plus{} 2a \plus{} b)^2} \leq\ \frac {1}{12}$[/quote]
Hint:
\[ (3a\plus{}2b\plus{}c)^2 \ge 4(a\plus{}2b)(c\plus{}2a)\]
:)

[quote="can_hang2007"][quote="nguoivn"]And the stronger:
$ \frac {1}{(3a \plus{} 2b \plus{} c)^2} \plus{} \frac {1}{(3b \plus{} 2c \plus{} a)^2} \plus{} \frac {1}{(3c \plus{} 2a \plus{} b)^2} \leq\ \frac {1}{12}$[/quote]
Hint:
\[ (3a \plus{} 2b \plus{} c)^2 \ge 4(a \plus{} 2b)(c \plus{} 2a)
\]
:)[/quote]
Nice! :lol:

We can also using Am-Gm with form: $ (3a \plus{} 2b \plus{} c)^2 \geq\ 2(a \plus{} b \plus{} c)(2a \plus{} b)$
By both of ways, we'll obtain:
$ 9(a \plus{} b \plus{} c) \leq\ (2a\plus{}b)(2b\plus{}c)(2c\plus{}a)$
It's equivalent to:
$ (a^2\plus{}2bc)(b^2\plus{}2ca)(c^2\plus{}2ab) \geq\ 9abc(a^2b\plus{}b^2c\plus{}c^2a)$
pqr works here
 :)


Acute $\triangle ABC,\ P\in [OH]\Longrightarrow PA+PB+PC\le 3R$.  (from [u]Caesar Lupu[/u])

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=19316 .

  darij


How would you evaluate something like $ \left(\frac {1}{2}\right)!$?

There's some sort of formula.  You wouldn't do it in the normal way.  Check wikipedia.

You would use the [[gamma function]].  There is no elementary definition of this function -- you need to know integral calculus to define it, and rather more to understand why it's important ;)

hm could someone show me how to evaluate that for the $ n\equal{}3/2$ case.

The major (elementary) property of the Gamma function $ \Gamma$ is that

\[ x! \equal{} \Gamma(x\plus{}1) \equal{} x \cdot \Gamma(x).\]

So $ (3/2)! \equal{} \Gamma(5/2) \equal{} 5/2 \cdot 3/2 \cdot \Gamma(1/2)$. Then use the fact that $ \Gamma(1/2) \equal{} \sqrt{\pi}$.


function
limit
real analysis
real analysis unsolved

Can anyone give or direct me to clear and concise proofs of the following four questions?

[img]http://i21.photobucket.com/albums/b256/lando-45/qs.jpg[/img]

These are all pretty straightforward applications of the definition of limit. So, what have you tried so far?

2) Let $ I$ be an interval containing the point $ a$. Let $ f$, $ g$, and $ h$ be functions defined on $ I$, except possibly at $ a$ itself. Suppose that for every $ x\in I$ not equal to $ a$, we have $ g(x) \leq f(x) \leq h(x)$.  Also suppose that $ \lim_{x \to a} g(x) \equal{} \lim_{x \to a} h(x) \equal{} L$.  Then $ \lim_{x \to a} f(x) \equal{} L$.

Proof
We first prove the special case where $ g(x) \equal{} 0$ for all $ x$ and $ L \equal{} 0$. In this case $ \lim_{x \to a} h(x) \equal{} 0$.  Let $ \epsilon>0$ be any fixed positive number. By the definition of the limit of a function, there is a $ \delta>0$ such that if $ 0<|x\minus{}a|<\delta$, then $ |h(x)|< \varepsilon$.  For any $ x\in I$ not equal to $ a$, $ 0\equal{}g(x)\leq f(x) \leq h(x)$ so that $ |f(x)| \leq |h(x)|$.  We conclude that if $ 0 < |x \minus{} a| < \delta$, then $ |f(x)| \leq |h(x)| < \varepsilon$.  This proves that $ \lim_{x \to a} f(x) \equal{} 0 \equal{} L$.

This completes the proof for the special case. Now, we prove the general theorem by letting $ g$ and $ L$ be arbitrary. For any $ x\in I$ not equal to $ a$, we have $ g(x) \leq f(x) \leq h(x)$.  Subtracting $ g(x)$ from each expression gives$ 0 \leq f(x) \minus{} g(x) \leq h(x) \minus{} g(x)$. As $ x \rightarrow a$, $ h(x) \rightarrow L$ and $ g(x) \rightarrow L$, so that $ h(x) \minus{} g(x) \rightarrow L \minus{} L \equal{} 0$.  The special case now shows that $ f(x) \minus{} g(x) \rightarrow 0$. We conclude that $ f(x) \equal{} (f(x) \minus{} g(x)) \plus{} g(x) \rightarrow 0 \plus{} L \equal{} L$.


[quote="rhbestrh"]We can multiply two numbers together in the following way. Write the two numbers down next to each other. Divide the first in half, rounding down to an integer and write this result below it. Double the second number, writing this result below it. We continue this halving/doubling until we are left with 1 in the first column. Cross out all those numbers in the second column that are opposite an even number and add the remaining numbers in this column together to get the product. Prove that this works.[/quote]This is simply binary multiplication in disguise. Write the two numbers in binary representation and everything will be clear.


[quote="scpajmb"]let  $x,y,z > 0,xyz \ge 1,p \ge 1$, Prove or disprove

$\frac{x}{{x^p + y + z}} + \frac{y}{{x + y^p + z}} + \frac{z}{{x + y + z^p }} \le 1$[/quote]
No,I don't think it is just yourself.Vasc has metioned it before ;)


calculus
derivative
function
inequalities
arithmetic sequence
real analysis
real analysis unsolved

Suppose $f:\mathbb{R}\to\mathbb{R}$ is a differentiable function for which the inequality $f'(x) \leq f'(x+\frac{1}{n})$ holds for every $x\in\mathbb{R}$ and every $n\in\mathbb{N}$.Prove that f is continiously differentiable

The condition tells us that for all $n$, the function $f_n(x)=f\left(x+\frac 1n\right)-f(x)$ is increasing.

Take any two reals $x<y$.

Assume first that $y-x\in\mathbb R\setminus\mathbb Q$. In this situation, because of the density of the set $\{n(y-x)\}$ in $(0,1)$, we can find a sequence $n_k\to\infty$ s.t. $\frac{y-x-\frac {m_k}{n_k}}{x+\frac{m_k+1}{n_k}-y}\to 0$ as $k\to\infty$, where $m_k$ is the largest positive integer for which $x+\frac{m_k}{n_k}<y$. A simple computation will then show that $\frac{f\left(x+\frac{m_k+1}{n_k}\right)-f\left(x+\frac{m_k}{n_k}\right)}{\frac 1{n_k}}\to f'(y)\ (*)$ as $k\to\infty$. On the other hand, $\frac{f\left(x+\frac 1{n_k}\right)-f(x)}{\frac 1{n_k}}\to f'(x)\ (**)$. From $(*),(**)$, and the fact that $f_{n_k}$ is increasing for all $k$, we find that $f'(x)\le f'(y)$.

If $y-x\in\mathbb Q$, it's even easier to show that $f'(x)\le f'(y)$: we just take $n_k\to\infty$ s.t. the denominator of $y-x$ divides all $n_k$.

We can thus show that $f'$ is increasing, and, since it has the intermediate value property, it must be continuous.

Since $f(x+1/n)-f(x)$ increases, we conclude that $f$ is convex on any arithmetic progression with rational denominator. Next, since $f$ is continious, $f$ is just convex. So, $f'$ increases (since if $a<b$, then $f'(a)\le \frac{f(b)-f(a)}{b-a}\le f'(b)$ by the very definition of a derivative and convexity). Then proceed as grobber: by Darboux theorem, $f'$ takes intermediate values and increases hence it is continious.


$ n,m$ are defined to be natural numbers. $ a$ is constant and defined to be $ 0 < a < 1$.

$ log_2 6 \equal{} m \plus{} \dfrac{1}{n \plus{} a}$

1.Calculate $ m,n$

2,Prove $ a > \dfrac{2}{3}$

1. $ n\in\mathbb{N},\ 0 < a < 1\Longrightarrow 0 < \frac {1}{n \plus{} a} < \frac {1}{n}\leq 1$.

$ \therefore m \equal{} [\log_2 6],\ \frac {1}{n \plus{} a} \equal{} \{\log_2 6\}$. Since $ 2^2 < 6 < 2^3\Longleftrightarrow 2 < \log_2 6 < 3\ \therefore \boxed{m \equal{} 2}$.

$ \therefore \frac {1}{n \plus{} a} \equal{} \log_2 6 \minus{} 2 \equal{} \log_2 \frac {3}{2}$, then $ \sqrt {2} < \frac {3}{2} < 2 \Longleftrightarrow \frac {1}{2} < \frac {1}{n \plus{} a} < 1$

$ \Longleftrightarrow 1 < n \plus{} a < 2$, yielding $ \boxed{n \equal{} 1}$.

2. $ a > \frac {2}{3}\Longleftrightarrow \frac {1}{1 \plus{} a} < \frac {3}{5}$.

Proof: $ \frac {3}{5} \minus{} \frac {1}{1 \plus{} a} \equal{} \frac {3}{5} \minus{} \log_2 3 \equal{} \frac {1}{5}\log_2 \frac {8\cdot 2^5}{3^5} > 0\ Q.E.D.$


How many pairs of positive integers (a,b) are there such that gcd (a,b) = 1 and

(a/b) + (14b/9a)

is an integer?

[hide="Solution"]$ 9a^2 \plus{} 14b^2 \equal{} 9abk$

$ 14b^2\equiv 0 \bmod a$, which is impossible unless $ 14\equiv 0\bmod a$, so $ a|14$

For the same reason, $ b|9$.

Therefore $ b\in \{1,3,9\}$ and $ a\in\{1,2,7,14\}$.

If $ b \equal{} 9$, the RHS is divisible by 81, but the LHS is $ 9\bmod 81$, which makes no sense.

If $ b \equal{} 1$, $ 9a^2 \plus{} 14 \equal{} 9ak$, which is impossible.

If $ b \equal{} 3$, $ 9a^2 \plus{} 14\cdot 9 \equal{} 9\cdot 3\cdot ak\implies a^2 \plus{} 14 \equal{} 3ak$. Checking $ \bmod 3$, all the $ a$ work, so there are $ \boxed{4}$ solutions.[/hide]

Edit: Also, there is a resources section for a reason (http://www.artofproblemsolving.com/Forum/viewtopic.php?p=767742)

[quote="21 7 15"]i got 3 instead of 4 but i'm still going through it[/quote]

[hide="I would bet you a million dollars that you are wrong..."]...mainly because 3 is not an answer choice.[/hide]


Lenght of sides of a cyclic pentagon are $ a, b, c, d, e$ given. How can we find the area of cyclic pentagon?

for example: Lenght of sides of cyclic pentagon are $ 7, 8, 9 , 10$ and $ 12$. What is its area? (Thank you for intrest)

Here comes some links:

[url]http://mathworld.wolfram.com/CyclicPentagon.html[/url]

[url]http://arxiv.org/PS_cache/math/pdf/0403/0403503v1.pdf[/url]


geometry
perimeter
area of a triangle
Heron\u0027s formula

Ok, does anyone know how to solve this? I might be looking at the question wrong, but anyways... tell me if anyone can get the answer to this:

wait wait, area is A and perimeter P and apotem r
$A=\frac{1}{2}rP$
$r=\frac{2A}{P}$
???
why two different answers?

The formula for the area $A$ of a triangle with semiperimeter $s$ and circle with radius $r$ inscribed in it, is $A=rs$, so we have $r=A/s$ and the problem is easy. :D

[quote="10000th User"]but why this is calculator problem "calculator icon"?
sides of triangle a,b,c
$a=c_1+r$
$b=c_2+r$
$c=c_1+c_2$
solve for r $2r=a+b-c$[/quote]

Where did you get this from? You should define your variables.

[quote="10000th User"]but why this is calculator problem "calculator icon"?
sides of triangle a,b,c
$a=c_1+r$
$b=c_2+r$
$c=c_1+c_2$
solve for r $2r=a+b-c$[/quote]
ok sry this is wrong, because 2-3-4 triangle is not right triangle  :oops:

[quote="donnyouyang"]Whoa... I'm lost.[/quote]

Oh sorry :blush: I don't know what level you're at, so I'll keep it nice and basic:

Let us denote the area of a triangle by $A$. The circle in the triangle is [i]inscribed[/i], meaning that it is tangent to all three sides of the triangle, and is contained inside the triangle. Let us call the radius of the inscribed circle $r$. (in this problem, however, it is called $x$.) Let us define the [i]semiperimeter[/i] $s$ of a polygon to be half of the perimeter, so in this case it is $s=\frac{2+3+4}{2}=\frac{9}{2}.$ A famous theorem you should learn states that:

\[A=rs\]

In this case, $2.9047=x \cdot (9/2) \Rightarrow x=\frac{2 \cdot 2.9047}{9}$. You can use a calculator to find this value.

Hope you understand now. :)

First to find the area of the triangle, apply Heron's formula, so you get:

$4.5(4.5-3)(4.5-2)(4.5-4)=8.4375$, Now square root it, so you get $2.90473751$ which is the area of the triangle. So then use $rs=A$ and so $r(4.5)=2.9$, so $r=.644...$. Yes, I did it the laziest way possible but atleast it's right :P

EDIT: Oh my god, I didn't even see that it allready gave the area of the triangle, feel stupid now :blush:

[quote="Iversonfan2005"]Yes, I did it the laziest way possible but atleast it's right :P[/quote]

All good mathematicians are lazy. Well, at least in math. Always look for the simplest solution.

[hide]Let $A$ be the Area of the Triangle, $r$ be the Radius of the Circle, and $s$ be the Semiperimeter of the Triangle. Thus, using chess64's Formula,
$r=\frac{A}{s}=\frac{2.9047}{4.5}\approx 0.645$[/hide]
Extension.
Prove the Heron's Formula.

Heron's formula can be proven by taking a triangle, drawing an altitude, letting it be h, letting the two segments created be x and c-x, and then solving for area in terms of h.

$\text{Heron's Formula}$ provides the area of the triangle. So the only thing we need to know is the length of the sides.

I'm not sure if you know Heron's formula or these other formulas, so let me try and explain so you can see where they come from.

First, let's label the points of the triangle: A, B, and C.  We'll call the center of the circle O.  If you draw lines from each of the vertices to O, you form 3 triangles within triangle ABC.  I'm sure you know the area of a triangle is $\frac{1}{2}bh$, where b is base and h is height or altitude.  To find the area of the whole triangle, you can sum up the areas of the smaller triangles.  And for the smaller triangles, the altitudes are the radii of the circle.

For this problem:

$A = \frac{2r}{2} + \frac{3r}{2} + \frac{4r}{2} = r \cdot \frac{2 + 3 + 4}{2} = \frac{9r}{2} = rs$

Then you can plug in and solve.

Yep, this is the easiest way to solve this problem. But if we only know the length of the 3 sides, then we can use Heron's formula to find the area, it states like that:
\[S=\sqrt{p(p-a)(p-b)(p-c)}\]
Where $p=\frac{1}{2}(a+b+c)$.

But we don't need Heron's formula for this question. I just accidently used it because I didn't see that it gave the area for the triangle already. ;)

Find the area of the small triangles and then equate them to the large are to give $x$ that is the simplest solution.


Shift the units to times between $ 0$ and $ 60$, let John's time be $ y$, and the train's arrival time be $ x$. Then the possible region is $ 0\le x,y\le60$, and the requirement is $ x\le y\le x\plus{}20$. Graphing this and finding the area of the successful region of a trapezoid, the probability is $ \frac{\frac12(60^2\minus{}40^2)}{60^2}\equal{}\frac{1000}{3600}\equal{}\boxed{\frac5{18}}$.

I don't get it. In the solution. How did we get 1000?

[hide]simplify out $ \frac{\frac12(60^2-40^2)}{60^2}$

we can do this as $\frac{\frac12(60\cdot 60-40\cdot 40}{60\cdot 60}$
and then distributing the $\frac12$ will get us

$\frac{60\cdot 30-40\cdot 20}{60\cdot 60}$.

Since the top has subtraction, we can't cross out stuff so we have to multiply it out.

We will then get $\frac{\bold{1000}}{3600}$

hope this helped[/hide]

[hide=Solution]We put the time the train arrives on the $y$-axis and the time John arrives on the $x$-axis, and shade in the region where John arrives while the train is there.

[asy]
draw((0,0)--(60,0));
draw((0,0)--(0,60)--(60,60)--(60,0));
label("2:00", (0,0), SW);
label("3:00", (60,0), S);
label("3:00", (0,60), W);
label("2:20",(20,0),S);
fill((0,0)--(60,60)--(60,40)--(20,0)--cycle, gray(.7));
[/asy]

The probability that John arrives while the train is at the station is the ratio of the shaded area to the whole square. If we divide the axes into 60 units, the shaded region can be split up into a triangle of area $20\cdot 20/2=200$ square units and a parallelogram of area $20\cdot 40=800$ square units, and the whole square has an area of 3600 square units. The ratio is $1000/3600=\boxed{5/18}$.[/hide]

[quote=GameBot]John needs to catch a train. The train arrives randomly some time between 2:00 and 3:00, waits for 20 minutes, and then leaves. If John also arrives randomly between 2:00 and 3:00, what is the probability that the train will be there when John arrives?[/quote]

[hide=slightly dodgy complementary events]the complementary event for both is that they aren't arriving in a 20 minute window, that covers $\frac 23$ of an hour so the product of these is $\frac 49$ so the complementary to that has $\frac 59$ but we notice half their not there time is the minimum overlap for not there and there are 2 events not 1 so we divide by 2 [/hide]

[hide=solution I only thought of after but it works ok]
If the train arrives from 2:00 to 2:40, John has a full 20-minute window to arrive.

Otherwise, the size of the time window John can arrive in shrinks linearly from 2:40 to 3:00, reaching 0 minutes at the end. Since it changes linearly, the average time window is just the average of the start and end: $\frac{20+0}2 = 10$.

Altogether, this gets

$\frac{40}{60}\cdot\frac{20}{60} + \frac{20}{60}\cdot\frac{10}{60} = \frac{2}{3}\cdot\frac{2}{6} + \frac{1}{3}\cdot\frac{1}{6}$

$= \frac{4}{18} + \frac{1}{18} = \boxed{\frac{5}{18}}.$
[/hide]

[quote=abc2142]aand my rating dropped from 99.6 to 99.1 again[/quote]

same[hide]5/18 is the answer[/hide]


The first phase of OBM be held on June 14, a Saturday leap year 2008. Hence the how many years on June 14 will be again on Saturday ?

Portuguese question (if necessary) :

A primeira fase da OBM se realiza no dia 14 de junho, um sábado do ano bissexto 2008. Daqui a quantos anos o dia 14 de junho será novamente no sábado ?

A) 4
B) 5
C) 6
D) 7
E) 8


calculus limit algebra function domain logarithms integration

In my calculus textbook, it gives the following proof for the division law of limits:

[quote]
$ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{L}{M}$ where $ L = \lim_{x \to a} f(x)$, $ M = \lim_{x \to a} g(x)$ and $ M \neq 0$

First let us show that

$ \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}$

To do this we must show that, given $ \epsilon >0$, there exists $ \delta >0$ such that

if $ 0 < |x-a| < \delta$ then $ \left| \frac{1}{g(x)} - \frac{1}{M} \right | < \epsilon$

Observe that $ \left| \frac{1}{g(x)} - \frac{1}{M} \right | = \frac{|M-g(x)|}{|Mg(x)|}$

We know that we can make the numerator small. But we also need to know that the denominator is not small when $ x$ is near $ a$. Since $ \lim_{x \to a} g(x) = M$, there is a number $ \delta_{1}$ such that, whenever $ 0 < |x-a| < \delta_{1}$, we have

$ |g(x)-M| < \frac{|M|}{2}$

and therefore $ |M| = |M - g(x) + g(x)| \leq |M-g(x)| + |g(x)| < \frac{|M|}{2} + |g(x)|$

This shows that

if $ 0 < |x-a| < \delta_{1}$ then $ |g(x)| > \frac{|M|}{2}$

and so, for these values of $ x$,

$ \frac{1}{|Mg(x)|} = \frac{1}{|M\parallel{}g(x)|} < \frac{1}{|M|} \cdot \frac{2}{|M|} = \frac{2}{M^{2}}$

Also, there exists $ \delta_{2}$ such that

if $ 0 < |x-a| < \delta_{2}$ then $ |g(x)-M| < \frac{M^{2}}{2}\epsilon$

Let $ \delta = \min \{ \delta_{1}, \delta_{2} \}$. Then, for $ 0 < |x-a| < \delta$, we have

${ \left| \frac{1}{g(x)} - \frac{1}{M} \right | = \frac{|M-g(x)|}{|Mg(x)|} < \frac{2}{M^{2}} \cdot \frac{M^{2}}{2}}\epsilon = \epsilon$

It follows that $ \lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M}$. Finally, using law 4 (the product rule for limits), we obtain

$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} f(x) \left( \frac{1}{g(x)} \right) = \lim_{x \to a} f(x) \cdot \lim_{x \to a} \frac{1}{g(x)} = L \cdot \frac{1}{M} = \frac{L}{M}$

[/quote]

I understand every aspect of the proof, except for the following: in the statements $ |g(x)-L| < \frac{|M|}{2}$ and $ |g(x)-L| < \frac{M^{2}}{2}\epsilon$, what is the purpose of the $ 2$? I do not understand why this number is introduced in the proof since it ends up being cancelled out in the third to last line of the proof. Can someone enlighten me here?

The best way to figure it out is to remove one or both of those $ 2$'s and try to make the same computations. What will you get?

Note that we had $ g(x)\to M$ and we have $ M\ne 0$ as an explicit condition. What is the enemy of the proof, the thing that must be conquered? The enemy is that we could be dividing by a number which is too small - that is, we have to guard against $ g(x)$ being too close to zero. Why can't $ g(x)$ be too close to zero? Because it's close to $ M$ instead.

That's what the $ \frac{|M|}2$ is doing. We [b]must[/b] pick a number like that that is in between $ 0$ and $ |M|.$ (It doesn't have to be exactly half way in between, but that's just a matter of convenience.)

Thanks for the help. I have another question though:

How does one go about proving $ \lim_{x \to a} f(x)^{g(x)} \equal{} L^{M}$ where $ L \equal{} \lim_{x \to a} f(x)$, $ M \equal{} \lim_{x \to a} g(x)$ and both $ f(x)$ and $ g(x)$ are positive over their respective domains (such as to prevent indeterminate forms like $ (\minus{}3)^{\pi}$ and $ 0^{0}$)? I can't get very far independently.

[quote]

Let $ \epsilon >0$ be given. We must find $ \delta >0$ such that

$ 0 < |x\minus{}a| < \delta \Longrightarrow |f(x)^{g(x)} \minus{} L^{M}| < \epsilon$

Notice

$ |f(x)^{g(x)} \minus{} L^{M}| \equal{} |f(x)^{g(x)\minus{}M} \cdot f(x)^{M} \minus{} L^{M}|$

[/quote]

I'm not sure how to introduce the $ f(x) \minus{}L$ term aside from maybe introducing a very ugly binomial expansion. Hints?

That's a matter of an entirely different character, and is completely out of place in the first few weeks of a first analysis course - the sort of place where you'd find the argument you started this topic with.

The fundamental issue: what is the definition of $ a^b?$ It's not an elementary law of arithmetic like $ ab$ or $ \frac ab.$

We wind up (with the restriction that $ a>0$ with no need to restrict $ b$) defining $ a^b$ as $ e^{b\ln a},$ and we can only do that after we've already defined the logarithm and exponential functions.

The outline of the argument you want goes something like this:

1. Define $ \ln x\equal{}\int_1^x\frac{dt}t.$ Note that the Fundamental Theorem of Calculus shows that this is a differentiable, hence continuous, function on its domain.

2. Argue that $ f(x)\equal{}\ln x$ is monotone, continuous, and maps $ (0,\infty)$ onto $ (\minus{}\infty,\infty).$ Then the Intermediate Value Theorem assures us that it has an inverse, and we can show that its inverse is continuous. Use the symbol $ g(x)\equal{}e^x$ to write that inverse.

3. Show that the usual laws of exponents hold.

Now, having set up all that, assume we have $ f(x)\to L$ and $ g(x)\to M$, where $ L>0$ and $ f(x)>0.$

a) Because the logarithm is a continuous function, we have $ \ln f(x)\to \ln L.$

b) We have previously proved that the limit of the product is the product of the limits. Hence $ g(x)\ln f(x)\to M\ln L.$

c) Since the exponential is a continuous function, then we have that $ e^{g(x)\ln f(x)}\to e^{M\ln L}.$

And that proves the theorem you want. There's really no pathway that's any more elementary than that - in particular, I can't imagine tackling it without having already proved a number of important theorems about continuous functions.

Again, thank you for the quick response. I'm not actually taking a college calculus course at the moment, but rather studying from a textbook and making some personal notes along the way that I have assembled in a pdf document. One of the things I wanted to cover that was never taught to me in class and hard to find on the internet is epsilon delta proofs for all the operations that can be applied to limits, so it was out of my own curiosity to ask about $ \lim_{x \to a} f(x)^{g(x)}$. Now that I've seen all the intricacies behind the proof, I can understand why my textbook didn't seem to cover it.

Anyways, my final question was this: Assuming I've adequately covered addition, subtraction, multiplication, division, exponentiation and composition of limits, what would be the remaining operations to cover? I see some sites list power and root laws as follows:

[quote]

${ \lim_{x \to a} \sqrt [n]{f(x)}} = \sqrt [n]{\lim_{x \to a} f(x)}$ where $ n \in \mathbb{Z}$ and $ n > 0$ for even $ n$.

$ \lim_{x \to a} f(x)^{n} = \left( \lim_{x \to a} f(x) \right)^{n}$ where $ n \in \mathbb{Z}$

[/quote]

But if I'm not mistaken, both of those laws are simply special cases of the composition law. What about logarithms?

If $ f(x)^{g(x)} = a$ then $ \log_{f(x)}(a) = g(x)$ but one could also write something such as $ \log_{g(x)}(f(x))$ and maybe look at whether ${ \lim_{x \to a}\log_{g(x)}(f(x)) = \log_{\lim_{x \to a} g(x)}} (\lim_{x \to a} f(x))$ holds true under specific conditions. Is there even much purpose at looking at the logarithm operation in such a way or is there no practical use for such a thing? Sorry if I'm not explaining myself well.

So, what exactly is the "exponential function"? (i.e., why does it exist?)

By the way, I reformatted your attachment description a bit to fit the screen better :)

[quote]So, what exactly is the "exponential function"? (i.e., why does it exist?) [/quote]

Actually I've (re-)reached the section on continuity in my book, and all it said concerning $ a^{x}$'s continuity is that "In Section 1.5 we defined the exponential function $ a^{x}$ so as to fill the holes in the graph of $ y \equal{} a^{x}$ where $ x$ is rational. In other words, the very definition of $ a^{x}$, in other words the very definition of $ a^{x}$ makes it a continuous function on $ \mathbb{R}$."

However, this doesn't satisfy me at all as it doesn't seem to give a very rigorous definition for the definition of $ a^{x}$ for irrational values. It simply gives an example of how $ 2^{\sqrt {3}}$ can be shown to be bound between progressively more accurate rational values, but it doesn't really go any further than that.

I myself have been thinking of the following idea. Let $ p_{1}, p_{2}, q_{1},q_{2}$ be integers and $ i$ be an irrational value, and $ \frac {p_{1}}{q_{1}} < i < \frac {p_{2}}{q_{2}}$. Thus, we'd have $ a^{\frac {p_{1}}{q_{1}}} < a^{i} < a^{\frac {p_{2}}{q_{2}}}$. I wanted to define $ a^{i}$ as follows:

$ \lim_{\frac {p_{1}}{q_{1}} \to i} a^{\frac { p_{1} } { q_{1} }} \equal{} \lim_{\frac {p_{2}}{q_{2}} \to i} a^{\frac {p_{2}}{q_{2}}} \equal{} a^{i}$

I'm not sure if $ \frac {p_{1}}{q_{1}} \to i$ properly conveys the meaning of what I'm trying to accomplish since $ p_{1}$ and $ q_{1}$ are increasing at different rates every time a better approximation is formed. So how would one go about defining $ a^{i}$ rigorously? Could I just state that the rationals are infinitely dense?

Alternatively, I know that a function $ f$ can be proved continuous if it satisfies $ \lim_{h \to 0} f(c \plus{} h) \equal{} f(c)$. Since $ a^{c \plus{} h} \equal{} a^{c}a^{h}$, wouldn't we just need to show that $ \lim_{h \to 0} a^{h} \equal{} 1$? That seems like a much more straightforward approach.

Edit: I tried using an epsilon delta proof for $ \lim_{h \to 0} a^{h} \equal{}1$ and came up with:

Let $ \epsilon>0$ be given. We must find a $ \delta>0$ such that:

$ 0 < |h| < \delta \Longrightarrow |a^{h} \minus{}1| < \epsilon$

Since $ |a^{h} \minus{}1| < a^{|h|} \plus{}1$, then setting $ \epsilon \equal{} a^{\delta} \plus{}1$ would work to complete the proof.


geometry
rectangle
combinatorics proposed
combinatorics

Consider all possible rectangles that can be drawn on a $ 8 \times 8$ chessboard, covering only whole cells. Calculate the sum of their areas. What formula is obtained if $ "8 \times 8"$ is replaced with $ "a \times b"$, where $ a,b$ are positive integers?

There are $ (a\minus{}m\plus{}1)(b\minus{}n\plus{}1)$ rectangles of length $ m$ and breadth $ n$. (look at it as the possible moves the smaller rectangle can make within the big board). This yields,

\[ \sum_{m\equal{}1}^{a} \sum_{n\equal{}1}^{b} (a\minus{}m\plus{}1)(b\minus{}n\plus{}1) mn\]

Another way to compute the same: Lets count "how many distinct rectangles include $ (k,l)$ th cell ($ 1\times 1$ component of the $ a\times b$ board) ?"

$ x$ and $ y$ components of the rectangle that cover the cell can be chosen in $ k(a\minus{}k\plus{}1)$ and $ l(b\minus{}l\plus{}1)$.


Points $ B_1$ and $ B_2$ lie on ray $ AM$, and points $ C_1$ and $ C_2$ lie on ray $ AK$. The circle with center $ O$ is inscribed into triangles $ AB_1C_1$ and $ AB_2C_2$. Prove that the angles $ B_1OB_2$ and $ C_1OC_2$ are equal.

$ \angle B_{1}OC_{1} \equal{} 90^{o} \plus{}\frac { \hat A} {2} \equal{} \angle B_{2}OC_{2}$ hence the conclusion follows.

Best regards,
sunken rock

isn't there the way of finding the relationship between nCr and nCr+1 and using that to find out the similars?


A convex quadrilateral $ ABCD$ inscribed in a circle $ k$ has the property that the rays $ DA,CB$ meet at a point $ E$ such that $ CD^2\equal{}AD\cdot ED$. The perpendicular to $ ED$ at $ A$ intersects $ k$ again at a point $ F$. Prove that the segments $ AD,CF$ are congruent if and only if the circumcenter of $ \triangle ABE$ lies on $ ED$.

To: radio

Attached, the proof.

Best regards,
sunken rock

A blind man sees better the details


Can everyone post their grade and the score they got on AIME?

I'll start.

9th grade... and  5.  :blush:

There are already multiple threads for this.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=77873]2006 AIME I Score Poll/Index[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78356]How did possible-floor-qualifiers score?[/url]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=78302]Usamo index poll for juniors and above[/url]

Right, so let's not start another one. Locked.


trigonometry
integration
calculus
calculus computations

[quote="perfect_radio"]Actually, the Wallis formula proves that the summand is $ O \left( \frac1{\sqrt k}\right)$, and that diverges. Maybe you meant something else?[/quote]

what i meant was to prove $ \text{equation}\;\;\#\textbf{\boxed{(14)}}$ of [url]http://mathworld.wolfram.com/InverseTangent.html[/url]

[quote="perfect_radio"]So your formula should be
\[ \frac{x}{1+x^{2}}\sum_{k=0}^\infty \left( \frac{x^{2}}{1+x^{2}}\right)^{k}\prod_{j=1}^{k}\frac{2j}{2j+1}= \arctan x . \]
By [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80515](3) from this link[/url], it suffices to prove that
\[ \frac{x}{1+x^{2}}\sum_{k=0}^\infty \left( \frac{x^{2}}{1+x^{2}}\right)^{k}\int_{0}^\frac{\pi}2 \sin^{2k+1}t \, dt = \arctan x . \]
Interchange sum and integral: $ \frac{x}{1+x^{2}}\int_{0}^\frac{\pi}2 \sin t \sum_{k=0}^\infty \left( \frac{x^{2}}{1+x^{2}}\cdot \sin^{2}t \right)^{k}\, dt = x \int_{0}^\frac{\pi}2 \frac{\sin t}{1+x^{2}\cos^{2}t}\, dt$.
Now substitute $ y = \cos t$ and the problem is over.[/quote]

actually, both your and the formula i had in my 1st post are [b]identical[/b]...
because in mine, $ \frac{x^{2}}{1+x^{2}}$ automatically gets included with every
term in the product ! :D

by the way, your solution -- nice !  :coolspeak:


Let $a$ be the greatest positive root of the equation $x^3-3x^2+1=0$
Prove that $[a^{1788}]$ and $[a^{1988}]$ are both divisible by $17$, where $[a]$ denotes the smallest integer less than or equal to $a$

This was a past ISL question I think.

Bomb

Hint: Find recurrence relation for roots.

That's right. It is from ISL 1988 :

http://www.kalva.demon.co.uk/short/soln/sh887.html

Pierre.


i need help in this home work ,some questions are easy and some can be confusing i will post them.

there are some stuff which are confusing ,i got some answer but some i dont have them ,i will post the answers later ,need some help!

1. In one year, an appliance company sold 231 washing
machines, 302 dryers, 467 stoves, 532 dishwashers, and 1,283
microwave ovens. How many appliances were sold altogether
that year?
A. 16,603 C. 2,815
B. 3,015 D. 1,815

2. In a restaurant, a man ordered soup for $3, an appetizer for
$4.25, dinner for $17.90, a beverage for $1.50, and dessert for
$3.75. He left the server a 15% tip. What was the total cost of
the man's meal, including the tip?
A.
$36.11 C. $30.40
B. $34.96 D. $4.56
3. Round 427,685 to the nearest thousand.
A. 400,000 C. 428,000
B. 427,000 D. 430,000
4. Round 0.8394263 to the nearest hundred-thousandth.
A. 0.840 C. 0.83942
B. 0.83943 D. 0.839
5. Add the fractions 2/3, 3/8, and 5/12 and reduce your answer to lowest terms.
A. 10/24 C. 1 11/24
B. 25/24 D. 1 1/2
6. In a mathematical equation, the coefficient is
A. the numeric value of the unknown that makes an equation true.
B. the term brought from one side of an equation to the other by performing the opposite
operation.
C. a number or quantity placed before another quantity to indicate multiplication.
D. the portion of an equation that's given.
7. Subtract the fraction 7 4/5 from the fraction 18 6/15 and reduce your answer to lowest terms.
A. 11 9/15 C. 10 12/15
B. 11 1/5 D. 10 3/5
8. Find the value of x in the equation x + 19 = 76.
A. x = 1,444 C. x = 57
B. x = 95 D. x = 4
9. What percent of 980 is 294?
A. 33 1/3% C. 3%
B. 30% D. 0.30%
10. Round the number 43,678.426785 to the nearest ten thousand.
A. 43,678.4279 C. 41,000
B. 43,678.4278 D. 40,000
11. Find the value of G in the equation G/5 - 13 = 42.
A. G = 275 C. G = 11
B. G = 135 D. G = 5 2/5
12. Which of the following would you use to record your checking account transactions?
A. A check stub or check book registry C. A signature card
B. A deposit slip D. Checks or drafts
13. Find the value of x in the equation 4(3x + 7) = 136.
A. x = 41 C. x = 10 3/4
B. x = 121/3 D. x = 9
14. Divide the fraction 14/27 by the fraction 2/3 and reduce your answer to lowest terms.
A. 2 25/28 C. 7/9
B. 1 2/7 D. 28/81
15. If a payee wants to transfer a check to another party, the payee would use a
A. blank endorsement. C. full endorsement.
B. restrictive endorsement. D. signature card.
16. A computer store sold a total of 300 items last month. The store sold six times as many
hard drives as they did CD-ROM drives, and half as many floppy drives as hard drives.
Based on this information, how many of each item were sold?
A. 30 hard drives, 90 floppy drives, and 180 CD-ROM drives
B. 40 hard drives, 20 floppy drives, and 240 CD-ROM drives
C. 180 hard drives, 90 floppy drives, and 30 CD-ROM drives
D. 240 hard drives, 20 floppy drives, and 40 CD-ROM drives
17. Change the fraction 38/8 to a mixed number and reduce your answer to lowest terms.
A. 4 3/4 C. 4
B. 19/4 D. 3 6/8
18. To calculate your adjusted bank balance, you'll need to locate all of your
A. outstanding checks and add them to the subtotal to get the statement balance.
B. outstanding checks and subtract them from the subtotal to get the statement balance.
C. deposits and subtract them from the subtotal.
D. service charges and add them to the checkbook balance to get a subtotal.
19. If two liters of soda cost
$0.80, how much would five liters cost?
A. $4.00 C. $2.00
B. $2.40 D. $1.60
20. Multiply the fraction 9/15 by the fraction 30/72 and reduce your answer to lowest terms.
A. 1/4 C. 1 11/25
B. 27/108 D. 5
21. A college professor has 370 papers to grade. In one evening, she grades 80 percent of
them. How many ungraded papers will remain for her to complete tomorrow?
A. 316 C. 82
B. 296 D. 74

22. A company employs a total of 527 people, and 47 of these people work in the operations
department. Which of the following is a ratio that compares the number of people in the
operations department to the total number of employees in the company?
A. 47 : 480 C. 480 : 47
B. 47 : 527 D. 527 : 47
23. Multiply the mixed number 4 2/7 by the mixed number 6 2/5 and reduce your answer to lowest
terms.
A. 30 C. 24 4/35
B. 27 3/7 D. 13 5/7
24. Subtract 476.234 from 538.4.
A. 47,085 C. 62.234
B. 4,708.50 D. 62.166
25. John and Paul work together at a music store. In one afternoon, they sold a total of 41
CDs. If Paul sold 9 more CDs than John, how many CDs did John sell?
A. 16 C. 32
B. 25 D. 34
26. The section of an invoice that lists the date of the payment and the cash discount is called
the
A. terms. C. description.
B. unit. D. F.O.B.
27. A flower shop buys flowerpots for $4.28 each and sells them for $6.50 each. What is the
amount of the markup on each flowerpot?
A. $10.78 C. $0.66
B. $2.22 D. $0.34
28. Calculate the simple interest on a loan of $18,500 at 11% interest for 18 months.
A.
$36,630.00 C. $2,543.75
B. $3,052.50 D. $763.13
29. The shipping term F.O.B. stands for
A. free on board. C. freight owed back.
B. free on bail. D. found on back.
30. What are the proceeds of a $9,500 note at a 12% discount rate for 8 months? Round your
answer to the nearest cent.
A. $760.00 C. $5,573.33
B. $912.00 D. $8,740.00

31. A lumberyard makes picnic tables for $87 and sells them for $127.89. What is the percent
markup on each picnic table based on cost?
A. 32% C. 47%
B. 41% D. 68%
32. On an invoice, the invoice total includes
A. merchandise only.
B. all other charges except merchandise.
C. insurance only.
D. merchandise plus all other charges.
33. Celina works at a crafts store where she earns $8.40 an hour and receives time-and-a-half
for overtime. She worked 43 hours last week. What was her gross pay for last week,
including overtime?
A. $399.00 C. $361.12
B. $373.80 D. $336.00
34. In a clothing store, a sweater that was originally priced at $73 was marked down and sold
for $54.75. What was the percent amount of the markdown?
A. 20% C. 33 1/3%
B. 25% D. 75%
35. An invoice for hairdressing supplies is shown in Figure A-1. Based on the information in
the invoice, calculate the invoice total.
A.
$585.19 C. $269.20
B. $569.81 D. $201.25
Examination, Lesson 1 35
CUSTOMER INVOICE
Stock
Number
Quantity Unit Merchandise
Description
Unit Price
22345 7 ea. Blow dryer $28.75
13672 5 doz.Permanent waves $53.84
44986 8 boxRubber gloves $12.42
Note: Shipping charges are $15.38.
36. Calculate the amount of interest on a loan of $3,200 at 6% interest for 60 days using the
ordinary interest method.
A. $3.16 C. $32.00
B. $31.56 D. $384.00
37. An amusement park purchased equipment from a manufacturer for a list price of
$42,000, minus trade discounts of 30/20/15. Calculate the amount of the trade discount
on the purchase.
A. $19,992 C. $23,520
B. $22,008 D. $29,400
38. A retail jeweler sells a pair of gold earrings for $213.90. The jeweler uses a 130% markup
based on cost. Based on this information, what was the wholesale cost of the earrings?
A. $93.00 C. $276.90
B. $164.53 D. $278.07
39. What would be the amount of compound interest on $6,000 invested for two years at 9%,
compounded semiannually? Round your answer to the nearest dollar.
A.
$7,155 C. $1,155
B. $2,160 D. $1,080
40. A store owner buys supplies from a vendor for $8,450. The terms of sale are 2/10, n/30.
What will be the net amount due if the owner pays the bill by the 10th day after he
receives the supplies?
A. $169 C. $8,281
B. $7,605 D. $8,450
41. Employers are required to match all FICA tax payments, which include
A. FUTA and SUTA. C. tuition reimbursement.
B. fringe benefits. D. Social Security and Medicare.
42. What is the effective rate of $18,600 invested for one year at 7.5%, compounded semiannually?
Round your answer to the nearest hundredth.
A. 7.64% C. 3.82%
B. 7.09% D. 3.75%
43. A software buyer is looking for a product to retail for
$230. He wants to apply a markup of
38% (based on selling price) to the product. Allowing for this markup, what is the maximum
amount the buyer can pay for the software?
A. $38.00 C. $142.60
B. $87.40 D. $370.97

44. The total amount of interest on a loan of $6,000 for 150 days is $210.50. Using the ordinary
interest method, what is the rate of interest on this loan? Round your answer to the
nearest hundredth.
A. 2.34% C. 8.53%
B. 8.42% D. 11.88%
45. Merchandise was purchased at list for $3,460 with a single trade discount of 23%. What
was the net price of this merchandise?
A.
$4,255.80 C. $795.80
B. $2,664.20 D. $772.80
46. Which of the following is a correct definition of the term present amount (or present
value)?
A. The total amount of principal and accumulated interest at the end of a loan or
investment
B. An amount of money that must be deposited today at compound interest in order to
provide a specific lump sum of money in the future
C. The idea that money “now,” or in the present, is worth more than money in the future
because it can be invested and earn interest as time goes by
D. The rate of return on investment
47. Darrell owns a consulting business and has an estimated annual income of $63,000. His
Social Security tax is 12.4%, Medicare is 2.9%, and his estimated federal income tax rate
is 22%. How much quarterly estimated tax must Darrell send to the IRS for the first
quarter?
A. $5,874.75 C. $15,750.00
B. $13,860.00 D. $23,498.99
48. On a promissory note, the payee is the
A. amount of money borrowed.
B. amount of money being paid back.
C. person borrowing the money and issuing the note.
D. person lending the money and receiving the payment.
49. Every week, a grocery store buys 30 gallons of milk at a cost of $1.58 per gallon. The
store owner anticipates a 15% spoilage rate. In order to achieve a 20% markup based on
selling price, what should be the selling price per gallon? Round your answer to the nearest
cent.
A.
$1.66 C. $2.19
B. $1.98 D. $2.32

50. EOM or end-of-month dating is the same as
A. F.O.B. dating. C. R.O.G. dating.
B. ordinary dating. D. proximo dating.

Could you tell which ones your having difficulty with, because most of these questions are simple arithmetic and basic algbra.

[quote]i am having some trouble with this one
number 42
what is the effective rate of 18600 invested for one year at 7.5 percent ,compounded semianually ,round the answer to the nearest handredth .

thanks alot !

[/quote]18600 invested for one year at 7.5 percent ,compounded semianually is
\[ {\text{18,600}}\left( {{\text{1 \plus{} }}\tfrac{{{\text{.075}}}}
{{\text{2}}}} \right)^{\left( {1 * 2} \right)}  \equal{} 20,021.16
\]
The effective rate is defined as (interest earned in one year)/(principal invested at the beginning of the year)
Therefore,
\[ \frac{{18,600\left( {1 \plus{} \tfrac{{.075}}
{2}} \right)^{\left( {1 * 2} \right)}  \minus{} 18,600}}
{{18,600}} \approx .07640625
\]

It seems to me you're just lazy to do your own homework.  :wink:


Fie $f,g: \mathbb{R}\to\mathbb{R}$, $f$ continua, $g$ monotona cu proprietatea ca pentru orice $a,b\in\mathbb{R}, a<b,$ exista $c, d\in (a,b)$ astfel incat $f(c)=g(d).$ Sa se demonstreze ca $f=g.$ ;)

EDIT: Imi cer scuze in loc de $g$ pusesem $f.$ :blush:

Pacat!!! Mie mi se pare o problema asa de draguta. :(

[quote="cezar lupu"]Fie $f,g: \mathbb{R}\to\mathbb{R}$, $f$ continua, $g$ monotona cu proprietatea ca pentru orice $a,b\in\mathbb{R}, a<b,$ exista $c, d\in (a,b)$ astfel incat $f(c)=f(d).$ Sa se demonstreze ca $f=g.$ ;)[/quote]
S-ar putea sa fie draguta ;) Ce conditie verifica $g$?


Solve the equation in $\mathbb{N}$
$12^m+1=5^n$

[hide]

All powers of 5 have 5 as last digit. So $12^m$ must have 4 as last digit, so $m=2+4k$. The equation becomes

$144^{2k+1}+1=5^n$

Look $\mod 13$: $144 \equiv 1 \mod 13 \implies 144^{2k+1} \equiv 1 \mod 13$. So $LHS \equiv 2 \mod 13$

But $5^n \not \equiv 2 \mod 13$ $\forall n \in \mathbb{N}$

[/hide]

[hide]$12^{m} \equiv 4 \bmod 5$. That gives $m = 4k + 2$.
$5^{n} \equiv 1 \bmod 3$. That gives $n = 2h$.
Hence $1 = (5^{h} - 12^{2k + 1})(5^{h} + 12^{2k + 1})$, that is obviously impossible.[/hide]


show your work....
[b]
CHEMISTRY
[/b]

in your chemistry lab, you have a bottle of 1% hydrochloric acid solution and a bottle of 5% bottle of hydrochloric acid solution. You need 100 milliliters of 3% hydrochloric acid solution for an experiment. Hom man liters of each solution do you need to mix together?

first one answer gets rated. good luck :ddr:

ooh, I get a RATED post  :D

[hide="systems of equations ftw"]Let the amount of the $ 1\%$ solution added be $ x$.  Then, $ 100\minus{}x$ milliliters of the $ 5\%$ are added.  Thus, the overall concentration is:

$ \frac{x\plus{}5(100\minus{}x)}{100}\equal{}3$
$ 500\minus{}4x\equal{}300$
$ x\equal{}50$

Thus, you add $ 50$ milliliters of both solutions.[/hide]

ty you get rated
butif someone post some other answer then i wold have to see if they are right or else ill change it

[quote="cf249"]OK, I'm just being picky here, but the question did ask
[quote]How many liters of each solution do you need to mix together? [/quote]
(with typos fixed).

So, the answer would be $ 0.05$ $ L$.[/quote]

Oh, sorry; I did not read that carefully.  You are correct.


Consider the following sequence of equalities:
$ 1\equal{}0\plus{}1$
$ 2\plus{}3\plus{}4\equal{}1\plus{}8$
$ 5\plus{}6\plus{}7\plus{}8\plus{}9\equal{}8\plus{}27$
$ 10\plus{}11\plus{}12\plus{}13\plus{}14\plus{}15\plus{}16\equal{}27\plus{}64$
$ ...$
Develop a general conjecture and prove that the conjecture is correct.


-p ^ (p v q),    truth table?

(-p ^ (p v q)) then q, truth table?[/quote]


vector
function
limit
calculus
calculus computations

What do you do if dy/dt and dx/dt of a vector or parametric function are both equal to zero and you need to make conclusions about whether there is asymptotic behavior at that point?


limit
real analysis
real analysis theorems

Let $ f(t) \equal{} e^{\minus{}a/t}\plus{}e^{\minus{}b/t}$ for t>0. Assume a>b>0.

It seems intuitively clear that as t goes to 0 from the right, the second term will "dominate" the second one. How do you make that precise?

The standard notation for that is that $ e^{ \minus{} a/t} \equal{} o(e^{ \minus{} b/t})$ as $ t\to 0^ \plus{}$, which means that $ \lim_{t\to 0^ \plus{}} \frac {e^{ \minus{} a/t}}{e^{ \minus{} b/t}} \equal{} 0$.


geometry
circumcircle
geometric transformation
homothety
geometry proposed

Be $ABC$ triangle with $T$ the touch point between its $NPC (O_9)$ and $(I_a)$ excircle of $A$. Produced $TG$ ($G$ is centroid of $ABC$) at $K$ on circumcircle of $ABC$. Prove that $KC=KA+KB$.

I'm having some trouble here: does all of this mean that $TG$ cuts the circumcircle of $ABC$ at $K$? Because if this is the case, then the assertion is not true. If $CA=AB$ and we make $A$ go really far away from $BC$, then $KB=KC\to \frac {BC}2$, while $KA\to\infty$.

Please, you consider ABC  acute angle and scalene triangle.

It doesn't matter that it's not exactly isosceles, you can make it "almost" isosceles (the difference between $AB$ and $AC$ can be made very small, but not $0$).

I am very sorry.
It is enough:AC greater than AB.

Am I the only one who thinks that this isn't true?

No, Grobber, the problem is true. It is not so well formulated (actually the point K could be simply defined as the image of T in the homothety with center G and factor -2, i. e. the anticomplement of T), but it is true. Here is my proof (well, not complete, since I leave out arrangement considerations):

If G is the centroid of triangle ABC, then the homothety with center G and factor -2 maps the midpoints $M_a$, $M_b$, $M_c$ of the sides BC, CA, AB of triangle ABC to the vertices A, B, C. Let this homothety map the point T to a point K'. Then, the point K' lies on the ray TG and on the circumcircle of triangle ABC (in fact, this is because the point T lies on the nine-point circle of triangle ABC, i. e. on the circumcircle of triangle $M_a M_b M_c$, and the homothety with center G and factor -2 maps the points $M_a$, $M_b$, $M_c$ to the points A, B, C, and thus also maps the circumcircle of triangle $M_a M_b M_c$ to the circumcircle of triangle ABC). Hence, the point K' coincides with our point K defined as the point of intersection of the ray TG with the circumcircle of triangle ABC. In other words, our point K is the image of the point T in the homothety with center G and factor -2. Since the points A, B, C are the images of the points $M_a$, $M_b$, $M_c$ in this homothety, we have $KA = 2\cdot TM_a$, $KB = 2\cdot TM_b$ and $KC = 2\cdot TM_c$. Hence, in order to prove KC = KA + KB, it is enough to show $TM_c = TM_a + TM_b$.

Now there is a nice theorem stating that if F is the point of tangency of the nine-point circle and the incircle of triangle ABC (i. e., the Feuerbach point of triangle ABC), then one of the segments $FM_a$, $FM_b$, $FM_c$ equals the sum of the two others. This actually holds not only for the point of tangency F of the nine-point circle and the incircle, but also for the point of tangency of the nine-point circle and any excircle (there are three such points). And since T is such a point of tangency, we see that one of the segments $TM_a$, $TM_b$, $TM_c$ equals the sum of the two others, i. e. we have either $TM_a = TM_b + TM_c$, or $TM_b = TM_c + TM_a$, or $TM_c = TM_a + TM_b$. In order to prove that exactly $TM_c = TM_a + TM_b$, we must consider some arrangements, and that's what I don't wanna do right now...

  Darij

[b]PS.[/b] Thanks, Juan Carlos, for the parallel tangents paper, I'll look it up.  Darij

Sorry for the negative comments: when juan said "$TG$ produced" I didn't think of the ray $(TG$, as I should have, but of the line $TG$, and then I labeled as $K$ the other point of intersection between $TG$ and the circumcircle.

I think there is a connection between this problem and the one from the Chinese TST, which has recently been posted in the Unsolved Problems section. Not a direct connection maybe, but the theorem you've mentioned shows that the Feuerbach point is $X(100)$ of the median triangle.


geometry
rectangle
advanced fields
advanced fields unsolved

[b]Def[/b] A collection $C$ is non-overlapping if distinct elements of $C$ have disjoint interior.

In $R^{n}$, given a countable collection $K$ of closed rectangles $(R_{i})$. Can we always create a non-overlapping countable collection $C$ of rectangles $(R'_{i})$ such that
    - For every $i$, $R'_{i}$ is a subset of $R_{k}$ for some $k$
    -  The union of all elements in $C$ is equal to that in $K$

The answer is "Yes", and this is a standard trick in measure theory. Just notice that the difference of any two rectangles is a union of several non-overlapping rectangles and that
\[R_{1}\cup R_{2}\cup R_{3}\cup\dots=R_{1}\cup(R_{2}\setminus R_{1})\cup ((R_{3}\setminus R_{1})\setminus R_{2})\cup\dots\]


Anyone know any good websites (besides mathcounts.org) for mathcounts/mathcounts like problems?

mathcounts.saab.org has mathcounts drills

And check the sticky, you can maybe trade,...

I believe the minnesota mathcounts homepage has some practice tests

There was another helpful site, but i think it got shut down :(

you can check the thread on MathCounts Resources which has people asking and giving problems etc..

[quote="PowerOfPi"][quote="IDKorean"]I believe the minnesota mathcounts homepage has some practice tests[/quote]
That went down for some reason.[/quote]

Yeah, I can only see the 2009 tests.

Tons of mock sets and competitions on my site [url=http://www.agmath.com/57427/index.html]here[/url] and [url=http://www.agmath.com/169763.html]here[/url].


Ok; I'm not quite sure what forum thisbelongs to; but it's a very easy non-=problem solving topic. My answer just deviated from the given answer so I want to know if it is right.

y = Arccos(1/x^2)

My answer:

2/x^3(sqrt(1-x^-4))
[/tex]

Answer is $\displaystyle \frac{dy}{dx}=\frac{2}{|x^3|}\frac{1}{\sqrt{1-x^{-4}}}=\frac{2}{|x|\sqrt{x^4-1}}$ but what do you mean 'deviated'?

Yeah, it's correct.  Multiply top and bottom by x^2 - does that give the 'answer in the back'?


What is the length of a diameter of a circle which has an area of $ 36\pi$ square units?


I am given $C_8 H_{10}$.

(a) If it is ethyl-benzene, then I must obtain $C_6 H_5 - C - CH_3$;
$--------------------- O$

(b) If it is 2-methyl-benzene (in para position - i.e. opposite), then I must obtain $HOOC -$ benzene $- COOH$;

both by using only substitutions.

I think I have solved the first, but I'm not really sure if it's doable.

i dont understand the second part!

but anyways, i think this would help;
[hide]
methyl group on benzene ring is an Ortho-Para directing group due to the sp3 benzilic carbon.[/hide]

In the second part, do you mean 1,4-dimethylbenzene (instead of 2-methylbenzene)?  As in,  para-dimethylbenzene?


Is this what you mean to write, marko avila?
$\frac{1}{1+a+b} + \frac{1}{1+b+c} + \frac{1}{1+a+c}\leq \frac{1}{2+a} + \frac{1}{2+b} + \frac{1}{2+c} $

We need some condition like $abc=1$

If instead we have $a+b+c=3$, if I'm not mistaken, the sign is flipped, at least if $a\geq b\geq c$ and $1\geq b$ by majorization.

I will try to solve if you clarify the condition.

With condition $abc=1$ it's [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Bulgaria&t=243]this[/url] peoblem.

And with the case $ a+b+c=1/a+1/b+1/c$ it is a mathlinks contest problem! :P

yes im very sorry abc=1  :blush:  mathlinks contest ??? no wonder i couldnt solve it¡¡¡ :P

this question is not from mathlinks contest. Well it has many solutions, like brutal force, mixing variables, etc.

well can you please tell me what is "mixing variables"? :blush:


[quote="cnyd"]it means $ abc \equal{} 1$ $ (a,b,c > 0)$

$ \sum{a^{6}b^{3}} \plus{} \sum{a^{3}b^{3}c^{3}}\geq \sum {a^{4}b^{4}c} \plus{} \sum{a^{5}b^{2}c^{2}}$[/quote]

It's just Schur applied to the numbers $ a^2b, b^2c, c^2a$   :lol:

[quote="kuing"][quote="cnyd"]it means $ abc \equal{} 1$ $ (a,b,c > 0)$

$ \sum{a^{6}b^{3}} \plus{} \sum{a^{3}b^{3}c^{3}}\geq \sum {a^{4}b^{4}c} \plus{} \sum{a^{5}b^{2}c^{2}}$[/quote]

the "$ \sum$" means  symmetric  or cyclic?...
 :|[/quote]

while  means  cyclic   $ \sum{a^{6}b^{3}} \plus{} \sum{a^{3}b^{3}c^{3}}\geq \sum {a^{4}b^{4}c} \plus{} \sum{a^{5}b^{2}c^{2}}$

holds only $ a,b,c>0$

I think it means  symmetric

Well, with

$ Schur(a^2b,b^2c,c^2a)$ we get indeed the cyclic version. Anyway it's only the term $ \sum a^6b^3$ that's not already symmetric...

And of course al this is valid only for $ a,b,c\ge0$.


Let $ \{x_k\},\{y_n\}$ are two real number sequences satisfy: $ x_1<x_2<...<x_k$,  $ y_1<y_2<...<y_n$ with $ k\geq5$ and $ n\geq5$, where $ x_i \neq y_j$ for all $ i\neq j$.
 Prove that: There exist an index $ n_0$ such that: $ |x_{n_0}\minus{}y_{n_0}|\geq 7$.

Are you sure that's correct? There are so many counterexamples to this result....  :ninja:

oh, sorry. There is a fault in supposition of the problem which i gave you!i post it again:
Let $ \{x_k\},\{y_n\}$ are two real number sequences satisfy: $ x_1 < x_2 < ... < x_k$,  $ y_1 < y_2 < ... < y_n$ with $ k\geq5$ and $ n\geq5$, where $ x_i \neq y_j$ for all $ i\equal{}1,2,...,k$ and $ j\equal{}1,2,...,n$.
 Prove that: There exist an index $ n_0$ such that: $ |x_{n_0} \minus{} y_{n_0}|\geq 7$.

Best! :)

ofcourse, $ k$ and $ n$ are tend to infinity!etc two sequences $ x_k$ and $ y_n$ are infinite element sequences.

But if you continue that pattern on both sequences, $ |x_{n_0} \minus{} y_{n_0}|\equal{}.5\not\geq7$.

ok.thanks you!i've checked suppostion of the above problem again!what a pity! i sent you wrongly problem.sorry you once again!
Here, i post correct problem:
Let $ \{x_k\},\{y_n\}$ are two integer number sequences satisfy: $ x_1 < x_2 < ... < x_k<...$,  $ y_1 < y_2 < ... < y_n<...$ with $ k\geq5$ and $ n\geq5$, where $ x_i \neq y_j$ for all $ i \equal{} 1,2,...,$ and $ j \equal{} 1,2,...,$.
 Prove that: There exist index $ n_0, m_0$ such that: $ |x_{n_0} \minus{} y_{m_0}|\geq 7$.

Best! :)

it seems that obviously,but you can prove?

because both n and $ k \geq 5$ so we consider the first 5 numbers from each sequence ,we call set $ M ( x_i |i\equal{}\bar{1,5})$ and set $ N (y_i |i\equal{}\bar{1,5})$ ; since they are all different so we can rearrange this 10 numbers into a increasing sequence which we called sequence (A), and without losing our generality we assume that $ x_1 < y_1$
now let assume that there is no pair $ x_{i}$ and $ y_{j}$ satisfying $ |x_i\minus{}y_{j}| \geq 7$
+)since $ x_1$ is the first number in (A) so the 8th ; 9th and 10th number in (A) must belong to set M  (because the difference between the 8th and the 1st $ \geq 7$)
+)since $ x_{10}$ is the 10th number so that the 2nd ; 3rd in (A) is belong to set M , because the difference between the 10th number and the 3rd number > 7 .
set M only cannot have 6 elements so we see the contradiction , QED

ok. Congratulation!your solution is very g :) ood!
By the way, may i ask you a question? are you vietnamese?
Best, Luan.

but we can solve this problem simply follow:
WLOG $ x_1<y_1$.
we also consider the case $ k\equal{}n\equal{}5$.
we have  $ x_1<x_2<...<x_5$ and $ y_1<y_2<...<y_5$
if $ x_3<y_5$ then $ x_2,x_3,y_1,y_2,y_3,y_4$ lie between $ x_1$ and $ y_5$ so $ y_5\minus{}x_1 \geq 7$.
Otherwise, we see that $ x_5\minus{}y_1 \geq 7$.  :)


Diophantine equation
number theory proposed
number theory

Find all integer solutions to the equation
\[ (a^2\plus{}ab\minus{}b^2)^2\equal{}1.\]

[hide="hint"]It is enough to describe all natural solutions  :wink:
Use the fact that if $ (a,b)$ is a solution then so is $ (b\minus{}a,a)$ and infinite descent.[/hide]

Can you please explain how we can find [b]all[/b] possible solutions in your way? Thanks a lot. :)

first you can determine all solutions where one of the variables is $ 0$. they are $ (0,\pm1),(\pm1,0)$.
Suppose $ (a,b)$ is a solution, then so are $ (\minus{}a,\minus{}b),(\minus{}b,a),(b,\minus{}a)$ so it is enough to solve it in positive integers.
If $ a\equal{}b$ the only solution is $ (1,1)$
Infinite descent tells us all such solutions can be retrieved by inversing the operation above a finite number of times from $ (1,1)$. We conclude $ (a,b)\equal{}(f_k,f_{k\plus{}1})$ ($ f_n$ is the $ n$'th Fibonacci number)
So all solutions are either of the above or $ (\minus{}f_k,\minus{}f_{k\plus{}1})$ or $ (\pm f_{k\plus{}1},\mp f_k)$  :wink:
I hope you can fill in the details

Thanks. What I actually intended to show was that all solutions to the above equation must be consecutive Fibonacci numbers in numerical values. I did something similar but I wasn't sure about the following.

[quote="Albanian Eagle"]Infinite descent tells us all such solutions can be retrieved by inversing the operation above a finite number of times from $ (1,1)$.[/quote]

So I rather turned it into a Pell equation ($ \pm 4 \equal{} (2a \plus{} b)^2 \minus{} 5b^2 \equal{} 5a^2 \minus{} (2b \minus{} a)^2$) and since all solutions to a Pell equation can be generated from the initial solution we can arrive at our conclusion.

Thanks for your help. :)