calculus
integration
algorithm
number theory
greatest common divisor
Euclidean algorithm

[quote="theax"]Please help!
"For which k is $ x^3 \plus{} y^3 \plus{} z^3 \plus{} kxyz$ divisible by $ x \plus{} y \plus{} z$?

Please provide the answer AS WELL as a solution.

Thanks in advance.[/quote]

For particular $ k$ which satisfies the condition, we have $ 1\plus{}1\plus{}(\minus{}2)\equal{}0 \Rightarrow 1^3 \plus{} 1^3 \plus{} (\minus{}2)^3 \plus{} k(1)(1)(\minus{}2)\equal{}0$
$ \Rightarrow k\equal{}\minus{}3$ which is a solution after checking.

So $ k\equal{}\minus{}3$ is the only solution.

You would have to check millions of real numbers if you go by that way.

For example, if you take $ x \equal{} 1, y \equal{} 2, z \equal{} 2$, we have $ 5|1 \plus{} 8 \plus{} 8 \plus{} 4k \implies k \equal{} 2 \minus{} 5n$ for all integral $ n$.

[hide="Solution"]$ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)$
$ \implies (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \plus{} x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2)$.

On the other hand, we also have $ (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \plus{} kxyz)$.

Hence, by Euclidean Algorithm, we have $ (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \minus{} kxyz)$.

Also, $ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$
$ \implies (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \plus{} 3xyz)$.

Applying Euclidean Algorithm once again, we have $ (x \plus{} y \plus{} z)|(k \plus{} 3)xyz$.

$ x \plus{} y \plus{} z$ won't be always able to divide $ xyz$, but we can make it always divide $ k \plus{} 3$, when $ k \plus{} 3 \equal{} 0$.

Hence, we will have $ k\plus{} 3 \equal{} 0\implies k \equal{} \minus{}3$.

Answer: $ k \equal{} \minus{}3$

[/hide]

[quote="FantasyLover"]
$ x \plus{} y \plus{} z$ won't be always able to divide $ xyz$, but we can make it always divide $ 3 \plus{} k$, when $ 3 \plus{} k \equal{} 0$.
[/quote]

On a competition, that must be proven. But, it's easy to construct a triple such that $ (xyz,x\plus{}y\plus{}z)\equal{}1$

Take $ x\equal{}y\equal{}z\equal{}1$

[quote="FantasyLover"]You would have to check millions of real numbers if you go by that way.

For example, if you take $ x \equal{} 1, y \equal{} 2, z \equal{} 2$, we have $ 5|1 \plus{} 8 \plus{} 8 \plus{} 4k \implies k \equal{} 2 \minus{} 5n$ for all integral $ n$.

[hide="Solution"]$ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2)$
$ \implies (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \plus{} x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2)$.

On the other hand, we also have $ (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \minus{} kxyz)$.

Hence, by Euclidean Algorithm, we have $ (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \minus{} kxyz)$.

Also, $ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(xy \plus{} yz \plus{} zx)$
$ \implies (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \plus{} 3xyz)$.

Applying Euclidean Algorithm once again, we have $ (x \plus{} y \plus{} z)|(3 \plus{} k)xyz$.

$ x \plus{} y \plus{} z$ won't be always able to divide $ xyz$, but we can make it always divide $ 3 \plus{} k$, when $ 3 \plus{} k \equal{} 0$.

Hence, we will have $ 3 \plus{} k \equal{} 0\implies k \equal{} \minus{} 3$.

Answer: $ k \equal{} \minus{} 3$

[/hide][/quote]

Hi,

I could be wrong, but, I was wondering, you said

"Hence, by Euclidean Algorithm, we have $ (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \minus{} kxyz)$."

And this came from the subtraction of $ x^3 \plus{} y^3 \plus{} z^3 \minus{} kxyz$ from $ x^3 \plus{} y^3 \plus{} z^3 \plus{} x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2$

But when this is done the $ \minus{} kxyz$ turns into a plus kxyz ie $ \plus{} kxyz$, so shouldnt it be

$ (x \plus{} y \plus{} z)|(x^2y \plus{} xy^2 \plus{} y^2z \plus{} yz^2 \plus{} z^2x \plus{} zx^2 \plus{} kxyz)$ ?

I apologize if im mistaken.

edit: I think the statement you made $ (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \minus{} kxyz)$ was ment to be $ (x \plus{} y \plus{} z)|(x^3 \plus{} y^3 \plus{} z^3 \plus{} kxyz)$. This would make my question obsolete. I think its just a typing error?

Nice solution by the way =)

@Bugi

First of all, I must point out that this is not a competition.

Second of all, even if this was a competition, my claim would be proven if I take just one counterexample.

@kurt.math

Yes, I have made a calculation mistake.

Thanks for pointing out. :)

EDIT: It turns out I did misread the problem...

I thought it was $ x^3 \plus{} y^3 \plus{} z^3 \minus{} kxyz$ instead of $ x^3 \plus{} y^3 \plus{} z^3 \plus{} kxyz$...

[quote="theax"]Thanks for the speedy responses!!!!!  :D

But can someone clarify FantasyLover's solution for me?
For example, I didn't really get how the first part $ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2})$ $ \implies (x \plus{} y \plus{} z)|(x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} x^{2}y \plus{} xy^{2} \plus{} y^{2}z \plus{} yz^{2} \plus{} z^{2}x \plus{} zx^{2})$

is used in the solution and where it came from. I also have difficulties understanding where the Euclidean algorithm comes into play (and how you use it) - isn't it used to find the gcd? I'm sorry if I'm mistaken.

In addition, is the first euclidean algorithm just a subtraction?

PLEASE HELP!!!!!![/quote]

Yeh i was thinking that about the "Euclidean Algorithm". It states

for $ \,\, x>y, GCD\left(x,y\right) \equal{} GCD\left(x\minus{}y, y\right)$

which basically means the greatest common divisor of two numbers is the same as the greatest common divisor of their difference. As all the divisors of a number are a divisor of their GCD, then we can say that any common divisors of two numbers are divisors of their difference (and also their sum).

But you could just state the following:

$ if\,\, a|b\,\,and\,\,a|c,\,\,then\,\,a|(b\minus{}c)\,\,and\,\,a|(b\plus{}c)$

And yes, the first application of the "euclidean algorithm" comes from

$ if\,\, a|b\,\,and\,\,a|c,\,\,then\,\,a|(b\minus{}c)$

which is as you said subtraction.

[quote="theax"]But can someone clarify FantasyLover's solution for me?
For example, I didn't really get how the first part $ (x \plus{} y \plus{} z)|(x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2})$ $ \implies (x \plus{} y \plus{} z)|(x^{3} \plus{} y^{3} \plus{} z^{3} \plus{} x^{2}y \plus{} xy^{2} \plus{} y^{2}z \plus{} yz^{2} \plus{} z^{2}x \plus{} zx^{2})$[/quote]
That results from expanding :D

[quote="FantasyLover"]You would have to check millions of real numbers if you go by that way.[/quote]
I don't know what you're trying to say.  stephencheng's solution is completely correct.

Er, okay.

Then consider my solution as a second one  :wink:


floor function
ceiling function
least common multiple
number theory proposed
number theory

Let $k$is fixed positive integer.With any n is positive integer,we assume $g(n)$ is least common multiple of$n+1,n+2,...,n+k$.
Find max$m_k$ and min$M_k$ s.t:
$m_k\le \frac{(n+1)(n+2)..(n+k)}{g(n)}\le M_k$.with all of $n$ is positive integer.

Let $(n+1) \ldots (n+k)=h(n)$, and $f(n)=\frac{h(n)}{g(n)}$. Note that $g(n)$ is the product of the primes dividing $(n+1) \ldots (n+k)$ with maximal exponent, i.e. $f(n)$ is just $(n+1) \ldots (n+k)$, with the maximal exponents taken away. We will now show that only a few primes matter to us, and that we can work them out separately.

Indeed, any prime $\geq k$ will divide at most one of $n+1, \ldots, n+k$. So the maximal exponent in the product will be the only one, and it will be taken away by $g(n)$. Then $f(n)$ will never contain any powers of primes $\geq k$. So we just consider the finite set of primes $\leq k-1$, that is, $2 \leq p_1< \ldots <p_l \leq k-1$.

Now, $f(n)$ is a product of powers of these primes. We will show that each prime works independently, i.e. if a prime $p_i$ has a set $S_i$ of possible exponents in $f(n)$, then we can achieve any value of $S_i$, no matter what exponents are required for the other primes (*).

An observation: for each prime $p_i$, there's an $E_i$ such that the product $h(n)$ contains at most one term divisible by $p_i^{E_i}$ (just take $E_i$ such that $p_i^{E_i} \geq k$). Then the exponent of $p_i$ in $f(n)$ is at most $(k-1)(E_i-1)$, i.e. it is bounded.

Now, for each $n$, let $e(n, j)$ the exponent of $p_j$ in $f(n)$. We will find $n_2$ such that $e(n_2, j)=e(n, j)$ for all $j \neq i$, and $p_i$ has any exponent in $S_i$ that we ask for.

For every $n$ and $j$, let $s(n, j)$ be the maximal exponent of $p_j$ in $h(n)$. Now take $p_j^{s(n, j)+1}=P_j, Q=\frac{P_1 \ldots P_l}{P_i}$. We use the numbers $n+Qx$ for $x=0, 1, \ldots, P_i-1$. These numbers are all congruent mod any $P_j$ with $j \neq i$; it is easy to see, then, that each term in $f(n)$ is fixed mod $P_j$, so the resulting exponents for each prime are fixed too. Also, we achieve all possible residues mod $P_i$, so one of our numbers is congruent to the integer that gives the exponent for $p_i$ that we're looking for. From this, (*) follows immediately.

Phew! Now we work with each prime. Obviously, it is enough to find the greatest and smallest possible power for each of them. Note that, in general, among $k$ consecutive numbers we have at least $\lfloor \frac{k}{a} \rfloor$ and at most $\lceil \frac{k}{a} \rceil$ multiples of $a$. Now, we will first work with multiples of $p$, then of $p^2$, and so on, as we do in the formula for the exponent of $p$ in $n!$.

For the minimum: we can get no less than $\sum \lfloor \frac{k}{p_i^j} \rfloor$ in $h(n)$. Since we're taking floor, we're forbidding powers $p^j$ such that $p^j>k$, that is, which might appear only once. So, let $F_i$ be the largest number such that $p_i^{F_i} \leq k$. With $g(n)$ we can drop no more than $F_i$, and it is easy to see that the above sum of floors minus $F_i$ is best possible. So the min exponent would be $\sum \lfloor \frac{k}{p_i^j} \rfloor-F_i$.

For the maximum, again we can assume that no powers $>F_i$ appear, since there would be at most one and it would be eaten by $g(n)$. Then the max is $\sum \lceil \frac{k}{p_i^j} \rceil-F_i$, with $i$ running only up to $F_i$.

Then $m_k=\frac{n!}{\prod p_i^{F_i}}$, and $M_k$ would be something similar, but less pleasant.


inequalities
ceiling function
function
real analysis
real analysis unsolved

If $ S: \equal{} \{\frac {1}{n} \minus{} \frac {1}{m}: m,n \in N\}$

I need to find the inf(S) and sup(S)

I know that they are -1 and 1. I know how to show that both of these are bounds. For example in the case of the Supremum, I don't know how to show that v<1 there exists a x>v for some x in S.

I tried using density of rationals to show v<$ \frac {1}{n} \minus{} \frac {1}{m}$<1 worked, but it doesn't.  Any help would be greatly appreciated.

[u]Infimum[/u]: To show $ \inf (S) \equal{} \minus{} 1$ it suffices to show two things: (1) $ \frac {1}{n} \minus{} \frac {1}{m}\geq \minus{} 1$ for all $ m,n\in \mathbb{N}$, and (2) for each $ \epsilon > 0$ there are $ m,n$ such that $ \frac {1}{n} \minus{} \frac {1}{m} < \minus{} 1 \plus{} \epsilon$.
(1): $ \frac {1}{n} > 0$ and $ \minus{} \frac {1}{m}\geq \minus{} 1$, so the inequality follows.
(2): Given any $ \epsilon > 0$, take $ m \equal{} 1$ and $ n \equal{} \left \lceil \frac {1}{\epsilon}\right \rceil$, and the inequality follows.
[u]Supremum[/u]: To show $ \sup (S) \equal{} 1$ it suffices to show two things: (1) $ \frac {1}{n} \minus{} \frac {1}{m}\leq 1$ for all $ m,n\in \mathbb{N}$, and (2) for each $ \epsilon > 0$ there are $ m,n$ such that $ \frac {1}{n} \minus{} \frac {1}{m} > 1 \minus{} \epsilon$.
The proofs of (1) and (2) are almost the same as for infimum above, so I leave them as an exercise. :)

Can $ \epsilon$ be any positive number?  If so wouldn't $ n$ no longer be a natural make the proof invalid?

$ \lceil \cdot \rceil$ denotes the [i]ceiling function[/i].  It returns the smallest integer that is greater than its argument, so $ n \equal{} \left \lceil \frac {1}{\epsilon}\right \rceil$ is always an integer, and $ \frac {1}{n} \leq \epsilon$.

Oh wow, sorry there Joe, I completely missed the ceiling of the function.

Last Question:

For part 2 of the infimum:
if $ \epsilon$ = 1
well then you would get get  1-1 < 1-1 in your infimum example, but that is a contradicting statement.

Clearly if $ \epsilon$ = 1 , you can adjust your m to show that there is an element less then 0, but I just wanted to make sure that example you gave would be for all $ \epsilon$ not equal to 1

You're right, I should have chosen $ m \equal{} 1$ and $ n \equal{} \left\lceil \frac {1}{\epsilon}\right \rceil \plus{} 1$, and the inequality would be strict.  For that matter, I could have chosen $ n \equal{} \left\lceil \frac {1}{\epsilon}\right \rceil \plus{} k$ for any $ k\in \mathbb{N}$, and the conclusion would still follow.

thanks, that's ultimately what I did, but I just wanted to make sure.


[b]4[/b]

yay jongao turned into a girl!!!! XD


I have a few quick coding questions:

How do I put spaces in between lines? I've been using "\\" every time I want a line break, and it's kind of inconvenient. If I don't use it, everything will be jammed onto one line (even if I press 'enter' several times in the code)

How do I put text within the dollar signs without it being squished and italicized?

How do I do the binomial sign?

How do I put the i=... at the bottom of the summation/product sign?

THanks a bunch.

1) Try using the \par command
2) the \text{} function

3) \binom{n}{r} (assumes amsmath package which you should always load)
4) use \limits or \displaystyle

What are packages? I used "\use{amsmath}" but the binom thing still doesn't work. Do I have to download the packages or are they automatically there?

It should be \usepackage{amsmath} in the preamble, that is before \begin{document}. amsmath package is almost certainly included in your Latex distribution whichever one you are using.

You can learn all about packages in [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuidePack.php]AoPS Guide[/url]. As it says there, you would probably include in all your documents the lines:
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}
so you get all the symbols and many of the facilities you might need.

Thanks for your help.

What I meant by the line break was that I noticed in other people's tex files (and on AoPS) that if you enter a line break in the code

like so

it appears in the output too. This doesn't happen when I do it in TeXnicCenter. Unless i put the "\\", everything will end up on one line, even if I put in a hundred spaces in the code.

You need not just a line break, you need a totally blank line inbetween.

So

[code]This is a test
of a line break[/code]

will appear all on one line, but

[code]This is a test

of a line break[/code]

will appear on two lines.

Yes, but even if I put 10 empty lines in between sentences in the code, the output will show them next to eachother, and the only way I can skip lines is to use "\\."

What happens if you try [code]\documentclass[12pt]{article}
\begin{document}
First line
Second line

Third line
\end{document}[/code] You should get [quote]First line Second line
Third line[/quote] as output. I have tried this in TexnicCenter. Do you get something different?

Ok, yes that is what I get. However, I notice that some people get line breaks in the output when they put them in the code.


Antonio's Jeep depreciates in value at a rate of 20% per year. If his jeep is currently valued at 9200 approximately how much will it be worth in three years. (round to the nearest dollar)
ok i think i understand bascially how to do it but would u multiply the orginal by 60% then subtract or multiply each new answer by 20%, or if im completely wrong can u explian to me how to do this

If it declines in value by 20 percent, the remaining value is 80 percent of the original (or $0.8$). Then the answer is $9200\times .8\times .8\times .8=(0.8)^3\times 9200=4710.4$.  Then just round that to $4710$ and you have your answer.


I was just wondering how synthetic division works. I know how to divide polynomials but synthetic division is new to me so i was just wondering if someone could help go through the process.

Ex. (4x^3+8x^2-x+1)/(2x+1)

most of the time it only is advantageous when its x+a for some integer a although fractions also work.  Basically you use -a and carry down the first coefficient and then multiply the -a by what was carried down.  You then subtract that from the next coefficient, that is assuming that the degree is one less than the first.  By repeating this you will eventually get your answer and a remainder.  The numbers you have are the coefficients beginning with 1 less than the degree of the original equation.  I hope this helps as it is hard to typeset this correctly.

Well, $x^2+x-2$ factors into $(x+2)(x-1)$ sooooo... I guess you could do synthetic division twice.

I don't know the general method though.

[quote="king_23"]I was wondering if you knew how to divide by a trinomial. ex. (x^2 + x -2)[/quote]

Synthetic division by a trinomial would be very cumbersome.  Just use long division, it works fine.

There is a short cut you can use. I would explain it but I dont know how to do it on the computer.

I will attempt to explain.

Since 2x+1 has 2 terms, you take the 1st 2 terms in the dividend. (4x^3+8x^2). So you divide the first term, 4x^3 by 2x, and get 2x^2. That is the first term of the quotient. Put that on top, and multiply 2x^2 by 2x+1. Subtract that from 4x^3+2x^2 and get 6x^2. Bring down the next term to get 6x^2-x. Again, do 6x^2/2x and get 3x. That's the 2nd term of the quotient. Multiply by 2x+1 to get 6x^2+3x. Subtract from 6x^2-x, then get -4x and bring down the 1. You now have -4x+1. since 4x/2x=2, the final term of the quotient is 2. multiplying and subtracting, you get 2x^2-3x-2 remainder 3.

I hope that made sense
I hope that helped
I hope that was actually what you were looking for.

Still, if I were to divide by a trinomial, I would still use long division to avoid getting confuse with synthetic div...


combinatorics solved
combinatorics
combinatorial geometry

What is the maximum number of regions a convex n-gon is divided into by its diagonals if no three diagonals go through a point ?

I think, answer is C(n,4)+n(n-3)/2+1.

Proof.
We have exactly 1 part if there are no diagonals.
Each inresection adds 1 additional part >> there are C(n,4)
Each diagonal adds 1 part except described above >> there are n(n-3)/2 diagonals.

It's right. Well done, Myth ! One can also write f(n) = Cn(n,4) + Cn(n-1,2)


algebra
function
domain
greatest common divisor
superior algebra
superior algebra solved

Let $\mathfrak{o}$ be a Dedekind domain.
Show that:

a) If there are only finitly many prime ideals over $\mathfrak{o}$, then all ideals in $\mathfrak{o}$ are generated by one element.

b) Every ideal of $\mathfrak{o}$ is generated by at most two elements.

For the second one (something very similar in nature can be done for the first one):

Factor an arbitrary ideal $\mathfrak o$ as $\mathfrak o=\mathfrak p_1^{a_1}\ldots\mathfrak p_k^{a_k}$. Now choose an element $\alpha$ of the domain for which $\mathfrak p_i^{a_i}\|(\alpha),\ \forall i$, let $\mathfrak p_i,\ i\in\overline{1,k}$ and $\mathfrak q_j,\ j\in\overline{1,t}$ be the prime ideals which appear in the decomposition of $(\alpha)$, and choose another element $\beta$ of the domain for which $\mathfrak p_i^{a_i}\|(\beta),\ \forall i$ and $\mathfrak q_j\not|(\beta),\ \forall j$.

The above implies that $\mathfrak o$ is the greatest common divisor of $(\alpha),(\beta)$, so $\mathfrak o=(\alpha,\beta)$.

Is it obvious that you can find an element whose ideal has the precise factorisation stated?

Yes it is: use chinese remainder theorem.


geometry
3D geometry
dodecahedron
probability

A regular dodecahedron is a convex polyhedron with 12 regular pentagonal faces and 20 vertices.  If two distinct vertices are chosen at random, what is the probability that the line connecting them lies inside the dodecahedron?

Hm...all I have is there are $ \binom{20}{2}$ ways to choose two distinct vertices...anyone wanna finish this off?

9 on the same face? I get 12 (i did it the same way though)

Wait why can't we do $\dfrac{\binom52 \cdot 12}{\binom{20}{2}}$?

What i did was to start off to find how many diagionals there were $\binom{20}{2} = 190$ and the i drew a vertex and the three pentagons around it counting that for one vertex there would be $9$ ways the diagional would just go over the surface and $10$ ways it would go through and you could do this on $20$ vertexes which equals $10 \cdot 20$ but here we counted it twice so its $\frac{20 \cdot 10}{2} = 100$ and this will equal  to $\frac{100}{190} = \boxed{\frac{10}{19}}$

I didn't do it the AoPS's way and also, I got this weird power hand-out. I think they should make the AoPS more based on skill than luck. Look attachment.

It is based on skill; it gives you power XP based on how hard of problems you solve. It gave me accuracy and stamina (Because I solve level 25's :-D )

[quote=jbatterson]I saw it differently than the solution given:

The first point does not matter.  There are 19 vertices left.  Of those 19, 9 are on the same face as the vertex that you have already selected (three pentagons share a vertex, a quick sketch makes this easy to see).  The other 10 are not.  10/19[/quote]

I did the same thing

[hide = Easy Solution]

Inspect a vertex on the dodecahedron. There are $9$ segments that connects this vertex to another one that lies on the dodecahedron.

There are $\frac{9\cdot 20}{2} = 90$ line segments that lie on the dodecahedron.

In total there are $\binom{20}{2} = 190$ segments, so our probability is

$$P(A) = 1- \frac{90}{190} = \boxed{\frac{10}{19}}$$

Solution 2 is quite clever!

I counted the number of ways to pick 2 vertices on each of the 12 faces, then accounted for each side being counted twice. 12(5C2) - 30.
(You can multiply the number of faces with the number of sides of each, then divide by two to get the total sides of the polygon: 12 * 5 / 2 = 30.)
Then divide that by 20C2 for the wanted answer.

An alternate way to find the number of edges is to use Euler's Polyhedron Formula, which states that $F+V-E=2$, where $F$ is the number of faces, $V$ is the number of vertices, and $E$ is the number of edges. We can substitute $F$ and $V$ to get $12+20-E=2\implies E=30$. Thus, there are $30$ edges.


Prove:

cos(-A)
_______=secA+tanA
1+sin(-A)

That's division. (Too many parenthese. COuld get messy)

Spoiler:

[hide]sin^2 A + cos^2 A = 1

cos^2 A = 1 - sin^2 A

(cosA)(cosA) = (1-sinA)(1+sinA)

cosA/(1-sinA) = (1+sinA)/cosA

cosA/(1-sinA) = 1/cosA + sinA/cosA

As cos -A = cos A, sin -A = - sin A,

cos(-A)/(1+sin(-A)) = sec(A) + tan(A)[/hide]

Nice

Sorry, bout posting the 2002 AIME questions tho.

BUt i wanted some new problems

But i gave u these trig problems to work on, see? :D

No, I like your posting the questions, since then I won't have to do it myself once Im done with actually taking the AIME.

haha
yea
i've already posted all the ones u might not be able to get

BTW #6 is nice.
Took me 7min to do it ritte now, but its nice.


Solve the equation:
$ \sin(\cot{x})\equal{}\tan(\cos{x})$


ABC is a triangle inscribed in a given circle.  From a variable point P on the circumference, perpendiculars PM and PN are dropped to AB and AC, respectively (M and N occur outside the circle for some positions of P).  Determine the position of P which makes the length of MN the greatest and find this maximum length.

Spolier:

[hide]

MN = AP sin(A) <= 2R sin (A) = BC So max length of MN is 2R sin (A) = BC  when AP is the diameter.

[/hide]

Solution with no trig.

[hide]A, P, M, N must form a cyclic quadrilateral at all times, with AP the diameter of the circle. I claim that AP / MN is constant. (This can be proven with similar triangles.) Therefore, MN is maximized when AP is maximized. AP is maximized when it is the diameter of the circle circumscribing triangle ABC; therefore, at that point, MN is maximized.

Proof of the claim:

Consider the following diagram. A, P, M, N are drawn according to the original problem, with AP the diameter. Angle MAN is fixed. Now, pick any point P' on AP or its extension. Construct circle O' with diameter AP'. The circle intersects AM at M' and AN at N'. Join M'N'. We want to prove that

AP / AP' = MN / M'N'.

Join ON and O'N'. AON and AO'N' are similar isosceles triangles, therefore AO / AN = AO' / AN'. Similarly, AO / AM = AO' / AM'. However, AP / AP' = (2AO) / (2AO') = AO / A'O'. Therefore, AM / AM' = AN / AN' = AP / AP'. Since angles MAN and M'AN' are congruent, we must have that triangles MAN and M'AN' are congruent, hence MN / M'N' = AN / A'N' = AP / AP'. QED.

(I don't know if this result is well known, so I proved it anyway. One can prove a more general result, in which AP doesn't have to be a diameter.)

Notice that the original problem is slightly different; in particular, we don't know that when one moves P around, the new MN we obtain will be M'N'. But since angle MAN's measure does not change, we can construct any chord subtended by an angle of that measure on circle O', and the chord's length will be equal to M'N'.[/hide]


[url=http://www.artofproblemsolving.com/Wiki/index.php/2002_AIME_I_Problems/Problem_4]AIME I Problem 4[/url]

In the solution it says that $ m\equal{}(29)(28)$ and $ n\equal{}28$, so the answer is $ 840$, but the question says that $ m<n$?

just switch them... the solution has $ m\minus{}n$ in the numerator, should be $ n\minus{}m$. everything should be fine with minor revisions.


function
advanced fields
advanced fields theorems

Hey there.

Sorry to bother you guys, and I am not sure whether I am right here. However I have just a problem writing some proofs together, or actually constructing a bijective mapping.

The Problem is as follows:

Assume that you know that for any set A it holds that \[ |P(A)| \geq |A| \], where the bars denote cardinality.

Show that the set \[ R \equal{} \{ \sum_{i\equal{}0}^{\infty}{a_i x^i} | a_i \in \{0,1\} \forall i \in \mathbb{N}_0 \} \]
of all series with coefficients 0 or 1 is not countable. For that, construct a bijective mapping from \[ R \to P(\mathbb{N}_0 ) \] and use what is given above.

I know that the statement  "any set A it holds that \[ |P(A)| \geq |A| \]" is derived using cantor's first diagonal argument, however I am just having trouble formulating a proper function which is such a bijective mapping.

I would say, that the series is entirely defined by a sequence of zeroes and ones, and I would say that my mapping has to do the following: if i have a series, e.g. (0,1,1,0,0...) (only zeroes follow), then my function would map this series into the element \[ a \in P(\mathbb{N}_0) \] which has the elements {1} and {2}.

But how can i write that formally correct and especially, how can i proof that this is indeed bijective.

Thanks a lot for your help,
regards
hmer

That construction is fine- just phrase it in terms of the values of the sequence: $ (a_i)\mapsto \{i: a_i\equal{}1\}$. To show that it's bijective, construct the inverse.

hey jmerry,

super, at least i got that part :-)

I m just thinking on the "correct" formulation.

I would argue like that:
Any series in R is uniquely defined by the sequence of coefficients $ (a_k)_{k \in \mathbb{N}_0} \equal{} (a_1,a_2,a_3,...)$, i.e. a sequence of zeroes and ones.

I was just wondering about the phrasing you suggested, as my function maps the series into the power set of the natural numbers, the function acutally has to be an "infinite" union. What do you say to this formulation?

We define our mapping:
$ f: R \minus{}> P(\mathbb{N}_0)$, with $ f: (R) \equal{} \bigcup_{i\equal{}0}^{\infty}{\{i: a_i \equal{} 1 \}}$.

This is bijective, the inverse would be:

$ f^{\minus{}1}: P(\mathbb{N}_0) \minus{}> R$, with $ f^{\minus{}1}(A) \equal{} \sum_{i\equal{}0}^{\infty}{a_i x^i}$, the $ A \equal{} {a_1,a_2,a_3,...} \in P(\mathbb{N}_0)$. However, I have to specifiy what the $ a_i$'s are. And this, I can only do, if A is finite, right? So only for real subsets of $ \mathbb{N}$. So should I define that inverse "piecewise"?

Do you have some hint on that one?

Thanks a lot for your help,

regards
hmer

All right- you don't understand how to construct the inverse. What you wrote is nonsense; the argument of $ f$ is a sequence and its value is a set, so $ f^{\minus{}1}$ should take sets to sequences. Try a few examples: what sequence gives the set of all even positive integers?

There is no difference between finite and infinite sets in this construction.


calculus
integration
calculus computations


We have a matrix X of order 2 and it has complex inputs. find out all X so that we have X^n=I_2

I_2=
1 0
0 1

How about this?[hide]

Let X =

a b

c d

Then we get the equations:

a2+bc=1

ab+bd=0

ac+cd=0

cb+d2=1

This second and third yield b=0 and c=0 or a=-d.

If b=0 and c=0, then a=-1 or 1 and d = -1 or 1.

If a= -d, we get (a, b, c, d) = (a, b, (1-a2)/b, -a),

where a, b are any complex number.

[/hide]--Dan


Can someone please answer the three title questions? :arrow:  :?:

See [url=http://www.artofproblemsolving.com/Wiki/index.php/Arml]this AoPS Wiki link[/url] and follow the link there for "How to participate."


A convex quadrilateral $ABCD$ with $AB\neq CD$ is cyclic. Quadrilaterals $AKDL$ and $CMBN$ are rhombs with side $a$. Prove that quadrilateral $KLMN$ is cyclic.

Let $ I$,$ J$ be midpoints of  $ AD$,$ BC$
$ KLMN$ cyclic  $ \Leftrightarrow$ $ OM.OK\equal{}OL.ON \Leftrightarrow OI^2\minus{}MI^2\equal{}OJ^2\minus{}NJ^2 \Leftrightarrow OI^2\minus{}(a^2\minus{}AI^2)\equal{}OJ^2\minus{}(a^2\minus{}BJ^2) \Leftrightarrow OA^2\equal{}OB^2\equal{}R^2$
Right!


algebra
difference of squares
special factorizations

Let $a+b=c,abc\neq 0$.Find the value of the following equation.

\[\frac{a^2+b^2-c^2}{2ab}+\frac{b^2+c^2-a^2}{3bc}+\frac{c^2+a^2-b^2}{4ca}\]

Source:2005 Aoyamagakuin University/Law,economics

Is the answer [hide]?[/hide]

It's fairly simple if you just factor the difference of squares and then do some substitution.  I might post the full solution later.

Alright, here's the solution:

[hide]

With all the helpful substitutions in the numerators of the fractions being taken from variations of .[/hide]


Could someone help me with this please.

(c) The line y = k meets the curve y = ax3 + bx2 + cx + d at least four times. Find all the possible values of a,
b, c, d and k.

Sorry the question came out ugly. It's actually this:
a(x)^3 + b(x)^2 + c(x) + d

(a times x cubed + b times x squared + c times x + d)

[hide="Solution"]
The problem states that $y = ax^3 + bx^2 + cx + d - k = 0$ has at least $4$ solutions.  This is a third-degree equation, clearly absurd.  Hence we must have $a = b = c = 0, d = k$.
[/hide]

If you don't mind, could you explain to me. I didn't quite follow. If you mind also, just a step by step solution.

[hide="Detailed solution"]
The lines

$y_1 = k$
$y_2 = ax^3 + bx^2 + cx + d$

Meet at at least four points.  This means that at at least four different values of $x$, $y_1 = y_2$, or

$k = ax^3 + bx^2 + cx + d$

Rearranging,

$ax^3 + bx^2 + cx + (d-k) = 0$

Which is a cubic equation that is supposed to have four solutions.  But because it is third degree, it can only have $1, 3$, or an infinite number of solutions.  Hence it must have an infinite number of solutions.  In other words, the equation is

$0 \cdot x^3 + 0 \cdot x^2 + 0 \cdot x + 0 = 0$

Which means that $a = b = c = 0, d = k$ as previously stated.  This equation has an infinite number of solutions.  For any other choices of $a, b, c, d, k$ we can only have a maximum of $3$ solutions.
[/hide]

[hide="Graphical solution"]
The graph of a cubic function has to go to $+ \infty$ on one side and $- \infty$ on the other.  In the meanwhile, it can "curve" a maximum of two times.  Any line drawn on the same graph as a cubic can only pass through it a maximum of $3$ times unless the line and the cubic are in fact the same graph.
[/hide]

Ah, now i understand you. Thanks a lot buddy!


Find all solutions to $ \sqrt{a}\plus{}\sqrt{b}\equal{}\sqrt{2009}$ in non-negative integers

$ \sqrt{a} \plus{} \sqrt{b} \equal{} \sqrt{2009}$:subtracting $ \sqrt{a}$ and squaring yields
$ b \equal{} 2009 \plus{} a \minus{}2\sqrt{2009a}$ , so clearly $ \sqrt{2009a} \equal{} \sqrt{7 \cdot 287 \cdot a}$ must be an integer.(someone check that 287 is prime and that 7*287 = 2009.) Then, if 287 is prime, $ a$ must have a factor of 287, and a factor of 7, which means a factor of 2009, which would make b=0 from the initial equation. So the only solutions are $ (a,b) \equal{} (0,2009) or (2009,0)$.

[hide="solution"]
$ \sqrt {a} \plus{} \sqrt {b} \equal{} \sqrt {2009} \Rightarrow b \equal{} 2009 \plus{} a \minus{} 2\cdot 7\sqrt {41a}$

Now $ \sqrt {41a}$ must be an intiger, so $ a \equal{} 41c^2$ where $ 0\le c \le 7$

Therefore $ b \equal{} 2009 \plus{} 82c^2 \minus{} 287c \equal{} 41(c \minus{} 7)^2$

So our paris $ (a,b) \equal{} (41c^2,41(c \minus{} 7)^2)$ with $ 0\le c \le 7$
[/hide]

Generally, all integer solutions to
\[ \sqrt a \plus{} \sqrt b \equal{} \sqrt c
\]
are given by $ (a,b,c) \equal{} (km^2,kn^2,k(m \plus{} n)^2)$, where $ k,m,n$ are integers such that $ \gcd(m,n) \equal{} 1$. Prove this! :wink:

I don't think $ \gcd(m,n)=1$ is necessary, e.g.
$ \sqrt{5\cdot 3^2}+\sqrt{5\cdot 6^2}=\sqrt{5\cdot 9^2} \longrightarrow \gcd(3,6)=3$

However, suppose $ m$ and $ n$ are coprime integers greater than zero that are irreducible in surd form.
We show that there are no solutions to
$ a\sqrt{m} + b{\sqrt{n} = c\sqrt{p}}$, for $ a,b,c,p \in \mathbb{N^+}$

Suppose it was true, then
$ a^2m + ab\sqrt{mn} + b^2n = c^2p \Longrightarrow c^2p -(a^2m +  b^2n)=ab\sqrt{mn}$
Now the LHS is an intiger and RHS is irrational, contradiction.

So the only solutions occurs when $ \sqrt{mn}$ is an intiger, i.e. $ m=n$
By a similar argument we find that $ p=m=n$

So it follows that the only solutions to our origional equation are
$ a\sqrt{m} + b\sqrt{m} = c\sqrt{m}$ where $ a+b=c$

The condition $ (m,n)\equal{}1$ isn't necessary, but sufficient. Because your example $ (5\cdot 3^2,5\cdot 6^2,5\cdot 9^2)$ is equivalent to $ (45\cdot 1^2,45\cdot 2^2,45\cdot 3^2)$. Hence $ (km^2,kn^2,k(m\plus{}n)^2)$ with $ (m,n)\equal{}1$ produces all solutions.


Billy Bob takes game theory. His teacher asks how he could find the number(s) with the greatest probability to roll if you roll n dice. He gives an example:

"For three dice, the numbers that you should expect to roll are 10 and 11."

Find the number(s) for n dice.

[hide]The expected value of rolling one die is:
$\frac{1}{6}\left(1+2+3+4+5+6\right) = 3.5$
For 3 die its $3.5+3.5+3.5 = 10.5$. So you expect $10$ or $11$.
For $n$ dice you expect $3.5\cdot(n)$.[/hide]

Computing the expected value (mean) is not enough to determine the numbers with greatest probability (modes).

I found a very unelegant solution that includes balls and urns formula, but I think I am on to a very very elegant solution.

To make sure I am on the right track, if n = 4, is the expected roll 14? and if n = 5, is the expected roll 17 or 18? and for n = 6, is it 21? and n = 7 its 24 or 25?

If so, then I'll post my solution. If not, drats.

IllCrow's solution isn't accurate but the answer is.

Try exploring a simpler case (such as coins - tails is 1, heads is 2, for example).

Both of your answers are right. This is the solution I had:

[hide="solution"]For 2 dice, the expected roll would be 7. For four dice, the expected roll is 14.

For one die, the expected roll is 3 or 4, so approxitmally 3.5 and so on

3.5n[/hide]

I don't get why my approach isn't right.  My approach is the same as 1=2.  The question says "For three dice the numbers you should   [b]expect[/b] to roll 10 or 11...". The expected roll for three dice is actually 10.5, non-integers rolls are impossible and 10.5 is equal distance from 10 or 11.  So you would expect a 10 or 11.  All I did was find the expected value for $n$ dice.

Why did you take the average of 1 through 6? Why not 7 through 14?

You didn't show any patterns...

illcrowflu, your approach is the same as that of 1=2, but neither is correct.  The original problem was to find the number(s) with the greatest probability.  (Later, the original post uses the term "expect", but it's better to avoid that word, because it is too close to expected value.)

Both of you correctly computed the expected value, which is a fancy term for average.  But the expected value is not necessarily the most frequent value (even if we allow rounding).  For example, suppose we have a biased die that comes up 6 with probability $50\%$, and comes up 1, 2, 3, 4, or 5 with probability $10\%$ each.  In one roll, the expected value is 4.5, but that doesn't mean the most frequent value is 4 or 5; the most frequent value is 6.

In the original problem about unbiased dice, it turns out that rounding the expected value does give the most frequent values.  But you need to prove it.

The reason that I think my approach is accurate is because if were dealing with one dice (that isn't biased) were working with a uniform distribution.  When working with $n$ dice ( as $n$ becomes arbitrarily large) the distribution approaches a normal distribution in which mean = median = mode.  Back to the example with three dice, the probability of rolling a 10 or 11 is high since there are several ways that this can occur( i.e. 1,6,4 - 4,6,1 - 6,1,4 - 3,3,4 - 4,3,3 etc...) whereas rolling values higher or lower have less combinations.

Are you sure it's $\frac{3}{2}n$?  For $n=1$, we get $3$ or $4$, but with $1$ die, we don't "expect" anything.


In a play of lottery from $ 01$ to $ 60$ are selected in a draw $ 6$ numbers. Find the probability of which in a draw there is a couple of consecutive numbers.

Note: $ 60$ and $ 01$ are not consecutive, and we want one and only one pair of consecutive numbers.

No. All six numbers must be different. There are $ 60$ balls. And one by one $ 6$ balls are drawn.

Ok, I guess this is a solution
[hide]
We wish to count the number of such occurrences divided by the number of total possible $ 6$ number combinations out of $ 60$. When we count, we can disregard the order which the balls are picked, which would still yield the same answer. Think of the $ 1$ to $ 60$ as 60 slots, and place an $ X$ at every respective slot for the number drawn. For example:
$ \minus{} X \minus{} \minus{} \minus{} \minus{} \minus{} X \minus{} \minus{} XX \minus{} \minus{} \minus{} \minus{} \minus{} X \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} \minus{} ........... \minus{} \minus{} \minus{} X \minus{} \minus{} \minus{} \minus{}$
We try to group 2 $ X$'s together only and try to count the number of such occurrences. If the $ 2$ $ X$'s aren't grouped at the very end, which there are $ 57$ instances of, there would be $ \binom{56}{4}$ ways to choose the $ 4$ remaining $ X$'s. But for the $ 2$ $ X$'s at the 2 ends, we have $ \binom{57}{4}$ ways to choose the $ 4$  remaining. So we have:

$ \frac {57\binom{56}{4} \plus{} 2\binom{57}{4}}{\binom{60}{6}} \equal{} \frac {1485}{3422}$

[/hide]

i got a different result:

[hide]I actually used the the same strategy as jimhu with the slots. But I simply choose $ 5$ slots out of the $ 55$ available slots for the drawn numbers. One of the selected slots contains $ 2$ numbers, the consecutive ones. In each case there are $ 5$ ways to choose this special double slot out of the $ 5$ slots. So I think there is nothing more to do and we get: $ \frac {5\binom{55}{5}} {\binom{60}{6}} \equal{} 0.34743...$   [/hide]

Please correct me if I am wrong!

wayne

[quote="Wayne10"]

[hide]I actually used the the same strategy as jimhu with the slots. But I simply choose $ 5$ slots out of the $ 55$ available slots for the drawn numbers. One of the selected slots contains $ 2$ numbers, the consecutive ones. In each case there are $ 5$ ways to choose this special double slot out of the $ 5$ slots. So I think there is nothing more to do and we get: $ \frac {5\binom{55}{5}} {\binom{60}{6}} \equal{} 0.34743...$   [/hide]

[/quote]

Why is there only 55 slots to choose from? And I dont think did you mentioned what would happen if two slots are together at both ends.

when you place the balls which were not drawn in a row there will be a row with $ 54$ balls. Now there are $ 53$ possibilities to place a drawn number (or the double pack) between two not-drawn numbers and $ 2$ possibilities to place them directly at an end. So all in all there are $ 55$ possible slots where you have to select $ 5$, $ 4$ of them get one ball and $ 1$ gets two. When you have chosen all $ 5$ slots you start to number the balls in the row (now there are $ 60$) beginning to count from one end to the other and I think that it's possible to get all outcomes.

For example:

$ \minus{}\minus{}\minus{}X\minus{}\minus{}\minus{}\minus{}\minus{}\minus{}X\minus{}\minus{}\minus{}X\minus{}\minus{}\minus{}\minus{}\minus{}...\minus{}\minus{}\minus{}\minus{}\minus{}X\minus{}\minus{}XX$

this selection of slots would define the drawn numbers to be $ 4, 11, 15, 56, 59, 60$

I think you're under-counting here. What happens if you have:

$ ...\minus{}XXX\minus{}...$

In this case, you have both a single and the double (or three singles) in the same slot. Since you assign each "ball" a distinct slot, this is impossible.

Jimhu's solution over-counts, because it counts each of these as separate cases:

$ ...\minus{}(XX)X\minus{}...$

$ ...\minus{}X(XX)\minus{}...$

I'm not sure if I'm right or how to correct for it if I am...

huh? Why do you have 3 $ X$'s in a row? I only considered at most 2 $ X$'s in a row..  :huh:

yes me too. are 3 consecutive numbers a "pair"?

All the X's are leading to confusion in this thread (even if it helps you personally solve the problem). So here's my attempt at an explanation with no X's:

Step 1: Choose 5 numbers from 1 through 55. There are $ \binom{55}5$ ways to do this.
Step 2: For each of the 5 chosen numbers EXCEPT the largest one, add a number in between that number and the next number. For example, if you chose 37, and its not the largest chosen number, add a little $ 37\frac 12$ between 37 and 38.
Step 3: Renumber you're 59 numbers with the numbers 1,2,...,59. You now have 5 chosen numbers from 1 through 59 with at least one other number in between any two chosen numbers.
Step 4: Pick one of the 5 chose numbers to be the SUPER number. There are FIVE ways to do this. Add an extra number after the SUPER number, but before the next number. For example if the SUPER number is 19, then add a $ 19\frac 12$ in between 19 and 20.
Step 5: Renumber you're 60 numbers with the numbers 1,2,...,60. If you count the number after the SUPER number as a chosen number, then you have 6 chosen numbers with one and only one pair of consecutive numbers.
NOTICE that there was only $ 5\binom{55}5$ ways for that to happen.
Thus, the probability that it does happen is $ \frac{5\binom{55}5}{\binom{60}6}$

Hopefully this is similar to Wayne10's train of thought :)

Wayne's solution doesn't count that possibility, since for each slot, he assigns ONLY 1 single/double. More than 1 ball can go in any slot.

For jimhu, the problem is that an $ X$ in a pair as you assign it is indistinguishable from an $ X$ on its own.

[quote]we want one and only one pair of consecutive numbers[/quote]
i'm not positive, but i think that if you have 3X's in a row, which represents 3 numbers in a row, then that means you have 2 pairs of consecutive numbers

if that's the case, then you don't need to count it

Whoops. Missed that part of the question.

Heh heh.


This is a problem I created out of a desire to see if I could create a circle with the golden ratio, rather than a rectangle.  I came up with an answer that I think is right, and I'll be interested in seeing if everyone else gets the same answer.  As far as I can tell, a graphing calculator (or an equivilent computer program) is needed to solve the problem, and the final answer must be rounded.

Problem:

In a circle of radius 1 drawn (not to scale) below, areas A and B are separated by a chord.  Point M is the midpoint of the chord and line segment LM (not drawn) is perpendicular to the chord.  The ratio of the area of A to the area of B is equal to the ratio of area B to the area of the entire circle.  What is the length of line segment LM?

[img]http://img.photobucket.com/albums/v223/BrianNight/MathChallengeDiagram.jpg[/img]

[hide="Solution"] Let the length of segment $LM$ be $x$.  Let the given chord be chord $CD$.  Let $CM = MD = y$.  Then

$(1-x)^{2}+y^{2}= 1 \implies y = \sqrt{2x-x^{2}}$
$[A] = \text{arctan }\frac{ \sqrt{2x-x^{2}}}{x}-(1-x) \sqrt{2x-x^{2}}$
$[B] = \pi-[A]$

We are given that

$\frac{[A]}{ \pi-[A] }= \frac{ \pi-[A] }{ \pi }\Leftrightarrow$
$\pi [A] = \pi^{2}-2 \pi [A]+[A]^{2}\Leftrightarrow$
$[A]^{2}-3 \pi [A]+\pi^{2}= 0 \Leftrightarrow$
$[A] = \pi \cdot \frac{ 3 \pm \sqrt{5}}{2}$

I don't think $x$ can be expressed in terms of elementary functions. [/hide]

I completely agree about the area of A, but I'm a bit confused about the first part.  How did you know that x^2 + y^2 = 1?

Also, I entered the equation you came up with for [A] and the value of [A] into the calculator and found the intersection, it yielded that [i]x[/i] is about equal to .976.  My answer (using an entirely different approach) was about[hide="answer"].814[/hide].  Of course, yours may very well be right, but could you explain how you came up with the equation for [A]?

Sorry, I messed up my variables there.  It should be correct now.

That makes more sense, but I'm still not sure how you got from finding [i]y[/i] in terms of [i]x[/i] to getting the equation for the area of A.  I know I'm probably not seeing something obvious, but if you wouldn't mind, could you explain to me how you came up with that equation?

[quote="Ad Infinitum"]That makes more sense, but I'm still not sure how you got from finding [i]y[/i] in terms of [i]x[/i] to getting the equation for the area of A. [/quote]

[hide="Explanation"] As before the chord is $CD$.  The area given by $[A]$ is the difference between the area of sector $CD$ (the area of a sector of measure $\theta$ in a circle of radius $r$ is $\frac{1}{2}r^{2}\theta$) and the area of $\triangle CBD$. [/hide]

Thanks, it makes a lot more sense now.  In fact, my solution was quite similar (using the fact that the triangle plus [A] equals the area of the sector) but I still got a different answer.

Since the equation for a sector has a one half in it, shouldn't that cancel with the 2 in the "2arctan"?  And also, did you mean (x-1) (the height of triangle CDB- B meaning the center of the circle) instead of x in the later parts of the equation?

Yes, you're right on both counts  :oops:


linear algebra
matrix
symmetry
geometry
angle bisector
geometry unsolved

Let $ ABC$ be an isosceles right triangle and $M$ be the midpoint of its hypotenuse $AB$. Points $D$ and $E$ are taken on the legs $AC$ and $BC$ respectively such that $AD=2DC$ and $BE=2EC$. Lines $AE$ and $DM$ intersect at $F$. Show that $FC$ bisects the $\angle DFE$.

$ N: AE \cap CM$, $ G: CF \cap BA$. For Menelaus on CMB with trasversal NE we obtaid $ \overline{CN}=\overline{NM}$ and for Ceva on ACM and point F we obtain $ \overline{AG}=2 \overline{GM}$ and so $ \overline{GB}=\overline{GA}$. For Menelaus on AEB and trasversal GF we have $ \frac{\overline{AF}}{\overline{FE}}= \frac{3}{2}$ and so $ \frac{\overline{AF}}{\overline{AE}}= \frac{3}{5}$. Let $ CA=CB=x$ so $ \overline{AD}=\frac{2}{3} x$ and $ \overline{AE}= \frac{\sqrt{10}}{3} x$ so $ \frac{\overline{AE}}{\overline{AD}} = \frac{\sqrt{10}}{2}= \frac{\overline{AC}}{\overline{AF}}$ so AFD and ACE are similar and $ AE \perp DM$, then DCEF is cyclic with DC = CE so C stay on the angle bisector of $ \angle DFE$

See that AM^2-AD^2=ME^2-DE^2, therefore AE and DM are perpendicular, CDFE cyclic with CD=CE, so CF is angle bisector of DFE.

Best regards,
sunken rock

Let $C(0,0)$, $B(0,a)$ and $A(a,0)$.
Easily, $D(\dfrac{a}{3},0)$, $E(0,\dfrac{a}{3})$, $M(\dfrac{a}{2},\dfrac{a}{2})$
Let $F(p,q)$. By solving a system,
$\begin{vmatrix}
p & q &1 \\
\dfrac{a}{2} & \dfrac{a}{2} & 1 \\
\dfrac{a}{3} & 0 & 1
\end{vmatrix}=0$

$\begin{vmatrix}
p & q &1 \\
0 & \dfrac{a}{3} & 1 \\
a & 0 & 1
\end{vmatrix}=0$

we easily get $p=\dfrac{2a}{5}$ and $q=\dfrac{a}{5}.$
$k_{EF}=-\dfrac{1}{3}$
$k_{CF}=\dfrac{1}{2}$
$k_{DF}=3$
$tg(\angle EFC)=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\dfrac{1}{3}}=1$
$tg(\angle CFD)=\dfrac{3-\dfrac{1}{2}}{1+3\cdot\dfrac{1}{2}}=1$
$\implies \angle EFC=\angle CFD$ and we are done.

Wait isnt this completely obvious by symmetry or am I messing up my diagram? Isosceles right triangle?

Because reviving the thread:
Take $N$ symmetrical of $B$ about $C; D$ is the centroid of $\triangle ABN\implies N-D-M$ collinear, so $\angle ADM=\angle CDN$, but $\angle CDN=\angle BDC$ by symmetry; also by symmetry $\angle AEC=\angle CDB$, so $\angle ADM=\angle FEC$ and $CDFE$ is cyclic; with $CD=CE$ we are done.

Best regards,
sunken rock


In the metric space $ (\mathbb R , d)$ , if we define :

1) $ d(x,y) \equal{} |exp(x) \minus{} exp(y)|$ , where $ x, y$ are real numbers .

2) $ d(x,y) \equal{} |{tan}^{ \minus{} 1}(x) \minus{} {tan}^{ \minus{} 1}(y)|$, where $ x, y$ are real numbers .

Show that $ (\mathbb R, d)$ w.r.t (1) and (2) are incomplete metric space . Also, what is the completion space of both $ (\mathbb R, d)$ w.r.t. (1) and (2).

I appreciate any help.

$ x_n\triangleq \minus{} n$ is a cauchy sequence in both metrics which does not converge.  The completion should be $ [\infty,\infty)$ in the first case and $ [\infty,\infty]$ in the second, both equipped with the metric $ \widetilde{d}$, where $ \widetilde{d}(x,y) \equal{} d(x,y)$ if neither $ x$ nor $ y$ are $ \pm\infty$, and where $ \widetilde{d}(x,y) \equal{} \infty$ if $ x$ or $ y$ are $ \pm\infty$.

e: sorry what i wrote is obviously wrong, in the first case we should have $ \widetilde{d}(x,\minus{}\infty) \equal{} e^x$, and the modifications are 'obvious' for the second one as well.

A metric is never infinite.

The metric on the completion is the usual one: $ \tilde{d}(x,y) \equal{} \lim_{n\to\infty}d(x_n,y_n)$ where $ x_n$ and $ y_n$ are cauchy-sequences in $ X$ converging to $ x,y\in\hat{X}$.

The completions are isometric to $ [0,\infty)$ and $ [\minus{}\frac{\pi}{2},\frac{\pi}{2}]$ with the absolute-value-metric.

You can let it have an infinite value by just setting $ \rho(x,y) \equal{} \frac{d(x,y)}{1\plus{}d(x,y)}$ instead

The metric $ \rho$ is quite different from $ d$, even though it defines the same topology.

Ohps, yeah. I forgot that other things are important too  :P


It has been posted many times before.

[hide="Hint"] Let $S$ equal the sum.  Creatively evaluate $2S - S$.  :) [/hide]


algorithm search trigonometry function articles

Well, you could always try Mathematica or Maple, if you can define the number.

There is a simple algorithm for computing square roots of numbers with known decimal representations.  (You lose half the digits each time, so it is best for square roots of rational numbers.)  A search for "square roots by hand" would probably be sufficient to find the algorithm, and it would be not-too-difficult to implement in any decently respectable programming language.  There are many, many series known that will give you values for $ \pi$, with varying rapidity of convergence.  (Taylor series of inverse trig functions I think give pretty quick convergence.)

[quote="BaBaK Ghalebi"]Hi every one,I dont know whether this is the right place to post this or not...  :oops:
anyway I am searching for a program which calculates some irrational numbers upto as many decimal numbers as possible...
such as $ \sqrt{2},\sqrt{3},\sqrt{1386},\pi,\frac{\sqrt{2}}{\pi},\frac{\pi}{\sqrt{2}}$
also if someone had a program and would just give me the representations,that would be soooo great and helpful and nice for me too...  :roll:
thanks...[/quote]

What is exactly your need : do you want to calculate decimals of $ \sqrt{2}$?, or do you want to have examples of decimals of some irrational numbers, whatever they could be ?

In this last case, you just to have to choose any sequence of decimals, with the only caution that these sequences are not periodic.

For example $ 0.1010010001000010000010000001\cdots$

well thanks guys for your helps...  :lol:
but actually I need to calculate the following numbers upto as much decimal numbers as possible...
$ \sqrt{2},\sqrt{3},\sqrt{1386},\pi,\frac{\sqrt{2}}{\pi},\frac{\pi}{\sqrt{2}}$
in other words,I need the decimal representations of the above numbers upto as much decimal numbers as possibel,for example upto 100 or 1000 decimal numbers...
and I need only the above numbers,not any arbitary,just the 6 numbers that I mentioned above,so if anyone could help me to find them (by givving the representation or to introduce a program itself) it would be very nice and helpful for me...  :roll:
I hope I explained it well...  :maybe:
you know many calculators,calculate $ \sqrt{2}$ upto only $ 16$ decimal numbers,but thats not enough for my work I need as much as possible,it should be calculated somewhere...  :maybe:
I remember there was an somehow artivcle which had computed $ \pi$ upto many many decimal numbers,it was like $ 2$ pages long...
where can I find that and also the $ 5$other numbers?
if only somebody could help me find them...  :roll:

I have searched the web and I have found a useful link,I need something like this:
[url]http://www.worldwideschool.org/library/books/sci/math/MiscellaneousMathematicalConstants/chap43.html[/url]

[quote="Kyron"]Try this Wikipedia article section for your square root ones

[url]http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Digit_by_digit_calculation[/url][/quote]
no,as I said before I need as many decimal numbers as possible,like 500 decimal numbers...
like the links mentiond by me or by JBL (2,3 posts above yours)...
thanks anyway its not necessary anymore... :)

[quote="BaBaK Ghalebi"][quote="JBL"]http://www.research.att.com/~njas/sequences/A002193
http://www.research.att.com/~njas/sequences/A000796

You can find many other constants there by putting in the first few digits and following them with the text "keyword:cons"[/quote]
woooooooooooow thank you very very much,that was great...  :lol:
I have found $ \sqrt{3}$ to,but ther are still 3 numbers which I couldnt find in the link...
$ \sqrt{1386},\frac{\pi}{\sqrt{2}},\frac{\sqrt{2}}{\pi}$
if I only could find these three numbers,it would be really greattttttt...[/quote]

I'll make a program and calculate it for you. Gimme a few days.

wow,thank you,that would be really greattttt...  :lol:  :roll:

wow,thanks thats awesome :w00t:
I really appreciate that

O ya, i just realized i didn't need to make the program for that.
There's one online:
http://web.ukonline.co.uk/home52365/bcalc.htm


Let $F_n$ be the Fibonacci sequence defined by $F_0 = F_1 = 1$ and $F_{n+2} = F_{n+1} + F_n$ for all $n \geq 0$.
Find all pairs $(a, b)$ of real numbers such that for each $n$, $aF_n + bF_{n+1}$ is a member of the sequence.
 Find all pairs $(u, v)$ of real numbers such that for each $n$, $u{F_{{n}}^2 + v{F_{n+1}}^2}$ is a member of the sequence.

Solution and discussion are in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=154386
I think moderator should close this


geometry
3D geometry
geometric transformation
rotation
function
polynomial
LaTeX

The game is basic.

Post #1: 1 in abstract algebra, is the multiplicative identity.

800th post!!!!!!!!!!
w00t

3 is one of the nicest numbers in number theory, even though it can get ugly.

4=100

The message is too small. Please make the message longer before submitting.

5 was my favorite number in...second grade.

6:

- a perfect number
- the product of the 2 smallest prime numbers
- the sum of the first 3 positive integers
- a factor of 42
- the last digit in my username
- how old i was 6 years ago

7 is a solution to the differential equation $ \frac{\text{dy}}{\text{dx}}\equal{}0$ :P

8: the first cube that is one less than a square.

9: The first square that is one more than a cube.

11: The first palindromic 2 digit number

12: 2 less than the smallest number whose square has not been proven to not be a Lychrel number

14: How old I will be in the middle of December.

15:How many points I got at the IMO this year  :D

751-Pepin The Short was elected as King..:P

$ \color{blue} \huge 752 \equal{} 2^{4} \times 47$
 :play_ball:  :play_ball:  :play_ball:

753: the sum of the first and last digits are double the middle digit :ninja:

760 is an area code in California.

760 is divisible by 10 & 76.
:|

[quote="aleph0"]169 equals 100+69, 69 looks like an astrological sign on its side
169 is relatively prime
169 is a :( number
169 is 13^2
169 backwards, 961, =31^2 and 31 is 13 backwards[/quote]

169 is also the largest number that cannot be expressed as the sum of 5 positive squares.

764 is half of 1528, which is the year that the first known Europeans set foot on the shores of modern day Texas.

765 is an odd number with a decreasing order of digits.


Write the matrix in row-echelon form. $ \left(\begin{array}{ccc}0 & 1 & 1\\ 1 & 2 & 3\\ 2 & 2 & 2\end{array}\right)$

What is row-echelon form?

Although I'm not sure how to do this myself:

http://en.wikipedia.org/wiki/Row_echelon_form
this might help

You use various matrix operations, like multiplying a line by some number and adding or subtracting it to another line.
You can also switch rows, and other stuff.
Basically, you reduce it to reduced row echelon or row echelon so its easier to solve it. You then can use Gaussian elimination or Gauss-Jordan.

ok, i learned this during the first few weeks of algebra 2 that i took this year... but i dont remember much (i have terrible memory)

if i understand you correctly (unless im mistaken)
u mean reduced row echelon form...
it is used to solve equations...
which is a useless method except for computers and calculators.
you try to get 1's on the main diagonal of the matrix, and 0's on all of the other stuff....

then the values that u have on the very right side are the values of the variables so that the equations balance (sorry if im doing a bad job at explaining by being too vague)

but isnt there supposed to be one more collumn than row?

(sorry if im completely wrong i have a terrible memory)


[quote="cezar lupu"]Let $a,b,c>0$.Is this true $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq (a+b+c)\sqrt\frac{a+b+c}{3abc}$? ;)[/quote]

Alas, it's wrong. Counterexample: a = 1, b = 46, c = 5.

  darij


If the number is $n$, is there any formula for finding the product of all positive integer divisors of $n$?

It depends on the number of divisors.  I think you can say that if there are $d$ divisors, then their product would be $n^{\frac{d}{2}}$.

[hide="hint"]
Think about symmetry... of divisors, of course.[/hide]

[quote="Magixmaniak"]unless n is as perfect square and then product would be n^((d-1)/2+1 )[/quote]Oh yeah, forgot about that.  I meant if the prime factorization is $p_{1}^{n_1}\times p_{2}^{n_2}\times p_{3}^{n_3}....$ it would be $n^{\frac{(n_1+1)(n_2+1)(n_3+1)....}{2}}$.


function
limit
calculus
topology
real analysis
calculus computations

Prove that a continuous function $f : \left[ a,b \right] \to \mathbb R$ is bounded and that we have $\sup \left( \textrm{Im} f \right) \in \textrm{Im} f$.

Let me give a try:

Suppose on the contrary that $f$ is not bounded and let $I_1=\left[a,b\right]$. Now form a decreasing sequence $I_n$ of subintervals $I_1\supset I_2\supset I_3\supset\cdots\supset I_n\supset I_{n+1}\supset\cdots$ such that $f$ is not bounded in $I_n$. Then take intersection and we have one limit number $\alpha=\bigcap_{n=1}^{\infty}I_n$. Because $\alpha$ is a limit and $f$ is continuous at $\alpha$, there is open set $O$ containing $\alpha$. So $f$ is bounded in $O$. But $O$ contains some interval $I_k$ because $\alpha\in I_k$ and we know $f$ is not bounded in $I_k$! Contradiction. So $f$ is bounded.

$\sup$ means least upper bound so if we do the same like before by letting $J_n$ be a decreasing sequence of subintervals of $J_1=\left[a,b\right]$, $J_1\supset J_2\supset J_3\supset\cdots\supset J_n\supset J_{n+1}\supset\cdots$, such that the least upper bound $U$ of $f$ is in each $J_n$, then there exists a number $\gamma\in\left[a,b\right]$ such that $\gamma=\bigcap_{n=1}^{\infty}J_n$ and then $f(\gamma)=U$. So $\sup(\textrm{Im }f)\in\textrm{Im }f$

well ,correct me if i'm wrong but I believe that there's this theorem , by weierstrass that states
"a continous function defined on a compact interval is bounded and reaches its bounds on that interval"
this pretty much solves the problem.... ;)

In fact this theorem was to be proved. :D

How do we prove that the "length" of $I_k$ goes to $0$ as $k$ goes to $\infty$?

actually i don't think we need to prove that...all we need is a common intersection of $I_k$

we can prove that it's bounded by assuming the contrary ...we get that the limit $x_n$ included in [a,b] isn't in this interval...but $x_n$ "belongs" to this interval and therefore has a convergent subsequence with its limit in [a,b]-> contradiction

[quote="blahblahblah"]1) The continuous image of a compact set is compact.  (this is easy to prove)

2) In $\mathbb{R}^n$, a set is compact iff it is closed and bounded.  We only need one direction - a compact set is closed and bounded.

3) A closed and bounded set in $\mathbb{R}^n$ contains its $\sup$ and $\inf$.[/quote]

[b]Definition[/b] A topological space is called compact iff all its open covers have a finite subcover.

I don't understand how this works. You say that $[0,1]$ is compact, right? How can it have an open cover?

$(-1,2)$ is an open cover of $[0,1]$ . for the case of a compact in $\mathbb{R}$, the proof for the fact that it is closed and bounded, is not hard.

[hide="Here is the Weierstrass - theorem and the its proof."]
\[ \mathrm {The\ level\ : \ for\ high\ school\ students\ !}  \][color=darkred][b]The Weierstrass' theorem.[/b] Let $f: [a,b]\rightarrow R$ be a continue function. Then the function $f$ is bounded and it touches the its (lower and upper) bounds.

[b]Proof[/b] ([i]in five steps[/i]).Denote the upper bound $s$ and the lower bound $i$, where

$s\equiv\sup_{a\le x\le b}f(x)=\sup Im(f)=\sup \{f(x)|a\le x\le b\}\in \overline {Im(f)}$
$i\equiv\inf_{a\le x\le b}f(x)=\inf Im(f)=\inf \{f(x)|a\le x\le b\}\in \overline {Im(f)}$

$\blacktriangleright$ The point $s$ is a adherent point for the set $Im(f)$
(definition)$\Longleftrightarrow$ exists a sequence $(x_n)_{n\in N^*}\subset [a,b]$
so that $f(x_n)\rightarrow s\in \overline R\equiv [-\infty,\infty ]$ (finite or infinite).

$\blacktriangleright$ The sequence $(x_n)_{n\in N^*}$ is bounded (from the [u]Cesaro's lemma[/u])$\Longrightarrow$
there is a convergent subsequence $(x'_n)_{n\in N^*}$ of the sequence $(x_n)_{n\in N^*}$,
i.e. $x'_n\rightarrow$ $x_s\in R$ (finite).

$\blacktriangleright$ The subsequence $(x'_n)_{n\in N^*}\subset [a,b]\Longrightarrow x_s\in [a,b]$.

$\blacktriangleright$ The function $f$ is continue on $[a,b]$ $\Longrightarrow$ the function $f$ is continue
in the point $x_s$ (definition) $\Longleftrightarrow$ $\lim_{n\to \infty} f(x'_n)=f(x_s)$.

$\blacktriangleright$ The sequence $\left(f(x'_n)\right)_{n\in N^*}$ is a subsequence of the sequence $\left(f(x_n)\right)_{n\in N^*}$ which has the limit $s$ $\Longrightarrow$ $s=f(x_s)$ $\Longleftrightarrow$ the function $f$ is superior bounded and it touches the its upper bound $s$, i.e. there is $x_s\in [a,b]$ so that for any point $x\in [a,b]$ we have $f(x)\le s=f(x_s)$.

Similarly prove that the function $f$ is inferior bounded and it touches the its lower bound $i$, i.e. there is a point $x_i\in [a,b]$ so that for any point $x\in [a,b]$ we have $f(x_i)=i\le f(x)$. Therefore, there are the points $\{x_i,x_s\}\subset [a,b]$ so that for any point $x\in [a,b]$ we have $f(x_i)=i\le f(x)\le s\le f(x_s)$, i.e. the function $f$ is bounded and it touches the its lower and upper bounds.

[u][b]Remark.[/b][/u] The [u]Weierstrass' theorem[/u], the [u]Fermat's theorem[/u] and the [u]Rolle's theorem[/u] are the most important theorems from the first part of the calculus (continuity and derivatives). Here I will prove soon and the last two theorems.[/color][/hide]

I did a quick read on open cover stuff for curiosity reason. My question below is not directly related to the question raised by the original poster. I apologize in advance  :roll:

Let $A = \{x : 0 < x < 1\}$. Define a collection of open sets $C = \bigcup_\alpha I_\alpha$ where $I_\alpha = \{1/\alpha < x < 2\}$ for all $\alpha \in A$, which covers $A$. It is claimed there is no finite subcollection of $C$ that can cover $A$. I don't quite see this. How about $I_\infty$? Doesn't that single set cover $A$?

[quote]How about $I_\infty?$ Doesn't that single set cover $A?$[/quote]
There is no such set. Your definition is a little garbled; it might be better to say that $I_n=\left(\frac1n,2\right),\,n\in\mathbb{N}.$

The union of any finite collection of the $I_n$ is $I_N$ for some $N.$ No such set can cover all of the original interval. We can copy the essential features of this example for any set that is not closed.

I just had a very long post disappear on me; I'm not going to try to reconstruct it. The heart of it was a cleaned-up version of 10000th User's argument, explicitly constructing a sequence of nested intervals by successive bisection.

As a pedagogical note: I would much prefer to proceed from the specific (perfect_radio's original question, which I call the Extreme Value Theorem) to the general (the notion of compactness.) Note that the open cover definition of compactness confuses nearly everyone the first time they see it - and what's it good for (at least as far as the beginning student can tell)? I would like to have several compactness related theorems already proved (the Extreme Value Theorem, the Bolzano-Weierstrass Theorem, that a continuous function on a closed bounded interval is uniformly continuous) before ever offering that general definition.

[quote="Kent Merryfield"] Your definition is a little garbled; it might be better to say that $I_n=\left(\frac1n,2\right),\,n\in\mathbb{N}.$[/quote]
Oh yes, admittedly I did not read what I wrote  :blush:

[quote="Kent Merryfield"][quote]How about $I_\infty?$ Doesn't that single set cover $A?$[/quote]
There is no such set.[/quote]
Then how would $\bigcup_{n=1}^\infty I_n$ be valid?

$\bigcup_{n=1}^{\infty}I_n$ does indeed include $A.$ That's what saying the collection is a cover of $A$ means. But that union is not itself any one of the $I_n$, and it is not contained in the union of any finite number of the $I_n.$

:idea:  I think I know what I just missed. It should be $\bigcup_{n \to \infty} I_n$  :D

To prove no finite collection:  Suppose we choose $m \in \mathbb{N}$ large enough ($m < \infty$) and claim that it covers $A$. But then there exists $\bar{x} \in A$ such that $0 < \bar{x} < \frac{1}{m}$. So $I_m$ does not cover $\bar{x} \in A$, contradicting our claim.


LaTeX
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MATHCOUNTS
Pythagorean Theorem
geometry
number theory
relatively prime

2 find Pythagorean triplits, pick a positive odd number. This #will be (A). square the #, but write it down 4 future reference. Now divide the new # by 2. so far the only important # is (A). after u divide the # by 2, subtract 0.5. the # u now have is (B). Now add 1. This #is (C). if u know how the Pythagorean theorem works, put these #s in and if u did it right. it will work. The #s will fit perfectly into the theorem.

You need to add [u]for all $ n > 1$[/u].

And not to disappoint you, but that formula is old; I knew it for over a year. It's not something you found by yourself :ninja: .

However, I wonder, how does that formula work?

The identity you've discovered is $ a^2 \plus{} \left( \frac{a^2 \minus{} 1}{2} \right)^2 \equal{} \frac{a^2 \plus{} 1}{2}$, and it is a special case of the general formula

$ (2mn)^2 \plus{} (m^2 \minus{} n^2)^2 \equal{} (m^2 \plus{} n^2)^2$

where $ m \equal{} a, n \equal{} 1$ and we divide by $ 4$.  Not to disappoint, but this formula has been known for approximately 2500 years.  :)

I'm probably one of the stupidest people on earth, but I don't know that formula.    :o

[quote="benjamin7xx"]This is complicated, mind someone explaining it more :(. I think it might be useful though.[/quote]

Hmm... I don't know how you can explain it more.

Expanding gives $ a^2\plus{}frac{a^4\minus{}2a^2\plus{}1}{4}\equal{}\frac{a^4\plus{}2a^2\plus{}1}{4}$ which is obviously true.

[quote="BOGTRO"][quote="benjamin7xx"]This is complicated, mind someone explaining it more :(. I think it might be useful though.[/quote]

Hmm... I don't know how you can explain it more.

Expanding gives $ a^2 \plus{} \frac{a^4 \minus{} 2a^2 \plus{} 1}{4} \equal{} \frac {a^4 \plus{} 2a^2 \plus{} 1}{4}$ which is obviously true.[/quote]

Fixed LaTeX

You can search for the derivation of the formula mentioned by t0r

I think MATHCOUNTS pulls that a lot, so it's pretty handy to know.

[quote="yankeeprez"]*Not claiming to have discovered this*

Another easy use for this idea is working backwards.
For example, if the question is something along the lines of:
Two legs of a triangle are 180 and 181, find the shorter leg.

What I would do, instead of squaring these large numbers, is add them together.
180+181=361

Does 361 look familiar? It should!
The square root of 361 is 19 and that is the answer!

Obviously, it doesn't work out with ever number (50 and 51) but for me, whenever I see a triangle problem with numbers that are one away, I just add them together and root them.
Hope that helped someone.[/quote]

Note that the two number $ m$ and $ n$ have to be relatively prime too.

Like a $ 6, 8, 10$ right triangle. Say the question gave you $ 10$ and $ 8$.  $ 10\plus{}8\equal{}18 \ne 6^2$

Simplifying gets us $ 5$ and $ 4$, and $ 5\plus{}4\equal{}3^2$.  Multiply everything by two to "unsimplify" it and you get $ 6$.

If a leg and the hypotenuse are given, they must have a difference of one to use this trick. :wink:

If the leg of a right triangle is an odd number greater than 1, than the the other leg and hypotenuse will be two consecutive integers that add up to the square of the odd integer.
Ex.: If the leg is 3, then the sum of the two other lengths will be 3^2=9. 4, and 5 are two consecutive integers that add up to 9. So the other leg is 4, and the hypotenuse is 5.
If the leg is 9, then the other two are 40, 41. (40+41=81)
Hope that helps.

[quote="5849206328x"][quote="megamath"]If the leg of a right triangle is an odd number greater than 1, than the other leg and hypotenuse will be two consecutive integers that add up to the square of the odd integer.[/quote]

This is just one possibility though :wink: For example, 9-12-15 doesn't follow this rule.

Even if we're only dealing with primitive triples, we still have triples such as 20-21-29[/quote]

When you make 9-12-15   3-4-5 and multiply by 3, it does.

[quote="5849206328x"][quote="megamath"]If the leg of a right triangle is an odd number greater than 1, than the other leg and hypotenuse will be two consecutive integers that add up to the square of the odd integer.[/quote]

This is just one possibility though :wink: For example, 9-12-15 doesn't follow this rule.

Even if we're only dealing with primitive triples, we still have triples such as 20-21-29[/quote]

True! :lol:


If water is running into a tank at the rate of 3 metres per second along a trough of rectangle cross-section 18cm wide and 15cm deep, find the amount of water that runs into the tank in 1 minute.

What do you mean by 3 meters per second? The flow of water should be measured in volume/time. If the tank is 18 cm wide and 15 cm deep, how long is it?

[hide="See after trying yourself!"]The amount of water that enters depend on the area of the cross section,time and speed of the water.

So, $ V \equal{} t\times A\times s$
$ V \equal{} 60\times 0.027\times 3$
$ V \equal{} 4.86m^3$  :D[/hide]


induction
combinatorics unsolved
combinatorics

The sum of $n$ real numbers is positive. Find the smallest possible number of their pairwise sums that must be positive.

I think the answer is $n-1$.
[b]Example[/b]: one of the numbers is very big and the others are <0.
[b]Proof that there are at least n-1[/b].Induction on n. WLOG there are no zeros. Let the number of
negative numbers is $a$ and the number of positive $b$. If there are some u>0 and v<0 such that $u+v>0$ then if we increase v and decrease u until v=0. We don't change the sum of the numbers and when we remove v=0 we get n-1 numbers satisfying the condition and by induction hypothesis there are at least n-2 pairwise positive sums. Since the sum u+v is positive we get n-1 positive sums.
If there are no such u and v then since the sum of all numbers i positive b>a
but now $\binom{b}{2}\geq\binom{\frac{n}{2}}{2}\geq n-1$

+)For n=3,4 .The answer is n-1
+)For n=5 .The answer is n-2( Example: 6,7,8.-9.-10 :) )
+)For$n\geq{6}$.We easily prove $min=n-1$ by[b]indution by n[/b]


MATHCOUNTS
probability
Alcumus
function
arithmetic sequence

Alright, so I just realized that I stink at counting questions on Mathcounts tests. The counting questions I am talking about are like "How many triangles are in the figure to the right?" and "How many numbers cannot be made adding the numbers 0,3,5,and 9 three times?" (these are just examples, I dont need an answer to them).

Anyways, which books would help a lot with that? Would the Intro to Counting and Prob. book be useful, or would something else be more useful for those types of pure counting problems.

Some of those are covered in Alcumus. Intro to counting and probability is also good. You should just do lots of problems, either from Alcumus or MATHCOUNTS (but some Alcumus problems are from MATHCOUNTS).

Order Introduction to CP.  While you're waiting for it to come, finish all the problems in Alcumus :D

I also fail at counting and probability, so I ordered intro to counting and prob and I am hoping to order intermediate counting and prob. Hope that helps! I really suggest learning to generate functions, it can be really help!

I suppose you mean generating functions?

Generating functions are not any more useful than simple counting techniques until you get to problems past MATHCOUNTS and AMC level.

Yes that is correct becausee to find the number of numbers the formula is:

$ \frac{highest\minus{}lowest}{interval} \plus{}1$

In the handbook next to each answer the problem is classified as to what type of problem it is.

[quote="Thunder365"][quote="maybach"]Anyway, here is a sample counting problem I saw from a textbook:  How many numbers are in the arithmetic sequence 3,10,17,24,31..........................2558?
I'm thinking the answer is 2,558-3=2,555/7=365+1=366.  Is 366 the right answer, or am I not even in the same ZIP Code? (I'm not very good at counting, but I'm terrible on combinatrics probability questions)[/quote]

See, these kind of counting questions Im good at. You can make a formula by y=mx+b and m=y2-y1/x2-x2. That gets m=7 and then b=-4. so 2558=7x-4, x=366.

The kind of counting I absolutely fail at is like "how many triangles are in the figure" and "How many ways can you make 5 dollars out of pennies, nickels and dimes"[/quote]
OK then.....here's a new question that I stunk at too.  It's 2009 Chapter Sprint Round Problem 24:  Janice bought 30 items each priced at 30 cents, 2 dollars, or 3 dollars.  If her total purchase price was $30, how many 30-cent items did she purchase.  I eventually got the answer(20) but I just guess-and checked and it took about 4 minutes to answer.  Is there a "trick" I can use for these problems?

for this problem, guessing and checking shouldn't take very long...IF you go through it systematically

the first thing you should notice is that the price of the item is 30 cents, and the others are in dollars
in addition, the amount she spent is in dollars
that should immediately tell you that the number of 30 cent items she bought is a multiple of 10 (why?)

thus we have reduced our choices to three answers: 10, 20, and 30

it can't be 30 (obviously), so we are left with 10 and 20

from here, we simply check to see which works

if you do it this way, it should only take you one minute, tops


A(zA),  B(zB),  C(zC)  ,D(zD)

  zA^2+zB^2=zB^2+zC^2=zC^2+zD^2=0

  Prove that  A,B,C,D  is a square.

[quote="tarta"]A(zA),  B(zB),  C(zC)  ,D(zD)

  zA^2+zB^2=zB^2+zC^2=zC^2+zD^2=0,(*)

  Prove that  A,B,C,D  is a square.[/quote]
Note: I will use $a=z_{A}$ ,$b=z_{B}$ ,$c=z_{C}$, $d=z_{D}$ because it's easier to write.

From $(*)$ we also obtain that  $a^{2}+d^{2}=0$ and $a^{2}=b^{2}=c^{2}=d^{2}$ $\leftrightarrow$ $|a|=|b|=|c|=|d|,(1)$
We have $(a+b)^{2}=a^{2}+b^{2}+2ab$ but $a^{2}+b^{2}=0$ $\rightarrow$ $(a+b)^{2}=2ab$ $\leftrightarrow$ $|a+b|^{2}=2|ab|$
From $a^{2}+b^{2}=0$ $\rightarrow$ $|a-b|^{2}=2|ab|$ therefore
$|a+b|=|a-b|$ $\rightarrow$ $\angle AOB=\frac{\pi}{2}$ ,where $O(0,0)$
Analogously we have $\angle BOC=\frac{\pi}{2}$ ,$\angle COD=\frac{\pi}{2}$ and $\angle DOA=\frac{\pi}{2}$ $(2)$
And from $(1)$ and $(2)$ the conclusion follows


geometry
analytic geometry
perpendicular bisector

Given square ABCD with side 8 feet. A circle is drawn through vertices A and D and tangent to side BC.  What is the radius of the circle in feet?

Connect the center to the perpendicular bisector of side AD. Use the right triangle 4,x, and 8-x. Solving for x gives you 3. Therefore, the radius is 5.

I got lazy and didn't type in latex.

i did it a long way, using coordinate geom... but got the same answer

basically i used the equation for a circle  (x-h)^2 + (y-k)^2 = r^2 , and set the square on the origin. Then i got a system of 3 equations, plugging the three locations that the circle hits the square -- (0,0) , (0,8) , and (8,4) -- and solved for easily for the center of the circle, which came out to h=3, k=4, so (3,4). So the radius is 8-3 = [b]5[/b].


conics parabola Asymptote geometry hyperbola

Hi y'all. I'm new here, and I have a question...

For a rational equation, like x/(x+2), the parabola can never touch the asymptote (in this case, x=-2) because division by zero is undefined. However, if you use geometry principles as opposed to algebra ones, the parabola HAS to touch and cross the asymptote at a infinite y value because the parabola is constantly curving toward the asymptote, right?

I think it becomes infinetly close, but never touches or crosses it.

But it always moves in the direction of the asymptote. I ask this because just because division by zero is "undefined" doesn't mean that it doesn't exist. Does it? :?

First, the name for that conic is [b]hyperbola[/b].

Second, it never crosses the asymptote, that's a property of hyperbola, due to division by zero you can say.

Third, when something is undefined, it's because nobody knows what that is, it is basically meaningless in some sense. So we should avoid it, there is no loss to it.


Looks like results have been mailed out, so I'm assuming we are free to discuss.

This contest seemed definitely harder than the last two.

And if you got owned by my problem, please don't blame me! I gave them $N = 400$, then Brian Dean came up with an $O(NlogN)$ and they set $N$ to $50000$.

Actually I thought Silver and Gold were both easier than previously, as in if I'd done Gold this time instead of Silver I'd have done about as well as my previous Silver scores... It only took me 1:30 to do all 3 problems and I only ran out of time on 4 cases. (I'm a Silver right now.)

Haha, I remember getting owned by one of your problems last year, in the Silver division.

And then when I saw your name, I got really scared.

I got owned by I/O on the first problem (fstream in C++...), got the second problem, and got owned on the third (darn you JSteinhardt... lol :P).

[quote="JSteinhardt"]Yeah, I/O seemed to hurt me on the first problem as well. Did you get the first 4 cases? That's what happened to me, though then again my solution still dies on the last case even with fscanf.

EDIT: I'd be more scared by Reid and Brian-type problems than anything I'd come up with. Then again, Brian Dean was the one who essentially made my problem extremely evil (not that it wasn't fairly challenging before), so I guess that sort of proves my point.

Though Reid's problems seem to be downright sadistic.[/quote]

Yeah I got the first 4, I didn't read the range-min query article though so I only had the $O(N+Q\sqrt{N})$ algorithm... the I/O thing probably messed things up because it made your algorithm (I'm assuming the $O(N\log{N}+Q)$ one) get the same number of points as mine.

I thought my algorithm was correct for the third one, just slow, but apparantly it was just plain wrong.

Yeah, the third problem seemed a little hard... I wonder how much it'll be weighted in the results.
By the way, JSteinhardt, what do you have to do to submit a problem for the contest? Do you just give them the problem statement, or do you also need a 'sample' solution, etc.?

EDIT: Wow, I just noticed that [url=http://ace.delos.com/traingate]this website[/url] exists. Is that the only way to submit problems for contests, or can anyone submit problems?

Also, I have a question about the analysis for problem schul:
[quote="USACO Analysis"]Suppose that for a particular value of D, using the teacher's choices gives a ratio of M, and Bessie wants to improve this ratio. If she thinks of every test as being M plus or minus some number of marks (not percent), then she will want to drop the tests with the minimum number of marks over M, and keep those with the most marks over M. If it happens that the teacher's choice is different, then she can switch some of the tests to get more than 0 over M and thus improve her score.[/quote]

I don't really understand what this means (particularly what M stands for and what they mean by 'M plus or minus some marks'). Could someone elaborate?

Thanks.

I got the first two, but was chasing a silly bug in #2 endlessly, so I ran out of time for the third.

I think #3 was a known problem (at least I've seen it a couple of times formulated in graph-terms), but nevertheless, very nice. Good job JSteinhardt :) I must say, though, Brian Dean's and Bruce Merry's solutions were awesome! ! !

PS: I was wondering, do you have to have an account at probgate to propose tasks for the contests? If so, how long after finishing usacogate do I have to wait? (it's been a while so far)

on the RMQ one, I used O(NlogN + Q) but it timed out on test case 8 for silver and all but the first four for gold :( (luckily i'm in silver :P)

I kinda hate myself for the first silver problem (flowers) because I abandoned the correct solution during the contest on the basis that I thought I'd found a very simple counterexample. In fact, my first submission for this problem would have gotten 7/10 test cases correct.

As for the last silver problem (tallest), I got pwnt by degenerate cases so I only got 4 test cases instead of all 10.

To answer several questions:

(1) http://ace.delos.com/traingate is reserved for people who have completed the training pages. If you have completed them but don't have an account, you should e-mail Rob Kolstad with USACO in the subject line and ask him for one. I am led to believe that the general purpose is for current participants to help the coaches develop Bronze and Silver problems, leaving them more time to come up with a challenging Gold problem set.

(2) http://ace.delos.com/probgate is for coaches only. It is where the contests themselves are assembled. I would assume that problems get transferred from traingate after they have been vetted by the members there, and other problems are added by the coaches. This is also where the Gold contests are created. I am not sure what the requirements are for becoming a coach, though I would not think that it would require anything more than having graduated from high school and a desire to help. Rob Kolstad would probably be the one to contact for this as well.

(3) I am guessing as weights 300ish for psolve and lineupg, and 400 for schul, though the test cases were a bit easy for schul and lineupg got a lot of people with the large input size, so that may affect some stuff. I would also surmise IOI grading based on the fact that only three people solved schul, but I could be wrong about that.

(4) In response to the question about the following explanation:

[quote]Suppose that for a particular value of D, using the teacher's choices gives a ratio of M, and Bessie wants to improve this ratio. If she thinks of every test as being M plus or minus some number of marks (not percent), then she will want to drop the tests with the minimum number of marks over M, and keep those with the most marks over M. If it happens that the teacher's choice is different, then she can switch some of the tests to get more than 0 over M and thus improve her score.[/quote]

I think that what they are saying is that if you have a score $\frac{p}{q}$, and the average is $M$, then the number of marks over $M$ is $(\frac{p}{q}-M)\cdot q$, as that is the number of points you earned over the average score of $M$. A slightly easier way to see this (though at the same time more difficult to come up with) is that

$\frac{n_{1}+\ldots+n_{k}}{d_{1}+\ldots+d_{k}}> M \iff (n_{1}-d_{1}M)+(n_{2}-d_{2}M)+\ldots+(n_{k}-d_{k}M) > 0$, and the given sum is equal to zero initially (letting $M$ be the score that the teacher's heuristic gives), so we just need to determine if there is a way to increase the sum, which is possible iff there exists a dropped test $i$ and an undropped test $j$ such that $n_{i}-d_{i}M > n_{j}-d_{j}M$.

(5) Anyone can submit problems for contests. I have submitted a bunch, two have been used, though I will admit that the two I submitted were the only two that were particularly good problems.

@solafidefarms: Maybe you're just getting better? Nice job on silver btw, I see you're joining us in gold now! Congrats!

[quote="Hamster1800"]Well I broke 400 this time :) Maybe by the end of freshman year I'll join gold (that's my goal anyway).

@nebula42: HOW IN THE WORLD DO YOU GO FROM 80 IN SILVER TO 733 IN GOLD IN ONE MONTH?

Is fscanf really that much faster than streams? Maybe that's why my lineup (when modified for lineupg in analysis mode) got the first four in <= 0.004 seconds and the all the rest in >2[/quote]

They're a lot faster, but I don't think the difference should be that great (it could be that the first four cases are just really small in comparison, I can't remember). My fifth test case went from just over 1 second to around .35 seconds by changing the streams to stdio.

Man...I changed my lineup to FILE*s and resubmitted it for silver and it got all 8 test cases...halved the running time on test case 8. Why didn't I know about these speed differences before?

I changed my lineupg to FILE*s as well and it went from
****t ttttt
to
***** **x*t

[quote="nebula42"][quote="Hamster1800"]@nebula42: HOW IN THE WORLD DO YOU GO FROM 80 IN SILVER TO 733 IN GOLD IN ONE MONTH?[/quote]
Actually, I'm not so sure myself.
It was probably that I learned a little more by doing some of the USACO training pages, though, since last month I didn't understand depth-first search and I had never heard of dynamic programming.
The funny thing though is that in November, I was in the Bronze division, but I did so badly I didn't even show up on the results.  :)
I just hope I can do well on next month's contest...[/quote]

Except that I've done more of the training pages than you...what's your secret to writing code correctly first time?

So wait, you went from bronze -> gold in two months? WOMGTWHTWDS

Correction: He went from the bottom of bronze to the top of gold in 2 months.

I do have to admit that was quite impressive. I suppose I'll be seeing you at camp then if you keep it up. As for error-free code, it's just important to be familiar with how to write all of the major data structures and algorithms on your own, and that should give you enough experience to do just about whatever you please, given a bit of practice.

[quote="Hamster1800"]One more thing. Did you (nebula42), since you were in silver last time, do the silver contest and move up during it?[/quote]
Yes, I also did the silver contest this month.
I suppose you're wondering about my silver results  :P , so here they are:

[code]============================== SUMMARY =============================

                     -- case number --
            1  2  3  4  5  6  7  8  9 10 11 12
flowers     *  *  *  *  *  *  *  *  *  *
lineup      *  *  *  *  *  *  *  t
lineupg     *  *  *  *  *  *  *  *  *  *
psolve      *  *  *  *  *  *  *  *  *  *
schul       *  *  x  x  x  x  x  x  x  *  *  t
tallest     *  *  *  *  *  *  *  *  *  *

. = no entry   t = time exceeded    *   = correct answer
x = wrong      s = signal           e   = bad exit status

=======================================================================[/code]

[quote="JSteinhardt"]Correction: He went from the bottom of bronze to the top of gold in 2 months.

I do have to admit that was quite impressive. I suppose I'll be seeing you at camp then if you keep it up.[/quote]
Well, I don't know about that. It seems like that would be a pretty high goal to try to reach.

By the way, what counts most for the selection of training camp attendees? Does the US Open count for almost all of it, or only a part of it?

[quote="nebula42"]
[quote="JSteinhardt"]Correction: He went from the bottom of bronze to the top of gold in 2 months.

I do have to admit that was quite impressive. I suppose I'll be seeing you at camp then if you keep it up.[/quote]
Well, I don't know about that. It seems like that would be a pretty high goal to try to reach.[/quote]

Well, if you keep doing as well you have been, I'm sure you'll make the camp.

[quote="nebula42"][quote="paladin8"][quote="nebula42"]By the way, what counts most for the selection of training camp attendees? Does the US Open count for almost all of it, or only a part of it?[/quote]

Pretty much every contest counts, and even some of your work on the training pages if I remember correctly. Plus, if you're a freshman, you get huge bonus points because there's a good chance that you'll be very good in the next few years and make the IOI.[/quote]
How would they count the contests if you were in a different division from Gold? (I was in Bronze and later Silver at the beginning of this year.) And about how much are the training pages weighted compared to the contests?[/quote]

I'm pretty sure they do selection subjectively, so they'll take it into account however they see fit. I'm not sure how the selection is exactly done or how the training pages are weighted.

[quote="USACO site"]  Criteria For 2007

We will use all contest results and work on the training materials as evaluation criteria for choosing attendees at the USACO training camp to be held in June, 2007.

		This means that candidates who show long-term strength (and/or growth) are more likely to be chosen for camp than those who have one lucky contest. We will be keeping a record of all  progress on the training materials.[/quote]

can any one please send me all the problems of bronze, silver and gold? i forgot to save them before february contest begins.

Google does wonders: http://infoman.musala.com/usaco/2006_07/jan/jan_gold_en.txt (you can get silver and bronze too)

Good luck on the Feb contest!


function
geometry
3D geometry
topology
advanced fields
advanced fields unsolved

It is a result due to Hlawka that the number of lattice points in $ B(0,\lambda)$, which we denote by $ N(\lambda)$, is
\[ N(\lambda) \equal{} |B(0,1)|\lambda^n \plus{} O(\lambda^{n\minus{}2\plus{}2/(n\plus{}1)})
\]
which essentially comes from using stationary phase to find asymptotics for the fourier transform of the characteristic function of the unit ball.

a)  Is there a simple argument (say, without fourier analytic methods) that shows that $ N(\lambda) \equal{} |B(0,1)|\lambda^n \plus{} O(\lambda^{n\minus{}1})$?

b)  Is there a simple argument that shows that we cannot get an estimate like $ N(\lambda) \equal{} |B(0,1)|\lambda^n \plus{} O(\lambda^{n\minus{}2})$?

Consider the Laplacian $ \Delta$ on the torus $ \mathbb{T}^n$.

c) Why is it the case that the eigenvalues of $ \minus{}\Delta$ are all integers

d) which are the sum of the squares of $ n$ integers?

e) What is a good reference for results on the eigenvalues of the Laplacian?

a) Divide the space into unit cubes and estimate how many of them lie within $ B(0,\lambda)$ or at least intersect it.

b) When $ n\equal{}2$, it suffices to prove that for any $ M$ there is $ \lambda$ such that $ \partial B(0,\lambda)$ contains at least $ M$ integer points. The latter follows from the fact that the unit circle contains infinitely many rational points. Not sure what to do when $ n>2$.

c)-d) Consider what the Laplacian does to the Fourier coefficients of a function.

e) "The Laplacian on a Riemannian manifold" by Rosenberg.


$x=4$ and $y=52$

i dont think that have anothers

Hint:

[hide]1. $x$  is even  [hide="because"]Otherwise , LHS is  (-1) mod 4 which is not possible.[/hide]
2.  $x<8$ [hide="because"]otherwise LHS is greater than ($(7^{x/2})^{2}$ and less than $(7^{x/2}+1)^{2}$  which can not be a perfect square[/hide]
Only have to try for x=2,4,6[/hide]

And without mod??

Just an elementay solution?


the other years I have known that TST USA starts 10-11/6.
I think may have started TST 2009.
Can anyone post them?

I think they are usually not made public for at least a little while.  Well the TST is still in progress, so in any case it would be a good idea to wait until it's entirely finished, at least.

The TST will very likely not be posted until after this year's IMO; this is the norm at least. In any case, it's not done being given anyway.

When USATST start???
I really want to know the time of the exam.


linear algebra
matrix
limit
algebra
polynomial
superior algebra
superior algebra solved

Let $a,b$ be constant numbers such that $a^2+b^2=1$  and  $A=\left(\begin{array}{cc} 4 & -2 \\ 3 & -1 \\ \end{array}\right)$.
Set $\left(\begin{array}{c} x_{n} & y_{n} \\ \end{array}\right)=A^n\left(\begin{array}{c} a & b \\ \end{array}\right)\ (n=1,2,\cdots)$.

Let $p_n=\frac{x_n}{\sqrt{{x_n}^2+{y_n}^2}},q_n=\frac{y_n}{\sqrt{{x_n}^2+{y_n}^2}}$. Find $\lim_{n\to\infty} p_n ,\lim_{n\to\infty} q_n$.

This is merely asking for a normalized eigenvector corresponding to the dominant eigenvalue. The characteristic polynomial of this matrix is $\lambda^2-3\lambda+2$ so the eigenvalues are 1 and 2. The dominant eigenvalue is 2, with eigenvector $\left[\begin{array}{c} 1 \\ 1 \end{array}\right],$ which normalizes to $\left[\begin{array}{c} \frac1{\sqrt{2}} \\ \frac1{\sqrt{2}} \end{array}\right].$

It's a straightforward question that I might ask as a fairly easy test or homework question in an introductory linear algebra course.

This is not so easy!

Can anyone solve this problem,completely? ;)

[quote="Kent Merryfield"]Ah, now I see it.  The key issue is the sign of the coefficient of $\left[\begin{array}{c} 1 \\ 1 \end {array}\right]$ in the representation of $\left[\begin{array}{c} a \\ b \end {array}\right]$  in the basis of eigenvectors.

Thus, based on the original values of $a$ and $b$:

If $3a-2b>0$ then the limit is $\left[\begin{array}{c} \frac1{\sqrt{2}} \\ \frac1{\sqrt{2}} \end {array}\right].$

If $3a-2b<0$ then the limit is $\left[\begin{array}{c} -\frac1{\sqrt{2}} \\ -\frac1{\sqrt{2}} \end {array}\right].$

If $3a-2b=0$ then the limit is $\left[\begin{array}{c} a \\ b \end {array}\right],$ which is either $\left[\begin{array}{c} \frac2{\sqrt{13}} \\ \frac3{\sqrt{13}} \end {array}\right]$ or $\left[\begin{array}{c} -\frac2{\sqrt{13}} \\ -\frac3{\sqrt{13}} \end {array}\right].$[/quote]

Thank you for your solution, Kent Merryfield.

Comment 1:To the third case, you need to classify the case of $b>a$ or $b<a$

Comment 2:The point is $\left(\begin{array}{c} a \\ b \end {array}\right)=(3a-2b)\left(\begin{array}{c} 1 \\ 1 \end {array}\right)+(b-a)\left(\begin{array}{c} 2 \\ 3 \end {array}\right)$,as you noticed.

kunny