email
algorithm
AMC
USA(J)MO
USAMO
abstract algebra

Can someone give me the java code for any one of the problems for the USACO training, or at least the first few lines of a code that contains the correct format? Even just the Hello World code to test submissions would be fine (so no, I'm not trying to cheat ;)).
The problem is that I'm not sure what's wrong with my submissions. Whenever I try to submit one of my answers (even for the Hello World program...), I get a "Fatal Syntax Error: / Found but BEGIN expected" and I'm not sure what's wrong, as the code works fine on my own computer.

Thank you[/i]

I'm fairly sure, actually, that Java is not available to use with the training pages. If you don't want to learn C or C++ (or Pascal) then my suggestion would be to bug Rob Kolstad about it -- it shouldn't be too hard for him to implement it, especially since the code is already there for the regular contests. This may be, though, a not-so-subtle hint that C/C++ is preferred and that Java is not acceptable at the IOI.

On a further note, I too have some questions about USACO.

I also use java, but lets for a second say that it is perfectly acceptable. I am used to (in class) providing a question in my programs, so that when my teacher would come over, he would read the question and either answer accordingly or try to "break" the program. But Im not so sure what to do for usaco being that there is no person there to read the question. Im assuming as soon as you submit, it gives the input, but im not too clear on this, and if that is the case, then im not so clear on how to control the input hehe.

Hello, maybe this is a little bit off topic, but which chapter or section of USACO are you guys on?

jelyman, you have to get the program to read in from a file.

BTW, I heard from my computer science teacher that in the near future, (next year or the year after), java will be accepted in IOI and all national contests as well - at least all Canadian contests.  Besides, java is allowed in topcoder.  The only thing they have to do I think is to set up a different standard and restrictions for java programs, since they are a lot slower than C and C++ programs.

[quote="anipoh"]BTW, I heard from my computer science teacher that in the near future, (next year or the year after), java will be accepted in IOI [/quote]

I don't think that's right, although I know that the issue of Java has been discussed by the executive committee at the past 2 IOIs, at least. I think there are technical details which have not been worked out.

Here's some discussion of the issues (scroll down to "Programming Languages and Compilers")
http://www.cs.uta.fi/~jyrki/itwg/

There is also a more informal discussion of this issue here:
http://ioi.ucsc.cmb.ac.lk/forum/forum/viewthread?thread=24

I myself is not perfect sure about java in IOI, and I don't care too much about using java or C++, C++ seems fine for me.  I think the reason some people want to use java is that the only programming language we learn in school is java.  (In Canadian schools, not sure about US schools)

Not that I can make it into IOI at all...

[quote="kidnappedturtle"]So java is allowed in the different contests throughout the year, but not in the training pages? :|
Is there any way to just see the different training problems then, even if I can't submit them and move on in the normal way?[/quote]

This is the first year that USACO has allowed Java for the internet contests. The training pages were developed 3 years ago and have had only minor changes since. I don't know what would be involved in making it so that the training pages could be done in Java, but I know that the organization is basically run by a small group of volunteers. There may not be anyone with both the time and Java expertise to do it. There also seems to be a feeling that C++ is superior to Java, even as changes are considered to account for the fact that US (and other countries') schools are moving to Java.

You can see all of the problems in a particular level of the training gate, so you can get a sense of what's there. You just can't advance to the next level until you finish the first one.

Right now, I would suggest that anyone who would like to be able to use Java on the training pages send email (usaco@lumeta.com) about it to show that there is demand for this. In the meantime, you may try to learn C++, which happens to be significantly faster than Java and might be easier for data structures, etc. due to pointers. Good resources for the type of code you would need are [url]http://www.cppreference.com[/url] and [url=http://www.topcoder.com]TopCoder[/url] to see how "contest" C++ works.

[quote="paradox"]Right now, I would suggest that anyone who would like to be able to use Java on the training pages send email (usaco@lumeta.com) about it to show that there is demand for this. [/quote]

I don't think the issue is whether or not there is demand. I think the issue is manpower to make the changes in the training gate, plus a belief that people seriously going after USACO/IOI would be better off learning C/C++ (or at least Pascal. That's what most of the Europeans use.).

Well, I can't imagine that the training pages grader differs significantly from the contest grader -- it can't be that hard to implement it. Besides, with the current switchover to Java in most schools in the country (and the AP course), it is vital to maintain the level of interest in USACO from new students.

You're right, of course, it's discouraging for beginners to have to learn a new language before they can even start on the training pages. But learning C/C++ on your own when you've learned Java in school is probably less of an issue for most USACO beginners than learning how to do the algorithms. I don't think that AP comp sci courses get into algorithms very much.

[quote="bugzpodder"]Microsoft supporting java?  thats a firsT!!!
guys interested in java AI development can also check out Robocode by IBM.  http://robocode.alphaworks.ibm.com/home/home.html

texas137 do you have link to that game?[/quote]

Oops, my bad. It's IBM, not Microsoft. I knew it was an important sponsor of ACM and IOI, but I picked the wrong one. (I've corrected my original post).

I don't have a link to the game, but you can see the screen and some code for it on the USACO site where they have daily camp reports. Here's the link to the photos:
http://oldweb.uwp.edu/academic/mathematics/usaco/2004/camp/CodeRuler/index.htm

Robocode looks great too. Beats learning Java by chugging through a technical manual.

[quote="probability1.01"]Any sample USACO problems?[/quote]

You can get an idea by signing up for the training gate:
http://ace.delos.com/usacogate

old IOI problems are here:
http://olympiads.win.tue.nl/ioi/tasks.html

here's a link to a bunch of problem archives for various computing contests:
http://oldweb.uwp.edu/academic/mathematics/usaco/home/links.htm

slightly different style but nevertheless a huge database: http://online-judge.uva.es/problemset/

What is a good site to learn C++? (I already know java, so it's fine if some stuff is sort of skeletal/bare--except for the stuff that's different in any case ;))

Also, what should I learn after an AP Computer Science course, if I'm not really interested in stuff like graphics and games? What [i]is[/i] there to learn anyways? Is it algorithms? What do you learn to do in algorithms courses anyways? I was looking at Cormen's Introduction to Algorithms and couldn't really get a good idea on what you actually do.

And in general, what do you actually learn in higher (theoretical (?)--not building computers and chips and stuff like that) computer science courses? Like in math there are higher math courses, but in computer science...? Is it sort of like foreign languages courses, where after you learn the essentials, you just keep on practicing? Or is there something else?

[quote="kidnappedturtle"]What do you learn to do in algorithms courses anyways? I was looking at Cormen's Introduction to Algorithms and couldn't really get a good idea on what you actually do.

And in general, what do you actually learn in higher (theoretical (?)[/quote]

I don't actually program myself, so I'm going to be passing along what I've been told. But my understanding is that algorithmic programming is sort of like USAMO problems. Each one is a little puzzle, and you have to first figure out how to think about it.  Actually writing the code (like writing up a USAMO problem) is kind of secondary. It's no accident that so many top USACO people have also done very well on USAMO.

I don't know the answer to your question about learning C++. But once you've learned the basics, the USACO training gate will teach you more than most college courses cover about algorithms.

I'm also trying to learn C++. I'm trying the first few problems on the USACO training site. I know how to structure the first few programs, but I learned (what little I know) C++ with the iostream library. I'm having trouble with the stdio library, especially with the reading and writing to a file in the way it needs to be sent to the site compiler. If anyone has the time to help me out I would really appreciate it.

Yeh, Id like to get in on that too. I dont really get how the program accepts the info and gives the response to the computer.

[quote="texas137"]Java programmers might want to watch for the IBM game Code Ruler, currently in final development. You write programs in Java to simulate storming and defending a castle. Then teams or individuals can run their programs against each other, sort of like BattleBot w/o the robots. It was tried out at the ACM World Finals a few weeks ago, and again this week at USACO training camp. Sounds pretty cool.[/quote]

Code Ruler was pretty fun -- Anders and I (mostly Anders, but I helped a bit with the strategic part of it) wrote a pretty good algorithm that beat a lot of the ACM people and got us in third place.

We were telling Rob that he should put Java on the training pages... I'm pretty sure they are going to add it pretty soon.

[quote="texas137"][quote="kidnappedturtle"]What is a good site to learn C++? [/quote]

There may not be one. Apparently this was a topic of discussion at the USACO camp. The organizers are considering developing something for the USACO site for people who want to learn C++, because they are not aware of anything already on the web that they think is good.

How do all of you know about USACO? It seems to be something individuals find out about from other individuals rather than finding out because their schools/teachers actively endorse it. It's a much better contest from a programming standpoint than ACSL, but does not seem nearly as well known.[/quote]

They are considering making a separate "beginner" set of training pages for people just learning to program.

What languages are allowed on the USACO?

[quote="kidnappedturtle"]What is a good site to learn C++? (I already know java, so it's fine if some stuff is sort of skeletal/bare--except for the stuff that's different in any case ;))[/quote]
Get a good book about learning C++. It'll always be there for you...

[quote="kidnappedturtle"]Also, what should I learn after an AP Computer Science course, if I'm not really interested in stuff like graphics and games? What [i]is[/i] there to learn anyways? Is it algorithms? What do you learn to do in algorithms courses anyways? I was looking at Cormen's Introduction to Algorithms and couldn't really get a good idea on what you actually do.[/quote]
Learn how to program in an obfuscated language, like [url=http://en.wikipedia.org/wiki/Brainfuck]Brainf***[/url]. Or learn C, or FORTRAN (whichever version you prefer), or B, or BCPL, or LOGO, or Lisp, or OCAML, or... Even better, learn how to make object code in hex. I'm not sure how you would go about doing it, though. But knowing hex and binary as you know base-10 should  be okay. You could even try making your own OS (kernel, shell, etc from scratch).

Or are you talking about concepts to learn? Have you learned about pointers and arrays? I hear they're quite useful.


inequalities
trigonometry
inequalities unsolved

Let $ \alpha$, $ \beta$, $ \gamma$ be the angles of acute-angled
triangle. Prove the following inequality
\[ \frac{\cos^{2}\alpha}{\cos^{2}\beta}\plus{}\frac{\cos^{2}\beta}{\cos^{2}\gamma}\plus{}
\frac{\cos^{2}\gamma}{\cos^{2}\alpha}\geq 4
(\cos^{2}\alpha\plus{}\cos^{2}\beta\plus{}\cos^{2}\gamma).\]

[quote="mkrnic"]Let $ \alpha$, $ \beta$, $ \gamma$ be the angles of acute-angled
triangle. Prove the following inequality
\[ \frac {\cos^{2}\alpha}{\cos^{2}\beta} \plus{} \frac {\cos^{2}\beta}{\cos^{2}\gamma} \plus{} \frac {\cos^{2}\gamma}{\cos^{2}\alpha}\geq 4 (\cos^{2}\alpha \plus{} \cos^{2}\beta \plus{} \cos^{2}\gamma).
\]
[/quote]
Let $ a^2 \plus{} b^2 \minus{} c^2 \equal{} z,$ $ a^2 \plus{} c^2 \minus{} b^2 \equal{} y$ and $ b^2 \plus{} c^2 \minus{} a^2 \equal{} x.$
Hence, your inequality is equivalent to
$ \sum_{cyc}(x^5y^2z^2 \plus{} 2x^5z^3y \plus{} x^5z^4\plus{}x^4y^4z \minus{} 3x^4y^3z^2 \minus{} 2x^4z^3y^2)\geq0,$ which is true by AM-GM.

[quote="mkrnic"]Let $ \alpha$, $ \beta$, $ \gamma$ be the angles of acute-angled
triangle. Prove the following inequality
\[ \frac {\cos^{2}\alpha}{\cos^{2}\beta} \plus{} \frac {\cos^{2}\beta}{\cos^{2}\gamma} \plus{} \frac {\cos^{2}\gamma}{\cos^{2}\alpha}\geq 4 (\cos^{2}\alpha \plus{} \cos^{2}\beta \plus{} \cos^{2}\gamma).
\]
[/quote]
Since $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$ ,we have: $ LHS \ge \frac{ cos^2{\alpha} \plus{}cos^2{\beta}\plus{}cos^2{\gamma}}{\sqrt[3]{cos^2 {\alpha}.cos^2{\beta}.cos^2{\gamma}}} \ge RHS$
(because: $ cos{\alpha}.cos{\beta}.cos{\gamma} \le \frac{1}{8}$)

Why  $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{a\plus{}b\plus{}c}{\sqrt[3]{abc}}$ holds?

[quote="mkrnic"]Why  $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$ holds?[/quote]
It's easy my friend...Try the substitution $ \sqrt[3]{a}\equal{}x$ and cyclic,and then it is Muirihead....

[quote="mkrnic"]Why  $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}$ holds?[/quote]
Because:
 $ \frac{a}{b}\plus{}\frac{a}{b}\plus{}\frac{b}{c} \ge 3\sqrt[3]{\frac{a^2}{bc}}\equal{}\frac{3a}{\sqrt[3]{abc}}$
 $ \frac{b}{c}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{3b}{\sqrt[3]{abc}}$
 $ \frac{c}{a}\plus{}\frac{c}{a}\plus{}\frac{a}{b} \ge \frac{3c}{\sqrt[3]{abc}}$
Then,we have Q.E.D :)


f is a polynomial function with nonnegative integer coefficients.  f(2)=23.  Find the number of different polynomials that f could be.

[hide="hint"] show it cannot be 5th degree[/hide]

[hide="hint"]factor[/hide]

[hide]The leading coefficient must be at least 1. This means that the polynomial's degree must be 1, 2, 3, or 4, since $1*2^5$, the smallest possible $f(2)$ with integer coefficients, is too big.

First degree polynomials are in the form $ax+b$, so we want solutions in the form $2a+b=23$. Since there are 11 even numbers less than 23, there are $11$ possible first degree polynomials.

Second degree polynomials are in the form $ax^2+bx+c$, so we want coefficients such that $4a+2b+c=23$. $c$ is odd, and can be from $1$ through $19$ inclusive, resulting in $30$ combinations of (a,b)

Third degree polynomials have coefficients satisfying $8a+4b+2c+d=23$, and now $d$ can be any odd number from $1$ through $15$ inclusive, resulting in $26$ possiblities.

Fourth degree polynomials have coefficients satisfying $16a+4c+2d+e=23$ ($b$ must be $0$). $e$ can be any odd number from $1$ to $7$, and there are $6$ ways to do this.

$\boxed{73}$ possibilities
[/hide]


[quote="flip2004"]If $a\le b\le c$ prove that   $\; \; \; (b+c-2a)^3(c-b)\ge (a-2b+c)^3(c-a)\; , \; \;$ with  equality iff $a=b$.[/quote]

put $a-b=x$
and $b-c=y$....

I set $b-a=x$ and $c-a=y$
Then you're trying to prove $(x+y)^3(y-x) \ge y(y-2x)^3$
$-x^4-2x^3y+2xy^3+y^4 \ge y^4-2xy^3+4x^2y^2-8x^3y$
$6x^3y+4xy^3-4x^2y^2-x^4 \ge 0$
$xy(x^2-4xy+4y^2)+x^3(5y-x) \ge0$
$xy(x-2y)^2+x^3(5y-x) \ge 0$
We know $y\gex\ge0$, so all of these terms are nonnegative and the inequality holds.


summer program
PROMYS
Ross Mathematics Program
MATHCOUNTS
Mathcamp
search
topology

Anyone around here planning on doing PROMYS?  Just wondering.. :)

look at the resources ---> programs section of this site.

it's a camp at BU and other sites too i think, but you would go to the one at Boston UNiversity.

AHEM, consider the ross program (the parent program to PROMYS) as well!

http://www.math.ohio-state.edu/ross/

[quote]I'm not going this year, but I went in 2002, so I can try to answer any questions you may have.[/quote]

How was it?

[quote]AHEM, consider the ross program (the parent program to PROMYS) as well! [/quote]

lol, I most certainly shall.. :)

I liked it.  You definitely should not go if you don't like math, though. I found it a little overwhelming at first because it wasn't quite clear what I was supposed to be doing, but after that it got better.

I do plan on submitting an application. Its the camp I most want to get into. Just curious though: How rigorous are the selections and did you have any outstanding achievements that made them recognize you when u applied?

The general consensus at PROMYS was that the program is not highly selective.  Some people joked that they didn't reject anybody.  Really, it was primarily self-selected--not everyone wants to spend six weeks doing math during the summer.  As long as you have some things that show you're interested in math and give the problems a good effort (which certainly does not mean answering all of the problems entirely correctly) you will probably get in.  When I applied, I had no math awards since MATHCOUNTS in Junior High (and even then, I was only fourth in my chapter).  I think that they mostly just want to make sure that the people who go there would actually get something out of the program.

I went last year.

Thing is you have to send in your app early! There were 8 math questions on the app, and since I sent in in so late I had to answer 7 1/2. I'm sure a lot of people that did go didn't answer that many.

Adding to the lateness was the influx of people from MOP 2002 (including myself), which added to the competitiveness of the selection.

This year, I'm predicting that it will probably be easier to get in than last year, since many people don't go back for a second year. I plan on returning, since the advanced classes looked very interesting but I only had time to do one on top of the required number theory homework for first year students.

The applications should be up soon... or so  they say.

Do you know if PROMYS accepts international students?  If they do, I'll apply next year.  I am applying to Mathcamp this year.

Nukular, how early do u mean? (I plan to send mine off March15ish).

wow... have you gotten recommendations and that stuff out of the way?  I'm going to ask my teachers for recommendations tomorrow...

I don't think it has to be done that early, though... it says admissions decisions will be made until June 1, so I suspect you could mail it in a couple weeks before that and it'd be fine.

They do rolling admissions.
March 15th or so is definitely fine.

But two weeks before June 1st, as I found out last year, means you better have all 8 problems answered.

[quote="beta"]How many problems do you have to do to get into PROMYS? also how much scholarship can you get?[/quote]

At this point in the game -- quite a few. (If not all)

They only accept 60 people(i didn't know that until today)!! I should have applied a lot earlier, I still haven't submit the application form. No way I am going to get accepted.

Has anyone not received a reply from PROMYS yet? (I haven't recieved one still)

i sent mine in about a month ago (little bit less) and havent heard from them...

Not only did I get accepted, but they offered me financial aids $1,000. I think I'm just lucky since my grades are so terrible.

I gave up waiting for PROMYS, because it's been 5 or 6 weeks since I sent it, and I got the registeration and stuff from Mathcamp a long time ago.  Besides, the Mathcamp registeration is due June 2.  I don't think the PROMYS information will even come then...

[quote="Yasha"]PROMYS? C'mon, people, [url=http://www.mathcamp.org]MATHCAMP[/url] (CANADA/USA) is so much better. (Though I shouldn't be talking since I've never been to PROMYS).[/quote]

Well, my daughter went to Mathcamp, and my son goes to PROMYS (it will be his 3rd year), and I wouldn't say that one is "so much better" than another. They are just different... Mathcamp gives more breadth, PROMYS and Ross give more depth, and maybe it would be a good idea to try both.

The problem is, after the first year everybody becomes so patriotic, they don't even consider going to another camp. Well, mostly so.

[quote="marma103"]I wouldn't say that one is "so much better" than another. They are just different... Mathcamp gives more breadth, PROMYS and Ross give more depth, and maybe it would be a good idea to try both.[/quote]

I'm looking ahead to the next few years and hope to be a parent of a kid who tries both. The issue I have to look at is which program has the tighter age limits on the young side of the age range.

[quote="marma103"]The problem is, after the first year everybody becomes so patriotic, they don't even consider going to another camp. Well, mostly so.[/quote]

How true. I guess I am excessively patriotic to Mathcamp.

[quote="marma103"]Well, we were lucky enough to catch the last year when PROMYS still could take kids under 14.[/quote]

LOL PROMYS sent me an invitation thingy (not acceptance, but an invitation to apply), and I'm < 14.

Yup, got my letter of acceptance today.  But I've already decided to go to Mathcamp.


limit
inequalities
calculus
integration
function
real analysis
real analysis unsolved

This is from an exam:

Let $(X, \mathcal{A}, \mu)$ be a finite measure space and let $1 < p < \infty$. Suppose that $(f_{n})$ is a bounded sequence in $L^{p}(X,\mu)$ (that is, $\sup_{n}\|f_{n}\|_{p}< \infty$), such that $f_{n}$ converges in measure to $f$.
a) Show that $f \in L^{p}(X,\mu)$
b) Show that $\lim_{n}\|f_{n}-f\|_{1}= 0$

I can do a) but not b).
[hide="a)"]
Take a subsequence $f_{n_{k}}$ that is convergent almost everywhere and apply Fatou's lemma to $|f_{n_{k}}|^{p}$.
[/hide]

In a finite measure space, the $p$-norm dominates the $1$-norm. Now that you have $f\in L^{p}$, look at $g_{n}=f_{n}-f$; this is a bounded sequence in $L^{p}$, which converges in measure to zero.. By the same Fatou argument, we can find a subsequence $g_{n(k)}$ of the $g_{n}$ with $\|g_{n(k)}\|_{p}\to 0$ (it's a subsequence of the subsequence which is convergent a.e).

Now, the final trick is a lemma I know as Urysohn's principle:
Suppose $a_{n}$ is a sequence in some topological space, and there is an element $a$ such that every subsequence of the $a_{n}$ has a further subsequence convergent to $a$. Then $a_{n}$ converges to $a$.
The proof (by contrapositive) is remarkably easy; if the sequence doesn't converge to $a$, we can find a subsequence that avoids some neighborhood.

Since convergence in measure passes to subsequence, we can apply the whole argument to any subsequence of the $g_{n}$, and Urysohn's principle tells us that $\|f_{n}-f\|_{p}\to 0$.

I don't understand how you want to apply Fatou's lemma to get the subsequence $g_{n(k)}$. Sorry  :oops:

Uh, oops. That's the direction of the inequality we don't have. My claim is false- consider the counterexample $f_{1}(x)=\begin{cases}4x&0\le x\le \frac12\\ 4(1-x)&\frac12<x\le 1\\0&\text{otherwise}\end{cases}$, $f_{n}(x)=n^{\frac1p}f(nx)$.

The way that counterexample works is actually something of a hint to the real argument; the integrals can only go bad because the function is large on a small set, and $p$-norm boundedness makes that go to zero in the $q$-norm for all $q<p$.

All of this is true for $p=1$ as well.  In fact working in $L^{p}$ ($p>1$) instead of $L^{1}$ is kind of a throwaway as long as you know that $L^{p}\supset L^{q}$ if $p<q$.

Since the imbedding of $L^{p}(\mu)$ into $L^{1}(\mu)$, $1<p$, is continuous (because $\mu(X)<\infty$), it suffices to show that $f_{n}\to f$ in $L^{p}(\mu)$.

Take arbitrary $\epsilon>0$, then since $f_{n}\to f$ in measure, the measure of $X_\epsilon^{(n)}: = \{x \in X : \epsilon \le |f_{n}(x)-f(x)| \}$ tends to 0 as $n\to\infty$. Thus, $\sup \|f_{n}\|_{p}< \infty$ imples that
\begin{eqnarray*}\int_{X}|f_{n}-f|^{p}& = & \int_{X_\epsilon^{(n)}}|f_{n}-f|^{p}+\int_{X \setminus X_\epsilon^{(n)}}|f_{n}-f|^{p}\\ & \le & \mathrm{const}\cdot \mu(X_\epsilon^{(n)})+\epsilon^{p}\mu(X \setminus X_\epsilon^{(n)}) \\ & \le & o(1)+\epsilon^{p}\mu(X) \quad \text{ as }n \to \infty, \end{eqnarray*}
yielding that $\limsup_{n\to\infty}\int |f_{n}-f|^{p}\le \epsilon^{p}\mu(X)$. Since $\epsilon$ can be choosen arbitrarily small, we obtain what desired.

[quote="jmerry"]No, it doesn't. If you have $L^{p}$ boundedness, you get $L^{q}$ convergence for all $q<p$, but you don't get $L^{p}$ convergence. Look at the example I gave- it has nonzero constant $p$-norm, and converges pointwise and in measure to zero. In particular, you don't get $L^{1}$ convergence if $p=1$.[/quote]

I was thinking of a similar exercise I had done with slightly different hypotheses, I feel kind of silly now :(

volodja- you're wrong, and you're the third person to make that mistake in this thread. It doesn't necessarily converge in $L^{p}$, as my example showed.

An argument that actually works:
Choose $1\le q<p$, and $\epsilon>0$. For each $n$, let $U_{n}=\{x: |f_{n}(x)-f(x)|<\epsilon\}$ and $V_{n}=\{x: |f_{n}(x)-f(x)|\ge\epsilon\}$. Now, $\int |f-f_{n}|^{q}=\int_{U_{n}}|f-f_{n}|^{q}+\int_{V_{n}}|f-f_{n}|^{q}$.
On $U_{n}$, we have the estimate that $\int_{U_{n}}|f-f_{n}|^{p}\le \int |f-f_{n}|^{p}\le M$. Since $x^{\frac{p}{q}}$ is convex for $x>0$, $\frac1{\mu(U_{n})}\int_{U_{n}}\left(|f-f_{n}|^{q}\right)^{p/q}\ge \left(\frac1{\mu(U_{n})}\int_{U_{n}}|f-f_{n}|^{q}\right)^{p/q}$ by Jensen. Solving for the integral we want to estimate,
$\int_{U_{n}}|f-f_{n}|^{q}\le \mu(U_{n})^{1-\frac{q}{p}}\int_{U_{n}}|f-f_{n}|^{p}\le \mu(U_{n})^{1-\frac{q}{p}}M$.

As $n\to\infty$, $\mu(U_{n})$ tends to zero, and thus so does this part of the integral. The other part is less than $\epsilon^{q}\mu(X)$. Since $\epsilon$ can be made arbitrarily small, we have $L^{q}$ convergence.

Very nice, jmerry. I think for $q=1$ it is easier to motivate the last inequality as a special case of Hölder's inequality  (writing $h = h\cdot 1$). I was actually using exactly that type of inequality in my attempted solution of the problem, but I think that I too messed up because I for some reason assumed that there would be $L^{p}$ convergence.


Let $ p$ is a prime number
     $ k$ is a natural number
Prove that exist $ (m,n)$ different from $ (1,1)$
$ \frac{m^{p}\plus{}n^{p}}{2}\equal{} (\frac{m\plus{}n}{2})^{k}$ if and only if  k=p.

Because there is still no answer, I guess it will be better to continue the discussion [hide="here"]http://www.mathlinks.ro/viewtopic.php?p=495854#495854[/hide]  :wink:


calculus
integration
logarithms
calculus computations

How to evaluate the following integrals without using formula table?

a) $\int_{-1}^2\frac{x}{\sqrt{x+2}}dx$

b) $\int_1^2\frac1{x+x^2}dx$

[quote="jelly119"][quote="jmerry"]1: Substitute $u=x+2$. Split the two terms apart.

2: Factor the denominator. This one calls for partial fractions.

In general: try stuff until something works. Don't just give up when your first idea doesn't work.[/quote]

Thanks. I've figured them out.  Please help me check my answers. :)

a) $\frac23$

b) $\ln\frac43$[/quote]

 :)


inequalities
trigonometry
geometry proposed
geometry

Given a covex quadrilateral $ ABCD$, where $ P$ is the midpoint of side $ BC$ and $ \angle APC\equal{}\frac{2\pi}{3}$. Prove that $ AB\plus{}\frac{BC}{2}\plus{}CD>\equal{}DA$.

Have $ AB\plus{}\frac{BC}{2}\plus{}CD$
$ \equal{}\sqrt{AP^2\plus{}BP^2\minus{}AP*BP}\plus{}BP\plus{}CD$ by Cosine Law
$ \equal{}\sqrt{AP^2\plus{}CP^2\minus{}AP*CP}\plus{}CP\plus{}CD$ as $ CP\equal{}BP$
Now, $ AD\le \sqrt{AP^2\plus{}(PC\plus{}CD)^2\plus{}(PC\plus{}CD)AP}$, because $ AD$ is maximized for fixed $ A,B,C$ when $ P,C,D$ collinear in this order (they cannot be collinear though because have a convex quad), and then if $ P,C,D$ collinear use Cosine Law in $ \triangle{APD}$
Then, if we can prove that:
$ \sqrt{AP^2\plus{}CP^2\minus{}AP*CP}\plus{}CP\plus{}CD\ge \sqrt{AP^2\plus{}(PC\plus{}CD)^2\plus{}(PC\plus{}CD)AP}$ we will be done. Square both sides:
$ AP^2\plus{}CP^2\minus{}AP*PC\plus{}2(PC\plus{}CD)\sqrt{AP^2\plus{}CP^2\minus{}AP*PC}\plus{}(PC\plus{}CD)^2\ge AP^2\plus{}(PC\plus{}CD)^2\plus{}(PC\plus{}CD)AP$
$ \Leftrightarrow CP^2\plus{}AP*PC\plus{}2(PC\plus{}CD)\sqrt{AP^2\plus{}CP^2\minus{}AP*PC}\ge (PC\plus{}CD)AP$
$ \Leftrightarrow CP^2\plus{}(PC\plus{}CD)(2\sqrt{AP^2\plus{}CP^2\minus{}AP*PC}\minus{}AP)\ge AP*PC$
$ \Leftarrow CP^2\plus{}PC(2\sqrt{AP^2\plus{}CP^2\minus{}AP*PC}\minus{}AP)\ge AP*CP$
$ \Leftrightarrow 2\sqrt{AP^2\plus{}CP^2\minus{}AP*PC}\ge 2AP\minus{}CP$  square both sides
$ \Leftrightarrow 4AP^2\plus{}4CP^2\minus{}4AP*PC\ge 4AP^2\plus{}CP^2\minus{}4AP*CP$
$ \Leftrightarrow 3CP^2\ge 0$ which is true and the result follows.

$ AB \plus{} \frac {BC}{2} \plus{} CD\ge DA$

$ \leftrightarrow AB \plus{} \frac {BC}{2}\ge DA \minus{} CD$. Since $ BC \equal{} 2AP,CA\ge DA \minus{} CD$ it is suffice to show $ AB \plus{} BP\ge CA$.

Let $ X$ be opposite side of $ BP$ as $ A$ such $ \triangle XBP$ is equilateral. $ PC \equal{} PB \equal{} PX,\angle APC \equal{} \angle APX \equal{} 120$. Hence $ AC \equal{} AX$. In triangle $ ABX,AB \plus{} BX\ge AX$. Hence $ AB \plus{} BP\ge AC$.


geometry
inequalities
number theory
national olympiad

Recently Baltic Way 2004 have held place and I'm just curious if any of you were there ?

Anyway, here are the problems:

http://bw2004.olimpiados.lt/images/bwprobl1.JPG
http://bw2004.olimpiados.lt/images/bwprobl2.JPG

PDF file: http://bw2004.olimpiados.lt/?x=12

They aren't looking hard (though I've not read'em all)

Here are the links to all the Baltic Way 2004 problems posted on the artofproblemsolving.com:

problem 1   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20210
problem 2   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5443
problem 3   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20171
problem 4   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20172
problem 5   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20212
problem 6   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20204
problem 7   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20207
problem 8   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20203
problem 9   http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20213
problem 10 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20214
problem 11 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20217
problem 12 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20215
problem 13 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20218
problem 14 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20173
problem 15 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20220
problem 16 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20167
problem 17 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20169
problem 18 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20168
problem 19 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20166
problem 20 http://www.artofproblemsolving.com/Forum/viewtopic.php?t=20174

cheers  :D

In [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20173]another thread[/url], Orl wrote:
__________________________________________________________________

@Darij and others

BTW is Christian also a math linker ?  :)  Hmm. Obviously Christian kills the Combinatorics problems, Darij the geometry ones. I don't know about Peter Scholze. Maybe number theory. The algebra problems could be a joint job.

Regarding the participants. If Saint-Petersburg participates you also may have one half or one third of the Russian IMO team....  ;)  :)
__________________________________________________________________

[quote="orl"]BTW is Christian also a math linker ?  :)  Hmm. Obviously Christian kills the Combinatorics problems, Darij the geometry ones. I don't know about Peter Scholze. Maybe number theory. The algebra problems could be a joint job.[/quote]

Today I was talking with Christian on the phone. Traditionally he is a combinatorican (how is this word correctly spelt in English? :) ), but on the Baltic Way he did the number theory problems - he did them quite well, as the statistics show -, and he had few time left for the combinatorial ones. If I would be there I would have probably done geometry and inequalities very quickly (I knew problems 2, 3, 17 and 20 and have solved problems 4, 16, 18 and 19 in altogether 20 minutes; problem 1 took 20 minutes more). For the other problems, however, I guess I wouldn't have got more than the current team (it was already a great surprise to me that I was able to solve four number theory problems, 5-8).

I don't know whether Christian Sattler is on MathLinks; however, I believe (not 100% sure) that Julia Steinberg is on MathLinks ([url=http://www.mathlinks.ro/Forum/profile.php?mode=viewprofile&u=1079]Julia1[/url], nickname Julia before the merge). If I remember correctly, the only German "land"s participating in the Baltic Way are Hamburg, Bremen, Schleswig-Holstein and Mecklenburg-Vorpommern (the rather northern parts of Germany, you see), hence you won't see most of the famous olympiad students among the Baltic Way participants (well, this would mostly change if Bavaria would take part  :D ).

Anyway, I find the Baltic Way problems a wonderful training for every mathematical contest, even for the IMO though the problems proposed there are mostly much more difficult. This is mainly the reason why I have started solving them. I posted them at MathLinks together with their solutions since I wanted to open threads on the one hand, and post solutions on the other hand. This doesn't make me a Lagrangia yet  :D .

Returning to what you said about Christian, Peter and me, I think Peter is not just a number theoretician. He would certainly solve the geometry problems and probably also the combinatorial ones.

Finally, if somebody is interested, here are the problems arranged accordingly to their difficulty, i. e. sorted by the total number of points achieved by all students at these problems:

[code]problem    : achieved points

problem 14 : 12
problem  8 : 22
problem 19 : 25
problem 16 : 27
problem 11 : 29
problem 20 : 30
problem  1 : 34
problem  3 : 34
problem 12 : 35
problem 10 : 38
problem  9 : 42
problem  4 : 45
problem 13 : 47
problem  5 : 48
problem  7 : 49
problem 18 : 50
problem  2 : 51
problem 15 : 52
problem 17 : 57
problem  6 : 60
[/code]

As you see, problem 14 is indeed the hardest one.

  Darij

I wonder why so few teams solve problem 3 ? - it is relatively easy inequality... :maybe:


Hello,

I just read something about Ptolemy's theorem but I find myself mixed up maybe from some weird notations so I decided to see if you have a better explaination for it and maybe one basic example  :D

Here's a very easy problem:

An isosceles trapezoid has side lengths $ 4, 9, 9,$ and $ 10$. Compute the length of the diagonals of the trapezoid.

Solution:

[hide]Isosceles trapezoids are cyclic quadrilaterals with diagonals of the same length. This is because their opposite angles are supplementary; this is obvious. Therefore, $ d^2 \equal{} 4 \times 10 \plus{} 9\times 9 \equal{} 121$, so $ d \equal{} 11$.[/hide]

What is the cyclic quadratail and noncyclic?

A cyclic quadrilateral is defined as a quadrilateral that can be inscribed in a circle. In other words, you can draw a circle that touches the quadrilateral only at its four vertices.

By definition, any quadrilateral that is not cyclic is noncyclic.

Could you explain me what is the noncyclic quadrilateral

Because I found this in the internet and I'm not sure if this is the noncyclic thing plus I can't imagine this quadrilateral

[quote]Let Ma, Mb and Mc be the midpoints of sides a, b, and c respectively. We will apply Ptolemy's theorem to the quadrilateral AMcBC.[/quote]

A [b]cyclic[/b] quadrilateral has the property that it can be inscribed in a circle. A noncyclic quadilareral is the opposite, it cannot be inscribed in a quadrilateral.

The problem stated makes no sense, as A-Mc-B-C is not a quadrilateral.

It is a proof of $ a \plus{} b \plus{} c > ma \plus{} mb \plus{} mc$ which ma and mb and mc are the medians of a triangle and a,b,c are its sides

I'm totally confused did I miss understand this proof?

[quote]Let Ma, Mb and Mc be the midpoints of sides a, b, and c respectively.  We will apply Ptolemy's theorem to the quadrilateral AMcBC. This  gives:

     AMc.BC + AC.BMc > CMc.AB
     c/2.a + b.c/2 > mc.c       [divide by c]
     a/2 + b/2 > mc

In the same way we find that b/2 + c/2 > ma and a/2 + c/2 > mb. Adding these three inequalities we get the desired result:

     a + b + c > ma + mb + mc[/quote]

ptolemy can be generalized to arbitrary quadrilaterals...

AB*CD+BC*DA >= AC*BD

with equality iff ABCD is cyclic.

OK can you read the weird quadrilateral post in my previous post please?

Thanks Altheman


geometry
rectangle
3D geometry
circumcircle
calculus
pigeonhole principle
analytic geometry

Ok.... im in,
so who wants to discuss solutions?

Here's some outlines of my solutions:

1) Put 2 tiles together to make 2x5 rectangle, use 10 of these to make 10x10 square.

2) (k+1)(k-1)=47p^2, eventually find out p=7 is only possibility.

3) Factor 10!, find out you can split into 2 groups with products 720, with 7 as only leftover, so answer is 1.

4) Find smallest odd n for p=2, and p^2+n isn't prime, n=5.

5) Use deviation on expression to find either x=y=z or 2 are negatives of the other. Use deviation to find minimum of f(x)=3x^4-4x^3, Answer:-1.

6) Divide into 5 groups: A{2}, B{1,8}, C{18}, D{3-13, except 8}, E{14-22, except 18}. Al first picks 2. After, if Bob picks number n from B, Al picks 9-n. IF Bob picks number n form D, Al picks 16-n. If Bob picks number n from E, Al picks 36-n. I had to prove a bunch of stuff that I'm too lazy to write out.

7) List out perfect cubes < 2008, list possibilities, not too hard. Unordered: (12, 6, 4), (10, 10, 2)

8) Um...too much stuff to write out. Assigned variables to side lengths, proved AC=BD if angle A= angle B, proved AC doesn't equal BD if angle A doesn't equal angle B.

9) Didn't do.

10) Divide diagonals (chords of circumcircle) into 4 types, then count. Answer: 60.

Order of difficulty that I did: 1, 3, 2, 4, 7, 10, 8, 5, 6.

Is there a way to do #5 without calculus?
#10 was way to easy...

I think in question 4, Carbon57's answer is wrong!
Here's my solution:
p=2 then choose n=2
p=3 then choose n=1
when p>=5 then p can be written in the form:4k+1 or 4k+3
In both case choose n=1 we will have: p^2+n is divisible by 4
Therefore, when p=2 then n=2 is the smallest value of n
                       p>=3, then n=1 is the smallest value of n
In number 5, simply using AM-GM for 4-variable as following:
x^4+y^4+z^4+1>=4xyz
Then x^4+y^4+z^4-4xyz>=-1
The equality occurs when x=y=z=1
After submitting my solution, I realize I got #9 wrong! :(  Until now, I don't have time to think of it! I'll try to find time to think of it later :wink:

[quote="Carbon57"]For #4, I think it's talking about the smallest possible value for n in general, not just for different cases...[/quote]
Yup, I think you're right :oops: . Another mistake because of carelessness :o

9:

Notice that 1, 5, and 7 are relatively prime with every other number, so you can arrange them last.

No matter what, you can only have 16 even numbers, and 16 3's and 9's total. (I drew a diagram). Let the even numbers be a combination of 2's, 4's, and 8's (such that no one number appear $ \geq 11$ times), which have no prime factor other than 2. We place the evens, and have imposed no other restrictions, so we place the 3's and 9's (again, such that no one number appear $ \geq 11$ times).

We can clearly place the 1, 5, and 7 in some orientation, but there are 32 spots unfilled! By Pigeonhole Principle, some number must appear 11 times.

This probably needs more rigor...

For #8, I used coordinates  :D  :P . I was aware that it could be done with the law of cosines, but I kinda wanted to irritate whoever was reading it (in my defense, my other 9 solutions were less than 1/2 page long).

I think my solution to #10 is easier...you consider the spaces between the three vertices, obviously we need one to have at least 5 dots in it...then you can just balls-and-urns it.

My solution for #6 was as follows: start with 9. If Bo does 7, do 18 and you win. The only other option for Bo is 16...then do 20, Bo has to do 5, then do 11, Bo has to do 14, then do 2, Bo has to do 7, win by doing 18.

distracted523, you may want to rigorously justify why there can only be 16 evens and 16 multiples of 3. You can do this by dividing the 8x8 into 16 2x2's.

My solution to #8 was ALL direct-proof, basic geometry. I called the intersection $ E$ and drew $ \overline{CE}$ and $ \overline{DE}$. Then I drew $ \overline{EF}$ and $ \overline{EG}$, the perpendiculars to $ \overline{AC}$ and $ \overline{BD}$. I used several triangle congruences (i.e. $ \bigtriangleup AFE \cong \bigtriangleup DFE$, $ \bigtriangleup CGE \cong \bigtriangleup BGE$, $ \bigtriangleup AEC \cong \bigtriangleup DEB$) and stuff.

I tried to be simple about #10 too. I fixed the vertex of the obtuse angle, and found 6 triangles for each possible vertex (1 with obtuse angle $ 144 \deg$, 2 with $ 126 \deg$, 3 with $ 108 \deg$ I think). $ 6*10\equal{}\fbox{60}$

But yeah, I need to work on my rigor.

yeah, number 10, find each vertex, and get 6 triangles, and multiply by 10!
60.

pretty easy... i got 7 or 8 right

How do you use AM-GM on number 5 since they're not necessarily nonnegative real numbers?

yeah they are...anything to the 4th is nonnegative and 1 is nonnegative

Oh...I forgot that you had to add in the 1.


Η παρακάτω άσκηση είναι εμπνευσμένη από αυτή που πρότεινε ο Μανώλης

( και να πώ ότι μου άρεσε πολύ Μανώλη και δεν είναι τόσο απλή όσο είπες στην αρχή ) :)

Άσκηση:

[b]Να αποδείξετε ότι υπάρχει μοναδικό ορθογώνιο τρίγωνο το οποίο έχει πλευρές με μήκη ακεραίους και η ακτίνα του εγγεγραμμένου κύκλου είναι ίση με 1.[/b]

[hide]

[url]http://www.mathlinks.ro/viewtopic.php?t=186030[/url]

Στο παραπάνω φαίνεται ότι το μοναδικό τρίγωνο με αυτά τα στοιχεία είναι αυτό με πλευρές με μήκη 3,4,5 (είναι και ορθογώνιο).[/hide]

Ναι βρε Μανιέ  :)  :lol:  ... από τη στιγμή που ενεργείς στο τυχαίο τρίγωνο έχοντας σαν δεδομένα 1. Μήκη ακέραια.   2. Ρ=1 και καταλήγεις στη μοναδική τριάδα τιμών 3,4,5 την έχεις αποδείξει τη μοναδικότητα.
Ζητάω μια διαφορετική λύση για τη μοναδικότητα ( Με επιπλέον δεδομένο αυτή τη φορά ότι το τρίγωνο είναι ορθογώνιο )
Τωρα θα μου πείς γιατί ανέβασα καινούργιο post ? Τέλος πάντων, καμια φορά σε κάποιες περιπτώσεις η επιστήμη σηκώνει τα χέρια  :lol:


Farz khonid $n\in \mathbb{N}$ va bozorg tar az $3$ mibashad. Neshan dahid majmoe mokaabate hameye adade haye kamtar az $n$ ke nesbat be $n$ aval hastannd bar $n$ bakhshpazir ast.

[quote]In masale vagean zibast ,na?[/quote]

 :P

[quote="Ramtin Madani"]$(n,a)=1 \Longleftrightarrow (n,n-a)=1$
$n|a^3+(n-a)^3$
hala hamaro jam kon... :lol:[/quote]

Oh yes it's very easy and useful


i've been having a lot of trouble entering calculator programs into my ti-84 plus silver edition. I really don't know what the symbols in the instructions mean. can someone please give me an extremely detailed explantion about entering any calculator program in. Thanks.

note that i will need to know which buttons to push.

 :juggle:  :nhl:  :omighty:

Try this website:
http://www.atomiclearning.com/ti_84


Make a Venn Diagram or use [url=http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion]PIE[/url] (basically the same thing).

So we have $ f\plus{}b\minus{}x\equal{}35\implies30\plus{}15\minus{}x\equal{}35\implies x\equal{}\boxed{10}$.

A different way:
There are $ 35$ girls in total. Subtract the $ 30$ who play football and the $ 15$ who play basketball from the $ 35$ to get $ \minus{}10$. However, the girls that play both were subtracted twice. If we were to add them twice, we would get the total number of girls who play both. This implies that if we added them once, we would work our way back up to $ 0$, and so there must be $ \boxed{10}$ girls who play both.
This is essentially taking the equation $ (f\plus{}b)\minus{}x\equal{}35$ and rearranging it to $ 35\minus{}f\minus{}b\equal{}\minus{}x$, then recognizing that adding $ 2x$ to both sides would give you $ x$.

This is basically a set theory problem that works off of the intersection of two sets, namely of girls who play basket ball and/or soccer. There is a formula for this, and its

$ |A \cup B| \equal{} |A| \plus{} |B| \minus{} |A \cap B|$

This is for intersecting sets

Note that A and B are the sets, ie basketball and soccer

What is easiest way to solve this equation using algebra?

Reduce to variables then solve?

A basket contains 25 marbles colored either red, yellow or orange. Ten are either red or yellow and 23 are either yellow or orange. How many are yellow?

I see it this way:

There are 10 marbles that are either Red or Yellow. Subtract that from the total amount to get the number of Orange marbles.
25 - 10 = 15 Orange Marbles

To find the number of Yellow marbles, we must subtract the number of Orange marbles from number of marbles that are either Yellow or Orange.
23 -15 = [b]8 Yellow Marbles[/b]

Then, if we wanted to find the number of Red marbles:
10 - 8 = 2 Red Marbles

15 Orange + 8 Yellow + 2 Red = 25 Total Marbles

The same could go with your first problem, too.

Look at these questions again today I'm amazed at how obvious the solutions are. This realization males it clear that the approach i use in solving  logic puzzles is crucial, otherwise I miss the obvious.

Is there a way to prepare you mind to think laterally for these questions?

"It's as easy as PIE" -- Richard Rusczyk.

Just think algebraically. For that problem, I would have just subtracted 23 and 10 from 25 to get -8. Then, because we've subtracted the yellows twice, adding them once would bring us up to 0, and so there must be 8 yellow ones.

[quote="learningfish"]In a class of 35 girls, all of whom play either football, basketball or both, 30 play football and 15 play basket ball. How many play both?

What is the easiest way to solve this using algebra?

35=f+b+x[/quote]

[hide="In my humble opinion ..."]

As other posters have already written, the PIE is perhaps the best method to solve a problem like this.  Though I thought it might be instructive to see how the PIE comes about in this case.

Let F be the set of girls who play football.

Let B be the set of girls who play basketball.

Then $ F \cup B$ is the set of all girls, since no girl doesn't play at least one sport.

We want $ |F\cap B|$. (The absolute value bars indicate the size of the set.)

The size of a union is easy to calculate if the sets that comprise it are disjoint. Disjoint sets have no elements in common by definition. Therefore the size of a union of disjoint sets is the sum of the sizes of the sets themselves. It follows that

$ |F\cup B| \equal{} |F| \plus{} |B\diagdown F|$     (These sets are disjoint because the intersection $ B\cap F$ is already accounted for in $ F$.)

$ |B\diagdown F|$ is read, "The size of the set $ B$ without $ F$."

$ |B\diagdown F| \equal{} |B| \minus{}|B\cap F|$

It follows that $ |F\cup B| \equal{} |F| \plus{} |B| \minus{} |B\cap F|$,

from which we obtain (by algebra), $ |B\cap F| \equal{} |F| \plus{} |B| \minus{} |B\cup F|$.

$ B\cap F\, \& \, F\cap B \text{ commute.}$

Likewise, $ B\cup F\, \& \, F\cup B \text{ also commute.}$

$ \therefore |F\cap B| \equal{} |F| \plus{} |B| \minus{} |F\cup B|$

$ \implies |F\cap B| \equal{} 30 \plus{} 15 \minus{} 35 \equal{} 10$.[/hide]


trigonometry
algebra
polynomial
sum of roots
product of roots

Greeting people, any bright ideas on how to tackle this question?

[img]http://img147.imageshack.us/img147/3598/10213374kd8.png[/img]

A gun G fires a shell with initial speed, v at an angle Θ above the horizontal. The shell hits a target T which is a distance d horizontally from G and a distance h below it. Given there is only one possible value for Θ for which this occurs, show that [b]v² = g ( - h + ( h² + √( h² + d² ) ) [/b]

[/img]

i think this has to do with parametric equations, which i know nothing about, but you can look it up. I will too just to see.

[quote="thetruthisfake"] [b]v² = g ( - h + ( h² + √( h² + d² ) ) [/b]

[/img][/quote]

I'm confused, what are the second pair of parenthesis for? Can't it be just
v² = g ( - h + h² + √( h² + d² ) )?

This is a physics question. The $ x$-component of the initial velocity is $ v\cos\theta$ and the $ y$-component of the initial velocity is $ v\sin\theta$. Place $ G$ at $ (0,h)$ and place $ T$ at $ (d,0)$. If $ G$ is at $ (0,h)$ at time $ t\equal{}0$, then:

$ x(t)\equal{}vt\cos\theta$
$ y(t)\equal{}\minus{}\frac 12 gt^2\plus{}vt\sin\theta\plus{}h$.

When $ x$ is at point $ d$, we have $ d\equal{}vt\cos\theta \implies t\equal{}\frac d{v\cos\theta}$. Plug this into the equation for $ y$, and we have $ \minus{}\frac 12 g\frac{d^2}{v^2\cos^2\theta}\plus{}d\tan\theta\plus{}h\equal{}0$.

[quote="Yongyi781"]This is a physics question. The $ x$-component of the initial velocity is $ v\cos\theta$ and the $ y$-component of the initial velocity is $ v\sin\theta$. Place $ G$ at $ (0,h)$ and place $ T$ at $ (d,0)$. If $ G$ is at $ (0,h)$ at time $ t \equal{} 0$, then:

$ x(t) \equal{} vt\cos\theta$
$ y(t) \equal{} \minus{} \frac 12 gt^2 \plus{} vt\sin\theta \plus{} h$.

When $ x$ is at point $ d$, we have $ d \equal{} vt\cos\theta \implies t \equal{} \frac d{v\cos\theta}$. Plug this into the equation for $ y$, and we have $ \minus{} \frac 12 g\frac {d^2}{v^2\cos^2\theta} \plus{} d\tan\theta \plus{} h \equal{} 0$.[/quote]

I continue this, rewrite the equation in $ \theta$, by $ 1 \plus{} \tan ^ 2 \theta \equal{} \frac {1}{\cos ^ 2 \theta}$,

we have $ \frac {gd^2}{2v^2}(1 \plus{} \tan ^ 2 \theta) \minus{} d\tan \theta \minus{} h \equal{} 0$

$ \Longleftrightarrow (d\tan \theta)^2 \minus{} \frac {2v^2}{g}(d\tan \theta) \plus{} d^2 \minus{} \frac {2h}{g}v^2 \equal{} 0$

Since $ 0 < \theta < \frac {\pi}{2},\ \tan \theta$ must be some positive real numbers, thus $ D/4\geq 0\Longleftrightarrow \left(\frac {v^2}{g}\right)^2 \minus{} \left(d^2 \minus{} \frac {2h}{g}v^2\right)\geq 0$ and

the sum of roots $ \equal{} \frac {2v^2}{g} > 0$ and the product of roots $ \equal{} d^2 \minus{} \frac {2h}{g}v^2\geq0$

$ \Longleftrightarrow \frac {v^2}{g} \geq \minus{} h \plus{} \sqrt {h^2 \plus{} d^2}$ and $ \frac{v^2}{g}\leq \frac{d^2}{2h}$, yielding $ \boxed{g( \minus{} h \plus{} \sqrt {h^2 \plus{} d^2})\leq v^2 \leq \frac {g}{2h}d^2}.$

[hide="Some thoughts..."]The horizontal component is unaffected.

Hence,

\begin{eqnarray}
tv\cos\theta = d \\
-h = tv\sin\theta - \frac{1}{2}gt^2
\end{eqnarray}

Eliminate $ \theta$ and $ t$. The problem is more of mathematical type and hence tedious.[/hide]


vector
analytic geometry
trigonometry
geometry
similar triangles

I was told that this is a famous problem. I need some help solving it, and I warn that it might require some knowledge of physics and differential equations. (Please tell me if the problem should be posted in a different forum.) Any ideas?

Problem:

A hound chases a fox which flees in a straight line with a constant velocity of u. The hound accelerates in the direction of the fox starting with a velocity v, and throughout the chase it maintains a speed of |v|. At time t = 0, u is perpendicular to v, and the distance between the fox and the hound is L.

1. What is the acceleration of the hound when t = 0?

2. How long does it take for the hound to reach the fox?

Let the two start with a distance of d between them.

[hide="1"] I believe it's uv/d...simply because after one second, the fox would be $u*s$ from its starting point, and we can then draw similar triangles with the vectors given to get the length of the parallel acceleration vector (because the fox doesn't change speed, the acceleration vector must always be perpendicular to the velocity vector)[/hide]

For part 2, or to generally solve the problem, it seems best to view everything from the fox's point of view and plot the hound's path in polar coordinates.  Given the angle the distance vector between them makes with the path of the fox, and the length of the distance vector itself, we can solve for everything.  So it seems to be a series of differential equations linking three pieces of information.  I'm now too lazy to work it all out.

there seems to be something wrong with this problem:

switch to the frame of the fox, then the velocity of the hound initially is u-v. notice that the acceleration is central. the vector, r x v is conserved. or, the angular momentum is conserved. the requirement is that the hound maintain a speed of |v|, so that the velocity of the hound is finite.

Assume that the angular momentum of the hound is non-zero initially, then the hound cannot reach the fox since if it does, the angular momentum will be zero (no longer conserved, a contradiction). A conclusion is that after some time, the hound will end up "orbiting" the fox (assume that the fox is a point).

the other possibility is that the hound travels in a straight line (zero angular momentum initially) and reaches the fox, but then it will be the trivial case.

What I suspect is that the actual question maybe:
a hound chasing a fox such that the velocity always points in the direction of the fox. That would make much more sense.

[quote="Tim_Lou"]
What I suspect is that the actual question maybe:
a hound chasing a fox such that the velocity always points in the direction of the fox. That would make much more sense.[/quote]

That's how I interpreted it.  Hoorah for not fully reading the question!  :D

well, he typed,
"The hound accelerates in the direction of the fox starting with a velocity v"

which to me implies an acceleration vector pointing in the direction of the fox. this is very different from letting the velocity vector pointing in the direction of the fox...

as for the solution... yeah, you probably wanna switch to the fox's frame, so that the the velocity of the hound is:
$\vec{r}'=-\vec{u}+v\vec{r}/r$

where v is the speed of the hound in the other frame. well, switch to polar coordinate and you can get rid of the r:
for x coordinate:
$-r\sin\theta*\omega+r'\cos\theta=-u+v\cos\theta$

and y coordinate:
$r\cos\theta*\omega+r'\sin\theta=v\sin\theta$

well, you can separate the variables in the second equation... well you can go from there...

If it is a famous problem, he undoubtedly means that the velocity points that way, since this is the infamous tractrix.

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

It actually isn't the tractrix...the leash is not constant length, but the dog is going at constant speed...

And to Tim, I was making a joke, ie for once, when I misread the question, I was actually right...

No, I don't mean that the velocity of the hound always points in the direction of the fox. If this were so, what would be the challenge of the problem? The hound starts with its velocity vector perpendicular to the velocity vector of the fox and accelerates always in the direction of the fox while keeping |v| (its speed) constant.

well, it might still be an interesting problem if it was the acceleration vector that points to the fox... hmmm, let's see the period of the orbits.... :huh:


geometry
geometric transformation
reflection

You have a 4x4 grid. (16 squares). all the squares have to be coloured either [color=red]red[/color], [color=blue]blue[/color] or [color=green]green[/color]. there are 4 rows, and 4 columms.
Givens:
1) All the boxes in column1 are the same colour.
2) all the boxes in row 1 are [color=red]red[/color].
3) there has to be an even number of [color=blue]blue[/color]
4) there has to be an even number of [color=green]green[/color]
5)[color=green]Green[/color] cannot have a common side or vertex with [color=red]red[/color].

What is the largest number of [color=green]green[/color] squares that u can have??

[hide]HINT: try to make a visual on a piece of paper[/hide]

[hide]

I got 2 also, here's my arrangement:

[color=red]R R R R[/color]
[color=red]R R[/color] [color=blue]B B[/color]
[color=red]R [/color][color=blue]B B B[/color]
[color=red]R[/color] [color=blue]B[/color] [color=green]G G[/color]

We have reds on the first column and row, so we need at least 5 blues to keep the greens from being adjacent. But we need an even number of blues (6), so we end up with 2 greens.[/hide]

there must be an even number of reds so plus another R, start with

RRRR
R
R
R

RRRR
RBRB
RBB
RB

RRRR
RRBB
RBB
RB

those two cases gives two G since all even

RRRR
RBBR
RBBR
RBBB

RRRR
RBBB
RBRB
RBBB

RRRR
RBBB
RBBR
RBXB

RRRR
RBBB
RBBB
RBBR

obviously none of those work...all cases covered (using reflection, etc)

This was a question on a PSAT in 2004 believe and I positive the answer is 2


If the radius of a circle is increased $100\%$, the area is increased:

$\textbf{(A)}\ 100\% \qquad
\textbf{(B)}\ 200\% \qquad
\textbf{(C)}\ 300\% \qquad
\textbf{(D)}\ 400\% \qquad
\textbf{(E)}\ \text{By none of these}$

[hide="solution"]If the radius is increased by 100%, that means it is doubled.
$ \pi r^{2}$ is the area of a circle.
Now if the radius is doubled
$ \pi (2r)^{2} \equal{} 4\pi r^{2}$
So the area is increased by 400%.

D. [/hide]

Edit: Thanks for the correction Pancho
\facepalm

[quote="Pancho Bigotes"]Hmmm. . . the original area was $ \pi r^2$, and the new area is $ 4\pi r^2$, so it was increased by $ 3\pi r^2$, shouldn't it then be a $ 300%
$ increase?[/quote]
Yeah, this question was on some Canadian contest too, missed it the first time.

[quote=Pancho Bigotes]Hmmm. . . the original area was $ \pi r^2$, and the new area is $ 4\pi r^2$, so it was increased by $ 3\pi r^2$, shouldn't it then be a $ 300%
$ increase?[/quote]

You are right, silly me  :wallbash_red:

This post is almost a decade old. Please don't bump it for the sake of 5 words.


complex analysis
function
trigonometry
geometry
rectangle
complex numbers
complex analysis unsolved

[quote="guiness"]This problem is from Schaum's (listed under the Rouche's Theorem heading):

Prove that z tan z = a, with a > 0 has only real roots.[/quote]
Rouché's Theorem applies to holomorphic functions, and tan(z) has singularities at odd multiples of ½π. So it seems like a good idea to rewrite the equation as $ z\sin z \minus{} a\cos z \equal{} 0$.

Now suppose that $ \text{re}(z) \equal{} \pm(n\plus{}\textstyle\frac12)\pi$ for some positive integer n. Then $ \sin z \equal{} \pm\cosh y$ and $ \cos z \equal{} \pm i\sinh y$, where y=im(z). If in addition we choose n large enough so that (n+½)π>a then $ |z\sin z| > |a\cos z|$.

What I want to do is to apply Rouché's Theorem to the region $ \Gamma \equal{} \{z: |\text{re}(z)|\leqslant (n\plus{}\textstyle\frac12)\pi\}$. This region is unbounded, and I'm not 100% sure that the therem applies. If it does, then $ z\sin z \minus{} a\cos z \equal{} 0$ has the same number of zeros as $ z\sin z \equal{} 0$ in Γ, namely 2(n+1). But it's easy to see that $ z\sin z \minus{} a\cos z \equal{} 0$ has 2(n+1) real roots in Γ. Hence all the roots must be real.

[quote="MysticTerminator"]I don't think Rouché's theorem should apply in unbounded regions -- consider, say, $ 2 \minus{} e^{iz^2}$ in the second quadrant; $ 2$ certainly doesn't have any roots but the difference does. Fortunately, for this particular case, I think the idea works -- just close off the rectangles by drawing sufficiently high and low horizontal lines to close off the rectangle. If they're drawn with the imaginary part fixed at $ \pm y$, the sine and cosine functions are both just essentially $ e^y/2$ so the factor of $ z$ as opposed to the constant factor of $ a$ should win.[/quote]
Thanks - I completely agree. (In fact, I came to the same conclusions, thinking about the problem overnight.)


MATHCOUNTS college Harvard Yale Princeton

The Manhattan chapter of MathCounts (a middle-school contest) held its competition last month.  There were 17 schools that participated, more than in previous years.  Each school can bring 4 team members and up to 4 individuals.  I was there as coach of Dalton's team (our first time participating).  Here are the scores of the top 6 schools (out of 66 points):

[code]
  1.  Hunter (60.25)
  2.  Lab School (58.75)
  3.  Dalton (56.50)
  4.  Salk (53.00)
  5.  Baruch (51.75)
  6.  Park East (49.75)
[/code]

  The top 3 teams (from Hunter, Lab School, and Dalton) move on to the state competition, which will be held next week on March 18 in upstate NY (RPI).

  Below is the list of the top 10 students.  I listed  the individual scores (out of 44) only when I know them.

[code]
  1.  Alex I. (Dalton) 44
  2.  Parker Z. (Dalton individual) 44
  3.  Taylor S. (Salk)
  4.  Paul H. (Hunter)
  5.  Evan C. (Lab School)
  6.  Emma Z. (Lab School)
  7.  Michael T. (Hunter)
  8.  Daniel Z. (Hunter)
  9.  Jesse S. (Dalton) 40
 10.  Yichi Z. (Salk) 39
[/code]

  In addition to the top 3 teams, Parker Z. (Dalton) and Taylor S. (Salk) will go to the state competition as high-scoring individuals.

  We at Dalton were happy to have the two highest individual scores.  Unfortunately, I did not place Parker on our team of 4, and so he did not contribute to our team score.

  I'll report the results of the state competition sometime next week.

[quote="Ravi B"]... I did not place Parker on our team of 4, and so he did not contribute to our team score ...[/quote]Actually it's a great move -- now you have 5 students in State instead of 4!  :D

That's true.  It was an, um, carefully designed, strategic move on my part...

I don't suppose you know if the Brooklyn chapter has more than 1 school in it nowadays, do you?

According to the MathCounts web site, four schools from Brooklyn registered: Berkeley Carroll, Brooklyn Amity, Cunningham IS 234, and Saint Ann's.  (The Rockward Park School from Howard Beach is listed as belonging to Brooklyn/Queens; I'm not sure which chapter they participated in.)  I imagine that three of these schools will go to the state competition.  If I see them listed in the state competition, I will let you know.

Did you participate in MathCounts back in the day?  How did you do?

I would love to see more schools from NYC participate in MathCounts.  From what I have observed, it seems to be an excellent stepping stone to higher math.  My daughter Meena, in 6th grade, is on Dalton's team; since making the team, she has been bugging me to learn more and more math.  It's quite a switch from when I used to nudge her to study math.

Yes, I went to St. Anns back in the day.  One year, we were the only team in the Brooklyn chapter, and the other year there was one other team (I don't remember from what school).

I believe we placed third in the state in 7th grade and 2nd (by a very slim margin -- perhaps one point?) in 8th grade.  (Our team+alternate was the same 5 students both years -- we were a really improbable collection of very mathy people in one class, and we conveniently overlapped with a teacher who was very into the competitions.  We actually won second place one semester in the Junior division of the NYCIML -- Mr. Linker introduced us as Saint Anns High School, and people laughed because we were this incredibly scrawny-looking group of kids, but then he re-announced us as Saint Anns Middle School and there wasn't any more laughter :-).)  I placed top 10 in 7th grade and 2nd in 8th grade.  (I beat out [url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=1210]Alison Miller[/url] on tiebreakers.)  I don't remember how we did at Nationals -- my individual ranking was 38th, I think.

Of my school team, two are at Harvard, one at Yale, one at Princeton and one at Michigan.  Of the state team, three of us are at Harvard.  I don't know what's happened to the 4th, although Alison should -- he was on her homeschool team.

Helping to start or to coach a Mathcounts team is definitely one of the things I'd like to do eventually.

Yes, Alison's mom was telling me that MathCounts played a key role in inspiring Alison.

Coaching a MathCounts team is cool -- you get to influence kids before it's too late.  Next year, my daughter will be transferring to Hunter, so I probably won't coach a team then, but I hope to be involved in other ways.  I'm hoping to have started enough momentum at Dalton for parents to insist on a MathCounts team in the coming years.


The value of $ x \plus{} x(x^x)$ when $ x \equal{} 2$ is:

$ \textbf{(A)}\ 10 \qquad\textbf{(B)}\ 16 \qquad\textbf{(C)}\ 18 \qquad\textbf{(D)}\ 36 \qquad\textbf{(E)}\ 64$

hello, we have $ x\plus{}x \cdot x^x\equal{}x\plus{}x^{x\plus{}1}$, inserting $ x\equal{}2$ we get $ 2\plus{}2^3\equal{}2\plus{}8\equal{}10$, so the answer is $ \text{A}$.
Sonnhard.

Inside the parentheses, we get $ 2^{2}\equal{}4$.  Now we multiply by 2 and get 8.  Finally, we add 2 and get 10.  Just follow the rules of PEMDAS (Parentheses, Exponent, Multiplication, Division, Addition, Subtraction).


Let $ f(n)=$ $ \frac{n-1}{2}$ if $ n$ is odd, or $ n+1$ if $ n$ is even. Let $ g(n+1,k)=f(g(n,k))$ for $ n\ge 1$ and $ g(1,k)=k$. How many integers $ 0\le k\le 1023$ are there such that $ g(2007,k)=0$?

[hide="hint"]Consider the binary representation of $ n$.[/hide]


what was your "MATHCOUNTS Index"? That is, what is your state score plus chapter score?

my index was terrible, i did okay at states, but i bombed chapter, so my index was 32+27=59.

oh yeah, I guess I should mention mine. I had 40 at chapter and 36 at state, so I had 76.

39 at Chapter
30 at State

so
69 index

Edit:
Shouldn't you include school (i know some don't take it)
just an idea

well i know that most people didn't take it...so it would kind of be an unfair advantage. Everyone took chapter and state though :)

Well the don't lie thing is kinda wrong because Kevin Chen did get a 92.

But I got 45+42=87.

omg, that's REALLY good (i just used "REALLY" twice within 2 min :D )

umm...well, mine's really bad

but if things went how they should've gone...about a 70-75

note that no one got like certain problems right in my state...kinda awkward...they must've screwed up on grading

jorian

lol blaming the graders is a terrible excuse.no offense.

isn't Kevin the smart guy who was the only 7th grader in the top 12 last year?

yep

(not taking a chance, so i'm adding some words)

[quote="kyyuanmathcount"]Shouldn't we weight it somehow?  Like... have chapter count as 3/4 and state as full?  After all, a point on chapter is easier to get than a point on state.

Anyways, Chapter 45 and state 42 so 87 under the kstan system.[/quote]

oh dude we have the exact same index
except you would have beaten me on both with tiebreakers

but your usamo index owned mine so yeah

39 chapter + 32 (ouch) state = $\boxed{71}$ (used a calculator just now  :lol: )

lol how can you be upset with that? are you just kidding?

I think he's saying that he made a dumb mistake

[quote="cognos599"]33+25=58  :([/quote]

cognos599, what r u so mad about?  my index was 59 and im going to nats.  then again, i bombed chapter, and my states wasn't much better.

80th Post  :agent:  :wow: :agent:

[quote="b-flat"]I don't know what it is with me and getting nervous at State (I scored 6 points be below my average), but I guess my index would be 43+37=80.

@leetspeak, what was your usamo index?[/quote]

His was 120+70=190.  :D  Mine was 141+110=251 :D.

Oh, haha, yeah, I probably did beat you on chapter tiebreaker because I got #5 wrong -__-.  And state I have to un-beast the target so I can win tiebreakers.

[quote="footballrocks41237"]I probably have the lowest index out of people going to Nationals.

21+32=53 LOL :mad:[/quote]

I bet not, ur just below me, and we got the same state score.  i know there r people out there that got lower than 53 as an index that r going to nats.

[quote="magixter"]What if your state didn't have chapters, like AR?[/quote]

Then I guess just say you didn't have chapters but make an "index" as 2*state score.

i don't know if this is a repeat, but i got 71. too bad it was my last year...

That would be great...I pwned my chapter with 2nd place written, 3rd CD, and I beat 5th place in my state at chapters (they had 6th at chapters, 5th at state o_O), but couldn't concentrate on the state sprint to due to nervousness and made really stupid mistakes (wrote 11 instead of 121 after wasting 5min with 100x + 10y + z and 10x  + z stuff until I thought, "what number gets divided by 11 (242/22) when the middle digit is removed" and memorizing the squares almost paid off because thinking 11 made me think of 121) and got a 26.

Even then, I'm in a tough state, and didn't prepare that well; I'd never really thought of the possibility of doing very well at states (if I even got there, my school never had before) until I did great at chapters and qualifyied in 2 ways for state.

[color=blue][b]My index:  31 + 26 = 57[/color][/b]

So I beat Watr at state, but his index was way better than mine.  By the way, do you know what 1st place's index is? (probably going to get boosted 32-46 pts by nats)

[quote="Walk Around The River"][quote="rd5493"]cognos599, what r u so mad about?  my index was 59 and im going to nats.  then again, i bombed chapter, and my states wasn't much better.[/quote]
man i had a 65 and i didnt make it.  i wish they used index scores.[/quote]

I had 78 and didn't make it-I wish I could switch my chapter and state scores-my team made states...

perfect 628, what state are you from?
I got a 78 as well and didn't qualify.
My chapter score was 41 (2nd place) and state was 37 (6th place).

umm...didn't you get lower in all of ca though?

cause ube got 6th with a 40.

jorian

Um excuse me but why do you refer to him as "ube?"
And no he didn't, he was sixth in Northern California. I believe math333 would be Evan O'Dorney.

sorry, too lazy to write ubemaya

oh, so math333 got 6th in s. cali and ube (sorry again, it just takes wayyyyyyyyyyyyyyy too long to type) got 6th in n. cali. now this makes sense.

ahh, i don't have 1337 posts anymore

jorian


let $ a,b,c>0$,$ a\plus{}b\plus{}c\equal{}1$;put $ abc\equal{}r$,$ ab\plus{}bc\plus{}ac\equal{}q$,prove that:
$ r \le \frac{{{q^2}(1 \minus{} q)}}{{2(2 \minus{} 3q)}}$


[quote="mathematikos"]Let $ a,b\in]0,\infty[$
find the minimum of:
$ A = (1 + \frac {a}{b})^{k} + (1 + \frac {b}{a})^{k}$
$ k\in{IN}$
I hope that be interesting  :)[/quote]
The minimum is $ 2^{k+1}.$ It's Austria 2001.


function
algebra
polynomial
complex analysis
complex analysis unsolved

Let $f(z)$ be entire function which is not linear, i mean that $f(z)$ we can't be represented as $f(z)=z+b$. Is it true that there exists $z \in \mathbb{C}$ such that $f(f(z))=z$ ?

If $f$ is a polynomial, everything is clear. If not, look at the function $g(z)=(f(f(z))-z)/(f(z)-z)$ and use Picard theorem. This function will be constant and from here it is easy to establish contradiction. By the way, I have a question (with no answer): can we caracterize the entire functions $f$ such that $f(f(z))=z$ implies $f(z)=z$?

Sorry for bumping this old topic, but how to proceed after knowing that $g$ is constant?

[quote="GreenKeeper"]Hint: show that $f$ must be injective.[/quote]
Excellent, thanks!


ratio
trigonometry
geometry proposed
geometry

[quote="Bugi"]In isosceles right triangle $ ABC$ a circle is inscribed. Let $ CD$ be the hypotenuse height ($ D\in AB$), and let $ P$ be the intersection of inscribed circle and height $ CD$. In which ratio does the circle divide segment $ AP$?[/quote]

[geogebra]cffcb969f3bf81e648aa52bfc167ff7ec1607a69[/geogebra]

We have $ r\equal{}\frac{[\triangle ABC]}{s}\equal{}\frac{x}{2\plus{}\sqrt{2}}$, where $ x\equal{}AC$ and $ s$ is the semiperimeter of $ \triangle ABC$.
WLOG, we assume that $ x\equal{}2 \sqrt{2}\plus{}2$, so $ P\equal{}(\minus{}1,\minus{}1)$, $ A\equal{}(\minus{}\sqrt{2},2 \sqrt{2}\plus{}2\minus{}\sqrt{2})\equal{}(\minus{}\sqrt{2}, \sqrt{2}\plus{}2)$

So $ PA\equal{}\sqrt{14\plus{}4\sqrt{2}}$

Let $ AP$ intersect the circle at $ Q$, then $ (AQ)(AP)\equal{}(\sqrt{2}\plus{}2)^2$

So $ \frac{AQ}{AP}\equal{}\frac{(\sqrt{2}\plus{}2)^2}{14\plus{}4\sqrt{2}}\equal{}\frac{3\plus{}2\sqrt{2}}{7\plus{}2\sqrt{2}}\equal{}\frac{13\plus{}8\sqrt{2}}{41}$

By $ O$ denote $ AP$ and $ \omega$ intersection point. Let also be $ \angle PAD = \alpha$. Since $ \triangle ABC$ is isosceles, $ PD$ is a diameter of $ \omega$. Hence $ \angle POD = 90 ^{\circ}$.
We must find $ \frac{AO}{OP}$: \[ \frac{AO}{OP} =  \frac{AD \cos{\alpha}}{PD \sin{\alpha}} = \cot^2{\alpha}\] So now we must find ${ \cot{\alpha}}$. \[ \cot{\alpha} = \frac{AD}{PD}\] But \[ PD = 2r = 2\cdot \frac{S}{p} = 2 \cdot \frac{AD^2}{AD(1+\sqrt{2})} = 2 \cdot \frac{AD}{1+\sqrt{2}}\] So \[ \cot{\alpha} = \frac{AD \cdot (1+\sqrt{2})}{AD \cdot 2} = \frac{1+\sqrt{2}}{2}\] and \[ \cot^2{\alpha} = \left( \frac{1+\sqrt{2}}{2} \right) ^2 = \frac{3+2\sqrt{2}}{4} = \frac{AO}{OP}\] $ q.e.d.$

[img]http://img.flashtux.org/upload/img1328ea23157cxb159bb5a.png[/img]

[hide="Solution in short notes"]

Very similar as jevgeniy's but you can use similarity of triangles or power of the point A.
Using same marks as jevgeniy:
$ AP\cdot AO\equal{}AD^2$
$ AP^2\equal{}AD^2\plus{}DP^2$

In every right triangle, $ r\equal{}\frac{a\plus{}b\minus{}c}2$, if $ c$ is the hypotenuse.

From these 3 facts, the problem is easily solved. [/hide]


[quote="litongyang"]What does $ P_{n \plus{} 5}$ mean?[/quote]

Oh! in my country $ P_{n\plus{}5} \equal{} (n\plus{}5)!$

It is equal to $ (n \plus{} 5)(n \plus{} 4)(n \minus{} k \plus{} 1)$ is not bigger than 60.


Let AOB be a right triangle with <O=90º. Choose a point F lying in AB and a point I lying in BO. Choose a point H inside the triangle. If F, H, I are collinear, show that :

   sin <A= ((HO^4)(IB^3)/(AH^4)(OI)(BF^2)) - ((HO)(IB)/(FA)(AH))

[quote="Pul de Algodoncito"]Let $ AOB$ be a triangle with $ OA\perp OB$. Choose a point $ F\in (AB)$, a point $ I\in (OB)$ and a point $ H$ inside the triangle. If $ F, H, I$ are collinear, show that $ \sin \widehat {OAB} \equal{} \frac {HO^4\cdot IB^3}{AH^4\cdot OI\cdot BF^2} \minus{} \frac {HO\cdot IB}{FA\cdot AH}\ .$[/quote]
I think that I wrote correctly what you wrote ugly, without LaTeX !

[b]Please, learn quickly LaTeX ![/b]

Sorry for the inconvenience. I'm on my way, it's just that it is hard for me to start.


function
real analysis
real analysis unsolved

Suppose $A,B\subset (0,1)$ are measurable subsets. Define $A\otimes B=\{0,a_1b_1a_2b_2...\mid 0,a_1a_2a_3...\in A,\,0,b_1b_2b_3...\in B\}$, where $0,a_1a_2a_3...$ and $0,b_1b_2b_3...$ are binary representations of real numbers.

The aim of this post is to discover the properties of $A\otimes B$.

1. Is $A\otimes B$ is a mesuarable set?
2. If $A\otimes B$ is a measurable set, then what can we say about $|A\otimes B|$ in terms of $|A|$ and $|B|$?

One little lemma:

Suppose $A=[a\cdot 2^{-j},(a+1)\cdot 2^{-j}]$ and $B=[b\cdot 2^{-j},(b+1)\cdot 2^{-j}]$ for integers $a$ and $b$, then

$A\otimes B=[c\cdot 2^{-j},(c+1)\cdot 2^{-2j}]$ where $c$ is an integer which is a function of $a$ and $b$. (Well, the sets are equal except possibly for some dyadic rational points, but there are at most countably many of those.)

In this particular case, $|A\otimes B|=|A|\cdot |B|.$

A probablistic interpretation:

Suppose $A$ and $B$ are (measurable) events that could occur in an infinite sequence of coin flips.

Perform an infinite sequence of (fair, independent) coin flips. Observer #1 transcribes only the results of the odd numbered flips, and observer #2 transcribes only the results of the even numbered flips.

Let $A'$ be the event that $A$ occurs within the sequence of flips transcribed by Observer #1, considered as an infinite sequence of coin flips in its own right. Similarly, $B'$ is the event that $B$ occurs with the sequence of flips transcribed by Observer #2.

Claims: $P(A')=P(A),\,P(B')=P(B),$ and $A'$ and $B'$ are independent.


calculus
derivative
inequalities solved
inequalities

Prove that  (3cos(x)/(1+2cos(x))<sin(x)/x<3/(4-cos(x)) for x in (0,pi/2)

I shall resort to calculus ^^ Not messy though.

For the right side, we must prove f(x) = 6x + sin 2x - 8sin x > 0.

The derivative is 6 + 2cos 2x - 8cos x = 4(t - 1)^2 \geq 0 where t = cos x. So f(x) is increasing in the interval (0, pi/2). f(0) = 0, hence result.

For the left side, we must prove (1 + 2cos x)/cos x > 3x/sin x or

f(x) = sin x/cos x + 2sin x - 3x > 0.

f'(x) = 1/cos^2 x + 2cos x - 3.

Proving f'(x) > 0 is equivalent to proving

2t^3 - 3t^2 + 1 > 0, where t = cos x.

But 2t^3 - 3t^2 + 1 = (2t + 1)(t - 1)^2, which is positive for x in the interval (0, pi/2).

Thus, f(x) is increasing in the interval (0, pi/2). f(0) = 0, hence result.


a) For $ x,y,z\in\mathbb{R}^\plus{}, x\neq y\plus{}z,y\neq z\plus{}x, z\neq x\plus{}y$, prove that:
$ 4(x\plus{}y\plus{}z)\plus{}5\sum_{cyc} \frac{x(x\plus{}y)^2}{(x\minus{}y\minus{}z)(3y\plus{}z\plus{}x)}\geq0$
b) If $ x,y,z\in\mathbb{R}$, does the inequality always stand? If yes, prove it. If no, give me a counterexample.

it fails for $ (x,y,z) \geq 0$ , take $ x\equal{}3$ , $ y\equal{}4$ , $ z\equal{}5$  :wink:


geometry
rectangle
inequalities
triangle inequality

I Understand the triangle inequality, but i dont seem to understand how its different with a quadrilateral(ok yes it isnt). When in the book AOPS v1, pg 182 #372  "three sides of a quadrilateral have lenghts 1, 2, and 5. The fourth side is an integer x, What is the sum of all the possible values of x?"
 OK well i figured that "any" quadrilateral could be split into two triangles. I drew a pictures, only to determine non-rigorously that the orientation didnt/doesnt matter. So then i did the very simple inequalities and determined that the range of values were 4-6 giving a sum of 15. Which is 1 off in both directions it seems. I'm frustrated.

Rather than blasting out a bunch of equations, here is a from-the-gut explanation of how the bounds will work. I leave it to you to formalize this logic. Consider it this way:

You begin by walking 5 miles due east. Then you pick a direction and travel 1 mile in that direction, and pick another direction and travel 2 miles in that direction. [i]What are the minimum and maximum distances you could be from where you started?[/i]

The maximum distance is if you keep going in the same direction - 5 miles east, another 1 mile east, another 2 miles east, for a total of 8 miles from where you started. That's as far as you can possibly be. However, your overall path, if you were to go back, would not be a quadrilateral, but a line segment, so we can't include 8. So, x < 8.

The minimum distance? Let's see - you're already 5 miles east, and you have 3 miles left to go. Your best bet is to start walking back to where you started. 5-3=2, so 2 miles is your lower bound. But again, this path isn't exactly a quadrilateral, so we can't count 2. This gives x > 2.

So, we have 2 < x < 8, so the possible values are 3, 4, 5, 6, 7, and their sum is 25.

Thanks the explanation clarified a way of looking at it. But is there a specific mathematical explanation why its not correct to look at it in a more basic(or actually less basic) fashion?

Since you didn't explain what you actually did, it's very difficult to show you what you did wrong.  If you would show everything you did to arrive at your incorrect answer, it would make it much easier to find the error in your reasoning.

Sorry i honestly thought there was only one way to interpret looking at a four sided figure as two three sided figures with a common side(not the unknown side clearly), and then looking at the figures individually with respect to the triangle inequality as to obtain an inequality for the quadrilateral

There are several different ways to order four edges and two different ways to divide a quadrilateral by a diagonal, so the fact that you think there's only one way might very well have something to do with your error.  Because you haven't actually shown what you've done (i.e., described (or provided a diagram of) how you set up the quadrilateral, the equations you derived from this and how you solved them), it is rather difficult to know for sure.


Euler
geometry
geometric transformation
reflection
geometry proposed

Let $ l_a,l_b,l_c$ be three parallel lines passing through $ A,B,C$ respectively. Let $ l_a'$ be reflection of $ l_a$ into $ BC$. $ l_b'$ and $ l_c'$ are defined similarly. Prove that $ l_a',l_b',l_c'$ are concurrent if and only if $ l_a$ is parallel to Euler line of triangle $ ABC$.

I couldn't find it , but it has been posted before in ML.

See Theorem 1 from [url=http://forumgeom.fau.edu/FG2008volume8/FG200806index.html]Cosmin Pohoata, On the Parry reflection point, Forum Geom., 8 (2008) 43--48.[/url]

Part [b]a)[/b] of http://www.mathlinks.ro/viewtopic.php?t=70770 .

  darij


inequalities
geometry
inradius
circumcircle
inequalities proposed

Is it true that in every triangle $ ABC$ with semiperimeter $ s$, inradius $ r$ and circumradius $ R$ there holds \[ s^2 (s^2 \plus{}r^2 \plus{}2Rr) \geq 48 Rr(R\plus{}r)^2\]

For right and equilateral triangles doesn`t hold

[quote="matha"]Is it true that in every triangle $ ABC$ with semiperimeter $ s$, inradius $ r$ and circumradius $ R$ there holds
\[ s^2 (s^2 \plus{} r^2 \plus{} 2Rr) \geq 48 Rr(R \plus{} r)^2\]
[/quote]

this ineq just hold in acute triangle.


induction
number theory
geometry unsolved
geometry

Let $A$ and $B$ points in the plane and $C$ a point in the perpendiclar bisector of $AB$. It is constructed a sequence of points $C_1,C_2,\dots, C_n,\dots$ in the following way: $C_1=C$ and for $n\geq1$, if $C_n$ does not belongs to $AB$, then $C_{n+1}$ is the circumcentre of the triangle $\triangle{ABC_n}$.

Find all the points $C$ such that the sequence $C_1,C_2,\dots$ is defined for all $n$ and turns eventually periodic.

Note: A sequence $C_1,C_2, \dots$ is called eventually periodic if there exist positive integers $k$ and $p$ such that $C_{n+p}=c_n$ for all $n\geq{k}$.

[quote="carlosbr"]if $C_n$ does not belongs to $AB$, then $C_{n+1}$ is the [b]circle[/b] of the triangle $\triangle{ABC_n}$.[/quote]

You probably mean "circumcentre"?

Consider $x_n=\frac{m\angle AC_nB}{\pi}$. Then the following reccurence holds: $x_{n+1}=2x_n$ if $x_n\le \frac 1 2$ and $x_{n+1}=1-x_n$ if $x_n\ge \frac 1 2$.

Now the problem is a number theory one.  :)  The answer is $x_1\in \mathbb{Q}$ . Firstly it is obvious that if $2\mid q$ then, the denominator will not contain any $2$ after few steps when $x_n$ is multiplied by 2. Hence is suficient to examine the case when $(q,2)=1$.  It is obvious that the denominator will be the same for all $x_i$. Denote by $p_n$ the denominator of $x_n$. If we prove that $p_n$ is eventually periodic, we are done, since the denominator is the same. But we can observe that at every step, $x_n\le 1$. Hence $p_n$ can take only finitely many values. Hence it is eventually periodic, since his reccurence depends only on the previous term.

After some steps the angle $AC_nB$ can be same with $AC_{n+p}B$ but $C_n$ and $C_{n+p}$ can be simetric wrt $AB$. But is easy to see that, then $C_{n+2p}\equiv C_n$.

I forgot.. the case when $x_n\in \mathbb{R}\setminus\mathbb{Q}$. We can prove by induction that $x_{n+k}=a_kx_n+b_k$ with $a_k,b_k$ integers. Hence, in order to be periodic we must have, $x_n=x_{n+k}=a_kx_n+b_k$. Hence $x_n\in \mathbb{Q}$. Hence $x_1\in \mathbb{Q}$

is "segment bisector" mean perpendicular bisector?


How many partitions of {1,2,3,4} are there that contains two sets?
I'm thinking that there's 5:
{{},{1,2,3,4}}
{{1},{2,3,4}}
{{1,2}.{3,4}}
{{1,2,3},{4}}
{{1,2,3,4},{}}

How many permutations of {1,2,3,4} are there that has no fixed point?
I have no clue what a fixed point is.
Can anybody explain this to me?

Thank you very much

[quote="MellowMelon"]You can think of a permutation as a bijection from a finite set to itself, so it becomes $ f : S \rightarrow S$ for $ S \equal{} \{1,2,3,4\}$. A fixed point is a number $ s \in S$ such that $ f(s) \equal{} s$.

For instance, if your permutation was $ (3,2,4,1)$, $ 2$ would be a fixed point (it's still in the $ 2$nd position).

For your first answer, I believe each set in the partition must be nonempty.[/quote]

So...
for the first one,
{{1},{2,3,4}}
{{1,2}.{3,4}}
{{1,2,3},{4}}
are the only ones?

and the second one:
it would be 3*3*3*3=81?

First question: yes.

Second question: There can be only one $ 1,2,3,4$ each in the permutation, so there's actually only $ 24$ possibilities before subtracting those with fixed points.

would the final answer be 20 then?
unfortunately, 20 isn't one of the answer choices. It only between 1 and 9...i think we're doing something wrong.

2143
3142
2413
4123
2341
4321
4312
3412
3421

That's all i can think of right now...any others that i missed?

I'm not sure why you only subtracted $ 4$ from $ 4!$. Remember, we can't have [i]any[/i] fixed points. There are more than four permutations that have some.

You have all of them listed. For more on these kinds of permutations, look here about [url=http://en.wikipedia.org/wiki/Derangement]derangements[/url].


trigonometry
vector
geometry
geometric transformation
rotation

[hide="Super hint"] Dot product! [/hide]

Given ...
$ a^2\plus{}b^2\equal{}1$ [b];[/b] $ c^2\plus{}d^2\equal{}1$ [b];[/b] $ ac\plus{}db\equal{}0$

[hide]$ d\equal{}\minus{}\frac{ac}{b}$
$ c^2\plus{}d^2\equal{}1\Longrightarrow c^2\plus{}\frac{a^2c^2}{b^2}\equal{}1\Longrightarrow \frac{b^2}{c^2}\equal{}a^2\plus{}b^2\equal{}1$
Thus we have $ \boxed{b^2\equal{}c^2}$ implying that $ a^2\plus{}c^2\equal{}1$ , $ b^2\plus{}d^2\equal{}1$ and $ ab\plus{}cd\equal{}0$[/hide]

Alternate solution:
[hide]
$ a\equal{}\cos \theta$
$ b\equal{}\sin \theta$
$ c\equal{}\cos \alpha$
$ d\equal{}\sin \alpha$
$ \cos\theta \cos \alpha \plus{} \sin \theta  \sin \alpha\equal{}\cos (\theta\minus{}\alpha)\equal{}0$
This means that $ \theta\minus{}\alpha\equal{}\frac{\pi}{2}$.
$ \cos (\frac{\pi}{2}\plus{}\alpha)\equal{}\cos \frac{\pi}{2} \cos \alpha\minus{}\sin \frac{\pi}{2}\sin \alpha\equal{}\minus{}\sin \alpha\equal{}\cos \theta$
So we have the results below following.[/hide]

[quote="igiul"][quote="CatalystOfNostalgia"]igiul: you have to make sure your denominators are not zero, but that's a trivial case.[/quote]
Can you elaborate? I checked for that stuff just after reading your comment but couldn't find anything. If you could point it out that would be very much appreciated.  :)[/quote]
He means make sure it's true for b=0/c=0.

[hide="With dot products"] Let $ \vec{v} \equal{} \left< a, b \right>, \vec{u} \equal{} \left< c, d \right>$ be vectors of length $ 1$.  We are given that their dot product is $ 0$.  Hence they are perpendicular.  Because they have the same length, one is a $ 90^{\circ}$ rotation of the other.  In particular, $ d \equal{} \minus{}a, b \equal{} c$ or $ d \equal{} a, b \equal{} \minus{}c$.  Now the problem is trivial. [/hide]

$ (a \plus{} ci)^2 \plus{} (b \plus{} di)^2 \equal{} (a^2 \plus{} b^2 \minus{} c^2 \minus{} d^2) \plus{} 2(ac \plus{} bd)i \equal{} 0$ so $ a \plus{} ci \equal{} \pm i(b \plus{} di)$ which gives the same result since we can equate real and imaginary parts.

$ a \plus{} ci \equal{} \pm i(b \plus{} di)$

$ a\plus{}ci\equal{}\mp d\pm bi$

$ a\pm bi\equal{}\pm d\minus{}ci$

$ a^2\minus{}b^2\pm 2abi\equal{}d^2\minus{}c^2\mp 2cd i$

hence $ ab\plus{}cd\equal{}0$ and $ a^2\plus{}c^2\equal{}b^2\plus{}d^2\equal{}t$...

But then $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1\plus{}1\equal{}2t$ so $ t\equal{}1$, as desired.

[quote="lena_siz"]a, b, c, d are real numbers so that
$ a^2 \plus{} b^2 \equal{} 1$
$ c^2 \plus{} d^2 \equal{} 1$
$ ac \plus{} db \equal{} 0$
Prove that:
$ a^2 \plus{} c^2 \equal{} 1$
$ b^2 \plus{} d^2 \equal{} 1$
$ ab \plus{} cd \equal{} 0$

Thanks![/quote]
Nice solutions to all!  Here is an alternate one to add (probably less elegant, though):
[hide="Alternate Solution"]

\[ a^2 \plus{} c^2 \equal{} k\implies b^2 \plus{} d^2 \equal{} (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2) \minus{} a^2 \minus{} c^2 \equal{} 2 \minus{} k\] \[ 1 \equal{} (1)(1) \minus{} 0^2 \equal{} (a^2 \plus{} b^2)(c^2 \plus{} d^2) \minus{} (ac \plus{} bd)^2\] \[ 1 \equal{} a^2d^2 \plus{} b^2c^2 \minus{} 2abcd \equal{} (ad \minus{} bc)^2\] \[ \minus{} 1(k \minus{} 1)^2 \equal{} k(2 \minus{} k) \minus{} 1 \equal{} (a^2 \plus{} c^2)(b^2 \plus{} d^2) \minus{} (ad \minus{} bc)^2\] \[ \minus{}1(k\minus{}1)^2\equal{} ab^2 \plus{} cd^2 \plus{} 2abcd \equal{} (ab \plus{} cd)^2\] \[ \minus{} 1(k \minus{} 1)^2\le 0\] \[ (ab \plus{} cd)^2\ge 0\] \[ ab \plus{} cd \equal{} 0 \equal{} \minus{} 1(k \minus{} 1)^2\] \[ ab \plus{} cd \equal{} 0\] \[ k \equal{} 1\] \[ a^2 \plus{} c^2 \equal{} 1\] \[ b^2 \plus{} d^2 \equal{} 2 \minus{} k \equal{} 2 \minus{} 1 \equal{} 1\]
[/hide]

Notice that this holds for even complex numbers.


Bad title! Locked! http://www.mathlinks.ro/Forum/viewtopic.php?t=144403


The Fibonacci sequence is defined by $ F_0\equal{}0, F_1\equal{}1$ and $ F_{n\plus{}2}\equal{}F_n\plus{}F_{n\plus{}1}$ for $ n \ge 0$. Prove that:
$ (a)$ The statement $ "F_{n\plus{}k}\minus{}F_n$ is divisible by $ 10$ for all $ n \in \mathbb{N}"$ is true if $ k\equal{}60$ but false for any positive integer $ k<60$.
$ (b)$ The statement $ "F_{n\plus{}t}\minus{}F_n$ is divisible by $ 100$ for all $ n \in \mathbb{N}"$ is true if $ t\equal{}300$ but false for any positive integer $ t<300$.

It has been posted before. But, thank you for showing the source. :D


calculus
geometry
3D geometry
USAMTS
function

The grading criteria for question 5 exclude the possibility of a completely correct calculus solution.  That is, a solution that shows the calculus, proves that it works correctly, and does not rely on any decimal approximations.  Of course, it is very possible to write such a solution, but by these criteria a calculus solution can only receive at most 4 points.  Why is this?

I think the criteria imply that if a student did all the calc by hand, and then showed that a nasty cube root thingy is less than 25/4rt6 by hand, they get 5/5. However, it seems sketchy to me that on a contest that is open to calculators and mathematica, they take off a point for utilizing mathematica (i.e. apparently decimal approximations are bad?)

Some people did get 5/5 with a calculus solution. I'd be grateful if he/she put up the solution so we can see?

On the USAMTS rules it explicitly states:
"Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose."

What I did was to find the maximum of my function in a certain interval (that made sense in the problem) using Mathematica.  Apparently, we have to prove that Mathematica can calculate the result correctly to receive full credit...... are we supposed to show exactly how Mathematica does this?  If so, that seems a bit harsh.

If anyone got full credit for proving this with Mathematica or protested and received full credit, could they please post their solution?

The last page of my attachment is a calculus solution that received full credit.

Yeah, that is a lot more intensive than some of the other calculus solutions.
If anyone gets a protest rewarded on #5 for a calc solution, can he/she mention it, please?

My protest was accepted.  My score was adjusted from a 4 to a 5 with a comment saying that the guy didn't know why I was docked the point in the first place.

My solution contains decimals, but all decimal values are exact and no approximation is used at any stage of the solution, even at the end.

[quote="Mewto55555"][quote="Jason27603"]On the USAMTS rules it explicitly states:
"Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and [b]must prove that the method used in the program does solve the problem[/b]. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose."
[/quote]

So apparently by this logic, me using a calculator to find $ 13^3$ on the family problem without proving it merited a one point deduction?[/quote]
"9. Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, [b]if a key step of a participant's solution uses these tools[/b], the participant must demonstrate that the tool completed the step correctly."

I think they only want you to prove it if it is a major step in the proof, and not if it is something obvious or easily verified (like $ 13^3\equal{}2197$).

I think that explaining why the nature of the graph is as it is (for example, why it is increasing, or why it is constant at an interval of 5 to 7) would be good enough. I am not sure, though.

Thanks Eric :wink:

EDIT:
[quote="EricMathPath09"]I think that explaining why the nature of the graph is as it is (for example, why it is increasing, or why it is constant at an interval of 5 to 7) would be good enough.[/quote]
Can anyone verify this?

Also, Im not sure if this counts as talking about the problem, as this is all hypothetical. If it does, a mod can just edit this post.

[quote="Thunder365"]This is a stupid question, but what if, hypothetically, you were to use Mathematica to graph? You wouldnt need to prove anything, right?[/quote]

Graphs aren't really rigorous proofs of anything, are they?


hello everybdy,
Suppose u have a transparent glass plate, refractive index $ n$ and thickness $ d$. Monochromatic light is incident on it( of a specific given wavelength) at an angle $ i$. For const. interference, what should b the reln involving the above terms.
i tried, but repeatedly got the $ cos$ $ r$ term in the denominator ( $ r$ being the angle of refraction) while irodov has it in the numerator.
 smbdy pls prove it 4 me, or give a link to its proof.
thanks in advance.


geometry
integration
function
calculus
calculus computations

How would you solve this problem as there are no real stated limits.

Find the area of this:

$f(x)=\sin(x)$

But as there is no limits i can do this integral.

$\int{\sin(x)dx=-\cos(x)}$

There are no boundary, so we can't find area. :|

Oh damn, now i understand the task. I have two functions:

$f(x)=sin x$

$g(x)=x^2-\pi x$

So i have to find the area these two functions make. That's what i misunderstood. I thought i had to find the area of both of them, but then i would need some limits, yes. OK, my fault :) Thx for the reply though.


Is there a better way of solving all the partitions for a number [i]X[/i] other than listing out everything?

There's the generating function thing, which I'm not that familiar with

In MATHCOUNTS, go with listing everything

I remember Ubemaya (or someone) started a whole thread about it.

I wonder if I can find it.

List in an organized manner.  Other methods of calculating the [url=http://mathworld.wolfram.com/PartitionFunctionP.html]partition function[/url] are much, much harder.

OK... thank you..... I think I'll stick with listing for now.... :lol:


Ring Theory
superior algebra
superior algebra unsolved

Suppose $ f: \mathbf Q_p \rightarrow \mathbf Q_q$ is a ring isomorphism, where $ p,q$ are rational primes ($ \mathbf Q_p$ is the p-adics).  Show $ p \equal{} q$ and $ f$ is trivial.

I know a way to do the problem but I think there should be a nicer way.  The point is that such a map should be continuous, then the result will follow.  I want to be able to show that the isomorphism restricts to the units $ \mathbf Z_p$, but the only way I know of doing that now is completely solving the problem in a fairly long-winded way (show that the roots of unity are different if $ p \neq q$, and if $ p \equal{} q$ show that the map is continuous by showing $ \mathbf Z_p^\times$ is sent to itself).

Take also a look at http://www.mathlinks.ro/viewtopic.php?t=100499 ($ p\equal{}q$ follows, as you said, from consideration of roots of unity).