logarithms
LaTeX
combinatorics proposed
combinatorics

$ T$ is a subset of $ {1,2,...,n}$ which has this property : for all distinct $ i,j \in T$ , $ 2j$ is not divisible by $ i$ . Prove that :
$ |T| \leq \frac {4}{9}n + \log_2 n + 2$

Ok here we go. First, it is clear that all the biggest odd divisors of elements of $ T$ are different. Let me say $ log_2\ n = x$ and define the set $ T_i$ for $ i = 0, 1,\ldots\,x$ such that $ T_i$ is the set of the biggest odd divisors of the elements of $ T$ which has $ 2^i$ as the biggest power of $ 2$ that divides them. Then for all even $ k$, I claim that for any two different element from the union $ T_k$ and $ T_{k + 1}$ any of them can't divide the other. (actually for all k of course, but i will use only this) This is straightforward from the condition.
Now, assume that we have a subset of the odd numbers from $ 1$ to $ t$ and for any two different element from this set, any of them can't divide the other. We will prove that this subset has at most $ \frac {t}{3}\ + 1$ elements. (Remember the problem that our set is the integers from $ 1$ to $ n$ and the other condition is the same. The method is absolutely the same.) Say this subset, $ S$. Now rewrite all the elements of $ S$ as $ 3^s\cdot\ p$ in which $ p$ is coprime with $ 3$. All the $ p$'s are different, otherwise the one with bigger $ s$ would divide the other. Moreover, all $ p$'s are from $ {1, 2,\ldots\ ,t}$; odd and coprime with $ 3$. Easy calculation shows that there are at most $ \frac {t}{3}\ + 1$ such numbers.
So for any even $ k$ the union of $ T_k$ and $ T_{k + 1}$ can consist at most $ \frac {n}{3\cdot\ 2^k} + 1$ elements.(since the elements of $ T_i$ are lesser than or equal to $ \frac {n}{2^k}$)  Summing all gives we can have lesser than $ \frac {4n}{9} + \frac {x + 1}{2}$ elements with only the possible exception of $ T_x$ which can have at most $ 1$ element (the sum of geometrical series) It's done.
Actually this is a better bound than the wanted one, but I couldn't see my mistake. Anyways, this is the spirit.

[quote="Love_Math1994"]I have a solution from a book of a math group:[/quote]

Is it really from a math group? It is totally the same as the offical solution given by Iranian (every year Chinese leader in IMO will publish some problems of some countries with offical solution). Which book is it?

It also from AMM:AMM E 3403 Proposed by Paul Erdos but have a different form:Prove that exist
$|T-\frac{4}{9}|<C \cdot \ln n$...I am finding it in my collection of AMMs:)

edit: Iran Mo 26:http://www.mediafire.com/?7w91n63sph7bf7o

[color=#008000]Moderator says: LaTeX knows \ln [/color]


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

Check the resources section: http://www.mathlinks.ro/Forum/resources.php
There you can find problems of lots of competitions

Here is the problem http://www.mathlinks.ro/Forum/viewtopic.php?p=357351#357351


THEOREM: Q (the set of all rational numbers) is countable.
Q can be defined as the set of all fractions a/b where a and b are integers and b > 0. This can be mapped onto the subset of ordered triples of natural numbers (a, b, c) such that b > 0, a and b are coprime, and c ∈ {0, 1} such that c = 0 if a/b ≥ 0 and c = 1 otherwise.
Can you tell where did $c$ came from?

[quote="mathnerd314"]In other words, c is 0 if a/b is positive and 1 if a/b is negative.[/quote]

Isn't it c is 0 a/b is NONNEGATIVE and 1 if a/b is negative? Srry for the correction, but I still don't know what the question is asking, "where did $c$ came from?".

Sorry but I'm new to this topic. What i was asking is that rational numbers are in the form of a/b, where b is not equal to zero. Then what is c, and why is it either 0 or 1?

O ok, I see. So is c like (-1) to the c times a/b to determine the sign of the rational number? when c=0, it is 1 times a/b which makes it nonnegative and when it is 1, it makes it negative.

[quote="Bictor717"][quote="guile"]...This can be mapped onto the subset of ordered triples of natural numbers (a, b, c) such that b > 0, a and b are coprime, and [b]c ∈ {0, 1} such that c = 0 if a/b ≥ 0 and c = 1 otherwise.[/b][/quote]
From here.
They tell you exactly what c can be and how to determine its value, right after it's introduced. $c\in{0,1}$ means c is in the set {0,1}, and the value of c is determined by the sign of a/b.
By using c, you are able to write a unique ordered triple of natural numbers for every rational number, which, I suppose, is necessary to prove the theorem.[/quote]

Ahh... you mean that for every rational number, there are 2 possible signs, that is, negative and positive and c use to represents it. Have I understood correctly?

Yup, that's what Bictor717 and I were trying to say.  :yup:

the one thing to understand is that it does not matter what two values c takes, it just matters that c can be 2 values (one for positive, and another for negative)


Let $a$,$b$,and $c$ denote three distinct integers, and let $P$ denote a polynomial
Having all integral coefficients. Show that it’s impossible that  $P(a)=b$ ,$P(b)=c$,
,and $P(c)=a$
 :ninja:

Using $x-y|P(x)-P(y)$ we get $(a-b)|(b-c)|(c-a)|(a-b)$, which is possible only if $a-b=\pm(b-c)$ and $b-c=\pm(c-a)$, but then all the possibilites give $a=b=c$. Contradiction.


calculus
integration
trigonometry
calculus computations

evaluate

int 0 to pi (sin6x/sinx)

can i have some  more  examples of this type?

put x=π - y and you will get I=-I means I=0 (I is our integral).
Ooops i see you want example. You can constract some ...if you have f(x)=-f(Π-x) then by integrating from 0 to π you can take result 0

In a similar vein, we have

$ I \equal{} \int_{0}^{a} x f(x) dx$

Let $ y \equal{} a \minus{} x$.

Then $ I \equal{} \int_{a}^{0} (a \minus{} y)f(a \minus{} y)( \minus{} dy)$

$ \equal{} \int_{0}^{a} (a \minus{} y)f(y)dy$

$ \equal{} a \int_{0}^{a}f(y) dy \minus{} I$,

so that $ I \equal{} \frac {1}{2} a\int_{0}^{a} f(y) dy \equal{} \frac {1}{2} a \int_{0}^{a} f(x) dx$.

Therefore, $ \int_{0}^{\pi} \frac {x}{1 \plus{} \cos \alpha \sin x}\ dx \equal{} \frac {\pi}{2} \int_{0}^{\pi} \frac {1}{1 \plus{} \cos \alpha \sin x}\ dx$.

Putting $ t\equal{}\tan (\frac{x}{2})$, we get

$ \frac{\pi}{2}\int_{0}^{\infty} \frac{2}{1\plus{}t^{2}\plus{} 2t \cos \alpha}\ dt$

$ \equal{}\pi \int_{0}^{\infty} \frac{1}{(t\plus{}\cos \alpha)^{2}\plus{}\sin^{2}\alpha}$

$ \equal{}\frac{\pi}{\sin \alpha}[\tan^{\minus{}1} \frac{t\plus{}\cos \alpha}{\sin \alpha}]_{0}^{\infty}$

$ \equal{} \frac{\pi}{\sin \alpha}(\frac{\pi}{2} \minus{} \tan^{\minus{}1}\cot \alpha)$

$ \equal{} \frac{\pi \alpha}{\sin \alpha}$.

to andrew tom,

can i have some more results of this type?

$ \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x \plus{} \sin x}\ dx$

$ \equal{}\int_{0}^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2} \minus{} x)}{\cos(\frac{\pi}{2}\minus{}x) \plus{} \sin(\frac{\pi}{2}\minus{}x)}\ dx$

$ \equal{}\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\sin x \plus{} \cos x}\ dx$

Therefore $ 2\int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x\plus{} \sin x}\ dx \equal{} \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x\plus{} \sin x}\ dx \plus{} \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos x\plus{} \sin x}\ dx$

$ \equal{}\int_{0}^{\frac{\pi}{2}} 1 \ dx \equal{} \frac{\pi}{2}$.

Therefore $ \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x\plus{} \sin x}\ dx \equal{} \frac{\pi}{4}.$

Putting $ x \equal{} a \minus{} u$ gives

$ \int_{\frac {a}{2}}^{a} f(x)\ dx \equal{} \minus{} \int_{\frac {a}{2}}^{0} f(a \minus{} u)\ du$

$ \equal{} \int_{0}^{\frac {a}{2}} f(a \minus{} u)\ du$

$ \equal{} \int_{0}^{\frac {a}{2}} f(a \minus{} x)\ dx$

Therefore $ \int_{0}^{a} f(x)\ dx \equal{} \int_{0}^{\frac {a}{2}} f(x) \plus{} f(a \minus{} x) \ dx$.

So,

$ \int_{0}^{\pi} x \sin^{n}x\ dx \equal{} \int_{0}^{\frac {\pi}{2}}x\sin^{n}x \plus{} (\pi \minus{} x)\sin^{n}(\pi \minus{} x)\ dx$

$ \equal{} \int_{0}^{\frac {\pi}{2}}x\sin^{n}x \plus{} (\pi \minus{} x) \sin^{n} x\ dx$

$ \equal{} \pi\int_{0}^{\frac {\pi}{2}} \sin^{n}x\ dx$

which may be evaluated using Wallis's formulas.

P.S. We can also use $ \int_{0}^{a} f(x)\ dx \equal{} \int_{0}^{a} f(a\minus{}x)\ dx$ to show that

$ \int_{0}^{\pi} x\sin x\  dx \equal{} \pi$, but perhaps you have enough examples by now.

If not, just ask.

Letting $ x\equal{}\frac{\pi}{2} \minus{}y$, in

$ I\equal{} \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}\plus{}\sqrt{\cos x}}\ dx$

we get $ 2I \equal{} \frac{\pi}{2}$ and so $ I \equal{} \frac{\pi}{4}$.

In fact, we can also show in the same way that

$ \int_{0}^{\frac{\pi}{2}} \frac{\sin^{m}x}{\sin^{m}x \plus{} \cos^{m}x}\ dx \equal{} \frac{\pi}{4}$

for all real values of $ m$.


linear algebra
matrix
invariant
function
quadratics

The function

$ f_{P}(A) = PAP^{-1}$

Preserves many properties of the matrix $ A$.  What about the function

$ g_{P}(A) = PAP^{T}$

?  (Subject to the conditions that $ \det(P) \neq 0$ and that $ A$ is square and symmetric)  Of course the determinant will no longer be the same, and I have doubts as to anything else, but what is invariant under this transformation?  Barring that, what properties does $ g_{P}(A)$ have?

Failing that:  If we have a pair of symmetric square matrices $ A, B$ which we then transform

$ (A, B) \to (g_{P}(A), g_{P}(B))$

What invariants or properties does this map have?  It's not hard to see that $ AB^{-1}$ is similar to $ g_{P}(A) g_{p}(B)^{-1}$, but what other invariants or properties are there?

My real question:  We have two transformations on two symmetric square matrices $ A, B$, which are

$ (A, B) \to \left( \alpha A+\beta B, \gamma A+\delta B \right), \alpha \delta-\beta \gamma \neq 0$
$ (A, B) \to (g_{P}(A), g_{P}(B))$

Where $ P$ is an arbitrary square matrix (of the same size).  What properties of the ordered pair $ (A, B)$ are invariant under both transformations?  It's also not hard to see that an invariant of the first operation is the set of matrices $ Q$ such that

$ Q = Ax+By, \det(Q) = 0$

But the second transformation takes $ Q$ to $ PQP^{T}$, and then I am back to my first question.

This is an important question which has arisen during the course of my research, and I freely admit to not having enough background to answer it  :oops:

This has a name: it's called [i]congruence[/i].

$ B$ is congruent to $ A$ iff $ B=P^{T}AP$ for invertible $ P.$
$ B$ is similar to $ A$ iff $ B=P^{-1}AP$ for invertible $ P.$

Roughly speaking, similarity preserves algebraic properties and congruence preserves geometric properties.

Congruence preserves rank.
Congruence preserves [i]signature[/i], which is the number of positive eigenvalues minus the number of negative eigenvalues.

Put these two together: Suppose $ A$ has $ a$ positive eigenvalues, $ b$ zero eigenvalues, and $ c$ negative eigenvalues. (Counted with multiplicity - and since you were assume $ A$ symmetric, hence diagonalizable, there is no ambiguity there.) Then if $ B$ is congruent to $ A,$ the same holds for $ B$ - $ a$ positive eigenvalues, $ b$ zero eigenvalues, and $ c$ negative eigenvalues.

Claim: the converse holds. If $ A$ and $ B$ are symmetric matrices with the same rank and signature, then $ A$ is congruent to $ B.$

Some vocabulary to look for:
Witt Cancellation
Sylvester's Law of Inertia
Hasse-Minkowski Theorem

Here's one step: suppose that $ a\ne0$ and there exists some $ x$ such that $ x^{T}Ax=a.$ (That is, $ a$ is in the range of the quadratic form.) Then the subspace $ D=\{y: x^{T}Ay=0\}$ must have dimension $ n-1.$ Furthermore, $ x\not\in D,$ so a set consisting of $ x$ and a basis for $ D$ would be $ n$-dimensional and be a basis for $ \mathbb{R}^{n}.$ Let $ P$ be the matrix whose first column is $ x$ and whose remaining columns are a basis for $ D.$ Then $ P$ is invertible.

$ P^{T}AP=\begin{pmatrix}a&0\\0&A'\end{pmatrix}$ where $ A'$ is an $ (n-1)\times(n-1)$ symmetric matrix.

[quote="Kent Merryfield"]Claim: the converse holds. If $ A$ and $ B$ are symmetric matrices with the same rank and signature, then $ A$ is congruent to $ B$.[/quote]

Wow, that's exactly the kind of result I was looking for.  Thanks for your informative reply  :)

Edit:  Does this result have a name / do you have a reference?  I may have to use this in my paper.  Thanks again  :oops:


I'm just wondering how people generally pronounce "AIME".

Come on, don't you think we'd have discussed this by now?

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24212

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=27374

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65668

I just say it a-i-m-e

Its easier I guess

Yes. There's a search button on teh top of the page. Otherwise known as http://www.artofproblemsolving.com/Forum/search.php

It's plain to see it's eh-YIME. (Though for the purposes of comprehension I usually just say aim.)

[quote="joml88"]How is that easier?  It's 4 syllables compared with either 1 or 2.[/quote]

aymee just sounds weird to me

Yeah me too. Makes it sound more dignified. :wink:

nah aymee sounds cooler. I also use A I M E, but I use aymee more

i usually say "aymee"

i started this poll because i hear my math teacher say "A I M E" all the time and it bothers me.

Everybody knows how to pronounced AIM. Since the preceding syllable is already long, the E is not silent. Hence AIME is pronounced Amy. I don't mind A-I-M-E though, it just gets outrageously cumbersome at times.

I've never even heard "aim" before. Maybe it's a state thing?

Oh yeah, and I pronounce AoPS... well... like it's written, where "ao" sounds like "cow" without the k. Funny thing is, I've never heard anybody else pronounce it like that.

[quote="mysmartmouth"]I pronounce it A-O-P-S, as do most people I know, but Joe (joml88) says "Ay-Ops" and I laugh whenever he says it.[/quote]
I used to say A-O-P-S until I was ridiculed.
I also said A-I-M so it is no surprise A-I-M-E.

I either say "Ay-ops" or just say the entire phrase out, "Art of Problem Solving". I don't ever spell out the abbreviation

As far as I know all Georgians say A-o-P-S, though one formerly unenlightened soul used to say A-SOP. I don't think I've heard A-OPS more than once in GA.

[quote="mysmartmouth"][quote="bpms"][quote="mysmartmouth"]I pronounce it A-O-P-S, as do most people I know, but Joe (joml88) says "Ay-Ops" and I laugh whenever he says it.[/quote]
hm, I was under the impression that most people say ay-ops. Perhaps I am mistaken though.[/quote]

Perhaps.  But we live in SC where everything is pronounced differently.  :wink:[/quote]

I don't understand...I pronounce it a-ops and I lived in SC whenever you heard me say it.

There's very rarely an "official" pronunciation for acronyms.  Regardless, most, if not all, of the 1337 AoPSers say ay-ops.  Kthx.

[quote="jb05"]Oh yeah, and I pronounce AoPS... well... like it's written, where "ao" sounds like "cow" without the k. Funny thing is, I've never heard anybody else pronounce it like that.[/quote]

Haha, like "owps"? I say it ay-ops like I thought everyone said it.

[quote="joml88"][/quote][quote="mysmartmouth"][quote="bpms"][quote="mysmartmouth"]I pronounce it A-O-P-S, as do most people I know, but Joe (joml88) says "Ay-Ops" and I laugh whenever he says it.[/quote]
hm, I was under the impression that most people say ay-ops. Perhaps I am mistaken though.[/quote]

Perhaps.  But we live in SC where everything is pronounced differently.  :wink:[/quote][quote="joml88"]

I don't understand...I pronounce it a-ops and I lived in SC whenever you heard me say it.

There's very rarely an "official" pronunciation for acronyms.  Regardless, most, if not all, of the 1337 AoPSers say ay-ops.  Kthx.[/quote]

Hm, well your a Bostonian now.  Y'all Yankees pronounce everything differently.  And you were the only SCer I ever heard say Ay-ops.  I even asked you about why you pronounced it that way one time, and you just kinda smiled and shrugged.

I say AIME as "Amy" and AoPS as "A-O-P-S".

i say "Amy" and A-O-P-S
i think the aops pronounciation depends on the area you live. in general, more people on the west say A-O-P-S, and east has more that say A-ops


How'd you do?

I got 85, and that is really horrible since I am an 8th grader and I should have done better.

As in total up to Round 4?
I only started doing USAMTS this round, and I sort of bombed it, I got 5,0,3,2,0 for a total of 10/25.

Then again, I was really busy, and only spent about an hour or so on it, so what did I expect?  :roll:

79.  I only devoted a half hour to round 4 because the maximum I could get was 95, which wouldn't get me gold.  Amazingly I'm still on the leaderboard, albeit barely.

Oh well, I'm happy :D

only took round 4
'
5 4 2 5 5

im trying to appeal problem 3, but i dont have my solutions with me


Here is a root formula which i cant derive.
[img]http://img337.imageshack.us/img337/9711/scan03000003uu0.png[/img][/img]

It seem easy but i cant derive. Can someone derive or give link to the solution?

[hide="Hint"]Try squaring both sides and then compare terms.[/hide]

Here's a verification of the two formulas:

OK, for number one:
$\sqrt{m+n+2\sqrt{mn}}= \sqrt{m}+\sqrt{n}$
Square both sides, giving
$m+n+2\sqrt{mn}= m+2\sqrt{m}\cdot\sqrt{n}+n$.
Rearranging and combining the square root,
$m+n+2\sqrt{mn}= m+n+2\sqrt{mn}$, which is of course true.

For number two:
$\sqrt{m+n-2\sqrt{mn}}= |\sqrt{m}-\sqrt{n}|$
Now, when squaring an absolute value, you can take the absolute value away (because the square will be positive anyway), so you get:
$m+n-2\sqrt{mn}= m+n-2\sqrt{mn}$, which is true.


Sorrry to lie but I need to get an attention...
Do anyone know the formula for permutation and combination?
Because I really need the formula...

Thank You...

[quote="Silverfalcon"]And can you prove that by this problem which was in state sprint round:

How many different combinations can you make with numbers 2, 3, 4, and 5 by using only three of them each time..
So, it would be like 234, 235, etc...[/quote]

This question confuses me because I'm not sure whether you're making a number, of just choosing three elements to put into a subset.

If it were choosing 3 and making a number, then it would actually be a permuation because order matters.  235 is a totally different number than 532.  Think of it as, you have 5 choices for the hundreds digit, 4 for the tens, and 3 for the units.  Multiply together to get 60.  I am also assuming that you can only use each digit once...

If you want to plug it into the formula of p=n!/(n-r)!, it would be 5!/(5-3)!=120/2=60

This problem is rather oddly, and I think there was more information given in the actual question.

I'll let someone else prove the identities for practice.

I think it's elements but what's the difference between
permutation and combination?

combinations don't care about order, permutations do.

An exmaple would be if you had to make numbers using the digits 1,2,3 only once, the number of cominations would be 1, because 123 utilizes all the digits.  The rest are all repititions.  If it were a permuation, it would be 123, 132, 213, 231, 312, 321.


function
complex numbers
complex analysis

Let $ f$ be analytic on an open connected set $ C$ and suppose $ D\equal{}\{z: |z|\le1\}\subset C$.  Supose $ f$ is real on $ \{z: |z|\equal{}1\}$, show that $ f$ must be constant in $ C$

Let $ f(z) \equal{} u(x,y) \plus{} iv(x,y)$  for real functions $ u,v$ where $ z \equal{} x \plus{} iy$.  Since $ f$ is analytic, $ v$ is harmonic.  Since $ v$ vanishes on the boundary of $ D$, by the maximum principle for harmonic functions $ v$ must vanish within $ D$.  The CR equations then demand
\[ \frac {\partial u}{\partial x} \equal{} \frac {\partial u}{\partial y} \equal{} 0
\]
so $ f$ is constant in $ D$, and hence also constant in $ C$.


I have a problem that is said to be very difficult and have just only one solution(by someone that I don't know :lol: ).
Here is it:
Let $xyz=1$ and x,y,z be positive number. Prove that:
$\frac{x}{y^{4}+2}+\frac{y}{z^{4}+2}+\frac{z}{x^{4}+2}\geq 1$
(source: http://www.mathnfriend.net)
(author: cậu bé quàng khăn đỏ).
I hope that some one could solve as soon as posiple. :P
I am waiting to have a solution :D

[quote="toetoe"]$xyz = 1$ ?[/quote]
Yes, thank you very much, I 've check the title.
NO ONE HAVE TIME TO SOLVE THIS PROBLEM?
I AM WAITING FOR YOU.

[quote="hpasten"][quote="timlaiminh"]
Let $xyz=1$ and a,b,c be positive number. Prove that:
$\frac{x}{y^{4}+2}+\frac{y}{z^{4}+2}+\frac{z}{x^{4}+2}\geq 1$
(source: http://www.mathnfriend.net)
(author: cậu bé quàng khăn đỏ).
I hope that some one could solve as soon as posiple. :P
I am waiting to have a solution :D[/quote]

sure about "a,b,c be positive number"? you mean x,y,z positive or you forget some a,b,c in the inequality?[/quote]

Ya, thank you very much, I have check all the mistake of the title. :P
AND HOW ABOUT A SOLUTION? OR ANY IDEA, IDEA ONLY IS POSSIPLE.  :huh:  :ninja:  :|  :sleeping:  :stink:  :what?:


LaTeX
logarithms
inequalities
calculus
induction
inequalities proposed

Let n>1 be an integer. Prove that

1/n+1/(n+1)+1/n^2 > 1.

Bomb

(EDIT - Sorry - is a +) NOw its right

There are many ways to prove it but how many are there in total????

Bomb

do you mean continued fraction or what? If not, I really can't see why changing - to + can correct the inequality.. :huh:

Sorry for my lack of Latex, but it is prove:

(1/n)+(1/(n+1))+(1/(n+2))+....+(1/(n^2)) > 1.

Bomb

Leepahkin, nice solution, good use of AMHM inequality and then pairing off. To nttu, your soln is basically calculus but provides a stronger result.
Others ways I 've found inc:
-induction,
-AMGM inequality
-direct Cauchyswcrhz or Jensens.
Are there any other ways....

Bomb


Hello to all.

I am very bad at math and I hope I am posting in the correct forum and posting enough information.

I want to build a wood mold for a planter based on a design I have drawn in a 3D program however I dont know how to determine the angles of the wood cuts.

I think it would be a trapezoid shape.  The info I have is

width 18" top
width 10" bottom
height 14"
I believe the top angle is 74º and bottom angle is 106º

Here is a pic of what I want to make. The pic shows a solid object however I am designing an empty box/mold for an outdoor container to hold plants/small trees. I will probably make the sides out of 3/4" plywood
[img]http://users.adelphia.net/~logiclab/Images/planter.gif[/img]

The link to a Cinema4D file is
[url]http://users.adelphia.net/~logiclab/Images/trapezoidPlanter_01.c4d[/url]

I need to find out the cutting angles where the pieces meet. After I do that, I am going to have to make a second smaller version that fits inside so I can pour cement inside the two molds. I need 1.5 inches of space between the two molds.

I hope I have explained this correctly.
Thank you all very much for your help.
[I wish I could figure this out for myself.]

Regards,
logiclab

calculate atan(3.5)=atan(2h/(B-b)), you will get 74.0546 degs, here h=height, B,b = bases of trapezoid.

Best regards,
sunken rock


number theory
prime numbers
Purple Comet

How many subsets of the set {1, 2, 3, …, 12} contain exactly one or two prime numbers?

I know there are 5 prime numbers.
So 2^5=32
What do I do next?

Well, there are $ 2^{12}$ subsets in the set but i got nothing though..

modmarc has a good start. now just subtract the ones made of all nonprimes which has 2^7 sets

the final answer is 2^12-2^7=3968

There are 10 ways to choose 2 prime numbers, and 5 ways to choose 1. Thus, 15 ways to choose the primes.

We have 2^7 ways to choose the nons. Thus, it is 15x128.

Mz94, your solution would be correct if it didnt specify EXACTLY one or two primes.

Isn't this question from Purple Comet! 2009 for middle school?

yup. I got it right!! :D  My partner didn't help though, and i did really boring casework. it took about 3 minutes though...

[quote="Mewto55555"]Mz94, your solution would be correct if it didnt specify EXACTLY one or two primes.[/quote]

i so read that part of the problem...hehe....

and yet i somehow got this right on the purple comet competition....


6. A cube of aluminum has side length 0.10 m. It is dropped into a deep swimming pool of water with density 1.00 x 103 kg/m3. The cube comes gently to rest at the bottom of the pool. If the density of the aluminum is 2.7 x 103 kg/m3, the magnitude of the buoyant force acting on the aluminum is:

A) 26.6 N B) 9.8 N C) 2.7 N D) 1.0 N E) 0 N

You must use Archimedes' principle: the buoyant force is equal to the weight of the displaced water. The weight of the displaced water then is simple its volume (the volume of the aluminum cube) times the water's density (careful: convert kg to N). The answer should be B) 9.8N.

By
http://www.walter-fendt.de/ph11e/buoyforce.htm
Answer = 9.8N
 :P


algebra
binomial theorem
algebra proposed

A simple calculator is given to you. (It contains 8 digits and only does the operations +,-,*,/,$ \sqrt{\mbox{}}$) How can you find $ 3^{\sqrt{2}}$ with accuracy of 6 digits.

Use the general/expanded binomial theorem
i.e. $ \displaystyle{{m \choose n} \equal{} \frac {m!}{n!(m \minus{} n)!} \equal{} \frac{m (m \minus{} 1) (m \minus{} 2) (m \minus{} 3) \cdot \cdot \cdot (m \minus{} n \plus{} 1)}{n!}}$
Then apply it to $ (a \plus{} b)^{\sqrt {2}}$

EDIT: wow, i need to type slower  :blush:

[quote="falconwing64"]
$ \displaystyle{{m \choose n} = \frac {m!}{(m - n)!} = m (m - 1) (m - 2) (m - 3) \cdot \cdot \cdot (m - n + 1)}$
[/quote]
$ \displaystyle{{m \choose n} = \frac {m!}{(m - n)!n!}}$ :huh:

Easier solution: Writed $ \sqrt 2$ in basis $ 2$ and we can can compute every number of form $ 2^{\frac{m}{2^n}}$.


[quote="neelnal"][quote="DaRKnEsS"]if  $[(p' \Rightarrow q) \vee r]' \equiv 1$
which is equivalent to $[(p\wedge q) \vee (p \Leftrightarrow r' ) ] \Rightarrow (p' \vee q)$

0,r,q,p, or p'?[/quote]

What is the difference between the double arrow and the single arrow? the V and the upside down V?

Also, your question itself is not clear at all.[/quote]question is true

neelnal, question does make sense, and those [i]things[/i] happen to be logical operators. See [url=http://www.rci.rutgers.edu/~cfs/305_html/Deduction/TTRef.html]this[/url].
(the prime here supposedly logical not).

First statement implies that $(p' \Rightarrow q) \vee r = 0$. This may happen only when $(p' \Rightarrow q) = 0$ and $r=0$.
Also, $(p' \Rightarrow q) = 0$ means $p'=1$ and $q=0$. Therefore,

$\boxed{p=q=r=0}$

Plugging these to the second statement, we find that answer is $1$. Among the choices, only non-zero thing is $p'$.


geometry
rectangle
limit
topology
3D geometry
prism
invariant

The concept of a [i]dimension print[/i] was introduced by C. A. Rogers in 1988, but I only learned about it today, from Falconer's book. It appears interesting, yet little-studied. Let's define it for subsets on $\mathbb R^{2}$ for simplicity. Given a rectangle $U\subset \mathbb R^{2}$, let $a(U)$ be its longer side and $b(U)$ the shorter side. (The sides need not be aligned along the axes.) For a set $E\subset\mathbb R^{2}$ and numbers $s,t\ge 0$, define
\[\mathcal{H}_{\delta}^{s,t}(E)=\inf\left\{\sum_{i}a(U_{i})^{s}b(U_{i})^{t}\colon E\subset\bigcup_{i}U_{i}, \ \mathrm{diam}(U_{i})\le\delta\right\}\]
and $\mathcal{H}^{s,t}(E)=\lim_{\delta\to 0}\mathcal{H}_{\delta}^{s,t}(E)$. The dimension print of $E$ is the set $\mathrm{print}(E): =\{(s,t)\colon \mathcal{H}^{s,t}(E)>0\}$.

This set encodes the Hausdorff dimension $\mathrm{dim}(E)$, because $\mathrm{dim}(E)=\sup\{s\colon (s,0)\in \mathrm{print}(E)\}$. But unlike $\mathrm{dim}(E)$, the dimension print can see the curvature of $E$. A line segment has dimension print $[0,1]\times \{0\}$, but smooth curves with nonzero curvature have dimension print $\{(s,t)\colon 2s+t\le 1\}$.  :!:

[b]Problem[/b] [from Falconer's book]: construct a set with a nonconvex dimension print.

Hmm, I'm inclined to believe that for the curve it should be $s+2t\le 1$  :maybe: . Anyway, wouldn't just the union of a line segment and the Cartesian square $K\times K$ of the standard Cantor set $K$ of dimension $\alpha\in \left(0,\frac{1}{2}\right)$ give you the desired example? My logic is the following: the line segment will give you $[0,1]\times\{0\}$. The Cantor square will never give anything with $s+t>2\alpha$, so the only chance for the dimension print of the union (which is the same as the union of dimension prints) to be convex is to coincide with $[0,1]\times\{0\}$. But the dimension print of the Cantor square contains the point $(\alpha,\alpha)$. Indeed, just take the measure $\nu=\mu\times \mu$ where $\mu$ lives on the standard one-dimensional Cantor set of dimension $\alpha$ and satisfies $\mu(I)\le |I|^\alpha$ for all intervals $I\subset\mathbb R$. Then, using the Fubini theorem, it is easy to show that $\nu(U)\le C a(U)^\alpha b(U)^\alpha$ for every rectangle $U$ (the sections of $U$ in at least one direction are intervals of length at most $\sqrt{2}\,b(U)$ and the projection of $U$ in the other direction is at most $\sqrt{2}\,a(U)$). Thus, $\mathcal H^{\alpha,\alpha}(K\times K)\ge C^{-1}\mu(K)^{2}>0$.

Yes, you are right on both counts. I should have spent more time on the problem before posting it.

It seems $\text{print}(E \cup F) \subset (\text{print}(E) \cap \text{print}(F))$. Can $\subset$ be replaced by $=$ ? If not, counterexample?

Similarly with $\text{print}(E \cap F) \supset (\text{print}(E) \cup \text{print}(F))$ I guess...

Correction: you probably meant $\text{print}(E \cup F) \subset (\text{print}(E) \cup \text{print}(F))$ and $\text{print}(E \cap F) \subset (\text{print}(E) \cap \text{print}(F))$.

The second inclusion is clear, because $E_{1}\subset E_{2}$ implies $\text{print}(E_{1}) \subset \text{print}(E_{2})$. We cannot expect equality, since the intersection set $E\cap F$ could even be empty.

The first inclusion is indeed an equality. Dimension prints are countably stable under unions, with the same proof as for the Hausdorff dimension.

It's clear that we could generalize this notion to any dimension (including 1, although there it would agree with Hausdorff measure) by covering with rectangular prisms and letting $a_{1}(U) \ge a_{2}(U) \ge \cdots \ge a_{n}(U)$ be the sides in decreasing order of length. Then the dimension print would be a subset of $\mathbb{R}^{n}$ too.

Then I'm pretty sure
$\text{print}(E_{1}\times E_{2}) = \text{print}(E_{1}) \times \text{print}(E_{2})$.

Also, this allows us to talk about projections... :D (I will post again soon if I can convince myself of anything!)

However, I asked Falconer about this idea and he told me that he thought it wasn't very useful and that dimension prints are too similar for many sets...  :maybe:

[quote="Xevarion"]
Then I'm pretty sure
$\text{print}(E_{1}\times E_{2}) = \text{print}(E_{1}) \times \text{print}(E_{2})$.
[/quote]
He-he, and why not $\text{print}(E_{2}) \times \text{print}(E_{1})$? ;)

Falconer may very well be right. The definition drew my attention primarily because I've been thinking of covering things with rectangular boxes. Curvature, or other non-biLipschitz-invariant properties, are relevant in this context. However, for my purposes a line segment and a smooth (or rectifiable) arc are pretty much the same: I'm trying to understand the measures that do not accumulate along rectifiable curves of any kind. Dimension prints don't seem to help much.
[hide="In case someone is interested"]here's the question. Let $\mu$ be a Radon measure on $\mathbb R^{n}$. Suppose that there is a constant $C$ such that $\frac{\mu(R_{1})}{\mu(R_{2})}\le C$ for any two congruent rectangular boxes $R_{1}$ and $R_{2}$ with $R_{1}\cap R_{2}\ne\varnothing$. Is it true that $\mu\ll\mathcal H^{s}$ for any $s<n$?
Known to be true for $s\le n-1$ (easy), and known to be false for $s=n$ (not so easy).[/hide]

[quote="fedja"]He-he, and why not $\text{print}(E_{2}) \times \text{print}(E_{1})$? ;)[/quote]
Ok yeah I guess there'd be some question of the arrangement of the coordinates. But we'd still know what the print of the product is.

---

I do still wonder about other properties of dimension print, though. For example, it's clear that if $(s_{0}, t_{0}) \in \text{print}(E)$, then $(s, t) \in \text{print}(E)$ for all $s \le s_{0}, t \le t_{0}$. But is that a sufficient condition for a set to be a dimension print? For example, does there exist a set $E$ such that $\text{print}(E) = \{(s, t) | s^{2}+t^{2}\le r < 1\}$?

I am pretty sure that if $C_\xi$ is a Cantor set with contraction ratio $\xi$ (so $\text{dim}(C_\xi) = \frac{\log(2)}{\log(1/\xi)}$), then $\text{print}(C_{\xi_{1}}\times C_{\xi_{2}})$ is a rectangle with $s$-length the larger dimension and $t$ length the smaller dimension. So I could approximate certain shapes by doing unions of these, but since only countable unions are legit I don't think this gets me anything more interesting than countable unions of rectangles...

[quote="mlok"]If I remember correctly, the dimension print of a product of Cantor sets is a trapezoid.[/quote]
 :o well I am mildly confused by that. What kind of rectangles do we want to try to cover with on the part that's sloped? I mean, the $\mathcal{H}^{s, t}$ - measure should be finite along the border, right? So there'd be some "most efficient" kind of covering. I thought if $t$ is too big then you would try to approximate squares, but apparently I was wrong (or I miscalculated what would happen in this case)...

Well I don't usually do math in my room. I'll work on this again tomorrow when I have a chance to go back to the math building where the chalkboards are...

[quote="mlok"]Known to be true for s\le n-1 (easy), and known to be false for s=n (not so easy).[/quote]
Let us consider the case $n=1$. Construct the following measure $\mu$ on $[0,1]$. Suppose that the ratio $\mu(I)/|I|=a$ for some 4-adic interval $I$ (in the beginning, we put $I=[0,1]$, $\mu(I)=1$). Then for the 4-adic subintervals $I_{1}$, $I_{2}$, $I_{3}$, $I_{4}$, make this ratio
$a$, $1.5a$, $0.5a$, and $a$ respectively. Now run this construction for all 4-adic intervals from the top down. You'll get some doubling measure that can be extended periodically to $\mathbb R$. Now, fix large $n$ and consider the 4-adic intervals of length $4^{-n}$, on which $\mu(I)/|I|\ge 1/n^{2}$. Clearly, the measure $\mu$ of their complement is at most $n^{-2}$. On the other hand, you can consider the logarithm of this ratio as a sum of $n$ independent random variables taking the values $0$, $\log 1.5$ and $\log 0.5$ with probabilities $1/2$, $1/4$, and $1/4$ respectively. Since the expectation of each such random variable is a negative constant and your variables are bounded, the probability that the deviation from the expectation is comparable to $n$ is of order $e^{-\delta n}$ for some absolute $\delta>0$. Hence, the number of such intervals is only $(4e^{-\delta})^{n}$, whence $\mu$ is supported by a set of Hausdorff dimension strictly less than $1$. I didn't think too much of the problem. Maybe the case $n=1$ is somehow exceptional :?

I forgot to mention that $n\ge 2$. :oops: In the one-dimensional case the condition on $\mu$ reduces to the usual doubling condition, and it is long known that a doubling measure can be supported on a set of arbitrarily small (but positive) Hausdorff dimension (although no geometric characterization of such sets is known, even in one dimension).

Yes, you are right: the case $n=2$ is different. Still, everything is not that bad. One possible idea would be to construct the required measure as a Riesz product $\prod_{n\ge 1}(1+\delta\cos\langle p_{n}, x\rangle)$. If $|p_{n}|$ grow "very fast" and the directions are "quite different", this yields a singular measure with the desired property. To ensure that the Hausdorff dimension be less than $2$, one needs $|p_{n}|$ to grow as a geometric progression. Then we shall have to choose the angles quite carefully. Perhaps, a reasonable choice for the angles would be $2\pi\{\sqrt 2 n\}$ (at least, I do not see why this choice should not be self-similar and this is one of the best self-similar choices to be made). Have you tried to look at such a construction, mlok? On the other hand, it is well past midnight where I am now, so, maybe, I'm saying some complete nonsense...

Yes, that is exactly what we (I+2 coauthors) [url=http://dx.doi.org/10.1007/s00209-007-0132-5]did[/url], except that the angles converge to zero in a geometric progression rather than go around in irrational multiples of $\pi$. (I considered irrational rotation too, but convinced myself it did not offer an advantage, which could well be wrong). Frustratingly, we could not limit the growth of $|p_{k}|$ to exponential: it ended up as something like $4^{k^{2}}$, hence the dimension of the singular measure turned out to be $n$.  :wallbash_red:
[hide="Technicalities"]Consider the restriction of $\cos\langle p_{k},x\rangle$ on a line segment $I$. The frequency of this trigonometric function is $f_{k}=|p_{k}|\cos \theta_{k}$, where $\theta_{k}$ is the angle between $I$ and the direction of $p_{k}$. If $p_{k}$ is rotating around, then $\cos\theta_{k}$ will be close to zero infinitely often. This may cause the sequence $f_{k}$ to jump down. Using the fact that quadratic irrationals are poorly approximable with rationals, we can make $f_{k}$ monotone (with a possible exception of one term) and lacunary, but only at the cost of making $|p_{k}|$ super-exponential.

If instead the angle of $p_{k}$ is $2^{-k}$, then we could achieve a lacunary series on $I$ with exponential growth of $|p_{k}|$. However, the number of "almost constant" terms in this series substantially depends on the orientation of $I$. Again, super-exponential growth of $|p_{k}|$ seems to be necessary to fix this.[/hide]

[quote="mlok"] hence the dimension of the singular measure turned out to be $n$.  :wallbash_red:
[/quote]
I've got the same :(. Well, this merely means that we should say farewell to cosines and try some other building blocks. Forgetting everything but linear terms in the product, we see that we need a Lipschitz function $w: \mathbb R^{2}\to [-1,1]$ with Lipschitz constant $1$ such that $|w|$ is about $1$ most of the time (say, $\int_{Q}w^{2}\ge 0.1$ for every unit square $Q$) and  $\sup_{|I|=R}\left|\int_{I}w\,d\ell\right|$ is much less than $R$ (preferrably, $R^{a}$ with $a<1$) where the supremum is taken over all intervals $I\subset \mathbb R^{2}$ of length $R>\!>1$. Do you know how to construct such a function?  :play_ball:

OK, mlok. Here is something you may want to consider. Check each statement separately; I believe I have them all right, but I haven't worked out all details.

1) If $F$ is a bounded Lipschitz function such that $\left|\int_{a}^{a+L}F(x)e^{i t x}\,dx\right|\le L^\alpha$ with some $\alpha<1$ for all $a\in\mathbb R$, $t\in\mathbb R$ and $L>\!>1$ (I'll call this property $(*)$), then the integral of $F(x)e^{iy}$ over any interval of length $R>\!>1$ is not greater than $R^\beta$ with $\beta<1$ depending only on $\alpha$.

2) We can construct by induction a sequence of functions $f_{j}$ such that all the functions $F(x)=f_{i_{1}}(2^{i_{1}-i_{n}}x)f_{i_{2}}(2^{i_{2}-i_{n}}x)f_{i_{n}}(x)$ enjoy the property $(*)$ for all $L>2^{i_{n}-i_{1}}$ with some fixed $\alpha$. The construction runs as follows. Consider double exponential dyadic lattice, i.e., the lattice of dyadic intervals with lengths $2^{2^{m}}$. It suffices to check the property $(*)$ for such intervals with $i_{n}-i_{1}$ going up to $3d$. Suppose that we have determined $f_{n}$ on the intervals of length $2^{d}$ in this lattice up to sign. Let us choose the signs for $2^{d}$ such intervals constituting the interval of length $2^{2d}$ at random. We have about $2^{100d}$ functions to integrate against and about $2^{100d}$ values of $t$ to check for each, so we get about $2^{200d}$ random sums consisting of $2^{d}$ terms each of which does not exceed $2^{d}$. With probability very close to $1$, all those sums will be bounded by $2^{d}2^{3d/4}$ (they are too few to beat the CLT tail).

3) Consider now the Riesz product $\prod_{n}(1+\delta f_{n}(2^{n}x)cos{2^{n}}y)$. Then the integral of this product over an interval of length $R$ through any point depends essentially only on the length of the interval and the point, but not on the direction of the interval

4) (This I do not see how to prove right away :( but I firmly believe it is true) The support of the resulting measure has Hausdorff dimension $<2$ for most random choices in the construction of $f_{j}$. Actually, for any fixed $x$, you have a Riesz product in $y$ (all right, I see now that I'd better change $2$ to $4$ everywhere, but the rest seems all right).

Again, this may be a complete nonsense because it is what I thought up during a party in short periods between entertaining the guests, but I decided to share it with you anyway :lol:


The function $ f : Q\rightarrow R$ satisfies $ f(x \plus{} y) \equal{} f(x) f(y) \minus{} f(xy) \plus{} 1$ for all $ x, y$ and $ f(1988)\neq f(1987)$. Show that $ f(\minus{}\frac{1987}{1988}) \equal{} \frac{1}{1988}$.

1.$ y\equal{}0$ give $ (f(x)\minus{}1)(1\minus{}f(0))\equal{}0$, because $ f(1997)\not \equal{}f(1998)\to f(0)\equal{}1$.
2. $ y\equal{}1$ give $ f(x\plus{}1)\equal{}f(x)f(1)\minus{}f(x)\plus{}1\equal{}af(x)\plus{}1$, were $ a\equal{}f(1)\minus{}1\not \equal{}0$ (else $ f(1)\equiv 1$).
Let $ g(x)\equal{}\frac{(a\minus{}1)f(x)\plus{}1}{a}$, then $ g(x\plus{}1)\equal{}ag(x),g(0)\equal{}1$. It give $ g(n)\equal{}a^n, n\in Z$.
3. $ x\equal{}\frac{1}{n}\minus{}1,y\equal{}n$ give $ f(\frac 1n \plus{}n\minus{}1)\equal{}f(\frac 1n \minus{}1)f(n)\minus{}f(1\minus{}n)\plus{}1$ or
$ (a\minus{}1)(ag(n\minus{}1\plus{}\frac 1n)\minus{}1)\equal{}(ag(\frac 1n \minus{}1)\minus{}1)(a^{n\plus{}1}\minus{}1)\minus{}(a\minus{}1)(a^{2\minus{}n}\minus{}1)\plus{}(a\minus{}1)^2$.
Let $ x_n\equal{}g(\frac 1n \minus{}1)$, then $ (a\minus{}1)(a^nx_n\minus{}1)\equal{}(ax_n\minus{}1)(a^{n\plus{}1}\minus{}1)\plus{}(a\minus{}1)(a\minus{}a^{2\minus{}n})$.
It give $ x_n\equal{}\frac{a\minus{}a^{n\plus{}1}\plus{}(a\minus{}1)(a\minus{}a^{2\minus{}n})}{a^{n\plus{}1}\minus{}a^n\minus{}a^{n\plus{}2}\plus{}a}.$


I was looking through my course booklet, and a lot of courses say they're offered through distance education... I'm wondering about the advantages & disadvantages of doing that..

Would there be a reason not to take courses like Physics and/or Chemistry through distance education??

Anyone else thinking of doing distance education??

i'd say the only advantage to distance education is the ability to work at your own pace (unless you really can't make the on-campus course)

i think you'd miss out on a lot of the hands-on activities (especially labs in science courses)
i know you get lab kits and stuff, but if you mess something up...

anyway, I still have one more year of high school so I haven't thought much about it

I took distance ed. courses before. I like them, but I would rather take the on campus course if I can. The distance ed. courses are normally for those who are not on campus, or if you got a time schedule conflict.

I took physics on campus; it was worthwhile.  However, I didn't take the lab, so really it would have been quite feasable to do it via DE.  But trust me; it's better to take the course on campus for those types of studies.

I did take Philosophy 145 via distance ed... this course I found very easy, and the lectures are pretty boring (they don't allow many jokes, dirty comments on the DE CDs, so the profs are forced to be pretty bland), and I truthfully believe that courses are more interesting on campus.  That said, if you just need an arts credit or two and don't want to waste your time going to lectures, DE is a great idea.  I just wouldn't recommend it for math or physics courses.

One trick that I used was the following... the lectures were pretty slow and boring, so to speed them up, I loaded the mp3 files into some of my DJing software, and sped everything up by about 40%.  Yea, the prof sounded a little bit like a chipmunk, but I learned everything 40% faster!


Fie a,b,c trei numere pozitive.Demonstrati ca este adevarata inegalitatea:
(-a+b+c)(a-b+c)(a+b-c)=<abc

Se disting 2 cazuri :

Pentru -a+b+c <0 atunci inegalitatea e clara..ca intr-o parte e negativ intr-una pozivitiv

Pentru -a+b+c > 0
Notam:
 * -a+b+c = x
 *  a-b+ c =y
 *  a+b-c = z

Rezulta ca:
 * a =(y+z)/2
 * b= (z+x)/2
 * c= (x+y)/2

Deci inegalitatea devine...8xyz<= (x+y)(y+z)(z+x), care este evidenta conform inegalitatii mediilor:
x+y > 2 sqrt xy..etc...

"end of story" :)


There is a 5 digit lock , digits ranging from 0 to 9. What is the probability that someone will get the right code within 20 tries?

First ask yourself, what is the probability that it [i]won't[/i] happen within 20 tries? (does it work better?)


Four bugs sit at the four corners of a square 12 cm on a side. Each bug is in love with the bug to the corner that is counterclockwise to it about the center of the square-- that is, if each bug is facing the center of the square, each one is in love to the bug at it's right. At time t=0, the bugs start moving toward their love at a rate of 1 cm/s.
How long before the bugs reach their love?

[hide]i dont think they ever do. if each bug is in love w/ the one counterclockwise from where it is, then theyre all in love w/ different bugs. that means they'll all keep moving counterclockwise at 1cm/s, and unless one stops to rest, theyll never catch up[/hide]

I was about to say this looks a lot like a calculus HMMT problem, because the bugs follow a certain curve from their respective corners to the center.

Well, to actually find the amount of time,

[hide] Consider each bug's speed radially towards the center of the square.[/hide]

Kent Merryfield's variation on that hint in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=89750 is (in my opinion) superior.

Thanks for your input.

quote:

"The situation is rotationally symmetric in that there is no one "distinguished" bug. If their starting configuration is that of a square, then they will always maintain that configuration. This is the key insight, believe it or not, and it is a very profitable one!"

So, if we generalize to an n-gon, i.e. n bugs are situated at the n corners of a regular n-gon with side 12 cm. If each bug is facing the center of the n-gon, each one turns to the bug at its right. At time t=0, the bugs start moving at the rate of of 1 cm/s. How long will it take them to reach the bug at its right?

It would be the same : 12 seconds, right?


The first four terms in an arithmetic sequence are x+y, x-y, xy, and x/y, in that order. What is the fifth term?
The answer is $ 123/40$, but how?

I didn't understand the solution and would like another good detailed solution, thanks.

In an arithmetic sequence, the difference between consecutive terms is constant.  Thus, $ x\minus{}y\minus{}(x\plus{}y)\equal{}xy\minus{}(x\minus{}y)\equal{}\frac{x}{y}\minus{}xy$.  Use this.


Well, the equation for a circle centered at $(h,k)$ is:

$(x-h)^2+(y-k)^2=r^2$

Now, note that when $(h,k)=(0,0)$ the origin, the equation becomes:

$x^2+y^2=r^2$

which is what you have. You can find $r$ the radius easily and then graph.

[quote="xfox"]Hi.

How do i graph a circle with the equation x2+y2=25

I dont need somone to graph it for me just to explain how you graph circles with equations in general?

Thank you in advance! :)[/quote]

Rewrite it as this to make it easier, then read the previous post.

(x - 0) :^2: + (y - 0) :^2: = 5 :^2: .

If you want to know how to graph it with a calculator there are a couple different options:

1.  Solve for $y$:

$y=+\sqrt{25-x^2}.$

and

$y=-\sqrt{25-x^2}.$

You have to punch in each separate equation to get the full circle.  This will produce a graph that appears to be broken near the x-axis.  Thus the other methods might be preferable.

(the next two are intermediate level...forgive me)

2.  Convert the equation to polar coordinates.  Using $x=r\cos \theta$ and $y=r\sin \theta$ we would get:

$r=5$ for the equation.

3.  Convert to parametric:

$x=5\cos t$
$y=5\sin t$

[quote="furious"]who would use a parametric function to graph a circle?[/quote]

I kinda already gave a reason why in my last post...

Pick up your calculator and try graphing $x^2+y^2=25$ in the usual function mode.  You have those stupid breaks near the x-axis.  If you use parametric or polar equations the circle appears as it should, i.e. continuous.

Also, parametric equations are basically breaking a function into its x and y components.  I'm sure there are many benefits for doing this just like breaking a vector into its components has many benefits.


Hi,
I want to show that if A(nxn) is symmetric and then q'aq=adj(b)*

What are you asking? You haven't defined anything.


linear algebra
matrix
quadratics
algebra
polynomial
symmetry
vector

Hello everybody.
Our math teacher wants us to solve the following exercise:

If   A is a n*n   matrix and every aij=1

Which are the eigenvalues and the eigenvectors??

I am pretty sure that this will be a piece of cake for you, but i am stuck and can't think of anything (all the solutions i came up with, are wrong according to my teacher). I would appreciate any help.
Thank you!

p.s. If i have posted the message in the wrong place, please correct me!

What is the rank of this matrix $ A$ (I usually call it $ J$)? How many nonzero eigenvalues does it have?

Two things to note for the purpose of understanding jmerry's hints:

1. (algebraic multiplicity) $ \ge$ (geometric multiplicity) for any eigenvalue. And jmerry's question about rank is connected to the geometric multiplicity of an eigenvalue by the rank-nullity theorem.

2. As for a way to get at the nonzero eigenvalue: the sum of all the eigenvalues (counted according to algebraic multiplicity) is the trace of the matrix (the sum of the elements on the main diagonal). $ J$ does have an easily computed trace.

An alternate way to get at that nonzero eigenvalue: Compute $ J^2.$ Can you find a quadratic polynomial that $ J$ satisfies? Remember that if a matrix satisfies a polynomial equation, than any eigenvalues of that matrix must satisfy the same polynomial equation.

well, $ [1,1,\ldots,1]^T$ certainly is an obvious candidate for an eigenvector by symmetry considerations.

First of all, thank you everybody for taking interest in my problem!

Regarding the rank and the nonzero eigenvalues, the teacher didn’t give us any clues. The whole concept was short of “figure it out yourselves…”

The only clue she gave me (as I was the only one who at least tried to do something), was that I should calculate the eigenvalues t, as if A-tI was 1*1, then 2*2, 3*3, 4*4 so on and supposably I would come up with a sequence.
We were not taught about traces so I cannot use them to solve the problem (we are an economics faculty and she is only teaching us some basic stuff).

Thank you.

Call the matrix $ J$ and the vector $ [1,1,\ldots,1]^T$ $ v$.  What is $ Jv?$

What do you get when you row-reduce $ J$.  How many nonzero rows are there?  How is the number of nonzero rows related to the multiplicity of the eigenvalue $ 0$?  In other words, what is the relation between the rank and the number of $ 0$ eigenvalues of a matrix?


A gaseous compound of disulfur difluoride is being considered:

The volume that 102 grams of this gas would occupy at a pressure of 700 mm of Hg and 25 degrees Celsius could be calculated from which of the following?

1.  (22.4 x 760 x 273)/ (700 x 298)
2.  (22.4 x 700 x 273)/ (760 x 298)
3.  (22.4 x 760 x 298)/ (700 x 273)
4. (102 x 760 x 298) / (700 x 273)
5. (102 x 700 x 273) / (760 x 298)

This seems a direct application of the ideal gas law. So, Mr. Math, where is the doubt?

Well, the choices of 1-5 do not seem to be related to the ideal gas law, and this might be a cause for confusion. Perhaps the problem means to find a similar expression in which the value equals the volume?

[hide="Solution"]
Disulfide difluoride has a molecular mass of 103.32, making the number of moles ($ n$) is .9872.

Using the ideal gas law, we have

\begin{eqnarray*}
PV &=& nRT \\
(0.921)(V) &=& (0.9872)(0.08206)(298) \\
(V) &=& 26.21 \text{L}
\end{eqnarray*}

From here, we go through the choices, and only $ \boxed{3}$ is the closest (with round-off error).
[/hide]


In the expansion of
\[ \left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{27}\right)\left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{14}\right)^2,
\]what is the coefficient of $ x^{28}$?

$ \textbf{(A)}\ 195 \qquad \textbf{(B)}\ 196 \qquad \textbf{(C)}\ 224 \qquad \textbf{(D)}\ 378 \qquad \textbf{(E)}\ 405$

[hide]Let $ A \equal{} \left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{14}\right)$ and $ B \equal{} \left(1 \plus{} x \plus{} x^2 \plus{} \cdots \plus{} x^{27}\right)$. We are expanding $ A \cdot A \cdot B$.
Since there are $ 15$ (not 14) terms in $ A$, there are $ 15^2 \equal{} 225$ ways to choose one term from each $ A$. The product of the selected terms is $ x^n$ for some integer $ n$ between $ 0$ and $ 28$ inclusive. For each $ n \neq 0$, there is one and only one $ x^{28 \minus{} n}$ in $ B$. Since there is only one way to choose one term from each $ A$ to get a product of $ x^0$, there are $ 225 \minus{} 1 \equal{} 224$ ways to choose one term from each $ A$ and one term from $ B$ to get a product of $ x^{28}$. Thus the coefficient of the $ x^{28}$ term is $ \boxed{224 \Rightarrow C}$.[/hide]

Let \[ a(x)\equal{}1\plus{}x\plus{}...\plus{}x^{27}\equal{}a_0\plus{}a_1x\plus{}...\plus{}a_{28}x^{27}\] \[ b(x)\equal{}(1\plus{}x\plus{}...\plus{}x^{14})^2\equal{}b_0\plus{}b_1x\plus{}...\plus{}b_{28}x^{28}\] Then the coefficient of $ x^{28}$ in $ a(x)\cdot b(x)$ is going to be \[ \sum_{k\equal{}0}^{28}a_{28\minus{}k}\cdot b_{k}\equal{}\sum_{k\equal{}1}^{28}b_k\equal{}b(1)\minus{}1\] since $ a_k\equal{}1$ for $ k\equal{}0,1,2,...27$ and $ a_k\equal{}0$ for $ k\equal{}28$. Hence the answer is $ 15^2\minus{}1\equal{}224$ or $ \boxed{C}$.

[hide="Generalization"] Given a sequence $ \{ a_k \}$, let $ A(x) \equal{} \sum_{k\equal{}0}^{\infty} a_k x^k$ be its generating function.  Let $ s_k \equal{} \sum_{i\equal{}0}^{k} a_i$ be the sequence of partial sums of $ \{ a_k \}$.  Then the generating function of $ s_k$ is

$ (1 \plus{} x \plus{} x^2 \plus{} ... ) A(x)$.

This problem is a finite version. [/hide]
I tried to find a cheap and dirty solution by factoring and working $ \bmod x^{29}$, but the solution doesn't seem to quite work.  Did anyone else find a cheap and dirty solution?

[hide]Let $ z$ be a primitive $ 28$th root of unity (that is $ n \equal{} 28$ is the minimal exponent that $ z^{28} \equal{} 1$). Suppose that the function in the problem is $ p(x)$. Then we have the sum of the coefficients in the form $ x^{28k}$ where $ k$ is an integer is
\[ \frac {1}{28}\cdot \sum_{k \equal{} 1}^{28}p(z^k) \equal{} a_0 \plus{} a_{28}
\]
by the roots of unity filter concept. For $ k \equal{} 1,2,...,27$, we have that
\[ z^{28} \equal{} 1\implies (z^{k})^{28} \minus{} 1 \equal{} 0\implies (z^k \minus{} 1)(1 \plus{} z^k \plus{} z^{2k} \plus{} ... \plus{} z^{27k}) \equal{} 0
\]
If $ z^k \equal{} 1$, then $ k$ is less than $ 28$ which is a contradiction to the fact that $ n \equal{} 28$ is the minimal exponent. Hence the second factor is zero, and so
\[ p(z^k) \equal{} (1 \plus{} z^k \plus{} z^{2k} \plus{} ... \plus{} z^{27k})(1 \plus{} z^k \plus{} ... \plus{} z^{14k})^2 \equal{} 0
\]
for $ k \equal{} 1,2,...,27$. Now for $ k \equal{} 28$, we have $ z^{28} \equal{} 1$, so the sum is $ \frac {1}{28}\cdot p(1) \equal{} 225$. We subract the constant coefficient which is one, for an answer of $ 224$ or $ \boxed{C}$.[/hide]

$ (1\plus{}x\plus{}x^2\plus{}\cdots\plus{}x^{14})^2 \equal{} 1\plus{}2x\plus{}3x^2\plus{}\cdots\plus{}14x^{13}\plus{}15x^{14}\plus{}14x^{15}\plus{}\cdots\plus{}x^{28}$
Sum the coefficients and subtract 1 (because there is no $ x^{28}$ in the first factor). Two ways to do this:

1) $ (1\plus{}1\plus{}1\plus{}\cdots\plus{}1)^2\minus{}1 \equal{} 15^2\minus{}1 \equal{} 224$
2) (what I did on the test) $ 2(1\plus{}2\plus{}\cdots\plus{}14)\plus{}15\minus{}1 \equal{} 2\binom{15}{2}\plus{}14 \equal{} 2\cdot 105\plus{}14 \equal{} 224$

Can this problem be solved using negative binomial coefficients? that's what i tried during the test. didnt work.. :(

[hide="With negative binomial coefficients"] Write the given expression as

$ \frac {1 \minus{} x^{28}}{1 \minus{} x} \cdot \left( \frac {1 \minus{} x^{15}}{1 \minus{} x} \right)^2 \equal{} (1 \minus{} x^{28})(1 \minus{} 2x^{15} \plus{} x^{30})(1 \minus{} x)^{ \minus{} 3}$.

If we only want to find the coefficient of $ x^{28}$, we can ignore all higher powers.  In other words, multiplying the first two terms and then ignoring higher powers we find that the above has the same $ x^{28}$ coefficient as

$ (1 \minus{} 2x^{15} \minus{} x^{28})(1 \minus{} x)^{ \minus{} 3}$.

The answer is then $ ( \minus{} 1)^{28} { \minus{} 3 \choose 28} \minus{} 2 ( \minus{} 1)^{13} { \minus{} 3 \choose 13} \minus{} 1 \equal{} 15 \cdot 29 \minus{} 15 \cdot 14 \minus{} 1 \equal{} 224$ as expected. [/hide]

[quote="t0rajir0u"][hide="With negative binomial coefficients"] Write the given expression as

$ \frac {1 \minus{} x^{28}}{1 \minus{} x} \cdot \left( \frac {1 \minus{} x^{15}}{1 \minus{} x} \right)^2 \equal{} (1 \minus{} x^{28})(1 \minus{} 2x^{15} \minus{} x^{30})(1 \minus{} x)^{ \minus{} 3}$.

If we only want to find the coefficient of $ x^{28}$, we can ignore all higher powers.  In other words, multiplying the first two terms and then ignoring higher powers we find that the above has the same $ x^{28}$ coefficient as

$ (1 \minus{} 2x^{15} \minus{} x^{28})(1 \minus{} x)^{ \minus{} 3}$.

The answer is then $ ( \minus{} 1)^{28} { \minus{} 3 \choose 28} \minus{} 2 ( \minus{} 1)^{13} { \minus{} 3 \choose 13} \minus{} 1 \equal{} 15 \cdot 29 \minus{} 15 \cdot 14 \minus{} 1 \equal{} 224$ as expected. [/hide][/quote]

Er... this doesn't seem to affect the answer, but isn't it $ (1\minus{}2x^{15}\textbf{\plus{}}x^{30})$?

[quote="Altheman"]Let
\[ a(x) \equal{} 1 \plus{} x \plus{} ... \plus{} x^{27} \equal{} a_0 \plus{} a_1x \plus{} ... \plus{} a_{27}x^{27}
\]

\[ b(x) \equal{} (1 \plus{} x \plus{} ... \plus{} x^{14})^2 \equal{} b_0 \plus{} b_1x \plus{} ... \plus{} b_{28}x^{28}
\]
Then the coefficient of $ x^{28}$ in $ a(x)\cdot b(x)$ is going to be
\[ \sum_{k \equal{} 0}^{28}a_{28 \minus{} k}\cdot b_{k} \equal{} \sum_{k \equal{} 1}^{28}b_k \equal{} b(1) \minus{} 1
\]
since $ a_k \equal{} 1$ for $ k \equal{} 0,1,2,...27$ and $ a_k \equal{} 0$ for $ k \equal{} 28$. Hence the answer is $ 15^2 \minus{} 1 \equal{} 224$ or $ \boxed{C}$.[/quote]

I fixed an error - should be $ a_{27}$ not $ a_{28}$ up there. By the way, the wiki has a really nice solution.
http://www.artofproblemsolving.com/Wiki/index.php/2008_AMC_12A_Problems/Problem_19

That is the same as the first solution in this thread  :wink:

Just out of curiosity, but could this expression be interpreted as how many ways there are to pick out $28$ boxes from $3$ distinct boxes with $27, 14$ and $14$ balls in them respectively? I assumed so, and ended up with $E$.

[hide=idk what I did wrong]$a+b+c=28$ in non-negative integers, where $b,c < 15$.

Answer is $\binom{30}{2}-2\binom{15}{2}$ by stars and bars, but I am getting $225$. What did I do wrong?[/hide]

Just did casework, let $a$ be exponent from the first factor, $b$ from the 2nd factor and $c$ from the 3rd factor (Split the squared term)

$a$ can be $0,1,2,3.....25,26,27,28$

$b$ can be $0,1,2,3.....11,12,13,14$

$c$ can be $0,1,2,3.....11,12,13,14$

Now just do casework with $a$ being the control, and you know for every $b$, there is ONLY one valid $c$

Now, go through all possible value of $a$, there's a pattern.... and you get $1+2+3.....15+14+13....+3+2=224$ and we are done!

The problem is equivalent to rolling a 27-sided die and 2 14-sided dice, and we want the number of ways that the sum of the dice can be 28. After some casework, we get $$\sum_{k=1}^{15} k+\sum_{k=2}^{14} k=\boxed{224}.$$

[b]Remark.[/b] This probably wasn't the best way to do the problem.

[hide]Let $F(x)$ be the polynomial in question. Then,
\[ F(x) = \frac{1 - x^{28}}{1-x}\cdot \frac{(1-x^{15})^2}{(1-x)^2} = \frac{(1-x^{28})(1-x^{15})^2}{(1-x)^3} = (1 - x^{28})(1-x^{15})(1-x^{15})\left(\binom{2}{2} + \binom{3}{2}x + \dots\right).\]
To compute the coefficient of $x^{28}$: If you include $-x^{28}$ from the first term, you have to take the constant term from every other factor, giving a contribution of $-1$. Otherwise, you can include the $x^{15}$ from the second/third factor and an $x^{13}$ coefficient from the last factor, contributing $-2\binom{15}{2}$, or you can just take $1$ from the second and third factors and just the $x^{28}$ term from the last factor, contributing $\binom{30}{2}$. Answer is $\binom{30}{2} - 2\binom{15}{2}-1 = \boxed{244}$.[/hide]

Ignore $1+x+...+x^{27}$, then there is a bijection between $x^{a+b}$ and that expression except when $a+b=0$. Thus, we simply have $15^2-1=\boxed{224}$.

probably MAA should've add the answer choice $225$ lol.

we can just sum the coefficients of (1+...+x^14)^2 excluding 1 because we can multply by x^(28-wtv) in (1+...+x^27) so ans is 15^2-1=224


first, we all know that girls require time and money.

Girls=time X money

And, we know time is money.

Girls=money X money
Girls=money squared.

Lastly, money is the root of all evil, so...

Girls=(sqrt(evil)) squared

Therefore, we are forced to conclude that...

Girls=evil.

heh, i've heard this one a thousand times. :rotfl: it doesn't make much sense, but, its still pretty funny. i've heard another one about how knowledge is corrupt.

Haha always good to hear again - I think I've posted this once before or something...

i wonder if girls came up with a similar one about guys................. :D

*the female AoPS community awakens and beats the heck out of buzzer11 for posting for the nth time a really really old, trite joke*
*some of the guys join in just for the fun of it*

eh an really old joke but funny nevertheless.  :D

:bomb:

I've never heard that before, it's okayish.

Yay girls are evil so now we can kill all of the guys and take over the world.!!! :coolspeak:  :D

THis is a good one (and it's true)  :rotfl:

:D never heard that joke before.  That is really funny!
Unfortunately i can disprove it :( Money is not the root of all evil. The [b]LOVE[/b] of money is the root of all evil :P

it's well known love = 1 (identity) so it actually comes to the same thing

here's another one i've heard: knowledge is power, power is corrupt, corruption is evil, therefore, knowledge is evil. :D

[quote="mathnerd314"]To err is human, but humans are intelligent; hence to err is intelligent. QED

Time is money, and knowledge is power. Power is work/time. Hence knowledge=work/money, and money=work/knowledge. So the less you know the more you have. QED

Don't believe either... :)[/quote]

 :rotfl:  so many false formulas  :rotfl:

NICE MATHNERD, our physics teacher derived that last year.

hmmm i'd venture to say that almost all of you are middle schoolers lol

very sorry lol...but no i didn't remember...i'll keep that in mind then...
unless you were jping...in which case i dunno

but anyhow i said "almost all"...so  :D

IT's the first time i heard this joke....

Sorry if it's too old


If a winter coat that normally costs $60 goes on sale, so that it now costs just $42, by
what percent has the price been reduced?


Well, this problem tha you read in the image is taken from Excalibur. I would be very happy, if someone could complete the proof without use of trigonemetry.I like this problem and a pure synthetic solution deserves some effort.
 Babis

This is a great problem from IMO SL 2002 . See the two best solutions here http://www.mathlinks.ro/Forum/viewtopic.php?t=14741 in the last two posts


The geometric mean of the area of two squares is 2987. Given that all sidelengths are integral and neither square is a unit square, find the perimeter of the larger square.

[hide]$ \sqrt{s_1^2 * s_2^2 }\equal{} s_1 * s_2 \equal{} 2987$
2987 factors as 103 * 29, so the side length of the larger square is 103, so the perimeter is 412[/hide]


Given two circles $ \omega_{1},\omega_{2}$ intersecting at $ A,B$. Let $ P_{1}P_{2}$ be a common tangent ($ P_{1}\in \omega_{1}, P_{2}\in \omega_{2}$). Let $ M_{1},M_{2}$ be the orthogonal projections of $ P_{1},P_{2}$ on the line $ O_{1}O_{2}$. The line $ AM_{1}$ cuts $ \omega_{1}$ again at $ N_{1}$, the line $ AM_{2}$ cuts $ \omega_{2}$ again at $ N_{2}$.

Prove that $ N_{1},B,N_{2}$ are collinear.


Let p a prime number and b a natural number such that (p,b)=1. Let
$p=\sum\limits_{i=0}^n p_ib^i$ in base b. Prove that the polynomial
$f=\sum\limits_{i=0}^n p_i X^i$ is ireducibile in Z[X]. :D

Pretty useless :): this seems to be the generalization of a BMO problem, where we had b=10.


function
integration
calculus
calculus computations

Let $ f$ be an integrable function on $ [a,b]$.  Suppose that $ f(x)\geq 0$ for all $ x$ and there is at least one point $ x_0\in [a,b]$ such that that $ f$ is continuous and strictly positive.  Show that $ \int_{a}^{b}f(x)dx>0$.

For every $ \epsilon>0$ there is $ \delta$ such that $ f(x_0)\minus{}\epsilon<f(x)<f(x_0)\plus{}\epsilon$ whenever $ |x\minus{}x_0|<\delta$. Use this with $ \epsilon\equal{}f(x_0)/2$.

[quote="mlok"]For every $ \epsilon > 0$ there is $ \delta$ such that $ f(x_0) \minus{} \epsilon < f(x) < f(x_0) \plus{} \epsilon$ whenever $ |x \minus{} x_0| < \delta$.[/quote]

How is this justified?

That's the definition of continuity at a point, which you assumed.


Find the integral value of

$ \sqrt [3] {1\plus{}\sqrt [3] {2\plus{}\sqrt [3]{3\plus{}\cdots\plus{}\sqrt[3] {1990\plus{}\sqrt [3] {1991}}}}}$

 :D

What does it mean to find the integral value of something that is plainly not an integer?

I believe it means $ \lfloor \sqrt [3]{1\plus{}\sqrt [3]{2\plus{}\sqrt [3]{3\plus{}\cdots\plus{}\sqrt[3]{1990\plus{}\sqrt [3]{1991}}}}} \rfloor$

[quote="PKMathew"]Find the integral value of

$ \sqrt [3] {1 \plus{} \sqrt [3] {2 \plus{} \sqrt [3]{3 \plus{} \cdots \plus{} \sqrt [3] {1990 \plus{} \sqrt [3] {1991}}}}}$

 :D[/quote]

Sorry , really find the integral part of the expression:
$ \sqrt [3] {1 \plus{} \sqrt [3] {2 \plus{} \sqrt [3]{3 \plus{} \cdots \plus{} \sqrt [3] {1990 \plus{} \sqrt [3] {1991}}}}}$

[hide="ans."]

$ 1< \sqrt [3] {1 \plus{} \sqrt [3] {2 \plus{} \sqrt [3]{3 \plus{} \cdots \plus{} \sqrt [3] {1990 \plus{} \sqrt [3] {1991}}}}} <2 //$

$ \therefore$  the integral part of the expression = $ 1$

[/hide]


geometry
inradius
incenter
circumcircle
power of a point
geometry unsolved

$ \triangle ABC$ with $ I$ (incenter), $ O$ (circumcenter), $ r$ (inradius)

$ OI\cap$ circumcircle = $ D$, $ E$

$ OI\cap$ incircle = $ F$, $ G$

$ D$, $ F$ - same side
$ G$, $ E$ - same side

Prove  $ \Longrightarrow$ $ r^{2} \equal{} DF\cdot EG$

May I ask you a favor ? Thank you  :)
I have no solution  :(

[hide]
\begin{eqnarray*}
r^2 &=& DF \cdot EG \\
&=& (DI-r)(IE-r) \\
&=& DI \cdot EI - (DI + IE)r + r^2 \\
&=& DI \cdot EI - 2Rr + r^2
\end{eqnarray*}
Let the line through $ I$ perpendicular to $ OI$ meet the circumcircle at $ P$ and $ Q$.  Then by Power of a Point, $ DI \cdot EI = PI \cdot QI = PI^2$.  Then from right triangle $ OIP$, we have $ PI^2 = OP^2 - OI^2$.  But $ OP=R$ and $ OI^2=R^2-2Rr$ (Euler).  Thus,
$ r^2 = DI \cdot EI - 2Rr + r^2 = R^2-(R^2-2Rr)-2Rr+r^2 = r^2$, which is true.[/hide]

Thanks !  archimedes1
It must be very simple.
:wallbash:


HCSSiM AMC USA(J)MO USAMO Putnam AIME geometry

This topic is for members of the NYC forum to introduce themselves.  You can mention where you go to school, your grade, what math class you are taking, what math contests you have participated in, and so on.  You don't need to reveal personal details such as your name.

I'll start things off.  My name is Ravi Boppana.  I am a parent of a 5th grader at Dalton, a private school in Manhattan.  I received a PhD in computer science at MIT.  After that, I was a professor of computer science at Rutgers and NYU for about 12 years.  I then joined the software industry for 4 years.  Now I work for a tutoring company as a math tutor and a developer of math materials.

I grew up in Maryland; the best part of high school for me was being part of a math team.  I attended a math summer program at Hampshire College in Massachusetts, which played an influential role in my pursuing the mathematical sciences.  I qualified for the USAMO twice, and received an honorable mention on the Putnam exam in college.

I have been an active member of the Art of Problem Solving site shortly after it started a year and a half ago.

Okay, now it's your turn...

My name is Joel Lewis.  I am, at the beginning of 2005, a sophomore math major.  I graduated Stuyvesant High School in 2003.  I was heavily involved in the Stuyvesant and New York City math teams and served as a captain on both in my junior and senior years.  I qualified for the AIME 5 times and the USAMO 4 times and am hoping that I have an HM on the Putnam this time around.  Upon graduation I will probably go to graduate school, most likely to pursue a career in academia or teaching high school.  I have been a member of AoPS for quite a while and at one time was the overall highest poster, although I have not visited the forum in quite some time except to assist in the online classes.

Hi. My name is Barry Weng. I am a senior at the Bronx High School of Science. I solve Rubik's cubes, play SET, and sleep in math class (all 3 of them). I went to the CCNY program after freshman year and HCSSiM after sophomore year. My time here has been wonderful so far, so I expect to stick around for a while.

Hi, Joel. I'm so glad to see you here :)

My name is Geehoon Hong. I am a senior at Benjamin Cardozo High School. I was New York City Interscholastic Math League Individual Top Scorer last year, and am involved in the New York City Math Team from sophomore year.

I attended CCNY program in the summer of 2003, met Joel there, and I attened the Canada/USA Mathcamp last summer.

I love math and am planning to major in math.

Hey Geehoon!  Good to see you here!  Also, I'll robably see you at NYSML (and maybe the practice session, depending when it's held.)

ayo!  petee here!  former stuyvesant math team captain ('99 - '00 school year).  i'm attending CCNY now, and will be working on a cure for topology this summer... yes, *cure*.  aside from that, inverted category theory is a MUST.

let it be known.  and hey joel! good job running this here show!

-petee outta the blue

wow.. Peter...

Hey on a different note.

Does anyone know where Kevin Zheng is or know of any means for me to contact him?????

He's dropped off the face of the earth.

hello, my name is Kamaldeep Gandhi aka KD to the G.

I was a Stuyvesant math team captain in the 1999-2000 year and was a USAMO winner that year as well. I believe I had a perfect score on the mandelbrot competition that year  and some other things which probably don't matter at this point. Anyhow, I'm currently a coach of the NYC math team, and help out with random math stuff at Hunter College HS.

Great to see ya'll here, including you Ming Jack! Really.

And yes, I've been looking for kevin too....any ideas? .... last I heard from him was some weeks back...

hey KD the great!!

nice to see you here  :)

how come people only know kd the great and not petee the great?  tsk tsk tsk.

hey Geehoon. Once again, congrats on both MIT and the USAMO. Hopefully I'll see you on fri. you taking the flushing bus?

petee// im sorry but i never had a chance to see you  :)

kd// yes i am taking the flushing bus

[quote="JackPo"][quote="JBL"]Upon graduation I will probably go to graduate school, most likely to pursue a career in academia or teaching high school. [/quote]

I know the REAL reason why Joel is pursuing a career in high school teaching ;)[/quote]Which is? :D

mister jack o' lantern, man, tsk tsk tsk for whatever it is you're trying to say.  joel is a good guy, unlike many of the inconsiderate people out there and i can testify to that... he was one of only 2 or maybe 3 (but i'm sure no more) people who cared to do anything in my math team back when i was captain.  the truth is, i should have been a lot tougher in order not to let those delinquents disillusion me the way they did; a large part was my fault, i guess, and i wish i could go back and change it.  if it weren't for joel and nikita, it would have been even worse, and i feel i let them down, because they really cared to learn :(

[quote="Romwell"]Oh, yes. Hi there Joel, Gehoon, Maria, Stan, etc. :)[/quote]

close enough...

Hi Roman, I'm glad to see you here.

Thanks for helping us at Stuy with power round practice, and i'm sure your friend Ralph is one of the "2" who is interested in math from your school.

I will be going to CCNY this summer, hopefully I'll see you there.
WELCOME!

Hi. I'm Jan Siwanowicz. I occasionally pretend I know math and help people out. If you have a question ask me. :)

Hey Jan!! You taught me the second pizza theorem in the practice.

Ilya went to Brooklyn Tech.

Greetings, everyone.  I'm Kevin.  No, not Zheng.

Anyway....I went to Stuy for the past four years.  Some of you may know me now as the coach for the C team.  I was a fairly mediocre student on the team, but apparently good enough to take captaincy last year.  Anyway...I went to NYSML and ARML for the past three years, so I thought I would come back.

Sho: Meeg!

Jan: I cannot come up with something clever to counter your statement.

Geehoon: You're doing a fantastic job keeping things running.

Peak: How are things running in 437?

I'd look for more, but I'm too lazy now.  See you all on Friday (Manhattan bus, at least)

[quote="Midnight"]Geehoon: You're doing a fantastic job keeping things running. [/quote]
Thank you. Welcome to NYC Forum! :welcomeani:

sweet jesus, what's going on here... who are these people... these faces... they look like caricatures of used car salesmen from brooklyn.

Hello everyone. I'm the math coach @ Flushing HS. Good luck to those taking the SAT this Saturday!

-Chew

good luck also to those who are taking the sat's for others!!

please, let's just sell more consciousness expansion.

hi all  :D
My name is Allan Zhao, I will be a sophomore in Brooklyn Tech in September. I am in the junior A team in Tech.
I participated in AIME once, and got 210 GRRR(stupid mistakes).

Good to see you all here.

Well i guess that revived an old topic, but it's for the better...

I'm Krishanu Sankar...going into junior year at Horace Mann school in Riverdale in two days.

Btw, I did PROMYS this year, and it was awesome.

AIME: Stupid careless mistakes again, a grand record of [b]5[/b] this time around (ie 2006)

well i could hope for hm next year maybe

Hi all

Rafaello here from Brooklyn
6th grade homeschooler who loves math
am on homeschool mathcounts team and have taken the AMC8 where i got an 18 out of 25.
:10: (i wish)


Basically, you post 3 words to continue the story that has been built by those who posted before you. Don't post twice in a row.

There once was

There have been at least three topics with this objective.

due to mcdonalds??
nvm

dnkywin, 1=2's friend

There once was a deaf dolphin who ate 1=2 and choked on his big head. A few days later the dolphin	found out that he lost weight. Dnkywin, 1=2's friend, I mean enemy goes and observes nowhere in particular. [b]Suddenly, dnkywin commits[/b]

There once was a deaf dolphin who ate 1=2 and choked on his big head. A few days later the dolphin found out that he lost weight. Dnkywin, 1=2's friend, I mean enemy goes and observes nowhere in particular. Suddenly, dnkywin commits nothing at all to his grades in health class. [b]The teacher was[/b]

i think we just skipped a lot that wuz posted, but w/e

and extremely stupid.

like Michael Jackson.

No offense, "ignorant" was getting on my nerves with Michael Jackson.

Locked for Repetition.

Hey, three words right?  :P

Seriously, there's another topic already.


quadratics
geometry
3D geometry
algebra
number theory

Find two positive integers, [i]x[/i] and [i]y[/i], that satisfy EITHER equation:

[b]x 2 - y 3 = 1 [/b]

[size=200]OR[/size]

[b]x 3 - y 2 = 1 [/b]

[hide]One pair is (using Equation 1) (2,3)[/hide]

Pretty Cool, right?  Wait, here's the cool part [hide][i][b][u]You're looking at an unsolved problem![/u][/b][/i][/hide]

So by either, you mean one or the other, but not both? If so, what's wrong with the solution you found... When you mean the problem is unsolved, is that to find all these integers, or just a single pair?

[quote="mysmartmouth"]
One pair is (using Equation 1) (2,3)
[/quote]

2,3 does not work in Equation one.

$2^2 - 3^3 \text { does not equal } 1$

i think you meant using equation 2.

If we take the last digit of a number to a certain power (represented by p):
if p=1: 1 2 3 4 5 6 7 8 9 0
if p=2: 1 4 9 6 5 6 9 4 1 0
if p=3: 1 8 7 4 5 6 3 2 9 0
So, there are a few cases where x^3-y^2=1.
x ends in 3, y ends in 6 or 4
x ends in 6, y ends in 5
x ends in 5, y ends in 8
x ends in 8, y ends in 1,9
x ends in 1, y ends in 0
x ends in 0, y ends in 3,7

We can test each case to see if its possible. It requires we get into the last two digits of numbers to a certain power, though.

One way to look for  integer solutions for $x^2 = y^3+1$ is to look at it as

$x^2 = (y+1)(y^2-y+1)$
 $=(y+1)((y+1)^2 - 3y)$
$ = (y+1)^2(y+1) -  \frac {3y}{y+1}))$
Now $(y+1)^2 $ is a compelete square, and  only integer solutions for $((y+1) - \frac{3y}{y+1}) $
would be  $y = 0 \ or \  2 $
(Similarlay you can tackle  the other case.

Hope this helps.

Why $(y+1) - \frac{3y}{y+1}$ should be integer?

So far, we know one set is  (3,2), but what could the other pair be.....

y+1-3y/(y+1) doesn't have to be an integer, because square roots on both sides give x=(y+1)*sqrt(y+1-3y/(y+1)) so sqrt(y+1-3y/(y+1)) can be any rational number (not necessarily completely reduced) with y+1 as its denominator.

Hello Myth and Goodyfresh:

Sorry for not crossing the "t" or dotting  the "i"   but you have to look at it more carefully.

The expression is :  $x^2 = (y+1)^2 ((y+1) - 3y/(y+1)) $

For integer solutions, the second factor has to be an integer (actually a square)  because:

If we start with the obvious conclusion  that it must be a square of a [b]rational number [/b]  because both $x^2$ and $(y+1)^2$  are perfect squares - we  can  easily show that then  above must be true ...

The details:

[hide]Let us , for the sake of simplicity write $z=y+1$ and so the second  factor is:

$z-\frac {3(z-1)}{z} $

let us assume $ z > 3$
 and note that  $z$  and $z-1$    have  no common factors:
We take the easy case when z is not a multiple of 3 first:
Now  since  $3(z-1)$  and $z$ have no common factors,
 the expression is  $z-\frac {3(z-1)}{z} =  \frac {z^2-3z+3}{z}$
Both $z$ and $z^2-3z+3$  have to be perfect squares.. but (Assuming z>3)  that's
[hide="impossible"]Perfect squares less than   $z^2$ are $(z-1)^2 = z^2 -2z+1$ or $(z-2)^2 = z^2 - 4z +4$ it will not work.  [/hide]
When  $z$ is a multiple of 3, Similar argument will prove that's
[hide="impssoble too"]If $z=3k$ (k>1) then we  will need  $3k(k-1) +1$  as a perfect square, or $3k(k-1) = 4a(a-1)$ etc ..[/hide][/hide]

[quote="Myth"]Hmm...
Have you deleted it, Gyan? Where is your reply I just saw?[/quote]

Heh..heh ... funny fingers..<smile>  din't realize the case when z is a factor of 3  was not that trivial ..:)

[quote="mysmartmouth"]Gyan:

Does that mean there is no solution other than (3,2)?

Can it be proved?[/quote]

Yes,  apart from  trivial (0,1)  and the one pair (3,2) there are no soulutions in fact for  ANY equation of the type:
$x^a - y^b = 1$

It  iwas  formally known as Catalan conjucture which is a theorem now  now (since a few years)  and have been proved,  by ( Mihailescu) I seem to recall a thread here in AoPS (do a search). In any case, Mathword (or google with Catalan as key word)   may have  more info: for example
[url=http://mathworld.wolfram.com/CatalansConjecture.html]Here[/url]

I seem to recall reading a   proof for $y^3-x^2 = 1$  from a chapter  in number theory in  Encyclopedia Britanica (This was when I used to like encycopedia's and bought copies for my sons - almost 20 years ago)

The proof goes something like this:  (AFAICR)

[hide]One uses Guasian integers and writes
$y^3 = x^2+1 = (x+i)(x-i)$
Now only common factors between  $(x+i)$ and $(x-i)$ can be  $2i$ but in that case $y$ would be even, and $x$ odd, and  LHS would be a multiple of  of 8 (0 mod 4)  while RHS is double of an odd number ( 2 Mod 4) so it will not work.

If (x+i) and (x-i) are prime to each other, then both of  these must be  perfect cubes.

so $(x+i) = (a+bi)^3$ which results into
$x=a^3-3ab^2$ and $1=b(3a^2-b^2)$
This gives $b=+-1 $ and $(3a^2 - 1) = +-1$
or $3a^2 = 0 \ or 2$  or $a=0$ means $x=0$[/hide]

[quote="chess64"][hide="Want to know..."]

$x^2 - y^3 = 1$ or $x^3 - y^2 = 1$.

$(3,2)$ is the only solution in positive integers. (It works in the first equation.)

[u][b]Proof[/b][/u]

[b]Case 1[/b]

If the first equation is true, then $y = \sqrt[3]{x^2 - 1}$, so $y^3 = k(k+2)$, where $k = x - 1$.  Then either $k = y$ and $k + 2 = y^2$, or $k + 2 = y$ and $k = y^2$. Solving for the positive root of the quadratic equation obtained by substitution in the first possibility, we see that $y = k = 2.$ The second possibility has no real roots. Thus the only positive integer solution to this equation is $(k+1,y) = (3,2).$

[b]Case 2[/b]

If the second equation is true, then $y = \sqrt{(x - 1)(x^2 + x + 1)}$. Then $y^2 = (x-1)(x^2+x+1)$, so $y = x - 1$ and $y = x^2 + x + 1$.  Setting these two values equal to each other,

\[x - 1 = x^2 + x + 1\]
\[0 = x^2 + 2.\]
\[x^2 = -2\]

Thus there are no real solutions for this case--the only positive integer solution for both equations is $(3,2).$

[/hide]...If the first equation is true, .... so $y^3 = k(k+2)$, where $k = x - 1$.  Then either $k = y$ and $k + 2 = y^2$, or $k + 2 = y$ and $k = y^2$[/quote]

hmmm..  why would be that? In fact, if  k is odd,  the two factors are prime to each other and can never have a common factor (let alone be equal to $y$ and $y^2$)

([i]A  good argument would be (in the case k is odd) that each factor must be  a   perfect cube, which is not possible etc ... hence there are no solutions_ ... you have to take the case when k is even a little deeper .. [/i]

(For example we do have  $216  = 6^3 = 12 \cdot 18 $ but   12 is not equal   to 6 (Nor 18 equal to 6^2)) .. Just to show that if a cube has two factors you can not automatically asume  that one is y and other is y^2) ..

Similary why:
[quote]hen $y^2 = (x-1)(x^2+x+1)$, so $y = x - 1$ and $y = x^2 + x + 1$. [/quote]

Hope it is helpful

Added later: (Such arguments can be valid when $y$ is prime and thus  $y^3$  can be factored only in  a finiite number of  ways and some concusions  such as above can be drawn ....)


[quote="Ji Chen"]If $ a,b,c,d$ be be positive numbers, then

$ \frac {c}{a}(8b \plus{} c) \plus{} \frac {d}{b}(8c \plus{} d) \plus{} \frac {a}{c}(8d \plus{} a) \plus{} \frac {b}{d}(8a \plus{} b)\geq 9(a \plus{} b \plus{} c \plus{} d),$

with equality if and only if $ a \equal{} c \wedge b \equal{} d$.[/quote]
By the AM-GM, we have
\[ \frac{bc}{a}\plus{}\frac{da}{c} \ge 2\sqrt{bd}\]
\[ \frac{cd}{b}\plus{}\frac{ab}{d} \ge 2\sqrt{ac}\]
Hence, it suffices to prove that
\[ \frac{c^2}{a}\plus{}\frac{a^2}{c}\plus{}16\sqrt{ac} \plus{}16\sqrt{bd} \ge 9(a\plus{}b\plus{}c\plus{}d)\]
To prove this inequality, it suffices to prove that
\[ \frac{x^4}{y^2}\plus{}\frac{y^4}{x^2} \plus{}16xy \ge 9(x^2\plus{}y^2) \quad \forall x,y \ge 0\]
\[ \Leftrightarrow \frac{x^4}{y^2}\plus{}\frac{y^4}{x^2} \minus{}x^2\minus{}y^2 \ge 8(x^2\plus{}y^2\minus{}2xy)\]
\[ \Leftrightarrow \frac{(x^2\minus{}y^2)^2(x^2\plus{}y^2)}{x^2y^2} \ge 8(x\minus{}y)^2\]
\[ \Leftrightarrow (x\plus{}y)^2(x^2\plus{}y^2) \ge 8x^2y^2\]
By the AM-GM Inequality, we get
\[ (x\plus{}y)^2 (x^2\plus{}y^2) \ge 4xy\cdot 2xy \equal{}8x^2y^2\]
We have done. :)


induction
graph theory
combinatorics unsolved
combinatorics

Let $A$ be a $K_n$ graph. Each turn you can remove an arbitrary edge from any cycle of $A$ of length $4$. What is the maximal number of turns?

Maximal number of turns is $f(n) = \frac {n(n-3)}2$, for $n \geq 3$, leaving a graph with exactly $n$ edges.

First, consider any vertex, say $X$, and let $M_1, \cdots , M_{n-1}$ be the remaining vertices. Using the $4$-cycles $XM_iM_1M_2$ we may delete the edges $XM_i$ for all $i \geq 2$. This uses $n-2$ operations and leaves a $K_{n-1}$ whose vertices are the $M_i$'s. It follows that $f(n) \geq f(n-1) + n-2.$
Clearly $f(3)=0$ so that $f(n) \geq 2+3+ \cdots + (n-2) = \frac {n(n-3)}2$.

Now, deleting an edge which belongs to a cycle in a connected graph leaves the connectedness unchanged. Thus, by an easy induction, after any deleting the remaining graph is connected. In particular, after a sequence of $f(n)$ deletions, we still have a connected graph. It follows that there are at least $n-1$ edges at the end. Since there are $\frac {n(n-1)} 2$ edges at the beginning, and that $ \frac {n(n-1)}2 - \frac {n(n-3)}2 = n$, all we have to do is to prove that the final graph is not a tree, that is it contains a cycle.

In fact, it is straightforward to prove that at each step the remaining graph always contains an odd cycle. But, I did prove it only by brute force (many many cases) and I did not see any cool argument  :?

Pierre.

He-he  :lol:
Pierre doesn't see easy arguments ;) :D
It is a noticeable day!

Well, let's try like that, which reduces the number of cases :

Assume that at some step, we want to delete edge $AB$ from $ABCD$, and assume (induction hypothesis) that the graph contains an odd non-degenerated (which means that the vertices are pairwise distincts) cycle $T$. Obviously, we may assume (*) that $AB$ is an edge from $T$ otherwise $T$ remains unchanged after the deletion...

- If neither $D$ nor $C$ belongs to $T$, then just consider the odd cycle $T'$ obtained from $T$ by replacing $AB$ by $ADCB$, which brings us back to (*), and we are done.

- If $D \in T$ and $C \notin T$ : Then, let $x,y$ be the number of vertices in $T$ between $A$ and $D$, $D$ and $B$ respectively after the deletion of $AB$. Since $T$ is odd, $x+y$ is even. If $x$ is odd then $AD$ is not an edge from $T$ (since it is non-degenerated) and consider the cycle $T'$ obtained from the path $AD$ from $T$ and adding the edge $AD$. Otherwise $y$ is even, and we can do the same with the path $BD$ from $T$ and adding $DCB$. Thus, we are done in that case.

- If $C,D \in T$ :
Let $x$ be the number of vertices between $C$ and $D$ in the remaining path after the deletion of $AB$. If $x$ is odd, then just add the edge $CD$ and we are done. Otherwise : If the path is $A - \cdots - D - \cdots - C - \cdots - B$, then since $T$ is odd, it means that the number $y$ of vertices between $A$ and $D$, or the number $z$ of vertices between $C$ end $B$ is odd, say $y$ is odd. Thus, $AD$ is not used in this path so that we may use this path and just add the edge $AD$.
If the path is $A - \cdots - C - \cdots - D - \cdots - B$, then neither $A-D-C$ nor $B-C-D$ is used in the path and, as above, the number of vertices between $A$ and $C$ or between $D$ and $B$ is odd. Say between $A$ and $C$. From the non-degenerated hypothesis, it follows that using this path and adding $A-D-C$ leads to the desired odd cycle.

Thus, in any case, we obtain a non-degenerated odd cycle in the remaining graph and we are done.

So, what's the cool point of view Myth?

Pierre.

I even don't want to read it ;) Too long for me...
There is an one-line argumentation.

[quote="pbornsztein"]
In fact, it is straightforward to prove that at each step the remaining graph always contains an odd cycle. But, I did prove it only by brute force (many many cases) and I did not see any cool argument  :?
[/quote]
A graph without odd cycles is bipartite, i.e., its vertices can be colored by 2 colors so that each edge connects vertices of different color. If you have a bipartite graph after deletion of an edge in a 4-cycle, you can easily see that your deleted edge connected two vertices of different color, so the graph was bipartite before your deletion as well. So, since the initial graph was not bipartite, the ending graph won't be bipartite either. In particular, it won't be a tree.

[quote="fedja"]If you have a bipartite graph after deletion of an edge in a 4-cycle, you can easily see that your deleted edge connected two vertices of different color[/quote]

Hmmm...I was sure that the cool argument was about bicoloration of a graph without odd cycle, but I didn't see that the two vertices had the same color  :?

Pierre.


Επειδή η συνδυαστική είναι μακράν από τα πιο στρυφνά κεφάλαια των μαθηματικών βρήκα δύο ασκήσεις Ολυμπιακού επιπέδου τις οποίες δεν κατάφερα να λύσω αλλά μου άρεσαν ιδιαιτέρως (διότι εκτός του ότι είναι εξαιρετικές ασκήσεις αγγίζουν τα ενδιαφέροντά μου). :wink:

ΑΣΚΗΣΗ 1

Κάθε ένας από ένα σύνολο 30 μαθηματικών με ταλέντο στο σκάκι έχει μια αγαπημένη βαριάντα (σειρά κινήσεων του ανοίγματος στο σκάκι) και μια αγαπημένη κλασική ανισότητα. Γίνεται μια έρευνα για αυτούς τους 30 που αποκαλύπτει ότι οι προτιμώμενες βαριάντες αυτές είναι ακριβώς 20 και οι ανισότητες ακριβώς 10. Επίσης, οι μαθηματικοί κατατάχτηκαν σε δύο κατηγορίες, στους μαθηματικούς Μ και στους υπόλοιπους. Για κάθε μαθηματικό Μ, η αγαπημένη του ανισότητα δηλώθηκε στην έρευνα περισσότερες φορές απ' ότι η αγαπημένη του βαριάντα. Να βρεθεί ο ελάχιστος αριθμός των μαθηματικών Μ.

(Στο πρόβλημα δινόταν και ζητούσαν την απόδειξη, έτσι είναι πιο δύσκολη). :wink:

Η δεύτερη είναι μάλλον λίγο πιο δύσκολη γι' αυτό θα τη βάλω αύριο μαζί με τη λύση της πρώτης, αν δεν έχει βρεθεί, που σίγουρα κάποιος θα την καταφέρει... :P

[quote="Protonios"]
 Για κάθε μαθηματικό Μ, η αγαπημένη του ανισότητα δηλώθηκε στην έρευνα περισσότερες φορές απ' ότι η αγαπημένη του βαριάντα. Να βρεθεί ο ελάχιστος αριθμός των μαθηματικών Μ. [/quote]

Μπορείς να γράψεις λίγο καλύτερα αυτό το σημείο ? Δεν είναι σαφές  ....
BTW , η άσκηση είναι καλή μην ανυσηχείς  :wink:

Ναι. Θα δώσω ένα παράδειγμα. :wink:  Έστω ένας από τους μαθηματικούς Μ. Ας πούμε ότι η αγαπημένη του βαριάντα είναι αγαπημένη βαριάντα άλλων 10 από τους μαθηματικούς. Δηλώθηκε δηλαδή 10 ακόμη φορές. Οπότε η αγαπημένη του ανισότητα δηλώθηκε από 11 φορές και πάνω. Αυτή η ιδιότητα ξεχωρίζει τους μαθηματικούς Μ από τους υπόλοιπους. Είναι κατανοητό? :?:

Ok  τώρα . Θα την προσπαθήσω μάλλον αύριο ...

Βασικά να δώσω μια υπόδειξη διότι δεν ξέρω κατά πόσον τη δυσκόλεψα αποκρύπτοντας ότι ο ζητούμενος αριθμός είναι

[hide]11[/hide]

Η άσκηση θυμίζει μια συνδυαστική από μια παλιά Shortlist (στη δεκατία του 80) (και οι προσεγγίσεις τους μοιάζουν)
Έστω $ v_{1},\ v_{2},...,v_{20}$ οι 20 διαφορ βαριάντες και $ i_{1},\ ..,i_{10}$ οι 10 διαφορετικές ανισότητες .
Έστω $ V_{i}$ το σύνολο των διαγωνιζομένων που διάλεξαν την βαριάντα $ v_{i}$ kai $ I_{j}$  το σύνολο των διαγωνιζομένων που διάλεξαν την anis;othta $ i_{j}$ .

Απο εδώ και στο εξής όπου γράφω σύνολο S θα εννοώ τον πληθικό αριθμό του S (giat;i to latex einai dyskolo  :P )
Δημιουργούμε τα ζεύγη $ (a_{M},\ b_{M})$ όπου $ a_{M}\equal{}\frac{1}{V_{i}}$ αν η αγαπημένη βαριάντα του Μ είναι h V_i kai
$ b_{M}\equal{}\frac{1}{I_{j}}$ αν η αγαπημένη anisotita του Μ είναι h I_j  . Θα ονομάζουμε ένα ζεύγος <<καλό >> αν $ a_{M}>b_{M}$

Προφανώς θέλουμε να βρούμε τον ελάχιστο αριθμό  των <<καλών>> ζευγών .
Τότε από τη συνθήκη έχουμε ότι
$ \sum_{M}\ (a_{M}\minus{}b_{M})\equal{}10$ . Και αφού $ a_{M}\minus{}b_{M}<1$ το άθροισμα που έγραψα πριν θα έχει τουλάχιστον 11 θετικούς όρους.

Άρα ο ελάχιστος αριθμός καλών ζευγών είναι 11 . :)

Ακριβώς έτσι!!!  :) Το κλειδί που δεν μπόρεσα να βρω ήταν το άθροισμα αυτών των αντιστρόφων ότι θα δίνει το 20 και το 10 (και προσπαθούσα να συνδυάσω το πλήθος των υποσυνόλων αυτών με τα νούμερα 20, 10 χωρίς επιτυχία). :(

Αύριο θα βάλω μια ελαφρώς πιο δύσκολη.

Δώδεκα μουσικοί Μ1, Μ2,....Μ12 συγκεντρώνονται σε ένα φεστιβάλ μουσικής δωματίου διάρκειας μιας βδομάδας. Κάθε μέρα υπάρχει μία προγραμματισμένη συναυλία στην οποία μερικοί μουσικοί συμμετέχουν ενεργά ενώ οι υπόλοιποι είναι θεατές. Για κάθε $ i \equal{} 1, 2,...12$ έστω $ A_i$ ο αριθμός των συναυλιών στις οποίες ο μουσικός Μi συμμετέχει ενεργά και έστω $ A \equal{} A_1 \plus{} A_2 \plus{} ...... \plus{} A_{12}$. Να προσδιοριστεί η μικρότερη τιμή του $ A$ έτσι ώστε να είναι δυνατό για κάθε μουσικό να ακούσει σαν θεατής όλους τους άλλους μουσικούς.

Προειδοποίηση: Είναι αρκετά έως πολύ στρυφνή. :(

Σε αντίθεση με την προηγούμενη αυτή είναι πολύ δύσκολη και brutal .Την βρηκα σε ένα παλιό TST της Κίνας τυχαία σε κάτι προετοιμασίες τους από το stfu .

Είναι όντως :(

Ποια λύση έχεις υπ' όψιν σου? :?:

Λύση: (εν συντομία γιατί είμαι κουρασμένος)
Κάποιες απλές αποδειξούλες τις αφήνω ως ερωτήματα για να μην είναι μασημένη τροφή... :wink:

Για να επιτευχθεί η συνθήκη παρατηρούμε ότι ανά $ 3$ μουσικοί πρέπει αναγκαστικά να συμμετέχουν σε τουλάχιστον $ 3$ συναυλίες (γιατί?). Επίσης, ανά $ 7$ ή παραπάνω πρέπει να συμμετέχουν σε τουλάχιστον $ 4$ συναυλίες (γιατί?). Τέλος, ανά $ 9$ πρέπει να συμμετέχουν σε τουλάχιστον $ 5$ συναυλίες (γιατί?)(αυτό είναι δύσκολο το αφήνω ως ξεχωριστό πρόβλημα)(hint: πρώτα αποδείξτε και χρησιμοποιήστε τα παραπάνω).

Υποθέτουμε τώρα ότι $ k$ μουσικοί συμμετέχουν μόνο σε μία συναυλία. Είναι αρχικά $ 0\leq k\leq 7$ (γιατί?). Επίσης οι συναυλίες αυτές είναι σόλο δηλαδή οι $ k$ συμμετέχουν μόνοι τους (γιατί?). Άρα $ k \equal{} 7$ απορρίπτεται. Οι υπόλοιποι $ 12 \minus{} k$ παίζουν όπως υποθέσαμε σε $ 2$ ή περισσότερες συναυλίες ο καθένας (άρα $ k \equal{} 6$ απορρίπτεται) και πρέπει να αλληλοπαρακολουθηθούν σε $ 7 \minus{} k$ συναυλίες.

Απορρίπτουμε τώρα εύκολα και $ k \equal{} 5$, $ k \equal{} 4$ και $ k \equal{} 3$ (γιατί?). Συνεπώς $ t\geq 22$ (γιατί?).

Μένει κάτι ακόμα (τι?).


email
number theory unsolved
number theory

Prove that for all natural $n$ the number:
$n(n+1)...(n+9)$ is not the square of positive integer

I would say that from a famous result of Erds and Selfridge, the product of any $k \geq 2$ consecutive positive integers is never a power, but I think it is not what you are expecting :D

Pierre.

[quote="pbornsztein"]I would say that from a famous result of Erd?nd Selfridge, the product of any $ k \geq 2$ consecutive positive integers is never a power, but I think it is not what you are expecting :D

Pierre.[/quote]
Can you post that statement here?

I believe this is the reference mentioned above:

P. Erdos and J.L. Selfridge, The product of consecutive integers is never a power, Illinois J. Math. 19 (1975), 292-301.

Can you send to me that paper? Here is my email: nttuan_qn@yahoo.com.vn  :wink:

Just for your notice, I think in the long run you will not be very happy posting your E-mail address in some forum ;)

NT TUAN,

Sorry, but I could not find a copy of the paper online.  I did find another older reference that you could also look into:

P. Erdos and V. Rigge (see P. Erdos, notes on the product of consecutive integers, I, II, J. London Math. Soc. 14(1939), pp. 194-198 and pp. 245-249).


Brown is known more for its applied math and computer science than pure math. The majority of my math professors were wonderful teachers, though. The writing courses are very popular and well supported, though. You can browse through some of the classes offered at mocha.cs.brown.edu. Sorry I can't say anything about the other schools.

I'm a first year at U. Chicago, so I'll put in what I can.

We have a good program in pure math, with a lot of emphasis on theory. I'm very happy with my math class. I don't know much about our creative writing program, but you can check out http://creativewriting.uchicago.edu/.

IIRC, Brown has an open curriculum. I think it's important to factor in the difference between Chicago's Core and Brown's open curriculum. I personally like the Core, but of course it's a matter of preference. I mainly like the Core because I think that (1) the analytic skills developed through the Core complement any field of study and (2) in the words of a friend, the Core forces me to read all the texts I wanted to read but may have set aside in favor of other passions like math.

Like the last poster, sorry I can't say anything about the other schools.

[quote="oab729"]Hi, all.  I'm kind of a weirdo in that I really love contest math, but I also really love English and creative writing, specifically poetry.  Of the following schools, which have strong undergrad pure math programs compared to say, U. Berkeley or Chicago?

Brown
Dartmouth
Williams
John Hopkins
U. Chicago
U. Michigan
UPenn
Swarthmore

Since five of my school choices are already taken up, I'd like to narrow this list to 3 or fewer.  Thanks in advance-TBAO[/quote]

Wow I thought I was the only one.  I love both math and poetry as well. :)


Mr. and Mrs. Zimmeran want to name their baby Zimmerman si that its monogram( first, middle, and last initials) will be in alphabetical order with no letters repeatd. How many such monograms are possible?

[quote="lisaliu"]Mr. and Mrs. Zimmeran want to name their baby Zimmerman so that its monogram( first, middle, and last initials) will be in alphabetical order with no letters repeated. How many such monograms are possible?[/quote]

Okay, notice how the initials are in the form $ ?,?,Z$.
So we know that there are $ 25$ letters that can be put in the first blank, but only $ 24$ that can be put in the second because we can't repeat.  First, let's put $ A$ in the first blank to get $ A,?,Z$.  We know that there are $ 25$ letters that are farther down in the alphabet than $ A$, but $ Z$ can't be one of those letters.  $ A$ has $ 24$ possible solutions, $ B$ has $ 23$, and so on, until we get to $ X$.  $ X$ has $ 1$ possible solution, and $ Y$ has $ 0$, since we can't use $ Z$ again.  So the answer is

$ 24 \plus{} 23 \plus{} ... \plus{} 1 \equal{} 300$

[hide]
Last letter is Z for sure. 25 letter left to choose from. Any combination we choose will have one orientation that we want (alphabetical). $ \binom{25}{2}\equal{}\boxed{300}$
[/hide]


calculus
integration
analytic geometry
calculus computations

So I came across the following integral from an old competition:

Evaluate
$ \iint_{D}\frac{x^{3}}{x^{2}\plus{}y^{2}}\ dA$ where $ D$ is the half disk given by $ (x\minus{}1)^{2}\plus{}y^{2}\leq 1,\ y\geq 0$.

I didn't have any trouble with this integral.  However, in the solutions, it says that in polar coordinates, the region sweeps from $ 0$ to $ \pi/2$.  Why does it not sweep to $ \pi$ if it is a half-disk?

Because its center is at a point different from $ (0,0)$.


Find all continuous functions $ f$ $ :$ $ R$ $ \longrightarrow$ $ R$ that for every $ x$ $ \in$ $ R$ :

$ f(1\minus{}f(x))$ $ \equal{}$ $ 1\minus{}x$

Any function whose graph is symmetric over the line $ y \equal{} 1 \minus{} x$ works. Rewriting the equation as
$ f(x) \plus{} f^{\minus{}1}(1\minus{}x) \equal{} 1$
also shows these are the only ones. Are those really all the conditions?