2n white balls and 2n black balls are arranged in a line. Show that however they are arranged it is possible to find 2n consecutive balls, just n of which are white.

look at the first 2n balls, if there are n white balls we are done, if there are less than n whites than the last 2n must more then n whites, or the other way around. Eitherway, shift the balls you are looking at over one, each time you shift over one the number of whites can only go up or down one, or stay the same, so as you go over to the other side, it must switch somewhere along the way so where it switches you have n white balls


1)
Aν $a,b,c,d \in \mathbb{N*}$

νδο $(\frac{a}{b}-\frac{c}{d})(\frac{d}{c}-\frac{b}{a})$ δεν ειναι ακεραιος!

2)
Να βρεθουν τα $a,b \in \mathbb{N}$ ωστε ο $\frac{\sqrt{2(a^2+b^2)}}{a+b}$ να ειναι ακεραιος!

 :)

Ωραια επιλογη Θαναση.ειδικα το πρωτο ερωτημα ειναι αρκετα ενδιαφερον.Για το α ερωτημα λοιπον μπορεις να κανεις πραξεις και να καταληξεις στην παρασταση:κ+1/κ-2οπου κ=αβ/γδ τις ασκησης. υποθετουμε οτι ειναι φυσικος αριθμος εστω  ν. και λυνοντας την ως δευτεροβαθμια εξισωση μεταβλητης κ παιρνεις διακρινουσα. εκει πρεπει ο κ^2-4 να ειναι τελειο τετραγωνο ομως κατι τετοιο δεν συμβαινει αφου χρειαζεται μονο η παρατηρηση οτι καθε τετραγωνο >του 4 απεχει απο το προηγουμενο του το διπλασιο + 1 δηλαδη 2ν+1. :jump: για αλλη λυση ... :)

Γεια!
[b]Για το1)[/b]Η τετραδα   (a,b,c,d)=(3,2,6,4) δινει 0 το οποιο ειναι ακεραιος!Μηπως λειπει κατι?Για παραδειγμα η συνθηκη (ad,bc)=1 δινει το ζητουμενο,αλλα μπορει να υπαρχει και πιο <σφιχτη>συνθηκη.
[b]Για το2)[/b]
[b]Α τροπος[/b]
Θετουμε ολο το κλασμα ισο με κ,θεωρουμε το τριωνυμο που προκυπτει ως προς α(η ως προς β) και επειδη ο α ειναι φυσικος ,απαιτουμε  η διακρινουσα να ειναι τελειο τετραγωνο ,το οποιο δινει κ^2-1=λ^2 με κ,λ ακεραιους που δινει (κ.λ)=(1,0),δηλαδη κ=1
[b]Β τροπος[/b]
Θετωντας ξανα ολο το κλασμα ισο με κ,και κανοντας ελαχιστες πραξεις παιρνουμε (α^2+β^2)(2-κ^2)=2κ^2αβ και επειδη ολα ειναι θετικα θα πρεπει 2-κ^2>0 <=>κ=0 η 1.Ομως το κ=0 δινει α=β=0 το οποιο ειναι ατοπο γιατι πρεπει α+β διαφορετικο του μηδενος.
Σε καθε περιπτωση λοιπον εχουμε κ=1 <=>(α-β)^2=0 <=>α=β.Αρα ειναι (α,β)=(t,t) με t ε Ν-{0}

ΡΑΛΛΗΣ :D

κατ'αρχην μια διορθωση στη λυση μου:κ=ad/bc.Ραλλη εχεις δικιο νομιζω οτι η συνθηκη που χρειαζεσαι ειναι ad/bc διαφ. του 1. αφου προκυπτει μια εξισωση της μορφης κ+ 1/κ -2=ακεραιος.(κ=ad/bc) για κ=0 =>...για κ διαφ.του 0 =>...


Suppose $A,B,C,D$ are for points on the circle $C$ with this order. Prove there exist 4 points $M$ which $\frac{MA}{MB}=\frac{MD}{MC}$ and these points make a quadrilateral which its diameters are perpendicular to each other.

We see $MA.MC=MB.MD$. We have:\[ \displaystyle MA.MC.AC=S_{ABC}.R \]\[ MB.MD.BD=S_{MBD}.R \] so we conclude $\frac{S_{MBD}}{S_{MAC}}=\frac{BD}{AC}$ so the distance of $M$ from lines $AC,BD$ must be equal. so $M$ is on the intersection of the circle with bisectors of $AC$ and $BD$.

You may have seen this problem:
"Find point $P$ on the circumcircle of $ABC$ such that $PA^2=PB.PC$"
Here is an interesting solution:
Let $L_1$ and $L_2$ be the lines $BC$ and the tangent at $A$. then $P$ lies on the bisectors of $L_1$ and $L_2$.
This can be proved by letting $C$ tend to $A$ in the solution of the previous problem.

another easy way:
use$bc=2Rh_a$(in every triangle $ABC$)

i dont know ,why  the two geometry problems in second round where too easy.

it was not good for me.because i have been pay attention to geometry more than the other ones. :(


Harvard
college
real analysis
linear algebra
complex analysis
rudin
real analysis theorems

If I wanted to get an intro to the material taught in Math 55 at Harvard, what books would be best?

Have you considered the books that Math 55 uses?  These are [url=http://linear.axler.net/]Axler[/url], [url=http://www.amazon.com/Naive-Theory-Undergraduate-Texts-Mathematics/dp/0387900926]Halmos[/url], and [url=http://www.amazon.com/Algebra-Michael-Artin/dp/0130047635]Artin[/url].

Thanks for that. I never knew they released the lists of what books they used. I'm going to be taking Linear Algebra and Diff. Equations in the summer, so do you think that would be enough for me to understand the texts you have linked? I started out with a book called Real and Complex analysis by Walter Rudin, but I didn't have enough of a background to understand it.

Halmos is self-contained, although I don't know much about it.  Axler has something of an ax to grind (pardon the pun); the way he does linear algebra is a little unorthodox, from what I've been told, so that book may or may not make sense after a normal linear algebra course.  Artin is wonderful, but you'll need to have a reasonably strong linear algebra background to appreciate it fully, and you'll need to be used to proof.  (Artin's also printing a new edition, so you might want to wait a bit before getting it.)

Big Rudin is very hard.  If you haven't taken an analysis course, there's no reason to expect it to make sense.

If you like, you can find the Math 55a problem sets [url=http://math.harvard.edu/~ctm/math55a/hw.html]here[/url].

Thanks for the link to the problems. The links to the solutions don't work, but they wouldn't help much if I don't even understand the problem lol.

Do you have any recommendations for an introductory book into this type of math?

Hm, I thought they used Rudin's [i]Principles of Mathematical Analysis[/i].  Did that change?

BTW, this book, often called baby Rudin, is as good an introduction to Analysis as any and you'd need to know much of the material in it before reading big Rudin.  It's also as challenging and mind-bending as any book I'm aware of.  It's pretty much self contained, little knowledge is absolutely required, but a high degree of mathematical maturity is.  I'm currently in the middle of it and I have less of a mathematical background than you, but I'm also probably older.  You'll have to ask yourself if you're ready for it.

No, 55b still uses baby Rudin.  I'm not particularly fond of it, though; it's far too dry.

What would you suggest for analysis? Not particularly to our liking, but just something rich and flavorful, etc...

Apostol's Mathematical Analysis, or [url=http://www.amazon.com/Real-Mathematical-Analysis-Charles-Chapman/dp/0387952977]Pugh[/url]


MATHCOUNTS
summer program
MathPath
algebra
binomial theorem
geometric sequence

[hide]
Plug in 1 for x and y and you get $2^5=32$
[/hide]

[quote="math92"]Find the sum of the coefficients in the expansion of $(x+y)^5$[/quote]

[hide]$\text{Using the binomial theorum:}$

$(x+y)^5$
$x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$\text{Therefore, the coefficients are}$
$1$
$5$
$10$[/hide]

1000th User, nice knowledge and slick solution, but that was my solution :D

Anyway, I'll stop with this nonsense coughcough(spam)coughcough.

How many distinct coefficients are there when $(x+x^7+x^{13})^{100}$ is multiplied out and simplified? Please show your work.

I fixed the questionable exponent, nat mc. Sorry :blush:

[quote="mathnerd314"]1000th User, nice knowledge and slick solution, but that was my solution :D

Anyway, I'll stop with this nonsense coughcough(spam)coughcough.

How many distinct coefficients are there when $(x+x^7+x^13)^{100}$ is multiplied out and simplified? Please show your work.[/quote]
Do you mean $(x+x^7+x^{13})^{100}$ or $(x+x^7+3x)^{100}$?

[quote="10000th User"][quote="236factorial"]Drunken_Math, nice knowledge, but no solution  :D

[hide]1+5+10+10+5+1 = 32. [/hide][/quote]
236factorial, nice knowledge and solution, but no slick solution :D

[hide]$(1+1)^5=32$[/hide][/quote]

Uh. I sorta knew that, from some high school contest,

or did I?  :P

[quote="mathnerd314"]How many distinct coefficients are there when $(x+x^7+x^{13})^{100}$ is multiplied out and simplified?[/quote]
This looks suspiciously identical to a problem posed by Dr. Paul Zeitz at MathPath.  I will leave the problem here, temporarily, but I insist that [b]all[/b] proposed solutions be at a MATHCOUNTS level.

I really think you should post this followup to one of the high-school forums, where a variety of solution-methods can be freely discussed.

[quote="math92"]Find the sum of the coefficients in the expansion of $(x+y)^5$[/quote]
For the record, most applications of the Binomial Theorem belong in higher-level forums.  The original post is within an acceptable level for this forum, but the solutions lack an explanation that a typical Chapter or State competitor could follow.

mathnerd / drunkenmath:  Maybe you guys could provide a solution using plain old algebra, and then show how that relates to the clever solutions you learned at MathPath.  Thanks.

Is the answer 201? If it is, I'll post the solution.

[quote="mathnerd314"]How many distinct coefficients are there when $(x+x^7+x^{13})^{100}$ is multiplied out and simplified? Please show your work.[/quote]

[hide="hint"]$x^{100}(1+x^6+x^{12})^{100}$. The $x^{100}$ won't affect the num of coefficients, so it's basically just $(1+x^6+x^{12})^{100}$[/hide]

[quote="10000th User"][quote="nat mc"][hide="hint maybe"]
$(1+x^6+x^{12})$ is a geometric sequence, so it is $\frac{(x^{18}-1)}{(x^6-1)}$
We keep factoring and get
$\frac{(x^9-1)(x^9+1)}{(x^3-1)(x^3+1)}$.
and again
$\frac{(x^3-1)(x^6+x^3+1)(x^3+1)(x^6-x^3+1)}{(x^3-1)(x^3+1)}$
cancel
$(x^6+x^3+1)(x^6-x^3+1)$
these are both geometric sequences.
I don't know if that helps
[/hide][/quote]
I think this better?
[hide="my hint"]Let $y=x^6$ and $(1+x^6+x^{12})^{100}=(1+y+y^2)^{100}$[/hide][/quote]
yeah thats much better, now the powers aren't as high

[hide="Possible solution..."]
$(1+y+y^2)^x$; Let's investigate what happens for small values of $x$...

$x=1 \rightarrow 3$
$x=2 \rightarrow (1+y+y^2)^2=1+2y+3y^2+2y^3+y^4 \rightarrow 5$
Hmm...maybe 7 is next? :P
$x=3 \rightarrow (1+y+y^2)^3=1+3y+6y^2+7y^3+6y^4+3y^5+y^6 \rightarrow 7$
So
$x=n \rightarrow 2n+1$
Maybe????
And
$x=100 \rightarrow \boxed {201}$
Maybe?????
[/hide]

Exactly how I did it chess64  :D

Do you know why? I think it involves introducing the lowest and highest powers... Nah, I'm going no where.

THink about $x=2$.

$(1+y+y^2)(1+y+y^2)=1(1+y+y^2)+y(1+y+y^2)+y^2(1+y+y^2)$

After multiplying by the $1$, there's already $3$ coefficients.

[hide="Wait..."]the problem says "distinct" coefficients. That means the answer should be $100$, not $201$[/hide]

Anyway...then, after adding the $y(1+y+y^2)$, you introduce $1$ new power/coefficient. So in all you introduce 2 new coefficients, so, $3+2=5$. This happens every time, so you get 2 new coefficients every time. However, the first half are the same as the last half, so you subtract 1 and divide by 2.

Maybe?????? :D

What ?

Can someone that went to mathpath tell us the solution?

[quote="JesusFreak197"]I would think that, assuming all coefficients are distinct up to the middle, there should be $101$. Think about it. We've proved that there are 201 coefficients, right? The first 100 are the exact same as the last 100, but the middle one is also distinct, meaning $\boxed{101}$ distinct coefficients.[/quote]

Yeah, I see what you're trying to say, so it's actually add one, then divide by 2?

oh sorry, I wasn't thinking about exponents in the beginning, I was thinking about exponents  :blush:

Still, according to this, there would only be 2 distinct coefficients if the power was 2. But there is still 3, isn't there?

[quote="JesusFreak197"]I would think that, assuming all coefficients are distinct up to the middle, there should be $101$. Think about it. We've proved that there are 201 coefficients, right? The first 100 are the exact same as the last 100, but the middle one is also distinct, meaning $\boxed{101}$ distinct coefficients.[/quote]

Yeah, that's what I meant.

Oh, sorry. I'm so stupid. I meant distinct [b]terms[/b]. :oops_sign: :blush:

But the way you've been solving it, I would think the answer would follow directly from what you've been doing.

Can anyone [b]prove[/b] that there would be such a pattern of the number of terms in the expansion of $(1+y+y^2)^x$? Because that's a key part of the problem.

[quote="236factorial"]Still, according to this, there would only be 2 distinct coefficients if the power was 2. But there is still 3, isn't there?[/quote]

$(1+y+y^2)^2=1+2y+3y^2+2y^3+y^4$, so three distinct coefficients.

[quote="236factorial"][quote="nat mc"][quote="mathnerd314"]Can anyone [b]prove[/b] that there would be such a pattern of the number of terms in the expansion of $(1+y+y^2)^x$? Because that's a key part of the problem.[/quote]
Thats pretty simple.  It is easy to see that $0$ is the smallest power, and $2x$ is the biggest power.  And since 0,1,2 don't have a common factor greater than 1, all powers from 0 to 2x will be in the expansion. Answer: 2x+1[/quote]

Yeah, I had a similar thinking process, but I couldn't really follow through on the thinking. So 201 is the answer?[/quote]
201 was not the answer to the one with distinct coefficients, it was 101, becuase it is symmetric. but if you want the number of terms, yup thats right

[quote="nat mc"][quote="236factorial"]Yeah, I had a similar thinking process, but I couldn't really follow through on the thinking. So 201 is the answer?[/quote]
201 was not the answer to the one with distinct coefficients, it was 101, becuase it is symmetric. but if you want the number of terms, yup thats right[/quote]

That was what I meant, since that was the original question  :)


group theory
abstract algebra
linear algebra
matrix
integration
vector
superior algebra

For (1b): Show that $G/N$ is isomorphic to the group of elements of the form $\begin{pmatrix}a&0\\0&d\end{pmatrix}$, which is obviously abelian.

For (2): More generally, let $H$ be a normal subgroup of $G$ and let $J$ be a characteristic subgroup of $H$ (fixed under all automorphisms of $H$). Then $J$ is a normal subgroup of $G$. Any subgroup of a cyclic group is characteristic.

Yes thats works James.. .thanks
but the exercises appear before we start learning Isomorphisms.

[quote="Beginner"]1)Let $G$ be the group of all $2 \times 2$ matrices $\left(\begin{array}{rc} a & b \\ 0 & d \end{array}\right)$ where $ad \not= 0$ under matrix multiplication.  Let $N = \left\{\left(\begin{array}{rc} 1 & b \\ 0 & 1 \end{array}\right)\right\}$ Prove that
(a) $N$ is a normal group of $G$.
(b) $G/N$ is abelian.

part a) is easy, but b) is not  for me.

2)If cyclic subgroup $T$ of $G$ is normal in $G$, then show that every subgroup of $T$ is normal in $G$.[/quote]

b) Let $A,B\in G$. Prove that $ABA^{-1}B^{-1}\in N$.
2)Let $t$ be a generator of $T$. Every subgroup of $T$ is cyclic generated by $t^n$ for some $n\in\mathbb N$.
Let $g\in G$. Because $T=<t>$ is normal in $G$, we know that $gtg^{-1}=t^k$ for some integer $k$. Then $gt^ng^{-1}=t^{kn}\in<t^n>$. This proves that $<t^n>$ is also normal in $G$.

About jmerry's isomorphism
       $G\simeq R^3$
             and
       $N\subset G$
              and
       $N \simeq R$

If we denote by $\hat{b}=\left\{\left(\begin{array}{rc} 1 & b \\ 0 & 1 \end{array}\right)\right\}$ the class of equivalence with representant b then

$\hat{b}=\left\{ \left(\begin{array}{rc} a & b \\ 0 & d \end{array}\right) with: a,b,d\in R \right\}$

But what is G/N ? it is said to be factor group but cant really understand
what G/N would be here.

after that...that high algebra stuff i dont really understand absolutely so ill try to see
what does amfulger means by that proposition for 1b) im working on it now

The definition of $G/N$ is the set of cosets $aN$, with the multiplication law $(xN)\cdot (yN)=xyN$. This is well-defined because $N$ is normal. In this case, we can choose representatives of the cosets which are themselves a subgroup; this subgroup is automatically isomorphic to $G/N$ by projection.

$G$ is [b]not[/b] isomorphic to $\mathbb{R}^3$; it's got multiplicative action on the main diagonal, and it's not abelian.

amfulger's method: by showing that the commutator of any two elements of $G$ is inside $N$, we have that the commutator of any two elements of $G/N$ is trivial: $(xN)(yN)(xN)^{-1}(yN)^{-1}=xNyNx^{-1}Ny^{-1}N=xyx^{-1}y^{-1}N$.

[quote="spx2"]...[/quote]

This problem is about groups, not about vector spaces.

uhmm...jmerry i would like it if you could be more explicit on what trivial is there ?
is it a property or is it trivial as in simple ?

but uhm should i pose myself the question now
 is $xyx^{-1}y^{-1}\in G$ ? oh...sorry i think it is

now what does this mean ?
$(xN)(yN)(xN)^{-1}(yN)^{-1}=xyx^{-1}y^{-1}N$
what does this mean related to the problem ?


oh , not hard but i haven't time . i'll post it in future


group theory
abstract algebra
superior algebra
superior algebra unsolved

sir
   If all the subgroups of a group has generators?i need examples?
   solve: If H is a proper subgroup of a group G,show that G=<G/H>

What do you mean by "has generators"?  Every group has generators - it is trivially generated by all of its elements.  Are you asking something along the lines of "are all the subgroups of a finitely generated group finitely generated?"  (The answer is no.)

[quote="barathisid"] If H is a proper subgroup of a group G, show that G=<G/H>. [/quote] First off, you need to recall the following fact:

[b][color=red]Lemma.[/color][/b] Let H and K be subgroups of the group G. Then $ H \cup K$ is a subgroup of G if and only if $ H \subseteq K$ or $ K \subseteq H$.

The proof of the original claim is now as follows.

[b]1.[/b] It is easy to show that $ G \equal{} H \mbox{ } \cup \mbox{ } < G \backslash H > .$
[b]2.[/b] As a consequence of [b]1[/b] and the above [b][color=red]lemma[/color][/b] we obtain that $ H \mbox{} \subseteq \mbox{} < G \backslash H >$. This implies at once that $ G \mbox{} \subseteq \mbox{} < G \backslash H >$ and the proof terminates.

Hope that these lines of mine are of some use to you.

Best of luck, :wink:


$ x_1,x_2,...x_m,y_1,y_2...y_n$ are non negative reals.denote $ X\equal{}\sum_{i\equal{}1}^{m}x_i$and  $ Y\equal{}\sum_{i\equal{}1}^{n}y_i$
Proof:$ 2XY\sum_{i\equal{}1}^{m}\sum_{j\equal{}1}^{n}|x_i\minus{}y_j|\ge X^2\sum_{i\equal{}1}^{n}\sum_{j\equal{}1}^{n}|y_i\minus{}y_j|\plus{}Y^2\sum_{i\equal{}1}^{m}\sum_{j\equal{}1}^{m}|x_i\minus{}x_j|$

This problem is extremely difficult.I think no more than ten students has worked it out :(

[quote="math10"]I think the completed test  China TST 2009
Can you post problem china TST 2009?[/quote]
China TST have 6 pre-exams and two Team selection test.This is the problem from the first pre-exam.
I'm very sorry that I can't post all of them for I failed to translate them into English :blush:

See [url=http://www.mathlinks.ro/viewtopic.php?p=1510361#1510361]here[/url] for a solution


Starting from point A, Treething walked 8 yards, turned right 45 degrees, walked another 8 yards, and turned right 45 degrees again. he then walked another 8 yards and turned left 45 degrees, walked 8 more yards, turned left 45 degrees again, and finally walked another 8 yards. If his final destination is point B, how many yards farther didh e walk than if he had walked a straight line path from A to B? Round your answer to the nearest thousandth.

[hide]
Treething is walking around the edges of an octagon with side 8.  Check if you don't believe me.  That means the straight line from A to B is the longest diagonal of the octagon.
Since the longest diagonal$^{2}$ = the medium diagonal$^{2}$ + a side$^{2}$ by pythagorean, all we have to do is find that medium diagonal.  Which turns out to be the base of an iso. trapezoid with 3 sides of 8.
That makes the medium diagonal $8+8\sqrt{2}$.  By pyth. thm, the longest diagonal is $\sqrt{256+128\sqrt{2}}$, so he saved 15.34 yards?
 :huh: I'm sure I did something wrong.[/hide]

Blarg, that makes no sense without my diagram...

HINT:

[hide]There are TWO octagons you are working with[/hide]

Answer for those who want to check:

[hide]6.548[/hide]

[quote="CircleSquared"]Oh, I read the question wrong.
However, my new solution requires an overly complicated and completely incomprehensible diagram.
I will only post an outline solution.
[hide]
Draw the diagram, draw the line from A to B, and you can immediately see a way to use pythagorean.  Because of the convenient 45 degree angles, two legs of the right triangle are $16+12\sqrt{2}$ and $4\sqrt{2}$.  The hypotenuse is the distance we are trying to find. After pythagorean, we get the answer to be $\sqrt{576+384\sqrt{2}}$, which turns out to be approximately $33.45$.  Subtract from 40, and the answer is $6.547675597$  :lol: [/hide][/quote]
[hide] I couldn't see the legs of the right triangle that you were talking about  :maybe: , but I tried doing it myself, and I got the legs to be $8\sqrt{2}+8$ and $8\sqrt{2}+16$. That was probably the long way to do it, because I got the same answer, but it seems kinda long compared to yours. [/hide]

I finally figured out how to upload pictures.

The legs are the lengths in red. :lol:

thank you very much for the picture at the end... otherwise i would have just kept staring at the diagram i drew like  :huh:
i couldnt understand why mine wasnt working until  i realized that i had drawn it from the wrong starting perspective and i didnt see that it would work like that until i saw that the leg should have been extended more..
again that illustration helped me understand.


inequalities
complex numbers
inequalities solved

z1,z2,z3 are vectors.
prove that
|z1+z2+z3|+|z1|+|z2|+|z3| \geq |z1+z2|+|z2+z3|+|z1+z3|

you can consider z1,z2,z3 as complex numbers as well.

I have a proof for the more generalized inequality:

(n-2) \sum |z_i | + | \sum z_i | \geq \sumi<j |z_i+z_j| for all complex numbers z_i, i=1,n.

I think it is called Hlawka's inequality.

To be rigourous, Hlawka's inequality is the particuliar case n = 3 of the inequality stated by Valentin. The general case is due to Adamovic.
This has also been generalized by Djokovic and others....

Pierre.

It seems to me the case n=3 was solved few months ago.

well, where is the solution? i cannot find it.

and where can i find the materials of its generalizations?


Find all triples of primes (p,q,r) such that $ p | q^r \plus{} 1, q|r^p \plus{} 1, r|p^q \plus{} 1.$

See here
http://www.mathlinks.ro/Forum/viewtopic.php?p=357351#357351

Daniel


AMC USA(J)MO USAMO AIME geometry induction inequalities

[quote="beta"][quote="pkerichang"]Let's say by some miracle I actually qualifies for USAMO, then how should I study for it??? I know it's very hard...[/quote]

If you haven't done any kind of preparation yet, it's probably already too late.[/quote]

If you're religious, pray.

...
I'd say just do problems, and give it your best shot.

It'd probably be easier sabotaging others.

The USAMO is very very very hard.  Maybe I'm a wimp, but it took me three years until I decided that I was going to really study for the USAMO.  The reason is that USAMO questions are a jump from short answer problems such as the AIME and AMC to proof problems.  You have to first learn what it means to prove something, learn the standard techniques of proofs, learn the much increased playbook of possible theorems or concepts you need to know, and then finally practice doing problems just like you would need to for short answer problems.  What I have to say is problably more relevant to trying to crush the USAMO.  If you want to study for a month beforehand, I would just take a peak at some old USAMO's to get a feel for what types of problems to expect.  Essentially they are all either geometry, number theory, algebra, or combinatorics.  And then learn the basic ways of proving things, ie WOP, induction, contradiction, case analysis.  There really is much to know.  I had a friend last year spend the couple of weeks before the USAMO just learning inequalities and getting better at them only to not have a single inequality show up on the USAMO.  So, my advice is to basically rely on your logical skills and short answer stuff to get the more low-tech problems, maybe like those games problems or something and expect to be left clueless on at least a couple of the problems.  I mean, looking at past years shows that most people get 1 or 2 problems total.

Good luck!  Hope I wasn't too discouraging.  It really is a lot of fun to think about interesting problems for 9 hours.

[quote="djshowdown2"]I wouldn't say its too late, but I would say to wait until the AIME's over with before assessing your USAMO situation since its just a couple days away.

A month isn't an insignificant amount of time for improving problem solving skills though.  I mean, its not like he's expecting to win it or anything.  Unless you are, in which case yeah, you might be a little too late  ;)[/quote]

Haha except I had been preparing for USAMO like since like last September(I'm only aiming for MOP and I probably won't make it). Actually I mean if you qualify for USAMO you should be able to solve 1 problem. For example last year's #1 and #4 were quite approachable.

You really can't make any significant improvement on solving USAMO-type problems in one month. What you can do:

1) Familiarize yourself with what types of problems show up on the USAMO by looking at past contests.
2) Learn to write proofs clearly.
3) Learn the basic geometry and inequality theorems. Just knowing these isn't likely to let you solve anything directly, but at least you won't be staring off into space.

That will allow you to fully tap into your own abilities... but as for actually increasing your abilities, I don't think there really is any short-term way.

[quote="JesusFreak197"]I've heard people say that USAMO is quite difficult. What makes it difficult? Is it just the level of the problems, or is it that you have to write formal proofs which are quite different than what most people do, or what? What do formal proofs look like? Would they be like the ones in my Calculus book? I really have no clue what USAMO is like, but I have a pretty good chance of qualifying for it this year, so I'm kind of wondering...[/quote]

USAMO problems require an extremely deep understanding of the topics, a vast knowledge of a wide variety of topics, creativity, and the ability to play with a problem for an extended period of time.  THEN, you have to write up the proof and make sure everything is logical.  Writing up the proof is obviously important, but I don't consider it to be what makes the USAMO as hard as it is.

[quote="JesusFreak197"]I've heard people say that USAMO is quite difficult. What makes it difficult? Is it just the level of the problems, or is it that you have to write formal proofs which are quite different than what most people do, or what? What do formal proofs look like? Would they be like the ones in my Calculus book? I really have no clue what USAMO is like, but I have a pretty good chance of qualifying for it this year, so I'm kind of wondering...[/quote]

Take a look at the problems :
[url=http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html]www.unl.edu/amc/a-activities/a7-problems/problemarchive.html[/url]

BTW Joe summarized it. But for USAMO the hardest part in my opinion is you really have to play around and experiment with ideas for a long time before getting a solution. Still that's my opinion you have to look at the problems yourself to judge what's hardest for you.

[quote="beta"]
BTW Joe summarized it. But for USAMO the hardest part in my opinion is you really have to play around and experiment with ideas for a long time before getting a solution. Still that's my opinion you have to look at the problems yourself to judge what's hardest for you.[/quote]

I should also mention that I'm not very experienced on USAMO matters.  But, yeah, they don't give you 1.5 hrs a problem without a good reason :P  Out of the very few Olympiad level problems I've done and written proofs for, I'd say that I spend usually an hour on them.  It takes a lot of time to think about the problem, experiment, think, get frustrated, try again etc. THEN you have to write the proof up which still can take some time.  Particularly, you have to work out all the minor details and stuff.  THe USAMO is like a marathon compared to the AMC and AIME.

Well think what professional mathematicians might do.  :D


trigonometry
calculus
integration
function
calculus computations

need some tips on how to apply parseval's identity to $f(t)=(1-|t|)_{\chi[-1,1]}$ to find
$\int_{-\infty}^\infty (\frac{sin(x)}{x})^{4}\,dx$
also, can the laplace transform be used for this integral?
thanks

Just answer two questions:

a) What is the Parceval identity?
b) What is the Fourier transform of $f$? (Yes, you'll have to actually make this boring computation ;))

for parsevals's identity
$\frac{1}{2\pi}\int_{-\pi}^\pi |f(t)|^{2}\,dt = \sum_{n=-\infty}^{\infty}|\hat{f}(n)|^{2}$

where $\hat{f}(n)$ is the fourier transform of $f(t)$

That's Parseval's identity for periodic functions. The theorem we actually want, although extremely closely related, usually goes by the name "Plancherel" rather than "Parseval." This is the identity relating $\int_{-\infty}^{\infty}|f(x)|^{2}\,dx$ to $\int_{-\infty}^{\infty}|\widehat{f}(\xi)|^{2}\,d\xi$ where $\widehat{f}(\xi)$ is the Fourier transform of $f.$

(I didn't write out the identity only because there's a detail there that depends on where you put the $2\pi$ in your definition of the Fourier transform.)

I would also add that something like (again, depending on where you put the $2\pi$) $\left(\frac{\sin\xi}{\xi}\right)^{2}$ is the Fourier transform of the "tent" function $(1-|x|)^{+}$

Laplace transforms are most at home on one-sided domains. Since the Laplace transform and Fourier transform are closely related, it would not be that big a surprise if a problem soluble using one tool is also soluble using the other tool - but this looks more naturally like a Fourier transform problem to me.

Another method for calculating this integral is to use the Gamma function as follows:

$\int_{-\infty}^{\infty}\left(\frac{\sin x}{x}\right)^{4}dx=2\int_{0}^{\infty}\left(\frac{\sin x}{x}\right)^{4}dx=\frac{2}{\Gamma(4)}\int_{0}^{\infty}t^{3}dt\int_{0}^{\infty}\sin^{4}x e^{-xt}dx$.

I used the fact that $\frac{1}{x^{4}}=\frac{1}{\Gamma(4)}\int_{0}^{\infty}e^{-xt}t^{3}dt$, and then interchange the order of integration.

Note that \[ \dfrac{\sin x}{x}\longrightarrow \begin{cases} 1\;\;\;\text{for}\;\;|t|<1\\ 0\;\;\text{else}\end{cases}\] and thus using multiplication property, we get
\[ \left(\dfrac{\sin x}{x}\right)^2\longrightarrow \frac{1}{\pi}\begin{cases} 1\minus{}|t|\;\;\;\text{for}\;\;|t|<2\\ 0\;\;\text{else}\end{cases}\]Thus \[ \int_{\minus{}\infty}^{\infty}\left(\dfrac{\sin x}{x}\right)^4dx\equal{}\frac{1}{2\pi^2}\int_{\minus{}2}^{2}\left(1\minus{}|t|\right)dt\]where we have used Parseval's relation. The above integral is easy to find out.
regards
bhupala


If $ log{}_1{}_0 2=k$, then $ log{}_1{}_0 625$ is equal to:

a) 4k   b) -4k    c) 4+k    d) (1-k)/4     e) 4-4k

I can't seem to find a suitable result. Help please :D

[hide="Solution"]The answer choice is E:
$ 4 \minus{} 4k \equal{} 4\left(1 \minus{} k\right) \\
\equal{} 4\left( 1 \minus{} \log 2 \right) \\
\equal{} 4\left( 1 \plus{} \log 2^{ \minus{} 1} \right) \\
\equal{} 4 \left( \log 10 \plus{} \log 2^{ \minus{} 1} \right) \\
\equal{} 4 \log 5  \equal{} \log 625 \Rightarrow \boxed{E}$[/hide]

ill write log base 10 as log
log625=log(5^4)=4log5
5=10/2
4log5 = 4log(10/2) = 4(log10-log2)=4-4k


Given a complete $ 2004\minus{}gon$ (all sides and diagonals drawn), delete some lines so that each vertex of the $ 2004\minus{}gon$ has degree at most $ 5$.

Prove that there exists a coloring, using only Red and Blue, of the verticies of the $ 2004\minus{}gon$ so that at least $ \frac{3}{5}$ of the remaining lines (sides and diagonals) have endpoints of different colors.

hello.
Can you send me problem of USA IMO Training, 2004(My gmail: math11a1@gmail.com).
I only have problems in year 2003.
Thank you very much.

does anyone have any ideas?

thank you  :)


Let $ p$ is a prime, $ p\leq n\in N$.Prove that:
$ \sum_{i\geq 1} {[\frac{n}{p^i}]}\leq[\frac{n}{p\minus{}1}]$

Do you understand why $ \left[\frac{n}{p\minus{}1}\right]\geq \left[\sum_{i\geq 1}\frac{n}{p^i}\right]$?
Do you understand why $ [a\plus{}b]\geq [a]\plus{}[b]$?

[quote="toanIneq"]Let $ p$ is a prime, $ p\leq n\in N$.Prove that:
$ \sum_{i\geq 1} {[\frac {n}{p^i}]}\leq[\frac {n}{p \minus{} 1}]$[/quote]

The LHS is the exponent of $ p$ in the prime decomposition of $ n!$. This is well-known to be also equal to $ \frac{n\minus{}s_p(n)}{p\minus{}1}$, where $ s_p(n)$ denotes the sum of the digits of $ n$ written in base $ p$.
The result follows easily  :D

Pierre.

$ \sum_{i \equal{} 1}^{inf} [ \frac {n}{p^i} ] \leq [ \sum_{i \equal{} 1}^{inf} \frac {n}{p^i} ] \equal{} [ \frac {n}{p \minus{} 1} ]$


calculus
integration
function
derivative
calculus computations

Let $ f(x)$ is continuos function .If $ f(1) \equal{} 1$ and

$ \int_0^x tf(2x \minus{} t)dt \equal{} \frac {1}{2}arctan(x^2)$ then Find $ \int_1^2 f(x)dx$

Differentiation and subsequenct product integration yields:

$ \frac{x}{1\plus{}x^2}\minus{}xf(x)\equal{}2\int_0^xtf'(2x\minus{}t)dt\equal{}\minus{}2xf(x)\plus{}2\int_0^xf(2x\minus{}t)dt$. This implies $ 2\int_x^{2x}f(w)dw\equal{}\frac{x}{1\plus{}x^2}\plus{}xf(x)$, in particular $ \int_1^2f(w)dw\equal{}\frac{3}{4}$


English and History are very time-consuming subjects for me, at least.

Math is by far my best subject, so no need to elaborate here.

Social studies consumes a lot of time and requires many things to be memorized. Despite having a very good memory and some interest in history, I find this subject to be quite boring. Luckily, this is not a required subject for me.

Science is probably my second best subject, as I find almost all aspects of it very interesting. My current teacher is much better than all prior ones except maybe last year's, but unfortunately not the most lenient on homework assignments.

English is regarded by most to be boring, especially with all the essays, books to read, and such things. I find writing essays quite interesting sometimes, presuming that the subject is at least somehow intriguing. I constantly read books anyway so there's no problem with those assignments, and all the other simple homework is easily manageable.

I find foreign language (in my case, Spanish) one of my best subjects, and one of the most interesting also. I don't have much homework yet (I'm sure I will, however), and my teacher really seems to know what he's doing, so this subject will not be a problem this year.

After listing all those things, I realize that I don't technically have a "worst subject", especially considering I get good grades in all of them. If a choice must be made, however, I would say social studies is the most boring.

English... Why do I need it? I'm not going to England very soon.

I'm sure there will be lots of people on this forum choosing math....

[hide="Math is by far my worst subject"]

Aside from math, French is my best and favorite subject. I find grammatical structure interesting in the same way math is. French also becomes very easy when you constantly speak it  :) . I'm also taking Latin, which I'll have to say I like too.

I'm good at science by default, but I haven't really enjoyed science classes in high school.

I can't say I'm bad at history and English, although I don't ease through them like languages and sciences. I love examining novels and thinking about the themes and character I find in them. With writing, I can be awkward at times, but after intense editing I often write very well.[/hide]

There should be a option called PE, because that is the worst subject in school. Although you could argue that is is non-academic.

The worst academic subject for me is English, that is usually one of my lowest grades (although still an A), I just think that English is too subjective for me, and I like the objective-style in math. Ironically, I'm in AP English Language, and was one of the people that my previous English teacher selected to automatically get into AP English Language regardless of the qualifying test score.

[quote="1/(ln x)"]There should be a option called PE, because that is the worst subject in school.[/quote]
I don't agree. In fact, I would rank PE in my top three, along with math and science. This year, our teacher is ten times better than all our previous teachers. Last year, for example, our teacher was extremely lenient about making us run, especially with the majority of the class being very lazy. Our current teacher, however, makes sure we do what we should be doing. If he sees anybody not putting enough effort into an activity, he takes off points. I really like this as it ensures that I won't be wasting several hours a week in PE, but instead doing something useful. The obvious problem: not all of the class appreciates his dedication to his work.

this seems like a typical result 4 people good at math. generally, if ur good at math, ur also good at science and foreign languages(not always speaking), but not necessarily english or social studies/history. i must say that i am that that statement personified. however, english is my worst and least favorite-i can't stand writing.

Based on last year's grades, history was my worst subject.

Hmm math as the hardest of my subjects?  :rotfl:

Nah.

English - definately. We're doing a bunch of Bible stuff and its annoying me because I'm an athiest and I lose participation points because I don't know, for example, what happened in Genesis  :| .

Social studies - it all depends on the teacher. Last year SS was one of my worst because my teacher graded us all unfairly. This year it's one of my best because my teacher is awesome :P

Science - meh middle school science sucks. By the time I truly appreciate a certain topic in our topic-jumping curriculm, we move on to something else. However, high school science is a lot better and since I'm taking biology, we focus on one general type of science. (Plus my teacher is cool and I have bio 95 minutes a day).

Spanish - we can't talk in English so that's a bit restraining but all in all I like spanish club.

1. Maths - My favwot. Nuff sed.
2. Science - Second favwot (as long as the teacher is always open to questions)
3. Chinese - Not bad, the syntax is almost like maths, and memorizing characters is just like memorizing pi.
4. English - I'm the best in my class, but compared to the nerds in other classes I feel Baaad. (im not in a noob class)
5. Social Studies - Ugh! I do not care for such things, therefore neither do my marks.

I don't really find history that interesting.

Love Mathematics, English and Science (Chemistry, Biology, Physics) and also Music!

Hate Social Studies (Geography, History, Classics) and Foreign Languages.

@7h3.D3m0n.117

But that's unfair because you don't subject to the logic behind Christian interpretations. I think your teacher should broaden the way the class views texts (e.g. from another point-of-view such as Marxism) so that 'special knowledge' (like the content of the Bible which some grew up with) is not localised to a specific group of people. I remember last year when we did Christian readings in class and I had no idea what was going on, so instead I adopted a Marxist/Psychoanalytic perspective instead.

The bible's not a particularly well written book either. The people who wrote them weren't exactly master poets.

Math and French are easily my favorite (and best) subjects. Math goes without saying, and French...well, c'est très interessant. I actually have a higher grade (A+ vs A) in Chemistry, but I'm not really too interested in it.
History, at least this semester, is my worst subject, but that's because our teacher requires us to do some intense thinking and stuff for his evil tests. And I'm bad at thinking.   :P But I kinda like the subject.
English is okay, it doesn't come as easily to me as math or French, but I can manage to pull off a decent grade, after editing my writing, etc.

@BanishedTraitor:

Lol I agree that yes, while reading the Christian literature may be an educational experiance, but I don't think I should get docked participation points because I didn't know what happened.

EDIT to my first post: Geometry is now one of my most boring subjects, partly because I don't pay attention, partly because it's boring (curriculm).

Really, PE is bad if you're physically disabled or near disabled like me.
Not really legally disabled, but just at the brink (higher than 10 minutes on the mile time  :rotfl: _
yeah...

[quote="mgao"](higher than 10 minutes on the mile time  :rotfl: _
yeah...[/quote]

You might be insulting some people on the forum...  :maybe:
Lol, my mile time is over 10 minutes, cuz I absolutely SUCK at PE. And I'm fat.

I don't do too well myself. I usually get about 7-8 minutes for the kilometre (which is significantly shorter than the mile).

But when I'm online, I literally go at the speed of light.  :pilot:

I'm pretty slow, I got ~8-9 minutes, but that was at the end of last year, so maybe a bit faster...?

History is my worst by far for me. I somehow take a lot of notes, but can not remember a lot of what the teacher says, and it's just too much, and is just not as good as science or math, etc.

By grades, it is also my worst subject.

I don't have a worst subject. I believe all subjects are brilliant: I just don't necessarily have a brilliant enough mind to be good at them all.

English is my worst subject because I am really slow, sloppy, and concrete with writing. I'm too concrete for digesting things that I read.

World Language (French) is my 2nd worst subject, with similar reasons as English...Also because it takes too much memorizing.

Math and social studies are my two favorite subjects out of those choices. Social studies is very interesting as long as we don't have to do a lot of writing work and relating stuff (i don't like deep stuff cuz as i said i'm concrete). Math is fun and doesn't require me to be deep.

Science is also a subject that I like, because it's got very interesting topics. It also doesn't require me to be deep.

Math...Probably a subject that I love, but could do some work on...I'm pretty good, so I'm not worrying about Math.

Social Studies is easy in Elementary school. Few dates to remember, Revolutionary War and Civil War fun to learn about. Some of you may be saying it's not possible...but I like history. I'm not worrying about it.

We are just starting Science, but it has been my worst subject ever...if you can say my work is bad. Science just doesn't amuse me...Space and learning about what worms like is entirely too boring. At the end of the year in 4th grade we started doing Eco Scouts, and that was fun, but then school ended and we couldn't do any more. I hope Science is more fun in 5th grade.

English is okay. I'm comfortable with writing and our teacher only grades us on Effort, Behavior, Attitude towards Working, responsibilities, etc. We rarely take tests or hand things in because as long as we put effort into it, creativity, and don't fool around, we get an A. Pretty easy.

Last is Foreign Language. You must be crazy to think we learn a Foreign Language in Elementary school! Well, we don't anyways. I tried looking at a Spanish book when I was in 2nd grade, and caught on to most of the stuff...But it seems hard memorizing and having a real conversation in Spanish. French is only harder..I want to learn both..and Latin!! But I'm horrible at Pig Latin anyway, so what difference does it make?

The worst subject listed right no is probably Science.. Since I haven't taken a real Foreign Language course.

French is my worst since my quarter grades are usually -A or +B, everything else is A or +A.

This reminds me, has anyone heard of EuroEnglish?


[quote="courtrigrad"]$x = \arctan \frac{4\cos 12+4\cos 36+1}{\sqrt{3}}$[/quote]
That's helpful.  :roll:

[hide="Hint"] [quote="Virgil Nicula"][color=darkblue][b]Lemma. [/b]$\boxed{\cos 36^{\circ}=\frac{1+\sqrt 5}{4}}\ .$
[b]The first proof.[/b] Let $ABC$ be a triangle for which $AB=AC=1$, $m(\widehat{BAC})=36^{\circ}$. Denote the middlepoint $E$ of the side $[AB]$ and the point $D\in (AC)$ for which $DA=DB$. Prove easily that $DA=DB=BC=x$, $DC=1-x$ and $ABC\sim BDC$. Therefore, $\frac{AB}{BD}=\frac{BC}{DC}$ $\Longrightarrow$ $\frac{1}{x}=\frac{x}{1-x}$ $\Longrightarrow$ $x^{2}+x-1=0\ ,\ x>0$ $\Longrightarrow$ $x=\frac{-1+\sqrt 5}{2}$. But $AB=2\cdot AE=2\cdot AD\cos 36^{\circ}$ $\Longrightarrow$ $1=2x\cos 36^{\circ}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1}{2x}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1}{\sqrt 5-1}$ $\Longrightarrow$ $\cos 36^{\circ}=\frac{1+\sqrt 5}{4}\ .$
[b]The second proof.[/b] Denote $u=36^{\circ}\ (5u=180^{\circ})$. Then $\sin 3u=\sin 2u$ $\Longleftrightarrow$ $\sin u(3-4\sin^{2}u)=2\sin u\cos u$ $\Longleftrightarrow$ $3-4(1-\cos^{2}u)=2\cos u$ $\Longleftrightarrow$ $4\cos^{2}u-2\cos u-1=0$ $\Longleftrightarrow$ $\cos u=\frac{1+\sqrt 5}{4}\ .$[/color][/quote] [/hide]


find a expression of general term to these sequence($ n\ge1$)
$ x_{1}\equal{}\frac{1}{2},x_{n}\equal{}2x_{n\minus{}1}\plus{}cos\frac{\pi}{7n}$
$ y_{1}\equal{}sin\frac{\pi}{11},y_{n}\equal{}y_{n\minus{}1}\plus{}sin\frac{\pi}{11n}$
thanks advanced

[quote="hoangvn.fix"]find a expression of general term to these sequence($ n\ge1$)
$ x_{1} \equal{} \frac {1}{2},x_{n} \equal{} 2x_{n \minus{} 1} \plus{} cos\frac {\pi}{7n}$
$ y_{1} \equal{} sin\frac {\pi}{11},y_{n} \equal{} y_{n \minus{} 1} \plus{} sin\frac {\pi}{11n}$
thanks advanced[/quote]

You basically have

$ x_n\equal{}2^{n\minus{}2}\plus{}\sum_{k\equal{}2}^n2^{n\minus{}k}\cos(\frac{\pi}{7k})$

$ y_n\equal{}\sum_{k\equal{}1}^n\sin(\frac{\pi}{11k})$

And I would be very surprised if some closed form existed for these two sums ... (this would imply closed forms for $ \cos(\frac{\pi}{7n})$ and $ \sin(\frac{\pi}{11n})$)

Are you sure about the problem statement ? where is it coming from ?

[quote="hoangvn.fix"]i also have the same result as you mr pco but i am stuck with counting 2 these sums,i think it is difficult because it contains variable at the numerator of the fraction.This problem comes from my school test( originally it requests counting $ x_{20}and y_{20}$approximately)
anyway thank you so much![/quote]
If the original problem was "compute an approximation of $ x_{20}$ and $ y_{20}$", please dont transform this in "give the general term of $ x_n$ and $ y_n$" !!!
That's quite different.
Or post it in  "open questions" since posting in "unsolved problem" means that you know there is an answer (although you dont know it), which is wrong !


Solve the system  in the integer numbers

$x^3-4x^2-16x+60=y$
$y^3-4y^2-16y+60=z$
$z^3-4z^2-16z+60=x$

Silouan,
 I think you must write  << Solve in the [u]set  Z of integers[/u] the ... system>>

 [u]Babis[/u]

 [hide]( I have it wrong in my book , too!!!)[/hide]

Yes, it should be solved in integers. I posted the problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=62219 .

  darij

I had the same solution sa Megus, but in thinking for another solution i get.
Adding the three equations we have

$(x+4)(x-3)(x-5)+(y+4)(y-3)(y-5)+(z+4)(z-3)(z-5)=0$

but I stopped there . Could we have solution from this?


induction
number theory unsolved
number theory

If $n(n+1)(n+2)=q^{2}$, then $n+1=r^{2}$. Therefore $n=u^{2},n+2=v^{2}$ or $n=2u^{2},n+2=2v^{2}$. It give contradition. I prove in Russian forum, that $n(n+1)...(n+k)\not =q^{2}$.

The general result (replace "3 integers" with "k integers" and "perfect square" with "perfect power") is a theorem due to Erdös and Selfridge.

As for the particular case where we have 3 consecutive integers, it's not hard to prove that it can never be a perfect power.
(indeed, $\gcd(n, n^{2}-1) = 1$; $n$ and $n^{2}-1$ are then perfect powers, conclusion)

[quote="icx"]Prove that for positive integers $n \in \mathbb{Z}$, $n(n+1)(n+2)$ isn't a perfect square.

Has induction something to do with this?[/quote]

we will prove that $(n-1)n(n+1)\neq k^{2}$. assume the contrary. we have
$n(n^{2}-1)=k^{2}$
so $n|k$. Let $k=nm$ so $n^{2}-1=nm^{2}$ or $n(n-m^{2})=1$, contradiction! since $n\ge 2$$\blacksquare$

ps. it took me less than a minute to get this. i don't believe it was a problem of an olympiad!


For all $n\in\mathbb{N}$   $r=?$ when $65^{6n-1}$is divided by $9$


There are 6 sets of connected paper rings consisting of 3 rings in 4 sets and 2 rings in 2 sets as depicted in the diagram below.  (These ringed sections are flexible.)  Your goal is to join them all into one larger unknotted ring by making cuts in rings, connecting ringed sections together, and then later taping back the cut rings (in order for it to hold together).  Making one cut will cut only one ring at a time.

What is the minimum number of cuts that you can make on the rings to accomplish this goal?

OOO

OOO

OOO

OOO

OO

OO

Could you explain your answer a bit more ihatepie? I don't quite get what you mean.

These sets of rings are connected chain-link style.

One cut cannot cut more than one ring at a time with one cut (as earlier mentioned).  After a ring is cut, it can be moved away from a ring.

Problem is OPEN still.

[hide]Would the answer be 6, because there are 6 places where it is needed to have two different chains taped together?[/hide]

No replies will be forthcoming from me on this puzzle at all.

[hide]
4 cuts
cut each ring in both OO , then you have 4 free O .
Join them to the 4 OOO , you get your ring [/hide]


quadratics
number theory unsolved
number theory

Let $x$ and $y$ be positive integers such that $xy$ divides $x^2 + y^2 + 1.$
Show that $\frac{x^2 + y^2 + 1}{xy} = 3.$

Suppose $xy \mid x^2+y^2+1$, then we have $x^2+y^2+1=nxy$, which can be written as $x^2+x(-ny)+(y^2+1)=0$ or $y^2+y(-nx)+(x^2+1)=0$. Now if $x=y$, we have $(n-2)x^2=1$ so that $n-2=1, x^2=1 \Rightarrow n=3, x=y=1$.

Suppose there's a solution to the equation with $x \neq y$, and assume $x>y$. Then the product of the roots of the quadratic in $x$ are $y^2+1$, so that $xx_2=y^2+1$. Since $x, y$ are positive, $x_2$ is also positive. If $x_2>y$, since $x>y$ we have $xx_2 \geq (y+1)^2>y^2+1$ for positive $y$, contradiction. So $x_2 \leq y<x$. So we get a solution $(x_2, y)$ such that $x_2+y<x+y$. Then we can again take the greater one of $x, y$ and repeat the process. If all the solutions thus obtained have $x \neq y$ we will have $x>y>x_2>y_2>x_3>...$ Contradiction, because all the $x_i, y_i$ must be positive. So at some point we'll get a solution with $x=y$. However, this implies $n=3$, as we've already seen.

 If $x_2>y$, since $x>y$ we have $xx_2 \geq (y+1)^2>y^2+1$ for positive $y$, contradiction.

Can you explain me why that means contradiction

It's the second example from the Vieta Jumping pdf see here: https://www.dropbox.com/sh/5k0y5y6xn70mbqf/AADLO-iLq733bdwCvi1uM5v7a/Vieta%20Jumping.pdf?dl=0


advanced fields
advanced fields theorems

In [b]Modular forms, Miyake [/b]has following text

''...A non-scalar element $ \alpha$ of $ GL_{2}^{+}(\mathbb{R})$ is called...'' (page 7)

i don't understand. What is ''[b] non-scalar [/b]element'' ?  :blush:

It means "not a multiple of the identity".

True ! I found that 20' before! Thank you very much!  :)


$ (1 \minus{} x^2)y'' \minus{} 2xy' \plus{} 6y \equal{} 0$ given that $ u(x)\equal{}3x^{2}\minus{}1$ is a solution of the equation and satisfies $ u(\minus{}1)\plus{}(1/3)u'(\minus{}1)\equal{}0$

find another solution $ v(x)$ of the equation satisying $ v(0)\equal{}0$

I tried:

$ v(x)\equal{}w(x)(3x^{2}\minus{}1)$

$ v'(x)\equal{}(3x^{2}\minus{}1)w'\plus{}6xw$
$ v''\equal{}(3x^{2}\minus{}1)w''\plus{}12xw'\plus{}6w$

how do i get to $ (3x^{2}\minus{}1)(1\minus{}x^2)w''\equal{}Xw'$ where X is some function. I will be able to go further once i know what X is.

Also when do you use the given boundary conditions?

Many Thanks

You introduce the initial conditions so that you can determine the constants at the end.

Write $ y\equal{}v(x)(3x^{2}\minus{}1)$.  Then $ y'\equal{}v'(3x^{2}\minus{}1)\plus{}6xv$ and $ y''\equal{}v''(3x^{2}\minus{}1)\plus{}12xv'\plus{}6v$.  Now we substitute these back into the original equation, to get $ (1\minus{}x^{2})(v''(3x^{2}\minus{}1)\plus{}12xv'\plus{}6v)\minus{}2x(v'(3x^{2}\minus{}1)\plus{}6xv)\plus{}v(3x^{2}\minus{}1)\equal{}0$.

I'm not expanding this; you can take it from here.


geometry
3D geometry
tetrahedron
sphere
geometry proposed

The tetrahedron $ABCD$ is inscribed in a sphere with center $O$, such that $O$ lies inside the tetrahedron. The lines $AO, BO, CO, DO$ intersect the sphere at $A', B', C', D'$, respectively. Denote $[XYZW]$ the volume of tetrahedron $XYZW$.
Prove that $2[ABCD] = [A'BCD] + [AB'CD] + [ABC'D] + [ABCD']$.

Assume AO,BO,CO,DO cut the planes (BCD) etc. in A",B",C",D" respectively. We have

$S = \sum \frac{\left[A^{\prime}BCD\right]}{\left[ABCD\right]} = \sum \frac{A^{\prime}A^{\prime\prime}}{AA^{\prime\prime}} = \sum \frac{R-OA^{\prime\prime}}{AA^{\prime\prime}}$.

On the other hand we have

$4 = \sum \frac{AA^{\prime\prime}}{AA^{\prime\prime}} = \sum \frac{R+OA^{\prime\prime}}{AA^{\prime\prime}}$ (1)

and

$1 = \sum \frac{\left[OBCD\right]}{\left[ABCD\right]} = \sum \frac{OA^{\prime\prime}}{AA^{\prime\prime}}$ (2).

By the subtraction (1) - 2 (2) we get what we want, which is S = 2.

The main idea of grobber's solution is the same as mine.
I thought out this problem when I saw the similar problem about triangle. And genalized to this one. Although it isn't very difficult, I love it.


Prove that if   $ x \plus{} y \plus{} z \ge xyz$

then,

$ x^{2} \plus{} y^{2} \plus{} z^{2} \ge xyz$

This inequality is Ireland 1997

With the same condition it can be prove that :

$ x^2\plus{}y^2\plus{}z^2\ge \sqrt{3}xyz$ (BMO 2001)

Let $ a\equal{}\frac{1}{yz}$ a.s.o. Then $ x\equal{}\sqrt{\frac{a}{bc}}$. we have $ a\plus{}b\plus{}c\ge 1$ and we need to prove

$ \frac{a}{bc}\plus{}\frac{b}{ca}\plus{}\frac{c}{ab}\ge \sqrt{\frac{3}{abc}}$

iff

$ (a^2\plus{}b^2\plus{}c^2)^2\ge 3abc$

Now $ (a^2\plus{}b^2\plus{}c^2)^2\ge 3(a\plus{}b\plus{}c)(abc)\ge 3abc$ is obvious.

If we cite Vasc Inequality in say, USAMO, will we be docked points?

[hide]
Not as nice but..
$ x\plus{}y\plus{}z \ge xyz \Longrightarrow \frac{1}{yz}\plus{} \frac{1}{zx}\plus{} \frac{1}{xy}\ge 1$

We need to show that $ \frac{x}{yz} \plus{} \frac{y}{zx}\plus{} \frac{z}{xy} \ge \sqrt{3}$

Since the inequality is symmetric assume $ x\ge y \ge z$, so by Chebyshev and our condition

$ \frac{x}{yz} \plus{} \frac{y}{zx}\plus{} \frac{z}{xy} \ge \frac{1}{3} (x\plus{}y\plus{}z)\left(\frac{1}{yz}\plus{} \frac{1}{zx}\plus{} \frac{1}{xy}\right) \ge \frac{1}{3}xyz \qquad(1)$

Also by AM-GM

$ \frac{x}{yz} \plus{} \frac{y}{zx}\plus{} \frac{z}{xy} \ge \frac{3}{\sqrt[3]{xyz}} \qquad(2)$

From (1) and (2) we always have one of

$ \frac{1}{3}xyz \ge \sqrt{3} \Longrightarrow xyz \ge 3\sqrt{3}$
or
$ \frac{3}{\sqrt[3]{xyz}} \ge \sqrt{3} \Longrightarrow 3\sqrt{3} \ge xyz$

Hence $ \frac{x}{yz} \plus{} \frac{y}{zx}\plus{} \frac{z}{xy} \ge \sqrt{3} \Longrightarrow x^2\plus{}y^2\plus{}z^2 \ge \sqrt{3}xyz$

Equality when $ x\equal{}y\equal{}z\equal{}\sqrt{3}$

[/hide]

[quote="ocha"][hide]
Not as nice but..
$ x \plus{} y \plus{} z \ge xyz \Longrightarrow \frac {1}{yz} \plus{} \frac {1}{zx} \plus{} \frac {1}{xy}\ge 1$

We need to show that $ \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy} \ge \sqrt {3}$

Since the inequality is symmetric assume $ x\ge y \ge z$, so by Chebyshev and our condition

$ \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy} \ge \frac {1}{3} (x \plus{} y \plus{} z)\left(\frac {1}{yz} \plus{} \frac {1}{zx} \plus{} \frac {1}{xy}\right) \ge \frac {1}{3}xyz \qquad(1)$

Also by AM-GM

$ \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy} \ge \frac {3}{\sqrt [3]{xyz}} \qquad(2)$

From (1) and (2) we always have one of

$ \frac {1}{3}xyz \ge \sqrt {3} \Longrightarrow xyz \ge 3\sqrt {3}$
or
$ \frac {3}{\sqrt [3]{xyz}} \ge \sqrt {3} \Longrightarrow 3\sqrt {3} \ge xyz$

Hence $ \frac {x}{yz} \plus{} \frac {y}{zx} \plus{} \frac {z}{xy} \ge \sqrt {3} \Longrightarrow x^2 \plus{} y^2 \plus{} z^2 \ge \sqrt {3}xyz$

Equality when $ x \equal{} y \equal{} z \equal{} \sqrt {3}$

[/hide][/quote]

Assuming $ xyz\sqrt {3}\ge x^2 \plus{} y^2 \plus{} z^2$ you get

$ xyz\sqrt {3}\ge x^2 \plus{} y^2 \plus{} z^2\ge 3\sqrt [3]{x^2y^2z^2}\Leftrightarrow xyz > 3\sqrt {3}$

Also :

$ \frac {x^2y^2z^2}{3}\le \frac {(x \plus{} y \plus{} z)^2}{3}\le x^2 \plus{} y^2 \plus{} z^2\le xyz\sqrt {3}\Rightarrow xyz < 3\sqrt {3}$


On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

$ \textbf{(A)}\ \frac {5}{11} \qquad
\textbf{(B)} \ \frac {10}{21} \qquad
\textbf{(C)}\ \frac {1}{2} \qquad
\textbf{(D)} \ \frac {11}{21} \qquad
\textbf{(E)}\ \frac {6}{11}$

[hide="Answer"]Looking at each case individually, we have a $\frac{1}{21}\cdot \frac{2}{6}=\frac{1}{63}$ chance if the dot on the $1$ is removed; $\frac{2}{21}\cdot \frac{4}{6}=\frac{4}{63}$ chance if one of the dots on the $2$ is removed; $\frac{3}{21}\cdot \frac{2}{6}=\frac{3}{63}$ chance if one of the dots on the $3$ is removed; $\frac{4}{21}\cdot \frac{4}{6}=\frac{8}{63}$ chance if one of the dots on the $4$ is removed; $\frac{5}{21}\cdot \frac{2}{6}=\frac{5}{63}$ chance if one of the dots on the $5$ is removed; $\frac{6}{21}\cdot \frac{4}{6}=\frac{12}{63}$ chance if one of the dots on the $6$ is removed. Adding these up, we have $\frac{1}{63}+\frac{4}{63}+\frac{3}{63}+\frac{8}{63}+\frac{5}{63}+\frac{12}{63}=\frac{33}{63}=\frac{11}{21}\Rightarrow \boxed{D}$.[/hide]

3/6 for even and same for odd.

1/6 for 1 dot removed from either parity.

3/6 * 1/21 + 3/6 * 20/21 = 1/2

actually..i think jesus' solution looks correct.

Yeah, I realized my error in thinking after posting that.

You didn't account for the fact that there are fewer total dots on odd faces.

[hide="Solution"]The probability of choosing a dot on a face with an even number of sides is $\frac{2+4+6}{21}=\frac{4}{7}$, and then the probability of choosing an odd-numbered face is $\frac{4}{6}=\frac{2}{3}$. The probability of choosing a dot on a face with an odd number of sides is $\frac{1+3+5}{21}=\frac{3}{7}$, and then the probability of choosing an odd-numbered face is $\frac{2}{6}=\frac{1}{3}$, so our total is
$\frac{4}{7}\cdot\frac{2}{3}+\frac{3}{7}\cdot\frac{1}{3}=\frac{8}{21}+\frac{3}{21}=\boxed{\frac{11}{21}}$.[/hide]

[hide="answer"]there is a 4/7 chance of picking a dot on an even face and 3/7 of picking an odd face.  Picking an even face creates 4 odd sides and 2 even and odd gives 4 even and 2 odd.  4/7 x 2/3 + 3/7 x 1/3= 11/21, so [color=red]D [/color]is the answer[/hide]

[hide="solution"]For the 1-dot removed, there is a $\frac{1}{21}*\frac{1}{3}=\frac{1}{63}$ chance that both situations will occur.  For the 2-dot side losing a dot, we have a $\frac{2}{21}*\frac{2}{3}=\frac{4}{63}$ chance that both situations will occur.  For the 3 dot side losing a dot, we have a $\frac{3}{21}*\frac{1}{3}=\frac{3}{63}$ chance that both situations will occur.  For the 4 dot side losing a dot, we have a $\frac{4}{21}*\frac{2}{3}=\frac{8}{63}$ chance that both situations will occur.  For the 5 dot side losing a dot, we have a $\frac{5}{21}*\frac{1}{3}=\frac{5}{63}$ chance that both situations will occur.  For the 6 dot side losing a dot, we have a $\frac{6}{21}*\frac{2}{3}=\frac{12}{63}$ chance that both situations will occur.  Overall, we add up the fractions, $\frac{1+4+3+8+5+12}{63}=\frac{33}{63}=\frac{11}{21}$  chance that it will happen, which is our answer (D)[/hide]


geometry
perimeter
combinatorics unsolved
combinatorics

Hi, everyone,

I am new on this site, so I apologize for any mistakes (placing the question in the wrong Forum, etc.)

I was asked about a solution to the following problem:

[i]A fly creeps clockwise along the perimeter of each face of a convex polyhedron (there are as many flies as faces). It is known that the velocities of all the flies at any moment are not less than 1 mm/h. Prove that at some moment some two flies will collide.[/i]

I don't remember the problem myself and was not able to come up with a solution after thinking for a while. I guess the solution is included into the second edition of Fomin's book, but I don't have it. So is there anyone who can help?

Thanks a lot in advance.

Just wanted to note that this has been posted before, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21916]here[/url]. I just wanted to link the two topics; there's no solution there either.


Will transcripts be available for this class? I signed up already, but I can only go to the Sunday class.

Yes, the transcripts will be available in the My Classes section within 24 hours of the end of each class.

I signed up for the AIME problems seminar but i can't figure out how to enter the virtual classroom.

Click on the Classroom button at left after 12:20 (after you've logged in).

I clicked on classroom, but only a page describing the classroom comes up.

Go there after 12:20; that is when the classroom will open.

I've tried to get in the classroom, but the computer continually freezes. Do you need a special plug-in to get in?

EDIT - Never mind. I got in.

um i cant seem to get in

i already clicked on "classroom"


geometry
3D geometry
tetrahedron
function
calculus
inequalities
induction

Let P be a point inside a regular tetrahedron whose edges have length 1. Show that the sum of distances from P to the vertices of the tetrahedron is at most 3.

This has already been discussed [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17028]here[/url].

This has already been posted and discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=17028 .

I gave the following solution: Let ABCD be our tetrahedron; then, the function $f:\mathbb{R}^3\rightarrow\mathbb{R}$, defined by f(P) = |PA| + |PB| + |PC| + |PD| is convex (since all of the functions $f_A\left(P\right) = \left|PA\right|$, $f_B\left(P\right) = \left|PB\right|$, $f_C\left(P\right) = \left|PC\right|$ and $f_D\left(P\right) = \left|PD\right|$ are convex), and since the maximum of a convex function inside a convex polyhedron is achieved at one of the polyhedron's vertices, the maximum of the function f(P) must be achieved at one of the vertices A, B, C, D of the tetrahedron ABCD. No matter which vertex we take, the value of f(P) for this vertex is clearly 3, so that 3 is the maximum of the function f(P). This solves the problem.

Here is, for the sake of completeness, a non-calculus proof of the (famous and helpful) fact that the maximum of a convex function inside a convex polyhedron is achieved at one of the polyhedron's vertices:

At first, we will use a lemma (whose proof is analogous to the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15431]proof of Lemma 1 in another post of mine[/url]):

[b]Lemma 1.[/b] If X and Y are two points in space, g is a convex function $g:\mathbb{R}^3\rightarrow\mathbb{R}$, and P is an arbitrary point on the segment XY, then at least one of the two inequalities $g\left(X\right) \geq g\left(P\right)$ and $g\left(Y\right) \geq g\left(P\right)$ holds.

Now, let g be our convex function, and let $A_1$, $A_2$, ..., $A_n$ be the vertices of our polyhedron. Among all these vertices, let $A_u$ be the one with the greatest value of g; in other words, let $A_u$ be the one such that $g\left(A_u\right) \geq g\left(A_i\right)$ for every $i\in\left\{1;\;2;\;...;\;n\right\}$. (If there is more than one vertex with the greastest value of g, take any of these vertices.)

We will prove that for every point P inside or on the boundary of the polyhedron, we have $g\left(A_u\right) \geq g\left(P\right)$. In fact, assume the contrary: Assume that $g\left(P\right) > g\left(A_u\right)$. We will now construct a contradiction.

Let an arbitrary line through the point P intersect the boundary of the polyhedron at two points U and V; then, the point P lies in the interior of the segment UV (thanks to the convexity of the polyhedron, and to the fact that the point P lies inside the polyhedron). Hence, since the function g is convex, we can apply Lemma 1 and obtain that at least one of the inequalities $g\left(U\right) \geq g\left(P\right)$ and $g\left(V\right) \geq g\left(P\right)$ holds. WLOG assume that the equation $g\left(U\right) \geq g\left(P\right)$ holds.

Now, the point U lies on the boundary of our polyhedron, i. e. inside one of its faces. This face is a convex polygon, and the vertices of this polygon are some of the vertices $A_1$, $A_2$, ..., $A_n$ of the polyhedron. Consider an arbitrary line passing through the point U and lying in the plane of this polygon; let this line meet the polygon in two points X and Y. Then, the point U lies in the interior of the segment XY (thanks to the convexity of the polygon, and to the fact that the point U lies inside the polygon). Hence, because of the convexity of the function g, we can use Lemma 1 again and see that at least one of the two inequalities $g\left(X\right) \geq g\left(U\right)$ and $g\left(Y\right) \geq g\left(U\right)$ holds. WLOG assume that the inequality $g\left(X\right) \geq g\left(U\right)$ holds.

Now, the point X lies on the boundary of our polygon, i. e. on an edge of our polyhedron. Let $A_iA_j$ be this edge. Since the point X lies in the interior of this edge, and since the function g is convex, Lemma 1 yields that at least one of the two inequalities $g\left(A_i\right) \geq g\left(X\right)$ and $g\left(A_j\right) \geq g\left(X\right)$ holds. Assume WLOG that the inequality $g\left(A_i\right) \geq g\left(X\right)$ holds.

Then,

  $g\left(A_i\right) \geq g\left(X\right) \geq g\left(U\right) \geq g\left(P\right) > g\left(A_u\right)$.

But this contradicts with our assumption that $g\left(A_u\right) \geq g\left(A_i\right)$ for every i. Hence, we have our contradiction, and the proof is complete.

Of course, by induction, this can be extended to n-dimensional space.

  Darij


function
topology
geometry
3D geometry
real analysis
real analysis unsolved

Let $\lambda : [a,b] \rightarrow \mathbb{R}^n$  continous and rectificable. Prove that $\lambda ([a,b])$ has measure zero (in $\mathbb{R}^n$ ).

PD: A set of Measure zero is a  set of points capable of being enclosed in intervals ( blocks in $\mathbb{R}^n$)   whose total length is arbitrarily small. $\lambda : [a,b] \rightarrow \mathbb{R}^n$ is rectificable if exists $C>0$ sucht that $\sum_{i=1}^{n} \| \lambda (t_i) - \lambda(t_{i-1}) \| < C$ for all $\{a=t_0, t_1, ... , t_n = b \}$ partition of $[a,b]$ .

$Tipe$

Take $n=1$ and $\lambda$ increasing. Does the image of $\lambda$ always have measure zero?

It's true if $n\ge 2$, I think. Basically, this says that curves with finite length cannot fill up volumes of dimension larger than $1$.

Yes: $n > 1$ , sorry, any idea?  :lol:

$Tipe$

i'd consider $l : = \inf \sum_{i=1}^{N} \| \lambda (t_i) - \lambda(t_{i-1}) \|$ taken over all partitions ${a=t_1 < t_2 < \dots < t_N = b}$ of $[a,b]$. of course $l<+\infty$, since curve is rectificable...
now, i think that's not difficult to show that $\mathcal{L}^n(\lambda([a,b]))$ can be stimated with $2l\varepsilon$ for some real $0< \varepsilon < \varepsilon_0$...
am i wrong?

What do you mean with $\mathcal{L}^n\left(\lambda(\left[a,b\right])\right)$ ?  :?

$Tipe$

Deleted

There were some serious mistakes here (see jmerry's comment below), and I don't have the time to correct them.

A correction: let the division be $a=t_0,t_1,t_2,\dots,t_m=b$. The volume of the body is less than $m\cdot(4\epsilon)^n$. If we choose the division so that each piece has length at least $\frac{\epsilon}2$ as well, then $m\le\frac{2L}{\epsilon}$ and everything works.

The global estimate has to depend on $\epsilon^{n-1}$, and the details are touchy.

Would this refined version work then?

Let $\lambda(t)=(\lambda^1(t),\ldots,\lambda_n(t))$. $\lambda$ has finite length $\ell$ iff $\lambda^i$ has finite length $\ell_i$ for all $i\in\overline{1,n}$. Just like before, pick some small $\varepsilon$, but this time choose a division $a=t_0<\ldots<t_m=b$ such that the lengths of $\lambda^i([t_{j-1},t_j])$ are $<\varepsilon$ for all $j\in\overline{1,m}$ and all $i\in\overline{1,n}$, with the additional condition that the lengths of $\lambda^n([t_{j-1},t_j])$ are equal to $\frac{\ell_n}m<\varepsilon$ for all $j\in\overline{1,m}$. If we can do this, we're done: every piece $\lambda([t_{j-1},t_j])$ lies in some parallelepiped with volume $2^n\cdot\frac{\ell_n}m\cdot\varepsilon^{n-1}$, and there are $m$ such parallelepipeds, so their total $n$-dimensional volume is $<C\cdot\varepsilon^{n-1}$ for a constant $C$, not depending on $\varepsilon$. All we need to do is prove that we can find such a partition $a=t_0<\ldots<t_m=b$.

The function $\gamma$ associating to $t$ the length of $\lambda^n([a,t])$ is an increasing homeomorphism of $[a,b]$ onto $[0,\ell_n]$, so it has a continuous inverse $\gamma^{-1}$. Given some $\varepsilon>0$, we can find $\delta=\delta(\varepsilon)$ such that whenever $x,y\in[a,b]$ with $|x-y|<\delta$ the lengths of $\lambda^i([x,y])$ are $<\varepsilon$ for all $i\in\overline{1,n}$. In turn, given $\delta>0$, we can find $u=u(\delta)<\varepsilon$ such that whenever $z,t\in[0,\ell_n]$ with $|z-t|<u$ we have $|\gamma^{-1}(z)-\gamma^{-1}(t)|<\delta$. Now simply divide $[0,\ell_n]$ into equal intervals of length $\frac{\ell_n}m<u<\varepsilon$ by means of a division $0=\tilde t_0<\ldots<\tilde t_m=\ell_n$ and take $t_j=\gamma^{-1}(\tilde t_j),\ \forall j\in\overline{0,m}$.

I hope there are no mistakes this time.

It probably works, but I'd prefer crude estimates and a simple argument. Partition the curve into $m$ pieces, each of length $\frac lm$. Since the length of the curve from $t=a$ to $t=c$ is an increasing and continuous function of $c$, we can do this. Now, at each $t_k$ ($0\le k<m$), take the cube $\|x-\lambda(t_k)\|_\infty\le \frac lm$. This cube has volume $2^n\left(\frac lm\right)^n$, and it clearly contains the curve between $t_k$ and $t_{k+1}$. The union of these $m$ cubes covers the curve and has volume at most $\frac{2^nl^n}{m^{n-1}}$, which can be made arbitrarily small if $n>1$.


vector
linear algebra
matrix
combinatorics unsolved
combinatorics

Hi,

I'm trying to deal with the following question...

Given a set of size n.
what is the maximal size for a family of subsets s.t.
each set is of size n/2 and the intersection of any two sets in the family does'nt exceed n/4.

thanks
Nir

Some fascinating results have been established by Ahlswede and Khachatrian especially the 'Complete intersection theorem' turns out to be a generalization of this.
citeseer.ist.psu.edu/ahlswede96complete.html

Thanks for the paper.

However, it seems I'm dealing with a different problem (maybe at first glance it seems dual but I think its not).

I'm interested in systems where the intersections are relatively small (any two sets "cover 3/4 of the total set) while the complete intersection theorems deal with intersections relatively large).

Am I missing something here?

Thanks again,
Nir

Make a vector out of each set $ A$ putting $ 1$ in the position $ j$ if $ j\in A$ and $ \minus{}1$ otherwise. You get several vectors with pairwise non-positive scalar products in $ \mathbb R^{n\minus{}1}$. But they cannot be more than $ 2(n\minus{}1)$ (simple induction). On the other hand, looking at $ n\minus{}1$ nontrivial rows in a Hadamard matrix, you can immediately get $ 2(n\minus{}1)$ such sets (positions of $ 1$'s and positions of $ \minus{}1$'s). So, normally, the answer is $ 2n\minus{}2$ (provided that $ n\times n$ Hadamard matrices exist; it is conjectured to be true whenever $ 4|n$ but it is not proved).


Find all positive number $ x,y,z$ such that $ xy\plus{}1$ is divisible by $ z$, $ yz\plus{}1$ is divisble by $ x$, $ zx\plus{}1$ is divisible by $ y$

Obvious d(x,y)=1= d(y,z)=d(z,x),  we can see   x divides xy+yz+zx+1   and so do y and z,  because of that x*y*z divides xy+yz+zx+1,  xy+yz+1=zx(x-1), w.l.g. we can say x>=y>=z, so  xyz=xy+yz+zx+1<=3xy +1
this means z=1,2,3. All this thre cases can be easily checked giving the solutions.

Only 2 solutions $ x>y>z$, there are $ (3,2,1)$ and $ (7,3,2)$.


i was talking to a friend... and thot about the "game of life" so to speak
this mite appeal to you game theorists out there or ppl who like this sort of thing

lets say you will end up meeting 2000 people of the opposite gender b4 ur 30
and your task is to find the "best person for you" b4 ur 30
or else ur single forever.

for simplification's purposes, what if after u meet a person you have 2 choices:
1. marry them
2. move on and meet more ppl and never meet that same person again

is there an optimal strategy for this?

what if the maximum number of people you'll meet is unknown?
what if you can meet that person again, but they mite b married already?
im sure there are otehr variations of this "game"

jus curious if ne one can think of an optimal strategy for this?
or any strategy? lol

 :P

This is very similar to the latest Regional Mandelbrot Team Question. (In fact, I think it may even be the same question.) I can't remember the color of the cards, but let's say we let the "best person for you" to be represented by a red card and all the other 1999 people to be represented by a black card.

The dealer has this stack of 2000 cards, and she will flip them one by one. Before each flip, you will have the opportunity to call the card (if you think it's red), or just let it go by without calling. If you call correctly, then congratulations. If you call (a red card) but a black card comes up, you lose. Or, if you don't call and a red card comes up, you also lose. So what's the optimal strategy here?

[hide="Hint"] You might want to set up a recursive relation. [/hide]

Um...just so we're clear, it's actually a little less random in real life. :-P

Um... every strategy should be the same... (unless the strategy is "never call").

Basically, you're just guessing which card is the correct card (unless you choose to never call). Any guess is equally likely to be correct.

Or am I missing something?

Yeah, actually, I think you're right.

Let's say the game changes a little, and there's more than one person that could make you extremely happy.

So there are now r red cards and b black cards.

But the other thing is you might have a spectrum of "red," and at any one point you see the card, but you don't know whether you have reached the "best possible" or not. So in that case it actually does matter when you stop.

Ah!  I've heard about this problem.  Not the one about the red and black cards, though.  In the one I heard about, you have a fixed number of dates, each with a quality ranking, but you don't know the total range of rankings.  To pick the best one, you watch a certain number of dates go by and then after a while you decide to pick the first one that's better than all the ones you've seen so far.  It turns out (no I don't know why) that the best strategy is to wait for 1/e of the dates to go by and then get ready to pick the next one you see that's better.

[quote]1/e? The transcendental number?[/quote]
yup.

i remember this from somewhere...

EDIT: [url=http://mathworld.wolfram.com/SultansDowryProblem.html]mathworld.wolfram.com/SultansDowryProblem.html[/url]

[quote]The problem is most commonly stated with  daughters, which gives the result that the commoner should wait until he has seen 37 of the daughters, then pick the first daughter with a dowry that is bigger than any preceding one. With this strategy, his odds of choosing the daughter with the highest dowry are surprisingly high: about $37.10$% (B. Elbows; Honsberger 1979, pp. 104-110, Mosteller 1987; Havil 2003, p. 136). As the number of daughters increases, this tends towards $1/e$ approximately $36.787...$%[/quote]

[quote="zeb"]How can 1/e dates go by?

And isn't 1/e less than 1?

So wouldn't you just be picking the first one?[/quote]

so apparently they meant 1/e percent.

not 1/e dates, and not 1/e percent, either.  I meant 1/e as a fraction of the total number of potential dates, like waiting for about 2000/e people in the problem in the original post.

In the formulation I've heard, one either estimates one's lifespan and waits until 1/e of one's life has passed, or estimates the total number of potential partners one is likely to meet over one's life (based on how many dates one currently has per year, perhaps) and waits until 1/e of those estimated potential partners have been met.

This is interesting, because I beleive that we make all signficant decision emotionally, then justify them rationally (this being the only rational part of signficant decision making).

Think about any significant decision you made in the past.  Is this description of decision making right for what you did?

In the end I think that when you are ready to get married, you will look at not 2,000 people, but closer to 20 or even 2 serious contenders.  If you are really interested in developing a strategy I suggest working with far fewer numbers of potentail mates.  Talk to those that have married and see if my thoughts are corrrect here.  It would be interesting to construct a poll and conduct it in a public place, such as a mall.

Try this, reject the first one then take the first one that is better than the immediate preceeding canidate.  I think you will be hard pressed to develop a better strategy than this.  (It also works in finding the cheapest gas station when traveling, where you don't know the discount stations.)

Have fun.

lol lots of interesting posts :D
haha... gjob ppl

Please, use words, not chatspeak.  Thanks.

If you really like the first one, then your choice is to abandon your strategy and . . .

I never thought that dating is such a complicated process  :lol:

Anything involving people can get very complicated very quickly.


[quote="ashrafmod"]Let a, b, and c be positive real numbers
such that a + b + c = 1.  Show that
a^2 + b^2  +  c^2  <=3(a^3 + b^3 + c^3).[/quote]

[hide]$a^{2}+b^{2}+c^{2}\le 3(a^{3}+b^{3}+c^{3})$
$\Leftrightarrow 3(a^{3}+b^{3}+c^{3})-(a+b+c)(a^{2}+b^{2}+c^{2})\ge 0$
$\Leftrightarrow 2(a^{3}+b^{3}+c^{3})\ge \sum_{sym}a^{2}b$

which follows from applications of AM-GM or Muirhead.[/hide]

From Cebashev we have $(a^{2}+b^{2}+c^{2})(a+b+c)\leq3(a^{3}+b^{3}+c^{3})$
The result follows :)


probability
limit
invariant
function
probability and stats

a funny exercice:
A opera singer is due to perform a long series of concerts. Having a fine artistic temperament, she is liable to pull out each night with probability $ \frac{1}{2}$. Once this has happened she will not sing again until the promoter convinces her of his high regard. This he does by sending flowers every day until she returns. Flowers costing $ x$ thousands dollars, $ 0<x<1$, brings about a reconciliation with probability $ \sqrt{x}$. The promoter stands to make $ 750$ from each succesful concert. How much should he spend on the flowers ?

Thanks!
Beautiful problem!
I get [hide="answer"]0.5[/hide]

In general settings i'v assumed discount factor $ 0 < \delta < 1$, $ s$ - profit of one concert (measured like $ x$ in thousand), $ p$ - probability of pulling out.
And [hide="generalyzed answer"] $ x \equal{} \left(\frac {\sqrt {1 \plus{} sa^{2}} \minus{} 1}{a}\right)^2$ where $ a \equal{} \frac {\delta}{1 \minus{} \delta(1 \minus{} p)}$.
When there are no discount $ \delta \equal{} 1$ and $ a \equal{} \frac {1}{p}$. In this case $ x \equal{} (\sqrt {p^{2} \plus{} s} \minus{} p)^{2}$.
[/hide]

Hi roah,

  I think that I did basically the same as you for the initial problem: I tried to optimize $ \lim_{n \to \infty} \frac{1}{n} \sum_1^n f(X_n)$ where $ X_n \in \{S,H\}$ represents the alternatives: the singer is $ S$inging or the singer is at $ H$ome, with $ f(S)\equal{}750$ and $ f(H)\equal{}\minus{}1000x$. This gives the same answer as yours, namely $ x\equal{}\frac{1}{2}$ is optimal.

Nevertheless, I am not sure about what you mean by "discount factor". As you, I tried to generalize as follows : what if we can borrow and put money in a bank with an interest rate equal to $ r$ (hence a discount factor $ \delta$). Do you mean that you optimized the quantity $ \sum_1^{\infty} \delta^n f(X_n)$ ? If so, what is the initial distribution of $ X_0$: you could take $ P[X_0\equal{}S]\equal{}1$ (the first day, the singer is singing) or you could take the invariant distribution of the appropriate Markov chain (we are interested in the long run), or any other initial distribution. The result depends on this choice, especialy if $ \delta$ is close to $ 0$ (very high interest rates).

Hi alekk,

why should optimal $ x$ depend on the initial condition?

I've supposed discount factor and infinite time horizont (the same function as your's). All situation where she is ok are similar, all situations where she is not ok are similar. So optimal investments in flowers are markov. That means that there is some unknown $ x^{*}$ that should be invested every grim day.

[quote="alekk"]a funny exercice:
A opera singer is due to perform a long series of concerts. Having a fine artistic temperament, she is liable to pull out each night with probability $ \frac {1}{2}$. Once this has happened she will not sing again until the promoter convinces her of his high regard. This he does by sending flowers every day until she returns. Flowers costing $ x$ thousands dollars, $ 0 < x < 1$, brings about a reconciliation with probability $ \sqrt {x}$. The promoter stands to make $ 750$ from each succesful concert. How much should he spend on the flowers ?[/quote]
Suppose that the opera singer is to perform $ n \in \mathbb{N}$ concerts.
We let $ a_i$ denote the money the promoter on average will get from the next $ i$ of the concerts if she is starting to sing in the first concert. In the same way let $ x_i$ be the money he should spend flowers, if $ i$ concerts is remaining, and she is not singing now.
Obviously $ a_1 \equal{} 750$. And $ a_{n \plus{} 1} \equal{} 750 \plus{} \frac {1}{2}a_n \plus{} \frac {1}{2}(\sqrt {x_n}a_n \minus{} 1000x_n)$ Since we want to maximise $ a_{n \plus{} 1}$ we see that $ \sqrt {x_n} \equal{} min(1, \frac {a_n}{2000}) \iff x_n \equal{} min(1, \frac {a_n}{2000})^2$
Let's first assume that $ a_n < 2000$, cause then we can replace $ x_n$ by $ \frac {a_n}{2000} ^2$
So $ a_{n \plus{} 1} \equal{} 750 \plus{} \frac {1}{2}a_n$
And $ a_n < 2000 \Rightarrow 750 \plus{} \frac {1}{2}a_n \equal{} a_{n \plus{} 1} < 1750$, so $ \frac {a_n}{2000} < 1$
But then $ a_n \equal{} (1 \plus{} \frac {1}{2} \plus{} \frac {1}{2^2} \plus{} \cdots \plus{} \frac {1}{2^{n \minus{} 1}})750 \equal{} \frac {2^n \minus{} 1}{2^{n \minus{} 1}}750$

So $ x_n \equal{} \frac {a_n^2}{2000^2} \equal{} (\frac {3(2^n \minus{} 1)}{2^{n \plus{} 2}})^2$

With infinitly many concerts left we see that $ \lim_{n \to \infty} x_n \equal{} \frac {9}{16}$
But you guys got $ x \equal{} 0.5$ and certanly $ 0.5 \neq \frac {9}{16}$ Can I see your solutions (well explained!!!), cause obviously either mine or yours is wrong...

Feel free to ask, if any questions.. Peace out :P

Edit: Sorry guys.. Just realised my mistake..


is the same of [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=214926]this[/url]: the circle with center A and radius p meet the halfline AB on K that is the tangency point of the A-excircle with AB, let a circle with radius $ r_a$ and center K that meet the perpendicolar line from K to AB on $ I_a$, s.t. $ I_a$ is on the same part of C. The circle with center $ I_a$ and radius $ r_a$ is the A-excircle $ \Gamma_a$. Let $ \Gamma$ be the circle with center A and radius $ h_a$. The tangent from A to $ \Gamma_a$ ($ \neq AB$) meet the internal tangent to $ \Gamma$ and $ \Gamma_a$ on C.

construct a circle with radius r(a) and draw tangents from a point with length p
let the point from which we draw the tangents be A and centre of circle be R
divide AR in the ratio h(a):r(a)
draw the tangent from the point of division
extend it to meet the other tangents at B,C
ABC is the required triangle


[hide="answer"]$max=2^{-\frac{n}{2}}$[/hide]

[hide="solution"]WLOG we can assume $\sum \tan x_{i}= n$
Let $L=\prod s_{i}-\lambda \sum t_{i}$ - Lagrange's function, where $s_{i}=\sin x_{i}, \ \ t_{i}=\tan x_{i}$
Then $\frac{\partial L}{\partial x_{i}}=\frac{\prod s_{i}}{t_{i}}-\lambda (1+t_{i}^{2})=0$
From strictly encreasing of $t(1+t^{2})$ it follows that all $t_{i}$ are equal one to another hence $t_{i}=1$
We are done because max evidently exist and we have only one point which is espected to give this max. [/hide]


Consider the pairwise distinct integers $ x_1, x_2,...,x_n$, $ n\geq 0$. Prove that
\[ x^2_1\plus{}x^2_2\plus{}...\plus{}x^2_n\geq x_1.x_2\plus{}x_2.x_3\plus{}...\plus{}x_n.x_1\plus{}2n\minus{}3.\]

[quote="mathvn82"]Consider the pairwise distinct integers $ x_1, x_2,...,x_n$, $ n\geq 0$. Prove that
\[ x^2_1 \plus{} x^2_2 \plus{} ... \plus{} x^2_n\geq x_1.x_2 \plus{} x_2.x_3 \plus{} ... \plus{} x_n.x_1 \plus{} 2n \minus{} 3.
\]
[/quote]
I think $ \sum_{i\equal{}1}^n(x_i^2\minus{}x_ix_{i\plus{}1})\geq\frac{n(n\minus{}1)}{2}$ is true too.


Are you sure you're not just breaking the length restriction?

[quote="moogra"]I can't have a link anymore when I try to edit my signature.
Am I one of the last people allowed to have a link?[/quote]You are not allowed to carry an external link in your signature anymore. If you previously had one, please edit your signature and remove it (or we will do it for you).

wait why aren't we allowed to carry a link anymore? I like my game...


hello, it is $ a_n\equal{}\frac{1}{4}\left(3\cdot2^{n\plus{}1}\minus{}4n\minus{}4\plus{}2^nn^2\minus{}2^nn\right)$.
Sonnhard.

The technique described [url=http://compgeom.cs.uiuc.edu/~jeffe/teaching/373/notes/recurrences.pdf]here[/url] works wonderfully.  Essentially, $ b_n \equal{} n \plus{} n \cdot 2^n$ is a sequence with characteristic polynomial $ (x \minus{} 2)^2 (x \minus{} 1)^2$, so $ a_n$ has characteristic polynomial $ (x \minus{} 2)^3 (x \minus{} 1)^2$ and the general solution should take the form

$ a_n \equal{} (An^2 \plus{} Bn \plus{} C) 2^n \plus{} (Dn \plus{} E)$

exactly as Dr Graubner has computed.

[quote="KenHungKK"]Suppose $ a_1 \equal{} 1$, $ a_{n \plus{} 1} \equal{} 2a_n \plus{} n\times(1 \plus{} 2^n)$ where $ n \equal{} 1,2,3,...$. Find the formula of $ a_n$.[/quote]

$ a_{n \plus{} 1} \equal{} 2a_n \plus{} n\times(1 \plus{} 2^n)$

$ \Longleftrightarrow \frac {a_{n \plus{} 1}}{2^{n \plus{} 1}} \minus{} \frac {(n \plus{} 1)^2}{4} \plus{} \frac {n \plus{} 2}{2^{n \plus{} 1}} \equal{} \frac {a_n}{2^n} \minus{} \frac {n^2}{4} \plus{} \frac {n \plus{} 1}{2^n} \minus{} \frac {1}{4}$

$ \therefore \frac {a_n}{2^n} \minus{} \frac {n^2}{4} \plus{} \frac {n \plus{} 1}{2^n} \equal{} \frac {a_1}{2^1} \minus{} \frac {1^2}{4} \plus{} \frac {1 \plus{} 1}{2^1} \minus{} \frac {1}{4}(n \minus{} 1)$

$ \Longleftrightarrow \frac{a_n}{2^n}\equal{} \frac {1}{4}(n^2 \minus{} n \plus{} 6) \minus{} \frac {n \plus{} 1}{2^n}$, yielding $ \boxed{a_n \equal{} (n^2 \minus{} n \plus{} 6)2^{n \minus{} 2} \minus{} n \minus{} 1\ (n\geq 1)}$.

[quote="KenHungKK"]Suppose $ a_1 \equal{} 1$, $ a_{n \plus{} 1} \equal{} 2a_n \plus{} n\times(1 \plus{} 2^n)$ where $ n \equal{} 1,2,3,...$. Find the formula of $ a_n$.[/quote]

$ a_{n \plus{} 1} \equal{} 2a_n \plus{} n\times(1 \plus{} 2^n)$

$ \Longleftrightarrow \frac {a_{n \plus{} 1}}{2^{n \plus{} 1}} \minus{} \frac {(n \plus{} 1)^2}{4} \plus{} \frac {n \plus{} 2}{2^{n \plus{} 1}}\minus{}\frac{1}{4} \equal{} \frac {a_n}{2^n} \minus{} \frac {n^2}{4} \plus{} \frac {n \plus{} 1}{2^n} \minus{} \frac {1}{4}$

$ \therefore \frac {a_n}{2^n} \minus{} \frac {n^2}{4} \plus{} \frac {n \plus{} 1}{2^n} \equal{} \frac {a_1}{2^1} \minus{} \frac {1^2}{4} \plus{} \frac {1 \plus{} 1}{2^1} \minus{} \frac {1}{4}(n \minus{} 1)$

$ \Longleftrightarrow \frac {a_n}{2^n} \equal{} \frac {1}{4}(n^2 \minus{} n \plus{} 6) \minus{} \frac {n \plus{} 1}{2^n}$, yielding $ \boxed{a_n \equal{} (n^2 \minus{} n \plus{} 6)2^{n \minus{} 2} \minus{} n \minus{} 1\ (n\geq 1)}$.


Prove that the line segments joining each vertex of a tetrahedron to the centroid of the opposite face are concurrent and divide each other in the ratio of $3:1$.

Let ABCD be a tetrahedron, and X be the midpoint of BC.  Take a plane through D, A, X.

Let G, H be the centroids of ABC and BCD.  We have (undirected lengths) XG/GA = XH/HD = 1/2.

Let DG and HA meet at K (its planar so they meet).  By menalaus, XA GK DH = AG KD HX implying the 3:1 property.

We have also shown two (arbitrary) midlines concur.  It is easy to show that if 4 midlines pairwise concur, they do so at a unique point.


Let A, B, C, D be four points on a line.
Prove that (A, B, C, D) = (B, A, D, C) = (D, C, B, A) = (C, D, A, B).
Starting off with a simple one.

Shouldn't this go into the Theorem's and Formulae section?