geometry
calculus
topology
integration
vector
set theory
linear algebra

I'm not sure if this is in the right forum and if it isn't, i apologize.

Anyways, right now i am currently a freshman student of Aerospace Engineering. I have been strongly considering doing a mathematics minor for the past year and i have been planning out my schedule for the next year. After much research, i have found that it is possible for me to finish a math minor by the end of next year. In addition to my other engineering classes and GE courses, this is what my schedule will look like in terms of mathematics courses. By the way, my school is on the quarter system.

Winter 07 (current) = MAT 115 Calculus/Analytic Geometry II

Spring 07 = MAT 116 Calculus/Analytic Geometry III

Summer 07= MAT 214 Calculus of Several Variables I
                & MAT 310 Introduction to Set Theory

Fall 07 = MAT 215 Calculus of Several Variables II
            & MAT 321 Introduction to Topology

Winter 08 = MAT 216 Differential Equations
              & MAT 208 Linear Algebra

Spring 08 = MAT 318 Mathematical Analysis of Engineering Problems & MAT 419 Abstract Linear Algebra

Basically i am taking the advanced classes RIGHT when i meet the prerequisites for them. In case this information is useful, here are the prerequisites:

Calc of Sev var I (Differential - MAT 214): Prereq = C or better in Calc III (MAT 116)

Calc of Sev var II (Integral - MAT 215): Prereq = C or better in Multivar Calc I (MAT 214)

Set Theory (MAT 310): Prereq = C or Better in Calc III (MAT 116)

Topology (MAT 321): Prereq = C or better in Set Theory (MAT 310)

Linear Algebra (MAT 208): Prereq = C or better in Multivar Calc I (MAT 214)

Differential Equations (MAT 216): Prereq = C or better in Calc III (MAT 116)

Analysis of Engr Prblms (MAT 318): Prereq = C or better in Multivar Calc II (MAT 215)

Abstract Linear Algebra: Prereqs = C or better in both Differential Equations (MAT 216) and Linear Algebra (MAT 208)

So here is my question. Is this doable for a sophomore in college or would this be too difficult? I can assure you that studying won't be an issue. I am an engineer with a love of mathematics and sometimes i study math on my own or read ahead in my current text when i am bored. Believe it or not, this has helped me learn math much better. Thanks for your input.

Whoa, two semesters of multivariable calculus.

  As someone who has taught a few aerospace engineering majors with math minors, I'll say: do the applied stuff as early as possible, and leave the abstract math for your junior year. In other words, do Calc/Diff.Eq/Lin.Alg (and Mat 318?) first, and take set theory/topology/abstract Lin.Alg. afterwards. If you don't decide to go for a business minor by then.  :roll:

  Don't underestimate the effort required by AERO courses: they can humble even the very best students.

Trying to match those course numbers to anything I've seen - Cal Poly Pomona?

There's a good evidence for that: http://www.csupomona.edu is the only Google match for "MAT 318 Mathematical Analysis of Engineering Problems".

You are correct. I go to Cal Poly Pomona.

And yes i would consider MAT 318 an applied class i guess. From what i have heard, that class reviews vector calculus and provides and introduction to complex analysis.

I don't know how heavy the courseload is in Aero, but from what I know about the courses you've listed, it should probably take you about 6-7 hours/week max which ought to be manageable.

Here is the thing though. I am on the 4 year pledge which means i am required to take at least 16 units every quarter (12 units is the minimum to be considered a full time student). The perk of the program is that you get to register for classes about a week or two before even the seniors who have accumulated the most units, so getting the classes you need is not an issue.


For all $ {n}\ge{2}$. let $ {a}_{n}$ be the  product of all coprime natural numbers less than $ n$. Prove that

(a) ${ {n}\mid{a}_{n}+1{\Leftrightarrow}{n}=2,4,{{p}^{\alpha}},{2{p}^{\alpha}}}$

(b) ${ {n}\mid{a}_{n}-1{\Leftrightarrow}{n}{\not=}2,4,{{p}^{\alpha}},{2{p}^{\alpha}}}$

Here $ p$ is odd prime number and$ {\alpha}\in{N}$

[quote="Litlle 1000t"]For all $ {n}\ge{2}$. let $ {a}_{n}$ be the  product of all coprime natural numbers less than $ n$[/quote]
I don't understand.

Coprime to $ n$ he means.
And it was posted before.

\[ a_n\equal{}\prod_{(x,n)\equal{}1}x.\]
Because $ \{x|(x,n)\equal{}1\}$ give group $ G_n$, for all x we have $ x'$, suth that $ xx'\equal{}1\mod n$.
Therefore \[ a_n\equal{}\prod_{x^2\equal{}1\mod n \ \ 1\le x<n}x\equiv c_n.\]
Therefore $ a_n^2\equal{}1\mod n$. Let $ G_n\equal{}\sum_{k\equal{}1}^m Z_k$, were $ Z_k$ are cyclic groups. Then we have $ 2^m$ solutions $ x^2\equal{}1\mod n$ and $ c_n\equal{}(\minus{}1)^{2^{m\minus{}1}}\mod n$. Therefore $ a_n\equal{}\minus{}1\mod n$ if $ G_n$ is cyclic or $ n\equal{}2,4,p^l,2p^l$ and $ a_n\equal{}\minus{}1\mod n$ for another case.


[code]\begin{tabular}{rcl}
case $x=-5$: & \hspace{1in} & $(-5)^3-216=0$ \\
&& $-125-216=0$ \\
&& false
\end{tabular}
[/code]

It won't work in the forum (because of the dollar signs), but it works in the TeXer.

tabular is the general column-alignment tool outside of math mode.  array works in math mode.

\begin{eqnarray*} is the same as \begin{array}{rcl}


geometry
probability
articles
rectangle
complex numbers

I used Google define in area because it's more glossological than mathematical :P Maybe you won't like the results:
area:Area is a quantity expressing the size of a figure in the plane
natural number (my definition :P) is a number that equals the sum of '1's. eg 4=1+1+1+1. I guess we can define it like this...
number=quantity...

[quote="basted"]
I guess we can define it like this... number=quantity...[/quote]

I dont think complex numbers can be classified as a quantity.

BTW good question.

I can only guess its meaning, but I believe that a number is just a symbol.

It's given meaning as a symbol varies.

If a number is a quantity, then what is a quantity?

I apologize in advance for posting in classroom math, but some clarifications need to be made here.

[quote="Z-coli"]Define "area,"[/quote]

The [url=http://en.wikipedia.org/wiki/Lebesgue_measure]Lebesgue measure[/url] on $ \mathbb{R}^2$.

[quote="Z-coli"]"natural number,"[/quote]

A natural number is an element of the set $ \{ 1, 2, 3, ... \}$.  To be more precise, it is the set of finite cardinalities of non-empty sets.

[quote="Z-coli"]and, yes, "number."[/quote]

This is an ambitious question.  In one sense, a number is a member of a [url=http://mathworld.wolfram.com/NumberField.html]number field[/url].  But when we speak of the "real numbers" or the "complex numbers," we might prefer to say that a "number" is a naive concept generalized by the notions of [url=http://mathworld.wolfram.com/Ring.html]ring[/url] and [url=http://mathworld.wolfram.com/Field.html]field[/url].

Edit:  Mathworld gives an even more general definition of [url=http://mathworld.wolfram.com/Number.html]number[/url].

[quote="Z-coli"]Define "area," "natural number," and, yes, "number."[/quote]

[url=http://dictionary.reference.com/browse/area]area[/url]
[url=http://dictionary.reference.com/browse/natural%20number]natural number[/url]

Naturals:

0 = {}
x' = x U {x}

0 = {}
1 = { { } }
2 = { {}, {{}} }

etc.

The naturals and the set of naturals are generally guarenteed to exist by axioms in set theory.

Integer: Ordered pair of natural numbers. (a, b) = (c, d) iff a + d = b + c.
Think of it as (a - b) = (c - d). Like, (5, 0) = 5, (3, 5) = -2, (0, n) = -n, etc.

Rational: Ordered pairs of two integers, where the second is nonzero. This should be self-explanatory. ((p, q) = p / q)!

Real: There are multiple definitions. One can say it is the set of all non-empty proper initial segments of Q. So for example, we can say $ 0 \equal{} \{x \in \mathbb{Q} : x < 0\}$, and $ \sqrt{2} \equal{} \{x \in \mathbb{Q} : x^2 < 2\}$. This is called a Dedekind cut. Or you can define it axiomatically as any ordered field such that every subset with an upper bound has a suprememum (least upper bound).

Complex: Ordered pairs of reals. This should be self-explanatory.

[quote="Zhero"]Integer: Ordered pair of natural numbers. (a, b) = (c, d) iff a + d = b + c.
Think of it as (a - b) = (c - d). Like, (5, 0) = 5, (3, 5) = -2, (0, n) = -n, etc.

Rational: Ordered pairs of two integers, where the second is nonzero. This should be self-explanatory. ((p, q) = p / q)![/quote]

These aren't particularly good definitions.  We should say that the integers are the completion of the natural numbers with respect to addition and additive inverse, and that the rationals are the completion of the integers with respect to multiplication and multiplicative inverse.  Alternately, the integers are the "smallest" ring containing the natural numbers and the rationals are the "smallest" field containing the integers.

The ring t0rajir0u talked about:

[img]http://id.mind.net/~zona/mmts/miscellaneousMath/typesOfNumbers/map10.gif[/img]

Well, no; I meant [url=http://mathworld.wolfram.com/Ring.html]ring[/url].

Number = quantity
area = the space in something

idk if i used the correct words for "area" but hope it helps :)

So define "quantity" and "space."  :)

In any case, the above diagram about different number systems is somewhat misleading:  there are in fact a lot more irrational numbers than rational numbers.  In fact, "almost" every real number is irrational:  in other words, if you threw a dart at the real number line, the probability that it hits a rational number is $ 0$.

See my link to the MathWorld article concerning the meaning of "number."

And you all should know well how to keep the content of this topic in check:

1) This is placed in a Middle School forum. Therefore, a basic 'definition' suffices.

2) If not, this topic should be moved to higher forums.

3) Concerning 'number', if this is a philosophical question, this topic shouldn't be here either.

Keep in mind that this forum should be the lowest difficulty level in AoPS/ML.

My apologies.  It seemed to me that the OP was pointing his questions in a direction not appropriate to the forum, but I don't like questions that "bait" people in this way, so I wanted to clear things up as rigorously as possible.

No no t0r, your experience, knowledge and insight are definitely welcomed  :D . But the point is ... let's roll the ball smoothly :)

OK, idk why everyone needs such complex answers to such simple questions. To put this simply...

Area is the measure of the space covered by a 2-dimensional figure. So a 6 by 4 rectangle covers an area of 24.

A natural number is what some people call a counting number. It is any integer greater than or equal to 1. The set runs: (1,2,3,...) and so on.

A number is any numerical or mathematical quantity or amount. It is hard to get much simpler than that really.

Anyways, hope this helps. :lol:

[quote="Temperal"]Are $ \mathbb{H}$ and $ \mathbb{O}$ considered "numbers"? After all, $ \mathbb{N}\subset \mathbb{Z} \subset \mathbb{Q}\subset \mathbb{R}\subset\mathbb{C}\subset\mathbb{H}\subset\mathbb{O}$, so it would follow natural progression.[/quote]

Wat is $ \mathbb{H}$ and $ \mathbb{O}$?

uh...yah ive never heard of them!!!!What are they? like $ N$ is like natural numbers right!!!


Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y:

E(X) = 50
E(Y) = 20
Var(X)= 50
Var(Y)= 30
Cov(X,Y)=10

100 people are randomly selected and observed for these three months. Let T be the total number of  hours these 100 people watch movies or sporting events during this three-month period.

Find P(T<7100).

My attempt:

T=100(X+Y)
E(T)=E[100(X+y)]= 100E(X+Y)= 100[E(X) +E(Y)]= 100 (50+20)= 100(70)=7,000
Var(T)= Var(100(X+Y))= 100^2 [Var(X+Y)]= 100^2[Var(X)+Var(Y) + 2Cov(X,Y)]= 100^2[50+30+2(10)]= 100^2[100]= 100^3.

The problem here is that the solution says the variance is of T is 100^2.  It multiplies Var(X+Y) by 100, when I thought that I was finding the Var(100(X+Y)) and not 100*Var(X+Y).

What am I doing wrong?

Your random variable is most emphatically not $ 100(X \plus{} Y).$ Rather, it is the sum of 100 independent identical copies of $ X \plus{} Y.$

The variance of $ NX$ is $ N^2\,\text{Var}\,X.$

The variance of $ \sum_{k \equal{} 1}^NX_k$ is $ N\,\text{Var}\,X.$

That's the difference.

There's one more thing going on: they didn't tell you the actual distributions for $ X$ and $ Y,$ just the first and second moments. But once you add up multiple independent copies of these, you invoke the Central Limit Theorem and claim that the sum is approximately normal.


topology
advanced fields
advanced fields unsolved

is it possible to obtain the fundamental group of the circle using the

Seifert Van Kampen theorem?

thx in advance

Probably not. In order to get any nontrivial fundamental group out of the theorem, you need an example of something which has a nontrivial fundamental group. The circle is always the first example.

Probably you are right but is it a complete answer for this problem.
I guess it is not.
Anyway appreciate your help

I wanna see any proof that for a given theorem, there is no proof using some other given theorem ;)

Yes you can use Van Kampen, just do the same as you do standardly when computing the fundamental group of the circle, then use Van Kampen on the open covering $\{A,B\}$ where $A$ is the circle itelf and $B$ is some small neighbourhood of some point. :P

The whole thing is, as said above, that we already need some nontrivial fundamental group to start with; and the simpliest one is that of the circle, so we circle around in circles when trying anything else than circles.
Well, you indeed could calculate the fundamental group of something else first by the covering-space-methode (this is the same as we do for the circle), but also then the problem occurs that we can't cover the circle by $\{A_1 , A_2 , ...\}$ in a good way when not already one of these is the circle itself because all other open connected subsets have trivial fundamental group.


Given an equilateral triangle, inside of this triangle is inscribed a square, with one side of the square lying on a side of the triangle. Inside of this square is inscribed a regular hexagon. What is the ratio of the area of the regular hexagon to the area of the equilateral triangle?

[hide="I hope there is a better way to do this"]Let a side of the equilaterial triangle be $1$.

Its altitude is then $\frac{\sqrt3}{2}$. Its incenter, which in an equilateral triangle, is 1/3 the altitude, is $\frac{\sqrt3}{6}$.

This incenter is also the center of the inscribed square, so half the side of the square is the length of the incenter. One side of the square is thus $\frac{\sqrt3}{3}$.

Let a side of the inscribed hexagon be $p$. If we draw a picture, we see that $p+\frac{p\sqrt2}{2}+\frac{p\sqrt2}{2}=p+p\sqrt2=\frac{\sqrt3}{3}$.

We find $p=\frac{\frac{\sqrt3}{3}}{1+\sqrt2}=\frac{\sqrt6-\sqrt3}{3}$.

The area of the hexagon is $\frac{3p^2\sqrt3}{2}=\frac{3(\sqrt6-\sqrt3)^2\sqrt3}{2(3^2)}=\frac{(9-6\sqrt2)\sqrt3}{6}=\frac{(3-2\sqrt2)\sqrt3}{2}$.

The area of the equilateral triangle is $\frac{\sqrt3}{4}$.

The ratio ($\sqrt3$ cancels) is $\frac{\frac{3-2\sqrt2}{2}}{\frac{1}{4}}=6-4\sqrt2$

I probably should've picked a higher number for the side of the equilateral triangle.[/hide]


geometry
analytic geometry
graphing lines
slope
trigonometry
similar triangles

Guy wires are strung from the top of each of two poles to the base of the other. What is the height from the ground where the two wires cross?

This is what I did...

[hide]I coordinatized by saying that the base of one pole was (0,0), top was (0,b), and the other pole had base of (d,0) and top of (d,b).

When i did point slope and set them equal and found x then y, i got y = d^2b/(d^2-b^2), but of course this doesn't work for d=b, and I'm starting to wonder if it's valid at all.

[/hide]

Any insight?

[hide]1. The distance between the two poles shouldn't matter.

2. Try looking at the question from a pure Euclidean (as opposed to coordinate) geometry perspective.

3. The heights of the poles aren't necessarily the same...[/hide]

hint [hide]drop an altitude from the point of intersection, and look for similar triangles.[/hide]

And if you knew the angles, trig would be helpful.

I have a 4-line solution to this problem, so I doubt trig is really necessary.

(Solution omitted for the benefit of others)

I have a proof of my answer using similar triangles but I don't wish to post it, it'd be more fun to see you all try proving it.

Answer In Spoiler:[hide]

1/x_1 + 1/x_2 = 1/x_3, where x_1 and x_2 are the heights of the two poles, and x_3 is the height of the intersection.[/hide]

No no... I just mean that if you HAD angles, that could be solved with trig.

Solution:

[hide]x/h=(x+y)/b

y/h=(x+y)/a

Add...

(x+y)/h=(x+y)(a+b)/ab

1/h=1/a+1/b[/hide]

Indeed.

(I just realised that a desperate engineer with only a marked ruler can use this diagram to figure out the resistance when two resistors are connected in parallel. :D )


I don't manage to finish this one...

Let $n \geq 2$ be an even integer.

Find all real numbers $a$ and $b$ satisfying the following conditions:

a) $b^n = 3a + 1$
b) $p(x) = (x^2 + x + 1)^n - x^n - a$ is divisible by $q(x) = x^3 + x^2 + x + b$.

[quote="Rafal"]hmm, $q(x)$ has only one root (and not a triple one) since $W(x)=x^3+x^2+x$ is a strictly increasing function.[/quote]
I am sorry, I do not understand what you meant.

[quote="enescu"]Answer: $a=0,b=1.$

Let $x_{1},x_{2},x_{3}$ be the roots of $q$; then $q$ divides $p$ iff $p(x_{1})=p(x_{2})=p(x_{3})=0.$

[/quote]

I was writing that $q(x)$ has only one root ( and not a triple one).

Now, $q$ is a polynomial of degree 3, so it must have three roots (not necessarily real, of course) and I denoted them by $x_1,x_2,x_3$. What is the problem?

Sorry enescu, I just thought you were talking about "real" roots. Of course, in this case I agree with you.


[i]Predict reactants and products for the following redox equation. Be sure to notice whether reactions occur in acidic or basic conditions. Also, all spectator ions must be eliminated. [/i]

Potassium permanganate solution is added to concentrated hydrochloric acid. $ KMnO_4 \plus{} HCl \xrightarrow{} KCl \plus{}
HMnO_4$
I can't seem to figure out the spectator ions. Please help. Thank you.

That isn't a redox reaction.

A redox reaction with those reactants with be

MnO4- + Cl- + H+ ---> Mn2+ + Cl2 + H2O

Here, the spectator ion is K+

So you think to yourself which compounds will dissociate in water (in other words, are aqueous).
All of these except for $ KCl\plus{}HMnO_{4}$ will dissociate because it is a weak acid.
So you write the ionic equation, and then cancel out the ions which are on both sides.

Also, this is not a redox equation, because there is no transfer of electrons.


Given $\triangle ABC$ with Fermat point $F$ prove that
$(FA.FB+FB.FC+FC.FA)(FA+FB+FC)=a^{2}.FA+b^{2}.FB+c^{2}.FC$

[quote="mecrazywong"]After a quick check, I suppose the result is only true when the largest angle of $\triangle ABC$ is smaller than $120^\circ$, right?[/quote]
Yes, you are right, because your suppose is the condition to exist Fermat point $F$.Can you post your proof?
To:SpongeBob Fermat point http://www.mathworld.wolfram.com/FirstFermatPoint.html

[quote="gemath"][quote="mecrazywong"]After a quick check, I suppose the result is only true when the largest angle of $\triangle ABC$ is smaller than $120^\circ$, right?[/quote]
Yes, you are right, because your suppose is the condition to exist Fermat point $F$.[/quote]
This is not the condition for the existence of Fermat's point. As far as I know, Fermat's point is defined as the the point which minimises the sum $AF+BF+CF$. The condition mentioned above is just to ensure the Fermat's point is at its "normal position" (the "120-120-120 configuration").
For the proof, I will post it later when I'm free.

What you need to do is to replace $a^{2}= FB^{2}+FC^{2}+FB.FC$.

Yes thank you for your solution, I found it when I found min value of
$\frac{(MA.MB+MB.MC+MC.MA)(MA+MB+MC)}{a^{2}MA+b^{2}MB+c^{2}MC}$ and I saw it is min when $M=F$ and with this equality then it is true!


analytic geometry
graphing lines
slope
quadratics
calculus
algebra

Let $p$ be a positive number. Let $C$ be the curve $y=2x^3$ and $P(p,2p^3)$ a point on C. Let $l_1$ be the tangent line at $P$ and $l_2$ be another tangent line of $C$ which passes through $P$.

1.) Express the slope of $l_2$ in terms of $p$.
2.) Find $tan\theta$, where $\theta$ is the angle formed by $l_1$ and $l_2$ and $0<\theta<\frac{\pi}{2}$
3.) Find the maximum value of $tan\theta$

[quote="willydavidjr"]Let $p$ be a positive number. Let $C$ be the curve $y=2x^3$ and $P(p,2p^3)$ a point on C. Let $l_1$ be the tangent line at $P$ and $l_2$ be another tangent line of $C$ which passes through $P$.

1.) Express the slope of $l_2$ in terms of $p$.
[/quote]

[hide]call $Q(q,2q^3)$ the point where $l_2$ is tangent to $C$. Clearly, it must be $q<0$. Now, the equation of $l_2$ is $y-2p^3=6q^2(x-p)$, because it has to pass through $P$. The slope is $6q^2$ $(f'(x)=6x^2)$, but it is also equal to $\frac{2p^3-2q^3}{p-q}$. So we arrive to solve the quadratic equation $2q^2-pq-p^2$ that has roots $p$ and $-\frac{1}{2}p$. but $q=p$ is not valid, because it has to be $q<0$. So $q=-\frac{1}{2}p$, and the slope is $6*(-\frac{1}{2}p)^2=\frac{3}{2}p^2$[/hide]

Hydro, how can you clearly say that q<0? It is not given on the problem? Or did you based it on the attachment?

I didn't remeber cause I did it a lot of time ago, but I think that since $f(x)$ is convex for $x \ge 0$, any tangent line to the curve has to be external to it, so it can't pass through P

This violates the no-calculus in Intermediate forum rule.

He didn't actually use $q < 0$ anywhere -- $q \neq p$ because when you have $q = p$, it's just the same tangent line you started with and you are explicitly looking for a different tangent line, and the only other option is $q = -\frac12 p$.
The fact that the positive half of the curve is convex does explain why it must be $q < 0$, but you didn't need that fact for the solution.


The Melbourne University Puzzle Hunt is a puzzle hunt ([url]http://en.wikipedia.org/wiki/Puzzlehunt[/url]) , and it starts this year on April 7th. I'm already on another team, but I'm just throwing this out there if anybody else on AoPS that are good at puzzles wants to make a team together (teams can have up to 10 people). Here is the website, which has previous puzzles posted on it: (somehow, I have a feeling that not many people will be interested  :P )

http://www.ms.unimelb.edu.au/%7Emums/puzzlehunt/

I would like to join, but can only do it on the Internet. Also, I'm just okay at puzzles.

[quote="mchoi815"]i haven't looked into this puzzle hunt much, but the MIT mystery hunt involves finding a coin.  can this specific one be done over the internet?[/quote]

Yes, this is internet only.

Edit: except the winning team who solves the meta-puzzle has to find something hidden inside the Melbourne university campus.

Well, it's almost over now.

The team I'm on is doing pretty good. :D

You can still take a look at the current puzzles here:
http://www.ms.unimelb.edu.au/~mums/puzzlehunt/puzzles.php


is it possible to obtain a [b]closed[/b] form for the sum of the following series...  :maybe:

$ \sum_{n=1}^\infty\;\left\{\;\textbf H_{\infty,2}\;-\;\textbf H_{n,2}\;\right\}$, i.e.

$ \sum_{n=1}^\infty\;\left\{\;\frac{\pi^{2}}{6}\;-\;\left(\;\sum_{m=1}^{n}\;\frac{1}{m^{2}}\;\right)\;\right\}$


calculus
integration
function
logarithms
real analysis
real analysis unsolved

Hi, I was wondering how would you compute the following integral:

$\int x^x dx$

I believe this is non-elementary (no known antiderivative)

That's right. It should be possible to represent it in some kind of series form, though.

As I recall, $\int_0^1 x^x dx = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^n}$, which you can estimate pretty accurately with just the first few terms.

So is it possible to evaluate this integral (mathematically) for limits of integration other than $[0,1]$?? For $[0,a]$ perhaps?

Probably, but I don't know how. You could always do it by Riemann sums, if you don't mind the method being somewhat inefficient.  :P

If I have time tonight I'll try to figure it out, or maybe Professor Merryfield will come by and clear things up (he's good at that).

[quote]As I recall, $\int_0^1 x^x dx = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^n}$, which you can estimate pretty accurately with just the first few terms.[/quote]

BTW, Xevarion I tried to derive this formula but could only get so far:

$\int_0^1 x^x dx = \int_0^1 e^{x\ln(x)} dx = \int_0^1 \sum_{k=0}^{\infty} \frac{{(x\ln(x))}^k}{k!} dx$

$= \sum_{k=0}^{\infty} \frac{1}{k!} \int_0^1 {(x\ln(x))}^k dx$

I am not quite sure what should I do next, could you please hlep me.  :?

If you know gamma function then $\int_0^1 x^k \ln^k x \, dx$ , subtitution $x=e^{-u}$

if you don't want to do with gamma function. Use integration by part to $\int_0^1 x^k \ln^k x \, dx$. and recognize the pattern.

I found the answer to this in another forum. Here it is:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=354567#p354567


Here is a little generlised form:
Let$\alpha_{1},\alpha_{2} \in \mathbb{R}$ and $n \in \mathbb{N}$. then prove that  there exist $a_1,a_2,b \in \mathbb{Z}$,$0 < b \le n$ such that :
$|\alpha_1- \frac{\alpha_1}{b}| \le \frac{1}{b\sqrt{n}}$ , , , ,   $|\alpha_2 - \frac{\alpha_2}{b}| \le \frac{1}{b\sqrt{n}}$

Any idea for this conjecture?
Let $\alpha_{i=1 \rightarrow n} \in \mathbb{R}$ and $n \in \mathbb{N}$.
Prove there exist numbers like$a_{i=1 \rightarrow n}, b \in \mathbb{Z}$
 $0 < b \le n$.
such that :
$|\alpha_i- \frac{\alpha_i}{b}| \le \frac{1}{b\sqrt[n]{n}}$

[quote="lomos_lupin"]This one is not funny at all, but it is challenging
Let $\alpha \in \mathbb{R}$ and $n \in \mathbb{N}$.prove that there exist integers numbers like $a,b$ such that $0 < b \le n$ and
$|\alpha -\frac{a}{b}| < \frac{1}{bn}$[/quote]

$\{\alpha\},\{2\alpha\},\ldots,\{n\alpha\}$ are $n$ points on the circle with unit circumference, so there must be two of them, $\{n_1\alpha\},\{n_2\alpha\},\ n_1<n_2$ with a distance of at most $\frac 1n$ between them, measured on the circle, in a counter-clockwise direction. This means that $0,\{(n_2-n_1)\alpha\}$ will have the same property, and $n_2-n_1$ is the $b$ we're looking for.

:o  :o
You are pretty good man.that problem was not too easy .
you solved it in less than 30 minutes,well done

The result was known to me :).

There's an $n$-dimensional analog, I believe (your second post is the $2$-dimensional analog), but I'll have to give that more thought :).

You can't imagine how many times it appeared on the forum, Lomos. It is classical very well known result.


[quote="motor angel"]1.Prove that
$ S \equal{} \frac1{9} \plus{} \frac1{25} \plus{} \frac1{49} \plus{} ... \plus{} \frac1{(2n \plus{} 1)^2} < \frac1{4}$
$ n \ge 1$ , $ n$ is a natural number
[/quote]

Note:  In the following steps, we can rearrange the order of terms,
because the expressions for $ S$ and $ \frac{{\pi}^2}{6}$ have terms
that converge absolutely.

If you allow for someone to know in advance that

$ \frac {{\pi}^2}{6} \equal{} 1 \plus{} \frac {1}{4} \plus{} \frac {1}{9} \plus{} \frac {1}{16} \plus{} ... \plus{} \frac {1}{{(2n})^2} \plus{} \frac {1}{{(2n \plus{} 1)}^2} \plus{} [\frac {1}{{(2n \plus{} 2)}^2} \plus{} \frac {1}{{(2n \plus{} 3)}^2} \plus{} ...]$,

then $ S \equal{} \frac {{\pi}^2}{6} \minus{} 1 \minus{} \frac14 \minus{} \frac {1}{16} \minus{} \frac {1}{36} \minus{} ... \minus{} \frac {1}{(2n)^2} \minus{} \frac {1}{(2n \plus{} 2)^2} \minus{} \frac {1}{(2n \plus{} 3)^2} \minus{} ...$

$ S \equal{} \frac {{\pi}^2}{6} \minus{} 1 \minus{} \frac14 \minus{} \frac {1}{16} \minus{} \frac {1}{36} \minus{} ... \minus{} [\frac {1}{(2n)^2} \plus{} \frac {1}{(2n \plus{} 2)^2} \plus{} \frac {1}{(2n \plus{} 4)^2} \plus{} ...]$ $ \minus{}$
$ [\frac {1}{(2n \plus{} 3)^2} \plus{} \frac {1}{(2n \plus{} 5)^2} \plus{} ...]$

$ \equal{} (\frac {{\pi}^2}{6} \minus{} 1) \minus{} \frac14(1 \plus{} \frac14 \plus{} \frac19 \plus{} ...) \minus{} [\frac {1}{(2n \plus{} 3)^2} \plus{} \frac {1}{(2n \plus{} 5)^2} \plus{} ...]$

$ \equal{} \frac {{\pi}^2}{6} \minus{} 1 \minus{} \frac14 (\frac {{\pi}^2}{6}) \minus{} [\frac {1}{(2n \plus{} 3)^2} \plus{} \frac {1}{(2n \plus{} 5)^2} \plus{} ...]$

$ \equal{} \frac {{\pi}^2}{8} \minus{} 1 \minus{} [\frac {1}{(2n \plus{} 3)^2} \plus{} \frac {1}{(2n \plus{} 5)^2} \plus{} ...]$

Suppose we use an appropriate slightly larger value of $ \pi$ here, so that if the left-hand side turns
out to be less than $ \frac14$, then using the exact value of $ \pi$ on the left-hand side will also be true.

$ \frac {(3.15)^2}{8} \minus{} 1 \minus{} [\frac {1}{(2n \plus{} 3)^2} \plus{} \frac {1}{(2n \plus{} 5)^2} \plus{} ...] \equal{}$

$ {{0.2403125}} \minus{} ($sum of fractions which are all positive$ )$ $ <$ $ \frac14$

For one, here's a simpler solution using this
[hide] $ \frac{1}{n^2}\le \frac{1}{(n\plus{}1)(n\minus{}1)}\equal{}\frac12(\frac{1}{n\minus{}1}\minus{}\frac{1}{n\plus{}1})$ [/hide]

[quote]2. Solve in real numbers
$ x_1^{1978} \plus{} x_2^{1978} \plus{} x_1^{1977} \plus{} x_2^{1977} \equal{} x_1^{1976} \plus{} x_2^{1976}$[/quote]
Are you sure the number of solutions is finite?


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

How can I find the Galois group of  x^12 + 2x +2?

SAGE outputs

Galois group Transitive group number 301 of degree 12 of the Number Field in a with defining polynomial x^12 + 2*x + 2

I don't know it this helps anything.

but is there a method to compute this galois group?

What do you think SAGE does¿  :wink:

There are general methods, but computing such a thing can get very messy. Depending on what the result is (don't know SAGE or what that output says) there may be a "nice" way to do it.

so ZetaX do you know a nice method to determine this galois group, then please let me know.


$f$ is pibonaci sequence let $p$ be a prime number and $p \not=2,5$ PROVE that:
$f_p\equiv 5^{\frac{p-1}{2}}$  $(\mod p)$

Have a look [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=fibonacci+quadratic&t=24981][u]here[/u][/url]


Does anyone know where I can find practice problems for the reactions section of the AP Chemistry exam? Also, where can I get help on that topic, e.g. synthesis, decomposition, double replacement, single replacement, etc?

You could try http://www.collegeboard.com
I don't remember how to get to them, but there is a section for every AP topic, each of which includes many problems for their respective topics. There is a book called Chemistry and Chemical Reactivity (Kotz and Treichel) with good discussions on reaction types. The book's front cover features a photo of a spoon full of decomposing ammonium chloride next to a model of the chemical's crystal structure. If you go to the Saunders College Publishing (the book's publisher - search for them on google) website and find the link for this book, there should be online tutorials for many topics in AP Chemistry, including reactions. If you can find a copy of the book's accompanying CD, I would highly recommend it to you. It has loads of great interactive tutorials and demos that really get the information across in a clear and student-friendly way. I have seen good deals around the web on used and new versions of the book/CD. Look around. see what you can find.

I hope these resources will aid you in your studies! If you want help with specific questions, you can pm me and I'll do my best to give you assistance.


search
number theory proposed
number theory

A sequence of primes $ p_1,p_2,p_3,....,p_n$ satisfies the condition that for $ n\ge 3$, $ p_n$ is the greatest prime divisor of $ p_{n\minus{}1} \plus{} p_{n\minus{}2} \plus{} 2000$. Prove that the sequence is bounded.

I searched for "bounded sequence of primes", and I got:

http://www.mathlinks.ro/viewtopic.php?search_id=695587681&t=151418
http://www.mathlinks.ro/Forum/viewtopic.php?p=372280#372280

Nice problem anyway.


Three numbers, a1,a2,a3, are drawn at random, without replacement, from the set {1, 2, 3, ... , 1000}. Three other numbers, b1, b2, and b3 are then drawn randomly and without replacement from the remaining set of 997 numbers. Find the probability that a brick with dimensions a1*a2*a3 can fit in a brick with dimensions b1*b2*b3 after a suitable rotation.

The answer is 1/4, but my question is, if the numbers were picked in order a1,b1,a2,b2,a3,b3 instead, would the answer be different (1/8)?

I've only given this a superficial thought, but it would appear to me that the answer would be different. After all, if you pick in order, then the b_k's can assume values that it couldn't under the original assumptions. Likewise, the a_k's could not assume certain values it would have under the original conditions.

I think the answer is still $\frac{1}{4}$

The numbers are chosen at random, so there is a bijection between the dimensions in the first problem and the dimensions in the second.

I'm just saying this because I originally got 1/8 for this problem by doing it the second way. My thinking was that the probability of a1<b1 = 1/2, a2<b2=1/2, and a3<b3 = 1/2, so it multiplies to 1/8. However, when you pick all the A's first, it's a little bit different because there are only 997 choices for b1, etc...


I am currently writing a mock AMC 10.

Please say I LOVE BOOGA if you are interested.

You will find out who BOOGA is later.  :D

We seem to have an influx of mock AMCs. I'll do it...

I'll do it, but I don't know if I am going to say - - - until I know just who this booga is. I smell a sinister plot :police:

[quote="13375P34K43V312"]Screw you I'll do it if I find the problems are somewhat decent when I first see them.[/quote]

oh.. put a sock in it  :lol:

I'm pretty sure booga is going to be some random person in one of the problems that does something stupid.

I LOVE BOOGA.

I'm going to guess that BOOGA is Go Around the Tree. But that's just a guess.

[quote="Xantos C. Guin"]Hopefully, BOOGA doesn't appear in all of them. That would be horrible.[/quote]

he might not be a person  :wink: and booga won't appear much... but will make a lasting impression

I remember in 4th grade when our teacher wrote a test , where every question had annoying characters like Spongebob and Patrick taking a bath, and all this junk. Haha don't worry it won't be like that.

I'm not going to say * **** *****....but this test better be good....!!!

[quote="anirudh"]I'll be out if it will be bad.[/quote]
You're one to talk :roll:

[quote="ragnarok23"][quote="anirudh"]I'll be out if it will be bad.[/quote]
I find this to be very funny[/quote]
Irony has it's place in life

I'm in.
[quote="now a ranger"]...Spongebob and Patrick taking a bath...[/quote]

Booga$\stackrel{?}{=}$Go Around the Tree
Now a ranger$\in$Go Around the tree

actually

booga is not who you think so HAH

Booga$\stackrel{?}{=}$Relative of Now a Ranger

[quote="anirudh"]Fine. I at least know that now a ranger$\stackrel{\in}{=}$Go Around the Tree.[/quote]
See Survivor AoPS island.

Um...There is no Booga in Survivor anymore; there used to be. :P

I sometimes post on now a ranger, sometimes on go around the tree.

It is because I now have two different computers. Therefore, one automatically signs on to now a ranger, the other to go around the tree.

So my posts to both are equal almost.  :lol:

ill do it if it's good....
forget the ****
you're trying to make me say/type.
who's Booga? :huh:

[quote="Go Around the Tree"]The Mock AMC 10: Booga will be posted On January 1, 2006!

I hope I made them well, and I hope you guys won't complain about how the problems don't RESEMBLE exactly the AMC 10 type problems. After all, AMC 10 consists of problems for 10th graders- Algebra 2 and under. No Trig and stuff. Who says the AMC 10 has to have blah blah blah and this kind of stuff on the test?

Hopefully it's about the right level.[/quote]
In actuallity, I think most of us prefer if the problems don't resemble AMC 10 problems at all. I personally prefer problems that are, to quote pakman2012, "fresh".


Are you a fan of Lego? I think that legos are cool!

i had them. my brother was obsessed with them so we had to throw them away.

I have liked Legos since I was in Kindergarten, and I still like them now.  The basic Legos are cool, but now, I use more advanced sets (and of course, the basic sets).  The MindStorm sets are really cool, because you build and program robots, which, in my opinion, is better than the traditional Legos.  I used to keep track of exactly how many Legos I had, but now, I just know that I have several thousand.

[quote="math92"][quote="PenguinIntegral"]Lego's big fault is their overuse of right angles.[/quote]
That would be nice if they had irregular shaped or trapezoidal legos.[/quote]

I thought they did.  :?

[quote="PenguinIntegral"]Lego's big fault is their overuse of right angles.[/quote]MOST lego pieces are shaped as rectangular prisms. I think this is primarily because the majority of the people who play with legos are little children and i suppose it would be easier for them if the pieces were simple. but i've seen irregular shapes.

The general thing is to use rectangular blocks for simplicity , but I'm quite sure I've seen a set of lego blocks being provided with motors and all that.....I guess lego's trying to get a customer base that's as wide as possible...


[b][color=blue]Pregunta 02:[/color]
Sabado 05 de mayo[/b]

Sean $a,b,c$ reales positivos, tales que: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$\\
Pruebe que:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}.\]

$\text{\LaTeX}{}ed$ by carlosbr

[quote="carlosbr"][b][color=blue]Pregunta 02:[/color]
Sabado 05 de mayo[/b]

Sean $a,b,c$ reales positivos, tales que: $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$\\
Pruebe que:
\[a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}. \]
$\text{\LaTeX}{}ed$ by carlosbr[/quote]
DEM.-
Supongamos lo contrario.
Existe una terna de números reales positivos $a,b,c$ tal que cumplen:
 $a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}...(1)$,
y tambien cumplen.
$a+b+c <\frac{3}{a+b+c}+\frac{2}{abc}....(2)$
luego de (1) y (2), $\frac{3}{a+b+c}+\frac{2}{abc}>(a+b+c \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$\to  (ab+bc+ca-2)/abc <3/(a+b+c)$, operando esto nos queda.
$2(a+b+c)>aab+bba+cca+aac+ccb+bbc..(3)$
multiplicando (1) y (3) y operando (no pongo todas las operaciones por falta de tiempo), nos quedara.
$aabbc+bbaac+aaccb+ccaab+bbcca+ccbba>aaabb+bbbaa+cccaa+aaacc+cccbb+bbbcc..(4)$, pero
$aaabb+bbbaa+cccaa+aaacc+cccbb+bbbcc\geq aabbc+bbaac+aaccb+ccaab+bbcca+ccbba..(5)$
de (4) y (5), $(\to <-)$.
porlo tanto, para toda terna de numeros reales $a,b,c$ que cumplan la desigualdad (1),  cumple que:
$a+b+c \geq \frac{3}{a+b+c}+\frac{2}{abc}.$
                                             $Q.E.D.$


[b]Setup:[/b] Let a vertex be a 0-face, an edge be a 1-face, a face be a 2-face, a cell be a 3-face, and in general, the k-dimensional extension be an k-face.

An [i]n[/i]-cube has side-length $ m\plus{}2$ for some positive integer $ m$, and is divided up into $ (m\plus{}2)^n$ unit [i]n[/i]-cubes.

Let the [i]order[/i] of such a unit [i]n[/i]-cube be the smallest nonnegative integer j such that at least one j-face of the unit is a subset of a j-face of the original. For example, in a 3-D cube of sidelength 5, there are 8 unit 3-cubes of order 0 (the vertexes, since a 0-face is a vertex).

[b]Problem:[/b] Show that the number of unit [i]n[/i]-cubes of order $ j$ is given by the coefficient of $ x^j$ in the expansion of $ (mx \plus{} 2)^n$.

[quote="The Zuton Force"]An [i]n[/i]-cube has side-length $ m \plus{} 2$ for some positive integer $ m$, and is divided up into $ m^n$ unit [i]n[/i]-cubes.[/quote]  This isn't possible in any of the dimensions you can draw pictures of, and it's not possible in any of the dimensions you [i]can't[/i] draw pictures of, either :P

Any takers?
[hide="Hint"]Use vertices and edges to define a particular n-face[/hide]


I was looking through the member lists and found about 5 people at the top with an insane amount of posts: around 1600000, one of them not joining but 2 days ago. The 6th person down had only 8000 posts. When I did a "Look at all of the posts" option, it showed only 2 posts for one person. All of these people are labelled as moderators, so I don't know if it is a glitch or not. Can someone explain?  :huh:

This problem has already been notified to the admins. It's a bug and it should be fixed soon  :wink: .

In fact, another thread on this is just a few lines down. It also appears to be fixed now.

It does tell us that the post count is a 24-bit unsigned integer.


topology
function
analytic geometry
graphing lines
slope
advanced fields
advanced fields unsolved

Is functionally Hausdorff same as Completely Hausdorff?

And If so, which one is finer than the other?

I read from some websites they different and other websites said they are same?

which one is true.?

Help please.

thank you in advanced

what are the definitions of these terms?

O.k. :

Here they are :

Functionally Hausdorff :

For any $ x\ne y \in X$ there exists a continuous function $ f: X\rightarrow [0,1]$ with $ f(x)\equal{}0$ and $ f(y)\equal{}1$.

Completely Hausdorf:

For any $ x\ne y \in X$ there exists two open set $ U,V$ s.t. $ x\in U$ , $ y\in V$ , and $ \bar{U}\cap \bar{V}\equal{}\phi$.

What do you think?

Thank you in Advanced.

In http://en.wikipedia.org/wiki/Completely_Hausdorff_space
Your "functionally Hausdorff" is called completely Hausdorff,
your "completely Hausdorff" is called Urysohn space.

It is claimed in the entry that completely Hausdorff => Urysohn (this is easily proved by looking at the inverse image of $ f^{\minus{}1}([0,1/3])$ and $ f^{\minus{}1}([2/3,1])$), but there exists Urysohn space that is not completely Hausdorff.

Fine, we can say functionally Hausdorff is finer than Hausdorff space and coarser than Regular space.

Am I true ?

About Urysohn Space..Could you tell me where should I put it?

I'm warping about Spaces Names.

Thank youuuuuuuuuuuu

My guess would be that these aren't always comparable (in terms of coarser and finer). If you assume $ T_1$, then functionally Hausdorff $ \Rightarrow$ regular $ \Rightarrow$ Hausdorff.

Hey guys;

I see
$ T_2\Leftarrow Urysohn \ Space \Leftarrow Functionally \ Hausdorff  \Leftarrow Regular\Leftarrow Completely \ regular$.

Hey HilbertThm90 , I have an example show that Functionally Hausdorff which s not regular .I think you mean as I said above.

Now could you check on that chain please.?

Thank You.

[quote="Soarer"] but there exists Urysohn space that is not completely Hausdorff.[/quote]

Could yo tell me what is it exactly please? And if  the example has name like " Irrational Slope topology " tell me what is it if I can find it in Counterexamples in General topology by Steen .( I have that book) .

Otherwise, Please write the example here.

Thank you.


linear algebra
matrix
algebra
polynomial
vector
linear algebra unsolved

i have to solve eigenvalues and eigenvectors of a matrix but i must not use the det(sI-A)=0 way. I have to use a differant but formal way.
The matrix is here.

0 1 1 1 1
1 0 1 1 1
1 1 0 1 1
1 1 1 0 1
1 1 1 1 0

the eigenvalues are 4,-1,-1,-1,-1

it suffices to find its minimal polynomial, which is not that hard.

Anyone who works often with matrices should regard the matrix $ J$ of all ones as a familiar old friend - nearly as familiar as $ I.$

Let $ J \equal{} J_{n\times n}.$ This is evidently a matrix of rank $ 1.$ By the rank-nullity theorem, its null space has dimension $ n \minus{} 1.$ That means that $ 0$ is an eigenvalue of $ J$ of geometric multiplicity $ n \minus{} 1.$ The algebraic multiplicity of an eigenvalue must be greater than or equal to its geometric multiplicity, so the algebraic multiplicity of $ 0$ as an eigenvalue of $ J$ must be at least $ n \minus{} 1.$ However, the sum of the eigenvalues (with their algebraic multiplicities) must be the trace of the matrix. The trace of $ J$ is $ n.$ The only way to make that work is for the eigenvalues of $ J$ to be $ n \minus{} 1$ copies of $ 0$ and one copy of $ n.$ Furthermore, the (unique up to scalar multiplication) eigenvector corresponding to the eigenvalue $ n$ is easy to spot: it's simply the column vector $ J_{n\times 1}.$

alekk would have you look for the minimal polynomial. That's easy enough, since $ J^2 \equal{} nJ.$ So that's the minimal polynomial right there: $ x^2 \minus{} nx.$

But the matrix ozlem81 started with wasn't exactly $ J.$ Instead, it was $ J \minus{} I.$ Adding or subtracting multiples of $ I$ is quite straightforward. If $ A$ has eigenvalues $ \lambda_i$ with corresponding eigenvectors $ v_i,$ then $ A \plus{} bI$ has eigenvalues $ \lambda_i \plus{} b$ and eigenvectors $ v_i.$ So if $ J$ has eigenvalues $ 0,0,0,0,5,$ it follows that $ J \minus{} I$ has eigenvalues $ \minus{} 1, \minus{} 1, \minus{} 1, \minus{} 1,4.$ And the eigenvector associated to the eigenvalue $ 4$ is the vector of all ones.

[color=green][For more on the "those who work often with matrices should regard $ J$ as a familiar old friend" idea, see jmerry's post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173012]here[/url].][/color]


limit
integration
logarithms
real analysis
calculus
calculus computations

For real numbers $a,\ b,$ let

\[x_{n}=\frac{1}{n^{b}}\left\{\frac{1}{n^{a}}+\frac{1}{(n+1)^{a}}+\cdots+\frac{1}{(2n-1)^{a}}\right\}, \ n=1,\ 2,\ 3,\ \cdots\cdots\]

Find the condition of $a,\ b$ for which $x_{n}$ converges when $n$ approaches infinity and the limit value.

$x_{n}=\frac1{n^{a+b-1}}\cdot\frac1n\sum_{k=0}^{n-1}\left( 1+\frac{k}{n}\right)^{-a}$

As a Riemann sum, $\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\left( 1+\frac{k}{n}\right)^{-a}=\int_{1}^{2}x^{-a}\,dx.$

Hence, we have the following:

If $a+b<1,$ then $\lim_{n\to\infty}x_{n}=\infty.$

If $a+b>1,$ then $\lim_{n\to\infty}x_{n}=0.$

If $a+b=1$ and $a\ne 1,$ then $\lim_{n\to\infty}x_{n}=\frac{2^{1-a}-1}{1-a}.$

If $a=1$ and $b=0,$ then $\lim_{n\to\infty}x_{n}=\ln 2.$

As you guys, the condition of converging $x_{n}$ is $a+b\geq 1.$


Check whether $ X(t)\equal{}exp[W_t\minus{}\frac 1 2 t]$ is Levy martingale
Assume that $ X_t$ is LM than $ X_t\equal{}\mathbb{E}[X|\mathcal{F}_t]$, where $ X\equal{}\lim_{t \to \infty} X_t$ a.s.
but due to Law of Iterated Logarithm we have that X=0 a.s..Contradiction.
Is this correct?
Can anyone offer another way of solving?

I claim that $ X(t)$ is a Martingale.
Let $ s<t$. Then
\begin{eqnarray*}E[X(t)|\mathcal{F}_s]&=&e^{-\frac{1}{2}t}\cdot E[e^{W_t}|\mathcal{F}_s]\\&=&e^{-\frac{1}{2}t}\cdot e^{W_s}\cdot E[e^{(W_t-W_s)}|\mathcal{F}_s]\\&=&e^{-\frac{1}{2}t+W_s+\frac{1}{2}(t-s)}\\&=&X(s)\end{eqnarray*}

terasenya what do you mean by a Levy Martingale exactly ?

It's clear that X_t is a martingale.
By Levy Martingale I mean uniformly integrable martingale.

Then it's not a Levy Martingale

Because the almost sure limit of $ W_t\minus{}1/2t$ is not defined

Remember that $ P(limsup W_t\equal{}\plus{}\infty)\equal{}1$ and $ P(liminf W_t\equal{}\minus{}\infty)\equal{}1$ (by for  example the law of iterated logarithm)

I've wrote the same in my first post. So my solution is OK.Thanks, [b]TheBridgeTwo[/b].

yes, but $ \lim W_t \minus{} \frac {t}{2} \equal{} \minus{} \infty$ so that $ M_t \equal{} \exp( W_t \minus{} \frac {t}{2})$ almost surely converges towards $ M_{\infty} \equal{} 0$. If it were uniformly integrable you would be able to apply the stopping theorem, and you would get $ M_t \equal{} E[M_{\infty} | \mathcal F_t] \equal{} E[0 | \mathcal F_t] \equal{} 0$, which is absurd :-)

[edit]: just noticed that it was what you wrote in your 1st post  :D


probability ratio geometry rectangle function circumcircle trigonometry

Three points are randomly selected within a square of side length 12. What is the probability that the three points all lie in a circle of radius 3? The circle is not predefined. Can you generalize this to a square of side m, a circle of radius r, and k points?

It's an easy math calculation.
The probability to hit one point, equals the ratio of (Area Circle):(Area Square).

Therefore your p = Area Circle / Area Square, which gets us to p = Pi r^2 / a^2.

And then your probability that all three points are within the circle:

P(three points in the circle) = p*p*p = Pi^3 r^6 / a^6.

(I hope i could help).

The circle's not predefined.
It can be centered anywhere you want it to be.
You define the circle after you define the three points.

It's not a problem, as long as the circle is full in the square.

It doesn't make a difference. If you define the points first, or you define the circle first, as only the both parts combined, give you a "question" in the first place.

It also doesn't matter where you place the circle first at, because the probability you hit is always the percentage which the area of rectangle and the circle intersect. There is also nothing said about "where" the k-points should lay, so i assumed they are absolutely random placed.
Otherwise you could place 4 points in the four corners of the square and you could never hit those corners by a circle which completely fits into the rectangle, only if you have a circle of radius 0 and you go into each corner, but then you no more have a circle by definition.

Who said anything about fitting into the rectangle?  You are not reading the question right.

Before anyone attempts it analytically, here's some matlab code which finds an approximate answer using monte carlo methods (i didn't use variance reduction or anything, nor did i feel like computing confidence intervals).

[code]
function P = triangleprob(s,r,n)
tic;
rand('twister', sum(100*clock));
points = s*rand(2,3*n);
distances = repmat(0,1,3*n);
k=1;
for i=1:n
    distances(k) = sqrt((points(1,k)-points(1,k+1))^2+(points(2,k)-points(2,k+1))^2);
    distances(k+1) = sqrt((points(1,k)-points(1,k+2))^2+(points(2,k)-points(2,k+2))^2);
    distances(k+2) = sqrt((points(1,k+1)-points(1,k+2))^2+(points(2,k+1)-points(2,k+2))^2);
    k=k+3;
end
cr = repmat(0,1,n);
count = 0;
for i=1:n
    cr(i) = circumradius(distances(3*i-2),distances(3*i-1),distances(3*i));
    if cr(i) < r
        count = count+1;
    end
end
P = [count/n , toc];
[/code]

With a million samples, we get a result of

[code]
>> triangleprob(12,3,1000000)

ans =

    0.1166   21.1341
[/code]

so the answer is 'about' 11.7%

n.b. the points fitting inside of a circle of radius three is the same as the circumradius of the triangle determined by the three points being at most three.  there's a useful formula for the circumradius of a triangle in terms of its sides (which i conveniently tucked away above in another m-file); this is probably useful in an attempted analytic solutions as well.

I suggest an intermediate problem: given a point $ P$ in the plane, consider the set of pairs of points $ (Q,R)$ such that $ P$, $ Q$, and $ R$ lie in some circle of radius $ r$.  This is a subset of $ \mathbb{R}^2\times \mathbb{R}^2$.  What is its volume?

Note that this is more or less the same as the original question, but ignoring the border of the rectangle.

[quote="blahblahblah"]the points fitting inside of a circle of radius three is the same as the circumradius of the triangle determined by the three points being at most three.[/quote]
No, it isn't. If the triangle is obtuse, it fits in a circle with the longest side as the diameter, but the circumradius is larger.

If we choose two points $ P$ and $ Q$ at distance $ d$, how likely is it that the third is close enough to fit in a circle of radius $ R$? Ignoring the sides of the rectangle, the area for which this works is $ R^2(3\theta\minus{}\sin\theta\plus{}\pi)$, where $ \theta$ is between $ 0$ and $ \pi$ with $ \cos\frac{\theta}{2}\equal{}\frac{d}{2R}$. If $ d>2R$, the area is zero.
Geometrically, this $ \theta$ is the angle $ O_1PO_2$, where $ O_1$ and $ O_2$ are the centers of the circles with radius $ R$ and $ PQ$ as a chord. The region is the union of four circular wedges: one each centered at $ P$ and $ Q$ with radius $ 2R$ and angle $ \theta$, and one each centered at $ O_1$ and $ O_2$ with radius $ R$ and angle $ \pi\minus{}\theta$. The overlap is the rhombus $ O_1PO_2Q$.

There is no reason to expect a nice answer.


geometry
geometric transformation
reflection

How many different bracelets have 3 identical orange beads, 3 identical black beads, and 1 teal bead, if rotating or flipping a bracelet does not change it?

[quote="Textangle"]Alt Solution:
Fix the bracelet along the teal bead. There are $ \frac{6!}{3!\cdot 3!}\equal{}20$ ways to group the rest of the beads, and then, dividing by two to account for reflections, gives $ \boxed{10}$ different bracelets.[/quote]
You could also say that there are $\dbinom{6}{3}$ ways.

Alt Alt Solution:
There are $\frac{7!}{3!3!1!}=140$ orderings, but we have to divide by $7\cdot2=14$ to account for rotations and reflections, and there are $\frac{140}{14}=\boxed{10\text{ bracelets.}}$

First, we find the amount of indistinguishable permutations there are (not accounting for rotations and flipping). We get $\frac{7!}{3!3!}=140$. However, because of the rotations, each case is counted 7 times. We only want it to be counted once, so we divide: $\frac{140}{7}=20$. Because each case is also counted twice (a flipped version) we must also divide by 2: $\frac{20}{2}=\boxed{10}$

[hide=my sol]
so basically we can do this in a letters way
it's $OOOBBBT$, so it's how many ways to arrange that which is $140$.

K, so now we divide by 7 because there are 7 beads and each bead can be rotated.

THen we divide by 2 because it has a 2 way reflectional symmetry

we get to $10$.

[hide = Solution]

Think of this as first arranging the beads in a line, and arranging them on a bracelet.

We have $\frac{7!}{3!} = 140$ ways to arrange them in a line.

However, on a bracelet there is a lot of symmetry: We must first divide the count by $7$, since there are $7$ rotations, and we must divide by $2$ since there are "flips".

Therefore, our answer is: $\frac{140}{7\cdot 2} = \boxed{10}$

[/hide]

[hide=Post #18 by fucongcong][url=aops.com/community/user/512453][b]fucongcong[/b][/url] · Nov 8, 2020, 9:06 PM [url=aops.com/community/p18784883](view)[/url][color=transparent]helo[/color]
What the I got this wrong and my rating went up

-----------
[color=#5b7083][aops]x[/aops] 18[color=transparent]hellloolo[/color] [aops]Y[/aops] 1 [color=transparent]hellloolo[/color] [/hide]

[tip=@fucongcong][img]https://avatar.artofproblemsolving.com/avatar_512453.png[/img]
AoPS User[/tip] what-

[quote=fucongcong]What the I got this wrong and my rating went up[/quote]

Are you sure about that? O_o

[quote=fucongcong]What the I got this wrong and my rating went up[/quote]

Wait what? Lol

[hide=sol]there are $\frac{7!}{3!3!1!}=140$ ways to arrange the beads, since flipping bracelets does not change it, divided by $2$ to get $70,$ then since there are $7$ different rotations, divide by $7$ to get $\boxed{10}$[/hide]

[quote=henryb][hide]Fix the bracelet along the teal bead. There are $ \frac{6!}{3!\cdot 3!}=20$ ways to group the rest of the beads, and then, dividing by two to account for reflections, gives $ \boxed{10}$ different bracelets.[/hide][/quote]

nice postfarm :)

And I have nothing to add to this thread, as there are plenty of solutions already. Please avoid posting any more solutions that are generally the same as previous ones, as it does not help the thread

[hide="wisely choose what to fix"]
Fix the teal bead, say, at the top. Then there are ${6 \choose 3}$ ways to order the rest. Reflections don't change it, so divide by 2 bc of mirroring to get $\frac{20}{2}=\boxed{10}$ ways.
[/hide]

richard did the key video and its basically the same thing right

Since there's no solution using combinations, I'll add one :)

[hide=solution with combinations]Imagine the bracelet has 7 "slots", each slot containing a bracelet. To place down the 3 orange beads in the 7 slots, we have $\dbinom{7}{3}$ ways to do that. Similarly, placing down the 3 black beads on the remaining four slots is $\dbinom{4}{3}$. Multiplying those together gives us 140, and accounting for division with rotations and reflections, we have $\dfrac{140}{7*2}=\boxed{10}$ solutions. [/hide]


[hide]If $x>1$ then $\text{LHS}>\text{RHS}$, if $0<x<1$ then $\text{LHS}<\text{RHS}$.[/hide]

Let $f(x)=2^{x^{2}+x}-2^{x+1}+log_{2}x$. Obviosly, that $f(x)<0$ when $0<x<1$ and $f'(x)>0$ when $x\ge 1$. Therefore $x=1$ is unique solution $f(x)=0$.

[hide]
It's equivalent to
\[\log_{2}x=2^{x+1}-2^{x^{2}+x}\]
But clearly this doesn't work if one side is positive and the other is negative, this rules out both $0<x<1$ and $x>1$. Therefore $\boxed{x=1}$ is only solution
[/hide]


Prove that if $ (a,b,c)\equal{}1$ then $ (ab,c)\equal{}(a,c).(b,c)$ where $ (a,b)$ is greatest common divisor of $ a$ and $ b$

[hide="Method 1"] If $ v_p(n)$ denotes the greatest integer $ k$ such that $ p^k | n$, verify that $ v_p(a, b) \equal{} \text{min}(v_p(a), v_p(b))$.  The first condition means that for any prime either $ v_p(a, c) \equal{} 0$ or $ v_p(b, c) \equal{} 0$. [/hide]
[hide="Method 2"] Write $ ap \plus{} cq \equal{} (a, c)$ and $ br \plus{} cs \equal{} (b, c)$.  Then

$ (a, c) (b, c) \equal{} ab(pr) \plus{} c (bqr \plus{} aps \plus{} cqs)$

so $ (ab, c) | (a, c) (b, c)$.  On the other hand, if $ (a, b, c) \equal{} 1$ then $ c$ cannot share a factor with both $ a$ and $ b$, so $ (a, c) | (ab, c), (b, c) | (ab, c) \implies (a, c) (b, c) | (ab, c)$. [/hide]


analytic geometry
rotation
calculus
inequalities
geometry
quadratics
number theory

just wrote the 2nd round of our national olympiad. the problems are quite nice compared to other years:

1) let n be an integer represented in the form $(a+b)^2+a-b$ where $a,b$ are positive integers.
a) prove n is even.
b) how many n are representable in that form in the range between 1 and 2005*2005?

2) let k be a circle and let g be a line which is a diameter of k. let P be a point on k and let the tangent through P to k meet g in S; let V be the midpoint of PS. prove that the orthogonal to g through V meets no inner points of k.

3) find all integers $x,y$ satisfying
$\frac{3+x}{6-x}+\frac{2+y}{9-y}=0$,
$\frac{4+x}{9-y}+\frac{1+y}{6-x}=0$.

4) andreas and ben play the following game: they take a regular n-gon($n\geq 4$) and alternatingly color edges or diagonals in their own color(andreas: blue, ben: red). if at the end, there is a triangle with all edges in one color, andreas wins. if not, ben wins. for which n has andreas a winning strategy if
a) andreas begins
b) ben begins?

The first round is posted [url=http://www.hh.schule.de/ifl/mathematik/mo-44-1.htm]here[/url] in German. :)

Quote:1) let n be an integer represented in the form  (a + b)^2 + a - b where a,b are positive integers.
a) prove n is even.
b) how many n are representable in that form in the range between 1 and 2005*2005?

An extremely long, unoriginal, stupid, and inelegent proof of 1a is in the spoiler below.

[hide]

If a is odd and b is even

(odd + even)^2 + odd - even

(odd)^2 + odd - even

odd + odd - even

even - even

even

If a is odd and b is odd

(odd + odd)^2 + odd - odd

(even)^2 + even

even + even

even

If a is even and b is odd

(even + odd)^2 + even - odd

(odd)^2 + odd

odd + odd

even

If a is even and b is even

(even + even)^2 + even - even

(even)^2 + even

even + even

even

[/hide]

Treething, if you know some modular arithmetic, you can approach 1a as follows:

1a:

Lemma: $x^2 - y^2$ has the same parity as $x - y$:

Proof:  $x - y \equiv x + y \bmod 2$ and $a^2 \equiv a \bmod 2$, so that:

$x^2 - y^2 \equiv (x-y)(x+y) \equiv (x-y)^2 \equiv x-y \bmod 2$ as desired.

-----

Problem: $(a+b)^2 + a - b \equiv 0 \bmod 2$:

Proof: $(a+b)^2 + a - b \equiv (a+b)^2 + a^2 - b^2 \equiv 2a^2 + 2ab \equiv 0 \bmod 2$

EDITED to tex

looking at (2), it looks like it's ridiculously easy to bash it with coordinates, as I will do right now because I suck at geometry.

We'll just scale and rotate the circle and line to the unit circle.  Take $(x_0,y_0)$ on the circle so that $y_0^2=1-x_0^2$.  Then $g$ cuts the x-axis at $(\frac {y_0^2}{x_0}+x_0,0)$ (I used calculus, but some similar triangles should do just as well).

WLOG, take $x,y>0$ (the other case is obtained by reflection).

It remains to show that:

$2x_0 + \frac {y_0^2}{x_0} \geq 1$

Substitute $y_0^2 = 1-x_0^2$ and we get that this is equivalent to

$x_0+\frac {1}{x_0}\geq 2$, but this is trivially true from AM-GM

The inequality is strict because $x$ can only approach, not equal, 1.  I have a feeling that there's a somewhat pretty coordinate solution with complex numbers.  I am sure that there is a much nicer solution for people who actually know geometry.

Plenty of time's been spent on 1a, but maybe the following little proof is worthwhile too - it's short with no modular arithmetic.

(a + b)^2 + a - b = a^2 + b^2 + 2ab + a - b = a(a + 1) + b(b - 1) + 2ab

All three terms there are obviously even, since one of two consecutive integers must be even; thus, their sum is even and, of course, n is even.

GBurnham, that's quite cute - mine is a bit overkill, but I wanted to show how it could be done without reducing to cases or even considering a(a+1) or b(b-1) (which you get quite easily if you assume a or b odd).

[hide]2004 numbers can be written under that form (unless I am making some small computational error. It is easy to check that if ,  then f is injective. Indeed, suppose that f(a,b)=f(c,d) => (a+b-c-d)(a+b+c+d) =-a+b+c-d =>

| a+b-c-d | (a+b+c+d) = |-a +b+c-d | < a+b+c+d => |a+b-c-d|<1 => a+b=c+d and also -a+b+c-d = 0 => -a+b=-c+d =>2b=2d => b=d and a=c.

All we need now if to check when  (since f(a,b) is even) that is (a+b-1/2)^2=(a+b)^2 -(a+b)+1/4 <(a+b)^2+a-b <2005^2, thus a+b < 2006. If a+b<2005, then |a-b|<2004, and thus (a+b)^2+a-b<2005^2, so all  pairs of positive integers (a,b) with a+b<2005 fulfill the conditions, and for each f(a,b) takes a different value.

Furthermore if a+b=2005, then a-b<0 is a required condition, which means that only 2005-1/2=1002 pairs verify, so in total we have 2003+1002=3006 possible values of n in the range 1,2005^2. [/hide]

[quote="Peter Scholze"]2) let k be a circle and let g be a line which is a diameter of k. let P be a point on k and let the tangent through P to k meet g in S; let V be the midpoint of PS. prove that the orthogonal to g through V meets no inner points of k.[/quote]

Here is a more geometric solution of this problem: In order to prove that the perpendicular to g through V meets no inner points of k, it is enough to show that the orthogonal projection W of the point V on g lies outside k (in fact, remember that g is a diameter of k). Now, if Q is the orthogonal projection of the point S on g, then the lines SQ and VW are parallel (since they are both perpendicular to g), and since V is the midpoint of PS, it follows that the line VW is a midparallel in triangle PSQ, so that the point W is the midpoint of PQ.

Now, since Q is the orthogonal projection of S on g, we have < OQS = 90; on the other hand, since the line PS is tangent to k at S, we have < OSP = 90, where O is the center of k. Thus, < OQS = < OSP. Clearly, we also have < QOS = < SOP. Thus, the triangles OQS and OSP are similar, and it follows that OQ : OS = OS : OP, so that $OP\cdot OQ = OS^2$. Since O is the center and OS is the radius of k, this equation yields that the point Q is the inverse of P in k. Hence, like any circle passing through two mutually inverse points with respect to k, the circle with diameter PQ must be orthogonal to the circle k. But the center of this circle with diameter PQ is the midpoint W of PQ. Hence, if r is the radius of k and s is the radius of the circle with diameter PQ, we have $OW^2 =r^2+s^2$, so that $OW^2>r^2$ and OW > r, and thus the point W lies outside k. Proof complete.

A nice problem, by the way (usually, DeMO geometry problems are far not that nice).

[quote="Peter Scholze"]3) find all integers $x,y$ satisfying
$\frac{3+x}{6-x}+\frac{2+y}{9-y}=0$,
$\frac{4+x}{9-y}+\frac{1+y}{6-x}=0$.[/quote]

The DeMO problem proposers are very successful in inventing ugly systems of equations. Anyway, this one can be easily solved for any [i]real[/i] numbers x and y, not just integers.

  Darij

[quote="darij grinberg"]

Here is a more geometric solution of this problem

...

[quote="Peter Scholze"]3) find all integers $x,y$ satisfying
$\frac{3+x}{6-x}+\frac{2+y}{9-y}=0$,
$\frac{4+x}{9-y}+\frac{1+y}{6-x}=0$.[/quote]

The DeMO problem proposers are very successful in inventing ugly systems of equations. Anyway, this one can be easily solved for any [i]real[/i] numbers x and y, not just integers.

  Darij[/quote]

Took some restraint to not laugh at my coordinates, huh?  :P

By the way, can I see your solution to the system of equations?  I ended up getting a system in the equations in the variables u = (x-1) and v = (x-4), or something like that that ended up being really ugly.

[quote="blahblahblah"]Took some restraint to not laugh at my coordinates, huh?  :P[/quote]

Hmm, I am almost sure that more than 50% of the students solved it using some kind of coordinates.

[quote="blahblahblah"]By the way, can I see your solution to the system of equations?  I ended up getting a system in the equations in the variables u = (x-1) and v = (x-4), or something like that that ended up being really ugly.[/quote]

Ah well, my solution is ugly, too. Just rewrite the equations

$\frac{3+x}{6-x}+\frac{2+y}{9-y}=0$;
$\frac{4+x}{9-y}+\frac{1+y}{6-x}=0$

in the forms

$-2xy+7x+3y+39=0$;
$-x^2-y^2+2x+8y+33=0$.

Subtracting, you get

$\left(-2xy+7x+3y+39\right) + \left(-x^2-y^2+2x+8y+33\right) = 0$,

what factors into

(y - x - 2) (y - x - 3) = 0.

Thus, either y = x + 2 or y = x + 3. In each of these two cases, plugging into one of the equations above yields a quadratic equation in one variable, so you can find the solution.

  Darij

@darij: i don't think 50% solved it using coordinates. it is a very low level olympiad and most of the students participating don't even know how to solve geometry problems with coordinates or at least are not used to doing so. i know noone doing it(and out of my school, about 50 participated).

Peter

[quote="Peter Scholze"]@valentin: learn counting ;). the pairs $(a,b)$ such that $a,b$ are positive integers and $a+b\leq 2004$ sure is not 2003 :)[/quote]ouch, I was thinking about the pairs that =2004 :D I knew I was making a mistake, was too early in the morning :P


I'm an officer for my math team this year and was wondering if I am allowed to create worksheets containing  AMC problems.  I won't be reproducing entire tests (like giving out the entire 2006 test, 2005 test, etc), but I want to make problem sets containing problems from various places, for example

Problem Set
Question 1: (2006 AMC 12)
Question 2: (1999 AIME)
Question 3: (2002 AMC 10)
Etc. etc.

Is this legal?
Can I also put the problems on our math team website, or is that a privelage that only AoPS can do?

I believe using the problems is legal, as long as they are cited properly. Posting on the internet, I'm not sure. I know people have posted old Mathcounts state competitions on the internet, but AMC might have a different policy...

Distributing the tests among your math team members is legal, though.

Does anyone know the rule about posting the problems on the internet?

bump.  On one of the AIME solution sheets, it says you cannot reproduce the answers, so does that mean AoPS is the only site that can host AMC and other contest questions + solutions?

The general rule is you can ignore copyrights for educational purposes.

The problem with posting stuff on the internet is that you no longer have any control over what they are used for, so you can no longer ignore the copyright. Presumably AoPS has permission from the AMC and other such organizations to have problems on the site (or maybe because AoPS is an educational website they can get around the copyright or something, I am not sure).


LaTeX
number theory unsolved
number theory

Find all terns (x,y,z) of positive integers such that:
1 + 4^x + 4^y=z^2

It should be $ (n,2n\minus{}1,2*4^{n\minus{}1}\plus{}1)$ and $ (n,2n\minus{}1,2*4^{n\minus{}1}\plus{}1)$, $ (2,3,5)$ is not a solution, $ (2,3,9)$ is.
[hide]WLOG $ x\le y$. z is obviously odd. So write:
$ (z\minus{}1)/2 * (z\plus{}1)/2 \equal{} 4^{x\minus{}1}(4^{y\minus{}x}\plus{}1)$
Let $ a\equal{}(z\minus{}1)/2$ and so $ a\plus{}1\equal{}(z\plus{}1)/2$. Note that $ gcd(a,a\plus{}1)\equal{}gcd(4^{x\minus{}1},4^{y\minus{}x}\plus{}1)\equal{}1$. So two cases:

Case 1) $ a\equal{}4^{y\minus{}x}\plus{}1, a\plus{}1\equal{}4^{x\minus{}1}$. Then $ 4^{y\minus{}x}\equal{}4^{x\minus{}1}\minus{}2$, no solutions.
Case 2) $ a\equal{}4^{x\minus{}1}, a\plus{}1\equal{}4^{y\minus{}x}\plus{}1$.
Then $ 4^{y\minus{}x}\equal{}4^{x\minus{}1}$, $ 2x\equal{}y\minus{}1$,$ y\equal{}2x\minus{}1$. Then plugging in we get $ z\equal{}2a\plus{}1\equal{}2*4^{x\minus{}1}\plus{}1$.
So $ (x,y,z)\equal{}(x,2x\minus{}1,2*4^{x\minus{}1}\plus{}1)$
Similarly if $ x\ge y$.
[/hide]

It isnt a right solution, because other tern might be (1;1;3), and then 2y-1=x , 4^x+ 4^y+1 always is a square=z^2. I proved that.  I'll try to download latex and maybe I will attach here my solution of course if i will understand how to work with latex. But my solution isnt finished and i dont know how to do that, whats why i ask her for someone help.

oh damn I haven't revealed your solution before, lol.. Thnx for help


Uglavnom, možda malo preuranjeno... ali, šta ćete... nemam šta raditi.

Ekipu Hrvatske čine (abecednim redom):
Ivana Antoliš, XV. gimnazija (MIOC), Zagreb
Ivo Božić, XV. gimnazija (MIOC), Zagreb
Borna Cicvarić, Gimnazija Andrije Mohorovičića (GAM), Rijeka
Matko Ljulj, XV. gimnazija (MIOC), Zagreb
Irma Telarović, XV. gimnazija (MIOC), Zagreb
Grgur Valentić, V. gimnazija, Zagreb

Ekipu Slovenije čine (valjda, čitam s dmfa.si):
Eva Breznik, I. gimnazija u Celju, Celje
Filip Kozarski, Gimnazija Bežigrad, Ljubljana
Matjaž Payrits, Gimnazija Bežigrad, Ljubljana
Tomaž Stepišnik Perdih, I. gimnazija u Celju, Celje

Sretno svima! :first:

(BTW, posebno skidam kapu ako netko pobijedi Poljake, Mađare ili Nijemce (zadnje dvije zemlje navodno sudjeluju ove godine))

Melkiore šta misliš koliko medalja ćemo uzet? Ja bih rekao dvije, najviše tri, zbog jače konkurencije.

Sretno;)

P. S. pa i nije prerano: MEMO je vec za 7 dana  :)

@hsiljak: da, nisam ni skužio da nisam napisao kad je natjecanje... uglavnom, koliko znam, odlazak iz ZG je u četvrtak

@djuro: ma barem 6 medalja (barem jedno zlato, dva srebra i tri bronce za Hrvatsku, da citiram...) Zezam se, naravno. Konkurencija je jaka, a ljudi ne trebaju razmišljati o medaljama - nakon svega, rad je ono što bi se trebalo cijeniti - ponekad rad urodi nekim konkretnim plodom kao što je medalja, ponekad ne... ali, rad je ono najbitnije. Dobivanje medalje ovisi o desecima faktora, a uloženi trud je, nažalost, samo jedan od njih.

Nam može ko objaviti zadatke, da se malo sprobamo.

Can somebody please post the problems and also the results, esp. for the German team! Is there also a web-page for it? Thanks!

Znam da Hrvati pojedinačno imaju 3 bronce (Ivo, Grga i Matko). Ekipno ništa, ali čuo sam da su zadaci na ekipnom bili loši jer su svi nagurani u malu bodovnu razliku.

Mislim da je granica za srebro bila 24, ako se ne varam. Uglavnom, granice su bile puno više nego prošle godine.

Čuo sam da je bronca bila na 16 bodova.

Meni se čini da su granice iste ko prošle godine plus jedan zadatak, a to je lagana teorija brojeva. Dakle rezultati ne odudaraju mnogo.

Izvinite zbog neznanja, ali šta je MEMO (da li je to Middle European Mathematical Olympiad=Centralno Evropska Matematička Olimpijada)?

Saznao sam nešto više...

Dakle, medalje su bile na redom 16, 24, 32 (zadatak nosi 8 bodova). Dakle na točno 2, 3, 4 zadatka. Primijetite da je za zlato trebalo imati sve bodove. Takvih ljudi je bilo 5.

Na ekipnom su bodovi išli otprilike 32, 32, 32, 28, 26, 25, 24, 22, 21, 8.
Dakle, tri pobjedničke ekipe i ostali jako blizu.

Zaključak: totalno promašeni zadaci.

Djuro, nije li bilo 9 ekipa (Poljska, Njemačka, Austrija, Švicarska, Češka, Slovačka, Mađarska, Slovenija, Hrvatska)?

Ako sam u pravu, onda imaš jednu ekipu viška na ekipnom poretku.  :D E, da, 21 i 8 bodova naspram maksimalnih 32 ne može biti tretirano kao "jako blizu". Ajde još 21, ali 8 ne. (Ne znam o koje se dvije ekipe radi, ali pretpostavljam.)

Dobro, nisu baš svi jako blizu. A ovih 10 score-ova, to je kopirano iz msn razgovora sa Grgom i prema tome vrlo neslužbeno :D

Evo par informacija koje sam povadio s foruma ÖMO (Austrijska matematička olimpijada) - naglašavam da ja [u]ne znam njemački[/u] na nekoj imalo dostojanstvenoj razini, pa tu može biti grešaka - ako se netko hoće mučiti s tim poganskim jezikom, be my guest  :D :

Njemačka, Poljska i Mađarska dijele 1. mjesto u ekipnom poretku s 32 boda (svi bodovi). Pojedinačno, Njemačka ima 1 zlato (dakle, ujedno i perfect score), 3 srebra i 1 broncu. Austrija ima 3 bronce pojedinačno i 26 bodova ekipno.

[quote="hsiljak"]Jes it iz  :)

Usput, evo i zadataka, da kompletiramo thread:

http://www.mathlinks.ro/Forum/resources.php?c=1&cid=114&year=2008[/quote]

 :rotfl:

Cestitam svima!!!


limit
integration
calculus
calculus computations

The limit is 1.

We have, since $\cot t^{2}$ decreases on $(0,\infty)$, that

$\frac{1}{x(x+1)\cot \frac{1}{(x+1)^{2}}}\geq I(x)\geq \frac{1}{x(x+1)\cot\frac{1}{x^{2}}}$, and hence the limit is 1.


geometry
parallelogram
circumcircle
geometry unsolved

A parallelogram ABCD with an acute angle BAD is given. The bisector of Angle BAD intersects CD at point L, and the line BC at point K. Let O be the circumcenter of Triangle LCK. Prove that quadrilateral DBCO is inscribed in a circle. - Bulgaria 1993

[hide]
It is easy to show that $ ADL$ and $ LCK$ are isosceles, and that since $ LCK$ is isosceles, $ LOC \cong COK$. Thus $ \angle DLO \equal{} \angle BCO$, $ DL \equal{} AD \equal{} BC$, and $ OL \equal{} OC$, so that $ DLO$ and $ BCO$ are congruent. Then $ \angle ODC \equal{} \angle ODL \equal{} \angle OBC$, so $ DBCO$ is cyclic.
[/hide]


calculus
derivative
geometry
calculus computations

[quote="flava22"]The area of a circle is decreasing at the rate of 2(pie) cm^2/s. At what rate is the radius of the circle decreasing when its area is 75(pie) cm^2?[/quote]

The area of a circle is $A(r)=\pi r^{2}$ and so $\frac{dA}{dt}=2\pi r\frac{dr}{dt}$ by the chain rule. Now this fact that "the area of a circle is [u]decreasing[/u] at the rate of $2\pi \frac{\mbox{cm}^{2}}{s}$" tells us that
\[\frac{dA}{dt}=-2\pi \frac{\mbox{cm}^{2}}{s}, \]
where the word decreasing indicated that this quantity is negative; we are asked, then, "At what rate is the radius of the circle decreasing when its area is $75\pi \mbox{cm}^{2}$?" that is, what is $\frac{dr}{dt}$ when $A(r)=75\pi \mbox{cm}^{2}$, the latter gives us the value of $r$ to be used, namely
\[A(r)=\pi r^{2}=75\pi \mbox{cm}^{2}\Rightarrow r=5\sqrt{3}\mbox{cm}, \]
now we solve for the desired value, namely $\frac{dr}{dt}$, from the above equations:
\[\frac{dA}{dt}=2\pi r\frac{dr}{dt}\Rightarrow-2\pi\frac{\mbox{cm}^{2}}{s}= 2\pi\left( 5\sqrt{3}\mbox{cm}\right) \frac{dr}{dt}\Rightarrow \frac{dr}{dt}=-\frac{1}{5\sqrt{3}}\frac{\mbox{cm}}{s}\approx-0.11547\frac{\mbox{cm}}{s}\]

neat related rates problem --Ben.


Let $ a(k)\equal{}\frac{(k\minus{}1)!\plus{}1}{k},b(m,n)\equal{}\frac{m^2\plus{}n^2}{mn\plus{}2},k,m,n\in N, \ P'\equal{}\{1,2,3,5,7,11,...\}$ set of primes and $ \{1\}$.
1. Prove, that $ a(k)\in N$ if and only if $ k\in P'$.
2. Prove, that $ b(m,n)\in N$ if and only if $ b(m,n)\equal{}2l^2$, if $ l\equal{}1$ if and only if $ |m\minus{}n|\equal{}2$, if $ l>1$ cfn be sojlved from solve Pell's equation.
3. Let $ Tw(m,n)\equal{}a(m)a(n)b(m,n)$. Prove, that if $ b(m,n)\in N, Tw(m,n)\in N$, then $ m,n\in P', |m\minus{}n|\equal{}2$ (m,n -are twins).
4. If $ Tw(m,n)\in N$ ((m,n) called pceudotwins) and $ m\ge n, (m,n)>1$, then $ n|m\equal{}kn, k|a(n)$. I prove, that in these case $ n>15$ and I don't know exist solution $ Tw(m,n)\in N, (m,n)>1$ or no.
5.  If (m,n) -pceudotwins, m>n, then or (m,n) twins or n is odd prime, $ m|a(n)$. I don't know had $ Tw(m,n)\in N$ solution  with not prime m or not.

1- Use wilson theorem !  http://en.wikipedia.org/wiki/Wilson's_Theorem  :)  then the other I haven't solved yet number 2 seems to be quite interesting
Nice problems
Daniel


Prove that $ \sum_{k\equal{}0}^m\binom{m}{k}\binom{n}{k}\binom{r\plus{}k}{m\plus{}n}\equal{}\binom{r}{m}\binom{r}{n}$

[quote="silouan"]Prove that $ \sum_{k \equal{} 0}^m\binom{m}{k}\binom{n}{k}\binom{r \plus{} k}{m \plus{} n} \equal{} \binom{r}{m}\binom{r}{n}$[/quote]

See http://www.mathlinks.ro/viewtopic.php?t=80551 .

  darij


quadratics
function
symmetry
algebra
domain
quadratic formula

[hide="general strategies"]
$ f(x) \equal{} ax^2 \plus{} bx \plus{} c$

Most of these can be seen from the quadratic formula

$ roots \equal{} \frac{\minus{}b \pm \sqrt{b^2 \minus{} 4ac}}{2a}$

or

$ roots \equal{} \frac{\minus{}b}{2a} \pm \frac{\sqrt{b^2\minus{}4ac}}{2a}$

The axis of symmetry is vertical. The plus or minus sign here makes it clear that the symmetric axis is $ x \equal{} \frac{\minus{}b}{2a}$

The vertex is the point on the axis of symmetry, so
$ (\frac{\minus{}b}{2a} , f(\frac{\minus{}b}{2a}) )$

x intecepts are simply the roots given above
the y intercept occurs when x is 0, so it is $ f(0)$ which equals $ c$

range can be found through the vertex and the concavity. You know the vertex is either the topmost point or the bottommost point. The vertex and concavity also determine where the function is increasing and decreasing.
[/hide]

Just know the definitions of what you're looking for and the problems should be done in 5 minutes at most.

And these seem way too much like homework problems

[quote="sharkman"]For each quadratic function given below, find its vertex, axis of symmetry, y-intercept and x-intercept, if any. Determine the domain and range of each function. Determine where the functions are increasing and decreasing.

(1) f(x) = x^2 - 4x

(2) f(x) = -3x^2 + 3x - 2[/quote]

[hide="vertex"]The vertex occurs at the average of the roots.(Why?)[/hide]
[hide="axis of symmetry"]It's vertical(Why?), and goes through the vertex.[/hide]
[hide="y intercept"]The y intercept occurs at x=0.[/hide]
[hide="x intercept"]The x intercept(s) occur at the root(s).[/hide]
[hide="domain"]This is pretty obvious: all real numbers.[/hide]
[hide="range"]The vertex is the only restriction. Try finding the range with that information.[/hide]
[hide="increasing/decreasing"]The vertex plays a crucial role in this.[/hide]


calculus
function
derivative
Asymptote
limit
real analysis
real analysis unsolved

[color=red] Please read the statement : [/color]

'' If function $f$ has derivative  for all real numbers and it's graph has an orizontal asymptote at $+\infty$ , then $\lim_{x\to \infty}f'(x) = 0$ ''

 I have seen in a book that this is true , if  $f$ has bounded second derivative(using Taylor's formula), but can this be true if $f$ is only once differentiable ?.I'm trying to find a counterexample for a negative answer , but I can find it.
[u] Babis[/u]

[quote="yevon"]Does your functoin have a horizontal asymptote at + infinite?[/quote]
 YES ! The horizontal asymptote is at $+\infty$


we all know that $x$ factorial is $x!$, or $x(x-1)(x-2)(x-3)...(2)(1)$.

let us say that $x$ divisoral is $x?$, or $(\frac{1}{x})(\frac{1}{x-1})(\frac{1}{x-2})...(\frac{1}{2})(\frac{1}{1})$.

if $\frac{x!(x-2)?}{((x-2)!)x?} = 16x^2$, what does $x$ equal?

please show your work. its easy once you get started.  ;)

[quote="fireemblem13"][hide]i think it's $x=0,5$ not sure though...
from what i understand, $x?=\frac{1}{x!}$
substituting this yields

$\frac{\frac{x!}{(x-2)!}}{\frac{(x-2)!}{x!}}=16x^2$

$\frac{x(x-1)}{\frac{1}{x(x-1)}}=16x^2$

$x^2(x-1)^2=16x^2$
$x(x-1)=4x$
$x^2-x=4x$
$x^2-5x=0$
$x=0,5$[/hide][/quote]

I think it is correct

From what I understand, $0$ cannot be an answer because it leads to an undefined negative factorial.

oh, right. so that would make it just $5$.

[quote="buzzer11"]we all know that $x$ factorial is $x!$, or $x(x-1)(x-2)(x-3)...(2)(1)$.

[/quote]

this is not true for x=0...
0!=1

[hide="solution"]x? = 1/(x!)

therefore 16x^2 = x!x!/[(x-2)!(x-2)!] = x^2(x-1)^2

then either 16 = (x-1)^2 ==> x = 5 or x = 0, if we adopt the convention that 0! = 0? = 1. x can't be -3 because facotrials (and likewise, divisorials) are undefined for negative integers.[/hide]

Note that x=0 leads to $(-2)!$ and $(-2)?$ so it's undefined as well.


Hey,

is there a significant difference between the two other than dates? Like... in terms of level of grading, difficulty, etc.

Ummm... why would there be? Supposedly they are pretty much the same in terms of everything except for dates, obviously. Oh yeah and maybe # of students taking them? But yeah they're the same I bet it's somewhere on http://www.unl.edu/amc

[quote="JesusFreak197"]For the AMC, not that I know of. For the AIME, the II is harder than the I. I'm pretty sure this is because the main reason for having two test dates for the AMC is so that some can take both the 10 and 12, but the reason for having two dates for the AIME is just in case anyone couldn't take the first, and it's only fair that they have a harder test for not taking the first one.[/quote]

Huh? Why does that make sense? Although for 2004 this seemed to be the case (and I don't remember anything about years before that), I don't think it was true for 2005. But anyways you should all consider taking my mock USAMO which I am going to make a thread about tomorrow (probably (most likely (I hope (I think (hopefully))))).

It makes sense because there is only supposed to be one test date for the AIME. They were gracious enough in adding a second date that whoever couldn't take the first one really shouldn't complain that it's harder.

I have to agree with probability1.01.

The level of "difficulty" between AMC-10/12 A and B should be same or similar and this applies to AIME A/B. According to you, JF, if your school had a snow day and you couldn't take the test, does that mean that it's YOUR FAULT that you couldn't take AIME A and as a punishment, you should take AIME B without complaining? I don't think so. Only thing is that you need to pay extra money (AIME A is FREE but AIME B requires 20 dollars of charges I think) but the level of difficulty should be same.

There is no way for people to make the two tests EQUALLY hard. Some may be little bit harder than other but the difference should be minor.

[quote="AMCDirector"]I can tell you from having worked and listened carefully in the committee meetings where the A and B versions of the AMC 10 and AMC 12 and the I and II versions of the AIME are hammered out, every effort is made to have the different date tests equivalent in terms of coverage and difficulty.

Because the problems are new and original every year, it is hard to estimate in advance what will be hard and what will be easy.  So your personal perceptions of which is "harder" may vary.  But the intent of the committees is to have contests which are equivalent in difficulty.

Steve Dunbar
AMC Director[/quote]

That's good news, our school is forcing us to take the alternate

One question - Is the USAMO cutoff different for the two different AIME's?

In practice, no.  The two AIMEs are intended to be equivalent, and the score distributions have worked out to be essentially identical since I have been working with them, demonstrating to my satisfaction that the two AIMEs are indeed equivalent.  As a result, I have not had to adjust USAMO qualification levels to account for differences between the two AIMEs.

Nevertheless, I do check for differences and I check for proportional representation on all combinations of qualification.  So far, in my experience it has not affected any of the qualification levels.

Could the qualification levels be different? Yes.  Has it happened yet? No.  Do I expect it to happen?  No, not if the committees continue to do the same excellent job of balancing difficulty they have been doing.

Steve Dunbar
AMC Director

My mistake; I'm sure I heard from someone that the AIME II is intended to be harder, or at least is harder despite the intention. It's possible that what I heard was just that individual person's assessment of the test. *shrug*


calculus
integration
trigonometry
logarithms
geometry
calculus computations

Let $x(t)=\tan t ,\ y(t)=-\ln \cos t\ \left(-\frac{\pi}{2}<t<\frac{\pi}{2}\right).$ Find the area surrounded by the curve $: x=x(t),\ y=y(t)$ and $x$ axis and the line $x=1.$

That's right, you can solve by elliminating prameter $t.$


geometry
trapezoid
parallelogram
trigonometry
geometry proposed

Let O be a point outside a given circle. Two lines OAB, OCD
through O meet the circle at A,B,C,D, where A,C are the midpoints of
OB,OD, respectively. Additionally, the acute angle θ between the lines is
equal to the acute angle at which each line cuts the circle. Find cos θ and
show that the tangents at A,D to the circle meet on the line BC.

It's enough to see that $ABDC$ is an isosceles trapezoid, thus the diagonal intersection $Q$ is collinear with the center $ K$ of the circle and $ O.$ If  tangents to the circle at $ B$ and $ C$ meet at $ P,$ then $ PO$ is the polar of $ Q$ WRT $ (K)$ $\Longrightarrow$ $ PO \parallel BD.$ Note that $ PB \parallel OD$ because of  $ \angle PBO \equal{}\angle BOD \equal{} \theta$ $\Longrightarrow$ $ PBDO$ is a parallelogram and since $ PD$ bisects $OB,$ it follows that $P,A,D$ are collinear. Tangents $p_a,p_d$ to $ (K)$ through $ A,D$ are the polars of $ A,D$ and $ BC$ is the polar of $ P$ WRT $ (K).$ If  $ P,A,D$ are collinear, then $ p_a,p_d,BC$ concur.

On the other hand, median $BC$ of $\triangle ODB$ is equal to its base $BD.$ Then

$ BD^2 \equal{} \frac {_1}{^2}(BD^2 \plus{} OB^2) \minus{} \frac {_1}{^4}OB^2 \ \Longrightarrow OB^2 \equal{} 2BD^2$

$\Longrightarrow \ \cos \theta \equal{} 1 \minus{} \frac {BD^2}{2OB^2} \equal{} 1 \minus{} \frac {1}{4} \equal{} \frac {3}{4}$


Triangle ABC has sides AC, BC and AB measuring 18, 24 and 30 units, respectively. If D is the midpoint of segment AB, what is the length of segment CD?

The angles at $ D$ add to $ 180^{\circ}$ so their cosines add to $ 0$.

Let $ d\equal{}CD$.

The cosine rule gives

$ 24^{2}\equal{} 15^{2}\plus{}d^{2} \minus{} 2 \cdot 15 \cdot d \cdot \cos \alpha$

$ 18^{2}\equal{} 15^{2} \plus{} d^{2} \minus{}2 \cdot 15 \cdot d \cdot \cos (180^{\circ} \minus{} \alpha)$

Adding $ d^{2} \equal{} \frac{24^{2}\plus{}18^{2}\minus{} 2 \cdot 15^{2}}{2} \equal{} 225$

$ d\equal{}15$


64 players play a tennis tournament.How many pairings are possible in the first round?

General version : http://www.mathlinks.ro/Forum/viewtopic.php?t=59921


I'm going to do a problem of the for the Basics forum like sirorange is doing for MOEMS.

So Problem for today:

It is curious that the "false" cancellation shown here gives the right answer $\frac{19}{95}$ cancel the 9's

Can you find a similar 'false equation' wiht the correct value [size=75]$ \frac{2}{5}$[/size]on the right hand side

Wow..only one person replied and that person(Pianoforte) was right

Well here's the next one:9/11OH!

A normal duck has two legs. A lame duck has 1 leg. A sitting dick has no legs. Ninety nine ducks have a total of a 100 legs. Given that these are half as many sitting ducks as normal ducks and lame ducks put together, find the number of lame ducks.

No one replied to my Problem of the Day today! That's very strange.

I couldn't figure it out.
^^^^^^^^^^^^^
Okay I redid it and finally got it. I guess the first time I was unorganized and got lost.
[hide]
$2n+l=100$
$n+l+s=99$
$2s=n+l$
^^^^^^^^^^^
$2n+2l+2s=198$
subtract$2s=n+l$
which is $n+l=198$
which equals$2n+2l=198$
subtract $2n+l=100$

gives you $l=98$
[/hide]

EDIT: I obviously did something wrong. :?  oh well. :P

luce  is correct  :)

Next problem of the day...note: I am going to stop posting problems unless more pwople start answering them! also if there is a reason you are not answering them please tell me!

Today's Problem:
8= 3+5 is the sum of two different prime numbers. What is the smalles whole number which can be written as the sum of two diferent whole numbers in two diffent ways????

[hide]
primes,
2,3,5,7,11,13

16? by guess and check
[/hide]

Yes you two are correct. I'll post a problem when I get back from school.

[quote="nonie"]Problem of the Day: 9/13/05

1990 is exactly 10 times a prime number. What is the next number after 1990 that is exactly 10 times a prime number??[/quote]

[hide]
next prime after 199 is 211

2110[/hide]

That is correct! :) I have decided that tomorrow will be the last day I will post on this thread unless 3 people asnwer to my Problem of The Day.

[hide]
$\frac{1990}{10}=199$

The next prime after 199 is 211 so we get

$211\cdot10=2110$[/hide]

sorry ppl, i forgot to hide my answer last time...

[hide]next prime after 199 is 211, 211x10=2110[/hide]

Both of you are right!  :)  remember atleast 3 people must answer this problem in order for me to continue posting on this thread.

Problem of the Day:

What is the largest prime factor of  2005?

[quote="luce86"][hide]$2005/5=401$[/hide][/hide][/quote]

I believe that is correct  :) ! Too bad I'm not going to be posting on this anymore..( not enough people replied to my last post.)


1. Every day Peter either works to school or takes a bus to school. The probability that he takes a bus to school is 0.25. If he takes a bus. the probability that he will be late is 2/3. If he walks to school, the probability that he will be late is 1/3. The probability that Peter will be on time for at least one out of two consecutive days is ____.

2. Five cards each have a single digit written on them. The digits are 9,9,8,7,6, respectively. The number of different 4-digit numbers that can be formed by placing four of the cards side by side is ___.

3. The number of arrangements of all eleven letters of the word M I S S I S S I P P I in which all the four letters I are consecutive is equal to ___.

[hide="1."]
Using complementary probability:

$ 1\minus{}\left(\frac{1}{4} \cdot \frac{2}{3} \plus{} \frac{3}{4} \cdot \frac{1}{3}\right)^2\equal{}\boxed{\frac{119}{144}}$[/hide]

[hide="2."]
Note that this is same as arranging all five cards and "reading" first four to get 4-digit number. Thus the number of different 4-digit numbers is  $ \; \frac{5!}{2!}\equal{}\boxed{60}$.[/hide]

[hide="3."]Consider all four I's as one object. Now we have 8 objects totally, therefore the number of arrangements is $ \frac{8!}{4!2!}\equal{}\fbox{840}$[/hide]


[quote="FISIKA"]I know that Vietnam is good in Physics, so can someone post the website for the National Physics Olympiad[/quote]
In Vietnamese forums for Physics, members rarely uses English documents. So all problems of National Physics Olympiads haven't been translated into other language, only Vietmese version.


symmetry
geometry
circumcircle
perpendicular bisector
geometry proposed

[hide](a) Let $ P$ be the intersection of $ A_2 A_{11}$ and $ A_3 A_{14}$.  Then $ \angle POA_3 \equal{} \angle PA_3 O \equal{} 20^\circ$, so $ P$ lies on the perpendicular bisector of $ OA_3$.  But triangles $ OA_3 A_6$ and $ OA_3 A_{18}$ are equilateral, so $ P$ lies on $ A_6 A_{18}$.  By symmetry, $ P$ also lies on $ A_1 A_8$ and $ A_4 A_{16}$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/7d7a959242484f29189587d9797109cd.png[/img]

(b) Let $ P$ be the intersection of $ A_3 A_{12}$ and $ A_1 A_6$, and let $ Q$ be the intersection of $ A_4 A_{11}$ and $ A_1 A_6$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/be7400da257b2c3a64b0c1987a72b7ac.png[/img]

Triangle $ QA_4 A_5$ is equilateral, so $ QA_4 \equal{} QA_5 \equal{} A_4 A_5 \equal{} s$, where $ s$ is the side length of the 18-gon.  Also, $ PA_3 \| QA_4$ and $ PQ \| A_3 A_4$, and $ QA_4 \equal{} s \equal{} A_3 A_4$, so quadrilateral $ PQ A_4 A_3$ is a rhombus.

Hence, $ \angle PA_4 A_3 \equal{} \angle QA_4 A_3/2 \equal{} 100^\circ/2 \equal{} 50^\circ$, so $ P$ lies on $ A_4 A_{16}$.  By symmetry, $ P$ also lies on $ A_2 A_8$ and $ A_5 A_{18}$.

It also follows that $ QA_3 A_4 \equal{} PA_3 A_4/2 \equal{} 80^\circ/2 \equal{} 40^\circ$, so $ Q$ lies on $ A_3 A_8$.  And, $ QA_5 \equal{} s \equal{} A_5 A_6$, so $ \angle A_5 QA_6 \equal{} \angle A_5 A_6Q \equal{} 40^\circ$, so $ Q$ lies on $ A_1 A_6$.  Thus, $ A_1 A_6$, $ A_3 A_8$, $ A_4 A_{11}$, and $ A_5 A_{16}$ concur.  This point of concurrency stands out somewhat, because unlike the other points, this point does not lie on a diameter (formed by a pair of the vertices).

(c) Let $ P$ be the intersection of $ A_1 A_{10}$ and $ A_4 A_{15}$.  Since $ OP \| A_{14} A_{15}$ and $ OA_{14} \| PA_{15}$, quadrilateral $ OPA_{15} A_{14}$ is a parallelogram.  Hence, $ OP \equal{} s$.

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/5fb8119bb15e5d11413f14f265bf6e3c.png[/img]

Furthermore, $ \angle POA_{18} \equal{} 20^\circ$ and $ OA_{18} \equal{} r$, where $ r$ is the circumradius of the 18-gon.  Also, $ A_1 A_{18} \equal{} s$, $ \angle A_{16} A_1 A_{18} \equal{} 20^\circ$, and $ A_1 A_{16} \equal{} r$, so triangles $ POA_{18}$ and $ A_{18} A_1 A_{16}$ are congruent.  Hence, $ \angle OA_{18} P \equal{} 10^\circ$, so $ P$ lies on $ A_8 A_{18}$.  By symmetry, $ P$ also lies on $ A_2 A_{12}$ and $ A_5 A_{16}$.

(d) This is the same diagram as part (c).

[img]http://www.artofproblemsolving.com/Admin/latexrender/pictures/5dd40202b3c6c9531389eabea248f9b3.png[/img][/hide]


geometry
3D geometry
probability
combinatorics proposed
combinatorics

just received the problems since i participated last year, but i won't participate this year again. anyway, one nice problem was there(ignoring the informatics part):
consider a cheese from suisse(meaning it has many holes). let it be a cube of 20x20x20 little cubes(or more generally, a x b x c), and any little cube is either cheese or empty(it can be split up in several parts), where cheese appears with probability $p$. now, let come water from above(assume the cheese lies on face down on a table and we get gravity perpendicular to the table). it flows through the cheese, but only from one little cube of air to another another if they shair a common side, not only an edge or a corner. we give water on it until there's no more water fitting in. what's the probability the water reaches the ground?

the original problem of the informatics olympiad is to write a program which approximates that probability by choosing some random examples and then to write some nonsense about the development of the probability when $p$ differs.

Peter

sorry for reviving this really old thread.
I'm bad at percolation theory.
Are you sure this problem actually has a closed answer?


let p ,q integers positives
prove that if p/q =1-1/2+1/3-1/4+..................-1/1334+1/1335 then p is divisible by 2003

just like imo 1979 obviously this can be generalised to whatever number

How do you use the Gaussian trick to conclude? I'm sorry, it's hard for my to jump steps...

$\frac{1}{668}+\frac{1}{1335} = \frac{2003}{\textrm{blahblahblah}}$, where $2003 \, \left( \textrm{a prime} \right) \nmid \textrm{blahblahblah}$. I think that's about it.

There is a more generalized problem
 If p is grater than 3, q=[2p/3] and let m and n be positive integers such that

m/n = 1- 1/2 +1/3 -1/4 + ... +(-1)exp(q-1) 1/q

then m is divisible by p

You just use Catalan Identity And then group since 1/([(q+1)/2]+k) + 1/(q-k)=p/(..)
(where k is 0,1,2...,(q-1)/2 )
And then p divides m

1335 = [2*2003/3]

I think that the Gaussian trick is $\sum_{i=1}^n x_i = \frac12 \sum_{i=1}^n \left( x_i + x_{n+1-i} \right)$. It's quite a well-known trick, but I found out just very recently that it has a name.


[quote]how long does it take for you to receive the result from AMC test? Mine is B at Feb. 16. When would I be abel to see my result? Like a week after it?[/quote]

Probably between 2 and 4 weeks, depending on your school. You're lucky its the B test - the A results come at the same time as the B, so it takes longer.

You can, however, check using the answers that will be provided on this page (and the AMC page) a few days after the contest.

The teachers are usually provided with answer keys so you would be able to calculate your score right after you took the test but it would not be official for another few weeks.