For some reason Latex doesn't want to build a PDF file. I select the Latex-->pdf option and when I press 'build and view', Latex opens Adobe, but with no document in it. If I check what he has build, I see that he creates my document as just a text-document, not a pdf-document (that's why it can't be opened in Adobe). He also doesn't create a .aux document.
In TeXniCenter it also says: "pdflatex.exe: is unknown."

Please tell me there's a simple solution for this, cause it drives me crazy!!

I have adobe 7.0. I allready tried to download everything again.
And the 'path of executables' is also right. It also doesn't work if I select: latex-->ps or latex-->dvi

Thanks!

[quote="marie2006"]In TeXniCenter it also says: "pdflatex.exe: is unknown."[/quote]That means TeXnicCenter can't find pdflatex.exe. I know you say that [quote="marie2006"]And the 'path of executables' is also right.[/quote] but are you absoutely sure? Press Alt+F7 and check that the top line really does point to the true location of pdflatex.exe. If it does and there's a space in the path (eg Program Files) try pressing Browse and getting the path that way. You could also try putting quotes round the path if there's a space in the path.

I checked the path again, and it's absolutely right. I browsed it myself and also tried putting quotes around it, but it still doesn't work.

In the Output Profile: Latex->pdf-box (in texniccenter) the following appears:

starting package maintenanct...
installation directory: C:\miktex
package repository: ftp://ftp.uvsq.fr/pub/Tex/CTAN/systems/win32/miktex/tm/packages/
DBlight digest: 0ba5cce9f6e86004e61b231855941d4e
pdflatex.exe: The package
pdflatex.exe:
pefdlatex.exe: ltxbase
pdflatex.exe:
pdflatex.exe: is unknown.
I couldn't open file name: 'C:\documents and settings\thuis\mijn documenten\marieline\latex\probeer.aux'
couldn't find input index file: 'C:\documents and settings\thuis\mijn documenten\marieline\latex\probeer nor 'C:\documents and settings\thuis\mijn documenten\marieline\latex\probeer.idx
usage: 'C:\Program Files\miktex\miktex\bin\makeindex.exe

I'm not sure that LaTeX is installed properly. Go to MiKTeX Package Manager then find ltxbase, right-click and choose Install and see what happens.

[quote="marie2006"]pdflatex.exe: The package
pdflatex.exe:
pefdlatex.exe: ltxbase
pdflatex.exe:
pdflatex.exe: is unknown.[/quote]
This looks misleading. It's not saying pdflatex.exe is unknown, but is a message from pdflatex to say ltxbase is unknown. Confusing isn't it?

[quote="marie2006"]So have you got any idea how I can make sure that he does find that package? I already downloaded miktex twice, so it should be there, i guess..

thanks a lot[/quote]MiKTeX has to go through 2 stages of installation. Once to set it up and download the packages and then again to install the packages. Have you done the second install?

so i went to my package manager, got packages from the internet and tried installing ltxbase. It partially did that and then gave me an error 'download failed for some reason'. Anyway, i tried compiling my file again and IT WORKS!!

It does give me a lot of warnings though.. like: 'font file missing'

maybe i should go through the whole installation procedure again..

thanks!


geometry
incenter
circumcircle
perpendicular bisector
geometry open

Dear All My Friends,
Two any points B, D are on perpendicular bisector of segment AC and X is any point on segment BC. Please prove that three incenters of triangles ABX, CDX, DAX and X are concyclic.
Thank you and best regards,
Bui Quang Tuan

[quote="armpist"]For a 4-liner Miquel theorem states that the centers of 4 Miquel circles are on a circle.

   M.T.[/quote]
Can you tell me more details?
I don't know "4-liner Miquel theorem "
Thank you,dear armpist :)

[quote="armpist"]See in

  http://mathworld.wolfram.com/MiquelsTheorem.html

    ... Furthermore, the centers of these four Miquel circles lie on a circle  (Johnson 1929, p. 139).

   My post is a note that both the problem and theorem have 4 centers concyclic.  Just speculation, nothing more.

    M.T.[/quote]
Thank you,dear armpist.I Know the theorem,but I don't know the name.
The problem tells about incenters,I can't see the relationship between our problem and the statement you mentioned

Dear All My Friends,
Thank you for interesting in this problem!
As I see, four our points are concyclic also if ABCD is one convex circumscribed quadrilateral.
Best regards,
Bui Quang Tuan

[quote="mihai miculita"][b]P interior point of the quadrilateral ABCD or on the bunday (ex: in [BC])![/b][/quote]
If "P interior point of the quadrilateral ABCD" your problem is false, you can check easily by computer.

[quote="plane geometry"][quote="encyclopedia"][quote="mihai miculita"][b]Sorry, in Miquel theorem the cercles circumcircles, and the four points circumcentres! [/b]

[b]Generalisation:[/b]
Let ABCD circumscriptible quadrilateral and P amy point in quadrilateral ABCD.
Denote $ I_{ab},I_{bc}, I_{cd}, I_{ad}$ the incentres of $ \Delta PAB, \Delta PBC, \Delta PCD, \Delta PAD$, respectively.
Prove that four points $ I_{ab},I_{bc}, I_{cd}, I_{ad}$ concyclic.[/quote]

What do you mean "P any point in quadrilateral ABCD "? If $ P$ is inside it then it is false but if $ P$ is in the sides of quadrilateral it is really a generalisation!

[b]Problem 1.[/b] Let $ ABCD$ be a circumscriptible quadrilateral $ P$ lie on $ AB$ prove that incenters of $ PAD,PDC,PCB$ and $ P$ concyclic.

And your problem is true if P is intersection of diagonals

[b]Problem 2.[/b] Let $ ABCD$ be a circumscriptible quadrilateral $ AC\cap BD \equal{} \{P\}$ prove that incenters of triangles $ PAB,PBC,PCA,PAB$ concyclic.[/quote]
Problem 2 has been solved by Darij Grinberg[/quote]

Could please you send the link?

[quote="encyclopedia"][/quote][quote="plane geometry"][quote="encyclopedia"][quote="mihai miculita"][b]Sorry, in Miquel theorem the cercles circumcircles, and the four points circumcentres! [/b]

[b]Generalisation:[/b]
Let ABCD circumscriptible quadrilateral and P amy point in quadrilateral ABCD.
Denote $ I_{ab},I_{bc}, I_{cd}, I_{ad}$ the incentres of $ \Delta PAB, \Delta PBC, \Delta PCD, \Delta PAD$, respectively.
Prove that four points $ I_{ab},I_{bc}, I_{cd}, I_{ad}$ concyclic.[/quote]

What do you mean "P any point in quadrilateral ABCD "? If $ P$ is inside it then it is false but if $ P$ is in the sides of quadrilateral it is really a generalisation!

[b]Problem 1.[/b] Let $ ABCD$ be a circumscriptible quadrilateral $ P$ lie on $ AB$ prove that incenters of $ PAD,PDC,PCB$ and $ P$ concyclic.

And your problem is true if P is intersection of diagonals

[b]Problem 2.[/b] Let $ ABCD$ be a circumscriptible quadrilateral $ AC\cap BD = \{P\}$ prove that incenters of triangles $ PAB,PBC,PCA,PAB$ concyclic.[/quote]
Problem 2 has been solved by Darij Grinberg[/quote][quote="encyclopedia"]

Could please you send the link?[/quote]
you can see here http://www.cut-the-knot.org/Curriculum/Geometry/CyclicIncenters.shtml

[quote="encyclopedia"]Thanks, that's nice, what do you think about idea of armpist, I don't understand.[/quote]
I have no idea……
I don't understand what's the relationship betwwen miquel theorem and  incenters

[quote="jayme"]Dear Quang tuan and Mathlinlkers,
is your problem solved? I think no.
Sincerely
Jean-Louis[/quote]

Dear Jean-Louis,

Of course, you are right! I don't know its solution now.
I am waiting for solution from MathLinkers.

Best regards,
Bui Quang Tuan

Dear Quang Tuan,
I have only explore your figure and give you my first conjecture.
Let 1, 2, 3 be the incircles of ABX, CDX, DAX, A', C' the centers of 1 and 2, T the second external tangent to 1 and 2.
I found that T is also tangent to 3 at point Y.
Let Z be the midpoint of A'C'; then ZXY is Z-isosceles.
Can you confirm these conjecture...
Sincerely.
Jean-Louis

I think this isn't an open problem, some one inspired it from [url=http://www.mathlinks.ro/viewtopic.php?t=215427]IMO 2008 problem 6[/url] and [url=http://www.mathlinks.ro/viewtopic.php?p=1186805#1186805]here[/url]

Here in this general problem, do you think three incircles have the same tangent?

Dear Quang Tuan and Mathlinkers,
a discussion with solution of the Quang tuan and Ignaccio Larrosa's Problem concerning a square can be found on
http://www.les-mathematiques.net/phorum/read.php?8,508015,page=1  (in French).
Sincerely
Jean-Louis

Dear Quang tuan and Mathlinkers,
the result I mentioned above can be seen in a more general developpement
http://perso.orange.fr/jl.ayme  vol.5,  le résultat de Larrosa Canestro, p.3
Sincerely
Jean-Louis


[quote="neelnal"]So, basically, now unfilled bubbles are only 1.5 points?

That invalidates the whole system though. Now you can, theoretically, guess and get the same score as leaving it blank.[/quote]

There are 5 choices, not 4.


hey all
this is my first shot at the national math olympiad(hope u won't b scandalized coz i didn't say NMO or watever)  :o
Well..i am reeely freaked and nervous..all u experienced cronies pleez tell me how it is, how it works,etc  :maybe:
i guess u all have "been there, done that" so pleez lend a hand. with all the copycats in my skool all i will learn is how to conceal ur answer paper.
my math teacher says u need an iq above zero to do well.(wonder how cum SHE couldn't solve the paper!lol  :lol: ). Is it really this easy?
pleez suggest reference material..watever...  :?:

Hello,
    I believe you are taking the RMO(Regional Mathematical Olympiad).In which state do you reside?To have an idea,you could look up the INMO Question Papers(though they are at a higher level of difficulty than RMO)at AOPS or at Kalva,John Scholes' home page http://www.kalva.demon.co.uk. Surely, don't have the idea that Olympiads are like your school exams.Especially Mathematics, it is highly challenging intellectually.
Sathej

Thanks Sathej
Yes, I hope it will be intellectually challenging  :)
I stay in Bangalore,Karnataka.
See, my dilemna occurs because my teacher says-"It's very easy.If ur IQ is above zero then you can do it."
And then she gets stuck when a girl asks her one of the questions from last year's paper, and she says"maybe you guys should go through IIT JEE papers to get an idea of what it is...such things are too high for a teacher's level"
AND SHE CALLS HERSELF AN MSc!!!
Anyway thanks.  :)
Anything else? this one says"error page cannot be found"

Hello,
   It will be definitely intellectually challenging.And your teacher has got it all wrong.The standard of an INMO Paper is much higher than an IIT JEE Paper in terms of mathematical content.The fact that you are asked to solve 6 problems in 4 hours speaks for itself.And in the IMO,it is 9 hours(on two days) for 6 problems.Sorry for the wrong link.I gave a full stop extra.Here is the correct one http://www.kalva.demon.co.uk
You can go to Google and type Kalva for the site,anyway.
  As for IQ,of course,if your IQ is negative,you cannot crack any exam,leave alone an RMO,INMO and IMO.And there are a lot of very good people related with the Olympiad in Bangalore at IISc like Prof.B.J Venkatachala,Prof C.R.Pranesachar and Shailesh Shirali of Rishi Valley School(I believe), to quote a few.
As for books,Challenges and Thrills of Pre College Mathematics by Pranesachar,Krishnamurthy,KN Ranganathan is very good for beginners.
Sathej

Why don't you post the problems  where you need help here. There are enough people to help you here. Best of luck.

(Hard/easy is a relative term, some may find the problems easy, some not so easy)

thanx ppl will surely do when i get time.It's on 24th August and i have tests right now..backlog of work..ya i guess i gotta tackle it positiverly..can u bleev it-they r giving us internal assessment marks on this...out of 20! i like math so it's ok but i feel sorry for my classmates.Will surely ask u guys 4 help..thanx a ton
Padma


I am thinking about buying one of the following books.
Either Problems Solving Through Problems or Problem Solving Strategies.
I am wondering what the major differences between these 2 books are, like do they include solutions to all of the problems. Also which is more like the AoPS books but not too much like ACoPS (I dont like this book that much because I dont feel that it does a good job of explaining techniques and strategies...but I do like the variety of questions it has at the end of each section).
so which book would you suggest.

[quote="t0rajir0u"]I have Problem-Solving Strategies in my backpack as we speak.  It covers more or less the same material as ACoPS (minus a few college-level topics, although it covers convexity as a useful problem-solving technique), but I think it does a little better job of explaining (although Engel is very terse most of the time).  Nearly all the problems have solutions - one or two that I've seen are listed with no solution (I can't imagine why), but I really enjoy reading the book because of occasional anecdotes and comments about the more difficult problems.

There's an anecdote about this problem:

$a, b, q \in \mathbb{N} : q = \frac{a^2 + b^2}{ab + 1} \implies q = c^2, c \in \mathbb{N}$

Where it was given to prestigious adult problem-solvers, notably George Szekeres and his wife, none of which were able to solve it within the IMO time limit.  It was then proposed at the IMO as the last problem anyway, and eleven students gave perfect solutions.  :)[/quote]

From what year????? Do you have the solution?

IMO 1998 #6, it's currently in the unsolved section of our contest database.

[quote="maokid7"]I should have mentioned before that I am trying to prepare for the Putnam exam.
If you do not think either of these book are the way to go then what would you suggest. also are the topics covered by the putnam exam about the same from year to year?[/quote]
1) I suggest a search to see what other people have said about preparing for the Putnam.  Looking in the Putnam sub-forum might help.
2) I suggest going to look at the Putnam to see what types of questions are asked on the Putnam.  Looking in the Putnam Archive might help.


[color=olive]Show that if a is an integer, then 3 divides (a to the power 3) - a[/color]

I don't know how to use the mathematical software to express the question, I apologize for any inconvenience. Thanks for helping me out.

[b]1)[/b] This is Fermat's little theorem for the case $p=3$.

[b]2)[/b] Working mod 3:

if $a \equiv 0 \mod 3$ then $a^{3}-a\equiv0-0=0 \mod 3$
if $a \equiv 1 \mod 3$ then $a^{3}-a\equiv1-1=0 \mod 3$
if $a \equiv 2 \mod 3$ then $a^{3}-a\equiv2-2=0 \mod 3$

[b]3)[/b] By induction:

Clearly $3|1^{3}-1$

Suppose that $3|k^{3}-k$

$(k+1)^{3}-(k+1)=k^{3}+3k^{2}+3k+1-k-1 =3(k^{2}+k)+(k^{3}-k)$

$3|3(k^{2}+k)$ and $3|k^{3}-k$ (by the inductive hypothesis)
So $3|(k+1)^{3}-(k+1)$

So by induction $3|a^{3}-a$ for all $a$.

simply $a^{3}-a=(a-1)a(a+1)$ is a product of three consecutive integers, so certainly one of them must be divisible by 3 ...

As mention before, not only for 3, the result is true for any prime number !!
 that is for any prime $p$

$a^{p}-a$ is divisibley by $p$
(This is called Fermat's Little Theorem )


Prove that:
Polynomial $f(x)=(5x^9+6x^8+3x^6+8x^5+9x^3+6x^2+8x+3)x^n+12$ irreduciple for all $n\geq0$

if $n$ is even, we have $f(x)=(x+1)q(x)$ where $q(x)$ has integer coefficients... maybe I am misunderstanding what irreducible means?  If so, can you define it?

Irreducible polynomial is ppolynomial suth that don't analysize two polynomial with lower degree.

[quote="Andreas"]The statment is not correct. $f(x)$ is [u]reducible[/u] for all $n \geq 0$  :D[/quote]
Could you post your solution?

Honestly I don't know how to approach it.  :(
I can say that it's reducible for all the non-negative integers because it's the statment of a problem I came up with.

either way we've shown it's reducible for even $n$ so the problem's not right

I think the above problem is very nice. But no solution has received up to now. Importantly how to construct such problem?

See a post of Vess about a result of Schinzel. I showed there how to construct them and how to solve this problem.

Where is the post of Vess about a result of Schinzel?


Prove that the zeta function
$\zeta(x) = \frac{1}{1^{x}}+\frac{1}{2^{x}}+\frac{1}{3^{x}}+...$
converges when $x\geq 2$.

P.S. When $x=1$, it becomes the harmonic series, and you can prove that this diverges in the High School Intermediate Topics forum.

In fact, it converges for $x > 1$  :wink:

[quote="vishalarul"]I was hoping that the generalization "$\zeta(x) = \frac{1}{1^{x}}+\frac{1}{2^{x}}+\frac{1}{3^{x}}+...$ converges for $x>1$" could be proved without calculus.[/quote]

I really doubt it.  It's not an elementary result.


Find primes number $ p,q$ such that $ p^{2}\minus{}p\plus{}1\equal{}q^{3}$

[quote="mathpk"]Find primes number $ p,q$ such that $ p^{2}\minus{}p\plus{}1 \equal{} q^{3}$[/quote]

The equation may be written $ p(p\minus{}1) \equal{} (q\minus{}1)(q^{2}\plus{}q\plus{}1)$. Let then $ a \equal{} q\minus{}1$, we have $ p(p\minus{}1) \equal{} a(a^{2}\plus{}3a\plus{}3)$
Since $ p$ is prime and $ p > q > a$, $ p|a^{2}\plus{}3a\plus{}3$ and it exists $ n\geq 1$ such that :
$ a^{2}\plus{}3a\plus{}3 \equal{} np$ and $ p\minus{}1 \equal{} na$
So $ p \equal{} na\plus{}1$ and we have $ a^{2}\plus{}3a\plus{}3 \equal{} an^{2}\plus{}n$ and so $ n^{2}\equal{} a\plus{}3\plus{}\frac{3\minus{}n}{a}$

From this, we can see that if $ n > 3$, $ a|n\minus{}3$ so $ a\leq n\minus{}3$, so $ n^{2}\equal{} a\plus{}3\minus{}\frac{n\minus{}3}{a}\leq n$, which is a contradiction. So $ n\leq 3$.

$ n \equal{} 1$ $ \implies$ $ a^{2}\plus{}3a\plus{}3 \equal{} a\plus{}1$ and no solution
$ n \equal{} 2$ $ \implies$ $ a^{2}\plus{}3a\plus{}3 \equal{} 4a\plus{}2$ and no solution
$ n \equal{} 3$ $ \implies$ $ a^{2}\plus{}3a\plus{}3 \equal{} 9a\plus{}3$ and $ a \equal{} 0$ or $ a \equal{} 6$

$ a \equal{} 0$ means $ q \equal{} 1$, not a prime.
$ a \equal{} 6$ means $ q \equal{} 7$ and $ p \equal{} na\plus{}1 \equal{} 3*6\plus{}1 \equal{} 19$

And indeed $ p \equal{} 19$ and $ q \equal{} 7$ is a solution, hence the only solution.


Let abcd=x. The first inequality becomes

\[
ac+ad+bc+bd \ge \sqrt{x}\cdot\left(\frac1a+\frac1b+\frac1c+\frac1d\right).
\]

The result follows quite easily from cauchy:

\[\left(\frac{bd}{a}+\frac{ac}{b}+\frac{ad}{c}+\frac{bc}{d}\right)\left(\frac1a+\frac1b+\frac1c+\frac1d\right)\cdot x\]

\[=\left(\frac{bd}{a}+\frac{ac}{b}+\frac{ad}{c}+\frac{bc}{d}\right)(abd+abc+acd+bcd)\]

\[\ge (bd+ac+ad+bc)^2\]

\[\ge x\cdot\left(\frac1a+\frac1b+\frac1c+\frac1d\right)^2,\]

and cancel the $x\cdot\left(\frac1a+\frac1b+\frac1c+\frac1d\right)$ to finish.

[quote="zabelman"]Since everything is homogeneous[/quote]

No it's not.

Wow, I must have been asleep when I wrote that!

And here is my solution.

It follows from Cauchy that

(1) $a+bcd \geq 2\sqrt{abcd},$
(2) $b+acd \geq 2\sqrt{abcd},$
(3) $c+abd \geq 2\sqrt{abcd},$
(4) $d+abc \geq 2\sqrt{abcd}.$

Summarizing these four inequalities we will have ($(1) \cdot bd + (2) \cdot ac + (3) \cdot ad + (4) \cdot bc$)

$(abd+abc+acd+bcd)+(b^2cd^2+a^2c^2d+a^2bd^2+ab^2c^2) \geq 2\sqrt{abcd}(bd+ac+ad+bc)=2\sqrt{abcd}(a+b)(c+d).$

Dividing this inequality by $abcd$ we will have

$\frac{bd}{a}+\frac{ac}{b}+\frac{ad}{c}+\frac{bc}{d} \geq 2\frac{(a+b)(c+d)}{\sqrt{abcd}} - \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right) \geq 1/a+1/b+1/c+1/d.$

So, the problem is solved.

[quote="Remike"]Let [tex]a[/tex], [tex]b[/tex], [tex]c[/tex], [tex]d[/tex] be positive real numbers such that [tex]\sqrt{\frac{bcd}{a}}+\sqrt{\frac{acd}{b}}+\sqrt{\frac{abd}{c}}+\sqrt{\frac{abc}{d}} \leq (a+b)(c+d).[/tex] Prove the inequality [tex]\frac{bd}{a}+\frac{ac}{b}+\frac{ad}{c}+\frac{bc}{d} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}.[/tex][/quote]
Let $ a,b,c,d$ be positive real numbers such that $\sqrt{\frac{bcd}{a}}+\sqrt{\frac{acd}{b}}+\sqrt{\frac{abd}{c}}+\sqrt{\frac{abc}{d}} \leq (a+b)(c+d).$ Prove the inequality\[\frac{bd}{a}+\frac{ac}{b}+\frac{ad}{c}+\frac{bc}{d} \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}.\]


Shouldn't this belong in the Puzzles and Brainteasers forum?

i guess it should...  :blush:

(edit)
I cant find the puzzles and brainteasers forum...

edit
okay i found it.

[hide]They didn't each pay 14 for the room, then 2 to the messenger, they payed 14 TOTAL each. So, the total was 42.00, not 44.00. They each are left with 1.00[/hide]

[quote="buzzer11"]this is a classic, with variations.

Explain the paradox. Three men rent a hotel room. They are charged 15 dollars each. Later the manager decides they should only have been charged 40 dollars for the room, so he gives 5 dollars to a messenger to give to the men. The messenger dishonestly keeps two dollars and gives each man 1 dollar. Each man has paid 14 dollars for the room and the messenger has 2 dollars, for a total of 44 dollars. However the men originally paid 45 dollars for the room.  Where's the other dollar?

Hide answers, please.[/quote]

[hide="Very Simple"]The total price is SUPPOSED to be $\$$40. Each person has paid $\$$14, for a total of $\$$42. Notice any money returned to the three men LOWERS the price each person pays. So instead of adding 2 to 42, we're supposed to subtract 2 from 42. For a total of$\$$40, the supposed price.
The 3 people should've gotten $\$13\frac{1}{3}$ back. But the messenger swiped the $\$\frac{2}{3}$ from each of them. The total is $3(\$13\frac{1}{3})=3(14)\boxed{-2}=\$40$[/hide]

haha. I couldn't get this.

*maybe the messenger took 3 dollars...*

 :blush:


(6√x^5) / (3√x^2)

6/3 * (√x^5)/(√x^2)

(√x^5)/(√x^2) reduces to √x^3

6/3 = 3

answer is 3√x^3

Did I do it correctly?

[quote="1=2"]You didn't simplify all the way, but you have the general idea.

How can you simplify further?[/quote]
At least I have it right so far. That's progress.

3√x^3 = 3√x * x * x

3x√x

√x^3

I don't really get how it can be reduced further.

??

[quote="Star00"]6/3 = 3 [/quote]

6/3 is 2.  :wink:

As for simplifying $ 2\sqrt{x^3}$, I think $ 2x\sqrt{x}$ would be the best way to write it, but I'm not sure.


Hi, I would love it if someone would help me set this equation up.  I know the answer, but I just can't set this up.

The question is:
A consignment of magazines was made up into 45 parcels each weighing either 5kg or 1kg. Calculate the number of 5kg parcels in the consignment if the total weight was 153kg.

To the one who can set this up, cheers!

Davis

This shouldn't be posted in Olympiad section.
$ \begin{cases}x\plus{}y\equal{}45 \\
5x\plus{}y\equal{}153 \end{cases}$, where $ x$ is no. of 5 kg, y is no. of 1 kg

Hey Stephen, thanks for the help.

I still don't understand how to merge those two equations in order to find the number of 5kg boxes... is there any way you could take it a step further and illustrate the next step so one can find the actual number of boxes?

Best wishes and thanks for the help,

Davis

srs? Remeber how you solve for a system of equations, either by substitution, elimination, addition, subtraction etc

I prefer substitution most of the time

(1): x + y = 45
(2): 5x + y = 153

from (1),

y = 45 - x

thus subbing into (2):

5x + (45 - x) = 153
4x + 45 = 153
4x = 108

x = 27, y = 18

since x is the number of 5kg parcels then there are 27 5kg parcels

If you don't know the systems of equation I prefer you search about it for more understanding  :)

Fantastic! Thanks so much for this help you all - absolutely cleared up the mental fog regarding this question!

Thanks again,

Davis


modular arithmetic
Digits
number theory
decimal representation
IMO
IMO 1978

Let $ m$ and $ n$ be positive integers such that $ 1 \le m < n$.  In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively, to the last three digits of $ 1978^n$.  Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.

[quote="orl"]Let $ m$ and $ n$ be positive integers such that $ 1 \le m < n$.  In their decimal representations, the last three digits of $ 1978^m$ are equal, respectively, so the last three digits of $ 1978^n$.  Find $ m$ and $ n$ such that $ m \plus{} n$ has its least value.[/quote]
We have $1978^m\equiv 1978^n\pmod {1000}$, or $978^m-978^n=1000k$ for some positive integer $k$ (if it is not positive just do $978^n-978^m=-1000k$). Hence $978^n\mid 1000k$. So dividing through by $978^n$ we get $978^{m-n}-1=\frac{1000k}{978^n}$. Observe that $2\nmid LHS$, so $2\nmid RHS$. So since $2|| 978^n$, clearly the minimum possible value of $n$ is $3$ (and then $489^n\mid k$). We will show later that if $n$ is minimal then $m$ is minimal. We have $978^{m-3}-1\equiv 0\pmod {125}\Leftrightarrow 103^{m-3}\equiv 1\pmod {125}$. Hence, $m-3\mid \varphi(125)\Rightarrow m-3\mid 100$. Checking by hand we find that only $m-3=100$ works (this also shows that minimality of $m$ depends on $n$, as claimed above). So $m=103$. Consequently, $m+n=106$ with $\boxed{(m,n)=(103,3)}$.

I want to add another solution for the 2nd part of the problem: to find a = n - 3, so that 1978^k is equivalent 1 (mod 125) without an "advanced" Theorem.

Step 1: modulo 5
1978 is equivalent 3 (mod 5), so 1978^4 is equivalent 3^4 (mod 5). 3^4 = 81 is equivalent 1 (mod 5).
-->a must be a multiply of 4

Step 2: modulo 25
1978^4 is equivalent 3^4 (mod 25). 3^4 = 81 is equivalent 6 (mod 25). 1978^4b is equivalent 6^b (mod 25). Because 6=5+1, it's easy to show that (5+1)^5 is equivalent 1 (mod 25)
-->b must be a multiply of 5 --> a is a multiply of 20

Final step: modulo 125
1978 is equivalent -22 (mod 125). 1978^20 is equivalent (-22)^20 equivalent 484^10 equivalent (-16)^10 equivalent 256^5 equivalent 6^5 (mod 125). Because 6=5+1, it's easy to show that (5+1)^5 is equivalent 5^2 + 1 (mod 125).
So 1978^20c is equivalent (5^2 + 1)^c (mod 125). It's also easy to show that (5^2 + 1)^5 is equivalent 1 (mod 125)
--> k = 20*5 = 100

P.S: I tried to post the full version with Latex, but new users can't post image. So I replace some calculations with "it's easy to show"


Prove that for integer $n$ we have:
\[n! \le \left( \frac{n+1}{2} \right)^n\]
[size=75][i](please note that the pupils in the competition never heard of AM-GM or alikes, it is intended to be solved without any knowledge on inequalities)[/i][/size]

There was also a similar problem in the greek junior MO,2002.
The problem was :

Prove that $2002! \le \left( \frac{2003}{2} \right)^{2002}$

I participated in that contest and solved it using this inequality.

Apart from AM-GM, this can be proven with the observation:

$k(n-k+1)\leq(\frac{n+1}{2})^2$...
 ;)

[quote="silouan"]To prove this that socrates said use that $xy\leq (\frac{x+y}{2})^2$
Construct $n$ similar equations and make a product.[/quote]And if $n$ is not even?

[quote="Peter VDD"]And if $n$ is not even?[/quote]

The middle term will be $\frac{n+1}{2}$ :)

[quote="Peter"]Prove that for integer $ n$ we have:
$ n! \le ( \frac {n + 1}{2} )^n$[/quote]

I proved this result by induction on $ n$

Prove the base case.

Assume for $ n\in \mathbb{N}$

Consider $ (n + 1)! = (n + 1)n! \le (n + 1)(\frac {(n + 1)^n}{2^n})$

It suffices to prove that $ (n + 1)(\frac {n + 1)^n}{2^n}) \le (\frac {n + 2}{2})^{n + 1})$

$ \Longleftrightarrow 2 \le (1 + \frac {1}{n + 1})^{n + 1}$ which is Bernoulli's inequality.

[quote="Peter"]Heu... I'm sorry, I read silouan's post as a standalone proof (and didn't get it) :oops:[/quote]

Note that $ xy \leq \left( \frac {x \plus{} y}{2} \right)^2$, which can be proven by expanding and simplifying to $ (x \minus{} y)^2 \geq 0$.

Now consider terms $ x \equal{} m$, $ y \equal{} n \plus{} 1 \minus{} m$. We know $ m (n \plus{} 1 \minus{} m) \leq \left( \frac {m \plus{} n \plus{} 1 \minus{} m}{2} \right)^2 \equal{} \left( \frac {n \plus{} 1}{2} \right)^2$

Now consider cases of $ m \equal{} 1,2,3, \ldots n/2$ for even $ n$, or cases $ m \equal{} 1,2,3, \ldots (n \minus{} 1)/2$ for odd $ n$, and multiply together all the resulting equations. For odd $ n$, you also need to multiply both sides by a final factor of $ (n \plus{} 1)/2$.

[hide="Concrete examples"]Here's a concrete example: $ n\equal{}5$
$ 1 \cdot 5 \leq 3^2$
$ 2 \cdot 4 \leq 3^2$
$ 3 \equal{} 3$

Multiply to get: $ 5! \leq 3^5$

Another example: $ n\equal{}6$
$ 1 \cdot 6 \leq (7/2)^2$
$ 2 \cdot 5 \leq (7/2)^2$
$ 3 \cdot 4 \leq (7/2)^2$

Multiply to get $ 6! \leq (7/2)^6$[/hide]


has anyone here taken the moems 2007 division M contests 1-4? if so, can you check if my answers are right through private message? Thanks

is division M the middle school division?

if so, i never took MOEMS. How do you enter it?

hmm...i would first like to check your location. As we know, the Moems is now an international math test, so every country has a different way of registration...yeah, it's middle school (division M)...and by the way, from what country are you?


If $\log_{12} 7 = .7831$ and $\log_{12} 4 = .5579$, then what is $\log_{12} 3$?

Thanks in advance!

     -- Pawel

he asked $\log_{12}{3}$, not $\log_{12}{4}$.
it should just simply be $\frac{4421}{5579}$

oops. my mistake. read it $\log_{4}{12}$. ;)


function
calculus
integration
ratio
absolute value
calculus computations

could someone please explain to me what is meant by absolute convergence?
the definition my calc book give is very vague and really isnt that informative.
also is there something special about sequences that converge absoluely?

I've been wondering what the point of having absolutely vs. conditionally convergent series/integrals is as well. I know that a series is absolutely convergent if the sum of the absolute value of the summand is finite, but what is the point in making the distinction?

well to answer your question

a series $u_{n}$ converges if

the sequence $\Sum_{i=1}^{n} u_{i}$ converges

it converges absolutely if $\Sum_{i=1}^{n} u_{i}$ converges, and you can prove this implies the latter

It is a very important condition.  For starters, if a series converges absolutely, you can permute the terms of the sequence, it will still converge and to the same limit.

There are many theorems for which you need it.

Basically, when you wanna see series as a sort of sum of numbers, absolute convergence lies closer to that intuitive perception, as here the sum of all things is not infinity.

It might be interesting to as what "unconditional convergence" would be.

That would mean this: there is a number $L$ such that not only is $\sum_{n=1}^{\infty}a_n=L$, but also any rearrangement would converge to the same limit.

That is, whenever $\sigma: \mathbb{N}\mapsto\mathbb{N}$ is a permuation - that is, a one-to-one function, then $\sum_{n=1}^{\infty}a_{\sigma(n)}=L$ as well.

And here's the big theorem: a (real or complex valued) series is unconditionally convergent if and only if it is absolutely convergent.

Hence, if the series converges absolutely and we rearrange it, it still converges to the same number. But if it converges without being absolutely convergent - that is, it is conditionally convergent - then we can rearrange it to converge to a different number or to diverge. For example:

$1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\cdots = \ln2\approx 0.69$ but

$1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16+\cdots$ converges, but to a number that is approximately $1.04.$

--

Another way to look at it: if you can argue that the series converges just by looking at the size of the terms, that's absolute convergence. All but one of the usual convergence tests of  a Calculus II (or BC) course are comparison tests. Comparison, limit comparison, the integral test, the ratio test, the root test - these are all comparison tests. (So is the condensation test.) All comparison tests are tests for absolute convergence.


function
calculus
derivative
calculus computations

let f'(x) be the function whose graph goes through the point (3,6) and whose derivative is given by f'(x) = (1+e^x)/(x^2)

use euler's method, starting at x=3 and use it to approximate f(3.1)
[b]use f''(x) to explain why this approximation is less than f(3.1)[/b]

please help me on the 2nd part of the ?  :idea:
 :lol:

Some friends of mine were working on this problem today.  Where do you live?   :rotfl:

Anyway, Euler's method basically uses a linear approximation (the tangent line).  If you graph the tangent line of an arbitrary curve, you'll see that it underestimates the actual value if $f(x)$ is concave up (convex), and overestimates the actual value if $f(x)$ is concave down (concave).

So, evaluate $f''(3)$, and you will know whether it is concave up or down in that area.


When considering the number of positive integer solutions to

\[\sum_{k=1}^{n}x_{k}=\alpha,\]
I understand that the "ball and urn" method is one way of taking care of this.  However, this would give $\binom{\alpha+n-1}{n-1}$ different ways of summing to $\alpha$ using the given sum, if I'm not mistaken.  This takes into account the order in which the terms are summed; for instance for $n=2$ and $\alpha =2$ we would have $3$ different ways of summing to $2$ by means of $0+2$, $1+1$, and $2+0$.  In order to find the number of solutions (not depending on order) to the same equation, would the answer simply be

\[\frac{1}{2}\left(\binom{\alpha+n-1}{n-1}-1\right)?\]
Also, how is it that the number of terms in the expanded (simplified) expression $\left(\sum_{k=1}^{n}x_{k}\right)^{\alpha}$ is the same as the number of solutions to $\sum_{k=1}^{n}x_{k}=\alpha$?

First of all $\binom{a+n-1}{n-1}$ is the number of solutions where $x_{1},...x_{n}$ are NONNEGATIVE integers (includes 0). $\binom{a-1}{n-1}$ is for POSITIVE integers (does not include 0).

I think you meant $\frac{1}{2}\left(\binom{\alpha+n-1}{n-1}+1\right)$, though this should only work if $n=2$

In your third question, think about what all the terms would be like (excluding coefficients). They would be like $x_{1}^\alpha$, $x_{1}^{1}x_{2}^{\alpha-1}$, $x_{1}^{2}x_{2}^{3}x_{4}^{\alpha-5}$. In each term, the sum of the exponents adds up to $\alpha$, which is why it's the same as the number of solutions to that equation.

$x_{1}=2,x_{2}=0$ and $x_{1}=0,x_{2}=2$ are not the same solution and should be counted separately. The question you're asking can't be solved with the same solution, as far as I can tell. There are complications like how you would handle three repeated addends and so forth.


There are 6 + 8 + 4 + 2 + 5 = 25 in total. Since 5 are E, we have $ 20$ percent.


Direct application of Minkowski's inequality.

$ \sqrt{a^2\plus{}b^2}\plus{}\sqrt{c^2\plus{}d^2}\ge \sqrt{(a\plus{}c)^2\plus{}(b\plus{}d)^2}$


For the classes that I would like to take: Mathcounts/AMC Counting and Probability and MATHCOUNTS/AMC Counting, I already know much of what is on the syllabus. My problem is that I cannot solve most problems that are "out-of-the-box." Basically, unless the problem is similar to a problem that I have already seen before, then I cannot do it. For example, I can do very difficult (content-wise)combinatorics questions, but I cannot do very difficult (problem-solving-wise) combinatorics questions.

Is it worth it for me to these classes? I am sure that I will strengthen my understanding of the subject matter after taking the class because there are some gaps in my knowledge, but I am not sure it will be worth it for me when I already know about 80% of the content.

What about the AIME/Olympiad Counting Class-- should I take that or should I completely and utterly master everything in the basics before moving on? Again, I already know most of what is on the syllabus, but, in this case, even a review would be nice because I am VERY shaky on the material, considering I have never had to do a problem that involved e.g. generating functions before except in the AoPS book.

Of course, I could take all of them and that way I'll have the best of all worlds, but money, time, and efficiency are at stake.

I say take the intro to counting (MC/AMC) class. If you like it, follow up with intermediate :-) (btw, generating functions is like the last few classes or intermediate counting)

I've never taken a subject class so far (but I have taken two problem series). Is the information and problems in the classes the same as the information and problems in the AoPS book? Right now, I am leaning towards AMC/AIME Algebra and AIME/Olympiad Counting.

I guess would just like to work on the problem sets if I took MC/AMC Counting and Probability because I have read the same information on basic counting described in the syllabus in three different books already. However, I need a good source of basic combinatorics questions. Anyone know of a good book for this? I've already done all the AoPS problems, and I want a book that has intermediate level competition type problems for combinatorics not textbook exercises.

I would consider the post test problems the real guide on the couting classes.  While you may have been exposed to permutations, combinatorics, and other counting concepts, these classes are about mastering those concepts to a level where you are confident in knocking out virtually any counting or probability problem you will ever see on the AMC-12.

If you are planning on enrolling in the independent study courses for two semesters I think it would be a superb idea to take the two counting and probability classes -- one each semester.  This would be a great foundation for combinatorics or probability classes you might take in college.

Additionally, even if you find the MATHCOUNTS/AMC Counting and Probability course a little easier than the other classes you take at AoPS, you will be taking a second class simultaneously as well as working through a broad body of material that we will offer through the independent study program.

Also, I'll add that if you're actually talking about taking these courses in the summer, then taking MATHCOUNTS/AMC Counting in the summer would set you up nicely to take the AIME/Olympiad Counting in the fall as part of the Independent Study program.

Isn't the AIME/Olympiad Counting class only in the summer, according to its description?

We expect to offer it again in the fall, probably the late fall (i.e. starting in Oct or early Nov).


calculus
integration
analytic geometry
trigonometry
calculus computations

I need to evaluate this double integral where the region is the triangle with the vertices (0,0)(0,1)(1,1). I'm not sure as to how to find the limits of integration. Since sin is involved I figured a good start would be with polar coordinates. Problem is polar coordinates confuse me. Thanks in advance.

Drawing a picture, we see that the boundary lines are $ x\equal{}0$ on the left, $ x\equal{}1$ on the right, $ y\equal{}x$ on the bottom, and $ y\equal{}1$.  This region is $ x$-simple so we define $ c\equal{}0$ and $ d\equal{}1$, and $ \psi_{1}\equal{}0$ and $ \psi_{2}\equal{}y$ which is just a consequence of the boundaries seen in the picture.  So we may write
$ \int_{0}^{1}\int_{0}^{y}\sin{y^{2}}\mbox{ }dx\mbox{ }dy$
Integrating,
$ \int_{0}^{1}y\sin{y^{2}}\mbox{ }dy\equal{}\frac{\sin{1}}{2}$
And we never used polar coordinates.

Thanks for the help, but what about sin when integrating? Wouldn't it become -cos?


inequalities
limit
calculus
calculus computations

Suppose that we have sequence $ \{a_n\}$
$ a_1\equal{}2;a_2\equal{}\minus{}2$
$ a_{n\plus{}1}\equal{}\frac{1}{2}(a_n^2\minus{}a_{n\minus{}1})$
Find $ lima_n$

We claim that $ a_n \to 0$ as $ n \to \infty$.

Put $ \gamma_n \equal{} \max \{ |a_{2n\minus{}1}|, |a_{2n}| \}$ and $ \alpha_n \equal{} \tfrac{1}{2}(1\plus{}\gamma_n)$. Then

$ |a_{2n \plus{} 1}| \leq \frac{1}{2}|a_{2n}|^2 \plus{} \frac{1}{2}|a_{2n\minus{}1}| \leq \frac{\gamma_n^2 \plus{} \gamma_n}{2} \equal{} \alpha_n \gamma_n.$

Suppose $ \gamma_n < 1$ so that $ \alpha_n < 1$. Then $ |a_{2n\plus{}1}| \leq \gamma_n$ and

$ |a_{2n \plus{} 2}| \leq \frac{1}{2}|a_{2n\plus{}1}|^2 \plus{} \frac{1}{2}|a_{2n}| \leq \frac{\gamma_n^2 \plus{} \gamma_n}{2} \equal{} \alpha_n \gamma_n.$

So $ \gamma_{n\plus{}1} \leq \alpha_n \gamma_n$ and $ \alpha_{n\plus{}1} \leq \alpha_n$.

Apply this inequality recursively to obtain $ \gamma_{n\plus{}k} \leq \alpha_n^k \gamma_n$. This yields $ \gamma_n \to 0$ as $ n \to 0$, thus $ \lim a_n \equal{} 0$.

Now it suffices to show that we have $ \gamma_n < 1$ for some $ n$. But this follows from direct calculation that $ \gamma_3 \equal{} 5/8 < 1$, which completes the proof.


Thomas stayed up very late before going to sleep. Though he is just a normal boy who needs just as much sleep as anyone else, staying up actually saved him from oversleeping the next day. If he had gone to bed when his parents told him to, he would have overslept.

How?

[quote="moogra"][hide="maybe?"]
Daylight Saving Time and Standard Time[/hide][/quote]

No, the boy is intelligent and would have remembered to set it forward staying up or not.

I agree with Bowei. Or as he was staying up, he realized that he had a late-night appointment, which he would have otherwise missed.

or maybe his parents live in a different time zone and they call him each night not realizing the time difference 0.o - not likely

[quote="bowei"][hide="maybe again"]
he did not set his alarm clock before his usual sleeping time
when he stayed up late, he remembered to set his alarm clock
and consequently did not oversleep the next day
[/hide][/quote]

No, the answer does not rely on an error on Thomas's part. Though the alarm clock part is correct, the reason is not.

[quote="hm29168"]This has nothing to do with alarm clocks, but an alternate solution could be that it was New Year's Eve and he was staying up for the ball drop. If he had gone to sleep, he would have missed it and his parents would be to blame.

I frankly have no idea what else it could be.[/quote]

nope, keep trying...

(the answer is pretty un-contrived)

[quote="lingomaniac88"][hide]Is it that his alarm clock takes a long time to set?[/hide][/quote]

Nope, just a standard alarm clock.

Did something happen to the clock? Maybe it stopped after parents told Thomas to go to sleep and no one would notice that if he wouldn't stay up so late?

[hide]Did he stay up all night. ie Never went to sleep until after the appointment?[/hide]

The above post was the first thing I thought of. Alternatively, there could have been a momentary power outage from a storm which caused his alarm clock to reset (this has actually happened to me before).

[quote="probability1.01"]The above post was the first thing I thought of. Alternatively, there could have been a[b] momentary power outage from a storm which caused his alarm clock to reset[/b] (this has actually happened to me before).[/quote]

Ding ding ding! Correct! (emphasis mine)


quadratics
MATHCOUNTS
algebra
polynomial
number theory
prime numbers
quadratic formula

[hide]
T) Solving using the quadratic formula, we find that the answer is one.

I) $2310 = 2 \times 3 \times 5 \times 7 \times 11$, so the sum of the prime factors is 28.

G) Primes between 12 and 25 are 13, 17, 19, 23. The sum of these is 72.

E) Looking at xxxxx's post, we find that E is 24.

R) Looking at the equation, we see x can be 0. Dividing both sides by x, we have $x^{2}= 1$, so x can also be 1 or -1. There are 3 values.

Plugging these values in, we have 1 (28 + 72) + (24 / 3)^2, which is 100 + 8^2, or 164.[/hide]

the multiple choices are:

A.160
B.8
C.12
D.18
E.2006

I need help with this one... :(

Uhhhhh......
Ubemaya, it should be 1 (28 + 72) + (24 / [b]3[/b])^2
so 100+64=164 which isn't a choice  :(

Oooops...my fault.  :blush:
The answers are supposed to be:

A. 160
B. 161
C. 162
D. 163
E. 164

[quote="ProtestanT"][quote="xxxxx"]Find $T\times(I+G)+(E\div R)^{2}$ if

T = number of positive solutions to $2x^{2}-11x-8=0$
I = the sum of the prime factors of 2310
G = sum of the prime numbers between 12 and 25
E = 24
R = the number of real numbers $x$, such that $x^{3}=x$[/quote]
for T just use the alternating signs rule.
for I just prime factor 2310 (2*3*5*7*11)
for G the prime numbers are 13, 17, 19, and 23
for E it is 24
and for R there should be 3 solutions.[/quote]
DesCartes law of signs is far beyond mathcounts, though this is a great application of it.

[quote="xxxxx"]"...Decartes' Rule of Signs says that for a polynomial P(x), the number of positive roots to the equation is equal to the number of sign changes in the coefficients of the polynomial, or is less than that number by a multiple of 2. The number of negative roots to the equation is the number of sign changes in the coefficients of P(-x), or is less than that by a multiple of 2."

I'm a little confused...can you explain to me?[/quote]
It, and it's proof are beyond mathcounts. You needn't worry about them. Anyways, here's a simpler explanation (I think): in f(x) then the number of times the sign of the number changes when you write it in descending order is the number of positive roots. the number can however, be less than that by a multiple of 2. For the number of negatives count the number of sign changes in f(-x).


integration
calculus
calculus computations

\[ y^3 dx \plus{} (x \plus{} y^2 )dy \equal{} 0\]

[quote="Xaenir"]
\[ y^3 dx + (x + y^2 )dy = 0\]
[/quote]

I'm getting $ xe^{-\frac{1}{2y^2}}=-\int\frac{1}{y}e^{-\frac{1}{2y^2}}dy$ (not able to integrate the rhs :( )


Prove:

If a = b (mod n), then a^k = b^k (mod n) for any positive integer k.

If a is an odd integer, then a^2 = 1 (mod 8)

For all integers a, a^3 = 0, 1, or 6 (mod 7)

The '=' signs are actally triple equal signs, don't know how to put them there.

I'm totally lost on the first and third questions. For the second, I have a proof for it already, but I'm not sure if mine covers all the cases. For the second I did

a^2 -1 = 4(K^2-K)

Then I did different cases for K is odd and K is even so I could write it as a multiple of 8. Is this Ok or is there a more elegant way?

For the first one, try induction on k.

For the third one, you can do 7 cases, and that'll work. I don't know of another easy way to do it.

[quote="calcsheep"]If a = b (mod n), then a^k = b^k (mod n) for any positive integer k.[/quote]

This can be done inductively, but there is a [hide="direct proof"] $a \equiv b \bmod n$ means $n | (a-b)$.  But since $a-b | a^{k}-b^{k}$ it follows $n | a^{k}-b^{k}$. [/hide]

[quote="calcsheep"]For all integers a, a^3 = 0, 1, or 6 (mod 7)[/quote]

This can be done with seven cases, but there is a [hide="cooler proof"] We know that $a^{6}\equiv 0, 1 \bmod 7$ by Fermat's Little Theorem.  Let $b = a^{3}$.  Then $b^{2}\equiv 0, 1 \bmod 7$, so $b \equiv 0, \pm 1 \bmod 7$ (I have taken the modular square root; consult my blog). [/hide]

[quote="calcsheep"]Thanks guys, I got all except the exponent proof so far.

t0rajir0u: I don't get the step (a-b) divides (a^k - b^k)

Can someone please explain?[/quote]

It's a well known factoring I believe

$a^{k}-b^{k}= (a-b)(a^{k-1}+a^{k-2}b+a^{k-3}b^{2}+...+a^{2}b^{k-3}+ab^{k-2}+b^{k-1})$

If I messed up the factoring, please tell me. (I think I got it right).

all of the proofs are essentially the same...but if you want another way to skirt around a rigirous proof then use the geometric series formula

[quote="calcsheep"]Thanks guys, I got all except the exponent proof so far.[/quote]

[hide="The inductive proof, for reference"] [b]Lemma:[/b]  If $a \equiv b \bmod m$ and $c \equiv d \bmod m$ then $ac \equiv bd \bmod m$.

Proof:  Let $b = mx+a$ and let $d = my+c$ where $x, y$ are integers.  Then

$bd = (mx+a)(my+c) = m^{2}xy+m(ay+cx)+ac \equiv ac \bmod m$

Inductively, since $a \equiv b \bmod m$ it follows $a \cdot a \equiv b \cdot b \bmod m$ by the lemma, and so forth. [/hide]


geometry
3D geometry
LaTeX
articles
probability
absolute value

Please, please, if you usually don't post on the math forums because the level of difficulty is too high, try some of these problems! Even if you can't completely finish a problem, [b]you can post what you've done so far[/b] and someone will be glad to help you finish it.

[b]1.[/b] Find the product of all the integers from -1000 to 1000, inclusive.

[b]2.[/b] If a cube's volume is equal to its surface area, find the length of an edge of the cube. (note: the fomula for the volume of a cube is [tex]s^3[/tex] and the formula for the surface area of a cube is [tex]6s^2[/tex], where [tex]\displaystyle s[/tex] is the length of an edge of the cube.)

[b]3.[/b] If [tex]x = \sqrt{x}[/tex], find the sum of all possible values of x.

[b]4.[/b] If D is the sum of all the integers from -10 to -1, inclusive, and Q is the sum of all the integers from 1 to 10, inclusive, find D + Q.

[b]5.[/b] If the first term of a sequence is 0, the second term of a sequence is 1, and any term after the second is the positive difference between the previous two terms, find:

a) The next 7 terms.

b) The 999th term.

c) The sum of the first 2004 terms.

[b]6.[/b] Find: [tex]\frac{1}{2} \times \frac{2}{3} \times \dots \times \frac{999}{1000}[/tex].

[b]7.[/b] (this is harder) Let a [i]yifanagon[/i] be a right triangle with side lengths that are consecutive integers. Prove that there is only one yifanagon, and find its side lengths.

here's what I got after a few minutes: ( please excuse me, as I'm a complete beginner and don't know any LaTeX whatsoever   )

[hide]1. 0? since 0 is included in that range of numbers.

2. s :^3:  - 6s :^2:  = 0

factor and solve, s = 0, 6, throw out the 0

3. only x values i can think of are 0 and 1, so the sum is 1?

x -  :sqrt: x = 0

 :sqrt: x ( :sqrt: x - 1) = 0

x = 0, x = 1

4. 0, b/c D = -Q?

5. the series goes like: 0, 1, 1, 0, 1, 1, 0, 1, 1, ...

a. 1, 0, 1, 1, 0, 1, 1

b. 1 because every third term is 1 ... how would u express this mathematically though, not just intuitively?

c. 2004 * (2/3) = 1336 nonzero terms, and since all of the nonzeroes are 1, then the sum is 1336

6. 1/1000

7. using pythagorean:

x :^2: + (x+1) :^2: = (x+2) :^2:

x :^2: + x :^2: + 2x + 1 = x :^2: + 4x + 4

x :^2: - 2x - 3 = 0

(x-3)(x+1) = 0

x = -1, 3, throw out the -1

so x = 3, the only positive integer root

[/hide]

as an aside, I feel that I'm very weak with proofs, which is mainly why I'm here. What do you suggest is the best way to get started in improving my proof-ing ability? I've browsed through this forum and I've seen recommendations for books, but which is the absolute best way? thanks for your help.

Well done, VeCTer! You solved them all correctly, but on #7 you have to make sure to read exactly what the question asks for. Besides that, though, good job!

Anyone else want to give these a shot?

[quote="VeCTer"]as an aside, I feel that I'm very weak with proofs, which is mainly why I'm here. What do you suggest is the best way to get started in improving my proof-ing ability? I've browsed through this forum and I've seen recommendations for books, but which is the absolute best way?[/quote]

The absolute best way to become better at proofs is to do as many as you can. There are many good ones to try on this forum (as well as a few on intermediate) that aren't too difficult.

Also, you might want to read the article on this site entitled "How to Write a Solution" which is pretty self-explanatory. You can find it here:

http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php

thanks for the help!

I forgot for #7, the sides are just [hide]3, 4, and 5[/hide]

I'll do number one because it took me less than 3 seconds:

[hide]Middleterm between -1000 and 1000 is 0. Anything multiply by 0 is 0, (identity property of multiplication), so answer is 0.[/hide]

Amazingly enough, I couldn't get #1 until I looked at Silverfalcon's solution. I'm seriously insane. But for the others:

2. [hide]If 6s^2=s^3, 6=(s^3)/(s^2)=s. Edge is 6 units long.[/hide]

3. [hide]1+0=1.[/hide]

4. [hide]-10-9-8-7-6-5-4-3-2-1+1+2+3+4+5+6+7+8+9+10. Every integer in there has an additive inverse, so the whole thing adds up to 0.[/hide]

5. [hide]a. The first seven terms are 0, 1, 1, 0, 1, 1, 0.

b. The repeating pattern (0-1-1) has three terms. Divide 999 by 3. If the remainder is 1, the term is 0. If the remainder is 2 or 0, the term is 1. 999/3 has a remainder of 0, so the 999th term is 1.

c. Two out of three times, a term is 1. Otherwise, it's 0. So the sum of 2004 terms is equal to 2004*2/3=1336. (Things would be a tiny bit more complicated if 2004 weren't divisible by 3.)[/hide]

6. [hide]All terms cancel except for 1 in the numerator and 1000 in the denominator. The product is 1/1000.[/hide]

7. [hide]In a yifanagon, with a being the shortest leg, a^2+(a+1)^2=(a+2)^2 from the Pythagorean Theorem. That has two solutions, -1 and 3, but the leg of a triangle can't have length -1 for obvious reasons. So there is really only one solution, a=3, and therefore only one yifanagon, with sides 3, 4, and 5.[/hide]

I still can't get over not being able to do #1...how could I have been so stupid...

[quote="JSRosen3"]7. (this is harder) Let a [i]yifanagon[/i] be a right triangle with side lengths that are consecutive integers. Prove that there is only one yifanagon, and find its side lengths.[/quote]
gee, what a cool name for a shape! :smile:
i would do a couple problems, but i don't want to risk destruction by the all mighty rosen

I might be wrong on this but here it is for number 6:

[hide]The nice thing in this problem is that except one and one thousand at the end, the every single number between them are being cancelled.  So, simply we're left with [tex]\frac {1}{1000}[/tex] , the answer.[/hide]

Another nice cancelling one:

[hide]4. If D is the sum of all the integers from -10 to -1, inclusive, and Q is the sum of all the integers from 1 to 10, inclusive, find D + Q. Same thing in here.  All the integers in set D are negative and Q's numbers are positive so the answer is 0.

[/hide]

Further approach question.

For the people who thought this was too easy, let me give you a nice easy one.

Find the absolute value of D+Q. Also, what's the difference of taking absolute value of all numbers in both set D and Q, and then adding compared to adding the numbers first and taking the absolute value of the result.  Now, think about that and it's not really hard so. Good luck!

Silverfalcon wrote:Find the absolute value of D+Q. Also, what's the difference of taking absolute value of all numbers in both set D and Q, and then adding compared to adding the numbers first and taking the absolute value of the result.  Now, think about that and it's not really hard so. Good luck!

[hide]Umm...since the whole thing is zero the absolute value is of course zero as well.

If you add up the absolute value of each number in D and Q, you have 2(1+2+3+...+10) which is 2*55 which is 110. If you take the absolute value of the sum, well, it's zero, like I've already said. For the first problem you have two sets of (1+2+3+...+10), and for the second you have one set of (1+2+3+...+10) and one of (-1-2-3-...-10).[/hide]

I should be ashamed of myself, not daring to take on problems any harder than this.

[quote="JSRosen3"]Try this one:

Square A has side length 1. A diagonal of square A is the side of square B. A diagonal of square B is the side of square C. The process is repeated until we arrive at square Q. A circle is inscribed in square Q. Find the probability that a point inside square Q is not inside the circle.[/quote]
it's not as hard as it seems...

[quote="JSRosen3"]I edited the problem since the probability is constant for all squares![/quote]
lol, shame on you!

JSRosen3 wrote:Try this one:

Square A has side length 1. A diagonal of square A is the side of square B. A diagonal of square B is the side of square C. The process is repeated until we arrive at square Q. A circle is inscribed in square Q. Find the area of this circle.

(edited)

spoiler:

[hide]

diagonal of 1st square w/side 1 =  :rt2:

diagonal of 2nd square w/side  :rt2:  = 2

3rd square diag = 2 :rt2:

4: 4

5: 4 :rt2:

...

for the nth square, the diagonal is of length  :rt2:^n

and since Q is the 17th letter in the alphabet, its diagonal is  :rt2: ^17 but we want its SIDE, which is just  :rt2: ^(n-1) which = :rt2: ^16 = 2^8 = 256

radius of inscribed circle = 256/2 = 128

so area of inscribed circle =  :pi: * 128 :^2:

[/hide]

BobofBobs wrote:Obviously its  [hide]4- :pi: /4[/hide].

There's another post about proving that up right now.

if you noticed, he changed the problem after he realized that. also, you should've noticed he got the idea for the new thread from this and he was the author of the new thread

If anybody wants to tackle the proof of the yifanagon, where there exists only one right triangle with consecutive integer side lengths, I would suggest it highly. It will help you become familiar with calculating primitive pythagorean triples, and methods for determining the next triplet. This is just a hint not an answer:

[hide]Remember that all primitive triplets can be expressed as ${(2m), (m^2+1),(m^2-1)}$. [/hide]

The answers and question # are hidden all in the same block.
[hide]
1. 0
2. s=6
4. 0
5. a) 0,1,0,0,1,0,0
    b) 0
    c) 668
6. 1/1000
7. The sides are 3,4,5 I don't have a proof.
[/hide]
P.S. How do you display LaTeX in posts?

[hide="ans. to 1"]
considering your also multiplying that by 0, your answer is obviously 0
>>0<<
[/hide]
[hide="ans. to 2"]
Area of Cube: s*s*s
Surface Area: 6*s*s
s^3=6s^2
hmmm, cancel out a s^2, and you get
s=6
so the side lengh must be 6!
>>6<<
[/hide]
[hide="ans. to 3"]
To only numbers that can do that are 1, 0, and -1.
1, and -1 cancel each other out so the answer is 0!
>>0<<[/hide]
[hide="ans. to 4"]
so the equation of D+Q=
(-10+-9+...+-1)+(1+2+...+10)
using the communative and associative properties, you can rearrange the numbers and you get,
-10+10+-9+9+...+-2+2+-1+1
they cancel each other out! Yay! 0 again!
>>0<<[/hide]
[hide="ans. to 5"]
so the sequence would be like this: (first seven terms for a)
A)0,1,1,0,1,1,0,...
B)999:equiv:0 (mod 3), so 1.
C)pattern repeats after 3, each 3 have a sum of 2, so 2/3 of 2004=1336
[/hide]
[hide="ans. to 6"]
all the terms cancel, except for the 1 on the numerator and the 1000 on the denominator, so the answer is 1/1000
[/hide]
all I have so far,, I'll come back here later

[quote="mathboyzld"]P.S. How do you display LaTeX in posts?[/quote]

You just put the LaTeX in between two dollar sign's.


AMC
AIME
calculus
invariant
geometry
search
graph theory

So i'm preparing for AIME now, but what chapters in ACoPS are essential or important for AIME (1-12)?  Also, do they send your home a letter if you qualified the AIME, what about those don't make it?

ACoPS is amazing. You're shortchanging the book by skipping anything. The calculus chapter is obviously unnecessary but otherwise everything else is pretty useful.

I don't have ACoPS anymore but I think that worthawhole bean is right. That is packed with good stuffs and really, there isn't too much you can take out.

Correct me if I'm wrong but isn't invariant and graph theory in there too? I don't think you need those for AIME though..

(Worthawholebean knows what he's talking about.)

I'd have to agree with the above two posts.  I haven't finished ACoPs yet, but from what I can see, the problem-solving chapters (#1-4) and the subject chapters excluding calculus (#5-8) are all pretty useful.  Obviously, if you're really good at geometry or something like that, then you should really just work some of the problems in that chapter to hone your skills and then focus more on the other subjects.  Also, I would [i]highly[/i] recommend Chapter 4, as techniques like using complex numbers are very useful in attacking certain AIME problems.

My biggest piece of advice: Do the problems! (ACoPS is especially good on problem choice.)

Can anyone tell me what's the full name for ACoPs and where can I buy it?

The Art and Craft of Problem Solving, by Paul Zeitz.  Google it to find sellers.  I bought mine from the publisher (Wiley)

[quote="cdymdcool"]Also, do they send your home a letter if you qualified the AIME[/quote]

I never received anything from the AMC people themselves last year about the AIME, my teacher notified me (about two weeks before the test).

You generally get notified by your teacher.

[quote="SnowEverywhere"]I do not have ACoPS and this is my first time writing the AMC 10 and making the AIME so I want to study. So far I have almost finished AOPS V1 and own AOPS V2. Are there any recommendations as to how I should prepare? I hope to get 5 or higher.[/quote]

By your usage of phrase "writing the AMC 10," am I not wrong that you have had some exposure to Canadian English? I know that Canadians websites discuss how to "write the test" when referring to "take the test."

My best suggestion is to search around the forum because this type of advice-seeking question has occurred. You can also try to look at my guide, which have helped a good number of students.

http://www.hiddendiscoveries.com

Also if you haven't done so, take some practice tests. As much as you can without overexerting your brain of course. And get a good night sleep day before.

@others: Keep in mind that ACoPS does not have solution available to the public. There is a solution, and yes, some people ( :roll: ) have it. But I can guarantee you that no one will give to anyone so just something to keep in mind.

Unfortunately I actually did not find ACoPS [i]all[/i] that helpful... maybe because I got it too late for it to be of much use. But finishing the book (except maybe the calculus chapter, and inversion) should be sufficient to get you into USAMO. But don't JUST use a book to prepare! [b]DOING PROBLEMS[/b] (from old tests, AoPS, and also books) is the best way! But you do need background knowledge etc., and that's where books (explanation books) come in. There's no substitute for doing actual competition problems though.


The two solutions of the equation $ x^2\plus{}bx\plus{}18\equal{}0$ are in the ratio of 2 to 1 for some values of $ b$. What is the largest possible value of $ b$?

[quote="GameBot"]The two solutions of the equation $ x^2 \plus{} bx \plus{} 18 \equal{} 0$ are in the ratio of 2 to 1 for some values of $ b$. What is the largest possible value of $ b$?[/quote]

$ (x\minus{}y)(x\minus{}2y)\equal{}x^2\plus{}bx\plus{}18$
$ x^2\minus{}3xy\plus{}2y^2\equal{}x^2\plus{}bx\plus{}18$
$ y\equal{}3, \minus{}3$
$ \pm 9x\equal{}bx$
$ b\equal{}9$.


You cannot go from this step to this:

(a+b)(a-b)=b(a-b)
(a+b)=b

because we assumed that a=b.  Dividing through by (a-b) is dividing through by 0, an illegal operation.

This very similar variation is the well known "proof" that any number is equal to zero. I use the inverted commas because, as already pointed out, dividing through by $ (a\minus{}b)$ is not allowed.

$ a\equal{}b$
$ a^2\equal{}ab$
$ a^2\minus{}b^2\equal{}ab\minus{}b^2$
$ (a\plus{}b)(a\minus{}b)\equal{}b(a\minus{}b)$
$ a\plus{}b\equal{}b$
$ a\equal{}0$

If you use Google, you can find many classic examples of similar mathematical fallacies, using anything from dividing by 0 to incorrectly differentiating.

This link just gives some:
[url]http://www.math.toronto.edu/mathnet/falseProofs/fallacies.html[/url]

[quote="Pie-R-not-squared"]I was surfing the web when I came upon a very interesting proof. The proof made sense to me, math wise, but not logically. Lets see who thinks that this proof is correct or incorrect.

 a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1[/quote]

we can't get $ a\plus{}b\equal{}b$ from $ (a\plus{}b)(a\minus{}b)\equal{}b(a\minus{}b)$ because $ a\minus{}b\equal{}0$

another example
let $ x\equal{}7$
then $ 3x\equal{}2x\plus{}7$
$ 3x\minus{}21\equal{}2x\plus{}7\minus{}21\equal{}2x\minus{}14$
$ 3(x\minus{}7)\equal{}2(x\minus{}7)$
$ 3\equal{}2$
obviously it is wrong


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

I'm not familiar with the theory of algebraical solvability, but I was asked by my friend the following question:
Is  the equation $ x^5\plus{} 5 x^3 \plus{}5 x \minus{}2\equal{}0$ algebraically solvable?
Thanks in advance.

The method you would use to solve this problem is to first find the splitting field $ E$ of $ f$ and then compute its Galois group and see if its solvable (just so you know, if it turns out $ Gal(E/Q)\cong S_5$, it will not be solvable). So to do this, there is probably something to do with computing the discriminant of $ E$ =\ I don't really know this too well though

[quote="leshik"]I'm not familiar with the theory of algebraical solvability, but I was asked by my friend the following question:
Is  the equation $ x^5 \plus{} 5 x^3 \plus{} 5 x \minus{} 2 \equal{} 0$ algebraically solvable?
Thanks in advance.[/quote]

A superior algebra way is to show that the Galois group of the polynomial has order 20 (it's the Frobenius group of order 20 in fact), and so it's solvable.

An inferior algebra way (haha) is to just solve the equation! The substitution $ x\equal{}y \minus{} \frac{1}{y}$ will do the job.  :wink:

[quote="ysharifi"][quote="leshik"]I'm not familiar with the theory of algebraical solvability, but I was asked by my friend the following question:
Is  the equation $ x^5 \plus{} 5 x^3 \plus{} 5 x \minus{} 2 \equal{} 0$ algebraically solvable?
Thanks in advance.[/quote]

A superior algebra way is to show that the Galois group of the polynomial has order 20 (it's the Frobenius group of order 20 in fact), and so it's solvable.[/quote]

Just to concur, the Galois group is order 20, and when it acts on the 5 roots of this polynomial it is the normalizer of a 5-cycle in S5, also known as the Frobenius group of order 20.

Most 5th degree polynomials do have Galois group S5, so the first response probably had a minor typo when they checked the polynomial.

I like the simple substitution.

Of course, simple substitution gives the result in no time, but he needs  ,,Galois'' argumentation :)
Thanks for your comments.


[quote="phamduyhiep"]Let $ABC$ be an acute triangle ($AC<AB$) with orthocentre $H$ and circumcentre $O$.
$CP \perp AB$ ($P \in AB$).$BH \cap OP=M$.The line passing through $P$ and parallel to $BC$ intersects the line $AB$ at $D$.
Show that $\widehat{MDP}=\widehat{BAC}$.[/quote]

Something is wrong with this problem. Did you mean http://www.mathlinks.ro/Forum/viewtopic.php?t=1133 ?

  Darij


Find the $ n^{th}$ term of the sequence
$ 1,3,7,13,21.......$

[hide="A little bit more explained"]

$ 3\minus{}1\equal{}2\equal{}a_2\minus{}a_1$
$ 7\minus{}3\equal{}4\equal{}a_3\minus{}a_2$
$ 13\minus{}7\equal{}6\equal{}a_4\minus{}a_3$
$ 21\minus{}13\equal{}8\equal{}a_5\minus{}a_4$
...

We connect the difference between the terms and the number of the smaller term in the difference:

2-1
4-2
6-3
8-4

We see that $ a_{n\plus{}1}\minus{}a_n\equal{}2n$, and now:

$ a_n\equal{}a_{n\minus{}1}\plus{}2(n\minus{}1)\equal{}a_{n\minus{}2}\plus{}2(n\minus{}1)\plus{}2(n\minus{}2)\equal{}\ldots\equal{}a_1\plus{}2((n\minus{}1)\plus{}(n\minus{}2)\plus{}\ldots\plus{}1)\equal{}a_1\plus{}n(n\minus{}1)\equal{}n^2\minus{}n\plus{}1$
[/hide]


Let $ a,b,c > 0$ such that   $ a \plus{} b \plus{} c \equal{} 1$ , prove

$ 1. \sum_{cyc}\frac {1}{a(1 \plus{} a \plus{} \ldots \plus{} a^{n \minus{} 1})} \ge \frac {2\cdot{3^{n \plus{} k}}}{3^{n} \minus{} 1}(a^k \plus{} b^k \plus{} c^k)(k \equal{} 1,2);$

$ 2.\sum_{cyc}{a(1 \plus{} b \plus{} \ldots \plus{} b^{n \minus{} 1})} \leq \frac {3^n \minus{} 1}{2\cdot{3^{n \minus{} k}}}(a^k \plus{} b^k \plus{} c^k)(k \equal{} 2,3,\ldots).$

And we have

$ \frac {3^n \minus{} 1}{2\cdot{3^{n \minus{} 2}}}(a^2\plus{}b^2\plus{}c^2)\geq\sum_{cyc}{a(1 \plus{} b \plus{} \ldots \plus{} b^{n \minus{} 1})} \geq \frac {3^n \minus{} 1}{2\cdot{3^{n \minus{} 2}}}(ab\plus{} bc \plus{} ca)$

in which $ a\plus{}b\plus{}c\equal{}1,a,b,c>0$.

Let $ n\to \infty$, then we have

\[ \frac {b \plus{} c}{a} \plus{} \frac {c \plus{} a}{b} \plus{} \frac {a \plus{} b}{c}\geq18(a^2 \plus{} b^2 \plus{} c^2);
\]
\[ \frac {9}{2}(a^2 \plus{} b^2 \plus{} c^2)\geq\frac {a}{a \plus{} c} \plus{} \frac {b}{b \plus{} a} \plus{} \frac {c}{c \plus{} b} \geq\frac {9}{2}(ab \plus{} bc \plus{} ca).
\]
in which  $ a \plus{} b \plus{} c \equal{} 1,a,b,c > 0.$

Generally, if $ x_1,x_2,\ldots,x_m > 0,x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m \equal{} 1$,prove(or disprove):

\[ \frac {m^{2 \minus{} n}(m^n \minus{} 1)}{m \minus{} 1}(x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_m^2)\geq\sum_{cyc}x_1(1 \plus{} x_2 \plus{} ... \plus{} x_2^{n \minus{} 1})
\]

\[ \geq \frac {m^{2 \minus{} n}(m^n \minus{} 1)}{m \minus{} 1}(x_1x_2 \plus{} x_2x_3 \plus{} \ldots \plus{} x_mx_1) ;
\]

\[ \frac {m^2}{m \minus{} 1}(x_1^2 \plus{} x_2^2 \plus{} \ldots \plus{} x_m^2)\geq(x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m)^2\sum_{i \equal{} 1}^m\frac {x_i}{x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m \minus{} x_{i \plus{} 1}}
\]

\[ \geq \frac {m^2}{m \minus{} 1}(x_1x_2 \plus{} x_2x_3 \plus{} \ldots \plus{} x_mx_1) .
\]

Further,if $ x_1,x_2,\ldots,x_m > 0,x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m \equal{} 1,k \equal{} 2,3,...,$ prove (or disprove)

\[ \frac {m^{2 \minus{} n}(m^n \minus{} 1)}{m \minus{} 1}(x_1^k \plus{} x_2^k \plus{} \ldots \plus{} x_m^k)\geq\sum_{cyc}x_1^{k \minus{} 1}(1 \plus{} x_2 \plus{} ... \plus{} x_2^{n \minus{} 1});
\]
\[ \frac {m^2}{m \minus{} 1}(x_1^k \plus{} x_2^k \plus{} \ldots \plus{} x_m^k)\geq(x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m)^2\sum_{i \equal{} 1}^m\frac {x_i^{k \minus{} 1}}{x_1 \plus{} x_2 \plus{} \ldots \plus{} x_m \minus{} x_{i \plus{} 1}}.
\]


algorithm
linear algebra
matrix
modular arithmetic
number theory unsolved
number theory

Let n be positive number such that n=$ {a}^{2}\plus{}{b}^{2}\plus{}{c}^{2}$(where a,b,c $ \in$$ N$*)and a,b,c are divisible by 3
     prove that there exist u,v,w $ \in$ $ N$* and u,v,w are not divisible by 3 such that n=$ {u}^{2}\plus{}{v}^{2}\plus{}{w}^{2}$.

[quote="Rust"]Consider $ u \equal{} (2b\plus{}2c\minus{}a)/3,v \equal{} (2a\plus{}2c\minus{}b)/3,w \equal{} (2a\plus{}2b\minus{}c)/3$.
If all u,v,w divisible to 3, repeat it.[/quote]

Let $ a\equal{}b\equal{}c\equal{}3$. how long must we repeat your algorithm?

Thank you pco.
If $ a\equal{}b\equal{}c$, and $ 3|a$, then work $ (3d,3d,3d)\to (5d,d,d)$ and repeat with $ (3a,3b,3c)\to (2b\plus{}2c\minus{}a,2a\plus{}2c\minus{}b,2a\plus{}2b\minus{}c)$.

[quote="Rust"]Thank you pco.
If $ a \equal{} b \equal{} c$, and $ 3|a$, then work $ (3d,3d,3d)\to (5d,d,d)$ and repeat with $ (3a,3b,3c)\to (2b\plus{}2c\minus{}a,2a\plus{}2c\minus{}b,2a\plus{}2b\minus{}c)$.[/quote]

Ok, and then what if $ (a,b,2b\minus{}a)$ ?; For example $ (3,27,51)$.

Your algorithm does not work in all the cases.

[quote="Rust"]Algoritm include  operations
1. $ (3a,3b,3c)\to (2b\plus{}2c\minus{}a,2a\plus{}2c\minus{}b,2a\plus{}2b\minus{}c)$
2. $ (3a,3b,6b\minus{}3a)\to (a,6b\minus{}a,3b\minus{}4a)$.
3. $ (3a,3b,6a\minus{}3b)\to (b,6a\minus{}b,3a\minus{}4b)$.
If first give same result or permutation, then work 2 or 3.[/quote]

Hello Rust.

Although you have nearly always very good ideas, I think it's not a good thing to just say "obviosly, here is the result" without giving a complete demo (users of this message board are generally more interested in good demos than in good results without demos and not interested at all in wrong results without demos).

In this case, you gave an incomplete result (so wrong) without demo.
On my first remark, you modified your position and gave a second incomplete result (so wrong) without demo.
On my second remark, you modified again your position and gave a third result without demo.

I'm not interested in verifying this third one (I've no obvious counter-example). Maybe you could give us a complete demo.

It's really a pity you never give demos since your ideas are very often very interesting.

[quote="Rust"]Algoritm include  operations
1. $ (3a,3b,3c)\to (2b\plus{}2c\minus{}a,2a\plus{}2c\minus{}b,2a\plus{}2b\minus{}c)$
2. $ (3a,3b,6b\minus{}3a)\to (a,6b\minus{}a,3b\minus{}4a)$.
3. $ (3a,3b,6a\minus{}3b)\to (b,6a\minus{}b,3a\minus{}4b)$.
If first give same result or permutation, then work 2 or 3.[/quote]

Wrong again : this does not work with $ (3,6,18)$

op2 and op3 can't be used. And op1 gives $ (15,12,0)$ which is not OK (non-zeroes values are required)

op1 again on $ (15,12,0)$ gives back $ (3,6,18)$

I'm really sorry, Rust. maybe an op4 is necessary.

[quote="Rust"]may be
2.' $ (3a,3b,6b\plus{}3a)\to (a,6b\plus{}a,3b\plus{}4a)$
3.' $ (3a,3b,6a\plus{}3b)\to (b,6a\plus{}b,3a\plus{}4b).$[/quote]

It does not fit with $ (3,6,18)$

I found ortogonal matrix for these case (when summ of 2  and third divides 9)$ B \equal{}\frac{1}{9}A$, were A:
\[ A\equal{}\begin{pmatrix}8& 1 & 4\\ \minus{}4 & 4 & 7\\ \minus{}1 &\minus{}8 & 4\end{pmatrix}\]
It give $ (3,6,18)B \equal{} (\minus{}2,\minus{}13, 14)$.

[quote="Rust"]I found ortogonal matrix for these case (when summ of 2  and third divides 9)$ B \equal{}\frac{1}{9}A$, were A:
\[ A\equal{}\begin{pmatrix}8& 1 & 4\\ \minus{}4 & 4 & 7\\ \minus{}1 &\minus{}8 &\minus{}4\end{pmatrix}\]
It give $ (3,6,18)B \equal{} (\minus{}2,\minus{}13,\minus{}2)$.[/quote]

I don't understand : $ 3^{2}\plus{}6^{2}\plus{}18^{2}\equal{}369\neq 2^{2}\plus{}13^{2}\plus{}2^{2}\equal{}177$ !

I am sorry, I edit $ \minus{}4\to 4$ for $ a_{33}$.
Let $ C\equal{}\frac{1}{3}D$, were \[ D\equal{}\begin{pmatrix}\minus{}1 & 2 & 2\\ 2 &\minus{}1 & 2\\ 2 & 2 &\minus{}1\end{pmatrix}\]
Only 2 operations work
1. $ (a,b,c)\to (a,b,c)C$
2. $ (a,b,c)\to (a,b,c)B$.
Let $ u\equal{}min(ord_{3}(a),ord_{3}(b),ord_{3}(c)),\ v\equal{}ord_{3}(a\plus{}b\plus{}c)$
If $ v\equal{}u$ we use first, else second.

[quote="Rust"]I am sorry, I edit $ \minus{}4\to 4$ for $ a_{33}$.
Let $ C \equal{}\frac{1}{3}D$, were
\[ D\equal{}\begin{pmatrix}\minus{}1 & 2 & 2\\ 2 &\minus{}1 & 2\\ 2 & 2 &\minus{}1\end{pmatrix}\]
Only 2 operations work
1. $ (a,b,c)\to (a,b,c)C$
2. $ (a,b,c)\to (a,b,c)B$.
Let $ u \equal{} min(ord_{3}(a),ord_{3}(b),ord_{3}(c)),\ v \equal{} ord_{3}(a\plus{}b\plus{}c)$
If $ v \equal{} u$ we use first, else second.[/quote]

$ (3,3,12)$ $ \implies$ $ \min(ord_{3}(3),ord_{3}(3),ord_{3}(12))\equal{}1\neq ord_{3}(3\plus{}3\plus{}12)\equal{}2$ $ \implies$ second $ \implies$ $ (0,\minus{}9,9)$

and now ?

Don't you think time is arrived to give a full demo?

Ok!
Let $ (a,b,c)S \equal{} (\minus{}a,b,c)$ if $ ord_{3}(b)\ge ord_{3}(a), ord_{3}(c)\ge ord_{3}(a)$ and
$ (a,b,c)S \equal{} (a,\minus{}b,c)$ if $ ord_{3}(a) > ord_{3}(b), ord_{3}(c)\ge ord_{3}(b)$ and
$ (a,b,c)S \equal{} (a,b,\minus{}c)$ if $ ord_{3}(a) > ord_{3}(c),ord_{3}(b) > ord_{3}(c)$ ($ ord_{3}(0) \equal{}\plus{}\infty$).
If $ ord_{3}(a\plus{}b\plus{}c) \equal{} v(a,b,c) > min(ord_{3}(a),ord_{3}(b),ord_{3}(c)) \equal{} u(a,b,c)$, then $ u((a,b,c)S) \equal{} v((a,b,c)S) \equal{} u(a,b,c)$.
Therefore $ (a,b,c)\to (a,b,c)SC$ give $ u((a,b,c)SC) < u(a,b,c)$.
I think it is full solution, and I use 2 operation $ C$ and $ S$.

[quote="Rust"]Ok!
Let $ (a,b,c)S = (-a,b,c)$ if $ ord_{3}(b)\ge ord_{3}(a), ord_{3}(c)\ge ord_{3}(a)$ and
$ (a,b,c)S = (a,-b,c)$ if $ ord_{3}(a) > ord_{3}(b), ord_{3}(c)\ge ord_{3}(b)$ and
$ (a,b,c)S = (a,b,-c)$ if $ ord_{3}(a) > ord_{3}(c),ord_{3}(b) > ord_{3}(c)$ ($ ord_{3}(0) =+\infty$).
If $ ord_{3}(a+b+c) = v(a,b,c) > min(ord_{3}(a),ord_{3}(b),ord_{3}(c)) = u(a,b,c)$, then $ u((a,b,c)S) = v((a,b,c)S) = u(a,b,c)$.
Therefore $ (a,b,c)\to (a,b,c)SC$ give $ u((a,b,c)SC) < u(a,b,c)$.
I think it is full solution, and I use 2 operation $ C$ and $ S$.[/quote]

Ok, Rust, thanks a lot. I really was looking for a rigorous solution and did not find it. i think now you did.
If you allow me, I'll re-write your demo in a longer but (I think) clearer way :

Let $ \mathbb{A}=\{(a,b,c)\in\mathbb{Z}^{3}-\{(0,0,0)\}$ such that $ a=b=c=0\pmod{3}\}$
Let $ ord_{3}(x)$ be the greatest power of $ 3$ which divides $ x$ (with convention that $ ord_{3}(0)=+\infty$)

[u]Lemma [/u]: $ \forall (a,b,c)\in\mathbb{A}$, it exists $ (u',v',w')\in\mathbb{A}$ such that $ a^{2}+b^{2}+c^{2}$ $ =u'^{2}+v'^{2}+w'^{2}$ and $ \min(ord_{3}(a),ord_{3}(b),ord_{3}(c))$ $ =ord_{3}(u'+v'+w')=ord_{3}(u')$ $ \leq ord_{3}(v')\leq ord_{3}(w'))$ :
==== Begin lemma ====
Let $ (u,v,w)$ a permutation of $ (a,b,c)$ such that $ ord_{3}(u)\leq ord_{3}(v)\leq ord_{3}(w)$ . Since $ (a,b,c)\neq(0,0,0)$, $ ord_{3}(u)$ is a finite number.

We have $ ord_{3}(u+v+w)\geq ord_{3}(u)$
If $ ord_{3}(u+v+w)= ord_{3}(u)$, then $ u'=u$, $ v'=v$, $ w'=w$ and the demonstration of the lemma is finished.
If $ ord_{3}(u+v+w)> ord_{3}(u)$, then $ 3^{ord_{3}(u)+1}$ divides $ u+v+w$ and so $ ord_{3}(v)=ord_{3}(u)$ (else $ 3^{ord_{3}(u)+1}$ would divide $ v+w$ but not $ u$, and so would not divide $ u+v+w$).
Then ${ ord_{3}(u-v+w})=ord_{3}(u)$ else $ 3^{ord_{3}(u)+1}$ would divide $ u+v+w$ and $ u-v+w$ and so $ v$, which is impossible since $ ord_{3}(v)=ord_{3}(u)$
So, take $ u'=u$, $ v'=-v$ and $ w'=w$ and the demonstration of the lemma is finished.
==== End lemma ====

[u]Demo itself [/u]:
From $ (a,b,c)\in\mathbb{A}$, take (with lemma above) $ (u',v',w')\in\mathbb{A}$ such that  $ a^{2}+b^{2}+c^{2}=u'^{2}+v'^{2}+w'^{2}$ and $ \min(ord_{3}(a),ord_{3}(b),ord_{3}(c))$ $ =ord_{3}(u'+v'+w')=ord_{3}(u')$ $ \leq ord_{3}(v')\leq ord_{3}(w'))$ .

Then let $ a'=2\frac{u'+v'+w'}{3}-u'$, $ b'=2\frac{u'+v'+w'}{3}-v'$ and  $ c'=2\frac{u'+v'+w'}{3}-w'$,
It's easy to verify that $ a^{2}+b^{2}+c^{2}=a'^{2}+b'^{2}+c'^{2}$
It is easy to verify that $ (a',b',c')\neq(0,0,0)$ (since $ a'^{2}+b'^{2}+c'^{2}=a^{2}+b^{2}+c^{2}>0$)
It is easy to verify that $ ord_{3}(a')=ord_{3}(b')=ord_{3}(c')=ord_{3}(u'+v'+w')-1$ $ =ord_{3}(u')-1<ord_{3}(u')$ $ =\min(ord_{3}(a),ord_{3}(b),ord_{3}(c))$

And so $ ord_{3}(a')=ord_{3}(b')=ord_{3}(c')$ $ =\min(ord_{3}(a'),ord_{3}(b'),ord_{3}(c'))=\min(ord_{3}(a),ord_{3}(b),ord_{3}(c))-1$

So, applying this transformation a finite number of times, we'll find $ 3$ numbers $ (x,y,z)$ such that :
$ a^{2}+b^{2}+c^{2}=x^{2}+y^{2}+z^{2}$ and $ 0=ord_{3}(x)=ord_{3}(y)=ord_{3}(z)$

Q.E.D.

Thanks Rust for your patience and your demo.


function
advanced fields
advanced fields theorems

Let $ f(X)\equal{}\sum_{i\equal{}1}^n a_iX_i \plus{}$(higher order terms) $ \in \mathbb{C}[[X]](\equal{}\mathbb{C}[[X_1,...,X_n]])$

and assume $ a_1 \ne 0$. Then $ f$ can be factored:

$ f(X)\equal{}u(X_1,...,X_n)(X_1\minus{}g(X_2,...,X_n))$ where $ u(0) \ne 0,g(0)\equal{}0$,

and every other power series $ k(X)$ can be written as uniquely:

$ k(X)\equal{}a(X_1,...,X_n)f(X_1,...,X_n)\plus{}b(X_2,...,X_n)$.

Please, prove this theorem.

I've never heard of this theorem before, and I don't know the answer, but I do have a guess for how to prove it.  In order to find $ g$, we "approximately" solve for $ X_1$ in the equation $ f(X)\equal{}0$ to get

$ X_1\equal{}\minus{}\frac{1}{a_1}\left(\sum_{i\equal{}2}^n a_i X_i\plus{}\text{higher order stuff}\right)$.

Note that RHS has no 1st order terms containing $ X_1$.  If you feed this formula into itself, you'll get an expression for $ X_1$ such that the RHS has no 1st or 2nd order terms containing $ X_1$.  Now feed the new expression into the RHS of the equation above, and lather, rinse, and repeat until you formally obtain a RHS that has no $ X_1$'s in it at all.  I conjecture that this is your desired $ g$.

yes this works. define a sequence of formal power series by $ g_0 \equal{}$ the formula for $ X_1$ above, and $ g_{n\plus{}1}\equal{}g_n(g_n(X_1,...,X_n),X_2,...,X_n)$. you can verify that $ g_n \to g$ for some $ g \in k[X_2,...,X_n]$ and that $ f(g(X_2,...,X_n),X_2,...,X_n)\equal{}0$.

this reminds me of:
http://www.uni-bonn.de/~habecker/dokumente/Henselsches_Lemma.pdf


You have made a rectangular stained glass window that is 2ft by 4ft. you have 7spuare feet of clear glass to create a border of uniform width around the window. what should be the width of the border if all of the clear glass is to be used?

(how do you set this up algebraically)

If the width of the border is $ w$, what would be the total width of the bordered window? And its height? How then will you express the area of the border itself?

see, i thought that if the height was 2, and the length was 4, then the area of the new rectangle made by the adding the border would be

$ (2x\plus{}4)(2x\plus{}2)\equal{}7\plus{}8$ where $ x$ is the width of the added border, but solving that out, your answers are not integers....

so im pretty positive that that's not the right equation

any ideas?

sorry, i didnt mean integers, i meant real numbers :oops:

(b/c the answer choices are fractions anyways  :P )

Do show your work on solving the equation, on my side it comes out very nicely.

ohh, i see.....when i was using the quadratic equation, i had an error in multiplication... :oops:

my final answer is $ \frac{1}{2}$   :D

is that correct?


geometry
incenter
geometric transformation
reflection
symmetry
homothety

[color=darkblue]Denote in a triangle $ABC$ : the incenter $I$, the middlepoint $A'$ of the side $[BC]$, the reflection $J_{a}$ of the point $I$ w.r.t. the point $A'$ and
the intersection $S_{a}\in BC\cap AJ_{a}$. Analogously define the points $S_{b}$, $S_{c}$. Prove that $\frac{S_{a}B}{S_{a}C}=\frac{a+b}{a+c}$ a.s.o. and the lines $AS_{a}$, $BS_{b}$, $CS_{c}$ are concurrently.[/color]

2nd part

$\triangle IB'A' \sim \triangle IS_{b}S_{a}$ by SAS~ ($IB'=B'S_{b}=IA'=A'S_{a}=r$) so $B'A' // S_{b}S_{a}$

$\triangle CB'A' \sim \triangle CAB$ by SAS~ ($2CB'=AC$, $2CA'=CB$) so $B'A' //AB$

then $S_{a}S_{b}// AB$

by symmetry other sides are parallel too, so $\triangle ABC$ is homothetic wrt $\triangle S_{a}S_{b}S_{c}$, then $AS_{a}$, $BS_{b}$, and $CS_{c}$ are concurrent at the center of homothety

[quote="Altheman"]2nd part

$\triangle IB'A' \sim \triangle IS_{b}S_{a}$ by SAS~ ($IB'=B'S_{b}=IA'=A'S_{a}=r$) so $B'A' // S_{b}S_{a}$[/quote]I think is $J_{a}$ and no $S_{a}$ a.s.o. The second point is trivially. The first is important in this problem.

[color=darkblue][b]Remark.[/b] Let $P$ be a inner point for the triangle $ABC$. Denote the intersections $D\in BC\cap AP$, $E\in CA\cap BP$, $F\in AB\cap CP$, the reflection $R$ of the point $P$ w.r.t. the middlepoint of the side $[BC]$, the intersection $S\in BC\cap AR$. Then there is the relation  $\boxed{\ \frac{SB}{SC}=\frac{c}{b}\cdot\frac{DC}{DB}\cdot \frac{AE}{AF}\ }\ .$ In the particular case $P: =I$ obtain $\frac{SB}{SC}=\frac{c}{b}\cdot \frac{b}{c}\cdot\frac{\frac{bc}{a+c}}{\frac{bc}{a+b}}=\frac{a+b}{a+c}\ .$[/color]


function
induction
vector
complex numbers
algebra open
algebra

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)\minus{}f(x)f(y)$ can be written as $ (1\minus{}f(x))(f(y)\minus{}1)\equal{}f(x\plus{}1)\minus{}1$. Defining a new function $ g(x)\equal{}f(x)\minus{}1$ we get $ \minus{}g(x)g(y)\equal{}g(x\plus{}y)$ . Inserting $ x\equal{}y\equal{}0$ in the latter we get $ g^{2}(0)\equal{}\minus{}g(0)$. Hence $ g(0)\equal{}0$ or $ g(0)\equal{}\minus{}1$.

If $ g(0)\equal{}0$, inserting $ y\equal{}0$, we get $ g(x)\equal{}0$ for all $ x\in R$.

If $ g(0)\equal{}\minus{}1$, inserting $ x\equal{}y$, we get $ g(2x)\equal{}\minus{}g^{2}(x)$. With induction we can prove that $ g(kx)\equal{}(\minus{}1)^{k\minus{}1}f^{k}(x)$ for all $ k\in N$. Denoting $ \frac{m}{n}\equal{}a$ or $ m\equal{}na$, we get $ g(m)\equal{}g(na)\equal{}(\minus{}1)^{m\minus{}1}g^{m}(1)\equal{}(\minus{}1)^{n\minus{}1}g(a)^{n}$. From the last we get $ g(a)\equal{}(\minus{}1)^{a\minus{}1}g(1)^{a}$ for all $ a\in Q^{\plus{}}$.

Here the problem appears, since $ (\minus{}1)^{a\minus{}1}$ is not real for all $ a\in Q^{\plus{}}$. $ (\minus{}1)^{a\minus{}1}\equal{}(cos(a\minus{}1)\pi\plus{}isin(a\minus{}1)\pi)$. Hence $ (\minus{}1)^{a\minus{}1}$ will be real when $ sin((a\minus{}1)\pi)\equal{}0$. So, $ (a\minus{}1)\pi\equal{}\frac{k\pi}{2}$ for $ k\in Z$. So, $ a\equal{}\frac{k\plus{}2}{2}$.

My question is, whether we can say, that $ g(x)$ and consequently $ f(x)$ maps from reals to complex numbers or must we limit it to $ x\equal{}\frac{k\plus{}2}{2}$? If we do not have to limit it, how do we extend the result to $ R^{\plus{}}$?

[quote="tmrfea"]$ f(x \plus{} y) \equal{} f(x) \plus{} f(y) \minus{} f(x)f(y)$ can be written as $ (1 \minus{} f(x))(f(y) \minus{} 1) \equal{} f(x \plus{} 1) \minus{} 1$. Defining a new function $ g(x) \equal{} f(x) \minus{} 1$ we get $ \minus{} g(x)g(y) \equal{} g(x \plus{} y)$ .... ?[/quote]

Take $ h(x)\equal{}1\minus{}f(x)$ and you have the well known equation $ h(x\plus{}y)\equal{}h(x)h(y)$

Continuous solutions are $ h(x)\equal{}0$ $ \forall x$ and $ h(x)\equal{}a^x$ for any $ a>0$ and so continuous solutions of your equation are :
$ f(x)\equal{}1$ $ \forall x$
$ f(x)\equal{}1\minus{}a^x$ $ forall a>0$

Many non continuous solutions exist too (with Axiom of Choice).

[quote="tmrfea"][quote="pco"]
Many non continuous solutions exist too (with Axiom of Choice).[/quote]

Can you explain this conclusion?[/quote]

We have $ h(x\plus{}y)\equal{}h(x)h(y)$ and so $ h(x)\equal{}h(\frac{x}{2})^2\geq 0$ and so, either $ h(x)\equal{}0$ $ \forall x$, either $ h(x)>0$ $ \forall x$ and we can define $ k(x)\equal{}\ln(h(x))$.

Then we have $ k(x\plus{}y)\equal{}k(x)\plus{}k(y)$ which is the well known additive Cauchy equation.

Consider then with Axiom of Choice, that exists a base of $ \mathbb{R}$ considered as a $ \mathbb{Q}$-vector space.

Each real $ x$ can be written as $ x\equal{}\sum_b x_bb$ where $ \{b\}$ is the vector-space base and $ x_b\in\mathbb{Q}$ and the sum is finite.

Then let $ \{y_b\}$ a set of rational numbers.

$ k(x)\equal{}\sum_bx_by_bb$ is solution of  $ k(x\plus{}y)\equal{}k(x)\plus{}k(y)$.

and $ f(x)\equal{}1\minus{}e^{k(x)}$ is a solution of the original equation. This solution is non-continuous if $ y_b$ are not all equal.

There is no need for $ \{y_b\}$ to be rationals.
And it's enough to have an upper bound of $ h(x)$ on some positive interval.
(or a positive lower bound)


What is the measure x of the smallest positive angle in radians, such that sin(x radians) = sin(x degrees)?
taken from USMC 10-12 which means NO CALC, AND express your answer in the simplest fraction with pi.

[hide]we know that sin(x) = sin( :pi:  - x). So sin(x radians) = sin(x degrees), after converting to radians, means that sin( :pi:  - x) = sin( :pi: x/180)  or  :pi:  - x =  :pi: x/180.

Solving, x =  :pi: /(1+ :pi:/180 )

However, converting to degrees, the statement can also mean that sin(180  -  180 x/ :pi: ) = sin x, which yields the solution x =  180/(1+180/:pi:).

They are the same exact solutions.

Thus, x = :pi:/(1+:pi:/180)[/hide][/hide]


i would be very gratefull if someone would show me how to proove cauchy-schwarz inequality, and AM>=GM

I think you should look at Theorem box to get it yourself.


Hey, my school has an awesome computer science department, but I missed out on it kinda...  What are some good books to learn Java from?  This is assuming complete beginner.

Thanks a lot!

Edit: Stuff in correlation with the AP Computer Science exam would be helpful.

Java Programming from the Beginning, by K.N. King is a really good book, and covers the AP cirruculum, but just the A part.  I don't really know any book that would cover the B part though.

For the AP test, I definitely recommend using the Barron's study guide.  It was very thorough - my teacher did not do a very good job of covering everything (it was his first year teaching, plus the other students weren't exactly phenomonal), and reading it and working through all the problems basically got me a 5 on the AP A level test.


geometry
inradius
circumcircle
trigonometry
congruent triangles

In case that I need to identify the source, it's this:

page. 172 of [i]Art of Problem Solving [/i] book one. chapter 13. polygons.

Anyway, I read this part of the book and I don't get what this means.

Please help:

Here it goes:

Just as with triangles, the perpendicular bisectors of the sides of a regular polygon all pass through a single point. (Can you use triangle congruence to prove this fact?) Furthermore, the angle bisectors of the interior angles also meet at this point.  Thus, we can construct both a circle when passes through the vertices of the polygon and a circle which is tangent to all the sides of the polygon.  We have drawn these circles below with radii R and r, respectively.

As shown in the figure (I won't put the figure in because I can't do it so I guess it means you have to look it up on the book), these two circles are concentric.  If the polygon has side length l and n sides, then we can determine the inradius (sometimes called the apothem(, r, the circumradius, R, and the area of the polygon by breaking the polygon into right triangles like triangle AXO at right. Remember that the radius of the circle is perpendicular to a tangent at the point of tangency, so angle     OXA = 90 degrees.  In this triangle, the hypotenuse is R, while the legs have length r and l/2. If we can determine  :theta: , we can find R and r using our basic trigonometric relations.  If we draw a radius like OX to all the sides of the polygon, we will form 2n congruent triangles like OXA.  All the angles at O together make up 360 degrees, so :theta: = 360 degrees/2n = 180 degrees/n.

Thus, we find r and R from the following trigonometric relations:

tan :theta: = tan (180 degrees/n) = $\frac{1/2}{r}$

sin :theta: = sin (180 degrees/n) = $\frac {1/2}{R}$

The area of the polygon is just 2n times the area of the OXA, or

Area = 2n[(l/2)(r)/2] = nlr/2.

--------------------------------------

Hoo... That was a lot of writing... maybe typing...

I have trouble on understanding what this is talking about. Please read through it and give me thorough explanation.

Thanks.

I don't understand either. Someone please give a reply.

Those 1's should be $l$'s, where $l$ is the length of a side.

Take a look at right triangle AXO.  $\sin \theta = \frac{AX}{AO}$.
Since $AX = l/2$ and $AO = R$, we have $\sin\theta = \frac{l/2}{R}$.

Similarly we can find $\tan\theta$.

Is this what you were asking about, or were you asking about why there's a circumcircle and an incircle?  Note that we can break up the regular polygon into a bunch of triangles like AOX which are all congruent.


calculus
integration
calculus computations

$ \int{\frac{e^{2x}\plus{}1}{e^{2x}\minus{}1}}dx$
I know you can change it to
$ \int{1\plus{}\frac{2}{e^{2x}\minus{}1}}dx$
But I don't know where to go from there.


Additive Number Theory
number theory
remainder
Divisibility
IMO
IMO 1977

Let $a,b$ be two natural numbers. When we divide $a^2+b^2$ by $a+b$, we the the remainder $r$ and the quotient $q.$ Determine all pairs $(a, b)$ for which $q^2 + r = 1977.$

[quote="orl"]Let $ a,b$ be two natural numbers. When we divide $ a^2 \plus{} b^2$ by $ a \plus{} b$, we the the remainder $ r$ and the quotient $ q.$ Determine all pairs $ (a, b)$ for which $ q^2 \plus{} r \equal{} 1977.$[/quote]

We note $ r < q$

Finding possible values of $ q$. Assume $ r \equal{} 0$.

Hence $ q_0 \equal{} \sqrt{1977}>\sqrt{1936}\equal{}44$

Therefore, the only value of $ q$ must be $ \leq 44$, since $ 1977\minus{}43^2>43$, $ q\equal{}44$

Therefore, $ r\equal{}41$

Therefore, we solve $ 44(a\plus{}b)\plus{}41\equal{}a^2\plus{}b^2$ to find $ (a,b)\equal{}(50, 7), (7,50), (50, 37), (37, 50)$ (I had to case bass that last bit, anyone got a better proof?)

why is $ r < q$ i was under the impression that the task states $ a^2 \plus{} b^2 \equal{} r\mod(a \plus{} b)$ not $ a^2 \plus{} b^2 \equal{} r\mod q$ and therefore only $ r < a \plus{} b$ not $ r < q$
you however got the right result :)

obivouslly $ q<45$
$ q^2\minus{}q(a\plus{}b)\plus{}a^2\plus{}b^2\minus{}1977\equal{}0$ therefore $ 7908\minus{}3a^2\minus{}3b^2\plus{}2ab$ must be a perfect square. so $ 3a^2\plus{}3b^2\minus{}2ab\equal{}3(a\plus{}b)^2 \minus{}8ab\leq7908$
By AM-GM we get $ 3(a\plus{}b)^2\minus{}8ab\geq3(a\plus{}b)^2\minus{}8(\frac{a\plus{}b}{2})^2\equal{}(a\plus{}b)^2>7908$ except if $ a\plus{}b\leq88$
and since $ r<a\plus{}b<89$ and $ 1977\minus{}43^2\equal{}128$ we can conclude $ q\equal{}44$ $ r\equal{}41$

[quote="Jure the frEEEk"]why is $ r < q$ i was under the impression that the task states $ a^2 \plus{} b^2 \equal{} r\mod(a \plus{} b)$ not $ a^2 \plus{} b^2 \equal{} r\mod q$ and therefore only $ r < a \plus{} b$ not $ r < q$[/quote]

Yep, thats a screw up on my part

[i]Solution.[/i] Using $r=1977-q^2$, we have $a^2+b^2=(a+b)q+1977-q^2$, or $q^2-(a+b)q+a^2+b^2-1977=0$, which implies $\Delta=7908+2ab-2(a^2+b^2)\ge 0$. If we now assume Wlog that $a\ge b$, it follows $a+b\le 88$. If $q\le 43$, then $r=1977-q^2\ge 128$, contradicting $r<a+b\le 88$. But $q\le 44$ from $q^2+r=1977$, thus $q=44$. It follows $r=41$, and we get $a^2+b^2=44(a+b)+41\Leftrightarrow (a-22)^2+(b-22)^2=1009\in \mathbb{P}$. By Jacobi's two squares theorem, we infer that $15^2+28^2=1009$ is the only representation of $1009$ as a sum of squares. This forces $\boxed{(a,b)=(37,50) , (7, 50)}$, and permutations. $\blacksquare$

Arkan


Fie O centrul cercului de raza R circumscris triunghiului ascutitunghic ABC. Cercul care trece prin O, B, C il notam Ca. In mod analog definim cercurile Cb si Cc. Fie OA intersectat cu Ca {M}, OB cu Cb {N} Oc cu Cc ={P}, OA int. cu BC={D}, OB int. cu AC = {E}, OC int. cu AB = {F}.

    Sa se arate ca:
a). OD*OE*OF*OM*ON*OP= R^6.
b). OD*OE*OF <= 1/8 * R^3.
c). OM*ON*OP >= 8 * R^3.

[hide="solutie a)"]
punctul $D$ se afla pe axa radicala a cercurilor $\mathcal{C}$ (cercul circumscris) si $\mathcal{C}_{A}$. deci $\rho_{\mathcal{C}}(D)=\rho_{\mathcal{C}_{A}}(D)$, adica $R^{2}-DO^{2}=DO\cdot DM$, sau $R^{2}=DO(DO+DM)$, adica $R^{2}=OD\cdot OM$. din celelalte doua relatii analoage rezulta concluzia. [/hide]
[hide="solutie b)"]
notez $OD=x$, etc.
deoarece $\frac{OD}{AD}=\frac{S_{OBC}}{S_{ABC}}$, rezulta ca : $\sum\frac{x}{R+x}=1$. (*)
presupun prin absurd ca $8xyz>R^{3}$. (**)
(*) este echivalent cu $\sum (xyz+R^{2}x+Rxy+Rxz)=(R+x)(R+y)(R+z)$, sau
$3xyz+R^{2}\sum x+2R\sum xy=R^{3}+R^{2}\sum x+R\sum xy+xyz$,
sau $2xyz+R\sum xy=R^{3}$.
din inegalitatea mediilor obtin $R^{3}=2xyz+R\sum xy\geq 2xyz+3R\sqrt[3]{(xyz)^{2}}$. (***)
fie $t=xyz$. atunci conform (**) :
$8t>R^{3}$
inlocuind in (***) obtin :
$4R^{3}\geq 4(2t+3Rt^{2/3})=8t+3R\cdot (8t)^{2/3}>R^{3}+3R\cdot R^{2}=4R^{3}$. contradictie
deci (**) este falsa, adica $8xyz\leq R^{3}$. [/hide]
[hide="solutie c)"]c) este evident din a) si b)[/hide]


If $ q$, $ r$, and $ s$ each represent a different integer 0-9, what is the value of $ s$?
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 \&$rrrr$ \\
+ \&$ssss$ \\
\hline
 \&$rqqqs$ \end{tabular}$

[quote="GameBot"]If $ q$, $ r$, and $ s$ each represent a different integer 0-9, what is the value of $ s$?
$ \begin{tabular}{l r} &qqqq\\
&rrrr\\
\plus{} &ssss\\
\hline &rqqqs\end{tabular}$[/quote]

Fixed.

r=1 or 2. Therefore q=9 pr 8. Also because q+r+s>10 we carry 1 and get q+r+s+1=q+10 so r+s=9

Since with this we carry 1 r=1 so s=$ 8$