analytic geometry
graphing lines
slope
integration
calculus
calculus computations

Okay, I was absent on day we went over this paper and I have no idea how to do this problem. I don't think we even learned it but anyway, here it goes. It's Free Response Question in AP Exam.

Consider the differential equation $\frac{dy}{dx}= x^{2}(y-1)$.

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the pink test booklet.)

(b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are positive.

(c) Find the particular solution $y = f(x)$ to the given differential equation with the initial condition $f(0) = 3$.

For (a), the axes they provide is like this. You have point O as center and the graph shows -1 and 1 to left and right, respectively. It doesn't show anything to bottom while it has upto 3 on the top. So, the twelve points they mean are (-1,0), (-1,1), (-1,2), (-1,3), (0,0), (0,1), (0,2), (0,3), (1,0), (1,1), (1,2), and (1,3).

Thanks for your help.. I appreciate it greatly.

(b) the slope is positive when $x^{2}(y-1)>1$. At which parts of the plane does this happen?

(c) I suppose you should separate $x$ and $y$ like this: $\frac{dy}{y-1}=x^{2}dx$ and then integrate both sides. $\int \frac{dy}{y-1}=\int x^{2}dx$. You'll get a formula relating $y$ and $x$ (with an indefinite constant $C$ on the right). The condition $y(0)=3$ will determine the value of $C$.

[hide="For bonus points"] Can anyone explain to me why this should be taught to schoolchildren?[/hide]


[hide]6y=x-1, divide both sides by 6 to get it in the form y=mx+b, where m is the slope. m=$\frac{1}{6}$[/hide]

[quote="math92"]Find the slope of this line:
6y = -1 + x[/quote]
[hide]
Let y=mx+b.  By definition m is the slope of the line.  So we have to convert the equation into this form.

This is $y=\frac{1}{6} x -1$.

So the slope is $\frac{1}{6}$[/hide]

[quote="math92"]Find the slope of this line:
6y = -1 + x[/quote]

[hide]$y=\frac{1}{6}x-\frac{1}{6}$
Therefore, the slope is $\boxed{\frac{1}{6}}$[/hide]

[hide="slope"]$\frac16$[/hide]
[hide="x int"]$1$[/hide]
[hide="y int"]$\frac{-1}{6}$[/hide]


There are 12 ordered pairs of integers $ (x,y)$ that satisfy $ x^2 \plus{} y^2 \equal{} 25$. What is the greatest possible sum $ x\plus{}y$?

[hide="solution"]
The integers can be positive or negative, but since we want to maximize their sum, they should both be positive. The solution that gives the greatest sum where both are positive integers is $ (3,4),$ and the sum is $ \boxed{7}.$[/hide]


calculus
integration
function
limit
Support
real analysis
real analysis unsolved

The following is a question that appeared on a recent exam.  Some of us have been debating the correct way to solve the problem and we haven't reached an agreement.

Let $ f(x)\equal{}x^{2}$ and $ \alpha(x)\equal{}x$ for $ x\in [0,1]$ and $ 3\minus{}x$ for $ x\in (1,2]$.  Compute $ \int_{0}^{2}f\,d\alpha$.

I believe your solution is wrong, function $ \alpha$ has a discontinuity at point $ 1$, and that doesn't mainfest in your solution. The idea is ok, writing the initial integral as a sum of two, but you have to add $ f(1)\big(\alpha(1) \minus{} \alpha(1 \minus{} )\big)$.

Edit: Please discard my comment, that is the entire post.

So Mr.Doe, should I consider your comment or do you believe it is wrong?

FibonacciFan, your solution is the one that myself, as well as a few other people produced, and we received full credit.  However, my friend made a claim similar to that of Mr.Doe and said that we need to add in $ f(1)\equal{}3$.  So which is it?

JRav: In the R-S integral $ \alpha$ should be a monotone increasing funtion. Isn't it?! But in your question  $ \alpha$ is not an increasing function.

$ \alpha$ doesn't have to be monotone increasing.

I *think* that Mr.Doe is right that you have to correct for the discontinuity. I get

$ \int_{1}^{2}x^2 d(\alpha(x)) \equal{} f(1)(\alpha(1\plus{}) \minus{} \alpha(1))\minus{} \frac{7}{3} \equal{} \minus{}\frac{4}{3}$

making the answer $ \minus{}1$. I could easily be making another mistake though.

I need to point out that I'm familiar with R-S integral only when function $ \alpha$ is monotone increasing and continous from right. My comment was pointed in that direction, but since that isn't the case here, I can't make any claims. It would be the best, if JRav ( in short ) explained how did they define R-S integral.

Indeed, in situations like this, [i]looking at the definitions[/i] is certainly a worthwhile endeavor :).  Wikipedia gives a definition that works fine for nonmonotone $ \alpha$, and which can be summarized as
\[ \int_a^b f \, d\alpha \equal{} \lim \sum f(c_i) (\alpha(x_{i \plus{} 1}) \minus{} \alpha(x_i))
\]
where $ x_i \leq c_i \leq x_{i \plus{} 1}$.  The question is whether we can disregard the part of the partition that contains $ 1$ and some point larger than $ 1$.  From the definition, it's clear that this part contributes in the limit $ f(1)(\alpha(1 \plus{} ) \minus{} \alpha(1)) \equal{} 1$.  In other words, if you split a R-S integral at a discontinuity of the integrating function, there better be something to indicate what you've done.

This basic point is well-illustrated by the functions in this post: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=261040

[quote="JRav"]FibonacciFan, your solution is the one that myself, as well as a few other people produced, and we received full credit.  However, my friend made a claim similar to that of Mr.Doe and said that we need to add in $ f(1) \equal{} 3$.  So which is it?[/quote]
Please compute $ \int_{[0,2]} x d\beta$ with your method, where
\[ \beta(x) \equal{} \begin{cases} 0 \quad (0 \le x < 1) \\ 1 \quad (1 \le x \le 2) \end{cases} \]
Then tell me if your answer makes sense.

In an effort to support what hpe is saying:

One of the best things about the Stieltjes integral is that it's the natural home of integration by parts. That is, as long as certain conditions are met (discontinuities at the same places cause trouble, but they're troublesome in general) we have
\[ \int_a^bf(x)\,dg(x)\equal{}f(x)g(x)\bigg{|}_a^b\minus{}\int_a^bg(x)\,df(x)\]

If you apply that to hpe's example, you get

$ \int_0^2x\,d\beta\equal{}2\cdot1\minus{}0\cdot0\minus{}\int_0^2\beta(x)\,dx\equal{}2\minus{}1\equal{}1.$


linear algebra
matrix
function
graph theory
linear algebra unsolved

What is the maximal number of 1's that a binary matrix whose square is also binary can have?

View the matrix as the adjacency matrix of a directed graph: $A_{ij} = 1$ is there is an edge from vertex i to vertex j.  The square represents the number of paths of length 2.  Look at the vertex in the middle of such a path.  If it has more than one edge in, it can have no edges out.  If it has more than one edge out, it can have no edges in.  So the vertices fall into three classes: $X$ sources (all edges out), $Y$ sinks (all edges in) and $Z$ with one out and one in.  This is a bipartite graph except some edges across are subdivided.  So the total number of edges is at most $XY + Z$.  Maximimising this subject to $X+Y+Z = N$ gives  $N^2/4$ or $(N^2-1)/4$.

I have only one thing to say :):

$\begin{pmatrix}0&0&0&1&1&1\\0&0&0&1&1&1\\0&0&0&1&1&1\\0&0&1&0&0&0\\0&1&0&0&0&0\\1&0&0&0&0&0\end{pmatrix}$

I'm sure it's obvious how to extend this example to th general case in order to obtain a matrix with $\frac{n^2}4+\frac n2\ 1$'s when $n$ is even (and something similar but with some floor functions thrown in here and there when $n$ is odd).

Quite right!  Il faut que j'y pense encore ...


Prove that if $ a, b$ and $ c$ are positive and satisfy $ a^{n} \plus{} b^{n} \equal{} c^{n}$ for some integer $ n\ge2$, then $ a, b$ and $ c$ are the sides of a triangle.

Fermat's Last Theorem states that this is impossible for $ n \ge 3$. Therefore the only posssibility by your conditions is that $ n \equal{} 2$, and by Pythagoras $ a,b,c$ are the sides of a right angled triangle.

EDIT: Oh wait, that only works if $ a,b,c$ are integers. Ignore this then.

Hi KrazyFK.

I think Fermat's Last Theorem is for integer solutions $ (a, b, c)$.

We have $ a < c$ and $ b < c$, so we need only show that $ a\plus{}b > c$.

For $ a, b > 0$ and $ n \ge 2$, the binomial theorem tells us that $ (a\plus{}b)^{n} > a^{n} \plus{} b^{n}$,

so if we assume that $ a\plus{}b \le c$, then $ a^{n} \plus{} b^{n} < (a\plus{}b)^{n} \le c^{n}$, contradicting $ a^{n} \plus{} b^{n} \equal{} c^{n}$.

Yes, I only realised that $ a,b,c$ weren't necessarily integers shortly after I posted.


Θυμηστε μου  το $ \int_{0}^{\pi/4}{\sqrt {1 \plus{} sin^2t}}\ dt$

hello, do you meant $ \int_{0}^{\frac{\pi}{4}}\sqrt{1\plus{}\sin(t)^2}dt$?
Sonnhard.

hello, your integral leads to an elliptic integral.
Sonnhard.


Let function $ f: \, \mathbb{R}\to \mathbb{R}$ satisfy the following two conditions:
$ (i)$ $ f(x+1)-f(x) = 2x+1$ for all $ x\in\mathbb{R}$.
$ (ii)$ $ |f(x)| \le 1$ for $ x \in [0, 1]$.
Prove that $ |f(x)| \le 2+x^{2}$ for all $ x\in\mathbb{R}$.


There are four different color eggs to work on.
How many combinations can we put 2 eggs in each basket?( I count it 10)
How many can we put 3 eggs in each basket? (I count it 20)
What about 4 eggs in a basket?(I count it 35)

What is the pattern?(please help here out)

10     20     35
    10     15
         5

The difference is 5 with a second difference of 5.

although i think jonathan knows what he's talking about, he's being quite confusing.

basically, the difference between terms increases by five each time..., so we increase by 10, then 15, then 20, then 25, ....

the sequence would be $ 10, 20, 35, 55, 80...$


algebra
polynomial
function
number theory proposed
number theory

Let $ a_n$ is the coefficient of $ x^n$ of polynomial $ (x^2\plus{}x\plus{}1)^n$ . If $ p>3$ is prime, prove that $ p^2 \mid a_{p}\minus{}1$

It is well known that algebraic integers, i.e. the roots of monic polynomials with integer coefficients, form a ring in which we have divisibility relation defined as always: $ a|b$ iff there exist an algebraic integer $ c$ such that $ b\equal{}ac$. It's very easy to prove that a rational number is an algebraic integer iff it's just an integer. So if we will prove that $ p^2|a_p \minus{} 1$ in ring of the algebraic integers we will be done, since the quotient $ \frac{a_p \minus{} 1}{p^2}$ is clearly rational.
$ (x^2\plus{}x\plus{}1)^p \equal{} (x\plus{}\omega)^p(x\plus{}\omega^2)^p$ where $ \omega$ is a third root of unity. Note that $ \omega$ is an algebraic integer. From the Viete's formula:
$ \minus{}a_p \equal{}\sum_{k\equal{}0}^{p} \binom{p}{k}^2\omega^k\omega^{2(p\minus{}k)}\equal{}\omega^p \plus{} \omega^{2p} \plus{} \sum_{k\equal{}0}^{p} \binom{p}{k}^2\omega^k\omega^{2(p\minus{}k)}$.
Since $ p|\binom{p}{k}$ for $ 1 \leq k \leq p\minus{}1$ each term in the sum is divisible by $ p^2$ and $ w^{p} \plus{} w^{2p} \equal{} w \plus{} w^2 \plus{} 1 \minus{} 1 \equal{} \minus{}1$ ($ p \ne 3$) so, it follows that $ p^2|a_p\minus{}1$.

[quote="Litlle 1000t"]Let $ a_n$ is the coefficient of $ x^n$ of polynomial $ (x^2 \plus{} x \plus{} 1)^n$ . If $ p > 3$ is prime, prove that $ p^2 \mid a_{p} \minus{} 1$[/quote]
[hide="Solution"]
Through a generating function argument, we see that $ a_p$ is the number of ordered partitions of $ p$ using the numbers $ 0$, $ 1$, and $ 2$ and only using $ p$ numbers.  Thus, $ a_p \minus{} 1$ is the number of ordered partitions of $ p$ using the numbers $ 0$, $ 1$, and $ 2$, using $ p$ numbers, but not including the partition using $ p$ ones.  Now, let a partition of $ p$ include $ k$ zeroes, $ i$ ones, and $ p \minus{} i \minus{} j$ twos.  This means that there are $ \frac {p!}{k!i!(p \minus{} i \minus{} j)!}$ ways to order this particular partition.  Since $ p$ is prime and $ k, i, p \minus{} i \minus{} j < p$, we see that $ p|(\frac {p!}{k!i!(p \minus{} i \minus{} j)!})$.  Let us say that there are $ l$ partitions of $ p$ (not including the $ p$ ones) where order doesn't matter.  Order these $ l$ partitions in any arbitrary order in a list.  Now, say that $ P_i$ is the $ i$th unordered partition of $ p$ in our list.  Say that there are $ x_i$ ways to order $ P_i$.  Thus, $ p|x_i$ for all $ i$ as proven above.  Notice that $ a_p \minus{} 1 \equal{} \sum_{i \equal{} 1}^{l} {x_i}\implies p|(a_p \minus{} 1)$ since $ p|x_i$ for all $ i$.  [/hide]

EDIT: My apologies, [b]ZetaX[/b] and [b]Litlle 1000t[/b], my solution is false.

It seems you only proved $ p|a_p\minus{}1$, but $ p^2|a_p\minus{}1$ was to be proven.


[hide]2/3

ummm, find the probabilty you chose a head/head coin and uncovered heads
that is 1/2

now, find the probabilty you chose a head/tail coin and uncovered heads
that is 1/4

(1/2) / (1/2+1/4) = 2/3[/hide]

Yes, that is correct.  But could you please edit ur post so the solution is hidden. Thanks

[hide]I got 2/3. This is because out of the three equally probable head you could select, two of them have a head on the other side of the coin. Simple.[/hide]

[hide] This is basically conditional probability. Out of four cases of coin face possibilities, three will be heads cases. Out of these 3, 2 will yield a head on the opposite face. Thus the answer is:

2/3[/hide]

[quote="G-UNIT"]A box contains two coins. One coin is heads on both sides and the other is heads on one side and tails on the other. One coin is selected from the box at random and the face of one side is observed. If the face is heads what is the probability that the other side is heads?[/quote]
[hide="Answer-solution"]
2/3, not 1/2.

There are three possibilites.
[/hide]


quadratics
pigeonhole principle
number theory

I already posted two selfmade problems in the "intermediate" section.
This is the third one :

All positive integers which can be written as the sum of two squares are called happy. For example, the number 5 = 2 + 1 is happy but the number 6 is unhappy since the equation 6 = p + q has no solutions in positive integers.

(a) Prove that the product of two numbers is a happy number.
(b) Prove that if a is a happy number then 7a is a unhappy number.

(a) (a+b)(x+y) = (ac-by) + (bx+ay) is know from complex numbers.

(b) n = a+b. 7n = c+d. I
t is easy to check (I did it by exhaustive computing) that x^2+y^2 = 0 mod 7 iff x=y=0 mod 7.

Then n = 7*(C+D) = a+b. Apply it again for a and b and so on. By infinite descent, we obtain a=b=c=d=0.

n=0 is the only happy number such as 7n is also happy.

-----------------

ARNE RULES  :D

Yeah but 0 is no positive integer so it is not happy !

The set of the positive integers is {1, 2, 3, ...}.

This can be done directly by using Fermat's thm:

a positive number n is what you called happy (love it! :D ) iff any prime divisor of n of the form 4k+3 appears in the decomposition of n at an even power. This follows quickly from here: if n is happy then, for any k>=0, (4k+3)*n isn't happy because 4k+3 must have a prime divisor of the form M4+3 at an odd power (I think  :) ).

For the first one: for example k=3 (like I said, I think all numbers of the form 4k+3 work for this one).

[quote="Arne"]There are even some generalisations.

FIRST ONE
Find another integer k such that if a is happy, then ka is unhappy.

SECOND ONE
Call a positive integer p lucky if p = a + nb for a fixed integer n and positive integers a and b. Is the product of two lucky numbers again a lucky number ? :D[/quote]
Alright, I'll take the lucky numbers. p=a:^2:+nb:^2:=|a+bi:sqrt:n|, q=c:^2:+nd:^2:=|c+di:sqrt:n|, pq=|a+bi:sqrt:n||c+di:sqrt:n|=|(a+bi:sqrt:n)(c+di:sqrt:n)|=
|(ac-bdn)+(bc+ad)i:sqrt:n|=(ac-bdn):^2:+n(ad+bc):^2:=lucky number yaaay!
Correct?

Of course MT !

How would you prove that each number of the form 4k + 3 works for the first generalisation Grobber ? Could be, but I'm wondering how someone could prove this ... by considering quadratic residues maybe, a bit of a slog. Then your method will be nicer Grobber but I never heard of it, probably I don't know enough number theory :(

Like I said, I think Fermat's thm can be used, although I don't know how to prove that! :D

[quote="Arne"]each number of the form 4k + 3 works for the first generalisation [/quote]

A little Number Theory will be a nice escape from my pigeon hole !

Let's prove first : if p=4k+3 is prime and p | a + b then p | a  and p | b.

Let's suppose p does not divide a.
Let c = ap-2. ac = 1 (mod p) by Fermat little theorem.

p | a + b, i.e. b = -a (mod p), (bc) = -1

So p does not divide bc. by Fermat little theorem again, (bc)p-1 = 1 = ((bc))(p-1)/2 = -1 because (p-1)/2 is odd. Contradiction.

Now it's easy to prove that p=4k+3 makes every happy number unhappy by infinite descente.

[quote="Arne"]Am I boring you with my pigeonhole problems ?[/quote]

[b]NOT AT ALL [/b]!!!!!

They are [b]very nice [/b]  :) but you can spend 2 hours on them without seeing even the start of a solution (like I did with the one with power of 4 :cry:)

Do you mean the one with the 4th power ? That's an IMO problem.

Hint : Look at squares instead of fourth powers first.


Find one (all you need is one) triple $ x, y, z$ that satisfies the following:

$ x, y, z$ are positive integers.

$ \sqrt{x^2 \plus{} y^2}, \sqrt{y^2 \plus{} z^2}, \sqrt{z^2 \plus{} x^2}$ are positive integers.

$ \sqrt{x^2 \plus{} y^2 \plus{} z^2}$ is a positive integer.

Kevin Hu is being very tricky and posting an open problem on High School Basics. [url]http://en.wikipedia.org/wiki/Euler_brick[/url]

Hmm, I'm not really sure where to put open problems.  I'll just lock it.


function
real analysis
real analysis unsolved

[quote="Slizzel"]We have $a,b\in\left(0,\frac{1}{2}\right)$ and $f: \mathbb{R}\to\mathbb{R}$ a continuous function such that $f(f(x))=a\cdot f(x)+b\cdot x$ for any real $x$. Show that $f(0)=0$.[/quote]

You forgot a space: $f(f(x))=a\cdot f(x)+b\cdot x$.

Sketch of the proof:

1) $f$ injective
2) $f$ have no foxed points, hence it is increasing.
3) If $f(x)\ge x$ then $fof(x)\ge f(x)\ge x$ hence $fof(x)\ge f(x)+x>a f(x)+bx$.
4) $0\le f(x)\le x\Rightarrow 0\le fof(x)\le (a+b)x$. Denote by $a_n=f^{(2n)}(x)$ is convergent to c and $fof(c)=c$ which implies that c=0 hence $fof(0)=0\Rightarrow f(0)=0$.

I think it works at a first sight. Good luck in putting pieces together.

Assuming there are no fixed points $f(x)>x$ (1), $f(x)<x$ (2) and $f$ is injective + continuous,m hence strictly monotonic. So from (1) $f$ is st. increasing  and from (2) $f$ is st. decreasing. We get some relations which are not true for any real $x$'es. (first $x=1$ then $x=-1$). So there is a real value for which $f(m)=m$. Then $f(f(m))=a\cdot f(m)+b \cdot m\Longleftrightarrow f(m)=a\cdot m +b\cdot m$ or $m\cdot (a+b-1)=0$ and this is true if $m=0$. Thus $f(0)=0$  :) .

Thanks xirti for the hints... I posted this because I think is a nice problem :yup:


http://spot.pcc.edu/~jkissick/winter/math112exam1review.pdf

Number 3 please.

I got f(c)=62,000 cos (pi/41) x + 138,000

which doesn't work!

Please help!
Thanks!


limit
induction
real analysis
real analysis unsolved

We define two sequences $(x_{n})_{n\geq 0}$, $(y_{n})_{n\geq 0}$ by $x_{0}>0,y_{0}>1$  and $x_{n+1}=\frac{x_{n}}{1+y_{n}x_{n}^2}$, $y_{n+1}=y_{n}(1+x_{n})^2$ for any natural $n$. Show that $(z_{n})_{n\geq 0}, z_{n}=x_{n}\cdot y_{n}$ converges and find $\lim_{n\rightarrow\infty}{z_{n}}$.

From easy induction we find out that ${x_n}$ is decrising and ${y_n}$ is increasing. There for $\lim_{n\to\infty}{x_n}$ exists.
We find this:  $x_{n+1}=\frac{x_{n}}{1+y_{n}x_{n}^2}<\frac{x_{n}}{1+x_{n}^2}$ and $\lim_{n\to\infty}{x_{n+1}}\leq \lim_{n\to\infty}\frac{x_{n}}{1+x_{n}^2}$ $\rightarrow\lim_{n\to\infty}x_n=0$. That is all that i could u till now :blush:

From the first recurrence one has that $\frac{1}{x_{n+1}}-\frac{1}{x_{n}}=x_{n}y_{n}$
and from the second that $y_{n+1}-y_{n}=2x_{n}y_{n}+y_{n}x_{n}^2$.
Hence by taking account that $\lim_{n\to\infty}\frac{1}{x_{n}}=\infty$ by Cesaro-Stolz we get
$\lim_{n\to\infty}\frac{y_{n}}{\frac{1}{x_{n}}}=\lim_{n\to\infty}\frac{y_{n+1}-y_{n}}{\frac{1}{x_{n+1}}-\frac{1}{x_{n}}}=\lim_{n\to\infty}\frac{2x_{n}y_{n}+y_{n}x_{n}^2}{x_{n}y_{n}}=\lim_{n\to\infty}(2+x_{n})=2$
and that's it.


From $ A\equal{}\frac{bh}{2}$, we have $ A\equal{}\frac{8(10)}{2}\equal{}\boxed{40}$.


Given that $P(x)=(x+1)(x+2)(x+3)\cdots(x+10)$ . Find the coefficient of $x^8$ .

Can generating function be used here ? If so can anyone show how ?

[quote="Altheman"]if you think of the degree of the term, that means that from each part there has to be 8 x's chosen, and the rest will be constant terms, and those will be muliplied, so like 15*14, etc, and each will pair up with the other so it will be like 15*14 +15*13+...+15*1, etc for each one so sum 1 to 15 = $\frac{15*16}{2} = 120$ then 120*15, then 120*14, etc, so it would be 120^2, but we have counted 14*15 and 15*14 so dividing by 2, we get $\boxed{7200}$ i think this is right, i may have made a mistake though[/quote]

But i use $\sum ab=\frac{(\sum a)^2-\sum a^2}{2}=\frac{55^2-385}{2}=1320$

Where could possibly gone wrong ?

 :?


From http://www.mugglenet.com:

"Book 6 title ... maybe!
The following was posted on a [url=http://boards.gamefaqs.com/gfaqs/genmessage.php?board=245&topic=14784493]message board[/url] today

Earlier today, someone got behind the door at JKRowling.com without hacking, and was greeted by a surprise. As we can see from the date, this information isn't supposed to be released until the first of July. It took the person to a book, on the left page was the dark mark, and on the right page was written the following:

    1st July 2004

    Erm it is perhaps time for me to reveal
    the title of Book 6! I am nearing completion of the
    first draft and have definitely decided upon a
    title. I had numerous possibilities in mind, but
    as I wrote, I realized that only one did the book
    justice.

    The more clever Potter fans may recognize that
    the title relates directly to both 'Order of the
    Phoenix' and 'Chamber of Secrets.' I will be
    highly amused if one of you can guess exactly
    what's going on!

    On a side note, the length of the book is
    growing so fast, I could swear Filius has put an
    Engorgement Charm on it; an editor's nightmare!
    I promised myself not to exceed the length of
    'Order of the Phoenix,' but every day that passes
    makes that undertaking more difficult.
    Incidentally, I have much more work ahead.

    Alas, the time has come! Here is the reward for
    you lot that so diligently figured out all the puzzles
    on the site. If you scroll down to the bottom of
    this page, I will reveal to you the title of Book 6!

    ...

    The title of Book 6 is:
    [u]Harry Potter and the Pillar of Storg[/u]

...The writing closely mirrors JKR's style...but, of course, we can't offer any sort of confirmation yet. Stay tuned.

For reference:

- Storg is pronounced Stor-gee.
- We had the book 5 title more than a year before its release.
- Storg is Greek for 'parental love'."

Oooh, this is interesting.

this is what our world needs more of. pillars that give you love.
it is a bit interesting, but i'm betting some people have already lost interest. 7 books just seems like a lot, and if somebody started to read the first book at 11, i'm not sure they'd read the 7th at 18 or 19.

Why the book 6 title is probably fake
Over the past 24 hours, over 2000 comments have been left regarding the possible title of Harry Potter and the Pillar of Storg for book six. Some of you have made strong cases for why the title may be fabricated:

For one, "she" uses American punctuation instead of British punctuation on this sentence: The more clever Potter fans may recognize that the title relates directly to both 'Order of the Phoenix' and 'Chamber of Secrets.'

In Britain, the ending quotation mark should go inside the period. As I was reading her biography on her site today, I noticed that she punctuated things the "British way," which leads to the assumption that either she made a typo in her book 6 post (the same problem appears in the next paragraph, except with a comma rather than a period) or that an American typed it up for her.

The first assumption seems incorrect because it is unlikely that she would make the same mistake twice in a row. The second doesn't seem likely either, because she affirmed that she is indeed the one writing the things on her website, and I'm sure Lightmaker made some easy system allowing her to update, so she could do it herself. Another reason the second assumption does not seem probable - the title of the sixth book is classified information, and she wouldn't let just any staff member at Lightmaker type that up, because then it would leak in no time.

Also, why would Jo put that information anywhere on her site prior to when she wanted to release it? She is the master of secrecy, and you think she'd be able to wait until July 1 to put it on her site. Although it's not in an easy-to-access area, it seems very non-JKRish to do that.

It is a bit strange that she would include a difficult-to-pronounce word such as Storg in the title of her book.

One more reason: The only way that some visitors can "hack" past the door is by downloading the file to their computer, and then opening it to find this link to a book with Latin characters on the pages. Once you open the file on your computer, you can modify it however you like, which includes modifying the text on the book.

Other reasons exist for why the title may be fabricated, these are just some.

Updates: A few more reasons...
1) When watching the video, the "help" tab disappears momentarily. While perusing her site, the "help" tab never disappears from the screen.
2) The Dark Mark (what Jo uses as a spoiler warning) is on the wrong side of the page. The Dark Mark should be on the page with the spoiler, and you then have to click on the Dark Mark to reveal the text. The Dark Mark then appears as a faded image underneath the text.
3) The font used for "Book 6" does not appear anywhere else on the site. Look at the words "Book" and "Biography" (on the main page) -- the B's don't match.
4) Although the text circulating the Internet said that the date on the post was July 1, a closer look at the video reveals that the date is "22.06.04" (June 22).

^from mugglenet.net

heh,  Smart Mathfiend.  I agree that it is probally fake and that a title might not have even been figured out.

Hmm... I just went to the site. I find the "Wall of Shame" very entertaining.

[quote="mathnovice"]random things and then meet Voldemorte at the end. "[/quote]

May I point out that in the third book, they did not meet Voldemort at all.

I think that the books have not been as near entertaining as they could have been, but they are still good.  book V was different.  WE found out somethings that I at least had been suspecting for awhile.  The books right now I think are setting up for something at the end of the series but I don't know.

Book V seemed to exist for the purpose of setting up the plot for the next two. I didn't quite like it as much as the others. The other books could be read by themselves, whereas the fifth one requires past and future books to complete it. All five of the books are very intriguing, though, so I am looking forward to the 6th book.

According to JKR's official site http://www.jkrowling.com book 6 will be called [u]Harry Potter and the Half Blood Prince.[/u]

[quote="mathnovice"]I have found that the Harry Potter books have been going downhill since Book III. The first part of Book IV was awful, and the entire Book V was mind-numbingly boring. Moreover, the plot has gotten rather tediously repetitive and predictable: random things and then meet Voldemorte at the end. JK Rowling should keep in mind the phrase "quality, not quantity."[/quote]
I always thought they were going steeply uphill ever since the first book...I thought the start of Book 4 was amazing and the entire Book 5 fast-paced and perfectly suited to readers like me...are you sure we're reading the same Harry Potter books?

Give JK some time. I like reading well-written long books. However, JK has come up with some totally gnarley new stuff although the fifth book was "different" in sort of bad ways. There needs to be more lust and hate. JK has made Harry a total looser with everybody, friends, family, and especially girls.

[quote="churchilljrhigh"]JK has made Harry a total looser with everybody, friends, family, and especially girls.[/quote]

Couldn't you call that a strategic maneuver to allow people like me to better relate?

[quote="hikari"]I'd have to agree that the way JKR wrote the fifth does make it easier to relate to. Everyone has those feelings and if you were going through the things Harry had to go through you would be even more likely to have them. He's had a hard life and his feelings show it well.[/quote]

Yeah, most people can relate to both of their parents as well as the only other parent-like adult in their life being murdered.

I think Book V was a fairly good book but it seems to rely too much on the previous and especially future books to finish up the plot.

[quote="Yasha"]I think Book V was a fairly good book but it seems to rely too much on the previous and especially future books to finish up the plot.[/quote]

Also a good marketing idea.  Increase sales of the first four in the series in addition to the fifth.

I just checked JKR's website, and found out the title

Spoiler:[hide] The title is Harry Potter and the Half Blood Prince[/hide]

For those of you having trouble opening the brick wall behind the door, this picture has the order in which to press the bricks.

I liked book five, and i think that while 7 books is a lot, at least it isn't totally arbitrary as it corresponds to the number of years.  It also has its precedents, like the Narnia series.  I also think that people will still read them even if they're older.  I will, anyway.  And I was 11 when I started, and by the time the last book comes out I probably will be 19 or even 20.  Scary thought.  I just hope the trend of increasing length stops.  I like reading, but it's kind of hard to carry around a 840-page book and I don't have unlimited bookshelves.

My 40-something-year-old friend stood in line to get the fifth book at midnight. She was not unusual (at least in that sense). I am definitely thinking there will be people who started off as young kids reading these when the last two (finally) come out.

my 82 or something year old grandmother stood in line to get the 5th book at midnight when it came out.  The books intrests a lot of people.  Like my dad who read LoTR when he was in highschool loves HP.

[quote="hikari"]I'd have to agree that the way JKR wrote the fifth does make it easier to relate to. Everyone has those feelings and if you were going through the things Harry had to go through you would be even more likely to have them. He's had a hard life and his feelings show it well.[/quote]

Yeah, but it's such an abrupt chnge from about a couple of weeks ago (in the story) when Harry Potter thought a whole lot less of himself, and cared a lot about Cedric's passing.

[quote="RobinHood3000"]I've seen people who don't mind Lord of the Rings loathe Harry Potter due to religious beliefs, though.  Perhaps I just don't get it, but it sounds like something of a double standard to me.[/quote]

The reason is that LoTR is a religious allegory, totally and completely. Tolkien was a lifelong Catholic, and that's reflected in his works. (If you don't get it from the LoTR trilogy, I suggest you read the Silmarillion and the Lost Tales of Numenor.)

No, I understand the religious convictions of the author.  I took a course on mythological structure in modern literature.  But the reason behind the shunning I witnessed of the Harry Potter license was because of the existence of magic.  From what I overheard, "Magic in Lord of the Rings is fine, because the movies and books were made by God-fearing people.  Magic in Harry Potter is bad because magic is not a God-given gift, and is the work of Satan."  I swear, on every scrap of honor I have, that that is what I heard.

I've heard some things similar to that as well. Honestly, I don't see how people like that survive in this world without a little imagination. It's sad how some people still want to live in the 1600s.

Though Rowling's books are dulling and only slightly amusing, the plot is still intricately woven - though not as clever and fascinating as some series. If anybody insists on dissing Harry Potter, please do it fashionably and flawlessly. People are righteous in disliking any book against their religion, but perhaps not all the world is similarly religious. Potter is a marvelous figure, who survives in a world not quite like our own, and Rowling has created a rather stunning tale. It may not be my favorite series - that award goes to Scott Card (Ender's Game and Ender's Shadow), but it deserves enough praise such that when the 6th comes out, I'll be first in line to get a glimpse.

you just simultaneously dissed and praised a book. congratulations, i think. very impressive.

It's not people disliking the book that bothers me, so long as they have a solid reason.  It's when people dislike it on the basis of religion with a double standard for similar series, simply because they refuse to suspend their disbelief.


calculus
derivative
calculus computations

I am starting into differentiation and I am stuck with:

Find all the solutions for:

$ \frac{dx}{dt} \equal{} |x|$.

$ f'(x)\equal{}1$ for $ x>0$ and $ f'(x)\equal{}\minus{}1$ for $ x<0$.

JRav, you're probably going to want to read that again.

Since $ |x|$ is continuous in $ t$ and Lipschitz in $ x$, the standard uniqueness theorem applies. We can easily find particular solutions given any initial condition, and these (given in V.V.'s post) are all solutions.

(Language note on the title: the English form is "equation", and qu is standard for that sound)


Revista de Matematică a elevilor din Timişoara are pagină web? Dacă da, atunci vă rog să-mi daţi şi mie adresa. Oricât am căutat-o cu Google, am găsit orice altceva, dar numai RMET nu.

Pai se numeste RMT, nu RMET.Cauta cu asta.Dar nu prea cred ca are pagina web. ;)

Bineînţeles că n-am căutat "RMET", ci Revista + Matematica + Timisoara şi alte combinaţii de genul acesta. Am folosit prescurtarea RMET în mesaj, ca să nu mai lungesc vorba. Întrebarea rămâne aceaşi: ştie cineva adresa de pe net pentru RMT? Credeţi-mă că am căutat-o singur până m-am săturat şi numai în disperare de cauză am lansat aici întrebarea.

Poti lua legatura cu redactia si sa intrebi daca are pagina de web.Incearca pe rmt.mate@yahoo.com sau sa iei legatura cu directorul RMT, pe dbirchi@xnet.ro Have a nice day, happy! :lol:


MATHCOUNTS
ratio
probability
geometry
rectangle
perimeter
3D geometry

I'll keep the contest open until 5pm EST tomorrow, which is Wednesday the 11th.  Sorry for my ignorance once again.

Try to take this within the 40 minute mark and with no calculator.

Sprint round:
1. Find the units digit of 2^59-7^33

2. How many 0s are at the end of 32! ?

3. Simplify: (sqrt48)/(sqrt9)+(sqrt54)/(sqrt8)

4. Pump A can fill a tank in 30 minutes. Pump B can fill the same tank in 50 minutes, and Pump C fill the same tank in 75 minutes. How long will it take all 3 to fill the tank together?

5. Right Triangle ABC is inscribed in circle N. C is a right angle, and AB=12in. What is the circumference of circle N in inches? Express your answer in terms of pi

6. Walter has 5 bottles of juice. He has cranberry and
orange juice. We know that he has at least two bottles of each type of juice, and the bottles must be at least 12 oz. and at most 20 oz. If he has a total of 80 oz. of juice, and has a 16 oz. bottle of orange juice, what is the maximum ratio of the amount of cranberry juice to the amount of orange juice? Express your answer as a common fraction.

7. The median of the set {5,7,7,10,11,15,19,21,x} is 10.
x is a positive integer. Find the sum of all possible values of x.

8. A point S is chosen inside the square MNPQ. What is the probability that the angle MSN is acute? Express your answer as a decimal to the nearest hundredths.

9. The square of a number n is eight more than twice n. What is the value of n?

10. What is the sum of all multiples of 7 less than 200 and
whose digits have a sum of 10?

11. At sunrise, the temperature in Avon, Connecticut was 50 degrees farenheit. At noon, the temperature was 72 degrees farenheit. If the temperature increases by 4 degrees every hour, then when was sunrise?

12. A train travels at a speed of 45 miles per hour and a plane travels at a speed of 200 miles per hour. If the train leaves 5 hours before the plane, and the plane has traveled 700 miles, how many miles has the train traveled?

13. Two different integers are randomly selected from the set of integers greater than 0 and less than 7 . What is the probability that their they have no common prime factors? Express your answer as a common fraction.

14. A waterbottle is 3/4 full. If Jill drinks 2/5 of the remaining contents, and then Joe drinks 1/3 of the remaining contents after Jill has drank, how much water is still in the water bottle compared to when it was full? Express as a common fraction.

15. Arty has 5 brushes. They are red, yellow, green, blue, and black. He places them in a specific order on the table. The yellow is not on either end, or the center. Counting from the left, the position blue is in is not a prime number. black is positioned between the red and yellow. Finally, the green brush is farthest away from blue, and closest to red. What brushes are to the right of black?

16. In triangle ABC, AB=BC, AC=10. D is the midpoint of AC, and all sides of ABC are integers. If the median from B is also an integer, find the value of it.

17. How many 5-digit numbers can be made with the digits 1,2,2,3,5?

18. What is the sum of all distinct roots of the equation x^5+5x^4+10x^3+10x^2+5x+1?

19. A rectangle is inscribed in a circle with a radius of 58 units long. The ratio of the dimensions of the rectangle is 20/21 . How many units are in the perimeter of the rectangle?

20. Find the area of a triangle with edges 3.25, 3.5 and 3.75. Express your answer as a common fraction.

21. Jeff the coat boy is serving at a party. He notices that every 2nd coat is black, every 5th coat is an overcoat, and every 7th coat is leather. If there have been 9 overcoats and 6 leather coats, how many black coats are there?

22. A slice is taken out of a circular pizza with a raduis of 9 inches. The slice has a central angle of 80 degrees. A rectangular pizza with side lengths 12 inches and 16 inches is cut into 20 equal pieces. What is the positive difference between the area of the circular slice and one of rectangular pieces?

23. Find the sum of the two roots of the equation x^2-14x+4=23

24. 3 tetrahedrons have the numbers 1-4 on their faces. If they are rolled at the same time, what is the probability of the sum of the faces facing down being prime?

25. A number is Jefrozen if it satisfies the inequality x^2-12x+36>40 and if it has 6 positive integral factors. What is the smallest Jefrozen number?

26. Brien takes naps of 1 hour each, one at 2:00 and another at 5:00. Jill takes naps of 30 minutes each, one at 1:45, another three hours after she started to her first one and another 45 minutes after she started her second one. Bob takes
naps of 90 minutes each, and takes naps 5 minutes after Jill falls asleep the first time and 10 minutes after Brien falls asleep for the second time. How long are all 3 of them asleep?

27. A sphere is inscribed in a cube with edge lengths of 50. A plane then crosses the sphere and they intersect to make a circle with diameter 14. How far away is the center of the circle from the center of the sphere?

28. Charlie, Evan and Gary play a game. First, Charlie rolls a dice. If he rolls a 4, then he wins and the game is over. If he doesn't roll a 4, then Evan tries to roll a 4. If he doesn't succeed, Gary tries. If he doesn't do it, then it goes back to Charlie and it keeps rotating. How much greater is Charlie's probability of winning than Gary's probability of winning?

29. A supermarket is having a huge sale. All food is discounted by 25%. All toys are discounted by 20% and everything else is discounted by 10%. If 80% of the store's products are food and 5% of the products are toys, then what is the average discount on an item? In other words, how many percent is the average item discounted?

30. In an 8x8 checkerboard, each side has a length of 16 inches. Coin A has a circumference of 1.2pi inches. Coin B has a circumference of 1.6pi inches. What is the probability that when both glided across the board, the edges of coin A don't touch any lines on the board, but the edges of coin B do?

[quote="Syntax Error"]
8. Let the [unparseable or potentially dangerous latex formula]b=a^b/(a*b) What is the value of (43? Express your answer as a common fraction.
[/quote]

I also said something about number 8...

[quote=RedFireTruck]BUMP!!!!![/quote]

You realize you can get a postban for bumping extremely old threads....


algebra
polynomial
superior algebra
superior algebra solved

When is $X^{n}+Y^{n}+Z^{2}$ irreducible over $k[X]$, where $k$ denotes a field ?

When $x^{n}+y^{n}+z^{2}= p(x,y,z)q(x,y,z)$ is some factoring, so is the image in $k[x,y,z]/(y-1)$, where the original polynomial is $x^{n}+1+z^{2}$. When $char(k) \neq 2$, then $z^{2}+1$ is either irreducible or factors into two different factors, thus by Eisensteins criterion (with the prime=one of the irreducible factors), we get that it is irreducible in $k[x,y,z]/(y-1)$, so the original one was in $k[x,y,z]$ (after some computation).

When $char(k)=2$ and $n=2k$ is even, we have $(x^{k}+y^{k}+z)^{2}= x^{n}+y^{n}+z^{2}$.
When $n$ is odd, we do the reduction $\mod (z-1)$ this time to get again irreducibility.

Hope it's at least correct...  :maybe:

[quote="ZetaX"]When $x^{n}+y^{n}+z^{2}= p(x,y,z)q(x,y,z)$ is some factoring, so is the image in $k[x,y,z]/(y-1)$[/quote]

Nice, I didn't think about it and I went all the way through brute force computations to get the result  :blush:


In right triangle $ ABC$ with right angle $ B$, $ \overline{AC}\equal{}\sqrt{12}$ units and $ \overline{BC}\equal{}\sqrt8$ units. How many units are in the length of $ \overline{AB}$?

$ \overline{AB}^2 \plus{} \overline{BC}^2 \equal{} \overline{CA}^2 \iff \overline{AB} \equal{} \sqrt {(\sqrt {12})^2 \minus{} (\sqrt {8})^2} \equal{} \boxed{2}$.


In this format:

Albert Gu
8
Redwood Middle School
8
10

Just curious.

Name: if you don't know already, I'm not about to tell you...you can easily find out with a little research anyway
Grade: smallest number greater than AIME score with the same parity as AIME score
School: see name
AIME score: see name
USAMO score: one more than twice AIME score

I have a strong urge to fill in bookaholic's information for her.

Hey I've used her name a lot on this forum.

Heh, please don't. I don't care if people find out my statistics, but they might as well have to work for it. It isn't hard at all.

I have no idea how I got that score, by the way. My best guess is that I got the majority of the points for #4 after all, which is really weird because I really don't think my so-called solution deserves it. Unless they gave a point for proving the equality case on #6 or something, which I somehow doubt they did.

Okay I will make people work for it by making them ask me on the forums before I post it, happy?

I think my score distribution was 7/0/0/3/0/0. I'm not sure how I got 3 on #4... probably one for the correct answer and 2 for the "proof", but I thought they would probably only give 1 for "proofs" so maybe I got 1 for the equality case for #6? Anyway that's my analysis of my score.

Hah, I told you I would fail! And you didn't believe me! And you kept making up bogus scores for my #1!

Hey, what was so bad about my three posts here that they all merit ratings of 1?

[quote]Hah, I told you I would fail! And you didn't believe me! And you kept making up bogus scores for my #1![/quote]
Should I even bother pointing out that I was [i]right[/i] about your score on #1 and you were spectacularly wrong? Darn, I just did. Oh well.

I haven't rated any posts anywhere on the forum yet, so don't look at me.

Since I didn't take or make USAMO, I'll put 0 for that slot.

[-cut for security reasons-] (Don't bother finding out my name)
8
[-cut for security reasons-] (Don't bother finding my school)
6
0

Lol, i'm also getting a strong urge to reveal certain information on here, vishalarul. Well, maybe not...

How about this. Nobody reveals who I am and I won't reveal anyone else (not that I know that many people)

[You don't want to know]
8
[Somewhere]
5 :oops:
N/A

a person
7
a school
3 :wallbash_red:  :wallbash_red:  :wallbash_red:
N/A


Points $D,M,N$ are chosen on sides $AC,AB,BC$ of triangle $ABC$. $AN,CM$ intersect at $P$, which lies on $BD$. Show that $BD$ is a median if and only if $AP: PN=CP: PM$

[hide="Large Hint"]
Ceva's, and area ratios.  :wink:  [/hide]

[hide="Another hint"]Use Menelaos theorem[/hide]

[hide="Sketch of Solution"]
The given condition is equivalent to
\[[ APM ] = [ CPN] \]
and since $[ADP]=[CDP]$ iff $BD$ is a median, we need to prove that $[BMP]=[BNP]$.
First, by the area formula on $\triangle AMP$ and $\triangle CNP$ we have that the condition
\[\iff AM \cdot h_{PC}= NC \cdot h_{PA}\iff \frac{AM}{NC}= \frac{h_{PA}}{h_{PC}}\]
where $h_{PA}$ is the distance from $P$ to side $A$, and likewise for $h_{PC}$.
But iff the lines are concurrent and $BD$ is a median we have by Ceva's that
\[\frac{AM}{MB}\cdot \frac{BN}{NC}\cdot \frac{CD}{DA}= \frac{AM}{MB}\cdot \frac{BN}{NC}=1 \implies \frac{BM}{BN}= \frac{AM}{NC}= \frac{h_{PA}}{h_{PC}}\]

\[BN \cdot h_{PA}= BM \cdot h_{PC}\implies [BMP]=[BNP] \]
Thus, by summing equal areas, we see that $[ABD]=[BDC] \iff \text{ BD is a median. }$ [/hide]


In a given triangle, prove that 2(r + R) = a(cot A) + b(cot B) + c(cot C).

$ a\cot A\equal{}2R\cos A$
So we need to prove $ r\equal{}2R(\cos A\plus{}\cos B\plus{}\cos C\minus{}1)$ which is a well known trigonometric identity.

[b]Bugi[/b], that one bugs me until now . . . . solution evades me

Sorry, made a typo, that 2 at end should disappear... (should be 1). I think it can be taken as a formula:

http://www.imomath.com/index.php?options=mbb|Bfa|f|Geo|For&p=0

2(r + R) = a cot A + b cot B + c cot C,
2(r + R) = a cot A + b cot B + c cot (180 - A - B),
2(r + R) = a (cos A/sin A) + b (cos B/sin B) - (c sin 2C)/(cos 2C - 1); what is next?

$ \text{Consider RHS}$

$ acotA\plus{}bcotB\plus{}ccotC\equal{}2R(cosA\plus{}cosB\plus{}cosC)$

                             $ \equal{}2R\left(2cos\left(\frac{A\plus{}B}{2}\right)cos\left(\frac{A\minus{}B}{2}\right)\plus{}1\minus{}2sin^2\left(\frac{C}{2}\right)\right)$

                            $ \equal{}2R\left(2sin\left(\frac{C}{2}\right)cos\left(\frac{A\minus{}B}{2}\right)\minus{}2sin^2\left(\frac{C}{2}\right)\plus{}1\right)$

                            $ \equal{}2R\left(2sin\left(\frac{C}{2}\right)\left(cos\left(\frac{A\minus{}B}{2}\right)\minus{}sin\left(\frac{C}{2}\right)\right)\plus{}1\right)$

                            $ \equal{}2R\left(2sin\left(\frac{C}{2}\right)\left(cos\left(\frac{A\minus{}B}{2}\right)\minus{}cos\left(\frac{A\plus{}B}{2}\right)\right)\plus{}1\right)$

                            $ \equal{}2R\left(4sin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)\plus{}1\right)$

                           $ \equal{}2\left(4Rsin\left(\frac{A}{2}\right)sin\left(\frac{B}{2}\right)sin\left(\frac{C}{2}\right)\plus{}R\right)$

                           $ \equal{}2(r\plus{}R)$

$ \text{Hence Proved}$

How did you jump from r = 4R(sin A/2)(sin B/2)(sinC/2)?

i can not see the logic. Is that a WELL known identity [b]picture74[/b]?

if it is, that will serve as your lemma. Do you have a proof of that?

Thanks


$ AB \equal{} AC$, $ \angle BAC \equal{} 80^\circ$. Let $ E$ be a point inside $ \triangle ABC$ such that $ AE \equal{} EC$ and $ \angle EAC \equal{} 10^\circ$. What is the measure of $ \angle EBC$?

$\textbf{(A)}\  10^\circ \qquad\textbf{(B)}\  15^\circ \qquad\textbf{(C)}\  20^\circ \qquad\textbf{(D)}\  25^\circ \qquad\textbf{(E)}\  30^\circ$

You messed up the diagram a little, E should be reflected across the altitude from A, not C, and $ \angle C \equal{} \angle B \equal{} 50$. But your idea was right. The answer is 20.


Suppose $ ABCDEF$ is a hexagon with all sides equal and opposite angles equal. Does it follow that it is regular? Explain your answer.

I've always thought that a regular polygon had all its sides (and angles equal). Am I missing the point?

I think what KrazyFX is saying is that if a hexagon has all sides equal but only [i]opposite[/i] angles equal, does it follow that all the angles are equal, therefore making the hexagon regular.

Thanks, I misunderstood. I should have realised that someone like KrazyFK would have a problem worthy of attack.

Pete Hein:

Problems worthy of attack
Prove their worth by hitting back.

No, Imagine it like the Following, but visualize it with equal sides. (I couldnt figure out the program... First time using it)

 [geogebra]2c8426db391afeea4fee3c7d838ce0f97af59d21[/geogebra] [/img]

minime1235813 is correct.

Contradictory example:

$ AB\equal{}BC\equal{}CD\equal{}DE\equal{}EF\equal{}FA$ ($ \equal{}1$ if you want), but

$ m\angle ABC\equal{}m\angle DEF\equal{}119^{\circ}$
$ m\angle BCD\equal{}m\angle EFA\equal{}121^{\circ}$
$ m\angle CDE\equal{}m\angle FAB\equal{}120^{\circ}$

Its actually kinda funny, in math the last year, We were taught this to be true until I showed my teacher (-:

Or, make it an hourglass shape:

 D\--------------/C
....\............../
......\.........../
......E\......../B
......../........\
......./...........\
....../..............\
...F/--------------\A

$ m\angle A \equal{} m\angle C \equal{} m \angle D \equal{} m \angle F$

$ m\angle B \equal{} m\angle E$ . . . (They are reflex angles.  That is, they are each greater than 180 degrees.)

Or "flip out" the sides and make angle B and angle E both obtuse for a
different type, again having two different numbers of degrees of angles
inside the hexagon.

I like that solution, because you can draw it on the board without being overly concerned that you measured the angles correctly, as I would have had to.


to Hienchuoi@ :
are you prove this proplem ? :D
 :no:

[quote]no one can? @mathboy_cntHDVN: you known this problem? give me the solution, thanks[/quote]
the solution very long
  after inequaliti i don't know it true or false :
 ${\sum_{i=1}^{n}a_{i}x_{i}x_{i=1}\leq\sum_{i=1}^{n}a_{i}}\sum_{i=1}^{n}x_{i}^{n}$


inequalities
rearrangement inequality
inequalities proposed

[quote="ductrung"]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\] [/quote]

[hide]$\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$

$\Leftrightarrow \sum_{\text{cyc}}a^{4}b+\sum_{\text{sym}}a^{4}b \ge \sum_{\text{sym}}a^{2}b^{2}c+\sum_{\text{cyc}}a^{2}b^{3}$

Since
$\sum_{\text{sym}}a^{4}b \ge \sum_{\text{sym}}a^{2}b^{2}c$ by muirhead
and
$\sum_{\text{cyc}}a^{4}b\ge \sum_{\text{cyc}}a^{2}b^{3}$ by rearrangement,

the given inequality is true.[/hide]

SplashD try $a \rightarrow 0$ and see that your rearrangement is wrong :!:

[quote="SplashD"][quote="ductrung"]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\] [/quote]

[hide]By Rearrangement,
$\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}$
$\le \frac{a(a+b)}{a+b}+\frac{b(b+c)}{b+c}+\frac{c(c+a)}{c+a}$
$=a+b+c$
$\le \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$ by cauchy[/hide][/quote]

You can't apply rearrangement ineq so, as even if you assume $a \geq b \geq c$, you cannot get the order of $a(a+b),b(b+c),c(c+a)$.

[quote="ductrung"]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\] [/quote]

$\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\leq (a+b+c)\leq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$

Ineq.
$\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\leq (a+b+c)$; after transfer is:

$b^{3}a+a^{3}c+c^{3}b\geq a^{2}bc+b^{2}ac+c^{2}ab$ which is true by  Muirhead ineq!!!

Ineq.
$(a+b+c)\leq \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$ after transfer is:
$a^{2}+b^{2}+c^{2}\geq ab+ac+bc$, wich is true by $AM\geq GM$

Hi SplashD: It is NOT true \[\sum_{cycl}a^{4}b\ge \sum_{cycl}a^{3}c^{2}\] Counter example can easily be found.

Hi Fitim: I think you make mistake because it can be proved that \[\frac{a^{k}(a+b)}{a+c}+\frac{b^{k}(b+c)}{b+a}+\frac{c^{k}(c+a)}{c+b}\ge a^{k}+b^{k}+c^{k}\] for $k=0,1,2$ (does not hold for $k=3$).

well if we proceeded as SplashD did we would have had no trouble noticing that :
$\sum_{cyc}a^{4}b+\sum_{sym}a^{4}b\geq \sum_{sym}a^{2}b^{2}c+\sum_{cyc}a{}^{2}{}^{b}{}^{3}$
$\leftrightarrow \sum_{cyc}ab(a+b)(a-b)^{2}+\sum_{cyc}(a^{4}b+b^{3}c^{2}-2a^{2}b^{2}c)\geq 0$ :wink:

[quote="SplashD"]I fixed it, I hope it's right this time.[/quote]

As Ductrung said, it's not true.

[quote="Fitim"]$b^{3}a+a^{3}c+c^{3}b\geq a^{2}bc+b^{2}ac+c^{2}ab$which is true by Muirhead ineq!!! [/quote]

This is true, but not by Muirhead. Muirhead inequality can be applied for only symmetric sums. For 3-variable cyclic sums, there was a post:
http://www.mathlinks.ro/Forum/viewtopic.php?t=103761
This can explain why SplashD's solution is wrong and fix Fitim's solution.

PS. $b^{3}a+a^{3}c+c^{3}b\geq a^{2}bc+b^{2}ac+c^{2}ab$ was posted and so were some solutions: http://www.mathlinks.ro/Forum/viewtopic.php?t=103282

yeag but in Fitim's proof if you take $a\rightarrow 0$ you will see that there is a mistake ...(in fact most probably he confused sides)

Dear Ductrung

is your solution with full expansion as Albanian Eagle did?

Thank you very much.

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\]
[/quote]

General problem:
Find the greatest constant $ k$ such that the following inequality holds for any non-negative real numbers $ a,b,c$:
$ \frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le \frac{k(a^{2}+b^{2}+c^{2})}{(a+b+c)}+\left(1-\frac{k}{3}\right)(a+b+c)$
I found $ k=2$

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\]
[/quote]

OK, following solution is for manlio, and Eagle. Thank you Trung for a very nice problem.

Clearly, the inequality is equivalent to
\[ \sum \frac{a(a+b)(a+b+c)}{a+c}\le 3(a^{2}+b^{2}+c^{2}). \]
Or, after reducing similar team, it becomes
\[ \frac{ab(a+b)}{a+c}+\frac{bc(b+c)}{b+a}+\frac{ca(c+a)}{c+b}\le 2(a^{2}+b^{2}+c^{2})-ab-bc-ca. \]
WLOG, assume that $ x,y,z$ is a permutation of $ a,b,c$ such that $ x\ge y\ge z.$ Since
\[ xy(x+y) \ge xz(x+z)\ge yz(y+z). \]
It is easy to refer that (by rearrangement inequality)
\[ \frac{ab(a+b)}{a+c}+\frac{bc(b+c)}{b+a}+\frac{ca(c+a)}{c+b}\le \]

\[ \frac{xy(x+y)}{y+z}+\frac{yz(y+z)}{y+x}+\frac{zx(z+x)}{z+x}=\frac{xy(x+y)}{y+z}+\frac{yz(y+z)}{y+x}+xz. \]
Now, we see that
\[ \frac{xy(x+y)}{y+z}+\frac{yz(y+z)}{y+x}-(xy+yz)=\frac{xy(x-z)}{y+z}+\frac{yz(z-x}{x+y}\]
\[ =\frac{y(x-z)^{2}(x+y+z)}{(x+y)(y+z)}\le (x-z)^{2}. \]
By this result, we can conclude that
\[ \frac{ab(a+b)}{a+c}+\frac{bc(b+c)}{b+a}+\frac{ca(c+a)}{c+b}\le (x-z)^{2}+(xy+yz+xz) \le 2(x^{2}+y^{2}+z^{2})-(xy+yz+zx). \]
As the desired result.

Actually, above solution shows the following stronger inequality
\[ \frac{ab(a+b)}{a+c}+\frac{bc(b+c)}{b+a}+\frac{ca(c+a)}{c+b}\le ab+bc+ca+\max\{(a-b)^{2},(b-c)^{2},(c-a)^{2}\}. \]

The equality holds for $ a=b=c$ or $ a=b,c=0$ or permutation.

Very nice solution, Hung. Thank you and keep posting more nice solutions  :lol:
To Manlio, mine is not as nice: I rewrite the inequality to the equivalent
\[ \sum_{cycl}(a-b)^{2}\left( 1-\frac{ac(a+b+c)}{(a+b)(b+c)(c+a)}\right) \ge 0\]

[quote="ductrung"]Very nice solution, Hung. Thank you and keep posting more nice solutions  :lol:
[/quote]

Hi, actually I just notice that a lot of nice inequalities from you have not any solution, or brute force. So this inspires me some attempt.

Very interesting solution Hungkhtn. :)

Congratulations.

[quote=ductrung]For any positive real numbers $a,b$ and $c$, \[\frac{a(a+b)}{a+c}+\frac{b(b+c)}{b+a}+\frac{c(c+a)}{c+b}\le\frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}\][/quote]

[img]http://s12.sinaimg.cn/middle/006ptkjAzy76sjkfLTl8b&690[/img]


geometry
circumcircle
geometric transformation
reflection
angle bisector
perpendicular bisector
power of a point

Hi!

Let $\triangle EFG$ be a triangle, and $EH$ is the internal bisector of $\angle FEG$ with $H\in FG$. $EH$ meets the circumcircle of $\triangle EFG$ in $I$. Let $J$ be a point of the circle with $EI$ as a diameter and such that $HJ\bot EI$. Prove that $IJ=IG$.

Let $K$ be the reflection of $J$ wrt to $EI$. Then we have $HK\cdot HJ=HI\cdot HE=HG\cdot HF$ by the power-of-a-point-theorem, thus $JGKF$ cyclic. In every triangle, an angle bisector, the perpendicular bisector of the opposite side and the circumcircle concur (does this theorem have a name?), thus $I$ lies on the perpendicular bisector of $FG$. $I$ also lies on the perpendicular bisector of $JK$ and thus $I$ is the circumcentre of $JGKF$.


geometry
3D geometry
probability
geometric transformation
linear algebra
matrix
tetrahedron

An ant is on the vertex of a cube. After every minute, the ant randomly chooses one of the three adjacent vertices to which it will move. What is the probability that, after six minutes, the ant is on the vertex on which it started?

And a follow-up: this is a "dithering" or "memoryless" ant, i.e., it might make a step and then immediately undo that step.  What if we have a "decisive" ant, i.e., one that never goes back down the edge it just came up?

I did this [the original] problem recursively, and I just want to make sure my answer is correct:

[hide="Answer?"]
-----
$ \frac {183}{729} \equal{} \boxed{\frac{61}{243}}$
-----
[/hide]

Can anyone confirm, either with recursion or some other method?

Yes, that's right, but it would have been much more interesting if you'd posted your solution :).

Here's a heuristic method to get that answer, by the way: [hide]after 6 moves, the ant should have had an opportunity to go pretty much everywhere, so the probability of it being in any given place should be about the same as its probability of being in any other given place.  But, one caveat: the cube is a bipartite graph, so actually after 6 steps there are only 4 vertices we could possibly be at (the vertex we started at and the 3 other vertices that could be joined to it by face diagonals of the cube), so the answer should be about $ \frac{1}{4}$.  But we also know that the answer must have a denominator of $ 3^6 \equal{} 729$, and $ 729/4 \equal{} 182.25$.  This means we have one vertex that should be $ 183/729$ and three vertices that should be $ 182/729$; the vertex we started at is "special," so we get the answer of $ 183/729$, as desired.[/hide]  This technique was taught to me by user ComplexZeta a very long time ago.  It isn't a proof, but it very frequently gives the right answer.

For those who are curious, the answer to my problem is
[hide]5/16 -- still close to 1/4, but not quite as close -- which agrees with my intuition, at least.[/hide]

[quote="JBL"]the vertex we started at is "special," so we get the answer of $ 183/729$, as desired.[/quote]

Hilarious.  :)

I only bring this up because I had this as a problem for my last math team meeting, and no one could understand my logic.  :( Here's what I did:

[hide="Solution."]
=====
At any vertex, we can only be $ 0$,$ 1$,$ 2$, or $ 3$ steps away from where we started. If we let $ p(n,i)$ be the number of paths of length $ n$ that end at a vertex $ i$ steps away from the start, then we can make the following observations:

$ p(n\plus{}1,0)\equal{}p(n,1)$, since any path of length $ n\plus{}1$ to a $ 0$ vertex is simply a path of length $ n$ to a $ 1$ vertex with a move to $ 0$ appended.

$ p(n\plus{}1,1)\equal{}3p(n,0)\plus{}2p(n,2)$, since any path of length $ n\plus{}1$ to a $ 1$ vertex is either a path of length $ n$ to the $ 0$ vertex with either of three moves to a $ 1$ vertex appended, or a path of length $ n$ to a $ 2$ vertex with either of two moves to a $ 1$ vertex appended.

$ p(n\plus{}1,2)\equal{}2p(n,1)\plus{}3p(n,3)$, using similar logic to the previous case.

$ p(n\plus{}1,3)\equal{}p(n,2)$, using similar logic to the first case.

Using this, and the fact that $ p(0,0)\equal{}1$ and $ p(0,1)\equal{}p(0,2)\equal{}p(0,3)\equal{}0$, the recursion spits out $ p(6,0)\equal{}183$, and since there are $ 3^6\equal{}729$ path of length $ 6$, we get the answer.
=====
[/hide]

In retrospect, I can see why everyone looked so confused. It was a very awful silence after I was done... Is there a better way to explain this?

I'm a big fan of methods like this, and that's how I would have solved the question, as well.  For explaining this solution to others, I think it may help to draw the cube so that the vertex you start at is on the bottom with the long diagonal pointing up -- then the "path distance" you're talking about becomes the height in the diagram, and it's relatively clear that each step either increases or decreases your height by 1.  To help explain this idea of attack, it might be easier to start with simpler graphs (e.g., a square instead of a cube) where you can see what's going on more easily.  In a smaller example, you can actually use a different variable for every vertex, but then explain that for the purposes of the problem, you can treat "equivalent" vertices in a single recursive equation.  (It may be that the tricky conceptual part is figuring out where those recursive equations come from -- when you have a separate equation for each vertex, it is a little bit clearer.)  Also, people might be getting lost in the translation from a probability problem to a combinatorics problem, so make sure that's explained very carefully.

Incidentally, we can solve that recursive system explicitly: I'm going to use $ a_n$, $ b_n$, $ c_n$ and $ d_n$ for your $ p(n, 0), \ldots, p(n, 3)$ because it makes it easier to keep track of what is what (and maybe this would make your explanation clearer, as well?): we have
$ a_{n \plus{} 1} \equal{} b_n$,
$ b_{n \plus{} 1} \equal{} 3a_n \plus{} 2c_n$,
$ c_{n \plus{} 1} \equal{} 2b_n \plus{} 3d_n$, and
$ d_{n \plus{} 1} \equal{} c_n$.
First thing we can do is eliminate $ d$ using the last equation, reducing the third equation to
$ c_{n \plus{} 1} \equal{} 2b_n \plus{} 3c_{n \minus{} 1}$.
Now subtract three times the second equation from itself with shifted index to get
$ b_{n \plus{} 1} \minus{} 3b_{n \minus{} 1} \equal{} 3(a_n \minus{} 3a_{n \minus{} 2}) \plus{} 2(c_n \minus{} 3c_{n \minus{} 2})$.
Use our new version of the third equation to rewrite this as
$ b_{n \plus{} 1} \minus{} 3b_{n \minus{} 1} \equal{} 3(a_n \minus{} 3a_{n \minus{} 2}) \plus{} 2(2b_{n \minus{} 1})$.
Finally, use the first equation to eliminate $ b$ and get
$ a_{n \plus{} 2} \minus{} 3a_{n} \equal{} 3(a_n \minus{} 3a_{n \minus{} 2}) \plus{} 2(2a_n)$
or equivalently
$ a_{n \plus{} 2} \minus{} 10a_n \plus{} 9a_{n \minus{} 2} \equal{} 0$.
(Note that this involves only every other term -- this has something to do with the bipartiteness I mentioned.)  We also need the initial values $ a_0 \equal{} 1$, $ a_1 \equal{} 0$, $ a_2 \equal{} 3$ and $ a_3 \equal{} 0$.  Then we get $ a_{2m \plus{} 1} \equal{} 0$ for all $ m$ and $ a_{2m} \equal{} a \cdot 9^m \plus{} b\cdot 1^m$, with $ a \plus{} b \equal{} 1$ and $ 9a \plus{} b \equal{} 3$.  Thus $ a \equal{} \frac {1}{4}$ (as expected!), $ b \equal{} \frac {3}{4}$ and we have $ a_{2m} \equal{} \frac {9^{m} \plus{} 3}{4}$ and in particular $ a_6 \equal{} \frac {9^3 \plus{} 3}{4} \equal{} \frac {732}{4} \equal{} 183$.

[quote="JBL"] I think it may help to [...].  To help explain this idea of attack, [...]  Also, people might be getting lost in the translation from a probability problem to a combinatorics problem, so make sure that's explained very carefully.[/quote]

Thanks for your suggestions, I will definitely use them next meeting, especially the simplification to a square. I don't think I want to bring up the derivation of the explicit formula, though. A few meetings ago was devoted to counting and probability problems, so I don't think that was the reason why they were confused... unless they forgot...

Thanks again!

[hide="Original Problem"]We have 6 moves _ _ _ _ _ _

We must fill these with letters U, D, L, R, F, and B (up, down, etc). For consistency, let's start at the lower-left-front vertex. Our moves must also satisfy the following properties:

1. There must be an even number of moves in each of the three classes, U/D, F/B, L/R
2. If we isolate the moves of each of these three classes, they must alternate (ie, we cannot move "back" twice without moving "forward" in between)
3. If we isolate the moves of a class, the first move must be the appropriate one of R, U, B.

So now, the question is easy. If we use exactly

1. One class , there are 3 ways of picking this, and there is 1 path in this dimension (ie, UDUDUD). Total of 3 ways.

2. Two classes (4 moves of one class, 2 moves of another class), there are 3*2=6 ways of choosing the two classes, and 6-choose-2 ways of picking the two moves to attribute to the second class. Total of 6*15=90 ways.

3. Three classes (2 moves of each class), there is 1 way of choosing the 3 classes (we need all of them), and there are 6!/(2!)^3 = 90 ways of attributing the moves among these classes. Total of 90 ways.

Adding, there are a total of 183 paths.[/hide]

jvenezuela716: glad to be of service.

TZF: Interesting approach, although it has the disadvantage of being extremely specialized to the cube.  (One could simplify slightly by not distinguishing between U & D or between L & R, etc.)  How much effort does it take to acquire the general solution from that method?

Right, I could have cast it as: [i]"You have three light switches, all of which are off. You make 6 flips at random. What is the probability that all the lights are off?"[/i], and then it would sound simpler.

Let us make $ 2a$ flips on switch A, $ 2b$ flips on switch B, and $ 2c$ flips on switch C. Then, for each solution in the non-negative integers to $ 2a\plus{}2b\plus{}2c\equal{}2n$, we must add the number of ways to arrange $ 2a$ A's, $ 2b$ B's, and $ 2c$ C's.

In other words, we want $ \sum_{a\plus{}b\plus{}c\equal{}n} \frac{(2n)!}{(2a)! \cdot (2b)! \cdot (2c)!}$

Any ideas on how to simplify this? It should simplify to $ \frac{3\plus{}3^{2n}}{4}$, which yeilds a probability of $ \frac{1}{4} \plus{} \frac{3}{4} \cdot 3^{\minus{}2n}$

[quote="TZF"]$ \sum_{a \plus{} b \plus{} c \equal{} n} \frac {(2n)!}{(2a)! \cdot (2b)! \cdot (2c)!}$

Any ideas on how to simplify this?[/quote]  I vote trinomial theorem and roots of unity sieve: think about $ (\pm 1 \pm 1 \pm 1)^{2n}$.  (Like my method, this can actually spit out a unified answer for the odd and even case involving both $ 3^n$ and $ (\minus{}3)^n$.)

How about an answer based on Markov chains?
Create a connected graph linking all vertices of a cube together, and label the vertices A - H.
Then create a matrix which expresses the probability of moving from one vertex to another. I've gone ahead and made this matrix. Note that the matrix has a diagonal filled with zeroes. This means that the ant will always move off of its current vertex (it doesn't sit still).

The columns of this matrix represent where the ant is coming from, and the rows represent where it is going to.
Each entry is the probability of a certain step happening.

Now the magic of the Markov chains come into play. Because this state diagram is the same no matter WHERE the ant is, we can simply multiply this matrix by itself as many times as we need, then read off the desired probability from the spot.
$ \left(
\begin{array}{cccccccc}
 0 & 1/3 & 1/3 & 1/3 & 0 & 0 & 0 & 0 \\
 1/3 & 0 & 0 & 0 & 1/3 & 1/3 & 0 & 0 \\
 1/3 & 0 & 0 & 0 & 1/3 & 0 & 1/3 & 0 \\
 1/3 & 0 & 0 & 0 & 0 & 1/3 & 1/3 & 0 \\
 0 & 1/3 & 1/3 & 0 & 0 & 0 & 0 & 1/3 \\
 0 & 1/3 & 0 & 1/3 & 0 & 0 & 0 & 1/3 \\
 0 & 0 & 1/3 & 1/3 & 0 & 0 & 0 & 1/3 \\
 0 & 0 & 0 & 0 & 1/3 & 1/3 & 1/3 & 0
\end{array}
\right)$

This is the desired Markov matrix $ M$, and
$ \left(
\begin{array}{cccccccc}
 \frac{61}{243} & 0 & 0 & 0 & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} & 0 \\
 0 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & 0 & 0 & 0 & \frac{182}{729} \\
 0 & \frac{182}{729} & \frac{61}{243} & \frac{182}{729} & 0 & 0 & 0 & \frac{182}{729} \\
 0 & \frac{182}{729} & \frac{182}{729} & \frac{61}{243} & 0 & 0 & 0 & \frac{182}{729} \\
 \frac{182}{729} & 0 & 0 & 0 & \frac{61}{243} & \frac{182}{729} & \frac{182}{729} & 0 \\
 \frac{182}{729} & 0 & 0 & 0 & \frac{182}{729} & \frac{61}{243} & \frac{182}{729} & 0 \\
 \frac{182}{729} & 0 & 0 & 0 & \frac{182}{729} & \frac{182}{729} & \frac{61}{243} & 0 \\
 0 & \frac{182}{729} & \frac{182}{729} & \frac{182}{729} & 0 & 0 & 0 & \frac{61}{243}
\end{array}
\right)$ is the $ M^6$ (generated from Mathematica).

Now, we arbitrarily decide that we want the probability of moving from A to A in six steps.
All we need to do here is read off the first entry ($ \frac{61}{243}$)

Note, interestingly, that this matrix to an ODD power will give a probability of ZERO for staying in the same place. This makes sense because, after moving an odd number of steps, you will never be back where you started.

That method is exactly equivalent to the recursive method, except you didn't take the simplifying step of identifying some of the vertices (which results in a smaller but less symmetric situation).  :)

@JBL: Your add-on problem seems simpler than the original.

[hide="second"]In the first move, you must be one unit away from the point of origin. The next move means you should be $ \sqrt{1^2\plus{}1^2}\equal{}\sqrt{2}$ away from the origin. There are two cases for the next move: $ \sqrt{3}$ from the origin or $ 1$ from the origin.

The first case is just like the first two moves: After two moves we are 1 from the origin, and the probability that we get back to the origin is $ \frac{1}{2}$. The probability of this case happening is $ \frac{1}{2}$, so the probability of getting back to the origin is $ \frac{1}{4}$.

The second case has a probability of $ \frac{1}{2}$ happening. Then we have a $ \frac{1}{2}$ probability that we don't go bak to the origin, because this ant is decisive, and if he goes back to the origin on the fourth move he won't be there on the 6th move. Then we have a $ \frac{1}{2}$ probability that the next move brings the ant to $ 1$ unit away from the origin, and a $ \frac{1}{2}$ probability that the and [i]does[/i] go to the origin on this move. The probability of all this happening is $ \frac{1}{16}$.

These are the only two cases, and they are independent probabilities, so we add: $ \frac{1}{4}\plus{}\frac{1}{16}\equal{}\boxed{\frac{5}{16}}$.[/hide]

The first problem has a few cases: First goes back to the origin on the 6th move, first goes back to the origin on the fourth move and comes back two moves later, goes back on the second move and takes four moves to get back there, or goes back on every other move. The last case is easy. The next two cases are compound, but they are essentially the same. The first case is the hardest.

Okay good -- can you solve it in the general case (for when 6 is replaced by $ n$)?

Here's yet another solution to the original problem.  After a phase of 2 steps, we have probability $ \frac{1}{3}$ of being back at the start, and probability $ \frac{2}{9}$ each of being at one of the 3 other points of the tetrahedron.  Another way to say that: With probability $ \frac{1}{9}$, we do the lazy thing of staying where we are; otherwise, we choose one of the 4 points of the tetrahedron equally at random.  If we ever choose the non-lazy thing, then we have lost memory of where we started, and so from then on will be equally likely to be at any tetrahedron point.  The probability of making all 3 phases lazy is $ \frac{1}{9^3}$.  So the probability of being back at the start after 6 steps is
\[ \frac{1}{9^3} \plus{} \left( 1 \minus{} \frac{1}{9^3} \right) \cdot \frac{1}{4}.
\]
That comes out to $ \frac{61}{243}$, agreeing with the previous answers.

This method generalizes to any number of steps, agreeing with TZF's formula.   This method is sometimes called coupling.

By the way, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125811&search_id=1178706192&start=26]here is another matrix approach[/url] for the original problem, from an old thread.

We can make a six-state model for Joel's problem by storing direction, slightly more complex than the 4-state model that can be used for the original problem. We define our states as follows:

0 - the origin, directed only toward (1A)
1A - any of the three points 1 step from the origin, directed only to points 2 steps from the origin (level 2B)
1B - 1 step from the origin, directed toward the origin and some point 2 steps from the origin (2B), with equal probability
2B - 2 steps from the origin, directed toward (3) and (1B) with equal probability
2A - 2 steps from the origin, directed only toward (1B)
3 - 3 steps from the origin, directed only toward (2A)

[hide="The relevant stochastic matrix is"]$ \left( \begin{array}{cccccc} 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 1/2 & 0 & 0 & 1/2 & 0 & 0 & 0 & 0 & 1/2 & 0 & 0 & 1/2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \end{array} \right)$
This can be translated into a recursion.[/hide]

Letting the probability sequences for those 6 states be $ a_n \ldots f_n$, we have the following [hide="system of equations"]$ \begin{array}{ccc} a_{n + 1} & = & c_n /2 & b_{n + 1} & = & a_n & c_{n + 1} & = & d_n /2 + e_n & d_{n + 1} & = & c_n /2 + b_n & e_{n + 1} & = & f_n & f_{n + 1} & = & d_n /2 \end{array}$[/hide]

1. Eliminate $ b$ and $ e$. [hide="This gives"]$ \begin{array}{ccc} a_{n + 1} & = & c_n /2 & c_{n + 1} & = & d_n /2 + f_{n - 1} & d_{n + 1} & = & c_n /2 + a_{n - 1} & f_{n + 1} & = & d_n /2 \end{array}$[/hide]
2. Eliminate $ c$ and $ d$ [hide="This gives"]$ \begin{array}{ccc} 2a_{n + 2} & = & f_{n + 1} + f_{n - 1} & 2f_{n + 2} & = & a_{n + 1} + a_{n - 1} & \end{array}$[/hide]
3. Eliminate $ f$ [hide="This gives"]$ 4a_{n + 2} = a_{n} + 2a_{n - 2} + a_{n - 4}$[/hide]

Now, we have a recurrence in $ a$, which we can solve to get the answer.

[quote="jvenezuela716"][quote="Ravi B"]Here's yet another solution to the original problem.  After a phase of 2 steps, we have probability $ \frac {1}{3}$ of being back at the start, and probability $ \frac {2}{9}$ each of being at one of the 3 other points of the tetrahedron.  Another way to say that: With probability $ \frac {1}{9}$, we do the lazy thing of staying where we are; otherwise, we choose one of the 4 points of the tetrahedron equally at random.  If we ever choose the non-lazy thing, then we have lost memory of where we started, and so from then on will be equally likely to be at any tetrahedron point.  The probability of making all 3 phases lazy is $ \frac {1}{9^3}$.  So the probability of being back at the start after 6 steps is
\[ \frac {1}{9^3} \plus{} \left( 1 \minus{} \frac {1}{9^3} \right) \cdot \frac {1}{4}.
\]
That comes out to $ \frac {61}{243}$, agreeing with the previous answers.

This method generalizes to any number of steps, agreeing with TZF's formula.   This method is sometimes called coupling.[/quote]

Ah, so instead of making six moves on a cube, you make three moves on a tetrahedron. That's clever!

[quote="DiscreetFourierTransform"]How about an answer based on Markov chains?[/quote]

Yes, I've used this method in my linear algebra class last year, but I only ever found it useful when the problem asked about the steady state, or what is expected to happen after an arbitrarily long time (IIRC, you have to find the eigenvalues of $ M$?).[/quote]

Yes, I think you're right about the eigenvalues. The general idea is that you're looking at the limit of the process $ \lim_{n \to \infty} M^{n}$.
If we call the limit matrix $ L$, then we have $ L*M \equal{} L$, and so it becomes $ L*(M \minus{} I) \equal{} 0$, and from there I believe you need an eigenvalue-argument.

[quote="jvenezuela716"]Yes, I've used this method in my linear algebra class last year, but I only ever found it useful when the problem asked about the steady state, or what is expected to happen after an arbitrarily long time (IIRC, you have to find the eigenvalues of $ M$?).[/quote]

When you find the eigenvalues of $ M$, you can do much more than just find the limit behavior. Consider the [url=http://tinurl.us/b32311]eigendecomposition of a matrix *[/url]:

In many cases, we can use the eigenvalues to write a matrix in the form $ A \equal{} VDV^{ \minus{} 1}$ for matrices $ V$ and $ D$, where $ D$ is a diagonal matrix. Then, $ A^n \equal{} VD^n V^{ \minus{} 1}$. But $ D^n$ is just $ D$ with the diagonal elements individually raised to the power $ n$. This allows us to write a closed-form expression for $ A^n$.

[i][b]*[/b] A more thorough explanation can be found [/i][url=http://tinyurl.com/673v4x]here[/url]

[quote="DiscreetFourierTransform"]Yes, I think you're right about the eigenvalues. The general idea is that you're looking at the limit of the process $ \lim_{n \to \infty} M^{n}$.
If we call the limit matrix $ L$, then we have $ L*M \equal{} L$, and so it becomes $ L*(M \minus{} I) \equal{} 0$, and from there I believe you need an eigenvalue-argument.[/quote]

This is not the most effective way of solving this. The solution set for that matrix equation includes not only the desired value $ L$, but also all matrices of the same dimensions whose rows are scalar multiples of the corresponding rows of the desired matrix. For example, if the desired limit is $ \binom{1/2 \; \; \; 1/2}{1/2 \; \; \; 1/2}$, then the matrices $ \binom{1 \; \; \; 1}{1 \; \; \; 1}$ and $ \binom{2 \; \; 2}{3 \; \; 3}$ would also be in the solution set for $ L$.

Also, we are not guaranteed that there is a limit in the first place.

A more reliable way of examining the end behavior is to find the eigendecomposition and take the limit as $ n \to \infty$.


Ring Theory
superior algebra
superior algebra solved

Hello,
Let R={a+bi|a,b are integers} be a ring. Show that M={x(2+i)|x is in R} is a maximal ideal of R.

I'd be thankful for your help.

$\mathbb{Z}[i]$ is a PID and $2+i$ is prime
this is the proof


inequalities
function
inequalities proposed

Let $a,b,c$ be nonnegative real numbers such that $a + b + c = 1$.
Prove that :
$\frac{1}{a+b^2+c^3}+\frac{1}{b+c^2+a^3}+\frac{1}{c+a^2+b^3} \leq \frac{81}{13}$

[quote="ashwath.rabindranath"]I think Jensen's inequality might work. I'm trying it out and posting it on the Forum in  a while....[/quote]

What function would you use?


Suppose G is a graph on 9 vertices such that, given any five of these vertices, there is some edge between some two of these vertices, what is the minimum number of edges that G can have in total? (Prove your answer)


MATHCOUNTS
AMC
AMC 10
AMC 12
AIME
AMC 8
AIME I

I know there are already several topics about this, sorry if this is spam, but I am taking AMC 8 for the 1st time this tuesday and I was wondering what the best ways are to prepare.

Do you know where I can find problems from the past? Unfortunately if I buy the books found on the AMC site, they will arrive too late. Are there any problems i can find online for free??

Thanks.

I recommend doing the problems, understanding the concepts, then doing the problems again but with complete work shown as if your going to explain to someone how you did it. This helps pick up on concepts easily, and the concepts could and HAVE BEEN repeated on such tests. That's the method i use, showing how the problem is done as if your explaining to someone, and it has helped me. Be sure you also attempt the understand most,if not all of the problems. Don't ignore one just because it's #25 on an AMC, see if you can understand all of it. If you dont, you have the privilege of posting it here, so take an advantage of that too. The only way you can get better is if you understand more concepts and the problems help you understand such concepts. Use the site xpmath gave to find problems.

I'm taking the AMC8 and AMC10 this year but my school won't let me take all 3. Is there a way to take the AMC12 in a different school district?

[quote="xpmath"][url=http://www.math.ksu.edu/main/handbook/ProblemSets]Try this.[/url][/quote]

Well, you could also use the AMC tests on this website...
There are a LOT of them.

[quote="xpmath"][url=http://www.math.ksu.edu/main/handbook/ProblemSets]Try this.[/url][/quote]

Wow...I had to buy the whole set because I didnt know they were here....thats unfair isnt it? :wallbash_red:  :mad:  :cursing:

Can you tell me all the sites with material I can possibly use? So I dont waste money again?

[quote="AwesomeToad"]How do I prepare for:
1) Mathcounts
2) AMC10
3) AMC12
4) AIME

I mean, I got 41 @ chapter mathcounts, then 25(!) :( at state.  :mad:  :mad:
Then I got 21:)onAMC8 but failed on AMC10(93)

I did this year practice AMC12, got a 90.

WHO SAID MAD FACES AND SAD FACES COUNTED AS SMILEYS???[/quote]
The best way to prepare is to do practice problems.


logarithms
trigonometry
integration
quadratics
continued fraction
calculus
calculus computations

If $ a \equal{} \pi\sqrt {2}$, does $ \sum_{k \equal{} 1}^{\infty}\frac {1}{(ak)ln(ak)sin(ak)}$ converge?

I think that it doesn't converge absolutely, but I can't figure out if it converges conditionally or not.

[quote="0714446459923"]

I think that it doesn't converge absolutely,.[/quote]
you are thinking right   :) . not only for
$ a\equal{}\pi \sqrt{2}$ main moment is that
serie will diverge absolutely if
$ a\equal{}\pi \beta$ where $ \beta$ is any algebraic numbers.

I just realized that I made a mistake. The log should be squared.

$ \sum_{k \equal{} 1}^{\infty}\frac {1}{k\ln^{2}(k)\sin(ak)}$

then if $ a \equal{} \beta n$ where $ \beta$ is algebraic number then :

serie absolutely  converge. but it's not so easy to show ..
I can't say nothing about conditionaly  :huh:

Why do you think so? later I will post full solution and essence of problems like this.
I try do it today  :wink:

I do some correction in my post and put full solution of this problem :
I will consider case : $ \sum \frac {1}{k\ln ^{a}k|\sin (\beta n \pi)|}$
where $ \beta$ is any algebraic number and investigate such $ a$ when serie absolutely converge.
few months ago I have already posted problem like this see here http://www.mathlinks.ro/viewtopic.php?t=166662
in this case we can do something like this :
$ %Error. "your" is a bad command.
sum ... \leq C + \sum _{k\geq 1}\frac {1}{2^{k}\ln^{a}2^{k}}\sum_{2^{k} < n\leq 2^{k + 1}}\frac {1}{|\sin(\pi \beta n)|}$
or our aim is to astimate sum $ \sum_{N/2 < n\leq N}\frac {1}{|\sin(\pi \beta n)|}$
as fedja wrote :
[quote="fedja"]The crucial point is that the distance from $ \sqrt 2 m$ to the nearest integer is at least $ c/m$ with some absolute $ c > 0$. Let us now look at the numbers $ n$ from $ N$ to $ 2N$ with $ |\sin\pi\sqrt 2 n| < 1/t$ for $ t > 0$. There are no such numbers for $ t > CN$ and the distance between two such consequtive numbers is at least $ ct$ for $ t < CN$, so we have at most $ CN/t$ of them. Thus
\[ \sum_{N\le n < 2N}\frac {1}{|\sin\pi\sqrt 2 n|}\le\int_{0}^{CN}\min\{N,CN/t\}\,dt\approx N\log N
\]
The rest is standard: just split your sum into dyadic blocks from $ 2^{k}$ to $ 2^{k + 1}$ and apply this estimate to each block.[/quote]
or upper bound is something like $ B N\ln N$ hence if we put $ N = 2^{k + 1}$ we will get
$ your sum ... \leq C + \sum_{k}\frac {D}{k^{a - 1}}$ and hence serie converge if $ a > 2$
Now I will try to explain fedjas estimate because for me his proof is  only sketch  and to understand his proof is not so easy.
I think that when he  proved during that moment he  was very got tired  :lol: .
but his proof is right .
now we will get more exact estimate :
denote:
$ S_{N}(\beta) = \sum _{N/2 < k\leq N}\frac {1}{|\sin (\beta \pi k)|}$
Lemma1.1 there exist $ \delta$ such as $ |\beta - \frac {m}{n}|\geq \frac {\delta}{n^{2}}$ for any algebraic number $ \beta$ and for any $ \frac {m}{n}$
(This was showen by K.F.Roth in 1955, an achievement which won him the Fields medal  :lol: )
case: if $ \beta = \sqrt {2}$ can be considered by Liouville Theorem , try to do prove it . fedja wrote it by next:
[quote="fedja"] ... distance from $ \sqrt 2 m$ to the nearest integer is at least $ c/m$ ... [/quote]  :lol: .
In my proof upper bound will be :
$ S_{N}(\beta) \leq 4 \frac {N}{\delta}\ln\frac {N}{\delta}$
Let
$ a_{n} = \min \limits_{m\in Z}|\beta - \frac {m}{n}|$
Lema1.2 (simplest lemma   ) $ a_{n} < \frac {1}{2n}$ and $ |\sin \pi n \beta |\geq 2na_{n}$
Hence
$ S_{N}(\beta)\leq \frac {1}{2}\sum_{N/2 < n\leq N}\frac {1}{na_{n}}$
after as fedja wrote $ S_{N}(\beta) \approx \int_{0}^{CN} ....$this is main moment and not so easy  as fedja wrote.
because :
Lemma1.3 : such $ n$ for which $ a_{n}\approx n^{ - 2}$ is like geometric progresion or in exact if we consider :
$ a_{n_{j}}\leq \frac {C}{n_{j}^{2}}$ then
$ \frac {n_{j + 1}}{n_{j}}\geq 1 + \frac {\delta}{2C}$ fedja wrote it as :
[quote="fedja"] ... distance between two such consequtive numbers is at least $ ct$ ... [/quote] :lol:
Consider : $ M_{\mu} = \{n \in N : \frac {2^{\mu - 1}}{n^{2}}\delta \leq a_{n} < \frac {2^{\mu}}{n^{2}}\delta \}$
We know that :
$ a_{n}\geq \frac {\delta}{n^{2}}$ therefore
\\

$ N = \bigcup\limits_{\mu \geq 1}M_{\mu}$
hence :
${ S_{N}(\beta}\leq \frac {1}{2}\sum_{\mu \geq 1} \sum \limits_{ n \in M_{\mu}, {N/2 < n \leq N} }\frac {1}{na_{n}}$
$ \leq \frac {1}{2}\sum_{\mu \geq 1} \sum \limits_{n \in M_{\mu}, {N/2 < n \leq N}} \frac {n}{2^{\mu}\delta}$
$ \leq \frac {N}{\delta}\sum_{\mu \geq 1}\frac {card (M_{\mu} \bigcap (\frac {N}{2}, N))}{2^{\mu}}$
suppose :
$ M_{\mu}\bigcap(\frac {N}{2},N] = \{ n_{1},..,n_{J}\}$
we know that :
$ \frac {n_{j + 1}}{n_{j}}\geq 1 + 2^{ - \mu + 1}$
then :
$ N\geq n_{J}\geq n_{1}(1 + 2^{ - \mu + 1})^{J - 1} > \frac {N}{2}(1 + 2^{ - \mu - 1})^{J - 1}$
therefore :
$ 1 > (J - 1)\log_{2}(1 + 2^{ - \mu - 1})\geq (J - 1)2^{\mu - 1}$ thus :
$ card (M_{\mu}\bigcap(\frac {N}{2},N]) = J < 1 + 2^{\mu + 1} = 1 + 2*2^{\mu}$
note that for big $ \mu$ , we have $ M_{\mu}\bigcap(\frac {N}{2},N]$ is empty.because :
$ \frac {2^{\mu - 1}}{n^{2}}\delta \leq a_{n} > \frac {1}{n}$
then $ 2^{\mu} < \frac {n}{\delta}\leq \frac {N}{\delta}$
hence after summating we have :
$ S_{N}(\beta) < \frac {N}{\delta}(1 + 2\log_{2}\frac {N}{\delta}) < 4\frac {N}{\delta}\ln \frac {N}{\delta}$ we are done

Note: never asked to professor : how to solve this problem, because you never understood his solution   :lol:. Anyway try to solve it yourself .

[quote="0714446459923"][quote="0714446459923"]how about to $ ln(ak)^{1.5}$? If my reasoning is right, it should grow at least as fast as ${ O(\sqrt {ln(k)})}$[/quote]
The errors of continued fraction approximations also have an upper bound of A/k. Also, the number of continued fractions with denominators less than k is at least $ Bln(k)$. Plug these in and it diverges[/quote]
I don't uderstand well your comment :
Continued fraction approximation for which number? you mean for $ \sum$? how do you construct continued fraction expansion?  :maybe:

anothering thing I was wondering: Why do you say it diverges for all algebraic b? I thought that the quadratic irrationals had at least as big error bound as any other number, so if it diverges for $ \sqrt{2}$, it diverges for every other number too

[quote="0714446459923"]anothering thing I was wondering: Why do you say it diverges for all algebraic b? I thought that the quadratic irrationals had at least as big error bound as any other number, so if it diverges for $ \sqrt {2}$, it diverges for every other number too[/quote]
yes. you are right.
I want to say:
 if $ a>2$ then it converge for any algebraic number $ \beta$ :)


Fie $DA$ perpendiculara pe planul triunghiului $ABC$ si $E$ un punct interior segmentului $(DA)$. Daca $M$,$N$,$P$,$Q$ sunt proiectiile lui $A$ pe $BD$, $CD$, $BE$, $CE$ sa se arate ca :
a) dreptele $MN$, $BC$ si $PQ$ au un punct comun.
b)patrulaterul $MNPQ$ este inscriptibil.

Well...it's quite easy

Luam planul (ABC) in care se afla si MN siBC=>MN si BC coplanare
Analog PQ si BC coplanare
Fie $X=MN \cap BC$ si $Y=PQ \cap BC$
Aplic Menelaus in triunghiul ABC cu transversala M-N-X si in triunghiul EBC cu transversala P-Q-Y.
Vedem ca CX=CY=> dreptele MN,BC,PQ au un punct comun

b)Vedem ca XM*XN=XP*XQ. Am calculat si iese asa => M,N,P,Q conciclice


How many equilateral triangles in the plane have two verticies in the set {(0,0),(0,1),(1,0),(1,1)}?

[hide]There are two triangles for any two points we can choose.
$2{4\choose 2}=12$[/hide]


calculus
integration
number theory unsolved
number theory

Find all non-negative integral solutions $(n_1,n_2,\ldots , n_{14})$ to \[ n_1^4 + n_2^4 + \cdots + n_{14}^4 = 1,599. \]

Note that a fourth power can only be 0 or 1 modulo 16. As 1599 is congruent to 15 modulo 16, there is no solution. :)


modular arithmetic
function
Olimpiada de matematicas

Poco tarde pero finalmente: :oops:

1. Hallar todas las funciones de los naturales en los naturales tales que $f(m^{2}+n^{2})=f^{2}(m)+f^{2}(n)$ para todo $(m,n)$ naturales.

(los naturales son $0,1,2...$)

2. Sea $S$ un conjunto de $n\geq3$ puntos en el plano. Pruebe  que existe un conjunto $T$ de $2n-5$ puntos tales que cualquier triangulo determinado por vértices en $S$ contiene  un punto de $T$.

3. Sea $ABC$ un triángulo acutángulo. Sea $M$ el punto medio del lado $AB$ y $D$ pie de la altura desde $C$. Una línea por $M$ intersecata a las líneas $CA$, $CB$ en $K$, $L$ respectivamente con $CK = CL$. Sea $O$ el cincuncentro del triángulo $CKL$. Demostrar que $OD = OM$.

4. Sean $p>5$ un primo y $n$ un entero par. Demostrar que

 Si $L_{n}\equiv 2 \pmod p$, entonces  $L_{n}\equiv 2 \pmod{p^{2}}$;

 Si $L_{n}\equiv-2 \pmod p$, entonces  $L_{n}\equiv-2 \pmod{p^{2}}$.

Solo una aclaración:
$L_{1}=1, L_{2}=3,L_{n+1}=L_{n}+L_{n-1}$ $(n\geq2)$
es la sucesión de Lucas.

$Tipe$

quedan pocos dias para enviar sus soluciones!!! donde estan todos???

se amplia el plazo a petición de varios participantes...

[color=blue]Gracias, en serio. Sucede que, en mi colegio, estamos en plena etapa de exámenes y recién el 20 de noviembre finaliza. Ya saben que cuando los estudios priman no hay mucho tiempo para resolver problemas matemáticos…[/color]

[quote="R.G.A.M."][color=blue]¿Significa [b][i]f[/i]²([i]x[/i])[/b] lo mismo que [b]([i]f[/i]([i]x[/i]))²[/b] o que [b][i]f[/i]([i]f[/i]([i]x[/i]))[/b]?

Por el momento, estuve utilizando la primera acepción, pero existen materiales que utilizan la otra.[/color][/quote]
bueno en algunos libros, consideran como aplicacion de la funcion $n$ veces, pero entonces indican, en el problema supongo que consideran la pimera acepcion.
chau, :D
PD: cuando vence el plazo??

[quote="Pascual2005"]3. Sea $ ABC$ un triángulo acutángulo. Sea $ M$ el punto medio del lado $ AB$ y $ D$ pie de la altura desde $ C.$ Una línea por $ M$ intersecta a las líneas $ CA,$ $ CB$ en $ K,$ $ L$ respectivamente con $ CK \equal{} CL.$ Sea $ O$ el cincuncentro del triángulo $ CKL.$ Demostrar que $ OD \equal{} OM.$[/quote]
Como $ MA \equal{} MB$ y $ CK \equal{} CL,$ al aplicar el teorema de Menelao en el $ \triangle ABC$ cortado por la transversal $ KML$ se tendrá $ AK \equal{} BL.$ Así $ E \equiv \odot(CKL) \cap \odot(ABC) \not\equiv C$ es el centro de la rotación que lleva $ AK$ en $ BL$ y como $ EA \equal{} EB,$ entonces $ E$ es el punto medio del arco $ AB$ de $ \odot(ABC).$ Desde luego, $ E$ yace en la bisectriz interior de $ \angle ACB$ $ \Longrightarrow$ $ O \in CE.$

Siendo $ CE$ el circundiámetro de $ \triangle CKL$ perpendicular a $ KL,$ se sigue que $ O$ es el punto medio de $ CE$ y si $ A',B'$ son los puntos medios de los lados $ CB,CA,$ entonces $ O$ es el corte de $ CE$ con la mediatriz de $ A'B'$ puesto que la homotecia positiva con centro $ C$ y coeficiente $ \frac {_1}{^2}$ transforma $ \overline{AB}$ en $ \overline{A'B'},$ $ E$ en $ O$ y la mediatriz $ EM$ de $ AB$ en la mediatriz de $ A'B'.$

Por último resta observar que el cuadrilátero $ A'B'DM$ es un trapecio isósceles en el cual $ O$ yace en su eje de simetría, ya que en efecto  $ \triangle MA'B'$ y  $ \triangle DB'A'$ son simétricos con respecto a la mediatriz de $ A'B',$ por tanto $ OD \equal{} OM.$


Given $ a, b, c$ are the [b]reals[/b] satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that: $ \frac{a^2\plus{}b^2c^2}{(b\minus{}c)^2}\plus{}\frac{b^2\plus{}c^2a^2}{(c\minus{}a)^2}\plus{}\frac{c^2\plus{}a^2b^2}{(a\minus{}b)^2} \ge 5$  :)

This strange ineq exists a very nice proof by Am-Gm  :lol:

It's also an interesting problem in new book "Inequalities With Beautiful Solutions" (Vasile Cirtoaje- Vo Quoc Ba Can - Tran Quoc Anh, GIL Publishing House 2009)  :)


Let ABCD is a quadrilateral that <ABC=<ADC and <ABC + <BCD < 180. If E=(CD,AB), prove that AC^2=CD.CE - AB.AE

Because <ABC + <BCD < 180, ray BA meets ray CD at E. The circle (ADE) meets the ray CA again at K. Then, $ \angle AKE\equal{}\angle ADC\equal{}\angle ABC$, so (EKBC) form a cyclic quadrilateral. Hence, $ AC^2\equal{}CA(CK\minus{}AK)\equal{}CA.CK\minus{}AC.AK\equal{}CD.CE\minus{}AB.AE$

We can Prove $ \angle ABC \equal{} \angle ADC$ if and only if $ AC^2 \equal{} CD.DE \minus{} AB.AE$
and it is 1996 Bugaria contest :D

[quote="farshid_lk_park"]MR Danh your i think your solution  is incorrect Because  <AKE= <ADE =180 - <ADC
[/quote]
I think mr.danh was right, you should note that he wrote E is the intersection of ray BA and ray CD. His proof completely fits to my figure.

Oh yes.I've made a mistake.Mr Danh your solution was really beautiful.thanks a lot. :roll:  :roll:


analytic geometry
graphing lines
slope
geometry
3D geometry
prism
area of a triangle

[quote="Scrambled"][quote="Silverfalcon"]I need everything or every similar types of questions for state..
I don't have a lot time and here are just few things I know:

Heron's Theorem
Pythogrem Theorem
45-45-90 Theorem
30-60-90 Theorem
Few algebra skills such as slopes and stuffs
... Um... I think that's all...
So, Please help...
i'm new here..[/b][/quote]
do you know how to find the diagonal of a rectangular prism?  How to find the area of a triangle by only knowing its three sides?(i think that showed up on a state before)[/quote]

I'm assuming Heron's Theorem was meant to be Heron's formula, which is that exactly..

diagonal of a rectangular prism? How to find the area of a triangle by only knowing its three sides?

I think I heard the diagnal of polygon's formula is:
n(n-1)/3? Is that right? I'm not sure so...
And how do you find the area of a triangle by only knowing three sides? Also, what's the formula for finding midpoint of two lines?


Let f(z) be an entire function and f(z) = 0 on a subinterval a < x < b of the real axis. Show that f(z) = 0 for all z in C (set of complex numbers). Conclude that if f and g are entire functions and that they agree on a segment of the real axis, then they must be the same.

How do I answer this question? Thanks for help in advance.

Since we know what the function is on a segment of the real axis, we can calculate what its Taylor series must be. Then, since an entire function is analytic and determined by its Taylor series, an entire function is determined by what it is on a segment of the real axis. Since the function $ f(z)\equal{}0$ is of course entire, any function that gives zero over a segment of the real axis must be zero everywhere.

This basic idea proves your statement for two functions, or you could just apply the result for zero to their difference (which is also entire).

This is a specific example of a more general phenomenon exhibited by Holomorphic functions. The property is as follows: let $ f(z)$ be holomorphic over an open set $ U$. If there exists a set of points $ \{a_n\}$ in $ U$ such that the limit of the series $ \{a_n\}$ is in $ U$ and $ f(z)$ vanishes on $ \{a_n\}$, then $ f(z)$ is zero on $ U$. This property guarantees us that analytic continuations are unique and that given functions which agree on a common open set. This is the historical origin for the theory of Riemann surfaces and cohomology of sheaves (which gives us a way to determine whether or not certain functions can be glued together to expand their domain). This "nice" behavior of holomorphic functions contrasts the behavior of real differentiable functions. There are many functions which are zero everywhere except a certain interval (functions with compact support). This is just one of the many reasons why complex functions behave much better than their real counterparts.

This typical property is sometimes known in the student community as [b]Identity Theorem[/b]


let n a positive integers if 4^n+2^n+1 is a prime than n is a power of 3 solve this problem man

[hide]Suppose $n=ab$ where $b$ is neither 1 nor a multiple of 3. Then let $d=4^a+2^a+1$.

$4^n+2^n+1=\frac{8^n-1}{2^n-1}$. The top is divisible by $d$. But the bottom is not because $8^a\equiv 1$ but $2^a\not\equiv 1\pmod{d}$. Let $r$ be the remainder when $b$ is divided by 3. Either $r$ is $1$ or 2. Then the bottom is congruent to $2^a-1$ or $4^a-1$, neither of which is divisible by $d$.

This means some factor of $d$ divides $4^n+2^n+1$, but since $b\neq 1$, $d<4^n+2^n+1$. Then $4^n+2^n+1$ is composite. Contradiction. It follows that $n$ must be a power of 3.[/hide]

If $\displaystyle\sum_{i=0}^{k}(2^n)^i$ is prime, $n$ must be a power of $k+1$.


A machine makes three kinds of cards: 50% are red, 30% are blue and 20% are green.
If you takes 5 cards, find the probability that it has 2 red, 2 blue and 1 green cards, rescp.

[hide="Fill in the blanks"]We consider the cards dealt in order.  There are ______ possible "good" orders in which the cards could be dealt.  The probability that any one of these occurs is ______.  Therefore, the total probability that a "good" hand is dealt is _______.[/hide]