ok so for everyone that doesnt know me, im Richard S. from Jenks

ok so i was doing some FTW a minute ago, and saw this question that i didnt know how to do

Given that 5^X+1 = 30, what is the vaule of 5^3x+1?

Simply use algebra and basic laws of exponents.

$ 5^X\equal{}29$, so $ 5^{3X}\plus{}1\equal{}29^3\plus{}1\equal{}\boxed{24390}$.

Richard!!! Hey. Wassup. What you doing on AoPS?

hey um...gauss?

i think i typed my question in wrong

its actually

if 5^(X+1)=30, then what is the value of 5^(3x+1)?

we had AoPS last year in Mathcounts, and so i thought i would get on again

[quote="jonathanchou711"]So.. Wouldn't it be like this:
$ 5^{x \plus{} 1} \equal{} 5^x*5$? So $ 5^x*5 \equal{} 30$. Divide both sides by $ 5$. So you get $ 5^x \equal{} 6$. So wouldn't $ 5^{3x} \equal{} 216$? I'm probably wrong. Good to have you back Richard.

P.S. New account? Only 2 posts?  :wink:[/quote]
You actually were right. :) Continuing from there, $ 5^{3X \plus{} 1} \equal{} 5^{3X}\times5 \equal{} 216\times5 \equal{} \boxed{1080}$.
There you go. :)

EDIT: Yay 2200th post! :D

[quote="yaofan"]jonathan

do u finally have a friend from oklahoma on aops?[/quote]

Actually, I already did. I have mr.k_74! And this other jack* guy.

EDIT: Well and dude25 and some other people, but they don't get on as much as I do.


function
trigonometry
calculus
derivative
symmetry
integration
calculus computations

Find the fourier coefficients of the following periodic function:

$ f(t)\equal{}
\begin{cases}
\sin\omega t, \mbox{   if   }  0<t<\frac{T}{4} \\
\minus{}\sin\omega t, \mbox{   if   } \frac{T}{4}<t<\frac{3T}{4} \\
\sin\omega t, \mbox{   if   }  \frac{3T}{4}<t<T
\end{cases}$

Here $ \omega \equal{} \frac{4\pi}{T}$

Some predictive observations about what the results of this calculation will be.

1. $ f$ is an odd function. This means that if you calculate the coefficients in complex for, $ \hat{f}(n)$ will be odd and pure imaginary; if we do this in real form - that is, as $  a_n\cos\left(\frac {2\pi nt}{T}\right) + b_n\sin\left(\frac {2\pi nt}{T}\right)$ - then all $ a_n = 0$ (including the constant term) and we need only compute $ b_n$ for a sine series.

2. $ f$ is symmetric about $ t = \frac {T}4.$ That is, $ f(t) = f\left(\frac {T}2 - t\right).$ As a result of this, we will have $ b_n = 0$ whenever $ n$ is even. Now we only need concentrate on computing $ b_n$ for $ n$ odd.

3. $ f$ is continuous. $ f'$ is bounded but not continuous. Considered as a weak derivative, $ f''$ contains Dirac deltas, added to what is otherwise a bounded function. Hence $ b_n = O(n^{ - 2})$ as $ n\to\infty.$ But this will also mean that the Fourier series will converge absolutely and uniformly.

Using all of the symmetry alluded to in the post above, we should be able to calculate the only nonzero coefficients as:

$ b_{2k \plus{} 1} \equal{} \frac {8}{T}\int_0^{\frac {T}{4}}\sin\left(\frac {4\pi t}{T}\right) \sin\left(\frac {2\pi (2k \plus{} 1)t}{T}\right)\,dt.$

If I did that computation correctly, I get that

$ b_{2k \plus{} 1} \equal{} \frac {4( \minus{} 1)^{k \plus{} 1}}{2\pi}\left(\frac1{2k \minus{} 1} \minus{} \frac1{2k \plus{} 3}\right) \equal{} \frac {8( \minus{} 1)^{k \plus{} 1}}{\pi(2k \minus{} 1)(2k \plus{} 3)}.$

And hence that

$ f(t) \equal{} \frac8{\pi}\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^{k \plus{} 1}}{(2k \minus{} 1)(2k \plus{} 3)}\,\sin \left(\frac {2\pi(2k \plus{} 1)t}{T}\right)$

$ \equal{} \frac8{\pi}\left[\frac13\sin\left(\frac {2\pi t}{T}\right) \plus{} \frac15\sin\left(\frac {6\pi t}{T}\right) \minus{} \frac1{21}\sin\left(\frac {10\pi t}{T}\right) \plus{} \frac1{45}\sin\left(\frac {14\pi t}{T}\right) \plus{} \cdots\right]$

This does meet my third prediction, that $ b_n \equal{} O(n^{ \minus{} 2})$ as $ n\to\infty.$

Added comment: after adding up 10 or so terms of this series numerically and viewing the graph, I do believe this computation is correct.


Partly to ask the question and partly to revive this forum after a respectful silence:

Has the schedule for Georgia's math competitions been posted yet?
Can anyone reccomend any that might not be on the list?

[url=http://www.gctm.org/Default.aspx?pageId=130271]Competition Schedule on GCTM Website[/url]

Some tournaments (e.g., Ga Tech) have also posted competition dates on their own websites, but not on the GCTM list.

For ones not on the list, I think I can recommend ARML local and ARML power round, Duke, Mandelbrot, and HMMT, Princeton, etc.  You might also try some nearby states, though you might want to check the style and quality of their past tournaments before you commit to a big trip.

Is Billy still gonna hold a math meet this year?  :P

how do we organize for ARML power contest?

is it just "hey everyone, lets meet at Rockdale at 11:00" or something like that for last year's arml members

Dr. Garner has organized it at least for the past two years.  It is not restricted to members of last year's team; anybody interested is permitted and encouraged to come.

[quote="Math Geek"]Is it going to take place this semester or the next?[/quote]
Yes---twice, in fact.  Once in November and once in February.

Mmkay.  So are they going to repost the website with [b]all[/b] the math tourneys, not just the GCTM-relevant ones?

They (GCTM) only post the ones in Georgia. And someone (the tournament organizers) has to tell them about a tournament before they post it on the web site. I don't know if they automatically post Ga. Tech, etc.

I can personally recommend the Furman tournament. It's a good one. The other nearby tournament that I've heard is good is the University of Alabama, but I've never been there.

So is Billy gonna hold a math meet this year?  I thought it was fun... :(

Billy's not doing his tournament this year. He's decided to concentrate on helping other groups in Henry County to get started with Math Team. Surely a worthy goal! (And one that Steve would heartily approved of.)

Separate question. Does anyone know if the University of Alabama is still hosting a tournament? I've heard that it is/was a good one, but haven't found a date yet.

Tom

Are enough people interested in going to the HMMT competition in February that a team has been set up? If so, who is it? I wanna go!

I'm also interested in going. Does anyone know the date for it?

I would be interested, too.

Santosh, check the website at http://web.mit.edu/hmmt/www/

I've already searched the site and I can't find the date for the 2009 competition.

Look at the Google cache for the HMMT site; it's supposed to be February 7 this year.  Apparently they recently took the date down; I hope this was accident, rather than change of date.

We were just trying to find out about the University of Alabama tournament this afternoon.  The website hasn't been updated in nearly seven years.


Given $ a,b,c>0$ and$ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
 $ \frac{\sqrt{a^3}}{\sqrt{b^2\plus{}c}}\plus{}\frac{\sqrt{b^3}}{\sqrt{c^2\plus{}a}}\plus{}\frac{\sqrt{c^3}}{\sqrt{a^2\plus{}b}} \ge \frac{3}{\sqrt{2}}$

[quote="quykhtn-qa1"]Given $ a,b,c > 0$ and$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
 $ \frac {\sqrt {a^3}}{\sqrt {b^2 \plus{} c}} \plus{} \frac {\sqrt {b^3}}{\sqrt {c^2 \plus{} a}} \plus{} \frac {\sqrt {c^3}}{\sqrt {a^2 \plus{} b}} \ge \frac {3}{\sqrt {2}}$[/quote]
Setting $ P\equal{}\sum {\frac{{\sqrt {a^3 } }}{{\sqrt {b^2  \plus{} c} }}}$
By Holder inequality, we have that:
$ P^2 .\sum {a^3 \left( {b^2  \plus{} c} \right)} .\left( {a^2  \plus{} b^2  \plus{} c^2 } \right) \ge \left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^4$
It suffices to prove that $ \sum {a^3 \left( {b^2  \plus{} c} \right)}  \le 6$ which is true as:
               $ a^3 c \plus{} b^3 a \plus{} c^3 b \le \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{3} \equal{} 3.$
         and $ a^3 b^2  \plus{} b^3 c^2  \plus{} c^3 a^2  \le 3$.(1)
(1) is in here:

Where can that theorem cited in the article above be found?

:)

[quote="leedt26"][quote="quykhtn-qa1"]Given $ a,b,c > 0$ and$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
 $ \frac {\sqrt {a^3}}{\sqrt {b^2 \plus{} c}} \plus{} \frac {\sqrt {b^3}}{\sqrt {c^2 \plus{} a}} \plus{} \frac {\sqrt {c^3}}{\sqrt {a^2 \plus{} b}} \ge \frac {3}{\sqrt {2}}$[/quote]
Setting $ P \equal{} \sum {\frac {{\sqrt {a^3 } }}{{\sqrt {b^2 \plus{} c} }}}$
By Holder inequality, we have that:
$ P^2 .\sum {a^3 \left( {b^2 \plus{} c} \right)} .\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^4$
It suffices to prove that $ \sum {a^3 \left( {b^2 \plus{} c} \right)} \le 6$ which is true as:
               $ a^3 c \plus{} b^3 a \plus{} c^3 b \le \frac {{\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 }}{3} \equal{} 3.$
         and $ a^3 b^2 \plus{} b^3 c^2 \plus{} c^3 a^2 \le 3$.(1)
(1) is in here:[/quote]
It is very nice and not nice,because,my solution is similar   :wink: :)

[quote="leedt26"][quote="quykhtn-qa1"]Given $ a,b,c > 0$ and$ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
 $ \frac {\sqrt {a^3}}{\sqrt {b^2 \plus{} c}} \plus{} \frac {\sqrt {b^3}}{\sqrt {c^2 \plus{} a}} \plus{} \frac {\sqrt {c^3}}{\sqrt {a^2 \plus{} b}} \ge \frac {3}{\sqrt {2}}$[/quote]
Setting $ P \equal{} \sum {\frac {{\sqrt {a^3 } }}{{\sqrt {b^2 \plus{} c} }}}$
By Holder inequality, we have that:
$ P^2 .\sum {a^3 \left( {b^2 \plus{} c} \right)} .\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \ge \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^4$
It suffices to prove that $ \sum {a^3 \left( {b^2 \plus{} c} \right)} \le 6$ which is true as:
               $ a^3 c \plus{} b^3 a \plus{} c^3 b \le \frac {{\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 }}{3} \equal{} 3.$
         and $ a^3 b^2 \plus{} b^3 c^2 \plus{} c^3 a^2 \le 3$.(1)
(1) is in here:[/quote]
can_hang 2007 wrote :$ a,b,c \ge  0$;$ a^2\plus{}b^2\plus{}c^2\equal{}3$.Prove that:
$ \frac{a}{ab\plus{}3}\plus{}\frac{b}{bc\plus{}3}\plus{}\frac{c}{ca\plus{}3} \le \frac{3}{4}$
.........................
This step: $ \sqrt{3(a^2\plus{}b^2\plus{}c^2)} \ge a\plus{}b\plus{}c\plus{}\frac{2(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)}{(\sqrt{3}\plus{}1)(a\plus{}b\plus{}c)}$ is true
                                                    $ \ge a \plus{}b\plus{}c\plus{}\frac{4(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)}{3(a\plus{}b\plus{}c)}$ is wrong
Then,next steps 's wrong.


ratio
geometry
trigonometry
geometry unsolved

Let's take the equailateral traingle $\ ABC$.A tangential line to the incircle of $\ ABC$ cut $\ AB$ and $\ AC$ in $\ D,E$.Prove that \[\frac{AD}{BD}+\frac{AE}{EC}=1\]

Let's take $AB=a$, $AD=x$ and $AE=y$. Quadriterial $BCDE$ has an inscribed circle so $DE=a-x-y$. From cosine theorem in triangle $ADE$ we have that $x^{2}+y^{2}-xy=(a-x-y)^{2}\Longleftrightarrow \frac{x}{a-x}+\frac{y}{a-y}=1$.

Let AF be a cevian that concurs with BE and CD at P. Since BDEC is a quadrilateral with inscribed circle, P must lie on the line connecting the two tangency points to AB and AC, so PF = PA. Then the ratio sum is:

$\frac{AD}{BD}+\frac{AE}{EC}= \frac{[APC]}{[BPC]}+\frac{[APB]}{[BPC]}= \frac{[ABPC]}{[BPC]}= 1$

And as you can see, we can calculate the ratio pretty easily with all triangles having AB = AC.

[color=darkblue][b]Remark.[/b] The point $P\in BE\cap CD$ belongs to the middle line w.r.t. the vertex $A$ in a equilateral triangle $ABC$.

[size=134][b]Generally[/b][/size]. For any triangle $ABC$ given are two points $M\in [AB]$ and $N\in [AC]$.

Then the line $MN$ is tangent to the incircle of a triangle $ABC\Longleftrightarrow\boxed{\ (p-b)\cdot\frac{MA}{MB}+(p-c)\cdot\frac{NA}{NC}=p-a\ }\ .$

[b]Remark.[/b] Choose conveniently the points $M\in [AB]$ and $N\in [AC]$ so that obtain interesting problems ! Here is [b]three examples[/b] :

$E1.\blacktriangleright$ For any triangle $ABC$ given are two points $M\in AB$ and $N\in AC$ so that the line $BC$ separates the point $A$ of the points $M$ and $N$.

Then the line $MN$ is tangent to the $A$-  exincircle  of a triangle $ABC\Longleftrightarrow \boxed{\ (p-c)\cdot\frac{MA}{MB}+(p-b)\cdot\frac{NA}{NC}=p\ }\ .$

$E2.\blacktriangleright$ Given are the triangle $ABC$ and two points $M\in [AB]$, $N\in [AC]$ so that the line $MN$ is tangent to the its incircle $C(r)$. Then

$MB+MC\ge \frac{4r}{\cos\frac{A}{2}}$ with equality if and only if $AB=AC$ and $MN\parallel BC\ .$

$E3.\blacktriangleright$ Given are the right triangle in the vertex $A$ and two points $M\in AB$, $N\in AC$.

Then the line $MN$ is tangent to the incircle of the triangle $ABC$ if and only if $\boxed{\ \frac{a-b}{MB}+\frac{a-c}{NC}=1\ }\ .$ \[\boxed{\ \mathrm{SUCCESS\ !}\ }\] [/color]


induction
number theory proposed
number theory

Let $ a,n\in \mathbb{N}^{*}$. Prove there exists $ m\in \mathbb{N}^{*}$ , for wich $ n|a^{m}-m$


[quote="Treething"]Jim has a pet goat named Lee.
She is a nice goat, see?

Lee produces four glasses of milk a day. If a seven glasses of milk can be exchanged at the store for a slice of pizza, six slices of pizza can be exchanged at the store for a backpack, and fourteen backpacks can be exchanged for a Meatspin Steak Sandwich, find how many values of x between 0 and 10 exist such that $\frac{2^{x}}{3}$ is an integer, and justify your answer.[/quote]

He never said x was an integer, just a value between 1 and 10. So I would say 341 values.

Oh.  :oops: Then it's
[hide]the values range continuously from $\frac{2^{0}}{3}=\frac{1}{3}$ to $\frac{2^{10}}{3}=341 \frac{1}{3}$. All integers from $1$ to $341$ occur in between, so the answer is $\boxed{341}$.[/hide]

[hide]The highest value of x gives $2^{10}/3=341.333...$, and the lowest, 1/3.

There are 341 integers in between those two numbers. [/hide]

That Meatspin Steak Sandwich costs quite a bit, doesn't it?

I want a meatspin sandwich!  Think of how long it must take the poor goat to produce enough milk to get one  :lol:

[hide="The solution"]If $0<x<10$, then $1<2^{x}<1024$.  The highest multiple of 3 less than 1024 is $1023=3\times341$.  The lowest multiple of 3 greater than 1 is $3=1\times3$.  Therefore, there are 341 values of $x$.[/hide]

That was a nice question!  The actual question was a trick question, and there was lots of useless information (unless, you go to the store that is mentioned in the problem).

[quote="b-flat"]Why was this moved to the Getting Started Forum?   :maybe:

The question does not actually involve higher level math, so unless someone can tell me otherwise, I am moving it back to the Mathcounts forum.  :)[/quote]
Actually it was first posted in the Getting Started forum, not moved there. Also, a question doesn't [i]have[/i] to have higher math to be put in the Getting Started forum. Many MathCounts level questions are also the appropriate level for the Getting Started forum. If you want me to, I can move this back to Getting Started.

Sorry, I had thought it was in the mathcounts forum.  You can do whatever you want.  It would probably be best to move it back to where it came from, though. :blush:

Wait, now it was moved back to mathcounts (and it wasn't me  :o ).  Why was it moved back?

How many times is this going to get moved.

all you have to do is do (2^10)/3 = 341.3333333333333333333...  and try to ignore the meatspin steak sandwiches... sand... meatspinsteak... wich... oh yeah, so the answer is 341 meatspin steak sandwiches.  I mean values of x.

[quote="nebula42"][quote="Treething"]Jim has a pet goat named Lee.
She is a nice goat, see?

Lee produces four glasses of milk a day. If a seven glasses of milk can be exchanged at the store for a slice of pizza, six slices of pizza can be exchanged at the store for a backpack, and fourteen backpacks can be exchanged for a Meatspin Steak Sandwich, find how many values of x between 0 and 10 exist such that $\frac{2^{x}}{3}$ is an integer, and justify your answer.[/quote]
[hide]This is [i]obviously[/i] a trick question  :P , and the answer is $\boxed{0}$ because $2^{x}$ can never have a factor of $3$ for whole number values of $x$.[/hide][/quote]
When I was 8, I wondered "How many times would you have to double one to get a number divisible by three." I knew the divisibility rule for three so it wasn't that bad. I was at 16384 when I just gave up. :rotfl:

[quote="Treething"]Neal is wrong =O, it's not 0[/quote]
I changed my answer a long time ago. Look at my second post.  :wink:
By the way, how are you doing on the AwesomeMath test so far?

[quote="Treething"]what awesomemath test[/quote]
Check your e-mail.  :wink:  If it's not there, maybe you don't have to do it.  :P

Oh that awesomemath test... it's pissing me off

[quote="nebula42"][quote="math92"][hide="my solution"]
$\frac{2x^{0}}{3}=\frac{1}{3}\\\\\frac{2x^{1}0}{3}=\frac{1024}{3}=341\frac{1}{3}$
which means that there are 341 integers [/hide][/quote]
Uhh... don't you mean that
[hide]$\frac{2^{0}}{3}=\frac{1}{3}$ and $\frac{2^{10}}{3}=\frac{1024}{3}=341\frac{1}{3}$? (You shouldn't have an $x$ in your expressions.) Also, if you want an expression like $x^{10}$, you have to write it like x^{10} (which is $x^{10}$) and not x^10 (which is $x^{1}0$).  :wink: [/hide][/quote]
yeah i did mean no x and i did mean 2^{10} instead of 2^10.  Sorry about that guys. I'm going to edit it right away.


[hide="Solution"]
Let $ m$ be the number of male students and let $ f$ be the number of female students.

We have:
$ m \plus{} f \equal{} 530$
$ f \minus{} 20 \equal{} \frac {8}{9}m$

$ m \equal{} 270$ and $ f \equal{} 260$.

So, there are $ 260$ female students in the school.[/hide]

[quote="sri340"]A certain school has 530 students, from which 20 female students and 1/9 of all male students left to attend a math competition. The remaining students are evenly divided between male and female. How many female students are there in the entire school?[/quote]
[hide]$ x$ amount of females, $ 530 \minus{} x$ amount of males. After those who went to the math competition: $ x \minus{} 20$ females left and $ \frac {4240 \minus{} 8x}{9}$ males left. $ x \minus{} 20 \equal{} \frac {4240 \minus{} 8x}{9}$---> $ x \equal{} 260$[/hide]


find $ f R \to R$ such that
$ f(n^2)\equal{}f(m\plus{}n)f(n\minus{}m)\plus{}m^2$ with all $ m;n \in R$

Put $ m \equal{} n \equal{} 0$ $ \Longrightarrow$ $ f(0) \equal{} 0,1$
Case 1. $ f(0) \equal{} 0$. Put $ m \equal{} n$ $ \Longrightarrow$ $ f(x) \equal{} x, \minus{} x$. Hence equation $ \Longleftrightarrow$ $ \pm n^2 \equal{} \pm (m \plus{} n)\pm (n \minus{} m) \plus{} m^2$ $ \Longrightarrow$ $ f(x) \equal{} x$
Case 2. $ f(0) \equal{} 1$. Put $ m \equal{} n$ $ \Longrightarrow$ $ f(m^2) \equal{} f(2m) \plus{} m^2$ (1). Put $ n \equal{} \minus{} m$ $ \Longrightarrow$ $ f(n^2) \equal{} f( \minus{} 2n) \plus{} n^2$ (2). From (1), (2) $ \Longrightarrow$ $ f(x) \equal{} f( \minus{} x)$. Put $ n \equal{} 0$ $ \Longrightarrow$ $ f(x) \equal{} \pm \sqrt {1 \minus{} m^2}$. Hence there is no solution for this case.


Find all prime numbers $ p$ for which $ 3^p\minus{}(p\plus{}2)^2$ is also prime.

Let's try with $ p \equal{} 2$, then $ 3^{2} \minus{} (2 \plus{} 2)^2 \equal{} 9 \minus{} 16 \equal{} \minus{} 7$, then $ p \equal{} 2$ works. Suppose then that $ p$ is odd.
$ 3^{p} \minus{} (p \plus{} 2)^{2} \equiv 1^{p} \minus{} p^{2} \minus{} 4p \minus{} 4 \equiv 1 \minus{} p^{2} \equiv 0 (mod 2)$ and hence it has to be $ 2$, but is easy to see that the equation $ 3^{p} \minus{} (p \plus{} 2)^{2} \equal{} 2$ has the only solution $ p \equal{} 3$. Thus the answer is that there are only two prime number which satisfies the problem, $ p \equal{} 2$ and $ p \equal{} 3$.


i have a question,not being able to think out a answer

a+b+c = 10
find number of whole number solutions to the above equation

i can find out the answer through balls n urns concept ie no of various solutions is equal to 12C2.

but suppose i dont make any distinction between a,b,c
and just wanna know the no of 3 numbers which add up to 10
for example
for the above equation 1+2+7 = 10 is a solution
n so is 2+1+7 = 10
but i dont wanna distinguish between the two cases n wanna treat them as 1 case only {1,2,7}
how many more sets are there..

plz help

Don't double post. A solution is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152694]here[/url].


The AoPS volume 2 is regarded as the "bible" for AIME level math and I have solved many problems with techniques I gained from it. However, why is it that none of the problems I have seen are from AIMEs? Also, just wanted to comment but I think I liked the old design for the cover of Vol. 1 better, though the new one for vol 2 is better (not that my opinion makes a difference).

But there are AIME problems in the Intro series...

We didn't have permission to use the AIMEs back in '93 when we wrote the original books.

[quote="bpms"]Another question, AHSME problems are quoted, but the thing is that some of them say "show that" something or another is true. Was the problem slightly paraphrased from an old AHSME problem, or did the AHSME use to have proofs.[/quote]
It's been a week, can someone please answer my question?

[quote="rrusczyk"]Some questions were paraphrased.[/quote]
Thanks for answering my question,
BPMS

also, the new version has alot of typo's in it.

It would be very helpful if you made a list of them and emailed it to me :)

[quote="rrusczyk"]I repeat, it would be helpful if you emailed them to us rather than simply stating they exist.[/quote]
How many copies of the first six printings combined did you guys sell of vol. 2?

[quote="now a ranger"]Is AoPS volume 2 only available on this website? I have never even seen it on Ebay or Amazon. I just see them from school because we use them as our textbooks in math team.[/quote]
Why would you want it from Ebay or Amazon? Here you can get solutions too...

To be honest, we've kinda lost track of exact sales figures, but we've definitely printed over 10,000 copies of Volume 2.  The actual number is probably closer to 15,000.  Volume 1 is a little bit higher than that.

We are the only regular distributor of the Art of Problem Solving series of books.

Also, I found 2 typos in one section in teh Intro to Number Theory book. Should I email them to you?

Um, yes.  That would be the logical thing to do after reading rrusczyk's posts in this thread...


Hello. I am Sarthak Kalani studying in class XI in Jaipur.I am an IIT aspirant and Have given the astronomy 2nd level recently. I am also an NTSE scholar (and now hoping to be a KVPY scholar too :D ). That's all about me.

sorry i shud have given u the link before...please reintroduce ur sel;f [url=http://www.mathlinks.ro/viewtopic.php?t=181606]here[/url] topic LOCKED


Let $f(x) = \frac{x+1}{x-1}$. Then for $x^2 \neq 1$, $f(-x)$ is

$ \textbf{(A)}\ \frac{1}{f(x)}\qquad\textbf{(B)}\ -f(x)\qquad\textbf{(C)}\ \frac{1}{f(-x)}\qquad\textbf{(D)}\ -f(-x)\qquad\textbf{(E)}\ f(x) $


Let be $ a>0$ and $ b,c\in [1,2)$ such that $ \frac{a\plus{}b}{b(1\plus{}c)}\plus{}\frac{a\plus{}c}{c(1\plus{}b)}\equal{}2$ . Prove that $ a,b,c$ can be the side lenghts of a triangle .

$ \frac {a \plus{} b}{b(1 \plus{} c)} \plus{} \frac {a \plus{} c}{c(1 \plus{} b)} \equal{} 2$

$ \frac {a \minus{} bc}{b(1 \plus{} c)} \plus{} \frac {a \minus{} bc}{c(1 \plus{} b)} \equal{} 0$

$ \therefore a \minus{} bc \equal{} 0$

$ \therefore 4 > b \plus{} c \ge 2\sqrt {bc} \equal{} 2\sqrt {a} > a$ Where the last inequality comes from $ 4 > 2\sqrt {a}$

Therefore $ b \plus{} c > a$

Also $ a \equal{} bc$ and $ b,c \ge 1$ implies that $ a \plus{} b > bc \ge c$. The same applies for $ a \plus{} c$

So we have $ b \plus{} c > a, a \plus{} b > c$ and $ a \plus{} c > b$. Therefore they are sides of a trianlge  :)


[hide="Answer"]$g(x)=\frac{\tan x+\sec x+\frac{1}{\tan x+\sec x}}{2}=\frac{\frac{\sin x+1}{\cos x}+\frac{\cos x}{\sin x+1}}{2}\Rightarrow$
$\frac{\frac{(\sin x+1)^2+\cos ^2x}{\cos x(\sin x+1)}}{2}=\frac{\sin x+1}{\cos x(\sin x+1)}=\frac{1}{\cos x}=\sec x\Rightarrow \boxed{E}$[/hide]

That's not what the answer says..... :noo:

he added instead of subtracted

when you subtract you get:

(D)

because you have
\[ \frac{1}{2}(\frac{\tan^2{x} + \sec^2{x} + 2\tan{x}\sec{x} - 1}{\tan{x} + \sec{x}}) = \tan{x} \]

[hide]$10^{f(x)}=10^{\displaystyle \log_{10}(\tan x + \sec x)} = \tan x +\sec x = \frac{1 + \sin x}{\cos x}$

$10^{-f(x)}=10^{\displaystyle \log_{10}(\tan x + \sec x)^{-1}} = \frac{1}{\tan x +\sec x} = \frac{\cos x}{1 + \sin x}$

$g(x) = \frac{ \displaystyle \frac{1+ \sin x}{\cos x} - \frac{\cos x}{1 + \sin x}}{2}$

$g(x)= \frac{2 \sin ^2 x + 2 \sin x}{2\cos x (1 + \sin x)}= \frac{2 \sin x (1 + \sin x)}{2\cos x (1 + \sin x)}= \tan x$[/hide]


cred ca folosim doar vandermonde :${{m+n}\choose k}=\sum_{l=0}^{\max\{k,n\}}{m\choose{k-l}}{n\choose l}$, cu $m=n=k$,iar  $n$ devine $n=i , i=\overline{0,n},$, si incep sa vina multe produse de 0 pe acolo, si s-ar putea sa iasa frumos.


If $ x^2 \plus{} y^2 \plus{}z^2 \equal{} 2x \plus{} 2y \plus{}4z$, what is the maximum possible value of $ x \plus{}y \plus{}z$?

[hide]$ (x^2 \minus{} 2x \plus{} 1) \plus{} (y^2 \minus{} 2y \plus{} 1) \plus{} (z^2 \minus{} 4z \plus{} 4) \equal{} 6$

$ (x \minus{} 1)^2 \plus{} (y \minus{} 1)^2 \plus{} (z \minus{} 2)^2 \equal{} 6$

$ x \plus{} y \plus{} z \equal{} (x \minus{} 1) \plus{} (y \minus{} 1) \plus{} (z \minus{} 2) \plus{} 4$

Letting $ a \equal{} x \minus{} 1$, $ b \equal{} y \minus{} 1$, $ c \equal{} z \minus{} 2$, our problem now becomes:

[i][color=darkblue]If $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 6$, what is the maximum value of $ a \plus{} b \plus{} c \plus{} 4$?[/color][/i][/hide]

Then BAM! Smack! Lagrange Multipliers.
[hide="Using TZFs Notation"]
Let $ W(a, b, c, \lambda) \equal{} a \plus{} b \plus{} c \plus{} 4 \plus{} \lambda(6 \minus{} c^2 \minus{} b^2 \minus{} a^2)$.
Taking partials with respect to a, b, and c, we get a series of symmetric equations that have the solutions
$ a \equal{} b \equal{} c \equal{} \sqrt{2}$, so the max is $ 3\sqrt{2} \plus{} 4$
[/hide]

Or...

[hide="A Staple Inequality"]$ \frac{a\plus{}b\plus{}c}{3} \leq \frac{|a|\plus{}|b|\plus{}|c|}{3} \leq \sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{3}} \equal{} \sqrt{2}$

[color=darkblue][i]Where the first part is just $ x \leq |x|$ and the second part is AM-QM on $ |a|,|b|,|c|$[/i][/color]

$ \therefore a\plus{}b\plus{}c\plus{}4 \leq 3 \sqrt{2} \plus{} 4$[/hide]

Always fun to overkill some of these things :)

But, nice AM-QM solution. The real work on this problem was the first few steps you took, anyways :)


function
floor function
number theory open
number theory

For $n \in \mathbb{N}$, let $\lambda(n)$ denote the minimum number of equations of the form

$n \equiv r_{k}\bmod p_{k}$

Where $p_{k}\in \mathbb{P}$, needed to uniquely identify $n$ as the smallest positive solution to the system

$n \equiv r_{1}\bmod p_{1}$
$n \equiv r_{2}\bmod p_{2}$

Etc.

For example, for a prime $p$ we have $\lambda(p) = 1$ since $p$ is the smallest positive solution to $n \equiv 0 \bmod p$.

1)  Can you tell me some properties of $\lambda(n)$?  Is it multiplicative in general?  In specific cases?

2)  Can you relate $\lambda(n)$ to other well-known arithmetic functions?

I don't actually know the answer to any of these questions, so feel free to tell me if what I ask is gibberish  :huh:

Let $n$ be arbitrary. Why can't we just take any prime $q>n$, then $x \equiv n \mod q$ has the smallest solution $x=n$¿
So when I didn't miss anything above: it's multiplicative  :lol:

:rotfl:

Would it be a little more interesting if I added the condition $p_{k}\le n \forall k$?  Again, this is just speculating about the properties of a random function.

Then $\lambda(n)=1$ iff $n$ is prime and $2$ otherwise:
The only thing left to prove is that it is $2$ when $n$ is not prime. For that it suffices to find $2$ different primes $p,q$ with $p,q<n$ and $pq>n$.
If $n$ is big ($>20$), then $k=\left\lfloor \sqrt{n}\right\rfloor+1 \leq \frac{n}{4}$. Thus by Bertrands postulate we find two primes $p \in [k,2k]$ and $q \in [2k,4k]$, each smaller than $n$ but their product greater.

This probably can be done without Bertrand, but it appeal to be used here...


Prove that phi(n) is even for any n greater than or equal to 3 (here phi(n) is the Euler phi-
function).

[hide="Solution"]
It is well known that $ \phi (n) \equal{} n(1 \minus{} \frac {1}{d_1})(1 \minus{} \frac {1}{d_2})...(1 \minus{} \frac {1}{d_k})$ where all $ d$ are all of the distinct divisors of $ n$.  This therefore simplifies to $ \phi (n) \equal{} c(d_1 \minus{} 1)(d_2 \minus{} 1)(d_3 \minus{} 1)...(d_k \minus{} 1)$, where $ c$ is $ \frac {n}{d_1d_2d_3...d_k}$.  Thus, $ \phi (n)$ is even if any one of $ d_i \minus{} 1$ is even and thus if $ d_i$ is odd.  So, all $ d_i$ are even for $ \phi (n)$ to be odd.  This implies that $ d_i \equal{} 2$, so all $ d_i$ are $ 2$ and thus $ n \equal{} 2^m$.  This, however gives that $ \phi (n) \equal{} 2^m(1 \minus{} \frac {1}{2}) \equal{} 2^{m \minus{} 1}$.  Notice that $ n > 2\implies m\ge 2$.  So, $ m \minus{} 1\ge 1\implies 2|2^{m \minus{} 1}$.  Thus, $ 2|\phi (n)$ either way.  So, $ \phi (n)$ is always even. [/hide]

EDIT: Ah, okay, thanks Brut3Forc3, edited.

You should probably be clear that the $ d_i$ are distinct prime divisors of $ n$.

@anncar: Please. Spell it "phi."

[hide="Or..."] If $ \gcd(n, a) \equal{} 1$ then $ \gcd(n, n\minus{}a) \equal{} 1$. [/hide]


quadratics
superior algebra
superior algebra unsolved

have this problem, it does not look difficult, but I do not seem to be able to bring n to one side to find N s.t. $ x_{n} \plus{}1$ < $ \frac{1}{10^2}$ for all n > N.

$ x_{n}\equal{}\frac{6\minus{}n^2}{n^2 \plus{} 5*n}$

$ \frac{6\minus{}n^2}{n^2 \plus{} 5*n} \plus{} 1$ < $ \frac{1}{10^2}$

$ \frac{6\plus{} 5*n}{n^2 \plus{} 5*n}$ < $ \frac{1}{10^2}$

$ \frac{(6/(5*n)) \plus{} 1}{(n/5) \plus{}1}$ < $ \frac{1}{10^2}$

would appreciate some help. thanx.

hell0, multiplying your inequalty $ \frac{6\plus{}5n}{n^2\plus{}5n}<\frac{1}{10^2}$ by $ 100$ we get $ 0<n^2\minus{}495n\minus{}600$,applying the formula for quadratic equations we have $ n>\frac{495}{2}\plus{}\frac{5}{2}\sqrt{9897}$.
Sonnhard.


[hide="Approach 1"]Use $2\arctan x=\arctan\frac{2x}{1-x^{2}}, \arctan x-\arctan y=\arctan\frac{x-y}{1+xy}$[/hide]

[hide="Approach 2"]Calculate $\arg\left\{\frac{(5+i)^{4}}{239+i}\right\}$[/hide]

The two facts on complex numbers:

1. $\arg\{a+bi\}=\arctan{b\over a}$

2. $\arg\{uv\}=\arg\{u\}+\arg\{v\},\arg\{{u\over v}\}=\arg\{u\}-\arg\{v\},\arg\{u^{n}\}=n\arg\{u\}$

[quote="me@home"]You are using $\text{arg()}$ as the angle made by a vector to the x axis in the complex plane?[/quote]I believe so, in $r \text{cis}(\theta), \ \theta$ is called the argument (abbreviated arg).

Yup, me@home - Hokkage gave you the answer.


Prove that for all positive integers n and all reals x, we have:

\sum0 \leq i \leq nC(n, i)(-1)i(x - i)n = n!.

i think it is usefull if u prove this... (maybe it's harder.. but today i learned it :D )
\sum {k=0,n+1} (-1)^kC(n+1,k)f(x-i) = 0
for every polynomial f of degree n with integer coeficients


[quote="DarkKnight"][quote="Elyot"]Here's a link to a few tracks I wrote and produced.  The samples and instruments were either made or recorded by me, or downloaded from various free sample sites.

[url=http://www.mp3unsigned.com/showband.asp?id=1315]http://www.mp3unsigned.com/showband.asp?id=1315[/url]

The genre is melodic trance... they're pretty heavy electronic tracks with big melodies and a lot of polyphony.  Each track is about 7-8 minutes, there are free downloads available, and DJs are free to use them in mixes and sets.

There's actually one track in there (Felicity) which has my girlfriend on vocals.[/quote]

Wow, they're great tracks. Listening to all of them right now.[/quote]

Thanks.  I actually did them all in a freeware program called Modplug Tracker (http://www.modplug.com).  I have more tracks but the site only lets me upload 3.

Nice tracks.  I like your style.

I loved the rendition of "God Rest Ye Merry Gentlemen."  Long intro!  But loved the melody over that backbeat once it got there.  Pretty original stuff there.

Liked the other two as well.  One suggestion, its tough to make out what your girlfriend says (other than "this is real")...just a little too whispery and too far back in the mix in "Felicity."  "Dream Dynamo" was good too, I really liked it once it got going about 20-30 seconds into it.

Nice work!  I passed it on to a friend of mine who DJs in Chicago.

I love music and I look forward to hearing other projects folks on this board are working on.  Keep it up...

[quote="DarkKnight"]Can you post something like a tutorial on how to use the MODPlug tracker? I have no idea how to use it.[/quote]

I think there are some tutorials that will help you on the modplug website.  The best thing to do is just play around a little.  Try taking existing songs and modifying them... playing with samples and instruments until you get the idea of what stuff does.  The help file isn't great, but it explains the effects pretty well.

The basic idea is that you import wav files in the samples section, create instruments, and then put together the music in the patterns section.  It takes a bit of figuring out, but once you get the basics, you can do almost anything with it.

AHSIMC, thanks for your comments.  I had never produced vocals before Felicity, so I don't think it's too bad for a first try.  Next time will undoubtedly be better.

I liked them alot.  I'm listening to them right now.

Really great. It is a powerful, emotional trance-like music. My dad likes it. WE are listening to it now!

hi...i'm new here and just reading criss-cross :)

i also 've got a band and we also recorded stuff... maybe you would like to hear 'em ? its:
http://mound.de
There's a section with downloadable mp3s... so please listen to that stuff and tell me your opinion..

thanks...

 :oops:

Hye man i liked your band.. i also play in a band..
http://www.mimamusica.com

thats not the site of my band.. but it is the iste of another band with who i often play .. we are doing our site now.. you can hear some music in that site..
the music we make is quite similar as the one that band does..but it is abit more experimental..

also noticing the style of your band i think you should look for a argentinan songwriter and composer called Gustavo Cerati. Im sure you'll like him, he is incredible..
also by your nickname I assume you like jazz!! I loooove john coltrane


1.the total number of degrees in the sum of the interior angles of two regular of two regular polygons is 1980.the sum of the number of diagonals in the two polygons is 34.what is the positive difference between the numbers of sides of the two polygons?

2.suppose a,b,c,and d are integers satisfying :a-b+c=3,b-c+d=4,c-d+a=1 and d-a+b=6.what is the value of a+b+c+d?

3.the variables a,b,c,d,e and f each represent exactly one of the integers 1 through 6.given the following facts,which integer is represented by e?
a+b=c      b-c=d    e+c=f

4.Groucho,Zippo,and Harpo have $65 altogether.Groucho and Zippo together have $1 more than Harpo has.and Groucho has $15 more than zippo.how many dollars does Groucho have?

5.on monday,30 girls and 0 boys in the Crescent Vally Band missed pratice, resulting in a 2:5 ratio of girls to boys.on tuesday,30 girls and 48 boys missed pratice,resulting in a 2:1 ratio of girls to boys.how many students are in the band?

P.S.plwase answer those questions~i'm so tired to typed that many questions~so please!!!thank you so much!!!   :wink:

                                                                                clover  :D

This is posted in the wrong forum (Middle School>>Classroom Math???). A quick check of the questions suggest that this belongs to HSB (High School Basics) forum.

[hide="Here are some steps"]
1. The sum of the interior angles of a regular n-gon is $ 180(n\minus{}2)$. The number of diagonals in an n-gon is $ \frac{n(n\minus{}3)}2$. Let the two polygons have sides $ n_1$ and $ n_2$. That should give you a system of equations with 2 unknowns.

2. Add them all.

3. If $ a\plus{}b\equal{}c$ and $ b\minus{}c\equal{}d$, $ d<0$. Please clarify and double check the statement of the problem.

4. Typical word problem. Set up equations just like what the problem wants... Let G=Groucho, Z=Zippo, H=Harpo. Then, G+Z+H=65, G+Z=1+H, G=15+Z. Now solve the system...

5. Let g=# of girls, b=# of boys. Then, $ \frac{g\minus{}30}{b\minus{}0}\equal{}\frac25$ and $ \frac{g\minus{}30}{b\minus{}48}\equal{}\frac21$. Again, solve the system...
[/hide]

BTW: For #5, the number of girls stays constant in both the ratio and the number who missed the practice. Thus 48 boys make up 4/5 of the boys, which means there are 60 boys, which means that 24 girls do not miss practice = 54 total, so 54+60=114.

[quote="qntty"]This forum isn't here to do your homework for you. You have enough hints, do it yourself.[/quote]okay sorry i will do it myself .....55555.....fine......i will .....it's my homework after all......thank you for reminder me .......

[quote="nurdyberdy"]I don't think number 3 is possible, because you have stated that
[quote]the variables a,b,c,d,e and f each represent exactly one of the integers 1 through 6
a+b=c
b-c=d
[/quote]

If you see, b<c, but b-c is a positive solution.[/quote]

Or alternatively,

\begin{eqnarray*}a + b &=& c \\ b - c &=& d \\ \hline\\ a &=& -d\end{eqnarray*}

Yet both $ a$ and $ d$ are positive integers. Stuff that makes you go [i]... hmm.[/i]


geometry
circumcircle
geometric transformation
homothety
linear algebra
matrix
incenter

Let $ \Omega$ be the circumcircle of triangle $ ABC$. Let $ D$ be the point at which the incircle of $ ABC$ touches its side $ BC$. Let $ M$ be the point on $ \Omega$ such that the line $ AM$ is parallel to $ BC$. Also, let $ P$ be the point at which the circle tangent to the segments $ AB$ and $ AC$ and to the circle $ \Omega$ touches $ \Omega$. Prove that the points $ P$, $ D$, $ M$ are collinear.

We deploy barycentric co-ordinates and a little thought.

Let $ \Gamma$ denote the circle tangent to $ AB$, $ AC$ and $ \Omega$, and $ \Psi$ the incircle.
$ P$ arises as the external centre of similitude of $ \Gamma$, $ \Omega$ and also lies on $ \Omega$.
Note that a positive homothety through $ A$ takes $ \Psi$ to $ \Gamma$, and the centres of two homotheties and the centre of their composition are collinear. We can thus characterise $ P$ as the intersection of line $ AX_{56}$ with $ \Omega$ ($ P \not= A$), where $ X_{56}$ is the external centre of similitude of the incircle and circumcircle, which has a standard barycentric characterisation in Kimberling's encyclopedia of triangle centres.

$ X_{56}(\frac{a^2}{b+c-a},\frac{b^2}{c+a-b},\frac{c^2}{a+b-c})$

So line $ AX_{56}$ can be parametrically represented as $ (\frac{a^2}{b+c-a},\frac{tb^2}{c+a-b},\frac{tc^2}{a+b-c})$.

Intersecting with the circumcircle $ a^2yz+b^2zx+c^2xy=0$, after cancelling $ a^2b^2c^2$ and clearing denominators we get
\[ t^2(b+c-a) + t(c+a-b+a+b-c)=t(t(b+c-a)+2a)=0\] which is obviously true (since $ 2s=a+b+c$).

Since we have $ A \not= P$, $ t\not=0$, so $ t=\frac{-2a}{b+c-a}$, yielding the co-ordinates of $ P$ as
\[ P(-a, \frac{2b^2}{c+a-b}, \frac{2c^2}{a+b-c}) = P(-a(s-b)(s-c), b^2(s-c), c^2(s-b))\].

The other two points are rather easier to obtain.

The point at infinity intersecting $ x=0$ is $ (0,-1,1)$ so the line through $ A$ parallel to $ BC$ contains $ (1,0,0),(0,-1,1)$, and is therefore $ (1,-t,t)$ ($ t=0,\infty$ will do it). Intersecting with the circumcircle, $ t(b^2-c^2)=a^2t^2$. Again, unless $ b=c$ we have $ M$ distinct from $ A$, so it corresponds to the root $ t=\frac{b^2-c^2}{a^2}$. Thus
\[ M(a^2, c^2-b^2, b^2-c^2)\].

Finally, $ D(0,s-c,s-b)$ is well known and obvious.

To show $ D,M,P$ are collinear it will thus suffice to show that the determinant of the 3x3 matrix with rows equal to these three triples of co-ordinates vanishes. This is given by the expression (taking DMP as the order in my matrix, not that it matters)
\[ (s-c)((b^2-c^2)(-a(s-b)(s-c)) - a^2c^2(s-b)) + (s-b)(a^2b^2(s-c)+a(c^2-b^2)(s-b)(s-c)\]
This vanishes (cancelling through $ a(s-b)(s-c)$) iff
\[ ab^2+(s-b)(c^2-b^2) = ac^2 + (s-c)(b^2-c^2)\]
Which is equivalent to $ a(b^2-c^2) = (s-c-s+b)(b^2-c^2)$. And this is clearly true by $ 2s=a+b+c)$, so we are done.

Let the circle which tangent to $ AB,AC$ and the circumcircle be $ w$ .
Let $ w$ touching the sides $ AB,AC$ at the points $ Q,R$ respectively . Let $ B',C'$ the midpoints of the arcs $ AC,AB$ which don't contain the points $ B,C$ .The the homothety with center $ P$ that sends $ w$ to the circumcircle. It's easy to see that $ Q$ goes to $ C'$ and
$ R$ goes to $ B'$ which means that the lines $ PQ,PR$ are bisectors of the angles
$ \angle{APB},\angle{APC}$ respectively .And also from pascal's theorem to the six points
$ A,B',C,P,B,C'$ we have that the points $ Q,I,R$ are collinear , where $ I$ is the incenter of $ ABC$ .Now from the bisector theorem we have $ \frac {QB}{QA} \equal{} \frac {PA}{PB}$
and $ \frac {RA}{RC} \equal{} \frac {PA}{PC}$ and dividing these two we have that
$ \frac {PC}{PB} \equal{} \frac {RC}{QB}$ (1). From the law of sines at the triangles $ PBD,PDC$ we find that $ \frac {\sin\angle{BPD}}{\sin\angle{DPC}} \equal{} \frac {CP\cdot BD}{BP\cdot CD}$  (2)

Now we have that since $ \angle{ARI} \equal{} 90 \minus{} \frac {A}{2}$ and $ \angle{RCI} \equal{} \frac {C}{2}$
we will have that $ \angle{RIC} \equal{} \frac {\angle{B}}{2}$ so from the law of sines at the triangle
$ RIC$ we have that $ RC \equal{} IC\cdot\frac {\sin\frac {B}{2}}{\cos\frac {A}{2}}$

and similarly we have $ BQ \equal{} IB\cdot\frac {\sin\frac {C}{2}}{\cos\frac {A}{2}}$

So $ \frac {RC}{QB} \equal{} \frac {IC}{IB}\cdot\frac {\sin\frac {B}{2}}{\sin\frac {C}{2}}$

But $ \frac {IC}{IB} \equal{} \frac {CD}{BD}\cdot\frac {\cos\frac {B}{2}}{\cos\frac {C}{2}}$

so $ \frac {RC}{QB} \equal{} \frac {CD}{BD}\cdot\frac {\sin B}{\sin C}$ (3)

So the relation (2) with the help of (1) and (3) becomes

$ \frac {\sin\angle{BPD}}{\sin\angle{DPC}} \equal{} \frac {\sin B}{\sin C}$

so $ \frac {\sin {(B \plus{} C \minus{} \angle{DPC})}}{\sin\angle{DPC}} \equal{} \frac {\sin (B \plus{} C \minus{} C)}{\sin C}$

But the function $ f(x) \equal{} \frac {\sin{(a \minus{} x)}}{\sin x}$ is monotonous so

$ \angle{BPD} \equal{} B$ and $ \angle{DPC} \equal{} C$ . Now if we let PD meets the circumcircle at $ M'$

we will have that $ \angle{M'PC} \equal{} C$ or $ \angle{M'AC} \equal{} C$ so the lines $ AM'$ and $ BC$
are parallel so $ M\equiv M'$ and we are done .

[b]LEMMA[/b] Given a triangle $ ABC$, where $ \alpha$ is circumcircle. $ P \in \alpha$, such that $ AP$ is external bisector. $ I$ incenter of $ ABC$.
$ \beta$ is the circle internally tangent to $ \alpha$,$ AB$ and $ AC$, $ \alpha$ touches $ \beta$ in $ S$.Then $ P,I,S$ are collinear.
----------------------------------------------------

In the problem,

[b]Lemma:[/b] In a triangle $ABC$ with circumcircle $\Omega$, let $J$ be the midpoint of arc $\widehat{BAC}$ and $I$ be the incenter of triangle $\triangle{ABC}$. If $P$ is the point such that the circle $\omega$ tangent to segments $AB$ and $AC$ and to $\Omega$ is tangent to $\Omega$, then $P$, $I$ and $J$ are collinear.

[i]Proof of Lemma.[/i] Let $R$ and $S$ denote the midpoints of arcs $\widehat{AB}$ and $\widehat{AC}$ not containing $C$ and $B$, respectively, and let $\omega$ be tangent to sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $\mathcal{H}$ be the homothety sending $\omega$ to $\Omega$. Note that $\mathcal{H}$ has center $P$. It follows that $\mathcal{H}(X)=R$ and $\mathcal{H}(Y)=S$ and hence that $P$, $X$ and $R$ are collinear and that $P$, $Y$ and $S$ are collinear, since $P$ is the center of homothety $\mathcal{H}$. Hence by Pascal's theorem $X$, $I$ and $Y$ are collinear, since $I$ is the intersection of $CR$ and $BS$. Further since $AX$ and $AY$ are tangents to $\omega$, it follows that $AX=AY$ and, since $AI$ bisects $\angle{XAY}$, it follows that $XY$ is perpendicular to $AI$, with $I$ the midpoint of $XY$. Now note that

\[\angle{IRS}=\angle{CRS}=\frac{1}{2}\angle{ABC}=90^\circ - \frac{1}{2}\angle{BAC} - \frac{1}{2}\angle{ACB}=\angle{JCB}-\angle{RCB}=\angle{RCJ}=\angle{RSJ}.\]

Hence $JS \| IR$ and, by the same argument, $JR \| IS$. Hence $JRIS$ is a parallelogram and if $K$ is the midpoint of $RS$, then $J$, $K$ and $I$ are collinear. It is well known that $R$ is the circumcenter of $\triangle{AIB}$ and that $S$ is the circumcenter of $\triangle{AIC}$. Hence $RS$ is the perpendicular bisector of $AI$. Now, since $AI \perp XY$ and $AI \perp RS$, it follows that $RS \| XY$. Since $I$ and $K$ are the midpoint of $XY$ and $RS$, respectively, it follows that $\mathcal{H}(I)=K$. Therefore, $K$, $I$ and $P$ are collinear, which implies that $P$, $I$ and $J$ are collinear. $\square$

Now let the midpoints of $BI$ and $CI$ be $U$ and $V$, respectively. Let the circumcircles of $\triangle{BUD}$ and $\triangle{CVD}$ intersect at $P' \neq D$. Note that $U$ and $V$ are the circumcenters of right triangles $\triangle{BID}$ and $\triangle{CID}$, respectively. Hence $\angle{UP'B}=\angle{UP'D}=\frac{1}{2}\angle{ABC}$ and $\angle{VP'D}=\angle{VP'C}=\frac{1}{2}\angle{ACB}$. Summing these angles yields that $\angle{BP'D}=180^\circ - \angle{BAC}$, which implies that $ABP'D$ is cyclic. Summing also yields that $\angle{UP'V}=90^\circ - \frac{1}{2}\angle{BAC}=180^\circ - \angle{BIC}=180^\circ - \angle{UBV}$, which implies that $IUP'V$ is cyclic. Since $U$ and $V$ are the midpoints of $BI$ and $CI$, it follows that $UV \| BC$ and hence that $\angle{IP'U}=\angle{IVU}=\frac{1}{2}\angle{ACB}$. Therefore $\angle{IP'B}=90^\circ - \angle{BAC}$, which implies that $J$, $I$ and $P'$ are collinear. By the lemma, since $P'$ lies on $\Omega$, this implies that $P'=P$. Now let $PD$ intersect $\Omega$ at $M'$. Since $PBUD$ is cyclic, $\angle{BPD}=\angle{ABC}$ which implies that the circumcircle of $PBUD$ is tangent to $AB$ and hence that $\angle{BDP}=180^\circ - \angle{PBA}=\angle{PM'A}$. Hence $M'A \| BD$ which implies that $M'=M$. Hence $P$, $D$ and $M$ are collinear. $\blacksquare$


Let $ f(a,b,c) \equal{} pa \plus{} qb \plus{} rc \minus{} 8abc$

Case 1)

 Presume that      $ q\geq r$,    $ a\geq b\geq c$  , then  $ a\geq \frac {1}{3}$

 $ f(a,b,c) \minus{} f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}) \equal{} (q \minus{} r)\frac {b \minus{} c}{2} \minus{} 8abc \plus{} 8a\left(\frac {b \plus{} c}{2}\right)^2\geq 0$

 But $ f(a,\frac {b \plus{} c}{2},\frac {b \plus{} c}{2}) \equal{} \frac {1}{2}[p(3a \minus{} 1) \plus{} (2a \minus{} 1)^2(1 \minus{} a)]\geq 0$ and inequality is prove.

  Equality occurs for
\[ p \equal{} 0 , ,a \equal{} 1 , b \equal{} c \equal{} 0
\]
Or
\[ p \equal{} 0 , a \equal{} \frac {1}{2} , b \equal{} c \equal{} \frac {1} {4} ,
\]
Case 2)  $ p\leq q\leq r$    ,  $ a\geq b\geq c$    :?:

[quote="Sylwek"]Let: $ a,b,c,p,q,r \geqslant 0$, and: $ a \plus{} b \plus{} c \equal{} p \plus{} q \plus{} r \equal{} 1$, and also: $ p,q,r \leqslant \frac {1}{2}$.

Prove: $ 8abc \leqslant pa \plus{} qb \plus{} rc$. When the equality occurs?[/quote]
$ p\plus{}q\minus{}r\equal{}1\minus{}2r\geq0.$
Let $ p\equal{}\frac{y\plus{}z}{2},$ $ q\equal{}\frac{x\plus{}z}{2},$  $ r\equal{}\frac{x\plus{}y}{2}$ and $ c\equal{}\max\{a,b,c\}.$
Hence, $ 8abc \leqslant pa \plus{} qb \plus{} rc\Leftrightarrow16abc\leq(1\minus{}x)a\plus{}(1\minus{}y)b\plus{}(1\minus{}z)c\Leftrightarrow$
$ \Leftrightarrow16abc\plus{}ax\plus{}by\plus{}cz\leq1\Leftrightarrow16abc\plus{}(a\minus{}c)x\plus{}(b\minus{}c)y\leq a\plus{}b.$
Thus, it remains to prove that $ a\plus{}b\geq16abc.$
But $ a\plus{}b\geq16abc\Leftrightarrow(a\plus{}b)(a\plus{}b\plus{}c)^2\geq16abc\Leftrightarrow$
$ (a\plus{}b)c^2\plus{}2(a^2\minus{}6ab\plus{}b^2)c\plus{}(a\plus{}b)^3\geq0,$ which is true because
$ (a^2\minus{}6ab\plus{}b^2)^2\minus{}(a\plus{}b)^4\equal{}\minus{}16ab(a\minus{}b)^2\leq0.$


[quote]
In problem 7.4. , the final words must be "is greater or equal than 8". Doesn't it?
[/quote]

Which problem?


ok, so i got this problem.
You have to find $ \ (x,y)$
$ \ x\minus{}y\equal{}2y\plus{}3$
$ \ x\minus{}y\equal{}5y\minus{}3$
I got $ \ (\frac {21}{5},\frac {2}{5})$
The book says no solutions.
Help?

I think you copied it wrong because the only question that looks like your's in AOPS vl 1 is:
\begin{align*}
x-y&=2x+3\\
x-2y&=5-3y\\
\end{align*}

And If you add these and rearrange you have $ 0=8$ So there are [b]no solutions[/b]

[hide="Solution"]Setting the left hand side of both equations equal (since the right hand sides are equal), we get:

$ 2y+3=5y-3 \\ \\
y=2$

and substituting gives:

$ x-2=2(2)+3 \\ \\
x=9$

and finally the ordered pair:

$ (x, y)=\boxed{(9, 2)}$[/hide]

Can you give us a page number or problem/exercise number?

i think you copied it wrong because if you solve this equation you get(7,2)

actually the answer is (9,2)
calcwiz u are wrong.
please check ur solution

AlphaBetaTheta and PowerOfPi are correct. Plug the numbers in
9-2=2(2)+3
7=7
and
9-2=5y-3
7=10-3
7=7


1)$ a_0\equal{}0$
$ a_1\equal{}1$
$ a_{n\plus{}2}\equal{}4a_{n\plus{}1}\minus{}a_n$
Prove that
$ 3^k|a_n\leftrightarrow 3^k|n$

It followed from $ a_n\equal{}\frac{(2\plus{}\sqrt 3 )^n\minus{}(2\minus{}\sqrt 3 )^n}{2\sqrt 3}$.


superior algebra
superior algebra unsolved

This seems to be a rather deep result: if $k\in K$ is a finite extension of fields and $K$ is an algebraic closure of $k$, then the extension has degree $2$. I have a very long and difficult proof of this result. Is there an easy approach?


geometry
3D geometry
number theory unsolved
number theory

Find all four number from 100 to 1000 such that the sum of the cubes of each digits equals to this number.
Thank you.

The four numbers are $ 153$, $ 370$, $ 371$, and $ 407$. I got this solution by finding some bounds on the sums of cubes depending on the values of the digits, but mostly it was just brute force and trial and error. I think there does not exist a nice solution to this problem.

Thank you. However, can you show me more the way you did?

Can anyone show me the way to reduce the possibility?
Thank you.


ratio
trigonometry
analytic geometry
graphing lines
slope
calculus
integration

A triangle has vertices $P=(-8,5)$, $Q=(-15,-19)$, and $R=(1,-7)$. The equation of the bisector of $\angle P$ can be written in the form $ax+2y+c=0$. Find $a+c$.

Solution posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=433508#p433508]here[/url] (post #91)

use the distance formula to find the lengths of the 3 sides of the triangle, then use angle bisector formula to find the lengths that the bisector divides QR into; use that ratio to find the point that it bisects QR at, and then use the 2 pts u now have to find the equation of the line... i don't see what's so hard about this problem, seems pretty straightforward to me (the numbers work out pretty nicely too)...

[hide="Alternate Solution"][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=652&sid=db10b9bf5177394d139fed02be513fc0[/img]

By the distance formula, we see that $PR=15$, $QR=20$, and $PQ=25$, so $\angle PRQ=90^\circ$. Therefore, we can extend $PR$ to point $I$ such that $\triangle PQI$ is isosceles. Some quick calculation yields that $I=(7,-15)$. Because $\angle P$ is the angle to the base of the triangle, its bisector bisects $IQ$ at $M$. By the midpoint formula, $M=(-4,-17)$. Using points $P$ and $M$, our equation is $11x + 2y + 78 = 0$, and $a+c=\boxed{089}$.[/hide]

[quote="Hikaru79"][quote="4everwise"]I typed up this solution to experiment with drawing diagrams on a graph. :D[/quote]
Neat diagram :D What exactly did you use to produce it?[/quote]Geometer's Sketchpad.  :)

[hide="Answer"]We have a 15-20-25 right triangle. Therefore, $\tan P=\frac{4}{3}$. We want $\tan \frac{P}{2}$. Using double-angle formula, we have $\tan P=\frac{4}{3}=\frac{2\tan \frac{P}{2}}{1-\tan ^2\frac{P}{2}}$. Simplifying, we have $2\tan ^2\frac{P}{2}+3\tan P-2=(\tan \frac{P}{2}+1)(2\tan \frac{P}{2}-1)=0$, so we have $\tan \frac{P}{2}=\frac{1}{2}$, as we cannot use a negative value. Therefore, if we label the point where the bisector touches $QR$ as $O$, we have $OR=\frac{15}{2}$. $PR$ has slope $\frac{-4}{3}$, so $OR$ has slope $\frac{3}{4}$. Solving this, we find $O\left(-5,-\frac{23}{2}\right)$. From there, we have that the slope of $OP$ is $\frac{-11}{2}$. Then, the $y$-intercept is $-39$, and we have $y=-\frac{11}{2}x-39\Rightarrow 11x+2y+78=0\Rightarrow 11+78=\boxed{89}$.[/hide]

JF197: $\tan (\theta/2)=\frac{1-\cos \theta}{\sin \theta}=\frac{\sin\theta}{1+\cos \theta}$.

[hide="Angle Bisector Solution"]We can easily find that $PQ=25$, $PR=15$, and $QR=20$.  Let H be the point of intersection of the angle bisector and $QR$. From the Angle Bisector Theorem, we know that $\frac{QH}{HR}=\frac{3}{5}$ and $\frac{QH}{QR}=\frac{3}{8}$.  From there, we can find that the point of intersection of the angle bisector and $QR$ is $(-5,-\frac{23}{2})$.  Then the slope of the angle bisector is $-\frac{11}{2}$.  We write the equation in point-slope form and manipulate it into $11x+2y+78=0$, and our answer is $11+78=\boxed{89}$.[/hide]

The equation of PR is $ 4x\plus{}3y\plus{}17\equal{}0$ and PQ is $ 24x\minus{}7x\plus{}227\equal{}0$

Then from the distance formula we have

$ \frac{|4x\plus{}3y\plus{}17|}5\equal{}\frac{|24x\minus{}7x\plus{}227|}{\sqrt{24^2\plus{}7^2}}$

$ 5|4x\plus{}3y\plus{}17|\equal{}|24\minus{}7x\plus{}227|$

From here, how do we determine what sign to take (i.e., +ve or -ve for the expression in the absolute value)?

I just tried the two different cases and it turned out that the one with one +ve and one -ve produces the $ 2$ as the coefficient for $ y$ (with the coefficients of other terms integral of course)... so we have

$ 5(4x\plus{}3y\plus{}17)\equal{}\minus{}(24\minus{}7x\plus{}227)$
$ 11x\plus{}2y\plus{}78$

Now in general how do we determine the signs (i got lucky in this problem since it hints that the coefficient of y must be 2, but what if i'm not so lucky?)?

I did the same thing [b]mathwizarddude[/b] did the first time I tried this problem. Does anyone know how to determine the sign?

[quote="mathwizarddude"]
Now in general how do we determine the signs (i got lucky in this problem since it hints that the coefficient of y must be 2, but what if i'm not so lucky?)?[/quote]

SIGN? I think if you graph it, use [ + ] sign if the ORIGIN (0,0) is inside the SPACE enclosed by the intersecting lines where the perpendicular bisector passes, otherwise  [ - ].

wow was not aware 19-7.5 was not equal to 12.5

[hide=Solution]Let the intersection of the bisector and $QR$ be $P$. Then from the Angle Bisector Theorem, we have $\frac{QP}{PR}=\frac{5}{3}$, so $H=(-5, -11.5)$. This means the equation is $11x+2y+78=0$, so $a+c=\boxed{89}$.[/hide]


[hide]
First, we will define $ f(x) \equal{} 1 \plus{} \frac {t^4}{4!} \plus{} \frac {t^8}{8!} \plus{} \cdots$,then we will find $ f(\frac {\pi}{2})$
It is known that :
$ {e}^{x} \equal{} 1 \plus{} x \plus{} \frac {x^2}{2!} \plus{} \frac {x^3}{3!} \plus{} \frac {x^4}{4!} \plus{} \cdots$
$ {e}^{ix} \equal{} 1 \plus{} ix \plus{} \frac {( \minus{} x^2)}{2!} \plus{} \frac {( \minus{} ix^3)}{3!} \plus{} \frac {x^4}{4!} \plus{} \cdots$
$ {e}^{ \minus{} x} \equal{} 1 \plus{} ( \minus{} x) \plus{} \frac {x^2}{2!} \plus{} \frac {( \minus{} x^3)}{3!} \plus{} \frac {x^4}{4!} \plus{} \cdots$
$ {e}^{ \minus{} ix} \equal{} 1 \plus{} ( \minus{} ix) \plus{} \frac {( \minus{} x^2)}{2!} \plus{} \frac {ix^3}{3!} \plus{} \frac {x^4}{4!} \plus{} \cdots$
Now add the forth equation to obtain that :
$ e^x \plus{} {e}^{ix} \plus{} {e}^{ \minus{} x} \plus{} {e}^{ \minus{} ix} \equal{} 4(1 \plus{} \frac {x^4}{4!} \plus{} \frac {x^8}{8!} \plus{} \cdots)$
$ f(x) \equal{} \frac {e^x \plus{} {e}^{ix} \plus{} {e}^{ \minus{} x} \plus{} {e}^{ \minus{} ix}}{4} \equal{} (1 \plus{} \frac {x^4}{4!} \plus{} \frac {x^8}{8!} \plus{} \cdots)$, $ f(\frac {\pi}{2}) \equal{} \frac {{e}^{\frac {\pi}{2}} \plus{} {e}^{\frac { \minus{} \pi}{2}}}{4}$
Then,
$ \frac {e^x \minus{} {e}^{ix} \plus{} e^x \minus{} {e}^{ \minus{} ix}}{4} \equal{} (\frac {x^2}{2!} \plus{} \frac {x^6}{6!} \plus{} \frac {{x}^{10}}{10!} \plus{} \frac {{x}^{14}}{14!} \plus{} \cdots)$
$ g(x) \equal{} \frac {e^x \minus{} {e}^{ix} \plus{} e^x \minus{} {e}^{ \minus{} ix}}{4x^2} \equal{} (\frac {1}{2!} \plus{} \frac {x^4}{6!} \plus{} \frac {{x}^{8}}{10!} \plus{} \frac {{x}^{12}}{14!} \plus{} \cdots)$. Then $ g(\frac {\pi}{2}) \equal{} \frac {2{e}^{\frac {\pi}{2}}}{\pi^2}$
Then the expression in the problem can be simplified in the form :
$ \frac {f(\frac {\pi}{2})}{g(\frac {\pi}{2})} \equal{} \pi^2.\frac {2{e}^{\frac {\pi}{2}} \plus{} {e}^{\frac { \minus{} \pi}{2}}}{4{e}^{\frac {\pi}{2}}} \equal{} \pi^2.\frac {1 \plus{} {e}^{ \minus{} \pi}}{2}$
[/hide]


geometry
integration
trigonometry
symmetry
calculus
rectangle
calculus computations

given that the line $y=c$ intersects the graph $y=sin(x)$ in the first quadrent for $x<\pi$, and that the two reginons formed by this intersection have equal area find that value of $k$.

sorry but i would provide a picture if i knew how.
assume that they intersect at $(a,c)$ and $(b,c)$
the regions of equal area are those to the left of $x=a$ between the two graphs and the one between $x=a$ and $x=b$ that is above $y=k$ and under $y=sin(x)$.

[hide="here is what i have so far"]
$cb=\displaystyle\int^b_0sin(x)dx$
$cb=1-cos(b)$
$b(sin(b))=1-cos(b)$
$b^2sin^2b=1-2cos(b)+cos^2b$
then i use $cosb=1-bsin(b)$ and get
$b^2sin^2b=1-2-2bsinb+1-sin^2b$
$b^2sin(b)+sin(b)+2b=0$ for $b>0$
and this is where i am stuck.
i am wondering if it is possible to find the exact value of $b$ by using this method.
any help would be greatly appreciated.
[/hide]

This comes down to

\[ \int_a^b(\sin x-c)\,dx=1 \]

By symmetry of the curve, $a+b=\pi$ and so $\sin a=\sin b$ and $\cos a=-\cos b$

But, from the graph $c=\sin a$

This reduces to the equation $2\cos a-(\pi-2a)\sin a=1$

[quote="maokid7"]could you please further explain?[/quote]Which part is unclear?

The area under the sine curve from 0 to $\pi$ is 2, so the divided area is 1.

Now, if you take $y=c$ as the 'base' line, the area under the sine curve and above line $y=c$ is the integral I wrote above.

You will need to do some simplication to arrive at the last equation I wrote. That last equation cannot be solved by elementary means. Use a calculator, that is the fastest way to obtain an approximate value of $a$.

o ok sorry.

but not all of the shaded regions are in the sine graph.
part of it is to the left and uner $y=c$ while the rest is in the graph and above $y=c$ and $y=sin(x)$.

I would gladly put up a picture if someone will tell me how i can do so  :)

Let me see if I understand the problem.
There is a sine curve, and the line y=c
Then y=c intersects the sine curve at (a,c) and (b,c)
The question is, given that these two areas are equal:
1. Area between [0,a] above the sine curve and below the line y=c
2. Area between [a,b] above the line y=c but below the sine curve
Find c.

If that's the question, then:
Call those areas A.
The whole rectangle from 0 to $\pi$ = $\pi$
But this is also equal to the area under the sine curve, plus the area of the rectangle from 1-c to 1, from 0 to pi (the upper rectangle), minus A (for the area between the two curves in the middle), plus 2A (for the area on the sides).

That gives the following equation:
$\pi=2+\pi(1-c)-A+2A$
$A=\pi c -2$

You can also find A by considering the first piece on the left.  (It is the rectangle minus the integral)
$A=ac-\int_0^a sin(t)dt=ac+cos(a)-1$

Setting A=A
$ac+cos(a)-1=\pi c -2$
And c=sin(a)
$(a-\pi)sin(a)+cos(a)+1=0$

Now I don't know how or if you can find the exact value, but I think a way to approximate this would be to use a Taylor series on sine and cosine.

I just put it in my calculator and I found the following:
a=.81047
c=sin(a)=.72461
A=.27643


Given integer n, how to check if it's prime or not?? I know that Erastotenes sieve can be used, but when the number is very big (for example 2^63-1), it takes ages to find out is it prime or not... Is there any algorithm to check if number is prime or not??

There was a recently discovered (past 2 years?) algorithm that runs is O(n^12) time where n is the number of digits.  If the Riemann Hypothesis is true, I think that it ends up running in O(n^3) or can be modified to do so.  I suggest, for your particular number, that you consider using more standard tricks such as looking for ways to factor numbers that have special forms.  Also, this is the kind of thing you could research online fairly easily.

Tytus -  You may like to  look here:
[url=http://www.utm.edu/research/primes/prove/index.html]Primality Proving ..[/url]

You may also like to search through AoPS messages  (Even I might have posted answers  to such messages  here before  :))

One of the  well known and easy method,  is using Ferma't last theorem... and check  if   $a^{n-1} \equiv 1 \bmod n$
Generally you can select  $a=2$   or something small   and  start taking the powers (by  repeated squaring - see the reference above)  you can find $a^n$ fairly fast for even large  $n$  ( For example to, check say a (40 digit number) you may need only  100-150  operations and may need only a few seconds of cpu time on  a desktop pc)

The problem is that  these methods  are not 100% accurate..If the number is shown as  a non-prime you are 100%  sure, but   this method  can falsly  detect some  numbers  as "psedo-prime"  which are not really  prime .. (Though you can modify the test  a little and in practice  you   can be more than  99.99999% correct .. Not 100% but fairly close to it  ...:))

[b]For special  type of numbers (eg 2^n  -1 ) these tests can be modified to provide definite answers.
[/b]
For example to test a number like  $2^{63} - 1$  if it is prime or not, one can use  a modern method  ( and this is   NOT hard to understand)   You do this by the following algorithm:

say $N = 2^{63}-1$

Start with  $S_0 = 4$

and consider the recurrence equation

  $S_n = S_{n-1}^2 -2   \bmod  N$

You will start getting numbers like  4, 14, 194 etc...
 and calculate  $S_{61}$  (That is 63-2)
 If  this is  0 mod $N$  the number is prime else it is not.
Since there are only  a few (about 60)  calculations are envolved .. it is not hard to do.

(If above is not clear do a google on Lucas Lehmer test and you may find many references)

Recently some undergrads from my alma mater  (IIT Kanpur, India)  with their young  prof found a method, which   can be used for any type of number  and that is  what JBL, I think,   is talking about. ..

If interested look here:

[url=http://www.cse.iitk.ac.in/news/primality.html]www.cse.iitk.ac.in/news/primality.html[/url]

(Side note: Prof  Agrawal  gave a talk  about this at Duke - about a year ago - and my son - an undergraduate - was ablle to follolw through - so the beauty of this method  is that it is not that advanced to understand)...

BTW,  to answer your question ,  yes the numbers like 2^63-1  can be tested  for primality fairly fast.  The test is easy  (see above)


How would you prove that (1+i) divides (a+bi) iff a is congruent to b (mod 2)?  I know that means that a and b would have to be either both positive or both negative but don't know where to take it for there.

$1+i$ divides $2$.  Hence if $a \equiv b \bmod 2$ then either $a, b \equiv 0 \bmod 2 \implies a, b \equiv 0 \bmod (1+i)$ (trivial) or $a, b \equiv 1 \bmod 2$ (in which case $a, b \equiv 1 \bmod (1+i)$ so $a+bi \equiv 1+i \bmod (1+i)$. )

By the way, it is not true that $a, b$ both have to be of the same sign.  For example,

$1+i |-2+2i$

What about the converse?  because you can't trade mod (1+i) for mod 2 since 2>(1+i)  correct?

Now, suppose $1+i$ divides $a+bi$.

Then, $\exists (c+di)$ such that $(c+di) * (1+i) = (c-d)+(c+d)i = a+bi$.

Therefore, $c-d = a$ and $c+d = b$.

So, $b-a = 2d \equiv 0 \, (\bmod \, 2)$.

Therefore, $b \equiv a \, (\bmod \, 2)$.

QED.


geometry
3D geometry
tetrahedron
pyramid
Pythagorean Theorem

[hide]A tetrahedron has an edge length $s$ and it is a pyramid, so the volume is $\frac13BH$, where $B$ is the area of the base and $H$ is the height to the base, $B$.  Each side is a equilateral triangle, which has an area of $\frac{s^2\sqrt3}{4}$.  The height forms a right traingle, when the altitude of one of the sides is drawn and part of the altitude from the point where the altitude meets the base is drawn to the base.  I do not have a picture, so I hope this is relatively clear.

The hypotenuse is going to be $\frac{s\sqrt3}2$, and the smaller leg is going to be $\frac{s}{2\sqrt3}=\frac{s\sqrt3}{6}$.  Now, use the pythagorean theorem to solve for the height:  $\left(\frac{s\sqrt3}2\right)^2-\left(\frac{2\sqrt3}{6}\right)^2=\frac{3s^2}4-\frac{3s^2}{36}=\frac{27s^2-3s^2}{36}=\frac{24s^2}{36}=\frac{2s^2}3$  Take the square root to get:$\frac{s\sqrt2}{\sqrt3}=\frac{s\sqrt6}{3}$

Now, the volume of the tetrahedron can be expressed as $\frac{3s^3\sqrt2}{36}=\frac{s^3\sqrt2}{12}$, which is 9.  Set them equal and solve:
$\frac{s^3\sqrt2}{12}=9$
$s^3\sqrt2=108$
$s^3=54\sqrt2$
$s=3\sqrt[3]{2\sqrt2}$
$s=3\sqrt[6]{8}$
EDOT:  simplifies to $3\sqrt2$
[/hide]

Yah, I just realized that.  I thought mine was a little messy.

[hide]
According to Wikipedia (muhahahaha), the volume of a [b]regular[/b] tetrahedron is $\frac{s^3\sqrt 2}{12}$.

Equating that to 9, we get $s^3=\frac{108}{\sqrt 2}=54\sqrt 2$.

So $s=\sqrt[3]{54\sqrt 2}$.

Edit: With a friendly reminder from b-flat:
$\sqrt[3]{54\sqrt 2}=3\sqrt[3]{2\sqrt 2}=3\sqrt{2}$

Woot. :w00t:
[/hide]

[quote="mathnerd314"][hide]
According to Wikipedia (muhahahaha), the volume of a [b]regular[/b] tetrahedron is $\frac{s^3\sqrt 2}{12}$.

Equating that to 9, we get $s^3=\frac{108}{\sqrt 2}=54\sqrt 2$.

So $s=\sqrt[3]{54\sqrt 2}$.

Unless I messed up...
[/hide][/quote]

You got the correct answer, just it is not simplified.  Look at my first post to see how to simplify that (and to figure out the formula for the volume of a tetrahedron instead of just trusting Wikipedia ;) )

[quote="Iron Fist"]A [b]regular[/b] tetrahedron has a volume of 9. Find the length of one edge.[/quote]

[hide]$3 \sqrt{2}$[/hide]


limit
real analysis
real analysis solved

We have the sequnces [tex]x_{n}[/tex] and [tex]y_{n} [/tex] which verify: [tex] (1+\sqrt{2})^{n}=x_{n}+y_{n}\sqrt{2}[/tex] forall n in N. Compute  [tex] \lim_{n\to \infty} \sum_{k=1}^{n} \arctan \frac{1}{6x_{k}^{2}y_{k}^{2}}  [/tex].

Just an idea,

Use $x_n -\sqrt2.y_n$ to an expression of ${x_n}$, ${y_n}$
Give an expression of $\arctan\frac{1}{6x_{k}^2y_{k}^2}$ like
 $arctan(u_{k+1}) - arctan(u_k)$ for the limit of the sum.

We claim that $\sum_1^n arctan(\frac{1}{6x_k^2y_k^2}) = arctan(\frac{x_ny_n}{x_{n+1}y_{n+1}})$

Proof: The result is true for $n=1$.  Now assume it is true for $n = j$.  Then $\sum_1^{j+1} arctan(\frac{1}{6x_k^2y_k^2}) = arctan(\frac{x_j y_j}{x_{j+1}y_{j+1}} + arctan(\frac{1}{6x_{j+1}^2y_{j+1}^2}) = arctan(\frac{x_{j+1}y_{j+1}}{x_{j+2}y_{j+2}}) (*)$, completing the induction.

Note: we can derive (*) using the collection formula $arctan( u) + arctan(v) = arctan(\frac{u+v}{1-uv})$ and the recurrence $x_{k+1} + sqrt(2) y_{k+1} = (1+sqrt(2))(x_k+sqrt(2) y_k)$.

Since $lim \frac{x_n}{x_{n+1}} = lim \frac{y_n}{y_{n+1}} = \frac{1}{1+sqrt(2)}$, $lim \sum_1^n arctan(\frac{1}{6x_k^2y_k^2}) = arctan(\frac{1}{(1+sqrt(2))^2}) = arctan(3-2sqrt(2))$.


Let $ a,b,c>0$. Prove that

$ \sum\sqrt{(a\plus{}c)(b\plus{}c)}\geq\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\plus{}a\plus{}b\plus{}c$

[quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \sum\sqrt {(a \plus{} c)(b \plus{} c)}\geq\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \plus{} a \plus{} b \plus{} c$[/quote]

$ \sqrt {(a \plus{} c)(b \plus{} c)}\ge \sqrt{ab}\plus{}c$

[quote="gigaman"][quote="Ligouras"]Let $ a,b,c > 0$. Prove that

$ \sum\sqrt {(a \plus{} c)(b \plus{} c)}\geq\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \plus{} a \plus{} b \plus{} c$[/quote]

$ \sqrt {(a \plus{} c)(b \plus{} c)}\ge \sqrt {ab} \plus{} c$[/quote]

Very nice... GIGAMAN thank you


geometry
incenter
analytic geometry
trigonometry
complex numbers
trig identities
Law of Sines

In a triangle $ ABC$, draw altitudes $ AD, BE, CF$ and let $ H$ be the orthocenter. Let $ O_1, O_2, O_3$ be the incenters of $ \triangle EHF, \triangle FHD, \triangle DHE$ respectively. Prove that $ AO_1, BO_2, CO_3$ are concurrent.

Construct the triangle $ \triangle A'BC$ outside of $\triangle ABC,$ such that $ \triangle A'BC \sim \triangle O_1EF$  and define analogously the points $ B',C'.$ Since $ EF$ is antiparallel to $ BC$ with respect to $ AB,$ $AC,$ the rays $ AO_1$ and $ AA'$ are isogonals with respect to $ AB,AC.$ Similarly, $ BB'$ and $CC'$ are the isogonals of $ BO_2$ and $CO_3.$  By angle chase, we have then

$ \angle CBA' \equal{} \angle ABC' \equal{} \frac {\pi \minus{} 2\angle B}{4} $

$ \angle BCA' \equal{} \angle ACB' \equal{} \frac {\pi \minus{} 2\angle C}{4}$

$ \angle CAB' \equal{} \angle BAC' \equal{} \frac {\pi \minus{} 2\angle A}{4}$

By Jacobi's theorem, the rays $ AA',BB',CC'$ concur at the Jacobi's perspector $ J$ relative to the triple $ \left (\frac {_{\pi \minus{} 2\angle A}}{^{4}} , \frac {_{\pi \minus{} 2\angle B}}{^{4}} , \frac {_{\pi \minus{} 2\angle C}}{^{4}} \right ).$ Therefore, the rays $ AO_1,BO_2,CO_3$ concur at the isogonal conjugate $ J'$ of $ J.$

It could also be proven in the following way making use of another theorem:

Let $ %Error. "trienagle" is a bad command.
ABC$ be a triangle, and let $ D,E,F$ be points on sides $ BC,CA,AB$, respectively, such that the cevians $ AD,BE,CF$ are concurrent. If $ M,N,P$ are points on $ EF, FD,DE$, respectively, then the lines $ AM, BN, CP$ are concurrent if and only if the lines $ DM, EN, FP$ concur. We can use this to solve the problem.

We know from triangle geometry that the orthocenter of a triangle is the incenter of its orthic triangle, therefore $ \angle EFO_1 = \angle DFO_2$, $ \angle FDO_2 = \angle EDO_3$, $ \angle DEO_3 = \angle FEO_1$.

Then, according to the same Jacobi theorem, the lines $ DO_1, EO_2, FO_3$ are concurrent, and because of the fact that the heights of any triangle are concurrent, the lines $ AO_1,BO_2,CO_3$ meet in one point by the theorem above.

Does anybody know if that theorem has a name? And/or has a proof that does involve barycentric coordinates or complex numbers?

By the way, it was the hardest problem of BW 2009. Only one team solved it.

[quote="Sassha"]It could also be proven in the following way making use of another theorem:

''Let $ \triangle ABC$ be a triangle, and let $ D,E,F$ be points on sides $ BC,CA,AB,$ respectively, such that the cevians $ AD,BE,CF$ are concurrent. If $ M,N,P$ are points on $ EF, FD,DE,$ respectively, then the lines $ AM, BN, CP$ are concurrent if and only if the lines $ DM, EN, FP$ concur.''

Does anybody know if that theorem has a name? And/or has a proof that does involve barycentric coordinates or complex numbers?[/quote]
Dear Sassha the theorem you mentioned is known as Cevian Nest Theorem, you can see a synthetic proof in [url]http://pagesperso-orange.fr/jl.ayme/[/url] (Volume 3).  But I'm afraid that this is not a direct application of the Cevian Nest Theorem as you claimed, since $ O_1,O_2,O_3$ do not lie on the sidelines of the orthic triangle $ \triangle DEF.$

I solved this using barycentric coordinates.

Given a triangle $ ABC$, let $ A_1 \equal{}(a_1,a_2,a_3),B_1\equal{}(b_1,b_2,b_3),C_1\equal{}(c_1,c_2,c_3)$. ($ a_1\plus{}a_2\plus{}a_3\equal{}b_1\plus{}b_2\plus{}b_3\equal{}c_1\plus{}c_2\plus{}c_3\equal{}1$) Then if the barycentric coordinate of $ P$ with regard to the triangle $ A_1B_1C_1$ is $ (x,y,z)$, the barycentric coordinate of $ P$ with regard to the triangle $ ABC$ is $ (xa_1\plus{}yb_1\plus{}zc_1,xa_2\plus{}yb_2\plus{}zc_2,xa_3\plus{}yb_3\plus{}zc_3)$. Since $ H\equal{}\frac{1}{\tan A\plus{}\tan B\plus{}\tan C}(\tan A, \tan B, \tan C)$, $ E\equal{}\frac{1}{\tan A\plus{}\tan C}(\tan A,0,\tan C)$, $ F\equal{}\frac{1}{\tan A\plus{}\tan B}(\tan A,\tan B,0)$, we can calculate the coordinate of $ O_1$, and hence the coordinate of $ P_1\equal{}AO_1 \cap BC$. Apply Ceva's theorem.

For a synthetic proof of the Jacobi's theorem, you can see
http://perso.orange.fr/jl.ayme  vol. 5  Le theoreme de Jacobi
Sincerely
Jean-Louis

There is also a succint solution using the trigonometric version of Ceva's theorem. First of all, we see that <DEF=180-2B, and because the altitude from B bisects this angle, we have that <FEH=90-B. Thus, <O_1EF=45-B/2, and <O_1FE=45-C/2.

Now, we see that [FAO_1]/[O_1AE] = AF*sin<FAO_1/AE*sin<EAO_1. However, we see that [FAO_1]/[O_1AE] is also equal to AF*FO_1*sin<AFO_1/AE*EO_1*sin<AEO_1. Thus, sin<FAO_1/sin<EAO_1= FO_1*sin<AFO_1/EO_1*sin<AEO_1. Now, because triangles AFE and ACB are similar <AFE=<ACB. Thus AFO_1=45-C/2+C=45+C/2.Similarly, <AEO_1=45+B/2. Using the law of sines in triangle FO_1E, we get that O_1E/O_1F=sin<45-C/2 / sin<45-B/2. Putting everything together, we have sin<FAO_1/sin<EAO_1 = sin(45-C/2)*sin(45+B/2)/sin(45-B/2)*sin(45+C/2). Taking the cyclic product of the LHS will then yield the desired result.


trigonometry
ratio
trig identities
Law of Cosines

Chords AE, AD, and AC in a given circle are all on the same side of diameter AB, and arcs AE, AD, and AC are 1/6, 1/4, and 1/3 as long as the circumference of the circle, respectively.  If chord CE meets chord AD in point V, and line segment CV is 3 units long, how long is DV?

[hide]I believe it is $\sqrt{6-3\sqrt{3}}$.

We know that $\angle{BOD} = 30^o$.

So if the radius is $x$ by Law of Cosines, $BD^2 = x^2(2-\sqrt{3}) \Rightarrow BD = x\sqrt{2-\sqrt{3}}$.

Because $\bigtriangleup{BVD} \approx \bigtriangleup{AVC}$, we use side ratios to find $DV$.

We know that $AC = x\sqrt{3}$.

So $\frac{DV}{3} = \frac{x\sqrt{2-\sqrt{3}}}{x\sqrt{3}}$, and so $DV = \sqrt{6-3\sqrt{3}}$.[/hide]


I edited the question, so hopefully it's better now.
[hide]Hint: what is $\sum_{i=k}^n\binom{i}{k}$?[/hide]


Let $Q= I- 2uu^T$, show that $Q^T= Q^{-1}$

I know that $Q^T=(I-2uu^T)^T= I-2uu^T$, but I dont know how to show it equals $Q^{-1}$, please help

Multiply it by $Q$. Show that you get the identity.


Solve an equation in $ x,y,z \in \mathbb{N}_0$:
$ 5^x\plus{}7^y\equal{}z^3$

note that $ z^3 \equiv 0,1,\minus{}1 (\mod 7)$ so we get that $ 5^x\equiv 0,1,\minus{}1 \mod 7$ therefore $ 3|x$
now let $ x\equal{}3a$therefore we have:
$ 7^y\equal{}(z\minus{}5^a)(z^2\plus{}z \times 5^a\plus{}5^{2a})$ if a prime number $ p$ divides $ z\minus{}5^a$ and $ z^2\plus{}z \times 5^a\plus{}5^{2a}$ then $ p|7$ and $ p|3\times z\times 5^a$ which is impossible,therefore
 $ \gcd(z\minus{}5^a,z^2\plus{}z \times 5^a\plus{}5^{2a})\equal{}1$ but both of them should be a power of $ 7$ hence one of them is equal $ 1$.but $ z^2\plus{}z \times 5^a\plus{}5^{2a}>1$ therefore $ z\equal{}5^a\plus{}1$
and:
$ z^2\plus{}z \times 5^a\plus{}5^{2a}\equal{}7^y\Rightarrow 3\times 5^a(5^a\plus{}1)\equal{}7^y\minus{}1$
therefore if $ a>0$ then:
$ 5|7^y\minus{}1 \Rightarrow 4|y \Rightarrow 16|7^y\minus{}1 \Rightarrow 16|5^a\plus{}1$ which is impossible!

[quote="ali666"]note that $ z^3 \equiv 0,1, \minus{} 1 (\mod 7)$[/quote]i didn't notice that obvious fact  :blush: , later the task is easy, thank you very much :)

P.S. I think, that it's better to use \cdot rather than \times:

$ z \cdot 5^a \\ z \times 5^a$


advanced fields
advanced fields unsolved

First of all, I have no idea what forum to put this in, so apologies if it is in the wrong place.

This may be a very simple question.

Let (V,<.,.>) be an inner product space. Show that if $(x^k)$ (a sequence) converges weakly to x (x in V), and $\left\| x^k \right\|\to\left\| x \right\|$ them $x^k \to x$ in V

I think I have a proof, but I just want to check something.

If I have

$\left\|x^k-x\right\|^2 = \left\| x^k \right\|^2-<x^k,x>-<x,x^k>+\left\| x \right\|^2$ is it allowable to take the limit as k goes to infinity of both sides? (In which case the results follows)

Yes. The limit of a sum is the sum of the limits, if all of the limits exist. You have enough conditions to say that all of these limits exist.

Thats what I thought.

A case of me thinking something was a little too easy

Thanks


Understood but how do I break this question down using the difference of cubes.

I have never really understood this type of factoring.

Thanks!

I appreciate your help but please provide the steps that SHOW ME how to answer this type of question on my own.

Thanks!

[quote="Interval"]I have never really understood this type of factoring.[/quote]

The general rule:

$a^{3}-b^{3}= (a-b)(a^{2}+ab+b^{2})$

So in this case,

$(ab)^{3}-(2x^{2}y^{2})^{3}= (ab-2x^{2}y^{2})( (ab)^{2}+(ab)(2x^{2}y^{2})+(2x^{2}y^{2})^{2})$

The last reply takes me through the main point that I needed broken down.

I thank you very much.


For [tex]x_1,x_2...x_n[/tex] and [tex]y_1,y_2,..,y_n[/tex] be non-negative numbers.Prove that:
[tex]\sum \min(x_ix_j,y_iy_j)\leq \sum \min(x_iy_j,x_jy_i) [/tex]

This is not an open question at all. See http://www.kalva.demon.co.uk/usa/usoln/usol006.html .

  darij

Oh,I known it.But it's so elaborate and so hard.I post it to see a shorter solution!! ;)


function
logarithms
inequalities
number theory open
number theory

I was playing around with the totient and pi functions, and thought, "wouldn't it be interesting to look at $\phi(x)-\pi(x)$?"  So, I did, and came up with this hypothesis, which I can see no easy way of proving:

If $n$ and $n+2$ are twin primes, $n>3$, then $\phi(n+1)-\pi (n+1)<0$.

The only "proof" that I have so far is that there are no counterexamples for n<50.  If anyone has any ideas, I'd love to hear them!

I think $n+1=72$ is a counterexample. It was really clear that there would be a counterexample because $\pi(n)\rightarrow \frac{n}{\log n}$ and $\phi(n)$ can be equal to $\frac{n}{6}$ and for big $n$ we have $\frac{n}{\log n}<\frac{n}{6}$. Perhaps your conjecture holds with the inequality the other way round for $n>c$ for a fixed $c$ but I am not sure :blush: .

Misha

Curses!  Oh well, it was still interesting...

using merten's theorem ( http://mathworld.wolfram.com/MertensTheorem.html ), it is easily seen that $\phi(n)$ grows much faster than $\pi(n)$, since it grows smallest in the sequence $p_1p_2\cdots p_k$ where $p_1<p_2<...<p_k$ are the primes. how it grows there is shown by merten's theorem. thus, misha's conjecture is correct(we can even drop the condition that $n$ and $n+2$ are twin primes).

Peter


Question:

If I take both the AMC 10 and the AMC 12 and get different scores on them, will my higher one be used to determine my USAMO index? Or my lower one? Or...random?

(especially if i take both the A and the B for both of them.)

I assume it's the one that gives you a better USAMO index.

The higher score is used because it yields the higher index.  This rule is implicit in Steps 3 and 4 of the selection algorithm.  It is also implicit the last sentence of Frequently Asked Question AMC 10/12 #21.  See also the Teachers' Manual, page 6, column 1.

[quote]especially if i take both the A and the B for both of them.)[/quote]

Note that you can take only one of the AMC 10 or the AMC 12 on a given official test date.  See Frequently Asked Question AMC 10/12 #12.  See also the Teachers' Manual, page 6.

Steve Dunbar
MAA Director, American Mathematics Competitions

take the 10, and both times...who cares if someone says that AMC 12 is a better contest and helps more, unless you can get 144+ on AMC 10, take 10 both times(if you get 144+ on 10A, you probably are safe in taking the 12B)

i wouldve been pretty mad last year had i not made it with an 8 on AIME because of that AMC 12 counting the same as AMC 10...so dont put yourself in that situation.

USAMO qualification isn't everything. It's not really something you should be basing this decision on. In fact, if the floor system worked as intended, it wouldn't even matter.

I went through high school when there was only the AHSME, but there's no way I would have taken the AMC 10 in tenth grade. At that point, I was taking calculus, and there's just no good reason to avoid the test that covers the full high school curriculum.

Don't forget that most students have access to only one test date, and AIME qualification is a far more common target. If you're just trying to reach the AIME, it's probably easier with the 12.

ok thanks!
i guess ill just take both 10's, since
1) I'm still in 9th grade and Algebra II
2) I'm trying to make Red MOP, not just the AIME
3) I qualified for USAMO last year and will feel stupid if I don't this year.


How many paths are there from $ C$ to $ D$ on the grid shown, if every step must be down or to the right?[asy]size(4cm,4cm);int w=5;int h=7;int i;for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("D", (w-1,0), SE);label("C", (0,h-1), NW);[/asy]

hello
if we take any path e.g. $ (RRRRDDDDDD)$
no. of ways=$ \frac{(4\plus{}6)!}{4!6!}$=$ 210$
thank u

Or you can do $ \binom{10}{4}$ to find the number of ways to choose the four moves when you go right.  This, when expanded, becomes $ \frac{10!}{4!6!}$ like you said, which is $ \boxed{210}$.

[hide=formula thingy]do 6+4 choose 4 = 210[/hide]


some one give me the proof of fermat's equation


The integer $ n$ is the smallest positive multiple of $ 15$ such that every digit of $ n$ is either $ 0$
or $ 2$. Compute $ \frac{n}{15}$.

the number ends in 0 (divisibility rule by 5)

so let n be the # of 2's in the number
n is a multiple of 3 (divisibility rule by 3)
and n is minimized when it is 3 and the # is minimized when it is 2220 and 2220/15=148


Does somebody know an [b]inequality on sides of triangle[/b] in [b]which equality occurs if the triangle is degenerated[/b]?

For example, $\frac a{a+b}+\frac b{c+a}+\frac c{a+b}< 2.$

Anybody?


Find all integers $a,b,c,d$ such that

(i) $1 \leq a \leq b \leq c \leq d$;

(ii) $ab + cd = a +b +c +d + 3$.

[quote="seamusoboyle"]This factorises as

$5=(a-1)(b-1)+(c-1)(d-1)$ Which generates infinate solutions... (Should it be positive integers?)[/quote]

 :D already  given that  a , b, c, d are more than 1 definetely positive
  and we get only finite solutions.
  a, b , c, d are integers  implies (a-1) (b-1) and (c-1) (d-1) also integers
 possibilities (a-1)(b-1) = 1  and (c-1)(d-1) = 4......proceede we get  sol
Brahmam

[b]shyong[/b], yes, you missed $(a,b,c,d)=(1,1,2,6)$

dondigo.. you missed it.... (1,1,2,6) is not a solution....

Take a-1 >=2 and a-1 = 0 and a-1 = 1 separately you will get the required answers

[quote=dondigo][b]felixx[/b], you're wrong $(1,1,2,6)$ works. Just check carefully
$1 \cdot 1+2 \cdot 6=1+1+2+6+3$[/quote]

Dondigo is right


vector
geometry
geometric transformation
linear algebra

I have some questions(i found them in an exam paper) and i can't answer them.The exercise is true or false but i can't understand why are true or fasle.Can anyone help me? Thanks in advance.

1)if $ A$ is an orthogonal matrice then $ A^ - ^1$ is also orthogonal

2)if the vectors $ x_1$ and $ x_2$ are solutions of the system $ Ax = b$ then also the vector $ 2/9x_1 + 7/9x_2$ is a solution of this system

3)if the vectors $ x_1$ , $ x_2$ ,$ x_3$ are linear dependent then also the vectors $ Av_1 , Av_2, Av_3$ where $ A\in R{}^n{}^x{}^n$ are linear dependent

4)if A is invertible and B non invertible matrices then AB is non  invertble

5)Let $ T: R^n - > R^m$ is a linear transformation and let that $ U$ is a subspace of $ R^m$. Is the invertible picture of $ U , T^ - 1 (U)$ a subspace of $ R^n$

1) $ A$ is orthogonal iff $ A^{ \minus{} 1} \equal{} A^T.$ Follow up on that idea.

2) The only thing that matters about $ \frac29$ and $ \frac79$ is that they sum to $ 1.$ In general, $ (1 \minus{} t)x_1 \plus{} tx_2$ is a solution to the equation. Use the linearity of multiplication by $ A.$

$ A((1 \minus{} t)x_1 \plus{} tx_2) \equal{} (1 \minus{} t)Ax_1 \plus{} tAx_2 \equal{} (1 \minus{} t)b \plus{} tb \equal{} b.$

3) What is the definition of linear dependence? There are constants $ c_1,c_2,c_3,$ not all zero, so that $ c_1v_1 \plus{} c_2v_2 \plus{} c_3v_3 \equal{} 0.$

Then $ c_1(Av_1) \plus{} c_2(Av_2) \plus{} c_3(Av_3) \equal{} A(c_1v_1 \plus{} c_2v_2 \plus{} c_3v_3) \equal{} A0 \equal{} 0.$

Hence, $ \{Av_1,Av_2,Av_3\}$ is dependent.

4. Let $ AB \equal{} C.$ If $ C$ were invertible, then $ C^{ \minus{} 1}(AB) \equal{} I,$ so $ (C^{ \minus{} 1}A)B \equal{} I$ and that makes $ B$ invertible.

5. Check that $ T^{ \minus{} 1}(U)$ is closed under addition and scalar multiplication. Let $ x,y\in T^{ \minus{} 1}(U)$ and $ \alpha,\beta$ be scalars. This means that $ Tx\in U$ and $ Ty\in U.$ Now consider $ z \equal{} \alpha x \plus{} \beta y.$ We have that $ Tz \equal{} T(\alpha x \plus{} \beta y) \equal{} \alpha Tx \plus{} \beta Ty\in U,$ since $ U$ is a subspace. Since $ Tz\in U,$ then $ z\in T^{ \minus{} 1}(U).$

All five statements are true, and in several cases true in more generality than stated. (For instance, in #3, we didn't have to restrict to a square matrix.)

Note on English translation: the idiomatic mathematical English would be "inverse image," not "invertible picture."

Thank you very much!
but i inderstand only 1,2 and 4.

in third question i don't understand the $ c_3v \plus{} 3$ (what is this?). maybe i make the mistake look again in my first post in three i made a change
and in 5 maybe you mean z = ax + βy and not z=ax + β or ????

thank you again

That was a typo- + instead of _. (adjacent keys) It should have been a subscript. On #5, your proposed correction is right.

[color=green][Typos now fixed. - K.M.][/color]


In the unit squares of an nxn chessboard are written n^2 different positive integers. In each column of the chessboard the unit square with the greatest number is colored red. A set S of n unit squares is called admissible if any two unit squares from S do not lie in the same column or in the same row of the chesscoard. Prove that the admissible set with the greatest sum of numbers written in its unit squares contains at least one red square.

well the idea is this:

let $A$ be the admisable set with bigger sum. And supose no element of $A$ is colored.  Now, let $B$ be the set of colored elements.  start at an element of $A$, it has a number on its column that belongs to $B$, and this number on $B$, has a number on its row belonging to $A$, so we eventually find a cycle whose elements alternate between elements from $A$ and $B$. now, taking out the elements of this cycle from $A$ and ading the elements of this cycle from $B$ to $A$ we get a new admisible set with bigger sum, contradiction.

[quote="Pascual2005"] start at an element of $A$, it has a number on its column that belongs to $B$, and this number on $B$, has a number on its row belonging to $A$,.[/quote]

well,you cannot start arbitrarily in $A$,because you may not be able to ensure that the resulting set after exchanging will correspond to a permutation.however we can modify the idea slightly as follows:assume that the colored squares occur along rows $r_1,r_2,..r_k(k \leq n)$.restrict to the elements of $A$ that lie in these rows.Consider a graph $G$ whose vertices are the cells of $A,a_1,a_2,..,a_k$ that lie on these $k$ rows.We put a directed edge from $a_i$ to $a_j$ if the colored edge of cell $a_i$ on the same row as cell $a_j$. Clearly each vertex has exactly one edge leaving it.but now we have a graph with k vertices and k edges with each vertex having exactly one edge leaving it.since the underlying undirected graph would have a cycle,so would the directed graph.by making all the exchanges among the vertices of the cycle,we are through.

well one starts in $A$ but one doesnt necesarily goes back into the starting square, however one gets a cycle!


function
trigonometry
induction
algebra unsolved
algebra

Find all $ f\in{C}^{2}(\mathbb{R},\mathbb{R})$ such that
\[f(x\plus{}y)f(x\minus{}y)\equal{}(f(x))^{2}\plus{}(f(y))^{2}\minus{}1\]
for all reals $x$ and $y$.

[quote="reason"]find all $ f\in{C}^{2}(\mathbb{R},\mathbb{R})$ such that:

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

thanx.[/quote]
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)f(x\minus{}y)\equal{}f(x)^2\plus{}f(y)^2\minus{}1$

$ f(x)\equal{}\plus{}1$ $ \forall x$ and $ f(x)\equal{}\minus{}1$ $ \forall x$ are solutions
So we'll now consider that $ \exists u$ such that $ f(u)\neq 1$

$ P(x,0)$ $ \implies$ $ f(0)^2\equal{}1$ and, since $ f(x)$ solution implies $ \minus{}f(x)$ solution, wlog consider $ f(0)\equal{}1$

1) $ f(u)>1$
$ P(x,u)$ $ \implies$ $ f(x\plus{}u)f(x\minus{}u)>0$ and so $ f(x)\neq0$ $ \forall x$ and so $ f(x)>0$ $ \forall x$ since $ f(x)$ is $ C^2$, so continuous.
$ P(x,x)$ $ \implies$ $ f(2x)\equal{}2f(x)^2\minus{}1$ and it is to see that if $ f(x)\in(0,1)$, then $ f(2^nx)<0$ sor some $ n$, which is impossible.
So $ f(x)\geq 1$ $ \forall x$
Let then $ f(a)\equal{}\cosh(b)$, $ b\geq 0$. From the above equation, we get $ f(\frac a{2^n})\equal{}\cosh(\frac b{2^n})$.
Setting $ n\to\plus{}\infty$, and using the fact that $ f(x)$ is $ C^2$, we get $ f'(0)\equal{}0$ and $ f''(0)\equal{}\frac{b^2}{a^2}$ and so $ \frac{b^2}{a^2}\equal{}$ constant and so $ f(x)\equal{}\cosh(cx)$
And it is easy to see that this function indeed is a solution.

2) $ f(u)<1$
If $ f(u)\leq 0$, we get that $ \exists v$ such that $ f(v)\equal{}0$
If $ f(u)>0$, $ P(x,x)$ $ \implies$ $ f(2u)\equal{}2f(u)^2\minus{}1$ and it is easy too to see that $ f(2^nu)<0$ sor some $ n$ and so  $ \exists v$ such that $ f(v)\equal{}0$.
Then let $ A\equal{}\{x>0$ such that $ f(x)\equal{}0$ and let $ w\equal{}\inf(A)$. Continuity gives that $ f(w)\equal{}0$ and $ f(x)>0$ $ \forall x\in[0,w)$
Let then $ a\in[0,w]$, so that $ f(a)\in[0,1]$ and let $ f(a)\equal{}\cos(b)$ with $ b\in[0,\frac{\pi}2]$
From $ f(2x)\equal{}2f(x)^2\minus{}1$ and $ f(x)\in(0,1]$ $ \forall x\in[0,w)$, we get that $ f(\frac a{2^n})\equal{}\cos(\frac b{2^n})$.
Setting $ n\to\plus{}\infty$, and using the fact that $ f(x)$ is $ C^2$, we get $ f'(0)\equal{}0$ and $ f''(0)\equal{}\minus{}\frac{b^2}{a^2}$ and so $ \frac{b^2}{a^2}\equal{}$ constant and so $ f(x)\equal{}\cos(cx)\equal{}\cos(\frac{\pi}{2w}x)$ $ \forall x\in[0,w]$

Then $ P(x,\minus{}x)$ $ \implies$ $ f(\minus{}x)\equal{}\pm f(x)$ and so, with continuity, $ f(x)\equal{}\cos(\frac{\pi}{2w}x)$ $ \forall x\in[\minus{}w,w]$

Last, $ P(x\plus{}w,w)$ $ \implies$ $ f(x\plus{}2w)f(x)\equal{}f(x\plus{}w)^2\minus{}1$ and so knowledge of $ f(x)$ on $ [\minus{}w,\plus{}w]$ gives knowledge on $ [w,2w]$, then on $ [2w,3w]$, ... and a simple induction gives $ f(x)\equal{}\cos(\frac{\pi}{2w}x)$ $ \forall x$

And it is easy to see that this function indeed is a solution.

3) synthesis of solutions :
$ f(x)\equal{}\minus{}1$
$ f(x)\equal{}1$
$ f(x)\equal{}\cosh(ax)$
$ f(x)\equal{}\minus{}\cosh(ax)$
$ f(x)\equal{}\cos(ax)$
$ f(x)\equal{}\minus{}\cos(ax)$

[i]Pco , The Lord of functional equations :lol:  [/i]


Five consecutive two-digit positive integers, each less than 30, are not prime. What is the largest of these five integers.

The two primes under 30 that are farthest apart are 23 and 29. So our best shot is the set of numbers in between: 24, 25, 26, 27 ,28. There are exactly 5 of them, so the largest number is 28.

The five consecutive two digit positive integers that are not prime under 30 are 24,25,26,27,28 and 28 is the largest in the group. This means that 28 is the answer and is therefore the largest number.


$\sqrt a - \frac{1}{2\sqrt a}=2 \Rightarrow \sqrt a = \frac{2+\sqrt 6}{2}, a = \frac{5+2\sqrt 6}{2}$

the first answer was wrong. It should be right now.


geometry
incenter
circumcircle
geometry proposed

Let $A'B'C'$ be a circumcevian triangle of the incenter $I$ of $ABC$ triangle. If $O$ is the circumcenter of $ABC$, prove that the circumcircles of the triangles $AOA',BOB', COC'$ are coaxials.

Let $\omega$, $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ be the circumcircles of triangles $ABC$, $AOA'$, $BOB'$, $COC'$, respectively.  Let $\omega_{1}'$, $\omega_{2}'$, $\omega_{3}'$ be the image of the inverses of $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ with respect to $\omega$.

Since $\omega_{1}$ passes through $O$, $\omega_{1}'$ is the line $AA'$.  Similarly, $\omega_{2}'$ is the line $BB'$, and $\omega_{3}'$ is the line $CC'$.  Hence, $\omega_{1}'$, $\omega_{2}'$, $\omega_{3}'$ concur at $I$.

Hence, circles $\omega_{1}$, $\omega_{2}$, $\omega_{3}$ all pass through $O$ and $I'$, the inverse of $I$ with respect to $\omega$.

Dear Mathlinkers,
without inversion, only with the thre chors theorem...
http://tech.groups.yahoo.com/group/Hyacinthos/message/13625

Sincerely
Jean-Louis

The links of Hyacinthos messages are changed by yahoo

This is the link to message 13625

https://groups.yahoo.com/neo/groups/Hyacinthos/conversations/messages/13625


is trues that the ovules contain the [b]microspore[/b] mother cell?
thanks for your help,

No, no, no...remember, if it has anything to do with ovules or ovaries in plants "mega-" is the key prefix because cells are much larger than their "male" analogies.

Egg: In the ovule, the megasporocyte undergoes meiosis resulting in four megaspores. Three degenerate and one is the surviving megaspore. In the ovule, it develops into the mature female gametophyte which consists of egg, antipodals, synergids, and polar nuclei (fuse with a sperm nucleus from the microgametophyte in order to form the 3n, triploid, endosperm.)

^^^Talking about in angiosperms here, just to be specific.


Hello,

I hav also post this but i can't find it anymore ??

[url=http://img350.imageshack.us/my.php?image=areatera3hp.jpg][img]http://img350.imageshack.us/img350/3731/areatera3hp.th.jpg[/img][/url]

Who can help? thanks.

Please post in LaTeX. Also, don't post attachments if that's not necessary.


Hi. I was maintaining my blog with the green color theme :D . However, everything disappeared about an hour ago :( . The blog options menu have all the options saved (i.e., color options, etc.), but it is not applied at all. Moreover, the title of the blog appears as some small dots; I assume it is the title but it is so small it cannot be read, and so does the other subject titles. I find nothing in blog options to adjust for that at all. The link to my blog is [url=http://www.mathlinks.ro/Forum/weblog.php?w=765]here[/url] or the equivalent in AoPS. I see the same thing happening to boxedexe's blog, which can be found [url=http://www.mathlinks.ro/Forum/weblog.php?w=770]here[/url]. FYI, I use Internet Explorer 6, on Windows XP, which is the most common. I don't know why this is happening......

Thanks :D

[quote="kohjhsd"]Internet Explorer 6, on Windows XP, which is the most common.[/quote]Not anymore, IE7 is the most common browser ;) But anyway, I'll look into this thing ...

on the subject of blogs, i have changed and added an image to the background. however, it doesn't show! also, ive changed the font to Cosmic Sans and it doesn't change!

BTW the URL of the background image is [url]http://www.sgeier.net/fractals/fractals/06/Theta%20Space.jpg[/url]

[quote="Valentin Vornicu"]That background is protected. You can't use it anyway.[/quote]

and what about the font?

[quote="#H34N1"]About the titles: I found a workaround:

get firefox
Then get that web developer toolbar
Go to your blog
Click on css>edit
then copy whatever is in the first tab and paste it in the second...[/quote]This only works for your browser, and only temporarily, AND the tabs get annoying...  :(

Problem should be fixed now. Continue reading at http://www.mathlinks.ro/Forum/viewtopic.php?t=145435


function
trigonometry
calculus
calculus computations

fina all functions $ f$ from $ C^{2}(R,R)$ such that,
for all réel $ x$ ,$ f''(x)\plus{}f(\minus{}x)\equal{}xcosx$

Just an opinion.

Start by considering f(x) as an even function, and so you have to solve $ y''\plus{}y \equal{} x\cos x$.
Next consider f(x) as an odd function, and now you have to solve $ y''\minus{}y \equal{} x\cos x$.

Both differential equations are trivial non-homogeneous linear differential equations.


function
topology
real analysis
real analysis unsolved

fk is a sequnce of measurable functions defined on E with m(E)<+inf.

For each x, fk is bounded, that is: Sup(|fk(x)|)<=Mx<+inf.

Prove: given e>0, we can find a finite M and a closed F in E, such that m(E-F)<e
and |fk(x)|<=M for all k and all x in F.

(I think this uses Egorov, but don't know how to do?)

Observe that $M(x): =\sup_{k}|f_{k}(x)|$ is a measurable function with finite values at every point. Thus there is $M$ such that the set $\{x\colon M(x)> M\}$ has measure less than $\epsilon$. It's not open in general, but is contained in an open set of measure less than $\epsilon$.


integration
trigonometry
geometry
trapezoid
calculus
function
calculus computations

The exact value of $ \int_{0}^{3\pi}\sin x\ dx \equal{} 2$. How large must $ n$ be for the Trapezoidal approximation $ T_{n}$ to satisfy $ |E_{T}|\leq (2\minus{}S_{n})$?

Here's a worst-case error estimate for the trapezoid rule: if $ M$ is the largest possible value of $ |f''(x)|,$ then the error is no larger than $ \frac{M(b\minus{}a)^{3}}{12n^{2}},$ where the integral is from $ a$ to $ b$ and we subdivide into $ n$ equal subintervals.

In this case, we may take $ M\equal{}1.$ That gives an error estimate of $ \frac{27\pi^{3}}{12n^{2}}.$

That's a worst-case estimate. The trapezoidal rule misses "to the inside of the curve" of a function. But this function has inflection points and curves back and forth; hence some of the trapezoidal errors will cancel out, and the actual error will be quite a bit smaller than that estimate.

---

As for lord12's question: as he has not explained what $ S_{n}$ is and I'm not a mind-reader, I can't even guess at an answer.

[quote="lord12"]$ S_{n}$ is Simmpson's Rule for $ n$ sub intervals.[/quote]
No need to start a new topic to say that. That belonged as a reply here.

Odd question. Very odd question, since you would expect, except for unusual coincidences, that the trapezoid error would be larger than the Simpson error, and that making $ n$ larger would only make that worse. Hence my guess as to the answer to your question would be "probably never, although there might be accidental exceptions."

If we were integrating from $ 0$ to $ 2\pi$, those "unusual coincidences" would happen- both rules would be exact for all meaningful $ n$. Going to $ 3\pi$, Simpson's rule is better for large enough $ n$.


geometry
vector
advanced fields
advanced fields unsolved

This is an easy problem but I want to double check to make sure I am right. your help is appreciated.

Let D be a constant distribution on R^n. (i.e: D(x)=V for all x in R^n, where V is a subspace of R^n.)
Let f(x)=Ax be a linear vector field on R^n.
Is the following true?
f belongs to D iff V is A invariant.

my solution:
No!
Consider A=I. then f(x) = x clearly then for any x1 in R^n such that x1 is not in V f is not in D since f(x1) is not in D(x1).

Looks right, but the question itself is kind of random.  I guess it's one of those questions that are asked just to get you to use the definition.