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Last summer, I was a student at AMSP. Now that I am coming back as a counselor, I feel that I have a lot of ideas about what made my time at AMSP enjoyable and enlightening. I'll update this as I think of things...I don't want a huge wall of text either...so it will get broken up that way.

I'm declaring it right now: [i]It is my goal to make your experience as good as possible. [/i]

I've made an effort to answer questions on this forum to the best of my ability. Right away, I think that the most important thing that you should do is communicate with the teachers and other students.

5/19/08
[i]Communication is paramount: if you don't understand something, ask for help.[/i]

AMSP is a community that learns together.

Don't be afraid of asking questions during lectures. Depending on your confidence level, you might find it easier to ask questions after class: go up to the teacher and ask for a clarification. Or if you prefer, assistants and other kids can lend their thoughts to the question as well.

[quote="Altheman"]
Don't be afraid of asking questions during lectures.[/quote]
This can't really be stressed enough, if you do not understand something, swallow your pride, and ask away.

Can someone give me tips on which courses I should take? Don't give me in terms of AMC/AIME scores... I didn't take those tests.  :( Right now I think I'm taking all the lowest level courses.

I hear AIME is a nice indicator test, so try working through some AIME problems (from the resources section) and see what level of problems you're most comfortable with. I bet whoever's in charge of course placement can help you out just fine.e

5/24/08

[i]What level am I at? What classes should I take? What should I do to challenge myself?
[/i]
These are always important questions to consider. I cannot give a full answer, but I'll give my opinion in the context of choosing classes for AwesomeMath.

I suggest that you think about the class choices before you come to camp. Dr. Andreescu and Dr. Feng put this together. This is the best advice that you can get. The first step is to read the document carefully:

[url]http://www.awesomemath.org/AMSP_dailysch.pdf[/url]

I suppose that I will put my own thoughts into the differences between the classes. Basically, there are four subjects, and three difficulties for each topic.

Choosing a Topic: Here is some advice that I like from last year: [i]Expand yourself with one topic, challeng yourself a course in a topic that you know well.[/i] Depending on your attitude towards failure (how do you feel when you cannot solve a problem?) you could take harder or easier classes.

Difficulty: I agree that using the AIME is a good measure of where you should be: the test has a wide difficulty gradient.

$ \cdot$ Introductory Level: The easiest AIME problems are an appropriate indicator of difficulty.
$ \cdot$ Intermediate Level: Mid-level AIME problems require a certain degree of adaptability and creativity in the problem solver.
$ \cdot$ Advanced Level: The hardest AIME problems introduce you to proof driven problems that require the solver to generate multiple non-obvious steps.

When you arrive at camp, I believe that you'll talk with one of the faculty (last year, Dr. Feng gave advice) and choose a course. If you want some concrete advice now, I'll discuss one of my old Mock AIMEs with you. Put yourself in a quiet place and do the problems that you can. Take extra time if necessary to at least consider each problem. Send me a pm with a some thoughts about each problem.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1045082941&t=126603]Mock AIME 5 for 2007[/url]

Thats enough for now.

[quote="Altheman"]I suggest that you think about the class choices before you come to camp.
...
When you arrive at camp, I believe that you'll talk with one of the faculty (last year, Dr. Feng gave advice) and choose a course. [/quote]
actually, this year we got a course request section in one of the forms of the precamp material stuffs

All the better. I was a bit caught off guard by the decision right after we got there. Anyway, I guess I'm here to give my thoughts about course placement if anybody wants them :D

Will we get to know how well we did on the admission tests, and how we could improve the proofs we've submitted? Does how well you did on the admission test influence the courses you can take?

I suppose that it is a good exercise to revise your proofs. I suppose you could bring that with you or something but I feel like the admissions test isn't that huge a factor in deciding your courses. So I feel like doing those corrections would just be for your own good. I feel like your personal decision is the largest factor. Then other scores could help determine your position as well. I suppose that I'm not totally sure I guess I'll ask.

Here's a slightly unrelated question:  Will we be able to access our dormitories at any time of the day during break?  I read somewhere that only the counselor will have a key, but in the packet I read that each person had a room card.

And also, what are the pianos like at AMSP?  Is there a concert hall with a grand piano in it?  (that would be nice  :D )  Lastly, how many people normally go to practice every day?  I'd like to know this so I don't hog the piano for too long...

 :oops:  My apologies if this is the wrong thread, but I'm afraid to revive any of the older ones.

Pianos o.o? I suck at piano.  Is AMSP held on a college campus or somewhere else? I never really got to reading about anything xD.

I've heard that there have been a few pianos, but I'm not sure...AMP is held at the University of Texas, and it's in Dallas.

[quote="CatalystOfNostalgia"][quote="pythag011"]How well should we do on the USAMO if we want to take advanced level?[/quote]

This is the type of question that can't be answered, and you can never expect a straight answer to this type of question. It really depends on how hard you want to work, how well you do in the class, and of course, how relevant next year's problems are to whatever you learn in the class.[/quote]
If you're goal is to get pretty much everything taught in the course, then you should be able to solve a USAMO #1 difficulty question in that subject more often than not, so 7 in a USAMO of that subject alone.

Although Advanced Number Theory was, well, freaking insane last year to be honest. I think some of the material would've troubled some blue MOPpers even, so if it's at all the same, I strongly dissuade you from taking it unless you're really really good at it.

CatalystofNostalgia said some stuff about pianos in the thread somewhere below started by zephyredx.... I'm so happy they have pianos  :) .....

My random advice: my room's mentor last year was named Oleg. Oleg would stay out partying all night, so we never ever saw him - I don't think he knew our names by the end of camp. Anyways, as a result of his partying, we could stay up as late as we wanted - usually we would stay up until ~3 in the morning. As a result, I would be extremely tired the next day and was unable to focus during class - in fact, on two occasions, I was over an hour late for my afternoon class as a result of falling asleep during lunch. Eventually, I fell so far behind in my class, Track 2, that I had to transfer to an easier class, Track 5. My point is, no matter how much fun you have with your roommates, you have to go to bed at a reasonable time, or you will miss out on a lot of learning.

Post #337
1000 more to go...

[quote="le00327146"]I got 9 on this year's AIME, but a 1 on the USAMO (I thought I would have gotten a 7 on one of them...). I don't really have any extremely significant strengths/weaknesses, although geometry is probably my weakest. What levels do you think I should go for?[/quote]
Make a list of classes where you think that you'll learn a lot, and be able to keep up.
Then pick the two that'll be hardest for you to teach yourselves.

I need help...I got 8 on AIME, and did really bad on USAMO...I can do almost all geo #1 and #4 on USAMO, and virtually no combinatorics or NT problems on USAMO, and my algebra is okay, i guess.

Should I take the medium NT and combo classes, or should I take like the Proof Geo class and one of those?

[quote="diophantient"]
Make a list of classes where you think that you'll learn a lot, and be able to keep up.
Then pick the two that'll be hardest for you to teach yourselves.[/quote]

Well...I'm currently stuck between the mid-level class and the advanced class. I'm pretty sure I'd be in the middle for geometry and number theory, but I'm not sure which way to lean on combinatorics and algebra.


function
Support
integration
geometry
3D geometry
sphere
calculus

I need a hint for the following.
Let $f: R^{n}\to R$ be a continous function. Show that $f$ is integrable if and only if for every $\epsilon>0$ there exists a function $g$ with compact support such that:

                     $\int_{R^{n}}^{}|f-g|$<$\epsilon$

If is obvious. Only if: this probably isn't the nicest way, but here's an idea. Cut $R^{n}$ into concentric spheres of radius $k$, $k = \{1, 2, ...\}$. The integral of $|f|$ over each shell of thickness 1 is finite because $f$ is cts and it won't harm us to include the boundaries on both sides for each shell. So, consider the sequence $I_{1}, I_{2}, ...$, the integral of $|f|$ over each shell. If the sum of that series does not converge, obviously the function is not integrable. But if it does, then there exists some large $m$ so that $\sum_{k=m}^\infty I_{m}< \epsilon$. So you need only cut and smooth $|f|$ at the boundary of the $m^{th}$ ball and you win.

Xevarion:
How is the if obvious?
Thanks a lot

Find a large ball that contains the support of $g$. Break the integral of $|f-g|$ into an integral over that ball and an integral over $R^{n}\setminus$ that ball. You will find that the integral of $f$ outside a large enough ball is small. ;)

Xevarion:
The function $g$ is not given. One has to show it exists.

You want to prove "if and only if". In my first post, I show how you can construct a $g$ given that $f$ is integrable. In my second post, I show how you can prove that $f$ is integrable, given the $g$.


Find every triples of positive integers $ (x,y,z)$ such that: $ 4^x\plus{}4^y\plus{}1\equal{}z^2$.

What's the source of the problem?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=305250]Same Problem[/url]


abstract algebra
superior algebra
superior algebra unsolved

R is a ring and consider commutative diagram of R -modules with exact rows.

http://images.planetmath.org:8080/cache/objects/5578/l2h/img1.png

then I want to show that Gamma is isomorphism if and only if

the sequence
0----->A_1----->A_2+B_1------>B_2----->0

here + is direct sum and teh first homomorphism sends a to \alpha(a),k_1(a)
and the second homomorphism sends (a_2,b) to  k_2(a_2)-\beta(b)

here k_1 is homomorphism from A_1 to B_1 and k_2 is homomorphism from
A_2 to B_2.

Use a diagram chase - which is a standard technique to solve problems like this.


function
algebra
functional equation
algebra unsolved

Find all functions $ f: \mathbb{Q} \rightarrow \mathbb{Q}$ such that:

$ f(x\plus{}f(y))\equal{}y\plus{}f(x)$ for all $ x,y \in \mathbb{Q}$.

[hide]Let $ P(x,y)$ be the statement $ f(x\plus{}f(y))\equal{}y\plus{}f(x)$.

$ P(0,0)$ yields: $ f(0\plus{}f(0))\equal{}0\plus{}f(0) \Rightarrow f(f(0)) \equal{} f(0)$

$ P(0,f(0))$ yields: $ f(0\plus{}f(f(0)))\equal{}f(0)\plus{}f(0) \Rightarrow f(f(0)) \equal{} 2f(0)$

Thus, $ f(0) \equal{} 2f(0) \Rightarrow f(0)\equal{}0$

$ P(0,y)$ yields: $ f(0\plus{}f(y))\equal{}y\plus{}f(0) \Rightarrow f(f(y)) \equal{} y$

$ P(x,f(y))$ yields: $ f(x\plus{}f(f(y)))\equal{}f(y)\plus{}f(x) \Rightarrow f(x\plus{}y) \equal{} f(x)\plus{}f(y)$

This is the Cauchy Functional Equation whose only solutions $ f: \mathbb{Q}\rightarrow\mathbb{Q}$ are $ f(x) \equal{} ax$ for some $ a \in \mathbb{Q}$.

Plugging this into the original equation yields:
$ f(x\plus{}f(y))\equal{}y\plus{}f(x)$
$ f(x\plus{}ay) \equal{} y\plus{}ax$
$ ax\plus{}a^2y \equal{} y\plus{}ax$
$ a^2 \equal{} 1$
$ a \equal{} \pm 1$

Thus, the only solutions are $ f(x) \equal{} x$ and $ f(x) \equal{} \minus{}x$. [/hide]

Slightly different but a very similar solution:

I claim that $f(x) = x$ and $f(x) = -x$ are the only solutions.

    Let $x = 0$. Then
    $$
    f(f(y)) = y + f(0).
    $$
    Setting $y = f(0 - f(0))$ gives
    $$
    f(x + f(f(0 - f(0)))) = f(x + 0 - f(0) + f(0)) = f(x) = f(0 - f(0)) + f(x),
    $$
    thus $f(0 - f(0)) = 0$, and $f(f(0 - f(0))) = 0 = f(0)$, so $f(0) = 0$.
    This implies that $f(f(y)) = y$.

    Finally, with $x = a$ and $y = f(b)$, we have
    $$
    f(a + f(f(b))) = f(a + b) = f(b) + f(a),
    $$
    which is Cauchy's equation, and over $\mathbb{Q}$ has only the solution $f(x) = kx$ for some $k \in \mathbb{Q}$.
    Substituting, we find that only $k = \pm 1$ works, so the only solutions are $f(x) = x$ and $f(x) = -x$, both of which are valid.

I think that we can extend the solution from $ f: \mathbb{Q} \rightarrow \mathbb{Q}$, to $ f: \mathbb{R} \rightarrow \mathbb{R}$:
1) If $x=0$, and $f(0)=a$, we get $f(f(y))=y+a$, and evident the function it is bijective
2) Swapping the variable, from $ f(x+f(y))=y+f(x)$, we get $f(y+f(x))=x+f(y)$. So we have $f(f(y+f(x))=f(x+f(y))=y+f(x)$, and from the surjectivity we have $f(f(t))=t$, and $f$ bijective, so $f(x)=x$, or $f(x)=-x$.
It is correct?

[quote = TuZo]...from the surjectivity we have $f(f(t))=t$, and $f$ bijective, so $f(x)=x$, or $f(x)=-x$.
It is correct?[/quote]
No. Any function $f: \mathbb{R} \to \mathbb{R}$ that satisfies $f(x+y)=f(x)+f(y)$ and $f(f(x))=x$ for all $x,y \in \mathbb{R}$ is a solution. There are infinitely many such functions.

But the $f$ it is bijective. Can you find the mistake in my attempt?


linear algebra
matrix
linear algebra solved

This was troubling me and I have a test on it tomorrow, so if someone could explain it, I would be REALLY grateful!  Thanks in advance!

If all entries of a (square) matrix A are integers and |A|= + or - 1, show that all entries of A^-1 are integers.


I play Flute and violin. But i love cello

do you know Jet Whistle of Villa-Lobos?! the cellos are.... woooww... too much good.

EDIT:

http://www.acplayers.com/music/lobos1.mp3

for the music. enjoy

No, but now I know. That was pretty cool.

YEAH, im playing ELgar and Shostakovich and popper, etdc

WHOA, you're playing elgar? what of shostakovich and popper are you playing? I've played popper's gavotte in D major, something else I can't remember and I'm slaving away at his etudes.

shostakovich concerto no 1, popper etudes, hungarian thing, concerto, etc...
i'm 13 rite now.

ahh, you prodigies.  :)

hmm, I'm working on Bloch's Prayer, any tips/suggestions/interpretations?

I play the cello too. And the name starcellist doesn't mean that I'm really good at cello; I actually just started about 2.5 years ago...
Um.. Right now I'm working on the Haydn C Major (1st mvmt.), Saint-Saens a minor (3rd mvmt.), Lots of Duport and Popper level etudes, and Suite No. 1 in G Major. I can't wait to learn the Shostakovich though!

I play cello and have been for two years.

No need to revive.

[color=#FF0000][size=89]No need to make a useless post contributing to the revival either. [/size][/color]


http://www.unl.edu/amc/a-activities/a7-problems/a7-1-archives/2003/AIMEpractice/03AIMEaltq.html

i cant really understand what #5 is asking about the logs wedges. is it like you cut thru the log so you get a half log, wwith a halfcircle for faces? then the second cut is directly down the middle at a 45 degree angle?

Basically just imagine starting at the upper left corner and sawing down to the right at a $45^\circ$ angle.  They want the volume of the smaller piece.

Just say that volume equals V. Now glue two of those pieces together to form a cylinder with volume 2V. You can easily find 2V. Then just divide by 2 to get V.


[hide="sol"]
$\sqrt{2n+1}=201$
Solve for $n$
[/hide]

You accidentally dropped a zero off the end, jensen, it should be $\boxed{20200}$

I change the left side to sumation form, i=1 to n, then simplify the thing to (1/2)(sqrt(2i+1)-sqrt(2i-1)), then break it out into (sqrt(3)+sqrt(5)+....+sqrt(2n-1)+sqrt(2n+1))-(1+sqrt(3)+sqrt(5)+....+sqrt(2n-1))=100
cancel out I got sqrt(2n+1)-1=100=>2n+1=10201=>n=5100
sorry, I don't know how to use the code thingy so :p


function
algebra
polynomial
email
Ring Theory
number theory
number theory open

Fermat's last theorem: There is no such solution of
Fermat equation $ x^n \plus{} y^n \equal{} z^n$ such that $ (x, y) \equal{} 1, (x, z) \equal{} 1, (y, z) \equal{} 1, n\geq5, n$ is odd prime, $ x, y, z\in\mathbb{Z}$.
\newline Proof: Suppose $ (a, b, c)$ is such solution. If $ a($ or $ b)$ is even, then let
$ (x, y, z) \equal{} (a, b, c)$. If $ c$ is even, since $ n$ is odd, we can
get $ c^n \plus{} ( \minus{} b)^n \equal{} a^n$, then let $ (x, y, z) \equal{} (c, \minus{} b, a)$. We can let
$ x$ is even. If $ (t, 2^{k \plus{} 1}) \equal{} 2^k$, then let the $ 2$-adic order
function $ \nu_2(t) \equal{} k$. Since $ x$ is even, let $ \nu_2(x) \equal{} k, k > 0$.
Hence $ \nu_2(z \minus{} y) \equal{} nk$. Considering cyclotomic field
$ \mathbb{Q}\{\zeta_n\}$. Since $ \bar{\zeta}_n^k \equal{} \zeta_n^{n \minus{} k}$,
let $ Re(\zeta_n^k) \equal{} \cos(k\alpha), \alpha \equal{} \frac {2\pi}{n}$ and we
have
$ x^n \equal{} z^n \minus{} y^n \equal{} (z \minus{} y)\prod_{k \equal{} 1}^{n \minus{} 1}(z \minus{} y\zeta_n^k) \equal{} (z \minus{} y)\prod_{k \equal{} 1}^{\frac {n \minus{} 1}{2}}(z^2 \plus{} y^2 \minus{} 2zy\cos(k\alpha))$.
Since $ \zeta_n^{n \minus{} k} \equal{} 2\cos(k\alpha) \minus{} \zeta_n^k$, we consider the
equivalence cyclotomic field $ \mathbb{Q}_1\{\zeta_n\}$ of
cyclotomic field $ \mathbb{Q}\{\zeta_n\}$, of which the power bases
are $ 1, \zeta_n^k, cos(k\alpha), k \equal{} 1, \cdots, \frac {n \minus{} 1}{2}$.
Since $ n\geq5, n$ is odd prime, $ cos(k\alpha), k \equal{} 1, \cdots, \frac {n \minus{} 1}{2}$ are irrational numbers $ p \equal{} 1 \plus{} \zeta_n$ is prime. Otherwise
the number of the power bases of cyclotomic field
$ \mathbb{Q}_1\{\zeta_n\}$ will be less than $ n$. We have $ p|2, (z^2 \plus{} y^2 \minus{} 2zy\cos(k\alpha),2p) \equal{} (z^2 \plus{} y^2 \minus{} zy(\zeta_n^k \plus{} 1 \plus{} \zeta_n^{n \minus{} k} \minus{} 1),2p) \equal{} (z^2 \plus{} y^2,2p) \equal{} 2$. As a result, we
can verify $ \nu_2(rightside) \equal{} nk \plus{} \frac {n \minus{} 1}{2}$ and
$ n|\nu_2(leftside)$ which yields a contradiction.

The root should make the equation to be an identity for any number field.
The power of a prime factor may vary from the change of number field,
but the ones of both sides of an identity equal to each other.

It would be nice for any mistake to be pointed out! Thank you!

Is there something wrong sensed by intuition? It would be nice for anything to be pointed out! Thank you!

[quote="Taoli"]The power of a prime factor [/quote] which should always be coprime to the other for these number fields considered in the proof.

$ p \equal{} 1 \plus{} \zeta_n$ is prime. Otherwise
the number of the power bases of cyclotomic field
$ \mathbb{Q}_1\{\zeta_n\}$ will be less than $ n$. We have $ p|2, (z^2 \plus{} y^2 \minus{} 2zy\cos(k\alpha),2p) \equal{} (z^2 \plus{} y^2 \minus{}zy(\zeta_n^k \plus{} 1 \plus{} \zeta_n^{n \minus{} k} \minus{} 1),2p) \equal{} (z^2 \plus{} y^2,2p) \equal{} 2$.

There are three changes in this version:
1、A lemma is added;
2、$ \nu_2(rightside)$ is recalculated;
3、The format is renewed.

Problem: Is there something about unique factorization theorem?

That's near to impossible to read...
Two things I saw:
- you claim that $ 1\minus{}\cos(k \alpha)$ is prime; but it isn't even an algebraic integer.
- $ p$ is not prime; in fact it's a unit and thus $ (p^2,2)\equal{}1$.

If $ e$ is a unit, then that $ 1\plus{}e,1\minus{}e$ are primes is followed since $ N_0(1\plus{}e)\equal{}N_0(1\minus{}e)\equal{}2$ is a rational prime.
-Since $ 2\cos(k\alpha)\equal{}\zeta_n^{n\minus{}k}\plus{}\zeta_n^k$, we can let $ \cos(k\alpha)$ be power bases as well as unit. Then we consider the equivalent cyclotomic field $ \mathbb{Q}_1\{\zeta_n\}$ of cyclotomic field $ \mathbb{Q}\{\zeta_n\}$.
-$ p|2\equal{}1\plus{}\zeta_n^n$.

I am sorry I have not practised professional English enough. Thank you!

[quote="Taoli"]If $ e$ is a unit, then that $ 1 \plus{} e,1 \minus{} e$ are primes is followed since $ N_0(1 \plus{} e) \equal{} N_0(1 \minus{} e) \equal{} 2$ is a rational prime.[/quote]
No, that norms are wrong, and thus is the whole reasoning. Also, the norm of $ 1\minus{}\zeta_n$ is $ n$, not $ 2$, and that of $ 1\plus{}\zeta_n$ is $ 1$ (a unit!).

I don't know what "power bases" or "equivalent fields" are supposed to be. But regardles of that, you can't call a number "prime" if it isn't even integral.

-The norm of an algebraic integer is again an integer, because it is equal (up to sign) to the constant term of the minimal polynomial.
-If $ \alpha$ is an algebraic integer then $ \beta\equal{}\alpha^{1/n}$ is another algebraic integer. A polynomial for $ \beta$ is obtained by substituting $ x^n$ in the polynomial for $ \alpha$.

These are based on wikipedia. I misunderstood a lemma I read before.
But without being called prime, those factors are something like prime. I will add the ignored details to the proof.

Just because the norm is an integer, it is not equal to $ 2$  :wink:
And no, these factors are not something like prime.

[b]Fermat's last theorem[/b]: There is no such
solution of Fermat equation $ x^n\plus{}y^n\equal{}z^n$ such that $ (x, y)\equal{}1, (x,
z)\equal{}1, (y, z)\equal{}1, n\geq5, n$ is odd prime, $ x, y, z\in\mathbb{Z}$.

[b]Idea[/b]: The root should make the equation to be an identity
for any number field. The power of a prime factor, which should
always be coprime to the other for these number fields considered
in the proof, may vary from the change of number field, but the
ones of both sides of an identity equal to each other.

[b]Lemma 1[/b]: Let $ k$ be a rational odd. Considering
cyclotomic field $ \mathbb{Q}\{\zeta_n\}$. We have $ (2,k)\equal{}1$.

[b]Proof[/b]: Let \[ (2,k)\equal{}a, 2\equal{}pa\equal{}\sum_ip_i\zeta_n^ia,
k\equal{}qa\equal{}\sum_jp_j\zeta_n^ja, (p,q)\equal{}1\]. Then we get
\[ k\sum_ip_i\zeta_n^i\equal{}2\sum_jp_j\zeta_n^j\]. Hence $ i\equal{}j, kp_i\equal{}2q_j,
kp\equal{}2q$. It yields that $ p\equal{}2, q\equal{}k, a\equal{}1$.

[b]Lemma 2[/b]: Considering cyclotomic field
$ \mathbb{Q}\{\zeta_n\}$. Let $ Re(\zeta_n^k)\equal{}\cos(k\alpha),
\alpha\equal{}\frac{2\pi}{n}, \beta_n^0\equal{}1, \beta_n^k\equal{}\frac{\zeta_n^k}{2},
 \beta_n^{n\minus{}k}\equal{}cos(k\alpha), k\equal{}1, \cdots, \frac{n\minus{}1}{2}$. Then
 \[ \mathbb{Z}\{\beta_n\}\equal{}\{x: x\equal{}a_0\plus{}\sum_{i\equal{}1}^{n\minus{}1}2a_i\beta_n^i, a_k\in\mathbb{N}\}\] is an algebraic integer
 ring.

[b]Proof[/b]: Since $ n\geq5, n$ is odd rational prime,
$ cos(k\alpha), k\equal{}1, \cdots, \frac{n\minus{}1}{2}$ are irrational numbers.
Otherwise the number of the power bases of cyclotomic field
$ \mathbb{Q}\{\zeta_n\}$ will be less than $ n$. We can verify the
fact from the definition.

[b]Proof[/b]: Suppose $ (a, b, c)$ is such solution. If $ a($ or $ b)$
is even, then let $ (x, y, z)\equal{}(a, b, c)$. If $ c$ is even, since $ n$
is odd, we can get $ c^n\plus{}(\minus{}b)^n\equal{}a^n$, then let $ (x, y, z)\equal{}(c, \minus{}b,
a)$. We can let $ x$ is even. If $ (t, 2^{i\plus{}1})\equal{}2^i$, then let the
$ 2$-adic order function $ \nu_2(t)\equal{}i$. Since $ x$ is even, let
$ \nu_2(x)\equal{}j, j>0$. Since
\[ \frac{x^n}{z\minus{}y}\equal{}\frac{z^n\minus{}y^n}{z\minus{}y}\] is a sum of $ n$ odds, we get
$ \nu_2(z\minus{}y)\equal{}nj$. Considering cyclotomic field
$ \mathbb{Q}\{\zeta_n\}$. Since $ \bar{\zeta}_n^k\equal{}\zeta_n^{n\minus{}k}$,
let $ Re(\zeta_n^k)\equal{}\cos(k\alpha), \alpha\equal{}\frac{2\pi}{n}$ and we
have\[ x^n\equal{}z^n\minus{}y^n\equal{}(z\minus{}y)\prod_{k\equal{}1}^{n\minus{}1}(z\minus{}y\zeta_n^k)\equal{}(z\minus{}y)\prod_{k\equal{}1}^{\frac{n\minus{}1}{2}}(z^2\plus{}y^2\minus{}2zy\cos(k\alpha))\].
Considering $ \mathbb{Z}\{\beta_n\}$, $ p\equal{}1\plus{}\zeta_n\equal{}1\plus{}2\beta_n,
p_k\equal{}1\minus{}\cos(k\alpha)\equal{}1\minus{}\beta_n^{n\minus{}k}$ are primes. Since $ (p^2,2)\equal{}p,
z\equiv y,a^2\equiv1,(mod8)$, if $ a$ is a odd, we have
\[ (z^2\plus{}y^2\minus{}2zy\cos(k\alpha),2)\equal{}2\] and
\[ (z^2\plus{}y^2\minus{}2zy\cos(k\alpha),p^2)\equal{}(2(1\minus{}\cos(k\alpha)),p^2)\equal{}p(p_k,p)\equal{}p\].
Hence
$ \nu_2(z^2\plus{}y^2\minus{}2zy\cos(k\alpha))\equal{}\nu_p(z^2\plus{}y^2\minus{}2zy\cos(k\alpha))\equal{}1$.
As a result, we can verify $ \nu_2(rightside)\equal{}nj\plus{}\frac{n\minus{}1}{2}$ and
$ n|\nu_2(leftside)$ which yields a contradiction.

Still, all that stuff concerning $ 1\plus{}\zeta_n$ is completely wrong. $ (p^2,2)$ is not $ 2$ or $ p$, but $ 1$. Also, neither $ \beta_n$ nor $ p_k$ are algebraic integers.

It would be nice for you to send a copy on the part of your algebraic number theory knowledge in order to make sure we do not misunderstang each other. My email is liangtao0517@zzu.edu.cn Thank you!

Any algebraic integer is referred to a verified algebraic number field or integer ring. For example, 1+i is integer over C but it does not exist over Z.

I don't understand what you want me to send, but sending all my knowledge (or my brain) is out of question  :D

The way you use some words is rather strange. $ 1\plus{}i$ is an algebraic integer [b]over[/b] $ \mathbb Z$ but not [b]in[/b] $ \mathbb Z$. I use exactly the standard definitions you can find in the Wikipedia or any book on that topics. I've no clue what a "verified algebraic integer" is supposed to be, and an algebraic integer isn't a ring, but an element of a number ring (in fact, of a lot of number rings) [a number ring is the ring of integers of some number field, which again is a finite extension of $ \mathbb Q$].
$ 1\plus{}i$ is as already said an algebraic integer over $ \mathbb Z$, but the numbers you use are more of type $ \frac{1\plus{}i}{2}$, which is [b]not[/b] an algebraic integer.

I am sorry I referred to "in" when I used "over" above.

My idea is that multiplying a rational integer factor to "power base" can derive a new algebraic number field from. Then we consider the algebraic integers in it.

[quote="Taoli"]Any algebraic integer is referred to a verified algebraic number field or integer ring. For example, $ 1\plus{}i$ is integer over $ C$ but it does not exist over $ Z$.[/quote] Any algebraic integer is referred to [b]being in [/b]a verified algebraic number field or integer ring. For example, $ 1\plus{}i$ is integer [b]in[/b] $ C$ but it does not exist [b]in[/b] $ Z$.

I havenot got a standard English vision of Algebraic Number Theory, which is my fault. Thank you!

[quote="ZetaX"]What is "power base"¿[/quote]

In my word, it is referred to the fact.
We can derive an algebraic number field over $ \mathbb{Q}$ from a set of algebraic numbers $ \alpha_n, n\in\mathbb{N}$.

Would it be better for me to use "base"?

[quote="ZetaX"]Ok, but what is the meaning of "verified number field"¿ The "verified" is a bit strange.[/quote]

specific.

sorry, I should train my vocabulary.

I will be back tomorrow. Thank you!

I hope I have not misunderstood the following facts. Any algebraic integer is referred to be in a specific algebraic number field. $ \beta_n$ is chosen to be power base of another algebraic number field, not to be integer in $ \mathbb{Q}\{\zeta_n\}$. It follows that $ \mathbb{Z}\{\beta_n\}$ is a ring and $ 1 \plus{} \zeta_n$ is not a unit any more for $ (1 \plus{} \zeta_n)^{ \minus{} 1}$ is not a integer in $ \mathbb{Q}\{\beta_n\}$.

My true idea is that if we use $ 1/2$ as the role of power base $ 1$, then $ n \plus{} 1/2$ is an "integer" multiple of power base $ 1/2$, under which $ n \plus{} 1/2$ is called an integer. Is it a well action?

Yes, base sounds more familiar than power base.
But still:
If you define "integer" dependent on some base, you cannot speak of "prime" etc. anymore: If you consider only sums of those numbers, you don't get a ring. If you consider the smallest subring containing it, then you get some localisation.

In an algebraic integer ring, If and only if (\alpha)(\alpha)^{-1}=1, it will act as a unity for every elment of this algebraic integer ring. It follows that some unity in the extension field of this algebraic integer ring can act as the role of a prime in this algebraic integer ring.

I think the definitions "ring" "field" I am using are referred to the ones in advanced algebra, not the specific in algebraic number theory.

[quote="Taoli"](\alpha)(\alpha)^{-1}=1[/quote]
That equation is always true. You want $ \alpha^{\minus{}1}$ to be in your ring.
And a unit is never prime (e.g. by definition). For the rest of the first part, I don't understand what you want to say.

And there is only one definition of ring/field. Number theory knows much more rings than number rings.

$ 1\equal{}(1\plus{}\zeta_n)(r/2), r$ is an algebraic integer but $ r/2$ is not.

You claimed $ 1\plus{}\zeta_n$ is unity above. [url]http://en.wikipedia.org/wiki/1_(number)[/url]

The "unity" I am using referred to $ \alpha^{\minus{}1}$ is an algebraic integer.

Yes I claimed that (and still believe it: I calculated it's norm to be $ 1$). Why do you think that the corresponding inverse is not an integer¿

The definition that if and only if the norm of an algebraic number is 1, the algebraic number is called unity is in algebraic number theory. I do define "integer" dependent on some base and also speak of the definitions such as "prime" etc, which follows. I have verified from the difinition of ring that I get a ring in lemma 2 which derive an extension field. And as a result I consider the "integer"s in the field.

The idea above follow that I think all of the coefficients of the corresponding inverse are not rational integers.

It's really time that you write down the defintion of that "ring" you use and prove that it is a ring.

I follow the definition of a ring to verify every item of the definition.
God bless me to have done a right ratiocination.

So you divide by two to multiply by 2 again¿ And writing $ \beta_n^k$ for something that isn't the $ k$-th power of $ \beta_n$ can be very confusing...

What's the "number of power bases"¿ If it's the dimension, then it's not $ n$ (but $ n\minus{}1$).

The reason for the first question is to keep the mark forms identical.
The reason for the second is a pure mark.

$ 1, \zeta_n^i, i \equal{} 1,...,n \minus{} 1$ is an example for my usage.

The "base" you give isn't lineary independent: their sum is $ 0$.

I think I have taught myself a wrong definition. Sorry.

If the "base" should be lineary independent, it's number is $ n\minus{}1$.

Thank you!

Are the attachments to say the acceptance of my paper?
What should I do in the following steps?
Thank you for you to advise me!

It is my hope to be your colleague and friend.


inequalities
function
inequalities proposed

$ {\text{( x , y ) }} \in \mathbb{R}$ Find All $ f,g: \mathbb{R} \to \mathbb{R}$

                                               $ f(x \plus{} yg(y)) \equal{} xf(x) \minus{} yf(x) \plus{} g(x)$

[quote="kunny"]$ \geq$ or $ \leq ?$[/quote]

Nothing $ \geq$ or $ \leq ?$ .  Functions  are  Equal  $ (\equal{})$

O.K. but, you shouldn't post the problem here. :wink:

Someone please move this topic to the Algebra forum.

If we let $ y\equal{}0$ we get $ f(x)\equal{}xf(x)\plus{}g(x)\implies g(x)\equal{}(1\minus{}x)f(x)\,\forall x$.

Thus our equation becomes $ f(x\plus{}y(1\minus{}y)f(y))\equal{}xf(x)\minus{}yf(x)\plus{}(1\minus{}x)f(x)$

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

In the last equality, $ y\equal{}1$ yields $ f(x)\equal{}0$. This implies $ g(x)\equal{}0$. Thus $ f\equiv g\equiv 0$.

I'm somewhat sure this isn't correct... :maybe:


geometry
trigonometry
area of a triangle
Heron\u0027s formula
complex numbers

The side lengths of a triangle are 15, rad165, and rad130. How do you find the area using Heron's Formula. I know what Heron's Formula is and how to use it with rational numbers, and I can actually do it with complex numbers too, but I can not figure out how to do it with irrationals.

Do you mean calculate it by hand?

By calculator, it's simple; just punch in the buttons.  :wink:

I know how to do it by calculator. I just don't know how to do it by hand. Sorry for not being so specific.  :)

In this case, this is all about how to add, subtract, multiply, and divided radicals. Check out your textbook on how to do this.

If you knew how to do arithmetic with complex numbers, you would know how to do arithmetic with irrational numbers too, since the latter is a subset of the former.  I'm also not sure why this question was specific to Heron's formula, since your problem is general.  Try this [url=http://en.wikibooks.org/wiki/Arithmetic/Radicals]Wikibook[/url].

[i]It helps to write Heron's Formula in the most convenient way. Writing Heron's Formula using semiperimeter is helpful if you are dealing with integers or with decimal approximations. With radicals, however, that is probably not too helpful; we can experiment with some other ways of writing Heron's Formula.[/i]

Let's multiply each of the four inside terms by $ 2$, thereby multiplying the interior of the squareroot by $ 16$, and compensate by dividing the whole interior by $ 16$. It looks like this:

$ K \equal{} \sqrt {\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)}{16}}$

You could also expand the interior:

$ K \equal{} \sqrt {\frac {2 \cdot \left( a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \right) \; \minus{} \; (a^4 \plus{} b^4 \plus{} c^4)}{16}}$

So now you have a formula in terms of the squares of the sides rather than the sides. This will get rid of those nasty radicals in your particular problem.

We can do a little manipulation in there to ease up the computation, too:

$ K \equal{} \sqrt {\frac {a^4 \plus{} b^4 \plus{}  c^4 \; \minus{}  | a^2 \minus{} b^2 |^2 \;  \minus{}| b^2 \minus{} c^2 |^2 \; \minus{}  |c^2 \minus{} a^2 | ^2 }{16}}$

This just makes some of the arithmetic lighter if you are dealing strictly with integers and squareroots. In this case, we have $ a^2 \equal{} 225$, $ b^2 \equal{} 130$, and $ c^2 \equal{} 165$, so:

$ K \equal{} \sqrt {\frac {225^2 \plus{} 130^2 \plus{} 165^2 \minus{} 95^2 \minus{} 35^2 \minus{} 60^2}{16}}$

if you know of the angles, then you can use the 1/2 sin something something formula....

Wickedestjr, in this problem, is referring to finding the area with [b]sides only[/b].

And what is the $ \frac12 \sin ...$ thing?

I believe he means $ \frac{ab\sin{C}}2$.

[quote="TZF"][i]It helps to write Heron's Formula in the most convenient way. Writing Heron's Formula using semiperimeter is helpful if you are dealing with integers or with decimal approximations. With radicals, however, that is probably not too helpful; we can experiment with some other ways of writing Heron's Formula.[/i]

Let's multiply each of the four inside terms by $ 2$, thereby multiplying the interior of the squareroot by $ 16$, and compensate by dividing the whole interior by $ 16$. It looks like this:

$ K \equal{} \sqrt {\frac {(a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b \plus{} c)( \minus{} a \plus{} b \plus{} c)}{16}}$

You could also expand the interior:

$ K \equal{} \sqrt {\frac {2 \cdot \left( a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2 \right) \; \minus{} \; (a^4 \plus{} b^4 \plus{} c^4)}{16}}$

So now you have a formula in terms of the squares of the sides rather than the sides. This will get rid of those nasty radicals in your particular problem.

We can do a little manipulation in there to ease up the computation, too:

$ K \equal{} \sqrt {\frac {a^4 \plus{} b^4 \plus{} c^4 \; \minus{} | a^2 \minus{} b^2 |^2 \; \minus{} | b^2 \minus{} c^2 |^2 \; \minus{} |c^2 \minus{} a^2 | ^2 }{16}}$

This just makes some of the arithmetic lighter if you are dealing strictly with integers and squareroots. In this case, we have $ a^2 \equal{} 225$, $ b^2 \equal{} 130$, and $ c^2 \equal{} 165$, so:

$ K \equal{} \sqrt {\frac {225^2 \plus{} 130^2 \plus{} 165^2 \minus{} 95^2 \minus{} 35^2 \minus{} 60^2}{16}}$[/quote]

Oh, I get it!!! So do you mean multiply the expressions, so that it is all powers of two, that way, you end up with rationals?

Yeah, that's what ends up happening. By "powers of 2", I assume you mean, "even powers".

Alright, I'll do that next time I have a problem similar to this one. Thanks for your help.  :)


logarithms
combinatorics unsolved
combinatorics

For any two people in a party of $ kn$ people, there is some person who knows both of them. Prove that if $ k<\frac{n}{2\ln n}$ then there are $ n$ people who all know one another.

This is not true.
Suppose that $ k \equal{} 3$ and $ n\geq 17$ (so that $ k < \frac {n}{2\ln n}$) and there are $ 3n$ people at the party, partitioned into $ n$ triples such that people from the same triple know each other but don't know any person from the other triples. Then there are no $ n$ people knowing each other.

[quote="maxal"]This is not true.
Suppose that $ k \equal{} 3$ and $ n\geq 17$ (so that $ k < \frac {n}{2\ln n}$) and there are $ 3n$ people at the party, partitioned into $ n$ triples such that people from the sa
me triple know each other but don't know any person from the other triples. Then there are no $ n$ people knowing each other.[/quote]
I think your constraction is not proper. "For any two people in a party of  kn people, there is some person who knows both of them." So who know both of person who are from different triples? :huh:
But i also think that problem is not true. See below :
[hide] Suppose that $ k \equal{} 3$ and $ n\geq 17$ (so that $ k < \frac {n}{2\ln n}$)(as maxal supposed) ,also suppose that n-is odd. So we have $ 3n$ people at the party. Let one of them(suppose he's person B) know $ 3n \minus{} 1$ people. and it's remained to maintain $ 3n \minus{} 1$ people. Let $ 3n \minus{} 1$ people partitioned into $ \frac {3n \minus{} 1}{2}$ pairs such that people from same pair know each other, but don't know any other person (except person B). So for every pair of people in the party there is someone who know both of them. But there is no $ n$ people who  know each other.[/hide]

SaYaT, thanks for correction and working counterexample.

I think there may be a typo in the original formulation - the statement [i]"then there are $ k$ people who all know one another"[/i] seems to be plausible.


The big square has a side length of 8. For each smaller square, the side of the square it is inscribed in is its diagonal. What's the area of the shaded region?
[img]http://img.photobucket.com/albums/v189/bigsister/mathdiagram.gif[/img]
(The black border is for style. Pretend it doesn't exist.)

It should be 48

I'm not sure... :maybe:

[hide]The area of the biggest square is 64

The are of the 2nd biggest square is (4* :sqrt: 2)^2 = 32

The are of the 3rd biggest square is 4^2 = 16

The area of the smallest square is (2* :sqrt: 2)^2 = 8

And you just subtract the 3rd buggest square from the 2nd and smallest squares.

32+8-16 = 24[/hide]

[hide]Right. Notice that the inner diagonal square has side 4rt2, so area 32, and the other inner inner shaded regionis half of half of 32 = 8, and the outer shaded region is 16. 16 + 8 = 24.[/hide]

[hide]Isn't it that the medium square is half the area of the biggest one and the smallest one is half the area of the medium. Then the non-shaded area is 1/4 of the large square's. Thus, 8^2*3/4=64*3/4=48[/hide]

Wait, is the area inside the border the largest square or is the largest one the bigger shaded square? I mean, which one has the side 8??

The white square with the border is the one with side 8. Sorry for not being clear, I was wondering how pakagawa got such a weird answer...

[hide]Or, you could just figure out the area of each triangle and not have to go through more complicated radicals, but it's all personal preference.[/hide]

In the diagram, if we pretend the four corners with area 4 don't exist, then we have a shape where the shaded region is exactly the same as the red portion. So (8^2-4*4)/2=24 is the answer.


Find those pairs of positive integers $(n,k),\ k\le n$ for which it is possible to fill an $n\times n$ board with reals such that the sum of all the numbers in the board is positive, but for any $k\times k$ (connected) square on the board, the sum of all the numbers written in it is negative.

Answer: any pair (n,k) such that k doesnt divide n.
Construction if k doesnt divide n: write 1 in columns 1,k+1,2k+1,..., write
x in cells (i,j) with k divide i and k divide j and write 0 in other cells.
We need x+k<0 and n([n/k]+1)+x[n/k]^2>0. Its easy to see that there
exists such x.
Is this correct?


geometry
geometric transformation
reflection
probability
rectangle
ratio
3D geometry

Some of these may not be too hard, but show the BEST way to find the solution! (StTg96)

1.  Angela attached numbers to the new lockers at the middle school. She used one stick-on number for each digit in the locker number. How many 5's did she use to number lockers 1 through 500 inclusive?

2.  Doug has two watches, one of which gains 1 second per hou and the other of which loses 6 seconds every 24 hours. He sets both to the correct time at 9:00pm. How many hours will pass before the positive difference in the times shown is 2 hours?

3. A cowboy agreed to work for one year for $ \$19,200$ plus a horse. He quit after 7 months and was paid $ \$10,450$ plus a horse. How many dollars was the horse worth, if the paid salary reflects the fair proportion of his yearly salary?

4.  A number can be written in the form ([i]m[/i] + 11)([i]n[/i]+22), where [i]m[/i] and [i]n[/i] are two digit numbers created by using each of the digits 2, 3, 4 and 5 exactly once. What is the smallest such number?

5.  A pair of dice must be rolled until a 7 occurs. What is the probability that a 7 is obtained on the fifth toss and not before? Express your answer as a decimal to the nearest hundredth.

6.  Find the smallest integer n > 1 so that the units digit of $ d^n$ is the same as the units digit of [i]d[/i] for all positive integers [i]d[/i].

7.  Rectangle ABCD, with AB = 2BC, is inscribed in a semicircle of radius 5 m. How many square meters, to the nearest whole number, are in the area of the shaded region?

8.  Cleo used 19¢ stamps and 29¢ stamps to mail a package. She spent exactly $ \$5.88$ (There is no tax on stamps.) How many 29¢ stamps did Cleo buy?

[quote="danb80"]Some of these may not be too hard, but show the BEST way to find the solution! (StTg96)

1.  Angela attached numbers to the new lockers at the middle school. She used one stick-on number for each digit in the locker number. How many 5's did she use to number lockers 1 through 500 inclusive?

2.  Doug has two watches, one of which gains 1 second per hoe and the other of which loses 6 seconds every 24 hours. He sets both to the correct time at 9:00pm. How many hours will pass before the positive difference in the times shown is 2 hours?

3. A cowboy agreed to work for one year for $ \$19,200$ plus a horse. He quit after 7 months and was paid $ \$10,450$ plus a horse. How many dollars was the horse worth, if the paid salary reflects the fair proportion of his yearly salary?

4.  A number can be written in the form ([i]m[/i] + 11)([i]n[/i]+22), where [i]m[/i] and [i]n[/i] are two digit numbers created by using each of the digits 2, 3, 4 and 5 exactly once. What is the smallest such number?

5.  A pair of dice must be rolled until a 7 occurs. What is the probability that a 7 is obtained on the fifth toss and not before? Express your answer as a decimal to the nearest hundredth.

6.  Find the smallest integer n > 1 so that the units digit of $ d^n$ is the same as the units digit of [i]d[/i] for all positive integers [i]d[/i].

7.  Rectangle ABCD, with AB = 2BC, is inscribed in a semicircle of radius 5 m. How many square meters, to the nearest whole number, are in the area of the shaded region?

8.  Cleo used 19¢ stamps and 29¢ stamps to mail a package. She spent exactly $ \$5.88$ (There is no tax on stamps.) How many 29¢ stamps did Cleo buy?[/quote]
[hide="hints"]
[hide="are you sure?"]
[hide="you did attempt the problem, right?"]

1) Be organized.
2) Think of it as one watch being exactly on time and the other gaining 5/4 seconds every hour
3) Standard algebra
4) I think $ m=20+d$, where $ d$ is a digit.  And the answer I got is the year I was born in...
5) Standard probability question
6) I tried proving the case without 10 cases
7) Use a calculator only for the last step.
8) Take mod 19 of both sides of $ 19a+29b=588$.  Remember you can divide in the congruence since 19 is prime.  Hey look, my first multi-line hint!

[/hide]
[/hide]
[/hide]

im kind of tired and a bit fuzzy, so i may have done some stuff completely wrong. im sorry, and please point it out for me.

1.
[hide]
i got 96

i counted how many 5's were in 1-99, inclusive, and got 19.
between 100-199, there are 19 5's.
between 200-299, there are 19 5's.
between 300-399, there are 19 5's.
between 400-499, there are 19 5's.

instead of counting all that, you can use your logic and do 19x5.

also, because 500 is included, so (19x5)+1
95+1=96
[/hide]

2. i think you meant 'hour' instead of 'hoe'......
[hide]
i got 560 hours, but that is ridiculously large. someone check my work.

one watch gains 1 sec per hour.
the other looses 1/4 sec per hour. (simple division, 6/24)

after an hour, the first watch has 1 hour and 1 sec past.
so that is 3601 seconds each 'hour'
after an hour, the second watch has 59 minutes, 59 and 3/4 seconds past.
so 3599 and 3/4 seconds each 'hour'
therefore, the difference between each watch per hour is 5/4 secs.

set up a ratio
5/4 secs: 1 hr = 7200 secs (2 hours in secs): x hr.
and solve
7200 : x5/4
5x = 28800
x =  560
[/hide]

3.
[hide]
1,800 dollars.

set up a ratio
12 months:(19,200+h) = 7 months:(10,450+h)
7(19,200+h) = 12(10,450+h)
134,400+7h = 125,400+12h
5h = 9,000
h = 1,800
[/hide]

4.
[hide]
2016

instead of punching all the possibilities in the calculator, its easier to use your logic.

to have the smallest number possible, you want your 4 and 5 as the units value of m or n. the possible choices are: 35 & 24, and 34 & 25.
after that, you just need to work it out.
so you get m=25, n=34.

check:
(25+11)(34+22)
56x36
2016
[/hide]

5. um... im a bit confused.

6.
[hide]
60

you are looking for a pattern in unit digit of numbers that had been squared, cubed, or whatever.

i.e.
2^1=2
2^2=4
2^3=8
2^4=16
2^5=32
2^6=64
and so on.
take a look at the units digit. 2, 4, 8, 6, 2, 4.... its an unending pattern.

so when 2 is put to the power of 5, the units digit starts to repeat.
all the numbers (1,2,3,4,5,6,7,8,9) have this pattern, though not necessarily the same power (i think many of you guys know this stuff). so what you are looking for is the exponent when the units digit of a number starts repeating.

2 starts repeating when the exponent is 5
3 starts repeating when the exponent is 5
4 starts repeating when the exponent is 3
5 has 5 as the units digit, always
6 has 6 as the units digit, always
7 starts repeating when the exponent is 4
8 starts repeating when the exponent is 5
9 starts repeating when the exponent is 3

5 and 4 has been repeated, and there is no reason to multiply them more than once. so 5x4x3=60
[/hide]

7.
[hide]
25(pi-2)/2

find the area of the semicircle first
5^2pi
25pi, but its a semicircle, so 25pi/2

now find the area of the rectangle.
(to make this easier, i'll call the center of the circle, E)
because AB/2=AD, if you cut the rectangle vertically in half, it becomes a square. the diagonal of the square is also the radius of the circle, so AE=5.

use the 45-45-90 theorum to find the length of AD.
AD= (5 radical 2)/2
because AB=2AD, AB= (5 radical 2)
now you know the length and the width, so the area of the rectangle is
(5 radical 2)/2 x (5 radical 2)= 25.

subtract the area of the rectangle from the area of the semicircle.
25pi/2 - 25
25pi - 50/2
simplify:25(pi-2)/2
[/hide]

8. um... im still confused.

#4, I got something smaller, and #6 has a single digit answer.

I'll check the rest later, I kinda have to do my english homework...

[hide="Spider"]
1.  You seem to have left out the double five

2.  Think about it some more.  Remember one is going back while the other goes forward...

3.  Correct!

4.  Smaller number possible

5.  Review your probability.  You might want to borrow the probability book from me.

6.  We're looking for [i]n[/i]...

7.  Read the question more carefully!

8.  You can figure this out using algebra![/hide]

Am I crazy, or did I get something much larger than you for #2?

I couldn't get on until now..stupid finals..

[hide="I think"][hide="Not sure on few"][hide="This is fun.."]
Casework:
5~95=20
105~195=[b]20[/b]
205~295=[b]20[/b]
305~395=[b]20[/b]
405~495=[b]20[/b]
500=[b]1[/b]
$ \boxed{101} \ 5'\text{s}$
I looked at it this way. one of which gains 24 seconds and the other loses 6 seconds per day.
That's 30 second diffence per day.
2 hours=7200(I hope I am right)seconds. Divide it by 30, and multiply by 24 to get $ \boxed{5760} \  \text{hours}$
Since #3 has been solved, I'll skip.
I keep getting 2116, but it's wrong.. BTW, alanchou, how did you come up with $ m=20+d$?
I am probably wrong on this.. but I tried.
I found that the probabillity of 7 occuring in one throw is $ \frac16$
Presence of 7 is absent for 4 times $ \rightarrow(\frac{5}{6})^4$
Then it happens in the fifth try$ \rightarrow \frac16$
I multiplied them to get $ \boxed{\approx{0.08}}$
The units digit repeats in pattern of 4, the smallest has to be $ \boxed 5$
I agree with spider's method, but she went off the course.. I get $ x= \frac{5\sqrt{2}}{2}$ and the area of the rectangle is $ 2x^2 = 25$
I get about $ \boxed{14}$
I modded out 19 just like alanchou told us to. That gave me $ a+10b=18$ and since $ a$ and $ b$ can't be negative, I get $ b=1$ so $ \boxed{1}$[/hide][/hide][/hide]

DK:  Good job.  Need to work on #'s 4 and 8

[hide="Hint for #4"]
Minimize your values for your two digit numbers [i]m[/i] and [i]n[/i], not the sums.

By the way, alanchou was not born after the year 2000.  He may be good, but not under 8 years old!  (See his hints)[/hide]

[hide="hinto for 8"]So like $ 19x\plus{}29y\equal{}588$, mod 19 gives $ 10y$ is 18 mod 19, keep adding 19 until you can divide by 10.[/hide]

[hide="following up from xpmaths hint"]
Alternatively, once you get the congruence, multiply both sides by 2, since $ 10\times 2\equiv 1$ (mod 19)
[/hide]

[quote="xpmath"][hide="hinto for 8"]So like $ 19x \plus{} 29y \equal{} 588$, mod 19 gives $ 10y$ is 18 mod 19, keep adding 19 until you can divide by 10.[/hide][/quote]
Since the last digit have to be 2 in order to make it divisible by 10 after adding it to 18, I think it should be 8.

How do you guys know these so well? I feel left behind.lol.

I know that number's have to be small. I know that in order to get the smallest product, the numbers have to be close to each other.
@alanchou: Can you explain how you got $ m\equal{}20\plus{}d$ for 4? I didn't get that.

Well, I just thought about it a little.

Yes, that is a true statement DK, as long as the larger number is still larger after subtracting.  Basically, you're saying that if $ a\geq b$ and by subtracting $ c$ from $ a$ and adding $ c$ to $ b$ so that $ a\minus{}c\geq b\plus{}c$, then the average remains the same while the product increases.  In fact, this a the lemma used in the proof of the AM-GM inequality.  Well, the proof that's in AoPS v2.

This practically gives away the answer, so only read if desperate and mods are confusing you...and you can't find another way to solve the problem
[hide="Hint on #8"]
Since mods are sometimes confusing, we can use a more expanded method. (Since this is a target question, you probably have time--and a calculator--to do more long hand work)

We know that some multiple of 19 plus some multiple of 29 adds to be 588
So lets list some of the multiple of 19 and 29
19, 38, 57, 76, 95, 114, 133, 152, 171, 190, 209--here the last digits start repeating, 228, 247, 266, 285...
29, 58, 87, 116, 145, 174, 203, 232, 261, 290, 319--here the last digits start repeating, 348, 377, 406, 435. 464, 493, 522, 551...
Now look for combinations that add to end in 8 (this saves a lot of unnecessary adding)[/hide]

:rotfl:  [hide]14 for problem 7[/hide]

[/list][/code]

Yep, but hide your answers, and explain your solution

[hide]I got $ \boxed{1995}$ for problem 4.  Is this correct?  I did $ (24 \plus{} 11)(35 \plus{} 22)$.

I don't get number 6.  Can't $ d$ keep changing?

I got 17 for number 8.  Is this correct?

Thanks.[/hide]

Your number 8 is correct, and your answer for #4 agrees with mine

[hide="6"]
The answer is $ \boxed{5}$.
[/hide]

Question 7
[hide]Q7. 14sqm.?[/hide]

I am just randomly doing questions, and I just, DK Y, did 7. It is WAY past my normal bedtime, so I am tired!!!!


Find the # of positive integers less than 1,000,000 where the sum of the digits is 31.

[quote="E^(pi*i)=-1"][hide]
Standard balls and urns gives $ \binom{36}{5}$.
[/hide][/quote]

How does that work? I looks like the number of solutions to $ a\plus{}b\plus{}c\plus{}d\plus{}e\plus{}f\equal{}31$, but that isn't really what we're after...

Remember, if the number is less than 6 digits we can fill in the beginning imaginary digits with 0s.

sure, but don't we still overcount some. For example, one solution would be $ 31\plus{}0\plus{}0\plus{}0\plus{}0\plus{}0 \equal{} 31$

But this doesn't represent any base 10 number, we need $ \{a,b,c,d,e,f\}\in[0,9]$

Very Nice!

But how did you find the coefficient in the last step?

[quote="m00li"][hide]
coeff of $ x^{31}$ in
$ (x^0 \plus{} x^1 \plus{} ... \plus{} x^{10})^6$
or $ (1 \minus{} x^{10})^6/(1 \minus{} x)^6$
or
$ \binom{36}{5} \minus{} \binom61\binom{26}{5} \plus{} \binom62\binom{16}{5} \minus{} \binom63\binom65$
[/hide][/quote]

How does the coefficients work?
$ (x^0 \plus{} x^1 \plus{} ... \plus{} x^{10})^6$

Isn't the function $ x^0 \plus{} x^1 \plus{} \cdots \plus{} x^9$? :maybe:


modular arithmetic
calculus
integration
AMC

[hide] Let the biking rate be $ b$, swimming rate be $ s$, jogging rate be $ j$, all in km/h.

We have $ 2b\plus{}3j\plus{}4s\equal{}74,2j\plus{}3s\plus{}4b\equal{}91$. Subtracting the second from twice the first gives $ 4j\plus{}5s\equal{}57$. Mod 4, we need $ s\equiv1\pmod{4}$. Thus, $ (j,s)\equal{}(13,1),(8,5),(3,9)$.

$ (13,1)$ and $ (3,9)$ give non-integral $ b$, but $ (8,5)$ gives $ b\equal{}15$. Thus, our answer is $ 15^{2}\plus{}8^{2}\plus{}5^{2}\equal{}\boxed{314}$[/hide]

[quote="The QuattoMaster 6000"][quote="E^(pi*i)=-1"]Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $ 74$ kilometers after biking for $ 2$ hours, jogging for $ 3$ hours, and swimming for $ 4$ hours, while Sue covers $ 91$ kilometers after jogging for $ 2$ hours, swimming for $ 3$ hours, and biking for $ 4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.[/quote]
[hide="Alternate Solution"]
Let $ b$, $ j$, and $ s$ be the biking, jogging, and swimming rates of the two people.  Hence, $ 2b \plus{} 3j \plus{} 4s \equal{} 74$ and $ 4b \plus{} 2j \plus{} 3s \equal{} 91$.  Subtracting gives us that $ 2b \minus{} j \minus{} s \equal{} 17$.  Adding three times this to the first equation gives that $ 8b \plus{} s \equal{} 125\implies b\le 15$.  Adding four times the previous equation to the first given one gives us that $ 10b \minus{} j \equal{} 142\implies b > 14\implies b\ge 15$.  This gives us that $ b \equal{} 15$, and then $ j \equal{} 8$ and $ s \equal{} 5$.  Therefore, $ b^2 \plus{} s^2 \plus{} j^2 \equal{} 225 \plus{} 64 \plus{} 24 \equal{} \boxed{314}$. [/hide][/quote]

you can do that??? i need to brush up on my algebra

Wow, I got crazy lucky on this problem and just estimated the rates, starting with 15kph for the biking, and got it from there, so I guess good old educated guess and check still works!

Cool, i did the same thing! I looked at the problem, left it blank, moved on. Came back later, moved on. Came back again with about 20 min left in the contest, started guessing numbers. Guessed b=15, and then it just came together. 314? no reference to $ \pi$? :P

[quote=The QuattoMaster 6000][quote="E^(pi*i)=-1"]Ed and Sue bike at equal and constant rates.  Similarly, they jog at equal and constant rates, and they swim at equal and constant rates.  Ed covers $ 74$ kilometers after biking for $ 2$ hours, jogging for $ 3$ hours, and swimming for $ 4$ hours, while Sue covers $ 91$ kilometers after jogging for $ 2$ hours, swimming for $ 3$ hours, and biking for $ 4$ hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.[/quote]
[hide="Alternate Solution"]
Let $ b$, $ j$, and $ s$ be the biking, jogging, and swimming rates of the two people.  Hence, $ 2b \plus{} 3j \plus{} 4s \equal{} 74$ and $ 4b \plus{} 2j \plus{} 3s \equal{} 91$.  Subtracting gives us that $ 2b \minus{} j \minus{} s \equal{} 17$.  Adding three times this to the first equation gives that $ 8b \plus{} s \equal{} 125\implies b\le 15$.  Adding four times the previous equation to the first given one gives us that $ 10b \minus{} j \equal{} 142\implies b > 14\implies b\ge 15$.  This gives us that $ b \equal{} 15$, and then $ j \equal{} 8$ and $ s \equal{} 5$.  Therefore, $ b^2 \plus{} s^2 \plus{} j^2 \equal{} 225 \plus{} 64 \plus{} 24 \equal{} \boxed{314}$. [/hide][/quote]

Hello, what is the motivation behind this, and how could I be able to recognize something like this -- these questions pop up on AIME ever so often that these are useful,
Thanks

[hide=Solution]
We convert the words into math to get that $74=2b+3j+4s$ and $91=2j+3s+4b$ where $b$, $j$, and $s$ are the speeds for biking, jogging, and swimming respectively.

We subtract to get that $17=2b-j-s$ so $51=6b-3j-3s$.

We add $51=6b-3j-3s$ to $74=2b+3j+4s$ to get that $125=8b+s$.

We add $51=6b-3j-3s$ to $91=2j+3s+4b$ to get that $142=10b-j$.

We see that the first equation tells us that $125\ge 8b$ so $15\ge b$.

We see that the second equation tells us that $142\le 10b$ so $15\le b$.

We see that $b=15$ so $s=125-8\cdot15=5$ and $j=10\cdot15-142=8$.

We calculate our answer to be $15^2+5^2+8^2=\boxed{314}$.
[/hide]

[hide=Solution]Let's say they bike at $x$ miles per hour, jog at $y$ miles per hour, and swim at $z$ miles per hour. Then we have $2x+3y+4z=74$ and $4x+2y+3z=91$, so $y=10x-142, z=125-8x$. This means $\frac{142}{10}<x<\frac{125}{8}$, so $x=15$. Then $y=8, z=5$, so the answer is $15^2+8^2+5^2=\boxed{314}$.[/hide]


advanced fields
advanced fields unsolved

show that every prime divisor of a^2+b^2 where mcd(a,b)=1 is in the form of c^2+d^2[/url]

by mcd i meant maximum common divisor but i remembered is called gcd.. i hope you figured it out...


[quote="alex__ib"]Prove that:

$3abc+3 \geq ab+bc+ac+a+b+c$

With a,b,c reals.

Thank you!  :)[/quote]

i dont think its true , just take $a=-1 ; b=2 ; c=2$ ,
i think there must have been more condition

Not true if $b,c$ approaching $0$ so that $abc<1$ and $a>6$.

Sorry,.....It is really annoying that i spent an hour on trying to prove it  :mad: . This can be locked.

It's equivalent to:
$\sum_{cyc}(a-1)(bc-1)\geq 0$
True for $a,b,c\geq 1$


Given a function $ f\colon \mathbb{R}\to\mathbb{R}$ defined for any real x and y such that  $ f(xf(y))\equal{}f(x)y$.Prove that$ f(xy)\equal{}f(x)f(y)$

___________________________
Azerbaijan Land of the Fire   :ninja:  :roll:  :P

The function $ f(x)\equal{}0$ for all $ x$ satisfies the equations, and then $ 0 \equal{} f(xy) \equal{} f(x)f(y) \equal{} 0$.

Now suppose there exist some $ a \in \mathbb{R}$ with $ f(a) \neq 0$.

Filling in $ x\equal{}1$ gives $ f(f(y)) \equal{} yf(1)$. Now,
$ f(1) x f(y) \equal{} f(f(xf(y)) \equal{} f(f(x)y) \equal{} f(y)x$,
Filling in $ y\equal{}a$ we get $ f(1) \equal{} 1$, so $ f(f(y)) \equal{} y$.
Filling in $ y\equal{}f(y)$ gives $ f(xy) \equal{} f(xf(f(y)) \equal{} f(x)f(y)$.


inequalities
trigonometry
geometry proposed
geometry

[quote="Ji Chen"]If $ a,b,c$ are the sides of a triangle,  $ m_a,m_b,m_c$ are the medians and  $ w_a,w_b,w_c$ are the angle-bisectors, then

$ (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\leq\left(\frac {m_a}{w_a} \plus{} \frac {m_b}{w_b} \plus{} \frac {m_c}{w_c}\right)^2,$

with equality if and only if the triangle is equilateral.[/quote]

$ m_a \ge \frac {1}{2}(b \plus{} c)\cos(\frac {\alpha}{2})$ and $ w_a \equal{} \frac {2bc\cos(\frac {\alpha}{2})}{b \plus{} c} \Rightarrow \frac {m_a}{w_a} \ge \frac {(b \plus{} c)^2}{4bc}\dots$

$ \frac {m_a}{w_a} \plus{} \frac {m_b}{w_b} \plus{} \frac {m_c}{w_c} \ge \frac{1}{4}(a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\plus{}\frac{3}{4}$

$ (a\plus{}b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c})\equal{}t \ge 9$

$ \Leftrightarrow (t\minus{}1)(t\minus{}9)\ge 0$

Nice proof, Sergic Primazon. $ \frac {m_a}{w_a} \ge \frac {(b \plus{} c)^2}{4bc}$ can be found in

G. Tsintsifas, L. Bankoff, L. Goldstone, Problem E 2471, Amer. Math. Monthly, 81 (1974) and 82 (1975), 523-524.

I approached it as follows :

$ 2\sum{\frac {m_{b}m_{c}}{w_{b}w_{c}}}\geq\sum{\frac {18a^4 \minus{} 3a^3(b \plus{} c) \minus{} a^2(17b^2 \minus{} 48bc \plus{} 17c^2) \minus{} a(b \plus{} c)(2b \minus{} c)(2c \minus{} b) \minus{} (b \minus{} c)^2(2b \plus{} c)(2c \plus{} b)}{4a(a \plus{} b \plus{} c)(c \plus{} a \minus{} b)(a \plus{} b \minus{} c)}}$

$ \equal{} (a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right) \minus{} \sum{\frac {m_a^2}{w_a^2}}.$

See also : http://www.mathlinks.ro/viewtopic.php?t=313282


Let $S$ be a square, let $C$ be the circle inscribed in $S$, and let $D$ be the circle circumscribed about $S$. Which statement most accurately describes the relationship between the areas of $C$ and $D$?

$[x] = \text{Area}$

A) $\frac{[D]}{[C]} = 2$
B) $\frac{[D]}{[C]} = 4$
C) $\frac{[D]}{[C]} = \frac{1}{2}$
D) $\frac{[D]}{[C]} = [S]$ for every square $S$
E) $\frac{[D]}{[C]}$ cannot be determined from the information given.

[quote="tarquin"]It's A because the ratio of their dimensions is $\sqrt 2$[/quote]

 :first: Nice job..

[quote="tarquin"]It's A because the ratio of their dimensions is $\sqrt 2$[/quote]

How do you find the area of a circle circumscribed around a square?

If you let the sidelength of the square be $x$ then the diagonal is $x \sqrt 2$ which is also the diameter of the circle so the area of the circle is $\left(\frac {x\sqrt 2}{2}\right)^2 \cdot \pi = \frac {x^2\pi}{2}$

This is funny because I just did a very similar problem a couple of days ago....
The key to this problem, as tarquin said, is the diagonal of the square. it is the diameter of the circle circumscribing the square.

Yeah, I figured out that method a while ago.


PLEASE, i do not know who to do any of these problems.

2.
An investment firm has 100,000 dollars to invest for a client and decides to invest it int two stocks A and B. The expected annual rate of return, or interest, for stock A is 15%, but there is some risk involved, and the client does not wish to invest more than 50,000 dollars in this stock. the annual rate of return on the more stable stock B is anticipated to be 10%. Determine how much should be invested at each rate so that the annual interest from these investments is 12,000.

4.
A radiator contains 8 quarts of a mixture of water and antifreeze. If 40% of the mixture is antifreeze, how much of the mixture should be drained and replaced by pure antifreeze so that resultant mixture will contain 60% antifreeze?

6.
A pharmacist is to prepare 15 ml of special eye drops for a glaucoma patient. The eye drop solution must have a 2% active ingredient, but the pharmacist only has 10% solution and 1% solution in stock. How much of each type of solution should be used to fill the prescription?

honestly, i do not know how to do these problems. i learn by viewing examples and their answers, then reproducing that with a different question and trying it myself.

so far ive gotten

0.1 V1 + 0.01 V2 = 0.3

lol yeah i know i havent gotten far, but i seriously do not know how to finish it[/hide]


Show that $ i^{n}$, when n is integral has one of four values $ \pm1$, $ \pmi$

[hide="Applying Rules for Exponents"]
$ i^{1}= i$
$ i^{2}=-1$
$ i^{3}=-i$
$ i^{4}= i^{2}\cdot i^{2}= 1$
$ i^{5}= i^{1}\cdot \left( i^{2}\right)^{2}=-i$
$ \vdots$

When continued those same four values $ \pm 1 \ , \ \pm i$ are repeated.
[/hide]

Can someone prove it more rigorously?

[hide]1. $ i^{1}=i$
2. $ i^{2}=-1$
3. $ i^{3}=-i$
4. $ i^{4}=1$
5. $ i^{n}=i^{n}\cdot 1=i^{n}\cdot i^{4}=i^{n+4}$
6. $ i^{n}=\frac{i^{n}}{1}=i^{n}i^{-4}=i^{n-4}$

Statements 1-5 cover all the positive integers.
Statements 1-4 and 6 cover all the negative integers and 0.[/hide]


Given 10^6 points in the space, show that the set of pairwise distances of given points has at least 79 elements

I thought: Suppose it is possible with 78 distances. Take a point, and of the remaining 999999 points at least $ \frac{999999}{78}$ of them are the same distance from this one point, and so lie on a sphere.

Take a point on the sphere, and of the (at least) $ \frac{999999}{78}\minus{}1$ others at least $ \frac{\frac{999999}{78}\minus{}1}{78}$ of are the same distance from this point, and so lie on another sphere centre this point. In fact they must lie on the intersection of the two spheres, which is a circle.

Take a point on this circle, and of the (at least) $ \frac{\frac{999999}{78}\minus{}1}{78}\minus{}1$ others, at least $ \frac{\frac{\frac{999999}{78}\minus{}1}{78}\minus{}1}{78}$ are the same distance from this point and so lie on another circle, and so on the intersection of the two circles. The intersection of two circles is two points, but as $ \frac{\frac{\frac{999999}{78}\minus{}1}{78}\minus{}1}{78} > 2$ there is a contradiction.

I am very unsure about the above argument as it seems too simple, and the same proof would work to prove that there are at least 80 distinct distances. Am I being very stupid, or is the bound in the question just weak?


There are 2007 stones in a pile. Two players alternately select a positive divisor of the number of stones currently in the pile and remove that number of stones. The player who removes the last stone loses. Find a winning strategy for one of the players.

The first player can remove an odd number of stones, because $ 2007$ is odd

The 2nd player has a strategy. It is : [u][b]Remove 1 stone[/b][/u]

So the 1st player has again an odd number of stones, and can remove odd, result is even, and so on

If the 2nd player wants to win quickly, he can remove more than 1 stone

[hide="How many?"]
Let $ n$ be the number of the stones when the 2nd player is called to play

Let $ n=2^{a}\cdot b$  (where $ b$ is odd)

If $ a>1$ then he can remove $ b$ stones $ \Rightarrow $ result $ =n-b=(2^{a}-1)b,$ odd

If $ a=1$ then he can remove $ x$ stones, where $ x$ is the larger divisor of $ b$ such that $ x<b$[/hide]

Generalization

The game is played with $ n$ stones, and taking the last one either wins or loses. Which rule is used is predetermined at the start. For which values of $ n$ does the result of the game change based on which rule is used? (assuming both players play optimally)


Let $ ABCDEF$ be a hexagon (which may not be regular), which circumscribes a circle $ S$. (That is, $ S$ is a tangent to each of the six sides of the hexagon.) The circle $ S$ touches $ AB, CD, EF$ at their midpoints $ P, Q, R$, respectively. Let $ X, Y, Z$ be the points of contact of $ S$ with $ BC, DE, FA$ respectively. Prove that $ PY, QZ, RX$ are concurrent.

It's easy to see that $ P$ is the middle of the arc $ XZ$, etc., hence $ PY, QZ$ and $ RX$ are the angle bisectors of $ \angle XYZ, \angle XZY$ and $ \angle ZXY$, consequently they are concurrent at the incenter of the $ \triangle XYZ$.

Best regards,
sunken rock


Ok I have been learning about motion in two dimensions and I just don't get the circular motion part. How did they derive that the acceleration=v^2/r? Also why is the acceleration inward? I don't just want to memorize the formulas...
THANKS!


Let $ n$ and $ k$ be positive integers. Please, find an explicit formula for
\[ \sum y_1y_2 \dots y_k,\]
where the summation runs through all $ k\minus{}$tuples positive integers $ (y_1,y_2,\dots,y_k)$ satisfying $ y_1\plus{}y_2\plus{}\dots\plus{}y_k\equal{}n$.

Consider a composition of $ n$ into $ k$ parts with the length of blocks from left being $ [l_1,l_2,\dots,l_k]$. Then we select one from each block, which can be done in $ l_1\cdots l_k$ ways. The summation in the problem counts this for every composition. Equivalently, we can find the number of compositions with $ 2k\minus{}1$ selections where an odd numbered selector gives the selected element of a block and an even numbered selector gives the end of the next block (right after that selector). But selecting the last element of a block shall lead to a pair of odd and even selector to lie on the same element. Thus we obtain, $ \sum_{i\equal{}0}^{k\minus{}1}{k\minus{}1\choose i}{n\choose 2k\minus{}1\minus{}i}$

Your sum simplifies to $ \binom{n \plus{} k \minus{} 1}{2k \minus{} 1}$.


[i]Can we find a pattern to calculate this sum ? [/i]

 $ S \equal{} \frac {1}{2} \plus{} \frac {1}{6 \cdot 8}\plus{}\frac {3}{40 \cdot 32}\plus{} \frac {5}{112 \cdot 128}\plus{} ...$

  [u]Babis [/u]

\begin{eqnarray*}
S & = & \dfrac{1}{2} + \dfrac{1}{6 \cdot 8} + \dfrac{3}{40 \cdot 32} + \dfrac{5}{112 \cdot 128} + \cdots \\
   & = & \dfrac{1}{2} + \dfrac{1}{\left( 2^1 \cdot 3 \right) \cdot 2^3} + \dfrac{3}{\left( 2^3 \cdot 5 \right) \cdot 2^5} + \dfrac{5}{\left( 2^4 \cdot 7 \right) \cdot 2^7} + \cdots
\end{eqnarray*}

I was not able to find a pattern :(

Unfortunately I did not find the general term of the series. I was given the problem by  two Physicians and I do not know if the source is higher physics or an olympiad competition.

 I will try to have again a contact.

Can you give at least $ 2$ extra terms to describe the series?

I don't know if anything can be brought out from this, but this is what I can see so far:
[hide]$ \frac {1}{2} \plus{} \frac {1}{6\times8} \plus{} \frac {1}{32\times40} \plus{} \frac {1}{112\times128} \plus{} ...$

$ \equal{} \frac {1}{2} \plus{} \left(\frac {1}{2}\right)^4\left(\frac {1}{3}\right) \plus{} \left(\frac {1}{2}\right)^8\left(\frac {3}{5}\right) \plus{} \left(\frac {1}{2}\right)^{11}\left(\frac {5}{7}\right) \plus{} ...$

With an unusual sequence $ 1, 4, 8, 11,...$ describing the exponents of the $ \frac{1}{2}$s.[/hide]

I don't understand how you got the final step

$ \text{arctanh } x \equal{} \frac{1}{2} \ln \left( \frac{1 \plus{} x}{1 \minus{} x} \right)$.  I would think that the confusing thing is the second-to-last step!


Is it very hard getting into PROMYS?
Did you have to answer all of the questions completely?

I left one or two blank.  I think you should just try to get good solutions to the problems you know how to do.


Find a formula for \[\frac{1}{1 \times 2 \times 3}+\frac{1}{2\times 3 \times 4}+\cdots+\frac{1}{n(n+1)(n+2)}\] Can this be generalized? Sorry if it's been posted before, but I did a search for "telescoping sums" and couldn't find anything  :maybe:

very nice Tim_Lou, that solution easily generalizes to any product of factors in the denominator!

Oops, should have seen that  :wallbash: Thanks for the solution  :)


Points $E$ and $F$ are located on square $ABCD$ so that $\Delta BEF$ is equilateral. What is the ratio of the area of $\Delta DEF$ to that of $\Delta ABE$?

[asy]
pair A=origin, B=(1,0), C=(1,1), D=(0,1), X=B+2*dir(165), E=intersectionpoint(B--X, A--D), Y=B+2*dir(105), F=intersectionpoint(B--Y, D--C);
draw(B--C--D--A--B--F--E--B);
pair point=(0.5,0.5);
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));[/asy]

$\textbf{(A)}\; \frac43\qquad \textbf{(B)}\; \frac32\qquad \textbf{(C)}\; \sqrt3\qquad \textbf{(D)}\; 2\qquad \textbf{(E)}\; 1+\sqrt3\qquad$

[hide]Let $AB=s$, $EF=x$, $DE=m$ and $AE=n$ the ratio is thus $\frac{\frac{1}{2}m^{2}}{\frac{1}{2}sn}=\frac{m^{2}}{(m+n)n}=\frac{m^{2}}{n^{2}+mn}$

Then we can note that $x^{2}=2m^{2}$ and $n^{2}+s^{2}=x^{2}$ however since $s=m+n$ then $x^{2}=2n^{2}+m^{2}+2mn$ setting both of these equations equal we get $2m^{2}=2n^{2}+2mn+m^{2}$ thus $m^{2}=2n^{2}+2mn$ which makes the ratio equivalent to $\frac{2(n^{2}+mn)}{n^{2}+mn}=2$ $\boxed{D}$

[/hide]


topology
search
real analysis
real analysis unsolved

(1). Does there exist a closed subset of $ \left[0,1\right]$ with an empty interior and positive measure?

(2). Construct a homeomorphism from $ \left[0,1\right]\cap\mathbb{Q}$ to $ \left(0,1\right)\cap\mathbb{Q}$.

For (1), I'm told that there is such a subset, but it seems to me that every closed set $ F$ of $ \left[0,1\right]$ with an empty interior has the property that for each $ \varepsilon > 0$, an interval of length $ \varepsilon /2^{k}$ can be placed around each point in $ F$ such that

\[ m\left(F\right)\leqslant\sum\varepsilon /2^{k} \equal{} \varepsilon ,\]
and therefore the subset would have measure $ 0$.

#1: Such a set was [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=231399]constructed[/url] here less than two weeks ago. It was very easy to search for.


group theory
abstract algebra
superior algebra
superior algebra solved

hi all,

here is a simple question: is there any isomorphism of group between
(GL(n,R),*) and (GL(m,R),*) where GL(m,R) is the group of invertible matrices m*m.

I wonder if the following is correct.

I want to prove that the order of a commutative subgroup of $GL_n(\mathbb R)$ for which all elements satisfy $X^2=I_n$ is at most $2^n$, and then get a contradiction. Take such a group. All of its elements are similar to some $diag (\pm 1,\pm 1,\ldots,\pm 1)$, and since they form a family of commuting diagonalizable matrices, it means that they're simultaneously diagonalizable, so our group turns out to be the conjugate of some subgroup of $G_n=(\{diag (\pm 1,\pm 1,\ldots,\pm 1)\},\cdot)$, so it has at most $2^n$ elements.

From that we can easily get a contradiction, since by an automorphism, such a group in $GL_n(\mathbb R)$ turns into a group satisfying the same properties in $GL_m(\mathbb R)$, and if we take $n>m$, then the image of $G_n$ will have to have $2^n>2^m$ elements, which is impossible.

Is it Ok?

you're right Grobber. Harazi posted a similar solution on another section.

What is your solution? I bet this problem has more than one approach...

almost the same as yours (cardinal of subgroups ..)


The length of each side of a cube is multiplied by 3.
What is the ratio of the volume of the larger cube to the volume of the smaller cube?

MY WORK:

3 times 6 sides of cube = 18

54/18 = 3/1 or 3 to 1.

But the book's answer is 27 to 1.

How can this be?

[hide]It's going to be the ratio of volumes, so it is $\frac{18^{3}}{6^{3}}=27$[/hide]

What made you divide 18^3 by 6^3?

What words in the question itself displayed that 18^3 needed to be divided by 6^3 in order to find the ratio 27/1 or 27 to 1?

[hide]okay. Let the original length of the side of the cube be a. The volume is a^3. The larger cube's side length is 3a. The volume is (3a)^2 = 27a^3. 27a^3/a^3 = 27 to 1.[/hide]

[quote="superwizard188"][hide]okay. Let the original length of the side of the cube be a. The volume is a^3. The larger cube's side length is 3a. The volume is [b](3a)^2 [/b]= 27a^3. 27a^3/a^3 = 27 to 1.[/hide][/quote]

That should be $(3a)^{3}$


Goodbye Billy, we'll all miss you!!!

As for everyone else, at this point, does it suddenly become morally wrong/insensitive to watch Billy Mays and the Orange Glowing Stuff or the like?

RIP Billy.

No, of course not. It makes sure his memory lives on.

Why not for Michael Jackson, "The King of Pop" ?

http://www.cnn.com/2009/SHOWBIZ/TV/06/28/mays.death/

It's really creepy, because at camp we were talking about both Billy Mays and Michael Jackson....[b]before[/b] they died...

a strange thing is that Michael Jackson and Billy Mays died at age 50

One of my friends said that the guy from Sham Wow was better. I was like You take that back!

[quote="dnkywin"]Yeah he hit his head on the airplane... and died... wtf[/quote]

The autopsy revealed his death was NOT caused by head trauma.

Billy Mays is believed to have died from heart disease, according to an article I read; I think the bump on the head was a coincidence.

[quote="Mewto55555"]Dude Vince Offer is a beast. He also got arrested in Florida recently.

Anyway, they finished teh autopsy. He had a blood clot.[/quote]

He talks too fast. He makes me want to slap him in the face with a smelly tuna fish.

billy is dead :'(

Now Pitchmen will be cancelled... why... Now I'll just hve to watch the reruns... WHY!!

[quote="bluecarneal"]or a wet sham wow[/quote]

Yea! give hiim a taste of his own medicine lol

Vince Offer was arrested in Florida recently, but was released.

[quote="pianoforte"][quote="1=2"]FOR THE LOVE OF GOD DON'T TALK ABOUT ME AT CAMP!!!!!![/quote]

XD I thought this was good.[/quote]

Piano, we were talking about you at camp!!!!

[quote="ernie"][quote="bluecarneal"]or a wet sham wow[/quote]

I thought ShamWow!-s dry quickly. :P[/quote]
well apparently only in commercials  :P
And also here's a nice little thing to have as ur signature:
[code][asy]import graph;
size(614);
draw(Circle((0,0),1));[/asy][/code]
Here is its output:
[asy]import graph;
size(614);
draw(Circle((0,0),1));[/asy]

lol. i still think my signature is awesome.

THE OXY-CLEAN COMMERCIAL REPLACED BILLY MAYS WITH A GIRL! HE'S NOT GONNA BE HAPPY ABOUT THIS!!!


logout is at the top

go to where it says
forum...latex help...classes

at the very end of that row is an icon that says resources

logout is right under there

MathLinks skin: Right at the top, under "Usergroups" and "Resources".
But it would be good if you send a picture how it looks for you.

in Aops skin, the logout button is at the top. But as zetaX said, it would help if you take a screen shot and post it. :)


Hi All,

I got a problem in solving the flux density equation for a cylindrical electrical machine equipped with permanent magnets. The coordinate system is cylindrical $(r,\theta,z)$.

The Maxwell equation for the flux density B in term of electromagnetic vector potential:

$\vec{B} = \nabla \times \vec{A}$

where in my case $\vec{A}=A_\theta (r,z) \vec{e}_\theta$

The  Laplace/Poisson equation is used for the governing equation:

$\nabla^2 \vec{A} = 0$ (in the free space)
$\nabla^2 \vec{A} = -\frac{1}{\mu_r} \nabla . \vec{M}$  (in the permanent magnets)

where $\vec{M}$ is magnetization

In the physical condition, it is usually used [b]B[/b] as boundary condition instead of [b]A[/b].
My quation is how to solve the above Laplace/Poisson equation (in [b]A[/b] terms) using the boundary condition (in [b]B[/b] terms). Solving [b]A[/b] from [b]B[/b] for the boundary condition will create new differential equations.

Your help would be highly appreciated.

Regards,
Wijono

I woud like to show you a part of what I have done (in the attachment, pdf file)

Thank you for your help ...


inequalities
calculus
derivative
algebra
polynomial
algebra unsolved

a.) Prove that there is no rational polynomial $f$ such that $\forall x$ both inequalities

\[f(x) > f^{''}(x), f^{'}(x) > f^{''}(x)\]

are satisfied.

b.) Prove or disprove that there is no rational polynomial $f$ such that $\forall x$ both inequalities

\[f(x) > f^{''}(x), f(x) > f^{''}(x)\]

are satisfied. Here the second inequality is different to a.)

[quote="Myth"]If he meant $f(x)>f'(x)$ in b) then answer is positive: $f(x)=x^2+3$.
BTW, what is rational polynomial?[/quote]
A rational polynomial is a polynomial having rational coefficients.

Check whether $f(x) = 2 - e^x + e^{-x}$ satisfies a.) and b.) .  :)


[quote="chemrock"]1. I think 3
2.Probably ring expansion would have caused it to lose optical activity or elimination of the quaternary amine or something like that!
3. Enna ketta athale avan settu erukka mattan avvalavu than.   :rotfl:
nanum avana kola pannala so not bothered about it :P[/quote]

Not commenting about your first 2 answers but the third one is the most brilliant question. Use chemistry da.

Onake overa ella ethavadu sollen da fool.

1. 3
2. probably bcoz the muscarine found in his body is not from the the mushroom grown in his locality( as it is opticaly active), and seeing that it has three chiral carbons, racemisation of the optically active muscarine in the body is highly improbable. the racemic mixture present in his body was probably prepared by someone to kill the Eugene guy.
3. in my opinion, he was murdered by someone

Ans 1 is right.

Ans 3 is "It is a case of mistaken identity - a chiral mushroom (of one typeconfiguration) may be useful and inverted configuration may be poisonous."

why coundt he have been murdered?
and how are we supposed to know whther both enantiomers are poisonous are not?
and what is answer 2? :(  :(

can you tell me the source of this problem, coz i came across a similar problem in morisson and boyd(stereochemistry).

no, the problem was slightly diff(did shreyas change it?)

i think it could have also been from career point test series  :)

but the answer doesnt match(morisson's and career point), and shreyas what is answer 2.


Find all positive integers $ x,y$ such that $ 3^x\plus{}4\equal{}7^y$

As is easily seen, $ x$, $ y$ are odd.
$ y\equiv 2(mod$ $ 3)$ since $ 7^y\equiv4(mod$ $ 9)$. Hence, $ 3^x\plus{}4\equal{}7^{3i\plus{}2}\equiv \minus{}8(mod$ $ 19)$. Hence, $ 3^x\equiv 7(mod$ $ 19)$. Since $ (7/19)\equal{}1$, $ x$ is even, which gives us the contradiction.

I will introduce my solution in order to solve the equation : $ 3^x \plus{} 4 \equal{} 7^y$          (1)
(1) is true when $ (x,y) \equal{} 1$.
When $ x,y\geq 2$,
By checking modulo 4, we have x and y are integers of the same parity. By checking modulo 10, we get x,y is odd.
$ y\equiv 2(mod3)$ since $ 7^y\equiv 4(mod9)$. Hence, $ y\equiv 5(mod 6)$ because y is odd.
$ 3(3^{x \minus{} 1} \minus{} 1)\equiv 0(mod7) \equal{} > 3^{x \minus{} 1}\equiv 1(mod7) \equal{} > x \minus{} 1\equiv 0(mod 3)$
Hence $ 3\equiv 3^{3m \plus{} 1}\equiv 7^y \minus{} 4\equiv7^{6n \plus{} 5} \minus{} 4\equiv ( \minus{} 1)^n.11 \minus{} 4 (mod13)$ , which gives us the contradiction.
Answer: $ x \equal{} y \equal{} 1$


for how many real number a does the quadratic equation x^2+ax+6a=0 have only integer roots for x?

[hide]$ x\equal{}\frac{\minus{}a\pm\sqrt{a^2\minus{}24a}}{2}$  A parity check shows that even and odd numbers are allowed.

$ a\equal{}24$ is trivial.
If $ a\equal{}n^3, a\minus{}24\equal{}n$, then $ (n\minus{}3)(n^2\plus{}3n\plus{}8)\equal{}0$, so $ a\equal{}27$
If $ a\equal{}bn, a\minus{}24\equal{}b$, then n is a perfect square and divides 24.  $ n\equal{}4, b\equal{}8, a\equal{}32$
If $ a\equal{}n^2, a\minus{}24\equal{}b^2$, then $ (a\plus{}k)^2\equal{}a^2\plus{}24\rightarrow 2ak\plus{}k^2\equal{}24\rightarrow a\equal{} \frac{24\minus{}k^2}{2k}\rightarrow k\equal{}2, 4\rightarrow a\equal{}25,49$.
That's 5 values for a.[/hide]

[hide="Hint"]Let the two integer roots be $ r,s$.

We then have that $ r\plus{}s\equal{}\minus{}a$ and $ rs\equal{}6a$, where $ a$ is unknown.

Equivalently, we can write $ rs\plus{}6r\plus{}6s\equal{}0 \implies (r\plus{}6)(s\plus{}6)\equal{}36$.[/hide]

another solution
[hide]
$ x\equal{}\frac{\minus{}a\pm\sqrt{a^2\minus{}24a}}{2}$
so $ \frac{\minus{}a\plus{}\sqrt{a^2\minus{}24a}}{2}$ and $ \frac{\minus{}a\minus{}\sqrt{a^2\minus{}24a}}{2}$ are both integral
add them and you get -a is integral (or a is integral) and subtract them to find $ a^2\minus{}24a$ is a perfect square
we know a is integral, so a^2-24a+144 is a perfect square
so 2 perfect squares differ by 144

define $ y\equal{}\sqrt{a^2\minus{}24a}$ and $ z\equal{}a\minus{}12$
so $ z^2\minus{}y^2\equal{}144 \implies (z\plus{}y)(z\minus{}y)$
we require y and x are both integers so z+y and z-y have the same parity (even)
and we find 5 factorizations of 144 into 2 even numbers (12*12 18*8 24*6 36*4 72*2)
we solve for a and we indeed do find 5 solutions : 24 25 30 35 49
[/hide]

Somehow a=0 was missed from everyone! :D

Yes, and so were several other values. I'm going to complete the solution I suggested earlier in this thread:

[hide="Solution"]Let the two integer roots be $ r,s$. We then have that $ r \plus{} s \equal{} \minus{} a$ and $ rs \equal{} 6a$, where $ a$ is unknown.

Equivalently, we can write $ rs \plus{} 6r \plus{} 6s \equal{} 0 \implies (r \plus{} 6)(s \plus{} 6) \equal{} 36$

To avoid redundancy, let $ r \geq s$. The solution set is $ (r,s) \in \{ ( \minus{} 12, \minus{} 12),( \minus{} 10, \minus{} 15),( \minus{} 9, \minus{} 18),( \minus{} 8, \minus{} 24),( \minus{} 7, \minus{} 42),(0, 0),(3, \minus{} 2),(6, \minus{} 3),(12, \minus{} 4),(30, \minus{} 5) \}$

The set of values of $ a \equal{} \minus{} (r \plus{} s)$ is then $ a \in \{ \minus{} 25, \minus{} 8, \minus{} 3, \minus{} 1, 0, 24, 25, 27, 32, 49 \}$, and there are [b][u]ten[/u][/b] possible values of $ a$.[/hide]


Holaa todos,
no se si conocen esta web.
pero  me sorprendio lo actualizado  q esta
y el nivel de detalle que tiene .. ;)

aki el link:
http://www.imo.org.yu/index.php?options=gl|imotres&p=02mL13
hagan click a un pais para tener detalle del pais, por año .. :)

entre a Peru:
http://www.imo.org.yu/index.php?options=gl|imotres&zemljarez=PER&p=02mL13

PD: Copiar el enlace en un navegador .. algo pasa con la direccion, creo q un caracter raro lo corta ..

prueben ...

Carlos Bravo :)
Lima -PERU

Muy buena página!!!!

Llena de información, muy buen link, gracias!

[color=blue]¡Ahora la I.M.O. tiene un [url=http://www.imo-official.com/?language=es]Sitio Oficial[/url] (ya sé que no es novedoso)!
¡Y admite varios idiomas![/color]


AMC
AIME
USAMTS
USA(J)MO
USAMO
AMC 12
AMC 10

For those of you who have done the USAMTS before, what should my score be to get into the AIME?

Thanks in Advance.

That isn't yet determined this year, but you should aim to be in the 70s after 3 rounds.  You should also take the AMC 10 and/or AMC 12.

you said that qualifying through the USAMTS would just give you a score of 100... this is only for the AMC 12 right? Wouldn't they give you a 120 for the AMC 10? because that is the qualifying score for that one.

[quote="Magnara"]If you want to qualify for USAMO, you should definately take the AMC.  Qualifying through USAMTS, no matter your scores, is equivalent to getting a 100 on AMC, which is the minimum to qualify for AIME.[/quote]

But for people who are 10th grade or below, the AMC score doesn't count for USAMO....

I believe that regardless of your grade, your contribution to your index is as if you scored 100 on the AMC 12.

[quote="anhnguyen"]Has the score been determined yet?[/quote]

I'm going to guess no, considering the fact that Round III has not been submitted or graded yet. From what I know (which may be wrong), the AMC folks take a look at people after the completion of 3 rounds and select the cutoff score, which, in the last few years has been 68.

[quote]so there are multiple ways to qualify for AIME??
I know that you need at least a 100 on AMC12 and between a 110 and 120 on AMC10 (this is based upon waht it has been in the past)
from what i read here, you can qualify through USAMTS...
does it matter which one, or do you have to do both?[/quote]

You have to qualify by one way, which can be any of these three ways. You can, however, qualify by several.

Is a 68 going to be enough this year(since that was the cutoff for last year)?

Generally, is this year harder or easier than last year?

I hope to have an answer to that by the end of the month.  We have to grade the 3rd round and talk with AMC before knowing for sure.

Please, please let the cutoff be 69 points...Darn. One trivial missed case might cost me 70.

[quote="rrusczyk"]The cut-off will be 68.[/quote]

yay! it's not 70!

[quote="sran"]Does this mean that the papers have been graded?[/quote]No.

In the profile for the usamts, I think it asked for a school code if you want to qualify for AIME. Do you have to have it or can they find out what school ur in without it?


Show that there exists no solutions in positive integers a,b,c satisfying:

k<1/a + 1/b + 1/c <1, where k = (pq+qr+rp)/pqr, where p,q,r are positive integers greater than 1.

Did you miss some condition?

Let $p,q,r \to \infty$ then $k \to 0$, and we can obviously find $a, b, c$.

you are right....looks like i can't just generalize a problem at will... :D

well, the original one is k = 41/42.

The one with $\frac {41} {42}$ is really easy, and has been proposed to the mathematic competition in Finland 2000/2001.
I've just sent a solution to Crux...

Pierre.

Hey, Pierre, try to see if the general case is also easy. If I'm not wrong, it caused some headaches to Erdos and some other people...  ;)

harazi, by the general case you mean k = (n-1)/n for sufficiently large positive integer n?

No, in fact the general problem is: what is the maximal value of  $\frac{1}{x_1}+...+\frac{1}{x_n}$ if $x_i$ are natural numbers such that $\frac{1}{x_1}+...+\frac{1}{x_n}<1$ ?[/code]


At a convention, each guest was given a ticket. $ 40\%$ of the guests received red tickets, $ \frac{3}{8}$ received blue tickets, 20 percent received green tickets, and 20 guests received yellow tickets. How many guests attending the convention received blue tickets?

Let the total number of guests be $ x$.
Then, $ \frac25x \plus{} \frac38x \plus{} \frac15x \plus{} 20 \equal{} x$
Solving, we get $ x \equal{} \boxed{800}$.

rubik, good try, but you are not done yet! Remember, we were asked to find the number guests that received blue tickets. Since we have $800$ guests, and $\frac{3}{8}$ of them received blue tickets, $\frac{3(800)}{8}=3(100)=\boxed{300}$.


Hello,
I'm trying to prove the following, and would like to get some hints:

Let $ \gamma$ be a closed regular (planar) curve, such that $ |\gamma(t)| < R$ for any t.
Prove that there exists a point P on $ \gamma$, such that $ |k(P)| \geq \frac {1}{R}$.
(Here k(P) is the curvature of this curve in P.)

Thanks!

Just take the point that is farthest from the origin. ;)

or just draw a circle which is smallest in the ex-tangent to curve and check at the tangent point

(Sorry for my poor English =o=;;)


if anyone has heard about the game of nim can anyone explain what it is.

Nim is a two-player mathematical game of strategy in which players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.

The rest of the article is here [url=http://en.wikipedia.org/wiki/Game_of_Nim]Nim - Wikipedia[/url]


[b]Problem.[/b] [i]Let $n$ be a positive integer, and let $S=\{ 1,\ldots, n\}$ Show that the number of permutations of $S$ with no fixed points and the number of permutations with exactly one fixed point differ by 1.[/i]

I have already solved the problem, but I am looking for other solutions.  Also, Could someone please tell me the source of this problem.  Beside the problem, it has written Canada 1983, but this makes absolutely no sense.  The following is my solution.

[b]Solution.[/b] The number of permutations with no fixed points is
\[n!-\left|\bigcup_{i=1}^{n}A_{i}\right|, \]
where $A_{i}$ denotes the set of all permutations with $i$ fixed.

The number of permutations with exactly one point fixed is
\[\sum_{i=1}^{n}|B_{i}|, \]
where $B_{i}$ is the set of permutations with only $i$ fixed.

Now, notice that
\[\sum_{i=1}^{n}|B_{i}|=\sum_{i=1}^{n}\left[(n-1)!-\left|\bigcup_{k=1}^{n-1}F_{k}\right|\right]=n!-n\left|\bigcup_{i=1}^{n-1}F_{i}\right|, \]
where $F_{i}$ is the set of permutations with fixed $i$ in the $(n-1)!$ permutations of the remaining $n-1$ elements.

So, we are to prove that
\[\left|\left(n!-\left|\bigcup_{i=1}^{n}A_{i}\right|\right)-\left(n!-n\left|\bigcup_{i=1}^{n-1}F_{i}\right|\right)\right|=1. \]
Using the [i]Boole-Sylvester Formula[/i] ([i]Sieve[/i])
\[\\ \left|\left(n!-\left|\bigcup_{i=1}^{n}A_{i}\right|\right)-\left(n!-n\left|\bigcup_{i=1}^{n-1}F_{i}\right|\right)\right|=\left| n\left|\bigcup_{i=1}^{n-1}F_{i}\right|-\left|\bigcup_{i=1}^{n}A_{i}\right|\right|\\ =\left| n\sum_{I\subseteq\{1,\ldots,n-1\},I\neq\emptyset}(-1)^{|I|+1}\left|\bigcap_{i\in I}F_{i}\right|-\sum_{I\subseteq\{1,\ldots,n\},I\neq\emptyset}(-1)^{|I|+1}\left|\bigcap_{i\in I}A_{i}\right|\right|\\ =\left|n\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n-1}{k}(n-k-1)!-\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n}{k}(n-k)!\right|\\ =\left|\sum_{k=1}^{n-1}(-1)^{k+1}(n-k-1)!\left(n\left[\binom{n-1}{k}-\binom{n}{k}\right]+k\binom{n}{k}\right)+(-1)^{n}\right|\\ =|(-1)^{n}|=1 \]
$\blacksquare$

all you have to prove is $D_{n}-nD_{n-1}=(-1)^{n}$, and this is easy using the sum definition of $D_{n}$, where $D_{n}$ is the number of derangements of $n$ objects

[quote="boxedexe"]That's exactly what I did above.  Do you have any other ideas?[/quote]

your proof is unnecessarily complicated...why call principle of inclusion exclusion the "boole-sylvester sieve"...also all the notation is difficult to read, and the derangement formula is common knowledge...

also a common recursion is $D_{n}= (n-1)(D_{n-1}+D_{n-2})$, can be proven with counting...the result becomes obvious through a little manipulation and an induction

[quote="boxedexe"]What I did was prove the derangement formula, so my proof uses the basics of counting.  I thought the notation would make it easier to understand since it is compact, especially since I proved the derangement formula.  Also, I thought that name was as well known as "Principle of Inclusion-Exclusion."  :maybe:  This is how I was introduced to Combinatorics by Titu Andreescu and Zuming Feng.[/quote]

i have not heard that name before...maybe it has something to do with living in seperate countries...also i don't see how some of those lines are supposed to be easy to understand...you need words to accompany them

like the whole going from $\sum_{I\subset\{1,2,...,n\}}(-1)^{|I|+1}\left|\bigcap_{i\in I}F_{i}\right|$ to
$\sum_{k=1}^{n-1}(-1)^{k+1}\binom{n-1}{k}(n-k-1)!$

i understand what it means...but it takes a while...i am not sure who you wrote this up for exactly, but supose it were for an olympiad, you want to write in a manner that reduces the amount of thinking for the grader

Indeed, the problem is equivalent to verify the well-known identity satisfied by the derangements: $D_{n}=nD_{n-1}+(-1)^{n}$. You can prove this directly using the derangement formula \[D_{n}=n!\sum_{k=0}^{n}\frac{(-1)^{i}}{i!}\] which can be deduced exactly as you did, that is, using the principle of inclusion-exclusion... And it's true, that contrived name, the "Boole-Sylvester formula" is due to Titu Andreescu and Zuming Feng, and appears in their book [i]A path to Combinatorics for undergraduates[/i]. I think it's an effort of the authors to be fair with the first mathematicians who discovered this theorem, but the name "principle of inclusion-exclusion" is far more standard. :)

We can give a proof of our identity with [color=orange][b]Altheman[/b][/color]'s formula, namely $D_{n}=(n-1)(D_{n-1}+D_{n-2})$, which is due to Euler. This formula implies that \[D_{n}-nD_{n-1}=-[D_{n-1}-(n-1)D_{n-2}]\] Iterating this relation we obtain \[D_{n}-nD_{n-1}=(-1)^{n-2}(D_{2}-2D_{1})\] But $D_{2}=1$, and $D_{1}=0$; hence \[D_{n}=nD_{n-1}+(-1)^{n}\]and we are done. :wink:


I am hereby announcing that the official Mathlinks chat has been created. This thread was posted in the Questions, Suggestions, & Announcements Forum, but nobody goes there so I will also start the topic here.
MODS DO NOT MOVE THIS PLEASE

So, here are the instructions:
[quote="ZetaX"]
First to all who want to chat:

So I proclamate now (later on we can still choose another one) EuIRC:
- irc.euirc.net::6665-6669
- channel: #mathlinks

For those new to that there is some html-client:
- go to http://html.chat.euirc.net/
- enter some nickname, leave the password and Username unchanged
- choose some random channel at the bottom (lets not bother with them Wink ), for example "#euirc - General information channel"
- click on "go"
- in the empty line you see type: "/join #mathlinks" and press enter.
- voila, there you are
[/quote]
Also:
One who doesn't want to chat is not forced to come there.
But everyone is invited to try chatting.

Please do not spam too much, but this forum is for chatting on Math-related subjects.
Please join and enjoy!

Have fun!

Oooh sweet remo.  I will come here to bug pro IMOers to help me ^_^


Let a, b, c be positive numbers such that $ \sqrt{ab} \plus{} \sqrt{bc} \plus{} \sqrt{ca} \equal{} 1$ . Prove that $ \frac{a^2 \plus{} 9c}{a \plus{} b} \plus{} \frac{b^2 \plus{} a}{b \plus{} c} \plus{} \frac{c^2 \plus{} 4b}{c \plus{} a}\ge \frac{9}{4}$ .

[quote="Ligouras"][quote="Ligouras"]Let a, b, c be positive numbers such that $ \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca} \equal{} 1$ . Prove that $ \frac {a^2 \plus{} 9c}{a \plus{} b} \plus{} \frac {b^2 \plus{} a}{b \plus{} c} \plus{} \frac {c^2 \plus{} 4b}{c \plus{} a}\ge \frac {9}{4}$ .[/quote]

Who can solve this?   :([/quote]

try

substitution  $ a \equal{} x^2,b \equal{} y^2,c \equal{} z^2$

and   substitution $ x \equal{} \frac {t}{1 \plus{} zt \plus{} z^2},y \equal{} \frac {1}{t \plus{} z}$

$ t> 0$


geometry
3D geometry
sphere
Putnam
college contests

What is the shortest distance between the plane $Ax + By + Cz + 1 = 0$and the ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$. What is the algebraic condition for the plane not to intersect the ellipsoid?

Let $\frac{x}{a}=X,\frac{y}{b}=Y,\frac{z}{c}=Z$, we can rewite the equations as
Plane $aAX+bBY+cCZ+1=0.$
Sphere $X^2+Y^2+Z^2=1.$
Thus  the algebraic condition for the plane not to intersect the ellipsoid is $\text{the distance between the center of the sphere, the origin and the plane}>\text{the radius of the sphere\ 1}$
$\Longleftrightarrow \frac{1}{\sqrt{(aA)^2+(bB)^2+(cC)^2}}>1,$ yielding $(aA)^2+(bB)^2+(cC)^2<1.$


Evaluate the sum: $\sum_{k = 2}^{\infty}\frac{k}{2^{k}+1}$.

I don't think it has a nice closed form.