LaTeX
combinatorics unsolved
combinatorics

can anyone show me an algebraic proof for:

(2n)Cn=0C0*2^n+2C1*2^(n-2)+...+(2k)Ck*2^(n-2k)+... where RHS stops when 2k>n

I know it, iura!
But I want mathlinkers express their problems in more clear way.

Did I miss something or this formula does not hold for $n=3$ ?

Pierre.

it does not hold, but i found a combinatoric proof for the same formula, but with the factor nC2k included in every term...in that case one can even let the sum rum till infinity :D

Peter

proof: just write a 1 if we chose that element of the 2n elements, and a 0 if not. then write the 2n elements in two rows of n elements. let k' be the number of columns for which the two numbers(0,1) written down are the same. since in the n-k' other columns, we get equally many zeroes and ones, it is clear that out of the k' columns, in $\frac{k'}{2}=:k$ there must be two 1's and in the other k columns two 0's. k is uniquely determined and lies between 0 and $\frac{n}{2}$. now given any k, there are nC2k ways to choose the columns with equal numbers(0,1), 2kCk to choose the columns with two 1's and $2^{n-2k}$ ways to arbritarily choose in the other n-2k columns whether we have a 0 or a 1 in the upper row.

Peter

[Edit: Latex formula looked horrible]

I'm not sure that it is an algebric proof as desired by Bugzpodder...
In another hand, since its statement is false, you have the right to answer as you please :D

Pierre.

no, i can't give him an algebraic proof...it's always hard to give proofs for wrong statements. but i didn't bother yet with giving an algebraic proof for the right formula, those proofs are never nice.

Peter

Peter, can you state more explicitely what formula didi you prove?

i proved $(2nCn)=\sum^{|\frac{n}{2}|}_{k=0}(nC2k)(2kCk)2^{n-2k}$. here (aCb) is the binomial coefficient $\frac{a!}{b!(a-b)!}$.

Peter

[Edit: sum indices]

Consider an array of n numbers and count the number of assigning to them 0 or 1 in the following way:
Choose 2k positions. Choose k from the chosen and assign them 1, to the other k assign 0. The other n-2k assign how we wish.
Now if a configuration has m ones and n-m zeros, in such a way it was counted
sum{mCk*(n-m)Ck) times (check!)
By Vandermonde this is equal to nCm.
So we must have sum{nCm*nCm} which is equal to 2nCn by same Vandermonde

I read now your solution Peter.
A really nice one!

Using the directly consequen from Ukraina 1996:Let M be set of numbers of 2n digits with n digits 1 and n digits 2.Let N be set of numbers of n digits with digits 1,2,3,4 fulfilling :the number of digits 1 equals to the number of digits 2.Then |M|=|N|.

Proof:We will construct a bijectivity relationship..
1.N->M
We "square" one number A of N becomes AA.Replace each 3->1 and 4->2 we have a number satisfying the condition of M.
2.M->N :
We "divide" the number A belongs to M into A'(the leader n digits) and A''(the last n digits) and write them into 2 columns to "add" accoring  as rules:
1+1=1,2+2=2,1+2=3,2+1=4


absolute value
combinatorics proposed
combinatorics

A certain sequence of $100$ real numbers has the property that, for every subsequence of $8$ of these numbers, there exists a subsequence of $9$ of these numbers with the same average as the $8$.  Show that the entire sequence is constant

First we reduce the problem to integer sequences:

Let $H$ be a Hamel basis of $\mathbb R$ over $\mathbb Q$. Every real $a$ can then be written as $\sum_{h\in H}c(a,h)h$. If we replace our sequence $(a_{n})_{n}$ with $(c(a_{n},h))_{n}$ for an arbitrary $h\in H$, and then multiply this new sequence by a suitable intger, the problem is reduced to integer sequences, as desired (because if we prove that for every $h$ the sequence $(c(a_{n},h))_{n}$ is constant, then $(a_{n})_{n}$ is constant as well).

Let now $(a_{n})_{n}$ be a finite integer sequence (that $100$ isn't really important, and $8$ and $9$ aren't particularly important either :)). We know that for any $8$ members of the sequence, say $a_{i_{1}},\ldots,a_{i_{8}}$, we have $9$ members $a_{j_{1}},\ldots,a_{j_{9}}$ such that $9\sum_{s=1}^{8}a_{i_{s}}=8\sum_{t=1}^{9}a_{j_{t}}$. In particular, this means that the sum of any $8$ members of our sequence is even. This clearly implies that they are all equal modulo $2$. If they are all even divide them all by $2$ and repeat the procedure, and if they are odd subtract $1$ from all of them, divide what you get by $2$, and repeat the procedure. This will decrease the maximal absolute value either until they are all $0$ or until they are all $-1$. Either way, they must all have been equal to begin with.


conics
ellipse
geometry
angle bisector
geometry open

This is an interesting problem in my opinion. I tried to solve it but without considerable success. A colleague of mine showed it few years ago. He said that the problem may have a practical application - for example for the spherical mirrors on the roads. I like this problem because it is combination between beauty of the math and its practical use.

[b]It is given a circumference k with center O(xo, yo) and radius R, and a segment AB it is defined by the points - A(x1, y1) and B(x2,y2). Find the locus of all the points M lying on k for which the line OM is an angle bisector of the angle AMB.[/b]

equivalent a generalizated problem:

ABC a triangle, find the locus of point P such that $ \angle APB \equal{} \angle APC$.

see [url=http://www.jstor.org/pss/2972061]here[/url] for the solution.

Thank you for your very valuable idea. But I still cannot find the solution  :( Sometimes I'm wondering how you find such interesting problems.

for your problem that is more simple:

A know lemma: Let and ellipse with focii E and F and a point P on it. Then the tangent on P to the ellipse is the external angle bisector of $ \angle EPF$.

If the angle bisector of $ \angle AMB$ pass trought O then the external angle bisector is tangent to the circle k, then the ellipse with focii A,B and passing trought M is tangent to k. Cleary the possibly point are 2: the point M the minimized AM+MB and the point M that maximized AM+MB.

Please, someone to post a full solution. That problem seems to be very difficult. It is not solved from long time ago on many forums.

Isnt the locus just the circle around AOB an the bisector of AB? But i didn draw a nice icture yet, just some with free hand.

Please, don't underestimate the problem.


Intr-un paralelipiped ABCDA'B'C'D' se cunosc muchiile $AB=2\sqrt{3},BC=2\sqrt{6},AA'=3\sqrt{2}$.Punctele M,N sunt mijlocele lui B'C' si A'D'.
a)Determinati masura planelor (AMB) si (CDN)
b)Demonstrati ca (AA'M)si(CC'N) sunt paralele.
c)Determinati masura unghiului dreptelor AM si CN.

*-scuzati ca dau probleme asa usoare

a)Dreapta comuna e $MN$. Se aplica teorema acoperisului(lema de paralelism) si se observa ca e paralela cu $DC$ si $AB$ etc.
b)$A'M$ e paralela cu $C'M$ si de aici concluzia.
c)Se ia $P$ mijlocul lui $AD$ etc.

P.S.: Multumesc ca le-ai postat.

[quote="Virgil Nicula"]Uite ca am auzit-o si pe asta : teorema acoperisului ... Exista si teorema podelei !?[/quote]

Manual pentru clasa a opta, ed. Sigma, pag.133
Si va asigur ca nu sunt singurul care foloseste aceasta teorema :mad:  :mad:  :mad:

domnule Nicula, nu am auzit de teorema podelei, dar am auzit de teorema fierastraului, des utilizata in rezolvarea problemelor de paralelism. Acelasi manual, aceeasi pagina.

P.S.: Prefer sa stiu teoreme cu 'nume din folclor' decat multe cu nume extravagante, pe care sa nu le folosesc niciodata.

[color=darkblue]Dragul meu, am glumit. Si eu am auzit de "problema fluturelui", chiar sunt doua : unu si doi. Le gasesti pe M.L. Insa in acest caz nu cred ca "teorema acoperisului" este de circulatie planetara ca "problema fluturelui". Daca ar fi asa, atunci cred ca ... a fost o proasta traducere. "Fluturele" este unic in timp ce "acoperisul" este de mai multe feluri (tavan drept, acoperis conic sau romboidal etc) ... Nu fac altceva decat sa-mi exprim o parere personala. Imi amintesc ca prof. Dan Barbilian la "multimea vida" spunea "multimea desarta" ... Intr-adevar, remarcabil si din folclor !

[b]P.S.[/b] Poti sa-mi dai un exemplu de o teorema cu un nume extravagant, mai ales ca ai spus ca sunt multe de acest fel ?[/color]


calculus
function
ARML
Putnam
complex numbers
complex analysis

I started to go through the section on complex numbers in Zeitz's book and I noticed that he referred to the book [cite]Visual Complex Analysis[/cite] by Tristan Needham.  So I wanted to here some of your opinions on the book.

I looked it up on Amazon.com.  I will probably wait a little while to buy it if I decide to buy it.  It looks like it goes very deep into complex numbers.  How far would this book cover?  Like which college class would it go up to?

Thanks

I borrowed the book from the library - and it goes very very deep indeed. Although much of the material is rather interesting, you probably should have at least some grounding in calculus before attempting to read it. It probably will not be very useful to you as a preparation for math competitions.

The book would cover roughly a course in "Complex Analysis" (or possibly "Complex Functions of One Variable,") which is calculus of functions on the complex rather than the real numbers..  I have never ever seen anything that looked even vaguely like complex analysis on a high school math contest.  That said, it's a really beautiful and interesting subject.  While I have not read the book myself, it was recommended to me by Larry Zimmerman, a wonderful teacher who wrote the ARML contest for many years, so I imagine it's pretty good.

Is there ever any complex analysis on Putnam?

I'm pretty sure that there isn't; although knowing complex numbers has certainly helped at times (IIRC, 2003 B4 or B5 was most easily solved using complex numbers).

"Visual Complex Analysis" is IMHO one of the best books you can get on the subject of complex analysis; alas, that won't help you in most pupils' math contests, since they mostly avoid the subject... But I still recommend you to learn it. Complex Analysis is one of the most beautyful subjects in maths, and once you start, you'll get addicted to it!

I don't know what you learn in which college class, but I think you won't see most of the contents of this book in any college class. So don't read it because you have to; read it because you want to understand complex numbers!

VCA is a fine book. Its aim is a bit different from most complex analysis books in that its main goal is for the reader to develop intuition about complex analysis rather than to be very rigorous. Therefore, if you want to learn complex analysis really well, VCA should be used in conjunction with another book, such as Ahlfors. But it will probably be much more interesting for you to read VCA than a standard complex analysis textbook.


MIT college AMC USA(J)MO USAMO AIME Support

I just got my letter today saying I was accepted for RSI 2007, so now I'm freaking out. I don't know what to do! I applied thinking I wasn't going to get in, so I just briefly looked up some math topics and BS'ed some stuff in my application. Now that I've actually gotten in, I don't know what I should do, because I'm not really good enough to get in!
Are we supposed to already have our own research all outlined and planned, or do we wait for the teachers there to give us directions? Is everyone there really, really smart and accomplished?
At the end, when everyone has to give their oral and written presentations, is it in front of all the students and the teachers?
Thanks!

1. Just wait until you get contacted by the people in charge, most likely Dr. Jenny Sendova. She'll set you up with your mentor and stuff. Basically, you get matched based on your application and the people (professors, grad students, professionals) who want to be mentors.

2. Most people are very smart and accomplished; that's how they get selected.

3. The oral presentations are given in front of about half of the students (two rooms go on at once, at least last year they did) and the counselors and probably some of their mentors. Your own mentor should be there as well.

Congratulations on getting in. I wouldn't worry too much about it. They'll take care of things :).

Exactly how rigorous is RSI? I can guess that because it's so prestigious that the students are expected to work really hard, but how far is the academic level above that of the average high school math student? I've looked on some of the example papers, and I am almost 100% sure I could never write something like that. If I get there and don't know anything, my mentor and the staff are going to hate me!

Haha, if you got in, I wouldn't worry about it. It's rigorous because you are basically concentrating your studies to a single subject for 6+ weeks. I'm sure that if you spent 6 weeks working on one thing with some guidance you could make a lot more progress than you think. For math in particular, it definitely depends on the project, but the projects I saw were clearly far above high school level, but if you're not expecting that I don't know why you want to go to RSI. I'm sure that if a problem proves too difficult, which happens a lot in math I would think, you can always change problems, prove special cases, find generalizations, etc.

[the projects I saw were clearly far above high school level, but if you're not expecting that I don't know why you want to go to RSI]
Let me explain my rationale: I really wanted to go to a math program, not only because I love math, but because I don't really have much to put on my resume in terms of productivity during the summer. But the problem was that every camp I was interested in was really, really expensive for me, so I kind of made halfhearted attempts and decided to focus on school year stuff. Then I found the MIT thing (it had been under the couch for who knows how long), and it was free and I like MIT, so I felt like if I didn't apply that would be really bad. So I applied partly to appease my conscience, knowing I wouldn't get in, and because my mom and my brother suggested I do it (my brother told me, after I got in, that he had reconsidered that suggestion but didn't tell me).
I've never been to ISEF or USAMO, my AIME score is pathetic, and I've only taken one college math course, and I think I only got a B. I'm perfectly willing to work my butt off, especially if it's for math, but I'm much more worried that I'm not going to be able to mentally keep up with everyone else, especially my mentor.
Are you an RSI alumnus? If yes, how was your experience there?

Yes, I am an alumnus; I went there last summer. The application process is extremely rigorous so basically anyone who gets in is fully capable of handling the work. The mentors do not expect the students to be on par with, say, graduate level students, though I'm sure some of them are. If your only concern is not being able to keep up, I would definitely say that it is not a sufficient reason for you to not go. There really isn't anything really demanding because you are doing your own research, so nobody can tell you what to do. It is a great opportunity to escape the pressures of college preparatory life, if anything.

RSI was one of the best experiences of my life by far. Not only do you get to work with a mentor who is an expert in whatever field you're doing, but you get to meet and perhaps even work with some of the best and brightest high school students this country AND other countries have to offer. It's difficult to emphasize how unique and valuable the whole experience is, but I doubt that there is a better summer program out there.

Of course it depends on what you want to accomplish as well. If your goal is to learn tons of math, going to a math camp would probably serve you better. If you want to have an open-ended problem to ponder for 6 weeks, RSI is the way to go. Several of the math people at RSI never qualified for the USAMO, but this by no means meant they were not perfectly capable of doing the research. You will be humbled by the awe-inspiring intelligence of many of the RSI students, but the best thing about it is that you'll fit right in (that's their hope, anyway). I definitely recommend that you attend RSI this summer; get to know some of the AoPSers who are also going, since that's always fun. I know for sure that t0rajir0u will be attending, and I bet several others are as well.

P.S. Sorry if my writing sounds incoherent or repetetive or anything, I never was the greatest of writers :P.

Thank you so much for the support, I'm sure I must sound so high-strung and immature.
How much off-time is there for extra studying (off time,  :oops: ) or reading? During the weeks that we're doing independent research, do we also attend classes? How much time do you spend with your mentor?
Thank you again!

No problem, I was pretty nervous myself before attending RSI. If you want to spend your off-time studying/reading, you will probably have plenty of opportunities for that (a good couple of hours per day if you don't sleep too much). You attend classes in the first week, but those are more or less pointless though some can be amusing. As for your mentor, it's really up to who your mentor is. My mentor, a EE professor, was very busy so I only met with them 2 or 3 times a week. A lot of people met with their mentors every day, while some very rarely did.

Are the mentors MIT professors, or TAs, or outside the university? Do they mentor several students at once, so you have work partners?

The mentors are a wide range of people, from MIT professors, to graduate students, to people working at companies like Verizon. As far as I know, most mentors only had a single student, though there were often students working in the same lab and occasionally students working with the same mentor on related but different projects.


Let $ ABCD$ be a convex quadrilateral.Let $ H$ be orthocenter of triangle $ ABC$.The line pass $ H$ paralell to $ DC,DA$ intesect $ BA,BC$ at $ M,N$ respectively.Prove that the line pass $ H$ perpendicular to $ MN$ pass the orthocenter of triangle $ \Delta{BDC}$. :D

I could guess some parts of the text, but what is the orthogonal of the triangle ?

Are you sure about this problem??? This is my asymptote sketch. :S


Prove:

If a and b are distinct primes, then for all integers n,

$ ab \mid (n^{ab} \minus{} n^a \minus{} n^b \plus{}n)$

can you explain a little more clear please

I try to explain
it's enough to show that
$ \frac {n^{ab} \minus{} n^{a} \minus{} n^{b} \plus{} n}{a} \in \mathbb{Z}$ (similary for $ b$)
this is true since :
\[ \frac {n^{ab} \minus{} n^{a} \minus{} n^{b} \plus{} n}{a} \equal{} \frac{n^{ab} \minus{} n^{b} }{a} \minus{} \frac {n^{a} \minus{} n}{a} \equal{} \frac {(n^{a} \minus{} n)(n^{a(b \minus{} 1)} \plus{} ... \plus{} n^{b \minus{} 1} )}{a} \minus{} \frac {n^{a} \minus{} n}{a}
\]
and by Fermat theorem
$ (n^{a} \minus{} n)/a \in \mathbb{Z}$


[quote="lowest_reynolds"]Hi,

I'm wondering what books would have problems strictly related in difficulty and style to AIME Geometry Problems.  I'm thinking between Geometry Revisited, and Challenging Problems in Geometry.  Any preferences or suggestions?[/quote]

Not so much in style, but more in depth of thought, I would recommend Challenging Problems in Geometry for this purpose.


Searching "Proizvolov" gives a nonempty set of results, but "*volov*" gives none. Why¿

Now, searching "hi" or "1" gives no results at all.

Forgetting the right star doesn't help anything  :wink:

I'm pretty sure it's that a star at the beginning is completely ignored. *tegrate came up with exactly one hit, in which "in tegrate" appeared for some reason.


Find, and give a proof of your answer, all positive integers n such that neither
n nor n^2 contain a 1 when written in base 3.

[hide="Solution"]
Consider the last digit of $ n$ in base 3. If it is a $ 1$, obviously it's not an answer. If it is a $ 0$, then $ 3\mid n$, and $ n$ is an answer only if $ n/3$ is an answer. So we only have to consider the case where $ n$ ends in $ 2$. Then $ n^{2}$ ends in a $ 1$. So no positive $ n$ works.
[/hide]

Strange solution set, though.

How about if neither $ n$ nor $ n^{2}$ contain a 2 in base 3?

[quote="MellowMelon"]How about if neither $ n$ nor $ n^{2}$ contain a 2 in base 3?[/quote]

Clearly, to get a new solution one can add an arbitrary number of zeros to the right of a solution and also delete any tail consisting of zeros (speaking in base 3).
So suppose the (base 3) number ends with a $ 1$. Then it's either $ 1$ (which is a solution) or of the form $ a100...01$, where a is a (possibly empty) string of 0's and 1's and we have $ k$ zeros inbetween the last two $ 1$'s. Now the latter number has value
$ 3^{k\plus{}2}a\plus{}3^{k\plus{}1}\plus{}1.$
Its square is:
$ 3^{2k\plus{}4}a^{2}\plus{}3^{2k\plus{}2}\plus{}1\plus{}2\cdot3^{2k\plus{}3}a\plus{}2\cdot3^{k\plus{}2}a\plus{}2\cdot3^{k\plus{}1}$
Whatever the value of k is, its base 3 representation contains always a $ 2$ due to the term $ 2\cdot3^{k\plus{}1}$.
So the only positive solutions are the nonnegative powers of 3.


find the area of the rectange with vertices 1,3 3,-4 and 1,-4

Did you try graphing it? If you do, it should be more obvious to find the fourth point. Then, you can find the area.

i tried don't get these type of probs...
get it know!!
thx a lot pascal!! :P

I'm not going to tell you the answer, but do the following:

1) Make a nice graph.
2) Plot the points you provided onto it.
3) Remember that the area of a rectangle is $ lw$?  You have the length and width available to you, just two of one.  Multiply them together and get the area.

Here is a quick graph.  See if you can solve it now.

[asy]draw((1,3)--(1,-4)--(3,-4));
dot((1,3));
dot((1,-4));
dot((3,-4));
label("$(1,3)$", (1,3), N);
label("$(1,-4)$", (1,-4), S);
label("$(3,-4)$", (3,-4), S);[/asy]

oh i see  :blush:
it's 2x6=12
thx a lot both of you!

Umm....not quite.  $ 7 \times 2 \equal{} \boxed{14}$

Uh yeah it should be 14... since it's, $ \minus{}4, \minus{}3, \minus{}2, \minus{}1, 0, 1, 2, 3$ which is 7 numbers, not 6... Perhaps you forgot to count zero?

uh you don't need to count them...

$ 4\minus{}(\minus{}3)\equal{}7$...

[quote="cgyao15"]oh yea mistyped.. :roll:[/quote]

are you sure? because you seemed to multiply wrong too ;)

Just graph it, use the distance formula, and you're done. Or, you could just count individually each unit square to find the length of the rectangle's sides.

-rts2007

sorry my test had a mistake :o it meant the area of a triangle with those vertices.  :roll: then it would make it 7.

[quote="saszs"]I definatly got 14.[/quote]

umm just saying next time dont make posts like this. make posts only if:

a) you are the first one to post, after which you should show your work for the person asking the question
b) you solved the problem using a different method

thank you

new problem of the day

2 people walked one mile.

personA's steps were 2 1/2 feet each
personB's steps were 2 1/3 feet each

how many more steps did personB take than person A.

ummm..
u already posted this in the MOEMS forum..
and this subject is about [b]one[/b] problem that I had. this is not a marathon :wink:

[quote="1=2"][[Shoelace Theorem]]

It's a cool theorem that you should TOTALLY look up.[/quote]

Yes, because we definitely have trouble finding the area of a rectangle...  :lol: pretty cool theorem, though


If $a,b,c \in R$ with
$b<c$
$b+c<a+1$
$a>1$

then prove that $b<a$

Since $b+c<a+1$ and $a>1$ then $b+c<2$. Since $b<c$ then $b< 1$. Since $a>1$ then $a>b$.

[quote="manuel"]Since $b+c<a+1$ and $a>1$ then $b+c<2$.[/quote]

Why? I don't think that this is true!

if b>a ,then b>1 then a+1>b+c>a+c,so c<1,but c>b>1,so it is wrong.
if b=a ,then b>1 then a+1>b+c=a+c,so c<1,but c>b>1,so it is wrong,too.
so b<a.

That's it.  :P

Another way is to add the first two ineqs and get $2b<a+1<2a$ because $a>1$
leading finally to $b<a$  :lol:

oops sorry for my stupid conclusion. !! :blush:


I have multiple tex files that I'd like to combine into one large pdf. I've used \include and \input in the past, but those commands required for the separate files to not have the preamble. Is there a command available to do the job even with the preamble in each of the smaller files?

For example:

inputfile1.tex and inputfile2.tex (preamble is present):
[code]\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{asymptote}

\begin{document}
content
\end{document}[/code]

rootfile.tex:
[code]\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{asymptote}
\usepackage{graphicx}
\usepackage[hang,small,bf]{caption}
\usepackage{multicol}
\usepackage[margin=1in]{geometry}

\begin{document}
\some_input_command{inputfile1}
\some_input_command{inputfile2}
\end{document}[/code]

I realize I could just exclude the preamble in the separate files and comment out \input if I ever need just a single file, but there are 27 files..

1. Write a program, in whatever language you are proficient in, to strip out the preamble etc in each of the 27 files. Since they are text documents that should be easy to do.
[i]or[/i]
2. Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=includex]includex[/url] or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=newclude]newclude[/url] packages which seem to do what you want. I had a quick try of includex's commands \includedoc, \includedocskip and it seemed to work.
[i]or[/i]
3. Compile all the files then use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=pdfpages]pdfpages[/url] package to include them in your final document. This is particularly useful if the files have different preambles or pagesizes or orientation.

The newclude package looks good. But I've been trying all morning to figure out how to work it. Reading the documentation, I think I'm supposed to put in arguments like \include[prehook]{filename}[posthook], but I don't know what the arguments are. The documentation talks about them "overriding the current values of \AtBeginInclude and \AtEndInclude," but what are they?

Using \includedoc like this:
[code]\documentclass[letterpaper,12pt]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage[margin=1in]{geometry}
\usepackage{times}
\usepackage{newclude}

\everymath{\displaystyle}

\begin{document}

\tableofcontents
\newpage

\includedoc{filename}

\end{document}[/code]
Says that \includedoc is an undefined control sequence. It said somewhere that includex was superseded by newclude - doesn't that mean \includedoc is also a part of newclude?

I've only tested includex; newclude does what includex does plus much more so is more complicated to use. I would try includex first and only go on to newclude if includex isn't good enough and you need more control.
So try
\usepackage{includex}
...
\includedoc{filename}
which should work - filename must not include the .tex extension

There's also \includedocskip. The difference is that \includedoc removes \documentclass, \usepackage and \begin{document} whereas \includedocskip removes the rest of the preamble as well.

Awesome - it works! :) Yes includex is good enough. Thanks for pointing out that package!

I tried includex, but it behaves strange: it does give errors about my declared newtheorems (saying they are already defined, even though he should not read that part!)
[code]\newtheorem{prp}{Problem}
\newtheorem{prkb}{B-K}
\newtheorem{rmk}{Comment}
\newtheorem{cas}{Case}
\newtheorem{cas1}{Case}
\newcommand{\proofend}{$\ \Box$}[/code]
even though they are before my \begin{document}. Any suggesitons what I should do?

Thanks!

Try the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=newclude]newclude package[/url] which is more complicated but has more customisable options.


logarithms
integration
calculus
calculus computations

Given $y_{1}= t$, find the other solution to $t^{2}y''+2ty'-2y = 0$ for $t>0$.

I am not too sure how to do this and attempting to learn on my own..  I let $y_{2}= At$ and eventually reduced the equation to
\[A''t^{2}+(t^{2}+2t)A' = 0\]
Then I let $Y = A'$ and reduced the equation to a first order linear:
$Y'+(1+\frac{2}{t})Y = 0$

I don't think it was legal to divide by $t^{2}$, but did not know how else to solve it.

Thanks!

My answer came out to
\[y = \frac{C}{e^{t+2\ln{t}}}\]

[quote="SnowStorm"]Given $y_{1}= t$, find the other solution to $t^{2}y''+2ty'-2y = 0$ for $t>0$.

I am not too sure how to do this and attempting to learn on my own..  I let $y_{2}= At$ and eventually reduced the equation to
$A''t^{2}+(t^{2}+2t)A' = 0$
.....
My answer came out to
$y = \frac{C}{e^{t+2\ln{t}}}$
[/quote]

You are wrong, the reduced equation is $A''t+4A'=0$ .. and so on ..
 :cool:

[quote="SnowStorm"]Given $y_{1}= t$, find the other solution to $t^{2}y''+2ty'-2y = 0$ for $t>0$.[/quote]

Use Liouville-Ostrogradky formula.

Other way: consider y=t^{\lambda}.

How do you deal with the $t$ that is attached to the $A''$ though?

First of all divide through by $t{^{2}}$ (we are allowed to do this) to get this equation:
$y''+\frac{2}{t}y'-\frac{2}{t^{2}}y = 0$
Then $P(t) = \frac{2}{t}$ and $Q(t) =-\frac{2}{t^{2}}$
Observe that the determinant is given by $D(t) = 2P'(t)+P(t){^{2}}-4Q(t)$. Solutions to an equation of the form $y''+P(t)y'+Q(t)y = 0$  are determined using the values you get after solving D, that is, call the value you will get after solving the equation D k, if k is zero then you have $u(t) = t$ so that $y_{2}= u(t)y_{1}= t^{2}$,
if k is greater than zero, then $u(t) = e^{Bt)}$ , where $B = \sqrt(k)$ ,
and finally if $k < 0$ , you have $u(t) = tan(At)$, where $A = \sqrt(\frac{k}{-4})$
Now you only have to find u(t), by solving D. And then use one of the three cases.

akech, I really don't know what you're doing there, but it doesn't lead to solutions of this differential equation. There will be no exponentials and no tangents.

It appears that SnowStorm probably attempted a method that would have worked except for a computational error. See Diogene's post.

Yes, the method I suggested works when the Discriminant (which I misktakenly called determinant in my previous post) is a constant. But the discriminant for the problem in question is a non - constant, it is, $D(t) = \frac{8}{t^{2}}$. But the original equation can be changed to make the the Discriminant equal to a constant by setting $x = ln(t)$ in the original equation. Once this is done, the method I suggested can be applied. I know there are a lot of standard ways than can handle this differential equation nicely, but it is fun to try something different (if it works).

[quote="Dvagner"][quote="V.V."]
Use Liouville-Ostrogradky formula.[/quote]
what's that?[/quote]

Consider Wronsky's determinant
$W(x)=\left|\begin{array}{cc}y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x)\end{array}\right|$.

Let $y_{1}$, $y_{2}$, ..., $y_{n}$~--- independent solutions of
$y^{(n)}+p_{1}(t)y^{(n-1)}+\ldots+p_{n-1}(t)y'+p_{n}(t)y=0$. Then
$W(x)=W(x_{0})e^{-\int_{t_{0}}^{t}p_{1}(s)ds}$. This is Liouville-Ostrogradky formula.

From Liouville-Ostrogradky formula for second order ODE
$y_{2}(t)=y_{1}(t)\int_{t_{0}}^{t}\frac{1}{y_{1}^{2}(s)}e^{\int_{t_{0}}^{s}p(\tau)d\tau}ds$.

In details in my book about ODE (in Russian; http://joker.botik.ru/~trushkov).


A body diagonal of a cube is a line through the centre joining oppsite vertices. Find the angles between the body diagonal of a cube.

answer: (not solution just answer)
[hide]71[/hide]

[hide]the diagonal of a cube is rt3

let that be the hypotenuse of the right rtrinagle.

the opposite side is 1

the adjacent side(diagonal of a square) is rt2

sin x=1/rt3

x=71[/hide]


For fixed [i]m [/i]and variable [i]k[/i], there are multipliers a and b such that f([i]m[/i]+[i]k[/i])=af([i]m[/i]+1)+bf([i]m[/i]), form a conjecture as to the nature of [i]a [/i]and [i]b [/i]in terms of [i]k[/i]. Prove this conjecture.

btw, i THINK that [i]a[/i]=f([i]k[/i]) and [i]b[/i]=f([i]k[/i]-1) but i cannot prove it

f(n) is the n th Fibonacci term where f(1)=f(2)=1


On a number line, the coordinate of point X is $ \minus{}18\frac{1}{2}$ and the coordinate of point Y is $ 21\frac{1}{2}$. What mixed number corresponds to the coordinate of point Z if the distance between X and Z is one-third of the distance between Z and Y ?

There are two possibilities:

Z is between XY. then XY is 40, so Z is -8 1/2

Z is not between XY. Z is -38 1/2

Ambiguous.

For this problem, we assume $Z$ is between $XY$.

Since the distance between $X$ and $Z$ is one-third of the distance between $Z$ and $Y$ and the length of $XY$ is $40$, we know that:

$\frac{1}{3}XY+XY=40$

$XY=30$

Coordinate $Z$ is $\frac{1}{3}XY$ away from $X$, hence $\boxed{-8 \frac{1}{2}}$


[quote="GameBot"]What is the sum of the fifth prime number, the sixth composite number, and the third perfect square?[/quote]

The fifth [url=http://www.google.com/search?q=prime+number&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]prime number[/url] is 11. The sixth [url=http://www.google.com/search?q=composite+number&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]composite number[/url] is 12. The third [url=http://www.google.com/search?hl=en&client=firefox-a&rls=org.mozilla%3Aen-US%3Aofficial&hs=iGX&q=perfect+square&btnG=Search]perfect square[/url] is 9. The sum is then $ \boxed{32}$.

what? i thought 0 is the first perfect square. in fact, i have never seen when 0 does not count as a perfect square

[quote="Wikipedia"]For a non-negative integer n, the nth square number is n^2, with 0^2 = 0 being the zeroth square[/quote]

[url]http://en.wikipedia.org/wiki/Square_number[/url]


function
linear algebra
linear algebra unsolved

Find all linear maps $f:M_n(K)->M_n(K)$  such that $detM=det f(M)$ for all $M)$.

I think there are other functions:
for example: $M \to AMB$ where $det(AB)=1$.
(and please, post proofs ...)

Moubinool, is it so difficult to post proofs of your statements? I mean, you know very well that solving a problem means more than guessing the answer. Anyway, I think Alekk knows the proof of this problem, since it follows easily from the first 2 parts of the written ENS exam in which the problem of linear maps preserving rank was discussed. But I wonder if there is easier solution.


Ok, so I'm curious.  Anyone else find it a bit weird how the Log1 takes into account, for the total school score, scores from every level of contest?  I mean, yes, I see that they want everybody, regardless of math class, to be competent with material from that class, but shouldn't they also encourage mathematical competence and skill throughout [i]all[/i] of mathematics (or at least mathematics accessible to a high-school student)?  I admit that I'm biased in that all of the top "mathletes" from my school are in Calculus or above (and therefore makes it impossible for us to do well in this contest, hehe), but I think that it should encourage more of an "Oh, I'm not in the top few scorers, I should study and learn more" - mentality than the idea that "I'm only required to know math from the course I'm taking right now, I need not 'expand my horizons' in pursuit of more mathematical knowledge."

Comments/ideas/opinions?

Sorry, I don't really have any opinions on your question, but I have a question for you!  :P

I swear I saw a Mu Alpha Theta forum on here before, but now I can't find it for my life... I am on the Mu Alpha Theta board of the chapter at my school, and we were thinking of some new things we could do next year so I went on the MAO website and found some information about this contest, but I'm still not really sure what it is. Could you give me some background information? How hard is it? how many students from a school can take it? How big of a hassle is registration/administration? Thanks!!


Also prove that this is also equal to 3cos²(36°) - sin²(36°).

EDIT: I have corrected the typo; sorry. (the cosine term is not squared)

----------------
also prove that sin 10◦ sin 50◦ sin 70◦ = cos 20◦ cos 40◦ cos 80◦

[hide]$ 36=x$ implies the left side is

\begin{align*}\frac{\sin 3x}{\sin x}&=\frac{\sin(2x+x)}{\sin x}\\
&=\frac{\sin 2x\cos x+\cos 2x\sin x}{\sin x}\\
&=\frac{2\sin x\cos^2x+\sin x\cos^2x-\sin^3x}{\sin x}\\
&=3\cos^2x-\sin^2x\\
&=3\cos^2 36^{\circ}-\sin^236^{\circ}\\
&=3\cos^2 36^{\circ}-\cos^236^{\circ}\\
&=2\cos^236^{\circ}\end{align*}[/hide]

[quote="nayel"][hide]$ 36 = x$ implies the left side is

\begin{align*}\frac {\sin 3x}{\sin x} & = \frac {\sin(2x + x)}{\sin x} \\
& = \frac {\sin 2x\cos x + \cos 2x\sin x}{\sin x} \\
& = \frac {2\sin x\cos^2x + \sin x\cos^2x - \sin^3x}{\sin x} \\
& = 3\cos^2x - \sin^2x \\
& = 3\cos^2 36^{\circ} - \sin^236^{\circ} \\
& = 3\cos^2 36^{\circ} - \cos^236^{\circ} \\
& = 2\cos^236^{\circ}\end{align*}[/hide][/quote]

How do you get $ \sin^236^\circ = \cos^236^\circ$, as you substituted on your second to last line?

In fact, I'm reasonably sure that $ \frac{\sin(108^\circ)}{\sin(36^\circ)}\neq 2\cos^2(36^\circ)$

[hide="Part 1"] $ \sin 108^{\circ} \equal{} \sin (180^{\circ} \minus{} 108^{\circ}) \equal{} \sin 72^{\circ} \equal{} 2 \sin 36^{\circ} \cos 36^{\circ}$ [/hide]

[hide="part 2"]$ \sin{x}\equal{}\cos{(90\minus{}x)}$.[/hide]

[hide="part 3"]We wish to prove that $ 3\cos^2{36}\minus{}\sin^2{36}\equal{}2\cos{36}$. $ 4\cos^2{36}\equal{}2\cos{36}\plus{}1$. $ \cos{36}\equal{}\dfrac{1\plus{}\sqrt{5}}{4}$, which is true.[/hide]

The second proof is a trivial application of the cofunction identity between sine and cosine. As for the first, we multiply both sides by $ \sin{36^{\circ}}$ and bank on the double angle identity that appears on the RHS. Thus, \[ sin{108^{\circ} = \sin{72^{\circ}}}\] Also, this is in the first/second quadrant so it suffices to show that 108-90=90-72 which is obviously true. Yes, the two are equal. edit: Geometrically, we have shown that there does indeed exist an isosceles triangle within the unit circle whose point lies on the origin with a unique angle of 108-72=36. This makes its other two angles 77 degrees each.

solutions only using double angle identity and cosx = sin(90-x)

[hide]for the first proof
cos(x) = sin(90 -x)
so sin(108) = sin 72
we now have to prove that:
sin 72 / sin 36 = 2cos36
 2sin36cos36 = sin 2(36) because sin2x = 2sinxcosx
so we have sin72 = sin72
LHS = RHS

for the second proof we again use cosx = sin(90-x)
so we have
cos20 x cos40 x cos80 = cos20 x cos40 x cos80
LHS = RHS
QED[/hide]

[quote="b-flat"][quote="nayel"][hide]$ 36 = x$ implies the left side is

\begin{align*}\frac {\sin 3x}{\sin x} & = \frac {\sin(2x + x)}{\sin x} \\
& = \frac {\sin 2x\cos x + \cos 2x\sin x}{\sin x} \\
& = \frac {2\sin x\cos^2x + \sin x\cos^2x - \sin^3x}{\sin x} \\
& = 3\cos^2x - \sin^2x \\
& = 3\cos^2 36^{\circ} - \sin^236^{\circ} \\
& = 3\cos^2 36^{\circ} - \cos^236^{\circ} \\
& = 2\cos^236^{\circ}\end{align*}[/hide][/quote]

How do you get $ \sin^236^\circ = \cos^236^\circ$, as you substituted on your second to last line?

In fact, I'm reasonably sure that $ \frac {\sin(108^\circ)}{\sin(36^\circ)}\neq 2\cos^2(36^\circ)$[/quote]

Sorry, you're right. I wasn't paying much attention..


tada voici quelque chose pour tous les gens qui sont extrêmement AWESOME à cause de parler français ou peut-être franglais

Oui, parlons Franglais.

Trigonometry1.01 you might as well ask Mystic Terminator for his phone number. Also, stop spamming.

DUDE WHAT

AND ALSO G&FF IS MEANT FOR SPAM

guys guys

THE HERALD OF DOOM IS NOT ARNaV

I MADE A BIG MISTAKE THANKS

I DON'T KNOW ANY LANGUAGE OF THE SET {THIS LANGUAGE} UNION {FRENCH}

Note: sorry, this is our computer which does not have accent capabilities :(

quoi de neuf, tlm?  ca fait longtemps que je ne viens pas ici :(.  C'est triste, n'est-ce pas?

Eh ben, qu'est-ce que vous allez faire pendant que vous ne devez pas aller a l'ecole?  Moi, je suis alle a New York pour jouer du violon a Carnegie Hall avec mon orchestra.  C'etait tres, tres amusant et j'y ai decouvert beaucoup de choses.  Par exemple, les taxis la-bas sont FOUS!!!!  Oh la la, en tout cas, la plupart du temps, on marchait et prenait le metro, mais une fois j'ai pris un taxi avec ma famille et le chauffeur nous conduisait trop vite, et y'avait beaucoup d'arretes aussi!  A Rockefeller Center, j'ai trouve un petit magasin qui vendait les livres francais, et j'y ai achete le livre "Harry Potter a l'ecole des Sorciers", malgre la prix de $20.  Quand je suis monte l'Empire State Building, il y avait presque 5 personnes qui parlaient francais devant moi, mais j'etait trop nerveux de leur parler (c'est toujours le probleme!).

Hmmm.... je vais egalement aller a Montreal en juillet avec ma famille et un copain.  Si j'ai de la chance, je pourrai pratiquer mon francais, mais si les gens sont comme la fois derniere, ils vont me dire "Vous savez, vous ne devez pas essayer de parler francais ici.  On parle anglais!"  Blegh, que c'est penible!

De toute facon, cet ete va etre vraiment amusant et je vais m'amuser beaucoup.  Je mangerai des crepes, prendrai le metro aux 2 villes, et parlerai a tous les gens que je verrai dans les rues (lol pas tout a fait, ce serait plutot dur, non?).

Wow that's kind of hard to read without accents.  Meh, whatever :)

lol what does that mean?

ou... je veux dire... qu'est-ce que ça signifie que cette phrase-là?


Euler
number theory unsolved
number theory

A very simple question . I have no idea how to solve

Find with proof . all the integers n with $phi$ $(n)=4$.( HINT: show that no prime greater than 5 can be a factor of n

NB: phi is the greek letter phi

Well, if $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}}...p_{k}^{\alpha_{k}}$, with $p_{i}$ prime, and $\alpha_{i}\geq{1}$ for each $i$, then
$\phi(n)=\phi(p_{1}^{\alpha_{1}})\phi(p_{2}^{\alpha_{2}})...\phi(p_{k}^{\alpha_{k}})$, but we have that $\phi(p_{i}^{\alpha_{i}})=p_{i}^{\alpha_{i}-1}(p_{i}-1)$, so for $p$ greater than 5, there isn't a solution for $\phi(n)=4$. So the prime divisors of n would be 2,3 or 5. If $n=2^{k}\Rightarrow{\phi(n)=4=2^{k-1}}\Rightarrow{n=2^3=8}$. If $n=3^{k}\Rightarrow{\phi(n)=4=3^{k-1}.2}\Rightarrow$ there isn't a solution. And if $n=2^{k}3^{j}\Rightarrow{\phi(n)=4=2^{k-1}3^{j-1}.2}\Rightarrow{n=12}$. If 5 divides $n$, $n=5$, or $n=10$.


algebra
function
domain
real analysis
real analysis theorems

The following "linear" properties hold when $ \mathcal{A}$ is a $ \sigma$-algebra, but do they hold then $ \mathcal{A}$ is merely an algebra?

if $ \mu, \nu$ are measures on $ \mathcal{A}$, $ t\in \mathbb{R}_{\plus{}}$ and $ A\in \mathcal{A}$, then the following are measures also:

1. $ \mu \plus{} \nu$ defined by $ (\mu\plus{}\nu)(E)\equal{}\mu(E) \plus{} \nu (E)$
2. $ t \nu$, defined by $ (t \mu)(E)\equal{}t\mu (E)$
3. $ \mu_A$, defined by $ \mu_A (E)\equal{}\mu (A \cap E)$.

My guess is [color=red]no[/color] and that these properties only hold when the domain is a $ \sigma$-algebra. Is that correct?

What do you mean? :? The measures are normally defined of $ \sigma$-algebras because you want complete freedom in uniting and intersecting countable families of sets but other than that the domains do not influence the properties of the measures: you can consider pre-measures (or whatever they are called) on arbitrary families of sets and demand that the countabe additivity holds whenever the union belongs to the family (not that it is very useful without any additional information but still it can be done and the theorem about the extension from a semi-ring starts exactly like that). Then the sum of such pre-measures is certainly a pre-measure, etc. It seems like you ask questions without understanding what exactly you are asking and I'm afraid other people have trouble understanding your questions as well :(.

Dear fedja,

yes, i am afraid that there is sometimes more mystery in my questions than i would like..i am just starting to get into measure theory...but i find recently that using more than one text will at least force myself to be at least a little clearer. Thank you for your answer--you give good advice, despite my not being clear at times! :0)

Los


For a particular peculiar pair of  dice,  the probabilities of rolling 1, 2, 3, 4, 5, and 6 on each
die are in the ratio $ 1: 2: 3: 4: 5: 6$.  What is the probability of rolling a total of 7 on
the two dice?

The solution says the answer is $ \frac{1*6\plus{}2*5\plus{}3*4\plus{}4*3\plus{}5*2\plus{}6*1}{21^2}$ I understand the numerator but wouldn't the denominator be $ 21^6$?

First, calculate each case separately and you will see why:

$ \frac{1}{21} \cdot {6}{21}\plus{}\frac{2}{21} \cdot \frac {5}{21} \plus{}\frac {3} {21} \cdot \frac{4}{21}\plus{}\frac{4}{21} \cdot \frac {3}{21} \plus{} \frac {5}{21} \cdot \frac{2}{21}\plus{}\frac{6}{21}\cdot \frac{1}{21}$

Which can be combined to make:

$ \frac{1 \cdot 6\plus{}2 \cdot 5\plus{}3 \cdot 4\plus{}4 \cdot 3\plus{}5 \cdot 2\plus{}6 \cdot 1}{21^{2}}$

OR

Since only two die are rolled, the exponent of $ 21$ must be $ 2$.

Wow I was going thru all of the AMC 10's and this morning I did the 2006 AMC 10 B, and look at that, this problem popped up on alcumus when I did it this morning.


probability
topology
Functional Analysis

Which courses whould you recommend me to take next year?

I have decided that I will take:Functional Analysis,Probability Theory,Topology and my big dillema is with Algebra. I have talked to my friend and they have told me that it would bad for me if I dont have it,no just because of the raw knowladge, but because of the way of thinking and methods used in Algebra. I have encountered a bit of Algebra in a course Combinatorial design, and I was blown away by the beauty of Algebra.
But I have a couple of problems with these courses; work load-if I take them all (including Algebra) I wll have about 32 houres of classes per week,so it might be a problem to master the all in my final year.
Next one is the fact that I am majoring in financial math, so I wont have most of the mentioned classes even I persue a phd degree( but one of the mathlinkers told me that taking these courses will help me get to a good university in case I want to get a phd )

Please help me make a wise decision, and please tell me about your experience!!

Thanks

More then a 100 views and not a single advice ? :(

I would take (abstract, modern) algebra before taking the other courses you mention.

functional analysis is more abstract than one might think. A good experience with abstract algebra will ease the pain

you should probably hold off the topology class for next year.

You can't take everything at once. Any schedule with more than three math courses on me strikes me as a likely overload. Even if you have no academic difficulties with a schedule with more math courses than that, you're likely to look back on it from a year or two later and discover that there's some course or two that you can't really remember.


linear algebra
matrix
real analysis
real analysis unsolved

We can use expression: $ \Gamma \left(n \plus{} \frac {1}{2}\right) \equal{} \sqrt {\pi}\cdot\frac {(2n \minus{} 1)!!}{2^n}$ and $ \Gamma (n \plus{} 1) \equal{} n!$. Now:
$ \sum_{n \equal{} 0}^\infty\left\{( \minus{} 1)^n\cdot\left(n \plus{} \dfrac{1}{4}\right)\cdot\left(\dfrac{\Gamma\left(n \plus{} \dfrac{1}{2}\right)}{\Gamma(n \plus{} 1)}\right)^3\right\} \equal{} \sum_{n \equal{} 0}^\infty\left\{( \minus{} 1)^n\cdot\left(n \plus{} \dfrac{1}{4}\right)\cdot\left(\dfrac{\sqrt {\pi}\cdot\frac {(2n \minus{} 1)!!}{2^n}}{n!}\right)^3\right\} \equal{}$
$ \equal{} \frac {\pi\sqrt {\pi}}{4}\cdot\sum_{n \equal{} 0}^\infty\left\{( \minus{} 1)^n\cdot(4n \plus{} 1)\cdot\left(\frac {(2n \minus{} 1)!!}{2^{n} n!}\right)^3\right\}$
And we should prove that: $ \sum_{n \equal{} 0}^\infty\left\{( \minus{} 1)^n\cdot (4n \plus{} 1)\cdot\left(\frac {(2n \minus{} 1)!!}{2^{n} n!}\right)^3\right\} \equal{} \frac {2}{\pi}$
To prove this, maybe, we sholud use hypergeometric series.

 :stretcher:

misan, ramanujan maybe was you relative  :rotfl:, you live just some miles to the north of his birthplace :D. In the internet I can find only results, not the proof of any his theorem :( :D.

[quote="Yuriy Solovyov"]To prove this, maybe, we sholud use hypergeometric series.
[/quote]
Yeah, this does work. We have to compute
\[ {}_{4}F_3\left(\left.\begin{matrix}\frac12 & \frac54 & \frac12 & \frac12 \\
&\frac14 & 1 & 1\end{matrix}\right| \minus{} 1\right).
\]
There's general formula
\[ {}_{4}F_3\left(\left.\begin{matrix}a &1\plus{} \frac a2 & d & e \\
& \frac a2 & 1 \plus{} a \minus{} d & 1 \plus{} a \minus{} e\end{matrix}\right| \minus{} 1\right) \equal{} \frac {\Gamma(1 \plus{} a \minus{} d)\Gamma(1 \plus{} a \minus{} e)}{\Gamma(1 \plus{} a)\Gamma(1 \plus{} a \minus{} d \minus{} e)}
\]
(cf. W.N.Bailey,Generalized Hypergeometric Series,$ \S4.4$,formula(3)).
Just put $ a \equal{} d \equal{} e \equal{} \frac12$.  :lol:


Was it intentional to miss ${\dbinom{6}{1}}^{2}$?
If so,
[hide]$\dbinom{12}{6}-{\dbinom{6}{1}}^{2}=888$[/hide]
If not,
[hide]$\dbinom{12}{6}=924$[/hide]

[quote="davidyko"]Was it intentional to miss ${\dbinom{6}{1}}^{2}$?[/quote]
If not, just subtract 36 at the end.  :)

Er...for the problem, as given, the solution is the first one in my post, the "if so" one.

thank you!  sorry i written wrong there will be C(6,1) too so answer is C(12,6)

[quote="rnwang2"][hide]
$\sum_{k=0}^{n}\binom{n}{k}^{2}=\binom{2n}{n}$[/hide][/quote]

Prove this.

[hide="Block walking"]
Basically, there are $\binom{n}{k}$ ways to get to the value $\binom{n}{k}$ on Pascal's Triangle.

And there are $\binom{n}{n-k}$ ways to get from $\binom{n}{k}$ to $\binom{2n}{n}$.

Thus there are a total of $\binom{n}{k}^{2}$ ways to get to $\binom{2n}{n}$.

This works for all $0\leq k\leq n$ (any value of k in the nth row of Pascal's Triangle).[/hide]


Let $a$, $b$, and $c$ be positive real numbers such that $abc=1$, prove that
\[\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}} \ge \sqrt{a}+\sqrt{b}+\sqrt{c}+3\]

$\frac {a}{2\sqrt{b}}+\frac {a}{2\sqrt{c}}\geq \sqrt{a}$

Sum this to get

$\frac {1}{2}LHS\geq \sum \sqrt{a}$

By AM-GMing each individual term, we also see that

$\frac {1}{2}LHS\geq \frac {6}{2}abc=3$

Sum and we're done.

Also see http://www.mathlinks.ro/Forum/viewtopic.php?t=22056 .

  darij


Mark earns 10 points for each correct answer on his math homework, but he loses 5 points for each incorrect answer. Mark earned a total of 10 points after doing 25 problems. How many problems did Mark answer correctly?

Assume he got everything wrong.  Then he gets -125 points.  So we need to add 135 points by adding in increments of 15 (for +10 and -5).  135/15=9 so the answer is $ \boxed{9}$.


Show that in any sequence of $ 2000$ integers each with absolute value not exceeding $ 1000$ such that the sequence has sum $ 1$, we can find a subsequence of one or more terms with zero sum.


induction
inequalities
geometry
algebra unsolved
graph theory

There are n circles with radii $r_1, r_2, ..., r_n$ connected together.
Show that it is possible to cover these circles with a cirlce of radius $r_1+r_2+..+r_n$.

what exactly do you mean by cnnected?(do you mean tangent, or....)

Assuming he means tangent, you can prove it by contradiction using the triangle inequality.  Assume that two points exist on the chain of smaller circles such that their distance is greater than the sum of the diameters.  The rest is left as an exercise for the reader :).

"Connected" means that any two circles among the whole set of circles are either tangent to each other or their intersection is non-empty.
Besides, the cirlces are not necessarily arranged in a chain.
This problem seems to be trivial but indeed it is not easy to solve.
For example, if we want to use induction to prove the statement, then we need to argue that after removing one circle, then the remaining circles are still connected. However, there is no clue to guarantee that the remaining circles are still connected.
Wish someone can help me.

Hmm.  Thinking constructively, describe the circles as $C(x_i,r_i)$, where the $x_i$'s are the respective centers and the $r_i$'s are the respective radii.  It seems like the point $x$ for which $\sum_i r_i\vert x-x_0\vert$ is minimal might work if we consider $C(x,r_1+r_2+\ldots +r_n)$.  We have existence issues and so on, but I'm going to give this a go.

When I said chain, I really meant a figure of connected circles.  Here is the proof:

I will use the variables given in the original problem statement.  Let $\displaystyle R = \sum_{i=1}^n r_i$.

Let the centers of small circles be $O_1, O_2, \ldots O_n$.  Given any two points $P_1$ and $P_2$ on the union of the circles, let them lie on a chain of circles with centers WLOG $O_1, O_2, \ldots O_k$, where $k \le n$, $P_1$ lies in circle $O_1$, and $P_2$ lies in circle $O_k$.  Then we have

$P_1P_2 \le P_1O_1+O_1O_2 + O_2O_3 + \ldots + O_{k-1}O_k + O_kP_2$

by the triangle inequality.  Note that $O_xO_{x+1} = r_x+r_{x+1}$.  We may then rewrite our inequality

$\displaystyle P_1P_2 \le P_1O_1 + O_kP_2 + 2\sum_{i=1}^k r_i - r_1 - r_k$

Since $P_1O_1 \le r_1$ and $O_kP_2 \le r_k$, we have

$\displaystyle P_1P_2 \le 2\sum_{i=1}^k r_i \le 2R$

Assume the figure of circles can not be covered by a circle of radius $R$.  Then there must exist two points in the figure such that their distance is greater than $2R$, but we have shown that this is not true.  By contradiction, the figure of circles can be covered by a circle of radius $R$.

It seems sort of understood (or maybe I'm just not thinking straight) but how would we show that given $n$ points, that if the maximum distance between two points is $2R$, then they lie in a circle of radius $R$?

You're right, that's not true :blush: .  Back to the drawing board!

I would like to see this problem moved - it's quite nice but this forum never gets any traffic.

Perhaps Algebra Unsolved Problems or Geometry Unsolved Problems?

It looks kind of similar to a problem in the MathLinks contest.

Edit--I think this can be proven by induction (I also think this was mentioned earlier somewhere)

[quote="luimichael"]"Connected" means that any two circles among the whole set of circles are either tangent to each other or their intersection is non-empty.[/quote]
"Any" is an ambiguous word.  Do you mean every pair of circles intersect, or some pair of circles intersect, or...

[quote="JBL"]Yes, induction would be very simple, but you need to first prove the assertation that it is always possible to find one disk such that removing it leaves the other n - 1 disks connected.  This is intuitively fairly clear but the proof is non-obvious, at least to me.[/quote]

If there's a disc that is connected just to one other disc, you can remove it and the set stays connected. If there's no such disc, then remove any disc and it stays connected. Or am I missing something?

First of all, you can start from one disk and add more.  Also, there is a graph theory proof.  I think my original idea can also be tweaked to work if you add some graph theory.

I have to apologize for making an incorrect explanation about the connectedness in my mind.
The connectedness should be referred to "simple connectedness".
Here is a link for the exact meaning:
http://mathworld.wolfram.com/SimplyConnected.html

I have also tried to solve the problem by myself and finally I needed to refer to some technical problems in graph theory.
We may try to denote the centers of the cirlces by C1, C2, ..., Cn. Next we put an edge between 2 centers of the corresponding circles if they are connected.
The crucial step in induction  is to guarantee the existence of a cirlce which can be deleted such that  the remaining circles are still connected.

After examining some cases for small values of n, I feel that we may divide the problem into to major cases:
1. There is a circle which is connected to exactly one of the other cirlces. ( [b]In the language of graph theory, deg(Cj) = 1 for some j[/b].)
The we may remove this particular cirlce and does not affect the connectedness of the remaining graph.
2. On the contrary, each circle is connected to at least two other circles. That is [b]deg(Cj) >=2 for all j.[/b]
I guess there could be a theorem to guarantee that such a graph is path-connected ?! If this is really true, we may choose a [b]path with maximum length[/b], then removing one circle at either ends of the path won't affect the connectedness of the remaining graph.

I am not familiar with graph theory, but I hope that my idea could asist to solve this problem.

That's basically it.  Now you can contradict--assume that the graph is no longer connected.  If the graph is not connected, then the point you removed must connect to at least two other vertices.  But then one of those points is one end of the diameter, and not your original point.  Well, this explanation is somewhat abbreviated.


There are five unmarked envelopes on a table, each with a letter for a different person.  If the mail is randomly distributed to these five people, with each person getting one letter, what is the probability that exactly four people get the right letter?

the answer is $ 0$, because if $ 4$ people have the right envelope, there is $ 1$ envelope, and $ 1$ person. they must match, and thus, the answer is $ \boxed 0$.

[quote="james4l"]Note that the problem is asking for [url=http://en.wikipedia.org/wiki/Rencontres_numbers]rencontres number[/url] $ D_{5,4} \equal{} {5 \choose 4} \cdot D_{1,0} \equal{} 5 \cdot 0 \equal{} \boxed 0$[/quote]

Well thanks for keeping this MC level...

Generally for n envelopes, it is impossible for n-1 people to have the right one.

Well it's to generalize for n envelopes and k people get their own envelope...

You pick n choose k to get their own envelope. Then take a derangement out of the last n-k people.

For the problem, since the derangement of 1 is 0 ({1} can be rearranged in one way: {1}, 0 which are derangements), our answer is 0*5=0.

[hide=Solution]Its impossible for 4 to get the right envelope without the 5th one also getting the right envelope, so its $0$.[/hide]


1.Let polynomial f(x)=x^n+a_1x^(n-1) +...+a_n with coefficients are integer and k,p are integer numbers .Show that if all follows numbers :f(k);f(k+1);...;f(k+p) are not divisible by p+1 then the equation f(x)=0 has no solution in integer.

2.Supose that n;p;q are positive integer numbers such that  n>p+q and a_0;a_1;....;a_n are integer numbers satisfying
     i)a_0=a_n=0
     ii) With each integer numbers i  (0<i<n+1) we have a_i - a_(i-1)=p or a_i -a_(i-1)=-q
  Prove that there exist one pair index (i;j)  (i<j and if i=0 then j <n ;if j=n then i>0) such that a_i=a_j

solution of 1. see here /http://www.mathlinks.ro/Forum/viewtopic.php?t=157202

$ \frac{solution\ of\ 2}{ }$
it's evident that:
$ a_{i}=a_{0}+b_{i}p-(i-b_{i})q=b_{i}(p+q)-iq$ where $ b_{i+1}=b_{i}\ or\ b_{i}+1,\ b_{0}=0$
and $ 0=a_{n}=b_{n}(p+q)-nq \Longrightarrow \ b_{n}(p+q)=nq$
then $ a_{i}-a_{j}=(b_{i}-b_{j})(p+q)+(j-i)q$

now we suppose that $ \forall (i,j)\not\in\{(0,n),(n,0)\}: \ a_{i}-a_{j}\neq 0$
then $ \forall (i,j)\in\{(0,n),(n,0)\}: \ (i-j)q\neq (b_{i}-b_{j})(p+q)$
$ \forall i\in\{0,1,...,n-(p+q)\}: \ (p+q)q\neq (b_{i+(p+q)}-b_{i})(p+q)\Longrightarrow b_{i+(p+q)}\neq b_{i}+q$
:arrow: i will continue after  :wink:


geometry
circumcircle
incenter
Euler
projective geometry
geometry unsolved

Dear Mathlinkers:

Let $ ABC$ be a triangle, $ O$ the circumcenter, $ I$ the incenter, $ X$ the tangency point of the incircle with $ BC, Fe$ the Feuerbach's point, $ P$ the intersection of $ OI$ and $ BC,$ and $ Q$ the intersection of $ AI$ and $ XFe.$ The result : $ PQ$ is perpendicular to $ AI.$

Sincerely
Jean-Louis Ayme

Dear Jean-Louis Ayme,

I like this problem a lot, because it seems to be possible to apply some very nice and recent results to prove it. In the following proof, I will use Antreas Hatzipolakis' result on the remarkable concurrence of three Euler lines at the Feuerbach point, which I think you know very well. If not, you can find this extremely nice result and some generalisations on Hyacinthos. Note that it can also be proved using some more classical arguments.

[b]Proof:[/b]

We project $ P$ orthogonally to $ AI$ to obtain a point $ Q'$. We try to prove that $ F_e, Q',X$ are collinear.

Clearly, $ PIQ'X$ is cyclic, so $ \measuredangle PXQ' \equal{} \measuredangle PIQ' \equal{} \measuredangle OIA$.
On the other hand, $ \measuredangle PXF_e \equal{}  \measuredangle BXF_e \equal{} \frac{1}{2} \measuredangle XIF_e$. We want to prove that these angles are indeed equal.

Now, call $ A_1,A_2$ the projections of $ A$ onto $ BI,CI$. An angle chase shows that $ \Delta AA_1A_2$ is similar to the circumcevian triangle of $ I$, and that $ A_1A_2$ and $ BC$ are parallel. We know that $ X'$, the antipode of $ X$ in the incircle, is the De Longchamps point of $ \Delta AA_1A_2$, and that $ X'F_e$ is the Euler line of $ \Delta AA_1A_2$. (For a proof of this result of Antreas Hatzipolakis, see for example Treegoner's proof in Hyacinthos message 10913).

We now see that $ \measuredangle XX'F_e \equal{} \frac{1}{2} \measuredangle XIF_e$ is the angle between the Euler line and the altitude from $ A$ in $ \Delta AA_1A_2$.

This angle is of course also the angle between Euler line and altitude of the circumcevian triangle of $ I$, because this triangle is similar to $ \Delta AA_1A_2$. The Euler line is in this case $ OI$, the altitude is $ AI$. This proves the statement.

[b]Comments:[/b]

The line $ PQ$ is the polar line of the intersection point of $ AI$ and the line connecting $ F_e$ to the midpoint of $ BC$. Proving this of course also implies the result, but the proof is considerably harder, and uses extensive angle chases and some far-fetched cyclic quadrilaterals.

[b]Figure:[/b]

[img]http://img504.imageshack.us/img504/9053/aymefeuerbachsj6.png[/img]

Best regards,
Jan Vonk

Dear Mathlinkers,
for more explanation, see for example
http://perso.orange.fr/jl.ayme   vol. 4  Symetriques de OI par rapport aux triangles de contact et médian
Sincerely
Jean-Louis


[color=yellow]If light particles do exist, then would they be smaller than the smallest particles currently recognised? Also if light contains particles, are the paticles of kinds of light, for example ultra violet vrs red, different, or would they probably be the same particles moving under different circumstances?[/list]

As far as I know, light has been classified through experiments to have what is (confusingly) wave-particle duality. Not sure how that works. I dont know how the particle part works, but waves of ultra violet are much shorter and higher frequency than that of red. Red, in contrast, is a much more stretched wave. I would assume then that particles of ultra violet would be smaller? Can anyone explain wave-particle duality?

well light shows both wave and particle properties in experiments. A gamma ray will record individual "click" in a GM tube, for example, indicating that light comes in discrete packages called photons. However it also shows wavelike properties such as interference - if you shine a coherent source of light through two narrow slits, the beams will diffract and interfere giving an interference pattern. So it is a particle sometimes and a wave other times.

Interestingly, other particles also show this property although the main distinction is that they don't travel at the speed of light and they have a mass. So an electron will also sometimes appear to be a particle (for example, in chemistry it is assumed I think) and again will cause a click on a GM tube, but when diffracted through graphite crystals interference patterns are again observed, which can only be explained by the peaks and troughts of a wave cancelling out and doubling up (intefering).


function
calculus
integration
number theory
prime factorization

Let $n=p_{1}^{e_{1}}\ldots p_{k}^{e_{k}}$ be a prime factorization of $n$. If $d(x)$ represents the number of positive divisors of the positive integer $x$, then prove that

${ \sum_{k|n}d(k) = \binom{e_{1}+2}{2}\ldots \binom{e_{k}+2}{2}}$.

i came up with this on my own. is this connected to any well known results?

[quote="Altheman"]the sum over the divisors of n of a multiplicitive function is, itself, a multiplicitive function, i.e.

s(pq)=s(p)s(q)

i think you expression needs some e_i 's there....[/quote]

oh of course! the e_i's are supposed to be where the p_i's are right now. stupid mistake, just like the AIME. it's fixed now  :)

[quote="Elemennop"]it's pretty obvious to prove it for $n=p^{e}$ and then use the multiplicative characteristic of $d(n)$ to show that it holds for any $n$ (just show that it's onto and one to one).[/quote]

that's a pretty quick way to do it. what about a proof not using the multiplicative property of $d$?

[quote="drunner2007"]what about a proof not using the multiplicative property of $d$?[/quote]

What for, if it is obvious consequence of simple statements?
Indeed
1) The set of multiplicative functions is closed under Dirichlet production $\circ$
2) The function $I(x)=1$ for all $x$ is multiplicative
3) $d=I\circ I$ and your LHS is $I\circ d=I\circ I \circ I$
4) [quote="Elemennop"]it's pretty obvious ...  [/quote]

To be fair:
I am too lazy even to think of combinatory interpretation althought I believe it exist of course.

Given $n = p_{1}^{e_{1}}p_{2}^{e_{2}}\dots p_{k}^{e_{k}}$, choose two $n$-tuples of non-negative integers $(u_{1}, u_{2}, \dots, u_{k})$ and $(v_{1}, v_{2}, \dots, v_{k})$ such that $0 \le u_{i}\le v_{i}\le e_{i}$ for all $i$.  Let $a = p_{1}^{u_{1}}p_{2}^{u_{2}}\dots p_{k}^{u_{k}}$ and $b = p_{1}^{v_{1}}p_{2}^{v_{2}}\dots p_{k}^{v_{k}}$.  This sets up a bijection between all possible pairs of $n$-tuples $(u_{1}, u_{2}, \dots, u_{k})$ and $(v_{1}, v_{2}, \dots, v_{k})$, and positive integers $a$ and $b$ such that $a|b$ and $b|n$.

Summing over $a$ and then $b$, the number of such pairs $(a,b)$ is $\sum_{b|n}d(b)$.  It is not hard to see that the number of such pairs of $n$-tuples $(u_{1}, u_{2}, \dots, u_{k})$ and $(v_{1}, v_{2}, \dots, v_{k})$ is the given formula.

Here's the way I thought of it.

Consider the number of positive integral 3 tuples $(x,y,z)$ such that $xyz=n=p_{1}^{e_{1}}\ldots p_{k}^{e_{k}}$.

Notice that when setting $x=k$, where $k$ is a divisor of $n$, the equation becomes $yz=n/k$, which gives $d(n/k)$ solutions. So the total number of solutions is $\sum_{k|n}d(n/k) = \sum_{k|n}d(k)$.

However, we can think of the number of solutions in another way. We need to appropriate the $p_{1}$'s, $p_{2}$'s, etc. among $x$, $y$, and $z$. It is not hard to see that there are $\binom{e_{1}+2}{2}$ ways to appropriate the $p_{1}$'s. In general, there are $\binom{e_{i}+2}{2}$ ways to appropriate $p_{i}$. So the total number of solutions is $\binom{e_{1}+2}{2}\ldots \binom{e_{k}+2}{2}$

[quote="Elemennop"]it's pretty obvious to prove it for $n=p^{e}$ and then use the multiplicative characteristic of $d(n)$ to show that it holds for any $n$ (just show that it's onto and one to one).[/quote]

when you refer to onto and one-to-one do you mean showing that it is onto by showing that n can be written as a product of powers of primes, and for one-to-one do you mean showing that n can be written uniquely in this form?


geometry
ratio
similar triangles
geometry proposed

Given two circles $k_1$ and $k_2$ which intersect at two different points $A$ and $B$. The tangent to the circle $k_2$ at the point $A$ meets the circle $k_1$ again at the point $C_1$. The tangent to the circle $k_1$ at the point $A$ meets the circle $k_2$ again at the point $C_2$. Finally, let the line $C_1C_2$ meet the circle $k_1$ in a point $D$ different from $C_1$ and $B$.

Prove that the line $BD$ bisects the chord $AC_2$.

Let $M$ be the point of intersection of the lines $BD$ and $AC_2$. We have to show that this point $M$ is the midpoint of the segment $AC_2$.
Now, $<C_1AB=<BC_2A=<BDC_2, <AC_1B=<BAC_2$, also: $MC_2B$ and $MDC_2$ are similar triangles: $MB.MD=MC_2^2$
As $AM$ is tangent to $k_1$: $AM^2=MB.MD$
Hence: $AM=MC_2$. QED.

This is a very nice proof.

Here is my solution (I know that there are many different solutions, I've myself found three or four, but this is the one I have submitted):

Let N and M be the midpoints of the chords $AC_1$ and $AC_2$, respectively. We will first show that the points A, B, N, M lie on one circle.

We will work with directed angles modulo 180°. Since the line $AC_2$ is the tangent to the circle $k_1$ at the point A, the tangent-chordal angle theorem yields $\measuredangle \left( AB;\;AC_{2}\right) =\measuredangle BC_{1}A$. In other words, $\measuredangle BAC_{2}=\measuredangle BC_{1}A$. Similarly, $\measuredangle BAC_{1}=\measuredangle BC_{2}A$, or, in other words, $\measuredangle BC_{2}A=\measuredangle BAC_{1}$. Together with $\measuredangle BAC_{2}=\measuredangle BC_{1}A$, this yields that the triangles $BAC_2$ and $BC_1A$ are directly similar. Now, corresponding points in directly similar triangles form equal angles. The points M and N are corresponding points in our two directly similar triangles $BAC_2$ and $BC_1A$, since M is the midpoint of the side $AC_2$ of the former triangle, while N is the midpoint of the side $C_1A$ of the latter one. Hence, $\measuredangle AMB=\measuredangle C_{1}NB$. But $\measuredangle C_{1}NB=\measuredangle ANB$. Thus, $\measuredangle AMB=\measuredangle ANB$, and it follows that the points A, B, N, M lie on one circle. This, in turn, yields $\measuredangle ABM=\measuredangle ANM$.

Now, since the points N and M are the midpoints of the sides $AC_1$ and $AC_2$ of triangle $AC_1C_2$, we have $NM\parallel C_{1}C_{2}$. Hence, $\measuredangle ANM=\measuredangle AC_{1}C_{2}$. Together with $\measuredangle ABM=\measuredangle ANM$, this gives $\measuredangle ABM=\measuredangle AC_{1}C_{2}$.

Now, since the points A, B, D, $C_1$ all lie on the circle $k_1$, we have $\measuredangle ABD=\measuredangle AC_{1}D$, so $\measuredangle ABD=\measuredangle AC_{1}D=\measuredangle AC_{1}C_{2}=\measuredangle ABM$. And this entails that the points B, D and M lie on one line. In other words, the line BD passes through the midpoint M of the chord $AC_2$, i. e. it bisects this chord. Qed..

By the way, this problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15908]has already been posted at MathLinks[/url].

  Darij

Let $d$ be the line through $D$ and parallel to $AC_2$. We have to show that $DC_1,d,DA,DB$ form a harmonic quadruple, and after working on this a bit we reduce it to $\frac{BC_1}{BA}=\frac{AC_2}{AC_1}$, which is easy to prove (it's in fact the ratio of the radii of the two circles).

Well, it's not that short, but this is the main idea :).


Yao is BALLIN'. He releases a basketball from a height of 10 feet moving in a perfect circular arc toward a vertical hoop 10 feet high, and the ball goes in perfectly. Find the area enclosed by Yao's release point, the hoop, the path of the ball, and the floor.

umm i think we need to know how far Yao is from the hoop.

Well yes, actually, but you can determine it from the information given.

Do you mean enclosed by Yao, his release point, the path, the hoop, the pole thing that supports the hoop, and the floor?

[hide="if so"]We can draw the diagram here.

[asy]draw((0,0)--(20,0)--(20,10)--(0,10)--cycle); path circ1; circ1=circle((10,10),10); draw(circ1);[/asy]
Since we know the path is a [b]perfect [/b]circular arc, we can extend it down to the floor which was done in the diagram (how do you draw semicircles?).  It becomes clear that the radius of the circle is 10, so the diameter is 20.  We have a semicircle of radius 10 and a rectangle with length 20 and width 10.  The total area is $\frac{100\pi}{2}+20\times10=50\pi+200$.[/hide]

I have to say indianamath is right.  Yao could be 1 foot away shooting in a circular path with diamter 1 ft, or 2 ft away shooting in a circular path with a diameter of 2 ft.  He also could be 10 feet away but shooting in a circular path with radius 37ft(anything > 10 ) it would just be a low shot and an arc less than half a circle.

Alan's solution, which may be what you were looking for, assumes 10 is the diameter, and that the ball and the hoop are at the ends of the diameter.

Maybe by perfect you mean semicircular arc, and goes through perpendicular to the plane the hoop defines?


Hi,

I have a problem and I can't find a good solution. Can you help me please?

Three spacecraft are landing on a spherical planet, their positions are chosen independently and distributed uniformly over the entire surface. Two spacecraft can not communicate with each other so that the angle of their positions relative to the center of the planet is less than $ \frac {\pi}{2}$. What is the probability that the three spacecraft can communicate with each other, possibly by using another spacecraft as the intermediary.

Thanks a lot

I was wrong. its easy.

answer: $ \frac{2\plus{}\pi}{4\pi}$


Hi All, how do I solve for n here?  The square root is for the entire fraction...

0.3 = 1.96*0.354√2/n

thanks in advacne

Pff... Please do not post such thing in the advanced section. There are more appropriate forums for that :mad:

Pierre.

I apologize but I thought i was posting in the "appropriate" section, as this is what is stated under the algebra link:

High-school Algebra. For College Algebra (groups, rings, matrices, etc.), please use the Superior Algebra forums.

If someone could please help me...I am a 4th year university student who has no choice but to take stats and I know this may be really easy for some of you, it's not for me.  Actually I'm in tears right now because I have wasted so much time trying to figure this out.

If someone could please have mercy on me.  Karma is a good thing.

If I have interpreted the question correctly...

$0.3=1.96 \times 0.354 \times \sqrt{\frac{2}{n}}$
$0.432376...=\sqrt{\frac{2}{n}}$ (dividing through by $1.96 \times 0.354$)
$0.186949...=\frac{2}{n}$ (squaring)
$0.186949... \times n = 2$
$n=\frac{2}{0.186949...}$

$n=10.69808...$

I think pbornsztein could have meant you used the high-school forum, perhaps the [url=http://www.mathlinks.ro/Forum/index.php?f=150]Intermediate Topics[/url] or [url=http://www.mathlinks.ro/Forum/index.php?f=149]High School Basics[/url] subforum.


Find $ \sqrt{1\plus{}\sqrt{1\plus{}\sqrt{1\plus{}\sqrt{1\plus{}...}}}}$ in simplest radical form

[hide]Let $ x\equal{}\sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} \sqrt {1 \plus{} ...}}}}$

Then $ x\equal{}\sqrt{1\plus{}x}$
$ x^2\equal{}1\plus{}x$
$ x^2\minus{}x\minus{}1\equal{}0$
$ x\equal{}\frac{1\pm\sqrt{5}}{2}$

However, $ x$ must be positive

So $ x\equal{}\frac{1\plus{}\sqrt{5}}{2}$[/hide]

I already told you in the MC PSS. The value of the square root is defined as positive.

obviously, in the original expression, all integers are positive.  Thus, the result must be positive.


AMC
USA(J)MO
USAMO
ARML
summer program
Mathcamp
MathPath

Hello, I am considering going to this camp this summer.  Has anyone here every attended or is planning to attend?  I'm not sure if I should go, considering that it is quite expensive and long.

[quote="eryaman"]Has anyone here every attended or is planning to attend?  I'm not sure if I should go, considering that it is quite expensive and long.[/quote]

One of my students went twice a few years ago and found it very worthwhile. He was a strong student (USAMO qualifier a couple of times). I also had a couple of (different) students attend the middle school version and say they had a good experience. A large portion of the Texas ARML team goes at some point or another. I haven't checked the current info, but in the past the price has actually been lower than other math camps because they received some sort of subsidy.

[quote="Rep123max"]I am definetly planning on applying. I just don't know if I will go to that or USA/Canada Math Camp (if I even have a chance at that). I went to the Middle School thing last year and it was lots of fun.[/quote]

Oh, Mathcamp is fabulous! That's the one my son has attended for the past 4 years. He's done a lot of summer programs (not the San Marcos one) and Mathcamp stands alone as his absolutely favorite place to be in the universe. It's going to be in Portland this year, which should be a good location.

Unless prices have changed, the San Marcos one is a lot cheaper than Mathcamp, and probably easier to get into. And for kids from Texas, there are likely to be people there they already know from various competitions.

yeah... I went to the middle school camp in San Marcos last year... probably will go the Honors camp in two years (summer after freshman year in highschool)

[quote="sran"]I've been trying to get to their website, but I keep getting the "this page cannot be displayed" message. Is their website down, or am I just googling the wrong terms? I tried "San Marcos Honors Math Camp" and "Texas Honors Math Camp".[/quote]

I got appropriate hits by googling "honors math camp" +"san marcos", but then I couldn't get anything to open. The url was http://www.mathworks.txstate.edu
Try again in a few days. Maybe they're having problems with their server.

[quote="EFuzzy"]Are you sure it's good? Because I when I took a look at the website, it seemed incredibly cheap, even for the residential program, and there was NO qualifying test of any kind. All you're doing is writing a paragraph. Mathpath seems to be better? Thanks.[/quote]

Mathpath is for middle schoolers. Are you a high schooler? The high school version is USA/Canada Mathcamp.

In spite of living w/in an hour of San Marcos, my son made the decision that flying across the country to Mathcamp would be better than the Honors camp, and for his purposes it has been. But the San Marcos program sounds very worthwhile from what I have heard and from the reputation of the director. It is not as selective as Mathcamp, PROMYS, or Ross. It attracts kids from the top 50 or so in Texas, even if it does not attract the top 5 in Texas, or very many kids from other states. It is a fairly recent program, and not as well known as some of the others.

The low price is because they are subsidized by outside funding, not because campers have to sleep in a tent or because the staff is on work-release from prison or anything. Participants are housed in a regular dorm. I've seen the campus, which is very nice. The campus and dorms are probably just as nice or nicer than the ones Mathcamp uses. San Marcos is on a river with a lot of greenspace and recreational activities. The program probably spends as much per person as any other residential camp, including MOP.

[quote="eryaman"]Hey texas137, where do you live?  You said you lived within one hour of San Marcos, I live in Austin, so we must be pretty close to each other.[/quote]

Sounds like it. I coach the Austin Area Homeschoolers team. You might know my son. Gee, you might even be on my team! :-)

Hmmm, I've heard of the AAHS, they used to do pretty well in MC when I was in middle school.

Their website is up and is simply [url=http://mathworks.txstate.edu/]mathworks.txstate.edu/[/url]

I went to the Texas State camp for a number of years and I highly recomend it.

I've ummm... visited the Mathworks math camp for the last three years... hence becoming the most well-known non-math camper there... and I have loads of fun every time I go.  Too bad I'd never actually get to go (as like a counselor... too old to be a camper).

TX State Mathcamp is one of the best things I've ever done.  And it's reasonably cheap if you have need.  I know tons of people that have gotten almost full or full scholarships due to financial aid.

oh man, math camp is reallllllly awesome. i sooo wanted to go to honors camp (i went to junior camp), but yeah, i say it was well worth every penny. if you have the opportunity to attend either the junior or honors camp, then by all means, do it.

btw, texas137, is your son alex? i think he's the only one i know from AAHS.

Hey Pranay!
It's sekhar. Pretty amazing there are other Mathworks people here.
To answer the question yes, Mathworks is an especially worthwhile camp. Don't be fooled by the application or website, it is a worthwhile camp. I'll be applying again this year. I attended the Junior (Middle School Version) Mathworks Camp for the past 2 summers. The director is also really friendly, and the counselors are all smart people.

I got my acceptance letter about 2 weeks ago, I'll probably be going this year.

haha, that IS a good reason. I think I might apply as well this year. Do you need to have taken junior to move up to honors?

[quote="Simfish"]For those who considered it worthwhile - can you please cite examples of why?[/quote]

It's a huge gathering of Texas math nerds. (Note that I do not consider myself one of them. ^-^)

[quote="MithsApprentice"]If you want massive amounts of rigorous work, rumor has it that PROMYS is looking for you. :-)[/quote]

Talk about it. I couldn't even get away with assuming things like a*0=0 ;)

Hmm.. PROMYS sounds good then. :)

And also for some additional info about the Texas Mathcamp: Hannah Chung, a Siemens semifinalist, got her inspiration from there.

[quote="Simfish"]Hmmm... Doesn't look like a rigorous program like the Summer Science program (i was rejected there probably due to poor essays so I decided to seek elsewhere) - more like CTY in that there is some academic material, but where socialization is pretty much the major part....[/quote]

The Texas State math camp sort of markets itself as more of a social experience than a math experience, but don't be fooled.  There's more to it than the packaging reveals.  The gist of this program is that you get an intensive, six-week course in number theory, including training in how to write mathematical proofs.  Even for students who are already proficient proof-writers, this is a really great introduction to how serious mathematics is done - a good change of pace from "speed" competitions like AMC and AIME.

I've found that the camp is particularly amazing in its ability to serve students at different ability levels (including average students all the way to extremely strong students).  The fastest students usually gripe the first week about how the material is too elementary, but that really tapers off at or around the second week as the problems get more challenging.  At that point, you pretty much go at whatever pace suits you.  You can rest assured that there will always be good enough people and problems there to give you a challenge if you're bored.  (After all, the King of Cody Bowl will be there, at least physically, for another summer.)

I'm very old-school in this regard, but I think that in order to call a certain program "rigorous" you need to have the opportunity to study a single subject for a period of several weeks.  There are math camps (PROMYS, San Marcos - cousins of the Ross program) that are great at this.  Others may or may not be.

Anyway, I've been trolling this topic for a while and wanted to take the opportunity to give my own take on it.  This camp is the best thing I've ever been involved in since I started in '98 and I wouldn't trade that away for any other program (even knowing some of the great ones that are available now).


hello
can you help by explaining this

How many different strings can be made from the letters in AARDV ARK

using seven or more letters?

[hide]There are 8 letters in Aardvark, so you there would be $8*7*6*5*4*3*2$ different strings if all of the letters were distinct. This has to be divided by 3!*2! because there will be some duplicates caused by the three A's and the 2 R's, so the final answer would be $8*7*5*4*2$ or 2240

edit: that was just counting 7-strings. There are 4480 altogether, because there are 2240 8 strings by the above logic.[/hide]


@king1: cool! my favorite number is 3 as well!! :D

yeah, basically wut king1 said: the beastliest number ever!!! :P

This is an extremely over-repeated thread, but anyway, by far, $ 382$.

22. Just 'cause it's awesome like that.

I also like 17. No one ever thinks about 17. 17 is lonely.  :(

I'd say my favorite number is either $ \pi$ or $ \infty$. If you want an integer, it would be $ 15$. :D

r15's favorite number reminds me of my email...

I like 9, 96, 56, and 22.

It only does now that I told you.  :wink

Hm, 56 was at one time my favorite number.


algebra
polynomial
inequalities proposed
inequalities

s = semiperimeter

Prove that

$ m_ah_a\plus{}m_bh_b\plus{}m_ch_c\leq s^2$


geometry
trigonometry
complex numbers
Pythagorean Theorem
trig identities
Law of Sines

If a1, a2, a3 be respectively the sides of a regular pentagon, hexagon, and decagon inscribed in a circle,                                         then PROVE that (a1)^2 = (a2)^2 + (a3)^2

$ sin^2(36)\minus{}sin^2(18)\equal{}(sin (36)\plus{} sin (18))(sin (36) \minus{} sin (18))\equal{}(2sin (27) cos (9))(2cos (27) sin (9))\equal{}(2sin (27) cos (27))(2sin (9) cos (9))\equal{}
sin (54) sin (18)\equal{}1/2 (cos (36)\minus{}cos 72)$.  $ x^5\plus{}1$ has 2 roots with real part $ cos (36)$ and 2 with real part $ \minus{}cos (72)$.  The last root is -1, so since the roots sum to 0, $ 2(cos 36\minus{}cos 72)\equal{}1$, so $ sin^2(36)\minus{}sin^2(18)\equal{}1/4\equal{}sin^2(30)$, so by the extended law of sines and pythagorean theorem, we are done.
[hide=" "]David, look under Mom's keyboard[/hide]


probability
advanced fields
advanced fields open

I am not successful (using Excel) to show that the skew of a lognormal distribution is (exp(s*s) +2)*sqrt(exp(s*s)-1) if the probability is given by
p(x) = exp(-0.5*(ln(x)/s)*(ln(x)/s)) / ( x*s*sqrt(2*pi()) ) for x > 0

I tried to take different s values and verify the computations by "discretizing" - that is
to consider values of x in increments of 0.1 or 0.01 or 0.001 etc.  Also I restricted the x values to have non-trivial p(x) (e.g p(x) > 0.0001).

The skew values ar not "close" -- I get 1.98 when I am looking for 3
I do not think it is to do with discretization.

Please help

thanks
ps: Is there a tool out there that can generate a "sample distribution" with the user specified skew and kurtosis?


LaTeX
number theory proposed
number theory

These are probably too easy for you,but im stucked at each one of them...

1.Find all integer solutions of 5n^2=36a^2+18b^2+6c^2

2.Tet p be prime odd number.Prove that there are unique m,n natural numbers such that m^2=n(n+p)

3.Let n be natural. In interval (n,n^2+n) we chose two different numbers a and b.Prove that in the interval we can't find number which divides ab,different from a and b.

4.let a,b,x,y be natural numbers with a^2+b^2 | ax+by .Prove that gcd(x^2+y^2,a^2+b^2)=1.

5.If p1,p2,...,pn different primes >3,prove that number 2^(p1*p2*p3.....*pn)+1 has at least 4^n dividors.

If somebody knows some of these write hint at least,plz  Anyway thanks!!!!

Please, force yourself to learn and use $ \text{\LaTeX}$, otherwise people will not read your post  :(

Add a dollar sign to the two sides of the code to $ LaTeX$ it , which is not difficult at all.

1.$ 5n^2\equal{}36a^2\plus{}18b^2\plus{}6c^2$ , obviously by infinite descent, we got $ n\equal{}a\equal{}b\equal{}c\equal{}0$ is the only solution.
2.Rewrite the equation as $ n^2\plus{}pn\minus{}m^2\equal{}0$, we get $ n\equal{}\frac{\minus{}p\plus{}\sqrt{p^2\plus{}4m^2}}{2}$, the value of $ m$ for which $ p^2\plus{}4m^2$ being a square of integer is unique , so done.

Add dollar signs before and after equations to LaTeX them.
1. Find all integer solutions of $ 5n^2 \equal{} 36a^2 \plus{} 18b^2 \plus{} 6c^2$
2. Let p be prime odd number.Prove that there are unique m,n natural numbers such that $ m^2 \equal{} n(n \plus{} p)$
3. Let n be natural. In interval $ (n,n^2 \plus{} n)$ we chose two different numbers a and b.Prove that in the interval we can't find number which divides ab,different from a and b.
4. Let $ a,b,x,y$ be natural numbers with $ a^2 \plus{} b^2|ax \plus{} by$ .Prove that $ gcd(x^2 \plus{} y^2,a^2 \plus{} b^2) \equal{} 1$.
5. If $ p_1,p_2,...,p_n$ different primes, each greater than 3,prove that number $ 2^{p_1\cdot p_2\cdot p_3\ldots\cdot p_n} \plus{} 1$ has at least $ 4^n$ divisors.

Ok i'm sorry for not using the latex,i'm new here, and still learning  :blush: @stephencheng at 1st i know its infinite descent,but which modulo?
Because i get 6 divides n,and after that 3 divides c,and after that nothing ....

Smilley wrote:
[quote]stephencheng at 1st i know its infinite descent,but which modulo?
Because i get 6 divides n,and after that 3 divides c,and after that nothing ....[/quote]

In detail:
After descent we get $ 10m^2 \equal{} 2x^2 \plus{} y^2 \plus{} 3z^2$

Case 1:not all of them are even
      Then $ y$ and $ z$ both odd or both even.$ mod\;\; 8$ both sides and we get no solution in this case.
Case 2:all of them are even
      Except $ (0,0,0,0)$, which is the only solution, for any other solution in this case, we can reduce it to Case 1, which has no solution.

Also for $ 3,4$ some conditions may be missed, $ 3: try \;\; n\equal{}3, a\equal{}4, b\equal{}6$; $ 4: try \;\; a\equal{}x, b\equal{}y$

[quote="Smilley"]for 3,im very sorry, the interval is (n^2,n^2+n),not (n,n^2+n) ,and i already solved it,so you don't have to write the solution.
For number 4. i missed all the numbers are different
Sorry again  :blush:[/quote]

I think there may be more condition missed in $ 4$.Try $ x\equal{}ka, y\equal{}kb$


Suppose you like to eat tacos every day, and you eat either 3, 4, or 5 at a time, once a day.  Suppose the probability of eating 3 tacos is 1/6, of eating 4 tacos is 1/3, and of eating 5 tacos is 1/2.  Now assume that you use taco sauce on each taco, such that you use 9 oz on 3 tacos, 14 oz on 4 tacos, and 20 oz on 5 tacos.  What is the probability that an 88 oz bottle of taco sauce will be consumed after 5 days?

[hide="solution"]
There are 3 cases to consider:

$ 1)$ every day you eat  $ 5$  tacos (then you consume $ 100$ oz bottles)

probability that this happens is $ \left(\frac{1}{2}\right)^5$

$ 2)$ 4 days you eat  $ 5$  tacos and the 5th day you eat either  $ 3$ or  $ 4$ (you consume $ 89$ or $ 94$ bottles)

probability that this happens is $ \binom 5 1 \cdot \left(\frac{1}{2}\right)^4 \cdot \left(\frac{1}{2}\right)^$

$ 3)$ 3 days you eat  $ 5$  tacos and 2 days  $ 4$ (you consume exactly $ 88$ bottles)

probability that this happens is $ \binom 5 2 \cdot \left(\frac{1}{2}\right)^3 \cdot \left(\frac{1}{3}\right)^2$

Adding all probabilities gives you $ \boxed{\frac{47}{144}}$[/hide]

Thank you very much for that solution Flame, but I am confused as to why you used the binomial expansion term in front of the product of the probabilities?  I've never seen probabilities calculated this way, and that is why I got the wrong solution..

Because there are different arrangements of days when you eat 3, 4, or 5 tacos and these are to be counted separately, thus the binomial term. Notice that there is only one possible arrangement in case 1.

You are welcome  :)


graph theory
combinatorics proposed
combinatorics

Suppose that $ \mathcal F$ is a family of subsets of $ X$. $ A,B$ are two subsets of $ X$ s.t. each element of $ \mathcal{F}$ has non-empty intersection with $ A, B$. We know that no subset of $ X$ with $ n \minus{} 1$ elements has this property. Prove that there is a representation $ A,B$ in the form $ A \equal{} \{a_1,\dots,a_n\}$ and $ B \equal{}  \{b_1,\dots,b_n\}$, such that for each $ 1\leq i\leq n$, there is an element of $ \mathcal F$ containing both $ a_i, b_i$.

There is a lot of lacking clarity. What does "We know that no subset of $ X$ with $ n \minus{} 1$ elements has this property" mean?

Taking $\mathcal{F}$ some family of subsets of $X$, all containing some element $x\in X$, and $A,B$ some subsets of $X$ also containing $x$. What given condition does this construction contradict?

I think, there is no other way of interpretation and the problem should be read as:

Suppose that $ \mathcal F$ is a family of subsets of $ X$. For a subset $X_1\subset X$ we say $X_1$ has the property $P$ if each element of $ \mathcal{F}$ has non-empty intersection with $X_1$. We know that no subset of $ X$ with $ n - 1$ elements has this property.
Let $ A,B$ be two subsets of $ X$, with $n$ elements each, such that they both have property $P$. Prove that $ A,B$ can be represented in the form $ A \equal{} \{a_1,\dots,a_n\}$ and $ B \equal{}  \{b_1,\dots,b_n\}$, such that for each $ 1\leq i\leq n$ there is an element of $ \mathcal F$ containing both $ a_i, b_i$.

This is a good exercise on the Hall's marriage theorem. Let $ A =\{a_1,\dots,a_n\} ,\, B =\{b_1,\dots,b_n\} $. Imagine a bipartite graph $G$ which parts are the vertices in $A$ and $B$, where $a_i$ and $b_j$ are connected iff there exists a set in $ \mathcal{F}$ containing both $ a_i, b_j$. Now prove that $G$ satisfies the marriage condition. It's easy, since otherwise (omitting some elements of $A$ and adding some elements of $B$) there will exist a set with fewer than $n$ elements also with property $P$.
Thus, there is a perfect matching in $G$ and we are done.


In quadrilateral $ABCD$, $AB=150$, $BC=200$, $CD=250$, and $DA=300$. There is a right angle at $B$. Let the circumcircles of $\triangle ABC$ and $\triangle ACD$ be $\omega_{1}$ and $\omega_{2}$, respectively. $\omega_{1}$ intersects $AD$ at $E$ and $CD$ at $F$. $\omega_{2}$ intersects $BC$ at $G$. Find the remainder when the area of quadrilateral $AEFG$ is divided by $1000$.

[dividing by 10] some of the results from the figure are nice...ABC = AEC = DEC, also CE, AF, DG are concurrent at H. [DG intersect AC at G'] It becomes a bunch of tedious length computation in triangle ACD. FC=7, GC=HC=8.75, DF=AG'=18, AF=24

area is AEF+AGF=1/2AFD+2AG'H+3HCF-FCG=1/2*1/2*18*24+2*1/2*15*5.25+3*1/2*7*5.25-1/2*7*8.75*sin 3arcsin3/5=21321/100...i shrank by 10 so, area is 21321...321 [barring lazyness]

ewwww...like <5mins with ti-89...on an aime....merr...


function
functional equation
algebra
IMO Shortlist

Let $ f(n)$ be a function defined on the set of all positive integers and having its values in the same set. Suppose that $ f(f(n) \plus{} f(m)) \equal{} m \plus{} n$ for all positive integers $ n,m.$ Find the possible value for $ f(1988).$

Eight years and one day later . . . the 500th view and
[hide="My Solution"]
\[ f \left( f(m) + f(n) \right) = m + n \]Substitute $ m = f(a) + f(b) $ and $ n = f(c) + f(d) $ to get \[ f \left( f \left( f(a) + f(b) \right) + f \left( f(c) + f(d) \right) \right) = f(a) + f(b) + f(c) + f(d). \]We know that $ f \left( f(a) + f(b) \right) = a + b $ and $ f \left( f(c) + f(d) \right) = c + d $. Thus: \[ f(a+b+c+d) = f(a) + f(b) + f(c) + f(d), \qquad \forall a, b, c, d \in \mathbb{N} \]If we substitue $a+1$ for $a$ and $b-1$ for $b$, we obtain \[ \begin {eqnarray*} f(a+b+c+d) &=& f(a+1) + f(b-1), \\ f(a) + f(b) &=& f(a+1) + f(b-1) \\ f(a+1) - f(a) &=& f(b) - f(b-1) \]Since this holds for all positive integers $a,b$, we conclude that $f$ is an arithmetic sequence. Hence, it is of the form $f(n) = \alpha \cdot n + \beta$.

We substitute this into our original function equation, and we get \[ \begin {eqnarray*} \alpha \cdot \left( \alpha \cdot m + \beta + \alpha \cdot n + \beta \right) &=& m + n \\ \implies \ \alpha^2 \cdot m + \alpha^2 \cdot n + 2 \alpha \beta + \beta &=& m + n. \]\[  \implies \left\{\begin{array}{lr} \alpha^2 = 1, \\ 2 \alpha \beta + \beta = 0.\end{array} \right. \]The second equation gives us $ \beta \cdot (2 \alpha + 1) = 0 $. If $ 2 \alpha + 1 = 0 $, then $ \alpha = -\dfrac{1}{2} $. But, $\alpha$ and $\beta$ are positive integers, so $ \beta = 0 $. From the first equation, we have $ \alpha = \pm 1 $, but $ \alpha $ must be a positive integer so it is $1$.

Therefore, $ f (n) = n $ and $ f(1988) = \boxed {1988} $.
[/hide]

The second solution from the IMO compendium gave that $f(f(1)+n+m)=f(f(1)+f(f(m)+f(n)))=1+f(m)+f(n)$, and hence concluded that $f(m)+f(n)$ is a function of $m+n$. How does this hold true?

If $f(x)$ is a function of $x$, then $f(x+a)+b$ is also a function of $x$ where $a,b$ are reals. Since $f(m+n)$ is obviously a function of $m+n$, then $f(m+n + f(1)) - 1 = f(m) + f(n)$ is also a function of $m+n$.

We claim that, $f(1988)=1988$.

We proceed as follows. Let $P(m,n)$ be the assertion given in the problem. Then, if there exists any $n_1,n_2$ with $f(n_1)=f(n_2)$, by inspecting $P(m,n_1)$ and $P(m,n_2)$, we would have deduced that $n_1=n_2$. Thus, $f(\cdot)$ is injective.

Next, notice that, for every $k\geq 2$, there exists a $k_0$ such that $f(k_0)=k$. We will now show that the values of $f(1)$ and $f(2)$ uniquely determines the rest of the sequence onwards. To show this, let $f(1)=a$ and $f(2)=b$. Then by considering $P(2,2)=P(1,3)$, together with injectivity, we have $f(1)+f(3)=2f(2)\implies f(3)=2b-a$. Now, $P(4,1)=P(3,2)\implies f(4)+a=3b-a\implies f(4)=3b-2a$. In particular, we claim that $f(k)=(k-1)b-(k-2)a$, for every $k$. For this, we'll rely on induction. Base cases are clear, and assume $f(k-1)=(k-2)b-(k-3)a$. Now, $P(k,1)=P(k-1,2)$ implies, $f(k)+a=(k-2)b-(k-3)a+b\implies f(k)=(k-1)b-(k-2)a$. In particular, we deduce that, $f(k)=uk+v$ for some $u,v\in\mathbb{Z}$, to be determined next.

Now, inserting $f(k)=uk+v$ into $P(m,n)$, we arrive at, $u=1$ and $v=0$; hence, $f(k)=k$. In particular, $f(1988)=1988$.

$f$ is the identity. So $f(1988)=1988.$ [url=https://artofproblemsolving.com/community/c146h150770p25259219]Proof for PEN[/url].


Distinct numbers are assigned
to regions $ A$, $ B$, and $ C$ in the
given rectangle. The numbers
shown in the circles represent the
sums of the numbers assigned
to the regions that border each
circle. What number is assigned
to region $ C$?

[asy]size(100); import graph; real length = 10, height = 5, cheight = .4height,radius = .55;

draw((0,0)--(length,0)--(length, height)--(0,height)--cycle);
draw((0,cheight)--(length,cheight));
draw((length/2,cheight)--(length/2,height));
label("$C$",(length/2,cheight/2)); label("$A$",(length/8,.5cheight+.5height)); label("$B$",(7length/8,.5cheight+.5height)); filldraw(Circle((.14length,cheight),radius),white); label("15",(.14length,cheight),fontsize(8pt)); filldraw(Circle(((1-.14)*length,cheight),radius),white); label("17",((1-.14)*length,cheight),fontsize(8pt));
filldraw(Circle((.5length,.6cheight + .4height),radius),white); label("22",(.5length,.6cheight + .4height),fontsize(8pt));[/asy]


I came up with these two ideas while playing around with two popular series for pi. Proofs or counterexamples are welcome!

(1) Let $S_k = 4 \sum_{x=0}^{10^k} \frac{(-1)^x}{2x+1}$, for positive integer $k$. Show that the first $2k$ digits after the decimal point in the decimal expansion of $\pi - S_k$ are $k$ $0s$ followed by $k$ $9s$.

(2) Let $T_k = \sqrt{6(\sum_{x=1}^{10^k}1/x^2)}$, for positive integer $k$. Show that the first $k$ digits after the decimal point in the decimal expansion of $\pi-T_k$ are $0s$ and the following digit is a $9$.

What about as the limit approaches infinity?

there is something even more strange:

For (1):
after your $2k$ numbers there are ($k>2$) $k$ more $0$'s followed by $74$, then $k-2$ $9$'s and one more time $74$, then $k-2$ $9$'s

For (2): when looking at the difference $\frac{\pi^2}{6}-\frac{T_k^2}{6}$ you get $k$ $0$'s, then $k$ $9$'s, after that a $5$ and then $(k-1)$ $0$'s and at last about $16666...$   !!!